3
$\begingroup$

Hello,

assuming Selberg's orthonormality conjecture, let's consider bijective maps $f$ from Selberg's class $\mathcal{S}$ to itself such as:

1) $f$ maps a primitive function of $\mathcal{S}$ to a primitive function of $\mathcal{S}$,

2) $f$ maps a function of degree $d$ to a function of degree $d$,

3) for every $F$ in $\mathcal{S}$, $n_{f(F)}=n_{F}$, where $n_{F}$ is the integer involved in Selberg's conjecture $A$,

4) if $F=F_{1}^{e_{1}}.F_{2}^{e_{2}}....F_{k}^{e_{k}}$, then $f(F)=f(F_{1})^{e_{1}}.f(F_{2})^{e_{2}}....f(F_{k})^{e_{k}}$ (so that $f$ is "strongly multiplicative"),

5) if $f$ verifies the above conditions, then so does the inverse of $f$.

The set of all such maps makes a group for the composition law. I would like to know whether this group is isomorphic to the absolute Galois group of the field of the rational numbers or not.

Thanks in advance.

Improve this question
$\endgroup$
3
  • $\begingroup$ What is the "dot" operation in 4)? Is it just multiplication of Dirichlet series? $\endgroup$
    – David Hansen
    Commented Jan 5, 2011 at 16:36
  • $\begingroup$ Yes indeed, it is just the product of the functions $F_i$. $\endgroup$
    – Sylvain JULIEN
    Commented Jan 15, 2011 at 12:13
  • 2
    $\begingroup$ While the Selberg class is a nice way of organizing the observed properties of L-functions, may I suggest that it is not much more than that, at least in the current state of the subject? Witness, for example, the immense effort required on the part of Kaczorowski-Perelli to prove that nothing in the Selberg class has degree $1<d<2$ (this appeared in the Annals of Math. recently). If a statement of such tempting simplicity requires Sisyphean expenditures,... $\endgroup$
    – David Hansen
    Commented Mar 14, 2011 at 6:28

4 Answers 4

Reset to default
7
$\begingroup$

Most elements in $\mathcal{S}$ probably don't have algebraic Dirichlet coefficients (even after scaling), e.g. this is conjectured for the $L$-functions of non-CM Maass forms. So I don't see that the Galois group in question acts on $\mathcal{S}$. I think it is fair to conjecture that the only automorphism of $\mathcal{S}$ is the identity.

Improve this answer
$\endgroup$
7
  • $\begingroup$ I'm not sure to understand your answer, but isn't the complex conjugation such an automorphism of $\mathcal{S}$ ? $\endgroup$
    – Sylvain JULIEN
    Commented Dec 30, 2010 at 18:56
  • $\begingroup$ You are right, this makes the dual $L$-function which enters the functional equation. So my guess is that this is the only nontrivial automorphism. My original answer tried to emphasize that most automorphic $L$-functions have highly transcendental Dirichlet coefficients, at least this is what we conjecture. Few automorphic $L$-functions (e.g. those of geometric type) have algebraic coefficients, but this is an exception. $\endgroup$
    – GH from MO
    Commented Dec 30, 2010 at 19:04
  • $\begingroup$ Could someone give a rigorous proof (always assuming Selberg's orthonormality conjecture) of the fact that there is no other automorphism than the identity and the complex conjugation? Thanks in advance. $\endgroup$
    – Sylvain JULIEN
    Commented Jan 30, 2011 at 20:36
  • $\begingroup$ I think this will be very hard to do. As far as I know we only have conjectures about the lack of algebraic relations between the Euler coefficients of a general automorphic L-function, e.g. an L-function associated to a Maass cusp form on SL(2,Z). $\endgroup$
    – GH from MO
    Commented Jan 31, 2011 at 9:45
  • $\begingroup$ But wouldn't it be possible to consider the conditions above as some kind of continuity assumptions that would make identity and complex conjugation the only possible maps such as condition 4) is true? A bit like the only continuous field automorphisms of $\mathbb{C}$ are namely the identity and the complex conjugation? $\endgroup$
    – Sylvain JULIEN
    Commented Feb 7, 2011 at 17:57
6
$\begingroup$

Doesn't twisting by (finite-order) Dirichlet characters satisfy all the properties you listed?

Improve this answer
$\endgroup$
6
  • $\begingroup$ Good point, but this might also be only conjectural. At least regarding the OP's requirement (4), Selberg says on page 50 of his paper "Old and new conjectures and results about a class of Dirichlet series": It is not unreasonable to conjecture that if they are not primitive, they will factor in the same way into primitive factors, in the sense that we would get the factorization of $F^\chi$ by putting the character $\chi$ into all the series that define the factors of $F$. $\endgroup$
    – GH from MO
    Commented Mar 13, 2011 at 22:06
  • 1
    $\begingroup$ Also, I don't see how the OP's requirement (5) works, i.e. how to "untwist". For example if $\chi$ is the principal character mod $N$, then twisting $\zeta(s)$ and $L(s,\chi)$ by $\chi$ results in the same function, namely $L(s,\chi)$. This shows that twisting is not bijective, some information is lost there. $\endgroup$
    – GH from MO
    Commented Mar 13, 2011 at 22:12
  • 2
    $\begingroup$ @GH: I agree that my proposal is conjectural. :) Regarding your second comment, L-functions with the "wrong" local factors at some finite set of places won't appear in the Selberg class, because they have the wrong functional equation. $\endgroup$
    – David Hansen
    Commented Mar 14, 2011 at 2:03
  • $\begingroup$ I agree that L-functions with the wrong local factors don't appear in the Selberg class, but doesn't that mean that twisting naively is not an automorphism? Perhaps you had in mind "twisting and then fixing the bad Euler factors"? $\endgroup$
    – GH from MO
    Commented Mar 15, 2011 at 1:18
  • $\begingroup$ @GH: Yes, that is what I meant, which (as you know) happens "automatically" in the automorphic world. :) $\endgroup$
    – David Hansen
    Commented Mar 15, 2011 at 19:07
5
$\begingroup$

The group in question really seems to be the direct product, over all $d$, of the full symmetric group of the set of primitive functions of degree $d$, extended by multiplicativity (condition 2 is automatic from the others because the integer $n_F$ is supposed to be the sum of squares of the multiplicities of the primitive factors). So this is a seriously huge group.

(About the last comment: the group can not act transitively, since it is supposed to preserve the degree; but the orbits are the same as the possible degrees).

Improve this answer
$\endgroup$
2
  • $\begingroup$ Very good. Is it known that every function in the Selberg class is a product of primitive functions in a unique fashion? I am not familiar with these things. $\endgroup$
    – GH from MO
    Commented Mar 13, 2011 at 22:20
  • $\begingroup$ By the way, is this group isomorphic to some subgroup of the general affine group of the affine plane over $\mathbb{R}$? $\endgroup$
    – Sylvain JULIEN
    Commented May 5, 2011 at 23:42
-1
$\begingroup$

In fact, I have quite good reasons to think that the above conjecture (the only automorphisms of $\mathcal{S}$ are the identity and the complex conjugation) is equivalent to the Riemann Hypothesis for the whole Selberg class. Certainly one would wish for a stricter proof here... :-)

EDIT: Considering David Hansen's answer, it appears that the group of automorphisms of $\mathcal{S}$ might be richer than I first expected. In fact I may have been a bit hasty saying "is equivalent to". An interesting question would be to ask whether the action of this group on $\mathcal{S}$ is transitive or not.

Improve this answer
$\endgroup$
6
  • $\begingroup$ But I guess that, even if $Aut \mathcal{S}$ is much bigger than we first expected, the only automorphisms that fix a given $F$ are the identity and the complex conjugation or the only identity (depending on whether $F(s)$ is real for $s$ real or not), am I wrong? $\endgroup$
    – Sylvain JULIEN
    Commented Mar 14, 2011 at 6:24
  • $\begingroup$ No, if you want to fix a given $F$, you can still permute arbitrarily all the other forms of the same degree, and all the other degrees. $\endgroup$
    – Denis Chaperon de Lauzières
    Commented Mar 14, 2011 at 7:56
  • $\begingroup$ By the way, would there be a natural way to define continuous automorphisms of $\mathcal{S}$? If so, what would the group of all such continuous autormophisms look like? $\endgroup$
    – Sylvain JULIEN
    Commented Mar 15, 2011 at 15:54
  • $\begingroup$ I might have an idea concerning my question above. Let's define the notion of isometric automorphism of $\mathcal{S}$ in such a way: $f$ is an isometric automorphism of $\mathcal{S}$ if and only if it verifies both of the following conditions: 1) $f$ is an automorphism of $\mathcal{S}$ 2) $\forall(F,G)\in\mathcal{S}^2$, for all positive integer $n$, $\vert a_{n}(F)-a_{n}(G)\vert=\vert a_{n}(f(F))-a_{n}(f(G))\vert$ where $a_{n}(F)$ is the coefficient of the Dirichlet series defining $F(s)$ for $\Re(s)>1$ Is the group of all isometric automorphisms of $\mathcal{S}$ of order 2? $\endgroup$
    – Sylvain JULIEN
    Commented Mar 22, 2011 at 17:39
  • $\begingroup$ Hello, would you please clarify more on the equivalence you mentioned? Because I think I have an answer to your original question. $\endgroup$
    – user48900
    Commented Mar 13, 2015 at 12:53

You must log in to answer this question.

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