Principal curvature, Cartan structure equation, and Bianchi identity

For a fundamental field $Z_P$, $\omega (Z_P)=Z$ is constant. The exterior-derivative formula therefore gives $d\omega (Z_P,Y)=Z_P(\omega (Y))-\omega ([Z_P,Y])=({\mathcal{L}}_{Z_P}\omega )(Y)$ by [F9]. Differentiating $R_{\exp (tZ)}^{\ast }\omega ={\mathrm{Ad}}_{\exp (-tZ)}\omega$ at $t=0$, [F5] yields ${\mathcal{L}}_{Z_P}\omega =-[Z,\omega ]$. The shuffle convention gives $\frac{1}{2}[\omega \wedge \omega ](Z_P,Y)=[Z,\omega (Y)]$, so ${\mathrm{\Omega }}_{\omega }(Z_P,Y)=0$. Every vertical vector is fundamental by the orbit-differential isomorphism, hence ${\mathrm{\Omega }}_{\omega }$ is horizontal.

Pullback commutes with $d$, and the fixed linear map ${\mathrm{Ad}}_{g^{-1}}$ preserves the Lie bracket by [F4]. Consequently, for fixed $g\in G$, $R_g^{\ast }{\mathrm{\Omega }}_{\omega }=d({\mathrm{Ad}}_{g^{-1}}\omega )+\frac{1}{2}[{\mathrm{Ad}}_{g^{-1}}\omega \wedge {\mathrm{Ad}}_{g^{-1}}\omega ]={\mathrm{Ad}}_{g^{-1}}{\mathrm{\Omega }}_{\omega }$. This is the claimed equivariance.

On an overlap, differentiate $s_j(x)=s_i(x)g_{ij}(x)$ along $X\in T_{x}M$. The derivative is the sum of $d(R_{g_{ij}(x)})d{s}_i(X)$ and the fundamental vertical vector at $s_j(x)$ generated by ${\theta }_L(g_{ij})(X)$, since $g_{ij}(x{)}^{-1}{g}_{ij}(x(t))$ has initial tangent ${\theta }_L(g_{ij})(X)$ at the identity. Applying $\omega$, its right equivariance and reproduction property give $A_j(X)={\mathrm{Ad}}_{g_{ij}^{-1}}{A}_i(X)+{\theta }_L(g_{ij})(X)$.

By [F7] and the graded-derivation rule in [F3], $d{\mathrm{\Omega }}_{\omega }=\frac{1}{2}d[\omega \wedge \omega ]=[d\omega \wedge \omega ]$, because graded skew-symmetry gives $[\omega \wedge d\omega ]=-[d\omega \wedge \omega ]$. Hence $D_{\omega }{\mathrm{\Omega }}_{\omega }=d{\mathrm{\Omega }}_{\omega }+[\omega \wedge {\mathrm{\Omega }}_{\omega }]=\frac{1}{2}[\omega \wedge [\omega \wedge \omega ]]=0$, the last equality being graded Jacobi from [F3] and the Jacobi identity in [F1]. Pulling back by each $s_i$ gives $D_{A_i}{F}_i=d{F}_i+[A_i\wedge F_i]=0$.

For horizontal vector fields $X,Y$, $\omega (X)=\omega (Y)=0$, so ${\mathrm{\Omega }}_{\omega }(X,Y)=d\omega (X,Y)=-\omega ([X,Y])$. Meanwhile $\omega (K_H(X,Y))=-\omega (v([X,Y]))=-\omega ([X,Y])$, since $\omega$ vanishes on the horizontal component. Both ${\mathrm{\Omega }}_{\omega }$ and $K_H$ depend only on the horizontal components of their inputs, so the equality extends to all $X,Y$: ${\mathrm{\Omega }}_{\omega }=\omega \circ K_H$. This fixes the sign relative to the imported vertical-valued curvature.

By [F7], pullback along $s_i$ commutes with the componentwise exterior derivative; the bracket is pointwise and pullback preserves its alternating arguments. Thus $F_i=s_i^{\ast }{\mathrm{\Omega }}_{\omega }=d{A}_i+\frac{1}{2}[A_i\wedge A_i]$. On the overlap, the same differentiated-section decomposition as in step 1.3 has a vertical term, which vanishes when inserted into ${\mathrm{\Omega }}_{\omega }$ by step 1.1. Its remaining term is transformed by the equivariance in step 1.2, yielding $F_j={\mathrm{Ad}}_{g_{ij}^{-1}}{F}_i$.

The statements include the zero-dimensional or abelian Lie-algebra cases: then the corresponding brackets vanish, and each displayed identity reduces to the ordinary exterior-derivative identity. A zero-dimensional base has no nonzero 2-forms, and the formulas remain valid. Only the declared ${\mathrm{A}\mathrm{C}}_{\omega }$ is inherited from [F1] and [F2]; no global section or full choice principle is assumed.