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I Is the circle a manifold?

  1. Jun 9, 2017 #1
    The answer to the question of the thread title is yes, according to what I found on web. Now a manifold is by definition a topological space that (aside from other conditions) is locally Euclidean.

    What does such condition means? Is it the same as saying that each of its points must have a homeomorphic neighborhood to ##\mathbb{R}^n## for a n-dimensional manifold?

    If so, how can we prove that it's possible to find an open neighborhood around each point of the circle? Could we use open spheres to do that?

    I have tried to prove this by considering the unit circle in ##\mathbb{R}^2## (with the usual metric) as
    $$\mathbb{S}^1: \big[x \in \mathbb{R}^2 \ | \ d(x,0) = 1 \big]$$ where ##0## is the point where the circle is centered in. Let ##P_0## be the center of an open sphere on the circle.
    We have ##d(0,x) = 1 = d(0,P_0)##

    Using the "triangle inequality" we have
    ##d(0,P_0) + d(P_0, x) \leq d(0,x)##

    The solution is ##d(P_0,x) = 0##. This says that ##P_0## and ##x## are the same point! That is, the open sphere contains only its center point. Let's call the open sphere ##A##.

    Now as this open sphere is in ##\mathbb{R}^2## a parametrization for its points could be ##\gamma: [0,\delta) \times [0, 2 \pi) \longrightarrow A## (the inverse mapping is the usual spherical polar coordinates). We would need both intervals in ##[0,\delta) \times [0, 2 \pi)## to be open in order to it to be homeomorphic to ##\mathbb{R}^2##. But if we do that for our open sphere ##A##, we will exclude its unique point, namely the central point ##P_0##. The solution would be that it doesn't contain any points at all. But then our proof that ##\mathbb{S}^1## is a manifold fails. How do we solve the problem?
     
    Last edited: Jun 9, 2017
  2. jcsd
  3. Jun 9, 2017 #2

    fresh_42

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    Yes.
    Yes.
    Why open spheres? The usual charts are the two stereographic projections from the poles.
    Sphere_small.jpg
    I don't understand what you want in ##\mathbb{R}^2##. The circle is one dimensional, so ##\mathbb{R}^1## is the Euclidean space of choice. Draw a line from the northpole through a point of the circle until you reach the ##x-##axis (##\mathbb{R}^1##) and you get on the ##x-##axis the homeomorphic Euclidean chart. For the points below the ##x-##axis do the same from the south pole. This gives you the two charts you need.
     
  4. Jun 9, 2017 #3
    This is a very easy way to show that ##\mathbb{S}^1## minus one of its points is homeomorphic to ##\mathbb{R}##. But is there a way of showing that by using open "1-dimensional spheres"?
    Just to see how the proof works if we use open spheres.
    I see
     
  5. Jun 9, 2017 #4

    fresh_42

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    A sphere isn't flat, Euclidean spaces are. Additionally why replace one sphere by another? With which goal? Of course you can as well use many small sections of tangents as charts, similar to ordinary atlases in the real world. This decreases the error to the price of more charts.
     
  6. Jun 9, 2017 #5

    WWGD

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    Just look at arcs about a point. They are twisted open intervals in the line. The twisting is " gentle-enough" to preserve (smooth, not just topological) equivalence with the Reals; no corners, nothing too nasty. EDIT: though the cricle is homeomorphic, though not diffeomorphic with a square (in standard embedding, differentiable structure)
     
  7. Jun 9, 2017 #6

    WWGD

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    On top of the local homeomorphism that must exist, this local homeomorphisms, must be "glued " to each other nicely-enough. You may say that a manifold is just a collection of Euclidean parts glued in/with "nice-enough" maps.
     
  8. Jun 10, 2017 #7

    lavinia

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    Here is another way to show that spheres are manifolds. Take the standard ##n##-sphere of radius 1 centered at the origin of ##R^{n+1}## Project the northern hemisphere onto the open unit disk in ##R^{n}## by dropping the last coordinate. (For the 2-sphere in ##R^3## one would drop the z coordinate.) This is a coordinate neighborhood of the north pole. Now take any other point on the sphere and rotate the sphere so that this point moves to the north pole. Follow this rotation by the same projection.

    - For the 2-sphere considered as the Riemann sphere, the two charts may be chosen to be ##z## and ##1/z## .

    - As you may know from calculus of several variables, the Implicit Function Theorem says that if ##f## is a smooth function defined on ##R^{n+1}## and its gradient is nowhere zero on the set ##f(x) = c## for some constant ##c##, then this set can be locally parameterized by a open region in ##R^{n}##. The inverses of these local parameterizations are coordinate charts.

    Consider the function defined on ##R^{n+1}##, ##f(x) = ||x||^2##. For instance in ##R^2## this function is ##f(x,y) = x^2+y^2##. A sphere of radius ##k## centered at the origin is the set of points ##f(x) = k^2##. Notice that the gradient of ##f## is nowhere zero on this set. So by the Implicit Function Theorem, the sphere is a manifold (actually a smooth manifold).

    More generally the Implicit Function Theorem works for a smooth function defined on any manifold not just Euclidean space. For instance stand a torus up on the plane and let ##f## be the height function of each point above the plane. ##f## is smooth and only has four critical points, the top and bottom of the torus - these are the absolute maximum and minimum - and the tops and bottom of the inner circle - these are the two saddle points. Everywhere else ##df## is non-singular. The Implicit Function Theorem guarantees that the points ##f(x) = c## is a 1 dimensional manifold whenever none of the four critical points are in ##f^{-1}(c)##.

    The Implicit Function Theorem is critically important. It would be useful to understand how it is used to find manifolds.
     
    Last edited: Jun 10, 2017
  9. Jun 10, 2017 #8
    Yes, this is a good way to see how it works, though it's a bit qualitative
    Cool
    I see

    @lavinia it's interesting to know about this way of finding a manifold. Thanks for pointing out.
     
  10. Jun 10, 2017 #9

    WWGD

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    Sorry, I can't think of a way of making this more formal without making use of e.g., tangent spaces and the tangent map. Have you seen them?
     
  11. Jun 10, 2017 #10
    Yes, I do. Go ahead. :smile:
     
  12. Jun 10, 2017 #11

    WWGD

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    Ok, a diffeomorphism gives rise to a(n) linear isomorphism between the tangent spaces ( of/between standard circle, square) . But the tangent space at the corners/edges of the square is not well-defined, so there can be no tangent space isomorphism, therefore no diffeomorphism.
     
  13. Jun 10, 2017 #12
    Is the Jacobian determinant equal to zero for those points?
     
  14. Jun 10, 2017 #13

    WWGD

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    Yes, but you would have to parametrize both and find a map between them. Once the result holds, it is independent of the choice of parametrization. EDIT: You can, e.g., inscribe the square in the circle and then map a point in the circle to the square by drawing a line from the center of the circle until it hits the square.
     
  15. Jun 10, 2017 #14
    Ah, ok. Thanks.
     
  16. Jun 10, 2017 #15

    WWGD

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    Maybe this is more precise: consider an arc in the circle mapping just left of the , say, (1,1) corner in the square, i.e., on (x(t),1) and another arc, mapping along (1, y(t)) and see what happens when ## x(t),y(t) \rightarrow 1 ## . You can use coordinates and compute the actual tangent map (i.e., the Jacobian) , which will not be invertible at that point.
     
  17. Jun 10, 2017 #16

    lavinia

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    You seem to be thinking of a smooth manifold where the composition of a chart with a local parameterization must be a diffeomorphism. For a general topological manifold there are no such requirements as far as I know.
     
  18. Jun 11, 2017 #17

    WWGD

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    Yes, you're right, I was assuming a smooth manifold, still, the gluing maps must be at least nice-enough to be homeomorphisms.
     
  19. Jun 11, 2017 #18

    lavinia

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    The gluing maps are automatically homeomorphisms.

    If one defines a manifold as a locally Euclidean space of a fixed dimension: That is there is a open neighborhood of any point that is homeomorphic to an open set in ##R^{n}## then the inverse of any such homeomorphism is by definition also a homeomorphism. So the composition of an inverse with another local homeomorphism must be a homeomorphism on the overlap of the two open sets.
     
  20. Jun 11, 2017 #19

    WWGD

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    Well, yes, this is what I said, that if we start with a collection of Euclidean open sets, we patch them together into ( topological, C^k, smooth) manifolds by using maps in the respective categories.
     
  21. Jun 11, 2017 #20

    lavinia

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    Right. But that is a different question. In this thread one starts with a topological space and asks when is it an n dimensional manifold. For this one only needs to say that each point has an open neighborhood that is homeomorphic to and open set in ##R^{n}##.Nothing needs to be said about overlapping domains. If one wants additional structure such as differentiability then one needs to say something about the coordinate transformations. But this is a restricted situation.

    If one starts with a collection of sets that are homeomorphic to open sets in ##R^{n}## then one only needs to say how they overlap. Nothing else about gluing is required to make a topological manifold.
     
    Last edited: Jun 11, 2017
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