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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 $\mathfrak{g}=\mathfrak{h}\oplus\mathfrak{p}$.

Warning: the phrase reductive homogeneous space as Sharpe uses it is not standard terminology. Most authors take that phrase to mean that $G$ is a reductive Lie group, in the sense of the theory of linear algebraic groups and that $G/H$ is a projective algebraic variety acted on by an algebraic group action. I would not encourage the use of Sharpe's terminology. Maybe just describe a Cartan geometry as having model with invariant connection instead of being reductive.

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 $\mathfrak{g}=\mathfrak{h}\oplus\mathfrak{p}$.

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 $\mathfrak{g}=\mathfrak{h}\oplus\mathfrak{p}$.

Warning: the phrase reductive homogeneous space as Sharpe uses it is not standard terminology. Most authors take that phrase to mean that $G$ is a reductive Lie group, in the sense of the theory of linear algeebraic groups. I would not encourage the use of Sharpe's terminology. Maybe just describe a Cartan geometry as having model with invariant connection.

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 $\mathfrak{g}=\mathfrak{h}\oplus\mathfrak{p}$.

Warning: the phrase reductive homogeneous space as Sharpe uses it is not standard terminology. Most authors take that phrase to mean that $G$ is a reductive Lie group, in the sense of the theory of linear algebraic groups and that $G/H$ is a projective algebraic variety acted on by an algebraic group action. I would not encourage the use of Sharpe's terminology. Maybe just describe a Cartan geometry as having model with invariant connection instead of being reductive.

Yes. Note that every point of $M$ is a coset $gH$ for some $g\in G$, so $M$ has a special point $m_0=1\cdot H$ and $G$ has a special point $1 \in G$. You have a linear map $\mathfrak{g}=T_1 G\to T_{m_0} M$ by quotienting by $\mathfrak{h}$, the tangent to $H$, the vertical. You split this somehow, in some manner which is $H$-equivariant. That splitting is $\mathfrak{g}=\mathfrak{h}\oplus\mathfrak{p}$. It has to be $H$-invariant because the Maurer--Cartan form transforms in the adjoint $H$-action, so splits into a sum in the two subspaces just when $\mathfrak{p}$ is $\operatorname{Ad} H$ invariant.

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 $\mathfrak{g}=\mathfrak{h}\oplus\mathfrak{p}$.

Warning: the phrase reductive homogeneous space as Sharpe uses it is not standard terminology. Most authors take that phrase to mean that $G$ is a reductive Lie group, in the sense of the theory of linear algeebraic groups. I would not encourage the use of Sharpe's terminology. Maybe just describe a Cartan geometry as having model with invariant connection.