 # Isometric versus affine

Throughout this article we let $$E$$ and $$F$$ denote real normed vector spaces. A map $$f : E \rightarrow F$$ is an isometry if $$\Vert f(x) – f(y) \Vert = \Vert x – y \Vert$$ for all $$x, y \in E$$, and $$f$$ is affine if $f((1-t) a + t b ) = (1-t) f(a) + t f(b)$ for all $$a,b \in E$$ and $$t \in [0,1]$$. Equivalently, $$f$$ is affine if the map $$T : E \rightarrow F$$, defined by $$T(x)=f(x)-f(0)$$ is linear.

First note that an isometry $$f$$ is always one-to-one as $$f(x) = f(y)$$ implies $0 = \Vert f(x) – f(y) \Vert = \Vert x- y \Vert$ hence $$x=y$$.

There are two important cases when every isometry is affine:

1. $$f$$ is bijective (equivalently surjective). This is Mazur-Ulam theorem, which was proven in 1932.
2. $$F$$ is a strictly convex space. Recall that a normed vector space $$(S, \Vert \cdot \Vert)$$ is strictly convex if and only if for all distinct $$x,y \in S$$, $$\Vert x \Vert = \Vert y \Vert =1$$ implies $$\Vert \frac{x+y}{2} \Vert <1$$. For example, an inner product space is strictly convex. The sequence spaces $$\ell_p$$ for $$1 < p < \infty$$ are also strictly convex.

However, an isometry need not be affine and we provide a counterexample.

To see this, let $$E$$ be the real line equipped with the absolute value $$\vert \cdot \vert$$ norm and let $$F$$ be the plane with the sup norm $$\Vert x \Vert = \sup (\vert x_1 \vert, \vert x_2 \vert)$$ for all $$x=(x_1,x_2) \in \mathbb R^2$$. Let $$g : \mathbb R \rightarrow \mathbb R$$ be a metric map, i.e. $$\vert g(s) – g(t) \vert \le \vert s – t \vert$$ for all $$s,t \in \mathbb R$$. For example $$g(t) = \vert t \vert$$, $$g(t) = \sin t$$ or $$g(t) = \arctan t$$. Now consider the map $$f : \mathbb R \rightarrow \mathbb R^2$$ defined by $$f(t) = (t, g(t))$$.

$$f$$ is an isometry as for $$x,y \in \mathbb R$$ we have $\Vert f(x) – f(y) \Vert = \sup(\vert x-y \vert,\vert g(x) – g(y) \vert) = \vert x – y \vert$

However $$f$$ is not an isometry for the maps $$g$$ mentioned above.