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?

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    $\begingroup$ 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) $\endgroup$ – user8268 Sep 18 '13 at 20:42
  • $\begingroup$ Perhaps someone should remark the horosphere will not be isometric to the Euclidean plane, only equivalent as a Riemannian manifold. $\endgroup$ – user641 Sep 18 '13 at 21:07
  • $\begingroup$ @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... $\endgroup$ – Moishe Kohan Sep 19 '13 at 8:35

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).

  • $\begingroup$ Okay, so the horosphere is not isometric to the Euclidean plane then as Steve D remarked? $\endgroup$ – Sid Sep 18 '13 at 21:17
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    $\begingroup$ Of course they are isometric, however, hyperbolic space is not isometric to the round sphere (they are not even homeomorphic). $\endgroup$ – Moishe Kohan Sep 19 '13 at 6:27
  • $\begingroup$ @studiosus: I thought that Hilbert's theorem says there's no immersion into $\Bbb R^3$, which is weaker than saying there's no isometric embedding because it requires infinite differentiability. Specifically, I thought that a $C^1$ embedding was proven to exist. $\endgroup$ – Mike Battaglia Nov 18 '15 at 23:12
  • $\begingroup$ @MikeBattaglia: I am assuming $C^2$-differentiability, which is the standard differential geometric assumption. Indeed, if one assumes only $C^1$, then the situation is totally different (Nash-Kuiper theorem). In the context of OP's question (he mentions Hilbert's theorem) the $C^2$ assumption makes sense. $\endgroup$ – Moishe Kohan Mar 15 '16 at 20:52

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