The essence of this question is:

Let $(M,g_M)$ and $(N,g_N)$ be Riemannian manifolds. How many different ways are there to embed $M$ isometrically in $N$?

In this context, I say the embedding $i_1$ is different from $i_2$ if there is no isometry $\varphi:N\to N$ suct that $i_2=\varphi\circ i_1$.

This question is rather easy for specific pairs of manifolds. For example, if $M$ is a line, any embedding has an arc length parametrization, which is just a different name for isometric embedding. Alternatively, let $M$ be Euclidean plane and $N$ Euclidean space. Then for every (smooth) function $f:\mathbb{R}\to\mathbb{R}$, the submanifold $$\{(x,y,z)|y=f(x)\}\subset\mathbb{R}^3,$$equipped with the Riemannian metric inherited from $\mathbb{R}^3$, is isometric to the plane. It is not hard to convince oneself that many of these submanifolds cannot be carried to one another by isometries of Euclidean space.

However, there are other simple examples for which I don't know the answer. For example, what happens when $M$ is $S^2$ with the unit sphere metric, and $N$ is Euclidean space? Somehow, the only way I can imagine $S^2$ embedded in $\mathbb{R}^3$ is by the standard embedding, but I can't find an argument to prove that. Intuitively speaking, it may have something to do with curvature. If so, I guess the answer to the question changes when allowing $C^1$ embeddings.

Anything regarding the last example, of the sphere and Euclidean space, would already make me happy. If, by any chance, someone has an answer to a more general case, that would be great. Insights about less smooth embeddings (e.g. $C^1$ embeddings) are also welcome, though I'm mostly interested in the smooth case.

  • $\begingroup$ I don't know what a complete answer to the general question could look like. Given a natural number $n$, I think I can give an example of a pair of manifolds so that there are exactly $n$ isometric embeddings, but this is hardly a full answer. Can you try to clarify what kinds of answers you are looking for, besides the example of the sphere? $\endgroup$ Commented Jul 15, 2015 at 14:30
  • $\begingroup$ @JoonasIlmavirta I'm looking for things that can be said. For example, what you're saying about $n$ different isometric embeddings sounds interesting, I'd like it if you could describe such a construction explicitly. $\endgroup$ Commented Jul 15, 2015 at 17:15

2 Answers 2


You should have a look at Berger's monumental A Panoramic View of Riemannian Geometry (specifically section 3.4 p. 131-142 and section 4.6, p. 216-218).

Two striking theorems are the following:

  1. Every Riemannian metric $g$ on $S^2$ with positive curvature is isometrically embeddable in $\mathbb R^3$, in a unique way. [Alexandrov, Weyl, Nirenberg, Pogorelov].

  2. A consequence of Nash's embedding theorem is that there are $C^1$ isometric embeddings of the standard $S^2$ in an arbitrarily small ball in $\mathbb R^3$. Obviously, this is quite far from the standard embedding...

  • $\begingroup$ Striking indeed. Thanks! $\endgroup$ Commented Jul 15, 2015 at 19:09

Here is one observation on isometric embeddings:

Claim: For any integer $n\geq0$ there is a pair of Riemannian manifolds $M$ and $N$ so that there is exactly $n$ ways to embed $M\to N$ smoothly and isometrically.

This depends on putting several connected components in $N$ and the following lemma. I do not have a proof for the lemma, but I believe it could be proven along the same lines as the result given by PseudoNeo. Let me denote by $S_r^n$ the sphere of radius $r$ in $\mathbb R^{n+1}$ with the usual round metric inherited from the ambient space.

Lemma: If $r<s$, there is exactly one way to embed $S_r^2$ into $S_s^3$ smoothly and isometrically.

Proof of the claim: If $n=0$, one can take $M=S_2^2$ and $N=S_1^2$, for example. Suppose then that $n\geq1$. Choose $M=S_{1/2}^2$ and let $N$ be the disjoint union $\coprod_{k=1}^n S_k^3$. Isometries of $N$ are isometries of the connected components and $M$ can be embedded in exactly one way to each component. Since there are $n$ components to choose from, there are exactly $n$ ways to embed $M$ isometrically and smoothly to $N$.


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