# Cohomologically trivial stacks

The following theorem of Serre is well-known:

A noetherian scheme $X$ is affine if and only if $H^i(X; \mathcal{F}) = 0$ for all quasi-coherent sheaves $\mathcal{F}$ on $X$ and all $i>0$. (Actually it is enough to have this for $i=1$ and all coherent ideal sheaves.)

I asked myself whether there is an extension of this theorem to (Artin/Deligne-Mumford) stacks. More precisely:

Question: Can one characterize the class of (Artin/Deligne-Mumford) stacks $X$ such that $H^i(X; \mathcal{F}) = 0$ for all quasi-coherent sheaves $\mathcal{F}$ on $X$ and all $i>0$?

It is certainly not true that affine schemes are here the only examples. For example, take a graded ring $A$ and consider $X = Spec A // \mathbb{G}_m$ (where the $\mathbb{G}_m$-action is induced by the grading). The category of quasi-coherent sheaves on $X$ is (by fpqc-descent) equivalent to that of graded $A$-modules and the global sections functor corresponds to taking the zeroth degree of such a graded module. This is clearly exact and thus all higher cohomology groups of all quasi-coherent sheaves vanish.

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More generally, the quotient of an affine scheme by a reductive group has this property (as the functor of invariants is exact). –  Sam Gunningham Nov 19 '12 at 22:57
@Sam: As Angelo points out below, you should replace "reductive" by "linearly reductive" or some other tameness condition. As is discussed, for instance, in the appendices of Mumford's "Geometric Invariant Theory", the functor of invariants is not exact for all reductive groups. –  Jason Starr Nov 20 '12 at 13:16

Suppose that $\mathcal X$ is an algebraic stack with finite inertia (for example, a separated Deligne-Mumford stack); then, by a well-known result of Keel and Mori, there exist a moduli space $\pi \colon \mathcal X \to M$. The stack $\mathcal X$ is called tame when $\mathrm R^i\pi_* F = 0$ for every quasi-coherent sheaf $F$ on $\mathcal X$ and every $i > 0$. From the definition it follows easily that tame stacks with affine moduli spaces have the property you require. In characteristic 0, an algebraic stack with finite diagonal is tame if and only if it is Deligne-Mumford.

There are several different characterizations of tame stacks; see the paper "Tame stacks in positive characteristic" by Dan Abramovich, Martin Olsson and myself. Using the results in that paper, it is not hard to show that a noetherian algebraic stack with finite inertia has the property you want if and only if it is tame with affine moduli space.

[Edit:] here is a proof that if a noetherian algebraic stack $\mathcal X$ with finite inertia has the property you want it is tame with affine moduli space. Let $\mathcal X \to M$ be the moduli space. Let $\mathcal G$ be the residual gerbe over a closed point of $M$; then $\mathcal G$ is closed in $\mathcal X$, so the cohomology of each quasi-coherent sheaf on $\mathcal G$ is trivial. The moduli space of $\mathcal G$ is the spectrum of a field, so $\mathcal G$ is tame. This implies that the automorphism group of an object of $\mathcal G$ is linearly reductive. One of the results in the paper implies that an open neighborhood of $\mathcal G$ in $\mathcal X$ is tame. Since every non-empty closed subset of $M$ contains a closed point of $M$, this implies that these open neighborhoods cover $\mathcal X$, so $\mathcal X$ is tame.

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Thanks for your answer! While I see why a tame stack with affine coarse moduli space has the property I asked for, I don't see the other direction. Can you give some hints? –  Lennart Meier Nov 20 '12 at 13:50

Let $\mathfrak{X}$ be a stack. Then the category $\mathrm{QCoh}(\mathfrak{X})$ (which you can define, say, as the homotopy limit of the categories of $R$-modules for every $R$-point of $\mathfrak{X}$, $R$ ranging over reasonably small rings -- this means that for each $R$-point of $\mathfrak{X}$ you get an $R$-module and these are compatible) is very good: it's presentable and filtered colimits are exact, and in particular (by a theorem of Grothendieck) it has enough injectives. It comes with a symmetric monoidal functor $\Gamma$ to the category of $\Gamma(\mathcal{O}_{\mathfrak{X}})$-modules, which preserves filtered colimits if $\mathfrak{X}$ is not too large.

Now you might want to say that $\Gamma$ is an equivalence. As I understand, a symmetric monoidal equivalence between quasi-coherent sheaves on sufficiently nice stacks (in particular, the diagonal should be affine) is necessarily induced by an equivalence of stacks. So under such hypotheses (which are anyway automatic if $\mathfrak{X}$ is indeed affine), the problem reduces to the following question: given an abelian category $\mathcal{A}$ and a functor $F: \mathcal{A} \to \mathrm{Mod}_R$ ($R$-modules for some ring $R$), when is it an equivalence?

This is a question you can solve using Morita theory. Namely, Morita theory says that essentially the defining property of the category of modules over a ring is that it has a compact, projective generator (for example, the ring itself). Whenever you have a (presentable) abelian category with such a compact projective generator $X$, there is induced an equivalence with the category of $R$-modules for some $R$: take $R = \hom(X, X)$ and the functor $\hom(X, \cdot)$ to induce the equivalence. Now, if $\mathfrak{X}$ is a noetherian stack with the property that all higher quasi-coherent cohomology vanishes, then the structure sheaf is compact and projective, and if $\mathcal{O}_{\mathfrak{X}}$ is a generator Morita theory tells you that $\Gamma$ is an equivalence. I believe that this argument is due to Knutson; this blog post of mine has a short exposition of it, although I just realized it leaves out an important piece (checking that the structure sheaf is a generator; Knutson seems to have an interesting argument for this which applies more generally to separated algebraic spaces at least).

Just take my example of $Spec A//\mathbb{G}_m$ for a graded ring $A$. The global sections corresponds to taking the zeroth-degree. Now there may be non-trivial graded $A$-modules which have vanishing zeroth degree. –  Lennart Meier Nov 20 '12 at 10:40