I am trying to find a more elementary proof of [GKKP-DIFFERENTIAL FORMS ON LOG CANONICAL SPACES, Corollary 13.4] for rational singularities, avoiding Du Bois pairs. When $X$ is a rational singularity the Corollary 13.4 should follow directly from the definition of a rational singularity (cf. [GKKP, page 37 (above Thm. 13.3)]). My idea was to use the short exact sequence

$ 0\rightarrow \mathcal O_Y(-E)\rightarrow \mathcal O_Y\rightarrow \mathcal O_E\rightarrow 0$

where $\pi:Y\rightarrow X$ is a resolution of singularities and $E$ is the reduced exceptional snc divisor. However, this only yields an isomorphism

$R^{n-2}\pi_*\mathcal O_E\cong R^{n-1}\pi_*\mathcal O_Y(-E)$.

Does someone have a different idea or does someone know how to improve my approach?

Thanks for your help. Sincerely,


  • 3
    $\begingroup$ it would be helpful if you wrote down the statement you are trying to prove $\endgroup$
    – pro
    Commented Aug 14, 2015 at 16:17

1 Answer 1


So, for those, who don't know all the statements of the cited paper by heart, the mentioned corollary states that in the situation described the following vanishing holds: $$R^{n-1}\pi_*\mathcal O_Y(-E)=0.$$

This is almost automatic for rational singularities of surfaces by your suggested approach. In that case you actually do not get the isomorphism that you claim but that $R^{n-1}\pi_*\mathcal O_Y(-E)\simeq \mathrm{coker}[\pi_*\mathscr O_Y \to \pi_*\mathscr O_E]$, but that's why it works, since the latter is indeed $0$.

As Karl points out in his comment it is also pretty easy for isolated rational singularities in arbitrary dimension. The essence of the argument is that if the singularity is a deformation retract of a neighbourhood, then the restriction map on (singular) cohomology from $Y$ to $E$ is surjective. Using Hodge theory this means that the same is true for the natural morphism $$ R^i\pi_*\mathscr O_Y \to R^i\pi_*\mathscr O_E $$ and you get the vanishing you need for your argument to work. For more details see Lemma 2.14 in Mixed Hodge structures associated with isolated singularities by Steenbrink.

One should note that here we are actually using the fact that normal crossings singularities are Du Bois (in the Hodge theoretic part), but I accept that this is indeed simpler than using Du Bois singularities in general.

In higher dimensions in general I am not sure how to make this work. Perhaps by some clever use of general hyperplane sections you can prove that the support of this sheaf is zero-dimensional, but I don't know how to do more.

There is a good reason we used Du Bois singularities in the paper and instead of trying to avoid them, you should perhaps embrace them. ;)

In higher dimensions ($n>2$) in order to prove what you want you would need $R^{n-2}\pi_*\mathcal O_E$ to vanish. The $n-2$ here doesn't seem to allow a special treatment (unlike the $n-1$) this is probably as hard as proving this for all $i>0$ (or at least $i>n-3$ which in dimension $3$ is the same thing). Given that you assume that $X$ has rational singularities this is equivalent to the natural morphism $$ R^i\pi_*\mathscr O_Y \to R^i\pi_*\mathscr O_E $$ being an isomorphism. Now, then if $X$ has only isolated singularities this is equivalent to having Du Bois singularities.

Of course, we actually have a simple proof for isolated singularities, but my feeling is that the truth is that this vanishing really works because of the Du Bois property, which in the isolated case can be grasped easily and so in that case we can seemingly avoid it, but it is there anyway.

  • $\begingroup$ Maybe it's worth pointing out that for isolated singularities, one has the injection $R^{i} \pi_* \mathcal{O}_Y(-E) \hookrightarrow R^{i} \pi_* \mathcal{O}_Y$, then the vanishing you want holds very easily. See for instance Mixed Hodge structures associated with isolated singularities by Steenbrink for a topological argument. One can also deduce this by the degeneration of the Hodge-to-De Rham spectral sequence. $\endgroup$ Commented Aug 15, 2015 at 3:31
  • $\begingroup$ Karl, you are right. When I started I meant to say this. Then I got interrupted and forgot what I had planned. I guess I'm getting old... I'll edit the answer accordingly. $\endgroup$ Commented Aug 15, 2015 at 4:06

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