REPRESENTATIONS OF BRAID

GROUPS VIA DETERMINANTAL RINGS

Stephen P. Humphries

Abstract. We construct representations for braid groups B_{n} via actions of B_{n} on a determinantal ring, thus mirroring the setting of the classical representation theory for GL_{n}. The respresentations that we construct fix a certain unitary form.

§1. Introduction

Let C be a commutative ring with identity. In this paper we attempt to do for the braid groups B_{n} what has been done for GL_{n}(C) relative to their representation theory. The point of view will be the following classical way of understanding the representation theory of GL_{n}(C): Let X = (a_{ij}) be a generic n × m matrix (where m ≥ n ≥ 1) with indeterminate entries and let R_{n} = C[a_{ij}, 1 ≤ i, j ≤ n] be the corresponding coordinate ring. Let G = GL_{n}(C) × GL_{m}(C). Then G acts on R as follows:

(A, B)(X) = AXB^{−1} for (A, B) ∈ G.

The representation theory of G is described using Young diagrams. Recall that a Young diagram is a finite subset σ of ℕ × ℕ such that (i, j) ∈ σ and 1 ≤ i′ ≤ i, 1 ≤ j′ ≤ j implies that (i′, j′) ∈ σ. The kth row of σ will be denoted by σ_{k}. There is a natural partial ordering of the σ. These diagrams correspond to irreducible representations of GL_{n}(C) as follows.

Given such a σ a tableau of shape σ is a function S : σ → {1, …, n}. The kth row of S will be denoted S_{k}. One writes σ = |S| and there is a natural partial ordering of tableau extending the above order of diagrams. A bitableau is a pair (S|T) of tableau of shape σ. Now to each bitableau we can associate a product of minors of the matrix X: for each row σ_{k} of σ = |S| = |T | we get the minor μ(S_{k}|T_{k}) corresponding to the entries a_{ij} where i ∈ S_{k}, j ∈ T_{k}. Then μ(S|T) is the product μ(S_{1}|T_{1})μ(S_{2}|T_{2}) … μ(S_{r}|T_{r}).

For each σ we let A_{σ} denote the subspace of R spanned by all tableau τ with τ ≥ σ and let A′_{σ} denote the subspace of R spanned by all tableau τ with τ > σ. Then A_{σ}/A′_{σ} is an irreducible G − module and this construction gives all of the irreducible representations of G. A fundamental role in this is given to the Plücker relations, these giving all relations among products of the monomials and a partial ordering to the minors. Details of this can be found in [BV, DEP1, DEP2]. This approach also allows the calculation of the ring of invariants and many other important properties, even in arbitrary characteristic [op. cit.].

We now describe an analogous setup for B_{n}, noting the following papers which contain results on the general representation theory of the braid groups [A2, At, B, BLM, BW, F, Iv, J, La, Le, Li]. However we should also note that the representation theory for braid

Typeset by AMS-TEX

groups is more involved than that for linear groups since, for example, the braid groups are not rigid (see Theorem 1.9 below).

Let B_{n} denote the (algebraic) braid group, with standard generators σ_{1}, σ_{2}, …, σ_{n−1} and with relations

σ_{j}σ_{j+1}σ_{j} = σ_{j+1}σ_{j}σ_{j+1}, σ_{j}σ_{i} = σ_{i}σ_{j}  (1.1)

for 1 ≤ i, j < n − 1 and |i − j| > 1. It is well known that B_{n} has a faithful representation in Aut(F_{n}), where F_{n} =< T_{1}, …, T_{n} > is a free group on n generators [Bi]. This comes from an action of B_{n} on the disc D_{n} with n punctures π_{1}, …, π_{n} as isotopy classes of diffeomorphisms of D_{n} each fixing the boundary of the disc, so that the action of B_{n} on the fundamental group π_{1}(D_{n}, p) for p on the boundary of D_{n} gives the monomorphism ϕ_{n} : B_{n} → Aut(F_{n}). The action of a generator σ_{i}, i < n is as follows:

ϕ_{n}(σ_{i})(T_{i}) = T_{i}T_{i+1}T_{i}^{−1}; ϕ_{n}(σ_{i})(T_{i+1}) = T_{i}; ϕ_{n}(σ_{i})(T_{j}) = T_{j} for j ≠ i, i + 1.

Artin characterised the image of ϕ_{n}: each ψ ∈ Aut(F_{n}) such that

(i) ψ(T_{j}) is a conjugate of some T_{k}; and (ii) ψ(T_{1})ψ(T_{2}) … ψ(T_{n}) = T_{1}T_{2} … T_{n} [Bi]. We will also need to note that there is an epimorphism Π_{n} : B_{n} → S_{n}, given by the permutation action of a braid on the punctures {π_{1}, …, π_{n}}, whose kernel is the pure braid group P_{n}.

We obtain an action of B_{n} on a finitely generated polynomial algebra as follows; this will be defined by representing the generators T_{j} of F_{n} as transvections, specifically we let

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 a matrix M is a transvection [A1] if M = I_{n} + A where I_{n} is the identity matrix, det(M) = 1, rank(A) = 1 and A^{2} = 0. In particular, conjugates of transvections are transvections. We let R_{n}^{(0)} = C[a_{ij}, 1 ≤ i ≠ j ≤ n] be the corresponding ring, so that T_{i} ∈ SL_{n}(R_{n}^{(0)}), and in this context it will be convenient to put a_{ii} = 0 for i = 1, …, n.

The fact that the group < T_{1}, T_{2}, …, T_{n} > generated by these transvections is a free group of rank n was noted in [Hu2, Lemma 2.5]. The action of B_{n} on R_{n}^{(0)} comes from the action of B_{n} on the trace algebra associated to the matrix group F_{n}: note that the element T_{i}T_{j}, i ≠ j, represents the conjugacy class of the simple closed curve containing the punctures π_{i}, π_{j} in its interior. Now

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

and, if A, B ∈ F_{n}, then, since AT_{r}A^{−1}, BT_{s}B^{−1} are transvections, it similarly follows that

trace(AT_{r}A^{−1}BT_{s}B^{−1}) = b_{rs}b_{sr} + n,

for some b_{rs}, b_{sr} ∈ R_{n}^{(0)}. For α ∈ B_{n} Artin’s characterisation of braids given above shows that α(T_{i}) = AT_{r}A^{−1}, α(T_{j}) = BT_{s}B^{−1} for some A, B ∈ F_{n}, where Π_{n}(i) = r, Π_{n}(j) = s, and so if trace(AT_{r}A^{−1}BT_{s}B^{−1}) = b_{rs}b_{sr} + n, then [Hu1, Hu2, Hu4] we may choose b_{rs}, b_{sr} such that

b_{rs} = a_{rs} + terms of higher order, b_{sr} = a_{sr} + terms of higher order.

The action of B_{n} on R_{n}^{(0)} is then given by α(a_{ij}) = b_{rs}.

This action is non-linear; on generators it is given as follows:

σ_{i}(a_{i i+1}) = a_{i+1 i}, σ_{i}(a_{i+1 i}) = a_{i i+1}, σ_{i}(a_{h i+1}) = a_{h i},
σ_{i}(a_{h i}) = a_{h i+1} + a_{h i}a_{i,i+1}, σ_{i}(a_{i+1 j}) = a_{i j},
σ_{i}(a_{i j}) = a_{i+1 j} − a_{i+1 i}a_{i j},  (1.2)

where h, j ≠ i, i + 1.

A quotient of the above representation (1.2) was found by Magnus [Ma] when he was looking at the action of Aut(F_{n}) on the character variety of 2 × 2 complex matrices. This character variety is essentially a polynomial ring with some quadratic relations; however Magnus noted (”somewhat surprisingly” [Ma p.100]) that, relative to a certain set of generators, the action of B_{n} ⊂ Aut(F_{n}) on this character variety lifted to an action on a polynomial algebra with n(n − 1)/2 generators. In [Hu1] we gave the above explanation for the existence of a lift of Magnus’s representation, extending it to act on the n(n − 1) indeterminates a_{ij}.

Let c_{ijk...rs} represent the cycle a_{ij}a_{jk} … a_{rs}a_{si} ∈ R_{n}^{(0)}. Then the cycles generate a subalgebra of R_{n}^{(0)} denoted Y_{n}^{(0)}. Then Y_{n}^{(0)} is the trace ring for the matrix group F_{n}. A cycle c_{ijk...rs} will be called simple if i, j, k, …, r, s are all distinct. It is easy to see that every cycle in R_{n}^{(0)} is a product of simple cycles. It is also clear from the above action (1.2) of B_{n} on R_{n}^{(0)} that if c_{I} is a cycle in R_{n}^{(0)} and α ∈ B_{n}, then α(c_{I}) is a sum of integral multiples of monomials, each of which is a cycle. Thus Y_{n}^{(0)} is a B_{n}-invariant subring of R_{n}^{(0)}.

It follows from [Hu1, Theorem 2.5 and Theorem 6.2] that the kernel of the action of B_{n} on R_{n}^{(0)} is the centre of B_{n} and that if B_{n} and R_{n}^{(0)} are thought of as sub-objects of B_{n+1} and R_{n+1}^{(0)} (respectively), then the action of B_{n} on R_{n+1}^{(0)} is faithful.

We note as in [Hu2] that there is a natural ring involution ∗ on R_{n}^{(0)} which commutes with the action of B_{n}: α(w^{∗}) = α(w)^{∗} for all α ∈ B_{n} and all w ∈ R_{n}^{(0)}. This involution is determined by its action on the generators a_{ij} of R_{n}^{(0)} which is as follows: a_{ij}^{∗} = −a_{ji}. This involution has the following property:

trace(A^{−1}) = trace(A)^{∗},

for all A ∈ F_{n}. The action that Magnus discovered in [Ma], and that we referred to above, was the action on the n(n − 1)/2 symbols a_{ij} + a_{ij}^{∗}.

It is clear from the above presentation of B_{n} that, for r < n, the subgroup

B_{r,n} =< σ_{r}, σ_{r+1}, …, σ_{n−1} >

of B_{n} is isomorphic to B_{n−r+1} with B_{1,n} = B_{n}. Now given n_{1}, n_{2}, …, n_{s} ≥ 1 we let

G = G_{n_{1},n_{2},...,n_{s}} = B_{1,n_{1}} × B_{n_{1}+1,n_{1}+n_{2}} × B_{n_{1}+n_{2}+1,n_{1}+n_{2}+n_{3}} × …
× B_{n_{1}+···+n_{s−1}+1,n_{1}+···+n_{s}}
≅ B_{n_{1}} × B_{n_{2}} × ⋯ × B_{n_{s}}.

Then by the above there is an action of G on R_{n_{1}+···+n_{s}+r}^{(0)} and on Y_{n_{1}+···+n_{s}+r}^{(0)} for any r ≥ 0. Let n = n_{1} + n_{2} + ⋯ + n_{s} + r and M_{n}^{(0)} = (a_{ij}) where we have a_{ii} = 0 for all i = 1, …, n. Then this action respects the minors of M_{n}^{(0)} as follows: for any subsequences S, T of {1, 2, …, n} of the same length (thought of as bitableau with a single row) we let (S|T)^{(0)} denote the minor of M_{n}^{(0)} having rows taken from S and columns taken from T. (If either of S or T is the empty sequence, then the corresponding determinant will be taken to be 0). The action of B_{n} (or the subgroup G) on R_{n}^{(0)} induces an action of B_{n} on the (S|T)^{(0)} which, for S, T with one row, is given on generators as follows:

σ_{r}(S|T)^{(0)} = t_{r}[(S|T)^{(0)} + (S|S_{r}^{r+1}T)^{(0)}a_{r+1r} − a_{rr+1}(S_{r}^{r+1}S|T)^{(0)}
− a_{rr+1}(S_{r}^{r+1}S|S_{r}^{r+1}T)^{(0)}a_{r+1r}].  (1.3)

Here t_{r} is the transposition (r, r + 1) ∈ S_{n} acting on the indices of the a_{ij} and S_{r}^{r+1} acts on the sequences S, T as follows: S_{r}^{r+1}T is the empty sequence unless r is in T, while if r is in T, then S_{r}^{r+1}T is T with r replaced by r + 1. We shall sometimes write (1.3) as

σ_{r}(S|T)^{(0)} = t_{r} [(S − a_{rr+1}S_{r}^{r+1}(S)|T + S_{r}^{r+1}(T)a_{r+1r})^{(0)}] ,  (1.4)

where linearity in the two entries is understood (we will later give a better account of the context in which this action occurs).

The above shows that there is an action of B_{n} on the determinantal ideals generated by the minors of M_{n}^{(0)}, the complication being that if α ∈ B_{n}, then α(S|T)^{(0)} is a sum of terms of the form w(S′|T′)^{(0)}, where w ∈ R_{n}^{(0)}, and so this does not result in a finite-dimensional representation over C. We will indicate below how this situation can be modified so as to produce a finite-dimensional representation.

The action of B_{n} on the (S|T)^{(0)} can be extended to an action on products

(S_{1}|T_{1})^{(0)} × ⋯ × (S_{r}|T_{r})^{(0)}

in the obvious way. These products can then be represented using bitableau, as in the GL_{n} case. The set of all minors corresponding to such bitableau of a given shape, generate an ideal of R_{n}^{(0)} which, by (1.3), is B_{n}-invariant. Thus we now have a way of assigning to each Young diagram σ a B_{n}-invariant ideal A_{σ}^{(0)} of R_{n}^{(0)}. Again the problem is that, since the action of B_{n} on the a_{ij} is non-linear, this ideal is not finitely generated as a C-module.

The first modification will be to show (in Proposition 3.1) that the action of B_{n} on the (S|T)^{(0)} lifts to an action on the R_{n}^{(0)}-module freely generated by products of abstract symbols (S|T)′ where S, T are tableau of shape σ taking values in {1, 2, …, n}, having the same length, the action on such symbols being given by the analogue of (1.3). We will also impose the conditions:

(i) (S|T)′ = 0 if either S or T contains repeated entries; (ii) if S′ is S with two entries interchanged, then (S′|T)′ = −(S|T)′ (with a similar condition for T); (iii) if S = (S_{1}, …, S_{k}) and T = (T_{1}, …, T_{k}), then (S|T)′ = (S_{1}|T_{1})′(S_{2}|T_{2})′ … (S_{k}|T′_{k}).

The second modification is that we think of the (S|T)′ as generating an R_{n}^{(0)}-algebra and then reduce the elements of R_{n}^{(0)} modulo a certain ideal that we now describe. Let u be another indeterminate and replace C by the field of rational functions C(u) (so that we will now have to assume that C is an integral domain). For S ⊆ {1, 2, …, n} we let I(S) be the ideal of R_{n}^{(0)} generated by the elements

a_{ij}a_{ji} − \frac{1}{u(u + 1)}, for i, j ∈ S, i ≠ j;

a_{ij}a_{jk} − \frac{1}{u}a_{ik}, if i, j, k ∈ S and (j − i)(k − i)(k − j) > 0;

a_{ij}a_{jk} − \frac{1}{u + 1}a_{ik} if i, j, k ∈ S and (j − i)(k − i)(k − j) < 0.  (1.5)

For disjoint subsets S_{1}, S_{2}, …, S_{r} ⊆ {1, 2, …, n} we let

I(S_{1}, S_{2}, …, S_{r}) =< I(S_{1}), I(S_{2}), …, I(S_{r}) > .

Then for example we see that the invariance of I({1, 2, …, m}) under the action of B_{m} implies the invariance of I({1, 2, …, n_{1}}, {n_{1} + 1, n_{1} + 2, …, n_{1} + n_{2}}, …) under the action of G_{n_{1},n_{2},...,n_{s}}.

Now fix n_{1}, …, n_{s} ≥ 1 and let n = ∑_{i=1}^{s} n_{i}. Choose a Young diagram σ for {1, …, n}. Let

I_{n_{1},...,n_{s}} = I({1, …, n_{1}}, {n_{1} + 1, …, n_{1} + n_{2}}, {n_{1} + n_{2} + 1, …, n_{1} + n_{2} + n_{3}}, …).

Let R_{n}(σ) be the R_{n}^{(0)}/I_{n_{1},...,n_{s}}-module generated by all (S|T)′ where |S| = |T | = σ and S_{1}, T_{1} ⊂ {1, …, n_{1}}, S_{2}, T_{2} ⊂ {n_{1} + 1, …, n_{1} + n_{2}}, . . .. If we set the degree of (S|T)′ to be 1 (and degree(a_{ij}) = 0), then R_{n}(σ) is a graded R_{n}^{(0)}/I_{n_{1},...,n_{s}}-algebra, which we write as

R_{n}(σ) = ⊕_{k=0}^{∞}R_{n}^{k}(σ).

Theorem 1.1. Each R_{n}^{k}(σ) is a finitely-generated free C(u)-module which is G_{n_{1},...,n_{s}}- invariant.

We will study the summands of these representations. We will show that the following contribute to the existence of such summands:

(i) Multiple Laplace expansions of the determinant of (a_{ij}). (ii) Ideals generated by the Plücker relations. (iii) The existence of invariant involutions. (iv) The existence of fixed forms for the action.

Examples 1.2. The bitableau (1, 2, …, n|1, 2, …, n) gives the trivial representation of G_{n_{1},...,n_{s}}. The bitableau (1, 2, …, n|n + 1, n + 2, …, 2n) gives the sign representation of G_{n_{1},...,n_{s}}.

If M is a matrix, then M^{t} will denote its transpose.

Theorem 1.3. The action of G_{n_{1},...,n_{s}} on each R_{n}^{k}(σ) fixes a non-degenerate form which is unitary relative to the involution ∗: for fixed n_{i}, k there is a basis {b_{i}} for R_{n}^{k}(σ) and a non-degenerate matrix E over C(u) such that if M is the matrix representing α ∈ G_{n_{1},...,n_{s}} relative to the basis {b_{i}} , then we have ME(M^{t})^{∗} = E. The matrix E satisfies E^{∗} = ±E^{t}.

Each R_{n}^{k}(σ) splits as a sum of G_{n_{1},...,n_{s}}-irreducible subrepresentations.

Theorem 1.4. In the representation R_{n}^{k}(σ) the matrix representing any of the generators σ_{i}, 1 ≤ i < n, is diagonalisable.

Here is a complete description of one case:

Theorem 1.5. The representation given by (1, 2, …, n|1, 2, …, î, …, n, n+1) splits as V_{1}⊕ V_{2} where V_{1} has dimension n and is irreducible and monomial, and where V_{2} is irreducible and has dimension n(n − 1).

We will give a branching law for the the restrictions Res_{B_{n−1}}^{B_{n}}V_{i}, i = 1, 2 of the above representations in Theorem 7.6

We see how the representation theory of S_{n} appears in the following case where we consider the action of B_{n} on R_{n}/I({1, 2, …, n}).

Theorem 1.6. Suppose that μ = ∏_{k} a_{i_{k}j_{k}} ∈ R_{n}^{(0)} is a monomial where {i_{k}}_{k} ∩ {j_{k}}_{k} = ∅. Then there is a partition λ = (λ_{1} ≥ λ_{2} ≥ ⋯ ≥ λ_{r}) of n such that the C(u)-module generated by the orbit B_{n}(μ) splits into B_{n}-invariant summands in exactly the same way as the representation of S_{n} induced from the trivial representation of the subgroup

S_{λ_{1}} × S_{λ_{2}} × ⋯ × S_{λ_{r}}

does. (The summands so obtained are not necessarily B_{n}-irreducible.)

Before describing how these representations split we will need to describe in greater detail the action of B_{n} on R_{n}^{(0)} (in §2), the action of B_{n} on the ideals I(S_{1}, …, S_{r}) (in §3) etc.

In §10 we will show that R_{n}(σ) is sometimes an algebra with straightening law (ASL). In [Hu5] we have shown that B_{4} acts on an ASL.

Theorem 1.7. Let 1 ≤ k ≤ n. Then for S = {1, …, k} the representation R_{n}^{1}(S|{n + 1, …, n + k\}) has dimension \binom{n}{k}^{2} and splits as E_{1} ⊕ E_{2} where E_{1} has dimension \binom{n}{k}\binom{n−1}{k} and is irreducible and E_{2} has dimension \binom{n}{k}\binom{n−1}{k−1}.

The Plücker relations (defined in §11) determine certain representation spaces also:

Theorem 1.8. Let U be a set of Plücker relations coming from Young diagrams with a single row of length k. Then there is an associated B_{n}-invariant finitely generated free C(u)-module < B_{n}(U) > associated to U. If n > 3 is odd, then < B_{n}(U) > has an irreducible summand of dimension n. The action of B_{n} on this representation is monomial.

A finitely generated group is rigid if it has only finitely many classes of irreducible complex representations in each dimension.

Theorem 1.9. For n ≥ 3 the braid group B_{n} and the braid commutator groups B′_{n} are not rigid.

The braid commutator groups B′_{n} have been studied by Gorin and Lin [GL] and play an important role in Lin’s study [L] of representations of B_{n}.

A result of Dyer, Formanek and Grossman [DFG] gives a connection between B_{4} and Aut(F_{2}). We use this to prove Theorem 1.10. The automorphism group Aut(F_{2}) is not rigid.

The question of whether Aut(F_{n}) is rigid is posed in the ‘Open problems in combinatorial group theory’ list [P, problem F5].

§2 Action of B_{n} on R_{n}

In this section we describe in greater detail the action of B_{n} on R_{n}^{(0)}. A coordinate free definition [A1] of a transvection in SL(Q^{n}) (for a commutative ring Q with identity) is as a pair T = (ϕ, d) where d ∈ Q^{n} and ϕ is an element of the dual space of Q^{n} satisfying ϕ(d) = 0. The action of T on Q^{n} is given by

T (x) = x + ϕ(x)d for all x ∈ Q^{n}.

Then we have [Hu1, Lemma 2.1]

Lemma 2.1. Let T = (ϕ, d) and U = (ψ, e) be two transvections. Then for all λ ∈ ℤ we have

U^{λ}T U^{−λ} = (ϕ − λϕ(e)ψ, U^{λ}(d)). □

Let T = {T_{1} = (ϕ_{1}, d_{1}), …, T_{n} = (ϕ_{n}, d_{n})} be a fixed set of transvections in SL((R_{n}^{(0)})^{n}) where ϕ_{i}(d_{j}) = a_{ij} for all 1 ≤ i ≠ j ≤ n as in the above. For any set of transvections

T′ = {T′_{1} = (ϕ′_{1}, e′_{1}), …, T′_{n} = (ϕ′_{n}, e′_{n})}

we let M(T′) denote the n × n matrix (ϕ′_{i}(e′_{j})) and we call M(T′) the M-matrix of the set of transvections T′.

Any monomial in R_{n}^{(0)} that can be written in the form a_{j_{1}j_{2}}a_{j_{2}j_{3}} … a_{j_{r−1}j_{r}} will be called a j_{1}j_{r}-word. Note that by (1.2) if α ∈ B_{n} and 1 ≤ i ≠ j ≤ n, then α(a_{ij}) is a sum of rswords, where α(T_{i}) is a conjugate of T_{r} and α(T_{j}) is a conjugate of T_{s}. Let α ∈ B_{n} where α(T_{i}) = w_{i}T_{j}w_{i}^{−1} in freely reduced form for i = 1, ..., n and where w_{i} = w_{i}(T_{1}, …, T_{n}). Then for i = 1, …, n we have w_{i}T_{i}w_{i}^{−1} = (ψ_{i}, f_{i}) for some ψ_{i}, f_{i} determined by Lemma 2.1, which result in fact shows that

ψ_{i} = q_{1}ϕ_{1} + ⋯ + q_{n}ϕ_{n} and f_{i} = p_{1}d_{1} + ⋯ + p_{n}d_{n},

where p_{1}, …, p_{n}, q_{1}, …, q_{n} ∈ R_{n}^{(0)}. Since the a_{ij} are algebraically independent the ϕ_{i} and d_{j} are linearly independent and so the above representation is unique. We define the action of B_{n} on R_{n}^{(0)} by

α(a_{ij}) = ψ_{i}(f_{j}).

One can check that this agrees with the previous definition. Thus the M-matrix is acted upon naturally by B_{n}:

α(M(T)) = M(α(T_{1}), …, α(T_{n})).

From Lemma 2.3 of [Hu1] we have:

Lemma 2.2. Let α ∈ B_{n} where α(T_{i}) = C_{1}T_{k}C_{1}^{−1}, α(T_{j}) = C_{2}T_{p}C_{2}^{−1}, with C_{1}, C_{2} ∈ < T_{1}, …, T_{n} > and let C = C_{1}^{−1}C_{2} = T_{j_{1}}^{q_{1}} … T_{j_{r}}^{q_{r}} be freely reduced with j_{r} ≠ p, j_{1} ≠ k, q_{s} ≠ 0 for s = 1, …, r and j_{s} ≠ j_{s+1}, for s = 1, …, r − 1. Then

α(a_{ij}) = ∑_{h=1}^{n} A_{h}a_{hp}

where A_{h} is equal to the sum of all the products of the form

q_{r_{1}}q_{r_{2}} … q_{r_{m}}a_{kj_{r_{1}}} a_{j_{r_{1}} j_{r_{2}}} … a_{j_{r_{m−1}} j_{rm}}

where 1 ≤ r_{1} < r_{2} < ⋯ < r_{m} ≤ r and j_{r_{m}} = h. If p ≠ j_{r}, then the summand of α(a_{ij}) of highest degree is unique and is equal to

±q_{1}q_{2} … q_{r}a_{kj_{1}}a_{j_{1}j_{2}} … a_{j_{r−1}j_{r}}a_{j_{r}p}. □

For example if α(T_{1}) = T_{3}T_{2}^{−1}T_{1}T_{2}T_{3}^{−1} and α(T_{2}) = T_{2}^{−1}T_{3}T_{2}, then we would have C = T_{2}T_{3}^{−1}T_{2}^{−1} and

α(a_{12}) = a_{13} + a_{13}a_{32}a_{23} + a_{12}a_{23}a_{32}a_{23}.

Note that (1.2) follows from Lemma 2.2 and the action of σ_{r} in (1.3) was already noted in [Hu4, Hu5].

§3 Lifting the determinantal representation

Here we prove:

Proposition 3.1. There is an action of G_{n_{1},...,n_{s}} on the R_{n}^{(0)}/I_{n_{1},...,n_{s}}-algebra generated by the (S|T)′ where the action of a generator σ_{r} is given by (1.3).

Proof. The proof consists in showing that the braid relations (1.1) are satisfied by the rule for the σ_{i} given in (1.3) i.e. we need to show that σ_{i}σ_{i+1}σ_{i}(S|T)′ = σ_{i+1}σ_{i}σ_{i+1}(S|T)′ for i < n and that σ_{i}σ_{j}(S|T)′ = σ_{j}σ_{i}(S|T)′ for |i − j| > 1. But, if S = (S_{1}, …, S_{k}) and T = (T_{1}, …, T_{k}), then (S|T)′ = (S_{1}|T_{1})′(S_{2}|T_{2})′ … (S_{k}|T_{k})′ and since the σ_{i} act as ring homomorphisms we need only consider the case k = 1.

The alternative formula (1.4) can be interpreted as giving actions of B_{n} on the S part and on the T part (which we put into some suitable category), namely:

σ_{r}(S) = t_{r}(S − a_{rr+1}S_{r}^{r+1}S), σ_{r}(T) = t_{r}(T + a_{r+1r}S_{r}^{r+1}T).  (3.1)

This is put in context in the following way: let V_{n} be a free R_{n}^{(0)}/I_{n_{1},...,n_{s}}-module with basis x_{1}, …, x_{n}. We will associate to every subsequence S = (s_{1}, …, s_{k}) of {1, …, n} the element x_{s_{1}} ∧ ⋯ ∧ x_{s_{k}} of the exterior algebra ⋀^{k} V_{n}. Then we will check that the first equation of (3.1) gives an action on ⋀^{k} V_{n} and the second equation gives a dual action (relative to the involution ^{∗}). This will suffice to prove Proposition 3.1. The fact that these are dual actions means that we need only do the S action for example. This then amounts to showing that the action of the σ_{i} on the S part of (3.1) satisfies the relations in the standard presentation (1.1).

Now the relation σ_{i}σ_{j} = σ_{j}σ_{i} for |i − j| > 1 is easy to check (since in this case t_{i} and t_{j} commute, as do t_{i} and S_{j}^{j+1} etc.). For the relation σ_{i}σ_{i+1}σ_{i} = σ_{i+1}σ_{i}σ_{i+1} we note that this is trivially true if S ∩ {i, i + 1, i + 2} = ∅. So we can assume that S ∩ {i, i + 1, i + 2} ≠ ∅. A further simplification is that we need only do the case i = 1. Also when i = 1 we may ignore the presence in S of any indices greater than 3 since these are unaffected by σ_{1} and σ_{2}. There are now 7 cases to consider: (i) S contains only 1 (out of 1, 2, 3); (ii) S contains only 2; (iii) S contains only 3; (iv) S contains only 1, 2; (v) S contains only 1, 3; (vi) S contains only 2, 3; (vii) S contains 1, 2, 3.

For (i) by the above remarks we may (since we are ignoring all indices of S greater than 3) write S = (1) and so we have

σ_{1}σ_{2}σ_{1}(1) = σ_{1}σ_{2}t_{1}((1) − a_{12}(2)) = σ_{1}σ_{2}((2) − a_{21}(1))
= σ_{1}((3) − a_{32}(2) − (a_{31} − a_{32}a_{21})(1))
= (3) − a_{31}(1) − (a_{32}((2) − a_{21}(1));
σ_{2}σ_{1}σ_{2}(1) = σ_{2}σ_{1}(1) = σ_{2}((2) − a_{21}(1))
= (3) − a_{32}(2) − (a_{31} − a_{32}a_{21})(1).

Thus σ_{1}σ_{2}σ_{1}(1) = σ_{2}σ_{1}σ_{2}(1). For (ii) we may similarly write S = (2) and we get:

σ_{1}σ_{2}σ_{1}(2) = σ_{1}σ_{2}(1) = σ_{1}(1) = (2) − a_{21}(1);
σ_{2}σ_{1}σ_{2}(2) = σ_{2}σ_{1}((3) − a_{32}(2)) = σ_{2}((3) − a_{31}(1)) = (2) − a_{21}(1).

For (iii) we write S = (3) and we get:

σ_{1}σ_{2}σ_{1}(3) = σ_{1}σ_{2}(3) = σ_{1}(2) = (1);
σ_{2}σ_{1}σ_{2}(3) = σ_{2}σ_{1}(2) = σ_{2}(1) = (1).

For (iv) we write S = (12) and we get:

σ_{1}σ_{2}σ_{1}(12) = σ_{1}σ_{2}(−(12)) = −σ_{1}((13) − a_{32}(12)) = −((23) − a_{21}(13) + a_{31}(12));
σ_{2}σ_{1}σ_{2}(12) = σ_{2}σ_{1}((13) − a_{32}(12))) = σ_{2}((23) − a_{21}(13) + a_{31}(12))
= (32) − (a_{31} − a_{32}a_{21})(12) + a_{21}((13) − a_{32}(12))
= −((23) − a_{21}(13) + a_{31}(12)).

For (v) we write S = (13) and we get:

σ_{1}σ_{2}σ_{1}(13) = σ_{1}σ_{2}((23) − a_{21}(13)) = σ_{1}((32) − (a_{31} − a_{32}a_{21})(12))
= (31) + a_{32}(12);
σ_{2}σ_{1}σ_{2}(13) = σ_{2}σ_{1}(12) = −σ_{2}(12) = −(13) + a_{32}(12).

For (vi) we write S = (23) and we get:

σ_{1}σ_{2}σ_{1}(23) = σ_{1}σ_{2}(13) = σ_{1}(12) = (21);
σ_{2}σ_{1}σ_{2}(23) = σ_{2}σ_{1}(32) = σ_{2}(31) = (21).

For (vii) we write S = (123) and we get:

σ_{1}σ_{2}σ_{1}(123) = σ_{1}σ_{2}(213) = σ_{1}(312) = (321);
σ_{2}σ_{1}σ_{2}(123) = σ_{2}σ_{1}(132) = σ_{2}(231) = (321).

This concludes the proof of all cases □

The proof of the above result immediately gives:

Corollary 3.2. For all 1 ≤ k ≤ n we have an action of B_{n} on ⋀^{k} V_{n}. □

§4 Invariant ideals

Lemma 4.1. Let S be the sequence (1, 2, …, n). Then

(i) The ideal I(S) is B_{n}-invariant. (ii) If c_{I} = a_{ij}a_{jk} … a_{rs} is an is-word where i, j, k, …, r, s ∈ S, then there are two cases:

a) if i = s, then c_{I} + I(S) = r + I(S), where r ∈ C(u);

b) if i ≠ s, then c_{I} + I(S) = ra_{is} + I(S) for r ∈ C(u). (iii) If i ≠ j, then a_{ij} + I(S) ≠ r + I(S) for r ∈ C(u).

Proof. (i) The proof is to check that if e = a_{ij}a_{ji} − \frac{u}{u+1}, e = a_{ij}a_{jk} − \frac{1}{u+1}a_{ik} or e = a_{ij}a_{jk} − \frac{1}{u}a_{ik} is one of the ideal generators (as in (1.5)), then σ_{r}(e) ∈ I(S). Note that if {i, j, k} ∩ {r, r + 1} = ∅, then σ_{r}(e) = e. Thus one may assume that {i, j, k} ∩ {r, r + 1} ≠ ∅ so that card({i, j, k}∪{r, r +1}) ≤ 4. One easily sees that in fact one may renumber so that i, j, k ∈ {1, 2, 3, 4} and r = 1, 2, 3, thus reducing the checking to a finite number of cases, as indicated below. (One may also use the invariance of the B_{n}-action under the involution ^{∗} to further reduce the number of cases to be checked.) For example,

σ_{1}(a_{12}a_{23} − \frac{1}{u}a_{13}) = (1 + \frac{1}{u})(a_{21}a_{13} − \frac{1}{u + 1}a_{23});

σ_{2}(a_{12}a_{23} − \frac{1}{u}a_{13}) = (a_{13}a_{32} − \frac{1}{1 + u}a_{12}) + a_{12}(a_{23}a_{32} − \frac{1}{u(1 + u)});

σ_{3}(a_{12}a_{23} − \frac{1}{u}a_{13}) = (a_{12}a_{24} − \frac{1}{u}a_{14}) + a_{34}(a_{12}a_{23} − \frac{1}{u}a_{13});

σ_{1}(a_{13}a_{32} − \frac{1}{u + 1}a_{12}) = (a_{23}a_{31} − \frac{1}{u}a_{21}) − a_{21}(a_{13}a_{31} − \frac{1}{u(u + 1)});

σ_{2}(a_{13}a_{32} − \frac{1}{u + 1}a_{12}) = \frac{u}{u + 1}(a_{12}a_{23} − \frac{1}{u}a_{13});

σ_{3}(a_{13}a_{32} − \frac{1}{u + 1}a_{12}) = (a_{14}a_{42} − \frac{1}{u + 1}a_{12}) + a_{13}(a_{34}a_{42} − \frac{1}{u}a_{32})

− a_{32}(a_{14}a_{43} − \frac{1}{u + 1}a_{13}) − a_{13}a_{32}(a_{34}a_{43} − \frac{1}{u(u + 1)});

σ_{1}(a_{12}a_{21} − \frac{1}{u(u + 1)}) = a_{12}a_{21} − \frac{1}{u(u + 1)};

σ_{2}(a_{12}a_{21} − \frac{1}{u(u + 1)}) = (a_{13}a_{31} − \frac{1}{u(u + 1)}) − a_{13}(a_{32}a_{21} − \frac{1}{u + 1}a_{31})

+ a_{31}(a_{12}a_{23} − \frac{1}{u}a_{13}) − (a_{12}a_{21} − \frac{1}{u(u + 1)})a_{23}a_{32} + \frac{1}{u(u + 1)}(a_{13}a_{31} − a_{23}a_{32}); etc.

Alternatively, for a fixed ring C = ℚ, 𝔽_{q}, one can do these calculations (faster) using a Gröbner basis algorithm, as implemented in, for example, Magma [MA], since, as we have already noticed, one only has to deal with the case n = 4.

(ii) Given a cycle c_{I} = a_{ij}a_{jk} … a_{rs} of degree d one can use the relations in I(S) to replace, for example, a_{ij}a_{jk} by a non-zero C(u)-multiple of a_{ik}, thus reducing the degree, while the resulting monomial of degree d − 1 is still an is-word. (ii) follows.

(iii) Define a ring homomorphism η = η_{S} : R_{n}^{(0)} → R_{n}^{(0)} by its action on generators:

η(a_{ij}) = \frac{1}{u} if i, j ∈ S and i < j; η(a_{ij}) = \frac{1}{u + 1} if i, j ∈ S and i > j.  (4.1)

Then one checks that η(I(S)) = 0. But clearly η(a_{ij}) ≠ 0 and this gives (iii) □

Lemma 4.2. Suppose that S, T have a single row. Let (S|T)^{(0)} be the minor of M_{n}^{(0)} with row indices from S and column indices from T. Then, when expanded out, each monomial of (S|T)^{(0)} is a product of rs-words and ii words for distinct choices of r ∈ S \T and s ∈ T \S.

If α ∈ B_{n} and T ∩{1, 2, …, n} = ∅, then each monomial of α(S|T)′ has the form w(S′|T)′ where (S′|T)′ has the same shape as (S|T)′ and where either w ∈ C or w ∈ R_{n}^{(0)} is a product of r_{i}s_{i}-words, i = 1, …, k, where the r_{i} and the s_{i} are all distinct and s_{i} ∈ S′ and r_{i} ∉ S′.

If α ∈ B_{n} and S ∩ {1, 2, …, n} = ∅, then each monomial of α(S|T)′ has the form (S|T′)′w′ where (S|T′)′ has the same shape as (S|T)′ and where either w′ ∈ C or w′ ∈ R_{n}^{(0)} is a product of r_{i}s_{i}-words, i = 1, …, k, where the r_{i} and the s_{i} are all distinct and r_{i} ∈ T′ and s_{i} ∉ T′.

If α ∈ B_{n}, then each monomial of α(S|T)′ has the form w(S′|T′)′w′ where (S′|T′)′ has the same shape as (S|T)′ and where w and w′ are as in the last two paragraphs.

Proof. The first statement follows from elementary properties of determinants. The rest follows from using (1.3) by induction on the length of α as a product of the standard generators. □.

Remark 4.3. Now most of the time we will reduce the monomials w, w′ referred to in Lemma 4.2 mod I_{n_{1},...,n_{s}} and only deal with representatives which are products of r_{i}s_{i}-words of smallest degree (see Lemma 4.1). We note that for w(S′|T′)′w′ as in the last paragraph of Lemma 4.2, Lemma 4.2 then places a bound on the degree of such monomials w, w′ (when so reduced). However one should note that there may well be further reductions for ww′ e.g. … a_{ij}(…, j, … | …, j, …)a_{jk} … could be reduced to … a_{ik}(…, j, … | …, j, …) . . .. Note that this latter form may not look like it has the form indicated in Lemma 4.2.

We note that if σ is a Young diagram with a single row of length k, then there are only a finite number of the (S|T)′ with |S| = |T | = k and by Lemma 4.2 there are only a finite number of monomials w(U|V)′ in the B_{n}-orbit of such (S|T)′. Thus R_{n}^{1}(σ) is a finite-dimensional free C(u)-module and so R_{n}^{m}(σ) is a finite-dimensional free C(u)-module since it is a quotient of the mth symmetric power of R_{n}^{1}(σ) where a basis consists of all w(S_{1}|T_{1})′ … (S_{m}|T_{m})′ with w satisfying conditions similar to those of Lemma 4.2. The case where σ has more than one row is similar. This proves Theorem 1.1. □

From (1.5) we see that if (j−i)(k−i)(k−j) > 0 then mod I_{n_{1},...,n_{s}} we have a_{ij}a_{jk} = \frac{1}{u}a_{ik}. Acting on this latter equation by ∗ we get a_{ji}a_{kj} = −\frac{1}{u^{∗}} a_{ki} and comparing this with (1.5) again we see that it is natural to define

u^{∗} = −(u + 1).

One then checks:

Lemma 4.4. For all w ∈ I_{n_{1},...,n_{s}} we have w^{∗} ∈ I_{n_{1},...,n_{s}}.

Proof. One need only consider the case where w is one of the generators of I_{n_{1},...,n_{s}} as in (1.5) and we have already done one case above. The rest are also easily checked. □

We now define the action of the involution ^{∗} on the generators (S|T)′, where S, T have a single row, by

((S|T)′)^{∗} = (T |S)′.

We extend this action naturally: (w_{1}S|w_{2}T)^{∗} = w_{1}^{∗}w_{2}^{∗}(T |S), and then C(u)-linearly over monomials. This now gives:

Lemma 4.5. For α ∈ B_{n} we have

α((S|T)′)^{∗} = α(((S|T)′)^{∗}).

In particular, for all x ∈ R_{n} and all α ∈ B_{n} we have α(x^{∗}) = α(x)^{∗}.

Proof. We need only prove the first statement, and this only in the case α = σ_{r}, 1 ≤ r < n. Using (1.4) we have:

(σ_{r}(S|T)′)^{∗} = (t_{r}(S − a_{rr+1}S_{r}^{r+1}S|T + S_{r}^{r+1}T a_{r+1r}))^{∗}
= (t_{r}S − a_{r+1r}t_{r}S_{r}^{r+1}S|t_{r}T + t_{r}S_{r}^{r+1}T a_{rr+1})^{∗}
= (t_{r}T − t_{r}S_{r}^{r+1}T a_{r+1r}|t_{r}S + a_{rr+1}t_{r}S_{r}^{r+1}S)
= t_{r}(T − S_{r}^{r+1}T a_{r+1r}|S + a_{rr+1}S_{r}^{r+1}S)
= σ_{r}(T |S)′ = σ_{r}(((S|T)′)^{∗}),

as required □

Lemma 4.6. Let S = (1, …, n) and choose distinct u_{1}, …, u_{r}, v_{1}, …, v_{r} ∈ {1, …, n}, r ≥ 2. Then for any ζ ∈ S_{n} there is c ≠ 0 ∈ C such that

a_{u_{1}v_{1}}a_{u_{2}v_{2}} … a_{u_{r}v_{r}} = c × a_{u_{1}v_{ζ1}}a_{u_{2}v_{ζ2}} … a_{u_{r}v_{ζr}} mod I(S).

Proof. It will suffice to do the case r = 2, since transpositions generate S_{n}. Now u_{1}, u_{2}, v_{1}, v_{2} are distinct and so there are c_{1}, c_{2} ∈ C(u) such that:

a_{u_{1}v_{1}}a_{u_{2}v_{2}} = c_{1}a_{u_{1}u_{2}}a_{u_{2}v_{1}}a_{u_{2}v_{2}} = c_{1}a_{u_{1}u_{2}}a_{u_{2}v_{2}}a_{u_{2}v_{1}} = c_{1}c_{2}a_{u_{1}v_{2}}a_{u_{2}v_{1}},

as required. □

Remark 4.7. For those who like their ring involutions to look like complex conjugation we can (in the situation where C = ℂ) put u = −\frac{1}{2} + iy, where i^{2} = −1.

Proposition 4.8. Suppose 2 is invertible in C and that V is a B_{n}-invariant subrepresentation of R_{n}^{k}(σ) with V^{∗} = V. Then V splits as V^{+} ⊕ V^{−} where

V^{±} = {b ∈ V |b^{∗} = ±b}.

Here V^{±} are both B_{n}-invariant.

Proof. Lemma 4.5 shows that each of V^{±} are invariant under the action of B_{n}. The rest follows since for b ∈ V we can write b = (b + b^{∗})/2 + (b − b^{∗})/2 = b^{+} + b^{−}, where b^{±} ∈ V^{±}. □

Let w(S|T) be a monomial where S, T have a single row. We assume that w is in normal form (see Lemma 4.2) so that w = w_{1}w_{2} with w_{1} = a_{r_{1}s_{1}} … a_{r_{z}s_{z}}, w_{2} = a_{p_{1}q_{1}} … a_{p_{y}q_{y}} with s_{i} ∈ S, p_{i} ∈ T, s_{i}, q_{i} ∉ S ∪ T. Then we let

E^{−} = (S ∪ {r_{1}, …, r_{z}}) \ {s_{1}, …, s_{z}}, E^{+} = (T ∪ {q_{1}, …, q_{y}}) \ {p_{1}, …, p_{z}}.

Lemma 4.9. Supoose that S, T have a single row. Let α ∈ B_{n} and let μ be a monomial in α(w(S|T)′). Then

|E^{+}(w(S|T)′) ∩ E^{−}(w(S|T)′)| = |E^{+}(μ) ∩ E^{−}(μ)|.

If σ is a Young diagram, then we have the following B_{n}-invariant splitting

R_{n}^{1}(σ) = ⊕_{i=0}^{|S|}W_{i}

where W_{i} is spanned by all monomials μ ∈ R_{n}^{1}(σ) with |E^{+}(μ) ∩ E^{−}(μ)| = i. Further we have W_{i} = W_{i}^{+} ⊕ W_{i}^{−}.

Proof. The last sentence follows from Proposition 4.8. The first sentence follows from (1.3) and (1.4): induct on the length of α as a word in the standard braid generators. The rest follows from the first sentence. □

§5 Invariant forms

We will first consider the case where the Young diagram has a single row.

By the above we have an action of B_{n} on the ring R_{n}^{(0)}. This can be extended to an action of B_{n} on a ring

R_{n}^{∞} = C[a_{ij}|i, j ≥ 1, a_{ii} = 0 for i ≤ n].

The action is still given by (1.2) so that α(a_{ij}) = a_{ij} for α ∈ B_{n} and i, j > n. As usual, we will think of the a_{ij} as entries in a matrix of sufficiently large degree.

Given finite subsequences S, T, U, V of ℕ of the same size and w_{1}, w_{2} ∈ R_{n} we define

< w_{1}(S|T)′, w_{2}(U|V)′ >′= w_{1}w_{2}^{∗}(S|U)^{(0)}(V |T)^{(0)} mod I_{n} = I({1, …, n}).

We will say that S, T, U, V and w_{1}, w_{2} are compatible if < w_{1}(S|T)′, w_{2}(U|V)′ >′ is in the subring C(u)[a_{ij}|i, j > n] of R_{n}^{∞} (so that it is fixed by the action of B_{n}). We define

< w_{1}(S|T)′, w_{2}(U|V)′ >= \{ \begin{matrix}< w_{1}(S|T)′, w_{2}(U|V)′ >′ if S, T, U, V, w_{1}, w_{2} are compatible; \\ 0 otherwise \end{matrix}

We extend <, > C(u)-linearly to act on R_{n}^{1}. For notational convenience we will sometimes use det(S|T) for (S|T)^{(0)}.

Example 5.1. If n = 4, S = [1], T = [5], then the only w(U|V)′ which are compatible with (S|T)′ are b_{1} = (S|T)′, b_{2} = a_{12}([2], [5])′, b_{3} = a_{13}([3], [5])′, b_{4} = a_{14}([4], [5])′ and the values of < b_{i}, b_{j} > are given in the following matrix (where we suppress the det([5], [5]) = a_{55} factor) :

( \begin{matrix}0 & −\frac{1}{u(u+1)} & −\frac{1}{u(u+1)} & −\frac{1}{u(u+1)} \\ \frac{1}{u(u+1)} & 0 & −\frac{1}{u(u+1)^{2}} & −\frac{1}{u(u+1)^{2}} \\ \frac{1}{u(u+1)} & −\frac{1}{u^{2}(u+1)} & 0 & −\frac{1}{u(u+1)^{2}} \\ \frac{1}{u(u+1)} & −\frac{1}{u^{2}(u+1)} & −\frac{1}{u^{2}(u+1)} & 0 \end{matrix} ) .

Proposition 5.2. The form <, > is C(u)-linear in both entries and for w ∈ R_{n}^{∞}, x, y ∈ R_{n}^{1} satisfies

< wx, y >= w < x, y >, < x, wy >= w^{∗} < x, y >, < x, y >^{∗}=< y, x > .

Further, <, > is B_{n}-invariant: for all α ∈ B_{n}, x, y ∈ R_{n}^{1} we have < α(x), α(y) >=< x, y >.

Proof. The linearity and the first two properties are clear. To show that < x, y >^{∗}=< y, x > we need only do the case where x = (S|T), y = (U|V). We need to note that det(S|T)^{∗} = (−1)^{|S|}det(T |S) and then we have:

< (S|T)′, (U|V)′ >^{∗} = (det(S|U)det(V |T))^{∗} mod I_{n}
= det(S|U)^{∗}det(V |T)^{∗} mod I_{n}
= det(U|S)det(T |V) mod I_{n}
=< (U|V)′, (S|T)′ > .

We now prove the invariance under the B_{n}-action, again noting that it suffices to check this for α = σ_{r}, 1 ≤ r < n, and x = (S|T)′, y = (U|V)′. First note that by (1.4) we have

σ_{r}(S) = t_{r}S − a_{r+1r}t_{r}S_{r}^{r+1}S, σ_{r}(U) = t_{r}U − a_{r+1r}t_{r}S_{r}^{r+1}U

and so

< σ_{r}(S|T), σ_{r}(U|V) > =< (σ_{r}S|σ_{r}T), (σ_{r}U|σ_{r}V) >
= det(σ_{r}S, σ_{r}U)det(σ_{r}V, σ_{r}T) mod I_{n}.

We will now prove that det(σ_{r}S, σ_{r}U) = σ_{r}det(S, U):

det(σ_{r}S, σ_{r}U) = det(t_{r}S − a_{r+1r}t_{r}S_{r}^{r+1}S, t_{r}U − a_{r+1r}t_{r}S_{r}^{r+1}U)
= det(t_{r}S, t_{r}U) − a_{r+1r}det(t_{r}S_{r}^{r+1}S, t_{r}U) − det(t_{r}S, a_{r+1r}t_{r}S_{r}^{r+1}U)
+ a_{r+1r}det(t_{r}S_{r}^{r+1}S, a_{r+1r}t_{r}S_{r}^{r+1}U)
= det(t_{r}S, t_{r}U) − a_{r+1r}det(t_{r}S_{r}^{r+1}S, t_{r}U) + a_{rr+1}det(t_{r}S, t_{r}S_{r}^{r+1}U)
− a_{r+1r}a_{rr+1}det(t_{r}S_{r}^{r+1}S, t_{r}S_{r}^{r+1}U)
= σ_{r}det(S|U).

We similarly have det(σ_{r}V, σ_{r}T)′ = σ_{r}det(V, T)′. Now combining these results we get

< σ_{r}(S|T), σ_{r}(U|V) > = det(σ_{r}S, σ_{r}U)det(σ_{r}V, σ_{r}T) mod I_{n}
= σ_{r}det(S, U)σ_{r}det(V, T) mod I_{n}
= σ_{r}(det(S, U)det(V, T)) mod I_{n}
= det(S, U)det(V, T) mod I_{n},

the last equality coming from the fact that S, T, U, V are compatible. □

Proposition 5.3. The form <, > is non-degenerate.

Proof. We will need:

Lemma 5.4. Let s_{0}(x, y) = 1 and for n > 0 let s_{n}(x, y) = x^{n} + x^{n−1}y + x^{n−2}y^{2} + ⋯ + y^{n} and

X_{n} = ( \begin{matrix}0 & \frac{1}{u} & \frac{1}{u} & … & \frac{1}{u} \\ \frac{1}{u+1} & 0 & \frac{1}{u} & … & \frac{1}{u} \\ \frac{1}{u+1} & \frac{1}{u+1} & 0 & … & \frac{1}{u} \\ ⋮ & ⋮ & ⋮ & ... & ⋮ \\ \frac{1}{u+1} & \frac{1}{u+1} & \frac{1}{u+1} & … & 0 \end{matrix} ) .

Then det(X_{n}) = (−1)^{n−1} \frac{1}{u(u+1)}s_{n−2}(\frac{1}{u}, \frac{1}{u+1}).

Proof. This follows directly from the last exercise in [Mu, §828, p. 764]. □

Now we have noted above that we may find a basis of R_{n}^{1} of the form {b_{i} = w_{i}(S_{i}|T_{i})′}_{i}. We may order this basis so that b_{1}, …, b_{N_{1}} are all compatible, b_{N_{1}+1}, …, b_{N_{1}+N_{2}} are all compatible (but not compatible with b_{1}), etc. In fact the number of b_{i} compatible with a given b_{j} is the same, so we have N_{i} = N_{j}. Relative to this basis the matrix representing the form <, > has block form

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

where each C_{i} is an N_{1} × N_{1} matrix. Thus to show that the form <, > is non-degenerate it suffices to show that each of the matrices C_{k} = (< b_{i}, b_{j} > |(k − 1)N_{1} ≤ i, j ≤ kN_{1}) is non-degenerate.

Fix (S|T)′ with |S| = |T | = k and consider all b_{i} which are compatible with (S|T)′. We will first consider the case of arbitrary S and T = (n+1, n+2, …, 2n). In fact there is little loss in this case in assuming that S = (1, 2, …, k). Now note that given the compatibility of each such b_{i} = w(S′|T)′ with (S|T)′ we see that S′ completely determines w (and viceversa). Further, for each S′ ⊂ {1, 2, …, n} with |S′| = k there is w′ such that w′(S′|T)′ is a basis element. It follows that there are exactly \binom{n}{k} of the b_{i} which are compatible with (S|T)′, one for each subset of \{1, 2, …, n\} of cardinality k. Thus N_{1} = \binom{n}{k}.

Recall the ring homomorphism η = η_{{1,...,n}} : R_{n}^{(0)} → C(u) defined in the proof of Lemma 4.1 (iii). We there showed that it satisfies: η(I_{n}) = 0. We can extend η as follows:

η(a_{ij}) = \frac{1}{u} if \{i, j\} ∩ \{1, …, n\} ≠ ∅ and i < j;

η(a_{ij}) = \frac{1}{u + 1} if \{i, j\} ∩ \{1, …, n\} ≠ ∅ and i > j;

η(a_{ij}) = a_{ij} if {i, j} ∩ {1, …, n} = ∅.

Lemma 5.5. For compatible b_{i} = w_{i}(S_{i}|T) and b_{j} = w_{j}(S_{j}|T) we have

< w_{i}(S_{i}|T), w_{j}(S_{j}|T) >= η(w_{i}w_{j}^{∗}det(S_{i}|S_{j})det(T |T)).

Proof. Since η(I(S)) = 0, and b_{i} and b_{j} are compatible we have

η(w_{1}w_{2}det(S_{1}, S_{2})det(T |T)) = η(< b_{i}, b_{j} >) =< b_{i}, b_{j} >,

as required. □

Now we wish to show that det(C_{k}) ≠ 0 where (C_{k})_{ij} is < w_{i}(S_{i}|T), w_{j}(S_{j}|T) >; but by Lemma 5.5 and the fact that η is a ring homomorphism, it suffices to show that the matrix E with i, j entry equal to η(< (S_{i}|T), (S_{j}|T) >) is non-degenerate. But since det(T |T) is a constant and non-zero factor this latter fact will follow if we can show that the matrix D with i, j entry equal to η(det(S_{i}|S_{j})) is non-degenerate.

Lemma 5.6. Fix 1 ≤ k ≤ n and let X_{n} be as in Lemma 5.4. Let S_{1}, …, S\binom{n}{k} be the subsets

of \{1, …, n\} of cardinality k and let D be the \binom{n}{k} × \binom{n}{k} matrix (η(det(S_{i}, S_{j}))). Then D is invertible.

Proof. Lemma 5.4 shows that X_{n} is invertible. We can think of X_{n} as acting on a C(u)- vector space V_{n} with basis x_{1}, …, x_{n}. Then by [Bo, Prop. 10 p. 529; see also Ex. 11 p. 640 (watch for the misprint!)] we see that the matrix D represents the action of X_{n} on the exterior algebra ⋀^{k} V_{n}. Since X_{n} is invertible we see that D is also. □

Conjecture 5.7. We conjecture that the determinants det(C_{i}) have the form \frac{(u+1)^{m}−u^{m}}{(u(u+1))^{p}} for some m, p. If this were the case and one solves det(C_{i}) = 0, then one obtains (u+1)^{m}−u^{m} = 0 and finds (over ℂ) that the solutions are:

u = −\frac{1}{2} − i\frac{sin(2kπ/m)}{1 − cos(2kπ/m)},

for 1 ≤ k < m. We compare these solutions with Remark 4.7.

Lemma 5.8. Assume that |S| = |T | = k. Then the action of B_{n} on (S|T)′ is the same as the action of B_{n} on the elements (S|[n + 1, …, n + k])′([n + 1, …, n + k]|T)′.

Proof. We need only check that for α ∈ B_{n} we have

α(S|T)′ = α(S|[n + 1, …, n + k])′α([n + 1, …, n + k]|T)′,

and in fact we need only check this for α = σ_{i}, i < n. However this latter fact in this case follows from (1.3). □

We now show how the above implies the non-degeneracy for general S, T.

Now the action of B_{n} on the w([n+1, …, n+k]|T) is dual to the action on the w^{∗}(S|[n+ 1, …, n + k]). Thus by the above the action of B_{n} on the tensor product (over C(u)) generated by all w(S|[n + 1, …, n + k]) ⊗ w′([n + 1, …, n + k]|T) is also a B_{n}-representation space with the B_{n} action fixing a non-degenerate B_{n}-invariant form; denote this space by U_{n} ⊗ U_{n}^{∗}. Then U_{n} ⊗ U_{n}^{∗} splits as a sum of B_{n}-irreducibles.

Now by Lemma 5.8 we see that the B_{n}-representation space that we are interested in is a quotient of this tensor product; denote it by Q. Thus, due to the above splitting property, this quotient can be identified with a summand of U_{n} ⊗ U_{n}^{∗} i.e. U_{n} ⊗ U_{n}^{∗} ≅ Q ⊕ Y. Then the form on U_{n} ⊗ U_{n}^{∗} restricts to a form on Q, which, since Q and Y are orthogonal relative to the form on U_{n} ⊗ U_{n}^{∗}, is also non-degenerate. This does the case where S, T have a single row. The general case follows by a similar argument since R_{n}^{1}(σσ′) is a quotient of R_{n}^{1}(σ) ⊗ R_{n}^{1}(σ′). □

§6 Diagonalisability

In this section we prove

Theorem 6.1. For all Young diagrams σ of n ≥ 2 and all 1 ≤ i < n the matrix representing the action of σ_{i} on R_{n}^{1}(σ) is diagonalisable over a finite extension of C(u).

Proof. It clearly suffices to prove the result in the case where σ has a single row. Now fix a monomial μ. We will show that the orbit O_{i}(μ) = {σ_{i}^{k}(μ)}_{k∈ℤ} is spanned by a certain finite set M_{i}(μ) of monomials and that the action on the subspace V_{i}(μ) spanned by these monomials is diagonalisable. Now by Lemma 4.2 we may assume that μ = w_{1}(S|T)′w_{2} with w_{1}, w_{2} as in Lemma 4.2. Since σ_{i}(a_{rs}) = a_{rs} for all r, s ≠ i, i + 1 we see that there are a finite number of cases to be checked, depending upon whether i, i + 1 occur in S or T or as subscripts of factors of w_{1}, w_{2}. Here, for example, a_{ji}(i|−) will indicate a monomial where a_{ji} is a divisor of w_{1}, j ≠ i, i + 1, (but none of a_{ij}, a_{i+1j}, a_{ji+1}, a_{ii+1}, a_{i+1i} are) and i ∈ S (but i + 1 ∉ S), and i, i + 1 ∉ T. The cases are: μ =

(i) (−|−), (ii + 1|−), (−|ii + 1); (ii) (i|−); (iii)(i + 1|−); (iv) (i|i); (v) (i|i + 1); (vi) (ii + 1|−); (vii) (ii + 1|i); (viii) (ii + 1|i + 1); (ix) a_{ji}(i|−); (x) a_{ji}(i|i + 1); (xi) a_{ji}(ii + 1|−); (xii) a_{ji}(ii + 1|i); (xiii)a_{ji}(ii + 1|i + 1); (xv) a_{ij}(|−); (xvi)a_{ij}(|i); (xvii) a_{ij}(−|i); (xviii) a_{i+1i}(i|−); (ixx) a_{ii+1}(i + 1|−); (xx) a_{i+1i}(i|i + 1);

(xxi) a_{ii+1}(i + 1|i); (xxii) a_{i+1i}(i + 1|i + 1); (xxiii) a_{i+1i}(i|i); (xxiv) a_{i+1i}^{2}(i|i + 1);

(xxv)a_{ii+1}^{2}(i + 1|i); (xxvi) a_{ji}(i|i + 1)a_{i+1i}; (xxvii) a_{ji+1}(i + 1|i + 1); (xxviii) a_{ji+1}(i + 1|i); (xxix) a_{ji+1}(i|i); (xxx) a_{ji}(i + 1|i + 1); a_{ji}(i|i); (xxxi) a_{ji+1}a_{ii+1}(i + 1|i); (xxxii) a_{ji}a_{i+1i}(i|i + 1); (xxxiii) a_{ji}a_{i+1k}(−|−); etc.

Here we have only indicated some of the cases, other cases will follow by duality.

We now indicate how each case can be checked. Of course σ_{i}(−|−) = (−|−) and similarly σ_{i}(ii + 1|−) = −(ii + 1|−) and so these cases are easy. For (ii) μ = (i|−), we have

M(μ) = {(i|−), (i + 1|−), a_{i+1i}(i|−), a_{ii+1}(i + 1|−)

and relative to this basis the matrix representing the action of σ_{i} is

M = ( \begin{matrix}0 & 1 & 0 & 0 \\ 1 & 0 & −1 & 0 \\ 0 & \frac{−1}{u(u+1)} & 0 & 1 \\ 0 & 0 & 1 & 0 \end{matrix} )

which has characteristic polynomial (z^{2} − \frac{u}{u+1})(z^{2} − \frac{u+1}{u}) and so M is diagonalisable over a finite extension of C(u). This is case (ii), but we note that this also takes care of cases (iii), (xviii) and (ixx).

For (iv) we have

M(μ) = {(i + 1|i + 1), (i|i), a_{i+1i}(i|i + 1); a_{ii+1}(i + 1|i)}

and relative to this basis the matrix representing the action of σ_{i} is

M = ( \begin{matrix}0 & 1 & 0 & 0 \\ 1 & \frac{−1}{u(u+1)} & −1 & 1 \\ 0 & \frac{−1}{u(u+1)} & 0 & 1 \\ 0 & \frac{1}{u(u+1)} & 1 & 0 \end{matrix} )

and this has characteristic polynomial (z − 1)^{2}(z + \frac{u}{u+1})(z + \frac{u+1}{u}). To obtain the result in this case we just need to note that (1, 1, 0, 1), (0, 0, 1, 1) span the 1-eigenspace. This does (iv) and also (xx) and (xxi).

For (v) we have

M(μ) = {(i + 1|i), (i|i + 1), a_{i+1i}(i + 1|i + 1),a_{i+1i}(i|i), a_{ii+1}(i + 1|i + 1),
a_{ii+1}(i|i); a_{i+1i}^{2}(i|i + 1), a_{ii+1}^{2}(i + 1|i)},

and relative to this basis the matrix representing the action of σ_{i} is

M = ( \begin{matrix}0 & 1 & 0 & 0 & 0 & 1 & 0 & 0 \\ 1 & 0 & 0 & −1 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 \\ 0 & \frac{−1}{u(u+1)} & 0 & 0 & 1 & \frac{−1}{u(u+1)} & 0 & 1 \\ 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 \\ \frac{1}{u(u+1)} & 0 & 1 & \frac{−1}{u(u+1)} & 0 & 0 & −1 & 0 \\ 0 & 0 & 0 & 0 & 0 & \frac{−1}{u(u+1)} & 0 & 1 \\ 0 & 0 & 0 & \frac{1}{u(u+1)} & 0 & 0 & 1 & 0 \end{matrix} ) .

This has characteristic polynomial

(z − 1)^{2}(z + 1)^{2}(z − \frac{u}{u + 1})(z + \frac{u}{u + 1})(z − \frac{u + 1}{u})(z + \frac{u + 1}{u}).

REPRESENTATIONS OF BRAID GROUPS VIA DETERMINANTAL RINGS 19

Here we note that the eigenspaces for the ±1 eigenvectors are generated (respectively) by (1, 0, u(u + 1), u(u + 1), u(u + 1), u(u + 1), 0, u(u + 1)),

(0, 1, −u(u + 1), −u(u + 1), −u(u + 1), −u(u + 1), u(u + 1), 0),
(1, 0, u(u + 1), u(u + 1), −u(u + 1), −u(u + 1), 0, −u(u + 1)),
(0, 1, u(u + 1), u(u + 1), −u(u + 1), −u(u + 1), −u(u + 1), 0).

This does this case and (xxii), (xxiii), (xxiv), (xxv).

For (ix) we have M(μ) = \{a_{ji}(i|−), a_{ji+1}(i+1|−)\} and the matrix is ( 0 1_{\frac{u+1}{u} \frac{−1}{u}} ) which has distinct eigenvalues.

For (x) we have

M(μ) = {a_{ji+1}(i + 1|i + 1), a_{ji+1}(i + 1|i),a_{ji+1}(i|i), a_{ji}(i + 1|i + 1), a_{ji}(i|i + 1), a_{ji}(i|i),
a_{ji}a_{i+1i}(i|i + 1), a_{ji+1}a_{ii+1}(i + 1|i)}.

Here the σ_{1} matrix is

( \begin{matrix}0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & \frac{1}{u} & 0 & 1 & 0 & 0 & 0 \\ 0 & \frac{1}{u} & 0 & 1 & 0 & \frac{−1}{u(u+1)} & −1 & 0 \\ 0 & 0 & \frac{u+1}{u} & 0 & 0 & 0 & 0 & 0 \\ 0 & \frac{u+1}{u} & 0 & 0 & 0 & \frac{−1}{u} & 0 & 0 \\ \frac{u+1}{u} & 0 & −\frac{1}{u^{2}} & 0 & 0 & 0 & 0 & \frac{u+1}{u} \\ 0 & 0 & −\frac{1}{u^{2}} & 0 & 0 & 0 & 0 & \frac{u+1}{u} \\ 0 & 0 & 0 & 0 & 0 & \frac{1}{u(u+1)} & 1 & 0 \end{matrix} ) .

This has characteristic polynomial

(z^{2} − \frac{u}{u + 1})(z^{2} − \frac{u + 1}{u})^{2}(z^{2} − \frac{(u + 1)^{3}}{u^{3}} )

and the eigenvectors for the squared factor are:

(−u\sqrt{(u + 1)u}, 0, −u\sqrt{(u + 1)u}, − (u + 1) u, \sqrt{(u + 1) u}, − (u + 1) u, 1, 0), ((u + 1) u, \sqrt{(u + 1) u}, (u + 1) u, \sqrt{(u + 1) u} (u + 1) , 0, \sqrt{(u + 1) u} (u + 1) , 0, 1), ((u + 1) u, −\sqrt{(u + 1) u}, (u + 1) u, −\sqrt{(u + 1) u} (u + 1) , 0, −\sqrt{(u + 1) u} (u + 1) , 0, 1), (u\sqrt{(u + 1) u}, 0, u\sqrt{(u + 1) u}, − (u + 1) u, −\sqrt{(u + 1) u}, − (u + 1) u, 1, 0).

This again shows diagonalisability for (x) and (xxvi)-(xxxii).

The rest of the cases are similarly checked, giving Theorems 6.1 and 1.4. □

§7 The (1, 2, …, n|1, 2, …, î, …, n, n + 1) representation.

In this section we prove Theorem 1.5. Let μ_{i} = μ_{i}^{(n)} = (1, 2, …, n|1, 2, …, î, …, n, n+1).

Lemma 7.1. For any 1 ≤ i ≤ n the C(u)-module V_{n} generated by the B_{n}-orbit of μ_{i} is freely generated by μ_{k}, a_{ij}μ_{j} for i, j, k = 1, …, n with i ≠ j. It has dimension n^{2}.

Proof. We note the following:

σ_{i}(μ_{i}) = −μ_{i+1}; σ_{i}(a_{ii+1}μ_{i+1}) = −a_{i+1i}μ_{i} − \frac{u}{u + 1}μ_{i+1};

σ_{i}(a_{ij}μ_{j}) = \frac{u}{u + 1}a_{i+1j}μ_{j}; σ_{i}(a_{i+1j}μ_{j}) = a_{ij}μ_{j}; σ_{i}(μ_{i+1}) = −μ_{i} − a_{ii+1}μ_{i+1};

σ_{i}(a_{ji}μ_{i}) = −\frac{(u + 1)}{u}a_{ji+1}μ_{i+1}; σ_{i}(a_{ji+1}μ_{i+1}) = −a_{ji}μ_{i} − \frac{1}{u}a_{ji+1}μ_{i+1}.  (7.1)

Here j ≠ i, i + 1. The first of these equations shows that all the ±μ_{j} are in the orbit of μ_{1}, for example. For i < j < n the 5th equation shows that we can get a_{jj+1}μ_{j+1} and then

σ_{i}σ_{i+1} … σ_{j−1}(a_{jj+1}μ_{j+1}) = a_{ij+1}μ_{j+1}.

We can similarly get all a_{ij}μ_{j} for i > j. □

Define the following vectors:

v_{1} = μ_{1} − ua_{12}μ_{2} + ua_{13}μ_{3} − ua_{14}μ_{4} + ua_{15}μ_{5} − ⋯ + (−1)^{n+1}ua_{1n}μ_{n} v_{2} = −(u + 1)a_{21}μ_{1} + μ_{2} − ua_{23}μ_{3} + ua_{24}μ_{4} − ua_{25}μ_{5} + ⋯ + (−1)^{n}ua_{2n}μ_{n}

v_{3} = (u + 1)a_{31}μ_{1} − (u + 1)a_{32}μ_{2} + μ_{3} − ua_{34}μ_{4} + ua_{35}μ_{5} − ⋯ + (−1)^{n+1}ua_{3n}μ_{n}; v_{4} = −(u + 1)a_{41}μ_{1} + (u + 1)a_{42}μ_{2} − (u + 1)a_{43}μ_{3} + μ_{4} − ua_{45}μ_{5} + ⋯ + (−1)^{n}ua_{4n}μ_{n};

⋮ v_{n} = ±(u + 1)a_{n1}μ_{1} ∓ (u + 1)a_{n2}μ_{2} ± (u + 1)a_{n3}μ_{3} ∓ (u + 1)a_{n4}μ_{4}

± (u + 1)a_{n5}μ_{5} ∓ ⋯ + μ_{n}.  (7.2)

We now note that σ_{i}(a_{jk}) = a_{jk} and σ_{i}(μ_{j}) = μ_{j} for all j, k ≠ i, i + 1. From (7.1) we see that for j ≠ i, i + 1 the only monomials in v_{j} which are not fixed are a_{ji}μ_{i} and a_{ji+1}μ_{i+1}, both having the same coefficients only differing in sign; so we have:

σ_{i}(a_{ji}μ_{i} − a_{ji+1}μ_{i+1}) = −\frac{(u + 1)}{u}a_{ji+1}μ_{i+1} + a_{ji}μ_{i} + \frac{1}{u}a_{ji+1}μ_{i+1} = a_{ji}μ_{i} − a_{ji+1}μ_{i+1},

showing that σ_{i}(v_{j}) = v_{j} for all j ≠ i, i + 1. We also have (for j ≠ i, i + 1):

σ_{i}(a_{ij}μ_{j}) = \frac{u}{u + 1}a_{i+1j}μ_{j}, and

σ_{i}(μ_{i} − ua_{ii+1}μ_{i+1}) = −μ_{i+1} + u(a_{i+1i}μ_{i} + \frac{1}{u(u + 1)}μ_{i+1}) = ua_{i+1i}μ_{i} − \frac{u}{u + 1}μ_{i+1}.

This shows that σ_{i}(v_{i}) = −\frac{u}{u+1}v_{i+1}. Similarly we have σ_{i}(a_{i+1j}μ_{j}) = a_{ij}μ_{j} for j ≠ i, i + 1 and

σ_{i}((u + 1)a_{i+1i}μ_{i} − μ_{i+1}) = −(u + 1)a_{ii+1}μ_{i+1} + μ_{i} + a_{ii+1}μ_{i+1} = μ_{i} − ua_{ii+1}μ_{i+1}.

Which shows that σ_{i}(v_{i+1}) = −v_{i}. Thus we get a monomial representation ρ of degree n where

ρ(σ_{1}) = ( \begin{matrix}0 & −\frac{u}{u+1} & 0 & 0 & … \\ −1 & 0 & 0 & 0 & … \\ 0 & 0 & 1 & 0 & … \\ 0 & 0 & 0 & 1 & … \\ ⋮ & ⋮ & ⋮ & ⋮ & ... \end{matrix} ) , ρ(σ_{2}) = ( \begin{matrix}1 & 0 & 0 & 0 & … \\ 0 & 0 & −\frac{u}{u+1} & 0 & … \\ 0 & −1 & 0 & 0 & … \\ 0 & 0 & 0 & 1 & … \\ ⋮ & ⋮ & ⋮ & ⋮ & ... \end{matrix} ) , etc. (7.3)

We recall that a monomial representation of a group G is a representation ρ : G → GL(V), where for each g ∈ G the matrix ρ(g) has only one entry in each row and each column; such a matrix is called a monomial matrix. General results about monomial groups and representations can be found in [O, Sc].

We now show that this representation V =< v_{1}, …, v_{n} > is irreducible. For suppose that W is an invariant subspace and let v ∈ W, 0 ≠ v = ∑_{i=1}^{n} λ_{i}v_{i}. Let r = r(v) = min{i|λ_{i} ≠ 0}. From the above we see that the action of σ_{i}^{2} is represented relative to the basis v_{1}, …, v_{n} by the diagonal matrix diag(1, …, 1, \frac{u}{u+1}, \frac{u}{u+1}, 1, …, 1), where the \frac{u}{u+1} entries are in the i and i + 1 positions. Thus if r = r(v) > 1, then the span of v and σ_{r−1}^{2}(v) contains v_{r}. Since V is a monomial representation whose corresponding permutation representation is transitive we see that v_{i} ∈ W for all i ≤ n and so W = V.

Similarly, if W is not 1-dimensional, then there is 0 ≠ v ∈ W with r(v) > 1 and so as in the above we are done. Thus we may assume that dim(W) = 1, W =< v > and r(v) = 1 and in this situation the span < v, σ_{1}^{2}(v) > contains an element of the form w = λ_{1}v_{1} +λ_{2}v_{2}. If λ_{1}λ_{2} = 0, then v_{i} ∈ W for some i = 1, 2; whereas if λ_{2} ≠ 0, then v_{2} ∈< w, σ_{2}^{2}(w) >. In either case we again see that W = V and we are done.

Remark 7.2. The action of B_{n} on < v_{1}, …, v_{n} > is not faithful since, for example, one can show that the images of σ_{1}^{2} and of σ_{2}^{2} are both diagonal and so commute. However it is well-known [Bi] that the subgroup < σ_{1}^{2}, σ_{2}^{2} > of B_{3} is free on the two given generators.

Now any μ_{i} can be evaluated as a minor of the (n + 1) × (n + 1) matrix (a_{rs}) and then we can look at this element mod I_{n+1}). This map we denote by I_{n+1}.

Lemma 7.3. (i) Let n ≥ 2 and 1 ≤ j ≤ n. Then

I_{n+1}(μ_{j}) = (−1)^{j+1} \frac{1}{(u + 1)^{j−1}u^{n−j}} a_{jn+1}.

(ii) For n ≥ 2 we have

I_{n+1}(det(a_{ij})_{n×n}) = (−1)^{n+1} (\frac{1}{u^{n−1}} − \frac{1}{(u + 1)^{n−1}} ) .

(iii) For n ≥ 2 and 1 ≤ i < j ≤ n we have

I_{n}((1, … î, …, n|1, … ĵ, …, n)) = (−1)^{n+i+j+1} \frac{1}{u^{n+i−j−1}(u + 1)^{j−i−1}} a_{ji}

Proof. We first show that (ii) for n follows from (i) for n − 1. Expanding det(a_{ij}) along the last row we get (remembering that a_{ii} = 0):

det(a_{ij}) = ∑_{i=1}^{n−1}(−1)^{n+i}a_{ni}(1, 2, …, n − 1|1, 2, …, î, …, n)

= ∑_{i=1}^{n−1}(−1)^{n+i}μ_{i}^{(n−1)} = ∑_{i=1}^{n−1}(−1)^{n+i}a_{ni}(−1)^{i+1} \frac{a_{in}}{(u + 1)^{i−1}u^{n−1−i}}

= (−1)^{n+1} ∑_{i=1}^{n−1} \frac{1}{(u + 1)^{i}u^{n−i}} = (−1)^{n+1} \frac{1}{u^{n}} \frac{u}{u + 1} ∑_{i=0}^{n−2} (\frac{u}{u + 1})^{i}

= (−1)^{n+1} \frac{1}{u^{n}} \frac{u}{u + 1} \frac{(1 − (\frac{u}{u+1})^{n−1})}{1 − \frac{u}{u+1}}

= (−1)^{n+1} (\frac{1}{u^{n−1}} − \frac{1}{(u + 1)^{n−1}} ) .

Proof of (i). This is by induction on n ≥ 2, the case n = 2 being easy to check. So assume that the lemma is true for n − 1 ≥ 2 and for all j ≤ n − 1. Then expanding along

REPRESENTATIONS OF BRAID GROUPS VIA DETERMINANTAL RINGS 23

the jth row we have:

μ_{j}^{(n)} = ∑_{i=1}^{j−1}(−1)^{j+i}a_{ji}(1, 2, …, ĵ, …, n|1, 2, …, î, …, ĵ, …, n, n + 1)

+ ∑_{i=j+1}^{n}(−1)^{j+i−1}a_{ji}(1, 2, …, ĵ, …, n|1, 2, …, ĵ, …, î, …, n, n + 1)

+ (−1)^{j+n+1}a_{jn+1}(1, 2, …, ĵ, …, n|1, 2, …, ĵ, …, n)

= ∑_{i=1}^{j−1}(−1)^{j+i}a_{ji}μ_{i}^{(n−1)} + ∑_{i=j+1}^{n}(−1)^{j+i−1}a_{ji}μ_{i−1}^{(n−1)} + (−1)^{j+n+1}a_{jn+1}det((a_{ij})_{(n−1)×(n−1)})

= ∑_{i=1}^{j−1}(−1)^{j+i}a_{ji} \frac{(−1)^{i+1}a_{in+1}}{(u + 1)^{i−1}u^{n−1−i}} + ∑_{i=j+1}^{n}(−1)^{j+i−1}a_{ji} \frac{(−1)^{i}a_{in+1}}{(u + 1)^{i−2}u^{n−i}}

+ (−1)^{j+n}a_{jn+1}(−1)^{n} (\frac{1}{u^{n−2}} − \frac{1}{(u + 1)^{n−2}} )

= (−1)^{j+1} ∑_{i=1}^{j−1} \frac{a_{jn+1}}{(u + 1)^{i}u^{n−i−1}} + (−1)^{j+1} ∑_{i=j+1}^{n} \frac{a_{jn+1}}{(u + 1)^{i−2}u^{n−i+1}}

+ (−1)^{j}a_{jn+1} (\frac{1}{u^{n−2}} − \frac{1}{(u + 1)^{n−2}} )

= (−1)^{j+1}a_{jn+1} (∑_{i=1}^{j−1} \frac{1}{(u + 1)^{i}u^{n−i−1}} + ∑_{i=j+1}^{n} \frac{1}{(u + 1)^{i−2}u^{n−i+1}} − \frac{1}{u^{n−2}} + \frac{1}{(u + 1)^{n−2}} )

= (−1)^{j+1}a_{jn+1} (\frac{1}{u^{n−1}} ∑_{i=1}^{j−1} (\frac{u}{u + 1})^{i} + \frac{(u + 1)^{2}}{u^{n+1}} ∑_{i=j+1}^{n} (\frac{u}{u + 1})^{i} − \frac{1}{u^{n−2}} + \frac{1}{(u + 1)^{n−2}} )

= (−1)^{j+1}a_{jn+1} (\frac{1}{u^{n−1}} \frac{(\frac{u}{u+1} − \frac{u^{j}}{(u+1)^{j}} )}{(1 − \frac{u}{u+1})} + \frac{(u + 1)^{2}}{u^{n+1}} \frac{(\frac{u^{j+1}}{(u+1)^{j+1}} − \frac{u^{n}}{(u+1)^{n}} )}{(1 − \frac{u}{u+1})} − \frac{1}{u^{n−2}} + \frac{1}{(u + 1)^{n−2}} )

= (−1)^{j+1}a_{jn+1}(\frac{u}{u^{n−1}} (1 − (\frac{u}{u + 1})^{j−1}) + \frac{(u + 1)^{3}}{u^{n+1}} \frac{u^{j+1}}{(u + 1)^{j+1}} (1 − (\frac{u}{u + 1})^{n−j})

− \frac{1}{u^{n−2}} + \frac{1}{(u + 1)^{n−2}} )

= (−1)^{j+1}a_{jn+1}(\frac{1}{u^{n−2}} − \frac{1}{(u + 1)^{j−1}u^{n−1−j}} + \frac{1}{u^{n−j}(u + 1)^{j−2}} − \frac{1}{(u + 1)^{n−2}}

− \frac{1}{u^{n−2}} + \frac{1}{(u + 1)^{n−2}} )

= (−1)^{j+1}a_{jn+1} \frac{1}{(u + 1)^{j−1}u^{n−j}} ,

as required for (i). (iii) is just a variation of (i). □

Note that as a C(u)-module the image of I_{n+1} has dimension n with generators a_{1n+1}, a_{2n+1}, …, a_{nn+1}. Now let K_{n} = {x ∈ V_{n}|I_{n+1}(x) = 0}. Then by Lemma 4.1 K_{n} is B_{n}- invariant and by Lemma 7.1 and the above remark it has dimension n(n − 1). Now we have

I_{n+1}(v_{1}) = I_{n+1}(μ_{1} − ua_{12}μ_{2} + ua_{13}μ_{3} − ua_{14}μ_{4} + ua_{15}μ_{5} − ⋯ + (−1)^{n+1}ua_{1n}μ_{n})

= \frac{1}{u^{n−1}} + \frac{u}{(u + 1)u^{n−2}} a_{12}a_{2n+1} + \frac{u}{(u + 1)^{2}u^{n−3}} a_{13}a_{3n+1} + …

+ \frac{u}{(u + 1)^{n−1}} a_{1n}a_{nn+1}

= (\frac{1}{u^{n−1}} + \frac{1}{(u + 1)u^{n−2}} + \frac{1}{(u + 1)^{2}u^{n−3}} + ⋯ + \frac{1}{(u + 1)^{n−1}} ) a_{1n+1},

which is clearly non-zero. One similarly (or even directly from this) sees that I_{n+1}(v_{j}) is a non-zero multiple of a_{jn+1} for all j ≤ n. Thus < v_{1}, …, v_{n} > ∩K_{n} = {0}. It follows that < v_{1}, …, v_{n} > is a complement to K_{n}, both being B_{n}-invariant.

We now show that K_{n} is an irreducible representation of B_{n}. For this we define the following basis: for 1 ≤ i ≠ j ≤ n let

γ_{ij} = a_{ij}μ_{j} − \frac{(−1)^{i+j}}{u} (\frac{u}{u + 1})^{j−i} μ_{i} if i < j;

γ_{ij} = a_{ij}μ_{j} − \frac{(−1)^{i+j}}{u + 1} (\frac{u + 1}{u})^{i−j} μ_{i} if i > j.

It will be convenient to put γ_{ii} = 0 for all i. Then using Lemma 7.3 we see that I_{n+1}(γ_{ij}) = 0 for all i ≠ j. Since the γ_{ij} are clearly independent, they form a basis for K_{n}. We will find it convenient to write γ_{ij} = a_{ij}μ_{j} − λ_{ij}μ_{i}, which thus defines the λ_{ij} ∈ C(u).

Lemma 7.4. (i) For 1 ≤ i < n and j ≠ i, i + 1 we have

σ_{i}(γ_{ii+1}) = −γ_{i+1i}; σ_{i}(γ_{i+1i}) = (λ_{i+1i} − 1)γ_{ii+1}; σ_{i}(γ_{ij}) = \frac{u}{u + 1}γ_{i+1j};

σ_{i}(γ_{i+1j}) = γ_{ij} + λ_{i+1j}γ_{ii+1}; σ_{i}(γ_{ji}) = −\frac{u + 1}{u}γ_{ji+1};

σ_{i}(γ_{ji+1}) = −γ_{ji} − \frac{1}{u}γ_{ji+1}.

(ii) For the action of σ_{i}^{2} we have:

σ_{i}^{2}(γ_{ii+1}) = \frac{u + 1}{u}γ_{ii+1}; σ_{i}^{2}(γ_{j}) = \frac{u}{u + 1}(γ_{ij} + λ_{i+1j}γ_{ii+1});

σ_{i}^{2}(γ_{ji}) = \frac{u + 1}{u}(γ_{ji} + \frac{1}{u}γ_{ji+1}); σ_{i}^{2}(γ_{i+1j}) = \frac{u}{u + 1}γ_{i+1j} − λ_{i+1j}γ_{i+1i};

σ_{i}^{2}(γ_{ji+1}) = \frac{1}{u} + \frac{u^{2} + u + 1}{u^{2}}γ_{ji+1}; σ_{i}^{2}(γ_{i+1i}) = \frac{1 + u}{u}γ_{i+1i}.

Proof. (i) follows from (1.2) and (7.1), and (ii) follows from (i). □

Lemma 7.5. Let 1 ≤ i < n. The matrix m_{i}^{2} representing the action of σ_{i}^{2} on K_{n} is diagonalisable and has eigenvalues 1, \frac{u}{u+1}, \frac{u+1}{u}, \frac{(u+1)^{2}}{u^{2}} .

Proof. We need only consider i = 1 and so we will give a basis for the eigenspaces of m_{1}^{2} corresponding to these eigenvalues. The dimensions will be seen to sum to n(n − 1) and so the result will follow.

The elements γ_{rs} for r, s ≠ i, i + 1 are 1 eigenvectors; there are (n − 2)(n − 3) of these. For i > 2 the elements γ_{i1} − γ_{i2} are also fixed; there are n − 2 of these.

The elements γ_{ii+1}, γ_{i+1i} are eigenvectors for the eigenvalue \frac{u+1}{u}.

For i > 2 the elements γ_{i1} + \frac{u+1}{u}γ_{i2} are eigenvectors for the eigenvalue \frac{(u+1)^{2}}{u^{2}} ; there are n − 2 of these.

For 1 < i < n the elements γ_{1i} − \frac{λ_{2j}}{λ_{2n}} γ_{1n} are eigenvectors for \frac{u}{u+1}; there are n−2 of these. For i ≠ 2, n the elements γ_{2i} − \frac{λ_{2j}}{λ_{2n}} γ_{2n} are eigenvectors for \frac{u}{u+1}; there are n − 2 of these. □

This last result shows that this representation (over ℂ say) has at least 4 eigenvalues for each σ_{i}. Since we will show that it is irreducible, it follows that it is not a summand of the Jones representation [J].

Let 0 ≠ b ∈ K_{n} and write b = ∑ c_{ij}γ_{ij} with c_{ij} ∈ C(u). Let i′ = min{i|c_{ij} ≠ 0 for some j}, j′ = min{j|c_{ij} ≠ 0 for some i}. Let r = min{i′, j′}. Assume r = i′ (the other case is similar). Let s = min{j|c_{rj} ≠ 0}. Note that if r > 1, then σ_{r−1}(γ_{rs}) = γ_{r−1s} + λ_{is}γ_{r−1r} and so we have a smaller r in σ_{r−1}(b). Thus we may assume that r = 1. Similarly, if s ≠ 2, then we can lower s by acting on b by σ_{s−1}. It follows that we may assume that c_{12} ≠ 0.

For this b we can now let b = ∑_{j=1}^{4} b_{j} where each b_{j} is an eigenvector for σ_{i}^{2}; namely σ_{1}^{2}(b_{1}) = b_{1}, σ_{1}^{2}(b_{2}) = \frac{u+1}{u}b_{2} etc. But c_{12} ≠ 0 shows that b_{2} ≠ 0 and so we see that some C(u)-combination of σ_{1}^{2k}(b) contains a non-zero element in the \frac{u+1}{u}-eigenspace of σ_{1}^{2}. This eigenspace is spanned by γ_{12} and γ_{21} and so we may assume that b = c_{1}γ_{12} + c_{2}γ_{21}. But one now checks that either γ_{12} or γ_{21} is a linear combination of b, σ_{2}^{2}(b), σ_{2}^{4}(b), σ_{2}^{6}(b) (use Lemma 7.4). Thus Lemma 7.4 shows that C(u)(B_{n}(b)) contains γ_{12}. But C(u)(B_{n}(γ_{12})) contains all the elements γ_{ij} and so K_{n} is irreducible. This proves Theorem 1.5. We will denote the K_{n} representation of B_{n} by V_{n,n^{2}−n}.

We now consider how these two irreducible B_{n}-representations split when considered as B_{n−1}-modules. First, the representation < v_{1}, …, v_{n} > clearly splits as < v_{1}, …, v_{n−1} > ⊕ < v_{n} >, both of which are irreducible B_{n−1} representations. We will denote the trivial representation of B_{n−1} by V_{n−11} and the < v_{1}, …, v_{n−1} > representation of B_{n−1} by V_{n−1n−1}.

For the K_{n} representation we note that the element ∑_{i=1}^{n−1}(−1)^{i}γ_{ni} is fixed by B_{n−1}. This gives a 1-dimensional summand. From the above we clearly see that the span of {γ_{ij}|1 ≤ i, j ≤ n − 1} is an irreducible B_{n−1}-module; it has dimension (n − 1)(n − 2).

Now let w_{i} = γ_{ni} + \frac{u+1}{u}γ_{ni+1} for 1 < i < n and W =< w_{2}, …, w_{n−1} >. We will show that W is an irreducible B_{n−1}-module. We first note:

σ_{i}(w_{i−1}) = σ_{i}(γ_{ni−1} + \frac{u + 1}{u}γ_{ni}) = γ_{ni−1} + \frac{u + 1}{u} (−\frac{u + 1}{u}γ_{ni+1})

= γ_{ni−1} + \frac{u + 1}{u}γ_{ni} − \frac{u + 1}{u} (γ_{ni} + \frac{u + 1}{u}γ_{ni+1}) = w_{i−1} − \frac{u + 1}{u}w_{i};

σ_{i}(w_{i}) = σ_{i}(γ_{ni} + \frac{u + 1}{u}γ_{ni+1}) = −\frac{u + 1}{u}γ_{ni+1} + \frac{u + 1}{u}(−γ_{ni} − \frac{1}{u}γ_{ni+1}) = −\frac{u + 1}{u}w_{i};
σ_{i}(w_{i+1}) = σ_{i}(γ_{ni+1} + \frac{u + 1}{u}γ_{ni+2}) = −γ_{ni} − \frac{1}{u}γ_{ni+1} + \frac{u + 1}{u}γ_{ni+1} = −w_{i} + w_{i+1}.

Thus the (n − 2) × (n − 2) matrices m_{i}, i < n − 1 representing σ_{i} relative to the basis w_{i} are (where t = \frac{u+1}{u}):

m_{1} = ( \begin{matrix}−t & 0 & 0 & 0 & … \\ −1 & 1 & 0 & 0 & … \\ 0 & 0 & 1 & 0 & … \\ 0 & 0 & 0 & 1 & … \\ ⋮ & ⋮ & ⋮ & ⋮ & ... \end{matrix} ) , m_{2} = ( \begin{matrix}1 & −t & 0 & 0 & … \\ 0 & −t & 0 & 0 & … \\ 0 & −1 & 1 & 0 & … \\ 0 & 0 & 0 & 1 & … \\ ⋮ & ⋮ & ⋮ & ⋮ & ... \end{matrix} ) ,

m_{3} = ( \begin{matrix}1 & 0 & 0 & 0 & … \\ 0 & 1 & −t & 0 & … \\ 0 & 0 & −t & 0 & … \\ 0 & 0 & −1 & 1 & … \\ ⋮ & ⋮ & ⋮ & ⋮ & ... \end{matrix} ) , etc.

We now show that this gives an irreducible representation of B_{n−1}. It will suffice to show that the action of these matrices on the row space is irreducible. Let U be a non-trivial subrepresentation and let w ∈ U. We first show that U contains some e_{i} = (0, …, 0, 1, 0, …, 0). This is certainly the case if wm_{i} ≠ w for some i. Now if wm_{i} = w for all i, then wM = 0, where

M = m_{1} + ⋯ + m_{n−2} − (n − 2)I_{n−2} = ( \begin{matrix}−t & −t & 0 & 0 & 0 & … \\ −1 & −t & −t & 0 & 0 & … \\ 0 & −1 & −t & −t & 0 & … \\ 0 & 0 & −1 & −t & −t & … \\ ⋮ & ⋮ & ⋮ & ⋮ & ⋮ & ... \end{matrix} ) .

But this matrix has non-zero determinant and so w = 0; thus e_{i} ∈ W for some i.

Now suppose that e_{i} ∈ W. Then e_{i+1} ∈< e_{i}, e_{i}m_{i+1} > and one easily sees that W =< e_{1}, …, e_{n−2} >. This gives the irreducibility. We will denote this representation of B_{n−1} by V_{n−1n−2}.

For the last representation we let

z_{1} = −uγ_{12} + uγ_{13} − uγ_{14} + uγ_{15} − …
z_{2} = −(u + 1)γ_{21} − uγ_{23} + uγ_{24} − uγ_{25} + …
z_{3} = (u + 1)γ_{31} − (u + 1)γ_{32} − uγ_{34} + uγ_{35−...}
…
z_{n−1} = ±(u + 1)γ_{n−11} ∓ (u + 1)γ_{n−12} ± (u + 1)γ_{n−13} ∓ (u + 1)γ_{n−14} …

exactly in analogy to how we defined the v_{i} in (7.2). Then the same argument used there shows that Z =< z_{1}, …, z_{n−1} > is an irreducible B_{n−1}-module with the (n − 1) × (n − 1) matrices given by (7.3). We denote this representation by V_{n−1,n−1}.

We have now proved that the restrictions of V_{n,n} and of V_{n,n^{2}−n} to B_{n−1} are multiplicity free:

Theorem 7.6. The restrictions Res_{B_{n−1}}^{B_{n}}V_{nn} and Res_{B_{n−1}}^{B_{n}}V_{nn^{2}−n} decompose according to the following diagram (branching law):

The restriction Res_{B_{n−1}}^{B_{n}}V_{n,n−1} decomposes as V_{n−1,1} ⊕ V_{n−1,n−2}. □

For results concerning the existence of branching laws for the classical groups see [GW, Ch. 8].

§8 The action of B_{n} on (S|{n + 1, …, n + |S|})

By Corollary 3.2 and Lemma 5.8 the action on the elements in the B_{n}-orbit of (S|{n + 1, …, n + |S|}) is the same as the action on the R_{n}^{(0)}-module ⋀^{|S|} V_{n}. By Lemma 4.2 we see that every monomial in the B_{n}-orbit of (S|{n + 1, …, n + |S|}) has the form a_{r_{1}s_{1}} … a_{r_{k}s_{k}}(S′|{n + 1, …, n + |S|}) where s_{i} ∈ S′ and r_{i} ∉ S′. Let V (S) denote the C(u)-module generated by all such elements.

Lemma 8.1. The dimension of V (S) is \binom{n}{|S|}^{2}.

Proof. We will need the following:

Lemma 8.2. Let a_{r_{1}s_{1}} … a_{r_{k}s_{k}} be given as above. Let π be any permutation of the set {r_{1}, …, r_{k}}. Then there is c ∈ C(u) such that

a_{π(r_{1})s_{1}} … a_{π(r_{k})s_{k}} = ca_{r_{1}s_{1}} … a_{r_{k}s_{k}}modI({1, 2, …, n}).

Proof. From the defining relations for I({1, …, n}) we see that for any distinct i, j, k, m there are non-zero c, c′ ∈ C(u) such that ca_{im}a_{mj} = a_{ij} and a_{km}a_{mj} = c′a_{kj}. Thus in R_{n}/I({1, …, n}) we have

a_{ij}a_{km} = ca_{im}a_{mj}a_{km} = ca_{im}a_{km}a_{mj} = cc′a_{im}a_{kj} = cc′a_{kj}a_{im}.

Thus in R_{n}/I({1, …, n}) we can interchange i and k in any product of the form a_{ij}a_{km}. The result easily follows. □

We count the number of elements of the form a_{r_{1}s_{1}} … a_{r_{k}s_{k}}(S′|{n + 1, …, n + |S|}) with the r_{i}, s_{i} as described above and with S′ fixed. Note that there exactly \binom{n}{|S|} of the (S′|{n + 1, …, n + |S|})s.

Now for fixed |S| ≥ k ≥ 0 there are \binom{n−s}{k} choices of the r_{i} and \binom{s}{k} choices of the s_{i} and so there are \binom{n−s}{k}\binom{s}{k} total such choices. Summing over the various k gives

dimV (S) = ∑_{k=0}^{|S|} \binom{n − s}{k}\binom{s}{k} = \binom{n}{|S|},

the last equality being a well-known binomial identity [R]. This proves Lemma 8.1. □

We will need the following construction. Let S be a subsequence of {1, 2, …, n} with |S| = k. For m > 0 we let N_{m} = [n + 1, n + 2, …, n + m]. Then to the element (S|N_{k})′ we associate

ω(S|N_{k})′ = (S, N_{n−k}|1, 2, …, n)′.

We extend the action of ω so as to obtain an R_{n}^{∞}-module map, also denoted by ω.

The action of B_{n} on (N_{n}|1, 2, … n)′ gives the sign permutation ϵ : B_{n} → S_{n} → {±1}. Thus the action of α ∈ B_{n} on the ω(S|N_{k})′ is given by

α(ω(S|N_{k})′) = ϵ(α)ω(α(S|N_{k})′).

Thus the representation theory for the (S|N_{k})′ is the same as for the ω(S|N_{k})′. Now we define a map J = J_{n+1} by

J_{n+1}(w(S|N_{k})′) = w × det(ω(S|N_{k})′) mod I_{n}.

Since the ideal I_{n} is B_{n}-invariant we see that the B_{n}-action commutes with J: for all α ∈ B_{n}, b ∈ R_{n}^{k} we have J α(b) = αJ (b). Thus the image of J is a B_{n}-representation space which is isomorphic to a direct sum of the B_{n}-irreducible summands of R_{n}^{1}. We will next show that J is not the zero homomorphism:

Lemma 8.3. For 1 ≤ k < n we have

J_{n+1}(([1, 2, …, k]|N_{k})′) = \frac{(−1)^{k}}{u^{k}} det(n + 1, n + 2, …, n|k + 1, k + 2, …, n).

Proof. Consider the matrix M_{k} = (1, 2, …, k, n + 1, n + 2, … n + (n − k)|1, 2, …, n). Then

M_{k} = ( \begin{matrix}0 & a_{12} & a_{13} & … & a_{1k} & a_{1k+1} & … & a_{1n} \\ a_{21} & 0 & a_{23} & … & a_{2k} & a_{2k+1} & … & a_{2n} \\ ⋮ & ⋮ & ⋮ & … & ⋮ & ⋮ & … & ⋮ \\ a_{k1} & a_{k2} & a_{k3} & … & 0 & a_{k,k+1} & … & a_{kn} \\ a_{n+11} & a_{n+12} & a_{n+13} & … & a_{n+1k} & a_{n+1k+1} & … & a_{n+1n} \\ a_{n+21} & a_{n+22} & a_{n+23} & … & a_{n+2k} & a_{n+2k+1} & … & a_{n+2n} \\ ⋮ & ⋮ & ⋮ & … & ⋮ & ⋮ & … & ⋮ \\ a_{n+n−k,1} & a_{n+n−k,2} & a_{n+n−k,3} & … & a_{n+n−k,k} & a_{n+n−k,k+1} & … & a_{n+n−k,n} \end{matrix} )

Now a_{12}a_{2p} = \frac{1}{u}a_{1p} for p > 2 and so adding −ua_{12} times the second row to the first row produces the matrix whose first row is (\frac{−1}{u+1}, a_{12}, 0, 0, …, 0). Similarly, adding −ua_{23} times the third row to the second row produces the matrix whose second row is (0, \frac{−1}{u+1}, a_{23}, 0, …, 0). Repeating this process k − 1 times and then adding −(u + 1)a_{kn+1} times the k + 1th row to the kth row produces the matrix

( \begin{matrix}\frac{−1}{u+1} & a_{12} & 0 & 0 & … & 0 & 0 & … & 0 \\ 0 & \frac{−1}{u+1} & a_{23} & 0 & … & 0 & 0 & … & 0 \\ 0 & 0 & \frac{−1}{u+1} & a_{34} & … & 0 & 0 & … & 0 \\ ⋮ & ⋮ & ⋮ & ⋮ & … & ⋮ & ⋮ & … & ⋮ \\ 0 & 0 & 0 & 0 & … & \frac{−1}{u+1} & a_{k−1,k} & … & 0 \\ \frac{−1}{u}a_{k1} & \frac{−1}{u}a_{k2} & \frac{−1}{u}a_{k3} & \frac{−1}{u}a_{k4} & … & \frac{−1}{u}a_{k,k−1} & \frac{−1}{u} & … & 0 \\ a_{n+11} & a_{n+12} & a_{n+13} & a_{n+14} & … & a_{n+1,k−1} & a_{n+1k} & … & a_{n+1n} \\ a_{n+21} & a_{n+22} & a_{n+23} & a_{n+24} & … & a_{n+2,k−1} & a_{n+2k} & … & a_{n+2n} \\ ⋮ & ⋮ & ⋮ & ⋮ & … & ⋮ & ⋮ & … & ⋮ \\ a_{n+n−k1} & a_{n+n−k2} & a_{n+n−k3} & a_{n+n−k4} & … & a_{n+n−k,k−1} & a_{n+n−k,k} & … & a_{n+n−kn} \end{matrix} )

One can now see that the determinant of this matrix is det(n + 1, n + 2, …, n + n − k|k + 1, k + 2, …, n) multiplied by the determinant of the principal k × k matrix. To find this latter determinant, which we prove is (−1/u)^{k}, we induct on k ≥ 1 the case k = 1 being clear (look at the (k,k) entry, not the (1,1) entry). For k > 1 we note that by the k − 1 case the (1, 1) entry of the adjoint matrix is (−1/u)^{k−1}; thus expanding along the first row we get:

\frac{−1}{u + 1} (\frac{−1}{u})^{k−1} − a_{12}a_{23}a_{34} … a_{k−1k} (\frac{−1}{u}a_{k1}) (−1)^{k−1}

= \frac{−1}{u + 1} (\frac{−1}{u})^{k−1} − \frac{1}{u^{k−2}} (\frac{−1}{u}) \frac{1}{u(u + 1)}(−1)^{k}

= \frac{(−1)^{k}}{u^{k−1}(u + 1)} (1 + \frac{1}{u}) = \frac{(−1)^{k}}{u^{k}} ,

as required. □

Proposition 8.4. The image and the kernel of J are non-trivial.

Proof. That the imnage is non-trivial follows from Lemma 8.3. Let η = η_{{1,...,n,n+1}} as in (4.1).

Lemma 8.5. A Gröbner basis for the ideal I_{n+1} relative to the ‘degree lexicographical’ order [AL, p. 19] consists of all elements of the following forms:

a_{ij}a_{ji} − \frac{1}{u(u + 1)}, for i ≠ j;

a_{ij}a_{jk} − \frac{1}{u}a_{ik}, if (j − i)(k − i)(k − j) > 0;

a_{ij}a_{jk} − \frac{1}{u + 1}a_{ik} if (j − i)(k − i)(k − j) < 0

a_{ij}a_{rs} − \frac{η(a_{ij}a_{rs})}{η(a_{is}a_{rj})}a_{is}a_{rj} if i, j, r, s are distinct and i < r, j < s.

Proof. One can check the the elements given satisfy the requirements for a Gröbner basis relative to the degree lexicographical order [AL, §1.6]. (Note that this order is called ‘glex’ in [MA]). □

Now note that if S is a subsequence of {1, …, n} with |S| = k, then J (S|N_{k}) is the determinant of a certain matrix, which determinant can be expanded along the row labelled n + 1:

J (S|N_{k})′ = ∑_{i=1}^{n}(−1)^{n+1+i}a_{n+1i}det(S, n + 2, … n + n − 1|1, …, î, …, n)

and then the monomials can be reduced mod I_{n+1}, so that each monomial looks like μ = a_{n+1i}b_{n+2c_{2}}b_{n+3c_{3}} … b_{n+n−kc_{n−k}} where b_{n+ic_{i}} either has the form a_{n+vj(v)}a_{j(v)e(v)} or the form a_{n+vuk(v)}. In the first case we will call the j(v) the middle indices of the monomial. We note that no two b_{n+ic_{i}} have the same middle indices.

We will say that such a monomial μ has end set {i, e(2), e(3), …, e(n − k)}. We note that for a monomial of J (S|N_{k}) as in the above we must have {i, e(2), e(3), …, e(n − k)} = {1, …, n} \ S. We can collect together all such terms and so are able to write

J_{n+1}((S|N_{k})′) = ∑_{i∉S} a_{n+1i}μ_{i}  (8.1)

where each μ_{i} is a sum of monomials all having the same end set, and each monomial of μ_{i} is in normal form a_{n+1i}b_{n+2,c_{2}}b_{n+3,c_{3}} … b_{n+n−k,c_{n−k}} (as in the above) relative to the Gröbner basis of Lemma 8.5.

It is now easy to see from Lemma 8.3 that J_{n+1}(([1, 2, …, k]|N_{k})′ ≠ 0. We will next show that the kernel of J is non-trivial.

Note that given any subsequence S ⊂ {1, …, n}, |S| = k and any end set E, |E| = n − k, there is w ∈ R_{n}^{(0)} such that all monomials in w × det(S, n + 1, … n + n − k|1, …, n) have the end set E.

Now note that if S, S′ are subsequences of {1, …, n} with |S| = |S′| = k, then J (S|N_{k})′ ≠ J (S′|N_{k})′ whenever S ≠ S′, as they have different end sets. Since there are \binom{n}{k} of the Ss we see that we must check for relations among the wJ (S|N_{k})′ only in the set of such which have the same end sets.

For S a subsequence of {1, …, n} and {1, …, n} \ S = {s_{1}, … s_{n−k}} let

δ_{S} = δ_{s_{1},s_{2},...,s_{n−k}} = det(n + 1, n + 2, …, n + n − k|s_{1}, s_{2}, …, s_{n−k}) mod I_{n+1}.

Then Lemma 8.3 shows that we have δ_{k+1,k+2,...,n} in the image of J. From the above we see that for any end set E and any sequence S ⊂ {1, …, n + 1} with |S| = k there is w ∈ R_{n}^{(0)} such that each monomial of the expanded form of wδ_{S} has the end set E.

We also see from (1.3) that

σ_{k}^{2}δ_{k+1,k+2,...,n} = σ_{k}δ_{k,k+2,...,n}
= δ_{k+1,k+2,...,n} + a_{k,k+1}δ_{k,k+2,...,n}.

Similarly we have

σ_{k−1}σ_{k}^{2}δ_{k+1,k+2,...,n} = δ_{k+1,k+2,...,n} + a_{k−1,k+1}δ_{k−1,k+2,...,n},
σ_{k−2}σ_{k−1}σ_{k}^{2}δ_{k+1,k+2,...,n} = δ_{k+1,k+2,...,n} + a_{k−2,k+1}δ_{k−2,k+2,...,n},

etc. Thus we can get all a_{hk+1}δ_{h,k+2,...,n} for h < k + 1. We also have

σ_{k}^{2}σ_{k+1}^{2}(a_{hk+1}δ_{(h,k+2,...,n)}) = \frac{u + 1}{u^{2}}a_{h,k+2}a_{k,k+1}δ_{h,k,k+3,...,n}

+ \frac{(u + 1)^{2}}{u^{2}} a_{h,k+1}δ_{h,k+2,...,n} + \frac{u + 1}{u^{2}}a_{h,k+2}δ_{h,k+1,k+3,...,n};

σ_{k}^{4}σ_{k+1}^{2}(a_{hk+1}δ_{(h,k+2,...,n)}) = \frac{2u^{2} + 2u + 1}{u^{3}}a_{h,k+2}a_{k,k+1}δ_{h,k,k+3,...,n}

+ \frac{(u + 1)^{3}}{u^{3}} a_{h,k+1}δ_{h,k+2,...,n} + \frac{u^{2} + u + 1}{u^{3}}a_{h,k+2}δ_{h,k+1,k+3,...,n},

from which we see that we can get a_{h,k+2}a_{k,k+1}δ_{h,k,k+3,...,n} and a_{h,k+2}δ_{h,k+1,k+3,...,n}. Continuing in this way we get the first sentence of:

Lemma 8.6. The C(u)-span of the B_{n}-orbit of δ_{k+1,...,n} contains all wδ_{S} with the end set {k + 1, …, n}. The C(u)-span of the B_{n}-orbit of δ_{k+1,...,n} contains all wδ_{S} with any end set.

Proof. For the second sentence we let V_{E} be the C(u)-span of all wδ_{S} with end set E. Then this follows from the fact that for α ∈ B_{n} we have α(V_{E}) = V_{Π_{n}(α)(E)}. □

Next we note that if E = {s_{1}, …, s_{n−k+1}} with s_{1} < s_{2} < ⋯ < s_{n−k+1}, then we can evaluate det(s_{1}, n + 1, n + 2, …, n + n − k|s_{1}, …, s_{n−k+1}) in two ways: (i) by Lemma 8.3 we see that it is equal to −\frac{1}{u}δ_{s_{2},...,s_{n−k+1}}; (ii) expanding along the row labeled s_{1} we have

det(s_{1}, n + 1, n + 2, …, n + n − k|s_{1}, …, s_{n−k+1}) = ∑_{i=1}^{n−k+1}(−1)^{i}a_{s_{1}s_{i}}δ_{s_{1},...,ŝ_{i},...s_{n−k+1}}

= ∑_{i=2}^{n−k+1}(−1)^{i}a_{s_{1}s_{i}}δ_{s_{1},...,ŝ_{i},...s_{n−k+1}}.

Thus we have the relation

∑_{i=2}^{n−k+1}(−1)^{i}a_{s_{1}s_{i}}δ_{s_{1},...,ŝ_{i},...s_{n−k+1}} + \frac{1}{u}δ_{s_{2},...,s_{n−k+1}} = 0,  (8.2)

among the wδ_{S}. This is a relation involving terms all having the same end set, namely {s_{2}, …, s_{n−k+1}}. However, given any end set E′ we can multiply such an expression (8.2) by some w ∈ R_{n}^{∞} so that the resulting product has end set E′. There are \binom{n}{k} end sets. This proves Proposition 8.4. □

We now show that there are exactly \binom{n−1}{k} of the wδ_{S} with |S| = k and all having the same end set E. We may clearly take E = {k + 1, …, n} and we count them according to the degree of w (where we always take w reduced as in the above). For deg(w) = 0, there is just one possibility. If deg(w) = 1, then w = a_{n+z,i}, where z = 2, … n − k, 1 ≤ i ≤ k and there are \binom{n−k−1}{1} × \binom{k}{1} possibilities. If deg(w) = 2, then w = a_{n+z_{1},i}a_{n+z_{2},j}, where z_{1}, z_{2} = 2, … n − k, z_{1} ≠ z_{2}, 1 ≤ i ≠ j ≤ k and there are \binom{n−k−1}{2} × \binom{k}{2} possibilities (by Lemma 8.5 or Lemma 4.6). Continuing in this way we see that the total number is

∑_{i=0}^{k} \binom{n − k − 1}{i} × \binom{k}{i} = \binom{n − 1}{k},

as required. It follows from Lemma 8.5 that these are all linearly independent over C(u). Thus the image of J_{n+1} has dimension \binom{n}{k}\binom{n−1}{k}, proving the first part of Theorem 1.7.

We now show that this image is an irreducible representation. We have the following actions:

σ_{i}(a_{n+z,i}) = \frac{u + 1}{u}a_{n+z,i+1}, σ_{i}(a_{n+z,i+1}) = a_{n+z,i},

σ_{i}(a_{n+z,j}) = a_{n+z,j} for j ≠ i, i + 1,

σ_{i}(δ_{s_{1},...,s_{m}}) = δ_{s_{1},...,s_{m}} if {s_{1}, …, s_{m}} ∩ {i, i + 1} = ∅,

σ_{i}(δ_{s_{1},...,s_{m}}) = −δ_{s_{1},...,s_{m}} if {s_{1}, …, s_{m}} ∩ {i, i + 1} = {i, i + 1},

σ_{i}(δ_{s_{1},...,s_{j}=i,...,s_{m}}) = δ_{s_{1},...,i+1,...,s_{m}} + a_{i+1,i}δ_{s_{1},...,i,...,s_{m}},

σ_{i}(δ_{s_{1},...,s_{j}=i+1,...,s_{m}}) = δ_{s_{1},...,i,...,s_{m}}.  (8.3)

Recall that a generator wδ_{S} is completely determined by its end set together with w (always chosen in normal form). Choose b ≠ 0, an element in an irreducible subrepresentation of C(u) < wδ_{S} >. We will show that the C(u)-span of the B_{n}-orbit of b contains some wδ_{S}; this will show irreducibility of C(u) < wδ_{S} >. Then by (8.3) we can assume that (some image of) b has a non-trivial monomial of the form wδ_{1,2,...,m}; and then that the span of b, σ_{m}^{2}(b), σ_{m}^{4}(b), … contains a single monomial. This proves the irreducibility. □

Conjecture 8.7. We conjecture that the kernel of J (having dimension \binom{n}{k}\binom{n−1}{k−1}) is also irreducible.

§9 Plücker relations

Here we prove Theorem 1.8. We first describe the Plücker relations. These give the relations between the minors of generic matrices:

Lemma 9.1. [BH Lemma 7.2.3] For every m × n matrix X, m ≤ n, with entries in a commutative ring A and all indices

a_{1}, …, a_{p}, b_{q}, …, b_{m}, c_{1}, …, c_{s} ∈ {1, 2, …, n}

such that s = m − p + q − 1 > m and t = m − p > 0, we have

∑_{\begin{matrix}i_{1}<···<i_{t} \\ i_{t+1}<···<i_{s} \\ \{i_{1},...,i_{s}\}=\{1,...,s\} \end{matrix}} σ(i_{1}, …, i_{s})[a_{1}, …, a_{p}, c_{i_{1}}, …, c_{i_{t}}][c_{i_{t+1}}, …, c_{i_{s}}, b_{q}, …, b_{m}] = 0.

Here if {1, …, s} = {i_{1}, …, i_{s}}, then σ(i_{1}, …, i_{s}) is the sign of the permutation determined by (i_{1}, …, i_{s}) and [x_{1}, …, x_{m}] is the minor of X using columns indexed by x_{1}, …, x_{m}.

Example 9.2. The simplest non-trivial Plücker relation is the following ‘Pfaffian’:

[1, 2][3, 4] − [1, 3][2, 4] + [1, 4][2, 3] = 0.

Now given any set U of Plücker relations we can look at the corresponding elements of (R_{n}/I_{n})[(S|N)′|S ⊂ {1, …, n}, |S| = k]. These are obtained by replacing each [S] in the Plücker relation by (S|N)′. For example for n = 4, S = [1, 2], N = [5, 6] and [1, 2][3, 4] − [1, 3][2, 4] + [1, 4][2, 3] ∈ U (as in 9.2) then the element of R_{4}^{2} would be

(1, 2|5, 6)′(3, 4|5, 6)′ − (1, 3|5, 6)′(2, 4|5, 6)′ + (1, 4|5, 6)′(2, 3|5, 6)′.

The orbit of such elements under the action of B_{n} would then generate a B_{n}-invariant C(u)-submodule < B_{n}(U) >⊂ R_{n}^{2}.

Lemma 9.3. If U is finite, then < B_{n}(U) > is a finitely-generated free C(u)-module.

Proof. We need only consider the case where U = {a} has a single element. But now each monomial in a has the same form and there are only a finite number of monomials of the same form as the monomials in a. The result follows. □

If π is a Plücker relation, then the corresponding ring element will be denotes by (π|N)′. In the next result we will give the action of B_{n} on these Plücker relations.

Lemma 9.4. Let S, T ⊂ {1, 2, …, n} with |S| = |T | = k and let N = N_{k}. Let π be the Plücker relation determined by (S|N)′(T |N)′ and let 1 ≤ r < n. Then

σ_{r}(π|N)′ = −(π|N)′ if r, r + 1 ∈ S ∪ T;
σ_{r}(π|N)′ = t_{r}(π|N)′ − a_{r+1r}(π|N)′ if r ∈ S ∪ T, r + 1 ∉ S ∪ T;
σ_{r}(π|N)′ = t_{r}(π|N)′ if r ∉ S ∪ T, r + 1 ∈ S ∪ T;
σ_{r}(π|N)′ = (π|N)′ if r, r + 1 ∉ S ∪ T.

Here t_{r} ∈ S_{n} is the transposition.

Proof. We consider the action of the generators σ_{r} on the monomial summands of (π|N). Let (S|N)(T |N) represent one of these monomial summands. First from (1.3) we note that

σ_{r}(S|N)′ = (S|N)′ if r, r + 1 ∉ S;
σ_{r}(S|N)′ = −(S|N)′ if r, r + 1 ∈ S;
σ_{r}(S|N)′ = (t_{r}S|N)′ − a_{r+1r}(S|N)′ if r ∈ S, r + 1 ∉ S;
σ_{r}(S|N)′ = (t_{r}S|N)′ if r + 1 ∈ S, r ∉ S.  (9.1)

Now consider the situation where r, r +1 ∈ S ∪T. Then there are four cases: (i) r, r +1 ∈ S; (ii) r, r + 1 ∈ T; (iii) r ∈ S, r + 1 ∈ T; (iv) r ∈ T, r + 1 ∈ S.

For (i) (9.1) shows that σ(S|N)′ = −(S|N)′, σ(T |N)′ = −(T |N)′ and so the result follows. (ii) is similar.

For (iii) (9.1) gives

σ_{r}(S|N)′ = (t_{r}S|N); σ_{r}(T |N)′ = (t_{r}T |N)′ − a_{r+1r}(T |N)′.

Now let U = t_{r}S, V = t_{r}T. Then the monomial summand (U|N)′(V |N)′ also occurs in (π|N)′, only with sign opposite to the sign of (S|N)′(T |N)′. Now we note that

σ_{r}(U|N)′ = (t_{r}U|N)′ and σ_{r}(V |N)′ = (t_{r}V |N)′ − a_{r+1r}(V |N)′

form which it follows that

σ_{r}((S|N)′(T |N)′ − (U|N)′(V |N)′) = −((S|N)′(T |N)′ − (U|N)′(V |N)′).

Thus σ_{r}(π|N)′ = −(π|N)′ as required. Case (iv) is similar (interchange S, T with U, V) and so this proves the first statement in Lemma 9.4. The rest is similar. □

Proposition 9.5. Let n = 2m + 1 > 3 be odd. Then there is an n-dimensional proper subrepresentation of the the Plücker representation which is B_{n}-invariant. This representation is irreducible and monomial.

Proof. For i ≤ m let π_{i} be the Plücker generator corresponding to (S_{i}|N)(T_{i}|N) where S_{i} ∩ T_{i} = ∅, S_{i} ∪ T_{i} = {1, 2, …, n} \ {i}, |S_{i}| = |T_{i}| = m. Thus, for example, when n = 5, we would have π_{5} = (1, 2|5, 6)′(3, 4|5, 6)′ − (1, 3|5, 6)′(2, 4|5, 6)′ + (1, 4|5, 6)′(2, 3|5, 6)′. Now for i ≤ n we let

v_{1} = \frac{1}{u}π_{1} − \frac{1 + u}{u}a_{21}π_{2} + \frac{1 + u}{u}a_{31}π_{3} − …

⋮

v_{i} = a_{1i}π_{1} − a_{2i}π_{2} + ⋯ + (−1)^{i+1} \frac{1}{u}π_{i} − (−1)^{i+1} \frac{1 + u}{u}a_{i+1i}π_{i+1} + …

⋮

v_{n} = a_{1n}π_{1} − a_{2n}π_{2} + ⋯ + (−1)^{n}a_{n−1n}π_{n−1} − (−1)^{n} \frac{1}{u}π_{n}.

Now using Lemma 9.4 one can now check the following actions:

σ_{i}(v_{i}) = −\frac{1 + u}{u}v_{i+1}; σ_{i}(v_{i+1}) = −v_{i}; and

σ_{i}(v_{j}) = −v_{j} for all j ≠ i, i + 1.

It is now clear that we have a representation and that the representation is monomial relative to the basis v_{1}, v_{2}, …, v_{n}. The irreducibility is proved in the same way that the matrices in (7.3) were proved to generate an irreducible representation. □

§10. Algebras with straightening law

Let R be a commutative ring, let A be an R-algebra and Π ⊂ A a finite subset with a partial order ≤. Then A is a graded algebra with straightening law (on Π, over R) (shortened to ASL most of the time) if we have:

(1) A = ⨁_{i≥0} A_{i} is a graded R-algebra such that A_{0} = R, the poset Π consists of homogeneous elements of positive degree which generate A as an R-algebra.

(2) The products ψ_{1} … ψ_{m}, m ∈ ℕ, ψ_{i} ∈ Π, such that ψ_{1} ≤ ψ_{2} ≤ ⋯ ≤ ψ_{m} are linearly independent. They are called the standard monomials.

(3) For all incomparable ψ, ν ∈ Π the product ψν has a representation

ψν = ∑ a_{μ}μ, a_{μ} ∈ R, μ a standard monomial

where the μ’s occuring in the above sum with non-zero coefficients each contain a factor ζ ∈ Π such that ζ ≤ ψ, ζ ≤ ν.

The standard monomials form a basis for A as an R-module. For a proof of this fact and general information about ASL’s see [BV]. ASL’s are called ordinal Hodge algebras in [DEP2].

In this section we will point out various situations where R_{n} is an ASL.

Lemma 10.1. Let n = 3 and (S|T) = (1, 2|4, 5). Putting (i, j) = (i, j|4, 5) then

b_{1} = (2, 3), b_{2} = (1, 3), b_{3} = (1, 2), b_{12} = a_{12}(2, 3), b_{13} = a_{13}(2, 3),
b_{21} = a_{21}(1, 3), b_{23} = a_{23}(1, 3), b_{31} = a_{31}(1, 2), b_{32} = a_{32}(1, 2)

are a basis for R_{3}^{1}(S|T). Further R_{n}(S|T) is an ASL with partial order as follows:

Proof. Standard monomials will be products of the b_{i}, b_{ij} where we do not have b_{ij} and b_{jk} in such a product. Note that for distinct 1 ≤ i, j, k ≤ 3 we see that b_{ij}b_{jk} ∈ b_{ik}b_{j}C(u) and that b_{ij}b_{ji} ∈ b_{i}b_{j}C(u). This shows that every monomial in the b_{i}, b_{jk} is a C(u)-multiple of a standard monomial. We have also thus shown that incompatible products satisfy (3) of the above definition. The rest is easy. □

§11 The R_{n}^{(0)}/I_{n} Representations

In this section we look at the situation where B_{n} acts on R_{n}^{(0)}/I_{n}.

We need to recall the standard generators for P_{n} [Bi p. 20]: For 1 ≤ i < j ≤ n we let

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

The action of the generators of B_{n} and P_{n} on R_{n}^{(0)}/I_{n} is given in

Lemma 11.1. (i) For all 1 ≤ i < n we have

σ_{i}(a_{i i+1}) = a_{i+1 i}, σ_{i}(a_{i+1 i}) = a_{i i+1}, σ_{i}(a_{h i+1}) = a_{h i},

σ_{i}(a_{h i}) = \frac{u + 1}{u}a_{h i+1}, σ_{i}(a_{i+1 j}) = a_{i j},

σ_{i}(a_{i j}) = \frac{u}{u + 1}a_{i+1 j}, σ_{i}(a_{h j}) = a_{h j},

where h, j ≠ i, i + 1.

(ii) For all 1 ≤ i < j ≤ n we have

A_{ij}(a_{ij}) = a_{ij}, A_{ij}(a_{ji}) = a_{ji}, A_{ij}(a_{ih}) = \frac{u}{u + 1}a_{ih}, A_{ij}(a_{jh}) = \frac{u}{u + 1}a_{jh},
A_{ij}(a_{hi}) = \frac{u + 1}{u}a_{hi}, A_{ij}(a_{hj}) = \frac{u + 1}{u}a_{hj}, A_{ij}(a_{rs}) = a_{rs},

for all r, s, h ≠ i, j.

(iii) Let 1 ≤ k < n. Then

(σ_{1}σ_{2} … σ_{k})^{k+1}(a_{ij}) = \frac{u^{k}}{(u + 1)^{k}} a_{ij}, (σ_{1}σ_{2} … σ_{k})^{k+1}(a_{ji}) = \frac{(u + 1)^{k}}{u^{k}} a_{ji}

for all 1 ≤ i ≤ k, j > k.

Proof. (i) follows immediately from (1.2) and the definition of I_{n}. (ii) follows from (i) by induction on j − i ≥ 1. For (iii) we use (ii) and the formula

(σ_{1}σ_{2} … σ_{k})^{k+1} = A_{12} × A_{13}A_{23} × ⋯ × A_{1 k+1}A_{2 k+1} … A_{k k+1}

found in [Bi, p.28] as follows. We note that for fixed 1 ≤ i < j ≤ n there are exactly k + 1 occurrences of the generators A_{rs} in this product where {r, s}∩{i, j} ≠ ∅, one of these being A_{ij}. The action of each such A_{rs} on a_{ij} is given in (ii), with the action of A_{ij} being trivial, and so we get the factor of \frac{u^{k}}{(u+1)^{k}} or its reciprocal. □

Now given any monomial μ ∈ R_{n}^{(0)} we let

I(μ) = {i|a_{ij} divides μ for some j ≤ n}, J(μ) = {i|a_{ji} divides μ for some j ≤ n}.

For I, J ⊆ {1, 2, …, n} we will say that μ has type IJ if I(μ) = I and J(μ) = J. For any b ∈ R_{n}^{(0)} we may write

b = ∑_{I,J⊆{1,2,...,n}} μ_{IJ}

where μ_{IJ} is a sum of C(u)-multiples of monomials of type IJ. Further, given such a monomial μ ∈ R_{n} there are c ∈ C(u) and μ′ ∈ R_{n} such that μ ≡ cμ′ mod I_{n} and where μ′ has type IJ with I ∩J = ∅; for if j ∈ I ∩J, then there are a_{ij} and a_{jk} both dividing μ and we can replace the product a_{ij}a_{jk} in μ by a C(u)-multiple of a_{ik}, thus reducing the degree of the monomial. In this section we will always assume that all monomials in R_{n}^{(0)}/I({1, 2, …, n}) are so represented.

Further, to any such monomial μ ∈ R_{n}^{(0)}/I({1, 2, …, n}) (or any non-trivial C(u)-multiple of μ) we may associate the directed, weighted graph Γ(μ) whose vertices are the numbers 1, 2, …, n and where we have an edge from i to j of weight k if a_{ij}^{k} divides μ (but a_{ij}^{k+1} doesn’t). We note that by Lemma 11.1 for α ∈ B_{n} the graphs Γ(μ) and Γ(α(μ)) are isomorphic as directed, weighted graphs; in fact we have Γ(α(μ)) = Π_{n}(α)Γ(μ) where S_{n} acts on the graphs Γ(μ) in the obvious way.

It is easily seen from Lemma 11.1 that the B_{n}-orbit of a monomial μ consists of C(u)- multiples of monomials μ′ such that Γ(μ) ≅ Γ(μ′) and that, conversely, if μ, μ′ are monomials with Γ(μ) ≅ Γ(μ′), then there is α ∈ B_{n} and c ∈ C(u) with α(μ) = cμ′. Thus the C(u)- module generated by all μ′ with Γ(μ) ≅ Γ(μ′) is B_{n}-invariant. We denote it by C(u)(B_{n}(μ)). Elementary group theory shows that

dim_{C(u)}(C(u)(B_{n}(μ))) = \frac{n!}{|Sym(Γ(μ))|}

where Sym(Γ(μ)) ⊆ S_{n} is the group of all symmetries of the directed, labeled graph Γ(μ).

Now the action of B_{n} on R_{n}^{(0)}/I_{n} thus splits as a sum of irreducible summands of these C(u)(B_{n}(μ)), which we now investigate.

Note that Γ(μ) is a bipartite graph with vertices being either sources or sinks (if a_{ij} divides μ, then the vertex i is a source and j is a sink). Now to each vertex i of Γ(μ) we associate its signed degree (the sum of the weights of the adjacent edges) which we denote by d_{i}(μ) = d_{i}(Γ(μ)). Here we let d_{i}(μ) be positive if the vertex i is a source, and negative otherwise. Note then that by Lemma 11.1 (ii) we have

A_{ij}^{w}(μ) = (\frac{u}{u + 1})^{w(d_{i}(μ)+d_{j}(μ))} μ  (11.1)

for all 1 ≤ i < j ≤ n and w ∈ ℤ.

For a monomial μ ∈ R_{n}/I({1, 2, …, n}) we let

W in(μ) = ∑_{i∈J(μ)} d_{i}(μ), W out(μ) = ∑_{i∈I(μ)} d_{i}(μ),

so that W in(μ) ≤ 0 and W out(μ) ≥ 0.

Lemma 11.2. Let 0 ≠ b ∈ R_{n}/I({1, 2, …, n}). Then there is k ≥ 1 such that in the C(u)-span of the B_{n}-orbit of b there is some b′ which is a sum of monomials all of type {1, 2, …, k}, {k + 1, …, n} and all having the same W in and W out values.

Proof. Choose a monomial μ in b such that (i) W out(μ) is maximal among all monomials of b; and (ii) among all monomials μ satisfying (i) we choose a μ with |J(μ)| smallest. For such a μ let k = |J(μ)|. Let I = I(μ), J = J(μ). Now the fact that B_{n} surjects onto the symmetric group S_{n}, together with Lemma 11.1 shows that there is α ∈ B_{n} such that α(μ) is a C(u)-multiple of a monomial μ′ of type {1, 2, …, k}, {k + 1, …, t} for some t ≤ n.

Now write b′ = α(b) = ∑ b_{I,J,r,s}, where b_{I,J,r,s} is the sum of all cμ′, c ∈ C(u), such that I(μ′) = I, J(μ′) = J, W out(μ′) = r, W in(μ′) = s. Let β = (σ_{1}σ_{2} … σ_{k})^{(k+1)w} and consider the action of β. Lemma 11.1 (iii) shows that b_{I,J,r,s} gets multiplied by \frac{u^{kw}}{(u+1)^{kw}} while every

other μ_{I′,J′,r′,s′} is multiplied by a strictly smaller power of \frac{u^{k}}{(u+1)^{k}}. Thus b_{I,J,r,s} is in the C(u)-span of the B_{n}-orbit of b. □

Now suppose that b is a C(u)-sum of monomials all of type {1, 2, …, k}, {k + 1, …, n}: b = ∑_{m=1}^{r} c_{m}μ_{m} where c_{m} ≠ 0, m = 1, …, r. We will call r = r(b) the length of b. Suppose that 1 ≤ i ≠ j ≤ n and that d_{i}(μ_{s}) + d_{j}(μ_{s}) ≠ d_{i}(μ_{t}) + d_{j}(μ_{t}) for some s ≠ t ≤ r. Then (11.1) above shows that the C(u)-linear span of the elements A_{ij}^{w}(b) contains a non-zero element b′ with r(b′) < r(b). We thus have the first part of

Lemma 11.3. For 0 ≠ b ∈ R_{n}^{(0)}/I_{n} and k as in Lemma 11.2 we see that there is b′ ∈ C(u)(B_{n}(b)) such that if b′ = ∑_{m=1}^{r} c_{m}μ_{m}, where 0 ≠ c_{m} ∈ C(u), then

d_{i}(μ_{s}) + d_{j}(μ_{s}) = d_{i}(μ_{t}) + d_{j}(μ_{t}),

for all s ≠ t ≤ r and for all 1 ≤ i ≠ j ≤ n. In particular, the μ_{m} satisfy I(μ_{s}) = I(μ_{t}) and J(μ_{s}) = J(μ_{t}) for all 1 ≤ s, t ≤ r.

Further, we may suppose that for all α ∈ P_{n} we have α(b′) ∈ C(u)b′.

Lastly, if d_{i}(μ_{s}) = d_{j}(μ_{s}), then d_{i}(μ_{t}) = d_{j}(μ_{t}) for all t.

Proof. We only need to prove the statement in the last paragraph. Since n ≥ 3 we can choose k ≠ i, j and from the above we obtain

d_{i}(μ_{s}) + d_{k}(μ_{s}) = d_{i}(μ_{t}) + d_{k}(μ_{t}) and
d_{j}(μ_{s}) + d_{k}(μ_{s}) = d_{j}(μ_{t}) + d_{k}(μ_{t}).

Thus if d_{i}(μ_{s}) = d_{j}(μ_{s}), then d_{i}(μ_{t}) = d_{j}(μ_{t}) as required. □

Now S_{n} clearly acts on the algebra R_{n}^{(0)} and so on monomials μ ∈ R_{n}^{(0)} and on the graphs Γ(μ). Given any monomial μ we let

Aut(μ) = {α ∈ S_{n}|α(μ) = μ}, DAut(μ) = {α ∈ S_{n}|d_{α(i)}(μ) = d_{i}(μ), for all i = 1, …, n}.

Note that if 0 ≠ b ∈ R_{n}^{(0)}/I_{n}, then Lemma 11.3 shows that there is 0 ≠ b′ ∈ C(u)(B_{n}(b)) with b′ = ∑_{m=1}^{r} c_{m}μ_{m} where I(μ_{m}) = {1, 2, …, k} for every m and J(μ_{s}) = J(μ_{t}) for all s, t. We also have: if d_{i}(μ_{s}) = d_{j}(μ_{s}), then d_{i}(μ_{t}) = d_{j}(μ_{t}) for all t. It follows that for 1 ≤ i ≤ r there is α ∈ DAut(μ_{1}) with α(μ_{1}) = μ_{i}. In particular we see that if r > 1, then DAut(μ_{1}) ≠ Aut(μ_{1}).

We now notice that if C(u)(B_{n}(b)), b = ∑_{m=1}^{r} c_{m}μ_{m}, is to give an irreducible representation of B_{n}, then all the monomials μ_{m} with c_{m} ≠ 0 are isomorphic under the action of S_{n} in the sense that there is some α ∈ S_{n} with α(μ) = μ′. For otherwise we could write b = b_{1} + b_{2} + ⋯ + b_{y}, y > 1, where each of the b_{i} is a C(u)-sum of isomorphic monomials, and this would give a splitting of C(u)(B_{n}(b)). We have:

Lemma 11.4. Let b = ∑_{m=1}^{r} c_{m}μ_{m}, c_{m} ≠ 0. If C(u)(B_{n}(b)) is an irreducible representation of B_{n}, then all the μ_{m} are isomophic monomials. □

Proposition 11.5. Let W be an irreducible subrepresentation of R_{n}^{(0)}/I_{n}. Then W is contained in C(u)(B_{n}(μ)) for some monomial μ ∈ R_{n}.

Let μ ∈ R_{n}^{(0)}/I_{n} be a monomial. Suppose that DAut(μ) = Aut(μ). Then C(u)(B_{n}(μ)) is an irreducible, monomial representation of B_{n} of degree n!/|Aut(μ)|.

Proof. The first statement follows directly from Lemma 11.4.

We note that for any monomial μ the representation C(u)(B_{n}(μ)) is a monomial representation (see (11.1)). Now suppose that there is a subrepresentation W of C(u)(B_{n}(μ)) and that 0 ≠ b ∈ W, b = ∑_{m=1}^{r} c_{m}μ_{m}, where c_{m} ≠ 0. Then by Lemma 11.3 and Lemma 11.4 we may assume that I(μ_{i}) = I(μ_{j}) and J(μ_{i}) = J(μ_{j}) for all i, j, and that if for some i, j, s we have d_{i}(μ_{s}) = d_{j}(μ_{s}), then d_{i}(μ_{t}) = d_{j}(μ_{t}) for all t. Now if r > 1, then there would be an element of DAut(μ) \ Aut(μ), a contradiction. The rest is standard. □

Now for a monomial μ and d ≥ 1 we let I_{d}(μ) be the set of vertices of Γ(μ) having degree d; similarly for J_{d}(μ). Now note that

DAut(μ) = ∑_{d} Sym(I_{d}(μ)) × ∑_{d} Sym(J_{d}(μ)).

Let N ⊂ ℕ be finite. We recall the basis facts about representations of symmetric groups Sym(N) [FH]. Given any tableau T with entries from N we let

P = P_{T} = {g ∈ S_{r}|g preserves each row of T }, Q = Q_{T} = {g ∈ S_{r}|g preserves each column of T };

a_{T} = ∑_{g∈P} g and b_{T} = ∑_{g∈Q} sgn(g)g.

The Young symmetriser is c_{T} = a_{T} b_{T} ∈ ℂSym(N), the fundamental fact being that a complex multiple of c_{T} is an idempotent. Further c_{T} ℂSym(N) is an irreducible representation of Sym(N), where different tableau corresponding to the same Young diagram give isomorphic representations.

Let μ be a monomial such that each of I_{d}(μ), J_{d}(μ) is an interval of distinct positive integers. Let X(μ) be the space generated by all μ′ such that I_{d}(μ′) = I_{d}(μ), J_{d}(μ′) = J_{d}(μ), d = 1, 2, . . .. Then X(μ) is an S_{n}-module. Further, if α ∈ P_{n} and x ∈ X(μ), then (11.1) shows that α(x) = c(α)x for some c(α) ∈ C(u). Thus the representation of P_{n} on X(μ) is 1-dimensional, and so any representation induced from it is monomial [S]. Define the following subgroup of B_{n}:

B_{n}(μ) =< σ_{i}|there is d ≥ 1 with i, i + 1 ∈ I_{d}(μ) or i, i + 1 ∈ J_{d}(μ) > .

By (11.1) we see that X(μ) is invariant under the action of P_{n}. Further, by Lemma 11.1 we see that X(μ) is invariant under the action of B_{n}(μ). Thus X(μ) is invariant under the action of B̄_{n}(μ) =< B_{n}(μ), P_{n} >. But the index of B̄_{n}(μ) in B_{n} is clearly finite and so we get an induced action Ind_{B̄_{n}(μ)}^{B_{n}} [S]. Now note that if Y is an S_{n}-invariant subspace of X(μ) with matrices M(α), α ∈ S_{n}, then Y is also a B̄_{n}(μ)-invariant subspace of W (see Lemma 11.1) and the matrices for the elements α′ ∈ B̄_{n}(μ) have the form M(π_{n}(α′))D(α′) where D(α′) is a diagonal matrix. Thus if Y is irreducible as an S_{n}-module, then Y is irreducible as a B̄_{n}(μ)-module.

Now given any irreducible B_{n}-invariant subspace W of V = C(u)(B_{n}(μ)) Lemmas 11.3 and 11.4 show that there is b′ = ∑_{m} c_{m}μ_{m} ∈ W where all the μ_{m} are isomorphic and where I(μ_{m}) = I(μ_{m′}), J(μ_{m}) = J(μ_{m′}) for all m, m′. We may also assume that each of I_{d}(μ), J_{d}(μ) is an interval of positive integers. Note that the I_{d}(μ), J_{d}(μ) determine a partition of n. Thus X(μ_{1}) ∩ W ≠ {0} and so the action of B_{n} on W is induced by a non-trivial action of B̄_{n}(μ_{1}). Putting this all together gives:

Proposition 11.6. Let μ ∈ R_{n}^{(0)}/I_{n} be a monomial. Then X(μ) is a Π_{n}(B̄_{n}(μ))-module and let X(μ) = ⊕_{i}X_{i} be a decomposition into irreducibles as a Π_{n}(B̄_{n}(μ))-module. Then, there is a corresponding decomposition of the action of B_{n} on V (μ) = C(u)(B_{n}(μ)), as V = ⊕_{i}V_{i} where V_{i} = Ind_{B̄_{n}(μ)}^{B_{n}}. □

Example 11.7. One situation that has received a lot of attention is that, in our context, where Aut(μ) = {id}, so that S_{λ} = Π_{n}(B̄_{n}(μ)) = S_{n_{1}} × S_{n_{2}} × ⋯ × S_{n_{r}}. Recalling that the irreducible representations of S_{n} correspond to Young diagrams, we see that the multiplicities of the irreducible components (corresponding to a Young diagram κ) of the induced representations Ind_{S_{λ}}^{S_{n}} are given by the Kostka numbers K_{λκ}. Here K_{λκ} is defined to be the coefficient of the monomial X^{λ} = x_{1}^{λ_{1}}x_{2}^{λ_{2}} … x_{r}^{λ_{r}} in the Schur polynomial S_{κ} [FH, p. 56]. Explicit bases for these irreducible subspaces are given in [E].

The situation where Aut(μ) ≠ {id} is more complicated.

§12 Rigidity

It is well-known that there is an epimorphism ψ_{n} : P_{n} → P_{n−1} obtained by “pulling out the nth string” [Bi, p. 23]. Composing ψ_{n}, ψ_{n−1}, …, ψ_{4} we see that there is an epimorphism ϕ_{n} : P_{n} → P_{3}. Now using the presentation for P_{3} given in [Bi, p. 20] or [Ha] with generators

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

for 1 ≤ i < j ≤ n we see that

P_{3} =< A_{12}, A_{13}, A_{23} >=< A_{12}, A_{23} > × < A_{23}A_{13}A_{12} >≅ F_{2} × ℤ.

Here z_{3} = A_{23}A_{13}A_{12} generates the centre of B_{3} [Bi, p.28]. This allows the construction of an epimorphism τ_{n} : P_{n} → F_{2} for each n ≥ 3. Now F_{2} has infinitely many irreducible representations in dimension 2; for example we could just choose the degree 2 irreducible representations of the dihedral groups D_{m} =< r, s|r^{m}, s^{2}, (rs)^{2} > of order 2m. These are described in [S]. For simplicity let us only consider the case where m = 2k is even. Let w = e^{2πi/m} and h ∈ ℤ. Then a representation ρ^{h} of D_{m} is defined by

ρ^{h}(r^{k}) = ( \begin{matrix}w^{hk} & 0 \\ 0 & w^{−hk} \end{matrix} ) , ρ^{h}(sr^{k}) = ( \begin{matrix}0 & w^{−hk} \\ w^{hk} & 0 \end{matrix} ) .

Then for 0 < h < m/2 the representation ρ^{h} is irreducible and these account for all such degree 2 irreducible representations of D_{m} [S]. Let W_{mh} be the corresponding representation space.

Now the induced representation Ind_{P_{n}}^{B_{n}}W_{mh} has degree 2 × n! and if the irreducibles so obtained (for varying m and h) were finite in number, then only a finite number of primes would show up in the orders of the matrix groups Ind_{P_{n}}^{B_{n}}W_{mh} so induced. But m is arbitrary and the order of Ind_{P_{n}}^{B_{n}}W_{mh} is clearly divisible by m. Thus we must have infinitely many distinct representations of B_{n}, all of degree at most 2 × n!. This proves the B_{n} case of Theorem 1.9.

Now for the B′_{n} case we note that Gorin and Lin [GL] show that B′_{n} is finitely generated and is perfect for n > 4. We have B_{n}/B′_{n} ≅ ℤ and so B′_{n} consists of those braids having zero exponent sum in the standard generators σ_{i}. Also [B′_{n} : B′_{n} ∩ P_{n}] = n!/2. Using the above we have maps

B′_{n} ∩ P_{n} ↪ P_{n} → F_{2} × ℤ → F_{2}.

Call this composite ζ_{n}. The generator of this central ℤ in F_{2} × ℤ is z_{3} = A_{23}A_{13}A_{12} which has exponent 6 in the standard generators. Thus any word w ∈ F_{2} =< A_{12}, A_{23} > of exponent a multiple of 6 is in the image of ζ_{n}. The set of such words is a subgroup of F_{2} of finite index and so the image of ζ_{n} is a finitely generated free group [MKS]. This now allows the construction of an epimorphism B′_{n} ∩ P_{n} → F_{2} and an argument similar to that used to prove that B_{n} is not rigid now proves the B′_{n} case of Theorem 1.9. □

We now explain the relationship between B_{4} and Aut(F_{2}) described in [DFG] that will allow us to prove Theorem 1.10. Recall Artin’s embedding of B_{n} in Aut(F_{n}). The fact that each α ∈ B_{n} ⊂ Aut(F_{n}) fixes T_{1} … T_{n} implies that there is a representation B_{n} → B_{n}^{∗} < Aut(F_{n−1}). The kernel of this representation is the centre Z(B_{n}) ≅ ℤ [DFG]. The connection between B_{4} and Aut(F_{2}) is now expressed as: B_{4}^{∗}(≅ B_{4}/Z(B_{4})) is isomorphic to a subgroup Aut^{+}(F_{2}) of Aut(F_{2}) of index 2. In [DFG] they use this result to show that B_{4} has a faithful representation over ℂ if and only if Aut(F_{2}) does. To prove Theorem 1.10 it will thus suffice to show that B_{4}/Z(B_{4}) is not rigid.

Now there is an epimorphism β : B_{4} → B_{3} given by its action on the generators σ_{i}:

β(σ_{1}) = σ_{1}, β(σ_{2}) = σ_{2}, β(σ_{3}) = σ_{1}.

We obtain by composition an epimorphism

P_{4} → P_{3} =< A_{12}, A_{13} > × < A_{23}A_{13}A_{12} >→< A_{12}, A_{13} >≅ F_{2}.

Call this α : P_{4} → F_{2} and note that α(A_{23}A_{13}A_{12}) = id. Now the cyclic generator of Z(B_{4}) = Z(P_{4}) is z_{4} = (σ_{1}σ_{2}σ_{3})^{4} = A_{34}A_{24}A_{23}A_{14}A_{13}A_{12} [Bi, p. 28] and so the above gives

β(z_{4}) = β((σ_{1}σ_{2}σ_{3})^{4}) = (σ_{1}σ_{2}σ_{1})^{4} = (z_{3})^{2}.

Thus α(z_{4}) = id, showing that α induces an epimorphism α : P_{4}/Z(P_{4}) → F_{2}. Since [B_{4}^{∗} : P_{4}/Z(P_{4})] = 24 we can now construct infinitely many distinct irreducible degree 2 representations of B_{4}^{∗} = B_{4}/ < z_{4} > as in the B_{n} case. This proves Theorem 1.10. □

Remarks 12.1. 1. The group H(n) of symmetric automorphisms of the free group F_{n} is the subgroup of automorphisms α such that α(x_{i}) is a conjugate of x_{j}, where i ↦ j is a permutation of {1, …, n}. A set of relations for H(n) is given by McCool [Mc]. Let P H(n) be those corresponding to the identity permutation. Then P H(n) is a subgroup of index n! in H(n) and there are epimorphisms P H(n) → P H(n − 1). From the presentation given by McCool one can see that H(3) is an extension of F_{3} by F_{3} and one easily uses ideas similiar to those used in the above to show that P H(n) and H(n) are not rigid for n ≥ 3.

2. In each of Theorem 1.9, 1.10 we obtained our infinite set of irreducible representations in some fixed degree by using the epimorphisms F_{2} → D_{m}. This meant that the degrees were necessarily bounded by a function of n. However we could also use epimorphisms F_{2} → SL_{r}(F_{p}) for any r ≥ 3 and prime p, since SL_{r}(F_{p}) is a 2-generator group. Using these representations the above methods show that there are an infinite number of degrees d such that B_{n}, B′_{n} and Aut(F_{2}) have an infinite number of irreducible representations of degree d.

Acknowledgement. All the calculations in writing this paper were made using MAGMA [MA].

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