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Determine the group of isometries of $\mathbb{R}^n$ equipped with the sup metric.

My wild guess is $(a_1,\ldots,a_n)\mapsto(\pm a_{\sigma(1)},\ldots,\pm a_{\sigma(n)})+(c_1,\ldots,c_n)$ where $(c_1,\ldots,c_n)$ is constant, the choice of $\pm$'s are the same for all elements in the domain, and $\sigma$ is a permutation of $(1,2,\ldots,n)$. What I note was that subtracting the function by a constant, changing the sign in one entry, and permuting the entries preserve the isometric property.

So we can assume $f(0,\ldots,0)=(0,\ldots,0)$. For any vector $(a_1,\ldots,a_n)$, $\|f(a_1,\ldots,a_n)-(0,\ldots,0)\|=\|(a_1,\ldots,a_n)-(0,\ldots,0)\|$, so $\|f(a_1,\ldots,a_n)\|=\max|a_i|$. Therefore, for some $k$, there are infinitely many vectors $(a_1,\ldots,a_n)$ with $\|f(a_1,\ldots,a_n)\|=|a_k|$. Among them, there are infinitely many $(a_1,\ldots,a_n)$ such that $\|f(a_1,\ldots,a_n)\|$ is the absolute value of its $j$-th position. By permuting the entries, we can assume $j=k=1$. These vectors have the property that the first entries of $f(a_1,\ldots,a_n)$ and $(a_1,\ldots,a_n)$ coincide in absolute value. Since we can change the sign in one entry, assume infinitely many of them coincide with the same sign.

I was thinking, from here we may be able to prove the same result for any vector, i.e., the first entries of $f(a_1,\ldots,a_n)$ and $(a_1,\ldots,a_n)$ are equal. Then, we can move on to the second, third, all the way up to the $n$-th entry. But I got stucked here.

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Think about what an isometry fixing $0$ has to do when restricted to the extremal points of the unit ball. Your wild guess is correct. – t.b. Mar 24 '12 at 12:03
I managed to prove that the isometry sends the extremal points into themselves (it's very difficult for me :P). But I cannot prove that it maps "correctly", and I don't know how to continue from here. – pola Mar 24 '12 at 13:01
Notice that every vector of the standard basis is uniquely a combination of $2^{n-1}$ of the extremal points and conversely, such a combination of norm $1$ of $2^{n-1}$ of the extremal points is $\pm$ a standard basis vector (prove this by induction). – t.b. Mar 24 '12 at 13:09
By combination I mean of course a convex combination. – t.b. Mar 24 '12 at 13:24
Can somebody write up an answer perhaps - @pola? It can be a CW one if you don't want credit. – Sharkos Jul 16 '13 at 21:17

1 Answer 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.

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