Let $G$ be a Lie group and $H$ be any Lie subgroup. Regard the Lie algebra $\mathfrak h$ as a point in the Grasmannian $Gr_k(\mathfrak g)$ where $k:=\dim _\mathbb R \mathfrak h$.
I want to understand why the adjoint representation $Ad :G\to GL_\mathbb R(\mathfrak g)$ yields a smooth action of $G$ on $Gr_k(\mathfrak g)$ and why the isotropy group $G_\mathfrak h$ is the normalizer
$$N_G(H^0):=\{g\in G; \ gH^0g^{-1}=H^0\}$$
where $H^0$ is the identity connected component.
Best Answer
Let $V$ be a (real) vector space, let $U\subset V$ be a linear subspace, and let $T:V\to V$ be a linear automorphism. Then $T$ carries $U$ to another linear subspace of the same dimension as $U$. This explains why a group action on $V$ yields an action of the same group on all the Grassmannians of $V$.
The isotropy group $G_\mathfrak{h}$ is defined by $$G_\mathfrak{h}:=\{g\in G\;|\;\mathrm{Ad}_g(\mathfrak{h})=\mathfrak{h}\}.$$ Now, for $g\in G$ and $\xi\in\mathfrak{h}$ we have $$\mathrm{exp}(\mathrm{Ad}_g(\xi))=g\cdot\mathrm{exp}(\xi)\cdot g^{-1}.$$ If $g\in N_G\left(H^0\right)$, then RHS is in $H^0$, which means that $\mathrm{Ad}_g(\xi)\in\mathfrak{h}.$ Conversely, if $g\in G_\mathfrak{h}$, then $\mathrm{Ad}_g(\xi)\in\mathfrak{h},$ and it follows from the above equality that $g\cdot\mathrm{exp}(\xi)\cdot g^{-1}\in H^0.$ The claim now follows from the fact that $H^0$ is generated by $\mathrm{exp}(\mathfrak{h}).$