In Proposition 7.6 of his paper "Higher Direct Images of Dualizing Sheaves", Kollár shows that if $X,Y$ are smooth complex projective varieties and $f:X\rightarrow Y$ is a proper surjective morphism with connected fibers, then there is an isomorphism $R^df_{\ast}\omega_X \simeq \omega_Y$, where $d=\dim X-\dim Y$.

I was wondering whether this is true in positive characteristic. By Grothendieck duality one has $$Rf_{\ast}\mathcal{O}_X \simeq Rf_{\ast}R\mathcal{H}om(\omega_X^{\bullet},\omega_X^{\bullet}) \simeq R\mathcal{H}om(Rf_{\ast}\omega_X^{\bullet}, \omega_Y^{\bullet})[-d]$$ so taking 0-th cohomology we get $$f_{\ast}\mathcal{O}_X \simeq \mathcal{E}xt^{-d}(Rf_{\ast}\omega_X^{\bullet}, \omega_Y^{\bullet})$$

From the spectral sequence $$\mathcal{E}xt^p(\mathcal{H}^{-q}(\mathcal{F}^{\bullet}), \mathcal{G}^{\bullet}) \Longrightarrow \mathcal{E}xt^{p+q}(\mathcal{F}^{\bullet},\mathcal{G}^{\bullet})$$ it follows that the RHS is isomorphic to $\mathcal{H}om(R^df_{\ast}\omega_X,\omega_Y)$ and the LHS is just $\mathcal{O}_Y$ by the connected fibers assumption so we have $$\mathcal{O}_Y \simeq \mathcal{H}om(R^df_{\ast}\omega_X,\omega_Y)$$

Finally since $R^df_{\ast}\omega_X$ is locally free we conclude that $R^df_{\ast}\omega_X \simeq \omega_Y$.

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    $\begingroup$ Where did you get the $[-d]$ in the first displayed equation? Or alternatively you should drop the $^\bullet$ from the dualizing complexes. If you do that, then you end up with either $\mathscr Ext^0$ or $^{-d}$, but without the $^\bullet$'s. (cont'd) $\endgroup$ Feb 17, 2015 at 17:48
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    $\begingroup$ Isn't such vanishing the result of Chatzistamatiou and Rülling in "Higher direct images of the structure sheaf in positive characteristic", ANT 2011? $\endgroup$ Feb 18, 2015 at 17:56
  • $\begingroup$ @MatthieuRomagny: No. They handle the case when $d=0$ and even pose this as an open problem. $\endgroup$ Feb 18, 2015 at 23:41
  • $\begingroup$ ... and the case of $d=0$ of this question is straightforward. $\endgroup$ Feb 18, 2015 at 23:43
  • $\begingroup$ The case of $d = 0$ is even known in mixed characteristic and goes back at least to Lipman in his work in the 70s. I do have something to say about the question though which is probably too big to fit in a comment. $\endgroup$ Feb 19, 2015 at 3:32

1 Answer 1


So a paper of Blickle, myself and Tucker handles some questions really closely related to this, see the arXiv version here

Ok, throw away the connected fibers hypothesis, I'm going to assume my base field is $F$-finite (although we can also work with integral $F$-finite separated schemes instead of varieties).

Suppose that $f : X \to Y$ is a proper surjective morphism of varieties in characteristic $p > 0$. Then we have a map $R^d \pi_* \omega_X \to \omega_Y$. This map is never zero (see Prop 2.13 in the aforementioned paper).

The nice thing is that the image of this morphism in $\omega_Y$ always contains the parameter test submodule $\tau(\omega_Y) \subseteq \omega_Y$ (see Prop 2.21 and Definition 2.33). For smooth varieties, $\tau(\omega_Y) = \omega_Y$ (and even for varieties with mild singularities). It's worth remarking that this is in some sense optimal: For non-smooth varieties, we also show that there always exists some $X \to Y$ such that $\tau(\omega_Y)$ is the image of $R^d \pi_* \omega_X \to \omega_Y$.

Anyway, the upshot of the above discussion is that if $Y$ is smooth, then $R^d \pi_* \omega_X \to \omega_Y$ is surjective. Obviously without the connected fibers hypothesis, it can't be injective.

EDIT: Ok, so then we need the injectivity under the hypothesis of connected fibers (or maybe also $X$ smooth). In an earlier version of this ansewr I was incorrectly speculating that maybe we could use something like the argument in the question to help here. As Sándor points out, that won't work...

  • $\begingroup$ Karl, I meant to post an answer with essentially the same conclusion (without appealing to your test submodule). I don't see how that spectral sequence would help with the injectivity. That is equivalent to saying that $R^d\pi_*\omega_X$ is torsion-free (which is also shown by Kollár in char=0). The OP claims that it is locally free... I don't see why. $\endgroup$ Feb 19, 2015 at 19:20
  • $\begingroup$ BTW, you exchanged $X$ and $Y$ everywhere where there is an $R^d\pi_*$... $\endgroup$ Feb 19, 2015 at 19:20
  • $\begingroup$ Hi Sándor, thanks, and I also fixed the $X$ to $Y$ stuff. $\endgroup$ Feb 20, 2015 at 1:35

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