TORSION-FREE QUOTIENTS OF BRAID GROUPS

Stephen P. Humphries

Department of Mathematics Brigham Young University Provo, UT 84602, U.S.A. email steve@math.byu.edu

Abstract. In this paper we show that the braid groups B_{n}, n ≤ 6, have proper torsion-free non-abelian quotients. We also show that for n ≤ 6 the homotopy braid groups are torsion-free.

AMS Classification : 20F36.

§1. Introduction.

In this paper we show that the braid groups B_{n}, 2 < n ≤ 6, have a proper, torsion-free, non-abelian factor group. This answers questions of Lin [L, §§0.9.e, f; see also Remark 7.3 which conjectures that there are no proper torsion-free quotients of the commutator group B′_{n} for n > 4]. This problem is also proposed in the ‘Open problems in combinatorial group theory’ list [P, problem B6]. The result is wellknown for n = 3, 4, but we also produce new examples in these cases. The method of proof will be to describe (following [Hu]) a representation of B_{n} on a polynomial algebra R_{n+1} and show that there is a B_{n}-invariant ideal I_{n+1} of R_{n+1} such that the action of B_{n} on R_{n+1}/I_{n+1} is torsion free, non-abelian and finite by nilpotent.

For R a commutative ring with identity we let R_{n}[a_{ij}] denote the polynomial algebra

R[a_{12}, a_{13}, …, a_{1n}, a_{21}, a_{23}, …, a_{2n}, …, a_{n1}, …, a_{n,n−1}]

where the a_{ij} are commuting indeterminates. We put a_{ii} = 0 for convenience. Let the group H(n) = < t_{ij}|1 ≤ i ≠ j ≤ n > act on R_{n}[a_{ij}] as follows: for λ ∈ ℤ, t_{kj}^{λ}(a_{rs}) = b_{rs}, where

b_{ij} = a_{ij} + λa_{ik}a_{kj}, if i ≠ j;
b_{ji} = a_{ji} − λa_{jk}a_{ki}, if i ≠ j;
b_{hi} = a_{hi},  if h ≠ j, i ≠ j.  (1.1)

Here t_{kj} = t_{kj}^{1} and (t_{kj})^{−1} = t_{kj}^{−1}. For i = 1, …, n we will put t_{ii} = id for convenience. Let the symmetric group Σ_{n} act on R_{n}[a_{ij}] by permutation of subscripts

Typeset by AMS-TEX

1991 Mathematics Subject Classification. 20F36, 13F50.

Key words and phrases. Braid groups.

and let Ĥ(n) = < H(n), Σ_{n} >. Then one checks that H(n) is normal in Ĥ(n) (in fact for σ ∈ Σ_{n} we have σ^{−1}t_{ij}σ = t_{(i)σ(j)σ}) and there is a split exact sequence

1 → H(n) → Ĥ(n) ⟶^{π} Σ_{n} → 1.

In fact Ĥ(n) is the group of symmetric automorphisms of the free group F_{n} = < x_{1}, …, x_{n} > of rank n ([Mc], [Hu]). In this situation (x_{j})t_{ij} = x_{i}x_{j}x_{i}^{−1} and (x_{k})t_{ij} = x_{k} for k ≠ j.

The action of Ĥ(n) on R_{n}[a_{ij}] extends to an action of Ĥ(n) on R_{n+r}[a_{ij}], r ≥ 0, in the obvious way. This action of H(n) on R_{n+r}[a_{ij}] is only faithful if r > 0, otherwise the kernel is the free normal subgroup Inn(n) generated by the inner automorphisms, t_{i1}t_{i2} … t_{in}, for (i = 1, …, n) [Hu, Theorem 2.5].

Note that the braid groups B_{n}, having standard generators σ_{1}, …, σ_{n−1} and presentation

σ_{i}σ_{i+1}σ_{i} = σ_{i+1}σ_{i}σ_{i+1}, for i = 1, …, n − 2;
σ_{i}σ_{j} = σ_{j}σ_{i},  for |i − j| > 1,

are subgroups of Ĥ(n) and the pure braid groups P_{n} = H(n) ∩ B_{n} = ker(π|_{B_{n}}) are subgroups of H(n) [Bi, p. 30]. In fact σ_{i} = t_{i+1i}(i, i + 1) if B_{n} acts on the right. In what follows we will also need to notice that this implies that π|_{B_{n}} : B_{n} → Σ_{n} is onto. The braid groups are well-known to be torsion free [FN,D]; see also [M] for other braid groups. In order to follow the arguments of this paper it is not necessary to understand where the representation of B_{n} in Aut(R_{n}) comes from (as described in [Hu]; but see also [M1, p. 197]). It is only necessary to realise that this is a representation, which fact is easily checked.

We consider the action of B_{n} on a quotient of R_{n+1}, as follows. To any sequence i_{1}, i_{2}, …, i_{r} ∈ {1, …, n+1} we associate the cycle a_{i_{1}i_{2}}a_{i_{2}i_{3}} … a_{i_{r−1}i_{r}}a_{i_{r}i_{1}} ∈ R_{n+1}. Such a cycle is simple if all the i_{j} are distinct. Let Y_{n+1} denote the subring generated by all simple cycles in R_{n+1}. Then one checks using (1.1) that Y_{n+1} is B_{n}-invariant. Let I_{n+1} be the ideal of Y_{n+1} generated by all cycles which are not simple i.e. I_{n+1} is generated by all elements y_{1}y_{2} where y_{1}, y_{2} are cycles. Then I_{n+1} is B_{n}-invariant and we let V_{n+1} = Y_{n+1}/I_{n+1}. Thus B_{n} acts on V_{n+1}, which is a finite-dimensional R_{n}-module. We further note that V_{n+1} has a B_{n}-invariant submodule which we denote by U_{n+1} and which is generated by all cycles which contain n + 1 exactly once as an index. Let ρ_{n} : B_{n} → GL(U_{n+1}) denote the representation so obtained.

Theorem 1.1. For n ≤ 6 the image of B_{n} in GL(U_{n+1}) is a torsion-free, nonabelian, finite by nilpotent group.

In fact what we prove is

Theorem 1.2. Let α ∈ B_{n} and let τ = π(α). If α ∈ P_{n}, then either ρ(α) = 1 or ρ(α) has infinite order. Assume that the order of the permutation τ is divisible by 2, 3 or 5. Then the image ρ(α) in GL(U_{n+1}) has infinite order.

Theorem 1.2 clearly imples Theorem 1.1.

The fact that B_{4} has such a proper torsion-free non-abelian image follows from the well-known fact that B_{3} is torsion-free and that there is an epimorphism B_{4} → B_{3} defined on generators by σ_{1} ↦ σ_{1}, σ_{2} ↦ σ_{2}, σ_{3} ↦ σ_{1}. Further, B̂_{3}, the three strand homotopy braid group (see §4) is such a quotient of B_{3}.

§2. Proof of Theorem 1.2

First we note that if α ∈ B_{n} and τ = π(α), then by (1.1) we have

(a_{ij})α = a_{(i)τ(j)τ} + terms of degree greater than 1.

Thus if y ∈ Y_{n+1} is a cycle, then each monomial of (y)α − (y)τ has degree at least 1 + degree(y). Choose a basis [b_{1}], …, [b_{r}] of classes of cycles for U_{n+1} so that the degrees are non-decreasing. Then the above gives:

Lemma 2.1. Relative to this basis each matrix ρ(α) is block upper triangular:

ρ(α) = ( \begin{matrix}X_{22} & X_{23} & … & X_{2 n+1} \\ 0 & X_{33} & … & X_{3 n+1} \\ ⋮ & ⋮ & ... & ⋮ \\ 0 & 0 & … & X_{n+1 n+1} \end{matrix} ) .  (2.1)

Here the X_{ii} give the permutation action of α on the cycles of length i in U_{n+1}. Further, if α ∈ P_{n}, then ρ(α) is upper-triangular with 1s on the diagonal. □

From Lemma 2.1 it follows that if β ∈ P_{n} and ρ(β) ≠ id, then ρ(β) has infinite order. For a group G let G_{k} denote the kth term of the lower central series of G so that G_{1} = G and G_{k+1} = [G, G_{k}] for k ≥ 1. Then we also have

Lemma 2.2. For all n the subgroup (P_{n})_{n} is in the kernel of ρ_{n}. □

Now suppose that π(α) = τ ≠ id. Then we wish to show that ρ(α)^{k} is not the identity for all k ≥ 1. Let q be the order of τ. Then by Lemma 2.1 it will suffice to show that ρ(α)^{q} is not the identity. Further, if q = pr is not prime (but p is), then we can clearly replace α by α^{r} so as to be able to assume that q is prime.

Assume first that q = 2 and that n ≥ 2 is arbitrary. Then by conjugating by an element of B_{n} we may assume that α = σ_{1}β, σ_{1}σ_{3}β, σ_{1}σ_{3}σ_{5}β, etc, where β ∈ P_{n}. Now note that if m ≥ 3 and if m is an index in a cycle y ∈ Y_{n+1}, then m or m ± 1 occurs as an index in every cycle of (y)α depending in whether m is odd or even. Thus we may consider the action of α on the quotient of U_{n+1} by the ideal generated by all cycles having an index m in them with 2 < m < n + 1. This amounts to pulling out all the mth braid strands for these values of m (or to looking at the submodule of U_{n+1} with indices 1, 2, n + 1). In particular we may now assume that α = σ_{1}β where β is now a pure braid on the strands 1, 2, n+1 i.e. β = σ_{1}^{2k} for some k ∈ ℤ. We show that the action of α on this ring has infinite order. By renumbering we may assume that we are in the case n + 1 = 3, and further that α = σ_{1}β where β = σ_{1}^{2k}. Thus ρ(α^{2}) = ρ(σ_{1}^{2+4k}) and one checks that the only way that ρ(σ_{1}^{m}) can be the identity is if m = 0. Thus ρ(α^{2}) ≠ id and this case is complete.

Now consider the case q = 3. Then by conjugating we may assume that α = (σ_{1}σ_{2})β, (σ_{1}σ_{2})(σ_{4}σ_{5})β, (σ_{1}σ_{2})(σ_{4}σ_{5})(σ_{7}σ_{8})β, etc, where β ∈ P_{n}. As in the above case we may reduce to the situation where α = σ_{1}σ_{2}β and n + 1 = 4. Let A_{ij}, 1 ≤ i < j ≤ n be the standard generators for P_{n} [Bi] so that

A_{ij} = σ_{i}σ_{i+1} … σ_{j−1}σ_{j}^{2}σ_{j−1}^{−1} … σ_{i+1}^{−1}σ_{i}^{−1}.

Then relative to the following basis

a_{14}a_{41}, a_{24}a_{42}, a_{34}a_{43}, a_{12}a_{24}a_{41}, a_{13}a_{34}a_{41}, a_{21}a_{14}a_{42}, a_{23}a_{34}a_{42}, a_{31}a_{14}a_{43},
a_{32}a_{24}a_{43}, a_{12}a_{23}a_{34}a_{41}, a_{31}a_{12}a_{24}a_{43}, a_{21}a_{13}a_{34}a_{42},
a_{13}a_{32}a_{24}a_{41}, a_{32}a_{21}a_{14}a_{43}, a_{23}a_{31}a_{14}a_{42}

the matrices that we get are:

ρ(σ_{1}) = ( \begin{matrix}0 & 1 & 0 & 1 & 0 & −1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 1 & 0 & −1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 1 & 0 & 0 & −1 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 \end{matrix} ) ,

ρ(σ_{2}) = ( \begin{matrix}1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 & 0 & 1 & 0 & −1 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & −1 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & −1 & 1 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 \end{matrix} ) .

If we let h = σ_{1}σ_{2} and z = A_{12}A_{23}A_{12}^{−1}A_{23}^{−1}, then we have the following relations (where a, b ∈ ℤ):

[z, h] = [z, A_{12}] = [z, A_{23}] = [z, A_{13}] = 1;
z = [A_{13}, A_{12}] = [A_{23}, A_{13}];
[A_{12}^{a}, A_{23}^{b}] = z^{ab}, [A_{13}^{a}, A_{12}^{b}] = z^{ab}, [A_{23}^{a}, A_{13}^{b}] = z^{ab}
hA_{12}h^{−1} = A_{23}, hA_{23}h^{−1} = A_{13}, hA_{13}h^{−1} = A_{12}.

Now we note that if I is the identity matrix and B_{ij} = ρ(A_{ij}) − I, then we have ρ(A_{ij})^{k} = I + kB_{ij} for all k ∈ ℤ and that similarly we have z^{k} = I + k(z − I) for all k. Using these relations we see that any element of ρ(P_{3}) can be represented as ρ(A_{12}^{a}A_{13}^{b}A_{23}^{c}z^{d}) for unique a, b, c, d ∈ ℤ. Thus we may now calculate as follows, where we indicate which powers of generators are being commuted by using parentheses:

ρ(βh)^{3} = ρ(z^{3d}A_{12}^{a}A_{13}^{b}A_{23}^{c}.h(A_{12}^{a}A_{13}^{b}A_{23}^{c})h^{−1}.h^{2}(A_{12}^{a}A_{13}^{b}A_{23}^{c})h^{−2}.h^{3})
= ρ(z^{3d}A_{12}^{a}A_{13}^{b}A_{23}^{c}.A_{23}^{a}A_{12}^{b}A_{13}^{c}.A_{13}^{a}A_{23}^{b}A_{12}^{c}.h^{3})
= ρ(z^{3d}A_{12}^{a}A_{13}^{b}(A_{23}^{a+c}A_{12}^{b})A_{13}^{a+c}(A_{23}^{b}A_{12}^{c}).h^{3})
= ρ(z^{3d−b(a+c)−bc}A_{12}^{a}(A_{13}^{b}A_{12}^{b})A_{23}^{a+c}(A_{13}^{a+c}A_{12}^{c})A_{23}^{b}.h^{3})
= ρ(z^{3d−b(a+2c)+b^{2}+c(a+c)}A_{12}^{a+b}A_{13}^{b}(A_{23}^{a+c}A_{12}^{c})A_{13}^{a+c}A_{23}^{b}.h^{3})
= ρ(z^{3d−b(a−b+2c)+c(a+c)−c(a+c)}A_{12}^{a+b}(A_{13}^{b}A_{12}^{c})(A_{23}^{a+c}A_{13}^{a+c})A_{23}^{b}.h^{3})
= ρ(z^{3d−b(a−b+2c)+bc+(a+c)^{2}}A_{12}^{a+b+c}A_{13}^{a+b+c}A_{23}^{a+b+c}.h^{3})
= ρ(z^{3d−b(a−b+c)+(a+c)^{2}}A_{12}^{a+b+c}A_{13}^{a+b+c}A_{23}^{a+b+c}.h^{3}).

Now we also have h^{3} = A_{12}A_{13}A_{23}z. Thus considering the 23 block of ρ(βh)^{3} (as in (2.1)) we see that we must have a + b + c + 1 = 0. With this condition the exponent of z becomes 3d + 3b^{2} + 3b + 1 and the non-diagonal entries of ρ(βh)^{3} are all multiples of 3d + 3b^{2} + 3b + 2 which clearly can never be zero. This completes the q = 3 case.

§3 The q = 5 case

We have already defined the generators A_{ij} if 1 ≤ i < j ≤ n. Note that hA_{ij}h^{−1} = A_{i+1 j+1} if j + 1 ≤ n. Thus if we define hA_{ij}h^{−1} = A_{i+1 j+1} (where indices are taken mod n as necessary) we have now defined A_{ij} for all i ≠ j that we shall need below. We will let B_{ij} = ρ(A_{ij}) for all i, j.

Define

B_{ijk} = [B_{ij}, B_{jk}] = B_{ij}B_{jk}B_{ij}^{−1}B_{jk}^{−1}; B_{ijkm} = [B_{ijk}, B_{km}];
and B_{ijkmr} = [B_{ijkm}, B_{mr}].

Define an ordered set of matrices:

G = {B_{12}, B_{23}, B_{34}, B_{45}, B_{51}, B_{13}, B_{24}, B_{35}, B_{41}, B_{52},
B_{123}, B_{234}, B_{345}, B_{451}, B_{512},
B_{124}, B_{235}, B_{341}, B_{452}, B_{513},
B_{1234}, B_{2345}, B_{3451}, B_{4512}, B_{5123},
B_{1345}, B_{2451}, B_{3512}, B_{4123}, B_{5234},
B_{12345}, B_{23451}, B_{34512}, B_{45123}, B_{51234}, z = [B_{124}, B_{235}]}.

We also let G_{i} denote the subgroup of < G > generated by z and all B_{I} ∈ G where I has length at least i + 1.

Lemma 3.1. (i) The element z is central and for all B_{J} ∈ G we have

B_{J}^{k} = I + k(B_{J} − I).

(ii) For every element β ∈ P_{n} there are integers e_{12}, e_{23}, e_{34}, …, e_{51234}, e such that we can write:

ρ(β) = B_{12}^{e_{12}}B_{23}^{e_{23}}B_{34}^{e_{34}} … B_{51234}^{e_{51234}}z^{e}.  (3.1)

(iii) For all x ∈ G_{i} and y ∈ G_{j} we have xyx^{−1}y^{−1} ∈ G_{i+j}. (iv) The integers e_{12}, e_{23}, e_{34}, …, e_{51234}, e are unique.

Proof. The proof is by constructing the representation ρ_{5} and checking that (i) and (iii) are true. But then (ii) follows from (i) and (iii). For (iv) we just need to check that the quotients G_{1}/G_{2}, G_{2}/G_{3}, G_{3}/G_{4} and G_{4} are free abelian groups of ranks 10, 10, 10, and 6 respectively. □

Property (ii) is called bounded generation. We call the representation (3.1) of ρ(β) the canonical representation. For example in what follows it will be useful to know that the element h^{5} has the following canonical representation (we just give the values of e_{12}, …):

1, 0, 0, 1, −1, 0, 0, 0, −1, −1, −1, 2, 1, 0, 0, 0, 1, 1,
0, 0, 0, −1, −1, −1, −1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, −1.

We will now consider a general element ρ(β) as in (3.1); i.e. we will consider the e_{J} as integer indeterminates. The goal will be to show that there is no such choice of integers e_{J} giving ρ(βh)^{5} = id.

Now note that if π(h) = τ = (1, 2, 3, 4, 5), then h acts on the set G such that B_{J}^{h} = hB_{J}h^{−1} = B_{(J)τ}. Consider the element

(βh)^{5} = ββ^{h}β^{h^{2}}β^{h^{3}}β^{h^{4}}.h^{5}.

Now the exponents of B_{12} and B_{13} in the canonical representation of this element are easily (since we can ignore all B_{J} where J has length greater than 2) seen to be multiples of

e_{12} + e_{23} + e_{34} + e_{45} + e_{51} + 1 and e_{13} + e_{24} + e_{35} + e_{41} + e_{52} + 1.

Thus these must both be zero if ρ((βh)^{5}) is to be the identity. One obtains the same conclusion if one looks at the coefficient of a_{12}a_{26}a_{61} and of a_{13}a_{36}a_{61} in the image of a_{16}a_{61} under the action of ρ((βh)^{5}). In fact if one also considers the coefficients of a_{12}a_{23}a_{36}a_{61}, a_{12}a_{24}a_{46}a_{61}, a_{12}a_{23}a_{34}a_{46}a_{61} and a_{12}a_{23}a_{34}a_{45}a_{56}a_{61} in this image, then one now has six polynomials and it can be checked that they generate the element 1 by using a Groebner basis calculation as implemented in, for example, MAGMA [MA]. In fact in these calculations one can work over the integers mod 5. Thus no integer solution of these six polynomials exists. All of the author’s computations of the representations ρ_{n} were carried out in MAGMA. This completes the proof of the q = 5 case. We note that there is an automorphism ∗ of the ring R_{n+1} which satisfies a_{ij}^{∗} = −a_{ji} and which is Ĥ(n)-invariant i.e. (r^{∗})α = (r)α^{∗} for all r ∈ R_{n+1} and all α ∈ Ĥ(n). This enables one to simplify calculations by, for example, reducing the degree of the representation ρ_{5} from 325 to 165.

§4 Connection with homotopy braid groups and string links

In this section we relate the representation ρ_{n} to the homotopy braid group B̂_{n} studied by Goldsmith [G] and to the group of link-homotopy classes of string links [HL]. We think of a braid as a set of n strings embedded in the D×I ,where D is the open unit disc with n distinguished points p_{1}, …, p_{n} and I = [0, 1]. The endpoints of these strings are (p_{i}, 0), (p_{i}, 1). Then we define two n-braids to be homotopic if one can be deformed to the other by simultaneous homotopies of the braid strings in D × I, keeping each end-point fixed, and allowing each arc to intersect (pass through) itself but not to intersect any other arc; this gives an equivalence relation ∼ on elements in B_{n}, and the resulting group B_{n}/ ∼ is the homotopy braid group B̂_{n}. Let ψ_{n} : B_{n} → B̂_{n} be the quotient map. Then in [G] Goldsmith (using results of Milnor [M2]) proves that the ker ψ_{n} is the normal subgroup of P_{n} generated by the commutators [A_{ij}, gA_{ij}g^{−1}], 1 ≤ i < j ≤ n, where g is in the free subgroup < A_{i,i+1}, ⋯, A_{in} >. We show

Proposition 4.1. For all n ≥ 3 we have ker(ρ_{n}) is contained in ker(ψ_{n}).

Proof. In what follows it will be convenient to denote a cycle c_{ii_{1}i_{2}...i_{r}} by c_{iS}. The action of P_{n} on the monomials of U_{n+1} is given in:

Lemma 4.2. (i) If 1 ≤ i < j ≤ n, and S does not contain any k with i < k ≤ j, then

(c_{iS})A_{ij} = c_{iS} + c_{jiS} − c_{ijS} and (c_{jS})A_{ij} = c_{jS} − c_{jiS} + c_{ijS};

(ii) If 1 ≤ i < k < j ≤ n, and S does not contain any k′ ≠ k with i < k′ ≤ j, then

(c_{kS})A_{ij} = c_{kS} + c_{kjiS} − c_{kijS} + c_{ijkS} − c_{jikS};

(iii) For all g ∈< A_{i,i+1}, ⋯, A_{in} > we have

(c_{S})g = c_{S} + ∑_{α} c_{iS_{α}} and (c_{S})g^{−1} = c_{S} − ∑_{α} c_{iS_{α}}.

(iv) For all S we have

(c_{S})A_{ij} = c_{S} + ∑_{β} c_{ijS_{β}} and (c_{S})A_{ij}^{−1} = c_{S} − ∑_{β} c_{ijS_{β}}.

Proof. (i) From (1.1) we see that

(a_{ir})A_{ij} = a_{ir} − a_{ij}a_{jr}, (a_{ri})A_{ij} = a_{ri} + a_{rj}a_{ji},
(a_{jr})A_{ij} = a_{jr} − a_{ji}a_{ir}, (a_{rj})A_{ij} = a_{rj} + a_{ri}a_{ij}.

Part (i) follows easily and (ii) is similar. Now (iii) follows from (i) and (ii) by induction on the length of g as a product of the generators A_{i,i+1}, ⋯, A_{in}. Lastly, (iv) follows from (1.1), (i) and (ii). □

For the proof of Proposition 4.1 we will show that

(c_{S})A_{ij}gA_{ij}g^{−1} = (c_{S})gA_{ij}g^{−1}A_{ij}

for all sequences S. Now note that for g as in Lemma 4.2 we have (c_{S})g = c_{S} if S contains i and if S contains both i and j, then (c_{S})A_{ij} = c_{S}. Thus using Lemma 4.2 again we obtain

(c_{S})A_{ij}gA_{ij}g^{−1} = (c_{S} + ∑_{β} c_{ijS_{β}})gA_{ij}g^{−1}

= (c_{S} + ∑_{α} c_{iS_{α}} + ∑_{β} c_{ijS_{β}})A_{ij}g^{−1}

= (c_{S} + ∑_{α} c_{iS_{α}}A_{ij} + 2 ∑_{β} c_{ijS_{β}})g^{−1}

= c_{S} + 2 ∑_{β} c_{ijS_{β}} + ∑_{α} c_{iS_{α}}A_{ij} − ∑_{α} c_{iS_{α}}.  (4.1)

On the other hand we have

(c_{S})gA_{ij}g^{−1}A_{ij} = (c_{S} + ∑_{α} c_{iS_{α}})A_{ij}g^{−1}A_{ij}

= (c_{S} + ∑_{α} c_{iS_{α}}A_{ij} + ∑_{β} c_{ijS_{β}})g^{−1}A_{ij}

= (c_{S} − ∑_{α} c_{iS_{α}} + ∑_{α} c_{iS_{α}}A_{ij} + ∑_{β} c_{ijS_{β}})A_{ij}

= c_{S} − ∑_{α} c_{iS_{α}}A_{ij} + ∑_{α} c_{iS_{α}}A_{ij}^{2} + 2 ∑_{β} c_{ijS_{β}}.  (4.2)

Now the reason that (4.1) and (4.2) are equal is that by Lemma 4.2 each monomial of ∑_{α} c_{iS_{α}}A_{ij} − ∑_{α} c_{iS_{α}} has both an i and a j in it and so this element is fixed by A_{ij}. This concludes the proof of Proposition 4.1. □.

Corollary 4.3. The homotopy braid group B̂_{n} is torsion-free for n = 3, 4, 5, 6.

Proof. Now the image of P_{n} in B̂_{n} is a torsion free nilpotent group [G, HL]. On the other hand, if α ∉ P_{n}, then by Theorem 1.1 we see that ρ(α) has infinite order hence so does ψ(α). □

Now let H(n) be the group of link-homotopy classes of n-string links [HL]. This we now explain. An n-string link is a smooth or piecewise linear proper imbedding of n disjoint arcs I_{i} into D × I such that the endpoints of the image of I_{i} are (p_{i}, 0) and (p_{i}, 1). Now two such string links σ, σ′ are link-homotopic if there is a homotopy of the strings in D × I fixing the endpoints and deforming σ to σ′ such that the images of different strings remain disjoint during the homotopy.

Corollary 4.4. For n = 3, 4, 5 the image ρ(P_{n}) and H(n) have the same abelian nilpotent quotients for the lower central series.

Proof. One calculates each case for ρ(P_{n})_{k−1}/ρ(P_{n})_{k} and checks this against the known values for H(n)_{k−1}/H(n)_{k} [HL, p.414]; see also [M2, p.194], namely

(k − 2)!\binom{n}{k}

for k ≥ 2 □

References

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[P] http://zebra.sci.ccny.cuny.edu/web/problems/probBr.html.