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Apr 1 at 19:46 vote accept 0207
Apr 1 at 18:59 history became hot network question
Apr 1 at 17:54 history edited LSpice CC BY-SA 4.0
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Apr 1 at 17:48 answer added mme timeline score: 11
Apr 1 at 16:56 comment added 0207 @mme Yes we can adopt this definition.
Apr 1 at 16:50 comment added mme Here is a formal statement which I think the OP would be happy with as a definition of "approximate". For any compact Lie group $G$ and any integer $n$, is there some closed manifold (of large dimension) $M$ which is $n$-connected and carries a free action of $G$?
Apr 1 at 16:47 comment added Moishe Kohan What exactly do you mean by "approximate?"
Apr 1 at 16:41 history edited 0207 CC BY-SA 4.0
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Apr 1 at 16:35 comment added 0207 @Achim Krause Thanks, this is an excellent answer. I just realized I'm more interested in the converse problem. Given a Lie group $G$ and its classfying space $BG$, can we always approximate it using boundaryless manifolds?
Apr 1 at 12:49 comment added Ryan Budney Which definition of $BG$ are you using to get a well-defined homeomorphism type? Usually $BG$ is only well-defined up to a homotopy-equivalence. So $BG = \{*\}$ and $BG=[0,1]$ are often seen as basically "the same".
Apr 1 at 4:25 comment added Achim Krause But I think $BG$ is basically never finite dimensional. For example for compact connected $G$, this is a nice exercise with the Serre spectral sequence.
Apr 1 at 4:23 comment added Achim Krause Any finite CW complex is homotopy equivalent to a manifold with boundary (by embedding into $\mathbb{R}^n$ and thickening), and so every finite type CW complex can be approximated by manifolds with boundary.
S Apr 1 at 3:44 review First questions
Apr 1 at 6:41
S Apr 1 at 3:44 history asked 0207 CC BY-SA 4.0