No Uncountable Polish Group Can be a Right-Angled Artin Group

TL;DR

This paper demonstrates that uncountable Polish groups cannot have a generator system with a length function satisfying specific properties, implying such groups cannot be automorphism groups of uncountable structures, extending previous results.

Contribution

It proves that uncountable Polish groups cannot be right-angled Artin groups, generalizing earlier results for free and free Abelian groups.

Findings

Uncountable Polish groups cannot admit certain length functions.

Automorphism groups of countable structures are not uncountable right-angled Artin groups.

Extends previous results from free and free Abelian groups.

Abstract

We prove that no uncountable Polish group can admit a system of generators whose associated length function satisfies the following conditions: (i) if $0 < k < \omega$, then $lg(x) \leq lg(x^k)$; (ii) if $lg(y) < k < \omega$ and $x^k = y$, then $x = e$. In particular, the automorphism group of a countable structure cannot be an uncountable right-angled Artin group. This generalizes results from [3] and [5], where this is proved for free and free Abelian uncountable groups.

Equations

$${\Gamma }_{\eta }=\{x_{n+1}^{\eta (n)}=x_n{g}_n:n<\omega \}.$$

$${\bar{b}}_{\eta }\in G^{\omega }\text{ and }\underset{n<\omega }{\bigwedge }{b}_{\eta ,n+1}^{\eta (n)}=b_{\eta ,n}{g}_n.$$

$$\pi (g_n)=g_n^{\prime },\pi (b_{\eta ,n})=c_{\eta ,n}\text{ and }{m}_{\ast }(\eta )=lg(c_{\eta ,0}).$$

$$\begin{array}[]{rcl}lg(c_{\eta,n})&\leqslant&lg(c_{\eta,n}^{\eta(m)})\\ &=&lg(c_{\eta,m}g^{\prime}_{m})\\ &\leqslant&lg(c_{\eta,m})+lg(g^{\prime}_{m})\\ &<&f_{2}(m)+f_{1}(m)\\ &=&f_{2}(n).\end{array}$$

$$\eta (n)>f_2(n+1)=f_2(n)+f_1(n)>lg(c_{\eta ,n})+lg(g_n^{\prime })⩾lg(c_{\eta ,n}{g}_n^{\prime }),$$

$$\eta (n)>f_2(n+2)⩾f_1(n+1)>lg(g_{n+1}^{\prime }).$$

$$c_{\eta ,n+2}^{\eta (n+1)}=c_{\eta ,n+1}{g}_{n+1}^{\prime }=g_{n+1}^{\prime }.$$

Full text

1. (i)

if $0<k<\omega$, then $lg(x)\leqslant lg(x^{k})$; 2. (ii)

if $lg(y)<k<\omega$ and $x^{k}=y$, then $x=e$.

In particular, the automorphism group of a countable structure cannot be an uncountable right-angled Artin group. This generalizes results from [3] and [5], where this is proved for free and free abelian uncountable groups.

Partially supported by European Research Council grant 338821. No. 1112 on Shelah’s publication list.

In a meeting in Durham in 1997, Evans asked if an uncountable free group can be realized as the group of automorphisms of a countable structure. This was settled in the negative by Shelah [3]. Independently, in the context of descriptive set theory, Becher and Kechris [1] asked if an uncountable Polish group can be free. This was also answered negatively by Shelah [4], generalizing the techniques of [3]. Inspired by the question of Becher and Kechris, Solecki [5] proved that no uncountable Polish group can be free abelian. In this paper we give a general framework for these results, proving that no uncountable Polish group can be a right-angled Artin group (see below for a definition). We actually prove more:

1. (i)

if $0<k<\omega$, then $lg(x)\leqslant lg(x^{k})$; 2. (ii)

if $lg(y)<k<\omega$ and $x^{k}=y$, then $x=e$.

Then $G$ is not isomorphic to $A$, in fact there exists a subgroup $G^{*}$ of $G$ of size $\mathfrak{b}$ (the bounding number) such that $G^{*}$ is not embeddable in $A$.

[TABLE]

By [4], for every $\eta\in\Lambda$, $\Gamma_{\eta}$ is solvable in $G$. Let $\bar{b}_{\eta}=(b_{\eta,n})_{n<\omega}$ witness it, i.e.:

[TABLE]

Let $G^{*}$ be the subgroup of $G$ generated by $\{g_{n}:n<\omega\}\cup\{b_{\eta,n}:\eta\in\Lambda,n<\omega\}$. Towards contradiction, suppose that $\pi$ is an embedding of $G^{*}$ into $A$, and let $S$ be a system of generators for $A$ whose associated length function $lg_{S}=lg$ satisfies conditions (i) and (ii) of the statement of the theorem. For $\eta\in\Lambda$ and $n<\omega$, let:

[TABLE]

Now, $m_{*}$ is a function from $\Lambda$ to $\omega$ and so there exists unbounded $\Lambda_{1}\subseteq\Lambda$ such that for every $\eta\in\Lambda_{1}$ the value $m_{*}(\eta)$ is a constant $m_{*}$. Fix such a $\Lambda_{1}$ and $m_{*}$, and let $f_{1},f_{2}\in\omega^{\omega}$ increasing satisfying the following:

1. (1)

$f_{1}(n)>lg(g^{\prime}_{n})$; 2. (2)

$f_{2}(n)=(m_{*}+1)+\sum_{\ell<n}f_{1}(\ell)$.

Claim 1.1*.* For every $\eta\in\Lambda_{1}$, $lg(c_{\eta,n})<f_{2}(n)$.

Proof. By induction on $n<\omega$. The case $n=0$ is clear by the choice of $f_{1}$ and $f_{2}$. Let $n=m+1$. Because of assumption (i) on $A$, the choice of $\Lambda_{1}$ and the choice of $f_{1}$ and $f_{2}$, we have:

[TABLE]

Now, by the choice of $\Lambda_{1}$, we can find $\eta\in\Lambda_{1}$ and $n<\omega$ such that $\eta(n)>f_{2}(n+2)$. Notice then that by the claim above and the choice of $f_{1}$ and $f_{2}$ we have:

[TABLE]

[TABLE]

Thus, by (1) and the fact that $c_{\eta,n+1}^{\eta(n)}=c_{\eta,n}g^{\prime}_{n}$, using assumption (ii) we infer that $c_{\eta,n+1}=e$. Hence,

[TABLE]

Furthermore, if $\eta(n+1)>lg(g^{\prime}_{n+1})$, then, again by assumption (ii), we have that $c_{\eta,n+2}=e$, and so $c_{\eta,n+2}^{\eta(n+1)}=g^{\prime}_{n+1}=e$, which contradicts the choice of $(g_{n})_{n<\omega}$. Hence, $\eta(n)<\eta(n+1)\leqslant lg(g^{\prime}_{n+1})$, contradicting (2). It follows that the embedding $\pi$ from $G^{*}$ into $A$ cannot exist. ∎

Thus, for $\Gamma$ a graph with no edges (resp. a complete graph) $A(\Gamma)$ is a free group (resp. a free abelian group).

1. (1)

$g$* can be written as $hfh^{-1}$ with $f$ cyclically reduced and $lg(g)=lg(f)+2lg(h)$;* 2. (2)

if $0<k<\omega$ and $f$ is cyclically reduced, then $lg(f^{k})=klg(f)$; 3. (3)

if $0<k<\omega$ and $g=hfh^{-1}$ is as in (1), then $lg(hfh^{-1})^{k}=klg(f)+2lg(h)$.

As well known, the automorphism group of a countable structure is naturally endowed with a Polish topology which respects the group structure, hence:

The situation is different for right-angled Coxeter groups, in fact the structure $M$ with $\omega$ many disjoint unary predicates of size $2$ is such that $Aut(M)=(\mathbb{Z}_{2})^{\omega}$, i.e. $Aut(M)$ is the right-angled Coxeter group on $K_{\mathfrak{c}}$ (a complete graph on continuum many vertices). Notice that in this group for any $a\neq b\in K_{\mathfrak{c}}$ we have:

1. (i)

$(ab)^{2}=1$; 2. (ii)

$lg(ab)=2<3$, $(ab)^{3}=ab$ and $ab\neq e$.

We hope to investigate realizability of uncountable right-angled Coxeter groups as groups of automorphisms of countable structures in a future work.