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 $\pi^* \mathcal{O}(1)$?

  • $\begingroup$ $\pi^* \mathcal{O}(1)$ 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. $\endgroup$ – 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 $(-1,-1,-1)$).


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.