Weil conjecture for algebraic surfaces - MathOverflow

Weil conjecture for algebraic surfaces

2010-07-15T05:16:28Z

Deligne's proof of the Weil conjecture is difficult.

On the other hand, there are some "simpler" proofs of the Weil conjecture in the case of algebraic curves.

For instance, in GTM52, one see it eventually reduced to the Hodge index theorem, which is the geometric input.

And there even exists some elementary proofs, by Bombieri or Stephanov.

So what I am asking is, will there be some "simple" proofs of the Weil conjecture for algebraic surfaces, at least for some special classes of them? And in that case, what can be the geometric inputs, without using the Standard Conjecture?

Answer by Tony Scholl for Weil conjecture for algebraic surfaces

2010-07-15T13:30:21Z

This is a somewhat subjective topic, but a lot of people believe that the answer is "no". There are various reasons why the RH for curves is much easier than the general case.

One is that for curves, one can replace $\ell$-adic cohomology with the Jacobian. In higher-dimensions the geometric objects underlying $\ell$-adic cohomology groups are motives.

Another is that in dimension 1, RH is equivalent to the estimate on the number of rational points $|\#X(\mathbb{F}_{q^r}) -q^r| = O(q^{r/2})$ . This is exactly what Stepanov and Bombieri proved. But in dimension d>1 the main error term comes from the cohomology in dimension $2d-1$, and so a point-counting estimate does not give you any more information than the Lang-Weil estimate $|\#X(\mathbb{F}_{q^r}) -q^{dr}| = O(q^{(2d-1)r/2})$ - which is proved by reduction to curves.

There are some special cases where RH is equivalent to a point-counting estimate - for example, (smooth projective) hypersurfaces, where the only interesting cohomology is in the middle dimension. Katz asked a long time ago whether one could give an elementary proof in this case, and Bombieri also thought about it. (I recently found how to deduce the general RH from the case of hypersurfaces, so a different proof of this special case would certainly be interesting.)

Taking a step back, you can also ask for "simpler" proofs of the other parts of the Weil conjectures. The proof of the rationality of the zeta function for curves is very simple, just using Riemann-Roch. As far as I know, there are no simple proofs in dimension > 1, although a while back Fesenko mentioned in a paper that his adelic methods would give a non-cohomological proof of rationality for surfaces.

Answer by Felipe Voloch for Weil conjecture for algebraic surfaces

2010-07-15T13:41:42Z

I've tried to extend my proof (with Stohr, Proc LMS 52, 1986) to smooth surfaces in P^3, where as Scholl pointed out in his answer, it's equivalent to a point counting inequality. Unfortunately I have only very limited success so far (see AMS Cont Math 324). I think it can be done.

Answer by Richard Borcherds for Weil conjecture for algebraic surfaces

2010-07-15T14:21:03Z

Deligne (La conjecture de Weil pour les surfaces $K3$. (French) Invent. Math. 15 (1972), 206--226 MR0296076 ) gave a proof for K3 surfaces shortly before his proof of the general case, but it is not exactly easy. Manin also proved a few specisl cases in higher dimensions using motives.

In general, one expects an easy proof whenever there is an easy description of the l-adic cohomology groups. The zeroth cohomology is trivial to describe, the first can be described in terms of the Picard variety, and the cohomology above the middle dimension can be reduced to the rest using Poincare duality. For curves this gives you all the cohomology, but in higher dimensions it does not, which is why the case of curves is so much easier. There are many cases in higher dimensions, such as abelian varieties, where one can somehow describe the cohomology, and in these cases there is again an elementary proof. There are also numerous trivial variations of known cases, such as products of curves or Kummer surfaces or rational surfaces, that can be done easily because the cohomology is known.

For surfaces the only "unknown" cohomology is the second cohmology. It is conceivable that one could get at this by mumbling something about the Brauer group in order to get a proof in this case, but this would probably end up as difficult as the general case.