To establish the Weil conjectures for $n$-dimensional projective space over a finite field is elementary. Does there exist a simple direct proof of the conjectures for finite field Grassmannians?
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$\begingroup$ More generally, as Ben's answer notes, it's easy to do the point counts for varieties that have nice cell decompositions. So generalized flag varieties and toric varieties are other examples where there should be direct proofs. Maybe someone who knows Weil's original paper can comment: I thought that homogeneous spaces were one of his motivating examples? $\endgroup$– GSCommented Jan 27, 2010 at 10:03
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$\begingroup$ Weil did this on the last paragraph of the paper where he states the Weil conjectures. $\endgroup$– Felipe VolochCommented Jan 27, 2010 at 16:00
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$\begingroup$ Reference ams.org/bull/1949-55-05/S0002-9904-1949-09219-4/home.html $\endgroup$– Felipe VolochCommented Jan 27, 2010 at 16:02
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2 Answers
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The first result on the google search "zeta function of grassmannian" seems to contain quite a direct and not too long derivation of the zeta function for a grassmannian over a finite field:
http://www.math.mcgill.ca/goren/SeminarOnCohomology/GrassmannVarieties%20.pdf
From the zeta you see that it is rational, of course get the zeros (which are none), but you don't immediately get confirmation of the functional equation. Though, from the very simple combinatoric representation of the zeta function, it might be easy to prove directly, I will try with pen and paper later.
I'm glad I searched this, I didn't know the zeta was so simple in this case as well
answered Nov 24, 2009 at 13:59
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$\begingroup$ A question that just occurred to me: There exist Grassamnians that are not projective varieties? if so, does the formulation of the Weil conjectures still trouble free? $\endgroup$ Commented Nov 24, 2009 at 14:05
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1$\begingroup$ @John: What do you mean? The usual Grassmanians (subspaces of a fixed dimension in a vector space of a given dimension) is always projective. $\endgroup$ Commented Nov 24, 2009 at 14:07
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1$\begingroup$ A projective variety is a closed subvariety of a projective space. $\endgroup$ Commented Nov 24, 2009 at 14:14
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2$\begingroup$ It's not obvious, but it's true. Proof 1: Grassmannian's are proper, so the image of the Plucker embedding is closed. Proof 2: There are classic, explicit equations called the Plucker relations which cut out the Grassmannian. For a detailed and elementary proof that the Grassmannian is exactly the zeroes of the Plucker equations, see Ezra Miller and Bernd Sturmfels' book. $\endgroup$ Commented Nov 24, 2009 at 14:17
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1$\begingroup$ @John: what Wikipedia means is that the simplest grassmanian which is not isomorphic to a projective space is $G(4,2)$, not that that is the simpler grassmanian which is not a projective variety (in which case we should correct it!) $\endgroup$ Commented Nov 24, 2009 at 18:19
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Yes. Both cohomology and number of points are readily determined by looking at the Schubert cells (a cell of dimension k contributes one dimension to $H^{2k}$, and $q^k$ to the point count) and they match.
In fact, it's very easy to check Weil's conjecture directly for any smooth variety which has a decomposition into cells.
answered Nov 24, 2009 at 13:55
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$\begingroup$ in the sense of a CW decomposition, I think, when you work over the complex numbers $\endgroup$ Commented Nov 24, 2009 at 16:47
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$\begingroup$ But doesn't every smooth manifold have a CW decomposition by Morse theory? $\endgroup$ Commented Nov 24, 2009 at 21:02
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1$\begingroup$ The cells are affine spaces inside the variety, which stratify it, not CW cells. For example $\mathbb{P}^n = \mathbb{C}^n \cup \mathbb{C}^{n-1} \cup \ldots \cup \mathbb{C}^0$. $\endgroup$– mdelandCommented Nov 24, 2009 at 21:34
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$\begingroup$ Kevin- Think about an elliptic curve. $\endgroup$– Ben Webster ♦Commented Nov 24, 2009 at 21:35