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.

 

[cianfa72]

I found an insight on Lee - Introduction to smooth manifolds, chapter 11 Ex 11.48.

In standard cartesian coordinates $\{ x,y \}$, he says that a computation shows that $\omega=d(\tan^{-1} y/x)$. The latter is defined for $x \neq 0$.

So it seems like $$\omega = \frac {-y} {x^2 + y^2} dx + \frac {x} {x^2 + y^2} dy
$$ is actually exact on $U= \mathbb R^2 \setminus \{0\}$ excluding the $y$-axis ($x \neq 0$), while using polar coordinates on $U$ excluding just an "half-line cut from the origin".

 

[Orodruin]

excluding the y-axis (x≠0),

So … not globally exact.

 

[cianfa72]

So … not globally exact.

Yes, but the point is: why we get different coverage's answers using different coordinate patches ?

Namely, $\{ x,y \}$ coordinates map $U$ on the same open region (identity map) on which $\tan^{-1} (y/x)$ is defined all but the entire $y$-axis. On the other hand, polar coordinates map $U$ minus an "half-line cut" to an open rectangle in $\{ r,\theta \}$ plane. This time the coordinate function $\theta$ that returns the one-form $\omega$ as differential, is actually differentiable on a broader region (i.e. $U$ minus just an "half-line cut").

 

[Orodruin]

As long as your coordinate patch is simply connected you will be able to find a function $f$ such that $\omega = df$ on that coordinate patch. The point is that you cannot find a global function such that this is the case.

 

[cianfa72]

As long as your coordinate patch is simply connected you will be able to find a function $f$ such that $\omega = df$ on that coordinate patch. The point is that you cannot find a global function such that this is the case.

Ah ok. A coordinate patch that comprises $U$ minus an "half-line cut from the origin" is simply connected. It maps to an open rectangle in $\{ r,\theta \}$ plane and the differential (i.e. exterior derivative) of the coordinate function $\theta$ gives $\omega=d\theta$ on that coordinate patch.

On the other hand, $U$ itself as (global) coordinate patch for $U$ isn't simply connected, hence it can't exist a global function $f$ (i.e. a function defined on such global coordinate patch) such that $\omega = df$ there.

 

[Orodruin]

Yes. Note however that there still exist forms which are globally exact, such as $dx$.

 

[cianfa72]

Yes. Note however that there still exist forms which are globally exact, such as $dx$.

Ok, let's try to recap: as long as the manifold's coordinate patch is simply connected, then any closed one-form defined on it is also exact on that coordinate patch. Nevertheless, even on a non simply connected coordinate patch, still exist (closed) one-forms exact on that entire coordinate patch. In case the latter is the entire manifold, those specific/particular (closed) one-forms will actually be globally exact (e.g $dx$).

 

[Orodruin]

Yes