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

Let $\phi$ be a $GL(2)$ automorphic form with Fourier coefficients $a(n)$ and $a(1)=1$. Obviously we have $L(s,\phi)=\sum \frac{a(n)}{n^s}$.

Shimura have the following formula $L(s, Ad\; \phi)=\zeta(2s)\sum\frac{a(n^2)}{n^s}$. [*]

I would like to see its generalization to $GL(3)$ or $GL(N)$. Let $\phi$ be a $GL(3)$ automorphic form with Fourier coefficients $a(m,n)$ and $a(1,1)=1$. We have $L(s,\phi)=\sum \frac{a(1,n)}{n^s}$. I would like to know a formula for $L(s, Ad\; \phi)$ like [*].

We shall note that we always have $L(s,Ad\;\phi)=L(s,\phi \times \overline{\phi})/\zeta(s)$ but that is not what I want.

share|improve this question
1  
The $n^2$ in Shimura's formula suggests half-integer weight forms are involved, so maybe there is something similar involving the metaplectic double cover of $GL(3)$? –  David Loeffler Jun 25 '13 at 10:41
2  
Shimura's formula is only valid for square-free level and trivial nebentypus. For the coefficients of $L(s,\phi \times \overline{\phi})$ there is a suitable formula, see Definition 12.1.1 in Goldfeld's Automorphic forms and L-functions for the group GL(n,R). Convolving these with the Möbius function gives the coefficients you need. In the case of $n=2$ the coefficients are pretty because $a(p^k)^2-a(p^{k-1})^2=a(p^{2k})$ for $p$ prime. –  GH from MO Jun 27 '13 at 23:14
add comment

Know someone who can answer? Share a link to this question via email, Google+, Twitter, or Facebook.

Your Answer

 
discard

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

Browse other questions tagged or ask your own question.