Question about external derivative

Question

I am new on computing external derivative, wedge product and pull-backs so I am having issues to understund some things about those things. For example, an excercise of my class notes is to prove that there is not any $f : \Bbb{R} - \{ 0 \} \rightarrow \Bbb{R}$ smooth that satisfies $df = \omega$ where $$\omega = \frac{-y \, dx + x \, dy}{x^2+y^2}$$

The problem is that, supposing that it exist such $f$ and integrating the partials of $f$, I obtained the following: $$ \frac{\partial f}{\partial x} = \frac{-y}{x^2+y^2} \Rightarrow f(x,y) = -\arctan (\frac{x}{y}) + c_1 \\ \frac{\partial f}{\partial y} = \frac{x}{x^2+y^2} \Rightarrow f(x,y) = \arctan (\frac{y}{x}) + c_2 $$

Both of them satisfies $df = \omega$ so, I do not know if I am missing something or do not know some theoretical result that helps to conclude. Any possible help would be appreciated :)

Answer 1

Exact differential 1-forms have vanishing integral along any closed path, so in order to prove such a $f$ does not exist it is sufficient to prove that, for some closed path $\gamma : [0,1] \to \mathbb{R}^2$, $\int_\gamma \omega \neq 0$.

For homotopical reasons (Poincaré Lemma), such a $\gamma$ must be searched among paths winding the origin (the "hole" in your domain), the most obvious one is the unitary circle parametrized by $$ \gamma(t) = \begin{bmatrix} \cos t\\ \sin t \end{bmatrix}, t \in [0,2\pi] $$ Therefore, $$ \int_\gamma \omega = \int_{[0,2\pi]} \gamma^* \omega = \int_0^{2\pi}\frac{\sin^2 t \, dt + \cos^2 t \, dt}{\sin^2 t + \cos^2 t} = \int_0^{2\pi} d t = 2\pi $$ and so $\omega$ can't be exact, id est there exists no smooth $f : \mathbb{R}^2 \setminus \{0\} \to \mathbb{R}$ such that $\omega = d f$.

This fact is quite deep, indeed the idea at the base of the De Rham cohomology is precisely to "count" how many differential forms are closed but not exact. Roughly speaking, it turns out that this "measures" how many "holes" there are in your domain.

The computation you made doesn't yield to a contradiction, indeed the expressions you found do not glue back together into a well defined function defined on $\mathbb{R}^2 \setminus \{0\}$. They just provide a primitive defined only on a half-plane.