Let $G$ be a (maybe Lie) group, and $M$ a space (perhaps a manifold). Then a principal $G$-bundle over $M$ is a bundle $P \to M$ on which $G$ acts (by fiber-preserving maps), so that each fiber is a $G$-torsor (a $G$-action isomorphic, although not canonically so, to the action of $G$ on itself by multiplication). A map of $G$-bundles is a bundle map that plays well with the actions.

Then I more-or-less know what the classifying space of $G$ is: it's some bundle $EG \to BG$ that's universal in the homotopy category of (principal) $G$-bundles. I.e. any $G$-bundle $P \to M$ has a (unique up to homotopy) map $P\to EG$ and $M \to BG$, and conversely any map $M\to BG$ (up to homotopy) determines a (unique up to isomorphism) bundle $P \to M$ and by pulling back the obvious square.

At least this is how I think it works. Wikipedia's description of $BG$ is here.

So, let $G$ be a Lie group and $M$ a smooth manifold. On a $G$-bundle $P \to M$ I can think about connections. As always, a connection should determine for each smooth path in $M$ a $G$-torsor isomorphism between the fibers over the ends of the path. So in particular, a bundle-with-connection is a (smooth) functor from the path space of $M$ to the category of $G$-torsors. But not all of these are connections: the value of holonomy along a path is an invariant up to "thin homotopy", which is essentailly homotopy that does not push away from the image of the curve. So one could say that a bundle-with-connection is a smooth functor from the thin-homotopy-path-space.

More hands-on, a connection on $P \to G$ is a ${\rm Lie}(G)$-valued one-form on $P$ that is (1) invariant under the $G$ action, and (2) restricts on each fiber to the canonical ${\rm Lie}(G)$-valued one-form on $G$ that takes a tangent vector to its left-invariant field (thought of as an element of ${\rm Lie}(G)$).

Anyway, my question is: is there a "space" (of some sort) that classifies $G$-bundles over $M$ with connections? By which I mean, the data of such a bundle should be the same (up to ...) as a map $M \to $ this space. The category of $G$-torsors is almost right, but then the map comes not from $M$ but from its thin-homotopy path space.

Please re-tag as desired.

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    $\begingroup$ Interesting question. It should be "principal" not principle. $\endgroup$
    – Q.Q.J.
    Feb 17, 2010 at 21:38
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    $\begingroup$ Oopa! Spelling fixed. $\endgroup$ Feb 18, 2010 at 1:47
  • $\begingroup$ This question seem to be written with super care.. I enjoy reading this question and its answers.. I read it at least two/three times in a month.. every time I read, I get some extra information.. thanks $\endgroup$ Sep 2, 2019 at 4:31
  • $\begingroup$ What, in the first place, is the classifying space of smooth G-bundles (G: Lie group)? $\endgroup$
    – Student
    Sep 18, 2019 at 0:24

3 Answers 3


There is a stupid answer which is equivalence classes of G-bundles with connection on M are the same as homotopy classes of maps $M \to BG$. That is as long as two G-bundles with connection are considered equivalent if they have the same underlying principal bundle. This isn't meant to be a serious answer, just point out that your question is not exactly well posed.

But more seriously, there is a stack which represents G-principal bundles with connections. It even has a nice form:

$$ Bun_G^\nabla = [ \Omega^1( - ; \mathfrak{g}) / G]$$

Maps from M to this stack are principal G-bundles with connection.

The problem with this stack is that it is not presentable. It is not covered by a manifold. It can be describe as a quotient stack, but thing you act on is the sheaf $\Omega^1(-; \mathfrak{g})$ of Lie algebra valued 1-forms. This is a sort of generalized manifold (in a loose sense), but this sheaf is not representable (great exercise!).

If it was a presentable stack, then we could take its classifying space (there are several ways to do this, e.g take the realization of the simplicial manifold obtained by iterated fiber products of the covering manifold). Homotopy classes of maps to this space could then be related to certain isomorphism classes of maps to the stack. But since $Bun^\nabla_G$ is not presentable we are kinda stuck.

You could ask, well what happens if I replace $\Omega^1(-; \mathfrak{g})$ with an honest topological space that is the best approximation to it (for maps into it). Well it turns out the space which best approximates $\Omega^1(-; \mathfrak{g})$ is the point. So you get the classifying space of the stack $[pt/G]$ which is just the usual BG.

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    $\begingroup$ Incidentally, Dan Freed this week gave a colloquium talk where he described various stacks (on manifolds, valued in simplicial sets), with a focus on $Bun^\nabla_G$. He had earlier explained some calculations with the stach $\Omega^1$; in particular, he had calculated the de Rham complex $\Omega^\bullet(\Omega^1) = Maps(\Omega^1 \to \Omega^\bullet)$ to be $\mathbb R \overset0\to \mathbb R \overset\sim\to \mathbb R \overset0\to \mathbb R \overset\sim\to \dots$. Then he argued that $Bun^\nabla_G = \Omega^1\otimes\mathfrak g/G$, and calculated $\Omega^\bullet(Bun^\nabla_G)$. $\endgroup$ May 13, 2011 at 14:53
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    $\begingroup$ Do you have a reference where I could read about this in more detail? Thanks in advance. $\endgroup$ Feb 6, 2015 at 2:33

To add to Chris' answer, since the space of connections is contractible, if you are looking for a space that classifies bundles-with-connections wrt homotopy classes of maps into it, then you're out of luck: either $BG$ if you want a proper space, or the stack if you're prepared to accept something more general.

But if you work in a different category, then you can get more. I have a vague memory of being told that $BG$ classifies bundles-with-connection if you work with the whole homotopy type of the mapping space $Map(X,BG)$ rather than just $\pi_0$ of it (which is what you get if you take homotopy classes of maps). I'm not quite sure how to make complete sense of that, but maybe some kind soul will step in in the comments and enlighten us.

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    $\begingroup$ Yeah, now that you mention it, I seem to remember something similar. I'm pretty sure this point of view was explained in the Hopkins-Singer paper on generalized smooth cohomology: arxiv.org/abs/math.AT/0211216 If memory serves it wasn't just the mapping space, but also a special filtration of the mapping space that allowed you to recover the correct category of prin. bundles with connection. I remember the machinery made heavy use of simplicial methods. $\endgroup$ Feb 17, 2010 at 21:53
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    $\begingroup$ What Andrew's referring to might be in Donaldson-Kronheimer (Geometry of 4-Manifolds, Section 5.1). They show that the classifying space for "framed families of connections" is weakly equivalent to Map^0 (M, BG)_P, so homotopy classes of maps X to Map^0 (M, BG)_P classify framed families of connections on P, parametrized by X. Here the mapping space is the space of base-point preserving maps which induce the bundle P. Not sure if this is the sort of answer Theo was looking for, though. $\endgroup$
    – Dan Ramras
    Feb 17, 2010 at 22:19
  • $\begingroup$ I have wondered what is the analogue for Map(M,BG), of dividing out by thin homotopy (as mentioned in the question). I sort of know what this might mean simplicially but not in this context. $\endgroup$
    – Tim Porter
    Feb 18, 2010 at 7:24

There should be an answer to Theo's question in terms of universal connections, but I don't know it.

This universal connection is a connection on the universal principal $G$-bundle over $BG$, such that every $G$-bundle with connection over $M$ is isomorphic (as a bundle with connection) to the pullback of $EG$ along some map $M \to BG$.

I've never found an answer to the following immediate question: what is the correct equivalence relation on the space of maps from $M$ to $BG$, such that equivalence classes of maps are in one-to-one correspondence with isomorphism classes of $G$-bundles with connection over $M$. Does anybody know this?

Also, I wanted to remark that $G$-bundles with connection are not the same as smooth functors on the thin path groupoid of $M$. Assuming global smoothness, you'll only get connections on trivializable bundles. The full story is here: http://arxiv.org/abs/0705.0452.

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    $\begingroup$ Oh, awesome. I'll definitely check out your paper. I guess I don't know what a "smooth functor" is, either. Anyway, so there is a universal connection on $EG \to BG$? That's cool. $\endgroup$ Feb 18, 2010 at 3:58
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    $\begingroup$ The reference is Roger Schlafly, "Universal connections", in Invent. Math. $\endgroup$ Feb 18, 2010 at 8:22
  • $\begingroup$ Your question probably is answered in page 27-28 of arxiv.org/pdf/math/0211216v2.pdf $\endgroup$
    – user40276
    Mar 4, 2016 at 13:27

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