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As pointed out by 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.

Given a Lie group $G$, is it possible to approximate $BG$ by using boundaryless manifolds? For example, $BS^1$ can be approximated by $\mathbb{CP}^N$. Is this a general phenomenon or special case for torus? (The classifying space of a torus supports a group structure, so it has to be boundaryless.)

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    $\begingroup$ 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. $\endgroup$
    – Achim Krause
    Commented Apr 1 at 4:23
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    $\begingroup$ 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. $\endgroup$
    – Achim Krause
    Commented Apr 1 at 4:25
  • $\begingroup$ 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". $\endgroup$
    – Ryan Budney
    Commented Apr 1 at 12:49
  • $\begingroup$ @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? $\endgroup$
    – 0207
    Commented Apr 1 at 16:35
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    $\begingroup$ 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 $G$? $\endgroup$
    – mme
    Commented Apr 1 at 16:50

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Yes, this is possible. The construction below is fairly standard and I'm not sure where I learned it.

The aim is to show that for each compact Lie group $G$, there is a closed $n$-connected manifold $M_n$ with a free action of $G$. Passing to a subsequence if necessary, one can equip these with a sequence of equivariant embeddings $M_n \hookrightarrow M_{n+1}$; then the colimit of $M_n/G$ is a model of $BG$ (and the colimit of the $M_n$ a model of $EG$).

The key observation is that as a consequence of the Peter--Weyl theorem, every compact Lie group embeds as a closed subgroup of $U(k)$ for some large $k$, so it suffices to construct $M_n$ for $U(k)$ and simply restrict the $U(k)$-action to $G$.

Now for $n \ge k$ we may take $M_n$ to be the Stiefel manifold $V_k(\Bbb C^{n+k})$ of lists $(v_1, \cdots, v_k)$ of linearly independent vectors in $\Bbb C^{n+k}$.

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  • $\begingroup$ The earliest example I see exploiting the embedding $G \hookrightarrow U(k)$ to give topological information about $BG$ is Venkov, 1959, which uses this and the Leray--Serre spectral sequence to argue that $H^*(BG)$ is finitely generated. This observation is mentioned in Dieudonne's history of topology (the final paragraph of 3.2.F), but as a part of the mathematical exposition and not a historical comment. It seems quite plausible from context it could have appeared earlier in foundational work of Whitney, but I'm not sure. $\endgroup$
    – mme
    Commented Apr 1 at 22:16
  • $\begingroup$ When $G$ is discrete, $BG$ is a $K(G, 1)$. Do you know if there is a model for $K(G, n)$ which is a colimit of closed manifolds? There is such a model for $K(\mathbb{Z}, 2)$ for example (namely $\mathbb{CP}^{\infty}$). $\endgroup$
    – Michael Albanese
    Commented Apr 27 at 1:45

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