Twisted conjugacy in braid groups Juan González-Meneses Universidad de Sevilla Paris, september Tresses à Paris Rencontres Parisiennes du GDR Tresses Joint with E. Ventura.

Introduction Conjugacy problem In a group G : Conjugacy Decision Problem: Conjugation: Conjugacy Search Problem: Determine whether two elements are conjugate. Find a conjugating element for two given conjugate elements. a ~ b

Introduction Twisted conjugacy problem In a group G : Twisted Conjugacy Decision 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 A G. Introduction Motivation Bogopolski, Martino, Ventura, H = f.g.-free f.g.-t.f.-hyperbolic … Solvable conj. problem. Solvable twisted conj. problem. F = f.g.-abelian f.g.-free … G has solvable conj. problem, A G < Aut ( F ) is orbit decidable Can we put braid groups here? ( Out B n is finite) ?

Braid groups Normal form B n : 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 B n : (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 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 } 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 B n conjugate, where every intermediate conjugate w has

Back to the conjugacy problem ElRifai-Morton’s solution u v c c 1 c 2  c r (can assume positive) (left normal form) u v c1c1 w1w1 c2c2 w2w2  w r -1 crcr Each c i is simple Then u and v can be joined through conjugations by simple elements, Theorem (Elrifai-Morton, 1988): Let u, v 2 B n 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 B n, x  p is positive for p big enough. Twst conj Positive! 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. First idea: Restrict to positive braids. The set of positive twisted-conjugates of x is infinite. (braid) 11 11 33 33 22 22 55 55 … … But…

Twisted conjugacy problem Solution A positive braid x is palindromic-free if it cannot be written as: Every positive braid is twisted conjugate to a palindromic-free one. 1232112321 232232 33 212212 11 3212332123 = Second idea: Restrict to positive, palindromic-free braids.

Twisted conjugacy problem Solution 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. 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. 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.

For usual conjugacy problem… Computing MPF( x ) u v c1c1 w1w1 c2c2 w2w2  w r -1 crcr …simple conjugations. For twisted conjugacy problem: 121121 11 212212 = 22 22 11 22 22 11 11 Simple twisted-conjugation: x y simple

Computing MPF( x ) Then u and v can be joined through simple twisted-conjugations, u v w1w1 w2w2  w r -1 Then u and v can be joined through conjugations by simple elements, Theorem (Elrifai-Morton, 1988): Let u, v 2 B n conjugate, where every intermediate conjugate w has Theorem (GM-Ventura, 2008): Let u, v 2 B n twisted conjugate, where every intermediate twisted-conjugate w has All palindromic-free

Computing MPF( x ) Ingredients of the proof: u v 1 Then use Elifai-Morton’s Theorem.

Conclusion H = f.g.-free f.g.-t.f.-hyperbolic … G has solvable conjugacy problem (decision & search) Since A G < Aut ( B n ) is orbit decidable, and B n has solvable twisted conjugacy problem,