1. Proving the Wang's theorem: (adapted from Kobayashi-Nomizu with step by step deductions)

First we prove that $ \lambda : J \rightarrow G$ is a homomorphism.

Given that $j_1 \in J$, $j_2 \in J$ and $ju_0 = u_0 \lambda(j)$ we have

$$u_0 \lambda(j_1j_2) = (j_1j_2)u_0 = j_1(j_2u_0)=j_1(u_0 \lambda(j_2))=(j_1u_0)\lambda(j_2)= (u_0\lambda(j_1))\lambda(j_2)$$

$$u_0 \lambda(j_1j_2) = u_0(\lambda(j_1)\lambda(j_2))$$

$$ \lambda(j_1j_2) = \lambda(j_1)\lambda(j_2)$$.

From $ju_0 = u_0 \lambda(j)$ and $eu_0=u_0e$ we deduce that $\lambda(e)=e$.

With $j_1 = j$ and $j_2 = j^{-1}$ we have that

$$\lambda(jj^{-1}) = \lambda(j)\lambda(j^{-1})$$

$$\lambda(e) = \lambda(j)\lambda(j^{-1})$$

$$e= \lambda(j)\lambda(j^{-1})$$

$$\lambda(j)^{-1}= \lambda(j^{-1})$$

We next prove that $\omega(X^*)$ is independent of the choice of $k$ and $a$. Let

$$u_0 = k u a = k_1 u a_1$$

where $k \in K$, $k_1 \in K$, $a \in G$ and $a_1 \in G$ . Then we have that $k^{-1}u_0a^{-1} = u$ and $k_1^{-1}u_0a_1^{-1} = u$; it is to say

$$k^{-1}u_0a^{-1} = k_1^{-1}u_0a_1^{-1} $$

which is equivalent to

$$k_1k^{-1}u_0 = u_0a_1^{-1}a$$.

Comparing the last equation with $ju_0 = u_0 \lambda (j)$ where $j \in J$ we deduce that

$$j = k_1k^{-1} \in J$$

and

$$\lambda (j) = a_1^{-1}a$$.

From these last equations we obtain that

$$k_1 = jk$$

and

$$a_1=a\lambda (j)^{-1}=a \lambda (j^{-1}) $$.

Using all these we have

$$k_1 \circ R_{a_1}X^* = (jk) \circ R_{a \lambda (j^{-1}) }X^* $$

$$k_1 \circ R_{a_1}X^* = (jk) \circ R_{ \lambda (j^{-1}) }( R_{a}X^* ) $$

$$k_1 \circ R_{a_1}X^* = j \circ R_{ \lambda (j^{-1}) }( k \circ R_{a}X^* ) $$

$$k_1 \circ R_{a_1}X^* = j \circ R_{ \lambda (j^{-1}) }( \hat{X}_{u_0} +A_{u_0}^*) $$

$$k_1 \circ R_{a_1}X^* = j \circ R_{ \lambda (j^{-1}) }( \hat{X}_{u_0} )+j \circ R_{ \lambda (j^{-1}) }(A_{u_0}^*) $$

$$k_1 \circ R_{a_1}X^* = j ( \hat{X}_{u_0 \lambda (j^{-1})} )+j \circ R_{ \lambda (j^{-1}) }(A_{u_0}^*) $$

$$k_1 \circ R_{a_1}X^* = j ( \hat{X}_{u_0 \lambda (j^{-1})} )+ R_{ \lambda (j^{-1}) }(jA_{u_0}^*) $$

$$k_1 \circ R_{a_1}X^* = j ( \hat{X}_{u_0 \lambda (j^{-1})} )+ R_{ \lambda (j^{-1}) }(A_{ju_0}^*) $$

$$k_1 \circ R_{a_1}X^* = j ( \hat{X}_{u_0 \lambda (j^{-1})} )+ R_{ \lambda (j^{-1}) }(A_{u_0\lambda(j)}^*) $$

$$k_1 \circ R_{a_1}X^* = \hat{Z}_{u_0}+ C^*_{u_0}$$

where

$$Z = ad(j)(X)$$

and

$$C = ad(\lambda (j) )(A)$$

Then we have that

$$ad(a_1)(\Lambda (Z) +C)= ad(a_1)[\Lambda (ad(j)(X) )+ad(\lambda (j) )(A)] $$

$$ad(a_1)(\Lambda (Z) +C)= ad(a_1)[ad(\lambda (j))(\Lambda(X))+ad(\lambda (j) )(A)] $$

$$ad(a_1)(\Lambda (Z) +C)= ad(a_1)[ad(\lambda (j))(\Lambda(X)+A)] $$

$$ad(a_1)(\Lambda (Z) +C)= ad(a_1)[ad(a_1^{-1}a)(\Lambda(X)+A)] $$

$$ad(a_1)(\Lambda (Z) +C)= a_1[ad(a_1^{-1}a)(\Lambda(X)+A)]a_1^{-1} $$

$$ad(a_1)(\Lambda (Z) +C)= a_1[(a_1^{-1}a)(\Lambda(X)+A)(a_1^{-1}a)^{-1}]a_1^{-1} $$

$$ad(a_1)(\Lambda (Z) +C)= a_1[(a_1^{-1}a)(\Lambda(X)+A)(a^{-1}a_1)]a_1^{-1} $$

$$ad(a_1)(\Lambda (Z) +C)= (a_1a_1^{-1})[a(\Lambda(X)+A)a^{-1}](a_1a_1^{-1} )$$

$$ad(a_1)(\Lambda (Z) +C)= e[a(\Lambda(X)+A)a^{-1}]e$$

$$ad(a_1)(\Lambda (Z) +C)= a(\Lambda(X)+A)a^{-1}$$

$$ad(a_1)(\Lambda (Z) +C)= ad(a)(\Lambda(X)+A)$$.

This proves that $\omega(X^*)$ depends only on $X^*$.

Now, we prove that $\omega$ is a connection one-form. Let $X^* \in T_u(P)$ and

$u_0 = kua$ . Let $b$ be an arbitrary element of the structure group $G$. We write

$$Y^* = R_b X^* \in T_v(P)$$

where $v = ub$ and for hence $u_0 =kua=ku(bb^{-1})a=k(ub)(b^{-1}a)= kv(b^{-1}a)$. Then we have that

$$k \circ R_{b^{-1}a}Y^* = k \circ R_{b^{-1}a}R_b X^* = k \circ R_{bb^{-1}a}X^* = k \circ R_aX^* =(\hat{X}_{u_0}+A^*_{u_0})$$.

From $k \circ R_aX^* =\hat{X}_{u_0}+A^*_{u_0}$ we write the definition of $\omega$ as

$$\omega(X^*)=ad(a)(\Lambda (X) +A)$$

then from $k \circ R_{b^{-1}a}Y^* =\hat{X}_{u_0}+A^*_{u_0}$ we write that

$$\omega(Y^*)=ad(b^{-1}a)(\Lambda (X) +A).$$

Then we have

$$\omega(R_b X^*)=\omega(Y^*)=ad(b^{-1}a)(\Lambda (X) +A)$$

$$\omega(R_b X^*)= (b^{-1}a)(\Lambda (X) +A)(b^{-1}a)^{-1}$$

$$\omega(R_b X^*)= (b^{-1}a)(\Lambda (X) +A)(a^{-1}b)$$

$$\omega(R_b X^*)= b^{-1}[a(\Lambda (X) +A)a^{-1}]b$$

$$\omega(R_b X^*)= ad(b^{-1})[a(\Lambda (X) +A)a^{-1}]$$

$$\omega(R_b X^*)= ad(b^{-1})[ad(a)(\Lambda (X) +A)]$$

$$\omega(R_b X^*)= ad(b^{-1})\omega(X^*).$$

Finally, we prove that $\omega$ is invariant by $K$. Let $X^* \in T_u(P)$ and $k_1$ be an arbitrary element of $K$. Then $k_1X^* \in T_{k_1u},(P)$ and we have that

$$\omega(k_1 X^*)=k_1(ad(a)(\Lambda (X) +A))$$

$$\omega(k_1 X^*)=ad(ak_1)(k_1(\Lambda (X) )+k_1(A))$$

$$\omega(k_1 X^*)=ad(ak_1)[ad(k_1^{-1})(\Lambda (X) )+ad(k_1^{-1})(A)]$$

$$\omega(k_1 X^*)=ad(ak_1)[ad(k_1^{-1})(\Lambda (X) )]+ad(ak_1)[ad(k_1^{-1})(A)]$$

$$\omega(k_1 X^*)=(ak_1)[ad(k_1^{-1})(\Lambda (X) )](ak_1)^{-1}+(ak_1)[ad(k_1^{-1})(A)](ak_1)^{-1}$$

$$\omega(k_1 X^*)=(ak_1)[k_1^{-1}\Lambda (X) k_1](k_1^{-1}a^{-1})+(ak_1)[k_1^{-1}Ak_1](k_1^{-1}a^{-1})$$

$$\omega(k_1 X^*)=a(k_1k_1^{-1})\Lambda (X)(k_1k_1^{-1})a^{-1}+a(k_1k_1^{-1})A(k_1k_1^{-1})a^{-1}$$

$$\omega(k_1 X^*)=ae\Lambda (X)ea^{-1}+aeAea^{-1}$$

$$\omega(k_1 X^*)=a\Lambda (X)a^{-1}+aAa^{-1}$$

$$\omega(k_1 X^*)=a[\Lambda (X)+A]a^{-1}$$

$$\omega(k_1 X^*)=ad(a)(\Lambda (X)+A)$$

$$\omega(k_1 X^*)=\omega( X^*)$$