Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

I'm trying to prove the following problem in the Deformation theory book by Hartshorne.

Any normalized vector bundle $\mathcal E$ of rank 2 degree 1 on an elliptic curve $\mathcal C$ can be written as a non-split extension

$0 \to \mathcal{O_C} \to \mathcal{E} \to \mathcal{O_C(p)} \to 0$

by a uniquely determined point p. (up to isomorphism)

It is easy to see that such data gives unique non-split extension. But the converse direction is not easy to show. I thought the proof of the classification of vector bundles on $\mathbb{P^1}$ may helps me, but I failed. How can I do this? I appreciate any helps or reference.

share|improve this question

1 Answer 1

up vote 4 down vote accepted

I guess you mean that any vector bundle with these properties and which is not a sum of line bundles can be written in that way.

The following standard proof can be found in Friedman's book "Algebraic surfaces and holomorphic vector bundles", page 35.

Since $\textrm{rank}(\mathcal{E})=2$ and $\det{\mathcal{E}}=1$, by Riemann-Roch we have $\chi(\mathcal{E})=1$, hence $h^0(\mathcal{E}) \geq 1$. This means that there is a non-zero map $\mathcal{O}_C \to \mathcal{E}$.

If this map vanishes at some point we have an exact sequence $$0 \to L_1 \to \mathcal{E} \to L_2 \to 0,$$ with $\deg L_1 =d \geq 1$ and $\det L_2=1-d$. Therefore $$\deg (L_1^{-1} \otimes L_2)= 1-2d <0,$$ hence $H^1(L_2^{-1} \otimes L_1)=H^0(L_1^{-1} \otimes L_2)=0$. This shows that $\mathcal{E}=L_1 \oplus L_2$, a contradiction.

Therefore the map $\mathcal{O}_C \to \mathcal{E}$ is never vanishing, so it gives a subbundle of $\mathcal{E}$ whose cokernel has determinant $1$. This precisely means that there exists a point $p \in C$ such that one has a non-split short exact sequence $$0 \to \mathcal{O}_C \to \mathcal{E} \to \mathcal{O}_C(p) \to 0.$$

share|improve this answer
Thank you so much! It's shame that I've almost forgot about general Riemann-Roch formula. By the way, I've just find out normalized condition prevents $\mathcal{E}$ from being a sum of invertible sheaves. So everything works just fine. –  Choa Apr 24 '12 at 16:21

Your Answer


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

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