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?
Grassmanians $Gr_k(\mathbb R^n) \cong Gr_{n-k}(\mathbb R^n)$
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general-topology
differential-topology
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0Hodge duality? Can you write the Grassmanians in terms of wedge products? The hodge dual connects $k$ and $n-k$ forms. This also requires a metric. – 2012-08-31
1 Answers
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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?