Single chart which covers entire [itex]S^1\times R[/itex] manifold

[stevendaryl]

Only if the range of $\theta$ is at least half-closed, i.e., it needs to include $\theta = 0$ or $\theta = 2 \pi$. But a chart is supposed to map open intervals to open intervals. If the range of $\theta$ is the open interval $(0, 2 \pi)$, then the image is missing an entire ray in the $x, y$ plane, from the origin to infinity along the positive $x$ direction.

The region S(A,B) on a cylinder with A < z < B and 0 \leq \theta \leq 2 \pi is an open set. To see this, note that it is the union of two open sets:

1. S_1(A,B) is the set of all points (z,\theta) with A < z < B and 0 < \theta < \pi
2. S_2(A,B) is the set of all points (z,\theta) with A < z < B and \frac{\pi}{2} < \theta < \frac{5 \pi}{2}

By definition of open, a union of two (or even countably many) open sets is another open set. If we take countably many such open sets, we get another open set:

S(0,\infty) = S(0, 1) \cup S(\frac{1}{2}, 2) \cup S(\frac{3}{2}, 3) \cup ...

Also, if the chart has to map open intervals to open intervals, the domain of the chart is not all of $S^1$; it must be missing at least one point (the one corresponding to $\theta = 0$ and/or $\theta = 2 \pi$). I always understood that this was the primary reason why closed manifolds could not be covered by a single chart.

The entire S^1 is an open interval, since it is the union of the interval 0 < \theta < \pi and \frac{\pi}{2} < \theta < \frac{5 \pi}{2}.

So it's a weird fact that S^1 cannot be covered by a single chart, but S^1 \times R can be.

 

[DrGreg]

Only if the range of $\theta$ is at least half-closed, i.e., it needs to include $\theta = 0$ or $\theta = 2 \pi$. But a chart is supposed to map open intervals to open intervals. If the range of $\theta$ is the open interval $(0, 2 \pi)$, then the image is missing an entire ray in the $x, y$ plane, from the origin to infinity along the positive $x$ direction.

Also, if the chart has to map open intervals to open intervals, the domain of the chart is not all of $S^1$; it must be missing at least one point (the one corresponding to $\theta = 0$ and/or $\theta = 2 \pi$). I always understood that this was the primary reason why closed manifolds could not be covered by a single chart.

If $\theta$ was a member of the real line $\mathbb{R}$, your objection would be valid. But it's not, it's a member of $S^1$, "real numbers mod 2$\pi$", so "$\theta = 0$" and "$\theta = 2 \pi$" are two descriptions of the same point in $S^1$, and of the same point in $\mathbb{R}^2$ under stevendaryl's mapping (for a given $z$).

 

[Cruz Martinez]

Only if the range of $\theta$ is at least half-closed, i.e., it needs to include $\theta = 0$ or $\theta = 2 \pi$. But a chart is supposed to map open intervals to open intervals. If the range of $\theta$ is the open interval $(0, 2 \pi)$, then the image is missing an entire ray in the $x, y$ plane, from the origin to infinity along the positive $x$ direction.

Also, if the chart has to map open intervals to open intervals, the domain of the chart is not all of $S^1$; it must be missing at least one point (the one corresponding to $\theta = 0$ and/or $\theta = 2 \pi$). I always understood that this was the primary reason why closed manifolds could not be covered by a single chart.

I think you're repeating some of the misconceptions you had on a previous thread here. For example, the reason that $\mathbb{S}^1$ cannot be covered by a single chart is not that it is a 'closed' manifold (All manifolds are closed).

 

[lavinia]

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?

S1xR is homeomorphic to an open subset of R2. Just take R2 minus the origin. So it is a manifold that can be covered by a single chart and is topologically nontrivial - by which I suppose you mean it is not contractible.

S2xR is homeomorphic to an open subset of R3.

 

[Ravi Mohan]

Thanks for explaining that. Now regarding my second question

when we define the homeomorphism from an open set of the manifold to $\mathbb{R}^n$, do we assume the topology of $\mathbb{R}^n$ to be usual topology?

 

[lavinia]

Thanks for explaining that. Now regarding my second question

yes. whenever one says Rn one means the usual topology.

 

[Ravi Mohan]

Ok that is great. I think I have got the answers to my satisfaction here.

 

[PeterDonis]

If $\theta$ was a member of the real line $\mathbb{R}$, your objection would be valid. But it's not, it's a member of $S^1$, "real numbers mod 2$\pi$", so "$\theta = 0$" and "$\theta = 2 \pi$" are two descriptions of the same point in $S^1$, and of the same point in $\mathbb{R}^2$ under stevendaryl's mapping (for a given $z$).

Ah, got it.

 

[PeterDonis]

I think you're repeating some of the misconceptions you had on a previous thread here.

Not sure what previous thread you're referring to (possibly you have confused me with another poster?), but see my response to DrGreg.

All manifolds are closed

Doesn't a closed manifold have to be compact? Not all manifolds are compact.

 

[lavinia]

Doesn't a closed manifold have to be compact? .

Yes. And it doesn't have a boundary.

 

[Cruz Martinez]

Not sure what previous thread you're referring to (possibly you have confused me with another poster?), but see my response to DrGreg.
Doesn't a closed manifold have to be compact? Not all manifolds are compact.

In which sense do you mean closed?

Edit: This thread //www.physicsforums.com/threads/general-relativity-and-curvilinear-coordinates.818066/page-6#post-5141351

 

[lavinia]

a closed manifold can not be covered by a single chart since no open subset of Rn is compact, So the circle can not be covered by a single chart - nor can a sphere or a doughnut.

- a manifold might not be closed even as a metric space, for instance Rn minus a point.

 

[Cruz Martinez]

a closed manifold can not be covered by a single chart since no open subset of Rn is compact, So the circle can not be covered by a single chart - nor can a sphere or a doughnut.

- a manifold my not be closed even as a metric space.

Which is exctly why i asked in which sense he means closed this time, because last time i saw he used it in a very different (wrong) way. With this of course i agree that the sphere cannot be covered by one chart since homeos preserve compactness as all continuous maps do.

 

[PeterDonis]

In which sense do you mean closed?

In the sense of a closed manifold. In the previous thread you linked to, it was pointed out (and I agreed once it was pointed out) that, topologically speaking, the entire space is always closed. But "closed" in the sense of a closed manifold is something more specific: a compact manifold without boundary. Not all manifolds meet that definition. And, as you appear to agree, such a manifold cannot be covered by a single chart. Which was what I said about $S^1$ earlier.

By the same definition, $S^1 \times R$ is not a closed manifold (it's not compact), so it can be covered by a single chart, as stevendaryl showed (and after his and DrGreg's correction, I see how that works, as I posted before).

 

[lavinia]

Which is exctly why i asked in which sense he means closed this time, because last time i saw he used it in a very different (wrong) way. With this of course i agree that the sphere cannot be covered by one chart since homeos preserve compactness as all continuous maps do.

You said that every manifold is closed. But whatever definition of closed you choose this is not true. The plane minus a point is not closed as a metric space and is not a closed manifold.

 

[lavinia]

A manifold that is not closed may still not be coverable with a single chart. An example is a torus minus a point.

 

[Cruz Martinez]

You said that every manifold is closed. But whatever definition of closed you choose this is not true. The plane minus a point is not closed as a metric space and is not a closed manifold.

I meant closed in the sense of 'a closed subset of its own topology' which is the sense in which peter tried to use the term last time. It is certainly true for any manifold that it is a closed subset of its own topology! I tried to interpret what peter said this time and i was simply wrong in that interpretation. This doesn't meant i don't know a closed manifold is compact without boundary, and i will leave it at that.