# Isometric embedding

I am confused by the term "Isometric Embedding". To my knowledge, this refers to a distance preserving map from a space to another (a mapping $f:(E,d_1) \to (F,d_2)$ such that $d_2(f(x_1), f(x_2)) = d_1(x_1, x_2) )$. But I have the following problem :

On one side, I see papers saying that an isometric embedding of a sphere (with its geodesic distance) to an euclidean space cannot exist; e.g., see The Sphere is not Flat by P. L. Robinson.

On the other side, I see the Nash embedding theorem which says that any surface can be embedded into $R^n$ for some $n$.

What didn't I understand ?

Thanks!

The Nash embedding theorem uses a different $d_2$. When you have a submanifold of a Riemann manifold there's an induced Riemann metric, which comes from the inner product on all the tangent spaces of the ambient manifold. Call this metric $d_3$. In the Nash embedding theorem, in your statement above, replace $(F,d_2)$ with $(f(E), d_3)$. That's the kind of isometric embedding this theorem refers to.

• thanks for the answer! It took me some time to understand it, but with the help of Henning's answer (which I thus upvoted), I think I got it... thank you both! – WhitAngl Dec 1 '11 at 22:11
• I am still a little confused. So when Nash talks about isometric embedding, Does he really mean that 2 points on a Riemannian manifold $M$ of dimension $m$, separated a (geodesic) distance $D$ before the embedding, will be now separated the same distance $D$ on $\mathbb{R}^n$, but following a straight line? Thanks Braco – user66877 Mar 15 '13 at 14:06
• @Braco: no, not at all. If a manifold is in Euclidean space it has a preferred Riemann metric (coming from the Euclidean metric). That induces a preferred path-metric (the infimum of the lengths of all paths between points). Nash's theorem refers to an abstract Riemann manifold embedding in Euclidean space so that the two path metrics (the one from the abstract Riemann metric and the one induced from Euclidean space) agree. – Ryan Budney Mar 15 '13 at 18:45

The usual 2-sphere exists naturally in $\mathbb R^3$, and in general the usual definition of $S^n$ is as a particular subset of $\mathbb R^{n+1}$ with the induced metric. In that case, the identity map is a locally metric-preserving embedding into $\mathbb R^2$, but it doesn't preserve the global distance. To wit, two diametrically opposed points have distance $2$ in $\mathbb R^3$ but distance $\pi$ along geodesics in the sphere itself.

Thus, the natural embedding works as an isometry when we view the two spaces as Riemannian manifolds, but not when we consider them directly as metric spaces. It appears that both kinds of maps can be called "isometric embeddings", but nonetheless they are different concepts.

• That's very helpful, thanks. The confusing part for me is that people describe Nash embedding theorem as saying that "every manifold can be isometrically embedded in some Euclidean space". To me it would be clearer if they instead said "every manifold can be isometrically embedded in a submanifold of some Euclidean space (with the ambient inner product)". – Călin May 17 '19 at 12:23

The confusion here comes from the fact that there are two different notions both called "metric" and two different notions, both called "an isometry" or "an isometric embedding."

(1). In the theory of metric spaces. A metric, or a distance function, is a map $$X\times X\to {\mathbb R}_+$$ satisfying certain conditions, such as symmetry, triangle inequality, etc. Accordingly, a map $$f: (X, d_X)\to (Y, d_Y)$$ is an isometry in the sense of metric geometry if $$d_Y(f(x_1), f(x_2))=d_X(x_1, x_2), \forall x_1, x_2\in X.$$

(2). In Riemannian geometry. A metric or, rather a Riemannian metric is a certain tensor (or a tensor field) on a differentiable manifold $$M$$. More precisely, a Riemannian metric $$g$$ on $$M$$ is a collection of inner products on tangent spaces $$T_pM, p\in M$$, satisfying certain smoothness condition. If $$M$$ happens to be connected then $$g$$ defines the Riemannian distance function $$d_g$$ on $$M$$: $$d_g(x,y)= \inf_{c} \int_0^1 g(c'(t), c'(t))^{1/2}dt,$$ where the infimum is taken over all (say, smooth) paths $$c: [0,1]\to M$$ connecting $$x$$ to $$y$$. Then $$(M, d_g)$$ is a metric space in the sense of (1).

Example. $$M={\mathbb R}^n$$. Then all tangent spaces $$T_xM$$ are canonically isomorphic to $${\mathbb R}^n$$. Then taking an inner product $$\langle \cdot, \cdot \rangle$$ on $${\mathbb R}^n$$ we obtain a Riemannian metric $$g_0$$ on $$M$$.

The corresponding notion of an isometric embedding in Riemannian geometry is:

Given two Riemannian manifolds $$(M_1, g_1), (M_2,g_2)$$, a smooth map $$f: (M_1, g_1)\to (M_2,g_2)$$ is called a (Riemannian) isometric embedding if it is a topological embedding and for every $$x\in M_1$$ and any two tangent vectors $$u, v\in T_xM_1$$, we have $$g_2(df_x(u), df_x(v))= g_1(u,v).$$

Now, Nash isometric embedding theorem is about Riemannian metrics and Riemannian isometric embeddings. It states:

Suppose that $$(M,g)$$ is a $$C^k$$-smooth $$m$$-dimensional Riemannian manifold, $$k\in [3,\infty]$$. Then there exists a $$C^k$$-smooth isometric embedding $$f: (M, g)\to ({\mathbb R}^n, g_0)$$ for $$n=m(3m+11)/2$$.

Note that if $$M$$ is connected, this theorem does not claim that $$f$$ preserves the Riemannian distance function, i.e. it does not hold, in general that $$|f(x)-f(y)|=d_g(x,g), x, y\in M.$$