Is the Stiefel manifold $V_k(\mathbb{R}^n)$ homeomorphic to $O(n)/O(n-k)$?

Question

Am I right that the Stiefel manifold $V_k(\mathbb{R}^n)$ (set of all orthonormal $k$-frames in $\mathbb{R}^n$) is 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.

Answer 1

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.

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

2. 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.