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Let $G$ be the compact Lie group $SO(n)$. There are some classical constructions of the classifying bundle of $G$ based upon on direct limits of Grassmann and Stiefel manifolds:

$$BG \simeq \underset{m \to \infty}{\lim} SO(m+n)/SO(m) \times SO(n)$$

$$EG \simeq \underset{m \to \infty}{\lim} SO(m+n)/SO(m)$$

with the evident $G$-bundle projection $EG \to BG$. One may give an explicit map $i: G \to \Omega BG$ which is a weak homotopy equivalence, by comparison of long exact homotopy sequences. Since $G$ and $\Omega BG$ have the homotopy types of CW complexes, the map $i$ is in fact a homotopy equivalence.

I am interested in whether an explicit homotopy inverse $h: \Omega BG \to G$ to $i$ can be given by taking holonomy of loops with respect to some suitably chosen connection on the universal bundle. Naturally the "manifold" $BG$ is infinite-dimensional, but I'm thinking it would suffice to work with the finite-dimensional $G$-bundles $V_m \to G_m$ (Stiefel to Grassmann) which approximate to the direct limits above, provided that the connections on each of the approximating bundles are compatible with respect to inclusion into the next, "compatible" here having an obvious sense in terms of $G$-valued holonomy.

Has anyone seen this idea worked out? Naturally I'm curious also about whether a similar idea works out for a general compact Lie group $G$.

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$\Omega BG$ being $A_\infty$ homotopy equivalent to G for G(F) is indeed ancient' - see my birth certificate! though it's more subtle than for smooth bundles with connection see my now updated version of parallel transport revisited' at the n-lab so what is it you'ld like to do with it –  Jim Stasheff Jan 24 '11 at 1:44
Jim, the question was not whether $\Omega BG$ is ($A_\infty$) homotopy equivalent to $G$ -- that much I indicated I already knew. It's a question of giving an explicit pair of maps which exhibits the equivalence. When I took algebraic topology as a graduate student, much of this type of thing was left in a black box: one can easily describe an appropriate map $i: G \to \Omega BG$, and give an abstract argument for why this is a homotopy equivalence, but as I've gotten older I like to get more concrete, giving a homotopy inverse explicitly. Thanks for the reference! I'll have a look. –  Todd Trimble Jan 24 '11 at 15:04
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3 Answers

up vote 10 down vote accepted

Yes, certainly. The model for BG as you described it has a canonical, universal connection for its $SO(n)$ bundle: just the induced Riemannian connection from Euclidean space.

As you move an $n$-dimensional plane in $\mathbb E^{n+m}$, the induced connection is the limit of compositions of orthogonal projections between nearby planes. In the limit, these become isometries. It doesn't matter what is the dimension of the ambient space, as long as the projection is defined. So, a loop in the Grassmanian gives an element of $G$. One example: if you hold a bicycle wheel by its axle, and move it around in a loop, when it comes back, it has rotated by some angle. That's the map.

It's a homotopy inverse of the map going the other way, namely, the classifying map for the bundle obtained by the suspension of an element of $G$. The suspension of a homomorphism has a canonical connection; its holonomy is $G$, so one composition of these two maps equals the identity. The other composition is homotopic to the identity, basically because the space of connections is contractible, and the bundle $EG$ over this model of $BG$ has a universal connection: Every connection on every $SO(n)$ bundle over a $CW$ complex $X$ is induced from a map $X \rightarrow BG$ (and this is also true in a relative form). This is a standard fact; I don't have a handy reference, but the proof is "soft".

However, to get a more immediate classifying space for connections that works for all Lie groups, just make a simplicial complex whose simplices are connections for a $G-bundles$ over the simplex. The $G$-bundle is specified by in terms of a trivialization associated with each vertex; the data needed is the chart-transition cocyle. In addition, give a connection; this amounts to specifying a connection form. Glue these simplices with $G$-bundles and connections together, to make a model for $BG$. Since the space of connections is contractible, this has the same homotopy type as the more usual model for $BG$ where just the cocycle is specified.

The same construction works to give an explicit homotopy inverse in a much more general context, e.g. the group of diffeomorphisms for a manifold.

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Thank you, Bill, for walking me through that! Your visual, concrete description in the case SO(n) is quite helpful, and your general description is easier than I was expecting it to be -- I really appreciate it. –  Todd Trimble Jan 20 '11 at 22:07
@Todd: I've "wasted" a lot of time muddling through formal descriptions of things and trying to come to terms with them intuitively, so I like passing them on if it might help save someone from either muddle or time or both. I'm curious now in what generality the notion of "connection" can work. I think it should work for locally contractible topological groups, but I wonder about topological groups that are not locally contractible. Maybe it works anyway, when the base is a CW complex. –  Bill Thurston Jan 20 '11 at 22:55
May I remind you, Bill, of Jim Stasheff's “Parallel” transport in fibre spaces, Bol. Soc. Mat. Mexicana (2), 11:68–84, 1966. and Parallel transport and classification of fibrations in Springer's LNM 428. –  David Roberts Jan 21 '11 at 1:03
@David: yeah. Stasheff's approach is more-or-less what I write about below. –  John Klein Jan 21 '11 at 12:52
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I'm going to change my original answer, since I interpreted the question wrongly. (I hope that's alright.)

Suppose $p: E\to B$ is a Hurewicz fibration, where $F = p^{-1}(\ast)$ is the fiber over the basepoint and $B$ is connected. Then one can cook up a map $\Omega B \times F \to F$ which might be called a "holonomy" in the algbraic topology sense.

The idea is this: Let $$ \Lambda_p = E\times_B B^I $$ be the space of path lifting problems for $p$ (this is the space of pairs $(e,\lambda)$ where $e\in E$ and $\lambda$ is a path starting at $p(e)$. There is a map $$ q: E^I \to \Lambda_p $$ by sending path $\lambda$ in $E$ to $(\lambda(0), p\circ \lambda)$. Then the condition that $p$ be a Hurewicz fibration is tantamount to saying that $q$ has a section. A choice of section might be regarded as parallel transport along a path in the algebraic topological sense. Choose such a section. This gives a way of associating to each path in $B$, starting at $x$ and ending at $y$, a map $E_x \to E_y$, where $E_x$ is the fiber at $x$. This map is a homotopy equivalence. (When $p$ is a fiber bundle, one can choose the section in such a way that each parallel transport is a homeomorphism of fibers.)

Evaluating the section when $x=y$ is the basepoint gives the holonomy operation $\Omega B \times F \to F$, or adjointly as $\Omega B \to G(F)$, where $G(F)$ is the topological monoid of self homotopy equivalences of $F$. If $p$ is a fiber bundle with structure group $G$, then the transport operation described above can be factored as $$ \Omega B \to G\to G(F) . $$ If we choose a basepoint in $F$, then the value of the operation on the basepoint gives a map $$ \Omega B \to F . $$ This map is well-known: it's the map sitting in the homotopy fiber sequence $$ \Omega B \to F \to E . $$ (this should be in any reasonable text on the subject).

So, in the particular case when $p: EG \to BG$ and $F = G$, then map $\Omega BG \to G$ will be a homotopy equivalence, using the above homotopy fiber sequence, since $E = EG$ is contractible. We have also seen this map as decribed by the orbit of a point in $G$ under the holonomy operation $\Omega BG \times G \to G$ as given above.

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Thank you, John. I follow you fine except for the point where you say that the transport operation can be factored through G, which I don't see since the preceding only used the fact that the bundle is a Hurewicz fibration. (Perhaps that particular point doesn't matter if all one is setting out to do is get the homotopy inverse I inquired about, but classical holonomy of a connection would give the extra information that the map $\Omega BG \to G$ takes path composition to group multiplication.) –  Todd Trimble Jan 21 '11 at 16:36
If the fibration is actually a fiber bundle with structure group G, then the section of $E^I \to \Lambda_p$ can be chosen so that the parallel transport along the path is a homeomorphism and is given by the action of $G$. How does one chose it? By the classical holonomy of course. In any case, the space of sections of the lifting problem is contractible, so whatever section you chose is homotopic to mine and so classical holonomy $\Omega B \to G$ followed by the map in $G(F)$ is homotopic to mine. Lastly, the map I constructed is a homomorphism in the $A_\infty$ sense. –  John Klein Jan 21 '11 at 20:19
Actually, I'm not sure the map I constructed is apriori an $A_\infty$ homomorphism. I will think about how to rectify that. –  John Klein Jan 21 '11 at 22:27
Yes, sure. I was simply saying that "the transport operation described above factors through G" does not hold for every choice of section (as "described above"). Otherwise I'm fine with what you had written (and thanks again; it was a good answer). –  Todd Trimble Jan 21 '11 at 23:31
@Todd, regarding my last comment: the issue of why the map $\Omega B \to G(F)$ is multiplicative is discussed in Whitehead's book. –  John Klein Jan 23 '11 at 2:26
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Let $\pi:P \to M$ be a smooth principal $G$-bundle on a Hilbert manifold. There exist smooth connections in this situation, so pick one. Pick a point $p \in P$, $x:=\pi(p)$. Let $\Omega_{\infty} M$ be the space of smooth loops based at $x$. Consider the lifting problem

$$ \xymatrix{ \Omega_{\infty} \times \{0\} \ar[d] \ar[r] & P \ar[d] \Omega_{\infty} M \times [0,1] \ar[r] \ar[ur]^{l} & M } $$

The bottom map is the evaluation, the top map is the constant map $p$. Now parallel transport along curves defines the lift $l$. Restriction of $l$ to $\Omega_{\infty} M \times \{1\}$ defines the holonomy $hol:\Omega_{\infty} M \to \pi^{-1}(x) =G$, the last identification depending on the point $p$.

Now specialize to a compact Lie group $G$. Take the Grassmann manifold $Gr_n$ of $n$-dimensional subspaces of the Hilbert space $\ell^2$. This is a model for $BSO(n)$ as a Hilbert manifold, the corresponding model for $EO(n)$ is the Stiefel manifold $V_n$ of orthonormal $n$-frames in $\ell^2$; this is a Hilbert manifold as well. If $G$ is compact, we can embed $G$ into $O(n)$ for some $n$ (Peter-Weyl Theorem). Then $V_n \to V_n/G=BG$ is a model for the universal $G$-bundle, in the context of Hilbert manifolds.

Now I claim that $hol: \Omega_{\infty} BG \to G$ is a weak homotopy equivalence.

Let $f:E \to B$ be a Hurewicz fibration with fibre $F=f^{-1}(x)$. There is the fibre transport map $T:\Omega B \to F$ obtained by lifting the paths. It is not hard to see that $\pi_{n+1}(B) \cong \pi_n (\Omega B) \stackrel{T}{\to} \pi_n (F)$ is the same as the connecting homomorphism in the long exact homotopy sequence of $F \to E \to B$. If the fibration is $EG \to BG$, you get a weak homotopy equivalence $\Omega BG \to G$. The fibre transport is defined only up to homotopy, but above, we have constructed one using a connection. Thus the holonomy is homotopic to the fibre transport and hence a weak homotopy equivalence.

I am running out of steam, but I think it can be shown along these lines that $hol$ is also a homotopy inverse to the natural map $G \to \Omega_{\infty} BG$ (which does not look so natural in this setting).

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So your use of sub-infty indicates smooth? suggest sub-smooth if you need a decoration as opposed to just saying that's what your Omega means –  Jim Stasheff Jan 24 '11 at 1:38
The use of the sub-infty was only an ad-hoc notation. –  Johannes Ebert Jan 24 '11 at 10:03
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