Selberg class definition and Riemann hypothesis

Question

Looking at the Selberg class definition on Wikipedia, under "Comment on definition", there is this paragraph:

"The condition that the real part of $\mu_i$ be non-negative is because there are known L-functions that do not satisfy the Riemann hypothesis when $\mu_i$ is negative. Specifically, there are Maass forms associated with exceptional eigenvalues, for which the Ramanujan–Petersson conjecture holds, and have a functional equation, but do not satisfy the Riemann hypothesis."

$\mu_i$ is a constant in the Gamma factor which is part of the functional equation.

I am looking for an article reference or book showing this particular Maass for which RH is wrong.

Below is the link to Wikipedia article where there is no reference supporting this result.

https://en.wikipedia.org/wiki/Selberg_class.

Answer 1

If such an $L$-function with gamma factors whose constants have negative real parts exists, then it will have at most finitely many zeros inside the critical region because of the poles of these gamma factors. And in order to be symmetric about the real axis by virtue of the functional equation, these gamma factors must have either a purely real constant or come in pairs with conjugate constants. See e.g. the discussion after Proposition 1.9.1 in Bump's book Automorphic forms and representations (MR1431508, Zbl 0868.11022).

The way I see it, these will still be trivial zeros (i.e. they can be spotted by inspection of the arguments of the gamma factors), despite being located inside the critical region.

Note that the Euler product and the non-vanishing of the $L$-function on $\Re(s)=1$ prevents the trivial zeros from being symmetric outside the critical region and on its boundaries. For instance, Dirichlet $L$-functions of even characters have a trivial zero at $s=0$, with the notable exception of the RZF whose simple pole at $s=1$ cancels that of the gamma function at $s=0$, thus removing the singularity there, which is the reason why $\zeta(0)=-1/2$.

EDIT: Now, if these poles of the gamma factors inside the critical region are reflecting those of the $L$-function (which are not cancelled by zeros of its constituent factors) so that the completed function is symmetric by virtue of the functional equation, and also assuming that the poles of the $L$-function correspond to those of the RZF (with suitably shifted arguments) as constituent factors of the $L$-function, then it would violate the RH because it will have infinitely many non-trivial zeros inside the critical region but not on the critical line. This, however, will contradict the assumption that the Ramanujan conjecture holds for this particular Maass form associated with exceptional eigenvalues. See e.g. the remarks on p.2 of the paper On the Selberg class of Dirichlet series: small degrees by Conrey & Ghosh.

EDIT 2: But the Maass forms under consideration are cusp forms, so no poles.

The exceptional eigenvalues are discussed at the end of section 3 on p.13 in Bump's paper Spectral theory and the trace formula (expanded text), in line with the discussions of the Ramanujan/Selberg conjectures in Sections 3.5 and 3.7 of his book. See also the remarks following the definition of the Selberg zeta function in Section 3 of Selberg's 1956 paper Harmonic analysis and discontinuos groups in weakly symmetric Riemannian spaces with applications to Dirichlet series, or similarly, Theorem 39 on p.28 in Bump's paper.

Answer 2

The only reference I managed to find is page 116 of Value Distribution of $L$-Functions, by Jörn Steuding (Springer, 2007).

If we assume the existence of an arithmetic subgroup of $\mathsf{SL}_2(\mathbb{R})$ together with a Maass cusp form that corresponds to an exceptional eigenvalue, and if we further suppose that all local roots are sufficiently small (more precisely, that the Ramanujan–Petersson conjecture holds), then the $L$-function associated with the Maass cusp form has a functional equation where the $\mu_j$ satisfy $\Re \mu_j < 0$, but this $L$-function violates Riemann’s hypothesis.

As it is written, it seems that the existence of the considered Maaß forms is only conjectural, so it would be interesting to figure this out.

Edited after Peter's comment above: it is more precisely conjectured that such Maaß forms do not exist.