Group of isometries of $\mathbb{R}^n$

Solution 1:

Theorem. The isometries of $\mathbb{R}^n$ with respect to the sup metric are precisely the functions of the form $$ f(p) \;=\; Ap + b $$ where $A\in\textit{O}(n)$ is a Euclidean symmetry of the unit cube $[-1,1]^n$ and $b\in\mathbb{R}^n$.

Proof: It is easy to see that maps of the given form are isometries. We must show that every isometry has this form.

Let $d$ denote the sup metric. We shall refer to balls in $\mathbb{R}^n$ with respect to $d$ as "cubes". Here are some definitions that we can make using only $d$:

• If $p,q\in\mathbb{R}^n$, a midpoint of $p$ and $q$ is a point $m\in\mathbb{R}^n$ for which $$ d(p,m) \;=\; d(q,m) \;=\; \frac{1}{2} d(p,q). $$

• Two points $p,q\in\mathbb{R}^n$ are diagonal from one another if they have a unique midpoint.

• A corner of a cube is a point on the boundary that is diagonal from the center point. Each cube has $2^n$ corners.

• Two cubes are adjacent if they have exactly $2^{n-1}$ corners in common. Each cube has precisely $2n$ different cubes adjacent to it.

• The intersection of two adjacent cubes is a facet of each. A face of a cube is any intersection of its facets.

Now, let $f$ be an isometry of $\mathbb{R}^n$. By composing with a translation, we may assume that $f$ fixes the origin. From the above discussion, it is clear that $f$ must permute the corners of the unit cube $[-1,1]^n$ in a face-preserving way. Composing with a Euclidean symmetry of the cube, we may assume that $f$ fixes each of the corners. At this point, we wish to show that $f$ is the identity map.

To do so, consider the subdivision of the unit cube $[-1,1]^n$ into $2^n$ subcubes of radius $1/2$. These subcubes are precisely those that have the origin and a corner of the large cube as opposite corners, and therefore $f$ must map each of these subcubes to itself. Furthermore, since $f$ fixes the facets of the large cube, $f$ must fix the corners of each of these subcubes. Repeating this process, we find that the $4^n$ canonical subcubes of radius $1/4$ are invariant under $f$, and so forth. But every point in the unit cube is the intersection of some nested sequence of these subcubes, and therefore $f$ fixes every point in $[-1,1]^n$.

To get outside the unit cube, observe that any cube $C$ adjacent to the unit cube must map to itself under $f$. Moreover, since $f$ fixes a facet of $C$, it must fix all of the corners of $C$, and therefore must fix $C$ pointwise by the argument above. Continuing in this fashion, we can prove that every cube in the integer lattice is fixed by $f$ pointwise, and therefore $f$ is the identity map.