I have a question about the spectral cover construction of Friedman, Morgan, and Witten (typically used to map a description in heterotic string theory into F-theory). I realise this is a highly specialised topic; I am asking in the hope that there might be some specialists around. I also realise this may be more appropriate in the mathematics StackExchange, and may post a question there later.

Specifically I have a question about the more mathematical FMW paper, `Vector bundles over elliptic fibrations': https://arxiv.org/abs/alg-geom/9709029

Consider the second short exact sequence on page 72, of sheaves on the elliptic fibration $Z$ (the one involving $V_{A,a}$ sheaves). The three sheaves appearing in the first two slots of the sequence are I think supposed to all be vector bundles (the $V_{A,a}$ sheaves should be vector bundles since they are supposed to correspond to the spectral cover description). But the sheaf in the third slot is, I think, zero (that is, not a trivial bundle, but actually just zero) away from the subspace $D$, and hence is not a vector bundle. How then could both the left and the middle sheaves in the sequence be vector bundles? Additionally, away from the subspace $D$, do we then have an isomorphism between the sheaves in the first two slots? And in the subspace $D$ (where the final sheaf is not just zero), do we just have the ordinary quotient relation between the three sheaves?


The short exact sequence in question is, $$0\to V_{A,a}(n)\to V_{A,a}(n-1)\oplus \pi^*L^a\to(\pi^*L^a)|D\to0\,.$$ The space where these bundles (or sheaves) live is an elliptic fibration $\pi:Z\to B$ (and $L^a$, where it appears, is being pulled back by $\pi^*$ to the space $Z$). The $|D$ denotes restriction to a subspace $D$ of $Z$. Here $V_{A,a}(m)$ is a sheaf, and specifically is I believe supposed to be a vector bundle for both values of $m$ seen in the sequence. The final term in the sequence is I think not a vector bundle (it is 0 away from the subspace $D$), and this gives rise to my questions above. (Essentially this is a question about exact sequences of sheaves, but as it appeared in a physics context I thought I'd try my luck finding someone on here who is familiar with this spectral cover construction.)

  • $\begingroup$ Please make your question self-contained - even if I was an expert in the area I would have no idea what the question is without opening the paper and reading page 72. At least write down the exact sequence you're talking about and the definition of the symbols in it. $\endgroup$ – ACuriousMind Jul 25 '16 at 10:21
  • $\begingroup$ Thanks @ACuriousMind, I've added a clarification that hopefully makes the question more narrow and self-contained. $\endgroup$ – diracula Jul 25 '16 at 11:43

The quotient of locally free sheaves is not necessarily locally free, i.e. the first two terms being locally free does not force the third to be locally free, cf. this math.SE post.

You have no guarantee that, given a map $f: E\to F$ of vector bundles, that the quotient $\mathrm{coker}(E\to F)$ (which we would also like to write as $F/\mathrm{im}(E)$) exists as a vector bundle. In other words, the category of vector bundles is not an Abelian category, it does not have all kernels and cokernels. To see this, simply choose any map of vector bundles whose rank on the fibers jumps somewhere. This shows that you should not expect to be able to lift an exact sequence of sheaves to a sequence of vector bundles unless you know all sheaves to be locally free, since the category of sheaves is Abelian so maps of two locally free sheaves will always have kernels and cokernels as sheaves that are not kernels and cokernels as locally free sheaves.

  • $\begingroup$ Thanks very much for the helpful answer. Would you possibly also be able to say something about the other part of the question? That is, away from the subspace $D$, can we say that the vector bundles in the first two slots of the sequence are isomorphic (but not isomorphic in $D$)? $\endgroup$ – diracula Jul 26 '16 at 12:59
  • $\begingroup$ @diracula: Yes - just restrict the sequence to a set that doesn't intersect $D$, then the sequence has zero in the third term. However, I don't think that's a very relevant observation, since all locally free sheaves of the same rank are locally isomorphic anyway. $\endgroup$ – ACuriousMind Jul 26 '16 at 13:11
  • $\begingroup$ Thanks for the reply. While locally free sheaves of the same rank will be isomorphic on a single open set, isn't it much more non-trivial to say that two locally free sheaves are isomorphic everywhere away from some subspace (in particular this is some codimension 1 subspace)? $\endgroup$ – diracula Jul 28 '16 at 0:37

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