Submanifolds on 2x2 matrices with trace=0. Let $S=\{X\in\text{Mat}_2(\mathbb{R}) : \text{tr}(X)=0\}$. The group $\text{SL}_2(\mathbb{R})$ acts on $S$ by conjugation, that is $g \cdot X = gXg^{-1}$.

*

*Describe the orbits of this action.

*When are the orbits submanifolds of S and what is their dimension?


My attempt thus far:
For 1, the orbits are given by matrices with the same Jordan Form, but I'm not sure how to approach 2.
My idea is that for any $A \in \text{Orb}(X)$, $\text{det}(A) = \text{det}(X)$, so the orbit is given by the submanifold defined by $f^{-1}(0)$, $f:S \rightarrow \mathbb{R}$, $f(A) = \text{det}(A)-\text{det}(X)$. Then in order for the orbit to be a submanifold, $Df_A$ must be of constant rank on $\text{Orb}(X)$, but I am unsure of how to show this.
 A: There are four types of orbits:

*

*The orbit of the zero matrix. This orbit consists only of the zero matrix and is a zero-dimensional sub-manifold.

*The orbit $O_{\lambda}$ of the matrix
$$ \begin{pmatrix} \lambda & 0 \\ 0 & -\lambda \end{pmatrix} $$
where $\lambda > 0$. These orbits are closed, two-dimensional, submanifolds. Indeed, if you define $f \colon S \rightarrow \mathbb{R}$ by $f(A) = \det(A)$ then any non-zero $\mu \in \mathbb{R}$ is a regular value of $f$. In particular, we have that $f^{-1}(\{ -\lambda^2 \}) = O_{\lambda}$ is an embedded, closed, two-dimensonal submanifold of $S$.

*The orbit $\hat{O}_{\lambda}$ of the matrix
$$ \begin{pmatrix} 0 & \lambda \\ -\lambda & 0 \end{pmatrix} $$
for $\lambda \neq 0$. Note that unlike in the previous case, here $\hat{O}_{\lambda}$ and $\hat{O}_{-\lambda}$ are not the same orbit as there is no $g \in \operatorname{SL}_2(\mathbb{R})$ which conjugates $O_{\lambda}$ to $O_{-\lambda}$. These orbits are distinct and are embedded, closed, two-dimensional submanifolds which you can see by noting that $f^{-1}( \{ \lambda^2 \}) = \hat{O}_{\lambda} \cup \hat{O}_{-\lambda}$.

*The orbits $\tilde{O}_{\pm}$ of the matrices
$$ \begin{pmatrix} 0 & +1 \\ 0 & 0 \end{pmatrix}, \begin{pmatrix} 0 & -1 \\ 0 & 0 \end{pmatrix}. $$
This case is "special" because the representative of the orbit is not diagonalizable (even over $\mathbb{C}$) and so the orbits are not closed. Note that while the two matrices above are conjugate by an element of $\operatorname{GL}_2(\mathbb{R})$, you can verify directly that they are not conjugate under an element of $\operatorname{SL}_2(\mathbb{R})$ so they indeed give you two distinct orbits. We have
$$\left( f|_{S \setminus \{ 0_{2 \times 2} \}}\right)^{-1}(0) = \tilde{O}_{+} \cup \tilde{O}_{-} $$
and so again, these are two-dimensional, locally closed, submanifolds.


If you want to "see" what is going on geometrically, note that you can identify
$$ S = \left \{ \begin{pmatrix} a & b \\ c & -a \end{pmatrix} \Big\vert a,b,c \in \mathbb{R} \right \}$$
with $\mathbb{R}^3 = \{ (a,v,w) \, | \, a, v, w \in \mathbb{R} \}$ using the change of variables
$$ a = a, b = v + w, c = v - w. $$
Under this change of variables, the determinant function $\det \colon S \rightarrow \mathbb{R}$ becomes $\det(a,v,w) = w^2 - a^2 - v^2$. The orbits in $S$ corresponds to the level set of $w^2 - a^2 - v^2$ in the following way. Note that the level sets of $w^2 - a^2 - v^2$ are of three types:

*

*The zero level set $w^2 - a^2 - v^2 = 0$ is a double cone. This is not a manifold but a union of three submanifolds: the upper cone, the lower cone and the origin. This corresponds to the union of the three orbits $\tilde{O}_{+}, \tilde{O}_{-}, \{ 0_{2 \times 2} \}$.

*The level set $w^2 - a^2 - v^2 = \lambda^2$ where $\lambda \neq 0$ is a two-sheeted hyperboloid. It is a disconnected submanifold and corresponds to the union of the two orbits $\hat{O}_{\lambda},\hat{O}_{-\lambda}$.

*The level set $w^2 - a^2 - v^2 = -\lambda^2$ where $\lambda > 0$ is a one-sheeted hyperboloid. It is a connected submanifold which is precisely the orbit $O_{\lambda}$.

You can see the various orbits in the following picture:

To see that any non-zero $\mu \in \mathbb{R}$ is a regular value of $f$, note that by Jacobi's formula we have
$$ df|_{A}(B) = \operatorname{tr} \left( \operatorname{adj}(A) B \right) = \left< \operatorname{adj}(A), B^T \right> $$
where $ \left< \cdot, \cdot \right>$ is the standard (Frobenius) inner product on $M_2(\mathbb{R})$. Note that for a $2 \times 2$ matrix we have
$$ \operatorname{tr}(\operatorname{adj}(A)^T) = \operatorname{tr}(\operatorname{adj}(A)) = \operatorname{tr}(A) $$
so if $A \in S$ then $\operatorname{adj}(A),\operatorname{adj}(A)^T \in S$ and so we have
$$ df|_A(\operatorname{adj}(A)^T) = \left< \operatorname{adj}(A), \operatorname{adj}(A) \right>. $$
Hence, if $df|_A = 0$ we must have $\operatorname{adj}(A) = 0$ which implies that $A = 0$. Hence, any non-zero $\mu \in \mathbb{R}$ is a regular value of $f$ and if you restrict $f$ to $S \setminus \{ O_{2 \times 2} \}$ then even $0$ is a regular value of $f$.
A: A more accurate answer to the first question can be given by listing all orbits. Here is a complete list of representatives of all orbits:
$$
\begin{pmatrix}
0 & 0\\
0 & 0
\end{pmatrix},\ 
\begin{pmatrix}
0 & 1\\
0 & 0
\end{pmatrix},\ 
\begin{pmatrix}
0 & -1\\
0 & 0
\end{pmatrix},\
$$
$$ 
\begin{pmatrix}
t & 0\\
0 & -t
\end{pmatrix},\ 
\begin{pmatrix}
0 & t\\
-t & 0
\end{pmatrix},\ 
\begin{pmatrix}
0 & -t\\
t & 0
\end{pmatrix},
$$
where $t\in\mathbb{R}$, $t>0$.
