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A symmetric idenpotent matrix of trace $k$ is nothing but a projection onto an $k$-dimensional space (obvious, or found in the textbook of 丘维声 (maybe)). Hence the Grassmannian $G_{n}(\mathbb F^m)$, consisting of $m$-dimensional linear subspaces of $\mathbb F^n$, is isomorphic to the manifold of symmetric idenpotent matrices in $\mathbb F^{n\times n}$ of trace $m$. Here $G_{n+1}(\mathbb F^1)=\mathbb P_{\mathbb F}^n$. |
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