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N-sphere as manifold without embedding

  1. Dec 29, 2015 #1
    I have been working through John Lee's "Introduction to Smooth Manifolds" recently. I am having trouble visualizing the unit n-sphere as a manifold in its own right, i.e. without an ambient space. It seems impossible to visualize it without embedding it in Euclidean space. I mean, the unit sphere is defined as the set of points whose (Euclidean) distance from the origin (a point of [itex] \mathbb{R}^n [/itex] that is NOT on the sphere), is equal to 1. So it seems impossible/useless to speak about the sphere without embedding it first into Euclidean space. Could someone please clarify where I may be wrong in my thinking, and also elaborate on the issue for me? Thank you.
     
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  3. Dec 29, 2015 #2

    mathwonk

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    yes that is the embedded definition. Here is the charts definition, in dimension 2. the 2-sphere, as a topological manifold, is the topological space obtained from the disjoint union of two copies of the complex numbers, by identifying each non zero point z in the first copy, with the point 1/z in the second copy. (with the quotient topology.) to give the smooth structure we must also say that a function on an open set is smooth if and only if its local pullbacks to both copies of the complex numbers are smooth where defined. I.e. if S is the sphere, we have by definition two maps C1-->S and C2-->S where C1 and C2 are the two copies of the complex numbers, and these maps are used to pull back a function f:S-->R to functions Cj-->S-->R.
     
    Last edited: Dec 29, 2015
  4. Dec 29, 2015 #3

    WWGD

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    I don't know if this would be a satisfactory answer but you could take a flat closed disk { ## x:||x|| \leq1 ## } and identify the boundary {## x: ||x||=1 ##} to a point.
     
  5. Dec 29, 2015 #4

    lavinia

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    Similar to the explanations already given, imagine cutting the sphere with a scissors in two ways. First along the Tropic of Cancer and second alone the Tropic of Capricorn. The pieces are partial spheres that overlap in a band around the equator. Each of these pieces is diffeomorphic to a disk in the plane. So the sphere can be viewed as two disks that overlap in an annulus. This construction is essentially the same as Mathwonk's construction.

    WWGD's example does not present the sphere as a manifold but it does present it as a topological space. One imagines closing a trash bag until it is tied off. One can also view the sphere as two disks glued together along their boundary, (two hemispheres pasted along the equator)In these two cases one needs to show that the resulting space is a manifold.

    In all of these examples, the sphere has no shape but is merely a topological space defined as a quotient with of one or of two disks. To give it a geometry like the unit sphere in Euclidean space you need to do more. You need to give it a Riemannian metric. This is easily done by parameterizing the hemispheres of the unit sphere.

    If you want to visualize the topological sphere then you need to do more than give it a Riemannian metric. You need to realize it a subset of Euclidean space.

    Higher dimensional spheres can be constructed as topological spaces by gluing n dimensional balls together along their boundary spheres. Or one can use WWGD's construction and identify the boundary of a single ball to a point.
     
    Last edited: Dec 29, 2015
  6. Dec 29, 2015 #5

    WWGD

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    Like Lavinia said, you need relatively minimal conditions to allow for a Riemannian metric in a manifold. Usually you can pull-it back in each chart ---the local diffeomorphisms preserve the positive -definiteness of the form defined in ##\mathbb R^n ## , and then you only need to glue the chart definitions together , so you need a partition of unity, which you can get through second countability and/or paracompactness.

    And Lavinia's map needed to glue the two disk homeomorphs seems pretty clearly smooth-enough so that there will be no problems with the overlaps . So just define the chart maps on each cutoff and then there is no problem in the points where two disks are glued together under a quotient map.
     
    Last edited: Dec 29, 2015
  7. Dec 30, 2015 #6
    Ah, so in order for me to visualize the sphere as it is usually visualized, it must be embedded in Euclidean space first. Makes sense. Now @lavinia, you brought up something interesting. You said that in these examples, the sphere has no shape, but is just a topological space. My question is: What is the point of studying the sphere merely as a topological space with no shape or geometry? It seems to me that the sphere would be interesting to work with precisely because its geometry.
     
  8. Dec 30, 2015 #7

    WWGD

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    You can start with the standard ##S^n##. Apply any homeomorphism from ##S^n ## to itself, you still have a sphere. Apply any diffeomorphism to any initial geometry, you still have the same sphere with the same geometry and same topology.
     
  9. Dec 30, 2015 #8
    I'm saying though, what's the point of studying a sphere as a manifold without embedding it into Euclidean space, as it loses all of its geometric properties, when those geometric properties were the interesting properties?
     
  10. Dec 30, 2015 #9

    WWGD

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    I am not sure the n-sphere has an intrinsic geometry; different choices of metric give rise to different types of geometry. But maybe you can say a special type of connection like the Levi -Civita one is the natural/intrinsic one? For low dimensions, for some dimensions, homeomorphic implies diffeomorphic, but not in dimension, e.g. 4 , so even within the same topology, you can have different types of geometry. You need additional conditions on, e.g., homology to have homeomorphic to imply diffeomorphic.
     
    Last edited: Dec 30, 2015
  11. Dec 30, 2015 #10
    @WWGD, sorry, I don't think I am wording my question properly. Let me try again. Is there a point in studying the sphere as a smooth manifold without embedding it in Euclidean space? When I think of a sphere, I think it is special because of its shape and because of it's (Euclidean) geometry. Without embedding it into Euclidean space, it no longer has this shape or geometry. So what's the point of studying it without embedding it first?
     
  12. Dec 30, 2015 #11

    WWGD

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    Sorry, I guess I misunderstood/misunderestimated your post, let me rethink.
     
  13. Dec 30, 2015 #12
    Sometimes abstractions simplify the matter in a useful way. One may or may not take in interest in shape or geometry.

    If all you are interested in is what is going on in the manifold, then you would not care about the surrounding space, or even whether or not it existed.

    Our Universe is thought to be a 4D manifold. We have no awareness of anything outside of it, so why complicate things by assigning assumed properties to the space in which it may or may not be embedded?
     
  14. Dec 30, 2015 #13
    Is there anything about the sphere, taken as a manifold in its own right (without an embedding), that is interesting? Also, could you please give me an example of a manifold where not embedding it simplifies things? I am trying to remove this thought that I have that a manifold must be embedded in Euclidean space in order to be useful/interesting to us. I know that my point of view is wrong (based on more knowledgeable people telling me so), but I need to be more convinced.
     
  15. Dec 30, 2015 #14

    WWGD

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    Actually, the usual perspective is that the embedding information is extra structure that distracts from the understanding of the intrinsic properties of the object itself, which are of more interest.
     
  16. Dec 30, 2015 #15
    @WWGD could you give me an example?
     
  17. Dec 30, 2015 #16

    WWGD

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    Sure, please give me some time to think of some good examples.
     
  18. Dec 30, 2015 #17
    Thanks!
     
  19. Dec 31, 2015 #18
    It is quite difficult to escape the reliance on a background or framework. The first to do so IMO was Grassman back in 1850 or so. He entered a geometry-with-no-framework contest and won. He was the only entry. No one else even tried.

    It could be that even Gauss had trouble with it. He had his student Reimann work on the problem. Reimann found it quite difficult, but Gauss was thrilled with the results.

    I'm told that Reimann's methods were the basis of the math used in general relativity. One of the main objectives of that theory was to get away from the more-or-less Euclidean geometry that limited special relativity. In general relativity the universe more or less creates the metric used to measure it in a recursive feedback process. Or so I gather.

    I have yet to learn these techniques, so I can't help. Suffice to say that thousands of students have been that way, and the trail is well marked. My best guess would be to start with description of a pendulum by a Lagrangian. That might help.
     
  20. Dec 31, 2015 #19

    mathwonk

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    an object, even a simple one like the sphere, has many properties, and it is useful to isolate them and ignore some of them in order to simplify the study of that object and understand which ones are more basic. The topology of the sphere is its most fundamental type of property. Adding on a smooth structure, or a metric defines still more refined geometric properties, but just the topology already determines some things of interest. E.g. the topology determines the euler characteristic, and this number tells you already how many zeroes a general smooth vector field will have (more precisely, the difference between the number of positively oriented zeroes and negatively oriented ones), no matter which smooth structure you give it. I.e. although a smooth vector field is not defined until you have a smooth structure, and a sphere can have more than one such smooth structure, the (weighted) number of zeroes in general is already determined by the topology. More refined still is a Riemannian metric, which allows to define the curvature, but the average curvature, the integral of the curvature over the manifold, is again determined by the euler characteristic, as lavinia has beautifully explained elsewhere.
     
    Last edited: Dec 31, 2015
  21. Jan 1, 2016 #20

    lavinia

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    The sphere does not necessarily lose all of its geometry if it is not embedded in 3 space. One can define its "intrinsic" geometry with a Riemannian metric on its tangent vectors. One starts with the sphere described soley as a smooth manifold - i.e. in terms of overlapping coordinate domains - then assigns a Riemannian metric to its tangent bundle. Once its has a Riemannian metric, it has a geometry.

    In order to retrieve the geometry of the embedded sphere, one needs to choose a particular Riemannian metric. Not all Riemannian metrics will work. Each choice of a Riemannian metric, gives this manifold a different intrinsic geometry. So the sphere as a manifold is the domain of many possible geometries. For instance, it can be given the geometry of an ellipsoid. It may even have geometries that can not be realized as a surface embedded in 3 space.

    Another example is a soap bubble floating in the air. At first, it wobbles around taking on various irregular shapes, and then finally becomes a round sphere (if it doesn't pop first). It is a manifold-sphere with many different geometries and only becomes the standard geometric sphere after it quiets down.

    The idea of giving a manifold an intrinsic geometry applies not only to the sphere-manifold but to all smooth manifolds. An important example is the Universe, interpreted as a smooth 4 dimensional manifold that has a geometry derived from a semi-Riemannian metric. This geometry is determined in part by the distribution of stars in the Universe, so as they move, the geometry of the Universe changes.
     
    Last edited: Jan 1, 2016
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