1. Proof of the proposition 6.3:

(b) First we prove that $\omega((R_a)_* X) = ad(a^{-1})\omega(X)$ where $a \in G \times H$ and $X \in (P+Q)_u$ with $u \in P + Q$. We assume that $a = a_G \times a_H$ where $a_G \in G$ , $a_H \in H$ and then $h_G (a) = a_G$ and $h_H (a) = a_H$ . Also we assume that $h_P (X) = X_P$ and $h_Q(X) = X _Q$. and then $\omega_P((R_{a_G})_* X_P) = ad(a_{G}^{-1})\omega_P(X_P)$ and $\omega_Q((R_{a_H})_* X_Q) = ad(a_{H}^{-1})\omega_Q(X_Q)$ . Then we have that

$$\omega((R_a)_* X) = (h^*_P \omega_P + h^*_Q \omega_Q)((R_a)_* X)$$

$$\omega((R_a)_* X) = (h^*_P \omega_P)((R_a)_* X) + (h^*_Q \omega_Q)((R_a)_* X)$$

$$\omega((R_a)_* X) = \omega_P(h_P((R_a)_* X))+ \omega_Q(h_Q((R_a)_* X))$$

$$\omega((R_a)_* X) = \omega_P((R_{h_G(a)})_* h_P(X))+ \omega_Q((R_{h_H(a)})_* h_Q(X))$$

$$\omega((R_a)_* X) = \omega_P((R_{h_G(a_G \times a_H)})_* h_P(X))+ \omega_Q((R_{h_H(a_G \times a_H)})_* h_Q(X))$$

$$\omega((R_a)_* X) = \omega_P((R_{a_G })_* h_P(X))+ \omega_Q((R_{a_H})_* h_Q(X))$$

$$\omega((R_a)_* X) = \omega_P((R_{a_G })_* X_P)+ \omega_Q((R_{a_H})_* X_Q)$$

$$\omega((R_a)_* X) = ad(a_{G}^{-1})\omega_P(X_P) +ad(a_{H}^{-1})\omega_Q(X_Q) $$

From other side we have that

$$ad(a^{-1})\omega(X)= a\omega(X)a^{-1}$$

$$ad(a^{-1})\omega(X)= (a_G \times a_H)\omega(X)(a_G \times a_H)^{-1}$$

$$ad(a^{-1})\omega(X)= (a_G \times a_H)\omega(X)(a_G^{-1} \times a_H^{-1})$$

$$ad(a^{-1})\omega(X)= (a_G \times a_H)[(h^*_P \omega_P + h^*_Q \omega_Q)(X)](a_G^{-1} \times a_H^{-1})$$

$$ad(a^{-1})\omega(X)= (a_G \times a_H)[(h^*_P \omega_P )(X)+ (h^*_Q \omega_Q)(X)](a_G^{-1} \times a_H^{-1})$$

$$ad(a^{-1})\omega(X)= (a_G \times a_H)[ \omega_P (h_P(X))+ \omega_Q(h_Q (X))](a_G^{-1} \times a_H^{-1})$$

$$ad(a^{-1})\omega(X)= (a_G \times a_H)[ \omega_P (X_P)+ \omega_Q( X_Q)](a_G^{-1} \times a_H^{-1})$$

$$ad(a^{-1})\omega(X)= (a_G \times a_H) \omega_P (X_P) (a_G^{-1} \times a_H^{-1}) + (a_G \times a_H)\omega_Q( X_Q)(a_G^{-1} \times a_H^{-1})$$

$$ad(a^{-1})\omega(X)= (a_G \omega_P (X_P)a_G^{-1}) ( a_H a_H^{-1} ) + (a_G a_G^{-1})(a_H\omega_Q( X_Q)a_H^{-1} \ )$$

$$ad(a^{-1})\omega(X)= (a_G \omega_P (X_P)a_G^{-1}) ( e_H ) + (e_G)(a_H\omega_Q( X_Q)a_H^{-1} \ )$$

$$ad(a^{-1})\omega(X)= a_G \omega_P (X_P)a_G^{-1} + a_H\omega_Q( X_Q)a_H^{-1} $$

$$ad(a^{-1})\omega(X)= ad(a_{G}^{-1})\omega_P(X_P) +ad(a_{H}^{-1})\omega_Q(X_Q) $$.

Second, we prove that $\omega(A^*) = A$ as follows.

$$\omega(A^*)= (h^*_P \omega_P + h^*_Q \omega_Q)(A^*)$$

$$\omega(A^*)= (h^*_P \omega_P)(A^*) + (h^*_Q \omega_Q)(A^*)$$

$$\omega(A^*)= \omega_P(h_P(A^*)) + \omega_Q( h_Q(A^*))$$

$$\omega(A^*)= \omega_P(A^*_P) + \omega_Q( A^*_Q)$$

$$\omega(A^*)= A_P+ A_Q$$

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

Finally, we prove that $\Omega = h^*_P \Omega_P + h^*_Q \Omega_Q$.

$$\Omega = d\omega+\frac{1}{2} [\omega,\omega]$$

$$\Omega = d(h^*_P \omega_P + h^*_Q \omega_Q)+\frac{1}{2} [(h^*_P \omega_P + h^*_Q \omega_Q),(h^*_P \omega_P + h^*_Q \omega_Q)]$$

$$\Omega = d(h^*_P \omega_P) + d(h^*_Q \omega_Q)+\frac{1}{2} [h^*_P \omega_P , h^*_P \omega_P ] +\frac{1}{2} [h^*_P \omega_P , h^*_Q \omega_Q] +$$

$$\frac{1}{2} [h^*_Q \omega_Q , h^*_P \omega_P] +\frac{1}{2} [h^*_Q \omega_Q , h^*_Q \omega_Q ] $$

$$\Omega = d(h^*_P \omega_P) + d(h^*_Q \omega_Q)+\frac{1}{2} [h^*_P \omega_P , h^*_P \omega_P ] +0 +0 +\frac{1}{2} [h^*_Q \omega_Q , h^*_Q \omega_Q ] $$

$$\Omega = d(h^*_P \omega_P) + d(h^*_Q \omega_Q)+\frac{1}{2} [h^*_P \omega_P , h^*_P \omega_P ] +\frac{1}{2} [h^*_Q \omega_Q , h^*_Q \omega_Q ] $$

$$\Omega = d(h^*_P \omega_P) + d(h^*_Q \omega_Q)+\frac{1}{2}h^*_P ( [ \omega_P , \omega_P ] ) +\frac{1}{2}h^*_Q( [ \omega_Q , \omega_Q ]) $$

$$\Omega =h^*_P (d \omega_P) + h^*_Q (d\omega_Q)+\frac{1}{2}h^*_P ( [ \omega_P , \omega_P ] ) +\frac{1}{2}h^*_Q( [ \omega_Q , \omega_Q ]) $$

$$\Omega =h^*_P (d \omega_P+\frac{1}{2} [ \omega_P , \omega_P ]) + h^*_Q (d\omega_Q+ +\frac{1}{2} [ \omega_Q , \omega_Q ]) $$

$$\Omega = h^*_P \Omega_P + h^*_Q \Omega_Q$$.