SOME LINEAR REPRESENTATIONS OF BRAID GROUPS

Stephen P. Humphries

Abstract. In this paper we exhibit a way of obtaining linear representations of the braid groups B_{n} over ℤ[t] by studying their action on the set of isotopy classes of sets of simple closed curves on a punctured disc. The cases n = 3, 4, 5 are shown to be very different from the cases n > 5. We show a connection between representations of B_{3} and Pascal’s triangle. We also show that there is a sequence of polynomials κ_{i}(t), i ≥ 0, related to polynomials P_{i}(t) defined by V.F.R. Jones all of whose roots give values of t for which these representations are not faithful.

§1. Introduction.

Let C = C_{n} denote the set of isotopy classes of positively oriented simple closed curves on a punctured disc D_{n} with n punctures, where each such simple closed curve surrounds two of the punctures. There is a standard way of representing B_{n} as diffeomorphisms of D_{n} [Bi, Ch.1]. We use this idea to exhibit a way of obtaining linear representations of the braid groups B_{n} over ℤ[t] by first finding a natural map from C to a polynomial algebra, and then considering the induced action of B_{n} on certain quotients of ideals in this polynomial algebra. This is described in greater detail in §2. We study these linear representations and show that some of these representations are not faithful at certain values of t which are roots of polynomials closely related to polynomials occuring in the work of Jones [J3]. We show a connection between some representations of B_{3} obtained in this way and Pascal’s triangle. In fact the polynomials of Jones referred to above are also related to Pascal’s triangle. We also show that these representations fix symmetric bilinear forms, which for certain values of t are positive definite.

We now descibe how we make this action of B_{n} on C into an action on a polynomial algebra. First let H(n) < Aut(F_{n}) be the group of pure symmetric automorphisms of the free group F_{n} =< x_{1}, …, x_{n} >. Thus α ∈ H(n) if and only if α ∈ Aut(F_{n}) and α(x_{i}) is a conjugate of x_{i} for i = 1, …, n. Now H(n) =< t_{ij}|1 ≤ i ≠ j ≤ n > where t_{ij}(x_{k}) = x_{k} if k ≠ j and t_{ij}(x_{j}) = x_{i}x_{j}x_{i}^{−1} [Mc]. We will put t_{ii} = id ∈ H(n) for convenience. Let the symmetric group S_{n} act on F_{n} by permutation of subscripts and let Ĥ(n) =< H(n), S_{n} >. Then there is a split exact sequence

1 → H(n) → Ĥ(n) → S_{n} → 1.

Ĥ(n) is the group of symmetric automorphisms of the free group F_{n} of rank n [Mc].

Typeset by AMS-TEX

1991 Mathematics Subject Classification. 20F36.

Key words and phrases. Braid groups, linear groups, simple closed curves.

From [H1] we see that Ĥ(n) acts as automorphisms of a polynomial algebra: 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 (algebraically independent) commuting indeterminates. Then the group H(n) =< t_{ij}|1 ≤ i ≠ j ≤ n > acts 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)

As indicated above, the origins of this action are explained in [H1], however in §2 we will give an alternative way of describing where this representation comes from.

Now the symmetric group S_{n} acts on R_{(n)}[a_{ij}] by permutation of subscripts. The action of H(n) and Ĥ(n) on R_{(n)}[a_{ij}] extends to an action of H(n) and Ĥ(n) on R_{(n+r)}[a_{ij}], r > 0 in the obvious way. In the case where R has characteristic zero this action of H(n) on R_{(n+r)}[a_{ij}] is faithful if and only if r > 0, otherwise the kernel is the free normal subgroup Inn(n) generated by the ‘inner’ automorphisms {J_{k} = t_{k1}t_{k2} … t_{kn}|k = 1, …, n} [H1, Theorem 2.5]. We note as in [H2] that there is a natural ring involution ∗ on R_{(n+r)}[a_{ij}] which commutes with the action of H(n) so that α(x^{∗}) = α(x)^{∗} for all α ∈ Ĥ(n) and all x ∈ R_{(n)}[a_{ij}]; this involution is determined by its action on the generators a_{ij} which is as follows: a_{ij}^{∗} = −a_{ji}.

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} are subgroups of H(n) [Bi, p. 30]. Recall the action of B_{n} by diffeomorphisms on the disc with punctures D_{n} referred to above. It turns out that each such diffeomorphism coming from a braid can be chosen to fix ∂D_{n}, the boundary of D_{n}. Let p ∈ ∂D_{n}. We will identify the fundamental group π_{1}(D_{n}, p) with the free group F_{n} = < x_{1}, …, x_{n} >. The x_{i} are simple closed cirves surrounding a single hole; see Figure 2 where we have identitfied x_{i} with T_{i} for i = 1, …, n. This now gives an action of B_{n} on F_{n} =< x_{1}, …, x_{n} >. Here the generators σ_{i}, i = 1, …, n − 1 act as ‘half-twist’ diffeomorphisms which interchange two adjacent punctures ; their action on F_{n} is as follows:

σ_{i}(x_{j}) = \{ \begin{matrix}x_{j} & if & j ≠ i, i + 1; \\ x_{i}x_{i+1}x_{i}^{−1} & if & j = i; \\ x_{i} & if & j = i + 1. \end{matrix}

Note that it follows from this that for α ∈ B_{n} and i = 1, …, n the element α(x_{i}) is a conjugate of some x_{j}, where i → j is a permutation of {1, …, n}. This latter condition, together with the condition that the boundary curve x_{1} … x_{n} is fixed by each element of B_{n}, are necessary and sufficient conditions for an element of Aut(F_{n}) to be in B_{n} [Bi, p. 30].

A presentation for H(n) is given by McCool [Mc]:

t_{ij}t_{rs} = t_{rs}t_{ij} if card ({i, j, r, s}) = 4
t_{ij}t_{is} = t_{is}t_{ij} if card ({i, j, s}) = 3; and
t_{ij}t_{is}t_{js} = t_{js}t_{ij}t_{is} if card ({i, j, s}) = 3

The action of B_{n} on R_{(n+r)}[a_{ij}], r ≥ 0, is not faithful if and only if r = 0, the kernel when r = 0 being exactly the cyclic centre of B_{n}; this is true even in the situation where R has positive characteristic [Hu1, Corollary 2.9].

For i, j, k, …, r, s ∈ {1, 2, …, n} let c_{ijk...rs} denote the cycle a_{ij}a_{jk} … a_{rs}a_{si} ∈ R_{(n)}[a_{ij}]. Then the cycles generate a subalgebra of R_{(n)}[a_{ij}] denoted Y_{n}. A cycle c_{ijk...rs} will be called simple if i,j,k,...,r,s are all distinct. The ring Y_{n} is generated by the (finite number of) simple cycles. By considering the action given in (1.1) one sees that B_{n} acts on the ring Y_{n}. In all examples considered below the rings on which B_{n} acts will be subrings and quotients of Y_{n}.

In section 2 we explain the construction of these representations in greater detail. In section 3 we find an irreducible linear 3 × 3 representation of B_{3} in the above manner. In section 4 we show that there is an infinite sequence of linear representations of B_{3} which are closely related to Pascal’s triangle (and to the 3×3 representation of B_{3} referred to above). In section 5 we describe how the above gives a linear 11 × 11 representation of B_{4} and prove

Theorem 1.1. For n = 4 the above method yields an 11 × 11 representation of B_{4} over ℚ[t] which splits into 3 × 3 and 8 × 8 irreducible representations of B_{4}. The 3×3 representation factors through B_{3}. The 11×11 representation is bi-orthogonal in that there are two linearly independent non-degenerate symmetric bilinear forms over ℚ[t] which are fixed by the 11 × 11 representation. One of these symmetric bilinear forms can be chosen so as to be positive definite for t close to −1/2.

In section 6 we briefly describe what happens when n = 5:

Theorem 1.2. For n = 5 the above method yields a 39 × 39 representation of B_{5} over ℚ[t]. This representation is also bi-orthogonal over ℚ[t]. One of these symmetric bilinear forms can be chosen so as to be positive definite for t close to −1.

In §8 we show why we are unable to obtain finite-dimensional representations for n > 5. In §9 we indicate a connection of this work with work of Jones [J2,J3]; specifically, we show that certain polynomials P_{k}(t) used by Jones to calculate dimensions of certain spaces related to the Jones representation occur naturally in our setting. In §10 we look at what happens when we specialise the value of t in our representations. For example we show that with t = 0 we always (for any n) obtain a finite-dimensional representation of B_{n} which is actually finite by (infinite) nilpotent.

§2 Construction of representations

Here is a geometric formulation of these representations. For n ≥ 1 let D_{n} denote the disc with n punctures p_{1}, …, p_{n}. For u= (u_{1}, …, u_{r}) ∈ ℕ^{n} where u_{1} ≥ u_{2} ≥ ⋯ ≥ u_{r} ≥ 1 and u_{1} + u_{2} + ⋯ + u_{r} ≤ n let γ_{u} denote the union of the oriented simple closed curves shown in Figure 1.

Let Ĉ_{nu} denote the set of all images of γ_{u} under the action of the braid group; thus Ĉ_{nu} consists of all collections of simple closed curves which are diffeomorphic to γ_{u}. If r = 1 and u= (u_{1}), then we will write Ĉ_{nu_{1}} instead of Ĉ_{nu}. Let C_{nu} denote the free R-algebra R[c|c ∈ Ĉ_{nu}] generated by the elements c ∈ C_{nu}. Let D_{nu} be the ideal of C_{nu} generated by all differences c − c′, where c, c′ ∈ C_{nu} so that C_{nu}/D_{nu} is isomorphic to R[t] for some indeterminate t.

Now there is a ring homomorphism ϕ = ϕ_{n} : C_{nu} → R_{(n)}[a_{ij}] which is defined as follows: For i = 1, …, n define the following n × n matrices (transvections)

T_{i} = ( \begin{matrix}1 & 0 & … & 0 & … & 0 & 0 \\ 0 & 1 & … & 0 & … & 0 & 0 \\ ⋮ & ⋮ & ... & ⋮ & … & ⋮ & ⋮ \\ a_{i1} & a_{i2} & … & 1 & … & a_{in−1} & a_{in} \\ ⋮ & ⋮ & … & ⋮ & ... & ⋮ & ⋮ \\ 0 & 0 & … & 0 & … & 1 & 0 \\ 0 & 0 & … & 0 & … & 0 & 1 \end{matrix} )

where the non-zero off-diagonal entries occur in the ith row. Here we define a transvection as a matrix M = I_{n} + A where I_{n} is the identity matrix, det(M) = 1 and A^{2} = 0. In particular, conjugates of transvections are transvections.

Then F_{n} =< T_{1}, …, T_{n} > is a free group of rank n. This follows from the fact that free groups are Hopfian [MKS] and from [H3, Theorem 1.2] which shows that even if we take a_{ij} = 6 for all 1 ≤ i ≠ j ≤ n we get a free group of rank n. This allows us to identify x_{i} and T_{i} for i = 1, …, n. See Figure 2 where we also define some simple arcs joining the punctures of D_{n}.

Now, as we noted above, B_{n} acts by automorphisms on F_{n} in such a way that α(T_{i}) is a conjugate of some T_{j} i.e. α(T_{i}) is also a transvection. This is the key fact that we will use later on.

Further, if c is a simple closed curve on D_{n}, then c represents a conjugacy class in F_{n} and so its trace is well-defined (the trace of the corresponding product of transvections). In fact one easily sees that trace(c) ∈ Y_{n} [H1, Lemma 2.3]. Further, if c ∈ Ĉ_{nu} is a union of simple closed curves c_{1}, …, c_{r}, then we define

trace(c) = trace(c_{1}) + ⋯ + trace(c_{r}).

Then ϕ is defined uniquely by its action on each c ∈ Ĉ_{nu}, which is ϕ(c) = trace(c)− rn.

Now trace(T_{i}T_{j}) = a_{ij}a_{ji}+n and in general if A, B ∈ F_{n}, then trace(AT_{i}A^{−1}BT_{j}B^{−1}) = b_{ij}b_{ji} + n where the b_{ij} ∈ R_{(n)}[a_{ij}] satisfy b_{ij} = −b_{ji}^{∗}. This can be proved by noting that we need only consider the case B = I_{n} and then the proof is by induction on the length of A. In fact in this way it is also easy to see that

b_{ij} = a_{ij} + terms of higher degree.

Now for α ∈ B_{n} the image α(T_{i}) is a conjugate AT_{j}A^{−1} for some A ∈ F_{n} and 1 ≤ j ≤ n. Here the action of α on the a_{ij} is given by ([H1])

trace(α(T_{i})α(T_{j})) = α(a_{ij})α(a_{ji}) + n,

so that it has the following naturality property (with respect to the action of B_{n} on F_{n}): for all 1 ≤ i, j, …, r, s ≤ n we have

trace(α(T_{i})α(T_{j}) … α(T_{r})α(T_{s})) = α(trace(T_{i}T_{j} … T_{r}T_{s})).

In Theorem 2.3 we show that for 3 ≤ n ≤ 5 and u = (2) the ring E_{nu} = ϕ(D_{nu})/ϕ(D_{nu})^{2} is a finitely generated free ϕ(C_{nu})/ϕ(D_{nu})-module. Now ϕ(C_{nu})/ϕ(D_{nu}) is in general a quotient of R[t] and, further, B_{n} acts trivialy on ϕ(C_{nu})/ϕ(D_{nu}). In fact we have:

Lemma 2.1. ϕ(C_{nu})/ϕ(D_{nu}) ≅ R[t].

Proof. To see this note that there is an R-module homomorphism ψ : Y_{n} → R[t] which is defined on cycles (monomials) μ as follows : ψ(μ) = 0 if degree(μ) > 2, and ψ(μ) = t if degree(μ) = 2. Clearly ψ is onto and ψ(ϕ(D_{nm})) = 0. We now need:

Lemma 2.2. If w ∈ F_{n} =< T_{1}, …, T_{n} > and w → T_{1}^{e_{1}} … T_{n}^{e_{n}} ∈ ℤ^{n} = F_{n}/F′_{n}, then trace(w) = n + ∑_{i,j=1}^{n} e_{i}e_{j}a_{ij}a_{ji} + higher degree terms.

Proof. The proof is by induction on the length of w when written as a product of the generators T_{i}, T_{i}^{−1}. □

It thus easily follows that the composite ψϕ : C_{nm} → R[t] is not the zero homomorphism. Thus we have ϕ(C_{nm})/ϕ(D_{nm}) ≅ R[t]. □

Remark. We note that we can generalise this construction as follows: one may consider (instead of u) any set of simple closed curves, whether they are disjoint or not.

If w ∈ F_{n}, then we let γ(w) ∈ C_{nu}, be an oriented closed curve representing w. We will also use the notation γ_{ij} for γ(x_{i}x_{j}).

If γ is a simple closed curve in D_{n} surrounding k holes, then we let Σ(γ) denote the spine of γ, so that Σ(γ) is an embedded tree in D_{n} ∪{p_{1}, …, p_{n}} whose valence 1 vertices are in {p_{1}, …, p_{n}} such that γ is isotopic to the boundary of a small tubular neighbourhood of Σ(γ) (with the correct orientation) . Thus Σ(γ) will have k end points which are punctures of D_{n}.

If d ∈ D_{nu}, then [d] will denote its class in E_{nu} and if c ∈ C_{nu}, then [[c]] will denote its class in C_{nu}/D_{nu}.

Theorem 2.3. For all 3 ≤ n ≤ 5 the quotient ϕ(D_{n2})/ϕ(D_{n2})^{2} is a finitely generated free ϕ(C_{n2})/ϕ(D_{n2})-module.

The proof is by construction and will be given in sections 3 − 6. We will denote the resulting representation (over ℚ[t]) by ρ_{n} = ρ_{n}^{(t)}. The rest of this section will shed more light on the properties of this representation in general.

Lemma 2.4. Let α, β_{1}, β_{2}, γ_{1}, γ_{2} ∈ Ĉ_{n2} be curves with spines as shown in Figure 3, where we assume that there are no punctures inside the heart-shaped diagram other than the three shown (p_{i}, p_{j} and p_{k}).

Then we have the following relation:

[γ_{1} − γ_{2}] + (1 + [[α]])[β_{1} − β_{2}] = 0.  (2.1)

In particular we have

[γ(x_{1}x_{2}x_{1}^{−1}x_{3}) − γ(x_{1}x_{2}x_{3}x_{2}^{−1})] = (1 + [[γ_{12}]])[γ_{23} − γ_{13}].  (2.2)

Proof. The first statement, which we refer to as the heart relation, follows immediately from the second, since B_{n} acts transitively on all sets of curves homeomorphic to those shown in Figure 3. The second relation is easily checked by calculating the traces and showing that in R_{(n)}[a_{ij}] we have

trace(γ(x_{1}x_{2}x_{1}^{−1}x_{3}))−trace(γ(x_{1}x_{2}x_{3}x_{2}^{−1})) = (1+a_{12}a_{21})(trace(γ_{23})−trace(γ_{13})). □

We note that in (2.2) we have γ(x_{1}x_{2}x_{1}^{−1}x_{3}) = σ_{1}^{2}(γ_{23}), γ(x_{1}x_{2}x_{3}x_{2}^{−1}) = σ_{1}^{−1}(γ_{23}) and γ(x_{1}x_{3}) = σ_{1}(γ_{23}). Thus we can write (2.2) as (σ_{1}^{2} − σ_{1}^{−1} − (1 + a_{12}a_{21})(1 − σ_{1})[γ_{23}] = 0. Upon acting on this equation by σ_{1} on the left this gives (since σ_{1}(a_{12}a_{21}) = a_{12}a_{21})

(σ_{1}^{3} − 1 − (1 + a_{12}a_{21})(σ_{1} − σ_{1}^{2}))[γ_{23}] = 0.  (2.3)

Let I_{n2} denote the ideal of Ẽ_{n2} = D_{n2}/D_{n2}^{2} generated by all the heart relations i.e. by the images of any one heart relation under the action of the braid group. Then I_{n2} is contained in ker(ϕ); the author has been unable to find any relations in ϕ(D_{n2})/ϕ(D_{n2}^{2}) (considered as a ϕ(C_{n2})/ϕ(D_{n2})-module) that are not consequences of the heart relation. We thus conjecture that

Ẽ_{n2}/I_{n2} ≅ ϕ(D_{n2})/ϕ(D_{n2}^{2}) = E_{n2},

i.e. that ker(ϕ) = I_{n2}. We will call this the heart conjecture. We will prove this conjecture in the cases n = 3, 4, 5.

Note that if we put t = −1 in the heart relation, then we get [γ_{1} − γ_{2}]|_{t=−1} = 0. This gives the motivation for the following definition: Suppose that s ∈ ℂ and that for some p ∈ ℤ^{+} we have [σ_{1}^{p}(γ_{23}) − γ_{23}]|_{t=s} = 0; we will let p(s) denote the smallest such (positive) integer p. Thus, for example, the heart relation shows that if s = −1, then p(s) = 3. In Corollary 9.2 we give a way of finding all s ∈ ℂ having finite p(s). For γ_{1}, γ_{2} ∈ C_{n2}, we will write γ_{1} ∼_{(s)} γ_{2} if

[γ_{1} − γ_{2}]|_{t=s} = 0.

This defines an equivalence relation on C_{n2}.

Lemma 2.5. If there are only a finite number k of equivalence classes for ∼_{(s)}, then k − 1 is a lower bound for the number of generators for ϕ(Ẽ_{n2})/I_{n2} as a ϕ(C_{n2})/ϕ(D_{n2})-module. If the heart conjecture is true, then a basis for ϕ(D_{n2})/D_{n2}^{2} as a ϕ(C_{n2})/ϕ(D_{n2})-module will have at least k − 1 elements.

Lastly, if E_{n2} has k −1 generators, then the heart conjecture is true for this value of n.

Proof. If c_{1}, c_{2}, c_{3}, …, c_{k} are distinct representatives for the equivalence classes, then

[c_{2} − c_{1}], [c_{3} − c_{1}], …, [c_{k} − c_{1}]

will be a linearly independent set of generators for a submodule of ϕ(D_{n2})/ϕ(I_{n2}) as a ϕ(C_{n2})/ϕ(D_{n2})-module. The rest is also easy. □

We now indicate how to calculate the number of equivalence classes for ∼_{(s)}. If γ ∈ C_{n2}, then there is α ∈ B_{n} with α(γ_{12}) = γ. Then we let

τ_{γ} = ασ_{1}α^{−1}

denote the positive half twist about γ.

Corollary 2.6. Let s ∈ ℂ, let p(s) be as above and let γ_{1}, γ_{2} ∈ C_{n2} be curves with spines meeting only at a puncture. Then

γ_{1} ∼_{(s)} τ_{γ_{2}}^{±p(s)}(γ_{1}).

Proof. This follows directly from the relation [σ_{1}^{p}(γ_{23}) − γ_{23}]|_{t=s} = 0 since B_{n} acts transitively on the set of curves in C_{n2} with spines meeting only at a puncture. □

We now need to find Stab_{B_{n}}(γ_{12}). For i ≤ j, following Gurzo [G], we let

A_{ij} = σ_{i}σ_{i+1} … σ_{j}, B_{ji} = σ_{j}σ_{j−1} … σ_{i}.

Proposition 2.7. Let L_{n} = Stab_{B_{n}}(γ_{12}). Then

L_{n} =< σ_{1}, σ_{3}, σ_{4}, …, σ_{n−1},B_{21}A_{12}, B_{21}B_{32}A_{23}A_{12}, B_{21}B_{32}B_{43}A_{34}A_{23}A_{12}, …,

B_{21}B_{32}B_{43} … B_{n−2,n−3}A_{n−3,n−2} … A_{34}A_{23}A_{12} > .

Proof. This ammounts to finding the centraliser (or normaliser) of σ_{1} in B_{n}. The result then follows from Theorem 1 of [G] which calculates the normaliser and shows that it is equal to the centraliser. □

Note that the braid (σ_{1}σ_{2} … σ_{k−1})^{k} is the Dehn twist [Bi] about the simple closed curve γ(x_{1}x_{2} … x_{k}). If G is a group, then we will let Z(G) denote the centre of G. We have Z(B_{n}) =< (σ_{1}σ_{2} … σ_{n−1})^{n} > [Bi].

Corollary 2.8. In B_{n}, n = 3, 4, 5 we have

L_{3} =< σ_{1}, σ_{3}, (σ_{1}σ_{2})^{3} >;
L_{4} =< σ_{1}, σ_{3}, (σ_{1}σ_{2})^{3}, (σ_{1}σ_{2}σ_{3})^{4} >;
L_{5} =< σ_{1}, σ_{3}, σ_{4}, (σ_{1}σ_{2})^{3}, (σ_{1}σ_{2}σ_{3})^{4}, (σ_{1}σ_{2}σ_{3}σ_{4})^{5} > .

For n = 6 in B_{6}/Z(B_{6}) we have

L_{6}/Z(B_{6}) =< σ_{1}, σ_{3}, σ_{4}, σ_{5}, (σ_{1}σ_{2})^{3}, (σ_{1}σ_{2}σ_{3})^{4}, (σ_{1}σ_{2}σ_{3}σ_{4})^{5} > .

Proof. This follows from Proposition 2.7, Lemma 33 of [G] and the following relations:

(σ_{1}σ_{2})^{3} = σ_{1}B_{21}A_{12}σ_{1};
(σ_{1}σ_{2}σ_{3})^{4} = σ_{3}σ_{1}B_{21}B_{32}A_{23}A_{12}σ_{1}σ_{3};
(σ_{1}σ_{2}σ_{3}σ_{4})^{5} = σ_{4}σ_{3}σ_{1}B_{21}B_{32}B_{43}A_{34}A_{23}A_{12}σ_{4}σ_{3}σ_{4}σ_{3}σ_{1}.  □

Proposition 2.9. Let s ∈ ℂ and let p(s) be as defined above. Let K_{n} = K_{n}^{(s)} =< L_{n}, σ_{2}^{p(s)} > . Then for any α ∈ K_{n} we have

α(γ_{12}) ∼_{(s)} γ_{12}.

Proof. Any element α ∈ K_{n} can be written as

α = α_{m}σ_{2}^{±p(s)}α_{m−1}σ_{2}^{±p(s)} … α_{2}σ_{2}^{±p(s)}α_{1},

where α_{i} ∈ Stab_{B_{n}}(γ_{12}) for i = 1, …, m. We prove the result by induction on m, the case m = 1 being trivial. Assume that the result is true for m − 1 and let

α = α_{m}σ_{2}^{±p(s)}α_{m−1}σ_{2}^{±p(s)} … α_{2}σ_{2}^{±p(s)}α_{1} = α_{m}σ_{2}^{±p(s)}α′.

Then by induction we have α′(γ_{12}) ∼_{(s)} γ_{12} and so

α(γ_{12}) = α_{m}σ_{2}^{±p(s)}α′(γ_{12}) ∼_{(s)} α_{m}σ_{2}^{±p(s)}(γ_{12}) = α_{m}σ_{2}^{±p(s)}α_{m}^{−1}(γ_{12}).

Now

α_{m}σ_{2}^{±p(s)}α_{m}^{−1} = τ_{α_{m}(γ_{23})}^{±p(s)}

and α_{m}(γ_{23}) is a curve whose spine meets the spine of γ_{12} in only the puncture p_{2}. Thus by Corollary 2.6 we have

α_{m}σ_{2}^{±p(s)}α_{m}^{−1}(γ_{12}) = τ_{α_{m}(γ_{23})}^{±p(s)}(γ_{12}) ∼_{(s)} γ_{12}.

The result follows by induction. □

Corollary 2.10. If for some s and n ≥ 3, the index [B_{n} : K_{n}^{(s)}] is finite, then there are at most k = [B_{n} : K_{n}^{(s)}] equivalence classes of elements of C_{n2} under the equivalence relation ∼_{(s)}.

Proof. Suppose that c_{1}, c_{2}, …, c_{k+1} ∈ C_{n2}. Then, since B_{n} acts transitively on C_{n2}, there are α_{1}, …, α_{k+1} ∈ B_{n} such that α_{i}(c_{i}) = γ_{12} for i = 1, …, k + 1. Since k + 1 > [B_{n} : K_{n}^{(s)}] there are 1 ≤ i ≠ j ≤ k + 1 such that α_{i}α_{j}^{−1} ∈ K_{n}^{(s)}. By Proposition 2.9 we see that α_{i}α_{j}^{−1}(γ_{12}) ∼_{(s)} γ_{12}, which implies that

c_{j} = α_{j}^{−1}(γ_{12}) ∼_{(s)} α_{i}^{−1}(γ_{12}) = c_{i}.

This shows that there are at most k such equivalence classes. □

Proposition 2.11. For n = 3, 4, 5 we have

[B_{3} : K_{3}^{(−1)}] = 4,  [B_{4} : K_{4}^{(−1)}] = 12,  [B_{5} : K_{5}^{(−1)}] = 40.

Proof. These indices can be calculated using the Todd-Coxeter algorithm as implemented in, for example, MAGMA [M]. □

In the following sections the author has further investigated these representations. From these computations it is clear (from Lemma 2.5) that the heart conjecture is true for n = 3, 4, 5 since the degrees of these representations are 3, 11 and 39 (respectively) and the only relations used in obtaining these representations are the heart relation and consequences thereof.

We have two obvious questions: 1) Is the heart conjecture true for n ≥ 6; and 2) What is [B_{6} : K_{6}^{(−1)}]? In §8 we show that [B_{n} : K_{n}^{(−1)}] = ∞ for n > 5.

§3. Action of B_{3} on C_{32}

As a basis for E_{32} as a ℤ[t]-module we can take

S ={ϕ(γ_{23} − γ(x_{1}x_{2}x_{1}^{−1}x_{3})) = a_{13}a_{32}a_{21} − a_{12}a_{23}a_{31} + a_{12}a_{21}a_{13}a_{31},
ϕ(γ_{12} − γ_{13}) = a_{12}a_{21} − a_{13}a_{31}, ϕ(γ_{12} − γ_{23}) = a_{12}a_{21} − a_{23}a_{32}}.

The matrices that we obtain are

σ_{1} → S_{1} = ( \begin{matrix}−1 − t & t & −t \\ 1 & 0 & 1 \\ 0 & 1 & 0 \end{matrix} ) , σ_{2} → S_{2} = ( \begin{matrix}−1 − t & −2t − t^{2} & 2t + t^{2} \\ −1 & −1 − t & t \\ −1 & −1 − t & 1 + t \end{matrix} ) .

Here S_{1}, S_{2} are the images of σ_{1}, σ_{2} under the representation. One can check that (for generic values of t) these matrices have three distinct eigenvalues. They are related to the representations of B_{3} described in [F].

This representation of B_{3}/Z(B_{3}) is faithful and irreducible. We show the faithfulness by noting that if we put t = −4, then the matrices r_{1}, r_{2} that we obtain are both unipotent. If we let c be the matrix

c = ( \begin{matrix}0 & 2 & −4 \\ 1 & 1 & 0 \\ 1 & 0 & 1 \end{matrix} ) ,

then

c^{−1}r_{1}c = ( \begin{matrix}1 & 1 & −1 \\ 0 & 1 & −2 \\ 0 & 0 & 1 \end{matrix} ) , c^{−1}r_{2}c = ( \begin{matrix}1 & 0 & 0 \\ −2 & 1 & 0 \\ −1 & 1 & 1 \end{matrix} ) .

Call these matrices X_{1}, X_{2}. In section 4 we will prove the following result as a specific case of a more general result.

Lemma 3.1. The matrices X_{1}^{2}, X_{2}^{2} generate a free group of rank 2.

Let w be the matrix diag(1, 1, −1). What we actually prove in §4 is that the squares of the matrices

(wX_{1}w)^{T} = ( \begin{matrix}1 & 0 & 0 \\ 1 & 1 & 0 \\ 1 & 2 & 1 \end{matrix} ) , (wX_{2}^{−1}w)^{T} = ( \begin{matrix}1 & 2 & 1 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{matrix} ) ,

generate a free group of rank 2. Lemma 3.1 will be used to prove

Proposition 3.2. The representation ρ_{3}^{(t)} : B_{3}/Z(B_{3}) →< S_{1}, S_{2} > is faithful.

Proof. By considering a presentation [Bi p. 20] for the pure braid group P_{3} one sees that the image of the subgroup < σ_{1}^{2}, σ_{2}^{2} > in P_{3}/Z(P_{3}) is the same as the image of all of P_{3}. Further, this image is a rank 2 free group. Since P_{3} is normal of finite index in B_{3}, if there were any kernel for the representation < σ_{1}, σ_{2} >→< S_{1}, S_{2} >, this kernel would meet P_{3}/Z(P_{3}) non-trivially. Since free groups are Hopfian [MKS], the group < X_{1}^{2}, X_{2}^{2} > is free if and only if the representation is faithful. □

In order to obtain a faithful representation of B_{3} from this representation we need only replace S_{1}, S_{2} by uS_{1}, uS_{2}, where u is an indeterminate (one could take u = t). This representation would then be over the ring ℤ[t, u, u^{−1}]. One could even replace X_{1}, X_{2} by 2X_{1}, 2X_{2} and obtain a faithful 3 × 3-representation of B_{3} over ℤ[1/2] in which σ_{1} and σ_{2} are represented by upper and lower triangular matrices (respectively).

Lastly in this section we note that the matrix group < S_{1}, S_{2} > fixes the nondegenerate symmetric bilinear form having the following matrix representation:

( \begin{matrix}−t & 0 & t/2 \\ 0 & 1 & 1/2 \\ t/2 & 1/2 & 1 \end{matrix} ) .

§4. Pascal’s triangle and representations of B_{3}

For fixed n ≥ 1 we let Σ_{1} = Σ_{1}(n) and Σ_{2} = Σ_{2}(n) (respectively) be the following (n + 1) × (n + 1) lower and upper (respectively) triangular matrices:

( \begin{matrix}1 \\ 1 & 1 \\ 1 & 2 & 1 \\ 1 & 3 & 3 & 1 \\ 1 & 4 & 6 & 4 & 1 \\ ⋮ & ⋮ & ⋮ & ⋮ & ⋮ & ... \\ \binom{n}{0} & \binom{n}{1} & \binom{n}{2} & \binom{n}{3} & \binom{n}{4} & … & \binom{n}{n} \end{matrix} ) , ( \begin{matrix}\binom{n}{n} & … & \binom{n}{4} & \binom{n}{3} & \binom{n}{2} & \binom{n}{1} & \binom{n}{0} \\ & ... & ⋮ & ⋮ & ⋮ & ⋮ & ⋮ \\ & & 1 & 4 & 6 & 4 & 1 \\ & & & 1 & 3 & 3 & 1 \\ & & & & 1 & 2 & 1 \\ & & & & & 1 & 1 \\ & & & & & & 1 \end{matrix} ) .

We will find it natural to index the rows and columns of all (n+1)×(n+1) matrices in this section starting at 0. Let E = E_{n} be the (n + 1) × (n + 1) permutation matrix corresponding to the permutation (0 n)(1 n − 1)(2 n − 2) …. We will show that σ_{1} → Σ_{1}, σ_{2} → Σ_{2}^{−1} gives a representation for B_{3}.

Lemma 4.1. We have EΣ_{1}E^{−1} = Σ_{2}. Further

Σ_{2}^{−1} = ( \begin{matrix}\binom{n}{n} & … & (−1)^{n−4}\binom{n}{4} & (−1)^{n−3}\binom{n}{3} & (−1)^{n−2}\binom{n}{2} & (−1)^{n−1}\binom{n}{1} & (−1)^{n}\binom{n}{0} \\ ⋮ & ... & ⋮ & ⋮ & ⋮ & ⋮ & ⋮ \\ 0 & … & 1 & −4 & 6 & −4 & 1 \\ 0 & … & 0 & 1 & −3 & 3 & −1 \\ 0 & … & 0 & 0 & 1 & −2 & 1 \\ 0 & … & 0 & 0 & 0 & 1 & −1 \\ 0 & … & 0 & 0 & 0 & 0 & 1 \end{matrix} ) .

There is a similar expression for Σ_{1}^{−1}.

Proof. The first statement is clear. We prove the second statement for Σ_{1}^{−1}. Note that row m of the proposed Σ_{1}^{−1} is (−1)^{m} (\binom{m}{0}, −\binom{m}{1}, \binom{m}{2}, −\binom{m}{3}, …, (−1)^{n}\binom{m}{n})

12 STEPHEN P. HUMPHRIES

while column j of Σ_{1} is (\binom{0}{j}, \binom{1}{j}, \binom{2}{j}, …, \binom{n}{j}). Now ∑_{i=0}(−1)^{i}\binom{m}{i}\binom{i}{j} = δ_{mj} is a standard identity among binomial coefficients [R p.4]. This gives the result. □ Lemma 4.2. We have (here, as in the above, a blank indicates a zero entry)

Σ_{1}Σ_{2}^{−1} = ( \begin{matrix}1 & −\binom{n}{1} & \binom{n}{2} & −\binom{n}{3} & \binom{n}{4} & … & 1 \\ ⋮ & ⋮ & ⋮ & ⋮ & ⋮ & ·^{·^{·}} \\ 1 & −4 & 6 & −4 & 1 \\ 1 & −3 & 3 & −1 \\ 1 & −2 & 1 \\ 1 & −1 \\ 1 \end{matrix} )

Proof. Row m of Σ_{1} is (\binom{m}{0}, \binom{m}{1}, \binom{m}{2}, \binom{m}{3}, …, \binom{m}{n}), while column j of Σ_{2}^{−1} is

(−1)^{j} (\binom{n − 0}{n − j}, −\binom{n − 1}{n − j}, \binom{n − 2}{n − j}, −\binom{n − 3}{n − j}, …, (−1)^{n}\binom{n − n}{n − j}) .

But by [R, p. 8 eq. (5)]

∑_{i=0}(−1)^{i}\binom{m}{i}\binom{n − i}{n − j} = ∑_{i=0}(−1)^{i}\binom{m}{i}\binom{n − i}{j − i} = \binom{n − m}{j}.

The result follows. □

Let G = G_{n} be the (n + 1) × (n + 1) matrix diag(1, −1, 1, −1, …)E_{n}. Lemma 4.3. We have Σ_{1}Σ_{2}^{−1}Σ_{1} = (−1)^{n}G_{n}. Proof. Row m of Σ_{1}Σ_{2}^{−1} is (\binom{n−m}{0}, −\binom{n−m}{1}, \binom{n−m}{2}, −\binom{n−m}{3}, …, (−1)^{n}\binom{n−m}{n}), while column j of Σ_{1} is (\binom{0}{0−j}, \binom{1}{1−j}, \binom{2}{2−j}, \binom{3}{3−j}, …, \binom{n}{n−j}) . Thus using [R, p. 3 (iv)] we obtain

∑_{i=0}(−1)^{i}\binom{n − m}{i}\binom{i}{i − j} = ∑_{i=0}(−1)^{i}\binom{n − m}{i}\binom{i}{j}

= ∑_{i=0}(−1)^{i}\binom{n − m}{j}\binom{n − m − j}{i − j}

= \binom{n − m}{j}δ_{(n−m−j) 0}

= (−1)^{j}δ_{(n−m−j) 0},

as required. □

For fixed n, let C_{n} denote the group generated by Σ_{1} and Σ_{2}. Let I_{n+1} denote the (n + 1) × (n + 1) identity matrix. Let Δ = σ_{1}σ_{2}σ_{1}. Then it is well-known [Bi] that Δ^{2} generates the infinite cyclic centre of B_{3}.

Theorem 4.4. The map ϕ : B_{3} → C_{n} induced by ϕ(σ_{1}) = Σ_{1}, ϕ(σ_{2}) = Σ_{2}^{−1} is a group homomorphism. In particular we have Σ_{1}Σ_{2}^{−1}Σ_{1} = Σ_{1}Σ_{2}^{−1}Σ_{1}. We also have (Σ_{1}Σ_{2}^{−1}Σ_{1})^{2} = (−1)^{n}I_{n+1}. Lastly, the subgroup generated by Σ_{1}^{2} and Σ_{2}^{2} is free of rank 2.

Proof. We know that Σ_{1}Σ_{2}^{−1}Σ_{1} = (−1)^{n}G_{n} and that EΣ_{1}E = Σ_{2}, EΣ_{2}E = Σ_{1}. We can also check that G_{n} = (−1)^{n}EG_{n}E. Thus

G_{n} = (−1)^{n}EG_{n}E = EΣ_{1}Σ_{2}^{−1}Σ_{1}E = Σ_{2}Σ_{1}^{−1}Σ_{2}.

Now we also have G_{n}^{−1} = (−1)^{n}G_{n}, which shows that

Σ_{1}Σ_{2}^{−1}Σ_{1} = G_{n} = (−1)^{n}G_{n}^{−1} = Σ_{2}^{−1}Σ_{1}Σ_{2}^{−1}.

This shows that ϕ gives a representation of B_{3}. We also have

(Σ_{1}Σ_{2}^{−1}Σ_{1})^{2} = G_{n}G_{n} = (−1)^{n}G_{n}G_{n}^{−1} = (−1)^{n}I_{n+1},

as required.

We now prove the last assertion. Let u and v be indeterminates and let

X = (u^{0}v^{n}, u^{1}v^{n−1}, u^{2}v^{n−2}, …, u^{n}v^{0}) .

Then for any r ∈ ℤ we have

Σ_{1}^{r}(X) = ((u + rv)^{0}v^{n}, (u + rv)^{1}v^{n−1}, (u + rv)^{2}v^{n−2}, …, (u + rv)^{n}v^{0})
Σ_{2}^{r}(X) = (u^{0}(v + ru)^{n}, u^{1}(v + ru)^{n−1}, u^{2}(v + ru)^{n−2}, …, u^{n}(v + ru)^{0}) .

Thus the above action of < Σ_{1}^{r}, Σ_{2}^{r} > on the vector X gives rise to an action of < Σ_{1}^{r}, Σ_{2}^{r} > on u and v, namely Σ_{1}(u) = u+v, Σ_{1}(v) = v, Σ_{2}(u) = u, Σ_{2}(v) = v+u. This gives a representation

Σ_{1} → ( \begin{matrix}1 & 1 \\ 0 & 1 \end{matrix} ) , Σ_{2} → ( \begin{matrix}1 & 0 \\ 1 & 1 \end{matrix} )

It is well-known that the matrices ( \begin{matrix}1 & 2 \\ 0 & 1 \end{matrix} ) , ( \begin{matrix}1 & 0 \\ 2 & 1 \end{matrix} ) generate a free group of rank

2 [LS p. 168]. □

The proof of Lemma 3.2 now gives the following

Corollary 4.5. The homomorphism ϕ : B_{3} → C_{n} induced by ϕ(σ_{1}) = Σ_{1}, ϕ(σ_{2}) = Σ_{2}^{−1} has kernel < Δ^{2} > if n is even and kernel < Δ^{4} > if n is odd. □

The results of the last two sections prove Theorem 2.3 for n = 3.

§5. Action of B_{4} on ℤ_{(4)}[a_{ij}]

The 11 × 11 representation that we obtain is reducible with irreducible subrepresentations of dimensions 3 and 8. There is a 2-parameter family of symmetric bilinear forms (over ℚ[t]) that is fixed by this representation i.e. there are two non-degenerate 11 × 11 symmetric matrices A and B over ℚ[t] such that for any matrix M in the image of this representation and any α, β ∈ ℂ[t] we have M(αA + βB)M^{T} = αA + βB (where T stands for transpose). The 3 × 3 irreducible subrepresentation factors through B_{3}. The 8×8 irreducible subrepresentation comes from the 8-dimensional B_{4}-submodule that is generated by the image of [γ_{12} − γ_{34}] under the action of B_{4}. In fact the following elements serve as a basis for this submodule:

[γ_{12} − γ_{34}] = a_{12}a_{21} − a_{34}a_{43},

[γ_{23} − γ_{14}] = a_{23}a_{32} − a_{14}a_{41},

[γ_{13} − γ(x_{1}x_{2}x_{1}^{−1}x_{4})] = a_{13}a_{31} − a_{24}a_{42} − a_{24}a_{41}a_{12} + a_{21}a_{14}a_{42} + a_{21}a_{14}a_{41}a_{12},

[γ_{24} − γ(x_{2}x_{3}x_{2}^{−1}x_{1})] = a_{24}a_{42} − a_{13}a_{31} − a_{31}a_{12}a_{23} + a_{32}a_{21}a_{13} + a_{32}a_{21}a_{12}a_{23},

[γ_{13} − γ(x_{3}x_{4}x_{3}^{−1}x_{2})] = a_{13}a_{31} − a_{24}a_{42} − a_{42}a_{23}a_{34} + a_{43}a_{32}a_{24} + a_{43}a_{32}a_{23}a_{34},

[γ_{24} − γ(x_{4}x_{1}x_{4}^{−1}x_{3})] = a_{24}a_{42} − a_{13}a_{31} − a_{13}a_{34}a_{41} + a_{14}a_{43}a_{31} + a_{14}a_{43}a_{34}a_{41},

[γ_{14} − γ(x_{1}x_{2}x_{1}^{−1}x_{4}^{−1}x_{3}x_{4})] = a_{14}a_{41} − a_{23}a_{32} − a_{31}a_{12}a_{23} + a_{32}a_{21}a_{13} + a_{21}a_{13}a_{31}a_{12} − a_{42}a_{23}a_{34} − a_{34}a_{23}a_{41}a_{12} + a_{34}a_{21}a_{13}a_{42} + a_{34}a_{21}a_{13}a_{41}a_{12}

+ a_{43}a_{32}a_{24} + a_{43}a_{24}a_{31}a_{12} − a_{43}a_{21}a_{14}a_{32} − a_{43}a_{21}a_{14}a_{31}a_{12}

+ a_{34}a_{43}a_{24}a_{42} + a_{34}a_{43}a_{24}a_{41}a_{12} − a_{34}a_{43}a_{21}a_{14}a_{42} − a_{34}a_{43}a_{21}a_{14}a_{41}a_{12},

[γ_{12} − γ(x_{2}x_{3}x_{2}^{−1}x_{1}^{−1}x_{4}x_{1})] = a_{21}a_{12} − a_{34}a_{43} − a_{42}a_{23}a_{34} + a_{43}a_{32}a_{24} + a_{32}a_{24}a_{42}a_{23} − a_{13}a_{34}a_{41} − a_{41}a_{34}a_{12}a_{23} + a_{41}a_{32}a_{24}a_{13} + a_{41}a_{32}a_{24}a_{12}a_{23}

+ a_{14}a_{43}a_{31} + a_{14}a_{31}a_{42}a_{23} − a_{14}a_{32}a_{21}a_{43} − a_{14}a_{32}a_{21}a_{42}a_{23}

+ a_{41}a_{14}a_{31}a_{13} + a_{41}a_{14}a_{31}a_{12}a_{23} − a_{41}a_{14}a_{32}a_{21}a_{13} − a_{41}a_{14}a_{32}a_{21}a_{12}a_{23}.

The action of the braid generators σ_{1}, σ_{2}, σ_{3} (respectively) on this submodule (relative to the generators given above) is given by the matrices

U_{1} = ( \begin{matrix}1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 \\ 0 & 1 + t & −1 − t & −1 & 0 & 0 & 0 & 0 \\ 0 & −1 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 1 + t & 0 & −1 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 \\ 0 & −1 − t & −1 & 0 & 1 & 1 + t & −1 − t & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 \end{matrix} ) ,

\begin{matrix}U_{2} = ( \begin{matrix}0 & 0 & 0 & −1 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 \\ −1 − t & 0 & 0 & −1 − t & −1 & 0 & 0 & 0 \\ 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ −1 − t & 0 & 1 & 0 & −1 & 0 & 0 & 0 \\ 1 + t & 0 & 0 & −2 − 3 t − t^{2} & −1 − t & 1 + t & 1 & −1 − t \\ 1 + t & 0 & 1 + t & −1 & 0 & 1 & 0 & −1 − t \end{matrix} ) , \\ U_{3} = ( \begin{matrix}1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & −1 & 0 & 0 & 0 \\ 0 & −1 − t & 0 & 1 & 0 & −1 & 0 & 0 \\ 0 & 2 + 2 t + t^{2} & 1 + t & −1 − t & −1 − t & 1 + t & 1 & 0 \\ 0 & −1 − t & 0 & 0 & −1 − t & −1 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 \\ 0 & −(1 + t)^{3} & −t − t^{2} & (1 + t)^{2} & −1 − t & −2 − 3 t − t^{2} & −1 − t & 1 \end{matrix} ) . \end{matrix}

Here U_{i} = ρ_{4}^{(t)}(σ_{i}) for i = 1, 2, 3. One can check that the matrices U_{i} are orthogonal in that they fix the following symmetric matrix A over ℚ[t] (so that for any matrix M ∈< U_{1}, U_{2}, U_{3} > we have MAM^{T} = A):

A = ( \begin{matrix}1 & 0 & 0 & 0 & 0 & 0 & 0 & c_{1} \\ 0 & 1 & 0 & 0 & 0 & 0 & −c_{1} & 0 \\ 0 & 0 & 1 & −1 − t & c_{1} & −1 − t & 0 & 0 \\ 0 & 0 & −1 − t & 1 & −1 − t & c_{1} & c_{2} & 0 \\ 0 & 0 & c_{1} & −1 − t & 1 & −1 − t & c_{2} & c_{2} \\ 0 & 0 & −1 − t & c_{1} & −1 − t & 1 & 0 & c_{2} \\ 0 & −c_{1} & 0 & c_{2} & c_{2} & 0 & 1 & c_{2} \\ c_{1} & 0 & 0 & 0 & c_{2} & c_{2} & c_{2} & 1 \end{matrix} ) .

where c_{1} = 1 + t + t^{2}/2 and c_{2} = −t(2 + t)(1 + t)/2.

The matrices for the 11 × 11 representation ρ_{4}^{(t)} : B_{4} → GL_{n}(ℤ[t]) are

σ_{1} → ( \begin{matrix}0 & 1 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 \\ 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 & 0 & 1 & 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 \\ t & −t & 0 & 0 & 0 & −1 − t & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & t & −t & 0 & 0 & −1 − t & 0 & 0 & 0 & 0 \\ 0 & 0 & −t & 0 & t & 0 & 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & t & −t & −t & 0 & 1 & −t & 1 & 0 \\ t^{2} & −t^{2} & 0 & 0 & 0 & −t^{2} & 0 & 0 & 2 t & −1 & 0 \\ −t^{2} & t^{2} & 0 & 0 & 0 & −t & −t^{2} − t & 0 & t & −1 & 1 \end{matrix} )

and

π_{4} → ( \begin{matrix}0 & −1 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & −1 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & −1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 1 & −1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & −1 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 \\ t & −t & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & −t & 0 & 0 & 1 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 \end{matrix} ) .

Here π_{4} = σ_{1}σ_{2}σ_{3} is the permutation braid of order 4 in B_{4}/Z(B_{4}) so that we have π_{4}σ_{1}π_{4}^{−1} = σ_{2} and π_{4}^{2}σ_{1}π_{4}^{−2} = σ_{3}. Thus π_{4} and σ_{1} generate B_{4}.

The 2 linearly independent symmetric matrices (q_{1} and q_{2}) that are fixed by this representation (as well as the above matrices) can be down-loaded from my WWW page. The address is

http : //www.math.byu.edu/P eople/F aculty/Humphries Stephen.html. The above matrices are calculated relative to the following elements (which we will shortly show is a basis): a_{12}a_{21} − a_{13}a_{31}, a_{12}a_{21} − a_{23}a_{32}, a_{12}a_{21} − a_{14}a_{41}, a_{12}a_{21} − a_{24}a_{42}, a_{12}a_{21} − a_{34}a_{43}, − a_{23}a_{31}a_{12} + a_{21}a_{13}a_{32} + a_{21}a_{13}a_{31}a_{12}, −a_{24}a_{41}a_{12} + a_{21}a_{14}a_{42} + a_{21}a_{14}a_{41}a_{12}, a_{24}a_{43}a_{32} − a_{23}a_{34}a_{42} + a_{23}a_{32}a_{24}a_{42}, a_{31}a_{14}a_{43} − a_{34}a_{41}a_{13} + a_{34}a_{43}a_{31}a_{13}, a_{34}a_{43}a_{21}a_{13}a_{31}a_{12} − a_{43}a_{32}a_{21}a_{14} + a_{43}a_{31}a_{12}a_{24} + a_{34}a_{42}a_{21}a_{13} − a_{34}a_{41}a_{12}a_{23}, a_{41}a_{14}a_{32}a_{24}a_{42}a_{23} − a_{43}a_{32}a_{21}a_{14} + a_{14}a_{42}a_{23}a_{31} + a_{41}a_{13}a_{32}a_{24} − a_{34}a_{41}a_{12}a_{23}.

Each of the above polynomials is a sum of elements of ϕ(D_{42}) as one can readily check.

Remark 5.1. We note that for values of t close to −1/2 and for c_{1}, c_{2} close to 1 the invariant symmetric matrix c_{1}q_{1} + c_{2}q_{2} is positive definite. Further, for t in the interval (−2, 0) the above 8 × 8 matrix A is also positive definite.

Lemma 5.2. The above set of polynomials is a free basis for ϕ(D_{42})/ϕ(D_{42})^{2} as a ℂ[t] ≅ ϕ(C_{42})/ϕ(D_{42})-module.

Proof. Note that for any n, since D_{n2} ⊂ C_{n2}, we see that there is a homomorphism

η = η_{n} : ϕ(D_{n2})/ϕ(D_{n2})^{2} → ϕ(D_{n2})/(ϕ(C_{n2})^{2} ∩ ϕ(D_{n2})) ↪ ϕ(C_{n2})/ϕ(C_{n2})^{2},

which is induced by this inclusion. Here ϕ(D_{n2})/(ϕ(C_{n2})^{2}∩ϕ(D_{n2})) is a ϕ(C_{n2})/ϕ(D_{n2}) ≅ ℂ[t]-module and ϕ(D_{n2})/(ϕ(C_{n2})^{2} ∩ ϕ(D_{n2})) and ϕ(C_{n2})/ϕ(C_{n2})^{2} are ℂ-modules, so that we can think of η acting on ℂ[t] by η(t) = 0. Now to show that ϕ(D_{42})/ϕ(D_{42})^{2} is a free ϕ(C_{42})/ϕ(D_{42})-module (on the above 11 generators) it will suffice to show that the images under η_{4} of these elements are linearly independent over ℂ. Now mod ϕ(C_{42})^{2} the above 11 elements are

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

Note that each element in this list is a sum of monomial summands each of which is a simple cycle. Note next that the only such simple cycle which occurs as a summand in more than one of these elements is a_{12}a_{21}. It is thus clear that these elements are linearly independent over ℂ. □

Lemma 5.3. The representations ρ_{n}^{(t)} are infinite except possibly when t has the form t = −2(cos(πa) + 1) where a ∈ ([1/2, 3/2] ∩ ℚ) + 2ℤ.

Proof. The characteristic polynomial of ρ_{4}(σ_{1}) is χ(z) = (z − 1)^{5} (z t + z^{2} + 2 z + 1)^{3}. If ρ_{4}(σ_{1}) is to have finite order, then we must have z = e^{iπa} where a ∈ ℚ. One solves to obtain t = −2(cos(πa) + 1). With this value of t we see that the characteristic polynomial of ρ_{4}^{(t)}(σ_{1}^{2}σ_{2}^{2}σ_{3}^{2}) is

(z − 1)^{3}(−z^{2} + 2 z − 1 + 16 cos(π a) z + 16 z cos^{2}(π a))(1 + 2 z + 4 cos(π a)z + z^{2})^{2}×

(16 z cos^{4}(π a) + 16 z cos^{3}(π a) + 4 z cos^{2}(π a) − z^{2} − 1 − 2 z).

The result follows by showing that the only roots of unity z satisfying the second factor of this equation correspond to the values of a given in the Lemma. □

If we put t = −1 in the above 8×8 representation for B_{4}, then we get an integral matrix representation of the Hessian group of order 216. If we put t = 0 in the 8 × 8 representation, then we obtain a representation which can be shown to be equivalent to a representation looking like

( \begin{matrix}A & 0 & 0 \\ B & C & 0 \\ D & E & F \end{matrix} )

where A gives a 2 × 2 representation of S_{3} and both C and F give a 3 × 3 representation of S_{4} (which are not equivalent). Thus the group with t = 0 fits in a short exact sequence

1 → N →< U_{1}, U_{2}, U_{3} > |_{t=0} → S_{3} × S_{4} × S_{4} → 1.

Here N is an infinite class 2 nilpotent group .

§6. Action of B_{5} on C_{52}

The situation here is very similar to that for n = 4. This action gives a representation of degree 39. The matrices, being large, will not be given here, but can be found on my WWW page. The address is given above. Also to be found there are two symmetric matrices w_{1} and w_{2} which represent symmetric forms that are fixed by this representation. There is a result analogous to Lemma 5.2 in this situation which shows that we also have a free ϕ(C_{52})/ϕ(D_{52})-module in this case. Checking all of this for the given matrices will constitute a proof of Theorem 1.2. If one puts t = −1 in this representation, then one obtains a representation of the simple group O(5, 3) of order 25920. This representation (with t = −1) splits into two irreducible representations of O(5, 3) of degrees 15 and 24.

Lastly we note that for values of t close to −1 and for c_{1}, c_{2} close to 1 the invariant symmetric matrix c_{1}w_{1} + c_{2}w_{2} is positive definite. In fact for t = −1 the matrix w_{1} + w_{2} is I_{39} + J_{39} where J_{39} is the 39 × 39 all 1 matrix, and I_{39} + J_{39} is easily seen to be positive definite.

§7. Action of B_{4} on C_{43}

There is a relation that we can use in C_{n3} analogous to the heart relation for C_{n2}:

Lemma 7.1. Let c = γ(x_{2}x_{3}x_{4}). Then we have

[(σ_{1} − σ_{1}^{3} + a_{12}a_{21}σ_{1} − a_{12}a_{21}σ_{1}^{2} − σ_{1}^{2} + 1)(c)] = 0.

Proof. This is easily checked, as in the proof of the heart relation, by taking traces. □

Remark 7.2. The element σ_{1} − σ_{1}^{3} + a_{12}a_{21}σ_{1} − a_{12}a_{21}σ_{1}^{2} − σ_{1}^{2} + 1 of the group algebra of B_{n} over ℤ[a_{12}a_{21}] has the property that

[(σ_{1} − σ_{1}^{3} + a_{12}a_{21}σ_{1} − a_{12}a_{21}σ_{1}^{2} − σ_{1}^{2} + 1)(c)] = 0

for any c = γ_{23...k} with k ≤ n. In fact for any c ∈ C_{n} there is a polynomial P_{c} = P_{c}(σ_{1}) over ℤ[a_{12}a_{21}] such that [P_{c}(c)] = 0. We shall not use this fact in this paper.

Lemma 7.1 indicates that we might obtain a finite-dimensional representation if we look at D_{n3}/D_{n3}^{2} as a C_{n2}/D_{n2}-module.

Recall that we are considering D_{n3}/D_{n3}^{2} as a C_{n2}/D_{n2}-module and that C_{n2}/D_{n2} ≅ ℂ[t]. We again note that if we put t = −1, then we get

[σ_{1}^{3}(γ(x_{2}x_{3}x_{4})) − γ(x_{2}x_{3}x_{4})]|_{t=−1} = 0.

Again we define an equivalence relation ∼^{3} by γ_{1} ∼^{3} γ_{2} if [γ_{1} − γ_{2}]|_{t=−1} = 0. An argument similar to that for the heart relation for C_{n2} in §2 shows that Proposition 7.3. a) Let Q_{n} =< Stab_{B_{n}}(γ_{123}), σ_{3}^{3} > . Then for any α ∈ Q_{n} we have

α(γ_{123}) ∼^{3} γ_{123}.

b) Stab_{B_{n}}(γ_{123}) =< σ_{1}, σ_{2}, σ_{4}, …, σ_{n−1}, (σ_{1}σ_{2})^{3}, (σ_{1}σ_{2}σ_{3})^{4}, …, (σ_{1}σ_{2} … σ_{n−1})^{n} > for n = 4, 5.

c) [B_{4} : Q_{4}] = 9; [B_{5} : Q_{5}] = 90.

d) [B_{n} : Q_{n}] = ∞ for n > 5.

Proof. The proof of a) is similar to the proof of Corollary 2.6. The proof of b) again follows from the results of Gurzo [G]. MAGMA calculates the indices for c). The proof of d) follows from the methods of section 8. □

Unfortunately we have not been able to show that in this case we obtain a finite-dimensional representation.

§8. The case n = 6

In this section we consider the case n = 6. If G is a group and S ⊂ G, then we let N(S) = N_{G}(S) denote the normal closure of S in G.

Theorem 8.1. a) The quotient B_{6}/N(σ_{1}^{3}, Z(B_{6})) is an infinite group.

b) The index [B_{6} : K_{6}] is infinite.

Proof. We first prove a). Let ω = (1 + i\sqrt{3})/2. Consider the representation Φ : B_{6} → GL(6, ℤ[ω]) induced by sending σ_{i} to M_{i}, where M_{i} are the following complex reflections of order 3:

M_{1} = ( \begin{matrix}0 & 1 & 0 & 0 & 0 & 0 \\ ω̄ & ω & 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 0 & 1 \end{matrix} ) , M_{2} = ( \begin{matrix}1 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 & 0 \\ 0 & ω̄ & ω & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 0 & 1 \end{matrix} ) ,

M_{3} = ( \begin{matrix}1 & 0 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & ω̄ & ω & 0 & 0 \\ 0 & 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 0 & 1 \end{matrix} ) , M_{4} = ( \begin{matrix}1 & 0 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & ω̄ & ω & 0 \\ 0 & 0 & 0 & 0 & 0 & 1 \end{matrix} ) ,

M_{5} = ( \begin{matrix}1 & 0 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 & ω̄ & ω \end{matrix} ) .

Now we let Q ∈ GL(6, ℤ[ω, ω̄]) be the following matrix:

Q = ( \begin{matrix}1 & ω & −ω̄ & −1 & −ω & ω̄ \\ 0 & −1 & 1 & 0 & 0 & 0 \\ 0 & −1 & 0 & 1 & 0 & 0 \\ 0 & −1 & 0 & 0 & 1 & 0 \\ 0 & −1 & 0 & 0 & 0 & 1 \\ 0 & 1 & 0 & 0 & 0 & 0 \end{matrix} ) ,

and τ_{i} = QM_{i}Q^{−1} for i = 1, …, 5. Then

τ_{1} = ( \begin{matrix}1 & 0 & 0 & 0 & 0 & 0 \\ −ω^{−1} & 1 + ω & −ω^{−1} & −1 & −ω & 0 \\ −ω^{−1} & ω & ω & −1 & −ω & 0 \\ −ω^{−1} & ω & −ω^{−1} & 0 & −ω & 0 \\ −ω^{−1} & ω & −ω^{−1} & −1 & ω^{−1} & 0 \\ ω^{−1} & −ω & ω^{−1} & 1 & ω & 1 \end{matrix} ) , τ_{2} = ( \begin{matrix}1 & 0 & 0 & 0 & 0 & 0 \\ 0 & −ω^{−1} & 0 & 0 & 0 & 0 \\ 0 & −1 & 1 & 0 & 0 & 0 \\ 0 & −1 & 0 & 1 & 0 & 0 \\ 0 & −1 & 0 & 0 & 1 & 0 \\ 0 & 1 & 0 & 0 & 0 & 1 \end{matrix} ) ,

τ_{3} = ( \begin{matrix}1 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 & 0 \\ 0 & ω^{−1} & ω & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 0 & 1 \end{matrix} ) , τ_{4} = ( \begin{matrix}1 & 0 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & ω^{−1} & ω & 0 & 0 \\ 0 & 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 0 & 1 \end{matrix} ) ,

τ_{5} = ( \begin{matrix}1 & 0 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & ω^{−1} & ω & 0 \\ 0 & 0 & 0 & 0 & 0 & 1 \end{matrix} ) .

Let T =< τ_{1}, …, τ_{5} >. Note that all of these matrices have 0’s in the (1, 2), (1, 3), …, (1, 6) and (2, 6), (3, 6), …, (5, 6) entries. This fact allows us to define a representation ψ as follows: for i = 1, …, 5 let μ_{i} be the 4 × 4 matrix obtained from τ_{i} by deleting rows 1 and 6 and columns 1 and 6. It is easy to see that this is a representation from the above fact. Thus we have an epimorphism

ψ : T → M =< μ_{1}, …, μ_{5} >, ψ(τ_{i}) = μ_{i}, i = 1, …, 5.

Using MAGMA one can show that the group M is a finite group of order 155, 520 having O(5, 3) as a quotient. The kernel of ψ is thus a subgroup of the group

E =< E_{21}(y), E_{31}(y), E_{41}(y), E_{51}(y), E_{61}(y), E_{62}(y), E_{63}(y), E_{64}(y), E_{65}(y) > .

Here y ∈ {1, ω, ω̄} and for 1 ≤ i ≠ j ≤ 6 E_{ij}(y) is the 6×6 matrix differing from the identity only in the ij entry which is equal to y. Note that the subgroup E is nilpotent of class 2 and so E/Z(E) is abelian. Here Z(E) =< E_{61}(ω), E_{61}(ω̄), E_{61}(1) >.

Now, the image of (σ_{1}σ_{2} … σ_{5})^{6}, which generates the centre Z(B_{6}), is

E_{61}(−ω̄) = ( \begin{matrix}1 & 0 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 & 1 & 0 \\ −ω̄ & 0 & 0 & 0 & 0 & 1 \end{matrix} ) .

Now the only τ_{i} with a non-zero entry in the (6, 1) entry is τ_{1} and so any element of < τ_{1}, …, τ_{5} > has (6, 1) entry which is a multiple of ω̄. Thus to prove a) we need only find an element of ker(ψ) which is not in < E_{61}(ω̄) > and which has infinite order. The shortest such element that we can find (using MAGMA) is

τ_{1}τ_{5}τ_{4}τ_{3}^{2}τ_{2}^{2}τ_{1}^{2}τ_{2}τ_{4}^{2}τ_{3}τ_{4}^{2}τ_{2}^{2}τ_{5}τ_{4}^{2}τ_{3}^{2}τ_{2}τ_{1}^{2}τ_{2}τ_{3}τ_{4}^{2}τ_{5}τ_{3}τ_{4}τ_{2}^{2}τ_{1}τ_{2}^{2}τ_{3}^{2}τ_{4}^{2}τ_{2}^{2}τ_{1}τ_{2}^{2}τ_{3}τ_{4}^{2}τ_{5}^{2}τ_{3}τ_{4}^{2}

τ_{2}^{2}τ_{1}τ_{2}^{2}τ_{3}τ_{4}^{2}τ_{2}^{2}τ_{1}τ_{2}^{2}τ_{3}τ_{4}^{2}τ_{2}τ_{4}τ_{3}^{2}τ_{2}τ_{1}^{2}τ_{2}τ_{4}τ_{3}^{2}τ_{2}τ_{1}^{2}τ_{2}τ_{4}τ_{3}^{2}τ_{5}τ_{4}τ_{3}^{2}τ_{2}τ_{1}^{2}τ_{2}τ_{4}τ_{3}τ_{2}τ_{1}^{2}τ_{2}τ_{4}^{2}τ_{3}^{2}τ_{5}^{2}

τ_{4}τ_{3}^{2}τ_{2}^{2}τ_{1}τ_{2}^{2}τ_{3}τ_{4}τ_{5}^{2}τ_{2}τ_{4}τ_{3}^{2}τ_{4}τ_{2}^{2}τ_{1}τ_{2}τ_{3}τ_{4}^{2}τ_{5}^{2}τ_{1}^{2}τ_{5}τ_{4}τ_{3}^{2}τ_{2}^{2}τ_{1}^{2}τ_{2}τ_{4}^{2}τ_{3}τ_{4}^{2}τ_{2}^{2}τ_{5}τ_{4}^{2}τ_{3}^{2}τ_{2}τ_{1}^{2}τ_{2}τ_{3}τ_{4}^{2}

τ_{5}τ_{3}τ_{4}τ_{2}^{2}τ_{1}τ_{2}^{2}τ_{3}^{2}τ_{4}^{2}τ_{2}^{2}τ_{1}τ_{2}^{2}τ_{3}τ_{4}^{2}τ_{5}^{2}τ_{3}τ_{4}^{2}τ_{2}^{2}τ_{1}τ_{2}^{2}τ_{3}τ_{4}^{2}τ_{2}^{2}τ_{1}τ_{2}^{2}τ_{3}τ_{4}^{2}τ_{2}^{2}τ_{4}τ_{3}^{2}τ_{2}τ_{1}^{2}τ_{2}τ_{4}τ_{3}^{2}τ_{2}τ_{1}^{2}

τ_{2}τ_{4}τ_{3}^{2}τ_{5}τ_{4}τ_{3}^{2}τ_{2}τ_{1}^{2}τ_{2}τ_{4}τ_{3}τ_{2}τ_{1}^{2}τ_{2}τ_{4}^{2}τ_{3}^{2}τ_{5}^{2}τ_{4}τ_{3}^{2}τ_{2}^{2}τ_{1}τ_{2}^{2}τ_{3}τ_{4}τ_{5}^{2}τ_{2}τ_{4}τ_{3}^{2}τ_{4}τ_{2}^{2}τ_{1}τ_{2}τ_{3}τ_{4}^{2}τ_{5}^{2}τ_{2}τ_{1}^{2}

τ_{2}^{2}τ_{3}τ_{2}τ_{1}τ_{5}τ_{4}τ_{3}^{2}τ_{2}^{2}τ_{1}^{2}τ_{2}τ_{4}^{2}τ_{3}τ_{4}^{2}τ_{2}^{2}τ_{5}τ_{4}^{2}τ_{3}^{2}τ_{2}τ_{1}^{2}τ_{2}τ_{3}τ_{4}^{2}τ_{5}τ_{3}τ_{4}τ_{2}^{2}τ_{1}τ_{2}^{2}τ_{3}^{2}τ_{4}^{2}τ_{2}^{2}τ_{1}τ_{2}^{2}τ_{3}τ_{4}^{2}τ_{5}^{2}τ_{3}τ_{4}^{2}

τ_{2}^{2}τ_{1}τ_{2}^{2}τ_{3}τ_{4}^{2}τ_{2}^{2}τ_{1}τ_{2}^{2}τ_{3}τ_{4}^{2}τ_{2}^{2}τ_{3}τ_{4}τ_{3}τ_{2}^{2}τ_{1}τ_{3}τ_{2}^{2}τ_{1}τ_{4}τ_{3}^{2}τ_{2}τ_{4}τ_{5}τ_{4}^{2}τ_{3}^{2}τ_{2}τ_{1}^{2}τ_{2}τ_{5}τ_{4}τ_{3}^{2}τ_{2}τ_{1}τ_{4}τ_{3}τ_{2}^{2}τ_{5}.

Call this element τ′. Then τ′ is the matrix

( \begin{matrix}1 & 0 & 0 & 0 & 0 & 0 \\ −2 − i\sqrt{3} & 1 & 0 & 0 & 0 & 0 \\ 1/2 − \frac{3 i\sqrt{3}}{2} & 0 & 1 & 0 & 0 & 0 \\ 5/2 − \frac{i\sqrt{3}}{2} & 0 & 0 & 1 & 0 & 0 \\ 5/2 − \frac{i\sqrt{3}}{2} & 0 & 0 & 0 & 1 & 0 \\ −25/2 − \frac{5 i\sqrt{3}}{2} & 5 + i\sqrt{3} & 1/2 + \frac{3 i\sqrt{3}}{2} & −9/2 + \frac{i\sqrt{3}}{2} & −3 − 2 i\sqrt{3} & 1 \end{matrix} ) .

By considering the representation of E that one obtains by looking at the principal 2 × 2 submatrix of any matrix in E we see that the above element has infinite order when considered as an element of B_{6}/N(σ_{1}^{3}, ζ_{6}). Here ζ_{n} is a generator for Z(B_{n}), so that for example ζ = ζ_{6} = (σ_{1}σ_{2}σ_{3}σ_{4}σ_{5})^{6} is a generator for Z(B_{6}). This proves a).

For b) we have

[B_{6} : N(σ_{1}^{3}, ζ_{6})] = [B_{6} :< K_{6}, N(σ_{1}^{3}, ζ_{6}) >][< K_{6}, N(σ_{1}^{3}, ζ_{6}) >: N(σ_{1}^{3}, ζ_{6})]
= [B_{6} :< K_{6}, N(σ_{1}^{3}, ζ_{6}) >][K_{6} : K_{6} ∩ N(σ_{1}^{3}, ζ_{6})].  (8.1)

We next show that [< K_{6}, N(σ_{1}^{3}, ζ) >: N(σ_{1}^{3}, ζ)] is finite. Now

K_{6} =< σ_{1}, σ_{2}^{3}, σ_{3}, σ_{4}, σ_{5}, ζ_{3}, ζ_{4}, ζ_{5}, ζ_{6} >,

Further, < σ_{1}, ζ_{3}, ζ_{4}, ζ_{5}, ζ_{6} > is in the centre of K_{6}/N(σ_{1}^{3}, ζ_{6}). One can check using MAGMA [M] that

[B_{4} : N_{B_{4}}(σ_{1}^{3})] = 216, [B_{5} : N_{B_{5}}(σ_{1}^{3})] = 25, 920.

Thus < ζ_{3}, ζ_{4}, ζ_{5}, ζ_{6} > /(< ζ_{3}, ζ_{4}, ζ_{5}, ζ_{6} > ∩ N(σ_{1}^{3}, ζ_{6})) is finite. Thus to see that [< K_{6}, N(σ_{1}^{3}, ζ_{6}) >: N(σ_{1}^{3}, ζ_{6})] is finite we need only note the isomorphism

< σ_{3}, σ_{4}, σ_{5} > /(N_{B_{6}}(σ_{1}^{3})∩ < σ_{3}, σ_{4}, σ_{5} >) ≅ B_{4}/N_{B_{4}}(σ_{1}^{3})

and the fact that B_{4}/N_{B_{4}}(σ_{1}^{3}) is finite by the above.

From (8.1) we now see that

[B_{6} : N(σ_{1}^{3}, ζ_{6})] = ∞ if and only if [B_{6} :< K_{6}, N(σ_{1}^{3}, ζ_{6}) >] = ∞.

From a) we thus get that [B_{6} :< K_{6}, N(σ_{1}^{3}, ζ_{6}) >] = ∞, which clearly implies that [B_{6} : K_{6}] = ∞. This concludes the proof of Theorem 8.1. □

We have noted (i) that E, and so ker(ψ), is nilpotent of class 2; (ii) that any τ ∈ T has (6, 1) entry which is an integer multiple of the (6, 1) entry of τ_{1}; (iii) the (6, 1) entry of τ_{1} is equal to the negative of the (6, 1) entry of the image of ζ_{6}. Denote the subgroup generated by Φ(ζ_{6}) by Z. From these facts we conclude that ker(ψ)/Z is a free abelian group and that T /Z fits in an exact sequence

1 → ℤ^{r} → T /Z → C → 1,

where C is a finite group.

§9. The heart relation revisited.

If we let g_{i} = [σ_{1}^{i}(γ_{23})] for i ≥ 0, then the heart relation can be restated as g_{3} = g_{0} +(1+t)(g_{1} −g_{2}). Thinking of the g_{i} as indeterminates which are related by this latter relation and the fact that g_{i+1} = σ_{1}g_{i} we find that we can easily express g_{i} for i ≥ 0 as a linear sum

g_{i} = c_{i0}g_{0} + c_{i1}g_{1} + c_{i2}g_{2}

where c_{ij} ∈ ℤ[t]. More precisely we have

Lemma 9.1. a) c_{00} = 1, c_{01} = 0, c_{02} = 0, c_{10} = 0, c_{11} = 1, c_{12} = 0, c_{20} = 0, c_{21} = 0, c_{22} = 1, c_{30} = 1, c_{31} = 1 + t, c_{32} = −1 − t and for i ≥ 3 we have c_{i+1 0} = c_{i2}, c_{i+1 1} = c_{i0} + c_{i2}(1 + t) and c_{i+1 2} = c_{i1} − c_{i2}(1 + t).

b) Let κ_{i} = gcd(c_{i0} − 1, c_{i1}, c_{i2}). Then we have

c_{i0} = 1 + (−1)^{i−1}κ_{i}κ_{i−3};
c_{i1} = \{ \begin{matrix}κ_{i}κ_{i−2} if i is odd; \\ −tκ_{i}κ_{i−2} if i is even; \end{matrix}
c_{i2} = (−1)^{i}κ_{i}κ_{i−1}.

c) For all i ≥ 1 we have

κ_{i}κ_{i−1} − κ_{i+1}κ_{i−2} = (−1)^{i}.

d) κ_{i} = (2 + t)κ_{i−2} − κ_{i−4} and

κ_{i} = \{ \begin{matrix}κ_{i−1} + κ_{i−2} if i iseven; \\ tκ_{i−1} + κ_{i−2} if i is odd. \end{matrix}

e) The degree of κ_{i} is [(i − 1)/2].

f) For all i ≥ 0 we have

κ_{i} = \frac{(t + \sqrt{t})(\sqrt{4 + t}/2 + \sqrt{t}/2)^{i}}{2t \sqrt{4 + t}} + \frac{(t − \sqrt{t})(\sqrt{4 + t}/2 − \sqrt{t}/2)^{i}}{2t \sqrt{4 + t}}

− \frac{(t + \sqrt{t})(−\sqrt{4 + t}/2 + \sqrt{t}/2)^{i}}{2t \sqrt{4 + t}} − \frac{(t − \sqrt{t})(−\sqrt{4 + t}/2 − \sqrt{t}/2)^{i}}{2t \sqrt{4 + t}} .

Proof. Since σ_{1}g_{i} = g_{i+1} and g_{3} = g_{0} +(1+t)(g_{1} −g_{2}) the proof of a) is straightforward. The proof of b), c), d) and e) is by simultaneous induction on i. The proof of f) follows by standard methods from d). □

Corollary 9.2. a) For i ≥ 1 we have the factorisation

g_{i} − g_{0} = κ_{i}K_{i}

where K_{i} ∈ ℤ[t, g_{0}, g_{1}, g_{2}].

b) At a value of t which is a root of κ_{i} we see that

[σ_{1}^{i}(γ_{23})] = g_{i} = g_{0} = [γ_{23}].

c) At a value of t which is a root of any of the κ_{i}, i > 0, the representation is not faithful.

Proof. Only the last statement is not completely obvious. Let ρ = ρ_{n}^{(t)} be the representation. For c), if κ_{i}(s) = 0, then by b) we have [σ_{1}^{i}(γ_{23})]|_{t=s} = [γ_{23}], from which it follows that ρ(σ_{1}^{i}σ_{2}σ_{1}^{−i}σ_{2}^{−1})|_{t=s} = id. But in B_{n} no power of σ_{1} can commute with σ_{2} (the proof is easy, but this fact also follows directly from the results of Gurzo [G]). Thus we have a non-trivial element in the kernel. □

Now note that we can put the coefficients of the κ_{i} (in decreasing powers of t) in a table as follows:

1 1

1  1
1  2
1  3  1
1  4  3
1  5  6  1
1  6  10  4
1  7  15  10  1
⋮ ⋮ ⋮ ⋮ ⋮ ⋮ ⋮ ⋮ ⋮

Note that this table is just a skewed version of Pascal’s triangle so that the entries in the kth row of this table are \binom{k}{0}, \binom{k−1}{1}, \binom{k−2}{2}, . . .. Here the first row in the table (corresponding to κ_{0}) is labelled the 0th row of the table. We further note that the polynomials κ_{i} are just a slightly disguised version of some polynomials P_{k}(t) defined by Jones in [J3]. These are polynomials defined recursively by P_{0} = 0, P_{1} = 1, P_{k}(t) = P_{k−1}(t) − tP_{k−2}(t) for k ≥ 2. The exact relationship between Jones’s polynomials and the κ_{i}(t) is given by

Lemma 9.3. For i ≥ 1 we have P_{i}(t) = (−t)^{[(i−1)/2]}κ_{i}(−1/t). □

This gives an indication of a link with the work of Jones [J3]. From the above lemma and [J3, Lemma 4.2.4] we obtain exact information about where the zeros of κ_{n}(t) are (which applies to Lemma 9.2):

Corollary 9.4. For fixed n the [(n−1)/2] roots of κ_{n}(t) are exactly all −4cos^{2}(mπ/n) for m ∈ ℤ, m > 0 with m/n < 1/2. □

Corollary 9.2 c) immediately gives

Corollary 9.5. For fixed n and t = −4cos^{2}(mπ/n) where m ∈ ℤ, m > 0 the representation is not faithful. □

Corollary 9.6. If i divides j, then κ_{i} divides κ_{j}.

Proof. By Corollary 9.4 κ_{i} has distinct roots and the result follows immediately since again Corollary 9.4 shows that a root of κ_{i} is also a root of κ_{j}. □

For s ∈ ℂ we let πρ_{n}^{(s)} denote the projectivised representation

B_{n} → ρ_{n}^{(s)}(B_{n}) → ρ_{n}^{(s)}(B_{n})/Z(ρ_{n}^{(s)}(B_{n})).

Proposition 9.7. Let n = 3, 4, 5 and s ∈ ℂ.

(i) If μ ∈ Core(K_{n}^{(s)}) and γ ∈ C_{n2}, then ρ_{n}^{(s)}(μτ_{γ}) = ρ_{n}^{(s)}(τ_{γ}μ). (ii) Core(K_{n}^{(s)}) ⊆ ker(πρ_{n}^{(s)}). (iii) The representation πρ_{n}^{(s)} factors through Core(K_{n}^{(s)}) and the representation ρ_{n}^{(t)} factors through ⋂_{s∈ℂ} Core(K_{n}^{(s)}).

Proof. First note that ρ_{n}^{(s)}(μτ_{γ}) = ρ_{n}^{(s)}(τ_{γ}μ) if and only if μ(γ) ∼_{(s)} γ. Also by Proposition 2.9 we see that if μ ∈ Core(K_{n}^{(s)}), then μ(γ_{12}) ∼_{(s)} γ_{12} and so ρ_{n}^{(s)}(μσ_{1}) = ρ_{n}^{(s)}(σ_{1}μ). For (i) we let μ ∈ Core(K_{n}^{(s)}) and γ ∈ C_{n2}. Then there is β ∈ B_{n} such that βτ_{γ}β^{−1} = σ_{1}. We have

ρ_{n}^{(s)}(βμτ_{γ}) = ρ_{n}^{(s)}(βμβ^{−1}σ_{1}β)
= ρ_{n}^{(s)}(σ_{1}βμβ^{−1}β) = ρ_{n}^{(s)}(σ_{1}βμ)
= ρ_{n}^{(s)}(ββ^{−1}σ_{1}βμ) = ρ_{n}^{(s)}(βτ_{γ}μ).

which gives (i).

(ii) If μ ∈ Core(K_{n}^{(s)}), then by (i) we have ρ_{n}^{(s)}(μτ_{γ}) = ρ_{n}^{(s)}(τ_{γ}μ), which shows that ρ_{n}^{(s)}(μ) is in the centre of ρ_{n}^{(s)}(B_{n}). Thus (ii) follows.

The first statement of (iii) is clear and the second follows since for n = 3, 4, 5 it is easy to show that the centre of ρ_{n}^{(t)}(B_{n}) is trivial. □

Remark 9.8. In order to show that ρ_{n}^{(t)} is faithful it would suffice to show (i) that ker(πρ_{n}^{(s)}) = Core(K_{n}^{(s)}) for all s ∈ ℂ; and (ii) ⋂_{s∈ℂ} Core(K_{n}^{(s)}) = Z(B_{n}).

Conjecture 9.9. We conjecture that the representation ρ_{n}^{(t)} of B_{n}/Z(B_{n}) is faithful for n = 4, 5.

§10. The representations ρ_{n}^{(t)} at some specialisations of t.

In this section all representations will be over the complex numbers. We have already indicated that puting t = −1 gives rise to representations of B_{n}, n = 3, 4, 5 which are groups of orders 24, 216 and 25920 (respectively), groups which also occur naturally in the work of Jones [J2].

The proof of Lemma 5.2 prompts us to look at the situation where we put t = 0. It was there indicated that doing so corresponds to looking at the representation of B_{n} on the quotient ϕ(D_{n2})/(ϕ(C_{n2})^{2} ∩ ϕ(D_{n2})) (as a vector space over ℂ). We will now in fact consider the more natural action of B_{n} on Y_{n}/Y_{n}^{2}. This latter vector space is easily seen to be finite dimensional since (as we have noted previously) there are only a finite number of simple cycles in Y_{n}.

Now the symmetric group S_{n} also acts on Y_{n}/Y_{n}^{2} and we can write this action as M_{1} ⊕ ⋯ ⊕ M_{κ(n)} where the M_{i} are irreducible representations of S_{n} such that M_{k} is a S_{n}-submodule generated by elements of degree d(k). We can also require that if i < j, then d(i) ≤ d(j). We will assume that the basis b_{1}, …, b_{β(n)} for Y_{n}/Y_{n}^{2} is a concatenation of the bases for the M_{k} in the same non-degree-decreasing order. We define the lowest degree ld(x) of an element x of a polynomial ring to be the smallest degree of any non-trivial monomial summand of x.

Note that if α ∈ B_{n} and b_{i} ∈ M_{k}, then α(b_{i}) = b + b′ where b ∈ M_{k} and b′ is a sum of monomials each having strictly greater lowest degree than the lowest degree of b_{i}. The above now gives

Theorem 10.1. The representation of B_{n} on Y_{n}/Y_{n}^{2} can be decomposed so that relative to the basis b_{1}, …, b_{β(n)} chosen above the matrices have the form

( \begin{matrix}A_{1} & 0 & 0 & … \\ B_{2} & B_{1} & 0 & … \\ C_{3} & C_{2} & C_{1} & … \\ ⋮ & ⋮ & ⋮ & ... \end{matrix} )

where A_{1}, B_{1}, C_{1}, … are irreducible representations of S_{n}. This representation is finite by nilpotent. □

At the end of §5 we gave the indicated decomposition for the action of B_{4} on ϕ(D_{n2})/ϕ(D_{n2})^{2} with t = 0 and found that all the non-abelian representations occurred as the associated representations of S_{4}. In fact this also happens for n = 5 with multiplicities 1 or 2.

For the specialisation t = −2 we can show that for n = 4 we get a decomposition of the representation looking like

( \begin{matrix}1 & 0 & 0 & 0 \\ ∗ & A & 0 & 0 \\ ∗ & ∗ & B & 0 \\ ∗ & ∗ & ∗ & C \end{matrix} ) .

Here A and C are 3×3 irreducible representations of S_{4} and B is a 4×4 irreducible representation of a group G of order 96 with centre C_{4} and quotient G/C_{4} ≅ S_{4}.

For t = −2 and n = 5 we get a similar block lower triangular decomposition with irreducible representations of sizes 1, 5, 8, 10 and 15. The first is the trivial representation, while the representations of sizes 5 and 10 are of a group G of order 1920 having a normal subgroup N ≅ C_{2}^{4} and quotient G/N ≅ S_{5}. The representation of size 8 is of a group G′ of order 7680 with centre C_{4} and quotient G′/centre(G′) ≅ G. The remaining representation is infinite.

For n = 4 and the specialisation where t is a root of the polynomial κ_{5}(t) = t^{2} +

3t+1 we get a decomposition of the representation having the form ( \begin{matrix}A & 0 \\ ∗ & B \end{matrix} ) where

A is a 3 × 3 irreducible representation of A_{5} and B is an irreducible representation of an infinite group.

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