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I'm trying to figure out some counterexamples of the following: If $F$ is an irrotational vector field then there exists some $U$ such that $F = \text{grad}(U)$ (or more generally, using the language of forms, I'm trying to find a closed form which is not exact).

As @peek-a-boo suggested I Googled a bit and found out the following example. In $\mathbb{R}^2 \setminus \{0\}$ the vector field $$F(x,y) := \left( -\frac{y}{x^2 + y^2},\ \frac{x}{x^2 + y^2} \right)$$ is closed (it has zero rotational) but not exact (it is not the gradient of any other function).

I would really appreciate if

  • Someone could give me more examples of such vector fields in $\mathbb{R}^2 \setminus \{0\}$.
  • More generally, someone could provide me with some examples of $k$-forms $\omega$ on n-dimensional manifolds $M$ such that its de Rham cohomology class $[\omega]$ is nonzero.

EDITED: I corrected the observations made in the first four comments, and also @Ted Shifrin's comment.

groupoid
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  • Did you search the site? Or google? There are several posts about closed non-exact forms. Btw you have a typo: there is no such thing as an exact form which is not closed. – peek-a-boo Mar 03 '23 at 18:47
  • the divergence take a vector field as entry and gives a function, so "$F = \operatorname{div} U$ for $F$ a vector field is nonsense. Did you mean the gradient of a function? – Didier Mar 03 '23 at 19:22
  • If you're only looking at subsets of $\mathbb{R}^2$, then you're limited to punching holes and creating vector fields that wind around them. That's not much "harder" than a single hole. What the vector field dual to $d\theta$ on the cylinder $S^1 \times \mathbb{R}$? or on the torus $S^1 \times S^1$? – Matthew Leingang Mar 03 '23 at 20:09
  • What a mess. As Didier already said, a vector field cannot be the divergence of a vector field, as the latter is a scalar function. And every $C^1$ exact form is automatically closed (because $d^2=0$), no matter the domain. Please rethink your question. Do you want to be thinking about a vector field that is not $C^1$ on a domain rather than a domain with some sort of hole? – Ted Shifrin Mar 03 '23 at 20:17
  • I apologize for this mess. I should've Googled a bit more, and I also got confused with names (rotational, divergence, exact, closed...). I believe now the question makes more sense. Thanks for all your corrections. – groupoid Mar 03 '23 at 21:27
  • At the end of this answer I made a few attempts to come up with other examples which do not just pull back that ubiquitous $\omega$ in OP. As you can see it was not successful. – Kurt G. Mar 03 '23 at 21:47
  • Because the cohomology of $\Bbb R^2-{0}$ comes from the circle sitting in there, up to diffeomorphic changes of coordinates, there are no other examples. For your final question, you say $n$-form on a manifold. Why specify $n$? Is the manifold $n$-dimensional? If so, there is a non-exact $n$-form if and only if the manifold is compact and orientable. – Ted Shifrin Mar 03 '23 at 22:22
  • @TedShifrin Thanks for this additional example! I guess this non-exact n-form is the volume form, am I right? When I wrote n-form, I meant an arbitrary form, n not necessarily being the dimension of the manifold. I'll edit this now. – groupoid Mar 05 '23 at 11:36

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