The electromagnetic gauge field is $A + d\theta$, where $\theta \colon \mathbb{R}^n \to \mathbb{R}$ comes from a gauge function, $e^{i\theta(\vec x)} $. Let's set $A=0$. The curvature form is $0$ since the gauge field is closed (in other words $d(d\theta) = 0$).
My question is partially mathematical in nature. I am also curious about physical interpretation if anyone can speak to it. $\theta(\vec x)$ appears to have no curvature, because of my reasoning (the connection it defines is a closed form). However... as a function it does not necessarily describe a flat manifold. And integral curves of the vector field created by the connection (for example, the connection coefficients $(\partial_{1} \theta(x_1)\ldots\partial_n \theta(x_n))$ after we dot $d\theta$ with a set of basis vectors) will have acceleration if the coefficient functions are not constants. So there's a set of contradictory observations for me about the role these calculations play in curvature. How do I combine these facts? Specifically, why am I calculating the curvature of the form is always zero when $\theta$ could be a curved surface (or $n$ dimensional generalization), and what does this have to do with not having a gauge force when the geodesics of that surface may curve?
Its possible I'm conflating a few different ideas and this is leading to my confusion, so any help straightening this out in my head would be really appreciated.
A connection $A$ on a bundle $E\to M$ determines a curvature $F$, which in some sense describes a local twisting picture of the bundle. But it is not locally intrinsic to the bundle or anything. One can always choose a connection in a local trivialization which has $F=0$.
There is also this special case, where you have a Riemannian metric $g$ on a manifold $M$, which uniquely determines a special connection $\nabla^{\text{LC}}$ defined on the tangent bundle $TM\to M$. The curvature of this connection is called the Riemann curvature $R$. It now makes sense to say that $R$ is tied to the geometry, or underlying shape, of your manifold $M$, because it is associated to $g$ which is giving $M$ that shape.