Poincare lemma for one-form on ##\mathbb R^2 \backslash \{ 0 \}##

[cianfa72]

TL;DR Summary

Poincare lemma applied to a one-form defined on the set $\mathbb R^2 \, \backslash \{ 0 \}$

Consider the following 1-form $\omega$ defined on $U = \mathbb R^2 \, \backslash \{ 0 \}$: $$\omega = \frac {y} {x^2 + y^2} dx + \frac {-x} {x^2 + y^2} dy$$
It is closed on $U$ since $\partial \left (\frac {y} {x^2 + y^2} \right) / \partial y = \partial \left (\frac {-x} {x^2 + y^2} \right ) / \partial x$, however it isn't globally exact on $U$ (i.e. there is a not a smooth function $f$ on $U$ such that its partial derivatives are the required two functions).

By Poincare lemma, however, $\omega$ is locally exact. Just to fix ideas pick the point $p=(1,1)$. Which is the function $\varphi$ defined in a neighborhood of $p$ such that locally $\omega=d\varphi$ ?

Thanks.

 

[Orodruin]

Hint: Polar coordinates.

 

[cianfa72]

In polar coordinates the set $U = U = \mathbb R^2 \setminus \{0\}$ is mapped into the $(r,\theta)$ open region $\{ -\pi \lt \theta \leq \pi, r \gt 0 \}$ and we get:
$$\omega = - d\theta$$
Btw, the coordinate function $\theta$ is actually globally defined on $U$ and not just locally. Why it is the case ?

 

[Orodruin]

In polar coordinates the set $U = \mathbb R^2 \, \backslash \{ 0 \}$ is mapped into the $(r,\theta)$ open region $\{ -\pi \lt \theta \leq \pi, r \gt 0 \}$ and we get:
$$\omega = - d\theta$$
Btw, the coordinate function $\theta$ is actually globally defined on $U$ and not just locally. Why it is the case ?

No, the coordinate function $\theta$ is not globally defined.

 

[cianfa72]

No, the coordinate function $\theta$ is not globally defined.

Ah ok, this is definitely the point I'm confused about.

In my previous post I made a mistake: by polar coordinates $U = U = \mathbb R^2 \setminus \{0\}$ is mapped into the region $\{ -\pi \lt \theta \leq \pi, r \gt 0 \}$ that is not open in $\mathbb R^2$ (indeed $U$ is topologically a cylinder that is not homeomorphic to the plane).

Coming back to the coordinate function $\theta$, I believe it is actually defined on all points of $U$ even though it is discontinuous at the half-axis $\{ y=0, x \lt 0 \}$. Is this the reason why you claim it is not globally defined on $U$ ?

 

[Orodruin]

Ah ok, this is definitely the point I'm confused about.

In my previous post I made a mistake: by polar coordinates $U = \mathbb R^2 \, \backslash \{ 0 \}$ is mapped into the region $\{ -\pi \lt \theta \leq \pi, r \gt 0 \}$ that is not open in $\mathbb R^2$ (indeed $U$ is topologically a cylinder that is not homeomorphic to the plane).

Coming back to the coordinate function $\theta$, I believe it is actually defined on all points of $U$ even though it is discontinuous at the half-axis $\{ y=0, x \lt 0 \}$. Is this the reason why you claim it is not globally defined on $U$ ?

Your problem is that your coordinate patch is not open, it has a closed boundary at $\theta = \pi$. The region $\mathbb R^2 \setminus \{0\}$ is an open subset of $\mathbb R^2$, but $(-\pi, \pi] \times (0,\infty)$ is not.

Using the $\theta$ of any maximal coordinate patch in polar coordinates, there is no way to make that function continuous across the cut. The form will be exact on the coordinate patch, but not globally.

 

[fresh_42]

Wikipedia has the formula how to construct it:
https://en.wikipedia.org/wiki/Poincaré_lemma
and a better readable on
https://de.wikipedia.org/wiki/Poincaré-Lemma

 

[cianfa72]

Your problem is that your coordinate patch is not open, it has a closed boundary at $\theta = \pi$. The region $\mathbb R^2 \setminus \{0\}$ is an open subset of $\mathbb R^2$, but $(-\pi, \pi] \times (0,\infty)$ is not.

Using the $\theta$ of any maximal coordinate patch in polar coordinates, there is no way to make that function continuous across the cut. The form will be exact on the coordinate patch, but not globally.

Ah ok, I think I got it.

You mean: any maximal coordinate patch on $U = \mathbb R^2 \setminus \{0\}$ is by definition open in the punctured topology on $U$. Any of them maps to an open set in $(r,\theta)$ plane, however, always leaves "an half-line cut" in $U$ unmapped (basically each of them picks a different "half-line cut").

Thus the 1-form $\omega = - d\theta$ will be exact only in any of those maximal coordinate patches.

If one insists on defining an onto one-to-one map $\gamma$ between $U$ and the $(r,\theta)$ plane, one will get a non open set in $\mathbb R^2$ and the coordinate function $\theta$ will not be continuous "on the boundary" (i.e. it will not be continuous on the mapped set in $(r,\theta)$ plane endowed with the topology defined there through $\gamma^{-1}$).

Is the above correct ? Thanks.