Last time we stated the following theorem.
Theorem 1. Let $X$ be a quasi-compact quasi-separated derived scheme over $\Lambda$. Then
- the category $\mathsf{QCoh}(X)$ is compactly generated,
- the compact objects and dualizable objects agree, and they are perfect sheaves on $X$,
- for any prestack $Y / \Lambda$, we have $$ \mathsf{QCoh}(X) \otimes_ \Lambda \mathsf{QCoh}(Y) \simeq \mathsf{QCoh}(X \times_ \Lambda Y). $$
Here is what we mean by perfect objects.
Definition 2. Let $\mathfrak{X}$ be any prestack. We define the full subcategory $$ \mathsf{Perf}(\mathfrak{X}) \subseteq \mathsf{QCoh}(\mathfrak{X}) $$ consisting of those $\mathscr{F}$ scuh that $f^\ast \mathscr{F} \in \mathsf{Mod}_ A$ is perfect for every $\Spec A \to \mathfrak{X}$. We say that a module $M \in \mathsf{Mod}_ A$ is perfect if it belongs to the smallest idempotent-complete stable subcategory that contains $A$.
Proof of the theorem#
Proposition 3. Parts (1) and (2) are true when $X = \Spec A$ is affine.
Proof.
First note the following general fact: if $\mathcal{C} \in \mathsf{CAlg}(\mathsf{LinCat}_ \Lambda)$ is a symmetric monoidal category, and $1_ \mathcal{C}$ is copmact, then dualizable objects are compact. This is because $\operatorname{Map}(x, -) = \operatorname{Map}(1, x^\vee \otimes -)$ and the right hand side can be seen to commute with filtered colimits. On the other hand, because $\mathsf{Mod}_ A$ is generated by the compact object $A$, it automatically follows that compact objects are the same as perfect objects. So it remains to show that perfect objects are dualizable.
This will follow from this general fact: if $\mathcal{C}$ is an idempotent-complete symmetric monoidal stable category, then the full subcategory of dualizable objects is stable under finite colimits and retracts. Finite colimits are made up of finite direct sums and cofibers. Finite direct susm are easy. For cofibers, say $X \to Y \to Z$ is a triangle and $X, Y$ are dualizable. Then one checks that the fiber $Z^\vee \to Y^\vee \to X^\vee$ is a dual. For retracts we do a similar thing.
Now we do the general case.
Lemma 4. Let $\mathcal{C} = \varprojlim_i \mathcal{C}_ i$ in $\mathsf{CAlg}(\mathsf{LinCat}_ \Lambda)$. Then $c \in \mathcal{C}$ is dualizable if and only if $c_i \in \mathcal{C}_ i$ is dualizable for all $i$.
This will imply that perfect objects are the same as dualizable objects for every prestack $\mathfrak{X}$. In one direction, since $C \to C_i$ is symmetric monoidal, it sends dualizable objects to dualizable objects. In the other direction, we use the following fact.Proof.
Definition 5. A commutative diagram $$ \begin{CD} C_{00} @>{f^\prime}>> C_{01} \br @V{g^\prime}VV @V{g}VV \br C_{10} @>{f}>> C_{11} \end{CD} $$ in $\mathsf{LinCat}_ \Lambda$ is called right adjointable if both $f, f^\prime$ admit continuous right adjoints $f^R, (f^\prime)^R$ and the induced Beck–Chevalley base change map $$ g^\prime \circ (f^\prime)^R \to f^R \circ f \circ g^\prime \circ (f^\prime)^R \simeq f^R \circ g \circ f^\prime \circ (f^\prime)^R \to f^R \circ g $$ is an isomorphism.
Remark 6. This “continuous right adjoint” condition should be thought of as the notion of right adjoint in the $2$-category $\mathsf{LinCat}_ \Lambda$. (Here, we only consider continuous functors.)
Let $I$ be a small category and let $$ \mathsf{Fun}^\mathrm{LAd}(I, \mathsf{LinCat}_ \Lambda) \subseteq \mathsf{Fun}(I, \mathrm{LinCat}_ \Lambda) $$ be the $1$-full subcategory consisting of
- objects only those with all $C_i \to C_j$ admiting condinuous right adjoints,
- morphisms only those such that all morphisms from $C_i \to C_j$ to $D_i \to D_j$ are right adjointable.
Similarly, one can define $\mathsf{Fun}^\mathrm{RAd}(I, \mathsf{LinCat}_ \Lambda)$.
Proposition 7. The categories $\mathsf{Fun}^\mathrm{LAd}(I, \mathsf{LinCat}_ \Lambda)$ and $\mathsf{Fun}^\mathrm{RAd}(I, \mathsf{LinCat}_ \Lambda)$ are both presentable, and moreover the inclusions $$ \mathsf{Fun}^\mathrm{LAd}, \mathsf{Fun}^\mathrm{RAd} \subseteq \mathsf{Fun} $$ preserve limits.
We will apply this to $I = [1]$. If we look at these diagrams $$ \mathsf{Mod}_ A \xrightarrow{\otimes c_A} \mathsf{Mod}_ A $$ these are all in both $\mathsf{Fun}^\mathrm{LAd}$ and $\mathsf{Fun}^\mathrm{RAd}$. Now we take the limit over all $\Spec A \to \mathfrak{X}$ and this can be computed in $\mathsf{Fun}$ to be $$ \mathsf{QCoh}(\mathfrak{X}) \xrightarrow{\otimes c} \mathsf{QCoh}(\mathfrak{X}). $$ This has to be in both $\mathsf{Fun}^\mathrm{RAd}$ and $\mathsf{Fun}^\mathrm{LAd}$, so this means that $- \otimes c$ has both a left adjoint and a right adjoint. From this we can deduce that $c$ is dualizable.
Proposition 8. Let $X$ be a quasi-compact quasi-separated scheme. Then $\mathscr{O}_ X$ is compact.
Proof.
For simplicity we assume that it is separated, and that it is $X = U \cup V$ with $U$, $V$, and $U \cap V$ affine. Then we see that $$ \Hom(\mathscr{O}_ X, -) = \operatorname{fiber}(\Hom(\mathscr{O}_ U, -) \oplus \Hom(\mathscr{O}_ V, -) \to \Hom(\mathscr{O}_ {U \cap V}, -)) $$ and finite limits commute with filtered colimits.
So now we know that perfect/dualizable objects are compact. For the other inclusion, we use the following fact.
Lemma 9. Let $f \colon \Spec A \to S$ be a morphism with $X$ quasi-compact quasi-separated, and $f_\ast = (f^\ast)^R$ continuous. Then $f^\ast$ preserves compact objects.
The dual of a category#
Theorem 10. Let $\mathcal{C}$ be a compactly generated category in $\mathcal{C}$. Then it is dualizable in $\mathsf{LinCat}_ \Lambda$.
This can be constructed int two ways: we can define $$ \mathcal{C}^\vee = \mathsf{Fun}^{\mathrm{L},\Lambda}(\mathcal{C}, \mathsf{Mod}_ \Lambda), $$ and more explicitly, we can take $$ \mathcal{C}^\vee = \operatorname{Ind}((\mathcal{C}^\omega)^\mathrm{op}). $$
The evaluation map is relatively easy to write down; we have $$ (\mathcal{C}^\omega)^\mathrm{op} \to \mathcal{C}^\omega \to \mathsf{Mod}_ \Lambda; \quad (c, d) \mapsto \Hom(c, d) $$ and then we can complete it to $$ \operatorname{Ind}((\mathcal{C}^\omega)^\mathrm{op}) \otimes_\Lambda \mathcal{C} \to \mathsf{Mod}_ \Lambda. $$ From this theorem it will follow that $\mathsf{QCoh}(X)$ is dualizable for $X$ quasi-compact quasi-separated.
Proposition 11. Let $\mathfrak{X}$ be a prestack such that $\mathsf{QCoh}(\mathfrak{X})$ is dualizable. Then for any $\mathfrak{Y}$, we have $$ \mathsf{QCoh}(\mathfrak{X}) \otimes \mathsf{QCoh}(\mathfrak{Y}) \to \mathsf{QCoh}(\mathfrak{X} \times_ \Lambda \mathfrak{Y}). $$
Proof.
We have tensoring with $\mathsf{QCoh}(\mathfrak{X})$ commuting with limits, becasue it is dualizable. So we can write $$ \begin{align} \mathsf{QCoh}(\mathfrak{X}) \otimes_ \Lambda \mathsf{QCoh}(\mathfrak{Y}) &= \varprojlim_{\Spec A \to \mathfrak{Y}} \mathsf{QCoh}(\mathfrak{X}) \otimes_ \Lambda \mathsf{Mod}_ A \br &= \varprojlim_{\Spec A \to \mathfrak{Y}} \varprojlim_{\Spec B \to \mathfrak{X}} (\mathsf{Mod}_ B \otimes_ \Lambda \mathsf{Mod}_ A) \br &= \varprojlim_{\Spec C \to \mathfrak{X} \times_ \Lambda \mathfrak{Y}} \mathsf{Mod}_ C. \end{align} $$ Here, we are using some cofinality in the last line.
Corollary 12. Let $X$ and $Y$ be quasi-compact and quasi-separated. Let $F \colon \mathsf{QCoh}(X) \to \mathsf{QCoh}(Y)$ be a continuous functor. Assume there is an equivalence $$ \mathbb{D}_ X \colon \mathsf{QCoh}(X)^\vee \simeq \mathsf{QCoh}(X) $$ satisfying some property. Then $F$ is of the form $$ F = (\mathrm{pr}_ Y)_ \ast (\mathscr{K} \otimes \mathrm{pr}_ X^\ast(-)). $$
Here, the condition we need is the following. We have a functor $$ \mathsf{Perf}(X) \otimes \mathsf{Perf}(X) \to \mathsf{Mod}_ \Lambda; \quad (\mathscr{F}, \mathscr{G}) \mapsto \Hom(\Delta_\ast \mathscr{O}_ X, \mathscr{F} \boxtimes \mathscr{G}). $$ We want this to induce $$ \mathbb{D}_ X \colon \mathsf{QCoh}(X) \simeq \mathsf{QCoh}(X)^\vee. $$