# A Polar coordinate neighbourhoods in manifolds

1. Aug 3, 2016

### Bas73

In my introduction to manifolds the following is stated:

Polar coordinates (r, phi) cover the coordinate neighborhood (r > 0, 0 < phi < 2pi); one
needs at least two such coordinate neighborhoods to cover R2.

I do not understand why two are needed. Any point in R2 can be described by polar coordinates.

Any idea anyone?

Thanks
Bas

2. Aug 3, 2016

### vanhees71

No, the origin is a coordinate singularity. Note that the Jacobian determiant between Cartesian and polar coordinates is $r$, which vanishes at $r$, and thus the polar coordinates are singular in this point.

In 3D the spherical coordinates are singular along the entire polar axis.

3. Aug 3, 2016

### Bas73

Hi vanhees71,

Thanks. As I understand you, zero forms the second neighborhood, its chart being the identity.

The same seems to happen with the Cylinder C2. In my book it is defined as (x,y) with (0 < x < 2pi, -oo < y < oo). Since x=0 is not included, this line needs a neighborhood as well. By why is it not included? (I'm new to this field, just in case you were wondering)

Bas

4. Aug 3, 2016

### DrSuage

Hi - welcome to differential geometry!

A key feature of manifolds is that transition functions in the overlaps between coordinate patches need to be smooth. This can be achieved with R2 by considering to sets of spherical polar coordinates where the origin of the second set is offset by some length l in any direction, so that each set of coordinates covers the other's origin.

You cannot have just the point zero as the second neighborhood - one point cannot contain an open set!

5. Aug 3, 2016

### Bas73

Thanks. It's a strange world though ;-)

For now, I am happy that I understand your answer at all and it seems logical.

The line as second neighborhood for the cylinder is no good either (not open). Since parameters need to be open intervals, I guess you could simply pick (0 < x < 2pi, -oo < y < oo) and (-pi < x < pi, -oo < y < oo) as your two neighborhoods.

Bas

6. Aug 3, 2016

### Orodruin

Staff Emeritus
Zero is not a neighbourhood. A neighbourhood is an open set. Using polar coordinates on the plane has two problems:
1. You do not include $r = 0$ since the coordinates need to be in an open subset of $\mathbb R^2$.
2. You cannot cover all angles $\phi$, again since the coordinates need to be in an open subset. You either miss at least one value of $\phi$ or you do not have a bijection from the coordinates to the manifold (the map would be several-to-one).

The same as reason 2 above. Note that you can cover the cylinder by a single coordinate chart though, just not that one.

7. Aug 7, 2016

### Ben Niehoff

I'm fairly sure you cannot cover the cylinder $\mathbb{R} \times S^1$ in a single chart. What chart do you have in mind?

8. Aug 7, 2016

### Orodruin

Staff Emeritus
The real plane with a single point removed works fine, but it is just one out of an infinite number of examples. Any subset of R^2 with the same topology works. It is fairly straightforward to construct an explicit map to the natural embedding of the cylinder in R^3. Note that there is no requirement that a chart must be simply connected.

Edit: To do the explicit construction now that I am on a laptop instead of a mobile device. Use $\xi$ and $\upsilon$ as coordinates on $\mathbb R^2$ and $x,y,z$ as coordinates in $\mathbb R^3$. We want to map the plane with the origin removed to the subset $x^2 + y^2 = 1$ of $\mathbb R^3$. Take the following map:
$x = \frac{\xi}{r}$
$y = \frac{\upsilon}{r}$
$z = \log(r)$
where $r = \sqrt{\xi^2+\upsilon^2}$.

Last edited: Aug 7, 2016
9. Aug 7, 2016

### Ben Niehoff

Ah, I forgot that a chart didn't have to be simply connected.