# When does the group of isometries of a norm determine the norm?

$\newcommand{\<}{\langle} \newcommand{\>}{\rangle}$

Let $V$ be a finite-dimensional normed space. Assume that $G=\text{ISO}(||\cdot ||_1) = \text{ISO}(||\cdot ||_2)$. When can we conclude that that $\| \|_1$ is a scalar multiple of $\| \|_2$?

Note:

There are two extreme cases:

(1) $G= \text{ISO}(\<,\>)$ for some inner product $\<,\>$ , it is known that the two norms are induced by inner products. Hence, since the isometry group of an inner product determines it up to scalar multiplication*, it follows that one norm is a scalar multiple of the other.

(2) $G=\{Id,-Id\}$, ie the norms are rigid. I think here it is not supposed to be hard to find two rigid norms which are not scalar multiples of each other. (However, I do not have a specific example in mind, and I would be happy to see one.)

Intuitively, the larger the isometry group, the more restrictions there are on the norm. One could ask for some "richness criterion" of the isometry group which defines some "richness threshold", such that every norm which has "enough" isometries is determined up to a scalar multiple by its isometry group.

Putting it differently, how symmetric must $\| \|_i$ must be in order for such a determination to hold? (In particular must $G$ be infinite?)

*This follows as a corollary from an argument given here which shows which inner products are preserved by a given linear automorphism.

Fix a norm $||\cdot||$ on a vector space $V$. In your previous question, you found that the isometry group $G$ of $||\cdot||$ also preserves an inner product $<\cdot,\cdot>$. Fix that inner product, then there is a dual norm $||\cdot||_*$ given by $$||v||_* = \sup_{||u|| = 1}<v,u>$$ This is really just identifying $V$ with its dual via the inner product. If $G$ preserves $||\cdot||$, then clearly it also preserves the dual norm as follows: $$||g\cdot v||_* = \sup_{||u|| = 1}<g\cdot v,u> = \sup_{||u|| = 1}<v,g^{-1}u> = \sup_{||g\cdot u|| = 1}<v,u> = ||v||_*$$
Hence $||\cdot||$ is only unique if it is equal to the dual norm. But then if this is the case, the norms are induced by the inner product because $$||v||_{Euc}^2 = <v,v> \leq ||v||\cdot||v||_* = ||v||^2$$ Fix $||v||=1$. If we choose $u$, such that $||u||=1$, and $<u,v> = 1$, we have $$1 = <u,v> \leq ||u||_{Euc}||v||_{Euc}\leq ||u||\cdot||v|| = 1\cdot 1$$ Hence equality holds, and we have $||v||=||v||_{Euc}$
• Your argument is very nice. However you only showed $ISO(\| \|) \subseteq ISO(\| \|_*)$, and in order to force these two norms to be scalar multiple of each other you need also to show the other direction of the inclusion. (i.e, to show they are equal, since the claim I stated assumes equality and not just containment between the two isometry groups). I guess this follows from the fact that the double dual is canonically isomorphic to the original space and the double dual norm isometric to the original norm, but this needs to be said explicitly. – Asaf Shachar Jun 22 '15 at 11:40
• The point of the edit was to clarify the point of up two scalar multiple. (I think there were two many one's and if some of them will be replaced by proportion constant $\lambda$ the solution will be clearer). – Asaf Shachar Jun 22 '15 at 11:54
A simple example of 2 norms where $$\text{ISO}(||\cdot ||_1) = \text{ISO}(||\cdot ||_2)$$ but $$||\cdot||_1 \not= ||\cdot||_2$$ would be the L3 and L4 norms. For a pair of rigid norms use the two norms where the unit "circles" are two (non-similar) parallelograms (not rectangles or rhombuses) centred at the origin.