1. Proof of the proposition 6.3:
(b) First we prove that ω((Ra)∗X)=ad(a−1)ω(X) where a∈G×H and X∈(P+Q)u with u∈P+Q. We assume that a=aG×aH where aG∈G , aH∈H and then hG(a)=aG and hH(a)=aH . Also we assume that hP(X)=XP and hQ(X)=XQ. and then ωP((RaG)∗XP)=ad(a−1G)ωP(XP) and ωQ((RaH)∗XQ)=ad(a−1H)ωQ(XQ) . Then we have that
ω((Ra)∗X)=(h∗PωP+h∗QωQ)((Ra)∗X)
ω((Ra)∗X)=(h∗PωP)((Ra)∗X)+(h∗QωQ)((Ra)∗X)
ω((Ra)∗X)=ωP(hP((Ra)∗X))+ωQ(hQ((Ra)∗X))
ω((Ra)∗X)=ωP((RhG(a))∗hP(X))+ωQ((RhH(a))∗hQ(X))
ω((Ra)∗X)=ωP((RhG(aG×aH))∗hP(X))+ωQ((RhH(aG×aH))∗hQ(X))
ω((Ra)∗X)=ωP((RaG)∗hP(X))+ωQ((RaH)∗hQ(X))
ω((Ra)∗X)=ωP((RaG)∗XP)+ωQ((RaH)∗XQ)
ω((Ra)∗X)=ad(a−1G)ωP(XP)+ad(a−1H)ωQ(XQ)
From other side we have that
ad(a−1)ω(X)=aω(X)a−1
ad(a−1)ω(X)=(aG×aH)ω(X)(aG×aH)−1
ad(a−1)ω(X)=(aG×aH)ω(X)(a−1G×a−1H)
ad(a−1)ω(X)=(aG×aH)[(h∗PωP+h∗QωQ)(X)](a−1G×a−1H)
ad(a−1)ω(X)=(aG×aH)[(h∗PωP)(X)+(h∗QωQ)(X)](a−1G×a−1H)
ad(a−1)ω(X)=(aG×aH)[ωP(hP(X))+ωQ(hQ(X))](a−1G×a−1H)
ad(a−1)ω(X)=(aG×aH)[ωP(XP)+ωQ(XQ)](a−1G×a−1H)
ad(a−1)ω(X)=(aG×aH)ωP(XP)(a−1G×a−1H)+(aG×aH)ωQ(XQ)(a−1G×a−1H)
ad(a−1)ω(X)=(aGωP(XP)a−1G)(aHa−1H)+(aGa−1G)(aHωQ(XQ)a−1H )
ad(a−1)ω(X)=(aGωP(XP)a−1G)(eH)+(eG)(aHωQ(XQ)a−1H )
ad(a−1)ω(X)=aGωP(XP)a−1G+aHωQ(XQ)a−1H
ad(a−1)ω(X)=ad(a−1G)ωP(XP)+ad(a−1H)ωQ(XQ)
.
Second, we prove that ω(A∗)=A as follows.
ω(A∗)=(h∗PωP+h∗QωQ)(A∗)
ω(A∗)=(h∗PωP)(A∗)+(h∗QωQ)(A∗)
ω(A∗)=ωP(hP(A∗))+ωQ(hQ(A∗))
ω(A∗)=ωP(A∗P)+ωQ(A∗Q)
ω(A∗)=AP+AQ
ω(A∗)=A
Finally, we prove that Ω=h∗PΩP+h∗QΩQ.
Ω=dω+12[ω,ω]
Ω=d(h∗PωP+h∗QωQ)+12[(h∗PωP+h∗QωQ),(h∗PωP+h∗QωQ)]
Ω=d(h∗PωP)+d(h∗QωQ)+12[h∗PωP,h∗PωP]+12[h∗PωP,h∗QωQ]+
12[h∗QωQ,h∗PωP]+12[h∗QωQ,h∗QωQ]
Ω=d(h∗PωP)+d(h∗QωQ)+12[h∗PωP,h∗PωP]+0+0+12[h∗QωQ,h∗QωQ]
Ω=d(h∗PωP)+d(h∗QωQ)+12[h∗PωP,h∗PωP]+12[h∗QωQ,h∗QωQ]
Ω=d(h∗PωP)+d(h∗QωQ)+12h∗P([ωP,ωP])+12h∗Q([ωQ,ωQ])
Ω=h∗P(dωP)+h∗Q(dωQ)+12h∗P([ωP,ωP])+12h∗Q([ωQ,ωQ])
Ω=h∗P(dωP+12[ωP,ωP])+h∗Q(dωQ++12[ωQ,ωQ])
Ω=h∗PΩP+h∗QΩQ
.