The irreducible Specht modules in characteristic 2

Gordon James and Andrew Mathas

In the representation theory of finite groups it is useful to know which ordinary irreducible representations remain irreducible modulo a prime p. For the symmetric groups 𝔖_{n} this amounts to determining which Specht modules are irreducible over a field of characteristic p. Throughout this note we work in characteristic 2, and in this case we classify the irreducible Specht modules, thereby verifying the conjecture in [3, p. 97].

Recall that a partition is 2–regular if all of its non–zero parts are distinct; otherwise the partition is 2-singular. The irreducible Specht modules S^{λ} with λ a 2–regular partition were classified in [2]. Let λ′ denote the partition conjugate to λ. If S^{λ} is irreducible then S^{λ′} is irreducible, since S^{λ′} is isomorphic to the dual of S^{λ} tensored with the sign representation. It turns out that if neither λ nor λ′ is 2–regular then S^{λ} is irreducible only if λ = (2, 2).

In order to state our theorem, for an integer k, we let l(k) be the least non–negative integer such that k < 2^{l(k)}.

Main Theorem Let λ = (λ_{1}, λ_{2}, …) be a partition. Then the Specht module S^{λ} is irreducible in characteristic 2 if and only if

(i) λ_{i} − λ_{i+1} ≡ −1 mod 2^{l(λ_{i+1}−λ_{i+2})} for all i ≥ 1; or, (ii) λ′_{i} − λ′_{i+1} ≡ −1 mod 2^{l(λ′_{i+1}−λ′_{i+2})} for all i ≥ 1, where λ′ = (λ′_{1}, λ′_{2}, …) is

the partition conjugate to λ; or, (iii) λ = (2, 2).

In part (i) of the Theorem, λ is necessarily 2–regular and the irreducible Specht modules S^{λ} for λ a 2–regular partition were classified in [2, Corollary 2.14]. Part (ii) therefore classifies the irreducible Specht modules S^{λ} when λ′ is 2–regular. For the proof of the Theorem we must show that (2, 2) is the only partition λ such that λ and λ′ are 2–singular and S^{λ} is irreducible.

If λ is 2-regular then S^{λ} modulo its radical is an irreducible module D^{λ}, and every irreducible module for the symmetric group arises in this way. If both λ and λ′ are 2–singular and λ ≠ (2, 2) we show that S^{λ} is reducible by exhibiting two distinct composition factors. One of these composition factors is D^{λ^{R}} where λ^{R} is the 2–regularization of λ, which we define next.

Let ⊵ be the usual dominance order on partitions; see [3, Definition 3.2]. Given partitions λ and μ write λ ∼ μ if, for all k, we have

# { (i, j) | 1 ≤ j ≤ λ_{i} and k = i + j } = # { (i, j) | 1 ≤ j ≤ μ_{i} and k = i + j } .

A.M.S. subject classification (1991): 20C30

It turns out that, given λ, there is a unique partition λ^{R} such that λ^{R} ⊵ μ for all partitions μ with μ ∼ λ.

Proposition 1 [1, Theorem A] The partition λ^{R} is 2–regular. Moreover, D^{λ^{R}} is a composition factor of S^{λ}.

In order to produce another composition factor of S^{λ}, we introduce some definitions.

Definition Let λ = (λ_{1}, λ_{2}, …) be a partition and suppose that r is minimal such that λ_{r+1} = 0. We define integers N_{i}(λ) for i ≥ 1 as follows.

(i) For i > r let N_{i}(λ) = 0. (ii) Let N_{r−1}(λ) be the largest integer of the form 2^{m} − 1 such that N_{r−1}(λ) ≤

λ_{r−1}. (iii) Next, define N_{i}(λ) for i = r − 2, r − 3, …, 1 to be the largest integer of the

form 2^{m} − 1 such that N_{i}(λ) ≤ λ_{i} − λ_{i+1} + N_{i+1}(λ).

(iv) Finally, let N_{r}(λ) = \{ \begin{matrix}min\{λ_{r}, N_{r−1}(λ)\}, & if r > 1, \\ λ_{1}, & otherwise. \end{matrix}

Lemma 1 Suppose that λ = (λ_{1}, λ_{2}, …, λ_{r}) is a partition with λ_{r} ≠ 0.

(i) If i ≥ 1 then N_{i}(λ) ≤ λ_{i} − λ_{i+1} + N_{i+1}(λ). (ii) If 1 ≤ i ≤ r − 1 then N_{i}(λ) is the largest integer of the form 2^{m} − 1 such

that N_{i}(λ) ≤ λ_{i} − λ_{i+1} + N_{i+1}(λ). (iii) If 1 ≤ i ≤ r − 1 then λ_{i} − λ_{i+1} + N_{i+1} ≤ 2N_{i}(λ). (iv) We have λ_{r−1} ≤ 2N_{r−1}(λ).

Proof. (i) This follows easily from the definition of N_{i}(λ).

(ii) If 1 ≤ i < r − 1 then the result is just the definition of N_{i}(λ). Next, note that N_{r−1}(λ) ≤ λ_{r−1} − λ_{r} + N_{r}(λ) ≤ λ_{r−1} since N_{r}(λ) = min{λ_{r}, N_{r−1}(λ)}. However, N_{r−1}(λ) is the largest integer of the form 2^{m} − 1 such that N_{r−1}(λ) ≤ λ_{r−1}. Therefore, part (ii) holds when i = r − 1.

(iii) This follows from the maximality of N_{i}(λ) in part (ii).

(iv) This follows from the definition of N_{r−1}(λ).

Observe, also, that λ_{i} ≥ N_{i}(λ) ≥ 1 for 1 ≤ i ≤ r and

λ_{1} − N_{1}(λ) ≥ λ_{2} − N_{2}(λ) ≥ · · · .

Therefore, (λ_{1} − N_{1}(λ), λ_{2} − N_{2}(λ), ⋯) is a partition.

Definition Let α(λ) = (λ_{1} − N_{1}(λ), λ_{2} − N_{2}(λ), ⋯).

For i ≥ 1 and j ≥ 1 let N_{i,j}(λ) = N_{i}(α^{i−1}(λ)).

Note that N_{i}(λ) = N_{i,1}(λ).

Example In the figures below, the number of j’s in row i is equal to N_{i,j}(λ) for λ = (11, 9, 5, 1) and for λ = (12, 10, 6, 2) respectively.

3 2 2 2 1 1 1 1 1 1 1 3 2 1 1 1 1 1 1 1 3 2 1 1 1 1  3 3 2 2 2 1 1 1 1 1 1 1 2 2 2 1 1 1 1 1 1 1 2 2 2 1 1 1 1 1

Lemma 2 Suppose that λ = (λ_{1}, λ_{2}, …, λ_{r}) is a partition with λ_{r} ≠ 0.

(i) If j ≥ 1 then N_{1,j}(λ) ≥ N_{2,j}(λ) ≥ N_{3,j}(λ) ≥ · · ·. (ii) If i ≥ 1 then N_{i,1}(λ) ≥ N_{i,2}(λ) ≥ N_{i,3}(λ) ≥ · · ·.

Proof. (i) By Lemma 1(iii), if 1 ≤ i ≤ r − 2 then N_{i+1,1}(λ) ≤ 2N_{i,1}(λ), and hence N_{i+1,1}(λ) ≤ N_{i,1}(λ) since both N_{i+1,1}(λ) and N_{i,1}(λ) are one less than a power of 2. Also, by definition, 0 = N_{r+1,1}(λ) < N_{r,1}(λ) ≤ N_{r−1,1}(λ); so we have now proved part (i) for j = 1. The general case follows by replacing λ by α^{j−1}(λ).

(ii) As in part (i), it suffices to show that N_{i,1}(λ) ≥ N_{i,2}(λ) for all i ≥ 1. If i > r then N_{i,2}(λ) = N_{i,1}(λ) = 0. Next, consider the case i = r. If N_{r,1}(λ) = λ_{r} then N_{r,2}(λ) = 0 < N_{r,1}(λ). If

N_{r,1}(λ) = N_{r−1,1}(λ) then λ_{r} ≤ λ_{r−1} ≤ 2N_{r−1,1}(λ), by Lemma 1(iv); therefore, N_{r,2}(λ) ≤ λ_{r} − N_{r,1}(λ) ≤ N_{r−1,1}(λ) = N_{r,1}(λ).

Now assume that 1 ≤ i < r. Then N_{i+1,1}(λ) ≥ N_{i+1,2}(λ) by induction. Therefore, N_{i,1}(λ) ≥ N_{i+1,2}(λ) since N_{i,1}(λ) ≥ N_{i+1,1}(λ) by part (i), and

λ_{i} − λ_{i+1} + N_{i+1,1}(λ) ≥ (λ_{i} − N_{i,1}(λ)) − (λ_{i+1} − N_{i+1,1}(λ)) + N_{i+1,2}(λ)
≥ N_{i,2}(λ), by Lemma 1(i).

If N_{i,2}(λ) has the form 2^{m}−1 for some m, then N_{i,2}(λ) ≤ N_{i,1}(λ) by Lemma 1(ii). If N_{i,2}(λ) does not have the form 2^{m} − 1 then N_{i+1,1}(λ) = λ_{i+1} and so λ_{i} ≤ 2N_{i,1}(λ) by Lemma 1(iii); therefore, N_{i,2}(λ) ≤ λ_{i} − N_{i,1}(λ) ≤ N_{i,1}(λ) as required.

Lemma 3 Suppose that λ = (λ_{1}, λ_{2}, …, λ_{r}) is a partition with λ_{r} ≠ 0 and that 1 ≤ i ≤ r − 1. Then

λ_{i} ≤ 2N_{i,1}(λ) + ∑_{j=i+1}^{r−1} N_{j,1}(λ),

with equality only if N_{j,2}(λ) = N_{j,1}(λ), for i ≤ j ≤ r − 1, and N_{r,2}(λ) = λ_{r} − N_{r,1}(λ).

Proof. By parts (iv) and (iii) of Lemma 1 we have

λ_{r−1} ≤ 2N_{r−1,1}(λ)
λ_{r−2} − λ_{r−1} ≤ 2N_{r−2,1}(λ) − N_{r−1,1}(λ)
λ_{r−3} − λ_{r−2} ≤ 2N_{r−3,1}(λ) − N_{r−2,1}(λ)
⋮
λ_{i} − λ_{i+1} ≤ 2N_{i,1}(λ) − N_{i+1,1}(λ).

Summing these inequalities we obtain

λ_{i} ≤ 2N_{i,1}(λ) + N_{i+1,1}(λ) + ⋯ + N_{r−1,1}(λ)

with equality only if we have equalities throughout.

Assume that we have equalities throughout. Recall that N_{j,2}(λ) = N_{j,1}(α(λ)) and write α(λ) = (α(λ)_{1}, α(λ)_{2}, …).

First, note that α(λ)_{r−1} = λ_{r−1} − N_{r−1,1}(λ) = N_{r−1,1}(λ) > 0.

Suppose that i ≤ j < r − 1. Then, using the equalities above,

α(λ)_{j} = λ_{j} − N_{j,1}(λ) = λ_{j+1} − N_{j+1,1}(λ) + N_{j,1}(λ) = α(λ)_{j+1} + N_{j,1}(λ).

Since α(λ)_{r−1} ≠ 0, Lemma 1(ii) implies that N_{j,2}(λ) is the largest integer of the form 2^{m} − 1 such that

N_{j,2}(λ) ≤ α(λ)_{j} − α(λ)_{j+1} + N_{j+1,2}(λ) = N_{j,1}(λ) + N_{j+1,2}(λ).

Therefore, N_{j,2}(λ) ≥ N_{j,1}(λ), and so N_{j,2}(λ) = N_{j,1}(λ), by Lemma 2(ii).

Next, consider N_{r−1,2}(λ). We have α(λ)_{r−1} = N_{r−1,1}(λ). Therefore, if α(λ)_{r} = 0 then

N_{r−1,2}(λ) = min{N_{r−1,1}(λ), N_{r−2,2}(λ)} = N_{r−1,1}(λ),

since we have already shown that N_{r−2,2}(λ) = N_{r−2,1}(λ), and N_{r−2,1}(λ) ≥ N_{r−1,1}(λ) by Lemma 2(i). Assume, now, that α(λ)_{r} ≠ 0. Then N_{r−1,2}(λ) is the largest integer of the form 2^{m} − 1 such that N_{r−1,2}(λ) ≤ α(λ)_{r−1} = N_{r−1,1}(λ). Therefore, N_{r−1,2}(λ) = N_{r−1,1}(λ).

Finally, note that N_{r,1}(λ) = min{λ_{r}, N_{r−1,1}(λ)}; therefore, λ_{r} − N_{r,1}(λ) ≤ λ_{r−1} − N_{r−1,1}(λ) = N_{r−1,1}(λ). Thus,

N_{r,2}(λ) = min{λ_{r} − N_{r,1}(λ), N_{r−1,2}(λ)} = λ_{r} − N_{r,1}(λ)

since N_{r−1,2}(λ) = N_{r−1,1}(λ).

Definition Suppose that λ is a partition of n. Let λ^{♭} = (λ_{1}^{♭}, λ_{2}^{♭}, …) where λ_{j}^{♭} = ∑_{i≥1} N_{i,j}(λ).

By Lemma 2(ii), λ^{♭} is a partition of n. It is easy to see that λ^{♭} ⊵ λ′. The point of the construction of λ^{♭} is the following.

Proposition 2 [3, Theorem 24.6] Suppose that λ is a partition of n and let F be a field of characteristic 2. Then Hom_{F 𝔖_{n}}(M^{λ^{♭}}, S^{λ}) ≠ 0, where M^{λ^{♭}} is the permutation representation of 𝔖_{n} on the Young subgroup for the partition λ^{♭}.

Proof. Let T be a λ^{♭}–tableau of type λ such that for all i and j the number of i’s in row j of T is N_{i,j}(λ). The definition of N_{i,j}(λ) implies that for all i, j we have

N_{i,j}(λ) ≡ −1 mod 2^{l(N_{i+1,j}(λ))}.

(Recall that l(k) is the least integer such that k < 2^{l(k)}.) By [3, Theorem 24.6] there exists a non–zero θ_{T} ∈ Hom_{F 𝔖_{n}}(M^{λ^{♭}}, S^{λ}). (In [3], the hypothesis that T is reverse semistandard is used only to show that ker θ_{T} ⊆ S^{λ^{♭}⊥}; this is not needed here.)

Corollary Suppose that λ^{R} ⋭ λ^{♭}. Then S^{λ} is reducible.

Proof. By [3, Theorem 12.1], if D^{μ} is a composition factor of M^{λ^{♭}} then μ ⊵ λ^{♭}. Therefore, by Proposition 2, S^{λ} has a composition factor D^{μ} with μ ⊵ λ^{♭}; in particular, μ ≠ λ^{R}. Hence, S^{λ} is reducible by Proposition 1.

The next three Lemmas apply this Corollary in cases which interest us.

Lemma 4 Suppose that λ = (λ_{1}, λ_{2}, …, λ_{r}) is a partition with λ_{r} ≠ 0. Assume that λ′_{1} ≥ λ_{1} and that there exists an i ≥ 2 such that λ′_{i} = λ′_{i+1} > 0. Then S^{λ} is reducible.

Proof. Take i to be minimal such that i ≥ 2 and λ′_{i} = λ′_{i+1} > 0. Let k = λ′_{i}. We claim that λ_{1}^{R} ≤ r + k. To see this, first note that

λ_{1}^{R} = max { λ′_{j} + j − 1 | λ′_{j} ≠ 0 } .

Suppose that λ′_{j} ≠ 0; then 1 ≤ j ≤ λ_{1}. If j ≥ i then λ′_{j} ≤ λ′_{i} so λ′_{j} + j − 1 ≤ λ′_{i} + λ_{1} − 1 ≤ k + r − 1. On the other hand, if j < i then, because i is minimal, we have λ′_{1} ≥ λ′_{2} > λ′_{3} > ⋯ > λ′_{j}; so λ′_{j} + j − 1 ≤ λ′_{1} + 1 = r + 1 ≤ r + k. This completes the proof of our claim that λ_{1}^{R} ≤ r + k.

Next, consider λ^{♭}. Since i ≥ 2 and λ′_{i} = λ′_{i+1} > 0 we have λ_{k} ≥ 3. In particular, if k = r then λ_{r} ≥ 3 and N_{k,1}(λ) ≥ 3. We claim that N_{k,1}(λ) ≥ 3 even when k < r. To prove this assume that k ≤ r − 1; then, by Lemma 1(iii), we have

λ_{k} − λ_{k+1} + N_{k+1,1}(λ) ≤ 2N_{k,1}(λ).

Now, λ_{k} −λ_{k+1} ≥ 2 and N_{k+1,1}(λ) ≥ 1 and N_{k,1}(λ) is one less than some power of 2. Therefore, N_{k,1}(λ) ≥ 3 as we claimed above.

Thus, for 1 ≤ j ≤ k we have N_{j,1}(λ) ≥ N_{k,1}(λ) ≥ 3 by Lemma 2(i). Also, N_{j,1}(λ) ≥ 1 for k < j ≤ r. Therefore,

λ_{1}^{♭} = ∑_{j=1}^{r} N_{j,1}(λ) ≥ 3k + r − k > r + k.

Hence, λ_{1}^{♭} > λ_{1}^{R}, so λ^{R} ⋭ λ^{♭}. Thus, S^{λ} is reducible by the Corollary to Proposition 2.

Lemma 5 Suppose that λ = (λ_{1}, λ_{2}, …, λ_{r}) is a partition with λ_{r} ≠ 0 and r ≥ 2. Assume that λ_{r−1} > λ_{r} and λ′_{1} = λ′_{2} and λ′_{i} > λ′_{i+1} for all i with 2 ≤ i ≤ λ_{1}. Then S^{λ} is reducible.

Proof. Since λ_{r−1} > λ_{r} ≥ 2, we have N_{r−1,1}(λ) ≥ 3; so N_{j,1}(λ) ≥ 3 for 1 ≤ j ≤ r−1 by Lemma 2(i). Therefore, λ_{1}^{♭} = N_{r,1}(λ)+∑_{j=1}^{r−1} N_{j,1}(λ) ≥ 2+3(r−1). On the other hand, λ_{1}^{R} = λ′_{1} + 1 = r + 1 because no two columns after the first two columns have the same length. Once more, λ^{R} ⋭ λ^{♭} and S^{λ} is reducible.

Lemma 6 Suppose that λ = (λ_{1}, λ_{2}, …, λ_{r}) is a partition with λ_{r} ≠ 0 and r ≥ 3. Assume that λ_{1} = λ_{2} and λ_{1}^{R} = λ_{1} + 1 and λ_{2} > λ_{2}^{R}. Then S^{λ} is reducible.

Proof. Since λ_{1} = λ_{2}, we have N_{1,1}(λ) = N_{2,1}(λ). By Lemma 3,

λ_{1} = λ_{2} ≤ 2N_{2,1}(λ) + N_{3,1}(λ) + ⋯ + N_{r−1,1}(λ) = λ_{1}^{♭} − N_{r,1}(λ),

since r ≥ 3. Thus, λ_{1}^{♭} ≥ λ_{1} + N_{r,1}(λ) ≥ λ_{1}^{R}.

If λ_{1}^{♭} > λ_{1}^{R} then λ^{R} ⋭ λ^{♭} and S^{λ} is reducible.

Assume, now, that λ_{1}^{♭} = λ_{1}^{R}. Since α(λ)_{1} = α(λ)_{2}, we have that

λ_{2}^{♭} = 2N_{2,2}(λ) + N_{3,2}(λ) + ⋯ + N_{r,2}(λ).

Since λ_{1}^{♭} = λ_{1}^{R}, we have λ_{1}^{♭} = λ_{1} + N_{r,1}(λ) and

λ_{2} = 2N_{2,1}(λ) + N_{3,1}(λ) + ⋯ + N_{r−1,1}(λ);

so, by Lemma 3, N_{j,2}(λ) = N_{j,1}(λ) for 2 ≤ j ≤ r−1 and N_{r,2}(λ) = λ_{r} −N_{r,1}(λ). Therefore,

λ_{2}^{♭} = 2N_{2,1}(λ) + N_{3,1}(λ) + ⋯ + N_{r−1,1}(λ) + λ_{r} − N_{r,1}(λ)
= λ_{2} + λ_{r} − N_{r,1}(λ).

Hence,

λ_{1}^{♭} + λ_{2}^{♭} = λ_{1}^{♭} + λ_{2} + λ_{r} − N_{r,1}(λ) = λ_{1} + λ_{2} + λ_{r} > λ_{1}^{R} + λ_{2}^{R},

since λ_{1}^{R} = λ_{1} + 1 and λ_{2} > λ_{2}^{R}.

Thus, λ^{R} ⋭ λ^{♭}, and S^{λ} is reducible.

Proof of the Main Theorem. By the comments which follow the statement of the Theorem, it remains to consider the Specht modules S^{λ} where λ and λ′ are 2–singular partitions.

Assume that λ and λ′ are 2–singular. We may also assume that λ_{1} ≥ λ′_{1} because S^{λ} is reducible if and only if S^{λ′} is reducible.

Suppose first that λ has only two non–zero parts; then λ = (k, k) for some k ≥ 2. By [3, Theorem 24.15] the Specht module S^{λ} is irreducible if and only if k = 2.

We may therefore assume that λ has at least three non–zero parts. If λ_{i} = λ_{i+1} > 0 for some i ≥ 2 then S^{λ}′ is reducible by Lemma 4. Since λ is 2– singular, we can now assume that λ_{1} = λ_{2} and λ_{i} > λ_{i+1} whenever i ≥ 2 and λ_{i} ≠ 0. If λ_{2} = λ_{3} + 1 then S^{λ}′ is reducible by Lemma 5. If λ_{2} ≠ λ_{3} + 1 then λ_{1} = λ_{2} > λ_{3} + 1 > λ_{4} + 1 > ⋯; in this case, λ_{1}^{R} = λ_{1} + 1 and λ_{2}^{R} = λ_{2} − 1, so S^{λ} is reducible by Lemma 6.

This completes the proof of the Theorem.

In conclusion, we remark that in characteristic p, with p odd, the irreducibility of Specht modules is much more complicated. We are unable even to conjecture which p–singular Specht modules are irreducible in characteristic p.

References

[1] G. D. James, On the decomposition matrices of the symmetric groups II, J. Algebra, 43 (1976), 45–54.

[2] , On a conjecture of Carter concerning irreducible Specht modules, Proc. Camb. Phil. Soc., 83 (1978), 11–17.

[3] , The representation theory of the symmetric groups, SLN, 682, Springer–Verlag, New York, 1978.

G.J. Imperial College, 180 Queen’s Gate, London SW7 2BZ. U.K.

       g.james@ic.ac.uk

A.M. University of Sydney, Sydney NSW 2006. Australia.

       a.mathas@maths.usyd.edu.au