# Galois group of an L-function

Let $M$ be a class of L-functions such that whenever $F$ and $G$ belong to $M$, then so do their product $F.G$ and their tensor product $F\otimes G$ defined by $F\otimes G : s\mapsto\sum_{n>0}\frac{a_{n}(F)a_{n}(G)}{n^s}$ for $\Re(s)>1$ if $F : s\mapsto\sum_{n>0}\frac{a_{n}(F)}{n^s}$ and $G : s\mapsto\sum_{n>0}\frac{a_{n}(G)}{n^s}$ if $\Re(s)>1$ . Suppose also that the constant map $s\mapsto 1$ and the Riemann Zeta function $\zeta$ belong to $M$ . An automorphism of $M$ is a bijection of $M$ that sends a primitive element (i.e irreducible for the product) to a primitive element and that commutes to both the usual and the tensor product.

Let's define for an element $F$ of $M$ and a field automorphism of $C$ denoted by $\sigma$ the map $\Psi_{\sigma} : F\mapsto F_{\sigma}=\sum_{n>0}\frac{\sigma(a_{n}(F))}{n^s}$ if $\Re(s)>1$ . $\Psi_{\sigma}$ is an automorphism of $M$ .

Let's now define the 'Galois group' of $F \in M\setminus\{1,\zeta\}$ as the group $\operatorname{Gal}(F)$ , under composition, of field automorphisms $\sigma$ of $C$ such that $F_{\sigma}=F$ . If $G\in M$ is such that there exists $\sigma$ such that $G=F_{\sigma}\neq F$ and $\operatorname{Gal(F)} =\operatorname{Gal}(G)$ then I managed to prove that this group is abelian. Is it finite ?

• What is an $L$-function, for the purposes of this question? – Kevin Buzzard Jan 12 '17 at 15:16
• Say, an automorphic L-function belonging to the Selberg class. The important thing is that the tensor product of two L-functions is required to be an L-function. – Sylvain JULIEN Jan 12 '17 at 15:35

Let $M$ be the set of finite products of Dirichlet $L$-functions. These surely form a class of $L$-functions as in the question. Now take some prime $p$ congruent to 1 mod 4 and let $\chi$ be one of the two Dirichlet $L$-functions of conductor $p$ and order 4 (the other one will then be $\overline{\chi}$). Let $F$ be $L(\chi,s)$ and let $G$ be $L(\overline{\chi},s)$. Then $Gal(F)=Gal(G)$ is the automorphisms of the complex numbers which leave $i$ fixed. This group is certainly not finite (indeed it is uncountably infinite). If $\sigma$ is complex conjugation then $G=F_\sigma\not=F$, so there is a counterexample.
• I accepted the answer, but something remains unclear to me. The degree of an element of $M$ can be any integer, so how do you prove that $F\otimes G$ belongs in $M$ whenever $F$ and $G$ do ? – Sylvain JULIEN Jan 12 '17 at 15:59
• You believe it for $F$ and $G$ irreducible because the product of two Dirichlet characters is a Dirichlet character. So then it's true in general because tensor is distributive over product. – Kevin Buzzard Jan 12 '17 at 16:00
• In representation theoretic terms, if $\rho$ and $\sigma$ are representations of a group and they're both a finite direct sum of irreducible 1-dimensional representations, then the same is true of $\rho\otimes\sigma$. Indeed if $\rho\cong\oplus_i\chi_i$ and $\sigma\cong\oplus_j\psi_j$ then $\rho\otimes\sigma\cong\oplus_{i,j}\chi_i\psi_j$. – Kevin Buzzard Jan 12 '17 at 16:02
• I've seen them before in work of Clozel. For an algebraic automorphic representation there should be a coefficient field $E$, a number field in $\mathbb{C}$, and your automorphism group is just the automorphisms of $\mathbb{C}$ that fix $E$. Clozel used them to prove cohomological representations were arithmetic if I remember correctly, but in some sense this is the standard trick involving the automorphisms of the complexes that everyone uses and it's a pretty coarse invariant. – Kevin Buzzard Jan 12 '17 at 16:09