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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.

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    $\begingroup$ Wikipedia is wrong, in the sense that we do not know of the existence of any such $L$-functions, and indeed the Selberg eigenvalue conjecture predicts that no such $L$-functions exist. $\endgroup$
    – Peter Humphries
    Jul 12, 2022 at 14:30
  • $\begingroup$ OK, I see, IF it exists then it would violate the RH, but why it would violate the RH, this possible violation is mentionned wihtout any explanation in the Book "Value Distribution of L-function", as if it was obvious... but it is not obvious for me, any insight ? $\endgroup$
    – Bertrand
    Jul 13, 2022 at 3:06

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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.

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  • $\begingroup$ Clear, thanks, this possible violation is mentionned without any explanation in the Book, as if it was obvious... but it is not obvious for me, any insight ? $\endgroup$
    – Bertrand
    Jul 13, 2022 at 3:07
  • $\begingroup$ I don't know sufficiently well the subject to provide such an insight but this may be related to the zeta function bearing his name he introduced in 1956. $\endgroup$
    – Sylvain JULIEN
    Jul 15, 2022 at 22:20
  • $\begingroup$ The answer to this former question of mine may actually be of help: mathoverflow.net/questions/247009/… $\endgroup$
    – Sylvain JULIEN
    Jul 15, 2022 at 22:24

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