Deciding endofixedness in free groupsepsem.upc.edu/~ventura/ventura/files/Newcastle-12-4-10.pdf · 2010. 4. 11. · 1.Some history Algorithmic results Needed tools The proof Train-tracks

1.Some history Algorithmic results Needed tools The proof

Deciding endofixedness in free groups

Enric VenturaDepartament de Matemàtica Aplicada III

Universitat Politècnica de Catalunya

Newcastle, April 12, 2010


Outline

1 Some history

2 Algorithmic results

3 Needed tools

4 The proof


Outline

1 Some history


3 Needed tools

4 The proof


Notation

A = {a1, . . . ,an} is a finite alphabet (n letters).A±1 = A ∪ A−1 = {a1,a−1

1 , . . . ,an,a−1n }.

Fn is the free group on A.Aut (Fn) ⊆ Mono (Fn) ⊆ End (Fn).I let endomorphisms φ : Fn → Fn act on the right, x 7→ xφ.Fix (φ) = {x ∈ Fn | xφ = x} 6 Fn.If S ⊆ End (Fn) thenFix (S) = {x ∈ Fn | xφ = x ∀φ ∈ S} = ∩φ∈SFix (φ) 6 Fn.


Notation


1 , . . . ,an,a−1n }.



Notation


1 , . . . ,an,a−1n }.



Notation


1 , . . . ,an,a−1n }.



Notation


1 , . . . ,an,a−1n }.



Notation


1 , . . . ,an,a−1n }.



Notation


1 , . . . ,an,a−1n }.



Fixed subgroups are complicated

φ : F3 → F3a 7→ ab 7→ bac 7→ ca2

Fixφ = 〈a,bab−1, cac−1〉

ϕ : F4 → F4a 7→ dacb 7→ c−1a−1d−1acc 7→ c−1a−1b−1acd 7→ c−1a−1bc

Fixϕ = 〈w〉,where...

w = c−1a−1bd−1c−1a−1d−1ad−1c−1b−1acdadacdcdbcda−1a−1d−1

a−1d−1c−1a−1d−1c−1b−1d−1c−1d−1c−1daabcdaccdb−1a−1.



φ : F3 → F3a 7→ ab 7→ bac 7→ ca2








φ : F3 → F3a 7→ ab 7→ bac 7→ ca2








φ : F3 → F3a 7→ ab 7→ bac 7→ ca2








φ : F3 → F3a 7→ ab 7→ bac 7→ ca2







What is known about fixed subgroups ?

Theorem (Dyer-Scott, 75)

Let G 6 Aut (Fn) be a finite group of automorphisms of Fn. Then,Fix (G) 6ff Fn; in particular, r(Fix (G)) 6 n.

Conjecture (Scott)

For every φ ∈ Aut (Fn), r(Fix (φ)) 6 n.

Theorem (Gersten, 83 (published 87))

Let φ ∈ Aut (Fn). Then r(Fix (φ)) <∞.

Theorem (Thomas, 88)

Let G 6 Aut (Fn) be an arbitrary group of automorphisms of Fn. Then,r(Fix (G)) <∞.





Conjecture (Scott)










Conjecture (Scott)










Conjecture (Scott)







Train-tracks

Main result in this story:

Theorem (Bestvina-Handel, 88 (published 92))

Let φ ∈ Aut (Fn). Then r(Fix (φ)) 6 n.

introducing the theory of train-tracks for graphs.

After Bestvina-Handel, live continues ...

Theorem (Imrich-Turner, 89)

Let φ ∈ End (Fn). Then r(Fix (φ)) 6 n.

Theorem (Turner, 96)

Let φ ∈ End (Fn). If φ is not bijective then r(Fix (φ)) 6 n − 1.


Train-tracks











Train-tracks











Description of fixed subgroups

There are three easy ways of building fixed points:

(Construction-1)

Let φ : Fn → Fn be an automorphism and Fix (φ) its fixed subgroup.Then, there are many ways of extending φ to φ′ : Fn ∗ Fm → Fn ∗ Fmsuch that Fix (φ′) = Fix (φ) (for example, invert all generators of Fm).

(Construction-2)

Let φ1 : Fn → Fn and φ2 : Fm → Fm be two automorphisms andFix (φ1) and Fix (φ2) their fixed subgroups. Then,φ1 ∗ φ2 : Fn ∗ Fm → Fn ∗ Fm has Fix (φ1 ∗ φ2) = Fix (φ1) ∗ Fix (φ2).

(Construction-3)

Let φ : Fn → Fn be an automorphism and Fix (φ) its fixed subgroup.Let h,h′ ∈ Fn be such that hφ = h′hh′−1. Then, the extensionφ′ : Fn ∗ 〈z〉 → Fn ∗ 〈z〉 defined by z 7→ h′hr z satisfiesFix (φ′) = Fix (φ) ∗ 〈z−1hz〉.




(Construction-1)


(Construction-2)


(Construction-3)





(Construction-1)


(Construction-2)


(Construction-3)





(Construction-1)


(Construction-2)


(Construction-3)




These are essentially the only possibilities:

Observation

A cyclic subgroup 〈w〉 6 Fn is the fixed subgroup of someφ ∈ Aut (Fn) if and only if w is not a proper power.

Theorem (Martino-V., 04)

Every automorphism φ : Fn → Fn and its fixed subgroup Fix (φ) canbe built from finitely many automorphisms φi : Fmi → Fmi (mi 6 n),i = 1, . . . , r , with cyclic fixed subgroup, r(Fix (φi)) = 1, by finitelymany applications of Constructions 1, 2 and 3.




Observation







Observation





Inertia

Definition

A subgroup H 6 Fn is called inert if r(H ∩K ) 6 r(K ) for every K 6 Fn.

Theorem (Dicks-V, 96)

Let G ⊆ Mon (Fn) be an arbitrary set of monomorphisms of Fn. Then,Fix (G) is inert; in particular, r(Fix (G)) 6 n.

Theorem (Bergman, 99)

Let G ⊆ End (Fn) be an arbitrary set of endomorphisms of Fn. Then,r(Fix (G)) 6 n.

Conjecture (V.)

Let φ ∈ End (Fn). Then Fix (φ) is inert.


Inertia

Definition






Conjecture (V.)



Inertia

Definition






Conjecture (V.)



Inertia

Definition






Conjecture (V.)



The four families

DefinitionA subgroup H 6 Fn is said to be

1-auto-fixed if H = Fix (φ) for some φ ∈ Aut (Fn),1-endo-fixed if H = Fix (φ) for some φ ∈ End (Fn),auto-fixed if H = Fix (S) for some S ⊆ Aut (Fn),endo-fixed if H = Fix (S) for some S ⊆ End (Fn),

Easy to see that 1-mono-fixed = 1-auto-fixed.


The four families





The four families





The four families





The four families





Relations between them

1− auto − fixed ⊆ 1− endo − fixed

∩| ∩|

auto − fixed ⊆ endo − fixed



1− auto − fixed⊆6= 1− endo − fixed

∩| ∩|

auto − fixed⊆6= endo − fixed

Example (Martino-V., 03; Ciobanu-Dicks, 06)

Let F3 = 〈a,b, c〉 and H = 〈b, cacbab−1c−1〉 6 F3. Then,H = Fix (a 7→ 1, b 7→ b, c 7→ cacbab−1c−1), but H is NOT the fixedsubgroup of any set of automorphism of F3.




∩| ‖ ? ∩| ‖ ?


Conjecture

For every S ⊆ End (Fn) (S ⊆ Aut (Fn)) there exists φ ∈ End (Fn)(φ ∈ End (Fn)) such that Fix (S) = Fix (φ).


Let S ⊆ End (Fn). Then, ∃φ ∈ 〈S〉 such that Fix (S) 6ff Fix (φ).

But... free factors of 1-endo-fixed (1-auto-fixed) subgroups need notbe even endo-fixed (auto-fixed).




∩| ‖ ? ∩| ‖ ?


Conjecture

For every S ⊆ End (Fn) (S ⊆ Aut (Fn)) there exists φ ∈ End (Fn)(φ ∈ End (Fn)) such that Fix (S) = Fix (φ).


Let S ⊆ End (Fn). Then, ∃φ ∈ 〈S〉 such that Fix (S) 6ff Fix (φ).

But... free factors of 1-endo-fixed (1-auto-fixed) subgroups need notbe even endo-fixed (auto-fixed).


Outline

1 Some history


3 Needed tools

4 The proof


Computing fixed subgroups

Proposition (Turner, 86)

There exists a pseudo-algorithm to compute fix of an endo.

Easy but is not an algorithm...

Theorem (Maslakova, 03)

Fixed subgroups of automorphisms of Fn are computable.

Difficult but it is an algorithm!

Conjecture

Fixed subgroups of endomorphisms of Fn are computable.









Conjecture










Conjecture



Deciding fixedness

In this talk, I’ll solve the two dual problems:

TheoremGiven H 6fg Fn, one can algorithmically decide whether

i) H is auto-fixed or not,ii) H is endo-fixed or not,

and in the affirmative case, find a finite family, S = {φ1, . . . , φm}, ofautomorphisms (endomorphisms) of Fn such that Fix (S) = H.

Conjecture

Given H 6fg Fn, one can algorithmically decide whetheri) H is 1-auto-fixed or not,ii) H is 1-endo-fixed or not,

and in the affirmative case, find one automorphism (endomorphism)φ of Fn such that Fix (φ) = H.


Deciding fixedness





Conjecture




Deciding fixedness





Conjecture




Outline

1 Some history


3 Needed tools

4 The proof


Fixed closures

DefinitionGiven H 6fg Fn, we define the (auto- and endo-) stabilizer of H,respectively, as

AutH(Fn) = {φ ∈ Aut (Fn) | H 6 Fix (φ)} 6 Aut (Fn)

andEndH(Fn) = {φ ∈ End (Fn) | H 6 Fix (φ)} 6 End (Fn)

DefinitionGiven H 6 Fn, we define the auto-closure and endo-closure of H as

a-Cl (H) = Fix (AutH(Fn)) > H

ande-Cl (H) = Fix (EndH(Fn)) > H


Fixed closures

DefinitionGiven H 6fg Fn, we define the (auto- and endo-) stabilizer of H,respectively, as

AutH(Fn) = {φ ∈ Aut (Fn) | H 6 Fix (φ)} 6 Aut (Fn)

andEndH(Fn) = {φ ∈ End (Fn) | H 6 Fix (φ)} 6 End (Fn)

DefinitionGiven H 6 Fn, we define the auto-closure and endo-closure of H as

a-Cl (H) = Fix (AutH(Fn)) > H

ande-Cl (H) = Fix (EndH(Fn)) > H


Main result

Theorem

For every H 6fg Fn, a-Cl (H) and e-Cl (H) are finitely generated andone can algorithmically compute bases for them.

Corollary

Auto-fixedness and endo-fixedness are decidable.

Observe that e-Cl (H) 6 a-Cl (H) but, in general, they are not equal.


Main result

Theorem


Corollary




Main result

Theorem


Corollary




Retracts

DefinitionA subgroup H 6 Fn is a retract if there exists a retraction, i.e. amorphism ρ : Fn → H which restricts to the identity of H.

Free factors are retracts, but there are more.

Observation

If H 6 Fn is a retract then r(H) 6 n (and, r(H) = n ⇔ H = Fn).

Observation (Turner)

It is algorithmically decidable whether a given H 6 Fn is a retract ornot.


Retracts



Observation





Retracts



Observation





The stable image

Definition

Let φ ∈ End (Fn). The stable image of φ is Fnφ∞ = ∩∞i=1Fnφ

i .


For every endomorphism φ : Fn → Fn,i) Fnφ

∞ is φ-invariant,ii) the restriction φ : Fnφ

∞ → Fnφ∞ is an isomorphism,

iii) Fnφ∞ is a retract.

iv) Fix (φ) 6 Fnφ∞.

Example: For φ : F2 → F2, a 7→ a, b 7→ b2, we have F2φ = 〈a,b2〉,F2φ

2 = 〈a,b4〉, F2φ3 = 〈a,b8〉, . . .. So, F2φ

∞ = 〈a〉 6ff F2.


The stable image

Definition


i .








2 = 〈a,b4〉, F2φ3 = 〈a,b8〉, . . .. So, F2φ

∞ = 〈a〉 6ff F2.


The stable image

Definition


i .








2 = 〈a,b4〉, F2φ3 = 〈a,b8〉, . . .. So, F2φ

∞ = 〈a〉 6ff F2.


Stallings’ graphs and intersections

Theorem (Stallings, 83)

For any free group Fn = F (A), there is an effectively computablebijection

{f.g. subgroups of Fn} ←→ {finite A-labeled core graphs}

TheoremGiven sets of generators for H,K 6fg Fn, one can algorithmicallycompute a basis for H ∩ K .


Stallings’ graphs and intersections

Theorem (Stallings, 83)

For any free group Fn = F (A), there is an effectively computablebijection

{f.g. subgroups of Fn} ←→ {finite A-labeled core graphs}

TheoremGiven sets of generators for H,K 6fg Fn, one can algorithmicallycompute a basis for H ∩ K .


Algebraic extensions

DefinitionAn extension of subgroups H 6 K 6 Fn is called algebraic, denotedH 6alg K , if H is not contained in any proper free factor of K . Write

AE(H) = {K 6 Fn | H 6alg K}.

Theorem (Takahasi, 51)

If H 6fg Fn then AE(H) is finite and computable (i.e. H has finitelymany algebraic extensions, all of them are finitely generated, andbases are computable from H).

TheoremEvery extension of subgroups H 6 K 6 Fn factors in a unique way asH 6alg L 6ff K 6 Fn.




AE(H) = {K 6 Fn | H 6alg K}.







AE(H) = {K 6 Fn | H 6alg K}.





Outline

1 Some history


3 Needed tools

4 The proof


The automorphism case

Theorem (McCool)

Let H 6fg Fn. Then AutH(Fn) is finitely generated (in fact, finitelypresented) and a finite set of generators (and relations) isalgorithmically computable from H.

Theorem

For every H 6fg Fn, a-Cl (H) is finitely generated and algorithmicallycomputable.

Proof. a-Cl (H) = Fix (AutH(Fn))= Fix (〈φ1, . . . , φm〉)= Fix (φ1) ∩ · · · ∩ Fix (φm). �



Theorem (McCool)


Theorem





Theorem (McCool)


Theorem




The endomorphism case

For the endomorphism case, a similar approach does not workbecause:

• we don’t know how to compute fix subgroups of endomorphisms

• H 6fg Fn does not imply that EndH(Fn) is finitely generated assubmonoid of End (Fn)













Example (Ciobanu-Dicks, 06)

Consider F3 = 〈a,b, c〉, the element d = ba[c2,b]a−1, and thesubgroup H = 〈a,d〉 6 F3. Clearly, the morphisms

ψ : F3 → F3 φ : F3 → F3 φnψ : F3 → F3a 7→ a a 7→ a a 7→ ab 7→ d b 7→ b b 7→ dc 7→ 1 c 7→ cb c 7→ dn

satisfy H 6 Fix (φnψ) for every n ∈ Z.With some computations, it can be shown that

EndH(F3) = {Id , φnψ | n ∈ Z}.

But, φmψ · φnψ = φmψ. Hence, EndH(F3) is not finitely generated.

Furthermore, a-Cl (H) = Fix (Id) = F3 and e-Cl (H) = Fix (ψ) = H.







































Theorem

For every H 6fg Fn, e-Cl (H) is finitely generated and algorithmicallycomputable.

Proof. Given H (in generators),Compute AE(H) = {H1,H2, . . . ,Hq}.Select those which are retracts, AE ret(H) = {H1, . . . ,Hr}(1 6 r 6 q).Write the generators of H as words on the generators of eachone of these Hi ’s, i = 1, . . . , r .Compute bases for a-Cl H1(H), . . . ,a-Cl Hr (H).Compute a basis for a-Cl H1(H) ∩ · · · ∩ a-Cl Hr (H).

Claim

a-Cl H1(H) ∩ · · · ∩ a-Cl Hr (H) = e-Cl (H).



Theorem



Claim




Theorem



Claim




Theorem



Claim




Theorem



Claim




Theorem



Claim




Theorem



Claim




Claim


Proof. Let us see that

r⋂i=1

⋂α ∈ Aut (Hi )H 6 Fix (α)

Fix (α) =⋂

β ∈ End (Fn)H 6 Fix (β)

Fix (β).

Take β ∈ End (Fn) with H 6 Fix (β).∃i = 1, . . . , r such that H 6alg Hi 6ff Fβ∞ 6ret F .Now, β restricts to an automorphism α : Hi → Hi .And, clearly, H 6 Fix (α) 6 Fix (β).Hence, we have "6”.



Claim



r⋂i=1


Fix (α) =⋂


Fix (β).




Claim



r⋂i=1


Fix (α) =⋂


Fix (β).




Claim



r⋂i=1


Fix (α) =⋂


Fix (β).




Claim



r⋂i=1


Fix (α) =⋂


Fix (β).




Claim



r⋂i=1


Fix (α) =⋂


Fix (β).




r⋂i=1


Fix (α) =⋂


Fix (β).

Take Hi ∈ AE ret(H), and α ∈ Aut (Hi) with H 6 Fix (α).Let ρ : F → Hi be a retraction, and consider the endomorphism,β : Fn

ρ→ Hiα→ Hi

ι↪→ Fn.

Clearly, H 6 Fix (α) = Fix (β).Hence, we have ">”. �



r⋂i=1


Fix (α) =⋂


Fix (β).


ρ→ Hiα→ Hi

ι↪→ Fn.




r⋂i=1


Fix (α) =⋂


Fix (β).


ρ→ Hiα→ Hi

ι↪→ Fn.




r⋂i=1


Fix (α) =⋂


Fix (β).


ρ→ Hiα→ Hi

ι↪→ Fn.




r⋂i=1


Fix (α) =⋂


Fix (β).


ρ→ Hiα→ Hi

ι↪→ Fn.



THANKS

Documents

Deciding endofixedness in free groupsepsem.upc.edu/~ventura/ventura/files/Newcastle-12-4-10.pdf · 2010. 4. 11. · 1.Some history Algorithmic results Needed tools The proof Train-tracks