On direct limit of Stiefel mainfold - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-20T11:07:51Z http://mathoverflow.net/feeds/question/118819 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/118819/on-direct-limit-of-stiefel-mainfold On direct limit of Stiefel mainfold Oscar1778 2013-01-13T15:44:47Z 2013-01-28T01:22:00Z <p>I'd like to build a model for the space $EU(n)$: the total space of universal bundle $\pi:EU(n) \rightarrow BU(n)$. $\;$ $EU(n)$ must be a conctractible space on which $U(n)$ acts freely. So I consider Stiefel manifold $V_{n}(\mathbb{C}^{k})$ of $\;$ $n-$frame in $\mathbb{C}^{k}$ and the associated fiber bundle $$V_{n-1}(\mathbb{C}^{k-1}) \rightarrow V_{n}(C^{k}) \rightarrow S^{2k-1}$$ The induced sequence in homotopy shows that all homotopy gruops vanish for $k$ large. Indeed, the natural action of $U(n)$ on $V_{n}(\mathbb{C}^{k})$ is free (the quotien is the Grassmannian manifold). So I can choose $$EU(n)=\lim_{k \to \infty} V_{n}(\mathbb{C}^{k}) = V_{n}(\mathbb{C}^{\infty})$$ the direct limit of $V_{n}(\mathbb{C}^{k})$. How could I prove that the action of $U(n)$ on $V_{n}(\mathbb{C}^{k})$ is still free? In other words, why is the limit of a free action free?</p> http://mathoverflow.net/questions/118819/on-direct-limit-of-stiefel-mainfold/118852#118852 Answer by Peter Michor for On direct limit of Stiefel mainfold Peter Michor 2013-01-14T00:38:40Z 2013-01-14T00:38:40Z <p>See section 47 of <a href="http://www.mat.univie.ac.at/~michor/apbookh-ams.pdf" rel="nofollow">(here)</a>.</p>