On the nature of the generating series of walks in the quarter plane

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 function coefficients.

Appendices

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 [24].

The appropriate setting for this Galois theory 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 $(R,\tau )$ where R is a ring and $\tau$ is an automorphism of R. A difference ideal $I\subset R$ is an ideal such that $\tau (I) \subset I$.

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 $R^\tau$ of a difference ring R are

$$\begin{aligned} R^\tau = \{ c \in R \ | \ \tau (c) = c\}. \end{aligned}$$

One can show that $R^\tau$ forms a ring and, if R is a field, then $R^\tau$ is also a field.

Example A.3

• 1. $(\mathbb {C}[x], \tau )$ where $\tau (x) = x+1$. The only difference ideals are $\mathbb {C}[x]$ and $\{0\}$. The constants of this ring are $\mathbb {C}$.

• 2. $(\mathbb {C}[x], \tau )$ where $\tau (x) = qx$, q not a root of unity. The difference ideals are $\mathbb {C}[x], \{0\}$ and the $(x^{k})$ for $k\in \mathbb {Z}_{\ge 1}$. The constants of this ring are $\mathbb {C}$.

• 3. $({\mathcal {M}}(\mathbb {C}), \tau )$ where ${\mathcal {M}}(\mathbb {C})$ is the field of meromorphic functions on $\mathbb {C}$ and $\tau (\omega ) = \omega +\omega _3$ for some $\omega _3 \in \mathbb {C}$. This is and the next example are difference fields. The constants of this field form the field of $\omega _3$-periodic meromorphic functions.

• 4. $(\mathbb {C}(\overline{E_t}), \tau )$ where $\mathbb {C}(\overline{E_t})$ is the field of meromorphic functions on $\overline{E_t}$ and for some fixed $P \in \overline{E_t} \ , \tau (f(X)) = f(X\oplus P)$ for all $f \in \mathbb {C}(\overline{E_t})$. If P is of infinite order, then the constants are $\mathbb {C}$ (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 $\tau (Y) = AY$ where $A\in \mathrm{GL}_n(K)$ where K is a difference field. Often one wants to deal with equations of the form form $L(y) = \tau ^n(y)+ a_{n-1}\tau ^{n-1}(y) + \cdots + a_0y = 0, \ a_i \in K$. If $a_0 = \ldots = a_{j-1} = 0, a_j \ne 0$, we can make a change of variables $z=\tau ^j(y)$ and assume $a_0 \ne 0$. One then sees that questions concerning solutions of $L(y) = 0$ can be reduced to questions concerning the system $\tau (Y) = A_LY$ where

$$\begin{aligned} A_L=\small {\begin{pmatrix} 0\quad &{}\quad 1\quad &{}\quad 0\quad &{}\quad \cdots \quad &{}\quad 0\\ 0\quad &{}\quad 0\quad &{}\quad 1\quad &{}\quad \ddots \quad &{}\quad \vdots \\ \vdots \quad &{}\quad \vdots \quad &{}\quad \ddots \quad &{}\quad \ddots \quad &{}\quad 0\\ 0\quad &{}\quad 0\quad &{}\quad \cdots \quad &{}\quad 0\quad &{}\quad 1\\ -a_0\quad &{}\quad -a_{1}\quad &{}\quad \cdots \quad &{}\quad \cdots \quad &{}\quad -a_{n-1} \end{pmatrix}} \in \mathrm{GL}_{n}(K). \end{aligned}$$

If z is a solution of $L(y) = 0$ in some difference ring containing K, then $(z, \tau (z), \ldots , \tau ^{n-1}(z))^T$ is a solution of $\tau (Y) = A_LY$.

In addition to considering individual solutions of $\tau (Y) = AY$, it is useful to consider matrix solutions and, in particular

Definition A.4

Let R be a difference ring and $A \in \mathrm{GL}_n(R)$. A fundamental solution matrix of $\tau (Y) = AY$ is a matrix $U \in \mathrm{GL}_n(R)$ such that $\tau (U) = AU$.

Note that if $U_1$ and $U_2$ are fundamental solution matrices of $\tau (Y) = AY$, then $\tau (U_1^{-1}U_2) = U_1^{-1}U_2$ so $U_2 = U_1 D$ where $D \in \mathrm{GL}_n(R^\tau )$.

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 $A \in \mathrm{GL}_n(K)$. We say that a k-algebra R is a Picard-Vessiot ring for $\tau (Y) = AY$ if

• (1) R is a simple difference ring extension of K (i.e., the only difference ideals of R are R and $\{0\}$).

• (2) $R = K[U, 1/\det (U)]$ for some fundamental solution matrix $U\in \mathrm{GL}_n(R)$ of $\tau (Y) = AY$.

It can be shown (cf. [24, Chapter 1.1]) that Picard-Vessiot rings always exist and if $K^\tau$ is algebraically closed they are unique up to k-difference isomorphisms. Furthermore, when $K^\tau$ is algebraically closed , we have $R^\tau = K^\tau$. Although some of the following results hold in more general situations, we will for simplicity assume from now on that

$K^\tau$ is algebraically closed and of characteristic zero.

Even under this assumption, the Picard-Vessiot ring need not be an integral domain. In [24, 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 $\tau (Y) = AY, \ A \in \mathrm{GL}_n(K)$. The Galois group G of R (or of $\tau (Y) = AY$) is

$$\begin{aligned} G = \{ \sigma :R\rightarrow R \ | \ \sigma \text { is a { K}-algebra automorphism of { R} and } \sigma \tau = \tau \sigma \} \end{aligned}$$

Using the notation of the definition, fix a fundamental solution matrix in $\mathrm{GL}_n(R)$ of the equation $\tau (Y) = AY$. If $\sigma \in G$, then

$$\begin{aligned} \tau (\sigma (U)) = \sigma (\tau (U)) = \sigma (AU) = A\sigma (U). \end{aligned}$$

Therefore, $\sigma (U)$ is again a fundamental solution matrix and so $\sigma (U) = U[\sigma ]_U$ where $[\sigma ]_U \in \mathrm{GL}_n(K^\tau )$. A key fact (cf. [24, Chapter 1.2]) forming the basis of the Galois theory of linear difference equations is

The map $\rho :G \rightarrow \mathrm{GL}_n(K^\tau )$ given by $\rho (\tau ) = [\tau ]_U$ is a group homomorphism whose image is a linear algebraic group.

A subgroup $G \subset \mathrm{GL}_n(K^\tau )$ is a linear algebraic group if it is a closed in the Zariski topology on $\mathrm{GL}_n(K^\tau )$, the topology whose closed sets are common solutions of systems of polynomial equations in $n^2$ variables.

Example A.7

1. Consider the equation

$$\begin{aligned} \tau (y) - y = b, \ b \in K. \end{aligned}$$

(A.1)

This equation is not a homogeneous linear difference equation but it is equivalent to the matrix equation

$$\begin{aligned} \tau (Y) = \begin{pmatrix} 1&{}\quad b\\ 0&{}\quad 1\end{pmatrix}Y. \end{aligned}$$

If z satisfies (A.1), then $U = \begin{pmatrix} 1 &{}\quad z\\ 0&{}\quad 1\end{pmatrix}$ is a fundamental solution of the matrix equation. The Picard-Vessiot extension of k is then given by $R = K[z]$. If $\sigma \in G$, then $y=\sigma (z)$ also satisfies (A.1). Therefore $\sigma (z) - z = d_\sigma \in K^\tau$. This implies that the Galois group of the matrix equation may be identified with a Zariski closed subgroup of

$$\begin{aligned} \left. \left\{ \begin{pmatrix} 1&{}\quad d\\ 0&{}\quad 1 \end{pmatrix} \ \right| \ d \in K^\tau \right\} . \end{aligned}$$

Note that this latter group is just the additive group $(K^\tau , +)$. The Zariski closed subgroups of this group are identified with $K^\tau$ and $\{0\}$.

2. Consider the system of equations

$$\begin{aligned} \tau (y_0) - y_0 = b_0, \ldots , \tau (y_n) - y_n = b_n\ b_0, \ldots , b_n \in K. \end{aligned}$$

(A.2)

As above, this system is equivalent to the matrix equation

$$\begin{aligned} \tau (Y) = \begin{pmatrix} B_0 &{}\quad 0 &{}\quad \ldots &{}\quad 0\\ 0&{}\quad B_1 &{}\quad \ldots &{}\quad 0\\ \vdots &{}\quad \vdots &{}\quad \vdots &{}\quad \vdots \\ 0 &{}\quad 0 &{}\quad \vdots &{}\quad B_n \end{pmatrix}Y, \ \text { where } B_i = \begin{pmatrix} 1 &{}\quad b_i \\ 0 &{}\quad 1 \end{pmatrix}\end{aligned}$$

The Picard-Vessiot extension of this equation is $R=K(z_0, \ldots , z_n)$ where $\tau (z_i) - z_i = b_i$ and the Galois group is a subgroup of

$$\begin{aligned} \left. \left\{ \begin{pmatrix} C_0 &{}\quad 0 &{} \quad \ldots &{}\quad 0\\ 0&{}\quad C_1 &{}\quad \ldots &{}\quad 0\\ \vdots &{}\quad \vdots &{}\quad \vdots &{}\quad \vdots \\ 0 &{}\quad 0 &{}\quad \vdots &{}\quad C_n \end{pmatrix} \ \right| \ \ C_i = \begin{pmatrix} 1 &{}\quad d_i \\ 0 &{}\quad 1 \end{pmatrix} d_i \in K^\tau , i = 0, \ldots , n\right\} .\end{aligned}$$

This latter group is just the direct sum of $n+1$ copies of the additive group $(K^\tau , +)$, that is $G \subset ((K^\tau )^{n+1}, +)$. The Zariski closed subgroups of $(K^\tau )^{n+1}$ 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 $(K^\tau )^{n+1}$ then there exist $c_0, \ldots , c_n \in K^\tau$, not all zero, such that

$$\begin{aligned} G \subset \left\{ (d_0, \ldots , d_n) \in (K^\tau )^{n+1} \ \left| \sum _{i=0}^n c_id_i = 0\right\} \right. . \end{aligned}$$

In [24, Chapter 1.3] a Galois correspondence and other basic facts are described. For our purposes, we only need

• (Gal 1) An element $z\in R$ is in K if and only if z is left fixed by all elements of G.

• (Gal 2) The ring R is an integral domain if and only if G is connected in the Zariski topology.

• (Gal 3) When G is connected, the dimension of G as an algebraic variety over $K^\tau$ 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 $z_0, \ldots , z_n \in R$ be solutions of this system. If $z_0, \ldots , z_n$ are algebraically dependent over K, then there exist $c_i \in K^\tau$, not all zero, and $g \in K$ such that

$$\begin{aligned} c_0 b_0 + \cdots + c_n b_n = \tau (g) - g. \end{aligned}$$

Proof

We follow ideas due to M. van der Put appearing in the appendix of [11]. As in Example A.7.2, the Galois group G is a subgroup of $(K^\tau )^{n+1}$ and so is connected. (Gal 2) implies that $R = K[z_0, \ldots , z_n]$ is a domain. Since $z_0, \ldots , z_n$ 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 $(K^\tau )^{n+1}$. Lemma A.8 implies that $G \subset \{(d_0, \ldots , d_n) \ | \sum _{i=0}^n c_id_i = 0\}$ for some $c_i \in K^\tau$. For any $\sigma \in G$, we have

$$\begin{aligned} \sigma \left( \sum _{i=0}^n c_iz_i\right) = \sum _{i=0}^n c_i(z_i + d_i) = \sum _{i=0}^n c_iz_i+ \sum _{i=0}^n c_id_i = \sum _{i=0}^n c_iz_i. \end{aligned}$$

From (Gal 1) we conclude that $\sum _{i=0}^n c_iz_i = g \in K$. Applying $\tau$ to this last equation and subtracting yields

$$\begin{aligned} \tau (g) - g = \tau \left( \sum _{i=0}^n c_iz_i\right) -\sum _{i=0}^n c_iz_i = \sum _{i=0}^n c_i(z_i+b_i) - \sum _{i=0}^n c_iz_i = \sum _{i=0}^n c_ib_i \end{aligned}$$

$\square$

We now turn to

Proof of Proposition 3.6

Let f satisfy $P(f, \delta (f),\ldots ,\delta ^n(f)) = 0$ for some polynomial P with coefficients in $K = \mathbb {C}(\overline{E_t})$. Since $\delta$ and $\tau$ commute, we have

$$\begin{aligned} \tau (f) - f = b, \tau (\delta (f)) - \delta (f) = \delta (b), \ldots , \tau (\delta ^n(f)) - \delta ^n(f) = \delta ^n(b). \end{aligned}$$

Let I be a maximal difference ideal in the difference ring $K[f, \delta (f), \ldots , \delta ^n(f)]$ and let $R = K[f, \delta (f), \ldots , \delta ^n(f)]/I$. The difference ring R is a simple difference ring of the form $K[z_0, \ldots , z_n]$ where $z_i$ is the image of $\delta ^i(f)$ and $\tau (z_i) - z_i = b_i, \ b_i = \delta (b)$. 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 $K[f, \delta (f), \ldots , \delta ^n(f)]^\tau = \mathbb {C}$. A computation shows that $\tau (L(f) -g) -(L(f) - g) =0$, where $L = a_n\delta ^n + a_{n-1}\delta ^{n-1} + \ldots + a_0$. Therefore $L(f) = g+ c, \text { for some } c \in \mathbb {C}$. If $g+c = 0$, this equation shows that f is holonomic over K. If $g+c \ne 0$, we can derive the same conclusion by considering $\delta (L(f)) - (\delta (g)/(g+c))L(f) = 0$. $\square$

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 [7], a proof of Proposition 3.6 was given based on the differential Galois theory of linear difference equations. This is a theory, presented in [14], 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 [15].

Appendix B: Telescopers and orbit residues

We have seen in Sect. 3.3 that the study of the hypertranscendance of $F^{1}(x,t)$ and $F^{2}(y,t)$ is intimately related to the study of equations of the form ${L(b)=\tau (g)-g}$ for some nonzero linear differential operator L with coefficients in $\mathbb {C}$ and some $b,g \in \mathbb {C}(\overline{E_t})$. In other contexts (cf. [6]), 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 $\tau :k(E)\rightarrow k(E)$ denote map corresponding to $Q\mapsto Q\oplus P$ on E, where $\oplus$ denotes the group law on E. We let $\Omega$ 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 $\delta$ of k(E) such that $d(f)=\delta (f) \Omega$ commutes with $\tau$.

We will prove the following

Proposition B.2

Let $b \in k(E)$. The following are equivalent.

• (1) There exist $g \in k(E)$ and a nonzero operator $L \in k[\delta ]$ such that ${L(b) = \tau (g) - g}$.

• (2) For all poles $Q_0$ of b, we have that

  $$\begin{aligned} h(X) =\sum _{i=1}^t b(X\oplus n_iP) \end{aligned}$$

  is regular at $X=Q_0$ where $Q_0\oplus n_1P, \ldots , Q_0\oplus n_tP$ are the poles of b that belong to $Q_0\oplus {\mathbb Z}P$.

This proposition allows one to give the following useful criteria guaranteeing when Condition 6.1 does or does not hold.

Corollary B.3

Let $b\in k(E)$ and assume that there exists $Q_0 \in E$ such that

• (1) b has a pole of order $m > 0$ at $Q_0$, and

• (2) b has no other pole of order $\ge m$ in $Q_0 \oplus {\mathbb Z}P$.

Then there is no nonzero $L \in k[\delta ]$ and $g\in k(E)$ such that $L(b) = \tau (g) - g$.

Proof

This follows easily from Proposition B.2 since the pole of b(X) at $Q_0$ cannot be cancelled by any pole of any $b(X\oplus nP)$ and so h(X) is not regular at $X=Q_0$. $\square$

Corollary B.4

Let $b \in k(E)$ and assume that there exists $Q_0 \in E$ such that

• (1) all poles of b occur in $Q_0 \oplus {\mathbb Z}P$, and

• (2) all poles of b are simple.

Then there exist $g \in k(E)$ and a nonzero operator $L \in k[\delta ]$ such that $L(b) = \tau (g) - g$.

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). $\square$

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 ${\mathcal {L}}(D)$ the finite dimensional k-space $\{ f \in k(E) \ | \ (f) + D \ge 0\}$, where (f) is the divisor of f. In Subsection 1 we will prove

Proposition B.5

Let $b \in k(E)$. The following are equivalent.

• (1) There exist $g \in k(E)$ and a nonzero operator $L \in k[\delta ]$ such that ${L(b) = \tau (g) - g}$.

• (2) There exists $Q\in E$, $e\in k(E)$ and $h \in {\mathcal {L}}(Q+(Q\ominus P))$⁴ such that

  $$\begin{aligned} b = \tau (e) - e+ h. \end{aligned}$$

To state the next equivalence, we need two definitions. Corresponding to each point $Q \in E$ there exists a valuation ring ${\mathfrak O}_Q \subset k(E)$. A generator $u_Q$ of the maximal ideal of ${\mathfrak O}_Q$ is called a local parameter at Q. Local parameters are unique up to multiplication by a unit of ${\mathfrak O}_Q$.

Definition B.6

Let $\mathcal {S} = \{ u_Q \ | \ Q\in E\}$ be a set of local parameters at the points of E. We say S is a coherent set of local parameters if for any

$$\begin{aligned} Q \in E, u_{Q\ominus P} =\tau (u_Q). \end{aligned}$$

We fix, once and for all, a coherent set of local parameters. All local parameters mentioned henceforth will be from this set.

Let $u_Q$ be a local parameter at a point $Q \in E$ and let $v_Q$ be the valuation corresponding to the valuation ring at Q. If $f \in k(E)$ has a pole at Q or order n, we may write

$$\begin{aligned} f = \frac{c_{Q,n}}{u_Q^n} + \cdots + \frac{c_{Q,2}}{u_Q^2} + \frac{c_{Q,1}}{u_Q} + \tilde{f} \end{aligned}$$

where $v_Q(\tilde{f}) \ge 0$. The following definition is similar to Definition 2.3 of [6].

Definition B.7

Let $f \in k(E)$ and $S =\{ u_Q \ | \ Q\in E\}$ be a coherent set of local parameters and $Q\in E$. For each $j \in {\mathbb N}_{>0}$ we define the orbit residue of order j at Q to be

$$\begin{aligned} \mathrm{ores}_{Q,j}(f) = \sum _{i \in {\mathbb Z}} c_{Q\oplus iP, j.} \end{aligned}$$

Note that if $Q' = Q \oplus tP$ for some $t \in {\mathbb Z}$, then $\mathrm{ores}_{Q',j}(f) = \mathrm{ores}_{Q,j}(f)$ for any $j \in {\mathbb N}_{>0}$. Furthermore $\mathrm{ores}_{Q,j}(f) = \mathrm{ores}_{Q,j}(\tau (f))$. We shall prove the next result in Subsection B.2.

Proposition B.8

Let $b \in k(E)$ and $S =\{ u_Q \ | \ Q\in E\}$ be a coherent set of local parameters. The following are equivalent.

• (1) There exists $Q\in E$, $e\in k(E)$ and $g \in {\mathcal {L}}(Q+(Q\ominus P))$ such that

  $$\begin{aligned} b = \tau (e) - e+ g. \end{aligned}$$

• (2) For any $Q \in E$ and $j \in {\mathbb N}_{>0}$

  $$\begin{aligned} \mathrm{ores}_{Q,j}(b) = 0. \end{aligned}$$

Proposition B.5, Proposition B.8 and the following lemma immediately imply Proposition B.2.

Lemma B.9

Let $b \in k(E)$. The following are equivalent

• (1) For all poles $Q_0$ of f, we have that

  $$\begin{aligned} h(X) =\sum _{i=1}^t b(X\oplus n_iP) \end{aligned}$$

  is regular at $X=Q_0$ where $Q_0\oplus n_1P, \ldots , Q_0\oplus n_tP$ are the poles of b that belong to $Q_0\oplus {\mathbb Z}P$.

• (2) For any $Q \in E$ and $j \in {\mathbb N}_{>0}$

  $$\begin{aligned} \mathrm{ores}_{Q,j}(b) = 0. \end{aligned}$$

Proof

If u is the local parameter at $Q_0$, we may write

$$\begin{aligned} h = \frac{c_n}{u^n} + \ldots + \frac{c_1}{u} + h' \end{aligned}$$

where $v_{Q_0}(h') \ge 0$. One easily sees that

$$\begin{aligned} c_j = \mathrm{ores}_{Q_0,j}(b). \end{aligned}$$

The conclusion now follows. $\square$

Remark B.10

Assume that $k=\mathbb {C}$. Then, one can consider the analytification $E^{an}$ of E. Instead of considering algebraic local parameters on E, one can consider analytic local parameters $\{u_{Q} \ \vert \ Q \in E\}$, i.e., for any $Q \in E$, $u_{Q}$ is a biholomorphism between a neighborhood of Q in $E^{an}$ and a neighborhood of 0 in $\mathbb {C}$. There is an obvious notion of coherent analytic local parameters, extending the notion introduced in Definition B.6, and a corresponding notion of $\mathrm{ores}_{Q,j}$. 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 ${\mathcal {L}}$ spaces (e.g. [13]). The proof that (2) implies (1) in Proposition B.5 is also constructive. Therefore given $b \in k(E)$ one can decide if there exist $g \in k(E)$ and a nonzero operator $L \in k[\delta ]$ such that $L(b) = \tau (g) - g$.

1.1 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 $\tau$. 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 $v_u$ be the associated valuation. Then $v_u(\delta (u)) = 0$ and, for any $f \in k(E)$ such that $v_u(f)\ne 0$, we have

• (1) if $v_u(f) \ge 0$ then $v_u( \delta (f)) \ge 0$;

• (2) if $v_u(f)<0$ then $v_u( \delta (u))= v_u(f)-1$.

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 ${\mathcal {L}}(D)$ then

$$\begin{aligned} l(D) = \text{ degree } \text{ of } D. \end{aligned}$$

This implies that if Q is a point on E, u is a local parameter at Q, $n\ge 2$, and $c_2, \ldots , c_n \in k$, then there exists an $f \in {\mathcal {L}}(nQ)$ and $c_1 \in k$ such that

$$\begin{aligned} f = \frac{c_n}{u^n} + \ldots + \frac{c_2}{u^2} + \frac{c_1}{u} + \tilde{f} \end{aligned}$$

where $v_u(\tilde{f}) \ge 0$. A priori, we have no control of the element $c_1$.

Finally we need some definitions:

Definition B.13

Let $f \in k(E)$ and $Q \in E$.

• (1) If Q is a pole of f, the polar dispersion of f at Q, $\mathrm{pdisp}(f,Q)$ is   the   largest   nonnegative   integer $\ell$ such that $Q\oplus \ell P$ is also a pole of f.

• (2) The polar dispersion of f, pdisp(f), is $\max \{\mathrm{pdisp}($f$,Q) \ | \ Q \text{ a } \text{ pole } \text{ of } f\}$.

• (3) The weak   polar   dispersion   of   f, wpdisp(f), is $\max \{\ell \ | \ \exists Q \in E s.t. f has \, a \, pole \, of \, order \, at \, least \, 2 \, at \, Q \, and Q\oplus \ell P \}$.

The following is an analogue of [14, Lemma 6.2].

Lemma B.14

Let $f \in k(E)$. There exist $f^*, g \in k(E)$ such that ${\mathrm{pdisp}(f^*) \le 1}$, wpdisp($f^*$) = 0 and $f = f^* + \tau (g) - g.$

Proof

We begin by showing that there exist $f^*, g \in k(E)$ such that wpdisp($f^*$) = 0 and $f = f^* + \tau (g) - g.$ We will then further refine $f^*$ so that $\mathrm{pdisp}(f^*) \le 1$ as well.

Let $N =$ wpdisp(f)$\ge 1$ and $n_f =$ the number of points $Q \in E$ such that f has poles of order at least two at Q and $Q\ominus NP$. Fix such a point Q and let u be a local parameter at Q. We may write

$$\begin{aligned} f = \sum _{i=1}^m \frac{a_i}{u^i} + h_f \end{aligned}$$

where $m \ge 2$ and $v_{Q}(h_f) \ge 0$. The Riemann-Roch Theorem implies that there exists a $\tilde{g} \in {\mathcal {L}}(mQ)$ such that

$$\begin{aligned} \tilde{g} = \sum _{i=1}^m \frac{b_i}{u^i} + h_{\tilde{g}} \end{aligned}$$

where $b_i = -a_i$ for $i = 2, \ldots , m$ and $v_{Q}(h_{\tilde{g}}) \ge 0$. Note that $\tau (\tilde{g})$ has a pole of order m at $Q\ominus P$. Letting $\tilde{f} = f - (\tau (\tilde{g}) - \tilde{g})$, one sees that $\tilde{f}$ has a pole of order at most 1 at Q. Therefore either the wpdisp(f) = wpdisp($\tilde{f}$) and $n_{\tilde{f}} < n_f$ or wpdisp(f) > wpdisp($\tilde{f}$). An induction allows us to conclude that there exist $f^*, g \in k(E)$ such that wpdisp($f^*$) = 0 and $f = f^* + \tau (g) - g.$

We may now assume that wpdisp(f) $= 0$ and let $\mathrm{pdisp}(f) = N \ge 2$. Let f have poles at both Q and $Q\oplus NP$. 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.

$\underline{f \,\hbox {has a pole of order 1 at} Q\oplus NP.}$ The Riemann-Roch Theorem implies that there exists a nonconstant $\tilde{g} \in {\mathcal {L}}((Q\oplus (N-1)P) + (Q \oplus NP))$. Note that $\tau (\tilde{g}) \in {\mathcal {L}}((Q\oplus (N-2)P) + (Q \oplus (N-1)P))$. For some $a \in k$, $\tilde{f} - (\tau (a\tilde{g}) - a\tilde{g})$ has no pole at $Q\oplus NP$ and so $\mathrm{pdisp}(f,Q) < N$. An induction finishes the proof.

$\underline{f \,\hbox {has a pole of order 1 at} Q.}$ The Riemann-Roch Theorem implies that there exists a nonconstant $\tilde{g} \in {\mathcal {L}}((Q\oplus P) + (Q \oplus 2P))$. Note that $\tau (\tilde{g}) \in {\mathcal {L}}((Q) + (Q \oplus P))$. For some $a \in k$, $\tilde{f} - (\tau (a\tilde{g}) - a\tilde{g})$ has no pole at Q and so $\mathrm{pdisp}(f,Q) < N$. 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 $\mathrm{pdisp}(b) \le 1$ and wpdisp(b) = 0 (here one uses the fact that $L\circ \tau = \tau \circ L$, since $L \in \mathbb {C}[\delta ]$ and $\tau \circ \delta = \delta \circ \tau$). We will first show that for any $Q \in E$, if b has a pole at Q then this pole must be simple and it has another pole of the same order at $Q\oplus P$ or at $Q\ominus P$. To see this note that if b has a pole at Q, then either g or $\tau (g)$ has a pole at Q. Assume that g has a pole at Q (the argument assuming $\tau (g)$ has a pole at Q is similar). Let r be the largest integer such that $Q\oplus rP$ is a pole of g and s be the largest integer such that $Q\ominus sP$ is a pole of g. We then have that $Q\oplus rP$ and $Q\ominus (s +1)P$ are both poles of $\tau (g)-g$ and therefore of L(b). Using Lemma 3.2 above, one sees that they must also be poles of b. Since $\mathrm{pdisp}(b) \le 1$, we have $r = s = 0$. In particular, the only pole of g in $Q\oplus {\mathbb Z}P$ is at Q, the only pole of $\tau (g)$ in $Q\oplus {\mathbb Z}P$ is at $Q\ominus P$ 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) = 0, 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 $\{Q_1, Q_1\ominus P\}, \ldots , \{Q_r, Q_r\ominus P\}$ where $(Q_i \oplus {\mathbb Z}P) \cap (Q_j \oplus {\mathbb Z}P) = \emptyset$ for $i \ne j$.

We will now show how one can construct an element e such that $b -(\tau (e) - e)$ has at most one pair of poles $\{Q, Q\ominus P\}$. This will yield (2) and our contention. Assume $r > 1$ and that b has simple poles at the pairs $\{Q_1, Q_1\ominus P\}$ and $\{Q_2, Q_2\ominus P\}$. Let $h\in k(E)$ be a nonconstant element of ${\mathcal {L}}(Q_1+Q_2)$. There exists an $a \in k$ such that $\tilde{b} = b - (\tau (ah)-ah)$ has no pole at $Q_1$. The element $\tilde{b}$ has only simple poles and $\text{ pdisp }(b) \le 1$. Therefore its poles occur at possibly $Q_1\ominus P, \{Q_2, Q_2\ominus P\}, \ldots , \{Q_r, Q_r\ominus P\}$. Since $\tilde{b}$ satisfies an equation of the form $L(\tilde{b}) = \tau (\tilde{g}) - \tilde{g}$ for some $\tilde{g} \in k(E)$ the poles of such an $\tilde{b}$ must occur in pairs. Therefore we have that $\tilde{b}$ has no pole at $Q_1\ominus P$. Continuing in this way we find an $e \in k(E)$ such that $b -(\tau (e) - e)$ has at most one pair of poles $\{Q, Q\ominus P\}$.

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 $\tau (u)$ is a local parameter at $Q\ominus P$.

Lemma B.15

If $g \in {\mathcal {L}}(Q + (Q\ominus m_1 P) + \cdots + (Q\ominus m_t P) )$ where $m_1, \ldots , m_t \in {\mathbb Z}\backslash \{0\}$ then

$$\begin{aligned} \mathrm{ores}_{Q,1}(g) = 0. \end{aligned}$$

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 $v_{u_Q}(\delta (u_Q)) = 0$ so we may write $\delta (u_Q)^{-1} = \alpha + \bar{u}$ where $0 \ne \alpha \in k$ and $\bar{u}$ is regular and zero at Q. For each $i \in {\mathbb Z}$ we write

$$\begin{aligned} g = \frac{c_{Q\ominus iP, -1}}{u_{Q\ominus iP}} + g_{Q\ominus iP} \end{aligned}$$

where $g_{Q\ominus iP}$ is regular and zero at $Q\ominus iP$. Now consider the differential $g\Omega$. Since for any $i \in {\mathbb Z}$, $\Omega =\delta (u_{Q\ominus iP})^{-1}du_{Q\ominus iP}$, we have

$$\begin{aligned} g\Omega= & {} \left( \frac{c_{Q\ominus iP, -1}}{u_{Q\ominus iP}} + g_{Q\ominus iP}\right) (\delta (u_{Q\ominus iP})^{-1}du_{Q\ominus iP})\\= & {} \left( \frac{c_{Q\ominus iP, -1}}{u_{Q\ominus iP}} + g_{Q\ominus iP}\right) (\tau ^i(\delta (u_Q)^{-1})du_{Q\ominus iP})\\= & {} \left( \frac{c_{Q\ominus iP, -1}}{u_{Q\ominus iP}} + g_{Q\ominus iP}\right) ((\alpha + \tau ^i(\bar{u}))du_{Q\ominus iP}) \end{aligned}$$

where the second equality follows from the fact that $u_{Q\ominus iP} = \tau ^i(u_Q)$ and $\tau \delta =\delta \tau$. Therefore the residue of $g\Omega$ at $Q\ominus iP$ is $\alpha c_{Q\ominus iP, -1}$. Since $\alpha \ne 0$ and the sum of the residues of a differential form is 0 we have $\mathrm{ores}_{Q,1}(g) = 0$.$\square$

Remark B.16

The proof of Lemma B.15 shows that if the poles of $g \in k(E)$ are simple and belong to $Q \oplus \mathbb {Z}P$, then there exists $0 \ne \alpha \in \mathbb {C}$ such that $\mathrm{ores}_{Q,1} (g)=\alpha \sum _{i \in \mathbb {Z}}{\text {Res}}_{Q \oplus iP} (g \Omega )$. Therefore, $\mathrm{ores}_{Q,1} (g)=0$ if and only if $\sum _{i \in \mathbb {Z}}{\text {Res}}_{Q \oplus iP} (g \Omega )=0$.

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 $f \in {\mathcal {L}}(Q + (Q\ominus m_1 P) + \cdots + (Q\ominus m_t P) )$ where $m_i = -n_i$. The residue of $h(X) = \sum _{i=1}^tf(X\oplus n_iP)$ at $X=P$ is $\mathrm{ores}_{Q,1}(f)$, so h(X) is regular at $Q_0$. 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 $\tilde{g}$ and a nonzero operator L such that $L(h) = \tau (\tilde{g}) -\tilde{g}$. Assume that we have done this. Then

$$\begin{aligned} L(b) = L(\tau (e) - e +h)= \tau (L(e)) - L(e) +\tau (\tilde{g}) - \tilde{g} = \tau (g) - g \end{aligned}$$

where $g = L(e) + \tilde{g}$.

If h is constant, then the result is obvious (take $L=\delta$ and $\tilde{g} =0$). We shall now assume that h is not constant.

To simplify notation, we write u for $u_Q$ and let $\delta (u) = u_0 + \bar{u}$, where $0 \ne u_{0} \in k$ and $\bar{u}$ is regular and zero at Q, and so $\delta (\tau (u)) = u_0 + \tau (\bar{u})$. Using Lemma B.15, one sees that

$$\begin{aligned} \delta (h)= & {} \frac{-u_0a}{u^2}(1+ h_u)\\= & {} \frac{u_0a}{\tau (u)^2}(1+ h_{\tau (u)}) \end{aligned}$$

where $v_u(h_u) > 0$ and $v_{\tau (u)}(h_{\tau (u)}) > 0$. Selecting an element $f \in {\mathcal {L}}(2Q)$ such that $f = u_0a/u^2 + \cdots ,$ we have

$$\begin{aligned} \delta (h) - (\tau (f) - f) \in {\mathcal {L}}(Q+(Q\ominus P)). \end{aligned}$$

Since $\{1,h\}$ forms a basis of ${\mathcal {L}}(Q+(Q\ominus P))$ (recall that h is not constant), there exist elements $c, d \in k$ such that

$$\begin{aligned} \delta (h) - (\tau (f) - f) -ch -d= 0. \end{aligned}$$

Therefore

$$\begin{aligned} \delta ^2(h) - c\delta (h) = \tau (\delta (f)) - \delta (f) \end{aligned}$$

and conclusion (2) holds for $L = \delta ^2 - c\delta$ and $\tilde{g} = \delta (f)$.$\square$

Remark B.18

One cannot weaken condition (2) in Proposition B.5, that is, for a general $b \in k(E)$, condition (1) of Proposition B.5 does not imply the following condition :

• (3) There exist $Q\in E$, $e\in k(E)$ and a constant $c \in k$ such that

  $$\begin{aligned} b = \tau (e) - e+ c. \end{aligned}$$

To see this, let b be a nonconstant element of ${\mathcal {L}}(Q+(Q\ominus P))$. Note that $\mathrm{pdisp}(b) =1$. We have just shown that b satisfies (1) of Proposition B.8. Now assume $b = \tau (e)-e +c$ for some $e\in k(E), c\in k$. Since $\mathrm{pdisp}(\tau (e) - e) = \mathrm{pdisp}(e) + 1$ if $e \notin k$, we have $\mathrm{pdisp}(e) =0$. Since b has no poles outside of $\{Q, Q\ominus P\}$, 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 $b \notin k$.

1.2 B.2 Proof of Proposition B.8

Proof that (1) implies (2) in Proposition B.8

For any $Q\in E$ and $j \in {\mathbb N}_{>0}$, we have $\mathrm{ores}_{Q,j}(e) = \mathrm{ores}_{Q,j}(\tau (e))$. Furthermore Lemma B.15 implies that $\mathrm{ores}_{Q,j}(g) = 0$. Therefore $\mathrm{ores}_{Q,j}(f) =\mathrm{ores}_{Q,j}(\tau (e)-e+g) =\mathrm{ores}_{Q,j}(\tau (e))-\mathrm{ores}_{Q,j}(e) +\mathrm{ores}_{Q,j}(g) = 0$.

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 $\mathrm{pdisp}(f)\le 1$ and $\mathrm {wdisp}(f) = 0$. Therefore if f has a pole of order $j \ge 2$ at some $Q \in E$, then Q is the only point in $Q + {\mathbb Z}P$ at which f has a pole. Since $\mathrm{ores}_{Q,j}(f) = 0$, we have that f has no poles of order greater than 1. Since we also have $\mathrm{pdisp}(f)\le 1$, we can conclude that that f has only poles of order 1 and the poles of f occur in pairs $\{Q_1, Q_1\ominus P\}, \ldots , \{Q_r, Q_r\ominus P\}$ where $(Q_i \oplus {\mathbb Z}P) \cap (Q_j \oplus {\mathbb Z}P) = \emptyset$ for $i \ne j$.

We will now show how one can construct an element e such that $f -(\tau (e) - e)$ has at most one pair of poles $\{Q, Q\ominus P\}$. This will yield condition 2. of the Proposition. We can assume that $r >1$. Let $h\in k(E)$ be a nonconstant element of ${\mathcal {L}}(Q_1+Q_2)$. There exists an $a \in k$ such that $\tilde{f} = f - (\tau (ag)-ag)$ has no pole at $Q_1$. The element $\tilde{f}$ has only simple poles and $\text{ pdisp }(f) \le 1$. Therefore its poles occur at possibly $Q_1\ominus P, \{Q_2, Q_2\ominus P\}, \ldots , \{Q_r, Q_r\ominus P\}$. Since $\mathrm{ores}_{Q_1,1}(f) = 0$, f cannot have a singe pole in $Q_1+{\mathbb Z}P$. Therefore we have that $\tilde{f}$ has no pole at $Q_1\ominus P$. Continuing in this way we find an $e \in k(E)$ such that $f -(\tau (e) - e)$ has at most one pair of poles $\{Q, Q\ominus P\}$.$\square$

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 $\tau :k(E)\rightarrow k(E)$ denote map corresponding to $Q\mapsto Q\oplus P$ on E, where $\oplus$ denotes the group law on E. Let $\iota _1$ and $\iota _2$ two involutions of E such that $\tau =\iota _2 \circ \iota _1$. We let $\Omega$ be a non zero regular differential form on E and we keep notation as in §B.

Lemma C.1

Let $b \in k(E)$ such that $\iota _1(b)=-b$. Let $Q \in E$ be a simple pole of b such that $Q \ne \iota _1(Q)$. Then, $\iota _{1}(Q)$ is a simple pole of b and the residue of $b \Omega$ at Q coincides with its residue at $\iota _1(Q)$.

Proof

The assertion follows from the fact that, since $\iota _1(b)=-b$ and ${\iota _1^*(\Omega )=-\Omega }$ (see [8, Lemma 2.5.1 and Proposition 2.5.2]), the form $\eta =b\Omega$ satisfies ${\eta = \iota _1^*(\eta )}$. Indeed, if $u_Q$ is a local parameter at Q, then we have $\eta = v du_Q$ with $v=\frac{c_Q}{u_Q}+\overline{v}$ where $c_Q \in \mathbb {C}$ is the residue of $\eta$ at Q and $\overline{v}$ is regular at Q. Hence, $\iota _1(u_Q)$ is a local parameter at $\iota _1(Q)$, and we have $\iota _1^*(\eta ) = \iota _1(v) d\iota _1(u_Q)$ with $\iota _1(v)=\frac{c_Q}{\iota _1(u_Q)}+\iota _1(\overline{v})$ where $\iota _1(\overline{v})$ is regular at $\iota _1(Q)$. So, ${\text {Res}}_{\iota _1(Q)} (\eta ) = {\text {Res}}_{\iota _1(Q)} (\iota _1^*(\eta )) = c_Q = {\text {Res}}_{Q} (\eta )$. $\square$

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 $\{u_{Q} \ \vert \ Q \in E \}$ on $E^{\mathrm {an}}$ such that $\iota _{1}(u_{Q})=-u_{\iota _{1}(Q)}$. Let $b \in k(E)$ such that $\iota _1(b)=-b$. For such a set of local parameters, if

$$\begin{aligned} b = \frac{c_{Q,n}}{u_Q^n} + \cdots + \frac{c_{Q,2}}{u_Q^2} + \frac{c_{Q,1}}{u_Q} + \tilde{f} \end{aligned}$$

where $v_Q(\tilde{f}) \ge 0$, then

$$\begin{aligned} b =\frac{c_{\iota _{1}(Q),n}}{u_{\iota _{1}(Q)}^n} + \cdots + \frac{c_{\iota _{1}(Q),2}}{u_{\iota _{1}(Q)}^2} + \frac{c_{\iota _{1}(Q),1}}{u_{\iota _{1}(Q)}} + \tilde{g} \end{aligned}$$

where $v_{\iota _{1}(Q)}(\tilde{g}) \ge 0$ and $c_{\iota _{1}(Q),j}=(-1)^{j+1}c_{Q,j}$. If follows that, if all the poles of b belong to the same $\tau$-orbit, then, for any even number j, we have $\mathrm{ores}_{Q,j}(b)=0$.

Proof

We first prove the existence of analytic local parameters with the desired properties. According to [8, p.35 and Remark 2.3.8], $\iota _{1}(P)=[-1]P\oplus P_{0}$ for some $P_{0} \in E$. By uniformization, it is equivalent to prove the following result : Consider a lattice $\Lambda \subset \mathbb {C}$ and two endomorphisms of the complex torus $\mathbb {C}/\Lambda$ given by $\iota _{1} : \overline{z} \mapsto \overline{-z+p_{0}}$ and $\tau : \overline{z} \mapsto \overline{z+q_{0}}$ for some $p_{0},q_{0} \in \mathbb {C}$. Then, there exists a set of analytic local parameters $\{u_{\overline{\omega }} \ \vert \ \overline{\omega } \in \mathbb {C}/\Lambda \}$ on the complex torus $\mathbb {C}/\Lambda$ such that $\iota _{1}(u_{\overline{\omega }})=-u_{\iota _{1}(\overline{\omega })}$ and $\tau (u_{\overline{\omega }})=u_{\tau (\overline{\omega })}$. Such local parameters are given by $u_{\overline{\omega }} : \overline{z} \mapsto z-\omega$ for z close to $\omega$. The rest of the Lemma is a direct consequence of the following easy computation. Indeed, applying $\iota _{1}$ to

$$\begin{aligned} b= \frac{c_{Q,n}}{u_Q^n} + \cdots + \frac{c_{Q,2}}{u_Q^2} + \frac{c_{Q,1}}{u_Q} + \tilde{f}, \end{aligned}$$

we get

$$\begin{aligned}&-b=\iota _{1}(b) = \frac{c_{Q,n}}{\iota _{1}(u_Q)^n} + \cdots + \frac{c_{Q,2}}{\iota _{1}(u_Q)^2} + \frac{c_{Q,1}}{\iota _{1}(u_Q)} + \iota _{1}(\tilde{f})\\&\quad = \frac{(-1)^{n}c_{Q,n}}{u_{\iota _{1}(Q)}^n} + \cdots + \frac{(-1)^{2}c_{Q,2}}{u_{\iota _{1}(Q)}^2} + \frac{(-1)^{1} c_{Q,1}}{u_{\iota _{1}(Q)}} + \iota _{1}(\tilde{f}) \end{aligned}$$

where $v_{\iota _{1}(Q)}(\iota _{1}(\tilde{f})) \ge 0$, as expected. $\square$

Lemma C.3

If $g \in {\mathcal {L}}(2Q + 2(Q\ominus m_1 P) + \cdots + 2(Q\ominus m_s P) )$ where $m_1, \ldots , m_s \in {\mathbb Z}\backslash \{0\}$ is such that $\mathrm{ores}_{Q,2}(g)=0$ then

$$\begin{aligned} \mathrm{ores}_{Q,1}(g) = 0. \end{aligned}$$

Proof

We may write $\delta (u_Q)^{-1} = \alpha + \beta u_{Q} + \bar{u}$ where $0 \ne \alpha \in k$, $\beta \in k$ and $\bar{u}$ is regular and has a zero of order 2 at Q. For each $i \in \{0, m_1, \dots ,m_s \}$ we write

$$\begin{aligned} g = \frac{c_{Q\ominus iP, 2}}{u_{Q\ominus iP}^{2}} + \frac{c_{Q\ominus iP, 1}}{u_{Q\ominus iP}} + g_{Q\ominus iP} \end{aligned}$$

where $g_{Q\ominus iP}$ is regular and zero at $Q\ominus iP$. Now consider the differential $g\Omega$. Since for any $i \in {\mathbb Z}$, $\omega =\delta (u_{Q\ominus iP})^{-1}du_{Q\ominus iP}$, we have

$$\begin{aligned} g\Omega= & {} \left( \frac{c_{Q\ominus iP, 2}}{u_{Q\ominus iP}^{2}} +\frac{c_{Q\ominus iP, 1}}{u_{Q\ominus iP}} + g_{Q\ominus iP}\right) (\delta (u_{Q\ominus iP})^{-1}du_{Q\ominus iP})\\= & {} \left( \frac{c_{Q\ominus iP, 2}}{u_{Q\ominus iP}^{2}} +\frac{c_{Q\ominus iP, 1}}{u_{Q\ominus iP}} + g_{Q\ominus iP}\right) (\tau ^i(\delta (u_Q)^{-1})du_{Q\ominus iP})\\= & {} \left( \frac{c_{Q\ominus iP, 2}}{u_{Q\ominus iP}^{2}} +\frac{c_{Q\ominus iP, 1}}{u_{Q\ominus iP}} + g_{Q\ominus iP}\right) ((\alpha + \beta u_{Q \ominus iP} + \tau ^i(\bar{u}))du_{Q\ominus iP}) \end{aligned}$$

where the second equality follows from the fact that $u_{Q\ominus iP} = \tau ^i(u_Q)$ and $\tau \delta =\delta \tau$. Note that $\tau ^{i}(\bar{u})$ is regular and has a zero of order 2 at $Q\ominus iP$. Therefore the residue of $g\omega$ at $Q\ominus iP$ is $\alpha c_{Q\ominus iP, 1} + \beta c_{Q\ominus iP, 2}$. Since the sum of the residues of a differential form is 0 we get $\alpha \mathrm{ores}_{Q,1}(g) + \beta \mathrm{ores}_{Q,2}(g)= 0$. Since $\alpha \ne 0$ and $\mathrm{ores}_{Q,2}(g)=0$, we get $\mathrm{ores}_{Q,1}(g)=0$. $\square$