most recent 30 from http://mathoverflow.net 2013-05-19T01:02:50Z http://mathoverflow.net/feeds/question/6685 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/6685/weil-conjectures-for-grassmannians John McCarthy 2009-11-24T13:34:27Z 2010-01-27T09:20:27Z

<p>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?</p>

http://mathoverflow.net/questions/6685/weil-conjectures-for-grassmannians/6687#6687 Ben Webster 2009-11-24T13:55:49Z 2009-11-24T15:30:41Z

<p>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 $H^k$, and $q^k$ to the point count) and they match.</p> <p>In fact, it's very easy to check Weil's conjecture directly for any smooth variety which has a decomposition into cells.</p>

http://mathoverflow.net/questions/6685/weil-conjectures-for-grassmannians/6688#6688 Dror Speiser 2009-11-24T13:59:32Z 2009-11-24T19:34:15Z

<p>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:</p> <ul> <li><a href="http://www.math.mcgill.ca/goren/SeminarOnCohomology/GrassmannVarieties.pdf" rel="nofollow">http://www.math.mcgill.ca/goren/SeminarOnCohomology/GrassmannVarieties.pdf</a></li> </ul> <p>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.</p> <p>I'm glad I searched this, I didn't know the zeta was so simple in this case as well.</p>