I know that if $U\subseteq\mathbb{R}^n$ is an open simply connected set, every closed 1-form $\omega\in\Omega^1(U)$ is also exact.
I was wondering: does the converse hold? So if every closed $\omega\in\Omega^1(U)$ is exact, is $U$ necessarily simply connected? Are there conditions under which this holds? If it doesn't hold in general, can you provide a counterexample together with the above-mentioned conditions if they exist?
I thought that given a $C^1$ path in $U$ we could first try to look for a point which isn't in any "finger-shaped" bit of the boundary of the set, parametrize the path so that it starts from that point, and use segments to squash every part of the path in any of those bits out of those bits so to get the path included in a big part of $U$ where we can use segments, but that obviously requires some formalization, and I still need to exclude there are holes in the big part.
Best Answer
No. "Every closed form is exact" is equivalent to the claim that the first de Rham cohomology $H^1_{dR}(U, \mathbb{R})$ vanishes. This means, equivalently, that there are no nontrivial homomorphisms from the fundamental group $\pi_1(U)$ to $\mathbb{R}$. But it does not imply that the fundamental group is trivial (which is equivalent to simply connected).
In fact any reasonable space (e.g. a manifold) is homotopy equivalent to an open subset of some $\mathbb{R}^n$ (embed it into $\mathbb{R}^n$ in a nice way, then take a tubular neighborhood), and lots of reasonable spaces have the property that their fundamental group is nontrivial but admits no nontrivial maps to $\mathbb{R}$. Maybe the simplest such space is the real projective plane $\mathbb{RP}^2$, which embeds nicely into $\mathbb{R}^4$ (I think) and has fundamental group $\mathbb{Z}_2$.
If you really want an explicit open subset of some $\mathbb{R}^n$, you can take $GL_3^{+}(\mathbb{R})$, the space of $3 \times 3$ matrices with positive determinant, which is homotopy equivalent to $SO(3)$, which is in turn the real projective space $\mathbb{RP}^3$, and which in particular again has fundamental group $\mathbb{Z}_2$. This is a connected open subset of $\mathbb{R}^9$.
In general, homomorphisms $\pi_1(U) \to \mathbb{R}$ correspond to homomorphisms $H_1(U, \mathbb{Z}) \to \mathbb{R}$, where $H_1$ is the first singular homology. If the fundamental group $\pi_1(U)$ is finitely generated (which is again true in all reasonable cases), every such homomorphism is trivial iff $H_1(U, \mathbb{Z})$ is torsion iff it is finite.