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.
 A: This is right, as has been pointed out in comments. There is something left to do: you've shown a bijection but not commented on the topology. You may or may not find this pedantic, but either way there are at least two ways to deal with it.


*

*Use the orbit–stabilizer bijection you've noted to port over a topology for $V_k(\mathbb R^n)$.

*a. Define your topology on $V_k(\mathbb R^n)$. This is most easily done by considering it as a subset of the $n \times k$ matrices, namely those for which $A^\top A = I_k$ is the $k \times k$ identity matrix. This subspace topology is Hausdorff because the product topology on $\mathbb{R}^{kn}$ is (in turn because $\mathbb R$ is). 
b. The map $O(n)/O(n-k) \to V_k(\mathbb R^n)$ is continuous because the map $O(n) \to V_k(\mathbb R^n)$ it descends from is a restriction of the matrix multiplication $\mathbb R^{k \times n} \times \mathbb R^{n \times n} \to \mathbb R^{k \times n}$.
c. The space $O(n)/O(n-k)$ is compact because $O(n)$ is (in turn, $O(n)$ is closed since it is defined by the polynomial equations $A^\top A = I_n$ and bounded since it is contained in, say, the cube $[-1,1]^{n \times n}$ ).
d. A continuous bijection from a compact space to a Hausdorff space is a homeomorphism.
