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A Polishchuk (2002)
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nils.matthes@maths.ox.ac.uk Department of Mathematics, We characterize Zagier’s generating series of extended period polynomials of Kyushu University, 744 Motooka, normalized Hecke eigenforms for PSL (Z) in terms of the period relations and existence Nishi-Ku, Fukuoka 819-0395, 2 Japan of a suitable factorization. For this, we prove a characterization of the Kronecker Full list of author information is function as the “fundamental solution” of the Fay identity. available at the end of the article Keywords: Periods of modular forms, Theta functions Mathematics Subject Classification: 11F67 (11F27) 1 Introduction 1.1 Zagier’s generating series of extended period polynomials To every cusp form f , one can attach its period polynomial r (X). This is an important arithmetic invariant whose coefficients are, up to elementary factors, the critical values of the completed L-function of f . Period polynomials were studied extensively by Eichler, Shimura and Manin (see, for example, [7, Chapter V] for a textbook account), and their definition was extended to all modular forms by Zagier [12]. Nowadays, extended period polynomials are well-studied objects in number theory which keep attracting attention. For example, they are related to zeta functions of real quadratic fields [6] and multiple zeta values [4]. More recently, an analogue of the Riemann hypothesis has been proved for them [2,5]. In [12], Zagier introduced a generating series C(X, Y ; τ; T) of extended period polyno- mials of normalized Hecke eigenforms (see Sect. 4.1 for its definition) and related it to the Kronecker function [10, Chapter VIII], θ (0)θ (u + v) F (u, v):= . θ (u)θ (v) τ τ Here θ (u) denotes Jacobi’s odd theta function (see Sect. 2.1 for the definition). More precisely, Zagier’s main result is as follows. Theorem 1.1 (Zagier [12], Eq. (17)) We have the following identity of formal series in variables X,Y,T: C(X, Y ; τ; T) = F (T, −XYT)F (XT, YT). τ τ Here and in what follows every modular form will be for the full modular group PSL (Z):= SL (Z)/{±1}. 2 2 © The Author(s) 2019. This article is distributed under the terms of the Creative Commons Attribution 4.0 International License (http://creativecommons.org/licenses/by/4.0/), which permits unrestricted use, distribution, and reproduction in any medium, provided you give appropriate credit to the original author(s) and the source, provide a link to the Creative Commons license, and indicate if changes were made. 0123456789().,–: volV 24 Page 2 of 13 Matthes Res Math Sci (2019) 6:24 As two applications, we mention the algorithmic computation (up to scalars) of the period polynomials of all cuspidal Hecke eigenforms, as well as a new proof of the Eichler–Selberg trace formula for PSL (Z)[11]. 1.2 Period relations and the main result The goal of this paper is to give another application of Zagier’s result, namely a purely algebraic characterization of C in terms of the period relations for extended period poly- nomials (1.1) r (X)|(1 + S) = r (X)|(1 + U + U ) = 0, f f 0 −1 1 −1 where S = , U = and | denotes the slash operator. Indeed, since C is defined 10 10 as a generating series of extended period polynomials it also satisfies a suitable version of the period relations; see Eq. (4.1). It turns out that the period relations alone do not suffice to characterize C, as for any k−2 modular form f of weight k the series C + r (X)T also satisfies the period relations. However, if we demand in addition the existence of a factorization as in Theorem 1.1, then the situation is much better and our main result is as follows. m n Theorem 1.2 For a formal Laurent series f (u, v) = a u v ∈ C((u, v)), m,n m,n> >−∞ assume that C (X, Y, T):= f (T, −XYT)f (XT, YT) ∈ C((X, Y, T)) satisfies the period relations (Eq. (4.1)). Then either: (i) there exist unique α, β ∈ C such that (αXY − β)(βX + αY ) C (X, Y, T) = C (X, Y, T) = , f P α,β 2 2 2 X Y T −1 −1 where P (u, v) = αu + βv or α,β (ii) there exist α, β, δ ∈ C and τ ∈ H ∪{i∞} such that T αY C (X, Y, T) = C (X, Y ,T ) = C(X, Y ; τ; T ), where T := ,Y := . α β Moreover, if α , β , δ , τ are a different choice of parameters as above, then there exists ab a matrix ∈ SL (Z) such that cd δ aτ + b (α , β , δ , τ ) = α(cτ + d), β(cτ + d), , . cτ + d cτ + d The basic idea is that, via Theorem 1.1, the period relations for C are equivalent to the Fay identity for F (u, v)(Eq.(2.1)) which is a consequence of Fay’s trisecant identity for theta functions [3, p. 34, Eq. (45)]. More precisely, C satisfies the period relations if and only if f satisfies the Fay identity; see Proposition 4.1. We are therefore led to characterize solutions of the Fay identity, for which we have the following result. m n Theorem 1.3 Assume that f (u, v) = a u v satisfies the Fay identity m,n m,n> >−∞ (Eq. (2.1)). Matthes Res Math Sci (2019) 6:24 Page 3 of 13 24 (i) If (a ,a ) = (0, 0), then there exist unique α, β, γ ∈ C such that 3,0 5,0 α β γ uv f (u, v) = e + . u v (ii) If (a ,a ) = (0, 0), then there exist α, β, δ ∈ C , γ ∈ C and τ ∈ H ∪{i∞} such that 3,0 5,0 u v γ uv f (u, v) = δe F , . α β Moreover, if α , β , γ , δ , τ are a different choice of parameters as above, then there ab exists ∈ SL (Z) such that cd c/2πi δ aτ + b (α , β , γ , δ , τ ) = α(cτ + d), β(cτ + d), γ − , , . αβ(cτ + d) cτ + d cτ + d The idea of proof is that the Fay identity implies recurrence relations for the coefficients of f such that every solution is uniquely determined by at most five of its coefficients (see Proposition 3.3 and Theorem 3.6 for the precise statements). Varying the parameters α, β, γ , δ, τ suitably, we can arrange that f is equal to one of the functions in Theorem 1.3. After a previous version of this manuscript had been submitted, the author was informed that the existence part of Theorem 1.3 is equivalent to a result of Polishchuk [8,Theorem 5]. To be precise, Polishchuk works with the (scalar) associative Yang–Baxter equation which is equivalent to the Fay identity [9] and classifies solutions in the space of mero- morphic functions defined in a neighborhood of (0, 0) instead of formal Laurent series. The two proofs are quite similar, but ours may still be of independent interest as it yields slightly more information about the coefficients of solutions to the Fay identity (see Propo- sition 3.8). This in turn sheds some light on how solutions to the associative Yang–Baxter equation can be constructed algorithmically. 1.3 Content Section 2 is preliminary; we recall the definition of the Kronecker function F (u, v)and state the Fay identity. This identity is then studied in detail in Sect. 3 which culminates in the proof of Theorem 1.3. In Sect. 4, we explain the relation between Fay identity and period relations, building on Zagier’s fundamental result, and finish by giving a proof of Theorem 1.2. Notation Given variables X , ... ,X and a ring R,wewilldenoteby R((X , ... ,X )) the 1 n 1 n R-algebra of formal Laurent series of the form m m a X ... X , with a ∈ R. m ,...,m m ,...,m 1 n n 1 n m ,...,m > >−∞ 1 n 2 The Kronecker function and its functional equations The reference for this section is [12]. 2.1 The Fay identity for the Kronecker function 1 1 1 n+ n+ u n 2 2 2 Consider the classical odd Jacobi theta function θ (u):= (−1) q e .It n∈Z is entire, has simple zeros exactly for u ∈ 2πi(Z + Zτ) and satisfies θ (u + 2πi(mτ + r)) = m+r − −mu (−1) q e θ (u), for m, r ∈ Z. τ 24 Page 4 of 13 Matthes Res Math Sci (2019) 6:24 Definition 2.1 The Kronecker function is the meromorphic function F : C → C, defined by θ (0)θ (u + v) F (u, v) = . θ (u)θ (v) τ τ From the properties of θ mentioned above, it follows that F (u, v) has simple poles if u or τ τ v is in 2πi(Z + Zτ), simple zeros for u + v ∈ 2πi(Z + Zτ) and satisfies F (u + 2πi(nτ + −mn −mu−nv s),v + 2πi(mτ + r)) = q e F (u, v), for m, n, r, s ∈ Z. Also, at the cusp i∞,the Kronecker function degenerates to a trigonometric function: 1 u v F (u, v)| = coth + coth . τ τ =i∞ 2 2 2 Remark 2.2 The above version of the Kronecker function is the one given in [12]. Other sources such as [1,8] use the function 2πiF (2πiu, 2πiv) instead. We are interested in functional equations satisfied by F .Since θ (−u) =−θ (u), the τ τ τ Kronecker function is antisymmetric, i.e., F (u, v) + F (−u, −v) = 0. More interestingly, τ τ it satisfies the following three-term functional equation [1], Proposition 5.(iii). Proposition 2.3 The Kronecker function satisfies the Fay identity, F (u ,v )F (u ,v ) + F (−u ,v − v )F (u + u ,v ) τ 1 1 τ 2 2 τ 2 1 2 τ 1 2 1 +F (−u − u , −v )F (u ,v − v ) = 0. (2.1) τ 1 2 2 τ 1 1 2 Proof We give a proof for convenience of the reader. Writing out the definition of F (u, v) and multiplying by the common denominator of the left-hand side of (2.1), we see that (2.1) is equivalent to 0 = θ (u + v )θ (u + v )θ (u + u )θ (v − v ) τ 1 1 τ 2 2 τ 1 2 τ 1 2 − θ (−u + v − v )θ (u + u + v )θ (v )θ (u ) (2.2) τ 2 1 2 τ 1 2 1 τ 2 τ 1 + θ (−u − u − v )θ (u + v − v )θ (u )θ (v ), τ 1 2 2 τ 1 1 2 τ 2 τ 1 where we also used θ (−u) =−θ (u). Now substituting τ τ α = u + u + v , α =−u + v − v , α =−α − α =−u − v , 0 1 2 2 1 2 1 2 2 0 1 1 1 β = u , β = v , β =−β − β =−u − v , 0 2 1 2 2 0 1 2 2 and again using antisymmetry of θ ,wecan write (2.2) in the more symmetric form θ (α )θ (β )θ (α + β )θ (α − β ) = 0, τ i τ i τ i−1 i+1 τ i+1 i−1 i∈Z/3Z and this is precisely [12, Proposition 5]. Slightly more generally, we have the following result. Note that our version is slightly different; see Remark 3.2 for more details. Matthes Res Math Sci (2019) 6:24 Page 5 of 13 24 Corollary 2.4 The functions α β u v γ uv γ uv e + and δe F , u v α β both satisfy the Fay identity, for all α, β, γ ∈ C in the first case and for all α, β ∈ C , γ , δ ∈ C and τ ∈ H ∪{i∞} in the second. γ u v γ u v 1 1 2 2 Proof The key observation is that the product e e is invariant under the following two linear transformations which occur in the Fay identity: (u ,v ,u ,v ) → (−u ,v − v ,u + u ,v ), 1 1 2 2 2 1 2 1 2 1 (u ,v ,u ,v ) → (−u − u , −v ,u ,v − v ). 1 1 2 2 1 2 2 1 1 2 The corollary then follows from partial fractions in the first case and from Proposition 2.3 in the second. 3 Algebraic structure of the Fay identity m n In this section, we always let f (u, v) = a u v ∈ C((u, v)) be a formal Laurent m,n m,n> >−∞ series. The goal of this section is to derive constraints on the coefficients a imposed m,n by the Fay identity. Our main results (Theorem 3.6 and Proposition 3.3) show that if f satisfies the Fay identity, it is uniquely determined by at most five of its coefficients. 3.1 Some basic implications of the Fay identity We begin by showing how the Fay identity implies the vanishing of many of the coefficients a . m,n Proposition 3.1 If f satisfies the Fay identity, then a a −1,0 0,−1 m n f (u, v) = + + a u v ,witha = 0,if m + n ∈ 2Z. m,n m,n u v m,n≥0 In particular, f (−u, −v) =−f (u, v).Moreover, if (a ,a ) = (0, 0),thenf = 0. −1,0 0,−1 Proof We first show that if f = 0, then f must have a pole at either u = 0or v = 0. Indeed, if f does not have a pole at v = 0, then the Fay identity implies f (u , 0)f (u , 0) + f (−u , 0)f (u + u , 0) + f (−u − u , 0)f (u , 0) = 0, 1 2 2 1 2 1 2 1 f (u ,v )f (u ,v ) + f (−u , 0)f (u + u ,v ) + f (−u − u , −v )f (u , 0) = 0. 1 1 2 1 2 1 2 1 1 2 1 1 −1 The first equation implies f (u, 0) = a u , as can be seen by comparing coefficients. −1,0 In particular, if f does not have a pole at u = 0, we must have f (u, 0) = 0, butthenthe second equation implies f (u, v) = 0. Now assume that f = 0and let M, N ∈ Z be the largest integers such that a = 0 −M,n for some n ∈ Z and a = 0 for some m ∈ Z.Since f has a pole at either u = 0or m,−N v = 0, we have M ≥ 1or N ≥ 1. Define M n N g(v):= (u f (u, v))| = a v ,h(u):= (v f (u, v))| u=0 −M,n v=0 n> >−∞ = a u . m,−N m> >−∞ 24 Page 6 of 13 Matthes Res Math Sci (2019) 6:24 Both g and h are well defined by construction of M and N.Now,if f has a pole at u = 0, we multiply the Fay identity by (u u ) and then set u = u = u = 0toget 1 2 1 2 M M −1 −1 g(v )g(v ) + g(v − v )g(v ) + g(−v )g(v − v ) = 0, 1 2 1 2 1 2 1 2 2 2 and it is straightforward to verify that this implies M = 1 and that g(v) = a . Likewise, −1,0 if f (u, v)has apoleat v = 0(i.e., N ≥ 1), then a similar argument yields N = 1and h(u) = a . This shows that 0,−1 a a −1,0 0,−1 m n f (u, v) = + + a u v , m,n u v m,n≥0 with (a ,a ) = (0, 0) if f = 0. −1,0 0,−1 It remains to prove that a = 0if m + n is even which is equivalent to antisymmetry m,n f (−u, −v) =−f (u, v). For this, we may clearly assume that f = 0. Taking the residues of the Fay identity at u = 0, respectively, at v = 0, gives 1 1 a f (u ,v ) + a f (−u , −v ) = 0, −1,0 2 2 −1,0 2 2 a f (u ,v ) + a f (−u , −v ) = 0, 0,−1 2 2 0,−1 2 2 and since (a ,a ) = (0, 0), the result follows. −1,0 0,−1 Remark 3.2 We have already mentioned that our version of the Fay identity is slightly different from the one in [1, Proposition 5]. In particular, the latter does not imply anti- symmetry. Indeed, for every α = 0 the function f (u, v):= α(coth(αu) + 1) satisfies the Fay identity as given in [1] (with u corresponding to ξ), butdoesnot satisfyEq. (2.1). On the other hand, antisymmetry together with the version of the Fay identity given in loc.cit. are equivalent to (2.1). Therefore, our version of the Fay identity subsumes antisymmetry as well which is the reason why we prefer to work with it. By Proposition 3.1, we already know that f = 0, if (a ,a ) = (0, 0). Therefore, the −1,0 0,−1 next proposition finishes the proof of Theorem 1.3 in the special case a a = 0. −1,0 0,−1 Proposition 3.3 Assume that f = 0 satisfies the Fay identity. (i) If a = 0, then f is uniquely determined by a and a ,and we have 0,−1 −1,0 0,1 γ uv a e a −1,0 0,1 f (u, v) = ,with γ = . u a −1,0 (ii) If a = 0, then f is uniquely determined by a and a ,and we have −1,0 0,−1 1,0 γ uv a e a 0,−1 1,0 f (u, v) = ,with γ = . v a 0,−1 Note that γ is well defined in both cases. Indeed, since f = 0wemusthave a = 0in −1,0 the first case and a = 0 in the second one, by Proposition 3.1. 0,−1 Matthes Res Math Sci (2019) 6:24 Page 7 of 13 24 Proof We only prove (i), the proof of (ii) is analogous. Since a = 0, Proposition 3.1 0,−1 implies that f (u, v) does not have a pole at v = 0 and the Fay identity with v := v = v 1 2 and u := u = u yields 1 2 f (u, v) − 2f (u, 0)f (2u, v) = 0. (3.1) −1 Writing out the Laurent expansion of f and using that f (u, 0) = a u (see the proof −1,0 of Proposition 3.1), Eq. (3.1) is equivalent to ⎛ ⎞ m n m+1 m−1 n ⎝ ⎠ a u v − a (2 − 2)a u v = 0. m,n −1,0 m,n m,n≥0 m≥1,n≥0 This implies a = 0 for m =−1, and then, by induction on n we get a = 0, if m,0 m,n m = n − 1. More generally, ⎛ ⎞ ⎜ ⎟ n−2 n m+1 m−1 n ⎝ a a ⎠ u v − a (2 − 2)a u v = 0, i−1,i j−1,j −1,0 m,n 1≤i,j≤n−1 m≥1 i+j=n for every n, showing that a is recursively determined by a and a , and therefore, n−1,n −1,0 0,1 f itself is uniquely determined by a and a . On the other hand, by Corollary 2.4 there −1,0 0,1 exists a solution to the Fay identity for any given values of a ∈ C and a ∈ C, −1,0 0,1 γ uv −1 namely αe u with α = a and γ = a /a , and this ends the proof. −1,0 0,1 −1,0 3.2 The ideal of Fay relations To study the Fay identity in more detail, it will be convenient to replace the coefficients a ∈ C by symbols A and accordingly to study a formal version of the Fay identity. m,n m,n More precisely, consider the polynomial C-algebra A := C[{A ,A }∪{A | m, n ≥ 0,m + n odd}] (3.2) −1,0 0,−1 m,n (the restriction on the indices (m, n) is justified by Proposition 3.1)and let m n −1 (u, v) = A u v ∈ (uv) A[[u, v]] m,n be the generic element where the sum is over all (m, n)asin (3.2). By definition, it satisfies (−u, −v) =− (u, v). Also, define F(u ,u ,v ,v ):= (u ,v ) (u ,v ) + (−u ,v − v ) (u + u ,v ) 1 2 1 2 1 1 2 2 2 1 2 1 2 1 (3.3) (−u − u , −v ) (u ,v − v ). 1 2 2 1 1 2 −1 A priori, it is contained in (u u v v (u + u )(v − v )) A[[u ,u ,v ,v ]]. 1 2 1 2 1 2 1 2 1 2 1 2 Lemma 3.4 The element F(u ,u ,v ,v ) is contained in A[[u ,u ,v ,v ]], i.e., 1 2 1 2 1 2 1 2 m m n n 1 2 1 2 F(u ,u ,v ,v ) = c u u v v , 1 2 1 2 m ,m ,n ,n 1 2 1 2 1 2 1 2 m ,m ,n ,n ≥0 1 2 1 2 for some c ∈ A. m ,m ,n ,n 1 2 1 2 24 Page 8 of 13 Matthes Res Math Sci (2019) 6:24 Proof By definition, (u, v) has simple poles exactly along u = 0and v = 0 with residues A and A , respectively. It is therefore sufficient to check that all residues of F −1,0 0,−1 along u = 0, u = 0, u + u = 0, v = 0, v = 0and v − v = 0 vanish, which is 1 2 1 2 1 2 1 2 straightforward and essentially only uses that (−u, −v) =− (u, v). Definition 3.5 Define J ⊂ A to be the ideal generated by the coefficients c of m ,m ,n ,n 1 2 1 2 F. Also, define A := A/J to be the corresponding quotient. The ideal J could be called the ideal of Fay relations. By definition, giving a formal m n Laurent series f (u, v) = a u v ∈ C((u, v)) which satisfies the Fay identity m,n m,n> >−∞ is equivalent to giving a homomorphism of C-algebras ϕ : A → C, A → a . m,n m,n Solutions to the Fay identity with a a = 0 correspond under this identification to −1,0 0,−1 −1 −1 ¯ ¯ ¯ ¯ ¯ homomorphisms ϕ : A → C, where A := A ⊗ C[A , A ]. By Proposition 3.3, f 0 0 C −1,0 0,−1 it is enough to classify the latter, and we shall therefore be interested in understanding ¯ ¯ ¯ the structure of A .Tothisend,let A ⊂ A,bethe C-subalgebra generated by the set ¯ ¯ ¯ ¯ {A }∪{A | m ∈{(−1, 1, 3, 5)}} and denote by ι : A → A the canonical inclusion. 0,−1 m,0 −1 −1 ¯ ¯ Extending scalars to C[A , A ], we get an induced map −1,0 0,−1 −1 −1 ¯ ¯ ¯ ¯ ¯ ¯ ι : A → A , where A := A ⊗ C[A , A ], 0 0 C 0 0 −1,0 0,−1 which is clearly injective. Theorem 3.6 The map ι is an isomorphism of algebras. More concretely, using the 1-1 correspondence f ↔ ϕ described above, Theorem 3.6 says that every solution f (u, v) ∈ C((u, v)) to the Fay identity with a a = 0 is uniquely −1,0 0,−1 determined by its coefficients a and a , for m =−1, 1, 3, 5. 0,−1 m,0 Remark 3.7 Although we will not need this, one can show that A is freely generated, −1 −1 ¯ ¯ ¯ ¯ as a C[A , A ]-algebra, by A and A for m =−1, 1, 3, 5; this follows from 0,−1 m,0 −1,0 0,−1 Theorem 3.6 since for every quintuple (z ,w ,z ,z ,z ) ∈ C such that z w = 0 −1 −1 1 3 5 −1 −1 m n there exists a solution f (u, v) = a u v to the Fay identity such that a = m,n 0,−1 m,n> >−∞ w and a = z for m =−1, 1, 3, 5 (see the proof of Theorem 1.3). −1 m,0 m Before we prove Theorem 3.6, we need to introduce some more notation. Let p ⊂ A be the ideal generated by all A with m, n ≥ 0. This is a homogeneous prime ideal of m,n A, the grading being defined by giving A degree m + n. Moreover, since the ideal J of m,n Fay relations is homogeneous, this grading descends to the quotient A. In general, given a homogeneous ideal I of either A or A,wewilldenoteby I its component of degree k. The following proposition gives explicit formulas for some of the coefficients c m ,m ,n ,n 1 2 1 2 and will be the key for proving Theorem 3.6. Proposition 3.8 We have the following formulas for the coefficients of F: c = 3A A − 3A A (3.4) 0,0,0,0 0,−1 0,1 −1,0 1,0 Matthes Res Math Sci (2019) 6:24 Page 9 of 13 24 and for k ≥ 2 even, c ≡ (k + 3)A A − 2A A mod p , (3.5) 0,0,0,k 0,−1 0,k+1 −1,0 1,k c ≡ 2A A − (k + 3)A A mod p , (3.6) 0,k,0,0 0,−1 k,1 −1,0 k+1,0 k + 2 k c ≡ + 1 A A − ( + δ )A A mod p , (3.7) 0,−1 −1,0 0,0,2,k−2 0,k+1 k,2 1,k 3 2 k + 2 k c ≡− + 1 A A + + δ A A mod p . k−2,2,0,0 −1,0 k+1,0 k,2 0,−1 k,1 3 2 (3.8) Finally, for 0 < m < k with k as above, we have c ≡ (k + 2 − m)A A − (m + 2)A A mod p . (3.9) 0,−1 −1,0 0,m,0,k−m m,k+1−m m+1,k−m r s Proof Since the Fay identity is homogeneous, for every monomial u v with r, s ≥ 0the 2 2 r s 2 r s coefficient of u v in (3.3) is congruent modulo p to the coefficient of u v in 2 2 r+s 2 2 ϕ (u ,v )ϕ (u ,v ) + ϕ (−u ,v − v )ϕ (u + u ,v ) r,s 1 1 r,s 2 2 r,s 2 1 2 r,s 1 2 1 + ϕ (−u − u , −v )ϕ (u ,v − v ), (3.10) r,s 1 2 2 r,s 1 1 2 −1 −1 r s+1 r+1 s where ϕ (u, v) = A u + A v + A u v + A u v . A straightforward r,s −1,0 0,−1 r,s+1 r+1,s k−m computation of the coefficient of u v in (3.10) now yields (3.4), (3.5), (3.6)and (3.9). 2 2 2 k−2 k−2 2 Similarly, for k ≥ 2 the coefficient of v v (respectively, of u u )in(3.3) is congruent 1 2 1 2 modulo p to the coefficient of the same monomial in (3.10) for (r, s) = (0,k) (respectively, for (r, s) = (k, 0)), and we get (3.7)and (3.8). Proof of Theorem 3.6 It is clearly enough to show that A ∈ A for all m, n.Weprove m,n ¯ ¯ ¯ this by induction on the degree d = m + n of A .For d = 1, we have A ∈ A by m,n 1,0 −1 ¯ ¯ ¯ ¯ ¯ definition and it follows from (3.4) that A = A A A ∈ A . Note that this also 0,1 0,−1 1,0 −1,0 0 2 2 2 ¯ ¯ ¯ ¯ implies that (p ) ⊂ A where p denotes the ideal generated by the image of p in A . 2 0 0 0 0 ¯ ¯ Now we use induction on d to show that A ∈ A for all m, n with d = m + n.(This m,n ¯ ¯ ¯ shows in particular that (p ) ⊂ A .) For d = 3or d = 5, since A ∈ A in that d−1 d,0 0 0 case, we see from (3.6) together with (p¯ ) ⊂ A (which follows from the induction d−1 0 0 ¯ ¯ hypothesis) that A ∈ A . Repeating the same argument, using (3.9) and finally (3.5), d−1,1 ¯ ¯ we obtain A ∈ A for all m + n = d,if d = 3or d = 5. m,n k+2 k If d ≥ 7, the crucial point is that the row vectors (−(k +3), 2) and (− −1, + δ ), k,2 3 2 where k = d − 1, are linearly independent. Therefore, from (3.6)and (3.8), we see that 2 2 ¯ ¯ ¯ ¯ A ≡ 0 mod (p ) , i.e., A ∈ (p ) . By our induction hypothesis, this shows d,0 d−1 d,0 d−1 0 0 ¯ ¯ ¯ ¯ A ∈ A and using a similar argument as before we get A ∈ A for all m + n = d. m,n d,0 0 0 3.3 Proof of Theorem 1.3 m n Let f (u, v) = a u v be a solution to the Fay identity. By Proposition 3.1, m,n m,n> >−∞ we have a a −1,0 0,−1 m n f (u, v) = + + a u v , with a = 0, if m + n ∈ 2Z. m,n m,n u v m,n≥0 24 Page 10 of 13 Matthes Res Math Sci (2019) 6:24 The case a a = 0 has already been taken care of by Propositions 3.1 and 3.3 so that −1,0 0,−1 we may assume a a = 0, i.e., a and a are both invertible. We distinguish −1,0 0,−1 −1,0 0,−1 between two cases. Case (i): (a ,a ) = (0, 0). Let 3,0 5,0 b b −1,0 0,−1 m n + + b u v , m,n u v m,n≥0 γ uv 3 be the Laurent expansion of e (α/u + β/v). It is straightforward to verify that setting α = a , β = a and γ = a /β,wehave a = b and a = b for −1,0 0,−1 1,0 0,−1 0,−1 m,0 m,0 m =−1, 1, 3, 5. Therefore, a = b for all m, n by Theorem 3.6. The uniqueness of m,n m,n α, β and γ is clear. Case (ii): (a ,a ) = (0, 0). For any choice of parameters α, β, γ , δ, τ as in the statement 3,0 5,0 of Theorem 1.3.(ii), we get the following Laurent expansion u v b b −1,0 0,−1 γ uv m n δe F , = + + b u v . τ m,n α β u v m,n≥0 It follows from [12, Theorem 3.(iv)] that b = αδ, b = βδ, and −1,0 0,−1 −m −(m+1) α α −2δ G (τ) =−2b G (τ),m = 1, m+1 −1,0 m+1 b = m! m! m,0 −1 −2 βγ δ − 2δα G (τ) = b γ − 2b α G (τ),m = 1, 2 0,−1 −1,0 2 where G (τ) is the Hecke-normalized Eisenstein series of weight k. (The normalization is in fact irrelevant here; we will only need that G (τ) is a modular form of weight k.) We now view b and b as functions of (α, τ) and claim that the map 3,0 5,0 × 2 C × H → C \{(0, 0)}, (α, τ) → (b ,b ), 3,0 5,0 where H := H ∪{i∞}, is surjective. Indeed, it is enough to prove that the map (α, τ) 3 2 (b ,b ) is surjective which in turn is equivalent to proving surjectivity of 3,0 5,0 1 3 2 H → P (C), τ → [b : b ]. 3,0 5,0 3 2 But this map is surjective because the quotient b /b is a non-constant modular function 3,0 5,0 3 2 1 (being proportional to G (τ) /G (τ) ), and every such function surjects onto P (C). 4 6 By what we have just proved, given any coefficients (a ,a ) ∈ C \{(0, 0)},wecan 3,0 5,0 choose α ∈ C and τ ∈ H such that we have an equality of Laurent coefficients b = a 3,0 3,0 and b = a . Now setting 5,0 5,0 −2 a a a + 2a α G (τ) −1,0 0,−1 1,0 −1,0 2 δ = , β = , γ = , α δ a 0,−1 we get a = b as well as a = b for m =−1, 1, 3, 5; hence, a = b for all 0,−1 0,−1 m,0 m,0 m,n m,n m, n by Theorem 3.6. Here and in the following, we suppress the dependence of the Laurent coefficients b on the parameters α, β, γ , δ, τ. m,n Matthes Res Math Sci (2019) 6:24 Page 11 of 13 24 On the other hand, assume that α , β , γ , δ , τ are different parameters such that u v γ uv 3 2 f (u, v) = δ e F , . Then, since the quotient b /b (which depends neither α β 3,0 5,0 ab on α, nor on α ) is a non-constant modular function, there exists ∈ SL (Z) such that cd aτ +b c/2πi τ = .Thefactthat α = α(cτ +d)and γ = γ − then follows by looking at the cτ +d αβ(cτ +d) coefficients b and using the modular transformation property of G (τ), if τ, τ = i∞. m,0 m+1 Finally, that δ = follows from the previous result using that αδ = b = α δ and −1,0 cτ +d similarly one shows β = β(cτ + d). This ends the proof of Theorem 1.3. 4 Proof of the main result We now give the proof of Theorem 1.2. First, we need to describe the relation between Fay identity and period relations. 4.1 Comparison with the period relations We again follow [12]. Consider the formal generating series k−2 (XY − 1)(X + Y ) T −2 C(X, Y ; τ; T) = T + c (X, Y ; τ) , 2 2 X Y (k − 2)! k=2 whose coefficients c (X, Y ; τ) are given by (r (X)r (Y ) − r (−X)r (−Y )) f f f f c (X, Y ; τ) = f (τ). k−3 (2i) f, f Here the sum is over all normalized Hecke eigenforms f of weight k, r (X) denotes the extended period polynomial of f and ·, · is the Petersson inner product. It follows from the definition of C and Eq. (1.1) that C satisfies the following variant of the period relations C(X, Y ; τ; T) + C − ,Y ; τ; XT = 0, (4.1) 1 1 C(X, Y ; τ; T) + C 1 − ,Y ; τ; XT + C ,Y ; τ;(1 − X)T = 0. X 1 − X Using the “slash operator” aX + b ab (C|γ )(X, Y ; τ; T):= C ,Y ; τ;(cX + d)T , γ = ∈ SL (Z), cX + d cd Eq. (4.1) can be expressed more concisely as C|(1 + S) = C|(1 + U + U ) = 0, where 0 −1 1 −1 S = and U = are generators of SL (Z)and | is extended linearly to the 10 10 group ring of SL (Z). Now recall from Theorem 1.1 that C(X, Y ; τ; T) = F (T, −XYT)F (XT, YT). In our τ τ context, this gives a relation between the Fay identity and the period relations, which we describe next. Given a formal Laurent series f (u, v) ∈ C((u, v)), we define C (X, Y, T) = f (T, −XYT)f (XT, YT) ∈ C((X, Y, T)). f 24 Page 12 of 13 Matthes Res Math Sci (2019) 6:24 Proposition 4.1 The series C satisfies the period relations (Eq. (4.1))ifand only if f satisfies the Fay identity (Eq. (2.1)). Proof By Proposition 3.1, every solution to the Fay identity is antisymmetric, so we may as well prove that the period relations are equivalent to Fay identity and antisymmetry. The equivalence of antisymmetry for f and the period relation C |(1+S) = 0 is straightforward. On the other hand, setting u := T, u :=−XT, v :=−XYT and v :=−YT, the period 1 2 1 2 relation C |(1 + U + U ) = 0is f (u ,v )f (−u , −v ) + f (−u ,v − v )f (−u − u , −v ) 1 1 2 2 2 1 2 1 2 1 (4.2) + f (−u − u , −v )f (−u ,v − v ) = 0, 1 2 2 1 2 1 which together with antisymmetry implies (2.1). Conversely, Fay identity and antisymme- try together imply (4.2). 4.2 Proof of Theorem 1.2 Combining Proposition 4.1 and Theorem 1.3, we know that C (X, Y, T) satisfies the period relations if and only if α β u v γ uv γ uv f (u, v) = e + , or f (u, v) = δe F , , u v α β for some α, β, γ ∈ C in the first case and for some α, β, δ ∈ C , γ ∈ C, τ ∈ H ∪{i∞} in the second. With the notation of Theorem 1.2, in the first case we get C (X, Y, T) = C (X, Y, T) and in the second case C (X, Y, T) = δC(X, Y ; τ; T )with δ := δ ,as P f α,β claimed. The uniqueness assertion follows from the corresponding statement in The- orem 1.3. Author details 1 2 Department of Mathematics, Kyushu University, 744 Motooka, Nishi-Ku, Fukuoka 819-0395, Japan, Present address: Mathematical Institute, University of Oxford, Andrew Wiles Building, Radcliffe Observatory Quarter, Woodstock Road, Oxford OX2 6GG, UK. Acknowledgements This work was done while the author was a JSPS Postdoctoral Fellow, partly supported by JSPS KAKENHI Grant No. 17F17020. The author would like to thank his academic host, Professor Masanobu Kaneko, for his support and fruitful discussions on the contents of this manuscript, as well as Ulf Kühn for useful comments. Final corrections were made while the author was a Postdoctoral Research Assistant at University of Oxford, partly supported by ERC Grant 724638. Received: 23 January 2019 Accepted: 21 May 2019 References 1. Brown, F., Levin, A.: Multiple elliptic polylogarithms. arXiv:1110.6917 2. 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Polishchuk, A.: Classical Yang–Baxter equation and the A constraint. Adv. Math. 168(1), 56–95 (2002) 9. Polishchuk, A.: Triple Massey products on curves, Fay’s trisecant identity and tangents to the canonical embedding. Mosc. Math. J. 3(1), 105–121, 256 (2003) Matthes Res Math Sci (2019) 6:24 Page 13 of 13 24 10. Weil, A.: Elliptic Functions According to Eisenstein and Kronecker. Ergebnisse der Mathematik und ihrer Grenzgebiete, Band 88, ii+93 p. Springer, Berlin (1976) 11. Zagier, D.: Hecke Operators and Periods of Modular Forms. Festschrift in Honor of I. I. Piatetski-Shapiro on the Occasion of His Sixtieth Birthday, Part II (Ramat Aviv 1989), pp. 321–336. Israel Mathematical Conference Proceedings, vol. 3, Weizmann, Jerusalem (1990) 12. Zagier, D.: Periods of modular forms and Jacobi theta functions. Invent. Math. 104(3), 449–465 (1991) Publisher’s Note Springer Nature remains neutral with regard to jurisdictional claims in published maps and institutional affiliations.
Research in the Mathematical Sciences – Springer Journals
Published: Jun 17, 2019
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