ScalingStacks

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

$$f^{+}(M)\simeq{\mathcal{H}om}_{\mathcal{O}_{X/G}}(f_{*}\mathcal{O}_{X},M)\simeq M\otimes_{\mathcal{O}_{X/G}}(f_{*}{\mathcal{O}_{X}})^{\vee}.$$

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

$$f^{*}(M\otimes_{\mathcal{O}_{X/G}}(f_{*}{\mathcal{O}_{X}})^{\vee})\simeq f^{*}(M)\otimes_{\mathcal{O}_{X}}q^{\prime*}g^{*}((g_{*}{\mathcal{O}_{k}})^{\vee})\simeq f^{*}(M)\otimes_{\mathcal{O}_{X}}(\mathcal{O}_{X}\otimes_{\mathcal{O}_{k}}\Gamma(G,\mathcal{O}_{G})^{\vee}).$$

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. ∎