On the orbit of a Fréchet Lie group action

Question

Suppose that $G$ is a Fréchet Lie group acting on a Fréchet manifold $X$. Fix $x\in X$ and let $\alpha(t)$ be a smooth path in $X$ such that $$ \begin{cases} \alpha(0)=x\\ \alpha(t)\in G\cdot x. \end{cases} $$ Also denote $\rho_{x}:G\rightarrow X:g\mapsto g\cdot x$. Is it true that $\alpha'(0)\in \operatorname{Im}(d_{e}\rho_{x})$?

In the finite dimensional setting, this is clearly the case: the orbit $G\cdot x$ is a weakly embedded submanifold of $X$, hence $\alpha(t)$ is also smooth as a curve in $G\cdot x$. Consequently $$\alpha'(0)\in T_{x}(G\cdot x)=\operatorname{Im}(d_{e}\rho_{x}).$$

In the case that is of interest to me, $G=\operatorname{Diff}(M)$ is the space of diffeomorphisms of a compact manifold $M$, and $X$ is the Fréchet space of rank-$k$ distributions $\Gamma(\operatorname{Gr}_{k}(M))$.

Answer 1

I'm not aware of a precise characterization of when the orbits are initial submanifolds in infinite dimensions. In fact, the manifold structure on the orbits is a hard problem even for $G$-actions on a coset space $G / H$, see Problem IX.3 in Neeb - Towards a Lie theory for locally convex groups.

Proposition 3.17 in Diez and Rudolph - Slice theorem and orbit type stratification in infinite dimensions shows that the orbits of Fréchet actions are embedded submanifolds if the action is proper, the stabilizer is a Lie subgroup and the linearization of the action has a tame inverse family. However, this proposition does not apply in your situation as the action is not proper in general (for example, if $k=0$, the zero distribution has the whole diffeomorphism group as stabilizer). Maybe, it works if you restrict attention to a subset of sufficiently rigid distributions.