Isometries of $\mathbb R^2$ with maximum norm.

Question

I have been asked to prove that all isometries of $\mathbb R^2$ with suprenum norm

$$ \|(x,y)\|_{\infty}=\max \{ |x|,|y| \} $$

Are $T(x,y)=(ax+by, cx+dy)$ where $b=c=0$ and $a, d \in \{1, -1\}$ or $ a=d=0$ and $b, c \in \{1, -1\}$.

It is obvious that these are isometries. Why are these all of them.

It seems that by computation we can show these are all of these are isometries. But can we use classic theorems to answer the question?

Answer 1

The concept of extreme points helps here, even though it does not involve the norm itself. The definition implies that if $p$ is an extreme point of a set $K$, and $T$ is an invertible linear map, then $T(p)$ is an extreme point of $T(K)$ .

Consider the closed unit ball of this space, which is a square $Q$. Its extreme points are $\{(\pm 1, \pm 1)\}$. Any isometry maps $Q$ onto itself, and therefore maps the extreme points onto extreme points. This leaves only a few choices:

• $(1, 1)$ can be mapped to $(1, 1)$, $(1, -1)$, $(-1, 1)$, or $(-1, -1)$.
• $(1, -1)$ has to be mapped to one of those points too, but it can't be the same as the image of $(1, 1)$; it can't be the negative of it, either.

Since $(1, 1)$ and $(1, -1)$ form a basis of the space, an isometry is determined by their images.

This leaves us with at most $4\times 2$ isometries, and you already found as many.