Recall that $A \in \mathsf{Alg}(\mathsf{LinCat}_ \Lambda)$ is rigid when
- the tensor functor $m \colon A \otimes A \to A$ admits a $(A \otimes A^\mathrm{rev})$-linear right adjoint,
- the object $1_A$ is compact.
Trace formula over a rigid monoidal category#
Example 1. If $A$ is compactly generated, then $A$ is rigid if and only if every compact object admits both a left and a right duals and $1_A$ is compact.
Proposition 2. Let $A$ be a rigid category.
- Let $F \colon M \to N$ be an $A$-linear functor of left $A$-modules. If $F^R$ is continuous, then it is $A$-linear.
- $A$ is self-dualizable as an object in $\mathsf{Mod}_ \Lambda$. Moreover it is $2$-dualizable as an algebra in $\mathsf{LinCat}_ \Lambda$.
- Let $M$ be a left $A$-module. Then $M$ is dualizable in $\mathsf{LinCat}_ \Lambda$ if and only if $M$ is left dualizable as an $A$-module.
Proof.
For the first part, we just look at the Beck–Chevalley map $$ \begin{CD} A \otimes M @>{\id_A \otimes F}>> A \otimes N \br @V{\mathrm{act}_ M}VV @V{\mathrm{act}_ N}VV \br M @>{F}>> N. \end{CD} $$
For dualizability of second part, we use the maps $$ \mathsf{Mod}_ \Lambda \xrightarrow{1_A} A \xrightarrow{m^R} A \otimes_ \Lambda A $$ and $$ A \otimes A \xrightarrow{m} A \xrightarrow{\Hom(1_A, -)} \mathsf{Mod}_ \Lambda. $$ For $2$-dualizability, we need to check that $A$ is left dualizable as a $\mathsf{Mod}_ \Lambda$-module and as a $A \otimes A^\mathrm{rev}$-module.
For the third part, we use the right adjoint of the map $$ N \otimes_A M \to N \otimes M $$ to get the unit $\mathsf{Mod}_ \Lambda \to N \otimes_A M$.
Corollary 3. Suppose $A$ is rigid.
- Let $F_1 \to F_2$ be an $A$-bilinear functor. Then the diagram $$ \begin{CD} F_1 @>>> HH(A, F_1) \br @VVV @VVV \br F_2 @>>> HH(A,, F_2) \end{CD} $$ is right adjointable.
- Suppose $F_1 \to F_2$ admits a continuous right adjoint. Then the diagram $$ \begin{CD} F_1 @>>> F_2 \br @VVV @VVV \br HH(A, F_1) @>>> H(A, F_2) \end{CD} $$ is right adjointable.
Proof.
Recall that $HH(A, F)$ is computed as the colimit of some simplicial diagram $$ \dotsb A \otimes F \rightrightarrows F. $$ But each arrow has a right adjoint and then we can compute this colimit as a limit of the right adjoints. For the second proof you use a similar fact.
Theorem 4. Assume $A$ is rigid and equipped with a monoidal endomorphism $\phi \colon A \to A$. Then $$ \End_{HH(A, {}^\phi A)}([1_A]_ {{}^\phi A}) \cong \tr(A, \phi)^\mathrm{rev} $$ as an $\Lambda$-algebra. More generally, let $M$ be a left dualizable $A$-module equipped with the functor $\phi \colon M \to {}^\phi A \otimes_A M = {}^\phi M$. Then $$ \Hom_{HH(A, {}^\phi A)}([1_A]_ {\phi_A}, [M, \phi]_ {\phi_A}) \cong \tr(M, \phi). $$
Proof.
We have the diagram $$ \begin{CD} {}^\phi A \otimes A @>>> {}^\phi A \br @VVV @V{[-]_ {{}^\phi A}}VV \br HH(A, {}^\phi A \otimes A) \cong {}^\phi A @>{[-]_ {{}^\phi A}}>> HH(A, {}^\phi A). \end{CD} $$ Then we can compute the endomorphism.
Theorem 5 (S=T). Let $A$ be rigid compactly generatedand $a \in A$ be compact. For simplicity, let $\phi = \id$. Then in $$ \Hom([1_A]_ A, [1_A]_ A) \cong \tr(A), $$ the Chern character $\mathrm{ch}(a) \in \tr(A)$ corresponds to the endomorphism $$ [1_A]_ {\phi_A} \to [a \otimes a^\vee]_ {\phi_A} \cong [a^\vee \otimes a]_ {\phi_A} \to [1_A]_ {\phi_A}. $$
Example 6. If we take $A$ the sheaves on $L^+G \backslash LG / L^+G$ with $\phi$ being the Frobenius action, then $\tr(A, \phi)$ is the space of functions on $C_c(G(\mathcal{O}) \backslash G(F) / G(\mathcal{O}))$.
Theorem 7. If $A$ is rigid and $F_1, F_2$ are two $A$-bimodules, both of which admit left duals. Suppose we are given an isomorphism $$ \alpha \colon F_1 \otimes_A F_2 \cong F_2 \otimes_A F_1. $$ Recall we have a trace formula $$ \tr(HH(A, F_1), HH(F_2, \alpha^{-1})) \cong \tr(HH(A, F_2), HH(F_1, \alpha)). $$ Let $M$ be a left-smooth and left-proper $A$-module, equipped with a commutative diagram $$ \begin{CD} F_1 \otimes_A M @<{\beta_1}<< M @>{\beta_2}>> F_2 \otimes_A M \br @V{\id \otimes \beta_2}VV @. @V{\id \otimes \beta_1}VV \br F_1 \otimes_A F_2 \otimes_A M @>{\alpha}>> @>>> F_2 \otimes_A F_2 \otimes_A M. \end{CD} $$ Assume $\beta_1, \beta_2$ both admit continuous right adjoints.
- There exists a natural transformation $$ (\eta_1 \colon [M, \beta_1]_ {F_1} \Rightarrow HH(F_2, \alpha^{-1}) \circ [M, \beta_1]_ {F_1}) \colon \mathsf{Mod}_ \Lambda \to HH(A, F_1) $$ and similarly $\eta_2$.
- The functors $[M, \beta_i]_ {F_i}$ are compact in $HH(A, F_i)$.
- We have $$ \operatorname{ch}([M, \beta_i]_ {F_1}, \tr(HH(F_2, \alpha^{-1}), \eta_1)) = \mathrm{ch}([M, \beta_2]_ {F_2}, \tr(HH(F_1, \alpha), \eta_2)). $$
Application to Deligne–Lusztig theory#
The first is the major example. Let $G / \mathbb{F}_ q$ be a reductive group and let $B \subseteq G$ be a Borel. Write $k = \bar{\mathbb{F}}_ q$. Take $$ A = \operatorname{Ind}(D_c^b((B \backslash G / B)_ k, \mathbb{Q}_ l)). $$
Proposition 8. The category $A$ is rigid, where the algebra structure is given by the correspondence $$ B \backslash G / B \times B \backslash G / B \leftarrow B \backslash (G \otimes^B G) / B \to B \backslash G / B. $$
Now let $\phi \colon A \to A$ be the pullback by the $q$-Frobenius. Then we can show that $$ \tr(A, \phi) = R\Gamma((B \backslash G / B)^\Frob, \mathbb{Q}_ l) = C_c(B(\mathbb{F}_ q) \backslash G(\mathbb{F}_ q) / B(\mathbb{F}_ q)). $$
Theorem 9. We have $$ HH(A, {}^\phi A) = A \otimes_{A \otimes A^\mathrm{rev}} {}^\phi $$ is the essential image of $$ \operatorname{Ind} D_c^b(B \backslash G / B) \to \operatorname{Ind} D_c^b(G / \mathrm{Ad}_ \Frob B) \to \operatorname{Ind} D_c^b(G / \mathrm{Ad}_ \Frob G) $$ induced by the correspondence $$ B \backslash G / B \leftarrow G / \mathrm{Ad}_ \Frob B \to G / \mathrm{Ad}_ \Frob G = [\ast / G(\mathbb{F}_ q)]. $$
If you think about this composite functor, this is nothing but Deligne–Lusztig induction. In particular, if $S_w$ is the Schubert cell in $B \backslash G / B$, then $$ j_{w,!} = j_! (\mathbb{Q}_ l)_ {S_w} \in D_c^b(B \backslash G / B) $$ has $$ [j_{w,!}]_ {{}^\phi A} $$ the compactly supported cohomology of the Deligne–Lusztig variety associated to $w$. Then for $w = e$, we see $$ [j_{e,!}]_ {\phi_A} \in C(G(\mathbb{F}_ q) / B(\mathbb{F}_ q)) $$ and then $$ \End_ {G(\mathbb{F}_ q)}(C(G(\mathbb{F}_ q) / B(\mathbb{F}_ q))) = C(B(\mathbb{F}_ q) \backslash G(\mathbb{F}_ q) / B(\mathbb{F}_ q)). $$