The adjoint action and its orbits

Question

Let $G$ be a Lie group and $H$ a closed subgroup of $G$. If we consider the action of $G$ on its Lie algebra $\frak g$ by the adjoint representation and regard $\frak h$ as a point in the Grassmannian $Gr_k(\mathfrak g)$ where $k=dim\frak\ h$. How to show that the orbit of this point is $G/N$ where $N$ is the normalizer in $G$ of the connected component $H^0$ of $H$?

Answer 1

I am not sure whether it is always true that the orbit is diffeomorphic to $G/N$, but general arguments bring you close to that. The map sending $g\in G$ to the subspace $Ad(g)(\mathfrak h)\subset\mathfrak g$ by definition descends to a smooth bijective map from the quotient of $G$ by the stabilizer subgroup $\{g\in G:Ad(g)(\mathfrak h)=\mathfrak h$. It is even known (for any smooth action of a Lie group on a manifold) that this is the embedding of an initial submanifold, but not a diffeomorphism onto its image in general. On the other hand, taking the normalizer $N$ of $H^0$ in $G$, it is clear that $N$ is contained in the stabilizer subgroup from above (just differentiate the defining property of the normalizer). On the other hand, assume that $g\in G$ is such that $Ad(g)(X)\in\mathfrak h$ for each $X\in\mathfrak h$. Then $\exp(tAd(g)(X))=g\exp(tX)g^{-1}$ lies in $H^0$ for all $t\in\mathbb R$, so $g\exp(X)g^{-1}\in H^0$ for all $X\in\mathfrak h$. Since the elements $\exp(X)$ for $X\in\mathfrak h$ generate $H^0$, it follows that $g\in N$.