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P.S. I asked the question on MSE more than a week ago, but didn't get any desired answer, so asking here.

Let $m < n \in \mathbb{N}$. Let us equip $\mathbb{R}^m, \mathbb{R}^n $ with their canonical Euclidean (Riemannian) metrics. How can we characterize the extrinsic distance preserving embeddings of $\mathbb{R}^m$ into $\mathbb{R}^n $? To be more precise, I'm looking for sufficiently regular, injective, distance preserving transformations from $\Phi: \mathbb{R}^m \to \mathbb{R}^n$, so that: $d_m(x, y)= d_n(\Phi(x), \Phi(y))$, where $d_m, d_n$ represents the distances in $m, n$ dimensional Euclidean spaces respectively, so $d_m(x,y):=||x-y||_{\mathbb{R}^m}, d_n(x,y):=||x-y||_{\mathbb{R}^n}$. Note that this means: we're looking to preserve the extrinsic distances between the two Euclidean spaces, we don't care about if the first one embeds isometricaly as a submanifold in the second.

I think the answer is:

$$x \mapsto A(Bx, O(Bx))$$ where $A$ is an Euclidean isometry of $\mathbb{R}^n$ (i.e. a rigid motion), $B$ is an Euclidean isometry of $\mathbb{R}^m$ (i.e. a rigid motion), and $O:\mathbb{R}^m \to \mathbb{R}^n$ is "zero padding $n-m$" times, namely: $O(x)= (x, 0,\dots 0)$.

If the above correct/incorrect, how do I go about proving it or characterizing the Euclidean embeddings?

Approaches that immediately come to mind are as follows:

(1) Assume $\phi: \mathbb{R}^m \to \mathbb{R}^n$ is such an isometric embedding. Then we can try to consider the canonical projection $\pi: \mathbb{R}^n \to \mathbb{R}^m$ and try to consider the map: $\phi \circ \pi: \mathbb{R}^n \to \mathbb{R}^n $, but this will not be an isometry (not one to one), which is a problem.

(2) Alternately, we can consider: $\pi \circ \phi: \mathbb{R}^m \to \mathbb{R}^m$. We can't guarantee that $\pi \circ \phi$ is an isometry of $\mathbb{R}^m$, which is again a problem.

(3) As someone suggested in the comments of the MSE question, there exists an isometry, say $A$, of $\mathbb{R}^n$ so that $A \circ \phi (\mathbb{R}^m)= \mathbb{R}^m \times \{0\}^{n-m}$. This intuitively seems correct, as the isometry group of $\mathbb{R}^n$ has dimension $\frac{n(n+1)}{2}$, which means it's "high dimensional enough", i.e. "there're enough elements in it" to render the last $n-m$ components to $0$. This could be made rigorous if $\phi$ was a linear or affine map, but if $\phi$ is nonlinear, I can't see how this argument goes through? In this case we need to show first that $\phi$ is affine, i.e. if $\phi: \mathbb{R}^m \to \mathbb{R}^n$ fixes $0$, then it has to be a linear map from $\mathbb{R}^m$ to $\mathbb{R}^n$.(?)

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  • $\begingroup$ This is incorrect, even for $m=1$ and $n-3$, as well as $m=2$ and $n-3$. In the first case, any arclength parameterization $c: \mathbb{R}\rightarrow\mathbb{R}^2$ is isometric. In the second case, any map of the form $(x,y) \mapsto (a(x),y,b(x))$, where $x\mapsto c(x)=(a(x),b(x))$ is an arclength parameterization of a curve in $\mathbb{R}^2$, is an isometric embedding of $\mathbb{R}^2$ in $\mathbb{R}^3$. $\endgroup$
    – Deane Yang
    Commented Mar 30, 2020 at 19:28
  • $\begingroup$ @DeaneYang Sorry but that's not the kind of isometry i'm looking for, as specified in the question. What you described is an isometry, if we consider the intrinsic distance on the image, not the extrinsic Euclidean distance. For example, if $m=1, n=2,$ then $d(\phi(t_1), \phi(t_2))= |t_1 - t_2|$ if $|\phi'|=1$, fir sure; but the intrisic distance $d(\phi(t_1), \phi(t_2)) \ne ||\phi(t_1) - \phi(t_2)||_{\mathbb{R}^2}$. $\endgroup$
    – Learning math
    Commented Mar 30, 2020 at 19:38
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    $\begingroup$ Oops. Sorry about that. So here's how the argument for your question goes. First show that, given inner product spaces $V$ and $W$ if a map $f: V \rightarrow W$ satisfies $|f(v_1)-f(v_2)| = |v_1-v_2|$, for all $v_1, v_2 \in V$, then $f(v_1)\cdot f(v_2) = v_1\cdot v_2$, for all $v_1, v_2 \in V$. This implies that $f$ maps an orthonormal basis to an orthonormal basis. You can prove that $f$ is linear and therefore a rotation. Finally, given an extrinsic distance preserving map, compose it with a translation so the origin is fixed, This map satisfies the assumptions of $f$ above. $\endgroup$
    – Deane Yang
    Commented Mar 30, 2020 at 21:51
  • $\begingroup$ By the way, no assumption on $n$ and $m$ is needed. That $m \le n$ is a consequence of the proof. $\endgroup$
    – Deane Yang
    Commented Mar 30, 2020 at 22:18
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    $\begingroup$ The trick is to observe that you can recover the coefficients of a vector with respect to an orthonormal basis by using the inner product. That plus the fact the inner product is preserved under $f$ implies the coefficients are preserve by $f$. $\endgroup$
    – Deane Yang
    Commented Mar 30, 2020 at 22:58

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The answer is yes. Since $\Phi$ is isometric in your sense, it sends a geodesic to a geodesic, i.e. a straight line to a straight line. Then, it is a classical result of affine geometry (see e.g. Artin's wonderful book "Geometric Algebra") that an injective map $R^m\to R^n$ sending straight lines to straight lines is affine (except maybe for $m=1$, but here you have moreover the distance being preserved).

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