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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.

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I'm having some difficulty seeing the obstruction to being an embedded submanifold. Isn't the tangent space at each point of an orbit naturally isomorphic to Lie($GL_n$)/Lie(centralizer)? – S. Carnahan May 30 '10 at 2:07
@Scott: strictly speaking, it doesn't make sense to say "tangent space at each point of an orbit" until one first addresses the issue of submanifold structure at that point for the orbit. More generally, if an analytic Lie group $G$ acts on the left on an analytic manifold $X$ then for $x \in X$ with stablizer $G_x$ the issue is whether or not $G/G_x \rightarrow X$ is an embedding. This must be proved somewhere in Chapter 3 of Bourbaki Lie Groups & Lie Algebras. As usual, Google Books seems to be missing the interesting relevant pages. – BCnrd May 30 '10 at 4:16
To Brian: I don't think this is true; the standard examples of actions of $\mathbb R$, or $\mathbb C$, on a torus with a dense orbit are analytic actions. However, it is certainly true for algebraic group actions on algebraic manifold (over $\mathbb C$, but from this case it is easy to deduce the real case). – Angelo May 30 '10 at 4:34
@Angelo: Whoops, I meant to say "immersion" rather than "embedding". It may be that one still needs some mild hypotheses for this case to be true (my copy of Bourbaki is far away), but my recollection is that orbits are immersed submanifolds. – BCnrd May 30 '10 at 4:56
Note also that the main matrix of interest is semisimple, so that its orbit is closed in Zariski topology and bad stuff doesn't happen. – Victor Protsak May 30 '10 at 6:03
up vote 4 down vote accepted

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 $\mathrm{M}_n(\mathbb R)$.

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Angelo, "loc. closed" aspect ok in general alg. case over $\mathbf{R}$, even if distinction of $\mathbf{R}$-orbits and $\mathbf{R}$-points of $\mathbf{C}$-orbits isn't nice as for ${\rm{GL}}_ n$: for $k = \mathbf{R}$, if $G$ is f.type $k$-gp acting on smooth sep'td f.type $k$-scheme $X$ then for $x \in X(k)$, the map $G(k)/G(k)_x \rightarrow X(k)$ is loc. closed submanifold. For proof, $G/G_x$ is smooth loc. closed subscheme of $X$, and $G(k)/G(k)_x$ is submanifold in $X(k)$, hence in $(G/G_x)(k)$. But submanifold $G(k)/G(k)_x \rightarrow (G/G_x)(k)$ has same dim., hence open submanifold! QED – BCnrd May 30 '10 at 13:02
Yes, I completely agree. I was just answering the specific question. – Angelo May 30 '10 at 13:15

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