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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?

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Maybe you are stuck because the homeomorphism is not natural? I think you need to pick an inner product so that you can form orthogonal complement of a subspace. Then you have a bijection between the two sets. I believe showing continuity should not be difficult since these spaces are metric. The inverse map should also be similar. This is essentially showing that the dual of the dual is primal. –  Tunococ Aug 31 '12 at 2:46
    
Hodge 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. –  James S. Cook Aug 31 '12 at 4:01

1 Answer 1

up vote 4 down vote accepted

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?

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