[proofplan] The reductive splitting gives two linear projections from $\mathfrak g$ onto $\mathfrak h$ and $\mathfrak m$, and applying these projections to the restricted connection form $\omega|_{TQ}$ gives the desired two summands. The principal connection axioms for $\omega$ imply the vertical normalization and equivariance properties after projection; the only point needing care is that $\operatorname{Ad}_H$ preserves both summands, so the projections commute with the $H$-adjoint action. Finally, preservation of the reduction is equivalent to saying that every $\omega$-horizontal lift at a point of $Q$ lies in $TQ$, which is equivalent to $\omega|_{TQ}$ being $\mathfrak h$-valued. [/proofplan] [step:Project the restricted connection form onto the reductive summands] Let $M$ denote the base manifold of the principal $G$-bundle, and let $\pi:P \to M$ denote its bundle projection. Let $\operatorname{pr}_{\mathfrak h}:\mathfrak g \to \mathfrak h$ denote the linear projection with kernel $\mathfrak m$, and let $\operatorname{pr}_{\mathfrak m}:\mathfrak g \to \mathfrak m$ denote the linear projection with kernel $\mathfrak h$. Define the two smooth $1$-forms $\omega_{\mathfrak h}:TQ \to \mathfrak h$ and $\omega_{\mathfrak m}:TQ \to \mathfrak m$ by $\omega_{\mathfrak h}(v)=\operatorname{pr}_{\mathfrak h}(\omega(v))$ and $\omega_{\mathfrak m}(v)=\operatorname{pr}_{\mathfrak m}(\omega(v))$, for every $q \in Q$ and every $v \in T_qQ$. These forms are smooth because they are obtained by composing the smooth restricted form $\omega|_{TQ}:TQ \to \mathfrak g$ with fixed linear maps between finite-dimensional vector spaces. Since every element of $\mathfrak g$ has a unique expression as a sum of an element of $\mathfrak h$ and an element of $\mathfrak m$, we have \begin{align*} \omega(v)=\omega_{\mathfrak h}(v)+\omega_{\mathfrak m}(v) \end{align*} for every $v \in TQ$. Thus \begin{align*} \omega|_{TQ}=\omega_{\mathfrak h}+\omega_{\mathfrak m}. \end{align*} The same direct-sum uniqueness proves uniqueness of the decomposition: if $\omega|_{TQ}=\alpha+\beta$ with $\alpha \in \Omega^1(Q;\mathfrak h)$ and $\beta \in \Omega^1(Q;\mathfrak m)$, then for each $v \in TQ$ the equality $\omega(v)=\alpha(v)+\beta(v)$ is the unique $\mathfrak h \oplus \mathfrak m$ decomposition of $\omega(v)$, so $\alpha=\omega_{\mathfrak h}$ and $\beta=\omega_{\mathfrak m}$. [/step] [step:Verify that the $\mathfrak h$ component is a principal $H$ connection] For $A \in \mathfrak h$, let $A_Q:Q \to TQ$ be the fundamental vector field of the principal $H$-bundle $Q \to M$, defined by \begin{align*} A_Q(q)=\left.\frac{d}{dt}\right|_{t=0} q\exp(tA). \end{align*} Because $Q \subset P$ is an $H$-subbundle, this is also the restriction to $Q$ of the fundamental vector field on $P$ determined by $A \in \mathfrak g$. Since $\omega$ is a principal $G$-connection form, \begin{align*} \omega(A_Q(q))=A. \end{align*} Applying $\operatorname{pr}_{\mathfrak h}$ gives \begin{align*} \omega_{\mathfrak h}(A_Q(q))=\operatorname{pr}_{\mathfrak h}(A)=A. \end{align*} Thus $\omega_{\mathfrak h}$ has the required vertical normalization. Now fix $h \in H$, and let $R_h:P \to P$ denote right multiplication by $h$, so $R_h(p)=ph$ for $p \in P$; its restriction $R_h:Q \to Q$ is the right action of $h$ on the principal $H$-bundle $Q$. Since $\mathfrak h$ is the Lie algebra of $H$, $\operatorname{Ad}_{h^{-1}}(\mathfrak h)\subset \mathfrak h$. Since the splitting is reductive, meaning that $\mathfrak m$ is $\operatorname{Ad}_H$-invariant, we have $\operatorname{Ad}_{h^{-1}}(\mathfrak m)\subset \mathfrak m$. Hence $\operatorname{Ad}_{h^{-1}}$ preserves both summands in $\mathfrak g=\mathfrak h\oplus\mathfrak m$, so \begin{align*} \operatorname{pr}_{\mathfrak h}\circ \operatorname{Ad}_{h^{-1}}=\operatorname{Ad}_{h^{-1}}\circ \operatorname{pr}_{\mathfrak h}. \end{align*} Using the principal $G$-connection equivariance of $\omega$, namely $(R_h)^*\omega=\operatorname{Ad}_{h^{-1}}\omega$, we obtain \begin{align*} (R_h)^*\omega_{\mathfrak h}=\operatorname{pr}_{\mathfrak h}((R_h)^*\omega)=\operatorname{pr}_{\mathfrak h}(\operatorname{Ad}_{h^{-1}}\omega)=\operatorname{Ad}_{h^{-1}}\operatorname{pr}_{\mathfrak h}(\omega)=\operatorname{Ad}_{h^{-1}}\omega_{\mathfrak h}. \end{align*} Therefore $\omega_{\mathfrak h}$ is a principal $H$-connection form on $Q$. [guided] We must check the two defining properties of a principal connection form on the principal $H$-bundle $Q \to M$: it must reproduce fundamental vertical vectors, and it must be equivariant for the right $H$-action. First take an element $A \in \mathfrak h$. The associated fundamental vector field on $Q$ is the map $A_Q:Q \to TQ$ defined by \begin{align*} A_Q(q)=\left.\frac{d}{dt}\right|_{t=0} q\exp(tA). \end{align*} Because $Q$ is an $H$-subbundle of the principal $G$-bundle $P$, this curve is also the curve in $P$ used to define the fundamental vector field for $A$ as an element of $\mathfrak g$. Since $\omega$ is a principal $G$-connection form, it reproduces fundamental vertical vectors: \begin{align*} \omega(A_Q(q))=A. \end{align*} Now $A$ already lies in $\mathfrak h$, so projecting to the $\mathfrak h$ summand does not change it: \begin{align*} \omega_{\mathfrak h}(A_Q(q))=\operatorname{pr}_{\mathfrak h}(\omega(A_Q(q)))=\operatorname{pr}_{\mathfrak h}(A)=A. \end{align*} This proves the vertical normalization condition for $\omega_{\mathfrak h}$. Next we verify $H$-equivariance. Fix $h \in H$, and let $R_h:P \to P$ be the right multiplication map $R_h(p)=ph$; since $Q$ is an $H$-subbundle, the same notation denotes its restriction $R_h:Q \to Q$. The principal connection form $\omega$ on $P$ satisfies \begin{align*} (R_h)^*\omega=\operatorname{Ad}_{h^{-1}}\omega. \end{align*} We want to project this identity to $\mathfrak h$. The reason this is legitimate is that the splitting is respected by the $H$-adjoint action. The subspace $\mathfrak h$ is preserved because it is the Lie algebra of $H$, and the subspace $\mathfrak m$ is preserved because the reductive hypothesis means precisely that $\mathfrak m$ is $\operatorname{Ad}_H$-invariant. Therefore $\operatorname{Ad}_{h^{-1}}$ preserves the two summands separately, and hence commutes with the projection onto $\mathfrak h$: \begin{align*} \operatorname{pr}_{\mathfrak h}\circ \operatorname{Ad}_{h^{-1}}=\operatorname{Ad}_{h^{-1}}\circ \operatorname{pr}_{\mathfrak h}. \end{align*} Applying this identity to the equivariance formula for $\omega$ gives \begin{align*} (R_h)^*\omega_{\mathfrak h}=\operatorname{pr}_{\mathfrak h}((R_h)^*\omega)=\operatorname{pr}_{\mathfrak h}(\operatorname{Ad}_{h^{-1}}\omega)=\operatorname{Ad}_{h^{-1}}\operatorname{pr}_{\mathfrak h}(\omega)=\operatorname{Ad}_{h^{-1}}\omega_{\mathfrak h}. \end{align*} Thus $\omega_{\mathfrak h}$ satisfies both defining axioms of a principal $H$-connection form. [/guided] [/step] [step:Verify that the $\mathfrak m$ component is horizontal and $H$ equivariant] Let $A \in \mathfrak h$ and let $A_Q:Q \to TQ$ be its fundamental vector field. As above, \begin{align*} \omega(A_Q(q))=A \end{align*} for every $q \in Q$. Applying the projection onto $\mathfrak m$ gives \begin{align*} \omega_{\mathfrak m}(A_Q(q))=\operatorname{pr}_{\mathfrak m}(A)=0. \end{align*} Thus $\omega_{\mathfrak m}$ vanishes on the vertical tangent bundle of $Q \to M$, so it is horizontal. For $h \in H$, the same invariance of the reductive splitting gives \begin{align*} \operatorname{pr}_{\mathfrak m}\circ \operatorname{Ad}_{h^{-1}}=\operatorname{Ad}_{h^{-1}}\circ \operatorname{pr}_{\mathfrak m}. \end{align*} Using $(R_h)^*\omega=\operatorname{Ad}_{h^{-1}}\omega$, we obtain \begin{align*} (R_h)^*\omega_{\mathfrak m}=\operatorname{pr}_{\mathfrak m}((R_h)^*\omega)=\operatorname{pr}_{\mathfrak m}(\operatorname{Ad}_{h^{-1}}\omega)=\operatorname{Ad}_{h^{-1}}\operatorname{pr}_{\mathfrak m}(\omega)=\operatorname{Ad}_{h^{-1}}\omega_{\mathfrak m}. \end{align*} Therefore $\omega_{\mathfrak m}$ is horizontal and $H$-equivariant. [/step] [step:Identify preservation of the reduction with vanishing of the $\mathfrak m$ component] For $q \in Q$, define the $\omega$-horizontal subspace at $q$ by $\mathcal H_q^\omega=\ker(\omega_q)\subset T_qP$. Since $\omega$ is a principal connection form, the differential $d\pi_q$ restricts to a linear isomorphism $\mathcal H_q^\omega \to T_{\pi(q)}M$; hence each vector in $T_{\pi(q)}M$ has a unique $\omega$-horizontal lift at $q$. We say that $\omega$ preserves $Q$ precisely when \begin{align*} \mathcal H_q^\omega \subset T_qQ \end{align*} for every $q \in Q$. Assume first that $\omega_{\mathfrak m}=0$. Let $q \in Q$, let $x=\pi(q)$, and let $v \in T_xM$. Since $Q \to M$ is a principal $H$-bundle, the differential \begin{align*} d\pi_q:T_qQ \to T_xM \end{align*} is surjective. Choose $u \in T_qQ$ with $d\pi_q(u)=v$. Since $\omega_{\mathfrak m}=0$, we have $\omega(u)=\omega_{\mathfrak h}(u)\in\mathfrak h$. Let $A=\omega_{\mathfrak h}(u)\in\mathfrak h$, and let $A_Q(q)\in T_qQ$ be the corresponding fundamental vertical vector. Define \begin{align*} u_0=u-A_Q(q). \end{align*} Then $u_0\in T_qQ$, $d\pi_q(u_0)=v$, and \begin{align*} \omega(u_0)=\omega(u)-\omega(A_Q(q))=A-A=0. \end{align*} Thus $u_0\in \mathcal H_q^\omega\cap T_qQ$. For each $v\in T_xM$, the $\omega$-horizontal lift at $q$ lies in $T_qQ$. Hence $\mathcal H_q^\omega\subset T_qQ$, so $\omega$ preserves $Q$. Conversely, assume that $\omega$ preserves $Q$. Let $q \in Q$ and let $u \in T_qQ$. Set $x=\pi(q)$ and $v=d\pi_q(u)\in T_xM$. By preservation, the unique $\omega$-horizontal vector $u_0\in \mathcal H_q^\omega$ with $d\pi_q(u_0)=v$ lies in $T_qQ$. Then $u-u_0$ is vertical for the principal $H$-bundle $Q \to M$, so there exists a unique $A\in\mathfrak h$ such that \begin{align*} u-u_0=A_Q(q). \end{align*} Using $\omega(u_0)=0$ and $\omega(A_Q(q))=A$, we get \begin{align*} \omega(u)=\omega(u_0)+\omega(A_Q(q))=A\in\mathfrak h. \end{align*} Projecting to $\mathfrak m$ gives $\omega_{\mathfrak m}(u)=0$. Since $q\in Q$ and $u\in T_qQ$ were arbitrary, $\omega_{\mathfrak m}=0$. Therefore $Q$ is preserved by $\omega$ if and only if $\omega_{\mathfrak m}=0$. [/step]