7.2 Conjugacy and Examples

Conjugacy

Consider two spaces $X$ and $Y$ and two maps $f : X \to X$ and $g : Y \to Y$ .

We say that $h : X \to Y$ forms a conjugacy from $f$ to $g$ if:

$h$ is a homeomorphism (i.e. bijection s.t. $h$ and $h^{- 1}$ are both continuous)
$h \circ (f (x)) = g \circ (h (x))$

\begin{array}{ccc} X & \overset{f}{\to} & X \\ ↓ h & ↓ h \\ Y & \overset{g}{\to} & Y \end{array}

If $h$ is not bijective then we have a semi-conjugacy

Let $h$ be a (semi-) conjugacy from $f : X \to X$ to $g : Y \to Y$ .

Let $x$ be a k-periodic point in $X$ under $f$ , then $y = h (x)$ is a k-periodic point in $Y$ .
If $x$ is a fixed point in $X$ under $f$ , then $y = h (x)$ is a fixed point in $Y$ under $g$ .
Let $x$ be a k-periodic point in $X$ under $f$ . The linear stability of $x$ is the same as the linear stability of $y = h (x)$ .
If $x$ is repelling/attracting, so is $y = h (x)$ .
If $x^{*}$ is such that $O_{f} (x^{*})$ is dense in $X$ , then $y^{*} = h (x^{*})$ is such that $O_{g} (y^{*})$ is dense in $Y$ .

D (x) = {\begin{cases} 2 x, & x \in [0, \frac{1}{2}] \\ 2 x - 1, & x \in [\frac{1}{2}, 1] \end{cases}

The binary expansion map $h : [0, 1] \to Σ_{+}$
any $x \in [0, 1]$ has a binary representation as

x = \sum_{i = 1}^{\infty} \frac{a_{i}}{2^{i}}

h (x) = a_{1} a_{2} a_{3} \dots

The binary map forms a semi-conjugacy between $D$ and $σ$ .

h (D (x)) = σ (h (x))

\begin{array}{ccc} [0, 1] & \overset{D}{\to} & [0, 1] \\ ↓ h & ↓ h \\ Σ_{+} & \overset{σ}{\to} & Σ_{+} \end{array}

\begin{aligned} h (D (x)) & = h (2 x \mod 1) \\ = h (2 (\sum_{k = 1}^{\infty} \frac{a_{k}}{2^{k}}) \mod 1) \\ = h (\sum_{k = 1}^{\infty} \frac{a_{k}}{2^{k - 1}} \mod 1) \\ = h (\frac{a_{1}}{2^{0}} + \sum_{k = 2}^{\infty} \frac{a_{k}}{2^{k - 1}} \mod 1) \\ = a_{2} a_{3} a_{4} a_{5} \dots \\ = σ (h (x)) \end{aligned}

Based on our theorems we are confident that $x$ is periodic in under $D$ $i f f$ $h (x)$ is periodic under $σ .$

First note that:

h (D^{(n)} (x)) = σ^{(n)} (h (x))

\begin{aligned} h (D \underset{y}{\underset{⏟}{(D (x))}}) & = h (D (y)) \\ (conjugacy) & = σ (h (y)) \\ = σ (h (D (y))) \\ = σ (σ (h (x))) \end{aligned}

Assume that $x$ is periodic in under $D$ :

\begin{aligned} D^{k} (x) = x & ⟹ h (D^{k} (x)) = h (x) \\ ⟹ σ^{k} (h (x)) = h (x) \end{aligned}

So $h (x)$ is periodic under $σ$

Assume that $h (x)$ is periodic under $σ$

\begin{aligned} σ^{k} (h (x)) = h (x) & ⟹ h (D^{k} (x)) = h (x) \\ ⟹ h^{- 1} (h (D^{k} (x))) = h^{- 1} (h (x)) \\ ⟹ D^{k} (x) = x \end{aligned}

we ignore dyadic numbers so that $h$ is bijective

We know that constructing $s^{*} = 0 1 00 11 01 10 000010 \dots$ creates a dense orbit:

O_{σ} (s^{*}) = {s_{k} = σ^{(k)} (s^{*}), k \in N}

in $Σ_{+}$

Take $x^{*}$ such that $h (x^{*}) = s^{*}$

\begin{aligned} h (D^{k} (x^{*})) & = σ^{k} (h (x^{*})) = σ^{k} (s^{*}) \\ ⟹ h (D^{k} (x^{*})) = σ^{k} (s^{*}) \\ applying h^{- 1} \\ ⟹ D^{k} (x^{*}) = (\frac{σ^{k} (s^{*})_{1}}{2} + \frac{σ^{k} (s^{*})_{2}}{2^{2}} + \dots) \end{aligned}

which is dense in $[0, 1]$

(D^{k})^{'} (x) = \underset{k times}{\underset{⏟}{2 \cdot 2 \cdot 2 \cdot \dots \cdot 2}} = 2^{k}

| f (x) - f (y) | \approx | f^{'} (x) | \cdot | x - y |

Let $x \neq y$ :

| D^{k} (x) - D^{k} (y) | = 2^{k} | x - y |