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 Krapanov (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?
Best Answer
I think you can prove this statement from (a weaker form of) the usual Serre duality statement using some compatibilities between derived functors, namely:
As a starting point consider the following statement, which is part of the usual Serre duality statement:
We generalize this statement to bounded complexes of coherent sheaves:
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:
[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:
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]