In Hartshorne's book, Section 3.5, the cohomology of projective spaces is computed. How to compute the cohomology of projective schemes? Maybe the general case is complicated, please look at the following simple case:

Let $R=k[x_1,\dotsc,x_n]$ with the usual grading, $f \in R$ be a homogeneous element of degree $d$, and $X=\mathrm{Proj}(R/(f))$. How to compute $H^i(X, \mathcal{O}_X(a))$ for each $a \in \mathbb{Z}$?

I'd also glad to see some other typical examples to compute the cohomology of projective schemes.

up vote 13 down vote accepted

The cohomology of projective schemes can be computed by resolving the structure sheaf by twists of $\mathcal O_{\mathbb P^n}$. In the case of hypersurfaces, the resolution is especially nice, and the cohomology can be found by writing long exact sequences.

In your example, we have an exact sequence $$ 0 \to \mathcal O_{\mathbb P^ n}(-d) \to \mathcal O_{\mathbb P} \to \mathcal O_X \to 0 $$

where the left map is multiplication by $f$. Then one can take cohomology of this to compute $H^i(X,\mathcal O_X)$. We get $$ ...\to H^{i}(\mathcal O_{\mathbb P}(-d)) \to H^i(\mathcal O_{\mathbb P}) \to H^i(\mathcal O_X) \to H^{i+1}(\mathcal O_{\mathbb P}(-d)) \to ... $$

If $0 < i < n$, then $H^i(\mathcal O_{\mathbb P}(l))=0$, so we get $H^i(\mathcal O_X(k))=0$ for $0 < i < n-1$. It remains to compute $H^0(\mathcal O_{X}(k))$ and $H^{n-1}(\mathcal O_X(k))$.

We have $$ 0 \to H^0(\mathcal O_{\mathbb P}) \to H^0(\mathcal O_{X}) \to 0 $$ so $H^0(\mathcal O_{X})=1$. But twisting will give $$ 0 \to H^0(\mathcal O_{\mathbb P}(-d+k)) \to H^0(\mathcal O_{\mathbb P}(k)) \to H^0(\mathcal O_X(k)) \to 0 $$ So for $k \geq d$, we get that $H^0(\mathcal O_X(k))$ is a difference of binomial coefficients. For $k < d$, we have $H^0(\mathcal O_X(k)) \simeq H^0(\mathcal O_{\mathbb P}(k))$.

To compute the top cohomology, just reverse the previous argument (in fancy language: use Serre duality on $\mathbb P^n$ and the adjunction formula on $X$).

In general, if $X$ is defined by more than one equation, you will have to split the resolution of $\mathcal O_X$ into short exact sequences and do more work. But it all follows from the cohomology of $\mathbb P^n$ and the fact that $h^0(\mathcal O_{\mathbb P}(d))=\binom{n+d}{d}$.

  • Dear Fredrik, everything is correct in your beautiful answer but actually we also have $H^0(X, \mathcal O_X(k)) \gt 1$ for any $k\geq 1$ (not only for $k\geq d$). Anyway: +1, of course. – Georges Elencwajg Nov 14 '14 at 10:04
  • @Georges: Thank you for your nice comment and for the correction. I've edited my answer. – Fredrik Meyer Nov 14 '14 at 10:07

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