It is well-known under what conditions the quotient of two finite-dimensional Lie groups forms a smooth manifold. It is also well-known what smooth structure the resulting quotient manifold might have. I am interested in generalizing results along these lines to infinite-dimensional Lie groups.
Dimension Independent Setup
Let $G_1$ be a Lie group and $G_2\subset G_1$ a generic subgroup thereof. Let $Q=G_1/G_2$ be the set of all of the $G_2$-cosets in $G_1$. For any $g\in G_1$, let $\bar{g}:Q\to Q$ be the group action given by $\bar{g}(h G_2) = (g h) G_2$. Collecting these maps together we have $\bar{g}\in\overline{G}_1\subset\text{Perm}(Q)$, a group of permutations on $Q$. Note that the stabilizer subgroup of $G_1$'s action (say at $e G_2$) is exactly $G_2$.
Theorem when $G_1$ is Finite Dimensional
If $G_1$ is finite-dimensional, then $G_2$ being a closed Lie subgroup of $G_1$ is both necessary and sufficient for the following: There exists a unique smooth structure $\mathcal{S}$ for $Q$ such that $\mathcal{M}=(Q,\mathcal{S})$ is a finite-dimensional smooth manifold for which the $\bar{g}:Q\to Q$ maps are diffeomorphisms, $\overline{G}_1\subset\text{Diff}(\mathcal{M})$.
I will sketch a proof of this result at the end of the post. I am interested in finding either necessary of sufficient conditions in the analogous claim when $G_1$ is allowed to be infinite-dimensional.
Schema for Infinite Dimensional Case
If $G_1$ is infinite-dimensional, then $G_2$ standing in some relation to $G_1$ is necessary and/or sufficient for the following: There exists a unique smooth structure $\mathcal{S}$ for $Q$ such that $\mathcal{M}=(Q,\mathcal{S})$ is a finite-dimensional smooth manifold for which the $\bar{g}:Q\to Q$ maps are diffeomorphisms, $\overline{G}_1\subset\text{Diff}(\mathcal{M})$.
I know that $G_2$ being closed is still necessary but I doubt that it is still sufficient. Moreover, $G_2$ obviously needs to be infinite-dimensional as well if it going to cancel off most of $G_1$.
Sketch of Finite Dimensional Proof
$G_2$ being a closed Lie subgroup of $G_1$ is necessary since otherwise the quotient space is not a T1-space (since there is a co-set in the quotient which cannot be separated from the identity by an open set).
Given that $G_2$ is a closed Lie subgroup of $G_1$ we can use the Quotient Manifold Theorem (Theorem 1 here) to transfer the smooth structure from the finite-dimensional manifold, $G_1$, down to its quotient $Q=G_1/G_2$. In particular, there is a unique smooth structure $\mathcal{S}_A$ for $Q$ which makes the projector $\pi:G\to Q$ into a smooth submersion. It is trivial to check that according to $\mathcal{M}_A=(Q,\mathcal{S}_A)$ we have $\overline{G}_1\subset\text{Diff}(\mathcal{M}_A)$. This proves the existence claim.
To see uniqueness, suppose that $\mathcal{M}_B=(Q,\mathcal{S}_B)$ also has $\overline{G}_1\subset\text{Diff}(\mathcal{M}_B)$. Noting that $G_1$ acts transitively over $\mathcal{M}_B$ we can reconstruct $\mathcal{M}_B$ up to diffeomorphism as a quotient of $G_1$ and its stabilizer subgroup $G_\text{fix}$. But this is just $G_\text{fix}=G_2$. This reconstruction of $\mathcal{M}_\text{B}\cong G_1/G_2$ uses the smooth structure coming from the quotient manifold theorem, see Theorem 5 here. Hence, we have $\mathcal{M}_\text{B}\cong G_1/G_2 \cong \mathcal{M}_\text{A}$.
Difficulty in Generalizing
The reason that this proof doesn't straight-forwardly generalize to the infinite-dimensional case is that we lack the Quotient Manifold Theorem. If $G_1$ is an infinite-dimensional Lie group it becomes much more difficult to talk about its smooth structure and transfer this down via a quotient. Note, however, that the theorem schema only talks about smooth structures for finite-dimensional manifolds.
Any help/thoughts would be much appreciated.