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Hello,

for any given function $F$ of the Selberg class $\mathcal{S}$, let $A_{F}$ be the set of coefficients $a_{n}$ of the Dirichlet series defining $F(s)$ for $\Re(s)>1$, and let $A=\bigcup_{F\in\mathcal{S}}A_{F}$. Is it true that $\mathbb{Q}(A)=\mathbb{C}$? Same question for $F$ running through $\mathbb{P_{\mathcal{S}}}$, where $\mathbb{P_{\mathcal{S}}}$ is the set of all primitive functions of $\mathcal{S}$. Thank you in advance.

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    $\begingroup$ Since there are only countably many automorphic $L$-functions, aren't there going to be only countably many functions in the Selberg class (at least conjecturally)? Hence won't the field $\mathbb Q(A)$ be countably generated over $\mathbb C$ (again, at least conjecturally)? $\endgroup$
    – Emerton
    Commented Jun 23, 2011 at 6:17

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Take any Dirichlet series $L(s)=\sum_{n \geq 1} \frac{a_n}{n^s}$ in the Selberg class. If $L(s)$ has no pole at $s=1$, then for any $\theta \in \mathbf{R}$, the additive twist $L_{\theta}(s)=L(s+i\theta) = \sum_{n \geq 1} \frac{a_n n^{-i\theta}}{n^s}$ still belongs to the Selberg class. It follows that your set $A$ contains all complex numbers of absolute value $1$ and that $\mathbf{Q}(A)=\mathbf{C}$.

So you probably want to ask your question in a more precise form, by considering Dirichlet series only up to additive twist.

As pointed out by Emerton, it is believed that all functions in the Selberg class come (in a sense which might not be formulated very precisely at the moment) from arithmetic automorphic forms, see for example the abstract of this preprint of J. B. Conrey and A. Ghosh. Since the set of all arithmetic automorphic forms is (provably) countable, the set of coefficients of the Dirichlet series which are (conjecturally) associated to them is countable, so cannot generate the field $\mathbf{C}$, whose transcendance degree over $\mathbf{Q}$ is the cardinality of the continuum.

But, as pointed out by Paul Garrett, the question in terms of elements the Selberg class is likely to be completely intractable (like many open questions on the Selberg class).

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  • $\begingroup$ By the way, would it be possible to establish the isomorphy of the groups $G=Gal(\mathbb{Q}(A)/\mathbb{Q}(A_{F})$ and $G'$ defined as the group of automorphisms of the Selberg class (see a previous question from myself whose title is precisely "Automorphisms of the Selberg class") that preserve a given primitive function $F$ belonging to $\mathcal{S}$? Maybe through categorical issues? $\endgroup$
    – Sylvain JULIEN
    Commented Jun 27, 2011 at 20:40
  • $\begingroup$ @Sylvain : I don't see how it could be possible. First $\mathbf{Q}(A)=\mathbf{C}$ so $G$ isn't a Galois group, but just an automorphism group. Second and more importantly, if you take an arbitrary Dirichlet series belonging to the Selberg class, you can't just act on the coefficients to get another element in the Selberg class. This works only in some cases when the coefficients are algebraic (ex. L-functions of modular forms) but certainly not in general. Finally there are plenty of elements in the Selberg class with rational coeffs, so the group would act trivially on all of them. $\endgroup$
    – François Brunault
    Commented Jun 27, 2011 at 22:01
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This is a very serious question, probably intractable.

The "arithmetic theory of automorphic forms", interpreted first as pertaining to holomorphic Siegel-Hilbert or other discrete-series-at-archimedean-places automorphic forms, does connect immediately to algebraic-geometric rationality notions...

The next larger regime is "cohomological" repns... at infinity, only, since finite primes seem to have no obvious constraint.

But... all Maass cuspforms for SL(2,Z)? I think no one has any expectation that associated data would be algebraic, etc. Not that there is "proof" to the contrary, but that there is no reason to think otherwise.

My own opinion is that we do not know the proper question about this, at this time.

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