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Let X be a smooth projective variety over the complex numbers, of dimension at least two. $D$ is an ample divisor on X. Then we know for $m>>0$, $H^i(mD)=0$. Now suppose $E$ is another divisor that is algebraiclly equivalent to $mD$, i.e. the line bundle $\mathcal{L}(E-mD)\in Pic^0(X)$ lies in the identity component of of Picard variety of X. Then is it true that even if they are not linearly equivalent, we still have $dim H^0(\mathcal{L}(E))=dim H^0(\mathcal{L}(mD))$, for $m>>0$ ?

Since the Euler character is a topological invariant, we know $\chi(\mathcal{L}(E))=\chi(\mathcal{L}(mD))$. Therefore if we know $H^i(\mathcal{L}(E))=0$ for $i>0$, we are done. However it is not obvious to me if that is true or not.

Some of my mumbling which may or may not be related:

For a general line bundle $\mathcal{L}\in Pic^0(X)$, if $\mathcal{L}\neq \mathcal{O}_X$, then $H^0(\mathcal{L})=0$ since for effective divisor in a projective variety we have a notion of degree, see Hartchorne chap II Exer 6.2. But I don't know if $H^i(\mathcal{L})=0$ or not.

In a series of paper by Green and Lazarsfeld, they looked at the case where X is compact Kahler, not necessarily projective, the behavior where $\mathcal{L}\in Pic^0(X)$ but $H^i(\mathcal{L})\neq 0$. see paper. But I don't know how to use that to construct an example where $E\sim_{alg}mD$ but $H^i(\mathcal{L}(E))\neq 0$ for $i>0$, or $H^0(\mathcal{L}(E))\neq H^0(\mathcal{L}(mD))$.

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1 Answer 1

up vote 4 down vote accepted

What you want follows easily from the Kodaira vanishing theorem:

If $m$ is sufficiently large then $mD - K$, where $K$ is the canonical divisor, is ample (this is true for $K$ replaced by any divisor). Since ampleness is preserved by algebraic equivalence, for example by Kleiman's criterion, it follows that $E - K$ is also ample if $E$ is algebraically equivalent to $mD$. Kodaira's vanishing theorem then implies that $H^i(\mathcal{L}(E)) = 0$ for $i>0$.

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Thanks, I realized this is a stupid question earlier today... – Ying Zhang Sep 9 '11 at 18:14

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