# Serre duality in derived category

Let $$X$$ be a smooth projective variety over a field $$k$$, and $$\omega_X$$ its canonical bundle. Denote by $$D^b(X):=D^b(\mathbf{Coh})$$ the bounded derived category of coherent sheaves on $$X$$.

Theorem: For any two comples $$\mathcal{E}^\bullet,\mathcal{F}^\bullet$$ in $$D^b(X)$$ there exists a functorial isomorphism $$\eta : Ext^i(\mathcal{E}^\bullet,\mathcal{F}^\bullet) \to Ext^{n-i}(\mathcal{F}^\bullet,\mathcal{E}^\bullet\otimes \omega_X)^*,$$ where $$^*$$ denotes the vector space dual.

This theorem can be found in D. Huybrechts's Fourier-Mukai Transforms in Algebraic Geometry (theorem 3.12). It basically says that the exact functor $$\cdot\otimes\omega_X[n] : D^b(X) \to D^b(X)$$ is a Serre functor in the sense of Bondal and Kapranov (see here for instance).

Now my question is: how can we prove this theorem? I know that this is a particular case of Grothendieck-Verdier duality, but is it possible to avoid it?

My attempts: I know the classical Serre Duality (Harthorne's Algebraic Geometry, III.7), which in particular says that the theorem is true when $$\mathcal{E}^\bullet = \mathcal{E}$$ is a locally free sheaf and $$\mathcal{F}^\bullet=\mathcal{F}$$ is any coherent sheaf. For the general case, I tried to use the definition: $$Ext^i(\mathcal{E}^\bullet,\mathcal{F}^\bullet) = H^i(RHom^\bullet(\mathcal{E}^\bullet,\mathcal{F}^\bullet)),$$ where $$Hom^n(A^\bullet,B^\bullet)=\bigoplus Hom(A^k,B^{k+n})$$ with $$d(f)=d_B\circ f - (-1)^nf\circ d_A$$. I tried to replace $$\mathcal{F}^\bullet$$ by a quasi-isomorphic complex of injectives (of quasi-coherent sheaves) and $$\mathcal{E}$$ by a quasi-isomorphic complex of locally free sheaves so I can use the classical Serre duality on each summand of the direct sum, but I'm not sure this is a way to start. If someone could give me some help I would really appreciate.

EDIT: Thanks to the answer of @mayer_vietoris, I tried the following. Suppose $$\mathcal{E}^\bullet$$ is a complex of locally free sheaves and $$\mathcal{F}$$ is a complex of injectives so that $$RHom^\bullet(\mathcal{E}^\bullet,\mathcal{F}^\bullet)=Hom^\bullet(\mathcal{E}^\bullet,\mathcal{F}^\bullet).$$ \begin{align*} Hom^i(\mathcal{E}^\bullet,\mathcal{F}^\bullet) &= \bigoplus_k Hom(\mathcal{E}^k,\mathcal{F}^{k+i}) \\ &= \bigoplus Ext^0(\mathcal{O}_X,(\mathcal{E}^k)^ \vee\otimes\mathcal{F}^{k+i})\\ &\simeq \bigoplus Ext^n(\mathcal{F}^{k+i},\mathcal{E}^k\otimes\omega_X)^* & \text{(by Serre Duality)}\\ &\simeq \bigoplus Hom(\mathcal{F}^{k+i},\mathcal{E}^k\otimes\omega_X[n])^*\\ &\simeq Hom^{n-i}(\mathcal{F}^\bullet,\mathcal{E}^\bullet\otimes\omega_X)^*. \end{align*} Up to a replacement of $$\mathcal{E}^\bullet\otimes\omega_X$$ by a complex of injectives, I would like to conclude by replacing $$Hom^i(\dots$$ by $$H^i(Hom^\bullet(\dots$$ and similarly with $$Hom^{n-i}$$. Thus I obtain the desired equality. Am I wrong somewhere?

• I think it works out. Maybe you could put it into its own answer? Also, it would have helped me to mention that $\operatorname{Ext}^k(A, B) = \operatorname{Hom}_{D^b(X)}(A, B[k])$ for all $A, B \in D^b(X)$. Commented Nov 19, 2019 at 12:02

I think you can prove this statement from (a weaker form of) the usual Serre duality statement using some compatibilities between derived functors, namely:

• For all $$\mathcal{F}^{\bullet}\in D^{b}(X)$$ we have a natural isomorphism $$R\Gamma(X,-)\circ R\mathcal{Hom}(\mathcal{F}^{\bullet},-)\cong R\operatorname{Hom}(\mathcal{F}^{\bullet},-).$$
• For all $$n\in \mathbb{N}$$ we have a natural isomorphisms $$\operatorname{Hom}(\mathcal{O}[-n],-)\cong H^{n}(R\Gamma(X,-)).$$
• For all $$\mathcal{F}^{\bullet},\mathcal{G}^{\bullet},\mathcal{E}^{\bullet}\in D^{b}(X)$$ we have an isomorphism $$\operatorname{Hom}(\mathcal{F}^{\bullet},\mathcal{E}^{\bullet}\otimes^{L}\mathcal{G}^{\bullet})\cong \operatorname{Hom}(R\mathcal{Hom}(\mathcal{E}^{\bullet},\mathcal{F}^{\bullet}),\mathcal{G}^{\bullet}).$$ This follows from formula (3.15) in Huybrechts' book after applying derived global sections and taking cohomology on degree zero. The formula (3.15) is functorial, because it is based on a sequence of canonical isomorphisms coming from exercise II.5.1 in Hartshorne's book. Hence the previous isomorphism is functorial as well.

As a starting point consider the following statement, which is part of the usual Serre duality statement:

For all line bundles $$\mathcal{L}$$ on $$X$$ we have a natural perfect pairing $$\operatorname{Hom}(\mathcal{L},\omega_{X})\times H^{n}(X,\mathcal{L})\to H^{n}(X,\omega_{X})\cong k$$

We generalize this statement to bounded complexes of coherent sheaves:

For all $$\mathcal{F}^{\bullet}\in D^{b}(X)$$ we have a natural perfect pairing $$\operatorname{Hom}(\mathcal{F}^{\bullet},\omega_{X})\times \operatorname{Hom}(\mathcal{O}_{X}[-n],\mathcal{F}^{\bullet})\to {\operatorname{Hom}(\mathcal{O}_{X}[-n],\omega_{X})}\cong k$$

Proof:

We have to check that the induced morphism $$\operatorname{Hom}(\mathcal{F}^{\bullet},\omega_{X})\to \operatorname{Hom}(\mathcal{O}_{X}[-n],\mathcal{F}^{\bullet})^{\vee}$$ is an isomorphism. We divide the proof into smaller steps:

1. The result is true for finite direct sums of line bundles, because all functors are additive and we know the result for line bundles already.
2. The result is true for all coherent sheaves (complexes concentrated on degree zero). To see this we use that every coherent sheaf $$\mathcal{F}$$ on a smooth projective variety admits a surjection from a finite direct sum of line bundles $$\mathcal{E}=\oplus_{i=1}^{l}\mathcal{L}_{i}$$. So we have a distinguished triangle $$\mathcal{K} \to \mathcal{E} \to \mathcal{F} \to \mathcal{K}[1]$$ Apply the cohomological functors $$\operatorname{Hom}(-,\omega_{X})$$ and $$\operatorname{Hom}(\mathcal{O}_{X}[-n],-)^{\vee}$$ to obtain a ladder diagram with first row the exact sequence $$0\to \operatorname{Hom}(\mathcal{F},\omega_{X})\to \operatorname{Hom}(\mathcal{E},\omega_{X})\to \operatorname{Hom}(\mathcal{K},\omega_{X})$$ and second row the exact sequence $$0\to \operatorname{Hom}(\mathcal{O}_{X}[-n],\mathcal{F})^{\vee}\to \operatorname{Hom}(\mathcal{O}_{X}[-n],\mathcal{E})^{\vee}\to \operatorname{Hom}(\mathcal{O}_{X}[-n],\mathcal{K})^{\vee}.$$ The zero on the first row corresponds to a negative ext group between coherent sheaves and the one on the second row follows from Grothendieck vanishing. By exactness of the rows and the result for finite direct sums of line bundles, the composition $$\operatorname{Hom}(\mathcal{F},\omega_{X})\to \operatorname{Hom}(\mathcal{O}_{X}[-n],\mathcal{E})^{\vee}$$ is injective. By commutativity of the diagram the map $$\operatorname{Hom}(\mathcal{F},\omega_{X})\to\operatorname{Hom}(\mathcal{O}_{X},\mathcal{F})^{\vee}$$ is injective as well. Since $$\mathcal{F}$$ was an arbitrary coherent sheaf on $$X$$, this is also true for the corresponding map for $$\mathcal{K}$$. Hence we can apply the 5-lemma (the version which is sometimes refered to as 4-lemma) to get the isomorphism that we wanted.
3. The result is true for all bounded complexes of coherent sheaves. This can be shown by induction on the length of the complex. The result is already known for complexes of length zero, so let $$\mathcal{F}^{\bullet}$$ be a complex of length $$n+1$$ for some integer $$n\geqslant 0$$. For simplicity assume that this complex is concentrated between degrees $$-n$$ and $$0$$. Let $$\mathcal{F}^{\bullet}_{\leqslant -1}$$ be the stupid truncation (not preserving the cohomology of the complex), so that we get a distinguished triangle $$\mathcal{F}^{\bullet}_{\leqslant -1}\to \mathcal{F}^{\bullet}\to \mathcal{F}^{0} \to \mathcal{F}^{\bullet}_{\leqslant -1}[1]$$ By induction hypothesis and arguing with the 5-lemma as before we obtain the desired isomorphism.

[End of proof]

Now we can use the compatibilities mentioned at the beginning to deduce the result in the form stated in Huybrechts' book. The conclusion is the following:

The category $$D^{b}(X)$$ has a Serre functor given by $$\mathcal{F}^{\bullet} \mapsto \mathcal{F}^{\bullet}\otimes \omega_{X}[n]$$

Proof:

For all $$\mathcal{F}^{\bullet}$$ and $$\mathcal{G}^{\bullet}$$ in $$D^{b}(X)$$ we have the following sequence of functorial isomorphisms: $$\operatorname{Hom}(\mathcal{F}^{\bullet},\mathcal{G}^{\bullet})$$ $$\cong \operatorname{Hom}(\mathcal{F}^{\bullet}\otimes \omega_{X},\mathcal{G}^{\bullet}\otimes \omega_{X})$$ $$\cong \operatorname{Hom}(R\mathcal{Hom}(\mathcal{G}^{\bullet},\mathcal{F}^{\bullet}\otimes \omega_{X}),\omega_{X})$$ $$\cong \operatorname{Hom}(\mathcal{O}_{X}[-n],R\mathcal{Hom}(\mathcal{G}^{\bullet},\mathcal{F}^{\bullet}\otimes \omega_{X}))^{\vee}$$ $$\cong \operatorname{Hom}(\mathcal{O}_{X},R\mathcal{Hom}(\mathcal{G}^{\bullet},\mathcal{F}^{\bullet}\otimes \omega_{X}[n]))^{\vee}$$ $$\cong H^{0}(R\Gamma(\mathcal{Hom}(\mathcal{G}^{\bullet},\mathcal{F}^{\bullet}\otimes \omega_{X}[n])))^{\vee}$$ $$\cong H^{0}(R\operatorname{Hom}(\mathcal{G}^{\bullet},\mathcal{F}^{\bullet}\otimes \omega_{X}[n]))^{\vee}$$ $$\cong \operatorname{Hom}(\mathcal{G}^{\bullet},\mathcal{F}^{\bullet}\otimes \omega_{X}[n])^{\vee}.$$ [End of proof]

I believe that you can find out what you look for explicitly in a very beautiful answer here, The idea behind the notion of dualizing sheaf.

• Thank you for your answer. Indeed the answer given in the other post is really interesting, but it does not deal with complexes (I mean, it only treats complexes concentrated in degree 0, i.e. coherent sheaves). Commented May 1, 2018 at 16:21

As proposed by red_trumpet, here is my solution, initially posted as an edit in the question:

Using resolution, we can assume that $$\mathcal{E}^\bullet$$ is a complex of locally free sheaves and $$\mathcal{F}^\bullet$$ is a complex of injectives so that $$RHom^\bullet(\mathcal{E}^\bullet,\mathcal{F}^\bullet)=Hom^\bullet(\mathcal{E}^\bullet,\mathcal{F}^\bullet).$$ Recall we have $$Hom_{D^b(X)}(A,B[i])\simeq Ext^i(A,B):=\mathcal{H}^i(RHom(A,B))$$ for any $$A,B\in D^b(X)$$ and this $$Ext$$ coincide with the usual one for coherent sheaves. We obtain: \begin{align*} Hom^i(\mathcal{E}^\bullet,\mathcal{F}^\bullet) &= \bigoplus_k Hom(\mathcal{E}^k,\mathcal{F}^{k+i}) \\ &= \bigoplus Ext^0(\mathcal{O}_X,(\mathcal{E}^k)^ \vee\otimes\mathcal{F}^{k+i})\\ &\simeq \bigoplus Ext^n(\mathcal{F}^{k+i},\mathcal{E}^k\otimes\omega_X)^* & \text{(by Serre Duality)}\\ &\simeq \bigoplus Hom(\mathcal{F}^{k+i},\mathcal{E}^k\otimes\omega_X[n])^*\\ &\simeq Hom^{n-i}(\mathcal{F}^\bullet,\mathcal{E}^\bullet\otimes\omega_X)^*. \end{align*} Now, one needs to show that these morphisms induce a complex morphism $$Hom^\bullet(\mathcal{E}^\bullet,\mathcal{F}^\bullet) \to Hom^\bullet(\mathcal{F}^\bullet,\mathcal{E}^\bullet\otimes\omega_X[n])^*$$ which I think comes from naturality of (classical) Serre Duality.

I realize that another way to prove $$Ext^i(\mathcal{E}^\bullet,\mathcal{F}^\bullet)\simeq Ext^{n-i}(\mathcal{F}^\bullet,\mathcal{E}^\bullet)$$ is to use the exact sequences $$E^{p,q}_2 = Ext^p(\mathcal{H}^{-q}(\mathcal{E}^\bullet),\mathcal{F}^\bullet) \Rightarrow Ext^{p+q}(\mathcal{E}^\bullet,\mathcal{F}^\bullet)$$ $$E^{p,q}_2 = Ext^p(\mathcal{E}^\bullet,\mathcal{H}^{q}(\mathcal{F}^\bullet)) \Rightarrow Ext^{p+q}(\mathcal{E}^\bullet,\mathcal{F}^\bullet)$$ but naturality of the isomorphisms still requires naturality of classical Serre Duality.