all 23 comments

sorted by: hot top controversial new old
[–] 50 points 6 months ago* (last edited 6 months ago) (3 children)

How on earth can an infinite dimensional sphere be able to agree to a contract, how would it even sign it?

  • source
  • hideshow 3 child comments
  • [–] 36 points 6 months ago (8 children)
  • [–] 18 points 6 months ago* (last edited 6 months ago) (4 children)

    I am not a topologist, but I can try..

    A space (shape) is contractible if you can "contract" (shrink) it to a point without cutting, pinching or punching through holes. For example, a mattress is contractible, since you can shrink it to the center - each point can follow the line to the center, continuously. Meanwhile, a doughnut, a circle or a hollow sphere are not contractible, you can never remove the inner "hole" to shrink to a point without cutting.

    In general, any dimensional sphere is not contractible... Until it is - infinite dimensional sphere is contractible. Somehow, it loses the "hollow space" inside.

  • source
  • parent
  • hideshow 4 child comments
  • [–] 16 points 6 months ago (1 child)

    For each finite dimension n (1, or 2, or 3, etc...), the sphere in dimension n can't be contracted because of that empty n-dimensional space it surrounds. But that same sphere is the "equator" of the sphere in the next higher dimension, n+1. There, the n-dimensional equator can contract along one of the hemispheres, to a pole. But then that whole (n+1)-dimensional sphere still isn't contractible, because of the (n+1)-dimensional space it surrounds.

    BUT the (n+1)-dimensional sphere can contract along one of the hemispheres in the (n+2)-dimensional sphere. And so on.

    For any particular finite dimension n, there is an n-dimensional obstruction to contracting the sphere in that dimension. But if you go all the way to infinitely-many dimensions, there is no obstruction that ever stops contractibility of the infinite-dimensional sphere.

  • source
  • parent
  • hideshow 1 child comment
  • [–] 8 points 6 months ago (5 children)

    Also isn’t an infinite dimensional sphere practically hollow?

    (If you were to integrate the sphere to calculate volume like you do for lower dimensional ones, you would sum the volume of shells—which is just their surface area times a thickness—making it up. With infinite dimensions, each shell becomes infinitely larger than the preceding shell no matter how fine you make the slices. This means the largest shell contains basically all the volume.)

  • source
  • hideshow 5 child comments
  • [–] 7 points 6 months ago* (last edited 6 months ago) (4 children)

    This reasoning is pretty weird, but the conclusion is basically right. That is, there is absolutely no way to extend the conventional notion of volume to Rinfinity, which is basically what most people would imagine is the infinite equivalent of our dimensional space. Edit: what I mean by Rinfinity is a bit ambiguous, but let's say for the purpose of a hypersphere we want something like l^2 hilbert space to ensure no vectors with infinite length appear, then we have a separable space and the proof is complete.

  • source
  • parent
  • hideshow 4 child comments
  • [–] 2 points 6 months ago (1 child)

    Wait, I thought the volume of a sphere approaches 0 as dimensions go to infinity, no? Thr general formula for the volume of any nth dimensional sphere has the gamma function in the denominator, which rises faster than whatever is in the numerator. At some point (5 dimensions, iirc) the volume starts decreasing

  • source
  • parent
  • hideshow 1 child comment
  • [–] 7 points 6 months ago
    [–] 5 points 6 months ago (1 child)

    How much do you need to contract an infinite sphere before it becomes finite?

  • source
  • hideshow 1 child comment
  • [–] 3 points 6 months ago

    There is no sphere but s p h e r e itself.

  • source