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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.

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  • $\begingroup$ 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 $\begin{pmatrix}a&0\\0&-a\end{pmatrix}$. $\endgroup$ – user647486 Apr 15 at 9:59
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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.

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  • $\begingroup$ 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. $\endgroup$ – Hongrui Li Apr 16 at 0:19
  • $\begingroup$ @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. $\endgroup$ – Joppy Apr 16 at 0:47

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