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Let $X$ be a separated connected scheme of characteristic $p > 0$. I am going to assume that $F : X \to X$ (the absolute Frobenius) is a finite map. This condition is called being $F$-finite.

Question Suppose that $X$ has a dualizing complex $\omega_X^{\bullet}$. Is $F^! \omega_X^{\bullet}$ quasi-isomorphic to $\omega_X^{\bullet}$? In other words, is $R Hom_R^{\bullet}(F_* O_X, \omega_X^{\bullet})$ isomorphic to $F_* \omega_X^{\bullet}$ as $F_* O_X$-modules?

Special Cases If $X$ is of finite type over an $F$-finite field $k$, then this is true. One has the commutative diagram involving the absolute Frobenius on the base field and $X$, and one notices that $F^! k = Hom_k(F_* k, k)$ is abstractly isomorphic to $F_* k$ and the result follows.

In general of course, $F^! \omega_X^{\bullet}$ is still a dualizing complex, so it agrees with the original up to shift on connected components or twist by line bundles (and it's obvious that there are no shifts because one can work locally/generically, so the only question is might one obtain a twist by a line bundle).

I don't know it in the case that $X$ is CM (in which case some statements become easier to digest). I would be interested even in the normal case, which should reduce to the CM case.

Remark According to Gabber, the assumption that $X$ has a dualizing complex is implied by $X$ being $F$-finite (see Remark 13.6 in O. Gabber: Notes on some t-structures, Geometric aspects of Dwork theory. Vol. I, II, Walter de Gruyter GmbH & Co. KG, Berlin, 2004, pp. 711–734). He provides a proof in the case that $X$ is affine.

Background I've talked to a few people about this question over the past couple years (in particular, Wenliang Zhang, who, iirc, first brought this question to my attention and has said he doesn't mind if I ask it here, as well as Gennady Lyubeznik, Joe Lipman, Shunsuke Takagi, and Manuel Blickle) but as far as I know, no one I've asked knows how to do it or anything like it. Has anyone ever run into this question or a situation that looks similar?

EDIT (Feb. 23, 2011): It's been a while since I asked this, but I thought I would point out the following. In general, there are choices of dualizing complexes $\omega_X^{\bullet}$ for $X$ such that $F^! \omega_X^{\bullet}$ is not isomorphic to $\omega_X^{\bullet}$. For example, choose any non-torsion line bundle $\mathcal{L}$, then $\omega_X^{\bullet}$ and $\omega_X^{\bullet} \otimes \mathcal{L}$ are both dualizing complexes. Then $$F^! (\omega_X^{\bullet} \otimes \mathcal{L}) = F^!(\omega_X^{\bullet}) \otimes F^* \mathcal{L} = F^!(\omega_X^{\bullet}) \otimes \mathcal{L}^p.$$ So in particular, $\omega_X^{\bullet} \otimes \mathcal{L}$ will NOT satisfy the desired conclusion for an arbitrary choice of $\mathcal{L}$. In particular, it should really only work for one dualizing complex up-to-shifting (philosophically speaking anyways).

I still have no idea how to do this but I thought maybe this might stir someone's memory?

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Karl, line bundle and shift come from this: if $R$ and $R'$ are dualizing & $X$ conn'd then $\mathbf{R}\mathcal{H}om(R,R')$ in $D^b_c(X)$ is isom. to $\mathcal{L}[n]$ for inv'tble $\mathcal{L}$ and integer $n$ (shifting is Zariski-local, could vary on conn'd comp's). In your setup (no shift), $\mathbf{R}\mathcal{H}om(R,F^{!}R)$ is inv'tble $O_X$-mod, want trivial. Apply $F_{\ast}H^0$ and finite duality to get $\mathcal{H}om(F_{\ast}R,R)$ is inv'tble $F_{\ast}O_X$-mod, want trivial. Since $R$ has finite inj. dim., seek $t:F_{\ast}R \rightarrow R$ generating Hom over completions. Candidate?? –  BCnrd Jul 21 '10 at 13:47
    
Brian, of course, I'm always thinking connected (i'll add a comment to that effect). You are right, I should seek such a $t$. In practice, I'm often working in a case where $X$ is finite type, so I can identify the map you suggest with the dual of Frobneius / ie the Cartier map (which as you point out, is equivalent to what I want). I'll have to think about finding such a map. Thanks! –  Karl Schwede Jul 21 '10 at 15:30
    
Karl, in case it's not yet irrelevant (nor too vague to be useful): what about working with a rigid dualizing complex a la Yekutieli-Zhang? –  Thomas Nevins Jun 10 '11 at 20:23
    
Tom, I'm not familiar with these. I'll take a look, thanks! –  Karl Schwede Jun 11 '11 at 1:50
    
If you are going to look at rigid dualizing complexes a la Y-Z, you might want to look at the work of Avramov-Iyengar-Lipman that (if I remember correctly) gives vast generalizations with simpler proofs. –  Sean Sather-Wagstaff Dec 1 '11 at 2:38
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