On extensions and branching rules for modules close to completely splittable

Vladimir Shchigolev

vkshch@vens.ru

Abstract

We describe the modules D^{λ}↓_{Σ_{n−1}} and D^{λ}↑^{Σ_{n+1}} for certain simple KΣ_{n}-modules (completely splittable and some close to) D^{λ}, where K is a field of characteristic p > 0 and Σ_{n} is the symmetric group of degree n. This result is based on an upper bound of the dimensions of the Ext^{1}-spaces between some simple modules.

1 Introduction

Fix a field K of characteristic p > 0. We denote by Σ_{n} the symmetric group of degree n. We shall assume the natural inclusion Σ_{n−1} ⊂ Σ_{n}. Calculation of the modules D^{λ}↓_{Σ_{n−1}} (if n > 0) and D^{λ}↑^{Σ_{n+1}}, where D^{λ} is the simple KΣ_{n}-module corresponding to a p-regular partition λ of n, is of great importance for the representation theory of the symmetric group. We have the following decomposition into blocks of KΣ_{n−1} and KΣ_{n+1} (see [CR, (55.2)] and [KSh1, §1]):

D^{λ}↓_{Σ_{n−1}} = ⨁_{α∈ℤ_{p}} Res_{α} D^{λ}, D^{λ}↑^{Σ_{n+1}} = ⨁_{α∈ℤ_{p}} Ind^{α} D^{λ}.

A lot of information about Res_{α} D^{λ} and Ind^{α} D^{λ} is contained in [K4] and [BK2]. For example, the socles of these modules are known. It is also known when a module arbitrarily chosen from these is simple. On the other hand, not all composition multiplicities of nonsimple modules Res_{α} D^{λ} and Ind^{α} D^{λ} are known in the general case.

However, all the above mentioned multiplicities can be explicitly found for some nonsimple block components. For example, the main result of [S2] allows us to prove in the present paper Theorems 9.4 and 9.5. Together with the known Proposition 9.1, they yield all composition multiplicities of

• D^{λ}↑^{Σ_{n+1}}, where D^{λ} is a completely splittable KΣ_{n}-module (see [K3, Definition 0.1]);

• D^{λ^{B}}↓_{Σ_{n}}, where D^{λ} is a completely splittable KΣ_{n}-module distinct from D^{(1^{p−1})} and B is the bottom λ-addable node.

Theorem 6 from [S2] prompts for what other modules one may hope to prove similar results. A partition λ is called big (for fixed p), if D^{λ} is completely splittable, λ has more than one nonzero parts and at least one rim p-hook. In that case, we denote by λ̃ the partition obtained from λ by moving all the nodes from the last row of the highest rim p-hook of λ to the first row. Big modules exist only for p > 2. In the present paper, we prove an upper bound (Theorem 8.2) of the dimensions of Ext_{Σ_{n}}^{1}(Dλ̃, D^{μ}), where λ̃ does not strictly dominate μ, similar to the bound of [S2, Theorem 6]. There are examples showing that this bound is not exact. A separate paper is planed to prove the exact formula. However, the bound we have obtained is enough to • prove the exact formula (Proposition 9.1 and Theorem 9.6) for composition multiplic-

ities of Dλ̃↑^{Σ_{n+1}}, where λ is a big partition of n such that h(λ) ⩾ \frac{p+3}{2}, λ_{1} + h(λ) ≠ 0 (mod p) and h_{2,1}(λ) ≠ p − 1;

• put forward conjectures on the composition multiplicities of some Ind^{α} D^{λ} and Res_{α} D^{λ} (Conjectures 9.7–9.9) confirmed by calculations within the known decomposition matrices.

In [BK2], the composition multiplicities of D^{μ} in Ind^{α} D^{λ} and Res_{α} D^{λ} are calculated for μ one node more or respectively less than λ. Applying the Mullineux map m, one can calculate the same composition multiplicities when m(μ) is one node more or respectively less than m(λ). However, to prove Conjectures 9.7–9.9, one must show that the required multiplicities equal 1 for some partitions μ having neither of the two mentioned forms. We conjecture that this can be done by the methods of [K4].

The paper is organized as follows. In §2, we introduce the main objects used in the text. The technique of the present paper differs from that of [S2] mainly in using abaci. The theory of abaci is presented in [JaKe]. We prove some inequalities in §3. In particular, the most useful inequalities of [S2, §3] are reproved by the methods standard for homological algebra. In §4, an inductive method of obtaining an upper bound of the dimensions of Ext_{Σ_{n}}^{1}(D^{λ}, D^{μ}) is described. With its help, we reprove the main result of [S2] but in a much simpler and more visual way due to using abaci. A slight, though necessary, modification of this method is used to prove Theorem 8.2. For a more precise bound in this theorem, an auxiliary upper bound obtained in §7 is needed. To sharpen this auxiliary bound, we use multiplication by the sign representation and the Mullineux map connected with it. The corresponding calculations are given in §6. Finally in §9, we prove the above mentioned results on the composition multiplicities of the induced and restricted modules.

2 Notation and definitions

2.1 Generalities. Throughout the paper, we fix a field K of positive characteristic p. All rings and modules are assumed finite dimensional over K. For n ∈ ℤ, let n̄ denote n + pℤ, which is an element of ℤ_{p} = ℤ/pℤ. For a pair n, m ∈ ℤ, where m > 0, let quo(n, m) and rem(n, m) denote the integers such that n = quo(n, m)m + rem(n, m) and 0 ⩽ rem(n, m) < m. For integers r and s, the following notation will be used:

\begin{matrix}[r, s] = \{i ∈ ℤ : r ⩽ i ⩽ s\}, & (r, s] = \{i ∈ ℤ : r < i ⩽ s\}, \\ [r, s) = \{i ∈ ℤ : r ⩽ i < s\}, & (r, s) = \{i ∈ ℤ : r < i < s\}. \end{matrix}

For an arbitrary assertion ρ, let [ρ] denote 1 if this assertion holds and denote 0 otherwise. For any set of integers S, we define its characteristic function by S̄(n) = [n ∈ S] for n ∈ ℤ. We define an ordered set ℤ′ = ℤ ∪ {+∞}, where +∞ > n holds for any n ∈ ℤ. For any module M and any simple module N, let [M : N] denote the composition multiplicity of N in M.

2.2 Partitions. Given a sequence a, |a| denotes its length. If a positive integer i is such that i ⩽ |a| in the case where |a| < +∞, then a_{i} stands for the i^{th} from the beginning element of a.

A partition of an integer n is an infinite nonincreasing sequence of nonnegative integers, whose elementwise sum equals n. To say that λ is a partition of n, the notation λ ⊢ n is used.

In practice we write only a finite initial part of a partition that is followed by zeros (not to be confused with finite sequences). For example, if λ is a partition and we write λ = (5, 3, 0), then λ_{1} = 5, λ_{2} = 3 and λ_{i} = 0 for i ⩾ 3. The height of a partition λ is the number h(λ) of its nonzero entries. A partition λ that does not contain p or more identical entries is called p-regular. Let ∑ λ denote the sum of all components of λ.

For a partition λ we define its Young diagram by the formula [λ] = {(i, j) ∈ ℤ × ℤ : 1 ⩽ i ⩽ h(λ), 1 ⩽ j ⩽ λ_{i}}. Elements of ℤ × ℤ are called nodes. For a node A = (i, j), we put r(A) = i and res A = \overline{j − i}. If the diagrams of partitions λ and μ contain the same number of nodes of each p-residue, then we write λ ∼ μ. Removable, addable, normal, good, conormal and cogood nodes of a partition λ are defined in [BK2]. We also use the notations λ_{A} and λ^{B}, where A is a removable and B is an addable node of λ, for partitions with diagrams [λ] \ {A} and [λ] ∪ {B} respectively. Let λ^{t} denote the partition, whose diagram is obtained by transposing [λ].

For a partition λ and an integer i, we put σ_{i}(λ) = ∑_{1⩽j⩽i} λ_{j}. A partition λ is said to dominate μ if σ_{i}(λ) ⩾ σ_{i}(μ) for any i ⩾ 1. This fact is denoted by λ ⊵ μ. The formula λ ▷ μ means that λ ⊵ μ and λ ≠ μ.

Let h_{i,j}(λ) denote the length of the hook of λ with base (i, j). We have h_{i,j}(λ) = λ_{i} + λ_{j}^{t} − i − j + 1. Rim, p-segment, p-edge, rim p-hook, p-core of a partition λ are defined in [JaKe] and [Mu]. Let e(λ) denote the number of nodes in the p-edge of a partition λ and φ(λ) denote the partition obtained from λ by removing its p-edge.

2.3 Modules. To each partition λ of n there corresponds a KΣ_{n}-module S^{λ}, which is called the Specht module (see, for example, [Ja, Definition 4.3]). We put D^{λ} = S^{λ}/ rad S^{λ}. The map λ ↦ D^{λ} defines a one-to-one correspondence between p-regular partitions of n and simple KΣ_{n}-modules.

For n ⩾ 0, we put ε_{n} = ((k + 1)^{d}, k^{p−1−d}), where n = k(p − 1) + d, k ∈ ℤ and 0 ⩽ d < p − 1. The following proposition plays an important role in the current paper.

Proposition 2.1. Let p > 2 and λ, μ be partitions of n such that h(λ) < p, μ is p-regular and λ ▷̸ μ. Then we have

Ext_{Σ_{n}}^{1}(S^{λ}, D^{μ}) ≅ \{ \begin{matrix}K & if λ = μ = ε_{n} and n ⩾ p; \\ 0 & otherwise. \end{matrix}

Proof is virtually the same as that of Theorem 2.9 from [KSh1]. One must only show that μ = ε_{n} or μ = κ_{n} (after the assumptions λ_{A} = ε_{n−1} and γ = ε_{n−1} are made) without using Lemma 1.6.

Indeed, μ_{C} ⊴ γ = ε_{n−1} = λ_{A} and μ_{C} ◁̸ λ_{A}, imply μ_{C} = ε_{n−1}. This means μ = κ_{n} or μ = ε_{n} or μ = ε_{n−1}^{D}, where D = (p, 1). However, the last case is impossible, since ε_{n−1}^{D} ◁ ε_{n}, κ_{n} and λ = ε_{n} or λ = κ_{n}. □

The following proposition is proved similarly to [KSh1, Theorem 2.10] but using Proposition 2.1 instead of [KSh1, Theorem 2.9].

Proposition 2.2. Let p > 2 and λ, μ be partitions of n such that h(λ) < p, μ is p-regular and λ ▷̸ μ. Then Ext_{Σ_{n}}^{1}(D^{λ}, D^{μ}) ≅ Hom_{Σ_{n}}(rad S^{λ}, D^{μ}).

Finally let us note the following proposition, which follows directly from [JaKe] and [KSh1, Theorem 2.10].

Proposition 2.3. If λ ≁ μ or if p > 2, λ = μ and h(λ) < p, then Ext_{Σ_{n}}^{1}(D^{λ}, D^{μ}) = 0.

The modules Ind^{α} M and Res_{α} M, where M is a KΣ_{n}-module, are defined, for example in §1 of [KSh1].

2.4 Abaci. We shall slightly modify the classical notion of abacus introduced in [JaKe] to make it more symmetrical and convenient to work both with removing and adding nodes. Everything what follows can be proved by the methods of [JaKe] and [FK].

An abacus is any map Λ : ℤ → {0, 1} for which there exists a number N such that Λ(n) = 1 for n ⩽ −N and Λ(n) = 0 for n ⩾ N. The shift of an abacus Λ is the limit shift(Λ) = lim_{x→−∞} x + ∑_{n⩾x} Λ(n) over integer x. Every abacus Λ defines an injective map node_{Λ} : ℤ → ℤ × ℤ by

node_{Λ}(a) = (1 + ∑_{n>a} Λ(n), ∑_{n⩽a}(1 − Λ(n))).

We have

c − r = a − shift(Λ), where node_{Λ}(a) = (r, c);
res node_{Λ}(a) = ā − \overline{shift(Λ)}.

(2.1)

Indeed, take any integer x such that Λ(n) = 1 for n < x. Then c−r = ∑_{x⩽n⩽a}(1−Λ(n))− 1 − ∑_{n>a} Λ(n) = a − x − ∑_{x⩽n} Λ(n) = a − shift(Λ).

An element a ∈ ℤ such that Λ(a) = 1 is called a bead of Λ, and an element b ∈ ℤ such that Λ(b) = 0 is called a space of Λ.

A bead a of an abacus Λ is called

• proper if there is a space b of Λ, strictly less than a;

• initial if Λ(a − 1) = 0;

• normal if it is initial and ∑_{0<k⩽s}(Λ(a + pk) − Λ(a − 1 + pk)) ⩾ 0 for any s > 0;

• good if it is the smallest normal bead of a given p-residue;

• movable up if Λ(a − p) = 0.

A space b of an abacus Λ is called

• initial if Λ(b − 1) = 1;

• conormal if it is initial and ∑_{0<k⩽s}(Λ(b − 1 − pk) − Λ(b − pk)) ⩾ 0 for any s > 0;

• cogood if it is the greatest conormal space of a given p-residue.

If an abacus contains at least one proper bead, then it is called proper. Otherwise it is called improper.

The first formula of (2.1) shows what partition should be assigned to an abacus Λ. Let a_{1}, a_{2}, … be all the beads of Λ written in descending order. We have node_{Λ}(a_{i}) = (i, λ_{i}) for some numbers λ_{i}. We define the partition P (Λ) = (λ_{1}, λ_{2}, …). Λ is said to be an abacus of P (Λ). Let h denote the number of proper beads in Λ. Clearly P (Λ) has height h. In this connection the number of proper beads of an abacus is called its height. We have h_{i,1}(P (Λ)) = a_{i} − b for 1 ⩽ i ⩽ h, where b is the smallest space of Λ, which is obviously equal to shift(Λ) − h.

Let Λ be an arbitrary abacus and m ∈ ℤ. For any n ∈ ℤ, we put (m+Λ)(n) = Λ(n+m). An elementary verification shows that

P (m + Λ) = P (Λ) and shift(m + Λ) = −m + shift(Λ).  (2.2)

Note that for every partition λ there is exactly one abacus Λ of a given shift such that λ = P (Λ). This fact and (2.2) imply that if P (Λ) = P (M) then shift(Λ)+Λ = shift(M)+M.

Proposition 2.4. node_{Λ} bijectively maps:

(1) the set of Λ-initial beads to the set of P (Λ)-removable nodes;

(2) the set of Λ-normal beads to the set of P (Λ)-normal nodes;

(3) the set of Λ-good beads to the set of P (Λ)-good nodes;

(4) the set of Λ-initial spaces to the set of P (Λ)-addable nodes;

(5) the set of Λ-conormal spaces to the set of P (Λ)-conormal nodes;

(6) the set of Λ-cogood spaces to the set of P (Λ)-cogood nodes.

Let c be an initial bead or an initial space of an abacus Λ. Denote by Λ_{c} in the former case and by Λ^{c} in the latter case the abacus, whose value at n is 1 − Λ(n) if n = c or n = c − 1 and is Λ(n) if n ≠ c and n ≠ c − 1. The operations of removing and adding nodes are connected with the transformations of abaci just described by

P (Λ_{c}) = P (Λ)_{node_{Λ}(c)} if c is an initial bead,
P (Λ^{c}) = P (Λ)^{node_{Λ}(c)} if c is an initial space.

One of the main causes to use abaci is that they help easily watch the removal of rim p-hooks and p-edges. Let a be a movable up bead of Λ. Denote by hook_{Λ}(a) the rim hook of P (Λ) with base (∑_{n⩾a} Λ(n), ∑_{n⩽a−p}(1 − Λ(n))). The abacus, whose value at n is 1 − Λ(n) if n = a or n = a − p and is Λ(n) if n ≠ a and n ≠ a − p, is said to be obtained from Λ by moving a one position up.

Proposition 2.5. The map hook_{Λ} is a bijection from the set of all movable up beads of Λ to the set of all rim p-hooks of P (Λ). Moreover, if Λ̄ is the abacus obtained from Λ by moving the bead a one position up, then P (Λ̄) is the partition obtained from P (Λ) by removing hook_{Λ}(a).

The above terminology is explained by the following way of representing abaci. Let Λ be an arbitrary abacus. Let the position (r, c) of the table T, where r ∈ ℤ and c = 0, …, p − 1, be occupied by · if Λ(pr + c) = 0 and by ∘ if Λ(pr + c) = 1. According to the definition of an abacus there are two numbers r_{1} and r_{2} such that row r of T is occupied solely by ∘ if r < r_{1} and is occupied solely by · if r > r_{2}. Let us draw only rows r of T with r_{1} ⩽ r ⩽ r_{2} and indicate the number of any row.

In this connection the set {n ∈ ℤ : n = i (mod p)}, where 0 ⩽ i ⩽ p − 1, is called the i^{th} runner.

Example. Let Λ = \overline{(−∞, 0] ∪ \{4, 8, 9, 11, 14, 17\}}. A representation of this abacus for p = 7 is

\begin{matrix}∘ & · & · & · & ∘ & · & · \\ · & ∘ & ∘ & · & ∘ & · & · \\ ∘ & · & · & ∘ & · & · & · \end{matrix}

where row 0 is the highest. We have P (Λ) = (11, 9, 7, 6, 6, 3).

Let us continue the functions φ and e defined in §2.2 to the set of all abaci so that P (φ(Λ)) = φ(P (Λ)) and e(Λ) = e(P (Λ)) for any abacus Λ.

Let Λ be an arbitrary abacus. Denote by a the greatest bead and by b the smallest space of this abacus.

We put φ(Λ) = Λ and e(Λ) = 0 for any improper abacus Λ. In the remaining part of this subsection we assume that Λ is proper. Following [FK, Definition 1.3], we consider the set of beads {m_{1}, …, m_{N}} of Λ (in the paper just cited it is called “set of r-movable beads”) that is as follows. We put m_{1} = a. Suppose the beads m_{1}, …, m_{i} are already chosen. If one of the following conditions holds:

• m_{i} − p is an improper bead;

• m_{i} − p is a space and there are no beads less than m_{i} − p, then we put N = i and we stop here. In the opposite case, m_{i+1} is equal to the greatest bead c of Λ such that c ⩽ m_{i} − p. It can be easily seen that all the beads m_{1}, …, m_{N} we have built are proper.

We shall build the abacus φ(Λ) as follows. If m_{N} − p is an improper bead, then we move m_{N} to position b. Otherwise we move m_{N} to position m_{N} − p. The remaining beads m_{N−1}, …, m_{1} must be moved one position up one after another starting with m_{N−1} and ending with m_{1}. Denoted by φ(Λ) the resulting abacus. Let

e(Λ) = ∑ P (Λ) − ∑ P (φ(Λ)).

2.5 Operation H_{ε}. In the sequel for any abacus Λ and a positive integer i, b^{Λ}(i) denotes the i^{th} bead of Λ counting from the greatest one and b_{Λ}(i), where i ⩽ h(Λ), denotes the i^{th} proper bead of Λ counting from the smallest one. If Λ is proper then we abbreviate b^{Λ} = b^{Λ}(1) and b_{Λ} = b_{Λ}(1).

Let ε be a finite set of integers. Consider the following map

Λ + ∑_{i=1}^{|ε|}(\overline{\{b^{Λ}(i) + pε_{i}\}} − \overline{\{b^{Λ}(i)\}}).  (2.3)

As we want to obtain an abacus again, we shall say that H_{ε} is applicable to Λ if and only if (2.3) defines an abacus. In that case we define H_{ε}(Λ) to be map (2.3). For a partition λ, we put H_{ε}(λ) = P (H_{ε}(Λ)), where λ = P (Λ). Clearly H_{ε}(λ) is well defined.

In other words, if H_{ε} is applicable to Λ, then H_{ε}(Λ) is obtained from Λ by moving each bead b^{Λ}(i), where 1 ⩽ i ⩽ |ε|, to position b^{Λ}(i) + pε_{i}.

Definition 2.6. An abacus (partition) is called ε-big, if it is completely splittable of height |ε| and H_{ε} is applicable to it.

We have the following evident assertion.

Lemma 2.7. Let Λ be an ε-big abacus and 1 ⩽ i ⩽ |ε| such that the runner containing b^{Λ}(i) + 1 contains no proper bead of Λ. Then Λ^{b^{Λ}(i)+1} is ε-big and H_{ε}(Λ^{b^{Λ}(i)+1}) = H_{ε}(Λ)^{b^{Λ}(i)+1+pε_{i}}.

3 Inequalities

We state the following standard fact for future reference.

Proposition 3.1. Let A, B, C be finite dimensional K-spaces and R be a K-algebra.

(a) If A → B → C is an exact sequence, then dim B ⩽ dim A + dim C.

(b) If A, B, C are R-modules and A is isomorphic to a submodule of B, then dim Hom_{R}(C, A) ⩽ dim Hom_{R}(C, B).

(c) If A, B, C are R-modules and A is a homomorphic image of B, then dim Hom_{R}(A, C) ⩽ dim Hom_{R}(B, C).

3.1 Case of restricted height. We define

ε(λ, μ) = \{ \begin{matrix}1 if λ = μ^{B} for a μ-conormal node B; \\ 0 otherwise. \end{matrix}

γ(μ, λ) = \{ \begin{matrix}1 if μ = λ_{A} for a λ-normal node A; \\ 0 otherwise. \end{matrix}

Lemma 3.2. Let p > 2, λ and μ be p-regular partitions of n > 0, h(λ) < p, λ ▷̸ μ and A be a μ-good node of residue α. Then we have

dim Ext_{Σ_{n}}^{1}(D^{λ}, D^{μ}) ⩽ dim Ext_{Σ_{n−1}}^{1}(Res_{α} D^{λ}, D^{μ_{A}}) + ε(λ, μ_{A}).

Proof. In view of Proposition 2.3, we can assume λ ∼ μ. Applying the functor Res_{α} to the exact sequence 0 → rad S^{λ} → S^{λ} → D^{λ} → 0, we get the exact sequence 0 → Res_{α} rad S^{λ} → Res_{α} S^{λ} → Res_{α} D^{λ} → 0. Applying Hom_{Σ_{n−1}}(−, D^{μ_{A}}) to the last sequence, we get the exact sequence

Hom_{Σ_{n−1}}(Res_{α} S^{λ}, D^{μ_{A}}) → Hom_{Σ_{n−1}}(Res_{α} rad S^{λ}, D^{μ_{A}}) → Ext_{Σ_{n−1}}^{1}(Res_{α} D^{λ}, D^{μ_{A}}). (3.1)

By Frobenius reciprocity, [BK2, Theorem E(iii)] and λ ∼ μ, we have

dim Hom_{Σ_{n−1}}(Res_{α} S^{λ}, D^{μ_{A}}) = dim Hom_{Σ_{n}}(S^{λ}, Ind^{α} D^{μ_{A}}) = ε(λ, μ_{A}).

By Frobenius reciprocity, the fact that D^{μ} is isomorphic to a submodule of Ind^{α} D^{μ_{A}}, Proposition 3.1(b) and Proposition 2.2, we get

dim Hom_{Σ_{n−1}}(Res_{α} rad S^{λ}, D^{μ_{A}}) = dim Hom_{Σ_{n}}(rad S^{λ}, Ind^{α} D^{μ_{A}}) ⩾
dim Hom_{Σ_{n}}(rad S^{λ}, D^{μ}) = dim Ext_{Σ_{n}}^{1}(D^{λ}, D^{μ}).

Now it remains to apply Proposition 3.1(a) to sequence (3.1). □ Dually we prove

Lemma 3.3. Let p > 2, λ and μ be p-regular partitions of n, h(λ) < p, λ ▷̸ μ and A be a μ-cogood node of residue α. Then we have

dim Ext_{Σ_{n}}^{1}(D^{λ}, D^{μ}) ⩽ dim Ext_{Σ_{n+1}}^{1}(Ind^{α} D^{λ}, D^{μ^{A}}) + γ(λ, μ^{A}).

3.2 Case of arbitrary height. In the absence of the restriction h(λ) < p, the estimation of the dimension of Ext_{Σ_{n}}^{1}(D^{λ}, D^{μ}) can be carried out in a different way.

Lemma 3.4. Let λ and μ be distinct p-regular partitions of n > 0 and A be a λ-good node of residue α. We have

dim Ext_{Σ_{n}}^{1}(D^{λ}, D^{μ}) ⩽ dim Ext_{Σ_{n−1}}^{1}(D^{λ_{A}}, Res_{α} D^{μ}) + dim Hom_{Σ_{n}}(rad Ind^{α} D^{λ_{A}}, D^{μ}).

Proof. Let I be the injective hull of D^{μ}. Since Hom_{Σ_{n}}(D^{λ}, I) = 0, we have Ext_{Σ_{n}}^{1}(D^{λ}, D^{μ}) ≅ Hom_{Σ_{n}}(D^{λ}, I/D^{μ}). Let M be a KΣ_{n}-submodule of I such that D^{μ} ⊂ M and M/D^{μ} ≅ soc(I/D^{μ}). Applying the functor Res_{α} to the exact sequence 0 → D^{μ} → M → M/D^{μ} → 0 and then applying Hom_{Σ_{n−1}}(D^{λ_{A}}, −), we get the exact sequence

Hom_{Σ_{n−1}}(D^{λ_{A}}, Res_{α} M) → Hom_{Σ_{n−1}}(D^{λ_{A}}, Res_{α}(M/D^{μ})) → Ext_{Σ_{n−1}}^{1}(D^{λ_{A}}, Res_{α} D^{μ}). (3.2)

By Frobenius reciprocity, the fact that D^{λ} is a homomorphic image of Ind^{α} D^{λ_{A}} and Proposition 3.1(c), we get

dim Hom_{Σ_{n−1}}(D^{λ_{A}}, Res_{α}(M/D^{μ})) = dim Hom_{Σ_{n}}(Ind^{α} D^{λ_{A}}, M/D^{μ}) ⩾
dim Hom_{Σ_{n}}(D^{λ}, soc(I/D^{μ})) = dim Hom_{Σ_{n}}(D^{λ}, I/D^{μ}) = dim Ext_{Σ_{n}}^{1}(D^{λ}, D^{μ}).

Apply Hom_{Σ_{n}}(−, M) to the exact sequence

0 → rad Ind^{α} D^{λ_{A}} → Ind^{α} D^{λ_{A}} → D^{λ} → 0

and get the exact sequence

Hom_{Σ_{n}}(D^{λ}, M) → Hom_{Σ_{n}}(Ind^{α} D^{λ_{A}}, M) → Hom_{Σ_{n}}(rad Ind^{α} D^{λ_{A}}, M).  (3.3)

The first term of this sequence is 0, since μ ≠ λ

Let us see what is the image of the last morphism of sequence (3.3). Take any φ ∈ Hom_{Σ_{n}}(Ind^{α} D^{λ_{A}}, M). Since M/D^{μ} is semisimple, we have rad Ind^{α} D^{λ_{A}} ⊂ Ker(π ∘ φ), where π : M → M/D^{μ} is the natural projection. Hence φ(rad Ind^{α} D^{λ_{A}}) ⊂ D^{μ}.

Therefore the last term of sequence (3.3) can be replaced by Hom_{Σ_{n}}(rad Ind^{α} D^{λ_{A}}, D^{μ}). Hence by Frobenius reciprocity

dim Hom_{Σ_{n−1}}(D^{λ_{A}}, Res_{α} M)= dim Hom_{Σ_{n}}(Ind^{α} D^{λ_{A}}, M)⩽ dim Hom_{Σ_{n}}(rad Ind^{α} D^{λ_{A}}, D^{μ}).

Now it remains to apply Proposition 3.1(a) to sequence (3.2). □ Dually we prove

Lemma 3.5. Let λ and μ be distinct p-regular partitions of n and A be a λ-cogood node of residue α. We have

dim Ext_{Σ_{n}}^{1}(D^{λ}, D^{μ}) ⩽ dim Ext_{Σ_{n+1}}^{1}(D^{λ^{A}}, Ind^{α} D^{μ}) + dim Hom_{Σ_{n}}(rad Res_{α} D^{λ^{A}}, D^{μ}).

Remark. It follows from [KSh1, Theorem 2.10] that Lemmas 3.4 and 3.5 remain true for λ = μ in the case p > 2 and h(λ) < p.

It turns out that the parameters dim Hom_{Σ_{n}}(rad Ind^{α} D^{λ_{A}}, D^{μ}) and dim Hom_{Σ_{n}}(rad Res_{α} D^{λ^{A}}, D^{μ}) arising in Lemmas 3.4 and 3.5 sometimes admit upper bounds.

Lemma 3.6. Let λ be a p-regular partition of n and M be a KΣ_{n}-module such that Ext_{Σ_{n}}^{1}(S^{λ}, rad M) = 0, Hom_{Σ_{n}}(S^{λ}, rad M) = 0 and head(M) ≅ D^{λ}. Then M is a KΣ_{n}-homomorphic image of S^{λ}.

Proof. Applying Hom_{Σ_{n}}(S^{λ}, −) to the exact sequence 0 → rad M → M → D^{λ} → 0, we get the exact sequence

0 = Hom_{Σ_{n}}(S^{λ}, rad M) → Hom_{Σ_{n}}(S^{λ}, M) → Hom_{Σ_{n}}(S^{λ}, D^{λ}) → Ext_{Σ_{n}}^{1}(S^{λ}, rad M) = 0

Hence Hom_{Σ_{n}}(S^{λ}, M) ≅ Hom_{Σ_{n}}(S^{λ}, D^{λ}) ≅ K and there exists a nonzero homomorphism φ : S^{λ} → M. Since Hom_{Σ_{n}}(S^{λ}, rad M) = 0, we have Im φ ⊄ rad M and thus Im φ = M. □

Lemma 3.7. Let λ and μ be distinct p-regular partitions of n > 0 such that Ext_{Σ_{n}}^{1}(S^{λ}, D^{μ}) = 0. Suppose that there exists a λ-good node A and a μ-good node B of residue α and that A is the unique λ-normal node of residue α and λ_{A} ≠ μ_{B}. Then dim Hom_{Σ_{n}}(rad Ind^{α} D^{λ_{A}}, D^{μ}) ⩽ ε(λ, μ_{B}).

Proof. In view of Proposition 2.3, we can assume λ ∼ μ. We put for brevity n = dim Hom_{Σ_{n}}(rad Ind^{α} D^{λ_{A}}, D^{μ}). Let V be a KΣ_{n}-submodule of rad Ind^{α} D^{λ_{A}} such that (rad Ind^{α} D^{λ_{A}}) /V ≅ ⨁ nD^{μ}. We put M = (Ind^{α} D^{λ_{A}}) /V. Applying the functor Res_{α} to the exact sequence 0 → rad M → M → D^{λ} → 0 and taking into account the equivalence Res_{α} D^{λ} ≅ D^{λ_{A}}, which follows from the uniqueness of A as a λ-normal node of residue α, we get the exact sequence

0 → Res_{α} rad M → Res_{α} M \underrightarrow{π} D^{λ_{A}} → 0.  (3.4)

By Frobenius reciprocity, we get

0 ≠ Hom_{Σ_{n}}(Ind^{α} D^{λ_{A}}, M) ≅ Hom_{Σ_{n−1}}(D^{λ_{A}}, Res_{α} M).

Let χ be any nonzero element of the last space. We have

soc Res_{α} rad M ≅ soc Res_{α} (⨁nD^{μ}) ≅ soc (⨁n Res_{α} D^{μ}) ≅ ⨁n soc Res_{α} D^{μ} ≅ ⨁nD^{μ_{B}}.

It follows from this formula and from λ_{A} ≠ μ_{B} that Im χ ∩ Res_{α} rad M = 0. Therefore χ splits the epimorphism π of sequence (3.4) and

Res_{α} M ≅ D^{λ_{A}} ⊕ Res_{α} rad M.  (3.5)

Since D^{μ_{B}} is a homomorphic image of Res_{α} D^{μ}, the module ⨁ nD^{μ_{B}} is a homomorphic image of Res_{α} rad M, which by (3.5) is a homomorphic image of Res_{α} M.

In view of head(M) ≅ D^{λ} and rad M ≅ ⨁ nD^{μ}, Lemma 3.6 is applicable to M. Therefore M is a homomorphic image of S^{λ}. Hence Res_{α} M is a homomorphic image of Res_{α} S^{λ}. As a result we get that ⨁ nD^{μ_{B}} is a homomorphic image of Res_{α} S^{λ}.

By Proposition 3.1(c), Frobenius reciprocity, [BK2, Theorem E(iii)] and λ ∼ μ, we get

n = dim Hom_{Σ_{n−1}}(⨁nD^{μ_{B}}, D^{μ_{B}}) ⩽ dim Hom_{Σ_{n−1}}(Res_{α} S^{λ}, D^{μ_{B}})
= dim Hom_{Σ_{n}}(S^{λ}, Ind^{α} D^{μ_{B}}) = ε(λ, μ_{B}).
□

Dually we prove Lemma 3.8. Let λ and μ be distinct p-regular partitions of n such that Ext_{Σ_{n}}^{1}(S^{λ}, D^{μ}) = 0. Suppose that there exists a λ-cogood node A and μ-cogood node B of residue α and that A is the unique λ-conormal node of residue α and λ^{A} ≠ μ^{B}. Then dim Hom_{Σ_{n}}(rad Res_{α} D^{λ^{A}}, D^{μ}) ⩽ γ(λ, μ^{B}).

3.3 Filtrations and self-duality. For the remainder of the section, we use the notation of [BK2, § 2].

Lemma 3.9. Fix λ ∈ X^{+}(n) and a residue α ∈ ℤ_{p}. Let s_{1} < ⋯ < s_{k} denote the set of all j such that j is conormal for λ and res(j, λ_{j} + 1) = α. Take γ ∈ X^{+}(n) such that

[Tr^{α} L_{n}(λ) : L_{n}(γ)] > 2 ∑_{i=2}^{k}[Δ_{n}(λ + ε_{s_{i}}) : L_{n}(γ)].

Then γ = λ + ε_{s_{1}}.

Proof. We put N = Tr^{α} L_{n}(λ). By [BK2, Theorem C], there exists a filtration 0 = N_{0} ⊂ N_{1} ⊂ ⋯ ⊂ N_{k} = N such that for 1 ⩽ j ⩽ k the module N_{i}/N_{i−1} is a nonzero homomorphic image of Δ_{n}(λ + ε_{s_{i}}).

Suppose that γ ≠ λ + ε_{s_{1}}. Then we have [N_{1} : L_{n}(γ)] = [rad N_{1} : L_{n}(γ)]. Since N is contravariantly self-dual, there exists a submodule M ⊂ N contravariantly dual to N/ rad N_{1}. Assumption N_{1} ⊂ M (actually even rad N_{1} ⊂ M) leads to a contradiction as follows:

[N : L_{n}(γ)] = [N/ rad N_{1} : L_{n}(γ)] + [rad N_{1} : L_{n}(γ)] ⩽ 2[M : L_{n}(γ)]

= 2[N/ rad N_{1} : L_{n}(γ)] = 2[N/N_{1} : L_{n}(γ)] ⩽ 2 ∑_{i=2}^{k}[Δ_{n}(λ + ε_{s_{i}}) : L_{n}(γ)].

Now the second isomorphism theorem yields 0 ≠ N_{1} + M/M ≅ N_{1}/N_{1} ∩ M, whence [N_{1} + M/M : L_{n}(λ + ε_{s_{1}})] = 1. Recall that [N : L_{n}(λ + ε_{s_{1}})] = 1 as follows from [BK2, Theorem B(iv)] and [rad N_{1} : L_{n}(λ + ε_{s_{1}})] ⩽ [rad Δ_{n}(λ + ε_{s_{1}}) : L_{n}(λ + ε_{s_{1}})] = 0. Now a contradiction follows from

\begin{matrix}[M : L_{n}(λ + ε_{s_{1}})] = [N : L_{n}(λ + ε_{s_{1}})] − [rad N_{1} : L_{n}(λ + ε_{s_{1}})] = 1, \\ [N_{1} + M/M : L_{n}(λ + ε_{s_{1}})] ⩽ [N : L_{n}(λ + ε_{s_{1}})] − [M : L_{n}(λ + ε_{s_{1}})] = 0. \end{matrix}

□

Lemma 3.10. Fix a p-regular partition λ ⊢ r and a residue α ∈ ℤ_{p}. Let B_{1}, …, B_{k} be all the λ-conormal nodes of residue α counted from bottom to top. Suppose that for some p-regular γ ⊢ r + 1 there holds

[Ind^{α} D^{λ} : D^{γ}] > 2 ∑_{i=2}^{k}[S^{λ^{B_{i}}} : D^{γ}].

Then γ = λ^{B_{1}}.

Proof. Choose any n > r. Then [BK1, Theorem 4.16] yields [Ind^{α} D^{λ} : D^{γ}] = [Tr^{α} L_{n}(λ^{t}) : L_{n}(γ^{t})] and [G, Lemma (6.6b)] yields [S^{λ^{B_{i}}} : D^{γ}] = [Δ_{n}((λ^{B_{i}})^{t}) : L_{n}(γ^{t})], where λ^{t}, γ^{t}, (λ^{B_{i}})^{t} are considered as elements of X^{+}(n). Now the desired result follows from Lemma 3.9. □

Remark. We conjecture that the result for Res_{α} dual to Lemma 3.10 also holds and follows (applying [BK2, Theorems C′, B′(iv)]) form the lemma for Tr_{α} reciprocal to Lemma 3.9, [BK2, Lemma 7.4] for a suitable choice of n and [K1, Theorem 2.11(v)].

4 Completely splittable partitions

4.1 General construction. Choose a set X, whose every element has the form (λ, μ), where λ and μ are p-regular partitions such that h(λ) < p, λ ▷̸ μ, λ ∼ μ and there exists at most one λ-normal node of each residue.

Define the map U : X → ℤ′ inductively as follows. We put U(∅, ∅) = 0. Now let (λ, μ) be a pair of nonempty partitions of X. For each μ-good node A, let m_{A}(λ, μ) equal the following number:

• ε(λ, μ_{A}) if there is no λ-good node of residue res A;

• U(λ_{B}, μ_{A}) + ε(λ, μ_{A}) if there is a λ-good node B of residue res A and (λ_{B}, μ_{A}) ∈ X.

• +∞ if there is a λ-good node B of residue res A and (λ_{B}, μ_{A}) ∉ X,

Define U(λ, μ) = min{m_{A}(λ, μ) : A is a μ-good node}.

Lemma 4.1. Let p > 2 and (λ, μ) ∈ X. Then dim Ext_{Σ_{n}}^{1}(D^{λ}, D^{μ}) ⩽ U(λ, μ), where λ, μ ⊢ n.

Proof is by induction on n applying Lemma 3.2. □ 4.2 Case of completely splittable partitions. Definition [K3, 0.1]. An irreducible KΣ_{n}-module D^{λ} is called completely splittable

if and only if the restriction D^{λ}↓_{Σ^{μ}} to any Young subgroup Σ^{μ} ⊂ Σ_{n} is semisimple.

For a partition λ, we put χ(λ) = 0 if λ = ∅ and χ(λ) = λ_{1} − λ_{h(λ)} + h(λ) otherwise. The following result yields the exact criterion for a module to be completely splittable.

Theorem [K3, 2.1]. The module D^{λ} is completely splittable if and only if χ(λ) ⩽ p. A partition λ and any abacus Λ of this partition in the case where D^{λ} is completely

splittable are also called completely splittable.

The first formula of (2.1) shows that for any proper abacus we have

χ(P (Λ)) = b^{Λ} − b_{Λ} + 1.  (4.1)

It follows from this formula that any proper abacus Λ is completely splittable if and only if b_{Λ} > b^{Λ} − p.

Definition 4.2. A (−1, 0^{n}, 1)-big abacus (partition), where n ⩾ 0, is simply called big. For any big abacus (partition) of height h, we put Λ̃ = H_{(−1,0^{h−2},1)}(Λ).

It is easy to see that an abacus Λ is big if and only if it is proper, b^{Λ} > b_{Λ} > b^{Λ} − p and b^{Λ} is movable up.

Let γ be a partition and C be a γ-removable node. Then we have

σ_{i}(γ_{C}) = \{ \begin{matrix}σ_{i}(γ) & if i < r(C); \\ σ_{i}(γ) − 1 & if i ⩾ r(C). \end{matrix}  (4.2)

Lemma 4.3.

(a) If λ is a completely splittable partition, μ is any partition, A is a μ-removable node, B is a λ-good node, res A = res B and λ ⋭ μ, then λ_{B} ⋭ μ_{A}.

(b) If ν is a partition not containing distinct removable nodes of the same residue, μ is an arbitrary partition, A is a μ-removable node, B is a ν-good node, res A = res B and ν ◁ μ, then ν_{B} ⋭ μ_{A}.

Proof. We put for brevity x = r(A) and y = r(B).

(a) Suppose λ_{B} ⊵ μ_{A}. Since B is λ-normal, we have λ_{1} − λ_{y} + y < p. It follows from (4.2) that σ_{i}(λ) ⩾ σ_{i}(μ) for i < x and i ⩾ y. If σ_{i}(λ) ⩾ σ_{i}(μ) also holds for each i such that x ⩽ i < y, then we have λ ⊵ μ, contrary to the hypothesis. Therefore we assume that there exists j such that x ⩽ j < y, σ_{j}(λ) < σ_{j}(μ) and σ_{i}(λ) ⩾ σ_{i}(μ) for all i < j. Since (λ_{B})_{1} ⩾ (μ_{A})_{1}, we have λ_{1} ⩾ μ_{1} − 1 ⩾ μ_{x} − x. Hence

p > λ_{1} − λ_{y} + y ⩾ (μ_{x} − x) − (λ_{y} − y) ⩾ (μ_{j} − j) − (λ_{y} − y).

Now prove that the right hand side of the last inequality is greater than 0. Suppose this is false. The inequalities σ_{j}(λ_{B}) ⩾ σ_{j}(μ_{A}), σ_{j}(λ) < σ_{j}(μ) and (4.2) imply σ_{j}(λ) = σ_{j}(μ) − 1 or in a different form λ_{j} + σ_{j−1}(λ) − μ_{j} − σ_{j−1}(μ) + 1 = 0. Hence

(σ_{j−1}(λ) − σ_{j−1}(μ))+(λ_{j} − λ_{y} + 1)+(y − j)⩽ 0.

This formula is a contradiction, as the expression in its first pair of brackets is nonnegative and the expressions in the other two are positive. Therefore, we have p > (μ_{x} − x) − (λ_{y} − y) > 0 and res A ≠ res B.

(b) Suppose ν_{B} ⊵ μ_{A}. It follows from (4.2) that σ_{i}(ν) ⩾ σ_{i}(μ) for i < x and i ⩾ y. If σ_{i}(ν) ⩾ σ_{i}(μ) also holds for each i such that x ⩽ i < y, then we have ν ⊵ μ, contrary to the condition ν ◁ μ. Therefore we assume that there exists j such that x ⩽ j < y, σ_{j}(ν) < σ_{j}(μ) and σ_{i}(ν) ⩾ σ_{i}(μ) for all i < j. Since ν ◁ μ, we have σ_{i}(ν) = σ_{i}(μ) for all i < j and thus ν_{i} = μ_{i} for all 0 ⩽ i < j. Similarly to part (a), we get σ_{j}(ν) = σ_{j}(μ) − 1, whence ν_{j} = μ_{j} − 1.

If x < j then A and B are distinct ν-removable nodes of the same residue, which is a contradiction. Therefore x = j and A is ν-addable. By the definition of a normal node there exists a ν-removable node of residue res B strictly between B and A. This is again a contradiction. □

Let us apply the construction described in §4.1 to the following set

X = {(λ, μ) : λ is completely splittable, μ is p-regular, λ ⋭ μ, λ ∼ μ},

which we fix until the end of this section. Note that X = ∅ if p = 2, as in that case the following conditions cannot hold simultaneously: λ is completely splittable , λ ⋭ μ, λ ∼ μ.

Lemma 4.4. Let (λ, μ) ∈ X. Then U(λ, μ) = 0 except the case μ = λ̃, in which we have U(λ, λ̃) ⩽ 1.

Proof. Induction on n = ∑ λ. For n = 0, by definition we have U(λ, μ) = 0.

Now let n > 0. Suppose that the theorem is true for partitions of numbers less than n. Choose some abaci M and Λ of the same shift such that μ = P (M) and λ = P (Λ). Let A be an arbitrary μ-good node. It exists because μ is nonempty and p-regular. Denote by a the bead of M corresponding to A.

First consider the case ε(λ, μ_{A}) = 1. Then Λ = (M_{a})^{c} for some conormal space c of M_{a}. Since λ ∼ μ and λ ⋭ μ, a and c are in the same runner and moreover a is below c. We have M = (Λ_{c})^{a}. This at once implies c = b_{Λ}, a = b_{Λ} + p, b^{Λ} + 1 = b_{Λ} + p (i.e. χ(λ) = p) and M = Λ̃ (i.e. μ = λ̃). Note that in this case the only bead of Λ from the same runner as a is c, which is not normal. Therefore, there is no λ-good node of residue res A and U(λ, μ) ⩽ m_{A}(λ, μ) = ε(λ, μ_{A}) = 1.

Now consider the case ε(λ, μ_{A}) = 0 If there is no λ-good node of residue res A, then U(λ, μ) = m_{A}(λ, μ) = 0. Therefore we assume that there is a λ-good node B of residue res A. Applying Lemma 4.3(a), we get λ_{B} ⋭ μ_{A}, whence (λ_{B}, μ_{A}) ∈ X. Thus U(λ, μ) ⩽ m_{A}(λ, μ) = U(λ_{B}, μ_{A}). Therefore we shall consider the case U(λ_{B}, μ_{A}) > 0.

Let b be the bead of Λ corresponding to B. By the inductive hypothesis, we get μ_{A} = λ̃_{B}, M_{a} = Λ̃_{b} and U(λ_{B}, μ_{A}) ⩽ 1. Hence s ⩽ b^{Λ_{b}} − p, where s is the smallest space of Λ_{b}. Therefore b ≠ s, as otherwise we would get b^{Λ} − b_{Λ} = b^{Λ_{b}} − s ⩾ p, contrary to the fact that Λ is completely splittable .

Recall that a and b are cogood spaces of M_{a} and Λ_{b} respectively, since a and b are good beads of M and Λ respectively. We have b = b^{Λ_{b}}(i) + 1 < b_{Λ_{b}} + p for 1 ⩽ i ⩽ h(Λ), since b_{Λ_{b}} + p is not a cogood space of Λ_{b}. Now Lemma 2.7 can be applied to Λ_{b}. This yields Λ̃ = M_{a}^{b+pε_{i}}, where ε = (−1, 0^{h(Λ)−2}, 1). Since a and b are in the same runner and the only cogood space of M_{a}, which is equal to Λ̃_{b}, being in the same runner as b is the space b + 1 + pε_{i}, then a = b + 1 + pε_{i} and M = Λ̃. □

4.3 Exact formula. The next lemma shows that Definition 4.2 of a big partition and the map λ ↦ λ̃ given in the current paper is equivalent to [S2, Definition 4].

Lemma 4.5. A partition λ is big if and only if λ is completely splittable of height more than one and h_{1,1}(λ) ⩾ p. In that case [λ̃] is obtained from [λ] by moving all the nodes of the rim p-hook with base (1, λ_{1} + h(λ) − p) from the last row to the first row.

Proof. The first part follows from h_{1,1}(λ) = b^{Λ} − c, where Λ is a proper abacus and c is its smallest space.

Now let λ be a big partition. We put for brevity h = h(λ) and j = λ_{1} + h − p. Since 1 ⩽ j ⩽ λ_{1}, the node (1, j) belongs to [λ]. We have λ_{j}^{t} ⩽ h. The inequality λ_{j}^{t} < h would imply λ_{h} < j and χ(λ) > p. Hence λ_{j}^{t} = h and h_{1,j}(λ) = λ_{1} − j + λ_{j}^{t} = p.

Since b^{Λ} − p < b_{Λ} and b^{Λ} < b_{Λ} + p, for each bead d of Λ such that b_{Λ} < d < b^{Λ}, the numbers of beads preceding d in Λ and Λ̃ coincide and the numbers of beads following d in Λ and Λ̃ also coincide. The beads b^{Λ} − p and b_{Λ} + p are respectively the smallest and the greatest beads of Λ̃. There are b^{Λ} − p − c = λ_{1} − 1 + h − p spaces in Λ̃ preceding the former bead and b_{Λ} + p − c − (h − 1) = λ_{h} − h + p + 1 spaces preceding the latter bead (c is the smallest space of Λ). Thus λ̃ = (λ_{h} − h + p + 1, λ_{2}, …, λ_{h−1}, λ_{1} + h − p − 1). This is exactly the partition, whose diagram is obtained from [λ] by moving all the nodes of the rim p-hook with base (1, j) from the last row to the first row. □

Theorem 4.6. ( [S2, Theorem 6] ) Let p > 2 and λ, μ be p-regular partitions of n such that D^{λ} is completely splittable and λ ▷̸ μ. Then

Ext_{Σ_{n}}^{1}(D^{λ}, D^{μ}) ≅ \{ \begin{matrix}K & if μ = λ̃; \\ 0 & otherwise. \end{matrix}

Proof. The equality Ext_{Σ_{n}}^{1}(D^{λ}, D^{μ}) = 0 in the case where λ is not big or μ ≠ λ̃ follows from Lemma 4.4 and Proposition 2.3.

Now suppose μ = λ̃. Lemma 4.5 implies that the length of the hook h_{i,j}(λ), where j = λ_{1} + h(λ) − p and 1 ⩽ i ⩽ λ_{j}^{t} = h(λ) is divisible by p if and only if i = 1. Since h(λ) > 1, the Carter conjecture proved in [JaM] implies that the module S^{λ} is not simple. By Lemma 4.4, a unique top composition factor of a nonzero module rad S^{λ} is Dλ̃. This fact and the second assertion of Lemma 4.4 yield dim Ext_{Σ_{n}}^{1}(D^{λ}, Dλ̃) = 1. □

4.4 Almost completely splittable partitions.

Definition 4.7. An abacus (partition) M is called almost completely splittable if M = Λ̃ for some completely splittable Λ, which is called the preimage of M.

It is important to notice that the preimage Λ is uniquely determined by M. Indeed, let a = b^{M}. If M is a completely splittable abacus, then denote by b its greatest improper bead. If M is not a completely splittable abacus, then put b = b_{M}. Then Λ is obtained from M by moving a up and moving b down one position.

A module D^{λ}, where λ is an almost completely splittable partition, is also called almost completely splittable .

It is convenient to define many abaci encountered in this paper with the help of the following construction. Take x ∈ ℤ and S ⊂ [0, p − 1]. Let x_{0}, …, x_{i}, … be all the elements of the set {n ∈ ℤ : n ⩾ x, rem(n, p) ∈ S} written in the ascending order. For i ⩾ 0, we put ⟨x, S, i⟩ = {x_{k} : i ⩽ k < i + |S|}. Note that if i > 0 then the set ⟨x, S, i − 1⟩ is obtained from ⟨x, S, i⟩ by replacing the greatest element a of the latter set with a − p (moving a one position up).

Example. Let p = 7 and S = {1, 3, 4, 6}. We have ⟨9, S, 5⟩ = {18, 20, 22, 24} and ⟨9, S, 6⟩ = {20, 22, 24, 25}.

Definition 4.8. Let h, i, x be integers such that 0 < i ⩽ h < p and x ⩾ 0. Define Λ^{(h,i,x)} = \overline{(−∞, 0) ∪ ⟨0, [0, i), x⟩ ∪ [i, h)} and λ^{(h,i,x)} = P (Λ^{(h,i,x)}).

Clearly, the partition λ^{(h,i,x)} has p-weight x and therefore is a partition of px. Moreover, if x > 0 the height of λ^{(h,i,x)} is h, core(λ^{(h,i,x)}) = ∅, the top removable node of λ^{(h,i,x)} is its only normal node and this node has residue −h̄. Obviously the equalities λ^{(h,i,0)} = ∅ and λ^{(h,i,x)} = λ^{(h,x,x)} (following from Λ^{(h,i,x)} = Λ^{(h,x,x)}) for x < i represent all nontrivial equalities between partitions λ^{(h,i,x)}. We put for brevity λ^{(H,x)} = λ^{(H,H,x)}. Explicitly Λ^{(h,i,x)} is written as

Λ^{(h,i,x)} = \overline{(−∞, 0) ∪ [(q + 1)p, (q + 1)p + r) ∪ [qp + r, qp + i) ∪ [i, h)},

where q = quo(x, i) and r = rem(x, i).

Remark. In [H], the partitions λ^{(H,x)} are called minimal.

Proposition 4.9. If λ is a completely splittable partition, then χ(core(λ)) < p.

Proof follows from (4.1) and the fact that core(λ) is completely splittable . □

Lemma 4.10. Let λ be a completely splittable partition, χ(λ) = p and the residue of any core(λ)-normal node is equal to the residue of the bottom λ-removable node. Then λ = λ^{(H,x)}, where 1 < H < p and H ∤ x.

Proof. Suppose that core(λ) is not empty. Proposition 4.9 implies that χ(core(λ)) < p and core(λ) contains only one removable node. Hence core(λ) = P (\overline{(−∞, 0) ∪ [b, a]}), where 0 < b ⩽ a < p. Therefore λ = P (\overline{(−∞, 0) ∪ [(q + 1)p + b, (q + 1)p + r) ∪ [qp + r, qp + a]}) for q and r such that q ⩾ 0 and b ⩽ r ⩽ a.

The case r = b is impossible, as we would get the contradiction χ(λ) = χ(core(λ)) < p. The case r > b is also impossible. Indeed, in this case the smallest initial bead of \overline{(−∞, 0) ∪ [(q + 1)p + b, (q + 1)p + r) ∪ [qp + r, qp + a]}, which is equal to qp + r, and the normal bead b of \overline{(−∞, 0) ∪ [b, a]} would belong to different runners. This is a contradiction.

Now the desired assertion follows from core(λ) = ∅ and χ(λ) = p. □

Proposition 4.11. Let λ be a nonempty partition and Λ be its abacus. Then λ−(1^{h(λ)}) = P (Λ′), where Λ′ is obtained from Λ by replacing its smallest space by the bead.

Corollary 4.12. Let λ and μ be partitions such that h(λ) = h(μ) and core(λ) = core(μ). Then core(λ − (1^{h(λ)})) = core(μ − (1^{h(μ)})).

Lemma 4.13. Let χ(λ) = p, core(λ) = core(μ) = ∅, h(μ) ⩽ h(λ) < p and the residue of any μ-normal node is equal to the residue of the bottom λ-removable node. Then h(μ) < h(λ).

Proof. Suppose h(μ) = h(λ). We put h = h(λ) and n = ∑ λ.

Case n > p. We have λ = P (Λ^{(h,x)}). We put α = res(h, λ_{h}), λ̄ = λ − (1^{h}) and μ̄ = μ − (1^{h}). In the case under consideration x > 1. Therefore λ_{h} > 1 and χ(λ̄) = p. By Corollary 5.3, for any core(μ̄)-normal node, there is a μ̄-normal node of the same residue. However all μ̄-normal nodes have residue α−1, which is equal to the residue of the bottom λ̄-removable node (h, λ_{h} − 1). Since core(μ̄) = core(λ̄) by Corollary 4.12, the residue of any core(λ̄)-normal node is equal to the residue of the bottom λ̄-removable node. By Lemma 4.10, we get core(λ̄) = ∅, which is a contradiction, since λ̄ is a partition of the number n − h not divisible by p.

Case n = p is reduced to the previous one by considering the pair of partitions λ̂ = λ + (p^{h}) and μ̂ = μ + (p^{h}). □

5 Removal of locally highest p-hooks

Definition 5.1. Let u be a movable up bead of Λ. The rim p-hook corresponding to it is called locally highest if u + 1 is not a movable up bead.

Lemma 5.2. Let λ be a partition and λ̄ be the partition obtained from λ by removing its locally highest rim p-hook R. Then for any α ∈ ℤ_{p} the number of λ̄-normal nodes of residue α does not exceed the number of λ-normal nodes of the same residue.

Proof. Let Λ and Λ̄ be some abaci of λ and λ̄ respectively having shifts of p -residue 1−α. Consider the decomposition hook_{Λ}^{−1}(R) = pi + j, where 0 ⩽ j < p.

By Proposition 2.4 and the second formula of (2.1), the theorem will be proved if we define an embedding ι of the set of normal beads of the first runner of Λ̄ into the set of normal beads of the first runner of Λ. If j > 1 then we can take the identity map for ι. Thus we assume 0 ⩽ j ⩽ 1.

First consider the case j = 0. Let a = px + 1 be a normal bead of Λ̄.

If x > i then a is obviously a normal bead of Λ. In that case we put ι(a) = a.

If x = i then a − p is a bead of Λ, as otherwise it would be possible to move the bead a one position up in Λ, which contradicts the fact that the rim p-hook we have removed is locally highest. Take any s > i − 1. We have

∑_{i−1<k⩽s}(Λ(pk + 1) − Λ(pk)) = ∑_{i<k⩽s}(Λ(pk + 1) − Λ(pk)) = ∑_{i<k⩽s}(Λ̄(pk + 1) − Λ̄(pk)) ⩾ 0.

Now it is clear that a − p is a normal bead of Λ. We put ι(a) = a − p.

The case x = i − 1 is impossible, as otherwise the bead a would not be initial in Λ̄. Finally let x < i − 1. For any s > x, we have ∑_{x<k⩽s} Λ̄(pk) ⩾ ∑_{x<k⩽s} Λ(pk) and

therefore

∑_{x<k⩽s}(Λ(pk + 1) − Λ(pk)) ⩾ ∑_{x<k⩽s}(Λ̄(pk + 1) − Λ̄(pk)) ⩾ 0.

Thus a is a normal bead of Λ. We put ι(a) = a.

Now consider the case j = 1. Define the parameter x_{0} as follows. Let px + 1 and py + 1 be normal beads of Λ̄ such that y < x < i − 1. We have

\begin{matrix}∑_{y<k⩽i−1}(Λ̄(pk + 1) − Λ̄(pk)) = \\ ∑_{y<k⩽x−1}(Λ̄(pk + 1) − Λ̄(pk)) + 1 + ∑_{x<k⩽i−1}(Λ̄(pk + 1) − Λ̄(pk)) ⩾ 1. \end{matrix}  (5.1)

Therefore there is at most one number x_{0} < i − 1 such that px_{0} + 1 is a normal bead of Λ̄ and

∑_{x_{0}<k⩽i−1}(Λ̄(pk + 1) − Λ̄(pk)) = 0.  (5.2)

If there is no such number at all, then we put x_{0} = +∞. Under this definition, we get

Basic property of x_{0}. Let px + 1 be a normal bead of Λ̄ such that x < i − 1. Then x ⩽ x_{0} and ∑_{x<k⩽i−1}(Λ̄(pk + 1) − Λ̄(pk)) > 0 if x ≠ x_{0}.

Indeed in the case x_{0} < +∞ this fact follows from (5.1). In the case x_{0} = +∞ the sum under consideration is not equal to zero by the definition of x_{0} and thus is strictly positive.

If x > i then similarly to the previous case we get that a is a normal bead of Λ. We put ι(a) = a.

The case x = i is impossible, as Λ̄(pi + 1) = 0 and Λ̄(px + 1) = 1.

If x = i − 1 then Λ(pi) = Λ̄(pi) = 0, because Λ̄(pi + 1) = 0 and a = p(i − 1) + 1 is a normal bead of Λ̄. Take any s > i. We have

∑_{i<k⩽s}(Λ(pk + 1) − Λ(pk)) = ∑_{i−1<k⩽s}(Λ̄(pk + 1) − Λ̄(pk)) ⩾ 0.

Now it is clear that pi + 1 is a normal bead of Λ. We put ι(a) = pi + 1.

If x = x_{0} then by property (5.2) we have

0 ⩽ ∑_{x_{0}<k⩽i}(Λ̄(pk + 1) − Λ̄(pk)) = −Λ(pi)

0 = ∑_{x_{0}<k⩽i−1}(Λ̄(pk + 1) − Λ̄(pk)) =

∑_{x_{0}<k⩽i−2}(Λ̄(pk + 1) − Λ̄(pk)) + 1 − Λ(p(i − 1)) ⩾ 1 − Λ(p(i − 1)).

Therefore Λ(pi) = 0 and Λ(p(i − 1)) = 1.

Take any s > i. We have

∑_{i<k⩽s}(Λ(pk + 1) − Λ(pk)) = ∑_{x_{0}<k⩽s}(Λ̄(pk + 1) − Λ̄(pk)) ⩾ 0.

For the last rearrangement, we used property (5.2) and the equalities Λ̄(pi) = Λ̄(pi+1) = 0. Therefore pi + 1 is a normal bead of Λ. We put ι(a) = pi + 1.

Finally let x < i − 1 and x ≠ x_{0}. Clearly, for an arbitrary s > x not equal to i − 1 we have

∑_{x<k⩽s}(Λ(pk + 1) − Λ(pk)) = ∑_{i<k⩽s}(Λ̄(pk + 1) − Λ̄(pk)) ⩾ 0.

On the other hand by the basic property of x_{0} we have

∑_{x<k⩽i−1}(Λ(pk + 1) − Λ(pk)) = ∑_{x<k⩽i−1}(Λ̄(pk + 1) − Λ̄(pk)) − 1 ⩾ 0.

Therefore a is a normal bead of Λ. We put ι(a) = a.

The preimage of a bead b belonging to the image of ι, is given by the following formulas:

Case j > 1: ι^{−1}(b) = b;
Case j = 0: ι^{−1}(b) = \{ \begin{matrix}b & if b ≠ p(i − 1) + 1; \\ pi + 1 & if b = p(i − 1) + 1, \end{matrix}

Case j = 1: ι^{−1}(b) = \{ \begin{matrix}b & if b ≠ pi + 1; \\ p(i − 1) + 1 & if b = pi + 1 and Λ(p(i − 1)) = 0; \\ px_{0} + 1 & if b = pi + 1 and Λ(p(i − 1)) = 1.  □ \end{matrix}

Corollary 5.3. Let λ be a partition. For any α ∈ ℤ_{p}, the number of core(λ)-normal nodes of residue α does not exceed the number of λ-normal nodes of the same residue.

Proof follows from the fact that core(λ) can be obtained by removing the highest rim p-hook at each step. □

6 Mullineux map of some partitions

To calculate the Mullineux map of a partition λ, we shall use the Mullineux symbol defined in [BeO], which is the array

G_{p}(λ) = \binom{A_{0} ⋯ A_{z}}{R_{0} ⋯ R_{z}},

where A_{j} = e(φ^{j}(λ)), R_{j} = h(φ^{j}(λ)) and φ^{z+1}(λ) = ∅. The product of such arrays is understood as follows:

\binom{A_{0} ⋯ A_{z}}{R_{0} ⋯ R_{z}}\binom{A′_{0} ⋯ A′_{z′}}{R′_{0} ⋯ R_{z′}} = \binom{A_{0} ⋯ A_{z}A′_{0} ⋯ A′_{z′}}{R_{0} ⋯ R_{z}R′_{0} ⋯ R_{z′}}.

Lemma 6.1. Let 1 < H < p, x > 0 and H ∤ x. We put Q = quo(x, H) and R = rem(x, H). If R > 1 then

G_{p} \binom{}{λ^{(H,x)}} = \binom{2p}{H}^{Q}\binom{p + R − 1}{H}\binom{p}{H − 1}^{x−2(Q+1)}\binom{p − R + 1}{H − R + 1}

and if R = 1 then

G_{p} \binom{}{λ^{(H,x)}} = \binom{2p}{H}^{Q}\binom{p}{H − 1}^{x−2Q}.

Proof. Let S = [0, H − 1] \ \{R − 1\}, Λ = { \atop Λ^{(H,x)}} and λ = { \atop λ^{(H,x)}}. It is easy to check that

φ^{j}(Λ) = \overline{(−∞, 0) ∪ ⟨0, \{R − 1\}, Q − j⟩ ∪ ⟨0, S, x − Q − j⟩} for 1 ⩽ j ⩽ Q.

Hence h(φ^{j}(λ)) = H and e(φ^{j}(λ)) = 2p for 0 ⩽ j < Q.

Case R > 1. Then h(φ^{Q}(λ)) = H. We have

φ^{Q+j}(Λ) = \overline{(−∞, 0] ∪ ⟨1, S, x − 2Q − 1 − j⟩} for 1 ⩽ j ⩽ x − 2Q − 1.

Hence e(φ^{Q}(λ)) = p + R − 1, h(φ^{Q+j}(λ^{(H,x)})) = H − 1, e(φ^{Q+j}(λ^{(H,x)})) = p for 1 ⩽ j < x − 2Q − 1 and h(φ^{x−Q−1}(λ^{(H,x)})) = H − R + 1. Finally φ^{x−Q}(λ^{(H,x)}) = ∅ and e(φ^{x−Q−1}(λ^{(H,x)})) = p − R + 1.

Case R = 1. Then h(φ^{Q}(λ)) = H − 1. We have

φ^{Q+j}(Λ) = \overline{(−∞, 0] ∪ ⟨1, S, x − 2Q − j⟩} for 0 ⩽ j ⩽ x − 2Q.

Hence h(φ^{Q+j}(λ)) = H − 1 for 1 ⩽ j < x − 2Q, φ^{x−Q}(λ) = ∅ and e(φ^{Q+j}(λ)) = p for 0 ⩽ j < x − 2Q. □

Definition 6.2. Let H and x be integers such that 1 < H < p, x > 0 and H ∤ x. Define

\begin{aligned}N^{(H,x)} & = & \overline{(−∞, H) ∪ ⟨H, S_{1}, Q⟩ ∪ ⟨p + H − R, S_{2}, x − Q⟩}, \\ ν^{(H,x)} & = & P (N^{(H,x)}), \end{aligned}

where Q = quo(x, H), R = rem(x, H), S_{1} = [0, p−1]\{H−R} and S_{2} = {H−R}∪[H, p−1].

To gain a better understanding of the structure of N^{(H,x)}, we introduce the following notation: let a_{0}^{(H,x)}, a_{1}^{(H,x)}, … be all the elements of the set {n ∈ ℤ : n ⩾ H, rem(n, p) ∈ S_{1}} and b_{0}^{(H,x)}, b_{1}^{(H,x)}, … be all the elements of the set {n ∈ ℤ : n ⩾ p+H −R, rem(n, p) ∈ S_{2}} written in ascending order. An easy verification shows that

\begin{matrix}a_{y}^{(H,x)} = H + y + [\frac{R−1+y}{p−1}] , \\ b_{y}^{(H,x)} = p + H − 1 + y + (H − R) [\frac{y}{p−H+1}] + (R − 1) [\frac{y−1}{p−H+1}] . \end{matrix}  (6.1)

Lemma 6.3. Let H and x be integers such that 1 < H < p, x > 0 and H ∤ x.

Then m({ \atop λ^{(H,x)}})= ν^{(H,x)}.

Proof. Let Q, R, S_{1}, S_{2} be as in Definition 6.2 and N = N^{(H,x)}, ν = ν^{(H,x)}. By the main result of [FK] and Lemma 6.1, it suffices to prove that if R > 1 then

G_{p}(ν) = \binom{2p}{2p − H}^{Q}\binom{p + R − 1}{p + R − H}\binom{p}{p − H + 1}^{x−2(Q+1)}\binom{p − R + 1}{p − H + 1}  (6.2)

and that if R = 1 then

G_{p}(ν) = \binom{2p}{2p − H}^{Q}\binom{p}{p − H + 1}^{x−2Q}.  (6.3)

Let us see first how the sets ⟨H, S_{1}, y⟩ and ⟨p + H − R, S_{2}, x − 2Q + y⟩ are situated with respect to one another. Denote by u_{y} the greatest element of the former set and by v_{y} the smallest element of the latter set. We have u_{y} = a_{y+p−2}^{(H,x)} and v_{y} = b_{x−2Q+y}^{(H,x)}.

For y ⩾ 0, it follows from (6.1) that

\begin{matrix}v_{y−1} − u_{y} = b_{x−2Q+y−1}^{(H,x)} − a_{y+p−2}^{(H,x)} = \\ x − 2Q + (H − R) [\frac{x−2Q+y−1}{p−H+1}] + (R − 1) [\frac{x−2Q+y−2}{p−H+1}] − [\frac{R+y+p−3}{p−1}] ⩾ \\ x − 2Q − 1 + [\frac{x−2Q+y+p−H}{p−H+1} ] − [\frac{R+y+p−3}{p−1}] + (R − 1) [\frac{x−2Q+y−2}{p−H+1}] . \end{matrix}  (6.4)

Since x − 2Q ⩾ R, the first and the last summands of the last sum are nonnegative for any y ⩾ 0. Moreover p − H + 1 ⩽ p − 1. If Q > 0 then we have x − 2Q + y + p − H = (H − 2)Q + R + y + p − H > p + y + R − 3 and therefore

[\frac{x−2Q+y+p−H}{p−H+1} ] ⩾ [\frac{R+y+p−3}{p−1}] .  (6.5)

For Q = y = 0, we get

[\frac{x+p−H}{p−H+1}] ⩾ 1 ⩾ [\frac{R+p−3}{p−1}] .

Thus we have proved that v_{y−1} ⩾ u_{y} for 0 ⩽ y ⩽ Q. Recall that if y ⩾ 0 then being moved one position up the greatest bead of ⟨p + H − R, S_{2}, x − 2Q + y⟩ takes position v_{y−1}.

Now one can clearly see that

φ^{j}(N) = \overline{(−∞, H) ∪ ⟨H, S_{1}, Q − j⟩ ∪ ⟨p + H − R, S_{2}, x − Q − j⟩} for 0 ⩽ j ⩽ Q.

Hence h(φ^{j}(ν)) = 2p − H and e(φ^{j}(ν)) = 2p for 0 ⩽ j < Q. Moreover h(φ^{Q}(ν)) = p + R − H.

If R > 1 then x − 2Q ⩾ 2 and

φ^{Q+j}(N) = \overline{(−∞, p + H − 1) ∪ ⟨p + H, S_{2}, x − 2Q − 1 − j⟩} for 1 ⩽ j ⩽ x − 2Q − 1.

Hence e(φ^{Q}(ν)) = p+R−1, e(φ^{Q+j}(ν)) = p for 1 ⩽ j < x−2Q−1, h(φ^{Q+j}(ν)) = p−H +1 for 1 ⩽ j ⩽ x − 2Q − 1, e(φ^{x−Q−1}(ν)) = p − R + 1 and φ^{x−Q}(ν) = ∅.

If R = 1 then

φ^{Q+j}(N) = \overline{(−∞, p + H − 1) ∪ ⟨p + H − 1, S_{2}, x − 2Q − j⟩} for 0 ⩽ j ⩽ x − 2Q.

Hence e(φ^{Q+j}(ν)) = p and h(φ^{Q+j}(ν)) = p − H + 1 for 0 ⩽ j < x − 2Q and φ^{x−Q}(ν) = ∅. □

Lemma 6.4. Let h, i, x be integers such that 0 < i ⩽ h < p and x ⩾ i. If i ⩾ h/2 and x ⩾ h then

G_{p}(λ^{(h,i,x)}) = \binom{A_{0} ⋯ A_{h−i−1}}{R_{0} ⋯ R_{h−i−1}}\binom{p}{i}^{x−2(h−i)}\binom{A_{0} − 2i ⋯ A_{h−i−1} − 2i}{R_{0} − i ⋯ R_{h−i−1} − i},

if i < h/2 and x ⩾ 2i then

G_{p}(λ^{(h,i,x)}) = \binom{A_{0} ⋯ A_{i−1}}{R_{0} ⋯ R_{i−1}}\binom{p}{i}^{x−2i}\binom{A_{0} − 2(h − i) ⋯ A_{i−1} − 2(h − i)}{R_{0} − (h − i) ⋯ R_{i−1} − (h − i)},

if x < 2i, h then

G_{p}(λ^{(h,i,x)})=\binom{A_{0} ⋯ A_{x−i−1}}{R_{0} ⋯ R_{x−i−1}}\binom{p}{i + h − x}^{2i−x}\binom{A_{0} − 2(i + h − x) ⋯ A_{x−i−1} − 2(i + h − x)}{R_{0} − (i + h − x) ⋯ R_{x−i−1} − (i + h − x)},

where A_{j} = p + h − 1 − 2j and R_{j} = h − j,

Proof. Let S = [0, i − 1], Λ = Λ^{(h,i,x)} and λ = λ^{(h,i,x)}.

Case i ⩾ h/2 and x ⩾ h. For 0 ⩽ j < h − i, we have x − j − 1 ⩾ x − h + i ⩾ i and therefore the first element of the set ⟨0, S, x − j − 1⟩ is not less than p. Hence

φ^{j}(Λ) = \overline{(−∞, j) ∪ ⟨0, S, x − j⟩ ∪ [i, h − j)} for 0 ⩽ j ⩽ h − i.

Therefore h(φ^{j}(λ)) = R_{j} for 0 ⩽ j ⩽ h − i and e(φ^{j}(λ)) = A_{j} for 0 ⩽ j < h − i. Next we have

φ^{h−i+j}(Λ) = \overline{(−∞, h − i) ∪ ⟨h − i, S, x − 2(h − i) − j⟩} for 0 ⩽ j ⩽ x − 2(h − i).

Hence e(φ^{h−i+j}(λ)) = p for 0 ⩽ j < x − 2(h − i), h(φ^{h−i+j}(λ)) = i for 1 ⩽ j < x − 2(h − i) and h(φ^{x−(h−i)}(λ)) = h − i. Finally

φ^{x−(h−i)+j}(Λ) = \overline{(−∞, i + j) ∪ [p, p + h − i − j)} for 0 ⩽ j ⩽ h − i.

Hence e(φ^{x−(h−i)+j}(λ)) = A_{j} − 2i for 0 ⩽ j < h − i and h(φ^{x−(h−i)+j}(λ)) = R_{j} − i for 1 ⩽ j ⩽ h − i.

Case i < h/2 and x ⩾ 2i. For 0 ⩽ j < i, we have x − j − 1 ⩾ i and therefore the first element of the set ⟨0, S, x − j − 1⟩ is not less than p. Hence we get

φ^{j}(Λ) = \overline{(−∞, j) ∪ ⟨0, S, x − j⟩ ∪ [i, h − j)} for 0 ⩽ j ⩽ i.

Therefore h(φ^{j}(λ)) = R_{j} and e(φ^{j}(λ)) = A_{j} for 0 ⩽ j < i. Next

φ^{i+j}(Λ) = \overline{(−∞, h − i) ∪ ⟨p, S, x − 2i − j⟩} for 0 ⩽ j ⩽ x − 2i.

Hence e(φ^{i+j}(λ)) = p and h(φ^{i+j}(λ)) = i for 0 ⩽ j < x − 2i. Finally

φ^{x−i+j}(Λ) = \overline{(−∞, h − i + j) ∪ [p, p + i − j)} for 0 ⩽ j ⩽ i.

Hence e(φ^{x−i+j}(λ)) = A_{j} − 2(h − i) for 0 ⩽ j < i and h(φ^{x−i+j}(λ)) = R_{j} − (h − i) for 0 ⩽ j ⩽ i.

Case x < 2i, h. For 0 ⩽ j < x − i, we have x − j − 1 ⩾ i and therefore the first element of the set ⟨0, S, x − j − 1⟩ is not less than p. Hence

φ^{j}(Λ) = \overline{(−∞, j) ∪ ⟨0, S, x − j⟩ ∪ [i, h − j)} for 0 ⩽ j ⩽ x − i

Therefore h(φ^{j}(λ)) = R_{j} for 0 ⩽ j ⩽ x − i and e(φ^{j}(λ)) = A_{j} for 0 ⩽ j < x − i. Next

φ^{x−i+j}(Λ) = \overline{(−∞, x − i) ∪ ⟨x − i, S, 2i − x − j⟩ ∪ [h, h − x + i)} for 0 ⩽ j ⩽ 2i − x.

Hence e(φ^{x−i+j}(λ)) = p for 0 ⩽ j < 2i − x and h(φ^{x−i+j}(λ)) = i + h − x for 1 ⩽ j < 2i − x. Finally

φ^{i+j}(Λ) = \overline{(−∞, h − x + i + j) ∪ [p, p + x − i − j)} for 0 ⩽ j ⩽ x − i.

Hence e(φ^{i+j}(λ)) = A_{j} − 2(i + h − x) for 0 ⩽ j < x − i and h(φ^{i+j}(λ)) = R_{j} − (i + h − x) for 0 ⩽ j ⩽ x − i. □

Definition 6.5. Let h, i, x be integers such that 0 < i ⩽ h < p and x ⩾ i. Define

\begin{matrix}M^{(h,i,x)} = \overline{(−∞, p) ∪ [p + h − m, p + h) ∪ ⟨p, S, x⟩}, \\ μ^{(h,i,x)} = P (M^{(h,i,x)}), \end{matrix}

where m = max{i, i + h − x} and S = [0, h − m) ∪ [h, p − 1].

Let c_{0}^{(h,i,x)}, c_{1}^{(h,i,x)}, … be all the elements of the set {n ∈ ℤ : n ⩾ p, rem(n, p) ∈ S} written in ascending order. An easy verification shows that

c_{y}^{(h,i,x)} = p + y + m + m [\frac{y+m−h}{p−m}] .  (6.6)

Lemma 6.6. Let h, i, x be integers such that 0 < i ⩽ h < p and x ⩾ i. Then m(λ^{(h,i,x)}) = μ^{(h,i,x)}.

Proof. Let m and S be as in Definition 6.5, A_{j} and R_{j} as in Lemma 6.4 and M = M^{(h,i,x)}, μ = μ^{(h,i,x)}, S_{j} = p − j.

Case i ⩾ h/2 and x ⩾ h. Then m = i. We have

2p > A_{0} > ⋯ > A_{h−i−1} > p > A_{0} − 2i > ⋯ > A_{h−i−1} − 2i > 0.

Therefore by the main result of [FK] and Lemma 6.4, it suffices to prove that

G_{p}(μ) = \binom{A_{0} ⋯ A_{h−i−1}}{S_{0} ⋯ S_{h−i−1}}\binom{p}{p − i}^{x−2(h−i)}\binom{A_{0} − 2i ⋯ A_{h−i−1} − 2i}{S_{0} − i ⋯ S_{h−i−1} − i}.

Note that x − 2(h − i) ⩾ 0. Therefore for 0 ⩽ j < h − i we have x − j − 1 ⩾ h − i and thus the first element of the set ⟨p, S, x − j − 1⟩ is not less than p + h. Hence we have

φ^{j}(M) = \overline{(−∞, p + j) ∪ [p + h − i, p + h − j) ∪ ⟨p, S, x − j⟩} for 0 ⩽ j ⩽ h − i.

Therefore e(φ^{j}(μ)) = A_{j} and h(φ^{j}(μ)) = S_{j} for 0 ⩽ j < h − i. Next

φ^{h−i+j}(M) = \overline{(−∞, p + i) ∪ ⟨p + h, S, x − 2(h − i) − j⟩} for 0 ⩽ j ⩽ x − 2(h − i).

Hence e(φ^{h−i+j}(μ)) = p and h(φ^{h−i+j}(μ)) = p − i for 0 ⩽ j < x − 2(h − i). We have

φ^{x−(h−i)+j}(M) = \overline{(−∞, p + i + j) ∪ [p + h, 2p + h − i − j)} for 0 ⩽ j ⩽ h − i.

Hence e(φ^{x−(h−i)+j}(μ)) = A_{j} − 2i and h(φ^{x−(h−i)+j}(μ)) = S_{j} − i for 0 ⩽ j < h − i and φ^{x}(μ) = ∅.

Case i < h/2 and x ⩾ 2i. We have

2p > A_{0} > ⋯ > A_{i−1} > p > A_{0} − 2(h − i) > ⋯ > A_{i−1} − 2(h − i) > 0.

Therefore by the main result of [FK] and Lemma 6.4, it suffices to prove that

G_{p}(μ) = \binom{A_{0} ⋯ A_{i−1}}{S_{0} ⋯ S_{i−1}}\binom{p}{p − i}^{x−2i}\binom{A_{0} − 2i ⋯ A_{i−1} − 2(h − i)}{S_{0} − i ⋯ S_{i−1} − (h − i)}.

For 0 ⩽ j < i, we have x − j − 1 ⩾ h − m. Therefore the first element of the set ⟨p, S, x − j − 1⟩ is not less than p + h. Hence we get

φ^{j}(M) = \overline{(−∞, p + j) ∪ [p + h − m, p + h − j) ∪ ⟨p, S, x − j⟩} for 0 ⩽ j ⩽ i. Therefore e(φ^{j}(μ)) = A_{j} and h(φ^{j}(μ)) = S_{j} for 0 ⩽ j < i. For 0 ⩽ j ⩽ x − 2i, we have

φ^{i+j}(M)= \{ \begin{matrix}\overline{(−∞, p + i) ∪ [p + x − i − j, p + h − i) ∪ [p + h, 2p + x − i − j)} & if x < h; \\ \overline{(−∞, p + i) ∪ ⟨p + i, S, x − 2i − j⟩} & if x ⩾ h. \end{matrix}

Hence e(φ^{i+j}(μ)) = p and h(φ^{i+j}(μ)) = p − i for 0 ⩽ j < x − 2i. Finally

φ^{x−i+j}(M) = \overline{(−∞, p + h − i + j) ∪ [p + h, 2p + i − 1 − j]} for 0 ⩽ j ⩽ i. Hence e(φ^{x−i+j}(μ)) = A_{j} −2(h−i), h(φ^{x−i+j}(μ)) = S_{j} −(h−i) for 0 ⩽ j < i and φ^{x}(μ) = ∅.

Case x < 2i, h. Then m = i + h − x. We have

2p > A_{0} > ⋯ > A_{x−i−1} > p > A_{0} − 2(i + h − x) > ⋯ > A_{x−i−1} − 2(i + h − x) > 0. Therefore by the main result of [FK] and Lemma 6.4, it suffices to prove that

G_{p}(μ)=\binom{A_{0} ⋯ A_{x−i−1}}{S_{0} ⋯ S_{x−i−1}}\binom{p}{p − (i + h − x)}^{2i−x} \binom{A_{0} − 2(i + h − x) ⋯ A_{x−i−1} − 2(i + h − x)}{S_{0} − (i + h − x) ⋯ S_{x−i−1} − (i + h − x)}.

For 0 ⩽ j < x − i, we have x − j − 1 ⩾ i > x − i and therefore the first element of the set ⟨p, S, x − j − 1⟩ is not less than p + h. Hence we get

φ^{j}(M) = \overline{(−∞, p + j) ∪ [p + x − i, p + h − j) ∪ ⟨p, S, x − j⟩} for 0 ⩽ j ⩽ x − i. Therefore e(φ^{j}(μ)) = A_{j} and h(φ^{j}(μ)) = S_{j} for 0 ⩽ j < x − i. Next

φ^{x−i+j}(M) = \overline{(−∞, p + i + h − x) ∪ ⟨p + h, S, 2i − x − j⟩} for 0 ⩽ j ⩽ 2i − x. Hence e(φ^{x−i+j}(μ)) = p and h(φ^{x−i+j}(μ)) = p − (i + h − x) for 0 ⩽ j < 2i − x. Finally φ^{i+j}(M) = \overline{(−∞, p + i + h − x + j) ∪ [p + h, 2p + x − i − 1 − j]} for 0 ⩽ j ⩽ x − i. Hence e(φ^{i+j}(μ)) = A_{j} − 2(i + h − x) and h(φ^{i+j}(μ)) = S_{j} − (i + h − x) for 0 ⩽ j < x − i and φ^{x}(μ) = ∅. □

It is interesting to look at the partitions ν(2, x), when p > 2 and x is an odd number greater than 2. By Lemmas 6.3 and 6.6, we get m({ \atop λ^{(2,x)}})= ν^{(2,x)} and m(λ^{(2,x)}) = μ^{(2,2,x)} = λ^{(p−2,x)}. We have h(ν^{(2,x)}) = 2p − 2 and h(λ^{(p−2,x)}) = p − 2. Hence ν^{(2,x)} ▷̸ λ^{(p−2,x)} and (ν^{(2,x)})^{t} ◁̸ (λ^{(p−2,x)})^{t}. Since simple Σ_{px}-modules are self-dual, by Theorem 4.6 (or [KSh2, Theorem 3.5(iv)]), [KN, Theorem 4.4(b)] and Propositions [J, II.2.14(4)], [D, 2.1f], we have

K ≅ Ext_{Σ_{px}}^{1}(D{ \atop λ^{(2,x)}}, D^{λ^{(2,x)}})≅ Ext_{S(N,px)}^{1}(L((ν^{(2,x)})^{t}), L((λ^{(p−2,x)})^{t}))

≅ Hom_{S(N,px)}(rad Δ((ν^{(2,x)})^{t}), L((λ^{(p−2,x)})^{t})),

where N ⩾ px and S(N, px) denotes the Schur algebra. Definition 6.2 and (6.1) show that ν^{(2,x+2(p−1))} = ν^{(2,x)} + (p^{2p−2}). Applying Corollary 5(4), Theorem 4(a) and Lemma 4 from [S1], we obtain a negative solution of Problem 2 from [S1] for the following values of the parameters: λ := ν^{(2,3)}, q_{i} := pi, n := 2p − 2, V_{i} := rad S^{λ+(q_{i}^{n})}.

In the remaining part of this section, we fix integers H, x, i such that 1 < H < p, H ∤ x and 0 < i ⩽ x, h, where h = H − rem(x, H). Let also Q = quo(x, H), R = rem(x, H), m = max{i, i+h−x}, S_{1} = [0, p−1]\{h}, S_{2} = {h}∪[H, p−1] and S = [0, h−m)∪[h, p−1].

Lemma 6.7.

(a) if c_{y}^{(h,i,x)} ⩾ b_{z}^{(H,x)} and y, z > 0, then c_{y−1}^{(h,i,x)} ⩾ b_{z−1}^{(H,x)};

(b) if a_{y}^{(h,i,x)} ⩽ z and y > 0, then a_{y−1}^{(h,i,x)} ⩽ z − 1;

(c) if a_{y}^{(h,i,x)} > z, and y ⩾ m − 1, 0 ⩽ m ⩽ p, then a_{y−j}^{(h,i,x)} ⩾ z − j for 0 ⩽ j < m.

Proof follows from the mutual situation of the sets S, S_{1}, S_{2}. □

Theorem 6.8. The following conditions are equivalent:

(1) m({ \atop λ^{(H,x)}})⊴ m(λ^{(h,i,x)});

(2) m [\frac{p−h+x−1}{p−m}] ⩾ H − Q − 1 + h [\frac{x−Q−1}{p−H+1}] + (R − 1) [\frac{x−Q−2}{p−H+1}] .

Proof. Since shift(N^{(H,x)}) = shift(M^{(h,i,x)}) = 2p (we added one more row in the definition of M^{(h,i,x)} just to ensure this equality), the first formula of (2.1) and Lemmas 6.3 and 6.6 show that condition (1) of the current theorem is equivalent to u ⊴ v, where

\begin{matrix}u = (b_{x−Q+p−H}^{(H,x)}, …, b_{x−Q}^{(H,x)}, a_{Q+p−2}^{(H,x)}, …, a_{Q}^{(H,x)}, H − 1 …, 0), \\ v = (c_{x+p−m−1}^{(h,i,x)}, …, c_{x}^{(h,i,x)}, p + h − 1, …, p + h − m, p − 1, …, 0). \end{matrix}

It is easy to see that condition (2) of the current theorem is equivalent to u_{1} ⩽ v_{1}. Therefore (1) implies (2).

Now suppose that condition (2) is satisfied. Note that m ⩽ h < H and thus p − m ⩾ p−H +1. By the equivalence of the previous paragraph and Lemma 6.7(a), we get u_{j} ⩽ v_{j} for 1 ⩽ j ⩽ p − H + 1.

Now let p − H + 1 < j ⩽ p − m. By (6.1) and (6.6), we have

\begin{matrix}v_{j} − u_{j} = c_{x+p−m−j}^{(h,i,x)} − a_{Q+2p−H−j}^{(H,x)} = x − Q + m [\frac{x+p−j−h}{p−m}] − [\frac{2p−1+Q−j−h}{p−1}] \\ = x − Q − 1 + m [\frac{x+p−j−h}{p−m}] − [\frac{p+Q−j−h}{p−1}] . \end{matrix}

Since m ⩾ 1, p − m ⩽ p − 1 and (x + p − j − h) − (p + Q − j − h) = x − Q > 0, we have

m [\frac{x+p−j−h}{p−m}] ⩾ [\frac{p+Q−j−h}{p−1}]

and v_{j} ⩾ u_{j}.

Case 1: m = 0 or m > 0, v_{p−m+1} ⩾ u_{p−m+1}. By Lemma 6.7(b), we get v_{j} ⩾ u_{j} for p−m < j ⩽ p. Thus we have proved v_{j} ⩾ u_{j} for 1 ⩽ j ⩽ p. We have v_{j} ⩽ u_{j} for p < j ⩽ 2p, since v_{j} is an improper bead of M^{(h,i,x)} for such j. Hence σ_{j}(v)−σ_{j}(u) ⩾ σ_{2p}(v)−σ_{2p}(u) = 0 for any p ⩽ j ⩽ 2p. Therefore u ⊴ v.

Case 2: m > 0 and v_{p−m+1} < u_{p−m+1}. By Lemma 6.7(c), we get v_{j} ⩽ u_{j} for p − m < j ⩽ p. We have v_{j} ⩽ u_{j} for p < j ⩽ 2p, since v_{j} is an improper bead of M^{(h,i,x)} for such j. Hence σ_{j}(v) − σ_{j}(u) ⩾ σ_{2p}(v) − σ_{2p}(u) = 0 for any p − m ⩽ j ⩽ 2p. Therefore u ⊴ v. □

Remark. The only fact we will need is that (1) implies (2). The reverse implication has been proved only to show the impossibility of improving the bounds by replacing a simpler condition (2) with a more complicated condition (1).

7 Auxiliary upper bound

7.1 Systems. Introduce the following staircase abaci and partitions. Let k ⩾ 0 and 0 < r_{2} < ⋯ < r_{k} < p, i_{1} > ⋯ > i_{k−1} ⩾ 0 be some integers. We put

St(r_{2}, …, r_{k}; i_{1}, …, i_{k−1}) =
\overline{(−∞, 0) ∪ [pi_{1}, pi_{1} + r_{2}) ∪ [pi_{2} + r_{2}, pi_{2} + r_{3}) ∪ ⋯ [pi_{k−1} + r_{k−1}, pi_{k−1} + r_{k})},
st(r_{2}, …, r_{k}; i_{1}, …, i_{k−1}) = P (St(r_{2}, …, r_{k}; i_{1}, …, i_{k−1})).

For k = 1, we assume St(∅; ∅) = \overline{(−∞, 0)} and st(∅; ∅) = ∅.

We have already met special cases of staircase abaci in § 6. Indeed, let h, i, x be integers such that 0 < i ⩽ h < p and x ⩾ 0. We put q = quo(x, i) and r = rem(x, i). Then

Λ^{(h,i,x)} = \{ \begin{matrix}St(i, h; q, 0) & if r = 0; \\ St(r, i, h; q + 1, q, 0) & if r > 0. \end{matrix}  (7.1)

Lemma 7.1. Every solution of the system

\{ \begin{matrix}h(λ) < p, \\ core(λ) = ∅, \\ res A = res B for any λ-normal nodes A and B. \end{matrix}  (7.2)

has the form st(r_{2}, …, r_{k}; i_{1}, …, i_{k−1}).

Proof. Clearly, these partitions satisfy the system. Prove the converse fact by induction on the number n of nodes in [λ]. This is obviously true for n = 0. Now let n > 0 and λ be a solution of system (7.2). Since λ ≠ ∅ = core(λ), there is at least one rim p-hook of λ. Denote by λ̄ the partition obtained from λ by removing the highest of these hooks. By Lemma 5.2, the partition λ̄ also satisfies system (7.2). By the inductive hypothesis, λ̄ = st(r_{2}, …, r_{k}; i_{1}, …, i_{k−1}).

Take the abacus Λ such that

λ = P (Λ) and shift(Λ) = shift(St(r_{2}, …, r_{k}; i_{1}, …, i_{k−1})).

We see that there is a movable down bead a of St(r_{2}, …, r_{k}; i_{1}, …, i_{k−1}) such that Λ is obtained from St(r_{2}, …, r_{k}; i_{1}, …, i_{k−1}) by moving a one position down.

Case k = 1. Since h(λ) < p, we have −p < a < 0 and λ = st(1, −a; 1, 0) for a < −1 and λ = st(1; 1) for a = −1.

Case k > 1. We can assume i_{1} > 0, as otherwise λ̄ = ∅ and we are under the conditions of the previous case. Since h(λ) < p, we have r_{k} − p < a. If a < −1 or a ⩽ −1 and i_{k} > 1, then Λ contains the following two normal beads in different runners: pi_{1} and a + p.

Therefore, the third equation of system (7.2) is violated for λ and this case is impossible. If a = −1 and i_{k} = 1, then λ = st(r_{2} + 1, …, r_{k} + 1; i_{1}, …, i_{k−1}) (k = 2, 3).

Now suppose that a ⩾ 0. Since the third equation of system (7.2) holds for λ, a can take only the following values: pi_{t} + r_{t}, where t = 1, …, k − 1 and we assume r_{1} = 0. Otherwise pi_{1} and a + p would be normal beads of Λ from different runners. Now directly from the definition of staircase abacus one can see that λ has the desired form. □

Let us calculate e(st(r_{2}, …, r_{k}; i_{1}, …, i_{k−1})). It is zero for k = 1. Therefore consider the case k > 1. Define the sequence 1 = a_{1} < ⋯ < a_{l} ⩽ k − 1 by the following rule: a_{j+1} = a_{j} + 1 if a_{j} + 1 ⩽ k − 1 and i_{a_{j}+1} < i_{a_{j}} − 1; a_{j+1} = a_{j} + 2 if a_{j} + 2 ⩽ k − 1 and i_{a_{j}+1} = i_{a_{j}} − 1. From the definition of § 2.4, we get

e(st(r_{2}, …, r_{k}; i_{1}, …, i_{k−1})) = \{ \begin{matrix}pl & if i_{a_{l}} > 0; \\ p(l − 1) + r_{k} − 1 & if i_{a_{l}} = 0. \end{matrix}

Therefore if e(λ) < 2p and λ is a staircase partition, then λ has one of the following forms: ∅, st(r_{2}, r_{3}; i_{1}, 0), st(r_{2}, r_{3}, r_{4},; i_{1}, i_{1} − 1, 0). This fact and (7.1) yield

Lemma 7.2. Every solution of the system

\{ \begin{matrix}h(λ) < p, \\ core(λ) = ∅, \\ res A = res B for any λ-normal nodes A and B, \\ h(λ) + h(m(λ)) < 2p. \end{matrix}  (7.3)

has the form λ^{(h,i,x)}, where 0 < i ⩽ h < p and x ⩾ 0.

The last inequalities can be strengthen by i ⩽ x.

7.2 Bound. We state the following known result.

Proposition 7.3. Let n > 2 and V be a KΣ_{n−1}-module. Then

(a) V ↑^{Σ_{n}} ⊗ sgn_{n} ≅ (V ⊗ sgn_{n−1})↑^{Σ_{n}};

(b) (Ind^{α} V) ⊗ sgn_{n} ≅ Ind^{−α}(V ⊗ sgn_{n−1}).

Proof. (a) The isomorphism is given by (σ_{i} ⊗ v) ⊗ u ↦ σ_{i} ⊗ (v ⊗ u), where σ_{1}, …, σ_{n} are representatives of the left cosets of Σ_{n} over Σ_{n−1} having the same sign, v ∈ V and u is a basis of the sign representation of Σ_{n}.

(b) follows from (a) and the arguments from the proof of [K2, Theorem 4.7] applied to the induction operator instead of the restriction operator. □

Definition 7.4. Let π(H, x, i) be satisfied if and only if

(1) H, x, i ∈ ℤ, 2 < H < p, H ∤ x, 0 < i ⩽ x, h and x > 2;

(2) m [\frac{p−h+x−1}{p−m}] ⩾H−Q−1 + h [\frac{x−Q−1}{p−H+1}] +(R−1) [\frac{x−Q−2}{p−H+1}] ,

where Q = quo(x, H), R = rem(x, H), h = H − R and m = max{i, i + h − x}.

Definition 7.5. Let H, x, i be integers for which π(H, x, i) is satisfied. Denote by ε(H, x, i) the sequence such that −R + Λ^{(h,i,x)} = H_{ε(H,x,i)}(Λ^{(H,x)}), where R = rem(x, H) and h = H − R.

Clearly the required sequence ε(H, x, i) exists and is given by

ε(H, x, i) = ((−Q − 1)^{R}, (−Q)^{h−i}, (q − Q)^{i−r}, (q + 1 − Q)^{r}),

where Q = quo(x, H), R = rem(x, H), h = H − R, q = quo(x, i), r = rem(x, i).

Definition 7.6. A pair (ν, μ) is called minimal if

• ν is an almost completely splittable partition;

• μ is a p-regular partition;

• ν ▷̸ μ and ν ∼ μ;

• for any μ-good node A there exists a ν-good node B such that res B = res A and ν_{B} is not an almost completely splittable partition.

It follows directly from the definition that if B is a ν-normal node, ν is an almost completely splittable partition and ν_{B} is not, then χ(λ) = p and B = (1, ν_{1}), where λ is the preimage of ν. Therefore if (ν, μ) is a minimal pair, then all μ-normal nodes have residue \overline{ν_{1} − 1} = \overline{λ_{1}}.

Lemma 7.7. Let (ν, μ) be a minimal pair of partitions of n. We put x = n/p and H = h(λ), where λ is the preimage of ν. Suppose that (ν, μ) is different from ((p^{2}, p^{2} −p), (2p^{2} − p)) and from ({ \atop λ^{(H,2)}}, λ^{(H−2,2)}) and that Ext_{Σ_{n}}^{1}(D^{ν}, D^{μ}) ≠ 0. Then ν = { \atop λ^{(H,x)}} and μ = λ^{(H−rem(x,H),i,x)}, where π(H, x, i) is satisfied.

Proof. Let α be the residue of all μ-normal nodes and A be a ν-good node of residue α. We have n ⩾ p > 2. Since h(ν) < p, by Proposition 2.2, we have

Hom_{Σ_{n}}(rad S^{ν}, D^{μ}) ≅ Ext_{Σ_{n}}^{1}(D^{ν}, D^{μ}) ≠ 0.

Hence ν ◁ μ and in particular h(μ) ⩽ h(ν). Let ν = λ̃, where λ is a big partition. Since ν_{A} is not almost completely splittable , we get χ(λ) = p and A is in the first row.

We have h(ν) ⩽ h(λ). Suppose h(ν) < h(λ). Then ν is completely splittable . By the hypothesis of the current theorem and Theorem 4.6, we get that ν is a big partition and μ = ν̃. The rightmost node A′ of the first row of [μ] is removable and therefore is normal. However res A′ is equal to the residue of the rightmost node of the last row of [ν], which is distinct from res A, as ν is completely splittable and h(ν) > 1. The resulting contradiction gives h(ν) = h(λ).

By Corollary 5.3, the residue of any core(μ)-normal node, and by λ ∼ μ also of any core(λ)-normal node, is res A, which in turn equals the residue of the bottom λ-removable node. By Lemma 4.10, we have λ = λ^{(H,x)}, 1 < H < p, x > 0 and H ∤ x. The case x = 1 is impossible, as we would have a contradiction h(ν) < h(λ). Thus x > 1. By Lemma 4.13, we get h(μ) < h(λ) = H. We put Q = quo(x, H), R = rem(x, H) and h = H − R.

If R > 1 we put b = (Q + 1)p and if R = 1 we put b = Qp. Since x > 1, we get that b is the only normal and thus good bead of { \atop Λ^{(H,x)}} belonging to runner zero . Denote by B the node of ν corresponding to b. By the second formula of (2.1), we have res B = −\overline{shift(Λ^{(H,x)})} = −H̄. By Lemma 3.4, we have

0 < dim Ext_{Σ_{n}}^{1}(D^{ν}, D^{μ}) ⩽
dim Ext_{Σ_{n−1}}^{1}(D^{ν_{B}}, Res_{−H̄} D^{μ}) + dim Hom_{Σ_{n−1}}(rad Ind^{−H̄} D^{ν_{B}}, D^{μ}).

Since α = R̄−H̄ ≠ −H̄, the hypothesis of the current theorem yields Res_{−H̄} D^{μ} = 0. Hence Hom_{Σ_{n−1}}(rad Ind^{−H̄} D^{ν_{B}}, D^{μ}) ≠ 0 and in particular [Ind^{−H̄} D^{ν_{B}} : D^{μ}] > 0. Multiplying the modules from the last inequality by sgn_{n}, we get by Proposition 7.3(b) that

[IndH̄ D^{m(ν_{B})} : D^{m(μ)}] > 0.  (7.4)

It follows from the proof of [K2, Theorem 4.7] that m(ν_{B}) = m(ν)_{B^{(m)}}, where B^{(m)} is the m(ν)-good node of residue H̄. By Lemma 6.3, we get m(ν) = P (N^{(H,x)}). For brevity until the end of the current proof we shall use the notation N = N^{(H,x)}, a_{i} = a_{i}^{(H,x)}, b_{i} = b_{i}^{(H,x)} (see Definition 6.2 and (6.1)). We shall prove that m(ν) ⊴ m(μ).

Case 1: R > 1. We have a_{Q+p−1} < b_{x−Q}. Indeed if Q > 0, then taking into account (6.5) the substitution y = Q + 1 into (6.4) yields b_{x−Q} − a_{Q+p−1} ⩾ x − 2Q − 1 ⩾ R − 1 > 0. On the other hand, if Q = 0 then a_{Q+p−1} = H + p < b_{x} = b_{x−Q}, since x > 1.

Denote by b^{(m)} the element of ⟨p + H − R, S_{2}, x − Q⟩ from runner H. Clearly, b^{(m)} is a normal bead of N. The only initial bead of N distinct from b^{(m)} and belonging to runner H is a_{Q} in the case where Q > 0 and p − 1 | Q. However this bead is not normal, as in this case a_{Q} + p − 1 = a_{Q+p−2} and a_{Q} + p = a_{Q+p−1} < b_{x−Q}, whence N(a_{Q} + p − 1) = 1 and N(a_{Q} + p) = 0. Therefore b^{(m)} is a good bead of N, the node B^{(m)} corresponds to this bead, and P (N_{b^{(m)}}) = m(ν)_{B^{(m)}} = m(ν_{B}).

All the initial spaces of N_{b^{(m)}} from runner H are: b^{(m)}; H if Q > 0; a_{Q+p−1} if p − 1 | Q. If in the last case Q > 0, then the space a_{Q+p−1} is not conormal. We put c = H if Q > 0 and c = H + p if Q = 0. We have c ⩽ a_{Q+p−1} < b_{x−Q} ⩽ b^{(m)}. Therefore all the conormal spaces of N_{b^{(m)}} from runner H are b^{(m)} and c.

Case 1.1: m(μ) ≠ P (N_{b^{(m)}}^{c}). Denote by C the addable node of m(ν_{B}) corresponding to the space c of N_{b^{(m)}}. If we suppose [S^{m(ν)} : D^{m(μ)}] = 0, then by (7.4), Lemma 3.10 applied to the following parameters:

r := n − 1, λ := m(ν_{B}), α := H̄, k := 2, B_{1} := C, B_{2} := B^{(m)}, γ := m(μ)

yields a contradiction m(μ) = m(ν_{B})^{C} = P (N_{b^{(m)}}^{c}). Thus [S^{m(ν)} : D^{m(μ)}] > 0, m(ν) ⊴ m(μ) and m(ν) ⊴ m(μ).

Case 1.2: m(μ) = P (N_{b^{(m)}}^{c}). If one of the conditions R = 2 or Q = 0 is violated, then N_{b^{(m)}}^{c} contains the following normal beads not belonging to runner H −R: b^{(m)}−1 if R > 2; the bead of ⟨H, S_{1}, Q⟩ from runner H − 1 if R = 2 and Q > 0. Therefore the partition m(μ) = P (N_{b^{(m)}}^{c}) contains normal nodes of residue different from H̄ − R̄ = −α, which is a contradiction.

We have x = 2, h(ν) = H > 2, ν = { \atop λ^{(H,2)}} and

\begin{aligned}N & = & \overline{(−∞, H − 2 + p) ∪ \{H − 1 + p\} ∪ (H + p, 2p) ∪ \{H − 2 + 2p, H + 2p\}}, \\ N_{b^{(m)}}^{c} & = & \overline{(−∞, H − 2 + p) ∪ [H + p, 2p) ∪ \{H − 2 + 2p, H − 1 + 2p\}}. \end{aligned} (7.5)

Hence m(μ) = (H^{2}, 2^{p−H}) = λ^{(p−H+2,2,2)}. By Lemma 6.6, we get μ = λ^{(H−2,2)}, which contradicts the hypothesis of the theorem.

Case 2: R = 1. Since x > 1, we have Q > 0. In the proof of Lemma 6.3, it was shown that a_{Q+p−2} ⩽ b_{x−Q−1} < b_{x−Q}. Denote by b^{(m)} the element of ⟨H, S_{1}, Q⟩ from runner H. Since Q > 0, this bead is normal in N. Clearly, there is no other normal bead of N in runner H. Therefore b^{(m)} is a good bead of N, the node B^{(m)} corresponds to this bead, and P (N_{b^{(m)}}) = m(ν)_{B^{(m)}} = m(ν_{B}).

All the initial spaces of N_{b^{(m)}} from runner H are: b^{(m)}; H; b_{x−Q+p−H+1} if p − H + 1 | x − Q − 1. In the last case the space b_{x−Q+p−H+1} is not conormal in N_{b^{(m)}}, since b^{(m)} ⩽ a_{Q+p−2} < b_{x−Q}. We put c = H. Since Q > 0, we have c < b^{(m)}. Therefore all the conormal spaces of N_{b^{(m)}} from runner H are b^{(m)} and c. Denote by C the addable node of m(ν_{B}) corresponding to the space c of N_{b^{(m)}}.

Case 2.1: m(μ) ≠ P (N_{b^{(m)}}^{c}) is similar to case 1.1.

Case 2.2: m(μ) = P (N_{b^{(m)}}^{c}). We have h(m(μ)) = 2p − H + 1 and h(μ) ⩽ H − 1. Hence e(N_{b^{(m)}}^{c}) ⩽ h(μ) + h(m(μ)) ⩽ 2p. Clearly, this can happen only if b^{(m)} = a_{Q+p−2} = H + p (arguments similar to the ones used in the proof of Lemma 6.3). In this case

N_{b^{(m)}}^{c} = \overline{(−∞, H − 1) ∪ [H, H + p) ∪ ⟨p + H − 1, S_{2}, x − Q⟩}  (7.6)

and m(μ) = P (N_{b^{(m)}}^{c}) is a p-singular partition, which is a contradiction.

Thus we have considered all possible cases and proved that m(ν) ⊴ m(μ). We have h(μ) < h(ν) < p and h(m(μ)) ⩽ h(m(ν)). By (6.2) and (6.3), we have

h(μ) + h(m(μ)) < h(ν) + h(m(ν)) = e(ν) + [p ∤ e(ν)] ⩽ 2p.

Therefore μ satisfies system (7.3) and Lemma 7.2 yields μ = λ^{(h′,i,x)}, where 0 < i ⩽ h′, x and h′ < p. Since the only normal bead of λ^{(h′,i,x)} has residue −h̄′, the hypothesis of the current theorem yields −h̄′ = α = R̄ − H̄. Since 0 < H − R < p, we have −p < H − R − h′ < p. But H − R − h′ is divisible by p, whence h′ = H − R = h.

From m(ν) ⊴ m(μ) and Theorem 6.8, it follows that part (2) of the condition π(H, x, i) is satisfied. We claim that part (1) of the condition π(H, x, i) is satisfied. It suffices to prove that H, x > 2.

Indeed, if H = 2 then h = 1, x is an odd number greater than 2 and ν = (\frac{x+1}{2}p, \frac{x−1}{2}p), μ = (px). By [KSh2, Lemma 3.5(iv)], we get x = 2p−1 and (ν, μ) = ((p^{2}, p^{2}−p), (2p^{2}−p)). This is a contradiction.

If x = 2 then part (2) of the condition π(H, x, i) is not satisfied, contrary to what was proved. □

Lemma 7.8.

(a) Ext_{Σ_{2p^{2}−p}}^{1}(D^{(p^{2},p^{2}−p)}, D^{(2p^{2}−p)}) ≅ K;

(b) Ext_{Σ_{2p}}^{1}(D{ \atop λ^{(H,2)}}, D^{λ^{(H−2,2)}}) ≅ K, where 2 < H < p.

Proof. (a) follows from [KSh2, Lemma 3.5(iv)].

(b) It follows from (7.5) and Lemma 6.3 that m({ \atop λ^{(H,2)}})= { \atop λ^{(p−H+2,2)}}. Hence by Theorem 4.6, we get

K ≅ Ext_{Σ_{2p}}^{1}(D^{λ^{(p−H+2,2)}}, D^{m({ \atop λ^{(H,2)}})}) ≅ Ext_{Σ_{2p}}^{1}(D{ \atop λ^{(H,2)}}, D^{m(λ^{(p−H+2,2)})}).  (7.7)

By Lemma 6.6, we have m(λ^{(p−H+2,2)}) = m(λ^{(p−H+2,2,2)}) = λ^{(H−2,2)} (see case 1.2 of the proof of the previous lemma). Substituting this value into (7.7), we get the desired equivalence. □

8 Upper bound

8.1 General construction. To make the approach of § 4.1 applicable to almost completely splittable partitions, we shall modify it as follows.

Let X be a set satisfying the same conditions as in § 4.1 and we somehow know a map ζ : X → ℤ′ such that

dim Ext_{Σ_{n}}^{1}(D^{ν}, D^{μ}) ⩽ ζ(ν, μ) for any pair (ν, μ) ∈ X of partitions of n.  (8.1)

Define the map U : X → ℤ′ by induction as follows. We put U(∅, ∅) = 0. Now let (ν, μ) be a pair of nonempty partitions of X. For any μ-good node A, let m_{A}(ν, μ) equal:

• ε(ν, μ_{A}) if there is no ν-good node of residue res A;

• U(ν_{B}, μ_{A}) + ε(ν, μ_{A}) if there is a ν-good node B of residue res A and (ν_{B}, μ_{A}) ∈ X;

• +∞ if there is a ν–good node B of residue res A and (ν_{B}, μ_{A}) ∉ X.

We put U(ν, μ) = min({ζ(ν, μ)} ∪ {m_{A}(ν, μ) : A is a μ-good node}).

Lemma 8.1. Let (ν, μ) ∈ X. Then dim Ext_{Σ_{n}}^{1}(D^{ν}, D^{μ}) ⩽ U(ν, μ), where ν, μ ⊢ n.

Proof is by induction on n applying Lemma 3.2. □ 8.2 Case of almost completely splittable partitions. Let us keep the following notation until the end of this section:

X = {(ν, μ) : ν is almost completely splittable , μ is p-regular, ν ▷̸ μ , ν ∼ μ}.

Define ζ as follows. Let (ν, μ) ∈ X be a pair of partitions of n. We put x = n/p and H = h(λ), where λ is the preimage of ν. We define ζ(ν, μ) = +∞ except the following cases:

(1) μ ▷̸ ν. We put ζ(ν, μ) = 0.

(2) μ ▷ ν and (ν, μ) is minimal.

(2.1) (ν, μ) = ((p^{2}, p^{2} − p), (2p^{2} − p)). We put ζ(ν, μ) = 1.

(2.2) (ν, μ) = ({ \atop λ^{(H,2)}}, λ^{(H−2,2)}), where H > 2. We put ζ(ν, μ) = 1.

(2.3) Cases (2.2) and (2.3) do not hold and there is no i such that ν = { \atop λ^{(H,x)}}, μ = λ^{(H−rem(x,H),i,x)} and π(H, x, i) is satisfied. We put ζ(ν, μ) = 0.

It follows from Proposition 2.2 and Lemmas 7.7, 7.8 that property (8.1) holds for ζ we have just defined.

Theorem 8.2. Let ν, μ be p-regular partitions of n such that ν is almost completely splittable and ν ▷̸ μ. Then Ext_{Σ_{n}}^{1}(D^{ν}, D^{μ}) = 0 except the case when μ = H_{ε}(λ), where λ is the preimage of ν, H = h(λ) and one of the following conditions holds:

(1) H = 2 and ε = (1, −1), (−p, p);

(2) H = 3 and ε = (0, −1, 1), (−1, 1, 0), (−1, −1, 2);

(3) H > 3 and ε = (0, −1, 0^{H−3}, 1), (−1, 0^{H−3}, 1, 0), (−1, −1, 0^{H−4}, 1, 1); (4) ε = ε(H, x, i) for x and i such that π(H, x, i) is satisfied.

In cases (1)-(3) the bound Ext_{Σ_{n}}^{1}(D^{ν}, D^{μ}) ⩽ 1 holds.

Proof. In view of Proposition 2.3, it suffices to assume ν ∼ μ. Therefore we must prove that U(ν, μ) = 0 for any pair (ν, μ) ∈ X except cases (1)-(4) and that U(ν, μ) ⩽ 1 in cases (1)-(3). Apply induction on n. If n = 0 then by definition we have U(ν, μ) = 0.

Now let n > 0 and suppose that theorem is true for partitions of numbers less than n. Choose some abaci M and Λ of the same shift such that μ = P (M) and λ = P (Λ).

Case 1: μ ▷̸ ν. Then U(ν, μ) = ζ(ν, μ) = 0.

Case 2: μ ▷ ν and the pair (ν, μ) is not minimal. Let A be a μ-good node such that either there is no ν-good node of residue res A or such a node B exists and ν_{B} is almost completely splittable . Denote by a the bead of M corresponding to A.

First consider the case ε(ν, μ_{A}) = 1. Then Λ̃ = (M_{a})^{c} for some conormal space c of M_{a}. Since ν ∼ μ and ν ◁ μ, we get that a and c belong to the same runner and moreover a is below c. We have M = (Λ̃_{c})^{a}.

If H = 2 then we get c = b^{Λ} − p, a = b^{Λ} + p and b^{Λ} = b_{Λ} + 1. Hence M = {b_{Λ} − p, b^{Λ} + p} = H_{(1,−1)}(Λ).

If H > 3 then either c = b^{Λ} − p, a = b^{Λ}, b^{Λ} = b^{Λ}(2) + 1 or c = b_{Λ}(2), a = b^{Λ}(2) + p, b_{Λ}(2) = b_{Λ} + 1. In the first case M = H_{(0,−1,0^{H−3},1)}(Λ) and in the second case M = H_{(−1,0^{H−3},1,0)}(Λ). Note that in all cases the only bead of Λ̃, belonging to the same runner as a is c, which is not normal. Therefore there is no ν-good node of residue res A and U(ν, μ) ⩽ m_{A}(ν, μ) = ε(ν, μ_{A}) = 1.

Now consider the case ε(ν, μ_{A}) = 0. If there is no ν-good node of residue res A, then U(ν, μ) = m_{A}(ν, μ) = 0. Therefore we assume that there is a ν-good node B of residue res A. By the choice of A, the partition ν_{B} is almost completely splittable . Applying Lemma 4.3(b), we get ν_{B} ▷̸ μ_{A}, whence (ν_{B}, μ_{A}) ∈ X. Hence U(ν, μ) ⩽ m_{A}(ν, μ) = U(ν_{B}, μ_{A}). Therefore in the sequel we will consider the case U(ν_{B}, μ_{A}) > 0.

Let b be the bead of Λ̃ corresponding to B. If χ(λ) = p then b is different from the greatest bead of Λ̃ (i.e., r(B) > 1), since otherwise the greatest bead of Λ̃_{b} equal to b − 1 is not movable up, which contradicts the fact that ν_{B} is almost completely splittable . Hence ν_{B} = λ̃_{D}, where D is a λ-good node (see the definition of H_{ε} or [S2, Lemma 8]).

We put H̄ = h(λ_{D}). By the inductive hypothesis, we get μ_{A} = H_{ε}(λ_{D}), where ε is the sequence described in cases (1)-(4) of the current theorem, where H is replaced by H̄. Since ∑H̄ ; i=1 ε_{i} = 0, we have M_{a} = H_{ε}(Λ_{d}), where d is a bead of Λ corresponding to D.

Recall that a and d are cogood spaces of M_{a} and Λ_{d} respectively, since a and d are cogood beads of M and Λ respectively. Since ε ≠ (0H̄), we have ε_{j} < 0 for some 1 ⩽ j ⩽ H̄. Hence e ⩽ b^{Λ_{d}}(j) − p, where e is the smallest space of Λ_{d}. Therefore d ≠ e, as otherwise we would get b^{Λ} = b^{Λ_{d}}, b_{Λ} = e, b^{Λ} − b_{Λ} ⩾ b^{Λ_{d}}(j) − e ⩾ p, which contradicts the fact that Λ is completely splittable . Hence H = H̄.

We have d = b^{Λ_{d}}(k)+1 < b_{Λ_{d}}+p, as b_{Λ_{d}}+p is not a conormal space of Λ_{d}. Obviously the only cogood space of M_{a}, which equals H_{ε}(Λ_{d}) from the same runner as d is b^{Λ_{d}}(k)+pε_{k}+1. Hence a = b^{Λ_{d}}(k) + pε_{k} + 1, M = H_{ε}(Λ) and U(ν, μ) ⩽ 1 in cases (1)-(3).

Case 3: μ ▷ ν and (ν, μ) is minimal. The desired assertion follows from the definitions of U and ζ. □

The general form of the sequences ε(H, x, i) is quite complicated. However we have the following assertion.

Lemma 8.3. Let H be an integer such that \frac{p+3}{2} ⩽ H < p. Then π(H, x, i) is satisfied if and only if x = QH + 1, Q ∈ ℤ, Q ⩾ 1 and i = H − 1. In this case ε(H, x, i) = (−Q − 1, q̄^{H−1−r}, (q̄ + 1)^{r}), where q̄ = quo(Q + 1, H − 1) and r = rem(Q + 1, H − 1).

Proof. Assume π(H, x, i) is satisfied. We shall use the notation of Definition 7.4. We put

δ = H − Q − 1 + h [\frac{x−Q−1}{p−H+1}] + (R − 1) [\frac{x−Q−2}{p−H+1}] − m [\frac{p−h+x−1}{p−m}] .

Suppose R > 2. Since m ⩽ h, we have

δ ⩾ h ([\frac{x−Q−1}{p−H+1}] − [\frac{x−1}{p−H+2}]) + ([\frac{x−Q−2}{p−H+1} + 1] − [\frac{x}{H} ]) + R − 2.

To obtain a contradiction with condition (2) of Definition 7.4, it suffices to prove that both differences in the outer brackets are nonnegative. This follows from the inequalities

\begin{matrix}\frac{x−Q−1}{p−H+1} − \frac{x−1}{p−H+2} = \frac{x−1−Q(p−H+2)}{(p−H+1)(p−H+2)} = \frac{Q(2H−(p+2))+R−1}{(p−H+1)(p−H+2)} > 0, \\ \frac{x−Q−2}{p−H+1} + 1 − \frac{x}{H} = \frac{(2H−(p+2))x+H(p−1−H)+R}{H(p−H+1)} > 0. \end{matrix}

Now suppose R = 2. Then Q ⩾ 1. We have

δ ⩾ h ([\frac{x−Q−1}{p−H+1}] − [\frac{x−1}{p−H+2}]) + ([\frac{x−Q−2}{p−H+1}] − [\frac{x}{H} ]) + R − 1.

Nonnegativity of the difference in the first pair of the brackets is shown just as above. We have

\frac{x−Q−2}{p−H+1} − \frac{x}{H} = \frac{(2H−(p+3))x+H(Q−2)+4}{H(p−H+1)}.

The last expression and thus the difference in the second pair of the brackets is nonnegative if Q ⩾ 2. If Q = 1 then x = H + 2 and we have

[\frac{x−Q−2}{p−H+1}] − [\frac{x}{H} ] = [\frac{H−1}{p−H+1}] − [\frac{H+2}{H}] = [\frac{H−1}{p−H+1}] − 1 ⩾ 0.

We have a contradiction with condition (2) of Definition 7.4.

Thus we have proved that R = 1. Hence x = QH + 1, Q ⩾ 1 and m = i. Suppose i ≠ H − 1. Then i ⩽ H − 2. We have

δ⩾H−Q−1+h [\frac{x−Q−1}{p−H+1}]−(H−2) [\frac{p−H+x}{p−H+2}]= h ([\frac{x−Q−1}{p−H+1}]−[\frac{x−2}{p−H+2}])+([\frac{x−2}{p−H+2}]−[\frac{x}{H} ])+1.

The difference in the first pair of the outer brackets of the right hand side is nonnegative, as

\frac{x−Q−1}{p−H+1} − \frac{x−2}{p−H+2} = \frac{x−2−(Q−1)(p−H+2)}{(p−H+1)(p−H+2)} = \frac{(Q−1)(2H−(p+2))+H−1}{(p−H+1)(p−H+2)} > 0.

We have

\frac{x−2}{p−H+2} − \frac{x}{H} = \frac{(2H−(p+2))x−2H}{H(p−H+2)} = \frac{(2H−(p+3))x+H(Q−2)+1}{H(p−H+2)}.

The last expression and thus the difference in the second pair of the brackets is nonnegative if Q ⩾ 2. If Q = 1 then x = H + 1 and we have

[\frac{x−2}{p−H+2}] − [\frac{x}{H} ] = [\frac{H−1}{p−H+2}] − [\frac{H+1}{H}] = [\frac{H−1}{p−H+2}] − 1 ⩾ 0.

We have a contradiction with condition (2) of Definition 7.4.

Finally for x = QH + 1 and i = H − 1, where Q is any positive integer, we have

δ = h ([\frac{x−1−Q}{p−H+1}] − [\frac{x−1}{p−H+1}]) − Q < 0,

whence it follows that π(H, x, i) is satisfied. Now it is clear that ε(H, x, i) is given exactly by the suggested formula.

□

9 Applications to branching rules

9.1 Preliminary facts. In the following proposition, which follows directly from [BK2, Theorems E, E′], Res_{α} 0 is understood as 0.

Proposition 9.1. Let λ be a p-regular partition and α ∈ ℤ_{p}.

• If there is no λ-normal (λ-conormal) node of residue α, then Res_{α} D^{λ} = 0 (Ind^{α} D^{λ} = 0).

• If there is exactly one λ-normal (λ-conormal) node A of residue α, then Res_{α} D^{λ} ≅ D^{λ_{A}} (Ind^{α} D^{λ} ≅ D^{λ^{A}}).

Lemma 9.2. Let M be a module such that head M ≅ soc M and N_{1}, …, N_{k} be mutually nonisomorphic simple modules such that [M : N_{i}] = 1, i = 1, …, k. Then either M ≅ ⨁_{i=1}^{k} N_{i} or there is a simple module N nonisomorphic to any of N_{1}, …, N_{k} such that Hom(M, N) ≠ 0.

Proof. Suppose there is no such a module N. Prove by induction on n = 0, …, k that there exists a subset S_{n} ⊂ {1, …, k} of cardinality n such that ⨁_{i∈S_{n}} N_{i} is isomorphic to a submodule of M. The set S_{0} = ∅ corresponds to the case n = 0. Now let 0 < n < k and ι : ⨁_{i∈S_{n}} N_{i} → M be an embedding of modules. Since |S_{n}| < k, we have Im ι ≠ M and there is a maximal submodule M_{0} of M containing Im ι. By our assumption, M/M_{0} ≅ N_{j} for some j ∈ {1, …, k} \ S_{n}. Since head M ≅ soc M, we can put S_{n+1} = S_{n} ∪ {j}.

We have S_{k} = {1, …, k}. By our assumption from the beginning of the proof, we get M ≅ ⨁_{i=1}^{k} N_{i}. □

Lemma 9.3. Let λ be a partition of height less than p and B be a λ-addable node such that λ^{B} is p-singular. Then λ = (1^{p−1}) and B = (p, 1).

Proof. Since h(λ^{B}) ⩽ p and λ^{B} is p-singular, we have λ^{B} = (k^{p}). Hence k = (λ^{B})_{p} = 1 and B = (p, 1). □

9.2 Inducing completely splittable modules. In what follows, the image of an R-module M in the Grothendieck group of R is denoted by [M].

Theorem 9.4. Let λ be a completely splittable partition of n and α ∈ ℤ_{p}. Suppose there are more than one λ-conormal nodes of residue α. Then there are exactly two such nodes. Denote them by A and B, where A is above B. Then (in the Grothendieck group of KΣ_{n+1})

[Ind^{α} D^{λ}] = \{ \begin{matrix}2[D^{λ^{A}}] + [D^{λ^{B}}] & if h_{1,1}(λ) ≠ p − 1 or p > 2, λ = (p − 1); \\ 2[D^{λ^{A}}] + [D^{λ^{B}}] + [Dλ̃^{A}] & if h_{1,1}(λ) = p − 1 and λ ≠ (1^{p−1}), (p − 1); \\ 2[D^{(2,1^{p−2})}] + [D^{(3,1^{p−3})}] & if p > 2 and λ = (1^{p−1}); \\ 2[D^{(2)}] & if p = 2 and λ = (1). \end{matrix}

Proof. There are exactly two such nodes, since all proper beads of any abacus of λ belong to different runners. We put ν = λ^{A}. Clearly, ν is completely splittable , A is a ν-good node and h(λ) = h(ν). Suppose the assertion of the current theorem does not hold.

First consider the case where h_{1,1}(λ) ≠ p−1 or p > 2, λ = (p−1). Then λ^{B} is p-regular by Lemma 9.3 and [Ind^{α} D^{λ} : D^{λ^{B}}] = 1. Applying Lemma 9.2 to the following parameters

M := rad Ind^{α} D^{λ}/ soc Ind^{α} D^{λ},  k := 1,  N_{1} := D^{λ^{B}},

and taking into account [Ind^{α} D^{λ} : D^{λ^{A}}] = 2, we get that there is a module D^{γ} isomorphic neither to D^{λ^{B}} nor D^{λ^{A}} such that Hom_{Σ_{n+1}}(rad Ind^{α} D^{λ}, D^{γ}) ≠ 0.

By Lemma 3.10, we have 0 < [Ind^{α} D^{λ} : D^{γ}] ⩽ 2[S^{λ^{A}} : D^{γ}] and thus ν = λ^{A} ◁ γ. Hence λ ≠ (p − 1) (and therefore h_{1,1}(λ) ≠ p − 1), h(ν) > 1 and p > 2. Since head Ind^{α} D^{λ} ≅ D^{ν}, we have Ext_{Σ_{n+1}}^{1}(D^{ν}, D^{γ}) ≠ 0. Theorem 4.6 implies γ = ν̃. We have h(λ) = h(ν) > 1 and p ⩽ h_{1,1}(ν) ⩽ h_{1,1}(λ) + 1 by Lemma 4.5. Taking into account h_{1,1}(λ) ≠ p − 1, we get h_{1,1}(λ) ⩾ p and thus λ is a big partition. Therefore ν̃ = λ̃^{C}, where C is a ν̃-good node of residue α. Since Ext_{Σ_{n+1}}^{1}(S^{ν}, Dν̃) = 0 by Proposition 2.1, we have by Lemma 3.7 that

1⩽ dim Hom_{Σ_{n+1}}(rad Ind^{α} D^{λ}, D^{γ})= dim Hom_{Σ_{n+1}}(rad Ind^{α} D^{ν_{A}}, Dν̃)⩽ε(ν, ν̃_{C})=ε(λ^{A}, λ̃).

Hence λ^{A} = λ̃^{D}. This is possible only if r(A) = 1 and χ(λ) = p. This contradicts the fact that A is a λ-conormal node.

Now consider the case where h_{1,1}(λ) = p − 1 and λ ≠ (1^{p−1}), (p − 1). We have p > 2, n ⩾ 2 and A = (1, λ_{1}). We put A^{t} = (λ_{1}, 1). We have (λ^{t})^{A^{t}} = (λ^{A})^{t}. Since λ is a p-core, by [Mu, Lemma 5.2], we get m(λ) = λ^{t}. By [S2, Lemma 11], we get m((λ^{A})^{t}) = λ̃^{A}. By [BK2, Theorem E(iv)] and Proposition 7.3, we get

1 = [D^{λ^{t}}↑^{Σ_{n+1}} : D^{(λ^{t})^{A^{t}}}] = [D^{λ^{t}}↑^{Σ_{n+1}} ⊗ sgn_{n+1} : D^{(λ^{A})^{t}} ⊗ sgn_{n+1}] =
[D^{m(λ^{t})}↑^{Σ_{n+1}} : D^{m((λ^{A})^{t})}] = [D^{λ}↑^{Σ_{n+1}} : Dλ̃^{A}] = [Ind^{α} D^{λ} : Dλ̃^{A}].

In the case under consideration λ^{B} is p-regular and [Ind^{α} D^{λ} : D^{λ^{B}}] = 1. Applying Lemma 9.2 to the following parameters

M := rad Ind^{α} D^{λ}/ soc Ind^{α} D^{λ},  k := 2,  N_{1} := D^{λ^{B}},  N_{2} := Dλ̃^{A}

and taking into account [Ind^{α} D^{λ} : D^{λ^{A}}] = 2, we get that there is a module D^{γ} isomorphic to none of D^{λ^{B}}, Dλ̃^{A}, D^{λ^{A}}, such that Hom_{Σ_{n+1}}(rad Ind^{α} D^{λ}, D^{γ}) ≠ 0. Since head Ind^{α} D^{λ} ≅ D^{λ^{A}}, we have Ext_{Σ_{n+1}}^{1}(D^{λ^{A}}, D^{γ}) ≠ 0. Similarly to the previous case Lemma 3.10 implies λ^{A} ◁ γ. Hence by Theorem 4.6, we get a contradiction γ = λ̃^{A}.

If p > 2 then multiplying [Ind\overline{−1} D^{(p−1)}] ≅ 2[D^{(p)}] + [D^{(p−1,1)}] by sgn_{p} and applying Proposition 7.3, we get [Ind¯1 D^{(1^{p−1})}] ≅ 2[D^{(2,1^{p−2})}] + [D^{(3,1^{p−3})}].

Finally the formula [Ind¯1 D^{(1)}] = 2[D^{(2)}] for p = 2 can be checked by dimension comparison. □

Theorem 9.5. Let λ be a completely splittable partition of n different from (1^{p−1}) and α ∈ ℤ_{p}. Denote by B the bottom λ-addable node (i.e. from the first column). Suppose there are more than one λ^{B}-normal nodes of residue α. Then except B there is only one such node A. We have Res_{α} D^{λ^{B}} ≅ Ind^{α} D^{λ_{A}}.

Proof. Let B_{1}, …, B_{k} be all λ-conormal nodes different from B. Since there are more than one λ^{B}-normal nodes of residue α, the residues res B, res B_{1}, …, res B_{k} are mutually distinct. Hence D^{λ}↑_{Σ_{n+1}} ≅ D^{λ^{B}} ⊕D^{λ^{B_{1}}} ⊕⋯⊕D^{λ^{B_{k}}}. Now it is clear that the only nonsimple indecomposable summand of D^{λ}↑_{Σ_{n+1}}↓_{Σ_{n}} is Res_{α} D^{λ^{B}}.

Let A_{1}, …, A_{l} be all the λ-normal nodes. The residues res A_{1}, …, res A_{l} are mutually distinct and D^{λ}↓_{Σ_{n−1}} ≅ D^{λ_{A_{1}}} ⊕ ⋯ ⊕ D^{λ_{A_{l}}}. The only nonsimple indecomposable summand of D^{λ}↓_{Σ_{n−1}}↑_{Σ_{n}} is Ind^{α} D^{λ_{A}}.

By the subgroup theorem [CR, (44.2)] applied to G := Σ_{n+1}, R = S := Σ_{n} and L := D^{λ}, we have D^{λ}↑_{Σ_{n+1}}↓_{Σ_{n}} ≅ D^{λ}↓_{Σ_{n−1}}↑_{Σ_{n}} ⊕ D^{λ}. By the Krull-Schmidt theorem, we have Res_{α} D^{λ^{B}} ≅ Ind^{α} D^{λ_{A}}. □

9.3 Inducing some almost completely splittable modules.

Theorem 9.6. Let λ be a big partition of n having height H ⩾ \frac{p+3}{2} and α ∈ ℤ_{p} such that h_{2,1}(λ) ≠ p − 1 and the condition \overline{λ_{1}} = −\overline{h(λ)} = α does not hold. Suppose there are more than one λ̃-conormal nodes of residue α. Then there are exactly two such nodes. Denote them by A and B, where A is above B. We have [Ind^{α} Dλ̃] = 2[Dλ̃^{A}] + [Dλ̃^{B}] (in the Grothendieck group of KΣ_{n+1}).

Proof. Arguing as in the proof of Theorem 9.4, we obtain that there exactly two such nodes. Let Λ be an abacus of λ and c be its minimal space. Denote by a and b the spaces of Λ̃ corresponding to A and B respectively. The case h(λ̃) < h(λ) is impossible, as we would have c = b^{Λ} − p, which contradicts the existence of more than one initial space of Λ̃ in the same runner. Therefore h(λ̃) = h(λ) and c = b. Hence λ̃ is not completely splittable and in particular λ̃ ≠ (1^{p−1}). By Lemma 9.3, λ̃^{B} is p-regular.

Since res A = res B and the condition \overline{λ_{1}} = −\overline{h(λ)} = α is violated, there exists i = 2, …, H such that the runner containing b^{Λ}(i) + 1 contains the space c and no proper bead of Λ. Hence λ̃^{A} = λ̃^{D}, where D is a λ-cogood node such that h(λ^{D}) = H. Therefore λ^{D} is big.

Suppose that the assertion of the current theorem does not hold. We put ν = λ̃^{A} and denote by d the space of Λ corresponding to D. Applying Lemma 9.2 to the parameters

M := rad Ind^{α} Dλ̃/ soc Ind^{α} Dλ̃,  k := 1,  N_{1} := Dλ̃^{B}

and taking into account [Ind^{α} Dλ̃ : Dλ̃^{A}] = 2, we get that there is a module D^{γ} isomorphic neither to Dλ̃^{B} nor Dλ̃^{A} such that Hom_{Σ_{n+1}}(rad Ind^{α} Dλ̃, D^{γ}) ≠ 0. By Lemma 3.10, we get 0 < [Ind^{α} Dλ̃ : D^{γ}] ⩽ 2[Sλ̃^{A} : D^{γ}] and thus ν = λ̃^{A} ◁ γ. Since head Ind^{α} Dλ̃ ≅ D^{ν}, we have Ext_{Σ_{n+1}}^{1}(D^{ν}, D^{γ}) ≠ 0. Theorem 8.2 and Lemma 8.3 imply γ = H_{ε}(λ^{D}) for ε equal to one of the following sequences: (0, −1, 0^{H−3}, 1); (−1, 0^{H−3}, 1, 0); (−1, −1, 0^{H−4}, 1, 1); (−Q − 1, q̄^{H−1−r}, (q̄+ 1)^{r}), where Q ⩾ 1, q̄ = quo(Q + 1, H − 1) and r = rem(Q + 1, H − 1).

If we suppose that h(γ) < H, then, taking into account the exact form of possible values of ε mentioned above, we get c = b^{Λ^{d}} − p or c = b^{Λ^{d}}(2) − p. However the former condition does not hold as i > 1 and the latter does not hold as h_{2,1}(λ) ≠ p − 1. Hence h(γ) = H, there exists a unique γ-normal node E of residue α and H_{ε}(λ) = γ_{E}.

Since Ext_{Σ_{n+1}}^{1}(S^{ν}, D^{γ}) = 0 by Proposition 2.1, we have by Lemma 3.7 that

1⩽ dim Hom_{Σ_{n+1}}(rad Ind^{α} Dλ̃, D^{γ})= dim Hom_{Σ_{n+1}}(rad Ind^{α} D^{ν_{A}}, D^{γ})⩽ε(λ̃^{A}, γ_{E}).

Hence λ̃^{A} = (γ_{E})^{F}, where res F = α. Clearly, either F = E or h((γ_{E})^{F}) > H. The former case is impossible as λ̃^{A} ◁ γ and the latter is impossible as h(λ̃^{A}) = H. □

Remark. The reader can by oneself formulate and prove the analog of Theorem 9.5 for almost completely splittable partitions.

9.4 Conjectures. The following conjectures are based on Theorem 8.2 and calculations within the known decomposition matrices.

Conjecture 9.7. Let λ be a big partition of n having height H and α ∈ ℤ_{p} such that \overline{λ_{1}} = −\overline{h(λ)} = α does not hold. Suppose there are more than one λ̃-conormal nodes of residue α. Then there are exactly two such nodes. Denote them by A and B, where A is above B. We have

[Ind^{α} Dλ̃] =

2[Dλ̃^{A}] + [Dλ̃^{B}]  if h_{2,1}(λ)≠p − 1;
2[Dλ̃^{A}] + [Dλ̃^{B}] + [D^{H_{(−2,2)}(λ̃^{A})}]  if h_{2,1}(λ)=p − 1 and H=2;
2[Dλ̃^{A}] + [Dλ̃^{B}] + [D^{H_{(0,−1,1)}(λ̃^{A})}] + [D^{H_{(1,−1,0)}(λ̃^{A})}]  if h_{2,1}(λ)=p − 1 and H=3;
2[Dλ̃^{A}] + [Dλ̃^{B}] + [D^{H_{(0,−1,0^{H−3},1)}(λ̃^{A})}] + [D^{H_{(0,−1,0^{H−4},1,0)}(λ̃^{A})}] if h_{2,1}(λ)=p − 1 and H>3.

Example. Let p = 7 and λ = (5, 4, 2, 2). Then λ̃ = (6, 4, 2, 1) and [Ind^{3} D^{(6,4,2,1)}] = 2[D^{(6,5,2,1)}] + [D^{(6,4,2,1,1)}] + [D^{(6,6,2)}] + [D^{(7,7)}]. We have λ_{1} = 5 ≠ −4 = −h(λ) (mod 7), h_{2,1}(λ) = 6, A = (2, 5), B = (5, 1), λ̃^{A} = (6, 5, 2, 1), λ̃^{B} = (6, 4, 2, 1, 1), H_{(0,−1,0,1)}(λ̃^{A}) = (6, 6, 2), H_{(0,−1,1,0)}(λ̃^{A}) = (7, 7).

Conjecture 9.8. Let λ be a completely splittable partition of height H such that χ(λ) = p and let α ∈ ℤ_{p}. Denote by A the top λ-addable node (i.e. from the first row). Suppose there are more than one λ^{A}-normal nodes of residue α. Then except A the only such node is the bottom λ-removable node B. We have

[Res_{α} D^{λ^{A}}] = \{ \begin{matrix}2[Dλ̃] + [D^{λ}] + [D^{H_{(−1,0^{H−3},1,0)}(λ)}] + x[D^{H_{(0,−1,0^{H−3},1)}(λ)}] & if H > 2; \\ 2[Dλ̃] + [D^{λ}] + x[D^{H_{(1,−1)}(λ)}] & if H = 2, \end{matrix}

where x = [h_{2,1}(λ) ⩾ p].

Example. Let p = 5 and λ = (5, 5, 3). Then A = (1, 6), λ^{A} = (6, 5, 3) and [Res_{0} D^{(6,5,3)}] = 2[D^{(6,5,2)}] + [D^{(5,5,3)}] + [D^{(9,2,2)}] + [D^{(6,6,1)}]. We have λ̃ = (6, 5, 2), H_{(−1,1,0)}(λ) = (9, 2, 2), H_{(0,−1,1)}(λ) = (6, 6, 1) and h_{2,1}(λ) = 6.

Conjecture 9.9. If λ is a completely splittable partition of height 3 such that h_{1,1}(λ) = 2p − 1, then [Ind^{−¯3} Dλ̃] = 2[Dλ̃^{A}] + [Dλ̃^{B}] + [D^{H_{(0,1,−1)}(λ̃^{A})}], where A = (3, λ̃_{3} + 1) and B = (4, 1).

Example. Let p = 5 and λ = (7, 6, 6). Then λ̃ = (9, 6, 4) and [Ind^{2} D^{(9,6,4)}] = 2[D(9, 6, 5)]+ [D(9, 6, 4, 1)] + [D(10, 10)]. We have h_{1,1}(λ) = 9, λ̃^{A} = (9, 6, 5), λ̃^{B} = (9, 6, 4, 1) and H_{(0,1,−1)}(λ̃^{A}) = (10, 10). Moreover with the help of the known decomposition matrices and the subgroup theorem [CR, (44.2)], it is easy to verify Conjecture 9.9 for p = 5.

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