Proof of the proposition 6.4.

First we prove that $\omega_Q((R_a)_* X) = ad(a^{-1})\omega_Q(X)$ where $a \in H$ and $X \in T_v(Q)$ with $v \in Q$. Given that $\omega_P$ is a connection one-form in the principal bundle $P(M,G)$; given that $v \in P$ , $X \in T_v(P)$ and $a \in G$; then it is verified that

$$\omega_P((R_a)_* X) = ad(a^{-1})\omega_P(X)$$.

Given that the last expression is valid for all $a \in G$, it is also valid for $a \in H$ because $H$ is a subgroup of $G$. Then for $X \in T_v(Q)$ and $a \in H$; then it is verified that

$$\omega_P((R_a)_* X) = ad(a^{-1})\omega_P(X)$$.

Now, let $\phi$ the $m$-component of $\omega_P$ restricted to the subbundle $Q$, then it is possible to write $\omega_P = \omega_Q + \phi$; and for hence we have that

$$(\omega_Q + \phi)((R_a)_* X) = ad(a^{-1})(\omega_Q + \phi)(X)$$

$$\omega_Q((R_a)_* X) + \phi((R_a)_* X) = ad(a^{-1})(\omega_Q(X) + \phi(X))$$

$$\omega_Q((R_a)_* X) + \phi((R_a)_* X) = a(\omega_Q(X) + \phi(X))a^{-1}$$

$$\omega_Q((R_a)_* X) + \phi((R_a)_* X) = a\omega_Q(X)a^{-1} + a\phi(X)a^{-1}$$

$$\omega_Q((R_a)_* X) + \phi((R_a)_* X) = ad(a^{-1})\omega_Q(X) + ad(a^{-1})\phi(X)$$.

The generators for the subalgebra $h$ are denoted by $ \hat{h}_{\alpha}$ and the generators for $m$ are denoted $ \hat{m}_{\beta}$; then we have that

$$[\omega_Q((R_a)_* X)]^{\alpha}\hat{h}_{\alpha} + [\phi((R_a)_* X) ]^{\beta}\hat{m}_{\beta}=$$

$$ad(a^{-1})([\omega_Q(X)]^{\alpha}\hat{h}_{\alpha} ) +ad(a^{-1})([\phi(X)]^{\beta}\hat{m}_{\beta}) $$

which is rewritten as

$$[\omega_Q((R_a)_* X)]^{\alpha}\hat{h}_{\alpha} + [\phi((R_a)_* X) ]^{\beta}\hat{m}_{\beta}=$$ $$[\omega_Q(X)]^{\alpha}ad(a^{-1})(\hat{h}_{\alpha} ) + [\phi(X)]^{\beta}ad(a^{-1})(\hat{m}_{\beta}) $$

and then we have that

$$[\omega_Q((R_a)_* X)]^{\alpha}\hat{h}_{\alpha} + [\phi((R_a)_* X) ]^{\beta}\hat{m}_{\beta}= [\omega_Q(X)]^{\alpha}E_{\alpha}^{\beta}\hat{h}_{\beta} + [\phi(X)]^{\beta}C_{\beta}^{\gamma}\hat{m}_{\gamma} $$

where $E_{\alpha}^{\beta}$ and $C_{\beta}^{\gamma}$ are structure constants.; and the Einstein summation convention was used.

Then, given that $\hat{h}_{\alpha}$ and $\hat{m}_{\beta}$ are linearly independent, we deduce that

$$[\omega_Q((R_a)_* X)]^{\alpha}\hat{h}_{\alpha} = [\omega_Q(X)]^{\alpha}E_{\alpha}^{\beta}\hat{h}_{\beta}$$

$$\omega_Q((R_a)_* X) = [\omega_Q(X)]^{\alpha}ad(a^{-1})(\hat{h}_{\alpha} )$$

$$\omega_Q((R_a)_* X) = ad(a^{-1})([\omega_Q(X)]^{\alpha}\hat{h}_{\alpha} )$$

$$\omega_Q((R_a)_* X) = ad(a^{-1})\omega_Q(X)$$.

Second, we prove that $\omega_Q(A^*) = A$, where $A\in h$ and $A^*$ is the fundamental vector field corresponding to $A$. Then, given that $\omega_P(A^*)= A$ and $\phi(A^*)=0$ we have that

$$\omega_P = \omega_Q + \phi$$

$$\omega_P(A^*) = \omega_Q( A^*)+ \phi(A^*)$$

$$A= \omega_Q( A^*)+ 0$$

$$A= \omega_Q( A^*)$$