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Hello,

This is a question in the spirit of Nonvanishing of central L-values of quadratic twists? and the application I have in mind is to p-adic L-functions a la Ash-Ginzburg.

The question is this:

Suppose $\pi$ is an irreducible unitary automorphic cuspidal representation of $GL(4)$, say, over ${\mathbb Q}$. Under what conditions on $\pi$, do we know the existence of a finite order $\chi$ such that $L(\frac{1}{2}, \pi \otimes \chi) \ne 0$? Here $L$ is the completed $L$-function.

I know that by a result of Luo from 2005 if $\Re (s ) \ne 1/2$, then there is $\chi$ such that $L(s, \pi \otimes \chi) \ne 0$.

Any help would be greatly appreciated.

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  • $\begingroup$ I bet this is an open problem. The issue is that, if $\chi$ has conductor $q$, then the approx. fun. eqn. gives $L(1/2,\pi \otimes \chi)$ as a sum of about $q^2$ terms, and if you average over $\chi$ and $q$ there's an obvious main term, but the error term just barely fails to be manageable (you can succeed on $GL(2)$ and $GL(3)$.) $\endgroup$
    – David Hansen
    Commented Apr 12, 2011 at 21:28
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    $\begingroup$ Hey Ramin, if you take a look at the discussion over at mathoverflow.net/questions/60387 this might apply to the application you're interested in. I don't know much about the $p$-adic $L$-functions for $GL(4)$, but if the Ash–Ginzburg $p$-adic $L$-function is interpolating the values at critical integers (and your central point is critical, like an "even weight" condition for $GL(2)$), then as long as the weight of your automorphic representation is sufficiently regular ($k>2$ for $GL(2)$), the non-vanishing of the $p$-adic $L$-function trivially follows from the non-vanishing of an ... $\endgroup$
    – Rob Harron
    Commented Apr 13, 2011 at 3:20
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    $\begingroup$ (cont'd) $L$-value in the range of convergence of the Euler product. The fact that the $p$-adic $L$-function is a $p$-adic measure will then imply that it has only finitely many zeros and hence there's some twist of the central value which is non-zero. Now, the $p$-adic $L$-functions in Ash–Ginzburg are just for trivial coefficients, i.e. a very small weight, so I imagine this doesn't apply. However, Rob Pollack's answer to the aforementioned mathoverflow question outlines an idea for using higher weight automorphic representations to prove the result for small weights. $\endgroup$
    – Rob Harron
    Commented Apr 13, 2011 at 3:21
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    $\begingroup$ (cont'd) However, this would involve constructing two-variable $p$-adic $L$-functions for $GL(4)$, and knowing something about their $\mu$-invariants, so that would be what they call "difficult". $\endgroup$
    – Rob Harron
    Commented Apr 13, 2011 at 3:21
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    $\begingroup$ Hi Rob, This is very helpful! Thank you. $\endgroup$
    – Ramin
    Commented Apr 13, 2011 at 14:59

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