I'm trying to verify some computations in a paper I'm reading and am feeling a little lost. In particular I haven't been able to properly compute curvature of a connection acting on a line bundle. Here's a specific example (modeled on the Dirac monopole):
Consider spherical coordinates related to Euclidean coordinates by $$(t,x,y) = (R\cos \theta, R\cos \psi \sin \theta, R \sin \psi \sin \theta)$$ Take the Hermitian line bundle $L_k$ over $\mathbb{R}^3$ defined by the transition function $g_{\theta 0} = e^{ik\psi}$ on the complement of the $t$-axis from $\theta \neq 0$ to $\theta \neq \pi$. Consider the connection $\bigtriangledown$ on $L_k$ defined by connection matrices $$A_0 = (ik/2)(1+\cos \theta) d\psi$$ $$A_\pi = (ik/2)(-1+\cos \theta)d\psi$$ where the first is defined on $\theta \neq 0$, the second on $\theta \neq \pi$. Let $\phi = ik/2R$. Then, $$\bigtriangledown \phi = d\phi = -(ik/2R^2) dR = -\star((ik/2)d(\cos \theta d \psi)) = \star F_{\bigtriangledown}$$
Its the last equality that troubles me. How does one compute the connection from the matrices? I've also only computed very easy examples of curvature so I'm lost on obtaining $F_\bigtriangledown$ as well. A detailed computation/explanation would be helpful. Thanks.
Locally the curvature of the connection is just $dA$, as in this case we have an abelian group as a fibre ($S^1$) otherwise you would have to add a term like $A \wedge A$. In this case then the curvature is given by $ik/2 \sin \theta d\theta d\psi$. From this is clear, I hope, the last equality.