Generalization of the flatness of R3

(see my own question and answer in MO)

Motivation

Consider the manifold $M={\mathbb{R}}^3$ with the natural vector bundle connection $\mathrm{\nabla }$. This connection, like any connection on a vector bundle, induces, or is induced by, a principal connection $\mathrm{\Theta }$ on the frame bundle $FM={\mathbb{R}}^3\times GL(3)$.
The curvature of this connection is, of course, zero:

The pair $({\mathbb{R}}^3,\mathrm{\Theta })$ can be obtained from the point of view of Cartan geometry. This way we can understand to what extent the flatness of ${\mathbb{R}}^3$ can be generalized to arbitrary homogeneous spaces.
We have the affine group $G=\mathrm{Aff}(3)={\mathbb{R}}^3⋊\mathrm{GL}(3)$ (we fix it as the "relevant" transformations on ${\mathbb{R}}^3$) together with the closed subgroup $H=\mathrm{GL}(3)$ (the "relevant" transformations fixing a point). The Klein geometry $(G,H)$ provides the base space $M$, and the Maurer-Cartan form $A$ of $G$ is a Cartan connection with, of course, zero curvature

because of the Maurer-Cartan equation for any Lie group.
Since this is a reductive Klein geometry, the Cartan connection $A$ splits

for every specific choice of the subspace $\mathfrak{p}\subset \mathfrak{g}$ such that $\mathfrak{g}=\mathfrak{h}\oplus \mathfrak{p}$. The part $A_{\mathfrak{h}}$ is a principal connection on the frame bundle $FM$, playing the same role as $\mathrm{\Theta }$, but not necessarily equal to $\mathrm{\Theta }$ (is a different connection). Why do we have in this case a natural choice of $\mathfrak{p}$ such that $A_{\mathfrak{h}}=\mathrm{\Theta }$?

Generalization

The goal is understand what do we have to require to an arbitrary Klein geometry $(G,H)$ so that it behaves in the same way that ${\mathbb{R}}^n$ "with respect to flatness". That is to say, in order to have $G/H$ be flat not only as a Cartan geometry, which is trivially true by virtue of Maurer-Cartan equations, but also to have that the principal bundle

have a canonical flat principal connection.
(By the way, this principal bundle is interpreted as the set of "$G$-frames" for the space $X=G/H$, and the principal connection is a way of deciding if an assignation of G-frames along a curve in $X$ is constant.)

And I think that the answer is that $G$ must have a normal subgroup $N$ such that $G$ is the semidirect product

The key would be the following proposition, which I hope is true,
Proposition
Let $G$ be a Lie group that is the semidirect product of a normal subgroup $N$ and a subgroup $H$, i.e., $G=N⋊H$. Then the Lie algebra $\mathfrak{g}$ of $G$ has a canonical decomposition as an $Ad(H)$-module given by

where $\mathfrak{n}$ and $\mathfrak{h}$ are the Lie algebras of the subgroups $N$ and $H$, respectively.$◼$

Sketch of the proof

• Show that $\mathfrak{g}=\mathfrak{n}\oplus \mathfrak{h}$ as vector spaces.
• $\mathfrak{g}$ is an $Ad(H)$-module, and of course $Ad(H)(\mathfrak{h})\subset \mathfrak{h}$. But also, since $N$ is normal, we have that $Ad(H)(\mathfrak{n})\subset \mathfrak{n}$
• The decomposition is canonical, once $N$ is given.$◼$

So, assuming this proposition is true, the requirement of being a semidirect product implies that we have a reductive Klein geometry $G/H$, and then the Maurer-Cartan form splits as $A=A_{\mathfrak{n}}+A_{\mathfrak{h}}$, being $A_{\mathfrak{h}}$ a principal connection on the principal bundle $G\to X$ (see this answer), in a "canonical way" (since $N$ is given).
Moreover, we have that $\mathfrak{n}$ is not merely a vector subspace of $\mathfrak{g}$, but a Lie algebra, so $[\mathfrak{n},\mathfrak{n}]\subset \mathfrak{n}$. And because of this we can prove that the flatness in the sense of Cartan geometry ($dA-[A,A]=0$) implies the flatness of the principal connection:

and since $d{A}_{\mathfrak{h}}-[A_{\mathfrak{h}},A_{\mathfrak{h}}]$ is $\mathfrak{h}$-valued and $d{A}_{\mathfrak{n}}-[A_{\mathfrak{n}},A_{\mathfrak{n}}]$ is $\mathfrak{n}$-valued we have