ScalingStacks

Proof. We briefly sketch the argument of [L1, Proposition 5.5.5]. The derived global sections functor $\Gamma$ preserves finite colimits. Thus it suffices to show $\Gamma$ preserves small coproducts: we must check that the natural map $\coprod_{\alpha}\Gamma(M_{\alpha})\rightarrow\Gamma(\coprod_{\alpha}M_{\alpha})$ is an equivalence, i.e., that the induced map on homotopy groups $\coprod_{\alpha}\pi_{*}\Gamma(M_{\alpha})\rightarrow\pi_{*}\Gamma(\coprod_{\alpha}M_{\alpha})$ is an equivalence. Since $\pi_{*}$ is colimit preserving, it suffices to check that the individual terms $\pi_{i}\Gamma$ each preserve small coproducts.

This is shown in two steps. First, one checks that for $M\in\qc(X)$ concentrated in a single degree, there exists $m$ such that $\pi_{n}\Gamma M$ is zero for all $n<m$. Thus to establish the assertion, it suffices to work in the subcategory $M\in(\qc(X))^{\leq n+m}_{\geq n}$ for which $\pi_{*}\Gamma M$ is concentrated in bounded degrees.

Next, one chooses a finite affine cover $U\rightarrow X$ giving the usual simplicial object $U_{*}\rightarrow X$, and thus an identification $\Gamma M\simeq\lim(M|_{U_{q}})$. The resulting Bousfield-Kan, or C̆ech, spectral sequence has $E_{1}$-term given by the $\pi_{p}(M|_{U_{q}})$, and converges to $\pi_{*}\Gamma M$. (The construction of the $E_{1}$-term evidently commutes with coproducts in $\qc(X)$, since pullback to an affine is colimit preserving.) Using the previous step, only finitely many terms in the spectral sequence may involve differentials affecting a particular group $\pi_{i}\Gamma M$. Therefore we have expressed $\pi_{i}\Gamma M$ as a finite limit of terms which preserved colimits in $M$ and so the result follows. ∎

Proof. It suffices to show that the following diagram of algebra objects is Cartesian

Let $u:U\coprod V\to X$ denote the cover. Since the restriction $u^{*}(-)\simeq(-)\otimes_{\mathcal{O}_{X}}(\mathcal{O}_{U}\times\mathcal{O}_{V})$ is conservative and preserves finite limits, it suffices to show that the restriction of the above diagram is Cartesian. But this is nothing more than the clearly Cartesian diagram

∎

Proof. The result for ordinary (non-derived) schemes $X$ is a theorem of Neeman [N2], extending ideas of Thomason [TT]. (In fact, Neeman proves that for quasi-compact, quasi-separated schemes, $\qc(X)$ is compactly generated, and dualizable and compact objects coincide. We assume $X$ has affine diagonal only because the definition of perfect stack requires it.)

A modified exposition of Neeman’s argument appears in the work of Bondal-Van den Bergh [BV], who in fact prove that $\qc(X)$ is generated by a single perfect object. One can translate the latter proof, which occupies [BV, Section 3.3], directly into the derived setting, substituting Lemma 3.17 above for its underived version [BV, Corollary 3.3.4], and using the natural identification $\qc(\Spec A)\simeq\Mod_{k}$ instead of [BV, Corollary 3.3.5]. (In the derived setting, there is no general notion of the abelian category of quasi-coherent sheaves, so we do not need to worry about the potential distinction between its derived category and the quasi-coherent derived category). In what follows, we sketch the argument for the reader’s convenience, keeping the notation from [BV].

The proof that $\qc(X)$ is generated by a single perfect (dualizable) object is an induction on the number of opens in an affine cover of $X$. The base case of an affine derived scheme is Lemma 3.5. For the inductive step, we write $X=Y\cup U$ with $Y$ open and $U$ affine (putting us in the context of Lemma 3.18), and assume that $\qc(Y)$ has a perfect generator $E$. By [BoN, Proposition 6.1], there is an explicit compact generator $Q$ for the kernel of the restriction from $U$ to the intersection $S=Y\cap U$. (One can think of $Q$ as a form of the structure sheaf of the closed complement $V=U\setminus S$). The key to the inductive step is Neeman’s abstract categorical form [N1, Theorem 2.1] of Thomason’s extension theorem for compact objects. This allows us to extend $E\oplus E[1]|_{S}$ to a compact (hence perfect) object on $U$, and then to glue the latter to $E\oplus E[1]$ to obtain a perfect object $P$ on all of $X$. (Note that we extend $E\oplus E[1]$ rather than $E$ itself since K-theoretic obstructions vanish for the former.) One then checks by a Mayer-Vietoris argument that the sum of $P$ and the pushforward of $Q$ (which is itself compact and perfect by support considerations) to $X$ generates all of $\qc(X)$.

By Lemma 3.17, we know that $\mathcal{O}_{X}$ is compact, and hence that dualizable complexes are compact. The assertion that compact objects are dualizable follows from [N1]: if a set $\mathcal{C}^{\circ}$ of compact objects generates $\mathcal{C}$, then all compact objects of $\mathcal{C}$ are summands of finite colimits of objects of $\mathcal{C}^{\circ}$ and their shifts. Since $\qc(X)$ is generated by a perfect object, we conclude that all compact objects are summands of perfect objects, which allows one to check locally that they are indeed perfect. ∎

Proof. First note that pushforward $p_{*}$ along the perfect morphism $p$ is colimit preserving, hence (by adjunction) the pullback $p^{*}M$ of a compact object $M\in\qc(Y)$ is compact. In particular we find that the structure sheaf (the monoidal unit) on $X$ is compact, and hence that all dualizable objects are compact. Note also that the pullback of a dualizable object is always dualizable.

Now suppose that $U\to Y$ is any affine mapping to $Y$, and consider the base change $X_{U}$ of $X$ to $U$. By the definition of a perfect morphism applied to $p$, this base change is itself a perfect stack. Let $q:X_{U}\to X$ denote the base change morphism, which is affine since $Y$ has affine diagonal. If $M\in\qc(X)$ is any compact object and $M_{U}=q^{*}M$ is its pullback to $U$, then $M_{U}$ is itself compact since $q_{*}$ preserves colimits:

Proof. By Lemma 3.20, we know that compact and dualizable objects in $\qc(X)$ coincide. Thus we need only to check that $\qc(X)$ is compactly generated. The pullbacks $p^{*}M$ of compact (hence dualizable) objects on $Y$ are dualizable, hence compact, as are the line bundles $\mathcal{L}$ in the given relatively ample family. We claim the $\infty$-category of compact objects $p^{*}M\otimes\mathcal{L}$ generates $\qc(X)$. The argument is as above in the case of an external product: let $N$ be right orthogonal to the compact objects, so that $\Hom(p^{*}M\otimes\mathcal{L},N)=0$. We first find by adjunction and the fact that $Y$ is compactly generated that $p_{*}{\mathcal{H}om}(\mathcal{L},N)=0$. Since the objects $\mathcal{L}$ form a relatively ample family of line bundles this forces $N=0$. ∎

Proof. First note that in characteristic zero, $BGL_{n}$ is clearly perfect: the compact and dualizable objects are both finite dimensional representations which generate.

If $G$ is an affine algebraic group, we can embed $G\hookrightarrow GL_{n}$ as a subgroup of $GL_{n}$ for some $n$. Thus we obtain a morphism $BG\to BGL_{n}$ with fiber $GL_{n}/G$. By a theorem of Chevalley [Ch], $GL_{n}/G$ is a quasi-projective variety, and so by Proposition 3.21, $BG$ itself is perfect.

Finally, for a quasi-projective derived scheme $X$ with a linear action of $G$, the morphism $X/G\to BG$ is quasi-projective, so applying Proposition 3.21 again, we conclude that $X/G$ is perfect. ∎

Proof. Let $f:A\to Y$ be a morphism from an affine derived scheme to the base, and $X_{A}\to A$ the base change of $X$. Since $Y$ has affine diagonal, $f$ is affine as is the base change morphism $f_{X}:X_{A}\to X$. In particular Proposition 3.21 (again in the basic case of an affine morphism) applies to $f_{X}$, so that the total space $X_{A}$ is perfect. ∎

Proof. The second assertion follows from the first and Proposition 3.21, since $X_{1}\times_{Y}X_{2}\to X_{1}\times X_{2}$ is an affine morphism for $Y$ with affine diagonal.

To prove the first assertion, note that $X=X_{1}\times X_{2}$ has affine diagonal. Applying Lemma 3.20 to the projection to a factor, we see that compact and dualizable objects of $\qc(X)$ coincide. Thus to confirm that $X$ is perfect, it suffices to show that $\qc(X)$ is compactly generated.

Let us first check that the external product of compact objects is again compact. By assumption, compact objects $M_{i}$ on the factors $X_{i}$ are dualizable, so $M_{1}\boxtimes M_{2}$ is dualizable (it is the tensor product of pullbacks, and both operations preserve dualizabie objects), and hence compact.

Now let us check that external products of compact objects generate $\qc(X)$. The argument is a modification of the argument of Bondal-Van den Bergh [BV] in the case of a single compact generator. Namely, let $N$ be right orthogonal to $\qc(X)^{c}$, so that in particular $\Hom(M_{1}\boxtimes M_{2},N)\simeq 0$ for all $M_{i}\in\qc(X_{i})^{c}$. By adjunction, we have

for all $M_{1},M_{2}$, so that $\pi_{1*}{\mathcal{H}om}(\pi_{2}^{*}M_{2},N)\simeq 0$ since such $M_{1}$ generate $\qc(X_{1})$. For any affines $U\to X_{1}$ and $V\to X_{2}$, we therefore have

for all $M_{2}$. Since the latter objects generate $\qc(X_{2})$ it follows (upon restricting to $V$) that $\Gamma(U\times V,N)\simeq 0$, whence (by affineness of $U\times V$) that $N|_{U\times V}\simeq 0$, and finally (since affines of the form $U\times V$ cover $X_{1}\times X_{2}$) that $N\simeq 0$. ∎

Proof. Let $\Sigma_{0}$ be the $0$-simplices of $\Sigma$. Since $X$ has affine diagonal, the natural projection $X^{\Sigma}\to X^{\Sigma_{0}}$ is affine. By Proposition 3.24, the product $X^{\Sigma_{0}}$ is perfect, and so by Proposition 3.21 (in the basic case of an affine morphism), the assertion follows. ∎

Proof. We leave to the reader the exercise of checking that $X/G$ has affine diagonal since $X$ has affine diagonal and $G$ is affine.

We will first prove that condition (1) implies $\qc(X/G)$ is generated by compact dualizable objects. Since $f:X\to X/G$ is affine, we have the identification $\qc(X)\simeq\Mod_{f_{*}\mathcal{O}_{X}}(\qc(X/G))$.

We claim that the algebra object $f_{*}\mathcal{O}_{X}$ is perfect (or equivalently, dualizable). To see this, consider the pullback square of derived stacks

Via base change, we obtain an equivalence $g^{*}g_{*}\mathcal{O}_{k}\simeq p_{*}p^{*}\mathcal{O}_{k}$, or in other words, an equivalence of $k$-algebras $g^{*}g_{*}\mathcal{O}_{k}\simeq\Gamma(G,\mathcal{O}_{G})$. By assumption, $\Gamma(G,\mathcal{O}_{G})$ is a perfect complex, and $g^{*}$ is conservative and preserves perfect complexes, so we conclude that the pushforward $g_{*}\mathcal{O}_{k}$ is perfect. Now consider the pullback square of derived stacks

Since $g_{*}\mathcal{O}_{k}$ is perfect, $q^{*}g_{*}\mathcal{O}_{k}$ is perfect. By base change, we have the equivalence $q^{*}g_{*}\mathcal{O}_{k}\simeq f_{*}q^{\prime*}\mathcal{O}_{k}$, and thus we conclude that $f_{*}\mathcal{O}_{X}$ is perfect.

Next observe that the right adjoint $f^{+}$ to the pushforward $f_{*}$ can be calculated explicitly by

It follows immediately that $f^{+}$ preserves colimits. It also follows that $f^{+}$ is conservative since a diagram chase with the above identities leads to the identity

The unit $e:\Spec k\to G$ gives a factorization of the identity map

and taking duals, a factorization of the identity map of $\mathcal{O}_{k}$ through the dual $\Gamma(G,\mathcal{O}_{G})^{\vee}$. Hence if $f^{+}(M)$ were trivial, then $f^{*}(M)$ would also be trivial, but $f^{*}$ is conservative. Thus we conclude $f_{*}$ takes a generating set of compact objects to a generating set of compact objects.

We now appeal to condition $(2)$ that the unit in $\qc(BG)$ is compact, hence so are all dualizables in $\qc(BG)$. In the case when $X$ is a point, the above arguments show that $\qc(BG)$ is compactly generated. Furthermore, it shows that all compacts are in fact dualizable (since $f_{*}\mathcal{O}_{k}$ is a compact dualizable generator), and hence $BG$ itself is perfect. The morphism $X/G\to BG$ is then a perfect morphism with perfect base. Thus by Lemma 3.20 compact and dualizable objects in $\qc(X/G)$ coincide. This implies (in combination with the compact generation of $\qc(X/G)$ above) that $X/G$ is perfect as asserted. ∎