Manifold embedding

  1. nash proved that any manifold can be embedded in R^3 in which the higher dimensional manifold gets crumpled and smoothness is lost.

    is it possible that 11 dimensional space has already crumpled into our three dimensional universe and that wormholes exist precisely as a direct result of the crumpling?

    cheers,
    phoenix
     
  2. jcsd
  3. Hurkyl

    Hurkyl 16,090
    Staff Emeritus
    Science Advisor
    Gold Member

    http://www.wikipedia.org/wiki/Nash_embedding_theorem

    The theorem says that any Riemann manifold can be (isometrically) embedded in Rn for some n. It doesn't say 3 specifically. In fact, n has to be at least as great as the dimension of the manifold.
     
  4. there was a big embedding theorem by nash, which is the one you're talking about, and a small embedding theorem, which is the one i'm talking about. it was mentioned in the "a beautiful mind," but i can't find the paper on the web.

    cheers,
    phoenix
     
  5. i did a little more research and found this quote from nash's autobiographical essay for winning the nobel:

    So as it happened, as soon as I heard in conversation at M.I.T. about the question of the embeddability being open I began to study it. The first break led to a curious result about the embeddability being realizable in surprisingly low-dimensional ambient spaces provided that one would accept that the embedding would have only limited smoothness. And later, with "heavy analysis", the problem was solved in terms of embeddings with a more proper degree of smoothness.

    so again i ask this: is it possible that the higher dimensional space has either fully or partially collapsed in the three dimensional space and that the nonsmoothness has resulted in wormholes?
     
  6. Hurkyl

    Hurkyl 16,090
    Staff Emeritus
    Science Advisor
    Gold Member

    I repeat, no. An n-dimensional manifold cannot be embedded in an m-dimensional manifold for m < n.
     
  7. arivero

    arivero 2,991
    Gold Member

    a lot more

    actually if you want a lorentzian metric in both manifold, the embedding rises, it needs a lot more of dimensions. About ninety or so, perhaps.
     
  8. marcus

    marcus 24,127
    Science Advisor
    Gold Member
    2014 Award

    Re: a lot more

    glad you are back, long time no see
    I need to know the Cartan subgroup of SL(2,C)
    (I am told there is just one and I suspect it is
    the diagonal matrices but am not sure)
     
  9. arivero

    arivero 2,991
    Gold Member

    Re: Re: a lot more

    Hi!

    I am back just on a errand for nuclear data. I am afraid I can not confirm your guess -it seems a good one- without browsing across manuals :-( Two years teaching computer science and you see, one loses the training.
     
  10. marcus

    marcus 24,127
    Science Advisor
    Gold Member
    2014 Award

    In the Archive section of this forum I just posted what I think is the Weyl group of SL(2,C). These are new ideas for me, they seem nice

    I think the normalizer of the (main) diagonal matrices in SL(2,C) consists of the union of the major and minor diagonal matrices

    and then N(H)/H the Weyl group comes down to Z_2

    which just flips the diagonal matrix to the other direction--major to minor and viceversa, but there is a minus sign in there too

    you cant fool me, you have not gotten all that rusty by teaching computer science. it could even give you ideas
     
Know someone interested in this topic? Share a link to this question via email, Google+, Twitter, or Facebook

Have something to add?
Similar discussions for: Manifold embedding
Loading...