Showing why a form is not exact

Question

Consider the differential form

$$\omega=\frac{y dx-x dy}{x^2+y^2}$$ on $\mathbb{R}^2-0$. I have proved the following.

1-This $1$-form is closed on $\mathbb{R}^2-0$.

2-If $f_n:S^1 \to \mathbb{R}^2-0$ is the map $$f_n(\theta)=(\cos(n \theta), \sin(n \theta))$$ Then $$\int_{S^1} f_n^* \omega= 2\pi n \neq 0$$

and this is where I am stuck.

3-Use the previous parts to show that the form $\omega$ is not exact.

My Thoughts- If $\omega=d\tau$ is exact, then $d \omega =d^2 \tau=0$ and thus $f_n^* d \omega = d(f_n^* \omega)=0$ . But now I am not sure what to do next? I am thinking about using stokes theorem and looking for guidance in that direction.

Answer 1

The comments give a clear explanation, which I'll re-present here.

Suppose for contradiction that $\omega$ is exact. Then $\omega = d\eta$ for some $0$-form $\eta$. Now $$f_1^* \omega = f_1^*(d \eta) = d (f_1^* \eta),$$ so by Stokes' Theorem $$2\pi = \int_{S^1} f_1^* \omega = \int_{S^1} d(f_1^* \eta) = \int_{\partial S^1} f_1^* \eta.$$ On the other hand, since $\partial S^1 = \varnothing$, this integral equals $0$. Contradiction!

Answer 2

For $$\tag{1} \omega=\frac{y\,dx-x\,dy}{x^2+y^2}=a\,dx+b\,dy $$ we have \begin{align} d\omega&=(\partial_ya)\,dx\wedge dy+(\partial_xb)\,dy\wedge dx\\ &=\frac{(x^2-y^2)\,dx\wedge dy+(x^2-y^2)\,dy\wedge dx}{(x^2+y^2)^2}=0\tag{2} \end{align} because $dx\wedge dy$ is antisymmetric.

• $\omega$ is defined only on the punctured, and therefore non-simply connected, domain $\mathbb R^2\setminus\{(0,0)\}\,.$ The Poincare lemma says that, in general, only on simply connected domains all closed $p$-forms, with $p=1\,,$ are exact. (For $p>1$, thanks to a comment by peek-a-boo one works with star-shaped domains in that case or assumes that the domain is contractible.)

• Because the vector field of which $\omega$ is the dual, namely, $$\tag{3} \frac{1}{x^2+y^2}\begin{pmatrix}y\\-x\end{pmatrix}\,, $$ is tangent to the circles around the origin it looks geometrically as if it should have rotation. But since $\omega$ is closed it doesn't! We have here an example of an irrotational vortex.

• The form (1) occurs as a local gauge transformation in the Wu-Yang monopole which was in 1975 the first solution of the Yang-Mills field equations [1]. This $\omega$ is also well-known as the generator of the first de-Rham cohomology group of the punctured plane $\mathbb R^2\setminus\{(0,0)\}\,.$

• Let's look for the function $\phi$ on any simply connected domain $D\subset\mathbb R^2\setminus\{(0,0)\}$ that satisfies $$\tag{4} \omega=d\phi\,. $$ It turns out that this function is given in polar coordinates by $$\tag{5} \phi(r,\theta)=-\theta\,. $$ The graph of this function is a helix. In Cartesian coordinates this is $$\tag{6} \phi(x,y)=-{\rm sign}( y)\,\arccos\Big(\textstyle\frac{x}{\sqrt{x^2+y^2}}\Big)\, $$ but we will not use that. Since, \begin{align}\tag{7} \begin{pmatrix}\partial_x\\\partial_y \end{pmatrix}=\begin{pmatrix} \cos\theta&-\sin\theta\\\sin\theta&\cos\theta\end{pmatrix}=\begin{pmatrix}\partial_r\\\frac{1}{r}\partial_\theta\end{pmatrix} \end{align} it is easy to see from $\partial_r\phi=0$ and $\partial_\theta\phi=-1$ that \begin{align}\tag{8} \partial_x\phi=\frac{r\sin\theta}{r^2}=\frac{y}{x^2+y^2}\,,\quad\partial_y\phi=\frac{-r\cos\theta}{r^2}=\frac{-x}{x^2+y^2} \end{align} which shows $$\tag{9} d\phi=(\partial_x\phi)\,dx+(\partial_y\phi)\,dy=\omega\,. $$

• The maximal domain on which this $\phi$ can be defined without becoming multivalued is $\mathbb R^2$ with a branch cut, say, $\mathbb R^2\setminus\{x\ge 0,y=0\}\,,$ similar to the complex logarithm.

• The example (1) is very popular in homework exercises. I am a bit surprised that the following closed form which is not exact is rarely seen: Instead of (4) and (5) one could take an arbitrary strictly decreasing function $f:[0,+\infty)\to\mathbb R$ and define $$\tag{10} \omega=d\phi=df(\theta)\,. $$ Clearly, $\omega$ is closed, $d\omega=dd\phi=0$ but its integral around the circle $\theta\in[0,2\pi)$ is $f(2\pi)-f(0)<0\,.$ For example when $f(\theta)=-\frac{1}{2}\theta^2$ then $\partial_r\phi=0$ and $\partial_\theta\phi=-\theta$ so that here, instead of (8), and using (6), \begin{align}\tag{11} \partial_x\phi&=\theta\frac{r\sin\theta}{r^2}={\rm sign}( y)\, \arccos\Big(\textstyle\frac{x}{\sqrt{x^2+y^2}}\Big)\frac{y}{\sqrt{x^2+y^2}}\,,\\[2mm] \partial_y\phi&=-\theta\frac{r\cos\theta}{r^2}={\rm sign}( y)\, \arccos\Big(\textstyle\frac{x}{\sqrt{x^2+y^2}}\Big)\frac{-x}{\sqrt{x^2+y^2}}\,.\tag{12} \end{align} This leads to $$\tag{13} \omega={\rm sign}( y)\,\arccos\Big(\textstyle\frac{x}{\sqrt{x^2+y^2}}\Big)\displaystyle\frac{y\,dx-x\,dy}{x^2+y^2}\,. $$ But that form is not continuous on $\mathbb R^2\setminus\{(0,0)\}$ for the same reason that the function $\phi$ from (5) and (6) was not continuous.

• In this related post it is further explained that there are essentially no other one-forms sharing the properties of $\omega$ from (1).

[1] M. Nakahara, Geometry, Topology and Physics