Serre functors

Another fundamental tool in the study of derived categories is the Serre functor. We work with a triangulated category $\mathcal T$ which is linear over a field $k$.

We will set $$V^\ast := \operatorname{Hom}_k(V,k)$$

Definition. A Serre functor is an auto-equivalence $S: \mathcal T \to \mathcal T$ for which there are natural isomorphisms $$\operatorname{Hom}(X,Y)^\ast \cong \operatorname{Hom}(Y,S(X))$$

Example. Let $A$ be a dg-algebra over $k$. We set $$(-)^\vee := \mathbf{R}\operatorname{Hom}_A(-,A)$$

There is natural map $$M^\vee \overset{\mathbf{L}}{\otimes}_A N \to \operatorname{Hom}(M,N)$$ The subcategory of pairs $(M,N)$ where this is an isomorphism

• contains $(A[l],A[s])$ for each $l,s \in A$,
• is triangulated in $M$ and $N$,
• is closed under retracts in both $M$ and $N$, and
• is closed under arbitrary coproducts in $N$.

Thus this is an isomorphism for any perfect $M$ and any $N$.

Using tensor-Hom adjunction, we have $$\operatorname{Hom}_k(M^\vee \overset{\mathbf{L}}{\otimes}_A N,k) \cong \operatorname{Hom}_A(N, \operatorname{Hom}_k(M^\vee,k))$$

Thus, if we have Serre functor, then we know its form must be $$S(M) := \left(M^\vee\right)^\ast$$ The question becomes whether $$M \mapsto \left(M^\vee\right)^\ast$$ is an auto-equivalence.

Assuming it is an auto-equivalence, we know the $S$ commutes with $\bigoplus$. However, it not reasonable to expect $$\bigoplus (M_t^\vee)^\ast \to \left(\left(\bigoplus M_t \right)^\vee \right)^\ast$$ to be an isomorphism for general $M$. One needs some finiteness in general, either for $M$ as an $A$-module or $H^\ast M$ as a $k$-module or both.

We will restrict ourselves now to assuming that $A$ is proper over $k$. Recall this means that $\dim H^\ast A < \infty$ and we will assume that $M$ and $N$ are perfect $A$-modules.

We have a natural map
$$M \overset{\mathbf{L}}{\otimes}_A A^\ast (M^\vee)^\ast$$ which one can check is isomorphism for perfect $M$.

Given that we can write $S$ as tensoring with the bimodule $A^\ast$, it makes sense to look for another bimodule $A^!$ with $$A^! \overset{\mathbf{L}}{\otimes}_A A^\ast \cong A$$ and $$A^\ast \overset{\mathbf{L}}{\otimes}_A A^! \cong A$$

Consider $$\mathbf{R}\operatorname{Hom}_{A \otimes_k A^{op}}(A, A \otimes_k A) \overset{\mathbf{L}}{\otimes}_A A^\ast$$ If we assume that $A$ is compact, we have $$\mathbf{R}\operatorname{Hom}_{A \otimes_k A^{op}}(A, A \otimes_k A) \overset{\mathbf{L}}{\otimes}_A A^\ast \to \mathbf{R}\operatorname{Hom}_{A \otimes_k A^{op}}(A, A \otimes_k A^\ast)$$ is an isomorphism. Since $H^\ast A$ is finite-dimensional, we have $$\mathbf{R}\operatorname{Hom}_{A \otimes_k A^{op}}(A, A \otimes_k A^\ast) \to \mathbf{R}\operatorname{Hom}_{A \otimes_k A^{op}}(A, \operatorname{Hom}_k(A,A))$$ is also an isomorphism. Now, we can use tensor-hom adjunction $$A \cong \mathbf{R}\operatorname{Hom}_{A \otimes_k A^{op}}(A \otimes_k A, A) \to \mathbf{R}\operatorname{Hom}_{A \otimes_k A^{op}}(A, \operatorname{Hom}_k(A,A))$$ to see that $$\mathbf{R}\operatorname{Hom}_{A \otimes_k A^{op}}(A, A \otimes_k A) \overset{\mathbf{L}}{\otimes}_A A^\ast \cong A$$ We get a similar isomorphism if we tensor in the other order.

Thus, if $A$ is a compact $A$-$A$ bimodule with $\dim H^\ast A < \infty$, we see that $$M \mapsto M \overset{\mathbf{L}}{\otimes}_A A^\ast$$ is the Serre functor on $\operatorname{perf} A$. Furthermore, when $A$ is a compact bimodule, $\operatorname{perf} A$ is equivalence to the category $D(\operatorname{mod}_{fd} A)$ of $A$-modules with total cohomology finite dimensional over $k$.

Lemma. If we have an endofunctor $S: \mathcal T \to \mathcal T$ with natural isomorphisms $$\operatorname{Hom}(X,Y)^\ast \cong \operatorname{Hom}(Y,S(X))$$ and each $\operatorname{Hom}(X,Y)$ is finite-dimensional, then $S$ is fully-faithful.

Proposition. Assume $F: \mathcal T \to \mathcal S$ is exact and both $\mathcal T$ and $\mathcal S$ have Serre functors. If $F$ has right adjoint $R : \mathcal S \to \mathcal T$, $F$ has a left adjoint $$S_{\mathcal T}^{-1} R S_{\mathcal S} \vdash F$$ If $F$ has a left adjoint $L$, then $F$ has a right adjoint $$F \vdash S_{\mathcal T}^{-1} L S_{\mathcal S}$$