> 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).