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Let $E\xrightarrow{p} \Sigma X$ be a principal G-bundle over a suspension. Write $\Sigma X= C_+X\cup_X C_-X$. Then there are trivialisations of the restrictions $E|_{C_+X}\cong C_+X\times G$, $E|_{C_+X}\cong C_-X\times G$, and the transition function between them over their intersection X is defined by a map $t:X\rightarrow G$. The homotopy class of this map completely characterises the bundle, and the process is in fact reversible. Given such a clutching function, one can construct a unique bundle over the suspension.

So if $f:\Sigma X\rightarrow BG$ is a map classifying the G-bundle E, how does this map relate to the clutching function $t:X\rightarrow G\simeq \Omega BG$ ? How does one go between one and the other?

The adjoint isomorphism $[\Sigma X, BG] = [X,\Omega BG]= [X,G]$ obviously has a part to play, but for all I have tried I cannot for the life of me craft a strong enough argument to really convince myself of their true relationship.

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  • $\begingroup$ The suspension is built out of loops parameterized by $X$. The classifying map send these to loops in $BG$. Loops in $BG$ are homotopy equivalent to points in $G$. Putting all that together you construct a map from $X$ to $G$. ... $\endgroup$
    – Kevin Walker
    Commented Aug 20, 2014 at 12:37
  • $\begingroup$ ... But perhaps your problem is that you don't see why the map in the previous comment is the clutching map? Consider the case where $X$ is a point. $\endgroup$
    – Kevin Walker
    Commented Aug 20, 2014 at 12:41
  • $\begingroup$ Well, I think that my problem was no so much in the visualisation, or in understanding the mechanics on a superficial level, but actually chasing the isomorphisms through to construct the map from X to G. I'm happy to generate a map from X to G, or map from $\Sigma X$ to BG in the reverse case... But showing these maps represent the correct homotopy classes is proving more tricky for me. $\endgroup$
    – Tyrone
    Commented Aug 21, 2014 at 14:12
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    $\begingroup$ The clutching function correspondence gives you one way of constructing an equivalence $G\simeq \Omega BG$, and you are asking whether that equivalence is the same as the equivalence $G\simeq \Omega BG$. But this is impossible to answer without specifying how the latter equivalence is constructed. $\endgroup$
    – Eric Wofsey
    Commented Aug 27, 2014 at 12:54
  • $\begingroup$ @EricWofsey, I guess I read the question differently: Tyrone describes a clutching map $X\to G$ associated to a bundle $E\to \Sigma X$, and wants to know how this map is related to the classifying map for $E$. $\endgroup$
    – Dan Ramras
    Commented Aug 28, 2014 at 2:54

1 Answer 1

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Here is one possible way of answering the question, using the simplicial model for $EG$. In this context, we can show very explicitly that the map $\Sigma X\to BG$ built (via adjointness) from the clutching function for $E$ classifies $E$ (and then if $X$ is a CW complex, any other classifying map for $E$ is homotopic to this one, by a standard cell-by-cell argument).

If $E\to \Sigma X$ is a principal $G$-bundle, then as explained in the question, $E$ is trivial over each half of $\Sigma X$. So $E$ is (isomorphic to) the quotient space $$(C_{-} X\times G) \cup (C_+ X \times G)\,/\,(x_-, g)\sim(x_+, c(x)g)$$ where $c: X\to G$ is the clutching function, and the notation for the relation is supposed to indicate that $x_-$ and $x_+$ are the copies of $x$ in the bases of the cones $C_-X$ and $C_+ X$ (respectively).

Let's assume that $G$ is a Lie group, so that $BG$ can be taken to mean the bar construction on $G$ (as in, say, "Segal's paper Classifying spaces and spectral sequences"). Then there is a natural map $\Sigma G \to BG$ given by the fact that $G$ is the space of 1-simplices in $BG$, and it has an adjoint $i: G\to \Omega BG$ (which sends $G$ to the loop that "runs around the 1-simplex labeled '$g$'," so to speak).

The function $i \circ c: X \to \Omega BG$ has an adjoint $f: \Sigma X \to BG$. It sends $x$ to $[c(x), t]$, which means "the point t units along the 1-simplex labeled $c(x)$." We want to prove that $f$ classifies $E$ - that is, we want to construct a map $\tilde{f}: E\to EG$ that is $G$-equivariant and covers $f$.

This can be done very explicitly. In Segal's setup, $EG$ is also a simplicial space, with 1-simplices parametrized by $G\times G$; the $G$-action on this (universal) principal bundle is induced by right-multiplication in $G$, and the map $EG\to BG$ sends the 1-simplex $(g,h)$ to $g h^{-1}$.

I want to think of $\Sigma X$ as $X\times I$ with both ends collapsed to points, so I'll take $C_-$ to be parametrized by $t\in [0,1/2]$ and $C_+$ to be parametrized by $t\in [1/2, 1]$.

We can define $\tilde{f}$ separately on the two halves of $E$: on the left half $$C_{-} X\times G = \left[(X\times [0,1/2])/(X\times 0)\right]\times G,$$ we set $\tilde{f}(x, t, g) = [(c(x)g, g), t]$. On the right half $$C_{+} X\times G = \left[(X\times [1/2, 1])/(X\times 1)\right]\times G,$$ we set $\tilde{f}(x, t, g) = [(g, c(x)^{-1}g), t]$. Note that in both cases, when we project down to $BG$ we get just $[c(x), t]$, and that the map is well-defined on $E$ because when $t=1/2$ (where the two cones meet) the first formula sends $(x_-, g) = (x, 1/2, g)$ to $[(c(x)g, g), 1/2]$ and the second formula sends the equivalent point $(x_+, c(x)g) = (x, 1/2, c(x)g)$ to $[(c(x)g, c(x)^{-1} c(x)g], 1/2] = [(c(x)g, g), 1/2]$.

I guess something similar probably works for any topological group $G$, if one uses Milnor's join construction instead of Segal's simplicial model, but I haven't thought through it.

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  • $\begingroup$ Eric's comment on the question made me realize I didn't say why the map $G\to \Omega BG$ in my answer is a homotopy equivalence. If $EG$ is a contractible principal $G$-bundle, then the contracting homotopy $EG\times I\to EG$ induces a map $EG\to P(EG)\to P(BG)$ where $P()$ is the (based) path-space. This induces a diagram relating the fibrations $EG\to BG$ and $P(BG)\to BG$ and by the 5-lemma, the map on fibers is a weak equivalence $G\to \Omega BG$. $\endgroup$
    – Dan Ramras
    Commented Aug 28, 2014 at 3:04
  • $\begingroup$ (cont'd) Segal's model for $EG$ comes from a (topological) category that is equivalent to the trivial category. If you choose this equivalence correctly, it gives a contracting homotopy of $EG$ that induces the map $G\to \Omega BG$ in my answer. $\endgroup$
    – Dan Ramras
    Commented Aug 28, 2014 at 3:07
  • $\begingroup$ Thanks, Dan! That's very helpful. I'm a lot more familiar with Milnor's construction, and am just about to run your brilliant ideas through it! $\endgroup$
    – Tyrone
    Commented Aug 28, 2014 at 9:32
  • $\begingroup$ I'd be interested to see what sort of analogous argument can be done with Milnor's construction. Personally, I like the simplicial models because formulas like the ones above are easy to write down (although it did take me a few tries to get the map $\tilde{f}$ correct...). $\endgroup$
    – Dan Ramras
    Commented Aug 28, 2014 at 17:22
  • $\begingroup$ I finished off the piece I was studying, and if I understand it correctly, runs through pretty similarly. In Milnor's construction one takes the total space as the infinite join $EG=G*G*\dots*G*\dots$ and represents the point (equivalence classes) as finite abstract sums $(t_1g_1+t_2g_2+\dots)$. Then the majority of your formulas piece in without much change, continuity of $\tilde{f}$ being checked using the projection maps of $EG$. Unless I'm very much mistaken... $\endgroup$
    – Tyrone
    Commented Aug 29, 2014 at 10:42

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