Blow up of $\mathbb{P}^2$ in a point and direct image sheaves

Question

I am trying to understand better direct image sheaves. To do so, I want to start working in a particular and easy example.

Let $\pi:X\rightarrow \mathbb{P}^2$ be the blow up of $\mathbb{P}^2$ in a point, and $E\subset X$ the exceptional curve. What can we say about $\pi_*\mathcal{O}_X(E)$ or $\pi_*\mathcal{O}_X$? Are they invertible sheaves? If so, which is the integer $n$ such that they coincide with $\mathcal{O}_{\mathbb{P}^2}(n)$?

More in general, we can ask the same questions about the sheaves $\pi_*\mathcal{O}_X(aL+bE)$, where $a,b\in \mathbb{Z}$ and $L$ is the pullback of a line in $\mathbb{P}^2$.

Answer 1

1. $\pi_*(\mathcal{O}_X) \cong \mathcal{O}_{\mathbb{P}^2}$. In general given $f:X\rightarrow Y$ a birational projective morphism of noetherian integral schemes, with $Y$ normal, then $f_*\mathcal{O}_X = \mathcal{O}_Y$. The proof is in Hartshorne III.11.4.
2. $\pi_*(\mathcal{O}_X(-E)) \cong \mathcal{I}_p$, where $\mathcal{I}_p$ is the ideal sheaf of $p$ (where here $p$ is the point you're blowing up). To see this, consider the exact sequence

$$0\rightarrow \mathcal{O}_X(-E) \rightarrow \mathcal{O}_X \rightarrow \mathcal{O}_p \rightarrow 0$$ Applying $\pi_*$, we have that

$$0\rightarrow \pi_*\mathcal{O}_X(-E) \rightarrow \pi_*\mathcal{O}_X \rightarrow \pi_* \mathcal{O}_E = \mathcal{O}_p \rightarrow 0$$

where here we note that even though $\pi_*$ is not right exact in general, it is here because you can check by hand that the map is surjective, hence the claim. You can repeat this argument for $\pi_*(\mathcal{O}_X(aE))$ for any $a<0$.

3. $\pi_*(\mathcal{O}_X(E)) \cong \pi_*(\mathcal{O}_X) \cong \mathcal{O}_{\mathbb{P}^2}$.

Tensor the above short exact sequence by $\mathcal{O}_X(E)$ and apply $\pi_*$. We have that the following sequence is exact:

$$0\rightarrow \pi_*\mathcal{O}_X \rightarrow \pi_*\mathcal{O}_X(E) \rightarrow \pi_* \mathcal{O}_E(E)$$

Now we claim that $\pi_*\mathcal{O}_E(E) = 0$. Since this is supported on $p$, this is the same thing as the vector space $H^0(E,\mathcal{O}_E(E))$. Now using the fact that the self intersection of $E$ is $-1$, $\mathcal{O}_E(E)$ is a line bundle of degree $-1$ on $E\cong \mathbb{P}^1$ which has no non-zero global sections. Hence we deduce that $\pi_*(\mathcal{O}_X(E)) \cong \pi_*(\mathcal{O}_X)$. You can repeat this argument for $\pi_*(\mathcal{O}_X(aE))$ for any $a>0$.

4. Finally for $\pi_*\mathcal{O}_X(aL+bE)$, you can compute this using the projection formula. Namely, you have $\pi_*(\mathcal{O}_X(aL) \otimes \mathcal{O}_X(bE)) = \pi_*(\pi^*\mathcal{O}_{\mathbb{P}^2}(aL) \otimes \mathcal{O}_X(bE)) = \mathcal{O}_{\mathbb{P}^2}(aL) \otimes \pi_*\mathcal{O}_X(bE)$.