Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

I would like to know which results have been obtained concerning Selberg's orthonormality conjecture. For example, has it been proven that for every pair of distinct primitive functions of the Selberg class $(F,G)$, $\displaystyle{\sum_{p\leq x}\frac{a_p(F)\overline{a_p(G)}}{p}=o(\log\log x)}$? Thank you in advance.

EDIT November 5th 2013: considering the answer given below, would the results contained in the following paper: http://arxiv.org/pdf/1311.0754.pdf be a progress toward establishing Selberg's orthonormality conjecture for the whole Selberg class?

share|improve this question

1 Answer 1

up vote 1 down vote accepted

Orthogonality has been proven for certain pairs of automorphic L-functions. The proof procedes like a proof of the Prime Number Theorem, replacing the logarithmic derivative of $\zeta(s)$ with that of $L(s,\pi\otimes\tilde\pi')$, where $\pi$ and $\pi'$ are automorphic representations.
The only proofs in print are for pairs of representations on $GL_m$ (not necessarily the same $m$ for both representations) over ${\mathbb Q}$. It looks like it applies to any pair where the Rankin-Selberg L-function satisfies:
1) Absolutely convergent Euler product for Re(s)$>1$.
2) The expected meromorphic continuation and functional equation, with at most one simple pole.
3) Some estimate towards the Ramanujan conjecture.
4) Nonvanishing for Re(s)$\ge 1$.
5) Order 1 (in the sense that $\log|f(z)|\ll |z|^{1+\epsilon}$), away from the possible pole.
It would be very exciting if nonvanishing (or order 1) could be proven from the Selberg Class axioms. Without them (i.e. proving orthogonality for the entire Selberg Class), the above proof fails completely, so something entirely different would be necessary.

Ye has the above proof on his website, assuming one of the representations is self-dual. The above proof without the assumption can be found here.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.