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?


1 Answer 1


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.

  • $\begingroup$ 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). $\endgroup$ Nov 25, 2011 at 1:33
  • $\begingroup$ 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. $\endgroup$ Nov 25, 2011 at 8:48
  • $\begingroup$ 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. $\endgroup$
    – tarosano
    Nov 25, 2011 at 10:00
  • 2
    $\begingroup$ 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 $R^1\mu_*\mathscr O_{\widetilde X}(-E)$. $\endgroup$ Nov 25, 2011 at 10:08

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