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Closed but infinite space?

  1. Feb 21, 2009 #1
    I wonder if Universe with a positive curvature everywhere can be infinite.

    As an example I see a spring: you make a full rotation, but you dont get into the same place, but another place.

    Spring is 1-dimensional object (line) so curvature is not defined for it. But is it possible for the higher dimensional surfaces?
     
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  3. Feb 21, 2009 #2

    HallsofIvy

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    The curvature of a line certainly is defined: it is 0. In any case, yes, it is possible for a 3 dimensional spaced, with positive curvature to have infinite volume. The curvature would have to have 0 as greatest lower bound.
     
  4. Feb 21, 2009 #3
    A simple example would be a parabaloid of revolution. Note that the Gauss curvature falls to zero as you move far away from the apex.
     
  5. Feb 21, 2009 #4
    Ha, this is correct!
    I have to change my original question, I need to add "positive curvature, constant everywhere"
     
  6. Feb 21, 2009 #5

    HallsofIvy

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    In that case the hyper-volume is finite and, in fact, can be given as a standard formula in terms of the curvature, [itex]\kappa[/itex], and dimension, n:
    [tex]V= \frac{\pi^{n/2}}{\Gamma(1+ n/2)\kappa^n}[/tex].

    Note that if n is odd, [itex]\Gamma(1+ n/2)[/itex] and [itex]\pi^{n/2}[/itex] will both involve a square root of [itex]\pi[/itex] so this equation always has [itex]\pi[/itex] to an integer power.
     
  7. Feb 21, 2009 #6
    Wait, wait. I understand the classical view.

    Curvature is positive everywhere, so surface must intersect with itself. Intersect, if we look at the manifold.

    But what if it does not intersect? And when we complete a full cycle around a sphere we end not in the same place, but in another place? (an analogy with a spiral spring)

    That set of points can not be put in any manifold... so may be it can not be called 'space'. But still.. what properties of space we need to sacrifice in order to make such object?

    In general, I am interested in how curvature defines the topology. As far as I know, I does not.... hence any facts about the topology of our curved space (General Relativity) are not falsifiable?
     
  8. Feb 21, 2009 #7

    Hurkyl

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    You can prove that it does. Here's a fun hands-on example that will (hopefully) convey what I think the main proof idea is.


    In this exercise, we will be drawing triangles. The sides of the triangle don't need to be straight, but they have to have three vertices and three edges.

    Your goal is to draw triangles so that exactly four triangles meet at a vertex. Of course, adjacent triangles at a vertex must share a side. So, initially you might have something that looks like this:
    Code (Text):

       *
      /|\
     / | \
    *--*--*
     \ | /
      \|/
       *
     
    How many more triangles can you add? I claim that the best you can do 6 vertices and 7 triangles, with exactly three edges lying on the outside of the diagram. (If you wanted, you could include the "outside" as an 8th triangle)

    Try the same example again, but this time have exactly 5 triangles meet at a vertex. How many triangles can you draw now? (12 vertices, 19 triangles, and the outside)



    The point of this exercise is how you're constrained by the geometry. The above two examples correspond to an octahedron and an icosahedron respectively: knowing nothing more than how many triangles could meet at a vertex, you were forced to assemble them into an 8-sided or a 20-sided figure, respectively.


    Maybe you still don't trust me, and want to try something more involved. Well, go nuts! Draw any sort of combination of triangles you like (again, you don't have to draw straight sides). But from the constant-curvature metric, you can prove two things:

    (1) The sum of the angles at any vertex is [itex]2 \pi[/itex] radians
    (2) If a triangle has angles [itex]\alpha[/itex], [itex]\beta[/itex] and [itex]\gamma[/itex] radians, then its area is [itex]\alpha + \beta + \gamma - \pi[/itex] steradians. (The scaling factor between steradian and actual area depends on the actual numerical value of the curvature)

    You'll find that no matter how you draw your triangles, if you assign angle measures to the angles of the triangles (including the outside 'triangle'), the area will always be no greater than [itex]4 \pi[/itex] steradians... the area of a sphere.

    (Note that the sum of the angles of the triangles must be strictly greater than [itex]\pi[/itex])


    In the two examples above, the angle measures of the triangles were [itex]\pi/2[/itex] and [itex]2\pi / 5[/itex] respectively.



    I believe you can exceed [itex]4 \pi[/itex] if you allow yourself to have edges. e.g. imagine starting with a strip with the geometry of the equator of a sphere, and cut a vertical line through it so it's no longer a loop. I believe there is no problem attaching it another duplicate copy of the strip along that cut edge, and you could repeat that process as much as you like. However, you will never be able to 'complete' your manifold: you will always be stuck with edges. This isn't really suitable for a description of the universe: you could fly off of the universe at an edge... or worse, any sort of random stuff could instantaneously appear at an edge.



    P.S. if you're dealing with anything that, locally, has the topology of Euclidean space, then it's a manifold. That's (essentially) the definition of manifold. :tongue:
     
    Last edited: Feb 21, 2009
  9. Feb 21, 2009 #8

    HallsofIvy

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    The surface MUST intersect but what if it does NOT? What if 4 were a prime number?
     
    Last edited: Feb 21, 2009
  10. Feb 21, 2009 #9
    I'm not sure I totally understand your point. I can see that you can define a trivial Riemann curvature tensor on a 1 manifold but I don't see how this extends when you view curvature as an obstruction to the integrability of horizontal spaces.
     
  11. Feb 22, 2009 #10
    Hurkyl, thank you
    Your example with triangles explains it, so the 'spiral spring' analogy wont work if nummber of dimensions is >1.
     
  12. Feb 22, 2009 #11
    You can make an infinitely spiralling surface that has Gauss curvature zero.
     
  13. Feb 22, 2009 #12
    But I wonder if such a surface can be complete.
     
  14. Feb 23, 2009 #13
    Here's a try. Start with an infinite spiral in the plane. At each of its points, place a small segment that curves away from the orientation of the spiral and slide it around to create a surface of positive curvature. As you wind inwards, scale the segment increasingly shallower so that it will not cause self intersection. This surface is not complete because it can not include the limit point of the spiral. Also it has a boundary.

    It on starts with a segment that curves towards the orientation of the spiral, one gets a surface of negative curvature - like an infinite sand collar.
     
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