## 57.3 Serre functors

The material in this section is taken from .

Lemma 57.3.1. Let $k$ be a field. Let $\mathcal{T}$ be a $k$-linear triangulated category such that $\dim _ k \mathop{\mathrm{Hom}}\nolimits _\mathcal {T}(X, Y) < \infty$ for all $X, Y \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{T})$. The following are equivalent

1. there exists a $k$-linear equivalence $S : \mathcal{T} \to \mathcal{T}$ and $k$-linear isomorphisms $c_{X, Y} : \mathop{\mathrm{Hom}}\nolimits _\mathcal {T}(X, Y) \to \mathop{\mathrm{Hom}}\nolimits _\mathcal {T}(Y, S(X))^\vee$ functorial in $X, Y \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{T})$,

2. for every $X \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{T})$ the functor $Y \mapsto \mathop{\mathrm{Hom}}\nolimits _\mathcal {T}(X, Y)^\vee$ is representable and the functor $Y \mapsto \mathop{\mathrm{Hom}}\nolimits _\mathcal {T}(Y, X)^\vee$ is corepresentable.

Proof. Condition (1) implies (2) since given $(S, c)$ and $X \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{T})$ the object $S(X)$ represents the functor $Y \mapsto \mathop{\mathrm{Hom}}\nolimits _\mathcal {T}(X, Y)^\vee$ and the object $S^{-1}(X)$ corepresents the functor $Y \mapsto \mathop{\mathrm{Hom}}\nolimits _\mathcal {T}(Y, X)^\vee$.

Assume (2). We will repeatedly use the Yoneda lemma, see Categories, Lemma 4.3.5. For every $X$ denote $S(X)$ the object representing the functor $Y \mapsto \mathop{\mathrm{Hom}}\nolimits _\mathcal {T}(X, Y)^\vee$. Given $\varphi : X \to X'$, we obtain a unique arrow $S(\varphi ) : S(X) \to S(X')$ determined by the corresponding transformation of functors $\mathop{\mathrm{Hom}}\nolimits _\mathcal {T}(X, -)^\vee \to \mathop{\mathrm{Hom}}\nolimits _\mathcal {T}(X', -)^\vee$. Thus $S$ is a functor and we obtain the isomorphisms $c_{X, Y}$ by construction. It remains to show that $S$ is an equivalence. For every $X$ denote $S'(X)$ the object corepresenting the functor $Y \mapsto \mathop{\mathrm{Hom}}\nolimits _\mathcal {T}(Y, X)^\vee$. Arguing as above we find that $S'$ is a functor. We claim that $S'$ is quasi-inverse to $S$. To see this observe that

$\mathop{\mathrm{Hom}}\nolimits _\mathcal {T}(X, Y) = \mathop{\mathrm{Hom}}\nolimits _\mathcal {T}(Y, S(X))^\vee = \mathop{\mathrm{Hom}}\nolimits _\mathcal {T}(S'(S(X)), Y)$

bifunctorially, i.e., we find $S' \circ S \cong \text{id}_\mathcal {T}$. Similarly, we have

$\mathop{\mathrm{Hom}}\nolimits _\mathcal {T}(Y, X) = \mathop{\mathrm{Hom}}\nolimits _\mathcal {T}(S'(X), Y)^\vee = \mathop{\mathrm{Hom}}\nolimits _\mathcal {T}(Y, S(S'(X)))$

and we find $S \circ S' \cong \text{id}_\mathcal {T}$. $\square$

Definition 57.3.2. Let $k$ be a field. Let $\mathcal{T}$ be a $k$-linear triangulated category such that $\dim _ k \mathop{\mathrm{Hom}}\nolimits _\mathcal {T}(X, Y) < \infty$ for all $X, Y \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{T})$. We say a Serre functor exists if the equivalent conditions of Lemma 57.3.1 are satisfied. In this case a Serre functor is a $k$-linear equivalence $S : \mathcal{T} \to \mathcal{T}$ endowed with $k$-linear isomorphisms $c_{X, Y} : \mathop{\mathrm{Hom}}\nolimits _\mathcal {T}(X, Y) \to \mathop{\mathrm{Hom}}\nolimits _\mathcal {T}(Y, S(X))^\vee$ functorial in $X, Y \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{T})$.

Lemma 57.3.3. In the situation of Definition 57.3.2. If a Serre functor exists, then it is unique up to unique isomorphism and it is an exact functor of triangulated categories.

Proof. Given a Serre functor $S$ the object $S(X)$ represents the functor $Y \mapsto \mathop{\mathrm{Hom}}\nolimits _\mathcal {T}(X, Y)^\vee$. Thus the object $S(X)$ together with the functorial identification $\mathop{\mathrm{Hom}}\nolimits _\mathcal {T}(X, Y)^\vee = \mathop{\mathrm{Hom}}\nolimits _\mathcal {T}(Y, S(X))$ is determined up to unique isomorphism by the Yoneda lemma (Categories, Lemma 4.3.5). Moreover, for $\varphi : X \to X'$, the arrow $S(\varphi ) : S(X) \to S(X')$ is uniquely determined by the corresponding transformation of functors $\mathop{\mathrm{Hom}}\nolimits _\mathcal {T}(X, -)^\vee \to \mathop{\mathrm{Hom}}\nolimits _\mathcal {T}(X', -)^\vee$.

For objects $X, Y$ of $\mathcal{T}$ we have

\begin{align*} \mathop{\mathrm{Hom}}\nolimits (Y, S(X))^\vee & = \mathop{\mathrm{Hom}}\nolimits (Y[-1], S(X))^\vee \\ & = \mathop{\mathrm{Hom}}\nolimits (X, Y[-1]) \\ & = \mathop{\mathrm{Hom}}\nolimits (X, Y) \\ & = \mathop{\mathrm{Hom}}\nolimits (Y, S(X))^\vee \end{align*}

By the Yoneda lemma we conclude that there is a unique isomorphism $S(X) \to S(X)$ inducing the isomorphism from top left to bottom right. Since each of the isomorphisms above is functorial in both $X$ and $Y$ we find that this defines an isomorphism of functors $S \circ  \to  \circ S$.

Let $(A, B, C, f, g, h)$ be a distinguished triangle in $\mathcal{T}$. We have to show that the triangle $(S(A), S(B), S(C), S(f), S(g), S(h))$ is distinguished. Here we use the canonical isomorphism $S(A) \to S(A)$ constructed above to identify the target $S(A)$ of $S(h)$ with $S(A)$. We first observe that for any $X$ in $\mathcal{T}$ the triangle $(S(A), S(B), S(C), S(f), S(g), S(h))$ induces a long exact sequence

$\ldots \to \mathop{\mathrm{Hom}}\nolimits (X, S(A)) \to \mathop{\mathrm{Hom}}\nolimits (X, S(B)) \to \mathop{\mathrm{Hom}}\nolimits (X, S(C)) \to \mathop{\mathrm{Hom}}\nolimits (X, S(A)) \to \ldots$

of finite dimensional $k$-vector spaces. Namely, this sequence is $k$-linear dual of the sequence

$\ldots \leftarrow \mathop{\mathrm{Hom}}\nolimits (A, X) \leftarrow \mathop{\mathrm{Hom}}\nolimits (B, X) \leftarrow \mathop{\mathrm{Hom}}\nolimits (C, X) \leftarrow \mathop{\mathrm{Hom}}\nolimits (A, X) \leftarrow \ldots$

which is exact by Derived Categories, Lemma 13.4.2. Next, we choose a distinguished triangle $(S(A), E, S(C), i, p, S(h))$ which is possible by axioms TR1 and TR2. We want to construct the dotted arrow making following diagram commute

$\xymatrix{ S(C)[-1] \ar[r]_-{S(h[-1])} & S(A) \ar[r]_{S(f)} & S(B) \ar[r]_{S(g)} & S(C) \ar[r]_{S(h)} & S(A) \\ S(C)[-1] \ar[r]^-{S(h[-1])} \ar@{=}[u] & S(A) \ar[r]^ i \ar@{=}[u] & E \ar[r]^ p \ar@{..>}[u]^\varphi & S(C) \ar[r]^{S(h)} \ar@{=}[u] & S(A) \ar@{=}[u] }$

Namely, if we have $\varphi$, then we claim for any $X$ the resulting map $\mathop{\mathrm{Hom}}\nolimits (X, E) \to \mathop{\mathrm{Hom}}\nolimits (X, S(B))$ will be an isomorphism of $k$-vector spaces. Namely, we will obtain a commutative diagram

$\xymatrix{ \mathop{\mathrm{Hom}}\nolimits (X, S(C)[-1]) \ar[r] & \mathop{\mathrm{Hom}}\nolimits (X, S(A)) \ar[r] & \mathop{\mathrm{Hom}}\nolimits (X, S(B)) \ar[r] & \mathop{\mathrm{Hom}}\nolimits (X, S(C)) \ar[r] & \mathop{\mathrm{Hom}}\nolimits (X, S(A)) \\ \mathop{\mathrm{Hom}}\nolimits (X, S(C)[-1]) \ar[r] \ar@{=}[u] & \mathop{\mathrm{Hom}}\nolimits (X, S(A)) \ar[r] \ar@{=}[u] & \mathop{\mathrm{Hom}}\nolimits (X, E) \ar[r] \ar[u]^\varphi & \mathop{\mathrm{Hom}}\nolimits (X, S(C)) \ar[r] \ar@{=}[u] & \mathop{\mathrm{Hom}}\nolimits (X, S(A)) \ar@{=}[u] }$

with exact rows (see above) and we can apply the 5 lemma (Homology, Lemma 12.5.20) to see that the middle arrow is an isomorphism. By the Yoneda lemma we conclude that $\varphi$ is an isomorphism. To find $\varphi$ consider the following diagram

$\xymatrix{ \mathop{\mathrm{Hom}}\nolimits (E, S(C)) \ar[r] & \mathop{\mathrm{Hom}}\nolimits (S(A), S(C)) \\ \mathop{\mathrm{Hom}}\nolimits (E, S(B)) \ar[u] \ar[r] & \mathop{\mathrm{Hom}}\nolimits (S(A), S(B)) \ar[u] }$

The elements $p$ and $S(f)$ in positions $(0, 1)$ and $(1, 0)$ define a cohomology class $\xi$ in the total complex of this double complex. The existence of $\varphi$ is equivalent to whether $\xi$ is zero. If we take $k$-linear duals of this and we use the defining property of $S$ we obtain

$\xymatrix{ \mathop{\mathrm{Hom}}\nolimits (C, E) \ar[d] & \mathop{\mathrm{Hom}}\nolimits (C, S(A)) \ar[l] \ar[d] \\ \mathop{\mathrm{Hom}}\nolimits (B, E) & \mathop{\mathrm{Hom}}\nolimits (B, S(A)) \ar[l] }$

Since both $A \to B \to C$ and $S(A) \to E \to S(C)$ are distinguished triangles, we know by TR3 that given elements $\alpha \in \mathop{\mathrm{Hom}}\nolimits (C, E)$ and $\beta \in \mathop{\mathrm{Hom}}\nolimits (B, S(A))$ mapping to the same element in $\mathop{\mathrm{Hom}}\nolimits (B, E)$, there exists an element in $\mathop{\mathrm{Hom}}\nolimits (C, S(A))$ mapping to both $\alpha$ and $\beta$. In other words, the cohomology of the total complex associated to this double complex is zero in degree $1$, i.e., the degree corresponding to $\mathop{\mathrm{Hom}}\nolimits (C, E) \oplus \mathop{\mathrm{Hom}}\nolimits (B, S(A))$. Taking duals the same must be true for the previous one which concludes the proof. $\square$

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