# Pushing forward vector bundles on a plane curve via projection from a point

Let $C \subset \mathbb{P}^2$ be a smooth plane curve, $P \in \mathbb{P}^2$ is point not on $C$, consider projection from this point $$\pi :\mathbb{P}^2 - \{P\} \to \mathbb{P}^1,$$ and restrict this map to the curve $C$ $$\pi : C \to \mathbb{P}^1.$$

This construction realizes curve $C$ as a (branched) covering over $\mathbb{P}^1$, degree of this covering is $\operatorname{deg}(\pi)=\operatorname{deg}(C)$, let use denote this degree by $d$.

If $F$ is vector bundle over $C$ of rank $r$, then $\pi_*F$ is a vector bundle over $\mathbb{P}^1$, but any vector bundle over $\mathbb{P}^1$ splits as direct sum of line bundles $$\pi_* F \cong \mathcal{O}(a_1) \oplus \mathcal{O}(a_2) \oplus \ldots \oplus \mathcal{O}(a_l).$$

I have two questions.

First. Is there a way to compute numbers $a_1, \ldots, a_l$? I understand that $\operatorname{rk}(\pi_* F)=dr$ i.e. $l=dr$. Moreover, Riemann-Roch formulas for $\mathbb{P}^1$ and $C$ gives $$c_1(\pi_* F)=c_1(F)+r(1-g(C)-d),$$ in other words $a_1+ \ldots + a_l=c_1(F)+r(1-g(C)-d)$.

But is it possible to determine numbers $a_l$ just knowing discreet parameters of $F$? If we need to know moduli of $F$, can one answer this question for elliptic curves where moduli space of vector bundle is again copy of an elliptic curve?

Second question. What is $\pi^* \mathcal{O}(1)$?

• $\pi^* \mathcal{O}(1)$ is the same line bundle as the one from the embedding of $C$ in $\mathbb{P}^2$. Projection from a point does not change the divisor class, it just restricts us to a smaller (incomplete) linear system. – Jake Levinson Oct 11 '14 at 3:42

Am not sure what you mean by discreet parameters, degree and rank? That will not be enough. As you said, take $C$ to be a smooth cubic curve. Then $\mathcal{O}_C$ and a line bundle $L$ of degree zero but $L\neq \mathcal{O}_C$ have the same discreet parameters. But $\pi_*\mathcal{O}_C=\mathcal{O}_{\mathbb{P}^1}\mathcal{O}_{\mathbb{P}^1}(-1)\mathcal{O}_{\mathbb{P}^1}(-2)$, while $\pi_*L$ has no $\mathcal{O}_{\mathbb{P}^1}$ as a direct summand. (In fact, it has splitting type $(-1,-1,-1)$).