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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,