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Now, among algebraic geometers, at least, it is well known that there is an equivalence between locally free $\mathcal{O}_X$-modules of rank $n$ and vector bundles of rank $n$. So, equivalently, principal $\mathrm{GL}(n,\mathbb{C})$-bundles are given by locally free sheaves of rank $n$.

So...what about other groups? I guess that $\mathrm{SL}(n,\mathbb{C})$ bundles are then locally free sheaves of rank $n$ with top exterior power trivial, but can we phrase everything in terms of the properties of a sheaf and a group?

My guess is that in this context, if we can do it, we'll end up with something that's not quite locally free sheaves of rank n for $\mathrm{GL}(n,\mathbb{C})$, but which will be equivalent.

Note: I'm aware that we could just say something like "the sheaf of local sections of a $G$-bundle" but I'm looking for something intrinsic, a set of properties of the sheaf without reference to the geometric bundle, which can be reconstructed from the sheaf description.

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3 Answers 3

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$\newcommand{\O}{\mathcal{O}}$ $\newcommand{\F}{\mathcal{F}}$

The way you get a locally free sheaf of rank $n$ from a $GL(n)$-torsor $P$ is by twisting the trivial rank $n$ bundle $\O^n$ (which has a natural $GL(n)$-action) by the torsor. Explicitly, the locally free sheaf is $\F=\O^n\times^{GL(n)}P$, whose (scheme-theoretic) points are $(v,p)$, where $v$ is a point of the trivial bundle and $p$ is a point of $P$, subject to the relation $(v\cdot g,p)\sim (v,g\cdot p)$. Conversely, given a locally free sheaf $\F$ of rank $n$, the sheaf $Isom(\O^n,\F)$ is a $GL(n)$-torsor, and this procedure is inverse to the $P\mapsto \O^n\times^{GL(n)}P$ procedure above. (Note: I'm identifying spaces over the base $X$ with their sheaves of sections, both for regarding $Isom(\O^n,\F)$ as a torsor and for regarding $\O^n\times^{GL_n}P$ as a locally free sheaf.)

Similarly, if you have a group $G$ and a representation $V$, then you can associate to any $G$-torsor $P$ a locally free sheaf of rank $\dim(V)$, namely $V\times^G P$. But I don't know of a characterization of which locally free sheaves of rank $\dim(V)$ arise in this way.

Operations with the locally free sheaf (like taking top exterior power, or any other operation which is basically defined fiberwise and shown to glue) correspond to doing that operation with the representation $V$, so I think you're right that in the case of $SL(n)$ you get exactly those locally free sheaves whose top exterior power is trivial (since $SL(n)$ has no non-trivial $1$-dimensional representations).

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    $\begingroup$ This is the closest to the sort of thing I was looking for. Since asking, I realized that there was an obvious description that was what I was looking for, for any fiber bundle with fiber F, a locally trivial F-bundle should be (it's in my scratch notebook, not fully proved but should work) equivalent to a sheaf of sets locally isomorphic to sheaf Hom(X,F). This recovers the equivalence for vector bundles, and does about what I want for G-bundles. $\endgroup$ Commented Oct 25, 2009 at 16:00
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    $\begingroup$ @Anton: Algebraic subgroups of $GL(n)$ can be described by matrices fixing some tensor; consequently, $G$-bundles can be described as a vector bundle + some tensors. $\endgroup$
    – ACL
    Commented Jan 30, 2014 at 7:44
  • $\begingroup$ @ACL: Can you give me a reference for that? At some point I tried describing what sorts of tensor relations give a vector space the structure of a representation of given a finite group, but I didn't get anything I was happy with. I didn't know that there was a general result along these lines for affine algebraic groups. $\endgroup$ Commented Jan 30, 2014 at 23:14
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    $\begingroup$ @AntonGeraschenko: I learnt this in Katz's paper on (Bulletin SMF, 1982, numdam.org/item?id=BSMF_1982__110__203_0). He refers to Chevalley's 1968 book on Lie groups. $\endgroup$
    – ACL
    Commented Jan 31, 2014 at 9:19
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    $\begingroup$ @LSpice, it is at the beginning of section III (Interlude: Review of algebraic groups), page 211. The statement of Chevalley's theorem is on page 212. The key word is “construction”. (I imagine it shortens the expression “construction tensorielle”.) $\endgroup$
    – ACL
    Commented Dec 2, 2019 at 8:52
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If G is an affine algebraic group, a G-bundle is the same as a monoidal functor from G-reps to coherent sheaves. The map one way is take associated bundle, the other involves reconstructing the structure sheaf of the G-bundle from the associated ones. Roughly, you think of the functions on the group as a ring ind-object in the category of representations, and take the corresponding ring object in quasi-coherent sheaves. The Spec of this sheaf of rings is the G-bundle.

For GL(n), you'e lucky, since its category has a simple description: it's (basically) the free monoidal category with a single generator of dimension n. Other groups are a little more complicated, but not much worse.

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    $\begingroup$ By the way, do you know a good reference for the reconstruction result? $\endgroup$
    – S. Carnahan
    Commented Oct 25, 2009 at 16:06
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    $\begingroup$ Denis Gaitsgory, personal communication. $\endgroup$
    – Ben Webster
    Commented Oct 25, 2009 at 18:12
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    $\begingroup$ math.harvard.edu/~gaitsgde/grad_2009/SeminarNotes/… are some of Gaitsgory's notes that cover this construction (although in the notes the functor takes values in vector bundles, not coherent sheaves). $\endgroup$ Commented Oct 25, 2009 at 22:28
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Adding it so that it's easily found. The thing I was looking for, which is generally not written out except in the case of vector bundles, is that the sheaf of sections of an F-bundle with fiber F is a sheaf of sets that is locally isomorphic in the etale topology to the sheaf hom(-,F) ranging over small enough open subsets of X.

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    $\begingroup$ actually, you should be incredibly careful with this definition. There are things you want to be principal G-bundles, which are not locally trivial in the Zariski topology (like the map z -> z^2 on A^1-{0}). In the etale/smooth/fppf topology, this is basically fine. $\endgroup$
    – Ben Webster
    Commented Oct 27, 2009 at 16:50
  • $\begingroup$ Fair enough. Edited to mention that I'm thinking etale. $\endgroup$ Commented Oct 27, 2009 at 17:20
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    $\begingroup$ Why is this answer blue? $\endgroup$ Commented Oct 27, 2009 at 21:35
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    $\begingroup$ I think it's because it's mine, and I asked the question. $\endgroup$ Commented Oct 27, 2009 at 23:19

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