# Bound of dimension of $H^1$ of certain line bundle

Let $X$ be a smooth variety over a field $k=\bar k$, $\rm char$ $k=0$. Assume that $\mathcal{L}$ is a nef and big line bundle. Is there some bound saying that $$h^1(\mathcal{L}^{\otimes n}) \le f(n)$$ where $f(n)$ is a function only depending on $n$? For example, a polynomial.

And how about the case when $\rm char$ $k>0$?

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I remember this kind of things was studied by Alex Kuronya. Did you check his papers? – Hailong Dao Oct 11 '11 at 2:25
I have not yet. But I will try to check it soon. Thanks! – Tong Oct 11 '11 at 2:32

## 1 Answer

In general, for any line bundle $L$ over $X$, you have $h^i(X, L^{\otimes n})\leqslant C m^n$ where $m=\dim X$. (write $L$ as the difference of two very ample line bundles, and use the usual restriction exact sequence)

In the case where $L$ is nef, then one can say more: $h^i(X,L^{\otimes n})\leqslant C m^{n-i}$ using Fujita's theorem (cf Lazarsfeld, Positivity in Algebraic Geometry, 1.2.29 "growth of cohomology").

But in the case $i=1$, you don't need Fujita, and this can be done more basically. (see e.g Debarre's "Higher dimensional Algebraic Geometry", Proposition 1.31 p21)

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Is this result independent of characteristic of the base field? – Tong Oct 11 '11 at 16:12
As this is a formal consequence of the restriction exact sequence and the definition of intersection numbers, I would tend to say yes. However, I almost never work over algebraically closed fields with positive characteristic, so I might be missing something. – Henri Oct 11 '11 at 22:02