The Duals of the 2-Modular Irreducible Modules of the Alternating Groups

We determine the dual modules of all irreducible modules of alternating groups over fields of characteristic 2.


Introduction and statement of the result
Let S n be the symmetric group of degree n ≥ 1 and let k be a field of characteristic p > 0. In [7,Theorem 11.5] G. James constructed all irreducible kS n -modules D λ where λ ranges over the p-regular partitions of n. Here a partition is p-regular if each of its parts occurs with multiplicity less than p.
As the alternating group A n has index 2 in S n , the restriction D λ ↓ An is either irreducible or splits as a direct sum of two non-isomorphic irreducible kA n -modules. Moreover, every irreducible kA n -module is a direct summand of some D λ ↓ An .
In this note we determine the dual of each irreducible kA n -module. Now D λ ↓ An is a self-dual kA n -module, as D λ is a self-dual kS n -module. So we only need to determine the dual of an irreducible kA n -module which is a direct summand of D λ ↓ An , when this module is reducible. This paper is a contribution to the Special Issue on the Representation Theory of the Symmetric Groups and Related Topics. The full collection is available at https://www.emis.de/journals/SIGMA/symmetric-groups-2018.html For example D (7,5,1) ↓ A 13 ∼ = S ⊕ S * , for a non self-dual irreducible kA 13 -module S, and D (5,4,3,1) ↓ A 13 decomposes similarly. On the other hand D (7,6) ↓ A 13 ∼ = S 1 ⊕ S 2 where S 1 and S 2 are irreducible and self-dual. In order to prove Theorem 1.2, we use the following elementary result, which requires the assumption that k has characteristic 2: Lemma 1.3. Let G be a finite group and let M be a semisimple kG-module which affords a non-degenerate G-invariant symmetric bilinear form B. Suppose that B(tm, m) = 0, for some involution t ∈ G and some m ∈ M . Then M has a self-dual irreducible direct summand.
Proof . We have M = n i=1 M i , for some n ≥ 1 and irreducible kG-modules M 1 , . . . , M n . Write m = m i , with m i ∈ M i , for all i. Then The last equality follows from the fact that char(k) = 2 and Without loss of generality B(tm 1 , m 1 ) = 0. Then B restricts to a non-zero G-invariant symmetric bilinear form 2 Known results on the symmetric and alternating groups

The irreducible modules of the symmetric groups
We use the ideas and notation of [7]. In particular for each partition λ of n, James defines the Young diagram [λ] of λ, and the notions of a λ-tableau and a λ-tabloid.
where R x is the row stabilizer of x. We regard {x} as an ordered set partition of {1, . . . , n}. The Z-span of the λ-tabloids forms the ZS n -lattice M λ , and the set of λ-tabloids is an S n -invariant Z-basis of M λ .
Recall from [7, Section 4] that corresponding to each tableau x there is a polytabloid e x := sgn(σ){σx} in M λ . Here σ ranges over the permutations in the column stabilizer C x of the tableau x. The Specht lattice S λ is defined to be the Z-span of all λ-polytabloids. In particular S λ is a ZS n -sublattice of M λ ; it has as Z-basis the polytabloids corresponding to the standard λ-tableaux (i.e., the numbers increase from left-to-right along rows, and from top-to-bottom along columns). Now James defines , to be the symmetric bilinear form on M λ which makes the tabloids into an orthonormal basis. As the tabloids are permuted by the action of S n , it is clear that , is S n -invariant.
Suppose now that λ is a strict partition and consider the unique permutation τ ∈ R x which reverses the order of the symbols in each row of the tableau x. In [7, Lemma 10.4] James shows that τ e x , e x = 1, as {x} is the only tabloid common to e x and e τ x (in fact James proves that τ e x , e x is coprime to p, if λ is p-regular, for some prime p). Set J λ := {x ∈ S λ | x, y ∈ 2Z, for all y ∈ S λ }. Then 2S λ ⊆ J λ and it follows from [7,Theorem 4.9] that D λ := (S λ /J λ ) ⊗ F 2 k is an absolutely irreducible kS n -module, for any field k of characteristic 2.

The real 2-regular conjugacy classes of the alternating groups
A conjugacy class of a finite group G is said to be 2-regular if its elements have odd order. R. Brauer proved that the number of irreducible kG-modules equals the number of 2-regular conjugacy classes of G [4]. Now Brauer's permutation lemma holds for arbitrary fields [3, footnote 19]. So it is clear that the number of self-dual irreducible kG-modules equals the number of real 2-regular conjugacy classes of G.
We review some well known facts about the 2-regular conjugacy classes of the alternating group. See for example [8,Section 2.5].
Corresponding to each partition µ of n there is a conjugacy class C µ of S n ; its elements consist of all permutations of n whose orbits on {1, . . . , n} have sizes {µ 1 , . . . , µ } (as multiset). So C µ is 2-regular if and only if each µ i is odd.
Let µ be a partition of n into odd parts. Then C µ ⊆ A n . If µ has repeated parts then C µ is a conjugacy class of A n . As C µ is closed under taking inverses, C µ is a real conjugacy class of A n . Now assume that µ has distinct parts. Then C µ is a union of two conjugacy classes C ± µ of A n . Set m := n− (µ) 2 and let z ∈ C µ . Then z is inverted by an involution t ∈ S n of cycle type (2 m , 1 n−2m ). Since C Sn (z) ∼ = Z/µ j Z is odd, t generates a Sylow 2-subgroup of the extended centralizer C * Sn (z) of z in S n . It follows that z is conjugate to z −1 in A n if and only if t ∈ A n . This shows that C ± µ are real classes of A n if and only if n− (µ) 2 is even. This and the discussion above shows:

Bressoud's bijection
We need a special case of a partition identity of I. Schur [9]. This was already used by Benson in his proof of Proposition 1.1: Proposition 3.1 (Schur, 1926). The number of strict partitions of n into odd parts equals the number of strict partitions of n into parts congruent to 0, ±1 (mod 4) where consecutive parts differ by at least 4 and consecutive even parts differ by at least 8. D. Bressoud [5] has constructed a bijection between the relevant sets of partitions. We describe a simplified version of this bijection.
Next we form the sequence of positive integers σ = (σ 1 , σ 2 , . . . , σ s ), where σ j is the sum of the parts in the j-th block of µ. Then the σ j are distinct, as the odd parts form a decreasing sequence, with minimal difference 4, and the even parts form a decreasing sequence, with minimal difference 8. Moreover, each even σ j is the sum of a pair of consecutive odd integers. So σ j ≡ 2 (mod 4), for all j > 0.
By construction, the minimal difference between the parts of γ is 4 and the minimal difference between the even parts of γ is 8. Moreover, γ j ≡ ζ τ j (mod 4). So γ j ≡ 2 (mod 4). Then µ → γ is Bressoud's bijection.
Finally form a strict partition λ of n which has 2s − 1 or 2s parts, by defining Then λ satisfies the constraints (i) and (ii) of Proposition 1.1. Conversely, it is easy to see that if λ satisfies these constraints, then λ is the image of some strict odd partition µ of n under the above sequence of operations. Proof . Each pair of consecutive parts λ 2j−1 , λ 2j of λ corresponds to a block B of µ. Moreover by our description of Bressoud's bijection, there are integers q 1 , . . . , q s , with q j = 0 such that

Proof of Theorem 1.2
Let D(n) be the set of strict partitions of n and let S(n) be the set of strict partitions of n which satisfy conditions (i) and (ii) in Proposition 1.1. So there are 2| S(n)| + | D(n)\ S(n)| irreducible kA n -modules. Next set S(n) + := {λ ∈ S(n) | λ 2j is even}. Then it follows from Lemmas 2.1 and 3.2 that the number of self-dual irreducible kA n -modules equals 2| S(n) + | + | D(n)\ S(n)|. Now D λ ↓ An is an irreducible self-dual kA n -module, for λ ∈ D(n)\ S(n). So we can prove Theorem 1.2 by showing that the irreducible direct summands of D λ ↓ An are self-dual for all λ ∈ S(n) + .
Suppose then that λ ∈ S(n) + . Let τ ∈ S n be the permutation which reverses each row of a λ-tableau, as discussed in Section 2.1. We claim that τ ∈ A n . For τ is a product of Since D λ is irreducible and the form , is non-zero, , is non-degenerate on D λ . Write D λ ↓ An = S 1 ⊕ S 2 , where S 1 and S 2 are non-isomorphic irreducible modules. As τ ∈ A n , it follows from Lemma 1.3 that we may assume that S 1 is self-dual. Now S * 2 ∼ = S * 1 ∼ = S 1 and S * 2 is isomorphic to a direct summand of D λ ↓ An . So S 2 is also self-dual. This completes the proof of the theorem.

Irreducible modules of alternating groups over fields of odd characteristic
We now comment briefly on what happens when k is a splitting field for A n which has odd characteristic p. Let sgn be the sign representation of kS n . So sgn is 1-dimensional but nontrivial. G. Mullineux defined a bijection λ → λ M on the p-regular partitions of n and conjectured that D λ ⊗ sgn = D λ M for all p-regular partitions λ of n. This was only proved in the 1990's by Kleshchev and Ford-Kleshchev. See [6] for details. Now D λ ↓ An ∼ = D λ M ↓ An , and D λ ↓ An is irreducible if and only if λ = λ M See [2] for details. Moreover D λ and D λ M are duals of each other, by [7,Theorem 6.6]. So D λ ↓ An is self-dual, if λ = λ M . However when λ = λ M , we do not know how to determine when the two irreducible direct summands of D λ ↓ An are self-dual.