# Single chart which covers entire $S^1\times R$ manifold

1. Jun 17, 2015

### Ravi Mohan

I am studying Carroll's notes on GR. He defines charts as maps from an open subset in manifold $M$ to open subset in $R^n$. He then writes

"We therefore see the necessity of charts and atlases: many manifolds cannot be covered with a single coordinate system. (Although some can, even ones with nontrivial topology. Can you think of a single good coordinate system that covers the cylinder, S1 × R?)"

So when we have defined the chart as a map whose domain is an open subset in the manifold then how can a single chart cover the entire manifold? Or am I missing something?

2. Jun 17, 2015

### Ibix

A "subset" U of a set S can include everything in S - the only restriction is that it cannot contain anything that is not in S. A "strict subset" or "proper subset" must be entirely within S, but exclude at least one thing.

I gather that this convention is not universal - some people use subset in the sense of "strict subset" (i.e., as you were reading it). Carroll is apparently not one of them.

3. Jun 17, 2015

### Ravi Mohan

Thanks for clearing that up.

4. Jun 17, 2015

### Cruz Martinez

Also remember that in a topological space S, the whole of S is an open set, for any topology defined on it. A manifold M is a topological space, therefore M itself is an open set, for any manifold.

5. Jun 17, 2015

### Ravi Mohan

Ah, I was wondering about it. So how exactly do we define open set in the manifold? In $R^n$ (the set to which a chart maps the manifold elements) we make open set from the union of open balls (by the open ball I mean set of all the points $y$ in $R^n$ such that $|x-y| < r$ for some fixed $x$ in $R^n$).

6. Jun 17, 2015

### Cruz Martinez

The coordinate charts are homeomorphisms from the coordinate neighborhoods to R^n, so an open set in the manifold is the image of an open set of R^n, under the inverse of the coordinate map.
Now, all of the subsets of M satisfying the above are open in M by definition, but strictly speaking, they only are a basis for a topology on M. A general open set in M is then any subset that can be expressed as a union of sets belonging to the basis we defined.

Last edited: Jun 17, 2015
7. Jun 17, 2015

### micromass

Staff Emeritus
I'm sure you're right, but I have never seen that convention before.

8. Jun 17, 2015

### micromass

Staff Emeritus
Usually, the concept of manifolds come equipped with a collection of open sets to use. So we don't have to define anything. If you're looking at specific manifolds like $S^1$, then the open sets come from the restriction of open sets in $\mathbb{R}^2$. With this collection, it becomes a manifold.

9. Jun 17, 2015

### Ibix

I'm wrong. I used Wikipedia to get the names for strict subsets, which had escaped me. Its page on subsets notes (citing Rudin's Real and Complex Analysis) that some authors use $\subset$ to mean strict subset and $\subseteq$ to mean subset, while some use $\subset$ to mean subset and don't use $\subseteq$. It doesn't say that they use the names that way.

10. Jun 18, 2015

### Ravi Mohan

Just to be clear, do you mean open sets in $\mathbb{R}$?

11. Jun 18, 2015

### Cruz Martinez

Nope, he means R^2. The circle S^1 is a subset of R^2 with the subspace topology. For the definition of the subspace topology, look here: https://en.wikipedia.org/wiki/Subspace_topology

12. Jun 18, 2015

### Ravi Mohan

Ok, that is interesting. Thank you very much Cruz and micromass.

13. Jun 18, 2015

### Ben Niehoff

You are correct. $S^1 \times \mathbb{R}$ cannot be covered with a single chart. Note that Carroll's notes may have small mistakes in them, which were corrected in the published book.

Since practically the definition of "non-trivial topology" is that a manifold requires multiple charts, that certainly seems like an odd statement to make.

14. Jun 18, 2015

### Ravi Mohan

I am little confused. As Cruz and micromass have noted, the manifolds are open sets, so the domain of a chart can be the whole manifold (which is a subset of itself). Had the manifolds been closed sets, then it would have been a problem because the open subset of a closed set can't cover the entire set.

15. Jun 18, 2015

### micromass

Staff Emeritus
You are misunderstanding the subset topology. It is true that a manifold is both open and closed IN ITSELF. Of course, the circle $S^1$ is not open in $\mathbb{R}^2$ (but it is open in itself), and the half-line $(0,+\infty)$ is a manfiold that is not closed in $\mathbb{R}$ (but it is closed in itself). The thing here is that open and closed are relative notions depending on the underlying space.

16. Jun 18, 2015

### Ibix

I understood that it could - I've certainly seen solutions on this forum that purport to do it. Were they wrong?

17. Sep 20, 2015

### Ravi Mohan

Even if we consider the manifold with a non-trivial topology, we do have an open set which is the entire set itself and we can simply (in principle) define a homeomorphism from it to $\mathbb{R}^n$.

But in this case ($S^1\times\mathbb{R}$), I am not able to find a suitable single homeomorphism to $\mathbb{R}^2$. Can you point to the solutions in this forum?

18. Sep 20, 2015

### micromass

Staff Emeritus
No, we can't. There is no homeomorphism from the cylinder to any $\mathbb{R}^n$.

19. Sep 20, 2015

### Ravi Mohan

Ok. I guess I formulated that statement incorrectly. I just wanted to justify Carroll's statement that a manifold with a non-trivial topology can be covered by a single chart. An example would be $\mathbb{R}^n$.

20. Sep 20, 2015

### micromass

Staff Emeritus
What does non-trivial topology even mean?