Let $F$ and $G$ be two primitive functions of the Selberg class, and let $\mathbb{A}$ be the set of values taken by the function which maps a prime number $p$ to $a_{p}(F)\overline{a_{p}(G)}$. $\mathbb{A}$ is finite or countable, so that there exists $I\subset\mathbb{N}^*$ such that $\displaystyle{\mathbb{A}=\bigcup_{i\in I}\{A_{i}\}}$, with $A_i\neq A_j$ whenever $i\neq j$. Now, let $\delta$ be the function that maps $i\in I$ to $\displaystyle{\lim_{x\to\infty}\frac{\#\{p\leq x, a_{p}(F)\overline{a_{p}(G)}=A_i\}}{\pi(x)}}$.

Does Selberg's orthonormality conjecture imply $\displaystyle{\sum_{i}\delta(i)A_i=\delta_{F,G}}$, where $\delta_{F,G}=1$ if $F=G$ and $\delta_{F,G}=0$ otherwise? If so, is such an implication in fact an equivalence? Thank you in advance.


No. (This is very unlikely to hold for anything except $L$-functions with only finitely many distinct coefficients; the simplest counter-example is to take an elliptic curve $E$ and its associated $L$-function $L(E,s)$, normalized to have critical line at $1/2$, and $F=G=L(E,s)$; for any non-zero real number $x>0$, due to the renormalization -- which by the way makes the question rather strange --, any prime $p$ for which $|a(p)|^2=x$, the Hasse-Weil coefficient $a_p$ of the elliptic curve must satisfy $a_p^2=px$; in particular $\sqrt{px}$ is an integer, and there are rather obviously only finitely many $p$ which can satisfy this for a given $x$. In other words, your function $\delta$ is identically zero, despite the fact that $\delta_{F,F}=1$.)

One can also of course just take the $L$-function of the Ramanujan delta function, replacing just squareroots with powers $11/2$ in the right places.

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  • $\begingroup$ Indeed I thought about this question considering the case of Dirichlet L-functions and wondering whether it would hold for the whole Selberg class or not. Thank you much for your answer. $\endgroup$ – Sylvain JULIEN Mar 27 '11 at 18:53

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