Why universal G-bundles are contractible?

by Mostafa   Last Updated August 14, 2019 06:20 AM

Let $G$ be a nice topological group and $E\to B$ a universal $G$-bundle. I'm interested in a proof of contractibility of $E$ using only the universal property of it. I also know that if there is a contractible $G$-bundle, then all of others are also contractible, but does there exist a direct proof not using a special construction of contractible universal $G$-bundles?

Answers 1

Here are two different sketches-- I believe the second is easily completed:

First note that the projection $\pi: E \rightarrow B$ is nullhomotopic by the universal property of $B$ since the pullback of a principal bundle by its projection is always trivial.

This implies that the long exact sequence of the fibration decomposes into short exact sequences $0 \rightarrow \pi_{n+1}(B) \rightarrow \pi_n(G) \rightarrow \pi_n(E) \rightarrow 0$.

The map $\pi_{n+1}(B) \rightarrow \pi_n(G)$ being surjective implies the total space is weakly contractible, as desired.

This map is surjective thanks to the following construction:

Given a map $S^n \rightarrow G$ we may use it to construct a $G$-bundle over $S^{n+1}$ by clutching together two copies of the space $D^{n+1} \times G$ along $S^n \times G$ according to our map. By the universal property, this bundle corresponds to a map $S^{n+1}$ into our space, which in a just world would be the element we are after.

Here is the second:

It follows from clutching and the loop-suspension adjunction that both $\Omega B$ and $G$ represent $G$-bundles over the suspension of your space. This implies that they represent $G$-bundles over the spheres which by the cofiber sequence/clutching implies they both represent $G$-bundles over the suspension for all CW complexes.

This means that we can use the following rectangle, commutative up to homotopy since the map $E \rightarrow B$ is nullhomotopic (formatting taken from https://mathoverflow.net/a/132340/134512) . $$\begin{matrix} G & \to & E & \to & B \\ \downarrow = && \downarrow * && \downarrow = \\ \Omega B & \to & PB & \to & B \end{matrix}$$

This square induces a morphism of long exact sequences which by the five lemma is an isomorphism, which implies that $E$ is weakly contractible.

Connor Malin
Connor Malin
August 14, 2019 05:46 AM

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