# I Mapping between two spaces

1. Jun 14, 2017

### davidge

Consider the mapping $f: \mathbb{D}^1 \longrightarrow [0,1] \times (0,2 \pi]$ where $\mathbb{D}^1$ is the unit disk. This is the familiar polar coordinate system.

The area element is $dx \wedge dy$ in $\mathbb{D}^1$ and $r dr \wedge d\theta$ in $[0,1] \times (0,2 \pi]$*. Now at $r = 0$ this vanishes. But $dx \wedge dy = r dr \wedge d\theta$ and that means $dx \wedge dy = 0$ as well.

We know that the polar system is singular at $r = 0$ because there cannot be a bijection: for every $\theta \in (0,2 \pi]$ we have the point $(0,0)$ in $\mathbb{D}^1$.

But what is the meaning of the vanishing area element $dx \wedge dy$?

*I know this is two dimensional and the area element as defined in vector calculus as the vector product of two vector requires a third vector, which means a third dimension. But I think this is not relevant here.

2. Jun 14, 2017

### Staff: Mentor

It means that you have allowed the RHS to be multiplied by zero and the LHS not. The equations doesn't hold for $r=0\,$.

3. Jun 14, 2017

### davidge

Does it mean we have to remove the point $(0,0)$ from the disk?

4. Jun 14, 2017

### Staff: Mentor

In order to do what? I consider $(dx,dy)$ as Cartesian coordinates. If we give them coefficients, zero shouldn't be a problem. To me it is as if you asked, whether the origin of the plane should be removed, because the standard basis vectors aren't zero, whereas the radius can be zero.

5. Jun 14, 2017

### davidge

Yes, this is what I was asking you
Me too
In order not to have a inconsistent equation when $r = 0$.

6. Jun 14, 2017

### Staff: Mentor

Well, the relation $(0,0) \longleftrightarrow \{0\} \times (0,2\pi]$ is obviously not one-to-one which causes problems. To remove a point is problematic, as it directly affects the functions defined on $\mathbb{D}^1$. I'd rather change the coordinate system depending on the task.

7. Jun 14, 2017

### davidge

Any examples of such coordinate systems?
This is not a specific problem, I was just thinking about what I mentioned in post #1.

8. Jun 14, 2017

### Staff: Mentor

I assume in most cases it doesn't make a difference. If so, I'd simply use the Cartesian version, but that's only because I'm no big fan of polar coordinates. It would be interesting to have an example, where it makes a difference to see, what exactly shouldn't work.

9. Jun 14, 2017

### WWGD

Basically, your coordinate change map is not defined for r=0:

We have:
$x=rcos\theta$
$y=rsin\theta$
But this holds for $r \neq 0$ only , as you said . Then the same is true for $dx \wedge dy = (\frac {\partial}{\partial \theta}d\theta+ \frac{\partial}{\partial r}dr )( rcos\theta+ ysin\theta)$. The wedge is not defined at 0 because the coordinate expression for r=0 is not defined.
EDIT: the one way I can see $dx \wedge dy$ turned into 0 is if the determinant of the Jacobian of a change of coordinate is 0. This is what $f(x)dx\wedge dy$ measures; how much f(x) stretches/shrinks the area element under a trasnformation.

Last edited: Jun 14, 2017
10. Jun 14, 2017

### davidge

Which is not the case we are considering, because for this case we as you said, $r=0$ is not on the domain, isn't?

11. Jun 14, 2017

### WWGD

Right, we are not really transforming the space, we are expressing it in different terms, i.e., changing coordinate systems, so no squashing ( nor stretching) is happenning. EDIT: By ( many) definition(s) , actually , a change of variable is supposed to be a diffeomorphism, meaning the ais invertible,and therefore its determinant is non-zero. Which kind of makes sense, that you would want that for a coordinate change.

Last edited: Jun 14, 2017
12. Jun 15, 2017

### Orodruin

Staff Emeritus
Am I the only one bothered by the fact that the set $[0,1]\times [0,2\pi)$ is not open and so the described mapping is not a chart? Polar coordinates can never give you a chart at $r=0$ and so it makes no sense to try to interpret coordinate expressions for tensors in polar coordinates at the origin.

13. Jun 15, 2017

### WWGD

But I think the map is just intended to be a (local) change of coordinates. Is this equivalent to $[0,1] \times [0,2\pi)$ being a chart?EDIT: I can see how a manifold is a way of assigning coordinates, but I had never seen things that way. Still, I agree, the coordinate change is a local diffeomorphism so it should preserve the topology of $\mathbb D^1$.

14. Jun 15, 2017

### Orodruin

Staff Emeritus
A chart is a homeomorphism from an open subset of the manifold to an open subset of $\mathbb R^n$. It does not need to cover the entire manifold. Remove $r=0$, $r=1$ and $\theta = 0$ and you have a chart that covers almost the entire disc.

15. Jun 15, 2017

### davidge

@Orodruin I've been considering $[0,1] \times [0,2 \pi)$ throughout this thread because that set covers the entire disk and I think it's the most familiar way of mapping points on the disk, although it's not homeomorphic to $\mathbb{R}^2$.
Is the area element and all other quantities defined only when we have charts? Is it what you mean here?

16. Jun 15, 2017

### Orodruin

Staff Emeritus
If you want to use the coordinate expressions at a point, you must use a chart that is valid at that point. Writing $r\, dr\wedge d\theta$ makes no sense at $r = 0$ because your coordinate system does not define a chart there. There are some charts that cover all of the disc, your coordinate system with the given set is not one of them because it is not a chart. If you use any chart that covers the origin (such as Cartesian coordinates), you will find that the area element is non-zero there.

17. Jun 15, 2017

### davidge

I got it. Thanks.

18. Jun 16, 2017

### lavinia

This mapping cannot be continuous.

In the other direction the map is continuous but singular.

Last edited: Jun 16, 2017
19. Jun 16, 2017

### davidge

Where it is singular?

20. Jun 16, 2017

### lavinia

The map in the other direction maps the entire half open interval ${0}×(0,2π]$ onto the center of the unit disk.