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

  1. Jun 17, 2015 #1
    I am studying Carroll's notes on GR. He defines charts as maps from an open subset in manifold [itex]M[/itex] to open subset in [itex]R^n[/itex]. 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. jcsd
  3. Jun 17, 2015 #2

    Ibix

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    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.
     
  4. Jun 17, 2015 #3
    Thanks for clearing that up.
     
  5. Jun 17, 2015 #4
    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.
     
  6. Jun 17, 2015 #5
    Ah, I was wondering about it. So how exactly do we define open set in the manifold? In [itex] R^n[/itex] (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 [itex] y[/itex] in [itex] R^n[/itex] such that [itex] |x-y| < r[/itex] for some fixed [itex]x[/itex] in [itex] R^n[/itex]).
     
  7. Jun 17, 2015 #6
    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
  8. Jun 17, 2015 #7

    micromass

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    I'm sure you're right, but I have never seen that convention before.
     
  9. Jun 17, 2015 #8

    micromass

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    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.
     
  10. Jun 17, 2015 #9

    Ibix

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    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 [itex]\subset[/itex] to mean strict subset and [itex]\subseteq[/itex] to mean subset, while some use [itex]\subset[/itex] to mean subset and don't use [itex]\subseteq[/itex]. It doesn't say that they use the names that way.

    I mis-read it. Apologies.
     
  11. Jun 18, 2015 #10
    Just to be clear, do you mean open sets in ##\mathbb{R}##?
     
  12. Jun 18, 2015 #11
    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
     
  13. Jun 18, 2015 #12
    Ok, that is interesting. Thank you very much Cruz and micromass.
     
  14. Jun 18, 2015 #13

    Ben Niehoff

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    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.
     
  15. Jun 18, 2015 #14
    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.
     
  16. Jun 18, 2015 #15

    micromass

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    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.
     
  17. Jun 18, 2015 #16

    Ibix

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    I understood that it could - I've certainly seen solutions on this forum that purport to do it. Were they wrong?
     
  18. Sep 20, 2015 #17
    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?
     
  19. Sep 20, 2015 #18

    micromass

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    No, we can't. There is no homeomorphism from the cylinder to any ##\mathbb{R}^n##.
     
  20. Sep 20, 2015 #19
    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##.
     
  21. Sep 20, 2015 #20

    micromass

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    What does non-trivial topology even mean?
     
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