The Stacks project

57.3 Serre functors

The material in this section is taken from [Bondal-Kapranov].

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 $X, Y \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{T})$.

Remark 57.3.3. Let $X^0 \to X^1 \to X^2 \to X^0[1]$ and $Y^0 \to Y^1 \to Y^2 \to Y^0[1]$ be distinguished triangles in a triangulated category. For $p \in \mathbf{Z}$ write $p = 3n + i$ with $i \in \{ 0, 1, 2\} $ and set we set $X^ p = X^ i[n]$. Simlarly for $Y^ q$. Consider the double complex with terms

\[ K^{p, q} = \mathop{\mathrm{Hom}}\nolimits (X^{-p}, Y^ q) \]

The differential $d_1 : K^{p, q} \to K^{p + 1, q}$ is given by the map $X^{-p - 1} \to X^{-p}$ (equal to the corresponding map in the first distinguished triangle up the a shift) and the differential $d_2 : K^{p, q} \to K^{p, q + 1}$ likewise by the map $Y^ q \to Y^{q + 1}$. From Derived Categories, Lemma 13.4.2 we see that the rows and columns of this double complex are exact. Furthermore, we see that $K^{p, q} = K^{p + 3, q - 3}$ and these equalities are compatible with the differentials. Finally, axiom TR3 implies one additional property: given $\alpha \in K^{p, q}$ and $\beta \in K^{p - 1, q + 1}$ such that $d_2 \alpha = d_1 \beta $, there exists a $\gamma \in K^{p - 2, q + 2}$ such that $d_1 \gamma = d_2 \beta $ in $K^{p - 1, q + 2}$ and $d_2 \gamma = d_1 \alpha $ in $K^{p - 2, q + 3} = K^{p + 1, q}$. (Hint: for $p = q = 0$ this is exactly the statement of TR3 and for other indices prove it by shifting.) A double complex with these properties is called a matress (see [Bondal-Kapranov]).

Remark 57.3.4. Let $k$ be a field and let $K^{\bullet , \bullet }$ be a double complex of finite dimensional $k$-vector spaces which is also a matress as in Remark 57.3.3. We claim that the dual double complex $L^{p, q} = \mathop{\mathrm{Hom}}\nolimits _ k(K^{-q, -p}, k)$ is also a matress. The exactness of rows and columns and the 3-periodicity are immediate. To see the additional condition holds, consider the linear map

\[ \partial : K^{p, q} \oplus K^{p - 1, q + 1} \oplus K^{p - 2, q + 2} \longrightarrow K^{p, q + 1} \oplus K^{p - 1, q + 2} \oplus K^{p - 2, q + 3} \]

sending $(\alpha , \beta , \gamma )$ to $(d_2 \alpha - d_1 \beta , d_2 \gamma - d_1 \alpha , d_1 \gamma - d_2 \beta )$ with identifications as in Remark 57.3.3. The condition of being a matress is that

\[ \mathop{\mathrm{Im}}(\partial ) \cap \left( 0 \oplus K^{p - 1, q + 2} \oplus K^{p - 2, q + 3} \right) = \mathop{\mathrm{Im}}(\partial |_{0 \oplus 0 \oplus K^{p - 2, q + 2}}) \]

Denoting ${}^\wedge $ the dual of a vector space or map, taking duals we get

\[ \mathop{\mathrm{Ker}}(\partial ^\wedge ) + \left(L^{-p, -q - 1} \oplus 0 \oplus 0 \right) = \mathop{\mathrm{Ker}}(\text{pr}_{L^{-p + 2, -q - 2}} \circ \partial ^\wedge ) \]

where the $\text{pr}$ term indicates projection. This implies that $\mathop{\mathrm{Im}}(\partial ^\wedge ) \cap (L^{-p, -q} \oplus L^{-p + 1, -q - 1} \oplus 0)$ is equal to $\mathop{\mathrm{Im}}(\partial ^\wedge |_{L^{-p, -q - 1} \oplus 0 \oplus 0})$ and this translates into the matress condition for $L^{\bullet , \bullet }$. Some details omitted.

Lemma 57.3.5. 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)[1])^\vee & = \mathop{\mathrm{Hom}}\nolimits (Y[-1], S(X))^\vee \\ & = \mathop{\mathrm{Hom}}\nolimits (X, Y[-1]) \\ & = \mathop{\mathrm{Hom}}\nolimits (X[1], Y) \\ & = \mathop{\mathrm{Hom}}\nolimits (Y, S(X[1]))^\vee \end{align*}

By the Yoneda lemma we conclude that there is a unique isomorphism $S(X[1]) \to S(X)[1]$ 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 $S \circ [1] \to [1] \circ S$.

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 $S(A[1]) \to S(A)[1]$ constructed above to identify the target $S(A[1])$ of $S(h)$ with $S(A)[1]$. 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

\[ \ldots \to \mathop{\mathrm{Hom}}\nolimits (X, S(A)) \to \mathop{\mathrm{Hom}}\nolimits (X, S(B)) \to \mathop{\mathrm{Hom}}\nolimits (X, S(C)) \to \mathop{\mathrm{Hom}}\nolimits (X, S(A)[1]) \to \ldots \]

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

\[ \ldots \leftarrow \mathop{\mathrm{Hom}}\nolimits (A, X) \leftarrow \mathop{\mathrm{Hom}}\nolimits (B, X) \leftarrow \mathop{\mathrm{Hom}}\nolimits (C, X) \leftarrow \mathop{\mathrm{Hom}}\nolimits (A[1], X) \leftarrow \ldots \]

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

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

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

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

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 $, let us form the matress of Remark 57.3.3 from the distinguished triangles $A \to B \to C \to A[1]$ and $S(A) \to E \to S(C) \to S(A)[1]$. By Remark 57.3.4 the $k$-linear dual of this is also a matress. Using that $S$ is a Serre functor, we see as above that this dual matress is the double complex made from the triangles (the second of which we do not yet know to be distinguished) $S(A) \to E \to S(C) \to S(A)[1]$ and $S(A) \to S(B) \to S(C) \to S(A)[1]$. But the matress condition exactly tells us that the maps $S(A) \to S(A)$ and $S(C) \to S(C)$ fit into a morphism of triangles, i.e., that we have a $\varphi $ making the diagram above commute. $\square$

Comments (2)

Comment #9991 by Jonas on

In the last step of the proof of Lemma 57.3.3, how exactly is TR3 used to obtain an element of mapping to both and ?

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