MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
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

Your Answer


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

Browse other questions tagged or ask your own question.