Can a homotopy inverse of the map from a Lie group to loops on its classifying space be given by holonomy?

Question

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 $G$.

Answer 1

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.

Answer 2

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.

Answer 3

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).