18
$\begingroup$

The Hilbert–Pólya conjecture is the name given to the idea that the "reason" or "explanation" for the collinearity of the non-trivial zeros of the Riemann zeta function $\zeta(s)$ is that they are the spectrum of some self-adjoint operator. The empirical evidence for the Montgomery pair-correlation conjecture provides some support for the idea that the zeros of $\zeta(s)$ are "spectral."

In the function field case, the Riemann hypothesis has been famously proved by Deligne. My question is, is the intuition behind the Hilbert–Pólya conjecture vindicated in the function-field case?

As I understand it, in the function field case, there is an interpretation of the Riemann zeros as the zeros of a linear operator (a Frobenius action on cohomology). However, it's not clear to me that this necessarily answers my question affirmatively. It seems to me that for the (original) Riemann hypothesis to be "explained" by the spectrum of a self-adjoint operator, it isn't enough to demonstrate the mere existence of such an operator; there should be some "natural reason" for the operator to be self-adjoint. To put it another way, suppose someone were to prove the Riemann hypothesis by some kind of seemingly ad hoc computation that just happened to come out right, and then after the fact, we were to artificially construct a self-adjoint operator to fit the zeros. This would prove the Hilbert–Pólya conjecture in some sense, but it would seem not to satisfy the original desire to "explain" the Riemann hypothesis spectrally.

In the function field case, does Deligne's proof seem to give a satisfying "spectral explanation" of the Riemann hypothesis, or does it feel more like a calculation that just happens to work out?

I realize that this is a vague question that may be largely opinion-based, but I'm holding out hope that people with a thorough understanding of Deligne's proof will find this to be a meaningful question.

$\endgroup$
6
$\begingroup$

In the function field setting, the most natural spectral explanation for the Riemann hypothesis might be expressing the eigenvalues of Frobenius as the eigenvalues of a unitary operator on a finite-dimensional vector space (times $\sqrt{q}$). This would be related to the Hilbert-Polya picture by a logarithm.

There are a formidable number of proofs of the Riemann hypothesis over finite fields. For curves, there are the two proofs of Weil, and one of Bombieri-Stepanov. For higher-dimensional varieties there are Deligne's proof in Weil I and his proof in Weil II, as well as later simplifications of these.

I think the short answer is that none of these proofs really construct a unitary operator prior to proving these inequalities.

In particular, for Deligne's method, note that all variants of Deligne's proof use crucially his inductive argument where a series of stronger bounds for a Frobenius eigenvalue is proved, converging on the correct one. It's hard to see how such a proof could be converted into a reasonable construction of a vector space and operator.

Stepanov's method has no linear algebra in characteristic zero at all, and works entirely with algebra over the finite field and inequalities.

For Weil's proofs this is the most subtle, as he works with the algebra of endomorphisms of the Jacobian variety of a curve (in the first proof) or the divisor class group of a surface (in the second proof). Both of these are integer lattices and so can be naturally embedded into a complex vector space, unlike anything in Deligne's proof, which is entirely $\ell$-adic. But both of these complex vector spaces do not play the role of the vector space, but rather of the algebra of operators on that vector space.

I am not too sure of this but it might be said that Weil's first proof naturally produces a representation of Frobenius as a unitary element of a $C^*$ algebra.

$\endgroup$
  • 2
    $\begingroup$ I've been meaning to write an answer which describes the same facts with the opposite spin and say "basically, yes". Here is the outline: Frobenius in the Weil conjectures should be like $\sqrt{q} \exp(i H)$ where $H$ is the hypothetical Hilbert-Polya operator. So the question is if $U:= \mathrm{Frob}/\sqrt{q}$ is like a unitary operator. It obeys $\langle U x, U y \rangle = \langle x, y \rangle$ where $\langle \ , \ \rangle$ is cup product on $H^1$ of a curve. The hard question is in what sense $\langle \ , \ \rangle$ is positive definite, but Weil's proofs in some sense answer that. $\endgroup$ – David E Speyer Sep 11 at 3:04
  • $\begingroup$ Does this make any sense? $\endgroup$ – David E Speyer Sep 11 at 3:06
  • $\begingroup$ @DavidESpeyer It certainly makes at least some sense. I was definitely somewhat biased by the fact that the question mentions only Deligne. The subtlety is that Weil's proofs don't directly prove a positivity property for the cup product but only for the pairing it induces on the endomorphisms. One can't literally make cup product into a positive definite pairing as, because there's no complex conjugation involved, one would have make $H^1$ into a real vector space, but this is impossible for supersingular elliptic curves. But one could say this is "close enough". $\endgroup$ – Will Sawin Sep 11 at 11:47
  • $\begingroup$ @DavidESpeyer Another approach would be to work in the setting of ordinary curves and use Deligne's parameterization, i.e. view the tangent space of a canonical lift as a complex vector space with Hermitian form. I think this does work, but only in the ordinary case. $\endgroup$ – Will Sawin Sep 11 at 11:50
  • $\begingroup$ I thought abelian varieties had canonical lifts but curves didn't. That is to say, my impression was that if $X$ is an ordinary curve over $\mathbb{F}_p$, then $\mathrm{Jac}(X)$ is a CM abelian variety, and we can consider an abelian variety of the same CM type over $\mathbb{Q}_p$, but that the result didn't have to be a Jacobian. But I could definitely be wrong! You seem like the sort of person who would know. $\endgroup$ – David E Speyer Sep 11 at 16:59

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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