Braid automorphisms of free groups

Braids and Braid Groups

1.5 Braid automorphisms of free groups

In this section we realize the braid groupB_n as a group of automorphisms of the free groupF_n onngeneratorsx₁, x₂, . . . , x_n.

1.5.1 Braid automorphisms ofFn

We say that an automorphismϕofFn is abraid automorphism if it satisfies the following two conditions:

(i) there is a permutationμ of the set{1,2, . . . , n} such thatϕ(x_k) is con- jugate inFn to x_μ(k) for allk∈ {1,2, . . . , n};

(ii) ϕ(x1x2· · ·xn) =x1x2· · ·xn.

To give examples of braid automorphisms ofFn, observe that an endomor- phism ofFn is entirely determined by its action on the generatorsx1, . . . , xn. It is straightforward to check that the following formulas define two mutually inverse braid automorphismsσiand σ_i⁻¹ ofFn fori= 1,2, . . . , n−1:

σi(xk) =

⎧⎪

⎨

⎪⎩

x_k+1 ifk=i, x⁻_k¹xk−1xk ifk=i+ 1,

xk otherwise,

σ_i⁻¹(xk) =

⎧⎪

⎨

⎪⎩

xkxk+1x⁻_k¹ ifk=i, xk−1 ifk=i+ 1,

xk otherwise.

Denote the set of braid automorphisms of Fn byBn. It follows from the definitions that the inverse of a braid automorphism and the composition of two braid automorphisms are again braid automorphisms. ThereforeBn is a group with respect to compositionϕψ=ϕ◦ψfor anyϕ, ψ∈B_n.

We now state the main theorem relating braids to braid automorphisms.

Theorem 1.31.The formula σi →σi withi= 1,2, . . . , n−1 defines a group isomorphism Bn→Bn.

The image ofβ ∈ Bn under the isomorphismBn → Bn will be denoted byβ. In the proof of Theorem 1.31 we shall give a direct definition ofβ. Yet another interpretation ofβwill be given in Section 1.6.

Theorem 1.31 gives a solution to the word problem in B_n. For a group defined by generators and relations, the word problem consists in finding an algorithm that allows one to decide whether a given word in the generators represents the neutral element of the group. By Theorem 1.31, a braidβ∈Bn

is equal to 1 if and only ifβ= id. The latter condition can be easily verified;

it suffices to check thatβ(xk) =xk for allk= 1,2, . . . , n.

Abelianizing the action ofBn ∼=Bn onFn, we obtain an action ofBn on the latticeFn/[Fn, Fn] =Zⁿ with basis ˙x1, . . . ,x˙n determined byx1, . . . , xn. The latter action is determined by the canonical projection π : Bn → Sn. Indeed, the automorphism of Zⁿ induced by σi is the transposition of the vectors ˙x_i,x˙_i+1. Therefore for anyβ ∈B_n, the automorphism ofZⁿ induced byβacts as the permutationπ(β) on the vectors ˙x₁, . . . ,x˙_n.

1.5.2 Proof of Theorem 1.31

The braid relations forσ₁, . . . ,σ_n−1∈B_n can be verified by a direct compu- tation (they follow also from further arguments in this paragraph). Therefore the formula σi → σi defines a group homomorphism Bn → Bn. We give now another definition of this homomorphism. Recall the natural inclusion ι : Bn → Bn+1, the group of pure braids Pn+1 ⊂ Bn+1, and the forgetting homomorphismfn+1:Pn+1→Pn. Ifβ∈Bnandu∈Un+1= Kerfn+1, then ι(β)u ι(β)⁻¹ ∈Pn+1 sincePn+1 is a normal subgroup ofBn+1. Moreover, it follows from the definition offn+1 that

ι(β)u ι(β)⁻¹ ∈U_n+1.

Therefore the formulau →ι(β)u ι(β)⁻¹ defines an automorphism of Un+1. We obtain thus a group homomorphism,ξ, from Bn to the group AutUn+1

of automorphisms ofUn+1. By Theorem 1.16, we can identify Un+1 withFn

by settingxk =Ak,n+1 ∈Un+1 fork= 1,2, . . . , n. Under this identification, ξ(β) = β for all β ∈ Bn. Indeed, it suffices to verify this equality for the generatorsσ1, . . . , σn−1ofBn. This amounts to checking the equalities

ι(σi)Ak,n+1ι(σi)⁻¹=

⎧⎪

⎨

⎪⎩

Ak+1,n+1 ifk=i,

A⁻_k,n+1¹ A_k−1,n+1A_k,n+1 ifk=i+ 1,

Ak,n+1 otherwise.

These equalities are verified by drawing the corresponding braid diagrams and checking that the diagrams on both sides present isotopic braids.

Let us prove the injectivity of the homomorphism β → β : Bn → Bn. Consider a braid β ∈ Bn such that β = 1. Abelianizing β, we obtain the identity automorphism of Un+1/[Un+1, Un+1]. Hence, π(β) = 1, so that

β ∈Pn ⊂Bn. By formula (1.6), β =β2β3· · ·βn, where βj ∈ Uj ⊂Pj ⊂Pn

forj= 2,3, . . . , n. Ifβ= 1, then take the largesti≤nsuch thatβi= 1. Then β = β₂β₃· · ·β_i. Since β = 1, we must haveξ(β) = 1, so thatι(β) ∈ P_n+1 commutes with all elements ofU_n+1and in particular withA_i,n+1. Note that β2, β3, . . . , βi−1 are braids on the leftmosti−1 strings. Therefore they com- mute with Ai,n+1. Hence βi commutes with Ai,n+1. By Corollary 1.23, the braids A1,i, . . . , Ai−1,i, Ai,i+1, . . . , Ai,n+1 are free generators of a free sub- group ofPn+1. We know thatβi commutes withAi,n+1 and lies in the group Ui ⊂Pi ⊂Pn+1 generated by A1,i, . . . Ai−1,i. This is possible only ifβi = 1, which contradicts the choice ofi. Hence,β = 1.

Let us prove the surjectivity of the homomorphism β → β: Bn → Bn. Letϕbe a nontrivial braid automorphism ofFn. Suppose that

ϕ(xk) =Akx_μ(k)A⁻_k¹,

wherek= 1,2, . . . , nandA_k is a word in the alphabetx^±₁¹, . . . , x^±_n¹. We can always chooseA_k so that the product A_kx_μ(k)A⁻_k¹ is reduced, that is, does not contain consecutive entriesx_rx⁻_r¹ or x⁻_r¹x_r. By the definition of a braid automorphism,

A1x_μ(1)A⁻₁¹A2x_μ(2)A⁻₂¹· · ·Anx_μ(n)A⁻_n¹=x1x2· · ·xn. (1.11) We claim that there existj∈ {1,2, . . . , n−1}and a wordA(possibly empty) inx^±₁¹, . . . , x^±_n¹satisfying one of the following two conditions:

(a) we have an equality of wordsAj =Aj+1x_μ(j+1)A, (b) we have an equality of wordsAj+1 =Ajx⁻_μ(j)¹ A.

This claim will imply thatϕlies in the image of the homomorphismβ→β. To see this, define thelength of ϕto be the sum of the letter lengths of the wordsA_kx_μ(k)A⁻_k¹overk= 1,2, . . . , n. If (a) holds, then the homomorphism

ϕσ_j=ϕ◦σ_j :F_n→F_n

can be computed as follows. Bothϕandϕσj have the same effect onxk for k=j, j+ 1 and

ϕσj(xj) =ϕ(xj+1) =Aj+1x_μ(j+1)A⁻_j+1¹ , ϕσj(xj+1) =ϕ(x⁻_j+1¹ xjxj+1)

=A_j+1x⁻_μ(j+1)¹ A⁻_j+1¹ A_jx_μ(j)A⁻_j¹A_j+1x_μ(j+1)A⁻_j+1¹

=A_j+1x⁻_μ(j+1)¹ A⁻_j+1¹

×Aj+1x_μ(j+1)A x_μ(j)A⁻¹x⁻_μ(j+1)¹ A⁻_j+1¹ Aj+1x_μ(j+1)A⁻_j+1¹

=Aj+1A x_μ(j)A⁻¹A⁻_j+1¹ .

The word Aj+1A is shorter than Aj = Aj+1xμ(j+1)A. Therefore ϕσj has shorter length than ϕ. Similarly, if (b) holds, then ϕσ_j⁻¹ has shorter length

thanϕ. This implies thatϕcan be reduced to the identity by repeated com- position with appropriateσj or σ_j⁻¹. Thus, ϕ is a power product of the σj. Henceϕlies in the image of the homomorphismβ → β.

It remains to prove the claim stated above. Let us call the term x_μ(k) appearing in the middle ofAkx_μ(k)A⁻_k¹special. Each letterx1, . . . , xnappears as a special term on the left-hand side of (1.11) exactly once. Equality (1.11) implies that the left-hand side of (1.11) reduces to the right-hand side after all possible free reductions (i.e., cancellationsxrx⁻_r¹=x⁻_r¹xr= 1). Suppose that a special termxμ(k) is canceled with a letter x⁻_μ(k)¹ during these reductions.

Thisx⁻_μ(k)¹ cannot come from the word Akx_μ(k)A⁻_k¹, which was assumed to be reduced. If thisx⁻_μ(k)¹ comes fromA⁻_k−¹₁, then we must have an equality of wordsA⁻_k−¹₁=B x⁻_μ(k)¹ A⁻_k¹for some wordB. Then (a) holds forj=k−1 and A=B⁻¹. If the letterx⁻_μ(k)¹ canceling the special termx_μ(k) comes from the right of the special termx_μ(k+1), then we must have (a) forj =k. Similarly, ifx⁻_μ(k)¹ comes fromA_k+1or from the left of the special termx_μ(k−1), then (b) holds. If the special terms on the left-hand side of (1.11) do not cancel with other letters, then we must haveμ(k) = k for all k, A1 and An are empty words, and each pairA⁻_k¹A_k+1 cancels out, so thatA_k =A_k+1 for allk. Then

ϕ= id, which contradicts our choice ofϕ.

Remark 1.32.Theorem 1.31 yields another proof of the residual finiteness ofB_n (Corollary 1.21). Indeed, by the Baumslag–Smirnov theorem [Bau63], [Smi63], the group of automorphisms of an arbitrary residually finite group is residually finite. SinceFn is residually finite, its group of automorphisms and all subgroups of this group are residually finite.

Exercise 1.5.1.For any integerr, letσ^(r)_i be the automorphism ofFndefined by the same formulas asσi with the only difference that

σ^(r)_i (xi+1) =x⁻_i+1^rxix^r_i+1.

Verify thatσ₁^(r),σ^(r)₂ , . . . ,σ_n^(r)₋₁satisfy the braid relations. (The resulting rep- resentationB_n→Aut(F_n) is faithful for allr= 0; see [Shp01].)

Exercise 1.5.2.Let F2n be the free group on 2n generators a1, . . . , an, b1, . . . , bn. Forj= 1, . . . ,2n+ 1, define an automorphismσ_jofF2n as follows.

For evenj = 2i, set σ_j(ai) =b⁻_i¹ai, σ_j(ak) =ak fork=i, andσ_j(bk) =bk

for all k. If j is odd, then σ_j(ak) = ak for all k. Also, σ₁(b1) = a1b1 and σ₁(bk) = bk for k > 1; σ_2n+1(bn) = bnan and σ_2n+1 (bk) = bk for k < n.

For other oddj= 2i+ 1, setσ_j(bi) =biaia⁻_i+1¹, σ_j(bi+1) =ai+1a⁻_i¹bi+1, and σ_j(bk) =bk fork =i, i+ 1. Verify thatσ₁, . . . , σ_2n+1 satisfy the braid rela- tions. Check that the corresponding group homomorphismB2n+2→Aut(F2n) sends the center ofB2n+2to the identity. (Forn= 1, we recover the formulas of Exercise 1.1.10.)