# What does it mean for a gauge field to have no gauge force?

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. $$\theta(\vec x)$$ 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.

Well, one simple confusion is that $$A$$ is the gauge potential whilst $$A+d\theta$$ is a gauge transformation.