# Isometries of the $l_1^n$ space

02 Dec 2016Let’s recall what distance is. You define a norm on a vector space that induces a metric. Note that metric does not require the underlying set to be a vector space, nor does it have to come from a norm. In this note, however, we will focus on finite-dimensional normed vector spaces. In particular, we want to find isometries of the Banach space $l_1^n$.

The symmetry group of the $l_2^n$ space—known as the Euclidean group—consisting of isometries of the Euclidean space, has as subgroups the translation group $T(n)$ and the orthogonal group $O(n)$. In other words, translations, rotations, and reflections preserve the Euclidean distance.

What is the symmetry group of the $l_1^n$ space? Translations of course preserve the $l_1^n$ metric, so $T(n)$ should be there. What else? Notice that the symmetry group of an $l_p^n$ space has the symmetry group of its unit ball as a subgroup. The symmetry group of the $l_1^n$ unit ball (i.e., the diamond) consists of signed permutations of its vertices. That group is known known as the Hyperoctahedral group. There are no other symmetries, so we may conclude that only translations and signed permutations preserve the $l_1^n$ metric.

If you don’t like handwavy arguments and would rather prefer a more technical proof, check out this very nice post on isometries of vector $p$-norms and read the forum discussion.