Given a morphism of algebraic stacks $f: X \to Y$ (possibly non-representable) what sufficient conditions will guarantee that the derived push-forward of the structure sheaf on $X$ is isomorphic to the structure sheaf on $Y$? I am looking for a stacky version of a result of Buch-Mihalcea (Thm 3.1 of Quantum K-theory of Grassmannians https://arxiv.org/abs/0810.0981, which itself is a variation of a result of Kollar)

Thm 3.1 of BM: Let $f : X \rightarrow Y$ be a surjective map of projective varieties with rational singularities. Assume that the general fiber of $f$ is rational, i.e. $f^{−1}(y)$ is an irreducible rational variety for all closed points in a dense open subset of $Y$. Then $f_∗[\mathcal{O}_X]=[\mathcal{O}_Y]$ in the K-theory of Y.

I am happy to assume, in any proposed generalization, that the stacks are proper DM stacks (even smooth) and that the general fiber is an irreducible rational variety.

  • 2
    $\begingroup$ The stack structure on $Y$ is irrelevant: if $Y$ is a smooth DM stack, then there exists a smooth surjective morphism from a smooth scheme to $Y$. You can base change your original morphism by this smooth cover. So assume that $Y$ is a scheme. Then the morphism $f$ factors through the coarse moduli space of $X$, $p:X\to |X|$. In characteristic $0$ (or assuming tame), $Rp_*\mathcal{O}_X$ is the structure sheaf of $|X|$. Then, if $|f| : |X|\to Y$ satisfies the hypotheses of Buch-Mihalcea, then $[Rf_*\mathcal{O}_X]$ equals $[R|f|_*\mathcal{O}_{|X|}]$, and this equals $[\mathcal{O}_Y]$. $\endgroup$ – Jason Starr Jan 15 '17 at 10:48
  • $\begingroup$ The example I wrote was wrong! I will try to find a correct example, e.g., a quartic surface with a worse singularity. $\endgroup$ – Jason Starr Jan 15 '17 at 19:33
  • $\begingroup$ Oops, there are no counterexamples . . . posting answer below. $\endgroup$ – Jason Starr Jan 15 '17 at 20:00

I am amplifying the above comments. The theorem of Buch-Mihalcea also holds for stacks.

Let $k$ be a field of characteristic $0$. Let $Y$ be a Deligne-Mumford stack that is finite type over $k$ and that is normal: there exists a surjective, étale morphism $h:\widetilde{Y}\to Y$ with $\widetilde{Y}$ a normal scheme. Let $f:X\to Y$ be a proper, surjective morphism of Deligne-Mumford stacks. Assume that $X$ is integral, and assume that $X$ has rational singularities: there exists a surjective, étale morphism $g:\widetilde{X}\to X$ such that $\widetilde{X}$ is a $k$-scheme that has rational singularities. Finally, assume that the geometric generic fiber of $f$ is isomorphic (as a stack) to a smooth proper variety that is rationally connected, or even just $\mathcal{O}$-acyclic.

Proposition. The natural map $\mathcal{O}_Y\to f_*\mathcal{O}_X$ is an isomorphism, and $R^q f_*\mathcal{O}_X$ equals $0$ for every $q>0$.

Because $h$ is flat, $h^*R^qf_*\mathcal{O}_X$ equals $R^q \widetilde{f} \mathcal{O}_{X\times_Y \widetilde{Y}}$, where $\widetilde{f}:X\times_Y \widetilde{Y} \to \widetilde{Y}$ is the projection. Thus, without loss of generality, assume that $Y$ is a normal scheme. As in the comment, there exists a coarse moduli space $p:X \to |X|$. The morphism $f$ factors as the composition of $p$ and a surjective, proper $k$-morphism, $$|f|:|X|\to Y.$$ Because the characteristic is $0$, $X$ is a tame stack. Thus, $\mathcal{O}_{|X|} \to p_*\mathcal{O}_X$ is an isomorphism and $R^qp_*\mathcal{O}_X$ equals $0$ for all $q>0$. Therefore, it suffices to prove that $\mathcal{O}_Y\to |f|_*\mathcal{O}_{|X|}$ is an isomorphism and $R^q|f|_*\mathcal{O}_{|X|}$ equals $0$ for all $q>0$. Via Buch-Mihalcea, it suffices to prove that $|X|$ has rational singularities. This follows from Proposition 5.13, p. 157 of "Birational Geometry of Algebraic Varieties" by Kollár and Mori.

  • $\begingroup$ Jason Starr, That's excellent, thanks! Did you drop the projectivity hypotheses in BM on purpose, or was that an oversight? $\endgroup$ – Chris Woodward Jan 15 '17 at 22:09
  • $\begingroup$ Because of Chow's lemma and existence of a projective resolution of singularities (in characteristic $0$), the version with $|f|$ proper follows from the version with $|f|$ projective. $\endgroup$ – Jason Starr Jan 15 '17 at 22:35
  • $\begingroup$ To be clear, I think Jason also meant to require that Y has rational singularities (as in the original question). $\endgroup$ – Nick Addington Jul 23 '19 at 23:30

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