Let $G$, be a Lie Group and $\mathfrak{g}$ be its Lie algebra ,i.e, $Lie(G)=\mathfrak{g}$. Let $\zeta=(\ X,F)\ \in \mathfrak{g}\oplus\mathfrak{g^\*}$. Here $X\in \mathfrak{g} $ and $F\in \mathfrak{g^\*}$ . So we can construct orbit of $\zeta$. by 


$\mathfrak{G}=\{ ( Ad(g)X,Ad^\*(g)F\ ): g\in G \}$. By the action of $g\in G$ to $\zeta $ . My question is can we suppose $G$ as fibre bundle $\mathfrak{G}$. with projection


$P\_\zeta :G\to \mathfrak{G}$.


$P\_\zeta (g)=( Ad(g)X,Ad^\*(g)F\ ) $ ?



fiber bundle on an orbit of $\mathfrak{g}\oplus\mathfrak{g^*}$
Yes. If a Lie group $G$ acts smoothly on a manifold $X$, and $H$ is the $G$-stabilizer of a point $x\_0 \in X$, then $H$ is a closed subgroup of $G$, clearly, and the orbit is identified with $G/H \cong Gx\_0$ by the map $gH \mapsto gx\_0$. Every Lie group $G$ is an $H$-bundle over any of its homogeneous spaces $G/H$: $H \to G \to G/H$. For proof, see Brocker and tom Dieck, **Representations of Compact Lie Groups**, p. 33, theorem 4.3.


