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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.

MickG
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1 Answers1

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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.

Qiaochu Yuan
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  • $\mathbb{RP}^2$ does (smoothly) imbed in $\mathbb{R}^4$ by Whitney's theorem. (By a characteristic class argument, it doesn't imbed in $\mathbb{R}^3$, though there is an immersion in that dimension.) – anomaly Mar 08 '16 at 22:26
  • Perhaps I was just thinking in $\mathbb{R}^2$ :). Does my claim work in $\mathbb{R}^2$? I.e., are there (or not) open connected subsets of $\mathbb{R}^2$ with nontrivial $\pi_1$ but trivial $H^1$? Also, how is the equivalence between trivial $H^1$ and trivial $\mathrm{Hom}(\pi_1,\mathbb{R})$ proven? And how does one prove that homotopy equivalence? I was thinking of a retraction: is that the case? – MickG Mar 08 '16 at 22:48
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    @Mick: yes, it follows from the (nonobvious) fact that the fundamental group of an open subset of $\mathbb{R}^2$ is free. The equivalence is proven using the universal coefficient theorem. The homotopy equivalence is a deformation retraction of the tubular neighborhood of a nice subspace onto the subspace. – Qiaochu Yuan Mar 08 '16 at 23:01
  • For the nonobvious fact, this is a good reference. I was wondering if $\mathbb R^3$ has subsets with fundamental groups containing nontrivial torsion elements, and it seems that is an open problem. In general, I'm curious as to what $n$ gives a minimal counterexample to the false statement my question asks about. Obviously it is at most 9. – MickG Mar 10 '16 at 09:48
  • @Mick: as anomaly says, $\mathbb{RP}^2$ smoothly embeds into $\mathbb{R}^4$, so it's at most 4. The fundamental group need not have any torsion; what's important is that its abelianization admits no nontrivial maps to $\mathbb{R}$, which in the finitely generated case is equivalent to it being entirely torsion. But it could also just vanish. – Qiaochu Yuan Mar 10 '16 at 16:58
  • In general yes, but the question says "open". $\mathbb{RP}^2$ is compact, so its image under the embedding will also be, hence it will be closed. So $\mathbb{RP}^2$ is not a valid way of lowering the minimal dimension for open subsets of $\mathbb{R}^n$ which fail to satisfy that statement. If we remove "open", then naturally it will be at most 4. With "open", it is at most 9 since in $\mathbb{R}^9$ we have that general linear group, open since it is the preimage of the open set $\mathbb{R}\smallsetminus{0}$ under the determinant which is continuous. – MickG Mar 10 '16 at 17:01
  • Compactness of $\mathbb{RP}^2$ due, of course, to the fact it is a quotient of $S^2$ which is compact since it is closed and bounded in $\mathbb{R}^3$. – MickG Mar 10 '16 at 17:04
  • @Mick: I don't mean $\mathbb{RP}^2$ is the counterexample; as stated in my answer, you can take a tubular neighborhood of $\mathbb{RP}^2$, which will be open and homotopy equivalent to it, and so will have the same fundamental group and will give a counterexample in $\mathbb{R}^4$. – Qiaochu Yuan Mar 10 '16 at 18:03
  • And which deformation retracts onto $\mathbb{RP}^2$, thus having the same fundamental group $\mathbb{Z}_2$. I'd missed that. So the question now remains whether it is 4 or less. With "open" we can say it will be 3 or 4, because… wait, perhaps 1. But a connected open subset of $\mathbb{R}^1$ is an interval which has trivial fundamental group. And in general connected subsets of $\mathbb{R}$ are intervals with trivial f.g., right? So open or not, the minimal counter example will have to be in at least 2 dimensions, and for open subsets 3 or 4. Better ask a new question. – MickG Mar 10 '16 at 18:07
  • Moved the minimal counterexample issue here. – MickG Mar 10 '16 at 18:24