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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.

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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 $$0 \to \mathcal{O}_C \to \mathcal{E} \to \mathcal{O}_C(p) \to 0.$$

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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
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