Orbits of the action of $SL(n,\mathbb R)$ on $\mathbb R^n$.

Question

We know that $GL(n,\mathbb R)$ acts on $\mathbb R^n$ via left multiplication. We can easily see that there are two orbits viz $\{0\}$ and $\mathbb R^n-\{0\}$. Now we also know that if $G$ acts on $X$ and $H$ is a subgroup of $G$ then $H$ acts on $X$ and the $H$-orbits are actually suborbits of the $G$-orbits. Now I want to know what are the orbits of the action of $SL(n,\mathbb R)$ on $\mathbb R^n$. Does it subdivide the orbit of non-zero vectors further into smaller suborbits?

Answer 1

Consider the action of $SL(n,\mathbb{R})$ in $\mathbb{R}^n$ via left multiplication.

If $n=1$, then $SL(1,\mathbb{R})\cong\{1\}$ and the action is trivial.

We will study $n=2$, the case $n>2$ is equal with some dots around. Take a generic vector $\begin{pmatrix} x\\y \end{pmatrix}\ne\begin{pmatrix} 0\\0 \end{pmatrix}$. We see this vector is in the same orbit of $\begin{pmatrix} 1\\0 \end{pmatrix}$.

If $y\ne0$ $$\begin{pmatrix} 1&0\\y&1 \end{pmatrix}\begin{pmatrix} 1\\0 \end{pmatrix}=\begin{pmatrix} 1\\y \end{pmatrix}$$ $$\begin{pmatrix} 1&\frac{x-1}{y}\\0&1 \end{pmatrix}\begin{pmatrix} 1\\y \end{pmatrix}=\begin{pmatrix} x\\y \end{pmatrix}.$$

If $y=0$, then $x\ne 0$ and $$\begin{pmatrix} 1&0\\1&1 \end{pmatrix}\begin{pmatrix} 1\\0 \end{pmatrix}=\begin{pmatrix} 1\\1 \end{pmatrix}$$

$$\begin{pmatrix} 1&x-1\\0&1 \end{pmatrix}\begin{pmatrix} 1\\1\end{pmatrix}=\begin{pmatrix} x\\1 \end{pmatrix} $$ $$\begin{pmatrix} 1&0\\-\frac{1}{x}&1 \end{pmatrix}\begin{pmatrix} x\\1 \end{pmatrix}=\begin{pmatrix} x\\0 \end{pmatrix} .$$

So if $n\ge2$ the orbits are exactly the orbits of $GL(n,\mathbb{R})$.

Those kind of matrices that allow you to do operations with the rows of a vector are called shear matrices or transvections. They differ from the identity matrix only for an element which doesn't lie in the diagonal and if $n\ge 2$ they generate $SL(n,\mathbb{F})$ in any field $\mathbb{F}$.