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

Let $X = (f=0) \subset \mathbb{C}^3$ be an isolated hypersurface singularity and $\mu: \tilde{X} \rightarrow X$ be a resolution of a singularity whose exceptional locus $E$ is simple normal crossing. Then

Question Is $H^1(\tilde{X}, \mathcal{O}_{\tilde{X}}(- E)) =0$ ?

I think it's OK if it is Du Bois. How about in general?

share|improve this question

1 Answer 1

up vote 4 down vote accepted

If $X$ is a seminormal surface, then actually this vanishing is equivalent to $X$ being Du Bois:

1) If $X$ is Du Bois, the vanishing is a direct consequence of Cor.6.2 of this paper, but in this case it is actually pretty easy to prove directly, see #2.

2) Let $P\in X$ be the singular point. There is a standard distinguished triangle $$ \underline\Omega_X^0 \to \mathscr O_P \oplus R\mu_*\mathscr O_{\widetilde X}\to R\mu_*\mathscr O_E\overset{+1}\to.$$

See 4.11 of Du Bois's paper for this. It is easy to see (or use Lemma 2.1 of this paper) that this implies that there exists a distinguished triangle $$R\mu_*\mathscr O_{\widetilde X}(-E)\to \underline\Omega_X^0 \to \mathscr O_P\overset{+1}\to$$

Now observe that $h^0(\underline\Omega_X^0)$ is the seminormalization of $\mathscr O_X$ and the outside terms have no cohomology higher than $h^1$, so the only non-trivial map is $$R^1\mu_*\mathscr O_{\widetilde X}(-E)\to h^1(\underline\Omega_X^0)$$ This is an isomorphism, because $\mathscr O_P$ has very little cohomology. It follows that $X$ is Du Bois if and only if $R^1\mu_*\mathscr O_{\widetilde X}(-E)=0$. $\quad\square$

Remark It is essential in this proof that $X$ is a surface. In higher dimensions it is true that Du Bois implies the desired vanishing, but I don't think the opposite implication holds. See #1 and my second comment below.

share|improve this answer
    
Just a quick comment, Steenbrink's paper Mixed hodge structures and isolated singularities also studies this condition a little (in fact, it is very close to how Steenbrink interpreted Du Bois singularities). –  Karl Schwede Nov 25 '11 at 1:33
    
Karl, yes. At first I thought of just quoting Steenbrink, but I don't think he explains why, just kind of says that it's true. I thought it might be more useful if I explain the simple proof. –  Sándor Kovács Nov 25 '11 at 8:48
    
Thank you very much for the quite helpful answer. Actually, I was reading that paper of Steenbrink a bit and I couldn't find the condition on $H^1$ is necessary and sufficient to be Du Bois. Maybe I should learn more. –  tarosano Nov 25 '11 at 10:00
2  
Steenbrink says that for an isolated normal singularity being Du Bois is equivalent to the restriction maps $R^i\mu_*\mathscr O_{\widetilde X}\to R^i\mu_*\mathscr O_E$ being isomorphisms for all $i>0$. In the case of a surface this is equivalent to the vanishing of $R^1\mu_*\mathscr O_{\widetilde X}(-E)$. –  Sándor Kovács Nov 25 '11 at 10:08

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.