Irina Kurkova
email: irina.kourkova@upmc.fr.

Kilian Raschel
email: kilian.raschel@lmpt.univ-tours.fr.

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

Keywords: Walks in the quarter plane, counting generating function, holonomy, group of the walk, Riemann surface, elliptic functions, uniformization, universal covering AMS 2000 Subject Classification: primary 05A15, secondary 30F10, 30D05

.

Introduction and main results

In the field of enumerative combinatorics, counting walks on the lattice Z 2 is among the most classical topics. While counting problems have been largely resolved for unrestricted walks on Z 2 and for walks staying in a half plane [START_REF] Bousquet-Mélou | Walks confined in a quadrant are not always D-finite[END_REF], walks confined to the quarter plane Z 2 + 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} 2 \ {(0, 0)}; for examples, see Figures 1 and10. In [START_REF] Bousquet-Mélou | Walks with small steps in the quarter plane[END_REF], Bousquet-Mélou and Mishna constructed a thorough classification of these 2 8 walks. After eliminating trivial cases and exploiting equivalences, they showed that 79 inherently different walks remain to be studied. Let q(i, j; n) denote the number of paths in Z 2 + having length n, starting from (0, 0) and ending at (i, j). Define the counting function (CF) as Figure 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 2 + are discarded Q(x, y; z) = i,j,n 0 q(i, j; n)x i y j z n .

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 [START_REF] Flajolet | Analytic combinatorics[END_REF]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 [START_REF] Bousquet-Mélou | Walks with small steps in the quarter plane[END_REF]) (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 (1.2) K(x, y; z) = xyz[ (i,j)∈S x i y j -1/z] 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 [START_REF] Malyshev | Random Walks, Wiener-Hopf Equations in the Quarter Plane, Galois Automorphisms[END_REF][START_REF] Malyshev | Positive random walks and Galois theory[END_REF][START_REF] Malyshev | An analytical method in the theory of two-dimensional positive random walks[END_REF]. This is the group (1.3) ξ, η of birational transformations of C(x, y), generated by (1.4) ξ(x, y) = x, 1 y (i,-1)∈S x i (i,+1)∈S x i , η(x, y) = 1 x

(-1,j)∈S y j (+1,j)∈S y j , y . Each element of ξ, η leaves invariant the jump function (i,j)∈S x i y j . Further, ξ 2 = η 2 = 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 [START_REF] Bousquet-Mélou | Walks with small steps in the quarter plane[END_REF].

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 [START_REF] Bousquet-Mélou | Walks with small steps in the quarter plane[END_REF] by means of certain (half-)orbit sums of the functional equation (1.1). For the 23rd model with finite group, known as Figure 2. The 5 singular walks in the classification of [START_REF] Bousquet-Mélou | Walks with small steps in the quarter plane[END_REF] Gessel's walk (see Figure 10), the CFs have been expressed by radicals in [START_REF] Bostan | The complete generating function for Gessel walks is algebraic[END_REF] thanks to a guessing-proving method using computer calculations; they were also found in [START_REF] Kurkova | Explicit expression for the generating function counting Gessel's walks[END_REF] by solving some boundary value problems. For the 2 walks with infinite group on the left in Figure 2, they have been obtained in [START_REF] Mishna | Two non-holonomic lattice walks in the quarter plane[END_REF], by exploiting a particular property shared by the 5 models of Figure 2 commonly known as singular walks. Finally, in [START_REF] Raschel | Counting walks in a quadrant: a unified approach via boundary value problems[END_REF], the problem (i) was resolved for all 54 remaining walks-and in fact for all the 79 models. For the 74 non-singular walks, this was done via a unified approach: explicit integral representations 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 meromorphic 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 [START_REF] Bousquet-Mélou | Walks with small steps in the quarter plane[END_REF] 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 [START_REF] Bostan | The complete generating function for Gessel walks is algebraic[END_REF]. Alternative proofs for the nature of the (bivariate) CF for these 23 walks were given in [START_REF] Fayolle | On the holonomy or algebraicity of generating functions counting lattice walks in the quarter-plane. Markov Process[END_REF]. For the remaining 56 walks with an infinite group, not much is known: in [START_REF] Mishna | Two non-holonomic lattice walks in the quarter plane[END_REF] 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 [START_REF] Flajolet | Analytic combinatorics[END_REF]Appendix B.4], is non-holonomic. Accordingly [START_REF] Flajolet | Analytic combinatorics[END_REF]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 [START_REF] Melczer | Singularity analysis via the iterated kernel method[END_REF][START_REF] Mishna | Two non-holonomic lattice walks in the quarter plane[END_REF]. As for the 51 non-singular walks with infinite group (all of them are pictured on Figure 17), Bousquet-Mélou and Mishna [START_REF] Bousquet-Mélou | Walks with small steps in the quarter plane[END_REF] 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 subsets H 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 stable by specialization of a variable [START_REF] Flajolet | Analytic combinatorics[END_REF]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 further analysis when z ∈ H, namely by the use of methods developed in [8, Chapter 4], see Remark 16. This important set H 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 construction 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 [START_REF] Raschel | Counting walks in a quadrant: a unified approach via boundary value problems[END_REF] in a certain domain of C admits a direct holomorphic continuation on C \ [x 3 (z), x 4 (z)]. Points x 3 (z), x 4 (z) are among four branch points x 1 (z), x 2 (z), x 3 (z), x 4 (z) of the two-valued algebraic function 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 meromorphic continuation on any open domain containing the segment [x 3 (z), x 4 (z)], but admits a meromorphic continuation along any path going once through [x 3 (z), x 4 (z)]. This way, we obtain a second (and different) branch of the function, which admits a direct meromorphic continuation on the whole cut plane

C \ ([x 1 (z), x 2 (z)] ∪ [x 3 (z), x 4 (z)]). Next, if the function x → Q(x, 0; z) is continued along a path in C \ [x 1 (z), x 2 (z)] crossing once again [x 3 (z), x 4 (z)], we come across its first branch. But its continuation along a path in C \ [x 3 (z), x 4 (z)] crossing once [x 1 (z), x 2 (z)] leads to a third branch of this CF, which is meromorphic on C \ ([x 1 (z), x 2 (z)] ∪ [x 3 (z), x 4 (z)]). Making loops through [x 3 (z), x 4 (z)]
and [x 1 (z), x 2 (z)], successively, we construct x → Q(x, 0; z) as a multi-valued meromorphic function on C with branch points x 1 (z), x 2 (z), x 3 (z), x 4 (z), and with (generically) 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 Riemann 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 1 (z) ∈ iR and ω 2 (z) ∈ R that are uniformization periods. These periods as well as a new important period ω 3 (z) are made explicit in (3.1) and (3.2). In Section 4 we lift CFs Q(x, 0; z) and Q(0, y; z) to some domain of T, and then to a domain on its universal covering. In Section 5, using a proper lifting of the automorphisms ξ and η defined in (1.4) as well as the 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 [START_REF] Malyshev | Random Walks, Wiener-Hopf Equations in the Quarter Plane, Galois Automorphisms[END_REF][START_REF] Malyshev | Positive random walks and Galois theory[END_REF][START_REF] Malyshev | An analytical method in the theory of two-dimensional positive random walks[END_REF], at that time to study the stationary probability generating functions for random walks with small steps in the quarter plane Z 2 + . It is presented in [START_REF] Fayolle | Random walks in the quarter-plane[END_REF]Chapter 3] for the case of ergodic random walks in Z 2 + , and applies directly for our Q(x, 0; 1/|S|) and Q(0, y; 1/|S|) if the drift vector ( (i,j)∈S i, (i,j)∈S j) has not two positive coordinates. In Sections 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 2 (z)/ω 3 (z) of 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 2 (z)/ω 3 (z) is rational, these functions are holonomic.

For 23 models of walks with finite group ξ, η , the ratio ω 2 (z)/ω 3 (z) turns out 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 approach 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 [START_REF] Bostan | The complete generating function for Gessel walks is algebraic[END_REF][START_REF] Bousquet-Mélou | Walks with small steps in the quarter plane[END_REF][START_REF] Fayolle | On the holonomy or algebraicity of generating functions counting lattice walks in the quarter-plane. Markov Process[END_REF][START_REF] Mishna | Two non-holonomic lattice walks in the quarter plane[END_REF].

Section 7 is devoted to 51 models with infinite group ξ, η . For all of them, the sets Proposition 14. These 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. By Theorem 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):

H = {z ∈]0, 1/|S|[: ω 2 (z)/ω 3 (z) is rational} and ]0, 1/|S|[\H = {z ∈]0, 1/|S|[: ω 2 (z)/ω 3 (z) is irrational} are proved to be dense in ]0, 1/|S|[, see
(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 3 (z), x 4 (z) and y 3 (z), y 4 (z), respectively; (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.

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

(2.1) xyz[ (i,j)∈S x i y j -1/z] = a(y)x 2 + b(y)x + c(y) = a(x)y 2 + b(x)y + c(x),
where

a(y) = zy (+1,j)∈S y j , b(y) = -y+zy (0,j)∈S y j , c(y) = zy (-1,j)∈S y j , a(x) = zx (i,+1)∈S x i , b(x) = -x+zx (i,0)∈S x i , c(x) = zx (i,-1)∈S x i .
With these notations we define

(2.2) d(y) = b(y) 2 -4 a(y) c(y), d(x) = b(x) 2 -4a(x)c(x).
If the walk is non-singular, then for any z ∈]0, 1/|S|[, the polynomial d (resp. d) has three or four roots, that we call y ℓ (resp. x ℓ ). They are such that

|y 1 | < y 2 < 1 < y 3 < |y 4 | (resp. |x 1 | < x 2 < 1 < x 3 < |x 4 |), with y 4 = ∞ (resp. x 4 = ∞) if d (resp. d)
b(y) + 2 a(y)x] 2 = d(y) or [b(x) + 2a(x)y] 2 = d(x)
. Consequently [START_REF] Jones | Complex Functions[END_REF], the algebraic functions X(y) and Y (x) defined by

(2.3) (i,j)∈S X(y) i y j -1/z = 0, (i,j)∈S x i Y (x) j -1/z = 0 have two branches, meromorphic on the cut planes C \ ([y 1 , y 2 ] ∪ [y 3 , y 4 ]) and C \ ([x 1 , x 2 ] ∪ [x 3 , x 4 ]), respectively-note that if y 4 < 0, [y 3 , y 4 ] stands for [y 3 , ∞[∪{∞}∪] -∞, y 4 ]; the same holds for [x 3 , x 4 ].
We fix the notations of the two branches of the algebraic functions X(y) and Y (x) by setting (2.4)

X 0 (y) = -b(y) + d(y) 1/2 2 a(y) , X 1 (y) = -b(y) -d(y) 1/2 2 a(y) ,
as well as

(2.5) Y 0 (x) = -b(x) + d(x) 1/2 2a(x) , Y 1 (x) = -b(x) -d(x) 1/2 2a(x) .
The following straightforward result holds.

Lemma 2. For all y ∈ C, we have

|X 0 (y)| |X 1 (y)|. Likewise, for all x ∈ C, we have |Y 0 (x)| |Y 1 (x)|.
Proof. The arguments (via the maximum modulus principle [START_REF] Jones | Complex Functions[END_REF]) given in [START_REF] Fayolle | Random walks in the quarter-plane[END_REF]Part 5.3] for z = 1/|S| also work for z ∈]0, 1/|S|[. 3). For this purpose we take two Riemann spheres C∪{∞}, say S 1 x and S 2 x , cut along the segments [x 1 , x 2 ] and [x 3 , x 4 ], and we glue them together along the borders of these cuts, joining the lower border of the segment [x 1 , x 2 ] (resp. [x 3 , x 4 ]) on S 1

x 1 x 1 x 2 x 2 x 3 x 3 x 4 x 4 S 1 x S 2 x x 1 x 1 x 2 x 2 x 3 x 3 x 4 x 4
x to the upper border of the same segment on S 2 x and vice versa, see 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 x : T → S.

In a standard way, we can lift the function Y (x) to T, by setting

Y (s) = Y ℓ (h x (s)) if s ∈ S ℓ
x ⊂ T, ℓ ∈ {1, 2}. Thus, Y (s) is single-valued and continuous on T. Furthermore, K(h x (s), Y (s)) = 0 for 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 gluing together two copies S 1 y and S 2 y of the sphere S along the segments [y 1 , y 2 ] and [y 3 , y 4 ]. It is again homeomorphic to a torus.

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 x , h y : T → S. Then, for s ∈ T, we set x(s) = h x (s) and y(s) = h y (s), and we will often represent a point s ∈ T 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)) = 0 is valid for any s ∈ T.

2.3.

Real points of T. Let us identify the set Φ of real points of T, that are the points s ∈ T where x(s) and y(s) are both real or equal to infinity. Note that for y real, X(y) 4)

x 1 y 1 x 2 y 2 Φ 1 Φ 0 y 3 x
Φ 0 = {s ∈ T : x(s) ∈ [x 2 , x 3 ]} = {s ∈ T : y(s) ∈ [y 2 , y 3 ]} and Φ 1 = {s ∈ T : x(s) ∈ [x 4 , x 1 ]} = {s ∈ T : y(s) ∈ [y 4 , y 1 ]
}, and homologically equivalent to a basic cycle on T-note, however, that the equivalence class containing Φ 0 and Φ 1 is disjoint from that containing the cycle h -1

x ({x ∈ C : |x| = 1}). 2.4. Galois automorphisms ξ, η. We continue Section 2 by introducing two Galois automorphisms. Define first, for ℓ ∈ {1, 2}, the incised spheres

S ℓ x = S ℓ x \ ([x 1 , x 2 ] ∪ [x 3 , x 4 ]), S ℓ y = S ℓ y \ ([y 1 , y 2 ] ∪ [y 3 , y 4 ]
). For any s ∈ T such that x(s) is not equal to a branch point x ℓ , there is a unique s ′ = s ∈ T such that x(s) = x(s ′ ). Furthermore, if s ∈ S 1

x then s ′ ∈ S 2

x and vice versa. On the other hand, whenever x(s) is one of the branch points x ℓ , 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 ℓ , there exists a unique s ′′ = s ∈ T such that y(s) = y(s ′′ ). If s ∈ S 1 y then s ′′ ∈ S 2 y and vice versa. On the other hand, if y(s) is one of the branch points y ℓ , we have 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)).

Define now the mappings ξ : T → T and η : T → T by

(2.6) ξs = s ′ if x(s) = x(s ′ ), ηs = s ′′ if y(s) = y(s ′′ ).
Following [START_REF] Malyshev | Random Walks, Wiener-Hopf Equations in the Quarter Plane, Galois Automorphisms[END_REF][START_REF] Malyshev | Positive random walks and Galois theory[END_REF][START_REF] Malyshev | An analytical method in the theory of two-dimensional positive random walks[END_REF], we call them Galois automorphisms of T. Then ξ 2 = η 2 = Id, and

(2.7)

y(ξs) = c(x(s)) a(x(s)) 1 y(s) , x(ηs) = c(y(s)) a(y(s)) 1 x(s)
. Any s ∈ T such that x(s) = x ℓ (resp. y(s) = y ℓ ) 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 Φ 0 where x(s) = x 2 and x 3 (resp. y(s) = y 2 and y 3 ); then points s and ξs (resp. s and ηs) can be drawn symmetric about this straight line.

x 3 y 3 x 2 y 2 x 4 y 4 x 1 y 1 ω 1 ω 1 /2 0 ω 3 /2 ω 2 /2 [ω 2 + ω 3 ]/2 ω 2
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 1 Z + ω 2 Z),
where ω 1 , ω 2 are complex numbers linearly independent on R, see [START_REF] Jones | Complex Functions[END_REF]. The set (2.8) can obviously be thought as the (fundamental) parallelogram ω 1 [0, 1] + ω 2 [0, 1] whose opposed edges are identified. Up to a unimodular transform, ω 1 , ω 2 are unique, see [START_REF] Jones | Complex Functions[END_REF]. In our case, suitable ω 1 , ω 2 will be found in (3.1).

If we cut the torus on Figure 4 along [x 1 , x 2 ] and Φ 0 , it becomes the parallelogram 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 1 , ω 2 (see above) and of ω 3 (to be defined below, in (3.2)).

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.

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 [START_REF] Jones | Complex Functions[END_REF]. This way, the surface T can be considered as the additive group C factorized by the discrete subgroup ω 1 Z + ω 2 Z, where the periods ω 1 , ω 2 are complex numbers, linearly independent on R. Any segment of length |ω ℓ | and parallel to ω ℓ , ℓ ∈ {1, 2}, is projected ) is then homological to the cycles of real points Φ 0 and Φ 1 ; see Figures 5 and7.

T + ω 1 T -ω 2 T T + ω 2 T -ω 1 Figure 6.
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)), that we have introduced in Section 2. First, the periods ω 1 , ω 2 are obtained in [START_REF] Fayolle | Random walks in the quarter-plane[END_REF]Lemma 3.3.2] 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), (3.1)

ω 1 = i x 2 x 1 dx [-d(x)] 1/2 , ω 2 = x 3 x 2 dx d(x) 1/2 .
We also need to introduce

(3.2) ω 3 = x 1 X(y 1 ) dx d(x) 1/2 .
Further, we define

g x (t) = d ′′ (x 4 )/6 + d ′ (x 4 )/[t -x 4 ] if x 4 = ∞, d ′′ (0)/6 + d ′′′ (0)t/6 if x 4 = ∞,
as well as

g y (t) = d ′′ (y 4 )/6 + d ′ (y 4 )/[t -y 4 ] if y 4 = ∞, d ′′ (0)/6 + d ′′′ (0)t/6 if y 4 = ∞,
and finally we introduce ℘(ω; ω 1 , ω 2 ), the Weierstrass elliptic function with periods ω 1 , ω 2 . Throughout, we shall write ℘(ω) for ℘(ω; ω 1 , ω 2 ). By definition, see [START_REF] Jones | Complex Functions[END_REF][START_REF] Watson | A course of modern analysis[END_REF], we have

℘(ω) = 1 ω 2 + (ℓ 1 ,ℓ 2 )∈Z 2 \{(0,0)} 1 (ω -ℓ 1 ω 1 -ℓ 2 ω 2 ) 2 - 1 (ℓ 1 ω 1 + ℓ 2 ω 2 ) 2 .
Then we have the uniformization [8, Lemma 3.3.1]

(3.3) x(λω) = g -1 x (℘(ω)), y(λω) = g -1 y (℘(ω -ω 3 /2))
.

ω x 4 + ω 1 ω y 4 + ω 1 ω y 4 + ω 1 + ω 2 ω x 3 ω y 3 ω x 2 ω y 2 ω x 3 + ω 2 ω x 4 ω y 4 ω x 1 ω y 1 ω x 4 + ω 2 L x 2 x 1 L y 2 y 1 L x 3 x 4 L y 3 y 4 Figure 7
. Important points and cycles on the universal covering From now on, whenever no ambiguity can arise, we drop the dependence w.r.t. λ, writing 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 ℓ ) = x(ω), y(ω + ω ℓ ) = y(ω), ∀ℓ ∈ {1, 2}, ∀ω ∈ C.
Furthermore,

x(0) = x 4 y(0) = Y (x 4 ) , x(ω 1 /2) = x 3 y(ω 1 /2) = Y (x 3 ) , x(ω 2 /2) = x 1 y(ω 2 /2) = Y (x 1 ) , x([ω 1 + ω 2 ]/2) = x 2 y([ω 1 + ω 2 ]/2) = Y (x 2 ) .
Let us denote the points 0,

ω 1 /2, ω 2 /2, [ω 1 + ω 2 ]/2 by ω x 4 , ω x 3 , ω x 1 , ω x 2
, respectively, see Figures 5 and7. Let

L x 3 x 4 = ω x 4 + ω 1 R, L x 2 x 1 = ω x 1 + ω 1 R. Then λL x 3
x 4 (resp. λL x 2 x 1 ) is the cut of T where S 1 x and S 2 x are glued together, namely, {s ∈ T :

x(s) ∈ [x 3 , x 4 ]} (resp. {s ∈ T : x(s) ∈ [x 1 , x 2 ]}).
Moreover, by construction we have (see again Figures 5 and7)

x(ω 3 /2) = X(y 4 ) y(ω 3 /2) = y 4 , x([ω 1 + ω 3 ]/2) = X(y 3 ) y([ω 1 + ω 3 ]/2) = y 3 , x([ω 2 + ω 3 ]/2) = X(y 1 ) y([ω 2 + ω 3 ]/2) = y 1 , and x([ω 1 + ω 2 + ω 3 ]/2) = X(y 2 ) y([ω 1 + ω 2 + ω 3 ]/2) = y 2 .
We denote the points

ω 3 /2, [ω 1 + ω 3 ]/2, [ω 2 + ω 3 ]/2, [ω 1 + ω 2 + ω 3 ]/2 by ω y 4 , ω y 3 , ω y 1 , ω y 2 , respectively. Let L y 3 y 4 = ω y 4 + ω 1 R, L y 2 y 1 = ω y 1 + ω 1 R. Then λL y 3 y 4 (resp. λL y 2 y 1 )
is the cut of T where S 1 y and S 2 y are glued together, that is to say

{s ∈ T : y(s) ∈ [y 3 , y 4 ]} (resp. {s ∈ T : y(s) ∈ [y 1 , y 2 ]}).
The distance between L x 3 x 4 and L y 3 y 4 is the same as between L x 2 x 1 and L y 2 y 1 ; it equals ω 3 /2. This continuation is not unique, but it will be unique if we fix some ζω 0 ∈ λ -1 ζλω 0 for a given point ω 0 ∈ C.

D x Γ 0 y Γ 1 x Γ 0 x D y Γ 1 y
According to [START_REF] Fayolle | Random walks in the quarter-plane[END_REF], we define ξ, η by choosing their fixed points to be ω x 2 , ω y 2 , respectively. Since any conformal automorphism of C is an affine function of ω [START_REF] Jones | Complex Functions[END_REF] and since ξ 2 = η 2 = Id, we have

(3.5) ξω = -ω + 2ω x 2 , ηω = -ω + 2ω y 2 .
It follows that η ξ and ξ η are just the shifts via the real numbers ω 3 and -ω 3 , respectively:

(3.6) η ξω = ω + 2(ω y 2 -ω x 2 ) = ω + ω 3 , ξ ηω = ω + 2(ω x 2 -ω y 2 ) = ω -ω 3 .
By (2.6) and (2.7) we have

(3.7) x( ξω) = x(ω), y( ξω) = c(x(ω)) a(x(ω)) 1 y(ω) , x( ηω) = c(y(ω)) a(y(ω)) 1 x(ω)
, y( ηω) = y(ω).

Finally, ξL x 2 x 1 = L x 2 x 1 , ξL x 3 x 4 = L x 3 x 4 + ω 2 and ηL y 2 y 1 = L y 2 y 1 , ηL y 3 y 4 = L y 3 y 4 + ω 2 .
4. Lifting of x → Q(x, 0) and y → Q(0, y) to the universal covering

∈ T such that x(s) ∈ [x 1 , x 2 ]. Since the function x → K(x, 0)Q(x, 0) is holomorphic in the unit disc, we can lift it to D x ⊂ T as r x (s) = K(x(s), 0)Q(x(s), 0), ∀s ∈ D x .
In the same way, the domain D y = {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 1 , y 2 ]}, see Figure 8, and which contains the latter. We lift the function y → K(0, y)Q(0, y) to D y ⊂ T as

r y (s) = K(0, y(s))Q(0, y(s)), ∀s ∈ D y .
It is shown in [START_REF] Raschel | Counting walks in a quadrant: a unified approach via boundary value problems[END_REF]Lemma 3] that for any z ∈]0, 1/|S|[ and any x such that |x| = 1, we have |Y 0 (x)| < 1 and |Y 1 (x)| > 1. Hence, the cycles that constitute the boundary of D x are

Γ 0 x = {s ∈ T : |x(s)| = 1, |y(s)| < 1}, Γ 1 x = {s ∈ T : |x(s)| = 1, |y(s)| > 1}. We thus have Γ 0 x ∈ D y and Γ 1 x / ∈ D y , see
Γ 0 y = {s ∈ T : |y(s)| = 1, |x(s)| < 1}, Γ 1 y = {s ∈ T : |y(s)| = 1, |x(s)| > 1}. Furthermore, Γ 0 y ∈ D x and Γ 1 y / ∈ D x , see Figure 8. It follows that D x ∩ D y = {s ∈ T : |x(s)| < 1, |y(s)| < 1}
is not empty, simply connected and bounded by Γ 0

x and Γ 0 y . Since for any s ∈ T, K(x(s), y(s)) = 0, and since the main equation

(1.1) is valid on {(x, y) ∈ C 2 : |x| < 1, |y| < 1}, we have (4.1) r x (s) + r y (s) -K(0, 0)Q(0, 0) -x(s)y(s) = 0, ∀s ∈ D x ∩ D y .

4.2.

Lifting to the universal covering C. The domain D lifted on the universal covering consists of infinitely many curvilinear strips shifted by ω 2 :

λ -1 D x = n∈Z ∆ n x , ∆ n x ⊂ ω 1 R+]nω 2 , (n + 1)ω 2 [,
and, likewise,

λ -1 D y = n∈Z ∆ n y , ∆ n y ⊂ ω 1 R + ω 3 /2+]nω 2 , (n + 1)ω 2 [.
Let us consider these strips for n = 0, that we rename

∆ x = ∆ 0 x , ∆ y = ∆ 0 y .
The first is bounded by

Γ 1 x ⊂ λ -1 Γ 1
x and by Γ 0 x ⊂ λ -1 Γ 0 x , while the second is delimited by

Γ 0 y ⊂ λ -1 Γ 0 y and by Γ 1 y ⊂ λ -1 Γ 1 y . Further, note that the straight line L x 2
x 1 (resp. L y 2 y 1 ) defined in Section 3 is invariant w.r.t. ξ (resp. η) and belongs to ∆ x (resp. ∆ y ).

Then, by the facts that ξΓ 1 x = Γ 0 x and ηΓ 1 y = Γ 0 y , and by our choice (3.5) of the definition of ξ and η on the universal covering, we have ξ Γ 1

x = Γ 0 x and η Γ 1 y = Γ 0 y . In addition,

(4.2) ξω ∈ ∆ x , ∀ω ∈ ∆ x , ηω ∈ ∆ y , ∀ω ∈ ∆ y . Moreover, since Γ 0 y ∈ D x , Γ 1 y / ∈ D x and Γ 0 x ∈ D y , Γ 1 x / ∈ D y , we have Γ 0 y ∈ ∆ x , Γ 1 y / ∈ ∆ x and Γ 0 x ∈ ∆ y , Γ 1 x / ∈ ∆ y . It follows that ∆ x ∩ ∆ y is a non-empty strip bounded by Γ 0
x and Γ 0 y , and that ∆ = ∆ x ∪ ∆ y is simply connected, as in Figure 9.

ω x 4 + ω 1 ω y 4 + ω 1 + ω 2 Γ 1 x L x 2 x 1 L y 2 y 1 Γ 1 y ω x 4 ω y 4 + ω 2 ∆ Figure 9. Location of ∆ = ∆ x ∪ ∆ y
Let us lift the functions r x (s) and r y (s) holomorphically to ∆ x and ∆ y , respectively: we put

(4.3) r x (ω) = r x (λω) = K(x(ω), 0)Q(x(ω), 0), ∀ω ∈ ∆ x , r y (ω) = r y (λω) = K(0, y(ω))Q(0, y(ω)), ∀ω ∈ ∆ y .
It follows from (4.1) and (4.3) that (4.4)

r x (ω) + r y (ω) -K(0, 0)Q(0, 0) -x(ω)y(ω) = 0, ∀ω ∈ ∆ x ∩ ∆ y .
Equation (4.4) allows us to continue functions r x (ω) and r y (ω) meromorphically on ∆: we put

(4.5) r x (ω) = -r y (ω) + K(0, 0)Q(0, 0) + x(ω)y(ω), ∀ω ∈ ∆ y , r y (ω) = -r x (ω) + K(0, 0)Q(0, 0) + x(ω)y(ω), ∀ω ∈ ∆ x .
Equation (4.4) is then valid on the whole of ∆. We summarize all facts above in the next result.

Theorem 3. The functions

r x (ω) = K(x(ω), 0)Q(x(ω), 0) if ω ∈ ∆ x , -K(0, y(ω))Q(0, y(ω)) + K(0, 0)Q(0, 0) + x(ω)y(ω) if ω ∈ ∆ y , and r y (ω) = K(0, y(ω))Q(0, y(ω)) if ω ∈ ∆ y , -K(x(ω), 0)Q(x(ω), 0) + K(0, 0)Q(0, 0) + x(ω)y(ω) if ω ∈ ∆ x , are meromorphic in ∆. Furthermore, (4.6) r x (ω) + r y (ω) -K(0, 0)Q(0, 0) -x(ω)y(ω) = 0, ∀ω ∈ ∆.
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 x (ω) and r y (ω) are meromorphic on ∆. We now continue these functions meromorphically from ∆ to the whole of C.

Theorem 4. The functions r x (ω) and r y (ω) can be continued meromorphically to the whole of C. Further, for any ω ∈ C, we have

r x (ω -ω 3 ) = r x (ω) + y(ω)[x(-ω + 2ω y 2 ) -x(ω)], (5.1) r y (ω + ω 3 ) = r y (ω) + x(ω)[y(-ω + 2ω x 2 ) -y(ω)], (5.2) r x (ω) + r y (ω) -K(0, 0)Q(0, 0) -x(ω)y(ω) = 0, (5.3) r x ( ξω) = r x (ω), r y ( ηω) = r y (ω), (5.4) r x (ω + ω 1 ) = r x (ω), r y (ω + ω 1 ) = r y (ω). (5.5)
For the proof of Theorem 4, we shall need the following lemma.

Lemma 5. We have

(5.6) n∈Z (∆ + nω 3 ) = C. Proof. It has been noticed in Section 4 that ξ Γ 1 x = Γ 0 x ∈ ∆ y . By (4.2), η Γ 0 x ∈ ∆ y ⊂ ∆, so that, by (3.6), Γ 1 x + ω 3 = η ξ Γ 1 x ∈ ∆. In the same way, Γ 1 y -ω 3 ∈ ∆. It follows that ∆ ∪ (∆ + ω 3
) is a simply connected domain, see Figure 9. Identity (5.6) follows.

Proof of Theorem 4. For any ω ∈ ∆, by Theorem 3 we have

(5.7) r x (ω) + r y (ω) -K(0, 0)Q(0, 0) -x(ω)y(ω) = 0.
For any ω ∈ ∆ close enough to the cycle Γ 1 x , we have that ξω ∈ ∆ y since ξ Γ 1 x = Γ 0 x ∈ ∆ y . Then ω + ω 3 = η ξω ∈ ∆ y by (4.2). We now compute r y ( η ξω) for any such ω. Equation (4.6), which is valid in ∆ ⊃ ∆ y , gives (5.8) r x ( ξω) + r y ( ξω) -K(0, 0)Q(0, 0) -x( ξω)y( ξω) = 0.

By (3.7), x( ξω) = x(ω). For our ω ∈ ∆ x , by (4.2) we have ξω ∈ ∆ x , so that Theorem 3 yields

r x ( ξω) = K(x( ξω), 0)Q(x( ξω), 0) = K(x(ω), 0)Q(x(ω), 0) = r x (ω).
If we now combine the last fact together with Equation (5.7), Equation (5.8) and identity x( ξω) = x(ω), we obtain that

r y ( ξω) = r y (ω) + x(ω)[y( ξω) -y(ω)].
Since ξω ∈ ∆ y , then by (4.2) we have η ξω ∈ ∆ y . Equation (3.7) and Theorem 3 entail

r y ( η ξω) = K(0, y( η ξω))Q(0, y( η ξω)) = K(0, y( ξω))Q(0, y( ξω)) = r y ( ξω).
Finally, for all ω ∈ ∆ close enough to Γ 1

x we have r y ( η ξω) = r y (ω) + x(ω)[y( ξω) -y(ω)].

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 Γ 1

x allows us to continue r y meromorphically from ∆ to the whole of C. Equation (5.2) therefore stays valid for any ω ∈ C. The function r y ( ηω) = r y (-ω + ω y 2 ) is then also meromorphic on C. Since these functions coincide in ∆ y , then by the principle of analytic continuation [START_REF] Jones | Complex Functions[END_REF] they do on the whole of C. In the same way, we prove Equation (5.1) for all ω ∈ ∆ y close enough to Γ 1 y . Together with Theorem 3 and Lemma 5 this allows us to continue r x (ω) meromorphically to the whole of C. By the same continuation argument, the identity r x (ω) = r x ( ξω) 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 1 -periodic, it follows from Theorem 3 that r x (ω) and r y (ω) are ω 1 -periodic in ∆. The vector ω 3 being real, by (5.1) and (5.2) these functions stay ω 1 -periodic on the whole of C.

5.2.

Branches of x → Q(x, 0) and y → Q(0, y). The restrictions of r x (ω)/K(x(ω), 0) on (5.9)

M k,ℓ = ω 1 [ℓ, ℓ + 1[+ω 2 [k/2, (k + 1)/2[ for k, ℓ ∈ Z provide all branches on C \ ([x 1 , x 2 ] ∪ [x 3 , x 4 ]) of Q(x, 0) as follows: (5.10) Q(x, 0) = {r x (ω)/K(x(ω), 0) : ω is the (unique) element of M k,ℓ such that x(ω) = x}.
Due to the ω 1 -periodicity of r x (ω) and x(ω), the restrictions of these functions on M k,ℓ do not depend on ℓ ∈ Z, and therefore determine the same branch as on M k,0 for any ℓ. Furthermore, thanks to (5.4), (3.5) and (3.7) the restrictions of r x (ω)/K(x(ω), 0) on M -k+1,0 and on M k,0 lead to the same branches for any k ∈ Z. Hence, the restrictions of r x (ω)/K(x(ω), 0) to M k,0 with k 1 provide all different branches of this function. The analogous statement holds for the restrictions of r y (ω)/K(0, y(ω)) on (5.11)

N k,ℓ = ω 3 /2 + ω 1 [ℓ, ℓ + 1[+ω 2 ]k/2, (k + 1)/2]
for k, ℓ ∈ Z, namely:

(5.12) Q(0, y) = {r y (ω)/K(0, y(ω)) : ω is the (unique) element of N k,ℓ such that y(ω) = y}.

The restrictions on N k,ℓ for ℓ ∈ Z give the same branch as on N k,0 . For any k ∈ Z + the restrictions on N -k+1,0 and on N k,0 determine the same branches. Hence, the restrictions of r y (ω)/K(0, y(ω)) on N k,0 with k 1 provide all different branches of y → Q(0, y). The rationality or irrationality of the quantity ω 2 /ω 3 is crucial for the nature of the functions x → Q(x, 0) and y → Q(0, y) for a given z. Indeed, the following theorem holds true.

Theorem 7. For any z ∈]0, 1/|S|[ such that ω 2 /ω 3 is rational, the functions x → Q(x, 0) 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 [START_REF] Fayolle | On the holonomy or algebraicity of generating functions counting lattice walks in the quarter-plane. Markov Process[END_REF], so here we just recall the main ideas. The proof actually consists in applying [8, Theorem 4.4.1], which entails that if ω 2 /ω 3 is rational, the function Q(x, 0) can be written as

Q(x, 0) = w 1 (x) + Φ(x)φ(x) + w(x)/r(x),
where w 1 and r are rational functions, while φ and w are algebraic. Further, in [9, Lemma 2.1] it is shown that Φ is holonomic. Accordingly, Q(x, 0) is also holonomic. The argument for Q(0, y) is similar. Notice that Theorem 4.4.1 in [START_REF] Fayolle | Random walks in the quarter-plane[END_REF] is proved for z = 1/|S| only, but in [START_REF] Fayolle | On the holonomy or algebraicity of generating functions counting lattice walks in the quarter-plane. Markov Process[END_REF] 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 2 /ω 3 is rational and 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 2 /ω 3 takes rational and irrational values on subsets H and ]0, 1/|S|[\H, respectively, which are dense on ]0, 1/|S|[, as it will be proved in Proposition 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.

Finite group case

Define the covariance of the model as (6.1)

(i,j)∈S ij -[ (i,j)∈S i][ (i,j)∈S j] = (i,j)∈S ij.
The equality above follows from the fact that for each of the 23 models with a finite group, (i,j)∈S i = 0 or (i,j)∈S j = 0, see [START_REF] Bousquet-Mélou | Walks with small steps in the quarter plane[END_REF]. Lemma 8 below is proved in [21, Proposition 5].

Lemma 8. For all 23 models with finite group (1.3), ω 2 /ω 3 is rational and independent of z. More precisely:

-For the walks with a group of order 4, ω 2 /ω 3 = 2; -For the walks with a group of order 6 and such that the covariance is negative (resp. positive), ω 2 /ω 3 = 3 (resp. 3/2);

-For the walks with a group of order 8 and a negative (resp. positive) covariance, ω 2 /ω 3 = 4 (resp. 4/3).

In the sequel, we note ω 2 /ω 3 = k/ℓ; then, 2k is the order of the group. Since kω 3 = ℓω 2 , we obviously always have

r x (ω + ℓω 2 ) -r x (ω) = 1 m k r x (ω + mω 3 ) -r x (ω + (m -1)ω 3 ).
It follows from (5.1) and from properties (3.5), (3.6) and (3.7) of the Galois automorphisms that

r x (ω + ℓω 2 ) -r x (ω) = 1 m k (xy)(ω + mω 3 ) -(xy)( η(ω + mω 3 )) = 1 m k (xy)(( η ξ) m ω) -(xy)( ξ( η ξ) m-1 ω)) = θ∈ ξ, η (-1) θ xy(θ(ω)), (6.2)
where (-1) θ is the signature of θ; in other words, (-1) θ = (-1) ℓ(θ) , where ℓ(θ) is the length of θ, i.e., the smallest ℓ such that we can write θ = θ 1 • • • • • θ ℓ , with θ 1 , . . . , θ ℓ equal to ξ or η. The same identity with the opposite sign holds for r y . The quantity (6.2) is the 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 [START_REF] Bousquet-Mélou | Walks with small steps in the quarter plane[END_REF].

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.

After the lifting to the universal covering done in Theorem 4, results of [START_REF] Bostan | The complete generating function for Gessel walks is algebraic[END_REF][START_REF] Bousquet-Mélou | Walks with small steps in the quarter plane[END_REF] 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 ([3, 5]). 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 ([5]

). 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) 2 = 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/p can be written in terms of ℘ as

℘(ω) + p-1 ℓ=1 [℘(ω + ℓ ω/p) -℘(ℓ ω/p)], ∀ω ∈ C.
(P3) We have the addition theorem:

℘(ω + ω) = -℘(ω) -℘( ω) + 1 4 ℘ ′ (ω) -℘ ′ ( ω) ℘(ω) -℘( ω) 2 , ∀ω, ω ∈ C.
(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 Φ which is ω-periodic and such that

Φ(ω + ω) = Φ(ω) -1, ∀ω ∈ C.
Proof. Properties (P1), (P3) and (P4) are most classical, and can be found, e.g., in [START_REF] Jones | Complex Functions[END_REF][START_REF] Watson | A course of modern analysis[END_REF]. For (P2) we refer to [24, page 456], and for (P5), see [START_REF] Fayolle | Random walks in the quarter-plane[END_REF]Equation (4.3.7)]. Note that the function Φ in (P5) can be constructed via the zeta function of Weierstrass.

Proof of Proposition 11. If the orbit-sum O(ω) is zero, Equation (6.2) implies that r x (ω) is ℓω 2 -periodic. In particular, the property (P4) of Lemma 13 entails that there exist two rational functions R and S such that

(6.3) r x (ω) = R(℘(ω; ω 1 , ℓω 2 )) + ℘ ′ (ω; ω 1 , ℓω 2 )S(℘(ω; ω 1 , ℓω 2 )).
Further, the property (P2) together with the addition formula (P3) of Lemma 13 gives that ℘(ω; ω 1 , ℓω 2 ) is an algebraic function of ℘(ω)-we recall that ℘(ω) denotes the Weierstrass function ℘(ω; ω 1 , ω 2 ). Due to Lemma 13 (P1), ℘ ′ (ω) is an algebraic function of ℘(ω) too, so that ℘ ′ (ω; ω 1 , ℓω 2 ) is also an algebraic function of ℘(ω). Thanks to (6.3), we get that r x (ω) is algebraic in ℘(ω). Since ℘(ω) is a rational function of x(ω), see (3.3), we finally obtain that r x (ω) is algebraic in x(ω). Then q x (ω) = r x (ω)/K(x(ω), 0) is algebraic in x(ω), and so is q y (ω) in y(ω).

Proof of Proposition 12. In this proof we have ℓ = 1, see Lemma 8. Thanks to Lemma 13 (P5), there exists a function Φ which is ω 1 -periodic and such that Φ(ω + ω 2 ) = Φ(ω) -1.

In particular, transforming (6.2) we can write

r x (ω + ω 2 ) + Φ(ω + ω 2 )O(ω + ω 2 ) = r x (ω) + Φ(ω)O(ω).
This entails that r x (ω) + Φ(ω)O(ω) is elliptic with periods ω 1 , ω 2 . In particular, for the 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). As for the function Φ(ω), it is proved in [8, page 71] that it is a non-algebraic function of x(ω). Moreover, it is shown in [9, Lemma 2.1] that it is holonomic in x(ω). Hence r x (ω) is holonomic in x(ω) but not algebraic. The same is true for q x (ω) = r x (ω)/K(x(ω), 0) in x(ω) and for q y (ω) in y(ω).

Infinite group case

It has been shown in [START_REF] Raschel | Counting walks in a quadrant: a unified approach via boundary value problems[END_REF]Part 6.2] that for all 51 models with infinite group, ω 2 /ω 3 takes irrational values for infinitely many z. The next proposition states a more complete result. Proposition 25 below gives the asymptotic of ω 2 /ω 3 as z → 0: for any of 51 models there exist some rational L > 0 and some L = 0 such that as z > 0 goes to 0,

ω 2 /ω 3 = L + L/ ln z + O((1/ ln z) 2 ).
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 Proposition 25) that ω 2 /ω 3 converges to a rational positive constant L as z → 0 for all 51 models. Indeed, then, since ω 2 /ω 3 necessarily takes irrational values for some z ∈]0, 1/|S|[ (see [START_REF] Raschel | Counting walks in a quadrant: a unified approach via boundary value problems[END_REF]Part 6.2]), there exists an interval within ]0, 1/|S|[ where the ratio ω 2 /ω 3 is not 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 2 /ω 3 is irrational, and we prove in particular that their 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. Let H = {z ∈]0, 1/|S|[: ω 2 /ω 3 is rational} and ]0, 1/|S|[\H = {z ∈ ]0, 1/|S|[: ω 2 /ω 3 is irrational}.
(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 Theorem 17 (iii) below as explained in the Introduction.

Remark 16. It follows from Remark 6 that H can be characterized as the set of z ∈]0, 1/|S|[ such that the group ξ, η restricted to the curve {(x, y) ∈ (C ∪ {∞}) 2 : K(x, y; z) = 0} is finite. Then methods developed in [8, 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. Poles of the set of branches of x → Q(x, 0) and y → Q(y, 0) for irrational ω 2 /ω 3 : an informal study. Let us fix z ∈]0, 1/|S|[ such that ω 2 /ω 3 is irrational. We first informally 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 ℜω and ℑω the real and imaginary parts of ω ∈ C, respectively. Let Π x and Π y be the parallelograms defined by (7.1)

Π x = M 0,0 ∪ M 0,1 = ω 1 [0, 1[+ω 2 [0, 1[, Π y = N 0,0 ∪ N 0,1 = ω 3 /2 + ω 1 [0, 1[+ω 2 ]0, 1],
with notations (5.9) and (5.11). Function r x (ω) (resp. r y (ω)) on Π x (resp. Π y ) defines the first (main) branch of x → Q(x, 0) (resp. y → Q(0, y)) twice via (5.10) (resp. (5.12)).

Denote by f y (ω) = x(ω)[y(-ω + 2ω x 1 ) -y(ω)] the function used in the meromorphic continuation procedure (5.2). Assume that at some ω 0 ∈ Π y , r y (ω 0 ) = ∞ and f y (ω 0 ) = ∞. Further, suppose that

(7.2) ∄ω ∈ Π y : ℑω = ℑω 0 , f y (ω) = ∞.
By (5.2), for any n 1 we have

(7.3) r y (ω 0 + nω 3 ) = r y (ω 0 ) + f y (ω 0 ) + n-1 k=1 f y (ω 0 + kω 3 ).
We have r y (ω 0 ) + f y (ω 0 ) = ∞ by our assumptions. If ω 2 /ω 3 is irrational, then for any k 1 there is no p ∈ Z such that ω 0 + kω 3 = ω 0 + pω 2 . Function f y being ω 2 -periodic, it follows from this fact and assumption (7.2) that f y (ω 0 + kω 3 ) = ∞ for any k 1. Hence by (7.3), r y (ω 0 + nω 3 ) = ∞ for all n 1. Due to irrationality of ω 2 /ω 3 , for any n 1 there exists a unique ω n (ω 0 ) ∈ Π y and p ∈ Z such that ω 0 + nω 3 = ω n (ω 0 ) + pω 2 , and the set {ω n (ω 0 )} n 1 is dense on the curve (7.4) I y (ω 0 ) = y({ω :

ℑω = ℑω 0 , ω ∈ Π y }) ⊂ C ∪ {∞}.
By definition (5.12), the set of poles of all branches of y → K(0, y)Q(0, y) is dense on the curve I y (ω 0 ). The number of zeros of y → K(0, y) being at most two, the same conclusion holds true for y → Q(0, y).

Let us now identify the points ω 0 in Π y where f y (ω 0 ) is infinite. They are (at most) six such points a 1 , a 2 , a 3 , a 4 , b 1 , b 2 ∈ Π y , which correspond to the following pairs (x(ω 0 ), y(ω 0 )):

(7.5) a 1 = (x ⋆ , ∞), a 4 = (x ⋆ , y ⋆ ), a 2 = (x ⋆ , ∞), a 3 = (x ⋆ , y ⋆ ), b 1 = (∞, y • ), b 2 = (∞, y • ).
Here by (2.4) and (2.5)

x ⋆ = lim y→∞ -b(y) + [ b(y) 2 -4 a(y) c(y)] 1/2 2 a(y) , x ⋆ = lim y→∞ -b(y) -[ b(y) 2 -4 a(y) c(y)] 1/2 2 a(y) , y ⋆ = lim x→x ⋆ -b(x) + [b(x) 2 -4a(x)c(x)] 1/2 2a(x) , y ⋆ = lim x→x ⋆ -b(x) + [b(x) 2 -4a(x)c(x)] 1/2 2a(x) , y • = lim x→∞ -b(x) + [b(x) 2 -4a(x)c(x)] 1/2 2a(x) , y • = lim x→∞ -b(x) -[b(x) 2 -4a(x)c(x)] 1/2 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 y could be compensated in the sum (7.3). Furthermore, it may happen for some of these points that not only f y (ω 0 ) = ∞ but also r y (ω 0 ) = ∞, and consequently f y (ω) + r y (ω) may have no pole at ω = ω 0 . For 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 2 /ω 3 . In addition to the notation (7.4), define the curve (7.6)

I x (ω 0 ) = x({ω : ℑω = ℑω 0 , ω ∈ Π x }) ⊂ C ∪ {∞}.
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 2 /ω 3 is irrational, the following statements hold.

(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 3 and x 4 (resp. y 3 and y 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 1 , b 1 defined in (7.5), it is dense on the following curves (see Figure 11 (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 2 /ω 3 is irrational, for any ω 0 ∈ C and any n ∈ Z + , there exists a unique ω y n (ω 0 ) ∈ Π y (resp. ω x n (ω 0 ) ∈ Π x ) as well as a unique number p y ∈ Z (resp. p x ∈ Z) such that ω 0 + nω 3 = p y ω 2 + ω y n (ω 0 ) (resp. ω 0 + nω 3 = p x ω 2 + ω x n (ω 0 )). With these notations we can state the following lemma.

Lemma 18. Let z be such that ω 2 /ω 3 is irrational.

(a) For all n = m, we have

ω x n (ω 0 ) = ω x m (ω 0 ) and ω y n (ω 0 ) = ω y m (ω 0 ). (b) The set {ω x n (ω 0 )} n∈Z + (resp. {ω y n (ω 0 )} n∈Z + ) is dense on the segment {ω ∈ Π x : ℑω = ℑω 0 } (resp. {ω ∈ Π y : ℑω = ℑω 0 }).
Proof. Both (a) and (b) are direct consequences of the irrationality of ω 2 /ω 3 .

In the next definition, we introduce a partial order in Π y .

= ω ′ + pω 2 .
If ω ≪ ω ′ (and if ω 2 /ω 3 is irrational), both n and p are unique and sometimes we shall write ω ≪ n ω ′ . In particular, for any ω ∈ Π y , we have ω ≪ ω, since ω ≪ 0 ω. Definition 20. If either ω ≪ ω ′ or ω ≪ ω ′ , we say that ω and ω ′ are ordered, and we write ω ∼ ω ′ .

Let us denote by f x and f y the (meromorphic) functions used in the meromorphic continuation procedures (5.1) and (5.2), namely, by using (3.5):

f x (ω) = y(ω)[x( ηω) -x(ω)], f y (ω) = x(ω)[y( ξω) -y(ω)].
The following lemma will be the key tool for the proof of Theorem 17.

ω x 4 + ω 1 b 2 a 2 -ω 2 a 4 a 2 ω x 2 ω y 2 a 1 -ω 2 a 3 a 1 b 1 ω x 4 ω y 4 ω x 1 ω y 1 ω 3 /2 ω 3 /2
(7.8) ∀ω ∈ ∞ k=0 N k,0 with ℑω = {ℑa 1 , ℑa 2 , ℑa 3 , ℑa 4 , ℑb 1 , ℑb 2 }, r y (ω) = ∞.
Function x(ω) being ω 2 -periodic, it has no pole at any ω such that 0 ℑω < ω 1 and ℑω / ∈ {ℑb 1 , ℑb 2 }. Then Equation ( 5.3) and the fact that

∪ ∞ k=1 M k,0 ⊂ ∪ ∞ k=0 N k,0 imply that (7.9) ∀ω ∈ ∞ k=1 M k,0 with ℑω = {ℑa 1 , ℑa 2 , ℑa 3 , ℑa 4 , ℑb 1 , ℑb 2 }, r x (ω) = ∞.
In order to prove Theorem 17 (i), we shall prove the following proposition. The proof of this proposition is postponed to the next subsection. By this proposition, (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 1 , x 2 , x 3 , x 4 (resp. y 1 , y 2 , y 3 , y 4 ). Let us recall that the function x → Q(x, 0) is initially defined as a series i,n 0 q(i, 0; n)x i z n . The elementary estimate i 0 q(i, 0; n) |S| n implies that for any z ∈]0, 1/|S|[ and x ∈ C with |x| 1 this series is absolutely is convergent. Since |x 1 | < 1, |x 2 | < 1, and since also K(x, 0) is a polynomial with (at most two) roots that are smaller or equal to 1 by absolute value, the only singularities of the first branch of x → Q(x, 0) are the branch points x 3 and x 4 , that are out of the unit disc. By the same arguments the analogous statement 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 ∈ ∪ p k=1 M k,0 (resp. ω ∈ ∪ p ℓ=1 N 0,ℓ ) where f x (ω) = ∞ (resp. f y (ω) = ∞), Theorem 17 (ii) immediately follows from the meromorphic continuation procedure of r x and r y done in Section 5, namely Equations (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 x (ω 0 ) (resp. I y (ω 0 )) with ω 0 ∈ {a 1 , a 2 , a 3 , a 4 , b 1 , b 2 }. For any of 51 models, the set of these curves 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 0 appropriately chosen among a 1 , a 2 , a 3 , a 4 , b 1 , b 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).

: ω ∈ L y 3 y 4 + ω 2 , 0 < ℑω < ω 1 } (resp. {ω : ω ∈ L x 3 x 4 + ω 2 , 0 < ℑω < ω 1 }
). These points are symmetric w.r.t. the center of the interval, namely ω 1 /2 + ω 2 + ω 3 /2 (resp. ω 1 /2 + ω 2 ). The points ω corresponding to a 3 and a 4 are on the open interval {ω : ω ∈ L y 3 y 4 + ω 2 -ω 3 , 0 < ℑω < ω 1 } and are symmetric w.r.t. the center ω 1 /2 + ω 2 -ω 3 /2 as well. Furthermore, a 4 + ω 3 = a 2 and a 3 + ω 3 = a 1 , so that a 4 ≪ 1 a 2 and a 3 ≪ 1 a 1 , see Figure 12. Finally, we have ℑa 4 = ℑa 2 = ℑa 1 = ℑ 3 , hence for any a ∈ {a 2 , a 4 } and any a ′ ∈ {a 1 , a 3 }, a ∼ a ′ (in the sense of Definition 20).

If

y 4 > 0 or y 4 = ∞ (resp. x 4 > 0 or x 4 = ∞), the point y = ∞ (resp. x = ∞) is on ]y 4 , ∞] ∪ {∞}∪]∞, y 1 [ (resp. ]x 4 , ∞] ∪ {∞}∪]∞, x 1 [).
Accordingly, the points a 1 , a 2 (resp. b 1 , b 2 ) and also a 3 , a 4 are on the segment ]ω 3 /2, ω 2 + ω 3 /2].Their location on this 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 4 and y 4 : these are those of the walks pictured on Figure 17 The following remark gives a geometric interpretation of this classification.

Remark 24. Let 1 (i,j) be 1 if (i, j) ∈ S, otherwise 0. Then x 4 > 0 (resp. < 0, = ∞) if and only if 1 2 (1,0) -41 (1,1) 1 (1,-1) > 0 (resp. < 0, = 0), see Equation (2.2). A symmetric statement holds for y 4 .

As an example, Remark 24 implies that x 4 < 0 if and only if (1, 1) ∈ S and (1, -1) ∈ S.

Case I: y 4 < 0, Subcase I.A: x 4 < 0. This assumption yields

x ⋆ = x ⋆ ; x ⋆ , x ⋆ = ∞; y = y • ; y • , y • = ∞; y ⋆ , y ⋆ = ∞.
The location of the six points a 1 , a 2 , a 3 , a 4 , b 1 , b 2 is already described above and is pictured on Figure 12.

We first show that for all ω ∈ {ω : ℑω = ℑa 3 , ω y 1 ℜω < ω y 4 +ω 2 }, we have r y (ω) = ∞. The proof consists in three steps. Introducing, for any ω 0 , the set

(7.11) O ∆ (ω 0 ) = {ω 0 -ω 3 , ω 0 -2ω 3 , . . . , ω 0 -n ω 0 ω 3 }, n ω 0 = inf{ℓ 0 : ω 0 -ℓω 3 ∈ ∆},
we can rewrite (7.10) as

(7.12) r y (a 3 ) = r y (a 3 -n a 3 ω 3 ) + ω∈O ∆ (a 3 ) f y (ω).
In (7.12), the quantity r y (a 3 -n a 3 ω 3 ) is defined thanks to Theorem 3. It may be infinite, but only if y(a 3 -n a 3 ω 3 ) = ∞. In this case we must have a 3 -n a 3 ω 3 = a 1 -ω 2 . But since a 3 + ω 3 = a 1 , we then have (n a 3 + 1)ω 3 = ω 2 which is impossible, due to irrationality of ω 2 /ω 3 . Hence r y (a 3 -n a 3 ω 3 ) = ∞. Further, we immediately have (see indeed Figure 12) that a 2 , a 4 , a 2 -ω 2 , a 4 -ω 2 / ∈ O ∆ (a 3 ). Moreover, since either ℑb 1 = ℑa 3 , or ℑb 1 = ℑa 3 but then a 3 +ω 3 /2 = b 1 (see again Figure 12), we also have that b

1 , b 2 , b 1 -ω 2 , b 2 -ω 2 / ∈ O ∆ (a 3 ). Finally a 1 = a 3 + ω 3 / ∈ O ∆ (a 3 ) and a 1 -ω 2 = a 3 + ω 3 -ω 2 / ∈ O ∆ (a 3 ), since ω 2 /ω 3 is irrational. Thus f y (a 3 -kω 3 ) = ∞ for any k ∈ {1, .
. . , n a 3 }. Accordingly, r y (a 3 ) = ∞ and by the same arguments, r y (a 4 ) = ∞.

Step 2. We now show that r y (a

1 -ω 2 )+f y (a 1 -ω 2 ) = ∞ and r y (a 2 -ω 2 )+f y (a 2 -ω 2 ) = ∞. By Equation (5.3), r y (a 1 -ω 2 ) = -r x (a 1 -ω 2 ) + K(0, 0)Q(0, 0) + x(a 1 -ω 2 )y(a 1 -ω 2 ) and f y (a 1 -ω 2 ) = x(a 1 -ω 2 )[y(a 4 ) -y(a 1 -ω 2 )]; hence (7.13) r y (a 1 -ω 2 ) + f y (a 1 -ω 2 ) = -r x (a 1 -ω 2 ) + K(0, 0)Q(0, 0) + x(a 1 -ω 2 )y(a 4 ).
It follows from Equation (5.3) and from the first step that r x (a 4 ) = ∞, since x(a 4 )y(a 4 ) = x ⋆ y ⋆ = ∞. Then, by (3.5) and (5.4) we get that r x (a 1 -ω 2 ) = r x ( ξa 4 ) = r x (a 4 ) = ∞. Furthermore, x(a 1 -ω 2 )y(a 4 ) = x ⋆ y ⋆ = ∞. Finally, thanks to (7.13), r y (a 1 -ω 2 ) + f y (a 1ω 2 ) = ∞ and by the same arguments, r y (a 2 -ω 2 ) + f y (a 2 -ω 2 ) = ∞.

Step 3. Let us now take any ω 0 in {ω : ℑω = ℑa 3 , ω y 1 ℜω < ω y 4 + ω 2 }. If ω 0 ∈ ∆, then ω 0 ∈ ∆ y . Indeed, it is proved in Section 4 that the domain ∆ y (resp. ∆ x ), which is bounded by Γ 0 y and Γ 1 y (resp. Γ 0 x and Γ 1 x ), is centered around

L y 2 y 1 (resp. L x 2 x 1 ). Furthermore, Γ 0 y ∈ ∆ x and Γ 1 y / ∈ ∆ x (resp. Γ 0 x ∈ ∆ y and Γ 1 x / ∈ ∆ y ). It follows that for any ω 0 ∈ ∆ \ ∆ y , ℜω 0 < ω y 1 . Then, by Theorem 3, r y (ω 0 ) = ∞. If ω 0 /
∈ ∆, with (7.11) and (7.12) we have

r y (ω 0 ) = r y (ω 0 -n ω 0 ω 3 ) + ω∈O ∆ (ω 0 ) f y (ω).
For the same reasons as in the first step, we have that a

2 , a 4 , a 2 -ω 2 , a 4 -ω 2 / ∈ O ∆ (ω 0 ). If ℜω 0 < ℜa 3 , for obvious reasons O ∆ (ω 0 ) cannot contain a 3 . If ℜa 3 ℜω 0 < ω y 4 +ω 2 , it can neither contain a 3 , since ω y 4 + ω 2 -ℜa 3 = ω 3 , and hence ℜω 0 -ω 3 < ℜa 3 . If ℑb 1 = ℑa 3 , or ℑb 1 = ℑa 3 and ℜω 0 < ℜb 1 , it cannot contain b 1 . If ℑb 1 = ℑa 3 and ℜb 1 ℜω 0 < ω y 4 + ω 2 , then ℜω 0 -ℜb 1 ω y 4 + ω 2 -ℜb 1 = ω 3 /2 < ω 3 , and b 1 / ∈ O ∆ (ω 0 ). If a 1 -ω 2 /
∈ O ∆ (ω 0 ), then we have r y (ω 0 -n ω 0 ω 3 ) = ∞ and f y (ω 0 -kω 3 ) = ∞ for all k ∈ {1, . . . , n ω 0 }, so that r y (ω 0 ) = ∞ by (7.12).

If a 1 -ω 2 ∈ O ∆ (ω 0 ), then for some j ∈ {1, . . . n ω 0 }, we have ω 0 -jω 3 = a 1 -ω 2 . Then r y (ω 0 -n ω 0 ω 3 ) + j+1 k=n f y (ω 0 -kω 3 ) = r y (a 1 -ω 2 ) and thus by (7.12),

r y (ω 0 ) = r y (a 1 -ω 2 ) + f y (a 1 -ω 2 ) + 1 k=j-1 f y (ω 0 -kω 3 ).
The first term here is finite by the second step and f y (ω 0 -kω 3 ) = ∞ for k ∈ {1, . . . , j -1} by all properties said above, so that r y (ω 0 ) = ∞.

So far we have proved that for all ω ∈ {ω : ℑω = ℑa 3 , ω y 1 ℜω < ω y 4 +ω 2 }, r y (ω) = ∞. In the same way, we obtain that r y (ω) = ∞ for ω ∈ {ω :

ℑω = ℑa 4 , ω y 1 ℜω < ω y 4 + ω 2 }.
Since by (3.5), We proceed with the proof of Proposition 23. Let us verify the assumptions of Lemma 21 for ω 0 = a 3 , a 4 , b 1 , b 2 . We have proved that r y (a 3 ), r y (a 4 ), r y (b 1 ), r y (b 2 ) = ∞, a 3 ≪ 1 a 1 , a 4 ≪ 1 a 2 and that the pairs {a 1 , a 3 } and {a 2 , a 4 } are not ordered. Let us now show that for any k ∈ {3, 4} and ℓ ∈ {1, 2}, it is impossible to have 12). From the irrationality of ω 2 /ω 3 , it follows that b ℓ ∼ a 1 , a 3 and in the same way b ℓ ∼ a 2 , a 4 . Then there is no other ω ∈ Π y except for a 1 (resp. a 2 ) such that a 3 ≪ ω (resp. a 4 ≪ ω) and f y (ω) = ∞. There is no ω ∈ Π y such that b ℓ ≪ ω and f y (ω) = ∞, ℓ = 1, 2. Hence, Lemma 21 could be applied to any of four points ω 0 = a 3 , a 4 , b 1 , b 2 if the assumption (B) of this lemma is satisfied for these points. It is then immediate that

η{ω : ℑω = ℑa 3 , ω y 1 ℜω < ω y 4 + ω 2 } = {ω : ℑω = ℑa 4 , ω y 4 < ℜω ω y 1 }, η{ω : ℑω = ℑa 4 , ω y 1 ℜω < ω y 4 + ω 2 } = {ω : ℑω = ℑa 3 , ω y 4 < ℜω ω y 1 }, Equation (5.
a k ∼ b ℓ . If ℑb ℓ = ℑa 3 , ℑa 4 , this is obvious. If ℑb ℓ = ℑa 3 , then it is enough to note that b ℓ -a 3 = ω 3 /2 and a 1 -b 1 = ω 3 /2 (see Figure
lim ω→b ℓ f y (ω) = lim ω→b ℓ x(ω)[y( ξω)-y(ω)] = ∞, ℓ ∈ {1, 2}, since x(ω) → ∞ and the other term converges to ±[y • -y • ] = 0. Let us verify that lim ω→a 3 {f y (ω) + f y (ω + ω 3 )} = ∞.
We have

lim ω→a 3 {f y (ω) + f y (ω + ω 3 )} = lim ω→a 3 {x(ω)[y( ξω) -y(ω)] + x( η ξω)[y( ξ η ξω) -y( η ξω)]} = lim ω→a 3 {x( η ξω)y( ξ η ξω) -x(ω)y(ω)} + lim ω→a 3 {x(ω)y( ξω) -x( η ξω)y( η ξω)}.
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 3 , then ξω → a 2 -ω 2 so that the first term in the product converges to y(a 2 -ω 2 ) = y(a 2 ) = ∞. The second term of this product converges to x(a 2 -ω 2 ) -x(a 1 ) = x ⋆ -x ⋆ which is different from 0 as x ⋆ = x ⋆ .
Then assumption (B) is satisfied for ω 0 = a 3 and in the same way for ω 0 = a 4 . Lemma 21 applies to any of the four points

ω 0 = a 3 , a 4 , b 1 , b 2 . But by (3.7), I x (a 3 ) = I x (a 4 ) = I x (a 1 ) = I x (a 2 ), I y (a 3 ) = I y (a 4 ) = I y (a 1 ) = I y (a 2 ), I x (b 1 ) = I x (b 2 ), I y (b 1 ) = I y (b 2
) so that poles of x → Q(x, 0) are dense on the curves I x (a 1 ) and I x (b 1 ) and those of y → Q(0, y) are dense on the curves I y (a 1 ) and I y (b 1 ). Proposition 23 is proved.

Case I: y 4 < 0, Subcase I.B: x 4 = ∞. This assumption implies that

x ⋆ = x ⋆ ; x ⋆ , x ⋆ = ∞; y • = y • = ∞; y ⋆ , y ⋆ = ∞.
The points a 1 , a 2 , a 3 , a 4 are located as in the previous case, see Figure 12. Consequently we have the following facts: r y (ω) = ∞ on the segments {ω ∈ Π y : ℑω = a 3 , a 4 }, except for their ends ω = a 1 , a 2 ; r x (ω) = ∞ on the segments {ω : ℑω = a 3 , a 4 , ω x 1 ℜω ω x 4 +ω 2 }. Lemma 21 applies to ω 0 = a 3 , a 4 as in the previous case, as x ⋆ = x ⋆ . Then the set of poles of all branches of x → Q(x, 0) (resp. y → Q(0, y)) is dense on I x (a 1 ) (resp. I y (a 1 )) where I x (a 1 ) = I x (a 2 ) = I x (a 3 ) = I x (a 4 ) (resp. I y (a 1 ) = I y (a 2 ) = I y (a 3 ) = I y (a 4 )).

Since x 4 = ∞, we have that b 1 = b 2 = ω x 4 + ω 2 . Take any ω 0 with ℑω 0 = 0 such that ω y 1 ℜω 0 ω y 4 + ω 2 . Then y(ω 0 ) = ∞. Let us show that r y (ω 0 ) = ∞. If ω 0 ∈ ∆, then by the same reasons as in Subcase I.A ω 0 ∈ ∆ y and r y (ω 0 ) = ∞. If ω 0 / ∈ ∆, consider the set O ∆ (ω 0 ) defined as in (7.11) and (7.12). Clearly b 1 -

ω 2 = ω x 4 / ∈ O ∆ (ω 0 ). Since b 1 + ω 3 /2 = ω y 4 + ω 2 , we have ω 0 -ω 3 ω y 4 + ω 2 -ω 3 < b 1 and then b 1 / ∈ O ∆ (ω 0 ). Hence O ∆ (ω 0 )
does not contain any point where y(ω) or f y (ω) is infinite. Thus by (7.12), r y (ω 0 ) = ∞.

We have η{ω : (5.4) and (5.5), we get that r y (ω) = ∞ for all ω ∈ Π y with ℑω = 0. The segment {ω : ℑω = 0, ω x 1 ω ω x 4 + ω 2 } does not contain any point with y(ω) = ∞. By (5.3) this gives r x (ω) = ∞ for all ω on this segment except for the points where x(ω) = ∞ (that is only at ω = ω x 4 + ω 2 = b 1 ), and this concludes the proof of Proposition 22.

ℑω = 0, ω y 1 ℜω 0 ω y 4 + ω 2 } = {ω : ℑω = ω 1 , ω y 4 ℜω 0 ω y 1 }. Then by
We have proved in particular that r y (ω 0 ) = ∞ for ω 0 = b 1 . Furthermore, there is no ω ∈ Π y such that b 1 ≪ ω and f y (ω) = ∞. Finally (7.14) Hence, for two of these roots b(y) = -2[ a(y) c(y)] 1/2 and then X(y) 0 (see (2.4)), and for the two others b(y) = 2[ a(y) c(y)] 1/2 and then X(y) 0. But X(y 2 ) and X(y 3 ) are on the segment [x 2 , x 3 ] ⊂]0, ∞[. Thus X(y 1 ) 0 and X(y

lim ω→b 1 f y (ω) = lim ω→b 1 x(ω)[y( ξω) -y(ω)] = lim ω→b 1 x(ω) [b(x(ω)) 2 -4a(x(ω))c(x(ω))] 1/2 a(x(ω)) , b 1 b 2 ω x 4 ω y 4 ω x 1 ω y 4 + ω 2 -ω 3 ω x 4 + ω 2 ω y 4 + ω 2
4 ) 0. Since x(b 1 ) = x(b 1 -ω 2 ) = ∞, x 4 = x(ω x 4 ) > 0 and X(y 4 ) = x(ω y 4 ) < 0, it follows that b 1 -ω 2 ∈]ω x 4 , ω y 4 [, in such a way that b 1 ∈]ω x 4 + ω 2 , ω y 4 + ω 2 [. Also, b 2 = ξ(b 1 -ω 2 ) -ω 1 = 2(ω x 4 + ω 2 ) -b 1 is symmetric to b 1 w.r.t. ω x 4 + ω 2 . Since x(ω y 1 ) = X(y 1 ) 0 and x(ω x 4 + ω 2 ) = x 4 > 0, it follows that ω y 1 < b 2 < ω x 4 + ω 2 , see
-ω 2 < ω y 1 ω 0 , it follows that b 1 -ω 2 < ω 0 -n ω 0 ω 3 (see (7.11)), so that b 1 -ω 2 / ∈ O ∆ (ω 0 ). Obviously b 2 -ω 2 / ∈ O ∆ (ω 0 ). Furthermore, since ω y 4 + ω 2 -b 2 = ω 3 /2+ ω x 4 + ω 2 -b 2 < ω 3 /2+ ω 3 /2 = ω 3 , it follows that ω 0 -ω 3 < b 2 < b 1 for any such ω 0 . Hence b 1 , b 2 /
∈ O ∆ (ω 0 ). Thus for any ω ∈ O ∆ (ω 0 ), y(ω) = ∞ and f y (ω) = ∞. By (7.12), r y (ω 0 ) = ∞. This implies, exactly as in Subcase I.B-by (5.4) and (5.5)-, that r y (ω) = ∞ for all ω ∈ Π y such that ℑω = 0. By (5.3) this gives r x (ω) = ∞ for all ω with ℑω = 0 and ω x 1 ℑω ω x 4 + ω 2 , except for points ω where x(ω) = ∞ (that happens for ω = b 2 only), and this concludes the proof of Proposition 22.

In particular, we proved that r y (b 1 ) = ∞ and also r y (b 2 ) = ∞. Since y • = y • , we have lim ω→b 1 f y (ω) = ∞ and also lim Case II: y 4 > 0, location of a 1 , a 2 , a 3 , a 4 . We first exclude Subcase II.D where x 4 = ∞ and Y (x 4 ) = ∞, and we locate the points a 1 , a 2 , a 3 , a 4 . In this case we have x ⋆ = x ⋆ . Note that x 1 , x 2 , x 3 , x 4 are the roots of the equation It follows from the above considerations that a 1 ∈]ω x 4 + ω 2 , ω y 4 + ω 2 [, see Figure 14. In particular, we have a 1 -ω 2 ∈]ω x 4 , ω y 4 [ and a 2 = ηa 1 -ω 1 = -(a 1 -ω 2 ) + 2ω y 4 . This means that a 1 -ω 2 and a 2 are symmetric w.r.t. ω y 4 . Furthermore a 2 ∈]ω y 4 , ω y 1 [, but since y 4 > 0 and Y (x 1 ) 0, we have a 2 ∈]ω y 4 , ω x 1 [. We must put a 3 = ξa 2 -ω 1 = -a 2 + 2ω x 1 , in such a way that the points a 2 and a 3 are symmetric w.r.t. ω x 1 . Finally,

d(x) = (b(x) -[a(x)c(x)] 1/2 )(b(x) + [a(x)c(x)] 1/2 ) = 0.
a 4 = ξ(a 1 -ω 2 ) -ω 1 = -(a 1 -ω 2 ) + 2ω x 1 = a 3 + a 2 -(a 1 -ω 2 ). Note that ω x 4 + ω 2 -a 4 = a 1 -ω 2 -ω x 4 > 0.
Furthermore, a 1 -a 3 = ω 3 and a 2 + ω 2 -a 4 = ω 3 , so that a 3 ≪ a 1 and a 4 ≪ a 2 .

Case II: y 4 > 0, Subcase II.A: x 4 < 0. In this case we have y Consider now r x (ω) and r y (ω) for ω with ℑω = 0. See Figure 14 for the location of the points a 1 , a 2 , a 3 , a 4 .

• = y • ; y • , y • = ∞; x ⋆ , x ⋆ = ∞; y ⋆ , y ⋆ = ∞.
We first prove that for (7.15) r y (ω 0 ) = ∞, ∀ω 0 ∈ {ω : ℑω = 0, ω y 1 ω 0 ω y 4 + ω 2 } \ {a 1 }.

The proof consists in three steps.

Step 1. We prove that r y (a 3 ) = ∞ and r y (a 4 ) = ∞.

If a 3 ∈ ∆, then necessarily r y (a 3 ) = ∞, since x ⋆ , y ⋆ = ∞. Otherwise |x ⋆ |, |y ⋆ | 1, and then a 2 = ξa 3 -ω 1 / ∈ ∆. Since a 2 < ω x 1 < a 3 , ω x 1 ∈ ∆ and a 2 , a 3 / ∈ ∆, it follows that any ω ∈ O ∆ (a 3 ) must be in ]a 2 , a 3 [, hence f y (ω) = ∞, x(ω) = ∞ and y(ω) = ∞. Thus by (7.12), r y (a 3 ) = ∞.

We now show that r y (a 4 ) = ∞. Suppose first that a 1 -ω 2 ∈ ∆. Then, since a 1 -ω 2 / ∈ Π y , we have a 1 -ω 2 ∈ ∆ x , so that r x (a 1 -ω 2 ) = ∞ by Theorem 3. Then by Equations (5.4) and ( 5

.5), r x (a 4 ) = r x ( ξ(a 1 -ω 2 ) -ω 1 ) = r x (a 1 ) = ∞. Since x ⋆ , y ⋆ = ∞, we also have r y (a 4 ) = ∞ by (5.3). Assume now that a 1 -ω 2 / ∈ ∆. Since a 1 -ω 2 < ω x 1 < a 4 , then a 1 -ω 2 / ∈ O ∆ (a 4 ). Furthermore a 2 / ∈ O ∆ (a 4 ) as a 4 + ω 3 = a 2 + ω 2 and ω 3 /ω 2 is irrational. Finally a 4 -a 3 < a 1 -a 3 = ω 3 , so that a 3 / ∈ O ∆ (a 4
). It follows that for any ω ∈ O ∆ (a 4 ), we have f y (ω) = ∞ and y(ω) = ∞. Hence r y (a 4 ) = ∞.

Step 2. We prove that r y (a 2 ) + f y (a 2 ) = ∞ and that r y (a

1 -ω 2 ) + f y (a 1 -ω 2 ) = ∞. By Equation (5.3) r y (a 2 ) + f y (a 2 ) = -r x (a 2 ) + K(0, 0)Q(0, 0) + x(a 2 )y(a 2 ) + x(a 2 )[y( ξa 2 ) -y(a 2 )] = -r x (a 2 ) + K(0, 0)Q(0, 0) + x(a 2 )y( ξa 2 ). (7.16)
Since r y (a 3 ) = ∞ and since x ⋆ , y ⋆ = ∞, it follows from Equation (5.3) that r x (a 3 ) = ∞. Then by (5.4) and (5.5), r x (a 2 ) = r x ( ξa 3 -ω 1 ) = r x (a 3 ) = ∞. Since x(a 2 ) = x ⋆ = ∞ and y( ξa 2 ) = y ⋆ = ∞, (7.16) is finite. By completely analogous arguments we obtain that r y (a 1 -ω 2 ) + f y (a 1 -ω 2 ) = ∞.

Step 3. Let us now show (7.15). If ω 0 ∈ ∆, then by the same arguments as in Subcase I.A ω 0 ∈ ∆ y and then r y (ω 0 ) = ∞ by Theorem 3. Otherwise, consider O ∆ (ω 0 ) defined in (7.11). Note that (7.17)

ω / ∈ O ∆ (ω 0 ), ∀ω ∈]ω x 4 + ω 2 -ω 3 /2, ω y 4 + ω 2 ],
since in this case ω 0 -ω 3 < ω. In particular, a 4 / ∈ O ∆ (ω 0 ). Furthermore, a 3 ∈ O ∆ (ω 0 ) implies that ω 0 = a 3 + ℓω 3 for some ℓ 1. But a 3 + ω 3 = a 1 = ω 0 and a 3 + ℓω 3 > ω y 4 + ω 2 for any ℓ 2. Hence a 3 / ∈ O ∆ (ω 0 ). Since a 2 -(a 1 -ω 2 ) < ω 3 , it is impossible that both a 2 and a 1 -ω 2 belong to O ∆ (ω 0 ). If none of them belongs to O ∆ (ω 0 ), then y(ω) = ∞ and f y (ω) = ∞ for any O ∆ (ω 0 ), and then r y (ω 0 ) = ∞. Suppose, e.g., that a 2 ∈ O ∆ (ω 0 ). Then for some ℓ 1, ω 0 = a 2 +ℓω 3 , and by (7.12), r y (ω 0 ) = r y (a 2 )+f y (a 2 )+ 1 k=ℓ-1 f y (ω 0 -kω 3 ). But r y (a 2 ) + f y (a 2 ) = ∞ by the second step, and obviously 1 k=ℓ-1 f (ω 0 -kω 3 ) = ∞ by all facts said above, so that r y (ω 0 ) = ∞. The reasoning is the same if a 1 -ω 2 ∈ O ∆ (ω 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 y (ω 0 ) = ∞ for all ω 0 = a 2 = ηa 1 with ℑω 0 = 0 and ω y 4

ℜω 0 ω y 1 as well. Next,

a 2 a 3 a 1 = a 4 ω x 4 ω y 4 ω x 1 ω x 4 + ω 2
(ω) = ∞. If a 3 , a 4 , b 1 , b 2 are ordered (e.g., a 3 ≪ b 1 ≪ a 4 ≪ b 2 , then immediately a 3 ≪ 1 a 1 ≪ b 1 ≪ a 4 ≪ 1 a 2 ≪ b 2 )
, there is a maximal point in the sense of this order. If the maximal element is b ℓ for some ℓ ∈ {1, 2}, then there is no ω ∈ Π y with f y (ω) = ∞ such that b ℓ ≪ ω. If the maximal element is a 3 (resp. a 4 ), then there is no ω ∈ Π y except for a 1 (resp. a 2 ) with f y (ω) = ∞ such that a 3 ≪ ω (resp. a 4 ≪ ω). Lemma 21 applies with ω 0 equal this maximal element since all assumptions (A), (B) and (C) are satisfied. If a 3 , a 4 , b 1 , b 2 are not all ordered, then it is enough to apply Lemma 21 to the maximal element of any ordered subset. Finally I

x (ω 0 ) = R\]x 1 , x 4 [ and I y (ω 0 ) = R\]y 4 , y 1 [ for any ω 0 ∈ {a 3 , a 4 , b 1 , b 2 }, hence the set of poles of x → Q(x, 0) (resp. y → Q(0, y)
) is dense on the announced intervals. Proposition 23 is proved.

Case II: y 4 > 0, Subcase II.D:

x 4 = ∞ and Y (x 4 ) = ∞. In this case x ⋆ = ∞; y ⋆ = ∞; x ⋆ , y ⋆ = ∞.
We have a 1 = ω x 4 + ω 2 , and a 2 = ηa 1 -ω 1 = ω x 4 + ω 3 is symmetric to a 1 -ω 2 w.r.t. ω y 4 . Further, since y 4 > 0 and Y (x 1 ) 0, a

2 ∈]ω y 4 , ω x 1 [. Then a 3 = ξa 2 -ω 1 = ω x 4 + ω 2 -ω 3 is symmetric to a 2 w.r.t. ω x 1 . Finally, a 4 = ω x 4 + ω 2 = a 1 , b 1 = b 2 = a 1 , a 3 + ω 3 = a 1 and a 1 + ω 3 = a 2 + ω 2 , so that a 3 ≪ 1 a 1 ≪ 1 a 2 , see Figure 15.
We prove (7.15). For this purpose, we first show that r y (a

3 ) = ∞. If a 3 ∈ ∆, x ⋆ , y ⋆ = ∞, then by Theorem 3, r y (a 3 ) = ∞ If a 3 / ∈ ∆, consider O ∆ (ω 0 ). Since a 3 +2ω 3 = a 2 +ω 2 , a 2 /
∈ O ∆ (ω 0 ) by the irrationality of ω 2 /ω 3 . Obviously a 1 -ω 2 = ω x 4 / ∈ ∆ and then it is not in O ∆ (ω 0 ). Hence r y (a 3 ) = ∞. Next we prove that r y (a 2 )+f y (a 2 ) exactly as in Subcase II.A by using that x ⋆ y ⋆ = ∞.

For any ω 0 = a 1 with ℑω 0 = 0 and ω y 1 ω 0 ω y 4 +ω 2 , the orbit O ∆ (ω 0 ) cannot contain a 3 , since a 3 + ω 3 = a 1 and a 3 + 2ω 3 > ω 0 . It can neither contain a 1 , since ω 0 -ω 3 < a 1 , nor obviously a 1 -ω 2 = ω x 4 . If it does not contain a 2 , then by (7.12), r y (ω 0 ) = ∞. If it does, then exactly as in Subcase II.A, using r y (a 2 ) + f y (a 2 ) = ∞, we prove that r y (ω 0 ) = ∞ as well. This finishes the proof of (7.15).

By Equations (5.4), (5.5) and (5.3), we derive as in Subcase II.A that r x (ω 0 ) = ∞ for all ω 0 ∈ {ω : ℑω = 0, ω x 1 ω 0 ω x 4 + ω 2 } except for ω 0 where x(ω 0 ) = ∞, that is for a 1 . This finishes the proof of Proposition 22 in this case.

a 3 = a 4 b 1 = b 2 a 1 = a 2 ω x 4 ω y 4 ω x 1 ω y 4 + ω 2 -ω 3 ω x 4 + ω 2 ω y 4 + ω 2 Figure 16. Location of a 1 , a 2 , a 3 , a 4 , b 1 , b 2 if y 4 = ∞, Case III
To prove Proposition 23, we would like to apply Lemma 21 with ω 0 = a 3 . We have shown that r y (a 3 ) = ∞, a 3 ≪ 1 a 1 = a 4 = b 1 = b 2 ≪ a 2 , so that there is no ω ∈ Π y \ {a 1 , a 2 } such that a 3 ≪ ω and f y (ω) = ∞. It remains to verify assumption (B) of Lemma 21 for ω 0 = a 3 , that is that f y (ω) + f y (ω + ω 3 ) + f y (ω + 2ω 3 ) converges to infinity if ω → a 3 . The 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( η ξ 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 η ξω) where we used (3.7). If ω → a 3 , then the first term in this sum converges to x ⋆ y ⋆ -x ⋆ y ⋆ = 0. Next, η ξω → a 1 , so that x( η ξω) → ∞. We can also compute the values of y( ξω) and y( ξ η ξω) as Since x ⋆ , y ⋆ = ∞, it follows from Equations (5.3), (5.4) and (5.5) that r x (a 1 -ω 2 ) = r x ( ξ(a 1 -ω 2 )) = r x ( ξ(a 1 -ω 2 ) -ω 1 ) = r x (a 3 ) = ∞. Then, by (5.3), r y (a 1 -ω 2 ) + f y (a 1ω 2 ) = -r x (a 1 -ω 2 ) + K(0, 0)Q(0, 0) + x(a 1 -ω 2 )y( ξ(a 1 -ω 2 )) = ∞.

(-b(x) ± [b(x) 2 -4a(x)c(x)] 1/2 )/(2a(x)) with x = x( η ξω).
Take any ω 0 with ℑω 0 = 0 and ω y 1 ω 0 < ω y 4 + ω 2 . If ω 0 ∈ ∆, then by the same arguments as in Subcase I.A, ω 0 ∈ ∆ y , so that r y (ω 0 ) = ∞. Otherwise, we notice that ω 0 -ω 3 < a 3 , so that no point-and in particular b 1 -of [ω y 4 + ω 2 -ω 3 , ω y 4 + ω 2 [ belongs to O ∆ (ω 0 ). Clearly b 1 -ω 2 = ω x 4 / ∈ O ∆ (ω 0 ). Since either a 1 -ω 2 / ∈ O ∆ (ω 0 ) or a 1 -ω 2 ∈ O ∆ (ω 0 ) but r y (a 1 -ω 2 ) + f y (a 1 -ω 2 ) = ∞, and by the same reasoning as in Subcase I.A, we derive that r y (ω 0 ) = ∞. The rest of the proof of Proposition 22 in this case goes along the same lines as in Case II. Now note that b 1 is not ordered with a 1 and a 3 . Indeed, since b 1 + ω 3 /2 = a 1 and b 1 -ω 3 /2 = a 3 , this would contradict the irrationality of ω 2 /ω 3 . Then there is no ω ∈ Π y such that b 1 ≪ ω and f y (ω) = ∞. We also have f y Proof. In order to prove (7.19), we shall use expressions of the periods ω 2 and ω 3 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,

K(k) = 1 0 dt [1 -t 2 ] 1/2 [1 -k 2 t 2 ] 1/2 , (7.20) F (w, k) = w 0 dt [1 -t 2 ] 1/2 [1 -k 2 t 2 ] 1/2 . (7.21)
Then the new expressions of ω 2 and ω 3 are (7. [START_REF] Raschel | Paths confined to a quadrant[END_REF])

ω 2 = M Ω 2 , ω 3 = M Ω 3 ,
where Ω 2 = K (x 4 -x 1 )(x 3 -x 2 ) (x 4 -x 2 )(x 3 -x 1 ) , (7.23) Ω 3 = F (x 4 -x 2 )(x 1 -X(y 1 )) (x 4 -x 1 )(x 2 -X(y 1 )) , (x 4 -x 1 )(x 3 -x 2 ) (x 4 -x 2 )(x 3 -x 1 ) , (7.24) and where (below, 1 (i,j) = 1 if (i, j) ∈ S, otherwise 0)

(7.25) M =            2 z 1 (1 (1,0) -41 (1,1) 1 (1,-1) )(x 3 x 4 -x 2 x 3 -x 1 x 4 + x 1 x 2 ) if x 4 = ∞, 2 (2z1 (1,0) + 4z 2 [1 (1,1) 1 (0,-1) + 1 (1,-1) 1 (0,1) ])(x 3 -x 1 ) if x 4 = ∞.
The expressions of ω 2 and ω 3 written in (7.22), (7.23), (7.24) and (7.25) are obtained from (3.1) and (3.2) by making simple changes of variables.

We are now in position to analyze the behavior of ω 2 /ω 3 (or equivalently, thanks to (7.22), that of Ω 2 /Ω 3 ) in the neighborhood of z = 0. First, with (2.2) and [START_REF] Stanley | Enumerative Combinatorics[END_REF]Proposition 6.1.8], we obtain that as z → 0, x 1 , x 2 → 0 and x 3 , x 4 → ∞. For this reason, (7.26) k = (x 4 -x 1 )(x 3 -x 2 ) (x 4 -x 2 )(x 3 -x 1 ) → 1.

The behavior of X(y 1 ) as z → 0 is not so simple as that of the branch points x ℓ (indeed, as z → 0, X(y 1 ) can converge to 0, to ∞ or to some non-zero constant), but we can show that for all 51 models, (7.27) w = (x 4 -x 2 )(x 1 -X(y 1 )) (x 4 -x 1 )(x 2 -X(y 1 )) → 1.

Due to (7.26) and (7.27), in order to determine the behavior of Ω 2 /Ω 3 near z = 0 it suffices to know (i) the expansion of K(k) as k → 1;

(ii) the expansion of F (w, k) as k → 1 and w → 1. Point (i) is classical, and is known as Abel's identity (it can be found, e.g., in [START_REF] Gerretsen | Lectures on the theory of functions of a complex variable II: Geometric theory[END_REF]): 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 [START_REF] Gerretsen | Lectures on the theory of functions of a complex variable II: Geometric theory[END_REF], 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: As for Point (ii), we proceed as follows. We have F (w, k) = K(k) -F (w, k), with

A(k) = (
F (w, k) = 1 w dt [1 -t 2 ] 1/2 [1 -k 2 t 2 ] 1/2 .
Then, introduce the expansion 1/([1 + t] 1/2 [1 + kt] 1/2 ) = ∞ ℓ=0 µ ℓ (k)(1 -t) ℓ , so that (7.28)

F (w, k) = ∞ ℓ=0 µ ℓ (k) 1 w dt [1 -t] 1/2-ℓ [1 -kt] 1/2 .
In Equation (7.28), all µ ℓ (k) as well as all integrals can be computed. As an example (that we shall use), we have

µ 0 (k) 1 w dt [1 -t] 1/2 [1 -kt] 1/2 = 1 [2k(1 + k)] 1/2 × × [ln{(1 -k)/k 1/2 } -ln{-[1 -(1 + k)w + kw 2 ] 1/2 + [(k + 1)/2 -kw]/k 1/2 }].
Moreover, it should be noticed that as k → 1 and w → 1, the speed of convergence to zero of the integrals in (7.28) increases with ℓ. 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 with S = {(-1, 0), (-1, 1), (0, 1), (1, -1)} (which belongs to Subcase II.D of Figure 17). For this model, we easily obtain from (2.2) that x 1 = z -2z 2 + 3z 3 + O(z 4 ), The latter proves (7.19), and thus Proposition 25, with L = 4 and L = -4 ln(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 Ω 2 /Ω 3 . A contrario, we could also be interested 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 2 /ω 3 is not constant in z.) To that aim, instead of (7.29) and (7.30), we just need two-terms expansions of k and w, say k = 1 + αz p + o(z p ) and w = 1 + βz q + o(z q ), with α, β = 0. Then with (i) and (ii) we deduce that Ω 2 = -(p/2) ln(z)+ o(ln(z)) and Ω 3 = -(q/2) ln(z)+ o(ln(z)), in such a way that L = p/q, which obviously is (non-zero and) rational.

Figure 3 . 2 . 2 .

 322 Figure 3. Construction of the Riemann surface

Figure 5 .

 5 Figure 5. The Riemann surface C/(ω 1 Z + ω 2 Z) and the location of the branch points

Figure 8 . 3 . 3 .

 833 Figure 8. Location of the domains D x and D y on the Riemann surface T

4. 1 .

 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 1 and x 2 ) are in the unit disc. For this reason, and by construction of the surface T, the set {s ∈ T : |x(s)| = 1} is composed of two cycles (one belongs to S 1x and the other to S 2 x ) homological to the cut {s ∈ T : x(s) ∈ [x 1 , x 2 ]}. The domain D x = {s ∈ T : |x(s)| < 1} is bounded by these two cycles, see Figure8, and contains the points s

Figure 8 .

 8 In the same way, for any z ∈]0, 1/|S|[ and any y such that |y| = 1, we have |X 0 (y)| < 1 and |X 1 (y)| > 1. Therefore, the cycles composing the boundary of D y are

Figure 10 .

 10 Figure 10. Three famous examples, known as Kreweras', Gessel's and Gouyou-Beauchamps' walks, respectively

Proposition 14 .

 14 For all 51 walks with infinite group, the setsH = {z ∈]0, 1/|S|[: ω 2 /ω 3 is rational} and ]0, 1/|S|[\H = {z ∈]0, 1/|S|[: ω 2 /ω 3 is irrational} are dense in ]0, 1/|S|[.Proof. The function ω 2 /ω 3 is clearly real continuous function on ]0, 1/|S|[. In fact, it has been noticed in [21, Part 6.2] that the function ω 2 /ω 3 is expandable in power 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.

  ): (iii.a) For the walks of Subcase I.A in Figure 17,: I x (a 1 ) and I x (b 1 ) for x → Q(x, 0); I y (a 1 ) and I y (b 1 ) for y → Q(0, y). (iii.b) For the walks of Subcases I.B and I.C in Figure 17: I x (a 1 ) and R\]x 1 , x 4 [ for x → Q(x, 0); I y (a 1 ) and [y 4 , y 1 ] for y → Q(0, y). (iii.c) For the walks of Subcase II.A in Figure 17: I x (b 1 ) and [x 4 , x 1 ] for x → Q(x, 0); I y (b 1 ) and R\]y 1 , y 4 [ for y → Q(0, y). (iii.d) For the walks of Subcases II.B, II.C, II.D and Case III in Figure 17: R\]x 1 , x 4 [ for x → Q(x, 0); R\]y 1 , y 4 [ for y → Q(0, y).

Figure 12 .

 12 Figure 12. Location of a 1 , a 2 , a 3 , a 4 , b 1 , b 2 if y 4 < 0 and x 4 < 0, i.e., Subcase I.A

Proposition 22 .

 22 For all 51 models, for any ω ∈ M 1,0 (resp. ω ∈ N 1,0 ) with x(ω) = ∞ (resp. y(ω) = ∞) and ℑω ∈ {ℑa 1 , ℑa 2 , ℑa 3 , ℑa 4 , ℑb 1 , ℑb 2 }, we have r x (ω) = ∞ (resp. r y (ω) = ∞).

7. 3 .

 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 1 , a 2 , a 3 , a 4 , b 1 , b 2 on Π y . It depends heavily on the signs of x 4 and y 4 , see Figures 12, 13, 14, 15 and 16. Let us recall that 0 < x 3 < ∞ and 0 < y 3 < ∞, see Section 2. If y 4 < 0 (resp. x 4 < 0), the point y = ∞ (resp. x = ∞) obviously belongs to the real cycle ]y 3 , ∞[∪{∞}∪]∞, y 4 [ (resp. ]x 3 , ∞[∪{∞}∪]∞, x 4 [) of the complex sphere S. By construction of the Riemann surface T and of its universal covering, the points a 1 , a 2 (resp. b 1 , b 2 ) then lie on the open interval {ω

  , Subcases I.A, I.B, I.C, II.A, II.B, II.C, II.D and Case III.

Step 1 . 1 k=n f y (a 3 -

 113 Let us first prove that r y (a 3 ) = ∞ and r y (a 4 ) = ∞. If |y ⋆ | < 1, then a 3 ∈ ∆ y (see Section 4 for the definition of ∆ y ) and it is immediate from Theorem 3 that r y (a 3 ) = ∞. If |y ⋆ | 1, then by Lemma 5 there exists n ∈ Z + such that a 3 -nω 3 ∈ ∆, and by Equation (5.2) of Theorem 4, (7.10) r y (a 3 ) = r y (a 3 -nω 3 ) + kω 3 ).

  4) implies that r y (ω) = ∞ on the segments {ω ∈ Π y : ℑω = a 3 , a 4 }, except for their ends a 1 , a 2 . The segments {ω : ℑω = a 3 , a 4 , ω x 1 ℜω ω x 4 + ω 2 } do not contain any point where y(ω) = ∞. It follows from Equation (5.3) that r x (ω) = ∞ on these segments except for points where x(ω) = ∞ if they exist. This last fact happens if and only if ℑb 1 = ℑa 3 and only at the ends b 1 , b 2 of the segments. If ℑb 1 = ℑa 3 , we can show exactly in the same way that r y (ω) = ∞ on the two segments {ω ∈ Π y : ℑω = b 1 , b 2 } and that r x (ω) = ∞ on the segments {ω : ℑω = b 1 , b 2 , ω x 1 ℜω ω x 4 + ω 2 }, except for their ends b 1 , b 2 . This concludes the proof of Proposition 22.

Figure 13 .

 13 Figure 13. Location of b 1 , b 2 if x 4 > 0, Subcases I.C and II.C

Figure 13 .

 13 Now we show that for any ω 0 with ℑω 0 = 0 and ω y 1 ℜω 0 ω y 4 + ω 2 , r y (ω 0 ) = ∞. Note that y(ω 0 ) = ∞. If ω 0 ∈ ∆, by the same arguments as in Subcase I.A, ω 0 ∈ ∆ y and r y (ω 0 ) = ∞. If ω 0 / ∈ ∆, then consider O ∆ (ω 0 ) with the notation (7.11). Note that b 1 -ω 2 / ∈ ∆. For this, it is enough to prove that b 1 -ω 2 / ∈ ∆ x and that b 1 -ω 2 / ∈ ∆ y . First, b 1 -ω 2 / ∈ ∆ x , since x(b 1 ) = x(b 1 -ω 2 ) = ∞. Furthermore, ∆ y is centered w.r.t. L y 2 y 1 , and ω y 4 / ∈ ∆ y (since |y 4 | > 1). Hence the point b 1 -ω 2 < ω y 4 cannot be in ∆ y . Since ∆∩{ω ∈ C : ℑω = 0} is an open interval containing ω y 1 , and since b 1

  ω→b 2 f y (ω) = ∞ by the same arguments as in Subcase I.A. If b 1 and b 2 are not ordered, Lemma 21 applies to both of these points. If b 1 ≪ b 2

Figure 14 .

 14 Figure 14. Location of a 1 , a 2 , a 3 , a 4 if y 4 > 0, (x 4 , Y (x 4 )) = (∞, ∞), Subcases II.A, II.B and II.C

  Hence for two of these roots b(x) = -2[a(x)c(x)] 1/2 and then Y (x) 0 (see (2.5)), and for two others b(x) = 2[a(x)c(x)] 1/2 and then Y (x) 0. But Y (x 2 ) and Y (x 3 ) are on the segment [y 2 , y 3 ] ⊂]0, ∞[. Hence Y (x 1 ) 0 and Y (x 4 ) 0. If in addition x 4 = ∞, then Y (x 4 ) equals 0 or ∞; note also that if Y (x 4 ) = ∞ then necessarily x 4 = ∞.But the case when Y (x 4 ) = ∞ and x 4 = ∞ is excluded from our consideration at this moment. It follows that ∞ ∈ [y 4 , Y (x 4 )[, and in fact ∞ ∈]y 4 , Y (x 4 )[, since the case y 4 = ∞ is excluded from Case II.

  The points b 1 , b 2 are located as in Subcase I.A, see Figure 12. Further, we can show as in Subcase I.A that r y (ω) = ∞ on the segments {ω : ℑω = b 1 , b 2 , ω y 1 ℜω ω y 4 + ω 2 }, and we deduce that r x (ω) = ∞ on {ω : ℑω = b 1 , b 2 , ω x 1 ℜω ω x 4 + ω 2 } except for their ends b 1 , b 2 . Consequently, the set of poles of all branches of x → Q(x, 0) (resp. y → Q(0, y)) is dense on the curve I x (b 1 ) = I x (b 2 ) (resp. I y (b 1 ) = I y (b 2 )), as claimed in Proposition 23.

Figure 15 . 1

 151 Figure 15. Location of a 1 , a 2 , a 3 , a 4 if y 4 > 0, x 4 = ∞, Y (x 4 ) = ∞, i.e., Subcase II.D

  Since for all of the 9 models composing Subcase II.D, deg a = deg b = 1 and deg c = 2, then y( ξω) and y( ξ η ξω) are of order O(|x( η ξω)| 1/2 ), and their difference |y( ξω) -y( ξ η ξω)| is not smaller than O(|x( η ξω)| 1/2 ) as ω → a 3 . Finally, x(ω), x( η ξ η ξω) → x ⋆ = ∞ as ω → a 3 .Then as ω → a 3 in the sum(7.18) the second term is of the order not smaller than O(|x( η ξω)| 3/2 ) while the first vanishes and the third has the order O(|x( η ξω)| 1/2 ). This proves the assumption (B) of Lemma 21 for ω 0 = a 3 . By this lemma the poles of x → 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 4 = ∞. It remains here exactly one case to study, see Figure17. It is such thaty 4 = ∞, x 4 = ∞ and X(y 4 ) = ∞. Then x ⋆ = x ⋆ = ∞; y • = y • = ∞; y ⋆ = y ⋆ = ∞.The points b 1 = b 2 = ω x 4 + ω 2 are located as in Subcase I.B, a 1 = a 2 = ω y 4 + ω 2 and a 3 = a 4 = ω y 4 + ω 2 -ω 3 . In particular, a 3 + ω 3 /2 = b 1 , b 1 + ω 3 /2 = a 1 , see Figure16.We start by showing that r y (a 3 ) = ∞. If a 3 ∈ ∆, this is true thanks to (5.3) and since x ⋆ , y ⋆ = ∞. If a 3 / ∈ ∆, consider the orbit O ∆ (a 3 ). It cannot contain a 1 -ω 2 since a 3 + ω 3 = a 1 , neither b 1 , nor b 1 -ω 2 = ω x 4 . It follows that r y (a 3 ) = ∞.

7 . 4 .

 74 (ω) = x(ω)[y( ξω) -y(ω)] = x(ω)[b 2 (x(ω)) -4a(x(ω))c(x(ω))] 1/2 /a(x(ω)). If ω → b 1 , then x(ω) → ∞and since deg a = 2 and deg b = deg c = 1, we have f y (ω) → ∞. Lemma 21 applies with ω 0 = b 1 and proves Proposition 23. Asymptotic of ω 2 (ω)/ω 3 (ω) as z → 0. It remains to prove the following result announced at the beginning of Section 7. Proposition 25. For any of 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 2 /ω 3 = L + L/ ln(z) + O(1/ ln(z)) 2 .

  3/2) ln(2) + ((k -1)/4)(1 -3 ln(2)) + O(k -1) 2 , B(k) = -1/2 + (k -1)/4 + O(k -1) 2 .

x 2 =

 2 z + 2z 2 + 5z 3 + O(z 4 ),x 3 = 1/(4z 2 ) -1 -2z -8z 3 + O(z 4 ), x 4 = ∞, X(y 1 ) = 0.Then, with (7.26) and (7.27), we reach the conclusion thatk = 1 -8z 4 -4z 5 + O(z 6 ), (7.29) w = 1 -2z + z 2 -(7/4)z 3 -(65/8)z 4 + (613/64)z 5 + O(z 6). (7.30) Then, using Points (i) and (ii) above, we obtainΩ 2 = -2 ln(z) -(1/4)z + (1/16)z 2 + O(z 3 ln(z)), Ω 3 = -(1/2) ln(z) -(1/2) ln(2) + (1/4)z + (57/16)z 2 + O(z 3 ln(z)),so that (7.31) Ω 2 /Ω 3 = 4 -4 ln(2)/ ln(z) + O(1/(ln(z)) 2 ).

  Location of the branch points and of the cycles Φ 0 and Φ 1 on the Riemann surface Tis real if y ∈ [y 4 , y 1 ] or y ∈ [y 2 , y 3 ], and complex if y ∈]y 1 , y 2 [ or y ∈]y 3 , y 4 [, see (2.2). Likewise, for real values of x, Y (x) is real if x ∈ [x 4 , x 1 ] or x ∈ [x 2 , x 3 ], and complex if x ∈]x 1 , x 2 [ or x ∈]x 3 , x 4[. The set Φ therefore consists of two non-intersecting closed analytic curves Φ 0 and Φ 1 , equal to (see Figure

	3
	y 4	x 4
	Figure 4.

  Informal construction of the universal covering onto a closed curve on T homological to one of the elements of the normal basis on the torus. We choose λ([0, ω 1 ]) to be homological to the cut [x 1 , x 2 ] (and hence also to all other cuts [x 3 , x 4 ], [y 1 , y 2 ] and [y 3 , y 4 ]); λ([0, ω 2 ]

Acknowledgments

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 ushave been very helpful in the elaboration of this paper. We also warmly thank R. Krikorian and J.-P. Thouvenot for encouraging discussions. Finally, we thank two anonymous referees for their careful reading and their remarks.

x 4 y 1 y 4

Figure 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 Lemma 21. Let z be such that ω 2 /ω 3 is irrational; let ω 0 ∈ Π y be such that r y (ω 0 ) = ∞, and let A(ω 0 ) = {ω ∈ Π y : ℑω = ℑω 0 , f y (ω) = ∞}. Assume that ω 0 ∈ A(ω 0 ) and that for some ω 1 , . . . , ω k ∈ A(ω 0 ):

Proof. By Equation (5.2) of Theorem 4, we have, for any n ∈ Z + and any ω ∈ Π y , (7.7) r y (ω

Let ω 0 be as in the statement of Lemma 21. Due to assumption (C), Lemma 18 (a) and the ω 2 -periodicity of f y , the set {ω 0 + nω 3 } n>n 1 +•••+n k does not contain any point ω where f y (ω) = ∞. Further, by the assumptions (A) and (C), Lemma 18 (a) and also by the ω 2 -periodicity of f y , the set {ω 0 + nω 3 } 0 n n 1 +•••+n k contains exactly k + 1 poles of f y that are ω 0 , ω 0 + n 1 ω 3 , . . . , ω 0 + n k ω 3 . Then, by (7.7), assumption (B) and the fact that r y (ω 0 ) = ∞, we reach the conclusion that for any n > n 1 + • • • + n k , the point ω 0 + nω 3 is a pole of r y (ω).

Due to Equation (5.3), any ω pole of r y such that x(ω)y(ω) = ∞ is also a pole of r x . Define now B, the set of (at most twelve) points in Π y where either

, the points ω 0 + nω 3 are poles of r x as well, and both K(x(ω 0 + nω 3 ), 0) and K(0, y(ω 0 + nω 3 )) are non-zero. By Lemma 18 (b) and definitions (5.10) and (5.12), Lemma 21 follows.

We are now ready to give the proof of Theorem 17.

Proof of Theorem 17. Functions f y (ω) and y(ω) being ω 2 -periodic, it follows that both of them have no pole at any ω with 0 ℑω < ω 1 and ℑω

exactly as in Subcase I.B, thanks to (5.3), we derive that r x (ω) = ∞ for all ω 0 such that ℑω = 0 and ω x 1 ℜω ω x 4 + ω 2 except possibly for points where x(ω) = ∞. But these 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 ω→a 4 {f y (ω) + f y (ω + ω 3 )} = ∞ and lim ω→a 3 {f y (ω) + f y (ω + ω 3 )} = ∞. Further, a 3 ≪ 1 a 1 and a 4 ≪ 1 a 2 . If in addition a 3 ≪ a 4 , then due to the fact that a 3 +ω 3 = a 1 we have

There is no point ω ∈ Π y \ {a 2 } such that a 4 ≪ ω and f y (ω) = ∞. Lemma 21 applies to ω 0 = a 4 . If a 4 ≪ a 3 , then also a 4 ≪ 1 a 2 ≪ a 3 ≪ 1 a 1 and this lemma applies to ω 0 = a 3 . If a 3 ∼ a 4 , Lemma 21 can be applied to both a 3 and a 4 . Since I x (a 3 ) = I x (a 4 ) = I x (a 1 ) = I x (a 2 ) = [x 1 , x 4 ] and I y (a 3 ) = I y (a 4 ) = I y (a 1 ) = I y (a 2 ) = R\]y 4 , y 1 [, the set of poles of x → Q(x, 0) (resp. y → Q(0, y)) is dense on the announced intervals and Proposition 23 is proved.

Case II: y 4 > 0, Subcase II.B: x 4 > 0 and exactly one of y • , y • is ∞. Assume, e.g., that

while a 1 , a 2 , a 3 , a 4 are pictured as previously, in 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 y (a 3 ) = ∞. Let us note that in this case a 1 -ω 2 / ∈ ∆, as x(a 1 -ω 2 ), y(a 1 -ω 2 ) = ∞. Next, we derive as in Subcase II.A that r y (a 4 ) = ∞ and that r y (a 2 ) + f y (a 2 ) = ∞, since x ⋆ y ⋆ = ∞. Finally, again by the same arguments as in Subcase II.A, we conclude that for any ω 0 = a 1 with ℑω 0 = 0 and ω y 1 ℜω 0 ω y 4 + ω 2 , the orbit O ∆ (ω 0 ) does not contain a 3 and a 4 . The orbit can neither contain a 1 -ω 2 , since a 1 -ω 2 / ∈ ∆ and a 1 -ω 2 < ω y 1 < ω 0 , where ω y 1 ∈ ∆. Since r y (a 2 ) + f y (a 2 ) = ∞, then as in Subcase II.A we have r y (ω 0 ) = ∞.

It follows from (5.4) and (5.5) that r y (ω 0 ) = ∞ for any ω 0 such that ℑω 0 = 0 and ω y 4 ω 0 ω y 4 + ω 2 , except for a 1 and ηa 1 -ω 1 = a 2 . By (5.3), r x (ω 0 ) = ∞ for any ω 0 such that ℑω 0 = 0 and ω x 1 ω 0 ω x 4 + ω 2 , except for points ω 0 where x(ω 0 ) = ∞ (this is ω 0 = a 4 in this case). This finishes the proof of Proposition 22 in this case.

Since x ⋆ = x ⋆ , the same reasoning as in Subcase I.A gives lim ω→a 4 {f y (ω)+f y (ω+ω 3 )} = ∞ and lim ω→a 3 {f y (ω) + f y (ω + ω 3 )} = ∞. The rest of the proof of Proposition 23 via the use of Lemma 21 with ω 0 = a 3 if a 4 ≪ a 3 or with ω 0 = a 4 if a 3 ≪ a 4 , or with indifferent choice of a 3 or a 4 if a 3 ∼ a 4 , is the same as in Subcase II.A.

Case II: y 4 > 0, Subcase II.C: x 4 > 0 and y • , y • = ∞, or x 4 = ∞ and Y (x 4 ) = ∞. In this case, we have y

The points a 1 , a 2 , a 3 , a 4 are pictured as in Subcases II.A and II.B, see Figure 14 Figure 17. Different cases considered in the proof of Theorem 17-they correspond to the 51 non-singular walks with infinite group, see [START_REF] Bousquet-Mélou | Walks with small steps in the quarter plane[END_REF]