(see also reductive Klein geometry)
A Cartan geometry $P$ of type $(G,H)$ on $M$ is called reductive if the Lie algebra $\mathfrak{g}$ seen as a $Ad(H)$-module (see this) can be decomposed
$$ \mathfrak{g}=\mathfrak{h}\oplus\mathfrak{p}, $$with $\mathfrak g$ and $\mathfrak h$ being the corresponding Lie algebras.
Let's denote $M=P/H$. In a reductive Cartan geometry we can identify $T_{\pi(p)}M$ with a subspace of $\mathfrak g$ complementary to $\mathfrak h$ and invariant under the adjoint action of $H$. So we have a kind of "horizontal" direction. Let's see.
You can fix a choice of $\mathfrak p$ and you get a projection $f:\mathfrak g \to \mathfrak h$. This way, the Cartan connection (or the MC form, if we are in a Klein geometry) can be split
$$ A=(f+id-f)A=A_{\mathfrak h}+A_{\mathfrak p} $$The 1-form $\omega=A_{\mathfrak h}$ is the 1-form of a principal connection on a principal bundle, and $\ker(A_{\mathfrak h})$ describe the horizontal subspaces. The map $d\pi_p:\ker(A_{\mathfrak h})\to T_{\pi(p)}M$ is a isomorphism, since it is surjective, and dimensions agree. Observe that if $V\in V_p P$ then $A_p(V)=\omega_p(V)$.
Personal belief
Conversely, given a Cartan geometry and a principal connection on its principal bundle such that $A_p(V)=\omega_p(V)$ we can define a projection $f:\mathfrak g \to \mathfrak h$ given by
$$ f:\mathfrak g \stackrel{A_e{-1}}{\longrightarrow} T_e P \stackrel{\omega_e}{\longrightarrow} \mathfrak h $$and then $\mathfrak g=\mbox{im}(f)\oplus \ker f=\mathfrak h \oplus \mathfrak p$, since the short exact sequence
$$ 0 \longrightarrow \mathfrak h \stackrel{i}{\longrightarrow} \mathfrak g \stackrel{}{\longrightarrow} \mathfrak g/\mathfrak h \longrightarrow 0 $$split. I think it can be proved that $\mathfrak p$ is a $Ad(H)$-module and that $f$ is well-defined, i.e., independent of $p\in P$. See my own question in MO.
$\blacksquare$
In any case, if we have a reductive Cartan geometry, we have a isomorphism between $T_{\pi(p)}M$ and $\mathfrak p$:
$$ T_{\pi(p)}(P/H) \stackrel{d\pi_p^{-1}}{\longrightarrow}\ker(A_{\mathfrak h})\stackrel{A_{\mathfrak p}}{\longrightarrow} \mathfrak p $$This means that reductive Cartan geometries are geometries in which a little displacement on the base manifold from $x:=\pi(p)\in M$ to a nearby point $x'\in M$ corresponds to a canonical element $v\in \mathfrak p \subset \mathfrak g$. A kind of "canonical translation" or transvection (see "Symmetric Space Cartan Connections and Gravity in Three and Four Dimensions" by Derek Wise page 3). That is, we obtain a one-parameter subgroup of the group of $G$ of privileged transformations of the geometry, associated to this little displacement. This is in total analogy with what happens in the case of affine space and usual translations. ^6297a6
Conversely, given a Klein geometry in which there is an isomorphism from $T_{\pi(p)}M$ to a subspace $\mathfrak p$ of $\mathfrak g$ then the short exact sequence
$$ 0 \longrightarrow \mathfrak h \stackrel{i}{\longrightarrow} \mathfrak g \stackrel{d\pi_e}{\longrightarrow} T_{\pi(e)} M \longrightarrow 0 $$splits, and $\mathfrak g=\mathfrak h \oplus \mathfrak p$. Of course, several properties have to be checked...
Continue with Jacob Erickson...
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Author of the notes: Antonio J. Pan-Collantes
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