Twisted conjugacy in braid groups

Juan González-Meneses

Universidad de Sevilla

Paris, 17-20 september 2008.

Tresses à ParisRencontres Parisiennes du GDR Tresses

Joint with E. Ventura.

Introduction Conjugacy problem

In a group G:

Conjugacy Decision Problem:

Conjugacy Decision Problem:

Conjugation:

Conjugacy Search Problem: Conjugacy Search Problem:

Determine whether two elements are conjugate.

Find a conjugating element for two given conjugate elements.

a ~ b a ~ b

Introduction Twisted conjugacy problem

In a group G:

Twisted Conjugacy Decision Problem: Twisted Conjugacy Decision Problem:

Twisted Conjugacy Search Problem:

Twisted Conjugacy Search Problem:

Conjugation:Twisted

Fixed automorphism.

Determine whether two elements are twisted conjugate.

Find a conjugating element for two given twisted conjugate elements.

Reidemeister (1936)

Conjugacy problem

Introduction Twisted conjugacy problem

If f : G G is an inner automorphism:

Twisted conjugacy problem

Just need to focus on representatives of Out(G) = Aut(G)/Inn(G).

One can determine, given x,y 2 F,whether x ~ f(y), for some f 2 AG.

Introduction Motivation

Bogopolski, Martino, Ventura, 2008.

H = f.g.-freef.g.-free f.g.-t.f.-hyperbolicf.g.-t.f.-hyperbolic …

Solvable conj. problem.Solvable twisted conj. problem.

F = f.g.-abelianf.g.-abelian f.g.-freef.g.-free …

G has solvable conj. problem , AG < Aut (F ) is orbit decidableG has solvable conj. problem , AG < Aut (F ) is orbit decidable

Can we put braid groups here? (Out Bn is finite)

?

Braid groups Normal form

Bn: Braid group on n strands.

Left normal form:

Each factor is a simple element. (permutation braid)

Canonical length = No. of factors.

Braid groups Automorphisms

Automorphisms of Bn: (Dyer-Grossman, 1981)

Just need to solve the twisted conjugacy problem for .

Braid groups Twisted conjugacy

Twisted conjugation for :

( c written backwards )

c

Braid groups Twisted conjugacy

Twisted conjugation for :

This is the twisted conjugation we will consider.

c

Braid groups Examples

Twst conj Twst conj

are conjugate,

are twisted conjugate.

How to solve the twisted conjugacy problems?

but not twisted conjugate.

Back to the conjugacy problem ElRifai-Morton’s solution

(ElRifai-Morton, 1988)Algorithm to solve the conjugacy problem.

Compute a finite set, invariant of the conjugacy class.

SSS(x) = { conjugates of x, of minimal canonical length}SSS(x) = { conjugates of x, of minimal canonical length}

One can compute SSS(x) using the following:

Then u and v can be joined through conjugations by simple elements,

Theorem (Elrifai-Morton, 1988): Let u,v 2 Bn conjugate,

where every intermediate conjugate w has

Back to the conjugacy problem ElRifai-Morton’s solution

u vcc1 c2 cr

(can assume positive)

(left normal form)

u vc1 w1

c2 w2 wr-1

cr

Each ci is simpleEach ci is simple

Then u and v can be joined through conjugations by simple elements,

Theorem (Elrifai-Morton, 1988): Let u,v 2 Bn conjugate,

where every intermediate conjugate w has

Back to the conjugacy problem ElRifai-Morton’s solution

If no new element is found, SSS(x) is computed.

Conjugate by all simple elemets...

…keeping elemets of minimal length.

x

SSS(x)

Computing SSS(x):

This solves the conjugacy problem.

Twisted conjugacy problem Solution

For every x2 Bn, xp is positive for p big enough.For every x2 Bn, xp is positive for p big enough.

Twst conjPositive!

Every braid is twisted conjugate to a positive braid.Every braid is twisted conjugate to a positive braid.

First idea: Restrict to positive braids.

Twisted conjugacy problem Solution

Every braid is twisted conjugate to a positive braid.Every braid is twisted conjugate to a positive braid.

First idea: Restrict to positive braids.

The set of positive twisted-conjugates of x is infinite. The set of positive twisted-conjugates of x is infinite.

(braid) 11 33 22 55 ……

But…

Twisted conjugacy problem Solution

A positive braid x is palindromic-free if it cannot be written as:A positive braid x is palindromic-free if it cannot be written as:

Every positive braid is twisted conjugate to a palindromic-free one.Every positive braid is twisted conjugate to a palindromic-free one.

12321 232 3

212 132123

=

Second idea: Restrict to positive, palindromic-free braids.

Twisted conjugacy problem Solution

Every positive braid is twisted conjugate to a palindromic-free one.Every positive braid is twisted conjugate to a palindromic-free one.

Second idea: Restrict to positive, palindromic-free braids.

The set of positive, palindromic-free twisted-conjugates of x can be infinite. The set of positive, palindromic-free twisted-conjugates of x can be infinite.

But…

Example:

k

k

These braids are palindromic-free, for all k.

They are twisted conjugate.

kk

Positive,MPF(x)= Twisted conjugates of x which are: Palindromic-free of minimal length.

Positive,MPF(x)= Twisted conjugates of x which are: Palindromic-free of minimal length.

Twisted conjugacy problem Solution

Third idea: Restrict to positive, palindromic-free braids, of minimal length

This is a finite set, invariant of the twisted-conjugacy class.

How to compute it?

Computing MPF(x), we solve the twisted.conjugacy problem.Computing MPF(x), we solve the twisted.conjugacy problem.

For usual conjugacy problem…

Computing MPF(x)

u vc1 w1

c2 w2 wr-1

cr

…simple conjugations.

For twisted conjugacy problem:

121

1

212 =

2

2

1 2

2 11

Simple twisted-conjugation:x y

simple simple

Computing MPF(x)

Then u and v can be joined through simple twisted-conjugations,

u vw1 w2 wr-1

Then u and v can be joined through conjugations by simple elements,

Theorem (Elrifai-Morton, 1988): Let u,v 2 Bn conjugate,

where every intermediate conjugate w has

Theorem (GM-Ventura, 2008): Let u,v 2 Bn twisted conjugate,

where every intermediate twisted-conjugate w has

All palindromic-free

Computing MPF(x)

Ingredients of the proof:

u v1

Then use Elifai-Morton’s Theorem.

Conclusion

H = f.g.-freef.g.-free f.g.-t.f.-hyperbolicf.g.-t.f.-hyperbolic …

G has solvable conjugacy problemG has solvable conjugacy problem

(decision & search)

Since AG < Aut (Bn) is orbit decidable,

and Bn has solvable twisted conjugacy problem,