Beiträge zur Algebra und Geometrie Contributions to Algebra and Geometry Volume 37 (1996), No. 2, 289-307.

Representations of the Generalized Symmetric Groups

Himmet Can^{∗}

Mathematics Department, University of Wales Aberystwyth, Dyfed, SY23 3BZ, Wales, UK

Introduction

The main aim of this paper is to construct a full set of irreducible, inequivalent representations of the generalised symmetric groups G(m, 1, n) in terms of m-sets of partitions of n and combinatorial concepts connected with generalised Young tableaux, etc. As a matter of fact, the irreducible representations of the generalised symmetric groups G(m, 1, n) were studied by Osima [5], Puttaswamaiah [7], Read [9] and Hughes [2], respectively. However, there are well-known constructions of irreducible representations and of irreducible modules, called Specht modules, for the symmetric groups S_{n} which are based on elegant combinatorial concepts connected with Young tableaux, etc. (see, e.g. [3]). Therefore, in this paper we show how the Specht approach to the irreducible representations of the symmetric groups can be extended to deal with the generalised symmetric groups G(m, 1, n). As a convention, throughout this paper, we assume that ξ is a primitive m-th root of unity.

1. Diagrams, Tableaux and Tabloids

The generalised symmetric group G(m, 1, n) is the group of all permutations σ of {ξ^{k}i | i = 1, …, n; k = 1, …, m} such that σ(ξ^{k}i) = ξ^{k}σ(i) for i = 1, …, n; k = 1, …, m. The order of G(m, 1, n) is m^{n}n!

Let S_{n} be the symmetric group on n symbols and C_{m} be the cyclic group of order m. Then G(m, 1, n) is the semi-direct product of C_{m}^{n} with S_{n}, written as C_{m}^{n} × S_{n}, where C_{m}^{n} is the direct product of n copies of C_{m}. The group G(m, 1, n) will sometimes be denoted by W_{n}^{m}. The group G(m, 1, n) has the following presentation (see [1]):

G(m, 1, n) = ⟨r_{1}, …, r_{n−1}, w_{1}, …, w_{n} | r_{i}^{2} = (r_{i}r_{i+1})^{3} = (r_{i}r_{j})^{2} = e, |i − j| ≥ 2,
w_{i}^{m} = e, w_{i}w_{j} = w_{j}w_{i}, r_{i}w_{i} = w_{i+1}r_{i}, r_{i}w_{j} = w_{j}r_{i}, j ≠ i, i + 1⟩.

∗The author was supported by the University of Erciyes. Permanent address:

Department of Mathematics; Faculty of Arts & Science, Erciyes University, 38039 Kayseri, Turkey 0138-4821/93 $ 2.50 \begin{matrix}c \\ ◯ \end{matrix} 1996 Heldermann Verlag, Berlin

We may identify r_{i} (i = 1, …, n − 1) with the transposition (i, i + 1) and therefore the group generated by {r_{1}, …, r_{n−1}} is the symmetric group S_{n}. The w_{i} (i = 1, …, n) may

be identified with the mapping ( \begin{matrix}i \\ ξi \end{matrix} ) : \{1, …, n\} → C\\{0\} defined by i → ξi and j → j,

j = 1, 2, …, i − 1, i + 1, …, n.

The above relations imply that if τ ∈ S_{n}, then τw_{i}τ^{−1} = w_{τ(i)}, i = 1, …, n. If C_{m}^{n} is the subgroup of G(m, 1, n) generated by {w_{1}, …, w_{n}}, then C_{m}^{n} ◁ G(m, 1, n) and G(m, 1, n) permutes the letters {1, …, n} as well as multiplying any number of them by some power of ξ. Furthermore, for the group G(m, 1, n) if we take m = 1 and 2 then we have the symmetric group S_{n} and the hyperoctahedral group O_{n} respectively (see [9]).

If τ = ( \begin{matrix}1 & 2 & … & n \\ τ(1) & τ(2) & … & τ(n) \end{matrix} ) ∈ S_{n}, let

σ = ( \begin{matrix}1 & 2 & … & n \\ ξ^{s_{1}}τ(1) & ξ^{s_{2}}τ(2) & … & ξ^{s_{n}}τ(n) \end{matrix} ) = τ ∏_{i=1}^{n} w_{i}^{s_{i}} ∈ G(m, 1, n),

where 1 ≤ s_{i} ≤ m. Define φ : G(m, 1, n) → S_{n} by φ(σ) = τ. Then φ is an epimorphism and ker φ = C_{m}^{n}. Define the sign of σ ∈ G(m, 1, n) to be sgn(σ) = sgn(φ(σ)).

Definition 1.1 An element σ ∈ G(m, 1, n) is even if φ(σ) is an even element of S_{n} and odd otherwise. In terms of the generators r_{i}, w_{i} given above, σ is even if and only if the number of r_{i} appearing in any expression for σ is even (see [8]).

Any element σ ∈ G(m, 1, n) may be written uniquely (up to reordering) as the product of disjoint cycles σ = θ_{1} … θ_{t}, where

θ_{i} = ( \begin{matrix}b_{i1} & b_{i2} & … & b_{ik_{i}} \\ ξ^{s_{i1}}b_{i2} & ξ^{s_{i2}}b_{i3} & … & ξ^{s_{ik_{i}}}b_{i1} \end{matrix} ) ,

b_{ij} ∈ {1, …, n}, s_{ij} ∈ {1, …, m}, k_{i} is the length of the cycle θ_{i}, i = 1, …, t.

Let f(θ_{i}) = ∑_{j=1}^{k_{i}} s_{ij}, and put f(σ) = ∑_{i=1}^{t} f(θ_{i}). Define a_{pq}(σ) to be the number of cycles

θ_{i} of σ of length q such that f(θ_{i}) ≡ p (mod m) for 1 ≤ p ≤ m, 1 ≤ q ≤ n. The m × n matrix (a_{pq}(σ)) is called the type of σ, denoted by T y(σ) (see [9]). Then it is well known that σ, π ∈ G(m, 1, n) are conjugate in G(m, 1, n) if and only if T y(σ) = T y(π) (see [4]). Let

π = ( \begin{matrix}1 & 2 & … & n \\ ξ^{r_{1}}γ(1) & ξ^{r_{2}}γ(2) & … & ξ^{r_{n}}γ(n) \end{matrix} ) = γ ∏_{i=1}^{n} w_{i}^{r_{i}} ∈ G(m, 1, n)

where γ = ( \begin{matrix}1 & 2 & … & n \\ γ(1) & γ(2) & … & γ(n) \end{matrix} ) ∈ S_{n}, 1 ≤ r_{i} ≤ m. Then we have

σπ = ( \begin{matrix}1 & 2 & … & n \\ ξ^{s_{1}}τ(1) & ξ^{s_{2}}τ(2) & … & ξ^{s_{n}}τ(n) \end{matrix} ) ( \begin{matrix}1 & 2 & … & n \\ ξ^{r_{1}}γ(1) & ξ^{r_{2}}γ(2) & … & ξ^{r_{n}}γ(n) \end{matrix} )
= ( \begin{matrix}1 & 2 & … & n \\ ξ^{(r_{1}+s_{γ(1)})}τγ(1) & ξ^{(r_{2}+s_{γ(2)})}τγ(2) & … & ξ^{(r_{n}+s_{γ(n)})}τγ(n) \end{matrix} )
= τγ ∏_{i=1}^{n} w_{i}^{(r_{i}+s_{γ(i)})}.

Hence, f(σπ) ≡ f(σ) + f(π) (mod m) for any σ, π ∈ G(m, 1, n). Furthermore, we assume that f(τ) = 0 for all τ ∈ S_{n}. Thus, f(σ^{−1}) = mk −f(σ) for any σ ∈ G(m, 1, n), where k ∈ Z.

Definition 1.2 An m-set of partitions of n, (λ^{(1)}, …, λ^{(m)}), consists of m partitions λ^{(1)}, …, λ^{(m)} such that λ^{(i)} = (λ_{1}^{(i)}, …, λ_{s_{i}}^{(i)}) is a partition of l_{i}, where

∑_{j=1}^{s_{i}} λ_{j}^{(i)} = l_{i} and ∑_{i=1}^{m} l_{i} = n.

For any i, if l_{i} = 0 then it is still necessary to write λ^{(i)} = 0. Let (λ^{(1)}, …, λ^{(m)}) be an m-set of partitions of n, then (λ^{(1)}, …, λ^{(m)}) will be abbreviated to λ^{[m]}. In the λ^{[m]}, the part corresponding to λ^{(i)} (1 ≤ i ≤ m) will be called the i-th constituent of the λ^{[m]}. Let P^{(m)}(n) denote the set of all m-sets of partitions of n. Saeed-ul-Islam ([11]) has proved that the number of conjugacy classes in G(m, 1, n) is given by

|P^{(m)}(n)| = ∑ p(l_{1}) … p(l_{m}),

where the summation is taken over all m-tuples {l_{1}, …, l_{m}} such that ∑_{i=1}^{m} l_{i} = n and p(l_{i}) denotes the number of partitions of l_{i}, i = 1, …, m.

Furthermore, Osima ([5]) gives that the number of ordinary irreducible representations of G(m, 1, n) is |P^{(m)}(n)|.

The group G(m, 1, n) has 2m one dimensional representations 1, σ_{1}, σ_{2}, …, σ_{m−1}, η, ϵ_{1}, …, ϵ_{m−1} defined by

(a) 1(g) = 1 for all g ∈ G(m, 1, n),
(b) σ_{k}(r_{i}) = 1 (i = 1, …, n − 1), σ_{k}(w_{j}) = ξ^{k} (j = 1, …, n),
(c) η(r_{i}) = −1 (i = 1, …, n − 1), η(w_{j}) = 1 (j = 1, …, n),
(d) ϵ_{k}(r_{i}) = −1 (i = 1, …, n − 1), ϵ_{k}(w_{j}) = ξ^{k} (j = 1, …, n),

where k = 1, …, m − 1 and ξ is a primitive m-th root of unity. Note that 1, η are real representations (called the trivial and sign representations respectively) and if m is even then σ\frac{m}{2} , ϵ\frac{m}{2} are also real representations.

Definition 1.3 Let λ^{[m]} ∈ P^{(m)}(n). The generalized Young diagram of λ^{[m]} is [λ^{[m]}] = ([λ^{(1)}], …, [λ^{(m)}]), where [λ^{(i)}] is the usual Young diagram of λ^{(i)}, i = 1, …, m. For any i, if λ^{(i)} = 0 then we write [λ^{(i)}] = ∅.

For example if λ^{[4]} = (3^{2}1, 21, 0, 2^{2}1) ∈ P^{(4)}(15), then we shall draw the generalized Young diagram of λ^{[4]} as in the following:

[λ^{[4]}] = ( \begin{matrix}• & • & • \\ • & • & • \\ • \end{matrix} , \begin{matrix}• & • \\ • \end{matrix} , ∅, \begin{matrix}• & • \\ • & • \\ • \end{matrix} )

Definition 1.4 Let λ^{[m]} ∈ P^{(m)}(n). A λ^{[m]}-tableau t is an array of the symbols {ξ^{k}i | i = 1, …, n; k = 1, …, m} obtained by replacing each node in [λ^{[m]}] by one of the symbols {ξ^{k}i | i = 1, …, n; k = 1, …, m} with the moduli of all the numbers appearing being distinct.

A λ^{[m]}-tableau t will sometimes be written (t_{λ^{(1)}}, …, t_{λ^{(m)}}). Clearly there are m^{n}n! λ^{[m]}- tableaux for any generalized Young diagram [λ^{[m]}].

Let t(i, j, k) = t_{λ^{(k)}}(i, j), where t_{λ^{(k)}}(i, j) stands for the entry of t_{λ^{(k)}} in position (i, j), as in a matrix, where k = 1, …, m. Thus, t(i, j, k) stands for the entry of t in position (i, j, k) where k = 1, …, m. In the tableau t, the part corresponding to t_{λ^{(k)}} (1 ≤ k ≤ m) will be called the k-th constituent of the tableau t.

The group G(m, 1, n) acts on a λ^{[m]}-tableau t = (t(i, j, k)) as follows: σt = (σt(i, j, k)) = (σt_{λ^{(1)}}(i, j), …, σt_{λ^{(m)}}(i, j)) for all σ ∈ G(m, 1, n).

For example, let λ^{[3]} = (21, 0, 2^{2}) ∈ P^{(3)}(7). Then

t = ( \begin{matrix}2 & ξ7 \\ ξ^{2}1 \end{matrix} , ∅, \begin{matrix}ξ^{2}5 & 6 \\ 3 & ξ^{2}4 \end{matrix} )

is a λ^{[3]}-tableau. If

σ = ( \begin{matrix}1 & 2 & 3 & 4 & 5 & 6 & 7 \\ ξ1 & 3 & ξ^{2}4 & 7 & 2 & 5 & ξ^{2}6 \end{matrix} ) ∈ G(3, 1, 7)

then

σt = σ ( \begin{matrix}2 & ξ7 \\ ξ^{2}1 \end{matrix} , ∅, \begin{matrix}ξ^{2}5 & 6 \\ 3 & ξ^{2}4 \end{matrix} ) = ( \begin{matrix}3 & 6 \\ 1 \end{matrix} , ∅, \begin{matrix}ξ^{2}2 & 5 \\ ξ^{2}4 & ξ^{2}7 \end{matrix} )

Definition 1.5 Let t be a λ^{[m]}-tableau. A row permutation of t is an element σ of W_{n}^{m} such that σ permutes the symbols in each row of t and multiplies any number of symbols in t_{λ^{(k)}} (k = 2, …, m) by a power of ξ. A column permutation of t is an element σ of W_{n}^{m} such that σ permutes the symbols in each column of t and multiplies any number of symbols in t_{λ^{(1)}} by a power of ξ.

We can now define the row stabilizer R_{t} and column stabilizer C_{t} of t as follows. Let R_{t} be the group of row permutations of t. Thus, R_{t} is isomorphic to

∏_{i=1}^{m} (W_{λ_{1}^{(i)}}^{m_{i}} × … × W_{λ_{s_{i}}^{(i)}}^{m_{i}}) ,

where

m_{i} = \{ \begin{matrix}1 & if i = 1 \\ m & if i = 2, …, m \end{matrix}

and is a subgroup of W_{n}^{m}.

Let C_{t} be the group of column permutations of t. Thus, C_{t} is isomorphic to

∏_{i=1}^{m} (W_{μ_{1}^{(i)}}^{m_{i}} × … × W_{μ_{r_{i}}^{(i)}}^{m_{i}}) ,

where

m_{i} = \{ \begin{matrix}m & if i = 1 \\ 1 & if i = 2, …, m \end{matrix} ,

(μ_{1}^{(i)}, …, μ_{r_{i}}^{(i)}) is the partition conjugate to (λ_{1}^{(i)}, …, λ_{s_{i}}^{(i)}), and C_{t} is a subgroup of W_{n}^{m}. Now, we define an equivalence relation ∼ on the set of λ^{[m]}-tableaux as follows.

Definition 1.6 Two λ^{[m]}-tableaux t_{1} and t_{2} are row equivalent, written t_{1} ∼ t_{2}, if there exists σ ∈ R_{t_{1}} such that t_{2} = σt_{1}. The equivalence class which contains the λ^{[m]}-tableau t is {t} and is called a λ^{[m]}- tabloid.

If λ^{[m]} ∈ P^{(m)}(n) with notation as in Definition 1.2, then the number of λ^{[m]}-tabloids is

m^{l_{1}} \frac{n!}{∏_{i=1}^{m} (∏_{j=1}^{s_{i}} λ_{j}^{(i)}!)}.

Since {t} = {t′ | t′ = σt for some σ ∈ R_{t}}, the number of λ^{[m]}-tableaux in any given equivalence class is

|R_{t}| = (∏_{i=1}^{s_{1}} λ_{i}^{(1)}!) (∏_{i=2}^{m} ∏_{j=1}^{s_{i}} m^{λ_{j}^{(i)}}λ_{j}^{(i)}!) .

Furthermore, it can be verified that if σ ∈ W_{n}^{m} then R_{σt} = σR_{t}σ^{−1} and C_{σt} = σC_{t}σ^{−1}.

Definition 1.7 If λ^{[m]} ∈ P^{(m)}(n), let M^{λ^{[m]}} be the vector space over the complex field C whose basis elements are the λ^{[m]}-tabloids.

Now, we define the action of W_{n}^{m} on the vector space M^{λ^{[m]}} as follows. Let τ ∈ S_{n} and

1 ≤ s_{i} ≤ m. If σ = τ ∏_{i=1}^{n} w_{i}^{s_{i}} ∈ W_{n}^{m}, then

σ{t} = ξ^{ϕ(σ,t)}{σt}

where ϕ(σ, t) = ∑_{i=1}^{n} k_{i}s_{i} if i appears in the k_{i}-th constituent of the tableau t. Note that if

σ ∈ S_{n} then since ϕ(σ, t) ≡ 0 (mod m) we have σ{t} = {σt}. Now we show that this is a group action on the vector space M^{λ^{[m]}}. Let

π = γ ∏_{i=1}^{n} w_{i}^{r_{i}} ∈ W_{n}^{m},

where γ ∈ S_{n} and 1 ≤ r_{i} ≤ m. Then we have

σπ = τγ ∏_{i=1}^{n} w_{i}^{(r_{i}+s_{γ(i)})} and so
(σπ){t} = ξ^{ϕ(σπ,t)}{σπt},

where ϕ(σπ, t) = ∑_{i=1}^{n} k_{i}(r_{i} + s_{γ(i)}) if i appears in the k_{i}-th constituent of the tableau t. (Note

that at this point, we have not used the relation ξ^{m} = 1). For π = γ ∏_{i=1}^{n} w_{i}^{r_{i}} ∈ W_{n}^{m} if i appears

in the k_{i}-th constituent of the tableau t then γ(i) appears in the same k_{i}-th constituent of the tableau πt.

Let γ(i) = j (1 ≤ i, j ≤ n). Then for σ = τ ∏_{j=1}^{n} w_{j}^{s_{j}} ∈ W_{n}^{m}, w_{j}^{s_{j}} = w_{γ(i)}^{s_{γ(i)}} and so

σ(π{t}) = σ (ξ^{ϕ(π,t)}{πt}) = ξ^{ϕ(π,t)}ξ^{ϕ(σ,πt)}{σπt},

where ϕ(σ, πt) = ∑_{i=1}^{n} k_{i}s_{γ(i)} and γ(i) appears in the k_{i}-th constituent of the tableau πt. Thus,

ϕ(σπ, t) = ϕ(π, t) + ϕ(σ, πt) and so
(σπ){t} = σ(π{t}).

This action is well-defined, for if {t_{1}} = {t_{2}} then there exists σ ∈ R_{t_{1}} such that t_{2} = σt_{1}. Then for any π ∈ W_{n}^{m}, πσπ^{−1} ∈ πR_{t_{1}}π^{−1} = R_{πt_{1}}, so

ξ^{ϕ(π,t_{1})}{πt_{1}} = ξ^{ϕ(π,σt_{1})}{πσt_{1}} = ξ^{ϕ(π,t_{2})}{πt_{2}}.

Extending this action to be linear on M^{λ^{[m]}} turns M^{λ^{[m]}} into a CW_{n}^{m}-module, and so we have the following theorem.

Theorem 1.8 The CW_{n}^{m}-module M^{λ^{[m]}} is isomorphic to the permutation module of W_{n}^{m} on the subgroup R_{t}. M^{λ^{[m]}} is a cyclic CW_{n}^{m}-module generated by any one λ^{[m]}-tabloid and

dim_{C}M^{λ^{[m]}} = m^{l_{1}} \frac{n!}{∏_{i=1}^{m} (∏_{j=1}^{s_{i}} λ_{j}^{(i)}!)}

2. Dominance Partial Ordering

Now we impose the following partial ≥ order on P^{(m)}(n), which we call the generalized dominance partial order.

Definition 2.1 Let λ^{[m]}, μ^{[m]} ∈ P^{(m)}(n). Then λ^{[m]} dominates μ^{[m]}, written λ^{[m]} ≥ μ^{[m]}, if one of the following conditions is satisfied:

(i) |λ^{(1)}| > |μ^{(1)}|, (ii) |λ^{(j)}| = |μ^{(j)}| (1 ≤ j ≤ k) and |λ^{(k+1)}| > |μ^{(k+1)}| for some k ∈ {1, 2, …, m − 2}, (iii) |λ^{(i)}| = |μ^{(i)}| and λ^{(i)} ≥ μ^{(i)} for all i = 1, …, m.

Lemma 2.2 Let t be a λ^{[m]}-tableau and t′ be a μ^{[m]}-tableau such that for all a, if a occurs in t_{λ^{(i)}} then ξ^{s_{i}}a occurs in t′_{μ^{(i)}}, where i ∈ {1, …, m − 1} and s_{i} ∈ {1, …, m}. Suppose that a, b belong to the same row of t′_{μ^{(i)}} implies that c, d belong to different columns of t_{λ^{(i)}} where c = ξ^{ρ_{1}}a, d = ξ^{ρ_{2}}b, (1 ≤ ρ_{1}, ρ_{2} ≤ m), and i ∈ {1, …, m}. Then λ^{[m]} ≥ μ^{[m]}.

Proof. By hypothesis, we have that |λ^{(i)}| = |μ^{(i)}| for all i = 1, …, m. Thus, by using a similar argument to that in the proof of the corresponding lemma (Lemma 2.2.4 in [10] for the symmetric groups) the result can be deduced. □

Corollary 2.3 Suppose |λ^{(i)}| = |μ^{(i)}| for all i = 1, …, m − 1 and λ^{[m]} ≱μ^{[m]}. Let t be a λ^{[m]}-tableau and t′ be a μ^{[m]}-tableau such that for all a, if a occurs in t_{λ^{(i)}} then ξ^{s_{i}}a occurs in t′_{μ^{(i)}}, where i ∈ {1, …, m − 1} and s_{i} ∈ {1, …, m}. Then there exist a, b in the same row of t′_{μ^{(i)}} and c, d in the same column of t_{λ^{(i)}}, where c = ξ^{ρ_{1}}a, d = ξ^{ρ_{2}}b, (1 ≤ ρ_{1}, ρ_{2} ≤ m), and i ∈ {1, …, m}.

Himmet Can: Representations of the Generalized Symmetric Groups 295

3. Specht Modules We now construct a complete set of irreducible modules of G(m, 1, n) over the complex field C; these are the so-called Specht modules, S^{λ^{[m]}}. Definition 3.1 Let t be a λ^{[m]}-tableau. Define κ_{t} ∈ CW_{n}^{m} by

κ_{t} = ∑_{σ∈C_{t}} ξ^{−f(σ)}(sgn σ)σ,

where f(σ) is defined as in section 1. The λ^{[m]}-polytabloid e_{t} associated with the tableau t is given by e_{t} = κ_{t}{t}. Lemma 3.2 Let t be a λ^{[m]}-tableau and π ∈ W_{n}^{m}. Then

(i) κ_{πt} = πκ_{t}π^{−1} (ii) πe_{t} = ξ^{ϕ(π,t)}e_{πt}, (iii) if π ∈ C_{t} then πe_{t} = ξ^{f(π)}(sgn π)e_{t}. Proof. (i) If π ∈ W_{n}^{m} then

κ_{πt} = ∑_{σ∈C_{πt}} ξ^{−f(σ)}(sgn σ)σ = ∑_{σ∈πC_{t}π^{−1}} ξ^{−f(σ)}(sgn σ)σ
= ∑_{τ∈C_{t}} ξ^{−f(πτπ^{−1})}sgn(πτπ^{−1})πτπ^{−1}
= πκ_{t}π^{−1}.

(ii) πe_{t} = πκ{t} = κ_{πt}π{t} = ξ^{ϕ(π,t)}κ_{πt}{πt} = ξ^{ϕ(π,t)}e_{πt}. (iii) If π ∈ C_{t} then

πe_{t} = ∑_{σ∈C_{t}} ξ^{−f(σ)}(sgn σ)πσ{t} = ∑_{τ∈C_{t}} ξ^{−f(π^{−1}τ)}sgn(π^{−1}τ)τ{t}
= ξ^{f(π)}(sgn π)e_{t}. □

We are now in a position to define the Specht modules. Definition 3.3 Let λ^{[m]} ∈ P^{(m)}(n). The Specht module S^{λ^{[m]}} for the m-set of partitions λ^{[m]} is the submodule of M^{λ^{[m]}} spanned by λ^{[m]}-polytabloids. Because of part (ii) of Lemma 3.2, we have the following. Theorem 3.4 S^{λ^{[m]}} is a cyclic CW_{n}^{m}-module generated by any one polytabloid. If ν = ∑_{i} α_{i}{t_{i}} ∈ M^{λ^{[m]}}, where α_{i} ∈ C, then we say that {t_{i}} appears in ν if α_{i} ≠ 0. Remark 3.5 The action of G(m, 1, n) on M^{λ^{[m]}} has already been defined in section 1. Now, we explain that this is a natural action because of the following reason. For the group G(m, 1, n) if we take n = 1 then

G(m, 1, 1) = C_{m}^{1} = ⟨w_{1} | w_{1}^{m} = e⟩ and P^{(m)}(1) consists of precisely m distinct m-sets of partitions of 1, i.e.

P^{(m)}(1) = {λ_{i}^{[m]} (i = 1, …, m) | 1 appears in the i − th constituent of λ_{i}^{[m]}}.

For λ_{1}^{[m]} = (1, 0, …, 0) we have

M^{λ_{1}^{[m]}} = ⟨{ξ^{k}1, ∅, …, ∅} | k = 1, …, m⟩ and
S^{λ_{1}^{[m]}} = ⟨e_{t} | e_{t} = ∑_{k=1}^{m}{ξ^{k}1, ∅, …, ∅}⟩.

For i = 2, 3, …, m,

M^{λ_{i}^{[m]}} = S^{λ_{i}^{[m]}} = ⟨{t} | 1 appears in the i−th constituent of t⟩.

Now define X_{i} : G(m, 1, 1) → GL(1, C) by X_{i}(w_{1}^{j}) = ξ^{ij} for i, j = 1, …, m. Then X_{1}, X_{2}, …, X_{m} = 1 are m distinct irreducible matrix representations of G(m, 1, 1) having degree 1. Let Y_{i} be the matrix representation of G(m, 1, 1) afforded by S^{λ_{i}^{[m]}} for i = 1, …, m. Then X_{i} and Y_{i} are equivalent irreducible representations of G(m, 1, 1) for i = 1, …, m.

The following lemma is a simple generalization of the corresponding result (Lemma 1.4 in [6]) for the symmetric groups.

Lemma 3.6 Let t and t′ be λ^{[m]}-tableaux. The following conditions are equivalent: (i) {t′} is contained in e_{t}, (ii) there exist ρ ∈ R_{t′}, π ∈ C_{t} such that ρt′ = πt, (iii) a, b belong to the same row of t′_{λ^{(i)}} implies that c, d belong to different columns of t_{λ^{(i)}},

where c = ξ^{ρ_{1}}a, d = ξ^{ρ_{2}}b, (1 ≤ ρ_{1}, ρ_{2} ≤ m), and i ∈ {1, …, m}.

Lemma 3.7 Let t be a λ^{[m]}-tableau and t′ be a μ^{[m]}-tableau. Suppose there exist a, b in the same row of t′_{μ^{(i)}} with c, d in the same column of t_{λ^{(i)}}, where c = ξ^{ρ_{1}}a, d = ξ^{ρ_{2}}b, (1 ≤ ρ_{1}, ρ_{2} ≤ m), and i ∈ {1, …, m}. Then C_{t} ∩ R_{t′} ≠ ∅.

Proof. Let a = ξ^{s_{1}}k and b = ξ^{s_{2}}l, where 1 ≤ s_{1}, s_{2} ≤ m, 1 ≤ k, l ≤ n and k ≠ l. Then we may assume without loss of generality that c = ξ^{ρ_{1}+s_{1}}k and d = ξ^{ρ_{2}+s_{2}}l, where 1 ≤ ρ_{1}, ρ_{2} ≤ m.

First of all, let i = 1. Then

(a, b) = (k, l)w_{k}^{m+s_{2}−s_{1}}w_{l}^{m+s_{1}−s_{2}} ∈ C_{t} ∩ R_{t′}.

Now let i ∈ {2, 3, …, m}. Then

(c, d) = (k, l)w_{k}^{m+(ρ_{2}+s_{2})−(ρ_{1}+s_{1})}w_{l}^{m+(ρ_{1}+s_{1})−(ρ_{2}+s_{2})} ∈ C_{t} ∩ R_{t′}.  □

Lemma 3.8 Let t be a λ^{[m]}-tableau and t′ be a μ^{[m]}-tableau. If there exist a, b in the same row of t′_{μ^{(i)}} such that c, d are in the same column of t_{λ^{(i)}}, where c = ξ^{ρ_{1}}a, d = ξ^{ρ_{2}}b, (1 ≤ ρ_{1}, ρ_{2} ≤ m), and i ∈ {1, …, m}. Then κ_{t}{t′} = 0.

Proof. In the first place, let i = 1. Then by Lemma 3.7, (a, b) ∈ C_{t}∩R_{t′} and (e−(a, b)){t′} = 0, and furthermore (a, b) generates a subgroup of order 2 of C_{t}. Select signed coset representatives σ_{1}, …, σ_{r} for this subgroup of C_{t}; then

κ_{t}{t′} = (∑_{i=1}^{r} ξ^{−f(σ_{i})}(sgn σ_{i})σ_{i}) (e − (a, b)){t′} = 0.

Secondly, let i ∈ {2, 3, …, m}. Then, by Lemma 3.7, (c, d) ∈ C_{t}∩R_{t′} and (e−(c, d)){t′} = 0, and so (c, d) generates a subgroup of order 2 of C_{t}. Select once again signed coset representatives σ_{1}, …, σ_{s} for this subgroup of C_{t}; then

κ_{t}{t′} = (∑_{i=1}^{s} ξ^{−f(σ_{i})}(sgn σ_{i})σ_{i}) (e − (c, d)){t′} = 0.  □

Lemma 3.9 Suppose |λ^{(i)}| = |μ^{(i)}| for all i = 1, …, m − 1 and λ^{[m]} ≱μ^{[m]}. Let t be a λ^{[m]}-tableau and t′ be a μ^{[m]}-tableau such that for all a, if a occurs in t_{λ^{(i)}} then ξ^{s_{i}}a occurs in t′_{μ^{(i)}}, where i ∈ {1, …, m − 1} and s_{i} ∈ {1, …, m}. Then κ_{t}{t′} = 0.

Proof. The result follows from Corollary 2.3 and Lemma 3.8. □

Lemma 3.10 Let t be a λ^{[m]}-tableau and t′ be a μ^{[m]}-tableau such that for all a, if a occurs in t_{λ^{(i)}} then ξ^{s_{i}}a occurs in t′_{μ^{(i)}}, where i ∈ {1, …, m − 1} and s_{i} ∈ {1, …, m}. Suppose that κ_{t}{t′} ≠ 0. Then λ^{[m]} ≥ μ^{[m]}.

Proof. Suppose that a and b are two elements in the same row of t′_{μ^{(i)}}, where i ∈ {1, …, m}. Then c and d cannot be in the same column of t_{λ^{(i)}}, where c = ξ^{ρ_{1}}a, d = ξ^{ρ_{2}}b, (1 ≤ ρ_{1}, ρ_{2} ≤ m) and i ∈ {1, …, m}, for if so, then by Lemma 3.8, κ_{t}{t′} = 0, contradicting our hypothesis. Thus, Lemma 2.2 yields λ^{[m]} ≥ μ^{[m]}. □

Definition 3.11 Let t and t′ be λ^{[m]}-tableaux. We say that t agrees with t′ if whenever a occurs in t_{λ^{(i)}} then ξ^{s_{i}}a occurs in t′_{λ^{(i)}} for some s_{i} ∈ {1, …, m}, for all a and i ∈ {1, …, m}.

With this definition, we can prove the following lemma and its corollary by using Lemma 3.6 and Lemma 3.8.

Lemma 3.12 Let t and t′ be λ^{[m]}-tableaux such that t agrees with t′.

(i) If {t′} is not contained in e_{t} then κ_{t}{t′} = 0. (ii) If {t′} is contained in e_{t} then κ_{t}{t′} = ±e_{t}.

Corollary 3.13 Let t be a λ^{[m]}-tableau and u ∈ M^{λ^{[m]}}. Suppose that u is a linear combination of all tabloids {t′} such that t agrees with each t′. Then κ_{t}u is a multiple of e_{t}.

Remark 3.14 Let t and t′ be λ^{[m]}-tableaux such that t agrees with t′. Since e_{t′} is a linear combination of all tabloids such that t agrees with all of them then by Corollary 3.13 we have κ_{t}e_{t′} = βe_{t}, where β ∈ C.

Lemma 3.15 Let t be a λ^{[m]}-tableau. Then  (i) κ_{t}κ_{t} = |C_{t}|κ_{t},  (ii) κ_{t}e_{t} = |C_{t}|e_{t}.

Proof. We know that κ_{t} = ∑_{σ∈C_{t}} ξ^{−f(σ)}(sgn σ)σ. Then

κ_{t}κ_{t} = ( ∑_{σ∈C_{t}} ξ^{−f(σ)}(sgn σ)σ) ( ∑_{π∈C_{t}} ξ^{−f(π)}(sgn π)π)
= ∑_{σ,π∈C_{t}} ξ^{−f(σ)}(sgn σ)ξ^{−f(π)}(sgn π)σπ
= ∑_{σ,γ∈C_{t}} ξ^{−f(σ)}(sgn σ)ξ^{−f(σ^{−1}γ)}sgn(σ^{−1}γ)γ, (where σπ = γ)
= ∑_{σ,γ∈C_{t}} ξ^{−f(γ)}(sgn γ)γ
= ∑_{σ∈C_{t}} ∑_{γ∈C_{t}} ξ^{−f(γ)}(sgn γ)γ
= |C_{t}|κ_{t},

thus proving (i), and (ii) follows from (i). □

Lemma 3.16 Let t be a λ^{[m]}-tableau and w_{i} ∈ W_{n}^{m}, 1 ≤ i ≤ n. The following conditions are equivalent:

(i) i occurs in the k-th constituent of t, (ii) i occurs in the k-th constituent of every tabloid appearing in e_{t}, (iii) w_{i}e_{t} = ξ^{k}e_{t}.

Proof. By definition of the column stabilizer C_{t} of t, the equivalence of (i) and (ii) follows directly from the formula

e_{t} = ∑_{σ∈C_{t}} ξ^{−f(σ)}(sgn σ)σ{t}.

(ii)⇒(iii): Assume (ii) so that i occurs in the k-th constituent of every tabloid appearing in e_{t}. Then by definition of the action of W_{n}^{m} on M^{λ^{[m]}} we have w_{i}e_{t} = ξ^{k}e_{t}. (iii)⇒(ii): Let w_{i}e_{t} = ξ^{k}e_{t}. Suppose that i occurs in the l-th constituent of every tabloid appearing in e_{t} such that k ≠ l. Then we have w_{i}e_{t} = ξ^{l}e_{t}, which is a contradiction. □

We now give the main theorem of this section.

Theorem 3.17 Let C be the field of complex numbers and λ^{[m]} ∈ P^{(m)}(n). Then

Hom_{CW_{n}^{m}}(S^{λ^{[m]}}, S^{λ^{[m]}}) ≅ C.

Proof. Let I be the identity function of S^{λ^{[m]}}. We show that every CW_{n}^{m}-endomorphism of S^{λ^{[m]}} is a scalar multiple of I.

Let θ be an endomorphism of S^{λ^{[m]}} and let t be a λ^{[m]}-tableau. Suppose that for all k = 1, …, m we have that ξ^{ρ_{1}^{(k)}}i_{1}^{(k)}, …, ξ^{ρ_{s_{k}}^{(k)}}i_{s_{k}}^{(k)} occur in the k-th constituent of the tableau t, where ρ_{j}^{(k)} ∈ {1, …, m} and i_{j}^{(k)} ∈ {1, …, n} for j = 1, 2, …, s_{k}. Suppose that e_{t^{∗}} is involved in θ(e_{t}) for some e_{t^{∗}} ≠ αe_{t}, where α ∈ C and t^{∗} is a λ^{[m]}-tableau. By Lemma 3.16, we have

w_{i_{j}^{(k)}}θ(e_{t}) = θ(w_{i_{j}^{(k)}}e_{t}) = ξ^{k}θ(e_{t})

for j = 1, 2, …, s_{k} and for all k = 1, …, m.

If there exists at least one entry, say ξ^{ρ_{j}^{(k)}}i_{j}^{(k)}, in the k-th constituent of t such that ξ^{ρ_{j}^{(k)}}i_{j}^{(k)} occurs in the l-th constituent of t^{∗}, where j = 1, 2, …, s_{k}, k = 1, …, m and k ≠ l, then by using Lemma 3.16, we have w_{i_{j}^{(k)}}e_{t^{∗}} = ξ^{l}e_{t^{∗}}. This implies that for w_{i_{j}^{(k)}} ∈ W_{n}^{m},

θ(w_{i_{j}^{(k)}}e_{t}) ≠ w_{i_{j}^{(k)}}θ(e_{t}).

This means θ does not preserve the action of W_{n}^{m}. Thus θ(e_{t}) is at most a linear combination of all λ^{[m]}-polytabloids e_{t′} such that t agrees with each t′, i.e.

θ(e_{t}) = ∑ α_{t′}e_{t′}

where the summation is taken over all λ^{[m]}-tableaux t′ such that t agrees with each one of them. Then by Lemma 3.15 (ii) and Remark 3.14 we have

|C_{t}|θ(e_{t}) = θ(κ_{t}e_{t}) = κ_{t}θ(e_{t}) = ∑ α_{t′}κ_{t}e_{t′} = ∑ α_{t′}βe_{t}

where β ∈ C. Thus, θ(e_{t}) = αe_{t} where α = \frac{1}{|C_{t}|} ∑ α_{t′}β, and θ = αI for some scalar α which proves the theorem. □

Corollary 3.18 Let C be the field of complex numbers and λ^{[m]} ∈ P^{(m)}(n). Then S^{λ^{[m]}} is irreducible.

We have constructed an irreducible Specht module S^{λ^{[m]}} for each λ^{[m]} ∈ P^{(m)}(n). We shall next show that they are mutually non-isomorphic.

As in the case of symmetric groups, we define an inner product ⟨·, ·⟩ on M^{λ^{[m]}} by setting

⟨\{t\}, \{t′\}⟩ = \{ \begin{matrix}1 & if \{t\} = \{t′\} \\ 0 & if \{t\} ≠ \{t′\} \end{matrix}

and extend to be bilinear on M^{λ^{[m]}}.

Proposition 3.19 Let λ^{[m]}, μ^{[m]} ∈ P^{(m)}(n) such that |λ^{(1)}| > |μ^{(1)}| or |λ^{(j)}| = |μ^{(j)}| (1 ≤ j ≤ k) and |λ^{(k+1)}| > |μ^{(k+1)}| for some k ∈ {1, …, m − 2}. Suppose the field of scalars is C and θ ∈ Hom(S^{λ^{[m]}}, S^{μ^{[m]}}). Then θ = 0.

Proof. Let θ ∈ Hom(S^{λ^{[m]}}, S^{μ^{[m]}}). Let t be a λ^{[m]}-tableau and t′ be a μ^{[m]}-tableau. Then by hypothesis there exists at least one entry, say ξ^{ρ_{1}}i, in the p-th constituent of t such that ξ^{ρ_{2}}i occurs in the q-th constituent of t′, where i ∈ {1, …, n}, ρ_{1}, ρ_{2} ∈ {1, …, m} and p ≠ q. Then by Lemma 3.16 we have

w_{i}e_{t} = ξ^{p}e_{t} and w_{i}e_{t′} = ξ^{q}e_{t′}.

This implies that θ(w_{i}e_{t}) ≠ w_{i}θ(e_{t}), which means that θ does not preserve the action of W_{n}^{m}. Thus there is no alternative but θ(e_{t}) = 0 for all λ^{[m]}-tableaux t. Therefore θ = 0 which implies that S^{λ^{[m]}} and S^{μ^{[m]}} are not isomorphic when |λ^{(1)}| > |μ^{(1)}| or |λ^{(j)}| = |μ^{(j)}| (1 ≤ j ≤ k) and |λ^{(k+1)}| > |μ^{(k+1)}| for some k ∈ {1, …, m − 2}. □

Proposition 3.20 Let λ^{[m]}, μ^{[m]} ∈ P^{(m)}(n) such that |λ^{(i)}| = |μ^{(i)}| for all i = 1, …, m − 1. Suppose the field of scalars is C and θ ∈ Hom(S^{λ^{[m]}}, M^{μ^{[m]}}) is non-zero. Thus λ^{[m]} ≥ μ^{[m]} and if λ^{[m]} = μ^{[m]} then θ is multiplication by a scalar.

Proof. Since θ ≠ 0, there is some basis vector e_{t} such that θ(e_{t}) ≠ 0. Because ⟨·, ·⟩ is an inner product with complex scalars, M^{λ^{[m]}} = S^{λ^{[m]}} ⊕ (S^{λ^{[m]}})^{⊥}. Thus we can extend θ to an

element of Hom(M^{λ^{[m]}}, M^{μ^{[m]}}) by setting θ ((S^{λ^{[m]}})^{⊥}) = 0. So

0 ≠ θ(e_{t}) = θ(κ_{t}{t}) = κ_{t}θ({t}) = κ_{t} (∑ α_{t′}{t′}) ,

where the summation is taken over all μ^{[m]}-tableaux t′ such that for all a, if a occurs in t_{λ^{(i)}} then ξ^{s_{i}}a occurs in t′_{μ^{(i)}}, where i ∈ {1, …, m − 1} and s_{i} ∈ {1, …, m}. (Here, we need this restriction otherwise θ does not preserve the action of W_{n}^{m}). By Lemma 3.10 we have λ^{[m]} ≥ μ^{[m]}.

In the case λ^{[m]} = μ^{[m]} the above summation is taken over all λ^{[m]}-tableaux t′ such that t agrees with each t′. Hence, Corollary 3.13 yields θ(e_{t}) = βe_{t} for some β ∈ C. So for any σ ∈ W_{n}^{m},

θ(e_{σt}) = θ(ξ^{m−ϕ(σ,t)}σe_{t}) = ξ^{m−ϕ(σ,t)}σθ(e_{t})
= ξ^{m−ϕ(σ,t)}σ(βe_{t}) = βe_{σt}.

Thus θ is multiplication by a scalar. □

Theorem 3.21 The S^{λ^{[m]}} (λ^{[m]} ∈ P^{(m)}(n)) give a complete set of irreducible CW_{n}^{m}-modules.

Proof. The S^{λ^{[m]}} are irreducible by Corollary 3.18. Since we have the right number of modules for a full set, it suffices to show that they are pairwise inequivalent. Now, let λ^{[m]}, μ^{[m]} ∈ P^{(m)}(n). If |λ^{(1)}| > |μ^{(1)}| or |λ^{(j)}| = |μ^{(j)}| (1 ≤ j ≤ k) and |λ^{(k+1)}| > |μ^{(k+1)}| for some k ∈ {1, …, m − 2} then by Proposition 3.19, S^{λ^{[m]}} and S^{μ^{[m]}} are not isomorphic. If |λ^{(i)}| = |μ^{(i)}| for all i = 1, …, m − 1 and S^{λ^{[m]}} ≅ S^{μ^{[m]}}, then there exists a non-zero homomorphism θ ∈ Hom(S^{λ^{[m]}}, M^{μ^{[m]}}), since S^{μ^{[m]}} ⊆ M^{μ^{[m]}}. Thus, by Proposition 3.20, λ^{[m]} ≥ μ^{[m]}. Similarly, μ^{[m]} ≥ λ^{[m]}, so λ^{[m]} = μ^{[m]}. □

4. A Basis for Specht Modules

As in the case of the symmetric group, generally the polytabloids that generate S^{λ^{[m]}} are not linearly independent. We now determine a subset which forms a basis for S^{λ^{[m]}}.

Definition 4.1 A λ^{[m]}-tableau t is said to be standard if the entries of t are all positive integers, and t_{λ^{(i)}} is a standard tableau for all i = 1, …, m. A tabloid {t} is a standard tabloid if there is a standard tableau in the equivalence class {t}. A polytabloid e_{t} is a standard polytabloid if t is a standard tableau.

Theorem 4.2 The set {e_{t} | t is a standard λ^{[m]} − tableau} is a C-basis for S^{λ^{[m]}}. The dimension of S^{λ^{[m]}} is the number of standard λ^{[m]}-tableaux.

We will spend the next two sections proving this theorem. First we shall establish that the e_{t} are linearly independent. For this case, we shall need a partial order on tabloids associated with the same m-set of partitions of n.

Let t = (t_{λ^{(1)}}, …, t_{λ^{(m)}}) be a λ^{[m]}- tableau. Let |t_{λ^{(k)}}| denote the modulus of t_{λ^{(k)}} for all k = 1, …, m, that is, if t_{λ^{(k)}} = (t_{λ^{(k)}}(i, j)), where t_{λ^{(k)}}(i, j)) stands for the entry of t_{λ^{(k)}} in position (i, j) then |t_{λ^{(k)}}| = (|t_{λ^{(k)}}(i, j)|) for all k = 1, …, m.

Let (s_{i}(t_{λ^{(1)}}))_{1≤i≤n} be the sequence associated with t_{λ^{(1)}} such that for all i, s_{i}(t_{λ^{(1)}}) =the power of the complex coefficient of the i if i occurs in t_{λ^{(1)}}, 0 if i is not contained in t_{λ^{(1)}}, or if i occurs in t_{λ^{(1)}} without complex coefficient. (Note that at this point, we do not use the relation ξ^{\frac{m}{2}} = −1 if m is even.)

For example, if λ^{[3]} = (31, 2^{2}, 21) ∈ P^{(3)}(11) and

t = ( \begin{matrix}ξ^{2}1 & ξ7 & 3 \\ ξ9 \end{matrix} , \begin{matrix}ξ^{2}6 & 4 \\ 11 & ξ5 \end{matrix} , \begin{matrix}8 & ξ2 \\ 10 \end{matrix} )

then

|t_{λ^{(1)}}| = 1 7 3 9 , |t_{λ^{(2)}}| = 6 4 11 5 , |t_{λ^{(3)}}| = 8 2 10

and (s_{i}(t_{λ^{(1)}}))_{1≤i≤11} = (2, 0, 0, 0, 0, 0, 1, 0, 1, 0, 0).

Given any tableau t = (t_{λ^{(1)}}, …, t_{λ^{(m)}}), let m_{ir}(t_{λ^{(k)}}) denote the number of entries less than or equal to i (i = 1, …, n) in the first r rows of |t_{λ^{(k)}}|, where k = 1, …, m.

Definition 4.3 Let λ^{[m]} ∈ P^{(m)}(n). Let {t} and {s} be λ^{[m]}-tabloids. Then we write

(a) {t_{λ^{(1)}}} ≤ {s_{λ^{(1)}}} if one of the following conditions is satisfied:

1. some entries of t_{λ^{(1)}} are complex (or negative) and the entries in s_{λ^{(1)}} are all positive integers (in this case {t_{λ^{(1)}}} ◁ {s_{λ^{(1)}}}),

2. |t_{λ^{(1)}}| = |s_{λ^{(1)}}| and s_{i}(t_{λ^{(1)}}) ≤ s_{i}(s_{λ^{(1)}}) for all i = 1, …, n,

3. |t_{λ^{(1)}}| ≠ |s_{λ^{(1)}}| and for all i (1 ≤ i ≤ n) and r m_{ir}(t_{λ^{(1)}}) ≤ m_{ir}(s_{λ^{(1)}}), and

(b) {t_{λ^{(k)}}} ≤ {s_{λ^{(k)}}} if for all i (1 ≤ i ≤ n) and r m_{ir}(t_{λ^{(k)}}) ≤ m_{ir}(s_{λ^{(k)}}), where k = 2, 3, …, m.

Hence, {s} dominates {t}, written {t} ≤ {s}, if {t_{λ^{(k)}}} ≤ {s_{λ^{(k)}}} for all k = 1, …, m.

Let {t} be a λ^{[m]}-tabloid. Suppose that a, b occur in t_{λ^{(j)}} such that a = ξ^{s_{1}}k, b = ξ^{s_{2}}l, k < l and a is in the p-th row and b is in the q-th row of t_{λ^{(j)}}, where 1 ≤ s_{1}, s_{2} ≤ m, k, l ∈ {1, …, n} and j ∈ {1, …, m}. Then the definition of m_{ir}(t_{λ^{(j)}}) gives

(∗)  m_{ir}((a, b)t_{λ^{(j)}}) − m_{ir}(t_{λ^{(j)}}) = \{ \begin{matrix}1 & if q ≤ r < p and k ≤ i < l \\ −1 & if p ≤ r < q and k ≤ i < l \\ 0 & otherwise \end{matrix}

where j ∈ {1, …, m}.

Just as for partitions, there is a dominance lemma for tabloids.

Lemma 4.4 (Dominance Lemma for Tabloids) Let {t} be a λ^{[m]}- tabloid. If there exist a, b in t_{λ^{(j)}} such that a = ξ^{s_{1}}k, b = ξ^{s_{2}}l, k < l and a appears in a lower row than b, where 1 ≤ s_{1}, s_{2} ≤ m, k, l ∈ {1, …, n} and j ∈ {1, …, m} then {t} ◁ (a, b){t}.

Proof. Let a and b be two elements in t_{λ^{(j)}} such that a = ξ^{s_{1}}k, b = ξ^{s_{2}}l, where 1 ≤ k < l ≤ n, 1 ≤ s_{1}, s_{2} ≤ m and j ∈ {1, …, m}. Then

(a, b) = (k, l)w_{k}^{m+s_{2}−s_{1}}w_{l}^{m+s_{1}−s_{2}} and (a, b){t} = {(a, b)t}.

Therefore the result follows from (∗). □

The dominance lemma puts a restriction on which tabloids can appear in a standard polytabloid.

Corollary 4.5 If t is standard and {s} appears in e_{t}, then {s} ≤ {t}.

Proof. Let s = πt, where π ∈ C_{t}, so that {s} appears in e_{t}. We induct on the number of column inversions in s. Let π = γτ ∈ C_{t} such that τ ∈ S_{n} and γ = ∏_{i} w_{i}^{s_{i}} ∈ C_{m}^{n}, where each i appears in t_{λ^{(1)}} and 1 ≤ s_{i} ≤ m. We now consider all the possible cases in terms of π ∈ C_{t}.

(i) Firstly, if τ = e then π = γ. Thus, since s = γt, γ multiplies any number of symbols in t_{λ^{(1)}} by a power of ξ. Hence, some entries of s_{λ^{(1)}} are complex (or negative) and the entries in s_{λ^{(j)}} do not change for all j = 2, 3, …, m. Then we have {s_{λ^{(1)}}} ◁ {t_{λ^{(1)}}} by part (a)(1) of Definition 4.3 and {s_{λ^{(j)}}} = {t_{λ^{(j)}}} for all j = 2, 3, …, m. Therefore, {s} ◁ {t}.

(ii) Secondly, if γ = e then π = τ. Since s = τt then in the same column of s_{λ^{(j)}} there are positive integers k < l such that k is in a lower row than l, where j ∈ {1, …, m}. Thus, by Lemma 4.4 {s} ◁ (k, l){s}. Since (k, l){s} is involved in e_{t} and (k, l){s} has fewer inversions than {s}, induction shows that (k, l){s} ≤ {t}. Therefore, {s} ◁ {t}.

(iii) Finally, if τ, γ ≠ e then π = γτ and s = γτt. By (i), some entries of s_{λ^{(1)}} are complex (or negative) and so we have at once {s_{λ^{(1)}}} ◁ {t_{λ^{(1)}}}. Furthermore, in the same column of s_{λ^{(j)}} there are entries a = ξ^{s_{1}}k and b = ξ^{s_{2}}l such that k < l and a appears in a lower row than b, where 1 ≤ s_{1}, s_{2} ≤ m, k, l ∈ {1, …, n} and j ∈ {1, …, m}. Thus, by Lemma 4.4 {s} ◁ (a, b){s}. Since (a, b){s} is involved in e_{t} and (a, b){s} has fewer inversions than {s}, induction shows that (a, b){s} ≤ {t}. Therefore, {s} ◁ {t}. Hence, we have the required result. □

The previous corollary says that {t} is the maximum tabloid in e_{t}.

Lemma 4.6 Let v_{1}, …, v_{k} be elements of M^{λ^{[m]}}. Suppose, for each v_{i}, we can choose a tabloid {t_{i}} appearing in v_{i} such that

1. {t_{i}} is maximum in v_{i}, and

2. the {t_{i}} are all distinct. Then v_{1}, …, v_{k} are linearly independent over C.

Proof. Choose the labels such that {t_{1}} is maximal among the {t_{i}}. Thus conditions 1 and 2 ensure that {t_{1}} appears only in v_{1}. (If {t_{1}} occurs in v_{i}, i > 1, then {t_{1}}◁{t_{i}}, contradicting the choice of {t_{1}}.) It follows that in any linear combination

c_{1}v_{1} + c_{2}v_{2} + ⋯ + c_{k}v_{k} = 0

we must have c_{1} = 0 since there is no other way to cancel {t_{1}}. By induction on k, the rest of the coefficients must also be zero. □

We now have all the ingredients to prove independence of the standard basis.

Proposition 4.7 The set {e_{t} | t is a standard λ^{[m]}-tableau} is linearly independent over C.

Proof. By Corollary 4.5, {t} is maximum in e_{t}, and by hypothesis they are all distinct. Thus Lemma 4.6 applies. □

5. The Garnir Relations

In this section we shall show that standard λ^{[m]}-polytabloids span S^{λ^{[m]}}; that is, if t is an arbitrary λ^{[m]}- tableau, then e_{t} is a linear combination of standard λ^{[m]}-polytabloids.

To determine the Garnir element of t associated with e_{t}, we use the following relations.

Lemma 5.1 Let t be a λ^{[m]}-tableau.

(i) Let a and b be two elements in the same column of t_{λ^{(j)}} such that a = ξ^{s_{1}}k, b = ξ^{s_{2}}l, where 1 ≤ s_{1}, s_{2} ≤ m, k, l ∈ {1, …, n} and j ∈ {1, …, m}. Then (a, b) = (k, l)w_{k}^{m+s_{2}−s_{1}}w_{l}^{m+s_{1}−s_{2}} generates a subgroup of order 2 of C_{t} and

(e + (a, b))e_{t} = 0 (alternacy relation).

(ii) For any entry a in the tableau t such that a = ξ^{s}i, where 1 ≤ s < m, 1 ≤ i ≤ n,

(e + αw_{i}^{m−s})e_{t} = 0 (primitive root change relation),

where α = −ξ^{−j(m−s)} if a appears in t_{λ^{(j)}} for j ∈ {1, …, m}.

Proof. (i) follows immediately from (a, b)e_{t} = −e_{t}. To prove (ii), we use the following fact

w_{i}^{m−s}e_{t} = ξ^{ϕ(w_{i}^{m−s},t)}e_{t},

where ϕ(w_{i}^{m−s}, t) = j(m − s) if a appears in t_{λ^{(j)}} for j ∈ {1, …, m}. □

Remark 5.2 The previous lemma says that we can find the elements of W_{n}^{m} which remove any complex (or negative) entries of t associated with e_{t} (primitive root change relation) and which make t column standard (alternacy relation), i.e. the tableau t associated with e_{t} may be reorganized so that no complex (or negative) entries remain and all columns are standard.

Now we shall find elements of the group algebra of G(m, 1, n) which annihilate the given polytabloid e_{t}. Let t be a λ^{[m]}-tableau associated with e_{t} such that the entries of t were reorganized by the primitive root change and alternacy relations. Suppose that t has entries t(i, j, k) and t(i, j + 1, k), where k ∈ {1, …, m}. Let A be the set of entries in column j of t below and including t(i, j, k), and let B be the set of entries in column j + 1 of t above and including t(i, j + 1, k). Suppose that A = {a_{i} : i ∈ I} and B = {b_{j} : j ∈ J}, where I and J are index sets. Let S_{A}, S_{B}, S_{A∪B} be the subgroups of G(m, 1, n). Let σ_{1}, …, σ_{r} be coset representatives for S_{A} × S_{B} in S_{A∪B}, and let

S_{A∪B} = ⨄_{j=1}^{r} σ_{j}(S_{A} × S_{B}) and G_{A,B} = ∑_{j=1}^{r}(sgn σ_{j})σ_{j}.

G_{A,B} is called a Garnir element associated with t.

Remark 5.3 (i) If the elements of A and B belong to t_{λ^{(1)}}, then we assume that S_{A}, S_{B}, S_{A∪B} are the subgroups of G(m, 1, n) generated by the elements in A, B, A ∪ B respectively. Let C_{A∪B} (⊂ C_{m}^{n}) be the subgroup of S_{A∪B} generated by {w_{a_{i}}, w_{b_{j}} | i ∈ I, j ∈ J}. Since C_{A∪B} is also a subgroup of S_{A} × S_{B}, all the coset representatives σ_{1}, …, σ_{r} are in S_{n}. (ii) If the elements of A and B belong to the t_{λ^{(j)}}, where j ∈ {2, 3, …, m} then we assume that S_{A}, S_{B}, S_{A∪B} are the subgroups of G(m, 1, n) generated by permutations on A, B, A ∪ B respectively. (iii) The coset representatives σ_{1}, …, σ_{r} are, of course, not unique, but for practical purposes note that we may take σ_{1}, …, σ_{r} so that σ_{1}t, …, σ_{r}t are all the tableaux which agree with t except in the positions occupied by A ∪ B, and whose entries increase vertically downwards in the positions occupied by A ∪ B.

Let H be any subset of W_{n}^{m}. Define

H̄ = ∑_{σ∈H} ξ^{−f(σ)}(sgn σ)σ

and if H = {σ} then we write σ̄ = ξ^{−f(σ)}(sgn σ)σ for H̄.

Lemma 5.4 Let H ≤ W_{n}^{m} be a subgroup. Let a = ξ^{s_{1}}k and b = ξ^{s_{2}}l, where 1 ≤ s_{1}, s_{2} ≤ m, and k, l ∈ {1, 2, …, n}, k ≠ l.

(i) If the element (a, b) = (k, l)w_{k}^{m+s_{2}−s_{1}}w_{l}^{m+s_{1}−s_{2}} ∈ H then we can factor H̄ = c(e−(a, b)), where c ∈ CW_{n}^{m}.

(ii) If t is a λ^{[m]}-tableau with a, b in the same row of t_{λ^{(j)}}, where j ∈ {1, 2, …, m}, and (a, b) ∈ H, then H̄{t} = 0.

Proof. (i) Consider the subgroup K = {e, (a, b)} of H. Select coset representatives σ_{1}, …, σ_{s}

for K in H and write H = ⨄_{i=1}^{s} σ_{i}K. But then

H̄ = (∑_{i=1}^{s} σ̄_{i}) (e − (a, b)),

as desired. (ii) By hypothesis, (a, b){t} = {(a, b)t} = {t}. Thus,

H̄{t} = c(e − (a, b)){t} = c({t} − {t}) = 0.  □

Proposition 5.5 Let t, A, B be as in the definition of a Garnir element. If |A ∪ B| is greater than the number of elements in column j of t then G_{A,B}e_{t} = 0.

Proof. Let

\overline{S_{A} × S_{B}} = ∑_{σ∈S_{A}×S_{B}} ξ^{−f(σ)}(sgn σ)σ and \overline{S_{A∪B}} = ∑_{σ∈S_{A∪B}} ξ^{−f(σ)}(sgn σ)σ.

Consider any σ ∈ C_{t}. If the elements of A and B are in the t_{λ^{(1)}} then by the hypothesis there must be positive integers a, b ∈ A ∪ B such that c, d belong to the same row of σt_{λ^{(1)}} where c = ξ^{s_{1}}a, d = ξ^{s_{2}}b, 1 ≤ s_{1}, s_{2} ≤ m. Thus,

(c, d) = (a, b)w_{a}^{m+s_{2}−s_{1}}w_{b}^{m+s_{1}−s_{2}} ∈ S_{A∪B},

by Remark 5.3 (i). If the elements of A and B are in the t_{λ^{(j)}}, where j ∈ {2, 3, …, m} then by the hypothesis there must be positive integers a, b ∈ A ∪ B such that a, b belong to the same row of σt_{λ^{(j)}}, where j ∈ {2, 3, …, m}. Therefore, (a, b) ∈ S_{A∪B} by Remark 5.3 (ii).

In terms of the elements of A and B appearing in the t_{λ^{(j)}}, where j ∈ {1, 2, …, m}, (c, d) or (a, b) belong to the same group S_{A∪B}. But then \overline{S_{A∪B}}\{σt\} = 0 by Lemma 5.4. Since this is true for every σ appearing in κ_{t}, we have \overline{S_{A∪B}}e_{t} = 0.

Now S_{A∪B} = ⨄_{j=1}^{r} σ_{j}(S_{A}×S_{B}) so \overline{S_{A∪B}} = G_{A,B}\overline{S_{A} × S_{B}} by Remark 5.3. Since S_{A}×S_{B} ⊂ C_{t}

then \overline{S_{A} × S_{B}} is a factor of κ_{t} and \overline{S_{A} × S_{B}}e_{t} = |S_{A} × S_{B}|e_{t}. Therefore,

0 = \overline{S_{A∪B}}e_{t} = |S_{A} × S_{B}|G_{A,B}e_{t}.

Thus, G_{A,B}e_{t} = 0. □

We shall now use the Garnir relations, primitive root change relation and alternacy relation to prove that any polytabloid can be written as a linear combination of standard polytabloids.

Theorem 5.6 The set {e_{t} | t is a standard λ^{[m]}−tableau} spans S^{λ^{[m]}}.

Proof. First we write [t] for the column equivalence class of t, that is

[t] = {t′ | t′ = σt for some σ ∈ C_{t}}.

The column equivalence classes are partially ordered in a way similar to the partial order in Definition 4.3 on the row equivalence classes.

Let t be any λ^{[m]}-tableau associated with e_{t}. Then we may assume that e_{t} may be written as a linear combination of column standard polytabloids by Lemma 5.1. Therefore, because of Remark 5.2, we may suppose that the entries in t are all positive integers and are in increasing order down the columns. Suppose that t is not standard. This means that some constituents of t are not standard, say r constituents are not standard where r ≤ m. Suppose that the k-th constituent of t is not standard. By induction, we may assume that e_{s} can be written as a linear combination of the polytabloids e_{t′} such that the k-th constituent of each t′ is standard when [s]▷[t] and prove the same result for e_{t} by considering the k-th constituent of t.

Since we have supposed that the k-th constituent of t is not standard then there must be some adjacent pair of columns in the k-th constituent of t, say the j-th and (j + 1)-th columns, with entries a_{1} < a_{2} < … < a_{p} and b_{1} < b_{2} < … < b_{q} with a_{i} > b_{i} for some i. Thus we have the following situation in the k-th constituent of t:

a_{1}  b_{1}
∧
a_{2}  b_{2}
∧
⋮  ⋮
∧
a_{i} > b_{i}
∧
a_{i+1}  ⋮
∧  b_{q}
⋮
∧
a_{p}

Take A = {a_{i}, a_{i+1}, …, a_{p}} and B = {b_{1}, b_{2}, …, b_{i}}, and consider the corresponding Garnir element G_{A,B} = ∑_{σ}(sgn σ)σ. By Proposition 5.5, we have G_{A,B}e_{t} = 0 so that

e_{t} = − ∑_{σ≠e}(sgn σ)e_{σt}.

But b_{1} < b_{2} < … < b_{i} < a_{i} < … < a_{p} implies that [σt] ≥ [t] for σ ≠ e by the column analogue of the dominance lemma for tabloids (Lemma 4.4). Since

e_{t} = − ∑_{σ≠e}(sgn σ)e_{σt},

the result follows from our induction hypothesis for the k-th constituent of t associated with e_{t}. Hence, if we repeat the above process (r − 1)-times then the result can be deduced. □

Acknowledgement. I would like to express my appreciation to Professor A. O. Morris who, as my supervisor, provided both assistance and encouragement.

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Received July 13, 1995