When can we extend distance-preserving maps between finite sets of points to isometries?

Question

Suppose I have $n$ points $x_1, \ldots ,x_n \in \mathbb{R}^d$ where the affine span of $\{x_1, \ldots ,x_n \}= \mathbb{R}^d$. If I have another set of points $y_1, \ldots, y_n$ where $\|y_i - y_j\| = \|x_i - x_j \|$ for all $i,j \in \{1, \ldots, n\}$ where $\| \cdot \|$ is the Euclidean norm then there exists an isometry $f : \mathbb{R}^d \rightarrow \mathbb{R}^d$ such that $f(x_i)=y_i$ for all $i = 1 , \ldots, n$. The same cannot be said for general normed spaces, for instances if we take $$x_1 =(0,0), ~ x_2 = (1,0), ~ x_3=(0,1)$$ in $(\mathbb{R}^2, \| \cdot \|_1)$ and $$y_1 = (0,0), ~ y_2 = (0.5,0.5), ~ y_3 = (0.5,-0.5).$$ Does anyone know of where to look into such results? My intuition say that for $\mathbb{R}^d$ with a smooth $\ell_p$ norm (i.e. $1<p<\infty$) we would need $2d$ points but I have no idea where to begin to look for papers in this area. Any ideas and/or references would be much appreciated.

Answer 1

This is an amazing property of Euclidean geometry. Given any subset $A \subseteq \mathbb R^n$ and an isometry of $A$ into $\mathbb R^n$, there is an extension to an isometry of the whole space. And if the affine span of $A$ is the whole space, then the extension is unique. Hyperbolic space also has this property. In dimension 2, this is classical (as in classical Greek times) and can be thought of as underpinning the theory of triangles.

Long ago I learned all about this, and much more, in the old book
Busemann & Kelley Projective Geometry and Projective Metrics
which is likely in your university library. They study many types of geometry; this extension property for isometry is sometimes seen, but it is rare.

Answer 2

I could not find the statement in the book referenced in the answer by GEdgar and proving the statement was easier than to continue looking. Here is my work for anyone looking in the future:

Def: Isotropies "extend" on $\mathcal{X}$

If someone comes up with better naming please tell me - I hate it

For a metric space $(\mathcal{X},d)$ with a set of isotropies $\Phi$, we say isotropies "extend", if:

For any $n\in\mathbb{N}$, $x_0,\dots, x_n, y_0,\dots, y_n\in \mathcal{X}$ with $$ d(x_i,x_j) = d(y_i, y_j) $$ there exists an isometry $\phi\in \Phi$, such that $\phi(x_i) = y_i$.

Example: Hilbert space

If $\mathcal{X}$ is a Hilbertspace, then the isotropies of the metric induced by the scalar product of the Hilbert space "extend".

Proof

Let $x_i, y_i\in \mathcal{X}$ as in the definition of extending isotropies and assume without loss of generality that $x_i\neq x_j$ for $i\neq j$ (in the case of equality we would also have $y_i = y_j$ due to positive definiteness of the metric). We define $$ \tilde{x}_i = x_i - x_0 $$ and similarly for $y$. In particular $\tilde{x}_0 = \tilde{y}_0=0$. Since $\tilde{x}_i$ and $\tilde{y}_i$ satisfy the requirements of the Lemma (Hilbertspace extension - below), there exists a linear isometry $\tilde{\phi}$ with $\tilde{\phi}(\tilde{x}_i) = \tilde{y}_i$. Then the isometry $$ \phi : x\mapsto \tilde{\phi}(x-x_0) + y_0 $$ does the job.

Example: $n$-Sphere

Let $\mathcal{X} = \mathbb{S}^{n-1} = \{x: \|x\|=1\}$ be the sphere in $\mathbb{R}^n$. Then isotropies extend.

Proof

The Lemma (Hilbertspace extension) can be applied to $x_i$ and $y_i$ viewed as members of $\mathbb{R}^n$ and the resulting isometry can be restricted to the sphere as linear isometries map spheres to spheres.

Lemma: Hilbertspace extension

Let $\mathcal{X}$ be a Hilbertspace, $x_i,y_i \in \mathcal{X}$ for $i=1,\dots, n$. If we have $$ \|x_i\|= \|y_i\| \quad \text{and}\quad \|x_i-x_j\| = \|y_i-y_j\| $$ for all $i,j$, then there exists a linear isometry $\phi$ with $$ \phi(x_i) = y_i. $$

Proof

By the polarization formula $$ \langle x_i, x_j\rangle = \frac{\|x_i\|^2 + \|y_i\|^2 - \|x_i - x_j\|^2}{2} = \langle y_i, y_j\rangle \qquad \forall i,j $$ We apply the Gram-Schmidt orthonormalization procedure to both $x_i$ and $y_i$ such that $$ U_{k_n} = \text{span}(u_1, \dots, u_{k_n}) = \text{span}(x_1,\dots, x_n) $$ for orthonormal $u_i$ where we skip $x_m$ if it is already in $U_{k_{m-1}}$ (resulting in $k_m = k_{m-1}$), and similarly $$ V_{k_n} = \text{span}(v_1, \dots, v_{k_n}) = \text{span}(y_1,\dots, y_n). $$ Since this procedure only uses scalar products, we inductively get $$ \langle x_k, u_j\rangle = \langle y_k, v_j\rangle \quad \forall k, j $$ We now extend $u_i$ and $v_i$ to orthonormal basis of $\mathcal{X}$ and define the linear mapping by its behavior on the basis elements $\phi: u_i \mapsto v_i$. Mapping an orthonormal basis to an orthonormal basis is an isometry and we have $$ \phi(x_k) = \phi\Bigl( \sum_{j=1}^k \langle x_k, u_j\rangle u_j \Bigr) = \sum_{j=1}^k \langle x_k, u_j\rangle \phi(u_j) = \sum_{j=1}^k \langle y_k, v_j\rangle v_j = y_k. $$