5
$\begingroup$

For a relative variety $X$ over a ring of integers $O_K$, we can define a zeta function. This zeta function is defined as the product of the zeta functions of the variety specialized to $O_K/\mathfrak{p}$ as $\mathfrak{p}$ runs over $Spec(O_K)$. In turn, zeta functions of varieties over finite fields are easy to define using the counting of rational points. As Grothendieck proved, these zeta functions can be expressed as a product of $L$-functions indexed by $i$, where the $i^{th}$ $L$-function is related to the $i^{th}$ (Weil) cohomology of $X_{O_K/\mathfrak{p}}$. The $i^{th}$ $L$-function of $X$ over $O_K$ is defined to be the product over $\mathfrak{p}$ of the $i^{th}$ $L$-function of $X_{O_K/\mathfrak{p}}$.

The Weil conjectures give us a lot of information about the zeta functions of varieties over finite fields, and in fact about their $L$-functions.

The Langlands program is about properties of $L$-functions of $X$ over $O_K$.

Is it possible to interpret the Weil conjectures as telling us something meaningful about the Langlands program?

$\endgroup$
0

1 Answer 1

8
$\begingroup$

If $\pi$ is a regular algebraic cuspidal automorphic representation of $GL_n/K$ with K totally real or CM, and $\pi$ satisfies a certain self-duality condition, then $\pi_v$ is tempered for all finite $v$. This monumental theorem is a vast generalization of Deligne's proof of the Ramanujan conjecture, and the proof ultimately appeals to the Weil conjectures, by proving an identity $L(s,\pi)=L(s,M(\pi))$ where $M(\pi)$ is (essentially) a submotive of the cohomology of a Shimura variety.

$\endgroup$
1
  • 3
    $\begingroup$ So what David is pointing out is that, because the L-functions of many Shimura varieties are now known to be related to automorphic forms, the parts of the Weil conjectures which give bounds for complex eigenvalues of Frobenius (Riemann hypo) can be used to deduce bounds for eg Satake parameters of certain automorphic representations, and these turn out to be profound results on the automorphic side. In particular it is "only" the RH parts playing a role here. $\endgroup$ Oct 26, 2011 at 6:51

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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