Explanation of this proof of $d\theta$ not being an exact form

Question

I have been reading Spivak's Introduction to Differential Geometry and I cannot get this. I have already seen other proofs of the following statement, and I get those, I just do not follow this one.

$d\theta$ is closed but it is not exact

I need some explanation, the proof is simple, but I do not quite get why is there a contradiction, the proof goes as it follows:

Proof: Let $\omega=d\theta$, and suppose there exists a $C^1$ function $f:\mathbb{R}^2\setminus \{ 0\} \rightarrow \mathbb{R}$ such that $\omega=df$, clearly this implies that $\partial f/ \partial x=\partial \theta / \partial x$ and $\partial f/ \partial y=\partial \theta / \partial y$, and hence $f=\theta + constant$ on $\mathbb{R}^2\setminus L$, which is impossible. $\square$

Note:

Here $\omega$ is defined (as always) as $$\omega = \frac{-y}{x^2+y^2}dx+\frac{x}{x^2+y^2}dy$$.

I do not get why is it impossible, some explanation would really be appreciation, thanks in advance.

Answer 1

This problem illustrates that $d\theta$ is a particularly misleading and hence poor notation for this $1$-form: It isn't the exterior derivative of anything!

(More generally the common notation $dV$ for a volume form on an oriented manifold suffers the same issue. For this reason I prefer a slightly more verbose notation like $\operatorname{vol}$ for a volume form.)

This notation also makes the provided proof possibly confusing. Here's a bit more detail:

If $\omega = df$ for some differentiable (and hence continuous) function $f$, then in particular that equality holds on the slit plane $$S := \Bbb R^2 \setminus L , \qquad L := \{(x, 0) \in \Bbb R^2 : x \geq 0\} ,$$ where in particular we can define the usual coordinates $(r, \theta) : S \to (0, \infty) \times (0, 2 \pi)$, so that $$df \vert_S = \omega\vert_S = d\theta .$$

(The mentioned abuse of notation $d\theta$ makes a sidebar important here: In the preceding equation, $d\theta$ is the exterior derivative of the function $\theta$, and hence is defined only $S$; written there it is not the form $\omega$, but rather its restriction to $S$.)

So, on $S$ we have $$0 = df\vert_S - d\theta = d(f\vert_S - \theta) ,$$ and in particular (since $S$ is connected) $f\vert_S - \theta$ is some constant function $C$, that is, as claimed, $$f\vert_S = \theta + C .$$

On the other hand, $f\vert_S$ cannot be the restriction of a continuous function $f$ on $\Bbb R^2 - \{(0, 0)\}$: At any point $(x, 0)$, $x > 0$, in the slit, the limit of $f$ at that point as you approach from above the $x$-axis is $C$, but the limit from below is $C + 2 \pi$, so $\lim_{p \to (x, 0)} f$ does not exist, hence $f$ is not continuous there, a contradiction.

A much faster proof is available once you have access to integration: By Stokes' Theorem (in this case, the Fundamental Theorem of Line Integrals, essentially), if $\omega$ were an exact form, then we would have $\int_\gamma \omega = 0$ for any curve $\gamma$. But parameterizing a convenient closed anticlockwise curve that encloses the origin once---say, the unit circle---and computing directly gives $\int_\gamma \omega = 2 \pi$, a contradiction.

Answer 2

The reason is that $\theta+const$ (for any constant) is a function that has a jump along $L$ (a $2\pi$-jump, to be precise). Therefore, no smooth function defined on $\mathbb{R}^2\setminus\{0\}$ can coincide with it on $\mathbb{R}^2\setminus L$.