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This question is inspired by, but is independent of: Sheaf Description of G-Bundles

Line bundles are classified by $H^1(X,\mathcal{O}^\times_X)$. We also know that in general that $H^1(X,G)$, where $G$ is a sheaf from open sets in $X$ to $Grps$, classifies $G$-torsors over X.

With this insight in mind: $\mathcal{O}^\times_X$-torsors should correspond to line bundles. Indeed, if $P$ is one, then $\mathcal{O} _X \times _{\mathcal{O} _X^\times}P$ gives the desired line bundle, and all line bundles are achieved this way (see the question I linked to).

My question is about the more mundane $H^1(X,\mathbb{C})$, which can be thought of as $H^1(X,\underline{\mathbb{C}})$ where $\underline{\mathbb{C}}$ is the constant sheaf $\mathbb{C}$ (which I think of as a sheaf going to $Grps$). These should supposedly correspond to $\mathbb{C}$-bundles over $X$. Which appears to be line bundles. But of course there's no reason for $H^1(X,\mathbb{C})$ to equal $H^1(X,\mathcal{O}^\times_X)$... My intuition is that this should correspond to the more naive version of fiber bundles that doesn't involve a structure group. Do you have any thoughts?

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I think i can answer this in algebraic topology land, but it would involve singular cohomology with integral coefficients, here H^2 is represented by BS^1, and BG classifies G-bundles over X ie, G-torsors. If this is indeed the type of thing that you are looking for I can put it as an answer, but it might not be... Perhaps this might help with intuition though..., and just in case BS^1=CP^inf –  Sean Tilson Apr 28 '10 at 23:20
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C does not act on the fibers of a line bundle by addition, so line bundles can't be C-torsors. One thing to observe is that line bundles come with a canonical choice of zero section, and are not necessarily trivial, unlike the situation with torsors where having a section automatically makes the bundle trivial. One thing you can say is that a line bundle is a "bundle of groups", but this is very different from being a torsor. –  Lucas Culler Apr 28 '10 at 23:33

3 Answers 3

up vote 9 down vote accepted

The general principle is: if you have some objects which are locally trivial but globally possibly not trivial then the isomorphism classes of such objects are classified by $H^1(X,\underline{Aut})$, where $\underline{Aut}$ is the sheaf of automorphisms of your objects.

So, if your objects locally are $U\times \mathbb A^n$ (i.e. vector bundles) or they are $\mathcal O_U^{\oplus n}$ (i.e. locally free sheaf) then either of these are classifed by $H^1(X, GL(n,\mathcal O))$. For $n=1$, you get $GL(1,\mathcal O)=\mathcal O^*$.

Now what is $\mathbb C$ an automorphism group of? Certainly not of line bundles (zero has to go to zero).

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Would it be fair to say: fiber bundles over X with fiber F, structure group G, and transition functions with property P (either: nothing, continuous, constant, etc.) is in correspondence with T-torsors, where T is the sheaf on X of P-function to G? –  Makhalan Duff Apr 29 '10 at 4:06
    
Which, in turn, corresponds to H^1(X, T)? –  Makhalan Duff Apr 29 '10 at 4:08
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@MD: Yes, that's fair. In fact it makes it easy to explain your point of confusion: you were conflating the fiber $F$ with the structure group $G$. As you can see, the latter is more important because it is used to classify the torsors, and you can always restrict to principal $G$-bundles. But a bundle with structure group $\mathbb{C}$ is not a line bundle! –  Pete L. Clark Apr 29 '10 at 5:52

I think the right thing to look at is $H^1(X, \mathbb C^\*)$. This classifies line bundles with a flat connection, or equivalently, line bundles with locally constant transition functions.

Now the natural embedding $\mathbb C^\* \to \mathbb O_X^\*$ induces on a map on cohomology $H^1(X, \mathbb C^\*) \to H^1(X, \mathbb O_X^\*)$ which is forgetting the flat connection.

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Your confusion seems to stem from the difference between topological bundles and algebraic/holomorphic bundles.

For a scheme $(X, \mathcal{O}_X),$ you say that $H^1(X, \mathcal{O}_X^{\times})$ classifies line bundles on $X$. This is true, as long as you mean algebraic line bundles. If, for example, $X$ is something like a complex algebraic variety with the analytic topology (which seems to be how you're thinking about it, perhaps), then there can certainly be plenty of topological line bundles on $X$ which aren't algebraic.

Also, as Lucas Culler notes in the comments to your question, a line bundle is not a torsor under the additive group $\mathbb{C}$. Instead, it is actually a $\mathbb{C}^{\times}$ torsor, if you remove the zero section. Algebraically, a torsor for the additive group $\mathbb{C}$ (which, by the way, is often denoted $\mathbb{G}_a$ to avoid confusion) is classified by all extensions of $\mathcal{O}_X$ by $\mathcal{O}_X$.

Anyway, it is true that (maybe with some mild assumptions on your space $X$) that $H^1(X,G)$ classifies topological $G$-bundles on $X$.

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I'm confused. Are you treating the constant sheaf C as G_a? But the first applied to Z (the integers) is C (additive), and the other would be the additive group Z. Are you saying H^1(X, C) (topological cohomology) is H^1(x, G_a)? –  Makhalan Duff Apr 29 '10 at 0:53
    
No, the constant sheaf $\mathbb{C}$ is not $\mathbb{G}_a,$ I was merely describing what an algebraic $\mathbb{C} = \mathbb{G}_a$ torsor is. This was meant to clarify your confusion over thinking that a $\mathbb{C}$ torsor is the same as a line bundle. –  Mike Skirvin Apr 29 '10 at 1:12
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One way to express the distinction is viewing $\mathbf{C}$ as a complex-analytic Lie group in two ways: with the usual topology or the discrete topology. The former gives rise to torsors classified by ${\rm{H}}^1(X, O_X)$ and the latter gives rise to torsors classified by ${\rm{H}}^1(X, \underline{\mathbf{C}})$. Neither case related to line bundles, and for connected $X$ the group ${\rm{H}}^1(X, \underline{\mathbf{C}}^{\times})$ classifies line bundles with locally constant transition functions (for some trivializing cover). –  BCnrd Apr 29 '10 at 1:46

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