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The same question was asked in Math StackExchange about 3 months ago. Since nobody has answered to it, I would like to post it here.


Weil's Basic Number Theory(denoted by BNT).

Bourbaki's Commutative Algebra(denoted by BCA).

Let $K$ be a topological ring with an identity. Suppose every non-zero element of $K$ is invertible. Let $K^* = K - \{0\}$ be the multipilcative group of $K$ If the map $x \rightarrow x^{-1}$ is continuous on $K^*$, we say $K$ is a topological division ring. Suppose the topological space $K$ is non-discrete, Hausdorff and locally compact. Then we say, by abuse of terminology, $K$ is a locally compact division ring.

Let $K$ be a locally compact division ring. Then the aditive group $K$ is a locally compact group. Hence there exists a Haar measure $\mu$ on $K$. Let $a$ be an element of $K^*$. Then the map $x \rightarrow ax$ is an automorphism of the locally compact group $K$. Hence the map $X \rightarrow \mu(aX)$ defines an invariant measure on $K$ where $X$ is any measurable subset of $K$. Therefore there exists a constant $c \gt 0$ such that $\mu(aX) = c\mu(X)$ for every measurable subset $X$ such that $0 \lt \mu(X) \lt \infty$. We denote $c$ by $mod(a)$. We define $mod(0) = 0$.

$mod(a)$ can also be defined by the map $x \rightarrow xa$(see BNT or BCA).

Clearly $mod(ab) = mod(a)mod(b)$ for all $a, b \in K$. The function $mod$ is continuous(see BNT or BCA). The subset $\{x \in K|\ mod(x) \le d\}$ is compact for every real number $d \gt 0$(see BNT).

Locally compact division rings are classified as follows(see BNT or BCA).

  1. The field of real numbers $\mathbb{R}$.

  2. The field of complex numbers $\mathbb{C}$.

  3. The field of Hamilton's quaternions $\mathbb{H}$.

  4. Finite division algebras over the field of $p$-adic numbers.

  5. Finite division algebras over the field of formal Laurent series over a finite field.

Here is my question. Is the following proposition true?

Proposition Let $K$ be a locally compact division ring. Let $\phi$ be a real valued function defined on $K$. Suppose $\phi$ satisies the following conditons.

  1. $\phi$ is continuous.

  2. $\phi(x) \gt 0$ for all $x \neq 0$ and $\phi(0) = 0$.

  3. $\phi(xy) = \phi(x)\phi(y)$ for all $x, y$.

Then there exists a real number $c \gt 0$ such that $\phi(x) = mod(x)^c$ for all $x$.

Remark Let $K$ be a locally compact division ring. Let $x \in K^*$.

  1. If $K = \mathbb{R}$, then $mod(x) = |x|$.

  2. If $K = \mathbb{C}$, then $mod(x) = |x|^2$.

  3. If $K = \mathbb{H}$, then $mod(x) = |x|^4$.

  4. If $K = \mathbb{Q}_p$, then $mod(x) = |x|_p$ where $|x|_p$ is the canonical absolute value, i.e. $|p|_p = 1/p$.

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I have given an answer over at MSE. The basic idea is to take advantage of the compactness of the kernel of the modulus function on $K^\ast$, and the fact that compact subgroups of $\mathbb{R}_{>0}$ are trivial. I would prefer to keep the discussion there for now, in view of recent meta discussions. – Todd Trimble Oct 20 '13 at 4:12

This is true (assuming as you do implicitly that all your division rings are nondiscrete). The reason is that for any nondiscrete locally compact division ring $\mathbf{K}$, the multiplicative group $\mathbf{K}^*$ of $\mathbf{K}$ has a unique maximal compact subgroup (namely the elements of modulus 1) and the quotient is either isomorphic to $\mathbf{R}$ (Archimedean case) or to $\mathbf{Z}$ (non-Archimedean case). Therefore $\mathrm{Hom}(\mathbf{K}^*,\mathbf{R})$ is 1-dimensional, so all continuous homomorphisms from $\mathbf{K}^*$ to $\mathbf{R}$ are proportional. Multiplicatively, this means that every continuous homomorphism $f:\mathbf{K}^*\to\mathbf{R}_{>0}$ has the form $f(x)=\mathrm{mod}(x)^c$ for some $c\in\mathbf{R}$. If you require in addition that $\lim_{x\to 0}f(x)=0$, then necessarily $c>0$.

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