Find orbits of adjoint representation of $SL(2,\mathbb{R})$ group I am studying group theory and I can't move further than understanding definitions.
So, I've got a group of two-dimensional matrixes with determinant 1 ( $ SL(2,\mathbb{R}))$.
Adjoint representation is defined by: $\operatorname{Ad}_g: h \rightarrow  ghg^{-1}, \forall g \in SL(2,\mathbb{R}), h \in sl(2, \mathbb{R}) $
Orbit of a group element $x$ is $G(x) = \{gx \in X: g \in G\}$ for each g in G.
Could you please explain at least key steps of finding orbits of adjoint representation of $ SL(2,\mathbb{R})$ ?
 A: The Lie algebra $\mathfrak{sl}(2,\mathbb{R})$ is the set of $2\times 2$ matrices with real entries and trace 0,
$$\mathfrak{sl}(2,\mathbb{R}) = \{ h\in M_2(\mathbb{R}): \text{tr}(h) = 0\}.$$
As you say, the adjoint action is by matrix conjugation:
$$Ad_g(h) = ghg^{-1}.$$
Let's take a simple orbit and think back to what you know about linear algebra. Take the traceless matrix
$$ h = \left( \begin{array}{cc} 1 & 0 \\ 0 & -1 \end{array}\right).$$
Can you describe the set of matrices of the form $ghg^{-1}$, $g\in SL(2, \mathbb{R})$?  This set is the orbit of $h$,
$$\text{orbit of }h = \left\{ ghg^{-1} : g\in SL(2,\mathbb{R})\right\}.$$
What about a different traceless matrix, such as 
$$ h = \left( \begin{array}{cc} 0 & 1 \\ 0 & 0 \end{array}\right)?$$
A: I think the following should be complete, if I am not mistaken somewhere:
$\mathfrak{sl}_2(\mathbb{R})$ are traceless $2\times 2$ matrices. One orbit is the zero orbit containing only the zero matrix, this one should be obvious. Then the orbit generated by the diagonal matrix 
\begin{pmatrix}
       a & 0 \\
       0 & -a
   \end{pmatrix}
should contain exactly the matrices with eigenvalues $(a,-a)$. For starters, anything in the same orbit will have said eigenvalues. Then, you can see that conjugation by 
\begin{pmatrix}
       1 & -c/2a \\
       0 & 1
\end{pmatrix}
will give you
\begin{pmatrix}
       a & c \\
       0 & -a
\end{pmatrix}
Similarly you can get matrices in the lower block. You can also exchange the position of $a,-a$ by conjugating with
\begin{pmatrix}
       0 & 1 \\
       1 & 0
\end{pmatrix}
I'll leave as an exercise how to get the rest of the matrices with same eigenvalues.
So we have one orbit (these are called semisimple orbits) for each $a\neq 0$. What about $a=0$? Here things become a little more interesting: You actually only have the zero matrix in the one orbit, but there is another orbit generated by 
\begin{pmatrix}
       0 & 1 \\
       0 & 0
\end{pmatrix}
This is a nilpotent orbit. By conjugation with one of the matrices previously mentioned (which one?) you can actually also get 
\begin{pmatrix}
       0 & 0 \\
       -1 & 0
\end{pmatrix}
which is therefore on the same orbit. What about 
\begin{pmatrix}
       0 & -1 \\
       0 & 0
\end{pmatrix}?
Well actually calculations will show you that this belongs in a different orbit (as long as you are in $\mathbb{R}$ you will get a square there when you conjugate!). 
To recap (please correct me if I forgot something):
We have one semisimple orbit for each $a\neq 0$. Every semisimple orbit is closed.
Two open nilpotent orbits. Their closure also contains zero.
The zero orbit.
By the way, notice that nilpotent orbits are smaller. Namely, the dimension of the semisimple orbits is 2, as you have a freedom in choosing $b,c$ in
\begin{pmatrix}
       a & b \\
       c & a
\end{pmatrix}
while the dimension of the open nilpotent orbits is 1.
