Equivalence of Definitions of Equivalent Division Ring Norms/Open Unit Ball Equivalent implies Norm is Power of Other Norm/Lemma 2/Lemma 2.3
Theorem
Let $R$ be a division ring.
Let $\norm {\, \cdot \,}_1: R \to \R_{\ge 0}$ and $\norm {\, \cdot \,}_2: R \to \R_{\ge 0}$ be norms on $R$.
Let $\norm {\, \cdot \,}_1$ and $\norm {\, \cdot \,}_2$ satisfy:
- $\forall x \in R:\norm x_1 < 1 \iff \norm x_2 < 1$
Let $x, y \in R \setminus 0_R$ such that $\norm x_1, \norm y_1 \ne 1$.
Let $\alpha = \dfrac {\log \norm x_1} {\log \norm x_2}$ and $\beta = \dfrac {\log \norm y_1} {\log \norm y_2}$.
Then:
- $\alpha = \beta$
Proof
Because $x, y \in R \setminus 0_R$:
- $\norm x_1 , \norm y_1, \norm x_2 , \norm y_2 > 0$.
Because $\norm{x}_1 , \norm {y}_1 \ne 1$, by Lemma 2:
- $\norm x_2 , \norm y_2 \ne 1$.
Hence:
- $\log \norm x_1 , \log \norm y_1, \log \norm x_2, \log \norm y_2 \ne 0$
and $\alpha, \beta$ are well-defined.
Let $r = \dfrac n m \in \Q$ be any rational number where $n, m \in \Z$ are integers and $m \ne 0$.
Then:
| \(\ds \norm y_1^n < \norm x_1^m\) | \(\leadstoandfrom\) | \(\ds \dfrac {\norm y_1^n} {\norm x_1^m} < 1\) | ||||||||||||
| \(\ds \) | \(\leadstoandfrom\) | \(\ds \norm y_1^n \norm {x^{-1} }_1^m\) | Norm of Inverse in Division Ring | |||||||||||
| \(\ds \) | \(\leadstoandfrom\) | \(\ds \norm {y^n}_1 \norm {\paren {x^{-1} }^m}_1 < 1\) | Norm Axiom $(\text N 2)$: Multiplicativity | |||||||||||
| \(\ds \) | \(\leadstoandfrom\) | \(\ds \norm {y^n \paren {x^{-1} }^m}_1 < 1\) | Norm Axiom $(\text N 2)$: Multiplicativity | |||||||||||
| \(\ds \) | \(\leadstoandfrom\) | \(\ds \norm {y^n \paren {x^{-1} }^m}_2 < 1\) | By assumption | |||||||||||
| \(\ds \) | \(\leadstoandfrom\) | \(\ds \norm {y^n}_2 \norm {\paren {x^{-1} }^m}_2 < 1\) | Norm Axiom $(\text N 2)$: Multiplicativity | |||||||||||
| \(\ds \) | \(\leadstoandfrom\) | \(\ds \norm y_2^n \norm {x^{-1} }_2^m\) | Norm Axiom $(\text N 2)$: Multiplicativity | |||||||||||
| \(\ds \) | \(\leadstoandfrom\) | \(\ds \dfrac {\norm y_2^n} {\norm x_2^m} < 1\) | Norm of Inverse in Division Ring | |||||||||||
| \(\ds \) | \(\leadstoandfrom\) | \(\ds \norm y_2^n < \norm x_2^m\) |
By Logarithm is Strictly Increasing:
- $\log \norm y_1^n < \log \norm x_1^m \iff \log \norm y_2^n < \log \norm x_2^m$
- $n \log \norm y_1 < m \log \norm x_1 \iff n \log \norm y_2 < m \log \norm x_2$
Because $m, \log \norm x_1, \log \norm x_2 \ne 0$:
- $r = \dfrac n m < \dfrac {\log \norm x_1} {\log \norm y_1} \iff r = \dfrac n m < \dfrac {\log \norm x_2} {\log \norm y_2}$
By Between two Real Numbers exists Rational Number:
- $\dfrac {\log \norm x_1} {\log \norm y_1} = \dfrac {\log \norm x_2} {\log \norm y_2}$
Because $\log \norm x_2, \log \norm y_2 \ne 0$:
- $\dfrac {\log \norm x_1} {\log \norm x_2} = \dfrac {\log \norm y_1} {\log \norm y_2}$
That is:
- $\alpha = \beta$
$\blacksquare$