1. Proving the Wang's theorem: (adapted from Kobayashi-Nomizu with step by step deductions)
First we prove that λ:J→G is a homomorphism.
Given that j1∈J, j2∈J and ju0=u0λ(j) we have
u0λ(j1j2)=(j1j2)u0=j1(j2u0)=j1(u0λ(j2))=(j1u0)λ(j2)=(u0λ(j1))λ(j2)
u0λ(j1j2)=u0(λ(j1)λ(j2))
λ(j1j2)=λ(j1)λ(j2)
.
From ju0=u0λ(j) and eu0=u0e we deduce that λ(e)=e.
With j1=j and j2=j−1 we have that
λ(jj−1)=λ(j)λ(j−1)
λ(e)=λ(j)λ(j−1)
e=λ(j)λ(j−1)
λ(j)−1=λ(j−1)
We next prove that ω(X∗) is independent of the choice of k and a. Let
u0=kua=k1ua1
where k∈K, k1∈K, a∈G and a1∈G . Then we have that k−1u0a−1=u and k−11u0a−11=u; it is to say
k−1u0a−1=k−11u0a−11
which is equivalent to
k1k−1u0=u0a−11a
.
Comparing the last equation with ju0=u0λ(j) where j∈J we deduce that
j=k1k−1∈J
and
λ(j)=a−11a
.
From these last equations we obtain that
k1=jk
and
a1=aλ(j)−1=aλ(j−1)
.
Using all these we have
k1∘Ra1X∗=(jk)∘Raλ(j−1)X∗
k1∘Ra1X∗=(jk)∘Rλ(j−1)(RaX∗)
k1∘Ra1X∗=j∘Rλ(j−1)(k∘RaX∗)
k1∘Ra1X∗=j∘Rλ(j−1)(ˆXu0+A∗u0)
k1∘Ra1X∗=j∘Rλ(j−1)(ˆXu0)+j∘Rλ(j−1)(A∗u0)
k1∘Ra1X∗=j(ˆXu0λ(j−1))+j∘Rλ(j−1)(A∗u0)
k1∘Ra1X∗=j(ˆXu0λ(j−1))+Rλ(j−1)(jA∗u0)
k1∘Ra1X∗=j(ˆXu0λ(j−1))+Rλ(j−1)(A∗ju0)
k1∘Ra1X∗=j(ˆXu0λ(j−1))+Rλ(j−1)(A∗u0λ(j))
k1∘Ra1X∗=ˆZu0+C∗u0
where
Z=ad(j)(X)
and
C=ad(λ(j))(A)
Then we have that
ad(a1)(Λ(Z)+C)=ad(a1)[Λ(ad(j)(X))+ad(λ(j))(A)]
ad(a1)(Λ(Z)+C)=ad(a1)[ad(λ(j))(Λ(X))+ad(λ(j))(A)]
ad(a1)(Λ(Z)+C)=ad(a1)[ad(λ(j))(Λ(X)+A)]
ad(a1)(Λ(Z)+C)=ad(a1)[ad(a−11a)(Λ(X)+A)]
ad(a1)(Λ(Z)+C)=a1[ad(a−11a)(Λ(X)+A)]a−11
ad(a1)(Λ(Z)+C)=a1[(a−11a)(Λ(X)+A)(a−11a)−1]a−11
ad(a1)(Λ(Z)+C)=a1[(a−11a)(Λ(X)+A)(a−1a1)]a−11
ad(a1)(Λ(Z)+C)=(a1a−11)[a(Λ(X)+A)a−1](a1a−11)
ad(a1)(Λ(Z)+C)=e[a(Λ(X)+A)a−1]e
ad(a1)(Λ(Z)+C)=a(Λ(X)+A)a−1
ad(a1)(Λ(Z)+C)=ad(a)(Λ(X)+A)
.
This proves that ω(X∗) depends only on X∗.
Now, we prove that ω is a connection one-form. Let X∗∈Tu(P) and
u0=kua . Let b be an arbitrary element of the structure group G. We write
Y∗=RbX∗∈Tv(P)
where v=ub and for hence u0=kua=ku(bb−1)a=k(ub)(b−1a)=kv(b−1a). Then we have that
k∘Rb−1aY∗=k∘Rb−1aRbX∗=k∘Rbb−1aX∗=k∘RaX∗=(ˆXu0+A∗u0)
.
From k∘RaX∗=ˆXu0+A∗u0 we write the definition of ω as
ω(X∗)=ad(a)(Λ(X)+A)
then from k∘Rb−1aY∗=ˆXu0+A∗u0 we write that
ω(Y∗)=ad(b−1a)(Λ(X)+A).
Then we have
ω(RbX∗)=ω(Y∗)=ad(b−1a)(Λ(X)+A)
ω(RbX∗)=(b−1a)(Λ(X)+A)(b−1a)−1
ω(RbX∗)=(b−1a)(Λ(X)+A)(a−1b)
ω(RbX∗)=b−1[a(Λ(X)+A)a−1]b
ω(RbX∗)=ad(b−1)[a(Λ(X)+A)a−1]
ω(RbX∗)=ad(b−1)[ad(a)(Λ(X)+A)]
ω(RbX∗)=ad(b−1)ω(X∗).
Finally, we prove that ω is invariant by K. Let X∗∈Tu(P) and k1 be an arbitrary element of K. Then k1X∗∈Tk1u,(P) and we have that
ω(k1X∗)=k1(ad(a)(Λ(X)+A))
ω(k1X∗)=ad(ak1)(k1(Λ(X))+k1(A))
ω(k1X∗)=ad(ak1)[ad(k−11)(Λ(X))+ad(k−11)(A)]
ω(k1X∗)=ad(ak1)[ad(k−11)(Λ(X))]+ad(ak1)[ad(k−11)(A)]
ω(k1X∗)=(ak1)[ad(k−11)(Λ(X))](ak1)−1+(ak1)[ad(k−11)(A)](ak1)−1
ω(k1X∗)=(ak1)[k−11Λ(X)k1](k−11a−1)+(ak1)[k−11Ak1](k−11a−1)
ω(k1X∗)=a(k1k−11)Λ(X)(k1k−11)a−1+a(k1k−11)A(k1k−11)a−1
ω(k1X∗)=aeΛ(X)ea−1+aeAea−1
ω(k1X∗)=aΛ(X)a−1+aAa−1
ω(k1X∗)=a[Λ(X)+A]a−1
ω(k1X∗)=ad(a)(Λ(X)+A)
ω(k1X∗)=ω(X∗)