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INVARIANT AND STATIONARY MEASURES FOR THE SL(2, R) ACTION ON MODULI SPACE by ALEX ESKIN and MARYAM MIRZAKHANI ABSTRACT We prove some ergodic-theoretic rigidity properties of the action of SL(2, R) on moduli space. In particular, we show that any ergodic measure invariant under the action of the upper triangular subgroup of SL(2, R) is supported on an invariant affine submanifold. The main theorems are inspired by the results of several authors on unipotent flows on homogeneous spaces, and in particular by Ratner’s seminal work. CONTENTS 1. Introduction ...................................................... 96 1.1. The main theorems .............................................. 98 2. Outline of the paper . ................................................. 99 2.1. Some notes on the proofs ........................................... 99 2.2. Notational conventions . ........................................... 101 2.3. Outline of the proof of Step 1 ........................................ 102 3. Hyperbolic properties of the geodesic flow ...................................... 108 4. General cocycle lemmas ............................................... 115 4.1. Lyapunov subspaces and flags ........................................ 115 4.2. Equivariant measurable flat connections .................................. 117 4.3. The Jordan canonical form of a cocycle ................................... 119 4.4. Covariantly constant subspaces ....................................... 119 4.5. Some estimates on Lyapunov subspaces ................................... 120 4.6. The cover X . ................................................. 124 4.7. Dynamically defined norms ......................................... 126 4.8 . Proof of Lemma 4.7 .............................................. 129 4.9 . Proof of Propositions 4.4 and 4.12 ...................................... 131 4.10 . Proof of Proposition 4.15 ........................................... 134 5. Conditional measure lemmas . ............................................ 137 5.1 . Proof of Lemma 5.2 .............................................. 140 5.2 . Proof of Proposition 5.3 ............................................ 141 6. Divergence of generalized subspaces ......................................... 144 6.1. Approximation of generalized subspaces and the map A(·,·,·,·) ..................... 158 6.2. The stopping condition . ........................................... 164 6.3 . Proof of Proposition 6.16 ........................................... 166 6.4 . Proof of Lemma 6.4 .............................................. 170 6.5 .Construction of the map A(q , u,, t) .................................... 173 6.6 . Proofs of Proposition 6.11 and Lemma 6.14 ................................ 180 7. Bilipshitz estimates . ................................................. 183 8. Preliminary divergence estimates ........................................... 185 8.1. The U -inert subspaces E(x) ......................................... 186 9. The action of the cocycle on E ............................................ 192 9.1. The Jordan canonical form of the cocycle on E(x) ............................. 192 9.2. Time changes ................................................. 194 9.3. The foliations F , F and the parallel transport R(x, y) .......................... 194 ij v 9.4. A maximal inequality ............................................. 196 10. Bounded subspaces and synchronized exponents .................................. 197 Research of the first author is partially supported by NSF grants DMS 0604251, DMS 0905912 and DMS Research of the second author is partially supported by the Clay foundation and by NSF grant DMS 0804136. https://doi.org/10.1007/s10240-018-0099-2 96 ALEX ESKIN, MARYAM MIRZAKHANI 10.1 . Bounded subspaces and synchronized exponents.............................. 199 10.2 . Invariant measures on X × P(L) ....................................... 207 11. Equivalence relations on W ............................................. 217 11.1 . Proof of Lemma 11.2 ............................................. 223 11.2 . Proof of Proposition 11.4 ........................................... 226 12. The inductive step ................................................... 241 13. Proof of Theorem 2.1 ................................................. 258 14. Random walks ..................................................... 260 15. Time changes and suspensions . ........................................... 266 16. The martingale convergence argument ....................................... 269 Acknowledgements ..................................................... 280 Appendix A: Forni’s results on the SL(2, R) action ................................... 280 A.1. The Hodge norm and the geodesic flow ................................... 280 A.2. The Kontsevich-Zorich cocycle ....................................... 281 Appendix B: Entropy and the Teichmüller geodesic flow ................................ 287 Appendix C: Semisimplicity of the Lyapunov spectrum ................................. 297 C.3. An ergodic lemma ............................................... 299 C.4. A zero-one law ................................................ 302 C.5. Proof of Theorem C.6 ............................................ 304 Appendix D: Dense subgroups of nilpotent groups ................................... 313 References ......................................................... 317 Index of Notation §1-§16 .................................................. 320 1. Introduction Suppose g ≥ 1, and let α = (α ,...,α ) be a partition of 2g − 2, and let H(α) be a 1 n stratum of Abelian differentials, i.e. the space of pairs (M,ω) where M is a Riemann sur- face and ω is a holomorphic 1-form on M whose zeroes have multiplicities α ...α .The 1 n form ω defines a canonical flat metric on M with conical singularities at the zeros of ω. Thus we refer to points of H(α) as flat surfaces or translation surfaces. For an introduction to this subject, see the survey [Zo]. The space H(α) admits an action of the group SL(2, R) which generalizes the action of SL(2, R) on the space GL(2, R)/SL(2, Z) of flat tori. In this paper we prove ergodic-theoretic rigidity properties of this action. In what follows, we always replace H(α) by a finite cover X which is a manifold. Such a cover can be found by e.g. considering a level 3 structure (see Section 3). However, in the introduction, we suppress this from the notation. Let ⊂ M denote the set of zeroes of ω.Let {γ ,...,γ } denote a sym- 1 k plectic Z-basis for the relative homology group H (M,, Z).Wecan defineamap : H(α) → C by (M,ω) = ω,..., ω . γ γ 1 k The map (which depends on a choice of the basis {γ ,...,γ }) is a local coordinate sys- 1 k tem on (M,ω). Alternatively, we may think of the cohomology class [ω]∈ H (M,, C) as a local coordinate on the stratum H(α). We will call these coordinates period coordinates. We can consider the measure λ on H(α) which is given by the pullback of the 1 k Lebesgue measure on H (M,, C) ≈ C .The measure λ is independent of the choice INVARIANT AND STATIONARY MEASURES 97 of basis {γ ,...,γ }, and is easily seen to be SL(2, R)-invariant. We call λ the Lebesgue or 1 k the Masur-Veech measure on H(α). The area of a translation surface is given by a(M,ω) = ω∧¯ω. A “unit hyperboloid” H (α) is defined as a subset of translation surfaces in H(α) of area on H (α) is defined by disintegration one. The SL(2, R)-invariant Lebesgue measure λ (1) 1 of the Lebesgue measure λ on H (α),namely dλ = cdλ da, (1) where c is a constant. A fundamental result of Masur [Mas1] and Veech [Ve1]isthat λ (H (α)) < ∞. In this paper, we normalize λ so that λ (H (α)) = 1(and so λ is (1) 1 (1) (1) 1 (1) a probability measure). For a subset M ⊂ H (α) we write 1 1 RM = (M, tω) | (M,ω) ∈ M , t ∈ R \{0} ⊂ H(α). 1 1 Definition 1.1. — An ergodic SL(2, R)-invariant probability measure ν on H (α) is called 1 1 affine if the following conditions hold: (i) The support M of ν is an immersed submanifold of H (α), i.e. there exists a 1 1 1 manifold N and a proper continuous map f : N → H (α) so that M = f (N ).The 1 1 self-intersection set of M , i.e. the set of points of M which do not have a unique preimage 1 1 under f , is a closed subset of M of ν -measure 0. Furthermore, each point in N has a 1 1 neighborhood U such that locally Rf (U) is given by a complex linear subspace defined over R in the period coordinates. (ii) Let ν be the measure supported on M = RM so that dν = dν da. Then each point in 1 1 N has a neighborhood U such that the restriction of ν to Rf (U) is an affine linear measure in the period coordinates on Rf (U), i.e. it is (up to normalization) the induced measure of the Lebesgue measure λ to the subspace Rf (U). Definition 1.2. — We say that any suborbifold M for which there exists a measure ν such 1 1 that the pair (M ,ν ) satisfies (i) and (ii) is an affine invariant submanifold. 1 1 We also consider the entire stratum H(α) to be an (improper) affine invariant submanifold. It follows from [EMiMo, Theorem 2.2] that the self-intersection set of an affine invariant manifold is itself a finite union of affine invariant manifolds of lower dimension. For many applications we need the following: Proposition 1.3. —Any stratum H (α) contains at most countably many affine invariant submanifolds. 98 ALEX ESKIN, MARYAM MIRZAKHANI Proposition 1.3 is deduced as a consequence of some isolation theorems in [EMiMo]. This argument relies on adapting some ideas of G. A. Margulis to the Te- ichmüller space setting. Another proof is given by A. Wright in [Wr1], where it is proven that affine invariant submanifolds are always defined over a number field. The classification of the affine invariant submanifolds is complete in genus 2 by the work of McMullen [Mc1][Mc2][Mc3][Mc4][Mc5]and Calta[Ca]. In genus 3 or greater it is an important open problem. See [Mö1], [Mö2], [Mö3], [Mö4], [BoM], [BaM], [HLM], [LN1], [LN2], [LN3], [Wr1], [Wr2], [MW], [NW], [ANW], [Fi1]and [Fi2] for some results in this direction. 1.1. The main theorems. — Let 1 t e 0 N = , t ∈ R , A = , t ∈ R , −t 01 0 e N = , t ∈ R t 1 cosθ sinθ Let r = ,and let SO(2) ={r | θ ∈[0, 2π)}.Then N, N, A and SO(2) are θ θ − sinθ cosθ subgroups of SL(2, R).Let P = AN denote the set of upper triangular matrices of deter- minant 1, which is a subgroup of SL(2, R). Theorem 1.4. —Let ν be any ergodic P-invariant probability measure on H (α). Then ν is SL(2, R)-invariant and affine. The following (which uses Theorem 1.4) is joint work with A. Mohammadi and is proved in [EMiMo]: Theorem 1.5. —Suppose S ∈ H (α). Then, the orbit closure PS = SL(2, R)S is an affine invariant submanifold of H (α). For the case of strata in genus 2, the SL(2, R) part of Theorems 1.4 and 1.5 were proved using a different method by Curt McMullen [Mc6]. The proof of Theorem 1.4 uses extensively entropy and conditional measure tech- niques developed in the context of homogeneous spaces (Margulis-Tomanov [MaT], Einsiedler-Katok-Lindenstrauss [EKL]). Some of the ideas came from discussions with Amir Mohammadi. But the main strategy is to replace polynomial divergence by the “exponential drift” idea of Benoist-Quint [BQ]. Stationary measures. — Let μ be an SO(2)-invariant compactly supported measure on SL(2, R) which is absolutely continuous with respect to Lebesgue measure. A measure ν on H (α) is called μ-stationary if μ ∗ ν = ν,where μ ∗ ν = (g ν)dμ(g). SL(2,R) INVARIANT AND STATIONARY MEASURES 99 Recall that by a theorem of Furstenberg [F1], [F2], restated as [NZ,Theorem 1.4], there exists a probability measure ρ on SL(2, R) such that ν → ρ ∗ ν is a bijection between ergodic P-invariant measures and ergodic μ-stationary measures. Therefore, Theorem 1.4 implies the following: Theorem 1.6. — Any ergodic μ-stationary measure on H (α) is SL(2, R)-invariant and affine. Counting periodic trajectories in rational billiards. — Let Q be a rational polygon, and let N(Q, T) denote the number of cylinders of periodic trajectories of length at most T for the billiard flow on Q. By a theorem of H. Masur [Mas2][Mas3], there exist c and c 1 2 depending on Q such that for all t > 1, 2t t 2t c e ≤ N Q, e ≤ c e . 1 2 Theorem 1.4 and Proposition 1.3 together with some extra work (done in [EMiMo]) imply the following “weak asymptotic formula” (cf. [AEZ]): Theorem 1.7. — For any rational polygon Q, there exists a constant c = c(Q) such that s −2s lim N Q, e e ds = c. t→∞ The constant c in Theorem 1.7 is the Siegel-Veech constant (see [Ve2], [EMZ]) associated to the affine invariant submanifold M = SL(2, R)S where S is the flat surface obtained by unfolding Q. It is natural to conjecture that the extra averaging on Theorem 1.7 is not neces- t −2t sary, and one has lim N(Q, e )e = c. This can perhaps be shown if one obtains a t→∞ classification of the measures invariant under the subgroup N of SL(2, R). Such a result is in general beyond the reach of the current methods. However it is known in a few very special cases, see [EMS], [EMM], [CW]and [Ba]. Other applications to rational billiards. — All the above theorems apply also to the mod- uli spaces of flat surfaces with marked points. Thus one should expect applications to the “visibility” and “finite blocking” problems in rational polygons as in [HST]. It is likely that many other applications are possible. 2. Outline of the paper 2.1. Some notes on the proofs. — The theorems of Section 1.1 are inspired by the re- sults of several authors on unipotent flows on homogeneous spaces, and in particular by 100 ALEX ESKIN, MARYAM MIRZAKHANI Ratner’s seminal work. In particular, the analogues of Theorems 1.4 and 1.5 in homoge- neous dynamics are due to Ratner [Ra4], [Ra5], [Ra6], [Ra7]. (For an introduction to these ideas, and also to the proof by Margulis and Tomanov [MaT] see the book [Mor]. See also the papers [Dan1], [Dan2], [Dan3], [Dan4], [DM1], [DM2], [DM3], [DM4], [Mar1], [Mar2], [Mar3], [Mar4], [Ra1], [Ra2], [Ra3], [MoSh]. The homogeneous ana- logue of the fact that P-invariant measures are SL(2, R)-invariant is due to Mozes [Moz] and is based on Ratner’s work. All of these results are based in part on the “polynomial divergence” of the unipotent flow on homogeneous spaces. However, in our setting, the dynamics of the unipotent flow (i.e. the action of N) on H (α) is poorly understood, and plays no role in our proofs. The main strategy is to replace the “polynomial divergence” of unipotents by the “exponential drift” idea in the recent breakthrough paper by Benoist and Quint [BQ]. One major difficulty is that we have no apriori control over the Lyapunov spectrum of the geodesic flow (i.e. the action of A). By [AV1] the Lyapunov spectrum is simple for the case of Lebesgue (i.e. Masur-Veech) measure, but for the case of an arbitrary P-invariant measure this is not always true, see e.g. [Fo2], [FoM]. In order to use the Benoist-Quint exponential drift argument, we must show that the Zariski closure (or more precisely the algebraic hull, as defined by Zimmer [Zi2]) of the Kontsevich-Zorich cocycle is semisimple. The proof proceeds in the following steps: Step 1. — We use an entropy argument inspired by the “low entropy method” of [EKL] (using [MaT] together with some ideas from [BQ]) to show that any P-invariant measure ν on H (α) is in fact SL(2, R) invariant. We also prove Theorem 2.1 which gives control over the conditional measures of ν . This argument occupies Sections 3–13 and is outlined in more detail in Section 2.3. Step 2. — By some results of Forni (see Appendix A), for an SL(2, R)-invariant measure ν , the absolute cohomology part of the Kontsevich-Zorich cocycle A : SL(2, R) × H (α) → Sp(2g, Z) is semisimple, i.e. has semisimple algebraic hull. For an exact statement see Theorem A.6. Step 3. — We pick an SO(2)-invariant compactly supported measure μ on SL(2, R) which is absolutely continuous with respect to Lebesgue measure, and work in the random walk setting as in [F1][F2]and [BQ]. Let B denote the space of infi- nite sequences g , g ,... ,where g ∈ SL(2, R). We then have a skew product shift map 0 1 i −1 T : B × H (α) → B × H (α) as in [BQ], so that T(g , g ,... ; x) = (g , g ,... ; g x). 1 1 0 1 1 2 Then, we use (in Appendix C) a modification of the arguments by Guivarc’h and Raugi [GR1], [GR2], as presented by Goldsheid and Margulis in [GM, §4–5], and an argument of Zimmer (see [Zi1]or[Zi2]) to prove Theorem C.5 which states that the Lyapunov spectrum of T is always “semisimple”, which means that for each SL(2, R)-irreducible component of the cocycle, there is a T-equivariant non-degenerate inner product on the INVARIANT AND STATIONARY MEASURES 101 Lyapunov subspaces of T (or more precisely on the successive quotients of the Lyapunov flag of T). This statement is trivially true if the Lyapunov spectrum of T is simple. Step 4. — We can now use the Benoist-Quint exponential drift method to show that the measure ν is affine. This is done in Sections 14–16. At one point, to avoid a prob- lem with relative homology, we need to use a result, Theorem 14.3 about the isometric (Forni) subspace of the cocycle, which is proved in joint work with A. Avila and M. Möller [AEM]. Finally, we note that the proof relies heavily on various recurrence to compact sets results for the SL(2, R) action, such as those of [EMa]and [Ath]. All of these results originate in the ideas of Margulis and Dani [Mar1], [Dan1], [EMM1], [EMM2]. 2.2. Notational conventions. — For t ∈ R,let e 0 1 t g = , u = . t t −t 0 e 01 Let A ={g : t ∈ R},N ={u : t ∈ R}.Let P = AN. t t Let X denote a finite cover of the stratum H (α) which is a manifold (see Sec- 0 1 ˜ ˜ tion 3). Let X denote the universal cover of X .Let π : X → X denote the natural 0 0 0 0 projection map. We will need at some point to consider a certain measurable finite cover X of X . This cover will be constructed in Section 4.6 below. Let X denote the “universal cover” of X, see Section 4.6 for the exact definition. We abuse notation by denoting the covering map from X to X also by the letter π . If f is a function on X or X we sometimes abuse notation by denoting f ◦ π by f and write f (x) instead of f (π(x)). A point of H(α) is a pair (M,ω),where M is a compact Riemann surface, and ω is a holomorphic 1-form on M. Let denote the set of zeroes of ω. The cohomology class of ω in the relative cohomology group 1 1 2 ∼ ˜ H (M,, C) = H (M,, R ) is a local coordinate on H(α) (see [Fo]). For x ∈ X ,let 1 2 V(x) denote a subspace of H (M,, R ). Then we denote by the image of V(x) under the affine exponential map, i.e. V[x]= y ∈ X : y − x ∈ V(x) . (For some subspaces V, we can define V[x] for x ∈ X as well. This will be explained in Section 4.6. Also, depending on the context, we sometimes consider V[x] to be a subset of X or X .) 1 1 Let p : H (M,, R) → H (M, R) denote the natural map. Let 1 1 (2.1)H (x) = v ∈ H (M,, R) : p(Re x) ∧ p(v) = p(Im x) ∧ p(v) = 0 , ⊥ 102 ALEX ESKIN, MARYAM MIRZAKHANI where we are considering the “real part map” Re and the “imaginary part map” Im as 1 1 2 1 maps from H (M,, C) H (M,, R ) to H (M,, R).Let 1 1 W(x) = R(Im x) ⊕ H (x) ⊂ H (M,, R), so that W(x) = v ∈ H (M,, R) : p(Im x) ∧ p(v) = 0 . − 1 Let π : W(x) → H (M,, R) denote the map (defined for a.e. x ∈ X ) − 1 (2.2) π (c Im x + v) = c Re x + v c ∈ R,v ∈ H (x), x ⊥ so that − 1 π W(x) = v ∈ H (M,, R) : p(Re x) ∧ p(v) = 0 . 1 2 2 1 We have H (M,, R ) = R ⊗ H (M,, R). For a subspace V(x) ⊂ W(x), we write + − − V (x) = (1, 0) ⊗ V(x), V (x) = (0, 1) ⊗ π V(x) . + − Then W [x] and W [x] play the role of the unstable and stable foliations for the action of g on X for t > 0, see Lemma 3.5. t 0 Starred subsections. — Some technical proofs are relegated to subsections marked with a star. These subsections can be skipped on first reading. The general rule is that no statement from a starred subsection is used in subsequent sections. 2.3. Outline of the proof of Step 1. — The general strategy is based on the idea of additional invariance which was used in the proofs of Ratner [Ra4], [Ra5], [Ra6], [Ra7] and Margulis-Tomanov [MaT]. The aim of Step 1 is to prove the following: Theorem 2.1. —Let ν be an ergodic P-invariant measure on X . Then ν is SL(2, R)- invariant. In addition, there exists an SL(2, R)-equivariant system of subspaces L(x) ⊂ W(x) such + + that for almost all x, the conditional measures of ν along W [x] are the Lebesgue measures along L [x], − − and the conditional measures of ν along W [x] are the Lebesgue measures along L [x]. + + In the sequel, we will often refer to a (generalized) subspace U [x]⊂ W [x] on which we already proved that the conditional measure of ν is Lebesgue. The proof of Theorem 2.1 will be by induction, and in the beginning of the induction, U [x]= Nx. (Note: generalized subspaces are defined in Section 6.) + + In this introductory subsection, let U (x) ⊂ W (x) denote the subspace { y − x : y ∈ U [x]}. (This definition has to be modified when we are dealing with generalized subspaces, see Section 6.) INVARIANT AND STATIONARY MEASURES 103 FIG. 1. — Outline of the proof of Theorem 2.1 Outline of the proof of Theorem 2.1.— Let ν be an ergodic P-invariant probability measure on X . Since ν is N-invariant, the conditional measure ν of ν along W 0 W is non-trivial. This implies that the entropy of A is positive, and thus the conditional measure ν − of ν along W is non-trivial (see e.g. [EL]). This implies that on a set of almost full measure, we can pick points q and q in the support of ν such that q and q are in the same leaf of W and d(q, q ) ≈ 1/100, see Figure 1. Let > 0 be a large parameter. Let q = g q and let q = g q .Then q and q are 1 1 1 1 very close together. We pick u ∈ U (q ) with u≈ 1/100, and pick (as described below) u ∈ U (q ). Consider the points uq and u q . With our choice of u , the points uq and 1 1 1 1 u q will be close, but they are no longer in the same leaf of W ,and we expect them to diverge under the action of g as t →+∞.Let t be chosen so that q = g uq and t 2 t 1 q = g u q be such that d(q , q ) ≈ ,where > 0is fixed. t 2 2 1 2 Consider the bundle (which we will denote for short H ) whose fiber above x ∈ 1 1 1 H(α) is H (M,, R). The presence of the integer lattice H (M,, Z) in H (M,, R) allows one to identify the fibers at nearby points. This defines a flat connection, called the Gauss-Manin connection on this bundle. The action of SL(2, R) and in particular the geodesic flow g on H(α),extends to an action on the bundle H , where the action on the fibers is by parallel transport with respect to the Gauss-Manin connection. The action on the bundle takes the form g (x,v) = g x, A(g ,v) , t t t where A : SL(2, R) × H (α) → GL(H (M,, R)) is the Kontsevich-Zorich cocycle. It is continuous (in fact locally constant) and log-integrable. Thus the multiplicative ergodic theorem can be applied. Let 1 1 1 1 1 = λ H >λ H ≥··· ≥ λ H >λ H =−1 1 2 k−1 k 104 ALEX ESKIN, MARYAM MIRZAKHANI denote the Lyapunov spectrum of the Kontsevich-Zorich cocycle. (The fact that λ < 1 is due to Veech [Ve1]and Forni[Fo].) We have 1 1 H (M,, R) = V H (x) i=1 1 1 where V (H )(x) is the Lyapunov subspace corresponding to λ (H ) (see Section 4). Note i i that V (H )(x) corresponds to the unipotent direction inside the SL(2, R) orbit. In the + 1 first step of the induction, U (x) = V (H )(x). + 1 In general, for y ∈ U [x], if we identify H at x and y using the Gauss-Manin connection, we have (see Lemma 4.1), 1 1 (2.3) V H (y) ⊂ V H (x). i j j≤i 1 + 1 + We say that the Lyapunov exponent λ (H ) is U -inert if for a.e. x, V (H )(x) ⊂ U (x) i i and also, for a.e. y ∈ U [x], 1 + 1 V H (y) ⊂ U (x) + V H (x). i i 1 + + (In other words, V (H )(x) is constant (modulo U )along U [x].) Note that in view of 1 + 1 (2.3), λ (H ) is always U -inert. We now assume for simplicity that λ (H ) is the only 2 2 U -inert exponent. We may write u q − uq = w + g (uq ) + w 1 + s 1 − + − where w ∈ W (uq ), w ∈ W (uq ),and s ∈ R. Furthermore, due to the assumption + 1 − 1 that λ is the only inert exponent, after possibly making a small change to u and u (see Section 6), we may write w = v + i i=2 where v ∈ V (H )(uq ), and furthermore, v /u q − uq is bounded from below. i i 1 2 1 Then, q − q will be approximately in the direction of V (H )(q ), see Section 8 for 2 2 2 the details. Let f (x) denote the conditional measure of ν along (V + V )(H )[x]. (This condi- 2 1 2 tional measure can be defined since ν is U -invariant.) Let q = g q and q = g q where 3 s 1 s 3 1 s > 0 is such that the amount of expansion along V (H ) from q to q is equal to the 2 1 3 amount of expansion along V (H ) from uq to q . Then, as in [BQ], 2 1 2 (2.4) f (q ) = A f (q ), and f q = A f q , 2 2 ∗ 2 3 2 2 2 ∗ 3 INVARIANT AND STATIONARY MEASURES 105 where A and A are essentially the same bounded linear map. But q and q approach each other, so that f (q ) ≈ f q . 2 3 2 Hence (2.5) f (q ) ≈ f q . 2 2 2 Taking a limit as →∞ of the points q and q we obtain points q˜ and q˜ in the same 2 2 2 2 leaf of (V + V )(H ) and distance apart such that 1 2 (2.6) f (q˜ ) = f q˜ . 2 2 2 This means that the conditional measure f (q˜ ) is invariant under a shift of size approx- 2 2 imately . Repeating this argument with → 0 we obtain a point p such that f (p) is invariant under arbitrarily small shifts. This implies that the conditional measure f (p) restricts to Lebesgue measure on some subspace U of (V + V )(H ), which is distinct new 1 2 + + from theorbitof N.Thus, we canenlarge U to be U ⊕ U . new TechnicalProblem #1.— The argument requires that all eight points q, q , q , q , q , q , q , q belong to some “nice” set K of almost full measure. We will give a very 2 3 2 3 rough outline of the solution to this problem here; a more detailed outline is given at the beginning of Section 5. We have the following elementary statement: Lemma 2.2. —If ν − is non-trivial, then for any δ> 0 there exist constants c(δ) > 0 and ρ(δ) > 0 such that for any compact K ⊂ X with ν(K)> 1 − δ there exists a compact subset K ⊂ K with ν(K )> 1 − c(δ) so that for any q ∈ K there exists q ∈ K ∩ W [q] with ρ(δ) < d q, q < 1/100. Furthermore, c(δ) → 0 as δ → 0. In other words, there is a set K ⊂ K of almost full measure such that every point q ∈ K has a “friend” q ∈ W [q],with q also in the “nice” set K, such that d q, q ≈ 1/100. Thus, q can be chosen essentially anywhere in X . (In fact we use a variant of Lemma 2.2, namely Proposition 5.3 in Section 5.) We also note the following trivial statement: 106 ALEX ESKIN, MARYAM MIRZAKHANI FIG.2.—(a)Wekeeptrack of therelativepositionofthe subspaces U [x] and U in part by picking a transversal Z(x) to + + U [x], and noting the distance between U [x] and U along Z[x].(b) If we apply the flow g to the entire picture in (a), we + + see that the transversal g Z[x] can get almost parallel to g U [x]. Then, the distance between g U [x] and g U along g Z[x] t t t t t may be much larger then the distance between g x ∈ g U [x] and the closest point in g U t t t Lemma 2.3. — Suppose ν is a measure on X invariant under the flow g .Let τˆ : X × 0 t 0 R → R be a function such that there exists κ> 1 so that for all x ∈ X and for t > s, −1 (2.7) κ (t − s)≤ˆτ(x, t)−ˆτ(x, s) ≤ κ(t − s). Let ψ : X → X be given by ψ (x) = g x. Then, for any K ⊂ X and any δ> 0, there exists t 0 0 t τ( ˆ x,t) 0 asubset E ⊂ R of density at least (1 − δ) such that for t ∈ E, −1 c 2 c ν ψ K ≤ κ /δ ν K . (We remark that the maps ψ are not a flow, since ψ is not in general ψ ◦ ψ .However, t t+s t s Lemma 2.3 still holds.) In Section 7 we show that roughly, q = ψ (q),where ψ is as in Lemma 2.3. 2 t t (A more precise statement, and the strategy for dealing with this problem is given at the beginning of Section 5.) Then,tomakesurethat q avoids a “bad set” K of small −1 measure, we make sure that q ∈ ψ (K) which by Lemma 2.3 has almost full measure. Combining this with Lemma 2.2, we can see that we can choose q, q and q all in an a priori prescribed subset K of almost full measure. A similar argument can be done for all eight points, see Section 12, where the precise arguments are assembled. TechnicalProblem #2.— Beyond the first step of the induction, the subspace U (x) may not be locally constant as x varies along W (x). This complication has a ripple effect on the proof. In particular, instead of dealing with the divergence of the points g uq and g u q we need to deal with the divergence of the affine subspaces U [g uq ] and t 1 t t 1 + + + + U [g u q ]. As a first step, we project U [g u q ] to the leaf of W containing U [g uq ], t t t 1 1 1 to get a new affine subspace U . One way to keep track of the relative location of U = + + U [g u q ] and U is (besides keeping track of the linear parts of U and U )topick atransversal Z(x) to U [x], and to keep track of the intersection of U and Z(x), see Figure 2. However, since we do not know at this point that the cocycle is semisimple, we cannot pick Z in a way which is invariant under the flow. Thus, we have no choice except to pick some transversal Z(x) to U (x) at ν -almost every point x ∈ X , and then deal with the need to change transversal. INVARIANT AND STATIONARY MEASURES 107 It turns out that the formula for computing how U ∩ Z changes when Z changes is non-linear (it involves inverting a certain matrix). However, we would really like to work with linear maps. This is done in two steps: first we show that we can choose the approximation U and the transversals Z(x) in such a way that changing transversals involves inverting a unipotent matrix. This makes the formula for changing transversals polynomial. In the second step, we embed the space of parameters of affine subspaces near U [x] into a certain tensor power space H(x) so that on the level of H(x) the change of transversal map becomes linear. The details of this construction are in Section 6. TechnicalProblem #3.— There may be more than one U -inert Lyapunov expo- nent. In that case, we do not have precise control over how q and q diverge. In particular the assumption that q − q is nearly in the direction of V (H )(q ) is not justified. Also 2 2 2 + + we really need to work with U [q ] and U [q ].Solet v ∈ H(q ) denote the vector corre- 2 2 + + sponding to (the projection to W (q ) of) the affine subspace U [q ]. (This vector v takes on the role of q − q .) We have no a-priori control over the direction of v (even though we know that v≈ ,and we know that v is almost contained in E(q ) ⊂ H(q ),where E(x) 2 2 is defined in Section 8 as the union of the Lyapunov subspaces of H(x) corresponding to the U -inert Lyapunov exponents). Theideaistovary u (while keeping q , q , fixed). To make this work, we need to define a finite collection of subspaces E (x) of H(x) (which actually only make sense [ij],bdd on a certain finite measurable cover X of X )suchthat (a) By varying u (while keeping q , q , fixed) we can make sure that the vector v becomes close to one of the subspaces E ,and [ij],bdd (b) For a suitable choice of point q = q = g q ,the map 3 3,ij s 1 ij (g ug ) E (q ) → E (q ) t −s ∗ [ij],bdd 3 [ij],bdd 2 ij is a linear map whose norm is bounded independently of the parameters. (c) Also, for a suitable choice of point q = q = g q ,the map s 1 3 3,ij ij (g ug ) E q → E q t −s ∗ [ij],bdd [ij],bdd 3 2 ij is a linear map whose norm is bounded independently of the parameters. For the precise conditions see Proposition 10.1 and Proposition 10.2. This construction is done in detail in Section 10. The general idea is as follows: Suppose v ∈ E (x) ⊕ E (x) i j where E (x) and E (x) are the Lyapunov subspaces corresponding to the U -inert (simple) i j Lyapunov exponents λ and λ . Then, if while varying u,the vector v does not swing i j towards either E or E , we say that λ and λ are “synchronized”. In that case, we consider i j i j the subspace E (x) = E (x) ⊕ E (x) and show that (b) and (c) hold. [i] i j The conditions (b) and (c) allow us to define in Section 11 conditional measures f on W (x) which are associated to each subspace E . In fact the measures are sup- ij [ij],bdd + + ported on the points y ∈ W [x] such that the affine subspace U [ y] maps to a vector in E (x) ⊂ H(x). [ij],bdd 108 ALEX ESKIN, MARYAM MIRZAKHANI Technical Problem #4. — More careful analysis (see the discussion following the state- ment of Proposition 11.4) shows that the maps A and A of (2.4)are notexactly thesame. Then, when one passes to the limit →∞ one gets, instead of (2.6), f (q˜ ) = P q˜ , q˜ f q˜ ij 2 2 ij 2 2 + + + where P : W (q˜ ) → W (q˜ ) is a certain unipotent map (defined in Section 4.2). Thus the conditional measure f (q˜ ) is invariant under the composition of a translation of size ij 2 and a unipotent map. Repeating the argument with → 0 we obtain a point p such that the conditional measure at p is invariant under arbitrarily small combinations of (translation + unipotent map). This does not imply that the conditional measure f (p) ij restricts to Lebesgue measure on some subspace of W , but it does imply that it is in the Lebesgue measure class along some polynomial curve in W . More precisely, for ν -a.e x ∈ X there is a subgroup U = U (x) of the affine group of W (x) such that new new the conditional measure of f (x) on the polynomial curve U [x]⊂ W [x] is induced ij new from the Haar measure on U . (We call such a set a “generalized subspace”.) The exact new definition is given in Section 6. Thus, during the induction steps, we need to deal with generalized subspaces. This is not a very serious complication since the general machinery developed in Section 6 can deal with generalized subspaces as well as with ordinary affine subspaces. Completion of the proof of Theorem 2.1.— Let L(x) ⊂ H (M,, R) be the smallest subspace such that ν − is supported on L (x). Roughly, the above argument can be W (x) iterated until we know the conditional measure ν + is Lebesgue on a subspace U [x], W (x) where U(x) ⊂ H (M,, R) contains L(x). (The precise condition for when the induc- tion stops is given by Lemma 6.15 and Proposition 6.16.) Then a Margulis-Tomanov style entropy comparison argument (see Section 13)shows that U(x) = L(x),and the − + conditional measures along L (x) are Lebesgue. Since U (x) contains the orbit of the unipotent direction N, this implies that L (x) contains the orbit of the opposite unipotent ¯ ¯ direction N ⊂ SL(2, R). Thus, the conditional measure along the orbit of N is Lebesgue, which means that ν is N-invariant. This, together with the assumption that ν is P = AN- invariant implies that ν is SL(2, R)-invariant, completing the proof of Theorem 2.1. 3. Hyperbolic properties of the geodesic flow The spaces X and X .— Let X be afinitecover of thestratum H (α) which is 0 0 0 1 a manifold. (Such a cover may be obtained by choosing a level 3 structure, i.e. a basis for the mod 3 homology of the surface.) Let X be the universal cover of X . Then the 0 0 fundamental group π (X ) acts properly discontinuously on X .Let ν be a P-invariant 1 0 0 ergodic probability measure on X . We recall the following standard fact: INVARIANT AND STATIONARY MEASURES 109 Lemma 3.1 (Mautner phenomenon). — Let ν be an ergodic P-invariant measure on a space Z. Then ν is A-ergodic. Proof.—Seee.g.[Moz]. + + Lemma 3.2. — For almost all x ∈ X , the affine exponential map from W (x) to W [x] is globally defined and is bijective, endowing W [x] with a global affine structure. The same holds for W [x]. − + Proof. — Since W and W play the role of the stable and unstable foliations for the action of g ∈ A (cf. Lemma 3.5), this follows from the Poincaré recurrence theorem. 1 1 The bundle H .— Let H denote the bundle whose fiber above x ∈ X is 1 1 H (M,, R). We denote the fiber above the point x ∈ X by H (x). The geodesic flow acts on H by parallel transport using the Gauss-Manin connec- tion (see Section 2.3). 1 1 1 1 The bundles H and H .— Let H denote the same bundle as H except that the + − + 1 t action of g on H includes an extra multiplication by e on the fiber. (In other words, if t 1 1 h (x,v) = (x, e v) and i : H → H is the identity map, then g ◦ i(x,v) = h ◦ i ◦ g (x,v).) t t t t 1 1 Similarly, let H denote the same bundle as H except that the action of g includes an −t extra multiplication by e on the fiber. 1 1 We use the notation H (x) and H (x) to refer to the fiber of the corresponding + − bundle above the point x ∈ X . (+) (−) (++) (−−) The bundles H , H , H , H and H .— In this paper, we will need to deal big big big big big with several bundles derived from the Hodge bundle H . It is convenient to introduce a bundle H so that every bundle we will need will be a subbundle of H .Let d ∈ N be big big a large integer chosen later (it will be chosen in Section 6 and will depend only on the Lyapunov spectrum of the Kontsevich-Zorich cocycle). Let j k−j d k 1 1 H (x) = H (x) ⊗ H (x) , big k=1 j=1 i=1 l=1 d k j k−j (+) 1 1 H (x) = H (x) ⊗ H (x) , big + + k=1 j=1 i=1 l=1 j k−j d k (−) 1 1 H (x) = H (x) ⊗ H (x) , big − − k=1 j=1 i=1 l=1 110 ALEX ESKIN, MARYAM MIRZAKHANI and let (+) (−) ˜ ˆ ˆ ˆ H (x) = H (x) ⊕ H (x) ⊕ H (x). big big big big Suppose L ⊂ L ⊂ H are g -invariant subbundles. We say that L /L is an admissi- 1 2 big t 2 1 ble quotient if the cocycle on L /L is measurably conjugate to a conformal cocyle (see 2 1 Lemma 4.3), and also L /L is maximal in the sense that if L ⊃ L and L ⊂ L are 2 1 2 1 2 1 g -invariant subbundles with the cocycle L /L measurably conjugate to a conformal co- 2 1 cycle, then L = L and L = L .Wethenlet denote the set of all admissible quotients 2 1 big 2 1 of H and let big H (x) = Q(x). big Q∈ big (+) (−) (+) (−) ˆ ˆ (We apply a similar operation to the bundles H and H to get bundles H and H .) big big big big The flow g acts on the bundle H in thenaturalway.Wedenotethe action on the t big (++) fibers by (g ) .Let H (x) denote the direct sum of the positive Lyapunov subspaces of t ∗ big (−−) H (x). Similarly, let H (x) denote the direct sum of the negative Lyapunov subspaces big big of H (x). big (++) Lemma 3.3. — The subspaces H (x) are locally constant along W [x], i.e. for almost all big (++) (++) x ∈ X and almost all y ∈ W [x] close to x we have H (y) = H (x). Similarly, the subspaces big big (−−) H (x) are locally constant along W [x]. big Proof. — Note that 1 (g ) v −t ∗ (++) H (x) = v ∈ H (x) : lim log < 0 big big t→∞ t v (++) Therefore, the subspace H (x) depends only on the trajectory g x as t →∞.How- −t big ever, if y ∈ W [x] then g y will for large t be close to g x, and so in view of the affine −t −t structure, (g ) will be the same linear map on H (x) and H (y). This implies that −t ∗ big big (++) (++) H (x) = H (y). big big The Avila-Gouëzel-Yoccoz norm. — The Avila-Gouëzel-Yoccoz norm on the relative cohomology group H (M,, R) is described in Appendix A. This then induces a norm which we will denote by · and then, as the projective cross norm, also on H .We Y big also use the notation · to denote the AGY norm at x ∈ X . Y,x 0 + + 1 The distance d (x, y).— Since the tangent space to W [x] is included in H (M, 1 + , R),the AGYnormon H (M,, R) defines a distance on W [x].Wedenotethis + + + distance by d (·,·).(Thus,for y, z ∈ W [x], d (y, z) is the length of the shortest path in W [x] connecting y and z, where lengths of paths are measured using the AGY norm.) INVARIANT AND STATIONARY MEASURES 111 + + + The ball B (x, r).— Let B (x, r) ⊂ W [x] denote the ball of radius r centered at x, in the metric d (·,·). The following is a rephrasing of [AG, Proposition 5.3]: Proposition 3.4. —For all x ∈ X ,x + v is well defined for v ∈ W (x) with v ≤ 1/2. 0 Y Also, for all y, z ∈ B (x, 1/50), we have y − z ≤y − z ≤ 2y − z , Y,y Y,z Y,y and y − z ≤ d (y, z) ≤ 2y − z . Y,y Y,y − − Note that we have a similar distance d (·,·) on W [x], and the analogue of Propo- sition 3.4 holds. The “distance” d (·,·).— Suppose x, y ∈ X are not far apart. Then, there exist + − unique z ∈ W [x] and t ∈ R such that g z ∈ W [ y].Wethendefine X + − d (x, y) = d (x, z) +|t|+ d (g z, y). + X + − X − 0 0 Thus, if y ∈ W [x] then d (x, y) = d (x, y), and if y ∈ W [x],then d (x, y) = d (x, y). We sometimes abuse notation by using the notation d (x, y) where x, y ∈ X .By this we mean d (x˜, y˜) where x˜ and y˜ are appropriate lifts of x and y. Choose a compact subset K ⊂ X with ν(K ) ≥ 5/6. Let K ={x ∈ X : 0 thick 0 thick thick d (x, K ) ≤ 1/100}. thick Lemma 3.5. — There exists α> 0 such that the following holds: (a) Suppose x ∈ X and t > 0 are such that the geodesic segment from x to g xspends at least 0 t half the time in K .Then, forall v ∈ W (x), thick −αt (g ) v ≤ e v . t ∗ Y Y (b) Suppose x ∈ X and t > 0 are as in (a). Then, for all v ∈ W (x), αt (g ) v ≥ e v . t ∗ Y Y (c) For every > 0 there exist a compact subset K ⊂ X with ν(K )> 1 − and thick thick (++) t > 0 such that for x ∈ K ,t > t and all v ∈ H (x), 0 0 thick big αt (g ) v ≥ e v . t ∗ Y Y 112 ALEX ESKIN, MARYAM MIRZAKHANI (d) For all v ∈ W (x),all x ∈ X and all t > 0, (g ) v ≥v . t ∗ Y Y Proof. — Parts (a), (b) and (d) follow from Theorem A.2. Part (c) follows immediately from the Osceledets multiplicative ergodic theorem. We also have the following simpler statement: Lemma 3.6. — There exists N > 0 such that for all x ∈ X ,all t ∈ R, and all v ∈ H (x), 0 big −N|t| N|t| e v ≤(g ) v ≤ e v . Y t ∗ Y Y For v ∈ W [x],wecan take N = 2. Proof. — This follows immediately from Theorem A.2. Proposition 3.7. — Suppose C ⊂ X is a set with ν(C)> 0,and T : C → R is a 0 0 − −1 measurable function which is finite a.e. Then we can find x ∈ X ,asubset C ⊂ W [x ]∩ π (C) 0 0 1 0 + + and for each c ∈ C asubset E [c]⊂ W [c] of diameter in the AGY metric at most 1/200 and a number t(c)> 0 such that if we let J = g E [c], c −t 0≤t<t(c) then the following holds: + + (a) E [c] is relatively open in W [c]. (b) π( J ) ∩ π( J ) =∅ if c = c . c c (c) π( J ) is embedded in X , i.e. if π(g x) = π(g x ) where x, x ∈ E [c] and 0 ≤ t < c 0 −t −t t(c), 0 ≤ t < t(c) then x = x and t = t . (d) π( J ) is conull in X . c 0 c∈C + + (e) For every c ∈ C there exists c ∈ C such that π(g E [c]) ⊂ π(E [c ]). 1 1 −t(c) (f) t(c)> T (c) for all c ∈ C . 0 1 Remark. — All the construction in Section 3 will depend on the choice of C and T , but we will suppress this from the notation. The set C and the function T will be finally chosen in Lemma 4.14. The proof of Proposition 3.7 relies on the following: Lemma 3.8. —Suppose C ⊂ X is a set with ν(C)> 0,and T : C → R is a measurable 0 0 − −1 function which is finite a.e. Then we can find x ∈ X ,asubset C ⊂ W [x ]∩ π (C) and for each 0 0 1 0 + + c ∈ C asubset E [c]⊂ W [c] of diameter in the AGY metric at most 1/200 so that the following hold: INVARIANT AND STATIONARY MEASURES 113 + + (0) E [c] is a relatively open subset of W [c]. + + (1) The set E = π( E [c]) is embedded in X , i.e. if π(x) = π(x ) where x ∈ E [c] c∈C and x ∈ E [c ],then x = x and c = c . (2) For some > 0, ν( g E)> 0. t∈(0,) + + + (3) If t > 0 and c ∈ C is such that π(g E [c]) ∩ E = ∅,then π(g E [c]) ⊂ π(E [c ]) 1 −t −t for some c ∈ C . (4) Suppose t, c, c are as in (3).Then t > T (c). Proof. — This proof is essentially identical to the proof of Lemma B.1,exceptthat we need to take care that (4) is satisfied. In this proof, for x ∈ C,wedenoteby ν the W [x] conditional measure of ν along W [x]∩ C . Choose T > 0so that if we let C ={x ∈ C : T (x)< T } then ν(C )>ν(C)/2. 1 4 0 1 4 Let X denote the union of the periodic orbits of g . By the P-invariance of ν and per t the ergodicity of g , ν(X ) = 0, and the same is true of the set X = W [x]. t per per x∈X per −1 − −1 Therefore there exists x ∈ π (C ) and a compact subset C ⊂ W [x ]∩ π (C ) with 0 4 3 0 4 ν (C )> 0such that for x ∈ C and 0 < t < T , π(g x)/ ∈ π(C ). Then, since W [x ] 3 3 1 −t 3 + + C is compact, we can find a small neighborhood V ⊂ W of the origin such that + + the set π( V [c]) is embedded in X and for x ∈ V [c] and 0 < t < T , 0 1 c∈C c∈C 3 3 π(g x)/ ∈ π( V [c]). −t c∈C There exists C ⊂ C with ν − (C )> 0and N > T such that for all c ∈ C and 2 3 W [x ] 2 1 2 all T > N, t ∈[0, T]: π(g c) ∈ K ≥ T/2. −t thick Then, for c ∈ C ,T > Nand x ∈ V [c], t ∈[0, T]: π(g x) ∈ K ≥ T/2. −t thick Let Y,x + + + M = sup : x ∈ V [c], y ∈ V [c], c ∈ C ,v ∈ W (x) Y,y 2 −αN Let α> 0 be as in Lemma 3.5, and choose N > Nsuch thatM e < 1/10. Then, for + + c ∈ C , x, y ∈ π(V [c]) and t > N such that g x ∈ π( V [c]), in view of Lemma 3.5 2 1 −t c∈C and Proposition 3.4, X X 0 0 d (g x, g y) ≤ d (x, y). −t −t Now choose C ⊂ C with ν − (C )> 0so that if we let Y = π( V [x]) then g Y∩ 1 2 W [x ] 1 −t 0 c∈C Y =∅ for 0 < t < max(T , N ), in other words, the first return time to Y is at least 1 1 max(T , N ). (This can be done e.g. by Rokhlin’s Lemma.) Condition (4) now follows 1 1 114 ALEX ESKIN, MARYAM MIRZAKHANI since T (c)< T for all c ∈ C . The rest of the proof is essentially the same as the proof of 0 1 1 Lemma B.1, applied to the first return map of g to Y. −t Proof of Proposition 3.7.—For x ∈ E, let t(x) ∈ R be the smallest such that g x ∈ E. By property (3), the function t(x) is constant on each set of the form π(E [c]). −t(x) Let F ={x ∈ E : t(x) = t}.(We have F =∅ if t < N .) By property (2) and the ergodicity t t 1 of g , up to a null set, −t X = g F . 0 −s t t>0 s<t Then properties (a)–(f) are easily verified. Notation. — For x ∈ X ,let J[x] denote the set π( J ) containing x.For x ∈ X ,let 0 c 0 −1 J[x] denote γ J where γ ∈ π (X ) is such that γ x ∈ J . c 1 0 c Lemma 3.9. — Suppose x ∈ X ,y ∈ W [x]∩ J[x].Thenfor any t > 0, g y ∈ J[g x]∩ W [g x]. −t −t −t Proof. — This follows immediately from property (e) of Proposition 3.7. Notation. — For x ∈ X ,let + −1 B [x]= π g J[g x˜]∩ W [g x˜] , where x˜ is any element of π (x). t −t t t Lemma 3.10. (a) For t > t ≥ 0, B [x]⊂ B [x]. t t (b) Suppose t ≥ 0, t ≥ 0,x ∈ X and x ∈ X are such that B [x]∩ B [x ]=∅.Then 0 0 t t either B [x]⊇ B [x ] or B [x ]⊇ B [x] (or both). t t t t Proof. — Part (a) is a restatement of Lemma 3.9. For (b), without loss of generality, we may assume that t ≥ t. Then, by (a), we have B [x]∩ B [x ]=∅. t t Suppose y ∈ B [x]∩ B [x ].Then g y ∈ B [g x] and g y ∈ B [g x ]. Since the sets t t t 0 t t 0 t B [z], z ∈ X form a partition, we must have B [g x]= B [g x ]. Therefore, B [x]= 0 0 0 t 0 t t B [x ],and thus,by(a), B x ⊂ B x = B [x]. t t t By construction, the sets B [x] arethe atomsofameasurablepartition of X sub- 0 0 ordinate to W (see Definition B.4). Then, let ν + denote the conditional measure of W [x] ν along the atom of the partition containing x. For notational simplicity, for E ⊂ W [x], we sometimes write ν +(E) instead of ν + (E). W W [x] INVARIANT AND STATIONARY MEASURES 115 Lemma 3.11. —Suppose δ> 0 and K ⊂ X is such that ν(K)> 1 − δ. Then there exists ∗ ∗ 1/2 ∗ asubset K ⊂ K with ν(K )> 1 − δ such that for any x ∈ K ,and any t > 0, 1/2 + + ν K ∩ B [x] ≥ 1 − δ ν B [x] . W t W t c ∗ Proof.—Let E = K ,so ν(E) ≤ δ.Let E denote the set of x ∈ X such that there exists some τ ≥ 0with 1/2 (3.1) ν + E ∩ B [x] ≥ δ ν + B [x] . W τ W τ ∗ 1/2 It is enough to show that ν(E ) ≤ δ .Let τ(x) be the smallest τ> 0so that (3.1) holds for x. Then the (distinct) sets {B [x]} cover E and are pairwise disjoint by τ(x) x∈E Lemma 3.10(b). Let F = B [x]. τ(x) x∈E Then E ⊂ F. For every set of the form B [ y],let (y) denote the set of distinct sets B [x] where x varies over B [ y]. Then, by (3.1) τ(x) 0 ν + F ∩ B [ y] = ν +(B ) W 0 W τ(x) (y) −1/2 −1/2 ≤ δ ν + E ∩ B [x] ≤ δ ν + E ∩ B [ y] . W τ(x) W 0 (y) −1/2 Integrating over y,weget ν(F) ≤ δ ν(E).Hence, ∗ −1/2 1/2 ν E ≤ ν(F) ≤ δ ν(E) ≤ δ . 4. General cocycle lemmas 4.1. Lyapunov subspaces and flags. — Let V (H )(x),1 ≤ i ≤ k denote the Lyapunov subspaces of the Kontsevich-Zorich cocycle under the action of the geodesic flow g ,and let λ (H ),1 ≤ i ≤ k denote the (distinct) Lyapunov exponents. Then we have for almost all x ∈ X , 1 1 H (M,, R) = V H (x) i=1 and for all non-zero v ∈ V (H )(x), 1 (g ) v t ∗ lim log = λ H , t→±∞ t v 116 ALEX ESKIN, MARYAM MIRZAKHANI where · is any reasonable norm on H (M,, R) for example the Hodge norm or the AGY norm defined in Section A.1. By the notation (g ) v we mean the action of the t ∗ geodesic flow (i.e. parallel transport using the Gauss-Manin connection) on the Hodge bundle H (M,, R). We note that the Lyapunov exponents of the geodesic flow (viewed as a diffeomorphism of X )are in fact 1 + λ ,1 ≤ i ≤ k and −1 + λ ,1 < i ≤ k. 0 i i We have 1 1 1 1 = λ H >λ H > ··· >λ H =−1. 1 2 k 1 1 1 It is a standard fact that dim V (H ) = dim V (H ) = 1, V (H ) corresponds to the 1 k 1 direction of the unipotent N and V (H ) corresponds to the direction of N. Let p : 1 1 H (M,, R) → H (M, R) denote the natural map. Recall that if x ∈ X denotes the pair (M,ω),then 1 1 H (x) = α ∈ H (M,, R) : p(α) ∧ Re(ω) = p(α) ∧ Im(ω) = 0 . Then k−1 1 1 H (x) = V H (x). i=2 We note that the subspaces H (x) are equivariant under the SL(2, R) action on X (since so is the subspace spanned by Reω and Imω). Since the cocycle preserves the symplectic form on p(H ),wehave 1 1 λ H =−λ H , 1 ≤ i ≤ k. k+1−i i Let i k 1 1 1 1 V H (x) = V H (x), V H (x) = V H (x). ≤i j ≥i j j=1 j=i Then we have the Lyapunov flags 1 1 1 1 {0}= V H (x) ⊂ V H (x) ⊂··· ⊂ V H (x) = H (M,, R) ≤0 ≤1 ≤k and 1 1 1 1 {0}= V H (x) ⊂ V H (x) ⊂··· ⊂ V H (x) = H (M,, R). >k >k−1 >0 We record some simple properties of the Lyapunov flags: INVARIANT AND STATIONARY MEASURES 117 Lemma 4.1. 1 + (a) The subspaces V (H )(x) are locally constant along W [x], i.e. for almost all x ∈ X , ≤i 0 + 1 1 for almost all y ∈ W [x] close to x we have V (H )(y) = V (H )(x) for all 1 ≤ i ≤ k. ≤i ≤i 1 1 (Here and in (b) we identify H (x) with H (y) using the Gauss-Manin connection.) 1 − (b) The subspaces V (H )(x) are locally constant along W [x], i.e. for almost all x ∈ X and ≥i 0 − 1 1 for almost all y ∈ W [x] close to x we have V (H )(y) = V (H )(x) for all 1 ≤ i ≤ k. ≥i ≥i Proof.—To prove(a),notethat 1 (g ) v −t ∗ 1 1 V H (x) = v ∈ H (M,, R) : lim log ≤−λ . ≤i i t→∞ t v Therefore, the subspace V (H )(x) depends only on the trajectory g x as t →∞.How- ≤i −t ever, if y ∈ W [x] then g y will for large t be close to g x, and so in view of the affine −t −t structure, (g ) will be the same linear map on H (M,, R) at x and y,asinSection 3. −t ∗ 1 1 This implies that V (H )(x) = V (H )(y). The proof of property (b) is identical. ≤i ≤i 1 1 1 1 The action on H and H .— Recall that the bundles H and H were defined in + − + − Section 3. All of the results of Section 4.1 also apply to these bundles. Also, 1 1 1 1 λ H = 1 + λ H ,λ H =−1 + λ H . i i i i + − Furthermore, under the natural identification by the identity map, for all x ∈ X , 1 1 1 V H (x) = V H (x) = V H (x). i i i + − (++) 4.2. Equivariant measurable flat connections. — Let L be a subbundle of H .Re- big call that by Lemma 3.2, typical leaves of W are simply connected. By an equivariant measurable flat W -connection on L we mean a measurable collection of linear “parallel transport” maps: F(x, y) : L(x) → L(y) defined for ν -almost all x ∈ X and ν + almost all y ∈ W [x] such that 0 W [x] (4.1)F(y, z)F(x, y) = F(x, z), and (4.2) (g ) ◦ F(x, y) = F(g x, g y) ◦ (g ) . t ∗ t t t ∗ For example, if L = W (x), then the Gauss-Manin connection (which in period local + 1 coordinates is the identity map) is an equivariant measurable flat W connection on H . + 1 However, there is another important equivariant measurable flat W -connection on H which we describe below. 118 ALEX ESKIN, MARYAM MIRZAKHANI + − 1 1 The maps P (x, y) and P (x, y).— Recall that V (H )(x) ⊂ H (x) are the Lyapunov subspaces for the flow g . Recall that the V (H )(x) are not locally constant along leaves t i + 1 + of W , but by Lemma 4.1, the subspaces V (H )(x) = V (W )(x) are locally con- ≤i j j=1 + + 1 stant along the leaves of W . Now suppose y ∈ W [x].Any vector v ∈ V (H )(x) can be written uniquely as 1 1 v = v + v v ∈ V H (y), v ∈ V H (y). i <i + 1 1 + Let P (x, y) : V (H )(x) → V (H )(y) be the linear map sending v to v .Let P (x, y) be i i + 1 the unique linear map which restricts to P (x, y) on each of the subspaces V (H )(x).We call P (x, y) the “parallel transport” from x to y. The following is immediate from the definition: Lemma 4.2. — Suppose x, y ∈ W [z]. Then + 1 1 (a) P (x, y)V (H )(x) = V (H )(y). i i + + −1 (b) P (g x, g y) = (g ) ◦ P (x, y) ◦ (g ) . t t t ∗ ∗ + 1 1 1 1 (c) P (x, y)V (H )(x) = V (H )(y). If we identify H (x) with H (y) using the Gauss- ≤i ≤i Manin connection, then the map P (x, y) is unipotent. + + + (d) P (x, z) = P (y, z) ◦ P (x, y). + 1 1 1 Note that the map P on H is thesameason H , provided we identify H with + + H via the identity map. The statements (b) and (d) imply that the maps P (x, y) define an equivariant mea- + 1 surable flat W -connection on H . This connection is in general different from the Gauss- Manin connection, and is only measurable. − − If y ∈ W [x],thenwecan defineasimilarmap whichwedenoteby P (x, y). This − 1 yields an equivariant measurable flat W -connection on H . + + Clearly the connection P (x, y) induces an equivariant measurable flat W -con- (++) nection on H . This connection preserves the Lyapunov subspaces of the g -action on big (++) H ,asinLemma 4.2(a). In view of Proposition 4.12 below, the connection P (x, y) also big induces an equivariant measurable flat W -connection on any g -equivariant subbundle (++) of H . big + + + Equivariant measurable flat U -connections. — Suppose U [x]⊂ W [x] is a g -equi- + + + variant family of algebraic subsets, with U [ y]= U [x] for y ∈ U [x]. In fact, we will only consider families compatible with ν as defined in Definition 6.2.Wedenotethe con- ditional measure of ν along U [x] by ν + . In the cases we will consider, these measures U [x] are well defined a.e. and are in the Lebesgue measure class, see Section 6. (++) By an equivariant measurable flat U -connection on a bundle L ⊂ H we mean big a measurable collection of linear maps F(x, y) : L(x) → L(y) satisfying (4.1)and (4.2), defined for ν -almost all x ∈ X and ν + -almost all y ∈ U [x]. 0 U [x] INVARIANT AND STATIONARY MEASURES 119 4.3. The Jordan canonical form of a cocycle. Zimmer’s amenable reduction. — The following is a general fact about linear cocycles over an action of R or Z. It is often called “Zimmer’s amenable reduction”. We state it only for the cases which will be used. (++) Lemma 4.3. — Suppose L is a g -equivariant subbundle of H . (For example, we could i t big have L (x) = V (H )(x).) Then, there exists a measurable finite cover σ : X → X such that for i i L L 0 + i i −1 σ (ν)-a.e x ∈ X there exists an invariant flag L i (4.3) {0}= L (x) ⊂ L (x) ⊂··· ⊂ L (x) = L (x), i,0 i,1 i,n i and on each L (x)/L (x) there exists a nondegenerate quadratic form ·,· and a cocycle λ : ij i,j−1 ij,x ij X × R → R such that for all u,v ∈ L (x)/L (x), L ij i,j−1 λ (x,t) ij (g ) u,(g ) v = e u,v . t ∗ t ∗ ij,x ij,g x −1 (Note: For each i, the pullback measures σ (ν) is uniquely defined by the condition that for almost all −1 −1 x ∈ X , the conditional of σ (ν) on the (finite) set σ (x ) is the normalized counting measure.) 0 0 0 L L i i Remark. — The statement of Lemma 4.3 is the assertion that on the finite cover X one can make a change of basis at each x ∈ X so that in the new basis, the matrix of the cocycle restricted to L is of the form ⎛ ⎞ C ∗ ... ∗ i,1 ⎜ ⎟ 0C ... ∗ i,2 ⎜ ⎟ (4.4) , ⎜ ⎟ . . . . . ⎝ ⎠ . . ∗ 00 ... C i,n where each C is a conformal matrix (i.e. is the composition of an orthogonal matrix and i,j a scaling factor λ ). ij We call a cocycle block-conformal if all the off-diagonal entries labeled ∗ in (4.4)are 0. Proof of Lemma 4.3.—See[ACO] (which uses many of the ideas of Zimmer). The statement differs slightly from that of [ACO, Theorem 5.6] in that we want the cocycle in each block to be conformal (and not just block-conformal). However, our statement is in fact equivalent because we are willing to replace the original space X by a finite cover X . 4.4. Covariantly constant subspaces. — The main result of this subsection is the fol- lowing: 120 ALEX ESKIN, MARYAM MIRZAKHANI Proposition 4.4. — Suppose L is a g -equivariant subbundle over the base X . We can write t 0 L(x) = L (x), where L (x) ≡ V (L)(x) is the Lyapunov subspace corresponding to the Lyapunov exponent λ . Suppose i i i there exists an equivariant flat measurable W -connection F on L, such that (4.5)F(x, y)L (x) = L (y). i i Suppose that M is a finite collection of subspaces of L which is g -equivariant. Then, for almost all x ∈ X and almost all y ∈ B [x], 0 0 F(x, y)M(x) = M(y), i.e. the collection of subspaces M is locally covariantly constant with respect to the connection F. Remark. — The same result holds if F is only assumed to be a measurable U -con- nection, and B [x] is replaced by B[x]. The following is a generalization of Lemma 4.1: Corollary 4.5. — Suppose M ⊂ H (M,, R) is a g -equivariant subbundle over the base X . t 0 Suppose also for a.e x ∈ X , V (x) ⊂ M(x) ⊂ V (x). Then (up to a set of measure 0), M(x) is 0 <i ≤i locally constant along W (x). Proof of Corollary 4.5. — By Lemma 4.1,L(x) ≡ V (x)/V (x) is locally constant ≤i <i along W [x].Let F(x, y) denote the Gauss-Manin connection (i.e. the identity map) on L(x). Note that the action of g on L(x) has only one Lyapunov exponent, namely λ . t i Thus, (4.5) is trivially satisfied. Then, by Proposition 4.4,M(x)/V (x) ⊂ L(x) is locally <i constant along W [x]. Since V (x) is also locally constant (by Lemma 4.1), this implies <i that M(x) is locally constant. Remark. — Our proof of Proposition 4.4 is essentially by reference to [L,Theo- rem 1]. It is given in Section 4.9 and can be skipped on first reading. For similar results in a partially hyperbolic setting see [AV2], [ASV], [KS]. 4.5. Some estimates on Lyapunov subspaces. — Let (V,· ) be a normed vector space. By a splitting E = (E ,..., E ) of V we mean a direct sum decomposition 1 n V = E ⊕ ··· ⊕ E 1 n Suppose E = (E ,..., E ) and E = (E ,..., E ) are two splittings of V, with dim E = 1 n i 1 n dim E for 1 ≤ i ≤ n. i INVARIANT AND STATIONARY MEASURES 121 We define D E, E = max sup inf : v + w ∈ E , and w ∈ E , 1≤i≤n v∈ E \{0} j≤i j>i j≤i and D E, E = max sup inf : v + w ∈ E , and w ∈ E . 1≤i≤n v∈ E \{0} Y j≥i j<i j≥i Note that D (E, E ) depends on E only via the flag E ,1 ≤ i ≤ n. Similarly, j≤i j − + D (E, E ) depends on E only via the flag E ,1 ≤ i ≤ n. Also D (E, E ) = j≥i j − + D (E, E ) = 0ifE = E ,and D (E, E ) =∞ if some E has non-trivial intersec- j≤i j tion with E . j>i In this subsection, we write V (x) for V (H )(x),etc.For almost all x in X ,wehave i i 0 the splitting H (x) = V (x) ⊕ ··· ⊕ V (x). 1 n GM For x, y ∈ X , we have the Gauss-Manin connection P (x, y), which is a linear 1 1 map from H (x) to H (y) (see Section 2.3). Let + + GM GM D (x, y) = D V (x),..., V (x) , P (y, x)V (y),..., P (y, x)V (y) . 1 n 1 n − − GM GM D (x, y) = D V (x),..., V (x) , P (y, x)V (y),..., P (y, x)V (y) . 1 n 1 n Distance between subspaces. — For a subspace V of H (x), let SV denote the intersec- tion of V with the unit ball in the AGY norm. For subspaces V , V of H (x),wedefine 1 2 (4.6) d (V , V ) = The Hausdorff distance between SV and SV Y 1 2 1 2 measured with respect to the AGY norm at x. Lemma 4.6. — There exists a continuous function C : X → R such that for subspaces 0 0 V , V of H (x) of the same dimension, 1 2 −1 C (x) d (V , V ) ≤ δ (V , V ) ≤ d (V , V ), 0 Y 1 2 Y 1 2 Y 1 2 where δ (V , V ) = max min v − v . Y 1 2 1 2 Y v ∈SV v ∈SV 1 1 2 2 122 ALEX ESKIN, MARYAM MIRZAKHANI Proof. — Sinced (V , V ) = max(δ (V , V ),δ (V , V )), the inequality on the Y 1 2 Y 1 2 Y 2 1 left follows immediately from the definition of the Hausdorff distance. To prove the inequality on the right it is enough to show that for some continuous function C : X → R , 0 0 −1 (4.7)C (x) δ (V , V ) ≤ δ (V , V ). 0 Y 2 1 Y 1 2 To prove (4.7), pick some arbitrary inner product ·,· on H (M,, R),and let · be 0 0 the associated norm. Then, there exists a continuous function C : X → R such that 1 0 for all v ∈ H (x), −1 C (x) v ≤v ≤ C (x)v . 1 0 Y 1 0 Let δ (·,·) and d (·,·) be the analogues of δ (·,·) and d (·,·) for the norm · . Then, 0 0 Y Y 0 it is enough to prove that there exists a constant c > 0 depending only on the dimension such that for subspaces V ,V of equal dimension, 1 2 (4.8) c δ (V , V ) ≤ δ (V , V ). 2 0 2 1 0 1 2 For subspaces U, V of equal dimension n,let u ,..., u and v ,...,v be orthonormal 1 n 1 n bases for U and V respectively. Then, we have n 1/2 n n 1/2 2 2 (4.9) inf u − v = n − u ,v i i j 0 0 v∈V i=1 i=1 j=1 Note that the expression on the left in (4.9) is independent of the basis for V, and the expression on the right of (4.9) is symmetric in U and V. Thus, the expression in (4.9) is independent of the basis for U as well, and thus defines a function d (U, V). (This function is called the Frobenius or chordal distance between subspaces, see e.g. [De], [WWF].) From the expression on the left of (4.9) it is clear that there exists a constant c depending only on the dimension so that −1 c d (V , V ) ≤ d (V , V ) ≤ c d (V , V ). 3 H 1 2 0 1 2 H 1 2 Since d (V , V ) = d (V , V ),(4.8) follows. H 1 2 H 2 1 Lemma 4.7. — There exists α> 0 depending only on the Lyapunov spectrum, and a function C : X → R finite almost everywhere such that the following holds: ˜ ˜ (a) For all t > 0,and all x ∈ X ,and all y ∈ X such that d (g x, g y) ≤ 1/100 for 0 0 s s 0 ≤ s ≤ t, we have, for all 1 ≤ i ≤ n, GM + −αt d V (g x), P (g y, g x)V (g y) ≤ min C(g x) 1 + D (x, y) e . Y ≤i t t t ≤i t s 0≤s≤t INVARIANT AND STATIONARY MEASURES 123 ˜ ˜ (b) For all t > 0,and all x ∈ X ,and all y ∈ X such that d (g x, g y) ≤ 1/100 for 0 0 −s −s 0 ≤ s ≤ t, we have, for all 1 ≤ i ≤ n, GM − −αt d V (g x), P (g y, g x)V (g y) ≤ min C(g x) 1 + D (x, y) e . Y ≥i −t −t −t ≥i −t −s 0≤s≤t The proof of Lemma 4.7 is a straightforward but tedious argument using the Osceledets multiplicative ergodic theorem. It is done in Section 4.8 . Lemma 4.8. — There exists a function C : X → R finite almost everywhere, such that for 3 0 − X + all x ∈ X ,all y ∈ W [x] with d (x, y)< 1/100 we have D (x, y) ≤ C (x)C (y). Similarly, 0 3 3 + X − for all x ∈ X ,all y ∈ W [x] with d (x, y)< 1/100 we have D (x, y) ≤ C (x)C (y). 0 3 3 Proof of Lemma 4.8.—For > 0, let K ⊂ X be a compact set with measure at least 1 − on which the functions x → V (x) are continuous. Then there exists ρ = ρ() −1 − −1 X + such that if x ∈ π (K ), y ∈ W [x]∩ π (K ) and d (x .y )<ρ then D (x , y )< 1. Then, by the Birkhoff ergodic theorem and Lemma 3.5, there exists a compact K ⊂ X −1 − with ν(K )> 1 − 2 and C = C () such that for all x ∈ π (K ),all y ∈ W [x]∩ 2 2 −1 X π (K ) with d (x, y)< 1/100 there exists C () < t < 2C () with g x ∈ K , g y ∈ K 2 2 t t X + + and d (x, y)<ρ().Thus, D (g x, g y)< 1, which implies that D (x, y)< C = C (). t t 2 2 Without loss of generality, we may assume that C ≥ 1and that K and C () both 2 2 decrease as functions of .Now for x ∈ X ,let ϒ(x) ={ : x ∈ K },and let C (x) = inf C () : ∈ ϒ(x) . The proof of the second assertion is identical. Corollary 4.9. — There exists a measurable function C : X → R finite a.e such that if 1 0 − X x ∈ X ,y ∈ W [x] with d (x, y)< 1/100, we have for all t > 0, − GM −αt (4.10) P (g x, g y)P (g y, g x) − I ≤ C (x)C (y)e , t t t t Y 1 1 where α> 0 depends only on the Lyapunov spectrum. Consequently, for almost all x ∈ X ,and almost all y ∈ W [x], − GM (4.11)lim P (g x, g y)P (g y, g x) − I = 0. t t t t Y t→∞ − + − + The same assertions hold if W is replaced by W ,g by g and P by P . t −t Proof of Corollary 4.9.—Let C (x) = C(x)C (x),where C(·) is as in Lemma 4.7 and 1 3 C (·) is as in Lemma 4.8. Then, by Lemmas 4.7 and 4.8, GM −αt d V (g x), P (g y, g x)V (g y) ≤ C (x)C (y)e . Y ≤i t t t ≤i t 1 1 124 ALEX ESKIN, MARYAM MIRZAKHANI GM Since by Lemma 4.1, V (x) = P (y, x)V (y),weget,for t > 0, ≥i ≥i GM −αt d V (g x), P (g y, g x)V (g y) ≤ C (x)C (y)e . Y i t t t i t 1 1 This, by the definition of P (x, y), implies that (4.10) holds as required. Even if we do not X − assume that d (x, y)< 1/100, then for almost all x and almost all y ∈ W [x],for t large enough d (g x, g y)< 1/100, and thus, in view of (4.10), (4.11) holds. t t 4.6. The cover X.— Let L = H viewed as a bundle over X .Let L = V (L).By big 0 i i Lemma 4.3, there exists a measurable finite cover X of X such that Lemma 4.3 holds on X for all the L . We always assume that the degree of the covering map σ : X → X i 0 0 is as small as possible. The set (x ).— For x ∈ X ,let (x ) denote the set of flags 0 0 0 i 0 −1 (x ) = {0}= L (x) ⊂ L (x) ⊂··· ⊂ L (x) = L (x) : x ∈ σ (x ) . i 0 i,0 i,1 i,n i 0 i 0 Let (x ) denote the Cartesian product of the (x ). Then, we can think of a point 0 i 0 x ∈ Xas a pair (x , F) where F ∈ (x ). 0 0 The measure ν on X.— We can use σ to define a pullback of the invariant measure ν on X to X, by requiring that the pushforward of the pullback measure by σ is ν,and 0 0 that the conditionals of the pullback measure on the fibers of σ are the (normalized) counting measure. We abuse notation by denoting the pullback measure also by ν . Lemma 4.10. — The measure ν is ergodic for the action of g on X. Proof. — Suppose E is a g -invariant set of X with ν(E)> 0. Then by the ergodicity −1 of the action of g on X , σ(E) is conull. Let N(x ) denote the cardinality of σ (x ) ∩ E. t 0 0 0 Then, again by the ergodicity of g ,N(x ) is constant almost everywhere. If E does not t 0 have full measure, then we have that N(x ) is smaller than the degree of the cover σ . 0 0 Then, we could replace X by E, contradicting the assumption that the degree of the covering map σ is as small as possible. ˜ ˜ ˜ The space X.— Recall that X is the universal cover of X .Let Xdenote the cover 0 0 of X corresponding to the cover σ : X → X . More precisely, 0 0 0 ˜ ˜ X = (x , F) : x ∈ X , F ∈ (x ) . 0 0 0 0 ˜ ˜ We denote the covering map from Xto X again by σ . 0 0 INVARIANT AND STATIONARY MEASURES 125 ˜ ˜ Stable and unstable manifolds for X and X.— Suppose x = (x , F) ∈ X. We define + + + (4.12) W [x]= y , F ∈ X : y ∈ W [x ], and F = P (x , y )F . 0 0 0 0 0 − − − (4.13) W [x]= y , F ∈ X : y ∈ W [x ], and F = P (x , y )F . 0 0 0 0 0 This definitions make sense, since by Proposition 4.4, + + P (x , y )(x ) = (y ) for y ∈ W [x ], 0 0 0 0 0 0 − − P (x , y )(x ) = (y ) for y ∈ W [x ]. 0 0 0 0 0 0 Remark. —Eventhough X itself does not have a manifold structure, for almost all + − x ∈ X, the sets W [x] and W [x] have the structure of an affine manifold (intersected with a set of full measure in X), see Lemma 3.2. Lemma 4.11 below asserts that these can be interpreted as the strong stable and strong unstable manifolds for the action of g on X. + − Notation. — If x ∈ X and V is a subspace of W (x) or W (x) we write V[x]= y ∈ W [x]: y − x ∈ V(x) . X + ˜ ˜ The “distance” d (·,·).— For x = (x , F) ∈ X, and y = (y , F ) ∈ Xand y ∈ W [x] 0 0 or W [x] define X X GM (4.14) d (x, y) = d (x , y ) + d F, P (y , x )F , 0 0 Y 0 0 where we extend the distance d between subspaces defined in (4.6) to a distance between flags. + X Lemma 4.11. —For almost all x ∈ X and almost all y ∈ W [x],d (g x, g y) → 0 as t t − X t →−∞. Similarly, for almost all x ∈ X and almost all y ∈ W [x], we have d (g x, g y) → 0 as t t t →∞. Proof. — This follows immediately from Corollary 4.9. Notational convention. — If f is an object on X ,and x ∈ X, we write f (x) instead of + + f (σ (x)). Thus, we can define V (H )(x) for x ∈ X, P (x, y) for x ∈ Xand y ∈ W [x], 0 i big etc. Also, if x ∈ X, we write f (x) instead of f (π ◦ σ (x)) etc. The partitions B of X.— Suppose x = (x , F) ∈ X. We define t 0 B [x]= y , F : y ∈ B [x ], F = P (x , y )F . t 0 0 t 0 0 0 126 ALEX ESKIN, MARYAM MIRZAKHANI Then B is a measurable partition of X subordinate to W .Inasimilarway,wecan −1 define sets J[x] for x ∈ Xand E [c] for c ∈ σ (C ),where C is as in Proposition 3.7. 1 1 Proposition 3.7 and all subsequent results of Section 3 apply to X as well as X . The following is an alternative version of Proposition 4.4 adapted to the cover X. Proposition 4.12. —Suppose L is a g -equivariant subbundle of H .For almost all x ∈ X, t big we can write L(x) = L (x), where L (x) is the Lyapunov subspace corresponding to the Lyapunov exponent λ . Suppose there exists i i an equivariant flat measurable W -connection F on L, such that F(x, y)L (x) = L (y), i i and that M ⊂ L is a g -equivariant subbundle. Then, (a) For almost all y ∈ B [x], F(x, y)M(x) = M(y), i.e. the subbundle M is locally covariantly constant with respect to the connection F. (b) For all i, the decomposition (4.3) of L is locally covariantly constant along W , i.e. for ν + -almost all y ∈ B [x],for all i ∈ I and for all 1 ≤ j ≤ n , W [x] 0 i (4.15)L (y) = F(x, y)L (x). ij ij Also, up to a scaling factor, the quadratic forms ·,· are locally covariantly constant i,j along W , i.e. for almost all y ∈ B [x],and for v,w ∈ L (x)/L (x), 0 ij i,j−1 (4.16) F(x, y)v, F(x, y)w = c(x, y)v,w . ij,x ij,y Proposition 4.12 will be proved in Section 4.9 . The proof also shows the following: Remark 4.13. —Proposition 4.12 applies also to U -connections, provided the measure along U [x] is in the Lebesgue measure class, and provided that in the state- ment, the set B [x] is replaced by B[x]= B [x]∩ U [x]. 0 0 4.7. Dynamically defined norms. — In this subsection we work on the cover X. We (++) define a norm on · on H , which has some advantages over the AGY norm · . big ++ Notation. — In Section 4.7 we let L denote the entire bundle H , write L for big V (L),and foreach i, consider the decomposition (4.3). i INVARIANT AND STATIONARY MEASURES 127 The function (x).— For x ∈ X, let 1/2 (x) = sup sup v,v : v ∈ L (x)/L (x), v = 1 , ij i,j−1 Y,x ij,x ij and let 1/2 (x) = inf inf v,v : v ∈ L (x)/L (x), v = 1 . ij i,j−1 Y,x ij,x ij Let + − (x) = (x)/ (x). −1 We have (x) ≥ 1for all x ∈ X. For x ∈ X ,wedefine (x ) to be max (x). 0 0 0 x∈σ (x ) Let d (·,·) be the distance between subspaces defined in (4.6). Let C ⊂ X Y 0 0 with ν(C )> 0and M ≥ 1 be chosen later. (We will choose them immediately before 0 0 Lemma 6.8 in Section 6.) Lemma 4.14. —Fix > 0 smaller than min |λ |, and smaller than min |λ − λ |, i i i=j i j (++) where the λ are the Lyapunov exponents of H . There exists a compact subset C ⊂ C ⊂ X with i 0 0 big ν(C)> 0 and a function T : C → R with T (x)< ∞ for ν a.e. x ∈ C such that the following 0 0 hold: (a) There exists σ> 0 such that for all c ∈ C , and any subset S of the Lyapunov exponents, d L (c), L (c) ≥ σ. Y i j i∈S j∈/S (b) There exists M > 1 such that for all c ∈ C , (c) ≤ M . −1 −1 (b ) There exists a constant M < ∞ such that for all x ∈ π (C),for all y ∈ π (C) ∩ + X GM W [x] with d (x, y)< 1/100, the Gauss-Manin connection P satisfies the estimate: GM P (x, y)v Y,y GM P (x, y) ≡ sup ≤ M . v=0 Y,x (c) For all c ∈ C,for all t > T (c) and for any subset S of the Lyapunov spectrum, −t d L (g c), L (g c) ≥ e . Y i −t j −t i∈S j∈/S + X Hence, for all c ∈ C and all t > T (c) and all c ∈ C ∩ W [g c] with d (g c, c )< 0 −t −t 1/100, P (g c, c )v −t Y,c −2 −t + −1 t (4.17)M ρ e ≤ P g c, c ≡ sup ≤ M ρ e , 1 −t 0 0 1 v=0 Y,g c −t where ρ = ρ (M ,σ, M , M )> 0. 1 1 0 128 ALEX ESKIN, MARYAM MIRZAKHANI (d) There exists ρ> 0 such that for all c ∈ C,for all t > T (c), for all i and all v ∈ L (c), 0 i −(λ +)t 2 −1 −2 −(λ −)t i i e ρ ρ v ≤g v ≤ ρ ρ e v . 1 Y,c −t Y,g c Y,c −t Proof. — Parts (a) and (b) hold since the inverse of the angle between Lyapunov subspaces and the ratio of the norms are finite a.e., therefore bounded on a set of almost full measure. To see (c), note that by the Osceledets multiplicative ergodic theorem, [KH, Theorem S.2.9 (2)] for ν -a.e. x ∈ X , lim log sin ∠ L (g x), L (g x) = 0. i −t j −t t→∞ i∈S j∈/S Also, (d) follows immediately from the multiplicative ergodic theorem. We now choose the set C and the function T of Proposition 3.7 and Lemma 3.8 to be as in Lemma 4.14. The main result of this subsection is the following: (++) Proposition 4.15. — For almost all x ∈ X thereexistsaninnerproduct ·,· on H (x) big (or on any bundle for which the conclusions of Lemma 4.14 hold) with the following properties: (a) For a.e. x ∈ X, the distinct eigenspaces L (x) are orthogonal. (b) Let L (x) denote the orthogonal complement, relative to the inner product ·,· of L (x) x i,j−1 ij (++) in L (x).Then, fora.e. x ∈ X,all t ∈ R and all v ∈ L (x) ⊂ H (x), ij ij big λ (x,t) ij (g ) v = e v + v , t ∗ where λ (x, t) ∈ R, v ∈ L (g x), v ∈ L (g x),and v =v. Hence (since v ij t i,j−1 t ij and v are orthogonal), λ (x,t) ij (g ) v≥ e v. t ∗ (c) There exists a constant κ> 1 such that for a.e. x ∈ X and for all t > 0, −1 κ t ≤ λ (x, t) ≤ κ t. ij (++) (d) There exists a constant κ> 1 such that for a.e x ∈ X and for all v ∈ H (x), and all big t ≥ 0, −1 κ t κ t e v≤(g ) v≤ e v. t ∗ (e) For a.e. x ∈ X,and a.e. y ∈ B [x] and all t ≤ 0, λ (x, t) = λ (y, t). ij ij INVARIANT AND STATIONARY MEASURES 129 (++) (f) For a.e. x ∈ X,a.e. y ∈ B [x],and any v,w ∈ H (x), big + + P (x, y)v, P (x, y)w =v,w . We often omit the subscript from ·,· and from the associated norm · . x x + −1 The inner product ·,· is first defined for x ∈ E [c] for c ∈ σ (C ) (in the nota- x 1 tion of Section 3, see also Section 4.6). We then interpolate between x ∈ E [c] and g x −t(c) (again in the notation of Section 3). The details of the proof of Proposition 4.15, which canbeskippedonfirstreading,are giveninSection 4.10 . The dynamical norm · on X .— The dynamical inner product ·,· and the 0 x dynamical norm · of Proposition 4.15 are defined for x ∈ X. For x ∈ X ,and v,w ∈ x 0 0 H (x ) we define big 0 1/2 (4.18) v,w = v,w , v =v,v . x x x 0 0 −1 0 |σ (x )| −1 x∈σ (x ) Remark 4.16. — The inner product and norm on X satisfy properties (a) and (d) of Proposition 4.15. Lemma 4.17. —For every δ> 0 there exists a compact subset K(δ) ⊂ X with ν(K(δ)) > (++) (−−) 1 − δ and a number C (δ) < ∞ such that for all x ∈ K(δ) and all v on H (x) or H (x), big big −1 C (δ) ≤ ≤ C (δ), 1 1 Y,x where · is the dynamical norm defined in this subsection and · is the AGY norm. x Y,x Proof. — Since any two norms on a finite dimensional vector space are equivalent, (++) there exists a function : X → R finite a.e. such that for all x ∈ Xand all v ∈ H (x), big −1 (x) v ≤v ≤ (x)v . 0 Y,x x 0 Y,x Since {x : (x)< N} is conull in X, we can choose K(δ) ⊂ Xand C = C (δ) so 0 1 1 N∈N that (x)< C (δ) for x ∈ K(δ) and ν(K(δ)) ≥ (1 − δ). 0 1 4.8 . Proof of Lemma 4.7.— We first prove (a). Note that the action of g commutes GM with P , i.e. GM GM P (g x, g y) ◦ g = g ◦ P (x, y). t t t t Let α = min |λ − λ |,where the λ = λ (H ). We will choose 0 < <α /100. For 0 i=j i j i i 0 every > 0 there exists a compact set K = K () ⊂ X with ν(K )> 1 − /4and 0 0 0 0 130 ALEX ESKIN, MARYAM MIRZAKHANI σ = σ() > 0 such that for any subset S of the Lyapunov exponents, −1 (4.19) d V (x), V (x) >σ for all x ∈ π (K ). Y i j 0 i∈S j∈/S By the multiplicative ergodic theorem and the Birkhoff ergodic theorem, there exists a set K = K() ⊂ K with ν(K)> 1 − /2and a constantC = C() such that such that for −1 all z ∈ π (K),all s ∈ R and all v ∈ V (z), −1/2 λ s−(/6)|s| 1/2 λ s+(/6)|s| i i (4.20)C() v e ≤g v ≤ C() v e , Y s Y Y and also for any interval I ⊂ R containing the origin of length at least 4 log C()/α ,and −1 any z ∈ π (K), (4.21) |{s ∈ I : g z ∈ K }| ≥ (1 − )|I|. s 0 Suppose the set {g x : 0 ≤ s ≤ t} intersects K. We will show that for all y ∈ X such that s 0 d (g x, g y) ≤ 1/100 for 0 ≤ s ≤ t, s s GM 2 + −αt (4.22) d V (g x), P (g y, g x)V (g y) ≤ C (x)C(σ)C() 1 + D (x, y) e , Y ≤i t t t ≤i t 0 where C (x) is as in Lemma 4.6.Let ϒ(x) ={ : x ∈ K()} and let C(x) = C (x) inf C(σ)C() : ∈ ϒ(x) . Since the union as → 0 of the sets K = K() is conull, (4.22) implies part (a) of the lemma. We now prove (4.22). We may assume that t > 4log C()/α , otherwise (4.22) trivially holds. Then, by (4.21), there exists (1 − )t < t ≤ t with g x ∈ K .Inviewof t 0 Lemma 3.6 the inequality (4.22)for t implies the inequality (4.22)for t (after replacing α by α − 4 ). Thus, we may assume without loss of generality that g x ∈ K . t 0 By assumption, there exists 0 < s < t such that g x ∈ K. Let z = g x. Then, applying s s (4.20)twice at z,weget,for all v ∈ V (x), −1 λ t−(/3)t λ t+(/3)t i i (4.23)C() v e ≤g v ≤ C()v e . Y t Y Y GM Let v ∈ P (g y, g x)V (g y) be such that v = 1and t t ≤i t Y GM d v , V (g x) = δ P (g y, g x)V (g y), V (g x) , Y ≤i t Y t t ≤i t ≤i t GM where δ (·,·) is as in Lemma 4.6. Then, v = g v for some v ∈ P (y, x)V (y).Wemay Y t ≤i write v = v + w, v ∈ V (x), w ∈ V (x). 0 0 ≤i >i INVARIANT AND STATIONARY MEASURES 131 We have, by the definition of D (·,·), w ≤ D (x, y)v . Y 0 Y Then, we have v = g v = g v + g w, t t 0 t and by (4.23), −1 (λ −/3)t g v ≥ C() e v , t 0 Y 0 Y and (λ +/3)t i+1 g w ≤ C()e w . t Y Y Thus, 2 + −(α −2/3)t g w ≤ C() D (x, y)e g v . t Y t 0 Y Since g v ∈ V (g x) and g w ∈ V (g x), this, together with (4.19) implies t 0 ≤i t t >i t 2 + −(α −2/3)t d v , V (g x) ≤ C(σ)C() 1 + D (x, y) e . Y ≤i t This, together with Lemma 4.6, completes the proof of (4.22). The proof of (b) is identical. 4.9 . Proof of Propositions 4.4 and 4.12.— The proof of Proposition 4.4 will essen- tially be by reference to [L, Theorem 1]. We recall the setup (in our notation): Let (X,ν) be a measure space, and let T : X → X be a measure preserving trans- formation. Let B be a σ -subalgebra of the σ -algebra of Borel sets on X, such that B −1 is T-decreasing (i.e. T B ⊂ B). Let B denote the σ -algebra generated by all the −∞ σ -algebras T B, n ∈ Z. Let V be a vector space, and let A : X → GL(V) be a log-integrable B-measurable function. Let (n) n−1 A (x) = A T x ... A(x) for n > 0 (0) A (x) = Id and (n) −1 n −1 −1 A (x) = A T x ... A T x for n < 0 We have a skew-product map T : X × V → X × Vgiven by T(x,v) = Tx, A(x)v , 132 ALEX ESKIN, MARYAM MIRZAKHANI and then, n n (n) T (x,v) = T x, A (x)v . Let (n) γ = lim logA (x)dν(x), n→∞ −1 (n) γ = lim − log A (x) dν(x), n→∞ where · is the operator norm. The limits exist by the subadditive ergodic theorem. The matrix A also naturally acts on the projective space P(V).Weuse thenotation T to denote also the associated skew-product map X × P(V) → X × P(V). We have the following: Theorem 4.18 (Ledrappier, [L, Theorem 1]). — Suppose (a) γ = γ . + − (b) x → ν is a family of measures on P(V) defined for almost every x such that A(x)ν = ν x x Tx andsuchthatthe map x → ν is B -measurable. x −∞ Then, x → ν is B-measurable. Proof of Proposition 4.4. — We first make some preliminary reductions. For x ∈ X , 1 k write M(x) ={M (x),..., M (x)}. Since M(x) is g -equivariant, for 1 ≤ j ≤ k, j j M (x) = M (x), M (x) ⊂ L (x). i i 1 k Let M (x) ={M (x),..., M (x)}. Thus, it is enough to show that i i F(x, y)M (x) = M (y). i i Without loss of generality, we may assume that for a fixed i,all the M have the same dimension. Suppose x ∈ J ,where J is as in Proposition 3.7. Then the sets {g c : 0 ≤ t ≤ c c −t t(c)} and B [x]= J ∩ W [x] intersect at a unique point x ∈ X . Then, we can replace 0 c 0 0 the bundle L(x) by L(x) ≡ F(x, x )L(x). Then, for y ∈ B [x], 0 0 ˜ ˜ L(y) = F(y, x )L(y) = F(y, x )F(x, y)L(x) = F(x, x )L(x) = L(x), 0 0 0 ˜ ˜ i.e. L(x) is locally constant along W (x). Also, by (4.2), the action of (g ) on L is locally t ∗ constant. Thus, without loss of generality, we may assume that F is locally constant (or else we replace L by L). Thus, it is enough to show that assuming the subspaces L (x) are almost everywhere locally constant along W , the set of subspaces M (x) is also almost everywhere locally constant along W . In other words, we assume that the functions INVARIANT AND STATIONARY MEASURES 133 x → L (x) are B -measurable, and would like to show that the functions x → M (x) are i 0 i B -measurable. Let T = g denote the time 1 map of the geodesic flow. Fix i and j,and let 1 k i d = dim M = ··· = dim M .Let V(x) = (L (x)/L (x)).Notethat V(x) is B -mea- i i i−1 0 i i surable and g -equivariant. We can write the action of (g ) (for t = 1) on the bundle V as t ∗ (g ) (x,v) = g x, A(x)v . 1 ∗ 1 Then, A(x) is B -measurable (where B is as in Section 3). Also, the condition γ = γ 1 t + − follows from the multiplicative ergodic theorem. (In fact, γ = γ = d λ ,where λ is the + − i i i Lyapunov exponent corresponding to L .) Let ν denote the Dirac measure on (the line through) v ∧··· ∧ v ,where 1 d {v ,...,v } is any basis for M (x),and let 1 d ν = ν . j=1 Then, since the M (x) are g -equivariant, the measures ν are T-invariant. Also note that i t x B is the partition into points. Thus, we can apply Theorem 4.18 (with B = B ). We −∞ 1 conclude that the function x → ν is B -measurable, which implies that the M (x) are x 1 i locally constant on atoms of B . Since the M (x) are g -equivariant, this implies that 1 i t the M (x) are also locally constant (in particular the function x → M (x) is B -mea- i i 0 surable). Proof of Proposition 4.12. — Note that (a) and also (4.15) follow immediately from Proposition 4.4. We now prove (4.16). After making the same reductions as in the proof of Propo- sition 4.4, we may assume that the L and F are locally constant. Let K ⊂ K denote ij a compact subset with ν(K)> 0.9where ·,· is uniformly continuous. Consider the ij points g x and g y,as t →−∞.Then d (g x, g y) → 0. Let t t t t −λ (x,t) −λ (x,t) ij ij v = e (g ) v, w = e (g ) w, t t ∗ t t ∗ where λ (x, t) is as in Lemma 4.3. Then, by Lemma 4.3,wehave ij (4.24) v ,w =v,w , v ,w = c(x, y, t)v,w , t t ij,g x ij,x t t ij,g y ij,y t t λ (x,t)−λ (y,t) ij ij where c(x, y, t) = e . Now take a sequence t →∞ with g x ∈ K, g y ∈ K (such a sequence exists for k t t k k ν -a.e. x and y with y ∈ B [x]). Then, since the L (x) and the connection F are assumed 0 ij to be locally constant, c(x, y, t ) is bounded between two constants. Also, v ,w −v ,w → 0. t t ij,g x t t ij,g y k k t k k t k k Now Equation (4.16) follows from (4.24). 134 ALEX ESKIN, MARYAM MIRZAKHANI 4.10 . Proof of Proposition 4.15.— To simplify notation, we assume that M = 1 (where M is as in Lemma 4.14). The inner products ·,· on E [c].— Note that the inner products ·,· and the ij ij R-valued cocycles λ of Lemma 4.3 are not unique, since we can always multiply ·,· ij ij,x by a scalar factor c(x), and then replace λ (x, t) by λ (x, t)+ log c(g x)− log c(x).Inviewof ij ij t (4.16)inProposition 4.12(b), we may (and will) use this freedom to make ·,· constant ij,x −1 + + on each set E [c],where c ∈ σ (C ) and E [c] is as in Section 3 (see also Section 4.6). The inner product ·,· on E [c].— Let (4.25) {0}= V ⊂ V ⊂··· ≤0 ≤1 (++) be the Lyapunov flag for H ,and foreach i,let big (4.26) V = V ⊂ V ⊂··· V = V ≤i−1 ≤i,0 i,1 ≤i,n ≤i be a maximal invariant refinement. (++) (++) Let L = V (H ) denote the Lyapunov subspaces for H .Thenwehavea i i big big maximal invariant flag {0}= L ⊂ L ⊂ ··· ⊂ L = L , i,0 i,1 i,n i where L = L ∩ V . ij i ≤i,j −1 + Let c ∈ σ (C ),E [c] be as in Section 3 and Section 4.6. By Lemma 4.14(b), we can (and do) rescale the inner products ·,· so that after the rescaling, for all v ∈ ij,c L (c)/L (c), ij i,j−1 −1 1/2 M v ≤v,v ≤ M v , Y,c Y,c ij,c where · is the AGY norm at σ (c) and M > 1 is as in Lemma 4.14.Wethen Y,c 0 choose L (c) ⊂ L (c) to be a complementary subspace to L (c) in L (c),sothatfor all ij i,j−1 ij ij v ∈ L (c) and all v ∈ L (c), i,j−1 ij v + v ≥ ρ max v ,v , Y,c Y,c Y,c and ρ > 0 depends only on the dimension. Then, ∼ ∼ L (c) L (c)/L (c) V (c)/V (c). = = ij i,j−1 ≤i,j ≤i,j−1 ij Let π : V → V /V be the natural quotient map. Then the restriction of π to ij ≤i,j ≤i,j ≤i,j−1 ij L (c) is an isomorphism onto V (c)/V (c). ≤i,j ≤i,j−1 ij INVARIANT AND STATIONARY MEASURES 135 (++) We can now define for u,v ∈ H (c), big u,v ≡ π (u ),π (v ) , c ij ij ij ij ij,c ij where u = u ,v = v , u ∈ L (c), v ∈ L (c). ij ij ij ij ij ij ij ij In other words, the distinct L (c) are orthogonal, and the inner product on each L (c) ij ij coincides with ·,· under the identification π of L (c) with V (c)/V (c). ij,c ij ≤i,j ≤i,j−1 ij (++) We now define, for x ∈ E [c],and u,v ∈ H (x) big + + u,v ≡ P (x, c)u, P (x, c)v , + + where P (·,·) is the connection defined in Section 4.2.Thenfor x ∈ E [c], the inner product ·,· induces the inner product ·,· on V (x)/V (x). x ij,x ≤i,j ≤i,j−1 Symmetric space interpretation. — We want to define the inner product ·,· for any x ∈ J[c] by interpolating between ·,· and ·,· ,where c is such that g c ∈ E [c ].To c c −t(c) define this interpolation, we recall that the set of inner products on a vector space V is canonically isomorphic to SO(V)\GL(V),where GL(V) is the general linear group of V (++) and SO(V) is the subgroup preserving the inner product on V. In our case, V = H (c) big with the inner product ·,· . (++) Let K denote the subgroup of GL(H (c)) which preserves the inner product big (++) ·,· .Let Q denote the parabolic subgroup of GL(H (c)) which preserves the flags big (4.25)and (4.26), and on each successive quotient V (c)/V (c) preserves ·,· .Let ≤i,j ≤i,j−1 ij,c (++) K A denote the point in K \GL(H (c)) which represents the inner product ·,· , i.e. c c c big u,v = A u, A v . Then, since ·,· induces the inner products ·,· on the space V (c )/V (c ) c ij,c ≤i,j ≤i,j−1 which is the same as V (g )/V (g ), we may assume that the matrix product ≤i,j −t(c)c ≤i,j−1 −t(c)c A (g ) is in Q. −t(c) ∗ Let N be the normal subgroup of Q in which all diagonal blocks are the identity, and let Q = Q/N .(We mayconsider Q to be the subgroup of Q in which all off- diagonal blocks are 0.) Let π denote the natural map Q → Q . Claim 4.19. — We may write A (g ) = A , −t(c) ∗ −λ t(c) where ∈ Q is the diagonal matrix which is scaling by e on L (c), A ∈ Q and A = t(c) O(e ). 136 ALEX ESKIN, MARYAM MIRZAKHANI Proof of claim. — Suppose x ∈ E [c] and t =−t(c)< 0where c ∈ C and t(c) is as in Proposition 3.7. By construction, t(c)> T (c),where T (c) is as in Lemma 4.14. Then, 0 0 the claim follows from (4.17) and Lemma 4.14(d). Interpolation. — We may write A = DA , where D is diagonal, and det A = 1. In 1 1 t t view of Claim 4.19, D= O(e ) and A = O(e ). We now connect K \A to the identity by the shortest possible path :[−t(c), 0]→ c 1 K \K Q, which stays in the subset K \K Q of the symmetric space K \SL(V).(We c c c c c parametrize the path so it has constant speed.) This path has length O(t) where the implied constant depends only on the symmetric space. Now for −t(c) ≤ t ≤ 0, let −t/t(c) (4.27)A(t) = (D) (t). Then A(0) is the identity map, and A(−t(c)) = A (g ) . Then, we define, for x ∈ E [c] −t(c) ∗ and −t(c) ≤ t ≤ 0, (g ) u,(g ) v = A(t)u, A(t)v . t ∗ t ∗ g x x Proof of Proposition 4.15. — Suppose first that x = c,where c and E [c] are as in Section 3 and Section 4.6. Then, by construction, (a) and (b) hold. Also, from the con- struction, it is clear that the inner product ·,· induces the inner product ·,· on c ij,c L (c)/L (c). ij i,j−1 + + Now by Proposition 4.12,for x ∈ E [c],P (x, c)L (x) = L (c),and for u¯,v¯ ∈ ij ij + + L (x)/L (x), u,v =P (x, c)u, P (x, c)v . Therefore, (a), (b), (e) and (f) hold ij i,j−1 ij,x ij,c + + for x ∈ E [c],and also for x ∈ E [c], the inner product ·,· induces the inner prod- uct ·,· on L (x)/L (x). Now, (a), (b), (e) and (f) hold for arbitrary x ∈ J[c] since ij,x ij i,j−1 A(t) ∈ Q. Let ψ : Q → R denote the homomorphism taking the block-conformal matrix ij + Q to the scaling part of block corresponding to L /L .Let ϕ = ψ ◦ π ;then ϕ : ij i,j−1 ij ij ij Q → R is a homomorphism. From (4.27), we have, for x ∈ E [c] and −t(c) ≤ t ≤ 0, λ (x, t) = logϕ A(t) = tλ + γ (x, t), ij ij i ij t/t(c) t/t(c) where tλ is the contribution of and γ (x, t) is the contribution of D (t).By i ij Claim 4.19,for all −t(c) ≤ t ≤ 0, (4.28) γ (x, t) = O() ij ∂ t where > 0 isasinClaim 4.19, and the implied constant depends only on the symmetric space. Without loss of generality, the function T (x) in Lemma 4.14 can be chosen large enough so that since t(c)> T (c), (c) holds. 0 INVARIANT AND STATIONARY MEASURES 137 The lower bound in (d) now follows immediately from (b) and (c). The upper bound in (d) follows from (4.28). 5. Conditional measure lemmas In Sections 5–8 we work on X (and not on X). Motivation. — We use notation from Section 2.3.Recallthat L (q) is the smallest linear subspace of W (q) containing the support of the conditional measure ν .For W (q) two (generalized) subspaces U and U and x ∈ X let hd (U , U ) denote the Hausdorff X X X 0 0 0 distance between U ∩ B (x, 1/100) and U ∩ B (x, 1/100),where B (x, r) denotes X X 0 0 { y ∈ X : d (x, y)< r}.For x ∈ X , we will sometimes write hd (U , U ) instead of 0 0 hd (U , U ) as long as the proper lift x˜ ∈ X of x is clear from the context. x˜ We can write X + + hd U q , U [q ] = Q q − q , 2 t q 2 where Q : L (q) → R is a map depending on q, u, ,and t.The map Q is essentially the t t composition of flowing forward for time , shifting by u ∈ U and then flowing forward X + + again for time t. We then adjust t so that hd (U [q ], U [q ]) ≈ ,where > 0is a q 2 priori fixed. In order to solve “technical difficulty #1” of Section 2.3, it is crucial to ensure that t does not depend on the precise choice of q (it can depend on q, u, ). The idea is to use the following trivial: Lemma 5.1. — For any ρ> 0 there is a constant c(ρ) with the following property: Let A : V → W be a linear map between Euclidean spaces. Then there exists a proper subspace M ⊂ V such that for any v with v= 1 and d(v, M)>ρ , we have A≥Av≥ c(ρ)A. Proof of Lemma 5.1.—Thematrix A A is symmetric, so it has a complete orthog- onal set of eigenspaces W ,..., W corresponding to eigenvalues μ >μ > ···μ .Let 1 m 1 2 m M = W . Now suppose the map Q : L (q) → R is of the form Q (v) =Q (v) where Q : t t t t L (q) → H(q ) is a linear map, and H(q ) a vector space. This in fact happens in the first 2 2 + + step of the induction where U is the unipotent N (and we can take H(q ) = W (q )/N). 2 2 We can then choose t, depending only on q, u and , such that the operator norm Q (v) Q ≡ sup = . − v v∈L (q) 138 ALEX ESKIN, MARYAM MIRZAKHANI Then,weneed to provethatwecan choose q ∈ L [q] such that q − q≈ 1/100, q avoids an a priori given set of small measure, and also q − q is at least ρ away from the “bad subspace” M = M (q,) of Lemma 5.1. (Actually, since we do not want the choice of q to depend on the choice of u, we want to choose q such that q − q avoids most of the subspaces M as u ∈ U varies over a unit box.) Then, for most u, c(ρ) ≤ Q q − q ≤ , t 2 and thus + + (5.1) c(ρ) ≤ hd U [q ], U q ≤ , q 2 2 2 as desired. In general we do not know that the map Q is linear, because we do not know the dependence of the subspace U (q) on q. To handle this problem, we can write Q q − q = A F q − F(q) t t (r) + where the map A : L [q] → W (q ) is linear (and can depend on q, u, ), and the t ext 2 − (r) measurable map F : L [q]→ L [q] depends only on q.(SeeProposition 6.11 below ext (r) for a precise statement.) The map F and the space L [q] are defined in this section, ext and the linear map A = A(q, u,, t) is defined in Section 6.1. We then proceed in thesameway.Wechoose t =ˆτ(q, u,,) so that A = . (A crucial bilipshitz type property of the function τˆ similar to (2.7)isprovedinSection 7.) In this section we prove Proposition 5.3, which roughly states that (for most q)wecan choose q ∈ L [q] while avoiding an a priori given set of small measure, so that F(q ) − (r) F(q)≈ 1/100 and also F(q )− F(q) avoids most of a family of linear subspaces of L [q] ext (which will be the “bad subspaces” of the linear maps A as u varies over U ). Then as above, for most u,(5.1) holds. We can then proceed using (a variant of) Lemma 2.3 as outlined in Section 2.3. In view of the above discussion, we need to keep track of the way U [ y] varies as y varies over W [x].InviewofProposition 4.12(a), all bundles equivariant with respect to the geodesic flow are, when restricted to W , equivariant with respect to the connection P (x, y) defined in Section 4.2. Thus, it will be enough for us to keep track of the maps − − P (x, y). However, this is a bit awkward, since P (x, y) depends on two points x and y. Thus, it is convenient to prove the following: (−) Lemma 5.2. — There exists a subbundle Y ⊂ H , locally constant under the Gauss- big Manin connection along W , and for almost all x ∈ X an invertible linear map P(x) : X → 0 0 Hom(Y(x), H (M,, R)), such that for almost all x, y, − −1 (5.2)P (x, y) = P(y) ◦ P(x) . The proof of Lemma 5.2 is simple, but notationally heavy, and is relegated to Section 5.1 . It may be skipped on first reading. INVARIANT AND STATIONARY MEASURES 139 − − − The spaces L (x) and L (x).— Let the subspace L (x) ⊂ W (x) be the smallest ext such that the conditional measure ν − is supported on L [x]. Since ν is invariant un- W [x] der N, the entropy of any g ∈ A is positive. Therefore for ν -almost all x ∈ X , L (x) = {0} t 0 (see Proposition B.5). In the same spirit, let L [x]⊂ Hom Y(x), H (M,, R) ext denote the smallest affine subspace which for almost every y ∈ W [x] contains the vector P(y). (This makes sense since Y(x) is locally constant along W [x].) We also set L (x) ext to be the vector space spanned by all vectors of the form P(y) − P(x) as y varies over W [x]. Then, L (x) = L [x]− P(x). ext ext Note that for almost all x and almost all y ∈ W [x], L [ y]= L [x]. ext ext (r) The space L (x) and the function F.— For a vector space V we use the notation ext ⊗m V to denote the m-fold tensor product of V with itself. If f : V → W is a linear map, ⊗m ⊗m ⊗m ⊗m ⊗m we write f for the induced linear map from V to W .Let j : V → V denote the map v → v ⊗ ··· ⊗ v (m-times). m ⊗k m Let V denote V .If f : V → W is a linear map, we write f for the k=1 m m induced linear map from V to W given by m ⊗1 ⊗2 ⊗m f (v) = f (v), f (v),... f (v) . m m Now if V and W are affine spaces, then we can still canonically define V and W ,and m m m an affine map f : V → W induces an affine map f : V → W . Let r be an integer to be chosen later. Let F : X → L [x] denote the diagonal 0 ext embedding F(x) = P(x) . Let (r) r L [x] ⊂ L (x) ext ext denote the smallest affine subspace which contains the vectors F(y) for almost all y ∈ W [x]. We also set (r) (r) L (x) = L [x] − F(x). ext ext − (r) (r) Note that for y ∈ W [x], L [ y] = L [x] . ext ext In this section, let (B,|·|) be a finite measure space. (We will use the following proposition with B ⊂ U is a “unit box”. The precise setup will be given in Section 6.) To carry out the program outlined at the beginning of Section 5, we need the following: 140 ALEX ESKIN, MARYAM MIRZAKHANI Proposition 5.3. —For every δ> 0 there exist constants c (δ) > 0, (δ) > 0 with 1 1 c (δ) → 0 and (δ) → 0 as δ → 0, and also constants ρ(δ) > 0, ρ (δ) > 0,and C(δ) < ∞ 1 1 such that the following holds: For any subset K ⊂ X with ν(K )> 1 − δ, there exists a subset K ⊂ K with ν(K)> 1 − c (δ) such that the following holds: suppose for each x ∈ X we have a measurable map from B to 1 0 (r) (r) proper subspaces of L (x) , written as u → M (x),where M (x) is a proper subspace of L (x) . ext u u ext Then, for any q ∈ K there exists q ∈ K with (5.3) ρ (δ) ≤ d q, q ≤ 1/100 and (5.4) ρ(δ) ≤ F q − F(q) ≤ C(δ) and so that (5.5) d F q − F(q), M (q) >ρ(δ) for at least 1 − (δ) -fraction of u ∈ B. Y u 1 This proposition is proved in Section 5.2 . The proof uses almost nothing about the maps F or the measure ν , other than the definition of L (x). It may be skipped on ext first reading. ∗ 1 1 5.1 . Proof of Lemma 5.2.— As in Section 4.1,let V (x) ≡ V (H )(x) ⊂ H (M, i i , R) denote the subspace corresponding to the (cocycle) Lyapunov exponent λ .Let Y(x) = V (x)/V (x), ≥i >i i=1 where V and V areasinSection 4.1.Let π : V (x) → V (x)/V (x) denote the ≥j >j i ≥i ≥i >i natural projection. For x ∈ X ,let P ∈ Hom(V (x)/V (x), H (M,, R)) denote the unique linear 0 i,x ≥i >i map such that for x¯ ∈ V (x)/V (x),P (x¯) ∈ V (H )(x) and π (P (x¯)) =¯ x.Notethat ≥i >i i,x i i i,x the P satisfy the following: i,x −1 (5.6)P = g ◦ P ◦ g , i,g x t i,x t t and (5.7)P (u¯) − P (u¯) ∈ V (x). i,x i,y >i Example. —The space V /V is one dimensional, and corresponds to the Lya- ≥1 >1 punov exponent λ = 1. If we identify it with R in the natural way then P : R → 1 1,x H (M,, R) is given by the formula (5.8)P (ξ) = (Im x)ξ 1,x where for x = (M,ω), we write Im x for the imaginary part of ω. INVARIANT AND STATIONARY MEASURES 141 Let P : X → Hom V (x)/V (x), H (M,, R) 0 ≥i >i i=1 be given by P(x) = (P ,... P ). 1,x k,x Then, we can think of P(x) as a map from Y(x) to H (M,, R) and (5.2) holds, where P (x, y) is as in Section 4.2. 5.2 . Proof of Proposition 5.3. The measure ν˜ .— Let ν˜ = F (ν − ) denote the pushforward of ν − under F. x x ∗ W [x] W (r) − Then ν˜ is a measure supported on L [x] . (Note that for y ∈ W [x], ν˜ =˜ν .) x ext x y Lemma 5.4. —For ν -almost all x ∈ X , for any > 0 (which is allowed to depend on x), (r) the restriction of the measure ν˜ to the ball B(F(x),) ⊂ L [x] is not supported on a finite union of x ext (r) proper affine subspaces of L [x] . ext Outline of proof. — Suppose not. Let N(x) be the minimal integer N such that for some = (x)> 0, the restriction of ν˜ to B(F(x),) is supported on N affine subspaces. (r) Note that in view of (5.6)and (5.7), the induced action on L (and hence on L )of g ext −t ext for t ≥ 0 is expanding. Then N(x) is invariant under g , t ≥ 0. This implies that N(x) −t (r) is constant for ν -almost all x, and also that the only affine subspaces of L [x] which ext contribute to N(·) pass through F(x). Then, N(x)> 1 almost everywhere is impossible. Indeed, suppose N(x) = k a.e., then pick y near x such that F(y) is in one of the affine subspaces through F(x); then there must be exactly k affine subspaces of non-zero mea- sure passing though F(y), but then at most one of them passes through F(x).Thus, the measure restricted to a neighborhood of F(x) gives positive weight to at least k + 1 sub- spaces, contradicting our assumption. Thus, we must have N(x) = 1 almost everywhere; but then (after flowing by g for sufficiently large t > 0) we see that for almost all x, ν˜ −t x (r) is supported on a proper subspace of L [x] passing through x, which contradicts the ext (r) definition of L (x) . ext Remark. — Besides Lemma 5.4, the rest of the proof of Proposition 5.3 uses only the measurability of the map F. The measure νˆ .— Let B be the analogue of the partition B constructed in Sec- x 0 tion 3 but along the stable leaves W . (The only properties we use here is that B is a measurable partition subordinate to W with atoms of diameter at most 1/100.) Let − − B [x]⊂ W [x] denote the atom of the partition B containing x. 0 0 142 ALEX ESKIN, MARYAM MIRZAKHANI Let νˆ = F (ν − | ), i.e. νˆ is the pushforward under F of the restriction of x ∗ W [x] x B [x] − − ν to B [x]. Then, for y ∈ B [x], νˆ =ˆν . Suppose δ> 0 is given. Since W [x] x y 0 0 (r) lim νˆ B F(x), C =ˆν L [x] , x x ext C→∞ there exists a function c(x)> 0 finite almost everywhere such that for almost all x, 1/2 (r) νˆ B F(x), c(x) > 1 − δ νˆ L [x] . x x ext 1/2 Therefore, we can find C = C(δ) > 0 and a compact set K with ν(K )> 1 − δ such δ δ that for each x ∈ K , 1/2 (r) (5.9) νˆ B F(x), C > 1 − δ νˆ L [x] for all x ∈ K . x x ext In the rest of Section 5.2 , C will refer to the constant of (5.9). Lemma 5.5. —For every η> 0 and every N > 0 there exists β = β (η, N)> 0, ρ = 1 1 1 ρ (η, N)> 0 and a compact subset K of measure at least 1 − η such that for all x ∈ K ,and 1 η,N η,N (r) any proper subspaces M (x),..., M (x) ⊂ L (x) , 1 N ext (5.10) νˆ B F(x), C \ Nbhd M (x),ρ ≥ β νˆ B F(x), C . x k 1 1 x k=1 Outline of proof. — By Lemma 5.4, there exist β = β (N)> 0and ρ = ρ (N)> 0 x x x x (r) such that for any subspaces M (x),... M (x) ⊂ L (x) , 1 N ext (5.11) νˆ B F(x), C \ Nbhd M(x),ρ ≥ β νˆ B F(x), C . x x x x k=1 Let E(ρ ,β ) be the set of x such that (5.10) holds. By (5.11), 1 1 ν E(ρ ,β ) = 1. 1 1 ρ >0 β >0 Therefore, we can choose ρ > 0and β > 0such that ν(E(ρ ,β )) > 1 − η. 1 1 1 1 Lemma 5.6. —For every η> 0 and every > 0 there exists β = β(η, )> 0, a compact 1 1 set K = K ( ) of measure at least 1 − η,and ρ = ρ(η, )> 0 such that the following holds: η η 1 1 (r) Suppose for each u ∈ B let M (x) be a proper subspace of L (x) .Let u ext E (x) = v ∈ B F(x), C : for at least (1 − )-fraction of u in B, good 1 d v − F(x), M (x) >ρ/2 . Y u Then, for x ∈ K , (5.12) νˆ E (x) ≥ βνˆ B F(x), C . x good x INVARIANT AND STATIONARY MEASURES 143 (r) Proof.—Let n = dim L [x] . By considering determinants, it is easy to show that ext for any C > 0 there exists a constant c = c (C)> 0 depending on n and C such that n n for any η> 0 and any points v ,...,v in a ball of radius C with the property that for 1 n all 1 < i ≤ n, v is not within η of the subspace spanned by v ,...,v ,then v ,...,v i 1 i−1 1 n are not within c η of any n − 1 dimensional subspace. Let k ∈ N denote the smallest n max max integer greater then 1 + n/ ,and let N = N( ) = .Let β , ρ and K be as in 1 1 1 1 η,N n−1 Lemma 5.5.Let β = β(η, ) = β (η, N( )), ρ = ρ(η, ) = ρ (η, N( ))/c ,K ( ) = 1 1 1 1 1 1 n η 1 K .Let E (x) = B(F(x), C) \ E (x). To simplify notation, we choose coordinates η,N( ) bad good so that F(x) = 0. We claim that E (x) is contained in the union of the ρ -neighborhoods bad 1 of at most N subspaces. Suppose this is not true. Then, for 1 ≤ k ≤ k we can induc- max tively pick points v ,...,v ∈ E (x) such that v is not within ρ of any of the subspaces 1 k bad j 1 spanned by v ,...,v where i ≤ ··· ≤ i < j . Then, any n-tuple of points v ,...,v i i 1 n−1 i i 1 n−1 1 n is not contained within ρ = c ρ of a single subspace. Now, since v ∈ E (x), there exists n 1 i bad U ⊂ B with |U |≥ |B| such that for all u ∈ U , d (v , M )<ρ/2. We now claim that i i 1 i Y i u for any 1 ≤ i < i < ··· < i ≤ k, 1 2 n (5.13)U ∩ ··· ∩ U =∅. i i 1 n Indeed, suppose u belongs to the intersection. Then each of the v ,...v is within ρ/2 i i 1 n of the single subspace M , but this contradicts the choice of the v . This proves (5.13). u i Now, k k max max k |B|≤ |U |≤ n U ≤ n|B|. 1 max i i i=1 i=1 This is a contradiction, since k > 1 + n/ . This proves the claim. Now (5.10) implies max 1 that νˆ E (x) ≥ˆν B F(x), C \ Nbhd M (x),ρ x good x k 1 k=1 ≥ βνˆ B F(x), C . Proof of Proposition 5.3.—Let − 1/2 − K = x ∈ X : ν − K ∩ B [x] ≥ 1 − δ ν − B [x] . 0 W [x] W [x] 0 0 1/2 By Lemma 3.11,wehave ν(K ) ≥ 1 − δ . We have, for x ∈ K , − 1/2 (r) (5.14) νˆ F K ∩ B [x] ≥ 1 − δ νˆ L [x] . x x ext Let β(η, ) be as in Lemma 5.6.Let 1/2 2 2 1/2 c(δ) = δ + inf η + : β(η, ) ≥ 8δ . 1 144 ALEX ESKIN, MARYAM MIRZAKHANI We have c(δ) → 0as δ → 0. By the definition of c(δ) we can choose η = η(δ) < c(δ) and 1/2 = (δ) < c(δ) so that β(η, ) ≥ 8δ . 1 1 1 Now suppose x ∈ K ∩ K . Then, by (5.9)and (5.14), − 1/2 (5.15) νˆ F K ∩ B [x] ∩ B F(x), C ≥ 1 − 2δ νˆ B F(x), C . x x By (5.12), for x ∈ K , 1/2 (5.16) νˆ E (x) ≥ 8δ νˆ B F(x), C . x good x 1/2 Let K = K ∩ K ∩ K ∩ K .Wehave ν(K) ≥ 1−δ − 2δ − c(δ),so ν(K) → 1as δ → 0. Also, if q ∈ K, by (5.15)and (5.16), F K ∩ B [q] ∩ E (q) ∩ B F(x), C = ∅. good Thus, we can choose q ∈ K ∩ B [q] such that F(q ) ∈ E (q) ∩ B(F(q), C).Then(5.5) good holds with ρ = ρ(η(δ), (δ)) > 0. Also the upper bound in (5.3) holds since B [q] has diameter at most 1/100, and the upper bound in (5.4) holds since F(q ) ∈ B(F(q), C). Since all M (q) contain the origin q, the lower bound in (5.4) follows from (5.5). Finally, the lower bound in (5.3) follows from lower bound in (5.4) since in view of (5.8), q − q is essentially a component of F(q) − F(q ). 6. Divergence of generalized subspaces 1 1 The groups G , G and G .— Recall that H (x) denotes H (M,, R).(In fact the + ++ dependence on x is superfluous, but we find it useful to consider H (x) as the fiber over 1 1 X of a flat bundle.) Let G(x) = (SL(H ) H )(x) which is isomorphic to the group of 1 1 affine maps of H (x) to itself. We can write g ∈ G(x) as a pair (L,v) where L ∈ SL(H (x)) and v ∈ H (x). We call L the linear part of g,and v the translational part. Let Q (x) denote the group of linear maps from H (x) to itself which preserve the 1 1 1 flag {0}⊂ V (H )(x) ⊂ ··· ⊂ V (H )(x) = H (x),and let Q (x) ⊂ Q (x) denote the ≤1 ≤k ++ + 1 1 unipotent subgroup of maps which are the identity on V (H )(x)/V (H )(x) for all i. ≤i <i Let G (x) denote the subgroup of G(x) in which the linear part lies in Q (x),and let + + G (x) denote the subgroup of G (x) in which the linear part lies in Q (x).Notethat ++ + ++ + 1 + G (x) is unipotent. Also, since W (x) = V (H )(x), G (x) preserves W (x). ++ ≤k−1 ++ GM 1 1 For y near x, we have the Gauss-Manin connection P (x, y) : H (x) → H (y). GM + This induces a map P (x, y) : G(x) → G(y). In view of Lemma 4.1,for y ∈ W [x], GM GM P (y, x)G (y) = G (x), P (y, x)Q (y) = Q (x), + + + + ∗ ∗ GM GM P (y, x)Q (y) = Q (x) and P (y, x)G (y) = G (x). ++ ++ ++ ++ ∗ ∗ INVARIANT AND STATIONARY MEASURES 145 We may consider elements of G (x) and G (x) as affine maps from W [x] to + ++ + + + W [x]. More precisely, g = (L,v) ∈ G(x) corresponds to the affine map W [x]→ W [x] given by: (6.1) z → x + L(z − x) + v. Then, Q (x) is the stabilizer of x in G (x).Wedenoteby Lie(G )(x) the Lie algebra ++ ++ ++ of G (x),etc. ++ We will often identify W (x) with the translational part of Lie(G )(x). Then, we ++ + + have an exponential map exp : W (x) → G (x), taking v ∈ W (x) to expv ∈ G (x). ++ ++ + + Then, expv : W [x]→ W [x] is translation by v. The maps Tr(x, y) and tr(x, y).— For h ∈ G(x),let Conj(h) to be the conjugation −1 map g → hgh ,and let Ad(h) : Lie(G)(x) → Lie(G)(x) be the adjoint map. Suppose y ∈ W [x].Let Tr(x, y) : G(x) → G(y) and tr(x, y) : Lie(G)(x) → Lie(G)(y) be defined as GM Tr(x, y) = P (x, y) ◦ Conj exp(x − y) , GM tr(x, y) = P (x, y) ◦ Ad exp(x − y) . The following lemma is clear from the definitions: Lemma 6.1. — Suppose y ∈ W [x]. Then the elements g ∈ G(x) and g ∈ G(y) correspond x y + + to the same affine map of W [x]= W [ y] (in the sense of (6.1)) if and only if g = Tr(x, y)g . y x Admissible partitions. — By an admissible measurable partition we mean any parti- tion B as constructed in Section 3 (with some choice of C and T (x)). 0 0 Generalized subspaces. — Let U (x) ⊂ G (x) be a connected Lie subgroup. We write ++ U [x]= ux : u ∈ U (x) and call U [x] a generalized subspace. We have U [x]⊂ W [x]. Definition 6.2. — Suppose that for almost all x ∈ X we have a distinguished subgroup U (x) of G (x). We say that the family of subgroups U (x) is compatible with ν if the following hold: ++ (i) The assignment x → U (x) is measurable and g -equivariant. (ii) For any admissible measurable partition B of X , the sets of the form U [x]∩ B [x] are a measurable partition of X . (iii) For any admissible measurable partition B of X ,for almost every x ∈ X , the conditional 0 0 + + measure of ν along U [x]∩ B [x] is a multiple of the unique U (x) invariant measure on + + + + + U [x] U (x)/(U (x) ∩ Q (x)). (Note that both U (x) and U (x) ∩ Q (x) ++ ++ are unimodular, since they are unipotent. Hence there is a well-defined Haar measure on the + + quotient U (x)/(U (x) ∩ Q (x)).) ++ 146 ALEX ESKIN, MARYAM MIRZAKHANI (iv) We have, for almost all x ∈ X and almost all u ∈ U (x), + + (6.2)U (ux) = Tr(x, ux)U (x). + + (This is motivated by Lemma 6.1 and the fact that we want U [ux]= U [x].) Thus, + + (6.3)Lie U (ux) = tr(x, ux) Lie U (x). + + (v) U (x) ⊃ exp N(x) where N(x) ⊂ W (x) is the direction of the orbit of the unipotent N ⊂ SL(2, R). Standing assumption. — We are assuming that for almost every x ∈ X thereisa + + distinguished subgroup U (x) of G (x) so that the family of subgroups U (x) is com- ++ patible with ν in the sense of Definition 6.2. This will be used as an inductive assumption in Section 12. We emphasize that U (x) is defined for x ∈ X . Using our notational conventions, + + for x ∈ X, we write U (x) for U (σ (x)) etc. The unipotent N as a compatible system of measures. — At the start of the induction + + we have U (x) = exp N(x) ⊂ G (x).Wenow verify that U (x) = exp N(x) is a fam- ++ ily of subgroups compatible with ν in the sense of Definition 6.2.Notethat N(x) = 1 1 + V (H )(x) = V (H )(x). In particular, by Lemma 4.1,for y ∈ W [x], ≤1 1 (6.4)N(y) = P (x, y)N(x). GM This implies (i) and (ii) of Definition 6.2. The subgroup U (x) = exp N(x) ⊂ G (x) consists of pure translations (i.e. ++ + + U (x) ∩ Q (x) is only the identity map). In particular, U [x]= N[x]. This, together ++ with the N-invariance of ν implies (iii) of Definition 6.2. + + Note that since U (x) consists of pure translations, for any y ∈ W [x],Conj(exp(y− + + x))(U (x)) = U (x). This, together with (6.4) implies (iv) of Definition 6.2. The sets B[x], B [x] and B(x).— Recall the partitions B [x] from Section 3.Let t t B [x]= U [x]∩ B [x].Wewillalsouse thenotation B[x] for B [x]. t t 0 For notational reasons, we will make the following construction: let + + B (x) = u ∈ U (x)/ U (x) ∩ Q (x) : ux ∈ B [x] . t ++ t We also write B(x) for B (x). 0 INVARIANT AND STATIONARY MEASURES 147 The Haar measure. — Let |·| denote the conditional measure of ν on B[x].(By our assumptions, this measure is U (x)-invariant where it makes sense.) We also denote the Haar measure (with some normalization) on B(x) by |·|. Unless otherwise specified, all statements will be independent of the choice of normalization. ThesameargumentasLemma 3.11 also proves the following: Lemma 6.3. — Suppose δ> 0, θ > 0 and K ⊂ X, with ν(K)> 1 − δ. Then there exists ∗ ∗ ∗ asubset K ⊂ K with ν(K )> 1 − δ/θ such that for any x ∈ K ,and any t > 0, |K ∩ B [x]| ≥ 1 − θ |B [x]|, t t and thus u ∈ B (x) : ux ∈ K ≥ 1 − θ |B (x)|. t t The “ball” B(x, r).— For notational reasons, for 0 < r ≤ 1/50, and x ∈ X we define + + + B(x, r) = u ∈ U (x)/ U (x) ∩ Q (x) : d (ux, x)< r , ++ where d (·,·) is as in Section 3.InviewofProposition 3.4, we will normally use the ball + + B(x, 1/100) ⊂ U (x)/(U (x) ∩ Q (x)). ++ Lyapunov subspaces. — Suppose W is a subbundle of H .Let λ (W)>λ (W)> big 1 2 ··· >λ (W) denote the Lyapunov exponents of the action of g on W, and for x ∈ X let n t 0 V (W)(x) denote the corresponding subspaces. Let V (W) = V (W). i ≤i i j=1 Notational convention. — In this subsection, we write V (x), V (x) and λ instead of i ≤i i V (Lie(G ))(x), V (Lie(G ))(x) and λ (Lie(G )). i ++ ≤i ++ i ++ Since Lie(U )(x) and Lie(Q )(x) are equivariant under the g action, we have ++ t + + Lie U (x) = Lie U (x) ∩ V (x), Lie(Q )(x) = Lie(Q )(x) ∩ V (x). ++ ++ i The spaces H (x) and H (x).— Let H (x) = Hom(Lie(U )(x), Lie(G )(x)). + ++ + ++ (Here, Hom means linear maps between vector spaces, not Lie algebra homomorphisms.) For every M ∈ H (x), we can write (6.5)M = M where M ∈ Hom Lie U (x) ∩ V (x), Lie(G )(x) ∩ V (x) . ij ij j ++ i ij 148 ALEX ESKIN, MARYAM MIRZAKHANI Let H (x) = M ∈ H (x) : M = 0if λ ≤ λ . ++ + ij i j Then, H is the direct sum of all the positive Lyapunov subspaces of the action of g ++ t on H . Parametrization of generalized subspaces. — Suppose M ∈ H (x) is such that (I + M) Lie(U )(x) is a subalgebra of Lie(G )(x). We say that the pair (M,v) ∈ H (x) × ++ + W (x) parametrizes the generalized subspace U if U = exp (I + M)u (x + v) : u ∈ Lie U (x) . (Thus, U is the orbit of the subgroup exp[(I + M) Lie(U )(x)] through the point x + v ∈ W [x].) In this case we write U = U(M,v). Remark. — In this discussion, U is a generalized subspace which passes near the + + point x ∈ X .However, U need not be U [x],oreven U [ y] for any y ∈ X . 0 0 Remark. — From the definitions, it is clear that any generalized subspace U ⊂ + + W [x] can be parametrized by a pair (M,v) ∈ H (x) × W (x). Also, if v = v and (6.6)I + M = I + M ◦ J, + + + where J : Lie(U )(x) → Lie(U )(x) is a linear map, then (M,v) ∈ H (x) × W (x) and (M ,v ) ∈ H (x) × W (x) are two parameterizations of the same generalized subspace U . Example 1. — We give an example of a non-linear generalized subspace. (The example does not satisfy condition (v) of Definition 6.2 but this is not relevant for the + + discussion.) Suppose for simplicity that W has two Lyapunov exponents λ (W ) and + + + + λ (W ) with λ (W ) = 2λ (W ).Let e (x) and e (x) be unit vectors so that V (W )(x) = 2 1 2 1 2 1 Re (x),and V (W )(x) = Re (x). 1 2 2 + 3 3 Let i : W (x) → R be the map sending ae (x) + be (x) → (a, b, 1) ∈ R .Weiden- 1 2 + 3 tify W (x) with its image in R under i. Then, we can identify ⎛ ⎞ ⎛ ⎞ 1 ∗∗ 0 ∗∗ ⎝ ⎠ ⎝ ⎠ G (x) = 01 ∗ , Lie G (x) = 00 ∗ . ++ ++ 00 1 00 0 Suppose ⎧ ⎫ ⎛ ⎞ ⎨ 1 t ⎬ ⎝ ⎠ U (x) = : t ∈ R , 01 t ⎩ ⎭ 00 1 INVARIANT AND STATIONARY MEASURES 149 ⎧ ⎫ ⎛ ⎞ 0 t 0 ⎨ ⎬ ⎝ ⎠ Lie U (x) = 00 t : t ∈ R . ⎩ ⎭ + t + Then, U [x] is the parabola {x + te (x) + e (x) : t ∈ R}⊂ W [x]. 2 1 Transversals. — Note that we have, as vector spaces, Lie(G )(x) = Lie(Q )(x) ⊕ W (x) ++ ++ whereweidentify W (x) with the subspace of Lie(G )(x) corresponding to pure trans- ++ lations. For each i,and each x ∈ X ,let Z (x) ⊂ W (x) ∩ V (x) ⊂ Lie(G )(x) ∩ V (x) be 0 i1 i ++ i a linear subspace so that Lie(G )(x) ∩ V (x) = Z (x) ⊕ Lie U + Lie(Q ) (x) ∩ V (x) . ++ i i1 ++ i Let Z (x) ⊂ Lie(Q )(x) ∩ V (x) be such that i2 ++ i + + Lie U + Lie(Q ) (x) ∩ V (x) = Lie U (x) ∩ V (x) ⊕ Z (x). ++ i i i2 Let Z (x) = Z (x) ⊕ Z (x),and let Z(x) = Z (x).Wealwaysassume that thefunction i i1 i2 i x → Z(x) is measurable. We say that Z(x) ⊂ Lie(G )(x) is a Lyapunov-admissible transversal ++ to Lie(U )(x). All of our transversals will be of this type, so we will sometimes simply use the word “transversal”. Note that Z (x) = Z(x) ∩ W (x) ∩ V (x). i1 i + + + Example 2. — Suppose U (x) is as in Example 1. Then (since λ (W )−λ (W ) = 1 2 λ (W )), + + λ ≡ λ Lie(G ) = λ W λ ≡ λ Lie(G ) = λ W , 1 1 ++ 1 2 2 ++ 2 ⎛ ⎞ 00 ∗ ⎝ ⎠ V ≡ V Lie(G ) = 00 0 , 1 1 ++ 00 0 ⎛ ⎞ 0 ∗ 0 ⎝ ⎠ V ≡ V Lie(G ) (x) = 00 ∗ , 2 2 ++ 00 0 ⎛ ⎞ 0 ∗ 0 ⎝ ⎠ Lie(Q ) ∩ V (x) = 000 , ++ 2 ⎧ ⎫ ⎛ ⎞ 0 t 0 ⎨ ⎬ ⎝ ⎠ Lie U ∩ V (x) = 00 t : t ∈ R , ⎩ ⎭ 000 150 ALEX ESKIN, MARYAM MIRZAKHANI and (Lie(U ) ∩ V )(x) = (Lie(Q ) ∩ V )(x) ={0}. Therefore, Z (x) ={0},and 1 ++ 1 12 ⎛ ⎞ ⎛ ⎞ 0 ∗ 0 00 ∗ ⎝ ⎠ ⎝ ⎠ Z (x) = 000 Z (x) = 00 0 , Z (x) ={0}. 22 11 21 000 00 0 We note that in this example, the transversal Z was uniquely determined (and is in fact invariant under the flow g ). This is a consequence of the fact that we chose an example with simple Lyapunov spectrum, and would not be true in general. Parametrization adapted to a transversal. — We say that the parametrization (M,v) ∈ H (x) × W (x) of a generalized subspace U = U(M,v) is adapted to the transversal Z(x) if v ∈ Z(x) ∩ W (x) and Mu ∈ Z(x) for all u ∈ Lie U (x). The following lemma implies that adapting a parametrization to a transversal is similar to inverting a nilpotent matrix. Lemma 6.4. — Suppose the pair (M ,v ) ∈ H (x) × W (x) parametrizes a generalized ++ subspace U.Let Z(x) be a Lyapunov-admissible transversal. Then, there exists a unique pair (M,v) ∈ H (x) × W (x) which parametrizes U and is adapted to Z(x). If we write ++ M = M ij ij as in (6.5),and v = v , where v ∈ W (x) ∩ V (x), then M = M and v = v are given by formulas of the form j ij i j ij i (6.7) v = L v + p v , M i i i (6.8)M = L M + p M ij ij ij ij where L is a linear map and p is a polynomial in the v and M which depends only on the v with i i j jk j λ <λ and the M with λ − λ <λ . Similarly, L is a linear map, and p is a polynomial which j i j k i ij ij jk depends on the M with λ − λ <λ − λ . k l i j kl If we assume in addition that (M ,v ) is adapted to another Lyapunov-admissible transversal Z (x), then L and L can be taken to be invertible linear maps (depending only on Z(x) and Z (x)). i ij INVARIANT AND STATIONARY MEASURES 151 The proof of Lemma 6.4 is a straightforward but tedious calculation. It is done in Section 6.4 . Z + The map S .— Suppose Z is a Lyapunov-admissible transversal to U (x). Then, Z + + let S : H (x) × W (x) → H (x) × W (x) be given by ++ ++ S M ,v = (M,v) where M and v are given by (6.8)and (6.7) respectively. Note that S is a polynomial, but is not a linear map in the entries of M and v . To deal with the non-linearity, we work with certain tensor product spaces defined below. ˆ ˜ Tensor products: the spaces H, H and the maps j.— As in Section 5,for avectorspace ⊗m m ⊗m m ⊗m m Vand a map f : V → W weuse thenotations V ,V , f , f , j , j . Let m be the number of distinct Lyapunov exponents on H ,and let n be the ++ number of distinct Lyapunov exponents on W .Let (α;β) = (α ,...,α ;β ,...,β ) be 1 m 1 n a multi-index, and let m n ⊗α ⊗β i j (α;β) + H (x) = V (H )(x) ⊗ V W (x) i ++ j i=1 j=1 and let m n (α;β) ⊗α + ⊗β i j H (x) = H (x) ⊗ W (x) . ++ i=1 j=1 (α;β) (α;β) (α;β) ˆ ˜ We have a natural map πˆ : H (x) → H (x) given by (α;β) πˆ Y ⊗ ··· ⊗ Y ⊗ Y ⊗··· ⊗ Y 1 m 1 n ⊗β ⊗β ⊗α 1 n 1 ⊗α = π (Y ) ⊗··· ⊗ π (Y ) ⊗ π Y ⊗··· ⊗ π Y , 1 m m 1 1 n n + + where π : H (x) → V (H )(x) and π : W (x) → V (W )(x) are the natural pro- i ++ i ++ j jections associated to the direct sum decompositions H (x) = V (H )(x) and ++ i ++ i=1 + + W (x) = V (W )(x). j=1 Let S be a finite collection of multi-indices (chosen in Lemma 6.6 below). Then, let (α;β) (α;β) ˜ ˜ ˆ ˆ (6.9) H (x) = H , H (x) = H 0 0 (α;β)∈S (α;β)∈S (α;β) (α;β) ˆ ˜ ˆ Let πˆ : H (x) → H (x) be the linear map with coincides with πˆ on each H . 0 0 152 ALEX ESKIN, MARYAM MIRZAKHANI (α;β) + (α;β) ˆ ˆ Let j : H (x) × W (x) → H (x) be the “diagonal embedding” ++ (α;β) j (M,v) = M ⊗ M··· ⊗ M ⊗ v ⊗··· ⊗ v, + (α;β) ˆ ˆ ˆ and let j : H (x) × W (x) → H (x) be the linear map j .Let ++ 0 (α;β)∈S (6.10) j : H (x) × W (x) → H (x) ++ 0 ˆ ˆ ˆ ˜ denote πˆ ◦ j.Let H(x) denote the linear span of the image of j,and let H(x) denote the linear span of the image of j. ˆ ˜ Induced linear maps on H(x) and H(x).— Suppose F : H (x) → H (y) and t ++ ++ + + F : W (x) → W (y) are linear maps. Let f = (F , F ). Then, f induces a linear map t t t t t ˆ ˆ ˆ f : H(x) → H(y).If F sends each V (H )(x) to each V (H )(y) and F sends each t t i ++ i ++ + + ˜ ˜ ˜ V (W )(x) to V (W )(y),then f also induces a linear map f : H(x) → H(y). j j t t (++) (++) ˜ ˆ Note that H(x) ⊂ H(x) ⊂ H (x) where H (x) is as in Section 3. big big + + Notation. — For an invertible linear map A : W (x) → W (y),let A : Lie(G )(x) → Lie(G )(y) denote the map ++ ++ −1 (6.11)A (Y) = A ◦ Y ◦ A + A ◦ Y ∗ 1 2 where for Y ∈ Lie(G )(x),Y is the linear part of Y and Y is the pure translation part. ++ 1 2 Lemma 6.5. — Suppose x ∈ X ,u ∈ U (x). Then, there exists a linear map u : H (x) × 0 ∗ ++ + + W (x) → H (ux) × W (ux) with the following properties: ++ (a) If (M ,v ) ∈ H (x) × W (x) parametrizes a generalized subspace U,then (M,v) = ++ u (M ,v ) parametrizes the same generalized subspace U . (b) If (M,v) = u (M ,v ), then M and v are given by formulas of the form (6.7) and (6.8). Proof. — In fact we claim that (6.12) u M ,v = tr(x, ux) ◦ M ◦ tr(ux, x), exp I + M Y x + v − exp(Y)x , where Y = log u. This can be verified as follows. Let U = U(M ,v ) denote the generalized subspace parametrized by (M ,v ),and let U = exp((I+ M ) Lie(U )(x)),sothat U is a subgroup of G (x). Then, for any w ∈ U , U = U w. Then, in view of Lemma 6.1 and (6.1), ++ U = Tr(x, ux)U ux + (w − ux) . Thus, (M,v) ∈ H (ux) × W (ux) parametrizes U if ++ (6.13)exp (I + M) Lie U (ux) = Tr(x, ux)U INVARIANT AND STATIONARY MEASURES 153 and (6.14) v = w − ux for some w ∈ U . Now let (M,v) be the right-hand-side of (6.12). We claim that (6.13)and (6.14) hold. Indeed, by (6.3), + + tr(ux, x) Lie U (ux) = Lie U (x), + + and furthermore, tr(ux, x)(Lie(U ) ∩ V )(ux) = (Lie(U ) ∩ V )(x).Now, ≤i ≤i Tr(x, ux)U = exp tr(x, ux) Lie U = exp tr(x, ux) I + M Lie U (x) = exp tr(x, ux) I + M tr(ux, x) Lie U (ux) = exp (I + M) Lie U (ux) , verifying (6.13). Also, let w = exp I + M Y x + v ∈ U = U M ,v . Therefore, since exp(Y)x = ux, w − ux = exp I + M Y x + v − exp(Y)x = v, and hence (6.14) holds. Thus, u (M ,v ) ∈ H (ux) × W (ux) as defined in (6.12) ∗ ++ parametrizes the same generalized subspace U as (M ,v ) ∈ H (x)× W (x). This com- ++ pletes the proof of part (a). It is clear from (6.12) that part (b) of the lemma holds. Lemma 6.6. — For an appropriate choice of S , the following hold: + Z(x) (a) Let Z(x) be a Lyapunov-admissible transversal to U (x). There exists a linear map S : ˜ ˜ H(x) → H(x) such that for all (M,v) ∈ H (x) × W (x), ++ Z(x) Z(x) S ◦ j (M,v) = j ◦ S (M,v). x x + + (b) Suppose u ∈ U (x),and let Z(ux) be a Lyapunov-admissible transversal to U (ux).Then, ˜ ˜ there exists a linear map (u) : H(x) → H(ux) such that for all (M,v) ∈ H (x) × ∗ ++ W (x), Z(ux) (u) ◦ j (M,v) = j ◦ S ◦ u (M,v), ∗ ∗ ux + + where u : H (x) × W (x) → H (ux) × W (ux) is as in (6.12). ∗ ++ ++ 154 ALEX ESKIN, MARYAM MIRZAKHANI Proof. — Part (a) formally follows from the universal property of the tensor prod- uct and the partial ordering in (6.7)and (6.8). We now make a brief outline: see also Example 3 below. Let H (x) and j be as in (6.9)and (6.10) with the dependence on S explicit. Let S denote the set of multi-indices of the form (0,..., 0, 1, 0,..., 0; 0,..., 0) or S + (0,..., 0; 0,..., 0, 1, 0,..., 0).Then j is an isomorphism between H (x) × W (x) ++ and H (x). Z(x) Let (M,v) = S (M ,v ).By(6.7), (6.8) and the universal property of the tensor ˜ ˜ product, there exists S ⊃ S and a linear map S : H (x) → H (x) such that 1 0 1 S S 1 0 S S 0 1 j (M,v) = S ◦ j M ,v . We now repeat this procedure to get a sequence S of multi-indices. More precisely, at each stage, for each (α;β) ∈ S , we may write, by (6.7), (6.8) and the universal property of the tensor product, (α;β) (α;β) (α;β) (α;β) (α ;β ) j (M,v) = L j (M,v) + S j M ,v , j+1 (α ;β )∈S(α;β) (α;β) (α;β) where L and S are linear maps; we then define S = S ∪ S(α;β). j+1 j j+1 (α;β)∈S Putting these maps together, we then get a linear map S such that S S j j+1 j (M,v) = S ◦ j M ,v . Because of the partial order in (6.7)and (6.8), we may assume that S(α;β) consists of multi-indices (α ;β ) where either α hasmorezeroentries than α or β hasmorezero entries than β . Therefore, this procedure eventually terminates, so that S = S for large j+1 j enough j.Wethendefine S to be the eventual common value of the S ;thenpart(a) of Lemma 6.6 holds. To prove part (b) of Lemma 6.6, note that part (b) of Lemma 6.5 and the proof ˜ ˜ of part (a) of Lemma 6.6 show that there exists a map u˜ : H(x) → H(ux) such that ˜ ˜ u˜ ◦ j = j ◦ u ,where u is as in (6.12). Now, we can define (u) : H(x) → H(ux) to be ∗ ∗ ∗ ∗ Z(ux) Z(ux) S ◦˜ u ,where S is as in (a). Thus (u) denotes the induced action of u on H(x). ∗ ∗ ux ux Example 3. — Suppose U is as in Examples 1 and 2.Let ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ 010 001 000 ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ F = 001 , E = 000 , E = 001 . 1 2 000 000 000 Then, (Lie(U ) ∩ V )(x) = RF, (Lie(G ) ∩ V )(x) = RE . Then, for M ∈ H (x),the 2 ++ 1 1 ++ only non-zero component is M ∈ Hom((Lie(U ) ∩ V )(x),(Lie(G ) ∩ V )(x)), which 12 2 ++ 1 INVARIANT AND STATIONARY MEASURES 155 is 1-dimensional. Let ∈ Hom Lie U ∩ V (x), Lie(G ) ∩ V (x) 2 ++ 1 denote the element such that F = E ,sothat H = R . 1 ++ With the choice of transversal Z given in Example 2, Equations (6.7)and (6.8) become: (6.15) v =−M v + v − v ,v = 0, M = M . 1 2 12 12 2 1 2 12 Then we can choose S ={(1; 0, 0),(0; 1, 0),(0; 0, 1),(1; 0, 1),(0; 0, 2)}, so that (drop- ping the (x)), + + + H = H ⊕ V W ⊕ V W ⊕ H ⊗ V W 0 ++ 1 2 ++ 2 + + ⊕ V W ⊗ V W . 2 2 ⊗0 (Sincefor anyvectorspace V, V = R, we have omitted such factors in the above for- Z(x) ˜ ˜ mula.) Let S = S . Then, the linear map S : H(x) → H(x) is given by S() = , S(E ) = E , S(E ) = 0, S( ⊗ E ) =−E , 1 1 2 2 1 S(E ⊗ E ) =−E . 2 2 1 Example 4. — We keep all notation from Examples 1–3. Suppose u = exp Y, where Y = tF. We now compute the map (u) . Note that by Lemma 4.1,wehave e (ux) = e (x). Also note that by Example 1, 1 1 at x, the tangent vector to U [x] coincides with e (x). Recall that we are assuming that the foliation whose leaves are U [x] is invariant under the geodesic flow. This implies that at the point ux, the tangent vector to the parabola U [x] is e (ux). Therefore, e (ux) = e (x), e (ux) = te (x) + e (x). 1 1 2 1 2 Therefore, + + P (x, ux)e (x) = e (ux), P (x, ux)e (x) = e (ux) = te (x) + e (x). 1 1 2 2 1 2 Suppose U is parametrized by (M ,v ),where M = M , v = v e (x) + v e (x). 1 2 12 1 2 Then ⎛ ⎞ ⎛ ⎞ 1 1 2 2 1 t t + M t 1 t t 2 2 ⎝ ⎠ ⎝ ⎠ exp I + M Y = 01 t , exp(Y) = 01 t . 00 1 00 1 Therefore, ⎛ ⎞ v + tv + tM 1 2 12 ⎝ ⎠ exp I + M Y x + v − exp(Y)x = v . 0 156 ALEX ESKIN, MARYAM MIRZAKHANI Let ∈ Hom((Lie(U ) ∩ V )(ux),(Lie(G ) ∩ V )(ux)) be the analogue of ,but at 2 ++ 1 the point ux. Then, u M ,v = u M ,v e (x) + v e (x) ∗ ∗ 1 2 12 1 2 = M , v + tv + tM e (x) + v e (x) 1 2 12 1 2 12 2 = M , v + tM e (ux) + v e (ux) 1 2 12 1 12 2 Z(ux) Then, in view of (6.15), (S ◦ u )(M ,v ) = (M ,v e (ux) + v e (ux)),where ∗ 12 1 1 2 2 ux v =−M v + v + tM − v ,v = 0, M = M . 1 2 12 12 2 1 12 2 12 ˜ ˜ Then, (u) : H(x) → H(ux) is given by (u) () = + tE,(u) (E ) = E,(u) (E ) = 0, ∗ 1 ∗ 1 1 ∗ 2 (u) ( ⊗ E ) =−E,(u) (E ⊗ E ) =−E . ∗ 2 1 ∗ 2 2 1 The dynamical system G .— Suppose we fix some Lyapunov-admissible transversal Z(x) for every x ∈ X . Suppose (M,v) ∈ H (x) × W (x) is adapted to Z(x).Let 0 ++ Z(g x) −1 + G (M,v) = S g ◦ M ◦ g ,(g ) v ∈ H (g x) × W (g x), t t t ∗ ++ t t g x t where (g ) on the right-hand side is g acting on W (x),and g on the right-hand t ∗ t t side is the natural map Lie(Q )(x) → Lie(Q )(g x), which maps Lie(U )(x) to ++ ++ t Lie(U )(g x). Then, if U is the generalized subspace parametrized by (M,v) then (M ,v ) = G (M,v) ∈ H (g x) × W (g x) parametrizes g U and is adapted to Z(g x). t ++ t t t t From the definition, we see that G = G ◦ G . t+s t s Also, it is easy to see that for (M,v) ∈ H (x) × W (x), ++ −1 −1 g Z(g x) G (M,v) = g ◦ M ◦ g ,(g ) v , where M ,v = S (M,v). t t t ∗ t x The bundle H(x).— Suppose we are given a Lyapunov adapted transversal Z(x) at each x ∈ X .Let Z(x) H(x) = S H(x) Z(x) + denote the image of H(x) under S . Then, if (M,v) ∈ H (x) × W (x) is adapted to ++ Z(x),then j(M,v) ∈ H(x). We can also consider (u) as defined in Lemma 6.6(b) to be a map (u) : H(x) → H(ux). ∗ INVARIANT AND STATIONARY MEASURES 157 The bundle H and the flow g .— Let Z(x) be an admissible transversal to U (x) for every x ∈ X .Let (g ) : H(x) → H(g x) be given by 0 t ∗ t Z(g x) −1 (6.16) (g ) = S ◦ f where f (M,v) = g ◦ M ◦ g ,(g ) v , t ∗ t t t t ∗ g x t ˜ ˜ f is the map induced by f on H ⊃ H, (g ) on the right-hand side is g acting on W (x), t t t ∗ t + + Z g on the right-hand side is the natural map Lie(U )(x) → Lie(U )(g x),and S is as in t t Lemma 6.6.Then (g ) is a linear map, and for (M,v) ∈ H (x) × W (x), t ∗ ++ (6.17) (g ) j(M,v) = j G (M,v) . t ∗ t Since G ◦ G = G , and the linear span of j(H (x)× W (x)) is H(x) ⊃ H(x), it follows t s t+s ++ from (6.17)that (g ) ◦ (g ) = (g ) . t ∗ s ∗ t+s ∗ Lemma 6.7. (a) Suppose u x = ux ∈ U [x] and v ∈ H(x).Then (u) v = (u ) v. ∗ ∗ + + (b) Suppose u ∈ U (g x). Then there exists u ∈ U (x) such that g u x = ug x. Furthermore, t t t for any choice of u satisfying g u x = ug x and any v ∈ H(x), we have (u) (g ) v = t t ∗ t ∗ (g ) (u ) v. t ∗ ∗ Proof. — It is enough to prove (a) for v = j(M,v) where (M,v) ∈ H (x)× W (x). ++ Let U be the generalized subspace parametrized by (M,v). Then, (u) v = j(M ,v ) where (M ,v ) ∈ H (ux) × W (ux) is the (unique) parametrization of U adapted to ++ Z(ux).But then (u ) v is also a parametrization of U adapted to Z(ux). Therefore (u ) v = (u) v. ∗ ∗ The proof of (b) is essentially the same. + Z(x) Choosing M and C .— For a.e. x ∈ X, let M (x) =S ,and let 0 0 − Z(x) M (x) = sup inf v: v ∈ H(x), S (v) = w . Z(x) w∈S (H(x)) + − Choose M > 1 sufficiently large so that C ≡{x ∈ X : max(M (x), M (x)) < M } has 0 0 0 0 positive measure. Let C ⊂ C and T : C → R be as in Lemma 4.14 (with this choice of 0 0 M , C ). 0 0 Adjusting the transversal Z(x).— For c ∈ C,let E [c], t(c) and J be as in Propo- + + sition 3.7.For x ∈ E [c] we define Z(x) = P (c, x) Z(c),and for 0 ≤ t < t(c),wedefine Z(g x) = g Z(x). This defines Z(y) for y ∈ J . From now on, we assume that the transver- −t −t c sal Z is obtained via this construction. 158 ALEX ESKIN, MARYAM MIRZAKHANI ˜ ˜ ˜ Lemma 6.8. —Let (g ) : H(x) → H(g x) and f : H(x) → H(g x) be as in (6.16). Then t ∗ t t t Z(x) the Lyapunov subspaces for (g ) at x are the image under S of the Lyapunov subspaces of f at x, and t ∗ t the Lyapunov exponents of g are those Lyapunov exponents of f whose Lyapunov subspace at a generic t t Z(x) point x is not contained in the kernel of S . ˜ ˜ Proof.—Let V (H)(x) and V (H)(x) denote the Lyapunov subspaces of the flow f i i t and g respectively, and let λ (H) and λ (H) denote the corresponding Lyapunov expo- t i i nents. Then, for v ∈ V (H), by the multiplicative ergodic theorem, for every > 0, Z(x) Z(g x) Z(g x) λ (H)t+|t| t t i ˜ ˜ g S v=S f v ≤S f v≤ C (x)C (g x)e . t t Y t 1 t x g x g x t t Z(x) Taking t →∞ and t →−∞ we see that λ (H) = λ (H) and S v ∈ V (H)(x). i i i The measurable flat connection P (x, y).— Recall that the measurable flat g -equi- + + 1 variant W -connection map P on H induces a measurable flat g -equivariant connec- (++) ˜ ˜ tion on H ,and thus on H. We will call this connection P (x, y). Then, we can define big + + a measurable flat W -connection P (x, y) : H(x) → H(y) by + Z(y) + + (6.18) P (x, y) = S ◦ P (x, y), y ∈ W [x]. Without loss of generality, we may assume that Lemma 4.3 applies to subbundles of H (++) as well as subbundles of H (or else we can replace X by a measurable finite cover). big Then, Proposition 4.12 applies to P . The dynamical inner product ·,· and the dynamical norm · on H.— Even though H x x (++) (++) is not formally a subbundle of H , H ⊂ H ⊂ H . Thus, the AGY norm makes sense big big + + in H. Note that by our choices of C and M ,(4.17) holds for P in place of P (and 1 in 0 0 place of M ). Then, the proof of Proposition 4.15 goes through. Thus, Proposition 4.15 also applies to H, with a norm which may be different from the norm obtained from (++) thinking of H as a subset of H . big 6.1. Approximation of generalized subspaces and the map A(·,·,·,·). Hausdorff distance between generalized subspaces. — For x ∈ X , and two generalized subspaces U and U ,let hd (U , U ) denote the Hausdorff distance using the metric X X X 0 0 0 d (·,·) defined in Section 3 between U ∩ B (x, 1/100) and U ∩ B (x, 1/100).(The balls B (·,·) are defined in Section 5.) Lemma 6.9. — Suppose x ∈ X , (M,v) ∈ H (x) × W (x),and 0 ++ X + hd U [x], U(M,v) ≤ 1/100. x INVARIANT AND STATIONARY MEASURES 159 (a) We have for some absolute constant C > 0, X + hd U [x], U(M,v) ≤ Cmax(v ,M ). Y Y Also if (M,v) ∈ H (x) × W (x) is adapted to Z(x), then there exists c(x)> 0 such ++ that X + hd U [x], U(M,v) ≥ c(x) max(v ,M ). Y Y (b) For some c (x)> 0, we have, for (M,v) ∈ H (x) × W (x) adapted to Z(x), 1 ++ X + −1 c (x)j(M,v) ≤ hd U [x], U(M,v) ≤ c (x) j(M,v) . 1 Y 1 Y Proof. — Part (a) is immediate from the definitions and Proposition 3.4. To see (b) note that part (a) implies that max(M ,v ) = O(1), and thus all the higher order Y Y terms in j(M,v) which are polynomials in M and v , have size bounded by a constant ij j multiple of the size of the first order terms, i.e. by max(M ,v ). Y Y We will be dealing with Hausdorff distances of particularly well-behaved sets (i.e. generalized subspaces parametrized by elements of H (x) × W (x)). For such sub- ++ spaces, the following holds: Lemma 6.10. — Suppose x ∈ X ,and U ⊂ W [x] is a generalized subspace. Then, (a) We have, for t ∈ R, −2|t| X + X + 2|t| X + 0 0 0 e hd U [x], U ≤ hd U [g x],(g ) U ≤ e hd U [x], U , t t ∗ x g x x provided the quantity on the right is at most 1/100. (The first inequality in the above line holds as long as the quantity in the middle is at most 1/100.) (b) Suppose that U is parametrized by an element of H (x) × W (x). There exists a func- ++ tion C : X → R finite almost everywhere and β> 0 depending only on the Lyapunov spectrum, such that, for t ≥ 0, −1 β t X + X + 0 0 C(x) e hd U [x], U ≤ hd U [g x],(g ) U , t t ∗ x g x provided the quantity on the right is at most 1/100.Also, for t < 0, X + −β|t| X + 0 0 hd U [g x],(g ) U ≤ C(x)e hd U [x], U , t t ∗ g x x provided the quantity on the right is at most 1/100. + X + Proof.—Recall that B (x, r) = B (x, r) ∩ W [x] denotes the ball of radius r in the metric d (·,·). Suppose t ≥ 0. Note that, by Lemma 3.5(d), for t > 0, + −1 + + B [x]≡ g B (g x, 1/100) ⊂ B (x, 1/100). t t 160 ALEX ESKIN, MARYAM MIRZAKHANI 2t Note that the action of g can expand in any direction by at most e , see also Lemma 3.6. Therefore, X + 2t X + + + 0 0 hd (g ) U [x],(g ) U ≤ e hd U [x]∩ B [x], U ∩ B [x] t ∗ t ∗ g x x t t 2t X + ≤ e hd U [x], U . This completes the proof of the second inequality in (a). The first inequality in (a) follows after renaming x to g x. We now begin the proof of (b). We assume t ≥ 0 (the proof for the case t < 0is identical). It is enough to show that for any δ> 0 there exists C = C(δ) < ∞ and a set K(δ) with measure at least 1 − δ such that for x ∈ K(δ) and t > 0, −1 β t X + X + 0 0 (6.19)C(δ) e hd U [x], U ≤ hd U [g x],(g ) U . t t ∗ x g x For any η> 0let K be the set where c (x)>η,where c (x) is as in Lemma 6.9. Choose η 1 1 η so that K has measure at least 1 − δ/4. By the Birkhoff ergodic theorem we may find a set K of measure at least 1 − δ/2and t > 0such that for x ∈ K and t > t , there exists 1 1 t ∈ R with |t − t | <t,and g x ∈ K . t η Let α> 0 be as in Lemma 3.5. Choose < α/2. By Lemma 3.5(c), we may find a set K ⊂ K of measure at least 1 − δ/2, and a constant t = t (δ) such that for all x ∈ K η 2 2 all t > t and all v ∈ H(x), αt (6.20) (g ) v ≥ e v . t ∗ Y Y Let K(δ) = K ∩ K ,and let t = max(t , t ).If 0 ≤ t <(1+)t ,then(6.19) holds in view 0 1 2 0 of Lemma 6.10(a). Suppose t >(1 + )t ,and let t be as in the definition of K . Since x ∈ K and g x ∈ K , by Lemma 6.9 and (6.20), η t η X + 2 αt X + 0 0 hd U [g x],(g ) U ≥ η e hd U [x], U . t t ∗ g x x Then, again using Lemma 6.10(a), we get X + −t X + 0 0 hd U [g x],(g ) U ≥ e hd U [g x],(g ) U . t t ∗ t t ∗ g x g x Now, (6.19) follows, with β = (α − ). Motivation. — We work in the universal cover X .Let q , q be as in Section 2.3, 0 1 so in particular, q ∈ W [q ]. Suppose u ∈ B(q , 1/100) and t > 0. Note that the gener- 1 1 + + alized subspace U [g q ]= U [g uq ] passes through the point g uq .If t is not too large, t 1 t 1 t 1 the generalized subspace U [g q ] will pass near g uq . These subspaces are not on the t t 1 + + + same leaf of W (even though the leaf W [g q ] containing U [g q ] gets closer to the leaf t t 1 1 + + + W [g q ]= W [g uq ] containing U [g uq ] as t →∞). It is convenient to find a way to t 1 t 1 t 1 + + “project” the part of U [g q ] near g uq to W [g uq ]. In particular, we want the projec- t t 1 t 1 tion to be again a generalized subspace (i.e. an orbit of a subgroup of G (g uq )). We also ++ t 1 INVARIANT AND STATIONARY MEASURES 161 want the projection to be exponentially close, in a ball of radius 1/100 about g uq ,tothe t 1 original generalized subspace U [g q ]. Furthermore, in order to carry out the program outlined in the beginning of Section 5, we want the pair (M ,v ) parameterizing the pro- jection to be such that j(M ,v ) ∈ H(g uq ) depends polynomially on P (q , q ).Thenit t 1 1 will depend linearly on F(q) − F(q ) since any fixed degree polynomial in P (q , q ) can (r) be expressed as a linear function of F(q) − F(q ) as long as r in the definition of L (q) ext is chosen large enough. More precisely, we need the following: Proposition 6.11. — Suppose α > 0 is a constant. We can choose r sufficiently large (de- pending only on α and the Lyapunov spectrum) so that there exists a linear map A(q , u,, t) : 3 1 (r) + L (g q ) → H(g uq ), defined for almost all q ∈ X , almost all u ∈ U [x],all ≥ 0 and all ext − 1 t 1 1 0 t ≥ 0, and a constant α > 0 depending only on α and the Lyapunov spectrum such that the following 1 3 hold: (i) We have (6.21) A q , u, + , t + t = g ◦ A(q , u,, t) ◦ g . 1 t 1 (ii) Suppose δ> 0,and is sufficiently large depending on δ. There exists a set K = K(δ) with ν(K)> 1−δ and constants C (δ) and C (δ) such that the following holds: Suppose 1 2 −1 −1 − q ∈ π (K).Let q = g q (see Figure 1). Suppose q ∈ π (K) ∩ W [q] satisfies 1 − 1 the upper bounds in (5.3) and (5.4) with the same constant δ, and write q = g q . For all −1 u ∈ B(q , 1/100) such that uq ∈ π (K),and any t > 0 such that 1 1 (6.22) t ≤ α , X + (6.23) d g uq , U g q ≤ 1/100, t 1 t and also −α X + + 1 0 (6.24)C (δ)e ≤ hd U [g uq ], U g q , 1 t 1 t g uq 1 t 1 we have −1 (6.25) C(g uq ) A(q , u,, t) F q − F(q) t 1 1 X + + ≤ hd U [g uq ], U g q t 1 t g uq 1 t 1 ≤ C(g uq ) A(q , u,, t) F q − F(q) , t 1 1 where C : X → R is a measurable function finite almost everywhere. (iii) Suppose δ, ,q,u,q ,q ,are as in (ii), and t satisfies (6.22) and (6.23). Then, we have (6.26) A(q , u,, t) F q − F(q) = j M ,v , 1 162 ALEX ESKIN, MARYAM MIRZAKHANI where the pair (M ,v ) ∈ H (g uq ) × W (g uq ) (which will be chosen in the proof) ++ t 1 t 1 is adapted to Z(g uq ) and parametrizes a generalized subspace U(M ,v ) ⊂ W (g uq ) t 1 t 1 satisfying X + −α 0 1 (6.27) hd U g q , U M ,v ≤ C (δ)e . t 3 g uq 1 t 1 Part (ii) of Proposition 6.11 is key to resolving “Technical Problem #1” of Sec- tion 2.3 (see the discussion at the beginning of Section 5). We claim part (ii) of Proposi- tion 6.11 follows easily from part (iii) of Proposition 6.11 and Lemma 6.9(b). Indeed, by the triangle inequality, X + + X + 0 0 (6.28) hd U [g uq ], U g q = hd U [g uq ], U M ,v t 1 t t 1 g uq 1 g uq t 1 t 1 X + + O hd U M ,v , U g q . g uq 1 t 1 The O(·) term on the right-hand-side of (6.28)isbounded by (6.27), and the size of the first term on the right-hand-side of (6.28) is comparable to j(M ,v ) by Lemma 6.9(b). Thus, (6.25) follows from (6.26). Lemma 6.12. —For any δ> 0, there exists K = K (δ) ⊂ X with ν(K )> 1 − c(δ) where c(δ) → 0 as δ → 0, and constants C (δ) > 0, C (δ) > 0 and C (δ) > 0 such that in 1 2 4 Proposition 6.11(ii) and (iii), the conditions (6.23) and (6.24) can be replaced by either (a) g uq ∈ K and t 1 −α (6.29)C (δ)e ≤ A(q , u,, t) F q − F(q) ≤ C (δ), 1 2 or by (b) −α X + (6.30)C (δ)e ≤ hd U [g uq ], U M ,v ≤ 1/400, t 1 4 g uq t 1 where U(M ,v ) is as in (6.26). Proof of Lemma 6.12.—Let c (x) be as in Lemma 6.9(b). There exists a compact K ⊂ X with ν(K )> 1 − c(δ),with c(δ) → 0as δ → 0, and a constant 1 < C (δ )< ∞ −1 with C (δ ) →∞ as δ → 0such that c (x) < C (δ ) for all x ∈ K . Then, in view of Lemma 6.9(b), there exist 0 < C (δ) < C (δ) and C (δ) > 0such that for t such that 1 2 4 g uq ∈ K and (6.29) holds, (6.30) also holds. Thus, it is enough to show that if for some t 1 t > 0(6.22)and (6.30) hold, then (6.23)and (6.24) also hold. + X + Let t = min{s ∈ R : d (g uq , U [g q ]) ≥ 1/100},sothatfor 0 ≤ t ≤ t max s 1 s max (6.23) holds. If t ≥ α ,thenfor t ∈[0,α ),(6.23) is automatically satisfied. Now as- max 3 3 sume t <α . Then, by the definition of t and Proposition 6.11(iii), (i.e. (6.26)and max 3 max (6.27)), and assuming is sufficiently large (depending on δ)wehave d g uq , U M ,v ≥ 1/200. t 1 max INVARIANT AND STATIONARY MEASURES 163 Let U = g U(M ,v ) ⊂ W [uq ].ByProposition 6.11(iii), for 0 ≤ t ≤ t , g U is 0 −t 1 max t 0 max parametrized by (M ,v ) satisfying (6.26). t t Suppose t > 0satisfies (6.22)and (6.30). Let + X t = max s ∈ R : d (g uq , g U ) ≤ 1/200 . 1 s 1 s 0 Since by Lemma 3.5(iv) the function s → d (g uq , g U ) is monotone increasing, we have s 1 s 0 t < t ≤ t . Thus, since t < t ,(6.23) holds. In particular, Proposition 6.11(iii) applies 1 max max and then, (6.27)and (6.30) (with a proper choice of C (δ)) imply (6.24). Corollary 6.13. — Suppose δ, ,q,u,q ,q , are as in Proposition 6.11(ii),and s ≥ 0 is such that (6.22), (6.23),and (6.24) hold for s in place of t. Suppose t ∈ R is such that 0 < t + s <α . Then, there exists C (δ) > 0 such that (a) We have, for t ∈ R such that 0 < t + s <α , −2|t| X + + −α e hd U [g uq ], U g q − C (δ)e s 1 s 4 g uq 1 s 1 X + + ≤ hd U [g uq ], U g q s+t 1 s+t g uq 1 s+t 1 2|t| X + + −α ≤ e hd U [g uq ], U g q + C (δ)e , s 1 s 4 g uq 1 s 1 provided the quantity on the right is at most 1/800. (The first inequality in the above line holds as long as the quantity in the middle is at most 1/800.) (b) There exists a function C : X → R finite almost everywhere and β> 0 depending only on the Lyapunov spectrum, such that, for t ≥ 0, −1 β t X + + −α C(g uq ) e hd U [g uq ], U g q − C (δ)e s 1 s 1 s 4 g uq 1 s 1 X + + ≤ hd U [g uq ], U g q , s+t 1 s+t g uq 1 s+t 1 provided the quantity on the right is at most 1/800.Also, for t < 0, X + + hd U [g uq ], U g q s+t 1 s+t g uq 1 s+t 1 −β|t| X + + −α ≤ C(g uq )e hd U [g uq ], U g q + C (δ)e , s 1 s 1 s 4 g uq 1 s 1 provided the quantity on the right is at most 1/800. Proof. — Suppose 0 ≤ t ≤ α ,and is sufficiently large depending on δ.Let U 3 t denote the generalized subspace of Proposition 6.11(iii). Then, by Proposition 6.11 if X + X 0 0 d (g uq , U [g q ])< 1/200, then d (g uq , U )< 1/100. Conversely, by (the proof of) t 1 t t 1 t X X + 0 0 Lemma 6.12(b), if d (g uq , U )< 1/400, then d (g uq , U [g q ])< 1/200. Also, by t 1 t t 1 t Proposition 6.11(iii) and Lemma 6.12(b), if either of these conditions holds, then (6.27) holds. Thus, the corollary follows from Lemma 6.10. 164 ALEX ESKIN, MARYAM MIRZAKHANI Proposition 6.11 is proved by constructing a linear map P (uq , q ) : W (uq ) → s 1 1 W (q ) with nice properties; then the approximating subspace U(M ,v ) is given by −1 + ∗ g P (uq , q ) U [q ]. The construction is technical, and is postponed to Section 6.5 . t s 1 1 1 Then, Proposition 6.11 is proved in Section 6.6 . From the proof, we will also deduce the following lemma (which will be used in Section 12): Lemma 6.14. —For every δ> 0 there exists > 0 and a compact set K ⊂ X with ν(K)> −1 1 − δ so that the following holds: Suppose < 1/100. Suppose q ∈ π (K), > 0 is sufficiently − −1 large depending on δ, and suppose q ∈ W [q]∩ π (K) is such that (5.3) and (5.4) hold. Let q = g q, q = g q (see Figure 1). Fix u ∈ B(q , 1/100), and suppose t > 0 is such that 1 1 X + + hd U [g uq ], U g q ≤ . t 1 t 0 g uq 1 t 1 −1 + Furthermore, suppose q ,q ,uq ,q ,and g uq all belong to π (K). Suppose x ∈ U [g uq ]∩ 1 1 t 1 t 1 1 1 B (g uq , 1/100).Let t 1 + X A = U [g uq ]∩ B (x, ), t t 1 0 + X A = U g q ∩ B (x, ). t 0 t 1 Then, |g A | |g A | −t t −t −1 t κ ≤ + + + + |U [q ]∩ B (q , 1/100)| |U [q ]∩ B (q , 1/100)| 1 1 1 1 |g A | −t t ≤ κ , + + |U [q ]∩ B (q , 1/100)| 1 1 where κ depends only on δ and the Lyapunov spectrum, the “Haar measure” |·| is defined at the beginning of Section 6, and the ball B (x, r) is defined in Section 3.Also, X −α hd g A , g A ≤ e , −t t −t where hd (·,·) denotes the Hausdorff distance, and α depends only on the Lyapunov spectrum. This lemma will also be proved in Section 6.6 . 6.2. The stopping condition. — We now state and prove Lemma 6.15 and Proposi- tion 6.16 which tell us when the inductive procedure outlined in Section 2.3 stops. Recall the notational conventions Section 2.2. − − − − The sets L (q) and L [q].— For a.e q ∈ X ,let L [q]⊂ W [q] denote the smallest real-algebraic subset containing, for some > 0, the intersection of the ball of radius with the support of the measure ν , which is the conditional measure of ν along W [q] − − − W [q]. Then, L [q] is g -equivariant. Since the action of g is expanding along W [q], t −t we see that for almost all q and any > 0, L [q] is the smallest real-algebraic subset of − − X − − W (q) such that L [q] contains support(ν − ) ∩ B (q,).Let L (q) = L [q]− q. W [q] INVARIANT AND STATIONARY MEASURES 165 + + + + − − The sets L (q) and L [q].— Let πˆ : W(x) → W (x) and πˆ : W(x) → W (x) denote the maps + − − πˆ (v) = (1, 0) ⊗ v, πˆ (v) = (0, 1) ⊗ π (v), q q q 1 1 1 − + + − −1 − + + where π is as in (2.2). Let L (q)=ˆπ ◦ (πˆ ) L (q),and let L [q]= q + L (q). q q q The automorphism h and the set S [x].— Let h denote the automorphism of the t t 2t affine group G (x) which is the identity on the linear part and multiplication by e on ++ the translational part. For x ∈ X ,let + + S [x]= h U [x]. t∈R + + + + It is clear from the definition that S [x] is relatively closed in W [x],S [x]⊂ U [x],and + + also S [x] is star-shaped relative to x (so that if x + v ∈ S [x],sois x + tv for all t > 0). Lemma 6.15. — The following are equivalent: + + (a) L [x]⊂ S [x] for almost all x ∈ X . + + (b) There exists E ⊂ X with ν(E)> 0 such that L [x]⊂ S [x] for x ∈ E. + + (c) There exists E ⊂ X with ν(E)> 0 such that L [x]⊂ U [x] for x ∈ E. + + Proof. — It is immediately clear that (a) implies (b). Also, since S [x]⊂ U [x], (b) immediately implies (c). It remains to prove that (c) implies (a). Now suppose (c) holds. Let ⊂ X be the set such that for q ∈ , g q spends a 0 1 t 1 positive proportion of the time in E. Then, by the ergodicity of g , is conull. For q ∈ , t 1 we have, for a positive fraction of t, + + L [g q ]⊂ U [g q ]. t 1 t 1 + t Let A(x, t) denote the Kontsevich-Zorich cocycle. Then g acts on W by e A(q , t) and t 1 − −t − −t − + acts on W by e A(q , t). Therefore, L (g q ) = e A(q , t)L (q ),and thus L (g q ) = 1 t 1 1 1 t 1 −t + + t + e A(q , t)L (q ). Also, we have U (g q ) = e A(q , t)U (q ). Thus, for a positive measure 1 1 t 1 1 1 set of t,wehave + 2t + + (6.31)L (q ) ⊂ e U (q ) = h U (q ), 1 1 t 1 where h is as in the statement of Proposition 6.16. Since both sides of (6.31)depend + + analytically on t, we see that (6.31) holds for all t. Then, L [q ]⊂ S [q ]. 1 1 Proposition 6.16. — Suppose the equivalent conditions of Lemma 6.15 do not hold. Then, there exist constants α > 0, α > 0 and α > 0 depending only on the Lyapunov spectrum, such that for 1 2 1 any δ> 0 and any sufficiently small (depending on δ) > 0, there exist (δ,) > 0 and a compact 0 166 ALEX ESKIN, MARYAM MIRZAKHANI K ⊂ X with ν(K)> 1 − δ such that for q ∈ K there exists a subset Q(q ) ⊂ B(q , 1/100) with 0 1 1 1 |Q(q )| >(1 − δ)|B(q , 1/100)|,suchthatfor > (δ,),for u ∈ Q(q ),and for t > 0 such 1 1 0 1 that (6.32) −α ≤ α t − α ≤ 0, 1 2 1 we have −α α t 1 2 (6.33) A(q , u,, t)≥ e e . Consequently, if > 0 is sufficiently small depending on δ, > (δ,),q ∈ K,u ∈ Q(q ),and 0 1 1 t > 0 is chosen to be as small as possible so that A(q , u,, t)= , then t < α ,where α = α /α depends only on the Lyapunov spectrum. 3 3 1 2 Remark. — The constant α constructed during the proof of Proposition 6.16 de- pends only on the Lyapunov spectrum. This value of α is then used in Proposition 6.11 to construct the function A(·,·,·,·), which is referred to in (6.33). 6.3 . Proof of Proposition 6.16. Lemma 6.17. — Suppose k ∈ N,and > 0. For every sufficiently small δ> 0,and every compact K with ν(K )> 1 − δ, there exists a constant β(, k,δ) > 0 and compact set K = K (, K , k,δ) ⊂ K with ν(K )> 1 − c (δ) where c (δ) → 0 as δ → 0 such that the following 1 1 holds: −1 − Suppose q ∈ π (K ) and H ⊂ L [q] is a connected, degree at most k, R-algebraic set which − −1 − X is a proper subset of L [q]. Then there exists q ∈ π (K ) ∩ L [q] with d (q , q)< and d q , H >β. Proof. — This argument is virtually identical to the proof of Lemma 5.4 and of Lemma 5.5. Lemma 6.18. — Suppose k ∈ N,m ∈ N,q ∈ X ,and U ⊂ W [q ] is the image of a 1 0 1 m + + polynomial map of degree at most k from R to W [q ]. Suppose furthermore that U [q ] is also the 1 1 m + image of a polynomial map of degree at most k from R to W [q ],and > 0 is such that there exists u ∈ B(q , 1/100) with d uq , U = . Suppose δ> 0.Then, foratleast (1 − δ)-fraction of u ∈ B(q , 1/100), d uq , U >β, where β> 0 depends only on k, m, δ and the dimension. INVARIANT AND STATIONARY MEASURES 167 Proof. — This is a compactness argument. If the lemma was false, we would (after passing to a limit) obtain polynomial maps whose images are Hausdorff distance > 0 apart, yet coincide on a set of measure at least δ. This leads to a contradiction. The following lemma is stated in terms of the distance d (·,·).However,inview of Proposition 3.4, it is equivalent to the analogous statement for the Euclidean distance on W [x]. Lemma 6.19. — There exists C : X → R finite a.e and α> 0 depending only on the + X ˜ 0 Lyapunov spectrum such that for all q ∈ X and all z ∈ L [x] with d (z, q )< 1/100, 1 0 1 X + X + + 0 0 d z, U [x] ≥ C(x)d z, U [x]∩ L [x] . Proof. — By the Łojasiewicz inequality [KuSp, Theorem 2] for any x ∈ X and any + + X k-algebraic sets U ⊂ W [x],L ⊂ W [x],and any z with d (z, x)< 1/100, X X X α 0 0 0 d (z, U) + d (z, L) ≥ c(U, L)d (z, U ∩ L) , where c(U, L)> 0and α> 0 depends only on k and the dimension. + + + In our case, U = U [x].L = L [x],and z ∈ L [x]. The lemma follows. + − Recall that for x near q , π + (x) is the unique point in W [q ]∩ AW [x].Let 1 W (q ) 1 1 τ n = ⊂ N ⊂ SL(2, R). Lemma 6.20. — Suppose q ∈ X ,q ∈ W [q ]. Then, we have 1 0 1 −1 −1 − π n q = n q + (1, 0) ⊗ τ(1 + cτ) πˆ q − q , W (q ) τ τ 1 1 1 1 q 1 −1 where c = p(v) ∧ p(Im q ),q − q = (0, 1) ⊗ v,and τ = (1 − c)τ(1 + cτ) . 1 1 Proof. — Abusing notation, we work in period coordinates. Since q ∈ W [q ],we can write q = q + (0, 1) ⊗ v,where p(v) ∧ p(Re q ) = 0. Then, 1 1 n q = (1, 0) ⊗ Re q + τ(Im q + v) + (0, 1) ⊗ (Im q + v). τ 1 1 1 Let −1 w = v + cτ(1 + cτ) Im q . Then, p(w) ∧ p(Re(n q )) = 0, and thus, (0, 1) ⊗ w ∈ W (n q ). Therefore, τ τ 1 1 n q − (0, 1) ⊗ w = (1, 0) ⊗ Re q + τ(Im q + v) τ 1 1 −1 − + (0, 1) ⊗ (1 + cτ) Im q ∈ W n q . 1 τ 1 168 ALEX ESKIN, MARYAM MIRZAKHANI −1 (1+cτ) 0 We have ∈ A. Therefore, 01+cτ −1 − (6.34) (1, 0) ⊗ (1 + cτ) Re q + τ(Im q + v) + (0, 1) ⊗ Im q ∈ AW n q . 1 1 1 τ It is easy to check that (6.34)isin W [q ]. Therefore, −1 π + n q = (1, 0) ⊗ (1 + cτ) Re q + τ(Im q + v) W (q ) τ 1 1 1 1 + (0, 1) ⊗ Im q −1 = q + (1, 0) ⊗ τ(1 + cτ) Im q + v , 1 1 where v ∈ H is such that v = c Re q + v . Also −1 n π + n q = π + n q − (1, 0) ⊗ τ Im q W (q ) τ W (q ) τ 1 1 1 1 1 −1 = π + n q − (1, 0) ⊗ (1 − c)τ(1 + cτ) Im q . W (q ) τ 1 1 1 Therefore, −1 −1 n π + n q = q + (1, 0) ⊗ τ(1 + cτ) c Im q + v . W (q ) τ 1 1 τ 1 Also, −1 −1 − − c Im q + v = π (v) = πˆ q − q . 1 1 q q 1 1 1 This completes the proof of the lemma. Proof of Proposition 6.16. — Suppose the equivalent conditions of Lemma 6.15 do − − + −1 + − − not hold. For x ∈ X ,let U (x)=ˆπ ◦ (πˆ ) U (x),and let U [x]= x + U (x). Then, x x − − − − for a.e x ∈ X ,L [x]⊂ U [x], and hence U [x]∩ L [x] is a proper algebraic subset of L [x]. By Lemma 6.17, there exists a K ⊂ X with ν(K )> 1 − δ/4and K ⊂ X with 0 0 −1 ν(K )> 1 − δ/2such that for any q ∈ π (K ) and any degree k proper real algebraic − − subset H of L [q], there exists q ∈ L [q] satisfying the upper bounds in (5.3)and (5.4) such that d (q , H)>β (δ). −1 Now assume that q ≡ g q ∈ π (K ). (We will later remove this assumption.) − 1 − − − −1 Then, we apply Lemma 6.17 with H = g (U [q ]∩ L [q ]) to get q ∈ L [q]∩ π (K ) − 1 1 satisfying the upper bounds in (5.3)and (5.4)and so that X − − d q , g U [q ]∩ L [q ] ≥ β (δ). − 1 1 In view of Lemma 3.6 and Proposition 3.4, there exists N > 0such that for all − X x ∈ X and all y ∈ W [x] with d (x, y)< 1/100 and all t > 1, X −Nt X 0 0 d (g x, g y)> e d (x, y). t t INVARIANT AND STATIONARY MEASURES 169 Let q = g q . Then, q ∈ L [q ],and 1 1 X − − −N d q , U [q ]∩ L [q ] ≥ β (δ)e . 1 1 + + − −1 Let z ∈ L [q ] be such that πˆ ◦ (πˆ ) (q ) = z. Then, we have q q 1 1 1 X + + −N d z, U [q ]∩ L [q ] ≥ β (δ)e , 1 1 and thus by Lemma 6.19, X + −αN (6.35) d z, U [q ] ≥ β(δ)β (δ)e . Let U = U [q ]. Then, U is a generalized subspace, and q ∈ U . Furthermore, 1 1 both U and U [q ] are invariant under the action of N ⊂ SL(2, R). Without loss of generality, we may assume that is large enough so that the con- −1 stant c in Lemma 6.20 satisfies c < 1/2. Now choose τ so that τ(1 + cτ) = 1, and let τ be as in Lemma 6.20. Let U = π (U). Then, since n q ∈ U , we have, by Lemma 6.20, W (q ) τ 1 1 n z = π + n q ∈ U . τ W (q ) τ 1 1 But, since U [q ] is N-invariant and (6.35) holds, we have X + −αN d n z, U [q ] >β (δ)e . τ 1 Thus (because of n z and Lemma 6.18), X + −αN hd U [q ], U >β (δ)e . Then, by Lemma 6.18,for (1 − δ)-fraction of u ∈ B(q , 1/100), X −αN (6.36) d uq , U >β (δ)e . By Lemma 3.5, and Proposition 3.4, there exists a compact set K of measure at least −1 (1 − δ) and λ depending only on the Lyapunov spectrum such that for x ∈ π (K ) min 2 and y ∈ W [x], X λ t X 0 min 0 d (g x, g y)> c(δ)e d (x, y), t t as long as t > 0and d (g x, g y)< 1/100. Let t > 0 be the smallest such that t t 0 X −1 d (g x, g U ) = 1/100. Therefore, assuming uq ∈ π (K ) in addition to (6.36)we t t 1 2 0 0 have, for 0 < t < t , X λ t−αN 0 min d g uq , g U > c(δ)β (δ)e . t 1 t Hence, for 0 < t < t , X + λ t−αN min hd U [g uq ], g U > c (δ)e , t 1 t 1 g uq t 1 170 ALEX ESKIN, MARYAM MIRZAKHANI and thus, in view of Proposition 6.11(iii) and Lemma 6.12(b), X + λ t−αN 0 min hd U [g uq ], g U > c (δ)e . t 1 t 2 g uq Let α = λ /2, α = 2αN, and let α = α /α .Let α > 0be as in Proposition 6.11 for min 3 1 2 1 1 2 this choice of α .Thenwecan choose α > 0 to be smaller than α ,sothatif(6.32) holds 3 1 and is sufficiently large then (6.24) holds. Hence, by Proposition 6.11(ii) if (6.32) holds, −1 0 < t < t , (and assuming that g uq ∈ π (K ) where K is a compact set of measure at 0 t 1 least 1 − δ), λ t−αN min A(q , u,, t) F q − F(q) ≥ c (δ)e 1 3 Then, for 0 < t < t satisfying (6.32), λ t−αN min A(q , u,, t)≥ c (δ)e 1 4 If t ≥ t satisfies (6.32), then λ t−αN min A(q , u,, t)≥A(q , u,, t )≥ c (δ) ≥ c (δ)e 1 1 0 5 5 Thus, for all t such that (6.32) holds, λ t−αN min A(q , u,, t)≥ c (δ)e . 1 6 −1 This implies (6.33), assuming that is sufficiently large (depending on δ), q ∈ π (K ) −1 and g uq ∈ π (K ). t 1 −1 For the general case (i.e. without the assumptions that q ∈ π (K ) and g uq ∈ t 1 −1 −1 π (K )), note that we can assume that g q ∈ π (K ) for a set of of density at least − 1 −1 (1 − 2δ),and also g uq ∈ π (K ) for a set of t of density at least (1 − 2δ).Now the t 1 general case of (6.33) follows from the special case, Proposition 6.11(i) and Lemma 3.6. ∗ + 6.4 . Proof of Lemma 6.4.— We can choose a subspace T(x) ⊂ Lie(U )(x),sothat Lie U (x) + Lie(Q )(x) = T(x) ⊕ Lie(Q )(x). ++ ++ + + (In particular, if Lie(U )(x) ∩ Lie(Q ) ={0},T(x) = Lie(U )(x).) Then, ++ Lie(G )(x) = Z(x) ∩ W (x) ⊕ T(x) ⊕ Lie(Q )(x). ++ ++ Thus, for any vector Y ∈ Lie(G )(x), we can write ++ (6.37)Y = π (Y) + π (Y) + π (Y), Q Z T where π (Y) ∈ Lie(Q )(x), π (Y) ∈ Z(x) ∩ W (x), π (Y) ∈ T(x). Q ++ Z T Suppose there exists u˜ ∈ T(x) such that (in W (x)) (6.38) x + v ≡ exp I + M u˜ x + v ∈ x + Z(x) ∩ W (x). INVARIANT AND STATIONARY MEASURES 171 Then there exists q ∈ Lie(Q )(x), z ∈ Z(x) ∩ W (x) such that in G (x), ++ ++ (6.39)exp I + M u˜ exp v = exp(z) exp(q). In this subsection, we write V (x) for V (Lie(G ))(x),and λ for λ (Lie(G )).Wealso i i ++ i i ++ i−1 write V (x) = V (x). <i j j=1 Write u˜ = u˜ ,where u˜ ∈ (Lie(U ) ∩ V )(x). Also, write q = q ,where i i i i i i q ∈ (Lie(Q ) ∩ V )(x), v = v ,where v ∈ (W ∩ V )(x),and z = z where i ++ i i i i i i i z ∈ Z (x) = Z(x) ∩ W (x) ∩ V (x). i i1 i For h ∈ G (x) we may write h = h h where h ∈ Q (x),and h ∈ W (x) is a ++ 1 2 1 ++ 2 pure translation. Let i(h) denote the element of Lie(G )(x) whose linear part is h − I ++ 1 and whose pure translation part is h . Then, i : G (x) → Lie(G )(x) is a bijective 2 ++ ++ g -equivariant map. Recall that our Lyapunov exponents are numbered so that λ >λ for i < j . Then, i j we claim that (6.40) i exp I + M u˜ exp v + V (x) <i ' ( ' ( =˜ u + v + i exp I + M u˜ exp v + V (x). i j <i i j j>i j>i Indeed, any term involving u˜ or v for j < i would belong to V (x) (since it would lie in a j <i subspace with Lyapunov exponent bigger than λ ). Also, for the same reason, any terms involving u˜ or v other than those written on the left-hand-side of (6.40) would belong to V (x). Similarly, <i (6.41) i exp(z) exp(q) + V (x) <i = z + q + i exp z exp q + V (x). i i j j <i j>i j>i We now apply i to both sides of (6.39), plug in (6.40)and (6.41), and compare terms in V (x).Weget equationsofthe form u˜ + v + p = z + q , i i i i where p is a polynomial in the u˜ and q for λ <λ , and in the M for λ − λ <λ . Then, i j j j i j k i jk the equation can be solved inductively, starting with the equation with i maximal (and thus λ minimal). Thus, Equation (6.38) can indeed be solved for u˜ and we get, u˜ =−π v + p , z = π v + p , q = π v + p , i T i i Z i i Q i i i i where π , π and π as in (6.37). This shows that v = exp(z)v has the form given in Q T Z (6.7). 172 ALEX ESKIN, MARYAM MIRZAKHANI Let U = exp((I + M ) Lie(U )(x)). By our assumptions, U is a subgroup of G . ++ Therefore, for u˜ as in (6.38), U = U · x + v = U exp − I + M u˜ · (x + v) = U · (x + v). Then, (M ,v) is also a parametrization of U.Tomake M adapted to Z(x) we proceed as follows: For u ∈ Lie(G )(x), we can write u = u + z ,where u ∈ Lie(U )(x) and z ∈ ++ Z + Z(x).Let π : Lie(G ) → Lie(U ) be the linear map sending u to u . + ++ + + In view of (6.6), we need to find a linear map J : Lie(U )(x) → Lie(U )(x),sothat if we define M via the formula (6.6), then M is adapted to Z(x).Write u = Ju. Then, u ∈ Lie(U )(x) must be such that u + M u = u + z,where z ∈ Z. Then, u + π M u = u, hence u = Ju must be given by the formula −1 u = I + π ◦ M u. Thus, in view of (6.6), we define M by −1 (6.42)M = I + M I + π ◦ M − I. Then for all u ∈ Lie(U )(x),Mu = (I + M)u − u = (I + M )u − u ∈ Z(x).Thus (M,v) is adapted to Z(x). Since M ∈ H (x), ++ Z Z π ◦ M = π ◦ M , + + U U ij i<j where M ∈ Hom(Lie(U ) ∩ V , Lie(G ) ∩ V ). Since Z(x) is a Lyapunov-admissible j ++ i ij Z + transversal, π takes Lie(G ) ∩ V to Lie(U ) ∩ V . Therefore, ++ j i Z + + π ◦ M ∈ Hom Lie U ∩ V , Lie U ∩ V . j i U ij Thus, π ◦ M is nilpotent. Then (6.8) follows from (6.42). This argument shows the existence of a pair (M,v) which parametrizes U and is adapted to Z(x). The uniqueness follows from the same argument. Essentially one shows that any (M,v) which parametrizes U and is adapted to Z(x) must satisfy equations whose unique solution is given by (6.7)and (6.8). INVARIANT AND STATIONARY MEASURES 173 6.5 . Construction of the map A(q , u,, t). − + + Motivation. — Suppose q ∈ X , q ∈ W [q ], u ∈ U (q ),so uq ∈ W [q ].Tocon- 1 0 1 1 1 1 struct the generalized subspace U = U(M ,v ) of Proposition 6.11,wefirstlet U = g U t 0 and construct the generalized subspace U ⊂ W [uq ].Let z = π + (uq ),sothat z is 0 1 W (q ) 1 + − + + the unique point in W [q ]∩ AW [uq ].Inparticular, W [q ]= W [z]. (Note that we 1 1 are not assuming any ergodic properties of z; in particular the Lyapunov subspaces at z may not be defined.) We will construct a π (X )-equivariant linear map P (uq , q ) : W (uq ) → 1 0 s 1 1 + −1 + + + W (z),and let U = P (uq , q ) U [q ]. (This makes sense since U [q ]⊂ W [q ]= 0 s 1 1 1 1 1 W [z].) We want P (uq , q ) to have the following properties: s 1 + + ˜ ˜ (P1) P (uq , q ) depends only on W [q ], i.e. for z ∈ W [q ],wehave P (uq , z ) = s 1 s 1 1 1 1 ˜ ˜ ˜ P (uq , q ). In particular, for any u ∈ U (q ), P (uq , q ) = P (uq , u q ). s 1 s 1 s 1 1 1 1 1 GM 1 1 (P2) For nearby x, y ∈ X ,let P (x, y) : H (x) → H (y) denote the Gauss-Manin connection. For u ∈ B(q , 1/100), u ∈ B(q , 1/50) and t ≥ 0with (6.43) d g uq , g u q < 1/100, t 1 t let z = π + (g uq ). Then, there exists α > 0 depending only on the Lya- W (g u q ) t 1 1 −1 GM −α punov spectrum such that P (g uq , g u q ) P (g uq , z ) − I = O(e ), s t 1 t t 1 Y for all t > 0such that (6.43) holds. (Also note that the points uq and u q satisfy d (g uq , g u q ) = O(1) for all 0 ≤ τ ≤ .) −τ 1 −τ X X 0 0 Note that as long as (6.43) holds, d (g uq , z ) = O(1) and d (z , g u q ) = t 1 t GM O(1) so that P (g uq , z ) connects nearby points. This would not be the t 1 + + case if we defined P (uq , q ) to be a linear map from W (uq ) to W (q ), s 1 1 1 1 since g uq and g q would quickly become far apart. t 1 t −1 (P3) The (entries of the matrix) P (uq , q ) are polynomials of degree at most s s 1 in (the entries of the matrix) P (q , q ). −1 + (P4) The generalized subspace U = P (uq , q ) U [q ] can be parametrized s 1 1 1 by (M ,v ) ∈ H (uq ) × W (uq ) (and not by an arbitrary element of ++ 1 1 H (uq ) × W (uq )). + 1 1 The construction will take place in several steps. Notation. — In this subsection, V (x) refers to V (H )(x). i i The map P(x, y).— There exists a set K of full measure such that each point x in K is Lyapunov-regular with respect to the bundle W , i.e. H (x) = V (x), i 174 ALEX ESKIN, MARYAM MIRZAKHANI where V (x) = V (H )(x) are the Lyapunov subspaces, and the multiplicative ergodic the- i i orem holds. We have the flag (6.44) {0}⊂ V (x) ⊂ ··· ⊂ V (x) = H (x), ≤1 ≤n + + where V (x) = V (x).Notethat V (x) = W (x).If y ∈ W [x] is also Lyapunov- ≤j i ≤n−1 i=1 regular, then the flag (6.44)at y agrees with the flag at x, provided we identify H (y) with H (x) using the Gauss-Manin connection. Thus, we may define (6.44) at any point x such that W [x] contains a regular point. Now suppose x and y are restricted to a subset where the V vary continuously. Then, for nearby x and y,wehave, foreach i, (6.45)H (x) = V (y) ⊕ V (x). ≤i j j=i+1 Let z = π (x),and let P : V (x) → H (z) be the map taking v ∈ V (x) to its V (y) W (y) i i i ≤i 1 1 component under the decomposition (6.45). Let P(x, y) : H (x) → H (z) be the linear ˆ ˆ map which agrees with P on each V (x).Notethat P(x, y) is defined for all nearby x, y i i + + such that (6.45) holds for all i.Let P[x, y] be the affine map from W [x] to W [ y] whose linear part is P(x, y) and such that x maps to z = π + (x). To simplify notation, we will W (y) ˆ ˆ denote P[x, y] also by P(x, y). We have ˆ ˆ P(g x, g y) = g ◦ P(x, y) ◦ g , t t t −t and GM (6.46) P(x, y)V (x) = P (y, z)V (y) = V (z). ≤i ≤i ≤i + GM (Since z ∈ W [ y],wecan define V (z) to be P (y, z)V (y) even if V (z) were not ≤i ≤i i originally defined.) The following lemma essentially states that the map P(uq , q ) has properties (P1) and (P2). Lemma 6.21. —Suppose δ> 0, α > 0 and is sufficiently large depending on δ and α . Suppose q ∈ X and q ∈ W [q] satisfy the upper bounds in (5.3) and (5.4).Let q = g q(see 3 0 1 Figure 1), and write q = g q . Then, for almost all u ∈ B(q , 1/100) and t with 0 < t <α such 1 3 that X + d g uq , U g q < 1/100, t 1 t the following holds: INVARIANT AND STATIONARY MEASURES 175 −1 + + ) ˆ ) Let U = P(uq , q ) (U [q ]). Then U ⊂ W [q ] is a generalized subspace, and 1 1 1 1 X + −α(t+) 0 ) hd g U, U g q ≤ C(q )C(uq )e , t t 1 1 g uq 1 t 1 where α> 0 depends only on α and the Lyapunov spectrum, and C : X → R is finite almost 3 0 everywhere. 1 1 Proof. — In this proof, we write V (x) for V (H )(x) and V (x) for V (H )(x).For i i ≤i ≤i convenience, we also choose u ∈ B(q , 1/50) with X X + 0 0 d g uq , g u q = d g uq , U g q ≤ 1/100. t 1 t t 1 t 1 1 (Nothing in the proof will depend on the choice of u .) Let q = g uq , q = g u q .Weclaim that 2 t 1 t 2 1 GM −α(t+) (6.47) d V (q ), P q , q V q ≤ C(q )C(uq )e , Y ≤i 2 2 ≤i 1 1 2 2 where α> 0 depends only on the Lyapunov spectrum, and C : X → R (which depends on δ) is finite a.e. We will apply Lemma 4.7 (with t + in place of t) to the points x = g q and −(t+) 2 y = g q . Thus, we need to bound D (x, y). In the following argument, we identify −(t+) 1 1 1 1 H (x),H (y),H (q) and H (q ) using the Gauss-Manin connection, while suppressing GM P from the notation. Suppose v ∈ V (y) realizes the supremum in the definition of D (x, y), i.e. v = ≤i v + w where v ∈ V (x), w ∈ V (x),and D (x, y) =w /v . ≤i >i Y Y Note that V (x) = V (q) and V (y) = V (q ).Thus, v ∈ V (q ). Also note that ≤i ≤i ≤i ≤i ≤i − − V (q ) = V (q) for all i,P (q , q)V (q ) = V (q), and by Lemma 4.2(c), P (q , q) is lower >i >i i i triangular and unipotent. By the upper bound in (5.4), P (q , q) ≤ C (δ).(In par- ticular, we have a lower bound, depending on δ, on the angles between the Lyapunov subspaces V (q ).) Hence we can write v = v + w v ∈ V (q), w ∈ V (q), w ≤ C(δ)v . ≤i >i Y Y Since V (x) = V (q),wehave v ∈ V (x). By Corollary 4.9 (applied with x = q , y = uq ≤i ≤i ≤i 1 1 and t = ) we can write w = v + w v ∈ V (x),w ∈ V (x), 2 2 2 ≤i 2 >i −α and v ≤ C (q )C (uq )e w . 2 Y 1 1 1 1 Y Thus, v = v + v,w = w . 2 2 176 ALEX ESKIN, MARYAM MIRZAKHANI If is bounded depending on C (q )C (uq ) and δ, then (in view of the condition t <α ), 1 1 1 1 3 the desired estimate (6.47) is trivially true. Thus, we may assume that is sufficiently large so that −α C (q )C (uq )e ≤ 1. 1 1 1 1 Then, w ≤w +v ≤ 2w ≤ 2C(δ)v . 2 Y Y 2 Y Y Y But, −α −α v ≤ C (q )C (uq )e w ≤ 2C(δ)C (q )C (uq )e v . 2 Y 1 1 1 1 2 Y 1 1 1 1 Y Arguing as above, we may assume, without loss of generality, that is sufficiently large so that v ≥v −v ≥ (1/2)v . Y Y 2 Y Y Then, 2 Y D (x, y) = ≤ 4C(δ). Hence, by Lemma 4.7,(6.47) follows. By Lemma 4.14(c), for any > 0 and any subset S of the Lyapunov exponents, −t −(t+) (6.48) d V (q ), V (q ) > C (uq )e > C (uq )e . Y i 2 j 2 1 1 i∈S j∈/S Choose < α/2, where α is as in (6.47). Then, by (6.48), (6.47), and the definition of ˆ ˆ P(q , q ) = P(g uq , g u q ), 2 t 1 t 2 1 −1 GM −α (+t) (6.49) P g uq , g u q P g uq , g u q − I ≤ C (uq )C (q )e , t 1 t t 1 1 1 1 t 1 where α = α − depends only on the Lyapunov spectrum, and C (·),C (·) are finite a.e. Also note that by the upper bound in (5.3) and Lemma 3.5,wehave −α d (uq , z) ≤ C (q )e , Y 1 1 and again by Lemma 3.5, −α t −α (t+) (6.50) d (g uq , g z)< C (uq )e d (uq , z) ≤ C (q )C (uq )e . Y t 1 t 1 Y 1 1 1 Note that U is the orbit of a subgroup Uof G(uq ) whose Lie algebra is −1 P uq , q Lie U q 1 1 ∗ INVARIANT AND STATIONARY MEASURES 177 (and we are using the notation (6.11)). By (6.46) and the fact that Lie(U )(q ) ∈ G (q ) ++ 1 1 ˆ ) we have Lie(U) ∈ G (uq ).Thus, U is a generalized subspace. ++ 1 + + + + Since U [q ] is a generalized subspace, for all u ∈ U (q ),U [q ]= U [u q ].We 1 1 1 1 have −1 −1 + + ˆ ˆ g U = g P uq , u q U u q = P g uq , g u q U g u q . t t 1 t 1 t t 1 1 1 1 Therefore, the lemma follows from (6.49)and (6.50). + − Motivation. — Suppose q ∈ X , u ∈ U (q ), q ∈ W [q ].InviewofLemma 6.21, 1 0 1 1 P(uq , q ) has properties (P1) and (P2). We claim that it does not in general have the properties (P3) and (P4). ˆ ˆ Let z = π + (uq ) so in particular P(uq , q ) = P(uq , z) and let W (q ) 1 1 1 −1 GM − + ˆ ˆ (6.51) Q uq ; q = P(uq , z) P q , z P q , q ◦ P (uq , q ), 1 1 1 1 1 1 1 1 so that GM − + ˆ ˆ (6.52) P(uq , z)Q uq ; q = P q , z P q , q P (uq , q ). 1 1 1 1 1 1 1 1 1 1 ˆ ˆ Then, Q(uq ; q ) : H (uq ) → H (uq ) and Q(uq ; q )V (uq ) = V (uq ),hence 1 1 1 1 ≤i 1 ≤i 1 1 1 + + ˆ ˆ Q(uq ; q ) ∈ Q (uq ). In particular Q(uq ; q )W (uq ) = W (uq ). 1 + 1 1 1 1 1 1 ˆ ˆ We now show how to compute P(uq , q ) and Q(uq ; q ) in terms of P = 1 1 1 1 + − − + P (uq , q ) and P = P (q , q ). In view of Lemma 4.2,P is upper triangular with 1’s 1 1 1 along the diagonal in terms of a basis adapted to V (uq ). Also by Lemma 4.2 applied i 1 − + − to P instead of P ,P is lower triangular with 1’s along the diagonal in terms of a ba- + + −1 − + sis adapted to V (q ). Therefore, since P takes V (uq ) to V (q ), (P ) P P is lower i 1 i 1 i 1 triangular with 1’s along the diagonal in terms of a basis adapted to V (uq ). i 1 ˆ ˆ ˆ ˆ ˆ ˆ Let P = P(uq , q ), Q = Q(uq ; q ). Then, in view of the definition of P, Pis lower 1 1 1 1 triangular with 1’s along the diagonal in terms of a basis adapted to V (uq ) (and we i 1 1 1 identify H (q ) with H (uq ) using the Gauss-Manin connection). Also, since Q preserves the flag V (uq ), Q is upper triangular in terms of the basis adapted to V (uq ).Thus, ≤i 1 i 1 (6.52) can we written as −1 − + + + − + ˆ ˆ (6.53) PQ = P P = P P P P Recall that the Gaussian elimination algorithm shows that any matrix A in neigh- borhood of the identity I can be written uniquely as A = LU where L is lower tri- ˆ ˆ angular with 1’s along the diagonal and U is upper triangular. Thus, P = P(uq , q ) ˆ ˆ and Q = Q(uq ; q ) are the L and U parts of the LU decomposition of the matrix − + + A = P (q , q )P (uq , q ). (Note that we are given A = U L where U = P is upper 1 1 1 + −1 − + triangular and L = (P ) P P is lower triangular, so we are really solving the equation LU = U L for L and U.) 178 ALEX ESKIN, MARYAM MIRZAKHANI Since the Gaussian elimination algorithm involves division, the entries of −1 + − P(uq , q ) are rational functions of the entries of P (uq , q ) and P (q , q ),but not 1 1 1 1 1 1 in general polynomials. This means that P(uq , q ) does not in general have property (P3). Also, the diagonal entries of Q(uq ; q ) are not 1. This eventually translates to the failure of the property (P4). Both problems are addressed below. ˆ ˜ ˆ The maps P (uq , q ) and P (uq , q ).— For s > 1, let Q (uq ; q ) be the order s s 1 s 1 s 1 1 1 1 Taylor approximation to Q(uq ; q ), where the variables are the entries of P (q , q ) (and 1 1 1 1 ˆ ˆ u, q and the entries of P (uq , q ) are considered constants). Then, Q = Q (uq ; q ) ∈ 1 1 1 s s 1 Q (uq ). We may write + 1 ˆ ˜ Q = D + Q , s s s ˜ ˜ where D preserves all the subspaces V (uq ) and Q = Q (uq ; q ) ∈ Q (uq ).Let s i 1 s s 1 ++ 1 ˜ ˜ P (uq , q ) = P (uq , z) be defined by the relation: s 1 s 1 −1 + − GM ˜ ˜ (6.54) P uq , q = Q uq ; q P (q , uq )P q , q P z, q . s 1 s 1 1 1 1 1 1 1 1 Motivation. — We will effectively show that for s sufficiently large (chosen at the end of the proof of Proposition 6.11)the map P (uq , q ) has the properties (P1), (P2), (P3) and s 1 (P4). We have, by (6.54), −1 −1 ˜ ˜ P uq , q V q = P uq , q V (z) = V (uq ). s 1 ≤i s 1 ≤i ≤i 1 1 1 1 As a consequence, −1 ˜ ˜ P uq , q ◦ Y ◦ P uq , q ∈ G (uq ) for all Y ∈ G q . s 1 s 1 ++ 1 ++ 1 1 1 −1 Thus, for any subalgebra L of Lie(G )(q ), it follows that P (uq , q ) (L) is a subalgebra ++ s 1 1 1 ∗ −1 of Lie(G )(uq ),where P (uq , q ) : Lie(G )(q ) → Lie(G )(uq ) is as in (6.11). ++ 1 s 1 ++ ++ 1 1 ∗ 1 The map i . u,q ,s Motivation. — For q ∈ X and u ∈ B(q , 1/100),wewant i : L (q ) → 1 0 1 u,q ,s ext 1 H (uq ) × W (uq ) to be such that ++ 1 1 i P q − P(q ) = (M ,v ), u,q ,s 1 s s where the pair (M ,v ) ∈ H (uq ) × W (uq ) parametrizes the approximation s s ++ 1 1 −1 + + P (uq , q ) U [q ] to U [q ] constructed above. Furthermore, we want i to be a s 1 u,q ,s 1 1 1 1 polynomial map of degree at most s in the entries of P(q ) − P(q ). 1 INVARIANT AND STATIONARY MEASURES 179 By Proposition 4.12(a), we have + − + + (6.55)Lie U q = P q , q ◦ P (uq , q ) Lie U (uq ) , 1 1 1 ∗ 1 1 1 −1 + where we used the notation (6.11). Let U = P (uq , q ) U [q ].Wefirstfind (M ,v ) ∈ s 1 s 1 1 s s H (q ) × W (q ) which parametrizes U .Let + 1 1 −1 + + v = P uq , q q ∈ U ⊂ W [q ]= W [uq ]. s s 1 1 1 1 1 s By (6.55), U = U · v where the subgroup U of G (uq ) is such that s s s ++ 1 −1 GM − Lie(U ) = P (uq , z) ◦ P q , z ◦ P q , q s s 1 1 ∗ 1 1 ∗ ∗ + + ◦ P (uq , q ) Lie U (uq ) . 1 1 ∗ 1 By (6.54), (6.56)Lie(U ) = Q uq ; q Lie U (uq ). s s 1 1 Let M = Q uq ; q − I. s s 1 Then (M ,v ) parametrizes U . Since Q (uq ; q ) ∈ Q (uq ),M ∈ H (q ). s s s 1 ++ 1 s ++ 1 s 1 Note that by (5.8), we can recover Im q from P(q ). Also, since q is considered 1 1 1 known and fixed here, knowing Im q is equivalent to knowing q since Re q = Re q . 1 1 1 Also, since by Proposition 4.12(a), for q ∈ W [q ], + − + −1 + (6.57)Lie U q = P q , q Lie U (q ) = P q ◦ P(q ) Lie U (q ), 1 1 1 1 1 1 1 ∗ ∗ + + we can reconstruct U (q ) if we know P(q ),U (q ) and P(q ).Now let i : 1 1 u,q ,s 1 1 1 L (q ) → H (uq ) × W (uq ) be the map taking P(q ) − P(q ) to (M ,v ).Inviewof ext 1 ++ 1 1 1 s s (6.56), this is a polynomial map, since Q is a polynomial, and both Im q and Lie(U )(q ) 1 1 can be recovered from P(q ) using (5.8)and (6.57). (Note that q is considered fixed here, so knowing P(q ) − P(q ) is equivalent to knowing P(q ).) 1 1 ⊗a ⊗a The maps (i ) and i .— For a ∈ N,let j : L (x) → L (x) be the “diag- u,q ,s ∗ u,q ,s ext ext 1 1 onal embedding” ⊗a j (v) = v ⊗ ··· ⊗ v, (a times) a a and let j denote the corresponding map L (x) → L (x) . ext ext Since i : L (q ) → H (uq )× W (uq ) is a polynomial map, by the universal u,q ,s ext 1 ++ 1 1 property of the tensor product, there exists a > 0and a linear map (i ) : L (q ) → u,q ,s ∗ ext 1 H (uq ) × W (uq ) such that ++ 1 1 i = (i ) ◦ j . u,q ,s u,q ,s ∗ 1 1 180 ALEX ESKIN, MARYAM MIRZAKHANI Furthermore, there exists r > a and a linear map i : L (q ) → H(uq ) such that u,q ,s ext 1 1 (6.58) j ◦ (i ) = i ◦ j , u,q ,s ∗ u,q ,s 1 1 where j is as in (6.10). Then i takes F(q ) − F(q ) ∈ L (q ) to j(M ,v ) ∈ H(uq ), u,q ,s 1 ext 1 s s 1 −1 + + where (M ,v ) is a parametrization of the approximation P (uq ; q ) U [q ] to U [q ]. s s s 1 1 1 1 Construction of the map A(q , u,, t).— Let s ∈ N be a sufficiently large integer to be chosen later. (It will be chosen near the end of the proof of Proposition 6.11, depending only on the Lyapunov spectrum.) Let r ∈ N be such that (6.58) holds. Suppose q ∈ X 1 0 and u ∈ B(q , 1/100).For > 0and t > 0, let (r) A(q , u,, t) : L (g q ) → H(g uq ), 1 ext − 1 t 1 be given by Z(uq ) r A(q , u,, t) = (g ) ◦ S ◦ˆπ ◦ i ◦ (g ) 1 t ∗ u,q ,s uq ∗ where (g ) : L (q) → L (g q) is given by ∗ ext ext −1 (g ) (P) = g ◦ P ◦ g . Then A(q , u,, t) is a linear map. Unraveling the definitions, we have, for P ∈ L (g q ), 1 ext − 1 r + Z(uq ) A(q , u,, t) j (P) = j G ◦ S ◦ (i ) ◦ (g ) (P) 1 u,q ,s ∗ t uq Thus, for q satisfying the upper bounds in (5.3)and (5.4), (6.59) A(q , u,, t) F(q) − F q = j M ,v , where (M ,v ) ∈ H (g uq ) × W (uq ) is a parametrization of the approximation ++ t 1 1 −1 g P uq , u q U u q t s 1 1 1 + + X to U [g u q ],where u q ∈ U [q ] is such that d (g uq , g u q )< 1/100. t t 1 t 1 1 1 1 6.6 . Proofs of Proposition 6.11 and Lemma 6.14. Proof of Proposition 6.11. — Note that Proposition 6.11(i) follows immediately from the definition of A(·,·,·,·).Wenow begin the proof of Proposition 6.11(iii). Let P = P(q ) − P(q) ∈ L (q).Let ext −1 P = (g ) (P) = g ◦ P ◦ g ∈ L (q ). 1 ∗ ext 1 Let (M ,v ) = i (P ). s s u,q ,s 1 1 INVARIANT AND STATIONARY MEASURES 181 * * Let U = U (M ,v ) be the generalized subspace parametrized by (M ,v ).Then s s s s s s −1 (6.60) U = P uq , q U q . s s 1 1 1 Let −1 −1 + + ) ) ˆ ˆ (6.61) U = P uq , q U q , U = P uq , q U [q ]. 1 s s 1 1 1 1 1 Suppose (6.23) holds. By Lemma 6.21, X + −α t 0 1 (6.62) hd U, U g u q = O e , t uq g uq 1 1 t 1 where α depends only on the Lyapunov spectrum. We have, in view of (5.3)and (5.4), for sufficiently large depending on δ, − GM −α (6.63) P q , q P q , q − I = O e 1 1 q 1 1 1 where α depends only on the Lyapunov spectrum. Therefore, X + + −α 0 2 hd U [uq ], U q = O e 1 q uq 1 1 ˆ ˆ To go from Qto Q we are doing order s Taylor expansion of the solution to (6.53) − GM in the entries of P (q , q )P (q , q ) − I. Thus, by (6.63), 1 1 1 1 −α (s+1) ˆ ˆ Q uq ; q − Q uq ; q = O e s 1 1 q ,uq 1 1 1 1 and thus, by (6.54), −1 −1 −α (s+1) ˆ ˆ (6.64) P uq , q − P uq , q = O e s 1 1 q ,uq 1 1 1 1 Then, by (6.61), X −α (s+1) 0 2 ) ) hd (U, U ) = O e . s q ,uq uq 1 1 Then, by Lemma 6.10(a), X −α (s+1)+2t 0 2 ) ) (6.65) hd (g U, g U ) = O e . t t s q ,uq g uq 1 1 t 1 Also, by (6.63), (6.51)and (6.49), we have −α Q uq ; q − I = O e , 1 q ,uq 1 1 and therefore −α Q uq ; q − I = O e . s 1 q ,uq 1 1 1 Y 182 ALEX ESKIN, MARYAM MIRZAKHANI Thus, −α ˜ ˆ D = Q uq ; q − Q uq ; q = O e s Y s 1 s 1 q 1 1 Therefore, since D preserves all the eigenspaces V , and the Osceledets multiplicative s i ergodic theorem, for sufficiently small > 0 (depending on the Lyapunov spectrum), −1 −α +t −(α /2) 2 2 g ◦ D ◦ g ≤ C (q )C (uq ,)e ≤ C (q )C (uq )e . t s Y 1 1 2 1 1 1 1 t 2 Thus, −1 −1 −(α /2) ˜ ˆ (6.66) P g uq , g u q − P g uq , g u q = O e s t 1 t s t 1 t uq 1 1 1 and hence by (6.60)and (6.61), X −1 −(α /2) 0 2 ) * (6.67) hd (g U , g U ) = O g ◦ D ◦ g = O e . t s t s uq t s Y uq 1 1 g uq t t 1 We now choose s so that α α (s + 1) − 3 >α . Then, by (6.22), (6.62), (6.65), and (6.67), 2 3 2 X + −α (6.68) hd g U , U g q ≤ C(q )C(uq )e , t s t 1 1 g uq 1 t 1 where α depends only on the Lyapunov spectrum. In view of (6.59), the pair (M ,v ) parametrizes g U . Therefore, (6.26) holds. Finally, (6.27) is an immediate consequence of t s (6.68). This completes the proof of Proposition 6.11(iii). (Note that is was shown imme- diately after the statement of Proposition 6.11 that Proposition 6.11(iii) implies Proposi- tion 6.11(ii).) Proof of Lemma 6.14. — In the proof of this lemma we normalize the measure |·| + + + on U [q ] so that |U [q ]∩ B (q , 1/100)|= 1 and similarly we normalize the measure 1 1 1 + + + |·| on U [q ] so that |U [q ]∩ B (q , 1/100)|= 1. As in the proof of Lemma 6.21,we 1 1 1 + + choose u ∈ B(q , 1/50) with V (g u q ) and U [g u q ]= U [g q ] defined and i t t t 1 1 1 1 X X + + 0 0 d g uq , g u q ≤ hd U [g uq ], U g q ≤ . t 1 t t 1 t 1 g uq 1 t 1 (Nothing in the proof will depend on the choice of u .) −1 ˜ ˜ ˜ Let A = g A ,A = g A .Let P be as in (6.54). Let A = P (g uq , g u q ) A . 0 −t t −t s t s t 1 t 0 t 1 t Then, −1 ˜ ˜ ˜ A ≡ g A = P uq , u q A . 0 −t t s 1 1 0 As in the proof of Proposition 6.11 (i.e. by combining (6.49), (6.64)and (6.66)), we have −1 GM −α P uq , u q P uq , u q − I = O e . s 1 1 1 1 −1 GM −α P g uq , g u q P g uq , g u q − I = O e . s t 1 t t 1 1 t 1 Y INVARIANT AND STATIONARY MEASURES 183 ˜ ˜ Hence, |A | is comparable to |A | and |A | is comparable to |A |. Thus, it is enough to t 0 t 0 show that |A | is comparable to |A |. 0 0 As in the proof of Proposition 6.11,let (M ,v ) be the pair parameterizing g U = t s −1 + + ˜ ˜ P (g uq , g u q ) U [g u q ].Let f : Lie(U )(g uq ) → g U be the “parametrization” map s t 1 t 1 t t t 1 t s f (Y) = exp I + M Y (g uq ) g uq + v . t t 1 t 1 + + Similarly, let f : Lie(U )(g uq ) → U [g uq ] be the exponential map t t 1 t 1 f (Y) = exp(Y)g uq . t t 1 Then, provided that is sufficiently small, we have −1 −1 −1 ˜ ˜ (6.69)0.5f (A ) ⊂ f (A ) ⊂ 2f (A ) t t t −1 −1 −1 + ˜ ˜ ˜ Let M = g ◦ M ◦ g , v = g v . Then, g ◦ f ◦ g = f ,where f : Lie(U )(uq ) → U 0 t 0 t t 0 0 1 s t t t is given by f (Y) = exp (I + M )Y (g uq )(g uq + v ). 0 0 t 1 t 1 0 −1 + + Similarly, g ◦ f ◦ g = f ,where f : Lie(U )(uq ) → U [uq ] is given by the exponential t t 0 0 1 1 map f (Y) = exp(Y)uq . 0 1 −1 Then, it follows from applying g to (6.69)that −1 −1 −1 ˜ ˜ (6.70)0.5f (A ) ⊂ f (A ) ⊂ 2f (A ) 0 0 0 0 0 0 −1 −1 ˜ ˜ Thus, |f (A )| is comparable to |f (A )|=|A |. But, since M ∈ H (g uq ) and v ∈ 0 0 0 ++ t 1 0 0 W (g uq ) are O(),M and v are exponentially small. Therefore, the map f is close t 1 0 0 0 −1 ˜ ˜ to f (and since Y is small, it is close to the identity). Therefore, |f (A )| is comparable 0 0 to |A |. The second assertion of the Lemma also follows from (6.70) and the fact that M 0 0 and v are exponentially small. 7. Bilipshitz estimates In this section, we continue working on X (and not X). Let · be the norm on (++) (++) H defined in (4.18). Since H ⊂ H , · is also a norm on H. We can also define big big (−−) (−−) (r) anormon H in an analogous way. Since L (x) ⊂ H (x), the norm · is also ext x big big (r) anormon L (x) .Let A(q , u,, t) =A(q , u,, t) where the operator norm is with ext 1 1 respect to the dynamical norms · at g q and g uq . In the rest of this section we assume − 1 t 1 that the equivalent conditions of Lemma 6.15 do not hold, and then by Proposition 6.16, (6.33) holds. 184 ALEX ESKIN, MARYAM MIRZAKHANI For 1/100 > > 0, almost all q ∈ X , almost all u ∈ B(q , 1/100) and > 0, let 1 0 1 τˆ (q , u,) = sup t : t > 0and A(q , u,, t) ≤ . () 1 1 Note that τˆ (q , u, 0) need not be 0. () 1 For x ∈ X ,let A (x, t) : H(x) → H(g x) denote the action of g on H as in (6.16). 0 + t t (r) (r) (r) Let A (x, s) : L (x) → L (g x) denote the action of g on L (x). − ext ext s s ext Lemma 7.1. — There exist absolute constants N > 0, α> 0 such that for almost all x, and t > 0, −αt −Nt αt Nt e ≥A (x, t)≥ e , e ≤A (x, t)≤ e , − + and, Nt αt −Nt −αt e ≥A (x,−t)≥ e , e ≤A (x,−t)≤ e . − + Proof. — This follows immediately from Proposition 4.15. Lemma 7.2. — Suppose 0 < < 1/100. There exists κ > 1 (depending only on the Lya- punov spectrum) with the following property: for almost all q ∈ X ,u ∈ B(q , 1/100), for all > 0 1 0 1 and s > 0, −1 τˆ (q , u, + s)> τˆ (q , u,) + κ s. () 1 () 1 Proof. — Note that by (6.21), A(q , u, + s, t + τ) = A (g uq ,τ)A(q , u,, t)A (g q , s). 1 + t 1 1 − −(+s) 1 Let t =ˆτ (q , u,),sothat A(q , u,, t) = . Therefore, () 1 1 A(q , u, + s, t + τ) ≤A (g uq ,τ)A(q , u,, t)A (g q , s) 1 + t 1 1 − −(+s) 1 Nτ−αs ≤ A (q uq ,τ)A (g q , s)≤ e , + t 1 − −(+s) 1 where we have used the fact that A(q , u,, t) = and Lemma 7.1.If t + τ = τˆ (q , u,+ s) then A(q , u,+ s, t +τ) = . It follows that Nτ −αs ≥ 0, i.e. τ ≥ (α/N)s. () 1 1 Hence, τˆ (q , u, + s)≥ˆτ (q , u,) + (α/N)s. () 1 () 1 Lemma 7.3. — Suppose 0 < < 1/100. There exists κ > 1 (depending only on the Lya- punov spectrum) such that for almost all q ∈ X , almost all u ∈ B(q , 1/100),all > 0 and all 1 0 1 s > 0, τˆ (q , u, + s)< τˆ (q , u,) + κ s. () 1 () 1 2 INVARIANT AND STATIONARY MEASURES 185 Proof.—We have A(q , u,, t) = A (g uq ,−τ)A(q , u, + s, t + τ)A (g q ,−s). 1 + t+τ 1 1 − − 1 Let t + τ =ˆτ (q , u, + s). Then, by Lemma 7.1, () 1 A(q , u,, t) ≤A (q uq ,−τ)A(q , u, + s, t + τ)A (g q ,−s) 1 + t+τ 1 1 − − 1 −ατ+Ns ≤ A (q uq ,−τ)A (g q ,−s)≤ e , + t+τ 1 − − 1 wherewehaveused thefactthat A(q , u, + s, t + τ) = . Since A(q , u,, t) = ,it 1 1 follows that −ατ + Ns > 0, i.e. τ<(N/α)s. It follows that τˆ (q , u, + s)< τˆ (q , u,) + (N/α)s () 1 () 1 Proposition 7.4. — There exists κ> 1 depending only on the Lyapunov spectrum, and such that for almost all q ∈ X , almost all u ∈ B(q , 1/100),any > 0 and any measurable subset 1 0 1 E ⊂ R , bad τˆ (q , u, E ) ∩ τˆ (q , u, 0),τˆ (q , u,) ≤ κ|E ∩[0,]| () 1 bad () 1 () 1 bad t ∈[0,]| : τˆ (q , u, t) ∈ E ≤ κ E ∩ τˆ (q , u, 0),τˆ (q , u,) . () 1 bad bad () 1 () 1 −1 Proof.—Let κ = max(κ ,κ ),where κ , κ areasinLemmas 7.2 and 7.3. Then, 2 1 2 for fixed q , u, τˆ (q , u,) is κ -bilipshitz as a function of . The proposition follows 1 () 1 immediately. 8. Preliminary divergence estimates In this section, we continue working on X (and not X). Motivation. — Suppose in the notation of Section 2.3, q and q are fixed, but u ∈ B(q , 1/100) and u ∈ B(q , 1/100) vary. Then, as u and u vary, so do the points q 1 2 + + and q , and thus the subspaces U [q ] and U [q ].Let U = U(M (u),v (u)) be the 2 2 approximation to U [q ] given by Proposition 6.11, and as in Proposition 6.11,let v(u) = j(M (u),v (u)) ∈ H(q ) be the associated vector in H(q ). 2 2 In this section we define a certain g -equivariant and (u) -equivariant subbundle t ∗ E ⊂ H such that, for fixed q , q ,for most u ∈ U [q ], v = v(u) is near E(q ) (see Propo- 1 1 2 sition 8.5(a) below for the precise statement). We call E the U -inert subbundle of H. The subbundle E is the direct sum of subbundles E ,where E is contained in the i-th i i Lyapunov subspace of H, and also each E is both g -equivariant and (u) -equivariant. i t ∗ 186 ALEX ESKIN, MARYAM MIRZAKHANI 8.1. The U -inert subspaces E(x).— We apply the Osceledets multiplicative ergodic theorem to the action on H(x) (see (6.16)). We often drop the ∗ and denote the action simply by g . In this section, λ denotes the i-th Lyapunov exponent of the flow g on the t i t bundle H. Let V (x) = V (H)(x), V (x) = V (H)(x), ≤i j <i j j≤i j<i V (x) = V (H)(x), V (x) = V (H)(x). ≥i j >i j j≥i j>i This means that for almost all x ∈ X and for v ∈ V (x) such that v ∈/ V (x), 0 ≤i <i 1 g v (8.1)lim log = λ , t→−∞ t v and for v ∈ V (x) such that v ∈/ V (x), ≥i >i 1 g v (8.2)lim log = λ . t→∞ t v By e.g. [GM, Lemma 1.5], we have for a.e. x ∈ X , (8.3) H(x) = V (x) ⊕ V (x). ≤i >i Let (8.4) F (x) = v ∈ H(x) : for almost all u ∈ B(x), (u) v ∈ V (ux) , ≥j ∗ ≥j where (u) is as in Lemma 6.6.Inother words, if v ∈ F (x), then for almost all u ∈ B(x), ∗ ≥j (8.5)limsup log (g ) (u) v≤ λ . t ∗ ∗ j t→∞ From the definition of F (x),wehave ≥j (8.6) {0}= F (x) ⊂ F (x) ⊂ F (x) ⊂··· F (x) ⊂ F (x) = H(x). ≥n+1 n ≥n−1 2 1 Let E (x) = F (x) ∩ V (x). j ≥j ≤j In particular, E (x) = V (x) = V (H)(x).Wemay have E (x) ={0} if j = 1. 1 ≤1 1 j INVARIANT AND STATIONARY MEASURES 187 Lemma 8.1. —For almost all x ∈ X the following holds: suppose v ∈ E (x)\{0}.Thenfor 0 j almost all u ∈ B(x), (8.7)lim log (g ) (u) (v)= λ . t ∗ ∗ j t→∞ Thus (recalling that V (H) denotes the subspace of H corresponding to the Lyapunov exponent λ ), we j j have for almost all x, using Fubini’s theorem, E (x) ⊂ V (H)(x). j j In particular, if i = j, E (x) ∩ E (x) ={0} for almost all x ∈ X . i j 0 Proof. — Suppose v ∈ E (x).Then v ∈ V (x). Since in view of (8.1), V (ux) = j ≤j ≤j (u) V (x) for all u ∈ U (x),wehavefor almost all u ∈ B(x), (u) v ∈ V (ux). It follows ∗ ≤j ∗ ≤j from (8.3) that (outside of a set of measure 0), (u) v ∈/ V (ux).Now (8.7) follows from ∗ >j (8.2). Lemma 8.2. — After possibly modifying E (x) and F (x) on a subset of measure 0 of X,the j ≥j following hold: (a) E (x) and F (x) are g -equivariant, i.e. (g ) E (x) = E (g x),and (g ) F (x) = j ≥j t t ∗ j j t t ∗ ≥j F (g x). ≥j t (b) For almost all u ∈ U (x), E (ux) = (u) E (x),and F (ux) = (u) F (x). j ∗ j ≥j ∗ ≥j Proof. — Note that for t > 0, g B[x]⊃ B[g x]. Therefore, (a) for the case t > 0 t t follows immediately from the definitions of E (x) and F (x). Since the flow {g } is j ≥j t t>0 ergodic, it follows that almost everywhere (8.4) holds with B[x] replaced by arbitrary large balls in U [x]. This implies that almost everywhere, F (x) = v ∈ H(x) : for almost all u ∈ U , (u) v ∈ V (ux) , ≥j ∗ ≥j where (u) v is as in Lemma 6.6. Therefore (b) holds. Then, (a) for t < 0 also holds, as long as both x and g x belong to a subset of full measure. By considering a transversal for the flow g , it is easy to check that it is possible to modify E (x) and F (x) on a subset t j ≥j of measure 0 of X in such a way that (a) holds for x in a subset of full measure and all t ∈ R. Lemma 8.3. —For x ∈ X ,let Q(v) = u ∈ B(x) : (u) v ∈ V (ux) . ∗ ≥j Then for almost all x, either |Q(v)|= 0,or |Q(v)|=|B(x)| (and thus v ∈ F (x)). ≥j 188 ALEX ESKIN, MARYAM MIRZAKHANI Proof. — For a subspace V ⊂ H(x),let Q(V) = u ∈ B(x) : (u) V ⊂ V (ux) . ∗ ≥j Let d be the maximal number such that there exists E ⊂ X with ν(E )> 0such that for x ∈ E there exists a subspace V ⊂ H(x) of dimension d with |Q(V)| > 0. For a fixed x ∈ E ,let W(x) denote the set of subspaces V of dimension d for which |Q(V)| > 0. Then, by the maximality of d,if V and V are distinct elements of W(x) then Q(V) ∩ Q(V ) has measure 0. Let V ∈ W(x) be such that |Q(V )| is maximal (among elements x x of W(x)). Let > 0 be arbitrary, and suppose x ∈ E . By the same Vitali-type argument as in the proof of Lemma 3.11, there exists t > 0 and a subset Q(V ) ⊂ Q(V ) ⊂ B(x) such 0 x x that for all u ∈ Q(V ) and all t > t , x 0 (8.8) |B (ux) ∩ Q(V )|≥ (1 − )|B (ux)|. t x t (In other words, Q(V ) are“pointsofdensity”for Q(V ), relative to the “balls” B .) Let x x t ∗ ∗ E = ux : x ∈ E , u ∈ Q(V ) . ∗ ∗ Then, ν(E )> 0. Let ={x ∈ X : g x ∈ E for an unbounded set of t > 0 }.Then 0 −t ν() = 1. Suppose x ∈ . We can choose t > t such that g x ∈ E .Notethat 0 −t (8.9) B[x]= g B [g x]. t t −t Let x = g x,and let V = (g ) V .Theninviewof(8.8)and (8.9), −t t,x t ∗ x |Q(V )|≥ (1 − )|B(x)|. t,x By the maximality of d (and assuming < 1/2), V does not depend on t.Hence,for t,x every x ∈ , there exists V ⊂ H(x) such that dim V = d and |Q(V)|≥ (1 − )|B(x)|. Since > 0 is arbitrary, for each x ∈ , there exists V ⊂ H(x) with dim V = d,and |Q(V)|=|B(x)|. Now the maximality of d implies that if v ∈/ V then |Q(v)|= 0. By Lemma 8.1, E (x) ∩ E (x) ={0} if j = k.Let j k = i : E (x) = {0} for a.e. x . Let the U -inert subbundle E be defined by E(x) = E (x). i∈ Then E(x) ⊂ H(x). INVARIANT AND STATIONARY MEASURES 189 In view of (8.5), (8.6) and Lemma 8.1,wehave F (x) = F (x) unless j ∈ . ≥j ≥j+1 Therefore if we write the elements of in decreasing order as i ,..., i we have the flag 1 m (consisting of distinct subspaces) (8.10) {0}= F ⊂ F (x) ⊂ F (x) ⊂··· F (x) ⊂ F (x) = H(x). ≥i ≥i ≥i ≥i ≥i m+1 m m−1 2 1 For a.e. x ∈ X ,and 1 ≤ r ≤ m,let F (x) be the orthogonal complement (using the inner product ·,· defined in Section 4.7)to F (x) in F (x). x ≥i ≥i r+1 r Lemma 8.4. —Given δ> 0 there exists a compact K ⊂ X with ν(K )> 1 − δ, 01 0 01 β(δ) > 0, β (δ) > 0,and forevery x ∈ K any j ∈ any v ∈ P(F )(x) asubset Q = 01 01 Q (x, v ) ⊂ B(x) with |Q | >(1 − δ)|B(x)| such that for any j ∈ any v ∈ F (x) and 01 01 any u ∈ Q , we can write (u) v = v + w , v ∈ E (ux), w ∈ V (ux), ∗ u u u j u >j with v ≥ β(δ)v ,and v >β (δ)w . u u u Proof. — This is a corollary of Lemma 8.3.Let ⊂ X be the conull set where (8.3) holds and where F (x) = F (x) for all i ∈/ . Suppose x ∈ . ≥i ≥i+1 Let F (x) ⊂ F (x) be the next subspace in the flag (8.10) (i.e. F ={0} if j is ≥k ≥j ≥k the maximal index in and otherwise we have k > j be minimal such that k ∈ ). Then F (x) = F (x). Since F (x) is complementary to F (x) we have that F (x) is ≥j+1 ≥k ≥k j j complementary to F (x). ≥j+1 By Lemma 8.2, F is g -equivariant, and therefore, by the multiplicative ergodic ≥j t theorem applied to F , F is the direct sum of its Lyapunov subspaces. Therefore, in ≥j ≥j view of (8.3), for almost all y ∈ X , (8.11) F (y) = F (y) ∩ V (y) ⊕ F (y) ∩ V (y) . ≥j ≥j ≤j ≥j >j Since F (x) ⊂ F (x), we have by Lemma 8.2, (u) v ∈ F (ux) for almost all u ∈ B(x). ≥j ∗ ≥j By the definition of F (x), since v ∈/ F (x), for almost all u if we decompose using ≥j+1 ≥j+1 (8.11), (u) v = v + w , v ∈ F (ux) ∩ V (ux), w ∈ F (ux) ∩ V (ux), ∗ u u u ≥j ≤j u ≥j >j then v = 0. Since by definition F (ux) ∩ V (ux) = E (ux) we have v ∈ E (ux).Let u ≥j ≤j j u j E (x) = v ∈ P F (x) : u ∈ B(x) :v ≥ v >(1 − δ/2)|B(x)| . Then the E (x) are an increasing family of open sets, and E (x) = P(F (x)). Since n n n=1 P(F (x)) is compact, there exists n(x) such that E (x) = P(F (x)). We can now choose n(x) j j 190 ALEX ESKIN, MARYAM MIRZAKHANI K ⊂ with ν(K )> 1 − δ/2such that for x ∈ K , n(x)< 1/β(δ). This shows that for 01 01 01 x ∈ K ,for any v ∈ P(F (x)),for (1−δ/2)-fraction of u ∈ B(x) we have v >β(δ)v . 01 j To prove the final estimate note that there exists a set K with ν(K )> 1 − δ/2 01 01 and a constant C(δ) such that for all x ∈ K and at least (1 − δ/2)-fraction of u ∈ B(x), we have (u) v ≤ C(δ)v .Let K = K ∩ K . Then, for at least (1 − δ)-fraction of ∗ 01 01 01 u ∈ B(x),wehave −1 w ≤(u) v ≤ C(δ)v ≤ C(δ)β(δ) v . u ∗ u Proposition 8.5. (a) For every δ> 0 there exists K ⊂ X of measure at least 1 − δ and a number L (δ) > 0 0 2 such that the following holds: Suppose x ∈ K, v ∈ H(x). Then, for any L > L (δ) there exists L < t < 2L such that for at least (1 − δ)-fraction of u ∈ B(g x), −t (g ) (u) (g ) v s ∗ ∗ −t ∗ −αt d , E(g ug x) ≤ C(δ)e , s −t (g ) (u) (g ) v s ∗ ∗ −t ∗ where s > 0 is such that (8.12) (g ) (u) (g ) v=v, s ∗ ∗ −t ∗ and α depends only on the Lyapunov spectrum. (b) There exists > 0 (depending only on the Lyapunov spectrum) and for every δ> 0 a compact set K with ν(K )> 1− c(δ) where c(δ) → 0 as δ → 0 such that the following holds: Suppose there exist arbitrarily large t > 0 with g x ∈ K so that for at least (1−δ)- −t fraction of u ∈ B(x), the number s > 0 satisfying (8.12), also satisfies (8.13) s ≥ 1 − t. Then v ∈ E(x). Proof.—Let > 0 be smaller than one third of the difference between any two Lyapunov exponents for the action on H. By the Osceledets multiplicative ergodic theo- rem, there exists a compact subset K ⊂ X with ν(K )> 1 − δ and L > 0such that for 1 0 1 x ∈ K and all j and all t > L, (λ +)t (g ) v≤ e v, v ∈ V (x) t ∗ ≥j and (λ −)t (g ) v≥ e v, v ∈ V (x). t ∗ ≤j ∗ ∗ ∗ By Fubini’s theorem there exists K ⊂ X with ν(K )> 1 − 2δ such that for x ∈ K , 1 1 1 u ∈ B(x) : ux ∈ K ≥ (1 − δ/2)|B(x)|. 1 INVARIANT AND STATIONARY MEASURES 191 Let K = K ∩ K ,where K is as in Lemma 8.4 (with δ replaced by δ/2). Let K, L (δ) 01 01 2 be such that for all x ∈ Kand allL > L , there exists t with L < t < 2L and g x ∈ K . 2 −t Write (8.14) (g ) v = v , v ∈ F (g x). −t ∗ −t j j j j∈ We have g x ∈ K ∩ K . Suppose u ∈ Q (g ) and ug x ∈ K . Then, by Lemma 8.4, −t 01 01 −tx −t 1 we have (8.15) (u) (g ) v = (v + w ), ∗ −t ∗ j j j∈ where v ∈ E (ug x), w ∈ V (ug x),and forall j ∈ , j j −t j >j −t (8.16) v ≥ β (δ)w . j j Then, (λ +)s j+1 (g ) w ≤ e w , s ∗ j j and, (λ −)s (λ −)s j j (8.17) (g ) v ≥ e v ≥ e β (δ)w . s ∗ j j j Thus, for all j ∈ , −(λ −λ +2)s −1 j j+1 (g ) w ≤ e β (δ) (g ) v . s ∗ j s ∗ j Since (g ) v ∈ E and using part (a) of Proposition 4.15, we get (a) of Proposition 8.5. s ∗ j To prove (b), suppose v ∈/ E(x). We may write v = vˆ , vˆ ∈ F (x) i i i∈ Let j be minimal such that vˆ ∈/ E (x).Let k > j be such that F (x) ⊂ F (x) is the sub- j j ≥k ≥j space preceding F (x) in (8.10). Then, F (x) = F (x) for k + 1 ≤ i ≤ j . ≥j ≥i ≥j Since vˆ ∈/ E (x), vˆ must have a component in V (H)(x) for some i ≥ j + 1. There- j j j i fore, by looking only at the component in V (H),weget −(λ +)t j+1 (g ) v≥ C(v)e . −t ∗ Also since F is g -equivariant we have F (x) = F (x) ∩ V (H). Note that by the ≥k t ≥k ≥k m −λ t multiplicative ergodic theorem, the restriction of g to V (H) is of the form e h ,where −t i t h = O(e ). Therefore (again by looking only at the component in V (H) and using t i Proposition 4.15(a)), we get −(λ +2)t j+1 d (g ) v, F (g x) ≥ C(v)e . −t ∗ ≥k −t 192 ALEX ESKIN, MARYAM MIRZAKHANI (Here and below, d(·,·) denotes the distance on H(x) given by the dynamical norm · .) Therefore (since (g ) v ∈ F (g x)), we see that if we decompose (g ) v as in (8.14), we −t ∗ ≥j −t −t ∗ get −(λ +2)t j+1 v ≥ C(v)e . We now decompose (u) (g ) v as in (8.15). Then, from (8.16)and (8.17), ∗ −t ∗ (λ −)s (λ −)s (λ −)s −(λ +2)t j j j j+1 (8.18) (g ) v ≥ e v ≥ e β(δ)v ≥ e β(δ)C(v)e . s ∗ j j If s satisfies (8.12), then (g ) v = O(1). Therefore, in view of (8.18), s ∗ j (λ −)s −(λ +2)t j j+1 e e ≤ c = c(v,δ). Therefore, (λ + 2)t + log c(v,δ) j+1 s ≤ . (λ − ) Since λ >λ , this contradicts (8.13)if is sufficiently small and t is sufficiently large. j j+1 9. The action of the cocycle on E In this section, we work on the finite cover X defined in Section 4.6.Recallthatif f (·) is an object defined on X ,thenfor x ∈ X we write f (x) instead of f (σ (x)) (where 0 0 σ : X → X is the covering map). 0 0 In this section and in Section 10, assertions will hold at best for a.e x ∈ X, and never for all x ∈ X. This will be sometimes suppressed from the statements of the lemmas. 9.1. The Jordan canonical form of the cocycle on E(x).— We consider the action of the cocycle on E. The Lyapunov exponents are λ , i ∈ .WenotethatbyLemma 8.2, the bundle E admits the equivariant measurable flat U -connection given by the maps (u) : E(x) → E(y),where (u) is as in Lemma 6.6. This connection satisfies the condition ∗ ∗ (4.5), since by Lemma 8.2, (u) E (x) = E (y).For each i ∈ , we have the maximal flag ∗ j j as in Lemma 4.3, (9.1) {0}⊂ E (x) ⊂ ··· ⊂ E (x) = E (x). i1 i,n i Let denote the set of pairs ij which appear in (9.1). By Proposition 4.12 and Re- mark 4.13,wehavefor a.e. u ∈ B(x), (u) E (x) = E (ux). ∗ ij ij Let · and ·,· denote the restriction to E(x) of the norm and inner product on x x H(x) defined in Section 4.7 and Section 6. (We will often omit the subscript from ·,· x INVARIANT AND STATIONARY MEASURES 193 and · .) Then, the distinct E (x) are orthogonal. For each ij ∈ let E (x) be the x i ij orthogonal complement (relative to the inner product ·,· )to E (x) in E (x). x i,j−1 ij Then, by Proposition 4.15, we can write, for v ∈ E (x), ij λ (x,t) ij (9.2) (g ) v = e v + v , t ∗ where v ∈ E (g x), v ∈ E (g x),and v =v. Hence (since v and v are orthogo- t i,j−1 t ij nal), λ (x,t) ij (g ) v≥ e v. t ∗ In view of Proposition 4.15 there exists a constant κ> 1such that for a.e x ∈ Xand for all v ∈ E(x) and all t ≥ 0, −1 κ t κ t (9.3) e v≤(g ) v≤ e v. t ∗ Lemma 9.1. — For a.e. x ∈ X and for a.e. y = ux ∈ B[x], the connection (u) : E(x) → + + E(y) agrees with the restriction to E of the connection P (x, y) induced from the map P (x, y) defined in Section 4.2. 1 1 1 Proof.—Let V (x) = V (H )(x) and V (x) = V (H )(x),where V (H )(x) and ≤i ≤i i i ≤i V (H )(x) areasinSection 4.1. Consider the definition (6.12)of u in Section 6.For a i ∗ + + + fixed Y = log u ∈ Lie(U )(x) and M ∈ H (x),let h : W (x) → W (ux) be given by ++ h(v) = exp (I + M)Y (x + v) − exp(Y)x. From the form of h, we see that h(V (x)) = V (ux),and also, h induces the identity map ≤i ≤i on V (x)/V (x) = V (ux)/V (ux).Thus, for v ∈ V (x), ≤i <i ≤i <i i h(v) ∈ P (x, ux)v + V (ux). <i Similarly, M ≡ tr(x, ux) ◦ M ◦ tr(ux, x) agrees with M up to higher Lyapunov exponents. Then, in view of (6.12), (6.18) and Lemma 6.8,for v ∈ E (x), (u) v ∈ P (x, ux)v + V (ux). ∗ <i But, for v ∈ E (x), (u) v ∈ E (ux) (and thus has no component in V (ux)). Hence, for all i ∗ i <i v ∈ E (x),wehave (u) v = P (x, ux)v. i ∗ 194 ALEX ESKIN, MARYAM MIRZAKHANI 9.2. Time changes. ij ij The flows g and the time changes τˆ (x, t).— We define the time changed flow g so that ij t t ij (after the time change) the cocycle λ (x, t) of (9.2) becomes λ t. We write g x = g x. ij i τˆ (x,t) t ij Then, by construction, λ (x,τˆ (x, t)) = λ t. We note the following: ij ij i Lemma 9.2. — Suppose y ∈ B [x]. Then for any ij ∈ and any t > 0, ij ij g y ∈ B g x . −t −t Proof. — This follows immediately from property (e) of Proposition 4.15,and the ij definition of the flow g . −t In view of Proposition 4.15,wehave (9.4) |t − t |≤ τˆ (x, t)−ˆτ x, t ≤ κ|t − t | ij ij where κ depends only on the Lyapunov spectrum. 9.3. The foliations F , F and the parallel transport R(x, y).— For x ∈ X, let ij v G[x]= g ug x : t ≥ 0, s ≥ 0, u ∈ B(g x) ⊂ X. s −t −t For y = g ug x ∈ G[x],let s −t R(x, y) = (g ) (u) (g ) . s ∗ ∗ −t ∗ Here (g ) is as in (6.16)and (u) : H(g x) → H(ug x) is as in Lemma 6.6.Itiseasyto s ∗ ∗ −t −t see using Lemma 6.7 that R(x, y) : H(x) → H(y) depends only on x, y and not on the choices of t, u, s. We will usually consider R(x, y) as a map from E(x) → E(y). In view of (9.2), Lemma 9.1 and Proposition 4.15(e) and (f), we have, for v ∈ E (x), ij and any y = g ug x ∈ G[x], s −t λ (x,y) ij (9.5)R(x, y)v = e v + v where v ∈ E (y), v ∈ E (y),and v =v.In(9.5), we have i,j−1 ij (9.6) λ (x, y) = λ (x,−t) + λ (ug x, s). ij ij ij −t INVARIANT AND STATIONARY MEASURES 195 Notational convention. — We sometimes use the notation R(x, y) when x ∈ X (instead of X) and y ∈ G[x]. For x ∈ Xand ij ∈ ,let F [x] denote the set of y ∈ G[x] such that there exists ij ≥ 0so that ij ij (9.7) g y ∈ B g x . − − By Lemma 9.2,if(9.7) holds for some , it also holds for any bigger . Alternatively, ij ij ij F [x]= g ug x : ≥ 0, u ∈ B g x ⊂ X. ij − − As above, when x ∈ X, we can think of the leaf of the foliation F [x] as a subset of X ij (not X). In view of (9.6), it follows that (9.8) λ (x, y) =0if y ∈ F [x]. ij ij We refer to the sets F [x] as leaves. Locally, the leaf F [x] through x is a piece of U [x]. ij ij More precisely, for y ∈ F [x], ij F [x]∩ B [ y]⊂ U [ y]. ij 0 ij Then, for any compact subset A ⊂ F [x] there exists large enough so that g (A) is ij ij contained in a set of the form B[z]⊂ U [z]. Then the same holds for g (A),for any −t t >. Recall (from the start of Section 6) that the sets B[x] support a “Lebesgue measure” + + |·|, namely the pushforward of the Haar measure on U (x)/(U (x) ∩ Q (x))(x) to ++ B[x] under the map u → ux.(Recall that Q (x) is the stabilizer of x in the affine group ++ G (x).) As a consequence, the leaves F [x] also support a Lebesgue measure (defined ++ ij up to normalization), which we also denote by |·|. More precisely, if A ⊂ F [x] and ij B ⊂ F [x] are compact subsets, we define ij ij |A| |g (A)| (9.9) ≡ , ij |B| |g (B)| ij ij where is chosen large enough so that both g (A) and g (B) are contained in a set of − − the form B[z], z ∈ X. It is clear that if we replace by a larger number, the right-hand- side of (9.9) remains the same. We define the “balls” F [x,]⊂ F [x] by ij ij ij ij (9.10) F [x,]= y ∈ F [x]: g y ∈ B g x . ij ij − − Lemma 9.3. — Suppose x ∈ X and y ∈ F [x]. Then, for large enough, ij F [x,]= F [ y,]. ij ij 196 ALEX ESKIN, MARYAM MIRZAKHANI ij ij Proof. — Suppose y ∈ F [x]. Then, for large enough, g y ∈ B[g x],and then ij − − ij ij B[g y]= B[g x]. − − v v The “flows” g .— Suppose x ∈ Xand v ∈ E(x).Let g x = g x, where the time τˆ (x,t) t t v change τˆ (x, t) is chosen so that v t g v = e v . v x ∗ g x (Note that we are not defining g y for y = x.) We have, for x ∈ X, v (g ) v v t ∗ g x = g g x. t+s s t By (9.3), (9.4) holds for τˆ instead of τˆ . v ij For y ∈ G[x] and ∈ R,let v,x (9.11) g˜ = g y, where w = R(x, y)v. − − (When there is no potential for confusion about the point x and the vector v used, we ij v,x v,x denote g˜ by g˜ .) Note that Lemma 9.2 still holds if g is replaced by g˜ . − −t −t The foliations F .— For v ∈ E(x) we can define the foliations F [x] and the “balls” v v ij v,x F [x,] as in (9.7)and (9.10), with g˜ replacing the role of g . −t −t For y ∈ F [x],wehave F [x]= F [ y], where w = R(x, y)v. v w We can define the measure (up to normalization) |·| on F [x,] as in (9.9). Lemma 9.3 holds for F [x] without modifications. The following follows immediately from the construction: Lemma 9.4. — For a.e. x ∈ X,any v ∈ E(x), and a.e. y ∈ F [x], we have R(x, y)v =v . y x 9.4. A maximal inequality. Lemma 9.5. — Suppose K ⊂ X with ν(K)> 1 − δ. Then, for any θ > 0 there exists a ∗ ∗ 2 ∗ subset K ⊂ X with ν(K )> 1 − 2κ δ/θ such that for any x ∈ K and any > 0, (9.12) |F [x,]∩ K| > 1 − θ |F [x,]|. ij ij INVARIANT AND STATIONARY MEASURES 197 Proof.—For t > 0let ij ij ij ij B [x]= g B g x ∩ U g x = B [x], t −t 0 t t τ ij ij where τ is such that g x = g x.Let s > 0 be arbitrary. Let K = g K. Then ν(K )> τ s s t −s 1 − κδ. Then, by Lemma 6.3, there exists a subset K with ν(K ) ≥ (1 − 2κδ/θ ) such s s that for x ∈ K and all t > 0, ij |K ∩ B [x]| ≥ 1 − θ /2 |K |. s t s ij ∗ ij ij ij ∗ Let K = g K , and note that g B [x]= F [g x, s − t]. Then, for all x ∈ K and all 0 < t ij s s s s s s s − t < s, |F [x, s − t]∩ K|≥ 1 − θ /2 |F [x, s − t]|. ij ij ∗ 2 ∗ We have ν(K ) ≥ (1 − 2κ δ/θ ). Now take a sequence s →∞,and let K be the set of points which are in infinitely many K . 10. Bounded subspaces and synchronized exponents Recall that indexes the “fine Lyapunov spectrum” on E. In this section we define an equivalence relation called “synchronization” on ; the equivalence class of ij ∈ is denoted by [ij] and the set of equivalence classes is denoted by .For each ij ∈ we define a g -equivariant and locally (u) -equivariant (in the sense of t ∗ Lemma 6.6(b)) subbundle E of the bundle E ≡ V (E) and we define ij,bdd i i E (x) = E (x). [ij],bdd kr,bdd kr∈[ij] In fact we will show that there exists a subset [ij] ⊂[ij] such that (10.1) E (x) = E (x). [ij],bdd kr,bdd kr∈[ij] Then, we claim that the following three propositions hold: Proposition 10.1. — There exists θ> 0 depending only on ν and n ∈ N depending only on the dimension of X such that the following holds: for every δ> 0 and every η> 0, there exists a subset K = K(δ,η) of measure at least 1−δ and L = L (δ,η) > 0 such that the following holds: Suppose 0 0 x ∈ X, v ∈ E(x), L ≥ L ,and n+1 |g K ∩ F [x, L]| ≥ 1 − (θ/2) |F [x, L]|. [−1,1] v v 198 ALEX ESKIN, MARYAM MIRZAKHANI Then, for at least (θ/2) -fraction of y ∈ F [x, L], R(x, y)v d , E (y) <η. [ij],bdd R(x, y)v ij∈ Proposition 10.2. — There exists a function C : X → R finite almost everywhere so that for all x ∈ X,for all y ∈ F [x], for all v ∈ E (x), ij [ij],bdd −1 −1 C (x) C (y) v≤R(x, y)v≤ C (x)C (y)v. 3 3 3 3 (Recall from Section 2.2 that by C (x) we mean C (π(x)).) 3 3 Proposition 10.3. — There exists θ> 0 (depending only on ν)andasubset ⊂ X with ν() = 1 such that the following holds: Suppose x ∈ , v ∈ H(x),and thereexists C > 0 such that for all > 0, and at least (1 − θ)-fraction of y ∈ F [x,], ij R(x, y)v≤ Cv. Then, v ∈ E (x). [ij],bdd Proposition 10.1 is what allows us to choose u so that there exists u such that the vector in H associated to the difference between the generalized subspaces U [g u q ] and U [g uq ] points close to a controlled direction, i.e. close to E (g uq ). This allows t 1 [ij],bdd t 1 us to address “Technical Problem #3” from Section 2.3. Then, Proposition 10.2 and Proposition 10.3 are used in Section 11 to define and control conditional measures f ij associated to each [ij]∈ , so we can implement the outline in Section 2.3.Wenotethat it is important for us to define a family of subspaces so that all three propositions hold. The number θ> 0, the synchronization relation and the subspaces E are de- ij,bdd ∗ ∗ fined in Section 10.1 . Also Proposition 10.1 is proved in Section 10.1 .Proposition 10.2 and Proposition 10.3 are proved in Section 10.2 . Both subsections may be skipped on first reading. Example. — To completely understand the example below, it necessary to read at least Section 10.1 . However, we include it here to give some flavor of the construction. Suppose we have a basis {e (x), e (x), e (x), e (x)} for E(x), relative to which the 1 2 3 4 cocycle has the form (for y ∈ G[x]): ⎛ ⎞ λ (x,y) e u (x, y) 00 λ (x,y) ⎜ 12 ⎟ 0 e 00 ⎜ ⎟ R(x, y) = . λ (x,y) ⎝ ⎠ 00 e 0 λ (x,y) 000 e INVARIANT AND STATIONARY MEASURES 199 Suppose E (x) = Re (x) ⊕ Re (x) (so e and e correspond to the Lyapunov expo- 1 1 2 1 2 nent λ ), E (x) = Re (x), E (x) = Re (x) (so that e and e correspond to the Lyapunov 1 3 3 4 4 3 4 exponents λ and λ respectively). Therefore the Lyapunov exponents λ and λ have 3 4 3 4 multiplicity 1, while λ has multiplicity 2. Then, we have E (x) = Re (x), E (x) = Re (x), E (x) = Re (x). 31,bdd 3 41,bdd 4 11,bdd 1 (For example, if y ∈ F [x] then λ (x, y) = 0, so that by (9.5), R(x, y)e =e .) 31 31 3 3 Now suppose that 31 and 41 are synchronized, but all other pairs are not synchro- nized. (See Definition 10.8 for the exact definition of synchronization, but roughly this means that |λ (x, y)| is bounded as y varies over F [x], but for all other distinct pairs ij 41 31 and kl , |λ (x, y)| is essentially unbounded as y varies over F [x].) Then, ij kl E (x) = Re (x) ⊕ Re (x). [31],bdd 3 4 Depending on the boundedness behavior of u (x, y) as y varies over F [x] we would 12 12 have either E (x) ={0} or E (x) = Re (x). 12,bdd 12,bdd 2 Since [11] ={11} and [12] ={12},wehave E (x) = E (x) and E (x) = [11],bdd 11,bdd [12],bdd E (x). 12,bdd ˜ ˜ 10.1 . Bounded subspaces and synchronized exponents. — For x ∈ X, y ∈ X, let |t| if y = g x, ρ(x, y) = ∞ otherwise. ˜ ˜ If x ∈ Xand E ⊂ X, we let ρ(x, E) = inf ρ(x, y). y∈E Lemma 10.4. —For every η> 0 and η > 0 there exists h = h(η ,η) such that the following holds: Suppose v ∈ E (x) and ij d , E (x) >η . i,j−1 Then if y ∈ F [x] and ρ y, F [x] > h ij then d R(x, y)v, E (y) ≤ ηv. i,j−1 200 ALEX ESKIN, MARYAM MIRZAKHANI Proof. — There exists t ∈ R such that y = g y ∈ F [x].Then t ij ρ y, F [x] = ρ y, y =|t| > h. ij We have the orthogonal decomposition v =ˆ v + w,where vˆ ∈ E (x) and w ∈ E (x). i,j−1 ij Then by (9.5) we have the orthogonal decomposition. λ (x,y ) ij R x, y vˆ = e v + w , where v ∈ E y , w ∈ E y , ˆ v=v . i,j−1 ij Since R(x, y )w ∈ E (y ),wehave i,j−1 2 2 2λ (x,y ) 2 2λ (x,y ) 2 ij ij R x, y v = e ˆ v + w + R x, y w ≥ e ˆ v . By (9.8), we have λ (x, y ) = 0. Hence, ij R x, y v ≥vˆ≥ η v. Since y ∈ F [x], R(x, y)v=v. Since |t| > h, we have either t > h or t < −h.If t < −h, then by (9.3) and Lemma 9.4, −1 −1 κ h κ h v=R(x, y)v= (g ) R x, y v ≥ e R x, y v ≥ e η v, −t ∗ which is a contradiction if h >κ log(1/η ). Hence we may assume that t > h.Wehave, λ (x,y) ij R(x, y)v = e v + w where v ∈ E (y) with v =vˆ,and w ∈ E (y).Hence, i,j−1 ij λ (x,y) λ (x,y) ij ij d R(x, y)v, E (y) = e ˆ v≤ e v. i,j−1 But, −1 λ (x, y) = λ x, y + λ y ,−t ≤−κ t ij ij ij by (9.8) and Proposition 4.15. Therefore, −1 −1 −κ t −κ h d R(x, y)v, E (y) ≤ e v≤ e v. i,j−1 INVARIANT AND STATIONARY MEASURES 201 The bounded subspace. — Fix θ> 0. (We will eventually choose θ sufficiently small depending only on the dimension.) Definition 10.5. — Suppose x ∈ X. A vector v ∈ E (x) is called (θ, ij)-bounded if there ij exists C < ∞ such that for all > 0 and for (1 − θ )-fraction of y ∈ F [x,], ij (10.2) R(x, y)v≤ Cv. Remark. — From the definition and (9.5), it is clear that every vector in E (x) is i1 (θ, i1)-bounded for every θ . Indeed, we have E = E ,and λ (x, y) = 0for y ∈ F [x], i1 i1 i1 i1 thus for y ∈ F [x] and v ∈ E (x), R(x, y)v=v. i1 i1 Lemma 10.6. —Let n = dim E (x) (for a.e x). If there exists no non-zero θ/n-bounded vector ij in E (x) \ E (x),weset E ={0}. Otherwise, we define E (x) ⊂ E (x) to be the linear ij i,j−1 ij,bdd ij,bdd ij span of the θ/n-bounded vectors in E (x). This is a subspace of E (x), and any vector in this subspace ij ij is θ -bounded. Also, (a) E (x) is g -equivariant, i.e. (g ) E (x) = E (g x). ij,bdd t t ∗ ij,bdd ij,bdd t (b) For almost all u ∈ B(x), E (ux) = (u) E (x). ij,bdd ∗ ij,bdd Proof.—Let E (x) ⊂ E (x) denote the linear span of all (θ/n, ij)-bounded vec- ij,bdd ij tors. If v ,..., v are any n (θ/n, ij)-bounded vectors, then there exists C > 1such that for 1 n 1 − θ fraction of y in F [x, L],(10.2) holds. But then (10.2) holds (with a different C) for ij any linear combination of the v . This shows that any vector in E (x) is (θ, ij)-bounded. i ij,bdd To show that (a) holds, suppose that v ∈ E (x) is (θ/n, ij)-bounded, and t < 0. In view ij ij ij of Lemma 8.2, it is enough to show that v ≡ (g ) v ∈ E (g x) is (θ/n, ij)-bounded. (This t ∗ ij t ij ij would show that for t < 0, (g ) E (x) ⊂ E (g x) which, in view of the ergodicity of ∗ ij,bdd ij,bdd t t the action of g , would imply (a).) ij Let x = g x.By(9.3), there exists C = C (t) such that for all z ∈ Xand all w ∈ t 1 1 E(z), ij −1 (10.3)C w≤ g w ≤ C w. t 1 ij ij Suppose y ∈ F [x, L] satisfies (10.2). Let y = g y.Then y ∈ F [x ].Let v = (g ) v. (See ij t ij t ∗ Figure 3.) Note that R x , y v = R y, y R(x, y)R x , x v = R y, y R(x, y)v hence by (10.3), (10.2), and again (10.3), R x , y v ≤ C R(x, y)v≤ C Cv≤ C Cv . 1 1 ij Hence, for y ∈ F [x, L] satisfying (10.2), y = g y ∈ F [x ] satisfies ij t ij (10.4) R x , y v < CC v . 1 202 ALEX ESKIN, MARYAM MIRZAKHANI FIG. 3. — Proof of Lemma 10.6(a) ij ij Therefore, since F [g x, L + t]= g F [x, L],wehavethatfor 1 − θ/n fraction of ij ij t t y ∈ F [x , L + t],(10.4) holds. Therefore, v is (θ/n, ij)-bounded. Thus, E (x) is ij ij,bdd g -equivariant. This completes the proof of (a). Then (b) follows immediately from (a) since Lemma 9.3 implies that F [ux, L]= F [x, L] for L large enough. ij ij Remark 10.7. — Formally, from its definition, the subspace E (x) depends on ij,bdd the choice of θ . It is clear that as we decrease θ , the subspace E (x) decreases. In view ij,bdd of Lemma 10.6, there exists θ > 0and m ≥ 0such that for all θ< θ and almost all 0 0 x ∈ X, the dimension of E (x) is m. We will always choose θ θ . ij,bdd 0 Synchronized exponents. Definition 10.8. — Suppose θ> 0. We say that ij ∈ and kr ∈ are θ -synchronized −1 if there exists E ⊂ X with ν(E)> 0,and C < ∞,suchthatfor all x ∈ π (E), for all > 0,for at least (1 − θ)-fraction of y ∈ F [x,], we have ij ρ y, F [x] < C. kr Remark 10.9. — By the same argument as in the proof of Lemma 10.6(a), if ij and kr are θ -synchronized then we can replace the set E in Definition 10.8 by g E. |s|<t Therefore, we can take E in Definition 10.8 to have measure arbitrarily close to 1. Remark 10.10. — Clearly if ij and kr are not θ -synchronized, then they are also not θ -synchronized for any θ <θ . Therefore there exists θ > 0 such that if any pairs ij and kr are not θ -synchronized for some θ> 0 then they are also not θ -synchronized. We will always consider θ θ , and will sometimes use the term “synchronized” with no modi- fier to mean θ -synchronized for θ θ .TheninviewofRemark 10.9, synchronization is an equivalence relation. We now fix θ min(θ ,θ ). If v ∈ E(x), we can write (10.5) v = v , where v ∈ E (x),but v ∈/ E (x). ij ij ij ij i,j−1 ij∈I v INVARIANT AND STATIONARY MEASURES 203 In the sum, I is a finite set of pairs ij where i ∈ and 1 ≤ j ≤ n .(Recall that denotes v i the Lyapunov spectrum of E.) Since for a fixed i the E (x) form a flag, without loss of ij generality we may (and always will) assume that I contains at most one pair ij for each i ∈ . For v ∈ E(x),and y ∈ F [x],let H (x, y) = supρ y, F [x] . v ij ij∈I Lemma 10.11. — There exists a set ⊂ X with ν() = 1 such that the following holds: Suppose x ∈ , C < ∞,and thereexists v ∈ E(x) so that for each L > 0, for at least (1−θ)-fraction of y ∈ F [x, L] H (x, y)< C. Then, if we write v = v as in (10.5),thenall {ij} are synchronized, and also for all ij ∈ I , ij ij∈I v ij∈I v v ∈ E (x). ij ij,bdd Proof.—Let = g E, where E is as in Definition 10.8. (In view of Re- t∈R mark 10.9, we may assume that the same E works for all synchronized pairs.) Suppose ij ∈ I and kr ∈ I .Wehavefor at least (1 − θ)-fraction of y ∈ F [x, L], v v v ρ y, F [x] < C,ρ y, F [x] < C. ij kr Let y ∈ F [x] be such that ρ(y, F [x]) = ρ(y, y ). Similarly, let y ∈ F [x] be such that ij ij ij ij kr kr ρ(y, F [x]) = ρ(y, y ).Wehave kr kr (10.6) ρ(y , y ) ≤ ρ(y , y) + ρ(y, y ) ≤ 2C. ij kr ij kr ij ij v,x Note that g˜ (F [x, L]) = g (F [x, L ]),where L is chosen so that g x = g x,where v ij −L −L −L −L the notation g˜ is as in (9.11). Hence, in view of (10.6)and (9.9), for any L > 0, for (1−θ)- fraction of y ∈ F [x, L ], ρ(y , F [x]) ≤ 2C. Then, for any t ∈ R,for any L > 0, for ij ij ij kr (1−θ)-fraction of y ∈ F [g x, L ], ρ(y , F [g x]) ≤ C(t). Since x ∈ , we can choose t so ij ij t ij kr t that g x ∈ E where E is as in Definition 10.8. This implies that ij and kr are synchronized. Recall that I contains at most one j for each i ∈ . Since R(x, y) preserves each E , v i and the distinct E are orthogonal, for all y ∈ G[x], 2 2 R x, y v = R x, y v . ij ij∈I Therefore, for each ij ∈ I ,and all y ∈ G[x], R x, y v ≤ R x, y v . ij 204 ALEX ESKIN, MARYAM MIRZAKHANI In particular, R(x, y )v ≤R(x, y )v. ij ij ij We have for (1 −θ)-fraction of y ∈ F [x, L ], ρ(y , y)< C, where y ∈ F (x).Wehave, by ij ij ij v Lemma 9.4, R(x, y)v=v, and hence, by (9.3), for (1 − θ)-fraction of y ∈ F [x, L], ij ij R(x, y )v≤ C v. ij 2 Hence, for (1 − θ)-fraction of y ∈ F [x, L ], ij ij R(x, y )v ≤ C v. ij ij 2 This implies that v ∈ E (x). ij ij,bdd We write ij ∼ kr if ij and kr are synchronized. With our choice of θ> 0, synchro- nization is an equivalence relation, see Remark 10.10. We write [ij]={kr : kr ∼ ij}. Let E (x) = E (x). [ij],bdd kr,bdd kr∈[ij] For v ∈ E(x), write v = v ,asin(10.5). Define ij ij∈I height(v) = (dim E) + j ij∈I The height is defined so it would have the following properties: • If v ∈ E (x) \ E (x) and w ∈ E (x) then height(w)< height(v). ij i,j−1 i,j−1 • If v = v , v ∈ E , v = 0, and w = w , w ∈ E , w = 0, and also the i i i i j j j j i∈I j∈J cardinality of J is smaller then the cardinality of I ,then height(w)< height(v). Let P (x) ⊂ E(x) denote the set of vectors of height at most k. This is a closed subset of E(x). Lemma 10.12. —For every δ> 0 and every η> 0 there exists a subset K ⊂ X of mea- sure at least 1 − δ and L > 0 such that for any x ∈ K and any unit vector v ∈ P (x) with d(v, E )>η and d(v, P (x)) > η, there exists 0 < L < L so that for at least θ -fraction [ij],bdd k−1 ij of y ∈ F [x, L ], R(x, y)v d , P (y) <η. k−1 R(x, y)v INVARIANT AND STATIONARY MEASURES 205 Proof. — Suppose C > 1 (we will later choose C depending on η). We first claim that we can choose K with ν(K)> 1 − δ and L > 0so that for every x ∈ g Kand every [−1,1] v ∈ P (x) such that d(v, E )>η there exists 0 < L < L so that for θ -fraction of k [ij],bdd ij y ∈ F [x, L ], (10.7)H (x, y) ≥ C. (Essentially, this follows from Lemma 10.11, but the argument given below is a bit more elaborate since we want to choose L uniformly over all v ∈ P (x) satisfying d(v, E )>η.) Indeed, let E ⊂ P (x) denote the set of unit vectors v ∈ P (x) [ij],bdd L k k ij such that for all 0 < L < L, for at least (1 − θ)-fraction of y ∈ F [x, L ],H (x, y) ≤ C. v v Then, the E are closed sets which are decreasing as L increases, and by Lemma 10.11, E ⊂ E (x) ∩ P (x). L [ij],bdd k L=1 ij∈ Let F denote the subset of the unit sphere in P (x) which is the complement of the η-neighborhood of E (x). Then the E are an open cover of F, and since F is [ij],bdd ij L compact, there exists L = L such that F ⊂ E .Now forany δ> 0 we can choose L so that L > L for all x in a set K of measure at least (1 − δ). Now suppose v ∈ F. Since F ⊂ E , v ∈/ E , hence there exists 0 < L < L (possi- bly depending on v) such that the fraction of y ∈ F [x, L ] which satisfies H (x, y) ≥ Cis v v greater than θ . Then, (10.7) holds. Now suppose (10.7) holds (with a yet to be chosen C = C(η)). Write v = v ij ij∈I as in (10.5). Let w = R(x, y)v, w = R(x, y)v . ij ij Since y ∈ F [x], by Lemma 9.4, w=v= 1. Let ij ∈ I be such that the supremum v v in the definition of H (x, y) is achieved for ij.If w <η/2 we are done, since w = v ij w has smaller height than v,and d(w, )<η. Hence we may assume that kr kr=ij w 1 ≥w ≥ η/2. ij Since d(v, P (x)) ≥ η,wehave k−1 d v , E (x) ≥ η ≥ ηv , ij i,j−1 ij where the last inequality follows from the fact that v ≤ 1. In particular, we have 1 ≥ ij v ≥ η. ij 206 ALEX ESKIN, MARYAM MIRZAKHANI Let y = g y be such that y ∈ F [x].Notethat t v ij 1 = R x, y v = R y, y w =(g ) w and 1 ≥w ≥ η/2. ij ij t ∗ ij ij Then, in view of (9.3), |t|≤ C (η), and hence R(y , y)≤ C (η). Let C = C (η) + h(η, η/C (η)),where h(·,·) is as in Lemma 10.4.Wenow 1 0 choose the constant C in (10.7)tobe C .If H (x, y)> C then, by the choice of ij , 1 v 1 ρ(y, F [x])> C . Since y = g y and |t|≤ C (η),wehave ij 1 t 0 ρ y , F [x] > C − C (η) = h η, η/C (η) . ij 1 0 Then, by Lemma 10.4 applied to v and y ∈ F [x], ij v ij 1 1 d R x, y v , E y ≤ η/C (η) v ≤ η/C (η). ij i,j−1 ij 0 0 2 2 Then, since w = R(y , y)R(x, y )v , ij ij d w , E (y) ≤ R y , y d R x, y v , E y ij i,j−1 ij i,j−1 ≤ R y , y η/C (η) ≤ . Let w be the closest vector to w in E (y),and let w = w + w .Then ij i,j−1 ij ij ij kr=ij d(w, )<η and w ∈ P . k−1 Proof of Proposition 10.1.—Let n denote the maximal possible height of a vector. Let δ = δ/n.Let η = η.Let L = L (δ ,η ) and K = K (δ ,η ) be chosen so that n n−1 n−1 n n−1 n−1 n Lemma 10.12 holds for k = n − 1, K = K ,L = L and η = η .Let η be chosen n−1 n−1 n n−1 so that exp(N(L + 1))η ≤ η ,where N isasinLemma 7.1. We repeat this process n−1 n−1 n until we choose L , η .Let L = L + 1. Let K = K ∩··· ∩ K .Then ν(K)> 1 − δ. 1 0 0 1 0 n−1 Let R(x, y)v E = y ∈ F [x, L]: d , P (y) ∪ E (y) <η , v k [ij],bdd k R(x, y)v ij∈ and let E =˜ g E , k −L so E ⊂ B[z],where z =˜ g x. Since E = F [x, L],wehave E = B[z].Let Q = k −L v n g˜ (g K ∩ F [x, L]). Then, by assumption, −L [−1,1] v n+1 (10.8) |Q|≥ 1 − (θ/2) |B[z]|. INVARIANT AND STATIONARY MEASURES 207 By Lemma 10.12, for every point uz ∈ (E ∩ Q) \ E there exists a “ball” B [uz] (where k k−1 t t = L − L and L is as in Lemma 10.12)suchthat (10.9) |E ∩ B [uz]| ≥ θ|B [uz]|. k−1 t t (When we are applying Lemma 10.12 we do not have v ∈ P but rather d(v/v, P )< k k η ; however by the choice of the η’s and the L’s this does not matter.) The collection of balls {B [uz]} as in (10.9)are acover of (E ∩ Q)\ E . These balls satisfy the t uz∈(E ∩Q)\E k k−1 k k−1 condition of Lemma 3.10(b); hence we may choose a pairwise disjoint subcollection which still covers (E ∩ Q) \ E .Weget |E |≥ θ|E ∩ Q|.Hence,by(10.8) and induction k k−1 k−1 k over k,wehave n−k |E |≥ (θ/2) |B[z]|. n n Hence, |E |≥ (θ/2) |B[z]|. Therefore |E |≥ (θ/2) |F [x, L]|. Since P =∅, the Propo- 0 v 0 sition follows from the definition of E . 10.2 . Invariant measures on X × P(L).— In this subsection we prove Proposi- tion 10.2. Recall that any bundle is measurably trivial. Lemma 10.13. —Suppose L(x) is an invariant subbundle or quotient bundle of H(x).(In fact the arguments in this subsection apply to arbitrary vector bundles.) Let μ˜ be the measure on X × P(L) defined by (10.10) μ˜ (f ) = f x, R(y, x)v dydρ (v)dν(x) |F [x,]| X P(L) ij F [x,] ij where ρ is the “round” measure on P(L).(In fact, ρ can be any measure on P(L) in the measure class 0 0 of Lebesgue measure, independent of x and fixed once and for all.) Let μˆ be the measure on X × P(L) defined by (10.11) μˆ (f ) = f y, R(x, y)v dydρ (v)dν(x). |F [x,]| X P(L) ij F [x,] ij Then μˆ is in the same measure class as μ˜ ,and dμˆ −2 2 (10.12) κ ≤ ≤ κ , dμ˜ where κ is as in Proposition 4.15. Proof.—Let F(x, y) = f x, R(y, x)v dρ (v). P(L) 208 ALEX ESKIN, MARYAM MIRZAKHANI Then, (10.13) μ˜ (f ) = F(x, y)dydν(x) |F [x,]| X ij F [x,] ij (10.14) μˆ (f ) = F(y, x)dydν(x) |F [x,]| X ij F [x,] ij ij −1 Let x = g x. Then, in view of Proposition 4.15, κ dν(x) ≤ dν(x ) ≤ κ dν(x). Then, 1 1 ij ij μ˜ (f ) ≤ F g x , g z dzdν x ≤ κμ˜ (f ), κ |B[x ]| X B[x ] and 1 1 ij ij μˆ (f ) ≤ F g z, g x dzdν x ≤ κμˆ (f ) κ |B[x ]| X B[x ] Let X consist of one point from each B[x]. In view of Definition 6.2(iii), we now disinte- grate dν(x ) = dβ(x )dz where x ∈ X , z ∈ B[x ]. ij ij F g x , g z dzdν x |B[x ]| X B[x ] ij ij = F g z , g z dz dzdβ x X B[x ]×B[x ] ij ij = F g z, g z dz dzdβ x X B[x ]×B[x ] ij ij = F g z, g x dzdν x . |B[x ]| X B[x ] Now (10.12) follows from (10.13)and (10.14). Lemma 10.14. —Let μ˜ be any weak-star limit of the measures μ˜ .Then, (a) We may disintegrate dμ˜ (x, v) = dν(x)dλ (v), where for each x ∈ X, λ is a measure ∞ x x on P(L). (b) For x ∈ X and y ∈ F [x], ij λ = R(x, y) λ , y ∗ x (where to simplify notation, we write λ and λ instead of λ and λ ). x y π(x) π(y) INVARIANT AND STATIONARY MEASURES 209 (c) Let w ∈ P(L) be a point. For η> 0 let B(w,η) = v ∈ P(L) : d(v, w) ≤ η . Then, for any t < 0 there exists c = c (t, w)> 0 and c = c (t, w)> 0 such that for 1 1 2 2 x ∈ X, λ B(g w, c η) ≥ c λ B(w,η) . g x t 1 2 x Consequently, for t < 0, the support of λ contains the support of (g ) λ . g x t ∗ x (d) For almost all x ∈ X there exist a measure ψ on P(L) such that λ = h(x)ψ x x for some h(x) ∈ SL(L), and also for almost all y ∈ F [x], ψ = ψ (so that ψ is constant ij y x on the leaves F ). The maps x → ψ and x → h(x) are both ν -measurable. ij x Proof.—If f (x, v) is independent of the second variable, then it is clear from the definition of μ˜ that μ˜ (f ) = fdν . This implies (a). To prove (b), note that R(y , y) = R(x, y)R(y , x). Then, λ = lim R y , y ρ dy y 0 k→∞ |F [ y, ]| ij k F [ y, ] ij k = R(x, y) lim R y , x ρ dy ∗ 0 k→∞ |F [ y, ]| ij k F [ y, ] ij k = R(x, y) lim R y , x ρ dy ∗ 0 k→∞ |F [x, ]| ij k F [x, ] ij k = R(x, y) λ ∗ x wheretopassfromthe secondline to thethirdweused thefactthat F [x,]= F [ y,] ij ij for large enough. This completes the proof of (b). We now begin the proof of (c). Let w(x) = w. Working in the universal cover, we define for y ∈ G[x], w(y) = R(x, y)w(x).Wedefine w (x) = v ∈ P L(x) : d v, w(x) ≤ η . (Here we are thinking of the space as X× P(L) and using the same metric on all the P(L) fibers.) ij ij Let x = g x, y = g y.Wehave t t R y , x = R x, x R(y, x)R y , y . 210 ALEX ESKIN, MARYAM MIRZAKHANI −1 −1 −1 Since R(x, x ) ≤ c ,where c depends on t,wehave R(x, x ) w (x ) ⊂ w (x). cη η Then, ρ v : R y , x v ∈ w x 0 cη −1 = ρ v : R(y, x)R y , y v ∈ R x, x w x 0 cη ≥ ρ v : R(y, x)R y , y v ∈ w (x) 0 η −1 = ρ R y, y u : R(y, x)u ∈ w (x) 0 η −1 = R y, y ρ u : R(y, x)u ∈ w (x) 0 η ≥ c ρ u : R(y, x)u ∈ w (x) . 0 η ij ij ij ij Note that for t < 0, g F [x,]⊂ F [g x,] and |g F [x,]| ≥ c(t)|F [g x,]|. Substitut- t ij ij t t ij ij t ing into (10.10) completes the proof of (c). To prove part (d), let M denote the space of measures on P(L).Recallthatby [Zi2, Theorem 3.2.6] the orbits of the special linear group SL(L) on M are locally closed. Then, by [Ef, Theorem 2.9 (13), Theorem 2.6(5)] there exists a Borel cross section φ : M/SL(L) → M. Then, let ψ = φ(π(λ )) where π : M → M/SL(L) is the x x quotient map. We also recall the following well known Lemma of Furstenberg (see e.g. [Zi2, Lemma 3.2.1]): Lemma 10.15. —Let L be a vector space, and suppose μ and ν are two probability measures on P(L).Suppose g ∈ SL(L) are such that g →∞ and g μ → ν . Then the support of ν is contained i i i in a union of two proper subspaces of L. In particular, if the support of a measure ν on P(L) is not contained in a union of two proper subspaces, then the stabilizer of ν in SL(L) is bounded. Lemma 10.16. — Suppose L is either a subbundle or a quotient bundle of H. Suppose that θ> 0, and suppose that for all δ> 0 there exists a set K ⊂ X with ν(K)> 1 − δ and a constant C < ∞, such that such that for all x ∈ K,all > 0 and at least (1 − θ)-fraction of y ∈ F [x,], 1 ij (10.15) R(x, y)v≤ C v for all v ∈ L. Then for all δ> 0 and for all > 0 there exists a subset K () ⊂ X with ν(K ()) > 1 − c(δ) where c(δ) → 0 as δ → 0,and thereexists θ = θ (θ,δ) with θ → 0 as θ → 0 and δ → 0 such that for all x ∈ K (), for at least (1 − θ )-fraction of y ∈ F [x,], ij −1 (10.16)C v≤R(x, y)v≤ C v for all v ∈ L. The “condition C” of [Ef] is satisfied since SL(L) is locally compact and M is Hausdorff. INVARIANT AND STATIONARY MEASURES 211 Proof.—Let f be the characteristic function of K × P(L).By(10.10), μ˜ (f ) ≥ (1 − δ). By Lemma 10.13 we have μˆ (f ) ≥ (1 − κ δ). Therefore, by (10.11) there exists 2 1/2 a subset K () ⊂ Xwith ν(K ()) ≥ 1 − (κ δ) such that such that for all x ∈ K (), 1/2 |F [x,]∩ K|≥ 1 − κ δ |F [x,]|. ij ij For x ∈ X, let Z [x ]= (x, y) ∈ F [x ,]× F [x ,]: 0 ij 0 ij 0 x ∈ K, y ∈ K, and (10.15) holds . 2 1/2 Then, if x ∈ K () and θ = θ + (κ δ) then, by Fubini’s theorem, |Z [x ]| ≥ 1 − θ |F [x ,]× F [x ,]|. 0 ij 0 ij 0 Let Z [x ] = (x, y) ∈ F [x ,]× F [x ,]: (y, x) ∈ Z [x ] . 0 ij 0 ij 0 0 Then, for x ∈ K (), |Z [x ]∩ Z [x ] |≥ 1 − 2θ |F [x ,]× F [x ,]|. 0 0 ij 0 ij 0 For x ∈ F [x ,],let ij 0 Y (x) = y ∈ F [x,]: (x, y) ∈ Z [x]∩ Z [x] . ij 1/2 Therefore, by Fubini’s theorem, for all x ∈ K () and θ = (2θ ) , (10.17) x ∈ F [x ,]:|Y (x)|≥ 1 − θ |F [x ,]| ≥ 1 − θ |F [x ,]|. ij 0 ij 0 ij 0 (Note that F [x ,]= F [x,].) Let ij 0 ij K () = x ∈ X :|Y (x)|≥ 1 − θ |F [x,]| . ij Therefore, by (10.17), for all x ∈ K (), |F [x ,]∩ K ()|≥ 1 − θ |F [x ,]|. ij 0 ij 0 Then, by the definition of μˆ , μˆ K () × P(L) ≥ 1 − θ ν K () ≥ 1 − 2θ , and therefore, by Lemma 10.13, ν K () =˜μ K () × P(L) ≥ 1 − 2κ θ . Now, for x ∈ K (),and y ∈ Y (x),(10.16) holds. 212 ALEX ESKIN, MARYAM MIRZAKHANI Lemma 10.17. — Suppose L(x) = E (x). Then there exists a -invariant function C : ij,bdd ˜ ˜ finite almost everywhere such that for all x ∈ X,all v ∈ L(x),and all y ∈ F [x], X → R ij −1 −1 C(x) C(y) v≤R(x, y)v≤ C(x)C(y)v, Proof.—Let μ˜ and μˆ be as in Lemma 10.13. Take a sequence →∞ such that μ˜ →˜μ ,and μˆ →ˆμ . Then by Lemma 10.14(a), we have dμ˜ (x, v) = dν(x)dλ (v) ∞ ∞ ∞ x k k where λ is a measure on P(L).Let E ⊂ Xbe such that for x ∈ E, λ is supported on at x x most two subspaces. We will show that ν(E) = 0. Suppose not; then ν(E)> 0, and for x ∈ E, λ is supported on F (x) ∪ F (x),where x 1 2 F (x) and F (x) are subspaces of L(x).Wealwayschoose F (x) and F (x) to be of minimal 1 2 1 2 dimension, and if λ is supported on a single subspace F(x) (of minimal dimension), we let F (x) = F (x) = F(x). Then, for x ∈ E, F (x) ∪ F (x) is uniquely determined by x. 1 2 1 2 After possibly replacing E by a smaller subset of positive measure, we may assume that dim F (x) and dim F (x) are independent of x ∈ E. 1 2 Let ={x ∈ X : g x ∈ Eand g x ∈ Efor some t > 0and s > 0}. t −s Then, ν() = 1. If x ∈ , then, by Lemma 10.14(c), (10.18) (g ) F (g x) ∪ (g ) F (g x) ⊂ suppλ ⊂ (g ) F (g x) ∪ (g ) F (g x). s ∗ 1 −s s ∗ 2 −s x −t ∗ 1 t −t ∗ 2 t Since F (g x) and F (g x) have the same dimension, the sets on the right and on the left i t i −s of (10.18) coincide. Therefore, E ⊃ (and so E has full measure) and the set F (x) ∪ F (x) is g -invariant. By Proposition 4.4 (see also the remark immediately following the 2 t Proposition) the set F (x) ∪ F (x) is also U -invariant. 1 2 Fix δ> 0 (which will be chosen sufficiently small later). Suppose > 0 is arbi- trary. Since L = E , there exists a constant C independent of and a compact subset ij,bdd 1 K ⊂ Xwith ν(K)> 1 − δ and for each x ∈ K a subset Y (x) of F [x,] with |Y (x)|≥ ij (1 − θ)|F [x,]|,suchthatfor x ∈ Kand y ∈ Y (x) ∩ Kwe have ij R(x, y)v≤ C v for all v ∈ L. Therefore by Lemma 10.16, there exists 0 <θ < 1/2, K () ⊂ Xand for each x ∈ K () a subset Y (x) ⊂ F [x,] with |Y (x)|≥ (1 − θ )|F [x,]| such that for x ∈ K () and ij ij y ∈ Y (x),(10.16) holds. Let Z(x,η) = v ∈ P(L) : d v, F (x) ∪ F (x) ≥ η . 1 2 We may choose η> 0 small enough so that there exists K ⊂ Xwith ν(K () ∩ K )> 0 such that for all x ∈ K , ρ Z(x, C η) > 1/2. 0 1 INVARIANT AND STATIONARY MEASURES 213 Let S(η) = (x, v) : x ∈ X, v ∈ Z(x,η) Let f denote the characteristic function of the set (x, v) : x ∈ K () ∩ K , v ∈ Z(x,η) ⊂ S(η). We now claim that for any , (10.19) μˆ (f ) ≥ ν K () ∩ K 1 − θ (1/2). Indeed, if we restrict in (10.11)to x ∈ K () ∩ K , y ∈ Y (x),and v ∈ Z(x, C η),then by (10.16), f (x, R(x, y)v) = 1. This implies (10.19). Thus, (provided δ> 0and θ> 0in Definition 10.5 are sufficiently small), there exists c > 0such that for all , μˆ (S(η)) ≥ c > 0. Therefore, by Lemma 10.13, μ˜ (S(η)) ≥ c /κ . 0 0 There exists compact K ⊂ Xwith ν(K )> 1 − c /(2κ ) such that the map x → 0 0 0 F (x)∩ F (x) is continuous on K .Let K ={(x, v) : x ∈ K }.Then S(η)∩ K is a closed 1 2 0 0 0 0 2 2 set with μ˜ (S(η) ∩ K ) ≥ c /(2κ ). Therefore, μ˜ (S(η) ∩ K )> c /(2κ )> 0, which is 0 ∞ 0 0 0 a contradiction to the fact that λ is supported on F (x) ∪ F (x). x 1 2 Thus, for almost all x, λ is not supported on a union of two subspaces. Thus the same holds for the measure ψ of Lemma 10.14(d). By combining (b) and (d) of Lemma 10.14 we see that for almost all x and almost all y ∈ F [x], ij R(x, y)h(x)ψ = h(y)ψ , x x −1 hence h(y) R(x, y)h(x) stabilizes ψ . Hence by Lemma 10.15, −1 h(y) R(x, y)h(x) ∈ K(x) where K(x) is a compact subset of SL(L),and R(x, y) is the image of R(x, y) under the −1 natural map GL(L) → SL(L).Thus, R(x, y) ∈ h(y)K(x)h(x) ,and thus (10.20) R(x, y)≤ C(x)C(y). −1 ¯ ¯ Since R(x, y) = R(y, x), we get, by exchanging x and y, −1 (10.21) R(x, y) ≤ C(x)C(y). Note that by Lemma 10.6, there exists v ∈ L(x) = E (x) ⊂ E (x) such that v ∈/ ij,bdd ij E (x). Then, (9.5) and the fact that λ (x, y) = 0for y ∈ F [x] shows that (10.20) i,j−1 ij ij and (10.21) must hold for R(x, y) in place of R(x, y). This implies the statement of the lemma. 214 ALEX ESKIN, MARYAM MIRZAKHANI Lemma 10.18. — Suppose that for all δ> 0 there exists a constant C > 0 and a compact subset K ⊂ X with ν(K)> 1 − δ and for each > 0 and x ∈ K asubset Y (x) of F [x,] with ij |Y (x)|≥ (1 − θ)|F [x,]|, such that for x ∈ K and y ∈ Y (x) we have ij (10.22) λ (x, y) ≤ C. kr Then, ij and kr are synchronized, and there exists a function C : X → R finite ν -almost everywhere such that for all x ∈ X,and all y ∈ F [x], ij (10.23) ρ y, F [x] ≤ C(x)C(y). kr Proof. — The proof is a simplified version of the proof of Lemma 10.17.Let L = E /E , L = E /E ,and L = L × L . ij i,j−1 2 kr k,r−1 1 2 We have, for y ∈ G[x],and (v¯, w¯ ) ∈ L, λ (x,y) λ (x,y) ij kr (10.24) R(x, y)(v¯, w¯ ) = e v¯ , e w ¯ , where ¯ v =v¯ and ¯ w =w¯ . Recall that λ (x, y) = 0for all y ∈ F [x]. Therefore, (10.22) implies that for all x ∈ K, all ij ij > 0and all y ∈ Y (x), R(x, y)(v¯, w ¯ )≤ C (v¯, w ¯ ). Therefore, by Lemma 10.16, there exists a subset K () ⊂ Xwith ν(K ()) > 1 − c(δ) where c(δ) → 0as δ → 0, and for each x ∈ K () a subset Y ⊂ F [x,] with |Y | > ij (1 − θ )|F [x,]| such that for all y ∈ Y , ij −1 C (v¯, w ¯ )≤R(x, y)(v¯, w ¯ )≤ C (v¯, w¯ ). This implies that for x ∈ K (), y ∈ Y (x), (10.25) |λ (x, y)|=|λ (x, y) − λ (x, y)|≤ C . kr ij kr 1 Let μ˜ and μˆ be as in Lemma 10.13. Take a sequence →∞ such that μ˜ → μ˜ ,and μˆ →ˆν .ThenbyLemma 10.14(a), we have dμ˜ (x, v) = dν(x)dλ (v) where ∞ ∞ ∞ x λ is a measure on P(L). We will show that for almost all x ∈ X, λ is not supported on x x L ×{0}∪{0}× L . 1 2 Suppose that for a set of positive measure λ is supported on (L ×{0})∪({0}× L ). x 1 2 Then, in view of the ergodicity of g and Lemma 10.14(c), λ is supported on (L ×{0}) ∪ t x 1 ({0}× L ) for almost all x ∈ X. Let Z(x,η) = (v¯, w ¯ ) ∈ L(x), (v¯, w ¯ )= 1, d(v¯, L ) ≥ η, d(w ¯ , L ) ≥ η , 1 2 and let S(η) = x,(v¯, w ¯ ) : x ∈ X,(v¯, w ¯ ) ∈ Z(x,η) . Then we have μ˜ (S(η)) = 0. Therefore, by Lemma 10.13, μˆ (S(η)) = 0. ∞ ∞ INVARIANT AND STATIONARY MEASURES 215 By (10.24)and (10.25), for x ∈ K ( ) and y ∈ Y (x), (10.26)R(x, y)Z(x, C η) ⊂ Z(y,η). Choose η> 0 so that there exists K = K ( ) ⊂ Xwith ν(K ( ) ∩ K )> 0such that m m for x ∈ K , ρ (Z(x, C η)) > (1/2).Let f be the characteristic function of S(η). Then, if 0 1 we restrict in (10.11)to x ∈ K ( ) ∩ K , y ∈ Y (x),and v ∈ Z(x, C η),thenby(10.26), m 1 f (x, R(x, y)v) = 1. This implies that for all m, μˆ S(η) ≥ ν K ( ) ∩ K 1 − θ (1/2). Hence μˆ (S(η)) > 0 which is a contradiction. Therefore, for almost all x, λ is ∞ x not supported on L ×{0}∪{0}× L . Thus the same holds for the measure ψ of 1 2 x Lemma 10.14(d). By combining (b) and (d) of Lemma 10.14 we see that for almost all x ∈ X and almost all y ∈ F [x], ij R(x, y)h(x)ψ = h(y)ψ , x x −1 hence h(y) R(x, y)h(x) stabilizes ψ .Notethatinviewof(10.24), −1 α(x,y) α (x,y) h(y) R(x, y)h(x)(v¯, w ¯ ) = e v¯ , e w ¯ , where α(x, y) ∈ R,α (x, y) ∈ R, ¯ v =v¯ and ¯ w =w¯ . For i = 1, 2 let Conf (L ) denote the subgroup of GL(L ) which preserves the inner prod- x i i uct ·,· up to a scaling factor. Let Conf (L) = Conf (L ) × Conf (L ). Then, by an x x x 1 x 2 elementary variant of Lemma 10.15, since ψ is not supported on L ×{0}∪{0}× L , x 1 2 we get −1 h(y) R(x, y)h(x) ∈ K(x) −1 where K(x) is a compact subset of Conf (L).Thus, R(x, y) ∈ h(y)K(x)h(x) ,and thus R(x, y)≤ C(x)C(y). −1 Note that by reversing x and y we get R(x, y) ≤ C(x)C(y). Therefore, by (10.24), |λ (x, y) − λ (x, y)|≤ C(x)C(y). ij kr This completes the proof of (10.23). For any δ> 0 we can choose a compact K ⊂ Xwith ν(K)> 1 − δ and N < ∞ such that C(x)< Nfor x ∈ K. Now, the fact that ij and kr are synchronized follows from applying Lemma 9.5 to K. Proof of Proposition 10.2. — This follows immediately from Lemmas 10.18 and 10.17. 216 ALEX ESKIN, MARYAM MIRZAKHANI Proof of Proposition 10.3. — Choose < /10, where is as in Proposition 8.5(b). By the multiplicative ergodic theorem, there exists a set K ⊂ Xwith ν(K )> 1 − θ and 1 1 T > 0, such that for x ∈ K and t > T, (10.27) |λ (x, t) − λ t| <t, ij i where λ (x, t) is as in (9.2). Then, by Fubini’s theorem there exists a set K ⊂ K with ij 2 1 ν(K )> 1 − 3θ such that for x ∈ K ,for (1 − θ)-fraction of u ∈ B(x), ux ∈ K . 2 2 1 Let K be as in Proposition 8.5(b) with δ = θ . We may assume that the conull set in Proposition 10.3 is such so that for x ∈ , g x ∈ K ∩ K for arbitrarily large t > 0. −t Suppose g x ∈ K ∩ K and y ∈ F [x]. We may write −t ij ij ij y = g ug x = g ug x. s −t t −t By the definition of F [x, t ], and since g x ∈ K ,wehave g x ∈ K and for at least ij −t −t 2 1 (1 − θ)-fraction of y ∈ F [x, t ],wehave ug x ∈ K ,and thus,inviewof(10.27), ij −t |s − λ t |≤ t and |t − λ t |≤ t. i i Therefore for (1 − θ)-fraction of y ∈ F [x, t ] or equivalently for (1 − θ)-fraction of u ∈ ij B(g x), −t (10.28) |s − t|≤ 2t. Now suppose v ∈ H(x).Notethatif R(x, y)v≤ Cv,and s is as in Proposition 8.5, then s > s − O(1) (where the implied constant depends on C). Therefore, in view of (10.28), for (1 − θ)-fraction of u ∈ B(g x),(8.13) holds. Thus, by Proposition 8.5(b), we −t have v ∈ E(x).Thus, we canwrite v = v kr kr∈I where the indexing set I contains at most one r for each k ∈ . Without loss of generality, is such that for x ∈ , g x satisfies the conclusions of Proposition 4.15 infinitely often. −t Note that for y ∈ F [x], ij λ (x,y) kr R(x, y)v≥R(x, y)v ≥ e v . kr kr By assumption, for all > 0 and for at least 1−θ fraction of y ∈ F [x,], R(x, y)v≤ C. ij Therefore, for all > 0 and for at least (1 − θ) fraction of y ∈ F [x,],(10.22) holds. ij Then, by Lemma 10.18,for all kr ∈ I , kr and ij are synchronized, i.e. kr ∈[ij]. Therefore, for at least (1 − 2θ)-fraction of y ∈ F [x,], kr R x, y v ≤ R x, y v ≤ C . kr Now, by Definition 10.5, v (x) ∈ E (x). Therefore, v ∈ E (x). kr kr,bdd [ij],bdd It follows from the proof of Proposition 10.3 that (10.1) holds. INVARIANT AND STATIONARY MEASURES 217 11. Equivalence relations on W Let GSpc denote the space of generalized subspaces of W .Let H (x) denote ++ the set of M ∈ H (x) such that (I + M) Lie(U )(x) is a subalgebra of Lie(G )(x).We ++ ++ have a map U : H (x) × W (x) → GSpc taking the pair (M,v) to the generalized x ++ −1 subspace it parametrizes. Let U denote the inverse of this map (given a Lyapunov- adapted transversal Z(x)). For ij ∈ ,let −1 E [x]= Q ∈ GSpc : j U (Q) ∈ E (x) . ij [ij],bdd Motivation. — In view of Proposition 10.2 and Lemma 6.9(b), for any sufficiently X + small > 0, the conditions that Q ∈ E [x] and hd (Q, U [x]) = O() imply the follow- ij ing: for “most” y ∈ F [x], ij X + hd R(x, y)Q, U [ y] = O(). A partition of W [x].— Let B denote the measurable partition constructed in Section 3 (see also Section 4.6). We denote the atom containing x by B [x],and let B (x) ={v ∈ W (x) : v + x ∈ B [x]}. In this section, the only properties of B we will 0 0 0 + + use is that it is subordinate to W , and that the atoms B [x] are relatively open in W [x]. Equivalence relations. — Fix x ∈ X. For x, x ∈ W [x ] we say that 0 0 x ∼ x if x ∈ B [x] and U x ∈ E [x]. ij 0 ij Proposition 11.1. — The relation ∼ is a (measurable) equivalence relation. ij The main part of the proof of Proposition 11.1 is the following: Lemma 11.2. — There exists a subset ⊂ X with ν() = 1 such that for any ij ∈ ,if X + x ∈ ,x ∈ ,x ∈ B [x ] (so in particular d (x , x )< 1/100), and U [x ]∈ E [x ],then 0 1 1 0 0 0 1 1 ij 0 E [x ]= E [x ]. ij 1 ij 0 War ning. — We will consider the condition x ∼ x to be undefined unless x and x ij both belong to the set of Lemma 11.2. Motivation. — In view of Proposition 10.1, we can ensure, in the notation of Sec- tion 2.3 that for some ij ∈ ,U [q ] is close to E [q ]; then in the limit we would have ij 2 U [˜ q ]∈ E [˜ q ],and thus q˜ ∼ q˜ . ij 2 ij 2 2 2 Proof of Proposition 11.1, assuming Lemma 11.2.—We have 0 ∈ E (x), therefore, [ij],bdd (11.1)U [x]∈ E [x]. ij Thus x ∼ x. ij 218 ALEX ESKIN, MARYAM MIRZAKHANI Suppose x ∼ x. Then, x ∈ B [x],and so x ∈ B [x ].By(11.1), U [x]∈ E [x],and ij 0 0 ij by Lemma 11.2, E [x ]= E [x]. Therefore, U [x]∈ E [x ],and thus x ∼ x . ij ij ij ij Now suppose x ∼ x and x ∼ x . Then, x ∈ B [x]. Also, U [x ]∈ E [x ]= E [x], ij ij 0 ij ij therefore x ∼ x. ij Remark. — By Lemma 11.2,for x, x ∈ , x ∼ x if and only if x ∈ B [x] and ij 0 E [x ]= E [x]. ij ij Outline of the proof of Lemma 11.2.— Intuitively, the condition U [x ]∈ E [x ] is the 1 ij 0 + + same as “F [x ] and F [x ] stay close”, and “U [x ] and U [x ] stay closeaswetravel ij 1 ij 0 1 0 along F [x ] or F [x ]”, which is clearly an equivalence relation. We give some more ij 0 ij 1 detail below. Throughout the proof we will be using Lemma 9.2, without mentioning it explicitly. Fix 1/100. Suppose x ∈ B [x ], so in particular d (x , x )< 1/100, and 1 0 0 0 1 suppose X + + hd U [x ], U [x ] = . 1 0 Then, by Lemma 6.9(b), −1 + j U U [x ] = O(). + −1 + We are given that U [x ]∈ E [x ],thus j(U (U [x ])) ∈ E (x ). Then, by Proposi- 1 ij 0 1 [ij],bdd 0 tion 10.2,for most y ∈ F [x ], 0 ij 0 −1 + R(x , y )j U U [x ] = O(). 0 0 1 We have −1 + −1 + R(x , y )j U U [x ] = j U U y , 0 0 1 x y 1 0 0 for some y ∈ G[x ]. Then, by Lemma 6.9(b), for most y ∈ F [x ], 1 0 ij 0 X + + hd U y , U [ y ] = O() for some y ∈ G[x ]. 0 1 y 1 1 It is not difficult to show that y is near a point y ∈ F [x ].Thus, formost y ∈ F [x ], 1 ij 1 0 ij 0 X + + (11.2) hd U [ y ], U [ y ] = O() for some y ∈ F [x ]. 1 0 1 ij 1 Thus, most of the time F [x ] and F [x ] remain close, and also that for most y ∈ F [x ], ij 0 ij 1 0 ij 0 + + U [ y ] and U [ y ] remain close, for some y ∈ F [x ]. 1 0 1 ij 1 Now suppose Q ∈ E [x ],and 1 ij 1 X + hd Q , U [x ] = O(). 1 1 1 INVARIANT AND STATIONARY MEASURES 219 −1 Then, j(U (Q )) ∈ E (x ),and thus,for most y ∈ F [x ], using Proposition 10.2 1 [ij],bdd 1 1 ij 1 and Lemma 6.9(b) twice as above, we get that for most y ∈ F [x ], 1 ij 1 X + (11.3) hd R(x , y )Q , U [ y ] = O(). 1 1 1 1 In our notation, R(x , y )Q is the same generalized subspace (i.e. the same subset of W ) 1 1 1 as R(x , y )Q for y ∈ F [x ] close to y . Then, from (11.2)and (11.3), for most y ∈ 0 0 1 0 ij 0 1 0 F [x ], ij 0 X + hd R(x , y )Q , U [ y ] = O(). 0 0 1 0 Thus, using Lemma 6.9(b) again, we get that for most y ∈ F [x ], 0 ij 0 −1 R(x , y )j U (Q ) = O(). 0 0 1 −1 By Proposition 10.3, this implies that j(U (Q )) ∈ E (x ),and thus Q ∈ E [x ]. 1 [ij],bdd 0 1 ij 0 Thus, E [x ]⊂ E [x ]. ij 1 ij 0 Conversely, if Q ∈ E [x ], then the same argument shows that Q ∈ E [x ].There- 0 ij 0 0 ij 1 fore, E [x ]= E [x ]. ij 0 ij 1 The (tedious) formal verification of Lemma 11.2 is given in Section 11.1 below. The equivalence classes C [x].— For x ∈ we define the equivalence class ij C [x]= x ∈ B [x]: x ∼ x . ij 0 ij Let C denote the σ -algebra of ν -measurable sets which are unions of the equivalence ij classes C [x]. We do not distinguish between σ -algebras which are equivalent mod sets of ij ν -measure 0, so we can assume that C is countably generated (see [CK, §1.2]). We now ij want to show that (away from a set of measure 0), the atoms of the σ -algebra C are the ij sets C [x]. More precisely, we want to show that the partition C whoseatoms arethe sets ij ij C [x] is a measurable partition in the sense of [CK, Definition 1.10]. ij To see this, note that each set E [x] is an algebraic subset of GSpc, and is thus ij parametrized by a finite dimensional space Y. Let ψ : X → Y be the map taking x to the ij parametrization of E [x]. We note that the functions ψ are measurable. Also, in view of ij ij Lemma 11.2,wehave x ∼ y if and only if y ∈ B [x] and ψ (y) = ψ (x). ij 0 ij ij By Lusin’s theorem, for each ij , there exists a Borel function ψ such that ν -almost every- ij where, ψ = ψ . Now the measurability of C follows from [CK, Theorem 1.14]. ij ij ij 220 ALEX ESKIN, MARYAM MIRZAKHANI Lemma 11.3. — Suppose t ∈ R,u ∈ U (x). (a) g C [x]∩ B [g x]∩ g B [x]= C [g x]∩ B [g x]∩ g B [x]. t ij 0 t t 0 ij t 0 t t 0 (b) uC [x]∩ B [ux]∩ uB [x]= C [ux]∩ B [ux]∩ uB [x]. ij 0 0 ij 0 0 [x] and E (x) are g -equivariant. Therefore, Proof. — Note that the sets U [ij],bdd t so are the E [x], which implies (a). Part (b) is also clear, since locally, by Lemma 8.2, ij (u) E (x) = E (ux). ∗ ij ij The measures f [x].— We now define f [x] to be the conditional measure of ν along ij ij the C [x].Inother words, f [x] is defined so that for any measurable φ : X → R, ij ij E(φ | C )(x) = φdf [x]. ij ij We view f [x] as a measure on W [x] which is supported on C [x]. ij ij + + The measures f (x).— We can identify W [x] with the vector space W (x),where x ij corresponds to the origin. Let f (x) be the pullback to W (x) of f [x] under this identifi- ij ij cation. We will also call the f (x) conditional measures. (The term “leaf-wise” measures ij is used in [EL] in a related context.) We abuse notation slightly and write formulas such as E(φ | C )(x) = φdf (x). ij ij The “distance” d (·,·).— Suppose E , E are open subsets of a normed vector ∗ 1 2 space V, with E ∩ E = ∅. Suppose that for i = 1, 2, μ is a finite measure on E ,with 1 2 i i μ (E ∩ E )> 0. Then, let d (μ ,μ ) denote the Kontorovich-Rubinstein distance be- i 1 2 ∗ 1 2 tween (the normalized versions of) μ¯ and μ¯ , i.e. 1 2 1 1 d (μ ,μ ) = sup fdμ − fdμ , ∗ 1 2 1 2 μ (E ∩ E ) μ (E ∩ E ) f 1 1 2 E ∩E 2 1 2 E ∩E 1 2 1 2 where the sup is taken over all 1-Lipshitz functions f : E ∩ E → R with sup |f (x)|≤ 1. 1 2 The only property of d (·,·) we will use is that it induces the topology of weak-* convergence on the domain of common definition of the measures, up to normalization. The following proposition is the rigorous version of (2.5)inSection 2.3: Proposition 11.4. — There exists 0 <α < 1 depending only on the Lyapunov spectrum, and for every δ> 0 there exists a compact set K ⊂ X with ν(K )> 1 − δ such that the following holds: 0 0 −1 Suppose ij ∈ , 1 < C < ∞, 0 < < C /100, C < ∞,t > 0,t > 0,and |t − t| < C. −1 − −1 X Furthermore suppose q ∈ π (K ) and q ∈ W [q]∩ π (K ) are such that d (q, q )< 1/100. 0 0 ij ij −1 Let q = g q, q = g q .Alsolet q = g q ,q = g q . Suppose q , q ,q ,q all belong to π (K ). 1 3 1 1 3 0 1 t 3 t 1 1 3 INVARIANT AND STATIONARY MEASURES 221 FIG.4.—Proposition 11.4 ij Suppose u ∈ B(q , 1/100),u ∈ B(q , 1/100).Let q = g uq . We write q = g uq for 1 2 2 τ 1 1 1 −1 −1 some τ> 0,and let q = g u q (see Figure 4). Also suppose uq ∈ π (K ),u q ∈ π (K ), τ 1 0 0 2 1 1 −1 −1 q ∈ π (K ),q ∈ π (K ) and 2 0 0 −1 X + + C ≤ hd U [q ], U q ≤ C and >α τ. 2 1 0 1 q 2 −1 −1 + In addition, suppose there exist q˜ ∈ π (K ) and q˜ ∈ π (K ) such that σ (q˜ ) ∈ W [σ (q˜ )], 2 0 0 0 0 2 2 2 X X and also d (q˜ , q )<ξ and d (q˜ , q )<ξ . Then, provided ξ is small enough and t is large enough 2 2 2 2 (depending on K ), (11.4) q˜ ∈ W [˜ q ]. Also, there exists ξ > 0 (depending on ξ , K and C and t) with ξ → 0 as ξ → 0 and t →∞ such that (11.5) d P q˜ , q˜ f (q˜ ), f q˜ ≤ ξ . ∗ 2 ij 2 ij 2 2 (In (11.5) we think of f (q˜ ) as a measure on B [˜ q ], P (q˜ , q˜ )f (q˜ ) as a measure on ij 0 2 ij 2 2 2 2 + + P (q˜ , q˜ )B [˜ q ], and we use the AGY norm · on W (q˜ ) for the norm in the definition of 2 0 2 Y 2 2 d (·,·).) Proposition 11.4 is proved in Section 11.2 . We give an outline of the argument below. Outline of the proof of Proposition 11.4.— The initial intuition behind the proof of Proposition 11.4 is that “one goes from q to q by nearly the same linear map as from 3 2 q to q ; since this map is bounded on the relevant subspaces, f (q ) should be related to 3 2 ij 2 222 ALEX ESKIN, MARYAM MIRZAKHANI f (q ) and f (q ) should be related to f (q ); since f (q ) and f (q ) are close, f (q ) should ij 3 ij ij 2 ij 3 ij ij 2 3 2 be related to f (q ).” ij 2 There are several problems with this argument. First, because of the need to change transversals, there is no linear map from the space GSpc(q ) of generalized sub- spaces near q to the space GSpc(q ) of generalized subspaces near q . This difficulty is 3 2 2 easily handled by working instead with the linear maps R(q , q ) : H(q ) → H(q ) and 3 2 3 2 R(q , q ) : H(q ) → H(q ). 3 2 3 2 The second difficulty is connected to the first. We would like to say that the two maps R(q , q ) and R(q , q ) are close, but the domains and ranges of the maps are 3 2 3 2 different. Thus we need “connecting” linear maps from H(q ) to H(q ),and also from H(q ) to H(q ). The first map is easy to construct: since q and q are in the same leaf 2 3 2 3 − − − of W , we can just use the linear map P (q , q ) induced by the “W -connection map” P (q , q ) defined in Section 4.2. Instead of constructing directly a map from H(q ) to H(q ) we construct, using the Z(q ) Z(q ) choice of transversal Z(·), linear maps P (q , q˜ ) : H(q ) → H(q˜ ) and P (q , q˜ ) : 2 2 2 2 2 2 H(q ) → H(q˜ ). Since q and q˜ are close, and also since q and q˜ are close, these maps 2 2 2 2 2 2 are in a suitable sense close to the identity. Then, since q˜ and q˜ are on the same leaf + + + + of W ,wehavethe map P (q˜ , q˜ ) induced by the W -connection map P (q˜ , q˜ ) of 2 2 2 2 Section 4.2. Thus, finally we have two maps from H(q ) to H(q˜ ): + Z(q ) A = P q˜ , q˜ ◦ P (q , q˜ ) ◦ R(q , q ) 2 2 2 3 2 and Z(q ) − A = P q , q˜ ◦ R q , q ◦ P q , q . 2 2 3 2 3 Even though A and A are defined on H(q ), in what follows we only need to consider their restrictions to E (q ) ⊂ H(q ); we will denote the restrictions by B and B re- [ij],bdd 3 3 spectively. We would like to show that B and B are close. By linearity, it is enough to show that the restrictions of B and B to each E (q ) ⊂ E (q ) are close. Note that by ij,bdd 3 [ij],bdd 3 Proposition 4.12(a), P (q , q )E (q ) = E (q ). Continuing this argument, we see 3 ij,bdd 3 ij,bdd 3 3 that the two subspaces BE (q ) and B E (q ) are close to E (q˜ ) (and thus are ij,bdd 3 ij,bdd 3 ij,bdd close to each other). Also, from the construction and Proposition 10.2, we see that both B and B are uniformly bounded linear maps. However, this is still not enough to conclude that B and B are close. In fact we also check that B and B are close modulo V (q˜ ). <i 2 (This part of the argument uses the assumptions on q, q , q , q , etc.) Then we apply the elementary Lemma 11.5 below with E = E (q ),L = H(q˜ ),F = E (q˜ ),V = ij,bdd 3 ij,bdd 2 2 V (q˜ ) to get <i (11.6) B − B → 0as ξ → 0. INVARIANT AND STATIONARY MEASURES 223 The final part of the proof of Proposition 11.4 consists of deducing (11.5)from(11.6)and the fact that B and B are uniformly bounded (Proposition 10.2). Lemma 11.5. — Suppose L is a finite-dimensional normed vector space, F and V are subspaces of L, with F∩ V ={0}.Let S denote the unit sphere in L,and let hd(·,·) denote the Hausdorff distance induced by the norm on L. Suppose E is another finite-dimensional normed vector space, and B : E → L and B : E → L are two linear maps each of norm at most C.Let π denote the projection L → L/V. Suppose ξ> 0 is such that (i) π ◦ B − π ◦ B ≤ ξ . V V (ii) hd(B(E) ∩ S, F ∩ S) ≤ ξ . (iii) hd(B (E) ∩ S, F ∩ S) ≤ ξ . Then, B − B ≤ ξ ,where ξ depends on ξ , C and the angle between V and F.Furthermore, ξ → 0 as ξ → 0 (and the other parameters remain fixed). In the course of the proof, we will prove the following lemma, which will be used in Section 12: Lemma 11.6. —For every δ> 0 there exists a compact set K ⊂ X with ν(K )> 1 − δ 0 0 −1 + + such that the following holds: Suppose x, x , y, y ∈ π (K ),y ∈ W [x],y ∈ W [x ] and x ∈ − X X 0 0 W [x]. Suppose further that d (x, y) ≤ 1/100,d (y, y ) ≤ 1/100, and that there exists s > 0 X X 0 0 such that for all |τ|≤ s, d (g x, g x ) ≤ 1/100 and d (g y, g y ) ≤ 1/100.Furthermore,suppose τ τ τ τ −1 X 0 <α < 1 and that 0 < t <α sis such that d (g y, g y )< 1/100,g y ∈ K and g y ∈ K . 0 t t t 0 t 0 Then, for all ij ∈ , (11.7) τˆ (y, t)−ˆτ y , t ≤ C, ij ij where C depends only on δ, α and the Lyapunov spectrum. 11.1 . Proof of Lemma 11.2.— Let θ > 0and δ> 0 be small constants to be cho- sen later. Let K ⊂ Xand C > 0be such that ν(K)> 1 − δ,for x ∈ K the Lemma 6.9(b) −1 holds with c (x)> C ,and forall x ∈ K, all v ∈ E (x) and all > 0, for at least 1 [ij],bdd (1 − θ ) fraction of y ∈ F [x,], 1 ij (11.8) R(x, y)v < Cv. ∗ ∗ 2 1/2 By Lemma 9.5 there exists a subset K ⊂ Kwith ν(K ) ≥ (1 − 2κ δ ) such that for ∗ 1/2 ∗ ∗ x ∈ K ,(9.12) holds with θ = δ . Furthermore, we may ensure that for x ∈ K ,K ∩ F [x] is relatively open in F [x]. (Indeed, suppose z ∈ F [x] is near x ∈ K . Then, there ij ij ij exists such that for > , F [x,]= F [z,] and thus (9.12) holds for z.For < , 0 0 ij ij 0 (9.12) holds for z sufficiently close to x by continuity.) Let ∗ 2 1/2 = x ∈ X : lim t ∈[0, T]: g x ∈ K ≥ 1 − 2κ δ . −t T→∞ 224 ALEX ESKIN, MARYAM MIRZAKHANI FIG. 5. — Proof of Lemma 11.2 Then ν() = 1. From its definition, is invariant under g . Since K ∩ F [x] is relatively t ij open in F [x], is saturated by the leaves of F . This implies that is (locally) invariant ij ij under U .Now,let K = x ∈ : ∗ 2 1/2 for all T > N, t ∈[0, T]: g x ∈ K ≥ 1 − 4κ δ T . −t 2 1/2 (We may assume that 4κ δ 1.) We have K = . Suppose x ∈ K , x ∈ B [x ]∩ K ,so d (x , x )< 1/100. For k = 0, 1, let Q ⊂ 0 N 1 0 0 N 0 1 k E [x ] be such that ij k X + hd Q , U [x ] ≤ 1/100, k k and the vector −1 v = j U (Q ) k 1−k satisfies v ≤ 1/100. −1 We claim that v ∈ H(x ).Indeed,wemay write U (Q ) = (M ,v ). Also k k 1−k 1−k 1−k 1−k −1 + we may write U (U [x ]) = (M ,v ). Then, Q is parametrized (from x )byapair 1−k 1−k k x k k (M ,w ) where w ∈ W (x ),and k k k M = (I + M ) ◦ I + M − I 1−k k k (This parametrization is not necessarily adapted to Z(x ).) Since M and M are both k 1−k in H ,M ∈ H (x ).Thus, v = S (j(M ,w )) ∈ H(x ). ++ ++ k k x k k k k k For C (N) sufficiently large, we can find C (N)< t < 2C (N) such that x ≡ 1 1 1 ij ij ij ∗ ∗ g x ∈ K , x ≡ g x ∈ K , see Figure 5. By Lemma 9.2, x ∈ B [x ].Let v = g v , −t 0 −t 1 0 −t k 1 1 0 k INVARIANT AND STATIONARY MEASURES 225 + + FIG. 6. — Proof of Lemma 11.2. In (b), the subspaces U [ y ] and U [ y ] stay close since x ∈ E (x ),and also for k ∈{0, 1}, ij 0 1 1 0 the subspaces R(x , y )Q and U [ y ] stay close since Q ∈ E (x ) ij,bdd k k k k k k ij Q = g Q , see Figure 6. By choosing C (N) sufficiently large (depending on N), we can k 1 −t ensure that X + + −3 X + −3 0 0 hd U x , U x ≤ C , hd Q , U x ≤ C . k 1−k k k x x k k By Lemma 6.9, since x ∈ K, −1 + −2 −1 −2 (11.9) j U U x ≤ C , j U Q ≤ C . 1−k k x x k k ij Let > 0 be arbitrary, and let be such that g F [x, ]= F [x,]. Then, for k = 0, 1, ij ij since x ∈ K , 1/2 y ∈ F x , : y ∈ K ≥ 1 − δ F x , . ij ij k k k −1 + + + Since U [x ]∈ E [x ],wehave U [x ]∈ E [x ],and thus j(U (U [x ])) ∈ E (x ). 1 ij 0 ij [ij],bdd 1 0 1 0 Since x ∈ K, we have by (11.8), for at least (1 − θ )-fraction of y ∈ F [x , ], 1 ij 0 0 0 −1 + −1 + −1 (11.10) R x , y j U U x ≤ C j U U x ≤ C , 0 0 1 1 x x 0 0 1/2 where we have used (11.9) for the last estimate. Let θ = 2θ + 2δ . Then, for at least 1 − θ /2fraction of y ∈ F [x , ], y ∈ Kand (11.10) holds. Therefore, by Lemma 6.9, ij 0 0 0 for at least (1 − θ /2)-fraction of y ∈ F [x , ],for asuitable y ∈ F [x , ], ij ij 0 0 1 1 + + (11.11) hd U y , U y ≤ 1/100. 0 1 −1 Also, since Q ∈ E [x ], Q ∈ E [x ],and thus j(U (Q )) ∈ E (x ).Hence,by(11.8), k ij k ij [ij],bdd k k k k for at least (1 − θ)-fraction of y ∈ F [x , ], ij k k −1 −1 −1 (11.12) R x , y j U Q ≤ C j U Q ≤ C , k k x k x k k k where we used (11.9) for the last estimate. Then, for at least (1 − θ /2)-fraction of y ∈ F [x , ], y ∈ Kand (11.12) holds. Therefore, by Lemma 6.9,for at least (1 − θ /2)- ij k k fraction of y ∈ F [x , ], ij k k 0 + hd U y , R x , y Q ≤ 1/100. y k k k k k 226 ALEX ESKIN, MARYAM MIRZAKHANI Therefore, by (11.11), for at least (1 − θ )-fraction of y ∈ F [x , ],for asuitable y ∈ ij k k 1−k F [x , ], ij 1−k 0 + (11.13) hd U y , R x , y Q ≤ 1/50. y k 1−k 1−k 1−k Let −1 w = j U R x , y Q = R x , y v . k 1−k 1−k 1−k k k k Then, assuming y ∈ Kand (11.13) holds, by Lemma 6.9, w ≤ C. ij Let y = g y ,and let k t w = R y , y w = R(x , y )v . k k k k k k k Then, for at least (1 − θ )-fraction of y ∈ F [x ,], R(x , y )v ≤ C (N). This implies, k ij k k k k 2 by Proposition 10.3,that v ∈ E (x ). (By making θ > 0and δ> 0 sufficiently small, k [ij],bdd k 1 we can make sure that θ <θ where θ> 0is as in Proposition 10.3.) −1 −1 Thus, for all Q ∈ E [x ] such that j(U (Q )) ≤ 1/100, we have j(U (Q )) ∈ k ij k k k x x 1−k 1−k −1 −1 E (x ). Since both U and j are analytic, this implies that j(U (Q )) ∈ [ij],bdd 1−k k x x 1−k 1−k E (x ) for all Q ∈ E [x ].Thus, for k = 0, 1, E [x ]⊂ E [x ]. This implies that [ij],bdd 1−k k ij k ij k ij 1−k E [x ]= E [x ]. ij 0 ij 1 11.2 . Proof of Proposition 11.4.— Let O ⊂ X be an open set contained in the fundamental domain, and let x → u ∈ U (x) be a function which is constant on each set of the form U [x]∩ O.Let T : O → X be the map which takes x → u x. u x Lemma 11.7. — Suppose E ⊂ O.Then ν(T (E)) = ν(E). Proof. — Without loss of generality, we may assume that T (O) ∩ O =∅.For each x ∈ O,let U[x] be a finite piece of U [x] which contains both U[x]∩O and T (U[x]∩O). We may assume that U[x] is the same for all x ∈ U[x]∩ O.Let U be the σ -algebra of functions which are constant along each U[x]. Then, for any measurable φ : X → R, φdν = E(φ | U)dν X X ˜ ˜ Now suppose φ is supported on O.Wehave E(φ ◦ T | U) = E(φ | U) since the con- + + ditional measures along U are Haar, and T restricted to O ∩ U [x] is a translation. Thus φ ◦ T dν = E(φ ◦ T | U)dν = E(φ | U)dν = φdν. u u X X X X We also recall the following standard fact: INVARIANT AND STATIONARY MEASURES 227 Lemma 11.8. — Suppose : X → X preserves ν , and also for almost all x, C [(x)]∩ ij B [(x)]∩ (B [x]) = (C [x]) ∩ B [(x)]∩ (B [x]).Then, 0 0 ij 0 0 f (x) ∝ f (x), ij ∗ ij in the sense that the restriction of both measures to the set B [(x)]∩ (B [x]) where both make 0 0 sense is the same up to normalization. Proof.—See[EL, Lemma 4.2(iv)]. Lemma 11.9. — We have (on the set where both are defined): f (g T g x) ∝ (g T g ) f (x). ij t u −s t u −s ∗ ij Proof. — This follows immediately from Lemma 11.7 and 11.8. + + The maps φ .— We have the map φ : W (x) → H (x) × W (x) given by x x ++ −1 + (11.14) φ (z) = U U [z] . −1 (Here U is defined using the transversal Z(x).) + + Suppose Z(x) is an admissible transversal to U (x). Since f (x) is Haar along U , ij we can recover f (x) from its restriction to Z(x). More precisely, the following holds: ij + + Let π : H (x) × W (x) → W (x) be projection onto the second factor. Then, 2 ++ for z ∈ Z(x), π (φ( z)) = z. Now, suppose Z is another transversal to U (x). Then, (f | )(x) = π ◦ S ◦ φ (f | ). ij Z 2 ij Z(x) The measures f (x).— Let ij f (x) = (j ◦ φ ) f (x). ij x ∗ ij Then, f (x) is a measure on H(x). ij Lemma 11.10. —For y ∈ F [x], we have (on the set where both are defined), ij f (y) ∝ R(x, y) f (x). ij ∗ ij ij ij Proof. — Suppose t > 0is such that x = g x and y = g y satisfy y ∈ B[x ]. Working −t −t ij in the universal cover, let Z[x]={z : z − x ∈ Z(x)}.Let Z[x ]= g Z[x],and let Z[ y ]= −t ij g Z[ y].For z ∈ Z[x ] near x ,let u be such that u z ∈ Z[ y ]. We extend the function −t z z z → u to be locally constant along U in a neighborhood of Z[x ]. Then, let ij ij = g ◦ T ◦ g . t u −t 228 ALEX ESKIN, MARYAM MIRZAKHANI Note that takes Z[x] into Z[ y], and by Lemma 11.9, (11.15) f (x) ∝ f (y). ∗ ij ij By the definition of u in Section 6,for z ∈ Z[x], −1 + −1 + R(x, y) ◦ j ◦ U U [z]= j ◦ U U (z) . x y Hence, by (11.14), (11.16) R(x, y) ◦ j ◦ φ (z) = (j ◦ φ ◦ )(z), x y where φ is relative to the transversal Z(y) and φ is relative to the transversal Z(x).(Here y x + + we have used the fact that (U [z]) = U [(z)] which follows from the equivariance of U . Also, in (11.16), R(x, y) is as in Section 9.3.) Now the lemma follows from (11.15) and (11.16). + − + Let P (x, y) and P (x, y) be as in Section 4.2.The maps P (x, y) : Lie(G )(x) → ∗ ++ Lie(G )(y) (where we use the notation (6.11)) are an equivariant measurable flat ++ W -connection on the bundle Lie(G ) satisfying (4.5). Then, by Proposition 4.12(a), ++ + + + (11.17)P (x, y) Lie U (x) = Lie U (y). + − + The maps P (x, y) and P (x, y).— In view of (11.17), the maps P (x, y) naturally ˜ ˜ ˜ induce a linear map (which we denote by P (x, y))from H(x) to H(y),sothatfor (M,v) ∈ H (x), ++ + + + −1 + P (x, y) ◦ j(M,v) = j P (x, y) ◦ M ◦ P (x, y) , P (x, y)v . + Z(y) + + Let P (x, y) = S ◦ P (x, y). Then the maps P (x, y) : H(x) → H(y) are an equivari- ant measurable flat W -connection on the bundle H satisfying (4.5). Then, by Proposi- tion 4.12(a), we have (11.18) P (x, y)E (x) = E (y). ij,bdd ij,bdd − − For y ∈ W [x],wehaveamap P (x, y) with analogous properties. Z Z The maps P (x, y) and P (x, y).— We also need to define a map between H(x) + − and H(y) even if x and y arenot on thesameleafof W or W .For every v ∈ V (x) ≡ i i V (H )(x),and i ∈ (where is the Lyapunov spectrum) we can write 1 1 v = v + v v ∈ V H (y), v ∈ V H (y). i i i i i i i j=i ! 1 1 1 Let P (x, y) : H (x) → H (y) be the linear map whose restriction to V (H )(x) sends ! 1 1 v to v . By definition, P (x, y) sends V (H )(x) to V (H )(y), but it is not clear that i i i i INVARIANT AND STATIONARY MEASURES 229 ! + + P (x, y) Lie(U )(x) = Lie(U )(y). To correct this, given a Lyapunov-adapted transver- sal Z(x),notethat(for y near x), ! −1 + Lie(G )(x) = P (x, y) Lie U (y) ⊕ Z(x). ++ Then, given v ∈ Lie(U )(x) ⊂ Lie(G )(x),wecan decompose ++ ! −1 + (11.19) v = v + v ,v ∈ P (x, y) Lie U (y), v ∈ Z(x). Define M(x; y) : Lie(U )(x) → Lie(G )(x) by ++ (11.20)Mv =−v . Then, since Z(x) is Lyapunov adapted, M(x; y) : Lie(U )(x) → Lie(G )(x) is the linear ++ map such that + ! −1 + (11.21) I + M(x; y) Lie U (x) = P (x, y) Lie U (y), and M(x; y)V (Lie(U ))(x) ⊂ Z (x),where Z (x) = Z(x) ∩ V (Lie(G ))(x) is as in Sec- i i i i ++ Z(x) tion 6. Then, let P (x, y) : H (x) → H (y) be the map taking f ∈ H (x) to ++ ++ ++ −1 Z(x) ! ! −1 P (x, y)f ≡ P (x, y) ◦ f ◦ I + M(x; y) ◦ P (x, y) ∈ H (y). ∗ ++ Then, since M(x; y)V (Lie(U ))(x) ⊂ V (Lie(G ))(x) we have for a.e. x, y, i i ++ Z(x) P (x, y)V (H )(x) = V (H )(y). i ++ i ++ Z(x) Z(x) + + Then P gives a map P (x, y) : H (x) × W (x) → H (y) × W (y) given by ++ ++ Z(x) Z(x) ! P (x, y)(f ,v) = P (x, y)f , P (x, y)v . Z(x) Therefore (after possibly composing with a change in transversal map S) P (x, y) in- Z(x) duces a map we will call P (x, y) between H(x) and H(y). This map satisfies Z(x) (11.22) P (x, y)V (H)(x) = V (H)(y), i i and has the equivariance property g Z(x) Z(x) −t P (g x, g y) = g ◦ P (x, y) ◦ g . −t −t −t t Lemma 11.11. —For y ∈ W [x], and any choice of Z(x), we have Z(x) + (11.23) P (x, y) = P (x, y). 230 ALEX ESKIN, MARYAM MIRZAKHANI + ! + Proof. — Suppose y ∈ W [x]. Then by Lemma 4.1,P (x, y) = P (x, y),thus ! −1 + + −1 + + P (x, y) Lie U (y) = P (x, y) Lie U (x) = Lie U (x) ∗ ∗ where for the last equality we used Proposition 4.12(a). Hence, M(x; y) = 0and (11.23) follows. Lemma 11.12. — For any δ> 0 there exists a compact subset K ⊂ X with ν(K)> −1 1 − δ/2 such that the following holds: Suppose x and y ∈ π (K),and s > 0 are such that for all |t| < s, d (g x, g y)< 1/100. Then, there exists α> 0 depending only on the Lyapunov spectrum, t t and C = C(δ) such that for all i, GM 1 1 −αs d P (x, y)V H (x), V H (y) ≤ C(δ)e . Y i i Proof. — There exists a compact subset K ⊂ X such that the functions x → 1 0 V (H )(x) are uniformly continuous. (Here we are using the Gauss-Manin connection to 1 −1 identify H (x) with H (y) for y near x.) Then, there exists σ> 0such that if x ∈ π (K ), 1 1 −1 X ( − y ∈ π (K ) and d (x, y)<σ then D x, y)< 1and D (x, y)< 1. (See Section 4.5 for the definition of D (·,·).) We also may assume that there exists a constant C (δ) such that C(x)< C (δ) for all x ∈ K ,where C(·) is as in Lemma 4.7. Then there exists a compact 0 1 subset K ⊂ Xwith ν(K)> 1 − δ,and t > 0such that for x ∈ K, for t > t ,for (1 − δ)- 0 0 fraction of t ∈[0, s], g x ∈ K , g x ∈ K also for at least half the fraction of t ∈[0, s], g x t 1 −t 1 t and g x belong to K where K is as in Lemma 3.5. −t thick thick −1 −1 Suppose x ∈ π (K),and y ∈ π (K). Then, by Lemma 3.5, there exists α > 0 depending only on the Lyapunov spectrum such that there exists t ∈[α s, s] with g x ∈ K , 1 t 1 X − g y ∈ K and d (g x, g y)<σ . Then, D (g x, g y)< 1. Then, by Lemma 4.7, applied to t 1 t t t t the points g x, g y,weget t t 1 1 −αt −αα s d V H (x), V H (y) ≤ C(δ)e = C(δ)e . Y ≥i ≥i Similarly, there exists t ∈[α s, s] with g x ∈ K and g y ∈ K . Then, we get 1 −t 1 −t 1 1 1 −αt −αα s d V H (x), V H (y) ≤ C(δ)e = C(δ)e . Y ≤i ≤i The lemma follows. For every δ> 0and every0 <α < 1 there exist compact sets K ⊂ K ⊂ Xwith ν(K )> 1 − δ such that the following hold: ! + 1 (K 1) The functions U (x), V (H )(x) and more generally, V (H )(x) for all i, i i big are uniformly continuous on K . ! ! (K 2) The functions Z(x) are uniformly continuous on K . ! ! (K 3) The functions E (x) are uniformly continuous on K . ij,bdd INVARIANT AND STATIONARY MEASURES 231 ! ! (K 4) The functions f (x) and f (x) are uniformly continuous on K (in the weak- ij ij * convergence topology). (K 5) There exists t > 0and < 0.25α min |λ − λ | such that for t > t , x ∈ 0 i=j i j 0 ! 1 K ,all i,and any v ∈ V (H )(x), (λ − )t (λ + )t i i e v≤(g ) v≤ e v t ∗ ! ! (K 6) The function C (·) of Proposition 10.2 is uniformly bounded on K . ! ! (K 7) E (x) and V (x) are transverse for x ∈ K . ij,bdd <i ! ! ! (K 8) K ⊂ K where K is as in Lemma 3.5(c). Also K ⊂ Kwhere Kis as thick thick in Lemma 11.12. ! ! (K 9) There exists c (δ) > 0with c (δ) → 0as δ → 0such that for all x ∈ K , 0 0 d (x,∂B [x])> c (δ) where B [x] is as in Section 3. 0 0 0 ! ! (K 10) There exists a constant C (δ) such that for all x ∈ K and all v ∈ H (x), 4 big −1 C (δ) v≤v ≤ C (δ)v. 4 Y 4 (K 11) There exists a constant C = C (δ) < ∞ such that for x ∈ K and all T > 1 1 0 C (δ) and all ij we have ij t ∈[C , T]: g x ∈ K ≥ 0.99(T − C ). 1 1 −t −1 + + Lemma 11.13. — Suppose x, x , y, y ∈ π (K ),y ∈ W [x],y ∈ W [x ] and x ∈ − X X 0 0 W [x]. Suppose further that d (x, y)< 1/100,d (y, y )< 1/100, and that there exists s > 0 X X 0 0 such that for all |t|≤ s, d (g x, g x )< 1/100 and d (g y, g y )< 1/100.Then, t t t t (a) There exists α depending only on the Lyapunov spectrum, such that ! GM −α s (11.24) P y, y P y , y − I = O e . (b) There exists α depending only on the Lyapunov spectrum such that + − GM + −α s (11.25) P x , y ◦ P x, x − P y, y ◦ P (x, y) = O e . Proof. — Note that part (a) follows immediately from Lemma 11.12, since we are assuming that d (g y, g y ) ≤ 1/100 for all t with |t|≤ s. t t To prove (b) we abuse notation by identifying H at all four points x, y, x , y using the Gauss-Manin connection. We write V (x) for V (H )(x). Since i i + − + −1 P x , y ◦ P x, x ◦ P (x, y) V (y) = V y , i i and by Lemma 11.12, −α s d V (y), V y = O e , Y i i it is enough to check that for v ∈ V (y), + − + −1 −α s (11.26) P x , y ◦ P x, x ◦ P (x, y) − I v + V (y) = O e v . <i Y But (11.26) follows from the following: 232 ALEX ESKIN, MARYAM MIRZAKHANI + −1 • P (x, y) is the identity map on V (y)/V (y) = V (x)/V (x). ≤i <i ≤i <i − − −α s • P (x, x )V (x) = V (x ) and by Lemma 11.12, P (x, x ) − I = O(e ). ≤i ≤i Y • P (x , y ) is the identity on V (x )/V (x ) = V (y )/V (y ). ≤i <i ≤i <i −α s • d (V (y), V (y )) = O(e ). Y ≤i ≤i This completes the proof of (11.26)and thus (11.25). Lemma 11.14. −1 ! X + X (a) Suppose x, x, ˜ y, y˜ all belong to π (K ),d (x, y)< 1/100, y˜ ∈ W [˜ x],d (x, x˜) ≤ ξ and d (y, y˜) ≤ ξ . Then + Z(x) Z(y) Z(x) P (x˜, y˜) ◦ P (x, x˜) − P (y, y˜) ◦ P (x, y)≤ ξ , where ξ → 0 as ξ → 0. −1 + + − (b) Suppose x, x , y, y ∈ π (K ),y ∈ W [x],y ∈ W [x ] and x ∈ W [x]. Suppose X X 0 0 further that d (x, y) ≤ 1/100,d (y, y ) ≤ 1/100, and that there exists s > 0 such X X 0 0 that for all |t|≤ s, d (g x, g x ) ≤ 1/100 and d (g y, g y ) ≤ 1/100.Furthermore, t t t t −1 suppose 0 <α < 1 and that 0 <τ <α sis such that d (g y, g y )< 1/100 and 0 τ τ g y ∈ K .Then, + − g Z(g y) + −αs −τ τ P x , y ◦ P x, x − P y, y ◦ P (x, y) = O e , where α depends only on the Lyapunov spectrum and α . + Z(x) + Proof of (a). — Since y ∈ W [x], by Lemma 11.11 we have P (x, y) = P (x, y). Z(x) ! ! Since P (x, y) depends continuously on x ∈ K and y ∈ K , part (a) follows from a compactness agreement. Proof of (b).—We first claimthat g Z(g y) GM −α s −τ τ (11.27) P y, y P y , y − I = O e , ∗ Y where α depends only on α and the Lyapunov spectrum. ! −1 ! −1 ! By (K 1) there exists > 0such that for x ∈ π (K ), y ∈ π (K ) with 0 1 1 X X + + ! 0 0 d (x , y )< , hd (U [x ], U [ y ])< 0.01. By (K 10) there exists t > s/2with 1 1 0 1 1 −1 ! −1 ! X g y ∈ π (K ), g y ∈ π (K ) and d (g y, g y )< 1/100. Therefore, by Lemma 3.5(c) and t t t t Proposition 3.4 we have 0 + + −α s hd U [ y], U y = O e , where α depends only on the Lyapunov spectrum. Therefore, we get −1 GM + + −α s d P y, y Lie U (y), Lie U y = O e . ∗ INVARIANT AND STATIONARY MEASURES 233 Then, by (11.24), −1 ! + + −α s (11.28) d P y, y Lie U y , Lie U (y) = O e where α depends only on the Lyapunov spectrum. −1 ! ! ! Since g y ∈ π (K ),by (K 1) and (K 2), d Z(g y) ∩ V Lie(G ) (g y), Lie U (g y) ∩ V Lie(G ) (g y) Y τ i ++ τ τ i ++ τ ≥ c K . By (K 5) (i.e. the multiplicative ergodic theorem), the restriction of g to V (Lie(G )) is τ i ++ λ τ τ e h ,where h = O(e ). Therefore, τ τ + − s (11.29) d g Z(g y) ∩ V Lie(G ) (y), Lie U (y) ∩ V Lie(G ) (y) ≥ ce Y −τ τ i ++ i ++ We may assume (since α> 0 in the choice of K is arbitrary), that <α /2. Then, it follows from (11.28), (11.29), (11.19)and (11.20)that −α s (11.30) M y; y = O e where M(·;·) is as in (11.21), and α depends only on α and the Lyapunov spectrum. 5 0 Now, (11.27) follows from (11.24)and (11.30). Combining (11.27), and (11.25)weget + − g Z(g y) + −α s −τ τ 6 P x , y ◦ P x, x − P y, y ◦ P (x, y) = O e . Now (b) of Lemma 11.14 follows immediately, see also (K 10). ! − ! X Lemma 11.15. — Suppose q ∈ K and q ∈ W [q]∩ K , are such that d (q , q )< 1 1 1 1 ! ! 1/100. Suppose u ∈ B(q , 1/100),u ∈ B(q , 1/100), with uq ∈ K ,u q ∈ K . We write q = 1 1 2 1 1 g uq for someτ> 0,and let q = g u q (see Figure 4). Suppose d (q , q )< 1/100, and also there τ 1 τ 2 2 1 2 exists α > 0 depending only on the Lyapunov spectrum such that for |t| <α τ,d (g uq , g u q )< 0 0 t 1 t 1/100. In addition, suppose there exist q˜ ∈ X and q˜ ∈ X with σ (q˜ ) ∈ W [σ (q˜ )] such that 2 0 0 2 2 2 X X ! d (q˜ , q )<ξ and d (q˜ , q )<ξ . Suppose further that q ,q , q˜ and q˜ all belong to K . 2 2 2 2 2 2 2 2 Then (assuming in (K 5) is sufficiently small depending on α and the Lyapunov spectrum), τ is sufficiently large and ξ is sufficiently small (both depending only on K ), we have q˜ ∈ W [˜ q ]. Proof. — In this proof, α is a generic constant depending only on α and the Lya- punov spectrum, with its value changing from line to line. By Lemma 11.13(a), −1 ! GM −ατ P uq , u q ◦ P uq , u q − I = O e . 1 1 1 1 Y 234 ALEX ESKIN, MARYAM MIRZAKHANI By Lemma 11.13(b), GM + + − −ατ P uq , u q ◦ P (q , uq ) − P q , u q ◦ P q , q = O e . 1 1 1 1 1 1 1 1 Thus, ! + + − −ατ (11.31) P uq , u q ◦ P (q , uq ) − P q , u q ◦ P q , q = O e . 1 1 1 1 1 1 1 1 Write u q = (σ (u q ), F ), uq = (σ (uq ), F) where F and F areasisinSection 4.6. 0 1 0 1 1 1 By Proposition 4.12 (see also (4.12)and (4.13)), + − + F = P q , u q ◦ P q , q ◦ P (uq , q )F. 1 1 1 1 1 1 Therefore, by (11.31), ! −ατ d F , P uq , u q F = O e , Y 1 where the distance d (·,·) between flags is as in Section 4.6. We now claim that ! −ατ (11.32) d g F , g P uq , u q F = O e . Y τ τ 1 Indeed to prove (11.32) it is enough to show that for each i, ! −ατ (11.33) d g F , g P uq , u q F = O e . Y τ τ 1 i i 1 But F ⊂ V (H )(u q ), F ⊂ V (H )(uq ),and i big i i big 1 i 1 P uq , u q V (H )(uq ) = V (H ) u q . 1 i big 1 i big 1 1 Thus, we have F ⊂ V (H ) u q , P uq , u q F ⊂ V (H ) u q i big 1 i i big i 1 1 1 λ τ The geodesic flow g restricted to V (H )(u q ) is of the form e h ,where h = τ i big τ τ Y τ ! O(e ).Thus, (11.33) and hence (11.32) follows. The equivariance property of P then implies that ! −ατ (11.34) d g F , P q , q g F = O e . Y τ 2 τ We have since the V are continuous on K and Lemma 4.1, GM ! + GM P q , q˜ ◦ P q , q − P q˜ , q˜ ◦ P (q , q˜ ) → 0, 2 2 2 2 2 2 2 2 as ξ → 0. Combining this with (11.34), we get GM + GM (11.35) d P q , q˜ g F , P q˜ , q˜ ◦ P (q , q˜ )g F → 0, Y τ 2 2 2 τ 2 2 2 as ξ → 0and τ →∞. INVARIANT AND STATIONARY MEASURES 235 Note that q = (σ (q ), g F), q = (σ (q ), g F ).Write q˜ = (σ (q˜ ), F), q˜ = 2 0 2 τ 0 τ 2 0 2 2 2 2 (σ (q˜ ), F ). Then, since d (q , q˜ ) → 0, in view of (4.14), 0 2 2 GM d P (q , q˜ )g F, F ≤ ξ Y 2 2 τ GM d P q , q˜ g F , F ≤ ξ Y τ 2 2 where ξ → 0as ξ → 0. Hence, by (11.35), ˜ ˜ d F , P q˜ , q˜ F →0as ξ → 0and τ →∞. Y 2 This implies that q˜ ∈ W [˜ q ] by (4.12). Z(·) Proof of Lemma 11.6. — Note that, by the construction of P (·,·),for all i ∈ , Z(g y) (11.36) P g y, g y V (H)(g y) = V (H) g y . t t i t i t However, even though for all ij ∈ , E (x) ⊂ V (H)(x),wemay have ij i Z(g y) P g y, g y E (g y) = E g y . t t ij t ij t Suppose v ∈ E (y),and that v is orthogonal to E (y) ⊂ E (y).Let ij i,j−1 ij + − + v = P x , y ◦ P x, x ◦ P (y, x)v. Then, by Proposition 4.12(a), v ∈ E (y ).By (K 1), and the fact that ij + − + P x , y ◦ P x, x ◦ P (y, x)E (y) = E y , i,j−1 i,j−1 we have −1 (11.37)C v≤ v + E y ≤ C v, i,j−1 1 where C depends only on K . By Lemma 11.14(b), 1 0 g Z(g y) −α t −t t P y, y v − v = O e v , where α depends only on α and the Lyapunov spectrum. By (11.36), g Z(g y) −t t P y, y v ∈ V (H) y . Then by the multiplicative ergodic theorem (see also (K 5)), Z(g y) −(α − )t (11.38) P g y, g y (g v) − g v = O e g v . t t t t t Since v is arbitrary, this implies that for all ij ∈ , Z(g y) −α t t 1 (11.39) d P g y, g y E (g y), E g y = O e , t t ij t ij t where α depends only on α and the Lyapunov spectrum. 1 0 236 ALEX ESKIN, MARYAM MIRZAKHANI ! ! By (K 1) and (K 2), Z(g y) P g y, g y ≤ C t t where C depends only on K . Therefore, by (11.38)and (11.39), −1 (11.40)C g v + E (g y)≤ g v + E g y ≤ C g v + E (g y), t i,j−1 t t i,j−1 t 2 t i,j−1 t where C depends only on K , α and the Lyapunov spectrum. 2 0 0 Note that g v + E (g y) g v + E (g y ) t i,j−1 t t i,j−1 t τˆ (y, t) = , τˆ y , t = . ij ij v v + E (y ) i,j−1 Now (11.7) follows from (11.37)and (11.40). Proposition 11.16. — Suppose α, ,s, ,t,t ,q,q , τ,q ,q ,q ,q ,u,u ,q ,q , q˜ , 1 3 2 2 1 3 2 + ! q˜ , C, C , ξ are as in Proposition 11.4. Suppose also q˜ ∈ W [˜ q ]. Then (assuming in (K 5)is 1 2 2 2 sufficiently small depending on α and the Lyapunov spectrum), (a) There exists ξ > 0 (depending on ξ , K and C and t) with ξ → 0 as ξ → 0 and t →∞ such that for v ∈ E (q ), [ij],bdd 3 Z(q ) − (11.41) P q , q˜ ◦ R q , q ◦ P q , q v 2 2 3 2 3 + Z(q ) − P q˜ , q˜ ◦ P (q , q˜ ) ◦ R(q , q )v ≤ ξ v. 2 2 2 3 2 (b) There exists ξ > 0 (depending on ξ , K , C and t) with ξ → 0 as ξ → 0 and t →∞ such that d P q˜ , q˜ f (q˜ ), f q˜ ≤ ξ . ∗ 2 ij 2 ij 2 2 Here d (·,·) is any metric which induces the weak-* convergence topology on the domain of common definition of the measures, up to normalization. Proof of (a). — Following the outline given after the statement of Proposition 11.4, the proof will consist of verifying conditions (i), (ii) and (iii) of Lemma 11.5,with E = E (q ),L = H(q˜ ),F = E (q˜ ),V = V (q˜ ),and B and B as the linear maps on the ij,bdd 3 ij,bdd <i 2 2 2 first and second line of (11.41). (We note that B and B are bounded by Proposition 10.2.) We start with (i). Note that by (9.4), we have −1 (11.42) κ τ ≤ t ≤ κτ, where κ depends only on the Lyapunov spectrum. Also, by assumption we have >α τ, where α depends only on the Lyapunov spectrum. 0 INVARIANT AND STATIONARY MEASURES 237 Suppose w ∈ E (q ). We now apply Lemma 11.14(b), with x = q , x = q , y = ij,bdd 1 1 uq , y = u q and τ = τ to get + − g Z(q ) + −τ 2 P u q , uq ◦ P q , q w − P uq , u q ◦ P (q , uq )w 1 1 1 1 1 1 1 1 −ατ = O e w . By Proposition 4.12(a), P (q , q )w ∈ E (q ) ⊂ E(q ). Therefore, by Lemma 9.1, this 1 ij,bdd 1 1 1 can be rewritten as − g Z(q ) −ατ −τ 2 u ◦ P q , q w − P uq , u q ◦ (u) w = O e w . 1 1 ∗ 1 1 Hence, − g Z(q ) −τ 2 (11.43) u ◦ P q , q w = P uq , u q ◦ (u) w + w 1 1 ∗ 1 1 where w ∈ H(u q ) satisfies −ατ −(λ +α− )τ (11.44) w = O e w = O e v , ij where we wrote w = g v for some v ∈ E (q ), and we have used (K 5), (11.42)and the ij 3 −t ij assumption |t − t | < C for the last estimate. We now apply g = g to both sides of (11.43) τ t and take the quotient mod V (q ).Weget <i (11.45) g ◦ u ◦ P q , q w + V q τ 1 <i 1 2 Z(q ) = P q , q ◦ g ◦ (u) w + g w + V q . 2 τ ∗ τ <i 2 2 We may write w = w , w ∈ V (H) u q . k k k Then, g w + V q = g w + V q = g w + V q , τ <i τ <i τ <i 2 k 2 k 2 k k≥i since for k < i, g w ∈ V (q ).By(K 5), for k ≥ i, τ <i k 2 (λ + )τ (λ + )τ −α τ k i 5 g w = O e w = O e w = O e v , k k k using (11.44)(andchoosing sufficiently small depending on α and the Lyapunov spec- trum). Therefore, substituting into (11.45), we get, for v ∈ E (q ), ij,bdd 3 R q , q ◦ P q , q v + V q 3 <i 3 2 3 2 Z(q ) −α τ 2 5 = P q , q ◦ R(q , q )v + O e v + V q . 2 3 2 <i 2 2 238 ALEX ESKIN, MARYAM MIRZAKHANI Z(q ) We now apply P (q , q˜ ) to both sides to get (using (11.22)) 2 2 Z(q ) − (11.46) P q , q˜ ◦ R q , q ◦ P q , q v + V q˜ 3 <i 2 2 3 2 3 2 Z(q ) Z(q ) −α τ 2 5 = P q , q˜ ◦ P q , q ◦ R(q , q )v + O e v + V q˜ . 2 3 2 <i 2 2 2 2 Since q , q˜ , q , q˜ all belong to K , we have by Lemma 11.14(a), 2 2 2 2 Z(q ) Z(q ) + Z(q ) 2 2 P q , q˜ ◦ P q , q − P q˜ , q˜ ◦ P (q , q˜ )≤ ξ , 2 2 2 2 3 2 2 2 2 where ξ → 0as ξ → 0. Therefore, substituting into (11.46), we get Z(q ) − P q , q˜ ◦ R q , q ◦ P q , q v + V q˜ 3 <i 2 2 3 2 3 2 + Z(q ) −α τ 2 5 = P q˜ , q˜ ◦ P (q , q˜ ) ◦ R(q , q )v + O e v + O(ξ v) 2 2 2 3 2 3 + V q˜ . <i This completes the verification of (i) of Lemma 11.5. We now verify (ii) of Lemma 11.5.For v ∈ E (q ),wehave R(q , q )v ∈ ij,bdd 3 3 2 E (q ),and then ij,bdd 2 + Z(q ) P q˜ , q˜ ◦ P (q , q˜ ) ◦ R(q , q )v 2 2 2 3 2 + Z(q ) ∈ P q˜ , q˜ ◦ P (q , q˜ )E (q ). 2 2 2 ij,bdd 2 ! ! X By (K 2) and (K 3), since d (q , q˜ )<ξ , 2 2 Z(q ) d P (q , q˜ )E (q ), E (q˜ ) <ξ , Y 2 2 ij,bdd 2 ij,bdd 2 0 where ξ → 0as ξ → 0. Then, using (11.18), + Z(q ) d P q˜ , q˜ ◦ P (q , q˜ )E (q ), E q˜ <ξ , Y 2 2 2 ij,bdd 2 ij,bdd 1 2 2 where ξ → 0as ξ → 0. This completes the verification of condition (ii) of Lemma 11.5. − − Also, by (11.18) (applied to P ), we have P (q , q )v ∈ E (q ). Then, R(q , q ) ◦ 3 ij,bdd 3 3 3 2 P (q , q )v ∈ E (q ),and 3 ij,bdd 3 2 Z(q ) − Z(q ) 2 2 P q , q˜ ◦ R q , q ◦ P q , q v ∈ P q , q˜ E q . 3 ij,bdd 2 2 3 2 3 2 2 2 ! ! By (K 2) and (K 3), Z(q ) d P q , q˜ E q , E q˜ <ξ , Y ij,bdd ij,bdd 2 2 2 2 2 where ξ → 0as ξ → 0. This completes the verification of condition (iii) of Lemma 11.5. Now (11.41) for arbitrary v ∈ E (q ) follows from Lemma 11.5. The gen- ij,bdd 3 eral case of (11.41) (i.e. for an arbitrary v ∈ E (q )) follows since E (q ) = [kr],bdd 3 [kr],bdd 3 E (q ) and all the maps on the left-hand-side of (11.41) are linear. ij,bdd 3 ij∈[kr] INVARIANT AND STATIONARY MEASURES 239 Proof of (b).—By (K 4), d P q , q f (q ), f q ≤ ξ , ∗ 3 ij 3 ij 1 3 3 where ξ → 0as and t →∞.Inviewofinviewofcondition (K 6), the assumption |t − t | < C and Proposition 10.2,that R(q , q ) is a linear map with norm bounded 3 2 depending only on K and C. It then follows from (a) that R(q , q ) is also a linear 3 2 ! ! map whose norm is bounded depending only on K and C. Furthermore, by (K 9) and Lemma 3.5 there exists a constant C (δ) such that if (11.47)C > t − t > C (δ), ij ij ij ij then if we write q = g ug q ,then g ug B [q ]∩ C [q ]⊃ B [q ]∩ C [q ]. Then, by 2 t 3 t 0 3 ij 3 0 2 ij 2 −t −t Lemma 11.10, f (q ) ∝ R(q , q ) f (q ) and f q ∝ R q , q f q . ij 2 3 2 ∗ ij 3 ij ij 2 3 2 3 In view of (K 11), we can assume that (11.47) holds: otherwise we can replace q and q ij ij ! ! by g q ∈ K and g q ∈ K where C (δ) < s < 2C (δ). (Without loss of generality we −s 3 −s 2 2 may assume that C > 2C (δ).) Hence, we have (11.48) d R q , q ◦ P q , q f (q ), f q ≤ ξ , ∗ 3 ij 3 ij 2 3 2 3 2 ! ! ! where ξ → 0as t →∞.Thus, by (K 1), (K 2), (K 3), Z(q ) d P q , q˜ f q , f q˜ ≤ ξ , ∗ ij ij 3 2 2 2 2 where ξ → 0as ξ → 0and t →∞.Hence, Z(q ) − (11.49) d P q , q˜ ◦ R q , q ◦ P q , q f (q ), f q˜ ≤ ξ , ∗ 3 ij 3 ij 4 2 2 3 2 3 2 where ξ → 0as ξ → 0and t →∞. Also, in view of (11.48), and since P (q˜ , q˜ ) is a 4 2 linear map whose norm is bounded depending only on K , + Z(q ) + (11.50) d P q˜ , q˜ ◦ P (q , q˜ ) ◦ R(q , q ) f (q ), P q˜ , q˜ f (q˜ )) ≤ ξ , ∗ 2 2 2 3 2 ij 3 2 ij 2 5 2 2 ∗ ∗ where ξ → 0as ξ → 0and t →∞. Now part (b) follows from (11.49), (11.50), and (11.41). Proof of Proposition 11.4. — Note that (11.4) follows from Lemma 11.15. We assume this from now on. Without loss of generality, and to simplify the notation, we may assume that Z(q˜ ) = P (q˜ , q˜ )Z(q˜ ). (Otherwise, we can further compose with a reparametrization 2 2 2 2 map at q˜ which will not change the result.) We have f (q˜ ) = (j ◦ φ ) f (q˜ ) ij 2 q˜ ∗ ij 2 2 240 ALEX ESKIN, MARYAM MIRZAKHANI and f q˜ = (j ◦ φ ) f q˜ ij q˜ ∗ ij 2 2 + + + As in Section 6,let P : H (q˜ ) × W (q˜ ) → H (q˜ ) × W (q˜ ) be given by ++ 2 2 ++ ∗ 2 2 −1 + + + + (11.51)P (M,v) = P q˜ , q˜ ◦ M ◦ P q˜ , q˜ , P q˜ , q˜ v . 2 2 2 ∗ 2 2 2 Then, + + (11.52) P q˜ , q˜ ◦ j(M,v) = j P (M,v) 2 ∗ We write A ≈ Bif d(A, B) → 0as ξ → 0and t →∞. Then, we have, by Proposi- ξ,t tion 11.16, (j ◦ φ ) f q˜ = f q˜ ≈ P q˜ , q˜ f (q˜ ) q˜ ∗ ij ij ξ,t 2 ij 2 2 2 2 2 ∗ = P q˜ , q˜ ◦ j ◦ φ f (q˜ ) 2 q˜ ij 2 2 2 By (11.52), (j ◦ φ ) f q˜ ≈ j ◦ P ◦ φ f (q˜ ). q˜ ∗ ij ξ,t q˜ ij 2 2 ∗ 2 ∗ Therefore, (φ ) f q˜ ≈ P ◦ φ f (q˜ ). q˜ ∗ ij ξ,t q˜ ij 2 2 ∗ 2 2 ∗ + + Let π : H (x)× W (x) → W (x) be projection onto the second factor. Then, applying 2 ++ π to both sides, we get (11.53) (π ◦ φ ) f q˜ ≈ π ◦ P ◦ φ f (q˜ ). 2 q˜ ∗ ij ξ,t 2 q˜ ij 2 2 ∗ 2 For z ∈ Z(q˜ ), π (φ (z)) = z, and thus in view of (11.51), 2 2 q˜ + + (11.54) π ◦ P ◦ φ (z) ≈ P q˜ , q˜ z. 2 q˜ ξ,t 2 ∗ 2 By assumption, we have Z(q˜ ) = P (q˜ , q˜ )Z(q˜ ). Then, similarly, for z ∈ Z(q˜ ) = 2 2 2 2 2 P (q˜ , q˜ )Z(q˜ ), 2 2 (11.55) (π ◦ φ )(z) = z. 2 q˜ Since f (x) is Haar along U ,wecan recover f (q˜ ) from its restrictions to Z(q˜ ) and f (q˜ ) ij ij 2 2 ij from its restriction to Z(q˜ ). It now follows from (11.53), (11.54)and (11.55)that f q˜ ≈ P q˜ , q˜ f (q˜ ). ij ξ,t 2 ij 2 2 2 ∗ INVARIANT AND STATIONARY MEASURES 241 12. The inductive step Proposition 12.1. — Suppose ν is a P-invariant measure on X . Suppose U (x) is a family − + of subgroups of G (x) compatible with ν in the sense of Definition 6.2.Let L [x] and L [x] be as ++ in Section 6.2, and suppose the equivalent conditions of Lemma 6.15 do not hold. Then, there exists a family of subgroups U (x) of G (x) compatible with ν in the sense of Definition 6.2 such that for ++ new + + almost all x, U (x) strictly contains U (x). new The rest of Section 12 will consist of the proof of Proposition 12.1. We assume − + + that L (x),L (x) and U (x) are as in Proposition 12.1, and the equivalent conditions of Lemma 6.15 do not hold. The argument has been outlined in Section 2.3, and we have kept the same notation (in particular, see Figure 1). Let f (x) be the measures on W (x) introduced in Section 11. We think of f as a ij ij function from X to a space of measures (which is metrizable). Let P (x, y) be the map introduced in Section 4.2.Proposition 12.1 will be derived from the following: + + − Proposition 12.2. —Suppose U , L , L are as in Proposition 12.1, and the equivalent conditions of Lemma 6.15 do not hold. Then there exists 0 <δ < 0.1,asubset K ⊂ X with 0 ∗ ν(K )> 1 − δ such that all the functions f ,ij ∈ are uniformly continuous on K ,and C > 1 ∗ 0 ij ∗ −1 )suchthatfor every 0 < < C /100 there exists a subset E ⊂ K with ν(E)> (depending on K ∗ ∗ −1 −1 δ ,suchthatfor every x ∈ π (E) there exists ij ∈ and y ∈ C [x]∩ π (K ) with 0 ij ∗ −1 X + + (12.1)C ≤ hd U [x], U [ y] ≤ C and (on the domain where both are defined) (12.2) f (y) ∝ P (x, y) f (x). ij ∗ ij We now begin the proof of Proposition 12.2. Choice of parameters #1. — Fix θ> 0as in Proposition 10.1 and Proposition 10.2. We then choose δ> 0 sufficiently small; the exact value of δ will we chosen at the end of this section. All subsequent constants will depend on δ. (In particular, δ θ;wewill make this more precise below.) Let > 0 be arbitrary and η> 0 be arbitrary; however we will always assume that and η are sufficiently small depending on δ. We will show that Proposition 12.2 holds with δ = δ/10. Let K ⊂ Xbe any 0 ∗ subset with ν(K )> 1 − δ on which all the functions f are uniformly continuous. It is ∗ 0 ij enough to show that there exists C = C(δ) such that for any > 0 and for an arbitrary compact set K ⊂ Xwith ν(K ) ≥ (1 − 2δ ), there exists x ∈ K ∩ K , ij ∈ and 00 00 0 00 ∗ y ∈ C [x]∩ K satisfying (12.1)and (12.2). Thus, let K ⊂ X be an arbitrary compact set ij ∗ 00 with ν(K )> 1 − 2δ . 00 0 242 ALEX ESKIN, MARYAM MIRZAKHANI We can choose a compact set K ⊂ K ∩ K with ν(K )> 1 − 5δ = 1 − δ/2so 0 00 ∗ 0 0 that Proposition 11.4 holds. In addition, there exists (δ) > 0such that for all x ∈ K , (12.3) d x,∂B [x] > (δ). (Here, d (·,·) is as in Section 3 and by ∂B [x] we mean the boundary of B [x] as a 0 0 subset of W [x].) Let κ> 1be as in Proposition 7.4,and so that (9.4) holds. Without loss of general- ity, assume δ< 0.01. We now choose a subset K ⊂ K ⊂ Xwith ν(K)> 1 − δ such that the following hold: • There exists a number T (δ) such that for any x ∈ Kand anyT > T (δ), 0 0 t ∈[−T/2, T/2]: g x ∈ K ≥ 0.9T. t 0 (This can be done by the Birkhoff ergodic theorem.) • Proposition 8.5(a) holds. • Proposition 10.1 holds. • There exists a constant C = C(δ) such that for x ∈ K, C (x) < C(δ) where C 3 3 is as in Proposition 10.2. • There is a constant C = C (δ) such that for x ∈ K, C(x)< C (δ) where C(x) is as in Lemma 6.10 or in Corollary 6.13. Also for x ∈ K, the function c (x) of −1 Lemma 6.9 is bounded from below by C (δ) . • Lemma 4.17 holds for K = K(δ) and C = C (δ). 1 1 • There exists a constant C = C (δ) such that for x ∈ K, C (x)< C ,C (x)< C 1 2 and C(x)< C where C (x),C (x) and C(x) areasinProposition 6.11. Also 1 2 K ⊂ K and also C (δ) < C ,C (δ) < C ,C (δ) < C and C (δ) < C where K , 1 2 4 C (δ),C (δ) and C (δ) are as in Lemma 6.12,and C (δ) is as in Corollary 6.13. 1 2 4 • Lemma 6.14 holds for K. • Proposition 11.4 and Lemma 11.6 hold for K (in place of K ). Let ˜ ˜ D (q ) = D (q , K ,δ,,η) ={t > 0 : g q ∈ K}. 00 1 00 1 00 t 1 For ij ∈ ,let −1 ˜ ˜ D (q ) = D (q , K ,δ,,η) = τˆ (q , t) : g q ∈ π (K), t > 0 . ij 1 ij 1 00 ij 1 t 1 Then by the ergodic theorem and (9.4), there exists a set K = K (K ,δ,,η) with D D 00 −1 ν(K ) ≥ 1 − δ and = (K ,δ,,η) > 0such that for q ∈ π (K ) and all ij ∈ D D D 00 1 D ˜ ˜ {00}∪ , D (q ) has density at least 1 − 2κδ for > .Let ij 1 D −1 E (q , u) = E (q , u, K ,δ,,η) = : g uq ∈ π (K) , 2 1 2 1 00 τˆ (q ,u,) 1 () 1 INVARIANT AND STATIONARY MEASURES 243 E (q , u) = E (q , u, K ,δ,,η) 3 1 3 1 00 = ∈ E (q , u) :∀ij ∈ , τˆ uq ,τˆ (q , u,) ∈ D (q ) . 2 1 ij 1 () 1 ij 1 Note that τˆ (uq ,τˆ (q , u,)) ∈ D (q ) if and only if ij 1 () 1 ij 1 −1 τˆ uq ,τˆ (q , u,) =ˆτ (q , s) and g q ∈ π (K). ij 1 () 1 ij 1 s 1 Claim 12.3. — There exists = (K ,δ,,η) > 0,aset K = K (K ,δ,,η) of 3 3 00 3 3 00 −1 measure at least 1 − c (δ) and for each q ∈ π (K ) asubset Q = Q (q , K ,,δ,,η) ⊂ 3 1 3 3 3 1 00 −1 B(q , 1/100) of measure at least (1 − c (δ))|B(q , 1/100)| such that for all q ∈ π (K ) and 1 1 1 3 −1 u ∈ Q ,uq ∈ π (K) and the density of E (q , u) (for > )isatleast 1 − c (δ), and we have 3 1 3 1 3 c (δ),c (δ) and c (δ) → 0 as δ → 0. 3 3 Proof of claim. — We choose K = K ∩ K ,and 2 D K = K ∩ x ∈ X : 3 2 u ∈ B(x, 1/100) : ux ∈ K >(1 − δ)|B(x, 1/100)| . −1 −1 Suppose q ∈ π (K ),and uq ∈ π (K ).Let 1 3 1 2 −1 c E = t : g uq ∈ π K . bad t 1 −1 Then, since uq ∈ π (K ),for > , the density of E is at most 2κδ.Wehave 1 D D bad E (q , u) = :ˆτ (q , u,) ∈ E . 2 1 () 1 bad Then, by Proposition 7.4,for >κ , the density of E (q , u) is at least 1 − 4κ δ. D 2 1 Let ˆ ˆ ˜ ˜ D(q , u) = D(q , u, K ,δ,,η) = t :∀ij ∈ , τˆ (uq , t) ∈ D (q ) . 1 1 00 ij 1 ij 1 −1 Since q ∈ π (K ),for each j,for > , the density of D (q ) is at least 1− 2κδ. Then, 1 D D ij 1 ˆ ˜ by (9.4), for >κ , the density of D(q , u) is at least (1 − 4||κ δ).Now D 1 E (q , u) = E (q , u) ∩ :ˆτ (q , u,) ∈ D(q , u) . 3 1 2 1 () 1 1 Now the claim follows from Proposition 7.4. Claim 12.4. — There exists a set D = D (K ,δ,,η) ⊂ R and a number = 4 4 00 4 (K ,δ,,η) > 0 so that D has density at least 1 − c (δ) for > ,and for ∈ D 4 00 4 4 4 4 asubset K () = K (, K ,δ,,η) ⊂ X with ν(K ()) > 1 − c (δ), such that for any 4 4 00 4 −1 q ∈ π (K ()) there exists a subset Q (q ,) ⊂ Q (q ,) ⊂ B(q , 1/100) with density at 1 4 4 1 3 1 1 −1 least 1 − c (δ), so that for all ∈ D ,for all q ∈ π (K ()) and all u ∈ Q (q ,), 4 1 4 4 1 (12.4) ∈ E (q , u) ⊂ E (q , u). 3 1 2 1 (We have c (δ),c (δ) and c (δ) → 0 as δ → 0.) 4 4 244 ALEX ESKIN, MARYAM MIRZAKHANI Proof of claim. — This follows from Claim 12.3 by applying Fubini’s theorem to X × R,where X ={(x, u) : x ∈ X, u ∈ B(x, 1/100)}. B B Suppose ∈ D . We now apply Proposition 5.3 with K = g K ().Wedenote 4 − 4 the resulting set K by K () = K (, K ,δ,,η). In view of the choice of ,wehave 5 5 00 1 ν(K ()) ≥ 1 − c (δ),where c (δ) → 0as δ → 0. 5 5 5 Let D = D and let K () = g K (). 5 4 6 5 Choice of parameters #2: choice of q, q ,q (depending on δ, ,q , ). — Suppose ∈ D 1 5 −1 −1 and q ∈ π (K ()).Let q = g q . Then, q ∈ π (K ()).Let A(q, u,, t) be as in 1 6 − 1 5 Section 6. (Note that following our conventions, we use the notation A(q , u,, t) for q ∈ X, even though A(q , u,, t) was originally defined for q ∈ X .) and for u ∈ Q (q ,) 1 1 1 0 4 1 let M be the subspace of Lemma 5.1 applied to the linear map A(q , u,,τˆ (q , u,)). u 1 () 1 − −1 By Proposition 5.3 and the definition of K (), we can choose q ∈ L [q]∩π (g K ()) 5 − 4 with ρ (δ) ≤ d (q, q ) ≤ 1/100 and so that (5.4)and (5.5)holdwith (δ) → 0as δ → 0. −1 Let q = g q .Then q ∈ π (K ()). 1 1 Standing assumption. — We assume ∈ D , q ∈ K () and q, q , q are as in Choice 5 1 6 of parameters #2. Notation. — For u ∈ B(q , 1/100), u ∈ B(q , 1/100),let τ(u)=ˆτ (q , u,), τ u =ˆτ q , u , . () 1 () The maps ψ and ψ .— For u ∈ B(q , 1/100),and u ∈ B(q , 1/100),let ψ(u) = g uq,ψ u = g u q . τ(u) 1 τ (u ) Claim 12.5. — We have −1 −1 (12.5) ψ Q (q ,) ⊂ π (K), and ψ Q q , ⊂ π (K). 4 1 4 Proof of claim. — Suppose u ∈ Q (q ,). Since q ∈ K and ∈ D , it follows from 4 1 1 4 4 (12.4)that ∈ E (q , u), and then from the definition of E (q , u) is follows that g uq ∈ 2 1 2 1 τ(u) 1 −1 −1 −1 π (K).Hence ψ(Q (q ,)) ⊂ π (K). Similarly, since q ∈ π (K ), ψ (Q (q ,)) ⊂ 4 1 4 4 1 1 −1 π (K), proving (12.5). The numbers t and t .— Suppose u ∈ Q (q ,), and suppose ij ∈ .Let t be de- ij 4 1 ij ij fined by the equation (12.6) τˆ uq ,τˆ (q , u,) =ˆτ (q , t ). ij 1 () 1 ij 1 ij INVARIANT AND STATIONARY MEASURES 245 Then, since ∈ D and in view of (12.4), we have ∈ E (q , u). In view of the definition 4 3 1 of E , it follows that −1 (12.7) g q ∈ π (K). t 1 ij Similarly, suppose u ∈ Q (q ,) and ij ∈ .Let t be defined by the equation 1 ij (12.8) τˆ u q ,τˆ q , u , =ˆτ q , t . ij () ij 1 1 1 ij Then, by the same argument, −1 (12.9) g q ∈ π (K). ij The map v(u) and the generalized subspace U(u).— For u ∈ B(q , 1/100),let (12.10) v(u) = v q, q , u,, t = A(q, u,, t) F(q) − F q where t =ˆτ (q , u,), F is as in Section 5 and A(·,·,·,·) is as in Section 6.1.ByProposi- () 1 tion 6.11, we may write v(u) = j(M ,v ),where (M ,v ) ∈ H (g uq )× W (g uq ). ++ τ(u) 1 τ(u) 1 Let U(u) ≡ U (M ,v ) denote the generalized affine subspace corresponding to v(u). g uq τ(u) 1 Thus, U(u) is the approximation to U [g q ] near g uq defined in Proposition 6.11. τ(u) τ(u) 1 Standing assumption. — We have C(δ) < 1/100 for any constant C(δ) arising in the course of the proof. In particular, this applies to C (δ) and C (δ) in the next claim. Claim 12.6. — There exists a subset Q = Q (q ,, K ,δ,,η) ⊂ Q (q ,) with 5 5 1 00 4 1 |Q |≥ (1 − c (δ))|B(q , 1/100)| (with c (δ) → 0 as δ → 0), and a number = (δ,) 5 1 5 5 5 5 such that for all u ∈ Q and > , 5 5 (12.11) τ(u)< α , where α > 0 is as in Proposition 6.16 and Section 6.1. In addition, X + + (12.12)C (δ) ≤ hd U [g uq ], U g q ≤ C (δ), 1 τ(u) 1 τ(u) 2 g uq 1 τ(u) 1 X + −α (12.13) hd U g q , U(u) ≤ C (δ)e , τ(u) 7 g uq 1 τ(u) 1 where α depends only on the Lyapunov spectrum. Also, (12.14)C (δ) ≤v(u)≤ C (δ), 1 2 and if u ∈ U [q ] is such that (12.15) d g uq , g u q < 1/100, τ(u) 1 τ(u) then u ∈ B(q , 1/100). 1 246 ALEX ESKIN, MARYAM MIRZAKHANI Proof of claim.—Let M be the subspace of Lemma 5.1 applied to the linear map A(q , u,,τˆ (q , u,)),where A(,,,) is as in Section 6.Let Q(q ) be as in Proposi- 1 () 1 1 tion 6.16,so |Q(q )|≥ (1 − δ)|B(q , 1/100)|.Let Q ⊂ Q ∩ Q(q ) be such that for all 1 1 4 1 u ∈ Q , d F(q) − F q , M ≥ β(δ) Y u where F is as in Section 5. Then, (12.11) follows from Proposition 6.16 and the fact that Q ⊂ Q . Also, by (5.5), 5 1 |Q |≥|Q |− δ + (δ) |B(q , 1/100)| 4 1 1 ≥ 1 − δ − (δ) − c (δ) |B(q , 1/100)|. 1 1 Then, let Q ={u ∈ Q : d(u,∂B(q , 1/100)) > δ},hence 5 1 |Q |≥ 1 − c (δ) − c (δ) − c δ |B(q , 1/100)|, 5 n 1 5 4 where c depends only on the dimension. −1 We have C(δ) ≤A(q , u,, t)≤ C(δ) by the definition of t =ˆτ (q , u,). 1 () 1 We now apply Lemma 5.1 to the linear map A(q , u,, t). Then, for all u ∈ Q , 1 5 c(δ)A(q , u,, t)≤ A(q , u,, t) F(q) − F q ≤A(q , u,, t). 1 1 1 Therefore, −1 C (δ) ≤ A(q , u,, t) F(q) − F q ≤ C (δ) This immediately implies (12.14), in view of the definition of v(u). We now apply Proposi- tion 6.11 and Lemma 6.12(a). (We assume is sufficiently small so that (6.29) holds. Also the condition (6.22)inProposition 6.11 holds in view of Proposition 6.16.) Now (12.12) follows from (6.25). Also (12.13) follows from (6.27). Finally, suppose u ∈ Q ,and u ∈ U (q ) is such that (12.15) holds. Then, by X −α Lemma 6.14,wehave d (uq , u q ) = O (e ). Then, assuming is sufficiently large 1 δ (depending on δ) and using Proposition 3.4,wehave u ∈ B(q , 1/100). Standing assumption. — We assume > . Claim 12.7. — Suppose u ∈ Q (q ,),u ∈ Q (q ,) and (12.15) holds. Then, there exists 5 1 4 C = C (δ) such that 0 0 (12.16) τˆ (q , u,)−ˆτ q , u , ≤ C (δ). () 1 () 0 1 INVARIANT AND STATIONARY MEASURES 247 Proof of claim.—Let t =ˆτ (q , u,), t =ˆτ (q , u ,). () 1 () By Proposition 6.11(ii) (with q and q reversed) and (5.4), = A q ,, u , t ≥ A q ,, u , t F q − F(q) 1 1 0 + + ≥ c(δ)hd U g u q , U [g uq ] t t 1 g u q t 1 −1 In view of Corollary 6.13(b), (12.11) and the fact that g u q ∈ π (K), this contradicts (12.12), unless t < t + C(δ). It remains give a lower bound on t .Let M denote the subspace as in Lemma 5.1 for A(q , u ,, t ). Note that by Proposition 5.3 (with the function u → M the constant function M ) we can choose q ∈ W [q] with d (F(q ) − F(q ), M )>ρ(δ), and also so the upper bounds in (5.3)and (5.4)holdwith q in place of q . Then, = A q ,, u , t ≤ c(δ) A q ,, u , t F q − F q . Write q = g q . Then, by Proposition 6.11(ii), and Lemma 6.12(a), 0 + + (12.17) hd U g u q , U g q ≥ c (δ). t t 2 g u q 1 1 −1 By Corollary 6.13(a), (12.11)and (12.12), since g u q ∈ π (K), 0 + + −β(t−t ) −α (12.18) hd U g u q , U [g uq ] ≤ C(δ)e + C (δ)e , t t 1 4 g u q 1 t 1 where α and β depend only on the Lyapunov spectrum. Then, by (12.17), (12.18), and the reverse triangle inequality, X + + −β(t−t ) −α (12.19) hd U [g uq ], U g q ≥ c (δ) − C(δ)e − C (δ)e . t 1 t 2 4 g uq 1 But, =A(q,, u, t)≥ c (δ) A(q,, u, t) F q − F(q) , and thus, by Proposition 6.11(ii) and Lemma 6.12(a), X + + hd U [g uq ], U g q ≤ c(δ) t 1 t g uq 1 t 1 −1 In view of Corollary 6.13(b) (and the fact that g uq ∈ π (K)) this contradicts (12.19) t 1 unless t > t − C (δ). We note the following trivial lemma: Lemma 12.8. — Suppose P and P are finite measure subsets of R with |P|=|P |,and we have N N P = P , P = P . j=1 j=1 248 ALEX ESKIN, MARYAM MIRZAKHANI Suppose there exists k ∈ N so that any point in P is contained in at most k sets P , and also any point in P is contained in at most k sets P . Also suppose Q ⊂ P and Q ⊂ P are subsets with |Q| >(1−δ)|P|, |Q | >(1 − δ)|P |. Suppose there exists κ> 1 such that for all 1 ≤ j ≤ N such that P ∩ Q = ∅, |P |≤ κ|P |. j j ˆ ˆ ˆ Then there exists Q ⊂ Q with |Q|≥ (1 − 2κ kδ)|P| such that if j is such that Q ∩ P = ∅, then Q ∩ P = ∅. Proof.—Let J ={j : P ∩ Q = ∅},and let J ={j : Q ∩ P = ∅},and let Q = x ∈ Q : for all j with x ∈ P ,wehave j ∈ J . Thus, if x ∈ Q \ Q, then there exists j ∈ Jwith x ∈ Q ∩ P but j ∈/ J . Then, |Q \ Q|≤ k |Q ∩ P |≤ k |P |≤ κ k |P |≤ κ k Q , j j j∈J\J j∈J\J j∈/J ˆ ˆ since if j ∈/ J then P ⊂ (Q ) .Thus, |Q \ Q|≤ κ kδ|P|,and so |Q|≥ (1 − 2κ kδ)|P|. The constant .— Let (δ) be a constant to be chosen later (we will choose (δ) 0 0 0 following (12.33)ofthe form (δ) = (δ)/C(δ)), where (δ) is as in (12.3). We will 0 0 always assume that < (δ) < (δ)/10. Claim 12.9. — There exists a subset Q (q ,) = Q (q ,, K ,δ,,η) ⊂ Q (q ,) 6 1 6 1 00 5 1 with |Q (q ,)| >(1 − c (δ))|B(q , 1/100)| and with c (δ) → 0 as δ → 0 such that for all 6 1 1 6 6 u ∈ Q (q ,) there exists u ∈ Q (q ,) such that 6 1 4 (12.20) d g uq , g u q < C(δ) (δ). τ(u) 1 τ(u) 0 Proof of claim. — Note that the sets {B [uq ]: u ∈ Q (q ,)} are a cover τ(u) 1 5 1 of Q (q ,)q . Then, since these sets satisfy the condition of Lemma 3.10(b), we can 5 1 1 find a pairwise disjoint subcover, i.e. find u ∈ Q (q ,),1 ≤ j ≤ N, with Q (q ,)q = j 5 1 5 1 1 B [u q ] and so that B [u q ] and B [u q ] are disjoint for j = k.Let τ(u ) j 1 τ(u ) j 1 τ(u ) k 1 j j k j=1 B ≡ g B [u q ]= B [g u q ]⊂ X j τ(u ) τ(u ) j 1 0 τ(u ) j 1 0 j j j In view of (12.3), Proposition 3.4, and the Besicovich covering lemma, there exists k, depending only on the dimension and points x ,..., x ⊂ B such that j,1 j,m(j) j m(j) −1 X + π (K) ∩ B ⊂ B x , (δ) ∩ U [g uq ], j j,m 0 τ(u ) 1 m=1 and also so that for a fixed j , each point is contained in at most k balls B (x , (δ)). j,m 0 Since (δ) < (δ)/10, in view of (12.3)and (12.21), the same is true without fixing j . 0 INVARIANT AND STATIONARY MEASURES 249 For 1 ≤ j ≤ Nand 1 ≤ m ≤ m(j),let P = u ∈ B(q , 1/100) : g uq ∈ B x , (δ) , j,m 1 τ(u ) 1 j,m 0 and let P = u ∈ B q , 1/100 : g u q ∈ B x , (δ) . τ(u ) j,m 0 j,m 1 1 By construction, each point is contained in at most k sets P ,and at most k sets P . j,m j,m By (12.12) applied to u , X + + (12.21) hd U [g u q ], U g q ≤ C (δ). τ(u ) j 1 τ(u ) 2 g u q j j 1 τ(u ) j 1 Suppose > 0 is sufficiently small (depending on δ) so that Lemma 6.14 holds with C (δ) in place of . Since for all x ∈ X , B [x]⊂ B (x, 1/200) we have 2 0 0 d (x , g u q )< 1/200, and j,m t j 1 X X 0 0 (12.22)B x , (δ) ⊂ B (g u q , 1/100). j,m 0 t j 1 By Lemma 6.14,for 1 ≤ j ≤ N, 1 ≤ m ≤ m(j),provided B ∩ Q (q ,) = ∅,we j 5 1 −1 have κ |P |≤|P |≤ κ|P |,where κ depends only on the Lyapunov spectrum, and j,m j j,m + + + we have normalized the measures |·| so that |U [q ]∩ B (q , 1/100)|=|U [q ]∩ 1 1 B (q , 1/100)|= 1. Let m(0) = 1and let N m(j) N m(j) P = B(q , 1/100) \ P , P = B q , 1/100 \ P . 0,1 1 j,m 0,1 1 j,m j=1 m=1 j=1 m=1 Then, m(j) m(j) N N B(q , 1/100) = P , B q , 1/100 = P . 1 j,m 1 j,m j=0 m=1 j=0 m=1 Then, applying Lemma 12.8 with P = B(q , 1/100),P = B(q , 1/100),Q = Q (q ,), 1 5 1 Q = Q (q ,), we get a set Q ≡ Q (q ,) with |Q (q ,)|≥ (1 − c (δ))|B(q , 1/100)| 4 6 1 6 1 1 1 6 where c (δ) → 0as δ → 0, so that, in view of (12.22) and the definitions of P and P , j,m 6 j,m for any u ∈ Q (q ,) there exists u ∈ Q (q ,) with uq ∈ B [u q ] and u ∈ Q (q ,) 6 1 j 5 1 1 τ(u ) j 1 4 j 1 with (12.23) d g uq , g u q ≤ (δ). τ(u ) 1 τ(u ) 0 j j 1 It remains to replace τ(u ) by τ(u) in (12.23). This can be done as follows: Since uq ∈ j 1 B [u q ],wehave, by (12.12) applied to u and Lemma 6.18, τ(u ) j 1 j −1 + + C (δ) ≤ hd U [g uq ], U g q ≤ C (δ) 2 g uq τ(u ) 1 τ(u ) 2 τ(u ) 1 j j 1 j 250 ALEX ESKIN, MARYAM MIRZAKHANI −1 Then, since g uq ∈ π (K),by(12.12), (12.13), (12.11) and Corollary 6.13,wehave τ(u) 1 (12.24) |τ(u) − τ(u )|≤ C (δ). j 1 Then, provided is small enough depending on δ,(12.20) follows from (12.23), (12.24), and Lemma 3.6. Claim 12.10. — There exists a constants c (δ) > 0 and c (δ) with c (δ) → 0 and 7 7 c (δ) → 0 as δ → 0 and a subset K () = K (, K ,δ,,η) with K () ⊂ K () and 7 7 00 7 6 −1 ν(K ()) > 1 − c (δ) such that for q ∈ π (K ()), 7 7 1 7 |B(q ) ∩ Q (q ,)|≥ 1 − c (δ) |B(q )|. 1 6 1 1 Proof of claim. — Recall that in view of Proposition 3.7, B(q ) ⊂ B(q , 1/100). 1 1 Given δ> 0, there exists c (δ) > 0with c (δ) → 0as δ → 0 and a compact set 7 7 −1 K ⊂ Xwith ν(K )> 1 − c (δ),suchthatfor q ∈ π (K ), |B(q ) ∩ B(q , 1/100)|≥ 1 1 1 7 7 7 7 1/2 −1 c (δ) |B(q , 1/100)|. Then, for q ∈ π (K ∩ K ), 1 1 6 6 7 c c |B(q ) ∩ Q (q ,) |≤|Q (q ,) |≤ c (δ)|B(q , 1/100)| 1 6 1 6 1 1 1/2 ≤ c (δ) |B(q )|. 1/2 Thus, the claim holds with c (δ) = c (δ) + c (δ) and c (δ) = c (δ) . 7 6 7 7 6 −1 Standing assumption. — We assume that q ∈ π (K ()). 1 7 The next few claims will help us choose u (once the other parameters have been chosen). Let Q (q ,) = B(q ) ∩ Q (q ,) 7 1 1 6 1 ∗ ∗ Claim 12.11. — There exists a subset Q (q ,) = Q (q ,, K ,δ,,η) ⊂ Q (q ,) 1 1 00 7 1 7 7 ∗ ∗ ∗ with |Q |≥ (1 − c (δ))|B(q )| such that for u ∈ Q and any > (δ) we have 1 7 7 7 7 |B (uq ) ∩ Q (q ,)|≥ 1 − c (δ) |B (uq )|, 1 7 1 1 where c (δ) → 0 as δ → 0. Proof. — This follows immediately from Lemma 6.3. Claim 12.12. — There exist a number = (K ,δ,,η) and a constant c (δ) with 8 8 00 8 c (δ) → 0 as δ → 0 and for every > asubset Q (q ,) = Q (q ,, K ,δ,,η) ⊂ B(q ) 8 8 8 1 8 1 00 1 with |Q (q ,)|≥ (1 − c (δ))|B(q )| so that for u ∈ Q (q ,) we have 8 1 8 1 8 1 v(u) −α (12.25) d , E(g uq ) ≤ C (δ)e , τ(u) 1 8 v(u) where v(u) is defined in (12.10) and α depends only on the Lyapunov spectrum. INVARIANT AND STATIONARY MEASURES 251 Proof of claim.—Let L > L (δ) be a constant to be chosen later, where L (δ) is as in 2 2 Proposition 8.5(a). Also let = (δ,, K ,η) be a constant to be chosen later. Suppose 8 8 00 ∗ −1 > , and suppose u ∈ Q (q ,), so in particular g uq ∈ π (K).Let t ∈[L , 2L ] be 8 1 τ(u) 1 such that Proposition 8.5(a) holds for v = v(u) and x = g uq . τ(u) 1 Let B ⊂ B(q ) denote B (uq )u,(where B (x) is defined in Section 6). u 1 τˆ (q ,u,)−t 1 t () 1 Suppose u ∈ B ∩ Q (q ,), and write 1 u 7 1 −1 g u q = g u g g uq . τ(u ) 1 1 s 2 τ(u) 1 1 t −1 Then, u ∈ B(g g uq ) and t ≤ 2L . 2 τ(u) 1 We now claim that (12.26) s ≤ κ t + C (δ) ≤ κ L + C (δ) 0 0 where κ depends only on the Lyapunov spectrum. Let + + U = U [g g uq ], U = U g g q . t −t τ(u) 1 −t τ(u) t 1 −1 By Corollary 6.13(b) applied at the point g uq ∈ π (K), τ(u) 1 X −β t −α hd U , U ≥ C(δ)e − c (δ)e , t 0 g g uq t −t τ(u) 1 where β depends only on the Lyapunov spectrum, and by Corollary 6.13(a) applied at −1 the point g u q ∈ π (K), τ(u ) 1 1 X −2s −α hd U , U ≤ c(δ)e + c (δ)e t 0 u g g uq t 2 −t τ(u) 1 where β also depends only on the Lyapunov spectrum. Also, by Lemma 6.18, X X −α 0 0 hd U , U ≥ c hd U , U − c (δ)e t 1 t 0 g g uq t u g g uq t −t τ(u) 1 2 −t τ(u) 1 where c is an absolute constant. Therefore, −β t −α −2s −α C(δ)e − c (δ)e ≤ c c(δ)e + c (δ)e . 0 1 0 This implies (12.26), assuming that is sufficiently large depending on . Since u ∈ Q (q ,),(12.12)and (12.13) hold. Therefore, 6 1 −1 + κ L −α hd g u g U(u), U g q = O e e , g u q s 2 τ(u ) τ(u ) 1 1 t 1 1 where κ and α depend only on the Lyapunov spectrum. Thus, using (12.13) at the point −1 g u q ∈ π (K), τ(u ) 1 1 −1 κ L −α hd g u g U(u), U(u ) = O e e . g u q s 2 1 τ(u ) 1 1 t 1 252 ALEX ESKIN, MARYAM MIRZAKHANI Therefore, −1 κ L −α (12.27) g u g v(u) − v(u ) = O e e . s 2 1 −1 In view of (12.14), v(u )≈ .Thus, (g u g ) v(u)≈ ,and 1 s 2 ∗ −1 (g u g ) v(u) v(u ) s 2 ∗ 1 t κ L −α − = O e . −1 v(u ) (g u g ) v(u) s 2 ∗ 1 −1 But, by Proposition 8.5(a), for 1 − δ fraction of u ∈ B(g g uq ), 2 τ(u) 1 (g u g ) v(u) s 2 −t ∗ −αL d , E(g u q ) ≤ C(δ)e . τ(u ) 1 1 (g u g ) v(u) s 2 −t ∗ Note that −1 B g g uq = g B . τ(u) 1 τˆ (q ,u,)−t u t () 1 Therefore, for 1 − δ fraction of u ∈ B , 1 u v(u ) κ L −α −αL (12.28) d , E(g u q ) ≤ C(,δ) e + e τ(u ) 1 1 v(u ) We can now choose L > 0to be α where α > 0 is a small constant depending only on the Lyapunov spectrum, and > 0so that for > the right-hand-side of the above 8 8 −α equation is at most e . The collection of balls {B } are a cover of Q (q ,). These balls satisfy u u∈Q (q ,) 1 1 7 the condition of Lemma 3.10(b); hence we may choose a pairwise disjoint subcollection which still covers Q (q ,). Then, by summing (12.28), we see that (12.25) holds for u in a subset Q ⊂ B[q ] of measure at least (1 − c (δ))|B[q ]| = (1 − δ)(1 − c (δ))|B[q ]|. 8 1 8 1 1 ∗ ∗ Claim 12.13. — There exists a subset Q (q ,) = Q (q ,, K ,δ,,η) ⊂ Q (q ,) 1 1 00 8 1 8 8 ∗ ∗ ∗ with |Q |≥ (1 − c (δ))|B(q )| such that for u ∈ Q and any t > (δ) we have 1 8 8 8 8 |B (uq ) ∩ Q (q ,)|≥ 1 − c (δ) |B (uq )|, t 1 8 1 t 1 where c (δ) → 0 as δ → 0. Proof. — This follows immediately from Lemma 6.3. Choice of parameters #3: choice of δ.— Let θ = (θ/2) ,where θ and n areasinPropo- sition 10.1. We can choose δ> 0so that (12.29) c (δ) < θ /2. 8 INVARIANT AND STATIONARY MEASURES 253 Claim 12.14. — There exist sets Q (q ,) = Q (q ,, K ,δ,,η) ⊂ Q (q ,) with 9 1 9 1 00 1 |Q (q ,)|≥ (θ /2)(1 − θ /2)|B(q )| and = (K ,δ,,η),suchthatfor > and 9 1 1 9 9 00 9 u ∈ Q (q ,), 9 1 v(u) (12.30) d , E (g uq ) < 4η. [ij],bdd τ(u) 1 v(u) ij∈ Proof of claim. — Suppose u ∈ Q (q ,). Then, by (12.25)and (12.14), we may write v(u) = v (u) + v (u), −α where v (u) ∈ E(g uq ) and v (u)≤ C(δ,)e . Arguing in the same way as in the τ(u) 1 proof of Claim 12.12, we see that for (1− O(δ))-fraction of y ∈ F [g uq , L],wehave v (u) τ(u) 1 y ∈ g K. Then, by Proposition 10.1 applied with L = L (δ,η) and v = v (u),weget [−1,1] 0 that for a at least θ -fraction of y ∈ F [g uq , L], v τ(u) 1 R(g uq , y)v (u) τ(u) 1 d , E (y) < 2η. [ij],bdd R(g uq , y)v (u) τ(u) 1 ij∈ κ L Note that by Proposition 4.15(d), for y ∈ F [g uq , L], R(g uq , y)≤ e ,where κ v τ(u) 1 τ(u) 1 is as in Proposition 4.15. Then, for at least θ -fraction of y ∈ F [g uq , L], v τ(u) 1 R(g uq , y)v(u) τ(u) 1 2κ L −α (12.31) d , E (y) < 3η + C(,δ)e e . [ij],bdd R(g uq , y)v(u) τ(u) 1 ij∈ Let B = B (uq )u.Inviewof(12.27)and (12.14) there exists C = C(,δ) such u τˆ (q ,u,)−L 1 () 1 that −1 F [g uq , L]∩ π (K) ⊂ g ψ(B ) and v τ(u) 1 [−C,C] u −1 ψ(B ) ∩ π (K) ⊂ g F [g uq , L]. u [−C,C] v τ(u) 1 −1 Then, by (12.31)and (12.29), for (θ /2)-fraction of u ∈ B , g u q ∈ π (K) and 1 u τ(u ) 1 1 R(g uq , g u q )v(u) τ(u) 1 τ(u ) 1 1 d , E (g u q ) [ij],bdd τ(u ) 1 1 R(g uq , g u q )v(u) τ(u) 1 τ(u ) 1 1 ij∈ 2κ L −α < C (,δ) 3η + e e . Then, by (12.27), for (θ /2)-fraction of u ∈ B , 1 u v(u ) 2κ L −α −α d , E (g u q ) < C (,δ) 3η + e e + e . [ij],bdd τ(u ) 1 1 2 v(u ) ij∈ 254 ALEX ESKIN, MARYAM MIRZAKHANI Hence, we may choose = (K ,,δ,η) so that for > the right-hand side of the 9 9 00 9 above equation is at most 4η.Thus, (12.30) holds for (θ /2)-fraction of u ∈ B . 1 u The collection of balls {B } are a cover of Q (q ,). These balls satisfy u u∈Q (q ,) 1 1 8 the condition of Lemma 3.10(b); hence we may choose a pairwise disjoint subcollection which still covers Q (q ,). Then, by summing over the disjoint subcollection, we see ∗ ∗ that the claim holds on a set E of measure at least (θ /2)|Q |≥ (θ /2)(1 − c (δ)) ≥ 8 8 (θ /2)(1 − θ /2). Choice of parameters #4: choosing ,q ,q,q ,q .— Choose > (K ,,δ,η).Now 1 9 00 choose q ∈ K (),and let q, q , q be as in Choice of Parameters #2. 1 7 Choice of parameters #5: choosing u, u ,q ,q ,ij, q ,q (depending on q ,q ,u, ). — 2 3,ij 1 2 3,ij 1 Choose u ∈ Q (q ,), u ∈ Q (q ,) so that (12.12)and (12.13) hold. We have ψ(u) = 9 1 4 −1 −1 g uq ∈ π (K) and ψ (u ) ∈ π (K).By(12.16), τ(u) 1 τˆ (q , u,)−ˆτ q , u , ≤ C (δ), () 1 () 0 therefore, −1 g u q ∈ π (g K), τ(u) [−C,C] where C = C(δ). By the definition of K we can find C (δ) and s ∈[0, C (δ)] such that 4 4 −1 −1 q ≡ g g uq ∈ π (K ), q ≡ g g u q ∈ π (K ). 2 s τ(u) 1 0 s τ(u) 0 2 1 In view of (12.12), (12.13), the fact that s ∈[0, C (δ)] and Corollary 6.13(a) we get X + + (12.32) ≤ hd U [q ], U q ≤ C(δ). q 2 C(δ) By (12.20), the fact that s ∈[0, C (δ)] and Lemma 3.6 we get X + (12.33) d q , q = d q , q ≤ C(δ) (δ). 2 2 0 2 2 We now choose (δ) so that C(δ) (δ) < (δ),where C(δ) is as in (12.33), and (δ) is 0 0 0 0 as in (12.3). Let ij ∈ be such that v(u) (12.34) d , E (g uq ) ≤ 4η. [ij],bdd τ(u) 1 v(u) By Lemma 11.6, τˆ uq ,τˆ (q , u,) −ˆτ u q ,τˆ (q , u,) ≤ C (δ). ij 1 () 1 ij () 1 1 4 INVARIANT AND STATIONARY MEASURES 255 Then, by (12.16)and (9.4), τˆ uq ,τˆ (q , u,) −ˆτ u q ,τˆ q , u , ≤ C (δ). ij 1 () 1 ij () 1 1 4 Hence, by Proposition 4.15(e) (cf. Lemma 9.2), (12.6)and (12.8), (12.35) |t − t |≤ C (δ). ij 5 ij Therefore, by (12.7)and (12.9), we have −1 −1 g q ∈ π (K), and g q ∈ π (g K). t 1 t [−C (δ),C (δ)] ij ij 1 5 5 By the definition of K, we can find s ∈[0, C (δ)] such that −1 −1 q ≡ g q ∈ π (K ), and q ≡ g q ∈ π (K ). 3,ij s +t 1 0 s +t 0 ij 3,ij ij 1 Let τ = s+ˆτ (q , u,), τ = s + t . Then, () 1 ij q = g uq , q = g u q , q = g q , q = g q . 2 τ 1 τ 3,ij τ 1 τ 2 1 3,ij 1 ij ij We may write q = g uq , q = g q . Then, in view of (12.35)and (9.4), 2 t 1 3,ij 1 |t − t |≤ C (δ). We note that by Proposition 6.16, >α τ,where α depends only on the Lyapunov 0 0 spectrum. Taking the limit as η → 0.— For fixed δ and , we now take a sequence of η → 0 (this forces →∞) and pass to limits along a subsequence. Let q˜ ∈ K be the limit of k 2 0 the q ,and q˜ ∈ K be the limit of the q . We may also assume that along the subsequence 2 0 2 2 ij ∈ is fixed, where ij is as in (12.34). By passing to the limit in (12.32), we get 0 + + (12.36) ≤ hd U [˜ q ], U q˜ ≤ C(δ). q˜ 2 C(δ) We now apply Proposition 11.4 (with ξ → 0as η → 0). By (11.4), q˜ ∈ W [˜ q ].By k 2 applying g to (12.34) and then passing to the limit, we get U [˜ q ]∈ E (q˜ ). Finally, it s ij 2 follows from passing to the limit in (12.33)that d (q˜ , q˜ ) ≤ (δ), and thus, since q˜ ∈ K 2 2 0 2 0 and q˜ ∈ K , it follows from (12.3)that q˜ ∈ B [˜ q ].Hence, 0 0 2 2 2 q˜ ∈ C (q˜ ). ij 2 Now, by (11.5), we have f (q˜ ) ∝ P q˜ , q˜ f q˜ . ij 2 2 ij 2 2 −1 −1 This concludes the proof of Proposition 12.2.Wehave q˜ ∈ π (K ) ⊂ π (K ∩ K ), 2 0 00 ∗ −1 and q˜ ∈ π (K ⊂ K ). 0 ∗ 2 256 ALEX ESKIN, MARYAM MIRZAKHANI Applying the argument for a sequence of ’s tending to 0.— Take a sequence → 0. We now apply Proposition 12.2 with = . After passing to a subsequence, we may assume ij is constant. We get, for each n a set E ⊂ K with ν(E )>δ and with the property that n ∗ n 0 for every x ∈ E there exists y ∈ C (x) ∩ K such that (12.1)and (12.2) hold for = .Let n ij ∗ n ∞ ∞ F = E ⊂ K n ∗ k=1 n=k (so F consists of the points which are in infinitely many E ). Suppose x ∈ F. Then there ex- + + ists a sequence y → x such that y ∈ C [x], y ∈/ U [x],and so that f (y ) ∝ P (x, y ) f (x). n n ij n ij n n ∗ ij Then (on the set where both are defined) f (x) ∝ (γ ) f (x), ij n ∗ ij where γ ∈ G (x) is the affine map whose linear part is P (x, y ) and whose translational n ++ n part is y − x. (Here we have used the fact that y ∈ C [x], and thus by the definition of con- n n ij + + ditional measure, f (y ) = (y − x) f (x),where (y − x) : W (x) → W (x) is translation ij n n ∗ ij n ∗ by y − x.) (x) denote the measure on G (x) given by Let f ij ++ ˜ ¯ f (x)(h) = hdf (x), ij ij W [x] where for a compactly supported real-valued continuous function h on G (x), h : ++ W [x]→ R is given by h(gx) = h(gq)dm(q), Q (x) ++ where m is the Haar measure on Q (x).(Thus, f (x) is the pullback of f (x) from ++ ij ij W [x] = G (x)/Q (x) to G (x).) Then, ++ ++ ++ ˜ ˜ (12.37) (γ ) f (x) ∝ f (x) n ∗ ij ij on the set where both are defined. For x ∈ X, let U (x) denote the maximal connected subgroup of G (x) such that ++ new for u ∈ U (x) (on the domain where both are defined), new ˜ ˜ (12.38) (u) f (x) ∝ f (x). ∗ ij ij + + By (12.37) and Proposition D.3,for x ∈ F, U (x) strictly contains U (x). new ˜ ˜ Suppose x ∈ F, y ∈ Fand y ∈ C [x]. Then, since f (y) = Tr(x, y) f (x),wehavethat ij ij ∗ ij (12.38) holds for u ∈ Tr(y, x)U (y) (see Lemma 6.1). Therefore, by the maximality of new U (x),for x ∈ F, y ∈ F ∩ C [x], ij new + + (12.39)Tr(y, x)U (y) = U (x). new new INVARIANT AND STATIONARY MEASURES 257 Suppose x ∈ F, t < 0and g x ∈ F. Then, since the measurable partition C is g -equivariant t ij t (see Lemma 11.3)wehavethat(12.38) holds for u ∈ g U (g x). Therefore, by the maxi- −t t mality of U (x),for x ∈ F, t < 0with g x ∈ Fwe have new + + (12.40) g U (g x) = U (x), −t t new new and (12.38)and (12.39) still hold. From (12.38), we get that for x ∈ Fand u ∈ U (x), new β (u) ˜ ˜ (12.41) (u) f (x) = e f (x), ∗ ij ij where β : U (x) → R is a homomorphism. Since ν(F)>δ > 0and g is ergodic, for x 0 t new almost all x ∈ X there exist arbitrarily large t > 0so that g x ∈ F. Then, we define U (x) −t new to be g U (g x). (This is consistent in view of (12.40).) Then, (12.41) holds for a.e. x ∈ X. t −t new It follows from (12.41)thatfor a.e. x ∈ X, u ∈ U (x) and t > 0, new (12.42) β (g ug ) = β (u). g x −t t x −t We can write β (u) = L (log u), x x where L : Lie(U )(x) → R is a Lie algebra homomorphism (which is in particular a linear map). Let K ⊂ X be a positive measure set for which there exists a constant C with L ≤ Cfor all x ∈ K. Now for almost all x ∈ Xand u ∈ U (x) there exists a new sequence t →∞ so that g x ∈ Kand g ug → e,where e is the identity element of j −t −t t j j j U . Then, (12.42) applied to the sequence t implies that β (u) = 0almost everywhere j x new (cf. [BQ, Proposition 7.4(b)]). Therefore, for almost all x ∈ X, the conditional measure of + + ν along the orbit U [x] is the push-forward of the Haar measure on U (x). new new The partition whose atoms are U [x] is given by the refinement of the measurable new partition C into orbits of an algebraic group. (For the atom C [x] this group is U (y) for ij ij new almost any y ∈ C [x];inviewof(12.39) and Lemma 6.1, this group, viewed as a group of ij affine maps of W [x] is independent of the choice of y.) Therefore the partition whose atoms are sets of the form U [x]∩ B [x] is a measurable partition. new In view of (12.39), and since for u near the identity, U [x]⊂ C [x] we have that ij new (6.2) holds for U . Then, it also holds for any u in view of g -equivariance. Finally, since new + + + + U (x) ⊃ U (x) and U (x) ⊃ exp N(x),wehave U (x) ⊃ N(x). new new Similarly, recall that the measure ν on X is the pullback of the measure on X such that the conditionals on the fibers of the covering map σ : X → X are the counting 0 0 measure. By (4.12) there exists a subset ⊂ X of full measure such that for any x ∈ , 0 0 0 0 −1 for any x ∈ σ (x ) we have an (almost-everywhere defined) identification σ between 0 x + + W [x]⊂ Xand W [x ]⊂ X and under this identification, the conditional measures 0 0 258 ALEX ESKIN, MARYAM MIRZAKHANI −1 coincide, i.e. (σ ) ν + = ν + . Suppose x ∈ and x ∈ σ (x ). After removing x ∗ W [x] W [x ] 0 0 0 0 0 from a set of measure 0, we may assume that Definition 6.2(iii) holds for x and U (x). new + −1 Therefore it also holds for x and σ ◦ U (x) ◦ σ ⊂ G (x ).Now for x ∈ define 0 x ++ 0 0 0 new x + + −1 U (x ) to be the group generated by all the groups σ ◦ U (x) ◦ σ where x varies 0 x new new x −1 over σ (x ). Then, Definition 6.2(iii) holds for x and U (x ).Inthe same way, allof 0 0 0 0 new + + the other parts of Definition 6.2 hold for x and U (x ) since they hold for x and U (x) 0 0 new new −1 for any x ∈ σ (x ). This completes the proof of Proposition 12.1. 13. Proof of Theorem 2.1 − + + Let L ,L ,S be as in Section 6.2. Apply Proposition 12.1 to get an equivari- ant system of subgroups U (x) ⊂ G (x) which is compatible with ν in the sense of ++ new Definition 6.2. We have that L [x] is smooth at x for almost all x ∈ X, see [AEM,§3].Let + + + T U (x) ⊂ W (x) denote the tangent subspace at x to the smooth manifold U [x],and − − − let T L (x) ⊂ W (x) denote the tangent subspace to L [x] at x. (This exists for almost all x.) + + If L [x]⊂ S [x] we can apply Proposition 12.1 again and repeat the process. When this process stops, the following hold: + + + (a) L [x]⊂ S [x]⊂ U [x]. In particular, + + −1 − + T L (x)≡ˆπ ◦ πˆ T L (x) ⊂ T U (x). R R R x x (b) The conditional measures ν + are induced from the Haar measure on U [x]. U [x] These measures are g -equivariant. + + (c) The subspaces T U (x) ⊂ W (x) is P = AN equivariant. (This follows from the fact that the N direction is contained in U (x),(6.2) and the fact that the N direction is in the center of G (x).) The subspaces T L (x) are g -equivariant. ++ R t (d) The conditional measures ν are supported on L [x]. W [x] 1 1 Let H denote the subspace of H (M,, R) which is orthogonal to the SL(2, R) or- bit, see (2.1). Let I denote the Lyapunov exponents (with multiplicity) of the cocycle in + 1 + 1 T U (x) ∩ H , J denote the Lyapunov exponents of the cocycle in T L (x) ∩ H .By R R ⊥ ⊥ (a), we have J ⊂ I. + 1 Since T U (x) ∩ H (x) is AN-invariant, by Theorem A.3 we have, (13.1) λ ≥ 0. i∈I INVARIANT AND STATIONARY MEASURES 259 We now compute the entropy of g .Wehave, by Theorem B.9(i) (applied to the flow in the reverse direction), (13.2) h(g ,ν) ≥ 2 + (1 + λ ) = 2 +|I|+ λ ≥ 2 +|I| t i i i∈I i∈I where the 2 comes from the direction of N, and for the last estimate we used (13.1). Also, by Theorem B.9(ii), (13.3) h(g ,ν) ≤ 2 + (1 − λ ), where the 2 is the potential contribution of N −t j j∈J ≤ 2 + (1 − λ ) since (1 − λ ) ≥ 0for all i i i i∈I ≤ 2 +|I| by (13.1) However, h(g ,ν) = h(g ,ν). Therefore, all the inequalities in (13.2)and (13.3)are in fact t −t equalities. In particular, I = J, i.e. + + (13.4)T L (x) = T U (x). R R + + + Since L [x]⊂ S [x] and S [x] is closed and star-shaped with respect to x, it follows that + + (13.5)T L [x]⊂ S [x]. + + Since S [x]⊂ U [x],weget,inviewof(13.4)and (13.5)that + + + T U [x]⊂ S [x]⊂ U [x]. + + Thus U [x] is an affine subspace of W [x]. Then, in view of (13.4), and the fact that + + + + + L [x]⊂ U [x],weget that L [x]= U [x].Thus, L [x] is an affine subspace, hence − − L (x) = L (x). We have h g , W = 2 + (1 − λ ). ν −t i i∈I By applying Theorem B.9(iii) to the affine subspaces L (x), this implies that the condi- ¯ ¯ tional measures ν (x) are Lebesgue, and that ν is N-invariant (where Nis as in Sec- tion 1.1). Hence ν is SL(2, R)-invariant. − − By the definition of L , the conditional measures ν are supported on L [x]. W [x] Thus, the conditional measures ν − are (up to null sets) precisely the Lebesgue mea- W [x] sures on L [x]. + + Let U [x] denote the smallest linear subspace of W [x] which contains the support of ν + . Since ν is SL(2, R)-invariant, we can argue by symmetry that the conditional W [x] 260 ALEX ESKIN, MARYAM MIRZAKHANI + + measures ν + are precisely the Lebesgue measures on U [x]. Since U [x] accounts W [x] + + + + for all the entropy of the flow, we must have U [x]= U [x]. Since U [x]= L [x], this completes the proof of Theorem 2.1. 14. Random walks In all of Sections 14–16, we work with the finite cover X (which is a manifold), and do not use the measurable cover X. We choose a compactly supported absolutely continuous measure μ on SL(2, R). We also assume that μ is spherically symmetric. Let ν be any ergodic μ-stationary prob- ability measure on X . By Furstenberg’s theorem [NZ, Theorem 1.4], 2π ν = (r ) ν dθ θ ∗ 0 2π where r is as in Section 1.1 and ν is a measure invariant under P = AN ⊂ SL(2, R). θ 0 Then, by Theorem 2.1, ν is SL(2, R)-invariant. Therefore the stationary measure ν is also in fact SL(2, R)-invariant. We can think of x ∈ X as a point in H (M,, C). For a subspace U(x) ⊂ H (M,, R) let U = C ⊗ U(x) denote its complexification, which is a subspace of H (M,, C). In all cases we will consider, U(x) will either contain the space spanned by Re x and Im x or will be symplectically orthogonal to that space. 1 1 Let area(x, 1) ⊂ H (M,, C) denote the set of y ∈ H (M,, C) such that x + y has area 1. We often abuse notation by referring to U (x) ∩ area(1, x) also as U (x).We C C also write U [x] for the corresponding subset of X . C 0 1 1 The map p : H (M,, R) → H (M, R) naturally extends to a map (also denoted 1 1 by p)from H (M,, C) → H (M, C). By Theorem 2.1, there is a SL(2, R)-equivariant family of subspaces U(x) ⊂ H (M,, R) containing Re x and Im x and such that the conditional measures of ν along U [x] are Lebesgue. Furthermore, for almost all x, the conditional measure of ν along + + − W [x] is supported on W [x]∩ U [x], and the conditional measure of ν along W [x] is supported on W [x]∩ U [x]. Lemma 14.1. — There exists a volume form d Vol(x) on U(x) which is invariant under the SL(2, R) action. This form is non-degenerate on compact subsets of X . Proof. — The subspaces p(U(x)) form an invariant subbundle p(U) of the Hodge bundle. By Theorem A.6(a) (after passing to a finite cover) we may assume that p(U) is a direct sum of irreducible subbundles. Then, by Theorem A.6(b), we have a decomposition p(U)(x) = U (x) ⊕ U (x) symp 0 INVARIANT AND STATIONARY MEASURES 261 where the symplectic form on U is non-degenerate, the decomposition is orthogonal symp with respect to the Hodge inner product, and U is isotropic. Then, by Theorems A.5 and A.4 the Hodge inner product on U is equivariant under the SL(2, R) action. Then we can define the volume form on p(U) to be the product of the appropriate power of the symplectic form on U and the volume form induced by the Hodge inner symp product on U . The subbundle U is clearly SL(2, R) equivariant. By [Fi1, Corol- 0 symp lary 5.4], applied to the section c ∧ ··· ∧ c where {c ,..., c } is a symplectic basis for 1 k 1 k U , it follows that the symplectic volume form on U agrees with the volume form symp symp induced by the Hodge inner product on U (which is non-degenerate on compact sets). symp This gives a volume form on p(U) with the desired properties. Since the Kontsevich-Zorich cocycle acts trivially on ker p, the normalized Lebesgue measure on ker p is well defined. Thus, the volume form on p(U) naturally induces a volume form on U. Remark. — In fact it follows from the results of [AEM]that U is trivial. ⊥ 1 Lemma 14.2. — There exists an SL(2, R)-equivariant subbundle p(U) ⊂ H (M, R) such that ⊥ 1 p(U)(x) ⊕ p(U) (x) = H (M, R). Proof. — This follows from the proof of Theorem A.6. The subbundles L .— By Theorem A.6 we have (14.1) p(U) (x) = L (x), k∈ ˆ ˆ where is an indexing set not containing 0, and for each k ∈ , L is an SL(2, R)- equivariant subbundle of the Hodge bundle. (In our notation, the action of the Kontsevich-Zorich cocycle may permute some of the L .) Note that L (x) is symplec- k k tically orthogonal to the SL(2, R) orbit of x. Without loss of generality, we may assume that the decomposition (14.1) is maximal, in the sense that on any (measurable) finite cover of X each L does not contain a non-trivial proper SL(2, R)-equivariant subbun- 0 k dle. (If this was not true, we could without passing to a finite cover, write a version of (14.1)withalarger k.) If U does not contain the kernel of p,thenwelet λ = 0, and let ˜ ˆ = ∪{0}. The Forni subbundle. — Let λ denote the top Lyapunov exponent of the geodesic flow g restricted to L .Let t k F(x) = L (x). {k : λ =0} k 262 ALEX ESKIN, MARYAM MIRZAKHANI We call F(x) the Forni subspace of ν . The subspaces F(x) form a subbundle of the Hodge bundle which we call the Forni subbundle. It is an SL(2, R)-invariant subbundle, on which the Kontsevich-Zorich cocycle acts by Hodge isometries. In particular, all the Lya- punov exponents of F(x) are 0. Let F (x) denote the orthogonal complement to F(x) in the Hodge norm. By Theorem A.9(b), F (x) = L (x). {k : λ =0} The following is proved in [AEM]: Theorem 14.3. — There exists a subset of the stratum with ν() = 1 such that for all x ∈ there exists a neighborhood U(x) such that for all y ∈ U(x) ∩ we have p(y − x) ∈ F (x). The backwards shift map. — Let B be the space of (one-sided) infinite sequences of elements of SL(2, R). (We think of B as giving the “past” trajectory of the random walk.) Let T : B → B be the shift map. (In our interpretation, T takes us one step into the past.) We define the skew-product map T : B × X → B × X by 0 0 −1 T(b, x) = Tb, b x , where b = (b , b ,...) 0 1 (Thus the shift map and the skew-product map are denoted by the same letter.) We define the measure β on B to be μ × μ··· . The skew product map T naturally acts on the bundle H (M, R),and thus on each L for k ∈ . For each k ∈ , by the multiplicative ergodic theorem we have the Lyapunov flag for this action (with respect to the invariant measure β ): (k) (k) (k) {0}= V ⊂ V (b, x) ⊂··· V (b, x) = L (x). ≤0 ≤1 ≤n By the multiplicative ergodic theorem applied to the action of SL(2, R) on R ,for β -almost all b ∈ B, σ = lim log b ... b 0 0 n n→∞ where σ > 0 is the Lyapunov exponent for the measure μ on SL(2, R). Then, the Lya- punov exponents of the flow g and the Lyapunov exponents of the skew-product map T differ by a factor of σ .Let λ denote the top Lyapunov exponent of T restricted to L . 0 k k The two-sided shift space. — Let B denote the two-sided shift space. We denote the measure ··· × μ × μ ×··· on Balso by β . INVARIANT AND STATIONARY MEASURES 263 Notation. — For a, b ∈ Blet (14.2) a ∨ b = (..., a , a , b , b ,...) ∈ B. 2 1 0 1 (Note that the indexing for a ∈ Bstarts at1not at0.) If ω = a ∨ b ∈ B, we think of the sequence ...,ω ,ω = ··· a , a −2 −1 2 1 as the “future” of the random walk trajectory. (In general, following [BQ], we use the symbols b, b etc. to refer to the “past” and the symbols a, a etc. to refer to the “future”.) The opposite Lyapunov flag. — Note that on the two-sided shift space B × X ,the map −1 T is invertible. Thus, for each a ∨ b ∈ B, we have the Lyapunov flag for T : (k) (k) (k) {0}= V ⊂ V (a, x) ⊂ ··· V (a, x) = L (x). ≥n −1 ≥0 ≥n k k (As reflected in the above notation, this flag depends only on the “future” i.e. “a”partof a ∨ b.) ˆ ˆ The top Lyapunov exponent λ .— Recall that λ ≥ 0 denotes the top Lyapunov expo- k k (k) nent in L . Then (since T steps into the past), for v ∈ V (b, x), ≤1 1 T (b, x) v (14.3)lim log =−λ . n→∞ n v n n In the above equation we used the notation T (b, x) to denote the action of T (b, x) on H (M, R). (k) Also, for v ∈ V (a, x),for some α> 0, >1 −n 1 T (a ∨ b, x) v lim log < λ − α. n→∞ n v Here, α is the minimum over k of the difference between the top Lyapunov exponent in L and the next Lyapunov exponent. The following lemma is a consequence of the zero-one law Lemma C.10(i): Lemma 14.4. —For every δ> 0 and every δ > 0 there exists E ⊂ X with ν(E )> good 0 good 1 − δ and σ = σ(δ,δ )> 0, such that for any x ∈ E , any k and any vector w ∈ P(L (x)), good k (k) (14.4) β a ∈ B : d w, V (a, x) >σ > 1 − δ >1 264 ALEX ESKIN, MARYAM MIRZAKHANI (In (14.4), d (·,·) is the distance on the projective space P(H (M, R)) derived from the AGY norm.) Proof. — It is enough to prove the lemma for a fixed k.For F ⊂ Gr (L (x)) (the n −1 k Grassmannian of n − 1 dimensional subspaces of L (x))let k k (k) (k) νˆ (F) = β a ∈ B : V (a, x) ∈ F , x >1 (k) and let νˆ denote the measure on the bundle X × Gr (L ) given by 0 n −1 k (k) (k) dνˆ (x, L) = dν(x)dνˆ (L). (k) Then, νˆ is a stationary measure for the (forward) random walk. For w ∈ P(L (x)) let I(w) ={L ∈ Gr (L (x)) : w ∈ L}.Let n −1 k (k) Z = x ∈ X :ˆν I(w) > 0for some w ∈ P L (x) . 0 k Suppose ν(Z)> 0. Then, for each x ∈ Z we can choose w ∈ P(L (x)) such that x k (k) νˆ (I(w )) > 0. Then, (k) (14.5) νˆ {x}× I(w ) > 0. x∈Z (k) Therefore, (14.5) holds for some ergodic component of νˆ . However, this contradicts Lemma C.10(i), since by the definition of L , the action of the cocycle on L is strongly k k c c irreducible. Thus, ν(Z) = 0and ν(Z ) = 1. By definition, for all x ∈ Z and all w ∈ L (x), (k) β a ∈ B : w ∈ V (a, x) = 0. >1 Fix x ∈ Z . Then, for every w ∈ P(L (x)) there exists σ (x,w,δ )> 0such that k 0 (k) β a ∈ B : d V (a, x),w > 2σ x,w,δ > 1 − δ . Y 0 >1 Let U(x,w) ={z ∈ P(L (x)) : d (z,w) < σ (x,w,δ )}. Then the {U(x,w)} k Y 0 w∈P(L (x)) form an open cover of the compact space P(L (x)), and therefore there exist w ,...w k 1 n with P(L (x)) = U(x,w ).Let σ (x,δ ) = min σ (x,w ,δ ). Then, for all x ∈ Z and k i 1 i 0 i i=1 all w ∈ P(L (x)), (k) β a ∈ B : d V (a, x),w >σ x,δ > 1 − δ . Y 1 >1 c 1 c c Let E (δ ) ={x ∈ Z : σ (x,δ )> }. Since E (δ ) = Z and ν(Z ) = 1, there N 1 N N=1 exists N = N(δ,δ ) such that ν(E (δ )) > 1 − δ.Let σ = 1/Nand letE = E . N good N INVARIANT AND STATIONARY MEASURES 265 Lyapunov subspaces and relative homology. — The following lemma is well known: Lemma 14.5. — The Lyapunov spectrum of the Kontsevich-Zorich cocycle acting on relative homology is the Lyapunov spectrum of the Kontsevich-Zorich cocycle acting on absolute homology, union n zeroes, where n = dim ker p. −1 1 Let L = p (L ) ⊂ H (M,, R). We have the Lyapunov flag k k (k) (k) (k) ¯ ¯ ¯ ¯ {0}= V ⊂ V (b, x) ⊂··· V (b, x) = L (x), ≤0 ≤1 ≤¯ n corresponding to the action on the invariant subspace L ⊂ H (M,, R). Also for each a ∈ B, we have the opposite Lyapunov flag (k) (k) (k) ¯ ¯ ¯ ¯ {0}= V ⊂ V (a, x) ⊂ ··· V (a, x) = L (x). ≥¯ n ≥¯ n −1 ≥0 k k Lemma 14.6. — Suppose λ = 0. Then for almost all (b, x), (k) (k) p V (b, x) = V (b, x), ≤1 ≤1 and p is an isomorphism between these two subspaces. Similarly, for almost all (a, x), (k) −1 (k) V (a, x) = p V (a, x) . >1 >1 ˆ ˆ Proof. — In view of Lemma 14.5 and the assumption that λ = 0, λ is the top k k Lyapunov exponent on both L and L .Notethat k k 1 T v¯ (k) ¯ ¯ (14.6) V = v¯ ∈ L : lim sup log ≤−λ . k k ≤1 t ¯v t→∞ Also, 1 T v (k) (14.7) V = v ∈ L : lim sup log ≤−λ . k k ≤1 t v t→∞ It is clear from the definition of the Hodge norm on relative cohomology (A.1)that p(v)≤ Cv for some absolute constant C. Therefore, it follows from (14.7)and (k) (k) (k) (k) ¯ ¯ (14.6)that p(V ) ⊂ V . But by Lemma 14.5,dim(V ) = dim(V ). Therefore, ≤1 ≤1 ≤1 ≤1 (k) (k) p(V ) = V . ≤1 ≤1 Remark. — Even though we will not use this, a version of Lemma 14.6 holds for all Lyapunov subspaces for non-zero exponents, and not just the subspace corresponding to the top Lyapunov exponent λ . k 266 ALEX ESKIN, MARYAM MIRZAKHANI The action on H (M,, C).— By the multiplicative ergodic theorem applied to the action of SL(2, R) on R ,for β -almost all b ∈ B there exists a one-dimensional subspace W (b) ⊂ R such that v ∈ W (b), + + −1 −1 lim log b ... b v=−σ . n 0 n→∞ Let + 1 W (b, x) = W (b) ⊗ H (M,, R) ∩ area(x, 1). 2 1 1 + Since we identify R ⊗ H (M,, R) with H (M,, C), we may consider W (b, x) as a subspace of H (M,, C). This is the “stable” subspace for T. (Recall that T moves into the past.) For a “future trajectory” a ∈ B, we can similarly define a 1-dimensional subspace W (a) ⊂ R such that lim log a ... a v=−σ for v ∈ W (a). n 1 0 − n→∞ 1 1 Let A : SL(2, R) × X → Hom(H (M,, R), H (M,, R)) denote the Kontsevich- Zorich cocycle. We then have the cocycle 1 1 A : SL(2, R) × X → Hom H (M,, C), H (M,, C) given by A(g, x)(v ⊗ w) = gv ⊗ A(g, x)w 1 2 1 andwehavemadethe identification H (M,, C) = R ⊗ H (M,, R). This cocycle can be thought of as the derivative cocycle for the action of SL(2, R). From the definition we see that the Lyapunov exponents of A are of the form ±σ + λ ,where the λ are the 0 i i Lyapunov exponents of A. 15. Time changes and suspensions There is a natural “forgetful” map f : B → B. We extend functions on B × X to B× X by making them constant along the fibers of f .The measure β ×ν is a T-invariant measure on B × X . The cocycles θ .— By Theorem A.6, the restriction of the Kontsevich-Zorich cocycle to each L is semisimple. Then by Theorem C.5, the Lyapunov spectrum of T on each L j j INVARIANT AND STATIONARY MEASURES 267 is semisimple, and the restriction of T to the top Lyapunov subspace of each L consists of a single conformal block. This means that there is a inner product , defined on j,b,x (j) W (b) ⊗ V (b, x) and a function θ : B × X → R such that for all u,v ∈ W (b) ⊗ + j 0 + ≤1 (j) V (b, x), ≤1 −1 −1 −θ (b,x) ˆ ˆ (15.1) A b , x u, A b , x v = e u,v . −1 j,b,x 0 0 j,Tb,b x To handle relative homology, we need to also consider the case in which the action of A(·,·) on a subbundle is trivial. We thus define an inner product , on R ,and a 0,b cocycle θ : B → R so that for u,v ∈ W (b), 0 + −1 −1 −θ (b) (15.2) b u, b v = e u,v . 0,b 0 0 0,Tb For notational simplicity, we let θ (b, x) = θ (b). 0 0 Switch to positive cocycles. — The cocycle θ corresponds to the A(·,·)-Lyapunov ex- ˆ ˆ ponent σ + λ ,where λ is the top Lyapunov exponent of A(·,·) in L . Since σ > 0and 0 j j j 0 λ ≥ 0, σ + λ = θ (b, x)dβ(b)dν(x)> 0. 0 j j B×X Thus, the cocycle θ has positive average on B × X . However, we do not know that θ is j 0 j positive, i.e. that for all (b, x) ∈ B × X , θ (b, x)> 0. This makes it awkward to use θ (b, x) 0 j j to define a time change. Following [BQ] we use a positive cocycle τ equivalent to θ . j j By [BQ, Lemma 2.1], we can find a positive cocycle τ : B × X → R and a measur- j 0 able function φ : B × X → R such that j 0 (15.3) θ − φ ◦ T + φ = τ j j j j and τ (b, x)dβ(b)dν(x)< ∞. B×X (j) For v ∈ W (b) ⊗ V (b, x) we define ≤1 φ (b,x) (15.4) v = e v , j,b,x j,b,x where the norm ·,· is as in (15.1)and (15.2). Then −1 −τ (b,x) (15.5) A b , x v = e v . 0 j,b,x j,T(b,x) 268 ALEX ESKIN, MARYAM MIRZAKHANI Suspension. — Let B = B × X × (0, 1].Recallthat β denotes the measure on B X X which is given by μ × μ···.Let β denote the measure on B given by β × ν × dt, where dt is the Lebesgue measure on (0, 1].In B we identify (b, x, 0) with (T(b, x), 1), X X X so that B is a suspension of T. We can then define a suspension flow T : B → B in the natural way. (Our suspensions are going downwards and not upwards, since we think of T as going into the past.) Then T preserves the measure β . ˜ ˜ Let B = B × X × (0, 1]. The suspension construction, the flow T , and the in- 0 t X X X variant measure β extend naturally from B to B . Let T (b, x, s) denote the action of T (b, x, s) on H (M,, C) (i.e. the derivative t ∗ t (j) cocycle on the tangent space). Then, for t ∈ Z and v ∈ W (b) ⊗ V (b, x) and 0 < s ≤ 1 ≤1 we have,inviewof(15.5), −τ (t,b,x) (15.6) T (b, x, s) v = e v , t ∗ j,T (b,x) j,b,x t−1 n X where τ (t, b, x) = τ (T (b, x)). We can extend the norm · from B × X to B by j j 0 n=0 j −(1−s)τ (b,x) v =v e . j,b,x,s j,b,x Then (15.6) holds for all t ∈ R provided we set for n ∈ Z and 0 ≤ s < 1, τ (n + s, b, x) = τ (n, b, x) + sτ T (b, x) . j j j The time change. — Here we differ slightly from [BQ] since we would like to have several different time changes of the flow T on the same space. Hence, instead of chang- ing the roof function, we keep the roof function constant, but change the speed in which one moves on the [0, 1] fibers. X X Let T : B → B be the time change of T where on (b, x) ×[0, 1] one moves at t t the speed 1/τ (b, x). More precisely, we set (15.7)T (b, x, s) = b, x, s − t/τ (b, x) , if 0 < s − t/τ (b, x) ≤ 1, t j j and extend using the identification ((b, x), 0) = (T(b, x), 1). Then T is the operation of moving backwards in time far enough so that the cocycle multiplies the direction of the top Lyapunov exponent in L by e .Infact, by (k) (15.6)and (15.7), we have, for v ∈ W (b) ⊗ V (b, x), ≤1 k − (15.8) T (b, x, s) v τ = e v . ∗ k j,b,x,s j,T (b,x,s) τ X τ k k The map T and the two-sided shift space. — On the space B ,T is invertible, and τ τ k k we denote the inverse of T by T . We write (15.9)T (a ∨ b, x, s) − INVARIANT AND STATIONARY MEASURES 269 1 k for the linear map on the tangent space H (M,, C) induced by T (a ∨ b, x, s).Inview (k) of (15.4)and (15.8), we have for v ∈ W (b) ⊗ V (b, x), ≤1 τ τ k k (15.10) T (a ∨ b, x, s) v= exp + φ (b, x, s) − φ T (a ∨ b, x, s) v. ∗ k k − − Here we have omitted the subscripts on the norm · and also extended the function k,b,x (k) φ (b, x, s) so that for all (b, x, s) ∈ B and all v ∈ W (b) ⊗ V (b, x), k + ≤1 φ (b,x,s) v = e v . k,b,x k,b,x,s τ ,X X Invariant measures for the time changed flows. — Let β denote the measure on B given by τ ,X dβ (b, x, t) = c τ (b, x)dβ(b)dν(x)dt, j j τ ,X X τ ,X j j where the c ∈ R is chosen so that β (B ) = 1. Then the measures β are invariant under the flows T . We note the following trivial: τ ,X X Lemma 15.1. —The measures β are all absolutely continuous with respect to β .For every δ> 0 there exists a compact subset K = K(δ) ⊂ B and L = L(δ) < ∞ such that for all j , τ ,X β (K)> 1 − δ, and also for (b, x, t) ∈ K, τ ,X X dβ dβ (b, x, t) ≤ L, (b, x, t) ≤ L. X τ ,X dβ dβ 16. The martingale convergence argument Standing assumptions. — Let + + W [b, x]= y : y − x ∈ W (b, x). Then, W [b, x] is the stable subspace for T. From the definition, for almost all b, (locally) the sets {W [b, x]: x ∈ X} form a measurable partition of X .Let + + + + U (b, x) = W (b, x) ∩ U (x), U [b, x]= W [b, x]∩ U [x]. C C − − + − We make the corresponding definitions for W (b, x),W [b, x],U [b, x] and U [b, x]. It follows from Theorem 2.1 applied to the flow r g r , using the fact that θ t −θ U [r x]= U [x],thatfor a.e. x, the conditional measures of ν along W [b, x] are sup- C θ C ± ± ported on U [b, x], and also that the conditional measures of ν along U [b, x] are Lebesgue. 270 ALEX ESKIN, MARYAM MIRZAKHANI X X Lemma 16.1. — There exists a subset ⊂ B with β () = 1 such that for all (b, x) ∈ , + + ∩ W [b, x]∩ ball of radius 1 ⊂ ∩ U [b, x]. Proof.—See[MaT]or[EL, 6.23]. The parameter δ.— Let δ> 0 be a parameter which will eventually be chosen suffi- ciently small. We use the notation c (δ) and c (δ) for functions which tend to 0 as δ → 0. In this section we use the notation A ≈ B to mean that the ratio A/B is bounded between two positive constants depending on δ. We first choose a compact subset K ⊂ ∩ with β (K )> 1 −δ> 0.999, the 0 0 conull set is as in Lemma 16.1, and the conull set is as in Theorem 14.3.Bythe multiplicative ergodic theorem and (14.3), we may also assume that there exists (δ) > 0 (k) such that for all (b, x, s) ∈ K all k and all v ∈ V (b, x) and all > (δ), 0 1 ≤1 −(λ /2) (16.1) T (b, x, s) v≤ e v. (Here, as in (14.3) the notation T (b, x, s) denotes the action on H (M,, R).) By the norm · in this section, we mean the AGY norm (see Section A.1). Lemma 16.2. —For every δ> 0 there exists K ⊂ B and C = C(δ) < ∞, β = β(δ) > 0 and C = C (δ) < ∞ such that (K1) For all L > C (δ), and all (b, x, s) ∈ K, χ T (b, x, s) dt ≥ 0.99. K t X τ ,X (K2) β (K)> 1 − c (δ). Also, for all j , β (K)> 1 − c (δ). 1 1 (K3) For all j and all (b, x, t) ∈ K, |φ (b, x, t)| < C,where φ is as in (15.3). j j (k) (K4) For all j , all (b, x, t) ∈ K all k = 0 and all v ∈ V (b, x), ≤1 (16.2) p(v)≥ β(δ)v. (K5) There exists C = C (δ) such that for all (b, x, s) ∈ K all j and all v ∈ W (b) ⊗ 0 0 + (j) −1 V (b, x), we have C v≤v ≤ C v. j,b,x 0 ≤1 0 Proof. — By the Birkhoff ergodic theorem, there exists K ⊂ B such that X X β (K )> 1 − δ/5 and (K1) holds for K instead of K. We can choose K ⊂ B and C = C(δ) < ∞ such that β (K )> 1 − δ/5 and (K3) holds for K instead of K. Let K = K(δ/5) and L = L(δ/5) be as in Lemma 15.1 with δ/5 instead of δ. Then choose τ ,X K ⊂ with β (K )> 1 − δ/(5d L),where d is the number of Lyapunov exponents. j j In view of Lemma 14.6 there exists K ⊂ X with β (K )> 1 − δ/5so that (16.2) 0 INVARIANT AND STATIONARY MEASURES 271 holds. Similarly, there exists a set K with K > 1 − δ/5 where (K5) holds. Then, let K = K ∩ K ∩ K ∩ K ∩ K ∩ K . The properties (K1), (K2), (K3) and (K4) are easily verified. War ning. — In the rest of this section, we will often identify K and K with their −1 X −1 X X X ˜ ˜ ˜ pullbacks f (K) ⊂ B and f (K ) ⊂ B where f : B → B is the forgetful map. τ ,X τ ,X j j The martingale convergence theorem. — Let B denote the σ -algebra of β measur- able functions on B .Asin[BQ], let τ ,X τ −1 j j τ ,X Q = T B . τ ,X (Thus if a function F is measurable with respect to Q , then F depends only on what happened at least time units in the past, where is measured using the time change τ .) Let τ ,X τ ,X j j Q = Q . >0 τ ,X The Q are a decreasing family of σ -algebras, and then, by the Martingale Conver- τ ,X X gence Theorem, for β -almost all (b, x, s) ∈ B , τ ,X τ ,X j j (16.3)lim E 1 | Q (b, x, s) = E 1 | Q (b, x, s) j K j K ∞ →∞ τ ,X where E denotes expectation with respect to the measure β . The set S .— In view of (16.3) and the condition (K2) we can choose S = S (δ) ⊂ B to be such that for all > ,all j,and all (b, x, s) ∈ S , τ ,X (16.4) E 1 | Q (b, x, s)> 1 − c (δ). j K 2 By using Lemma 15.1 as in the proof of Lemma 16.2 we may assume that (by possibly making larger)wehavefor all j , τ ,X (16.5) β S > 1 − c (δ). The set E .— By Lemma 14.4 we may choose a subset E ⊂ B (which is actu- good good ally of the form B × E for some subset E ⊂ X ×[0, 1]),with β (E )> 1 − c (δ), good 3 good good and a number σ(δ) > 0such that for any (b, x, s) ∈ E ,any j and any unit vector good w ∈ L (b, x), (j) (16.6) β a ∈ B : d w, V (a, x) >σ(δ) > 1 − c (δ). >1 3 272 ALEX ESKIN, MARYAM MIRZAKHANI We may assume that E ⊂ K. By the Osceledets multiplicative ergodic theorem and good Lemma 14.6, we may also assume that there exists α> 0 (depending only on the Lya- punov spectrum), and = (δ) such that for (b, x, s) ∈ E , > , at least 1 − c (δ) 0 0 good 0 (j) measure of a ∈ B, and all v¯ ∈ V (a, x), >1 (1−α) (16.7) T (a ∨ b, x, s) v¯≤ e ¯v. The sets .— In view of (16.5) and the Birkhoff ergodic theorem, for every ρ> 0 there exists a set = (δ) ⊂ B such that ρ ρ (1) β ( )> 1 − ρ . (2) There exists = (ρ) such that for all > ,and all (b, x, s) ∈ , 0 0 0 t ∈[−,]: T (b, x, s) ∈ S ∩ E ≥ 1 − c (δ) 2. t good 5 Lemma 16.3. — Suppose the measure ν is not affine. Then there exists ρ> 0 so that for every δ > 0 there exist (b, x, s) ∈ , (b, y, s) ∈ with y − x <δ such that p(y − x) ∈ p(U) (x), ρ ρ (16.8) d y − x, U (x) > y − x and (16.9) d y − x, W (b, x) > y − x (so y − x is in general position with respect to W (b, x)). Remark. — In view of Theorem 14.3, it follows that for (b, x, s), (b, y, s) satisfying the conditions of Lemma 16.3, p(y − x) is orthogonal to the complexification F (x) of the Forni subspace F(x). 1/2 Proof. — By Fubini’s theorem, there exists a subset ⊂ Xwith ν( ) ≥ 1 − ρ ρ ρ such that for x ∈ , 1/2 (16.10) (β × dt) (b, s) : (b, x, s) ∈ ≥ 1 − ρ . Let K be an arbitrary compact subset of X with ν(K)> 1/2, and let K denote its lift ˜ ˜ to X .Let π : X → X denote the natural map. We have 0 0 0 1/2 (16.11) ν ≥ 1 − 2ρ ν(K). In view of Lemma 14.1 we can find finitely many sets J ⊂ K ⊂ X and constants α α 0 N > 0and δ > 0 such that the following hold: 0 INVARIANT AND STATIONARY MEASURES 273 (i) For all α,K is diffeomorphic to an open ball, and the restriction of π to K α α is injective. (ii) The sets J are disjoint, and up to a null set π(K) = π( J ). α α (iii) Any point belongs to at most N of the sets π(K ). (iv) Recall that for x ∈ X ,U [x] denotes the (infinite) affine space whose tangent 0 C space is U (x).Wehave, for ν -almost all x ∈ J , C α (16.12)Vol U [x]∩ K ≥ δ , C α 0 where Vol(·) is as in Lemma 14.1. Let 1/4 (16.13) = x ∈ J : ν ∩ K ≥ 1 − ρ ν (K ) . α U (x) α U (x) α ρ C ρ C In the above equation, ν is the conditional measure of ν along U [x] (whichisinfact U (x) C a multiple of the measure Vol of Lemma 14.1). By (16.11), properties (ii), (iii) and Fubini’s 1/4 theorem, ν( ) ≥ (1 − 2Nρ )ν(K). In particular, is conull in K. ρ ρ>0 ρ Note that by the definition of ,if x ∈ ∩ J then U [x]∩ J ⊂ . It follows α C α ρ ρ ρ that we may write, for some indexing set I (ρ), ∩ J = U [x]∩ J . α C α x∈I (ρ) Suppose that for all α and all ρ> 0, I (ρ) is countable. Then, for a positive measure set of x ∈ X , x has an open neighborhood in U [x] whose ν -measure is positive. Then 0 C by ergodicity of the geodesic flow, this holds for ν -almost all x ∈ X and without loss of generality, for all x ∈ I (ρ). The restriction of ν to U [x] is a multiple of the measure Vol of Lemma 14.1, therefore there exists a constant ψ(x) = 0such that for E ⊂ U [x], ν(E) = ψ(x) Vol(E). Since both ν and Vol are invariant under the SL(2, R) action, ψ(x) is invariant, and thus by ergodicity ψ is constant almost everywhere. Let I = I (ρ).For x, y ∈ I write x ∼ y if U [x]∩ J = U [ y]∩ J ,and let α C α C α α ρ>0 α I ⊂ I be the subset where we keep only one member of each ∼-equivalence class. Note α α that by properties (i) and (iv), for distinct x, y ∈ I ,U [x]∩ K and U [ y]∩ K are disjoint C α C α up to a set of measure 0. Then (16.12) implies that for each α, ν(K ) ≥ ν U [x]∩ K = ψ Vol U [x]∩ K ≥ ψδ |I |, α C α C α 0 x∈I x∈I α α where |·| denotes the cardinality of a set. Since ν is a finite measure, we get that each I is finite. Since for a fixed K, there are only finitely many sets K , this implies that the support of restriction of ν to K is contained in a finite union of “affine pieces” each of the form U [x ]∩ K for some x ∈ K, and the measure ν restricted to each affine piece C j α j 274 ALEX ESKIN, MARYAM MIRZAKHANI coincides with ψ Vol. It follows from the ergodicity of g that the affine pieces fit together to form an (immersed) submanifold. Thus, ν is affine. Thus, we may assume that there exist α and ρ> 0such thatI (ρ) is not countable. Then we can find x ∈ I (ρ) and y ∈ I (ρ) such that 1 α n α lim hd U [x ]∩ K , U [ y ]∩ K = 0, C 1 α C n α n→∞ where hd denotes Hausdorff distance between sets (using the distance d defined in Sec- tion 3). Let f : p(U) [ y ]→ p(U) [x ] denote the function taking z ∈ p(U) [ y ] to the n C n C 1 C n unique point in p(U) [x ]∩ p(U) [z]. Then, for large n,the map f is almost measure C 1 n preserving, in the sense that for V ⊂ p(U) (y ), C n (0.5)|V|≤|f (V)|≤ 2|V|, where |·| denotes Lebesgue measure. Then, in view of the definition (16.13)of ,for sufficiently large n, there exist x ∈ U [x ]∩ and y ∈ U [ y ]∩ such that p(y − x) ∈ C 1 C n ρ ρ p(U) (x),and y − x <δ . Then, by the definition (16.10)of , we can choose (b, s) C ρ so that (b, x, s) ∈ , (b, y, s) ∈ ,and (16.8)and (16.9) holds. ρ ρ Standing assumption. — We fix ρ = ρ(δ) so that Lemma 16.3 holds. The main part of the proof is the following: Proposition 16.4. — There exists C(δ) > 1 such that the following holds: Suppose for every δ > 0 there exist (b, x, s),(b, y, s) ∈ with x − y≤ δ ,p(x − y) ∈ p(U) (x), and so that (16.8) and (16.9) hold. Then for every > 0 there exist (b , x , s ) ∈ K , (b , y , s ) ∈ K ,such 0 0 that y − x ∈ U (x ), ≤y − x ≤ C(δ), C(δ) (16.14) d y − x , U x ≥ y − x , C(δ) (16.15) d y − x , W b , x <δ , where δ depends only on δ ,and δ → 0 as δ → 0. ˜ ˆ ˆ Proof.—Let ⊂ denote the subset {k : λ = 0}.Wemay decompose (16.16) p(U) (x) = L (x) F(x) k∈ as in Section 14.For j ∈ ,let π denote the projection to L , using the decomposition j j (16.16). Note that by Theorem 14.3, the projection of p(y − x) to F(x) is always 0. INVARIANT AND STATIONARY MEASURES 275 FIG. 7. — Proof of Proposition 16.4. In the figure, going “up” corresponds to the “future”. The map T for m > 0takes one m steps into the “past” For m ∈ R , write (see Figure 7) b , x , s = T (b, x, s), b , y , s = T (b, y, s), m m and let w (m) = π x − y . j j (We will always have m small enough so that the above equation makes sense.) Let (m) be such that (m) e w (m)= . We also need to handle the relative homology part (where the action of the Kontsevich- Zorich cocycle is trivial). Set (m) to be the number such that (m) e x − y = . Choose 0 <σ λ where 0 <λ = min λ . We will be choosing m so that min min ˜ j j∈ (16.17) | log y − x| ≤ m ≤ σ | log y − x|. In view of (16.9)and Theorem A.1, (after some uniformly bounded time), w (m) is an −t increasing function of m (since the factor of e from the geodesic flow beats the contribu- tion of the Kontsevich-Zorich cocycle). Therefore, (m) is a decreasing function of m. For a bi-infinite sequence b ∈ Band x ∈ X ,let G (b, x, s) = m ∈ R : T T (b, x, s) ∈ S . j + m − (m) Let G (b, x, s) = G (b, x, s) ∩{m : T (b, x, s) ∈ E }. all j m good j 276 ALEX ESKIN, MARYAM MIRZAKHANI Lemma 16.5. —For (b, x, s) ∈ and N sufficiently large, |G (b, x, s) ∩[0, N]| all ≥ 1 − c (δ). Proof. — We can write T T = T . By definition, m −g (m) − (m) j m ∈ G (b, x, s) if and only if T (b, x, s) ∈ S . j −g (m) Since (m) is a decreasing function of m,sois g , and in fact, for all m > m j j 2 1 g (m ) − g (m )> m − m . j 1 j 2 2 1 This implies that −1 −1 (16.18) g (m ) − g (m )< m − m . 1 2 1 2 j j Let F ={t ∈[0, g (N)]: T (b, x)/ ∈ S }. By condition (2), for N large enough, |F|≤ j −t c −1 (1 − c (δ))g (N).Notethat G ∩[0, N]= g (F). Then, by (16.18), 5 j j j c −1 |G ∩[0, N]| = |g (F)|≤|F|≤ c (δ)g (N) ≤ c (δ)N, 5 j 6 j j where as in our convention c (δ) → 0as δ → 0. We now continue the proof of Proposition 16.4. We may assume that δ is small enough so that the right-hand-side of (16.17) is smaller then the N of Lemma 16.5. Sup- pose (b, x, s) ∈ , (b, y, s) ∈ . By Lemma 16.5,wecan fix m ∈ G (x) such that (16.17) ρ ρ all holds. Write = (m).Let j j b , x , s = T (b, x, s), b , y , s = T (b, y, s). m m For j ∈ ,let τ τ j j (b , x , s ) = T b , x , s,(b , y ,¯ s ) = T b , y , s . j j j j j j − (m) − (m) j j Since m ∈ G (b, x, s),wehave (b , x , s ) ∈ S , (b , y ,¯ s ) ∈ S . Then, by (16.4), for all j , all j j j j j j τ ,X E 1 | Q (b , x , s )> 1 − c (δ) , j K j j j 2 τ ,X E 1 | Q (b , y ,¯ s )> 1 − c (δ) . j K j j j 2 Since T (b , x , s ) = (b , x , s ),by[BQ, (7.5)] we have j j j τ ,X τ j j E 1 | Q (b , x , s ) = 1 T a ∨ b , x , s dβ(a), j K j j j K j j B INVARIANT AND STATIONARY MEASURES 277 where the notation a ∨ b is as in (14.2). Thus, for all j ∈ , (16.19) β a : T a ∨ b , x , s ∈ K > 1 − c (δ). Similarly, for all j ∈ , β a : T a ∨ b , y , s ∈ K > 1 − c (δ). Let w = x − y ,and let w = π (w). We can write j j (16.20) w=¯ w + w¯ 0 j j∈ where w¯ ∈ ker p,and for j > 0, w¯ are chosen so that π (w¯ ) = w ,and also ¯ w ≈w . 0 j j j j j j For any a ∈ B, we may write w = ξ (a) + v (a), j j j (j) where ξ (a) ∈ W (b ) ⊗ V (b , x ),and j + ≤1 (j) v (a) ∈ W (b) ⊗ V a, x + W (a) ⊗ L b , x . j + − j >1 This decomposition is motivated as follows: if we consider the Lyapunov decomposition C ⊗ L (x) = V (a ∨ b, x) j k then ξ (a) belongs to the subspace V (a ∨ b, x) corresponding to the top Lyapunov ex- j ≤1 ponent σ + λ for the action of T ,and v ∈⊕ V (a ∨ b, x) will grow with a smaller 0 j −t j k≥2 k Lyapunov exponent under T .Then v (a) will also grow with a smaller Lyapunov expo- −t j nent then ξ (a) under T . Since m ∈ G (b, x, s),wehave (b , x , s ) ∈ E . Then, by (16.6), for at least 1 − all good c (δ) fraction of a ∈ B, (16.21) v (a)≈ξ (a)≈w ≈ e , j j j where the notation A ≈ Bmeans that A/B is bounded between two constants depending only on δ. Since (b , x , s ) ∈ E ⊂ K, by condition (K3) we have |φ (b , x , s )|≤ C(δ). good j Also by (16.19), for at least 1 − c (δ) fraction of a ∈ B, we have T (a ∨ b , x , s ) ∈ K, so again by condition (K3) we have φ T a ∨ b , x , s ≤ C(δ). Thus, by (16.21), (15.10)and (16.7), we have, for all j ∈ ,and at least 1 − c (δ) fraction of a ∈ B, τ τ j j −α (16.22) T a ∨ b , x , s ξ (a) ≈ , and T a ∨ b , x , s v (a) = O e , j j − − j ∗ j ∗ 278 ALEX ESKIN, MARYAM MIRZAKHANI where α> 0 depends only on the Lyapunov spectrum. (The notation in (16.22)isdefined in (15.9).) Hence, for at least 1 − c (δ) fraction of a ∈ B, T a ∨ b , x , s w ≈ . Since λ ≥ 0(and by Theorem 14.3,if λ = 0then j = 0, and w¯ ∈ ker p where the action j j 0 of the Kontsevich-Zorich cocycle is trivial), we have for at least 1 − c (δ) fraction of a ∈ B, (16.23) T a ∨ b , x , s w¯ ≈ . Let t (a) = sup t > 0 : T a ∨ b , x , s w¯ ≤ , j −t j ˜ ˜ and let j(a) denote a j ∈ such that t (a) is as small as possible as j varies over . Then, if j = j(a),thenby(16.23), (16.24) T a ∨ b , x , s w¯ ≈ T a ∨ b , x , s w¯ ≈ . −t (a) j j j − ∗ j ∗ Also, for at least 1 − c (δ)-fraction of a ∈ B, if j = j(a) and k = j,thenby(16.23), (16.25) T a ∨ b , x , s w¯ ≤ C (δ), k 1 j ∗ where C (δ) depends only on δ. Therefore, by (16.20), (16.24), and (16.25), for at least 1 − c (δ)-fraction of a ∈ B, if j = j(a), (16.26) T a ∨ b , x , s y − x ≈ . j ∗ We now choose δ> 0so that c (δ) + 2c (δ) < 1/2, and using (16.19) we choose a ∈ Bso 4 2 that (16.26) holds, and also τ τ j j T a ∨ b , x , s ∈ K, T a ∨ b , y , s ∈ K. − − j j We may write T a ∨ b , x , s = T a ∨ b, x , s , −t T a ∨ b , y , s = T a ∨ b, y , s −t Then, |t − t|≤ C(δ). Therefore by condition (K1), there exists t with |t − t|≤ C(δ) such that b , x , s = T a ∨ b , x , s ∈ K , −t 0 b , y , s = T a ∨ b , y , s ∈ K . −t 0 Since w≈ e ,and →∞ as δ → 0, we have w=x − y → 0as − − δ → 0. Since T does not expand the W components, the W component of x − y is −t INVARIANT AND STATIONARY MEASURES 279 − − bounded by the W component of x − y . Thus, the size of the W component of x − y tends to0as δ → 0. Thus (16.15) holds. It remains to prove (16.14). If (16.27) p y − x ≥ y − x C(δ) then (16.14) holds since p(y − x ) ∈ p(U) (x ). This automatically holds for the case where ||= 1 (and thus, in particular, there are no marked points). If not, we may write y − x = w +¯ w + 0 where w ≤ c(δ)¯ w and w¯ ∈ ker p. We will need to rule out the case where w¯ is + 0 0 0 very close to U (x ) ∩ ker p. We will show that this contradicts the assumption (16.8). Let w , w¯ be such that + 0 w = T a ∨ b, x , s w , w¯ = T a ∨ b, x , s w¯ . −t −t + + 0 0 ∗ ∗ Then y − x = w +¯ w andinviewof(16.1)and (16.21), + 0 −λ t /2 −λ t /2 min min w ≤ e ¯ w ≈ e y − x . + 0 Applying T (b, x , s ) to both sides we get −m y − x = w +¯ w , + 0 where w¯ ∈ ker p,and λ t min 2m 2m− w ≤ e w ≤ e x − y. λ t λ t min min By (16.17), 2m − ≤− .Thus, w ≤ (1/100)y − x. Therefore, by (16.8), we 2 4 have d w¯ , ker p ∩ U (x) > w . 0 C 0 Since the action of the cocycle on ker p is trivial (and we have shown that in our situation the component in ker p dominates throughout the process), this implies 1 1 d w¯ , ker p ∩ U x > w ≥ y − x . 0 0 20 40 This, together with the assumption that (16.27) does not hold, implies (16.14). Proof of Theorem 1.4. — It was already proved in Theorem 2.1 that ν is SL(2, R)- invariant. Now suppose ν is not affine. We can apply Lemma 16.3, and then iter- ate Proposition 16.4 with δ → 0and fixed and δ. Taking a limit along a subse- quence we get points (b , x , s ) ∈ K and (b , y , s ) ∈ K such that x − y ≈ , ∞ ∞ ∞ 0 ∞ ∞ ∞ 0 ∞ ∞ + ⊥ + y ∈ W (b , x ) and y ∈ (U ) (b , x ). This contradicts Lemma 16.1 since K ⊂ . ∞ ∞ ∞ ∞ ∞ ∞ 0 Hence ν is affine. 280 ALEX ESKIN, MARYAM MIRZAKHANI Acknowledgements We thank Amir Mohammadi for many useful discussions relating to all aspects of this project. In particular some of the ideas for the proof of Theorem 2.1 came during discussions with Amir Mohammadi. We also thank Vadim Kaimanovich and Emmanuel Breuillard for their insights into the work of Benoist and Quint and Elon Lindenstrauss for his helpful comments. We also thank the anonymous referee for his truly extraordinary effort and his numerous detailed and helpful comments. The paper has vastly improved as a result of his contribution. Appendix A: Forni’s results on the SL(2, R) action In this appendix, we summarize the results we use from the fundamental work of Forni [Fo]. The recent preprint [FoMZ] contains an excellent presentation of these ideas and also some additional results which we will use as well. A.1 The Hodge norm and the geodesic flow Let M denote the moduli space of genus g curves. Fix a point S in H(α); then S is a pair (M,ω) where M ∈ M and ω is a holomorphic 1-form on M. Let · denote the g H,t Hodge norm (see e.g. [ABEM]) at the surface M = π(g S).Here π : H(α) → M is the t t g natural map taking (M,ω) to M. We recall that the Hodge norm is a norm on H (M, R). The following fundamental result is due to Forni [Fo,§2]: Theorem A.1. — For any λ ∈ H (M, R) and any t ≥ 0, λ ≤ e λ . H,t H,0 If in addition λ is orthogonal to ω, and for some compact subset K of M , the geodesic segment [S, g S] g t −1 spends at least half the time in π (K), then we have (1−α)t λ ≤ e λ , H,t H,0 where α> 0 depends only on K. The Hodge norm on relative cohomology. — Let denote the set of zeroes of ω.Let 1 1 p : H (M,, R) → H (M, R) denote the natural map. We define a norm · on the relative cohomology group H (M,, R) as follows: (A.1) λ =p(λ) + (λ − h) , z,w (z,w)∈× INVARIANT AND STATIONARY MEASURES 281 where · denotes the Hodge norm on H (M, R), h is the harmonic representative of the cohomology class p(λ) and γ is any path connecting the zeroes z and w. Since z,w p(λ) and h represent the same class in H (M, R), Equation (A.1) does not depend on the choice of γ . z,w Let · denote the norm (A.1) on the surface M . Then, up to a fixed multiplicative constant, the analogue of Theorem A.1 holds, for · ,aslongas S ≡ (M,ω) and g S belong to a fixed compact set. This assertion is essentially Lemma 4.4 from [AthF]. For a self-contained proof in this notation see [EMR,§8]. The Avila-Gouëzel-Yoccoz (AGY) norm. — The Hodge norm on relative cohomology behaves badly in the thin part of Teichmüller space. Therefore, we will use instead the Avila-Gouëzel-Yoccoz norm · defined in [AGY], some properties of which were further developed in [AG]. The norms · and · are equivalent on compact subsets of the strata H (α), and therefore the decay estimates on · in the style of Theorem A.1 also apply to the Avila-Gouëzel-Yoccoz norm. Furthermore, we have the following: Theorem A.2. —Suppose S = (M,ω) ∈ H(α).Let · denote the Avila-Gouëzel-Yoccoz (AGY) norm on the surface g S.Then, (a) For all λ ∈ H (M,, R) and all t > 0, λ ≤ e λ . t 0 (b) Suppose for some compact subset K of M , the geodesic segment [S, g S] spends at least g t −1 1 half the time in π (K).Suppose λ ∈ H (M,, R) with p(λ) orthogonal to ω. Then we have (1−α)t λ ≤ Ce λ , t 0 where α> 0 depends only on K. A.2 The Kontsevich-Zorich cocycle We recall that X denotes a finite cover of a stratum which is a manifold (see Section 3). In the sequel, a subbundle L of the Hodge bundle is called isometric if the action of the Kontsevich-Zorich cocycle restricted to L is by isometries in the Hodge metric. We say that a subbundle is isotropic if the symplectic form vanishes identically on the sections, and symplectic if the symplectic form is non-degenerate on the sections. A subbundle is irreducible if it cannot be decomposed as a direct sum, and strongly irreducible if it cannot be decomposed as a direct sum on any (measurable) finite cover of X . Theorem A.3. —Let ν be a P-invariant measure on X , and suppose L is a P-invariant ν -measurable subbundle of the Hodge bundle. Let λ ,...,λ be the Lyapunov exponents of the restriction 1 n 282 ALEX ESKIN, MARYAM MIRZAKHANI of the Kontsevich-Zorich cocycle to L.Then, λ ≥ 0. i=1 Proof. — Let the symplectic complement L of L be defined by (A.2)L (x) = v : v ∧ u = 0for all u ∈ L(x) . Then, L is a P-invariant subbundle, and we have the short exact sequence † † 0 → L ∩ L → L → L/ L ∩ L → 0. The bundle L/(L ∩ L ) admits an invariant non-degenerate symplectic form, and there- fore, the sum of the Lyapunov exponents on L/(L ∩ L ) is ≥ 0. Therefore, it is enough to show that the sum of the Lyapunov exponents on the isotropic subspace L ∩ L is 0. Thus, without loss of generality, we may assume that L is isotropic. Let {c ,..., c } be a Hodge-orthonormal basis for the bundle L at the point 1 n S = (M,ω), where M is a Riemann surface and ω is a holomorphic 1-form on M. For g ∈ SL(2, R),let V (g) denote the Hodge norm of the polyvector c ∧ ··· ∧ c at the S 1 n point gS, where the vectors c are transported following a path from the identity to g using the Gauss-Manin connection. (The result does not depend on the path since the Gauss-Manin connection is flat, and X has no orbifold points). Since V (kg) = V (g) for 0 S S k ∈ SO(2), we can think of V as a function on the upper half plane H. From the defini- tion of V and the multiplicative ergodic theorem, we see that for ν -almost all S ∈ X , S 0 log V (g ) S t (A.3)lim = λ , t→∞ i=1 where the λ are as in the statement of Theorem A.3. Let denote the hyperbolic Laplacian operator (along the Teichmüller disk). hyp By [FoMZ, Lemma 2.8] (see also [Fo, Lemma 5.2 and Lemma 5.2 ]) there exists a non- negative function : X → R such that for all S ∈ X and all g ∈ SL(2, R), 0 0 ( log V )(g) = (gS). hyp S We now claim that the Kontsevich-Forni type formula (A.4) λ = (S)dν(S) i=1 holds, which clearly implies the theorem. The formula (A.4)isprovedin[FoMZ](and for the case of the entire stratum in [Fo]) under the assumption that the measure ν INVARIANT AND STATIONARY MEASURES 283 is invariant under SL(2, R). However, in the proofs, only averages over “large cir- cles” in H = SO(2)\SL(2, R) are used. Below we show that a slightly modified ver- sion of the proof works under the a-priori weaker assumption that ν is invariant under P = AN ⊂ SL(2, R). This is not at all surprising, since large circles in H are approxi- mately horocircles (i.e. orbits of N). We now begin the proof of (A.4), following the proof of [FoMZ, Theorem 1]. Since (A.3) holds for ν -almost all S and ν is N-invariant, (A.3) also holds for almost all S ∈ X and almost all S ∈ S ,where 0 0 N 0 1 s = :|s|≤ 1 ⊂ N. −1 We identify SO(2)\SL(2, R)S with H so that SO(2)gS corresponds to g · i.Then 0 0 −4t S corresponds to the horizontal line segment connecting −1 + i to 1 + i.Let = e . N 0 Then, g S corresponds to the line segment connecting −1 + i to 1 + i . t N 0 Let f (z) = log V (SO(2)z).Notethat ∇ f is bounded (where ∇ is the gradient S hyp hyp with respect to the hyperbolic metric on H). Then, (A.3) implies that for almost all x ∈ [−1, 1], −2T T f (x + ie ) − f (x + i) 1 ∂ −2t λ = lim = lim f x + ie dt T→∞ T→∞ T T ∂ t i=1 Integrating the above formula from x =−1to x = 1, we get (using the bounded conver- gence theorem), T 1 1 ∂ −2t λ = lim f x + ie dx dt T→∞ T ∂ t 0 −1 i=1 −2t −2t Let R denote the rectangle with corners at −1 + ie ,1 + ie ,1 + i and −1 + i, see Figure 8.Wenow claimthat ∂ ∂ f −2t −4t −4t (A.5) f x + ie dx = e + O te , ∂ t ∂ n −1 ∂ R ∂ f where denotes the (outgoing) normal derivative of f with respect to the hyperbolic ∂ n metric. Indeed, the integral over the bottom edge of the rectangle R on the left hand −4t side of (A.5) coincides with the right hand side of (A.5) (the factor of e appears because ∂ f 2 −4t the hyperbolic length element is dx/y = e dx.) The partial derivative is uniformly ∂ n bounded, and the hyperbolic lengths of the other three sides of ∂ R are O(t). Therefore (A.5) follows. Now, by Green’s formula (in the hyperbolic metric), ∂ f = f = . hyp ∂ n ∂ R R R t t t 284 ALEX ESKIN, MARYAM MIRZAKHANI FIG. 8. — Proof of Theorem A.3 We get, for almost all S , −4t λ = lim e dt ≥ 0. T→∞ 0 R i=1 This completes the proof of the Theorem. It is also easy to conclude (by integrating over S )that(A.4) holds. Theorem A.4. —Let ν be an ergodic SL(2, R)-invariant measure, and suppose L is an SL(2, R)-invariant ν -measurable subbundle of the Hodge bundle. Suppose all the Lyapunov exponents of the restriction of the Kontsevich-Zorich cocycle to L vanish. Then, the action of the Kontsevich-Zorich cocycle on L is isometric with respect to the Hodge inner product, and the orthogonal complement L of L with respect to the Hodge inner product is also an SL(2, R)-invariant subbundle. Proof. — The first assertion is the content of [FoMZ, Theorem 3]. The second assertion then follows from [FoMZ, Lemma 4.3]. Theorem A.5. —Let ν be an ergodic SL(2, R)-invariant measure, and suppose L is an SL(2, R)-invariant ν -measurable subbundle of the Hodge bundle. Suppose L is isotropic. Then all the Lyapunov exponents of the restriction of the Kontsevich-Zorich cocycle to L vanish (and thus Theorem A.4 applies to L). Proof. — For a point x ∈ X and an isotropic k-dimensional subspace I ,let (x, I ) 0 k k k be as in [FoMZ, (2.46)] (or [Fo, Lemma 5.2’]). We have from [FoMZ, Lemma 2.8] that (x, I ) ≤ (x, I ) if i < j and I ⊂ I . k k j j k j Let λ ≥ ··· ≥ λ be the Lyapunov exponents of the restriction of the Kontsevich-Zorich 1 n cocycle to L. Let V (x) denote the direct sum of all the Lyapunov subspaces correspond- ≤j ing to exponents λ ≥ λ . By definition, V (x) = L(x). Suppose j = n or λ = λ . Then, i j n j j+1 INVARIANT AND STATIONARY MEASURES 285 by [FoMZ, Corollary 3.1] the following formula holds: λ + ··· + λ = x, V (x) dν(x) 1 j j ≤j (This formula is proved in [Fo]for thecasewhere ν is Lebesgue measure and L is the entire Hodge bundle.) We will first show that all the λ have the same sign. Suppose not, then we must have λ < 0 but not all λ < 0. Let k be maximal such that λ = λ .Then n j k n λ + ··· + λ = x, V (x) dν(x) 1 k k k and λ + ··· + λ = x, L(x) dν(x) 1 n n But (x, V (x)) ≤ (x, L(x)) since V (x) ⊂ L(x).Thus, k k n k (A.6) λ + ··· + λ ≥ 0. k+1 n But by the choice of k,all theterms in (A.6) are equal to each other. This implies that λ ≥ 0, contradicting our assumption that λ < 0. Thus all the λ ,1 ≤ j ≤ n have the same n n j sign. Since ν is assumed to be SL(2, R)-invariant, and any diagonalizable g ∈ SL(2, R) is conjugate to its inverse, we see that e.g. the λ cannot all be positive. Hence, all the Lyapunov exponents λ are 0. Algebraic hulls. — The algebraic hull of a cocycle is defined in [Zi2]. We quickly recap the definition: Let G be a group acting on a space X, preserving an ergodic mea- sure ν . Suppose H is an R-algebraic group, and let A : G × X → H be a measurable cocycle. We say that the R-algebraic subgroup H of H is the algebraic hull of A if H is the smallest R-algebraic subgroup of H such that there exists a ν -measurable map C : X → Hsuch that −1 C(gx) A(g, x)C(x) ∈ H for almost all g ∈ Gand ν-almost all x ∈ X. It is shown in [Zi2] (see also [MZ, Theorem 3.8]) that the algebraic hull exists and is unique up to conjugation. Theorem A.6. —Let ν be an ergodic SL(2, R)-invariant measure. Then, (a) The ν -algebraic hull H of the Kontsevich-Zorich cocycle is semisimple. (b) Every ν -measurable SL(2, R)-invariant irreducible subbundle of the Hodge bundle is either symplectic or isotropic. 286 ALEX ESKIN, MARYAM MIRZAKHANI Remark. — The fact that the algebraic hull is semisimple for SL(2, R)-invariant measures is key to our approach. Proof. — Suppose L is an invariant subbundle. It is enough to show that there exists an invariant complement to L. Let the symplectic complement L of L be defined as in † † (A.2). Then, L is also an SL(2, R)-invariant subbundle, and K = L ∩ L is isotropic. By Theorem A.5, K is isometric, and K is also SL(2, R)-invariant. Then, ⊥ † † ⊥ L = K ⊕ L ∩ K , L = K ⊕ L ∩ K , and 1 ⊥ † ⊥ H (M, R) = K ⊕ L ∩ K ⊕ L ∩ K † ⊥ Thus, L ∩ K is an SL(2, R)-invariant complement to L. This proves (a). To prove (b), let L be any irreducible SL(2, R)-invariant ν -measurable irreducible subbundle of the Hodge bundle, and let K = L ∩ L . Since K ⊂ L and L is irreducible, we have either K = 0 (so L is symplectic), or K = L and so L is isotropic. The same could be done on any finite cover. The Forni subspace. Definition A.7 (Forni subspace). — Let −1 R (A.7)F(x) = g Ann B , gx g∈SL(2,R) where for ω ∈ X the quadratic form B (·,·) is as defined in [FoMZ, (2.33)]. Remark. — It is clear from the definition, that as long as its dimension remains constant, F(x) varies real-analytically with x. Theorem A.8. — Suppose ν is an ergodic SL(2, R)-invariant measure. Then the subspaces F(x) where x varies over the support of ν form the maximal ν -measurable SL(2, R)-invariant isometric subbundle of the Hodge bundle. Proof.—Let F(x) be as defined in (A.7). Then, F is an SL(2, R)-invariant sub- bundle of the Hodge bundle, and the restriction of B to F(x) is identically 0. Then, by [FoMZ, Lemma 1.9], F is isometric. Now suppose M is any other ν -measurable isometric SL(2, R)-invariant subbun- dle of the Hodge bundle. Then by [FoMZ, Theorem 2], M(x) ⊂ Ann B . Since M is SL(2, R)-invariant, we have M ⊂ F. Thus F is maximal. INVARIANT AND STATIONARY MEASURES 287 Theorem A.9. —Let ν be an ergodic SL(2, R)-invariant measure on any finite cover of X . (a) For ν -almost all x ∈ X , the Forni subspace F(x) is symplectic, and its symplectic comple- † ⊥ ment F (x) coincides with its Hodge complement F (x). (b) Any ν -measurable SL(2, R)-invariant subbundle of F is symplectic, and the restriction of the Kontsevich-Zorich cocycle to any invariant subbundle of F has at least one non-zero Lyapunov exponent. ⊥ ⊥ ⊥ † Proof. — Suppose the subspace F is not symplectic. Let L = F ∩(F ) . Then L is isotropic, and therefore by Theorems A.5 and A.4,Lis an SL(2, R)-invariant isometric ⊥ ⊥ subspace. Hence L ⊂ Fby Theorem A.8.As L ⊂ F we get L = 0. Therefore F is symplectic. Let M be an irreducible subbundle of F . Then, in view of Theorem A.4 and the maximality of F, M must have at least one non-zero Lyapunov exponent. In particular, in view of Theorem A.5, M cannot be isotropic, so it must be symplectic in view of Theorem A.6(b). This proves the statement (b). ⊥ ⊥ † ⊥ Since F is symplectic, (F ) is SL(2, R)-invariant and complementary to F . Note that F is also SL(2, R)-invariant and complementary to F . In order to conclude ⊥ † that (F ) = F, it is enough to show that there is a unique SL(2, R)-invariant complement to F . Note that another complement to F would be the graph of an equivariant linear map A : F → F . If A is nonzero, then an invariant complement of its kernel in F exists by Theorem A.6, and it even contains an irreducible subbundle M . Then A induces an equivariant isomorphism between M and its image, an irreducible subbundle M of F . 2 1 Now, to get a contradiction, it is enough to show that for any irreducible subbundles M ⊂ F and M ⊂ F, the algebraic hulls H (M ) of the restriction of the Kontsevich- 1 2 i Zorich cocycle to M are not isomorphic to each other. But the later statement is clear, since H (M ) is compact and H (M ) is not (since it has at least one non-zero Lyapunov 2 1 ⊥ † ⊥ exponent by (b)). Thus, (F ) = F. Since we already showed that F is symplectic, this implies that so is F, which completes the proof of (a). Appendix B: Entropy and the Teichmüller geodesic flow The contents of this section are well-known, see e.g. [LY], [MaT]and also [BG]. How- ever, for technical reasons, the statements we need do not formally follow from the results of any of the above papers. Our setting is intermediate between the homogeneous dy- namics setting of [MaT] and the general C -diffeomorphism on a compact manifold setup of [LY], but it is closer to the former than the latter. What follows is a lightly edited but almost verbatim reproduction of [MaT, §9], adapted to the setting of Teichmüller space. It is included here primarily for the convenience of the reader. The (minor) dif- ferences between our presentation and that of [MaT] are related to the lack of uniform 288 ALEX ESKIN, MARYAM MIRZAKHANI hyperbolicity outside of compact subsets of the space, and some notational changes due to the fact that our space is not homogeneous. Notation. — We recall some notation from Section 2.2.Let X denote the finite cover of H (α) defined in Section 3 (which has no orbifold points). Let g denote the 1 t Teichmüller geodesic flow. In this section, ν is an ergodic g -invariant probability measure 1 2 on X .Let V(x) denote a subset of H (M,, R ).Thenwedenote V[x]= y ∈ X : y − x ∈ V(x) . This makes sense in a neighborhood of x. Let d (·,·) denote the AGY distance on X ,definedinSection 3. Fix a point p ∈ X (so p is not an orbifold point), and such that every neighborhood of p in X has 0 0 positive ν -measure. Fix relatively compact neighborhoods C (p) and Q(p) of 0 in W (p) and R respectively. Let C = g C [p]. t∈Q(p) For each c ∈ C choose a relatively compact neighborhood B (c) of 0 in W (c) with diam- eter in the AGY distance at most 1/200 so that the B (c) vary continuously with c.For c ∈ C, let B [c]= c + v : v ∈ B (c) , D = B [c]. c∈C We assume that C (p),Q(p) and the B (c) are sufficiently small so that D is open and contractible. Lemma B.1 (Cf. [MaT, Lemma 9.1]). — There exists s > 0, C ⊂ C and for each c ∈ C 1 1 there exists a subset E[c]⊂ W [c] such that (1) E[c]⊂ B [c]. (2) E[c] is open in W [c], and the subset E ≡ E[c] satisfies ν(E)> 0. c∈C (3) Let T = g denote the time s map of the geodesic flow. Then whenever T E[c]∩ E = 0, c ∈ C , n > 0, we have T E[c]⊂ E. Proof. — Fix a compact subset K ⊂ X ,with ν(K )< 0.01. Then by the Birkhoff 1 0 ergodic theorem, for every δ> 0 there exists R > 0 and a subset E with ν(E )> 1 − δ 1 1 such that for all x ∈ E and all N > R, n ∈[1, N]: g x ∈ K ≥ (1/2)N. n 1 INVARIANT AND STATIONARY MEASURES 289 By choosing δ> 0 small enough, we may assume that ν(D ∩ E )> 0. Let C = c ∈ C : c + v ∈ D ∩ E for some v ∈ B (c) . 1 1 Then there exists a compact K ⊃ K such that for all c ∈ C and all x ∈ B [c], 1 1 n ∈[1, N]: g x ∈ K ≥ (1/2)N. and all x ∈ B [c], By Lemma 3.5 there exists α> 0such that for all c ∈ C d (x, c) if n ≤ R d (g x, g c) ≤ n n X −α(n−R) d (x, c)e if n > R Therefore we may choose s > 0such that if we let T = g denote the time s map of the geodesicflow,thenfor all c ∈ C and all x ∈ B [c], X X 0 0 d (Tx, Tc) ≤ d (x, c). There exists a > 0so that for all c ∈ C ,B [c] contains the intersection with W [c] of a ball in the AGY metric of radius a and centered at c.Let a = Let B [c]⊂ W [c] denote the ball in the AGY metric of radius a and centered at c.Let (0) E [c]= B [c],and for j > 0let (j) (j−1) E [c]= E [c] n n (j−1) ∪ T B c : c ∈ C , n > 0and T B c ∩ E [c]= 0 . 0 0 Let (j) E[c]= E [c], and E = E[c]. j≥0 c∈C It easily follows from the above definition that E[c] has the properties (2) and (3). To show (1), it is enough to show that for each j , X (j) (B.1) d (x, c)< a/2, for all x ∈ E [c]. This is done by induction on j.The case j = 0 holds since a = a/10 < a/2. Suppose (B.1) (j) (j−1) holds for j − 1, and suppose x ∈ E [c]\ E [c]. Then there exist c = c, c ,..., c = x in 0 1 j C and non-negative integers n = 0,..., n such that for all 1 ≤ k ≤ j , 1 0 j n n k k−1 (B.2)T B [c ] ∩ T B [c ] = ∅. k k−1 0 0 290 ALEX ESKIN, MARYAM MIRZAKHANI Let 1 ≤ k ≤ j be such that n is minimal. Recall that B [ y]∩ B [z]=∅ if y = z, y ∈ C , k 1 −n z ∈ C . Therefore, in view of the inductive assumption, n ≥ 1. Applying T to (B.2)we 1 k get k−1 n −n n −n i k i k T B [c ] ∩ B [c ]=∅, and T B [c ] ∩ B [c ]=∅. i k i k 0 0 0 0 i=1 i=k+1 (j) Therefore, in view of (B.2), and the definition of the sets E [c], n −n (k−1) n −n (j−k) i k i k T B [c ] ⊂ E [c ], and T B [c ] ⊂ E [c ] i k i k 0 0 i=1 i=k (k−1) (j−k) By the induction hypothesis, diam(E [c ])< a/2, and diam(E [c ])< a/2. There- k k fore, n −n i k diam T B [c ] ≤ a. i=1 Then, applying T we get, diam T B [c ] ≤ i=1 Since diam(B [c]) ≤ a/10, we get j j n n i i diam T B [c ] ≤ diam B [c ] + diam T B [c ] i 0 i 0 0 0 i=0 i=1 a a a ≤ + < . 10 10 2 But the set on the left-hand-side of the above equation contains both c = c and x = c . 0 j Therefore d (c, x)< a/2, proving (B.1). Thus (1) holds. Lemma B.2 (Mañé). — Let E be a measurable subset of X , with ν(E)> 0.If ν is a compactly supported measure on E and q : E → (0, 1) is such that logqis ν -integrable, then there exists a countable partition P of E with entropy H(P)< ∞ such that, if P(x) denotes the atom of P containing x, then diam P(x)< q(x). Proof.—See[M1]or[M2, Lemma 13.3]. Let V(x) be a system of real-algebraic subsets of W (x). INVARIANT AND STATIONARY MEASURES 291 Definition B.3. — The system V(x) is admissible if it is T-equivariant and also for almost all x ∈ X , x is a smooth point of V[x]. Definition B.4. — We say that a measurable partition ξ of the measure space (X ,ν) is subordinate to an admissible system of real-algebraic subsets V(x) ⊂ W (x) if for almost all (with respect to ν)x ∈ X , we have (a) ξ[x]⊂ V[x] where ξ[x] denotes, as usual, the element of ξ containing x. (b) ξ[x] is relatively compact in V[x]. (c) ξ[x] contains a neighborhood of x in V[x]. Let η and η be measurable partitions of (X ,ν). We write η ≤ η if η[x]⊃ η [x] for −1 almost all (with respect to ν ) x ∈ X . We define a partition Tη by (Tη)[x]= T(η[T (x)]). Proposition B.5. — Assume that ν is T-ergodic (where T is as in Lemma B.1(3)). Then there exists a measurable partition η of the measure space (X ,ν) with the following properties: (i) η is subordinate to W . (ii) η is T-invariant, i.e. η ≤ Tη. (iii) The mean conditional entropy H(Tη | η) is equal to the entropy h(T,ν) of the automor- phism x → Tx of the measure space (X ,ν). Proof.—Let E[c] and E be as in Lemma B.1.Denoteby π : E → C the natural projection (π(x) = c if x ∈ E[c]). We set η[x]= E(π(x))for every x ∈ E. We claim that it is enough to find a countable measurable partition ξ of (X ,ν) − − −n such that H(ξ) < ∞ and η[x]= ξ [x] for almost all x ∈ Ewhere ξ = T ξ is the n=0 −n product of the partitions T ξ,0 ≤ n < ∞. Indeed, suppose the claim holds. Then it is clear that η is T-invariant. The set of x ∈ X for which properties (a) and (b) (resp. (c)) in the definition of a subordinate partition −1 are satisfied is T -invariant (resp. T-invariant) and contains E. But ν(E)> 0and ν is T-ergodic. Therefore, η is subordinate to W . To check the property (iii) it is enough to show that the partition ξ = T ξ is the partition into points, see [R,§9],or[KH, k=−∞ −n §4.3]. By [Fo]or[ABEM, Theorem 8.12] ξ (x) ={x} if T x ∈ E for infinitely many n. (Recall that by the construction of E, any such geodesic will spend at least half the time in the compact set K.) But ν(E)> 0and ν is T-ergodic. Hence ξ [x]={x} for almost all x, which completes the proof of the claim. Let us construct the desired partition ξ.For x ∈ E, let n(x) be the smallest positive integer n such that T x ∈ E. We have the classical Kac formula [Ka] (B.3) n(x)dν(x) = 1. E 292 ALEX ESKIN, MARYAM MIRZAKHANI Define a probability measure ν on C by −1 ν(π (F)) (B.4) ν (F) = , F ⊂ C . ν(E) Property (3) of the family {E[c]: c ∈ C } implies that n(x) is constant on every E[c], c ∈ C . Therefore, in view of (B.3)and (B.4), n(c)dν (c)< ∞. By Lemma 3.6, there exists κ> 1such that for all x, y ∈ X , X X 0 0 d (Tx, Ty) ≤ κ d (x, y). −2n(c) Since the function n(c) is ν -integrable, one can find a positive function q(c)<κ , c ∈ C such that log q is ν -integrable, and the ν -essential infimum ess inf q(c) is 0. 1 c∈C After replacing, if necessary, C (p),Q(p) and the B (c) for c ∈ C by smaller subsets we can find > 0 such that the minimum distance between lifts of E is at most /10 and also X X 0 0 (a) d (x, y)< 2d(π(x),π(y)) whenever x, y ∈ Eand d (x, y)<,and (b) if x, y ∈ C then d (x, y)< . Since the function log q(c) is ν -integrable, there exists a countable measurable parti- tion P of C such that H(P)< ∞ and diam P(x)< q(x) for almost all x ∈ C (see 1 1 Lemma B.2). After possibly replacing P by a countable refinement, we may assume that the function x → n(x) is constant on the atoms of P . Now we define a countable measur- able partition ξ of X by −1 π (P(π(x))) if x ∈ E ξ(x) = X \Eif x ∈/ E. Since H(P)< ∞ we get using (B.4)that H(ξ) < ∞. It remains to show that ξ [x]= η[x] for almost all x ∈ E. It follows from the property (3) of the family {E[c]} that η[z]⊂ ξ [z]. − − Let x and y be elements in E with ξ [x]= ξ [ y]. Since η[z]⊂ ξ[z], we can assume that x, y ∈ C .Then d (x, y)<.Set x = x, y = y and define by induction 1 1 1 n(x ) n(y ) k k x = T x , y = T y . k+1 k k+1 k Then, the sequence {x } (resp. { y } ) is the part of the T-orbit of x (resp. T-orbit of y) k k∈N k k∈N which lies in E. Let x˜ , y˜ be the lifts of x = x and y = y to Teichmüller space, and let x˜ , y˜ be 1 1 1 1 k k defined inductively by n(x ) n(y ) k k x˜ = T x˜ , y˜ = T y˜ . k+1 k k+1 k INVARIANT AND STATIONARY MEASURES 293 Then x˜ and y˜ are lifts of x and y respectively. We now claim that for all k ≥ 0, k k k k (B.5) d (x˜ , y˜ )<q π(x ) . k k k If k = 1, the inequality (B.5) is true because diam P(x)< q(π(x)) and P(x) = P(y). Assume that (B.5)isprovedfor k.Then X X n(x ) n(x ) n(x ) X 0 0 k k k 0 d (x˜ , y˜ ) = d T x˜ , T y˜ ≤ κ d (x˜ , y˜ ) k+1 k+1 k k k k n(x ) ≤ κ q π(x ) ≤ . Then since x and y belong to the same element of the partition ξ (because ξ [x]= k+1 k+1 ξ [ y])and diam(P(x )) ≤ qπ(x ), we get from condition (b) in the definition of k+1 k+1 > 0that (B.5) is true for k + 1. Since the measure ν is T-ergodic and ess inf q(c) = 0 we may assume that lim inf q(π(x )) = 0 (since this holds for almost all x ∈ E). Then (B.5) implies that k→∞ k lim inf d (x˜ , y˜ ) = 0. k k k→∞ By the definition of x˜ , y˜ , there exists a sequence m →+∞ such that x˜ = T x˜, y˜ = k k k k k T y˜.Thus, X m m 0 k k d T x˜, T y˜ = 0. 0+ But, by construction x˜ and y˜ are on the same leaf of W . This contradicts the non- contraction property of the Hodge distance [ABEM, Theorem 8.2], unless x˜ =˜ y.Thus we must have x = y. Lemma B.6 (See [LS, Proposition 2.2]). — Let T be an automorphism of a measure space (X ,ν), ν(X )< ∞, and let f be a positive finite measurable function defined on X such that 0 0 0 f ◦ T − 1 − log ∈ L (X,ν), where log (a) = min(log a, 0). Then f ◦ T log dν = 0. − − Suppose V (x) ⊂ W (x) is an admissible T-equivariant family of real-algebraic − − − subsets. Let (T V )(x) ⊂ W (x) denote the tangent space to smooth manifold V [x] − − at x. (Recall that since V is admissible, for almost every x,V [x] is smooth at x.) − − Definition B.7 (Margulis property). — Suppose V (x) ⊂ W (x) is an admissible T-equi- variant family of real-algebraic subsets. Let τ = τ(x) be a measure on each V [x]. We say that τ has the Margulis Property if for almost all x, τ(x) is in the Lebesgue measure class on V [x],and also d T τ(x) T τ(x) agrees with τ(Tx) up to normalization. (In other words the Radon-Nykodym derivative dτ(Tx) is locally constant along V [x].) 294 ALEX ESKIN, MARYAM MIRZAKHANI − − Proposition B.8. —Let T = g as in Lemma B.1(iii). Let V (x) ⊂ W (x) be a T-equivariant family of real-algebraic subsets. Suppose there exists a T-invariant measurable partition η of (X ,ν) subordinate to V . Then the following hold: (a) We have H(Tη | η) ≤ s V , where H(Tη | η) is the mean conditional entropy, and V = (1 − λ ), i∈I(V) where I(V) are the Lyapunov subspaces in T V (counted with multiplicity), and λ are the R i corresponding Lyapunov exponents of the Kontsevich-Zorich cocycle. (b) Suppose that for almost all x there exists a measure τ = τ(x) on each V [x] with the Margulis property. Then (b1) If the conditional measures of ν along V [x] agree with τ(x) (up to normalization), then H(Tη | η) = s V − − (b2) If H(Tη | η) = s(V ) then the conditional measures of ν along V [x] agree with τ(x) (up to normalization). Proof. — Since η ≤ Tη for almost all x ∈ X we have a partition η of η[x] such 0 x that η [ y]= (Tη)[ y] for almost all y ∈ η[x].Let τ(x) be a measure on V (x) in the Lebesgue measure class. (To simplify notation, we will sometimes denote τ(x) simply by τ .) (Here we pick some normalization of the Lebesgue measure on the connected components of the intersections of the leaves of V with a fixed fundamental domain.) Since η[x]⊂ V [x], τ induces a measure on η[x] which we will denote also by τ.Let J(x) denote the Jacobian of the restriction of the map T to V [x] at x (with respect to − − the Lebesgue measure class measures τ on V [x] and V [Tx]). Then, by the Osceledets multiplicative ergodic theorem, for almost all x ∈ X , N−1 −N 1 d(T τ)(x) 1 − −n −s V = lim log =− lim log J T x . N→∞ N→∞ N dτ(x) N n=0 Integrating both sides over X ,weget (B.6) − log J(x)dν(x) = s V . Put L(x) = τ(η[x]) and τ = τ/L(x), x ∈ X .Notethaton η[x] we have a conditional x 0 probability measure ν induced by ν.Put p(x) = τ (η [x]) and r(x) = ν (η [x]). x x x x x INVARIANT AND STATIONARY MEASURES 295 Let (B.7) η = η ∨ Tη ∨ ··· ∨ T η. Then, η is also T-invariant, and H(Tη | η ) = H(Tη | η). Thus, we can replace η by η . Suppose > 0 is given. Then, we can choose k large enough in (B.7)sothat(after replacing η by η ), on a set of measure at least (1 − ),wehave p(x)L(x) (B.8) (1 − ) ≤ ≤ (1 + ) −1 −1 J(T x)L(T x) From its definition, p(x) ≤ 1. Also (B.9) − log r(x)dν(x) = H(Tη | η). Let Y (x),1 ≤ i < ∞ denote the elements of the countable partition η of η[x]. i x Then we have τ (Y (x)) x i (B.10) log p(y)dν (y) − log r(y)dν (y) = log ν Y (x) . x x x i ν (Y (x)) x i η(x) η(x) i=1 We have that (B.11) τ Y (x) ≤ 1, x i i=1 and (B.12) ν Y (x) = 1. x i i=1 (In (B.11), we can have strict inequality because apriori it is possible that the measure τ of η[x]\ Y (x) is positive.) From (B.10), (B.11)and (B.12), using the convexity of log i=1 we get that log p(y)dν (y) ≤ log r(y)dν (y), x x η(x) η(x) and the equality holds if and only if p(y) = r(y) i.e. τ (η [ y]) = ν (η [ y]) for all y ∈ η[x]. x x x x Now using integration over the quotient space (X ,ν)/η of the measure space (X ,ν) 0 0 by η,weget from (B.9)that (B.13)H(Tη | η) ≤− log p(x)dν(x), and the equality holds if and only if τ ((Tη)[x]) = ν ((Tη)[x]) for almost all x ∈ X . x x 0 296 ALEX ESKIN, MARYAM MIRZAKHANI In view of (B.8) and the fact that p(x) ≤ 1, − log p(x)dν(x) ≤ 2 − log J(x)dv(x) X X 0 0 −1 + log L T x /L(x) dν(x). The last term vanishes by Lemma B.6. Since > 0 is arbitrary, we have, by (B.13)and (B.6) that (a) holds. −1 Now suppose that τ is as in (b). Then since η [x]= T(η[T x]) one easily sees that −1 −1 p(x) = J(T x)L(T x)/L(x). Therefore, by (B.6) and Lemma B.6, − log p(x)dν(x) = s V . If the conditional measures of ν along V coincide with τ ,then p(x) = r(x) and therefore equality in (B.13) holds. This proves (b1). Conversely, assume that H(Tη | η) = s(V ). k − Then H(T η | η) = ks(V ) for every k > 0. Using thesameargumentasabove and k k k replacing T by T ,weget that τ ((T η)[x]) = ν ((T η)[x]) for any k > 0 and almost all x x − − x ∈ X . On the other hand since η is subordinate to V and T is contracting on V , we have that T η is the partition into points. Hence the conditional measures of ν k=1 along V agree with τ . This proves (b2). Theorem B.9. —Let T = g denote the time s map of the geodesic flow. Assume that T acts ergodically on (X ,ν).Let V (x) be an admissible T-equivariant system of real-algebraic subsets of − − W (x),and let (V ) be as in Proposition B.8. (i) Suppose V has a system of measures τ with the Margulis property, and suppose that for al- most all x, the conditional measures of ν along V [x] agree with τ(x) up to normalization. Then, h(T,ν) ≥ s(V ). (ii) Assume that there exists a subset ⊂ X with ν -measure 1 such that ∩ W [x]⊂ − − V [x] for every x ∈ .Then h(T,ν) ≤ s(V ). (iii) Assume that there exists a subset ⊂ X with ν -measure 1 such that ∩ W [x]⊂ − − V [x] for every x ∈ . Also assume that V has a system of measures τ with the Margulis property, and that h(T,ν) = s(V ). Then, for almost all x, the conditional measures of ν along V [x] agree with τ(x) up to normalization. Proof. — According to Proposition B.5, there exists a measurable T-invariant par- tition η of (X ,ν), subordinate to W ,suchthat H(Tη | η) = h(T,ν). By Lemma 3.2, − − we may assume that the affine exponential map W (x) → W [x] is one-to-one and onto, − − and thus W [x] has an affine structure. Set η (x) = V [x]∩ η[x]. Suppose the assumptions of (i) hold. Then, (B.14) h(T,ν) ≥ H Tη | η . INVARIANT AND STATIONARY MEASURES 297 By Proposition B.8(b1), H(Tη | η ) = s(V ). This, together with (B.14) implies the conclusion of (i). Now suppose the assumptions of (ii) or (iii) hold. Then η and η coincide on , i.e. η[x]∩ = η [x]∩ .Hence H(Tη | η) = H(Tη | η ).ByProposition B.5(iii), h(T,ν) = H(Tη | η). Using Proposition B.8(a) we obtain (ii), and using Proposition B.8(b2) we ob- tain (iii). Appendix C: Semisimplicity of the Lyapunov spectrum In this section we work with a bit more generality than we need. Let X be a space on which SL(2, R) acts. Let μ be a compactly supported probability measure on SL(2, R) and let ν be an ergodic μ-stationary probability measure on X. Let L be a finite dimen- sional real vector space, and suppose A : SL(2, R) × X → SL(L) is a cocycle, such that for any g ∈ SL(2, R),the map x → log A(g, x) is in L (X,ν).Let H be the algebraic hull of the cocycle A (see Section A.2 for the definition). We may assume that a basis at every point is chosen so that for all g ∈ SL(2, R) and all x ∈ X, A(g, x) ∈ H . Definition C.1. —Wesay that ameasurablemap W : X → L is an invariant system of subspaces for A(·,·) if for μ-a.e. g ∈ SL(2, R) and ν -a.e. x ∈ X, A(g, x)W(x) = W(gx). Definition C.2 (Strongly irreducible). — We say that A is strongly irreducible if on any measurable finite cover of X there is no nontrivial proper invariant system of subspaces for A(·,·). Remark. — If a cocycle is strongly irreducible, then its algebraic hull is a simple Lie group. Let B be the space of (one-sided) infinite sequences of elements of SL(2, R).We define the measure β on B to be μ × μ··· .Let T : B × X → B × X be the forward shift, with β × ν as the invariant measure. We denote elements of B by the letter a (following the convention that these refer to “future” trajectories). If we write a = (a , a ,...) then 1 2 T(a, x) = (Ta, a x) (and we use the letter T to denote the shift T(a , a ,...) = (a , a ,...)). By the Osceledets 1 2 2 3 multiplicative ergodic theorem, for β × ν almost every (a, x) ∈ B × X there exists a Lya- punov flag (C.1) {0}= V (a, x) ⊂ V (a, x) ⊂··· ⊂ V (a, x) = L. ≥k ≥k−1 ≥0 Definition C.3. — The map T : B × X → B × X has semisimple Lyapunov spectrum if (after passing to a measurable finite cover), the algebraic hull of the cocycle Z × (B × X) → SL(L) given by (n, a, x) → A(a ... a , x) n 1 298 ALEX ESKIN, MARYAM MIRZAKHANI is block-conformal, see Section 4.3. In other words, T has semisimple Lyapunov spectrum if all the off-diagonal blocks labelled ∗ in (4.4) are 0. In Appendix C our aim is to prove the following general fact: Theorem C.4. — Suppose A is strongly irreducible and ν is μ-invariant. Then T has semisim- ple Lyapunov spectrum. Furthermore, the restriction of T to the top Lyapunov subspace V /V consists ≥1 >1 of a single conformal block, i.e. for β × ν almost every (a, x) there exists an inner product ·,· on a,x V (a, x)/V (a, x) and a function λ : B × X → R such that for all u,v ∈ V (a, x)/V (a, x), ≥1 >1 ≥1 >1 (C.2) a u, a v = λ(a , x)u,v . 1 1 (Ta,ax) 1 a,x If the algebraic hull H is all of SL(L), then all the Lyapunov subspaces consist of a single conformal block, i.e. for all 1 ≤ i ≤ k − 1 onecan defineaninnerproduct ·,· on V (a, x)/V (a, x) so that a,x ≥i >i (C.2) holds for some function λ = λ . The backwards shift. — We will actually use the analogue of Theorem C.4 for the backwards shift. Let T : B × X → B × X be the (backward) shift as in Section 14,with β as defined in [BQ, Lemma 3.1] as the invariant measure. By the Osceledets multiplicative ergodic theorem, for β almost every (b, x) ∈ B × X there exists a Lyapunov flag (C.3) {0}= V (b, x) ⊂ V (b, x) ⊂ V (b, x) ⊂ V (b, x) = L. ≤0 ≤1 ≤2 ≤k We need the following: Theorem C.5. — Suppose A is strongly irreducible and ν is μ-invariant. Then T has semisim- ple Lyapunov spectrum. Furthermore, the restriction of T to the top Lyapunov subspace V consists of a ≤1 single conformal block, i.e. for β almost every (b, x) there exists an inner product ·,· on V (b, x) b,x ≤1 and a function λ : B × X → R such that for all u,v ∈ V (b, x), ≤1 −1 −1 (C.4) b u, b v = λ(b , x)u,v . −1 0 b,x 0 0 (Tb,b x) If the algebraic hull H is all of SL(L), then all the Lyapunov subspaces consist of a single conformal block, i.e. for all 1 ≤ i ≤ k − 1 one can define an inner product ·,· on V (b, x)/V (b, x) so that b,x ≤i <i (C.4) holds for some function λ = λ . The two-sided shift. — As in Section 14,let B be the space of bi-infinite sequences of elements of SL(2, R), and we consider the two-sided random walk as a shift map on B × X. We abuse notation by using the same letter T both for the backwards shift and the bi-infinite shift. We denote a point in Bby a ∨ b where a denotes the “future” of the ˜ ˜ trajectory and b denotes the “past”. Let β denote the T-invariant measure on B × X which projects to the measure β × ν on the future trajectories, and to the measure β on the past trajectories. Then, at β almost all points (a ∨ b, x) we have both the flags INVARIANT AND STATIONARY MEASURES 299 (C.1)and (C.3). The two flags are generically in general position (see e.g. [GM, Lemma 1.5]) and thus we can intersect the flags to define the (shift-invariant) Lyapunov subspaces V (a ∨ b, x) so that i m V (b, x) = V (a ∨ b, x), V (a, x) = V (a ∨ b, x). ≤i j ≥i j j=1 j=i Then ∼ ∼ (C.5) V (b, x)/V (b, x) = V (a ∨ b, x) = V (a, x)/V (a, x). ≤i <i i ≥i >i We will prove the following: Theorem C.6. — Suppose A is strongly irreducible and ν is μ-invariant. Then T has semisim- ple Lyapunov spectrum. Furthermore, the restriction of T to the top Lyapunov subspace V consists of ≤1 a single conformal block, i.e. for β almost every (a ∨ b, x) there exists an inner product ·,· on a∨b,x V (a ∨ b, x) and a function λ : B × X → R such that for all u,v ∈ V (a ∨ b, x), 1 1 (C.6) a u, a v = λ(a ∨ b, x)u,v . 1 1 (T(a∨b),a x) a∨b,x If the algebraic hull H is all of SL(L), then all the Lyapunov subspaces consist of a single conformal block, i.e. for all 1 ≤ i ≤ k − 1 one can define an inner product ·,· on V (b, x) so that (C.6) a∨b,x i holds for some function λ = λ . Remark 1. — The proof of Theorems C.4–C.6 we give is essentially taken from [GM], and is originally from [GR1]and [GR2]. For most of the proof, we assume only that ν is μ-stationary (and not necessarily μ-invariant). The exceptions are Lemma C.10 and Claim C.14. We follow [GM] and present the proof of Theorems C.4–C.6 for the easier to read case where the algebraic hull H of the cocycle A is all of SL(L). The general case of semisimple H is treated in [EMat]. Remark 2. — It is possible to define semisimplicity of the Lyapunov spectrum in the e 0 context of the action of g = ⊂ SL(2, R) (instead of the random walk). Then the t −t 0 e analogue of Theorems C.4–C.6 remains true; the proof would use an argument similar to the proof of Proposition 4.12. Since we will not use this statement we will omit the details. C.3 An ergodic lemma We recall the following well-known lemma: 300 ALEX ESKIN, MARYAM MIRZAKHANI Lemma C.7. —Let T : → be a transformation preserving a probability measure β.Let F : → R be an L function. Suppose that for β -a.e. x ∈ , lim inf F T x =+∞. i=1 Then Fdβ> 0. Proof. — This lemma is due to Atkinson [At] and Kesten [Ke]. See also [GM, Lemma 5.3], and the references quoted there. We will need the following variant: Lemma C.8. —Let T : → be a transformation preserving an ergodic probability mea- sure β.Let F : → R be an L function. Suppose there exists K ⊂ with β(K )> 0 such that for β -a.e. x ∈ , i n (C.7) lim inf F T x : T x ∈ K =+∞. i=1 Then Fdβ> 0. Proof. — After passing to the natural extension, we may assume that T is invertible. We can choose a subset K ⊂ K with β(K)> 0, and C > 0such that for all x ∈ K, we have |F(x)| < C. Since K ⊂ K ,(C.7) holds with K replaced by K. Let A ={x : x ∈/ K},A ={x : x ∈ K, Tx ∈ K},and for n ≥ 0, −1 0 n n+1 A = x : x ∈ K, Tx ∈/ K,..., T x ∈/ K, T x ∈ K . n+1 Also let A = A . Note that by the ergodicity of T, for almost every x ∈ , n=−1 i : i ≥ 0, T (x) ∈ K =∞ (∗). Define G : → R defined on A (which has full measure) by • G(x) = 0if x ∈ A . −1 • G(x) = F(x) if x ∈ A . • G(x) = F(x) + F(Tx) + ··· + F(T x) if x ∈ A . n+1 We now claim the following hold: INVARIANT AND STATIONARY MEASURES 301 (1) For almost every x ∈ we have (C.8)lim G(x) + G(Tx) + ··· + G T x =∞. n→∞ , , (2) |G|dβ ≤ |F|dβ< ∞. , , (3) G(x)dβ(x) = F(x)dβ(x). Proof of (1). — Note that almost every x ∈ satisfies (C.7) (with K replaced by K). Also, we have, m−1 n i G(x) + G(Tx) + ··· + G T x = F T x , i=m k k where m = inf{k : T x ∈ K},and m = inf{k : k ≥ n, T x ∈ K}.Thus, n m m −1 j i i m G T x = F T x − F T x − F T x . j=0 i=1 i=0 Since m is independent of n,T x ∈ Kand for every x ∈ K, we have |F(x)| < C, Equation (C.7) implies (C.8). Proof of (3) assuming (2). — By the definition of G we can use the dominated con- vergence theorem, and get that Gdβ = Fdβ + F T x dβ(x) K A i=1 where A = A .Then j≥i i i i i−1 T A = T K − K ∪··· T K . i i i i j j i i Also K ∪ T A has full measure in ,and for i = j,T A ∩ T A and K ∩ T A have i=1 i −i i i measure zero. Note that A = T (T A ). Since β is T invariant, we have F T x dβ(x) = F(x)dβ(x), i i i A T A and hence Gdβ = Fdβ + F(x)dβ(x) = Fdβ. i i K T A i=1 Proof of (2). — This follows by applying (3) to |F| instead of F, and then using the triangle inequality. 302 ALEX ESKIN, MARYAM MIRZAKHANI Proof of Lemma C.8. — Now by (1), and (2), the function G satisfies the assumptions , , of Lemma C.7.Hence we have Fdβ = Gdβ> 0. C.4 A zero-one law Lemma C.9. — Suppose h is a bounded non-negative μ-subharmonic function, i.e. for ν -almost all x ∈ X, (C.9) h(x) ≤ h(gx)dμ(g). Then h is constant ν -almost everywhere. Proof. — By the random ergodic theorem [Fu, Theorem 3.1], for ν -almost all x ∈ X, N−1 lim h(gx)dμ (g) = hdν N→∞ G X n=0 Therefore, by (C.9), for ν -almost all x ∈ X, (C.10) h(x) ≤ hdν. Let s ≥ 0 denote the essential supremum of h, i.e. s = inf s ∈ R : ν {h > s} = 0 . Suppose > 0 is arbitrary. We can pick x ∈ Xsuch that (C.10) holds and h(x)> s − . Then, s − ≤ h(x) ≤ hdν ≤ s . 0 0 Since > 0 is arbitrary, hdν = s .Thus h(x) = s for ν -almost all x. 0 0 Let ν be an ergodic stationary measure on X. Fix 1 ≤ s < dim(L),and let Gr denote the Grassmannian of s-dimensional subspaces in L. Let X = X × Gr .Wethen have an action of SL(2, R) on X, by g · (x, W) = gx, A(g, x)W . ˆ ˆ Let νˆ be a μ-stationary measure on X which projects to ν under the natural map X → X. We may write dν( ˆ x, U) = dν(x)dη (U), where η is a measure on Gr . x s INVARIANT AND STATIONARY MEASURES 303 Let m = dim(L). For a subspace W of L, let I(W) = U ∈ Gr : dim(U ∩ W)> max 0, m − dim(U) − dim(W) Then U ∈ I(W) if and only if U and W intersect more than general position subspaces of dimension dim(U) and dim(W). Lemma C.10 (Cf. [GM, Lemma 4.2], [GR1, Theorem 2.6]). (i) Suppose the cocycle is strongly irreducible on L. Then for almost all x ∈ X, and any 1-dimensional subspace W ⊂ L, η (I(W )) = 0. x x x (ii) Suppose the algebraic hull H of the cocycle is SL(L). Then for almost all x ∈ X, for any nontrivial proper subspace W ⊂ L, η (I(W )) = 0. x x x Proof of Lemma C.10. — We give the proof under the extra assumption that ν is μ-invariant (and not just μ-stationary). The general case is proved in [EMat]. Suppose there exists a subset E ⊂ Xwith ν(E)> 0and for all x ∈ E, a nontrivial subspace W ⊂ Lsuch that η (I(W )) > 0. Let W = (W ,..., W ) denote a finite col- x x x 1 k lection of subspaces of L. If the assumptions of (i) hold, we are requiring the W to be one-dimensional; if the assumptions of (ii) hold, the W are allowed to be any dimension. Write I(W) = I(W ) ∩ ··· ∩ I(W ). 1 k # # # For x ∈ E, let S denote the set of I(W ) such that for any W so that I(W ) is a proper x x x x # # # subset of I(W ),wehave ν (I(W )) = 0. For x ∈ E, S is non-trivial since the subsets I(W) x x x are algebraic and thus there cannot be an infinite descending chain of them. For W ∈ S , let f (x) = η I(W) . I(W) Since νˆ is μ-stationary and ν is assumed to be μ-invariant, we have (C.11) f (x) = f (gx)dμ(g) # # I(W) I(A(g,x)W) # # # Let S(x) ={I(W) ∈ S : f (x)> 0}. Then, for I(W ) ∈ S(x),I(W ) ∈ S(x), x 1 2 I(W) # # η I(W ) ∩ I(W ) = 0. x 1 2 Thus f (x) ≤ 1. I(W) I(W)∈S(x) 304 ALEX ESKIN, MARYAM MIRZAKHANI Therefore S(x) is at most countable. Let (C.12) f (x) = max f (x). I(W) I(W)∈S(x) Applying (C.11)tosome I(W) for which the max is achieved, we get f (x) ≤ f (gx)dμ(g) i.e. f is a subharmonic function on X. By Lemma C.9, f is constant almost everywhere. Now substituting again into (C.11) we get that the cocycle A permutes the finite set of I(W) where the maximum (C.12) is achieved. Therefore the same is true for the algebraic hull H . If the assumptions of (ii) hold, this is a contradiction since H acts transitively on subspaces of L. If the assumptions of (i) hold then, for W = (W ,..., W ), since the W 1 k i are 1-dimensional, we have I(W) ≡ I(W ) ∩ ... I(W ) 1 k ={subspaces M ⊂ LsuchthatW +··· + W ⊂ M}. 1 k Since H must permute some finite set of I(W) it must thus permute a finite set of sub- spaces of L which contradicts the strong irreducibility assumption. C.5 Proof of Theorem C.6 Recall that we are assuming that the algebraic hull of the cocycle is SL(L) for some vector space L. Let m = dim L. Definition C.11 ((,δ)-regular). — Suppose > 0 and δ> 0 are fixed. A measure η on Gr (L) is (,δ)-regular if for any subspace U of L, η Nbhd I(U) <δ. Lemma C.12. — Suppose g ∈ GL(L) is a sequence of linear transformations, and η is a n n sequence of uniformly (,δ)-regular measures on Gr (L) for some k. Suppose δ 1. Write g = K(n)D(n)K (n), where K(n) and K (n) are orthogonal relative to the standard basis {e ,... e },and D(n) = 1 m diag(d (n),..., d (n)} with d (n) ≥ ··· ≥ d (n). 1 m 1 m (a) Suppose d (n) (C.13) →∞ d (n) k+1 INVARIANT AND STATIONARY MEASURES 305 Then, for any subsequential limit λ of g η there exists a subspace W ∈ Gr (L) such that n n k (C.14)K(n) span{e ,..., e }→ W, 1 k and λ({W}) ≥ 1 − δ. (b) Suppose g η → λ where λ is some measure on Gr (L). Suppose also that there exists a n n k subspace W ∈ Gr (L) such that λ({W})> 5δ. Then, as n →∞, (C.13) holds. As a consequence, by part (a), (C.14) holds and λ({W}) ≥ 1 − δ. Proof of (a). — This statement is standard. Suppose g η → λ. Without loss of gen- n n erality, K (n) is the identity (or else we replace η by K (n)η ). By our assumptions, for n n j < ··· < j , 1 k g (e ∧ ··· ∧ e ) n j j 1 k → 0 unless j = i for 1 ≤ i ≤ k. g (e ∧ ··· ∧ e ) n 1 k Therefore, if U ∈/ I(span{e ,..., e }), k+1 m d g U, K(n) span{e ,..., e } → 0, n 1 k where d(·,·) denotes some distance in Gr (L). After passing to a further subsequence, we may assume that for some W ∈ Gr (L),(C.14) holds. It follows from the (,δ)-regularity of η that λ(W) ≥ 1 − δ. Since δ< 1/2, W is uniquely determined by λ, and therefore (C.14) holds without passing to a further subsequence (but only assuming g η → λ). n n Proof of (b). — This is similar to [GM, Lemma 3.9]. Suppose d (n)/d (n) does k k+1 not go to ∞. Then, there is a subsequence of the g (which we again denote by g ) n n that K(n) → K and that for every j , either d (n)/d (n) converges as n →∞ or ∗ j j+1 d (n)/d (n) →∞ as n →∞. Also without loss of generality we may assume that K (n) j j+1 is the identity (or else we replace η by K (n)η ). n n Let 1 ≤ s ≤ k < r ≤ m be such that s is as small as possible, r is as large as possible, and d (n)/d (n) is bounded for s ≤ j ≤ r − 1. Then, for j < ··· < j , j j+1 1 k g (e ∧ ··· ∧ e ) n j j 1 k (C.15) → 0 unless j = i for 1 ≤ i ≤ s − 1 g (e ∧ ··· ∧ e ) n 1 k and s ≤ j ≤ r for s ≤ i ≤ k. Let V = span{e ,..., e }, V = span{e ,..., e }. − 1 s−1 + 1 r Let D = diag(d (1),..., d (m)) be any diagonal matrix such that for s ≤ j ≤ r − 1, ∗ ∗ ∗ d (j)/d (j + 1) = lim d (n)/d (n). ∗ ∗ j j+1 n→∞ 306 ALEX ESKIN, MARYAM MIRZAKHANI ⊥ ⊥ Then, in view of (C.15), for U such that U ∈/ I(V ) ∪ I(V ), if along some subsequence + − g U → U ,wehave K V ⊂ U ⊂ K V . ∗ − ∗ + −1 ⊥ ⊥ Therefore, we must have V ⊂ K W ⊂ V . Furthermore, for U ∈/ I(V ) ∪ I(V ), − + ∗ + − −1 −1 ⊥ ⊥ if g U → WthenU ∈ I D K W ∩ V + V . ∗ ∗ − + But, since η is (,δ)-regular, ⊥ ⊥ −1 −1 ⊥ ⊥ η Nbhd I V ∪ I V ∪ I D K W ∩ V + V < 3δ. + − ∗ ∗ − + Therefore λ({W})< 3δ which is a contradiction. Thus d (n)/d (n) →∞.Now by k k+1 part (a) (C.14) holds, and λ({W}) ≥ 1 − δ. Let F = F(L) denote the space of full flags on L. Let X = X × F . The cocycle A satisfies the cocycle relation A(g g , x) = A(g , g x)A(g , x). 1 2 1 2 2 The group SL(2, R) acts on the space Xby (C.16) g · (x, f ) = gx, A(g, x)f . Let νˆ be an ergodic μ-stationary measure on X which projects to ν under the natural map X → X. (Note there is always at least one such: one chooses νˆ to be an extreme point among the measures which project to ν.If νˆ =ˆν +ˆν where the νˆ are 1 2 i μ-stationary measures then ν = π (ν) ˆ = π (νˆ ) + π (νˆ ). Since ν is μ-ergodic, this im- ∗ ∗ 1 ∗ 2 plies that π (νˆ ) = π (νˆ ) = ν,hence the νˆ also project to ν . Since νˆ is an extreme point ∗ 1 ∗ 2 i among such measures, we must have νˆ =ˆν =ˆν.Thus νˆ is μ-ergodic.) 1 2 Lemma C.13 (Furstenberg). — For 1 ≤ s ≤ dim L,let σ¯ : SL(2, R) × X → R be given by A(g, x)ξ (f ) σ¯ (g, x, f ) = log ξ (f ) where ξ (f ) is the s-dimensional component of the flag f . (The norms in the above equation are on (V), and here and in the following we make sense of such expressions by picking the same basis for the ξ (f ) in the numerator and denominator.) Then, we have λ + ··· + λ = σ¯ (g, x, f )dν( ˆ x, f )dμ(g), 1 s s SL(2,R) X where λ is the i’th Lyapunov exponent of the cocycle A. i INVARIANT AND STATIONARY MEASURES 307 Proof. — See the proof of [GM, Lemma 5.2]. We may disintegrate dν( ˆ x, f ) = dν(x)dη (f ). Note that Lemma C.10 applies to the projections of the measures η to the various Grass- mannians which are components of F . For a ∈ B, let the measures ν , νˆ be as defined in [BQ, Lemma 3.2], i.e. a a −1 ν = lim (a ... a ) ν a n 1 n→∞ −1 νˆ = lim (a ... a ) ν. ˆ a n 1 n→∞ The limits exist by the martingale convergence theorem. We disintegrate dνˆ (x, f ) = dν (x)dη (f ). a a a,x k k For 1 ≤ k ≤ m,let η = (ξ ) η and η = (ξ ) η ,where ξ : F(L) → Gr (L) is the k ∗ x k ∗ a,x k k x a,x k k natural projection. Then, η and η are measures on Gr (L). x a,x Claim C.14. —Onasetof β × ν full measure, −1 lim (a ... a ) η = η . n 1 a ...a x a,x n 1 n→∞ Equivalently, using (C.16), −1 lim A (a ... a ) , a ... a x η = η . n 1 n 1 a ...a x a,x n 1 n→∞ Proof of claim. — In this claim, we use the invariance of ν.Let C ⊂ Xand D ⊂ F be measurable, and let χ denote the characteristic functions of C. Recall that dν( ˆ x, z) = dν(x)dη (z) is μ-stationary, so that η (D)dν(x)=ˆν(C × D) = (μ∗ˆν)(C × D) = χ (gy)A(g, y)η (D)dν(y)dμ(g) C y −1 = χ (x)A g, g x η −1 (D)dν(x)dμ(g) C g x −1 = A g, g x η −1 (D)dμ(g) dν(x) g x C G 308 ALEX ESKIN, MARYAM MIRZAKHANI Since C and D are arbitrary, we see that −1 η = A g, g x η −1 dμ(g) x g x −1 Therefore (replacing x by a ... a x and g by a ), we have n−1 1 −1 η = A a , a ... a x η dμ(a ). a ...a x n 1 a ...a x n n−1 1 n n 1 −1 Multiplying both sides on the left by A((a ... a ) , a ... a x) and using the cocycle n−1 1 n−1 1 identity −1 A (a ... a ) , a ... a x n 1 n 1 −1 −1 = A (a ... a ) , a ... a x A a , a ... a x , n−1 1 n−1 1 n 1 we get −1 (C.17) A (a ... a ) , a ... a x η n−1 1 n−1 1 a ...a x n−1 1 −1 = A (a ... a ) , a ... a x η dμ(a ). n 1 n 1 a ...a x n n 1 In view of (C.17), the expression −1 A (a ... a ) , a ... a x η n 1 n 1 a ...a x n 1 is a (measure-valued) martingale. Therefore, the claim follows from the martingale con- vergence theorem. If the Lyapunov spectrum is simple, we expect the measures η to be supported a,x at one point. In the general case, let λ ≥ λ ≥ ··· ≥ λ 1 2 m denote the Lyapunov exponents, and let I ={1 ≤ r ≤ m − 1 : λ = λ }. r r+1 Then, by the multiplicative ergodic theorem, Lemmas C.10 and C.12(a), for r ∈/ I, we m−r have η is supported at one point. (This point is the part of the flag (C.1) corresponding a,x to the Lyapunov exponents λ ,...,λ .) r+1 m INVARIANT AND STATIONARY MEASURES 309 Claim C.15. —For any r ∈ I and β × ν -almost all (a, x), for any subspace W(a, x) ∈ m−r Gr (L), we have η ({W(a, x)}) = 0. m−r a,x Proof of claim. — Suppose there exists δ> 0so that for some r ∈ I for a set (a, x) of positive measure, there exists W(a, x) ∈ Gr (L) with η ({W(a, x)})>δ. Then this m−r a,x happens for a subset of full measure by ergodicity. Note that by the cocycle relation, −1 −1 A g , gx = A(g, x) . Therefore, −1 −1 A (a ... a ) , a ... a x = A(a ... a , x) . n 1 n 1 n 1 Hence, on a set of β × ν -full measure, −1 lim A(a ... a , x) η = η . n 1 a ...a x a,x n 1 n→∞ In view of Lemma C.10 (cf. the proof of Lemma 14.4), there exists > 0and a compact K ⊂ Xwith ν(K )> 1 − δ such that the family of measures {η } is uni- δ δ x x∈K formly (,δ/5)-regular. Let N (a, x) ={n ∈ N : a ... a x ∈ K }. δ n 1 δ Write −1 (C.18)A(a ... a , x) = K (a, x)D (a, x)K (a, x) n 1 n n where K and K are orthogonal, and D is diagonal with non-increasing entries. We also n n write ¯ ¯ ¯ (C.19)A(a ... a , x) = K (a, x)D (a, x)K (a, x), n 1 n n ¯ ¯ ¯ where K and K are orthogonal, and D is diagonal with non-increasing entries. Let n n ¯ ¯ d (n, a, x) ≥ ··· ≥ d (n, a, x) be the entries of D (a, x),and let d (n, a, x) ≥ d (n, a, x) ≥ 1 m n 1 2 ¯ ¯ d (n, a, x) be the entries of D (a, x). Then, m n −1 d (n, a, x) = d (n, a, x), m+1−j (C.20) −1 −1 −1 ¯ ¯ K (a, x) = w K (a, x) w , K (a, x) = w K (a, x) w , 0 n n 0 0 n 0 n −1 where w = w is the permutation matrix mapping e to e . Then, by 0 j m+1−j m−r Lemma C.12(b), for β × ν almost all (a, x), η ({W(a, x)}) ≥ 1 − δ (and thus W(a, x) is a,x unique) and as n →∞ along N (a, x) we have: d (n, a, x)/d (n, a, x) →∞, m−r m+1−r 310 ALEX ESKIN, MARYAM MIRZAKHANI and (C.21)K (a, x) span{e ,..., e }→ W(a, x), n 1 m−r where the e are the standard basis for L. Then, by (C.20), ¯ ¯ (C.22) d (n, a, x)/d (n, a, x) →∞, r r+1 and −1 K (a, x) span{e ,..., e }→ w W(a, x) r+1 m 0 Therefore for any > 0 there exists a subset H ⊂ B × Xof β × ν -measure at least 1 − such that the convergence in (C.22)and (C.21) is uniform over (a, x) ∈ H .Hence there exists M > 0such that for any (a, x) ∈ H ,and any n ∈ N (a, x) with n > M, −1 (C.23) K (a, x) span{e ,..., e }∈ Nbhd w W(a, x) . r+1 m 0 n 1 By Lemma C.10 (cf. the proof of Lemma 14.4) there exists a subset H ⊂ Xwith ν(H )> 1 − c ( ) with c ( ) → 0as → 0such that for all x ∈ H ,and any 2 1 2 1 1 1 1 U ∈ Gr (L), m−r η Nbhd I(U) < c ( ), 2 3 1 where c ( ) → 0as → 0. Let 3 1 1 (C.24)H = (a, x, f ) : (a, x) ∈ H , x ∈ H and d ξ (f ), I w W(a, x) > 2 . r 0 1 1 1 Then, (β ׈ν)(H )> 1 − − c ( ) − c ( ),hence (β ׈ν)(H ) → 1as → 0. 1 2 1 3 1 1 1 1 Furthermore, by (C.23) and the definition of H ,for (a, x, f ) ∈ H and n ∈ N (a, x) with 1 1 n > M, we have −1 d ξ (f ), I K (a, x) span{e ,..., e } > . r r+1 m 1 Therefore, in view of (C.19) there exists C = C( ),suchthatfor any (a, x, f ) ∈ H ,any n ∈ N (a, x) with n > M, A(a ... a , x)ξ (f ) 1 n 1 r −1 (C.25)C > d (n, a, x) > ξ (f ) C i=1 (cf. [GM, Lemma 5.1]). Note that for all (a, x, f ) ∈ B × X, all n ∈ N and j = r − 1or j = r + 1we have A(a ... a , x)ξ (f ) n 1 j (C.26) ≤A(a ... a , x) j ≤ d (n, a, x). n 1 i (L) ξ (f ) i=1 INVARIANT AND STATIONARY MEASURES 311 Then, in view of (C.25)and (C.26), for all (a, x, f ) ∈ H ,as n →∞ in N (a, x), (A(a ... a , x))ξ (f ) ξ (f ) n 1 r r−1 (C.27) log ξ (f ) (A(a ... a , x))ξ (f ) r n 1 r−1 ξ (f ) d (n, a, x) r+1 r × ≥ log →∞ (A(a ... a , x))ξ (f ) d (n, a, x) n 1 r+1 r+1 Since (β ׈ν)(H ) → 1as → 0, (C.27) holds as n →∞ along N (a, x) for β ׈ν 1 δ almost all (a, x, f ) ∈ B × X. For 1 ≤ s ≤ m,let σ : B × X → R be defined by σ (a, x, f )=¯σ (a , x, f ),where σ¯ s s s 1 is as in Lemma C.13. Then, the left hand side of (C.27)isexactly n−1 (2σ − σ − σ ) T (a, x, f ) . r r−1 r+1 j=0 Also, we have n ∈ N (a, x) if and only if T (a, x) ∈ K . Then, by Lemma C.8, δ δ (2σ − σ − σ )(q)d(β ׈ν)(q)> 0. r r−1 r+1 B×X By Furstenberg’s formula Lemma C.13, the left hand side of the above equation is λ − λ .Thus λ >λ , contradicting our assumption that r ∈ I. This completes the proof r+1 r r+1 of the claim. Proof of Theorem C.6. — Pick an orthonormal basis at each point of X, and let C(a ∨ b, x) : L → L be a map which makes the subspaces V (a ∨ b, x) orthonormal. Let A denote the cocycle obtained by −1 A(n, a ∨ b, x) = C T (a ∨ b, x) A(a ... a , x)C(a ∨ b, x). n 1 Then A is cohomologous to A. Let η( ˆ a ∨ b, x) = C(a ∨ b, x) η , η˜ = C(a ∨ b, x) η . ∗ x a∨b,x ∗ a,x We have, on a set of β full measure, −1 n η˜ = lim A(n, a ∨ b, x) ηˆ T (a ∨ b, x) . a∨b,x n→∞ In view of Lemma C.10 there exists > 0and a compact K ⊂ B × Xwith β (K )> δ δ 1 − δ such that the family of measures {ˆη(a ∨ b, x)} is uniformly (,δ/5)-regular. (a∨b,x)∈K Write −1 A(n, a ∨ b, x) = K (a ∨ b, x)D (a ∨ b, x)K (a ∨ b, x) n n n 312 ALEX ESKIN, MARYAM MIRZAKHANI where K and K are orthogonal, and D is diagonal with non-increasing entries. Let n n d (n, a ∨ b, x) ≥ ··· ≥ d (n, a, x) be the entries of D (a ∨ b, x). 1 m n m−r By Claim C.15,for r ∈ I and almost all (a ∨ b, x) η˜ has no atoms. It follows a∨b,x that for every δ> 0 there exists K = K (δ) ⊂ B × Xand = (δ) > 0, such that for 1 1 1 1 m−r (a ∨ b, x) ∈ K , η a ∨ b, x gives measure at most δ to the -neighborhood of any point. 1 1 Then, by Lemma C.12(a), there exists C = C (δ) such that if (a ∨ b, x) ∈ K (δ) and 1 1 1 T (a ∨ b, x) ∈ K then for r ∈ I (C.28) d (n, a ∨ b, x)/d (n, a ∨ b, x) ≤ C . m−r m+1−r 1 Note that the matrix of A(n, a ∨ b, x) is block diagonal. We can write each block as a scaling factor times a determinant one matrix which we denote by A (n, a ∨ b, x).(Thus A (n, a ∨ b, x) is, up to a scaling factor, a conjugate of the restriction of A(n, a ∨ b, x) to V (a ∨ b, x).) Since the subspaces defining the blocks are by construction orthogonal, the −1 KAK decomposition of A(n, a ∨ b, x) is compatible with the KAK decompositions of −1 each A (n, a ∨ b, x) . Then, (C.28)for all r ∈ I implies that for all (a ∨ b, x) ∈ K (δ) such i 1 that T (a ∨ b, x) ∈ K we have A (n, a ∨ b, x)≤ C (δ) for all i. It follows that for all n ∈ Z ˜ ˜ β (a ∨ b, x) ∈ B × X :A (n, a ∨ b, x) > C (δ) ≤ 2δ. Since δ> 0 is arbitrary, this means (by definition) that the cocycle A is bounded in the sense of Schmidt, see [Sch]. It is proved in [Sch] that any bounded cocycle is conjugate to a cocycle taking values in an orthogonal group. Therefore the same holds for the determinant one part of the cocycle A| . Proof of Theorems C.4 and C.5.—To proveTheorem C.4, for the case where the algebraic hull is all of SL(L), it is enough to show that for almost all (a, x), the inner product ·, does not depend on b. The proof is similar to the proof of (4.16). a∨b,x For any > 0 exists a compact set K ⊂ B × X of measure 1 − such that the map (a ∨ b, x) →·,· is uniformly continuous on K. Then there exists ⊂ B × Xsuch a∨b,x X n that β () = 1and T (a ∨ b, x) ∈ K for set of n of asymptotic density at least 1/2. For (a ∨ b, x) ∈ B × Xand v,w ∈ V (a, x)/V (a, x),let ≥i >i v,w i,(a∨b,x) [v,w] = i,(a∨b,x) 1/2 1/2 v,v w,w i,(a∨b,x) i,(a∨b,x) Now suppose (a ∨ b, x) ∈ ,and (a ∨ b , x) ∈ . Consider the points T (a ∨ b, x) n n n and T (a ∨ b , x),as n →∞.Then d(T (a ∨ b, x), T (a ∨ b , x)) → 0. Let v = A(a ... a )v, w = A(a ... a )w. n n 1 n n 1 INVARIANT AND STATIONARY MEASURES 313 Then, by Theorem C.6,wehave (C.29) [v ,w ] n =[v,w] , [v ,w ] n =[v,w] . n n i,T (a∨b,x) i,x n n i,T (a∨b ,x) i,(a∨b ,x) n n Now take a sequence n →∞ with T (a ∨ b, x) ∈ K, T (a ∨ b , x) ∈ K(such a sequence exists by the definition of ). Then, [v ,w ] −[v ,w ] n → 0. n n i,T (a∨b,x) n n i,T (a∨b ,x) k k k k Nowfrom(C.29), we get [v,w] =[v,w] . i,(a∨b,x) i,(a∨b ,x) Therefore, for all v,w ∈ V (a, x)/V (a, x) ≥i >i v,w = c a, b, b , x v,w , i,(a∨b,x) i,(a∨b ,x) where c(a, b, b , x) ∈ R . We can (measurably) choose, for almost all (a, x) some b ∈ Bso that (a ∨ b , x) ∈ , and then replace ·,· by 0 i,(a∨b,x) v,w =v,w . i,a∨b ,x i,(a,x) 0 Then ·,· satisfies all the conditions of Theorem C.4. This concludes the proof of i,(a,x) Theorem C.4 for the case where the algebraic hull is all of SL(L). The proof of Theorem C.5 is identical. Appendix D: Dense subgroups of nilpotent groups The aim of this appendix is to prove Proposition D.3 which is used in Section 12. Let N be a nilpotent Lie group. For a subgroup ⊂ N, let denote the topological ¯ ¯ closure of ,and let denote the connected component of containing the identity e of N. Let B(x,) denote the ball of radius centered at x in some left-invariant metric on N. Lemma D.1. — Suppose N is a Lie group, and S ⊂ N is a subset. For > 0,let denote the subgroup generated by S ∩ B(e,). Then there exists > 0 and a connected closed Lie subgroup N of N such that for < , = N . 1 1 1 Proof. — By Cartan’s theorem (see e.g. [Kn, §0.4]), any closed subgroup of a Lie 0 0 ¯ ¯ group is a closed Lie subgroup. Let > 0 bearbitrary. Sincewehave ⊂ for < , there exists > 0such that for ≤ , the dimension of the Lie algebra of (and thus 0 0 itself) is independent of . Thus there exists a connected closed subgroup N ⊂ Nsuch that for ≤ , = N . In particular, 0 1 (D.1) ⊃ N . 1 314 ALEX ESKIN, MARYAM MIRZAKHANI From the definition it is immediate that is a closed subgroup of N. Thus, by ¯ ¯ Cartan’s theorem, and N = are closed submanifolds of N. Therefore, there exists < such that 1 0 ¯ ¯ B(e, ) ∩ = B(e, ) ∩ = B(e, ) ∩ N . 1 1 1 1 Then, for < < , 1 0 ∩ B(e, ) ⊂ ∩ B(e, ) ⊂ N . 1 1 1 Therefore, ⊂ N , and hence ⊂ N .Inviewof(D.1), the lemma follows. 1 1 Lemma D.2. —Suppose N is a simply connected nilpotent Lie group, and let S ⊂ N be an (infinite) subset. For each > 0 let ⊂ N denote the subgroup of N generated by the elements γ ∈ S ∩ B(e,). Suppose that for all > 0, is dense in N. Then, for every > 0 there exist 0 <θ < (depending on and S) such that for every γ ∈ with d(γ, e)<θ there exists n ∈ N and for 1 ≤ i ≤ nelements γ ∈ S with (D.2) γ = γ ...γ n 1 and for each 1 ≤ j ≤ n, (D.3) d(γ ...γ , e)<. j 1 Proof. — We will proceed by induction on dim N. Let N =[N, N] k −1 For k ∈ N,let S be the product of at most k elements in (S ∪ S ) ∩ B(e,).Let k k k T =[S , S ]. This decreases with , so a variant of Lemma D.1 shows that, for small enough , the closure of the group generated by T is a closed connected group N (and N is independent of for small enough). Since N increases with k, it is constant for k k large k.Fix k so that N = N . We will show that N = N . k k+2 k k −1 First, we show that N is normal. For a, b ∈ S and s ∈ S ,wehave s[a, b]s = −1 −1 k+2 k −1 k+2 [sas , sbs ]∈ T .So, sT s ⊂ T . Taking the closure of the generated groups, we −1 get sN s ⊂ N = N .Hence, N is normalized by S . Since S generates a dense k k+2 k k subset of N, N is normal. −1 We have [ab, c]= a[b, c]a [a, c]. This shows that, if [a, c] and [b, c] both belong to N ,then [ab, c] also belongs to N ,bynormality.For x, y ∈ S ,wehave [x, y]∈ N . k k k Taking products, and since S generates a dense subgroup of N, we get [z, y]∈ N for all z ∈ N. Doing the same argument with the other variable, we finally have [z, z ]∈ N for all z, z ∈ N, and therefore N = N as desired. Let S = T ⊂ N .For δ> 0let denote the subgroup of N generated by /4k δ S ∩ B(e,δ). Since (for sufficiently small δ) [B(e,δ), B(e,δ)]⊂ B(e,δ),wehave, for δ< /4k, ⊃ the subgroup generated by T = N . δ δ/4k INVARIANT AND STATIONARY MEASURES 315 Therefore, S ⊂ N satisfies the conditions of the Lemma. Let > 0be such that (D.4)B e, ⊂ B(e,/100). Since dim N < dim N, by the inductive assumption there exist 0 <θ < such that for any γ ∈ with d(γ , e)<θ , there exist γ ∈ S such that (D.2) holds, and (D.3) holds θ i with in place of . Suppose > η > 0. By construction, N/N is abelian. Note that N is connected and simply connected. Then, since = N, there exists a finite set S ≡{λ ,...,λ }⊂ ∩ S 0 1 k η with d(λ , e)<η for 1 ≤ i ≤ k so that λ N ,...,λ N form a basis over R for the vector i 1 k space N/N .Let denote the subgroup generated by the λ ,and let F ⊂ N/N denote the parallelogram centered at the origin whose sides are parallel to the vectors λ N .Then F is a fundamental domain for the action of on N/N ,and diam F = O(η). Let N be a local complement to N in N near the identity e. We can choose N to be a 0 0 smooth manifold transversal to N (N need not be a subgroup). Let π : N → N/N be −1 −1 the natural map, and let π : N/N → N be the inverse. Let F = π (F ).Wecan now choose η sufficiently small so that F ⊂ B(e,ρ),where θ >ρ >η > 0is such that B(e,ρ) ∩ N = B(e,ρ)B(e,ρ)B(e,ρ)B(e,ρ)B(e,ρ) ∩ N ⊂ B e,θ ∩ N . We now choose θ> 0so thatB(e,θ) ⊂ FO where O ⊂ N ∩ B(e,ρ) is some neigh- borhood of the origin. We now claim that for any x ∈ FO and any s ∈ B(e,θ), there exist −1 λ ∈ S ∪ S and γ ∈ such that γ λ sx ∈ FO. Indeed, since B(e,θ)N ⊂ FN ,for any 0 θ x ∈ FN , B(x,θ)N ⊂ λB(x,θ)N . −1 λ∈S ∪S −1 Thus, we can find λ ∈ S ∪ S such that λ sx ∈ FN . Since is dense in N , there exists 0 θ γ ∈ such that γ λ sx ∈ FO, completing the proof of the claim. Now suppose γ ∈ and γ ∈ B(e,θ) ⊂ FO. Then, we have γ = s ... s , where s ∈ S ∩ B(e,θ). n 1 i −1 Note that s ∈ FO.Wenow defineelements λ ∈ S ∪ S and γ ∈ inductively as 1 0 j 0 j θ follows. At every stage of the induction, we will have x ≡ γ λ s ...γ λ s ∈ FO. Suppose j j 1 j j 1 1 316 ALEX ESKIN, MARYAM MIRZAKHANI −1 γ ,...,γ and λ ,...λ have already been chosen. Now choose λ ∈ S ∪ S and 1 j−1 1 j−1 j 0 γ ∈ so that x = γ λ s x ∈ FO.Such λ and γ exist by the claim. j j j−1 j θ j j j j Note that −1 −1 −1 γ = x x s λ j j−1 j j −1 −1 −1 5 ∈ (FO)(FO) B(e,θ) S ∪ S ⊂ B(e,ρ) ⊂ B e,θ . Since x = λ γ s ...λ γ s ∈ FN ,wehave λ s ...λ s ∈ FN . Also γ = s ... s ∈ n n 1 n 1 n 1 n n 1 1 n 1 B(x,θ) ⊂ FN . Since FN is a fundamental domain for the action of on N/N , λ ...λ ∈ N .Thus, n 1 (D.5) γ = γ γ λ s ...γ λ s , n 1 n n 1 1 where γ ∈ N .Wehave −1 −1 γ = γ x ∈ B(e,θ)(FO) ⊂ B e,θ . For notational convenience, denote γ by γ . By the inductive assumption, for 1 ≤ i ≤ n+1 n + 1, we can express γ = s ... s such that s ∈ S ∩ B(e,θ ) and so that for all i, j , i i1 in ij d s ... s , e ≤ . ij i1 We now substitute this into (D.5). Finally, we express each s as a commutator of a product ij of at most k elements of S ∩ B(e,/4k). Then, in view of (D.4), the resulting word satisfies (D.3). Proposition D.3. — Suppose N is a simply connected nilpotent Lie group, O a neighborhood of the identity in N,and μ ameasure on N supported on O. Suppose S ⊂ N is a subset containing elements arbitrarily close to (and distinct from) e, and suppose for each γ ∈ S, (D.6) γ μ ∝ μ −1 on O ∩ γ O where both sides make sense. Then, there exists a nontrivial connected subgroup H of −1 N and a neighborhood O of the identity in H such that for all h ∈ O ,h μ ∝ μ on O ∩ h O. Furthermore, if U is a connected subgroup of N and S contains arbitrarily small elements not contained in U, then H is not contained in U. Proof.—Let N and be as in Lemma D.1. By our assumptions on S, N is non- 1 1 1 trivial (and also N is not contained in U). Now suppose > 0is such thatB(e,) ⊂ O, and let θ> 0 beasinLemma D.2, with N replaced by N . Without loss of generality, we may assume that θ< .Let be the subgroup of N generated by S ∩ B(e,θ). 1 θ 1 Since θ< , is dense in N . Now suppose γ¯ ∈ N ,and d(γ, ¯ e)<θ . Then, there 1 θ 1 1 exists γ ∈ such that γ → γ ,and d(γ , e)<θ . We can write each γ = γ ...γ as k θ k k k k,n k,1 in Lemma D.2. Then, by applying (D.6) repeatedly, we get that (γ ) μ ∝ μ. 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Index of Notation §1-§16 A, 98, 101 B [x], 114 A(q , u,, t), 161 B, 262 A (x, t), 184 B (x, r), 111 A (x, s), 184 B , 268 A, 266 B (x, r), 137 A(g ,v), 103 C [x], 219 t ij A(q , u,, t), 183 Conj , 145 A , 152 D (E, E ), 121 Ad , 145 D (x, y), 121 B, 262 D (E, E ), 121 B(x), 146 D (x, y), 121 B(x, 1/100), 147 E, 185 B(x, r), 147 E (x), 193 ij B[x], 126, 146 E(x), 188 B (x), 146 E (x), 185, 186 t j B [x], 146 E (x), 197 t [ij],bdd B , 141 E (x), 197, 201 ij,bdd 0 INVARIANT AND STATIONARY MEASURES 321 (−−) E (x), 192 H (x), 110 ij big E [x], 217 Im , 102 ij E [c], 112, 126 J[x], 114, 126 F, 139, 261 J , 112 F (x), 189 K , 111 j thick F (x), 186 ≥j K , 111 thick F [x,], 196 K , 111 thick F [x], 196 v Lie(U ), 146 F [x,], 195 ij Lie(G )(x), 145 ++ F [x], 195 ij L , 265 F, 124 L(x), 108 GSpc, 217 L (x), 108, 139 G(x), 144 L (x), 261 G (x), 144 L (x), 139 ext (r) G (x), 144 ++ L (x) , 139 ext (r) G[x], 194 L [x] , 139 ext G , 156 L (q), 165 H(x), 156 L [q], 165 H (x), 217 ++ L (q), 164 H (x), 147 L [q], 164 H (α),97 M, 96, 101 H (x), 148 ++ M(x; y), 229 H(x), 152 N, 98, 101 H (x), 151 N, 98 H(x), 152 N(x), 146 H (x), 151 P, 98, 101 H(α),96 ! P (x, y), 228 H , 109 + P (x, y), 158, 193 H (M,, R), 101 − P (x, y), 222 H (x), 109, 144 Z(x) P (x, y), 222, 229 H , 109 P (x), 204 H (x), 109 P, 138, 141 H , 109 P(x, y), 174 H (x), 109 P (uq , q ), 173 s 1 H (x), 101 P (x, y), 228 H (x, y), 203 v + P (x, y), 118 H , 109 big − P (x, y), 118 H (x), 110 big GM P (x, y), 144 (+) H (x), 110 GM big P , 144 (++) Z(x) H (x), 110 big P (x, y), 229 (−) H (x), 110 P , 140 i,x big 322 ALEX ESKIN, MARYAM MIRZAKHANI V (H )(x), 116 Q(uq ; q ), 177 ≤i ˆ V (W), 147 ≤i Q (uq ; q ), 178 s 1 V (x), 147 Q (x), 144 ≤i (k) τ ,X V , 262 ≤j Q , 271 V[x], 101, 125 Q (x), 144 ++ ⊗m V , 139 R(x, y), 194 V , 139 Re , 102 Z(x) W(x), 102 S , 153 W (b, x), 266 SO(2),98 W [b, x], 269 SV, 121 W [x], 102, 125 S [x], 165 W (b, x), 269 S , 151 W [b, x], 269 T, 262 τ − W [x], 102, 125 T , 268 W (b), 266 T , 268 W (a), 266 T (b, x, s) , 268 t ∗ X, 101, 124 Tr(x, y), 145 X, 101, 124 U(M,v), 148 X , 108 U , 217 0 X , 101, 108 U (x), 145 0 Y , 138 U [x], 145 Z(x), 149 U(x), 260 Z (x), 149 U (b, x), 269 Z (x), 149 U (x), 102, 145 i1 Z (x), 149 U [b, x], 269 i2 (x ), 124 U [x], 102, 145 (x ), 124 U [b, x], 269 i 0 U , 260 , 261 Vol(x), 260 , 197, 261 V (x), 186 , 192 <i V (x), 186 , 192 >i V (x), 186 , 272 ≥i ρ V (x), 186 , 217 ≤i (k) V , 265 , 96, 101 ≤j (x), 127 V (H )(x), 115 (x), 129 V (H )(x), 117 α,96 V (H )(x), 117 β , 262 V (W)(x), 147 V (H)(x), 158 β , 268 τ ,X V (x), 147 β , 269 V (H )(x), 116 δ (·,·), 121 ≥i Y INVARIANT AND STATIONARY MEASURES 323 (δ), 248 τ , 267 0 j exp, 145 θ , 201, 202 B [x], 125 θ , 202 0 0 B [x], 125 θ , 202 πˆ , 165 θ , 267 πˆ , 165 B , 268 Z(x) ˆ ˜ λ , 262 P (x, y), 229 λ , 261 a ∨ b, 263 λ , 147 area(x, 1), 260 λ (H ), 115 d(·,·), 192 λ (H ), 117 d (·,·), 110 λ (H ), 117 d (x, y), 125 λ (W), 147 d (·,·), 111 λ (H), 158 d (·,·), 220 i ∗ λ (x, t), 128, 193 d (V , V ), 121 ij Y 1 2 λ (x, y), 194 f (x), 227 ij ij ⊗m , , 267 f , 139 j,b,x ·,· , 128, 192 f , 139 ·,·, 129, 193 f (x), 220 ij ·,· , 119 f [x], 220 ij ij,x g,96 ·,· , 129 v,x g˜ , 196 μ, 260 g˜ , 196 μˆ , 207 g , 196 μ˜ , 207 ij ν , 102, 103, 124 g , 194 ν , 114 g , 101 W [x] t ν , 114 (g ) , 110, 116, 157 W t ∗ ω, 96, 101 h , 165 φ (z), 227 hd (·,·), 137, 158 π , 101 height(v), 204 (i ) , 179 πˆ , 151 u,q ,s ∗ i , 180 π , 102 u,q ,s i , 178, 179 π (X ), 108 1 0 u,q ,s π + , 173 ij ∼ kr, 204 W (q ) ρ(x, E), 199 [ij], 204 ρ(x, y), 199 j, 152 σ , 124, 262 j, 152 ⊗m ∼ , 217 j , 139 ij τˆ (x, t), 196 p, 101, 116, 260 τˆ (q , u,), 184 r, 139 () 1 τˆ (x, t), 194 r ,98 ij θ 324 ALEX ESKIN, MARYAM MIRZAKHANI t(c), 112 |·|, 147, 195, 196 ·, 126, 193, 270 tr(x, y), 145 · , 110 (u) , 153, 156 · , 129, 192 u , 152 · , 110 Y,x u , 101 · , 129 t x A. E. Department of Mathematics, University of Chicago, Chicago, IL 60637, USA eskin@math.uchicago.edu M. M. Department of Mathematics, Stanford University, Stanford, CA 94305, USA mmirzakh@math.stanford.edu Manuscrit reçu le 17 février 2014 Manuscrit accepté le 12 février 2018 publié en ligne le 17 avril 2018.
Publications mathématiques de l'IHÉS – Springer Journals
Published: Apr 17, 2018
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