Suppose $G$ is a Topological group then classification theorem of Principal $G$ bundles says that

there is a Principal $G$ bundle $EG\rightarrow BG$ such that any principal $G$ bundle over a decent topological space $X$ has to be pullback of a continuous map $f:X\rightarrow BG$.

Can we replace Topological group by Lie group and Topolgical space by Smooth manifold. Do we get all Principal $G$ bundles over smooth manifold in this case? Is $BG$ a smooth manifold??

In his book Fiber bundles, Dale Husemoller does not say anything (I could not see anything) about smooth version of that classification result. Now I have a doubt if that Milnor constriction $BG$ for a Lie group $G$ gives a smooth manifold or is this classification only for topological Principal $G$ bundles.

In similar way, when doing classification of vector bundles we construct what is called Grassmannian $G_n$ for each $n$ and a topological vector bundle $E_n\rightarrow G_n$ and say that for a decent topological space $X$, any rank $n$ vector bundle (in Topological sense, not smooth sense) over $X$ should be pullback of a continuous map $X\rightarrow G_n$. Here also we are classifying only topological vector bundles, not smooth vector bundles, right? I was thinking $G_n$ is a manifold and it classifies all smooth vector bundles but then realise I am thinking wrong.

Is the classification only restricted to Topological set up?

Is there similar classification in smooth set up? Like classifying smooth Principal $G$ bundles and classifying smooth vector bundles?

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    $\begingroup$ A Lie group is a topological group, a manifold is a decent topological space. In general $BG$ is not a smooth finite dimensional manifold. There are Hilbert manifold models for reasonable $G$. (Embed $G$ into some $GL(n)$ which you can embed into $GL(\mathbb{H})$ which is contractible by Kuipers Theorem. The group $G$ acts freely on $GL(\mathbb{H})$ by left multiplication. Take the quotient which is a Hilbert manifold for reasonable $G$.) $\endgroup$
    – Thomas Rot
    Commented Sep 10, 2018 at 11:58
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    $\begingroup$ For vector bundles: You can approximate the classifying map with a smooth map (take a finite dimensional approximation of the universal grassmannian to stay in the setting of finite dimensional manifolds). This shows that topological vector bundles are isomorphic to a smooth vector bundle. $\endgroup$
    – Thomas Rot
    Commented Sep 10, 2018 at 12:01
  • $\begingroup$ @ThomasRot A Lie group is a topological group a manifold is a decent topological space... I have no problem with that... My question is, in this set up, do we get smooth principal bundle? Or a smooth vector bundle? I think it is a no $\endgroup$ Commented Sep 10, 2018 at 12:01
  • $\begingroup$ @ThomasRot can you give some reference where this is done $\endgroup$ Commented Sep 10, 2018 at 12:05
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    $\begingroup$ @PraphullaKoushik In my experience, people just say "in any continuous homotopy class there is a smooth representative" and then they wave their hands a lot. $\endgroup$ Commented Sep 10, 2018 at 12:39

1 Answer 1


Here is a nice paper where the authors compare smooth and topological principal bundles: https://edoc.hu-berlin.de/bitstream/handle/18452/11495/214.pdf?sequence=1

Their comparison result is very general it works for infinite dimensional Lie Groups (such as diffeomorphisms groups).

  • $\begingroup$ Thanks thanks.. I will read. :) $\endgroup$ Commented Mar 27, 2019 at 1:56

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