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? 



$H^1$ of a certain line bundle on the resolution of  a 2-dimensional hypersurface singularity
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](https://projecteuclid.org/journals/kyoto-journal-of-mathematics/volume-51/issue-1/Du-Bois-pairs-and-vanishing-theorems/10.1215/0023608X-2010-020.full), 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](http://www.numdam.org/item/?id=BSMF_1981__109__41_0) for this.
It is easy to see (or use Lemma 2.1 of [this paper](https://www.ams.org/journals/jams/2010-23-03/S0894-0347-10-00663-6/home.html)) 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.


