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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.2012-03-24
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    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.2012-03-24
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    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).2012-03-24
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    By combination I mean of course a convex combination.2012-03-24
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    What is a combination of norm 1? Does it mean the arithmetic mean? Anyway, I haven't been able to prove it, but can you give a direction on how to complete the proof after this? I'm not sure how to make use of it, because the norm function is not linear.2012-03-24
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    Yes, you can say an arithmetic mean instead. If an arithmetic mean of $2^{n-1}$ extremal points has norm one then it is $\pm$ a basis vector. Now this tells you that every standard basis vector is sent to a standard basis vector. And since you certainly assume that your isometry is surjective, it is linear if it fixes zero, so it is determined by its action on the standard basis.2012-03-24
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    So, there's a theorem of Mazur-Ulam (I didn't know this theorem, just did a quick search). It makes everything much easier. Thanks a lot!2012-03-24
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    No problem. By the way, this result on the isometry group holds for all $\ell^p$ norms on $\mathbb{R}^n$ for $p \neq 2$ by a theorem of Banach and Lamperti.2012-03-24
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    Can somebody write up an answer perhaps - @pola? It can be a CW one if you don't want credit.2013-07-16

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