On certain permutation representations of the braid group. Part II
aa r X i v : . [ m a t h . G R ] A p r On certain permutation representations of thebraid group. Part II
Valentin Vankov IlievSection of Algebra,Institute of Mathematics and Informatics,Bulgarian Academy of Sciences, 1113 Sofia, Bulgariae-mail: [email protected] 20, 2018
Abstract
In [1] we find certain finite homomorphic images of Artin braid groupinto appropriate symmetric groups, which a posteriori are extensions ofthe symmetric group on n letters by an abelian group. The main theoremof this paper characterizes completely the extensions of this type that aresplit. Key words : Artin braid group, permutation representation, split extension
This paper is a natural continuation of [1] and we use freely the terminologyand notation used there.In [1, Theorem 9,(iii)] we prove that the braid-like permutation group B n ( σ )is an extension of the symmetric group Σ n by the abelian group A n ( q ) from[1, (3)], and, moreover, give a sufficient condition for this extension to split.Theorem 2 is the final version of [1, Theorem 9, (iii)], and proves a necessaryand sufficient condition for B n ( σ ) to be a semi-direct product of the symmetricgroup Σ n and the abelian group A n ( q ). Making use of [1, Theorem 9,(iii)], we have that the group B n ( σ ) is an extensionof the symmetric group Σ n by the abelian group A n ( q ) from [1, (3)], where q = q , and the corresponding monodromy homomorphism is defined in [1,Proposition 11].Any (set-theoretic) section ρ : Σ n → B n ( σa ) of the surjective homomorphism π from the short exact sequence in the proof of [1, Theorem 9,(iii)] produces an1lement ( a , . . . , a n − ) of the direct sum ` n − s =1 A n ( q ) via the rule ρ ( θ s ) = σ s a s , s = 1 , . . . , n − A be a (left)Σ n -module, and let m : Σ n → Aut ( A ) , θ s ι s , s = 1 , . . . , n − , be the corresponding structure homomorphism. Let us set J s = ι s + 1 , s = 1 , . . . , n − ,I r,r +1 = ι r,r +1 + ι r + ι r +1 , r = 1 , . . . , n − , where ι r,r +1 = ι r ι r +1 ι r = ι r +1 ι r ι r +1 . Note that J s and I r,r +1 are endomor-phisms of the abelian group A .Let G be a braid-like group, let g , . . . , g n − be a generating set of G thatsatisfies the braid relations, and let γ : B n → G be the corresponding surjectivehomomorphism. Let ν : B n → Σ n be the analogous surjective homomorphism,corresponding to the set of generators θ . . . , θ n − of the symmetric group Σ n .Moreover, let G be an extension of Σ n by a Σ n -module A ,0 −−−−→ A i −−−−→ G π −−−−→ Σ n −−−−→ , (1)where ν = π ◦ γ . We identify A with the normal abelian subgroup i ( A ) ≤ G and for any s = 1 , . . . , n − g s ∈ A by f s . Lemma 1
The extension (1) splits if and only if the linear system J s ( a s ) = − f s s = 1 , . . . , n − I r,r +1 ( a r − a r +1 ) = 0 r = 1 , . . . , n − , (2) has a solution a , . . . , a n − ∈ A . Proof:
The extension (1) splits if and only if there exists a section ρ of π , whichis a homomorphism of groups. By [2, Theorem 4.1], this comes to showing thatthere exist a , . . . , a n − ∈ A , such that the elements g s a s , s = 1 , . . . , n − G are involutions, and satisfy the braid relations. The equalities( g s a s ) = 1 are equivalent to g s ι s ( a s ) a s = 1 . (3)The equalities g r a r g r +1 a r +1 g r a r = g r +1 a r +1 g r a r g r +1 a r +1 are equivalent to g r g r +1 g r ( ι r ι r +1 )( a r ) ι r ( a r +1 ) a r = g r +1 g r g r +1 ( ι r +1 ι r )( a r +1 ) ι r +1 ( a r ) a r +1 . Since g s , s = 1 , . . . , n −
1, satisfy the braid relations, we have( ι r ι r +1 )( a r ) ι r ( a r +1 ) a r = ( ι r +1 ι r )( a r +1 ) ι r +1 ( a r ) a r +1 . (4)2riting (3) and (4) additively, we have that the elements a , . . . , a n − ∈ A satisfy the linear system J s ( a s ) = − f s ,s = 1 , . . . , n − ι r ι r +1 )( a r ) + ι r ( a r +1 ) + a r = ( ι r +1 ι r )( a r +1 ) + ι r +1 ( a r ) + a r +1 ,r = 1 , . . . , n − . After substituting a r and a r +1 from the first group equations into the secondone for r = 1 , . . . , n −
2, we obtain the linear system (2).Now, we introduce some notation. Let Z n → ( Z / ( q )) n , a ¯ a, be the canonical surjective homomorphism of abelian groups. Note that theelements ¯ f , . . . , ¯ f n − , ¯ f n , form a basis for the Z / ( q )-module ( Z / ( q )) n . Theelements ¯ f , . . . , ¯ f n − , ¯ g , . . . , ¯ g n − , generate A n ( σ ) as a subgroup of ( Z / ( q )) n ,and the group A n ( σ ) contains the elements ¯ h , . . . ,¯ h n . Moreover, there exists anisomorphism of abelian groups A n ( σ ) ≃ Z ¯ f / ( q ) ¯ f a · · · a Z ¯ f n − / ( q ) ¯ f n − a Z ¯ h n / ( q )¯ h n . On the other hand, the group ( Z / ( q )) n has a structure of Σ n -module, obtainedvia the isomorphism ( Z / ( q )) n → h τ i ( n ) , and A n ( σ ) is its Σ n -submodule. Throughout the end of the paper we consider J s and I r,r +1 as endomorphisms of the Z / ( q )-module ( Z / ( q )) n . Theorem 2
The extension from [1, Theorem 9, (iii)] splits if and only if doesnot divide q . Proof:
We consider several cases.Case 1. q is an odd number. We set a s = − ¯ f s , s = 1 , . . . , n −
1, where istaken modulo q . Then the first n − r = 1 , . . . , n −
2, we have I r,r +1 ( a r − a r +1 ) = − I r,r +1 ( ¯ f r − ¯ f r +1 ) , and I r,r +1 ( ¯ f r ) = I r,r +1 ( ¯ f r +1 ) = ¯ f r + ¯ f r +1 + ¯ g r . Therefore, the last n − q ≡ q = 2 q , where q is an odd number. Let us consider the linearsystem (2) over the Z / ( q )-module ( Z / ( q )) n . We are looking for solutions of (2)in the abelian group A n ( σ ) considered as a Z / ( q )-submodule of ( Z / ( q )) n . Thereduction of A n ( σ ) modulo q is the Z / ( q )-module ( Z / ( q )) n , and in accordwith Case 1, the reduction of the linear system (2) modulo q is consistent.Now, let us reduce the system (2) modulo 2. The reduction of A n ( σ ) modulo 2is the Z / (2)-linear space M = Z / (2)) n − , and a r = P n − t = r ¯ f t r = 1 , . . . , n − a n − = ¯ f n − a n − = ¯ f n − . is a solution of the linear system (2) in M . Therefore the reduction of the linearsystem (2) modulo 2 is consistent over A n ( σ ). Thus, the linear system (2) overthe Z / ( q )-module ( Z / ( q )) n is consistent, too.Case 3. q ≡ a = P n − t =1 x t ¯ f t + y ¯ h n be a generic element of the abelian group A n ( σ ).We have J ( a ) = (2 x + x ) ¯ f + 2 n − X t =3 [( − t − x + 2 x t ] f t + [( − n − x + 2 y ]¯ h n , and the equality J ( a ) = − f yields 2 x + x +1 ≡ q ), and 2( − n − x +4 y ≡ q ), which is a contradiction. References [1] V. V. Iliev, On certain permutation representations of the braid group,arXiv:0910.1727v2 [math.GR].[2] Ch. Kassel, V. Turaev,