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This may be a trivial question. If so, apologies in advance.

Let a $1$-form $\omega$ be given. Suppose that for any line segments $C_1=[i,j]$, $C_2=[j,k]$, $C_3=[i,k]$, we have that $\int_{C_1} \omega + \int_{C_2} \omega = \int_{C_3} \omega$.

My question is whether this implies that $\omega$ is exact. If this were true for any curves $C_1,C_2,C_3$, I believe the answer would be affirmative. I want to make sure that restricting attention to finitely (or countably) many line segments is without loss of generality.

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    $\begingroup$ Math.StackExchange is more appropriate for elementary math questions. $\endgroup$
    – Piotr Hajlasz
    Apr 28, 2018 at 16:09
  • $\begingroup$ May I ask why you need this? $\endgroup$
    – Avi Steiner
    Apr 28, 2018 at 16:42
  • $\begingroup$ I'm reading a paper, whose assumptions basically correspond to path-independence over line segments. I'm trying to obtain an easier way of proving their result leveraging equivalence between closedness/exactness. For that I need to establish that path independence over finitely many line segments suffices for exactness. $\endgroup$
    – Ozzy
    Apr 28, 2018 at 16:59
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    $\begingroup$ Have you already made any initial effort to prove the result or find a counterexample? Sometimes one or the other is reasonably easy, as long as you keep an open mind about whether it's true or not. $\endgroup$
    – Deane Yang
    Apr 28, 2018 at 17:22
  • $\begingroup$ The question is not clearly stated anyway. Are $C_i$ line segments or curves? Just check any textbook in Calculus 3, look for path independent line integrals and potential functions. $\endgroup$
    – Piotr Hajlasz
    Apr 28, 2018 at 17:26

2 Answers 2

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The answer is no. This answer is similar in the spirit to that of Ben McKay, but with all details provided. Consider the form $$ \omega = \frac{-y}{x^2+y^2}\, dx + \frac{x}{x^2+y^2}\, dy $$ on the annulus $\Omega=\{ (x,y):\, 1<x^2+y^2<1.1\}$. This form is closed, but not exact, because the integral along a circle around zero equals $2\pi$.

Any triangle in $\Omega$ will have the interior contained in $\Omega$, because the annulus $\Omega$ is so thin that the triangle cannot go around the origin. Since the form is closed, it is exact in a neighborhood of the triangle and hence the integral along the triangle equals zero showing that $$ \int_{C_1} \omega + \int_{C_2} \omega = \int_{C_3} \omega. $$

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Take the differential form $\omega$ to be $d\theta$ in polar coordinates, so that $\omega$ extends to a unique smooth 1-form on the punctured plane (punctured at the origin). Remove from the plane a horizontal line segment centred at the origin and of length 2. Remove a similar vertical one. Then remove the diagonal lines, except for segments centred at the origin of length 1/2. If you draw the picture, I think you can see that there is no triangle around the origin lying in the resulting domain, and that any triangle in that domain lies in a half space, so has integral zero. Of course, this is not a proof; there are some details left to work out.

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