# Continued Fractions, Attempt 3, Part 4: Equivalence Transformation

## 1. Equivalence Transformation of Finite Continued Fractions

For any $k\in\N$, let $t_k\in\C^\times$ be an arbitrary constant. Define $\tau:\cC\to\cC$ by $\tau:((a_0,b_0),(a_1,b_1),\ldots,(a_{n-1},b_{n-1}))\mapsto((t_0a_0,t_1t_0b_0),(t_1a_1,t_2t_1b_1),\ldots,(t_{n-1}a_{n-1},t_nt_{n-1}b_{n-1})).$ We call this the tau transformation of finite continued fractions. Apparently, $\tau$ preserves regularity of finite continued fractions.

Lemma 3.1 We have $\left\lang\overline{\tau(\gamma)}\right\rang=\lang\tau(\overline\gamma)\rang.$

Proof. When $\gamma$ is regular, it is trivial. Assume $\gamma=((a_0,b_0),\ldots,(a_{n-1},b_{n-1}))$ is degenerate and $\overline\gamma=((a_0,b_0),\ldots,(a_{k_0},1))$. Then, \begin{aligned} \left\lang\overline{\tau(\gamma)}\right\rang&=\lang(t_0a_0,t_1t_0b_0),\ldots,(t_{k_0}a_{k_0},1)\rang \\ &=\lang(t_0a_0,t_1t_0b_0),\ldots,(t_{k_0}a_{k_0},t_{k_0+1}t_{k_0})\rang \\ &=\lang\tau((a_0,b_0),\ldots,(a_{k_0},1))\rang \\ &=\lang\tau(\overline\gamma)\rang. \end{aligned}

Lemma 3.2 For any $k\in\N$, define $T_k\in\GL(2,\C)$ by $T_k=\M{t_k}{0}{0}{1}.$ Then, $M(t_{k-1}a_{k-1},t_kt_{k-1}b_{k-1})T_k=t_kT_{k-1}M(a_{k-1},b_{k-1}).$ Proof. $\M{t_{k-1}a_{k-1}}{t_kt_{k-1}b_{k-1}}{1}{0}\M{t_k}{0}{0}{1}=\M{t_kt_{k-1}a_{k-1}}{t_kt_{k-1}b_{k-1}}{t_k}{0}=t_k\M{t_{k-1}}{0}{0}{1}\M{a_{k-1}}{b_{k-1}}{1}{0}.$

Theorem 3.3 For any $\gamma\in\cC$, $\lang\tau(\gamma)\rang=t_0\lang\gamma\rang.$

Proof. First, assume $\gamma$ is regular. Noting that $\left[T_n\V{1}{0}\right]=\B{1}{0}$, we have \begin{aligned} \lang\tau(\gamma)\rang&=\phy\left(\left[M(t_0a_0,t_1t_0b_0)\cdots M(t_{n-1}a_{n-1},t_nt_{n-1}b_{n-1})T_n\V{1}{0}\right]\right) \\ &=\phy\left(\left[T_0M(a_0,b_0)\cdots M(a_{n-1},b_{n-1})\V{1}{0}\right]\right) \\ &=t_0\lang\gamma\rang. \end{aligned}

Next, assume $\gamma$ is degenerate. Then, $\lang\tau(\gamma)\rang=\left\lang\overline{\tau(\gamma)}\right\rang=\lang\tau(\overline\gamma)\rang=t_0\lang\overline\gamma\rang=t_0\lang\gamma\rang.$

$\square$

Because of this proposition, when $t_0=1$, $\tau$ is called the equivalence transformation of finite continued fractions.

## 2. Equivalence Transformation of Infinite Continued Fractions

Define $\tau:\cC_\infty\to\cC_\infty$ by $\tau:((a_0,b_0),(a_1,b_1),(a_2,b_2),\ldots)\mapsto((t_0a_0,t_1t_0b_0),(t_1a_1,t_2t_1b_1),(t_2a_2,t_3t_2b_2),\ldots).$ We call this the tau transformation of infinite continued fractions. Apparently, $\tau$ preserves regularity of infinite continued fractions.

Theorem 3.4 For any $\gamma\in\cC_\infty$, $\tau(\gamma)$ has a value in $\whC$ if and only if $\gamma$ has a value in $\whC$, and when they have, the values have the relation $\lang\tau(\gamma)\rang=t_0\lang\gamma\rang.$

Proof. Let $c_n$ be the $n$-th convergent of $\gamma$, and $c'_n$ the $n$-th convergent of $\tau(\gamma)$. By Theorem 3.3, we have $c'_n=t_0c_n$ and therefore $\lang\tau(\gamma)\rang=\lim_{n\to\infty}c'_n=t_0\lim_{n\to\infty}c_n=t_0\lang\gamma\rang.$

$\square$

Because of this proposition, when $t_0=1$, $\tau$ is called the equivalence transformation of infinite continued fractions.