The action of $GL(\mathbb{R}^{n})\otimes GL(\mathbb{R}^{m})$ on $\mathbb{R}P^{(mn-1)}$

Question

Motivated by the following RG question we ask a related question as follows:

We identify $\mathbb{R}^{n} \otimes \mathbb{R}^{m}$ with $\mathbb{R}^{mn}$. We define $GL(\mathbb{R}^{n})\otimes GL(\mathbb{R}^{m})$ as a subgroup of $GL(\mathbb{R}^{n} \otimes \mathbb{R}^{m})$ which consist simple tensors $T\otimes S\;\; \text{where}\;\;T\in GL(\mathbb{R}^{n}),\;S\in GL(\mathbb{R}^{m})$.

Is the action of $GL(\mathbb{R}^{n})\otimes GL(\mathbb{R}^{m})$ on $\mathbb{R}P^{(mn-1)}$ a transitive action? If not, what is the topological description of the quotion $\mathbb{R}P^{(mn-1)}/GL(\mathbb{R}^{n})\otimes GL(\mathbb{R}^{m})$?

The above action is the restriction of the natural action of $GL(\mathbb{R}^{n} \otimes \mathbb{R}^{m})$

Note: "$GL(\mathbb{R}^{n})\otimes GL(\mathbb{R}^{m})$" is just a notation so we do not confuse it with the tensor product of two groups or some thing else. After that I post this question, I heard from Todd Trimble that this group is isomorphic to $\frac{GL(\mathbb{R}^{n}) \times GL(\mathbb{R}^{m})} {\mathbb{R}^{*}}$ with obvious action $\lambda.(A,B)=(\lambda A, \lambda^{-1}B)$

Answer 1

By the evident isomorphism of $\mathbb R^m\otimes \mathbb R^n$ with $M_{m\times n}(\mathbb R)$, the orbits of the action of $Gl(\mathbb R^m)\otimes Gl(\mathbb R^n)$ corresponds to orbits of the action of $Gl(\mathbb R^m)\times Gl(\mathbb R^n)$ on $M_{m\times n}(\mathbb R)$ by left and right multiplication: $$(A,B)\cdot X = AXB^t.$$

And the orbit of an element in this action is determined only by its rank. It's because for a matrix $X\in M_{m\times n}(\mathbb R)$ its orbit under right multiplication is only determined by span of its column vectors, and two subspaces of $\mathbb R^m$ of the same dimension are conjugate under automorphisms of $\mathbb R^m$.