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A closed $1$-form in a simply connected set in $R^n$ is exact. I would like a similar condition (with a reference) on sets in $R^n$ that closed $2$-forms are exact. De Rham cohomology gives an algebraic answer, but I am interested in a condition like simply connected with geometric appeal.

The condition should not exclude "too many" sets. By analogy, closed $1$-forms on contractible sets are exact. But this condition excludes too many sets, e.g., the simply connected set $R^3 - \{{0\}}$.

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    A simply connected compact manifold $M\subset \mathbb R^n$ of dimension $3$ has the property you require, by Poincaré duality. Unfortunately some Russian guy has proved that these manifolds are fairly rare , so I'm not making this an answer.2012-05-31
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    @GeorgesElencwajg Why 'unfortunately'? ;-)2012-05-31
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    Dear @Thomas: oh, I meant it's unfortunate for me, because my class of examples is consequently not a very large one. So you have heard of that Russian too?2012-05-31

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You can appeal to the universal coefficient theorem to say that "every closed surface in $S$ bounds a 3-manifold in $S$" is a condition which implies that closed 2-forms on $S$ are exact. It's not quite fully general, but it's pretty broad and sounds geometrically appealing to me.

If you really want an if-and-only-if condition, you could say something like "every closed surface in $S$ has a multiple which bounds a 3-manifold in $S$", but anyone who is actually happy with that is probably also fine with talking in terms of de Rham cohomology. :)

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    Thanks, Micah Does the "every closed surface ... " condition apply to $R^n$, or just $R^3$?2012-06-09
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    This is a continuation of my last comment -- I pushed the wrong button. I want to state the result in a second year undergraduate text. Without questioning the accuracy of your reply, do you have a citation that states the result directly?2012-06-09
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    Micah, They keep cutting me off!! \\ Last part. I've read that "Every closed surface in S can be contracted to a point" also suffices. True? Citation?2012-06-09
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    @AlanMacdonald: That'd work too (and both of them work in $\mathbb{R}^n$). Both your condition and mine will ensure that the second *homology* group $H_2(S)$ vanishes (much as being simply connected ensures that the first homology group vanishes). Once you know that, the universal coefficient theorem allows you to connect homology to cohomology and de Rham's theorem gives you the rest. Unfortunately, I don't know of any single place where you can find all of this written down explicitly...2012-06-09
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    @AlanMacdonald ...my bookshelf doesn't even seem to contain any one book that states both the universal coefficient theorem and de Rham's theorem!2012-06-09