Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I read the answers to this very interesting question and saw that we can in fact embed the Euclidean plane into hyperbolic 3-space using what is called a horosphere. However, as Hilbert showed us, the reverse is not true; we cannot embed the hyperbolic plane into Euclidean 3-space. This made me interested in considering the other non-Euclidean geometry: elliptic geometry. We can embed the plane from elliptic geometry into Euclidean 3-space - the result is spherical geometry - but:

a) is the reverse true: can we embed the Euclidean plane into elliptic 3-space?

b) Furthermore, is it possible to embed the elliptic plane into hyperbolic 3-space?

c) What about the reverse: can we embed the hyperbolic plane into elliptic 3-space?

I'm really curious about this.

share|improve this question
1  
For elliptic in hyperbolic, take any sphere (in the hyperbolic metric) in the hyperbolic space (the relation between radius and curvature is of course different than in the Euclidean space) –  user8268 Sep 18 '13 at 20:42
    
Perhaps someone should remark the horosphere will not be isometric to the Euclidean plane, only equivalent as a Riemannian manifold. –  user641 Sep 18 '13 at 21:07
    
@SteveD: There are two notions of an isometric embedding $f: (M_1,g_1)\to (M_2,g_2)$ in Riemannian geometry. One just requires $f^*(g_2)=g_1$. The second requires an isometric embedding of the associated metric spaces. It seems clear that OP was asking for the first one (since this is what Hilbert's theorem is about), but one can never be completely sure, of course... –  studiosus Sep 19 '13 at 8:35
add comment

1 Answer 1

up vote 2 down vote accepted

Here is the detailed answer to eliminate the confusion. Let $H^n, R^n, S^n$ denote the $n$-dimensional spaces of sectional curvature $-1, 0$ and $1$ respectively. Then the following hold (Items 1 and 2 are immediate, but items 3 and 4 are not):

  1. For every $n$, $S^n$ embeds isometrically in $R^{n+1}$ and $H^{n+1}$ as a metric sphere of certain radius.

  2. $R^n$ isometrically embeds in $H^{n+1}$ as a horosphere.

  3. $H^2$ does not isometrically embed in $R^3$ (Hilbert's theorem). However, $H^2$ does embed (isometrically) in $R^6$.

  4. $R^n$ and $H^n$ do not isometrically embed in $S^k$ for any $n$ and $k$.

David Brander wrote a UPenn thesis in 2003 summarizing the results on isometric embeddings between various constant curvature spaces. See http://davidbrander.org/penn.pdf for details (in particular, he explains what happens if one considers other dimensions and other constant curvature values, including embeddings between spaces with the same curvature sign).

share|improve this answer
    
Okay, so the horosphere is not isometric to the Euclidean plane then as Steve D remarked? –  Sid Sep 18 '13 at 21:17
1  
Of course they are isometric, however, hyperbolic space is not isometric to the round sphere (they are not even homeomorphic). –  studiosus Sep 19 '13 at 6:27
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.