# Norms on $\mathbb{R}^d$ whose linear isometries are the hypercube group

It is a known fact that for any $2\neq p\in[1,\infty]$, the linear isometries for the corresponding norm $\|\cdot\|_p$ on $\mathbb{R}^d$ is the set of all square-matrices with entries in $\{-1,1,0\}$, with at exactly one nonzero entry for each line and each column, that is : the hypercube group.

Question : what are the norms that precisely have this set of matrices as linear isometry group ?

When trying to exhibit such norms $N$, the most general construction that I could guess is the following : $$N(x) = \int_1^{+\infty} \|x\|_p \,\mathrm{d}\mu,$$ where $\mu$ is a finite measure over $[1,+\infty)$. I expect that this should cover all possible such norms, but was not able to prove it.

One track would be to use some Choquet theory for which the previous formula would describe $N$ as a generalized linear combination of extreme points (the norms $\|\cdot\|_p$), but I am not much acquainted with such point of view.

I welcome any reference or suggestion on this question !

Thanks,

Ayman

• I guess you want to consider norms that are 1 on the standard basis vectors. The set of all such norms, considered as functions on $\mathbb{R}^d$ and endowed with the pointwise convergence topology is indeed a convex compact set on which the hypercube group $G$ acts by automorphisms. The fixed point set of $G$, $C$, is a sub-convex compact space, consisting of $G$ invariant norms. However, for these norms the group of isometries might be bigger than $G$ (as $p=2$ which is a limit point of $p\neq 2$ shows). Your question makes more sense without the word "precisely". Anyway, nice question! – Uri Bader Feb 15 '17 at 10:25
• Dear Uri, first thanks for answering so quickly. I do understand why we can focus on norms that are 1 on the canonical basis. But why isn't it adequate to search for norms that have exactly this group as isometries ? Concerning the second formulation, could you detail a little bit ? Is it equivalent to " the group of isometries of $\| \cdot \|$ contains $G$ ", or " the group of isometries of $\| \cdot \|$ is $G$ " ? Thanks ! – Ayman Moussa Feb 15 '17 at 10:39
• Sorry, Ayman, your question makes perfect sense as formulated. I meant to say that while trying to apply Choquet theory, in order to gain compactness, it makes sense to relax the condition "the group of isometries is exactly $G$" to "the group of isometries contains $G$". The latter is a closed condition. – Uri Bader Feb 15 '17 at 10:41
• The group of isometry of any norm is a compact linear group, thus conjugated into the orthogonal group $O(d)$. I guess that any closed subgroup of $O(d)$ that contains (a conjugate of) $G$ is either $O(d)$ or $G$. The only norm fixed by $O(d)$ is $\|\cdot\|_2$, thus classifying norms that their isometry group contains $G$ is equivalent (up to throwing away $\|\cdot\|_2$) to classfying norms with the precise group $G$. You are left with the task of finding the extreme points of the set $C$ described above. – Uri Bader Feb 15 '17 at 10:52
• Reading the answer by @Guillaume it occurred to me: take any linear functional $\theta$ on $\mathbb{R}^d$ and define $N(x)=c\cdot\sum_g |\theta(g^{-1}x)|$. Then $N(x)$ is a $G$ invariant norm which is not strictly convex, so if it is an integral as above than $\mu$ must be an atomic measure at $\{1,\infty\}$. Varying $\theta$ it is easy to arrange also that $N$ is not a combination of $\ell_1$ an $\ell_\infty$, thus not an integral as above. – Uri Bader Feb 15 '17 at 14:43

Here is a complete answer in dimension 2.

Among all the norms on $\mathbb{R}^2$ which are invariant under the dihedral group $D_8$, the extremal rays are norms whose unit ball are octagons with the required symmetries ($\ell_1$ and $\ell_{\infty}$ appear as octagons which degenerate into squares).

The reason is specific to dimension 2: any norm on $\mathbb{R}^2$ is of form $$\|x\| = \int_{S^1} | \langle x , \theta \rangle | d \mu(\theta)$$ for a positive finite measure $\mu$ on $S^1$. (This measure is unique if we require that it is even.) In other words, any 2-dimensional real space embeds into $L^1$. This fails for the $\ell_{\infty}$ norm on $\mathbb{R}^3$.

Under the required symmetries, the measure $\mu$ is uniquely defined by its restriction to the west-northwest part of the unit circle. Among those, extreme rays are Dirac masses, leading to octagons.

In higher dimensions you get examples of such norms using Orlicz norms. I don't know what are the extreme rays there.

Edit : here are some references, A Class of Convex Bodies Ethan D. Bolker Transactions of the American Mathematical Society Vol. 145 (Nov., 1969), pp. 323-345. Schneider, Rolf, and Wolfgang Weil. "Zonoids and related topics." Convexity and its Applications. Birkhäuser Basel, 1983. 296-317.

• Guillaume, could you explain a little bit how to prove your representation formula ? In the reference of Yost you give below, the embedding in $L^1$ seems to be proven in a somehow different way. I tried to compute this measure for the $\|\cdot\|_p$ norms, but apart from the easy cases $p\in\{1,2,\infty\}$, I did not succeed. Do you have any idea for these specific cases ? Btw : you claim this measure to be unique, but for $\|\cdot\|_1$, if $(e_1,e_2)$ is the canonical basis, it seems that both measures $\delta_{e_1} + \delta_{e_2}$ and $\delta_{-e_1}+\delta_{-e_2}$ do the job. – Ayman Moussa Feb 16 '17 at 16:10
• You are right, the measure if unique "up to the $\theta \mapsto -\theta$ symmetry", for example there is a unique even measure. – Guillaume Aubrun Feb 17 '17 at 8:15
• If you consider two independent random variables X,Y with a p-stable distribution (en.wikipedia.org/wiki/Stable_distribution), then $\| s X + t Y \|_{L^1}$ is proportional to $\|(s,t)\|_p$. I'm not sure how explicit will be the measure $\mu$ in this case. – Guillaume Aubrun Feb 17 '17 at 8:20
• To obtain the measure $\mu$ from the embedding in $L_1$, see Theorem 6.1 in the paper by Bolker I mentioned. – Guillaume Aubrun Feb 17 '17 at 8:25

This is not answer, rather a summarize of some remarks.

Let $G$ be the hypercube group. A first step towards classifying all norms with isometry group $G$ is to classify all norms with isometry group which contains $G$. This is what we discuss below.

Let $D$ be the set of all seminorms on $\mathbb{R}^d$ for which the average value on the standard basis vector is 1 (this is a natural normalization). Observe that $D$ is a convex set, compact wrt the pointwise convergence topology. Observe that $G$ acts naturally on $D$ and let $P:D\to D$ be the associated averaging operator $N(\cdot)\mapsto \frac{1}{|G|}\Sigma_g N(g^{-1}\cdot)$. Denoting by $C$ the set of $G$ invariant seminorms (in fact: norms, as $G$ acts irreducibly on $\mathbb{R}^d$) in $D$, $P$ is a projection on $C$. The $P$-preimage of an extreme point in $C$ is a face in $D$, hence contains an extreme point of $D$.

By this observation, it appears that our task is reduced to classifying the set $E$ consisting of all extreme points in $D$.

Given a non-zero linear functional $\theta\in (\mathbb{R}^d)^*$, up to a suitable normalization we have $|\theta(\cdot)|\in D$ and it is easy to see that this is an extreme point. Let us denote by $F\subset E$ the space of extreme points coming from functionals as above. Guillaume Aubrun made in his answer the following observation:

Observation: A norm in $N\in D$ is in the closed convex hull of $F$ iff $(\mathbb{R}^d,N)$ is embedable in $L^1$.

Proof: There exists a probability measure $\mu\in \text{Prob}(F)$ s.t for all $x\in \mathbb{R}^d$, $N(x)=\int |\theta(x)|d\mu(\theta)$ $\Leftrightarrow$ There exists a linear map $\mathbb{R}^d\to L^1(\mu)$, $x\mapsto \theta(x)$.

It then follows that for $d=2$ indeed $E=F$, as every norm on $\mathbb{R}^2$ is embedable in $L^1$ (right? reference?). However, for $d>2$ there are norms which are not embedable in $L^1$, e.g $\|\cdot\|_\infty$ (also $\|\cdot\|_p$, $p>2$ - correct?), thus $F\subsetneq E$. What are the other extreme points of $D$?

• You must have other extreme points than the $|\theta(\cdot)|$'s since all the norms constructed from them will embed isometrically in $L^1$, while $(\mathbb{R}^3,\|\cdot\|_{\infty})$ does not. – Guillaume Aubrun Feb 15 '17 at 16:56
• A related question: can you write in a non-trivial way the $\ell_{p}$ norm ($p > 2$) on $\mathbb{R}^d$ ($d >2$) as a sum of 2 norms ? I would bet no ... – Guillaume Aubrun Feb 15 '17 at 17:25
• By the way, norms whose isometries contain a copy of $G$ are precisely those norm which admit a 1-symmetric basis. – Tomasz Kania Feb 15 '17 at 19:54
• A self-contained reference (this was certainly known before) D. Yost, L1 contains every two-dimensional normed space, Ann. Polonici Math. 49 (1988), 17–19. – Guillaume Aubrun Feb 16 '17 at 10:28
• The fact that for $p>2$, $\ell_p$ of dimension at least $3$ does not embed in $L^1$ was proved in Dor L. E., Potentials and isometric embeddings in L, Israel J. Math 24 (1976), no. 3-4, 260–268. link.springer.com/article/10.1007/BF02834756 – Guillaume Aubrun Feb 16 '17 at 10:55