HOLOMORPHIC SOLUTIONS OF E-OPERATORS
T. RIVOAL AND J. ROQUESAbstract. We solve the problem of describing the solutions of E-operators of order µ ≥ 1admitting at z = 0 a basis over C of local solutions which are all holomorphic at z = 0.
P`
We prove that the components of such a basis can be taken of the form j=1 Pj (z)eβj z ,×where ` ≤ µ, β1 , . . . , β` ∈ Q , and P1 (z), . . . , P` (z) ∈ Q[z].

1. Introduction
We fix an enbedding of Q into C. An E-function is a power series∞Xan nz ∈ Q[[z]]f (z) =n!
n=0such that:(1) f (z) satisfies a non-zero linear differential equation with coefficients in Q(z);(2) there exists C > 0 such that(a) for any σ ∈ Gal(Q/Q), we have |σ(an )| ≤ C n+1 ;(b) there exists a sequence of positive integers dn such that dn ≤ C n+1 and dn am
is an algebraic integer for all m ≤ n.
P
P∞ an nn
A G-function at z = 0 is a power series f (z) = ∞n=0 an z ∈ Q[[z]] such thatn=0 n! z
1is an E-function. Both classes of functions have been first introduced by Siegel ( ). Much
work has been devoted to the study of the Diophantine nature of the values of thesefunctions, and the properties of the differential equations they satisfy proved to be crucial;
we refer to [1, 4, 6, 10] for surveys of these results.
dA theorem of Chudnovkii [5] says that if L ∈ Q[z, dz] \ {0} is the minimal differential
operator annihilating a G-function at z = 0, then all its solutions at any point α ∈ Q∪{∞}are essentially G-functions of the variable z−α or 1/z if α = ∞. Such a differential operator
is called a G-operator. Throughout the paper, by “solution of a differential operator L”,
it must be understood “solution of the differential equation Ly(z) = 0”.
dRecently, André [2, 3] defined an E-operator as a differential operator in Q[z, dz] suchthat its Fourier-Laplace transform is a G-operator. Recall that the Fourier-Laplace transddform Lb ∈ Q[z, dz] of an operator L ∈ Q[z, dz] is the image of L by the automorphism of
ddd
the Weyl algebra Q[z, dz ] defined by z 7→ − dz and dz
7→ z. Any E-function is solution ofDate: March 20, 2016.
1His definition was slightly less restrictive, but it is now believed that both definitions define the sameclass of functions.
12dbut is minimal for thean E-operator, which is not necessarily minimal for the degree in dz
degree in z. André proved that the only possible singular points of an E-operator are 0 and∞, 0 being a regular singularity. In general, ∞ is an irregular singularity but the slopes of
the Newton polygon of L are in {0, 1}. We emphasize that the solutions of E-operators are
not holomorphic at z = 0 in general, as they may have a non-trivial monodromy around 0.
This note grew out from the observation that the solutions of “non-trivial” E-operators
we have found in the literature have a non-trivial monodromy. Our main result shows that
this is actually a general fact.
dTheorem 1. Consider an E-operator L ∈ Q[z, dz] of order µ having a basis over C ofholomorphic solutions at z = 0. Then, L has a basis over C of solutions of the form

P1 (z)eβ1 z + · · · + P` (z)eβ` z(1.1)

×for some integer ` ≤ µ, some β1 , . . . , β` ∈ Q , and some P1 (z), . . . , P` (z) ∈ Q[z].
Remark. In [4], Bertrand gives two new proofs of the Siegel-Shidlovskii Theorem using
Laurent’s interpolation determinants method. The second proof given in Section 5 of [4]works under the assumption that the solutions of the underlying E-operator are holomorphic at z = 0. He then observes that “cette hypothèse n’est en pratique vérifiée que dansle cas du théorème de Lindemann-Weierstrass” (2). Theorem 1 shows that Bertrand’s
observation is the general situation: his second proof is indeed only a new proof of theLindemann-Weierstrass Theorem, as it does not cover any other E-function than those ofthe form (1.1).
As a by-product of the proof of Theorem 1, we also obtain the following result.
dTheorem 2. Given an E-operator L ∈ Q[z, dz], if the equation Ly(z) = 0 has a non×zero local solution F (z − α) at some α ∈ Q such that F (z) is an E-function, then theconclusion of Theorem 1 holds.

One of the steps in the proof of Theorem 1 is the following result, which is of independentinterest.
Pd j
Proposition 1. Consider a differential operator L = µj=0 Aj (z) dz
with Aj (z) ∈ Q[z],and Aµ (z) 6= 0. Let∞
XanF (z − α) =(z − α)n ∈ Q[[z − α]]n!
n=0be a local solution of L at an algebraic point α such that Aµ (α) 6= 0. Let dn denote thesmallest positive integer such that a0 , a1 , . . . , an are algebraic integers. Then, there exists
a positive integer C such that, for all n ≥ 0, dn divides C n+1 .
Pσ(an ) nMoreover, if F σ (z) := ∞n=0 n! z is an entire function for any σ ∈ Gal(Q/Q), and if
the slopes of the Newton polygon of L at ∞ are ≤ 1, then F (z) is an E-function.
2in practice, this assumption is satisfied only in the case of the Lindemann-Weierstrass Theorem.

3

2. Proof of Proposition 1Up to replacing L by mL for some non zero integer m, we can and will assume that
A0 (z), . . . , Aµ (z) are polynomials with algebraic integers coefficients. The vector function
Y (z) = t F (z − α), F 0 (z − α), . . . , F (µ−1) (z − α) satisfies the differential system Y 0 = M Y
where

010 ···0.. ..
 0.
01.  .
.
..
..
 ∈ Mµ (Q(z)).
.
.
M =.
.
.
.
0
 00··· 01 Aµ−1A1A0− Aµ − Aµ · · · · · · − AµWe define the sequence of square matrices (Mn )n≥0 by M0 = Iµ and, for all n ≥ 0,Mn+1 = Mn M + Mn0 .
Then, we haveY (z) =∞X
Mn (α)
n=0n!

!
(z − α)n Y (α).

f(See [8, p. 93].) By induction on n, we see that Mn is of the form A−nµ Mn for somefn ∈ Mµ (Q[z]) whose entries are polynomials with algebraic integers coefficients, and thatM
fn is ≤ Bn.
there exists B > 0 such that the degree of each entry of M
Let α be an algebraic number such that Aµ (α) 6= 0. Let u and v be non-zero integers
such that uα and v/Aµ (α) are algebraic integers. Then, for any n ≥ 0, the entries offn (α)(v/Aµ (α))n are algebraic integers. Let also w be a
the matrix uBn v n Mn (α) = uBn M
non-zero integer such that the vector wY (α) has algebraic integers components. Thenthe n-th Taylor coefficient of each component of Y (z) becomes an algebraic integer after
multiplication by w(uB v)n n!. This proves the first part of the proposition, and in fact amore precise result.
Let us now assume that, for all σ ∈ Gal(Q/Q), F σ (z) is an entire function and that theslopes of the Newton polygon of L at ∞ are ≤ 1. Note that, for all σ ∈ Gal(Q/Q), theslopes of the Newton polygon of Lσ at ∞ are ≤ 1 as well, and, hence, the solutions of Lσhave at most exponential growth at ∞ in any sector of bounded aperture. In particular,F σ (z − σ(α)) and, hence, F σ (z) have at most exponential growth of order 1 at ∞, i.e.
|F σ (z)|  eκ|z| for some κ > 0 as z → ∞. Using the Cauchy formulaZn!
F σ (z)σ(an ) =dz
2iπz n+1|z|=nand Stirling’s formula, we then deduce thatlim sup |σ(an )|1/n ≤ eκ−1 < ∞.
n→+∞

4Together with the above bound for the denominator of the Taylor coefficients of F (z), thisimplies that F (z) is an E-function.
P
anRemark. The proof shows the following. If the local solution F (z − α) = ∞n=0 n! (z − α) isentire and of order of growth ≤ 1, then it is a quasi E-function in the following sense: F (z)satisfies conditions (1), (2b) and a weak version of (2a) for σ = idQ only (a priori), wherethese labels refer to the definition of an E-function in the Introduction. The assumption on
F σ (z) in the second part of Proposition 1 forces (2a) for all σ ∈ Gal(Q/Q). We don’t know
if a quasi E-function is automatically an E-function. A similar problem can be formulatedfor G-functions.
3. Proof of Theorem 1Using Theorem 4.3 (iii) in [2], we see that L has a basis over C of solutions E1 , . . . , Eµwhich consists of E-functions. Then, for any σ ∈ Gal(Q/Q), the E-functions E1σ , . . . , Eµσform a basis over C of solutions of Lσ . Indeed, if E1σ , . . . , Eµσ were C-linearly dependent,
they would also be Q-linearly dependent, and using σ −1 on the induced Q-relations of Taylor coefficients, this would imply that E1 , . . . , Eµ are Q-linearly dependent. In particular,for any σ ∈ Gal(Q/Q), any solution of Lσ is an entire function.
×Let us consider α ∈ Q . Theorem 4.3 (iv) of [2] ensures that the slopes of the Newton
polygon of L at ∞ are ≤ 1. Using Proposition 1, we infer that the differential operator Ladmits at z = α a basis of solutions of the form F1 (z − α), . . . , Fµ (z − α), where each Fj (z)is an E-function. We now fix j and we set F (z) := Fj (z). We have
Lα F (z) = 0
where Lα is the shifted operator (of order µ) obtained bychanging z to z + α in L.
P∞
an nWerecallthattheLaplacetransformofF(z)=z is defined by g(z) :=n!
R ∞ −zt
Pn=0∞
ane F (t)dt. Locally around z = ∞, we have g(z) = n=0 zn+1 , which is a G-function
0of the variable 1/z. Denoting by Lbα the Fourier-Laplace transform of Lα , we have µdLbα g(z) = 0.
dz
b −αz , the above equality can(This is a general property, see [2, p. 716].) Since Lbα = eαz Le
be rewritten as follows:Lb e−αz g(z) = e−αz P (z)(3.1)
for some P (z) ∈ Q[z] of degree ≤ µ − 1. Let S be the (finite) set of singularities of g overP1 (C). Consider [γ] ∈ π1 (P1 (C) \ S, z0 ) (where z0 ∈ P1 (C) \ S is arbitrary) and denote by[γ]g the analytic continuation of g along [γ]. We set var[γ] (g) = [γ]g − g. Then, we deducefrom (3.1) that
Lb e−αz var[γ] (g) = 0.
Therefore, we havevar[γ] (g)(z) = eαz ϕ(z)

5b We claim that ϕ = 0. Indeed, assume at the contrary thatfor some solution ϕ of L.
ϕ 6= 0. Note that Lb is a G-operator and, hence, a fuchsian operator. Thus, ϕ has moderate
growth at ∞, so that eαz ϕ(z) has exponential growth at ∞ along some direction. But, g
is a G-function hence it has moderate growth near each s ∈ S along any sector with finite
aperture. It follows that var[γ] (g) has moderate growth at ∞ along any sector with finite
aperture. Whence a contradiction.
Therefore, we have var[γ] (g) = 0 i.e. [γ]g = g. In other words, the monodromy of g istrivial. Since g has moderate growth at each s ∈ S, we get g ∈ Q(z).
To recover F (z) from g(z), we use the inverse Laplace transform formula. Let δ be any
real number larger than 0 and than all the real parts of the singularities of g(z). Then,by [7, p. 61, Théorème 1], we haveR z
δ+iTZ
 0 F (u)du if z > 0g(x) xz
1lime dx =T →+∞ 2iπx0 if z < 0.
δ−iTNow, since g(z) is a rational function, the integral
1limT →+∞ 2iπ

δ+iTZg(x) xze dxx

δ−iTis easily
by the residues theorem; see [9] for a similar computation. It is equal
P computed
ρzto ρ Qρ (z)e when z > 0, where the summation is over the poles of g(x)/x and where
each Qρ (z) ∈ Q[z]. Hence, by differentiation, for z > 0, we haveX
Q0ρ (z) + ρQρ (z) eρz .
(3.2)
F (z) =
ρ

By analytic continuation, this identity holds for any z ∈ C. Since F (z) represents any localsolution of Ly(z) = 0 at z = α, it follows that L has a basis of solutions of the form (3.2).
This completes the proof of the theorem.
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T. Rivoal, Institut Fourier, Université Grenoble 1, CNRS UMR 5582, 100 rue des Maths,
BP 74, 38402 Saint-Martin d’Hères cedex, France.
Tanguy.Rivoal@univ-grenoble-alpes.fr
J. Roques, Institut Fourier, Université Grenoble 1, CNRS UMR 5582, 100 rue des Maths,
BP 74, 38402 Saint-Martin d’Hères cedex, France.
Julien.Roques@ujf-grenoble.fr