Why the orbit is of dimension $12$? Let $SL_3$ acts on the variety consisting of all nilpotent $3$ by $3$ matrices over $\mathbb{C}$ by conjugation. Let $S_p$ be the orbit of the matrix $$ a=\left( \begin{matrix} 0 & 1 & 0  \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{matrix} \right). $$ Why the orbit $S_p$ over $\mathbb{R}$ is of dimension $12$?
I think that $S_p$ consisting of all matrices $g a g^{-1} $, $g \in SL_3$. But how to show that $S_p$ over $\mathbb{R}$ is of dimension $12$? Thank you very much.
 A: By orbit-stabilizer, the dimension of the orbit should be the dimension of the group minus the dimension of the stabilizer, so we should compute the dimension of the stabilizer. Suppose
$$\begin{pmatrix} a & b & c \\ d & e & f \\ g & h & i\end{pmatrix} \begin{pmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0\end{pmatrix} = \begin{pmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0\end{pmatrix} \begin{pmatrix} a & b & c \\ d & e & f \\ g & h & i\end{pmatrix}.$$
Multiplying the matrices, this tells us
$$\begin{pmatrix} 0 & a & b \\ 0 & d & e \\ 0 & g & h \end{pmatrix} = \begin{pmatrix} d & e & f \\ g & h & i \\ 0 & 0 & 0 \end{pmatrix}.$$
Therefore $d,g,h=0$ and $a=e=i$ and $b=f$. So a stabilizing matrix looks like
$$\begin{pmatrix}a & b & c \\ 0 & a & b \\ 0 & 0 & a\end{pmatrix}.$$
In order to be in ${\rm SL}_3$ it must satisfy $a^3=1$, so $a$ is a cube root of unity. This leaves two degrees of freedom ($b$ and $c$), so the stabilizer has complex dimension $2$. The group ${\rm SL}_3(\Bbb C)$ has complex dimension equal to $3^2-1=8$. Therefore the orbit has complex dimension $8-2=6$, which means it has real dimension equal to $2\cdot 6=12$.
