If $(P,\pi, M)$ is a principal $G$-bundle, then given a left $G$-space $F$, using the $G$-product we can create a new bundle $(P_F, \pi_F, M)$ that is said to be associated to the first. Also, if we have a certain vector bundle, if we pick the principal bundle of bases we can see that this vector bundle is associated to the latter.

So in the case of vector bundles we have the converse: every vector bundle is associated to some principal bundle. Now, I wonder if this holds for more general bundles. Indeed, the book I'm reading makes it seems that it does:

A crucial idea is that of a 'principal fibre bundle', which is a bundle whose fibre is a Lie group in a certain special way. These have the important property that all non-principal bundles are associated with an underlying principal bundle. Furthermore, the twists in a bundle associated with a particular principal bundle are uniquely determined by the twists in the latter, and hence the topological implications of fibre-bundle theory are essentially coded into the theory of principal fibre bundles.

This sounds like every non-principal bundle is associated with some principal bundle. Is this true? If it is, how can we show it? I don't have yet an idea of a proof, but if it is true I imagine it has something to do with the transition functions between overlapping local trivializations. Is that right?


1 Answer 1


Yes, it's true in general. Given transition functions taking values in the structure group, you define a principal bundle as the quotient of the disjoint union of trivializing neighborhoods $U_i$ crossed with the group, modulo the relation imposed by the transition functions: probably you've already seen this construction of a vector bundle.


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