szymon g lab˘ - politechnika Śląska · szymon g lab˘ (and filip strobin) large free subgroups...

introduction – general problemFraısse limit

the main resultother results

Szymon G lab

Large free subgroups of automorphisms group of ultrahomogeneousspaces

with Filip Strobin

Institute of Mathematics, Lodz University of Technology

Szymon G lab (and Filip Strobin) Large free subgroups of automorphisms group of ultrahomogeneous spaces

Large free groups in groups of automorphism

Groups of automorphisms

Let A be a countable structure (in fact, we should write A = (A,F ,R, C)). ByAut(A) we denote the group of automorphisms of A.

general problem

Detect those countable structures A, whose groups of auomorphisms Aut(A)contains a large free group.

Macperson (1986)

If A is ω-categorical, then Aut(A) contains a dense free subgroup of ωgenerators.Automorphism group of a random graph contains a dense free subgroup of 2generators.

Melles and Shelah (1994)

If A is a saturated model of a complete theory T with |A| = λ > |T |, thenAut(A) has a dense free subgroup with cardinality 2λ.

general problem

Macperson (1986)

general problem

Macperson (1986)

general problem

Macperson (1986)

Gartside and Knight (2003)

If A is ω-categorical and Kn = {(g1, . . . , gn) ∈ Aut(A)n : g1, . . . , gn are freegenerators}, then Kn is comeager in Aut(A)n for every n.

Shelah (2003)

Aut(A) cannot be a free uncountable group where A is a countable structure

Shelah (2011)

Any uncoutable Polish group cannot be free.

H.D. Macpherson, A survey of homogeneous structures. Discrete Math. 311(2011), no. 15, 1599-1634.

Shelah (2003)

Shelah (2011)

Shelah (2003)

Shelah (2011)

Shelah (2003)

Shelah (2011)

ultrahomogneous structures, c-largeness

ultrahomogeneous structure

We say that a countable structure A is ultrahomogeneous, if each isomorphismbetween finitely generated substructures of A can be extended to anautomorphism of A. Additionally we assume that finitely generatedsubstructures of A are finite.A is ultrahomogeneous iff A is a Fraısse limit.

Question

For which countable ultrahomogeneous structures A there exists a familyH ⊂ Aut(A) of c-many free generators? Such groups Aut(A) will be calledc-large.

Question

Two examples

positive example

Let A = ω. Then Aut(A) = S∞ - the group of all bijections of ω.S∞ is c-large.

negative example

Let A = (ω, {Rn : n ∈ ω}), where Rn are unary relations such that

x ∈ Rn iff x ∈ {2n, 2n + 1}.

Then for every f ∈ Aut(A), f ◦ f = id, so Aut(A) does not contain anynonempty family of free generators.

Two examples

positive example

Let A = ω. Then Aut(A) = S∞ - the group of all bijections of ω.S∞ is c-large.

negative example

Let A = (ω, {Rn : n ∈ ω}), where Rn are unary relations such that

x ∈ Rn iff x ∈ {2n, 2n + 1}.

Then for every f ∈ Aut(A), f ◦ f = id, so Aut(A) does not contain anynonempty family of free generators.

Hereditary property, Joint embedding property and Amalgamation property

Let K be a class of finitely generated L-structures. K is called age if it has

Hereditary property (HP): if A ∈ K and B is finitely generatedsubstructure of A, then B ∈ K.

Joint embedding property (JEP): if A,B ∈ K , then there is C ∈ K suchthat A and B are embeddable in C .

Amalgamation property (AP): if A,B,C ∈ K and e : A→ B andf : A→ C , then there are D ∈ K and g : B → D and h : C → D suchthat ge = hf .

remark

In general JEP is not a special case of AP. Consider fields!

Strong amalgamation property (SAP)

AP + rng g ∩ rng h = rng ge(= rng hf ).

remark

Fraısse theorem

ultrahomogeneous structures

A is ultrahomogeneous, if each isomorphism between finitely generatedsubstructures of A can be extended to an automorphism of A.

Age of ultrahomogeneous structure

Let K be a family of all finitely generated substructures of ultrahomogeneousstructure A. Then K is an age (of A).

Fraısse theorem

Let L be a countable language and K be a countable age of L-structures.Then there is L-structure A, unique up to isomorphism, such that

A is countable.

K is an age of A.

A is ultrahomogeneous.

A is called a Fraısse limit of K.

Fraısse theorem

A is countable.

K is an age of A.

Fraısse theorem

A is countable.

K is an age of A.

Fraısse theorem

A is countable.

K is an age of A.

Fraısse theorem

A is countable.

K is an age of A.

Some examples of Fraısse limits

finite linear orders

If K = {finite linear orders}, then (Q,≤) is a Fraısse limit of K.

finite graphs

Random graph G is a Fraısse limit of K = {finite graphs}.

finite groups

Hall’s universal locally finite group is a Fraısse limit of {finite groups}.

finite groups where every element has order 2⊕i∈N Z2 is a Fraısse limit of {finite groups where every element has order 2}.

finitely generated torsion-free abelian groups⊕i∈N Q is a Fraısse limit of {finitely generated torsion-free abelian groups}.

finite graphs

finite groups

finite graphs

finite groups

finite graphs

finite groups

finite graphs

finite groups

(∗) condition

(x0, ..., xm)-function

Let x0, . . . , xm be pairwise distinct elements of A. A shift on {x0, . . . , xm} is apartial function ϕ : {x0, . . . , xm} → A such that ϕ(xi ) = xi−1 for i = 1, . . . ,m(ϕ is a left-shift) or ϕ(xi ) = xi+1 for i = 0, . . . ,m − 1 (ϕ is a right-shift).An (x0, . . . , xm)-function is a partial function g :

⋃ki=1 Ii → A such that:

(i) I1, . . . , Ik are pairwise disjoint;

(ii) each Ii is of the form {xp, xp+1, . . . , xq} for some 0 ≤ p < q ≤ m;

(iii) each restriction g � Ii is a shift.

(∗) condition

For any finitely generated substructures B1,B2 ⊂ A and any m ∈ N, there existpairwise distinct x0, ..., xm ∈ A \ (B1 ∪ B2) such that for any embeddingf : B1 → B2, and for any (x0, ..., xm)-function g , there exists an embeddingfg : gen(B1 ∪ dom(g))→ A such that f , g ⊂ fg .

(∗) condition

(∗) condition and Theorem 1

Theorem 1

Assume that A satisfies (∗) and each finitely generated substructure of A isfinite. Then Aut(A) is c-large.

The structure ω satisfies (∗) and every finitely generated substructure is finite.

S∞ group

The group S∞ of all bijections of ω is c-large.

(Q,≤)

the structure (Q,≤) satisfies (∗) and every finitely generated substructure isfinite.

countable atomless Boolean algebra

A countable atomless Boolean algebra B satisfies (∗) and finitely generatedsubalgebras are finite.

Theorem 1

S∞ group

(Q,≤)

Theorem 1

S∞ group

(Q,≤)

Theorem 1

S∞ group

(Q,≤)

Theorem 1

S∞ group

(Q,≤)

ω-independence

A relational structure A is ω-independent if for any finitely generatedsubstructures B1,B2 of A, and for any m, there is a set D ⊂ A \ (B1 ∪ B2)consisting of m + 1 elements such that, for any embedding f : B1 → B2, andany partial permutation g of D, the function f ∪ g is an embedding ofB1 ∪ dom(g) into A.

rational Urysohn space

A rational Urysohn space is a countable metric space U with rational distances,such that every finite metric space with rational distances has an isometriccopy in U.A rational metric space is ω-independent.

random graph

A random graph G is a countable graph such that for every finite X ,Y ⊂ G,there is a vertex with edges going to each vertex from X , and no edge going toa vertex of Y .A random graph is ω-independent.

ω-independence

random graph

ω-independence

random graph

ω-independence

random Kn-free graph

Let Gn be the random Kn-free graph, that is the ultrahomogeneous countablegraph which omits Kn, the complete graph on n vertices. Gn is ω-independent.

ELet E be a countable equivalence relation with infinitely many infiniteequivalence classes. E is ω-independent.

Let En be a countable equivalence relation with n many infinite equivalenceclasses. E is ω-independent.

DLet D be the universal countable ultrahomogeneous partially ordered set. It is aFraısse limit of all finite partially ordered sets. D is ω-independent.

ω-independence

ordered structures

Assume that A is a Fraısse limit of a class K0. Let

K = K0 ? LO := {〈B,≤〉 : B ∈ K0 and ≤ is a linear ordering on B}.

If K0 is a Fraısse class with the strong amalgamation property, then so is K.We will denote the Fraısse limit of K by A≤.

Proposition

Let A be ω-independent relational structure such that A is a Fraısse limit of aFraısse class with the strong amalgamation property. Then A≤ satisfies (∗).

U, G, Gn, E and En have strong amalgamation property. Their orderedcounterparts - ordered rational Urysohn space U≤, ordered random graph G≤,ordered Kn-free random graph Gn

≤ and ordered relations E≤ and En≤ - satisfy

(∗).

ordered structures

Proposition

(∗).

ordered structures

Proposition

(∗).

ordered structures

Proposition

(∗).

The proof of Theorem 1

partial order PPart(A) = all partial isomorphisms of A.P = the set of pairs (n, p) where p : {0, 1}n → Part(A) and dom(p(s)) is ann-element substructure of A for every s ∈ {0, 1}n.P is ordered in the following way:(n, p) ≤ (k, q) if and only if k ≤ n and q(t) ⊂ p(s) provided t ≺ s.

dense subsets of PD ⊂ P is dense if for any (k, q) there is (n, p) ∈ D with (n, p) ≤ (k, q).

filters in PG ⊂ P is a filter if(i) for any (k1, q1), (k2, q2) ∈ G there is (n, p) ∈ G with (n, p) ≤ (ki , qi ),(ii) if (n, p) ∈ G and (n, p) ≤ (k, q), then (k, p) ∈ G .

Rasiowa-Sikorski Lemma

If D1,D2, . . . are dense in P, then there is a filter G which has nonemptyintersection with each Di .

For every k ∈ A, the set

Dk := {(n, p) ∈ P : ∀s ∈ {0, 1}n k ∈ dom(p(s)) ∩ rng(p(s))}

is dense in P.

Generic filter G

If G ∩ Dk 6= ∅ is a filter, then for α ∈ {0, 1}N put

g(α) =⋃{p(α � n) : (n, p) ∈ G}.

g(α) ∈ Aut(A).

is dense in P.

Generic filter G

g(α) =⋃{p(α � n) : (n, p) ∈ G}.

g(α) ∈ Aut(A).

is dense in P.

Generic filter G

g(α) =⋃{p(α � n) : (n, p) ∈ G}.

g(α) ∈ Aut(A).

is dense in P.

Generic filter G

g(α) =⋃{p(α � n) : (n, p) ∈ G}.

g(α) ∈ Aut(A).

Dense sets Ds1,...,smw

Assume that A has the property (∗). For any nonempty word w(y1, . . . , ym)and any pairwise distinct finite sequences s1, . . . , sm of 0’s and 1’s of the samelength, the set

Ds1,...,smw = {(n, p) : |s1| ≤ n and for every t1, . . . , tm ∈ {0, 1}n with si ≺ ti

we have w(p(t1), . . . , p(tm)) 6= id}

is dense in P.

Let G be a filter such that G ∩ Dk 6= ∅ and G ∩ Ds1,...,smw 6= ∅. For α ∈ {0, 1}N

g(α) =⋃{p(α � n) : (n, p) ∈ G}.

Then {g(α) : α ∈ {0, 1}N} is a family of free generators.

we have w(p(t1), . . . , p(tm)) 6= id}

is dense in P.

g(α) =⋃{p(α � n) : (n, p) ∈ G}.

we have w(p(t1), . . . , p(tm)) 6= id}

is dense in P.

g(α) =⋃{p(α � n) : (n, p) ∈ G}.

we have w(p(t1), . . . , p(tm)) 6= id}

is dense in P.

g(α) =⋃{p(α � n) : (n, p) ∈ G}.

Extending free groups

problem

Let F ⊂ Aut(A) be a family of free generators. Does there exists a family H ofcardinality c such that F ∪H is a family of free generators?

Theorem 2

Let F ⊂ S∞ be a countable family of free generators. Then there is a familyH ⊂ S∞ of cardinality c such that F ∪H is a family of free generators.

problem

Let F ⊂ Aut(A) be a family of free generators. Does there exists a family H ofcardinality c such that F ∪H is a family of free generators?

Theorem 2

Let F ⊂ S∞ be a countable family of free generators. Then there is a familyH ⊂ S∞ of cardinality c such that F ∪H is a family of free generators.

New dense sets

For any k, l ∈ N, any word w(y1, . . . , ym) with k + l = m, any free generatorsf1, . . . , fk , and any pairwise different sequences s1, . . . , sl of 0’s and 1’s of thesame length, the set

Ds1,...,slw,f1,...,fk

= {(n, p) : n ≥ |s1| and for every t1, . . . , tl ∈ {0, 1}n such that si ≺ ti

we have w(f1, . . . , fk , p(t1), . . . , p(tl)) 6= id)}

is dense in P.

How many new dense sets are there?

If F ⊂ Aut(A) has cardinality κ, then there are κ many dense sets Ds1,...,slw,f1,...,fk

New dense sets

For any k, l ∈ N, any word w(y1, . . . , ym) with k + l = m, any free generatorsf1, . . . , fk , and any pairwise different sequences s1, . . . , sl of 0’s and 1’s of thesame length, the set

Ds1,...,slw,f1,...,fk

= {(n, p) : n ≥ |s1| and for every t1, . . . , tl ∈ {0, 1}n such that si ≺ ti

we have w(f1, . . . , fk , p(t1), . . . , p(tl)) 6= id)}

is dense in P.

How many new dense sets are there?

If F ⊂ Aut(A) has cardinality κ, then there are κ many dense sets Ds1,...,slw,f1,...,fk

Martin’s Axiom

ccc posets

Let (P,≤) be a partially ordered set. P has ccc if any set (antichain) C ⊂ Psuch that any two distinct x , y ∈ C do not have common extension in P (i.e.there is no z ∈ P with z ≤ x , z ≤ y) is countable.

Martin’s Axiom

MA(κ): for any ccc partially ordered set (P,≤) and any family {Dα : α < κ}of dense subsets of P there is a filter G in P which has nonempty intersectionwith every Dα.MA(ω) is true in ZFC. MA(c) is false.Martin’s Axiom = MA(κ) for every κ < c.

Martin’s Axiom

ccc posets

Martin’s Axiom

ccc posets

Martin’s Axiom

ccc posets

Martin’s Axiom

a result under Martin’s Axiom

Theorem 2

Assume MA. Then for every family F ⊂ S∞ of less than c many freegenerators, there exists a family H ⊂ S∞ of cardinality c such that H ∪ F is afamily of free generators.

M = σ-ideal of meager subsets of R.

min{κ : ’ MA(κ) for countable posets’ fails} = cov(M).

Corollary

For any family of free generators F of cardinality less than cov(M), there is afamily of free generators F ′ of cardinality c such that F ∪ F ′ is a family of freegenerators.

Theorem 2

Corollary

Theorem 2

Corollary

Theorem 2

Corollary

Product of countable groups

Theorem 3

Let Gn, n ∈ N, be groups.

(i) If for any nonempty word w(y1, ..., ym) there are infinitely many n’s forwhich there are gn,1, . . . , gn,m ∈ Gn with w(gn,1, . . . , gn,m) 6= en, then∏∞

n=1 Gn contains a free group of c generators.

(ii) If every Gn is countable and for some nonempty word w(y1, . . . , ym) andfor almost every n and every gn,1, . . . , gn,m ∈ Gn we havew(gn,1, . . . , gn,m) = en, then

∏∞n=1 Gn does not contain any free group of

uncountably many generators.

Corollary

Let Gn, n ∈ N, be countable groups. Then either∏

n∈N Gn contains freesubgroups of c generators or it does not contain free subgroup of uncountablymany generators.

Theorem 3

Corollary

Theorem 3

Corollary

Theorem 3

Corollary

Let n ≥ 2. There exists a family F = {fα : α < c} of functions from{0, 1, . . . , n − 1}N such that for any α0 < α1 < · · · < αn−1 < c there is k ∈ Nsuch that fαi (k) = i .

Let {pk : k ∈ N} be an enumeration of all subsets of N of cardinality n.Enumerate pk as {pk(0), . . . , pk(n − 1)}.Let {Uα : α < c} be an independentfamily of N (i.e. any finite boolean combination of Uα’s is nonempty).For anyα we define fα : N→ {0, 1, . . . , n − 1} as follows: Fix k ∈ N. If there is i < nsuch that pk(i) ∈ Uα and pk(j) /∈ Uα for every j 6= i , then put fα(k) = i ;otherwise put fα(k) = 0.Let α0 < α1 < · · · < αn−1. Pick mi ∈ Uαi \

⋃j 6=i Uαj and put p(i) = mi for

i < n. There is k ∈ N with p = pk . Then fαi (k) = i .

Theorem 3(i)

If for any nonempty word w(y1, ..., ym) there are infinitely many n’s for whichthere are gn,1, . . . , gn,m ∈ Gn with w(gn,1, . . . , gn,m) 6= en, then

∏∞n=1 Gn

contains a free group of c generators.

Assume that for any word w(y1, ..., ym) there are infinitely many n’s for whichthere are gw

n,1, . . . , gwn,m ∈ Gn with w(gw

n,1, . . . , gwn,m) 6= en.

Ew = {n ∈ N : there are gwn,1, . . . , g

wn,m ∈ Gn with w(gw

n,1, . . . , gwn,m) 6= en}.

Then {Ew : w is a nonempty word} is a countable family of infinite sets.Let{E ′w : w = w(y1, ..., ym) is a nonempty word} be a disjoint refinement of thisfamily, i.e. a family of pairwise disjoint infinite sets with E ′w ⊂ Ew for anynonempty word w .For any α < c, define fα ∈

∏Gn as follows. Let w be a word.

If w = w(y1, . . . , ym) then using Lemma, we can find a family {f wα : α < c}

such that for any α1 < · · · < αm there is n ∈ E ′w with f wαi

(n) = gwn,i for i ≤ m.

Finally, let fα(n) = f wα (n) if n ∈ E ′w , and fα(n) = en, otherwise.

Theorem 3(i)

∏∞n=1 Gn

n,1, . . . , gwn,m) 6= en.

n,1, . . . , gwn,m) 6= en}.

Theorem 3(i)

∏∞n=1 Gn

n,1, . . . , gwn,m) 6= en.

n,1, . . . , gwn,m) 6= en}.

Theorem 3(i)

∏∞n=1 Gn

n,1, . . . , gwn,m) 6= en.

n,1, . . . , gwn,m) 6= en}.

Theorem 3(i)

∏∞n=1 Gn

n,1, . . . , gwn,m) 6= en.

n,1, . . . , gwn,m) 6= en}.

Theorem 3(i)

∏∞n=1 Gn

n,1, . . . , gwn,m) 6= en.

n,1, . . . , gwn,m) 6= en}.

Theorem 3(i)

∏∞n=1 Gn

n,1, . . . , gwn,m) 6= en.

n,1, . . . , gwn,m) 6= en}.

Theorem 3(i)

∏∞n=1 Gn

n,1, . . . , gwn,m) 6= en.

n,1, . . . , gwn,m) 6= en}.

Theorem 3(ii)

If every Gn is countable and for some nonempty word w(y1, . . . , ym) and foralmost every n and every gn,1, . . . , gn,m ∈ Gn we havew(gn,1, . . . , gn,m) = en, then

Let w(y1, . . . , ym) be a word such that there is N with w(gn,1, . . . , gn,m) = enfor n ≥ N and all gn,1, . . . , gn,m ∈ Gn.Suppose

∏∞n=1 Gn contains a free group of

uncountably many generators, say {fα : α < ω1}. Then for every distinctα1, . . . , αm < ω1 and there is n < N, depending on αi ’s, withw(fα1 (n), . . . , fαm (n)) 6= en.Since the groups Gn are countable, there are twodistinct m-element sets {α1, . . . , αm} and {β1, . . . , βm} of ordinals less than ω1

such thatw(fα1 (n), . . . , fαm (n)) = w(fβ1 (n), . . . , fβm (n))

for every n < N. Then

w(fα1 (n), . . . , fαm (n))w−1(fβ1 (n), . . . , fβm (n)) = en

for every n ∈ N. This contradicts the fact that {fα : α < ω1} are freegenerators.

Theorem 3(ii)

Thank you for your attention

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