Localization sequences#
Let $A \in \mathsf{Alg}(\mathsf{LinCat})$. We made the following definition.
Definition 1. A sequence $$ M \xrightarrow{F} C \xrightarrow{G} N $$ of $A$-linear categories is called a localization sequence if
- both $F$ and $G$ admit $A$-linear right adjoints $F^R$ and $G^R$,
- the unit and counits $\id_M \to F^R \circ F$ and $G \circ G^R \to \id$ are equivalences,
- $G \circ F = 0$,
- for each $c \in C$ the sequence $$ F(F^R(c)) \to c \to G^R(G(c)) $$ is a fiber sequence.
If $G^R$ moreover admits an $A$-linear right adjoint, then we say that the images of $F$ and $G^R$ form a semi-orthogonal decomposition of $C$.
Remark 2. For $A = \mathsf{Mod}_ \Lambda$, and $M, N, C$ are compact, then the existence of $F^R, G^R$ implies that $F, G$ sends compacts to compacts. For a semi-orthogonal decomposition, similarly $G^R$ sends compacts to compacts. Then we get for a localization sequence $$ K_0(M^\omega) \to K_0(C^\omega) \to K_0(N^\omega), $$ and for a semi-orthogonal decomposition, we get a splitting $$ K_0(C^\omega) = K_0(M^\omega) \oplus K_0(N^\omega). $$ Here, we are using that $F$ is fully faithful and colimit-preserving, so compactness can be detected after applying $F$.
Lemma 3. Let $M \to C \to N$ be a localization sequence of dualizable categories. Then $$ M^\vee \xrightarrow{F^\circ} C^\vee \xrightarrow{G^\circ} N^\vee $$ is also a localization sequence.
Proposition 4. Let $M \to C \to N$ be a localization sequence of dualizable categories. Then $$ (F \otimes F^\circ) u_M \to u_C \to (G^R \otimes (G^\circ)^R) u_N $$ is a fiber sequence in $C \otimes_A C^\vee$.
Proof.
We first have the sequence $$ (F \otimes \id) (F^R \otimes \id) u_C \to u_C \to (G^R \otimes \id) (G \otimes \id) u_C $$ in $C \otimes C^\vee$. We also have $$ (\id \otimes F^\circ) (\id \otimes (F^\circ)^R) (G \otimes \id) u_C \to (G \otimes \id) u_C \to (\id \otimes (G^\circ)^R) (\id \otimes G^\circ) (G \otimes \id) u_C $$ and $$ (\id \otimes F^\circ) (\id \otimes (F^\circ)^R) (F^R \otimes \id) u_C \to (F^R \otimes \id) u_C \to (\id \otimes (G^\circ)^R) (\id \otimes G^\circ) (F^R \otimes \id) u_C. $$ But if we think about it, the first term in the second sequence has $$ (\id \otimes (F^\circ)^R) (G \otimes \id) u_C = (G \otimes \id) (\id \otimes F^\vee) u_C (G \otimes \id) (F \otimes \id) u_M = 0. $$ Similarly, the last term in the third sequence is zero.
So the last term in the first sequence can be rewritten as $$ (G^R \otimes \id) (\id \otimes (G^\circ)^R) (\id \otimes G^\circ) (G \otimes \id) u_C. $$ Now we just note that $$ (\id \otimes G^\circ) (G \otimes \id) u_C = (\id \otimes G^\circ) (\id \otimes (G^\circ)^R) u_N = u_N. $$ A similar thing can be done on the first term.
Proposition 5. Let $M \to C \to N$ be as above. Suppose there exists an $A$-linear functor $$ \phi_C \colon C \to C, $$ and define $$ \phi_M = F^R \circ \phi_C \circ F, \quad \phi_N = G \circ \phi_C \circ G^R. $$ This induces $$ \eta \colon F \circ \phi_M \Rightarrow \phi_C \circ F, \quad \delta \colon G \circ \phi_C \Rightarrow \phi_N \circ G. $$ Then $$ \tr(M, \phi_M) \xrightarrow{\tr(F, \eta)} \tr(C, \phi_C) \xrightarrow{\tr(G, \delta)} \tr(N, \phi_N) $$ is a fiber sequence in $A$. Moreover, if $(F(M), G^R(N))$ form a semi-orthogonal decomposition, then we have a canonical splitting $$ \tr(C, \phi_C) \cong \tr(M, \phi_M) \oplus \tr(N, \phi_N). $$
Proof.
Note that trace is just applying the evaluation map to $u_C$. From the previous proposition, we get $$ e_C(\phi_C \otimes 1)(F \otimes F^\circ) u_M\to \tr(C, \phi_C) \to e_C(\phi_C \otimes 1)(G^R \otimes (G^\circ)^R) u_N. $$ If you think about it, $\tr(F, \eta)$ is the composition $$ \begin{align} \tr(M, \phi_M) &= e_M (\phi_M \otimes \id) u_M \to e_C(F \circ \phi_M \otimes F^\circ) u_M = e_C(\phi_C \circ F \otimes F^\circ) u_M \br &= e_C(\phi_C \otimes 1)(F \otimes F^\circ) u_M \to \tr(C, \phi_C). \end{align} $$ Here, we can check that this one map is also an isomorphism, because $F$ is fully faithful.
Application to Chern characters#
Let $A = \mathsf{Mod}_ \Lambda$. Recall we defined a map $$ \mathrm{ch} \colon C^\omega \to \tr(C, \id_C). $$
Proposition 6. This map induces a map $$ K_0(C^\omega) \to \tr(C, \id_C). $$
Proof.
Given $c^\prime \to c \to c^{\prime\prime}$, we need to show that the Chern characters add up. Let $$ S_2 C \subseteq \mathsf{Fun}(\Lambda_1^2, C) $$ be the full subcategory of fiber sequences. This turns out to be also a compact $A$-linear category with compact objects being those sequences with all objects compact.
If we look at the functor $$ F \colon C \to S_2 C; \quad c \mapsto (c \to c \to 0) $$ then this has a right adjoint $$ F^R(c^\prime \to c \to c^{\prime\prime}) = c^\prime. $$ On the other hand, if we look at $$ G \colon S_2 C \to C; \quad (c^\prime \to c \to c^{\prime\prime}) \mapsto c^{\prime\prime} $$ with $G^R(c) = (0 \to c \to c)$, we can check that $$ C \xrightarrow{F} S_2 C \xrightarrow{G} C $$ is a semi-orthogonal decomposition.
Now we can check that once we have this semi-orthogonal decomposition, we get $$ \mathrm{ch}(C, c^\prime) \to \mathrm{ch}(S_2 C, c^\prime \to c \to c^{\prime\prime}) \to \mathrm{ch}(C, c^{\prime\prime}). $$ At this point, we check that $S_2 C \to C$ given by $(c^\prime \to c \to c^{\prime\prime}) \mapsto c$ gives the same Chern character.
Theorem 7 (Grothendieck–Riemann–Roch). If $C$ and $D$ are $\Lambda$-linear categories, and $F \colon C \to D$ admits a continuous right adjoint, then $$ \begin{CD} K_0(C^\omega) @>{\mathrm{ch}}>> H^0(\tr(C)) \br @VVV @VVV \br K_0(D^\omega) @>{\mathrm{ch}}>> H^0(\tr(D)) \end{CD} $$ commutes.
Example 8. For $C = \mathsf{QCoh}(X)$ for $X / \mathbb{C}$ smooth, we get $$ H^0 \tr(C) = \bigoplus H^i(X, \Omega^i). $$ So the Chern character lands here.
Example 9. Let $C = \mathsf{Rep}^\mathrm{sm}(G)$, where $G$ is a $p$-adic Lie group. Here the Chern character lands in $$ H^0(\mathscr{H} \otimes_{\mathscr{H} \otimes \mathscr{H}^\mathrm{rev}} \mathscr{H}) = \mathscr{H} / [\mathscr{H}, \mathscr{H}] = \bar{\mathscr{H}} $$ where $\mathscr{H}$ is the Hecke algebra. Then we can apply the theorem to parabolic induction and get interesting results.
Admissible objects#
Recall the following definition.
Definition 10. Let $C$ be an $A$-linear category. An object $c \in C$ is $A$-compact if $F_c = (-) \otimes c \colon A \to C$ admits an $A$-linear right adjoint, and $A$-admissible if $F_c$ admits a $A$-linear left adjoint.
Now suppose that $C$ is dualizable. Let $c$ be an $A$-admissible object. Then we have an $A$-linear left adjoint $$ F_c^L \colon C \to A, $$ and then we get a map $$ (\tr(C) \xrightarrow{\tr(F_c^L)} 1_A) \in Z(C, S_C). $$
What is this map? Explicitly, let $c^\vee = (F_c^L)^\vee (1_A) \in C^\vee$. Then this is the composition of $$ u_C \to c \boxtimes_A c^\vee, \quad e_C(c^\vee \boxtimes_A c) \to 1_A. $$ The second map corresponds to a map $c^\vee \boxtimes_A c \to K_{S_C}$. Now the composed map $$ \begin{align} c &= 1_A \boxtimes_A c \cong (\id_C \otimes e_C) (u_C \otimes \id_C) (1_A \boxtimes_A c) \cong (\id_C \otimes e_C) (u_C \boxtimes_A c) \br &\to (\id_C \otimes e_C) (c \boxtimes_A c^\vee \boxtimes_A c) \to c \boxtimes_A 1_A \cong c \end{align} $$ is the identity. A similar thing holds for $c^\vee$.
Example 11. Let $C = \mathsf{Rep}^\mathrm{sm}(G)$ for $G$ a $p$-adic group. Then $\pi$ is admissible if and only if $\Hom(\operatorname{cInd}_ K^G \Lambda, \pi)$ is perfect for every $K$ open compact. Then this will be the character $$ \Theta_\pi \colon \bar{\mathscr{H}} = \mathscr{H}/[\mathscr{H}, \mathscr{H}] \to \Lambda. $$