5.1. The elliptic hypergeometric functions. — We shall now introduce the elliptic hypergeometric functions following [Spi16]. Consider p, q ∈ C∗ such that |p| < 1,
|q|<1, andqZ∩pZ={1}. Define
Definition5.1. — The elliptic hypergeometric functionfε(z) is defined by the fol-lowing formula
fε(z) := V(q/cε1, . . . , q/cε5, cz, c/z, cε8;p, q)
Γ(c2z/ε8;p, q) Γ(z/ε8;p, q) Γ(c2/zε8;p, q) Γ(1/zε8;p, q).
Remark5.2. — As explained in [Spi16], the functionV(t;p, q)defined in (5.1) can be extended by analytic continuation, so thatQ
16j<k68(tjtk;p, q)∞V(t;p, q)is holomor-phic fort1, . . . , t8∈C∗. We should also mention for completeness that, as explained in [Spi16], in Definition 5.1 it is initially necessary to impose the constraints (ex-pressed in terms of the old parametrization) p
|pq| < |tj| < 1 for j = 1, . . . ,5 and p|pq|<|q±1tj|<1forj = 6,7,8, which can then be relaxed by analytic continuation.
These important but subtle considerations will not play a role in what follows.
5.2. The elliptic hypergeometric equation. — The elliptic hypergeometric function fε(z)satisfies the following equation
(5.4) A(z)(y(qz)−y(z)) +A(z−1)(y(q−1z)−y(z)) +νy(z) = 0,
It is easily seen that A(pz) = A(z), so that the previous equation has coefficients in Mp,1. Replacingzbyqz in (5.4), we obtain the following equation:
(5.6) σ2(y) +aσ(y) +by= 0, with
a=ν−A(qz)−A(q−1z−1)
A(qz) ,
b= A(q−1z−1)
A(qz) ∈Mp,1.
Remark5.3. — Note that the new parametersε1, . . . , ε8used in the definition offε(z) are not defined to be free independent parameters, since they are defined in terms of the old parameterst1, . . . , t8(which are free parameters save for the balancing condi-tionQ8
j=1tj =p2q2), and in fact one of the equations in the reparametrization (5.2) is equivalent toε8=ε7q.
On the other hand, the elliptic hypergeometric equation (5.4) is defined for arbi-trary parametersε1, . . . , ε8∈C∗, subject only to the balancing condition (5.3), which is equivalent to imposing that the coefficientsA(z)andA(z−1)actually belong to the field of elliptic functionsMp,1.
For this reason, we prove two related but distinct results on differential transcen-dence:
(A) differential transcendence of solutions of the elliptic hypergeometric equation (5.4), where we think of the εj as free parameters subject only to the balancing condition (5.3) and without imposing the additional constraintε8=ε7q; and
(B) differential transcendence of the elliptic hypergeometric functionsfε(z)where theεj are defined in terms of thetj as in (5.2), and where in particular we do impose the additional constraintε8=ε7q.
Note that in case (B) above the balancing condition (5.3) for the remaining inde-pendent parametersε1, . . . , ε7becomes
(5.7)
6 Y
j=1
εj
ε27=p2q.
In the next lemma we show that in case (B) there are no universal relations among the parameters ε1, . . . , ε7 induced from the reparametrization (5.2), save for formal algebraic consequences of the balancing condition (5.7). This result ensures that the hypothesis in case (B) of Theorem 5.6 and Theorem 5.7 below is not vacuous.
Lemma5.4. — Assume that case (B) holds. Every multiplicative relation among the ε1, . . . , ε7, p, qis induced by(5.7), in the sense that if there are integersα1, . . . , α7, m, n such that
7
Y
j=1
εαjj =pmqn,
thenα1=· · ·=α6=α=nandm=α7= 2αfor someα∈Z.
Proof. — Let us begin to writec and theεj in terms of the tj. We have c =√ t6t7, and
ε6= c2p4 ε8
=cp4t8=p4√
t6t7t8, ε7= ε8 q = c
qt8
=
√t6t7 qt8
. Assume now that there are integersα1, . . . , α8, m, n such that
7
Y
j=1
εαjj =pmqn.
Let us write this equality in terms of thetj. The relationQ7
j=1εαjj =pmqn gives (5.8)
5 Y
j=1
qαj (t6t7)αj/2tαjj
p4α6(t6t7)(α6+α7)/2tα86−α7q−α7=pmqn.
Using the balancing conditionQ8
j=1tj=p2q2, we obtain the existence of an integerα such that
α1=· · ·=α5=α.
Furthermore, regarding the terms inq, p,tj,j= 6,7, andt8 respectively, we find n−5α+α7=−2α, m−4α6=−2α,
−5α/2 +α6/2 +α7/2 =−α, α6−α7=−α.
If we put the equality of the fourth relation α6 =α7−αinto the third, we obtain 2α=α7. With α6 =α7−α, we find α=α6. Finally from the first and the second
equality, we deducen=αandm= 2α.
Remark5.5. — Assume that case (B) holds. With Lemma 5.4 and the relationε8= qε7 it follows that if there are integersα1, . . . , α8, m, n such that
8
Y
j=1
εαjj =pmqn
thenα1=· · ·=α6=α=n−α8 andα7+α8= 2α=mfor someα∈Z.
5.3. Irreducibility of theσ-Galois group of the elliptic hypergeometric function From now on, we denote byGtheσ-Galois group of (5.6) overK(with respect to some σ-PV ring).
Theorem5.6. — Assume one of the two hypotheses (A)or (B)below.
(A) Every multiplicative relation among the ε1, . . . , ε8, p, q is induced by (5.3), in the sense that if there are integers α1, . . . , α8, m, nsuch that
8
Y
j=1
εαjj =pmqn
thenα1=· · ·=α8=:αandm=n= 2αfor someα∈Z.
(B) ε8=ε7qand every multiplicative relation among theε1, . . . , ε7, p, q is induced
Proof. — To the contrary, assume thatGis reducible. According to Theorem 4.6, the following Riccati equation has a solutionu∈Mp,2 :
(5.9) uσ(u) +au+b= 0.
First, note thatu∈Mp,2 is a solution of (5.9) if and only if
v(σ(v) +σ−1(a)) +σ−1(b) = 0 withv=σ−1(u)∈K.
Then to simplify the expression of the divisors ofaandb, we may replace them by σ−1(a) =ν−A(z)−A(z−1)
We now considerr0, r1, r2∈Θ2as in Theorem 4.6. Fori= 1,2, let where the second equality is obtained from property (iv) of Theorem 4.6. After taking fourth powers we see that
Suppose first that we are in case (A). Since every multiplicative relation among the ε1, . . . , ε8, p, q is induced by (5.3), it follows from (5.10) that there existsα∈Nsuch that2αj+ 2α0j=αfor everyj∈ {1, . . . ,8}andm= 2 deg2(r2) +γ+ 2 deg2(r0) = 2α.
In particular, we have that 2 deg2(r2) 6 2α. On the other hand, it follows from properties (i) and (ii) of Theorem 4.6, respectively, thatα1+· · ·+α86deg2(r1)and α01+· · ·+α086deg2(r2). We note that, by property (iii) of Theorem 4.6,2 deg2(r2) = deg2(r1) + deg2(r2). Putting together these inequalities we obtain
4α=
Now suppose we are in case (B). Since every multiplicative relation among the ε1, . . . , ε8, p, q is induced by (5.7), it follows from (5.10) that there existsα∈Nsuch
It follows from this thatα= deg2(r1) = deg2(r2) = 0. Hence,r1/r2 is constant and ω2(r1/r2) = 1 =√
qdeg2(r0) modpZ
by property (iv) of Theorem 4.6. SincepZ∩qZ={1}, we see thatdeg2(r0) = 0also.
It follows from the above in either of the cases (A) or (B) thatv∈C∗ is constant.
Therefore (5.9) can be rewritten as
(5.11) v2A(z) +v(ν−A(z)−A(z−1)) +A(z−1) = 0, i.e.,
(5.12) (v2−v)A(z) +vν= (v−1)A(z−1).
But since√
q−1 is a pole ofA(z)but not ofA(z−1)and, on the other hand,√ qis a pole ofA(z−1)but not ofA(z), we obtain thatv2−v=v−1 =vν = 0. So we must have ν = 0. On the other hand, we see from the definition of ν in (5.5) that ν = 0 if and only ifεjε8=qp` for some` ∈Zand j = 1, . . . ,6, which is ruled out by our hypotheses in both cases (A) and (B). This contradiction concludes the proof thatG
is irreducible.
5.4. Differential transcendence of the elliptic hypergeometric functions
We may equip(K, σ)with the classical derivationδ:=zd/dzas in [DHRS18, §3.1].
Note that δ commutes with σ. Let Ce be the δ-closure of C. Following Lemma 3.1, we may consider L := Frac(K ⊗CCe) and we have Lσ = Ce. Recall that fε(z) is meromorphic on C∗, and note that the field of meromorphic functions on C∗ is a (σ, δ)-extension ofK.
Theorem5.7. — Assume one of the two hypotheses (A)or (B)below.
(A) Every multiplicative relation among the ε1, . . . , ε8, p, q is induced by (5.3), in the sense that if there are integers α1, . . . , α8, m, nsuch that
8
Y
j=1
εαjj =pmqn
thenα1=· · ·=α8=:αandm=n= 2αfor someα∈Z.
(B) ε8=ε7qand every multiplicative relation among theε1, . . . , ε7, p, q is induced by (5.7), in the sense that if there are integersα1, . . . , α8, m, n such that
8
Y
j=1
εαjj =pmqn
thenα1=· · ·=α6=α=n−α8 andα7+α8= 2α=m for someα∈Z. Then any non-zero solution to (5.4)is differentially transcendental overK.
Proof. — We apply the criteria of Theorem 3.5. We proved in Theorem 5.6 thatGis irreducible, which by [DR15, Lem. 13] is equivalent to the non-existence of a solution u∈Kto the Riccati equation
uσ(u) +au+b= 0.
It remains to show that there is no nonzero linear differential operator L in δ with coefficients inCandg∈Ksuch that
L(δb/b) =σ(g)−g.
Letk∈N∗such thatg∈Mp,kand considerbas an element ofMp,k. Letω∈C∗k/pZ be a zero or a pole ofb. Then it is a pole ofδb/b. Since L has constant coefficients, we get that ω is also a pole of L(δb/b). Therefore, ω is a pole of σ(g)−g and hence also a pole ofσ(g)or ofg. Furthermore,σ(g)−g has at least two distinct poles ω0, ω00∈C∗k/pZsuch thatω≡ω0≡ω00 modqZk, whereqk ∈C∗kis as in (4.2). Theseω0 andω00 are poles ofδb/b, and hence zeros or poles ofbhas well. We have proved that, for every ω∈C∗k/pZ that is a pole or zero ofb, there exists`∈Z6=0 such thatωqk` is a pole or zero ofb.
Let us now considerbas an element ofMp,1. From the preceding, we deduce that for every ω ∈C∗/pZ, pole or zero of b, there exists `∈ Z6=0 such that ωq` is a pole or zero of b. We will use this to find a contradiction. Note that the set of zeros or poles of
b= θ(q2z2;p)θ(q3z2;p) θ(q−2z−2;p)θ(q−1z−2;p)×
8
Y
j=1
θ(εjq−1z−1;p) θ(εjqz;p) , seen as an element ofMp,1, is included in
S ={q−1ε±11 , . . . , q−1ε±18 ,±q−1/2,±q−1/2√
p,±q−3/2,±q−3/2√
p} modpZ. Let us prove that the elements ofS are all distinct. To see this, note that if any two elements ofS were the same modulopZthen we would find a non-trivial multiplicative relation satisfied by at most four elements amongp, q, ε1, . . . , ε8. This contradicts the hypothesis in both cases (A) and (B). Therefore, no simplifications occur and S is exactly the set of zeros or poles of b. It suffices to show that for all ` ∈ Z6=0, we haveS ∩ {q`q−1ε1 modpZ}=∅. Let`∈Zsuch thatS ∩ {q`q−1ε1 modpZ} 6=∅. If`6= 0, then we again find a non-trivial multiplicative relation satisfied by at most four elements amongp, q, ε1, . . . , ε8. In either case (A) or case (B) this contradiction
to the hypothesis concludes the proof.
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Manuscript received 20th September 2019 accepted 5th January 2021
Carlos E. Arreche, The University of Texas at Dallas, Mathematical Sciences FO 35 800 West Campbell Road, Richardson, TX 75024, USA
E-mail :arreche@utdallas.edu
Url :https://personal.utdallas.edu/~arreche/
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 E-mail :dreyfus@math.unistra.fr
Url :https://sites.google.com/site/thomasdreyfusmaths/
Julien Roques, Univ Lyon, Université Claude Bernard Lyon 1, CNRS UMR 5208, Institut Camille Jordan
43 blvd. du 11 novembre 1918, F-69622 Villeurbanne cedex, France E-mail :roques@math.univ-lyon1.fr
Url :http://math.univ-lyon1.fr/~roques/