# Differential 1-form of line

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? • If you consider $\omega(x)dx$ as the differential form, what can you say about $f(x):=\int_0^x \omega(t)dt$? – Caffeine Feb 1 '19 at 15:57
• $dy-dx$ is not a differential form on the line, it is a differential form on the plane – Caffeine Feb 3 '19 at 15:15

## 1 Answer

$$dy-dx$$ 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

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