Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Let $X$ be a smooth complex surface of general type and $m>0$ an integer and $\eta\in Pic^0(X)$. If $X'$ is a smooth surface birational to $X$, then it is easy to show that $h^0(mK_X+\eta)=h^0(mK_{X'}+\eta)$ (I'm using the fact that $Pic^0(X)$ and $Pic^0(X')$ are canonically isomorphic). If one takes $X'$ to be the minimal model of $X$, then for $m>1$ by Kawamata-Viehweg vanishing $h^0(mK_X+\eta)=\chi(mK_X)$ is independent of $\eta\in Pic^0(X)$.

This argument does not work if $\dim X>2$ because minimal models in general are not smooth. My question is whether it is nevertheless true that $h^0(mK_X+\eta)$ is independent of $\eta$ for $m>1$.

I'm interested mainly in the case when the Albanese map of $X$ is generically finite.

ADDED: I would like to add some motivation. If $X$ is a variety with generically finite Albanese map and such that $\chi(K_X)>0$ the {\em paracanonical system} of $X$ (i.e. the family of effective divisors algebraically equivalent to $K_X$) contains a {\em main component}, that dominates a component of $Pic(X)$. This main component is a classical object of study.

As explained above, in the case of surfaces the analogous construction for $mK_X$, $m>1$ is ``boring'', since it gives a $\mathbb P^k$-bundle over acomponent of $Pic(X)$. So I am wondering whether for $\dim X>2$ things are similar, or, conversely, there is some interesting geometry also for $m>1$.

share|improve this question

1 Answer 1

Actually, I think the above argument always works because the minimal model has mild singularities (or one can reason directly on $X$; see ref below).

Let $X$ be a smooth variety of general type and $X'=Proj R(K_X)$ its canonical model (which exists by BCHM). Since $X'$ has rational singularities we have a natural identification $Pic ^0(X)\cong Pic ^0(X')$. For any $m\geq 1$ it is easy to see that $H^0(mK_X+\eta)\cong H^0(mK_{X'}+\eta)$ (after replacing $X$ by a birational model, we may assume that $f:X\to X'$ is a morphism and $K_{X}=f^*K_{X'}+E$ where $E\geq 0$ is an effective exceptional $\mathbb Q$-divisor....). By Kawamata-Viehweg vanishing (2.17 of arXiv:9601026 Singularities of Pairs by J. Koll\'ar) we have $h^i(mK_{X'}+\eta)=0$ for any $i>0$, $m\geq 2$ and $\eta \in Pic ^0(X')$ so that $h^0(mK_{X'}+\eta )=\chi (mK_{X'}+\eta)=\chi (mK_{X'})$. See 2.12 of Hacon-Pardini J. reine angew. Math. 546 (2002) 177-199 for a related statement in any Kodaira dimension where the proof is done on the smooth variety $X$.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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