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Throughout "curve" means smooth projective curve over an algebraically closed field.

Motivation and Background
I read somewhere that Atiyah has classified vector bundles on elliptic curves. My understanding is that the story is roughly: every vector bundles breaks up as a direct some of indecomposable vector bundles. The indecomposable vector bundles are further divided by their degree and rank. The set $Ind(d,r)$ of isomorphism classes of indecomposable vector bundles of a fixed degree $d$ and rank $r$ has the structure of variety isomorphic to the Jacobian of the curve; namely the curve itself. In fact $Ind(d,r) \cong Ind(0, gcd(d,r) )$ so its enough to consider the degree $0$ vector bundles.

Now for $V,V'$ vector bundles on any curve $C$ its true that $\deg V \otimes V' = \deg V\cdot rk(V') + \deg V'\cdot rk(V)$. So in the case of an elliptic curve the set $Ind(0,r)$ is a torsor for $Pic^0(C)$. Additionally, there is a unique isomorphism class $V_r \in Ind(0,r)$ characterized by the fact that $h^0(V_r) \ne 0$ (in fact $h^0(V_r)=1$).

The Question
Since curves of low genus are usually "simple enough" to easily describe explicitly, I would like to see how explicit I can be with a description of the $V_r$, or at least $V_2$. So my question is simply, given an explicit elliptic curve in $\mathbb{P}^2$, say $zy^2 - x(x-z)(x+z)$can you reasonably show how to construct $V_2$; i.e. give cocyles for it $\phi_{ij} \colon U_{ij} \to GL_2(k)$, or produce a graded module for it?

Initial Thoughts
$V_2$ should correspond to the unique nontrivial extension in $Ext^1(\mathcal{O}, \mathcal{O}) \cong H^1(\mathcal{O}) \cong k$.On a curve $C$, $0 \to \mathcal{O}_C \to K(C) \to K(C)/\mathcal{O}_C \to 0$ is a flasque resolution of the structure sheaf and an element in $H^1(\mathcal{O}_C)$ corresponds to a map $\alpha \colon \mathcal{O}_C \to K(C)/\mathcal{O}_C$. Then the desired extension should be the pullback of $0 \to \mathcal{O}_C \to K(C) \to K(C)/\mathcal{O}_C \to 0$ via $\alpha$.

The trouble with this is that I can't seem to pin down what $\alpha$ is. I now that, via Serre duality, it corresponds to a global section of $H^0(\omega_C)$ which I can explicitly describe as a differential on the curve. Using the example I mentioned above, on the affine patch where $z \ne 0$, it is: $\frac{dx}{2y} = -\frac{dy}{3x^2-1}$ but I guess I don't understand Serre duality well enough to determine $\alpha$ from this. Also I'm not even sure if this will lead to a reasonable way of getting at the cocylces of $V_2$. And this is so close to "just using the definitions" I have to imagine there might be a better way to go about this...

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Maybe you can find this useful: O. Iena, "Vector bundles on elliptic curves and factors of automorphy", arXiv:1009.3230 –  Francesco Polizzi Oct 30 '10 at 19:55
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1 Answer

up vote 12 down vote accepted

A description via cocycle usually is not convenient to work with. From my point of view, a description as an extension is much more useful. But if you want something else, I would advice the following. Since your curve $C$ is given as a double covering of $P^1$, that is as a relative spectrum of the sheaf of algebras $A = O \oplus O(-2)$ on $P^1$ (with $O$ summand being generated by the unit and with multiplication $O(-2)\otimes O(-2) \to O$ given by the ramification divisor $D$), the category of coherent sheaves on $C$ is identified with the category of sheaves on $P^1$ with a structure of a module over this algebra.

Spelling this out, a sheaf on $C$ is a sheaf $F$ on $P^1$ with a morphism $\phi:F(-2) \to F$ such that the composition $\phi\circ\phi(-2):F(-4) \to F(-2) \to F$ coincides with the morphism given by $D$. The structure sheaf $O_C$ corresponds to $A$. Hence the extension of $O_C$ with $O_C$ corresponds to an extension of $A$ with $A$, that is we have an exact sequence $$ 0 \to O \oplus O(-2) \to F \to O \oplus O(-2) \to 0. $$ Since $Ext^1_A(A,A) = Ext^1_O(O,A) = H^1(A) = H^1(O \oplus O(-2))$, we see that the nontrivial extension corresponds to the extension of $O$ in the right term by $O(-2)$ in the left term. Thus $F = O \oplus O(-1)^2 \oplus O(-2)$ and the above sequence can be rewritten as $$ 0 \to O \oplus O(-2) \to O \oplus O(-1) \oplus O(-1) \oplus O(-2) \to O \oplus O(-2) \to 0 $$ The maps are given by matrices $\left[\begin{smallmatrix}1 & 0\cr 0 & x\cr 0 & y\cr 0 & 0 \end{smallmatrix}\right]$ and $\left[\begin{smallmatrix} 0 & y & -x & 0 \cr 0 & 0 & 0 & 1 \end{smallmatrix}\right]$. The map $\phi:F(-2) \to F$ providing $F$ with a structure of an $A$-module is given by the matrix $\left[\begin{smallmatrix} 0 & g & h & -f \cr x & xy & -x^2 & h \cr y & y^2 & -xy & -g \cr 0 & y & -x & 0 \end{smallmatrix}\right]$. Here $f$ is the polynomial in $x$ and $y$ of degree $4$ such that $div(f) = D$, and $g,h$ are polynomials such that $f = gx +hy$.

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This is really cool! Thanks for this answer. But can you clarify a little about $D$ giving you a map $O(-2)\otimes O(-2) \to O$? I understand $D$ to be the effective divisor consisting of the 4 points in $\mathbb{P}^1$ where $C \to \mathbb{P}^1$ is ramified. So $O(D) \cong O(4)$ and a section would give the desired section, but which section? Intuitively this map should correspond to the relation given by the curve $Y^2 = f(X)$ where $\deg f = 3$. Also... –  solbap Oct 31 '10 at 18:00
    
... oh also the $f$ in $Y^2 = f(X)$ is not the same $f$ as yours, but in any case your $f$ seems determined up to a scalar by $D$; if $D = [0:1] + [1:1] + [-1:1] + [1:0]$ then $f \in k \cdot x(x-y)(x+y)y$, but $g,h$ don't seem unique: $f = x(g_1 = x^2y) + y(h_1 = y^2x) = x(g_2 = y^3) +y(h_2 = x^3)$. This construction should work for any choices of $g,h$? –  solbap Oct 31 '10 at 18:04
    
typo in my first comment, should read: So $O(D) \cong O(4)$ and a section would give the desired MAP, but which section? –  solbap Oct 31 '10 at 18:07
    
I'm also a little curious about how you determined the $A$-module structure. This 4x4 matrix somehow seems a little mysterious to me... –  solbap Oct 31 '10 at 18:10
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The section of $O(D)$ you need is the one which has zeros precisely in $D$. Certainly, different choices of $g$ and $h$ give you an isomorphic object. As for the 4x4 matrix --- if you right down the conditions that the maps in the exact sequence commute with the $A$-module structure, and also that the square of the matrix is $f$, you will obtain precisely this answer. –  Sasha Oct 31 '10 at 19:19
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