isometry

An *isometry* is a function that preserves a metric, either in the sense of a metric space or in the sense of a Riemannian manifold.

An **isometry** $f\colon (X,d) \to (X',d')$ between metric spaces is a function $f\colon X \to X'$ between the underyling sets that respects the metrics in that $d = f^* d'$. More explicitly, $d'(f(a),f(b)) = d(a,b)$ for any points $a,b$ in $X$.

The same idea holds for extended quasi-pseudo-generalisations of metric spaces.

An **isometry** $f\colon (X,g) \to (X',g')$ between Riemannian manifolds is a morphism $f\colon X \to X'$ between the underlying manifolds that respects the metrics in that $g = f^* g'$. More explicitly, $g'(f_*v,f_*w) = g(v,w)$ for any tangent vectors $v,w$ on $X$.

Global isometries are the isomorphisms of metric spaces or Riemannian manifolds. An isometry is **global** if it is a bijection whose inverse is also an isometry. Between metric spaces, isometries are necessarily injections and bijective isometries necessarily have isometries as inverses, so global isometries between metric spaces are also called *surjective isometries*; this does not work for Riemannian manifolds (where the inverse of an isometry need not be a morphism of manifolds), nor does it work for pseudometric spaces (where an isometry need not be injective).

In practice, isometries $E \to F$ between normed vector spaces tend to be affine maps. The following theorem gives a precise meaning to this.

A norm on a vector space is **strictly convex** if, whenever ${\|u\|} = 1 = {\|v\|}$, we have ${\|t u + (1-t)v\|} \lt 1$ for some (hence all!) $t$ in the range $0 \lt t \lt 1$. In brief, no sphere contains a line segment. Examples of strictly convex spaces include spaces of type $L^p$ for $1 \lt p \lt \infty$.

Let $f \colon E \to F$ an isometry between normed vector spaces, and suppose $F$ is strictly convex. Then $f$ is affine.

To say that $f \colon E \to F$ is affine means that $f$ preserves linear combinations of the form $t x + (1-t)y$. It suffices to consider only the case where $0 \lt t \lt 1$ and, by continuity considerations, only the case of dyadic rationals between $0$ and $1$. Continuing this train of thought, it suffices to prove that $f(\frac1{2}(x + y)) = \frac1{2}(f(x) + f(y))$ for all $x, y$.

In the case of strict convexity, midpoints $\frac1{2}(u+v)$ are determined in terms of the norm, as the unique point $w$ such that

${\|w - u\|} = \frac1{2}{\|u-v\|} = {\|w-v\|}.$

The midpoint satisfies these equations for any normed vector space, but the uniqueness is a consequence of strict convexity. For if there were two such points $w, w'$, then for some point $w''$ on the line segment between them, we would have ${\|w'' - u\|} \lt \frac1{2}{\|u-v\|}$, and ${\|w''-v\|} \leq \frac1{2}{\|u-v\|}$ by ordinary convexity. But these two inequalities taken together would violate the triangle inequality.

As a result, since $f$ is an isometry, $w = f(\frac1{2}(x+y))$ is forced to be the midpoint between $f(x)$ and $f(y)$ if $F$ is strictly convex. This completes the proof.

If $F$ is not strictly convex, then isometries need not be affine. For example, consider $E = \mathbb{R}$, and $F = \mathbb{R}^2$ equipped with the $l_\infty$ (max) norm. For any contractive map $\phi \colon \mathbb{R} \to \mathbb{R}$, e.g., any smooth function with ${|\phi'|} \leq 1$, the map $E \to F$ sending $x$ to $(x, \phi(x))$ is easily seen to be an isometry.

If however $f$ is a *surjective* isometry between normed vector spaces, then $f$ is affine, by the Mazur-Ulam theorem.

Revised on May 8, 2015 12:26:59
by Urs Schreiber
(195.113.30.252)