In what sense is the L2 norm the canonical norm on $\mathbb{R}^n$ but the max-norm is the canonical norm on $\mathbb{Q}_p^n$?

Question

I understand that the L2 norm is natural to consider on $\mathbb{R}^n$ both because of its geometric intuition, and because it is induced by the dot product on $\mathbb{R}^n$. I also understand that a max-norm is convenient to use on $\mathbb{Q}_p^n$ because it extends the nonarchimedean triangle inequality to the vector space. I further understand that every norm on a vector space over a normed field is topologically equivalent.

However, geometrically, different norms can have very different structures. Is there a natural criterion of "niceness" which, when applied to real vector fields, singles out an L2-norm (with respect to some basis), and when applied to $p$-adic vector spaces, singles out the max-norm (with respect to some basis)?

If it helps, this question was motivated by some comments in Shai Haran's "Mysteries of the Real Prime", where he noted that (following Tate) the factors of the completed Riemann zeta function can be viewed as integrals over the places of $\mathbb{Q}$, and in fact the integrand may be viewed as a limit of a function induced by a norm on $\mathbb{Q}_p^2$ or $\mathbb{R}^2$. But norm Haran used on $\mathbb{Q}_p^2$ was the max-norm, and the norm he used on $\mathbb{R}^2$ was the L2 norm. I am searching for a principled justification for this choice beyond "Well, it gave us the result we wanted."

Answer 1

• Euclidean norm on $\Bbb{R}^n$ : invariant under $O_n(\Bbb{R})$ a maximal compact subgroup of $GL_n(\Bbb{R})$

• $\sup $ norm on $\Bbb{Q}_p^n$ : invariant under $GL_n(\Bbb{Z}_p)$ a maximal compact subgroup of $GL_n(\Bbb{Q}_p)$

A few claims : all maximal compact subgroups are conjugates. If $\|.\|$ is a norm and $G$ (with Haar measure $\mu$) is a maximal compact subgroup of $GL_n$ then $\|x\|_G=\int_G \|g^{-1}(x)\|d\mu(g)$ is $G$-invariant and $\|.\|_G$ is unique up to a scalar.

Answer 2

$\mathbb{R}$ and $\mathbb{Q}_p$ already have completely different topologies, so at this point, the L2 norm and sup norm are incomparable between them; it's apples to oranges. All finite dimensional vector spaces over complete valued fields are equivalent, so we can freely choose any norm we like. But, why should we choose the L2 norm over the sup norm? The sup norm is not only easy to use, but it also obeys the ultrametric inequality while the L2 norm does not.