Is there anything special about principal $G$-bundles, such that $G$- abelian group? Are there some interesting consequenes of this property, which do not take place in general?
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1$\begingroup$ If $G$ is an abelian group, one can write them in terms of cohomology with coefficients in $G$, defined appropriately, so we can fit it into a larger cohomology theory, where if $G$ is not abelian we cannot do this so nicely. $\endgroup$– Will SawinCommented Sep 30, 2012 at 5:25
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1$\begingroup$ @Will Sawin: there is of course a first non-abelian Cech cohomology group that classifies non-abelian principal bundles.It just doesn't fit into a cohomology theory. $\endgroup$– Konrad WaldorfCommented Sep 30, 2012 at 8:13
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1$\begingroup$ General principal bundles form a sheaf of groupoids. Abelian principal bundles form a sheaf of Picard groupoid, i.e. monoidal groupoids with duals. $\endgroup$– Konrad WaldorfCommented Sep 30, 2012 at 8:15
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$\begingroup$ Konrad's last comment is the answer I would give to the question: principal $G$ bundles for $G$ abelian have a "multiplication", and so form a Picard stack $\endgroup$– Eugene LermanCommented Sep 30, 2012 at 12:16
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$\begingroup$ Thank you for comments. Unfortunately I am not quite an expert in Picard groupoids but I'll try to get some understanding of it. $\endgroup$– AxelCommented Oct 3, 2012 at 6:53
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In easy detail, the classifying space $BG$ for principal $G$-bundles ($G$ a topological abelian group) is itself a topological abelian group, just because the usual classifying space functor preserves products and the multiplication and inverse on $G$ are continuous homomorphisms. Since the set of principal $G$-bundles over any $X$ (assuming CW homotopy type) is in natural bijective correspondence with the set of homotopy classes of maps $X\to BG$, it naturally forms an abelian group. If $G$ is discrete, $BG$ is a $K(G,1)$ and this set is $H^1(X;G)$. In general, one can define continuous cohomology so that the conclusion remains true.
answered Sep 30, 2012 at 14:10