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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?

Praphulla Koushik
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    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 $G$.) – Thomas Rot Sep 10 '18 at 11:58
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    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. – Thomas Rot Sep 10 '18 at 12:01
  • @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 – Praphulla Koushik Sep 10 '18 at 12:01
  • @ThomasRot can you give some reference where this is done – Praphulla Koushik Sep 10 '18 at 12:05
  • There are some discrete groups that have finite-dimensional smooth classifying spaces (eg $\mathbb{Z}$, or surface groups), but in general one needs infinite-dimensional classifying spaces. For Hermitian vector bundles, the classifying space can be constructed as an infinite-dimensional manifold eg https://en.wikipedia.org/wiki/Classifying_space_for_U(n)#Construction_as_an_infinite_Grassmannian – David Roberts Sep 10 '18 at 12:39
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    @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. – Vít Tuček Sep 10 '18 at 12:39
  • I am not completely sure what should be a infinite dimensional classifying space mean... @DavidRoberts By infinite dimensional manifold you mean Banach manifold right?? – Praphulla Koushik Sep 10 '18 at 12:59
  • @VítTuček Hehe :D I do not know why they do that.. they can say little more clearly... – Praphulla Koushik Sep 10 '18 at 13:00
  • I dont know a reference – Thomas Rot Sep 10 '18 at 13:16
  • @PraphullaKoushik no. It makes sense for any space of the homotopy type of a CW complex to say it is infinite dimensional, and for a Lie group $G$, $BG$ can be constructed as such a space. There could also be a algebraic topology characterisation, for instance: having nontrivial cohomology in unboundedly-many degrees. eg https://mathoverflow.net/questions/79741/why-is-bg-infinite-dimensional-for-g-finite and https://mathoverflow.net/questions/247006/infinite-grassmannian-does-not-have-the-homotopy-type-of-a-finite-dimensional-co – David Roberts Sep 10 '18 at 20:43
  • @PraphullaKoushik note the examples mentioned here https://mathoverflow.net/questions/234901/does-mathbb-c-mathbb-p-infty-have-a-group-structure – David Roberts Sep 11 '18 at 00:29
  • @DavidRoberts Ok... Yes, You can say a CW complex is infinite dimensional... and by infinite dimensional it means as a CW complex here.. If a space is homotopic to a infinite dimensional CW complex you also call that to be infinite dimensional space... For a Lie group $G$ we construct $BG$... Ok Ok.. So, there is no smooth structure as such in $BG$.. Got it... Thanks for the links... – Praphulla Koushik Sep 11 '18 at 03:39

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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).

David C
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