Re: bounding cohomology of tensor products

On 11.04.2008 16:47, Fabrice Rosay wrote:
Hello,
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?

I think it is difficult to say something like this for a pair (F,G) of 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 latter assumption is equivalent to that there is an ample line bundle L on X. Now consider the sequence F(n) = F (x) L^{(x)n} of the n-th twist of the coherent O_X-module F w.r.t. L (n>=0) ["(x)" denotes the tensor product over the structure sheaf O_X of X].

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 \chi(F) is equal to dim(X) and the leading coefficient of f_F is a the form rk(F).deg(L)/(d!).

Here rk(F) denotes the rank of F, i.e. the dimension of the stalk of F in the generic point of X. Note that rk(F(x)G) = rk(F).rk(G) . If F is locally free - as you do -, rk(F) is simply the rank of F in every point of X. deg(L) is a positive integer associated to the line bundle L, called the degree of L.

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 (F(x)G)(n).

HTH.

--
Best wishes,
J.

.

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