Let $L=\mathbb{R}^2\times U(1)$ be the trivial $U(1)$-bundle over $\mathbb{R}^2$. Define a connection $\nabla=d+A$ where $A=fdx+gdy$ is an $\mathbb{R}$ valued $1$-form on $L$. That is, $\nabla$ gives a distribution $\mathcal{H}$ on $L$ - the horizontal distribution.

The distribution $\mathcal{H}$ is obtained as the graph of $-A$ as a linear map from $\mathbb{R}^2\rightarrow\mathbb{R}$. A horizontal lift $\tilde{X}$ of a vector field $X$ on $\mathbb{R}^2$ is given by $\tilde{X}=(X,-A(X))$.

Let $\alpha$ be the projection onto the vertical direction on $L$ i.e. $\ker(\alpha)=\mathcal{H}$, and define the curvature $2$-form $\Omega_{\nabla}$ of the connection $\nabla$ by

$$\Omega_{\nabla}(X,Y)=\alpha([\tilde{X},\tilde{Y}])$$

The following is expected to be true
$$\Omega_{\nabla}=-dA?$$

Here is my confusion:

Let $z$ be the local vertical coordinate, $X=\partial_x$ and $Y=\partial_y$. Then $\tilde{X}=-f\partial_z+\partial_x$ and $\tilde{Y}=-g\partial_z+\partial_y$. And
$$-dA(X,Y)=-X(A(Y)+Y(A(X))+A([X,Y])=-\partial_xg+\partial_yf$$
while
$$[\tilde{X},\tilde{Y}]=(f(\partial_zg)-g(\partial_zf)-\partial_xg+\partial_yf)\partial_z$$
therefore
$$\Omega_{\nabla}(X,Y)=\alpha([\tilde{X},\tilde{Y}])=f(\partial_zg)-g(\partial_zf)-\partial_xg+\partial_yf.$$
Where is the mistake? Thank you.

This post imported from StackExchange Physics at 2016-07-29 20:36 (UTC), posted by SE-user Student