On Tits Alternative and its Application

Wayne Peng

joint with J. Bell, K. Huang, T. Tucker, and S.-F. Tsai

Workshop on Dynamical System

Dichotomy
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In terms of growth rate

The existence of any relation between generators results in a reduction in the count of distinct words within $\Sigma^*_n$, leading to the inequality $\#\Sigma^*_n\leq (\#\Sigma)^n$.

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

  • For $n=2$, $\#\Sigma_n^*=n+1$.
  • For $n>2$, $\#\Sigma_n^*=H^m_n$.

We say the growth rate of the semigroup $\langle f,g\rangle$ is $f(n)$ if $\#\Sigma^*_n=O(f(n))$. A polynomially bounded growth corresponds to $f(n)$ being a polynomial, while an exponentially growth corresponds to $f(n)$ being an exponential function.

A Tits alternative for endomorphisms of the projective line

(Bell - Huang - :) - Tucker, upcoming on JEMS)

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).

  • We answered a conjecture posed by Cabrera and Makienko. If the measures of maximal entropy (julia set) for $f$ and $g$ are distinct, then there is a $j$ such that $\langle f^j, g^j\rangle$ is a free semigroup on two generators.
  • Let $A$ be an abelian variety. Then $\text{End}(A)$ (finite morphisms from $A$ to itself) satisfies tits alternative.

Let $f,g\in\mathcal{C}(x)$ be two endormophisms of $\mathbb{P}^1_{\mathbb{C}}$, each having degree greater than $1$. Then the followings are equivalent:

  1. $\text{Prep}(f)\cap\text{Prep}(g)$ is infinite;
  2. $\text{Prep}(f)=\text{Prep}(g)$;
  3. for any $w_1$, $w_2\in\langle f,g\rangle$, we have $\text{Prep}(w_1)=\text{Prep}(w_2)$;
  4. $\langle f,g\rangle$ has polynomial growth;
  5. $\langle f,g\rangle$ does not contain a nonabelian free semigroup;
  6. for any $l>0$, the semigroup $\langle f^l,g^l\rangle$ is not free on two generators.

Algorithm:$\langle f,g\rangle$ contains free semigroup

Input: polarized morphisms $f$ and $g$, and an integer $n$

Output: A Boolean indicating whether $f$ and $g$ have a composition relation, or "indeterminable" if the algorithm cannot determine by searching on a binary tree up to level $n$.

Idea of Algorithm

  1. For simplicity, we assume $\deg f=\deg g$.
  2. Composition has the property of right cancellation, i.e. $$ h_1\circ f = h_2\circ f\implies h_1=h_2. $$

    Therefore, to verify a nontrivial composition relation, we can simply assume that the rightmost functions are distinct.

  1. If there exists some $\alpha$ such that $$ B\left(\frac{h(A(\alpha))}{d^n},\dfrac{C}{(d-1)d^n}\right)\bigcap B\left(\frac{h(B(\alpha))}{d^n},\dfrac{C}{(d-1)d^n}\right)=\varnothing $$ where $C=\max\{C_f, C_g\}$, $A=A_n\circ\cdots\circ A_1$ and $B=B_n\circ\cdots\circ B_1$, then any composition that starts with $A$ won't be equal to one that starts with $B$.




Algorithm:$\langle f,g\rangle$ doesn't contains free semigroup

  • If one of the functions $f$ or $g$ in $\mathbb{C}(x)$ is not special, and $\langle f,g\rangle$ exhibits polynomial bounded growth, then it ultimately demonstrates linear growth.
  • Linear growth implies that for some positive integer n, $f^n=g^n$.
  • Thus, cases that require special attention are when both $f$ and $ g$ are special.
  • 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 $$ \langle \zeta_1x^{n_1},\zeta_2x^{n_2},\ldots,\zeta_mx^{n_m}\rangle. $$

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, none of which is generated by the other:

  1. $gf^2g=fg^2f$
  2. $fgfg=g^2f^2$
  3. $fg^4=g^4f$
  4. $f^2g^3=g^2fgf$
  5. $f^3g^2=gfgf^2$
  1. $f^4g=gf^4$
  2. $fgf^3g=gf^3gf$
  3. $fg^3fg=gfg^3f$
  4. $f^2g^2fg=g^3f^3$

A finitely generated semigroup $G=\Sigma^*/\mathcal{R}$ with the following properties:

  1. 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 $s\in\Sigma$ (If two composition are the same, then they use the same number of $f$ and $g$.)
  2. 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.)
  3. The semigroup $G$ has one-side cancellation property.

:) - Tsai (2023)

If $G$ 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.

Thank you

Takeaways:

  • We introduce the Banach-Tarski paradox, which inspired von Neumann to propose the von Neumann conjecture.
  • Tits confirmed the von Neumann conjecture for linear groups, a result now known as the Tits alternative.
  • We introduce the ping-pong lemma and provide a classical example.
  • We outline the proof and conclusion of our main result.
  • We demonstrate two subsequent research studies.