Self-duality of schemes

Theorem 1. Let $$ \begin{CD} X^\prime @>{g^\prime}>> X \br @V{f^\prime}VV @V{f}VV \br Y^\prime @>{g}>> Y \end{CD} $$ be a Cartesian diagram of quasi-compact quasi-separated schemes. Then
the natural Beck–Chevalley map $$ g^\ast \circ f_\ast \xrightarrow{\cong} (f^\prime)_ \ast (g^\prime)^\ast $$ is an isomorphism,
the projection formula $$ f_\ast(\mathscr{F}) \otimes \mathscr{G} \xrightarrow{\cong} f_\ast(\mathscr{F} \otimes f^\ast \mathscr{G}) $$ holds.

Proof.

For the first part, in the case when everything is affine, this is clear. Now asssume that $X \to Y$ is affine. In this case, we can write $\mathsf{QCoh}(X) \to \mathsf{QCoh}(Y)$ as a limit of $\mathsf{Mod}_ D \to \mathsf{Mod}_ C$, and everything satisfies the Beck–Chevalley condition. So by something we discussed last time, this is fine. Then we have to make the argument when $X \to Y$ is separated, and so on.

First, to get a natrual transformation, we use adjunction on $$ f^\ast(f_\ast \mathscr{F} \otimes \mathscr{G}) \cong f^\ast f_\ast \mathscr{F} \otimes f^\ast \mathscr{G} \to \mathscr{F} \otimes f^\ast \mathscr{G}. $$ Now because all functors commute with colimits, we may assume that $\mathscr{F}$ and $\mathscr{G}$ are both perfect. Using (1), we can assume that $\mathscr{G}$ is $\mathscr{O}_ Y$, in which case this is clear.

Self-duality of schemes#

Theorem 2. Let $\phi \colon \mathsf{QCoh}(X) \to \mathsf{QCoh}(X)$ be a $\Lambda$-linear continuous functor, where $X$ is a quasi-compact quasi-separated scheme over $\Lambda$. Then there exists a $\mathscr{K}_ \phi \in \mathsf{QCoh}(X \times_ \Lambda X)$ such that $$ \phi \simeq (p_2)_ \ast (p_1^\ast(-) \otimes \mathscr{K}_ \phi), $$ where $p_1, p_2 \colon X \times_\Lambda X \to X$ are the projections.

In general, let $\mathcal{C}$ be a dualizable $\Lambda$-linear category. Then we will have $$ \mathcal{C}^\vee = \mathsf{Fun}^L(\mathcal{C}, \mathsf{Mod}_ \Lambda) $$ the colimit-preserving functors. There is a natural pairing $$ e_\mathcal{C} \colon \mathcal{C}^\vee \otimes_\Lambda \mathcal{C} \to \mathsf{Mod}_ \Lambda, $$ which will be the counit map. The unit will be given by some $$ u_\mathcal{C} \colon \mathsf{Mod}_ \Lambda \to \mathcal{C} \otimes \mathcal{C}^\vee, $$ which corresponds to some object $u_\mathcal{C} \in \mathcal{C} \otimes \mathcal{C}^\vee$.

So in general, if we have a continuous functor $$ F \colon \mathcal{C} \to \mathcal{D}, $$ this gives rise to an element $$ K_F = (\id_{\mathcal{C}^\vee} \otimes F)(u_\mathcal{C}) \in \mathcal{C}^\vee \otimes \mathcal{D}. $$ Conversely, a $K_F$ gives rise to a continuous functor $$ \mathcal{C} \simeq \mathcal{C} \otimes \mathsf{Mod}_ \Lambda \xrightarrow{\id_ \mathcal{C} \otimes K_F} \mathcal{C} \otimes \mathcal{C}^\vee \otimes \mathcal{D} \xrightarrow{e_ \mathcal{C} \otimes \id_ \mathcal{D}} \mathcal{D}. $$

Recall that if $\mathcal{C}$ is compactly generated, then $\mathcal{C}$ is dualizable, with dual given by $$ \mathcal{C}^\vee = \operatorname{Ind}((\mathcal{C}^\omega)^\mathrm{op}). $$ In the case of $\mathsf{QCoh}(X)$, we can actually make this quite explicit. We can take $$ \mathsf{Qcoh}(X) \otimes \mathsf{QCoh}(X) \simeq \mathsf{QCoh}(X \times X) \to \mathsf{Mod}_ \Lambda; \quad \mathscr{F} \mapsto R\Gamma(\mathscr{F} \vert_{\Delta_X}), $$ and also $$ \mathsf{Mod}_ \Lambda \to \mathsf{QCoh}(X \times X); \quad \Lambda \mapsto \Delta_\ast \mathscr{O}_ X. $$

Once we have this, the object $$ \mathscr{K}_ \phi \in \mathsf{QCoh}(X \times X) $$ corresponds to the functor $$ \mathscr{F} \mapsto \mathscr{F} \boxtimes \mathscr{K}_ \phi \mapsto (p_2)_ \ast \Delta_{12}^\ast(\mathscr{F} \boxtimes \mathscr{K}_ \phi) \cong p_{2\ast} (p_1^\ast \mathscr{F} \otimes \mathscr{K}_ \phi). $$

The Serre functor#

Example 3. Suppose $\mathcal{C}$ is dualizable. The Serre functor is the functor $$ S_ \mathcal{C} \colon \mathcal{C} \to \mathcal{C} $$ corresponding to the kernel $$ K_S = (e_ \mathcal{C}^R)(\Lambda) \in \mathcal{C}^\vee \otimes \mathcal{C} $$ where $e_ \mathcal{C} \colon \mathcal{C}^\vee \otimes \mathcal{C} \to \mathsf{Mod}_ \Lambda$ always has a right adjoint by continuity.

Note that $e_ \mathcal{C}^R$ need not by continuous.

Example 4. When $\mathcal{C} = \mathsf{QCoh}(X)$, we will have $$ e \colon \mathsf{QCoh}(X \times X) \xrightarrow{\Delta^\ast} \mathsf{QCoh}(X) \xrightarrow{p_\ast} \mathsf{Mod}_ \Lambda $$ and so $K_S = \Delta_\ast \omega_X$. Then $S_\mathcal{C} = (-) \otimes \omega_X$.

Example 5. If $\mathcal{C}$ is compactly generated, then we will have for $c \in \mathcal{C}^\omega$ the formula $$ \Hom(d, S_\mathcal{C}(c)) = \Hom(c, d)^\ast. $$

The trace of a dualizable category#

Let $\mathcal{C}$ be dualizable and let $\Phi \colon \mathcal{C} \to \mathcal{C}$ be a continuous functor. We defined the trace $\tr(\Phi, \mathcal{C})$ as the continuous functor $$ \mathsf{Mod}_ \Lambda \xrightarrow{u_\mathcal{C}} \mathcal{C} \otimes \mathcal{C}^\vee \xrightarrow{\Phi \otimes \id_{\mathcal{C}^\vee}} \mathcal{C} \otimes \mathcal{C}^\vee \simeq \mathcal{C}^\vee \otimes \mathcal{C} \xrightarrow{e_\mathcal{C}} \mathsf{Mod}_ \Lambda. $$ But this can be thought of as a single object in $\mathsf{Mod}_ \Lambda$. We can also think of it as $$ \tr(\Phi, \mathcal{C}) = e_\mathcal{C}(\mathscr{K}_\Phi). $$ This is also called the Hochschild homology of $\mathcal{C}$.

Remark 6. We can also define a Hochschild cohomology $$ \Hom_{\mathcal{C} \otimes \mathcal{C}^\vee}(u_\mathcal{C}, \mathscr{K}_ \Phi) \in \mathcal{C} \otimes \mathcal{C}^\vee. $$ In particular, if $\Phi = \id$ then we will get $$ Z(\mathcal{C}) = \End_{\mathcal{C}^\vee \otimes \mathcal{C}}(u_\mathcal{C}) = \End_{\mathsf{Fun}^L(\mathcal{C}, \mathcal{C})}(\id_\mathcal{C}). $$

Example 7. For $\mathcal{C} = \mathsf{QCoh}(X)$ for $X$ quasi-compact and quasi-separated, consider $f \colon X \to X$ inducing $\Phi = f^\ast$. Then we see that the trace is given by pull-back/push-forward along $$ \mathrm{pt} \xleftarrow{p} X \xrightarrow{\Delta} X \times X \xleftarrow{f \times \id} X \times X \xleftarrow{\Delta_X} X \xrightarrow{p} \mathrm{pt}. $$ So we will have $$ \tr(f^\ast, \mathsf{QCoh}(X)) = R\Gamma(X^f, \mathscr{O}_ X), $$ thanks to the unconditional base change isomorphism we had.

Example 8. For $f = \id$ and $X / \Lambda$ is smooth, we have $$ H^i \mathscr{O}_ {X \times_{X \times X} X} = \Omega_X^{-i}. $$

Example 9. Let $\mathcal{C} = \mathsf{LMod}_ A$ for $A$ an associative algebra. This is dualizable with its dual being $\mathcal{C}^\vee = \mathsf{LMod}_ {A^\mathrm{rev}}$. Its unit is given by $$ A \in \mathsf{LMod}_ {A \otimes A^\mathrm{rev}}, $$ and its counit is $$ \mathcal{C}^\vee \otimes \mathcal{C} \to \mathsf{Mod}_ \Lambda; \quad M \boxtimes N \mapsto M \otimes_A N. $$ (More generally, we should take $M \mapsto M \otimes_{A \otimes A^\mathrm{rev}} A$. Then $$ \tr(\mathcal{C}, \id_\mathcal{C}) = A \otimes_{A \otimes A^\mathrm{rev}}^\mathbb{L} A = HH_\ast(A) $$ is really Hochschild homology.

Remark 10. In the beginning of the course, we defined this category $\mathsf{Morita}$. This is a full subcategory $$ \mathsf{Morita}_ \Lambda \hookrightarrow \mathsf{LinCat}_ \Lambda $$ and the inclusion is symmetric monoidal.

Smooth and proper categories#

Recall we have continuous functors $$ \mathsf{Mod}_ \Lambda \xrightarrow{u_\mathcal{C}} \mathcal{C} \otimes \mathcal{C}^\vee, \quad \mathcal{C}^\vee \otimes \mathcal{C} \xrightarrow{e_\mathcal{C}} \mathsf{Mod}_ \Lambda. $$

Definition 11. We say that $\mathcal{C}$ is smooth when $u_\mathcal{C}$ admits a continuous right adjoint.

This is equivalent to saying that $u_\mathcal{C} \colon \mathcal{C} \otimes \mathcal{C}^\vee$ is compact. This happens when $\mathcal{C} = \mathsf{QCoh}(X)$ for $X$ smooth.

Definition 12. We say that $\mathcal{C}$ is proper when $e_\mathcal{C}$ admits a continuous right adjoint.

When $\mathcal{C}$ is compactly generated, this is equivalent to $\Hom(c, d)$ being perfect for every $c, d$ compact.