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

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

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?

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?

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

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?

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

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. For general $\mathbb{R}^n$ the case is even "worse".

Is there some way to classify these equivalence classes?

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

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?

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

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. For general $\mathbb{R}^n$ the case is even "worse".

Is there some way to classify these equivalence classes?

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

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. For general $\mathbb{R}^n$ the case is even "worse".

Is there some way classify these equivalence classes?

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

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. For general $\mathbb{R}^n$ the case is even "worse".

Is there some way to classify these equivalence classes?

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