Generalization of the flatness of $\mathbb R^3$

To get $A_{\mathfrak{h}}$ to be an affine connection, you need precisely that $\mathfrak{p}\subseteq\mathfrak{g}$ is an $H$-invariant linear subspace complementary to $\mathfrak{h}\subseteq\mathfrak{g}$. To make it flat, you need precisely that $\mathfrak{p}\subseteq\mathfrak{g}$ is bracket closed, so a Lie subalgebra complementary to $\mathfrak{h}\subseteq\mathfrak{g}$. For Euclidean space, this is precisely the fact that translations of Euclidean space are defined independent of any choice of basis, i.e. form a normal subgroup in the affine group, hence invariant under affine transformations, so invariant under linear transformations (the group $H$), and bracket closed, so a flat connection.

Ben's answer has been useful to me in order to understand things. But I think it is a bit incomplete, so I am going to write here my conclusions in case they are useful for someone else in the future. I want to point out that some of the statements I am going to do are based on my intuition, and I don't have a proof, so I hope anybody can teach me if something is wrong.

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

• 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.$\blacksquare$

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 $$ dA_{\mathfrak{h}}-[A_{\mathfrak{h}},A_{\mathfrak{h}}]=0. $$