I am currently trying to understand Serre duality from a more formal point of view, with the goal of eventually looking at Grothendieck duality. I'm following Hartshorne III.6-7. For our sake, let's let $X$ be a smooth projective scheme of dimension $n$ over an algebraically closed field with dualizing sheaf $\omega_{X}$. I would like to formalise the statement of Serre duality as something like
The Yoneda pairing, $$\text{Ext}^{n-i}(\mathcal{O}_{X}, \mathcal{F}) \otimes \text{Ext}^{i}(\mathcal{F}, \omega_{X}) \longrightarrow \text{Ext}^{n}(\mathcal{O}_{X}, \omega_{X}),$$
is perfect for any coherent sheaf $\mathcal{F}$.
Is this actually a correct formulation? Of course it depends on the well known fact that $\text{Ext}^{m}(A, B)$ groups can be interpreted as (equivalence classes of) extensions,
$$
0 \rightarrow B \rightarrow E^{1} \rightarrow E^{2} \rightarrow \cdots E^{m} \rightarrow A \rightarrow 0.
$$
So the main question I wanted to ask was, how does this correspond to our normal notion of sheaf cohomology? In particular, we have that $H^{p}(X, \mathcal{F}) = \text{Ext}^{p}(\mathcal{O}_{X}, \mathcal{F})$. So given a $p$-extension of $\mathcal{F}$,
$$
0 \rightarrow \mathcal{F}\rightarrow E^{1} \rightarrow E^{2} \rightarrow \cdots E^{p} \rightarrow \mathcal{O}_{X} \rightarrow 0,
$$
how do we think of this as a $p$-cocycle as in Cech cohomology? I feel like this would massively help me bridge the gap between the more geometric formulation of duality, and the more abstract formal presentation. Any help is appreciated.
Best Answer
I think that conceptually the simplest reason why $\operatorname{Ext}^p(X,\mathcal F) = H^p(X,\mathcal F)$ is that they are both derived functors of the same functor $\operatorname{Hom}(\mathcal O, \bullet) = \Gamma(X,\bullet)$, but this is not very explicit.
Explicitly, let us start with a Čech cocycle. Take an open affine cover $X= \bigcup U_i$, and let us abbreviate the terms in the Čech complex by $\mathcal F^{(p)} = \bigoplus \mathcal F|_{U_{i_1}\cap\cdots \cap U_{i_{p+1}}}$. Take $\eta\in \mathcal F^{(p)}$.
We can use $\eta$ to construct a map of complexes from $\mathcal O_X[-p]$ to the Čech resolution of $\mathcal F$ with respect to this open cover, by mapping $1\in \mathcal O_X$ to $\eta$. $$ \require{AMScd} \begin{CD} @.\cdots @>>> 0 @>>> \mathcal O_X@>>> 0@>>> \cdots \\ @.@. @VVV @VVV @VVV \\ \mathcal F @>>> \cdots @>>> \mathcal F^{(p-1)} @>>> \mathcal F^{(p)} @>>> \mathcal F^{(p+1)} @>>> \cdots \end{CD} $$ Note that $\eta$ being a cocycle is the condition we need for the above to be a map of complexes. The Yoneda extension we are looking for is $$ 0\to \mathcal F \to \mathcal F^{(0)} \to \cdots \to \mathcal F^{(p-2)} \to \mathcal F^{(p-1)}\times_{\mathcal F^{(p)}} \mathcal O_X \to \mathcal O_X\to 0. $$ You can check that this is exact using that the Čech resolution is exact. Another way to see it is that this $\operatorname{Ext}$ class comes from the Yoneda pairing: let $Z^{p} = \ker d:\mathcal F^{(p)}\to \mathcal F^{(p+1)}$. Then the truncation of the Čech complex is a class in $\operatorname{Ext}^p(Z^{p},\mathcal F)$, which we are composing with $\eta\in \operatorname{Ext}^0(\mathcal O_X,Z^{p})$ to get the desired extension class.
Going in the opposite direction seems much harder to me. I don't know how to do it avoiding an injective resolution of $\mathcal F$, which likely defeats the purpose of trying to make things explicit.
Another way to go is to split the Čech resolution into short exact sequences: $$ 0\longrightarrow Z^p \longrightarrow \mathcal F^{(p)}\longrightarrow Z^{p+1}\longrightarrow 0.$$ Since for $p\ge 0$, $\mathcal F^{(p)}$ is a pushforward from affine open sets, $\operatorname{Ext}^i(\mathcal O_X,\mathcal F^{(p)})=0$ for $i>0$. Then we have that $\mathcal F = Z^0$, and taking the $\operatorname{Ext}(\mathcal O_X,\bullet)$ long exact sequences for all the short exact sequences above, we obtain: $$ \operatorname{Ext}^p(\mathcal O_X, Z^0) \cong \operatorname{Ext}^{p-1}(\mathcal O_X, Z^1) \cong \cdots \cong \operatorname{Ext}^{1}(\mathcal O_X, Z^{p-1})\cong \frac{\operatorname{Hom}(\mathcal O_X,Z^p)}{\operatorname{Hom}(\mathcal O_X,\mathcal F^{(p-1)})}. $$ And the last term is exactly the $p$th Čech cohomology group.
In conclusion, I think I answered your question only partially, and I don't know if it is possible to do better. I think this topic is a very good motivation to use derived categories, where all these functors amount to $\operatorname{Hom}_{D(X)}(\mathcal O_X,\bullet)$ and the Yoneda pairing is composition of Hom's.