Beckman–Quarles theorem

Remarks

No continuity, injectivity or surjectivity of $f$ is assumed; preserving the single distance $1$ is enough. The theorem was published by F. S. Beckman and D. A. Quarles in 1953. The proof uses rigid configurations made of unit segments, such as unit equilateral triangles and simplices. They show that $f$ also preserves other distances, and then that it preserves all distances. Since an isometry of Euclidean space is affine, $f$ is then of the form $x \mapsto Ax + b$ with $A$ orthogonal.

The hypothesis $n \ge 2$ is necessary. On $\R$, let $h \colon [0, 1) \to [0, 1)$ be any bijection, for instance $h(t) = t^2$, and set $f(x) = \floor{x} + h(x - \floor{x})$. If $\abs{x - y} = 1$, then $x$ and $y$ have the same fractional part and their integer parts differ by $1$, so $\abs{f(x) - f(y)} = 1$. But $f(0) = 0$ and $f(1/2) = 1/4$, so $f$ is not an isometry.

Going deeper

For a student

The theorem is a rigidity result in Euclidean distance geometry: one preserved distance forces the whole metric to be preserved. It contrasts with the classical characterization of isometries, which assumes that all distances are preserved. It is connected with unit distance graphs: a unit-distance-preserving map is a graph homomorphism of the unit distance graph of $\R^n$ to itself. The one-dimensional counterexample shows that the plane's richer supply of unit triangles is what makes the result work.