Equivalence classes of norms on $R^n$ under symmetries

Let $G \leq {\bf GL}_n$ be a symmetry group on $\mathbb{R}^n$. For simplicity, we can consider the case $G = {\bf GL}_n$.

Define two norms $\|\cdot\|_1$ and $\| \cdot\|_2$ to be equivalent under $G$ if there exists $A \in G$ such that

$\ \forall \ x \ \| A x \|_1 = \| x\|_2$

It is trivial to show that this defines equivalence classes on the set of all norms on $\mathbb{R}^n$.

Now, for $\mathbb{R}^1$, there is only a single class of norms but for $\mathbb{R}^2$ there are infinitely many classes, specifically all $p$ norms live in different equivalence classes (except for the fact that the $1$-norm and the $\infty$-norm are equivalent). For general $\mathbb{R}^n$ the case is even "worse".

Is there some way to classify these equivalence classes?

• I'm not sure that the case $n=2$ is much easier as you can always define a norm $\|x\|=\| \|x\|_1,\ldots,\|x\|_n\|_{\alpha}$ where $\|\cdot\|_i$ are norms on $\Bbb R^2$ and $\|\cdot \|_{\alpha}$ is a norm on $\Bbb R^n$. – Surb Sep 20 '17 at 9:02
• You can also do $\|x\| = \sum_i c_i\|x\|_i$ for any norms, so there certainly are a few things that can be done, but perhaps we can classify that "a norm is given by combinations of $p$ norms under these operations", or something similar. Anothing thing we may classify is some set of "fundamental" norms, that cannot be derived from any other norms. – Jonas Adler Sep 20 '17 at 9:04
• Note that the norm you mention is the special case where $\|\cdot\|_{\alpha}$ is a weighted $1$ norm on $\Bbb R^n$. Anyway, don't get me wrong, although I believe what you are asking is a very difficult question, I think it is of high interest. – Surb Sep 20 '17 at 9:09
• The question seems to be essentially to classify (balanced, bounded and absorbing) convex shapes in a linear space, which stated that way seems a bit hopeless. On the other hand, there could be much interesting to say about the topology or some other structure (and remarkable points, etc.) of the space of all norms. – Gro-Tsen Sep 20 '17 at 9:15
• @Surb Maybe I misunderstood what you are saying, but in general the expression $\lVert \lVert\cdot\rVert_1, \ldots\rVert_{\alpha}$ is not a norm. – gsa Sep 20 '17 at 15:59

When $G={\bf GL}_n$, the set of classes of equivalent norms over ${\mathbb R}^n$ is a compact Hausdorff space when equipped with the Banach-Mazur metric $$d(N_1,N_2)=\log\delta(N_1,N_2)$$ where $$\delta(N_1,N_2)=\inf_{T\in{\bf GL}_n}\|T^{-1}\|_{N_2\rightarrow N_1}\cdot\|T\|_{N_1\rightarrow N_2}\,.$$ See this page.
Of course, the smaller the group $G$, the bigger the set of equivalence classes.
The following paper describes some invariants of the $GL(n)$ action.