Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $X$ denote the blowup of $\mathbb P^2$, $E$ the exceptional divisor, and $H$ the pullback of the hyperplane class. How can I compute $H^0(X,mE+nH)$, $H^1(X,mE+nH)$, and $H^2(X,mE+nH)$ for $m,n \in \mathbb Z$? If I'm working analytically, how can I think of these geometrically (e.g. "an element of $H^1(X,2E)$ is equivalent to a $1$-form on $\mathbb P^2$ such that...")?

share|cite|improve this question
This is not really an answer, but rather a strategy. To compute the cohomology groups you are interested, I think there is no escaping simplying doing the Cech cohomology calculation directly. This will probably involve a bit of annoying combinatorics, but should be quite doable, since $X$ has a nice open cover by four affines, each of which is easily understood. You might be able to save a bit of labor by using Serre duality and the results on blowups of surfaces in Chapter V of Hartshorne. – Sam Lichtenstein Aug 25 '10 at 3:42
A remark: if you only wanted the Euler characteristics of these divisors, you could probably get away without computing anything. Use Riemann-Roch to express $\chi(D)$ in terms of the canonical divisor $K$ and the intersection pairing on $Pic(X)$. But this pairing is easy to compute geometrically: $H$ and $E$ freely generate $Pic(X)$ and the pairing is given by $E^2=-1, H^2=0,E\cdot H = 1$. And we know $K$ because we know the canonical divisor on $\mathbf{P}^2$, using results in Hartshorne, Ch. V (section 3, I believe). – Sam Lichtenstein Aug 25 '10 at 3:46
Unfortunately, I don't believe there is an easy "geometric" interpretation of these Zariski cohomology groups in degrees $1$, for example in terms of differential forms. The reason there is for curves $C$, you might say, is because by Serre duality $H^1(C,\mathcal{O}(D))$ can be related to 1-forms. But in dealing with surfaces $X$, 19th c. algebraic geometers considered $H^1(X,\mathcal{O}(D))$ to be sort of an "error term" in Riemann-Roch (which is naturally expressed using Euler characteristics), the so-called "superabundance". But $H^0$ and $H^2$ have "geometric" description by Serre dualty. – Sam Lichtenstein Aug 25 '10 at 3:51
All is not lost, however. Holomorphic differentials do capture cohomological information about a variety, the so-called "Algebraic de Rham cohomology" defined vaguely analogously to the way it is in diff. geom. But to compute it you actually need to use a spectral sequence with $E_2$ page $H^p(X,\Omega^q_X)$. On a smooth surface $X$, $\Omega^1_X$ is a vector bundle of rank 2, so without doing some more work, you can't immediately get at is cohomology using only the cohomology of divisors. – Sam Lichtenstein Aug 25 '10 at 3:57

I assume you're only blowing up at one point, so here's at least a nice geometric description for $H^0$.

If $p$ is the point blown up to $E$, then $dH-E$ is the system of plane curves of degree $d$ passing through $p$, $dH-2E$ are those that have a double point at $p$, etc. This works with any number of blownup points, and a good exercise is using this interpretation to find all 27 lines on a cubic surface (which is the blowup at 6 points)

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.