ELEMENTARY DIVISORS OF SPECHT MODULES

MATTHIAS KÜNZER AND ANDREW MATHAS

1. Introduction and statement of main results

The irreducible representations of the symmetric groups and their Iwahori-Hecke algebras have been classified and constructed by James [6] and Dipper and James [2], yet simple properties of these modules, such as their dimensions, are still not known. Every irreducible representation of these algebras is constructed by quotienting out the radical of a bilinear form on a particular type of module, known as a Specht module. The bilinear forms on the Specht modules are the objects of our study.

One way of determining the dimension of the simple modules would be to first find the elementary divisors of its Gram matrix over ℤ[q, q^{−1}] and then specialize. This would also give the dimensions of the subquotients of the Jantzen filtrations of the Specht modules over an arbitrary field; see [7]. In general, such an approach is not possible because, as Andersen has shown, Gram matrices need not be diagonalizable over ℤ[q, q^{−1}]; see [1, Remark 5.11]. We also give some examples of non–diagonalizable Specht modules in section 7.

Let G(λ) be the Gram matrix of the Specht module S(λ). Then the first result in this paper shows that G(λ) is diagonalizable if and only if G(λ′) is diagonalizable, where λ′ is the partition conjugate to λ. Moreover, if G(λ) is divisibly diagonalizable (that is, G(λ) is equivalent to a diagonal matrix diag(d_{1}, …, d_{m}) such that d_{i} divides d_{i+1}, for 1 ≤ i < m), then so is G(λ′). In this case we can speak of elementary divisors and we show how the elementary divisors of G(λ) and G(λ′) determine each other. This is a q–analogue of the corresponding result for the symmetric group [8].

We next consider the elementary divisors for the hook partitions. We show that when λ = (n − k, 1^{k}), for 0 ≤ k < n, the Gram matrix G(λ) is always divisibly diagonalizable over ℤ[q, q^{−1}], and we determine the elementary divisors. Again, this is a q–analogue of the corresponding result for the symmetric groups [8], however, the proof in the Hecke algebra case is more involved and requires some interesting combinatorics.

2. The Hecke algebra and permutation modules

Fix a positive integer n and let 𝔖_{n} be the symmetric group of degree n.

Let R be a commutative domain and let q be an invertible element in R.

2000 Mathematics Subject Classification. 20C08, 20G05, 33D80.

Key words and phrases. Specht modules, elementary divisors, hook partitions.

The Iwahori–Hecke algebra of 𝔖_{n} with parameter q is the unital associative algebra ℋ with generators T_{1}, T_{2}, …, T_{n−1} and relations

(T_{i} − q)(T_{i} + 1) = 0  for 1 ≤ i < n,
T_{i}T_{j} = T_{j}T_{i}  for 1 ≤ i < j − 1 < n − 1,
T_{i}T_{i+1}T_{i} = T_{i+1}T_{i}T_{i+1} for 1 ≤ i < n − 1 .

Let r_{i} = (i, i + 1), for i = 1, 2, …, n − 1. Then {r_{1}, r_{2}, …, r_{n−1}} generate 𝔖_{n} (as a Coxeter group). If w ∈ 𝔖_{n} then w = r_{i_{1}} ⋯ r_{i_{k}} for some i_{j} with 1 ≤ i_{j} < n. The word w = r_{i_{1}} … r_{i_{k}} is reduced if k is minimal; in this case we say that w has length k and we define ℓ(w) = k.

If r_{i_{1}} … r_{i_{k}} is reduced then we set T_{w} = T_{i_{1}} ⋯ T_{i_{k}}. Then T_{w} is independent of the choice of reduced expression for w; see, for example, [10, 1.11]. Furthermore, ℋ is free as an R–module with basis { T_{w} | w ∈ 𝔖_{n} }.

A composition μ of n is a sequence of non–negative integers (μ_{1}, μ_{2}, …) that sum to n. If, in addition, μ_{1} ≥ μ_{2} ≥ …, then μ is a partition of n.

Let μ be a composition of n and let 𝔖_{μ} be the associated Young subgroup. Then ℋ (𝔖_{μ}) = ⟨ T_{w} | w ∈ 𝔖_{μ} ⟩ is a subalgebra of ℋ . Given a (right) ℋ (𝔖_{μ})–module V, we define the induced ℋ –module

Ind_{ℋ (𝔖_{μ})}^{ℋ} (V) = V ⊗_{ℋ (𝔖_{μ})} ℋ .

Let D_{μ} = { d ∈ 𝔖_{n} | ℓ(dr_{i}) > ℓ(d) for all r_{i} ∈ 𝔖_{μ} } be the set of distinguished right coset representatives of 𝔖_{μ} in 𝔖_{n}. Then, as an R-module,

Ind_{ℋ (𝔖_{μ})}^{ℋ} (V) ≅ ⨁_{d∈D_{μ}} V ⊗ T_{d}

by [2, Theorem 2.7].

Let x_{μ} = ∑_{w∈𝔖_{μ}} T_{w}. Then T_{w}x_{μ} = x_{μ}T_{w} = q^{ℓ(w)}x_{μ} for all w ∈ 𝔖_{μ}. The trivial representation of ℋ (𝔖_{μ}) is the free R–module 1_{μ} = Rx_{μ}.

Let y_{μ} = ∑_{w∈𝔖_{μ}}(−q)^{−ℓ(w)}T_{w}. Then T_{w}y_{μ} = y_{μ}T_{w} = (−1)^{ℓ(w)}y_{μ} for all w ∈ 𝔖_{μ}. The sign representation of ℋ (𝔖_{μ}) is the free R–module E_{μ} = Ry_{μ}.

For any composition μ we define the permutation module M(μ) = Ind_{ℋ (𝔖_{μ})}^{ℋ} (1_{μ}) ≅ x_{μ}ℋ . Then M(μ) is free as an R–module of rank [𝔖_{n} : 𝔖_{μ}] with basis { x_{μ}T_{d} | d ∈ D_{μ} }. The ℋ –action on M(μ) is determined by

x_{μ}T_{d}T_{i} = \begin{cases}qx_{μ}T_{d}, & if ℓ(dr_{i}) > ℓ(d) and dr_{i} ∉ D_{μ}, \\ x_{μ}T_{dr_{i}}, & if ℓ(dr_{i}) > ℓ(d) and dr_{i} ∈ D_{μ}, \\ qx_{μ}T_{dr_{i}} + (q − 1)x_{μ}T_{d}, & otherwise. \end{cases}

Note that if ℓ(dr_{i}) < ℓ(d) then dr_{i} ∈ D_{μ}.

Let ∗: ℋ ⟶ ℋ be the R–linear map on ℋ determined by T_{w}^{∗} = T_{w^{−1}}, for all w ∈ 𝔖_{n}. This defines an R-algebra anti–automorphism on ℋ of order 2.

The module M(μ) carries a symmetric bilinear form ⟨ , ⟩_{μ} given by

⟨x_{μ}T_{a}, x_{μ}T_{b}⟩_{μ} = \begin{cases}q^{ℓ(a)}, & if a = b, \\ 0, & otherwise, \end{cases}

for a, b ∈ D_{μ}. It follows from the formulae above that the form ⟨ , ⟩_{μ} is associative in the sense that

⟨xh, y⟩_{μ} = ⟨x, yh^{∗}⟩_{μ}

for all x, y ∈ M(μ) and all h ∈ ℋ .

We will need two dualities on the category of right ℋ –modules. Both of them come from involutions on ℋ . The first duality comes from the involution ∗ defined above. The second is induced from the automorphism # : ℋ ⟶ ℋ which is the R–linear map on ℋ determined by T_{w}^{#} = (−q)^{l(w)}T_{w^{−1}}^{−1}, for all w ∈ 𝔖_{n}. It is straightforward to check that # preserves the relations in ℋ and, hence, that it is an R-algebra automorphism of order 2. Note that the involutions # and ∗ commute.

If V is an ℋ -module let V^{∗} be its R-linear dual. Then V^{∗} becomes an ℋ – module by letting (ϕ · ξ)(v) := ϕ(vξ^{∗}), where ϕ ∈ V^{∗}, v ∈ V and ξ ∈ ℋ . With the according operation on morphisms, this defines a contravariant self–equivalence on the category of ℋ -modules.

If V is an ℋ -module let V^{#} the ℋ -module with underlying R-module V and operation v ·_{#} ξ := v ·ξ^{#}, where v ∈ V and ξ ∈ ℋ . With the identical operation on morphisms, this defines a covariant self–equivalence on the category of ℋ -modules.

3. Specht modules

We recall some well-known facts due to Dipper and James [2].

Let λ = (λ_{1}, λ_{2}, …) be a composition of n. The diagram of λ is the set [λ] = { (i, j) ∈ ℕ^{2} | 1 ≤ j ≤ λ_{i} }. We identify the diagram of λ with an array of boxes in the plane. For example, if λ = (4, 3, 2) then

[λ] = .

The conjugate of λ is the partition λ′ = (λ′_{1}, λ′_{2}, …), where λ′_{j} = # { i ≥ 1 | λ_{i} ≥ j } for all j; that is, λ′ is the partition of n whose diagram is obtained by interchanging the rows and columns of the diagram of λ.

Formally, a λ–tableau is a bijection 𝔱: [λ]⟶{1, 2, …, n}; however, we will think of a λ–tableau as a labelling of the diagram of λ by the numbers 1, 2, …, n. Accordingly, we will speak of the rows and columns of a tableau. For example,

1 2 3 4 5 ,  1 3 5 2 4 ,  1 4 5 2 3  and  2 3 4 1 5

are all (3, 2)–tableaux.

A tableau is row standard if in each row its entries increase from left to right. A tableau is standard if it is row standard and in each column its entries increase from top to bottom. Let Std(λ) be the set of standard λ-tableaux.

All of the tableaux above are row standard; however, only the first two tableaux are standard.

The initial λ–tableau 𝔱^{λ} is the standard λ–tableau which has the numbers 1, 2, …, n entered in order from left to right, and then top to bottom, along its rows. The terminal λ–tableau 𝔱_{λ} is the standard λ–tableau which has the numbers 1, 2, …, n entered in order from top to bottom, and then left to right, along its columns. Of the (3, 2)–tableaux above, the first is 𝔱^{(3,2)} and the second is 𝔱_{(3,2)}.

The symmetric group 𝔖_{n} acts from the right on the set of λ–tableaux by permuting their entries. If 𝔱 is a λ–tableau let d(𝔱) be the unique permutation such that 𝔱 = 𝔱^{λ}d(𝔱). In particular, we set w_{λ} = d(𝔱_{λ}).

We remark that D_{μ} = { d(𝔱) | 𝔱 is a row standard μ–tableau } .

Suppose that λ is a partition of n and let z_{λ} = x_{λ}T_{w_{λ}}y_{λ′}. The Specht module is the submodule S(λ) = z_{λ}ℋ of M(λ).

Let S(λ)^{⊥} = { x ∈ M(λ) | ⟨x, y⟩ = 0 for all y ∈ S(λ) }. As ⟨ , ⟩_{λ} is associative, S(λ)^{⊥} is an ℋ -submodule of M(λ). More precisely, S(λ)^{⊥} is the kernel of the ℋ -linear map

M(λ) ^{\overline{}^{δ_{λ}}} S(λ)^{∗} ; x_{λ}h \overline{}⟨x_{λ}h, −⟩_{λ} ,

where h ∈ ℋ .

By restricting the bilinear form ⟨ , ⟩_{λ} on M(λ) we obtain a bilinear form on S(λ). If R is a field then D(λ) = S(λ)/S(λ) ∩ S(λ)^{⊥} is either zero or absolutely irreducible. Moreover, all of the irreducible ℋ –modules arise uniquely in this way [2, Theorem 5.2].

Before we can give a basis of S(λ) we need some more notation. If 𝔱 is a λ– tableau let 𝔱′ be the λ′–tableau obtained by interchanging the rows and columns of 𝔱. For example, (𝔱^{λ})′ = 𝔱_{λ′} and (𝔱_{λ})′ = 𝔱^{λ′}. Finally, if 𝔱 is a standard λ–tableau let v_{𝔱} = z_{λ}T_{d(𝔱′)}.

3.1 (Dipper–James [2, Theorem 5.6]) The Specht module S(λ) is free as an R– module with basis { v_{𝔱} | 𝔱 ∈ Std(λ) }.

We call { v_{𝔱} | 𝔱 ∈ Std(λ) } the Dipper–James basis of S(λ). Let n_{λ} = # Std(λ) be the number of standard λ–tableaux. Then, as an R-module, S(λ) is free of rank n_{λ}.

Fix an ordering of Std(λ) and let

G(λ) = (⟨v_{𝔰}, v_{𝔱}⟩_{λ})_{𝔰,𝔱∈Std(λ)}

be the Gram matrix of the bilinear form ⟨ , ⟩_{λ}, with respect to the Dipper–James basis. The matrix G(λ) depends on the choice of ordering on Std(λ); however, all of the quantities that we are interested in will be independent of this choice. We remark that det G(λ) has been explicitly computed by Dipper and James [3, Theorem 4.11].

4. Diagonalizability and elementary divisors

Given an integer m ≥ 1, an m×m matrix A with coefficients in R is diagonalizable if there exist matrices S and T in GL_{m}(R) such that SAT is a diagonal matrix. The matrix A is divisibly diagonalizable if SAT = diag(d_{1}, …, d_{m}) is a diagonal matrix such that d_{i} divides d_{i+1} in R, for 1 ≤ i < m. If A is divisibly diagonalizable and SAT = diag(d_{1}, …, d_{m}) satisfies this condition, then we call d_{1}, …, d_{m} the elementary divisors of A.

Given A ∈ R^{m×m}, we let I_{k}(A) be the ideal of the k × k minors of A, for 1 ≤ k ≤ m. Note that for B ∈ R^{m×m}, we have I_{k}(AB) ⊆ I_{k}(A) and I_{k}(BA) ⊆ I_{k}(A). Hence for S, T ∈ GL_{m}(R), we have I_{k}(A) = I_{k}(SAT). Therefore, if A is divisibly diagonalizable with resulting diagonal elements d_{1}, …, d_{m}, then I_{k}(A) = I_{k}(diag(d_{1}, …, d_{m})) is the principal ideal generated by d_{1}d_{2} ⋯ d_{k}. This shows that the resulting diagonal entries are independent, up to multiplication by units, of the choice of the diagonalizing matrices. In other words, the elementary divisors of a divisibly diagonalizable matrix are well–defined modulo units.

Whether or not A is diagonalizable, the ideals I_{k}(A) ⊆ R are invariant under the equivalence relation A ∼ SAT. It would be interesting to consider the equivalence classes within { A ∈ R^{m×m} | I_{k}(A) = J_{k} for 1 ≤ k ≤ m } for a fixed tuple (J_{i}) of ideals of R.

If R is a principal ideal domain then every matrix A ∈ R^{m×m} is divisibly diagonalizable by the elementary divisor theorem. The resulting diagonal matrix is known as the Smith normal form.

Now the Laurent polynomial ring ℤ[q, q^{−1}] is not a principal ideal domain and, in fact, there are strict inclusions of the set of divisibly diagonalizable matrices in the set of diagonalizable matrices, and of the set of diagonalizable matrices in all matrices with coefficients in ℤ[q, q^{−1}]. For example, the matrix A = ( \begin{matrix}q−1 & 0 \\ 0 & q+1 \end{matrix} ) is diagonalizable, but not divisibly diagonalizable because I_{1}(A), the ideal of R generated by the entries of A, is not principal.

Proving that a matrix is not diagonalizable is slightly harder. For example, we claim that the matrix B = ( \begin{matrix}q+1 & 2 \\ 0 & q+1 \end{matrix} ) is not diagonalizable over ℤ[q, q^{−1}]. To see this, notice that over ℚ[q, q^{−1}] the matrix B has elementary divisors 1 and (q +1)^{2}. Therefore, if B is diagonalizable over ℤ[q, q^{−1}] then one of these diagonal entries must be a unit in ℚ[q, q^{−1}]; that is, of the form aq^{b} with a, b ∈ ℤ. Reducing modulo 2 this shows that one of the elementary divisors of B over 𝔽_{2}[q, q^{−1}] is zero or a unit. However, this is a contradiction because the elementary divisors of B over 𝔽_{2}[q, q^{−1}] are q + 1 and q + 1.

In proving that certain Gram matrices G(λ) are divisibly diagonalizable over ℤ[q, q^{−1}], we shall make use of the following simple lemma.

4.1. Lemma. Let A be an m × m matrix with coefficients in R, and suppose that there exist invertible matrices S, T ∈ GL_{m}(R) such that

SAT = (\begin{matrix}d_{1} & b_{12} & … & b_{1m} \\ 0 & d_{2} & … & b_{2m} \\ ⋮ & ... & ... & ⋮ \\ 0 & ⋯ & 0 & d_{m} \end{matrix}) ,

where d_{1} | d_{2} | … | d_{m} and d_{i} divides b_{ij} for all j. Then A is divisibly diagonalizable and d_{1}, d_{2}, …, d_{m} are the elementary divisors of A.

Proof. The matrix SAT can be written as the product of diag(d_{1}, …, d_{m}) with a matrix in GL_{m}(R). □

As we saw with the non–diagonalizable matrix ( \begin{matrix}q+1 & 2 \\ 0 & q+1 \end{matrix} ) above, the requirement that d_{i} divides b_{ij} for all j is not superfluous.

5. Elementary divisors for conjugate partitions

Let R = ℤ[q, q^{−1}]. Let λ be a partition of n. In this section we relate the Gram matrices G(λ) and G(λ′). We start with some mild generalizations of some results about Specht modules which were proved by Dipper and James [2] over a field.

Recall that if Y is a submodule of an R–free module X then Y is pure if the quotient module X/Y is R-free.

5.1. Lemma. Suppose that λ is a partition. Then the Specht module S(λ) is a pure submodule of M(λ).

Proof. Using the Dipper–James basis of S(λ), and the basis { x_{λ}T_{d} | d ∈ D_{λ} } of M(λ), suitably ordered, the matrix representing the embedding S(λ) \overline{}M(λ) ℤ[q, q^{−1}]-linearly becomes triangular with 1s on the diagonal [2, Theorem 5.8]. □

5.2. Corollary. The map M(λ) ^{\overline{}^{δ_{λ}}} S(λ)^{∗} is surjective.

Proof. The map δ_{λ} is the composition of the map

M(λ) ^{\overline{}^{∼}} M(λ)^{∗} ; ξ \overline{}⟨ξ, −⟩_{λ}

with the dual of the inclusion map S(λ) \overline{}M(λ). This is surjective by Lemma 5.1. □

Let α(λ) = ∑_{i≥1}(i − 1)λ_{i} = ∑_{i≥1} \binom{λ′_{i}}{2}. Note that α(λ) = ℓ(w_{0,λ′}), where w_{0,λ′} is the unique element of longest length in 𝔖_{λ′}. The next lemma is well known; see, for example, [13, Prop. 2.2]. We include a proof for completeness.

Recall that automorphism #, and the corresponding operation on the module category of ℋ , were defined at the end of section 2.

5.3. Lemma. We have x_{λ}^{#} = q^{α(λ′)}y_{λ} and y_{λ}^{#} = q^{−α(λ′)}x_{λ}.

Proof. As # is an involution the two equalities are equivalent, so we prove only the first. For any integer i, with 1 ≤ i < n, we have x_{λ}^{#}T_{i} = (x_{λ}T_{i}^{#})^{#} = −x_{λ}^{#}. Write x_{λ}^{#} = ∑_{w∈𝔖_{λ}} a_{w}T_{w}, for some a_{w} ∈ ℤ[q, q^{−1}]. Comparing coefficients on both sides of the equation x_{λ}^{#}T_{i} = −x_{λ}^{#} shows that a_{wr_{i}} = (−q)a_{w} for each w that has a reduced expression ending in r_{i}; compare [10, Cor. 1.7]. Hence, x_{λ}^{#} is a scalar multiple of y_{λ}. Then T_{w_{0,λ}}^{#} = (−1)^{ℓ(w_{0,λ})}T_{w_{0,λ}} plus a linear combination of T_{v} where v ∈ 𝔖_{λ} and ℓ(v) < ℓ(w_{0,λ}). Therefore, comparing the coefficient of T_{w_{0,λ}} in x_{λ}^{#} and y_{λ} gives the result. □

Recall that S(λ) = z_{λ}ℋ , where z_{λ} = x_{λ}T_{w_{λ}}y_{λ′}. The importance of z_{λ}, and the irreducibility of S(λ) in the semisimple case, follow from the following simple fact. 5.4 (Dipper–James [2, Lemma 4.1]) Suppose that w ∈ 𝔖_{n}. Then

x_{λ}T_{w}y_{λ′} = \begin{cases}±q^{a}z_{λ}, & if w ∈ 𝔖_{λ}w_{λ}𝔖_{λ′}, \\ 0, & otherwise, \end{cases}

for some integer a.

The proof of this result amounts to the observation that 𝔖_{λ} ∩ w𝔖_{λ′}w^{−1} = {1} if and only if w ∈ 𝔖_{λ}w_{λ}𝔖_{λ′}.

5.5. Lemma (The Submodule Theorem). If U is a pure submodule of M(λ), then S(λ) ⊆ U or U ⊆ S(λ)^{⊥}.

Proof. For all u ∈ U, we have uy_{λ′} = α_{u}z_{λ} for some α_{u} ∈ ℤ[q, q^{−1}] by (5.4).

Case 1: α_{u} = 0 for all u ∈ U. Therefore, if u ∈ U and h ∈ ℋ then we have ⟨u, z_{λ}h⟩_{λ} = ⟨uh^{∗}y_{λ′}, x_{λ}T_{w_{λ}}⟩_{λ}, since y_{λ′}^{∗} = y_{λ′}. But uh^{∗} ∈ U, so uh^{∗}y_{λ′} = 0 and u ∈ S(λ)^{⊥}. Hence, U ⊆ S(λ)^{⊥}.

Case 2: α_{u} ≠ 0 for some u ∈ U. Now U ∋ uy_{λ′} = α_{u}z_{λ} implies z_{λ} ∈ U since U ⊆ M^{λ} is a pure submodule. Therefore, S(λ) ⊆ U. □

Note that the right ideal y_{λ′}T_{w_{λ}}^{−1}x_{λ}ℋ is isomorphic to S(λ)^{#} via ξ \overline{}ξ^{#}. Composing left multiplication by y_{λ′}T_{w_{λ}}^{−1} with this isomorphism, and using Lemma 5.3, we obtain a surjective ℋ -linear map

M(λ) = x_{λ}ℋ ^{\overline{}^{θ_{λ}}} S(λ′)^{#} ; x_{λ}h \overline{}z_{λ′} ·_{#} h ,

where h ∈ ℋ .

5.6. Lemma. We have Kern θ_{λ} = S(λ)^{⊥}. This induces an isomorphism

S(λ′)^{#} \overline{}^{ψ_{λ}}_{∼} S(λ)^{∗} ; z_{λ′} ·_{#} h \overline{}⟨x_{λ}h, −⟩_{λ} ,

where h ∈ ℋ .

Proof. Both Kern θ_{λ} and S(λ)^{⊥} are pure submodules of M(λ). Over ℤ[q, q^{−1}], both S(λ′)^{#} and S(λ)^{∗} are free of rank n_{λ}, so it suffices to prove that Kern θ_{λ} ⊆ S(λ)^{⊥}. By (5.5) this is equivalent to showing that S(λ) ⊈ Kern θ_{λ}. So it is enough to show that z_{λ}θ_{λ} ≠ 0. The bilinear form ⟨ , ⟩_{λ′} is associative, so

⟨z_{λ}θ_{λ}, x_{λ′}⟩_{λ′} = q^{−α(λ)}⟨z_{λ′}T_{w_{λ′}}^{−1} x_{λ′}, x_{λ′}⟩_{λ′}

= q^{−α(λ′)}( ∑_{w∈𝔖_{λ′}} q^{ℓ(w)})⟨z_{λ′}, x_{λ′}T_{w_{λ}}^{−1}⟩_{λ′} .

Now, z_{λ′} = ∑_{v∈𝔖_{λ}}(−q)^{−ℓ(v)}x_{λ′}T_{w_{λ′} v}, where each w_{λ′}v is a distinguished coset representative for 𝔖_{λ′}. In contrast, T_{w_{λ}}^{−1} is equal to T_{w_{λ′}} plus a linear combination of terms T_{u}, where u ∈ 𝔖_{n} with ℓ(u) < ℓ(w_{λ′}). Thus ⟨z_{λ′}, x_{λ′}T_{w_{λ}}^{−1}⟩_{λ′} = ⟨x_{λ′}T_{w_{λ′}} , x_{λ′}T_{w_{λ′}} ⟩_{λ} = q^{ℓ(w_{λ′})}. Hence, ⟨z_{λ}θ_{λ}, x_{λ′}⟩_{λ′} ≠ 0.

A comparison of the short exact sequences

0 \overline{}Kern θ_{λ} \overline{}M(λ) ^{\overline{}^{θ_{λ}}} S(λ′)^{#} \overline{}0

and

0 \overline{}S(λ)^{⊥} \overline{}M(λ) ^{\overline{}^{δ_{λ}}} S(λ)^{∗} \overline{}0

yields the isomorphism ψ_{λ}. □

For each node (i, j) ∈ [λ], we let h_{i,j} = (λ_{i} −j)+(λ′_{j} −i)+1 be the corresponding hook length and set h_{λ}(q) = ∏_{(i,j)∈[λ]}[h_{i,j}]_{q}. The next lemma follows from results of Murphy [11].

5.7. Lemma. We have z_{λ}T_{w_{λ}}^{−1}z_{λ} = q^{n−α(λ)}h_{λ}(q)z_{λ}.

Proof. For the purpose of this proof, we may assume R = ℚ(q). By [11, p. 510– 511], there exists an element Ψ_{𝔱_{λ}}^{∗} = T_{w_{λ′}} +∑_{ℓ(v)<ℓ(w_{λ′})} r_{v}T_{v} ∈ ℋ , for some r_{v} ∈ R, such that

z_{λ}Ψ_{𝔱_{λ}}^{∗} = q^{n−α(λ)+ℓ(w_{λ′})}h_{λ}(q)E_{λ} ,

where E_{λ} is a primitive idempotent such that E_{λ}ℋ = z_{λ}ℋ = S(λ). In particular E_{λ}z_{λ} = z_{λ}. (Note that z_{λ} = z_{λt} in Murphy’s notation; see [11, p. 496, p. 498].)

Note that T_{w_{λ}}^{−1} = q^{−ℓ(w_{λ})}T_{w_{λ′}} + ∑_{ℓ(v)<ℓ(w_{λ′})} r_{v}T_{v}, for some r_{v} ∈ R. Now, if ℓ(v) < ℓ(w_{λ′}) then v ∉ 𝔖_{λ′}w_{λ′}𝔖_{λ}, so y_{λ′}T_{v}x_{λ} = (x_{λ}T_{v}^{∗}y_{λ′})^{∗} = 0 by (5.4). Consequently, z_{λ}T_{v}z_{λ} = 0. Therefore,

z_{λ}T_{w_{λ}}^{−1}z_{λ} = q^{−ℓ(w_{λ})}z_{λ}T_{w_{λ′}} z_{λ}
= q^{−ℓ(w_{λ})}z_{λ}Ψ_{𝔱_{λ}}^{∗}z_{λ}
= q^{n−α(λ)}h_{λ}(q)E_{λ}z_{λ}
= q^{n−α(λ)}h_{λ}(q)z_{λ} .
□

Consider the ℋ -linear map

S(λ) ^{\overline{}^{γ_{λ}}} S(λ)^{∗} ; ξ \overline{}⟨ξ, −⟩_{λ} .

5.8. Lemma. The composition

S(λ) ^{\overline{}^{γ_{λ}}} S(λ)^{∗} \overline{}^{ψ_{λ}^{−1}}_{∼} S(λ′)^{#} ^{\overline{}^{γ_{λ′}^{#}}} S(λ′)^{∗,#} \overline{}^{(ψ_{λ′}^{#} )^{−1}}_{∼} S(λ)

is equal to scalar multiplication by (−q)^{ℓ(w_{λ′})}q^{n−α(λ)−α(λ′)}h_{λ}(q).

Proof. The element z_{λ} is mapped via γ_{λ} to ⟨z_{λ}, −⟩_{λ}, which is mapped via ψ_{λ}^{−1} to z_{λ′} ·_{#} T_{w_{λ}}y_{λ′} = z_{λ′}T_{w_{λ}}^{#}y_{λ′}^{#}, which in turn goes to ⟨z_{λ′}T_{w_{λ}}^{#}y_{λ′}^{#}, −⟩_{λ} via γ_{λ′}^{#}, and finally to

\begin{aligned}z_{λ} ·_{#} T_{w_{λ′}} y_{λ}T_{w_{λ}}^{#}y_{λ′}^{#} & = & (−q)^{ℓ(w_{λ′})}q^{−α(λ′)}z_{λ}T_{w_{λ}}^{−1}z_{λ} \\ & = & (−q)^{ℓ(w_{λ′})}q^{n−α(λ)−α(λ′)}h_{λ}(q)z_{λ} \end{aligned}

via (ψ_{λ′}^{#})^{−1}, by Lemma 5.7. □

Let I_{m} be the m×m identity matrix. Recall that n_{λ} = # Std(λ) is the dimension of the Specht module S(λ).

5.9. Proposition. Suppose that λ is a partition of n.

(1) There exist invertible matrices A, B ∈ GL_{n_{λ}}(ℤ[q, q^{−1}]) such that

G(λ) · A · G(λ′) · B = h_{λ}(q) · I_{n_{λ}} .

(2) G(λ) is diagonalizable to the diagonal matrix D if and only if G(λ′) is diagonalizable to the diagonal matrix h_{λ}(q)D^{−1}.

(3) G(λ) is divisibly diagonalizable if and only if G(λ′) is divisibly diagonalizable. In this case, the product of the i^{th} elementary divisor of G(λ) and the (n_{λ} + 1 − i)^{th} elementary divisor of G(λ′) is equal to h_{λ}(q).

Recall that elementary divisors are only well defined up to a unit in ℤ[q, q^{−1}]; the same is true of their product in (3).

Proof. (1) The R-linear map γ_{λ} is represented by the matrix G(λ) with respect to the Dipper–James basis and its dual basis. Thus the assertion follows by (5.8).

(2) If G(λ) = SDT with S, T ∈ GL_{n_{λ}}(ℤ[q, q^{−1}]) and D ∈ ℤ[q, q^{−1}]^{n_{λ}×n_{λ}} is a diagonal matrix, then G(λ′) = A^{−1}T^{−1}(h_{λ}(q)D^{−1})S^{−1}B^{−1}. Since G(λ′) has coefficients in ℤ[q, q^{−1}], so does h_{λ}(q)D^{−1}.

(3) Repeat the argument of (2). □

We remark that all of the results in this section hold more generally when the Hecke algebra ℋ is defined over an integral domain R such that ℋ ⊗_{R} Q is semisimple, where Q is the field of fractions of R. (We need semisimplicity over Q only when we apply Murphy’s results in the proof of Lemma 5.7.) In particular, Proposition 5.9 holds when R = F [q, q^{−1}] and F is any field. Notice that G(λ) is always diagonalizable in this case because F [q, q^{−1}] is a principal ideal domain.

6. The elementary divisors for hook partitions

Throughout this section we fix an integer k, with 0 ≤ k < n, and consider the Specht module S(λ), where λ = (n − k, 1^{k}). We will show that G(λ) is divisibly diagonalizable by explicitly constructing two bases of S(λ) which transform G(λ) into an upper triangular matrix satisfying the requirements of Lemma 4.1. In particular, this will allow us to determine the elementary divisors of S(λ).

The Specht module S(λ) is defined as a submodule of the permutation module M(λ); however, to compute the elementary divisors we will work inside a different permutation module.

By definition, S(λ) = x_{λ}T_{w_{λ}}y_{λ′}ℋ = x_{(n−k,1^{k})}T_{w_{(n−k,1^{k})}}y_{(k+1,1^{n−k−1})}. We first need to understand the permutation w_{λ} = w_{(n−k,1^{k})} a little better. This requires some new notation. For integers non–negative i and j define

r_{i,j} = \begin{cases}1, & if i = 0 or j = 0, \\ r_{i}r_{i+1} … r_{j}, & if 0 < i ≤ j, \\ r_{i}r_{i−1} … r_{j}, & if i > j > 0, \end{cases}

and set T_{i,j} = T_{r_{i,j}}. Next, let a and b be non–negative integers such a + b ≤ n. If either a = 0 or b = 0 then set w_{a,b} = 1. If both a and b are non–zero then define w_{a,b} = (r_{a+b−1,1})^{b}; then one can check that, in two–line notation,

w_{a,b} = ( \begin{matrix}1 & 2 & … & a & a + 1 & a + 2 & … & a + b \\ b + 1 & b + 2 & … & a + b & 1 & 2 & … & b \end{matrix} ).

It is not hard to see that w_{b,a} = w_{a,b}^{−1} and that w_{a,b} = r_{a,a+b−1}w_{a−1,b} and ℓ(w_{a,b}) = ℓ(r_{a,a+b−1}) + ℓ(w_{a−1,b}); see [4]. Consequently,

w_{a,b} = r_{a,a+b−1}r_{a−1,a+b−2} … r_{1,b} = r_{1,a}r_{2,a+2} … r_{b,a+b−1}

with the lengths adding in both cases. Hence, ℓ(w_{a,b}) = ab.

The permutation w_{(n−k,1^{k})} is essentially one of these permutations because

w_{(n−k,1^{k})} = ( \begin{matrix}1 & 2 & ⋯ & n − k & n − k + 1 & ⋯ & n \\ 1 & k + 2 & ⋯ & n & 2 & ⋯ & k + 1 \end{matrix} ).

Hence, w_{(n−k,1^{k})} = r_{n−k,n−1}r_{n−k−1,n−2} … r_{2,k+1}, with the lengths adding. So

T_{w_{(n−k,1^{k})}} = T_{n−k,n−1} … T_{2,k+1},

Notice also that T_{w_{n−k.k}} = T_{w_{(n−k,1^{k})}}T_{1,k}.

If w ∈ 𝔖_{(k,n−k)} ≅ 𝔖_{k} × 𝔖_{n−k} then we write w = (u, v), where u ∈ 𝔖_{k} and v ∈ 𝔖_{(1^{k},n−k)} are the unique permutations such that w = uv = vu. Set

x_{(k|n−k)} = y_{(k,1^{n−k})}x_{(1^{k},n−k)} = ∑_{(u,v)∈𝔖_{(k,n−k)}}(−q)^{−ℓ(u)}T_{uv}.

Then it is easy to see that Rx_{(k|n−k)} is an ℋ (𝔖_{μ})–module on which the subalgebras ℋ (𝔖_{k}) and ℋ (𝔖_{(1^{k},n−k)}) act via their sign and trivial representations, respectively. Let

M(k|n−k) = Ind_{ℋ (𝔖_{(k,n−k)})}^{ℋ (𝔖_{n})} (Rx_{(k|n−k)}) ≅ x_{(k|n−k)}ℋ .

As in section 2, the induced module M(k|n−k) is free as an R–module with basis { x_{(k|n−k)}T_{d} | d ∈ D_{(k,n−k)} }. Furthermore, M(k|n−k) possesses a natural non– degenerate associative bilinear form ⟨ , ⟩_{(k|n−k)} which is determined by

⟨x_{(k|n−k)}T_{u}, x_{(k|n−k)}T_{v}⟩_{(k|n−k)} = \begin{cases}q^{ℓ(u)}, & if u = v, \\ 0, & otherwise, \end{cases}

for u, v ∈ D_{(k,n−k)}. Donkin [5] calls M(k|n−k) a trivial source module.

Let y′_{k+1} = 1 + ∑_{j=1}^{k}(−q)^{j−k−1}T_{k,j} = 1 − q^{−1}T_{k} + q^{−2}T_{k,k−1} + ⋯ + (−q)^{−k}T_{k,1}. This is a sum over the right coset representatives of 𝔖_{k} in 𝔖_{k+1}. Consequently, it follows that y_{(k+1,1^{n−k−1})} = y_{(k,1^{n−k})}y′_{k+1}. The reason for introducing the module M(k|n−k) is the following result.

Given a non–negative integer k > 1 let [k]_{q} = 1 + q + ⋯ + q^{k−1} and [k]_{q}^{!} = [1]_{q}[2]_{q} ⋯ [k]_{q}. Notice that if q = 1 then [k]_{1} = k and [k]_{1}^{!} = k!.

6.1. Proposition. Let λ = (n − k, 1^{k}). The map

π_{k} : S(λ) ⟶ M(k|n−k) ; z_{λ}h ⟼ x_{(k|n−k)}y′_{k+1}h

is an injective ℋ –module homomorphism. Moreover,

⟨x, y⟩_{λ} = q^{\frac{k}{2} (2n−3k−1)}[k]_{q}^{!}⟨π(x), π(y)⟩_{(k|n−k)},

for all x, y ∈ S(λ).

Proof. By definition, S(λ) = x_{(n−k,1^{k})}T_{w_{λ}}y_{(k+1,1^{n−k−1})}ℋ . As remarked above, w_{n−k,k} = w_{(n−k,1^{k})}r_{1,k} with the lengths adding. Therefore, since r_{1,k} ∈ 𝔖_{λ′},

x_{(n−k,1^{k})}T_{w_{(n−k,1^{k})}}y_{(k+1,1^{n−k−1})} = (−1)^{k}x_{(n−k,1^{k})}T_{w_{(n−k,1^{k})}}T_{1,k}y_{(k+1,1^{n−k−1})}
= (−1)^{k}x_{(n−k,1^{k})}T_{w_{n−k,k}}y_{(k+1,1^{n−k−1})}
= (−1)^{k}T_{w_{n−k,k}}x_{(1^{k},n−k)}y_{(k+1,1^{n−k−1})}
= (−1)^{k}T_{w_{n−k,k}}x_{(1^{k},n−k)}y_{(k,1^{n−k})}y′_{k+1}
= (−1)^{k}T_{w_{n−k,k}}x_{(k|n−k)}y′_{k+1}.

Therefore, π(x) = (−1)^{k}T_{w_{n−k,k}}^{−1}x, for all x ∈ S(λ). As T_{w_{n−k,k}} is invertible, the first claim now follows.

To prove the second claim we first suppose that R = ℤ[q, q^{−1}]. If x, y ∈ S(λ) then, by extending scalars, we may assume that x and y are elements of S(λ)_{ℚ(q)} = S(λ)_{ℤ[q,q^{−1}]} ⊗ ℚ(q). Now S(λ)_{ℚ(q)} ≅ π(S(λ)_{ℚ(q)}) is irreducible so, up to a scalar, there is a unique associative bilinear form on S(λ)_{ℚ(q)}. To determine this scalar it is enough to compare the two inner products on z_{λ} and π(z_{λ}). Using associativity,

⟨z_{λ}, z_{λ}⟩_{λ} = ⟨x_{λ}T_{w_{λ}}y_{λ′}, x_{λ}T_{w_{λ}}y_{λ′}⟩_{λ} = ⟨x_{λ}T_{w_{λ}}y_{λ′}^{2}, x_{λ}T_{w_{λ}}⟩_{λ}

= q^{−\binom{k+1}{2}}[k + 1]_{q}^{!}⟨x_{λ}T_{w_{λ}}y_{λ′}, x_{λ}T_{w_{λ}}⟩_{λ}

= q^{−\binom{k+1}{2}}[k + 1]_{q}^{!} ∑_{v∈𝔖_{λ′}}(−q)^{−ℓ(v)}⟨x_{λ}T_{w_{λ}v}, x_{λ}T_{w_{λ}}⟩_{λ}

= q^{ℓ(w_{λ})−\binom{k+1}{2}}[k + 1]_{q}^{!}

= q^{\frac{k}{2} (2n−3k−3)}[k + 1]_{q}^{!}.

Similarly,

⟨π(z_{λ}), π(z_{λ})⟩_{(k|n−k)} = ⟨x_{(k|n−k)}y′_{k+1}, x_{(k|n−k)}y′_{k+1}⟩_{(k|n−k)} = q^{−k}[k + 1]_{q}

This proves that ⟨x, y⟩_{λ} = q^{\frac{k}{2} (2n−3k−1)}[k]_{q}^{!}⟨π(x), π(y)⟩_{(k|n−k)}, for all x, y ∈ S(λ) when R = ℤ[q, q^{−1}]. The general case now follows by specialization. □

6.2. Corollary. Suppose that λ = (n − k, 1^{k}). Then [k]_{q}^{!} divides ⟨x, y⟩_{λ}, for all x, y ∈ S(λ).

Let S′(λ) = π(S(λ)) = x_{(k|n−k)}y′_{k+1}ℋ . Then S′(λ) ≅ S(λ) by the Proposition. We will work with S′(λ) in what follows rather than working with S(λ) directly.

As a first step we need a basis of S′(λ). For any λ-tableau 𝔱 define

v′_{𝔱} = π(v_{𝔱}) = x_{(k|n−k)}y′_{k+1}T_{d(𝔱′)}.

The Dipper–James basis of S(λ), (3.1), combined with Proposition 6.1, give us the following.

6.3. Corollary. The module S′(λ) is R–free with basis { v′_{𝔱} | 𝔱 ∈ Std(λ) }.

In order to exploit this basis we introduce another type of tableaux. For our purposes we could get by using (k, n − k)–tableaux; however, we use the notation from the theory of trivial source modules.

The diagram of (k|n−k) is the ordered pair of diagrams [k|n − k] = ([k], [n − k]). A (k|n−k)–tableau is a bijection from [k|n − k] to {1, 2, …, n}. Once again, we will think of a (k|n−k)–tableau as being a labelling of [k|n − k]. Accordingly, we will write a (k|n−k)–tableau as an ordered pair (𝔞|𝔟), where 𝔞 and 𝔟 are suitable labellings of the diagrams of the partitions (k) and (n − k) respectively. We refer to 𝔞 and 𝔟 as the first and second components of (𝔞|𝔟).

A (k|n−k)–tableau (𝔞|𝔟) is (row) standard if the entries in 𝔞 increase from left to right and the entries in 𝔟 increase from left to right. Let Std(k|n − k) be the set of standard (k|n−k)–tableaux. For example, the standard (1|3)–tableaux are

( 1 | 2 3 4 ), ( 2 | 1 3 4 ), ( 3 | 1 2 4 ) and ( 4 | 1 2 3 ).

Let 𝔱^{(k|n−k)} be the standard (k|n−k)–tableau with 1, …, k entered in order, from left to right, in the first omponent and the numbers k + 1, …, n in the second. The first of the tableaux above is 𝔱^{(1|3)}.

Two (k|n − k)–tableaux (𝔞|𝔟) and (𝔰|𝔱) are row equivalent if 𝔞 and 𝔰 contain the same entries up to reordering (in which case, 𝔟 and 𝔱 also contain the same set of entries). As with ordinary tableaux, the symmetric group acts from the right on the set of (k|n−k)–tableaux. If (𝔞|𝔟) is a (k|n−k)–tableau we define d(𝔞|𝔟) to be the unique permutation such that (𝔞|𝔟) = 𝔱^{(k|n−k)}d(𝔞|𝔟). Then (𝔞|𝔟) and (𝔰|𝔱) are row equivalent if and only if d(𝔞|𝔟) = wd(𝔰|𝔱) for some w ∈ 𝔖_{(k,n−k)}. Consequently,

D_{(k,n−k)} = { d(𝔞|𝔟) | (𝔞|𝔟) ∈ Std(k|n−k)}.

So the standard (k|n − k)–tableaux index a basis of M(k|n−k).

For furture reference, notice that if (𝔞|𝔟) is a standard (k|n − k)–tableau then ℓ(d(𝔞|𝔟)) is equal to the number of pairs of integers (i, j) where i appears in 𝔞, j appears in 𝔟 and i > j. This follows because if w ∈ 𝔖_{n} then ℓ(w) is equal to the number of pairs a < b with i = a^{w} > j = b^{w}, and the entries in 𝔞 are the images of 1, …, k under d(𝔞|𝔟), whereas the entries in 𝔟 are the images of k + 1, …, n.

For any (k|n−k)–tableau (𝔞|𝔟) define x_{(𝔞|𝔟)} = x_{(k|n−k)}T_{d(𝔞|𝔟)}. Here we do not assume that (𝔞|𝔟) is standard. The following lemma is easily verified.

6.4. Lemma. Suppose that 0 ≤ k < n.

(i) M(k|n−k) is free as an R–module with basis { x_{(𝔞|𝔟)} | (𝔞|𝔟) ∈ Std(k|n−k) }. (ii) Suppose that (𝔞|𝔟) ∈ Std(k|n−k) and 1 ≤ i < n. Then

x_{(𝔞|𝔟)}T_{i} = \begin{cases}−x_{(𝔞|𝔟)}, & if i and i + 1 are both contained in 𝔞, \\ qx_{(𝔞|𝔟)}, & if i and i + 1 are both contained in 𝔟, \\ x_{(𝔞_{i}|𝔟_{i})}, & if i is in 𝔞 and i + 1 is in 𝔟, \\ qx_{(𝔞_{i}|𝔟_{i})} + (q − 1)x_{(𝔞|𝔟)}, & otherwise, \end{cases}

where (𝔞_{i}|𝔟_{i}) = (𝔞|𝔟)r_{i}.

The action of ℋ on M(k|n − k) is completely determined by (ii).

We now show how to write the basis {v′_{𝔱}} of S′(λ) in terms of this basis of M(k|n − k). To do this, if 𝔱 is a λ–tableau and (𝔞|𝔟) is a (k|n−k)–tableau write (𝔞|𝔟) ≺ 𝔱 if (𝔞|𝔟) is standard and all of the entries in 𝔞 are contained in the first column of 𝔱. Finally, if (𝔞|𝔟) ≺ 𝔱 we set I_{𝔱}(𝔞|𝔟) = i, the index of (𝔞|𝔟) in 𝔱, if the number in row i of 𝔱 does not appear in 𝔞.

6.5. Lemma. Suppose that 𝔱 is a standard λ–tableau. Then

v′_{𝔱} = ∑_{(𝔞|𝔟)≺𝔱}(−1)^{k+1−I_{𝔱}(𝔞|𝔟)}q^{ℓ(d(𝔱′))−ℓ(d(𝔞|𝔟))}x_{(𝔞|𝔟)}.

Proof. First consider v_{𝔱_{λ}}. Looking at the definitions we see that

v′_{𝔱_{λ}} = x_{(k|n−k)}y′_{k+1} = x_{𝔱^{(k|n−k)}}(1 − q^{−1}T_{k} + q^{−2}T_{k,k−1} − ⋯ + (−q)^{−k}T_{k,1})

= ∑_{(𝔞|𝔟)≺𝔱_{λ}}(−q)^{−ℓ(d(𝔞|𝔟))}x_{(𝔞|𝔟)}.

As ℓ(d(𝔱_{λ})′) = ℓ(d(𝔱^{λ′})) = 0 and ℓ(d(𝔞|𝔟)) = k + 1 − I_{𝔱_{λ}}(𝔞|𝔟), when (𝔞|𝔟) ≺ 𝔱_{λ}, the Lemma follows in this case.

Now suppose that 𝔱 is an arbitrary standard λ–tableaux. If 𝔱 ≠ 𝔱_{λ} then we can find another standard λ–tableau 𝔰 and an integer i in the first column of 𝔰 such that 𝔱 = 𝔰r_{i} and ℓ(d(𝔱)) = ℓ(d(𝔰)) − 1. (That is, 𝔱 ⊵ 𝔰 where ⊵ is the dominance order on tableaux; see, for example, [10].) Therefore, by induction,

v′_{𝔱} = v′_{𝔰}T_{i} = ∑_{(𝔞|𝔟)≺𝔰}(−1)^{k+1−I_{𝔰}(𝔞|𝔟)}q^{ℓ(d(𝔰′))−ℓ(d(𝔞|𝔟))}x_{(𝔞|𝔟)}T_{i}.

Since 𝔰 and 𝔱 are standard, i is in the first column of 𝔰 and the first row of 𝔱 and i+1 is in the first row of 𝔰 and the first column of 𝔱. Therefore, if (𝔞|𝔟) ≺ 𝔰 then the entries in the first component of (𝔞|𝔟)r_{i} are still in increasing order and the entries in the second component are in increasing order unless i and i+1 both appear in 𝔟. So, ℓ(d(𝔞|𝔟)r_{i}) = ℓ(d(𝔞|𝔟)) + 1 and by Lemma 6.4(ii) we have

x_{(𝔞|𝔟)}T_{i} = \begin{cases}qx_{(𝔞|𝔟)}, & if i and i + 1 both appear in 𝔟, \\ x_{(𝔞|𝔟)r_{i}}, & otherwise. \end{cases}

In the first case, when i and i + 1 both appear in 𝔟, we have that (𝔞|𝔟) ≺ 𝔱. Also, ℓ(d(𝔱′)) − ℓ(d(𝔞|𝔟)) = ℓ(d(𝔰′)) − ℓ(d(𝔞|𝔟)) + 1 and I_{𝔱}(𝔞|𝔟) = I_{𝔰}(𝔞|𝔟), so x_{(𝔞|𝔟)} has the required coefficient in v′_{𝔱}.

In the second case, i appears in 𝔞 and i+1 appears in 𝔟, so (𝔞r_{i}|𝔟r_{i}) = (𝔞|𝔟)r_{i} ≺ 𝔱, I_{𝔱}(𝔞r_{i}|𝔟r_{i}) = I_{𝔰}(𝔞|𝔟) and ℓ(d(𝔱′)) − ℓ(d(𝔞|𝔟)r_{i}) = ℓ(d(𝔰′)) − ℓ(d(𝔞|𝔟)). Hence, once again, x_{(𝔞|𝔟)r_{i}} has the predicted coefficient in v′_{𝔱}.

As there are exactly k standard (k|n − k)–tableaux (𝔞|𝔟) satisfying (𝔞|𝔟) ≺ 𝔱, this completes the proof. □

In order to compute the elementary divisors of S(λ) we need a second basis of S′(λ). Let

x_{n−k} = 1 + T_{1} + ⋯ + T_{1,n−k−1} = ∑_{j=0}^{n−k−1} T_{1,j}.

(Note that r_{1,0} = 1.) As with y′_{k+1}, we have x_{(n−k,1^{k})} = x_{(1,n−k−1,1^{k−1})}x_{n−k}. Now, for any standard (n − k, 1^{k})–tableau 𝔱 we define

w′_{𝔱} = \begin{cases}v′_{𝔱(1,n)}, & if n appears in row 1 of 𝔱, \\ v′_{𝔱^{λ}}x_{n−k}T_{d(𝔱)}, & otherwise. \end{cases}

We remark that it is not obvious that the set of elements { w′_{𝔱} | 𝔱 ∈ Std(λ) } is a basis of S′(λ). We will prove this below.

Lemma 6.5 gives an explicit description of the basis {v′_{𝔱}}. We need to do the same for the basis {w′_{𝔱}}, and for this we need some more notation. If 𝔱 is a standard λ–tableau let 𝔱^{∗} = 𝔱(1, n). If (𝔞|𝔟) is a (k|n−k)–tableau write (𝔞|𝔟) ≺_{n} 𝔱 if (𝔞|𝔟) ≺ 𝔱 and n is contained in 𝔞. Finally, if 1 appears in the first row of 𝔱 then we define (𝔞_{𝔱}^{∗}|𝔟_{𝔱}^{∗}) to be the unique standard (k|n − k)–tableau such that (𝔞_{𝔱}^{∗}|𝔟_{𝔱}^{∗}) ≺ 𝔱^{∗} and n appears in 𝔟_{𝔱}^{∗}. So (𝔞_{𝔱}^{∗}|𝔟_{𝔱}^{∗}) ≺ 𝔱^{∗} and (𝔞_{𝔱}^{∗}|𝔟_{𝔱}^{∗}) ≺ 𝔱.

6.6. Lemma. Suppose that 𝔱 is a standard λ–tableau and that n appears in the first row of 𝔱. Then

w′_{𝔱} = (−1)^{k}q^{2n−2k−3}x_{(𝔞_{𝔱}^{∗}|𝔟_{𝔱}^{∗})} + ∑_{(𝔞|𝔟)≺_{n}𝔱^{∗}} r_{𝔞𝔟}x_{(𝔞|𝔟)},

for some scalars r_{𝔞𝔟} ∈ ℤ[q, q^{−1}].

Proof. We now argue by downwards induction on 𝔱 beginning with 𝔱 = 𝔱_{λ}, this is an unpleasant calculation. Now,

w′_{𝔱_{λ}} = x_{(k|n−k)}y′_{k+1}T_{1,n−1}T_{n−2,1} = (−1)^{k}x_{(k|n−k)}y′_{k+1}T_{k+1,n−1}T_{n−2,1}

since x_{(k|n−k)}y′_{k+1} = x_{(1^{k},n−k)}y_{(k+1,1^{n−k−1})}. Therefore, using the definitions together with the braid relations,

w′_{𝔱_{λ}} = (−1)^{k}x_{(k|n−k)}y′_{k+1}T_{n−1} … T_{k+2}T_{k+1,n−1}T_{k,1}
= (−1)^{k}x_{(k|n−k)}T_{n−1} … T_{k+2}y′_{k+1}T_{k+1,n−1}T_{k,1}
= (−1)^{k}q^{n−k−2}x_{(k|n−k)}y′_{k+1}T_{k+1,n−1}T_{k,1}

= (−1)^{k}q^{n−k−2}x_{(k|n−k)}{1 + ∑_{j=1}^{k}(−q)^{j−k−1}T_{k,j}}T_{k+1,n−1}T_{k,1}

= (−1)^{k}q^{n−k−2}x_{(k|n−k)}{q^{n−k−1} + ∑_{j=1}^{k}(−q)^{j−k−1}T_{k,j}T_{k+1,n−1}}T_{k,1}

= (−1)^{k}q^{n−2k−3}x_{(k|n−k)}{q^{n}T_{k,1} + ∑_{j=1}^{k}(−q)^{j}T_{k,j}T_{k+1,1}T_{k+2,n−1}}

= (−1)^{k}q^{n−2k−3}x_{(k|n−k)}{q^{n}T_{k,1} + ∑_{j=1}^{k}(−q)^{j}T_{k+1,1}T_{k+1,j+1}T_{k+2,n−1}}

= (−1)^{k}q^{n−2k−3}x_{(k|n−k)}{q^{n}T_{k,1} − ∑_{j=1}^{k}(−q)^{j+1}T_{k,1}T_{k+1,j+1}T_{k+2,n−1}}

= (−1)^{k}q^{n−2k−3}x_{(k|n−k)}T_{k,1}{q^{n} − ∑_{j=2}^{k+1}(−q)^{j}T_{k+1,j}T_{k+2,n−1}}.

Now, 𝔱^{(k|n−k)}r_{k,1} = ( 2 ⋯ k + 1 | 1 k + 2 ⋯ n ) = (𝔞_{𝔱_{λ}}^{∗}|𝔟_{𝔱_{λ}}^{∗}) and, consequently,

𝔱^{(k|n−k)}r_{k,1}r_{k+1,j}r_{k+2,n−1} = ( 2 ⋯ j − 1 j + 1 ⋯ k + 1 n | 1 j k + 2 ⋯ n − 1 ), for

j = 2, …, k + 1. This completes the proof for w′_{𝔱_{λ}}.

Now suppose that 𝔱 is an arbitrary standard λ–tableau which has n in its first row. Then d(𝔱′) ∈ 𝔖_{(1,n−2,1)} so d(𝔱′) and (1, n) commute and ℓ(d(𝔱′)(1, n)) = ℓ(d(𝔱′)) + ℓ(1, n). Therefore, w′_{𝔱} = w′_{t_{λ}}T_{d(𝔱′)}. To complete the proof now argue by induction, as in the proof of Lemma 6.5; we leave the details to the reader. (Indeed, this shows that r_{𝔞𝔟} = ±q^{a} for some integer a.) □

For convenience we now write ⟨ , ⟩ = ⟨ , ⟩_{(k|n−k)}. In terms of the standard basis of M(k|n − k), the bilinear form ⟨ , ⟩ on M(k|n − k) is determined by

⟨x_{(𝔞|𝔟)}, x_{(𝔰|𝔱)}⟩ = \begin{cases}q^{ℓ(d(𝔞|𝔟))}, & if (𝔞|𝔟) = (𝔰|𝔱), \\ 0, & otherwise, \end{cases}

for standard (k|n − k)–tableaux (𝔞|𝔟) and (𝔰|𝔱).

6.7. Corollary. Suppose that 𝔰 and 𝔱 are standard λ–tableaux which have n in their first row. Then

⟨w′_{𝔱}, v′_{𝔰}⟩ = \begin{cases}q^{2n−2k−3+ℓ(d(𝔱′))}, & if 𝔰 = 𝔱, \\ 0, & otherwise. \end{cases}

Proof. By Lemma 6.5 and Lemma 6.6 we have

v′_{𝔰} = ∑_{(𝔞_{𝔰}|𝔟_{𝔰})≺𝔰}(−1)^{k+1−I_{𝔰}(𝔞_{𝔰}|𝔟_{𝔰})}q^{ℓ(d(𝔰′))−ℓ(d(𝔞_{𝔰}|𝔟_{𝔰}))}x_{(𝔞_{𝔰}|𝔟_{𝔰})}

and, by Lemma 6.6,

w′_{𝔱} = (−1)^{k}q^{2n−2k−3}x_{(𝔞_{𝔱}^{∗}|𝔟_{𝔱}^{∗})} + ∑_{(𝔞_{𝔱}|𝔟_{𝔱})≺_{n}𝔱^{∗}} r_{𝔞_{𝔱}𝔟_{𝔱}}x_{(𝔞_{𝔱}|𝔟_{𝔱})}.

Now, all of the tableaux appearing in w′_{𝔱} have 1 appearing in their second component. In contrast, the only tableau in v′_{𝔰} which has 1 in its second component is the tableau (𝔞_{𝔰}^{∗}|𝔟_{𝔰}^{∗}). Therefore,

⟨w′_{𝔰}, v′_{𝔱}⟩ = (−1)^{2k+1−I_{𝔰}(𝔞_{s}^{∗}|𝔟_{s}^{∗})}q^{2n−2k−3+ℓ(d(𝔰′))−ℓ(d(𝔞_{𝔰}^{∗}|𝔟_{𝔰}^{∗}))}⟨x_{(𝔞_{𝔰}^{∗}|𝔟_{𝔰}^{∗})}, x_{(𝔞_{𝔱}^{∗}|𝔟_{𝔱}^{∗})}⟩

= \begin{cases}(−1)^{1+I_{𝔱}(𝔞_{𝔱}^{∗}|𝔟_{𝔱}^{∗})}q^{2n−2k−3+ℓ(d(𝔱′))}, & if 𝔰 = 𝔱, \\ 0, & otherwise. \end{cases}

Finally, the sign vanishes when 𝔰 = 𝔱 because I_{𝔱}(𝔞_{𝔱}^{∗}|𝔟_{𝔱}^{∗}) = 1 . □

We need one more result before we can produce the elementary divisors of S(λ).

6.8. Lemma. Let (𝔞_{λ}^{+}|𝔟_{λ}^{+}) be the unique standard (k|n − k)–tableau which has the numbers n − k + 1, …, n in 𝔞_{λ}^{+}. Then

w′_{𝔱^{λ}} = (−1)^{k}q^{ℓ(w_{λ′})}{q^{−ℓ(d(𝔞_{λ}^{+}|𝔟_{λ}^{+}))}[n − k]_{q}x_{(𝔞_{λ}^{+}|𝔟_{λ}^{+})}

+ ∑_{\begin{matrix}(𝔞|𝔟)≺𝔱^{λ} \\ (𝔞|𝔟)≠(𝔞_{λ}^{+}|𝔟_{λ}^{+}) \end{matrix}}(−1)^{1−I_{𝔱^{λ}} (𝔞|𝔟)}q^{−ℓ(d(𝔞|𝔟))} ∑_{j=0}^{n−k−1} x_{(𝔞|𝔟)r_{1,j}}\}.

Proof. By definition w′_{𝔱^{λ}} = v′_{𝔱^{λ}}x_{n−k}. Also, d((𝔱^{λ})′) = d(𝔱_{λ′}) = w_{λ′} so, by Lemma 6.5,

w′_{𝔱^{λ}} = v′_{𝔱^{λ}}x_{n−k} = ∑_{(𝔞|𝔟)≺𝔱^{λ}}(−1)^{k+1−I_{𝔱^{λ}} (𝔞|𝔟)}q^{ℓ(w_{λ′})−ℓ(d(𝔞|𝔟))}x_{(𝔞|𝔟)}x_{n−k}

= ∑_{(𝔞|𝔟)≺𝔱^{λ}}(−1)^{k+1−I_{𝔱^{λ}} (𝔞|𝔟)}q^{ℓ(w_{λ′})−ℓ(d(𝔞|𝔟))}x_{(𝔞|𝔟)}(1 + T_{1} + ⋯ + T_{1,n−k−1}).

Let (𝔞|𝔟) be one of the tableaux appearing in this sum. If (𝔞|𝔟) ≠ (𝔞_{λ}^{+}|𝔟_{λ}^{+}) then 1 is contained in 𝔞 and all of the numbers 2, 3, …, n − k are contained in 𝔟. Therefore, (𝔞|𝔟)r_{1,j} is standard and x_{(𝔞|𝔟)}T_{1,j} = x_{(𝔞|𝔟)r_{1,j}}, for 0 ≤ j ≤ n − k − 1. On the other hand, x_{(𝔞_{λ}^{+}|𝔟_{λ}^{+})}T_{1,j} = q^{j}x_{(𝔞_{λ}^{+}|𝔟_{λ}^{+})}, for 0 ≤ j ≤ n − k − 1. This completes the proof of the Lemma. □

This result has two useful Corollaries.

6.9. Corollary. Suppose that 𝔱 ≠ 𝔱^{λ} is a standard (k|n − k)–tableau. Then

⟨w′_{𝔱^{λ}}, v′_{𝔱}⟩ = 0.

Proof. By Lemma 6.8, if x_{(𝔞|𝔟)} appears in w′_{𝔱^{λ}} then all but one of the entries in 𝔞 are contained in {1, n − k + 1, …, n}. On the other hand, by Lemma 6.5, if x_{(𝔞|𝔟)} appears in v′_{𝔱} then all of the entries in 𝔞 are contained in the first column of 𝔱.

Suppose now that 𝔱 ≠ 𝔱^{λ}. Then, by the last paragraph, x_{(𝔞_{λ}^{+}|𝔟_{λ}^{+})} cannot appear in v′_{𝔱} and the only way that the inner product ⟨w′_{𝔱^{λ}}, v′_{𝔱}⟩ can be non–zero is if the set of numbers in the first column of 𝔱 is of the form T = {1, j, n − k + 1, …, n} \ {m}, for some integers j and m with 1 < j ≤ n − k and n − k < m ≤ n. Let (𝔞|𝔟) be the standard (k|n − k)–tableau whose first component contains exactly the numbers in T \ {j} and let (𝔞′|𝔟′) = (𝔞|𝔟)r_{1,j−1}. Then (𝔞|𝔟) ≺ 𝔱^{λ}, I_{𝔱}(𝔞|𝔟) = 2 and I_{𝔱}(𝔞′|𝔟′) = 1. Also ℓ(d(𝔞′|𝔟′)) = ℓ(d(𝔞|𝔟)) + j − 1, so x_{(𝔞|𝔟)}T_{1,j−1} = x_{(𝔞′|𝔟′)}. Therefore, by Lemma 6.5 and Lemma 6.8 and the remarks above,

v′_{𝔱} = (−1)^{k}q^{ℓ(d(𝔱′))}(q^{−ℓ(d(𝔞′|𝔟′))}x_{(𝔞′|𝔟′)} − q^{−ℓ(d(𝔞|𝔟))}x_{(𝔞|𝔟)}) + other standard terms

and

w′_{𝔱^{λ}} = q^{ℓ(w_{λ′})−ℓ(d(𝔞|𝔟))}(x_{(𝔞′|𝔟′)} + x_{(𝔞|𝔟)}) + other standard terms,

where none of the “other standard terms” appear both in v′_{𝔱} and in w′_{𝔱^{λ}}. Consequently, ⟨w′_{𝔱^{λ}}, v′_{𝔱}⟩ = 0. Hence, ⟨w′_{𝔱^{λ}}, v′_{𝔱}⟩ = 0 whenever 𝔱 ≠ 𝔱^{λ} as claimed. □

6.10. Corollary. Suppose that 𝔱 is a standard (n − k, 1^{k})–tableau and that n does not appear in the first row of 𝔱. Then ⟨w′_{𝔱}, v′_{𝔱}⟩ = q^{k(n−k−2)}[n]_{q}.

Proof. Recall that if 𝔱 is a standard λ–tableau then d(𝔱′)d(𝔱)^{−1} = w_{λ′}, with the lengths adding; this is well–known and is easily proved by induction on the dominance order for tableaux. Therefore,

⟨w′_{𝔱}, v′_{𝔱}⟩ = ⟨w′_{𝔱^{λ}}T_{d(𝔱)}, x_{(k|n−k)}y′_{k+1}T_{d(𝔱′)}⟩ = ⟨w′_{𝔱^{λ}}, x_{(k|n−k)}y′_{k+1}T_{d(𝔱′)}T_{d(𝔱)}^{∗}⟩
= ⟨w′_{𝔱^{λ}}, x_{(k|n−k)}y′_{k+1}T_{d(𝔱^{λ}′)}⟩ = ⟨w′_{𝔱^{λ}}, v′_{𝔱^{λ}}⟩.

Hence, it is enough to consider the case where 𝔱 = 𝔱^{λ}.

Suppose that 𝔱 = 𝔱^{λ}. Then, by Lemma 6.5 and Lemma 6.8,

⟨w′_{𝔱^{λ}}, v′_{𝔱^{λ}}⟩ = q^{2ℓ(w_{λ′})}\{q^{−ℓ(d(𝔞_{λ}^{+}|𝔟_{λ}^{+}))}[n − k]_{q} + ∑_{\begin{matrix}(𝔞|𝔟)≺𝔱^{λ} \\ (𝔞|𝔟)≠(𝔞_{λ}^{+}|𝔟_{λ}^{+}) \end{matrix}} q^{−ℓ(d(𝔞|𝔟))}\}.

Using the remarks before Lemma 6.4 it is not hard to see that ℓ(d(𝔞_{λ}^{+}|𝔟_{λ}^{+})) = k(n − k) and that ℓ(d(𝔞|𝔟)) = (k − 1)(n − k) + 2 − I_{𝔱}(𝔞|𝔟), whenever (𝔞|𝔟) ≺ 𝔱^{λ} and (𝔞|𝔟) ≠ (𝔞_{λ}^{+}|𝔟_{λ}^{+}). Therefore,

⟨w′_{𝔱^{λ}}, v′_{𝔱^{λ}}⟩ = q^{2ℓ(w_{λ′})}{q^{−k(n−k)}[n − k]_{q} + ∑_{i=2}^{k+1} q^{−(k−1)(n−k)−2+i}}

= q^{2ℓ(w_{λ′})−k(n−k)}{[n − k]_{q} + q^{n−k} ∑_{j=0}^{k−1} q^{j}}

= q^{2ℓ(w_{λ′})−k(n−k)}[n]_{q}.

As ℓ(w_{λ′}) = k(n − k − 1) the result follows. □

Finally, we can prove the main result of this section.

6.11. Proposition. Suppose that λ = (n−k, 1^{k}), for some k with 0 ≤ k < n. Then the Gram matrix G(λ) of S(λ) is divisibly diagonalizable over ℤ[q, q^{−1}] with \binom{n−2}{k} elementary divisors equal to [k]_{q}^{!} and with the remaining \binom{n−2}{k−1} elementary divisors being equal to [k]_{q}^{!}[n]_{q}.

Proof. By Proposition 6.1 the Gram matrix G(λ) of S(λ) is equal to [k]_{q}^{!} times the Gram matrix of S′(λ). Therefore, by Lemma 4.1 it is enough to show that there is an invertible diagonal matrix D such that

G′(λ) = (⟨w′_{𝔰}, v′_{𝔱}⟩)_{𝔰,𝔱∈Std(k|n−k)} = D · ( \begin{matrix}I & ∗ \\ 0 & [n]_{q}U \end{matrix} ) ,

where I is a \binom{n−2}{k} × \binom{n−2}{k} identity matrix and U is a \binom{n−2}{k−1} × \binom{n−2}{k−1} upper triangular matrix with 1’s down its diagonal. Here we order the rows and columns lexicographically with respect to the entries in the first columns of 𝔰 and 𝔱. Because D is invertible its non–zero entries must all be of the form ±q^{m}, for some integer m.

By Corollary 6.7, the rows of G′(λ) which are indexed by those tableaux which have n in their first row have the required form. This accounts for the identity matrix in the top half of the Gram matrix G′(λ).

Next, suppose that 𝔰 is a standard (k|n − k)–tableau and that n does not appear in the first row of 𝔰. If 𝔰 = 𝔱^{λ} then ⟨w′_{𝔰}, v′_{𝔱}⟩ = 0, for all 𝔱 ≠ 𝔰, by Corollary 6.9. If 𝔰 ≠ 𝔱^{λ} then there exists an integer i, 1 < i < n, such that ℓ(d(𝔰)r_{i}) < ℓ(d(𝔰)). Therefore,

⟨w′_{𝔰}, v′_{𝔱}⟩ = ⟨w′_{𝔰r_{i}}T_{i}, v′_{𝔱}⟩ = ⟨w′_{𝔰r_{i}}, v′_{𝔱}T_{i}⟩.

By expanding v′_{𝔱}T_{i} and using induction, it follows that ⟨w′_{𝔰}, v′_{𝔱}⟩ = 0 if 𝔱 appears before 𝔰 in our chosen ordering of Std(λ). Similarly, if 𝔱 does not appear before 𝔰 then [n]_{q} divides ⟨w′_{𝔰}, v′_{𝔱}⟩ by Corollary 6.10. □

Notice, in particular, that the Gram matrix calculation in the proof of the Proposition implies that { w′_{𝔱} | 𝔱 ∈ Std(λ) } is indeed a basis of S′(λ).

We give one application of Proposition 6.11.

Let π: S(λ) ⟶ S′(λ) be the isomorphism of Proposition 6.1 and for each standard λ–tableau 𝔱 let w_{𝔱} = π^{−1}(w′_{𝔱}). Then { w_{𝔱} | 𝔱 ∈ Std(λ) } is a basis of S(λ). Then, in the case where S(λ) is not irreducible, the proof of Proposition 6.11 also gives a basis for the simple module D(λ). More precisely, we have the following.

6.12. Corollary. Suppose that R is a field, that [k]_{q}^{!} ≠ 0 and that [n]_{q} = 0. Then S(λ) is not irreducible and a basis of D(λ) = S(λ)/(S(λ)^{⊥} ∩ S(λ)) is given by

{ w_{𝔱} + (S(λ)^{⊥} ∩ S(λ)) | 𝔱 ∈ Std(λ) and n in first row of 𝔱 } ,

and a basis of S(λ)^{⊥} ∩S(λ) is given by { w_{𝔱} | 𝔱 ∈ Std(λ) and n is in first row of 𝔱 } .

7. Some counterexamples

Let R = ℤ[q, q^{−1}]. We write the m^{th} cyclotomic polynomial in q as Φ_{m} = Φ_{m}(q). Andersen remarked that in general the Gram matrix G(λ) is not diagonalizable

[1, Remark 5.11]. We give two examples of this kind.

Note that G(λ) is divisibly diagonalizable over ℤ_{(p)}[q, q^{−1}] for all but finitely many primes p. In fact, it suffices to exclude the primes occurring in the denominators of the entries of the matrices used to diagonalize G(λ) over ℚ[q, q^{−1}].

We record the elementary divisors in “jump notation”. That is, we write

\overline{}^{f_{1}} m_{1} ^{\overline{}^{f_{2}}} m_{2} ^{\overline{}^{f_{3}}} m_{3} ^{\overline{}^{f_{4}}} ⋯ ^{\overline{}^{f_{s}}} m_{s}

to indicate that the matrix has the elementary divisor f_{1} with multiplicity m_{1}, the elementary divisor f_{1}f_{2} with multiplicity m_{2}, …, and the elementary divisor f_{1} ⋯ f_{s} with multiplicity m_{s}.

7.1. Example. Let λ = (3, 3, 2). The elementary divisors of G(3, 3, 2) over ℚ[q, q^{−1}] are given by

\overline{}^{Φ_{2}^{2}} 1 ^{\overline{}^{Φ_{4}}} 20^{\overline{}^{Φ_{3}Φ_{5}}} 20 ^{\overline{}^{Φ_{4}}} 1 ;

over 𝔽_{2}[q, q^{−1}] they are given by

\overline{}^{Φ_{2}^{3}} 1 ^{\overline{}^{Φ_{2}}} 20^{\overline{}^{Φ_{3}Φ_{5}}} 20 ^{\overline{}^{Φ_{2}}} 1 ;

and, putting q = 1, over ℤ they are given by

\overline{}^{2^{3}} 21 ^{\overline{}^{3·5}} 21 .

We claim that G(3, 3, 2) is not diagonalizable over ℤ_{(2)}[q, q^{−1}]. To see this suppose that it is diagonalizable. Then, considered as an element of ℤ_{(2)}[q, q^{−1}], any resulting diagonal entry must contain the factor (q + 1) with exponent 2. Considered as an element of 𝔽_{2}[q, q^{−1}] the factor (q + 1) can occur only with even exponent in such a diagonal entry. But this is not the case, so we have a contradiction.

This claim in particular implies that G(3, 3, 2) is not diagonalizable over ℤ[q, q^{−1}]. We remark that the comparison of the elementary divisors over ℚ[q, q^{−1}] and

over ℤ yields a contradiction to diagonalizability over ℤ_{(2)}[q, q^{−1}], too.

7.2. Example. Let λ = (4, 2, 1, 1). The elementary divisors of G(4, 2, 1, 1) over ℚ[q, q^{−1}] are given by

\overline{}^{Φ_{2}} 14 ^{\overline{}^{Φ_{2}}} 1 ^{\overline{}^{Φ_{4}}} 30 ^{\overline{}^{Φ_{7}}} 30 ^{\overline{}^{Φ_{4}}} 1 ^{\overline{}^{Φ_{2}}} 14 ;

over 𝔽_{2}[q, q^{−1}] they are given by

\overline{}^{Φ_{2}} 14 ^{\overline{}^{Φ_{2}^{2}}} 1 ^{\overline{}^{Φ_{2}}} 30 ^{\overline{}^{Φ_{7}}} 30 ^{\overline{}^{Φ_{2}}} 1 ^{\overline{}^{Φ_{2}^{2}}} 14 ;

over 𝔽_{3}[q, q^{−1}] they are given by

\overline{}^{Φ_{2}} 13 ^{\overline{}^{Φ_{2}}} 2 ^{\overline{}^{Φ_{4}}} 30 ^{\overline{}^{Φ_{7}}} 30 ^{\overline{}^{Φ_{4}}} 2 ^{\overline{}^{Φ_{2}}} 13 ;

and, putting q = 1, over ℤ they are given by

\overline{}^{2} 14 ^{\overline{}^{2^{2}}} 31 ^{\overline{}^{7}} 31 ^{\overline{}^{2^{2}}} 14 .

We claim that G(4, 2, 1, 1) is not diagonalizable over ℤ_{(2)}[q, q^{−1}]. Again, by way of contradiction suppose that it is diagonalizable. In 𝔽_{2}[q, q^{−1}], 14 of the resulting diagonal entries contain the factor (q+1) with exponent 1. Therefore, in ℤ_{(2)}[q, q^{−1}], 14 of them contain the factor (q + 1) with exponent 1 and the factor (q^{2} + 1) with exponent 0. Similarly, in 𝔽_{2}[q, q^{−1}], 14 of the resulting diagonal entries contain the factor (q + 1) with exponent 7. Thus in ℤ_{(2)}[q, q^{−1}], 14 of them contain the factor (q+1) with exponent 3 and the factor (q^{2} +1) with exponent 2. Hence in 𝔽_{2}[q, q^{−1}], no other diagonal entry can contain (q + 1) with odd exponent. But in 𝔽_{2}[q, q^{−1}], there is a diagonal entry containing (q + 1) to the power 3 and another containing it to the power 5 so, again, we have a contradiction.

We claim that G(4, 2, 1, 1) is not diagonalizable over ℤ_{(3)}[q, q^{−1}]. Assume it to be diagonalizable. In ℤ_{(3)}[q, q^{−1}], 14 of the resulting diagonal entries contain (q + 1) with exponent 1. This contradicts the fact that in 𝔽_{3}[q, q^{−1}], only 13 of them contain (q + 1) with exponent 1.

Both claims independently imply that G(4, 2, 1, 1) is not diagonalizable over ℤ[q, q^{−1}].

We remark that the comparison of the elementary divisors over ℚ[q, q^{−1}] and over ℤ yields a contradiction to diagonalizability over ℤ_{(2)}[q, q^{−1}], too.

Finally, we give a (non-exhaustive) list of elementary divisors of some divisibly diagonalizable Gram matrices for non-hooks, calculated using Gap 3 [12] and Magma [9]. We omit the respective conjugate partition; compare (5.9).

n λ  Elementary divisors of G(λ)
4 (2, 2)  ^{\overline{}^{Φ_{2}}} 1 ^{\overline{}^{Φ_{3}}} 1
5 (3, 2)  ^{\overline{}^{1}} 1 ^{\overline{}^{Φ_{3}}} 3 ^{\overline{}^{Φ_{4}}} 1
6 (4, 2)  ^{\overline{}^{1}} 4 ^{\overline{}^{Φ_{4}}} 1 ^{\overline{}^{Φ_{2}}} 3 ^{\overline{}^{Φ_{5}}} 1
(3, 3)  ^{\overline{}^{Φ_{2}}} 1 ^{\overline{}^{Φ_{3}}} 3 ^{\overline{}^{Φ_{4}}} 1
(3, 2, 1)  ^{\overline{}^{1}} 4 ^{\overline{}^{Φ_{3}}} 4 ^{\overline{}^{Φ_{5}}} 4 ^{\overline{}^{Φ_{3}}} 4
7 (5, 2)  ^{\overline{}^{1}} 8 ^{\overline{}^{Φ_{5}}} 5 ^{\overline{}^{Φ_{3}Φ_{6}}} 1
(4, 3)  ^{\overline{}^{1}} 1 ^{\overline{}^{Φ_{3}}} 7 ^{\overline{}^{Φ_{4}}} 5 ^{\overline{}^{Φ_{5}}} 1
(3, 3, 1)  ^{\overline{}^{Φ_{2}}} 6 ^{\overline{}^{Φ_{3}}} 2 ^{\overline{}^{Φ_{5}}} 12 ^{\overline{}^{Φ_{4}}} 1
8 (6, 2)  ^{\overline{}^{1}} 13 ^{\overline{}^{Φ_{3}Φ_{6}}} 1 ^{\overline{}^{Φ_{2}}} 5 ^{\overline{}^{Φ_{7}}} 1
(5, 3)  ^{\overline{}^{1}} 8 ^{\overline{}^{Φ_{4}}} 6 ^{\overline{}^{Φ_{2}}} 7 ^{\overline{}^{Φ_{5}}} 6 ^{\overline{}^{Φ_{6}}} 1
(4, 4)  ^{\overline{}^{Φ_{2}}} 1 ^{\overline{}^{Φ_{3}}} 7 ^{\overline{}^{Φ_{4}}} 5 ^{\overline{}^{Φ_{5}}} 1
9 (7, 2)  ^{\overline{}^{1}} 19 ^{\overline{}^{Φ_{7}}} 7 ^{\overline{}^{Φ_{4}Φ_{8}}} 1
(6, 3)  ^{\overline{}^{1}} 21 ^{\overline{}^{Φ_{5}}} 19 ^{\overline{}^{Φ_{6}}} 1 ^{\overline{}^{Φ_{3}}} 6 ^{\overline{}^{Φ_{7}}} 1
(5, 4)  ^{\overline{}^{1}} 1 ^{\overline{}^{Φ_{3}}} 15 ^{\overline{}^{Φ_{4}}} 18 ^{\overline{}^{Φ_{5}}} 7 ^{\overline{}^{Φ_{6}}} 1

We do not know an example of a Gram matrix G(λ) that is diagonalizable over ℤ[q, q^{−1}], but not divisibly diagonalizable.

For a general partition λ, we can not decide whether G(λ) is diagonalizable over ℤ[q, q^{−1}].

Acknowledgements

The authors would like to thank Adrian Williams for pointing out some errors in an earlier version of this manuscript.

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Abteilung Reine Mathematik,

    Universität Ulm, D-89069 Ulm.  kuenzer@mathematik.uni-ulm.de
    Germany.  www.mathematik.uni-ulm.de/ReineM/kuenzer

School of Mathematics and Statistics F07

University of Sydney, Sydney N.S.W. 2006. mathas@maths.usyd.edu.au

    Australia.  www.maths.usyd.edu.au/u/mathas/