2
$\begingroup$

Norimatsu's lemma says that on a smooth projective complex variety $X$ of dimension $n$, then we have $H^i(X,\mathcal O_X(-A-E))=0$ for $i < n$ when $A$ is an ample divisor and $E$ is a simple normal crossings (SNC) divisor. Does this statement remain true if $E$ is an effective divisor with SNC support? In other words, if $\sum E_i$ is SNC, do we still have these vanishings for $\mathcal O_X(-A-\sum a_i E_i)$ where $a_i >0$? Or is there a counterexample?

$\endgroup$

2 Answers 2

5
$\begingroup$

Here is a counterexample.

Let $X$ be a smooth cubic surface in $\mathbf{P}^3$ and $E$ a line on it; moreover take $A=-K_X$.

Then by Serre duality $$H^1(X, -A-aE)=H^1(X, K_X + A + aE)=H^1(X, aE).$$

On the other hand $h^0(X, aE)=1$ since $E^2=-1$ and $h^2(X, aE)=h^0(X, K_X-aE)=0$ since $K_X$ is not effective.

Now Riemann-Roch yields $$\chi(X, aE)=\frac{aE(aE-K_X)}{2}+\chi(\mathcal{O}_X)=\frac{-a^2+a}{2}+1,$$ hence $$h^1(X, aE)=\frac{a(a-1)}{2}$$ which is not zero for $a \geq 2$.

$\endgroup$
0
7
$\begingroup$

Let $X$ be an arbitrary smooth projective surface, $A$ an arbitrary ample divisor and $E\subset X$ a smooth proper curve. Consider the short exact sequence $$ 0\to \mathscr O_X(-A-(a+1)E)\to \mathscr O_X(-A-aE)\to \mathscr O_X(-A-aE)|_E\to 0 $$ If $H^0(X, \mathscr O_X(-A-aE))=0$, then $H^1(X, \mathscr O_X(-A-(a+1)E))= 0$ only if $H^0(E, \mathscr O_X(-A-aE)|_E)= 0$. Therefore, if $(A+aE)\cdot E\ll 0$, then the desired vanishing will fail.

So, for example if $E$ is such that $E^2<0$, then this will happen for $a\gg 0$.

If, as in Francesco's example, $X$ is a Del Pezzo surface, $A=-K_X$, and $E$ is an exceptional curve, then $(-A-aE)\cdot E=K_X\cdot E -aE^2=-1+a$ and hence $\mathscr O_X(-A-aE)|_E\simeq \mathscr O_E(a-1)$, so if $a>0$, then $H^0(E, \mathscr O_X(-A-aE)|_E)\neq 0$ and hence this gives another proof that in Francesco's example the desired vanishing fails as soon as the coefficient of $E$ is at least $2$.


A similar example works in higher dimensions as long as $-E|_E$ is ample, for instance if it is the exceptional divisor of the blow up of a point (probably even under more general conditions). In other words, you cannot expect the vanishing you would like unless $E$ has some semi-positivity property (say $E$ is nef), but in that case you probably get it directly (since for example if $E$ is nef, then $A+aE$ is ample).

$\endgroup$
0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.