Let $r<n$ be two positive integers and $G=GL(n,\mathbb{R}).$ If $Gr(k,\mathbb{R}^n)$ is the set of all $k$-subspaces, then show that the $G$-sets $Gr(k,\mathbb{R}^n)$ and $G/P_k$ is isomorphic, given $P_k$ is the subgroup formed by all blockwise triangular matrices in the form $\begin{array}{l}\quad\left(\begin{array}{ll}A & B \\ 0 & D\end{array}\right) \\ \text { where } A \in G L(k, \mathbb{R}), D \in G L(n-k, \mathbb{R}), \text { and } B \in M_{k, n-k}(\mathbb{R})\end{array}$.

Should I make a homomorphism such that $P_k$ is the kernel of it, then use the 1st isomorphism theorem to prove it? Or should I find an isomorphism from these two $G$-sets directly?


Note that $P_k$ is not normal and the Grassmannian is not a group. Instead, argue that $GL(n)$ acts transitively on $Gr(k)$ with stabilizer of one point $X\in Gr(k)$ (which one?) equal to $P_k$. Then construct a continuous bijection $GL(n)/P_k\to Gr(k)$, $g\mapsto gX$, $g\in GL(n)$. Then argue that the inverse is also continuous.

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  • $\begingroup$ Should X be the trivial subspace ? $\endgroup$ – Tifsir Sep 13 at 10:12
  • $\begingroup$ @Tifsir: Not at all, the stabilizer o $\{0\}$ is the entire $GL(n)$. Try to work out the case $n=2$ first. $\endgroup$ – Moishe Kohan Sep 13 at 10:20

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