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

Abstract

We prove that if G is a Polish group and A a group admitting a system of generators whose associated length function satisfies: (i) if $0<k<\omega$, then $lg\left(x\right)\le lg\left(x^k\right)$; (ii) if $lg\left(y\right)<k<\omega$ and $x^k=y$, then $x=e$, then there exists a subgroup $G^{*}$ of G of size $\mathfrak{b}$ (the bounding number) such that $G^{*}$ is not embeddable in A. In particular, we prove that the automorphism group of a countable structure cannot be an uncountable right-angled Artin group. This generalizes analogous results for free and free abelian uncountable groups.

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 [1]. Independently, in the context of descriptive set theory, Becher and Kechris [2] asked if an uncountable Polish group can be free. This was also answered negatively by Shelah [3], generalizing the techniques of [1]. Inspired by the question of Becher and Kechris, Solecki [4] 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 Definition 1). We actually prove more:

Theorem 1.

Let $G=(G,d)$ be an uncountable Polish group and A a group admitting a system of generators whose associated length function satisfies the following conditions:

(i) 

if $0<k<\omega$, then $lg\left(x\right)\le lg\left(x^k\right)$;

(ii) 

if $lg\left(y\right)<k<\omega$ 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.

After the authors proved Theorem 1, they discovered that the impossibility to endow groups A as in Theorem 1 with a Polish group topology follows from an old important result of Dudley [5]. In fact, Dudley’s work implies more strongly that we cannot even find a homomorphism from a Polish group G into A. Apart from the fact that the claim about $G^{*}$ in Theorem 1 is of independent interest and not subsumed by Dudley’s work, our focus here is on techniques; i.e., the crucial use of the Compactness Lemma of [3]. This powerful result has a broad scope of applications, and is used by the authors in a work in preparation [6] to deal with classes of groups not covered by Theorem 1 or Dudley’s work, most notably the class of right-angled Coxeter groups (see Definition 1).

Proof of Theorem 1.

Let $\zeta ={\left({\zeta }_n\right)}_{n<\omega }\in {\mathbb{R}}^{\omega }$ be such that ${\zeta }_n<2^{-n}$, for every $n<\omega$, and $\overline{g}={\left(g_n\right)}_{n<\omega }\in G^{\omega }$ such that $g_n\ne e$ and $d(g_n,e)<{\zeta }_n$, for every $n<\omega$. Let $\mathrm{\Lambda }$ be a set of power $\mathfrak{b}$ of increasing functions $\eta \in {\omega }^{\omega }$ which is unbounded with respect to the partial order of eventual domination. For transparency, we also assume that for every $\eta \in \mathrm{\Lambda }$ we have $\eta \left(0\right)>0$. For $\eta \in \mathrm{\Lambda }$, define the following set of equations:

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

By (3.1, [3]), for every $\eta \in \mathrm{\Lambda }$, ${\mathrm{\Gamma }}_{\eta }$ is solvable in G. Let ${\overline{b}}_{\eta }={\left(b_{\eta ,n}\right)}_{n<\omega }$ witness it; i.e.,

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

Let $G^{*}$ be the subgroup of G generated by $\{g_n:n<\omega \}\cup \{b_{\eta ,n}:\eta \in \mathrm{\Lambda },n<\omega \}$. 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 $l{g}_S=lg$ satisfies conditions (i) and (ii) of the statement of the theorem. For $\eta \in \mathrm{\Lambda }$ and $n<\omega$, let:

$$\pi \left(g_n\right)=g_n^{\prime },\pi \left(b_{\eta ,n}\right)=c_{\eta ,n}\mathrm{and}{m}_{*}\left(\eta \right)=lg\left(c_{\eta ,0}\right).$$

Now, $m_{*}$ is a function from $\mathrm{\Lambda }$ to $\omega$ and so there exists unbounded ${\mathrm{\Lambda }}_1\subseteq \mathrm{\Lambda }$ such that for every $\eta \in {\mathrm{\Lambda }}_1$ the value $m_{*}\left(\eta \right)$ is a constant $m_{*}$. Fix such a ${\mathrm{\Lambda }}_1$ and $m_{*}$, and let $f_1,f_2\in {\omega }^{\omega }$ increasing satisfying the following:

(1)

$f_1\left(n\right)>lg\left(g_n^{\prime }\right)$;

(2)

$f_2\left(n\right)=(m_{*}+1)+{\sum }_{\ell <n}{f}_1\left(\ell \right)$.

Claim 1.

For every $\eta \in {\mathrm{\Lambda }}_1$, $lg\left(c_{\eta ,n}\right)<f_2\left(n\right)$.

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 ${\mathrm{\Lambda }}_1$, and the choice of $f_1$ and $f_2$, we have:

$$\begin{array}{lll}lg\left(c_{\eta ,n}\right)& \le & lg\left(c_{\eta ,n}^{\eta \left(m\right)}\right)\\ & =& lg\left(c_{\eta ,m}{g}_m^{\prime }\right)\hfill \\ & \le & lg\left(c_{\eta ,m}\right)+lg\left(g_m^{\prime }\right)\hfill \\ & <& f_2\left(m\right)+f_1\left(m\right)\hfill \\ & =& f_2\left(n\right).\hfill \end{array}$$

☐

Now, by the choice of ${\mathrm{\Lambda }}_1$, we can find $\eta \in {\mathrm{\Lambda }}_1$ and $n<\omega$ such that $\eta \left(n\right)>f_2(n+2)$. Notice then that by the claim above and the choice of $f_1$ and $f_2$, we have:

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

(1)

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

(2)

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

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

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

Definition 1.

Given a graph $\mathrm{\Gamma }=(E,V)$, the right-angled Artin group $A\left(\mathrm{\Gamma }\right)$ is the group with presentation:

$$\mathrm{\Omega }\left(\mathrm{\Gamma }\right)=\langle V\mid ab=ba:aEb\rangle .$$

If in the presentation $\mathrm{\Omega }\left(\mathrm{\Gamma }\right)$, we ask in addition that all the generators are involutions, then we speak of right-angled Coxeter groups $C\left(\mathrm{\Gamma }\right)$.

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

Definition 2.

Let $A\left(\mathrm{\Gamma }\right)$ be a right-angled Artin group and $lg$ its associated length function. We say that an element $g\in A\left(\mathrm{\Gamma }\right)$ is cyclically reduced if it cannot be written as $g=hf{h}^{-1}$ with $lg\left(g\right)=lg\left(f\right)+2$.

Fact 1.

Let $A\left(\mathrm{\Gamma }\right)$ be a right-angled Artin group, $lg$ its associated length function, and $g\in A\left(\mathrm{\Gamma }\right)$. Then:

(1) 

g can be written as $hf{h}^{-1}$ with f cyclically reduced and $lg\left(g\right)=lg\left(f\right)+2lg\left(h\right)$;

(2) 

if $0<k<\omega$ and f is cyclically reduced, then $lg\left(f^k\right)=klg\left(f\right)$;

(3) 

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

Proof. 

Item (1) is proved in (Proposition on p. 38, [7]). The rest is folklore. ☐

Corollary 1.

No uncountable Polish group can be a right-angled Artin group.

Proof. 

By Theorem 1 it suffices to show that for every right-angled Artin group $A\left(\mathrm{\Gamma }\right)$ the associated length function $lg$ satisfies conditions (i) and (ii) of the theorem, but by Fact 1, this is clear. ☐

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

Corollary 2.

The automorphism group of a countable structure cannot be an uncountable right-angled Artin group.

As already mentioned, 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 $Aut\left(M\right)={\left({\mathbb{Z}}_2\right)}^{\omega }$; i.e., $Aut\left(M\right)$ is the right-angled Coxeter group on $K_{\mathfrak{c}}$ (a complete graph on continuum many vertices). Notice that in this group for any $a\ne b\in K_{\mathfrak{c}}$, we have:

(i) 

${\left(ab\right)}^2=1$;

(ii) 

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