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

Question

Original setup

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 $FM=\mathbb R^3 \times \operatorname{GL}(3)$. The curvature of this connection is, of course, zero: $$ \Omega=d\Theta-[\Theta,\Theta]=0. $$

Cartan geometry

I want to obtain the pair $(\mathbb R^3,\Theta)$ from the point of view of Cartan geometry, to 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). Moreover, even if $A$ has zero curvature I think we cannot deduce that $A_{\mathfrak{h}}$ has zero curvature.

My questions are:

1. What data am I missing from the original setup to be able to fix a $\mathfrak{p}$ such that $A_{\mathfrak{h}}=\Theta$?
2. In general, what additional requirement do we have to impose to a reductive Klein geometry $(G,H)$ in such a way that we have a canonical choice of $\mathfrak{p}$ giving rise to a principal connection 1-form $\Theta:=A_{\mathfrak{h}}$ which is flat?

The motivation for my questions is that I have read that the Klein geometries are the flat models for the Cartan geometries. But even if I understand that they are flat, in the sense that their natural Cartan connections has zero curvature, I want to know which among them are "really flat", in the sense that they have a natural principal connection with zero curvature.

Answer 1

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.

Answer 2

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