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Let $X$ be a proper algebraic variety. $X$ is said to have polynomial point count if there is a polynomial $P$ such that for all finite fields $\mathbb F_q$ with $q$ elements, $|X(𝔽_q)|=P(q)$.

If in addition $X$ is smooth, then by the Weil conjectures one can derive that $X$ has no odd cohomology.

My question: Is there a weaker hypothesis than smooth under which the same conclusion can be reached?

Please don't answer with the condition $X$ has a paving by affine cells. Because I know about this condition, and know about it I should, since many (but not all) varieties that one comes across in real life (by which I provocatively and rather tounge-in-cheek mean my own mathematical experience) have affine pavings.

It wouldn't surprise me if there is no answer to this question as stated. So I would also like to issue a call for examples. In particular, what is the best-behaved variety you know which has polynomial point count and odd cohomology?

The example I know is the nodal plane cubic $$Y^2Z=X^2(X+Z)$$. It has $q$ points over $𝔽_q$ and one-dimensional $H^1$. Can you do better?

(If you are at all concerned that I haven't defined what base scheme $X$ is over or what cohomology theory I'm using, feel free to interpret such things in any sensible way you wish)

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2 Answers 2

You can do a bit better than a paving: it's enough to check that the cohomology is generated by algebraic cycles (or any other way of showing the Hodge filtration is pure).

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(I deleted my original answer since I realize I had misinterpreted the question a bit).

The cheekiest/simplest example that I know of a polynomial count variety with odd cohomology is $\mathbb{G}_m$. This example can be bootstrapped (standard `scissor formalism') to produce more complicated examples of such varieties. For instance intersection of Schubert cells with opposite Schubert cells, Mirkovic-Vilonen slices in the affine Grassmanian are two of my favorite nice examples of such varieties (the count polynomials are Kazhdan-Lusztig R-polynomials and Hall-Littlewood polynomials respectively).

As pointed out, the Weil conjectures yield that complete smooth polynomial count varieties have no odd cohomology. If one tries to relax smooth, then one thing we can replace both complete and smooth with is pure (which is the content of Ben's answer I believe). I don't know how more general of a class this really is. Certainly it contains examples that are not smooth (for instance rationally smooth but not smooth varieties, and all your other favorite examples of complete varieties whose intersection cohomology agrees with ordinary cohomology). But I don't think that's very helpful, since it can be a pain to show purity in specific examples without getting it for free out of the yoga of weights.

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But G_m is not proper. If you intersect a Schubert variety with an opposite Schubert variety, do you know how to compute the cohomology? –  Peter McNamara Jun 7 at 0:43
    
@PeterMcNamara: Sorry, didn't realize that you only wanted complete/projective examples. Apparently, I misinterpreted the question even more than I thought. I do not know how to compute the cohomology of the intersections (simply the weight polynomials, i.e., R-polynomials), I think it would be fantastic if someone explained how to compute the full cohomology (see mathoverflow.net/questions/116348/…). –  Reladenine Vakalwe Jun 10 at 13:22

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