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I'm trying to reconcile two results on the classification of principal bundles. First, we have $\mathrm{Prin}_G(X)$ (the equivalence classes of $G$-bundles on $X$) is isomorphic to $H^1(X;G)$ (the first Cech cohomology group of X -- I'm taking $G$ to be abelian). Second, we have $\mathrm{Prin}_G(X)$ is isomorphic to $[X,BG]$, the set of homotopy equivalences of maps from $X$ to $BG$, the classifying space of $G$. If we now take $G=\mathbb{R}$, the real line viewed as an additive group, the first result seems to say we can have non-trivial bundles, while the latter seems to contradict that (since we can take $BG$ to be a point). How do I reconcile these? Does one result not apply in this case?

Thanks in advance!

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The cohomology group $H^1(X,\mathbb{R})$ is the Cech cohomology group of the sheaf of differentiable functions over $X$, since there exist partition of unity, $H^1(X,\mathbb{R})=0.$

What is the right version of "partitions of unity implies vanishing sheaf cohomology"

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  • $\begingroup$ I think I was confused by notation, then. My understanding was that Cech cohomology groups and singular/de Rham cohomology groups are isomorphic on CW complexes, but I guess that when people say this they mean something different by $H^1(X,\mathbb{R})$. Is this right -- that sometimes this notation corresponds to the sheaf of differentiable functions while other times it corresponds to constant real-valued functions? $\endgroup$ – oggius Feb 21 '18 at 10:09
  • $\begingroup$ @oggius The point is that normally when one writes $H^1_{cech}(X, A)$ with $A$ an abelian group it is understood that $A$ carries the discrete topology, so that the cocycles are all locally constant functions. This group is isomorphic to singular cohomology if $X$ has the homotopy type of a CW complex. But normally when one writes $H^1_{cech}(X, G)$ where $G$ is a Lie group it is understood that the cocycles are continuous functions with values in $G$ equipped with its Lie group topology - this is the object that classifies principal $G$-bundles, and it is indeed zero if $G = \mathbb{R}$. $\endgroup$ – Paul Siegel Feb 22 '18 at 0:42
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The resolution, as others have said, is simply that $H^1_{cech}(X, \mathbb{R}) = 0$ if $\mathbb{R}$ is the additive group with the standard topology.

The reason you implicitly thought that this cohomology group is nonzero might have been due to the following "proof", which tripped me up in the past:

Suppose $H^1_{singular}(X, G)$ is nonzero. This group classifies maps from $X$ to $K(G,1)$ up to homotopy, and $BG$ is a $K(G,1)$. But $H^1_{cech}(X,G)$ classifies maps from $X$ to $BG$ up to homotopy, so the groups are the same and thus the latter is nonzero.

The flaw is in the step "$BG$ is a $K(G,1)$": this is only true if the topology on $G$ is discrete!

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Principal $(\mathbb R,+)$-bundles are indeed all trivial (Existence of a section suffices which exists for every fiber bundle with contractible fiber). You should consider principal $(\mathbb R\setminus\{0\},\cdot)$-bundles.

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