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It is well known and understood that linear connections, as holonomy functors, are composition-preserving mappings from the path groupoid to a structure group $G$. This extends the idea of a 1-form line integral to a non-Abelian setting.

It is known, although still a work in progress (http://ncatlab.org/nlab/show/connection+on+a+2-bundle), that using higher category theory we can do the very same for 2-forms, 3-forms, etcetera, raising the degree of the groups.

The geometrical picture is very nice. I was wondering, though, if there were such a picture for curvature on a good old bundle (vector or principal) of degree 1. That is: can we see curvature as a mapping between 2-dimensional surfaces and a group $G$ (the holonomy group)? How can we make this mapping not Abelian, to prevent what happens in the classic second homotopy group?

Is there a reference, maybe in nLab, that I missed about this?

I have some personal ideas about how to do such a thing, but I'd first rather knowing if it has been already done nicely elsewhere.

Thanks!

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This question is tightly linked to this very interesting one: math.stackexchange.com/questions/49583/…. –  geodude Jul 26 '13 at 9:35

1 Answer 1

If G=U(1), then the holonomy around some circle C bounding some disk D is the exponent of the integral of curvature over D. Thus the integral of curvature over D can be recovered as the logarithm of the holonomy around C.

The ambiguity in the logarithm can be resolved by observing that the integral of curvature varies smoothly as D varies smoothly, and any disk D can be smoothly deformed to a point, for which the integral of curvature is known to be zero.

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Yes! That is exactly where I started from. The question is, what is the general case? –  geodude Oct 31 '13 at 20:01
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@geodude: If the “general case” refers to principal bundles with nonabelian structure groups, then the answer is given by the nonabelian Stokes theorem, see Section 3.2, Theorem 3.4 and the displayed formula on top of page 48 in arxiv.org/abs/0802.0663. –  Dmitri Pavlov Oct 31 '13 at 22:07
    
There, that is exactly what I was looking for. Thanks! –  geodude Nov 1 '13 at 10:52

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