Groupes de discussion : sci.math.research
De : Jannick Asmus <jannick.n...@web.de>
Date : Sat, 12 Apr 2008 12:56:29 +0200
Date/heure locale : Sam 12 avr 2008 06:56
Objet : Re: bounding cohomology of tensor products
On 11.04.2008 16:47, Fabrice Rosay wrote:
> Hello, I think it is difficult to say something like this for a pair (F,G) of > I would like to know if it's possible to bound the dimension of the > cohomology vector spaces of a tensor product of two locally free > sheaves (on an algebraic variety) using only the dimension of the > cohomology vector spaces of the two sheaves and eventually invariants > of the variety? locally free sheaves only. But there is a way-out if you associate a sequence of locally free modules to each of F and G as follows. Let's assume that the k-variety X is irreducible and projective. The Then is is known that \chi(F)(n) := dim_k \Gamma(X,F(n)) is a polynomial in n (for n>>0). Since Supp(F)=X, the degree d of Here rk(F) denotes the rank of F, i.e. the dimension of the stalk of F With this in mind you get in your situation for n>>0 \chi(F(x)G)(n) = rk(F).rk(G).deg(L)/(d!) n^d + f(n) where f(X) is a polynomial of degree at most d-1. This can give you an upper bound of the dimensions for the sequence HTH. -- Vous devez vous connecter pour pouvoir envoyer des messages.
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