# Manfold embeddings

1. Jun 17, 2011

### TrickyDicky

One usually needs multiple local charts to cover a manifold ( no global coordinate system), the manifold in general is covered by local patches with overlapping areas.
But if we embed a certain manifold in a higher dimension space, like for instance embedding a 2-sphere in a Euclidean space, then from this space I guess we can cover the whole original manifold with a global coordinate system. If this was the case, the benefit seems clear, would there be any drawback?

2. Jun 17, 2011

### micromass

Staff Emeritus
Hi TrickyDicky!

Let's say we embed the 2-sphere in Euclidean space. What would be the global coordinate system you propose?

3. Jun 17, 2011

### quasar987

The number of coordinate charts required to cover a manifold has nothing to do with where it is embedded. It is an intrinsic property of the manifold governed by its global topology.

Assume the sphere has a global coordinate chart. Then it is homeomorphic to R^2. But this isn't so, as we know, because for instance, S² is compact and R² is not.

4. Jun 18, 2011

### TrickyDicky

I hope I'm not saying something completely wrong, maybe "global coordinate system" is not a very rigorous notion, but I guess it would be that of the Euclidean space, (x,y,z) in cartesian coordinates or (r,theta,phi) in spherical ones. So that all the points in the surface of a 2-sphere are defined in this Euclidean 3-space coordinates of the ambient space of the sphere.

Given this, I am thinking that a metric of this ambient space can be obtained in cartesian cordinates (x,y,z):

$$ds^2=dx^2+dy^2+dz^2$$

What would be the metric for a sphere in this coordinate system, in spherical coord. (r,theta,phi)?
I guess since we are in a flat Euclidean space if we consider the distance to an arbitrary origin of coordinates as r, we can describe a 2-sphere of radius r as the one which curvature radius is r^2:

$$ds^2=dr^2+r^2(d\theta^2+sin^2\theta d\phi)$$
I guess in this coordinate system the surface of the sphere would be coordinate-singularity free unlike if we constrain the line element to two coordinates.

5. Jun 18, 2011

### micromass

Staff Emeritus
OK, how would you then state that a map $\phi:S^2\rightarrow \mathbb{R}$ is differentiable, using this new coordinate system?

6. Jun 18, 2011

### TrickyDicky

Hmm..good point, I'm not sure at all of this but I vaguely recall that all topological manifolds up to maybe two or three dimensions is differentiable, so my guess is then in this case you wouldn't need it to be specifically stated?

7. Jun 18, 2011

### micromass

Staff Emeritus
I'm not talking about which manifolds are differentiable or not. I'm talking about which maps are differentiable. Certainly there are continuous maps $\phi:S^2\rightarrow \mathbb{R}$ which are continuous but not differentiable. The same applies for maps $\phi:S^2\rightarrow S^2$ for example.

In general, I'm not asking which manifolds are differentiable, but which maps between manifolds are differentiable!

8. Jun 18, 2011

### TrickyDicky

Ok, excuse my confusion, so you are saying there is no way to state that the maps are differentiable.

9. Jun 18, 2011

### micromass

Staff Emeritus
Indeed, I don't think there is an easy way to define differentiability. But I may be wrong of course.

Your coordinate system does have it's advantages. For example, it's much easier to define tangent spaces and all these things.

10. Jun 18, 2011

### TrickyDicky

Aha, so the drawback would be that you wouldn't have a way to assure differentiability of S^2 maps from this embedding so there could be just continuous maps to R, we could state differentiability of the embedding manifold but not of S^2, right?

11. Jun 18, 2011

### micromass

Staff Emeritus
That would be my criticism against such a global coordinate system.

12. Jun 18, 2011

### TrickyDicky

Thanks, micromass.

13. Jun 19, 2011

### George Jones

Staff Emeritus
In this example, isn't an equation of constraint needed? In Cartesian coordinates, this constraint equation is
$$0 = x^2 + y^2 +z^2 - R^2,$$
while in spherical coordinates, the constraint equation is just
$$0 = r - R,$$
where $R$ is constant. Then, the metric induced on the 2-sphere by spherical coordinates is
$$ds^2 = R^2 \left( d\theta^2 + sin^2\theta d\phi \right).$$
What happens when $\theta = 0$?

14. Jun 19, 2011

### TrickyDicky

Yes, that is the usual procedure, to use the sphere equation to express one of the coordinates as a function of the other two + the radius parameter, then you get the line element you wrote with R acting as parameter.
What happens when $\theta = 0$? Exactly what I was trying to avoid by using the coordinates of the embedding space, you find a limitation intrinsic to using a local chart, a coordinate-singularity at the equator in this example.
However if you found a singularity in the embedding manifold you would know is not a coordinate singularity, it is not very useful in the case of the 2-sphere since we know beforehand that it is singularity free surface. But in the case one didn't know, maybe this was a good strategy to ascertain if the points corresponding to the surface S^2 that look singular in 2-D coordinate system are truly singular or not in the embedding 3-space .

15. Jun 19, 2011

### Hurkyl

Staff Emeritus
If M is a submanifold of N, and N has a smooth structure, then M inherits that structure (unless it has singularities).

It becomes difficult to tell the difference between properties of your manifold, properties of your higher dimensional space, and properties of the embedding.

As I recall the history, this drawback was the primary reason that manifolds were invented.

16. Jun 19, 2011

### Hurkyl

Staff Emeritus
As an aside, (global) spherical coordinates aren't actually coordinates -- they are similar to your idea of embedding the manifold in some other space, but the other way around: covering your manifold with some other space.

Spherical coordinates are a coordinate chart on some "larger" three-dimensional space where each triple $(r, \theta, \phi)$ picks out a distinct point, together with a surjection from this space to $\mathbb{R}^3$ that sends $(r, \theta, \phi)$ to the appropriate point in Euclidean space.

Restricting to r=R and the corresponding sphere, Spherical coordinates are a mapping $\mathbb{R}^2 \to S^2$.

17. Jun 19, 2011

### micromass

Staff Emeritus
That doesn't really help since many subsets of a smooth manifolds won't be smooth manifolds themselfs. Indeed, it is only the submanifolds that inherirt the structure, but the OP doesn't want to define manifolds...

18. Jun 19, 2011

### George Jones

Staff Emeritus
Even at the level of topological manifolds, isn't
false because of what quasar987 posted? And couldn't there be other topological impediments to using coordinates for R^n to construct global coordinate systems on manifolds embedded in R^n?

19. Jun 19, 2011

### TrickyDicky

I think this is the key problem to my approach.

Let's go the opposite way, let's say we have a well defined 3-dimensional manifold N, but we are actually interested in a 2-dimensional submanifold M embedded in N, say because we believe this submanifold M models some physical entity.
It would be very hard to discern the properties of M (like curvature, singularities, covariance up to diffeomorphism, etc) from the properties of N, right?

20. Jun 19, 2011

### lavinia

the submanifold inherits a geometry from the manifold it is embedded in.