Examples of traces#
Let’s recall this category $\mathsf{Morita}(\mathsf{Vect}_ k^\heartsuit)$ where every object is dualizable. For any associative $k$-algebra $A$, we can take $A^\mathrm{rev}$ as the dual, with the bimodules $$ {}_ {A \otimes A^\mathrm{rev}} A_k, \quad {}_ k A_{A^\mathrm{rev} \otimes A}. $$ For any endomorphism of $A$, which is a bimodule ${}_ A M_A$, we can consider $$ \tr({}_ A M_A \vert A) = A \otimes_{A \otimes A^\mathrm{rev}} (M \otimes A^\mathrm{rev}) \otimes_{A \otimes A^\mathrm{rev}} A \in \mathsf{Vect}_ k^\heartsuit. $$ This can be thought of as “tensoring over a circle.” Then this trace can be identified as $$ A \otimes_{A \otimes A^\mathrm{rev}} M = M / \operatorname{span}\lbrace am - ma \rbrace. $$ Instead of $\mathsf{Vect}_ k^\heartsuit$ we can use $\mathsf{Vect}_ k$ and then we obtain Hochschild homology.
Similarly, we had the category of correspondences $\mathsf{Corr}(\mathsf{Sch}_ {/S})$. Here, every $X$ is again dualizable with dual $X$. The counit and unit maps be $$ X \times_S X \xleftarrow{\Delta} X \to S, \quad S \leftarrow X \xrightarrow{\Delta} X \times_S X. $$ Given any $f \colon X \to X$, we can consider the endomorphism $$ X \xleftarrow{f} X = X, $$ and the trace will be $$ \tr(f \vert X) = X^f \in \mathsf{Sch}_ {/S} = \End(S). $$
Let’s look at the cohomological correspondences $\mathsf{CohCorr}(\mathsf{Sch}_ {/S}^\mathrm{ft})$ where $S/k$ is of finite type and coefficients are in $\mathbb{F}_ l$ where $l$ is invertible in $k$. The objects are pairs $(X, \mathscr{F})$ where $X \in \mathsf{Sch}_ {/S}$ and $\mathscr{F} \in D(\mathsf{Shv}(X_\mathrm{et}, \mathbb{F}_ l)^\heartsuit)$. Here, not every object is dualizable.
Theorem 1. An object $(X, \mathscr{F})$ is dualizable if and only if $\mathscr{F}$ is universally locally acyclic with respect to $X \to S$.
Now let’s consider the category $\mathsf{LinCat}_ k$. A category of the form $$ \mathcal{C} = D(\mathsf{QCoh}(X)^\heartsuit) $$ for $X$ a quasi-compact quasi-separated scheme, is dualizable with the dual being $\mathcal{C}$ itself. One functor will be $$ \mathcal{C} \otimes \mathcal{C} \to \mathsf{Vect}_ k; \quad \mathscr{F} \boxtimes \mathscr{G} \mapsto R\Gamma(X, \mathscr{F} \otimes \mathscr{G}), $$ and the other functor will involve Grothendieck duality.
Any endomorphism $f \colon X \to X$ of schemes defines a functor $$ Lf^\ast \colon D(\mathsf{QCoh}(X)^\heartsuit) \to D(\mathsf{QCoh}(X)^\heartsuit). $$ What is the trace of this functor? We will see that $$ \tr(Lf^\ast \vert D(\mathsf{QCoh}(X)^\heartsuit)) = R\Gamma(X^f, \mathscr{O}) \in \mathsf{Vect}_ k, $$ where $X^f$ is to be understood in terms of derived geometry.
Later we will see that any sheaf theory gives a symmetric monoidal functor from $\mathsf{Corr}(\mathsf{Sch}_ {/S}) \to \mathsf{LinCat}_ k$. Then we can compute these things by functoriality of trace.
Monoidal categories#
Definition 2. A semigroup (non-unital monoidal) category is a triple $(\mathcal{C}, \otimes ,a)$ where
- $\mathcal{C}$ is a category,
- $\otimes \colon \mathcal{C} \times \mathcal{C} \to \mathcal{C}$ is a functor,
- the diagram $$ \begin{CD} \mathcal{C} \times \mathcal{C} \times \mathcal{C} @>{\otimes \times \id}>> \mathcal{C} \times \mathcal{C} \br @V{\id \times \otimes}VV @V{\otimes}VV \br \mathcal{C} \times \mathcal{C} @>{\otimes}>> \mathcal{C} \end{CD} $$ has $a$ as a natural equivalence between them, i.e., there is a functorial $a_{X,Y,Z} \colon (X \otimes Y) \otimes Z \cong X \otimes (Y \otimes Z)$,
- satisfying the property that the pentagonal diagram $$ \begin{CD} ((X \otimes Y) \otimes Z) \otimes W @>{\cong}>> (X \otimes Y) \otimes (Z \otimes W) @>{\cong}>> X \otimes (Y \otimes (Z \otimes W)) \br @VVV @. @AAA \br (X \otimes (Y \otimes Z)) \otimes W @>{\cong}>> \longrightarrow @>{\cong}>> X \otimes ((Y \otimes Z) \otimes W) \end{CD} $$ commutes.
Definition 3. A unit in a non-unital monoidal category $\mathcal{C}$ is a pair $(1, u)$ where
- $1 \in \mathcal{C}$ is an object,
- $u \colon 1 \otimes 1 \cong 1$ is an isomorphism,
such that both functors $$ 1 \otimes - \colon \mathcal{C} \to \mathcal{C}, \quad - \otimes 1 \colon \mathcal{C} \to \mathcal{C} $$ are fully faithful.
Proposition 4. Let $(1, u)$ be a unit in $(\mathcal{C}, \otimes, a)$.
- There are natural equivalences $$ l \colon 1 \otimes - \simeq \id_\mathcal{C}, \quad r \colon - \otimes 1 \simeq \id_\mathcal{C}. $$
- The following diagrams are commutative. $$ \begin{CD} (X \otimes 1) \otimes Y @>{\cong}>> X \otimes (1 \otimes Y) \br @V{r_X \otimes \id_Y}VV @V{\id_X \otimes l_Y}VV \br X \otimes Y @= X \otimes Y \end{CD} $$ $$ \begin{CD} (X \otimes Y) \otimes 1 @>{\cong}>> X \otimes (Y \otimes 1) \br @V{r_{X \otimes Y}}VV @V{\id_X \otimes r_Y}VV \br X \otimes Y @= X \otimes Y \end{CD} $$ $$ \begin{CD} (1 \otimes X) \otimes Y @>{\cong}>> 1 \otimes (X \otimes Y) \br @V{l_X \otimes \id_Y}VV @V{l_{X \otimes Y}}VV \br X \otimes Y @= X \otimes Y \end{CD} $$
- $l_1 = r_1 = u$.
- The monoid $\End_\mathcal{C}(1)$ is a commutative monoid.
- There are no nontrivial automorphisms of $(1, u)$.
- If $(1^\prime, u^\prime)$ is another unit, then there exists a unique $\epsilon \colon (1, u) \cong (1^\prime, u^\prime)$.
Proof.
For (1), we note that there is a natural isomorphism $$ 1 \otimes (1 \otimes X) \cong (1 \otimes 1) \otimes X \cong 1 \otimes X, $$ and by fully faithfulness of $1 \otimes -$, we get $1 \otimes X \cong X$.
For the last two diagrams of (2), we can just tensor $1$ on the left or right, check that they commute using the definition, and use fully faithfulness.
For the first diagram of (2), we somehow need to use the pentagon diagram where the two terms in the middle are $1$.
For (3), we note that we have the diagram $$ \begin{CD} 1 \otimes (1 \otimes 1) @>{\cong}>> (1 \otimes 1) \otimes 1 \br @V{\id \otimes l_1}VV @V{u \otimes \id}VV \br 1 \otimes 1 @= 1 \otimes 1 \end{CD} $$ is commutative, and then by the first diagram of (2), we get $u = r_1$.
For (4), say we have $f, g \colon 1 \otimes 1$. We have a diagram $$ \begin{CD} 1 \otimes 1 @>{f \otimes \id}>> 1 \otimes 1 \br @V{\id \otimes g}VV @V{g \otimes \id}VV \br 1 \otimes 1 @>{f \otimes \id}>> 1 \otimes 1 \end{CD} $$ that commutes. If we look at the lower left path, then the diagram $$ \begin{CD} 1 \otimes 1 @>{\id \otimes g}>> 1 \otimes 1 @>{f \otimes \id}>> 1 \otimes 1 \br @V{u}VV @V{u}VV @V{u}VV \br 1 @>{g}>> 1 @>{f}>> 1 \end{CD} $$ is commutative because of the second and third parts. Doing the same thing with the upper right path, we get $f \circ g = g \circ f$.
For (5), suppose that $f \colon 1 \to 1$ gives an automorphism of $(1, u)$, meaning that the outer square of $$ \begin{CD} 1 \otimes 1 @>{u}>> 1 \br @V{\id \otimes f}VV @V{f}VV \br 1 \otimes 1 @>{u}>> 1 \br @V{f \otimes \id}VV @| \br 1 \otimes 1 @>{u}>> 1 \end{CD} $$ commutes. On the other hand, the upper small square is always commutative. So the lower square must also be commutative, and by fully faithfulness, we get $f = \id_1$.
For (6), uniqueness was covered above, so we only need to check existence. We define $$ \epsilon \colon 1 \xrightarrow{r^\prime_1} 1 \otimes 1^\prime \xleftarrow{l_{1^\prime}} 1^\prime. $$ Then we can check that this indeed respects $u, u^\prime$.
Definition 5. A monoidal category is a non-unital monoidal category that has a unit.