1
$\begingroup$

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 ?

Improve this question
$\endgroup$
2
  • 2
    $\begingroup$ What is an $L$-function, for the purposes of this question? $\endgroup$
    – Kevin Buzzard
    Commented Jan 12, 2017 at 15:16
  • $\begingroup$ 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. $\endgroup$
    – Sylvain JULIEN
    Commented Jan 12, 2017 at 15:35

1 Answer 1

Reset to default
4
$\begingroup$

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.

Improve this answer
$\endgroup$
8
  • $\begingroup$ 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 ? $\endgroup$
    – Sylvain JULIEN
    Commented Jan 12, 2017 at 15:59
  • 1
    $\begingroup$ 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. $\endgroup$
    – Kevin Buzzard
    Commented Jan 12, 2017 at 16:00
  • $\begingroup$ 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 $\rho\otimes\sigma\cong\oplus_{i,j}\chi_i\psi_j$. $\endgroup$
    – Kevin Buzzard
    Commented Jan 12, 2017 at 16:02
  • $\begingroup$ It's amazing to see how fast you're able to type...Thank you much anyway. P.S. do you think notions related to this question can be of interest ? I'm becoming doubtful about it. $\endgroup$
    – Sylvain JULIEN
    Commented Jan 12, 2017 at 16:06
  • $\begingroup$ 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. $\endgroup$
    – Kevin Buzzard
    Commented Jan 12, 2017 at 16:09

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .