Proc. Lond. Math. Soc.(3), 74 (1997), 241–274.

A q–analogue of the Jantzen–Schaper theorem

Gordon James and Andrew Mathas^{∗}

Abstract

In this paper we prove an analogue of Jantzen’s sum formula for the q–Weyl modules of the q–Schur algebra and, as a consequence, derive the analogue of Schaper’s theorem for the q–Specht modules of the Hecke algebras of type A. We apply these results to classify the irreducible q-Weyl modules and the irreducible (e–regular) q– Specht modules, defined over any field. In turn, this allows us to identify all of the ordinary irreducible representations of the finite general linear group GL_{n}(q) which remain irreducible modulo a prime p not dividing q.

1 Introduction

One of the most important tools for calculating decomposition numbers for the symmetric group 𝔖_{r} is the theorem of Schaper [22], which is derived from Jantzen’s sum formula [16, 18]. For example, Schaper’s Theorem tells us precisely which simple modules are composition factors of the Specht module S^{λ}, provided that the composition factors (together with multiplicities), are known for all Specht modules S^{μ} with μ ▷ λ (here “▷” is the usual dominance order on partitions). A good account of such matters appears in [2]. Here we derive theorems for q–Schur algebras and Hecke algebras of type A which reduce to the results of Jantzen and Schaper when q is set equal to 1.

We base our results upon a formula for the determinant of the Gram matrix of the q– Weyl module W^{λ} for the q–Schur algebra. Now, W^{λ} may also be regarded as a Weyl module for quantum GL_{n} (see, for example, [3, Theorem 3.3.2]). Consequently, it is possible to construct a proof of the formula we need using [1, Prop. 10.5] together with the techniques of Schaper’s thesis [22, pp. 42–55]. The proof of [1, Prop. 10.5] is geometric, and relies upon quantum analogues of Kempf’s vanishing theorem [1, Theorem 5.7] and Serre duality [1, Theorem 7.3]. Here, we take a more direct and purely algebraic approach by introducing generalisations of the Murphy operators [19, 6] for the q–Schur algebra and using them to construct an orthogonal basis for the q–Weyl modules W^{λ}. By combining the properties of these Murphy operators with the combinatorial results in section 2 we arrive at the formula we seek in Theorem 3.26.

∗Supported in part by SERC grant GR/J37690

As an application of our results, we derive, for the first time, a necessary and sufficient condition for an ordinary irreducible representation of the finite general linear group GL_{n}(q) to remain irreducible modulo a prime number which does not divide q.

We have made this paper largely self contained, by quoting little beyond basic facts about q–Weyl modules [8]. In particular, our results include those of Jantzen and Schaper as special cases.

2 Combinatorial results involving tableaux

In this section we develop the combinatorial machinery necessary to describe the determinant of the Gram matrix, defined with respect to the semistandard basis, of the q–Weyl module (these objects will be defined in section 3).

Throughout this paper we fix two natural numbers n and r with n ≥ r. We regard a composition of r as a sequence μ = (μ_{1}, μ_{2}, …) of non–negative integers whose sum is r, and μ_{i} = 0 for all i > n. We write μ ⊧ r if μ is a composition of r. If, moreover, μ_{1} ≥ μ_{2} ≥ ⋯ ≥ μ_{n}, then μ is a partition of r and we write μ ⊢ r. We often suppress all zeros which are at the end of a composition.

Suppose now that λ ⊧ r and μ ⊧ r. The diagram [λ] consists of all ordered pairs of integers (i, j) with 1 ≤ i ≤ n and 1 ≤ j ≤ λ_{i}. We refer to the elements of [λ] as nodes and we illustrate diagrams as follows.

If λ = (4, 2, 1) then [λ] = .

A λ–tableau of type μ is obtained from [λ] by mapping each node to an integer between 1 and n in such a way that, for all i, the number of nodes mapped to i is μ_{i}. Let 𝒯 (λ, μ) denote the set of all λ–tableaux of type μ.

Suppose that t ∈ 𝒯 (λ, μ). Then t is row–standard if the entries in t are non–decreasing along each row. If λ ⊢ r, then t is semistandard if it is row–standard and the entries in t are increasing down each column. Let 𝒯_{0}(λ, μ) denote the set of semistandard λ–tableaux of type μ. In the case where t ∈ 𝒯_{0}(λ, (1^{r})), we say t is a standard λ–tableau; if t is a standard tableau then all of the entries in t are distinct.

Of the three (4, 2, 1)–tableaux of type (2, 3, 1, 1) illustrated below, only the first two are row standard, and only the first is semistandard.

1 1 2 2 2 3 4  1 1 2 4 2 3 2  1 2 3 2 4 1 2

Let t_{ij} denote the entry in row i and column j of t. We define a relation ⊵ on 𝒯 (λ, μ) as follows.

A q–analogue of the Jantzen–Schaper theorem 3

(2.1) Let s, t ∈ 𝒯 (λ, μ). Then s ⊵ t if and only if for all u and all k

|{ (i, j) ∈ [λ] : j ≤ k and t_{ij} ≤ u }| ≤ |{ (i, j) ∈ [λ] : j ≤ k and s_{ij} ≤ u }|.

Write s ▷ t if s ⊵ t and s ≠ t.

When λ ⊢ r, the restriction of ⊵ to 𝒯_{0}(λ, μ) is easy to understand, as we explain next. Assume that s and t belong to 𝒯_{0}(λ, μ) and consider the tableau obtained from s by omitting all entries from s which are greater than u; since s is semistandard, this tableau has the shape of a partition, which we call s_{|u}. Then, s ⊵ t if and only if s_{|u} ⊵ t_{|u} for all u, where the second symbol “⊵” denotes the usual dominance order on partitions. For example,

1 1 1 2 2 2 3 3 3  ▷ 1 1 1 2 3 2 2 3 3  ▷ 1 1 1 3 3 2 2 2 3 .

Let t^{λ} be the λ–tableau of type (1^{r}) which has the numbers 1, 2, …, r inserted in order along the rows. For example, if λ = (4, 2, 1) then

t^{λ} = 1 2 3 4 5 6 7 .

The symmetric group 𝔖_{r} acts on 𝒯 (λ, (1^{r})) from the right by letter permutations, in the usual way. Let 𝔖_{λ} denote the row stabilizer of t^{λ}. Then 𝔖_{λ} is a standard Young subgroup of 𝔖_{r}.

The basic transpositions (m, m + 1), with 1 ≤ m ≤ r − 1, generate 𝔖_{r}. If w ∈ 𝔖_{r}, we denote by ℓ(w) the shortest length of an expression for w as a product of basic transpositions. Each right coset 𝔖_{λ}w of 𝔖_{λ} in 𝔖_{r} contains a unique element of minimal length; this element is called a distinguished right coset representative of 𝔖_{λ}. The distinguished right coset representatives of 𝔖_{λ} are precisely those permutations d such that t^{λ}d is row– standard. Let 𝒟_{λ} denote the set of distinguished right coset representatives of 𝔖_{λ} in 𝔖_{r}, and let 𝒟_{λμ} = 𝒟_{λ} ∩ 𝒟_{μ}^{−1}. Then the elements of 𝒟_{λμ} form a complete set of 𝔖_{λ}–𝔖_{μ} double coset representatives in 𝔖_{r}. If d ∈ 𝒟_{λμ} then d is the unique element of minimal length in 𝔖_{λ}d𝔖_{μ}. Indeed, the following holds (see, for example, [5, 1.6]).

(2.2) Each w ∈ 𝔖_{r} can be written in a unique way as a product w = udv with u ∈ 𝔖_{λ}, d ∈ 𝒟_{λμ}, and v ∈ 𝒟_{ν} ∩ 𝔖_{μ} where ν is the composition of r which corresponds to the standard Young subgroup d^{−1}𝔖_{λ}d ∩ 𝔖_{μ} of 𝔖_{r}; moreover, ℓ(w) = ℓ(u) + ℓ(d) + ℓ(v).

We next relate this result to the row–standard tableaux in a way which will be useful later.

(2.3) Definition Suppose that λ, μ ⊧ r and that t is a row–standard tableau in 𝒯 (λ, (1^{r})). Let μ(t) be the row–standard tableau in 𝒯 (λ, μ) which is obtained from t by replacing each entry i by u if i belongs to row u of t^{μ}.

So, t ⟼ μ(t) gives a mapping from the set of row–standard tableaux in 𝒯 (λ, (1^{r}))

into 𝒯 (λ, μ). For example, if t = 1 2 5 8 9 3 4 7 6 10 and μ = (3, 3, 2, 2), then t^{μ} = 1 2 3 4 5 6 7 8 9 10 , and

μ(t) = 1 1 2 3 4 1 2 3 2 4 .

(2.4) Definition Suppose that λ, μ ⊧ r and that s is a row–standard tableau in 𝒯 (λ, μ). Then μ^{−1}(s) is defined to be the set of row–standard tableaux of type (1^{r}) which are obtained from s by replacing the 1’s in s by 1, 2, …, μ_{1}, then the 2’s in s by μ_{1} + 1, …, μ_{1} + μ_{2}, and so on. If, at each stage, we make the replacements so that the successive numbers are as high as possible, then we obtain a tableau which we call first(s). On the other hand, if we place successive numbers as low as possible, we get a tableau which we call last(s).

For example, if s = 1 1 1 2 3 3 5 2 2 3 4 , then first(s) = 1 2 3 4 7 8 11 5 6 9 10 and

last(s) = 1 2 3 6 8 9 11 4 5 7 10.

Note that if s is semistandard, then every tableau t in μ^{−1}(s) is standard, and satisfies first(s) ⊵ t ⊵ last(s).

We now relate the row–standard tableaux in 𝒯 (λ, μ) to the 𝔖_{λ}–𝔖_{μ} double cosets.

(2.5) Suppose that s is a row–standard tableau in 𝒯 (λ, μ) and that t = first(s). Then t = t^{λ}d where d is the unique permutation of minimal length in the double coset 𝔖_{λ}d𝔖_{μ}, and

𝔖_{λ}d𝔖_{μ} = ⋃_{t^{λ}d′∈μ^{−1}(s)^{d′}} 𝔖_{λ}d′.

Moreover, this gives a 1–to–1 correspondence between the set of row–standard tableaux in 𝒯 (λ, μ) and the set of 𝔖_{λ}–𝔖_{μ} double cosets.

The Young subgroup 𝔖_{ν} which appears in (2.2) is the intersection of the row stabilizer of

first(s) with 𝔖_{μ}. For example, if s = 1 1 1 2 2 3 4 2 3 3 , then first(s) = 1 2 3 4 5 7 10 6 8 9

and ν = (3, 2, 1, 1, 2, 1).

We shall use the following special case of (2.5).

(2.6) Suppose that i < j and that x is the last number in row i of t^{λ} and y is the first number in row j of t^{λ}. Then (x, y) is the permutation of minimal length in the double coset 𝔖_{λ}(x, y)𝔖_{λ} and

𝔖_{λ}(x, y)𝔖_{λ} = ⋃_{d′} 𝔖_{λ}d′,

where d′ runs over the elements of 𝒟_{λ} with the property that the entries in t^{λ}d′ agree with those in t^{λ} except that some pair of numbers have been swapped between row i and row j.

For example, if λ = (3, 2, 2) then t^{λ} = 1 2 3 4 5 6 7 , and, to calculate 𝔖_{λ}(3, 6)𝔖_{λ}, we

apply (2.5) to the case where s = 1 1 3 2 2 1 3 , so first(s) = t^{λ}(3, 6). We find that 𝔖_{λ}(3, 6)𝔖_{λ} =

⋃_{d′} 𝔖_{λ}d′ where d′ runs over

{(3, 6), (3, 6)(2, 3), (3, 6)(2, 3)(1, 2), (3, 6)(6, 7), (3, 6)(2, 3)(6, 7), (3, 6)(2, 3)(1, 2)(6, 7)}.

The next result is a straightforward computation in a similar vein.

(2.7) Assume that i < j and that x belongs to row i of t^{λ}, and y belongs to row j of t^{λ}. Suppose that there are a numbers to the right of x in row i of t^{λ}, and b numbers to the left of y in row j. Then (x, y) = wd where w ∈ 𝔖_{λ}, d ∈ 𝒟_{λ} and ℓ(w) = a + b.

For example, if t^{λ} = 1 2 3 4 5 6 7 8 9 then (2, 9) = (2, 3)(3, 4)(8, 9)d where t^{λ}d = 1 3 4 9 5 6 7 2 8 .

We now wish to introduce various rational functions associated with tableaux. To this end, let q be an indeterminate. For each non–negative integer m, let

[m]_{q} = ∑_{a=0}^{m−1} q^{a}  and  {m}_{q} = ∏_{a=1}^{m}[a]_{q}.

Note that [0]_{q} = 0 and {0}_{q} = 1. Also, setting q = 1, [m]_{q} and {m}_{q} reduce to m and m! respectively.

In the next section, we shall define vectors f_{s}, indexed by the semistandard tableaux s in 𝒯_{0}(λ, μ), in a certain inner product space (namely, the μ–weight space of the q–Weyl module W^{λ}). The square of the norm of f_{s} turns out to be the rational function γ_{s} ∈ ℤ(q), which we define next.

(2.8) Definition Let λ ⊢ r, μ ⊧ r, and s ∈ 𝒯_{0}(λ, μ). For each (i, j) ∈ [λ], let

Γ_{s}(i, j) = { (k, l) ∈ [λ] : l < j, s_{kl} < s_{ij}, and s_{k′l} > s_{ij} for all k′ > k },

and define

γ_{s} = ∏_{(i,j)∈[λ]} ∏_{(k,l)∈Γ_{s}(i,j)} \frac{[j − i + k − l + 1]_{q}}{[j − i + k − l]_{q}} .

Thus Γ_{s}(i, j) is the set of nodes (k, l) ∈ [λ] such that the l–th column of s lies to the left of the j–th column of s and contains no node labelled s_{ij} and, in addition, s_{kl} must be the largest entry in the l–th column which is less that s_{ij}. Moreover, the factor that a node (k, l) ∈ Γ_{s}(i, j) contributes to γ_{s} is precisely \frac{[ρ+1]_{q}}{[ρ]_{q}} where ρ is the so–called axial distance from (k, l) to (i, j) (see [13, p. 123]).

In order to help understand these definitions we illustrate them with several examples, many of which will be needed later in the paper.

(2.9) Example (i) If s = 1 1 2 2 3 then γ_{s} = [2]_{q}[2]_{q}.

(ii) If s = 1 1 3 2 2 then γ_{s} = \frac{[4]_{q}}{[3]_{q}} \frac{[3]_{q}}{[2]_{q}} .

(2.10) Example If t = t^{λ} then γ_{t} = ∏_{i=1}^{∞}{λ_{i}}_{q}.

For example, if t = 1 2 3 4 5 6 7 8 then the contributions to γ_{t} from the (1, j)–nodes are

[1]_{q},  [2]_{q},  \frac{[3]_{q}}{[2]_{q}} \frac{[2]_{q}}{[1]_{q}} ,  and  \frac{[4]_{q}}{[3]_{q}} \frac{[3]_{q}}{[2]_{q}} \frac{[2]_{q}}{[1]_{q}} ,

for 1 ≤ j ≤ 4 respectively, and the product of these terms is {4}_{q}.

(2.11) Example (cf. [19, 4.1], [6, 4.10]) Suppose that t and t′ are in 𝒯_{0}(λ, (1^{r})) and that t′ = t(m, m + 1), for some 1 ≤ m ≤ r − 1. Assume that t_{ij} = m + 1, t_{kl} = m where i < k; then j > l, since t and t′ are standard tableaux. Then,

γ_{t} = \frac{[ρ + 1]_{q}[ρ − 1]_{q}}{[ρ]_{q}^{2}} γ_{t′},

where ρ = j − i + k − l.

Proof: To see this, first note that Γ_{t′}(a, b) = Γ_{t}(a, b) for (a, b) ≠ (i, j); in particular, Γ_{t}(k, l) = Γ_{t′}(k, l). Also, the only difference between the two sets Γ_{t}(i, j) and Γ_{t′}(i, j) is that (k, l) ∈ Γ_{t}(i, j) and (k − 1, l) ∈ Γ_{t′}(i, j); these nodes contribute factors of

\frac{[ρ + 1]_{q}}{[ρ]_{q}}  and  \frac{[ρ]_{q}}{[ρ − 1]_{q}}

to γ_{t} and γ_{t′} respectively.

Recall the map t ⟼ μ(t) from Definition 2.3. For any standard tableau t, we want to relate the two functions γ_{t} and γ_{μ(t)}; for this we need the next definition.

(2.12) Definition Assume that λ ⊢ r, μ ⊧ r, and t ∈ 𝒯_{0}(λ, (1^{r})). For each (i, j) ∈ [λ], let

P_{t}^{μ}(i, j) = { (k, l) ∈ [λ] : l < j, t_{kl} < t_{ij}, and μ(t)_{kl} = μ(t)_{ij} },

and let

π_{t}^{μ} = ∏_{(i,j)∈[λ]} ∏_{(k,l)∈P_{t}^{μ}(i,j)} \frac{[j − i + k − l + 1]_{q}}{[j − i + k − l]_{q}} .

(2.13) Example Assume that μ = (2, 2).

(i) If t = 1 2 3 4 , then π_{t}^{μ} = [2]_{q}.

(ii) If t = 1 2 4 3 , then π_{t}^{μ} = \frac{[4]_{q}}{[3]_{q}} \frac{[2]_{q}}{[1]_{q}}.

(2.14) Example Suppose that t, t′ ∈ 𝒯_{0}(λ, (1^{r})) with μ(t) = μ(t′) and t′ = t(m, m + 1) for some m, with 1 ≤ m ≤ r − 1. Suppose that t_{ij} = m + 1, t_{kl} = m, and i < k. Then, by an argument similar to that used in Example 2.11, we have

π_{t}^{μ} = \frac{[ρ + 1]_{q}}{[ρ]_{q}} π_{t′}^{μ},

where ρ = j − i + k − l.

(2.15) Example Suppose that s ∈ 𝒯_{0}(λ, μ) and let t = first(s). Let d ∈ 𝒟_{λ} be given by t = t^{λ}d and let ν be the composition such that 𝔖_{ν} = d^{−1}𝔖_{λ}d ∩ 𝔖_{μ} (see (2.2)). Then,

π_{t}^{μ} = ∏_{i=1}^{∞}{ν_{i}}_{q}.

Proof: Notice that if (i, j) ∈ [λ], then P_{t}^{μ}(i, j) = { (k, l) ∈ [λ] : l < j and s_{kl} = s_{ij} }. Let x_{ij} = |P_{t}^{μ}(i, j)|, then

∏_{(k,l)∈P_{t}^{μ}(i,j)} \frac{[j − i + k − l + 1]_{q}}{[j − i + k − l]_{q}} = \frac{[x_{ij} + 1]_{q}}{[x_{ij}]_{q}} \frac{[x_{ij}]_{q}}{[x_{ij} − 1]_{q}} … \frac{[2]_{q}}{[1]_{q}} = [x_{ij} + 1]_{q}.

Therefore,

π_{t}^{μ} = ∏_{(i,j)∈[λ]}[x_{ij} + 1]_{q} = ∏_{i=1}^{∞} ∏_{u=1}^{∞}{c_{iu}}_{q},

where c_{iu} is the number of entries in the i–th row of s which are equal to u. However, the parts of ν are precisely the numbers c_{iu}, as i and u vary, by the remark following (2.5).

(2.16) Example Suppose that s ∈ 𝒯_{0}(λ, μ) and let t = last(s). Comparing the definitions of γ_{t} and γ_{s} we see that γ_{t} = π_{t}^{μ}γ_{s}.

Next, we give a name to a product of the rational functions γ_{s}. In the next section we will show that this product is (essentially) the Gram determinant of the μ–weight space of the q–Weyl module W^{λ}.

(2.17) Definition Let λ ⊢ r and μ ⊧ r. Define Δ_{μ}(λ) = ∏_{s∈𝒯_{0}(λ,μ)} γ_{s}.

The remainder of this section is devoted to establishing an explicit formula for Δ_{μ}(λ). Let ν be a partition of r and let ν′ = (ν′_{1}, ν′_{2}, …) denote the partition conjugate to ν.

That is, ν′ is the partition whose diagram is obtained by interchanging the rows and columns in [ν]. Recall that n is an integer with n ≥ r. If β_{i} = ν′_{i} + n − i for 1 ≤ i ≤ n then β_{1}, β_{2}, …, β_{n} is a sequence of β–numbers for ν′ (see [13, p. 77]). Note that β_{1} + ⋯ + β_{n} = r + n(n − 1)/2. Hereafter, whenever we deal with integers β_{1}, …, β_{n}, we shall implicitly assume that β_{1} + ⋯ + β_{n} = r + n(n − 1)/2.

(2.18) Definition Given integers β_{1}, β_{2}, …, β_{n} and a composition μ ⊧ r, define an integer d_{μ}(β_{1}, β_{2}, …, β_{n}) as follows.

(i) If β_{1} > β_{2} > ⋯ > β_{n} ≥ 0, then let d_{μ}(β_{1}, β_{2}, …, β_{n}) = |𝒯_{0}(ν, μ)| where ν′ is the partition whose β–numbers are β_{1}, β_{2}, …, β_{n}.

(ii) If β_{i} = β_{j} for some i ≠ j, or β_{i} < 0 for some i, then let d_{μ}(β_{1}, β_{2}, …, β_{n}) = 0. (iii) If w ∈ 𝔖_{n}, then let d_{μ}(β_{1w}, β_{2w}, …, β_{nw}) = sign(w)d_{μ}(β_{1}, β_{2}, …, β_{n}).

Now we fix our notation so that λ ⊢ r and μ ⊧ r. Let α_{1} > α_{2} > ⋯ > α_{n} ≥ 0 be β–numbers for λ′, and suppose that z is the largest integer such that μ_{z} ≠ 0.

Every s in 𝒯_{0}(λ, μ) has μ_{z} entries equal to z, and these entries are at the feet of distinct columns. If we delete these entries from s, then we obtain a tableau in 𝒯_{0}(\overline{λ}, \overline{μ}), say, where \overline{λ} is a partition of r − μ_{z} and \overline{μ} = (μ_{1}, …, μ_{z−1}). The β–numbers for \overline{λ}′ are \overline{α}_{1} > \overline{α}_{2} > ⋯ > \overline{α}_{n} ≥ 0 where for precisely z subscripts i = i_{1}, …, i_{z} we have \overline{α}_{i} = α_{i} − 1, and for the remaining subscripts j ≠ i_{1}, …, i_{z} we have \overline{α}_{j} = α_{j}. Since d_{μ}(α_{1}, …, α_{n}) = |𝒯_{0}(λ, μ)| and d\overline{μ}(\overline{α}_{1}, …, \overline{α}_{n}) = |𝒯_{0}(\overline{λ}, \overline{μ})|, it follows from our conventions in Definition 2.18 that

d_{μ}(α_{1}, …, α_{n}) = ∑_{i_{1}<···<i_{z}} d\overline{μ}(α_{1}, …, α_{i_{1}} − 1, …, α_{i_{z}} − 1, …, α_{n}).

And now, further thought shows that in fact we have the following.

(2.19) For all integers β_{1}, β_{2}, …, β_{n}

d_{μ}(β_{1}, …, β_{n}) = ∑_{i_{1}<···<i_{z}} d\overline{μ}(β_{1}, …, β_{i_{1}} − 1, …, β_{i_{z}} − 1, …, β_{n}).

Next, we extend Definition 2.17 of Δ_{μ}(λ), replacing λ by a sequence β_{1}, …, β_{n} of integers.

(2.20) Definition Suppose that β_{1} ≥ β_{2} ≥ ⋯ ≥ β_{n} are integers. If β_{1} > β_{2} > ⋯ > β_{n} ≥ 0 then let Δ_{μ}(β_{1}, …, β_{n}) = Δ_{μ}(ν) where β_{1}, …, β_{n} are a sequence of β–numbers for ν′; otherwise, let Δ_{μ}(β_{1}, …, β_{n}) = 1.

We now embark upon the task of evaluating Δ_{μ}(λ) in earnest. Recall the definition of γ_{s} from Definition 2.8. We have,

(2.21)  Δ_{μ}(λ) = ∏_{s∈𝒯_{0}(λ,μ)} γ_{s} = ∏_{s∈𝒯_{0}(λ,μ)} ∏_{(i,j)∈[λ]} ∏_{(k,l)∈Γ_{s}(i,j)} \frac{[j − i + k − l + 1]_{q}}{[j − i + k − l]_{q}} .

In order to relate Δ_{μ}(λ) to Δ\overline{μ}(\overline{λ}), where \overline{λ} and \overline{μ} are as above, notice that

(2.22)  \begin{matrix}∏_{s∈𝒯_{0}(λ,μ)} & ∏_{s_{ij}<z^{(i,j)∈[λ]}} & ∏_{(k,l)∈Γ_{s}(i,j)} & \frac{[j − i + k − l + 1]_{q}}{[j − i + k − l]_{q}} \\ = & ∏_{i_{1}<···<i_{z}} Δ\overline{μ}(α_{1}, …, α_{i_{1}} − 1, …, α_{i_{z}} − 1, …, α_{n}), \end{matrix}

since this calculation involves only those semistandard tableaux obtained by removing all entries labelled z from some semistandard tableau in 𝒯_{0}(λ, μ). The remaining contributions to Δ_{μ}(λ) come from those s ∈ 𝒯_{0}(λ, μ) and (i, j) ∈ [λ] for which s_{ij} = z. For such s and (i, j), the set Γ_{s}(i, j) consists of those nodes (k, l) which lie at the foot of a column in [λ] with l < j and s_{kl} ≠ z; so we have j − i + k − l = α_{l} − α_{j}. Therefore,

(2.23)  \begin{matrix}∏_{s∈𝒯_{0}(λ,μ)} & ∏_{s_{ij}=z^{(i,j)∈[λ]}} & ∏_{(k,l)∈Γ_{s}(i,j)} & \frac{[j − i + k − l + 1]_{q}}{[j − i + k − l]_{q}} \\ = & ∏_{i_{1}<···<i_{z}} ( ∏_{l<j ; {l∉\{i_{1},...,iz\} \atop j∈\{i_{1},...,iz\}}} \frac{[α_{l} − α_{j} + 1]_{q}}{[α_{l} − α_{j}]_{q}} )_{.}^{d\overline{μ}(α_{1},...,α_{i_{1}}−1,...,α_{iz} −1,...,α_{n})} \end{matrix}

Let X_{1} be the numerator on the right hand side of this equation. Then

(2.24)  \begin{aligned}X_{1} & = & ∏_{h=1}^{∞} ∏_{i_{1}<···<i_{z}} ∏_{l} ∏_{j}_{l∉\{i_{1},...,iz\} ; {j∈\{i_{1},...,iz\} \atop α_{l}−α_{j} +1=h}} [h]_{q}^{d\overline{μ}(α_{1},...,α_{i_{1}}−1,...,α_{iz} −1,...,α_{n})} \\ & = & ∏_{h=1}^{∞} ∏_{i_{1}<···<i_{z−1}} ∏_{l} ∏_{j} ; α_{l}−α_{j}+1=h^{\{l,j\}∩\{i_{1},...,i_{z−1}\}=∅}[h]_{q}^{f_{1}}, \end{aligned}

where f_{1} = d\overline{μ}(α_{1}, …, α_{i_{1}} − 1, …, α_{j} − 1, …, α_{i_{z−1}} − 1, …, α_{n}). Interchanging the entries α_{l} and α_{j} − 1 in f_{1}, so that α_{l} goes in position j and α_{j} − 1 goes in position l, we obtain

(2.25) \begin{aligned}f_{1} & = & −d\overline{μ}(α_{1}, …, α_{i_{1}} − 1, …, α_{l}, …, α_{j} − 1, …, α_{i_{z−1}} − 1, …, α_{n}) \\ & = & −d\overline{μ}(α_{1}, …, α_{i_{1}} − 1, …, α_{j} − 1 + h, …, α_{l} − h, …, α_{i_{z−1}} − 1, …, α_{n}) \end{aligned}

since α_{l} − α_{j} + 1 = h.

Let X_{2}^{−1} be the denominator on the right hand side of (2.23). Thus,

X_{2} = ∏_{h=1}^{∞} ∏_{i_{1}<···<i_{z}} ∏_{l} ∏_{j}_{l∉\{i_{1},...,iz\} ; {j∈\{i_{1},...,iz\} \atop α_{l}−α_{j} =h}} [h]_{q.}^{−d\overline{μ}(α_{1},...,α_{i_{1}}−1,...,α_{iz} −1,...,α_{n})}

An argument similar to that which we just used for X_{1} shows that

(2.26)  X_{2} = ∏_{h=1}^{∞} ∏_{i_{1}<···<i_{z−1}} ∏_{l} ∏_{j} ; α_{l}−α_{j}=h^{{l,j}∩{i_{1},...,i_{z−1}}=∅}[h]_{q}^{f_{2}},

where f_{2} = d\overline{μ}(α_{1}, …, α_{i_{1}} − 1, …, α_{j} + h, …, α_{l} − h − 1, …, α_{i_{z−1}} − 1, …, α_{n}).

Gathering together the formulae from (2.21) through to (2.26), we obtain a “branching rule” for Δ_{μ}(λ).

(2.27)  Δ_{μ}(λ) = X_{1}X_{2} ∏_{i_{1}<···<i_{z}} Δ\overline{μ}(α_{1}, …, α_{i_{1}} − 1, …, α_{i_{z}} − 1, …, α_{n}),

where X_{1} and X_{2} are as in (2.24) and (2.26).

We may now derive our formula for Δ_{μ}(λ).

(2.28) Theorem Suppose that λ ⊢ r and μ ⊧ r. Let α_{1} > α_{2} > ⋯ > α_{n} ≥ 0 be β–numbers for λ′. Then

Δ_{μ}(λ) = ∏_{h=1}^{∞} \begin{matrix}∏_{l} & ∏_{j} \\ α_{j}>α_{l}−h \end{matrix}[h]_{q}^{−d_{μ}(α_{1},...,α_{j}+h,...,α_{l}−h,...,α_{n})}.

Note that in this formula, we do not require that j < l, merely that α_{j} > α_{l} − h. Thus, in the exponent, we add h to α_{j} and subtract h from α_{l}, whether or not j < l.

Proof: First, observe that if α_{1}, …, α_{n} are integers with α_{1} ≥ α_{2} ≥ ⋯ ≥ α_{n} and α_{i} = α_{j}, for some i ≠ j, or α_{n} = −1, then both Δ_{μ}(λ) and the product given in the statement of the theorem are equal to 1. This remark allows us to apply induction on r.

As before, assume that μ_{z} is the last non–zero part of μ, and let \overline{μ} = (μ_{1}, …, μ_{z−1}). By (2.27) and induction on r, we have Δ_{μ}(λ) = X_{1}X_{2}Y_{1}Y_{2}Y_{3}Y_{4}, where X_{1} and X_{2} are as above and Y_{1}, Y_{2}, Y_{3}, and Y_{4} are given below and correspond to the four ways in which {l, j} and {i_{1}, i_{1}, …, i_{z}} can intersect.

Y_{1} = ∏_{h=1}^{∞} ∏_{i_{1}<···<i_{z}} ∏_{l} ∏_{j} ; α_{j}>α_{l}−h^{{l,j}∩{i_{1},...,iz}=∅}[h]_{q}^{g_{1}},

where g_{1} = −d\overline{μ}(α_{1}, …, α_{i_{1}} − 1, …, α_{j} + h, …, α_{l} − h, …, α_{i_{z}} − 1, …, α_{n});

Y_{2} = ∏_{h=1}^{∞} ∏_{i_{1}<···<i_{z−1}} ∏_{l} ∏_{j} ; α_{j}−1>α_{l}−h^{{l,j}∩{i_{1},...,i_{z−1}}=∅}[h]_{q}^{g_{2}},

where g_{2} = −d\overline{μ}(α_{1}, …, α_{i_{1}} − 1, …, α_{j} − 1 + h, …, α_{l} − h, …, α_{i_{z−1}} − 1, …, α_{n});

Y_{3} = ∏_{h=1}^{∞} ∏_{i_{1}<···<i_{z−1}} ∏_{l} ∏_{j} ; α_{j}>α_{l}−1−h^{{l,j}∩{i_{1},...,i_{z−1}}=∅}[h]_{q}^{g_{3}},

where g_{3} = −d\overline{μ}(α_{1}, …, α_{i_{1}} − 1, …, α_{j} + h, …, α_{l} − 1 − h, …, α_{i_{z−1}} − 1, …, α_{n}); and

Y_{4} = ∏_{h=1}^{∞} ∏_{i_{1}<···<i_{z−2}} ∏_{l} ∏_{j} ; α_{j}>α_{l}−h^{{l,j}∩{i_{1},...,i_{z−2}}=∅}[h]_{q}^{g_{4}},

where g_{4} = −d\overline{μ}(α_{1}, …, α_{i_{1}} − 1, …, α_{j} − 1 + h, …, α_{l} − 1 − h, …, α_{i_{z−2}} − 1, …, α_{n}). Now, g_{2} is equal to f_{1} by (2.25); so by (2.24) and the definition of Y_{2},

X_{1}Y_{2} = ∏_{h=1}^{∞} ∏_{i_{1}<···<i_{z−1}} ∏_{l} ∏_{j} ; α_{j}>α_{l}−h^{{l,j}∩{i_{1},...,i_{z−1}}=∅}[h]_{q}^{g_{2}}.

Also, g_{3} = −f_{2}. Hence,

X_{2}Y_{3} = ∏_{h=1}^{∞} ∏_{i_{1}<···<i_{z−1}} ∏_{l} ∏_{j} ; α_{j}>α_{l}−h^{{l,j}∩{i_{1},...,i_{z−1}}=∅}[h]_{q}^{g_{3}}.

Now, applying (2.19), we obtain

\begin{aligned}Δ_{μ}(λ) & = & Y_{1}(X_{1}Y_{2})(X_{2}Y_{3})Y_{4} \\ & = & ∏_{h=1}^{∞} \begin{matrix}∏_{l} & ∏_{j} \\ α_{j}>α_{l}−h \end{matrix}[h]_{q}^{−d_{μ}(α_{1},...,α_{j}+h,...,α_{l}−h,...,α_{n})}, \end{aligned}

as required.

(2.29) Example Suppose that λ = (3, 2). Then (6, 5, 3, 1, 0) are β–numbers for λ′, and

\begin{aligned}Δ_{μ}(λ) & = & [2]_{q}^{−d_{μ}(4,7,3,1,0)}[3]_{q}^{−d_{μ}(9,2,3,1,0)}[4]_{q}^{−d_{μ}(2,5,7,1,0)−d_{μ}(2,9,3,1,0)} \\ & = & [2]_{q}^{|𝒯_{0}((3,1^{2}),μ)|} [3]_{q}^{|𝒯_{0}((1^{5}),μ)|} [4]_{q.}^{|𝒯_{0}((2^{2},1),μ)|−|𝒯_{0}((1^{5}),μ)|} \end{aligned}

Now let μ = (2^{2}, 1). Then |𝒯_{0}((3, 1^{2}), μ)| = 1, |𝒯_{0}((1^{5}), μ)| = 0, and |𝒯_{0}((2^{2}, 1), μ)| = 1; so Δ_{μ}(λ) = [2]_{q}[4]_{q}. This agrees with Example 2.9 where we calculated γ_{s} for the two elements s of 𝒯_{0}(λ, μ).

There is another useful way to describe Δ_{μ}(λ); in order to state it we let h_{ij} = λ_{i} + λ′_{j} − i − j + 1 denote the length of the (i, j)–th hook in [λ] (see [10, §18]).

(2.30) Corollary (cf. [15, pp. 224, 233]) Suppose that λ ⊢ r and μ ⊧ r. Let α_{1} > α_{2} > ⋯ > α_{n} ≥ 0 be β–numbers for λ′. Then

Δ_{μ}(λ) = ∏_{a,b,c ; (a,b)∈[λ] ; {(a,c)∈[λ] \atop b<c}} (\frac{[h_{ab}]_{q}}{[h_{ac}]_{q}} )^{d_{μ}(α_{1},...,α_{b}+h_{ac},...,α_{c}−h_{ac},...,α_{n})} .

Proof: If we subtract h from α_{l} in (α_{1}, α_{2}, …, α_{n}) then we usually obtain an n–tuple where some integer is repeated or some integer is negative. The only exceptions occur when h = h_{al} for some (a, l) ∈ [λ]; see [13, 2.7.13]. Therefore, by Theorem 2.28,

Δ_{μ}(λ) = ∏_{(a,l)∈[λ]} ∏_{α_{j}>α_{l}−h_{al}^{j=1}}^{n}[h_{al}]_{q}^{−d_{μ}(α_{1},...,α_{j}+h_{al},...,α_{l}−h_{al},...,α_{n})}

= ∏_{(a,l)∈[λ]} ∏_{j<l}[h_{al}]_{q}^{−d_{μ}(α_{1},...,α_{j}+h_{al},...,α_{l}−h_{al},...,α_{n})}

× ∏_{(a,l)∈[λ]} ∏_{α_{j}>α_{l}−h_{al}^{j>l}}[h_{al}]_{q}^{−d_{μ}(α_{1},...,α_{l}−h_{al},...,α_{j}+h_{al},...,α_{n})}.

In the product where j > l we have

−d_{μ}(α_{1}, …, α_{l} − h_{al}, …, α_{j} + h_{al}, …, α_{n})
= d_{μ}(α_{1}, …, α_{j} + h_{al}, …, α_{l} − h_{al}, …, α_{n})
= \{ \begin{matrix}d_{μ}(α_{1}, …, α_{l} + h_{aj}, …, α_{j} − h_{aj}, …, α_{n}), & if (a, j) ∈ [λ], \\ 0, & if (a, j) ∉ [λ], \end{matrix}

since α_{l} − α_{j} = h_{al} − h_{aj} if (a, j) ∈ [λ]. However, if (a, l) ∈ [λ] and j < l then (a, j) ∈ [λ], so

Δ_{μ}(λ) = ∏_{(a,l)∈[λ] ; {(a,j)∈[λ] \atop j<l}} [h_{al}]_{q}^{−d_{μ}(α_{1},...,α_{j}+h_{al},...,α_{l}−h_{al},...,α_{n})} ∏_{(a,l)∈[λ] ; {(a,j)∈[λ] \atop j>l}} [h_{al}]_{q}^{d_{μ}(α_{1},...,α_{l}+h_{aj},...,α_{j}−h_{aj},...,α_{n})},

which is the same as the expression in the statement of the corollary.

The exponent which appears in Corollary 2.30 has the following interpretation (cf. [15, p. 223]). Given (a, b) and (a, c) ∈ [λ], with b < c, unwrap the skew (a, c)–hook from [λ] and wrap it back onto the resulting diagram so that the hand of the new skew hook is in column b. Suppose that by doing this we get a partition ν. Then

d_{μ}(α_{1}, …, α_{b} + h_{ac}, …, α_{c} − h_{ac}, …, α_{n}) = ±|𝒯_{0}(ν, μ)|;

where the sign is positive if and only if the sum of the leg lengths of the unwrapped and wrapped skew hooks is even.

(2.31) Example Consider, once again, the partition λ = (3, 2). The pairs of nodes in the same row which contribute to Δ_{μ}(λ) are as shown (the nodes (a, b) and (a, c) in [λ] are circled and we have labelled the nodes in the (a, c)–hook by c).

\begin{matrix}× ◯ & \begin{matrix}c \\ ◯ \end{matrix} & c \\ × & c \end{matrix} ⟶ × ; × ; c ; c ; c  giving (\frac{[4]_{q}}{[3]_{q}} )^{−|𝒯_{0}((1^{5}),μ)|}

\begin{matrix}× ◯ & × & \begin{matrix}c \\ ◯ \end{matrix} \\ × & × \end{matrix} ⟶ × × ; × × ; c  giving (\frac{[4]_{q}}{[1]_{q}} )^{|𝒯_{0}((2^{2},1),μ)|}

\begin{matrix}× & × & × \\ × ◯ & \begin{matrix}c \\ ◯ \end{matrix} \end{matrix} ⟶ × × × ; × ; c  giving (\frac{[2]_{q}}{[1]_{q}} )^{|𝒯_{0}((3,1^{2}),μ)|}

The product of these three quotients agrees with the answer for Δ_{μ}(λ) which was obtained in Example 2.29.

3 The Hecke algebra and the q–Schur algebra

Let R be an integral domain and let q be a unit in R. The Hecke algebra ℋ = ℋ_{R,q}(𝔖_{r}) is the R–free R–algebra with basis { T_{w} : w ∈ 𝔖_{r} }, where multiplication is defined by the rule that

T_{w}T_{v} = \{ \begin{matrix}T_{wv}, & if ℓ(wv) = ℓ(w) + 1, \\ qT_{wv} + (q − 1)T_{w}, & otherwise, \end{matrix}

where v is a basic transposition in 𝔖_{r} and w ∈ 𝔖_{r}.

If μ ⊧ r, then let

(3.1)  x_{μ} = ∑_{w∈𝔖_{μ}} T_{w}  and  y_{μ} = ∑_{w∈𝔖_{μ}}(−q)^{−ℓ(w)}T_{w}.

We are interested in homomorphisms defined on the right ideals x_{μ}ℋ of ℋ .

For λ, μ ⊧ r and d ∈ 𝒟_{λμ}, the map φ_{λμ}^{d} is defined by

φ_{λμ}^{d}(x_{μ}h) = ∑_{w∈𝔖_{λ}d𝔖_{μ}} T_{w}h, (h ∈ ℋ ).

Then φ_{λμ}^{d} ∈ Hom_{ℋ} (x_{μ}ℋ , x_{λ}ℋ ), and it is easy to see that { φ_{λμ}^{d} : d ∈ 𝒟_{λμ} } is a basis for this R–module.

Recall from (2.5) that the elements of 𝒟_{λμ} are indexed by the row–standard λ–tableaux of type μ. If d ∈ 𝒟_{λμ} corresponds to a row–standard tableau s in 𝒯 (λ, μ) we define 1_{s} = d^{−1} ∈ 𝒟_{μλ}. Then φ_{μλ}^{1_{s}} is an element of Hom_{ℋ} (x_{λ}ℋ , x_{μ}ℋ ) and

{ φ_{μλ}^{1_{s}} : s ∈ 𝒯 (λ, μ) is row–standard }

is a basis for this R–module.

(3.2) Example Hereafter, ω denotes the partition (1^{r}). Then x_{ω}ℋ = ℋ . For each h′ ∈ ℋ , the map which sends h ∈ ℋ to h′h is an ℋ –endomorphism of ℋ considered as a right ℋ –module. Moreover, { φ_{ωω}^{x} : x ∈ 𝔖_{r} } is a basis of Hom_{ℋ} (ℋ , ℋ ); so we identify ℋ with Hom_{ℋ} (ℋ , ℋ ) via the map T_{x} ⟼ φ_{ωω}^{x} (x ∈ 𝔖_{r}).

Let Λ(n, r) denote the set of compositions μ of r such that μ_{i} = 0 for all i > n. The algebra

S_{R}(n, r) = End_{ℋ} ( ⨁_{μ∈Λ(n,r)} x_{μ}ℋ )

is the q–Schur algebra [7, 8]. The element φ_{μλ}^{1_{s}} of Hom_{ℋ} (x_{λ}ℋ , x_{μ}ℋ ) is regarded as an element of S_{R}(n, r) by defining its action on x_{ν}ℋ to be zero for any composition ν ≠ λ. Then

{ φ_{μλ}^{1_{s}} : λ, μ ∈ Λ(n, r) and s ∈ 𝒯 (λ, μ) is row–standard }

is a basis of S_{R}(n, r).

In order to define the q–Weyl modules, recall from [8] that q–tensor space is the left ideal S_{R}(n, r) φ_{ωω}^{1} of the q–Schur algebra. An R–basis of q–tensor space is

{ φ_{μω}^{1_{t}} : μ ∈ Λ(n, r) and t ∈ 𝒯 (μ, (1^{r})) }.

Given λ ⊢ r, let w_{λ} be unique permutation such that t^{λ}w_{λ} has the numbers 1, 2, …, r

entered in order down its columns. So, for example, if t = 1 2 3 4 5 then t^{λ}w_{λ} = 1 3 5 2 4 ; so

w_{λ} = (2, 3, 5, 4).

Let z_{λ} = φ_{λω}^{1}T_{w_{λ}}y_{λ′}, where, as usual, λ′ is the partition conjugate to λ and y_{λ′} is defined in (3.1). Then z_{λ} is an element of q–tensor space and the right ℋ –module generated by z_{λ} is isomorphic to the right ideal of ℋ generated by x_{λ}T_{w_{λ}}y_{λ′}, which is called the q–Specht module S^{λ}. The left ℋ –module generated by φ_{ωλ}^{1}z_{λ} is isomorphic to the left ideal of ℋ generated by x_{λ}T_{w_{λ}}y_{λ′} and is the dual q–Specht module S̃^{λ} (in [6] this module is denoted S̃^{λ′}). The left ideal of the q–Schur algebra generated by z_{λ} is the q–Weyl module W^{λ}.

For each s ∈ 𝒯_{0}(λ, μ), let e_{s} = φ_{μλ}^{1_{s}}z_{λ}. In [8, 8.1] it is proved that W^{λ} is a free R–module with basis

(3.3)  { e_{s} : μ ∈ Λ(n, r) and s ∈ 𝒯_{0}(λ, μ) }.

This basis is called the semistandard basis of W^{λ}.

Now, { φ_{μμ}^{1} : μ ∈ Λ(n, r) } is a set of orthogonal idempotents in S_{R}(n, r) whose sum is the identity element of S_{R}(n, r). If U is a left S_{R}(n, r)–module, then φ_{μμ}^{1}U is the μ– weight space of U corresponding to the q–weight μ. We are interested in the μ–weight spaces W_{μ}^{λ} = φ_{μμ}^{1}W^{λ} of the q–Weyl module W^{λ}; these spaces also have a semistandard basis.

(3.4)  { e_{s} : s ∈ 𝒯_{0}(λ, μ) } is a basis of W_{μ}^{λ}.

It is the main aim of this section to calculate the determinant of the Gram matrix, with respect to the semistandard basis, of W_{μ}^{λ} (we define an inner product on W^{λ} below).

Before we begin this calculation we introduce two results which will be needed in the next section. The first of these is taken from [8, 8.8,8.9].

(3.5) Suppose that R is a field. Then every q-Weyl module W^{λ} has a unique top composition factor F^{λ}, and { F^{λ} : λ ⊢ r } is a complete set of non–isomorphic irreducible S_{R}(n, r)–modules. Moreover, F^{λ} is a composition factor of W^{λ} with multiplicity 1, and every composition factor of W^{λ} is of the form F^{ν} with λ ⊵ ν.

Next, we generalize a well known property of the ordinary Weyl modules. By way of notation, write

U ⟷ ∑_{λ⊢r} a_{λ}W^{λ} (a_{λ} ∈ ℤ)

to mean that

U ⊕ ⨁_{a_{λ}<0^{λ⊢r}}(−a_{λ})W^{λ}  and  ⨁_{a_{λ}>0^{λ⊢r}} a_{λ}W^{λ}

have the same composition factors.

(3.6) Lemma Assume that R is a field and that U is a left S_{R}(n, r)–module. Suppose that for all μ ⊢ r,

dim_{R} φ_{μμ}^{1}U = ∑_{λ⊢r} a_{λ} dim_{R} W_{μ}^{λ}.

Then U ⟷ ∑_{λ⊢r} a_{λ}W^{λ}.

Proof: From (3.5) we deduce that U ⟷ ∑_{λ⊢r} c_{λ}W^{λ} for some integers c_{λ}; so, for all μ ⊢ r,

0 = ∑_{λ⊢r}(a_{λ} − c_{λ}) dim_{R} W_{μ}^{λ} = ∑_{λ⊢r}(a_{λ} − c_{λ})|𝒯_{0}(λ, μ)|.

Now, |𝒯_{0}(λ, λ)| = 1 and |𝒯_{0}(λ, μ)| = 0 if λ ⋭ μ, so the matrix of coefficients |𝒯_{0}(λ, μ)|, as λ and μ run over all partitions of r, is unitriangular. Therefore, a_{λ} = c_{λ} for every λ ⊢ r.

Now we begin our investigation of the weight spaces of the q–Weyl module W^{λ}. Of these, the ω-weight space, where ω = (1^{r}), is the most easy to understand so we study maps between W_{μ}^{λ} and W_{ω}^{λ} for all μ ⊧ r. In order to do this we first consider the maps φ_{ωμ}^{1} and φ_{μω}^{1}. The reader may find it helpful to refer back to the results and definitions of (2.2)–(2.5).

(3.7) Lemma Suppose that s ∈ 𝒯 (λ, μ) is row–standard. Then

φ_{ωμ}^{1}φ_{μλ}^{1_{s}} = ∑_{t∈μ^{−1}(s)} φ_{ωλ}^{1_{t}}.

Proof: For all h ∈ ℋ ,

φ_{ωμ}^{1}φ_{μλ}^{1_{s}}(x_{λ}h) = ∑_{w∈𝔖_{μ}1_{s}𝔖_{λ}} T_{w}h = ∑_{t∈μ^{−1}(s)} ∑_{w∈1_{t}𝔖_{λ}} T_{w}h,

since 𝔖_{μ}1_{s}𝔖_{λ} = ∪_{t∈μ^{−1}(s)}1_{t}𝔖_{λ} by (2.5). Therefore, φ_{ωμ}^{1}φ_{μλ}^{1_{s}} = ∑_{t∈μ^{−1}(s)} φ_{ωλ}^{1_{t}} as claimed.

For the next result we recall from [8, 1.11], that there is an antiautomorphism ∗ of S_{R}(n, r) which is given by

(3.8)  (φ_{λμ}^{d})^{∗} = φ_{μλ}^{d^{−1}},  (λ, μ ∈ Λ(n, r), and d ∈ 𝒟_{λμ}).

Note that T_{w}^{∗} = T_{w^{−1}}; consequently, x_{μ}^{∗} = x_{μ} for all μ ⊧ r.

(3.9) Lemma Suppose that t ∈ 𝒯 (λ, (1^{r})) is row–standard and let s = μ(t). Write t = t^{λ}dw, where t^{λ}d = first(s) and w ∈ 𝔖_{μ} and let ν = (ν_{1}, ν_{2}, …) be the composition of r given by 𝔖_{ν} = d^{−1}𝔖_{λ}d ∩ 𝔖_{μ}. Then

φ_{μω}^{1}φ_{ωλ}^{1_{t}} = (q^{ℓ(w)} ∏_{i=1}^{∞}{ν_{i}}_{q}) φ_{μλ}^{1_{s}}.

Proof: We have that ℓ(dw) = ℓ(d) + ℓ(w) by (2.2), and T_{w}x_{μ} = q^{ℓ(w)}x_{μ} since w ∈ 𝔖_{μ}. Therefore,

x_{λ}T_{dw}x_{μ} = x_{λ}T_{d}T_{w}x_{μ} = q^{ℓ(w)}x_{λ}T_{d}x_{μ}.

If I = {i + 1, i + 2, …, i + j} is a subset of a row in t^{λ}, then x_{λ} ∑_{w∈𝔖(I)} T_{w} = {j}_{q}x_{μ}. Hence, by (2.2),

x_{λ}T_{d}x_{μ} = ∏_{i=1}^{∞}{ν_{i}}_{q} ∑_{w∈𝔖_{λ}d𝔖_{μ}} T_{w}.

Therefore,

x_{μ}T_{1_{t}}x_{λ} = (x_{λ}T_{dw}x_{μ})^{∗} = q^{ℓ(w)} ∏_{i=1}^{∞}{ν_{i}}_{q} ∑_{w∈𝔖_{μ}1_{t}𝔖_{λ}} T_{w} = q^{ℓ(w)} ∏_{i=1}^{∞}{ν_{i}}_{q} ∑_{w∈𝔖_{μ}1_{s}𝔖_{λ}} T_{w}.

This shows that x_{λ} has the same image under the two homomorphisms which appear in the statement of the lemma, so the homomorphisms are equal.

We use Lemma 3.9 to show that φ_{μω}^{1} acts as a projection operator, mapping W_{ω}^{λ} onto W_{μ}^{λ}.

(3.10) Lemma Suppose that t ∈ 𝒯_{0}(λ, (1^{r})), and let s = μ(t). Write t = t^{λ}dw, where t^{λ}d = first(s) and w ∈ 𝔖_{μ}, and let ν = (ν_{1}, ν_{2}, …) be the composition of r which is given by 𝔖_{ν} = d^{−1}𝔖_{λ}d ∩ 𝔖_{μ}.

(i) If s ∈ 𝒯_{0}(λ, μ) then φ_{μω}^{1}e_{t} = (q^{ℓ(w)} ∏_{i=1}^{∞}{ν_{i}}_{q}) e_{s}.

(ii) If s ∉ 𝒯_{0}(λ, μ) then φ_{μω}^{1}e_{t} is a linear combination of terms e_{s′} where s′ ∈ 𝒯_{0}(λ, μ) and s′ ▷ s.

Proof: By definition, φ_{μω}^{1}e_{t} = φ_{μω}^{1}φ_{ωλ}^{1_{t}}z_{λ}, so

φ_{μω}^{1}e_{t} = q^{ℓ(w)} ∏_{i=1}^{∞}{ν_{i}}_{q}φ_{μλ}^{1_{s}}z_{λ}

by Lemma 3.9. If s ∈ 𝒯_{0}(λ, μ), then φ_{μλ}^{1_{s}}z_{λ} = e_{s}. If s ∉ 𝒯_{0}(λ, μ), then the construction of the semistandard basis of W^{λ}, given in [8], shows that φ_{μλ}^{1_{s}}z_{λ} is a linear combination of the terms e_{s′} where s′ ∈ 𝒯_{0}(λ, μ) and s′ ▷ s.

Next, we introduce some elements of the q–Schur algebra which are generalizations of the Murphy operators [19, 6].

(3.11) Definition Let μ = (μ_{1}, μ_{2}, …) be a composition of r and suppose that μ_{m} > 0. For each i with μ_{i} > 0, let x_{i} be the last entry in the i–th row of t^{μ}. Let y be the first entry in the m–th row of t^{μ}, and note that (x_{i}, y) ∈ 𝒟_{μμ} by (2.6). Define L_{μ,m} by

L_{μ,m} = ∑_{μ_{i}>0^{i=1}}^{m−1} q^{−(1+μ_{i+1}+μ_{i+2}+···+μ_{m−1})}φ_{μμ}^{(x_{i},y)} + ([1]_{q} + [2]_{q} + ⋯ + [μ_{m} − 1]_{q})φ_{μμ}^{1}.

(3.12) Example Let μ = (4, 4, 3). Then t^{μ} = 1 2 3 4 5 6 7 8 9 10 11 and

\begin{aligned}L_{μ,2} & = & q^{−1}φ_{μμ}^{(4,5)} + ([1]_{q} + [2]_{q} + [3]_{q})φ_{μμ}^{1}, \\ L_{μ,3} & = & q^{−5}φ_{μμ}^{(4,9)} + q^{−1}φ_{μμ}^{(8,9)} + ([1]_{q} + [2]_{q})φ_{μμ}^{1}. \end{aligned}

Note that L_{ω,m} is the q–Murphy operator L_{m} of ℋ [6]. That is,

L_{ω,m} = L_{m} = q^{−(m−1)}T_{(1,m)} + q^{−(m−2)}T_{(2,m)} + ⋯ + q^{−1}T_{(m−1,m)}.

Our next result relates the operators L_{μ,m} and L_{m}; the rationale being that we understand the action of the q–Murphy operators on the (1^{r})–weight space of W^{λ}, so we can understand how L_{μ,m} acts on W_{μ}^{λ} by pulling the calculation back to W_{ω}^{λ}.

(3.13) Lemma Suppose that μ ⊧ r and that μ_{m} > 0. Let y be the first entry in the m–th row of t^{μ} and let k = μ_{m}. Then

φ_{μω}^{1}(L_{y} + L_{y+1} + ⋯ + L_{y+k−1}) = L_{μ,m}φ_{μω}^{1}.

Proof: For this proof we adopt the notation that if w ∈ 𝔖_{r} then \overline{w} is the unique element of minimal length in the coset 𝔖_{μ}w.

Suppose that 0 ≤ u ≤ k − 1. The definition of L_{y+u} gives

x_{μ}L_{y+u} = ∑_{j=1}^{y+u−1} q^{−(y+u−j)}x_{μ}T_{(j,y+u)}.

Define ρ(j) to be the row index of j in t^{μ} and let f(j) be the number of entries to the right of j in this row.

Assume first that ρ(j) < m. Then, by (2.7), x_{μ}T_{(j,y+u)} = q^{u+f(j)}x_{μ}T\overline{(j,y+u)}, so

q^{−(y+u−j)}x_{μ}T_{(j,y+u)} = q^{j+f(j)−y}x_{μ}T\overline{(j,y+u)} = q^{−(1+μ_{ρ(j)+1}+···+μ_{m−1})}x_{μ}T\overline{(j,y+u)},

since y = μ_{1} + ⋯ + μ_{m−1} + 1 and j + f(j) = μ_{1} + ⋯ + μ_{ρ(j)}.

Next, suppose that ρ(j) = m and j ≤ y + u − 1. Then q^{−(y+u−j)}x_{μ}T_{(j,y+u)} = q^{y+u−j−1}x_{μ} since ℓ(j, y + u) = 2y + 2u − 2j − 1. Therefore,

∑_{j=y}^{y+u−1} q^{−(y+u−j)}x_{μ}T_{(j,y+u)} = [u]_{q}x_{μ}.

So the action of φ_{μω}^{1}(L_{y} + ⋯ + L_{y+k−1}) on an element h of ℋ is given by

h ⟼ \{∑_{i=1}^{m−1} ∑_{a∈row i} ∑_{b∈row m} q^{−(1+μ_{i+1}+···+μ_{m−1})}x_{μ}T\overline{(a,b)} + ([1]_{q} + [2]_{q} + ⋯ + [k − 1]_{q})x_{μ}\} h,

where a ∈ row i means that ρ(a) = i and b ∈ row m means that ρ(b) = m. By (2.6) and the definitions, φ_{μμ}^{(x_{i},y)} is the map sending

x_{μ}h ⟼ ∑_{a∈row i} ∑_{b∈row m} x_{μ}T\overline{(a,b)}h,

so we deduce that φ_{μω}^{1}(L_{y} + L_{y+1} + ⋯ + L_{y+k−1}) = L_{μ,m}φ_{μω}^{1}, as required.

(3.14) Corollary Let μ be a composition of r. Then, for all a and b for which L_{μ,a} and L_{μ,b} are defined, L_{μ,a} and L_{μ,b} commute.

Proof: By [6, 2.2] the q–Murphy operators L_{m} commute. Therefore, by Lemma 3.13, L_{μ,a}L_{μ,b}φ_{μω}^{1} = L_{μ,b}L_{μ,a}φ_{μω}^{1}, and hence L_{μ,a} and L_{μ,b} commute.

From here until Theorem 3.26 we take our ground ring R to be 𝔽(q), where 𝔽 is an arbitrary field and q is an indeterminate. Thus we work with the Hecke algebra ℋ_{𝔽(q),q} and the corresponding q–Schur algebra. The case of Theorem 3.16 below, when R is any integral domain, follows from the case where R = 𝔽(q); however, in Definition 3.18 we will need to avoid certain rings.

As in [6, p. 71], define the q-residue of the (i, j)–th node of a diagram to be [j − i]_{q} if j ≥ i and −q^{j−i}[i − j]_{q} if i > j. The pattern of q–residues in a diagram is illustrated below.

0  1  1 + q  1 + q + q^{2} …
−q^{−1}  0  1  1 + q  …
−q^{−1} − q^{−2} −q^{−1}  0  ⋮
⋮  ⋮

If s ∈ 𝒯 (λ, μ) and m is an entry in s, let r_{s}(m) be the sum of the q–residues of the nodes which m occupies in s.

For example, if

s = 1 1 2 3 2 3 4 4 ,

then r_{s}(1) = 1, r_{s}(2) = −q^{−1} + 1 + q, r_{s}(3) = 1 + q + q^{2}, and r_{s}(4) = −2q^{−1} − q^{−2}. It is an easy exercise to prove the following result.

(3.15) If s and s′ are distinct tableaux in 𝒯_{0}(λ, μ) then r_{s}(m) ≠ r_{s′}(m) for some m.

Given s ∈ 𝒯_{0}(λ, μ), an expression such as ∑_{u▷s} a_{u}e_{u}, with a_{u} ∈ 𝔽(q), denotes a linear combination of terms e_{u} with u ∈ 𝒯_{0}(λ, μ) and u ▷ s.

(3.16) Theorem Let s ∈ 𝒯_{0}(λ, μ) and suppose that μ_{m} > 0. Then

L_{μ,m}e_{s} = r_{s}(m)e_{s} + ∑_{u▷s}a_{u}e_{u},

for some scalars a_{u} ∈ 𝔽(q).

Proof: The special case where μ = (1^{r}) is a version of the main property of the q–Murphy operators. Although this property is normally stated in terms of the q–Specht modules (cf. [6, 3.15] and [20, 4.6]), similar arguments prove the result for the dual q–Specht modules, and this is the version which we use here.

Let t = first(s). Then, by Lemma 3.10(i), e_{s} = π^{−1}φ_{μω}^{1}e_{t}, where π = ∏_{i=1}^{∞}{ν_{i}}_{q} for some composition ν of r. Let y and k be defined as in Lemma 3.13; then

L_{μ,m}e_{s} = π^{−1}L_{μ,m}φ_{μω}^{1}e_{t} = π^{−1}φ_{μω}^{1}(L_{y} + ⋯ + L_{y+k−1})e_{t}.

Now, L_{y+i}e_{t} = r_{t}(y + i)e_{t}+ a linear combination of terms e_{v} where v ∈ 𝒯_{0}(λ, (1^{r})) and v ▷ t because, as we remarked above, we know that the Theorem holds when t ∈ 𝒯_{0}(λ, (1^{r})). But, r_{t}(y) + ⋯ + r_{t}(y + k − 1) = r_{s}(m), so

L_{μ,m}e_{s} = π^{−1}φ_{μω}^{1} (r_{s}(m)e_{t} + ∑_{v▷t}b_{v}e_{v})

for some b_{v} ∈ 𝔽(q). Now, if v ▷ t then μ(v) ▷ μ(t); so, by applying Lemma 3.10, we obtain the result.

There is an inner product < , > defined on q–tensor space by

<φ_{μω}^{d}, φ_{νω}^{d′}> = \{ \begin{matrix}q^{ℓ(d)}, & if μ = ν and d = d′, \\ 0, & otherwise. \end{matrix}

Here μ, ν ∈ Λ(n, r), d ∈ 𝒟_{μ} and d′ ∈ 𝒟_{ν}. The inner product behaves nicely with respect to the antiautomorphism ∗ defined in (3.8). Namely,

(3.17)  <s^{∗}u, v> = <u, sv>  and  <uh^{∗}, v> = <u, vh>,

for all s ∈ S_{R}(n, r) and h ∈ ℋ , where u and v are elements of q–tensor space (see [8, 4.1]).

We now use Theorem 3.16 to transform the semistandard basis of W^{λ} into a basis of orthogonal vectors (cf. [19]).

(3.18) Definition Assume that R = 𝔽(q) and let s ∈ 𝒯_{0}(λ, μ). Define

E_{s} = ∏_{m with μ_{m}≠0} ∏_{r_{s′}(m)≠r_{s}(m)^{s′∈𝒯_{0}(λ,μ)}} \frac{L_{μ,m} − r_{s′}(m)}{r_{s}(m) − r_{s′}(m)},

and let f_{s} = E_{s}e_{s}.

Since L_{μ,a} commutes with L_{μ,b} by Corollary 3.14, there is no need to specify the order of the products in the definition of E_{s}.

(3.19) Lemma Assume that R = 𝔽(q) and that s and s′ are distinct elements of 𝒯_{0}(λ, μ). Then the following hold.

(i) Ł_{μ,m}f_{s} = r_{s}(m)f_{s}; (ii) f_{s} = e_{s} + ∑_{u▷s} a_{u}e_{u} for some a_{u} ∈ 𝔽(q); (iii) E_{s′}e_{s} = 0 if s ▷ s′; (iv) E_{s}f_{s} = f_{s} and E_{s}f_{s′} = 0.

Proof: The stated results follow from Theorem 3.16 by straightforward techniques (see [19, (3.4)–(3.11)] and [6, 4.6]). In order to prove that E_{s}f_{s′} = 0, it is important that if s ≠ s′ then r_{s}(m) ≠ r_{s′}(m) for some m, as we noted in (3.15). We leave the reader to check the details.

Now, L_{μ,m}^{∗} = L_{μ,m} where ∗ is the antiautomorphism of (3.8); so E_{s}^{∗} = E_{s} for all s in 𝒯_{0}(λ, μ). Therefore, by (3.17) and Lemma 3.19(iv), if s and s′ are distinct semistandard tableaux then

<f_{s}, f_{s′}> = <E_{s}f_{s}, f_{s′}> = <f_{s}, E_{s}f_{s′}> = 0.

Combining this observation with Lemma 3.19(ii), we deduce the following.

(3.20) { f_{s} : s ∈ 𝒯_{0}(λ, μ) } is an orthogonal basis of W_{μ}^{λ}.

Our next aim is to find the inner products <f_{s}, f_{s}> for all s ∈ 𝒯_{0}(λ, μ).

(3.21) Lemma Suppose that t and t′ ∈ 𝒯_{0}(λ, (1^{r})) and that t′ = t(m, m + 1). Assume that t_{ij} = m + 1, t_{kl} = m and that i < k. Then

T_{(m,m+1)}f_{t′} = \frac{−1}{[ρ]_{q}} f_{t′} + f_{t},

where ρ = j − i + k − l.

Proof: The proof of this result is almost identical to that of [6, 4.9], so we omit it. The coefficient here is different from that in [6, 4.9] because the ordering of the tableaux has been reversed.

For the next lemma, recall the rational function π_{t}^{μ} of Definition 2.12. In this lemma we study the projection of the orthogonal basis of W_{ω}^{λ} onto the orthogonal basis of W_{μ}^{λ}.

(3.22) Lemma Suppose that t ∈ 𝒯_{0}(λ, (1^{r})) and that s = μ(t) ∈ 𝒯_{0}(λ, μ). Then φ_{μω}^{1}f_{t} = π_{t}^{μ}f_{s}.

Proof: We show first that φ_{μω}^{1}f_{t} is a scalar multiple of f_{s}. Using the notation in the statement of Lemma 3.13, we have, for all m with μ_{m} > 0,

L_{μ,m}φ_{μω}^{1}f_{t} = φ_{μω}^{1}(L_{y} + ⋯ + L_{y+k−1})f_{t}  by Lemma 3.13
= φ_{μω}^{1}(r_{t}(y) + ⋯ + r_{t}(y + k − 1))f_{t}, by Lemma 3.19(i),
= r_{s}(m)φ_{μω}^{1}f_{t}.

Hence, E_{s}φ_{μω}^{1}f_{t} = φ_{μω}^{1}f_{t}. On the other hand, φ_{μω}^{1}f_{t} is a linear combination of terms f_{u} with u ∈ 𝒯_{0}(λ, μ) by (3.20); so Lemma 3.19(iv) now implies that φ_{μω}^{1}f_{t} is a multiple of f_{s}.

Write φ_{μω}^{1}f_{t} = b_{t}f_{s}. Then it remains to show that b_{t} = π_{t}^{μ}. To start with, consider the case where t = first(s). Then, by Lemma 3.19(ii), f_{t} = e_{t}+ a linear combination of terms e_{v} with μ(v) ▷ μ(t). Note also that φ_{μω}^{1}e_{t} = π_{t}^{μ}e_{s} by Lemma 3.10(i) and Example 2.15. Therefore, Lemma 3.19(ii) implies that

φ_{μω}^{1}f_{t} = π_{t}^{μ}e_{s} + ∑_{u▷s}a_{u}e_{u},

for some a_{u} ∈ 𝔽(q). However, since we already know that φ_{μω}^{1}f_{t} is a multiple of f_{s}, Lemma 3.19(ii) shows that φ_{μω}^{1}f_{t} = π_{t}^{μ}f_{s} in this case.

Now suppose that t ≠ first(s). Then there exists an integer m such that if t′ = t(m, m+1) then t′ ▷ t and μ(t) = μ(t′). So, if t_{ij} = m+1 and t_{kl} = m then i < k and j > l. By induction, φ_{μω}^{1}f_{t′} = π_{t′}^{μ}f_{s}. Also, because m and m + 1 are in the same row of t^{μ}, the map φ_{μω}^{1}T_{(m,m+1)} sends an element h in ℋ to x_{μ}T_{(m,m+1)}h = qx_{μ}h. Therefore, φ_{μω}^{1}T_{(m,m+1)} = qφ_{μω}^{1}.

Let ρ = j −i+k−l. Then, by Lemma 3.21, T_{(m,m+1)}f_{t′} = \frac{−1}{[ρ]_{q}} f_{t′} +f_{t}. Writing φ_{μω}^{1}f_{t} = b_{t}f_{s} as above, we have

qπ_{t′}^{μ}f_{s} = φ_{μω}^{1}T_{(m,m+1)}f_{t′} = φ_{μω}^{1} (\frac{−1}{[ρ]_{q}} f_{t′} + f_{t}) = (\frac{−π_{t′}^{μ}}{[ρ]_{q}} + b_{t}) f_{s}.

So, b_{t} = \frac{[ρ+1]_{q}}{[ρ]_{q}} π_{t′}^{μ} = π_{t}^{μ}; the last equality following from Example 2.14.

In contrast with Lemma 3.22, our next step is easy.

(3.23) Lemma Suppose that s ∈ 𝒯_{0}(λ, μ) and let t = last(s). Then there exists scalars a_{u} ∈ 𝔽(q) such that

φ_{ωμ}^{1}f_{s} = f_{t} + ∑_{u▷t}a_{u}f_{u}.

Proof: Using the definitions and Lemma 3.7, we see that

φ_{ωμ}^{1}e_{s} = φ_{ωμ}^{1}φ_{μλ}^{1_{s}}z_{λ} = ∑_{u∈μ^{−1}(s)} φ_{ωλ}^{1_{u}}z_{λ} = ∑_{u∈μ^{−1}(s)} e_{u} = e_{t} + terms e_{u} with u ▷ t,

since t = last(s). Therefore, there exist coefficients b_{v}, b′_{u}, and a_{u} ∈ 𝔽(q) such that

φ_{ωμ}^{1}f_{s} = φ_{ωμ}^{1}(e_{s} + ∑_{v▷s}b_{v}e_{v}), by Lemma 3.19(ii)
= e_{t} + ∑_{u▷t}b′_{u}e_{u}
= f_{t} + ∑_{u▷t}a_{u}f_{u},  by Lemma 3.19(ii).

Before we come to the main results of this section, we contract the inner product < , > to the left ideal of S_{R}(n, r) which is generated by y_{λ′}. To do this we define

(3.24)  ≪uy_{λ′}, vy_{λ′}≫ = <u, vy_{λ′}>,

for all u, v ∈ S_{R}(n, r) (cf. [9, § 5.5], [8, 4.3]).

Now recall the rational function γ_{s} from Definition 2.8.

(3.25) Theorem Suppose that s ∈ 𝒯_{0}(λ, μ). Then ≪f_{s}, f_{s}≫ = q^{a_{s}}γ_{s} for some a_{s} ∈ ℤ.

Proof: We consider first the case where μ = (1^{r}); here the result is essentially the same as in [6, 4.10], but we give the proof.

Let t = t^{λ} ∈ 𝒯_{0}(λ, (1^{r})). Then f_{t} = e_{t} = φ_{ωλ}^{1_{t}}x_{λ}T_{w_{λ}}y_{λ′} = x_{λ}T_{w_{λ}}y_{λ′}, and

≪f_{t}, f_{t}≫ = <x_{λ}T_{w_{λ}}, x_{λ}T_{w_{λ}}y_{λ′}> = <x_{λ}T_{w_{λ}}, x_{λ}T_{w_{λ}}>,

using (2.2) and the fact that w_{λ}^{−1}𝔖_{λ}w_{λ} ∩ 𝔖_{λ′} = {1}. Therefore,

≪f_{t}, f_{t}≫ = q^{ℓ(w_{λ})}<x_{λ}, x_{λ}> = q^{ℓ(w_{λ})} ∏_{i=1}^{∞}{λ_{i}}_{q} = q^{ℓ(w_{λ})}γ_{t},

by Example 2.10.

Now assume that t ∈ 𝒯_{0}(λ, (1^{r})) but t ≠ t^{λ}. Then there exist integers m and m + 1 such that if t_{ij} = m + 1 and t_{kl} = m then i < k and j > l. Let t′ = t(m, m + 1). By induction, ≪f_{t′}, f_{t′}≫ = q^{a_{t′}}γ_{t′} for some a_{t′} ∈ ℤ. By Lemma 3.21, T_{(m,m+1)}f_{t′} = \frac{−1}{[ρ]_{q}} f_{t′} + f_{t}, where ρ = j − i + k − l. Therefore, noting that ≪f_{t′}, f_{t}≫ = 0 by (3.20),

\frac{1}{[ρ]_{q}^{2}} ≪f_{t′}, f_{t′}≫ + ≪f_{t}, f_{t}≫ = ≪T_{(m,m+1)}f_{t′}, T_{(m,m+1)}f_{t′}≫
= ≪f_{t′}, T_{(m,m+1)}^{2}f_{t′}≫  by (3.17),
= q≪f_{t′}, f_{t′}≫ + (q − 1)≪T_{(m,m+1)}f_{t′}, f_{t′}≫
= q≪f_{t′}, f_{t′}≫ − \frac{1}{[ρ]_{q}} (q − 1)≪f_{t′}, f_{t′}≫.

Therefore,

≪f_{t}, f_{t}≫ = \frac{q[ρ]_{q}^{2} − (q − 1)[ρ]_{q} − 1}{[ρ]_{q}^{2}}≪f_{t′}, f_{t′}≫
= \frac{q[ρ + 1]_{q}[ρ − 1]_{q}}{[ρ]_{q}^{2}} q^{a_{t′}}γ_{t′}
= q^{a_{t′}+1}γ_{t};

the last equality coming from Example 2.11. This completes the proof of the theorem in the case where μ = (1^{r}).

Assume now that μ ≠ (1^{r}), let s ∈ 𝒯_{0}(λ, μ), and set t = last(s). Then Lemma 3.22, (3.17), and Lemma 3.23 give

≪f_{t}, f_{t}≫ = ≪φ_{ωμ}^{1}f_{s}, f_{t}≫ = ≪f_{s}, φ_{μω}^{1}f_{t}≫ = π_{t}^{μ}≪f_{s}, f_{s}≫.

Since ≪f_{t}, f_{t}≫ = q^{a_{t}}γ_{t}, Example 2.16 shows that ≪f_{s}, f_{s}≫ = q^{a_{t}}γ_{s} as required.

We have now reached the main technical result of the paper. Choose a total order < on 𝒯_{0}(λ, μ) such that s < s′ if s′ ▷ s where s, s′ ∈ 𝒯_{0}(λ, μ). Let 𝒢_{λμ} be the matrix whose entries are the inner products ≪e_{s}, e_{s′}≫ , where the rows and columns of 𝒢_{λμ} are ordered by <. That is to say, 𝒢_{λμ} is the Gram matrix of W_{μ}^{λ} with respect to the semistandard basis { e_{s} : s ∈ 𝒯_{0}(λ, μ) }.

We show that det 𝒢_{λμ} is given by the rational function Δ_{μ}(λ) which was introduced in Definition 2.17, and calculated in Theorem 2.28 and Corollary 2.30.

(3.26) Theorem Assume that R = 𝔽(q) and that λ ⊢ r and μ ⊧ r. Then

det 𝒢_{λμ} = q^{a}Δ_{μ}(λ)

for some integer a.

Proof: By Lemma 3.19(ii), the transition matrix from the semistandard basis {e_{s}} of W_{μ}^{λ} to the orthogonal basis {f_{s}} is unitriangular and hence has determinant 1. Therefore, because the vectors f_{s} are orthogonal,

det 𝒢_{λμ} = ∏_{s∈𝒯_{0}(λ,μ)} ≪f_{s}, f_{s}≫ = q^{a} ∏_{s∈𝒯_{0}(λ,μ)} γ_{s}

for some a ∈ ℤ by Theorem 3.25. The Theorem now follows from the Definition 2.17 of Δ_{μ}(λ).

(3.27) Corollary The rational function Δ_{μ}(λ) belongs to ℤ[q]. If R is an arbitrary integral domain, and q is an invertible element of R, then the determinant of the Gram matrix 𝒢_{λμ} is equal to q^{a}Δ_{μ}(λ), for some a ∈ ℤ, regarded as an element of R.

Proof: Assume first that R = ℚ(q), where q is transcendental. It is clear from the definitions that ≪e_{s}, e_{s′}≫ ∈ ℤ[q, q^{−1}] for all s, s′ ∈ 𝒯_{0}(λ, μ). Therefore Δ_{μ}(λ) ∈ ℤ[q, q^{−1}]. However, Δ_{μ}(λ), as given in Theorem 2.28, can be calculated when q = 0 (where its value is 1), so Δ_{μ}(λ) ∈ ℤ[q].

Now suppose that R is an arbitrary integral domain, and that \overline{q} is an invertible element in R. We work with ℋ_{R,\overline{q}}(𝔖_{r}). For each f ∈ ℤ[q, q^{−1}], let \overline{f} denote the element of R obtained from f by replacing q by \overline{q}, and 1 by the identity of R. If g_{s,s′} is the (s, s′)–th entry in the Gram matrix 𝒢_{λμ} defined over ℚ(q), where s, s′ ∈ 𝒯_{0}(λ, μ), then \overline{g_{s,s′}} is the (s, s′)–th entry in the Gram matrix \overline{𝒢_{λμ}} defined over R. Therefore, the determinant of the Gram matrix over R is \overline{q^{a}Δ_{μ}(λ)}, as we wished to show.

We do not know any direct way to prove that the rational function Δ_{μ}(λ), given by Theorem 2.28 or Corollary 2.30 is, in fact, a polynomial.

4 The main results

A technique, due to Jantzen, now enables us to derive information about the composition factors of the q–Weyl modules and q-Specht modules from the formulae we have obtained for the determinant of the Gram matrix 𝒢_{λμ} of W_{μ}^{λ}.

Suppose that R is a principal ideal domain and that 𝔭 is a prime in R. Note that R/𝔭R is a field and let ν_{𝔭}: R \ {0}⟶ℤ be the natural valuation on R associated with (R, 𝔭); thus, for x ∈ R, ν_{𝔭}(x) = i where i is the largest integer such that 𝔭^{i} divides x.

Let M be a free R–module of finite rank, equipped with a non–singular bilinear form ≪ , ≫. Denote by M(i) the R–submodule of M which consists of all elements m such that 𝔭^{i} divides ≪m, m′≫ for all m′ ∈ M. If U is an R–submodule of M, let \overline{U} denote the R/𝔭R–submodule (U + 𝔭U)/𝔭U of \overline{M} = M/𝔭M. Then

\overline{M} = \overline{M(0)} ≥ \overline{M(1)} ≥ \overline{M(2)} ≥ ⋯

gives a filtration of \overline{M} with \overline{M(i)} = 0 for all sufficiently large i.

Let e_{1}, …, e_{d} be an R-basis of M, and let 𝒢 be the Gram matrix whose (i, j)–th entry is ≪e_{i}, e_{j}≫. Then Jantzen’s [17, Lemma 3] key observation is that

(4.1)  ν_{𝔭}(det 𝒢) = ∑_{i>0} dim_{R/𝔭R} \overline{M(i)}.

The principal ideal domains R which are of particular interest for the q–Schur algebra are the following.

(4.2) (i) R = 𝔽[q, q^{−1}], where 𝔽 is an arbitrary field, q is transcendental over 𝔽, and 𝔭 is

a cyclotomic polynomial. (ii) R = 𝔽[q, q^{−1}], where 𝔽 is an arbitrary field, q is transcendental over 𝔽, and 𝔭 = q − \overline{q}

for some non–zero element \overline{q} of 𝔽. (iii) R = ℤ[q^{−1}], where q is an integer, and 𝔭 is a prime number which does not divide q.

In any case, we henceforth assume that R is a principal ideal domain in the definition of S_{R}(n, r), that 𝔭 is a prime in R, and that q is an invertible element in R such that ν_{𝔭}(q) = 0 and [a]_{q} ≠ 0 for all 1 ≤ a ≤ r. The last condition on q ensures that ≪ , ≫ is non–singular on W_{μ}^{λ} by Theorem 3.26.

To simplify the statements of our main results, we adopt conventions for the q–Weyl modules like those in Definition 2.18. Namely, given a sequence β_{1}, β_{2}, …, β_{n} of integers we define the (virtual) S_{R/𝔭R}(n, r)–module W (β_{1}, β_{2}, …, β_{n}) as follows.

(i) W (β_{1}, β_{2}, …, β_{n}) = W^{ν} if β_{1} > β_{2} > ⋯ > β_{n} ≥ 0 are β–numbers for the partition ν′; (ii) W (β_{1}, β_{2}, …, β_{n}) = 0 if β_{i} = β_{j}, for some i ≠ j, or β_{i} < 0 for some i; (iii) W (β_{1w}, β_{2w}, …, β_{nw}) = sign(w)W (β_{1}, β_{2}, …, β_{n}) for all w ∈ 𝔖_{n}.

We can now prove a q–analogue of Jantzen’s sum formula concerning the Weyl modules of general linear groups.

(4.3) Theorem Let λ ⊢ r and let α_{1} > α_{2} > ⋯ > α_{n} ≥ 0 be β–numbers for λ′. Then, as S_{R/𝔭R}(n, r)–modules,

∑_{i>0} \overline{W^{λ}(i)} ⟷ ∑_{(a,b)∈[λ] ; {(a,c)∈[λ] \atop b<c}} (ν_{𝔭}([h_{ab}]_{q}) − ν_{𝔭}([h_{ac}]_{q}))W (α_{1}, …, α_{b} + h_{ac}, …, α_{c} − h_{ac}, …, α_{n}).

Proof: Let μ ⊢ r and recall that 𝒢_{λμ} is the Gram matrix of W_{μ}^{λ} with respect to its semistandard basis. By Corollary 3.27, det 𝒢_{λμ} = q^{a}Δ_{μ}(λ) for some integer a. Therefore, applying (4.1) in the case where M = W_{μ}^{λ},

∑_{i>0} dim_{R/𝔭R} \overline{W_{μ}^{λ}(i)} = ν_{𝔭}(det 𝒢_{λμ}) = ν_{𝔭}(q^{a}Δ_{μ}(λ)) = ν_{𝔭}(Δ_{μ}(λ)),

since ν_{𝔭}(q^{a}) = 0. However, by Corollary 2.30,

ν_{𝔭}(Δ_{μ}(λ)) = ∑ (ν_{𝔭}([h_{ab}]_{q}) − ν_{𝔭}([h_{ac}]_{q})) dim (φ_{μμ}^{1}W (α_{1}, …, α_{b} + h_{ac}, …, α_{c} − h_{ac}, …, α_{n})) ,

where the sum is over all a, b, c as in the statement of the Theorem. The desired result now follows by Lemma 3.6.

The importance of Theorem 4.3 lies in the fact that \overline{W^{λ}(1)} is the unique maximal submodule of W^{λ} (defined over R/𝔭R) – see [8, 4.6]. Thus,

W^{λ} ⟷ F^{λ} + \overline{W^{λ}(1)},

and Theorem 4.3 gives us partial information about the composition factors F^{ν} of W^{λ} for ν ≠ λ.

(4.4) Example Suppose that λ = (2^{2}, 1). Then the hook lengths in [λ] are given in the diagram

4 2 3 1 1 .

Thus, Theorem 4.3 gives

∑_{i>0} \overline{W^{λ}(i)} = (ν_{𝔭}([4]_{q}) − ν_{𝔭}([2]_{q}))W^{(1^{5})} + ν_{𝔭}([3]_{q})W^{(21^{3})}.

Now, [4]_{q} = (1 + q^{2})[2]_{q}, so W^{(2^{2},1)} is irreducible unless 𝔭 divides (1 + q^{2}) or (1 + q + q^{2}). Assume for example that R = ℚ[q, q^{−1}] and that 𝔭 is a cyclotomic polynomial. Then

W^{(2^{2},1)} ⟷ F^{(1^{5})} + F^{(2^{2},1)}, if 𝔭 = 1 + q^{2},

since W^{(1^{r})} = F^{(1^{r})} for all r. Also,

W^{(2^{2},1)} ⟷ F^{(2,1^{3})} + F^{(2^{2},1)}, if 𝔭 = 1 + q + q^{2},

since Theorem 4.3 shows that W^{(2,1^{3})} = F^{(2,1^{3})} in this case.

(4.5) Theorem The q–Weyl module W^{λ}, defined over R/𝔭R, is irreducible if and only if ν_{𝔭}([h_{ab}]_{q}) = ν_{𝔭}([h_{ac}]_{q}) for all nodes (a, b) and (a, c) in [λ].

Proof: The Theorem follows at once from Theorem 4.3 since W^{λ} = F^{λ} if and only if \overline{W^{λ}(1)} = 0.

Next, we convert our theorems on the q–Weyl modules W^{λ} into results about the dual q–Specht modules S̃^{λ}. Note that φ_{ωω}^{1} is an idempotent in S_{R}(n, r) and that

(4.6)  φ_{ωω}^{1} S_{R}(n, r) φ_{ωω}^{1} = ℋ  and  φ_{ωω}^{1}W^{λ} = S̃^{λ}.

We may therefore apply the “eSe–argument” as in [9, § 6.2] to convert S_{R}(n, r)–modules into ℋ –modules. In particular, if we work over a field, then

{ φ_{ωω}^{1}F^{λ} : φ_{ωω}^{1}F^{λ} ≠ 0 }

is a complete set of irreducible ℋ –modules.

Now, by applying (4.6), we deduce from Theorem 4.3 the q–analogue of Schaper’s Theorem [22, Satz 4.2] for the dual q–Specht modules (we extend the definition of the relation ⟷, and our conventions on indexing modules by β–numbers, to the dual q–Specht modules in the obvious way).

(4.7) Theorem Let λ ⊢ r and let α_{1} > α_{2} > ⋯ > α_{n} ≥ 0 be β–numbers for λ′. Then, as ℋ_{R/𝔭R}–modules,

∑_{i>0} \overline{S̃^{λ}(i)} ⟷ ∑_{(a,b)∈[λ] ; {(a,c)∈[λ] \atop b<c}} (ν_{𝔭}([h_{ab}]_{q}) − ν_{𝔭}([h_{ac}]_{q}))S̃(α_{1}, …, α_{b} + h_{ac}, …, α_{c} − h_{ac}, …, α_{n}).

In order to be able to say something about which dual q–Specht modules are irreducible, we first need to investigate the relationship between the simple modules of the q–Schur algebra S_{R/𝔭R}(n, r) and the simple modules of the Hecke algebra ℋ_{R/𝔭R}(𝔖_{r}). Many properties of both of these algebras do not depend directly on q, but rather upon the related integer e (possibly infinite), which is defined next.

(4.8) Definition Let e be the smallest positive integer such that 𝔭 divides [e]_{q}; let e = ∞ if no such integer exists.

A partition λ = (λ_{1}, λ_{2}, …) is e–singular if and only if there is an integer i such that λ_{i+1} = λ_{i+2} = ⋯ = λ_{i+e} > 0; otherwise λ is e–regular. By convention, all partitions are ∞–regular.

The following is known (cf. [5, 7.6]).

(4.9) Suppose that the Hecke algebra ℋ is defined over a field. Then there exist ℋ – modules D̃^{λ} such that { D̃^{λ} : λ′ is e–regular } is a complete set of irreducible ℋ –modules. Moreover, if ν′ is an e-regular partition then D̃^{ν} is a composition factor of S̃^{ν}, appearing with multiplicity 1, and D̃^{ν} is a composition factor of S̃^{λ} only if λ ⊵ ν.

Comparing (3.5), (4.6), and (4.9), we see that

(4.10)  φ_{ωω}^{1}F^{λ} = \{ \begin{matrix}D̃^{λ}, & if λ′ is e–regular , \\ 0, & otherwise. \end{matrix}

(4.11) Lemma The irreducible S_{R/𝔭R}(n, r)–module F^{λ} is isomorphic to a submodule of S_{R/𝔭R}(n, r) only if λ′ is e–regular.

Proof: Let U be an irreducible submodule of S_{R/𝔭R}(n, r), and let u be a non–zero vector in U. Then φ_{μμ}^{1}u ≠ 0 for some μ ⊧ r. Hence, φ_{ωμ}^{1}φ_{μμ}^{1}u ≠ 0 (cf. [8, 2.5]), and so φ_{ωω}^{1}U ≠ 0. If F^{λ} ≅ U, it follows from (4.10) that λ′ is e–regular.

This allows us to prove the promised result concerning the irreducible dual q–Specht modules.

(4.12) Theorem Let λ be a partition of r. Then λ′ is e-regular and the dual q–Specht module S̃^{λ}, defined over R/𝔭R, is irreducible if and only if ν_{𝔭}([h_{ab}]_{q}) = ν_{𝔭}([h_{ac}]_{q}) for all (a, b) and (a, c) in [λ].

Proof: If λ′ is not e–regular then λ′_{i+1} = ⋯ = λ′_{i+e} > λ′_{i+e+1} for some i; hence, for a = λ′_{i+1} we have ν_{𝔭}([h_{a,i+1}]_{q}) = ν_{𝔭}([e]_{q}) > 0; but, [h_{a,i+e}]_{q} = 1. Thus, ν_{𝔭}([h_{ab}]_{q}) = ν_{𝔭}([h_{ac}]_{q}) for all (a, b) and (a, c) ∈ [λ] if and only if λ′ is e–regular and W^{λ} is irreducible. These conditions in turn imply that S̃^{λ} is irreducible by (4.6).

Next, assume that λ′ is e–regular and that S̃^{λ} is irreducible. Let F^{ν} be a bottom composition factor of W^{λ}. Then ν′ is e–regular, by Lemma 4.11. Therefore, D̃^{ν} is a composition factor of S̃^{λ}; so ν = λ and W^{λ} = F^{λ} which is irreducible. Hence, the Theorem follows.

Now, S̃^{λ} is mapped to S^{λ′} (or, rather to a left ideal of ℋ which corresponds to S^{λ′}), under the outer automorphism # of ℋ which is given by

T_{(i,i+1)}^{#} = (q − 1)T_{1} − T_{(i,i+1)},

for 1 ≤ i < r; see [7]. Also, the bilinear form ≪ , ≫, of (3.24), on the one–sided ideal of S_{R}(n, r) generated by y_{λ′} converts into a bilinear form on the one–sided ideal generated by x_{λ}. Therefore, Theorem 4.7 gives the following theorem on q–Specht modules.

(4.13) Theorem Let λ be a partition of r and suppose that, as ℋ_{R/𝔭R}–modules,

∑_{i>0} \overline{S̃^{λ}(i)} ⟷ ∑_{μ⊢r} a_{μ}S̃^{μ},

where the coefficients a_{μ} are given by Theorem 4.7. Then

∑_{i>0} \overline{S^{λ′}(i)} ⟷ ∑_{μ⊢r} a_{μ}S^{μ′}.

(4.14) Example As in Example 2.31, the calculations in Theorem 4.13 have a pictorial description in terms of moving hooks in the diagram [λ]; the difference being that hooks h_{ac} and h_{bc} in the same column must be compared and moved. Let λ = (2^{2}, 1). Then, by taking conjugates in Example 2.31, we see that

∑_{i>0} \overline{S^{λ}(i)} ⟷ ν_{𝔭}([2]_{q})S^{(3,1^{2})} + ν_{𝔭}([4]_{q})S^{(3,2)} − (ν_{𝔭}([4]_{q}) − ν_{𝔭}([3]_{q}))S^{(5)}.

We obtain from Theorem 4.12 a corresponding result about irreducible q–Specht modules.

(4.15) Theorem Let λ be a partition of r. Then λ is e–regular and the q–Specht module S^{λ}, defined over R/𝔭R, is irreducible if and only if ν_{𝔭}([h_{ac}]_{q}) = ν_{𝔭}([h_{bc}]_{q}) for all nodes (a, c) and (b, c) in [λ].

It was shown in [6] that the determinant of the Gram matrix, with respect to the standard basis of S^{λ}, is given by q^{a}Δ_{(1^{r})}(λ′), for some a ∈ ℤ. From this it is immediate that the condition given in Theorem 4.15 is sufficient to ensure that S^{λ} is irreducible; until now, we knew of no way to prove the converse (except in the case where 𝔽 is a field of characteristic zero and 𝔭 = q + 1 in (4.2); see [14]).

The e–regular partitions which satisfy the conditions of Theorem 4.15 have a particularly nice form. In order to describe this, we attach a prime number p (or ∞), to (R, 𝔭) in the definition below.

If m ∈ ℕ, then m1_{R} denotes the element 1_{R} + 1_{R} + ⋯ + 1_{R} (m times), of R.

(4.16) Definition (i) If char R = 0 and ν_{𝔭}(m1_{R}) ≠ 0 for some m ∈ ℕ, then let p be the smallest positive integer with ν_{𝔭}(p1_{R}) ≠ 0. In this case, p is a prime number and 𝔭 = u(p1_{R}) for some unit u in R.

(ii) If char R = 0 and ν_{𝔭}(m1_{R}) = 0 for all m ∈ ℕ, then let p = ∞. (iii) If char R ≠ 0, then let p = char R.

If p is a prime number then ν_{p} denotes the usual p–adic valuation on ℕ. We adopt the convention that if p = ∞ then ν_{p}(m) = 0 for all m ∈ ℕ.

The next lemma investigates the relationship between 𝔭, p, and e.

(4.17) Lemma Let a and b be integers with 1 ≤ a ≤ r and 1 ≤ b ≤ r. Then ν_{𝔭}([a]_{q}) = ν_{𝔭}([b]_{q}) if and only if either

(i) e ∤ a and e ∤ b; or (ii) e|a and e|b and ν_{p}(a) = ν_{p}(b).

Proof: By definition, e is the smallest positive integer such that 𝔭 divides [e]_{q}. Therefore, if m ∈ ℕ then 𝔭 divides [m]_{q} if and only if e divides m. It follows that ν_{𝔭}([a]_{q}) = ν_{𝔭}([b]_{q}) only if either (i) e ∤ a and e ∤ b, or (ii) e|a and e|b.

If e ∤ a and e ∤ b then ν_{𝔭}([a]_{q}) = ν_{𝔭}([b]_{q}) = 0, so we may assume that e|a and e|b.

First consider the case where p = ∞. If m ∈ ℕ then [me]_{q} = [e]_{q}[m]_{q^{e}} and [m]_{q^{e}} ≡ m1_{R} (mod 𝔭) since 𝔭 divides q^{e} − 1. Therefore, ν_{𝔭}([me]_{q}) = ν_{𝔭}([e]_{q}) since 𝔭 ∤ m1_{R}. Hence, ν_{𝔭}([a]_{q}) = ν_{𝔭}([b]_{q}) = ν_{𝔭}([e]_{q}) and the lemma follows in this case.

Now consider the case where p is a prime number. Suppose first that q = 1_{R}. If char R ≠ 0 then e = char R and so [e]_{q} = 0. However, in (4.2), we have assumed that [m]_{q} ≠ 0 for all 1 ≤ m ≤ r. Since e ≤ r, it follows that char R = 0 if q = 1_{R}. Thus, [a]_{q} = a1_{R} and the lemma holds in this case also.

Finally, suppose that p is a prime number and that q ≠ 1_{R}. Write a = eup^{x} and b = evp^{y} where p does not divide u or v. Then, since q ≠ 1_{R},

(4.18)  \begin{aligned}ν_{𝔭}([a]_{q}) = ν_{𝔭}([b]_{q}) & ⟺ & ν_{𝔭}(q^{a} − 1_{R}) = ν_{𝔭}(q^{b} − 1_{R}) \\ & ⟺ & ν_{𝔭}((q^{ep^{x}} − 1_{R})[u]_{q^{ep^{x}}}) = ν_{𝔭}((q^{ep^{y}} − 1_{R})[v]_{q^{ep^{y}}}) \\ & ⟺ & ν_{𝔭}(q^{ep^{x}} − 1_{R}) = ν_{𝔭}(q^{ep^{y}} − 1_{R}), \end{aligned}

the last line following because [u]_{q^{ep^{x}}} ≡ u1_{R} (mod 𝔭), since 𝔭 divides q^{e} − 1_{R}, and u1_{R} ≢ 0 (mod 𝔭) (and similarly for [v]_{q^{ep^{y}}}). Now, if y > x then q^{ep^{y}} − 1_{R} = (q^{ep^{x}} − 1_{R})[p^{y−x}]_{q^{ep^{x}}}. However, [p^{y−x}]_{q^{ep^{x}}} ≡ p^{y−x}1_{R} ≡ 0 (mod 𝔭), so this forces ν_{𝔭}(q^{ep^{y}} − 1_{R}) > ν_{𝔭}(q^{ep^{x}} − 1_{R}). Therefore,

ν_{𝔭}([a]_{q}) = ν_{𝔭}([b]_{q}) ⟺ ν_{p}(a) = ν_{p}(b),

completing the proof of the lemma.

Recall the definition of the e–core of a partition from [13, § 2.7].

(4.19) Example Suppose that λ = (14, 12, 7, 2^{2}, 1^{2}) and e = 3. We illustrate [λ], marking those nodes in its e–core by ×, and circling those nodes which have hook lengths divisible by e.

× × × × ⊗ × × × · ⊙ ⋯
× × × × ⊗ × ⋯ ⊙ · ·
× × × × ⊙ · ·
× ×
× ×

×

×

We see then from Theorem 4.15 and Lemma 4.17 that S^{λ} is irreducible for e = 3 if and only if p ≠ 2, 3.

Given partitions σ = (σ_{1}, σ_{2}, …) and τ = (τ_{1}, τ_{2}, …) and m ∈ ℕ, we define σ + τ to be the partition (σ_{1} + τ_{1}, σ_{2} + τ_{2}, …) and mτ to be the partition (mτ_{1}, mτ_{2}, …).

We can now improve upon Theorem 4.15. The reader might find it useful to refer to the previous example during the proof,

(4.20) Theorem Let λ be a partition of r. Then λ is e-regular and the q–Specht module S^{λ}, defined over R/𝔭R, is irreducible if and only if there exist partitions σ^{(1)}, σ^{(2)}, … and τ^{(1)}, τ^{(2)}, … such that

(i) λ = σ^{(1)} + eτ^{(1)}, σ^{(1)} is an e–core, and σ_{i}^{(1)} = σ_{i−1}^{(1)} + e − 1 for all i with τ_{i}^{(1)} ≠ 0; and (ii) for all j ≥ 1, τ^{(j)} = σ^{(j+1)} + pτ^{(j+1)}, σ^{(j+1)} is a p–core, and σ_{i}^{(j+1)} = σ_{i−1}^{(j+1)} + p − 1 for

all i with τ_{i}^{(j+1)} ≠ 0.

Proof: For the duration of the proof, let h_{ab}^{ν} denote the hook length of the (a, b)–node in the diagram [ν].

If e = ∞, then S^{λ} is always irreducible; in this case take σ^{(1)} = λ and let the partitions σ^{(2)}, …, τ^{(1)}, τ^{(2)}, … be partitions of 0. Thus, we may assume that e ≠ ∞.

By Theorem 4.15 and Lemma 4.17, λ is e–regular and S^{λ} is irreducible if and only if the following is true.

(4.21) For all nodes (a, c) and (b, c) in [λ] either e ∤ h_{ac}^{λ} and e ∤ h_{bc}^{λ}, or e|h_{ac}^{λ} and e|h_{bc}^{λ} and ν_{p}(h_{ac}^{λ}) = ν_{p}(h_{bc}^{λ}).

We shall prove that (4.21) is equivalent to (i) and (ii) from the statement of the theorem.

Draw the diagram [λ] and circle those nodes which have hook lengths divisible by e, as in Example 4.19. The condition that for all (a, c) and (b, c) ∈ [λ] either e ∤ h_{ac}^{λ} and e ∤ h_{bc}^{λ}, or e|h_{ac}^{λ} and e|h_{bc}^{λ} is readily seen to be equivalent to condition (i) in Theorem 4.20 (remove a skew e–hook from the last row in [λ] which contains a circled node and apply induction). Assume therefore that condition (i) holds.

Suppose that the (a, c)–node of [λ] is circled (ie. e divides h_{ac}^{λ}), and that (a, c) lies in the k–th column of circled nodes. Then h_{ac}^{λ} = eh_{ak}^{τ^{(1)}}. Therefore, ν_{p}(h_{ac}^{λ}) = ν_{p}(h_{bc}^{λ}) for all b with (b, c) ∈ [λ] if and only if ν_{p}(h_{ak}^{τ^{(1)}}) = ν_{p}(h_{bk}^{τ^{(1)}}) for all b with (b, k) ∈ [τ^{(1)}]. We now iterate this argument, replacing λ by τ^{(1)}, and e by p, to deduce that (4.21) is equivalent to (i) and (ii) in the statement of the theorem.

Now, the proof of [6, Theorem 2.14], which states that the centre of the Hecke algebra consists of symmetric polynomials in the q–Murphy operators, contains a gap. Although we believe that this result is correct, a consequence of the gap is that the proof of the Nakayama conjecture for Hecke algebras [6, Theorem 4.13] is incomplete. We therefore provide a proof of the Nakayama conjecture in Theorem 4.29 below.

First we classify the blocks of the q–Schur algebra. Now, by (3.5) the following holds.

(4.22) If ∑_{ν⊢r} a_{ν}W^{ν} ⟷ 0 then every integer a_{ν} is zero.

Therefore, by applying a suitable sum of block idempotents, we deduce the next result.

(4.23) Suppose that U is a non–zero S_{R/𝔭R}(n, r)–module with U ⟷ ∑_{ν⊢r} a_{ν}W^{ν} and suppose that all of the composition factors of U lie in the same block. Then, W^{ν} and U lie in the same block if a_{ν} ≠ 0.

(4.24) Theorem Suppose that λ and μ are partitions of r. Then the S_{R/𝔭R}(n, r)–modules W^{λ} and W^{μ} belong to the same block if and only if λ and μ have the same e–core.

Proof: If [ν] is a diagram which is obtained by unwrapping a skew hook of length divisible by e from [λ], and wrapping the skew hook on somewhere else, then λ and ν have the same e– core. Therefore, by Theorem 4.3, the composition factors of W^{λ}, besides F^{λ}, are composition factors of q–Weyl modules W^{ν} where λ and ν have the same e–core. (See the remark which follows Corollary 2.30.) Thus, W^{λ} and W^{μ} belong to the same block only if λ and μ have the same e–core.

Now suppose that ρ is an e–core of some partition of r, say ρ ⊢ r−we. Let γ = (γ_{1}, γ_{2}, …) be the partition of r for which γ_{1} = ρ_{1} + we and γ_{i} = ρ_{i} for i > 1. Then γ has e–core ρ.

Assume that μ ⊢ r and that μ has e–core ρ. To prove the theorem it suffices to show that W^{μ} and W^{γ} are in the same block. Suppose that μ ≠ γ. Then, for some (a, b) the (a, b)–skew hook of [μ] has length e and has its foot node in a row later than the first row. Remove this skew hook from [μ] and add it back at the end of the first row of [μ], to obtain the diagram [λ], say. Then the expression which appears in Theorem 4.3 for ∑_{i>0} \overline{W^{λ}(i)} includes W^{μ}. Therefore, W^{λ} and W^{μ} belong to the same block by (4.23). By induction, W^{λ} and W^{γ} are in the same block, so W^{μ} and W^{γ} are also in the same block.

To prove that the ℋ -modules S^{λ} and S^{μ} belong to the same block if and only if λ and μ have the same e–core is more tricky because we cannot replace q–Weyl modules by q–Specht modules in (4.22). For example, it is known [12, 6.5], that

(4.25)  ∑_{x=0}^{e−1}(−1)^{x}S^{(e−x,1^{x})} ⟷ 0.

However, (4.9) shows that the following is true.

(4.26) Suppose that ∑_{ν⊢r} a_{ν}S^{ν} ⟷ 0 where a_{ν} = 0 for every e–singular ν. Then a_{ν} = 0 for all ν.

Therefore, as in (4.23), we obtain the following.

(4.27) Suppose that U is a non–zero ℋ_{R/𝔭R,q}(𝔖_{r})–module with U ⟷ ∑_{ν⊢r} a_{ν}S^{ν} and suppose that all of the composition factors of U lie in the same block and that a_{ν} = 0 for every e–singular partition ν. Then S^{ν} and U lie in the same block if a_{ν} ≠ 0.

The following generalization of (4.25) comes from applying the Littlewood–Richardson Rule to combine the partition μ with the linear combination of partitions of e which appear in (4.25) (cf. [10, Remark p. 84]).

(4.28) Let μ be a partition of r − e. Then ∑_{ν} ±S^{ν} ⟷ 0, where the sum is over all partitions ν where [ν] is obtained by wrapping a skew e–hook onto [μ], and the sign is positive if and only if the leg length of the wrapped on skew hook is even.

We now prove the Nakayama conjecture for q–Specht modules.

(4.29) Theorem Suppose that λ and μ are partitions of r. Then the ℋ_{R/𝔭R}(𝔖_{r})–modules S^{λ} and S^{μ} belong to the same block if and only if λ and μ have the same e–core.

Proof: It follows from (4.6) and Theorem 4.24 that S^{λ} and S^{μ} belong to the same block only if λ and μ have the same e–core. Indeed, this implication is proved in [6, 4.4]. We concentrate upon the converse.

Suppose that ρ is an e–core of some partition of r, say ρ ⊢ r − we. Let γ = (γ_{1}, γ_{2}, …) be the partition of r for which γ_{1} = ρ_{1} + we and γ_{i} = ρ_{i} for i > 1. Then γ has e–core ρ.

Assume that λ ⊢ r has e–core ρ, where λ ≠ γ. We prove that S^{λ} and S^{γ} are in the same block, which will complete the proof of the theorem. Note that if β is a partition of r with e–core ρ then γ ⊵ β. Therefore, by induction, it suffices to show that there exists a partition β with e–core ρ such that S^{λ} and S^{β} are in the same block and β ▷ λ.

First, suppose that S^{λ} is reducible. Then S^{λ} shares a composition factor with some S^{μ} where μ is e–regular and μ ▷ λ. Moreover, μ has e-core ρ because S^{μ} and S^{λ} are in the same block; so we may take β = μ.

Next, if λ is e–singular and S^{λ} is irreducible then S^{λ} = D^{μ} for some e–regular partition μ such that μ ▷ λ. Therefore, λ and μ have the same e–core and again we are done.

It remains to consider the case where λ is e–regular and S^{λ} is irreducible. Such partitions are described explicitly in Theorem 4.20; in particular, for all (a, b) and (b, c) ∈ [λ], either e ∤ h_{ac} and e ∤ h_{bc}, or e|h_{ac} and e|h_{bc}. (A typical partition of this kind is illustrated in Example 4.19; the reader may care to keep this example in mind for the remainder of the proof.)

Let a be the largest integer such that e|h_{ac} for some c, and let b = λ_{a} − e + 1. Then h_{ab} = e and e|h_{ib} for 1 ≤ i ≤ a; so

(4.30)  λ_{i} ≡ λ_{i+1} − 1 (mod e)  for 1 ≤ i ≤ a − 1.

Let [μ] be the diagram obtained from [λ] by removing the (a, b)–skew hook; so μ_{i} = λ_{i} for i ≠ a and μ_{a} = λ_{a} − e. (In Example 4.19, a = 3 and μ = (14, 12, 4, 2^{2}, 1^{2}).)

As in (4.28), we now wrap a skew e–hook onto [μ] in all possible ways. Our first aim is to show that (except in one special case), the only one way to do this and obtain an e–singular diagram is to add a vertical e–hook to the end of first column of [μ].

First, recall that the only way to add a skew e–hook to an e–core to obtain an e–singular diagram is to add a vertical e–hook to the end of the first column (cf. [21]). A more general result is the following, which can be proved by removing the first part from [τ].

(4.31) Suppose that σ is an e–core and that τ is a partition such that τ_{1} = σ_{1} + ke, for some non–negative integer k, and τ_{i} = σ_{i} for all i > 1. Suppose that ν is an e–singular partition whose diagram is obtained by wrapping a skew e–hook onto [τ]. Then, either the skew e–hook was added vertically to the end of the first column of [τ], or e = 2, τ = (2), and ν = (2, 2).

We return now to the situation in hand. Assume that ν is an e–singular partition which is obtained by wrapping a skew e–hook into [μ], and suppose that the hand of the wrapped on skew e–hook lies in row x. We have that μ_{i} ≡ μ_{i+1} − 1 (mod e) for 1 ≤ i ≤ a − 1 by (4.30), and this implies that x ≥ a. Let τ = (μ_{a}, μ_{a+1}, …). Then τ is a partition of the kind considered in (4.31); also, μ_{a} +e < μ_{a−1}. Therefore, either the skew e–hook which was added to [μ] to obtain [ν] consisted of e nodes, placed vertically below the first column of [μ], or e = 2, λ_{a} = 4, λ_{a+1} = 0 and ν = (λ_{1}, …, λ_{a−1}, 2, 2). So, we have proved the following.

(4.32) Unless e = 2, λ_{a} = 4, and λ_{a+1} = 0, there is exactly one way to wrap a skew e–hook onto [μ] to obtain an e-singular diagram.

Assume, for the moment, that we are not in the case where e = 2, λ_{a} = 4, and λ_{a+1} = 0. Then (4.28) and (4.32) give us a linear combination of q–Specht modules ∑_{ν} a_{ν}S^{ν} such that there is a unique e–singular partition, σ say, with a_{σ} ≠ 0. Moreover, a_{λ} = 1 and a_{β} = 1 where β = (λ_{1} +e, λ_{2}, …, λ_{a} −e, …). We can therefore apply (4.27), with U = S^{σ}, to deduce that S^{λ} and S^{β} are in the same block. Since β ▷ λ, this completes the proof in this case.

Finally, suppose that e = 2, λ_{a} = 4, and λ_{a+1} = 0; we again show that λ and β, where β is as above, are in the same block. Here, we can add a vertical 2–hook to [μ] in only one way; we obtain the diagram [δ] where δ = (λ_{1}, …, λ_{a−1}, 2, 1^{2}). Therefore, (4.28) gives ∑_{ν} a_{ν}S^{ν} ⟷ 0, where a_{δ} = −1, a_{λ} = 1, a_{β} = 1, and a_{ν} ≥ 0 for all other partitions ν. Hence, S^{λ} and S^{δ} have a common composition factor, as do S^{β} and S^{δ}. This again implies that S^{λ} and S^{β} are in the same block, and so completes the proof.

In conclusion, we classify the ordinary irreducible representations of the finite general linear group GL_{n}(q) which remain irreducible modulo a prime number p which does not divide q. To state and prove our theorem we adopt the notation of [4]. In particular, the ordinary irreducible GL_{n}(q)–modules have the form S(I) where I is an (n, ∞)–index [4, Definition 2.4].

(4.33) Theorem Let S(I) be an ordinary irreducible GL_{n}(q)–module, where

I = ( \begin{matrix}d_{1} & … & d_{N} & k_{1} & … & k_{N} \\ s_{1} & … & s_{N} & λ^{(1)} & … & λ^{(N)} \end{matrix} ) .

Suppose that p is a prime number which does not divide q. Then the reduction of S(I) modulo p is irreducible if and only if the following two conditions hold.

(i) If i ≠ j and d_{i} = d_{j} then the p–regular parts of s_{i} and s_{j} are roots of distinct irreducible polynomials over GF (q).

(ii) For all i and for all nodes (a, c) and (b, c) in [λ^{(i)}], ν_{p}([h_{ac}]_{q^{d_{i}}}) = ν_{p}([h_{bc}]_{q^{d_{i}}}).

Proof: First consider the GL_{d_{i}k_{i}}(q)–modules S(I_{i}) where I_{i} = ( \begin{matrix}d_{i} & k_{i} \\ s_{i} & λ^{(i)} \end{matrix} ). The module

S(I_{i}) is irreducible modulo p if and only if the decomposition numbers d_{μ′λ^{(i)′}} of the q^{d_{i}}–Schur algebra are zero for all μ ≠ λ^{(i)} by [8, Theorem 4.6]. Thus, Theorem 4.5, applied to the case where R = ℤ[q^{−d_{i}}], shows that condition (ii), in the statement of the theorem, is equivalent to the statement that all of the modules S(I_{i}), with 1 ≤ i ≤ N, are irreducible modulo p.

Let I_{(p)} be the (n, p)-index which is given by

I_{(p)} = ( \begin{matrix}d_{1} & … & d_{N} & k_{1} & … & k_{N} \\ s′_{1} & … & s′_{N} & λ^{(1)} & … & λ^{(N)} \end{matrix} ) ,

where s′_{i} is the p–regular part of s_{i}. Then S(I_{(p)}) is the reduction modulo p of S(I). The module S(I_{(p)}) is irreducible only if S(I_{i}) is irreducible modulo p for all i; that is, condition (ii) holds.

Assume now that condition (ii) holds. Let e(d_{i}) be the smallest positive integer such that p divides [e(d_{i})]_{q^{d_{i}}}. Then the partition λ^{(i)} is e(d_{i})–regular, for all i, and S(I_{(p)}) = D(I_{(p)}) in the notation of [4]. If condition (i) holds, then I_{(p)} is a special foot index [4, 2.9]; so D(I_{(p)}) is irreducible by [4, 4.9]. If condition (i) does not hold, then we may use the Littlewood– Richardson Rule to express S(I_{(p)}) as a linear combination of modules S(J) where more than one (n, p)–index J occurs in the expression (cf. case 1 of the proof of [4, 6.2]); in particular, S(I_{(p)}) is reducible.

An example where condition (i) of Theorem 4.33 fails is provided by GL_{6}(2) with s_{1} a root of 1 + x + x^{3}, s_{2} a root of 1 + x^{2} + x^{3}, and

I = ( \begin{matrix}3 & 3 & 1 & 1 \\ s_{1} & s_{2} & (1) & (1) \end{matrix} ) .

If p = 7 then 1 is the p–regular part of s_{1} and s_{2}; so S(I) is an ordinary irreducible GL_{6}(2)– module which is reducible modulo p.

We note that the special case of Theorem 4.33 where

I = ( \begin{matrix}1 & n \\ 1 & λ \end{matrix} )

(so that S(I) is a unipotent module), was conjectured in [11, 20.5]; the numerical condition which appears there is equivalent to the condition that ν_{p}([h_{ac}]_{q}) = ν_{p}([h_{bc}]_{q}), for all nodes (a, c) and (b, c) in [λ], by Lemma 4.17.

Acknowledgements

We thank the referee and Jie Du for their comments on the manuscript. Also, our understanding of the action of the operators L_{μ,m} in the case where q = 1 was greatly enhanced by computations made using the computer algebra package Gap [23].

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Department of Mathematics, Imperial College, Queen’s Gate, London SW7 2BZ

                     e-mail:  g.james@ic.ac.uk and a.mathas@ic.ac.uk