If $\omega$ is closed on $\mathbb R ^2 - 0$ and $\text d \omega =0$, then $\omega = \text d g+ \lambda \text d \theta$.

Question

I'm trying to solve problem 4-30 from “Calculus on manifolds”, which is the one in the title, where $$\text d \theta = -\dfrac{y}{x^2+y^2}\text d x+\dfrac{x}{x^2+y^2}\text d y.$$ I think I'm on the right track, but I need some help concluding.

The problem is preceded by this simpler one:

If $\omega=f\text d x$ is a one form on $[0,1]$, there exists $g\colon [0,1]\to \mathbb R$ with $g(1)=g(0)$ and $\lambda\in \mathbb R$ such that $\omega = \text d g+ \lambda \text d x$.

and gives this hint:

Hint: if $C_R^*\omega = \lambda _R\text d t+ \text d (g_R),$ show that all numbers $\lambda _R $ are equal.

Here $C_R(t)=Re^{i2\pi t}$.

Ok, so by problem 4-29 follows that $C_R ^*\omega$ is of the form given in the hint, and using closeness of $\omega$ I've been able to prove that all $\lambda$'s are equal. Since $C_R ^* \text d \theta = 2\pi \text d t$, it follows that, if we redefine $\lambda \mapsto 2\pi \lambda$: $$C_R ^* (\omega -\lambda \text d \theta ) =g'_R(t)\text d t$$ for all $R$ and for some function $g_R\colon [0,1]\to \mathbb R$.

I'm stuck here. I thought, for example, that $$g'_R\text d t=C_R ^*[ (g'_R\circ C_R ^{-1})\text d \theta ],$$ but I don't know if this fact can help (I'm not even sure that it can make sense, since $C_R ^{-1}$ is discontinuous on the $x$ axis).

Any help or hint is much appreciated.

Answer 1

You're 90% done. You've shown that $\omega - \lambda d \theta$, pulled back by $C_R$, gives you $dg_R$. All you really need to do is build the function "g" on the punctured plane. And your individual $g_R$ functions are what you need to do that.

Basically, you'd like to say "my function $g$, at a point $P$ of the punctured plane, is just...well, if the polar coordinates of $P$ are $(R, \theta)$, then $g(P) = g_R(\dfrac{\theta}{2\pi})$, or something like that."

The problems have have left are

(1) cleaning up that statement a bit, and

(2) worrying about the $x$-axis and continuity and well-defined-ness there. As you observe, if your point $P$ is on the $x$-axis, then you could use either $g_R(0)$ or $g_R(1)$ to define $g(P)$. How can you be sure that they're the same? [Hint: does problem 4-30 have an additional conclusion about $g$? I seem to recall, from the last time I did this problem, about 40 years ago, that it does...]

Answer 2

To complete my proof above.

Let $g(r,\theta)=g_r(\frac{\theta}{2\pi})$, $r>0$, $\theta \in [0,2\pi]$. Denote with $p=(r,\theta)$ the polar coordinates on $\mathbb R ^2-{0}$ and let $G=g\circ p$ (by construction of $g$ in problem 4-29 it can be shown that $g^{(k)}(0)=g^{(k)}(1)$ if $f^{(k-1)}(0)=f^{(k-1)}(1)$, so $G$ is differentiable on the $x$ axis). Now: $$C_R^*\text d G=C_R^*[(\dfrac{\partial g}{\partial r }\circ p)\text d r+(\dfrac{\partial g}{\partial \theta }\circ p)\text d \theta]=g_r'(t)\text d t,$$ so:$$C_R ^* (\omega -\lambda \text d \theta -\text d G)=0,$$ or$$\omega =\lambda \text d \theta + \text d G+H\, \text d r$$ for some function $H=h\circ p$ ($\text d r$ is the differential of $\sqrt{x^2+y^2}$).

Since $\omega$ is closed, we have $$\dfrac{\partial h}{\partial \theta}=0,$$ so $$h(r,\theta)=k(r).$$

Let $g_1=g+\intop _1 ^r k$, $G_1=g_1\circ p$, and we are done.