# Conjugacy class in a group Lie

I have one question: Let $G$ is a group Lie and $H$ is closed subgroup. Let $M=\cup g^{-1}Hg < G$. Is it true that $M$ is manifold? What is the dimension of $M$?

upd. Let $G$ is a compact. I have a hypothesis that, in this case $M$ is manifold, and $\dim \, M = \dim \, G + \dim \, H - \dim \, N_G(H)$.

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Well, in the case $G$ is abelian it's obvious that $M=H$ hence $dim(M)=dim(H)$. Perhaps this continues to be true when the group operation is more interesting. The fact that $H$ is a manifold is given by Cartan's Theorem according to the Wikipedia article on Lie Groups. – James S. Cook Mar 1 '13 at 18:14
@JamesS.Cook The only mention of "Cartan's Theorem" that I see in that Wikipedia article refers to the result that a closed subgroup of a Lie group is a Lie group. This doesn't seem relevant, since the $M$ in the question is in general neither closed nor a subgroup. – Andreas Blass Mar 1 '13 at 19:12
@AndreasBlass but the question stated begins by assuming $H$ is a closed subgroup. As I point out, in the special case $G$ is Abelian $M=H$ so in fact $M$ is a closed Lie subgroup.Your example is nonabelian since rotations need not commute. – James S. Cook Mar 3 '13 at 4:44
@JamesS.Cook Sorry. I thought your mention of Cartan's theorem was in connection with the immediately preceding sentence in your comment (about "when the group operation is more interesting") rather than about the abelian situation. I overlooked that you applied Cartan's theorem to $H$, not $M$, so you intended to refer to the abelian case. – Andreas Blass Mar 4 '13 at 2:12

I assume that the union, in the definition of $M$, is over all elements $g$ of $G$. Let $G$ be the group of orientation-preserving rigid motions of the Euclidean plane (i.e., translations and rotations), and let $H$ be the subgroup of rotations about a certain point $P$. Then, unless I've made a mistake, $M$ consists of all rotations about all points, including the identity element of the group, but not including any non-trivial translations. This $M$ is not a manifold, because no neighborhood of the identity looks like Euclidean space.
I think in this example, $M$ is a manifold. If $P = 0$, then $M$ is just $SO(n)$. It's true that no neighborhood of the identity (in $M$) looks like Euclidean space of the same dimension of $G$, but it looks like a smaller dimensional Euclidean space. – Jason DeVito Mar 1 '13 at 19:49
@JasonDeVito If $P=0$ then $H$ is $SO(n)$, but $M$ is the union of all the conjugates of $SO(n)$. If you conjugate $SO(n)$ by a motion $g$ that sends $0$ to some other point $Q$ then you get the group of rotations fixing $Q$. (I'm composing motions from left to right here; if you prefer right-to-left, let $g$ move $Q$ to $0$). So the union $M$ of all the conjugates contains the rotations about all points, not just about $0$. – Andreas Blass Mar 1 '13 at 21:01
Ok, if I understand correctly, Lie group of this example is not compact. But if $G$ is compact, what we can say? – Alex-omsk Mar 2 '13 at 5:36
So, but what about $M \setminus \{id\}$? – Alex-omsk Mar 6 '13 at 3:16