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

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.

• Yes, this is correct. Voting as too localized, because no further answer seems necessary. Oct 23, 2011 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. Mar 19, 2014 at 0:30

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 $[-1,1]^{n \times n}$ ).