Computing sheaf cohomology on ${\mathbb{P}}^n$

Last time we saw that computing $$H^i({\mathbb{P}}_k^n,\mathcal{O}(j)):={\mathbf{R}}^i\Gamma ({\mathbb{P}}_k^n,\mathcal{O}(j))$$ could be achieved by computing the cohomology of the (augmented) Čech complex $$0\to k[x_0,\dots ,x_n]\to \underset{i=0}{\overset{n}{⨁}}k[x_0,\dots ,x_n{]}_{x_i}\to \cdots \to \underset{|I|=n}{⨁}k[x_0,\dots ,x_n{]}_{x_I}\to k[x_0,\dots ,x_n{]}_{x_0\cdots x_n}\to 0$$

Let's look at the case $n=0$ to get a foothold. This reduces to $$0\to k[x_0]\to k[x_0,x_0^{-1}]\to 0$$ Since $x_0$ is regular, there is no $H^0$. We have $$H^1=k[x_0,x_0^{-1}]/k[x_0]$$ which as a $k[x_0]$-module has a filtration $$0\subset \mathrm{Ann}(x_0)\subset \cdots \subset \mathrm{Ann}(x_0^m)\subset \mathrm{Ann}(x_0^{m+1})\subset \cdots$$ whose associated graded pieces are $$\frac{\mathrm{Ann}(x_0^{m+1})}{\mathrm{Ann}(x_0^m)}\cong \frac{k}{x_0^{m+1}}$$ In particular, $H^1$ is a free $k$-module. It is also graded with $|x_0^{-1}|=-1$.

The general complex is isomorphic to the tensor product of complexes $$\underset{i=0}{\overset{n}{⨂}}\left(k[x_i]\to k[x_i{]}_{x_i}\right)$$ over $k$. Since the homologies of the complexes are flat over $k$, we have $$H^s\left(\underset{i=0}{\overset{n}{⨂}}\left(k[x_i]\to k[x_i{]}_{x_i}\right)\right)\cong \underset{i_0+\cdots +i_n=s}{⨁}{H}^{i_j}(k[x_j]\to k[x_j{]}_{x_j})$$ From our computation above we see that $$H^s=\{\begin{array}{ll}0& s\ne n\\ \frac{k[x_0,\dots ,x_n{]}_{x_0\cdots x_n}}{{\sum }_{|I|=n}k[x_0,\dots ,x_n{]}_{x_I}}& s=n\end{array}$$

Theorem. We have $${\mathrm{Ext}}_{{\mathbb{P}}_k}^l(\mathcal{O}(i),\mathcal{O}(j))\cong H^l({\mathbb{P}}_k^n,\mathcal{O}(j-i))\cong \{\begin{array}{ll}k[x_0,\dots ,x_n{]}_{j-i}& l=0\\ k[x_0,\dots ,x_n{]}_{i-j-n-1}& l=n\\ 0& \text{otherwise}\end{array}$$

Fixing the basis vector $1/x_0\cdots x_n$ gives an isomorphism $$H^n({\mathbb{P}}_k^n,\mathcal{O}(-n-1))\cong k$$ From our computations, we see that composition $${\mathrm{Ext}}^i(\mathcal{O}(i),\mathcal{O}(j)){\otimes }_k{\mathrm{Ext}}^{n-i}(\mathcal{O}(j),\mathcal{O}(i-n-1))\to {\mathrm{Ext}}^n(\mathcal{O}(i),\mathcal{O}(i-n-1))\cong H^n({\mathbb{P}}_k^n,\mathcal{O}(-n-1))$$ is given by multiplication $$f,\frac{1}{g}\mapsto \frac{f}{g}$$ up to isomorphism.

Corollary. Composition $${\mathrm{Ext}}^i(\mathcal{O}(i),\mathcal{O}(j)){\otimes }_k{\mathrm{Ext}}^{n-i}(\mathcal{O}(j),\mathcal{O}(i-n-1))\to H^n({\mathbb{P}}_k^n,\mathcal{O}(-n-1))$$ is a perfect pairing. In other words, this induces an isomorphism $${\mathrm{Ext}}^i(\mathcal{O}(i),\mathcal{O}(j){)}^{\vee }\cong {\mathrm{Ext}}^{n-i}(\mathcal{O}(j),\mathcal{O}(i-n-1))$$

For a scheme $X\to \mathrm{Spec}k$, we can ask if the functor $$\mathbf{R}{\mathrm{Hom}}_k(\mathbf{R}{\mathrm{Hom}}_X(E,F),k)$$ is quasi-isomorphic to $$\mathbf{R}{\mathrm{Hom}}_X(F,S(E))$$ for some $k$-linear auto-equivalence $S$. In the case $k$ is simply a field, then this is just a Serre functor. In this statement, it is an enhanced version of a Serre functor.

Proposition. For any objects $E,F\in D^b(\mathrm{coh}{\mathbb{P}}_k^n)$ there are natural quasi-isomorphisms $$\mathbf{R}{\mathrm{Hom}}_k(\mathbf{R}\mathrm{Hom}(E,F),k)\cong \mathbf{R}\mathrm{Hom}(F,S(E))$$ where $S(E)=E(-n-1)[n]$.