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I Do these manifolds have a boundary?

  1. May 3, 2017 #1
    Any open subset of ##\mathbb{R}^{n}##;
    The n-Sphere, ##\mathbb{S}^n##;
    The Klein Bottle.

    I guess they don't have a boundary, as a neighborhood of any point of them is homeomorphic to ##\mathbb{R}^n##.
    I'd like to know whether my guess is correct and whether the reason I'm giving for them not to have a boundary is valid.

    (I actually found on web that the Klein Bottle does not have a boundary.)
     
    Last edited: May 3, 2017
  2. jcsd
  3. May 3, 2017 #2

    lavinia

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    None of these manifolds have a boundary and your reason is correct.
     
  4. May 3, 2017 #3
    Thnx
     
  5. May 3, 2017 #4
    Can you explain that a bit more ? I understand what homeomorphism is from Wikipedia - https://en.wikipedia.org/wiki/Homeomorphism but why would being homeomorphic result in not having a boundary ?
     
  6. May 3, 2017 #5

    lavinia

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    What definition of the boundary of a manifold are you thinking of?
     
  7. May 3, 2017 #6
    I think he is looking for an answer of why manifolds without boundary have to be homeomorphic to ##\mathbb{R}^n##.
     
  8. May 3, 2017 #7

    martinbn

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    They don't have to be, for example the sphere.
     
  9. May 3, 2017 #8
    I meant the points on the manifold have to be a neighborhood homeomorphic to ##\mathbb{R}^n##. Is this not so?
     
  10. May 3, 2017 #9

    lavinia

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    yes but what do you mean by a boundary? There is a difference between a manifold and a manifold with boundary.

    For instance a sphere has no boundary but a closed ball does.
     
  11. May 3, 2017 #10

    martinbn

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    May be, depends on what you mean by "the points". Do you mean all points i.e. the whole manifold? Then, no, it doesn't have to be homeomorphic to ##\mathbb R^n##.
     
  12. May 3, 2017 #11
    Please see below.

    Well, I mean something that is in agreement with our notion of a end point or a barrier, for familiar geometrical objects, like the Earth, a sphere, a disk. At the same time, this definition has to be correct for any kind of manifold we want to apply it. So, given
    a manifold ##M## and a point ##x \in M##,
    an open subset ##U## of ##M## containing ##x##;
    Suppose we can find an homeomorphism ##\Psi: U \rightarrow V \subseteq \mathbb{R}^n##. If this can happen for every ##x \in M## we say ##M## have no boundaries.

    For the case when it's possible to define only a half region around ##x## where there are a homeomorphism as above, then we say ##x## is in the boundary of ##M##. Thus ##M## have a boundary. All points on the boundary will get mapped to ##y \in \mathbb{R}^n = (0,y_1,...,y_n)## in the local mapping.

    We see that by this way we can recover our notions of a boundary for the objects that I mentioned above (The sphere, etc...)
    I've constructed a diagram showing these concepts for the case of the closed 1-ball, namely the disk. I'm not sure if that diagram is actually correct in everything, however.

    I wrote some notes in my language, not in English. So I'm going to translate it

    1 - Here we see the disk from far away. The disk consists of all points that lie within the red circle.

    dC5UB2G.png

    2 - Now we zoom into the border of the disk and we see something like this

    ioSPzDN.png

    3 - Here is the result of the map of several points of the border of the disk in ##\mathbb{H}^2##

    ZdJSr9P.jpg

    Of course objects have physical limitations, so the above diagram is only a representation.
     
  13. May 3, 2017 #12

    lavinia

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    For a manifold each point has to have a neighborhood that is homeomorphic to an open subset of ##R^{n}## as the OP said in the original post. For a manifold with boundary each point has to have a neighborhood that is homeomophic to an open subset of ##R^{n+}##.

    It is important to understand that a point can not both be an interior point and a boundary point according to this definition.
     
    Last edited: May 3, 2017
  14. May 3, 2017 #13
    "[..] In technical language, a manifold with boundary is a space containing both interior points and boundary points. [..]", from https://en.wikipedia.org/wiki/Manifold#Manifold_with_boundary
     
  15. May 3, 2017 #14

    PeterDonis

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    That's not the correct way to state it. The correct way to state it is the way lavinia stated it in post #12. "Only a half region where there is a homeomorphism" doesn't really make sense.
     
  16. May 3, 2017 #15
    "Each point has to have a neighborhood that is homeomophic to an open subset of ##\mathbb{R}^{n \ +}## (which is the same as ##\mathbb{H}^2##) as she stated, isn't the same thing as "there's some half-like-region around each point ##x \in M## that is homeomorphic to ##\mathbb{H}^2##"?
     
  17. May 3, 2017 #16

    PeterDonis

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    No, because "half-region" implies that the neighborhood that is homeomorphic to ##\mathbb{H}^2## is "half" of something. It isn't. It's just a neighborhood that is homeomorphic to ##\mathbb{H}^2##.
     
  18. May 3, 2017 #17
    Ok. Just to know, in my diagram above, is it correct to picture the region around the point as that circular red disk?
     
  19. May 3, 2017 #18

    PeterDonis

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    As the semicircular red region. Plus, the region includes the straight line boundary (since the point ##x## lies on it), but not the circular boundary (since ##\mathbb{H}^2## doesn't include that, it's open there).
     
  20. May 3, 2017 #19
    Ok, thanks
     
  21. May 3, 2017 #20
    I think this is the subtlety I was looking for. The difference between a sphere and a closed ball.
    https://en.wikipedia.org/wiki/Ball_(mathematics)
    That helps !
     
  22. May 4, 2017 #21

    lavinia

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    The sphere has no boundary but it is the boundary of the closed ball. One might ask more generally when a manifold that has no boundary is itself the boundary of another - one higher dimensional manifold. For instance a torus has no boundary but it is the boundary of a solid torus - a doughnut filled with strawberry jelly. The Klein bottle is also the boundary of a solid Klein bottle. However, the projective plane is a surface without boundary yet it is not the boundary of any manifold. There is no way to solidify it.

    The question of when a compact manifold without a boundary is the boundary of a one higher dimensional manifold can be asked for any dimension.
     
  23. May 4, 2017 #22
    So in my case the earth(land + ocean + atmosphere) has no boundary am I correct ?
     
  24. May 4, 2017 #23

    lavinia

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    The sold earth does have a boundary. It is the surface of the earth - approximately a sphere.
     
  25. May 4, 2017 #24
    What if you include the atmosphere ? I mean the earth as a sphere or an ellipsoid.
     
  26. May 5, 2017 #25

    WWGD

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    Same issue: if you include the outer layer, it is the same as the earth with its boundary.
     
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