Einstein Institute of Mathematics, The Hebrew University of Jerusalem, Edmond J. Safra Campus, Givat Ram, Jerusalem 91904, Israel
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.
Sidney A. Morris
Received: 28 March 2017 / Accepted: 4 May 2017 / Published: 11 May 2017
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 , then ; (ii) if and , then , then there exists a subgroup of G of size (the bounding number) such that 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.
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 . Independently, in the context of descriptive set theory, Becher and Kechris  asked if an uncountable Polish group can be free. This was also answered negatively by Shelah , generalizing the techniques of . Inspired by the question of Becher and Kechris, Solecki  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:
Let be an uncountable Polish group and A a group admitting a system of generators whose associated length function satisfies the following conditions:
if , then ;
if and , then .
Then G is not isomorphic to A; in fact, there exists a subgroup of G of size (the bounding number) such that 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 . 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 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 . This powerful result has a broad scope of applications, and is used by the authors in a work in preparation  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).
Let be such that , for every , and such that and , for every . Let be a set of power of increasing functions which is unbounded with respect to the partial order of eventual domination. For transparency, we also assume that for every we have . For , define the following set of equations:
By (3.1, ), for every , is solvable in G. Let witness it; i.e.,
Let be the subgroup of G generated by . Towards contradiction, suppose that is an embedding of into A, and let S be a system of generators for A whose associated length function satisfies conditions (i) and (ii) of the statement of the theorem. For and , let:
Now, is a function from to and so there exists unbounded such that for every the value is a constant . Fix such a and , and let increasing satisfying the following:
For every , .
By induction on . The case is clear by the choice of and . Let . Because of assumption (i) on A, the choice of , and the choice of and , we have:
Now, by the choice of , we can find and such that . Notice then that by the claim above and the choice of and , we have:
Thus, by (1) and the fact that , using assumption (ii), we infer that . Hence,
Furthermore, if , then again by assumption (ii), we have that , and so , which contradicts the choice of . Hence, , contradicting (2). It follows that the embedding from into A cannot exist. ☐
Given a graph , the right-angled Artin group is the group with presentation:
If in the presentation , we ask in addition that all the generators are involutions, then we speak of right-angled Coxeter groups .
Thus, for , a graph with no edges (resp. a complete graph) is a free group (resp. a free abelian group).
Let be a right-angled Artin group and its associated length function. We say that an element is cyclically reduced if it cannot be written as with .
Let be a right-angled Artin group, its associated length function, and . Then:
g can be written as with f cyclically reduced and ;
if and f is cyclically reduced, then ;
if and is as in (1), then .
Item (1) is proved in (Proposition on p. 38, ). The rest is folklore. ☐
No uncountable Polish group can be a right-angled Artin group.
By Theorem 1 it suffices to show that for every right-angled Artin group the associated length function 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:
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 many disjoint unary predicates of size 2 is such that ; i.e., is the right-angled Coxeter group on (a complete graph on continuum many vertices). Notice that in this group for any , we have:
, and .
Partially supported by European Research Council grant 338821. No. 1112 on Shelah’s publication list.
Gianluca Paolini and Saharon Shelah contribute equally to this manuscript.
Conflicts of Interest
The authors declare no conflict of interest.
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