Orbits form a manifold?

Question

A prominent example are the coadjoint orbits $O_x = \{Ad_u^*(x);u \in G\}$ where $x \in \mathfrak{g}$ and $G$ a Lie group with Adjoint map $Ad.$

Could anybody give me an easy argument why $O_x$ is a smooth manifold?

Answer 1

It follows from a much more general fact: Whenever a Lie group $G$ acts smoothly on a smooth manifold $M$, its orbits are immersed smooth manifolds. This can be derived from the following observations:

1. For each $p\in M$, the isotropy group $G_p = \{g\in G: g\centerdot p=p\}$ is a closed subgroup of $G$.
2. The quotient space $G/G_p$ has a unique smooth manifold structure, and $G$ acts transitively on this smooth manifold by left multiplication.
3. The orbit map $\mathscr O_p\colon G\to M$ given by $\mathscr O_p(g) = g\centerdot p$ descends smoothly to the quotient $G/G_p$ to give a smooth map $F_p\colon G/G_p\to M$ that is a bijection from $G/G_p$ onto the orbit $G\centerdot p$.
4. The map $F_p$ is equivariant with respect to the left actions of $G$ on $G/G_p$ and $M$, and therefore it has constant rank.
5. Because it is a constant-rank injection, $F_p$ is a smooth immersion, and therefore its image $G\centerdot p$ is a smoothly immersed submanifold of $M$.

I don't know if this would qualify as an "easy argument," but it's based on standard facts from elementary differential geometry, most of which can be found in Chapter 21 of my Introduction to Smooth Manifolds (2nd ed.).