Re: bounding cohomology of tensor products

On 12.04.2008 13:18, Jannick Asmus wrote:
On 12.04.2008 12:56, Jannick Asmus wrote:
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).

Additionally, this implies that, for n >> 0,

dim_k \Gamma(X,(F(x)G)(n)) = rk(G) . dim_k \Gamma(X,F(n))

This holds for any coherent sheaves F and G with Supp(F) = Supp(G) = X or, equivalently, rk(F)>0 and rk(G)>0.

Thus if you allow F to be substituted by the n-th twist w.r.t. L and n high enough, I think you will have achieved what you wanted.

One more addition: Note that if L is ample, the homology vector spaces h^i(X,F(n)) (i>0) vanish for n>>0.

--
Best wishes,
J.

.

Relevant Pages

  • Re: bounding cohomology of tensor products
    ... 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 ... cohomology vector spaces of the two sheaves and eventually invariants ... I think it is difficult to say something like this for a pair of locally free sheaves only. ...
    (sci.math.research)
  • Re: bounding cohomology of tensor products
    ... 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 ... cohomology vector spaces of the two sheaves and eventually invariants ... I think it is difficult to say something like this for a pair of locally free sheaves only. ...
    (sci.math.research)
  • Re: Tensor Products of Vector Spaces: Need Help in Understanding
    ... >> Given Vector Spaces U and V of dimension m and n respectively, ... >> Defines the Tensor Product U*V as the Dual of the Vector Space of all ... >> Bilinear Forms defined on the Vector Space Direct Sum of U and V. I'am ... Given x belonging to U ...
    (sci.math)
  • bounding cohomology of tensor products
    ... I would like to know if it's possible to bound the dimension of the ... sheaves (on an algebraic variety) using only the dimension of the ... cohomology vector spaces of the two sheaves and eventually invariants ...
    (sci.math.research)
  • Re: Tensor Products of Vector Spaces: Need Help in Understanding
    ... > Given Vector Spaces U and V of dimension m and n respectively, ... > Defines the Tensor Product U*V as the Dual of the Vector Space of all ... > Bilinear Forms defined on the Vector Space Direct Sum of U and V. I'am ... Given x belonging to U ...
    (sci.math)
Loading