# Conjugacy classes in space of trace zero 2*2 matrices

I'm trying to find the orbits when $$SL_2$$ operates by conjugation on $$\mathfrak{sl}_2=Lie(SL_2)=\{A|\operatorname{tr} A=0\}$$.

I have tried to write $$X\in sl_2$$ and corresponding $$AXA^{-1}$$ for random $$A\in SL_2$$ in form of $$xE_1+yE_2+zE_3$$, where $$E_1= \left[ {\begin{array}{cc} 1 & 0 \\ 0 & -1 \\ \end{array} } \right]$$,$$E_2= \left[ {\begin{array}{cc} 0 & 1 \\ 0 & 0 \\ \end{array} } \right]$$,$$E_3= \left[ {\begin{array}{cc} 0 & 0 \\ 1 & 0 \\ \end{array} } \right]$$, and find that for fixed $$x_0$$, all $$\left[ {\begin{array}{cc} x_0 & y \\ z & -x_0 \\ \end{array} } \right]$$ where $$yz\leq x_0^2$$ are in one orbit, but I can$$`$$t see how to deal with the rest.

• Note that the orbit of $X$ contains its Jordan form. Since the trace is zero, this would be $\begin{pmatrix}a&1\\0&-a\end{pmatrix}$ or $\begin{pmatrix}a&0\\0&-a\end{pmatrix}$. – user647486 Apr 15 at 9:59

The orbits of $$\operatorname{GL}_2$$ on $$\mathfrak{sl}_2$$ are labelled by the Jordan type of the matrices, which (because we are acting on traceless matrices) fall into two families: $$\begin{pmatrix}\lambda & 0 \\ 0 & -\lambda\end{pmatrix}$$ for any $$\lambda \in \mathbb{C}$$, and also $$\begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}$$. The orbits of $$SL_2$$ should be the same.
• Thank you for the answer, i tried to restrict to real matrices, so non-real λ are imaginary here and orbits with determinants $-λ^2=constant$ will be preserved under restriction. – Hongrui Li Apr 16 at 0:19
• @HongruiLi You should point out what field you're working over! In the real case, you get $\begin{pmatrix} \lambda & 0 \\ 0 & -\lambda\end{pmatrix}$ for all $\lambda \in \mathbb{R}$, then also $\begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}$, then also $\begin{pmatrix} 0 & \mu \\ -\mu & 0 \end{pmatrix}$ for all $\mu \in \mathbb{R}$ which correspond to those purely imaginary eigenvalue matrices. – Joppy Apr 16 at 0:47