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If $D$ is a simply connected set in $R^2$ or $R^3$, then closed $1$-forms on D are exact. This fact is suitable for the elementary vector calculus course. I have been unable to find similarly suitable sufficient conditions on a domain $D$ in $R^n$ that closed $1$-forms are exact. Or that closed $2$-forms are exact.

Closed forms on a contractible set are exact. However, "contractible" is not a suitable condition, as the simply connected set $R^3 - \{0\}$ is not contractible.

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  • $\begingroup$ Search for more information on de Rham cohomology. I'm not sure, but it may also be that your question would be more on-topic at math.stackexchange.com ? $\endgroup$
    – Jyrki Lahtonen
    Commented May 26, 2012 at 18:42
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    $\begingroup$ I am aware of de Rham cohomology. But I asked for something "suitable for the elementary vector calculus course". Thanks for the tip about math.stackexchange.com. I'll try there if I don't get an answer here within a couple of days. AM. $\endgroup$
    – Alan Macdonald
    Commented May 26, 2012 at 19:10
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    $\begingroup$ "Simply connected" still implies that closed 1-forms are exact, independent of dimension. $\endgroup$
    – Lee Mosher
    Commented May 26, 2012 at 19:15
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    $\begingroup$ Well, what I would say is any book on de Rham cohomology --- I like Bott-Tu --- which might not be what you have in mind. However, if you already have a proof in mind for the case of open sets in $R^2$ and $R^3$, the exact same proof should work in higher dimensions. $\endgroup$
    – Lee Mosher
    Commented May 26, 2012 at 19:51
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    $\begingroup$ If your intention is to teach this to undergraduates, just tell them that the primitive of a closed 1-form $\omega$ is obtained by integrating $\omega$ along paths starting from some point $p_0\in D$. Then explain why integral is independent of the choice of the path connecting $p_0$ to $p$. This, in view of simple connectivity assumption on $D$ that you are making, amounts to verifying path-independence when $D$ is the square, which, I presume, you already know how to do. This is how one proves that irrotational vector fields are conservative in a vector calculus class. $\endgroup$
    – Misha
    Commented May 27, 2012 at 1:10

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See http://en.wikipedia.org/wiki/De_Rham_cohomology and in particular De Rham's fundamental theorem. "Closed = exact" for a domain says the De Rham cohomology group is trivial in the appropriate degree; and the condition comes down to normal topological calculations.

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  • $\begingroup$ I am aware of de Rham cohomology. But I asked for something "suitable for the elementary vector calculus course". $\endgroup$
    – Alan Macdonald
    Commented May 26, 2012 at 19:10
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    $\begingroup$ Well, it's a matter of paraphrase, then. The second para of your question suggests you would like something more than just a sufficient condition on the topology. Taking the complement of a point in three-space, the topologist might divide it into two pieces that were contractible and overlapped. $\endgroup$
    – Charles Matthews
    Commented May 26, 2012 at 21:32
  • $\begingroup$ @Alan, you can say every closed loop is contractible, which is the same as simply connected. $\endgroup$
    – Xiaolei Wu
    Commented May 27, 2012 at 4:59
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The answer "every closed form is exact if and only if the appropriate de Rham cohomology group vanishes" carries no information whatever, since you are (in effect) just giving the definition of said group. You can say that "this property is sufficiently interesting that wise men (Poincare, de Rham, et al) have built a powerful machine to figure out when a space has the property and more general versions thereof. If you are interested, please go on to study Bott-Tu [or your favorite other reference), and I will be happy to supervise you in an independent study course". I believe that giving dumbed-down reasons is counterproductive, since the students will believe that your sufficient condition is necessary and sufficient, so the next bridge they build will fall down.

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