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

[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?

 

[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.

 

[Ravi Mohan]

Thanks for clearing that up.

 

[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.

 

[Ravi Mohan]

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.

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 $R^n$).

 

[Cruz Martinez]

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 $R^n$).

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.

 

[micromass]

some people use subset in the sense of "strict subset"

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

 

[micromass]

Ah, I was wondering about it. So how exactly do we define open set in the manifold?

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.

 

[Ibix]

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

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.

I mis-read it. Apologies.

 

[Ravi Mohan]

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.

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

 

[Cruz Martinez]

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

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

 

[Ravi Mohan]

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

 

[Ben Niehoff]

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?

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.

 

[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.

 

[micromass]

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.

 

[Ibix]

You are correct. $S^1 \times \mathbb{R}$ cannot be covered with a single chart.

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

 

[Ravi Mohan]

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

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 $\mathbb{R}^n$.

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

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?

 

[micromass]

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 $\mathbb{R}^n$.

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

 

[Ravi Mohan]

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

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 $\mathbb{R}^n$.

 

[micromass]

What does non-trivial topology even mean?

 

[micromass]

Anyway, it is true that $S^1\times \mathbb{R}$ is homeomorphic to an open subset of $\mathbb{R}^2$. Maybe this is what you meant?

 

[Ravi Mohan]

What does non-trivial topology even mean?

Well I can't give reference to it, but I "guess" every topology which is not trivial is non-trivial topology (like usual or discrete topology). Ok so my justification may not be valid.

Anyway, it is true that $S^1\times \mathbb{R}$ is homeomorphic to an open subset of $\mathbb{R}^2$. Maybe this is what you meant?

Oh that is great. Can you please define that homeomorphism?

 

[micromass]

Oh that is great. Can you please define that homeomorphism?

I really don't feel like writing it out, but maybe you can "see" that it is possible to define a homeomorphism with this https://en.wikipedia.org/wiki/Annulus_(mathematics) Note the second sentence which basically says what you want.

The open annulus is topologically equivalent to both the open cylinderS1 × (0,1) and the punctured plane.

 

[stevendaryl]

Well I can't give reference to it, but I "guess" every topology which is not trivial is non-trivial topology (like usual or discrete topology). Ok so my justification may not be valid.

Oh that is great. Can you please define that homeomorphism?

Let original coordinates for $S^1 \times R$ be $(\theta, z)$, where $\theta$ measures the distance along $S^1$ and $z$ measures the distance along $R$. Then we map to the $x-y$ plane via:

$x = e^{z} cos(\theta)$
$y = e^{z} sin(\theta)$

The image of the mapping is the entire plane minus the point at $x=0, y=0$.

 

[Ben Niehoff]

Do charts need to be contractible? For some reason I had thought that was part of the definition, but it is not mentioned on the Wiki article for manifolds.

 

[micromass]

Do charts need to be contractible? For some reason I had thought that was part of the definition, but it is not mentioned on the Wiki article for manifolds.

I guess it depends on the definition. But in most math texts I own (eg Lee, smooth manifolds), contractibility is not assumed.

 

[Ben Niehoff]

OK, so clearly, if charts need not be contractible, then the cylinder can be covered in one chart, while the circle cannot. Interesting...

 

[Ravi Mohan]

I really don't feel like writing it out, but maybe you can "see" that it is possible to define a homeomorphism with this https://en.wikipedia.org/wiki/Annulus_(mathematics) Note the second sentence which basically says what you want.

I don't "see" it as you are, but I think I certainly have something to discuss with my graduate friends in the mathematics department .

Ok I see it now. stevendaryl's solution is essentially the annulus (the open annulus).

 

[Ravi Mohan]

OK, so clearly, if charts need not be contractible, then the cylinder can be covered in one chart, while the circle cannot. Interesting...

My thoughts exactly.

Just one more 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?

 

[PeterDonis]

The image of the mapping is the entire plane minus the point at $x=0, y=0$.

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.

 

[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?