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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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    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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    @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