What are the orbits of $SL(2, R)$ on $R^2 \times R^2$? When $SL(2, R)$ acts on $R^2$, we get two orbits: 0 and $R^2 - 0$.
What happens when $SL(2, R)$ acts on $R^2 \times R^2$ by $g \cdot (u, v) = (g \cdot u, g \cdot v)$?
 A: Apart of $(0,0)$, one orbit generated by $(0,v)$ and one by $(v,0)$ for $v\neq 0$, all other orbits contain an element $((1,0),v)$, $v\neq 0$. The stabilizer of $(1,0)$ on $SL$ is the subgroup $\left\{\begin {pmatrix}1 & x\\ 0 & 1\end{pmatrix}|x\in\mathbb{R}\right\}$; these matrices send  $((1,0), (a,b))$ to $((1,0), (a+bx,b))$. It follows that
(1) The orbit of $((1,0),(a,0))$ contains no other element of the form $((1,0),v)$; you get one such orbit for each $a\neq 0$,
(2) For each $b\neq 0$, you have another orbit containing $((1,0), (0,b))$ (it also ontains all elements $((1,0), (*,b))$; this is another infinite family of orbits parametrized by $b$.
Clearly, each $((1,0),v)$ for $v\neq 0$ is contained in exactly one orbit of type (1) or (2).
A: $\mathrm{SL}(2,\mathbb{R})$ acts diagonally on $\mathbb{R}^2\times\mathbb{R}^2$ the same way it acts on $M_2(\mathbb{R})$ by left multiplication.
We may split elements of $M_2(\mathbb{R})$ into three classes, depending on rank.
Rank $2$. These are invertible matrices, i.e. elements of $\mathrm{GL}(2,\mathbb{R})$. Every matrix with determinant $d\ne0$ may be uniquely written in the form $$A\begin{bmatrix}\pm\sqrt{|d|} & 0 \\ 0 & \sqrt{|d|}\end{bmatrix}$$ (the sign matches that of $d$) where $A\in\mathrm{SL}(2,\mathbb{R})$. This parametrizes regular orbits by $\mathbb{R}\setminus0$.
Rank $1$. These matrices are all of the form $uv^T$ with $u,v\in\mathbb{R}^2$ nonzero column vectors. Without loss of generality $\|v\|=1$ and $v$ makes an angle in $[0,\pi)$ with the positive $x$-axis. To see this, take $v$ to be in the kernel's orthogonal complement and let $u$ be the image of $v$. This uniquely determines $u$ and $v$ (subject to the restriction on $v$), so these orbits are parametrized by $(\mathbb{R}^2\setminus0)\times[0,\pi]$ modulo $(u,0)\sim(u,\pi)$, which is $\simeq(\mathbb{R}^2\setminus0)\times S^1\simeq\mathbb{R}\times S^1\times S^1$. The stabilizer of any $u\in\mathbb{R}^2\setminus0$ will be conjugate to the stabilizer of $(1,0)^T$ (since $\mathrm{SL}(2,\mathbb{R})$ acts transitively), which is the subgroup of unitriangular matrices. 
Rank $0$. This is just the zero matrix. Obviously this is a fixed point.
There is an Iwasawa decomposition $\mathrm{SL}(2,\mathbb{R})=KAN$, where $K=\mathrm{SO}(2)\simeq S^1$, $A$ is the subgroup of positive diagonal matrices $\simeq\mathbb{R}$, and $N$ the subgroup of unitriangular matrices $\simeq\mathbb{R}$; topologically (but not group-theoretically), $\mathrm{SL}(2,\mathbb{R})$ is a direct product of these three subgroups. This can be used to determine the shape of all the orbits too.
A: $SL_2(\mathbb{R})$ acts on $x \in M_2(\mathbb{R})$ (the $2 \times 2$ matrices) as $(g,x) \mapsto gx$
Let $M_2(\mathbb{R}) = GL_2(\mathbb{R}) \cup E$ where $ E= \{x \in M_2(\mathbb{R}), \det(x) = 0\}$  


*

*$GL_2(\mathbb{R})/SL_2(\mathbb{R}) = \mathbb{R}^*$ which means that $x,y\in GL_2(\mathbb{R})$ are in the same orbit iff $\det(x) = \det(y)$

*If $x \in E$, then $x = (Au,Bu)$ with $u \in \mathbb{R}^2 \setminus \{\scriptstyle\begin{pmatrix}0 \\ 0\end{pmatrix}\}$ and $A,B \in \mathbb{R}$, 
so that $gx = (Av,Bv)$ where $v = gu$.
And since for any $u,v \in \mathbb{R}^2 \setminus \{\scriptstyle\begin{pmatrix}0 \\ 0\end{pmatrix}\}$ there is a  $g\in SL_2(\mathbb{R})$  such that $gu = v$, you get that the orbit of $x = (Au,Bu)$ is $\{(Av,Bv), v \ne \begin{pmatrix}0 \\ 0\end{pmatrix}\}$
