# Abelian flat connection maps to zero

In an abelian gauge theory, a flat connection is an $$A_\mu$$ such that $$F_{\mu\nu}=0$$. I have to prove that these connections are equivalent to $$0$$, i.e. there is a gauge transformation that maps the flat connection to zero.

Here's my attempt:

$$A_{\mu}~~|~~F_{\mu\nu}=0$$

$$0=A'_\mu=A_\mu+\partial_\mu\alpha~~\Rightarrow~~A_{\mu}=-\partial_\mu\alpha,$$ so I have to prove that $$\exists~\alpha~|~A_{\mu}=-\partial_\mu\alpha$$.

In three dimensions, if a vector field satisfies $$\vec{\nabla}\times\vec{V}=0$$, then it can be written as the gradient of a scalar function $$\vec{V}=\vec{\nabla}\phi$$. I would generalize this to $$d$$ dimensions this way:

$$\epsilon^{\mu_1\mu_2\cdot\cdot\cdot\mu_d}\partial_{\mu_1}A_{\mu_2}=0~~\Rightarrow~~A_\mu=-\partial_\mu\alpha$$

But $$F_{\mu\nu}=0$$ implies $$\epsilon^{\mu_1\mu_2\cdot\cdot\cdot\mu_d}\partial_{\mu_1}A_{\mu_2}=0$$ so it is proven.

My question is if the generalization to $$d$$ dimensions is right, and how could I prove it.

Hint: Given an abelian flat connection $$F=\mathrm{d}A=0$$, from Poincare Lemma we know that there exists a locally$$^1$$ defined $$0$$-form gauge transformation $$\alpha$$, so that 1-form gauge potential $$A=\mathrm{d}\alpha$$.
$$^1$$ The corresponding global problem can have topological obstructions.