Grassmanians $Gr_k(\mathbb R^n) \cong Gr_{n-k}(\mathbb R^n)$

Question

I am trying to prove that the Grassmanians $Gr_{n-k}(\mathbb R^n)$ and $Gr_{k}(\mathbb R^n)$ are homeomorphic. Intuitively, this makes sense; specifying a $k$-dimensional subspace is equivalent to specifying its $n-k$-dimensional orthogonal complement. But, I am not quite sure how to prove this formally. Can someone explain?

Answer 1

You do not need to pick a metric or anything. The correct coordinate-invariant form of the isomorphism is that $\text{Gr}_k(V)$ is canonically homeomorphic (moreover, isomorphic as a projective variety over an arbitrary field) to $\text{Gr}_{n-k}(V^{\ast})$, where $V$ is an $n$-dimensional vector space and $V^{\ast}$ is its dual. The canonical map sends a subspace $W$ of $V$ to its annihilator

$$\text{Ann}(W) = \{ v^{\ast} \in V^{\ast} : v^{\ast}(W) = 0 \}.$$

Can you fill in the rest from here?