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In my struggle to understand Cartan/Klein geometries, I have the intuition that reductive Klein geometries are the link to connect the "classical" differential geometry approach with this "modern" approach. So I want to understand them better.

A Klein geometry $(G,H)$ with $M=G/H$ is called reductive if the Lie algebra $\mathfrak{g}$ seen as a $Ad(H)$-module can be decomposed $$ \mathfrak{g}=\mathfrak{h}\oplus\mathfrak{p}, $$ with $\mathfrak g$ and $\mathfrak h$ being the corresponding Lie algebras.

I interpret a reductive Klein geometry like one in which there is a canonical sense of translation. Let's focus on a point $x\in M$, we know that a vector $v\in \mathfrak g$ gives rise to a one-parameter subgroup of $G$ and, in particular, a fundamental vector field on $M$ that we can understand like a "little displacement" of $x$ to $x'\in M$. But several different elements of $\mathfrak g$ can give rise to the same "little displacement" at $x$.

If the Klein geometry is reductive you can fix a choice of $\mathfrak p$ and get a projection $f:\mathfrak g \to \mathfrak h$. This way, the Maurer-Cartan form $A$ of $G$ 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 the principal bundle $\pi:G\to G/H$, and $\ker(A^{\mathfrak h})$ describe the horizontal subspaces. For $\pi(p)=x$, the map $d\pi_p:\ker(A^{\mathfrak h}_p)\to T_{x}M$ is a isomorphism, since it is surjective, and dimensions agree. So we have the isomorphism$$ T_{x}M \stackrel{d\pi_p^{-1}}{\longrightarrow}\ker(A^{\mathfrak h}_p)\stackrel{A^{\mathfrak p}_p}{\longrightarrow} \mathfrak p $$

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The meaning of this isomorphism is to assign a "canonical" one-parameter group of transformations to a little displacement from $x$ to $x'\in M$.

This is clearly visualized for the Euclidean plane: given an infinitesimal translation (a tangent vector) of a point, there are "lots" of transformations tangent to this vector: the translation itself and an infinite number of rotations of different radius. But we have a canonical euclidean transformation associated.

My question is: is this what characterize the reductive Klein geometries? That is, given a Klein geometry in which there are isomorphism from every tangent space of the base manifold to a fixed subspace of $\mathfrak g$, and assuming that these isomorphisms are consistent (in a sense that can't precise yet), can we conclude it is a reductive Klein geometry? Can we say, loosely speaking, that reductive Klein geometries are those in which little displacement on $M$ can be associated canonically with special one-parameter subgroups of $G$?

Added: Once I know the answer is yes, I wonder if these "special one-parameter subgroup" will eventually be the geodesics of the geometry (whatever they are in this context). Should I expect geodesics to exist only if we are in a reductive geometry?

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    $\begingroup$ Projective space has an obvious notion of geodesic: projective lines. It is a homogeneous space, hence any Cartan geometry modelled on it has geodesics. Such a Cartan geometry is called a projective connection. But these are unparameterized curves. To have a parameter well defined up to affine transformations, you need to have an affine connection, not just a projective connection. If this affine connection arises from the model, the model has to be reductive. $\endgroup$
    – Ben McKay
    Oct 31, 2022 at 14:16

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Yes. Recall that the Cartan geometry on $M=G/H$ has principal right $H$-bundle $E$ defined to be $E=G$ with bundle map $\pi\colon g \in G\mapsto gH\in G/H=M$.

Recall that any submersion $\pi\colon E\to M$ has a vertical bundle $V\subset TE$, the vector bundle whose each fiber $V_x$ at $x\in E$ is the kernel of $\pi'(x)\colon T_x E\to T_m M$ where $m=\pi(x)$. Recall that a connection on a principal $H$-bundle $\pi\colon E \to M$ is precisely an $H$-invariant splitting $TE=V\oplus W$.

Suppose that we have a principal right $H$-bundle $\pi\colon E\to M$, with an action of a Lie group $G$ by bundle automorphisms, acting transitively on $E$. Any connection $W$ is completely determined by the linear subspace $W_{x_0}\subset T_{x_0}E_H$ at any one given point $x_0\in E$, by $G$-invariance. Let $m_0=\pi(x_0)$. Let $K\subseteq G$ be the stabilizer of $m_0$. Each $k\in K$ moves $x_0$ to some point of the fiber $E_{m_0}$, say to $kx_0$. The group $H$ acts transitively on the fibers of $E\to M$, so some $h\in H$ has inverse $h^{-1}$ which takes $kx_0$ back to $x_0=kx_0h^{-1}$. This maps $K\to H$, easily seen to be a Lie group injection $\phi\colon K\to H$. We thus act on $E$ by a new $K$-action, $k\cdot x=kx\phi(k)^{-1}$, fixing $x_0$. Hence a $G$-invariant connection $W$ is precisely determined by one linear subspace $W_{x_0}$, which must be invariant under the new $K$-action, since it is both left $G$ and right $H$ invariant. Reversing the steps, any $K$-invariant linear subspace $W_{x_0} \subset TE_{x_0}$ complementary to $V_{x_0}$ determines a unique $G$-invariant connection.

Taking the special case of $M=G/H$, $x_0=1\in G$, $m_0=1\cdot H\in G/H$, we find $K=H$, $\phi\colon H\to H$ is the identity map, so invariant connections are identified with $\operatorname{Ad}H$ invariant complements $\mathfrak{p}=W_{x_0}$ to $\mathfrak{h}=V_{x_0}$ in $TE_{x_0}=\mathfrak{g}$. So a reductive homogeneous space, in the sense of Sharpe's book, is precisely a homogeneous space with invariant connection on the bundle $G\to G/H$, i.e. precisely a homogeneous space with an $H$-module decomposition $\mathfrak{g}=\mathfrak{h}\oplus\mathfrak{p}$.

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  • $\begingroup$ Thank you for your answer. Given a reductive Klein geometry, the $A^{\mathfrak h}$ part of the MC-form is a principal bundle connection, which is "very compatible" with the MC form (a Cartan connection). It is the converse true? That is, if we have a Klein geometry and together with the MC form (or the Cartan connection if we have a Cartan geometry instead) we have a principal bundle connection with this "compatibility", then is it a reductive Klein geometry? $\endgroup$ Oct 23, 2022 at 17:17
  • $\begingroup$ Wow, great answer with the edition. I am really learning a lot thanks to you. Only to confirm: in this kind of geometries (reductive, in Sharpe's sense) the invariant complement $\mathfrak p$ or, equivalently, the admitted invariant connection, is not canonically given a priori, isn't it? That is, you can have several valid decompositions of $\mathfrak g$... $\endgroup$ Oct 25, 2022 at 17:47
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    $\begingroup$ @A.J.Pan-Collantes: the choice of $\mathfrak{p}$ is not given a priori: rather, every $G$-invariant connection on $M=G/H$ determines one such $\mathfrak{p}$ and vice versa, so the choices of $\mathfrak{p}$ are arbitrary $\operatorname{Ad} H$-invariant complements to $\mathfrak{h}\subseteq\mathfrak{g}$. $\endgroup$
    – Ben McKay
    Oct 26, 2022 at 16:46
  • $\begingroup$ Little by little I am putting all the pieces of the puzzle. Thank you $\endgroup$ Oct 26, 2022 at 16:53
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    $\begingroup$ I'm not sure that I agree that Sharpe's usage is not standard. A quick search for the phrase "reductive homogeneous space" turns up more papers using it in Sharpe's sense than the other as well as Wikipedia, Encylopedia of mathematics and other posts on stack exchange/overflow all using Sharpe's version. $\endgroup$
    – Callum
    Oct 31, 2022 at 14:07

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