Can every Riemannian manifold be embedded in a sphere?

The famous Nash embedding theorem asserts that every closed Riemannian manifold can be isometrically embedded in Euclidean space $\mathbb{R}^n$ for $n$ sufficiently large.

Is it true that we can replace $\mathbb{R}^n$ with the round sphere $\mathbb{S}^n$?

What about $\mathbb{H}^n$ (Hyperbolic space)? or $\mathbb{T}^n$ (Torus)?

(i.e I am asking whether any Riemannianm manifold can be embedded in one of this spaces when allowing the ambient space to be of arbitrary dimension)

Of course, by the Nash embedding theorem, it's enough to check whether the Euclidean space can be embedded in these manifolds.

• My guess is that $\Bbb R^n\to\Bbb S^{2n}$ should work, but I'm only certain for $n=1$. – Hagen von Eitzen May 18 '16 at 21:08
• are you assuming compact domain? – user99914 May 18 '16 at 21:15
• No, I am not. However, I suspected there might be some trivial obstruction I have missed because the fact $\mathbb{S}^n$ is compact. Do you see one? (If so, I guess the question is more interesting for $\mathbb{H}^n$) – Asaf Shachar May 18 '16 at 21:36
• Horospheres are Euclidean, so an affirmative answer in the hyperbolic case follows immediately from Nash. – Andrew D. Hwang May 18 '16 at 21:58
• Since (rectangular) n dimensional flat torus embeds in a round sphere you get a positive answer at least in the case of compact manifolds. – Moishe Kohan May 18 '16 at 23:23

$\newcommand{\Reals}{\mathbf{R}}\newcommand{\Int}{\mathbf{Z}}$The answer is "yes" in all cases:

Hyperbolic Space:

Euclidean space $\Reals^{n}$ embeds isometrically in $\mathbf{H}^{n+1}$ as a horosphere.

Spheres and Tori:

The decomposition $\Reals^{2m} \simeq (\Reals^{2})^{m}$ gives an orthogonal product of $m$ circles of radius $\frac{1}{\sqrt{m}}$ in the unit sphere $S^{2m-1}$. The Euclidean line $\Reals$ can be isometrically embedded into a flat $S^{1} \times S^{1}$ (with circles of equal, arbitrary positive radius $r$). For example, take an arclength parametrization of the path $$t \mapsto (t, r\arctan t),$$ whose image lies in $\Reals \times (-\frac{1}{2} \pi r, \frac{1}{2} \pi r) \subset \Reals \times \Reals$: Divide by the square lattice $2\pi r\Int \times 2\pi r\Int$ to get an isometric embedding into a square torus. (The square in the preceding diagram is a fundamental domain for the lattice.) It follows that Euclidean $\Reals^{n}$ can be isometrically embedded in a $(2n)$-dimensional torus $(S^{1} \times S^{1})^{n}$ whose circles have equal radius, and a "suitably small" torus of this type embeds isometrically into the unit sphere $S^{4n - 1}$.

• Right, with one terminological quip: Usually, an "embedding" is understood as a homeomorphism to its image. The maps you are constructing will not be homeomorphisms to their images (in the sphere) if the Riemannian manifold is noncompact (at least in general). What you getting are isometric injective immersions. One needs to work a bit more to get embeddings. The trick is to first construct an isometric embedding $R\to T^2$ (not given by a line with irrational slope). – Moishe Kohan May 19 '16 at 21:11
• @studiosus: Yes, that was sloppy of me. You're thinking of, e.g., taking an arc-length reparametrization of $t \mapsto (t, \arctan t)$ and wrapping it around a product of unit circles, so the ends of $\Reals$ accumulate on a pair of parallel circles? – Andrew D. Hwang May 19 '16 at 22:00
• Andrew: That would work. I was thinking about a double spiral which is an isometric embedding of $R$ into $R^2$ with bounded image (contained in a small disk), like an Euler spiral. Then embed the small disk in the 2-torus. – Moishe Kohan May 19 '16 at 22:08
• Thanks for your comments! I have tried to implement carefully your ideas, but got stuck at some point. I have two difficulties; 1) We need to prove that the image of the Euler curve can be arbitrarilly small. (This is because when our circles gets smaller radii, their volume decreases, hence the disks we can isometrically embedd in them also decrease in diameter). "See my attempted proof of Lemma1 below" 2) I am having hard time convincing myself formally this is actually an embedding (and not just a smooth isometric immersion). – Asaf Shachar May 20 '16 at 16:27
• Since this is all a bit of a diversion from the original question, I have opened a new one here: math.stackexchange.com/questions/1793159/…. Can you help? Thanks! – Asaf Shachar May 20 '16 at 16:27

$\newcommand{\til}{\tilde}$ $\newcommand{\al}{\alpha}$

$(1)$ A product of isometric embeddings is an isometric embedding:

More precisely, assume $f:(M,g) \to (N,h)\, \,, \, \,\til f:(\til M,\til g) \to (\til N,\til h)$ are isometric embeddings. Then the map $f \times \til f:(M \times \til M,g \oplus \til g) \to (N \times \til N,h \oplus \til h)$ is an isometric embedding (where $g \oplus \til g,h \oplus \til h$ are the standard product metrics).

Of course, this remains true for any finit number of products.

$(2)$ A product of a finite number of flat Tori, is isometric to a flat Torus:

Proof:

The flat torus $\mathbb{T}^n=\mathbb{R}^n / \mathbb{Z}^n$ is isometric to $(\mathbb{S}^1_{\frac{1}{2\pi}})^n$ (product of $n$ copies of circles with radius $\frac{1}{2\pi}$ ). Hence $\mathbb{T}^n \times \mathbb{T}^m \cong (\mathbb{S}^1_{\frac{1}{2\pi}})^n \times (\mathbb{S}^1_{\frac{1}{2\pi}})^m \cong (\mathbb{S}^1_{\frac{1}{2\pi}})^{n+m}=\mathbb{T}^{n+m}$

Observations $(1),(2)$ imply that if $\mathbb{R}$ is isometrically embeddable in a flat torus, then so is $\mathbb{R}^n$.

We also need the following two results:

Lemma (1): $\mathbb{R}$ can be isometrically embedded into $\mathbb{S}^1_a \times \mathbb{S}^1_a$ for arbitrary positive $a$. (We at least need $a=\frac{1}{\sqrt{2n}}$ for every natural $n$ and $a=\frac{1}{2\pi}$).

Corollary (1): $\mathbb{R}$ can be embedded in a flat torus $\mathbb{T}^2$ (The case of $a=\frac{1}{2\pi}$). Also, $\mathbb{R}^n$ can be embedded in a flat torus $\mathbb{T}^{2n}$.

Lemma (2): A product of $m$ circles of radius $\frac{1}{\sqrt m}$ can be isometrically embedded in the unit sphere $\mathbb{S}^{2m−1}$;

Proof of Lemma (2):

The inclusion $\mathbb{S}^1_{\frac{1}{\sqrt m}} \subseteq \mathbb{R}^2$ is an isometric immersion (by definition). Now take the $m$-product of this embedding (and use observation $(1)$). Note that the image is contained in the unit sphere $\mathbb{S}^{2m−1} \subseteq \mathbb{R}^{2m}$.

Proposition (1): $\mathbb{R}^n$ can be isometrically embedded in $\mathbb{S}^{4n-1}$

Using Lemma1 (with $a= \frac{1}{\sqrt {2n}}$), we get that $\mathbb{R}$ is isometrically embedded in $\mathbb{S}^1_\frac{1}{\sqrt {2n}} \times \mathbb{S}^1_\frac{1}{\sqrt {2n}}$, hence (observation $(1)$ again) $\mathbb{R}^{n}$ is isometrically embedded in $\big( \mathbb{S}^1_\frac{1}{\sqrt {2n}} \times \mathbb{S}^1_\frac{1}{\sqrt {2n}} \big)^n \cong \big( \mathbb{S}^1_\frac{1}{\sqrt {2n}} \big) ^{2n}$ which by Lemma (2) (with $m=2n$) can be isometrically embedded in $\mathbb{S}^{4n-1}$.

Partial proof of Lemma(1):

We will use a version of Euler's spiral. Let $c >0$ and define

$\al(t)= \frac{1}{c} (\int_0^{ct} \cos(s^2)ds,\int_0^{ct} \sin(s^2)ds)$,

$\al$ is clearly a smooth isometric immersion. In fact, it's an isometric embedding $\mathbb{R} \to \mathbb{R}^2$ with bounded image. (Actually this is not trivial, see this question. For another construction for an isometric embedding of $\mathbb{R}$ in $\mathbb{R}^2$ with bounded image see here).

Since the integrals defining $\al$ are bounded,by choosing $c$ to be arbitrarily large , we can get the image to be contained in an arbitrarily small disk.

Since $\mathbb{S}^1_a \times \mathbb{S}^1_a$ is flat, it is locally isometric to a disk in $\mathbb{R}^2$. By the discussion above, no matter what is the size of the disk, we can think of our $\alpha$ (for a suitable value $c$) as an isometric embedding into that disk, hence into $\mathbb{S}^1_a \times \mathbb{S}^1_a$, as required.