Lifting of curves with respect to a Lie group action Let $G$ be a complex Lie group. Let $M$ a complex $G$-manifold.  Given any point $x\in M$, the orbit $G\cdot x$ is a complex submanifod of $M$.
My question is the following: given an analytic curve $\gamma: \mathbb{C} \to G\cdot x$ with $\gamma(0)=x$, does there exists an analytic curve $\tilde{\gamma}: \mathbb{C} \to G$ such that $\tilde{\gamma}(0)=e_G$ and $\tilde{\gamma}(t)\cdot x = \gamma(t)$ for all $t\in\mathbb{C}$.
Thanks a lot!
 A: The answer is positive, you can even replace the complex line with an arbitrary open Riemann surface $S$ (the most natural one besides ${\mathbb C}$ would be ${\mathbb C}^*$). Here is a proof. Let $H< G$ denote the stabilizer of $x\in M$; it is a complex Lie subgroup of $G$. Then $G/H$ is equivariantly biholomorphic to the submanifold $Gx\subset M$. Thus, the problem reduces to the lifting problem for holomorphic maps $f: S\to B=G/H$. The projection $\xi: G\to B$ is a  principal holomorphic $H$-bundle. Let $\eta=f^*\xi$ be its pull-back to $S$: The bundle $\eta: E\to S$ is again a holomorphic bundle (with the fiber $H$) over $S$. Recall that $E$ can be identified with the submanifold
$$
\{(x,y)\in S\times G: f(x)= \xi(y)\}
$$
By a theorem due to Grauert and Rohrl (see here), every holomorphic bundle over an open Riemann surface is holomorphically trivial. (Famously, this theorem fails if the base is a Stein manifold of dimension $\ge 2$.) Let $\sigma: S\to E$ be a holomorphic section of $\eta$. Then composing $\sigma$ with the projection 
$E\to G$ we obtain the required holomorphic lift $\tilde{f}: S\to G$ of the mapping $f: S\to B$.  
