# Rank 2 vector bundle on a product of elliptic curves

Let $E$, $F$ be two complex elliptic curves, and $A=E \times F$. Let us denote by

$\pi_E \colon A \to E, \quad \pi_F \colon A \to F$

the natural projections. For all $p \in F$ let us write $E_p$ instead of $\pi_F^*(p)$.

Now let us fix $p \in F$ and consider the unique indecomposable rank $2$ vector bundle $\mathcal{F}$ on $A$ defined by the extension

$0 \to \mathcal{O}_A \to \mathcal{F} \to \mathcal{O}_A(-E_p) \to 0$

My main question is the following:

What is the restriction of $\mathcal{F}$ to a fibre of type $E_x$, for $x \in F$ general?

Related to this, another question is

What are $\pi_{F*} \mathcal{F}$ and $R^1 \pi_{F*} \mathcal{F}$?

Observe that the restriction of $\mathcal{F}$ to $E_x$ is given by an extension

$0 \to \mathcal{O}_{E_x} \to \mathcal{F}|_{E_x} \to \mathcal{O}_{E_x} \to 0$,

so by Atiyah's classification of vector bundles over an elliptic curve we have that $\mathcal{F}|_{E_x}$ is either $\mathcal{O}_{E_x} \oplus \mathcal{O}_{E_x}$ or the unique non-trivial extension of $\mathcal{O}_{E_x}$ with itself.

But I cannot decide what happens generically.

Motivation. I met this problem when I started the investigation of some triple covers $f \colon X \to A$ such that

$f_*\mathcal{O}_X=\mathcal{O}_A \oplus \mathcal{F}^{\vee}$.

It is worth noticing that the vector bundle $\mathcal{F}$ is the easiest example of indecomposable vector bundle on $A$ which is not simple (in fact, $\textrm{End}(\mathcal{F})=\mathbb{C} \oplus \mathbb{C}$).

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It's the nontrivial extension. Here is why:

Let $\pi=\pi_F$ and consider

$$0\to \pi_*\mathscr O_A \to \pi_*\mathscr F \to \pi_*\mathscr O_A(-E_p)\to R^1\pi_*\mathscr O_A \to \dots$$

Now

1) $\pi_*\mathscr O_A\simeq \mathscr O_F$ and $\pi_*\mathscr O_A(-E_p)\simeq\mathscr O_F(-p)$

2) $R^1\pi_*\mathscr O_A\simeq R^1\pi_*\omega_A$ is torsion-free by Kollár's theorem (I guess it is also isomorphic to $\omega_F\simeq \mathscr O_F$ so you don't even need Kollár here), so

3) if $\pi_*\mathscr F\to \pi_*\mathscr O_A(-E_p)$ is not surjective, then it is zero.

4) $\pi_*\mathscr F$ is torsion free and hence locally free.

Suppose that the restriction of your sequence to any fiber is a trivial extension. Then $\pi_*\mathscr F$ has to have rank $2$ and hence by 1) and 3) the map $\pi_*\mathscr F\to \pi_*\mathscr O_A(-E_p)\simeq \mathscr O_F(-p)$ is surjective.

Next, take the short exact sequence $$0\to \mathscr O_F\to \pi_*\mathscr F \to \mathscr O_F(-p) \to 0.$$ This is necessarily the trivial extension, but also by construction it is the push-forward of your original sequence.

Then the pull-back of this sequence maps functorially to the original sequence and you get an isomorphism on the sides, so you get an isomorphism in the middle as well. That implies that $\mathcal F$ is a trivial extension, which is an obvious contradiction.

Now as a side result we get that $\mathscr O_F\simeq \pi_*\mathscr O_A\simeq \pi_*\mathscr F$ from the beginning of the long exact sequence and $$\begin{multline} 0\to \pi_*\mathscr O_A(-E_p)\simeq\mathscr O_F(-p)\to R^1\pi_*\mathscr O_A\simeq\mathscr O_F\to R^1\pi_*\mathscr F \to \\ \to R^1\pi_*\mathscr O_A(-E_p)\simeq \mathscr O_F(-p)\otimes R^1\pi_*\omega_A\simeq \mathscr O_F(-p)\to 0 \end{multline}$$ so $\pi_*\mathscr F\simeq \mathscr O_F$ and $R^1\pi_*\mathscr F\simeq \mathscr O_p\oplus \mathscr O_F(-p)$.

EDIT: added the last exact sequence and corrected conclusion according to Francesco's comment.

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Thank you! But following your computations (that seems to me correct) I obtain $\pi_* \mathcal{F}=\mathcal{O}_F$ and $R^1 \pi_* \mathcal{F}= \mathcal{O}_F(-p) \oplus \mathcal{O}_p$. Am I missing something? – Francesco Polizzi Nov 2 '10 at 18:03
Francesco, you are right, I forgot to put the term coming from the previous part of the long exact sequence. It is corrected now. – Sándor Kovács Nov 2 '10 at 21:27
Ok. In my opinion this is a nice example showing that $R^1 \pi_∗F$ may have a torsion part even in the "very good situation" where $F$ a locally free sheaf and $\pi$ is a smooth morphism. – Francesco Polizzi Nov 3 '10 at 7:52
___Yes, indeed. – Sándor Kovács Nov 3 '10 at 8:41