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We know that $BO(1) =\mathbb{R}P^\infty$ has closed, finite-dimensional manifold approximations $\mathbb{R}P^1\subset \mathbb{R}P^2\subset\cdots.$

Similarly $BO(2)$ can be approximated by closed, finite-dimensional manifolds, as follows. There is a fibration $$ BSO(2) \to BO(2) \to BO(1), $$ where the projection is given by the determinant. Both the base space $BO(1)=\mathbb{R}P^\infty$ and the fibre $BSO(2)=\mathbb{C}P^\infty$ can be approximated by finite-dimensional manifolds. Hence the total space $BO(2)$ can be approximated by the Dold manifolds $$ P(n,m) = \mathbb{C}P^n \times_{\mathbb{Z}/2} S^m,$$ where the generator of $\mathbb{Z}/2$ acts on the sphere by $-1$ and on $\mathbb{C}P^n$ by complex conjugation.

Do similar closed, finite-dimensional manifold approximations exist for $BO(3)$?

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  • $\begingroup$ In your examples, the approximating manifolds are closed. For $BO(3)$, do you want your manifolds to be closed? There are approximations of classifying spaces of Lie groups by open manifolds that are used, for instance, in constructing equivariant Chow theory. $\endgroup$
    – Jason Starr
    Commented Jul 31, 2015 at 13:49
  • $\begingroup$ @JasonStarr: Yes, I intended the manifolds to be closed, thanks for this. I'll edit. $\endgroup$
    – Mark Grant
    Commented Jul 31, 2015 at 13:54
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    $\begingroup$ For $BO(3)$ one can choose the Grassmannian of 3-planes in ${\mathbb R}^\infty$, and this is in some sense approximated by the finite-dimensional Grassmann manifolds of 3-planes in ${\mathbb R}^n$. What kind of approximation do you have in mind? $\endgroup$
    – Allen Hatcher
    Commented Jul 31, 2015 at 13:56
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    $\begingroup$ @user51223: In response to your first comment, because we can view $BO(2)$ as the total space of the Borel fibration $BSO(2)\times_{\mathbb{Z}/2} S^\infty$. $\endgroup$
    – Mark Grant
    Commented Aug 3, 2015 at 9:58
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    $\begingroup$ $BSpin(3)$ has the very nice representation $\mathbb H\mathbb P^\infty$. The fiber sequence $BSpin(3)\to BSO(3)\to K(\mathbb Z/2,2)$ suggests that $BSO(3)$ is on par with $K(\mathbb Z/2,2)$. ($BO(n)$ reduces to $BSO(n)$ in a way independent of $n$.) $\endgroup$
    – Ben Wieland
    Commented Aug 6, 2015 at 1:59

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