On the nature of the generating series of walks in the quarter plane
Thomas Dreyfus, Charlotte Hardouin, Julien Roques, Michael F., Singer

TL;DR
This paper introduces a novel Galois-theoretic approach to analyze the generating series of walks in the quarter plane, establishing new hypertranscendency results that show these series do not satisfy certain algebraic differential equations.
Contribution
It applies Galois theory of difference equations to study these generating series, providing new hypertranscendency results beyond previous findings.
Findings
Recovered recent results on generating series
Proved certain series are hypertranscendental
Established nonexistence of specific algebraic differential equations
Abstract
In the present paper, we introduce a new approach, relying on the Galois theory of difference equations, to study the nature of the generating series of walks in the quarter plane. Using this approach, we are not only able to recover many of the recent results about these series, but also to go beyond them. For instance, we give for the first time hypertranscendency results, {\it i.e.}, we prove that certain of these generating series do not satisfy any nontrivial nonlinear algebraic differential equation with rational coefficients.
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\sloppy
On the nature of the generating series of walks in the quarter plane
Thomas Dreyfus
Institut de Recherche Mathématique Avancée, U.M.R. 7501 Université de Strasbourg et C.N.R.S. 7, rue René Descartes 67084 Strasbourg, FRANCE
,
Charlotte Hardouin
Université Paul Sabatier - Institut de Mathématiques de Toulouse, 118 route de Narbonne, 31062 Toulouse.
,
Julien Roques
Université Grenoble Alpes, Institut Fourier, CNRS UMR 5582, 100 rue des Maths, BP 74, 38402 St Martin d’Hères
and
Michael F. Singer
Department of Mathematics, North Carolina State University, Box 8205, Raleigh, NC 27695-8205, USA
Abstract.
In the present paper, we introduce a new approach, relying on the Galois theory of difference equations, to study the nature of the generating series of walks in the quarter plane. Using this approach, we are not only able to recover many of the recent results about these series, but also to go beyond them. For instance, we give for the first time hypertranscendency results, i.e., we prove that certain of these generating series do not satisfy any nontrivial nonlinear algebraic differential equation with rational coefficients.
Key words and phrases:
Random walks, Difference Galois theory, Elliptic functions, Transcendence
2010 Mathematics Subject Classification:
05A15,30D05,39A06
This project has received funding from the European Research Council (ERC) under the European Union’s Horizon 2020 research and innovation programme under the Grant Agreement No 648132. The second author would like to thank the ANR-11-LABX-0040-CIMI within the program ANR-11-IDEX-0002-0 for its partial support. The second author’s work is also supported by ANR Iso-Galois. The work of the third author has been partially supported by the LabEx PERSYVAL-Lab (ANR-11-LABX-0025-01) funded by the French program Investissement d’avenir. The work of the fourth author was partially supported by a grant from the Simons Foundation (#349357, Michael Singer). All authors received funding from NSF grant DMS-1606334 to attend the DART VII conference where significant progress concerning these results was made. We thank Alexey Ovchinnikov and Alice Medvedev for making this possible. We would like to thank Mireille Bousquet-Mélou and Kilian Raschel for discussions and comments of this work. The second author would like to thank Marcello Bernardara, Thomas Dedieu and Stephane Lamy for many discussions and references on elliptic surfaces. In addition, we would like to thank the anonymous referees for many useful comments and suggestions concerning this article.
Contents
- 1 Introduction
- 2 Fundamental properties of the walks with small steps
- 3 Hypertranscendancy Criteria
- 4 Preliminary results on the elliptic curve Et
- 5 Hypertranscendance of generating series of the walks in Wtyp
- 6 Hyperalgebraicity of the generating series of the walks in Wex
- 7 Nonholonomicity in the exceptional cases
- 8 Some comments concerning singular and weighted walks
- A Galois Theory of Difference Equations
- B Telescopers and orbit residues
- C Some computation of orbit residues
1. Introduction
In the recent years, the nature of the generating series of walks in the quarter plane Z≥02 has attracted the attention of many authors, see [BMM10, BvHK10, FIM99, KR12, MM14, Ras12] and the references therein.
To be concrete, let us consider a walk with small steps in the quarter plane Z≥02. By “small steps” we mean that the set of authorized steps, denoted by D, is a subset of {0,±1}2\{(0,0)}. For i,j,k∈Z≥0, we let qD,i,j,k be the number of walks in Z≥02 with steps in D starting at (0,0) and ending at (i,j) in k steps and we consider the corresponding trivariate generating series
[TABLE]
The main questions considered in the literature are:
- •
is QD(x,y,t) algebraic over Q(x,y,t) ?
- •
is QD(x,y,t) holonomic, i.e., is QD(x,y,t) holonomic in each of the variables x,y,t. Here, holonomic in the variable x means that the Q(x,y,t)-vector space spanned by the partial derivatives ∂xi∂iQD(x,y,t), i∈Z≥0, of QD(x,y,t) is finite dimensional.
- •
is QD(x,y,t) x-hyperalgebraic (resp. y-hyperalgebraic), i.e., is QD(x,y,t), seen as a function of x, a solution of some nonzero (possibly nonlinear) polynomial differential equations with coefficients in Q(x,y,t)? In case of a negative answer, we say that QD(x,y,t) is x-hypertranscendental (resp. y-hypertranscendental).
We shall now make a brief overview of some recent works on these questions. Random walks in the quarter plane were extensively considered in [FIM99]. These authors attached a group to any such walk and introduced powerful analytic tools to study the generating series of such a walk. In [BMM10], Bousquet-Mélou and Mishna give a detailed study of the various walks with small steps in the quarter plane and make the conjecture that such a walk has a holonomic generating series if and only if the associated group is finite. They prove that, if the group of the walk is finite, then the generating series is holonomic, except, maybe, in one case, which was solved positively by Bostan, van Hoeij and Kauers in [BvHK10] (see also [FR10]). In [MR09] Mishna and Rechnitzer showed that two of the walks with infinite groups have nonholonomic generating series. Kurkova and Raschel proved in [KR12] (see also [BRS14, Ras12]) that for all of the 51 nonsingular walks with infinite group (see Section 2.3) the corresponding generating series is not holonomic. This work is very delicate and technical, and relies on the explicit uniformization of certain elliptic curves. Recently, Bernardi, Bousquet-Mélou and Raschel [BBMR16] have shown that the generating series for 9 of the nonsingular walks satisfy nonlinear differential equations despite the fact that they are not holonomic.
In the present paper, we introduce a new, more algebraic approach, relying on Galois theory of difference equations, to study the nature of the generating series of walks. Using this approach, we are not only able to recover the above mentioned remarkable results, but also to go beyond them. For instance, the following theorem, proved in Section 5, is one of the main results of this paper.
Theorem 1**.**
Except for the 9 walks considered in [BBMR16], the generating series of all nonsingular walks with infinite group are x- and y-hypertranscendental. In particular, they are nonholonomic.
We shall now briefly explain our proof of this result and the reason why the Galois theory of difference equations comes into play. Our starting point is the well-known fact that the generating series QD(x,y,t) satisfies a functional equation of the form
[TABLE]
where
[TABLE]
for some KD(x,y,t)∈C[x,y,t] and some ϵ∈{0,1}.
Fix t∈C. The equation KD(x,y,t)=0 defines a curve Et⊂P1(C)×P1(C) whose Zariski-closure Et happens to be an elliptic curve in all the situations considered in this paper.
Using the fact that the series QD(x,y,t) converges for ∣x∣<1, ∣y∣<1 and ∣t∣<1/∣D∣, one can see FD1(x,t) and FD2(y,t) as analytic functions on some small pieces of Et. Using the above functional equation, one can prove that these functions can be extended to multivalued meromorphic functions on the whole of Et. Uniformizing Et with Weierstrass functions, we can lift these extended functions to meromorphic functions rx,ry on the universal covering C of Et. These are the functions that will be found to satisfy difference equations. Indeed, intersecting Et with horizontal, resp. vertical lines, we define two involutions ι1,ι2 of Et. Their compositum τ:=ι2∘ι1 is a translation on Et. We lift these three mappings to C, keeping the same notations ι1, ι2, τ for the lifted mappings. Then, it can be proved that rx and ry satisfy difference equations of the form
[TABLE]
for some explicit elliptic functions b1 and b2. Now, the “nature” of QD(x,y,t) is tackled as followed: its hypertranscendency with respect to the derivation d/dx, resp. d/dy, is found to be equivalent to the hypertranscendency of FD1(x,t), resp. FD2(y,t), and then in turn to the hypertranscendency of rx, resp. ry. Results from the Galois theory of difference equations allow one to reduce the question of showing that rx and ry are hypertranscendental to showing that a certain linear differential equation defined on the elliptic curve Et has no solutions in the function field of that curve [Har08, HS08]. It turns out that the last question can be answered by some elementary considerations about the polar divisor of some elliptic functions. Note that the fact that difference equations come into play in the present context is classical (see [FIM99, KR12, Ras12]). The novelty of our approach consists in the algebraic way we exploit these functional equations, in the light of the Galois theory of difference equations.
Our techniques also allow us to study the 9 exceptional cases and to recover some of the results of [BBMR16], namely :
Theorem 2**.**
In the 9 exceptional cases treated in [BBMR16], the generating series QD(x,y,t) is x- and y-hyperalgebraic.
It is very likely that our method can be used to study the generating series of weighted walks in the quarter-plane as well as singular walks. This is explained in more details in Section 8. We hope to come back on this in future publications.
The paper is organized as follows. In Section 2 we review several useful facts and ideas that form the basis of this paper as well as previous investigations concerning the generating series of walks in the quarter plane: the functional equation, the elliptic curve associated to this equation together with certain involutions and automorphisms and the method by which one reduces the question of hypertranscendence to a similar question for a multivalued meromorphic function on the associated curve. In Section 3, we present the criteria based on the Galois theory of difference equations which we will use to determine if a function is hypertranscendental. A brief introduction to the Galois theory of difference equations is given in Appendix A, as well as a proof of the above mentioned criteria. In Sections 4 and 5 we present the calculations that show that for all but the nine exceptional cases, the generating series of nonsingular walks are hypertranscendental. In Section 6 we show that for the nine exceptional cases, the generating series have specializations that are hyperalgebraic and deduce Theorem 2 from this. In Section 7 we show these series are not holonomic. Appendix B contains useful necessary and sufficient conditions for certain linear differential equations on elliptic curves (similar to the telescopers appearing in the verification of combinatorial identities) to have solutions in the function field of the curve. These criteria involve some “orbit residues” and Appendix C provides useful results about them. These results, together with the results of Section 3, are the mains tools used in Sections 4 and 5 to determine hypertranscendency.
2. Fundamental properties of the walks with small steps
We start by recalling some basic facts about random walks in the quarter-plane, see [BMM10, FIM99, KY15, MM14] for more details.
2.1. The generating series
We consider a walk with small steps in the quarter plane Z≥02. The set of authorized steps D is a subset of {0,±1}2\{(0,0)}. For i,j,k∈Z≥0, we let qD,i,j,k be the number of walks in Z≥02 with steps in D starting at (0,0) and ending at (i,j) in k steps and we consider the corresponding trivariate generating series
[TABLE]
The obvious fact that ∣qD,i,j,k∣≤∣D∣k ensures that QD(x,y,t) converges for all (x,y,t)∈C2×R such that ∣x∣≤1, ∣y∣≤1 and 0<t<1/∣D∣.
2.2. Kernel and functional equation
The generating series QD(x,y,t) satisfies a functional equation that we shall now recall. The Kernel of the walk is defined by
[TABLE]
where
[TABLE]
with di,j is equal to 1 if (i,j)∈D, and to [math] otherwise. (Note that we slightly diverge from the notation of [BMM10, Lemma 4] where the Kernel of the walk is by definition equal to xyKD(x,y,t).)
One can consider SD(x,y) as a Laurent polynomial in x with coefficients that are Laurent polynomials in y and vice versa. Using the notations of [BMM10, KY15], we write
[TABLE]
where AD,i(x)∈x−1Q[x] and BD,i(y)∈y−1Q[y]. The generating series QD(x,y,t) satisfies the following functional equation (see [BMM10, Lemma 4]):
[TABLE]
where
[TABLE]
2.3. Classification of the walks with small steps
There are a priori 28=256 possible walks with small steps in the quarter plane Z≥02, but, as explained in [BMM10, §2], only 138 of them are truly worthy of consideration. Moreover, taking into account natural symmetries, we are finally left with 79 inherently different walks to study; see [BMM10, Figures 1 to 4].
Following [BMM10, Section 3] or [KY15, Section 3], we attach to any walk in the quarter plane its group, which is by definition the group ⟨i1,i2⟩ generated by the involutive birational transformations of C2 given by
[TABLE]
These transformations leave SD(x,y,t) and, hence, xyKD(x,y,t) invariant. Amongst the 79 walks mentioned above, 23 have a finite group and 56 have an infinite group; see [BMM10, Theorem 3].
In the finite group case, the generating series QD(x,y,t) is holonomic. This has been proved in [BMM10] for 22 walks, and in [BvHK10] for the remaining walk, the so-called Gessel walk (its generating series is actually algebraic; see also [FR10]).
Amongst the walks having an infinite group, we distinguish the singular and nonsingular walks, that is, those walks whose associated curves Et, defined below in Section 2.4, are singular or nonsingular. In the latter case, Et is an elliptic curve; see Proposition 2.1 below. Amongst the 56 walks under consideration, there are 5 singular walks and 51 nonsingular walks. The set of these nonsingular walks, on which this paper focuses, is denoted by W and is described in Figure 1. This Figure reproduces the table [KR12, Figure 17] and uses notations compatible with loc. cit. for the convenience of the reader and for the ease of reference. In [KR12], the authors show that the generating series for all walks in W are nonholonomic. In [BBMR16], the authors show that 9 of these are hyperalgebraic. The subset of W formed by these 9 walks, denoted by Wex, is described in Figure 2 with references to both [BBMR16, Tab. 2] and Figure 1. We will refer to the elements of Wex as the “exceptional walks”. As noted in [BMM10, § 2.2], interchanging x and y in the steps leads to equivalent counting problems. The notation “wIIB.2 (after x↔y)” refers to the walk wIIB.2 with the x and y axes interchanged. The complement of Wex in W is denoted by Wtyp. We will refer to the elements of Wtyp as the “typical walks”.
2.4. The algebraic curve Et defined by the kernel K(x,y,t)
We fix D a set of authorized steps and 0<t<1/∣D∣. We will omit the subscript D in our notations; for instance, the kernel KD will be denoted by K. We denote by K the homogenized polynomial
[TABLE]
We let Et be the algebraic curve in P1(C)×P1(C) defined by K, i.e.,
[TABLE]
The intersection of Et with (P1(C)∖{∞})×(P1(C)∖{∞}), where ∞=[1:0], will be denoted by Et and identified with a subset of C×C, i.e.,
[TABLE]
Such curves have been studied in detail by Duistermaat in [Dui10, Chapter 2].
2.5. Smoothness and genus of Et
We fix 0<t<1/∣D∣. For any [x0:x1] and [y0:y1] in P1(C), we denote by Δ[x0:x1]x and Δ[y0:y1]y the discriminants of the degree 2 homogeneous polynomials given by y↦K(x0,x1,y,t) and x↦K(x,y0,y1,t) respectively, i.e.,
[TABLE]
and
[TABLE]
Proposition 2.1** ([Dui10, §2.4.1, especially Proposition 2.4.3]).**
The following properties are equivalent :
- •
the curve Et is smooth;
- •
the discriminant Δ[x0:x1]x has simple roots in C2∖{(0,0)};
- •
the discriminant Δ[y0:y1]y has simple roots in C2∖{(0,0)}.
Moreover, if Et is smooth, then it has genus 1, i.e., it is an elliptic curve.
Remark 2.2**.**
In [FIM99, Section 2.3.2], other explicit equivalent conditions on the di,j are given ensuring that the discriminants have simple roots.
Corollary 2.3**.**
For any walk in W, the algebraic curve Et is an elliptic curve.
Proof.
This is a direct consequence of Proposition 2.1 in virtue of [KR12, Section 2.1] and [FIM99, Part 2.3]. ∎
Remark 2.4**.**
We work in P1(C)×P1(C) instead of the projective plane over C in order to get a smooth curve. Moreover the smoothness criteria of Proposition 2.1 can be encoded as follows. Following [Dui10, Proposition 2.4.3], the curve Et is smooth if and only if the Eisenstein invariants Fx (resp. Fy) of Δ[x0:x1]x (resp. Δ[y0:y1]y) are non zero (see [Dui10, Section 2.3.5]). The Eisenstein invariants Fx and Fy are given by explicit polynomial formulas in t and the di,j. Furthermore, if Et is smooth then it is an elliptic curve with modulus J that can be explicitly computed thanks to the Eisenstein invariants (see [Dui10, 2.3.23]).
2.6. The involutions ι1 and ι2 and the QRT mapping τ of Et
In this section, we fix 0<t<1/∣D∣ and we assume that Et is an elliptic curve.
We let ι1 and ι2 be the involutions of Et induced by i1 and i2, i.e.,
[TABLE]
These formulas define ι1 and ι2 as rational maps from Et to itself, but, since Et is a smooth projective curve, they are actually endomorphisms of Et. They are nothing but the vertical and horizontal switches of Et, i.e., for any P=(x,y)∈Et, we have
[TABLE]
Lemma 2.5**.**
A point P=([x0:x1],[y0:y1])∈Et is fixed by ι1 (resp. ι2) if and only if Δ[x0:x1]x=0 (resp. Δ[y0:y1]y=0).
Proof.
Since ι1 is the vertical switch of Et, P is fixed by ι1 if and only if y↦K(x0,x1,y,t) has a unique root in P1(C). The last property is equivalent to Δ[x0:x1]x=0. The proof for ι2 is similar. ∎
The automorphism τ of Et given by
[TABLE]
which will play a central role in this paper, is called the QRT mapping of Et. According to [Dui10, Proposition 2.5.2], τ is the addition by a point of the elliptic curve Et.
We denote by Gt the subgroup of Aut(Et), that is the group formed of the automorphisms of Et, generated by ι1 and ι2. Note that Gt has finite order if and only if τ has finite order.
If the group of the walk is finite then Gt is finite as well. However, if the group of the walk is infinite, there are a priori no reason why Gt should be infinite for all 0<t<1/∣D∣ (and this is false in general). However, we have the following result.
Proposition 2.6**.**
For any walk with infinite group, the set of t∈C such that Gt is infinite, i.e., such that τ has infinite order, has denumerable complement in C.
Proof.
This follows from [KR12, Remark 6, Proposition 14]. However, we give an alternate more elementary proof herebelow.
So, we assume that the group of the walk under consideration is infinite and we must show that the set of t∈C such that τ has finite order on Et is denumerable. To do this it suffices to show that, for each positive integer n, the set of t∈C such that τn is the identity on Et is finite. We know that the walks under consideration have an infinite group, i.e., that the birational transformation of C2 given by
[TABLE]
has infinite order. Fix a value of n>0. Using the formulas for ι1 and ι2 in Section 2.3, one sees that
[TABLE]
where p1,p2,q1,q2∈Q[x,y] are independent of t. This birational map is well defined on the complement of the curve Z⊂C2 defined by p2(x,y)q2(x,y)=0. For any t∈C, Bézout’s theorem ensures that either Et and Z have an irreducible component in common or Et∩Z is finite. Let S be the set of t∈C such that Et and Z have an irreducible component in common. We claim that S is finite. Assume not. For t∈S, K(x,y,t)=xy(1−tS(x,y)) and p2(x,y)q2(x,y) have a nonconstant factor in common. Since S is infinite, there are two values of t, t1 and t2, such that K(x,y,t1)=xy(1−t1S(x,y)) and K(x,y,t2)=xy(1−t2S(x,y)) will both be divisible by the same nonconstant factor d(x,y)∈C[x,y] of p2(x,y)q2(x,y). This implies that d(x,y) is a factor of xy and xyS(x,y), and, hence, of K(x,y,t). But, for the walks we consider, none of the factors of xy is a factor of K(x,y,t) (i.e., neither x nor y is a factor of K(x,y,t) in C[x,y,t]), whence a contradiction. Therefore, S is finite.
Let Xn be the set of (x,y,t)∈C3 such that
- •
t∈/S,
- •
(x,y)∈Et,
- •
p2(x,y)q2(x,y)=0,
- •
fn(x,y)=τn(x,y)=(x,y).
One sees that Xn is a constructible set111A subset of Cm is constructible if it lies in the boolean algebra generated by the Zariski closed sets.. The projection of a constructible set onto a subset of its coordinates is again constructible [CLO05, Ch.5.6, Cor.2, Ex.1] so the set
[TABLE]
is also constructible. If t∈Yn, then there is a point (x,y)∈Et such that fn is defined at this point and fn(x,y)=τn(x,y)=(x,y). Since τn is given by the addition of a point on Et, it must leave all of Et fixed. Conversely, if t∈/S and τn is the identity on Et then for some (x,y)∈C2, (x,y,t)∈Xn and so t∈Yn. We will have completed the proof once we show that Yn is finite. If Yn is not finite then, since constructible subsets of C are either finite or cofinite, it must contain an open set U. The set of points ∪t∈YnEt will then contain an open subset V of C×C such that fn leaves V pointwise fixed. Therefore, fn would be the identity on C×C, and this contradicts the fact that f has infinite order. Therefore Yn is finite. ∎
2.7. Functional equations satisfied by F1(x,t) and F2(y,t)
In this section, following [KR12], we describe how one identifies the formal series F1(x,t) and F2(y,t) with meromorphic functions on a suitable domain and give the functional equations satisfied by these functions for the walks in W.
In this section, we consider a walk in W. We recall that the corresponding generating series QD(x,y,t) satisfies the functional equation (2.1). This equation is formal yet, but for ∣x∣,∣y∣<1 and 0<t<∣D∣1, the series QD(x,y,t), FD1(x,t) and FD2(y,t) are convergent. Therefore we have
[TABLE]
for all x,y∈V:=Et∩{∣x∣,∣y∣<1}. This V is a non empty open subset of Et as explained in [KR12, Section 4.1]. In particular, F1(x,t) and F2(y,t) yield analytic functions on some small pieces of the elliptic curve Et. Thanks to uniformization, we can identify Et with C/Zω1+Zω2 via a map
[TABLE]
where q1,q2 are rational functions of p and its derivative dp/dω, p the Weirestrass function associated with the lattice Zω1+Zω2 (cf. [KR12, Section 3.2]). Therefore one can lift the functions F1(x,t) and F2(y,t) to functions rx(ω)=F1(q1(ω),t) and ry(ω)=F2(q2(ω),t), each defined on a suitable open subset of C. We summarize our constructions with the following diagrams :
[TABLE]
and
[TABLE]
where pr1 and pr2 denote the projections on the first and second coordinates respectively.
Note that the involutions ι1,ι2 and the map τ can also be lifted to the universal cover C of Et. We shall abuse notation and again denote these functions as ι1(ω),ι2(ω) and τ(ω). Furthermore, we have that τ(ω)=ω+ω3 on C where no nonzero integer multiple of ω3 belongs to the lattice Zω1+Zω2 if τ has infinite order. For any meromorphic function f on a Riemann surface X and for any automorphism σ of X, we set σ(f)=f∘σ. This notation is widely used in the theory of difference equations. The notation σ∗(f) is also classical but will not be used. Be careful, if σ,β are automorphisms of X, then (σ∘β)(f)=β(σ(f)).
Remark 2.7**.**
In [KR12, Section 3.2], explicit expressions for ω1,ω2, and ω3 are given. Furthermore, ω1 is a purely imaginary number, whereas ω2 and ω3 are real numbers.
One can be deduce from [KR12, Theorems 3 and 4], that the functions rx(ω) and ry(ω) can be continued meromorphically as univalent functions on the universal cover C. Furthermore, for any ω∈C, we have
[TABLE]
Remark 2.8**.**
1. In the statement of [KR12, Theorem 4] Kurkova and Raschel give equations that are different from equations (2.7) and (2.7). The above equations are presented in the proof of [KR12, Theorem 4] and are shown to be equivalent to those in the statement of their theorem.
2. The functions rx and ry are not ω2 periodic and therefore only define multivalued functions on Eˉt.
3. Equation (2.7) gives meaning to the formula
[TABLE]
and (2.7) gives meaning to the formula
[TABLE]
Although the right hand sides of these equations are well defined on Et, further analysis is needed to give meaning to “τ(F1(x,t))” and “τ(F2(y,t))”. This is one reason for lifting the various functions to the universal cover of this curve.
From now on and until the end of the paper, we shall make the following assumptions.
Assumption 2.9**.**
The walk is in W and 0<t<1/∣D∣ is such that :
- (1)
the group Gt is infinite, 2. (2)
t* is not algebraic over Q.*
Using the fact that the group of any walk in W is infinite (see Section 2.3) and Proposition 2.6, we see that, once we have fixed a walk in W, the set of t such that the assumptions 2.9 are satisfied has denumerable complement in ]0,1/∣D∣[.
Remark 2.10**.**
*Actually, the crucial properties used in the rest of the paper are the following : *
- (1)
the curve Et is an elliptic curve, 2. (2)
the group Gt is infinite, 3. (3)
t* is not algebraic over Q,* 4. (4)
the functions x↦F1(x,t) and y↦F2(y,t), each analytic on some open subset of Et, can be lifted and continued to functions rx and ry meromorphic on the universal cover of Et such that these functions satisfy equations (2.7), (2.7), (2.8), and (2.9).
These properties are automatically satisfied under the assumptions 2.9 (indeed, for (1) see Corollary 2.3 and for (4) see the beginning of the present Section).
3. Hypertranscendancy Criteria
In this section, we derive hypertranscendency criteria for F1(x,t) and F2(y,t). The related functions rx and ry satisfy “difference” equations of the form τ(Y)−Y=b. Galois theoretic methods to study the differential properties of such functions have been developed in [HS08] and [DHR15] (see also [Har16]). In this section we describe a consequence of this latter theory and how it will be used to show that the walks in Wtyp have hypertranscendental generating series. We will begin by making precise the differential situation.
3.1. A derivation on Et commuting with τ
We assume that Et is an elliptic curve. We denote its function field by C(Et). The space of meromorphic differential forms on Et forms a one dimensional vector space over C(Et) and the space of regular differential forms is a one dimensional vector space over C. We let Ω be a nonzero regular differential form on Et. In the following Lemma, we prove the existence of a derivation δ on Et commuting with the automorphism τ of Et.
Lemma 3.1**.**
The derivation δ of C(Et) such that d(f)=δ(f)Ω commutes with τ, that is
[TABLE]
Proof.
According to [Dui10, Lemma 2.5.1, Proposition 2.5.2] or to [Sil09, Proposition III.5.1], we have τ∗(Ω)=Ω, where τ∗ is the map induced by τ on the space of regular differential forms on Et. It follows that, for any f in C(Et), we have
[TABLE]
Whence the equality δ(τ(f))=τ(δ(f)). ∎
Lemma 3.2**.**
Let P∈Et and let vP be the associated valuation on C(Et). Then, for any f∈C(Et), we have
- •
if vP(f)≥0 then vP(δ(f))≥0;
- •
if vP(f)<0 then vP(δ(f))=vP(f)−1.
Proof.
We recall that ω has valuation [math] at any point of Et (see [Sil09, Proposition III.1.5]; see [Sil09, § II.4] for the definition of the valuation of differential forms). Let u be a local parameter of Et at P. Since d(u)=δ(u)Ω and since both du and Ω have valuation [math] at P, we get that vP(δ(u))=0. The result follows clearly from this and from the fact that, for f=∑i=vu(f)+∞aiui∈C(Et), we have δ(f)=∑i=vu(f)+∞aiiδ(u)ui−1. ∎
Remark 3.3**.**
In Section 2.7, we discussed the universal covering space map C→Et and used it to lift functions on Et to C. In particular the elements of C(Et) lift to doubly periodic meromorphic functions on C and so we can consider C(Et)⊂M(C) where M(C) is the field of meromorphic functions on C. One then sees that τ corresponds to the map ω↦ω+ω3, Ω corresponds (up to constant multiple) to the regular differential form dω and the τ-invariant derivation δ corresponds to dωd.
3.2. Hypertranscendancy criteria
We will reduce questions concerning the hypertranscendence of F1(x,t) and F2(y,t) to questions about the differential behavior of elements of C(Et). Criteria derived by using the Galois theory of difference equations allow us to do this. We will start with an abstract formulation but quickly specialize to the present situation.
Definition 3.4**.**
A δτ-field is a triple (K,δ,τ) where K is a field, δ is a derivation on K, τ is an automorphism of K and δ and τ commute on K. The τ-constants Kτ* of K is the set {c∈K ∣ τ(c)=c}.*
The triples (C(Et),δ,τ) and (M(C),dωd,τ:ω→ω+ω3) are examples (see Lemma 3.1 and Remark 3.3). We say that a δτ-field is a subfield of another δτ-field if the derivation and automorphism of the smaller field are just the restrictions of the derivation and automorphism of the larger field. If τ has infinite order as an automorphism of C(Et), then C(Et)τ=C. This follows from the fact that τ(f(X))=f(X⊕P) where P is a point of infinite order. If τ(f)=f, then f(Q)−f(Q⊕nP)=0 for all n∈Z and Q a regular point of f. This implies that f(X)=f(Q) on a Zariski-dense subset of Et and so f(X) must be constant on Et.
The following formalizes the notion of holonomic, hyperalgebraic and hypertranscendental.
Definition 3.5**.**
Let (E,δ)⊂(F,δ) be δ-fields We say that f∈F is hyperalgebraic over E if it satisfies a non trivial algebraic differential equation with coefficients in E, i.e., if for some m there exists a nonzero polynomial P(y0,…,ym)∈E[y0,…,ym] such that
[TABLE]
We say that f is honolomic over E if in addition, the equation is linear. We say that f is hypertranscendental over E if it is not hyperalgebraic.
Other terms have been used for the above concepts: hypotranscendental or differentially algebraic or δ-algebraic for hyperalgebraic and differentially transcendental or transcendentally transcendental for hypertranscendental.
Proposition 2.6 of [DHR15] gives criteria for hypertranscendence in this general setting. Here, we only state this result in our situation. As noted above, in Remark 3.3, we may consider (C(Et),δ,τ) as a subfield of (M(C),dωd,τ:ω→ω+ω3). Given f∈M(C) we denote by C(Et)<f>δτ the smallest subfield of M(C) containing C(Et) and {τi(δj(f)) ∣ i∈Z,j∈Z≥0}. Note that this is a δτ-field.
Proposition 3.6**.**
Let b∈C(Et) and f∈M(C) and assume that
[TABLE]
If f is hyperalgebraic over C(Et), then, there exist an integer n≥0, c0,…,cn−1∈C and g∈C(Et) such that
[TABLE]
Conversely, if b satisfies such an equality and if (C(Et)<f>δτ)τ=C, then f is holonomic over C(Et).
For the convenience of the reader, Appendix A contains a brief introduction to the Galois theory of difference equations and a self-contained proof of Proposition 3.6.
The condition (C(Et)<f>δτ)τ=C is not superfluous. We will reconfirm (cf. [BBMR16, Theorem 11]) in Section 6 that for the nine exceptional walks in Wex, specializations of the generating series are hyperalgebraic but we know that they are nonetheless not holonomic. This situation arises because when one tries to apply the last part of Proposition 3.6 one is forced to add new τ-constants. This is precisely the situation that occurs in the proof of Proposition 6.2
In the appendix, we give necessary and sufficient conditions on the poles of b to guarantee the existence of an equation of the form of (3.1). For instance, the following result is a particular case of Corollary B.3 from the Appendix.
Corollary 3.7**.**
Assume that b has a pole P∈Et of order m≥1 such that none of the τk(P) with k∈Z∖{0} is a pole of order ≥m of b. Then, f is hypertranscendental.
In Sections 5 and 6, we will verify conditions like this to show that various generating series are hypertranscendental.
3.3. Applications to F1(x,t) and F2(y,t)
We assume that assumptions 2.9 are satisfied. We shall now apply the results of Section 3.2 to F1(x,t) and F2(y,t).
We begin by considering F1(x,t) as a formal power series with coefficients in C. We wish to show that F1(x,t) does not satisfy a polynomial differential equation
[TABLE]
where P∈C[x,Y0,…,Yn]. Let us assume that such an equation existed. The derivation dxd extends uniquely to a derivation on C(Et) which we again denote by dxd. So, we have the following commutative diagram:
[TABLE]
where (pr1)∗ is the map induced on the function fields by the first projection pr1:Et→P1(C). The derivations on C(Et) form a one dimensional vector space over C(Et). Therefore δ on Et can be written as δ=hdxd for some h∈C(Et). In particular this implies that when we consider F1(x,t) as an analytic function on an open set of Et it will satisfy a polynomial differential equation P~(F1(x,t),δ(F1(x,t)),…,δn(F1(x,t)))=0 where P~ has coefficients in C(Et). When we lift this equation to the universal cover, we see that rx∈M(C) is hyperalgebraic over C(Et). By assumption 2.9 and Section 2.7, rx satisfies τ(rx)−rx=b1 where b1∈C(Et). We can therefore apply Proposition 3.6 and conclude that there exist an integer n≥0, c0,…,cn−1∈C and g∈C(Et) such that
[TABLE]
Therefore to show F1(x,t) is hypertranscendental, it is enough to show that such an equation does not exist. Notice that this last condition only involves elements in C(Et). Similar reasoning (where we replace dxd with dyd) shows that F2(y,t) is hypertranscendental over C(Et) if there is no relation such as (3.2) with b1 replaced by b2. Therefore we have
Proposition 3.8**.**
Let i∈{1,2}. Assume that there does not exist an integer n≥0, c0,…,cn−1∈C and g∈C(Et) such that
[TABLE]
Then, the function Fi is hypertranscendental over C(Et).
Once again, using results from the appendix (namely Corollary B.3), we have
Corollary 3.9**.**
Let i∈{1,2}. Assume that bi has a pole P∈Et of order m≥1 such that none of the τk(P) with k∈Z∖{0} is a pole of order ≥m of bi. Then, Fi is hypertranscendental over C(Et).
The following additional result will also be useful.
Proposition 3.10**.**
The function F1(x,t) is hypertranscendental over C(Et) if and only if F2(y,t) is hypertranscendental over C(Et).
Proof.
We note that b1+b2=τ(xy)−xy. So, for all k∈Z≥0,
[TABLE]
It follows that an equation of the form (3.3) holds true for i=1 if and only if such an equation holds true for i=2. ∎
4. Preliminary results on the elliptic curve Et
In this and the next section we will show that all walks in Wtyp have generating series that are hypertranscendental. As we have indicated, an examination of the poles of the bi,i=1,2, and of the orbits of these poles under τ will yield these results.
To get a sense of our techniques, we will outline in the following example how we show that F2(y,t) is hypertranscendental for the walk wIA.1.
Example 4.1*.*
For the walk wIA.1, we have d−1,1=d1,1=d1,−1=d0,−1=1 and all other di,j=0. The curve Et is defined by
[TABLE]
Recalling that x=x0/x1 and y=y0/y1 one sees that the poles of b2=x(ι1(y)−y) are among the poles of x, y, and ι1(y), that is, among the points
[TABLE]
where
[TABLE]
One sees that b2 has a pole at P1. We claim that b2 has no pole of the form τk(P1) with k∈Z\{0}. Once this is shown the result will follow from Corollary 3.9. Therefore we must show that τk(P1)=P2 for any k∈Z\{0} and τk(P1)=Q1,Q2 for any k∈Z.
To verify this, first note that P1,P2,Q1,Q2 all have coordinates in L=Q(t,−1). Let σ be the automorphism of L∣Q(t) defined by σ(−1)=−−1. A calculation shows that for any point Q∈Et(L) one has that ι1(Q),ι2(Q),τ(Q)∈Et(L) and that ιk∘σ=σ∘ιk,k=1,2, and τσ=στ (cf. Proposition 4.8). Therefore all points in S2 have coordinates in L. Furthermore P2=σ(P1) and Q2=σ(Q1).
If P2=τk(P1), then P1=σ(P2)=σ(τk(P1))=τk(σ(P1))=τk(P2)=τ2k(P1). Since τ corresponds to addition by a point of infinite order, we get a contradiction if k=0.
If Q1=τk(P1) for some k∈Z, then τk(P2)=σ(τk(P1))=σ(Q1)=Q2. Since P1=ι1(P2) and τ=ι2ι1, we have that
[TABLE]
We have used the fact that ι1τι1=τ−1. Therefore τ−2k+1(P1)=P1, again a contradiction, since k∈Z. The proof that Q2=τk(P1) for any k∈Z is similar.
A similar proof combining arithmetic (e.g., Galois theory) with geometry (the behavior of points under ι1,ι2,τ) shows that b1 or b2 for the walks wIA.*,222Notation such as wIA.* refers to all the walks wIA.1,…,wIA.9. wIB.*, wIC.*, and wIIA.* has a pole that is unique in its τ-orbit. In the remaining cases, the arguments become more complicated. The poles lie in Q(t) and so we do not have a field automorphism at our disposal. In addition we may need to look more carefully at which poles lie in which orbits333We note that given points Q1 and Q2 on an elliptic curve Et, a general procedure is given in [Mas88] to determine if there is an integer n such that Q1=Q2⊕nP where τ(Q)=Q⊕P. Our more elementary, direct approach is independent of [Mas88]. and consider their orders as well as the expansions at these poles. Nonetheless, the interplay of arithmetic and geometry will be the key to the arguments.
In this section we examine points that are possible poles of b1 and b2 and develop the properties needed in the next sections. In Section 5 we will apply these together with results from the appendix and Proposition 3.8 to conclude hypertranscendence for any walk in Wtyp.
As before, we suppose that the assumptions 2.9 are satisfied.
4.1. On the base points
Definition 4.2**.**
The points P=([x0:x1],[y0:y1]) of Et such that x0x1y0y1=0 will be called the base points of Et.
Remark 4.3**.**
This terminology comes from the fact that these points are the base points of a natural pencil of elliptic curves.
Let us recall that the notation [a:b]∈P1(C) represents a ray of points of the form {(αa,αb) ∣ 0=α∈C}. Let L be a subfield of C. We say that [a:b]∈P1(L) if there is some element of this ray with coordinates in L. A similar notation concerns points in Et(L), the elements of which will be called the L-points of Et.
Lemma 4.4**.**
Any base point of Et belongs to Et(Q).
Proof.
Let P=([x0:x1],[y0:y1]) be a base point of Et. Let us assume for instance that x0=0 (the other cases being similar). Then, we obviously have [x0:x1]=[0:1]∈P1(Q). Moreover, we have
[TABLE]
which is equal to [math] if and only if ∑j=02d−1,j−1y0jy12−j=0. Since at least one of the d−1,j−1 is nonzero, we get [y0:y1]∈P1(Q). ∎
Lemma 4.5**.**
The following arrays describe precisely what are the possible base points fixed by ι1 or ι2, the corresponding conditions on the di,j and the walks in W satisfying these conditions:
[TABLE]
[TABLE]
Proof.
Taking into consideration obvious symmetries, we see that it is sufficient to prove the Lemma for a base point of the form P=([0:1],[β0:β1]).
Assume that P is fixed by ι1. By Lemma 2.5, this is equivalent to
[TABLE]
Since the di,j belong to {0,1}, the latter condition is equivalent to the equality d−1,0=0=d−1,−1d−1,1.
If d−1,0=d−1,−1=0, then we have d−1,1=0 and the fact that P belongs to Et simply means that d−1,1β12=0. Therefore, we have P=([0:1],[1:0]).
If d−1,0=d−1,1=0, then we have d−1,−1=0 and the fact that P belongs to Et simply means that d−1,−1β02=0. Therefore, we have P=([0:1],[0:1]).
This is precisely the first line of the first array.
Assume now that P is fixed by ι2. Since P and ι2(P) belong to the curve Et, the x-coordinates of both P and ι2(P) must satisfy the homogeneous equation of degree 2 in x0 and x1 given by x02A+Bx0x1+Cx12=0 with
[TABLE]
Since the x-coordinate of P is equal to [0:1], we see that C=0. Moreover, the fact that ι2(P)=P ensures that [0:1] is the only solution in P1(C) of the homogeneous equation x02A+Bx0x1+Cx12=x02A+Bx0x1=0. This ensures that B=0. Thus, we have obtained the equalities
[TABLE]
But, according to Lemma 4.4, [β0:β1] belongs to P1(Q) and, by hypothesis, t is transcendental. Therefore, equation (4.2) is equivalent to
[TABLE]
i.e., β0=0 and d0,−1=0 or β1=0 and d0,1=0.
If β0=0 and d0,−1=0, then P=([0:1],[0:1]). This point belongs to Et if and only if d−1,−1=0.
If β1=0 and d0,1=0, then P=([0:1],[1:0]). This point belongs to Et if and only if d−1,1=0.
This gives the first line of the second array. ∎
Lemma 4.6**.**
The following properties hold true :
- •
if ι1(P)=P then P=([0:1],[0:1]) and d−1,0=d−1,−1=0; the last condition corresponds to the walks wIA.1,wIA.2,wIC.2,wIIB.7;
- •
*if ι2(P)=P then P=([0:1],[0:1]) and d0,−1=d−1,−1=0; the last condition corresponds to the walks wIA.3,wIA.5,wIIA.2,wIB.3,wIID.4. *
Proof.
Taking into consideration obvious symmetries, we see that it is sufficient to prove the first statement. Since t is transcendental, we can and will identify Q(t) with a field of rational functions. Assume that P=([α0:1],[β0:1])∈Et(Q(t)) is fixed by ι1. By Lemma 2.5, we must have Δ[α0:1]x=0, i.e.,
[TABLE]
Step 1: Case α0∈Q.
If α0∈Q, then, comparing the t-adic valuations of both sides of (4.4), we get α0=0, i.e., P=([0:1],[β0:1]). In this case, equation (4.4) is simply d−1,02=4d−1,1d−1,−1. Since di,j∈{0,1}, we get d−1,0=0=d−1,1d−1,−1. This leads us to consider the two cases d−1,0=d−1,−1=0 and d−1,0=d−1,1=0.
If d−1,0=d−1,−1=0, then we have d−1,1=0 and the fact that P belongs to Et simply means that d−1,1β02=0. Therefore, we have P=([0:1],[0:1]), as desired.
If d−1,0=d−1,1=0, the fact that P belongs to Et simply means that d−1,−1=0. But for every walks under consideration, d−1,−1=d−1,0=d−1,1=0 never occur so this is impossible.
We shall now prove that α0∈Q(t)∖Q is impossible. A key fact that is used several times in the proof is that, for any walks in W, each set of steps contain elements lying on both sides of each of the lines: i=0,i+j=0,j=0,i−j=0. For example, the following condition d−1,0=d−1,1=d0,1=0 is never realized since this condition would imply all steps lie on or below the line i−j=0. We argue by contradiction. Equation (4.4) ensures that α0 must have either a pole or a zero at t=0. Indeed, otherwise, the left hand side of (4.4) would have a pole at t=0 but not the right hand side.
Let us write α0=tlP(t) with l∈Z∗ and P(t)∈Q(t) without zero and pole at t=0.
Step 2: Case P(t) constant.
If P(t)=c∈Q∗ then (4.4) becomes
[TABLE]
If l<0, then, equating the coefficients of t0 in (4.5), we find the equality d−1,02=4d−1,1d−1,−1. This gives d−1,0=0 and d−1,1d−1,−1=0.
If d−1,0=0 and d−1,1=0, then d−1,−1=0. Moreover, equation (4.5) simplifies as follows.
[TABLE]
On the right hand side of (4.6), we find the monomial 4cd0,1d−1,−1tl, but the non trivial monomials appearing on the left hand side have degree 2l−2, 3l−1 or 4l and none of them is equal to l since l<0. Therefore, we must have d0,1d−1,−1=0 and, hence, d0,1=0. However, the condition d−1,0=d−1,1=d0,1=0 is never realized so this is impossible.
If d−1,0=0 and d−1,−1=0, then d−1,1=0. Moreover, equation (4.5) simplifies as follows.
[TABLE]
On the right hand side of (4.7), we find the monomial 4cd−1,1d0,−1tl, but the non trivial monomials appearing on the left hand side have degree 2l−2, 3l−1 or 4l and none of them is equal to l since l<0. Therefore, we must have d−1,1d0,−1=0 and, hence, d0,−1=0. However, the condition d−1,0=d−1,−1=d0,−1=0 is never realized so this is impossible.
If l>0, then, equating the coefficients of t4l in (4.5), we find the equality d1,02=4d1,1d1,−1. This gives d1,0=0 and d1,1d1,−1=0. We claim that this is impossible.
We first assume that d1,0=0 and d1,−1=0. In every walks under consideration, we must have d1,1=0. Equation (4.5) simplifies as follows:
[TABLE]
On the right side of (4.8), we find the monomial 4c3d1,1d0,−1t3l. But, since l>0, we have 3l=0,l−1,2l−2. It follows that d1,1d0,−1=0 and, hence, d0,−1=0. However, the condition d1,0=d1,−1=d0,−1=0 is never realized so this is impossible.
We now assume that d1,0=0 and d1,1=0. In every walks under consideration, we must have d1,−1=0. Equation (4.5) simplifies as follows:
[TABLE]
On the right side of (4.9), we find the monomial 4c3d0,1d1,−1t3l. But, since l>0, we have 3l=0,l−1,2l−2. It follows that d0,1d1,−1=0 and, hence, d0,1=0. However, the condition d1,0=d1,1=d0,1=0 is never realized so this is impossible.
So, P(t) is not constant and, hence, has at least one zero or pole at some t0∈C∗.
Step 3: Poles of P(t).
Assume that P(t) has a pole t0∈C∗ of order κ≥1. Multiplying both sides of (4.4) by (t−t0)4κ and evaluating at t0, we find that d1,02=4d1,1d1,−1, which implies that d1,0=0=d1,1d1,−1.
We first assume that d1,0=0 and d1,1=0. Then, (4.4) simplifies as follows:
[TABLE]
Multiplying both sides of (4.10) by (t−t0)3κ and evaluating at t0, we find that the term in α03 of the right hand side of (4.10) must be equal to [math], i.e., d0,1d1,−1=0. So, we have either d1,0=d1,1=d0,1=0 or d1,0=d1,1=d1,−1=0. However, these conditions are never realized so this is impossible.
We now assume that d1,0=0 and d1,−1=0. Then, (4.4) simplifies as follows:
[TABLE]
Multiplying both sides of (4.11) by (t−t0)3κ and evaluating at t0, we find that the term in α03 of the right hand side of (4.11) must be equal to [math], i.e., d1,1d0,−1=0. So, we have either d1,0=d1,−1=d1,1=0 or d1,0=d1,−1=d0,−1=0. However, these conditions are never realized so this is impossible. So P(t) has no poles in C.
Step 4: Zeros of P(t).
Assume that P(t) has a zero t0∈C∗ of order ν≥1. Evaluating (4.4) at t0, we get d−1,02=4d−1,1d−1,−1, which implies d−1,0=0=d−1,1d−1,−1.
We first assume that d−1,0=0 and d−1,1=0. Then (4.4) simplifies as follows:
[TABLE]
Since the walk is non degenerate, we have d−1,−1=0. Let α=0 be the value of (t−t0)−να0 at t0. Dividing both sides of (4.12) by (t−t0)ν and evaluating at t0 we find 0=4αd0,1d−1,−1, which implies d0,1=0. So, we have obtained d−1,0=d−1,1=d0,1=0. However, these conditions are never realized so this is impossible.
We now assume that d−1,0=0 and d−1,−1=0. Then (4.4) simplifies as follows:
[TABLE]
Since the walk is non degenerate, we have d−1,1=0. Let α=0 be the value of (t−t0)−να0 at t0. Dividing both sides of (4.13) by (t−t0)ν and evaluating at t0 we find 0=4αd−1,1d0,−1, which implies d0,−1=0. So, we have obtained d−1,0=d−1,−1=d0,−1=0. However, these conditions are never realized so this is impossible. So P(t) has no zeros in C. Since it is not constant and it has no poles in C, we find that α0∈Q(t)∖Q is impossible. This concludes the proof of Lemma 4.6. ∎
In the following lemma, we focus our attention on the base points ([x0:x1],[y0:y1]) of Et corresponding to the equation x1y1=0, namely:
[TABLE]
We will use the following notations :
[TABLE]
Lemma 4.7**.**
The points P1 and P2 (resp. Q1 and Q2) are Lx-points (resp. Ly-points) of Et. They are Q-points of Et if and only if Δ[1:0]x/t2 (resp. Δ[1:0]y/t2) is a square in Q. Moreover, the following properties hold true :
- •
Δ[1:0]x/t2* is a square in Q if and only if d1,−1d1,1=0; moreover :*
- –
d1,1=0* if and only if there exist i,j∈{1,2} such that Pi=Qj=([1:0],[1:0]); this corresponds to the walks wIIB.*, wIID.*;*
- –
d1,−1=0* if and only if there exists i∈{1,2} such that Pi=([1:0],[0:1]); this corresponds to the walks wIB.*, wIC.*, wIIC.*, wIII;*
- •
Δ[1:0]y/t2* is a square in Q if and only if d−1,1d1,1=0; moreover :*
- –
d1,1=0* if and only if there exist i,j∈{1,2} such that Pi=Qj=([1:0],[1:0]); this corresponds to the walks wIIB.*, wIID.*;*
- –
d−1,1=0* if and only if there exists j∈{1,2} such that Qj=([0:1],[1:0]); this corresponds to the walks wIIA.*, wIIB.1-6, wIIC.*, wIID.1, wIID.2, wIID.5,wIII.*
Proof.
Taking into consideration the obvious symmetry in x and y, we see that it is sufficient to prove the result for P1 and P2 and the first of the last two statements of the Lemma.
The y-coordinates [β0:β1] and [β0′:β1′] of P1 and P2 are the roots in P1(C) of the homogeneous polynomial in y0 and y1 given by
[TABLE]
Therefore, [β0:β1] and [β0′:β1′] belong to P1(Lx). Moreover, we see that they belong to P1(Q) if and only if Δ[1:0]x/t2=d1,02−4d1,−1d1,1 is a square in Q. Since the di,j are in {0,1}, we have that d1,02−4d1,−1d1,1 is a square in Q if and only if d1,−1d1,1=0.
The fact that d1,1=0 is equivalent to the fact that [1:0] is a root of equation (4.14) and this is equivalent to the fact that there exist i,j∈{1,2} such that Pi=Qj=([1:0],[1:0]).
Similarly, the fact that d1,−1=0 is equivalent to the fact that [0:1] is a root of equation (4.14) and this is equivalent to the fact that P1=([1:0],[0:1]) or P2=([1:0],[0:1]). ∎
4.2. Galois action on Et
Let Q(t)⊂L⊂C be a field extension. For any L-point P=([x0:x1],[y0:y1]) of Et, with x0,x1,y0,y1∈L, and any σ∈Aut(L/Q(t)), we set
[TABLE]
Since Et is defined over Q(t), σ(P) is an L-point of Et.
Proposition 4.8**.**
Let Q(t)⊂L⊂C be a field extension and let σ∈Aut(L/Q(t)). Let P be a L-point of Et. Then, the following properties hold true :
- •
ι1(P),ι2(P)* and, hence, τn(P) for any n∈Z, are L-points of Et;*
- •
for any k∈{1,2}, ιk∘σ=σ∘ιk on Et(L) and, hence, τ∘σ=σ∘τ on Et(L).
Proof.
We only prove the assertions concerning ι1. The proofs for ι2 are similar and the assertions concerning τ follow from those about ι1 and ι2 since τ=ι2∘ι1.
We set P=([a0:a1],[b0:b1])∈Et(L) (with a0,a1,b0,b1∈L) and ι1(P)=([a0:a1],[b0′:b1′]). The point [b0′:b1′] is characterized by the fact that [b0:b1] and [b0′:b1′] are the roots in P1(C) of the homogeneous polynomial in y0 and y1 given by
[TABLE]
where A(a0,a1)=d−1,1a12+d0,1a0a1+d1,1a02, B(a0,a1)=d−1,0a12−t1a0a1+d1,0a02 and C(a0,a1)=d−1,−1a12+d0,−1a0a1+d1,−1a02 (these are not all [math]). Since (4.15) has coefficients in L and b0,b1∈L, we can assume that b0′,b1′∈L as well. Hence, [b0′:b1′]∈P1(L) and ι1(P)∈Et(L), as desired.
Moreover, [σ(b0):σ(b1)] and [σ(b0′):σ(b1′)] are the roots in P1(C) of
[TABLE]
Therefore, ι1(σ(P))=([σ(a0):σ(a1)],[σ(b0′):σ(b1′)])=σ(ι1(P)). ∎
4.3. On the τ-orbits
Definition 4.9**.**
We define an equivalence relation ∼ on Et by
[TABLE]
If P∼Q is not true, we shall write P≁Q. An equivalence class for ∼ will be called a τ-orbit.
Lemma 4.10**.**
*We set Mx=Q(t)(Δ[1:0]x) and My=Q(t)(Δ[1:0]y). The following properties hold true : *
- •
if Q(t)⊊Mx or Q(t)⊊My then, for all i,j∈{1,2}, we have Pi≁Qj;
- •
if Q(t)⊊Mx (resp. Q(t)⊊My) then Pi≁Pj (resp. Qi≁Qj) for i=j.
Proof.
We recall that due to the assumption 2.9, τ has infinite order (so that it corresponds to a translation by a non torsion point). Let us prove the first assertion. Suppose to the contrary that, for instance, P1∼Q1 and that Q(t)⊊Mx, the other cases being similar. So, there exists n∈Z such that τn(P1)=Q1. The fact that P1=([1:0],[β0:β1]) belongs to Et means that
[TABLE]
Since Q(t)⊊Mx, we have that Δ[1:0]x/t2=d1,02−4d1,1d1,−1 is not a square in Q(t). It follows that d1,1d1,−1=0 and that P1∈Et(Mx)∖Et(Q(t)). On the other hand, the fact that Q1=([α0:α1],[1:0]) belongs to Et means that
[TABLE]
So, Q1 belongs to Et(My). Since τ−n(Q1)=P1, Proposition 4.8 ensures that P1∈Et(My) as well. Therefore, P1∈(Et(Mx)∖Et(Q(t)))∩Et(My). In particular, Mx∩My is not reduced to Q(t). Since Mx and My are fields extensions of degree at most 2 of Q(t), we get Mx=My. Let σ∈Gal(Mx/Q(t))=Gal(My/Q(t)) be an element of order 2. We obviously have σ(P1)=P2 and σ(Q1)=Q2. Using Proposition 4.8, it follows that τn(P2)=τn(σ(P1))=σ(τn(P1))=σ(Q1)=Q2. Therefore, ι2τn(P2)=ι2(Q2)=Q1. But, we have ι2τn=τ−n+1ι1 (because τ=ι2ι1 and the ιk are involutions). Then, we find τ−n+1(P1)=τ−n+1ι1(P2)=ι2τn(P2)=Q1=τn(P1). This gives τ−2n+1(P1)=P1. Since τ is a translation by a non torsion point of Et, this implies that −2n+1=0. This yields a contradiction because n∈Z.
We shall now prove the second assertion. Assume that Q(t)⊊Mx. In particular Δ[1:0]x=0 and, hence, P1=ι1(P1)=P2. Suppose to the contrary that P1∼P2, that is, that there exists n∈Z∗ such that τn(P1)=P2. Let σ∈Gal(Mx/Q(t)) be an element of order 2. We obviously have σ(P1)=P2. Using Proposition 4.8, we get τn(P2)=τn(σ(P1))=σ(τn(P1))=σ(P2)=P1. Therefore, τ2n(P1)=P1. Since τ is a translation by a non torsion point of Et, this implies that n=0 and hence, P1=P2. This yields a contradiction. ∎
4.4. The poles of b1 and b2
Lemma 4.11**.**
The set of poles of b1=ι1(y)(τ(x)−x) in Et is contained in
[TABLE]
Similarly, the set of poles of b2=x(ι1(y)−y) in Et is contained in
[TABLE]
Moreover, we have
[TABLE]
Proof.
The proof of the assertions about the localization of the poles of b1 and b2 are straightforward. Let us prove (4.17). By definition, ι1(y1y0) and y1y0 are the two roots of the polynomial y↦K(x0,x1,y,t). The square of their difference equals to the discriminant divided by the square of the leading term. Then, we have
[TABLE]
Therefore, we find
[TABLE]
∎
5. Hypertranscendance of generating series of the walks in Wtyp
The treatments of the walks in Wtyp are finalized in the following subsections; more precisely :
- •
for wIA.*, wIB.*, wIC.* and wIIA.*, see Section 5.1, Theorem 5.2;
- •
for wIIB.4, wIIB.5, wIIB.8, wIIB.9, wIIB.10, see Section 5.2.1, Theorem 5.3;
- •
for wIID.*, see Section 5.2.2, Theorem 5.4;
- •
for wIII, see Section 5.2.3, Theorem 5.8;
- •
For wIIC.3, see Section 5.2.4, Theorem 5.10.
5.1. Generic cases
The following proposition gives a diophantine criteria for the hypertranscendency of F1(x,t) and F2(y,t).
Proposition 5.1**.**
We assume that the assumptions 2.9 are satisfied. If Δ[1:0]x/t2=d1,02−4d1,−1d1,1 or Δ[1:0]y/t2=d0,12−4d−1,1d1,1 is not a square in Q then F1(x,t) and F2(y,t) are hypertranscendental over C(Et).
Proof.
Assume for instance that Δ[1:0]x/t2 is not a square in Q, the other case being similar. Combining Corollary 3.9 and Proposition 3.10, we see that it is sufficient to prove that P1 is a pole of b2 and that it is the only pole of b2 of the form τn(P1) with n∈Z. The fact that P1 is a pole of b2 is clear (indeed, on the one hand, P1 is a pole of x and, on the other hand, the y-coordinates of P1 and ι1(P1) are distinct because Δ[1:0]x=0, and, hence, P1 is not a zero of ι1(y)−y). Moreover, Lemma 4.10 implies that P1≁P2 and P1≁Qi for i=1,2. The latter also implies that P1≁τ−1(Qi) for i=1,2. But, Lemma 4.11 ensures that the set of poles of b2 is included in {P1,P2,Q1,Q2,τ−1(Q1),τ−1(Q2)}. So, P1 is the only pole of b2 of the form τn(P1) with n∈Z, as desired. ∎
Theorem 5.2**.**
For any of the walks wIA.*, wIB.*, wIC.* or wIIA.* and for any t∈]0,1/∣D∣[∖Q such that Gt is infinite, the generating series F1(x,t) and F2(y,t) are hypertranscendental over C(Et).
Proof.
According to Proposition 5.1, it is sufficient to prove that either the discriminant Δ[1:0]x/t2=d1,02−4d1,−1d1,1 or Δ[1:0]y/t2=d0,12−4d−1,1d1,1 is not a square in Q. This is true because, according to Lemma 4.7, they are both squares in Q if and only if the walk we consider is among wIIB.*,wIIC.*,wIID.*,wIII. ∎
5.2. Non generic Cases
In this subsection, we shall focus our attention on the walks in Wtyp which are not covered by Theorem 5.2, i.e., on the walks in Wtyp among wIIB.*, wIIC.*, wIID.* and wIII. This gives 16 walks, namely wIIB.4, wIIB.5, wIIB.8, wIIB.9, wIIB.10, wIIC.3, wIID.1, wIID.2, wIID.3, wIID.4, wIID.5, wIID.6, wIID.7, wIID.8, wIID.9, wIII. In each of these cases, we will show that the corresponding generating series is hypertranscendental.
For the convenience of the reader, we have included in Figure 3 a table of the the values of the di,j for the various walks considered in this section.
5.2.1. The walks wIIB.*
In that situation, we have
[TABLE]
The following properties hold :
- •
according to Lemma 4.7, there exist i,j∈{1,2} such that Pi=Qj; up to renumbering, we can and will assume that P1=Q1=([1:0],[1:0]);
- •
P1=P2=ι1(P1) because Δ[1:0]x=d1,02−4d1,1d1,−1=d1,02=0;
- •
P1=Q1=([1:0],[1:0])=Q2=ι2(Q1) in virtue of Lemma 4.5 (or simply because Δ[1:0]y=0);
- •
Q1=ι1(Q2) because Q1=Q2 so Q1 and Q2 do not have the same x-coordinates.
In particular, we see that
- •
P1=Q1=([1:0],[1:0]);
- •
ι1(P1)=P2=([1:0],[β1′:β2′]) with [β1′:β2′]=[1:0];
- •
ι1(P2)=P1=([1:0],[1:0]);
- •
Q2=([α0′:α1′],[1:0]) with [α0′:α1′]=[1:0];
- •
ι1(Q2)=P1,P2.
Theorem 5.3**.**
For any of the walks wIIB.4, wIIB.5, wIIB.8, wIIB.9, wIIB.10 and any t∈]0,1/∣D∣[ ∖Q such that Gt is infinite, F1(x,t) and F2(y,t) are hypertranscendental.
Proof.
Using Lemma 4.11, we see that the set of poles of b2 is included in {P1,P2,Q1,Q2,ι1(Q1),ι1(Q2)}={P1,P2,Q2,ι1(Q2)} and that:
- •
P1 is a pole of order 2 of b2 because
- –
P1 is a pole of order 1 of x0/x1;
- –
P1 is a pole of order 1 of y0/y1;
- –
P1 is not a pole of ι1(y0/y1).
- •
P2 is a pole of order 2 of b2 because
- –
P2 is a pole of order 1 of x0/x1;
- –
P2 is not a pole of y0/y1;
- –
P2 is a pole of order 1 of ι1(y0/y1).
There are at least two double poles. Since x0/x1, y0/y1 and ι1(y0/y1) have at most two poles counted with multiplicities, we find that b2=x0/x1(ι1(y0/y1)−y0/y1) has at most 6 poles, counted with multiplicities. So there are at most 3 double poles (in fact P1 and P2 are the only double poles in this situation but this fact will not be used).
If P1 and P2 are the only double poles, combining Corollary 3.9 and Proposition 3.10, we see that, in order to conclude, it is sufficient to show that P1≁P2. Assume that there exists a third double pole P3 and that P1∼P2. Then there should exists j∈{1,2} such that P3∼Pj. Combining Corollary 3.9 and Proposition 3.10, we see that we have the conclusion in this case. So it is sufficient to prove that P1∼P2.
Suppose to the contrary that P1∼P2, i.e., that there exists n∈Z such that τn(P1)=P2.
On the one hand, since P1∈Et(Q)⊂Et(Q(t)), Proposition 4.8 implies that τj(P1)∈Et(Q(t)) for any j∈Z. Moreover, Proposition 4.8 implies that P2=ι1(P1)∈Et(Q(t)) is such that, for any j∈Z, τj(ι2(P2))∈Et(Q(t)).
On the other hand, it is easily seen that the equality τn(P1)=P2, together with the fact that τ=ι2∘ι1 is the composition of two involutions, imply,
- •
if n=2k, then τk(P1)=τ−k(P2) and thus
[TABLE]
- •
if n=2k+1, then τk+1(P1)=τ−k(P2) and thus
[TABLE]
For n=2k, we get that τk(P1) is fixed by the involution ι1. Lemma 4.5 ensures that for the walks under considerations none of the base points of Et is fixed by ι1. Therefore, τk(P1) is not a base point. By Lemma 4.6, we conclude that τk(P1)∈/Et(Q(t)). This yields a contradiction. For n=2k+1, we get that τk(ι2(P2)) is fixed by the involution ι2. Lemma 4.5 ensures that for the walks under consideration none of the base points of Et is fixed by ι2. Therefore, τk(ι2(P2)) is not a base point. By Lemma 4.6, we conclude that τk(ι2(P2))∈/Et(Q(t)). This yields a contradiction. ∎
5.2.2. The walks wIID.*
In that situation, we have
[TABLE]
In particular, Δ[1:0]x=0 and, hence, P1=P2. Moreover, Lemma 4.7 ensures that Pi=Qj=([1:0],[1:0]) for some i,j∈{1,2}. Up to renumbering, we can and will assume that P1=Q1. Since Δ[1:0]y=d0,12=0, we have P1=Q1=Q2. Since P1=ι1(P1), we also have P1=Q1=ι1(Q2). Last, using Lemma 4.5, we see that Q2=ι1(Q2).
Theorem 5.4**.**
For any of the walks wIID.* and any t∈]0,1/∣D∣[∖Q such that Gt is infinite, F1(x,t) and F2(y,t) are hypertranscendental.
Proof.
We recall the formula (4.17)
[TABLE]
Since the curve Et is nonsingular, the point P1 is a simple zero of Δ[x0:x1]x seen as a rational function on P1(C) (cf Proposition 2.1 and Corollary 2.3). Using additionally d0,1=1, it is easily seen that the polar divisor of b2 is of the form
[TABLE]
for some i,j∈{0,1,2}. Thus, P1 is the only pole of b2 of order ≥3. The result is now a consequence of Corollary 3.9 and Proposition 3.10. ∎
5.2.3. The walk wIII
This walk is symmetric in the sense of the following definition.
Definition 5.5**.**
We say that a walk is symmetric if di,j=dj,i for all i,j∈{0,±1}.
Note that the walk is symmetric if and only if
[TABLE]
Therefore, the involutive morphism s of P1×P1 defined by
[TABLE]
induces an involutive morphism of Et in the symmetric case, still denoted by s. Note that
[TABLE]
Indeed, on the one hand, for any point P=(x,y)∈Et, we have that {P,ι1(P)}=Et∩({x}×P1(C)). and, hence, {s(P),s(ι1(P))}=Et∩(P1(C)×{x}). On the other hand, we find {s(P),ι2(s(P))}=Et∩(P1(C)×{x}). Whence the desired equality s(ι1(P))=ι2(s(P)).
Similarly, we have
[TABLE]
It follows that
[TABLE]
Lemma 5.6**.**
Assume that the walk under consideration is symmetric. Consider R1,R2∈Et(Q(t)) such that s(R1)=R2. If R1∼R2 then there exists R3∈Et(Q(t)) such that s(R3)=R3.
Proof.
We have to prove that, if there exists ℓ∈Z such that τℓ(R1)=R2 for some R1,R2∈Et(Q(t)) such that s(R1)=R2, then there exists R3∈Et(Q(t)) such that s(R3)=R3. Up to interchanging R1 and R2, we can assume that ℓ≥0. We argue by induction on ℓ≥0. The result is obvious for ℓ=0. Let us assume that the result is true for some ℓ≥0. The equality τℓ(R1)=R2 ensures that sτℓ(R1)=s(R2)=R1. Using (5.3), we get τ−ℓs(R1)=R1. Using the equality τ−ℓ=ι1τℓ−1ι2, we get ι1τℓ−1ι2s(R1)=τ−ℓs(R1)=R1. Apply ι1 in the both sides of the equality gives τℓ−1ι2s(R1)=ι1(R1). With (5.1), we obtain τℓ−1sι1(R1)=ι1(R1). Note that ι1(R1) belongs to Et(Q(t)) in virtue of Proposition 4.8. The induction hypothesis leads to the desired result. ∎
Lemma 5.7**.**
We assume that t∈C∖Q. For the walk wIII, there are no R1,R2∈Et(Q(t)) such that s(R1)=R2 and R1∼R2.
Proof.
According to Lemma 5.6, it is sufficient to prove that s does not have fixed points in Et(Q(t)). Suppose to the contrary that there exists P∈Et(Q(t)) such that s(P)=P. So, P=(x,x) for some x=[x0:x1]∈P1(Q(t)) such that
[TABLE]
If x0=0, then we can assume that x1=1 and we see that (5.4) is impossible. Assume that x0=0. Then, we can and will assume that x0=1 and we have
[TABLE]
Since t is transcendental, we can and will identify Q(t) with a field of rational functions. If x1 has a pole of order μ≥1 at some t=t0∈Q, then t(1+2x13+x14) has a pole of order 4μ or 4μ−1 at t0 (depending on whether t0 is equal to [math]) whereas x12 has a pole of order 2μ at t0. Equation (5.5) yields 2μ=4μ or 4μ−1, whence a contradiction.
If x1 vanishes at some t=t0∈Q, then the equality (5.5) specialized at t=t0 gives 0=t0.
Therefore, we have proved that x1=ctm for some c∈Q× and m∈Z≥0. Equation (5.5) becomes
[TABLE]
Equating the degrees of both sides of this equation, we get m=0. Now, the equality c2=t(1+2c3+c4) and t∈C∖Q implies that c=0, whence a contradiction.
∎
Theorem 5.8**.**
For the walk wIII and for any t∈]0,1/∣D∣[∖Q such that Gt is infinite, F1(x,t) and F2(y,t) are hypertranscendental.
Proof.
Here, we have (using Lemma 4.5 for instance):
[TABLE]
The formula (4.17) applied in this setting gives
[TABLE]
Since the curve Et is nonsingular, the point P1 is a simple zero of Δ[x0:x1]x seen as a rational function on P1(C) (see Proposition 2.1 and Corollary 2.3). Then, it is easily seen that the polar divisor of b2 is [P1]+[Q1]+[ι1(Q1)]=[P1]+[Q1]+[τ(Q1)]. Using Corollary 3.9 and Proposition 3.10, we see that, in order to conclude the proof, it is sufficient to prove that P1≁Q1. Since s(P1)=Q1, this follows from Lemma 5.7. ∎
5.2.4. The walk wIIC.3
We shall exploit the fact that this walk is symmetric in the sense of Definition 5.5.
Lemma 5.9**.**
We assume that t∈C∖Q. For the walk wIIC.3, there are no R1,R2∈Et(Q(t)) such that s(R1)=R2 and R1∼R2.
Proof.
According to Lemma 5.6, it is sufficient to prove that s does not have fixed points in Et(Q(t)). Suppose to the contrary that there exists P∈Et(Q(t)) such that s(P)=P. So, P=(x,x) for some x=[x0:x1]∈P1(Q(t)) such that
[TABLE]
If x1=0, then we can assume that x0=1 and we see that (5.7) is impossible. Assume that x1=0. Then, we can and will assume that x1=1 and we have
[TABLE]
As we can see in the proof of Lemma 5.7, there are no such x0∈Q(t), whence a contradiction. ∎
Theorem 5.10**.**
For the walk wIIC.3 and for any t∈]0,1/∣D∣[∖Q such that Gt is infinite, we have that F1(x,t) and F2(y,t) are hypertranscendental.
Proof.
In this case, we have
- •
P1=([1:0],[0:1]);
- •
P2=([1:0],[−1:1])=ι1(P1);
- •
ι2(P1)=P1;
- •
Q1=([0:1],[1:0]);
- •
Q2=([−1:1],[1:0])=ι2(Q1);
- •
ι1(Q1)=([0:1],[1:0])=Q1;
- •
ι1(Q2)=([−1:1],[−t:t+1]).
The polar divisor of b2 is
[TABLE]
The poles Q2 and ι1(Q2) belong to the same τ-orbit:
[TABLE]
Similarly, the poles P2 and P1 belong to the same τ-orbit:
[TABLE]
Since s(Q1)=P1, Lemma 5.9 implies that the τ-orbits of ι1(Q2) and P2 are distinct. We refer to Figure 4 for a summary of these facts.
Since ι1(b2)=−b2, Lemma C.1 ensures that the residue of b2ω at P1 and P2 are equal (and they are non zero). Therefore, the sum of the residues of b2(ω) on the τ-orbit of P2 is nonzero. Lemma B.9 together with Remark B.16 and Proposition B.2 lead to the desired result. ∎
In conclusion, we have shown that the series F1(x,t) and F2(y,t) for the walks in Wtyp are hypertranscendental. The next section is devoted to the study of the walks in Wex.
6. Hyperalgebraicity of the generating series of the walks in Wex
We will show below that the following is true for any walk in Wex :
Condition 6.1**.**
For i=1 or i=2 there exist an integer n≥0, c0,…,cn−1∈C and g∈C(Et) such that
[TABLE]
As we have noticed in the proof Proposition 3.10, if Condition 6.1 is satisfied for i=1 or i=2, then it is satisfied for both values. Condition 6.1 precludes the possibility of using Proposition 3.8 to show that the corresponding generating series are hypertranscendental. As we have already mentioned, this does not immediately imply that the series F1 and F2 are hyperalgebraic. Nonetheless, we can use this data together with properties of the related rx and ry to show that they are indeed hyperalgebraic. We are going to prove the following result.
Proposition 6.2**.**
If a walk satisfies Assumption 2.9 and Condition 6.1 then the corresponding F1(x,t) and F2(y,t) are hyperalgebraic over C.
Using Proposition 6.2, we are now able to prove Theorem 2.
- **Proof of Theorem 2. **
We will show below, in Subsections 6.1, 6.2 and 6.3, that the nine walks in Wex satisfy Condition 6.1. Proposition 6.2 together with Assumption 2.9 then imply that for these walks the corresponding series F1(x,t)=QD(x,0,t) and F2(y,t)=QD(0,y,t) are hyperalgebraic with respect to x and y. The functional equation (2.1) for QD(x,y,t) can be rewritten as
[TABLE]
Each term on the right hand side of this equation is both x- and y-hyperalgebraic. Since the property of hyperalgebraicity is closed under field operations, QD(x,y,t) is also x- and y-hyperalgebraic. □
We will need the following lemmas to prove Proposition 6.2.
Lemma 6.3**.**
Let U,V be open subsets of C and f:V→C,g:U→V be functions analytic in their domains. If f(x) and g(x) are hyperalgebraic over C, then so is f(g(x)).
Proof.
One easily checks that a function h is hyperalgebraic over C if and only if the field C(h(x),h′(x),…,h(n)(x),…) has finite transcendence degree over C. Since f is hyperalgebraic, the field C(f(g(x)),f′(g(x)),…,f(n)(g(x)),…) has finite transcendence degree over C. Faà di Bruno’s formula [Jor65, p. 33] for the derivative of a composite function shows that for all m
[TABLE]
which is of finite transcendence degree over C. ∎
We note that similar techniques show that if f and g are hyperalgebraic over C then so is any element of C(f,f′,…,g,g′,…).
Lemma 6.4**.**
Let h:U→V be a biholomorphism between open subsets of C. If h is hyperalgebraic over C, then h−1 is hyperalgebraic over C.
Proof.
We know that C(h,h′,…,h(k),…) has finite transcendence degree over C, so C(h∘h−1,h′∘h−1,…,h(k)∘h−1,…) has finite transcendence degree over C as well. But, the successive derivatives of h−1 are rational fractions in the h(k)∘h−1 (k∈Z≥0), so C((h−1)′,…,(h−1)(k),…) is a subfield of C(h′∘h−1,…,h(k)∘h−1,…) and, hence, has finite transcendence degree over C. ∎
- **Proof of Proposition 6.2. **
We will prove this for F1(x,t); the other case is similar. We set
[TABLE]
Let g∈C(Et) be as in Condition 6.1. We then have
[TABLE]
Therefore L(rx)−g is τ-invariant. From Assumption 2.9, we have that rx is invariant under ω↦ω+ω1 and since g∈C(Et) the same is true for g. Therefore L(rx)−g=R(p1,3,p1,3′) where R is a rational function of two variables and p1,3 is the Weierstrass p-function with periods ω1 and ω3. Since p1,3 is hyperalgebraic over C, we have that L(rx)−g is hyperalgebraic over C. The element g is a rational function of a Weierstass p-function p1,2 with periods ω1 and ω2 and its derivative, so it is also hyperalgebraic over C. Therefore L(rx) is hyperalgebraic over C and thus the same holds for rx. By definition, for some open set U, we have that rx(ω)=F1(q(ω),t) where q is a rational function of p1,2 and p1,2′. Let U′⊂U and V′ be nonempty open subsets of C such that p1,2 induces a biholomorphism U→V. Then, on V, we have F1(x,t)=rx(p1,2−1(x)) and we deduce from Lemma 6.4 and Lemma 6.3 that F1(x,t) is hyperalgebraic over C. □
The remainder of this section is devoted to showing that the walks in Wex satisfy Condition 6.1 for b2. These proofs rely heavily on the alternate characterizations of Condition 6.1 for b2 given in the appendix (i.e., Proposition B.2, Corollary B.4, Lemma B.9 and Remark B.16). These characterizations are just in terms of the τ-orbits of the poles of b2.
We list here the sections where the treatments of the walks in Wex are finalized:
- •
for wIIC.1, wIIC.2 and wIIC.4, see Section 6.1, Theorem 6.5;
- •
for wIIB.1, wIIB.2, wIIB.3, wIIB.6 and wIIB.7, see Section 6.2, Theorem 6.6;
- •
for wIIC.5, see Section 6.3, Theorem 6.8.
In all cases we assume that t has been chosen such that Assumption 2.9 holds.
6.1. The walks wIIC.1, wIIC.2 and wIIC.4
Theorem 6.5**.**
The walks wIIC.1, wIIC.2 and wIIC.4 satisfy Condition 6.1.
Proof.
We shall first give a detailed proof for wIIC.1. In this case, we have
[TABLE]
and
[TABLE]
We see that
- •
P1 is the only pole of x and it has order 2;
- •
the poles of y are Q1 and Q2 and they have order 1;
- •
the poles of ι1(y) are ι1(Q1)=([−1:1],[t+1t:1]),ι1(Q2)=([0:1],[0:1]), and they have order 1;
- •
since Q1,Q2,ι1(Q1),ι1(Q2) are two by two distinct, these four points are the poles of ι1(y)−y and they all have order 1;
- •
we have, see (4.17), (b2)2=x12(x0+x1)2Δ[x0:x1]x, but P1 is a zero of order 2 of x04Δ[x0:x1]x and a zero of order 4 of (x0x1)2, so P1 is a pole of order 2 of (b2)2, and hence of order 1 of b2;
- •
Q2 and ι1(Q2) are zeros of x;
- •
Q1, and ι1(Q1) are not zeros of x.
Finally, the polar divisor of b2 is [P1]+[Q1]+[ι1(Q1)] and P1,Q1,ι1(Q1) are two by two distinct.
Moreover, the first 4 elements of the orbit of ι1(Q1) by the iterated action of τ are given by :
[TABLE]
Therefore, all the poles of b2 belong to the same τ-orbit. In summary, b2 has only simple poles, and they all belong to the same τ-orbit. The result is now a direct consequence of Corollary B.4.
The other cases are similar. The polar divisor of b2 and the first few terms of the τ-orbit of one of the poles of b2 in the remaining cases are listed in Figure 5. ∎
6.2. The walks wIIB.1, wIIB.2, wIIB.3, wIIB.6 and wIIB.7
Theorem 6.6**.**
The walks wIIB.1, wIIB.2, wIIB.3, wIIB.6 and wIIB.7 satisfy Condition 6.1.
Proof.
In the 5 cases,
- •
([1:0],[1:0]) is a double pole of b2,
- •
there exists α∈{0,−1} such that ([1:0],[α:1]) is a double pole of b2,
- •
b2 has at most two simple poles.
Furthermore, every pole belong to the same τ-orbit, see Figure 5. We consider a set of analytic local parameters as given by Lemma C.2 and we use the same notations. Since ι1(b2)=−b2, Lemma C.2 ensures that ores([1:0],[1:0]),2(b2)=0. Moreover, since every pole of b2 belong to the same τ-orbit, Lemma C.3 ensures that oresQ,1(b2)=0 for all Q. Lemma B.9 together with Proposition B.2 lead to the desired result. ∎
6.3. The walk wIIC.5
In this case, unlike the previous cases, the poles of b2 form two distinct τ-orbits. To proceed, we will need the following
Lemma 6.7**.**
Let R1,R2∈Et(Q(t)) be such that ι1(R1)=R2 and R1∼R2. Then, there exist j∈{1,2} and R∈Et(Q(t)) such that ιj(R)=R.
Proof.
The reasonning is similar to a part of the proof of Theorem 5.3.
It is easily seen that the equality τn(R1)=R2, together with the fact that τ=ι2∘ι1 is the composition of two involutions, imply, if n=2k, that ι1(τk(R1))=τk(R1), and, if n=2k+1, that ι2(τk(ι2(R2)))=τk(ι2(R2)). Proposition 4.8 ensures that both τk(R1) and τk(ι2(R2)) belong to Et(Q(t)). Whence the desired result. ∎
Theorem 6.8**.**
The walk wIIC.5 satisfies Condition 6.1.
Proof.
We have
- •
P1=([1:0],[0:1]);
- •
P2=([1:0],[−1:1]);
- •
Q1=([0:1],[1:0]);
- •
Q2=([−1:1],[1:0]);
- •
ι1(Q1)=([0:1],[0:1]);
- •
ι1(Q2)=([−1:1],[t:2t+1]).
The polar divisor of b2 is
[TABLE]
The poles P1 and ι1(Q2) belong to the same τ-orbit:
[TABLE]
Similarly, the poles P2 and Q2 belong to the same τ-orbit:
[TABLE]
Moreover, the τ-orbits of ι1(Q2) and P2 are distinct. Indeed, otherwise, since we have
[TABLE]
Lemma 6.7 would imply the existence of R∈Et(Q(t)) such that ιj(R)=R for some j∈{1,2}. But, Lemma 4.6 ensures that R is a base point and Lemma 4.5 ensures that none of the base points are fixed by ιj, whence a contradiction.
Lemma C.1 ensures that the residues of b2ω at P1 and P2=ι1(P1) are equal and will be denoted by a. Similarly, the residues of b2ω at Q2 and ι1(Q2) are the same and will be denoted by b. Since the sum of the residues over Et of b2(ω) is equal to [math], we have 2a+2b=0, so a+b=0. Therefore, the sum of the residues of b2(ω) on any τ-orbit is equal to [math]. Lemma B.9 together with Remark B.16 and Proposition B.2 lead to the desired result. ∎
7. Nonholonomicity in the exceptional cases
Theorem 7.1**.**
For each walk in Wex the series F1(x,t) and F2(y,t) are not holonomic, i.e., they do not satisfy any nontrivial linear differential equation with coefficients in C(x) and C(y) respectively.
Proof.
We only present the proof for F2(y,t), the proof for F1(x,t) being similar. Assume that F2(y,t) is holonomic. Then, F2(y,t) could be analytically continued to a multivalued meromorphic function on Et\{\mboxafinitesetofpoints}. This implies that ry would be a meromorphic function on the universal cover of Et whose singular points form a finite set modulo the lattice Zω1+Zω2. For each walk in Wex, we have that F2(y,t) satisfies Condition 6.1. Therefore as in the proof of Proposition 6.2, there is a g∈C(Et) such that τ(L(ry)−g)=L(ry)−g. Note that the poles of L(ry)−g also form a finite set modulo the lattice Zω1+Zω2. Since L(ry)−g is τ-periodic, this set is left invariant by ω↦ω+ω3. Using the fact that the reduction of ω3 modulo Zω1+Zω2 has infinite order, we get that the set of poles of L(ry)−g is empty. Using Condition 6.1 again, we see that L(ry)−g is ω1-periodic as well as being ω3 periodic. Since it has no poles, we must have that L(ry)−g=c∈C.
We want to prove that this last fact leads to a contradiction. To do this we will use some notation from [KR12, Sec. 4.2]. Let Δx be the set of ω in ω1R+]0,ω2[ corresponding to points on the elliptic curve with ∣x∣<1 and let Δy be the set of ω in ω1R+ω3/2+]0,ω2[ corresponding to points on the elliptic curve with ∣y∣<1. Let us state and prove two lemmas.
Lemma 7.2**.**
The function ry has no poles in Δx∪Δy.
Proof of Lemma 7.2.
From [KR12, Theorem 3], one sees that ry has no poles on Δy and that rx has no poles on Δx.
Using formula (4.5) in [KR12] and the fact that rx has no poles on Δx, we find that the poles of ry on Δx are poles of xy. It is therefore sufficient to prove that xy does not have poles for ∣x∣<1. Note that a pole of xy with ∣x∣<1 is a pole of y.
We claim that ([0:1],[1:0]) is the only possible pole of xy with ∣x∣<1. We recall that the poles of y are Q1 and Q2. Let α1,α2∈P1(C) be the x-coordinates of Q1 and Q2 respectively. To prove the claim it is sufficient to prove that for j∈{1,2}, ∣αj∣<1 implies αj=0. The x-coordinates {α1,α2} are the two roots in P1(C), counted with multiplicities, of the polynomial d−1,1+d0,1X+d1,1X2. For the walks in Wex, we have d0,1=1. The following array summarizes the possible values of the pair (α1,α2) in the four situations:
[TABLE]
In the four situations, we see that ∣αj∣<1 implies αj=0, proving our claim.
Recall that we are interested in the poles of xy with ∣x∣<1. Thanks to the claim just proved we have to determine whether ([0:1],[1:0]) is a pole of xy. We have seen in the proof of the claim that for the walks in Wex, Q1=Q2 since their x-coordinates are different. So y has at most a simple pole at ([0:1],[1:0]). This shows that ([0:1],[1:0]) is not a pole of xy. Combined with the claim, this shows that xy has no poles for ∣x∣<1, proving the lemma.
∎
Lemma 7.3**.**
The function g has at least one pole in Δx∪Δy, that is, one pole at a point on the elliptic cuve in {∣x∣<1}∪{∣y∣<1}.
Proof of Lemma 7.3.
According to Figures 5 and 6, there exists a pair (Q,n) with n>0 such that Q and τn(Q) are poles of b2, and for all ℓ∈]−∞,−1]∪[n+1,∞[, τℓ(Q) is not a pole of b2. Since Condition 6.1 is satisfied, one sees that τn(Q) and τ(Q) should be poles of g. Using Figures 5 and 6, we see that for all walks in Wex except wIIB.7, this implies that g should have a pole for an ω∈C that corresponds to {∣x∣<1}∪{∣y∣<1} in the elliptic curve, proving the lemma for those cases.
It remains to treat wIIB.7. In this case, b2 has two double poles Q,τ4(Q) and two simples poles τ(Q),τ3(Q) with Q=([1:0],[−1:1]). The operator L in Condition 6.1 have coefficients in C. Let n be its order. With Lemma 3.2, we find that L(b2) have only two poles of order n+2, that are Q,τ4(Q), and no poles of higher order. With L(b2)=τ(g)−g, we find that g should have a pole of order n+2 at τ4(Q). So τ(g) have a pole of order n+2 at τ3(Q). Since L(b2)=τ(g)−g and L(b2) have no poles of order n+2 or higher at τ3(Q), g should have a pole of order n+2 at τ3(Q). But the y-coordinates of τ3(Q) is [0:1], proving the lemma in this case. ∎
Let us complete the proof of the theorem. From Lemma 7.2 and Lemma 7.3, we see that there exists ω0∈C, such that g has a pole at ω0 and such that ry is analytic at ω0. Since L has coefficients in C, L(ry) is analytic at ω0. This contradicts L(ry)−g∈C. ∎
8. Some comments concerning singular and weighted walks
8.1. Singular walks
If the walk is singular, then Et is a rational curve, i.e., is birational to P1(C). The difference equations on elliptic curves involved in the nonsingular case should be replaced by finite difference or q-difference equations on a rational curve. It seems plausible that our Galoisian methods can be used in order to study the generating series of the singular walks as well.
8.2. Weighted walks
We consider a walk with small steps in the quarter plane Z≥02. Following [FIM99, KY15], the step (i,j)∈{0,±1}2 is weighted by some di,j∈Q≥0. Let D be the set {di,j ∣ (i,j)∈{0,±1}2}. For i,j,k∈Z≥0, we let qD,i,j,k be the number of walks in Z≥02 with weighted steps in D starting at (0,0) and ending at (i,j) in k steps and we consider the corresponding trivariate generating series
[TABLE]
We can then ask for these weighted walks the same questions as for the unweighted walks considered in the present paper. It turns out that in the weighted context, we have generalizations of the basic tools used in the unweighted case (generalizations of the kernel, of the functional equation (2.1), etc). Moreover, explicit conditions can be deduced from [FIM99], Section 2.3.2, Corollary 4.2.11 and Theorem 3.2.1 in order to have the properties listed in Remark 2.10 satisfied. This should be the starting point to apply our technics in this context.
Appendix A Galois Theory of Difference Equations
In this section we describe some basic facts concerning the Galois theory of linear difference equations and indicate how these lead to a proof of Proposition 3.6. Many of these facts we state without proof but proofs can be found in [vdPS97].
The appropriate setting for this Galois is difference algebra, that is, the study of algebraic objects endowed with an automorphism, so we begin with
Definition A.1**.**
A difference ring is a pair (R,τ) where R is a ring and τ is an automorphism of R. A difference ideal I⊂R is an ideal such that τ(I)⊂I.
One can define difference subring, difference homomorphism, difference field, etc. in a similar way. A difference subring of particular importance in any difference ring is given in the following definition.
Definition A.2**.**
The constants Rτ of a difference ring R are
[TABLE]
One can show that Rτ forms a ring and, if R is a field, then Rτ is also a field.
Example A.3*.*
1. (C[x],τ) where τ(x)=x+1. The only difference ideals are C[x] and {0}. The constants of this ring are C.
2. (C[x],τ) where τ(x)=qx, q not a root of unity. The difference ideals are C[x],{0} and the (xk) for k∈Z≥1. The constants of this ring are C.
3. (M(C),τ) where M(C) is the field of meromorphic functions on C and τ(ω)=ω+ω3 for some ω3∈C. This is and the next example are difference fields. The constants of this field form the field of ω3-periodic meromorphic functions.
4. (C(Et),τ) where C(Et) is the field of meromorphic functions on Et) and for some fixed P∈Et ,τ(f(X))=f(X⊕P) for all f∈C(Et). If P is of infinite order, then the constants are C (see the argument following Definition 3.4).
When considering linear difference equations, it is most convenient to consider first order matrix equations, that is, equations of the form τ(Y)=AY where A∈GLn(K) where K is a difference field. Often one wants to deal with equations of the form form L(y)=τn(y)+an−1τn−1(y)+…+a0y=0, ai∈K. If a0=…=aj−1=0,aj=0, we can make a change of variables z=τj(y) and assume a0=0. One then sees that questions concerning solutions of L(y)=0 can be reduced to questions concerning the system τ(Y)=ALY where
[TABLE]
If z is a solution of L(y)=0 in some difference ring containing K, then (z,τ(z),…,τn−1(z))T is a solution of τ(Y)=ALY.
In addition to considering individual solutions of τ(Y)=AY, it is useful to consider matrix solutions and, in particular
Definition A.4**.**
Let R be a difference ring and A∈GLn(R). A fundamental solution matrix of τ(Y)=AY is a matrix U∈GLn(R) such that τ(U)=AU.
Note that if U1 and U2 are fundamental solution matrices of τ(Y)=AY, then τ(U1−1U2)=U1−1U2 so Uc=U1D where D∈GLn(Rτ).
The usual Galois theory of polynomial equations is cast in terms of a splitting field of the polynomial and a group of automorphisms of this field. For linear difference equations, the following takes the place of the splitting field.
Definition A.5**.**
Let K be a difference field and A∈GLn(K). We say that a k-algebra R is a Picard-Vessiot ring for τ(Y)=AY if
- (1)
R* is a simple difference ring extension of K (i.e., the only difference ideals of R are R and {0}).* 2. (2)
R=K[U,1/det(U)]* for some fundamental solution matrix U∈GLn(R) of τ(Y)=AY.*
It can be shown (cf. [vdPS97, Chapter 1.1]) that Picard-Vessiot rings always exist and if Kτ is algebraically closed they are unique up to k-difference isomorphisms. Furthermore, when Kτ is algebraically closed , we have Rτ=Kτ. Although some of the following results hold in more general situations, we will for simplicity assume from now on that
Kτ is algebraically closed and of characteristic zero.
Even under this assumption, the Picard-Vessiot ring need not be an integral domain. In [vdPS97, Corollary 1.16] a precise description of its structure is given but we will only use some basic facts listed below and not delve further.
We can now define the Galois group.
Definition A.6**.**
Let R be the Picard-Vessiot ring of τ(Y)=AY, A∈GLn(K). The Galois group G of R (or of τ(Y)=AY) is
[TABLE]
Using the notation of the definition, fix a fundamental solution matrix in GLn(R) of the equation τ(Y)=AY. If σ∈G, then
[TABLE]
Therefore, σ(U) is again a fundamental solution matrix and so σ(U)=U[σ]U where [σ]U∈GLn(Kτ). A key fact (cf. [vdPS97, Chapter 1.2]) forming the basis of the Galois theory of linear difference equations is
The map ρ:G→GLn(Kτ) given by ρ(τ)=[τ]U is a group homomorphism whose image is a linear algebraic group.
A subgroup G⊂GLn(Kτ) is a linear algebraic group if it is a closed in the Zariski topology on GLn(Kτ), the topology whose closed sets are common solutions of systems of polynomial equations in n2 variables.
Example A.7*.*
- Consider the equation
[TABLE]
This equation is not a homogeneous linear difference equation but it is equivalent to the matrix equation
[TABLE]
If z satisfies (A.1), then U=(10z1) is a fundamental solution of the matrix equation. The Picard-Vessiot extension of k is then given by R=K[z]. If σ∈G, then y=σ(z) also satisfies (A.1). Therefore σ(z)−z=dσ∈Kτ. This implies that the Galois group of the matrix equation may be identified with a Zariski closed subgroup of
[TABLE]
Note that this latter group is just the additive group (Kτ,+). The Zariski closed subgroups of this group are identified with Kτ and {0}.
- Consider the system of equations
[TABLE]
As above, this system is equivalent to the matrix equation
[TABLE]
The Picard-Vessiot extension of this equation is R=K(z0,…,zn) where τ(zi)−zi=bi and the Galois group is a subgroup of
[TABLE]
This latter group is just the direct sum of n+1 copies of the additive group (Kτ,+), that is G⊂((Kτ)n+1,+). The Zariski closed subgroups of (Kτ)n+1 are the vector subspaces and are all connected in the Zariski topology. From this we can deduce
Lemma A.8**.**
Using the notation of Example A.7.2, if G is a proper linear algebraic subgroup of (Kτ)n+1 then there exist c0,…,cn∈Kτ, not all zero, such that
[TABLE]
In [vdPS97, Chapter 1.3] a Galois correspondence and other basic facts are described. For our purposes, we only need
- (Gal 1)
An element z∈R is in K if and only if z is left fixed by all elements of G. 2. (Gal 2)
The ring R is an integral domain if and only if G is connected in the Zariski topology. 3. (Gal 3)
When G is connected, the dimension of G as an algebraic variety over Kτ is equal to the transcendence degree of the quotient field of R over K.
Concerning (Gal 3), when G is not connected then the dimension equals the Krull dimension of R.
We now present the main tool used in proving Proposition 3.6.
Proposition A.9**.**
Let R be the Picard-Vessiot extension for the system (A.2) and z0,…,zn∈R be solutions of this system. If z0,…,zn are algebraically dependent over K, then there exist ci∈Kτ, not all zero, and g∈K such that
[TABLE]
Proof.
We follow ideas due to M. van der Put appearing in the appendix of [Har08]. As in Example A.7.2, the Galois group G is a subgroup of (Kτ)n+1 and so is connected. (Gal 2) implies that R=K[z0,…,zn] is a domain. Since z0,…,zn are algebraically dependent, the transcendence degree of the quotient field of R is less than n+1. Therefore (Gal 3) implies that G is a proper subgroup of (Kτ)n+1. Lemma A.8 implies that G⊂{(d0,…,dn) ∣∑i=0ncidi=0} for some ci∈Kτ. For any σ∈G, we have
[TABLE]
From (Gal 1) we conclude that ∑i=0ncizi=g∈K. Applying τ to this last equation and subtracting yields
[TABLE]
∎
We now turn to
Proof of Proposition 3.6. Let f satisfy P(f,δ(f),…,δn(f))=0 for some polynomial P with coefficients in K=C(Et). Since δ and τ commute, we have
[TABLE]
Let I be a maximal difference ideal in the difference ring K[f,δ(f),…,δn(f)] and let R=K[f,δ(f),…,δn(f)]/I. The difference ring R is a simple difference ring of the form K[z0,…,zn] where zi is the image of δi(f) and τ(zi)−zi=bi, bi=δ(b). Therefore, R is a Picard-Vessiot ring for a system of the form (A.2). Applying Proposition A.9, yields the first conclusion of Proposition 3.6.
Now assume that K[f,δ(f),…,δn(f)]τ=C. A computation shows that τ(L(f)−g)−(L(f)−g)=0, where L=anδn+an−1δn−1+…+a0. Therefore L(f)=g+c, for some c∈C. If g+c=0, this equation shows that f is holonomic over K. If g+c=0, we can derive the same conclusion by considering δ(L(f))−(δ(g)/(g+c))L(f)=0. □
The above proof very much depends on the fact that we are considering systems of first order scalar difference equations of the form (A.2). In [DHR15], a proof of Proposition 3.6 was given based on the differential Galois theory of linear difference equations. This is a theory, presented in [HS08], that allows one to describe differential properties of general linear difference equations. A general introduction to this theory as well as an elementary introduction to the Galois theory of linear differential equations and the analytic theory of q-difference equations can be found in the articles in [HSS16].
Appendix B Telescopers and orbit residues
We have seen in Section 3.3 that the study of the hypertranscendance of F1(x,t) and F2(y,t) is intimately related to the study of equations of the form L(b)=τ(g)−g for some nonzero linear differential operator L with coefficients in C and some b,g∈C(Et). In other contexts (cf. [CS12]), L is referred to as a telescoper for b and g as a witness. The aim of this appendix is to study in more details these equations.
Let E be an elliptic curve defined over an algebraically closed field k of characteristic zero and k(E) be its function field. Let P be a non-torsion point on E and let τ:k(E)→k(E) denote map corresponding to Q↦Q⊕P on E, where ⊕ denotes the group law on E. We let Ω be a non zero regular differential form on E. A straightforward generalization of Lemma 3.1 shows the following result :
Lemma B.1**.**
The derivation δ of k(E) such that d(f)=δ(f)Ω commutes with τ.
We will prove the following
Proposition B.2**.**
Let b∈k(E). The following are equivalent.
- (1)
There exist g∈k(E) and a nonzero operator L∈k[δ] such that L(b)=τ(g)−g. 2. (2)
For all poles Q0 of b, we have that
[TABLE]
is regular at X=Q0 where Q0⊕n1P,…,Q0⊕ntP are the poles of b that belong to Q0⊕ZP.
This proposition allows one to give the following useful criteria guaranteeing when Condition 6.1 does or does not hold.
Corollary B.3**.**
Let b∈k(E) and assume that there exists Q0∈E such that
- (1)
b* has a pole of order m>0 at Q0, and* 2. (2)
b* has no other pole of order ≥m in Q0⊕ZP.*
Then there is no nonzero L∈k[δ] and g∈k(E) such that L(b)=τ(g)−g.
Proof.
This follows easily from Proposition B.2 since the pole of b(X) at Q0 cannot be cancelled by any pole of any b(X⊕nP) and so h(X) is not regular at X=Q0. ∎
Corollary B.4**.**
Let b∈k(E) and assume that there exists Q0∈E such that
- (1)
all poles of b occur in Q0⊕ZP, and 2. (2)
all poles of b are simple.
Then there exist g∈k(E) and a nonzero operator L∈k[δ] such that L(b)=τ(g)−g.
Proof.
Basically, this is true because the sum of the residues of a differential form on a compact Riemann surface is zero. More precisely, using Lemma B.15 below, one can show that the hypotheses of Corollary B.4 imply condition (2) of Proposition B.5 and therefore that the conclusion holds (see the remark following Lemma B.15).∎
To prove Proposition B.2 we shall prove two ancillary results, Propositions B.5 and B.8. These results give conditions equivalent to the conditions in Proposition B.2.
Before proceeding, we recall the following standard notation. If D is a divisor of E, we will denote by L(D) the finite dimensional k-space {f∈k(E) ∣ (f)+D≥0}, where (f) is the divisor of f. In Subsection B.1 we will prove
Proposition B.5**.**
Let b∈k(E). The following are equivalent.
- (1)
There exist g∈k(E) and a nonzero operator L∈k[δ] such that L(b)=τ(g)−g. 2. (2)
There exists Q∈E, e∈k(E) and h∈L(Q+(Q⊖P))444The symbol “+” represents the formal sum of divisors. We will use ⊕ and ⊖ for addition and subtraction of points on the curve. such that
[TABLE]
To state the next equivalence, we need two definitions. Corresponding to each point Q∈E there exists a valuation ring OQ⊂k(E). A generator uQ of the maximal ideal of OQ is called a local parameter at Q. Local parameters are unique up to multiplication by a unit of OQ.
Definition B.6**.**
Let S={uQ ∣ Q∈E} be a set of local parameters at the points of E. We say S is a coherent set of local parameters if for any Q∈E,
[TABLE]
We fix, once and for all, a coherent set of local parameters. All local parameters mentioned henceforth will be from this set.
Let uQ be a local parameter at a point Q∈E and let vQ be the valuation corresponding to the valuation ring at Q. If f∈k(E) has a pole at Q or order n, we may write
[TABLE]
where vQ(f~)≥0. The following definition is similar to Definition 2.3 of [CS12].
Definition B.7**.**
Let f∈k(E) and S={uQ ∣ Q∈E} be a coherent set of local parameters and Q∈E. For each j∈N>0 we define the orbit residue of order j at Q to be
[TABLE]
Note that if Q′=Q⊕tP for some t∈Z, then oresQ′,j(f)=oresQ,j(f) for any j∈N>0. Furthermore oresQ,j(f)=oresQ,j(τ(f)). We shall prove the next result in Subsection B.2.
Proposition B.8**.**
Let b∈k(E) and S={uQ ∣ Q∈E} be a coherent set of local parameters. The following are equivalent.
- (1)
There exists Q∈E, e∈k(E) and g∈L(Q+(Q⊖P)) such that
[TABLE] 2. (2)
For any Q∈E and j∈N>0
[TABLE]
Proposition B.5, Proposition B.8 and the following lemma immediately imply Proposition B.2.
Lemma B.9**.**
Let b∈k(E). The following are equivalent
- (1)
For all poles Q0 of f, we have that
[TABLE]
is regular at X=Q0 where Q0⊕n1P,…,Q0⊕ntP are the poles of b that belong to Q0⊕ZP. 2. (2)
For any Q∈E and j∈N>0
[TABLE]
Proof.
If u is the local parameter at Q0, we may write
[TABLE]
where vQ0(h′)≥0. One easily sees that
[TABLE]
The conclusion now follows.∎
Remark B.10**.**
Assume that k=C. Then, one can consider the analytification Ean of E. Instead of considering algebraic local parameters on E, one can consider analytic local parameters {uQ ∣ Q∈E}, i.e., for any Q∈E, uQ is a biholomorphism between a neighborhood of Q in Ean and a neighborhood of [math] in C. There is an obvious notion of coherent analytic local parameters, extending the notion introduced in Definition B.6, and a corresponding notion of oresQ,j. Lemma B.9 remains true in this context, with the same proof.
Remark B.11**.**
The proof that (2) implies (1) in Proposition B.8 is constructive. One only needs a constructive method for finding the bases of certain L spaces (e.g. [Hes02]). The proof that (2) implies (1) in Proposition B.5 is also constructive. Therefore given b∈k(E) one can decide if there exist g∈k(E) and a nonzero operator L∈k[δ] such that L(b)=τ(g)−g.
B.1. Proof of Proposition B.5
In the following lemma, we will collect some facts concerning the local behavior of functions under the actions of τ. Its proof is a straightforward generalization of the proof of Lemma 3.2.
Lemma B.12**.**
Let u be a local parameter of k(E) and let vu be the associated valuation. Then vu(δ(u))=0 and, for any f∈k(E) such that vu(f)=0, we have
- (1)
if vu(f)≥0 then vu(δ(f))≥0; 2. (2)
if vu(f)<0 then vu(δ(u))=vu(f)−1.
We will also need a consequence of the Riemann-Roch Theorem for elliptic curves: If D is a positive divisor on E and l(D) is the dimension of the space L(D) then
[TABLE]
This implies that if Q is a point on E, u is a local parameter at Q, n≥2, and c2,…,cn∈k, then there exists an f∈L(nQ) and c1∈k such that
[TABLE]
where vu(f~)≥0. A priori, we have no control of the element c1.
Finally we need some definitions:
Definition B.13**.**
Let f∈k(E) and Q∈E.
- (1)
If Q is a pole of f, the polar dispersion of f at Q, pdisp(f,Q) is the largest nonnegative integer ℓ such that Q⊕ℓP is also a pole of f. 2. (2)
The polar dispersion of f, pdisp(f), is max{pdisp(f,Q) ∣ Q\mboxapoleoff}. 3. (3)
*The weak polar dispersion of f, wpdisp(f), is *
\max\{\ell\ |\ \exists Q\in E\mbox{ s.t. fhasapoleoforderatleast2atQandQ\oplus\ell P}\}.
The following is an analogue of [HS08, Lemma 6.2].
Lemma B.14**.**
Let f∈k(E). There exist f∗,g∈k(E) such that pdisp(f∗)≤1, wpdisp(f∗) = [math] and f=f∗+τ(g)−g.
Proof.
We begin by showing that there exist f∗,g∈k(E) such that wpdisp(f∗) = [math] and f=f∗+τ(g)−g. We will then further refine f∗ so that pdisp(f∗)≤1 as well.
Let N= wpdisp(f)≥1 and nf= the number of points Q∈E such that f has poles of order at least two at Q and Q⊖NP. Fix such a point Q and let u be a local parameter at Q. We may write
[TABLE]
where m≥2 and vQ(hf)≥0. The Riemann-Roch Theorem implies that there exists a g~∈L(mQ) such that
[TABLE]
where bi=−ai for i=2,…,m and vQ(hg~)≥0. Note that τ(g~) has a pole of order m at Q⊖P. Letting f~=f−(τ(g~)−g~), one sees that f~ has a pole of order at most 1 at Q. Therefore either the wpdisp(f) = wpdisp(f~) and nf~<nf or wpdisp(f) > wpdisp(f~). An induction allows us to conclude that there exist f∗,g∈k(E) such that wpdisp(f∗) = [math] and f=f∗+τ(g)−g.
We may now assume that wpdisp(f) =0 and let pdisp(f)=N≥2. Let f have poles at both Q and Q⊕NP. Since wpdisp(f) =0, f has a pole of order greater than one at no more than one of these two points. We deal with the two cases separately.
f has a pole of order 1 at Q⊕NP. The Riemann-Roch Theorem implies that there exists a nonconstant g~∈L((Q⊕(N−1)P)+(Q⊕NP)). Note that τ(g~)∈L((Q⊕(N−2)P)+(Q⊕(N−1)P)). For some a∈k, f~−(τ(ag~)−ag~) has no pole at Q⊕NP and so pdisp(f,Q)<N. An induction finishes the proof.
f has a pole of order 1 at Q. The Riemann-Roch Theorem implies that there exists a nonconstant g~∈L((Q⊕P)+(Q⊕2P)). Note that τ(g~)∈L((Q)+(Q⊕P)). For some a∈k, f~−(τ(ag~)−ag~) has no pole at Q and so pdisp(f,Q)<N. An induction again finishes the proof.∎
We now turn to the
- **Proof that (1) implies (2) in Proposition B.5. **
Applying Lemma B.14, we may assume that pdisp(b)≤1 and wpdisp(b) = [math] (here one uses the fact that L∘τ=τ∘L, since L∈C[δ] and τ∘δ=δ∘τ). We will first show that for any Q∈E, if b has a pole at Q then this pole must be simple and it has another pole of the same order at Q⊕P or at Q⊖P. To see this note that if b has a pole at Q, then either g or τ(g) has a pole at Q. Assume that g has a pole at Q (the argument assuming τ(g) has a pole at Q is similar). Let r be the largest integer such that Q⊕rP is a pole of g and s be the largest integer such that Q⊖sP is a pole of g. We then have that Q⊕rP and Q⊖(s+1)P are both poles of τ(g)−g and therefore of L(b). Using Lemma 3.2 above, one sees that they must also be poles of b. Since pdisp(b)≤1, we have r=s=0. In particular, the only pole of g in Q⊕ZP is at Q, the only pole of τ(g) in Q⊕ZP is at Q⊖P and they must have the same orders. Once again, Lemma 3.2 above implies that b has poles at these points of equal orders. Since wpdisp(b) = [math], the orders of these poles must be 1.
We can therefore conclude that b has only poles of order 1 and the poles of b occur in pairs {Q1,Q1⊖P},…,{Qr,Qr⊖P} where (Qi⊕ZP)∩(Qj⊕ZP)=∅ for i=j.
We will now show how one can construct an element e such that b−(τ(e)−e) has at most one pair of poles {Q,Q⊖P}. This will yield (2) and our contention. Assume r>1 and that b has simple poles at the pairs {Q1,Q1⊖P} and {Q2,Q2⊖P}. Let h∈k(E) be a nonconstant element of L(Q1+Q2). There exists an a∈k such that b~=b−(τ(ah)−ah) has no pole at Q1. The element b~ has only simple poles and \mboxpdisp(b)≤1. Therefore its poles occur at possibly Q1⊖P,{Q2,Q2⊖P},…,{Qr,Qr⊖P}. Since b~ satisfies an equation of the form L(b~)=τ(g~)−g~ for some g~∈k(E) the poles of such an b~ must occur in pairs. Therefore we have that b~ has no pole at Q1⊖P. Continuing in this way we find an e∈k(E) such that b−(τ(e)−e) has at most one pair of poles {Q,Q⊖P}. □
In the proof that (2) implies (1) in Proposition B.5 we will need the following technical lemma. Let u be a local parameter at Q. Note that τ(u) is a local parameter at Q⊖P.
Lemma B.15**.**
If g∈L(Q+(Q⊖m1P)+…+(Q⊖mtP)) where m1,…,mt∈Z\{0} then
[TABLE]
Proof.
This result will follow from the fact that the sum of the residues of a differential form on a compact Riemann surface must be zero. We start by noting that Lemma 3.2 states that vuQ(δ(uQ))=0 so we may write δ(uQ)−1=α+uˉ where 0=α∈k and uˉ is regular and zero at Q. For each i∈Z we write
[TABLE]
where gQ⊖iP is regular and zero at Q⊖iP. Now consider the differential gΩ. Since for any i∈Z, Ω=δ(uQ⊖iP)−1duQ⊖iP, we have
[TABLE]
where the second equality follows from the fact that uQ⊖iP=τi(uQ) and τδ=δτ. Therefore the residue of gΩ at Q⊖iP is αcQ⊖iP,−1. Since α=0 and the sum of the residues of a differential form is [math] we have oresQ,1(g)=0.∎
Remark B.16**.**
The proof of Lemma B.15 shows that if the poles of g∈k(E) are simple and belong to Q⊕ZP, then there exists 0=α∈C such that oresQ,1(g)=α∑i∈ZResQ⊕iP(gΩ). Therefore, oresQ,1(g)=0 if and only if ∑i∈ZResQ⊕iP(gΩ)=0.
Remark B.17**.**
Lemma B.15 and Proposition B.2 imply Corollary B.4. To see this note that for f as in Corollary B.4 we have that f∈L(Q+(Q⊖m1P)+…+(Q⊖mtP)) where mi=−ni. The residue of h(X)=∑i=1tf(X⊕niP) at X=P is oresQ,1(f), so h(X) is regular at Q0. Applying Proposition B.2 yields the conclusion of Corollary B.4.
- **Proof that (2) implies (1) in Proposition B.5. **
Let us assume that condition (2) holds. We claim that it is enough to find an element g~ and a nonzero operator L such that L(h)=τ(g~)−g~. Assume that we have done this. Then
[TABLE]
where g=L(e)+g~.
If h is constant, then the result is obvious (take L=δ and g~=0). We shall now assume that h is not constant.
To simplify notation, we write u for uQ and let δ(u)=u0+uˉ, where 0=u0∈k and uˉ is regular and zero at Q, and so δ(τ(u))=u0+τ(uˉ). Using Lemma B.15, one sees that
[TABLE]
where vu(hu)>0 and vτ(u)(hτ(u))>0. Selecting an element f∈L(2Q) such that f=u0a/u2+…, we have
[TABLE]
Since {1,h} forms a basis of L(Q+(Q⊖P)) (recall that h is not constant), there exist elements c,d∈k such that
[TABLE]
Therefore
[TABLE]
and conclusion (2) holds for L=δ2−cδ and g~=δ(f). □
Remark B.18**.**
One cannot weaken condition (2) in Proposition B.5, that is, for a general b∈k(E), condition (1) of Proposition B.5 does not imply the following condition :
- (3)
There exist Q∈E, e∈k(E) and a constant c∈k such that
[TABLE]
To see this, let b be a nonconstant element of L(Q+(Q⊖P)). Note that pdisp(b)=1. We have just shown that b satisfies (1) of Proposition B.8. Now assume b=τ(e)−e+c for some e∈k(E),c∈k. Since pdisp(τ(e)−e)=pdisp(e)+1 if e∈/k, we have pdisp(e)=0. Since b has no poles outside of {Q,Q⊖P}, we would have that e has at most one pole and this pole would be simple. Therefore e must be constant. A contradiction with the fact that b∈/k.
B.2. Proof of Proposition B.8
- **Proof that (1) implies (2) in Proposition B.8. **
For any Q∈E and j∈N>0, we have oresQ,j(e)=oresQ,j(τ(e)). Furthermore Lemma B.15 implies that oresQ,j(g)=0. Therefore oresQ,j(f)=oresQ,j(τ(e)−e+g)=oresQ,j(τ(e))−oresQ,j(e)+oresQ,j(g)=0. □
- **Proof that (2) implies (1) in Proposition B.8. **
The proof of this implication is similar to the proof that (1) implies (2) in Proposition B.5. Lemma B.14 implies that we may assume that pdisp(f)≤1 and wdisp(f)=0. Therefore if f has a pole of order j≥2 at some Q∈E, then Q is the only point in Q+ZP at which f has a pole. Since oresQ,j(f)=0, we have that f has no poles of order greater than 1. Since we also have pdisp(f)≤1, we can conclude that that f has only poles of order 1 and the poles of f occur in pairs {Q1,Q1⊖P},…,{Qr,Qr⊖P} where (Qi⊕ZP)∩(Qj⊕ZP)=∅ for i=j.
We will now show how one can construct an element e such that f−(τ(e)−e) has at most one pair of poles {Q,Q⊖P}. This will yield condition 2. of the Proposition. We can assume that r>1. Let h∈k(E) be a nonconstant element of L(Q1+Q2). There exists an a∈k such that f~=f−(τ(ag)−ag) has no pole at Q1. The element f~ has only simple poles and \mboxpdisp(f)≤1. Therefore its poles occur at possibly Q1⊖P,{Q2,Q2⊖P},…,{Qr,Qr⊖P}. Since oresQ1,1(f)=0, f cannot have a singe pole in Q1+ZP. Therefore we have that f~ has no pole at Q1⊖P. Continuing in this way we find an e∈k(E) such that f−(τ(e)−e) has at most one pair of poles {Q,Q⊖P}.
□
Appendix C Some computation of orbit residues
Let E be an elliptic curve defined over an algebraically closed field k of characteristic zero and k(E) be its function field. Let P be a non-torsion point on E and let τ:k(E)→k(E) denote map corresponding to Q↦Q⊕P on E, where ⊕ denotes the group law on E. Let ι1 and ι2 two involutions of E such that τ=ι2∘ι1. We let Ω be a non zero regular differential form on E and we keep notation as in §B.
Lemma C.1**.**
Let b∈k(E) such that ι1(b)=−b. Let Q∈E be a simple pole of b such that Q=ι1(Q). Then, ι1(Q) is a simple pole of b and the residue of bΩ at Q coincides with its residue at ι1(Q).
Proof.
The assertion follows from the fact that, since ι1(b)=−b and ι1∗(Ω)=−Ω (see [Dui10, Lemma 2.5.1 and Proposition 2.5.2]), the form η=bΩ satisfies η=ι1∗(η). Indeed, if uQ is a local parameter at Q, then we have η=vduQ with v=uQcQ+v where cQ∈C is the residue of η at Q and v is regular at Q. Hence, ι1(uQ) is a local parameter at ι1(Q), and we have ι1∗(η)=ι1(v)dι1(uQ) with ι1(v)=ι1(uQ)cQ+ι1(v) where ι1(v) is regular at ι1(Q). So, Resι1(Q)(η)=Resι1(Q)(ι1∗(η))=cQ=ResQ(η). ∎
For the notion of coherent analytic parameters used below, we refer to Appendix B, especially to Remark B.10.
Lemma C.2**.**
There exists a coherent set of analytic local parameters {uQ ∣ Q∈E} on Ean such that ι1(uQ)=−uι1(Q). Let b∈k(E) such that ι1(b)=−b. For such a set of local parameters, if
[TABLE]
where vQ(f~)≥0, then
[TABLE]
where vι1(Q)(g~)≥0 and cι1(Q),j=(−1)j+1cQ,j. If follows that, if all the poles of b belong to the same τ-orbit, then, for any even number j, we have oresQ,j(b)=0.
Proof.
We first prove the existence of analytic local parameters with the desired properties. According to [Dui10, p.35 and Remark 2.3.8], ι1(P)=[−1]P⊕P0 for some P0∈E. By uniformisation, it is equivalent to prove the following result : Consider a lattice Λ⊂C and two endomorphisms of the complex torus C/Λ given by ι1:z↦−z+p0 and τ:z↦z+q0 for some p0,q0∈C. Then, there exists a set of analytic local parameters {uω ∣ ω∈C/Λ} on the complex torus C/Λ such that ι1(uω)=−uι1(ω) and τ(uω)=uτ(ω). Such local parameters are given by uω:z↦z−ω for z close to ω. The rest of the Lemma is a direct consequence of the following easy computation. Indeed, applying ι1 to
[TABLE]
we get
[TABLE]
where vι1(Q)(ι1(f~))≥0, as expected. ∎
Lemma C.3**.**
If g∈L(2Q+2(Q⊖m1P)+…+2(Q⊖msP)) where m1,…,ms∈Z\{0} is such that oresQ,2(g)=0 then
[TABLE]
Proof.
We may write δ(uQ)−1=α+βuQ+uˉ where 0=α∈k, β∈k and uˉ is regular and has a zero of order 2 at Q. For each i∈{0,m1,…,ms} we write
[TABLE]
where gQ⊖iP is regular and zero at Q⊖iP. Now consider the differential gΩ. Since for any i∈Z, ω=δ(uQ⊖iP)−1duQ⊖iP, we have
[TABLE]
where the second equality follows from the fact that uQ⊖iP=τi(uQ) and τδ=δτ. Note that τi(uˉ) is regular and has a zero of order 2 at Q⊖iP. Therefore the residue of gω at Q⊖iP is αcQ⊖iP,1+βcQ⊖iP,2. Since the sum of the residues of a differential form is [math] we get αoresQ,1(g)+βoresQ,2(g)=0. Since α=0 and oresQ,2(g)=0, we get oresQ,1(g)=0. ∎
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