Let $\theta: S^1 \rightarrow \mathbb{R}:(x,y) \mapsto \arctan(\dfrac{y}x)$. Prove that $d\theta$ is a closed 1-form which is not exact.
I managed to prove that it was closed but how can it possibly not be exact? Isn't it implicit in the notation that it is? It would be like asking me to prove that $2n$ isn't an even number. Is there a mistake in the question or what am I missing here?
Cheers.
Best Answer
If $\omega$ is an exact differential form on $\Bbb R^2$, and $c_1$ and $c_2$ are two arcs with the same starting/ending points, then $$ \int_{c_1} \omega = \int_{c_2} \omega. $$
With that in mind, consider the two paths from $(+1, 0)$ to $(-1, 0)$ that go, respectively, around the top semicircle and the bottom semicircle of the unit circle.
Let me amplify a little bit, because this problem, in more or less this form, reappears constantly, and it's a pity it's not stated better.
There's a function $\theta(x, y) = arctan(y/x)$, defined for points $(x, y)$ in the plane with $x \ne 0$. Let's denote by $U$ the set $\{ (x, y) \mid x \ne 0 \}$, so $U$ is the domain of $\theta$.
The exterior derivative of $\theta$ is also defined on $U$ (because $\theta$ is differentiable). A little computation shows that this exterior derivative has the formula (for points $(x, y)$ that are actually in $U$) \begin{align} d\theta (x, y) &= \frac{\partial \theta} {\partial x}(x, y) dx + \frac{\partial \theta} {\partial y}(x, y) dy \\ &= \frac{1} {1 + (y/x)^2}(\frac{-y}{x^2}) dx + \frac{1}{1 + (y/x)^2} \frac{1}{x} dy \\ &= \frac{-y} {x^2 + y^2} dx + \frac{x}{x^2 + y^2} dy. \end{align}
This 1-form is defined on $U$, and is an exact $1$-form on $U$; it's the exterior derivative of the function $\theta$, after all!
But algebraically, if you look at it, it makes sense for any pair of numbers $(x, y)$ except $(x, y) = (0, 0)$; in other words, it admits a continuous extension to the set $$ V = \Bbb R^2 - \{ (0,0) \}. $$ Indeed, because $U$ is a dense subset of $V$, there's at most one continuous extension to $V$, and hence it must be the one given by the algebraic formula above.
Now here's the stupid part: even though the function $\theta$ is defined on $U$ rather than $V$, and hence $d\theta$ is only defined on $U$, this unique continuous extension to $V$ is also called "$d\theta$". I can't fix that, but I can ignore it for now, and I'm going to. I'm going to define a $1$-form on $V$ called $\eta$, by the formula:
\begin{align} \eta (x, y) &= \frac{-y} {x^2 + y^2} dx + \frac{x}{x^2 + y^2} dy. \end{align}
You'll notice that (1) $\eta$ is a nice smooth 1-form. (2) On the subset $U$, we have $\eta(x, y) = d\theta(x, y)$.
Because $S^1$ is a subset of $V$, we can restrict $\eta$ to be a 1-form on $S^1$ as well, if we like.
The thing that the problem is asking you to show (hard though it may be to believe, given that they left out all this stuff) is that there's no 0-form (i.e., differentiable function) $f$ on either $V$ or $S^1$, with the property that $df = \eta$.
There are two approaches you can take.
The first is to use the observation I made above, and note that the integral of $\eta$ over the upper and lower semicircle paths from $(+1, 0)$ to $(-1, 0)$ are different --- one is $+\pi$; the other is $-\pi$.
The second is to note that if such a function $f$ exists on $V$, then on the right open half-plane, which I'll call $V^+$, it must be the same as $\theta$, up to an additive constant, $c_1$ because they have the same partial derivatives; the same is true on the left half-plane, $V^-$, up to a possibly different additive constant $c_2$. So we would have, on $V^+$, that $$ f(x, y) = \theta(x, y) + c_1. $$ Fixing $y = 1$ and letting $x$ approach $0$ from above, we'd be taking arctan of large positive numbers, and we'd find that $$ f(0, 1) = \frac{\pi}{2} + c_1. $$
On the other hand, fixing $y = 1$ and letting $x$ approach $0$ from below, we're taking $\arctan$ of large negative numbers, and conclude that $$ f(0, 1) = -\frac{\pi}{2} + c_2. $$ so that $c_2 - c_1 = \pi$.
Doing the same analysis at the point $(0, -1)$, we would find that $c_1 - c_2 = \pi$. And that's a contradiction.