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

Gianluca Paolini; Saharon Shelah

doi:10.3390/axioms6020013

and

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.

Axioms2017, 6(2), 13;https://doi.org/10.3390/axioms6020013

This article belongs to the Collection Topological Groups

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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 < ω

, then

l g (x) \leq l g (x^{k})

; (ii) if

l g (y) < k < ω

and

x^{k} = y

, then

x = e

, then there exists a subgroup

G^{*}

of G of size

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 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 $0 < k < ω$ , then $l g (x) \leq l g (x^{k})$ ;
(ii): if $l g (y) < k < ω$ 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

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

ζ = {(ζ_{n})}_{n < ω} \in R^{ω}

be such that

ζ_{n} < 2^{- n}

, for every

n < ω

, and

\bar{g} = {(g_{n})}_{n < ω} \in G^{ω}

such that

g_{n} \neq e

and

d (g_{n}, e) < ζ_{n}

, for every

n < ω

. Let

Λ

be a set of power

b

of increasing functions

η \in ω^{ω}

which is unbounded with respect to the partial order of eventual domination. For transparency, we also assume that for every

η \in Λ

we have

η (0) > 0

. For

η \in Λ

, define the following set of equations:

Γ_{η} = {x_{n + 1}^{η (n)} = x_{n} g_{n} : n < ω} .

By (3.1, [3]), for every

η \in Λ

,

Γ_{η}

is solvable in G. Let

{\bar{b}}_{η} = {(b_{η, n})}_{n < ω}

witness it; i.e.,

{\bar{b}}_{η} \in G^{ω} and \underset{n < ω}{⋀} b_{η, n + 1}^{η (n)} = b_{η, n} g_{n} .

Let

G^{*}

be the subgroup of G generated by

{g_{n} : n < ω} \cup {b_{η, n} : η \in Λ, n < ω}

. Towards contradiction, suppose that

π

is an embedding of

G^{*}

into A, and let S be a system of generators for A whose associated length function

l g_{S} = l g

satisfies conditions (i) and (ii) of the statement of the theorem. For

η \in Λ

and

n < ω

, let:

π (g_{n}) = g_{n}^{'}, π (b_{η, n}) = c_{η, n} and m_{*} (η) = l g (c_{η, 0}) .

Now,

m_{*}

is a function from

Λ

to

ω

and so there exists unbounded

Λ_{1} \subseteq Λ

such that for every

η \in Λ_{1}

the value

m_{*} (η)

is a constant

m_{*}

. Fix such a

Λ_{1}

and

m_{*}

, and let

f_{1}, f_{2} \in ω^{ω}

increasing satisfying the following:

(1): $f_{1} (n) > l g (g_{n}^{'})$ ;
(2): $f_{2} (n) = (m_{*} + 1) + \sum_{ℓ < n} f_{1} (ℓ)$ .

Claim 1.

For every

η \in Λ_{1}

,

l g (c_{η, n}) < f_{2} (n)

.

Proof.

By induction on

n < ω

. 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

Λ_{1}

, and the choice of

f_{1}

and

f_{2}

, we have:

\begin{array}{l} l g (c_{η, n}) & \leq & l g (c_{η, n}^{η (m)}) \\ = & l g (c_{η, m} g_{m}^{'}) \\ \leq & l g (c_{η, m}) + l g (g_{m}^{'}) \\ < & f_{2} (m) + f_{1} (m) \\ = & f_{2} (n) . \end{array}

☐

Now, by the choice of

Λ_{1}

, we can find

η \in Λ_{1}

and

n < ω

such that

η (n) > f_{2} (n + 2)

. Notice then that by the claim above and the choice of

f_{1}

and

f_{2}

, we have:

η (n) > f_{2} (n + 1) = f_{2} (n) + f_{1} (n) > l g (c_{η, n}) + l g (g_{n}^{'}) \geq l g (c_{η, n} g_{n}^{'}),

(1)

η (n) > f_{2} (n + 2) \geq f_{1} (n + 1) > l g (g_{n + 1}^{'}) .

(2)

Thus, by (1) and the fact that

c_{η, n + 1}^{η (n)} = c_{η, n} g_{n}^{'}

, using assumption (ii), we infer that

c_{η, n + 1} = e

. Hence,

c_{η, n + 2}^{η (n + 1)} = c_{η, n + 1} g_{n + 1}^{'} = g_{n + 1}^{'} .

Furthermore, if

η (n + 1) > l g (g_{n + 1}^{'})

, then again by assumption (ii), we have that

c_{η, n + 2} = e

, and so

c_{η, n + 2}^{η (n + 1)} = g_{n + 1}^{'} = e

, which contradicts the choice of

{(g_{n})}_{n < ω}

. Hence,

η (n) < η (n + 1) \leq l g (g_{n + 1}^{'})

, contradicting (2). It follows that the embedding

π

from

G^{*}

into A cannot exist. ☐

Definition 1.

Given a graph

Γ = (E, V)

, the right-angled Artin group

A (Γ)

is the group with presentation:

Ω (Γ) = ⟨ V ∣ a b = b a : a E b ⟩ .

If in the presentation

Ω (Γ)

, we ask in addition that all the generators are involutions, then we speak of right-angled Coxeter groups

C (Γ)

.

Thus, for

Γ

, a graph with no edges (resp. a complete graph)

A (Γ)

is a free group (resp. a free abelian group).

Definition 2.

Let

A (Γ)

be a right-angled Artin group and

l g

its associated length function. We say that an element

g \in A (Γ)

is cyclically reduced if it cannot be written as

g = h f h^{- 1}

with

l g (g) = l g (f) + 2

.

Fact 1.

Let

A (Γ)

be a right-angled Artin group,

l g

its associated length function, and

g \in A (Γ)

. Then:

(1): g can be written as $h f h^{- 1}$ with f cyclically reduced and $l g (g) = l g (f) + 2 l g (h)$ ;
(2): if $0 < k < ω$ and f is cyclically reduced, then $l g (f^{k}) = k l g (f)$ ;
(3): if $0 < k < ω$ and $g = h f h^{- 1}$ is as in (1), then $l g {(h f h^{- 1})}^{k} = k l g (f) + 2 l g (h)$ .

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 (Γ)

the associated length function

l g

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

ω

many disjoint unary predicates of size 2 is such that

A u t (M) = {(Z_{2})}^{ω}

; i.e.,

A u t (M)

is the right-angled Coxeter group on

K_{c}

(a complete graph on continuum many vertices). Notice that in this group for any

a \neq b \in K_{c}

, we have:

(i): ${(a b)}^{2} = 1$ ;
(ii): $l g (a b) = 2 < 3$ , ${(a b)}^{3} = a b$ and $a b \neq e$ .

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.

References

Shelah, S. A Countable Structure Does Not Have a Free Uncountable Automorphism Group. Bull. Lond. Math. Soc. 2003, 35, 1–7. [Google Scholar] [CrossRef]
Becker, H.; Kechris, A.S. The Descriptive Set Theory of Polish Group Actions; London Math. Soc. Lecture Notes Ser. 232; Cambridge University Press: Cambridge, UK, 1996. [Google Scholar]
Shelah, S. Polish Algebras, Shy From Freedom. Israel J. Math. 2011, 181, 477–507. [Google Scholar] [CrossRef][Green Version]
Solecki, S. Polish Group Topologies. In Sets and Proofs; London Math. Soc. Lecture Note Ser. 258; Cambridge University Press: Cambridge, UK, 1999. [Google Scholar]
Dudley, R.M. Continuity of Homomorphisms. Duke Math. J. 1961, 28, 34–60. [Google Scholar] [CrossRef]
Paolini, G.; Shelah, S. Polish Topologies for Graph Products of Cyclic Groups. In Preparation.
Servatius, H. Automorphisms of Graph Groups. J. Algebra 1989, 126, 34–60. [Google Scholar] [CrossRef]

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

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