An Azumaya variety over a field is by definition a pair $(X,\mathcal A_X)$, where $X$ is an algebraic variety of finite type over that field and $\mathcal A_X$ is a sheaf of Azumaya algebras, namely it is locally free of finite rank as an $\mathcal O_X$-module, and the canonical map of algebras \begin{equation} \mathcal A_X \otimes_{\mathcal O_X} \mathcal A_X^{\operatorname{op}} \to \mathcal End_{\mathcal O_X}(\mathcal A_X) \end{equation} is an isomorphism.

Given an Azumaya variety $(X,\mathcal A_X)$, there is a canonical "structure morphism" $\pi_X \colon (X,\mathcal A_X) \to (X,\mathcal O_X)$, which is defined by the map of sheaves of rings $\mathcal O_X \to \mathcal A_X$ which endows $\mathcal A_X$ with the $\mathcal O_X$-algebra structure. This morphism induces a pushforward functor \begin{equation} {\pi_X}_* \colon \operatorname{Mod}(\mathcal A_X) \to \operatorname{Mod}(\mathcal O_X) \end{equation} which is just a "restriction of scalars": it takes a right $\mathcal A_X$-module and maps it to itself viewed as an $\mathcal O_X$-module.

In Hyperplane sections and derived categories, Lemma 10.4, it is proved that:

A sheaf $F \in \operatorname{Coh}(X,\mathcal A_X)$ is locally projective over $\mathcal A_X$ in the Zariski topology if and only if ${\pi_X}_*F \in \operatorname{Coh}(X)$ is locally free, where $\pi_X \colon (X,\mathcal A_X) \to X$ is the structure morphism.

Setting $\mathcal D(X,\mathcal A_X)$ to be the derived category of the category of coherent modules $\operatorname{Coh}(X, \mathcal A_X)$, Definition 10.17 in the same article says:

An object $F \in \mathcal D(X,\mathcal A_X)$ is called a perfect complex if it is locally quasi-isomorphic to a bounded complex of [locally?] projective $\mathcal A_X$-modules of finite rank.

Question 1: given that perfect objects in $\mathcal D(X,\mathcal O_X)$ are precisely the complexes locally isomorphic to bounded complexes of locally free $\mathcal O_X$-modules of finite rank, is it reasonable to expect that the derived pushforward \begin{equation} \mathbb R {\pi_X}_* \colon \mathcal D(X,\mathcal A_X) \to \mathcal D(X,\mathcal O_X) \end{equation} has the property that $\mathcal P \in \mathcal D(X,\mathcal A_X)$ is perfect if and only if $\mathbb R {\pi_X}_* \mathcal P \in \mathcal D(X,\mathcal O_X)$ is perfect? Also, since in reasonable situations perfect objects coincide with compact objects, one would expect that $\mathcal K \in \mathcal D(X,\mathcal A_X)$ is compact if and only if $\mathbb R {\pi_X}_* \mathcal K \in \mathcal D(X,\mathcal O_X)$ is compact. How would one prove either statement?

Question 2: in a more "algebraic" setting, let $R$ be a commutative ring and let $A$ be an algebra over $R$. The restriction of scalars functor $\pi_* \colon \operatorname{Mod}(A) \to \operatorname{Mod}(R)$ induces a functor at the level of derived categories: \begin{equation} \mathbb R \pi_* \colon \mathcal D(\operatorname{Mod}(A)) \to \mathcal D(\operatorname{Mod}(R)). \end{equation} Which hypotheses can we put on $A$ in order that $M \in \mathcal D(\operatorname{Mod}(A))$ is perfect (resp. compact) if and only if $\mathbb R \pi_* M \in \mathcal D(\operatorname{Mod}(R))$ is perfect (resp. compact)? What if $A$ is Azumaya over $R$?

Remark: I've read that the perfect complexes are characterised as the dualisable objects, namely $M \in \mathcal D(\operatorname{Mod}(B))$ is dualisable if for all $N$ there is a natural isomorphism \begin{equation} \mathbb R \operatorname{Hom}(M, N) \cong \mathbb R \operatorname{Hom}(M,B) \otimes^{\mathbb L} N. \end{equation} How could I prove (and under which hypotheses) that, in the setting of Question 2, $M$ is dualisable if and only if $\mathbb R \pi_* M$ is dualisable?

Note/motivation: The aim of this kind of results would be, more in general, to understand perfect and compact objects of, say, $\mathcal D(\operatorname{Mod}(A)$ in terms of perfect and compact objects of $\mathcal D(\operatorname{Mod}(R)$. The ideal pattern would be: we know that perfects and compacts are the same in $\mathcal D(\operatorname{Mod}(R))$, we know that the results sketched above hold, then we deduce that perfects and compacts are the same in $\mathcal D(\operatorname{Mod}(A))$.

  • $\begingroup$ Certainly one needs that the $R$-algebra $A$ is a compact $R$-module. $\endgroup$ – Artur Jackson Apr 14 '17 at 7:18

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