Principal connection & curvature Let $(P, \pi, B)$ be a principal $G$-bundle over $B$ and $\omega$ a principal connection. Then the curvature is defined as 
$$ \Omega_\omega = d \omega + \frac{1}{2} \omega \wedge \omega$$
With the d being the standard differentiation. Let $p \in P$ and $\alpha, \beta \in T_p P $ be horizontal vectors (ie $\omega(v)=0$). Let $\hat{\alpha},\hat{\beta}$ be horizontal vector fields extending $\alpha, \beta$ near $p$. Then 
$$ \Omega_\omega(\alpha,\beta)_p = \omega([\hat{\alpha}, \hat{\beta} ])_p $$
How do we show this? I'm pretty sure I'm getting tripped up on something easy. 
 A: You mentioned that you have found an answer in the above comment, but since you never wrote it up, I hope you don't mind if I have a shot!
You have a manifold $B$, a Lie group $G$ and a principal $G$-bundle $\pi : P \to B$. We can define a connection on $P$ using a 1-form $ \omega \in \Omega^1(P) \otimes \mathfrak{g}$ satisfying
$$ R_g^* \omega = {\rm Ad}(g^{-1}) \omega $$
$$ \omega(e \exp(tX)) = X$$
Firstly, there is something wrong with your formula for the curvature. You are adding sections of different vector bundles:
$$ d \omega \in \Omega^2(P) \otimes \mathfrak{g} $$
$$ \omega \wedge \omega \in \Omega^2(P) \otimes \mathfrak{g}^{\otimes 2}$$
I think that you meant to write the formula
$$ \Omega = d\omega + \frac{1}{2} [\omega,\omega] $$
where $[\omega,\omega]$ is obtained from $\omega \wedge \omega$ by applying the Lie bracket $[-,-] : \mathfrak{g}^{\otimes 2} \to \mathfrak{g}$. We have the following formulas:
$$ [\omega,\omega](X,Y) = 2 [\omega(X),\omega(Y)]$$
$$ dw(X,Y) = X(\omega(Y)) - Y(\omega(X)) - \omega([X,Y])$$
The first is just unraveling the definitions. The second is one of the standard Cartan formulas, but for vector valued forms instead of real valued forms. When $X$ and $Y$ are horizontal, they are killed the $ \omega$, so we get that
$$ \Omega(X,Y) = - \omega([X,Y])$$
Im guessing that this is what you had in mind, but it is good to have questions like this answered on the site!
