Question about decomposing a space into orbits The question is:

If  the group $SL_2(\mathbb{R})$ operates by conjugation on the space of trace-zero matrices, then how can we decompose this space into orbits? 

(I know that this space is indeed the Lie algebra of $SL_2(\mathbb{R})$, and this operation is related to the adjoint representation.)
Any ideas or hints are welcome!
 A: This can be done using rational canonical form, which generalizes Jordan normal form to the non-algebraically closed case. Rational canonical form tells you that the orbits under the action of $GL_2(\mathbb{R})$ can be divided up into the following cases depending on the behavior of the eigenvalues:


*

*Nonzero real eigenvalues $\pm r$: these matrices are diagonalizable so there is one orbit for every positive real $r$, with diagonal representative $\left[ \begin{array}{cc} r & 0 \\ 0 & -r \end{array} \right]$.

*Zero eigenvalues: these matrices are conjugate to the Jordan block $\left[ \begin{array}{cc} 0 & 1 \\ 0 & 0 \end{array} \right]$, or are zero themselves. 

*Nonzero purely imaginary eigenvalues $\pm si$: these matrices are conjugate to companion matrices of their characteristic polynomials, which take the form $\left[ \begin{array}{cc} 0 & -s^2 \\ 1 & 0 \end{array} \right]$, although these are in turn conjugate to the matrices $\left[ \begin{array}{cc} 0 & -s \\ s & 0 \end{array} \right]$ which have the benefit of being rotations-and-scalings. 


These orbits can be visualized quite explicitly as subsets of $\mathbb{R}^3$: writing an element of $\mathfrak{sl}_2(\mathbb{R})$ as $X = \left[ \begin{array}{cc} x & y+z \\ y-z & -x \end{array} \right]$, they can be distinguished by the (negative of) the value of the determinant
$$-\det(X) = x^2 + (y+z)(y-z) = x^2 + y^2 - z^2$$
which is equal, in the three cases, to $r^2, 0, -s^2$ respectively. This is more or less the same as distinguishing based on the Killing form as Moishe suggests in the comments. 


*

*In the first case the orbit $x^2 + y^2 = z^2 + r^2$ is a one-sheeted hyperboloid,.

*In the second case the orbit $x^2 + y^2 = z^2$ is a cone. 

*In the third case the orbit $x^2 + y^2 + s^2 = z^2$ is a two-sheeted hyperboloid.


This picture of the three cases from Wikipedia might help you visualize how these orbits partition $\mathbb{R}^3$. It's lovely how everything fits together.

If we think of ourselves as working in $(2+1)$-dimensional Minkowski spacetime, these correspond to timelike, lightlike, and spacelike vectors respectively. In particular we can interpret the cone in the middle as a light cone. 
We can also describe pretty explicitly the corresponding one-parameter subgroups of $SL_2(\mathbb{R})$:


*

*$\exp \left[ \begin{array}{cc} r & 0 \\ 0 & -r \end{array} \right] = \left[ \begin{array}{cc} e^r & 0 \\ 0 & e^{-r} \end{array} \right]$ gives the hyperbolic elements, which are squeeze mappings, and whose action on $\mathbb{R}^2$ has orbits hyperbolas, in this case the (connected components of the) hyperbolas $xy = c$.

*$\exp \left[ \begin{array}{cc} 0 & t \\ 0 & 0 \end{array} \right] = \left[ \begin{array}{cc} 1 & t \\ 0 & 1 \end{array} \right]$ gives the parabolic elements, which are shear mappings, and whose action on $\mathbb{R}^2$ has orbits lines, in this case the lines $y = c$ plus a separate orbit for each point on the $x$-axis.

*$\exp \left[ \begin{array}{cc} 0 & -s \\ s & 0 \end{array} \right] = \left[ \begin{array}{cc} \cos s & - \sin s \\ \sin s & \cos s \end{array} \right]$ gives the elliptic elements, which are rotations, and whose action on $\mathbb{R}^2$ has orbits ellipses, in this case the circles $x^2 + y^2 = c$. 


Now, this entire time I've been describing the orbits under the action of $GL_2(\mathbb{R})$, but we wanted the orbits under the action of $SL_2(\mathbb{R})$. This is the same as asking for the orbits under the action of $GL_2^{+}(\mathbb{R})$, the matrices with positive determinant, because we can multiply a matrix in $SL_2(\mathbb{R})$ by the scalar matrix $\left[ \begin{array}{cc} \sqrt{d} & 0 \\ 0 & \sqrt{d} \end{array} \right]$ to get a matrix with determinant $d$, which doesn't affect how it conjugates. What this means is that rather than allow ourselves to change basis arbitrarily, we can only change basis in an orientation-preserving way; equivalently, we lose the freedom to conjugate by a matrix of determinant $-1$. 
This has the following effects in each of the three cases above:


*

*Nonzero real eigenvalues $\pm r$: in this case the one sheet of the hyperboloid remains a single orbit. 

*Zero eigenvalues: we now need to distinguish the two Jordan blocks $\left[ \begin{array}{cc} 0 & 1 \\ 0 & 0 \end{array} \right]$ and $\left[ \begin{array}{cc} 0 & 0 \\ 1 & 0 \end{array} \right]$, which are no longer conjugate; geometrically this disconnects the "past" and "future" light cone.

*Nonzero purely imaginary eigenvalues $\pm si$: we now need to distinguish $\left[ \begin{array}{cc} 0 & -s \\ s & 0 \end{array} \right]$ and its negative, which are no longer conjugate; geometrically this disconnects the two sheets of the hyperbola.
