Let $G$ be a Lie group (finite dimensional or Banach), and let $H$ be a Lie subgroup (in the Banach case we assume that $H$ is a submanifold which is also a Lie group). Let $\text{U}(1) \rightarrow \hat{G} \xrightarrow{\pi} G$ be a central extension of $\hat{G}$.
Suppose that $\hat{G}$ trivializes over $H$.
Write $f: G \rightarrow G/H$ for the projection map.



> 
> Is this sufficient to conclude that there exists a unique principal $\text{U}(1)$-bundle $\mathcal{L} \rightarrow G/H$, with the property that the pullback of $\mathcal{L}$ to $G$ is $\hat{G}$?
> 
> 
> 


I think the answer is yes, essentially using the construction in the proof of Proposition 5.3.1 in Brylinski's book "Loop spaces, characteristic classes and geometric quantization". I think that technique can be applied essentially because a trivialization $\sigma: H \rightarrow \hat{G}|\_{H}$ yields a descent isomorphism $\phi: p\_{1}^{\*} \hat{G} \rightarrow p\_{2}^{\*}\hat{G}$. (Here $p\_{1}$ and $p\_{2}$ are the projections $G \times\_{f} G \rightarrow G$.)
Specifically we set $\phi(\hat{g}\_{1},g\_{1},g\_{2}) = (\hat{g}\_{1} \sigma(g\_{1}^{-1}g\_{2}), g\_{1},g\_{2})$.


The problem that I have is that I'm not sure if this works, because Brylinski assumes that the projection map $G \rightarrow G/H$ is a local homeomorphism, which is definitely not true for us. (Also, I'm not sure if Brylinski works in the Banach setting.)


Brylinski then defines an equivalence relation on $\hat{G}$ as follows. Say that $\hat{g}\_{1} \sim \hat{g}\_{2}$ if $g\_{1} = \pi(\hat{g}\_{1})$ and $g\_{2} \pi(\hat{g}\_{2})$ satisfy $f(g\_{1}) = f(g\_{2})$, and moreover, if $\phi(\hat{g}\_{1},g\_{1},g\_{2}) = (\hat{g}\_{2},g\_{1},g\_{2})$.
Brylinski then claims that $\hat{G}/ \sim \rightarrow G/H$ does the trick.


This leads me to the second question:



> 
> How does one equip $\hat{G}/\sim$ with the structure of smooth manifold?
> 
> 
> 



Descending central extensions to homogeneous spaces
First up, $\sigma\colon G\times H \to G\times\_f G$ sending $(g,h)\mapsto (g,gh)$ is a diffeomorphism. For this, all you need is that $G\to G/H$ is a locally trivial $H$-bundle (so, with care, this works for beyond the Banach setting). Then since you have your descent data $\phi\colon p\_1^\*\hat{G}\to p\_2^\*\hat{G}$ over $G\times\_f G$, pulling this back along $\sigma$ gives an isomorphism $p\_1^\* \hat{G} \simeq (m^\*\hat{G})\big|\_{G\times H}$ where $m\colon G\times G \to G$ is multiplication. Since $\hat{G} \to G$ is an extension of groups, the multiplication $\hat{m}$ on $\hat{G}$ covers $m$. 


Since $U(1)\times U(1) \to U(1)$ is a homomorphism, you can change structure group for the $U(1) \times U(1)$-bundle $\hat{G}\times \hat{G}$ along it to get what is written $p\_1^\*\hat{G} \otimes p\_2^\*\hat{G}$, a $U(1)$-bundle on $G\times G$. As a result $\hat{m}$ descends to give a map of $U(1)$-bundles $p\_1^\*\hat{G} \otimes p\_2^\*\hat{G} \to \hat{G}$ covering $m$. It follows from this that $m^\*\hat{G} \simeq p\_1^\*\hat{G} \otimes p\_2^\*\hat{G}$ over $G\times G$.


Then $(m^\*\hat{G})\big|\_{G\times H} \simeq (p\_1^\*\hat{G} \otimes p\_2^\*\hat{G})\big|\_{G\times H}\simeq p\_1^\*\hat{G} \otimes p\_2^\*(\hat{G}\big|\_H)$ over $G\times H$, the latter isomorphism coming from the fact forming $\otimes$ commutes with pullback. But we have a trivialisation $\hat{G}\big|\_H \simeq \hat{G}\times U(1)$, and it's a general result that $\otimes$ with a trivial $U(1)$-bundle changes nothing, so that we have $p\_1^\*\hat{G} \otimes p\_2^\*(\hat{G}\big|\_H) \simeq p\_1^\*\hat{G} = \hat{G}\times H$.


Putting these bits together we get a map $\hat{G}\times H \to \hat{G}$ covering $G\times H\to G$, and the cocycle identity for the descent isomorphism ensures that this is an $H$-action, hence that $\hat{G}\to G$ is an $H$-equivariant bundle.


Then you can take $\hat{G}/\sim\, := \hat{G}/H$, the quotient by a Lie group action. All of this is quite general theory, and doesn't rely on assumptions past local triviality, which for non-Banach Lie groups (and subgroups) are built into the definitions rather than arising from the splitting on tangent space. If you are in the Banach setting, for simplicity, then charts are modelled on tangent spaces, and in particular, $G/H$ is modelled on $\mathfrak{m}:=\mathfrak{g}/\mathfrak{h}$, $G$ is modelled on $\mathfrak{g} \simeq \mathfrak{h}\oplus \mathfrak{m}$ and $\hat{G}$ is modelled on $\mathfrak{u}(1)\oplus \mathfrak{g} \simeq \mathfrak{u}(1)\oplus \mathfrak{h}\oplus \mathfrak{m}$. Then $\hat{G}/H$ is modelled on $\mathfrak{u}(1)\oplus \mathfrak{m}$. 


