# Question Regarding a form that is closed but not exact, and smooth angle function on the punctured plane

In Jack Lee's introduction to smooth manifold. Example 11.48 in Chapter 11.

It discusses that the 1 form $$\omega = \frac{x dy - y dx}{x^2+y^2}$$ is a closed 1 form but not exact (because it is not conservative).

Then Jack goes on to discuss why this form cannot be exact intuitively, here's what he says

"On the other hand, if we restrict the domain to the right half-plane U = {(x,y) | x>0}, a computation shows that $$\omega = d (\tan^{-1}\frac{y}{x})$$ . This can be seen more clearly in polar coordinates, where $$\omega = d(\theta)$$. The problem, of course, is that there is no smooth (or even continuous) angle function on all of $$\mathbb{R}^2 \backslash (0,0)$$, which is a consequence of the “hole” in the center."

What is a smooth angle function defined on a smooth manifold, and why is the hole in the center a problem, shouldn't the removal of the origin actually takes care of the problem of not having an angle defined at $$(0,0)$$?

I am at the beginning of the De Rham Cohomology Chapter, and I am trying to understand the relationship between the shape of a smooth manifold and what it has to do with differential forms not being exact, and this first example rather baffles me.

• If you go in a circle around $(0,0)$, the angle increases by $2\pi$. So what is the angle at $(1,0)$; is it $0$ or $2\pi$? The function must be multi-valued or discontinuous. Commented Jul 28, 2019 at 12:17
• @mr_e_man This I understand, but I don't understand why that has anything to do with the hole at the center, the problem of not being able to define the angle at (1,0) seems to be a problem even if you have the origin right? Commented Jul 28, 2019 at 16:50
• @Ecotistician When $(0,0)$ is missing, there's a well-defined angle function in a suitable neighborhood of any point, but no global angle function. When $(0,0$ is present, there isn't even a local angle function in a neighborhood of $(0,0)$. The closed-vs.-exact issue (also known as cohomology) is really about having things available locally but running into inconsistencies when you try to extend them globally. Commented Jul 29, 2019 at 0:06
• @Andreas Blass, I see, so perhaps I should interpret the statement as " When there's a hole in the plane, because closed form are not required to be exact (by Poincare lemma of star shaped space), it is possible to have a closed form that is not exact? Commented Jul 29, 2019 at 0:36
• *correction, not by Poincare Lemma, I meant to say Poincare lemma for star shape space does not apply here. Commented Jul 29, 2019 at 5:50

Identify $$\mathbb R^2$$ with $$\mathbb C$$. The trick is to define the angle function implicitly: an angle function on a subset $$U \subset S^1$$ is a continuous function $$\theta : U \to \mathbb R$$ such that $$e^{iθ(p)} = p,$$ for all $$p \in U.$$ This definition makes sense intutively because if $$e^{iθ(p)}=\cos \theta(p)+i\sin \theta (p)=p$$, then $$\theta$$ must map $$p$$ to the angle that the line segment from the origin to $$p$$ makes with the positive $$x$$ axis. Of course, there is the question of well-definedness and continuity of $$\theta$$, which is addressed by the following standard result:

There exists an angle function on an open subset $$U\subset S^1$$ if and only if $$U\neq S^1$$.

Suppose $$\theta :U\to \mathbb R$$ is an angle function and $$U = S^1 \theta$$ is continuous and $$S^1$$ is compact and connected, so $$\theta(S^1)$$ is an interval closed interval. And since $$\theta$$ is injective (why?), it is in fact, a homeomorphism from $$S^1$$ onto its $$\theta (S^1)$$. But, removing one point from $$\theta (S^1)$$ disconnects it, whereas removing one point from $$S^1$$ does not. Contradiction.

On the other hand, Suppose $$U \nsubseteq S^1$$ is open in $$S^1$$. Define $$\varphi : [0, 2\pi]\to S^1$$ by $$\varphi(t) = e^{it}$$ . Continuity of $$\varphi$$ implies that $$V =\varphi^{-1}(U)$$ is open in $$[0, 2\pi].$$ It follows that $$\varphi$$ is injective on $$V$$ (why?). So the restriction of $$\varphi$$ to $$V$$ is a continuous bijection $$:V\to U$$. Now. define $$\theta = (\varphi|_V )^{-1}$$. Then $$e^{iθ(p)} = p$$, for all $$p \in U$$ and if $$I\subset V$$ is an open interval then $$\theta^{-1}(I)=\varphi (I)$$ so $$\theta$$ is continuous.

Now that we have an angle function, we can relate it to the $$1$$-form $$\omega = \frac{x dy - y dx}{x^2+y^2}$$:

Without loss of generality, $$(0,1)\notin U$$. Then, if $$p=(x,y)\in S^1$$, we may take $$\theta(x,y)=\tan^{-1}(y/x)$$

and then because

$$d\theta \left (\frac{\partial}{\partial x}\right)=\frac{\partial \theta}{\partial x}=\frac{-y}{y^2+x^2}=\omega\left (\frac{\partial}{\partial x}\right)$$

and

$$d\theta \left (\frac{\partial}{\partial y}\right)=\frac{\partial \theta}{\partial y}=\frac{x}{y^2+x^2}=\omega\left (\frac{\partial}{\partial y}\right)$$

it follows that $$\omega=d\theta.$$