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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?

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The limit is a union, ie a direct limit of injective maps. If a point had isotropy in the union then it would have isotropy in the first Stiefeld manifold it belongs to. – Fernando Muro Jan 13 '13 at 15:58
I think that the OP is using "free" for a topological group action in a strong sense: not only is there no isotropy but the map to the orbit space is a (principal) bundle. I don't know how generally it is true that a $U(n)$-action with trivial isotropy groups is free in the strong sense, but in this case you can explicitly make trivializing open sets: just think of how you make charts in the (finite-dimensional) Grassmannians by looking at where a given set $n$ of the $k$ coordinates are nonzero. It still works with $k$ infinite. – Tom Goodwillie Jan 13 '13 at 16:43
@Tom, if you mean locally trivial when you say "strong", then it holds under very general topological assumtions satisfied by this example, see eg – Fernando Muro Jan 13 '13 at 21:15

See section 47 of (here).

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