Are coadjoint orbits symplectic manifolds?

Question

Let $G$ be a Lie group and $\mathfrak{g}$ be it's Lie algerba. Consider the coadjoint orbit for an element $x \in \mathfrak{g}: O_x$. It is well known that $$O_x \simeq G/G_x$$ where $G_x$ is the stabilizer of $x$. Moreover it is well known that there is a symplectic form on each orbit. However is it true to say that the orbit is actually a symplectic manifold?

Answer 1

Yes, it is. Firstly, the coadjoint orbit passes through a point $\mu\in\mathfrak{g}^*$ (not $x\in\mathfrak{g}$, as you stated). A tangent vector to the coadjoint orbit at $\nu\in\mathcal{O}_{\mu}$ (i.e. $\nu = \mathrm{Ad}_{g^{-1}}^*\mu$ for $g\in G$) is then a vector of the form $-\mathrm{ad}_\xi^*\nu$ for some $\xi\in\mathfrak{g}$. The symplectic form(s) on $\mathcal{O}_\mu$ is called the Kostant-Kirillov-Souriau form, and is defined by $$ \omega_\nu^\pm (-\mathrm{ad}_\xi^*\nu,-\mathrm{ad}_\zeta^*\nu) := \pm \nu([\xi,\zeta]) $$ for $\xi,\zeta\in\mathfrak{g}$.