# Stiefel Manifold

Am I right that Stiefel manifold $V_k(\mathbb{R}^n)$ (set of all orthonormal k-frames in $\mathbb{R}^n$) homeomorphic to $O(n)/O(n-k)$?

Sketch of proof: any element of $V_k(\mathbb{R}^n)$ can be obtained from standard basis $\mathbb{R}^n$ $e_1,\ldots , e_n$ by action group $O(n)$ on set $\{e_1,\ldots , e_k\}$, i.e.: $$V_k(\mathbb{R}^n)=\{(e_1,\ldots,e_k)A~|~A\in O(n)\}.$$ And $A\sim B$ iff $(e_1,\ldots,e_k)A=(e_1,\ldots,e_k)B$ $\Leftrightarrow$ $(e_1,\ldots,e_k)AB^{-1}=(e_1,\ldots,e_k)$ $\Leftrightarrow$ $AB^{-1}\in O(n-k)$, and hence $$V_k(\mathbb{R}^n)\approx O(n)/O(n-k).$$ $\blacksquare$

Thanks.

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Yes, this is correct. Voting as too localized, because no further answer seems necessary. –  Rasmus Oct 23 '11 at 8:27
I don't agree that "This question is unlikely to help any future visitors". Also, closing it prevents any better or more detailed proofs from being posted. –  user127096 Mar 19 at 0:30