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In how many ways can you embed a Riemannian manifold in R^n?

  1. Jan 14, 2014 #1
    By the well-known Whitney embedding theorem, any manifold can be embedded in [itex]\mathbb R^n[/itex].

    You might have also heard the Nash embedding theorem, which basically says that this is still true for Riemannian manifolds (i.e. now we demand the metric is induced from [itex]\mathbb R^n[/itex]).

    So fine, any Riemannian manifold can be seen as a submanifold of [itex]\mathbb R^n[/itex]. But my question is: in how many ways can one do this?

    For example, embedding a sphere in [itex]\mathbb R^3[/itex], it doesn't matter where we do it (translation symmetry) or how we orient it (rotational symmetry). I.e. the only freedom in embedding it is the symmetry of [itex]\mathbb R^3[/itex]. So an alternative formulation of the question is: are there isometric submanifolds of [itex]\mathbb R^n[/itex] which are not the same up to rotation and translation?
     
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  3. Jan 14, 2014 #2

    WWGD

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    Do you mean how many ways up to what; up to isotopy of embeddings, or up to something else ?
     
  4. Jan 14, 2014 #3
    I'm not sure myself what the best concept is. I think up to rotations and translations(?) The spirit of the question is: I think that if you have two isometric embeddings of the sphere in [itex]\mathbb R^3[/itex], then you can map one to the other using rotations and translations. But is this true for any embedded Riemannian manifold in [itex]\mathbb R^n[/itex]?
     
  5. Jan 14, 2014 #4

    jgens

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    Is this your question:

    Let E+(n) be the group of isometries defined by x→Ax+b where A is an element of SO(n) and b is an element of Rn. Suppose M is a Riemannian manifold and that we have two isometric embeddings f,g:M→Rn. Is there some element φ in E+(n) such that g = φf?


    If so, then the answer is clearly no. Just consider the standard embedding of the (n-1)-sphere and the embedding given by composing the standard one with a negative determinant element of O(n). If you expand the group E+(n) to the group E(n) consisting of isometries defined by x→Ax+b where A is an element of O(n) and b is an element of Rn, then my example fails. My hunch is that counter-examples still abound, but an explicit example eludes me at the moment.

    P.S. When you ask a question like "How many ways can you embed a Riemannian manifold in Rn?" you really should state up to which equivalence. For one it gives everyone else an indication of what you are trying to ask. It can also help members point you to relevant sources. For example trying to classify embeddings up to isotopy is a fairly classical problem and there is a lot of material written about it. If this were your question, then even though I know quite little about it, I could still direct you to some sources that might help with your question.
     
  6. Jan 14, 2014 #5
    Nope I'm not trying to classify up to isotopy. Actually your formulation hits the nail on the head, thanks! And you're right, inversion is a counter-example, but then I realize that means "E(n)" (as you say) is the proper group to ask the question about. So yes, thank you, my question is:


    Let E(n) be the group of isometries defined by x→Ax+b where A is an element of O(n) and b is an element of Rn. Suppose M is a Riemannian manifold and that we have two isometric embeddings f,g:M→Rn. Is there some element φ in E(n) such that g = φf?
     
  7. Jan 14, 2014 #6

    WWGD

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    Just to make the general point that a weaker result holds: if f: N-->M and g: N-->M are two injective immersions of N in M , i.e., the images are equal as sets of points, you can find a diffeomorphism h with f=goh
    http://diffgeom.subwiki.org/wiki/Submanifold_(differential_sense [Broken]). Of course, h is not necessarily an isometry.
     
    Last edited by a moderator: May 6, 2017
  8. Jan 14, 2014 #7

    Office_Shredder

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    It seems to me the interesting question isn't whether a rotation/translation can be applied, but any isometry at all.

    So if [itex] f:M\to \mathbb{R}^n[/itex] and [itex] g:M\to \mathbb{R}^n [/itex] are two embeddings of a Riemannian manifold,

    1.) Is there necessarily an isometry h on Rn such that [itex] g = h\circ f [/itex] (if there is then embeddings are boring in a certain sense)

    2.) Is there possibly a function [itex]h:\mathbb{R}^n \to \mathbb{R}^n [/itex] which is NOT an isometry, but still [itex] g = h \circ f [/itex] and h restricted to f(M) IS an isometry.
     
  9. Jan 14, 2014 #8

    jgens

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    In that case you might want to look up some literature on the Euclidean group. This is just the group E(n) I mentioned in my post above.

    Unless I am mistaken the isometry group of Euclidean space is just the group E(n). So this looks like the OPs modified question in post #5.

    This question is a bit more complicated. By adding some additional hypotheses---something like closed embeddings and not onto should probably work, but you might need something slightly stronger---the question becomes fairly simple, however, since you can start with an isometry between f(M) and g(M) and apply classical extension results from point-set topology combined with smooth approximation results to get loads of maps like this. In any case the OP would probably be more interested in maps h:RnRn with the properties you stated and the additional hypothesis that h|f(M) is not the restriction of some member in E(n).
     
    Last edited: Jan 14, 2014
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