# Computing sheaf cohomology on $P_{n}$

Last time we saw that computing $H_{i}(P_{k},O(j)):=R_{i}Γ(P_{k},O(j))$ could be achieved by computing the cohomology of the (augmented) Čech complex $0→k[x_{0},…,x_{n}]→i=0⨁n k[x_{0},…,x_{n}]_{x_{i}}→⋯→∣I∣=n⨁ k[x_{0},…,x_{n}]_{x_{I}}→k[x_{0},…,x_{n}]_{x_{0}⋯x_{n}}→0$

Let's look at the case $n=0$ to get a foothold. This reduces to $0→k[x_{0}]→k[x_{0},x_{0}]→0$ Since $x_{0}$ is regular, there is no $H_{0}$. We have $H_{1}=k[x_{0},x_{0}]/k[x_{0}]$ which as a $k[x_{0}]$-module has a filtration $0⊂Ann(x_{0})⊂⋯⊂Ann(x_{0})⊂Ann(x_{0})⊂⋯$ whose associated graded pieces are $Ann(x_{0})Ann(x_{0}) ≅x_{0}k $ In particular, $H_{1}$ is a free $k$-module. It is also graded with $∣x_{0}∣=−1$.

The general complex is isomorphic to the tensor product of complexes $i=0⨂n (k[x_{i}]→k[x_{i}]_{x_{i}})$ over $k$. Since the homologies of the complexes are flat over $k$, we have $H_{s}(i=0⨂n (k[x_{i}]→k[x_{i}]_{x_{i}}))≅i_{0}+⋯+i_{n}=s⨁ H_{i_{j}}(k[x_{j}]→k[x_{j}]_{x_{j}})$ From our computation above we see that $H_{s}=⎩⎨⎧ 0∑_{∣I∣=n}k[x_{0},…,x_{n}]_{x_{I}}k[x_{0},…,x_{n}]_{x_{0}⋯x_{n}} s=ns=n $

**Theorem**. We have
$Ext_{P_{k}}(O(i),O(j))≅H_{l}(P_{k},O(j−i))≅⎩⎨⎧ k[x_{0},…,x_{n}]_{j−i}k[x_{0},…,x_{n}]_{i−j−n−1}0 l=0l=notherwise $

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**Proof**. (Expand to view)

We have an isomorphism of functors $Hom(O(i),F)≅Γ(P_{k},F(−i))$ which gives the isomorphims of derived functors $Ext_{l}(O(i),F)≅R_{i}Γ(P_{k},F(−i)).$

Now we use the computation above. Recall that
$H_{s}(P_{k},O(d))≅H_{s}(0→i=0⨁n R_{x_{i}}→⋯→R_{x_{0}⋯x_{n}}→0)_{d}$
where $R=k[x_{0},…,x_{n}]$. Thus, we have
$H_{0}(P_{k},O(j−i))≅R_{j−i}$
and we get the vanishing for $0<s<n$. For $s=n$, from the discussion above,

$x_{0}⋯x_{n}1 ,a_{0},…,a_{n}≥1$
forms a $k$-basis for $H_{n}$. Rewriting this as
$x_{0}⋯x_{n}1 =x_{0}⋯x_{n}1 x_{0}⋯x_{n}1 $
we get a bijection between $H_{d}$ and $R_{−d−n−1}$.

Fixing the basis vector $1/x_{0}⋯x_{n}$ gives an isomorphism $H_{n}(P_{k},O(−n−1))≅k$ From our computations, we see that composition $Ext_{i}(O(i),O(j))⊗_{k}Ext_{n−i}(O(j),O(i−n−1))→Ext_{n}(O(i),O(i−n−1))≅H_{n}(P_{k},O(−n−1))$ is given by multiplication $f,g1 ↦gf $ up to isomorphism.

**Corollary**. Composition
$Ext_{i}(O(i),O(j))⊗_{k}Ext_{n−i}(O(j),O(i−n−1))→H_{n}(P_{k},O(−n−1))$
is a perfect pairing. In other words, this induces an isomorphism
$Ext_{i}(O(i),O(j))_{∨}≅Ext_{n−i}(O(j),O(i−n−1))$

For a scheme $X→Speck$, we can ask if the functor $RHom_{k}(RHom_{X}(E,F),k)$ is quasi-isomorphic to $RHom_{X}(F,S(E))$ 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 $E,F∈D_{b}(cohP_{k})$ there
are natural quasi-isomorphisms
$RHom_{k}(RHom(E,F),k)≅RHom(F,S(E))$
where $S(E)=E(−n−1)[n]$.

##
**Proof**. (Expand to view)

Some care is required for a general $k$ which we won't take. From composition we have $RHom(E,F)⊗L _{k}RHom(F,S(E))≅RHom(RHom(E,E),O(−n−1)[n])$ Using the natural map $O→RHom(E,E)$ lands us in \mathbf{R}\operatorame{Hom}(\mathcal O, \mathcal O(-n-1)[n]) which is quasi-isomorphic to $H_{n}(P_{k},O(−n−1))≅k$.

This gives a natural map $RHom(F,S(E))→RHom_{k}(RHom(E,F),k)$ which we see is a quasi-isomorphism for a collection of generators $O(j)$. Therefore, it is for all objects of $D_{b}(cohP_{k})$.

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