[proofplan] We prove the identity pointwise by decomposing tangent vectors into their horizontal and vertical components. Since both sides are $\mathfrak{g}$-valued alternating two-forms, it is enough to compare them on horizontal-horizontal, horizontal-vertical, and vertical-vertical pairs. The horizontal-horizontal case is exactly the defining formula for the curvature form, the mixed case follows by differentiating principal equivariance of the connection form, and the vertical-vertical case is computed directly on each principal fibre using left-invariant vector fields. [/proofplan] [step:Decompose tangent vectors into horizontal and fundamental vertical parts] Fix a point $p \in P$. Let $\mathfrak{X}(P)$ denote the real [vector space](/page/Vector%20Space) of smooth vector fields on $P$. Let $H_pP := \ker \omega_p \subset T_pP$ denote the horizontal subspace at $p$, and let $V_pP := \ker d\pi_p \subset T_pP$ denote the vertical subspace. Let $\exp: \mathfrak{g} \to G$ denote the Lie group exponential map. Since $\omega$ is a principal connection form, the restriction \begin{align*} \omega_p|_{V_pP}: V_pP \to \mathfrak{g} \end{align*} is inverse to the infinitesimal action map $\mathfrak{g} \to V_pP$ defined by $\xi \mapsto \xi_P(p)$, where $\xi_P \in \mathfrak{X}(P)$ is the fundamental vector field \begin{align*} \xi_P(q) = \frac{d}{dt}\bigg|_{t=0} q\exp(t\xi). \end{align*} Thus every vector $X_p \in T_pP$ has a unique decomposition \begin{align*} X_p = X_p^{\mathrm{hor}} + \xi_P(p), \end{align*} with $X_p^{\mathrm{hor}} \in H_pP$ and $\xi = \omega_p(X_p) \in \mathfrak{g}$. The horizontal distribution $H P := \ker \omega \subset TP$ is a smooth subbundle, so every vector in $H_pP$ extends locally to a smooth horizontal vector field. Fundamental vertical vector fields are globally defined by elements of $\mathfrak{g}$. Because $d\omega + \frac{1}{2}[\omega,\omega]$ and $\Omega$ are alternating bilinear forms on $T_pP$, their values at $p$ depend only on the two tangent vectors at $p$. Hence it is enough to check the identity on pairs consisting of horizontal local extensions and fundamental vertical vector fields. [/step] [step:Compare both sides on two horizontal vectors] Let $X,Y \in \mathfrak{X}(P)$ be smooth horizontal vector fields, so $\omega(X)=0$ and $\omega(Y)=0$. We use the definition of curvature as the horizontal [exterior derivative](/theorems/1525) of the connection form: for vector fields $A,B \in \mathfrak{X}(P)$, \begin{align*} \Omega(A,B) = d\omega(A^{\mathrm{hor}},B^{\mathrm{hor}}), \end{align*} where $A^{\mathrm{hor}}$ and $B^{\mathrm{hor}}$ denote the horizontal components determined by $\omega$. Since $X$ and $Y$ are already horizontal, \begin{align*} \Omega(X,Y) = d\omega(X,Y). \end{align*} The bracket term vanishes because \begin{align*} \frac{1}{2}[\omega,\omega](X,Y) = [\omega(X),\omega(Y)]_{\mathfrak{g}} = [0,0]_{\mathfrak{g}} = 0. \end{align*} Therefore \begin{align*} \Omega(X,Y) = \left(d\omega + \frac{1}{2}[\omega,\omega]\right)(X,Y) \end{align*} on horizontal-horizontal pairs. [/step] [step:Show the mixed horizontal-vertical terms vanish] Let $X \in \mathfrak{X}(P)$ be a smooth horizontal vector field and let $\xi \in \mathfrak{g}$. We compare the two sides on the pair $(X,\xi_P)$. Since $\Omega$ is horizontal by definition, \begin{align*} \Omega(X,\xi_P) = 0. \end{align*} For the bracket term, \begin{align*} \frac{1}{2}[\omega,\omega](X,\xi_P) = [\omega(X),\omega(\xi_P)]_{\mathfrak{g}} = [0,\xi]_{\mathfrak{g}} = 0. \end{align*} It remains to show that $d\omega(X,\xi_P)=0$. The exterior derivative of a one-form gives \begin{align*} d\omega(X,\xi_P) = X(\omega(\xi_P)) - \xi_P(\omega(X)) - \omega([X,\xi_P]). \end{align*} Here $\omega(\xi_P)$ is the constant function with value $\xi$, so $X(\omega(\xi_P))=0$. Also $\omega(X)=0$, so $\xi_P(\omega(X))=0$. It remains to check $\omega([X,\xi_P])=0$. For $g \in G$, define the right-translation map $R_g: P \to P$ by $R_g(q)=qg$, and let $(dR_g)_q: T_qP \to T_{qg}P$ denote its differential at $q$. The flow of $\xi_P$ is the right action by $\exp(t\xi)$: \begin{align*} \Phi_t^{\xi_P}(q)=R_{\exp(t\xi)}(q)=q\exp(t\xi). \end{align*} Let $\operatorname{Ad}: G \to GL(\mathfrak{g})$ denote the adjoint representation of $G$ on its Lie algebra. The connection form satisfies the principal equivariance identity \begin{align*} R_g^*\omega = \operatorname{Ad}_{g^{-1}}\omega \end{align*} for every $g \in G$. Pulling back by the flow and differentiating at $t=0$ gives the infinitesimal equivariance identity \begin{align*} \mathcal{L}_{\xi_P}\omega = -\operatorname{ad}_{\xi}\circ \omega, \end{align*} where $\operatorname{ad}_{\xi}: \mathfrak{g} \to \mathfrak{g}$ is the [linear map](/page/Linear%20Map) $\eta \mapsto [\xi,\eta]_{\mathfrak{g}}$. Evaluating this identity on the horizontal field $X$ gives \begin{align*} 0 - \omega([\xi_P,X]) = -[\xi,\omega(X)]_{\mathfrak{g}} = 0. \end{align*} Thus $\omega([\xi_P,X])=0$, and by antisymmetry of the Lie bracket of vector fields, $\omega([X,\xi_P])=0$. Hence $d\omega(X,\xi_P)=0$. Consequently \begin{align*} \left(d\omega+\frac{1}{2}[\omega,\omega]\right)(X,\xi_P)=0=\Omega(X,\xi_P). \end{align*} [guided] Let $X \in \mathfrak{X}(P)$ be horizontal and let $\xi \in \mathfrak{g}$. We want to evaluate the proposed identity on $(X,\xi_P)$, where $\xi_P$ is the fundamental vertical vector field defined by \begin{align*} \xi_P(q)=\frac{d}{dt}\bigg|_{t=0}q\exp(t\xi). \end{align*} The curvature form is horizontal by definition: it only sees the horizontal components of its inputs. Since the vertical vector field $\xi_P$ has horizontal component $0$, \begin{align*} \Omega(X,\xi_P)=0. \end{align*} Now compute the bracket term. Because $X$ is horizontal, $\omega(X)=0$. Because $\omega$ is a principal connection form, $\omega(\xi_P)=\xi$. Therefore \begin{align*} \frac{1}{2}[\omega,\omega](X,\xi_P) = [\omega(X),\omega(\xi_P)]_{\mathfrak{g}} = [0,\xi]_{\mathfrak{g}} = 0. \end{align*} So the only remaining point is the exterior derivative term. The formula for the exterior derivative of a $\mathfrak{g}$-valued one-form gives \begin{align*} d\omega(X,\xi_P) = X(\omega(\xi_P))-\xi_P(\omega(X))-\omega([X,\xi_P]). \end{align*} The first term is zero because $\omega(\xi_P)$ is the constant $\mathfrak{g}$-valued function with value $\xi$. The second term is zero because $\omega(X)=0$. Thus we must prove $\omega([X,\xi_P])=0$. This is exactly where principal equivariance enters, but we need its infinitesimal form rather than only the statement that right translations preserve horizontal subspaces. For $g \in G$, define $R_g: P \to P$ by $R_g(q)=qg$, and let $(dR_g)_q: T_qP \to T_{qg}P$ be its differential. The flow of $\xi_P$ is given by \begin{align*} \Phi_t^{\xi_P}=R_{\exp(t\xi)}, \end{align*} where $\exp: \mathfrak{g} \to G$ is the Lie group exponential map. The connection form satisfies the principal equivariance identity \begin{align*} R_g^*\omega=\operatorname{Ad}_{g^{-1}}\omega. \end{align*} Here $\operatorname{Ad}: G \to GL(\mathfrak{g})$ is the adjoint representation. Therefore \begin{align*} (\Phi_t^{\xi_P})^*\omega = \operatorname{Ad}_{\exp(-t\xi)}\omega. \end{align*} Differentiating at $t=0$ gives \begin{align*} \mathcal{L}_{\xi_P}\omega = -\operatorname{ad}_{\xi}\circ \omega, \end{align*} where $\operatorname{ad}_{\xi}: \mathfrak{g} \to \mathfrak{g}$ is the linear map $\eta \mapsto [\xi,\eta]_{\mathfrak{g}}$. Now evaluate this one-form identity on the horizontal vector field $X$. The Lie derivative of a one-form satisfies \begin{align*} (\mathcal{L}_{\xi_P}\omega)(X)=\xi_P(\omega(X))-\omega([\xi_P,X]). \end{align*} Since $\omega(X)=0$, the left side also equals $-\omega([\xi_P,X])$. The infinitesimal equivariance identity gives \begin{align*} -\omega([\xi_P,X]) = -[\xi,\omega(X)]_{\mathfrak{g}} = 0. \end{align*} Hence $\omega([\xi_P,X])=0$. Since $[X,\xi_P]=-[\xi_P,X]$, we also have \begin{align*} \omega([X,\xi_P])=0. \end{align*} Substituting into the exterior derivative formula gives \begin{align*} d\omega(X,\xi_P)=0. \end{align*} Combining the three computations, \begin{align*} \left(d\omega+\frac{1}{2}[\omega,\omega]\right)(X,\xi_P)=0=\Omega(X,\xi_P). \end{align*} [/guided] [/step] [step:Compute the vertical-vertical terms on each fibre using left-invariant vector fields] Let $\xi,\eta \in \mathfrak{g}$. We compare both sides on $(\xi_P,\eta_P)$. Since both arguments are vertical, their horizontal components are zero, and hence \begin{align*} \Omega(\xi_P,\eta_P)=0. \end{align*} Define the fibre parametrisation $\iota_p: G \to P$ by $\iota_p(g)=pg$. This map identifies $G$ with the fibre $\pi^{-1}(\pi(p))$. Define the pulled-back one-form $\theta := \iota_p^*\omega \in \Omega^1(G;\mathfrak{g})$. For $\zeta \in \mathfrak{g}$, let $\zeta^L \in \mathfrak{X}(G)$ denote the left-invariant vector field $\zeta^L_g=(dL_g)_e\zeta$, where $L_g: G \to G$ is left translation by $g$. Since $(d\iota_p)_g(\zeta^L_g)=\zeta_P(pg)$, the connection-form normalization gives \begin{align*} \theta(\zeta^L)=\zeta. \end{align*} Apply the exterior derivative formula to the left-invariant vector fields $\xi^L$ and $\eta^L$. The functions $\theta(\xi^L)$ and $\theta(\eta^L)$ are constant with values $\xi$ and $\eta$, and the Lie bracket of left-invariant vector fields satisfies $[\xi^L,\eta^L]=[\xi,\eta]_{\mathfrak{g}}^L$. Hence \begin{align*} d\theta(\xi^L,\eta^L) = -\theta([\xi^L,\eta^L]) = -[\xi,\eta]_{\mathfrak{g}}. \end{align*} Also, \begin{align*} \frac{1}{2}[\theta,\theta](\xi^L,\eta^L) = [\theta(\xi^L),\theta(\eta^L)]_{\mathfrak{g}} = [\xi,\eta]_{\mathfrak{g}}. \end{align*} Therefore \begin{align*} \left(d\theta+\frac{1}{2}[\theta,\theta]\right)(\xi^L,\eta^L)=0. \end{align*} Pulling this equality through $\iota_p$ gives \begin{align*} \left(d\omega+\frac{1}{2}[\omega,\omega]\right)(\xi_P,\eta_P)=0. \end{align*} Thus the desired equality holds on vertical-vertical pairs. [/step] [step:Conclude by bilinearity] At the fixed point $p \in P$, every tangent vector is a sum of a horizontal vector and a fundamental vertical vector. We have shown that \begin{align*} \Omega_p = \left(d\omega+\frac{1}{2}[\omega,\omega]\right)_p \end{align*} on horizontal-horizontal, horizontal-vertical, and vertical-vertical pairs. Since both sides are alternating bilinear maps \begin{align*} T_pP \times T_pP \to \mathfrak{g}, \end{align*} the equality holds on all pairs of tangent vectors at $p$. As $p \in P$ was arbitrary, \begin{align*} \Omega = d\omega+\frac{1}{2}[\omega,\omega] \end{align*} on all of $P$. [/step]