most recent 30 from http://mathoverflow.net 2013-05-21T09:47:17Z http://mathoverflow.net/feeds/user/7935 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/105058/tates-thesis-for-varieties-over-finite-fields Tom Price 2012-08-19T21:55:34Z 2012-08-21T02:52:46Z

<p>Tate showed that the functional equation for zeta functions of number fields can be proven with fourier-analytic methods on the adele ring. Can the same be done for zeta functions of varieties over finite fields?</p>

http://mathoverflow.net/questions/52479/effect-of-tensor-product-on-euler-characteristic-of-line-bundles Tom Price 2011-01-19T06:48:03Z 2011-01-19T13:33:54Z

<p>Suppose X is a curve.</p> <p>Under sufficiently nice conditions we have that every line bundle on X corresponds to an equivalence class of divisors modulo principal divisors, with tensor product of bundles corresponding to addition of divisors.</p> <p>Given a line bundle L on X, I will call the Euler Characteristic of L minus the Euler characteristic of the trivial line bundle the cohomological degree of L. By the first half of the Riemann Roch theorem, we have that this is equal to the degree of the equivalence class of divisors corresponding to L. Thus cohomological degree is a homomorphism from the Picard group of X to Z.</p> <p>Is there a more direct proof of the fact that cohomological degree is a homomorphism, that does not go through divisors and the Riemann-Roch theorem? Hopefully this is just a matter of homological algebra.</p> <p>Is there a similar, cohomological definition of degree that works in higher dimensions?</p> <p>Thank you.</p>

http://mathoverflow.net/questions/33476/when-is-the-yoneda-product-graded-commutative Tom Price 2010-07-27T05:57:40Z 2010-07-29T20:45:08Z

<p>Sometimes, given an object A in an Abelian category, the Yoneda product on Ext(A, A) is graded-commutative, for example in cases where it coincides with the cup-product in singular cohomology. Are there any nice theorems about when the Yoneda product is graded-commutative in general? Thanks in advance.</p>