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*Axioms*
**2017**,
*6*(2),
13;
https://doi.org/10.3390/axioms6020013

Article

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

^{1}

Einstein Institute of Mathematics, The Hebrew University of Jerusalem, Edmond J. Safra Campus, Givat Ram, Jerusalem 91904, Israel

^{2}

Department of Mathematics, The State University of New Jersey, Hill Center-Busch Campus, Rutgers, 110 Frelinghuysen Road, Piscataway, NJ 08854-8019, USA

^{*}

Author to whom correspondence should be addressed.

Academic Editor:
Sidney A. Morris

Received: 28 March 2017 / Accepted: 4 May 2017 / Published: 11 May 2017

## 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.

Keywords:

descriptive set theory; polish group topologies; right-angled Artin groupsIn 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.,

$${\overline{b}}_{\eta}\in {G}^{\omega}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{4.pt}{0ex}}\mathrm{and}\phantom{\rule{4.pt}{0ex}}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\underset{n<\omega}{\bigwedge}{b}_{\eta ,n+1}^{\eta \left(n\right)}={b}_{\eta ,n}{g}_{n}.$$

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:

$$\pi \left({g}_{n}\right)={g}_{n}^{\prime},\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\pi \left({b}_{\eta ,n}\right)={c}_{\eta ,n}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{4.pt}{0ex}}\mathrm{and}\phantom{\rule{4.pt}{0ex}}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}{m}_{*}\left(\eta \right)=lg\left({c}_{\eta ,0}\right).$$

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:

$$\eta \left(n\right)>{f}_{2}(n+1)={f}_{2}\left(n\right)+{f}_{1}\left(n\right)>lg\left({c}_{\eta ,n}\right)+lg\left({g}_{n}^{\prime}\right)\ge lg\left({c}_{\eta ,n}{g}_{n}^{\prime}\right),$$

$$\eta \left(n\right)>{f}_{2}(n+2)\ge {f}_{1}(n+1)>lg\left({g}_{n+1}^{\prime}\right).$$

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 $ab\ne e$.

## Acknowledgments

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

## Author Contributions

Gianluca Paolini and Saharon Shelah contribute equally to this manuscript.

## Conflicts of Interest

The authors declare no conflict of interest.

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