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This problem is from V.I Arnold's book Mathematics of Classical Mechanics. Q) Show that every differential 1-form on line is differential of some function

  1. Relevant equations The differential of any function is $$df_{x}(\psi): TM_{x} \rightarrow R$$
  2. The attempt at a solution

The tangent to line is line itself. The differential 1-form is $dy-dx=0$. Here I am struct. I don't know how to find out the differential. Can anyone help?enter image description here

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  • $\begingroup$ If you consider $\omega(x)dx$ as the differential form, what can you say about $f(x):=\int_0^x \omega(t)dt$? $\endgroup$ – Caffeine Feb 1 '19 at 15:57
  • $\begingroup$ $dy-dx$ is not a differential form on the line, it is a differential form on the plane $\endgroup$ – Caffeine Feb 3 '19 at 15:15
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$dy-dx$ is not a differential form on $\mathbb{R}$, since $\mathbb{R}$ has only one coordinate for every chart you choose. It is instead a differential form on $\mathbb{R}^2$, where is the differential of $f(x,y)=y-x$.

About Arnold's exercise, I would insted prove it in this way: Let $\omega(x)dx $ be the general differential form on $\mathbb{R}$, where $\omega(x)$ is $C^{\infty}[\mathbb{R}]$. Then, let $f:=\int_0^x\omega(t)dt$. By definition of differential and by the fundamental theorem of calculus, we obtain that: $df=\partial_{x}( \int_0^x \omega(t)dt)dx=\omega(x)dx$ QED

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  • $\begingroup$ @Abhi7731756 Is there anything wrong or unclear with the answer? $\endgroup$ – Caffeine Feb 4 '19 at 20:43

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