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Let $X\subset\mathbb R^d$ be a nonempty finite set and let $f:X\longrightarrow\mathbb R^m$ preserve all distances. Then there is an affine isometric embedding

$$F:\mathbb R^d\longrightarrow\mathbb R^{m+d} \quad\text{with}\quad F(x)=(f(x),0)\quad(x\in X).$$

In particular, a prescribed projection point $y\in\mathbb R^d$ can be carried along when $X$ is embedded into a larger transitive configuration.

Complete proof. Fix $x_0\in X$ and let $V=\operatorname{span}\{x-x_0:x\in X\}$. Polarization gives

$$\langle x-x_0,x'-x_0\rangle =\tfrac12\bigl(\|x-x_0\|^2+\|x'-x_0\|^2-\|x-x'\|^2\bigr).$$

Thus the vectors $f(x)-f(x_0)$ have exactly the same Gram matrix. The map $x-x_0\mapsto f(x)-f(x_0)$ extends linearly to a well-defined isometry $L:V\to\mathbb R^m$: a linear combination of the original vectors has norm zero exactly when the corresponding combination of image vectors does. Choose any linear isometry $J:V^\perp\to\mathbb R^d$. Decompose $u-x_0=v+w$ with $v\in V$, $w\in V^\perp$, and put

$$F(u)=(f(x_0)+Lv,Jw).$$

Its two linear summands are orthogonal, so it preserves distances. For $u\in X$ the second summand vanishes, giving the required extension. $\square$

This compilation lemma supplies the enclosure step omitted from the short proof of Corollary 4, source p. 6. It permits a base that is merely subsoluble; it does not assume that the full symmetry group of the base itself is soluble. Its use is explicit in Corollary 4.

Bears on. #174.