Proof that some Grassmannians are homeomorphic

Question

It is written in the wikipedia article on Grassmannians that $Gr_n(\mathbb{R}^k)$ is homeomorphic to $Gr_{k-n}(\mathbb{R}^k)$ via the map taking the orthogonal complement, could anyone give a proof/reference for this fact?

Answer 1

The orthogonal group $O(k) = \{A \in GL(k, \Bbb R) : A^{\top} A = I\}$ acts continuously and transitively on $Gr_{\ell}(\Bbb R^k)$ for all $\ell$, so we may identify $Gr_{n}(\Bbb R^k)$ as a homogeneous space $O(k) / H$, where $H$ is the stabilizer in $O(k)$ of an elected element of $Gr_n(\Bbb R^k)$. If we take the span of the first $n$ standard basis vectors, then $$H = \left\{\pmatrix{\ast&\ast\\0&\ast} \in O(k)\right\},$$ and the condition $A^{\top} A = I$ implies that all elements of $H$ actually have the form $\pmatrix{\ast&0\\0&\ast}$. But by symmetry the set of such elements is the stabilizer in $O(k)$ of the span of the last $k - n$ standard basis elements, so we can also identify $O(k) / H$ with $Gr_{k - n}(\Bbb R^k)$. (We don't need this fact, but applying the definition of the orthogonal group gives that $H \cong O(n) \times O(k - n)$.)