Infinite cylinder covered by a single chart

1. Jul 17, 2016

Figaro

1. The problem statement, all variables and given/known data
This is a problem from Spacetime and Geometry by Carroll,

Just because a manifold is topologically nontrivial doesn't necessarily mean it can't be covered with a single chart. In contrast to the circle $S^1$, show that the infinite cylinder $RxS^1$ can be covered with just one chart, by explicitly constructing the map.

2. Relevant equations

3. The attempt at a solution
Based on my understanding, a chart is a mapping from an open subset $U$ of a set $M$ to $R^n$ such that the image of $U$ is an open set in $R^n$.

From my searches, I've came across that I can use the annulus as the open subset because it is homeomorphic to the infinite cylinder, but how do I define/construct the mapping? I think there are a lot of ways?

2. Jul 17, 2016

andrewkirk

Map each line on the cylinder that runs parallel to the cylinder's axis of rotation, to a radial line in the annulus. Then set units along that radial line by parameterizing it in such a way that one approaches but never reaches the outer (inner) circumference of the annulus as $t\to\infty$ ($t\to-\infty$). The parameter $t$ can represent signed distance along the line on the cylinder from an arbitratily chosen starting point.

3. Jul 18, 2016

Figaro

Thanks for that but I just want to know, since the cylinder is made out of infinite circles stacked together, why is that the $S^1$ (circle) can at least be covered by two charts while the cylinder can be covered by one chart? Aren't they similar? For the circle, one round trip [0, 2π] certainly will be a closed set, so some people use two charts with domain (-π, π) and (0, 2π). Shouldn't the annulus method be valid also for the circle? So is there a general way of identifying how to construct a mapping such that, say one chart (as in this case) can cover the manifold (aside from the annulus method)?

4. Jul 18, 2016

andrewkirk

Because $S^1$ is one-dimensional whereas the cylinder is two-dimensional. The chart has to map to a subset of Euclidean space of the same number of dimensions, and Euclidean 2-space contains circular subsets whereas Euclidean 1-space does not.
No, because the annulus method maps the (cylindrical) manifold to a subset of $\mathbb R^2$ and a chart of $S^1$ must map into $\mathbb R^1$ instead.
No. Classifying manifolds is hard. There is no simple, general rule.

5. Jul 18, 2016

Figaro

Can I set the inner radius to 0 such that I have the map $φ: U(θ, z) → R^2~~~$ where $φ(U) = (θ, e^z), ~~~ -∞ < z < ∞, ~0≤θ<2π$

But I know $0≤θ<2π$ is not open, what do you think?

6. Jul 18, 2016

andrewkirk

That works. The image of that chart is a pierced plane not an annulus, but that's OK because Carroll's question doesn't specifically ask for an annulus. If you really wanted an annulus you'd need to use a function that is bounded above. One that works nicely is arctan (the logistic function is another).
It doesn't matter that $[0,2\pi)$ is not open. What has to be open is the domain of the chart, not necessarily all coordinate slices thereof.

7. Jul 18, 2016

Figaro

So you mean the map $φ: U(θ, z) → R^2~~~$ where $~φ(U) = (θ, arctan(z)), ~~~ -∞ < z < ∞, ~0≤θ<2π$
But doesn't that mean any function would work as long as it is bounded above and below?

So the cartesian product of a half open and open set is open? Hmmm...

8. Jul 18, 2016

andrewkirk

No.
The Cartesian product specifies only the sets, not the topology, so that statement has no meaning. One can put either a topology on the Cartesian product that makes the set open, or one that makes it half open. The product topology would make it half open, but the product topology is different from the subspace topology that the image inherits from $\mathbb R^2$ in this case.

9. Jul 19, 2016

Figaro

$[0, 2π )~$ doesn't cover a whole circle, so if the given problem is a $S^1$ (circle), this would pose a problem since it doesn't cover the whole so we need to add another chart just to cover the "hole". But in this case, it is an infinite cylinder, so $[0, 2π )~$ being not open doesn't matter because $-∞<z<∞$ is open, so the domain will be open? Is this what you are pointing out?

10. Aug 5, 2016

haruspex

It doesn't?

11. Aug 9, 2016

Figaro

Yes, it doesn't, 2π is not included.

12. Aug 9, 2016

andrewkirk

The circle is covered by the following map $f$, whose domain is $[0,2\pi)$. Here the circle $S^1$ is considered embedded in $\mathbb R^2$ as the circle of unit radius centred on the origin.

$f$ is the map from $[0,2\pi)$ to $S^1$ such that $f(\theta)$ is the point in $\mathbb R^2$ whose polar coordinates are $(1,\theta)$. This map is surjective onto $S^1$ and in that sense 'covers' it. The point that would be $f(2\pi)$ if $2\pi$ were included in the domain is already covered by $f(0)$.

The reason why $S^1$ is not homeomorphic to $[0,2\pi)$ is not that there is no 'covering' map but rather that the inverse of $f$ is not continuous.

13. Aug 10, 2016

haruspex

How is the 0 point on the circle different from the 2π point?

14. Aug 10, 2016

Figaro

Sorry, what I mean to say, it is not going to return to the original point, it just keeps getting closer but not exactly so in a sense it doesn't cover all? It is like running in an Olympic circle starting from the START but when you almost ran a full circle, you don't reach the START but just keep getting closer.

15. Aug 10, 2016

haruspex

In that analogy, the runner does not need to reach the start line since the runner has already been there.
All that is required is that for any given point on the circle there is a point in [0, 2π) that maps to it.

16. Aug 11, 2016

Figaro

Oh, fair enough, thanks for the clarification.