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Embedding hyperbolic constant-time hypersurface in Euclidean space.

  1. Sep 30, 2012 #1

    andrewkirk

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    In Bernard Schutz's 'A first course in General Relativity', p325 (1st edition) he says

    " [the constant-time hypersurface of a FLRW spacetime with k=-1 (hyperbolic)] is not realisable as a three-dimensional hypersurface in a four- or higher-dimensional Euclidean space."

    On the face of it, this appears to contradict the Strong Whitney Embedding Theorem, which states that :

    "any smooth m-dimensional manifold (required also to be Hausdorff and second-countable) can be smoothly embedded in Euclidean 2m-space, if m>0. "

    So it should be possible to embed the constant-time hypersurface in Euclidean space of six dimensions, if not less.

    Have I misunderstood either Schutz or Whitney here, or is Schutz just wrong?
     
  2. jcsd
  3. Sep 30, 2012 #2

    Ben Niehoff

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    The Whitney theorem only proves that a smooth embedding exists. Schutz is probably talking about an isometric embedding, which in addition to smoothness requires that the metric tensors agree. That is, the pullback along the embedding map of the flat metric tensor on R^n should equal the (already-given) metric tensor on the submanifold.

    Hyperbolic spaces are not isometrically embeddable in R^n. They are, however, isometrically embeddable in R^(n,1).
     
  4. Sep 30, 2012 #3

    andrewkirk

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    Thanks Ben. It looks like I chose the wrong embedding theorem.
    Perhaps the Nash embedding theorem (C1, or Nash-Kuiper, version) gives the general result needed, as that proves the existence of a C1 isometric map from the n-manifold to a hypersurface in En+1.

    Schutz's claim would still be wrong, even assuming he meant isometric (which I think he did), wouldn't it, because he says the hyperbolic hypersurface cannot be [isometrically] embedded in any higher dimensional Euclidean space?
     
  5. Oct 1, 2012 #4

    Ben Niehoff

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    Yeah, I was wondering that as well. And yet I've seen this claim in many places that hyperbolic spaces cannot be embedded in R^n.

    It seems reasonable to me, for example, that the hyperbolic plane should embed isometrically in R^3, as an infinite sheet that gets more and more wrinkly as you go further from the origin. You can certainly make tesselations of the hyperbolic plane that embed isometrically in R^3, such as the hyperbolic soccer ball.

    I think claims such as Schutz's are maybe asking for even more additional requirements. In the case of hyperbolic spaces, I can think of one: symmetry. Hyperbolic spaces embed in R^(n,1) in such a way that all the Killing vectors of the submanifold correspond with Killing vectors of R^(n,1). But embeddings into R^n have "wrinkles" that prevent this. However, this requirement cannot be extended in any natural way to spaces with less symmetry.
     
  6. Oct 2, 2012 #5
    Thanks for the very interesting comments, Ben. Could you elaborate a little on what the problem is with defining both a hyperbolic space and a euclidean space on the same manifold, where the spaces are one (maybe two) dimensions less than the manifold? Would the Euclidean space then just be charts for the hyperbolic space? Would the atlas represent the hyperbolic space?
     
    Last edited: Oct 2, 2012
  7. Oct 2, 2012 #6

    Ben Niehoff

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    I'm not sure what you're asking, because I've already elaborated as much as I know.

    Second, "Euclidean" vs. "Hyperbolic" is not a matter of charts and atlases. Topologically, R^n and H^n are the same space. But we are talking in the category of Riemannian manifolds, which have a local metric structure.

    H^2 can be mapped to the interior of a disk, so it actually embeds in R^2! But this is not an isometric embedding.
     
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