On the functions counting walks with small steps in the quarter plane

Irina Kurkova; Kilian Raschel

doi:10.1007/s10240-012-0045-7

On the functions counting walks with small steps in the quarter plane

Kurkova, Irina; Raschel, Kilian 2012-10-25 00:00:00 ON THE FUNCTIONS COUNTING WALKS WITH SMALL STEPS IN THE QUARTER PLANE by IRINA KURKOVA and KILIAN RASCHEL ABSTRACT Models of spatially homogeneous walks in the quarter plane Z with steps taken from a subsetS of the set of jumps to the eight nearest neighbors are considered. The generating function (x, y, z) → Q(x, y; z) of the numbers q(i, j; n) of such walks starting at the origin and ending at (i, j) ∈ Z after n steps is studied. For all non-singular models of walks, the functions x → Q(x, 0; z) and y → Q(0, y; z) are continued as multi-valued functions on C having inﬁnitely many meromorphic branches, of which the set of poles is identiﬁed. The nature of these functions is derived from this result: namely, for all the 51 walks which admit a certain inﬁnite group of birational transformations of C ,the interval ]0, 1/|S|[ of variation of z splits into two dense subsets such that the functions x → Q(x, 0; z) and y → Q(0, y; z) are shown to be holonomic for any z from the one of them and non-holonomic for any z from the other. This entails the non-holonomy of (x, y, z) → Q(x, y; z), and therefore proves a conjecture of Bousquet-Mélou and Mishna in Contemp. Math. 520:1–40 (2010). 1. Introduction and main results In the ﬁeld of enumerative combinatorics, counting walks on the lattice Z is among the most classical topics. While counting problems have been largely resolved for unrestricted walks on Z and for walks staying in a half plane [3], walks conﬁned to the quarter plane Z still pose considerable challenges. In recent years, much progress has been made for walks in the quarter plane with small steps, which means that the set S of possible steps is included in {−1, 0, 1} \{(0, 0)}; for examples, see Figures 1 and 10. In [2], Bousquet-Mélou and Mishna constructed a thorough classiﬁcation of these 2 walks. After eliminating trivial cases and exploiting equivalences, they showed that 79 in- herently different walks remain to be studied. Let q(i, j; n) denote the number of paths in Z having length n, starting from (0, 0) and ending at (i, j). Deﬁne the counting function (CF) as i j n Q(x, y; z) = q(i, j; n)x y z . i,j,n≥0 There are then two key challenges: (i) Finding an explicit expression for Q(x, y; z); (ii) Determining the nature of Q(x, y; z): is it holonomic (i.e., see [7, Appendix B.4], is the vector space over C(x, y, z)—the ﬁeld of rational functions in the three variables x, y, z—spanned by the set of all derivatives of Q(x, y; z) ﬁnite dimensional)? And in that event, is it algebraic, or even rational? The common approach to treat these problems is to start from a functional equation for the CF, which for the walks with small steps takes the form (see [2]) DOI 10.1007/s10240-012-0045-7 70 I. KURKOVA, K. RASCHEL FIG. 1. — Example of model (with an inﬁnite group) considered here—on the boundary, the jumps are the natural ones: those that would take the walk out Z are discarded (1.1) K(x, y; z)Q(x, y; z) = K(x, 0; z)Q(x, 0; z) + K(0, y; z)Q(0, y; z) − K(0, 0; z)Q(0, 0; z) − xy, where i j (1.2) K(x, y; z) = xyz x y − 1/z (i,j)∈S is called the kernel of the walk. This equation determines Q(x, y; z) through the boundary functions Q(x, 0; z),Q(0, y; z) and Q(0, 0; z). Known results regarding both problems (i) and (ii) highlight the notion of the group of the walk, introduced by Malyshev [13–15]. This is the group (1.3) ξ, η of birational transformations of C(x, y), generated by i j 1 1 (i,−1)∈S (−1,j)∈S (1.4) ξ(x, y) = x, ,η(x, y) = , y . i j y x x y (i,+1)∈S (+1,j)∈S i j 2 2 Each element of ξ, η leaves invariant the jump function x y . Further, ξ = η = (i,j)∈S Id, and ξ, η is a dihedral group of order even and larger than or equal to four. It turns out that 23 of the 79 walks have a ﬁnite group, while the 56 others admit an inﬁnite group, see [2]. For 22 of the 23 models with ﬁnite group, CFs Q(x, 0; z),Q(0, y; z) and Q(0, 0; z)—and hence Q(x, y; z) by (1.1)—have been computed in [2]bymeans of cer- tain (half-)orbit sums of the functional equation (1.1). The 23rd model with ﬁnite group is known as Gessel’s walk (see Figure 10); see [9, 11] and references therein for literature on this quite interesting model. For it, the CFs have been expressed by radicals in [1] thanks to a guessing-proving method using computer calculations; they were also found in [12] by solving some boundary value problems. For the 2 walks with inﬁnite group on the left in Figure 2, they have been obtained in [17], by exploiting a particular property shared by the 5 models of Figure 2 commonly known as singular walks. Finally, in [18], the problem (i) was resolved for all 54 remaining walks—and in fact for all the 79 models. For ON THE FUNCTIONS COUNTING WALKS IN THE QUARTER PLANE 71 FIG. 2. — The 5 singular walks in the classiﬁcation of [2] the 74 non-singular walks, this was done via a uniﬁed approach: explicit integral repre- sentations were obtained for CFs Q(x, 0; z),Q(0, y; z) and Q(0, 0; z) in certain domains, by solving boundary value problems of Riemann-Carleman type. In this article we go further, since both functions x → Q(x, 0; z) and y → Q(0, y; z) are computed on the whole of C as multi-valued functions with inﬁnitely many meromor- phic branches, that are made explicit for all z ∈]0, 1/|S|[. This result gives not only the most complete continuation of these CFs on their complex planes along all paths, but also permits to establish the nature of these functions, i.e., to solve Problem (ii). Problem (ii) is actually resolved for only 28 of the 79 walks. All 23 ﬁnite group models admit a holonomic CF. Indeed, the nature of Q(x, y; z) was determined in [2] for 22 of these walks: 19 walks turn out to have a holonomic but non-algebraic CF, while for 3 walks Q(x, y; z) is algebraic. As for the 23rd—again, Gessel’s model—, the CF is algebraic [1]. Alternative proofs for the nature of the (bivariate) CF for these 23 walks were given in [5]. For the remaining 56 walks with an inﬁnite group, not much is known: in [17] it was shown that for 2 singular walks (namely, the 2 ones on the left in Figure 2), the function z → Q(1, 1; z) has inﬁnitely many poles and, as a consequence [7, Appendix B.4], is non-holonomic. Accordingly [7, Appendix B.4], the trivariate function Q(x, y; z) is non-holonomic as well. It is reasonable to expect that the same approach would lead to the non-holonomy of all 5 singular walks, see [16, 17]. As for the 51 non- singular walks with inﬁnite group (all of them are pictured on Figure 17), Bousquet-Mélou and Mishna [2] conjectured that they also have a non-holonomic CF. In this article we prove the following theorem. Theorem 1. — For any of the 51 non-singular walks with inﬁnite group (1.3), the set ]0, 1/|S|[ splits into subsetsH and ]0, 1/|S|[\H that are both dense in ]0, 1/|S|[ and such that: (i) x → Q(x, 0; z) and y → Q(0, y; z) are holonomic for any z ∈H; (ii) x → Q(x, 0; z) and y → Q(0, y; z) are non-holonomic for any z ∈]0, 1/|S|[\H. Theorem 1(ii) immediately entails Bousquet-Mélou and Mishna’s conjecture: the trivariate function (x, y, z) → Q(x, y; z) is non-holonomic since the holonomy is sta- ble by specialization of a variable [7, Appendix B.4]. Further, Theorem 1(i) goes beyond it: it suggests that Q(x, y; z), although being non-holonomic, still stays accessible for fur- ther analysis when z ∈H, namely by the use of methods developed in [6,Chapter 4], 72 I. KURKOVA, K. RASCHEL see Remark 16. This important setH will be characterized in two different ways, see Corollary 15 and Remark 16 below. The proof of Theorem 1 we shall do here is based on the above-mentioned con- struction of the CF x → Q(x, 0; z) (resp. y → Q(0, y; z)) as a multi-valued function, that must now be slightly more detailed. First, we prove in this article that for any z ∈]0, 1/|S|[, the integral expression of x → Q(x, 0; z) given in [18]inacertaindo- main of C admits a direct holomorphic continuation on C \[x (z), x (z)]. Points x (z), x (z) 3 4 3 4 are among four branch points x (z), x (z), x (z), x (z) of the two-valued algebraic func- 1 2 3 4 tion x → Y(x; z) deﬁned via the kernel (1.2) by the equation K(x, Y(x; z); z) = 0. These branch points are roots of the discriminant (2.2) of the latter equation, which is of the second order. We refer to Section 2 for the numbering and for some properties of these branch points. We prove next that function x → Q(x, 0; z) does not admit a direct mero- morphic continuation on any open domain containing the segment [x (z), x (z)],but 3 4 admits a meromorphic continuation along any path going once through [x (z), x (z)]. This way, 3 4 we obtain a second (and different) branch of the function, which admits a direct mero- morphic continuation on the whole cut plane C \ ([x (z), x (z)]∪[x (z), x (z)]).Next, if 1 2 3 4 the function x → Q(x, 0; z) is continued along a path in C \[x (z), x (z)] crossing once 1 2 again [x (z), x (z)], we come across its ﬁrst branch. But its continuation along a path in 3 4 C \[x (z), x (z)] crossing once [x (z), x (z)] leads to a third branch of this CF, which is 3 4 1 2 meromorphic on C \ ([x (z), x (z)]∪[x (z), x (z)]). Making loops through [x (z), x (z)] 1 2 3 4 3 4 and [x (z), x (z)], successively, we construct x → Q(x, 0; z) as a multi-valued meromor- 1 2 phic function on C with branch points x (z), x (z), x (z), x (z), and with (generically) 1 2 3 4 inﬁnitely many branches. The analogous construction is valid for y → Q(0, y; z). In order to prove Theorem 1(ii), we then show that for any of the 51 non-singular walks with inﬁnite group (1.3), for any z ∈]0, 1/|S|[\H, the set formed by the poles of all branches of x → Q(x, 0; z) (resp. y → Q(0, y; z)) is inﬁnite—and even dense in certain curves, to be speciﬁed in Section 7 (see Figure 11 for their pictures). This is not compatible with holonomy. Indeed, all branches of a holonomic one-dimensional function must verify the same linear differential equation with polynomial coefﬁcients. In particular, the poles of all branches are among the zeros of these polynomials, and hence they must be in a ﬁnite number. The rest of our paper is organized as follows. In Section 2 we construct the Rie- mann surface T of genus 1 of the two-valued algebraic functions X( y; z) and Y(x; z) deﬁned by K X( y; z), y; z = 0, K x, Y(x; z); z = 0. In Section 3 we introduce and study the universal covering of T. It can be viewed as the complex plane C split into inﬁnitely many parallelograms with edges ω (z) ∈ iR and ω (z) ∈ R that are uniformization periods. These periods as well as a new important period ω (z) are made explicit in (3.1)and (3.2). In Section 4 we lift CFs Q(x, 0; z) and 3 ON THE FUNCTIONS COUNTING WALKS IN THE QUARTER PLANE 73 Q(0, y; z) to some domain of T, and then to a domain on its universal covering. In Sec- tion 5, using a proper lifting of the automorphisms ξ and η deﬁned in (1.4)aswellasthe independence of K(x, 0; z)Q(x, 0; z) and K(0, y; z)Q(0, y; z) w.r.t. y and x, respectively, we continue these functions meromorphically on the whole of the universal covering. All this procedure has been ﬁrst carried out by Malyshev in the seventies [13–15], at that time to study the stationary probability generating functions for random walks with small steps in the quarter plane Z . It is presented in [6, Chapter 3] for the case of ergodic random walks in Z , and applies directly for our Q(x, 0; 1/|S|) and Q(0, y; 1/|S|) if the drift vector ( i, j) has not two positive coordinates; see also [4]. In Sec- (i,j)∈S (i,j)∈S tions 3, 4 and 5 we carry out this procedure for all z ∈]0, 1/|S|[ and all non-singular walks, independently of the drift. Then, going back from the universal covering to the complex plane allows us in Subsection 5.2 to continue x → Q(x, 0; z) and y → Q(0, y; z) as multi-valued meromorphic functions with inﬁnitely many branches. For given z ∈]0, 1/|S|[, the rationality or irrationality of the ratio ω (z)/ω (z) of 2 3 the uniformization periods is crucial for the nature of x → Q(x, 0; z) and y → Q(0, y; z). Namely, Theorem 7 of Subsection 5.3 proves that if ω (z)/ω (z) is rational, these func- 2 3 tions are holonomic. For 23 models of walks with ﬁnite group ξ, η,the ratio ω (z)/ω (z) turns out 2 3 to be rational and independent of z, see Lemma 8 below, that implies immediately the holonomy of the generating functions. In Section 6 we gather further results of our ap- proach for the models with ﬁnite group concerning the set of branches of the generating functions and their nature. In particular, we recover most of the results of [1, 2, 5, 17]. Section 7 is devoted to 51 models with inﬁnite group ξ, η.For allofthem, the setsH ={z ∈]0, 1/|S|[: ω (z)/ω (z) is rational} and ]0, 1/|S|[\H ={z ∈]0, 1/|S|[: 2 3 ω (z)/ω (z) is irrational} are proved to be dense in ]0, 1/|S|[, see Proposition 14. These 2 3 sets can be also characterized as those where the group ξ, η restricted to the curve {(x, y) : K(x, y; z) = 0} is ﬁnite and inﬁnite, respectively, see Remark 6.ByTheorem 7 mentioned above, x → Q(x, 0; z) and y → Q(0, y; z) are holonomic for any z ∈H,that proves Theorem 1(i). In Subsections 7.1, 7.2 and 7.3, we analyze in detail the branches of x → Q(x, 0; z) and y → Q(0, y; z) for any z ∈]0, 1/|S|[\H and prove the following facts (see Theorem 17): (i) The only singularities of the ﬁrst (main) branches of x → Q(x, 0; z) and y → Q(0, y; z) are two branch points x (z), x (z) and y (z), y (z), respectively; 3 4 3 4 (ii) All (other) branches have only a ﬁnite number of poles; (iii) The set of poles of all these branches is inﬁnite for each of these functions, and is dense on certain curves; these curves are speciﬁed in Section 7, and in particular are pictured on Figure 11 for all 51 walks given on Figure 17; (iv) Poles of branches out of these curves may be only at zeros of x → K(x, 0; z) or y → K(0, y; z), respectively. It follows from (iii) that x → Q(x, 0; z) and y → Q(0, y; z) are non-holonomic for any z ∈]0, 1/|S|[\H. 74 I. KURKOVA, K. RASCHEL 2. Riemann surface T In the sequel we suppose that z ∈]0, 1/|S|[, and we drop the dependence of the different quantities w.r.t. z. 2.1. Kernel K(x, y).— The kernel K(x, y) deﬁned in (1.2) can be written as i j 2 2 (2.1) xyz x y − 1/z = a( y)x + b( y)x + c( y) = a(x)y + b(x)y + c(x), (i,j)∈S where j j j a( y) = zy y , b( y) =−y + zy y , c( y) = zy y , (+1,j)∈S (0,j)∈S (−1,j)∈S i i i a(x) = zx x , b(x) =−x + zx x , c(x) = zx x . (i,+1)∈S (i,0)∈S (i,−1)∈S With these notations we deﬁne 2 2 (2.2) d( y) = b( y) − 4 a( y) c( y), d(x) = b(x) − 4a(x)c(x). If thewalkisnon-singular, then forany z ∈]0, 1/|S|[, the polynomial d (resp. d)has three or four roots, that we call y (resp. x ). They are such that |y | < y < 1 < y < |y | (resp. 1 2 3 4 |x | < x < 1 < x < |x |), with y =∞ (resp. x =∞)if d (resp. d ) has order three: the 1 2 3 4 4 4 arguments given in [6, Part 2.3] for the case z = 1/|S| indeed also apply for other values of z. Now we notice that the kernel (1.2) vanishes if and only if [b( y) + 2 a( y)x] = d( y) or [b(x) + 2a(x)y] = d(x). Consequently [10], the algebraic functions X( y) and Y(x) deﬁned by i j i j (2.3) X( y) y − 1/z = 0, x Y(x) − 1/z = 0 (i,j)∈S (i,j)∈S have two branches, meromorphic on the cut planes C \ ([y , y ]∪[y , y ]) and C \ 1 2 3 4 ([x , x ]∪[x , x ]), respectively—note that if y < 0, [y , y ] stands for [y , ∞[ ∪ {∞} ∪ 1 2 3 4 4 3 4 3 ]−∞, y ]; the same holds for [x , x ]. 4 3 4 We ﬁx the notations of the two branches of the algebraic functions X( y) and Y(x) by setting 1/2 1/2 −b( y) + d( y) −b( y) − d( y) (2.4) X ( y) = , X ( y) = , 0 1 a( y) 2 a( y) as well as 1/2 1/2 −b(x) + d(x) −b(x) − d(x) (2.5) Y (x) = , Y (x) = . 0 1 2a(x) 2a(x) ON THE FUNCTIONS COUNTING WALKS IN THE QUARTER PLANE 75 FIG. 3. — Construction of the Riemann surface The following straightforward result holds. Lemma 2. —For all y ∈ C,wehave |X ( y)|≤|X ( y)|. Likewise, for all x ∈ C, we have 0 1 |Y (x)|≤|Y (x)|. 0 1 Proof. — The arguments (via the maximum modulus principle [10]) given in [6, Part 5.3] for z = 1/|S| also work for z ∈]0, 1/|S|[. 2.2. Riemann surface T.— We now construct the Riemann surface T of the alge- braic function Y(x) introduced in (2.3). For this purpose we take two Riemann spheres 1 2 C ∪{∞},say S and S , cut along the segments [x , x ] and [x , x ], and we glue them 1 2 3 4 x x together along the borders of these cuts, joining the lower border of the segment [x , x ] 1 2 1 2 (resp. [x , x ])on S to the upper border of the same segment on S and vice versa, see 3 4 x x Figure 3. The resulting surface T is homeomorphic to a torus (i.e., a compact Riemann surface of genus 1) and is projected on the Riemann sphere S by a canonical covering map h : T → S. In a standard way, we can lift the function Y(x) to T, by setting Y(s) = Y (h (s)) if s ∈ S ⊂ T, ∈{1, 2}.Thus, Y(s) is single-valued and continuous on T. Furthermore, K(h (s), Y(s)) = 0for any s ∈ T. For this reason, we call T the Riemann surface of Y(x). In a similar fashion, one constructs the Riemann surface of the function X( y),by 1 2 gluing together two copies S and S of the sphere S along the segments [y , y ] and 1 2 y y [y , y ]. It is again homeomorphic to a torus. 3 4 76 I. KURKOVA, K. RASCHEL FIG. 4. — Location of the branch points and of the cycles and on the Riemann surface T 0 1 Since the Riemann surfaces of X( y) and Y(x) are equivalent, we can work on a single Riemann surface T, but with two different covering maps h , h : T → S. Then, x y for s ∈ T, we set x(s) = h (s) and y(s) = h (s), and we will often represent a point s ∈ T x y by the pair of its coordinates (x(s), y(s)). These coordinates are of course not independent, because the equation K(x(s), y(s)) = 0is valid for any s ∈ T. 2.3. Real points of T.— Let us identify the set of real points of T,thatare the points s ∈ T where x(s) and y(s) are both real or equal to inﬁnity. Note that for y real, X( y) is real if y ∈[y , y ] or y ∈[y , y ], and complex if y ∈]y , y [ or y ∈]y , y [, see (2.2). 4 1 2 3 1 2 3 4 Likewise, for real values of x,Y(x) is real if x ∈[x , x ] or x ∈[x , x ], and complex if 4 1 2 3 x ∈]x , x [ or x ∈]x , x [. The set therefore consists of two non-intersecting closed 1 2 3 4 analytic curves and , equal to (see Figure 4) 0 1 = s ∈ T : x(s) ∈[x , x ] = s ∈ T : y(s) ∈[y , y ] 0 2 3 2 3 and = s ∈ T : x(s) ∈[x , x ] = s ∈ T : y(s) ∈[y , y ] , 1 4 1 4 1 and homologically equivalent to a basic cycle on T—note, however, that the equivalence −1 class containing and is disjoint from that containing the cycle h ({x ∈ C :|x|= 1}). 0 1 2.4. Galois automorphisms ξ, η.— We continue Section 2 by introducing two Ga- lois automorphisms. Deﬁne ﬁrst, for ∈{1, 2}, the incised spheres S = S \ [x , x ]∪[x , x ] , S = S \ [y , y ]∪[y , y ] . 1 2 3 4 1 2 3 4 x x y y For any s ∈ T such that x(s) is not equal to a branch point x , there is a unique s = s ∈ T 1 2 such that x(s) = x(s ). Furthermore, if s ∈ S then s ∈ S and vice versa. On the other x x hand, whenever x(s) is one of the branch points x , s = s . Also, since K(x(s), y(s)) = 0, y(s) and y(s ) give the two values of function Y(x) at x = x(s) = x(s ). By Vieta’s theorem and (2.1), y(s)y(s ) = c(x(s))/a(x(s)). Similarly, for any s ∈ T such that y(s) is different from the branch points y ,there 1 2 exists a unique s = s ∈ T such that y(s) = y(s ).If s ∈ S then s ∈ S and vice versa. y y ON THE FUNCTIONS COUNTING WALKS IN THE QUARTER PLANE 77 On the other hand, if y(s) is one of the branch points y ,wehave s = s . Moreover, since K(x(s), y(s)) = 0, x(s) and x(s ) the two values of function X( y) at y(s) = y(s ). Again, by Vieta’s theorem and (2.1), x(s)x(s ) = c( y(s))/ a( y(s)). Deﬁnenow themappings ξ : T → T and η : T → T by ξ s = s if x(s) = x(s ), (2.6) ηs = s if y(s) = y(s ). 2 2 Following [13–15], we call them Galois automorphisms of T.Then ξ = η = Id, and c(x(s)) 1 c( y(s)) 1 (2.7) y(ξ s) = , x(ηs) = . a(x(s)) y(s) a( y(s)) x(s) Any s ∈ T such that x(s) = x (resp. y(s) = y ) is a ﬁxed point for ξ (resp. η). To illustrate and to get some more intuition, it is helpful to draw on Figure 4 the straight line through the pair of points of where x(s) = x and x (resp. y(s) = y and y ); then points s and 0 2 3 2 3 ξ s (resp. s and ηs) can be drawn symmetric about this straight line. 2.5. The Riemann surface T viewed as a parallelogram whose opposed edges are identiﬁed. — Like any compact Riemann surface of genus 1, T is isomorphic to a certain quotient space (2.8) C/(ω Z + ω Z), 1 2 where ω ,ω are complex numbers linearly independent on R, see [10]. The set (2.8) 1 2 can obviously be thought as the (fundamental) parallelogram ω [0, 1]+ ω [0, 1] whose 1 2 opposed edges are identiﬁed. Up to a unimodular transform, ω ,ω are unique, see [10]. 1 2 In our case, suitable ω ,ω will be found in (3.1). 1 2 If we cut the torus on Figure 4 along [x , x ] and , it becomes the parallelogram 1 2 0 on the left in Figure 5. On the right in the same ﬁgure, this parallelogram is translated to the complex plane, and all corresponding important points are expressed in terms of the complex numbers ω ,ω (see above) and of ω (to be deﬁned below, in (3.2)). 1 2 3 3. Universal covering 3.1. An informal construction of the universal covering. — The Riemann surface T can be considered as a parallelogram whose opposite edges are identiﬁed, see (2.8)and Figure 5. The universal covering of T can then be viewed as the union of inﬁnitely many such parallelograms glued together, as in Figure 6. 78 I. KURKOVA, K. RASCHEL FIG.5.—TheRiemann surface C/(ω Z + ω Z) and the location of the branch points 1 2 FIG. 6. — Informal construction of the universal covering 3.2. Periods and covering map. — We now give a proper construction of the universal covering. The Riemann surface T being of genus 1, its universal covering has the form (C,λ),where C is the complex plane and λ : C → T is a non-branching covering map, see [10]. This way, the surface T can be considered as the additive group C factorized by the discrete subgroup ω Z + ω Z, where the periods ω ,ω are complex numbers, 1 2 1 2 linearly independent on R. Any segment of length |ω | and parallel to ω , ∈{1, 2},is projected onto a closed curve on T homological to one of the elements of the normal basis on the torus. We choose λ([0,ω ]) to be homological to the cut [x , x ] (and hence 1 1 2 also to all other cuts [x , x ], [y , y ] and [y , y ]); λ([0,ω ]) is then homological to the 3 4 1 2 3 4 2 cycles of real points and ; see Figures 5 and 7. 0 1 Our aim now is to ﬁnd the expression of the covering λ. We will do this by ﬁnding, for all ω ∈ C, the explicit expressions of the pair of coordinates (x(λω), y(λω)),thatwe have introduced in Section 2. First, the periods ω ,ω are obtained in [6, Lemma 3.3.2] 1 2 for z = 1/|S|. The reasoning is exactly the same for other values of z, and we obtain that with d as in (2.2), x x 2 3 dx dx (3.1) ω = i ,ω = . 1 2 1/2 1/2 [−d(x)] d(x) x x 1 2 ON THE FUNCTIONS COUNTING WALKS IN THE QUARTER PLANE 79 FIG. 7. — Important points and cycles on the universal covering We also need to introduce dx (3.2) ω = . 1/2 d(x) X( y ) Further, we deﬁne d (x )/6 + d (x )/[t − x ] if x = ∞, 4 4 4 4 g (t) = d (0)/6 + d (0)t/6if x =∞, as well as d ( y )/6 + d ( y )/[t − y ] if y = ∞, 4 4 4 4 g (t) = d (0)/6 + d (0)t/6if y =∞, and ﬁnally we introduce ℘(ω; ω ,ω ), the Weierstrass elliptic function with periods 1 2 ω ,ω . Throughout, we shall write ℘(ω) for ℘(ω; ω ,ω ). By deﬁnition, see [10, 20], 1 2 1 2 we have 1 1 1 ℘(ω) = + − . 2 2 2 ω (ω − ω − ω ) ( ω + ω ) 1 1 2 2 1 1 2 2 ( , )∈Z \{(0,0)} 1 2 Then we have the uniformization [6, Lemma 3.3.1] −1 x(λω) = g (℘ (ω)), (3.3) −1 y(λω) = g (℘ (ω − ω /2)). From now on, whenever no ambiguity can arise, we drop the dependence w.r.t. λ, writ- ing x(ω) and y(ω) instead of x(λω) and y(λω), respectively. The coordinates x(ω), y(ω) deﬁned in (3.3) are elliptic: (3.4) x(ω + ω ) = x(ω), y(ω + ω ) = y(ω), ∀ ∈{1, 2}, ∀ω ∈ C. 80 I. KURKOVA, K. RASCHEL Furthermore, x(ω /2) = x x(ω /2) = x x(0) = x 1 3 2 1 , , , y(0) = Y(x ) 4 y(ω /2) = Y(x ) y(ω /2) = Y(x ) 1 3 2 1 x([ω + ω ]/2) = x 1 2 2 y([ω + ω ]/2) = Y(x ) 1 2 2 Let us denote the points 0,ω /2,ω /2, [ω + ω ]/2by ω ,ω ,ω ,ω , respectively, see 1 2 1 2 x x x x 4 3 1 2 Figures 5 and 7.Let x x 3 2 L = ω + ω R, L = ω + ω R. x 1 x 1 x 4 x 1 4 1 x x 1 2 3 2 Then λL (resp. λL )isthe cutof T where S and S are glued together, namely, {s ∈ x x x x 4 1 T : x(s) ∈[x , x ]} (resp. {s ∈ T : x(s) ∈[x , x ]}). 3 4 1 2 Moreover, by construction we have (see again Figures 5 and 7) x(ω /2) = X( y ) x([ω + ω ]/2) = X( y ) 3 4 1 3 3 , , y(ω /2) = y y([ω + ω ]/2) = y 3 4 1 3 3 x([ω + ω ]/2) = X( y ) 2 3 1 y([ω + ω ]/2) = y 2 3 1 and x([ω + ω + ω ]/2) = X( y ) 1 2 3 2 y([ω + ω + ω ]/2) = y 1 2 3 2 We denote the points ω /2, [ω + ω ]/2, [ω + ω ]/2, [ω + ω + ω ]/2by ω , ω , ω , 3 1 3 2 3 1 2 3 y y y 4 3 1 ω , respectively. Let y y 3 2 L = ω + ω R, L = ω + ω R. y 1 y 1 y 4 y 1 4 1 y y 1 2 3 2 Then λL (resp. λL )isthe cutof T where S and S are glued together, that is to say y y y y 4 1 {s ∈ T : y(s) ∈[y , y ]} (resp. {s ∈ T : y(s) ∈[y , y ]}). 3 4 1 2 x y x y 3 3 2 2 The distance between L and L is the same as between L and L ; it equals x y x y 4 4 1 1 ω /2. 3.3. Lifted Galois automorphisms ξ, η.— Any conformal automorphism ζ of the sur- −1 face T can be continued as a conformal automorphism ζ = λ ζλ of the universal cover- −1 ing C. This continuation is not unique, but it will be unique if we ﬁx some ζω ∈ λ ζλω 0 0 for a given point ω ∈ C. 0 ON THE FUNCTIONS COUNTING WALKS IN THE QUARTER PLANE 81 According to [6], we deﬁne ξ, η by choosing their ﬁxed points to be ω ,ω ,re- x y 2 2 spectively. Since any conformal automorphism of C is an afﬁne function of ω [10]and 2 2 since ξ = η = Id, we have (3.5) ξω =−ω + 2ω , ηω =−ω + 2ω . x y 2 2 It follows that η ξ and ξ η are just the shifts via the real numbers ω and −ω , respectively: 3 3 η ξω = ω + 2(ω − ω ) = ω + ω , y x 3 2 2 (3.6) ξ ηω = ω + 2(ω − ω ) = ω − ω . x y 3 2 2 By (2.6)and (2.7)wehave c(x(ω)) 1 x(ξω) = x(ω), y(ξω) = , a(x(ω)) y(ω) (3.7) c( y(ω)) 1 x( ηω) = , y( ηω) = y(ω). a( y(ω)) x(ω) x x x x y y y y 2 2 3 3 2 2 3 3 Finally, ξ L = L , ξ L = L + ω and ηL = L , ηL = L + ω . 2 2 x x x x y y y y 1 1 4 4 1 1 4 4 4. Lifting of x → Q(x, 0) and y → Q(0, y) to the universal covering 4.1. Lifting to the Riemann surface T.— We have seen in Section 2 that for any z ∈ ]0, 1/|S|[, exactly two branch points of Y(x) (namely, x and x ) are in the unit disc. For 1 2 this reason, and by construction of the surface T, the set {s ∈ T :|x(s)|= 1} is composed 1 2 of two cycles (one belongs to S and the other to S ) homological to the cut {s ∈ T : x x x(s) ∈[x , x ]}. The domain D ={s ∈ T :|x(s)| < 1} is bounded by these two cycles, 1 2 x see Figure 8, and contains the points s ∈ T such that x(s) ∈[x , x ]. Since the function 1 2 x → K(x, 0)Q(x, 0) is holomorphic in the unit disc, we can lift it to D ⊂ T as r (s) = K x(s), 0 Q x(s), 0 , ∀s ∈ D . x x In thesameway,the domain D ={s ∈ T :|y(s)| < 1} is bounded by {s ∈ T : |y(s)|= 1}, which consists in two cycles homological to the cut {s ∈ T : y(s) ∈[y , y ]}, 1 2 see Figure 8, and which contains the latter. We lift the function y → K(0, y)Q(0, y) to D ⊂ T as r (s) = K 0, y(s) Q 0, y(s) , ∀s ∈ D . y y It is shown in [18, Lemma 3] that for any z ∈]0, 1/|S|[ and any x such that |x|= 1, we have |Y (x)| < 1and |Y (x)| > 1. Hence, the cycles that constitute the boundary of 0 1 D are = s ∈ T : x(s) = 1, y(s) < 1 , = s ∈ T : x(s) = 1, y(s) > 1 . x 82 I. KURKOVA, K. RASCHEL FIG. 8. — Location of the domains D and D on the Riemann surface T x y 0 1 We thus have ∈ D and ∈ / D , see Figure 8.Inthe same way, forany z ∈]0, 1/|S|[ y y x x and any y such that |y|= 1, we have |X ( y)| < 1and |X ( y)| > 1. Therefore, the cycles 0 1 composing the boundary of D are = s ∈ T : y(s) = 1, x(s) < 1 , = s ∈ T : y(s) = 1, x(s) > 1 . 0 1 Furthermore, ∈ D and ∈ / D , see Figure 8. x x y y It follows that D ∩ D ={s ∈ T :|x(s)| < 1, |y(s)| < 1} is not empty, simply con- x y 0 0 nected and bounded by and . Since for any s ∈ T,K(x(s), y(s)) = 0, and since the x y main equation (1.1)isvalid on {(x, y) ∈ C :|x| < 1, |y| < 1},wehave (4.1) r (s) + r (s) − K(0, 0)Q(0, 0) − x(s)y(s) = 0, ∀s ∈ D ∩ D . x y x y 4.2. Lifting to the universal covering C.— The domain D lifted on the universal cov- ering consists of inﬁnitely many curvilinear strips shifted by ω : −1 n n λ D = , ⊂ ω R + nω ,(n + 1)ω , x 1 2 2 x x n∈Z and, likewise, −1 n n λ D = , ⊂ ω R + ω /2 + nω ,(n + 1)ω . y 1 3 2 2 y y n∈Z Let us consider these strips for n = 0, that we rename 0 0 = , = . x y x y 1 −1 1 0 −1 0 The ﬁrst is bounded by ⊂ λ and by ⊂ λ , while the second is delimited by x x x x 0 −1 0 1 −1 1 ⊂ λ and by ⊂ λ y y y y x y 2 2 Further, note that the straight line L (resp. L ) deﬁned in Section 3 is invariant x y 1 1 w.r.t. ξ (resp. η) and belongs to (resp. ). x y ON THE FUNCTIONS COUNTING WALKS IN THE QUARTER PLANE 83 FIG.9.—Location of = ∪ x y 1 0 1 0 Then, by the facts that ξ and η , and by our choice (3.5)ofthe x x y y 1 0 1 0 deﬁnition of ξ and η on the universal covering, we have ξ and η .In x x y y addition, (4.2) ξω ∈ , ∀ω ∈ , ηω ∈ , ∀ω ∈ . x x y y 0 1 0 1 0 1 Moreover, since ∈ D , ∈ / D and ∈ D , ∈ / D ,wehave ∈ , ∈ / and x x y y x x y y x x y y 0 1 0 0 ∈ , ∈ / . It follows that ∩ is a non-empty strip bounded by and y y x y x x x y and that = ∪ x y is simply connected, as in Figure 9. Let us lift the functions r (s) and r (s) holomorphically to and , respectively: x y x y we put r (ω) = r (λω) = K(x(ω), 0)Q(x(ω), 0), ∀ω ∈ , x x x (4.3) r (ω) = r (λω) = K(0, y(ω))Q(0, y(ω)), ∀ω ∈ . y y y It follows from (4.1)and (4.3)that (4.4) r (ω) + r (ω) − K(0, 0)Q(0, 0) − x(ω)y(ω) = 0, ∀ω ∈ ∩ . x y x y Equation (4.4) allows us to continue functions r (ω) and r (ω) meromorphically on :we x y put r (ω) =−r (ω) + K(0, 0)Q(0, 0) + x(ω)y(ω), ∀ω ∈ , x y y (4.5) r (ω) =−r (ω) + K(0, 0)Q(0, 0) + x(ω)y(ω), ∀ω ∈ . y x x Equation (4.4) is then valid on the whole of . We summarize all facts above in the next result. 84 I. KURKOVA, K. RASCHEL Theorem 3. — The functions ⎪ K(x(ω), 0)Q(x(ω), 0) if ω ∈ , −K(0, y(ω))Q(0, y(ω)) + K(0, 0)Q(0, 0) r (ω) = + x(ω)y(ω) if ω ∈ , and ⎪ K(0, y(ω))Q(0, y(ω)) if ω ∈ , r (ω) = −K(x(ω), 0)Q(x(ω), 0) + K(0, 0)Q(0, 0) + x(ω)y(ω) if ω ∈ , are meromorphic in .Furthermore, (4.6) r (ω) + r (ω) − K(0, 0)Q(0, 0) − x(ω)y(ω) = 0, ∀ω ∈ . x y 5. Meromorphic continuation of x → Q(x, 0) and y → Q(0, y) on the universal covering 5.1. Meromorphic continuation. — In Theorem 3 we saw that r (ω) and r (ω) are x y meromorphic on . We now continue these functions meromorphically from to the whole of C. Theorem 4. — The functions r (ω) and r (ω) can be continued meromorphically to the whole x y of C.Further,for any ω ∈ C, we have (5.1) r (ω − ω ) = r (ω) + y(ω) x(−ω + 2ω ) − x(ω) , x 3 x y (5.2) r (ω + ω ) = r (ω) + x(ω) y(−ω + 2ω ) − y(ω) , y 3 y x (5.3) r (ω) + r (ω) − K(0, 0)Q(0, 0) − x(ω)y(ω) = 0, x y r (ξω) = r (ω), x x (5.4) r ( ηω) = r (ω), y y r (ω + ω ) = r (ω), x 1 x (5.5) r (ω + ω ) = r (ω). y 1 y For the proof of Theorem 4, we shall need the following lemma. Lemma 5. — We have (5.6) ( + nω ) = C. n∈Z ON THE FUNCTIONS COUNTING WALKS IN THE QUARTER PLANE 85 1 0 0 Proof. — It has been noticed in Section 4 that ξ ∈ .By(4.2), η x x x ⊂ ,sothat, by (3.6), 1 1 + ω = ηξ ∈ . x x In the same way, − ω ∈ . It follows that ∪ ( + ω ) is a simply connected domain, 3 3 see Figure 9. Identity (5.6) follows. Proof of Theorem 4.—Forany ω ∈ ,byTheorem 3 we have (5.7) r (ω) + r (ω) − K(0, 0)Q(0, 0) − x(ω)y(ω) = 0. x y 1 1 0 For any ω ∈ close enough to the cycle ,wehavethat ξω ∈ since ξ ∈ . y y x x x Then ω + ω = ηξω ∈ by (4.2). We now compute r ( ηξω) for any such ω. Equation 3 y y (4.6), which is valid in ⊃ ,gives (5.8) r (ξω) + r (ξω) − K(0, 0)Q(0, 0) − x(ξω)y(ξω) = 0. x y By (3.7), x(ξω) = x(ω). For our ω ∈ ,by(4.2)wehave ξω ∈ , so that Theorem 3 x x yields r (ξω) = K x(ξω), 0 Q x(ξω), 0 = K x(ω), 0 Q x(ω), 0 = r (ω). x x If we now combine the last fact together with Equation (5.7), Equation (5.8) and identity x(ξω) = x(ω),weobtainthat r (ξω) = r (ω) + x(ω) y(ξω) − y(ω) . y y Since ξω ∈ ,thenby(4.2)wehave ηξω ∈ . Equation (3.7)and Theorem 3 entail y y r ( ηξω) = K 0, y( ηξω) Q 0, y( ηξω) = K 0, y(ξω) Q 0, y(ξω) = r (ξω). Finally, for all ω ∈ close enough to we have r ( ηξω) = r (ω) + x(ω) y(ξω) − y(ω) . y y Using (3.6), we obtain exactly Equation (5.2). Thanks to Theorem 3 and Lemma 5, this equation shown for any ω ∈ close enough to allows us to continue r meromor- phically from to the whole of C. Equation (5.2) therefore stays valid for any ω ∈ C. The function r ( ηω) = r (−ω + ω ) is then also meromorphic on C. Since these func- y y y tions coincide in , then by the principle of analytic continuation [10] they do on the whole of C. In the same way, we prove Equation (5.1)for all ω ∈ close enough to Together with Theorem 3 and Lemma 5 this allows us to continue r (ω) meromorphi- cally to the whole of C. By the same continuation argument, the identity r (ω) = r (ξω) x x 86 I. KURKOVA, K. RASCHEL is valid everywhere on C. Consequently Equation (5.3), which a priori is satisﬁed in , must stay valid on the whole of C. Since x(ω) and y(ω) are ω -periodic, it follows from Theorem 3 that r (ω) and r (ω) are ω -periodic in .The vector ω being real, by (5.1) x y 1 3 and (5.2) these functions stay ω -periodic on the whole of C. 5.2. Branches of x → Q(x, 0) and y → Q(0, y).— The restrictions of r (ω)/ K(x(ω), 0) on (5.9) M = ω [, + 1[+ ω k/2,(k + 1)/2 k, 1 2 for k, ∈ Z provide all branches on C \ ([x , x ]∪[x , x ]) of Q(x, 0) as follows: 1 2 3 4 (5.10) Q(x, 0) = r (ω)/K x(ω), 0 : ω is the (unique) element of M x k, such that x(ω) = x . Due to the ω -periodicity of r (ω) and x(ω), the restrictions of these functions on M 1 x k, do not depend on ∈ Z, and therefore determine the same branch as on M for k,0 any . Furthermore, thanks to (5.4), (3.5)and (3.7) the restrictions of r (ω)/K(x(ω), 0) on M and on M lead to the same branches for any k ∈ Z. Hence, the restrictions −k+1,0 k,0 of r (ω)/K(x(ω), 0) to M with k ≥ 1 provide all different branches of this function. x k,0 The analogous statement holds for the restrictions of r (ω)/K(0, y(ω)) on (5.11) N = ω /2 + ω , + 1[+ ω ]k/2,(k + 1)/2 k, 3 1 2 for k, ∈ Z, namely: (5.12) Q(0, y) = r (ω)/K 0, y(ω) : ω is the (unique) element of N y k, such that y(ω) = y . The restrictions on N for ∈ Z give the same branch as on N .For any k ∈ Z the k, k,0 + restrictions on N and on N determine the same branches. Hence, the restrictions −k+1,0 k,0 of r (ω)/K(0, y(ω)) on N with k ≥ 1 provide all different branches of y → Q(0, y). y k,0 5.3. Ratio ω /ω 2 3 Remark 6. —For any z ∈]0, 1/|S|[ the value ω /ω is rational if and only if the 2 3 group ξ, η restricted to the curve {(x, y) ∈ C ∪{∞} : K(x, y) = 0} is ﬁnite, see [6, Section 4.1.2] and [18, Proof of Proposition 4]. The rationality or irrationality of the quantity ω /ω is crucial for the nature of 2 3 the functions x → Q(x, 0) and y → Q(0, y) for a given z. Indeed, the following theorem holds true. ON THE FUNCTIONS COUNTING WALKS IN THE QUARTER PLANE 87 FIG. 10. — Three famous examples, known as Kreweras’, Gessel’s and Gouyou-Beauchamps’ walks, respectively Theorem 7. —For any z ∈]0, 1/|S|[ such that ω /ω is rational, the functions x → Q(x, 0) 2 3 and y → Q(0, y) are holonomic. Proof. — The proof of Theorem 7 is completely similar to that of Theorems 1.1 and 1.2 in [5], so here we just recall the main ideas. The proof actually consists in applying [6, Theorem 4.4.1], which entails that if ω /ω is rational, the function Q(x, 0) can be 2 3 written as Q(x, 0) = w (x) + (x)φ (x) + w(x)/r(x), where w and r are rational functions, while φ and w are algebraic. Further, in [5, Lemma 2.1] it is shown that is holonomic. Accordingly, Q(x, 0) is also holonomic. The argument for Q(0, y) is similar. Notice that Theorem 4.4.1 in [6]isprovedfor z = 1/|S| only, but in [5] it is observed that this result also holds for z ∈]0, 1/|S|[. For all 23 models of walks with ﬁnite group (1.3), the ratio ω /ω is rational and 2 3 independent of z. This fact, which is speciﬁed in Lemma 8 below, implies the holonomy of the functions x → Q(x, 0) and y → Q(0, y) for all z ∈]0, 1/|S|[ by Theorem 7,and also leads to some more profound analysis of the models with a ﬁnite group. This analysis is the topic of the Section 6. For all 51 non-singular models of walks with inﬁnite group, ω /ω takes rational 2 3 and irrational values on subsetsH and ]0, 1/|S|[\H, respectively, which are dense on ]0, 1/|S|[,asitwillbeprovedinProposition 14 below. For any z ∈H, x → Q(x, 0) and y → Q(0, y) are holonomic by Theorem 7.For all z ∈]0, 1/|S|[\H, properties of the branches of x → Q(x, 0) and y → Q(0, y) (in particular, the set of their poles) will be studied in detail in Section 7; the non-holonomy will be derived from this analysis. 6. Finite group case Deﬁne the covariance of the model as (6.1) ij − i j = ij. (i,j)∈S (i,j)∈S (i,j)∈S (i,j)∈S 88 I. KURKOVA, K. RASCHEL The equality above follows from the fact that for each of the 23 models with a ﬁnite group, i = 0or j = 0, see [2]. Lemma 8 below is proved in [18, Proposition 5]. (i,j)∈S (i,j)∈S Lemma 8. — For all 23 models with ﬁnite group (1.3), ω /ω is rational and independent 2 3 of z. More precisely: – For the walks with a group of order 4, ω /ω = 2; 2 3 – For the walks with a group of order 6 and such that the covariance is negative (resp. positive), ω /ω = 3 (resp. 3/2); 2 3 – For the walks with a group of order 8 and a negative (resp. positive) covariance, ω /ω = 4 2 3 (resp. 4/3). In the sequel, we note ω /ω = k/; then, 2k is the order of the group. Since 2 3 kω = ω , we obviously always have 3 2 r (ω + ω ) − r (ω) = r (ω + mω ) − r ω + (m − 1)ω . x 2 x x 3 x 3 1≤m≤k It follows from (5.1) and from properties (3.5), (3.6)and (3.7) of the Galois automorphisms that (6.2) r (ω + ω ) − r (ω) = (xy)(ω + mω ) − (xy) η(ω + mω ) x 2 x 3 3 1≤m≤k m m−1 = (xy) ( η ξ) ω − (xy) ξ( η ξ) ω ) 1≤m≤k = (−1) xy θ(ω) , θ ∈ξ, η θ θ (θ ) where (−1) is the signature of θ;inother words, (−1) = (−1) ,where (θ ) is the length of θ , i.e., the smallest such that we can write θ = θ ◦···◦ θ ,with θ ,...,θ equal 1 1 to ξ or η. The same identity with the opposite sign holds for r . The quantity (6.2)isthe orbit-sum of the function xy under the group ξ, η, and is denoted by O(ω). It satisﬁes the property hereunder, which is proved in [2]. Lemma 9. — In the ﬁnite group case, the orbit-sum O(ω) is identically zero if and only if the covariance (6.1) is positive. We therefore come to the following corollary. Corollary 10. — In the ﬁnite group case, the functions x → Q(x, 0) and y → Q(0, y) have a ﬁnite number of different branches if and only if the covariance (6.1) is positive. ON THE FUNCTIONS COUNTING WALKS IN THE QUARTER PLANE 89 After the lifting to the universal covering done in Theorem 4, results of [1, 2] concerning the nature of the functions x → Q(x, 0) and y → Q(0, y) in all ﬁnite group cases can now be established by very short reasonings. For the sake of completeness, we show how this works. Proposition 11 ([1, 2]). — For all models with a ﬁnite group and a positive covariance (6.1), x → Q(x, 0) and y → Q(0, y) are algebraic. Proposition 12 ([2]). — For all models with a ﬁnite group and a negative or zero covariance (6.1), x → Q(x, 0) and y → Q(0, y) are holonomic and non-algebraic. Proofs of both of these propositions involve the following lemma. Lemma 13. —Let ℘ be a Weierstrass elliptic function with certain periods ω, ω. (P1) We have ℘ (ω) = 4 ℘(ω) − ℘(ω/2) ℘(ω) − ℘ [ω + ω]/2 × ℘(ω) − ℘( ω/2) , ∀ω ∈ C. (P2) Let p be some positive integer. The Weierstrass elliptic function with periods ω, ω/pcan be written in terms of ℘ as p−1 ℘(ω) + ℘(ω + ω/p) − ℘( ω/p) , ∀ω ∈ C. =1 (P3) We have the addition theorem: 1 ℘ (ω) − ℘ ( ω) ℘(ω + ω) =−℘(ω) − ℘( ω) + , ∀ω, ω ∈ C. 4 ℘(ω) − ℘( ω) (P4) For any elliptic function f with periods ω, ω, there exist two rational functions R and S such that f (ω) = R ℘(ω) + ℘ (ω)S ℘(ω) , ∀ω ∈ C. (P5) There exists a function which is ω-periodic and such that (ω + ω) = (ω) − 1, ∀ω ∈ C. Proof. — Properties (P1), (P3) and (P4) are most classical, and can be found, e.g., in [10, 20]. For (P2) we refer to [20, page 456], and for (P5), see [6, Equation (4.3.7)]. Note that the function in (P5) can be constructed via the zeta function of Weierstrass. 90 I. KURKOVA, K. RASCHEL Proof of Proposition 11. — If the orbit-sum O(ω) is zero, Equation (6.2) implies that r (ω) is ω -periodic. In particular, the property (P4) of Lemma 13 entails that there exist x 2 two rational functions R and S such that (6.3) r (ω) = R ℘(ω; ω ,ω ) + ℘ (ω; ω ,ω )S ℘(ω; ω ,ω ) . x 1 2 1 2 1 2 Further, the property (P2) together with the addition formula (P3) of Lemma 13 gives that ℘(ω; ω ,ω ) is an algebraic function of ℘(ω)—we recall that ℘(ω) denotes the 1 2 Weierstrass function ℘(ω; ω ,ω ). Due to Lemma 13(P1), ℘ (ω) is an algebraic function 1 2 of ℘(ω) too, so that ℘ (ω; ω ,ω ) is also an algebraic function of ℘(ω). Thanks to (6.3), 1 2 we get that r (ω) is algebraic in ℘(ω). Since ℘(ω) is a rational function of x(ω), see (3.3), we ﬁnally obtain that r (ω) is algebraic in x(ω).Then q (ω) = r (ω)/K(x(ω), 0) is x x x algebraic in x(ω), and so is q (ω) in y(ω). Proof of Proposition 12. — In this proof we have = 1, see Lemma 8. Thanks to Lemma 13(P5), there exists a function which is ω -periodic and such that (ω + ω ) = 1 2 (ω) − 1. In particular, transforming (6.2) we can write r (ω + ω ) + (ω + ω )O(ω + ω ) = r (ω) + (ω)O(ω). x 2 2 2 x This entails that r (ω) + (ω)O(ω) is elliptic with periods ω ,ω . In particular, for the x 1 2 same reasons as in the proof of Proposition 11, this is an algebraic function of x(ω). The function O(ω) is obviously also algebraic in x(ω).Asfor thefunction (ω),itis proved in [6, page 71] that it is a non-algebraic function of x(ω). Moreover, it is shown in [5, Lemma 2.1] that it is holonomic in x(ω).Hence r (ω) is holonomic in x(ω) but not algebraic. The same is true for q (ω) = r (ω)/K(x(ω), 0) in x(ω) and for q (ω) in y(ω). x x y 7. Inﬁnite group case It has been shown in [18, Part 6.2] that for all 51 models with inﬁnite group, ω /ω 2 3 takes irrational values for inﬁnitely many z. The next proposition states a more complete result. Proposition 14. — For all 51 walks with inﬁnite group, the setsH ={z ∈]0, 1/ |S|[: ω /ω is rational} and ]0, 1/|S|[\H ={z ∈]0, 1/|S|[: ω /ω is irrational} are dense 2 3 2 3 in ]0, 1/|S|[. Proof.—Thefunction ω /ω is clearly real continuous function on ]0, 1/|S|[.In 2 3 fact, it has been noticed in [18, Part 6.2] that the function ω /ω is expandable in power 2 3 series in a neighborhood of any point of the interval ]0, 1/|S|[. Thus it sufﬁces to ﬁnd just one segment within ]0, 1/|S|[ where this function is not constant. ON THE FUNCTIONS COUNTING WALKS IN THE QUARTER PLANE 91 Proposition 25 below gives the asymptotic of ω /ω as z → 0: for any of 51 models 2 3 there exist some rational L > 0and some L = 0such that as z > 0goes to0, ω /ω = L + L/ ln z + O (1/ ln z) . 2 3 This immediately implies that this function is not constant on a small enough interval in a right neighborhood of 0 and concludes the proof. Note however that there is another way to conclude the proof that does not need the full power of Proposition 25: it is enough to show (as done in the proof of Proposi- tion 25)that ω /ω converges to a rational positive constant L as z → 0for all51models. 2 3 Indeed, then, since ω /ω necessarily takes irrational values for some z ∈]0, 1/|S|[ (see 2 3 [18, Part 6.2]), there exists an interval within ]0, 1/|S|[ where the ratio ω /ω is not 2 3 constant. In Subsections 7.1, 7.2 and 7.3 we thoroughly analyze the branches of x → Q(x, 0) and y → Q(0, y) for z such that ω /ω is irrational, and we prove in particular that their 2 3 set of poles is inﬁnite and dense on the curves given on Figure 11, see Theorem 17.Then the following corollary is immediate. Corollary 15. —LetH ={z ∈]0, 1/|S|[: ω /ω is rational} and ]0, 1/|S|[\H ={z ∈ 2 3 ]0, 1/|S|[: ω /ω is irrational}. 2 3 (i) For all z ∈H,x → Q(x, 0) and y → Q(0, y) are holonomic; (ii) For all z ∈]0, 1/|S|[\H,x → Q(x, 0) and y → Q(0, y) are non-holonomic. Proof. — The statement (i) follows from Theorem 7, and (ii) comes from Theo- rem 17(iii) below as explained in the Introduction. Remark 16. — It follows from Remark 6 thatH can be characterized as the set of z ∈]0, 1/|S|[ such that the group ξ, η restricted to the curve {(x, y) ∈ (C ∪{∞}) : K(x, y; z) = 0} is ﬁnite. Then methods developed in [6, Chapter 4] speciﬁcally for the ﬁnite group case should be efﬁcient for further analysis of x → Q(x, 0) and y → Q(0, y) for any ﬁxed z ∈H. The analysis of the poles being rather technical, we start ﬁrst with an informal study. 7.1. Polesofthe setofbranchesof x → Q(x, 0) and y → Q( y, 0) for irrational ω /ω : 2 3 an informal study. — Let us ﬁx z ∈]0, 1/|S|[ such that ω /ω is irrational. We ﬁrst infor- 2 3 mally explain why the set of poles of all branches of x → Q(x, 0) and y → Q(0, y) could be dense on certain curves in this case. We shall denote by ω and ω the real and 92 I. KURKOVA, K. RASCHEL imaginary parts of ω ∈ C, respectively. Let and be the parallelograms deﬁned by x y = M ∪ M = ω [0, 1[+ ω [0, 1[, x 0,0 0,1 1 2 (7.1) = N ∪ N = ω /2 + ω [0, 1[+ ω ]0, 1], y 0,0 0,1 3 1 2 with notations (5.9)and (5.11). Function r (ω) (resp. r (ω))on (resp. ) deﬁnes the x y x y ﬁrst (main) branch of x → Q(x, 0) (resp. y → Q(0, y))twice via(5.10) (resp. (5.12)). Denote by f (ω) = x(ω)[y(−ω + 2ω ) − y(ω)] the function used in the mero- y x morphic continuation procedure (5.2). Assume that at some ω ∈ , r (ω ) = ∞ and 0 y y 0 f (ω ) =∞. Further, suppose that y 0 (7.2) ω ∈ :ω =ω , f (ω) =∞. y 0 y By (5.2), for any n ≥ 1we have n−1 (7.3) r (ω + nω ) = r (ω ) + f (ω ) + f (ω + kω ). y 0 3 y 0 y 0 y 0 3 k=1 We have r (ω ) + f (ω ) =∞ by our assumptions. If ω /ω is irrational, then for any y 0 y 0 2 3 k ≥ 1 there is no p ∈ Z such that ω + kω = ω + pω .Function f being ω -periodic, it 0 3 0 2 y 2 follows from this fact and assumption (7.2)that f (ω + kω ) = ∞ for any k ≥ 1. Hence y 0 3 by (7.3), r (ω + nω ) =∞ for all n ≥ 1. Due to irrationality of ω /ω ,for any n ≥ 1there y 0 3 2 3 exists a unique ω (ω ) ∈ and p ∈ Z such that ω + nω = ω (ω ) + pω , and the set n 0 y 0 3 n 0 2 {ω (ω )} is dense on the curve n 0 n≥1 (7.4)I (ω ) = y {ω :ω =ω ,ω ∈ } ⊂ C ∪{∞}. y 0 0 y By deﬁnition (5.12), the set of poles of all branches of y → K(0, y)Q(0, y) is dense on the curveI (ω ).The number of zerosof y → K(0, y) being at most two, the same conclusion y 0 holds true for y → Q(0, y). Let us now identify the points ω in where f (ω ) is inﬁnite. They are (at 0 y y 0 most) six such points a , a , a , a , b , b ∈ , which correspond to the following pairs 1 2 3 4 1 2 y (x(ω ), y(ω )): 0 0 a = x , ∞ , a = x , y , a = x , ∞ , a = x , y , 1 4 2 3 (7.5) ◦ • b = ∞, y , b = ∞, y . 1 2 ON THE FUNCTIONS COUNTING WALKS IN THE QUARTER PLANE 93 Here by (2.4)and (2.5) 2 1/2 −b( y) +[b( y) − 4 a( y) c( y)] x = lim , y→∞ a( y) 2 1/2 −b( y) −[b( y) − 4 a( y) c( y)] x = lim , y→∞ a( y) 2 1/2 −b(x) +[b(x) − 4a(x)c(x)] y = lim , x→x 2a(x) 2 1/2 −b(x) +[b(x) − 4a(x)c(x)] y = lim , 2a(x) x→x 2 1/2 −b(x) +[b(x) − 4a(x)c(x)] y = lim , x→∞ 2a(x) 2 1/2 −b(x) −[b(x) − 4a(x)c(x)] y = lim . x→∞ 2a(x) where a, b, c, a, b, c are introduced in (2.1). For most of 51 models of walks, assumption (7.2) holds true for none of these points, so that the previous reasoning does not work: some poles of f could be compensated in the sum (7.3). Furthermore, it may happen for some of these points that not only f (ω ) =∞ y 0 but also r (ω ) =∞, and consequently f (ω) + r (ω) mayhavenopoleat ω = ω .For y 0 y y 0 these reasons we need to inspect more closely the location of these six points for each of the 51 models and their contribution to the set of poles via (7.3). 7.2. Functions x → Q(x, 0) and y → Q(0, y) for irrational ω /ω .— In addition to 2 3 the notation (7.4), deﬁne the curve (7.6)I (ω ) = x {ω :ω =ω ,ω ∈ } ⊂ C ∪{∞}. x 0 0 x We now formulate the main theorem of this section. Theorem 17. — For all 51 non-singular walks with inﬁnite group (1.3) given on Figure 17 and any z such that ω /ω is irrational, the following statements hold. 2 3 (i) The only singularities on C of the ﬁrst branch of x → Q(x, 0) (resp. y → Q(0, y))are the branch points x and x (resp. y and y ). 3 4 3 4 (ii) Each branch of x → Q(x, 0) (resp. y → Q(0, y)) is meromorphic on C with a ﬁnite number of poles. (iii) The set of poles on C of all branches of x → Q(x, 0) (resp. y → Q(0, y)) is inﬁnite. With the notations (7.6), (7.4) above and points a , b deﬁned in (7.5), it is dense on the 1 1 following curves (see Figure 11): 94 I. KURKOVA, K. RASCHEL FIG. 11. — For walks pictured on Figure 17, curves where poles of the set of branches of x → Q(x, 0) and y → Q(0, y) are dense (iii.a) For the walks of Subcase I.A in Figure 17:I (a ) andI (b ) for x → Q(x, 0); x 1 x 1 I (a ) andI (b ) for y → Q(0, y). y 1 y 1 (iii.b) For the walks of Subcases I.B and I.C in Figure 17:I (a ) and R\]x , x [ for x 1 1 4 x → Q(x, 0);I (a ) and [y , y ] for y → Q(0, y). y 1 4 1 ON THE FUNCTIONS COUNTING WALKS IN THE QUARTER PLANE 95 (iii.c) For the walks of Subcase II.A in Figure 17:I (b ) and [x , x ] for x → Q(x, 0); x 1 4 1 I (b ) and R\]y , y [ for y → Q(0, y). y 1 1 4 (iii.d) For the walks of Subcases II.B, II.C, II.D and Case III in Figure 17: R\]x , x [ 1 4 for x → Q(x, 0); R\]y , y [ for y → Q(0, y). 1 4 (iv) Poles of branches of x → Q(x, 0) and y → Q(0, y) out of these curves may be only at zeros of K(x, 0) and K(0, y), respectively. Before giving the proof of Theorem 17, we need to introduce some additional tools. If the value of ω /ω is irrational, for any ω ∈ C and any n ∈ Z , there exists a unique 2 3 0 + y x ω (ω ) ∈ (resp. ω (ω ) ∈ )aswellasaunique number p ∈ Z (resp. p ∈ Z)such 0 y 0 x y x n n y x that ω + nω = p ω + ω (ω ) (resp. ω + nω = p ω + ω (ω )). With these notations we 0 3 y 2 0 0 3 x 2 0 n n can state the following lemma. Lemma 18. — Let z be such that ω /ω is irrational. 2 3 x x y y (a) For all n = m, we have ω (ω ) = ω (ω ) and ω (ω ) = ω (ω ). 0 0 0 0 n m n m x y (b) The set {ω (ω )} (resp. {ω (ω )} ) is dense on the segment {ω ∈ :ω = 0 n∈Z 0 n∈Z x + + n n ω } (resp. {ω ∈ :ω =ω }). 0 y 0 Proof. — Both (a) and (b) are direct consequences of the irrationality of ω /ω . 2 3 In the next deﬁnition, we introduce a partial order in . Deﬁnition 19. — For any ω, ω ∈ , we write ω ω if for some n ∈ Z and some p ∈ Z, y + ω + nω = ω + pω . 3 2 If ω ω (and if ω /ω is irrational), both n and p are unique and sometimes we shall 2 3 write ω ω . In particular, for any ω ∈ ,wehave ω ω, since ω ω. n y 0 Deﬁnition 20. — If either ω ω or ω ω ,wesay that ω and ω are ordered, and we write ω ∼ ω . Let us denote by f and f the (meromorphic) functions used in the meromor- x y phic continuation procedures (5.1)and (5.2), namely, by using (3.5): f (ω) = y(ω) x( ηω) − x(ω) , f (ω) = x(ω) y(ξω) − y(ω) . x y The following lemma will be the key tool for the proof of Theorem 17. Lemma 21. — Let z be such that ω /ω is irrational; let ω ∈ be such that r (ω ) = ∞, 2 3 0 y y 0 and let A(ω ) = ω ∈ :ω =ω , f (ω) =∞ . 0 y 0 y 1 k Assume that ω ∈A(ω ) and that for some ω ,...,ω ∈A(ω ): 0 0 0 96 I. KURKOVA, K. RASCHEL 1 k (A) ω ω ··· ω ; 0 n n n 1 2 k (B) lim {f (ω) + f (ω + n ω ) + f (ω + n ω ) +· · · + f (ω + n ω )}=∞; ω→ω y y 1 3 y 2 3 y k k (C) there is no other ω ∈A(ω ) such that ω ω. 0 0 Then the set of poles of all branches of x → Q(x, 0) (resp. y → Q(0, y)) is dense on the curveI (ω ) x 0 (resp.I (ω ))deﬁnedin(7.6)(resp.(7.4)). y 0 Proof. — By Equation (5.2)ofTheorem 4,wehave, forany n ∈ Z and any ω ∈ , + y (7.7) r (ω + nω ) = r (ω) + f (ω) + f (ω + ω ) + f (ω + 2ω ) + ··· y 3 y y y 3 y 3 + f ω + (n − 1)ω . y 0 3 Let ω be as in the statement of Lemma 21. Due to assumption (C), Lemma 18(a) and the ω -periodicity of f , the set {ω + nω } does not contain any point ω where 2 y 0 3 n>n +···+n 1 k f (ω) =∞. Further, by the assumptions (A) and (C), Lemma 18(a) and also by the ω - y 2 periodicity of f , the set {ω + nω } contains exactly k + 1 poles of f that are y 0 3 0≤n≤n +···+n y 1 k ω ,ω + n ω ,...,ω + n ω . Then, by (7.7), assumption (B) and the fact that r (ω ) = ∞, 0 0 1 3 0 k 3 y 0 we reach the conclusion that for any n > n +· · · + n , the point ω + nω is a pole of r (ω). 1 k 0 3 y Due to Equation (5.3), any ω pole of r such that x(ω)y(ω) = ∞ is also a pole of r . Deﬁne nowB, the set of (at most twelve) points in where either x(ω) =∞, x y y(ω) =∞,K(x(ω), 0) = 0orK(0, y(ω)) = 0. Introduce also M = max{m ≥ 0 : ω 0 m ω for some ω ∈B}—with the usual convention M =−∞ if ω ω for none ω ∈B.If n > max(M, n + ··· + n ), the points ω + nω are poles of r as well, and both K(x(ω + 1 k 0 3 x 0 nω ), 0) and K(0, y(ω + nω )) are non-zero. By Lemma 18(b) and deﬁnitions (5.10)and 3 0 3 (5.12), Lemma 21 follows. We are now ready to give the proof of Theorem 17. Proof of Theorem 17.—Functions f (ω) and y(ω) being ω -periodic, it follows that y 2 both of them have no pole at any ω with 0 ≤ω< ω and ω/∈{a , a , a , a , 1 1 2 3 4 b , b }. Then, by (5.2), 1 2 (7.8) ∀ω ∈ N with ω = {a , a , a , a , b , b }, r (ω) = ∞. k,0 1 2 3 4 1 2 y k=0 Function x(ω) being ω -periodic, it has no pole at any ω such that 0 ≤ω< ω and 2 1 ∞ ∞ ω/∈{b , b }. Then Equation (5.3) and the fact that M ⊂ N imply 1 2 k,0 k,0 k=1 k=0 that (7.9) ∀ω ∈ M with ω = {a , a , a , a , b , b }, r (ω) = ∞. k,0 1 2 3 4 1 2 x k=1 In order to prove Theorem 17(i), we shall prove the following proposition. ON THE FUNCTIONS COUNTING WALKS IN THE QUARTER PLANE 97 FIG. 12. — Location of a , a , a , a , b , b if y < 0and x < 0, i.e., Subcase I.A 1 2 3 4 1 2 4 4 Proposition 22. — For all 51 models, for any ω ∈ M (resp. ω ∈ N ) with x(ω) = ∞ 1,0 1,0 (resp. y(ω) = ∞)and ω ∈{a , a , a , a , b , b },wehave r (ω) = ∞ (resp. 1 2 3 4 1 2 x r (ω) = ∞). The proof of this proposition is postponed to the next subsection. By this propo- sition, (7.8)and (7.9), the only singularities of the ﬁrst branches of K(x, 0)Q(x, 0) (resp. K(0, y)Q(0, y)) may be only among the branch points x , x , x , x (resp. y , y , y , y ). Let 1 2 3 4 1 2 3 4 i n us recall that the function x → Q(x, 0) is initially deﬁned as a series q(i, 0; n)x z . i,n≥0 The elementary estimate q(i, 0; n) ≤|S| implies that for any z ∈]0, 1/|S|[ and i≥0 x ∈ C with |x|≤ 1 this series is absolutely is convergent. Since |x | < 1, |x | < 1, and since 1 2 also K(x, 0) is a polynomial with (at most two) roots that are smaller or equal to 1 by ab- solute value, the only singularities of the ﬁrst branch of x → Q(x, 0) are the branch points x and x , that are out of the unit disc. By the same arguments the analogous statement 3 4 holds true for y → Q(0, y). This ﬁnishes the proof of Theorem 17(i). Since for any p ∈ Z , there exist only ﬁnitely many ω ∈ M (resp. ω ∈ + k,0 k=1 N )where f (ω) =∞ (resp. f (ω) =∞), Theorem 17(ii) immediately follows from 0, x y =1 the meromorphic continuation procedure of r and r done in Section 5, namely Equa- x y tions (5.1)and (5.2) as well as the deﬁnitions (5.10)and (5.12). The following proposition proves Theorem 17(iii). Proposition 23. — The poles of x → Q(x, 0) (resp. y → Q(0, y)) are dense on the six curves I (ω ) (resp.I (ω )) with ω ∈{a , a , a , a , b , b }.For anyof 51 models, the set of these curves x 0 y 0 0 1 2 3 4 1 2 coincide with the one claimed in Theorem 17(iii). The proof of this proposition is postponed to the next subsection as well, it will be based on Lemma 21 with ω appropriately chosen among a , a , a , a , b , b . 0 1 2 3 4 1 2 The last statement (iv) of the theorem follows immediately from (7.8)and (7.9), Proposition 23 and Deﬁnitions (5.10)and (5.12). 7.3. Proof of Propositions 22 and 23.— To start with the proofs of Propositions 22 and 23, we need to study closer the location of points (7.5) a , a , a , a , b , b on .It 1 2 3 4 1 2 y 98 I. KURKOVA, K. RASCHEL depends heavily on the signs of x and y , see Figures 12, 13, 14, 15 and 16. Let us recall 4 4 that 0 < x < ∞ and 0 < y < ∞, see Section 2. 3 3 If y < 0 (resp. x < 0), the point y =∞ (resp. x =∞) obviously belongs to the real 4 4 cycle ]y , ∞[ ∪{∞}∪]∞, y [ (resp. ]x , ∞[ ∪{∞}∪]∞, x [) of the complex sphere S.By 3 4 3 4 construction of the Riemann surface T and of its universal covering, the points a , a 1 2 (resp. b , b ) then lie on the open interval {ω : ω ∈ L + ω , 0 < ω< ω } (resp. {ω : 1 2 2 1 ω ∈ L + ω , 0 < ω< ω }). These points are symmetric w.r.t. the center of the interval, 2 1 namely ω /2 + ω + ω /2 (resp. ω /2 + ω ). The points ω corresponding to a and a are 1 2 3 1 2 3 4 on the open interval {ω : ω ∈ L + ω − ω , 0 < ω< ω } and are symmetric w.r.t. the 2 3 1 center ω /2 + ω − ω /2 as well. Furthermore, a + ω = a and a + ω = a ,sothat 1 2 3 4 3 2 3 3 1 a a and a a , see Figure 12. Finally, we have a =a = a = ,hence for 4 1 2 3 1 1 4 2 1 3 any a ∈{a , a } and any a ∈{a , a }, a a (in the sense of Deﬁnition 20). 2 4 1 3 If y > 0or y =∞ (resp. x > 0or x =∞), the point y =∞ (resp. x =∞)is 4 4 4 4 on ]y , ∞] ∪ {∞} ∪ ]∞, y [ (resp. ]x , ∞] ∪{∞}∪]∞, x [). Accordingly, the points a , a 4 1 4 1 1 2 (resp. b , b )and also a , a are on the segment ]ω /2,ω + ω /2].Their location on this 1 2 3 4 3 2 3 segment will be speciﬁed latter. Therefore, Propositions 22 and 23 must be proved separately for eight subclasses of 51 models according to the signs of x and y : these are those of the walks pictured on 4 4 Figure 17, Subcases I.A, I.B, I.C, II.A, II.B, II.C, II.D and Case III. The following remark gives a geometric interpretation of this classiﬁcation. Remark 24. —Let 1 be 1 if (i, j) ∈S , otherwise 0. Then x > 0 (resp. < 0, =∞) (i,j) 4 if and only if 1 − 41 1 > 0 (resp. < 0, = 0), see Equation (2.2). A symmetric (1,1) (1,−1) (1,0) statement holds for y . As an example, Remark 24 implies that x < 0 if and only if (1, 1) ∈S and (1, −1) ∈S . Case I:y < 0,Subcase I.A:x < 0.— This assumption yields x = x ; x , x = ∞; 4 4 • ◦ • y = y ; y , y = ∞; y , y = ∞. The location of the six points a , a , a , a , b , b is already 1 2 3 4 1 2 described above and is pictured on Figure 12. We ﬁrst show that for all ω ∈{ω :ω =a ,ω ≤ω< ω + ω },wehave 3 y y 2 1 4 r (ω) = ∞. The proof consists in three steps. Step 1. — Let us ﬁrst prove that r (a ) = ∞ and r (a ) = ∞.If |y | < 1, then y 3 y 4 a ∈ (see Section 4 for the deﬁnition of ) and it is immediate from Theorem 3 that 3 y y r (a ) = ∞.If |y |≥ 1, then by Lemma 5 there exists n ∈ Z such that a − nω ∈ ,and y 3 + 3 3 by Equation (5.2)ofTheorem 4, (7.10) r (a ) = r (a − nω ) + f (a − kω ). y 3 y 3 3 y 3 3 k=n ON THE FUNCTIONS COUNTING WALKS IN THE QUARTER PLANE 99 Introducing, for any ω , the set O (ω ) ={ω − ω ,ω − 2ω ,...,ω − n ω }, 0 0 3 0 3 0 ω 3 (7.11) n = inf{ ≥ 0 : ω − ω ∈ }, ω 0 3 we can rewrite (7.10)as (7.12) r (a ) = r (a − n ω ) + f (ω). y 3 y 3 a 3 y ω∈O (a ) In (7.12), the quantity r (a − n ω ) is deﬁned thanks to Theorem 3. It may be inﬁnite, y 3 a 3 but only if y(a − n ω ) =∞. In this case we must have a − n ω = a − ω . But since 3 a 3 3 a 3 1 2 3 3 a + ω = a ,wethenhave (n + 1)ω = ω which is impossible, due to irrationality of 3 3 1 a 3 2 ω /ω .Hence r (a − n ω ) = ∞. Further, we immediately have (see indeed Figure 12) 2 3 y 3 a 3 that a , a , a − ω , a − ω ∈ / O (a ). Moreover, since either b = a ,or b =a 2 4 2 2 4 2 3 1 3 1 3 but then a + ω /2 = b (see again Figure 12), we also have that b , b , b − ω , b − ω ∈ / 3 3 1 1 2 1 2 2 2 O (a ). Finally a = a + ω ∈ / O (a ) and a − ω = a + ω − ω ∈ / O (a ), since ω /ω 3 1 3 3 3 1 2 3 3 2 3 2 3 is irrational. Thus f (a − kω ) = ∞ for any k ∈{1,..., n }. Accordingly, r (a ) = ∞ and y 3 3 a y 3 by the same arguments, r (a ) = ∞. y 4 Step 2. — We now show that r (a − ω ) + f (a − ω ) = ∞ and r (a − ω ) + y 1 2 y 1 2 y 2 2 f (a − ω ) = ∞. By Equation (5.3), r (a − ω ) =−r (a − ω ) + K(0, 0)Q(0, 0) + x(a − y 2 2 y 1 2 x 1 2 1 ω )y(a − ω ) and f (a − ω ) = x(a − ω )[y(a ) − y(a − ω )];hence 2 1 2 y 1 2 1 2 4 1 2 (7.13) r (a − ω ) + f (a − ω ) =−r (a − ω ) + K(0, 0)Q(0, 0) y 1 2 y 1 2 x 1 2 + x(a − ω )y(a ). 1 2 4 It follows from Equation (5.3) and from the ﬁrst step that r (a ) = ∞, since x(a )y(a ) = x 4 4 4 x y = ∞. Then, by (3.5)and (5.4)weget that r (a − ω ) = r (ξ a ) = r (a ) = ∞. x 1 2 x 4 x 4 Furthermore, x(a − ω )y(a ) = x y = ∞. Finally, thanks to (7.13), r (a − ω ) + 1 2 4 y 1 2 f (a − ω ) = ∞ and by the same arguments, r (a − ω ) + f (a − ω ) = ∞. y 1 2 y 2 2 y 2 2 Step 3. — Let us now take any ω in {ω :ω =a ,ω ≤ω< ω + ω }.If 0 3 y y 2 1 4 ω ∈ ,then ω ∈ . Indeed, it is proved in Section 4 that the domain (resp. ), 0 0 y y x 0 1 0 1 y x 2 2 which is bounded by and (resp. and ), is centered around L (resp. L ). y y x x y x 1 1 0 1 0 1 Furthermore, ∈ and ∈ / (resp. ∈ and ∈ / ). It follows that for any x x y y y y x x ω ∈ \ , ω <ω . Then, by Theorem 3, r (ω ) = ∞.If ω ∈ / ,with(7.11)and 0 y 0 y y 0 0 (7.12)wehave r (ω ) = r (ω − n ω ) + f (ω). y 0 y 0 ω 3 y ω∈O (ω ) 0 100 I. KURKOVA, K. RASCHEL For the same reasons as in the ﬁrst step, we have that a , a , a − ω , a − ω ∈ / O (ω ). 2 4 2 2 4 2 0 If ω < a , for obvious reasons O (ω ) cannot contain a .If a ≤ω <ω + ω , 0 3 0 3 3 0 y 2 it can neither contain a , since ω + ω −a = ω , and hence ω − ω < a .If 3 y 2 3 3 0 3 3 b = a ,or b =a and ω < b , it cannot contain b .If b =a and b ≤ 1 3 1 3 0 1 1 1 3 1 ω <ω + ω ,then ω −b ≤ ω + ω −b = ω /2 <ω ,and b ∈ / O (ω ). 0 y 2 0 1 y 2 1 3 3 1 0 4 4 If a − ω ∈ / O (ω ),thenwehave r (ω − n ω ) = ∞ and f (ω − kω ) = ∞ for 1 2 0 y 0 ω 3 y 0 3 all k ∈{1,..., n },sothat r (ω ) = ∞ by (7.12). ω y 0 If a − ω ∈ O (ω ),thenfor some j ∈{1,... n },wehave ω − jω = a − ω . 1 2 0 ω 0 3 1 2 j+1 Then r (ω − n ω ) + f (ω − kω ) = r (a − ω ) and thus by (7.12), y 0 ω 3 y 0 3 y 1 2 0 k=n r (ω ) = r (a − ω ) + f (a − ω ) + f (ω − kω ). y 0 y 1 2 y 1 2 y 0 3 k=j−1 Theﬁrsttermhereisﬁnitebythe secondstepand f (ω − kω ) = ∞ for k ∈{1,..., j − 1} y 0 3 by all properties said above, so that r (ω ) = ∞. y 0 So far we have proved that for all ω ∈{ω :ω =a ,ω ≤ω< ω + ω }, 3 y y 2 1 4 r (ω) = ∞. In the same way, we obtain that r (ω) = ∞ for ω ∈{ω :ω =a ,ω ≤ y y 4 y ω< ω + ω }. y 2 Since by (3.5), η{ω :ω =a ,ω ≤ω< ω + ω } 3 y y 2 1 4 ={ω :ω =a ,ω < ω ≤ ω }, 4 y y 4 1 η{ω :ω =a ,ω ≤ω< ω + ω } 4 y y 2 1 4 ={ω :ω =a ,ω < ω ≤ ω }, 3 y y 4 1 Equation (5.4) implies that r (ω) = ∞ on the segments {ω ∈ :ω = a , a },exceptfor y y 3 4 their ends a , a . The segments {ω :ω = a , a ,ω ≤ω ≤ ω + ω } do not contain 1 2 3 4 x x 2 1 4 any point where y(ω) =∞. It follows from Equation (5.3)that r (ω) = ∞ on these seg- ments except for points where x(ω) =∞ if they exist. This last fact happens if and only if b =a and only at the ends b , b of the segments. 1 3 1 2 If b = a , we can show exactly in the same way that r (ω) = ∞ on the two seg- 1 3 y ments {ω ∈ :ω = b , b } and that r (ω) = ∞ on the segments {ω :ω = b , b ,ω ≤ y 1 2 x 1 2 x ω ≤ ω + ω },exceptfor theirends b , b . This concludes the proof of Proposition 22. x 2 1 2 We proceed with the proof of Proposition 23. Let us verify the assumptions of Lemma 21 for ω = a , a , b , b .Wehaveprovedthat r (a ), r (a ), r (b ), r (b ) = ∞, 0 3 4 1 2 y 3 y 4 y 1 y 2 a a , a a and that the pairs {a , a } and {a , a } are not ordered. Let us now show 3 1 1 4 1 2 1 3 2 4 that for any k ∈{3, 4} and ∈{1, 2},itisimpossibletohave a ∼ b .If b = a , a , this k 3 4 is obvious. If b =a ,thenitisenoughtonotethat b − a = ω /2and a − b = ω /2 3 3 3 1 1 3 (see Figure 12). From the irrationality of ω /ω , it follows that b a , a and in the same 2 3 1 3 way b a , a . Then there is no other ω ∈ except for a (resp. a )suchthat a ω 2 4 y 1 2 3 ON THE FUNCTIONS COUNTING WALKS IN THE QUARTER PLANE 101 (resp. a ω)and f (ω) =∞. There is no ω ∈ such that b ω and f (ω) =∞, 4 y y y = 1, 2. Hence, Lemma 21 could be applied to any of four points ω = a , a , b , b 0 3 4 1 2 if the assumption (B) of this lemma is satisﬁed for these points. It is then immedi- ate that lim f (ω) = lim x(ω)[y(ξω) − y(ω)]=∞, ∈{1, 2}, since x(ω) →∞ ω→b y ω→b ◦ • and the other term converges to ±[y − y ] = 0. Let us verify that lim {f (ω) + ω→a y f (ω + ω )}=∞.Wehave y 3 lim f (ω) + f (ω + ω ) = lim x(ω) y(ξω) − y(ω) y y 3 ω→a ω→a 3 3 + x( η ξω) y(ξ η ξω) − y( η ξω) = lim x( η ξω)y(ξ η ξω) − x(ω)y(ω) ω→a + lim x(ω)y(ξω) − x( η ξω)y( η ξω) . ω→a The ﬁrst term above converges to x y − x y .By(3.7) the second term equals the limit of the product y(ξω)[x(ξω) − x( η ξω)].If ω → a ,then ξω → a − ω so that 3 2 2 the ﬁrst term in the product converges to y(a − ω ) = y(a ) =∞. The second term 2 2 2 of this product converges to x(a − ω ) − x(a ) = x − x which is different from 0 2 2 1 as x = x .Thenassumption (B)issatisﬁedfor ω = a and in the same way for 0 3 ω = a . Lemma 21 applies to any of the four points ω = a , a , b , b .But by (3.7), 0 4 0 3 4 1 2 I (a ) =I (a ) =I (a ) =I (a ),I (a ) =I (a ) =I (a ) =I (a ),I (b ) =I (b ), x 3 x 4 x 1 x 2 y 3 y 4 y 1 y 2 x 1 x 2 I (b ) =I (b ) so that poles of x → Q(x, 0) are dense on the curvesI (a ) andI (b ) y 1 y 2 x 1 x 1 and those of y → Q(0, y) are dense on the curvesI (a ) andI (b ).Proposition 23 is y 1 y 1 proved. Case I:y < 0,Subcase I.B:x =∞.— This assumption implies that x = x ; 4 4 ◦ • x , x = ∞; y = y = ∞; y , y = ∞. The points a , a , a , a are located as in the previous case, see Figure 12. Conse- 1 2 3 4 quently we have the following facts: r (ω) = ∞ on the segments {ω ∈ :ω = a , a }, y y 3 4 except for their ends ω = a , a ; r (ω) = ∞ on the segments {ω :ω = a , a ,ω ≤ω ≤ 1 2 x 3 4 x ω + ω }. Lemma 21 applies to ω = a , a as in the previous case, as x = x . Then the x 2 0 3 4 set of poles of all branches of x → Q(x, 0) (resp. y → Q(0, y)) is dense onI (a ) (resp. x 1 I (a ))whereI (a ) =I (a ) =I (a ) =I (a ) (resp.I (a ) =I (a ) =I (a ) =I (a )). y 1 x 1 x 2 x 3 x 4 y 1 y 2 y 3 y 4 Since x =∞,wehavethat b = b = ω + ω .Takeany ω with ω = 0such 4 1 2 x 2 0 0 that ω ≤ω ≤ ω + ω .Then y(ω ) = ∞. Let us show that r (ω ) = ∞.If ω ∈ , y 0 y 2 0 y 0 0 1 4 then by the same reasons as in Subcase I.A ω ∈ and r (ω ) = ∞.If ω ∈ / ,consider 0 y y 0 0 the set O (ω ) deﬁned as in (7.11)and (7.12). Clearly b − ω = ω ∈ / O (ω ). Since 0 1 2 x 0 b + ω /2 = ω + ω ,wehave ω − ω ≤ ω + ω − ω < b and then b ∈ / O (ω ). 1 3 y 2 0 3 y 2 3 1 1 0 4 4 Hence O (ω ) does not contain any point where y(ω) or f (ω) is inﬁnite. Thus by (7.12), 0 y r (ω ) = ∞. y 0 102 I. KURKOVA, K. RASCHEL FIG. 13. — Location of b , b if x > 0, Subcases I.C and II.C 1 2 4 We have η{ω :ω = 0,ω ≤ω ≤ ω + ω }={ω :ω = ω ,ω ≤ω ≤ ω }. y 0 y 2 1 y 0 y 1 4 4 1 Then by (5.4)and (5.5), we get that r (ω) = ∞ for all ω ∈ with ω = 0. The segment y y {ω :ω = 0,ω ≤ ω ≤ ω + ω } does not contain any point with y(ω) =∞.By(5.3) x x 2 1 4 this gives r (ω) = ∞ for all ω on this segment except for the points where x(ω) =∞ (that is only at ω = ω + ω = b ), and this concludes the proof of Proposition 22. x 2 1 We have proved in particular that r (ω ) = ∞ for ω = b . Furthermore, there is y 0 0 1 no ω ∈ such that b ω and f (ω) =∞. Finally y 1 y (7.14) lim f (ω) = lim x(ω) y(ξω) − y(ω) ω→b ω→b 1 1 2 1/2 [b(x(ω)) − 4a(x(ω))c(x(ω))] = lim x(ω) , ω→b 1 a(x(ω)) where x(ω) →∞ as x → b . For all models in Subcase I.B deg a(x) = 2, deg b(x) = 1 1/2 and deg c(x) = 1, so that (7.14)isofthe order O(|x(ω)| ). Thus lim f (ω) =∞.By ω→b y Lemma 21 with ω = b , the poles of x → Q(x, 0) and those of y → Q(0, y) are dense on 0 1 I (b ) =I (b ) andI (b ) =I (b ), respectively. They are the intervals of the real line x 1 x 2 y 1 y 2 claimed in Theorem 17(iii). Proposition 23 is proved. Case I:y < 0,Subcase I.C:x > 0.— The statements and results about a , a , a , a 4 4 1 2 3 4 are the same as in Subcases I.A and I.B, see Figure 12 for their location. We now locate b , b . By deﬁnition (see Section 2), the values y , y , y , y are the 1 2 1 2 3 4 roots of 1/2 1/2 d( y) = b( y) − 2 a( y) c( y) b( y) + 2 a( y) c( y) = 0. 1/2 Hence, for two of these roots b( y) =−2[ a( y) c( y)] and then X( y) ≥ 0 (see (2.4)), and for 1/2 the two others b( y) = 2[ a( y) c( y)] and then X( y) ≤ 0. But X( y ) and X( y ) are on the 2 3 segment [x , x ]⊂ ]0, ∞[.Thus X( y ) ≤ 0and X( y ) ≤ 0. Since x(b ) = x(b − ω ) =∞, 2 3 1 4 1 1 2 x = x(ω )> 0and X( y ) = x(ω )< 0, it follows that b − ω ∈]ω ,ω [,insuchaway 4 x 4 y 1 2 x y 4 4 4 4 that b ∈]ω + ω ,ω + ω [. Also, b = ξ(b − ω ) − ω = 2(ω + ω ) − b is symmetric 1 x 2 y 2 2 1 2 1 x 2 1 4 4 4 to b w.r.t. ω + ω . Since x(ω ) = X( y ) ≤ 0and x(ω + ω ) = x > 0, it follows that 1 x 2 y 1 x 2 4 4 1 4 ω < b <ω + ω , see Figure 13. y 2 x 2 1 4 Nowweshowthatfor any ω with ω = 0and ω ≤ω ≤ ω + ω , r (ω ) = ∞. 0 0 y 0 y 2 y 0 1 4 Note that y(ω ) = ∞.If ω ∈ , by the same arguments as in Subcase I.A, ω ∈ and 0 0 0 y r (ω ) = ∞.If ω ∈ / ,thenconsider O (ω ) with the notation (7.11). y 0 0 0 Note that b − ω ∈ / .For this,itisenoughtoprove that b − ω ∈ / and that 1 2 1 2 x b − ω ∈ / .First, b − ω ∈ / , since x(b ) = x(b − ω ) =∞. Furthermore, is 1 2 y 1 2 x 1 1 2 y ON THE FUNCTIONS COUNTING WALKS IN THE QUARTER PLANE 103 FIG. 14. — Location of a , a , a , a if y > 0, (x , Y(x )) = (∞, ∞), Subcases II.A, II.B and II.C 1 2 3 4 4 4 4 centered w.r.t. L ,and ω ∈ / (since |y | > 1). Hence the point b − ω <ω cannot be y y 4 1 2 y y 4 4 in . Since ∩{ω ∈ C :ω = 0} is an open interval containing ω , and since b − ω < y 1 2 ω ≤ ω , it follows that b − ω <ω − n ω (see (7.11)), so that b − ω ∈ / O (ω ). y 0 1 2 0 ω 3 1 2 0 1 0 Obviously b − ω ∈ / O (ω ). Furthermore, since ω + ω − b = ω /2 + ω + ω − b < 2 2 0 y 2 2 3 x 2 2 4 4 ω /2 + ω /2 = ω , it follows that ω − ω < b < b for any such ω .Hence b , b ∈ / 3 3 3 0 3 2 1 0 1 2 O (ω ).Thusfor any ω ∈ O (ω ), y(ω) = ∞ and f (ω) = ∞.By(7.12), r (ω ) = ∞. 0 0 y y 0 This implies, exactly as in Subcase I.B—by (5.4)and (5.5)—, that r (ω) = ∞ for all ω ∈ such that ω = 0. By (5.3) this gives r (ω) = ∞ for all ω with ω = 0and ω ≤ω ≤ y x x ω + ω , except for points ω where x(ω) =∞ (that happens for ω = b only), and this x 2 2 concludes the proof of Proposition 22. ◦ • In particular, we proved that r (b ) = ∞ and also r (b ) = ∞. Since y = y ,we y 1 y 2 have lim f (ω) =∞ and also lim f (ω) =∞ by the same arguments as in Sub- ω→b y ω→b y 1 2 case I.A. If b and b are not ordered, Lemma 21 applies to both of these points. If b b 1 2 1 2 (resp. b b ), then there is no ω ∈ such that ω = b (resp. ω = b ), f (ω) =∞ and 2 1 y 2 1 y b ω (resp. b ω). Hence Lemma 21 applies to ω = b (resp. ω = b ). Thus the set 2 1 0 2 0 1 of poles of all branches of x → Q(x, 0) (resp. y → Q(0, y)) is dense on the curvesI (b ) x 2 andI (b ) (resp.I (b ) andI (b )). We conclude the proof of Proposition 23 with the y 2 x 1 y 1 observation thatI (b ) =I (b ), whileI (b ) =I (b ) are intervals of the real line as x 2 x 1 y 2 y 1 claimed in Theorem 17(iii). Case II:y > 0, location of a , a , a , a .— We ﬁrst exclude Subcase II.D where x = 4 1 2 3 4 4 ∞ and Y(x ) =∞, and we locate the points a , a , a , a . In this case we have x = x . 4 1 2 3 4 Note that x , x , x , x are the roots of the equation 1 2 3 4 1/2 1/2 d(x) = b(x) − a(x)c(x) b(x) + a(x)c(x) = 0. 1/2 Hence for two of these roots b(x) =−2[a(x)c(x)] and then Y(x) ≥ 0 (see (2.5)), and 1/2 for two others b(x) = 2[a(x)c(x)] and then Y(x) ≤ 0. But Y(x ) and Y(x ) are on the 2 3 segment [y , y ]⊂]0, ∞[.Hence Y(x ) ≤ 0and Y(x ) ≤ 0. If in addition x =∞,then 2 3 1 4 4 Y(x ) equals 0 or ∞;notealsothatif Y(x ) =∞ then necessarily x =∞.But thecase 4 4 4 when Y(x ) =∞ and x =∞ is excluded from our consideration at this moment. It 4 4 follows that ∞∈[y , Y(x )[, and in fact ∞∈ ]y , Y(x )[, since the case y =∞ is excluded 4 4 4 4 4 from Case II. It follows from the above considerations that a ∈]ω + ω ,ω + ω [, see Fig- 1 x 2 y 2 4 4 ure 14. In particular, we have a − ω ∈]ω ,ω [ and a = ηa − ω =−(a − ω ) + 2ω . 1 2 x y 2 1 1 1 2 y 4 4 4 104 I. KURKOVA, K. RASCHEL This means that a − ω and a are symmetric w.r.t. ω . Furthermore a ∈]ω ,ω [, 1 2 2 y 2 y y 4 4 1 but since y > 0and Y(x ) ≤ 0, we have a ∈]ω ,ω [. We must put a = ξ a − ω = 4 1 2 y x 3 2 1 4 1 −a + 2ω , in such a way that the points a and a are symmetric w.r.t. ω .Fi- 2 x 2 3 x 1 1 nally, a = ξ(a − ω ) − ω =−(a − ω ) + 2ω = a + a − (a − ω ).Notethat 4 1 2 1 1 2 x 3 2 1 2 ω + ω − a = a − ω − ω > 0. Furthermore, a − a = ω and a + ω − a = ω ,so x 2 4 1 2 x 1 3 3 2 2 4 3 4 4 that a a and a a . 3 1 4 2 ◦ • ◦ • Case II:y > 0,Subcase II.A:x < 0.— In this case we have y = y ; y , y = ∞; 4 4 x , x = ∞; y , y = ∞. The points b , b are located as in Subcase I.A, see Figure 12. 1 2 Further, we can show as in Subcase I.A that r (ω) = ∞ on the segments {ω :ω = b , b ,ω ≤ω ≤ ω + ω }, and we deduce that r (ω) = ∞ on {ω :ω = b , b ,ω ≤ 1 2 y y 2 x 1 2 x 1 4 1 ω ≤ ω + ω } except for their ends b , b . Consequently, the set of poles of all branches x 2 1 2 of x → Q(x, 0) (resp. y → Q(0, y)) is dense on the curveI (b ) =I (b ) (resp.I (b ) = x 1 x 2 y 1 I (b )), as claimed in Proposition 23. y 2 Consider now r (ω) and r (ω) for ω with ω = 0. See Figure 14 for the location of x y the points a , a , a , a . 1 2 3 4 We ﬁrst prove that for (7.15) r (ω ) = ∞, ∀ω ∈{ω :ω = 0,ω ≤ ω ≤ ω + ω }\{a }. y 0 0 y 0 y 2 1 1 4 The proof consists in three steps. Step 1. — We prove that r (a ) = ∞ and r (a ) = ∞. y 3 y 4 If a ∈ , then necessarily r (a ) = ∞, since x , y = ∞. Otherwise |x |, |y |≥ 1, 3 y 3 and then a = ξ a − ω ∈ / . Since a <ω < a , ω ∈ and a , a ∈ / , it follows that 2 3 1 2 x 3 x 2 3 1 1 any ω ∈ O (a ) must be in ]a , a [,hence f (ω) = ∞, x(ω) = ∞ and y(ω) = ∞.Thusby 3 2 3 y (7.12), r (a ) = ∞. y 3 We now show that r (a ) = ∞. Suppose ﬁrst that a − ω ∈ . Then, since y 4 1 2 a − ω ∈ / ,wehave a − ω ∈ ,sothat r (a − ω ) = ∞ by Theorem 3.Thenby 1 2 y 1 2 x x 1 2 Equations (5.4)and (5.5), r (a ) = r (ξ(a − ω ) − ω ) = r (a ) = ∞. Since x , y = ∞,we x 4 x 1 2 1 x 1 also have r (a ) = ∞ by (5.3). Assume now that a − ω ∈ / . Since a − ω <ω < a , y 4 1 2 1 2 x 4 then a − ω ∈ / O (a ). Furthermore a ∈ / O (a ) as a + ω = a + ω and ω /ω is 1 2 4 2 4 4 3 2 2 3 2 irrational. Finally a − a < a − a = ω ,sothat a ∈ / O (a ). It follows that for any 4 3 1 3 3 3 4 ω ∈ O (a ),wehave f (ω) = ∞ and y(ω) = ∞.Hence r (a ) = ∞. 4 y y 4 Step 2. — We prove that r (a ) + f (a ) = ∞ and that r (a − ω ) + f (a − ω ) = ∞. y 2 y 2 y 1 2 y 1 2 By Equation (5.3) (7.16) r (a ) + f (a ) =−r (a ) + K(0, 0)Q(0, 0) + x(a )y(a ) y 2 y 2 x 2 2 2 + x(a ) y(ξ a ) − y(a ) 2 2 2 =−r (a ) + K(0, 0)Q(0, 0) + x(a )y(ξ a ). x 2 2 2 ON THE FUNCTIONS COUNTING WALKS IN THE QUARTER PLANE 105 Since r (a ) = ∞ and since x , y = ∞, it follows from Equation (5.3)that r (a ) = ∞. y 3 x 3 Then by (5.4)and (5.5), r (a ) = r (ξ a − ω ) = r (a ) = ∞. Since x(a ) = x = ∞ and x 2 x 3 1 x 3 2 y(ξ a ) = y = ∞,(7.16) is ﬁnite. By completely analogous arguments we obtain that r (a − ω ) + f (a − ω ) = ∞. y 1 2 y 1 2 Step 3. — Let us now show (7.15). If ω ∈ , then by the same arguments as in Subcase I.A ω ∈ and then r (ω ) = ∞ by Theorem 3. Otherwise, consider O (ω ) 0 y y 0 0 deﬁned in (7.11). Note that (7.17) ω/ ∈ O (ω ), ∀ω ∈]ω + ω − ω /2,ω + ω ], 0 x 2 3 y 2 4 4 since in this case ω − ω <ω. In particular, a ∈ / O (ω ). Furthermore, a ∈ O (ω ) 0 3 4 0 3 0 implies that ω = a + ω for some ≥ 1. But a + ω = a = ω and a + ω >ω + ω 0 3 3 3 3 1 0 3 3 y 2 for any ≥ 2. Hence a ∈ / O (ω ). Since a − (a − ω )<ω ,itisimpossiblethatboth a 3 0 2 1 2 3 2 and a − ω belong to O (ω ). If none of them belongs to O (ω ),then y(ω) = ∞ and 1 2 0 0 f (ω) = ∞ for any O (ω ),and then r (ω ) = ∞. Suppose, e.g., that a ∈ O (ω ).Then y 0 y 0 2 0 for some ≥ 1, ω = a + ω ,and by (7.12), r (ω ) = r (a ) + f (a ) + f (ω − 0 2 3 y 0 y 2 y 2 y 0 k=−1 kω ).But r (a ) + f (a ) = ∞ by the second step, and obviously f (ω − kω ) = ∞ 3 y 2 y 2 0 3 k=−1 by all facts said above, so that r (ω ) = ∞. The reasoning is the same if a − ω ∈ O (ω ). y 0 1 2 0 This concludes the proof of (7.15). Applying (5.4)and (5.5) exactly as in Subcase I.B, we now reach the conclusion that r (ω ) = ∞ for all ω = a = ηa with ω = 0and ω ≤ω ≤ ω as well. Next, y 0 0 2 1 0 y 0 y 4 1 exactly as in Subcase I.B, thanks to (5.3), we derive that r (ω) = ∞ for all ω such that x 0 ω = 0and ω ≤ω ≤ ω + ω except possibly for points where x(ω) =∞. But these x x 2 1 4 points are absent on this segment in this case. This concludes the proof of Proposition 22. For the same reason as in Subcase I.A, the fact that x = x gives lim {f (ω) + ω→a y f (ω + ω )}=∞ and lim {f (ω) + f (ω + ω )}=∞. Further, a a and a a .If y 3 ω→a y y 3 3 1 1 4 1 2 in addition a a , then due to the fact that a + ω = a we have a a a a . 3 4 3 3 1 3 1 1 4 1 2 Thereisnopoint ω ∈ \{a } such that a ω and f (ω) =∞. Lemma 21 applies to y 2 4 y ω = a .If a a ,thenalso a a a a and this lemma applies to ω = a .If 0 4 4 3 4 1 2 3 1 1 0 3 a a , Lemma 21 can be applied to both a and a . SinceI (a ) =I (a ) =I (a ) = 3 4 3 4 x 3 x 4 x 1 I (a ) =[x , x ] andI (a ) =I (a ) =I (a ) =I (a ) = R\]y , y [, the set of poles of x 2 1 4 y 3 y 4 y 1 y 2 4 1 x → Q(x, 0) (resp. y → Q(0, y)) is dense on the announced intervals and Proposition 23 is proved. ◦ • Case II:y > 0,Subcase II.B:x > 0 and exactly one of y , y is ∞.— Assume, e.g., 4 4 ◦ • • that y =∞ and y = ∞.Then x =∞; y = y = ∞; x = x =∞; y = ∞. It follows that b = a and b = ξ b − ω − ω = a , while a , a , a , a arepicturedaspreviously, in 1 1 2 1 1 2 4 1 2 3 4 Subcase II.A, see Figure 14. We ﬁrst derive (7.15). By the same reasoning as in Subcase II.A, we reach the conclusion that r (a ) = ∞. Let us note that in this case a − ω ∈ / ,as x(a − ω ), y(a − y 3 1 2 1 2 1 106 I. KURKOVA, K. RASCHEL ω ) =∞. Next, we derive as in Subcase II.A that r (a ) = ∞ and that r (a ) + f (a ) = ∞, 2 y 4 y 2 y 2 since x y = ∞. Finally, again by the same arguments as in Subcase II.A, we conclude that for any ω = a with ω = 0and ω ≤ω ≤ ω + ω , the orbit O (ω ) does 0 1 0 y 0 y 2 0 1 4 not contain a and a . The orbit can neither contain a − ω , since a − ω ∈ / and 3 4 1 2 1 2 a − ω <ω <ω ,where ω ∈ . Since r (a ) + f (a ) = ∞,thenasinSubcaseII.Awe 1 2 y 0 y y 2 y 2 1 1 have r (ω ) = ∞. y 0 It follows from (5.4)and (5.5)that r (ω ) = ∞ for any ω such that ω = 0and y 0 0 0 ω ≤ ω ≤ ω + ω ,exceptfor a and ηa − ω = a .By(5.3), r (ω ) = ∞ for any ω y 0 y 2 1 1 1 2 x 0 0 4 4 such that ω = 0and ω ≤ ω ≤ ω + ω , except for points ω where x(ω ) =∞ (this 0 x 0 x 2 0 0 1 4 is ω = a in this case). This ﬁnishes the proof of Proposition 22 in this case. 0 4 Since x = x , the same reasoning as in Subcase I.A gives lim {f (ω) + f (ω + ω→a y y ω )}=∞ and lim {f (ω) + f (ω + ω )}=∞. The rest of the proof of Proposition 23 3 ω→a y y 3 via the use of Lemma 21 with ω = a if a a or with ω = a if a a ,orwith 0 3 4 3 0 4 3 4 indifferent choice of a or a if a a , is the same as in Subcase II.A. 3 4 3 4 ◦ • Case II:y > 0,Subcase II.C:x > 0 and y , y = ∞,or x =∞ and Y(x ) = ∞.— In 4 4 4 4 ◦ • this case, we have y , y = ∞; x , x = ∞; x = x , y , y = ∞. The points a , a , a , a are pictured as in Subcases II.A and II.B, see Figure 14, 1 2 3 4 while b , b are pictured as in Subcase I.B (where b = b = ω ) or I.C, see Figure 13. 1 2 1 2 x +ω 4 2 They are such that b = a and b = a . In particular, b , b ∈]ω + ω − ω ,ω + ω [ and 1 1 2 4 1 2 x 2 3 x 2 4 4 are symmetric w.r.t. ω + ω ; b = b is in the middle of this interval if and only if x =∞. x 2 1 2 4 Hence for any ω with ω = 0and ω ≤ω ≤ ω + ω ,wehave ω − ω < b < b ,so 0 0 y 0 y 2 0 3 2 1 1 4 that b , b ∈ / O (ω ). Furthermore, by the same arguments as in Subcase I.C, b − ω ∈ / 1 2 0 1 2 O (ω ).Hence (7.15)provedinSubcaseII.Astays validinthis case andby(5.4)and (5.5), r (ω) = ∞ for all ω with ω = 0and ω ≤ω ≤ ω + ω ,exceptfor ω = a , a . y y y 2 1 2 4 4 By the identity (5.3), r (ω) = ∞ for all ω with ω = 0and ω ≤ω ≤ ω + ω ,except x x x 2 1 4 for points ω where x(ω) =∞,namely ω = b . This concludes the proof of Proposition 22 and proves in particular that r (ω ) = ∞ for any ω ∈{a , a , b , b }. y 0 0 3 4 1 2 Using x = x , we verify as in Subcase I.A that lim {f (ω) + f (ω + ω )}=∞ ω→a y y 3 ◦ • and lim {f (ω) + f (ω + ω )}=∞.If x > 0, since y = y ,weverifyasinSubcaseI.A ω→a y y 3 4 that lim f (ω) =∞ and lim f (ω) =∞.If x =∞,then b = b and we verify as ω→b y ω→b y 4 1 2 1 2 in Subcase I.B that lim f (ω) =∞.If a , a , b , b are ordered (e.g., a b a ω→b y 3 4 1 2 3 1 4 b , then immediately a a b a a b ), thereisa maximal point in the sense 2 3 1 1 1 4 1 2 2 of this order. If the maximal element is b for some ∈{1, 2}, then there is no ω ∈ with f (ω) =∞ such that b ω. If the maximal element is a (resp. a ), then there is no y 3 4 ω ∈ except for a (resp. a )with f (ω) =∞ such that a ω (resp. a ω). Lemma 21 y 1 2 y 3 4 applies with ω equal this maximal element since all assumptions (A), (B) and (C) are satisﬁed. If a , a , b , b are not all ordered, then it is enough to apply Lemma 21 to 3 4 1 2 the maximal element of any ordered subset. FinallyI (ω ) = R\]x , x [ andI (ω ) = x 0 1 4 y 0 R\]y , y [ for any ω ∈{a , a , b , b }, hence the set of poles of x → Q(x, 0) (resp. y → 4 1 0 3 4 1 2 Q(0, y)) is dense on the announced intervals. Proposition 23 is proved. ON THE FUNCTIONS COUNTING WALKS IN THE QUARTER PLANE 107 FIG. 15. — Location of a , a , a , a if y > 0, x =∞,Y(x ) =∞, i.e., Subcase II.D 1 2 3 4 4 4 4 Case II:y > 0,Subcase II.D:x =∞ and Y(x ) =∞.— In this case x =∞; y = 4 4 4 ∞; x , y = ∞. We have a = ω + ω ,and a = ηa − ω = ω + ω is symmetric to a − ω 1 x 2 2 1 1 x 3 1 2 4 4 w.r.t. ω . Further, since y > 0and Y(x ) ≤ 0, a ∈]ω ,ω [.Then a = ξ a − ω = y 4 1 2 y x 3 2 1 4 4 1 ω + ω − ω is symmetric to a w.r.t. ω . Finally, a = ω + ω = a , b = b = a , x 2 3 2 x 4 x 2 1 1 2 1 4 1 4 a + ω = a and a + ω = a + ω ,sothat a a a , see Figure 15. 3 3 1 1 3 2 2 3 1 1 1 2 We prove (7.15). For this purpose, we ﬁrst show that r (a ) = ∞.If a ∈ , x , y = y 3 3 ∞, then by Theorem 3, r (a ) = ∞ If a ∈ / ,consider O (ω ). Since a + 2ω = a + ω , y 3 3 0 3 3 2 2 a ∈ / O (ω ) by the irrationality of ω /ω . Obviously a − ω = ω ∈ / and then it is not 2 0 2 3 1 2 x in O (ω ).Hence r (a ) = ∞.Nextweprove that r (a ) + f (a ) exactly as in Subcase II.A 0 y 3 y 2 y 2 by using that x y = ∞. For any ω = a with ω = 0and ω ≤ ω ≤ ω + ω , the orbit O (ω ) 0 1 0 y 0 y 2 0 1 4 cannot contain a , since a + ω = a and a + 2ω >ω . It can neither contain 3 3 3 1 3 3 0 a , since ω − ω < a , nor obviously a − ω = ω . If it does not contain a ,then 1 0 3 1 1 2 x 2 by (7.12), r (ω ) = ∞. If it does, then exactly as in Subcase II.A, using r (a ) + y 0 y 2 f (a ) = ∞,weprove that r (ω ) = ∞ as well. This ﬁnishes the proof of (7.15). y 2 y 0 By Equations (5.4), (5.5)and (5.3), we derive as in Subcase II.A that r (ω ) = ∞ x 0 for all ω ∈{ω :ω = 0,ω ≤ ω ≤ ω + ω } except for ω where x(ω ) =∞,thatis 0 x 0 x 2 0 0 1 4 for a . This ﬁnishes the proof of Proposition 22 in this case. To prove Proposition 23, we would like to apply Lemma 21 with ω = a .Wehave 0 3 shown that r (a ) = ∞, a a = a = b = b a , so that there is no ω ∈ \{a , a } y 3 3 1 1 4 1 2 2 y 1 2 such that a ω and f (ω) =∞. It remains to verify assumption (B) of Lemma 21 for 3 y ω = a , that is that f (ω) + f (ω + ω ) + f (ω + 2ω ) converges to inﬁnity if ω → a .The 0 3 y y 3 y 3 3 last quantity is the sum of x(ω) y(ξω) − y(ω) + x( η ξω) y(ξ η ξω) − y( η ξω) + x( η ξ η ξω) y(ξ η ξ η ξω) − y( ηξ η ξω) , which equals (7.18) x( η ξ η ξω)y(ξ η ξ η ξω) − x(ω)y(ω) + x( η ξω) y(ξ η ξω) − y( η ξω) + x(ω)y(ξω) − x( η ξ η ξω)y(ξ ηξω) where we used (3.7). If ω → a , then the ﬁrst term in this sum converges to x y − x y = 0. Next, η ξω → a ,sothat x( η ξω) →∞. We can also compute the values of y(ξω) and 2 1/2 y(ξ η ξω) as (−b(x) ±[b(x) − 4a(x)c(x)] )/(2a(x)) with x = x( η ξω). Since for all of 108 I. KURKOVA, K. RASCHEL FIG. 16. — Location of a , a , a , a , b , b if y =∞, Case III 1 2 3 4 1 2 4 the 9 models composing Subcase II.D, deg a = deg b = 1and deg c = 2, then y(ξω) and 1/2 y(ξ η ξω) are of order O(|x( η ξω)| ), and their difference |y(ξω) − y(ξ η ξω)| is not 1/2 smaller than O(|x( η ξω)| ) as ω → a . Finally, x(ω), x( η ξ η ξω) → x = ∞ as ω → a . 3 3 Then as ω → a in the sum (7.18) the second term is of the order not smaller than 3/2 1/2 O(|x( η ξω)| ) while the ﬁrst vanishes and the third has the order O(|x( η ξω)| ). This proves the assumption (B) of Lemma 21 for ω = a . By this lemma the poles of x → 0 3 Q(x, 0) and y → Q(0, y) are dense on the intervals of the real line, as announced in the proposition. Case III:y =∞.— It remains here exactly one case to study, see Figure 17.It ◦ • is such that y =∞, x =∞ and X( y ) = ∞.Then x = x = ∞; y = y = ∞; y = 4 4 4 y = ∞. The points b = b = ω + ω are located as in Subcase I.B, a = a = ω + ω 1 2 x 2 1 2 y 2 4 4 and a = a = ω + ω − ω . In particular, a + ω /2 = b , b + ω /2 = a , see Figure 16. 3 4 y 2 3 3 3 1 1 3 1 We start by showing that r (a ) = ∞.If a ∈ , this is true thanks to (5.3)and y 3 3 since x , y = ∞.If a ∈ / , consider the orbit O (a ). It cannot contain a − ω since 3 3 1 2 a + ω = a , neither b ,nor b − ω = ω . It follows that r (a ) = ∞. 3 3 1 1 1 2 x y 3 Since x , y = ∞, it follows from Equations (5.3), (5.4)and (5.5)that r (a − ω ) = x 1 2 r (ξ(a − ω )) = r (ξ(a − ω ) − ω ) = r (a ) = ∞. Then, by (5.3), r (a − ω ) + f (a − x 1 2 x 1 2 1 x 3 y 1 2 y 1 ω ) =−r (a − ω ) + K(0, 0)Q(0, 0) + x(a − ω )y(ξ(a − ω )) = ∞. 2 x 1 2 1 2 1 2 Take any ω with ω = 0and ω ≤ ω <ω + ω .If ω ∈ , then by the same 0 0 y 0 y 2 0 1 4 arguments as in Subcase I.A, ω ∈ ,sothat r (ω ) = ∞.Otherwise,wenoticethat 0 y y 0 ω − ω < a , so that no point—and in particular b —of [ω + ω − ω ,ω + ω [ belongs 0 3 3 1 y 2 3 y 2 4 4 to O (ω ). Clearly b − ω = ω ∈ / O (ω ). Since either a − ω ∈ / O (ω ) or a − ω ∈ 0 1 2 x 0 1 2 0 1 2 O (ω ) but r (a − ω ) + f (a − ω ) = ∞, and by the same reasoning as in Subcase I.A, 0 y 1 2 y 1 2 we derive that r (ω ) = ∞. The rest of the proof of Proposition 22 in this case goes along y 0 thesamelinesasinCaseII. Now note that b is not ordered with a and a . Indeed, since b + ω /2 = a 1 1 3 1 3 1 and b − ω /2 = a , this would contradict the irrationality of ω /ω . Then there is no 1 3 3 2 3 ω ∈ such that b ω and f (ω) =∞.Wealsohave f (ω) = x(ω)[y(ξω) − y(ω)]= y 1 y y 2 1/2 x(ω)[b (x(ω)) − 4a(x(ω))c(x(ω))] /a(x(ω)).If ω → b ,then x(ω) →∞ and since deg a = 2and deg b = deg c = 1, we have f (ω) →∞. Lemma 21 applies with ω = b y 0 1 and proves Proposition 23. 7.4. Asymptotic of ω (ω)/ω (ω) as z → 0.— It remains to prove the following re- 2 3 sult announced at the beginning of Section 7. ON THE FUNCTIONS COUNTING WALKS IN THE QUARTER PLANE 109 Proposition 25. — For all 51 non-singular walks having an inﬁnite group, there exist a rational constant L > 0 and a constant L = 0 such that (7.19) ω /ω = L + L/ ln(z) + O 1/ ln(z) . 2 3 Proof.—In ordertoprove (7.19), we shall use expressions of the periods ω and ω different from that given in (3.1)and (3.2). To that purpose, deﬁne the complete and incomplete elliptic integrals of the ﬁrst kind by, respectively, dt (7.20) K(k) = , 2 1/2 2 2 1/2 [1 − t ] [1 − k t ] dt (7.21) F(w, k) = . 2 1/2 2 2 1/2 [1 − t ] [1 − k t ] Then the new expressions of ω and ω are 2 3 (7.22) ω = M,ω = M , 2 2 3 3 where (x − x )(x − x ) 4 1 3 2 (7.23) = K , (x − x )(x − x ) 4 2 3 1 (x − x )(x − X( y )) (x − x )(x − x ) 4 2 1 1 4 1 3 2 (7.24) = F , , (x − x )(x − X( y )) (x − x )(x − x ) 4 1 2 1 4 2 3 1 and where (below, 1 = 1if (i, j) ∈S , otherwise 0) (i,j) 2 1 if x = ∞, ⎪ 4 (1 − 41 1 )(x x − x x − x x + x x ) (1,0) (1,1) (1,−1) 3 4 2 3 1 4 1 2 (7.25) M = ⎪ if x =∞. (2z1 + 4z [1 1 + 1 1 ])(x − x ) (1,0) (1,1) (0,−1) (1,−1) (0,1) 3 1 The expressions of ω and ω written in (7.22), (7.23), (7.24)and (7.25) are obtained from 2 3 (3.1)and (3.2) by making simple changes of variables. We are now in position to analyze the behavior of ω /ω (or equivalently, thanks to 2 3 (7.22), that of / ) in the neighborhood of z = 0. First, with (2.2)and [19,Proposition 2 3 6.1.8], we obtain that as z → 0, x , x → 0and x , x →∞. For this reason, 1 2 3 4 (x − x )(x − x ) 4 1 3 2 (7.26) k = → 1. (x − x )(x − x ) 4 2 3 1 110 I. KURKOVA, K. RASCHEL The behavior of X( y ) as z → 0 is not so simple as that of the branch points x (indeed, as z → 0, X( y ) can converge to 0, to ∞ or to some non-zero constant), but we can show that for all 51 models, (x − x )(x − X( y )) 4 2 1 1 (7.27) w = → 1. (x − x )(x − X( y )) 4 1 2 1 Due to (7.26)and (7.27), in order to determine the behavior of / near z = 0it 2 3 sufﬁces to know (i) the expansion of K(k) as k → 1; (ii) the expansion of F(w, k) as k → 1and w → 1. Point (i) is classical, and is known as Abel’s identity (it can be found, e.g., in [8]): there exist two functions A and B, holomorphic at z = 0, such that K(k) = A(k) + ln(1 − k)B(k). Both A and B can be computed in an explicit way, see [8], and from all this we can deduce an expansion of K(k) as k → 1 up to any level of precision. For our purpose, it will be enough to use the following: A(k) = (3/2) ln(2) + (k − 1)/4 1 − 3ln(2) + O(k − 1) , B(k) =−1/2 + (k − 1)/4 + O(k − 1) . As for Point (ii), we proceed as follows. We have F(w, k) = K(k) − F(w, k),with dt F(w, k) = . 2 1/2 2 2 1/2 [1 − t ] [1 − k t ] 1/2 1/2 Then, introduce the expansion 1/([1 + t] [1 + kt] ) = μ (k)(1 − t) ,sothat =0 dt (7.28) F(w, k) = μ (k) . 1/2− 1/2 [1 − t] [1 − kt] =0 In Equation (7.28), all μ (k) as well as all integrals can be computed. As an example (that we shall use), we have dt μ (k) 1/2 1/2 [1 − t] [1 − kt] 1/2 1/2 2 = ln (1 − k)/k − ln − 1 − (1 + k)w + kw 1/2 [2k(1 + k)] 1/2 + (k + 1)/2 − kw /k . ON THE FUNCTIONS COUNTING WALKS IN THE QUARTER PLANE 111 FIG. 17. — Different cases considered in the proof of Theorem 17—they correspond to the 51 non-singular walks with inﬁnite group, see [2] Moreover, it should be noticed that as k → 1and w → 1, the speed of convergence to zero of the integrals in (7.28) increases with . This way, we can write an expansion of F(w, k)—and thus of F(w, k)—up to any level of precision. Unfortunately, the end of the proof cannot be done simultaneously for all 51 models, but should be done model by model. For the sake of shortness, we choose to present the details only for one model, namely for the model withS = {(−1, 0), (−1, 1), (0, 1), (1, −1)} (which belongs to Subcase II.D of Figure 17). For this 112 I. KURKOVA, K. RASCHEL model, we easily obtain from (2.2)that 2 3 4 x = z − 2z + 3z + O z , 2 3 4 x = z + 2z + 5z + O z , 2 3 4 x = 1/ 4z − 1 − 2z − 8z + O z , x =∞, X( y ) = 0. Then, with (7.26)and (7.27), we reach the conclusion that 4 5 6 (7.29) k = 1 − 8z − 4z + O z , 2 3 4 5 6 (7.30) w = 1 − 2z + z − (7/4)z − (65/8)z + (613/64)z + O z . Then, using Points (i) and (ii) above, we obtain 2 3 =−2ln(z) − (1/4)z + (1/16)z + O z ln(z) , 2 3 =−(1/2) ln(z) − (1/2) ln(2) + (1/4)z + (57/16)z + O z ln(z) , so that (7.31) / = 4 − 4ln(2)/ ln(z) + O 1/ ln(z) . 2 3 Thelatterproves(7.19), and thus Proposition 25,with L = 4and L =−4ln(2). Making expansions of higher order of k and w in (7.29)and (7.30), we could obtain more terms in the expansion (7.31)of / . A contrario, we could also be interested 2 3 in obtaining the ﬁrst term only (the constant term L) in (7.31). (Indeed, we saw in the proof of Proposition 14 that it was sufﬁcient for our purpose, i.e., for proving that in the inﬁnite group case, the ratio ω /ω is not constant in z.) To that aim, instead of 2 3 p p (7.29)and (7.30), we just need two-terms expansions of k and w,say k = 1 + αz + o(z ) q q and w = 1 + β z + o(z ),with α, β = 0. Then with (i) and (ii) we deduce that = −(p/2) ln(z) + o(ln(z)) and =−(q/2) ln(z) + o(ln(z)), in such a way that L = p/q, which obviously is (non-zero and) rational. Acknowledgements K. Raschel’s work was partially supported by CRC 701, Spectral Structures and Topological Methods in Mathematics at the University of Bielefeld. We are grateful to E. Lesigne: his mathematical knowledge and ideas—that he generously shared with us— have been very helpful in the elaboration of this paper. We also warmly thank R. Kriko- rian and J.-P. Thouvenot for encouraging discussions. Finally, we thank two anonymous referees for their careful reading and their remarks. ON THE FUNCTIONS COUNTING WALKS IN THE QUARTER PLANE 113 REFERENCES 1. A. BOSTAN and M. KAUERS, The complete generating function for Gessel walks is algebraic, Proc. Am. Math. Soc., 432 (2010), 3063–3078. 2. M. BOUSQUET-MÉLOU and M. MISHNA, Walks with small steps in the quarter plane, Contemp. Math., 520 (2010), 1–40. 3. M. BOUSQUET-MÉLOU and M. PETKOVSEK, Walks conﬁned in a quadrant are not always D-ﬁnite, Theor. Comput. Sci., 307 (2003), 257–276. 4. G. FAYOLLE and R. IASNOGORODSKI, Two coupled processors: the reduction to a Riemann-Hilbert problem, Z. Wahrscheinlichkeitstheor. Verw. Geb., 47 (1979), 325–351. 5. G. FAYOLLE and K. RASCHEL, On the holonomy or algebraicity of generating functions counting lattice walks in the quarter-plane, Markov Process. Relat. Fields, 16 (2010), 485–496. 6. G. FAYOLLE,R.IASNOGORODSKI,and V. MALYSHEV, Random Walks in the Quarter-Plane, Springer, Berlin, 1999. 7. P. FLAJOLET and R. SEDGEWICK, Analytic Combinatorics, Cambridge University Press, Cambridge, 2009. 8. J. GERRETSEN and G. SANSONE, Lectures on the Theory of Functions of a Complex Variable II: Geometric Theory,Wolters- Noordhoff Publishing, Groningen, 1969. 9. I. 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MISHNA and A. RECHNITZER, Two non-holonomic lattice walks in the quarter plane, Theor. Comput. Sci., 410 (2009), 3616–3630. 18. K. RASCHEL, Counting walks in a quadrant: a uniﬁed approach via boundary value problems, J. Eur. Math. Soc., 14 (2012), 749–777. 19. R. STANLEY, Enumerative Combinatorics, vol. 2, Cambridge University Press, Cambridge, 1999. 20. G. WATSON and E. WHITTAKER, A Course of Modern Analysis, Cambridge University Press, Cambridge, 1962. I. K. Laboratoire de Probabilités et Modèles Aléatoires, Université Pierre et Marie Curie, 4 Place Jussieu, 75252 Paris Cedex 05, France irina.kourkova@upmc.fr 114 I. KURKOVA, K. RASCHEL K. R. CNRS and Université de Tours, Faculté des Sciences et Techniques, Parc de Grandmont, 37200 Tours, France kilian.raschel@lmpt.univ-tours.fr Manuscrit reçu le 4 octobre 2011 Manuscrit accepté le 11 octobre 2012 publié en ligne le 25 octobre 2012. http://www.deepdyve.com/assets/images/DeepDyve-Logo-lg.png Publications mathématiques de l'IHÉS Springer Journals http://www.deepdyve.com/lp/springer-journals/on-the-functions-counting-walks-with-small-steps-in-the-quarter-plane-MPTf0P2Qti

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Publisher: Springer Journals
Copyright: Copyright © 2012 by IHES and Springer-Verlag Berlin Heidelberg
Subject: Mathematics; Analysis; Mathematics, general; Number Theory; Geometry; Algebra
ISSN: 0073-8301
eISSN: 1618-1913
DOI: 10.1007/s10240-012-0045-7
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Abstract

ON THE FUNCTIONS COUNTING WALKS WITH SMALL STEPS IN THE QUARTER PLANE by IRINA KURKOVA and KILIAN RASCHEL ABSTRACT Models of spatially homogeneous walks in the quarter plane Z with steps taken from a subsetS of the set of jumps to the eight nearest neighbors are considered. The generating function (x, y, z) → Q(x, y; z) of the numbers q(i, j; n) of such walks starting at the origin and ending at (i, j) ∈ Z after n steps is studied. For all non-singular models of walks, the functions x → Q(x, 0; z) and y → Q(0, y; z) are continued as multi-valued functions on C having inﬁnitely many meromorphic branches, of which the set of poles is identiﬁed. The nature of these functions is derived from this result: namely, for all the 51 walks which admit a certain inﬁnite group of birational transformations of C ,the interval ]0, 1/|S|[ of variation of z splits into two dense subsets such that the functions x → Q(x, 0; z) and y → Q(0, y; z) are shown to be holonomic for any z from the one of them and non-holonomic for any z from the other. This entails the non-holonomy of (x, y, z) → Q(x, y; z), and therefore proves a conjecture of Bousquet-Mélou and Mishna in Contemp. Math. 520:1–40 (2010). 1. Introduction and main results In the ﬁeld of enumerative combinatorics, counting walks on the lattice Z is among the most classical topics. While counting problems have been largely resolved for unrestricted walks on Z and for walks staying in a half plane [3], walks conﬁned to the quarter plane Z still pose considerable challenges. In recent years, much progress has been made for walks in the quarter plane with small steps, which means that the set S of possible steps is included in {−1, 0, 1} \{(0, 0)}; for examples, see Figures 1 and 10. In [2], Bousquet-Mélou and Mishna constructed a thorough classiﬁcation of these 2 walks. After eliminating trivial cases and exploiting equivalences, they showed that 79 in- herently different walks remain to be studied. Let q(i, j; n) denote the number of paths in Z having length n, starting from (0, 0) and ending at (i, j). Deﬁne the counting function (CF) as i j n Q(x, y; z) = q(i, j; n)x y z . i,j,n≥0 There are then two key challenges: (i) Finding an explicit expression for Q(x, y; z); (ii) Determining the nature of Q(x, y; z): is it holonomic (i.e., see [7, Appendix B.4], is the vector space over C(x, y, z)—the ﬁeld of rational functions in the three variables x, y, z—spanned by the set of all derivatives of Q(x, y; z) ﬁnite dimensional)? And in that event, is it algebraic, or even rational? The common approach to treat these problems is to start from a functional equation for the CF, which for the walks with small steps takes the form (see [2]) DOI 10.1007/s10240-012-0045-7 70 I. KURKOVA, K. RASCHEL FIG. 1. — Example of model (with an inﬁnite group) considered here—on the boundary, the jumps are the natural ones: those that would take the walk out Z are discarded (1.1) K(x, y; z)Q(x, y; z) = K(x, 0; z)Q(x, 0; z) + K(0, y; z)Q(0, y; z) − K(0, 0; z)Q(0, 0; z) − xy, where i j (1.2) K(x, y; z) = xyz x y − 1/z (i,j)∈S is called the kernel of the walk. This equation determines Q(x, y; z) through the boundary functions Q(x, 0; z),Q(0, y; z) and Q(0, 0; z). Known results regarding both problems (i) and (ii) highlight the notion of the group of the walk, introduced by Malyshev [13–15]. This is the group (1.3) ξ, η of birational transformations of C(x, y), generated by i j 1 1 (i,−1)∈S (−1,j)∈S (1.4) ξ(x, y) = x, ,η(x, y) = , y . i j y x x y (i,+1)∈S (+1,j)∈S i j 2 2 Each element of ξ, η leaves invariant the jump function x y . Further, ξ = η = (i,j)∈S Id, and ξ, η is a dihedral group of order even and larger than or equal to four. It turns out that 23 of the 79 walks have a ﬁnite group, while the 56 others admit an inﬁnite group, see [2]. For 22 of the 23 models with ﬁnite group, CFs Q(x, 0; z),Q(0, y; z) and Q(0, 0; z)—and hence Q(x, y; z) by (1.1)—have been computed in [2]bymeans of cer- tain (half-)orbit sums of the functional equation (1.1). The 23rd model with ﬁnite group is known as Gessel’s walk (see Figure 10); see [9, 11] and references therein for literature on this quite interesting model. For it, the CFs have been expressed by radicals in [1] thanks to a guessing-proving method using computer calculations; they were also found in [12] by solving some boundary value problems. For the 2 walks with inﬁnite group on the left in Figure 2, they have been obtained in [17], by exploiting a particular property shared by the 5 models of Figure 2 commonly known as singular walks. Finally, in [18], the problem (i) was resolved for all 54 remaining walks—and in fact for all the 79 models. For ON THE FUNCTIONS COUNTING WALKS IN THE QUARTER PLANE 71 FIG. 2. — The 5 singular walks in the classiﬁcation of [2] the 74 non-singular walks, this was done via a uniﬁed approach: explicit integral repre- sentations were obtained for CFs Q(x, 0; z),Q(0, y; z) and Q(0, 0; z) in certain domains, by solving boundary value problems of Riemann-Carleman type. In this article we go further, since both functions x → Q(x, 0; z) and y → Q(0, y; z) are computed on the whole of C as multi-valued functions with inﬁnitely many meromor- phic branches, that are made explicit for all z ∈]0, 1/|S|[. This result gives not only the most complete continuation of these CFs on their complex planes along all paths, but also permits to establish the nature of these functions, i.e., to solve Problem (ii). Problem (ii) is actually resolved for only 28 of the 79 walks. All 23 ﬁnite group models admit a holonomic CF. Indeed, the nature of Q(x, y; z) was determined in [2] for 22 of these walks: 19 walks turn out to have a holonomic but non-algebraic CF, while for 3 walks Q(x, y; z) is algebraic. As for the 23rd—again, Gessel’s model—, the CF is algebraic [1]. Alternative proofs for the nature of the (bivariate) CF for these 23 walks were given in [5]. For the remaining 56 walks with an inﬁnite group, not much is known: in [17] it was shown that for 2 singular walks (namely, the 2 ones on the left in Figure 2), the function z → Q(1, 1; z) has inﬁnitely many poles and, as a consequence [7, Appendix B.4], is non-holonomic. Accordingly [7, Appendix B.4], the trivariate function Q(x, y; z) is non-holonomic as well. It is reasonable to expect that the same approach would lead to the non-holonomy of all 5 singular walks, see [16, 17]. As for the 51 non- singular walks with inﬁnite group (all of them are pictured on Figure 17), Bousquet-Mélou and Mishna [2] conjectured that they also have a non-holonomic CF. In this article we prove the following theorem. Theorem 1. — For any of the 51 non-singular walks with inﬁnite group (1.3), the set ]0, 1/|S|[ splits into subsetsH and ]0, 1/|S|[\H that are both dense in ]0, 1/|S|[ and such that: (i) x → Q(x, 0; z) and y → Q(0, y; z) are holonomic for any z ∈H; (ii) x → Q(x, 0; z) and y → Q(0, y; z) are non-holonomic for any z ∈]0, 1/|S|[\H. Theorem 1(ii) immediately entails Bousquet-Mélou and Mishna’s conjecture: the trivariate function (x, y, z) → Q(x, y; z) is non-holonomic since the holonomy is sta- ble by specialization of a variable [7, Appendix B.4]. Further, Theorem 1(i) goes beyond it: it suggests that Q(x, y; z), although being non-holonomic, still stays accessible for fur- ther analysis when z ∈H, namely by the use of methods developed in [6,Chapter 4], 72 I. KURKOVA, K. RASCHEL see Remark 16. This important setH will be characterized in two different ways, see Corollary 15 and Remark 16 below. The proof of Theorem 1 we shall do here is based on the above-mentioned con- struction of the CF x → Q(x, 0; z) (resp. y → Q(0, y; z)) as a multi-valued function, that must now be slightly more detailed. First, we prove in this article that for any z ∈]0, 1/|S|[, the integral expression of x → Q(x, 0; z) given in [18]inacertaindo- main of C admits a direct holomorphic continuation on C \[x (z), x (z)]. Points x (z), x (z) 3 4 3 4 are among four branch points x (z), x (z), x (z), x (z) of the two-valued algebraic func- 1 2 3 4 tion x → Y(x; z) deﬁned via the kernel (1.2) by the equation K(x, Y(x; z); z) = 0. These branch points are roots of the discriminant (2.2) of the latter equation, which is of the second order. We refer to Section 2 for the numbering and for some properties of these branch points. We prove next that function x → Q(x, 0; z) does not admit a direct mero- morphic continuation on any open domain containing the segment [x (z), x (z)],but 3 4 admits a meromorphic continuation along any path going once through [x (z), x (z)]. This way, 3 4 we obtain a second (and different) branch of the function, which admits a direct mero- morphic continuation on the whole cut plane C \ ([x (z), x (z)]∪[x (z), x (z)]).Next, if 1 2 3 4 the function x → Q(x, 0; z) is continued along a path in C \[x (z), x (z)] crossing once 1 2 again [x (z), x (z)], we come across its ﬁrst branch. But its continuation along a path in 3 4 C \[x (z), x (z)] crossing once [x (z), x (z)] leads to a third branch of this CF, which is 3 4 1 2 meromorphic on C \ ([x (z), x (z)]∪[x (z), x (z)]). Making loops through [x (z), x (z)] 1 2 3 4 3 4 and [x (z), x (z)], successively, we construct x → Q(x, 0; z) as a multi-valued meromor- 1 2 phic function on C with branch points x (z), x (z), x (z), x (z), and with (generically) 1 2 3 4 inﬁnitely many branches. The analogous construction is valid for y → Q(0, y; z). In order to prove Theorem 1(ii), we then show that for any of the 51 non-singular walks with inﬁnite group (1.3), for any z ∈]0, 1/|S|[\H, the set formed by the poles of all branches of x → Q(x, 0; z) (resp. y → Q(0, y; z)) is inﬁnite—and even dense in certain curves, to be speciﬁed in Section 7 (see Figure 11 for their pictures). This is not compatible with holonomy. Indeed, all branches of a holonomic one-dimensional function must verify the same linear differential equation with polynomial coefﬁcients. In particular, the poles of all branches are among the zeros of these polynomials, and hence they must be in a ﬁnite number. The rest of our paper is organized as follows. In Section 2 we construct the Rie- mann surface T of genus 1 of the two-valued algebraic functions X( y; z) and Y(x; z) deﬁned by K X( y; z), y; z = 0, K x, Y(x; z); z = 0. In Section 3 we introduce and study the universal covering of T. It can be viewed as the complex plane C split into inﬁnitely many parallelograms with edges ω (z) ∈ iR and ω (z) ∈ R that are uniformization periods. These periods as well as a new important period ω (z) are made explicit in (3.1)and (3.2). In Section 4 we lift CFs Q(x, 0; z) and 3 ON THE FUNCTIONS COUNTING WALKS IN THE QUARTER PLANE 73 Q(0, y; z) to some domain of T, and then to a domain on its universal covering. In Sec- tion 5, using a proper lifting of the automorphisms ξ and η deﬁned in (1.4)aswellasthe independence of K(x, 0; z)Q(x, 0; z) and K(0, y; z)Q(0, y; z) w.r.t. y and x, respectively, we continue these functions meromorphically on the whole of the universal covering. All this procedure has been ﬁrst carried out by Malyshev in the seventies [13–15], at that time to study the stationary probability generating functions for random walks with small steps in the quarter plane Z . It is presented in [6, Chapter 3] for the case of ergodic random walks in Z , and applies directly for our Q(x, 0; 1/|S|) and Q(0, y; 1/|S|) if the drift vector ( i, j) has not two positive coordinates; see also [4]. In Sec- (i,j)∈S (i,j)∈S tions 3, 4 and 5 we carry out this procedure for all z ∈]0, 1/|S|[ and all non-singular walks, independently of the drift. Then, going back from the universal covering to the complex plane allows us in Subsection 5.2 to continue x → Q(x, 0; z) and y → Q(0, y; z) as multi-valued meromorphic functions with inﬁnitely many branches. For given z ∈]0, 1/|S|[, the rationality or irrationality of the ratio ω (z)/ω (z) of 2 3 the uniformization periods is crucial for the nature of x → Q(x, 0; z) and y → Q(0, y; z). Namely, Theorem 7 of Subsection 5.3 proves that if ω (z)/ω (z) is rational, these func- 2 3 tions are holonomic. For 23 models of walks with ﬁnite group ξ, η,the ratio ω (z)/ω (z) turns out 2 3 to be rational and independent of z, see Lemma 8 below, that implies immediately the holonomy of the generating functions. In Section 6 we gather further results of our ap- proach for the models with ﬁnite group concerning the set of branches of the generating functions and their nature. In particular, we recover most of the results of [1, 2, 5, 17]. Section 7 is devoted to 51 models with inﬁnite group ξ, η.For allofthem, the setsH ={z ∈]0, 1/|S|[: ω (z)/ω (z) is rational} and ]0, 1/|S|[\H ={z ∈]0, 1/|S|[: 2 3 ω (z)/ω (z) is irrational} are proved to be dense in ]0, 1/|S|[, see Proposition 14. These 2 3 sets can be also characterized as those where the group ξ, η restricted to the curve {(x, y) : K(x, y; z) = 0} is ﬁnite and inﬁnite, respectively, see Remark 6.ByTheorem 7 mentioned above, x → Q(x, 0; z) and y → Q(0, y; z) are holonomic for any z ∈H,that proves Theorem 1(i). In Subsections 7.1, 7.2 and 7.3, we analyze in detail the branches of x → Q(x, 0; z) and y → Q(0, y; z) for any z ∈]0, 1/|S|[\H and prove the following facts (see Theorem 17): (i) The only singularities of the ﬁrst (main) branches of x → Q(x, 0; z) and y → Q(0, y; z) are two branch points x (z), x (z) and y (z), y (z), respectively; 3 4 3 4 (ii) All (other) branches have only a ﬁnite number of poles; (iii) The set of poles of all these branches is inﬁnite for each of these functions, and is dense on certain curves; these curves are speciﬁed in Section 7, and in particular are pictured on Figure 11 for all 51 walks given on Figure 17; (iv) Poles of branches out of these curves may be only at zeros of x → K(x, 0; z) or y → K(0, y; z), respectively. It follows from (iii) that x → Q(x, 0; z) and y → Q(0, y; z) are non-holonomic for any z ∈]0, 1/|S|[\H. 74 I. KURKOVA, K. RASCHEL 2. Riemann surface T In the sequel we suppose that z ∈]0, 1/|S|[, and we drop the dependence of the different quantities w.r.t. z. 2.1. Kernel K(x, y).— The kernel K(x, y) deﬁned in (1.2) can be written as i j 2 2 (2.1) xyz x y − 1/z = a( y)x + b( y)x + c( y) = a(x)y + b(x)y + c(x), (i,j)∈S where j j j a( y) = zy y , b( y) =−y + zy y , c( y) = zy y , (+1,j)∈S (0,j)∈S (−1,j)∈S i i i a(x) = zx x , b(x) =−x + zx x , c(x) = zx x . (i,+1)∈S (i,0)∈S (i,−1)∈S With these notations we deﬁne 2 2 (2.2) d( y) = b( y) − 4 a( y) c( y), d(x) = b(x) − 4a(x)c(x). If thewalkisnon-singular, then forany z ∈]0, 1/|S|[, the polynomial d (resp. d)has three or four roots, that we call y (resp. x ). They are such that |y | < y < 1 < y < |y | (resp. 1 2 3 4 |x | < x < 1 < x < |x |), with y =∞ (resp. x =∞)if d (resp. d ) has order three: the 1 2 3 4 4 4 arguments given in [6, Part 2.3] for the case z = 1/|S| indeed also apply for other values of z. Now we notice that the kernel (1.2) vanishes if and only if [b( y) + 2 a( y)x] = d( y) or [b(x) + 2a(x)y] = d(x). Consequently [10], the algebraic functions X( y) and Y(x) deﬁned by i j i j (2.3) X( y) y − 1/z = 0, x Y(x) − 1/z = 0 (i,j)∈S (i,j)∈S have two branches, meromorphic on the cut planes C \ ([y , y ]∪[y , y ]) and C \ 1 2 3 4 ([x , x ]∪[x , x ]), respectively—note that if y < 0, [y , y ] stands for [y , ∞[ ∪ {∞} ∪ 1 2 3 4 4 3 4 3 ]−∞, y ]; the same holds for [x , x ]. 4 3 4 We ﬁx the notations of the two branches of the algebraic functions X( y) and Y(x) by setting 1/2 1/2 −b( y) + d( y) −b( y) − d( y) (2.4) X ( y) = , X ( y) = , 0 1 a( y) 2 a( y) as well as 1/2 1/2 −b(x) + d(x) −b(x) − d(x) (2.5) Y (x) = , Y (x) = . 0 1 2a(x) 2a(x) ON THE FUNCTIONS COUNTING WALKS IN THE QUARTER PLANE 75 FIG. 3. — Construction of the Riemann surface The following straightforward result holds. Lemma 2. —For all y ∈ C,wehave |X ( y)|≤|X ( y)|. Likewise, for all x ∈ C, we have 0 1 |Y (x)|≤|Y (x)|. 0 1 Proof. — The arguments (via the maximum modulus principle [10]) given in [6, Part 5.3] for z = 1/|S| also work for z ∈]0, 1/|S|[. 2.2. Riemann surface T.— We now construct the Riemann surface T of the alge- braic function Y(x) introduced in (2.3). For this purpose we take two Riemann spheres 1 2 C ∪{∞},say S and S , cut along the segments [x , x ] and [x , x ], and we glue them 1 2 3 4 x x together along the borders of these cuts, joining the lower border of the segment [x , x ] 1 2 1 2 (resp. [x , x ])on S to the upper border of the same segment on S and vice versa, see 3 4 x x Figure 3. The resulting surface T is homeomorphic to a torus (i.e., a compact Riemann surface of genus 1) and is projected on the Riemann sphere S by a canonical covering map h : T → S. In a standard way, we can lift the function Y(x) to T, by setting Y(s) = Y (h (s)) if s ∈ S ⊂ T, ∈{1, 2}.Thus, Y(s) is single-valued and continuous on T. Furthermore, K(h (s), Y(s)) = 0for any s ∈ T. For this reason, we call T the Riemann surface of Y(x). In a similar fashion, one constructs the Riemann surface of the function X( y),by 1 2 gluing together two copies S and S of the sphere S along the segments [y , y ] and 1 2 y y [y , y ]. It is again homeomorphic to a torus. 3 4 76 I. KURKOVA, K. RASCHEL FIG. 4. — Location of the branch points and of the cycles and on the Riemann surface T 0 1 Since the Riemann surfaces of X( y) and Y(x) are equivalent, we can work on a single Riemann surface T, but with two different covering maps h , h : T → S. Then, x y for s ∈ T, we set x(s) = h (s) and y(s) = h (s), and we will often represent a point s ∈ T x y by the pair of its coordinates (x(s), y(s)). These coordinates are of course not independent, because the equation K(x(s), y(s)) = 0is valid for any s ∈ T. 2.3. Real points of T.— Let us identify the set of real points of T,thatare the points s ∈ T where x(s) and y(s) are both real or equal to inﬁnity. Note that for y real, X( y) is real if y ∈[y , y ] or y ∈[y , y ], and complex if y ∈]y , y [ or y ∈]y , y [, see (2.2). 4 1 2 3 1 2 3 4 Likewise, for real values of x,Y(x) is real if x ∈[x , x ] or x ∈[x , x ], and complex if 4 1 2 3 x ∈]x , x [ or x ∈]x , x [. The set therefore consists of two non-intersecting closed 1 2 3 4 analytic curves and , equal to (see Figure 4) 0 1 = s ∈ T : x(s) ∈[x , x ] = s ∈ T : y(s) ∈[y , y ] 0 2 3 2 3 and = s ∈ T : x(s) ∈[x , x ] = s ∈ T : y(s) ∈[y , y ] , 1 4 1 4 1 and homologically equivalent to a basic cycle on T—note, however, that the equivalence −1 class containing and is disjoint from that containing the cycle h ({x ∈ C :|x|= 1}). 0 1 2.4. Galois automorphisms ξ, η.— We continue Section 2 by introducing two Ga- lois automorphisms. Deﬁne ﬁrst, for ∈{1, 2}, the incised spheres S = S \ [x , x ]∪[x , x ] , S = S \ [y , y ]∪[y , y ] . 1 2 3 4 1 2 3 4 x x y y For any s ∈ T such that x(s) is not equal to a branch point x , there is a unique s = s ∈ T 1 2 such that x(s) = x(s ). Furthermore, if s ∈ S then s ∈ S and vice versa. On the other x x hand, whenever x(s) is one of the branch points x , s = s . Also, since K(x(s), y(s)) = 0, y(s) and y(s ) give the two values of function Y(x) at x = x(s) = x(s ). By Vieta’s theorem and (2.1), y(s)y(s ) = c(x(s))/a(x(s)). Similarly, for any s ∈ T such that y(s) is different from the branch points y ,there 1 2 exists a unique s = s ∈ T such that y(s) = y(s ).If s ∈ S then s ∈ S and vice versa. y y ON THE FUNCTIONS COUNTING WALKS IN THE QUARTER PLANE 77 On the other hand, if y(s) is one of the branch points y ,wehave s = s . Moreover, since K(x(s), y(s)) = 0, x(s) and x(s ) the two values of function X( y) at y(s) = y(s ). Again, by Vieta’s theorem and (2.1), x(s)x(s ) = c( y(s))/ a( y(s)). Deﬁnenow themappings ξ : T → T and η : T → T by ξ s = s if x(s) = x(s ), (2.6) ηs = s if y(s) = y(s ). 2 2 Following [13–15], we call them Galois automorphisms of T.Then ξ = η = Id, and c(x(s)) 1 c( y(s)) 1 (2.7) y(ξ s) = , x(ηs) = . a(x(s)) y(s) a( y(s)) x(s) Any s ∈ T such that x(s) = x (resp. y(s) = y ) is a ﬁxed point for ξ (resp. η). To illustrate and to get some more intuition, it is helpful to draw on Figure 4 the straight line through the pair of points of where x(s) = x and x (resp. y(s) = y and y ); then points s and 0 2 3 2 3 ξ s (resp. s and ηs) can be drawn symmetric about this straight line. 2.5. The Riemann surface T viewed as a parallelogram whose opposed edges are identiﬁed. — Like any compact Riemann surface of genus 1, T is isomorphic to a certain quotient space (2.8) C/(ω Z + ω Z), 1 2 where ω ,ω are complex numbers linearly independent on R, see [10]. The set (2.8) 1 2 can obviously be thought as the (fundamental) parallelogram ω [0, 1]+ ω [0, 1] whose 1 2 opposed edges are identiﬁed. Up to a unimodular transform, ω ,ω are unique, see [10]. 1 2 In our case, suitable ω ,ω will be found in (3.1). 1 2 If we cut the torus on Figure 4 along [x , x ] and , it becomes the parallelogram 1 2 0 on the left in Figure 5. On the right in the same ﬁgure, this parallelogram is translated to the complex plane, and all corresponding important points are expressed in terms of the complex numbers ω ,ω (see above) and of ω (to be deﬁned below, in (3.2)). 1 2 3 3. Universal covering 3.1. An informal construction of the universal covering. — The Riemann surface T can be considered as a parallelogram whose opposite edges are identiﬁed, see (2.8)and Figure 5. The universal covering of T can then be viewed as the union of inﬁnitely many such parallelograms glued together, as in Figure 6. 78 I. KURKOVA, K. RASCHEL FIG.5.—TheRiemann surface C/(ω Z + ω Z) and the location of the branch points 1 2 FIG. 6. — Informal construction of the universal covering 3.2. Periods and covering map. — We now give a proper construction of the universal covering. The Riemann surface T being of genus 1, its universal covering has the form (C,λ),where C is the complex plane and λ : C → T is a non-branching covering map, see [10]. This way, the surface T can be considered as the additive group C factorized by the discrete subgroup ω Z + ω Z, where the periods ω ,ω are complex numbers, 1 2 1 2 linearly independent on R. Any segment of length |ω | and parallel to ω , ∈{1, 2},is projected onto a closed curve on T homological to one of the elements of the normal basis on the torus. We choose λ([0,ω ]) to be homological to the cut [x , x ] (and hence 1 1 2 also to all other cuts [x , x ], [y , y ] and [y , y ]); λ([0,ω ]) is then homological to the 3 4 1 2 3 4 2 cycles of real points and ; see Figures 5 and 7. 0 1 Our aim now is to ﬁnd the expression of the covering λ. We will do this by ﬁnding, for all ω ∈ C, the explicit expressions of the pair of coordinates (x(λω), y(λω)),thatwe have introduced in Section 2. First, the periods ω ,ω are obtained in [6, Lemma 3.3.2] 1 2 for z = 1/|S|. The reasoning is exactly the same for other values of z, and we obtain that with d as in (2.2), x x 2 3 dx dx (3.1) ω = i ,ω = . 1 2 1/2 1/2 [−d(x)] d(x) x x 1 2 ON THE FUNCTIONS COUNTING WALKS IN THE QUARTER PLANE 79 FIG. 7. — Important points and cycles on the universal covering We also need to introduce dx (3.2) ω = . 1/2 d(x) X( y ) Further, we deﬁne d (x )/6 + d (x )/[t − x ] if x = ∞, 4 4 4 4 g (t) = d (0)/6 + d (0)t/6if x =∞, as well as d ( y )/6 + d ( y )/[t − y ] if y = ∞, 4 4 4 4 g (t) = d (0)/6 + d (0)t/6if y =∞, and ﬁnally we introduce ℘(ω; ω ,ω ), the Weierstrass elliptic function with periods 1 2 ω ,ω . Throughout, we shall write ℘(ω) for ℘(ω; ω ,ω ). By deﬁnition, see [10, 20], 1 2 1 2 we have 1 1 1 ℘(ω) = + − . 2 2 2 ω (ω − ω − ω ) ( ω + ω ) 1 1 2 2 1 1 2 2 ( , )∈Z \{(0,0)} 1 2 Then we have the uniformization [6, Lemma 3.3.1] −1 x(λω) = g (℘ (ω)), (3.3) −1 y(λω) = g (℘ (ω − ω /2)). From now on, whenever no ambiguity can arise, we drop the dependence w.r.t. λ, writ- ing x(ω) and y(ω) instead of x(λω) and y(λω), respectively. The coordinates x(ω), y(ω) deﬁned in (3.3) are elliptic: (3.4) x(ω + ω ) = x(ω), y(ω + ω ) = y(ω), ∀ ∈{1, 2}, ∀ω ∈ C. 80 I. KURKOVA, K. RASCHEL Furthermore, x(ω /2) = x x(ω /2) = x x(0) = x 1 3 2 1 , , , y(0) = Y(x ) 4 y(ω /2) = Y(x ) y(ω /2) = Y(x ) 1 3 2 1 x([ω + ω ]/2) = x 1 2 2 y([ω + ω ]/2) = Y(x ) 1 2 2 Let us denote the points 0,ω /2,ω /2, [ω + ω ]/2by ω ,ω ,ω ,ω , respectively, see 1 2 1 2 x x x x 4 3 1 2 Figures 5 and 7.Let x x 3 2 L = ω + ω R, L = ω + ω R. x 1 x 1 x 4 x 1 4 1 x x 1 2 3 2 Then λL (resp. λL )isthe cutof T where S and S are glued together, namely, {s ∈ x x x x 4 1 T : x(s) ∈[x , x ]} (resp. {s ∈ T : x(s) ∈[x , x ]}). 3 4 1 2 Moreover, by construction we have (see again Figures 5 and 7) x(ω /2) = X( y ) x([ω + ω ]/2) = X( y ) 3 4 1 3 3 , , y(ω /2) = y y([ω + ω ]/2) = y 3 4 1 3 3 x([ω + ω ]/2) = X( y ) 2 3 1 y([ω + ω ]/2) = y 2 3 1 and x([ω + ω + ω ]/2) = X( y ) 1 2 3 2 y([ω + ω + ω ]/2) = y 1 2 3 2 We denote the points ω /2, [ω + ω ]/2, [ω + ω ]/2, [ω + ω + ω ]/2by ω , ω , ω , 3 1 3 2 3 1 2 3 y y y 4 3 1 ω , respectively. Let y y 3 2 L = ω + ω R, L = ω + ω R. y 1 y 1 y 4 y 1 4 1 y y 1 2 3 2 Then λL (resp. λL )isthe cutof T where S and S are glued together, that is to say y y y y 4 1 {s ∈ T : y(s) ∈[y , y ]} (resp. {s ∈ T : y(s) ∈[y , y ]}). 3 4 1 2 x y x y 3 3 2 2 The distance between L and L is the same as between L and L ; it equals x y x y 4 4 1 1 ω /2. 3.3. Lifted Galois automorphisms ξ, η.— Any conformal automorphism ζ of the sur- −1 face T can be continued as a conformal automorphism ζ = λ ζλ of the universal cover- −1 ing C. This continuation is not unique, but it will be unique if we ﬁx some ζω ∈ λ ζλω 0 0 for a given point ω ∈ C. 0 ON THE FUNCTIONS COUNTING WALKS IN THE QUARTER PLANE 81 According to [6], we deﬁne ξ, η by choosing their ﬁxed points to be ω ,ω ,re- x y 2 2 spectively. Since any conformal automorphism of C is an afﬁne function of ω [10]and 2 2 since ξ = η = Id, we have (3.5) ξω =−ω + 2ω , ηω =−ω + 2ω . x y 2 2 It follows that η ξ and ξ η are just the shifts via the real numbers ω and −ω , respectively: 3 3 η ξω = ω + 2(ω − ω ) = ω + ω , y x 3 2 2 (3.6) ξ ηω = ω + 2(ω − ω ) = ω − ω . x y 3 2 2 By (2.6)and (2.7)wehave c(x(ω)) 1 x(ξω) = x(ω), y(ξω) = , a(x(ω)) y(ω) (3.7) c( y(ω)) 1 x( ηω) = , y( ηω) = y(ω). a( y(ω)) x(ω) x x x x y y y y 2 2 3 3 2 2 3 3 Finally, ξ L = L , ξ L = L + ω and ηL = L , ηL = L + ω . 2 2 x x x x y y y y 1 1 4 4 1 1 4 4 4. Lifting of x → Q(x, 0) and y → Q(0, y) to the universal covering 4.1. Lifting to the Riemann surface T.— We have seen in Section 2 that for any z ∈ ]0, 1/|S|[, exactly two branch points of Y(x) (namely, x and x ) are in the unit disc. For 1 2 this reason, and by construction of the surface T, the set {s ∈ T :|x(s)|= 1} is composed 1 2 of two cycles (one belongs to S and the other to S ) homological to the cut {s ∈ T : x x x(s) ∈[x , x ]}. The domain D ={s ∈ T :|x(s)| < 1} is bounded by these two cycles, 1 2 x see Figure 8, and contains the points s ∈ T such that x(s) ∈[x , x ]. Since the function 1 2 x → K(x, 0)Q(x, 0) is holomorphic in the unit disc, we can lift it to D ⊂ T as r (s) = K x(s), 0 Q x(s), 0 , ∀s ∈ D . x x In thesameway,the domain D ={s ∈ T :|y(s)| < 1} is bounded by {s ∈ T : |y(s)|= 1}, which consists in two cycles homological to the cut {s ∈ T : y(s) ∈[y , y ]}, 1 2 see Figure 8, and which contains the latter. We lift the function y → K(0, y)Q(0, y) to D ⊂ T as r (s) = K 0, y(s) Q 0, y(s) , ∀s ∈ D . y y It is shown in [18, Lemma 3] that for any z ∈]0, 1/|S|[ and any x such that |x|= 1, we have |Y (x)| < 1and |Y (x)| > 1. Hence, the cycles that constitute the boundary of 0 1 D are = s ∈ T : x(s) = 1, y(s) < 1 , = s ∈ T : x(s) = 1, y(s) > 1 . x 82 I. KURKOVA, K. RASCHEL FIG. 8. — Location of the domains D and D on the Riemann surface T x y 0 1 We thus have ∈ D and ∈ / D , see Figure 8.Inthe same way, forany z ∈]0, 1/|S|[ y y x x and any y such that |y|= 1, we have |X ( y)| < 1and |X ( y)| > 1. Therefore, the cycles 0 1 composing the boundary of D are = s ∈ T : y(s) = 1, x(s) < 1 , = s ∈ T : y(s) = 1, x(s) > 1 . 0 1 Furthermore, ∈ D and ∈ / D , see Figure 8. x x y y It follows that D ∩ D ={s ∈ T :|x(s)| < 1, |y(s)| < 1} is not empty, simply con- x y 0 0 nected and bounded by and . Since for any s ∈ T,K(x(s), y(s)) = 0, and since the x y main equation (1.1)isvalid on {(x, y) ∈ C :|x| < 1, |y| < 1},wehave (4.1) r (s) + r (s) − K(0, 0)Q(0, 0) − x(s)y(s) = 0, ∀s ∈ D ∩ D . x y x y 4.2. Lifting to the universal covering C.— The domain D lifted on the universal cov- ering consists of inﬁnitely many curvilinear strips shifted by ω : −1 n n λ D = , ⊂ ω R + nω ,(n + 1)ω , x 1 2 2 x x n∈Z and, likewise, −1 n n λ D = , ⊂ ω R + ω /2 + nω ,(n + 1)ω . y 1 3 2 2 y y n∈Z Let us consider these strips for n = 0, that we rename 0 0 = , = . x y x y 1 −1 1 0 −1 0 The ﬁrst is bounded by ⊂ λ and by ⊂ λ , while the second is delimited by x x x x 0 −1 0 1 −1 1 ⊂ λ and by ⊂ λ y y y y x y 2 2 Further, note that the straight line L (resp. L ) deﬁned in Section 3 is invariant x y 1 1 w.r.t. ξ (resp. η) and belongs to (resp. ). x y ON THE FUNCTIONS COUNTING WALKS IN THE QUARTER PLANE 83 FIG.9.—Location of = ∪ x y 1 0 1 0 Then, by the facts that ξ and η , and by our choice (3.5)ofthe x x y y 1 0 1 0 deﬁnition of ξ and η on the universal covering, we have ξ and η .In x x y y addition, (4.2) ξω ∈ , ∀ω ∈ , ηω ∈ , ∀ω ∈ . x x y y 0 1 0 1 0 1 Moreover, since ∈ D , ∈ / D and ∈ D , ∈ / D ,wehave ∈ , ∈ / and x x y y x x y y x x y y 0 1 0 0 ∈ , ∈ / . It follows that ∩ is a non-empty strip bounded by and y y x y x x x y and that = ∪ x y is simply connected, as in Figure 9. Let us lift the functions r (s) and r (s) holomorphically to and , respectively: x y x y we put r (ω) = r (λω) = K(x(ω), 0)Q(x(ω), 0), ∀ω ∈ , x x x (4.3) r (ω) = r (λω) = K(0, y(ω))Q(0, y(ω)), ∀ω ∈ . y y y It follows from (4.1)and (4.3)that (4.4) r (ω) + r (ω) − K(0, 0)Q(0, 0) − x(ω)y(ω) = 0, ∀ω ∈ ∩ . x y x y Equation (4.4) allows us to continue functions r (ω) and r (ω) meromorphically on :we x y put r (ω) =−r (ω) + K(0, 0)Q(0, 0) + x(ω)y(ω), ∀ω ∈ , x y y (4.5) r (ω) =−r (ω) + K(0, 0)Q(0, 0) + x(ω)y(ω), ∀ω ∈ . y x x Equation (4.4) is then valid on the whole of . We summarize all facts above in the next result. 84 I. KURKOVA, K. RASCHEL Theorem 3. — The functions ⎪ K(x(ω), 0)Q(x(ω), 0) if ω ∈ , −K(0, y(ω))Q(0, y(ω)) + K(0, 0)Q(0, 0) r (ω) = + x(ω)y(ω) if ω ∈ , and ⎪ K(0, y(ω))Q(0, y(ω)) if ω ∈ , r (ω) = −K(x(ω), 0)Q(x(ω), 0) + K(0, 0)Q(0, 0) + x(ω)y(ω) if ω ∈ , are meromorphic in .Furthermore, (4.6) r (ω) + r (ω) − K(0, 0)Q(0, 0) − x(ω)y(ω) = 0, ∀ω ∈ . x y 5. Meromorphic continuation of x → Q(x, 0) and y → Q(0, y) on the universal covering 5.1. Meromorphic continuation. — In Theorem 3 we saw that r (ω) and r (ω) are x y meromorphic on . We now continue these functions meromorphically from to the whole of C. Theorem 4. — The functions r (ω) and r (ω) can be continued meromorphically to the whole x y of C.Further,for any ω ∈ C, we have (5.1) r (ω − ω ) = r (ω) + y(ω) x(−ω + 2ω ) − x(ω) , x 3 x y (5.2) r (ω + ω ) = r (ω) + x(ω) y(−ω + 2ω ) − y(ω) , y 3 y x (5.3) r (ω) + r (ω) − K(0, 0)Q(0, 0) − x(ω)y(ω) = 0, x y r (ξω) = r (ω), x x (5.4) r ( ηω) = r (ω), y y r (ω + ω ) = r (ω), x 1 x (5.5) r (ω + ω ) = r (ω). y 1 y For the proof of Theorem 4, we shall need the following lemma. Lemma 5. — We have (5.6) ( + nω ) = C. n∈Z ON THE FUNCTIONS COUNTING WALKS IN THE QUARTER PLANE 85 1 0 0 Proof. — It has been noticed in Section 4 that ξ ∈ .By(4.2), η x x x ⊂ ,sothat, by (3.6), 1 1 + ω = ηξ ∈ . x x In the same way, − ω ∈ . It follows that ∪ ( + ω ) is a simply connected domain, 3 3 see Figure 9. Identity (5.6) follows. Proof of Theorem 4.—Forany ω ∈ ,byTheorem 3 we have (5.7) r (ω) + r (ω) − K(0, 0)Q(0, 0) − x(ω)y(ω) = 0. x y 1 1 0 For any ω ∈ close enough to the cycle ,wehavethat ξω ∈ since ξ ∈ . y y x x x Then ω + ω = ηξω ∈ by (4.2). We now compute r ( ηξω) for any such ω. Equation 3 y y (4.6), which is valid in ⊃ ,gives (5.8) r (ξω) + r (ξω) − K(0, 0)Q(0, 0) − x(ξω)y(ξω) = 0. x y By (3.7), x(ξω) = x(ω). For our ω ∈ ,by(4.2)wehave ξω ∈ , so that Theorem 3 x x yields r (ξω) = K x(ξω), 0 Q x(ξω), 0 = K x(ω), 0 Q x(ω), 0 = r (ω). x x If we now combine the last fact together with Equation (5.7), Equation (5.8) and identity x(ξω) = x(ω),weobtainthat r (ξω) = r (ω) + x(ω) y(ξω) − y(ω) . y y Since ξω ∈ ,thenby(4.2)wehave ηξω ∈ . Equation (3.7)and Theorem 3 entail y y r ( ηξω) = K 0, y( ηξω) Q 0, y( ηξω) = K 0, y(ξω) Q 0, y(ξω) = r (ξω). Finally, for all ω ∈ close enough to we have r ( ηξω) = r (ω) + x(ω) y(ξω) − y(ω) . y y Using (3.6), we obtain exactly Equation (5.2). Thanks to Theorem 3 and Lemma 5, this equation shown for any ω ∈ close enough to allows us to continue r meromor- phically from to the whole of C. Equation (5.2) therefore stays valid for any ω ∈ C. The function r ( ηω) = r (−ω + ω ) is then also meromorphic on C. Since these func- y y y tions coincide in , then by the principle of analytic continuation [10] they do on the whole of C. In the same way, we prove Equation (5.1)for all ω ∈ close enough to Together with Theorem 3 and Lemma 5 this allows us to continue r (ω) meromorphi- cally to the whole of C. By the same continuation argument, the identity r (ω) = r (ξω) x x 86 I. KURKOVA, K. RASCHEL is valid everywhere on C. Consequently Equation (5.3), which a priori is satisﬁed in , must stay valid on the whole of C. Since x(ω) and y(ω) are ω -periodic, it follows from Theorem 3 that r (ω) and r (ω) are ω -periodic in .The vector ω being real, by (5.1) x y 1 3 and (5.2) these functions stay ω -periodic on the whole of C. 5.2. Branches of x → Q(x, 0) and y → Q(0, y).— The restrictions of r (ω)/ K(x(ω), 0) on (5.9) M = ω [, + 1[+ ω k/2,(k + 1)/2 k, 1 2 for k, ∈ Z provide all branches on C \ ([x , x ]∪[x , x ]) of Q(x, 0) as follows: 1 2 3 4 (5.10) Q(x, 0) = r (ω)/K x(ω), 0 : ω is the (unique) element of M x k, such that x(ω) = x . Due to the ω -periodicity of r (ω) and x(ω), the restrictions of these functions on M 1 x k, do not depend on ∈ Z, and therefore determine the same branch as on M for k,0 any . Furthermore, thanks to (5.4), (3.5)and (3.7) the restrictions of r (ω)/K(x(ω), 0) on M and on M lead to the same branches for any k ∈ Z. Hence, the restrictions −k+1,0 k,0 of r (ω)/K(x(ω), 0) to M with k ≥ 1 provide all different branches of this function. x k,0 The analogous statement holds for the restrictions of r (ω)/K(0, y(ω)) on (5.11) N = ω /2 + ω , + 1[+ ω ]k/2,(k + 1)/2 k, 3 1 2 for k, ∈ Z, namely: (5.12) Q(0, y) = r (ω)/K 0, y(ω) : ω is the (unique) element of N y k, such that y(ω) = y . The restrictions on N for ∈ Z give the same branch as on N .For any k ∈ Z the k, k,0 + restrictions on N and on N determine the same branches. Hence, the restrictions −k+1,0 k,0 of r (ω)/K(0, y(ω)) on N with k ≥ 1 provide all different branches of y → Q(0, y). y k,0 5.3. Ratio ω /ω 2 3 Remark 6. —For any z ∈]0, 1/|S|[ the value ω /ω is rational if and only if the 2 3 group ξ, η restricted to the curve {(x, y) ∈ C ∪{∞} : K(x, y) = 0} is ﬁnite, see [6, Section 4.1.2] and [18, Proof of Proposition 4]. The rationality or irrationality of the quantity ω /ω is crucial for the nature of 2 3 the functions x → Q(x, 0) and y → Q(0, y) for a given z. Indeed, the following theorem holds true. ON THE FUNCTIONS COUNTING WALKS IN THE QUARTER PLANE 87 FIG. 10. — Three famous examples, known as Kreweras’, Gessel’s and Gouyou-Beauchamps’ walks, respectively Theorem 7. —For any z ∈]0, 1/|S|[ such that ω /ω is rational, the functions x → Q(x, 0) 2 3 and y → Q(0, y) are holonomic. Proof. — The proof of Theorem 7 is completely similar to that of Theorems 1.1 and 1.2 in [5], so here we just recall the main ideas. The proof actually consists in applying [6, Theorem 4.4.1], which entails that if ω /ω is rational, the function Q(x, 0) can be 2 3 written as Q(x, 0) = w (x) + (x)φ (x) + w(x)/r(x), where w and r are rational functions, while φ and w are algebraic. Further, in [5, Lemma 2.1] it is shown that is holonomic. Accordingly, Q(x, 0) is also holonomic. The argument for Q(0, y) is similar. Notice that Theorem 4.4.1 in [6]isprovedfor z = 1/|S| only, but in [5] it is observed that this result also holds for z ∈]0, 1/|S|[. For all 23 models of walks with ﬁnite group (1.3), the ratio ω /ω is rational and 2 3 independent of z. This fact, which is speciﬁed in Lemma 8 below, implies the holonomy of the functions x → Q(x, 0) and y → Q(0, y) for all z ∈]0, 1/|S|[ by Theorem 7,and also leads to some more profound analysis of the models with a ﬁnite group. This analysis is the topic of the Section 6. For all 51 non-singular models of walks with inﬁnite group, ω /ω takes rational 2 3 and irrational values on subsetsH and ]0, 1/|S|[\H, respectively, which are dense on ]0, 1/|S|[,asitwillbeprovedinProposition 14 below. For any z ∈H, x → Q(x, 0) and y → Q(0, y) are holonomic by Theorem 7.For all z ∈]0, 1/|S|[\H, properties of the branches of x → Q(x, 0) and y → Q(0, y) (in particular, the set of their poles) will be studied in detail in Section 7; the non-holonomy will be derived from this analysis. 6. Finite group case Deﬁne the covariance of the model as (6.1) ij − i j = ij. (i,j)∈S (i,j)∈S (i,j)∈S (i,j)∈S 88 I. KURKOVA, K. RASCHEL The equality above follows from the fact that for each of the 23 models with a ﬁnite group, i = 0or j = 0, see [2]. Lemma 8 below is proved in [18, Proposition 5]. (i,j)∈S (i,j)∈S Lemma 8. — For all 23 models with ﬁnite group (1.3), ω /ω is rational and independent 2 3 of z. More precisely: – For the walks with a group of order 4, ω /ω = 2; 2 3 – For the walks with a group of order 6 and such that the covariance is negative (resp. positive), ω /ω = 3 (resp. 3/2); 2 3 – For the walks with a group of order 8 and a negative (resp. positive) covariance, ω /ω = 4 2 3 (resp. 4/3). In the sequel, we note ω /ω = k/; then, 2k is the order of the group. Since 2 3 kω = ω , we obviously always have 3 2 r (ω + ω ) − r (ω) = r (ω + mω ) − r ω + (m − 1)ω . x 2 x x 3 x 3 1≤m≤k It follows from (5.1) and from properties (3.5), (3.6)and (3.7) of the Galois automorphisms that (6.2) r (ω + ω ) − r (ω) = (xy)(ω + mω ) − (xy) η(ω + mω ) x 2 x 3 3 1≤m≤k m m−1 = (xy) ( η ξ) ω − (xy) ξ( η ξ) ω ) 1≤m≤k = (−1) xy θ(ω) , θ ∈ξ, η θ θ (θ ) where (−1) is the signature of θ;inother words, (−1) = (−1) ,where (θ ) is the length of θ , i.e., the smallest such that we can write θ = θ ◦···◦ θ ,with θ ,...,θ equal 1 1 to ξ or η. The same identity with the opposite sign holds for r . The quantity (6.2)isthe orbit-sum of the function xy under the group ξ, η, and is denoted by O(ω). It satisﬁes the property hereunder, which is proved in [2]. Lemma 9. — In the ﬁnite group case, the orbit-sum O(ω) is identically zero if and only if the covariance (6.1) is positive. We therefore come to the following corollary. Corollary 10. — In the ﬁnite group case, the functions x → Q(x, 0) and y → Q(0, y) have a ﬁnite number of different branches if and only if the covariance (6.1) is positive. ON THE FUNCTIONS COUNTING WALKS IN THE QUARTER PLANE 89 After the lifting to the universal covering done in Theorem 4, results of [1, 2] concerning the nature of the functions x → Q(x, 0) and y → Q(0, y) in all ﬁnite group cases can now be established by very short reasonings. For the sake of completeness, we show how this works. Proposition 11 ([1, 2]). — For all models with a ﬁnite group and a positive covariance (6.1), x → Q(x, 0) and y → Q(0, y) are algebraic. Proposition 12 ([2]). — For all models with a ﬁnite group and a negative or zero covariance (6.1), x → Q(x, 0) and y → Q(0, y) are holonomic and non-algebraic. Proofs of both of these propositions involve the following lemma. Lemma 13. —Let ℘ be a Weierstrass elliptic function with certain periods ω, ω. (P1) We have ℘ (ω) = 4 ℘(ω) − ℘(ω/2) ℘(ω) − ℘ [ω + ω]/2 × ℘(ω) − ℘( ω/2) , ∀ω ∈ C. (P2) Let p be some positive integer. The Weierstrass elliptic function with periods ω, ω/pcan be written in terms of ℘ as p−1 ℘(ω) + ℘(ω + ω/p) − ℘( ω/p) , ∀ω ∈ C. =1 (P3) We have the addition theorem: 1 ℘ (ω) − ℘ ( ω) ℘(ω + ω) =−℘(ω) − ℘( ω) + , ∀ω, ω ∈ C. 4 ℘(ω) − ℘( ω) (P4) For any elliptic function f with periods ω, ω, there exist two rational functions R and S such that f (ω) = R ℘(ω) + ℘ (ω)S ℘(ω) , ∀ω ∈ C. (P5) There exists a function which is ω-periodic and such that (ω + ω) = (ω) − 1, ∀ω ∈ C. Proof. — Properties (P1), (P3) and (P4) are most classical, and can be found, e.g., in [10, 20]. For (P2) we refer to [20, page 456], and for (P5), see [6, Equation (4.3.7)]. Note that the function in (P5) can be constructed via the zeta function of Weierstrass. 90 I. KURKOVA, K. RASCHEL Proof of Proposition 11. — If the orbit-sum O(ω) is zero, Equation (6.2) implies that r (ω) is ω -periodic. In particular, the property (P4) of Lemma 13 entails that there exist x 2 two rational functions R and S such that (6.3) r (ω) = R ℘(ω; ω ,ω ) + ℘ (ω; ω ,ω )S ℘(ω; ω ,ω ) . x 1 2 1 2 1 2 Further, the property (P2) together with the addition formula (P3) of Lemma 13 gives that ℘(ω; ω ,ω ) is an algebraic function of ℘(ω)—we recall that ℘(ω) denotes the 1 2 Weierstrass function ℘(ω; ω ,ω ). Due to Lemma 13(P1), ℘ (ω) is an algebraic function 1 2 of ℘(ω) too, so that ℘ (ω; ω ,ω ) is also an algebraic function of ℘(ω). Thanks to (6.3), 1 2 we get that r (ω) is algebraic in ℘(ω). Since ℘(ω) is a rational function of x(ω), see (3.3), we ﬁnally obtain that r (ω) is algebraic in x(ω).Then q (ω) = r (ω)/K(x(ω), 0) is x x x algebraic in x(ω), and so is q (ω) in y(ω). Proof of Proposition 12. — In this proof we have = 1, see Lemma 8. Thanks to Lemma 13(P5), there exists a function which is ω -periodic and such that (ω + ω ) = 1 2 (ω) − 1. In particular, transforming (6.2) we can write r (ω + ω ) + (ω + ω )O(ω + ω ) = r (ω) + (ω)O(ω). x 2 2 2 x This entails that r (ω) + (ω)O(ω) is elliptic with periods ω ,ω . In particular, for the x 1 2 same reasons as in the proof of Proposition 11, this is an algebraic function of x(ω). The function O(ω) is obviously also algebraic in x(ω).Asfor thefunction (ω),itis proved in [6, page 71] that it is a non-algebraic function of x(ω). Moreover, it is shown in [5, Lemma 2.1] that it is holonomic in x(ω).Hence r (ω) is holonomic in x(ω) but not algebraic. The same is true for q (ω) = r (ω)/K(x(ω), 0) in x(ω) and for q (ω) in y(ω). x x y 7. Inﬁnite group case It has been shown in [18, Part 6.2] that for all 51 models with inﬁnite group, ω /ω 2 3 takes irrational values for inﬁnitely many z. The next proposition states a more complete result. Proposition 14. — For all 51 walks with inﬁnite group, the setsH ={z ∈]0, 1/ |S|[: ω /ω is rational} and ]0, 1/|S|[\H ={z ∈]0, 1/|S|[: ω /ω is irrational} are dense 2 3 2 3 in ]0, 1/|S|[. Proof.—Thefunction ω /ω is clearly real continuous function on ]0, 1/|S|[.In 2 3 fact, it has been noticed in [18, Part 6.2] that the function ω /ω is expandable in power 2 3 series in a neighborhood of any point of the interval ]0, 1/|S|[. Thus it sufﬁces to ﬁnd just one segment within ]0, 1/|S|[ where this function is not constant. ON THE FUNCTIONS COUNTING WALKS IN THE QUARTER PLANE 91 Proposition 25 below gives the asymptotic of ω /ω as z → 0: for any of 51 models 2 3 there exist some rational L > 0and some L = 0such that as z > 0goes to0, ω /ω = L + L/ ln z + O (1/ ln z) . 2 3 This immediately implies that this function is not constant on a small enough interval in a right neighborhood of 0 and concludes the proof. Note however that there is another way to conclude the proof that does not need the full power of Proposition 25: it is enough to show (as done in the proof of Proposi- tion 25)that ω /ω converges to a rational positive constant L as z → 0for all51models. 2 3 Indeed, then, since ω /ω necessarily takes irrational values for some z ∈]0, 1/|S|[ (see 2 3 [18, Part 6.2]), there exists an interval within ]0, 1/|S|[ where the ratio ω /ω is not 2 3 constant. In Subsections 7.1, 7.2 and 7.3 we thoroughly analyze the branches of x → Q(x, 0) and y → Q(0, y) for z such that ω /ω is irrational, and we prove in particular that their 2 3 set of poles is inﬁnite and dense on the curves given on Figure 11, see Theorem 17.Then the following corollary is immediate. Corollary 15. —LetH ={z ∈]0, 1/|S|[: ω /ω is rational} and ]0, 1/|S|[\H ={z ∈ 2 3 ]0, 1/|S|[: ω /ω is irrational}. 2 3 (i) For all z ∈H,x → Q(x, 0) and y → Q(0, y) are holonomic; (ii) For all z ∈]0, 1/|S|[\H,x → Q(x, 0) and y → Q(0, y) are non-holonomic. Proof. — The statement (i) follows from Theorem 7, and (ii) comes from Theo- rem 17(iii) below as explained in the Introduction. Remark 16. — It follows from Remark 6 thatH can be characterized as the set of z ∈]0, 1/|S|[ such that the group ξ, η restricted to the curve {(x, y) ∈ (C ∪{∞}) : K(x, y; z) = 0} is ﬁnite. Then methods developed in [6, Chapter 4] speciﬁcally for the ﬁnite group case should be efﬁcient for further analysis of x → Q(x, 0) and y → Q(0, y) for any ﬁxed z ∈H. The analysis of the poles being rather technical, we start ﬁrst with an informal study. 7.1. Polesofthe setofbranchesof x → Q(x, 0) and y → Q( y, 0) for irrational ω /ω : 2 3 an informal study. — Let us ﬁx z ∈]0, 1/|S|[ such that ω /ω is irrational. We ﬁrst infor- 2 3 mally explain why the set of poles of all branches of x → Q(x, 0) and y → Q(0, y) could be dense on certain curves in this case. We shall denote by ω and ω the real and 92 I. KURKOVA, K. RASCHEL imaginary parts of ω ∈ C, respectively. Let and be the parallelograms deﬁned by x y = M ∪ M = ω [0, 1[+ ω [0, 1[, x 0,0 0,1 1 2 (7.1) = N ∪ N = ω /2 + ω [0, 1[+ ω ]0, 1], y 0,0 0,1 3 1 2 with notations (5.9)and (5.11). Function r (ω) (resp. r (ω))on (resp. ) deﬁnes the x y x y ﬁrst (main) branch of x → Q(x, 0) (resp. y → Q(0, y))twice via(5.10) (resp. (5.12)). Denote by f (ω) = x(ω)[y(−ω + 2ω ) − y(ω)] the function used in the mero- y x morphic continuation procedure (5.2). Assume that at some ω ∈ , r (ω ) = ∞ and 0 y y 0 f (ω ) =∞. Further, suppose that y 0 (7.2) ω ∈ :ω =ω , f (ω) =∞. y 0 y By (5.2), for any n ≥ 1we have n−1 (7.3) r (ω + nω ) = r (ω ) + f (ω ) + f (ω + kω ). y 0 3 y 0 y 0 y 0 3 k=1 We have r (ω ) + f (ω ) =∞ by our assumptions. If ω /ω is irrational, then for any y 0 y 0 2 3 k ≥ 1 there is no p ∈ Z such that ω + kω = ω + pω .Function f being ω -periodic, it 0 3 0 2 y 2 follows from this fact and assumption (7.2)that f (ω + kω ) = ∞ for any k ≥ 1. Hence y 0 3 by (7.3), r (ω + nω ) =∞ for all n ≥ 1. Due to irrationality of ω /ω ,for any n ≥ 1there y 0 3 2 3 exists a unique ω (ω ) ∈ and p ∈ Z such that ω + nω = ω (ω ) + pω , and the set n 0 y 0 3 n 0 2 {ω (ω )} is dense on the curve n 0 n≥1 (7.4)I (ω ) = y {ω :ω =ω ,ω ∈ } ⊂ C ∪{∞}. y 0 0 y By deﬁnition (5.12), the set of poles of all branches of y → K(0, y)Q(0, y) is dense on the curveI (ω ).The number of zerosof y → K(0, y) being at most two, the same conclusion y 0 holds true for y → Q(0, y). Let us now identify the points ω in where f (ω ) is inﬁnite. They are (at 0 y y 0 most) six such points a , a , a , a , b , b ∈ , which correspond to the following pairs 1 2 3 4 1 2 y (x(ω ), y(ω )): 0 0 a = x , ∞ , a = x , y , a = x , ∞ , a = x , y , 1 4 2 3 (7.5) ◦ • b = ∞, y , b = ∞, y . 1 2 ON THE FUNCTIONS COUNTING WALKS IN THE QUARTER PLANE 93 Here by (2.4)and (2.5) 2 1/2 −b( y) +[b( y) − 4 a( y) c( y)] x = lim , y→∞ a( y) 2 1/2 −b( y) −[b( y) − 4 a( y) c( y)] x = lim , y→∞ a( y) 2 1/2 −b(x) +[b(x) − 4a(x)c(x)] y = lim , x→x 2a(x) 2 1/2 −b(x) +[b(x) − 4a(x)c(x)] y = lim , 2a(x) x→x 2 1/2 −b(x) +[b(x) − 4a(x)c(x)] y = lim , x→∞ 2a(x) 2 1/2 −b(x) −[b(x) − 4a(x)c(x)] y = lim . x→∞ 2a(x) where a, b, c, a, b, c are introduced in (2.1). For most of 51 models of walks, assumption (7.2) holds true for none of these points, so that the previous reasoning does not work: some poles of f could be compensated in the sum (7.3). Furthermore, it may happen for some of these points that not only f (ω ) =∞ y 0 but also r (ω ) =∞, and consequently f (ω) + r (ω) mayhavenopoleat ω = ω .For y 0 y y 0 these reasons we need to inspect more closely the location of these six points for each of the 51 models and their contribution to the set of poles via (7.3). 7.2. Functions x → Q(x, 0) and y → Q(0, y) for irrational ω /ω .— In addition to 2 3 the notation (7.4), deﬁne the curve (7.6)I (ω ) = x {ω :ω =ω ,ω ∈ } ⊂ C ∪{∞}. x 0 0 x We now formulate the main theorem of this section. Theorem 17. — For all 51 non-singular walks with inﬁnite group (1.3) given on Figure 17 and any z such that ω /ω is irrational, the following statements hold. 2 3 (i) The only singularities on C of the ﬁrst branch of x → Q(x, 0) (resp. y → Q(0, y))are the branch points x and x (resp. y and y ). 3 4 3 4 (ii) Each branch of x → Q(x, 0) (resp. y → Q(0, y)) is meromorphic on C with a ﬁnite number of poles. (iii) The set of poles on C of all branches of x → Q(x, 0) (resp. y → Q(0, y)) is inﬁnite. With the notations (7.6), (7.4) above and points a , b deﬁned in (7.5), it is dense on the 1 1 following curves (see Figure 11): 94 I. KURKOVA, K. RASCHEL FIG. 11. — For walks pictured on Figure 17, curves where poles of the set of branches of x → Q(x, 0) and y → Q(0, y) are dense (iii.a) For the walks of Subcase I.A in Figure 17:I (a ) andI (b ) for x → Q(x, 0); x 1 x 1 I (a ) andI (b ) for y → Q(0, y). y 1 y 1 (iii.b) For the walks of Subcases I.B and I.C in Figure 17:I (a ) and R\]x , x [ for x 1 1 4 x → Q(x, 0);I (a ) and [y , y ] for y → Q(0, y). y 1 4 1 ON THE FUNCTIONS COUNTING WALKS IN THE QUARTER PLANE 95 (iii.c) For the walks of Subcase II.A in Figure 17:I (b ) and [x , x ] for x → Q(x, 0); x 1 4 1 I (b ) and R\]y , y [ for y → Q(0, y). y 1 1 4 (iii.d) For the walks of Subcases II.B, II.C, II.D and Case III in Figure 17: R\]x , x [ 1 4 for x → Q(x, 0); R\]y , y [ for y → Q(0, y). 1 4 (iv) Poles of branches of x → Q(x, 0) and y → Q(0, y) out of these curves may be only at zeros of K(x, 0) and K(0, y), respectively. Before giving the proof of Theorem 17, we need to introduce some additional tools. If the value of ω /ω is irrational, for any ω ∈ C and any n ∈ Z , there exists a unique 2 3 0 + y x ω (ω ) ∈ (resp. ω (ω ) ∈ )aswellasaunique number p ∈ Z (resp. p ∈ Z)such 0 y 0 x y x n n y x that ω + nω = p ω + ω (ω ) (resp. ω + nω = p ω + ω (ω )). With these notations we 0 3 y 2 0 0 3 x 2 0 n n can state the following lemma. Lemma 18. — Let z be such that ω /ω is irrational. 2 3 x x y y (a) For all n = m, we have ω (ω ) = ω (ω ) and ω (ω ) = ω (ω ). 0 0 0 0 n m n m x y (b) The set {ω (ω )} (resp. {ω (ω )} ) is dense on the segment {ω ∈ :ω = 0 n∈Z 0 n∈Z x + + n n ω } (resp. {ω ∈ :ω =ω }). 0 y 0 Proof. — Both (a) and (b) are direct consequences of the irrationality of ω /ω . 2 3 In the next deﬁnition, we introduce a partial order in . Deﬁnition 19. — For any ω, ω ∈ , we write ω ω if for some n ∈ Z and some p ∈ Z, y + ω + nω = ω + pω . 3 2 If ω ω (and if ω /ω is irrational), both n and p are unique and sometimes we shall 2 3 write ω ω . In particular, for any ω ∈ ,wehave ω ω, since ω ω. n y 0 Deﬁnition 20. — If either ω ω or ω ω ,wesay that ω and ω are ordered, and we write ω ∼ ω . Let us denote by f and f the (meromorphic) functions used in the meromor- x y phic continuation procedures (5.1)and (5.2), namely, by using (3.5): f (ω) = y(ω) x( ηω) − x(ω) , f (ω) = x(ω) y(ξω) − y(ω) . x y The following lemma will be the key tool for the proof of Theorem 17. Lemma 21. — Let z be such that ω /ω is irrational; let ω ∈ be such that r (ω ) = ∞, 2 3 0 y y 0 and let A(ω ) = ω ∈ :ω =ω , f (ω) =∞ . 0 y 0 y 1 k Assume that ω ∈A(ω ) and that for some ω ,...,ω ∈A(ω ): 0 0 0 96 I. KURKOVA, K. RASCHEL 1 k (A) ω ω ··· ω ; 0 n n n 1 2 k (B) lim {f (ω) + f (ω + n ω ) + f (ω + n ω ) +· · · + f (ω + n ω )}=∞; ω→ω y y 1 3 y 2 3 y k k (C) there is no other ω ∈A(ω ) such that ω ω. 0 0 Then the set of poles of all branches of x → Q(x, 0) (resp. y → Q(0, y)) is dense on the curveI (ω ) x 0 (resp.I (ω ))deﬁnedin(7.6)(resp.(7.4)). y 0 Proof. — By Equation (5.2)ofTheorem 4,wehave, forany n ∈ Z and any ω ∈ , + y (7.7) r (ω + nω ) = r (ω) + f (ω) + f (ω + ω ) + f (ω + 2ω ) + ··· y 3 y y y 3 y 3 + f ω + (n − 1)ω . y 0 3 Let ω be as in the statement of Lemma 21. Due to assumption (C), Lemma 18(a) and the ω -periodicity of f , the set {ω + nω } does not contain any point ω where 2 y 0 3 n>n +···+n 1 k f (ω) =∞. Further, by the assumptions (A) and (C), Lemma 18(a) and also by the ω - y 2 periodicity of f , the set {ω + nω } contains exactly k + 1 poles of f that are y 0 3 0≤n≤n +···+n y 1 k ω ,ω + n ω ,...,ω + n ω . Then, by (7.7), assumption (B) and the fact that r (ω ) = ∞, 0 0 1 3 0 k 3 y 0 we reach the conclusion that for any n > n +· · · + n , the point ω + nω is a pole of r (ω). 1 k 0 3 y Due to Equation (5.3), any ω pole of r such that x(ω)y(ω) = ∞ is also a pole of r . Deﬁne nowB, the set of (at most twelve) points in where either x(ω) =∞, x y y(ω) =∞,K(x(ω), 0) = 0orK(0, y(ω)) = 0. Introduce also M = max{m ≥ 0 : ω 0 m ω for some ω ∈B}—with the usual convention M =−∞ if ω ω for none ω ∈B.If n > max(M, n + ··· + n ), the points ω + nω are poles of r as well, and both K(x(ω + 1 k 0 3 x 0 nω ), 0) and K(0, y(ω + nω )) are non-zero. By Lemma 18(b) and deﬁnitions (5.10)and 3 0 3 (5.12), Lemma 21 follows. We are now ready to give the proof of Theorem 17. Proof of Theorem 17.—Functions f (ω) and y(ω) being ω -periodic, it follows that y 2 both of them have no pole at any ω with 0 ≤ω< ω and ω/∈{a , a , a , a , 1 1 2 3 4 b , b }. Then, by (5.2), 1 2 (7.8) ∀ω ∈ N with ω = {a , a , a , a , b , b }, r (ω) = ∞. k,0 1 2 3 4 1 2 y k=0 Function x(ω) being ω -periodic, it has no pole at any ω such that 0 ≤ω< ω and 2 1 ∞ ∞ ω/∈{b , b }. Then Equation (5.3) and the fact that M ⊂ N imply 1 2 k,0 k,0 k=1 k=0 that (7.9) ∀ω ∈ M with ω = {a , a , a , a , b , b }, r (ω) = ∞. k,0 1 2 3 4 1 2 x k=1 In order to prove Theorem 17(i), we shall prove the following proposition. ON THE FUNCTIONS COUNTING WALKS IN THE QUARTER PLANE 97 FIG. 12. — Location of a , a , a , a , b , b if y < 0and x < 0, i.e., Subcase I.A 1 2 3 4 1 2 4 4 Proposition 22. — For all 51 models, for any ω ∈ M (resp. ω ∈ N ) with x(ω) = ∞ 1,0 1,0 (resp. y(ω) = ∞)and ω ∈{a , a , a , a , b , b },wehave r (ω) = ∞ (resp. 1 2 3 4 1 2 x r (ω) = ∞). The proof of this proposition is postponed to the next subsection. By this propo- sition, (7.8)and (7.9), the only singularities of the ﬁrst branches of K(x, 0)Q(x, 0) (resp. K(0, y)Q(0, y)) may be only among the branch points x , x , x , x (resp. y , y , y , y ). Let 1 2 3 4 1 2 3 4 i n us recall that the function x → Q(x, 0) is initially deﬁned as a series q(i, 0; n)x z . i,n≥0 The elementary estimate q(i, 0; n) ≤|S| implies that for any z ∈]0, 1/|S|[ and i≥0 x ∈ C with |x|≤ 1 this series is absolutely is convergent. Since |x | < 1, |x | < 1, and since 1 2 also K(x, 0) is a polynomial with (at most two) roots that are smaller or equal to 1 by ab- solute value, the only singularities of the ﬁrst branch of x → Q(x, 0) are the branch points x and x , that are out of the unit disc. By the same arguments the analogous statement 3 4 holds true for y → Q(0, y). This ﬁnishes the proof of Theorem 17(i). Since for any p ∈ Z , there exist only ﬁnitely many ω ∈ M (resp. ω ∈ + k,0 k=1 N )where f (ω) =∞ (resp. f (ω) =∞), Theorem 17(ii) immediately follows from 0, x y =1 the meromorphic continuation procedure of r and r done in Section 5, namely Equa- x y tions (5.1)and (5.2) as well as the deﬁnitions (5.10)and (5.12). The following proposition proves Theorem 17(iii). Proposition 23. — The poles of x → Q(x, 0) (resp. y → Q(0, y)) are dense on the six curves I (ω ) (resp.I (ω )) with ω ∈{a , a , a , a , b , b }.For anyof 51 models, the set of these curves x 0 y 0 0 1 2 3 4 1 2 coincide with the one claimed in Theorem 17(iii). The proof of this proposition is postponed to the next subsection as well, it will be based on Lemma 21 with ω appropriately chosen among a , a , a , a , b , b . 0 1 2 3 4 1 2 The last statement (iv) of the theorem follows immediately from (7.8)and (7.9), Proposition 23 and Deﬁnitions (5.10)and (5.12). 7.3. Proof of Propositions 22 and 23.— To start with the proofs of Propositions 22 and 23, we need to study closer the location of points (7.5) a , a , a , a , b , b on .It 1 2 3 4 1 2 y 98 I. KURKOVA, K. RASCHEL depends heavily on the signs of x and y , see Figures 12, 13, 14, 15 and 16. Let us recall 4 4 that 0 < x < ∞ and 0 < y < ∞, see Section 2. 3 3 If y < 0 (resp. x < 0), the point y =∞ (resp. x =∞) obviously belongs to the real 4 4 cycle ]y , ∞[ ∪{∞}∪]∞, y [ (resp. ]x , ∞[ ∪{∞}∪]∞, x [) of the complex sphere S.By 3 4 3 4 construction of the Riemann surface T and of its universal covering, the points a , a 1 2 (resp. b , b ) then lie on the open interval {ω : ω ∈ L + ω , 0 < ω< ω } (resp. {ω : 1 2 2 1 ω ∈ L + ω , 0 < ω< ω }). These points are symmetric w.r.t. the center of the interval, 2 1 namely ω /2 + ω + ω /2 (resp. ω /2 + ω ). The points ω corresponding to a and a are 1 2 3 1 2 3 4 on the open interval {ω : ω ∈ L + ω − ω , 0 < ω< ω } and are symmetric w.r.t. the 2 3 1 center ω /2 + ω − ω /2 as well. Furthermore, a + ω = a and a + ω = a ,sothat 1 2 3 4 3 2 3 3 1 a a and a a , see Figure 12. Finally, we have a =a = a = ,hence for 4 1 2 3 1 1 4 2 1 3 any a ∈{a , a } and any a ∈{a , a }, a a (in the sense of Deﬁnition 20). 2 4 1 3 If y > 0or y =∞ (resp. x > 0or x =∞), the point y =∞ (resp. x =∞)is 4 4 4 4 on ]y , ∞] ∪ {∞} ∪ ]∞, y [ (resp. ]x , ∞] ∪{∞}∪]∞, x [). Accordingly, the points a , a 4 1 4 1 1 2 (resp. b , b )and also a , a are on the segment ]ω /2,ω + ω /2].Their location on this 1 2 3 4 3 2 3 segment will be speciﬁed latter. Therefore, Propositions 22 and 23 must be proved separately for eight subclasses of 51 models according to the signs of x and y : these are those of the walks pictured on 4 4 Figure 17, Subcases I.A, I.B, I.C, II.A, II.B, II.C, II.D and Case III. The following remark gives a geometric interpretation of this classiﬁcation. Remark 24. —Let 1 be 1 if (i, j) ∈S , otherwise 0. Then x > 0 (resp. < 0, =∞) (i,j) 4 if and only if 1 − 41 1 > 0 (resp. < 0, = 0), see Equation (2.2). A symmetric (1,1) (1,−1) (1,0) statement holds for y . As an example, Remark 24 implies that x < 0 if and only if (1, 1) ∈S and (1, −1) ∈S . Case I:y < 0,Subcase I.A:x < 0.— This assumption yields x = x ; x , x = ∞; 4 4 • ◦ • y = y ; y , y = ∞; y , y = ∞. The location of the six points a , a , a , a , b , b is already 1 2 3 4 1 2 described above and is pictured on Figure 12. We ﬁrst show that for all ω ∈{ω :ω =a ,ω ≤ω< ω + ω },wehave 3 y y 2 1 4 r (ω) = ∞. The proof consists in three steps. Step 1. — Let us ﬁrst prove that r (a ) = ∞ and r (a ) = ∞.If |y | < 1, then y 3 y 4 a ∈ (see Section 4 for the deﬁnition of ) and it is immediate from Theorem 3 that 3 y y r (a ) = ∞.If |y |≥ 1, then by Lemma 5 there exists n ∈ Z such that a − nω ∈ ,and y 3 + 3 3 by Equation (5.2)ofTheorem 4, (7.10) r (a ) = r (a − nω ) + f (a − kω ). y 3 y 3 3 y 3 3 k=n ON THE FUNCTIONS COUNTING WALKS IN THE QUARTER PLANE 99 Introducing, for any ω , the set O (ω ) ={ω − ω ,ω − 2ω ,...,ω − n ω }, 0 0 3 0 3 0 ω 3 (7.11) n = inf{ ≥ 0 : ω − ω ∈ }, ω 0 3 we can rewrite (7.10)as (7.12) r (a ) = r (a − n ω ) + f (ω). y 3 y 3 a 3 y ω∈O (a ) In (7.12), the quantity r (a − n ω ) is deﬁned thanks to Theorem 3. It may be inﬁnite, y 3 a 3 but only if y(a − n ω ) =∞. In this case we must have a − n ω = a − ω . But since 3 a 3 3 a 3 1 2 3 3 a + ω = a ,wethenhave (n + 1)ω = ω which is impossible, due to irrationality of 3 3 1 a 3 2 ω /ω .Hence r (a − n ω ) = ∞. Further, we immediately have (see indeed Figure 12) 2 3 y 3 a 3 that a , a , a − ω , a − ω ∈ / O (a ). Moreover, since either b = a ,or b =a 2 4 2 2 4 2 3 1 3 1 3 but then a + ω /2 = b (see again Figure 12), we also have that b , b , b − ω , b − ω ∈ / 3 3 1 1 2 1 2 2 2 O (a ). Finally a = a + ω ∈ / O (a ) and a − ω = a + ω − ω ∈ / O (a ), since ω /ω 3 1 3 3 3 1 2 3 3 2 3 2 3 is irrational. Thus f (a − kω ) = ∞ for any k ∈{1,..., n }. Accordingly, r (a ) = ∞ and y 3 3 a y 3 by the same arguments, r (a ) = ∞. y 4 Step 2. — We now show that r (a − ω ) + f (a − ω ) = ∞ and r (a − ω ) + y 1 2 y 1 2 y 2 2 f (a − ω ) = ∞. By Equation (5.3), r (a − ω ) =−r (a − ω ) + K(0, 0)Q(0, 0) + x(a − y 2 2 y 1 2 x 1 2 1 ω )y(a − ω ) and f (a − ω ) = x(a − ω )[y(a ) − y(a − ω )];hence 2 1 2 y 1 2 1 2 4 1 2 (7.13) r (a − ω ) + f (a − ω ) =−r (a − ω ) + K(0, 0)Q(0, 0) y 1 2 y 1 2 x 1 2 + x(a − ω )y(a ). 1 2 4 It follows from Equation (5.3) and from the ﬁrst step that r (a ) = ∞, since x(a )y(a ) = x 4 4 4 x y = ∞. Then, by (3.5)and (5.4)weget that r (a − ω ) = r (ξ a ) = r (a ) = ∞. x 1 2 x 4 x 4 Furthermore, x(a − ω )y(a ) = x y = ∞. Finally, thanks to (7.13), r (a − ω ) + 1 2 4 y 1 2 f (a − ω ) = ∞ and by the same arguments, r (a − ω ) + f (a − ω ) = ∞. y 1 2 y 2 2 y 2 2 Step 3. — Let us now take any ω in {ω :ω =a ,ω ≤ω< ω + ω }.If 0 3 y y 2 1 4 ω ∈ ,then ω ∈ . Indeed, it is proved in Section 4 that the domain (resp. ), 0 0 y y x 0 1 0 1 y x 2 2 which is bounded by and (resp. and ), is centered around L (resp. L ). y y x x y x 1 1 0 1 0 1 Furthermore, ∈ and ∈ / (resp. ∈ and ∈ / ). It follows that for any x x y y y y x x ω ∈ \ , ω <ω . Then, by Theorem 3, r (ω ) = ∞.If ω ∈ / ,with(7.11)and 0 y 0 y y 0 0 (7.12)wehave r (ω ) = r (ω − n ω ) + f (ω). y 0 y 0 ω 3 y ω∈O (ω ) 0 100 I. KURKOVA, K. RASCHEL For the same reasons as in the ﬁrst step, we have that a , a , a − ω , a − ω ∈ / O (ω ). 2 4 2 2 4 2 0 If ω < a , for obvious reasons O (ω ) cannot contain a .If a ≤ω <ω + ω , 0 3 0 3 3 0 y 2 it can neither contain a , since ω + ω −a = ω , and hence ω − ω < a .If 3 y 2 3 3 0 3 3 b = a ,or b =a and ω < b , it cannot contain b .If b =a and b ≤ 1 3 1 3 0 1 1 1 3 1 ω <ω + ω ,then ω −b ≤ ω + ω −b = ω /2 <ω ,and b ∈ / O (ω ). 0 y 2 0 1 y 2 1 3 3 1 0 4 4 If a − ω ∈ / O (ω ),thenwehave r (ω − n ω ) = ∞ and f (ω − kω ) = ∞ for 1 2 0 y 0 ω 3 y 0 3 all k ∈{1,..., n },sothat r (ω ) = ∞ by (7.12). ω y 0 If a − ω ∈ O (ω ),thenfor some j ∈{1,... n },wehave ω − jω = a − ω . 1 2 0 ω 0 3 1 2 j+1 Then r (ω − n ω ) + f (ω − kω ) = r (a − ω ) and thus by (7.12), y 0 ω 3 y 0 3 y 1 2 0 k=n r (ω ) = r (a − ω ) + f (a − ω ) + f (ω − kω ). y 0 y 1 2 y 1 2 y 0 3 k=j−1 Theﬁrsttermhereisﬁnitebythe secondstepand f (ω − kω ) = ∞ for k ∈{1,..., j − 1} y 0 3 by all properties said above, so that r (ω ) = ∞. y 0 So far we have proved that for all ω ∈{ω :ω =a ,ω ≤ω< ω + ω }, 3 y y 2 1 4 r (ω) = ∞. In the same way, we obtain that r (ω) = ∞ for ω ∈{ω :ω =a ,ω ≤ y y 4 y ω< ω + ω }. y 2 Since by (3.5), η{ω :ω =a ,ω ≤ω< ω + ω } 3 y y 2 1 4 ={ω :ω =a ,ω < ω ≤ ω }, 4 y y 4 1 η{ω :ω =a ,ω ≤ω< ω + ω } 4 y y 2 1 4 ={ω :ω =a ,ω < ω ≤ ω }, 3 y y 4 1 Equation (5.4) implies that r (ω) = ∞ on the segments {ω ∈ :ω = a , a },exceptfor y y 3 4 their ends a , a . The segments {ω :ω = a , a ,ω ≤ω ≤ ω + ω } do not contain 1 2 3 4 x x 2 1 4 any point where y(ω) =∞. It follows from Equation (5.3)that r (ω) = ∞ on these seg- ments except for points where x(ω) =∞ if they exist. This last fact happens if and only if b =a and only at the ends b , b of the segments. 1 3 1 2 If b = a , we can show exactly in the same way that r (ω) = ∞ on the two seg- 1 3 y ments {ω ∈ :ω = b , b } and that r (ω) = ∞ on the segments {ω :ω = b , b ,ω ≤ y 1 2 x 1 2 x ω ≤ ω + ω },exceptfor theirends b , b . This concludes the proof of Proposition 22. x 2 1 2 We proceed with the proof of Proposition 23. Let us verify the assumptions of Lemma 21 for ω = a , a , b , b .Wehaveprovedthat r (a ), r (a ), r (b ), r (b ) = ∞, 0 3 4 1 2 y 3 y 4 y 1 y 2 a a , a a and that the pairs {a , a } and {a , a } are not ordered. Let us now show 3 1 1 4 1 2 1 3 2 4 that for any k ∈{3, 4} and ∈{1, 2},itisimpossibletohave a ∼ b .If b = a , a , this k 3 4 is obvious. If b =a ,thenitisenoughtonotethat b − a = ω /2and a − b = ω /2 3 3 3 1 1 3 (see Figure 12). From the irrationality of ω /ω , it follows that b a , a and in the same 2 3 1 3 way b a , a . Then there is no other ω ∈ except for a (resp. a )suchthat a ω 2 4 y 1 2 3 ON THE FUNCTIONS COUNTING WALKS IN THE QUARTER PLANE 101 (resp. a ω)and f (ω) =∞. There is no ω ∈ such that b ω and f (ω) =∞, 4 y y y = 1, 2. Hence, Lemma 21 could be applied to any of four points ω = a , a , b , b 0 3 4 1 2 if the assumption (B) of this lemma is satisﬁed for these points. It is then immedi- ate that lim f (ω) = lim x(ω)[y(ξω) − y(ω)]=∞, ∈{1, 2}, since x(ω) →∞ ω→b y ω→b ◦ • and the other term converges to ±[y − y ] = 0. Let us verify that lim {f (ω) + ω→a y f (ω + ω )}=∞.Wehave y 3 lim f (ω) + f (ω + ω ) = lim x(ω) y(ξω) − y(ω) y y 3 ω→a ω→a 3 3 + x( η ξω) y(ξ η ξω) − y( η ξω) = lim x( η ξω)y(ξ η ξω) − x(ω)y(ω) ω→a + lim x(ω)y(ξω) − x( η ξω)y( η ξω) . ω→a The ﬁrst term above converges to x y − x y .By(3.7) the second term equals the limit of the product y(ξω)[x(ξω) − x( η ξω)].If ω → a ,then ξω → a − ω so that 3 2 2 the ﬁrst term in the product converges to y(a − ω ) = y(a ) =∞. The second term 2 2 2 of this product converges to x(a − ω ) − x(a ) = x − x which is different from 0 2 2 1 as x = x .Thenassumption (B)issatisﬁedfor ω = a and in the same way for 0 3 ω = a . Lemma 21 applies to any of the four points ω = a , a , b , b .But by (3.7), 0 4 0 3 4 1 2 I (a ) =I (a ) =I (a ) =I (a ),I (a ) =I (a ) =I (a ) =I (a ),I (b ) =I (b ), x 3 x 4 x 1 x 2 y 3 y 4 y 1 y 2 x 1 x 2 I (b ) =I (b ) so that poles of x → Q(x, 0) are dense on the curvesI (a ) andI (b ) y 1 y 2 x 1 x 1 and those of y → Q(0, y) are dense on the curvesI (a ) andI (b ).Proposition 23 is y 1 y 1 proved. Case I:y < 0,Subcase I.B:x =∞.— This assumption implies that x = x ; 4 4 ◦ • x , x = ∞; y = y = ∞; y , y = ∞. The points a , a , a , a are located as in the previous case, see Figure 12. Conse- 1 2 3 4 quently we have the following facts: r (ω) = ∞ on the segments {ω ∈ :ω = a , a }, y y 3 4 except for their ends ω = a , a ; r (ω) = ∞ on the segments {ω :ω = a , a ,ω ≤ω ≤ 1 2 x 3 4 x ω + ω }. Lemma 21 applies to ω = a , a as in the previous case, as x = x . Then the x 2 0 3 4 set of poles of all branches of x → Q(x, 0) (resp. y → Q(0, y)) is dense onI (a ) (resp. x 1 I (a ))whereI (a ) =I (a ) =I (a ) =I (a ) (resp.I (a ) =I (a ) =I (a ) =I (a )). y 1 x 1 x 2 x 3 x 4 y 1 y 2 y 3 y 4 Since x =∞,wehavethat b = b = ω + ω .Takeany ω with ω = 0such 4 1 2 x 2 0 0 that ω ≤ω ≤ ω + ω .Then y(ω ) = ∞. Let us show that r (ω ) = ∞.If ω ∈ , y 0 y 2 0 y 0 0 1 4 then by the same reasons as in Subcase I.A ω ∈ and r (ω ) = ∞.If ω ∈ / ,consider 0 y y 0 0 the set O (ω ) deﬁned as in (7.11)and (7.12). Clearly b − ω = ω ∈ / O (ω ). Since 0 1 2 x 0 b + ω /2 = ω + ω ,wehave ω − ω ≤ ω + ω − ω < b and then b ∈ / O (ω ). 1 3 y 2 0 3 y 2 3 1 1 0 4 4 Hence O (ω ) does not contain any point where y(ω) or f (ω) is inﬁnite. Thus by (7.12), 0 y r (ω ) = ∞. y 0 102 I. KURKOVA, K. RASCHEL FIG. 13. — Location of b , b if x > 0, Subcases I.C and II.C 1 2 4 We have η{ω :ω = 0,ω ≤ω ≤ ω + ω }={ω :ω = ω ,ω ≤ω ≤ ω }. y 0 y 2 1 y 0 y 1 4 4 1 Then by (5.4)and (5.5), we get that r (ω) = ∞ for all ω ∈ with ω = 0. The segment y y {ω :ω = 0,ω ≤ ω ≤ ω + ω } does not contain any point with y(ω) =∞.By(5.3) x x 2 1 4 this gives r (ω) = ∞ for all ω on this segment except for the points where x(ω) =∞ (that is only at ω = ω + ω = b ), and this concludes the proof of Proposition 22. x 2 1 We have proved in particular that r (ω ) = ∞ for ω = b . Furthermore, there is y 0 0 1 no ω ∈ such that b ω and f (ω) =∞. Finally y 1 y (7.14) lim f (ω) = lim x(ω) y(ξω) − y(ω) ω→b ω→b 1 1 2 1/2 [b(x(ω)) − 4a(x(ω))c(x(ω))] = lim x(ω) , ω→b 1 a(x(ω)) where x(ω) →∞ as x → b . For all models in Subcase I.B deg a(x) = 2, deg b(x) = 1 1/2 and deg c(x) = 1, so that (7.14)isofthe order O(|x(ω)| ). Thus lim f (ω) =∞.By ω→b y Lemma 21 with ω = b , the poles of x → Q(x, 0) and those of y → Q(0, y) are dense on 0 1 I (b ) =I (b ) andI (b ) =I (b ), respectively. They are the intervals of the real line x 1 x 2 y 1 y 2 claimed in Theorem 17(iii). Proposition 23 is proved. Case I:y < 0,Subcase I.C:x > 0.— The statements and results about a , a , a , a 4 4 1 2 3 4 are the same as in Subcases I.A and I.B, see Figure 12 for their location. We now locate b , b . By deﬁnition (see Section 2), the values y , y , y , y are the 1 2 1 2 3 4 roots of 1/2 1/2 d( y) = b( y) − 2 a( y) c( y) b( y) + 2 a( y) c( y) = 0. 1/2 Hence, for two of these roots b( y) =−2[ a( y) c( y)] and then X( y) ≥ 0 (see (2.4)), and for 1/2 the two others b( y) = 2[ a( y) c( y)] and then X( y) ≤ 0. But X( y ) and X( y ) are on the 2 3 segment [x , x ]⊂ ]0, ∞[.Thus X( y ) ≤ 0and X( y ) ≤ 0. Since x(b ) = x(b − ω ) =∞, 2 3 1 4 1 1 2 x = x(ω )> 0and X( y ) = x(ω )< 0, it follows that b − ω ∈]ω ,ω [,insuchaway 4 x 4 y 1 2 x y 4 4 4 4 that b ∈]ω + ω ,ω + ω [. Also, b = ξ(b − ω ) − ω = 2(ω + ω ) − b is symmetric 1 x 2 y 2 2 1 2 1 x 2 1 4 4 4 to b w.r.t. ω + ω . Since x(ω ) = X( y ) ≤ 0and x(ω + ω ) = x > 0, it follows that 1 x 2 y 1 x 2 4 4 1 4 ω < b <ω + ω , see Figure 13. y 2 x 2 1 4 Nowweshowthatfor any ω with ω = 0and ω ≤ω ≤ ω + ω , r (ω ) = ∞. 0 0 y 0 y 2 y 0 1 4 Note that y(ω ) = ∞.If ω ∈ , by the same arguments as in Subcase I.A, ω ∈ and 0 0 0 y r (ω ) = ∞.If ω ∈ / ,thenconsider O (ω ) with the notation (7.11). y 0 0 0 Note that b − ω ∈ / .For this,itisenoughtoprove that b − ω ∈ / and that 1 2 1 2 x b − ω ∈ / .First, b − ω ∈ / , since x(b ) = x(b − ω ) =∞. Furthermore, is 1 2 y 1 2 x 1 1 2 y ON THE FUNCTIONS COUNTING WALKS IN THE QUARTER PLANE 103 FIG. 14. — Location of a , a , a , a if y > 0, (x , Y(x )) = (∞, ∞), Subcases II.A, II.B and II.C 1 2 3 4 4 4 4 centered w.r.t. L ,and ω ∈ / (since |y | > 1). Hence the point b − ω <ω cannot be y y 4 1 2 y y 4 4 in . Since ∩{ω ∈ C :ω = 0} is an open interval containing ω , and since b − ω < y 1 2 ω ≤ ω , it follows that b − ω <ω − n ω (see (7.11)), so that b − ω ∈ / O (ω ). y 0 1 2 0 ω 3 1 2 0 1 0 Obviously b − ω ∈ / O (ω ). Furthermore, since ω + ω − b = ω /2 + ω + ω − b < 2 2 0 y 2 2 3 x 2 2 4 4 ω /2 + ω /2 = ω , it follows that ω − ω < b < b for any such ω .Hence b , b ∈ / 3 3 3 0 3 2 1 0 1 2 O (ω ).Thusfor any ω ∈ O (ω ), y(ω) = ∞ and f (ω) = ∞.By(7.12), r (ω ) = ∞. 0 0 y y 0 This implies, exactly as in Subcase I.B—by (5.4)and (5.5)—, that r (ω) = ∞ for all ω ∈ such that ω = 0. By (5.3) this gives r (ω) = ∞ for all ω with ω = 0and ω ≤ω ≤ y x x ω + ω , except for points ω where x(ω) =∞ (that happens for ω = b only), and this x 2 2 concludes the proof of Proposition 22. ◦ • In particular, we proved that r (b ) = ∞ and also r (b ) = ∞. Since y = y ,we y 1 y 2 have lim f (ω) =∞ and also lim f (ω) =∞ by the same arguments as in Sub- ω→b y ω→b y 1 2 case I.A. If b and b are not ordered, Lemma 21 applies to both of these points. If b b 1 2 1 2 (resp. b b ), then there is no ω ∈ such that ω = b (resp. ω = b ), f (ω) =∞ and 2 1 y 2 1 y b ω (resp. b ω). Hence Lemma 21 applies to ω = b (resp. ω = b ). Thus the set 2 1 0 2 0 1 of poles of all branches of x → Q(x, 0) (resp. y → Q(0, y)) is dense on the curvesI (b ) x 2 andI (b ) (resp.I (b ) andI (b )). We conclude the proof of Proposition 23 with the y 2 x 1 y 1 observation thatI (b ) =I (b ), whileI (b ) =I (b ) are intervals of the real line as x 2 x 1 y 2 y 1 claimed in Theorem 17(iii). Case II:y > 0, location of a , a , a , a .— We ﬁrst exclude Subcase II.D where x = 4 1 2 3 4 4 ∞ and Y(x ) =∞, and we locate the points a , a , a , a . In this case we have x = x . 4 1 2 3 4 Note that x , x , x , x are the roots of the equation 1 2 3 4 1/2 1/2 d(x) = b(x) − a(x)c(x) b(x) + a(x)c(x) = 0. 1/2 Hence for two of these roots b(x) =−2[a(x)c(x)] and then Y(x) ≥ 0 (see (2.5)), and 1/2 for two others b(x) = 2[a(x)c(x)] and then Y(x) ≤ 0. But Y(x ) and Y(x ) are on the 2 3 segment [y , y ]⊂]0, ∞[.Hence Y(x ) ≤ 0and Y(x ) ≤ 0. If in addition x =∞,then 2 3 1 4 4 Y(x ) equals 0 or ∞;notealsothatif Y(x ) =∞ then necessarily x =∞.But thecase 4 4 4 when Y(x ) =∞ and x =∞ is excluded from our consideration at this moment. It 4 4 follows that ∞∈[y , Y(x )[, and in fact ∞∈ ]y , Y(x )[, since the case y =∞ is excluded 4 4 4 4 4 from Case II. It follows from the above considerations that a ∈]ω + ω ,ω + ω [, see Fig- 1 x 2 y 2 4 4 ure 14. In particular, we have a − ω ∈]ω ,ω [ and a = ηa − ω =−(a − ω ) + 2ω . 1 2 x y 2 1 1 1 2 y 4 4 4 104 I. KURKOVA, K. RASCHEL This means that a − ω and a are symmetric w.r.t. ω . Furthermore a ∈]ω ,ω [, 1 2 2 y 2 y y 4 4 1 but since y > 0and Y(x ) ≤ 0, we have a ∈]ω ,ω [. We must put a = ξ a − ω = 4 1 2 y x 3 2 1 4 1 −a + 2ω , in such a way that the points a and a are symmetric w.r.t. ω .Fi- 2 x 2 3 x 1 1 nally, a = ξ(a − ω ) − ω =−(a − ω ) + 2ω = a + a − (a − ω ).Notethat 4 1 2 1 1 2 x 3 2 1 2 ω + ω − a = a − ω − ω > 0. Furthermore, a − a = ω and a + ω − a = ω ,so x 2 4 1 2 x 1 3 3 2 2 4 3 4 4 that a a and a a . 3 1 4 2 ◦ • ◦ • Case II:y > 0,Subcase II.A:x < 0.— In this case we have y = y ; y , y = ∞; 4 4 x , x = ∞; y , y = ∞. The points b , b are located as in Subcase I.A, see Figure 12. 1 2 Further, we can show as in Subcase I.A that r (ω) = ∞ on the segments {ω :ω = b , b ,ω ≤ω ≤ ω + ω }, and we deduce that r (ω) = ∞ on {ω :ω = b , b ,ω ≤ 1 2 y y 2 x 1 2 x 1 4 1 ω ≤ ω + ω } except for their ends b , b . Consequently, the set of poles of all branches x 2 1 2 of x → Q(x, 0) (resp. y → Q(0, y)) is dense on the curveI (b ) =I (b ) (resp.I (b ) = x 1 x 2 y 1 I (b )), as claimed in Proposition 23. y 2 Consider now r (ω) and r (ω) for ω with ω = 0. See Figure 14 for the location of x y the points a , a , a , a . 1 2 3 4 We ﬁrst prove that for (7.15) r (ω ) = ∞, ∀ω ∈{ω :ω = 0,ω ≤ ω ≤ ω + ω }\{a }. y 0 0 y 0 y 2 1 1 4 The proof consists in three steps. Step 1. — We prove that r (a ) = ∞ and r (a ) = ∞. y 3 y 4 If a ∈ , then necessarily r (a ) = ∞, since x , y = ∞. Otherwise |x |, |y |≥ 1, 3 y 3 and then a = ξ a − ω ∈ / . Since a <ω < a , ω ∈ and a , a ∈ / , it follows that 2 3 1 2 x 3 x 2 3 1 1 any ω ∈ O (a ) must be in ]a , a [,hence f (ω) = ∞, x(ω) = ∞ and y(ω) = ∞.Thusby 3 2 3 y (7.12), r (a ) = ∞. y 3 We now show that r (a ) = ∞. Suppose ﬁrst that a − ω ∈ . Then, since y 4 1 2 a − ω ∈ / ,wehave a − ω ∈ ,sothat r (a − ω ) = ∞ by Theorem 3.Thenby 1 2 y 1 2 x x 1 2 Equations (5.4)and (5.5), r (a ) = r (ξ(a − ω ) − ω ) = r (a ) = ∞. Since x , y = ∞,we x 4 x 1 2 1 x 1 also have r (a ) = ∞ by (5.3). Assume now that a − ω ∈ / . Since a − ω <ω < a , y 4 1 2 1 2 x 4 then a − ω ∈ / O (a ). Furthermore a ∈ / O (a ) as a + ω = a + ω and ω /ω is 1 2 4 2 4 4 3 2 2 3 2 irrational. Finally a − a < a − a = ω ,sothat a ∈ / O (a ). It follows that for any 4 3 1 3 3 3 4 ω ∈ O (a ),wehave f (ω) = ∞ and y(ω) = ∞.Hence r (a ) = ∞. 4 y y 4 Step 2. — We prove that r (a ) + f (a ) = ∞ and that r (a − ω ) + f (a − ω ) = ∞. y 2 y 2 y 1 2 y 1 2 By Equation (5.3) (7.16) r (a ) + f (a ) =−r (a ) + K(0, 0)Q(0, 0) + x(a )y(a ) y 2 y 2 x 2 2 2 + x(a ) y(ξ a ) − y(a ) 2 2 2 =−r (a ) + K(0, 0)Q(0, 0) + x(a )y(ξ a ). x 2 2 2 ON THE FUNCTIONS COUNTING WALKS IN THE QUARTER PLANE 105 Since r (a ) = ∞ and since x , y = ∞, it follows from Equation (5.3)that r (a ) = ∞. y 3 x 3 Then by (5.4)and (5.5), r (a ) = r (ξ a − ω ) = r (a ) = ∞. Since x(a ) = x = ∞ and x 2 x 3 1 x 3 2 y(ξ a ) = y = ∞,(7.16) is ﬁnite. By completely analogous arguments we obtain that r (a − ω ) + f (a − ω ) = ∞. y 1 2 y 1 2 Step 3. — Let us now show (7.15). If ω ∈ , then by the same arguments as in Subcase I.A ω ∈ and then r (ω ) = ∞ by Theorem 3. Otherwise, consider O (ω ) 0 y y 0 0 deﬁned in (7.11). Note that (7.17) ω/ ∈ O (ω ), ∀ω ∈]ω + ω − ω /2,ω + ω ], 0 x 2 3 y 2 4 4 since in this case ω − ω <ω. In particular, a ∈ / O (ω ). Furthermore, a ∈ O (ω ) 0 3 4 0 3 0 implies that ω = a + ω for some ≥ 1. But a + ω = a = ω and a + ω >ω + ω 0 3 3 3 3 1 0 3 3 y 2 for any ≥ 2. Hence a ∈ / O (ω ). Since a − (a − ω )<ω ,itisimpossiblethatboth a 3 0 2 1 2 3 2 and a − ω belong to O (ω ). If none of them belongs to O (ω ),then y(ω) = ∞ and 1 2 0 0 f (ω) = ∞ for any O (ω ),and then r (ω ) = ∞. Suppose, e.g., that a ∈ O (ω ).Then y 0 y 0 2 0 for some ≥ 1, ω = a + ω ,and by (7.12), r (ω ) = r (a ) + f (a ) + f (ω − 0 2 3 y 0 y 2 y 2 y 0 k=−1 kω ).But r (a ) + f (a ) = ∞ by the second step, and obviously f (ω − kω ) = ∞ 3 y 2 y 2 0 3 k=−1 by all facts said above, so that r (ω ) = ∞. The reasoning is the same if a − ω ∈ O (ω ). y 0 1 2 0 This concludes the proof of (7.15). Applying (5.4)and (5.5) exactly as in Subcase I.B, we now reach the conclusion that r (ω ) = ∞ for all ω = a = ηa with ω = 0and ω ≤ω ≤ ω as well. Next, y 0 0 2 1 0 y 0 y 4 1 exactly as in Subcase I.B, thanks to (5.3), we derive that r (ω) = ∞ for all ω such that x 0 ω = 0and ω ≤ω ≤ ω + ω except possibly for points where x(ω) =∞. But these x x 2 1 4 points are absent on this segment in this case. This concludes the proof of Proposition 22. For the same reason as in Subcase I.A, the fact that x = x gives lim {f (ω) + ω→a y f (ω + ω )}=∞ and lim {f (ω) + f (ω + ω )}=∞. Further, a a and a a .If y 3 ω→a y y 3 3 1 1 4 1 2 in addition a a , then due to the fact that a + ω = a we have a a a a . 3 4 3 3 1 3 1 1 4 1 2 Thereisnopoint ω ∈ \{a } such that a ω and f (ω) =∞. Lemma 21 applies to y 2 4 y ω = a .If a a ,thenalso a a a a and this lemma applies to ω = a .If 0 4 4 3 4 1 2 3 1 1 0 3 a a , Lemma 21 can be applied to both a and a . SinceI (a ) =I (a ) =I (a ) = 3 4 3 4 x 3 x 4 x 1 I (a ) =[x , x ] andI (a ) =I (a ) =I (a ) =I (a ) = R\]y , y [, the set of poles of x 2 1 4 y 3 y 4 y 1 y 2 4 1 x → Q(x, 0) (resp. y → Q(0, y)) is dense on the announced intervals and Proposition 23 is proved. ◦ • Case II:y > 0,Subcase II.B:x > 0 and exactly one of y , y is ∞.— Assume, e.g., 4 4 ◦ • • that y =∞ and y = ∞.Then x =∞; y = y = ∞; x = x =∞; y = ∞. It follows that b = a and b = ξ b − ω − ω = a , while a , a , a , a arepicturedaspreviously, in 1 1 2 1 1 2 4 1 2 3 4 Subcase II.A, see Figure 14. We ﬁrst derive (7.15). By the same reasoning as in Subcase II.A, we reach the conclusion that r (a ) = ∞. Let us note that in this case a − ω ∈ / ,as x(a − ω ), y(a − y 3 1 2 1 2 1 106 I. KURKOVA, K. RASCHEL ω ) =∞. Next, we derive as in Subcase II.A that r (a ) = ∞ and that r (a ) + f (a ) = ∞, 2 y 4 y 2 y 2 since x y = ∞. Finally, again by the same arguments as in Subcase II.A, we conclude that for any ω = a with ω = 0and ω ≤ω ≤ ω + ω , the orbit O (ω ) does 0 1 0 y 0 y 2 0 1 4 not contain a and a . The orbit can neither contain a − ω , since a − ω ∈ / and 3 4 1 2 1 2 a − ω <ω <ω ,where ω ∈ . Since r (a ) + f (a ) = ∞,thenasinSubcaseII.Awe 1 2 y 0 y y 2 y 2 1 1 have r (ω ) = ∞. y 0 It follows from (5.4)and (5.5)that r (ω ) = ∞ for any ω such that ω = 0and y 0 0 0 ω ≤ ω ≤ ω + ω ,exceptfor a and ηa − ω = a .By(5.3), r (ω ) = ∞ for any ω y 0 y 2 1 1 1 2 x 0 0 4 4 such that ω = 0and ω ≤ ω ≤ ω + ω , except for points ω where x(ω ) =∞ (this 0 x 0 x 2 0 0 1 4 is ω = a in this case). This ﬁnishes the proof of Proposition 22 in this case. 0 4 Since x = x , the same reasoning as in Subcase I.A gives lim {f (ω) + f (ω + ω→a y y ω )}=∞ and lim {f (ω) + f (ω + ω )}=∞. The rest of the proof of Proposition 23 3 ω→a y y 3 via the use of Lemma 21 with ω = a if a a or with ω = a if a a ,orwith 0 3 4 3 0 4 3 4 indifferent choice of a or a if a a , is the same as in Subcase II.A. 3 4 3 4 ◦ • Case II:y > 0,Subcase II.C:x > 0 and y , y = ∞,or x =∞ and Y(x ) = ∞.— In 4 4 4 4 ◦ • this case, we have y , y = ∞; x , x = ∞; x = x , y , y = ∞. The points a , a , a , a are pictured as in Subcases II.A and II.B, see Figure 14, 1 2 3 4 while b , b are pictured as in Subcase I.B (where b = b = ω ) or I.C, see Figure 13. 1 2 1 2 x +ω 4 2 They are such that b = a and b = a . In particular, b , b ∈]ω + ω − ω ,ω + ω [ and 1 1 2 4 1 2 x 2 3 x 2 4 4 are symmetric w.r.t. ω + ω ; b = b is in the middle of this interval if and only if x =∞. x 2 1 2 4 Hence for any ω with ω = 0and ω ≤ω ≤ ω + ω ,wehave ω − ω < b < b ,so 0 0 y 0 y 2 0 3 2 1 1 4 that b , b ∈ / O (ω ). Furthermore, by the same arguments as in Subcase I.C, b − ω ∈ / 1 2 0 1 2 O (ω ).Hence (7.15)provedinSubcaseII.Astays validinthis case andby(5.4)and (5.5), r (ω) = ∞ for all ω with ω = 0and ω ≤ω ≤ ω + ω ,exceptfor ω = a , a . y y y 2 1 2 4 4 By the identity (5.3), r (ω) = ∞ for all ω with ω = 0and ω ≤ω ≤ ω + ω ,except x x x 2 1 4 for points ω where x(ω) =∞,namely ω = b . This concludes the proof of Proposition 22 and proves in particular that r (ω ) = ∞ for any ω ∈{a , a , b , b }. y 0 0 3 4 1 2 Using x = x , we verify as in Subcase I.A that lim {f (ω) + f (ω + ω )}=∞ ω→a y y 3 ◦ • and lim {f (ω) + f (ω + ω )}=∞.If x > 0, since y = y ,weverifyasinSubcaseI.A ω→a y y 3 4 that lim f (ω) =∞ and lim f (ω) =∞.If x =∞,then b = b and we verify as ω→b y ω→b y 4 1 2 1 2 in Subcase I.B that lim f (ω) =∞.If a , a , b , b are ordered (e.g., a b a ω→b y 3 4 1 2 3 1 4 b , then immediately a a b a a b ), thereisa maximal point in the sense 2 3 1 1 1 4 1 2 2 of this order. If the maximal element is b for some ∈{1, 2}, then there is no ω ∈ with f (ω) =∞ such that b ω. If the maximal element is a (resp. a ), then there is no y 3 4 ω ∈ except for a (resp. a )with f (ω) =∞ such that a ω (resp. a ω). Lemma 21 y 1 2 y 3 4 applies with ω equal this maximal element since all assumptions (A), (B) and (C) are satisﬁed. If a , a , b , b are not all ordered, then it is enough to apply Lemma 21 to 3 4 1 2 the maximal element of any ordered subset. FinallyI (ω ) = R\]x , x [ andI (ω ) = x 0 1 4 y 0 R\]y , y [ for any ω ∈{a , a , b , b }, hence the set of poles of x → Q(x, 0) (resp. y → 4 1 0 3 4 1 2 Q(0, y)) is dense on the announced intervals. Proposition 23 is proved. ON THE FUNCTIONS COUNTING WALKS IN THE QUARTER PLANE 107 FIG. 15. — Location of a , a , a , a if y > 0, x =∞,Y(x ) =∞, i.e., Subcase II.D 1 2 3 4 4 4 4 Case II:y > 0,Subcase II.D:x =∞ and Y(x ) =∞.— In this case x =∞; y = 4 4 4 ∞; x , y = ∞. We have a = ω + ω ,and a = ηa − ω = ω + ω is symmetric to a − ω 1 x 2 2 1 1 x 3 1 2 4 4 w.r.t. ω . Further, since y > 0and Y(x ) ≤ 0, a ∈]ω ,ω [.Then a = ξ a − ω = y 4 1 2 y x 3 2 1 4 4 1 ω + ω − ω is symmetric to a w.r.t. ω . Finally, a = ω + ω = a , b = b = a , x 2 3 2 x 4 x 2 1 1 2 1 4 1 4 a + ω = a and a + ω = a + ω ,sothat a a a , see Figure 15. 3 3 1 1 3 2 2 3 1 1 1 2 We prove (7.15). For this purpose, we ﬁrst show that r (a ) = ∞.If a ∈ , x , y = y 3 3 ∞, then by Theorem 3, r (a ) = ∞ If a ∈ / ,consider O (ω ). Since a + 2ω = a + ω , y 3 3 0 3 3 2 2 a ∈ / O (ω ) by the irrationality of ω /ω . Obviously a − ω = ω ∈ / and then it is not 2 0 2 3 1 2 x in O (ω ).Hence r (a ) = ∞.Nextweprove that r (a ) + f (a ) exactly as in Subcase II.A 0 y 3 y 2 y 2 by using that x y = ∞. For any ω = a with ω = 0and ω ≤ ω ≤ ω + ω , the orbit O (ω ) 0 1 0 y 0 y 2 0 1 4 cannot contain a , since a + ω = a and a + 2ω >ω . It can neither contain 3 3 3 1 3 3 0 a , since ω − ω < a , nor obviously a − ω = ω . If it does not contain a ,then 1 0 3 1 1 2 x 2 by (7.12), r (ω ) = ∞. If it does, then exactly as in Subcase II.A, using r (a ) + y 0 y 2 f (a ) = ∞,weprove that r (ω ) = ∞ as well. This ﬁnishes the proof of (7.15). y 2 y 0 By Equations (5.4), (5.5)and (5.3), we derive as in Subcase II.A that r (ω ) = ∞ x 0 for all ω ∈{ω :ω = 0,ω ≤ ω ≤ ω + ω } except for ω where x(ω ) =∞,thatis 0 x 0 x 2 0 0 1 4 for a . This ﬁnishes the proof of Proposition 22 in this case. To prove Proposition 23, we would like to apply Lemma 21 with ω = a .Wehave 0 3 shown that r (a ) = ∞, a a = a = b = b a , so that there is no ω ∈ \{a , a } y 3 3 1 1 4 1 2 2 y 1 2 such that a ω and f (ω) =∞. It remains to verify assumption (B) of Lemma 21 for 3 y ω = a , that is that f (ω) + f (ω + ω ) + f (ω + 2ω ) converges to inﬁnity if ω → a .The 0 3 y y 3 y 3 3 last quantity is the sum of x(ω) y(ξω) − y(ω) + x( η ξω) y(ξ η ξω) − y( η ξω) + x( η ξ η ξω) y(ξ η ξ η ξω) − y( ηξ η ξω) , which equals (7.18) x( η ξ η ξω)y(ξ η ξ η ξω) − x(ω)y(ω) + x( η ξω) y(ξ η ξω) − y( η ξω) + x(ω)y(ξω) − x( η ξ η ξω)y(ξ ηξω) where we used (3.7). If ω → a , then the ﬁrst term in this sum converges to x y − x y = 0. Next, η ξω → a ,sothat x( η ξω) →∞. We can also compute the values of y(ξω) and 2 1/2 y(ξ η ξω) as (−b(x) ±[b(x) − 4a(x)c(x)] )/(2a(x)) with x = x( η ξω). Since for all of 108 I. KURKOVA, K. RASCHEL FIG. 16. — Location of a , a , a , a , b , b if y =∞, Case III 1 2 3 4 1 2 4 the 9 models composing Subcase II.D, deg a = deg b = 1and deg c = 2, then y(ξω) and 1/2 y(ξ η ξω) are of order O(|x( η ξω)| ), and their difference |y(ξω) − y(ξ η ξω)| is not 1/2 smaller than O(|x( η ξω)| ) as ω → a . Finally, x(ω), x( η ξ η ξω) → x = ∞ as ω → a . 3 3 Then as ω → a in the sum (7.18) the second term is of the order not smaller than 3/2 1/2 O(|x( η ξω)| ) while the ﬁrst vanishes and the third has the order O(|x( η ξω)| ). This proves the assumption (B) of Lemma 21 for ω = a . By this lemma the poles of x → 0 3 Q(x, 0) and y → Q(0, y) are dense on the intervals of the real line, as announced in the proposition. Case III:y =∞.— It remains here exactly one case to study, see Figure 17.It ◦ • is such that y =∞, x =∞ and X( y ) = ∞.Then x = x = ∞; y = y = ∞; y = 4 4 4 y = ∞. The points b = b = ω + ω are located as in Subcase I.B, a = a = ω + ω 1 2 x 2 1 2 y 2 4 4 and a = a = ω + ω − ω . In particular, a + ω /2 = b , b + ω /2 = a , see Figure 16. 3 4 y 2 3 3 3 1 1 3 1 We start by showing that r (a ) = ∞.If a ∈ , this is true thanks to (5.3)and y 3 3 since x , y = ∞.If a ∈ / , consider the orbit O (a ). It cannot contain a − ω since 3 3 1 2 a + ω = a , neither b ,nor b − ω = ω . It follows that r (a ) = ∞. 3 3 1 1 1 2 x y 3 Since x , y = ∞, it follows from Equations (5.3), (5.4)and (5.5)that r (a − ω ) = x 1 2 r (ξ(a − ω )) = r (ξ(a − ω ) − ω ) = r (a ) = ∞. Then, by (5.3), r (a − ω ) + f (a − x 1 2 x 1 2 1 x 3 y 1 2 y 1 ω ) =−r (a − ω ) + K(0, 0)Q(0, 0) + x(a − ω )y(ξ(a − ω )) = ∞. 2 x 1 2 1 2 1 2 Take any ω with ω = 0and ω ≤ ω <ω + ω .If ω ∈ , then by the same 0 0 y 0 y 2 0 1 4 arguments as in Subcase I.A, ω ∈ ,sothat r (ω ) = ∞.Otherwise,wenoticethat 0 y y 0 ω − ω < a , so that no point—and in particular b —of [ω + ω − ω ,ω + ω [ belongs 0 3 3 1 y 2 3 y 2 4 4 to O (ω ). Clearly b − ω = ω ∈ / O (ω ). Since either a − ω ∈ / O (ω ) or a − ω ∈ 0 1 2 x 0 1 2 0 1 2 O (ω ) but r (a − ω ) + f (a − ω ) = ∞, and by the same reasoning as in Subcase I.A, 0 y 1 2 y 1 2 we derive that r (ω ) = ∞. The rest of the proof of Proposition 22 in this case goes along y 0 thesamelinesasinCaseII. Now note that b is not ordered with a and a . Indeed, since b + ω /2 = a 1 1 3 1 3 1 and b − ω /2 = a , this would contradict the irrationality of ω /ω . Then there is no 1 3 3 2 3 ω ∈ such that b ω and f (ω) =∞.Wealsohave f (ω) = x(ω)[y(ξω) − y(ω)]= y 1 y y 2 1/2 x(ω)[b (x(ω)) − 4a(x(ω))c(x(ω))] /a(x(ω)).If ω → b ,then x(ω) →∞ and since deg a = 2and deg b = deg c = 1, we have f (ω) →∞. Lemma 21 applies with ω = b y 0 1 and proves Proposition 23. 7.4. Asymptotic of ω (ω)/ω (ω) as z → 0.— It remains to prove the following re- 2 3 sult announced at the beginning of Section 7. ON THE FUNCTIONS COUNTING WALKS IN THE QUARTER PLANE 109 Proposition 25. — For all 51 non-singular walks having an inﬁnite group, there exist a rational constant L > 0 and a constant L = 0 such that (7.19) ω /ω = L + L/ ln(z) + O 1/ ln(z) . 2 3 Proof.—In ordertoprove (7.19), we shall use expressions of the periods ω and ω different from that given in (3.1)and (3.2). To that purpose, deﬁne the complete and incomplete elliptic integrals of the ﬁrst kind by, respectively, dt (7.20) K(k) = , 2 1/2 2 2 1/2 [1 − t ] [1 − k t ] dt (7.21) F(w, k) = . 2 1/2 2 2 1/2 [1 − t ] [1 − k t ] Then the new expressions of ω and ω are 2 3 (7.22) ω = M,ω = M , 2 2 3 3 where (x − x )(x − x ) 4 1 3 2 (7.23) = K , (x − x )(x − x ) 4 2 3 1 (x − x )(x − X( y )) (x − x )(x − x ) 4 2 1 1 4 1 3 2 (7.24) = F , , (x − x )(x − X( y )) (x − x )(x − x ) 4 1 2 1 4 2 3 1 and where (below, 1 = 1if (i, j) ∈S , otherwise 0) (i,j) 2 1 if x = ∞, ⎪ 4 (1 − 41 1 )(x x − x x − x x + x x ) (1,0) (1,1) (1,−1) 3 4 2 3 1 4 1 2 (7.25) M = ⎪ if x =∞. (2z1 + 4z [1 1 + 1 1 ])(x − x ) (1,0) (1,1) (0,−1) (1,−1) (0,1) 3 1 The expressions of ω and ω written in (7.22), (7.23), (7.24)and (7.25) are obtained from 2 3 (3.1)and (3.2) by making simple changes of variables. We are now in position to analyze the behavior of ω /ω (or equivalently, thanks to 2 3 (7.22), that of / ) in the neighborhood of z = 0. First, with (2.2)and [19,Proposition 2 3 6.1.8], we obtain that as z → 0, x , x → 0and x , x →∞. For this reason, 1 2 3 4 (x − x )(x − x ) 4 1 3 2 (7.26) k = → 1. (x − x )(x − x ) 4 2 3 1 110 I. KURKOVA, K. RASCHEL The behavior of X( y ) as z → 0 is not so simple as that of the branch points x (indeed, as z → 0, X( y ) can converge to 0, to ∞ or to some non-zero constant), but we can show that for all 51 models, (x − x )(x − X( y )) 4 2 1 1 (7.27) w = → 1. (x − x )(x − X( y )) 4 1 2 1 Due to (7.26)and (7.27), in order to determine the behavior of / near z = 0it 2 3 sufﬁces to know (i) the expansion of K(k) as k → 1; (ii) the expansion of F(w, k) as k → 1and w → 1. Point (i) is classical, and is known as Abel’s identity (it can be found, e.g., in [8]): there exist two functions A and B, holomorphic at z = 0, such that K(k) = A(k) + ln(1 − k)B(k). Both A and B can be computed in an explicit way, see [8], and from all this we can deduce an expansion of K(k) as k → 1 up to any level of precision. For our purpose, it will be enough to use the following: A(k) = (3/2) ln(2) + (k − 1)/4 1 − 3ln(2) + O(k − 1) , B(k) =−1/2 + (k − 1)/4 + O(k − 1) . As for Point (ii), we proceed as follows. We have F(w, k) = K(k) − F(w, k),with dt F(w, k) = . 2 1/2 2 2 1/2 [1 − t ] [1 − k t ] 1/2 1/2 Then, introduce the expansion 1/([1 + t] [1 + kt] ) = μ (k)(1 − t) ,sothat =0 dt (7.28) F(w, k) = μ (k) . 1/2− 1/2 [1 − t] [1 − kt] =0 In Equation (7.28), all μ (k) as well as all integrals can be computed. As an example (that we shall use), we have dt μ (k) 1/2 1/2 [1 − t] [1 − kt] 1/2 1/2 2 = ln (1 − k)/k − ln − 1 − (1 + k)w + kw 1/2 [2k(1 + k)] 1/2 + (k + 1)/2 − kw /k . ON THE FUNCTIONS COUNTING WALKS IN THE QUARTER PLANE 111 FIG. 17. — Different cases considered in the proof of Theorem 17—they correspond to the 51 non-singular walks with inﬁnite group, see [2] Moreover, it should be noticed that as k → 1and w → 1, the speed of convergence to zero of the integrals in (7.28) increases with . This way, we can write an expansion of F(w, k)—and thus of F(w, k)—up to any level of precision. Unfortunately, the end of the proof cannot be done simultaneously for all 51 models, but should be done model by model. For the sake of shortness, we choose to present the details only for one model, namely for the model withS = {(−1, 0), (−1, 1), (0, 1), (1, −1)} (which belongs to Subcase II.D of Figure 17). For this 112 I. KURKOVA, K. RASCHEL model, we easily obtain from (2.2)that 2 3 4 x = z − 2z + 3z + O z , 2 3 4 x = z + 2z + 5z + O z , 2 3 4 x = 1/ 4z − 1 − 2z − 8z + O z , x =∞, X( y ) = 0. Then, with (7.26)and (7.27), we reach the conclusion that 4 5 6 (7.29) k = 1 − 8z − 4z + O z , 2 3 4 5 6 (7.30) w = 1 − 2z + z − (7/4)z − (65/8)z + (613/64)z + O z . Then, using Points (i) and (ii) above, we obtain 2 3 =−2ln(z) − (1/4)z + (1/16)z + O z ln(z) , 2 3 =−(1/2) ln(z) − (1/2) ln(2) + (1/4)z + (57/16)z + O z ln(z) , so that (7.31) / = 4 − 4ln(2)/ ln(z) + O 1/ ln(z) . 2 3 Thelatterproves(7.19), and thus Proposition 25,with L = 4and L =−4ln(2). Making expansions of higher order of k and w in (7.29)and (7.30), we could obtain more terms in the expansion (7.31)of / . A contrario, we could also be interested 2 3 in obtaining the ﬁrst term only (the constant term L) in (7.31). (Indeed, we saw in the proof of Proposition 14 that it was sufﬁcient for our purpose, i.e., for proving that in the inﬁnite group case, the ratio ω /ω is not constant in z.) To that aim, instead of 2 3 p p (7.29)and (7.30), we just need two-terms expansions of k and w,say k = 1 + αz + o(z ) q q and w = 1 + β z + o(z ),with α, β = 0. Then with (i) and (ii) we deduce that = −(p/2) ln(z) + o(ln(z)) and =−(q/2) ln(z) + o(ln(z)), in such a way that L = p/q, which obviously is (non-zero and) rational. Acknowledgements K. Raschel’s work was partially supported by CRC 701, Spectral Structures and Topological Methods in Mathematics at the University of Bielefeld. We are grateful to E. Lesigne: his mathematical knowledge and ideas—that he generously shared with us— have been very helpful in the elaboration of this paper. We also warmly thank R. Kriko- rian and J.-P. Thouvenot for encouraging discussions. Finally, we thank two anonymous referees for their careful reading and their remarks. ON THE FUNCTIONS COUNTING WALKS IN THE QUARTER PLANE 113 REFERENCES 1. A. BOSTAN and M. KAUERS, The complete generating function for Gessel walks is algebraic, Proc. Am. Math. Soc., 432 (2010), 3063–3078. 2. M. BOUSQUET-MÉLOU and M. MISHNA, Walks with small steps in the quarter plane, Contemp. Math., 520 (2010), 1–40. 3. M. BOUSQUET-MÉLOU and M. PETKOVSEK, Walks conﬁned in a quadrant are not always D-ﬁnite, Theor. Comput. Sci., 307 (2003), 257–276. 4. G. FAYOLLE and R. IASNOGORODSKI, Two coupled processors: the reduction to a Riemann-Hilbert problem, Z. Wahrscheinlichkeitstheor. Verw. Geb., 47 (1979), 325–351. 5. G. FAYOLLE and K. RASCHEL, On the holonomy or algebraicity of generating functions counting lattice walks in the quarter-plane, Markov Process. Relat. Fields, 16 (2010), 485–496. 6. G. FAYOLLE,R.IASNOGORODSKI,and V. MALYSHEV, Random Walks in the Quarter-Plane, Springer, Berlin, 1999. 7. P. FLAJOLET and R. SEDGEWICK, Analytic Combinatorics, Cambridge University Press, Cambridge, 2009. 8. J. GERRETSEN and G. SANSONE, Lectures on the Theory of Functions of a Complex Variable II: Geometric Theory,Wolters- Noordhoff Publishing, Groningen, 1969. 9. I. GESSEL, A probabilistic method for lattice path enumeration, J. Stat. Plan. Inference, 14 (1986), 49–58. 10. G. JONES and D. SINGERMAN, Complex Functions, Cambridge University Press, Cambridge, 1987. 11. M. KAUERS,C.KOUTSCHAN,and D. ZEILBERGER, Proof of Ira Gessel’s lattice path conjecture, Proc. Natl. Acad. Sci. USA, 106 (2009), 11502–11505. 12. I. KURKOVA and K. RASCHEL, Explicit expression for the generating function counting Gessel’s walks, Adv. Appl. Math., 47 (2011), 414–433. 13. V. MALYSHEV, Random Walks, Wiener-Hopf Equations in the Quarter Plane, Galois Automorphisms, Lomonossov Moscow Uni- versity Press, Moscow, 1970 (in Russian). 14. V. MALYSHEV, Positive random walks and Galois theory, Usp. Mat. Nauk, 26 (1971), 227–228. 15. V. MALYSHEV, An analytical method in the theory of two-dimensional positive random walks, Sib. Math. J., 13 (1972), 1314–1329. 16. S. MELCZER and M. MISHNA, Singularity analysis via the iterated kernel method (2011, in preparation). 17. M. MISHNA and A. RECHNITZER, Two non-holonomic lattice walks in the quarter plane, Theor. Comput. Sci., 410 (2009), 3616–3630. 18. K. RASCHEL, Counting walks in a quadrant: a uniﬁed approach via boundary value problems, J. Eur. Math. Soc., 14 (2012), 749–777. 19. R. STANLEY, Enumerative Combinatorics, vol. 2, Cambridge University Press, Cambridge, 1999. 20. G. WATSON and E. WHITTAKER, A Course of Modern Analysis, Cambridge University Press, Cambridge, 1962. I. K. Laboratoire de Probabilités et Modèles Aléatoires, Université Pierre et Marie Curie, 4 Place Jussieu, 75252 Paris Cedex 05, France irina.kourkova@upmc.fr 114 I. KURKOVA, K. RASCHEL K. R. CNRS and Université de Tours, Faculté des Sciences et Techniques, Parc de Grandmont, 37200 Tours, France kilian.raschel@lmpt.univ-tours.fr Manuscrit reçu le 4 octobre 2011 Manuscrit accepté le 11 octobre 2012 publié en ligne le 25 octobre 2012.

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On the functions counting walks with small steps in the quarter plane

On the functions counting walks with small steps in the quarter plane

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On the functions counting walks with small steps in the quarter plane

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