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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...")?

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  • $\begingroup$ 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. $\endgroup$ Aug 25, 2010 at 3:42
  • $\begingroup$ 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). $\endgroup$ Aug 25, 2010 at 3:46
  • $\begingroup$ 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. $\endgroup$ Aug 25, 2010 at 3:51
  • $\begingroup$ 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. $\endgroup$ Aug 25, 2010 at 3:57

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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)

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Let $X$ be a smooth surface and $\pi: \widetilde{X}\rightarrow X$ be the blow up of $X$ at one point. By Hartshorne, Algebraic Geometry, Proposition V.3.4, $$R^{i}\pi_{*}\mathcal{O}_{\widetilde{X}}=0$$ for all $i>0$ and $\pi_*\mathcal{O}_\widetilde{X}\cong\mathcal{O}_X$. By the projection formula, for any divisor $D\subset X$, we have $$H^{i}(\widetilde{X},\pi^{*}D)=H^{i}(X,D).$$

Now we come back to your question. If $X$ is the blow up of $\mathbb{P}^{2}$, $H^{i}(X,nH)=H^{i}(\mathbb{P}^{2},\mathcal{O}(n))$. $$0\rightarrow \mathcal{O}_{X}(nH+(m-1)E)\rightarrow \mathcal{O}_{X}(nH+mE)\rightarrow \mathcal{O}_{E}(nH+mE)\rightarrow 0$$ Since $E(nH+mE)=mE^2=-m$, $\mathcal{O}_{X}(nH+mH)\cong \mathcal{O}_{\mathbb{P}^2}(-m)$. Then by taking the long exact sequence, we may relate the cohomology of $nH+(m-1)E$ and $nH+mE$. Since we already know the cohomology of $nH$, we can inductively get the cohomology of $nH+mE$ for any $m,n\in \mathbb{Z}$.

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