(see my own question and answer in MO
Consider the manifold $M=\mathbb R^3$ with the natural vector bundle connection $\nabla$. This connection, like any connection on a vector bundle, induces, or is induced by, a principal connection $\Theta$ on the frame bundle $FM=\mathbb R^3 \times GL(3)$.
The curvature of this connection is, of course, zero:
$$ \Omega=d\Theta-[\Theta,\Theta]=0. $$The pair $(\mathbb R^3,\Theta)$ 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 $G=\operatorname{Aff}(3)=\mathbb{R}^3 \rtimes \operatorname{GL}(3)$ (we fix it as the "relevant" transformations on $\mathbb R^3$) together with the closed subgroup $H=\operatorname{GL}(3)$ (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
$$ dA-[A,A]=0, $$because of the Maurer-Cartan equation for any Lie group.
Since this is a reductive Klein geometry, the Cartan connection $A$ splits
$$ A=A_{\mathfrak{h}}+A_{\mathfrak{p}} $$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 $\Theta$, but not necessarily equal to $\Theta$ (is a different connection). Why do we have in this case a natural choice of $\mathfrak{p}$ such that $A_{\mathfrak{h}}=\Theta$?
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
$$ G\to G/H $$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
$$ G=N\rtimes H. $$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., $G = N \rtimes H$. Then the Lie algebra $\mathfrak{g}$ of $G$ has a canonical decomposition as an $Ad(H)$-module given by
$$ \mathfrak{g} = \mathfrak{n} \oplus \mathfrak{h}, $$where $\mathfrak{n}$ and $\mathfrak{h}$ are the Lie algebras of the subgroups $N$ and $H$, respectively.$\blacksquare$
Sketch of the proof
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:
$$ 0=dA-[A,A]= $$ $$ =dA_{\mathfrak{n}}+dA_{\mathfrak{h}}-[A_{\mathfrak{n}}+A_{\mathfrak{h}},A_{\mathfrak{n}}+A_{\mathfrak{h}}]= $$ $$ =dA_{\mathfrak{h}}-[A_{\mathfrak{h}},A_{\mathfrak{h}}]+dA_{\mathfrak{n}}-[A_{\mathfrak{n}},A_{\mathfrak{n}}], $$and since $dA_{\mathfrak{h}}-[A_{\mathfrak{h}},A_{\mathfrak{h}}]$ is $\mathfrak{h}$-valued and $dA_{\mathfrak{n}}-[A_{\mathfrak{n}},A_{\mathfrak{n}}]$ is $\mathfrak{n}$-valued we have
$$ dA_{\mathfrak{h}}-[A_{\mathfrak{h}},A_{\mathfrak{h}}]=0. $$________________________________________
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Author of the notes: Antonio J. Pan-Collantes
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