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As is well known, there is classification of line bundles on the moduli stack of elliptic curves over a nearly arbitrary base scheme in the paper The Picard group of $M_{1,1}$ by Fulton and Olsson: every line bundle is isomorphic to a tensor power of the line bundle of differentials $\omega$ and $\omega^{12}$ is trivial.

I am now interested in a similar classification scheme for higher dimensional vector bundles (i.e. etale-locally free quasi-coherent sheaves of finite rank). I am especially interested in the prime 3, so you may assume, 2 is inverted, or even that we work over $\mathbb{Z}_{(3)}$. I found really very little in the literature on these questions. I know only of two strategies to approach the topic:

1) I think, I can prove that every vector bundle $E$ on the moduli stack over $\mathbb{Z}$ localized at $p$ for $p>2$ is an extension of the form $0\to L \to E \to F$ where $L$ is a line bundle and $F$ a vector bundle of one dimension smaller than $E$ (this may hold also for $p=2$, but I haven't checked). In a paper of Tilman Bauer (Computation of the homotopy of the spectrum tmf) Ext groups of the so called WeierstraƟ Hopf algebroid are computed, which should amount to a computation of the Ext groups of the line bundles on the moduli stack of elliptic curves if one inverts $\Delta$. It follows than that every vector bundle on the moduli stack of elliptic curves over $\mathbb{Z}_{(p)}$ is isomorphic to a sum of line bundles for $p>3$ if I have not made a mistake. But for $p=3$, there are many non-trivial Ext groups and I did not manage to see which of the occuring vector bundles are isomorphic.

2) One can try to find explicit examples of a non-trivial higher dimensional vector bundles. A candidate was suggested to me by M. Rapoport: for every elliptic curve $E$ over a base scheme $S$ we have an universal extension of $E$ by a vector bundle. Take the Lie algebra of this extension and we get a canonical vector bundle over $S$. As explained in the book Universal Extensions and One Dimensional Crystalline Cohomology by Mazur and Messing, this is isomorphic to the deRham cohomology of $E$. This vector bundle is an extension of $\omega$ and $\omega^{-1}$ and lies in a non-trivial Ext group. But I don't know how to show that this bundle is non-trivial.

I should add that I am more a topologist than an algebraic geometer and stand not really on firm ground in this topic. I would be thankful for any comment on the two strategies or anything else concerning a possible classification scheme.

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How did you prove that every vector bundle is an extension? – Tyler Lawson Apr 9 '10 at 12:55
Since the coarse space is affine and the automorphism groups of the geometric points have order a power of 2 times a power of 3, and the Ext-group (or $\omega^{-1}$ by $\omega$, to be precise) is identified with degree-1 coherent cohomology (of $\omega^2$), its nontriviality is also not visible over $\mathbf{Z}[1/6]$. So can you also briefly indicate how you know it is nontrivial? – BCnrd Apr 9 '10 at 14:49
@Brian: The computation of the Ext-groups of tensor powers of $\omega$ on this moduli stack is written up in the Bauer paper that was linked to under part (1). – Tyler Lawson Apr 9 '10 at 15:07
You can also deduce that $Ext^1(O, \omega^2)$ is non-trivial from the fact that there are non-trivial "mod p" modular forms of weight 2 when p=2 or 3; these are computed (for instance) in Deligne's note on "formulas, after Tate" in Antwerp 4. – Charles Rezk Apr 9 '10 at 15:33
The fact that vector bundles on $\mathcal M_{1,1}$ split as sums of line bundles in characteristic larger than 3 can also be seen by the classical description of $\mathcal M_{1,1}$ as an open substack of the weighted projective stack $\mathbb P(4,6)$, coming from the Weierstrass form. Any locally free sheaf on $\mathcal M_{1,1}$ extends to a reflexive sheaf on $\mathbb P(4,6)$, which is locally free, because $\mathbb P(4,6)$ is regular of dimension 1. It is not hard to prove that any locally free sheaf on a weighted projective stack $\mathbb P(m,n)$ splits as a direct sum of line bundles. – Angelo Apr 9 '10 at 17:34

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