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I Boundary and homeomorphism

  1. May 21, 2017 #1
    The 2-sphere ##\mathbb{S}^2## can be expressed as the product ##\mathbb{S}^1 \times \mathbb{S}^1##
    Now can we express ##\mathbb{S}^1## as ##\mathbb{S}^1 \subset (-a,a)##, where ##(-a,a)## is some open interval of ##\mathbb{R}##? If so, then (I think) ##\mathbb{S}^1## is homeomorphic to ##\mathbb{R}##. Also, it's homemomorphic to ##\mathbb{R}^2## up to four coordinate charts covering it in ##\mathbb{R}^2##.

    If so, by the same reasoning ##\mathbb{S}^2 \subset \mathbb{R}^2## is homeomorphic to ##\mathbb{R}^2##. Also, it's homeomorphic to ##\mathbb{R}^3##, for we can define an embedding from it to ##\mathbb{R}^3##.

    Finally, if the above is correct, ##\mathbb{S}^2 \times \mathbb{R}## is homeomorphic to ##\mathbb{R}^3##, though it's not compact.

    I'm trying to get this because I'm interested in knowing whether ##\mathbb{S}^2 \times \mathbb{R}## as a manifold has a boundary or not.

    I'll appreciate any help.
     
    Last edited: May 21, 2017
  2. jcsd
  3. May 22, 2017 #2
    I was taught long ago that ##S^1\times S^1## is two dimensional torus not sphere.
     
    Last edited: May 22, 2017
  4. May 22, 2017 #3
    Oh, I thought the 2-sphere could be represented as ##\mathbb{S}^1 \times \mathbb{S}^1## because each one of the coordinates of a point ##x \in \mathbb{S}^2##, say ##(\theta, \phi)##, obeys a "circle equation".
     
  5. May 22, 2017 #4

    TeethWhitener

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    Also, ##\mathbb{S}^1## is not homeomorphic to ##\mathbb{R}## (because it's not homeomorphic to an open interval of ##\mathbb{R}##). Think about the values ##\theta## ranges over.
     
  6. May 22, 2017 #5
    So why can we map the entire circle using four coordinate charts from an open interval of ##\mathbb{R}##? Can you check out my work below and say whether it's correct or not?

    monlm7f.png
     
  7. May 22, 2017 #6

    fresh_42

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    This is locally (in a neighborhood of each point) true, but not globally (on the entire sphere), because you will have to leave out one of the poles.
     
  8. May 22, 2017 #7

    TeethWhitener

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    I can't really read what's going on here, but the easiest way to show that ##\mathbb{S}^1## is not homeomorphic to an open interval of ##\mathbb{R}## is to note that ##\mathbb{S}^1## is compact whereas an open interval of ##\mathbb{R}## is not. Since homeomorphisms preserve compactness, we immediately get that the two spaces are not homeomorphic.
     
  9. May 22, 2017 #8
    What if we choose a closed interval instead of a open one. A closed interval is compact.
    Then letting the "end points", namely the poles out of consideration is forbidden?
     
  10. May 22, 2017 #9
    yes, but it's precisely what I was trying to say in post number 1. Pick a closed interval. There's no way of homeomorphicaly map the entire circle to ##\mathbb{R}^2## using an closed interval (at least in that way I did in post #5). So you have to "remove the end points" of your closed interval. Then now you have an open interval that is homeomorphic to ##\mathbb{R}##.

    Can you point out where I'm wrong on the above ?
     
  11. May 22, 2017 #10

    fresh_42

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    It's not forbidden, it simply makes the difference between local and global. As entire set ##S^1 \times S^1 \ncong S^2## as a torus (on the left) isn't a sphere (on the right), because one has a hole and the other has not. But on a small patch you can always apply coordinates like on earth. You simply cannot find a single chart describing both. As manifolds they all have local flat charts patched to a whole atlas, but this doesn't make them equal.
     
  12. May 22, 2017 #11

    TeethWhitener

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    I think you answered your own question. Look here:
    https://www.physicsforums.com/threads/why-the-circle-cant-be-homeomorphic-to-a-real-interval.537897/
    http://mathhelpforum.com/differenti...-not-homeomorphic-any-interval-real-line.html
    Combining with what I said before: an open interval on ##\mathbb{R}## is not homeomorphic to ##\mathbb{S}^1## because compactness isn't preserved. A half-open interval won't work for the same reason. And as you said, there's no homeomorphism between a closed interval of ##\mathbb{R}## and ##\mathbb{S}^1##. The links above answer why: choose a closed real interval ##[a,b]## and assume a (edit: bijection homeomorphism) exists ##f: [a,b] \mapsto \mathbb{S}^1##. Then deleting any point ##p## (not equal to ##a## or ##b##) from the real interval gives another homeomorphism: ##g: [a,p) \cup (p,b] \mapsto \mathbb{S}^1 - \{f(p)\}##. But the new real interval ##[a,p)\cup(p,b]## is not connected, whereas ##\mathbb{S}^1 - f(p)## is connected. Since connectedness is a topological property, ##g## can't be a homeomorphism (and by extension, neither can ##f##).

    EDIT: one important note. The set ##\mathbb{R} \cup \{\infty\}## is homeomorphic to the circle. This is sometimes called the "point at infinity." It's a special case of one point compactification:
    https://en.wikipedia.org/wiki/Alexandroff_extension
     
    Last edited: May 22, 2017
  13. May 22, 2017 #12
    So can we apply the definition of connectness to intervals like the one above? I thought the definition could be applied only to open intervals. Also, doing the same as you did, but with an open interval ##(a,b)##, the same reasoning would led us to conclude that ##\mathbb{S}^1## and ##(a,b)## are not homeomorphic. Cool!
     
  14. May 22, 2017 #13
    There is another funny proof that ##[a,b]## is not homeomorphic to ##S^1##. Indeed, any continuous function ##f:[a,b]\to[a,b]## has a fixed point. But there is a continuous function of ##S^1## to itself that does not have a fixed point.
     
  15. May 22, 2017 #14

    WWGD

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    There is a result, I think by Ulam, that ## \mathbb S^n ## cannot be embedded in ## \mathbb R^n ## -or-lower. Still, it would be interesting to see whether ## \mathbb S^n ## " Is a square root" for some n , meaning whether it is homeomorphic to the product ## Y \times Y ## of some topological space ## Y ## . There is an argument showing ## \mathbb R^{2n+1} ## is not a square root in this sense.
     
    Last edited: May 22, 2017
  16. May 22, 2017 #15

    WWGD

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    Nice. And there is also the connectivity one that any point p separates [a,b] , i.e., for any p in [a,b] , [a,b] -{p} is disconnected, while the same is not true for ## \mathbb S^1 ## - or higher-dimensions. I don't know the formal name for this, though ; it comes down to that if X is homeo to Y through h, then X-{p} is homeo to Y-{h(p)}. EDIT 2, besides, with heavier machinery, you also have contractibility, etc. which is preserved under homotopy equivalence alone. EDIT 3:
     
    Last edited: May 22, 2017
  17. May 22, 2017 #16

    fresh_42

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    Nice idea. Or one can take ##S^1 \times S^1 \cong \mathbb{P}^2(1,\mathbb{R}) \cong \mathbb{P}(2,\mathbb{R}) \ncong \mathbb{P}(1,\mathbb{C}) \cong S^2\;##, which indicates the problem with the poles.
     
    Last edited: May 22, 2017
  18. May 22, 2017 #17
    what do you mean by having a fixed point?
     
  19. May 22, 2017 #18

    WWGD

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    Yet another way :
    It means there is a point that maps to itself, i.e., f(p)=p.
     
  20. May 22, 2017 #19
    Thanks
     
  21. May 22, 2017 #20

    WWGD

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    No problem, consider re, Zwierz' post and fixed points the (continuous) map from ##\mathbb S^1 ## to itself that rotates every point by a fixed amount that is not an integer multiple of ## 2\pi ## .
     
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