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 ?

$\endgroup$
  • 2
    $\begingroup$ What is an $L$-function, for the purposes of this question? $\endgroup$ – Kevin Buzzard Jan 12 '17 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 Jan 12 '17 at 15:35
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.

$\endgroup$
  • $\begingroup$ 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 Jan 12 '17 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 Jan 12 '17 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 Jan 12 '17 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 Jan 12 '17 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 Jan 12 '17 at 16:09

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.