Let $M\_{n\times n}$ denote the set of $n\times n$ real matrices and let $GL\_n$ be the subgroup of invertible matrices. $GL\_n$ acts on $M\_{n\times n}$ smoothly by conjugation, which means that each conjugacy class (which is an orbit of this action) is an immersed submanifold of $M\_{n\times n}$. However, the action is not proper (e.g. the isotropy groups are not compact) so the orbits may not be embedded submanifolds.


My question is if there are nice conditions on a matrix that guarantee that its conjugacy class is or is not an embedded submanifold. My interest in this question actually comes from trying to understand the space of all complex structures on a real vector space: it can be shown that the set of all complex structures is the conjugacy class of the block matrix $\begin{pmatrix} 0 & -I \\\ I & 0\end{pmatrix}$ and I was wondering if this is an embedded submanifold. So an answer to this question (if the above doesn't have a nice answer) would also be appreciated.



When is a conjugacy class of matrices an embedded submanifold?
If $G$ is an algebraic group (say over $\mathbb C$), acting on an algebraic variety $X$, the orbits are always locally closed smooth algebraic subvarieties of $X$. This is standard, and easy to prove. If $x$ is a point of $X$, and $G \to X$ is the morphism sending $g$ to $gx$, consider the orbit $Gx$ as a subset of its Zariski closure $\overline{Gx}$. By Chevalley's theorem, $Gx$ contains a dense open subset of $\overline{Gx}$. Hence, by homogeneity it is open in $\overline{Gx}$, and smooth.


From this it is easy to deduce that all the orbits of $\mathrm{GL}\_n(\mathbb R)$ in $\mathrm{M}\_n(\mathbb R)$ are embedded submanifolds. The point is that if two real matrices are conjugate as complex matrices, then they are conjugate as real matrices. These means that the orbits of the action of $\mathrm{GL}\_n(\mathbb R)$ in $\mathrm{M}\_n(\mathbb R)$ are of the form $\Omega(\mathbb R)$, where $\Omega$ is an orbit of $\mathrm{GL}\_n(\mathbb C)$ in $\mathrm{M}\_n(\mathbb C)$; since $\Omega$ is a smooth algebraic variety, it follows from the implicit function theorem that $\Omega(\mathbb R)$ is an embedded submanifold of $\mathrm{M}\_n(\mathbb R)$.


