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

Why is the described map bijective? The statement '$A \sim B \iff AB^T \in O(n - k)$ so $V_k(\mathbb{R}^n) \cong O(n)/O(n - k)$' seems to appear from nowhere to me.

Let's define $f: O(n) \to V_k(\mathbb{R}^n), A \mapsto (e_1, ..., e_k) A$. We have that $f(A) = f(B)$ for $A$ and $B$ in $\{ (e_1, ..., e_k)A = (e_1, ..., e_k) \ | \ A \in O(n) \} \cong O(n - k)$, so it descends to a map $\tilde{f}: O(n) \to V_k(\mathbb{R}^n)$. As $f$ is surjective, so is $\tilde{f}$, but why is $\tilde{f}$ injective? I think this is the case iff we have that $f(A) = f(B) \implies A \sim B \iff A, B \in O(n - k)$, but $f(A) = f(A^T)$ for any $A$ in $O(n)$, so this is not the case.

Rodrigo de Azevedo
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user388557
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  • The question you point to says $A \sim B \iff AB^{-1} \in O(n-k)$ (not $AB^{T}$) and it shows some intermediate steps to demonstrate the equivalence. It is not true that $f(A) = f(A^{T})$ ($f$ maps $A$ to the first $k$ rows of the matrix representation of $A$ and that is not invariant under transposition). – Rob Arthan Apr 13 '20 at 21:17
  • Isn't $B^{-1} = B^T$ if $B$ is orthogonal? When you say it like that, I understand that $f(A)$ is not $f(A^T)$, but I don't see why 'as $AA^T = I$ we have that $(e_1, ..., e_k) = (e_1, ..., e_k)AA^T$, so by right multiplying with $A^T$ we get that $(e_1, ..., e_k) A^T = (e_1, ..., e_k)A' is wrong. Can you shed some light on that? – user388557 Apr 14 '20 at 09:45
  • Sorry I was being a bit picky about you not quoting from the other question accurately. Sure $A^T = A^{-1}$ if $A$ is orthogonal. What the other question says is that $f(A) = f (B)$ iff $AB^{-1} \in O(n-k)$ (by which they mean the copy of $O(n-k)$ that acts on the last $n-k$ standard basis vectors). So $f(A) = f(A^T)$ iff $(A^T)^{-1} = A^2 \in O(n-k)$. (I think you substituted $A$ for $B$ when you should have substituted $A^T$.) – Rob Arthan Apr 14 '20 at 19:42

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Consider the right action of ${\rm O}(n)$ on $V_k(\Bbb R^n)$ given by $$V_k(\Bbb R^n)\times {\rm O}(n)\ni(\mathfrak{v},A)\mapsto \mathfrak{v}A\in V_k(\Bbb R^n).$$The stabilizer of a given $\mathfrak{v}\in V_k(\Bbb R^n)$ is isomorphic to ${\rm O}(n-k)$ since an element in said stabilizer is uniquely determined by it's action on ${\rm span}(\mathfrak{v})^\perp$. And the action is transitive, so the orbit-stabilizer theorem gives $V_k(\Bbb R^n)\cong {\rm O}(n)/{\rm O}(n-k)$.

As a bonus, note that we can also establish that the Grassmannian ${\rm Gr}(k,n)$ is a homogeneous space. We have the canonical map ${\rm span}:V_k(\Bbb R^n)\to {\rm Gr}(k,n)$ and a left action $${\rm O}(n)\times {\rm Gr}(k,n)\ni (A,W)\mapsto A[W]\in {\rm Gr}(k,n),$$for which the stabilizer of a given $W \in {\rm Gr}(k,n)$ is isomorphic to ${\rm O}(k)\times {\rm O}(n-k)$ (due to a block diagonal decomposition). It is also transitive, so the orbit-stabilizer theorem gives ${\rm Gr}(k,n)\cong {\rm O}(n)/({\rm O}(k)\times {\rm O}(n-k))$.

Ivo Terek
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