Curvature of $K$-invariant connection (principal bundles) Here is a proposition from Kobayashi & Nomizu's Foundations of Differential Geometry. 
I don't understand how they obtain the final line of the proof. 
They write:
\begin{align}
2\Omega(\tilde{X},\tilde{Y})&=\tilde{X}(\omega(\tilde{Y}))-\tilde{Y}(\omega(\tilde{X}))-\omega([\tilde{X},\tilde{Y}])+[\omega(\tilde{X}),\omega(\tilde{Y})].
\end{align}
They then claim that
\begin{align*}
\tilde{X}(\omega(\tilde{Y}))&=\omega([\tilde{X},\tilde{Y}])\\
\tilde{Y}(\omega(\tilde{X}))&=\omega([\tilde{Y},\tilde{X}])
\end{align*}
and,
\begin{align*}
\omega_{u_0}([\tilde{X},\tilde{Y}])&=\Lambda([X,Y]).
\end{align*}
It is also true by definition that 
\begin{align*}
[\omega(\tilde{X}),\omega(\tilde{Y})]=[\Lambda(X),\Lambda(Y)].
\end{align*}
So then, I believe the first equation, when evaluated at $u_0$ should be
\begin{align*}
2\Omega_{u_0}&=\omega([\tilde{X},\tilde{Y}])-\omega([\tilde{Y},\tilde{X}])-\Lambda([X,Y])+[\Lambda(X),\Lambda(Y)]\\
&=2\omega([\tilde{X},\tilde{Y}])+[\Lambda(X),\Lambda(Y)]-\Lambda([X,Y]).
\end{align*}
Either I have made a mistake, or  $2\omega([\tilde{X},\tilde{Y}])=0$ and I cannot see why this would be the case. 
Any help would be much appreciated.
 A: I have been stuck on this for a long time as well, but apparently the proof contains an error. (See errata of the 1996 edition.) The step $\omega_{u_0}([\bar{X},\bar{Y}]) = \lambda([X,Y])$ is wrong. We should note that $[\bar{X},\bar{Y}] = -\bar{[X,Y]}$, which leads to $$\omega_{u_0}([\bar{X},\bar{Y}]) = \omega_{u_0}(-\bar{[X,Y]}) = -\lambda([X,Y])$$
This correction should solve your problem.
The equality $[\bar{X},\bar{Y}] = -\bar{[X,Y]}$ follows from the fact that we have a left group action on our principal fibre bundle by the group $K$. Take a look at proposition 4.1, chapter I and switch right action for left action. This will change the Lie algebra homomorphism $\sigma$ into an anti-homomorphism.
If you do not have proposition 4.1 with you, I will try to explain it in short. The vector field $\bar{X}$ is induced by $X$ from $(\sigma_u)_*:k \rightarrow \chi(P)$, with $k$ the Lie algebra of $K$ and $\sigma_u: k \mapsto ku$. Let $x_t = \mathrm{exp}(t X)$, then $$[\bar{X},\bar{Y}] = \frac{d}{dt}_{t=0} (L_{x_t^{-1}})_* \bar{Y}_{x_t}$$
Next, we note that $$(L_{x_t^{-1}})_* \bar{Y}_{x_t} = (L_{x_t^{-1}})_* (\sigma_{x_t u})_* Y = \frac{d}{ds}_{s=0} x_t^{-1} \mathrm{exp}(s Y) x_t u = (\sigma_u)_* (\mathrm{Ad}(x_t^{-1}) Y)$$
This leads us to the equation $$\frac{d}{dt}_{t=0} (\sigma_u)_* (\mathrm{Ad}(x_t^{-1}) Y) = (\sigma_u)_* (-[X,Y]) = -\bar{[X,Y]}$$
