Why is it so hard to find any composition relation between two rational functions?

In this talk, our $f, g, ...$ are degree $d>1$ rational morphisms on $\mathbb{P}^n(K)$ where $K$ is a globel field.

A composition relation of $f$ and $g$ is an equation like: $f\circ f\circ g=g\circ f\circ f\text{ or }f\circ g=g\circ f$ (Omit $\circ$ if there is no dilemma)

Question! Given $f$ and $g$ , how can we determin if a relation between $f$ and $g$ exists?

A Lattès map is a rational map $f = \theta L\theta^{−1}$ from the complex sphere to itself such that $\theta$ is a holomorphic map from a complex torus to the complex sphere and $L$ is an affine map $z\mapsto az + b$ from the complex torus to itself.

If $\zeta$ is a primitive fifth root of unity, then the semigroup generated by $f=x^2$ and $g=\zeta x^3$ has the following nine relations $\mathcal{R}'$ , none of which is generated by the others

$gf^2g=fg^2f$
$fgfg=g^2f^2$
$fg^4=g^4f$
$f^2g^3=g^2fgf$
$f^3g^2=gfgf^2$

$f^4g=gf^4$
$fgf^3g=gf^3gf$
$fg^3fg=gfg^3f$
$f^2g^2fg=g^3f^3$

A Tits alternative for endomorphisms of the projective line

(Bell - Huang - :) - Tucker, JEMS, 2023)

Let $\mathcal{S}\subset\mathbb{C}(x)$ be a finitely generated semigroup of endomorphisms of $\mathbb{P}^1$ . Then either $\mathcal{S}$ has polynomially bounded growth or contains a nonabelian free semigroup (exponential growth).

If $\mathcal{S}=\langle f,g\rangle$ , then the free semigroup mentioned here is of the form $\langle f^j, g^j\rangle$ for some positive integer $j$ .

Question! $j=1$ ?

characters, words, and language

Consider a set of characters $\Sigma=\{f_1,f_2,\ldots,f_n\}$ . We define:

A word is a formal string of characters.

A language

$\Sigma^*$ is the collection of (finite and infinite) words.

$\Sigma^*_{<\infty}$ is the collection of finite words.

Let

$\Sigma^*_n$ denote the set of words of the length

$n$ , and

$\Sigma^*_{\leq n}$ denote the set of words of the length at most

$n$

let

$\Sigma^*_\infty$ be the set of infinite words.

The difference between $\langle\Sigma\rangle$ and $\Sigma^*_{<\infty}$ is that the later one is the collection of formal strings.

Growth rate

Let $\mathcal{R}$ be the set of composition relations. The existence of any relation between generators results in a reduction in the count of distinct elements within $\Sigma^*_n$ , leading to the inequality $\#\Sigma^*_n/\mathcal{R}\leq (\#\Sigma)^n$ .

$\mathcal{R}'$ is the smallest set of composition relations that generate $\mathcal{R}$ . We call it a set of basic relations

Let $\Sigma=\{f_1,\ f_2,\ \ldots,\ f_m\}$ , and assume $f_if_j=f_jf_i$ for all $i$ and $j$ .

For $n=2$ , $\#\Sigma_n^*/\mathcal{R}=n+1$ .
For $n>2$ , $\#\Sigma_n^*/\mathcal{R}=H^m_n$ .

The Ping-Pong lemma

sketch of the proof: Note that $f^{*}g^{*}\cdots f^{*}:A\to B$ cannot be an identity.

case ( $w=g^{*}f^{*}\cdots g^{*}$ ): we consider $fwf^{-1}$ .
case ( $w=f^ag^{*}\cdots g^{*}$ ): for $n\neq a$ and $0$ , we consider $f^{n}wf^{-n}$ .
case ( $w=g^*f^*\cdots f^a$ ): for $n\neq a$ and $0$ , we consider $f^{n}wf^{-n}$ .

To comprehend the essence of our proof for the main result, perform the following steps:

Use a naive height function $h:\mathbb{Q}^\times\to\mathbb{R}_{\geq 0}$ .
Let $f$ and $g$ be rational functions. Define both the dynamical canonical height of $f$ and the general multi-system dynamical canonical height $h_w$ , where $w\in \Sigma^*$ .
Given $\alpha\in\mathbb{Q}$ , define $X$ as the set of all $h_w(\alpha)$ where $w\in \Sigma^*$ .
Explain the action of $f$ and $g$ on $X$ .
Identify $A$ and $B$ , which are disjoint subsets of $X$ , and apply the ping-pong lemma.

Write a nonzero rational number $\frac{a}{b}$ in lowest terms. The naive height is $\begin{align*} h(\alpha) &= \log(\max\{|a|,|b|\})\\ &= \log(\max\{1,|\frac{a}{b}|\})+\sum_{p\text{ prime}}\log(\max\{1,|\frac{a}{b}|_p\}), \end{align*}$ e.g. $h(\frac{1}{2})=\log(2)$ and $h(\frac{1001}{2000})=\log(2000)$ .

A well-known inequality

Let $f$ be a rational function. Let $d$ be the degree of $f$ . Then, we have a constant $C_f$ , dependent only on $f$ , such that $|h(f(\alpha))-dh(\alpha)| < C_f$ We further define $C_f=\sup_{P\in\mathbb{P}^N(\bar{\mathbb{Q}})}\left|\dfrac{h(f(P))}{d}-h(P)\right|.$

Dynamical Canonical height

$h_f(\alpha)=\lim_{n\to\infty}\frac{h(f^n(\alpha))}{d^n}$

Key argument: For $m>n$ , $\left|\frac{h(f^m(\alpha))}{d^m} - \frac{h(f^n(\alpha))}{d^n}\right|$
$\begin{align*} &= \left|\frac{h(f^m(\alpha))}{d^m} - \frac{h(f^{m-1}(\alpha))}{d^{m-1}}+\frac{h(f^{m-1}(\alpha))}{d^{m-1}} - \frac{h(f^{m-2}(\alpha))}{d^{m-2}} + \cdots + \frac{h(f^{n+1}(\alpha))}{d^{n+1}} - \frac{h(f^n(\alpha))}{d^n}\right|\\\\ &\leq \left|\frac{h(f^m(\alpha))}{d^m} - \frac{h(f^{m-1}(\alpha))}{d^{m-1}} \right| + \left|\frac{h(f^{m-1}(\alpha))}{d^{m-1}} - \frac{h(f^{m-2}(\alpha))}{d^{m-2}}\right| + \cdots + \left|\frac{h(f^{n+1}(\alpha))}{d^{n+1}} - \frac{h(f^n(\alpha))}{d^n}\right|\\\\ &< \frac{C}{d^{m}} + \frac{C}{d^{m-1}} + \cdots + \frac{C}{d^{n+1}} = \frac{\frac{C}{d^{m+1}}-\frac{C}{d^{n+1}}}{\frac{1}{d}-1} \end{align*}$

Multi-system dynamical canonical height

Consider a sequence $\{f_1(\alpha),\ f_2f_1(\alpha),\ f_3f_2f_1(\alpha),\ \cdots \}$ where $f_i$ is equal to $f$ or $g$ . Let $w_n= f_n\ldots f_2f_1$ , and let $w$ be the formal composition $\ldots f_3f_2f_1$ ( $w\in \Sigma^*_\infty$ ). We define $h_w(\alpha) = \lim_{n\to\infty}\frac{h(w_n(\alpha))}{\deg(w_n)}.$

Proof of $j=1$

Let $d_f$ be the degree of $f$ . We have $\dfrac{C_f}{d_f}\left(1-\dfrac{1}{d_f-1}\right)<|h_f(\alpha)-h(\alpha)|<\dfrac{C_f}{d_f-1}$ and $\left|h_f(\alpha)-\dfrac{h(f^i(\alpha))}{d_f^i}\right|<\dfrac{C_f}{d_f^i(d_f-1)}$ for $P\in\mathbb{P}^N(\bar{Q})$ .

To show that $\langle f^j, g^j\rangle$ is free, we need to show that, for some $P\in\mathbb{P}^N(\bar{\mathbb{Q}})$ , $|\hat{h}_f(P)-\hat{h}_g(P)|\geq \left|\hat{h}_f(P)-\dfrac{h(f^j(P))}{d_f^j}\right|+\left|\hat{h}_g(P)-\dfrac{h(g^j(P))}{d_g^j}\right|.$ By the lemma, we have $\begin{align*} |\hat{h}_f(P)-\hat{h}_g(P)|&\geq |\hat{h}_f(P)-h(P)|-|h(P)-\hat{h}_g(P)|\\ &\geq \dfrac{C_f}{d_f}\left(1-\dfrac{1}{d_f-1}\right)-\frac{C_g}{d_g-1}. \end{align*}$ It follows that $j=1$ if the above difference is greater than $0$ , and we need ...

$\begin{align} C_f>\frac{d_f(d_f-1)(d_g+1)}{(d_f-3)d_g(d_g-1)}C_g. \end{align}$ It implies $\begin{align*} \dfrac{C_f}{d_f}&\left(1-\dfrac{1}{d_f-1}\right)-\frac{C_g}{d_g-1} = \frac{(d_f-2)d_g(d_g-1)C_f-d_f(d_f-1)d_g C_g}{d_f(d_f-1)d_g(d_g-1)}\\ =& \frac{d_g(d_g-1)C_f+d_f(d_f-1)C_g+d_g(d_g-1)(d_f-3)C_f-d_f(d_f-1)(d_g+1)C_g}{d_f(d_f-1)d_g(d_g-1)}\\ >& \frac{C_f}{d_f(d_f-1)}+\frac{C_g}{d_g(d_g-1)}. \end{align*}$ Therefore, we conclude that $j=1$ .

Caution! There are maybe infinitely many $f$ and $g$ satisfying the inequality on the top.

For any $B>0$ , integer $d_f>1$ , and a number field $K$ , we have $C_f>B$ for all but finitely many degree $d_f$ rational maps $f:\mathbb{P}^1(K)\to\mathbb{P}^1(K)$ .

draft of proof:

We know there are only finitely many rational function satisfies

$h(f)< B$ (Northcott's Theorem).

When

$f$ is a polynomial, it is a consequence of Vandermond matrices.

When

$f$ is a rational function on

$\mathbb{P}^1(K)$ , we need to cleverly applying a change of coordinate such that we can use the result from

$f$ being a polynomial.

An Effective Tits alternative for endomorphisms of the projective line

(:) - Tsai - Tucker, Under construction 🔨)

Given a morphism $g:\mathbb{P}^1\to\mathbb{P}^1$ of degree $d_g>1$ , the semigroup $\langle f,g\rangle$ is free for all but finitely many $f:\mathbb{P}^1\to\mathbb{P}^1$ of degree $d_f>3$ . In particular, the semigroup $\langle f^2, g\rangle$ is free for all but finitely many $f$ of degree $d_f$ .

Monomial $x^n$ : The Julia set is the unit circle centered at the origin (0,0).
Chebyshev polynomial: The Julia set is the closed interval $[-2,2]$
Lattès map: The Julia set is the entire space.

The only remaining case is when we have a semigroup generated by monomials like $\langle \zeta_1x^{n_1},\zeta_2x^{n_2},\ldots,\zeta_mx^{n_m}\rangle.$

A finitely generated semigroup $S=\Sigma^*_{<\infty}/\mathcal{R}$ is a semigroup with the following properties:

Let equivalent classes $[w_1]$ and $[w_2]$ belong to $G$ . If $[w_1]=[w_2]$ , then $|w_1|_s = |w_2|_s$ (counting the number of characters to be $s$ ) for all $s\in S$ (If two composition are the same, then they use the same number of $f$ and $g$ .)
There exists a positive integer $N$ such that $\#\mathcal{U}_I/\mathcal{R}\leq N$ for all indices $I$ (Number of possible composition result is bounded.)
The semigroup $G$ has one-side cancellation property.

The dimension theorem for finite represented semigroup

(:) - Tsai - Tucker, Under construction 🔨)

If $S$ is a finitely generated semigroup that meets the three specified conditions, then the cardinality of $\mathcal{R'}$ is finite and does not depend on our choice of relations. Furthermore, the process of finding relations can be finished in a finite number of steps.

Attempt at defining nilpotent-by-finite semigroup

We say a group $G$ is nilpotent if $G$ has a series of normal subgroup $G=G_0\triangleright G_1\triangleright G_2\triangleright\cdots \triangleright G_n=\{1\}$ where $G_i=[G,G_{i-1}]$ .

We say a cancellative seimgroup $S$ is nilpotent if the group of fraction of $M$ is nilpotent.

A semigroup is cancellative if it satisfies $\text{both } ac=bc\text{ and }ca=cb\implies a=b.$

For $x$ , $y$ , $u_1$ , $u_2$ , $\ldots\in S$ , we define $x_0=x$ , $y_0=y$ , $x_i=y_{i-1}u_ix_{i-1}\text{ and }y_i=x_{i-1}u_iy_{i-1}.$ We say a semigroup $M$ is nilpotent if for any $x$ , $y$ , $u_1$ , $u_2$ , $\ldots\in S$ , there exists an integer $n$ such that $x_n=y_n$ .

We say a sequence $\{u_i\}\subset M$ is $(x,y,n)-stable$ if $x_n=y_n$ . If $\{u_i\}\subseteq S_{\leq n}$ , we say $\{u_i\}$ is $(x,y,m,n)$ -stable.

The problem for our situation is that composition is not cancellative.