[proofplan] We compare the horizontal spaces of the $G$-connection on $P$ with the tangent spaces of the reduced subbundle $Q$. If the $G$-horizontal distribution is tangent to $Q$, then every tangent vector to $Q$ splits into its horizontal part plus an $H$-vertical fundamental vector, forcing the restricted connection form to be $\mathfrak h$-valued and to satisfy the connection axioms for $H$. Conversely, if the restriction is an $H$-connection form, then its kernel inside $T_qQ$ is contained in the $G$-horizontal space and has the same dimension, so the full $G$-horizontal space at $q$ lies in $T_qQ$. [/proofplan]

[step:Define the restricted form and the vertical fundamental vectors] Let $\iota: Q \to P$ denote the inclusion map. Define \begin{align*} \omega_Q := \iota^*\omega \in \Omega^1(Q;\mathfrak g). \end{align*} Equivalently, for $q \in Q$ and $v \in T_qQ$, \begin{align*} (\omega_Q)_q(v) = \omega_q(d\iota_q(v)). \end{align*} Since $Q$ is an embedded principal $H$-subbundle of $P$, we identify $T_qQ$ with its image $d\iota_q(T_qQ) \subset T_qP$. Let $\exp_G: \mathfrak g \to G$ and $\exp_H: \mathfrak h \to H$ denote the Lie group exponential maps of $G$ and $H$, respectively. For each $\xi \in \mathfrak g$, let $\xi_P: P \to TP$ denote the fundamental vector field on $P$ generated by the right $G$-action. At $p \in P$, it is given by \begin{align*} (\xi_P)_p = \frac{d}{dt}\Big|_{t=0} p\exp_G(t\xi). \end{align*} For each $\eta \in \mathfrak h$, let $\eta_Q: Q \to TQ$ denote the fundamental vector field on $Q$ generated by the right $H$-action. At $q \in Q$, it is given by \begin{align*} (\eta_Q)_q = \frac{d}{dt}\Big|_{t=0} q\exp_H(t\eta). \end{align*} Because the $H$-action on $Q$ is the restriction of the $G$-action on $P$ and the differential of the inclusion $H \hookrightarrow G$ identifies $\mathfrak h$ with a Lie subalgebra of $\mathfrak g$, we have \begin{align*} d\iota_q((\eta_Q)_q) = (\eta_P)_q \end{align*} for every $q \in Q$ and every $\eta \in \mathfrak h$. [claim:Identify vertical tangent spaces with fundamental vectors] For every $q \in Q$, the maps \begin{align*} \theta_{P,q}: \mathfrak g \to \ker(d\pi_q), \quad \zeta \mapsto (\zeta_P)_q \end{align*} and \begin{align*} \theta_{Q,q}: \mathfrak h \to \ker(d(\pi|_Q)_q), \quad \eta \mapsto (\eta_Q)_q \end{align*} are linear isomorphisms. Moreover $d\iota_q \circ \theta_{Q,q} = \theta_{P,q}|_{\mathfrak h}$. [/claim] [proof] This is the standard vertical-space identification for a principal bundle. For $P$, the orbit map $G \to \pi^{-1}(\pi(q))$, $g \mapsto qg$, is a diffeomorphism because the right $G$-action is free and transitive on each fiber. Its differential at the identity element $e_G \in G$ identifies $T_{e_G}G = \mathfrak g$ with $T_q(\pi^{-1}(\pi(q))) = \ker(d\pi_q)$, and this differential is exactly $\theta_{P,q}$. The same argument applied to the principal $H$-bundle $Q \to M$ gives the isomorphism $\theta_{Q,q}: \mathfrak h \to \ker(d(\pi|_Q)_q)$. Finally, the equality $\iota(qh) = \iota(q)h$ for $h \in H$ differentiates at $h = e_H$ in the direction $\eta \in \mathfrak h$ to give $d\iota_q((\eta_Q)_q) = (\eta_P)_q$, which is the asserted intertwining identity. [/proof] [/step]

[step:Restrict a preserved $G$-connection to an $H$-connection on $Q$]Assume that $Q$ is preserved by $\omega$, so that \begin{align*} \ker(\omega_q) \subset T_qQ \end{align*} for every $q \in Q$. Fix $q \in Q$ and $v \in T_qQ$. Since $\omega$ is a principal $G$-connection form, the tangent space $T_qP$ decomposes as \begin{align*} T_qP = \ker(\omega_q) \oplus \ker(d\pi_q), \end{align*} where $\ker(d\pi_q)$ is the vertical tangent space of the principal $G$-bundle $P \to M$. Hence there exists a unique horizontal vector $v_{\mathrm{hor}} \in \ker(\omega_q)$ and a unique element $\xi \in \mathfrak g$ such that \begin{align*} v = v_{\mathrm{hor}} + (\xi_P)_q. \end{align*} By preservation, $v_{\mathrm{hor}} \in T_qQ$. Since also $v \in T_qQ$, subtraction in the [vector space](/page/Vector%20Space) $T_qP$ gives \begin{align*} (\xi_P)_q = v - v_{\mathrm{hor}} \in T_qQ. \end{align*} Moreover $(\xi_P)_q$ is vertical for $\pi: P \to M$, so it lies in \begin{align*} T_qQ \cap \ker(d\pi_q) = \ker(d(\pi|_Q)_q). \end{align*} By the vertical-fundamental-vector identification proved in the first step, \begin{align*} \ker(d(\pi|_Q)_q) = \{(\eta_Q)_q : \eta \in \mathfrak h\}. \end{align*} Thus there exists $\eta \in \mathfrak h$ such that \begin{align*} (\xi_P)_q = d\iota_q((\eta_Q)_q) = (\eta_P)_q. \end{align*} By the vertical-fundamental-vector identification proved in the first step, the map $\theta_{P,q}: \mathfrak g \to \ker(d\pi_q)$ is a linear isomorphism, so $\xi = \eta \in \mathfrak h$. Therefore \begin{align*} (\omega_Q)_q(v) = \omega_q(v) = \omega_q(v_{\mathrm{hor}}) + \omega_q((\xi_P)_q) = 0 + \xi = \xi \in \mathfrak h. \end{align*} Since $q$ and $v$ were arbitrary, $\omega_Q$ is $\mathfrak h$-valued. It remains to verify the principal connection form axioms for the principal $H$-bundle $Q \to M$. For $\eta \in \mathfrak h$ and $q \in Q$, \begin{align*} (\omega_Q)_q((\eta_Q)_q) = \omega_q((\eta_P)_q) = \eta, \end{align*} using the reproduction axiom for the principal $G$-connection $\omega$. For $h \in H$, the right action $R_h^Q: Q \to Q$ is the restriction of $R_h^P: P \to P$. Hence \begin{align*} (R_h^Q)^*\omega_Q = \iota^*((R_h^P)^*\omega) = \iota^*(\operatorname{Ad}_{h^{-1}}\omega) = \operatorname{Ad}_{h^{-1}}\omega_Q. \end{align*} Because $H$ is a Lie subgroup and $\mathfrak h$ is its Lie algebra, $\operatorname{Ad}_{h^{-1}}(\mathfrak h) \subset \mathfrak h$ for every $h \in H$. Thus $\omega_Q \in \Omega^1(Q;\mathfrak h)$ satisfies the reproduction and $H$-equivariance axioms, so it is a principal $H$-connection form on $Q$.[/step]

[guided]Assume that $Q$ is preserved by $\omega$, meaning that every $G$-horizontal vector at a point of $Q$ is already tangent to $Q$. Fix a point $q \in Q$ and a tangent vector $v \in T_qQ$. We want to prove first that $\omega_q(v)$ actually lies in the smaller Lie algebra $\mathfrak h$, not merely in $\mathfrak g$. Since $\omega$ is a principal $G$-connection form, it splits every tangent vector in $T_qP$ into a horizontal part and a vertical fundamental part. Thus there are unique elements \begin{align*} v_{\mathrm{hor}} \in \ker(\omega_q) \end{align*} and \begin{align*} \xi \in \mathfrak g \end{align*} such that \begin{align*} v = v_{\mathrm{hor}} + (\xi_P)_q. \end{align*} The preservation hypothesis now matters: because $q \in Q$ and $v_{\mathrm{hor}}$ is $G$-horizontal, preservation gives $v_{\mathrm{hor}} \in T_qQ$. Since $v$ was already chosen in $T_qQ$, the difference \begin{align*} (\xi_P)_q = v - v_{\mathrm{hor}} \end{align*} also lies in $T_qQ$. This vector is vertical for $\pi: P \to M$, because every fundamental vector field of the principal action is tangent to the fiber. Therefore it lies in the intersection \begin{align*} T_qQ \cap \ker(d\pi_q). \end{align*} The vertical-fundamental-vector identification proved in the first step is exactly the tool needed here. It says that \begin{align*} T_qQ \cap \ker(d\pi_q) = \ker(d(\pi|_Q)_q) = \{(\eta_Q)_q : \eta \in \mathfrak h\}. \end{align*} Hence there exists $\eta \in \mathfrak h$ such that \begin{align*} (\xi_P)_q = d\iota_q((\eta_Q)_q) = (\eta_P)_q. \end{align*} The same identification says that $\theta_{P,q}: \mathfrak g \to \ker(d\pi_q)$ is a linear isomorphism. Therefore $\xi = \eta$, and in particular $\xi \in \mathfrak h$. Now apply $\omega_q$ to the decomposition of $v$: \begin{align*} (\omega_Q)_q(v) = \omega_q(v) = \omega_q(v_{\mathrm{hor}}) + \omega_q((\xi_P)_q). \end{align*} The first term is zero because $v_{\mathrm{hor}} \in \ker(\omega_q)$, and the second term is $\xi$ by the reproduction axiom for a principal $G$-connection form. Therefore \begin{align*} (\omega_Q)_q(v) = \xi \in \mathfrak h. \end{align*} Since $q \in Q$ and $v \in T_qQ$ were arbitrary, $\omega_Q$ is $\mathfrak h$-valued. It remains to check that this $\mathfrak h$-valued form satisfies the principal connection form axioms for the principal $H$-bundle $Q \to M$. Let $\eta \in \mathfrak h$. The fundamental vector field $\eta_Q$ on $Q$ maps under $d\iota$ to the fundamental vector field $\eta_P$ on $P$, so the reproduction axiom for $\omega$ gives \begin{align*} (\omega_Q)_q((\eta_Q)_q) = \omega_q((\eta_P)_q) = \eta. \end{align*} Next let $h \in H$. The right action of $h$ on $Q$ is the restriction of the right action of $h$ on $P$, so the $G$-equivariance of $\omega$ restricts to \begin{align*} (R_h^Q)^*\omega_Q = \operatorname{Ad}_{h^{-1}}\omega_Q. \end{align*} Because $H$ is a Lie subgroup, its adjoint action preserves its Lie algebra: $\operatorname{Ad}_{h^{-1}}(\mathfrak h) \subset \mathfrak h$. Thus the restricted form is $\mathfrak h$-valued, reproduces $H$-fundamental vector fields, and is $H$-equivariant. These are exactly the axioms for a principal $H$-connection form on $Q$.[/guided]

[step:Use the restricted $H$-connection to force the $G$-horizontal space into $TQ$] Conversely, assume that \begin{align*} \omega_Q \in \Omega^1(Q;\mathfrak h) \end{align*} and that $\omega_Q$ is a principal $H$-connection form on $Q$. Fix $q \in Q$. Define \begin{align*} K_q := \ker((\omega_Q)_q) \subset T_qQ. \end{align*} Since $(\omega_Q)_q$ is the restriction of $\omega_q$ to $T_qQ$, we have \begin{align*} K_q \subset \ker(\omega_q) = \mathcal H_q^\omega. \end{align*} We compare dimensions. Since $\omega$ is a principal $G$-connection form, its restriction to the vertical tangent space $\ker(d\pi_q)$ is the isomorphism induced by fundamental vector fields, so $\omega_q: T_qP \to \mathfrak g$ is surjective. Therefore \begin{align*} \dim \mathcal H_q^\omega = \dim T_qP - \dim \mathfrak g. \end{align*} Because $P \to M$ is a principal $G$-bundle, \begin{align*} \dim T_qP = \dim M + \dim G = \dim M + \dim \mathfrak g, \end{align*} and hence \begin{align*} \dim \mathcal H_q^\omega = \dim M. \end{align*} Similarly, because $\omega_Q$ is a principal $H$-connection form, $(\omega_Q)_q: T_qQ \to \mathfrak h$ is surjective. Thus \begin{align*} \dim K_q = \dim T_qQ - \dim \mathfrak h. \end{align*} Since $Q \to M$ is a principal $H$-bundle, \begin{align*} \dim T_qQ = \dim M + \dim H = \dim M + \dim \mathfrak h, \end{align*} so \begin{align*} \dim K_q = \dim M. \end{align*} We have an inclusion $K_q \subset \mathcal H_q^\omega$ between finite-dimensional vector spaces of the same dimension. Therefore \begin{align*} K_q = \mathcal H_q^\omega. \end{align*} Since $K_q \subset T_qQ$, it follows that \begin{align*} \mathcal H_q^\omega \subset T_qQ. \end{align*} The point $q \in Q$ was arbitrary, so the horizontal distribution of $\omega$ is tangent to $Q$. Hence $Q$ is preserved by $\omega$. [/step]

[step:Conclude the equivalence] The first implication proves that preservation of the reduction forces $\omega|_{TQ}$ to be an $\mathfrak h$-valued principal $H$-connection form. The second implication proves that any $\mathfrak h$-valued principal $H$-connection form obtained by restriction has kernel equal to the $G$-horizontal space along $Q$, and hence the $G$-horizontal distribution is tangent to $Q$. Therefore $Q$ is preserved by $\omega$ if and only if $\omega|_{TQ}$ takes values in $\mathfrak h$ and is a principal $H$-connection form on $Q$. [/step]