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Why there cannot be a single chart for n-sphere?

  1. Mar 17, 2012 #1
    My question is about a general comment made in differential geometry books:
    There cannot be a single chart for the manifold n-sphere.
    I have been trying a proof for long but ...
    Can you help me out?
  2. jcsd
  3. Mar 17, 2012 #2


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    What attempts at proof have you tried?
  4. Mar 17, 2012 #3
    I haven't really tried to prove that there cannot be a single chart. But the usual charts that we have, like the 2 charts using stereographic projection, the 6 charts using hemispheres...and many others---all produce atleast 2 charts.
  5. Mar 17, 2012 #4
    What happened was I was reading a book where they have discussed the construction of charts for the figure-8: {(sin t, sin 2t): t ε (0,2π)}. Using the subspace topology of R^2, you cannot make it into a manifold. But there are topologies that admit its manifold structure, and there is a topology that has a single chart for figure-8.
    From this example, we were trying to prove the existence of single charts for S^1: {(cos t, sin t): t ε [0,2π)}.
  6. Mar 17, 2012 #5
    There are several equivalent definitions of coordinate chart, and one says that a map is a chart if it is a homeomorphism from open subset of manifold to wholen. So if there was a single chart for a sphere or for a circle, it would have to be homeomorphic to ℝn.

    There are problems with this "chart". For example, it's range is not open. And even if you wanted to remedy this situation by saying that circle is a manifold with boundary (so that [0,2π) is open in H1), it still doesn't satisfy one other crucial condition to count as a chart.
  7. Mar 17, 2012 #6
    Could you please share that topology? I frankly don't see how to do it, unless you have different definition of (topological) manifold.
  8. Mar 17, 2012 #7
    I agree. But is there a proof that there CANNOT be ANY single chart for Sn?
  9. Mar 17, 2012 #8
    In stead of taking the subspace topology of ℝ^2, consider open intervals in (0,2π) to get mapped to the figure-8. This is indeed a homeomorphism, as in case of subspace topology of ℝ^2, it was not (because of the presence of the "cross" at the middle which was making our life hard).
  10. Mar 17, 2012 #9
    Well, if you want it formaly, I would do proof by contradiction.

    Suppose there is a single chart for all Sn. Then, by definition, there is a homeomorphism between Rn and Sn. But Rn and Sn are not homeomorphic. So contradiction.
  11. Mar 17, 2012 #10
    Sorry, still don't see it. How exactly do you map open intervals in (0,2π) to figure 8?

    There still have to be open set containing the middle point of figure 8 homeomorphic to, say, (0, 1), but IMHO that's impossible.

    Also, every connected 1-manifold is homeomorphic to either circle or real line, so it can be embedded into ℝ^2, so it has to have inherited topology from R^2.
  12. Mar 17, 2012 #11
    map the nbd of π in (0, 2π) to one branch at the middle of fig8, right nbd of 0 to one portion of the remaining branch near the middle of fig8 and left nbd of 2π to the remaining portion of the remaining branch near the middle of fig8. Thats how u can cover the whole of fig8.
  13. Mar 17, 2012 #12
    Well, but is inverse of this function really continuous? Homomorphism is bijection that is continuous and has continuous inverse.
  14. Mar 17, 2012 #13
    Inverse will be continuous if you look at fig8 as a separate object not embedded in R^2. If you consider it to be embedded in R^2, then it will not be a homeomorphism.
  15. Mar 17, 2012 #14
    R^2 doesn't really need to be considered. For this map to qualify as a chart, it would need to be homeomorphic to (0, 1) or (0, 2π) or R with usual euclidean topology (by definition of chart). What book are you reading?
  16. Mar 17, 2012 #15
    Also see Locally Euclidean.

    It is also possible to to define manifold using only coordinate charts without any reference to underlying topology, with that the topology can be inherited. So with manifolds, your are really restricted when it comes to choosing underlying topology. See this example of nonmanifold.
  17. Mar 17, 2012 #16
    A course in differential geometry-s.kumaresan.
  18. Mar 17, 2012 #17
    Try to make fig-8 with a piece of open string. The map is trivial.
  19. Mar 17, 2012 #18
    this is your answer, one is compact, the other is not.
  20. Mar 17, 2012 #19
    Ok, to convince you 8 is not a manifold, try this: since you say it is a manifold, there has to be open nbd around every point homeomorphic to (0, 1). So find me homeomorphism between figure X, which is a ndb of middle point of 8, and (0, 1). The (0, 1) interval is with usual euclidean topology.
  21. Mar 17, 2012 #20
    note carefully my figure 8: {(sin t, sin 2t): t ε (0, 2π)}. It is really not a 8. The middle portion is not a "cross". If t ε [0, 2π], then it is certainly not a manifold. By my "figure 8" IS a manifold and I am sure you know how. :)
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