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A theorem by Kollár asserts that if $X$ and $Y$ are projective varieties with $X$ smooth, and $f : X \to Y$ is a surjective map, then the higher direct images $R^if_*\omega_X$ vanish for $i$ greater than the generic fiber dimension. I'd like to know if one can weaken the assumption that $X$ is smooth, for example, to $X$ having quotient singularities, or something even weaker (whereby the sheaf $\omega_X$ represents an appropriate dualizing sheaf in such a case).

Kollár's original proof uses the fact that $X$ is smooth to prove that the higher direct images $R^if_*\omega_X$ are torsion-free which I suspect may fail for $X$ having singularities.

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  • $\begingroup$ There are definitely situations where the vanishing still holds if $X$ and $Y$ are not smooth. An easy example is when $f$ is flat. But maybe you are interested only in assumptions on $X$ that make this true for every $f \colon X \to Y$ with $Y$ smooth projective (say)? $\endgroup$
    – R. van Dobben de Bruyn
    Commented Feb 7, 2018 at 0:47
  • $\begingroup$ Thanks, Remy. For my purposes I'd really like to take $Y$ to be singular, and let $X$ be a partial resolution of the singularities of $Y$, so that in particular, $f$ is a birational map, but $X$ may still admit singularities itself. $\endgroup$
    – Mitchell Faulk
    Commented Feb 7, 2018 at 1:01
  • $\begingroup$ Good job on correcting the spelling! ;-) $\endgroup$
    – Sándor Kovács
    Commented Feb 9, 2018 at 17:13

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If $X$ has rational singularities, then this vanishing (and the torsion-freeness as well) follows almost trivially. Let $g:Z\to X$ be a resolution of singularities. Then $Rg_*\omega_Z\simeq \omega_X$ and hence $R^if_*\omega_X\simeq R^i(f\circ g)_*\omega_Z=0$ for $i>\dim X-\dim Y$ by Kollár's theorem (I suppose that's what you meant, not what you wrote, right?! ;-)

Quotient singularities are rational, so this includes for what you are hoping. Also, since rational singularities are Cohen-Macaulay there is indeed a dualizing sheaf and it plays nice.

p.s.: If I remember correctly, Kollár also mentions this in his paper(s).

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  • $\begingroup$ Also, sorry for the follow-up question, but your parenthetical remark suggests I may have formulated my original statement incorrectly? I was trying to follow as closely as possible the formulation written by Koll\'{a}r in Theorem 2.1(ii) of ``Higher Direct Images of Dualizing Sheaves I.'' $\endgroup$
    – Mitchell Faulk
    Commented Feb 7, 2018 at 16:03
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    $\begingroup$ Your formulation was fine. I just meant that Kollar$\neq$Kollár. :) $\endgroup$
    – Sándor Kovács
    Commented Feb 7, 2018 at 20:42

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