A PROOF OF THE MULLINEUX CONJECTURE

BEN FORD AND ALEXANDER KLESHCHEV

Dedicated to the memory of Professor Brian Hartley

Introduction

A partition λ of a positive integer n is a sequence (λ_{1} ≥ λ_{2} ≥ ⋯ ≥ λ_{m} > 0) of integers such that ∑ λ_{i} = n. For a positive integer p, a partition λ = (λ_{1} ≥ λ_{2} ≥ ⋯ ≥ λ_{m}) (or its Young diagram) is called p-regular if it does not have p or more equal parts, i.e. if there does not exist t ≤ m−(p−1) with λ_{t} = λ_{t+1} = ⋯ = λ_{t+p−1}.

Let F be a field of characteristic p > 0. It is well known that irreducible representations of the symmetric group S_{n} over F are naturally parametrized by p-regular partitions of n (cf. for example [8, 11]). If λ is such a partition we denote the corresponding irreducible module by D^{λ}.

Let sgn_{n} be the one-dimensional sign representation of S_{n} over F; i.e., sgn_{n} ≅ F as a vector space and g · f = sgn(g)f for any g ∈ S_{n}, f ∈ F. Here sgn(g) is just the sign of the permutation g.

It is clear that for any irreducible D^{λ}, the tensor product D^{λ} ⊗ sgn_{n} is also irreducible. The problem, usually called the problem of Mullineux, is to find the p-regular partition μ such that D^{λ} ⊗ sgn_{n} ≅ D^{μ}. Put

D^{λ} ⊗ sgn_{n} ≅ D^{b_{n}(λ)}.

In this way a bijection b_{n} on the set P_{n} of p-regular partitions of n is defined for each positive integer n, and the problem is:

Problem 1. Find b_{n}.

The analogous question in characteristic zero is easily answered. The corresponding bijection on the set of all partitions of n is given by λ ↦ λ′, where λ′ is the conjugate partition to λ (the partition whose Young diagram is the transpose of the Young diagram of λ). So the hope is to find some sort of “conjugation modulo p”.

1991 Mathematics Subject Classification. 20C05.

The authors thank the organizers, headed by Professor Hartley, of the August 1994 NATO Advanced Study Institute on Finite and Locally Finite Groups, at which this collaboration began.

The second author was supported by the Royal Society and the International Science Foundation.

The problem is important in the representation theory of the alternating group A_{n}. In addition, the module D^{λ} ⊗sgn_{n} appears in a natural way in many situations. For example, if λ′ is p-regular then:

(1) the socle of the Specht module S^{λ} is D^{λ′} ⊗ sgn_{n};

(2) the Young module Y^{λ} is the projective cover of D^{λ′} ⊗ sgn_{n} (cf. [10]);

(3) the image of the Schur functor applied to the irreducible polynomial representation F_{λ} of GL_{n}(F) is D^{λ′} ⊗ sgn_{n} (cf. [7, Section 6]).

We refer the reader to [4, 5, 24] for other cases of the occurrence of b_{n}.

In 1979, Mullineux ([20]) proposed a quite unmotivated algorithm which defines a bijection m_{n} : P_{n} → P_{n} (see Section 1 below for a detailed description) and conjectured that

Conjecture 2 (Mullineux). b_{n} = m_{n}.

In a subsequent paper ([21]), Mullineux proved that for any (p-regular) λ, b_{n}(λ) and m_{n}(λ) have the same p-core. That result is used below.

Of the more recent results on the problem we would like to mention the following: Martin ([17, 18]) proved that b_{n} = m_{n} if n < 3p. Andrews and Olsson [1] have shown that the numbers of fixed points of m_{n} and b_{n} coincide! Finally, Bessenrodt and Olsson ([2]) have proved that the Conjecture of Mullineux agrees with the (since proved) conjecture of Jantzen and Seitz (cf. [12, 13, 6, 14]).

Branching rules obtained by the second author made possible the approach we use here to prove Conjecture 2. In [15], the notion of a good node of a p-regular Young diagram is introduced (see Section 8 below) and it is proved that

soc(D^{λ} ↓_{S_{n−1}}) ≅ ⨁{D^{λ_{A}} | A is good}

where λ_{A} means λ \ {A} and soc(M) stands for the socle of the module M. From an evident isomorphism

soc((D^{λ} ⊗ sgn_{n}) ↓_{S_{n−1}}) ≅ soc(D^{λ} ↓_{S_{n−1}}) ⊗ sgn_{n−1}

we immediately get that for any good node A of λ there exists a good node B of b_{n}(λ) such that

b_{n−1}(λ_{A}) = b_{n}(λ)_{B}.

So if Conjecture 2 is true, then this equation must hold for m in place of b. Thus Conjecture 2 implies the following.

Conjecture 3. For any p-regular partition λ of any integer n ≥ 1 there exist a good node A of λ and a good node B of m_{n}(λ) such that m_{n−1}(λ_{A}) = m_{n}(λ)_{B}.

The following result is proved in [16] and is of crucial importance.

Theorem 4. Conjectures 2 and 3 are equivalent.

Thus, representation theory is completely eliminated and we have to consider a purely combinatorial question about m_{n}.

We would like to mention that [16] actually contains a solution of Problem 1 different from that proposed by Mullineux. However, the Mullineux Conjecture seems to be important not only from a representation theoretic but also from a combinatorial point of view. Also, the Mullineux algorithm is more convenient to compute by hand. The two algorithms may prove to be useful in different situations. For example, the algorithm from [16] is convenient in inductive arguments, while the Mullineux construction allows one to easily determine the number of parts of the partition b_{n}(λ).

Many questions about the combinatorial nature of both algorithms remain unclear, and we believe they will draw attention in the future.

The main result of the paper is:

Main Theorem. Conjecture 2 holds.

Actually, we prove Conjecture 3 and apply Theorem 4.

Remark. The classical Mullineux Conjecture proved here has a q-analog sometimes called the quantum Mullineux Conjecture (cf. [19, Section 7.6]). We are grateful to Gordon James for pointing out the following:

Let q be a root of unity in a field F, with q ≠ 1 if the characteristic p of F is zero. Let e be the least positive integer such that 1+q +q^{2} +⋯+q^{e−1} = 0, and let H_{q} be the Hecke algebra of type A over F with parameter q. If {T_{w} | w ∈ S_{n}} is the usual basis of H_{q} then there exists an outer automorphism ^{#} of H_{q}, due to Goldman, given by T_{v}^{#} = (q − 1)T_{1} − T_{v}, where v is any basic transposition in S_{n}. It is shown in [3] that irreducible H_{q}-modules are parametrized by the e-regular partitions of n. If λ is such a partition, let D_{q}^{λ} be the corresponding irreducible module. One may define (D_{q}^{λ})^{#} as the H_{q}-module with the twisted action h ∘ v = #(h)v, for h ∈ H_{q} and v ∈ D_{q}^{λ}. Then (D_{q}^{λ})^{#} is also irreducible, so (D_{q}^{λ})^{#} ≅ D_{q}^{λ^{∗}} for some e-regular λ^{∗}. The q-analog of the Mullineux Conjecture claims that λ^{∗} = m_{n}(λ) where m_{n} is the Mullineux map on the set of the e-regular partitions of n. Matthew Richards has proved in [23] that the permutation ^{∗} on the e-regular partitions depends only on e, not on p and q. If p > 0 and q = 1 then e = p, H_{q} ≅ F S_{n}. So the classical Mullineux Conjecture implies the quantum version for prime e. For general e, the q-analog of the Mullineux Conjecture remains open.

For our purposes, the abacus notation for partitions (cf. [11]) turned out to be effective. In Section 1 we reformulate for abaci the notions introduced in [15] of removable, good, normal, etc. nodes, and determine what the Mullineux algorithm looks like in this setting. In Section 2 we investigate the effect of adding a p-edge to a partition on runner sizes of the corresponding abacus. Sections 3 and 4 present some lemmas on normal and degenerate beads appearing in the abacus, leading to the necessary and sufficient condition given in Section 5 for the occurence of a degnerate bead. In Sections 6 and 7 we finally compare λ and m_{n}(λ), proving that a degenerate bead occurs in λ if and only if one occurs in m_{n}(λ); this leads shortly to the proof of the Main Theorem, presented in Section 8.

The reader who is interested in the main structure of the proof rather than the details should begin with Section 8.

1. Preliminaries

In this section we introduce the notion of an abacus for a partition and explain the Mullineux procedure in this setting. More detailed information concerning the language of the abacus can be found in [11, Section 2.7] or [22].

Throughout the paper p is a fixed prime and x ≡ y means x ≡ y (mod p). If a, b ∈ ℤ we denote

[a, b] = {c ∈ ℤ | a ≤ c ≤ b},  (a, b] = {c ∈ ℤ | a < c ≤ b},
[a, b) = {c ∈ ℤ | a ≤ c < b},  (a, b) = {c ∈ ℤ | a < c < b}.

If a ∈ ℤ and a = pq + r, q ∈ ℤ, r ∈ (0, p], we write \overline{a} for r.

Let λ = (λ_{1} ≥ λ_{2} ≥ ⋯ ≥ λ_{m} > 0) be a partition of n. We write h(λ) for m (the height of λ). Gathering together equal parts of λ we write it in the form

λ = (ν_{1}^{α_{1}}, …, ν_{k}^{α_{k}}), ν_{1} > ⋯ > ν_{k} > 0, α_{i} > 0. (1)

Throughout the paper we assume that λ is p-regular, i.e. α_{1}, …, α_{k} < p. To explain the Mullineux algorithm we have to introduce some notation.

The rim of a Young diagram λ is its south-east border — in other words, a node in row i and column j of λ belongs to its rim if and only if the node in row i + 1 and column j + 1 does not belong to λ.

Example. Let λ = (6, 4^{2}, 2, 1). Then the rim of λ contains the nodes represented by numbers in the following picture.

[figure: 3; 2; 1; 4; 7; 6; 5; 9; 8; 10]

Let us number the nodes of the rim moving from the “top-right” to the “leftbottom” (see the picture above). Define the first p-segment of the rim as the set consisting of the nodes whose numbers do not exceed p. If the last (i.e. having the largest number) node B of the first p-segment is in the last row of λ then λ has only one p-segment. If not, let r be the row containing B. The first node of the second p-segment is the node which has the smallest number, say δ, among the nodes of the rim lying in row r + 1. The second p-segment is now defined as the set consisting of the nodes whose numbers i satisfy δ ≤ i ≤ δ + p − 1. Repeating this procedure sufficiently many times we reach the bottom row of the diagram. It is clear that all p-segments except possibly the last one contain p nodes. The p-edge is defined as the union of the p-segments.

Example. Let λ = (6, 4^{2}, 2, 1), p = 5. The nodes of the p-edge (which consists of two p-segments) are coloured in black in the following picture.

[figure]

Now define diagrams λ^{(0)}, λ^{(1)}, …, λ^{(z+1)} as follows. Put λ^{(0)} = λ, and for i ≥ 1 put

λ^{(i)} = λ^{(i−1)} \ {p − edge of λ^{(i−1)}}.

We choose z to be maximal with respect to λ^{(z)} ≠ ∅; so λ^{(z+1)} = ∅. We call the p-edge of λ^{(j)} the j-th p-edge of λ. The Mullineux symbol for λ (introduced in [2]) is an array

G_{p}(λ) = (\begin{matrix}A_{0} & A_{1} & … & A_{z} \\ R_{0} & R_{1} & … & R_{z} \end{matrix})

where A_{j} is the number of nodes (or the length) of the j-th p-edge of λ, and R_{j} = h(λ^{(j)}) is the height of λ^{(j)}.

Example. Let λ = (6, 4^{2}, 2, 1), p = 5. The j-th p-edge contains the nodes labelled j in the following picture.

3 2 1 0 0 0 2 2 1 0 1 1 1 0 0 0 0

Thus λ^{(1)} = (3^{3}), λ^{(2)} = (2^{2}), λ^{(3)} = (1), z = 3, and

G_{p}(λ) = (\begin{matrix}8 & 5 & 3 & 1 \\ 5 & 3 & 2 & 1 \end{matrix})

For i ∈ [0, z] put ε_{i} = 0 if p|A_{i} and ε_{i} = 1 otherwise. The following proposition is proved in [20].

Proposition 1.1. The entries of G_{p}(λ) satisfy

ε_{i} ≤ R_{i} − R_{i+1} < p + ε_{i}, 0 ≤ i < z;

1 ≤ R_{z} < p + ε_{z};

R_{i} − R_{i+1} + ε_{i+1} ≤ A_{i} − A_{i+1} < p + R_{i} − R_{i+1} + ε_{i+1}, 0 ≤ i < z; R_{z} ≤ A_{z} < p + R_{z};

∑_{i=0}^{z} A_{i} = n.

Moreover, if A_{0}, …, A_{z}, R_{0}, …, R_{z} are positive integers such that these inequalities are satisfied then there exists exactly one p-regular partition λ of n such that

G_{p}(λ) = (\begin{matrix}A_{0} & A_{1} & … & A_{z} \\ R_{0} & R_{1} & … & R_{z} \end{matrix}) .

In fact, λ can easily be reconstructed from its Mullineux symbol by starting from the empty partition λ^{(z+1)} and adding the j-th p-edges for j = z, z − 1, …, 0. More precisely, if j ∈ [0, z] and λ^{(j+1)} = (l_{1}, …, l_{x}) has already been constructed, then to construct λ^{(j)} one has to add the j-th p-edge which consists of A_{j} nodes in R_{j} rows. It is convenient to begin adding each p-edge from the bottom of the diagram. The following description is rather more confusing than the actual process; this will be clear if one simply tries to reconstruct λ from its Mullineux symbol for a few sample partitions λ.

If R_{j} = R_{j+1} then p|A_{j} by Proposition 1.1. Let A_{j} = tp; that is, we have to add t p-segments. We start adding from position (x, l_{x} + 1) (i.e. the position in row x and column (l_{x} + 1)), and add p nodes one after another according to the following rule: Let B be the last added node. If the position above it is not occupied by a node of λ^{(j+1)} then we add the next node at this position. Otherwise we add the next node at the position to the right of B. In this way we add the bottom p-segment of the j-th p-edge. Assume that we have already added w < t segments from the bottom. Let u be the row in which the last node was added. Then we find the column v such that there is a node of λ^{(j+1)} in position (u − 1, v − 1) but none in position (u − 1, v). Add the first node of segment w + 1 at position (u − 1, v) and add the remaining p − 1 nodes of this segment following the rule described above.

If R_{j} > R_{j+1} the only difference from the previous case is that we have to start adding the nodes of the bottom p-segment, which consists of \overline{A_{j}} ∈ (0, p] nodes, from position (R_{j}, 1).

Here is the bijection which Mullineux defined in [20]:

Definition 1.2. Let λ have Mullineux symbol

G_{p}(λ) = (\begin{matrix}A_{0} & A_{1} & … & A_{z} \\ R_{0} & R_{1} & … & R_{z} \end{matrix}) .

Define a p-regular partition m_{n}(λ) of n via

G_{p}(m_{n}(λ)) = (\begin{matrix}A_{0} & A_{1} & … & A_{z} \\ S_{0} & S_{1} & … & S_{z} \end{matrix})

where S_{j} = A_{j} + ε_{j} − R_{j}.

The proof of the fact that the symbol (\begin{matrix}A_{0} & A_{1} & … & A_{z} \\ S_{0} & S_{1} & … & S_{z} \end{matrix}) does in fact correspond

to a p-regular partition of n can be found in [20]; Proposition 1.1 ensures that m_{n}(λ) is well defined.

We shall often write m(λ) instead m_{l}(λ), and we will use the evident equality m(λ^{(i)}) = m(λ)^{(i)} without comment.

Example. Let λ = (6, 4^{2}, 2, 1), p = 5. As we saw in the previous example,

G_{p}(λ) = (\begin{matrix}8 & 5 & 3 & 1 \\ 5 & 3 & 2 & 1 \end{matrix}) .

So

G_{p}(m(λ)) = (\begin{matrix}8 & 5 & 3 & 1 \\ 4 & 2 & 2 & 1 \end{matrix}) .

We reconstruct m(λ) from its Mullineux symbol according to the rule described above:

3 2 1 1 1 1 0 2 2 1 0 0 0 0 0 0 0

Now we will describe the abacus notation. We consider sequences

{s_{0}, s_{1}, …, s_{v}, … }

where each s_{v} is either “·” (space) or “×” (bead) (one could call them 0 and 1 as well but, having in mind an abacus, we prefer this more representative terminology). For λ = (ν_{1}^{α_{1}}, …, ν_{k}^{α_{k}}) written in the form of (1) and R ≥ h(λ) we construct the sequence s_{R}(λ) as follows: The first R − h(λ) members of s_{R}(λ) are beads. They are followed by ν_{k} spaces, then α_{k} beads, then ν_{k−1} − ν_{k} spaces, then α_{k−1} beads, …, then ν_{1} − ν_{2} spaces, then α_{1} beads, and then spaces. (Informally speaking, we move along the rim of λ from bottom to top, putting beads in the sequence when we move up and leaving spaces when we move to the right).

Example. Let λ = (6, 4^{2}, 2, 1). Then

s_{5}(λ) = · × · × · · × × · · ×,
s_{6}(λ) = × · × · × · · × × · · ×,
s_{7}(λ) = × × · × · × · · × × · · ×.

(we do not draw the spaces after the last bead).

Each sequence of spaces and beads with a finite number of beads is s_{R}(λ) for exactly one λ and exactly one R, but there are infinitely many sequences corresponding to a fixed partition λ, namely s_{h(λ)}, s_{h(λ)+1}, . . ..

It is convenient to draw the elements s_{0}, s_{1}, … in the array

\begin{matrix}s_{0} & s_{1} & … & s_{p−1} \\ s_{p} & s_{p+1} & … & s_{2p−1} \\ s_{2p} & s_{2p+1} & … & s_{3p−1} \\ ⋮ & ⋮ & … & ⋮ \end{matrix}

and for any residue i (mod p) we say that the elements s_{v} with v ≡ i form the i-th runner of the abacus. If s_{v} is a space we say that the v-th position of the abacus is unoccupied. We often denote beads by letters k, l, m, . . .. If s_{v} = k we say that bead k occupies the v-th position of the abacus.

Warning. Denoting beads by letters is convenient, but it can be misleading, because we do not distinguish between abaci having the same bead configuration. For example, the abaci (p = 5)

· k · l ·

and

· l · k ·

are identified.

If the abacus is constructed from a sequence s_{R}(λ) we say that it is an abacus for λ, and if it is constructed from s_{h(λ)}(λ) we call it the canonical abacus for λ. One can easily see that the only abacus for λ with position 0 unoccupied is the canonical one.

Abaci for λ are in 1–1-correspondence with the so called sequences of β-numbers for λ (cf. [11, 9]). If (β_{1}, β_{2}, …, β_{r}) is such a sequence then we obtain the corresponding abacus by placing beads in the abacus at positions β_{1}, β_{2}, …, β_{r}.

Example. Let λ = (6, 4^{2}, 2, 1), p = 5. Then

\begin{matrix}· & × & · & × & · \\ · & × & × & · & · \\ × \end{matrix} ,  \begin{matrix}× & · & × & · & × \\ · & · & × & × & · \\ · & × \end{matrix} ,  and  \begin{matrix}× & × & · & × & · \\ × & · & · & × & × \\ · & · & × \end{matrix}

are abaci for λ, the first being canonical.

If it is clear which abacus is used for a partition λ, we shall often denote this abacus by the same letter λ. For a bead k of an abacus λ we denote by P_{λ}(k) the position occupied by k in λ. If there is an unoccupied position α < P_{λ}(k) we call k proper; otherwise k is improper. All the beads of a canonical abacus are proper.

Now we want to describe the deletion and addition of a p-edge in the abacus notation. Informally speaking, if Λ is an abacus for λ^{(j)} then to construct an abacus for λ^{(j+1)} one should find the last bead of Λ and try to push it up one position. If this position is occupied by some proper bead l then the latter bead is pushed out by k and we try to move l one position up as we moved k. If the position above k is unoccupied then after moving k to this position we look for the bead m having the maximal position less than the new position of k and try to move m up similarly to k. Finally if k is in the top row of the abacus or the position above k is occupied by an improper bead then k moves to the smallest unoccupied position (this position will be in a different runner than that containing k in Λ) and the process of removing the p-edge is complete. We continue until the process completes. The procedure of adding p-edge is the reverse; we want to describe both operations formally.

Definition 1.3. Let Λ be an abacus for a nontrivial p-regular partition. We define the set {m_{1}, m_{2}, …, m_{N}} of r-movable beads of Λ and the abacus φ(Λ) as follows:

Let m_{1} be the proper bead such that P_{Λ}(m_{1}) is maximal (in other words, m_{1} is the last bead of the abacus).

Assume that m_{1}, …, m_{i} have been defined. If either • P_{Λ}(m_{i}) < p (i.e. m_{i} is in the first row of the abacus);

• P_{Λ}(m_{i}) ≥ p and P_{Λ}(m_{i}) − p is occupied by an improper bead; or

• P_{Λ}(m_{i}) ≥ p, P_{Λ}(m_{i}) − p is unoccupied, and there does not exist a proper

bead m with P_{Λ}(m) < P_{Λ}(m_{i}) − p;

then we put N = i. Otherwise, we define m_{i+1} to be the proper bead whose position is maximal with respect to the condition P_{Λ}(m_{i+1}) ≤ P_{Λ}(m_{i}) − p.

Put

α = \begin{cases}P_{Λ}(m_{N}) − p, if P_{Λ}(m_{N}) ≥ p and position P_{Λ}(m_{N}) − p is unoccupied in Λ; \\ the minimal unoccupied position in Λ, otherwise. \end{cases}

Now we define φ(Λ) to be the abacus with the same set of beads as Λ, but with P_{φ(Λ)}(k) = P_{Λ}(k) for k ≠ m_{1}, m_{2}, …, m_{N}, P_{φ(Λ)}(m_{N}) = α, and P_{φ(Λ)}(m_{i}) = P_{Λ}(m_{i}) − p for all i ∈ [1, N). (In other words, to obtain φ(Λ) from Λ, one has to move beads m_{1}, …, m_{N−1} one position up, and move m_{N} to position α.)

Definition 1.4. Let λ be a p-regular partition with Mullineux symbol G_{p}(λ) =

(\begin{matrix}A_{0} & A_{1} & … & A_{z} \\ R_{0} & R_{1} & … & R_{z} \end{matrix}) (possibly z = −1, which is interpreted as λ = ∅). Let A and

R be positive integers, and let ε = 0 if p|A and ε = 1 otherwise. Assume that • ε ≤ R − R_{0} < p + ε, R − R_{0} + ε_{0} ≤ A − A_{0} < p + R − R_{0} + ε_{0}, if z ≥ 0;

• 1 ≤ R < p + ε, R ≤ A < p + R, if z = −1.

Let Λ be an abacus for λ containing Q ≥ R beads. We define the set n_{1}, …, n_{D} of a-movable beads and a new abacus ψ(Λ) = ψ_{A,R}(Λ) as follows:

If z ≥ 0 and R = R_{0} then p|A, and we let n_{1} be the first proper bead of Λ (i.e. the proper bead such that P_{Λ}(n_{1}) is minimal), and put P_{ψ(Λ)}(n_{1}) = P_{Λ}(n_{1}) + p = P_{Λ}(n_{1}) + \overline{A}.

If R > R_{0} or z = −1, we let n_{1} be the (necessarily improper) bead of Λ with P_{Λ}(n_{1}) = Q − R, and put P_{ψ(Λ)}(n_{1}) = P_{Λ}(n_{1}) + \overline{A}.

Assume that n_{1}, …, n_{i} and P_{ψ(Λ)}(n_{1}), …, P_{ψ(Λ)}(n_{i}) have been defined. If P_{Λ}(k) < P_{ψ(Λ)}(n_{i}) for all beads k of Λ then we put D = i. Otherwise, we define n_{i+1} to be the bead minimal with respect to P_{Λ}(n_{i+1}) ≥ P_{ψ(Λ)}(n_{i}), and define P_{ψ(Λ)}(n_{i+1}) = P_{Λ}(n_{i+1}) + p.

Finally, we define ψ(Λ) as the abacus with the same set of beads as λ, but P_{ψ(Λ)}(k) = P_{Λ}(k) if k ≠ n_{1}, …, n_{D} and P_{ψ(Λ)}(n_{i}) defined as above. (In other words, to obtain ψ(Λ) from Λ one should move n_{1} to the position P_{Λ}(n_{1}) + \overline{A}, and beads n_{2}, …, n_{D} down one row).

Part (i) of the following lemma is proved in [21, Section 3] (in the language of β-numbers), and (ii) follows from (i).

Lemma 1.5. Let λ be a p-regular partition with Mullineux symbol

G_{p}(λ) = (\begin{matrix}A_{0} & A_{1} & … & A_{z} \\ R_{0} & R_{1} & … & R_{z} \end{matrix}) , 0 ≤ j ≤ z.

(i) If Λ^{(j)} is an abacus for λ^{(j)}, then

φ(Λ^{(j)})

is an abacus for λ^{(j+1)} (see Definition 1.3). (ii) If Λ^{(j+1)} is an abacus for λ^{(j+1)} with at least R_{j} beads, then

ψ_{A_{j},R_{j}}(Λ^{(j+1)})

is an abacus for λ^{(j)} (see Definition 1.4).

Example. Let λ = (6, 4^{2}, 2, 1), p = 5. Let Λ be the canonical abacus for λ, i.e.

Λ = \begin{matrix}· & a & · & b & · \\ · & c & d & · & · \\ e \end{matrix}

(We use letters rather than × to denote the beads so that it will be easier to follow their movement). Then

φ(Λ) = \begin{matrix}b & a & · & · & · \\ e & c & d \end{matrix} is an abacus for λ^{(1)},

φ^{2}(Λ) = \begin{matrix}b & a & d & · & · \\ e & c \end{matrix} is an abacus for λ^{(2)},

φ^{3}(Λ) = \begin{matrix}b & a & d & c & · \\ e \end{matrix} is an abacus for λ^{(3)}, and
φ^{4}(Λ) = b a d c e is an abacus for λ^{(4)} = ∅.

Recall from the example following Definition 1.2 that m(λ) has Mullineux symbol

G_{p}(m_{n}(λ)) = (\begin{matrix}A_{0} & A_{1} & … & A_{z} \\ S_{0} & S_{1} & … & S_{z} \end{matrix}) .

To reconstruct m(λ) we should start with an abacus for λ^{(4)} = ∅ which has at least S_{0} = 4 beads. So we begin with

M^{(4)} = r s t u ·

Then

M^{(3)} = ψ_{1,1}(Λ^{(4)}) = r s t · u
M^{(2)} = ψ_{3,2}(Λ^{(3)}) = \begin{matrix}r & s & · & · & u \\ t \end{matrix}

M^{(1)} = ψ_{5,2}(Λ^{(2)}) = \begin{matrix}r & s & · & · & · \\ t & · & · & · & u \end{matrix}

M^{(0)} = ψ_{8,4}(Λ^{(1)}) = \begin{matrix}· & s & · & r & · \\ · & · & · & · & u \\ t \end{matrix}

It is easily seen that M^{(i)} is an abacus for m(λ^{(i)}) = m(λ)^{(i)}.

Definition 1.6. Let k be a proper bead of an abacus λ. Then k is called a beginning of λ if P_{λ}(k) − 1 is unoccupied in λ. It is called an end if P_{λ}(k) + 1 is unoccupied in λ. If k is a beginning we denote by λ(k) the abacus obtained from λ by moving k to P_{λ}(k) − 1.

Definition 1.7. A bead k in runner a of an abacus λ is called normal if and only if

(i) it is a beginning; and (ii) if l_{1}, …, l_{r} are all the ends in runner a − 1 of λ with P_{λ}(l_{v}) > P_{λ}(k) − 1

then there exist r distinct beginnings k_{1}, …, k_{r} in runner a of λ such that P_{λ}(l_{v}) + 1 > P_{λ}(k_{v}) > P_{λ}(k) for all v ∈ [1, r].

A bead k of λ is called good if it is the top normal bead in its runner.

Remark. Ends and beginnings correspond to indent and removable nodes for the corresponding partition. Normal and good beads correspond to normal and good removable nodes, respectively. If the beginning k corresponds to a removable node A then λ(k) is an abacus for λ_{A} (cf. [16] and Section 8).

Example. As we saw above, the canonical abacus for (6, 4^{2}, 2, 1) for p = 5 is as follows:

\begin{matrix}· & k & · & l & · \\ · & m & q & · & · \\ t \end{matrix}

Here k, l, m, t are beginnings; k, l, q, t are ends. Of the beginnings, k is normal because for the end t there exists the beginning m as in 1.7; l is not normal because of the end q; m is not normal because of the end t; finally, t is normal. Both normal beads k and t turn out to be good in this example since they belong to distinct runners.

Definition 1.8. A good bead k of an abacus λ is called degenerate if and only if φ(λ(k)) = φ(λ). Otherwise k is called non-degenerate.

In the example above, k is degenerate and t is non-degenerate.

Definition 1.9. If k and m are beads in the same runner of an abacus λ, then we write k ≻ m (in λ) if k is above m (i.e. P_{λ}(k) < P_{λ}(m)).

Notation. We denote by (a)_{λ} the set of the proper beads in runner a of λ, and write |a|_{λ} for the order of (a)_{λ}.

Definition 1.10. Let L = {l_{1} ≻ ⋯ ≻ l_{r}} ⊆ (a − 1)_{λ}, K = {k_{1} ≻ ⋯ ≻ k_{s}} ⊆ (a)_{λ}. We say that K compensates L if s ≥ r and P_{λ}(k_{v}) ≤ P_{λ}(l_{v}) + 1 for all v ∈ [1, r]. We say that L is compensated if there exists a subset K ⊆ (a)_{λ} which compensates L.

The following lemma follows immediately from Definition 1.7.

Lemma 1.11. A bead k ∈ (a)_{λ} is normal (for λ) if and only if the set

K = {k′ ∈ (a)_{λ} | k′ ≺ k}

compensates the set

L = {l ∈ (a − 1)_{λ} | P_{λ}(l) ≥ P_{λ}(k) − 1}.

From now on we assume that λ is an arbitrary but fixed p-regular partition of n with Mullineux symbol

G_{p}(λ) = (\begin{matrix}A_{0} & A_{1} & … & A_{z} \\ R_{0} & R_{1} & … & R_{z} \end{matrix})

and we denote by the same letter λ the canonical abacus for λ (which contains R_{0} beads). We denote by λ^{(j)} the abacus φ^{j}(λ) which is a (non-canonical in general) abacus for the partition λ^{(j)}, according to Lemma 1.5 (j ∈ [0, z +1]). Thus all abaci λ^{(j)} contain R_{0} beads. For example λ^{(z+1)} is the abacus with beads in positions 0, 1, …, R_{0} − 1. We shall write P_{j}(k), |a|_{j}, and (a)_{j} for P_{λ^{(j)}}(k), |a|_{λ^{(j)}}, and (a)_{λ^{(j)}}, respectively.

Similarly, m(λ) is the canonical abacus for the partition m(λ) (with S_{0} beads), and we denote by m(λ)^{(j)} = m(λ^{(j)}) the abacus φ^{j}(m(λ)) which is a (non-canonical in general) abacus for the partition m(λ)^{(j)} = m(λ^{(j)}). All abaci m(λ)^{(j)} have S_{0} beads. We write P^{j}(k), |a|^{j}, (a)^{j}, etc. for P_{m(λ)^{(j)}}(k), |a|_{m(λ)^{(j)}}, (a)_{m(λ)^{(j)}}, etc.

Lemma 1.12. Let 0 ≤ j ≤ z.

(i) Let l be a bead of λ^{(j+1)} such that position P_{j}(l) is unoccupied in λ^{(j+1)}. Then position P_{j}(l) + 1 is unoccupied in λ^{(j)}.

(ii) Let k be a bead of λ^{(j)} which is proper for λ^{(j+1)}, and such that position P_{j+1}(k) is unoccupied in λ^{(j)}. Then position P_{j+1}(k) − 1 is unoccupied in λ^{(j+1)}.

Proof. (i) Since P_{j}(l) is unoccupied in λ^{(j+1)}, we have P_{j}(l) ≠ P_{j+1}(l), i.e. l is an a-movable bead of λ^{(j+1)}, or l = n_{w} for some w ∈ [1, D] (see 1.4). If P_{j}(l) + 1 is unoccupied in λ^{(j+1)} it will remain so in λ^{(j)} because P_{j}(n_{v}) < P_{j}(l) for v < w and P_{j+1}(n_{v}) ≥ P_{j}(l) for v > w by definition. If P_{j}(l) + 1 is occupied in λ^{(j+1)} then it is occupied necessarily by n_{w+1} and P_{j}(n_{w+1}) > P_{j+1}(n_{w+1}) = P_{j}(l) + 1. So position P_{j}(l) + 1 is unoccupied in λ^{(j)} in this case, too. The proof of (ii) is similar.

Definition 1.13. The subset M = {m_{1} ≻ ⋯ ≻ m_{r}} ⊆ (b)_{j} is called dense (in λ^{(j)}) if m_{1} ≻ m ≻ m_{r} implies m = m_{v} for some v ∈ (1, r).

Lemma 1.14. Let L = {l_{1} ≻ ⋯ ≻ l_{r}} ⊆ (a − 1)_{j+1}, K = {k_{1} ≻ ⋯ ≻ k_{s}} ⊆ (a)_{j+1}. Assume that K is dense and compensates L in λ^{(j+1)}.

(i) If there does not exist a bead k such that P_{j}(k) = P_{j+1}(k_{1}), then K compensates L in λ^{(j)}.

(ii) If there is a bead k such that P_{j}(k) = P_{j+1}(k_{1}), then K′ = {k, k_{1}, …, k_{r−1}}

compensates L in λ^{(j)}. (iii) If P_{j+1}(l_{1}) > P_{j+1}(k_{1}) − 1, then K compensates L in λ^{(j)}.

Proof. (i) and (ii). By definition r ≤ s and P_{j+1}(k_{v}) ≤ P_{j+1}(l_{v}) + 1, v ∈ [1, r]. If K does not compensate L in λ^{(j)}, choose w to be minimal such that P_{j}(k_{w}) > P_{j}(l_{w})+1. Then P_{j+1}(k_{w}) = P_{j+1}(l_{w}) + 1, P_{j}(l_{w}) = P_{j+1}(l_{w}), and P_{j}(k_{w}) = P_{j+1}(k_{w}) + p. By Lemma 1.12(ii) there exists a bead k with P_{j}(k) = P_{j+1}(k_{w}) (k_{w} is pushed down by k). If w > 1 then, since K is dense, k = k_{w−1}. Since P_{j}(l_{w−1}) < P_{j}(l_{w}) = P_{j}(k_{w−1}) − 1, we have P_{j}(k_{w−1}) > P_{j}(l_{w−1}) + 1 which contradicts the choice of w. This proves (i). To prove (ii) notice that if w = 1 then we get P_{j}(k) = P_{j}(l_{1}) + 1, P_{j}(k_{v}) ≤ P_{j}(l_{v+1}) + 1, because P_{j+1}(k_{v}) < P_{j+1}(l_{v+1}) + 1 for v ∈ [1, r).

(iii). Notice that P_{j+1}(l_{1}) ≥ P_{j+1}(k_{1})−1+p. Therefore P_{j+1}(l_{v}) ≥ P_{j+1}(k_{1})−1+vp for v ∈ [1, r]. Let y ≥ 1 be such that P_{j+1}(k_{v}) = P_{j+1}(k_{1})+(v −1)p for all v ∈ [1, y], and P_{j+1}(k_{1}) + yp is unoccupied in λ^{(j+1)}. Then P_{j+1}(l_{v}) ≥ P_{j+1}(k_{v}) − 1 + p, hence P_{j}(l_{v}) ≥ P_{j}(k_{v}) − 1 for all v ∈ [1, min(r, y)], i.e. {k_{1}, …, k_{min(r,y)}} compensates {l_{1}, …, l_{min(r,y)}} in λ^{(j)}. If r > y, then {k_{y+1}, …, k_{r}} compensates {l_{y+1}, …, l_{r}} in λ^{(j)} by part (i).

Corollary 1.15. Let i ≤ j and let L ⊆ (a−1)_{j} be a dense set which is compensated in λ^{(j)}. Then L is compensated in λ^{(i)}.

Proof. This follows from Lemma 1.14, parts (i) and (ii).

Lemma 1.16. Let L = {l_{1} ≻ ⋯ ≻ l_{r}} ⊆ (a − 1)_{j}, K = {k_{1} ≻ ⋯ ≻ k_{s}} ⊆ (a)_{j+1}(⊆ (a)_{j}). Assume that L is dense, there does not exist a bead l such that P_{j+1}(l) = P_{j}(l_{r}), and K compensates L in λ^{(j)}. Then L ⊆ (a − 1)_{j+1} and K compensates L in λ^{(j+1)}.

Proof. We first prove that L ⊆ (a − 1)_{j+1}. If not, then l_{1} ∉ (a − 1)_{j+1}. We have P_{j}(k_{1}) ≤ P_{j}(l_{1}) + 1.

If P_{j}(k_{1}) < P_{j}(l_{1}) + 1 then P_{j}(k_{1}) ≤ P_{j}(l_{1}) + 1 − p, and since P_{j+1}(l_{1}) ≥ P_{j}(l_{1}) − p we get P_{j+1}(k_{1}) ≤ P_{j+1}(l_{1}) + 1. Now since k_{1} is proper for λ^{(j+1)} (by assumption), so is l_{1}; i.e. l_{1} ∈ (a − 1)_{j+1}.

If P_{j}(k_{1}) = P_{j}(l_{1}) + 1, let w ≥ 1 be minimal with respect to the conditions that P_{j}(l_{v}) = P_{j}(l_{1}) + (v − 1)p, v ∈ [1, w], and P_{j}(l_{1}) + wp is unoccupied in λ^{(j)}. Then, since K compensates L in λ^{(j)}, one has P_{j}(k_{v}) = P_{j}(l_{v}) + 1 for v ∈ [1, w]. Now it follows from Lemma 1.12(i) that P_{j+1}(l_{v}) = P_{j}(l_{v}) for v ∈ [1, w]. In particular, P_{j+1}(l_{1}) = P_{j}(l_{1}). Therefore P_{j+1}(k_{1}) ≤ P_{j}(k_{1}) ≤ P_{j}(l_{1}) + 1 = P_{j+1}(l_{1}) + 1, and then the fact that k_{1} is proper for λ^{(j+1)} implies l_{1} is also.

Now let us prove that K compensates L in λ^{(j)}. If not, choose w ≤ r to be maximal with respect to P_{j+1}(l_{w}) < P_{j+1}(k_{w}) − 1. Then P_{j}(l_{w}) = P_{j}(k_{w}) − 1, P_{j+1}(l_{w}) = P_{j}(l_{w}) − p, and P_{j+1}(k_{w}) = P_{j}(k_{w}). Now it follows from Lemma 1.12(i) that there exists a bead l with P_{j+1}(l) = P_{j}(l_{w}). By assumption, w < r. Then, in view of the density of L, l = l_{w+1} and P_{j+1}(l_{w+1}) = P_{j+1}(k_{w}) − 1 < P_{j+1}(k_{w+1}) − 1, which contradicts the choice of w.

Definition 1.17. If k is a proper bead of λ we denote by St(k) (“St” for step) the integer j such that k is proper for λ^{(j)} and is improper for λ^{(j+1)}. If j = St(k) we call k a new (proper) bead of λ^{(j)}.

Remark. We shall often use the following evident facts: If k, m ∈ (a)_{j} and k ≻ m then St(k) ≤ St(m). If k, m ∈ (a)_{j} and St(k) < St(m) then k ≻ m.

Lemma 1.18. A bead k is a new bead of λ^{(j)} if and only if R_{0} − R_{j} ≤ P_{j+1}(k) ≤ R_{0} − R_{j+1} − 1.

Proof. This is an immediate consequence of Definitions 1.17 and 1.4 and Lemma 1.5.

The following constants measure the change from λ^{(j+1)} to λ^{(j)} (resp., from m(λ)^{(j+1)} to m(λ)^{(j)}) in the difference between the sizes of runners a and a − 1.

Definition 1.19. For all j ∈ [0, z] define

d_{j}(a) = (|a|_{j} − |a|_{j+1}) − (|a − 1|_{j} − |a − 1|_{j+1}),
d^{j}(a) = (|a|^{j} − |a|^{j+1}) − (|a − 1|^{j} − |a − 1|^{j+1}).

2. Runner sizes

In this section we investigate the changes that can occur in the size of a runner a, and in the difference of the sizes of two adjacent runners a and a − 1, with the removal or addition of a p-edge to the abacus. We always assume that 0 ≤ j ≤ z, and R_{z+1} is interpreted as 0.

Lemma 2.1. |a|_{j+1} ≤ |a|_{j} ≤ |a|_{j+1} + 2.

Proof. If k ∈ (a)_{j+1} then k ∈ (a)_{j}. This implies the first inequality. If there are no improper beads in runner a of λ^{(j+1)} then |a|_{j} ≤ |a|_{j+1} +1 because at most one bead can move to runner a from another runner with the addition of the p-edge. Let m be the lowest (i.e. with P_{j+1}(m) maximal) improper bead in runner a of λ^{(j+1)}. If m is the first row of λ^{(j+1)}, we obtain |a|_{j} ≤ |a|_{j+1} + 2 as above. If not, let l be the bead with P_{j+1}(l) = P_{j+1}(m) − p. It is enough to show that l is improper for λ^{(j)}. By 1.18, only the beads occupying positions from R_{0} − R_{j} to R_{0} − R_{j+1} − 1 in λ^{(j+1)} are new proper beads for λ^{(j)}. So l is improper for λ^{(j)} in view of the inequality R_{j} − R_{j+1} ≤ p given in Proposition 1.1.

Lemma 2.2. |a|_{j} = |a|_{j+1} + 2 if and only if A_{j} ≢ 0, a ≡ R_{0} − R_{j} + A_{j}, and \overline{A_{j}} < R_{j} − R_{j+1}. Moreover, the two new proper beads in runner a of λ^{(j)} are the beads m and k occupying the positions R_{0} − R_{j} and R_{0} − R_{j} + \overline{A_{j}} in λ^{(j+1)}, respectively, and P_{j}(m) = P_{j+1}(k), P_{j}(k) = P_{j+1}(k) + p. (In other words, m moves from R_{0} − R_{j} to R_{0} − R_{j} + \overline{A_{j}} and pushes k down one row from this last position.)

Proof. If A_{j} ≢ 0, a ≡ R_{0}−R_{j}+A_{j}, and \overline{A_{j}} < R_{j}−R_{j+1}, let m be the (improper) bead occupying position R_{0} − R_{j} in λ^{(j+1)}. Since \overline{A_{j}} < R_{j} − R_{j+1}, position R_{0} − R_{j} + \overline{A_{j}} is occupied in λ^{(j+1)} by some improper bead k (cf. 1.18. Notice that k is in runner a because a ≡ R_{0} − R_{j} + A_{j}. When the j-th p-edge is added, m moves to position R_{0} − R_{j} + \overline{A_{j}} and pushes k out, to position R_{0} − R_{j} + \overline{A_{j}} + p. Notice that k and m are proper for λ^{(j)} because P_{j+1}(m) is unoccupied in λ^{(j)}. Hence |a|_{j} ≥ |a|_{j+1} + 2. So Lemma 2.1 implies |a|_{j} = |a|_{j+1} + 2.

In the other direction, let |a|_{j} = |a|_{j+1} + 2. Then there are some improper beads in runner a of λ^{(j+1)} because otherwise |a|_{j} ≤ |a|_{j+1} + 1 (recall that at most one bead changes runners with the addition of a p-edge). Let k be the lowest improper bead in this runner. Then |a|_{j} = |a|_{j+1} + 2 implies that k becomes proper for λ^{(j)} and some new bead comes to runner a from another runner. Therefore A_{j} ≢ 0, a ≡ R_{0} − R_{j} + A_{j}, and \overline{A_{j}} < R_{j} − R_{j+1}.

Corollary 2.3. If |a|_{j} = |a|_{j+1} + 2 then |b|_{j} ≤ |b|_{j+1} + 1 for any b ≢ a.

Proof. This follows from the Lemmas 2.2, 2.1.

Lemma 2.4. |a|_{j} = |a|_{j+1} + 1 if and only if one of the following conditions holds.

(i) a ≡ R_{0} − R_{j} + x with 0 < x < R_{j} − R_{j+1}, x ≢ A_{j}. In this case the bead occupying position R_{0} − R_{j} + x in λ^{(j+1)} remains at the same position and becomes a new proper bead in runner a of λ^{(j)}.

(ii) a ≡ R_{0} − R_{j} + A_{j}, \overline{A_{j}} ≥ R_{j} − R_{j+1} > 0. In this case the bead occupying position R_{0} − R_{j} in λ^{(j+1)} moves to position R_{0} − R_{j} + \overline{A_{j}} and becomes a new proper bead in runner a of λ^{(j)}.

Proof. The “if” direction is straightforward. For example, if a ≡ R_{0} − R_{j} + A_{j} and \overline{A_{j}} ≥ R_{j} − R_{j+1} > 0, then the bead k occupying position R_{0} − R_{j} in λ^{(j+1)} moves to position R_{0} − R_{j} + \overline{A_{j}} in runner a with the addition of the j-th p-edge (if A_{j} ≡ 0 we need to use R_{j} − R_{j+1} > 0). Moreover, k is improper for λ^{(j+1)} and proper for λ^{(j)}. It follows from \overline{A_{j}} ≥ R_{j} − R_{j+1} and Lemma 1.18 that k is the only new proper bead in runner a of λ^{(j)}.

Let us prove “only-if”. First, note that R_{j} − R_{j+1} > 0 because some new proper beads must appear.

If A_{j} ≡ 0, then by Lemma 1.18 the new proper beads of λ^{(j)} are the bead k which has moved from position R_{0} − R_{j} in λ^{(j+1)} to position R_{0} − R_{j} + p in λ^{(j)}, and the beads occupying positions R_{0} −R_{j} +1 through R_{0} −R_{j+1} −1 in both λ^{(j+1)} and λ^{(j)}. So since |a|_{j} > |a|_{j+1}, we must have a ≡ R_{0} − R_{j} + x for some x ∈ [0, R_{j} − R_{j+1}). If a ≡ R_{0} −R_{j} we are in case (ii), as A_{j} ≡ 0; if a ≡ R_{0} −R_{j} +x with 0 < x < R_{j} −R_{j+1} we are in case (i).

If A_{j} ≢ 0 and \overline{A_{j}} ≥ R_{j} − R_{j+1}, then by Lemma 1.18 the new proper beads of λ^{(j)} are the bead k which has moved from position R_{0} − R_{j} in λ^{(j+1)} to position R_{0} − R_{j} + \overline{A_{j}} in λ^{(j)}, and the beads occupying positions R_{0} − R_{j} + 1 through R_{0} − R_{j+1} − 1 in both λ^{(j+1)} and λ^{(j)}. Again, since |a|_{j} > |a|_{j+1}, we must have a ≡ R_{0} − R_{j} + \overline{A_{j}} or a ≡ R_{0} − R_{j} + x for x ∈ (0, R_{j} − R_{j+1}). If a ≡ R_{0} − R_{j} + \overline{A_{j}}, we have (ii); and if a ≡ R_{0} − R_{j} + x for x ∈ (0, R_{j} − R_{j+1}), we have (i).

Finally if A_{j} ≢ 0 and \overline{A_{j}} < R_{j} − R_{j+1} we find in a similar manner that we are in case (i).

Lemma 2.5. −2 ≤ d_{j}(a) ≤ 2.

Proof. This follows from Lemma 2.1 (and Definition 1.19).

Lemma 2.6. d_{j}(a) = 2 if and only if A_{j} ≡ 1, a ≡ R_{0} − R_{j} + 1, and R_{j} − R_{j+1} > 1. Moreover, in this case the bead m which occupies position R_{0} − R_{j} in λ^{(j+1)} moves to position R_{0} − R_{j} + 1 and pushes down by one row the bead k which occupies R_{0} − R_{j} + 1 in λ^{(j+1)}. Beads k and m are the two new proper beads in runner a of λ^{(j)}, and runner a − 1 of λ^{(j)} does not have new proper beads.

Proof. By Lemma 2.1, d_{j}(a) = 2 if and only if |a|_{j} = |a|_{j+1} + 2, |a − 1|_{j} = |a − 1|_{j+1}. According to Lemma 2.2, the first equation holds if and only if A_{j} ≢ 0, a ≡ R_{0} − R_{j} + A_{j}, R_{j} − R_{j+1} > \overline{A_{j}}. Now if A_{j} ≢ 1, it follows from Lemma 2.4(i) that |a − 1|_{j} = |a − 1|_{j+1} + 1. If A_{j} ≡ 1, it follows from 2.4, 2.2, and 2.1 that |a − 1|_{j} = |a − 1|_{j+1}. Thus, the first part of the Lemma is proved. The second part follows from the second part of Lemma 2.2.

Lemma 2.7. d_{j}(a) = 1 if and only if one of the following conditions holds.

(i) a ≡ R_{0} − R_{j} + A_{j}; R_{j} − R_{j+1} > 0; \overline{A_{j}} > R_{j} − R_{j+1} if R_{j} − R_{j+1} > 1 and \overline{A_{j}} ≥ 1 if R_{j} − R_{j+1} = 1. In this case the bead k occupying position R_{0} − R_{j} in λ^{(j+1)} moves to position R_{0} − R_{j} + \overline{A_{j}} and becomes the new proper bead in runner a of λ^{(j)}. Runner a − 1 of λ^{(j)} does not have new proper beads.

(ii) \overline{A_{j}} ≢ 0, 1, a ≡ R_{0} − R_{j} + A_{j}, and \overline{A_{j}} < R_{j} − R_{j+1}. In this case the bead m occupying position R_{0} − R_{j} in λ^{(j+1)} moves to R_{0} − R_{j} + \overline{A_{j}} and pushes a bead k down one row from the latter position. Beads k and m are the new proper beads in runner a of λ^{(j)}, and the bead l occupying R_{0} − R_{j} + \overline{A_{j}} − 1 in both λ^{(j+1)} and λ^{(j)} is the new proper bead in runner a − 1 of λ^{(j)}.

(iii) \overline{A_{j}} ≢ 0, 1, a ≡ R_{0} − R_{j} + 1, and R_{j} − R_{j+1} > 1. In this case the bead occupying R_{0} − R_{j} + 1 in both λ^{(j+1)} and λ^{(j)} becomes the new proper bead in runner a of λ^{(j)}. Runner a − 1 of λ^{(j)} does not have new proper beads.

Proof. By Lemma 2.1, d_{j}(a) = 1 if and only if one of the following conditions holds. (a) |a|_{j} = |a|_{j+1} + 2, |a − 1|_{j} = |a − 1|_{j+1} + 1;

(b) |a|_{j} = |a|_{j+1} + 1, |a − 1|_{j} = |a − 1|_{j+1}.

By 2.2 and 2.4, (a) holds if and only if A_{j} ≢ 0, 1, a ≡ R_{0} − R_{j} + A_{j}, and \overline{A_{j}} < R_{j} − R_{j+1}; thus we are in case (ii).

By 2.4, |a|_{j} = |a|_{j+1} + 1 if and only if one of the following conditions holds.

(α) a ≡ R_{0} − R_{j} + x, 0 < x < R_{j} − R_{j+1}, x ≢ A_{j}.
(β) a ≡ R_{0} − R_{j} + A_{j}, \overline{A_{j}} ≥ R_{j} − R_{j+1} > 0.

If (α) holds, then |a − 1|_{j} = |a − 1|_{j+1} if and only if x ≡ 1 and A_{j} ≢ 0, in view of 2.2 and 2.4. Thus we are in case (iii).

If (β) holds, then |a−1|_{j} = |a−1|_{j+1} if and only if \overline{A_{j}} > R_{j}−R_{j+1} for R_{j}−R_{j+1} > 1 and \overline{A_{j}} ≥ 1 for R_{j} − R_{j+1} > 1, according to 2.2 and 2.4. This gives (i).

Lemma 2.8. d_{j}(a) = −1 if and only if one of the following conditions holds. (i) A_{j} ≡ 0, R_{j} − R_{j+1} > 0, and a ≡ R_{0} − R_{j+1}; (ii) A_{j} ≢ 0, A_{j} ≢ R_{j} − R_{j+1}, A_{j} ≢ R_{j} − R_{j+1} − 1, and a ≡ R_{0} − R_{j+1}; (iii) A_{j} ≢ 0, A_{j} ≢ R_{j} − R_{j+1} − 1, and a ≡ R_{0} − R_{j} + A_{j} + 1. Moreover, in cases (i) and (ii), runner a of λ^{(j)} does not have new proper beads,

and the bead l occupying position R_{0} − R_{j+1} − 1 in λ^{(j+1)} is the new proper bead in runner a − 1 of λ^{(j)}. In case (iii) the bead l occupying R_{0} − R_{j} in λ^{(j+1)} moves to position R_{0} − R_{j} + \overline{A_{j}} in runner a − 1 of λ^{(j)}. If \overline{A_{j}} > R_{j} − R_{j+1} − 1 then l is the only new proper bead in runner a−1 of λ^{(j)}, and runner a of λ^{(j)} has no new proper beads. If \overline{A_{j}} < R_{j} − R_{j+1} − 1 then l pushes down a bead m occupying R_{0} − R_{j} + \overline{A_{j}} in λ^{(j+1)}, and the beads l and m are the new proper beads in runner a − 1 of λ^{(j)}; while the bead k occupying R_{0} −R_{j} +\overline{A_{j}} +1 in both λ^{(j)} and λ^{(j+1)} is the new proper bead in runner a of λ^{(j)}.

Proof. This follows from from 2.1, 2.2, and 2.4 in a fashion similar to the proofs of Lemmas 2.6 and 2.7.

Lemma 2.9. d_{j}(a) = −2 if and only if A_{j} ≢ 0, A_{j} ≡ R_{j} − R_{j+1} − 1, and a ≡ R_{0} − R_{j+1}. In this case the bead l occupying position R_{0} − R_{j} in λ^{(j+1)} moves to position R_{0} −R_{j+1} −1 (in runner a−1) occupied by an improper bead m, and pushes m down one row. Beads m and l are the two new proper beads in runner a − 1 of λ^{(j)}. Runner a of λ^{(j)} has no new proper beads.

Proof. Again, this follows from from 2.1, 2.2, and 2.4 in a fashion similar to the proofs of Lemmas 2.6 and 2.7.

3. Some lemmas on normal beads

Throughout the section 0 ≤ j ≤ z.

Lemma 3.1. Let k be a normal bead for λ^{(j)}. Assume that either k is proper for λ^{(j+1)} or A_{j} ≡ 0. Then position P_{j}(k) − 1 is unoccupied in λ^{(j+1)}.

Proof. Assume on the contrary that l_{1} is the bead with P_{j+1}(l_{1}) = P_{j}(k) − 1. Then since k is a beginning in λ^{(j)}, P_{j}(l_{1}) ≠ P_{j}(k) − 1. But since either k is proper for λ^{(j+1)} or A_{j} ≡ 0, the only way that l_{1} can move is one row up. So we have P_{j}(l_{1}) = P_{j}(k)−1+p. Let l_{1}, …, l_{w} be the beads such that P_{j}(l_{v}) = P_{j}(k)−1+vp for v ∈ [1, w] and P_{j}(k)−1+(w+1)p is unoccupied in λ^{(j)}. Since k is normal, there exist beads k_{1}, …, k_{w} with P_{j}(k_{v}) = P_{j}(k) + vp, v ∈ [1, w]. Therefore P_{j+1}(l_{v}) = P_{j}(l_{v}) for v ∈ [1, w] by 1.12(i). Taking v = 1 gives a contradiction.

Lemma 3.2. Let k ∈ (a)_{j+1} be a normal bead for λ^{(j+1)}, and let y ≥ 0 be the minimal integer such that position P_{j+1}(k) − 1 + (y + 1)p is unoccupied in λ^{(j+1)}. Then there are beads k_{v} with P_{j+1}(k_{v}) = P_{j+1}(k) + vp, v ∈ [1, y]. Moreover,

(i) Assume that k is a beginning for λ^{(j)}. Then k is normal for λ^{(j)}, and if k is not good for λ^{(j+1)} then it is not good for λ^{(j)}.

(ii) Assume that k is not a beginning for λ^{(j)}. Then y ≥ 1, k_{y} is normal for λ^{(j)} and if k is not good for λ^{(j+1)} then k_{y} is not good for λ^{(j)}.

Proof. Let

L = {l_{1} ≻ ⋯ ≻ l_{r}} = {l ∈ (a − 1)_{j+1} | P_{j+1}(l) ≥ P_{j+1}(k) − 1},
K = {k_{1} ≻ ⋯ ≻ k_{s}} = {k′ ∈ (a)_{j+1} | k′ ≺ k}.

Then according to Lemma 1.11, K compensates L in λ^{(j+1)}. By assumption, for v ∈ [1, y] we have P_{j+1}(l_{v}) = P_{j+1}(k) − 1 + vp. So P_{j+1}(k_{v}) = P_{j+1}(k) + vp for v ∈ [1, y].

Let x be minimal with respect to P_{j+1}(k_{v}) = P_{j+1}(k) + vp for all v ∈ [1, x] and P_{j+1}(k) + (x + 1)p is unoccupied in λ^{(j+1)}. Then x ≥ y.

Assume that k is a beginning for λ^{(j)}. This means that either P_{j}(k) = P_{j+1}(k) or P_{j}(k) = P_{j+1}(k) + p and y = 0. In both cases we have

{l ∈ (a − 1)_{j} | P_{j}(l) ≥ P_{j}(k) − 1} = L,
{k′ ∈ (a)_{j} | k′ ≺ k} = K.

So in view of Lemma 1.11, the k is normal for λ^{(j)} if and only if K compensates L in λ^{(j)}. But the latter fact follows from Lemma 1.14, parts (i) and (iii).

Assume that k is not a beginning in λ^{(j)}. Then y ≥ 1 and P_{j}(k) = P_{j+1}(k) + p. Hence P_{j}(l_{v}) = P_{j+1}(l_{v}) and P_{j}(k_{v}) = P_{j+1}(k_{v})+p for v ∈ [1, y]. So k_{y} is a beginning for λ^{(j)}. The proof of the normalcy of k_{y} for λ^{(j)} is similar to that of the normalcy of k above.

Now assume that k is not good for λ^{(j+1)}, i.e. there exists a normal bead m ∈ (a)_{j+1} with m ≻ k (in λ^{(j+1)}). Applying the above arguments to m, one of the following happens:

(1) m is normal in λ^{(j)}. Then k is not good for λ^{(j)} since m ≻ k in λ^{(j)}.

(2) there exists y′ ≥ 1 and beads f_{v}, e_{v} such that P_{j+1}(f_{v}) = P_{j+1}(m) − 1 + vp and P_{j+1}(e_{v}) = P_{j+1}(m) + vp for v ∈ [1, y′], with P_{j+1}(m) − 1 + (y′ + 1)p unoccupied in λ^{(j+1)} and with e_{y′} normal for λ^{(j)}. But since none of e_{1}, …, e_{y′} is a beginning in λ^{(j+1)}, it follows that k ≺ e_{y′} in λ^{(j+1)}, hence in λ^{(j)}. Thus k is not good in λ^{(j)}.

Corollary 3.3. If there exists a good bead in runner a of λ^{(j+1)}, then there exists a good bead in runner a of λ^{(j)}.

Proof. By the previous lemma if there exists a normal bead in runner a of λ^{(j+1)} then there exists a normal bead in runner a of λ^{(j)}. Now it suffices to recall that acccording to Definition 1.7, any runner which contains a normal bead also contains a good bead.

Lemma 3.4. Let k ∈ (a)_{j} be a normal bead for λ^{(j)} and assume that there exist u ≥ 0 and beads e_{1}, …, e_{u} such that P_{j}(e_{v}) = P_{j}(k) − vp for v ∈ [1, u], P_{j}(k) − (u + 1)p > 0, and position P_{j}(k) − (u + 1)p is unoccupied in λ^{(j)}. Suppose that k is proper for λ^{(j+1)} if u = 0 and that e_{u} is proper for λ^{(j+1)} if u > 0.

(i) Let k be a beginning for λ^{(j+1)}. Then k is normal for λ^{(j+1)}. Moreover, if k is good for λ^{(j)} then it is so for λ^{(j+1)}.

(ii) Assume that k is not a beginning for λ^{(j+1)}. Then u > 0 and e_{u} is normal for λ^{(j+1)}. Moreover, if k is good for λ^{(j)} then e_{u} is so for λ^{(j+1)}.

Proof. If u > 0 then k is proper for λ^{(j+1)} since e_{u} is so. If u = 0 then k is proper for λ^{(j+1)} by assumption. Thus k is proper for λ^{(j+1)} in any case.

Set

L = {l ∈ (a − 1)_{j} | P_{j}(l) ≥ P_{j}(k) − 1}

and

K = {k′ ∈ (a)_{j} | k′ ≺ k}
= {k′ ∈ (a)_{j+1} | k′ ≺ k}.

By Lemma 3.1 we have that position P_{j}(k) − 1 is unoccupied in λ^{(j+1)}, so k is not a beginning for λ^{(j+1)} if and only if P_{j+1}(k) = P_{j}(k) − p, u > 0, and position P_{j}(k) − p − 1 is occupied (in λ^{(j+1)}) by some bead l. So if k is a beginning for λ^{(j+1)} then

L = {l ∈ (a − 1)_{j+1} | P_{j+1}(l) ≥ P_{j+1}(k) − 1}.

By Lemma 1.11, K compensates L in λ^{(j)}, so K compensates L in λ^{(j+1)} in view of 1.16. Now k is normal in λ^{(j+1)} by virtue of 1.11.

Next assume that k is not a beginning for λ^{(j+1)}. By Lemma 3.1 as above, we have P_{j+1}(e_{v}) = P_{j}(e_{v}) − p for v ∈ [1, u]. Since e_{u} is proper for λ^{(j+1)}, Lemma 1.12(ii) implies that P_{j+1}(e_{u}) − 1 is unoccupied in λ^{(j+1)}, i.e. e_{u} is a beginning for λ^{(j+1)}; thus e_{u} ≠ k and u > 0.

In view of Lemma 1.11, to prove the normalcy of e_{u} in λ^{(j+1)} it suffices to show that

K′ = {e_{1}, …, e_{u−1}} ∪ {k} ∪ K

compensates

{l′ ∈ (a − 1)_{j+1} | P_{j+1}(k) − 1 ≥ P_{j+1}(l′) ≥ P_{j+1}(e_{u}) − 1} ∪ L.

But K compensates L in λ^{(j+1)} by Lemma 1.16 and {e_{1}, …, e_{u−1}} ∪ {k} clearly compensates {l′ ∈ (a − 1)_{j+1} | P_{j+1}(k) − 1 ≥ P_{j+1}(l′) ≥ P_{j+1}(e_{u}) − 1}.

The remaining statements of the lemma (those regarding good beads) follow from 3.2.

4. Degenerate beads

Difficulties arise in the proof of the Main Theorem when the abacus for λ has a degenerate bead. In the next section we show that λ has a degenerate bead if and only if m(λ) does; here are some lemmas towards that end.

Lemma 4.1. Assume that there exists a bead k of λ with P_{0}(k) = 1. Then λ(k)^{(1)} = λ^{(1)}.

Proof. It follows from the definition that for any bead m with P_{1}(m) > P_{0}(k) we have P_{1}(m) = P_{λ(k)^{(1)}}(m). If P_{1}(k) = 0 then P_{1}(m) > P_{0}(k) for any bead m ≠ k. So it suffices to observe that P_{λ(k)^{(1)}}(k) = 0 since P_{λ(k)}(k) = 0.

If P_{1}(k) = 1 then there exists a bead f with P_{1}(f) = 0. It is easily seen that P_{1}(m) = P_{λ(k)^{(1)}}(m) for all m ≠ k, f, P_{λ(k)^{(1)}}(f) = P_{1}(k), and P_{λ(k)^{(1)}}(k) = P_{1}(f), i.e. λ(k)^{(1)} = λ^{(1)}.

The hypothesis of the next lemma can be depicted schematically as

f_{x} e_{x}
… ⋮ ⋮ …
f_{1} e_{1}
· k

if a ≢ 0 and as

· . . . f_{x}
e_{x} … f_{x−1}
⋮ … ⋮
e_{2} … f_{1}
e_{1} . . . ·
k …

if a ≡ 0, with f_{x} in the first row of the abacus in both cases. Notice that if a ≡ 1, P_{0}(k) > 1, then the hypotheses of 4.2 cannot hold because λ is a canonical abacus, i.e. position 0 is unoccupied.

Lemma 4.2. Assume that bead k of λ is r-movable (i.e. P_{1}(k) < P_{0}(k)) and k is a beginning for λ. Let x be maximal with respect to P_{0}(k)−1−xp ≥ 0, and assume that there are beads f_{v} ∈ (a − 1)_{0}, e_{v} ∈ (a)_{0} such that P_{0}(f_{v}) = P_{0}(k) − 1 − vp, P_{0}(e_{v}) = P_{0}(k) − vp for v ∈ [1, x]. Then λ(k)^{(1)} = λ^{(1)}.

Proof. Note that

P_{1}(k) = min(0, P_{0}(k) − p)
P_{1}(e_{v}) = min(0, P_{0}(e_{v}) − p)
P_{1}(f_{v}) = P_{0}(f_{v})
P_{λ(k)^{(1)}}(k) = min(0, P_{0}(k) − 1 − p)
P_{λ(k)^{(1)}}(e_{v}) = P_{0}(e_{v})
P_{λ(k)^{(1)}}(f_{v}) = min(0, P_{0}(f_{v}) − p)
P_{λ(k)^{(1)}}(m) = P_{1}(m)

for any v ∈ [1, x] and any bead m ∉ {k, e_{1}, …, e_{x}, f_{1}, …, f_{x}}. To prove that λ(k)^{(1)} = λ^{(1)} it now suffices to observe that P_{λ(k)^{(1)}}(k) = P_{1}(k) = 0 if x = 0, and

P_{λ(k)^{(1)}}(k) = P_{1}(f_{1})
P_{λ(k)^{(1)}}(e_{1}) = P_{1}(k)
P_{λ(k)^{(1)}}(e_{v}) = P_{1}(e_{v−1}) for v ∈ [2, x]
P_{λ(k)^{(1)}}(f_{v}) = P_{1}(f_{v+1}) for v ∈ [1, x)
P_{λ(k)^{(1)}}(f_{x}) = P_{1}(e_{x})

if x > 0.

Lemma 4.3. Assume that k ∈ (a)_{0} is a good bead for λ, and μ is the partition such that λ(k) is an abacus for μ.

(i) If k is non-degenerate then there exists a bead g ∈ (a)_{1} such that g is good for λ^{(1)} and λ(k)^{(1)} = λ^{(1)}(g).

(ii) The bead k is degenerate if and only if at least one of the following conditions

holds.

(a) P_{0}(k) = 1;

(b) k is r-movable and if x is maximal with respect to P_{0}(k) − 1 − xp ≥ 0, then for v ∈ [1, x], there exist beads f_{v}, e_{v} with P_{0}(f_{v}) = P_{0}(k) − 1 − vp, P_{0}(e_{v}) = P_{0}(k) − vp.

(iii) μ is p-regular, the first column of the Mullineux symbol G_{p}(μ) is (\begin{matrix}A_{0} \\ R_{0} \end{matrix}) if

k is non-degenerate, and the first column of the Mullineux symbol G_{p}(μ) is

(\begin{matrix}A_{0} − 1 \\ R_{0} − ξ \end{matrix}) with ξ ∈ \{0, 1\} if k is degenerate. Moreover, ξ = 1 if P_{0}(k) = 1

and ξ = 0 otherwise. (iv) Let k be degenerate, with ξ as in (iii). Then A_{0} ≡ 0 implies ξ = 0, and

A_{0} ≡ 1 implies ξ = 1.

Proof. (i) and (ii). Lemmas 4.1 and 4.2 tell us that if either (a) or (b) holds, then k is degenerate. So we assume that neither of (a), (b) holds, and prove that k is non-degenerate and λ(k)^{(1)} = λ^{(1)}(g) for some bead g ∈ (a)_{1} which is good for λ^{(1)}.

Let y be maximal such that for all v ∈ [1, y] there exist beads f_{v} with P_{0}(f_{v}) = P_{0}(k)−1−vp (as usual we put y = 0 if P_{0}(k)−1−p is unoccupied or P_{0}(k)−1−p < 0). Then y ≤ x. Let w be maximal such that there exist beads e_{v} with P_{0}(e_{v}) = P_{0}(k) − vp for v ∈ [1, w]. Then w ≤ x.

Notice that w ≤ y since otherwise e_{y+1} would be a normal bead above k, which is impossible since k is good.

Assume first that k is not r-movable. Then k is good for λ^{(1)} by Lemma 3.4. Now the fact that P_{0}(k) − 1 is unoccupied in λ^{(1)} implies P_{λ(k)^{(1)}}(m) = P_{1}(m) for any bead m ≠ k, and P_{λ(k)^{(1)}}(k) = P_{1}(k) − 1; i.e. λ(k)^{(1)} = λ^{(1)}(k).

Next assume that k is r-movable. Then w < x since otherwise w = y = x and (b) holds. So P_{1}(k) = P_{0}(k) − p, and P_{0}(e_{v}) = P_{0}(e_{v}) − p for v ∈ [1, w].

If w = 0 then k is a beginning for λ^{(1)} by 1.12(ii). So it follows from 3.4 that k is good for λ^{(1)}.

As in the case in which k is not r-movable, we get P_{λ(k)^{(1)}}(m) = P_{1}(m) for any bead m ≠ k, and P_{λ(k)^{(1)}}(k) = P_{1}(k) − 1; i.e. λ(k)^{(1)} = λ^{(1)}(k).

Let w > 0. Then e_{w} is good for λ^{(1)} by 3.4(ii), and we have λ(k)^{(1)} = λ^{(1)}(e_{w}). (iii). Assume that μ is not p-regular. Since λ is p-regular, this means that there

are proper beads m_{1}, …, m_{p−1} with P_{0}(m_{v}) = P_{0}(k) − 1 − v for v ∈ [1, p − 1], and P_{0}(k) − 1 − p is unoccupied. But then m_{p−1} is normal for λ which contradicts the fact that k is good for λ (because m_{p−1} ≻ k in λ).

If k is non-degenerate then P_{0}(k) > 1 by (ii). So λ(k) has R_{0} proper beads. Moreover, λ^{(1)} is a partition of n − A_{0}, so λ(k)^{(1)} = λ^{(1)}(g) implies that μ^{(1)} is a partition of n − A_{0} − 1. Notice that μ is a partition of n − 1. Summarizing, the p-edge of μ has length A_{0}. The remainder of (iii) is obtained similarly.

(iv) Let A_{0} ≡ 0. If P_{0}(k) = 1 then by Lemma 3.1, position P_{0}(k)−1 is unoccupied in λ^{(1)}. Therefore R_{1} = R_{0}. If ξ = 1 we get by (iii) that

G_{p}(μ) = (\begin{matrix}A_{0} − 1 & A_{1} & … & A_{z} \\ R_{0} − 1 & R_{1} & … & R_{z} \end{matrix})

giving a contradiction to 1.1. So P_{0}(k) ≠ 1, and ξ = 0 by (iii).

Let A_{0} ≡ 1. If P_{0}(k) ≠ 1 then (ii) implies that k is non-degenerate since position 0 is unoccupied in λ. Since k is degenerate, P_{0}(k) = 1; hence ξ = 1 by (iii).

Proposition 4.4. Let (a − 1)_{1} = {l_{1} ≺ ⋯ ≺ l_{r}} = L. Then runner a of λ contains a degenerate (good) bead if and only if at least one of the following conditions holds.

(i) d_{0}(a) > 0 and L is compensated in λ^{(1)}; (ii) d_{0}(a) = 2 and {l_{1}, …, l_{r−1}} is compensated in λ^{(1)}.

Proof. Let K = (a)_{1} = {k_{1} ≺ ⋯ ≺ k_{s}}. Assume that runner a of λ contains a degnerate (good) bead k. Then in view of 4.3 at least one of the following conditions holds.

(a) P_{0}(k) = 1;

(b) k is r-movable and if x is maximal with respect to P_{0}(k) − 1 − xp ≥ 0, then for v ∈ [1, x], there exist beads f_{v}, e_{v} with P_{0}(f_{v}) = P_{0}(k) − 1 − vp, P_{0}(e_{v}) = P_{0}(k) − vp.

Suppose (a) holds. Then a ≡ 1. Notice that k is improper for λ^{(1)}: Otherwise, R_{0} = R_{1}, which would imply p|A_{0} in view of 1.1, and then by 1.3 there exists a bead m_{N} with P_{1}(m_{N}) = 0.

Thus |a|_{0} ≥ |a|_{1} + 1.

By Lemma 3.1, A_{0} ≢ 0. So in view of Lemmas 2.2, 2.4, and 2.1, we have |a − 1|_{0} = |a − 1|_{1}, i.e. d_{0}(a) = |a|_{0} − |a|_{1} ≥ 1.

If d_{0}(a) = 2 then P_{1}(k) = 0 and for some bead m we have P_{1}(m) = 1, P_{0}(k) = P_{1}(m), and P_{0}(m) = p + 1, according to Lemma 2.6. ¿From the normalcy of k in λ and Lemma 1.11 we get that {m} ∪ K compensates L in λ. Hence K compensates {l_{1}, …, l_{r−1}} in λ, and so K compensates {l_{1}, …, l_{r−1}} in λ^{(1)} by virtue of 1.16.

If d_{0}(a) = 1 then by 2.7 either P_{1}(k) = 0 and position 1 is unoccupied in λ^{(1)}, or P_{1}(k) = 1 and there exists a bead m with P_{1}(m) = 0, P_{0}(m) ∈ (0, p). In both cases the normalcy of k and Lemmas 1.11, 1.16 imply that K compensates L in λ^{(1)}.

Now assume that (b) holds.

If x = 0 then |a|_{0} = |a|_{1} + 1, |a − 1|_{0} = |a − 1|_{1}, P_{1}(k) = 0, P_{0}(k) = \overline{A_{0}} since k is r-movable, and P_{0}(k) − 1 is unoccupied in both λ and λ^{(1)}. So d_{0}(a) = 1 and

L = {l ∈ (a − 1)_{0} | P_{0}(l) ≥ P_{0}(k) − 1},
K = {k′ ∈ (a)_{0} | k′ ≺ k}.

By 1.11 the normalcy of k implies that K compensates L in λ, and so K compensates L in λ^{(1)} according to Lemma 1.16.

Let x > 0 and assume that f_{x} is proper for λ^{(1)}. Then e_{x} is the only new proper bead in (a)_{0}, and runner a − 1 of λ has no new proper beads. Thus d_{0}(a) = 1 and

(a − 1)_{1} = (a − 1)_{0} = {f_{1}, …, f_{x}} ∪ L′,
K = {e_{1}, …, e_{x−1}} ∪ {k} ∪ K′

where

L′ = {l ∈ (a − 1)_{1} | l ≺ f_{1}},
K′ = {k′ ∈ (a)_{1} | k′ ≺ k}.

The normalcy of k in λ and Lemma 1.11 imply that K′ compensates L′ in λ, hence in λ^{(1)} in view of 1.16. Also, {e_{1}, …, e_{x−1}} ∪ {k} clearly compensates {f_{1}, …, f_{x}} in λ^{(1)}.

If x > 0 and f_{x} is improper for λ^{(1)}, we see that f_{x} is the only new proper bead in runner a − 1 of λ, and e_{x}, e_{x−1} (or e_{x}, k if x = 1) are the new proper beads in runner a of λ. So d_{0}(a) = 1. The proof of the fact that (a)_{1} compensates (a − 1)_{1} in λ^{(1)} is similar to the proof in the previous case, using 1.11 and 1.16.

In the other direction, assume that d_{0}(a) = 2 and {l_{1}, …, l_{r−1}} is compensated in λ^{(1)}. By Lemma 2.6, a ≡ 1 and there are beads k, m with P_{1}(k) = 0, P_{1}(m) = 1, P_{0}(k) = 1, and P_{0}(m) = p + 1. Let g_{v} = k_{s−(r−1−v)} for v ∈ [1, r − 1]. Then g_{1} ≺ ⋯ ≺ g_{r−1} are the top r − 1 beads in (a)_{1}, and G = {g_{1}, …, g_{r−1}} compensates L′ = {l_{1}, …, l_{r−1}} in λ^{(1)}. If P_{1}(g_{r−1}) > 1 + p then G compensates L′ in λ by Lemma 1.14(i). Otherwise, G compensates L′ in λ by Lemma 1.14(iii) because P_{1}(l_{r−1}) > P_{1}(l_{r}) ≥ p. In addition, m compensates l_{r} in λ since P_{0}(l_{r}) ≥ p. In view of Lemma 1.11, k is normal for λ (hence good); thus (a) holds. By Lemma 4.3 this means that k is degenerate for λ.

Assume d_{0}(a) = 1 and L is compensated in λ^{(1)}. Let g_{v} = k_{s−(r−v)}, v ∈ [1, r]. Then G = {g_{1}, …, g_{r}} compensates L in λ^{(1)}. Since d_{0}(a) = 1 there are some new beads in λ, hence R_{0} − R_{1} > 0 and position 0 is occupied in λ^{(1)}. Let m be the bead with P_{1}(m) = 0.

In view of Lemma 2.7, one of the following occurs.

(α) a ≡ A_{0}, P_{0}(m) = \overline{A_{0}}, and m is the only new proper bead in (a)_{0}; (β) a ≡ A_{0} ≢ 0, 1, P_{0}(m) = \overline{A_{0}}, and there are improper beads l and k with P_{0}(m)−1 = P_{1}(l), P_{0}(m) = P_{1}(k), P_{0}(l) = P_{1}(l), and P_{0}(k) = P_{0}(m)+p. In this case m and k are the new proper beads in (a)_{0} and l is the new proper bead in (a − 1)_{0}; (γ) a ≡ 1, P_{0}(m) ∈ (1, p) and there is a bead k with P_{1}(k) = 1 = P_{0}(k) which is the only new proper bead in (a)_{0}.

Consider case (α). If P_{0}(m) − 1 is unoccupied in λ^{(1)} then K compensates L in λ by Lemma 1.14, parts (i) and (iii), and m is good for λ by Lemma 1.11. So (b) holds, and by Lemma 4.3, k is degenerate for λ.

If P_{1}(l_{r}) = P_{0}(m) − 1 then P_{1}(g_{r}) = P_{0}(m) since G compensates L in λ^{(1)}. Let u ∈ [1, r] be such that P_{1}(l_{v}) = P_{0}(l_{v+1}) + p for v ∈ [u, r) and position P_{1}(l_{u}) − p is unoccupied in λ^{(1)}. Then since G compensates L in λ^{(1)} we have P_{1}(g_{v}) = P_{1}(l_{v})+1 for all v ∈ [u, r]. Then e_{u} is the top beginning in (a)_{0}. Moreover, e_{u} is good in view of Lemma 1.14, parts (i) and (iii), and so (b) holds. By Lemma 4.3, e_{u} is degenerate for λ.

In case (γ) we immediately see that k is good, so (a) holds, and by Lemma 4.3, k is degenerate for λ. Case (β) is similar to (α).

5. Compensation and “steps”

The main result of this section is Corollary 5.8, giving a necessary and sufficient condition for the occurence of a degenerate bead in runner a of λ, in terms of the d_{j}(a). Throughout the section 0 ≤ j ≤ z.

Lemma 5.1. Let l ∈ (a − 1)_{j}.

(i) If |a|_{j} = |a|_{j+1} + 1 and k ∈ (a)_{j} is a new bead of λ^{(j)} then {k} compensates {l}.

(ii) If |a|_{j} = |a|_{j+1} + 2 and k′, k ∈ (a)_{j}, k′ ≺ k are new proper beads of λ^{(j)} then {k} compensates {l} and {k, k′} compensates any subset L ⊆ (a − 1)_{j} with |L| = 2.

Proof. It follows from Lemmas 2.2 and 2.4 that P_{j}(l) ≥ P_{j}(k) − 1 for any l ∈ (a)_{j}, giving (i). To prove (ii) we have also to notice that P_{j}(k′) = P_{j}(k) + p in view of Lemma 2.2.

Lemma 5.2. Assume that d_{j}(a) < 0, and let L be a subset of (a − 1)_{j} with |L| ≥ |a − 1|_{j} + d_{j}(a) + 1. Then L is not compensated in λ^{(j)}.

Proof. Let (a − 1)_{j} = {l_{1} ≺ ⋯ ≺ l_{r}}.

Assume that d_{j}(a) = −2. We must show that if |L| ≥ r − 1, then L is not compensated in λ^{(j)}. It is clearly sufficient to prove that

L′ = {l_{1} ≺ ⋯ ≺ l_{r−1}}

is not compensated in λ^{(j)}.

It follows from Lemma 2.9 that l_{r−1} and l_{r} are the new proper beads of (a − 1)_{j}, and that (a)_{j} does not have new proper beads, i.e. (a)_{j} = (a)_{j+1}. Moreover, by the same lemma, l_{r−1} is an improper bead in runner a − 1 of λ^{(j+1)}, and

P_{j+1}(l_{r}) < P_{j}(l_{r}) = P_{j+1}(l_{r−1}) = P_{j}(l_{r−1}) − p.

Notice that position P_{j+1}(l_{r−1}) + 1 is unoccupied in λ^{(j+1)}, since if P_{j+1}(k) = P_{j+1}(l_{r−1}) + 1 then k would be a new proper bead in (a)_{j}. Therefore P_{j+1}(l_{r−1}) + 1 is also unoccupied in λ^{(j)}.

Let w be the smallest integer such that

P_{j+1}(l_{v}) = P_{j+1}(l_{v+1}) + p for v ∈ [w, r − 1), P_{j+1}(l_{w}) + p is unoccupied in λ^{(j)}.

Then P_{j}(l_{v}) = P_{j+1}(l_{v}) + p, v ∈ [w, r − 1] and P_{j}(l_{w}) + 1 is unoccupied in λ^{(j)} by Lemma 1.12(i). Now it follows that L′ is not compensated in λ^{(j)} since

|{l ∈ L′ | P_{j}(l) ≤ P_{j}(l_{w})}| = |{l_{w}, …, l_{r−1}}| = r − w,
|{k ∈ (a)_{j} | P_{j}(k) ≤ P_{j}(l_{w}) + 1}| ≤ r − w − 1.

If d_{j}(a) = −1 we have to prove that {l_{1} ≺ ⋯ ≺ l_{r}} is not compensated in λ^{(j)}. This is similar to the case d_{j}(a) = −2 above, using Lemma 2.8 instead of Lemma 2.9.

Notation. For the remainder of the section we assume that

(a − 1)_{0} = {l_{1} ≺ ⋯ ≺ l_{s}}

and put j_{v} = St(l_{v}) for v ∈ [1, s]. Then j_{1} ≥ ⋯ ≥ j_{s}. (cf. Definition 1.17 and the remark following it.)

Definition 5.3. We define an integer M ∈ [0, s], integers h_{1}, …, h_{M}, and beads k_{1} ≺ ⋯ ≺ k_{M} ∈ (a)_{0} inductively as follows.

If St(k) > j_{1} for all k ∈ (a)_{0} we set M = 0. Otherwise let k_{1} be the lowest bead in (a)_{0} such that St(k_{1}) ≤ j_{1}.

Assume we have already defined k_{1}, …, k_{t}. If either (1) t = s or (2) t < s but St(k) > j_{t+1} for all k ∈ (a)_{0} with k ≻ k_{t}, then we put M = t. Otherwise, let k_{t+1} be the lowest bead in (a)_{0} such that St(k_{t+1}) ≤ j_{t+1} and k_{t+1} ≻ k_{t}.

Finally put h_{v} = St(k_{v}) for v ∈ [1, M].

Proposition 5.4. Assume r ≤ s and j ≤ j_{r} (i.e. l_{r} ∈ (a − 1)_{j}). Then L = {l_{1}, …, l_{r}} is compensated in λ^{(j)} if and only if M ≥ r and h_{r} ≥ j (i.e. k_{r} ∈ (a)_{j}).

Proof. We proceed by induction on r. Let r = 1.

Assume {l_{1}} is compensated in λ^{(j)}. We may suppose that either (1) j = j_{1} or (2) j < j_{1} but {l_{1}} is not compensated in λ^{(j+1)}. We must show that M ≥ 1 and h_{1} ≥ j. Assume not.

We claim that St(k) > j for any k ∈ (a)_{j}. Indeed, if M = 0 then by Definition 5.3, St(k) > j_{1} ≥ j for any k ∈ (a)_{0} ⊇ (a)_{j}. If M = 1, and h_{1} < j, then k_{1} ∉ (a)_{j}. But St(k) = j for some k ∈ (a)_{j} implies k ≺ k_{1} in λ and St(k) = j ≤ j_{1}, which contradicts Definition 5.3.

Thus |a|_{j} = |a|_{j+1}. Assume that j = j_{1}. Then |a−1|_{j+1} = 0 (since l_{1} is the lowest proper bead in runner a), and |a − 1|_{j} > 0. thus d_{j}(a) = −|a − 1|_{j} < 0 and {l_{1}} is not compensated in λ^{(j)} by Lemma 5.2, contrary to our assumption.

Now assume that j < j_{1} and {l_{1}} is not compensated in λ^{(j+1)}. Let k ∈ (a)_{j} be a bead such that {k} compensates {l_{1}} in λ^{(j)}. If k ∈ (a)_{j+1} then {k} would compensate {l_{1}} in λ^{(j+1)} by Lemma 1.16. Hence St(k) = j. Therefore M ≥ 1 and h_{1} ≥ j since k_{1} ⪯ k by definition.

In the other direction, let M ≥ 1, h_{1} ≥ j. Since j_{1} ≥ h_{1} then l_{1} ∈ (a − 1)_{h_{1}}. Moreover, by Lemma 5.1, {l_{1}} is compensated in λ^{(h_{1})}. By Corollary 1.15, {l_{1}} is compensated in λ^{(j)}.

Now for the inductive step: Let r > 1 and assume that the proposition has been proved for all subsets {l_{1}, …, l_{r′}} of (a − 1)_{j′} for all j′ and all r′ < r.

Assume that L is compensated in λ^{(j)}. We may suppose that either j_{r} = j (i.e. L ⊈ (a−1)_{j+1}) or j_{r} > j but L is not compensated in λ^{(j+1)}. Since L is compensated in λ^{(j)} then it is compensated by the set of the top r beads, say m_{1} ≺ ⋯ ≺ m_{r}, of (a)_{j}.

By induction we have M ≥ r − 1, h_{r−1} > j. Assume that M = r − 1, or M ≥ r but h_{r} < j.

Note that either m_{r} = k_{r−1} or St(m_{r}) > j_{r} (or both) since otherwise we would have M ≥ r, h_{r} ≥ j by Definition 5.3.

If St(m_{r}) > j then {m_{1}, …, m_{r}} ⊆ (a)_{j+1}. So in view of Lemma 1.16, L ⊆ (a − 1)_{j+1} and {m_{1} …, m_{r}} compensates L in λ^{(j+1)}. Thus St(m_{r}) = j and m_{r} = k_{r−1}.

We consider three cases.

(a) j_{r−1} > j and {l_{1}, …, l_{r−1}} is compensated in λ^{(j+1)}. By the inductive hypothesis, h_{r−1} ≥ j + 1. But h_{r−1} = St(k_{r−1}) = St(m_{r}) = j, giving a contradiction.

(b) j_{r−1} > j and {l_{1}, …, l_{r−1}} is not compensated in λ^{(j+1)}. Now if St(m_{r−1}) > j then {m_{1}, …, m_{r−1}} ⊆ (a)_{j+1} and so {m_{1}, …, m_{r−1}} compensates {l_{1}, …, l_{r−1}} by Lemma 1.16. So St(m_{r−1}) = j. We claim that r ≥ 3 and m_{r−1} = k_{r−2}. Indeed, if r = 2 we have St(m_{1}) = St(m_{2}) = j, St(l_{1}) ≥ j, and St(l_{2}) ≥ j. Hence k_{1} ⪯ m_{1} ≺ m_{2} = k_{1}, giving a contradiction. If r ≥ 3 and k_{r−2} ≺ m_{r−1}, then St(m_{r−1}) = St(m_{r}) = j implies k_{r−1} ⪯ m_{r−1}, giving a contradiction since k_{r−1} = m_{r}. Thus k_{r−2} = m_{r−1}. Therefore h_{r−2} = j < j + 1. By the inductive hypothesis, {l_{1}, …, l_{r−2}} is not compensated in λ^{(j+1)}. But {m_{1}, …, m_{r−2}} ⊆ (a)_{j+1} and so {m_{1}, …, m_{r−2}} compensates {l_{1}, …, l_{r−2}} in λ^{(j+1)} by Lemma 1.16.

(c) j_{r−1} = j. Then also j_{r} = j. Thus we get |a − 1|_{j} = |a − 1|_{j+1} + 2. Hence d_{j}(a) < 0 because at most one runner can grow by two beads with the addition of a p-edge, cf. Corollary 2.3. By Lemma 5.2, L is not compensated in λ^{(j)}.

In the other direction, assume that M ≥ r and h_{r} ≥ j. By Corollary 1.15 we may suppose that j = h_{r} (taking into account that j_{r} ≥ h_{r}).

We have j_{r−1} ≥ j_{r} ≥ j. If j_{r−1} = j then j_{r} = j; and h_{r−1} ≥ h_{r} = j, h_{r−1} ≤ j_{r−1} by definition. So h_{r−1} = h_{r} = j which contradicts Corollary 2.3. So j_{r−1} > j, i.e. {l_{1}, …, l_{r−1}} ⊆ (a − 1)_{j+1}.

We consider two cases.

(a) h_{r−1} = j. If r = 2 then by Lemma 5.1(ii), {k_{1}, k_{2}} compensates L in λ^{(j)}. Let r > 2. Since |a|_{j} ≤ |a|_{j+1} + 2 by 2.1, we have h_{r−2} > j. By the inductive hypothesis L′ = {l_{1}, …, l_{r−2}} is compensated in λ^{(j+1)}. Let N = {n_{1} ≺ ⋯ ≺ n_{r−2}} be the set of the top r − 2 beads in (a)_{j+1}. Then N compensates L′ in λ^{(j+1)}. We noted above that l_{r−1} is a proper bead in runner a − 1 of λ^{(j+1)}. So either P_{j+1}(n_{r−2}) < P_{j+1}(l_{r−2}) + 1, or position P_{j+1}(n_{r−2})−p is unoccupied in λ^{(j+1)} (since n_{r−2} is the top (proper) bead of (a)_{j+1}). Now N compensates L′ in λ^{(j)} by Lemma 1.14(iii) in the first case and by Lemma 1.14(i) in the second. Now it suffices to notice that {k_{r−1}, k_{r}} compensates {l_{r−1}, l_{r}} in λ^{(j)} by Lemma 5.1, and {k_{r−1}, k_{r}} ∩ N = ∅ since St(k_{r−1}) = St(k_{r}) = j.

(b) h_{r−1} > j. Then by the inductive hypothesis L′ = {l_{1}, …, l_{r−1}} is compensated in λ^{(j+1)}. Let N = {n_{1} ≺ ⋯ ≺ n_{r−1}} be the set of the top r − 1 beads in (a)_{j+1}. If P_{j+1}(l_{r−1}) > P_{j+1}(n_{r−1})−1 or there does not exist a bead n with P_{j}(n) = P_{j+1}(n_{r−1}) then N compensates L′ in λ^{(j)} by Lemma 1.14, parts (i) and (iii), and the top new bead in (a)_{j} compensates {l_{r}} in λ^{(j)} by Lemma 5.1. So we may assume that P_{j+1}(l_{r−1}) = P_{j+1}(n_{r−1}) − 1 and P_{j}(n) = P_{j+1}(n_{r−1}) for some necessarily improper bead n of λ^{(j+1)}. Note that l_{r} ∉ (a − 1)_{j+1} since otherwise P_{j+1}(n_{r−1}) − p would be unoccupied in λ^{(j+1)}, giving a contradiction. For the same reason we have that either P_{j+1}(l_{r−1}) < p or P_{j+1}(l_{r−1}) − p is occupied by an improper bead. Notice that P_{j}(l_{r−1}) = P_{j+1}(l_{r−1}) because otherwise n_{r−1} would not be amovable in λ^{(j+1)}. Since l_{r} ≻ l_{r−1}, l_{r} ∈ (a − 1)_{j} we get P_{j}(l_{r−1}) > p. So position P_{j+1}(l_{r−1}) − p is occupied by the improper bead l_{r} in λ^{(j+1)} (l_{r} cannot be in a runner of λ^{(j+1)} other than a because that would force P_{j}(l_{r}) = P_{j+1}(l_{r−1}), contradicting P_{j}(l_{r−1}) = P_{j+1}(l_{r−1})); that is, P_{j+1}(l_{r}) = P_{j+1}(l_{r−1}) − p. Since P_{j}(l_{r−1}) = P_{j+1}(l_{r−1}), we have also P_{j}(l_{r}) = P_{j+1}(l_{r}). We claim that P_{j+1}(n) = P_{j+1}(n_{r−1}) − p. Indeed, P_{j+1}(n) ≥ P_{j}(n) − p = P_{j+1}(n_{r−1}) − p, and if n belongs to a runner different from a in λ^{(j+1)} then l_{r} is not proper for λ^{(j)}. Since l_{r} is proper for λ^{(j)} we also conclude that there exists some bead m with P_{j+1}(m) < P_{j+1}(l_{r}), and P_{j}(m) = P_{j+1}(n) (then m ∈ (a)_{j} since P_{j}(m) > P_{j}(l_{r})).

Now {n, n_{2}, …, n_{r−1}} compensates L′ in λ^{(j)} by Lemma 1.14(ii) and {m}

compensates {l_{r}} in λ^{(j)} by Lemma 5.1.

Recall that z was introduced in Section 1 as the length of the Mullineux symbol for λ, and the integers j_{v} = St(l_{v}) were defined before 5.3.

Lemma 5.5. Let (a−1)_{j} = {l_{1} ≺ ⋯ ≺ l_{r}} and q ≤ r. The subset L = {l_{1}, …, l_{q}} ⊆ (a − 1)_{j} is compensated in λ^{(j)} if and only if for every h ∈ [j_{q}, z],

|{l ∈ L | St(l) ≤ h}| ≤ |{k ∈ (a)_{j} | St(k) ≤ h}| (2)

Proof. Assume that L is compensated in λ^{(j)}. Then by Proposition 5.4, M ≥ q and h_{q} ≥ j (hence k_{q} ∈ (a)_{j}). Let h ∈ [j_{q}, z]. If w is minimal with respect to St(l_{w}) ≤ h, then the set on the left hand side of (2) is {l_{w}, …, l_{q}}. However the set on the right hand side of (2) contains {k_{w}, …, k_{q}} since St(k_{v}) ≤ St(l_{v}) by definition.

Conversely, assume that (2) holds for any h ∈ [j_{q}, z]. We want to show that M ≥ q and h_{q} ≥ j, and then to apply Proposition 5.4.

We may assume q ≥ 1. By induction on t = 1, 2, …, q, we show that M ≥ t, h_{t} ≥ j: Put h = j_{1} ∈ [j_{q}, z]. It follows from (2) that M ≥ 1, h_{1} ≥ j. Suppose t ≤ q and we have already proved that M ≥ t − 1, h_{t−1} ≥ j. Put h = j_{t} ∈ [j_{q}, z]. Then it follows from (2) that M ≥ t and h_{t} ≥ j (cf. Definition 5.3).

Notation. Let (a − 1)_{j} = {l_{1} ≺ ⋯ ≺ l_{r}}, q ≤ r and h ∈ [j_{q}, z]. We denote

L(h) = L_{j}(q, h) = {l ∈ {l_{1}, …, l_{q}} | St(l) ≤ h};
K(h) = K_{j}(h) = {k ∈ (a)_{j} | St(k) ≤ h}.

Lemma 5.6. |K(h)| − |L(h)| = (r − q) + ∑_{i=j}^{h} d_{i}(a).

Proof. Note that

|K(h)| = |a|_{j} − |a|_{h+1} and
|L(h)| = |a − 1|_{j} − |a − 1|_{h+1} − (r − q).

So

|K(h)| − |L(h)| = |a|_{j} − |a|_{h+1} − (|a − 1|_{j} − |a − 1|_{h+1}) + (r − q).

To complete the proof of the lemma it suffices to recall that d_{i}(a) = |a|_{i} − |a|_{i+1} − (|a − 1|_{i} − |a − 1|_{i+1}).

Corollary 5.7. Let (a−1)_{j} = {l_{1} ≺ ⋯ ≺ l_{r}}, q ≤ r. The subset L = {l_{1}, …, l_{q}} ⊆ (a − 1)_{j} is compensated in λ^{(j)} if and only if for every h ∈ [j, z],

(r − q) + ∑_{i=j}^{h} d_{i}(a) ≥ 0.

Proof. In view of Lemmas 5.5 and 5.6 we have only to prove that if L is compensated in λ^{(j)}, then (r − q) + ∑_{i=j}^{h} d_{i}(a) ≥ 0 for every h ∈ [j, j_{q}). For such h write

(r − q) + ∑_{i=j}^{h} d_{i}(a) = (r − q) + |a|_{j} − |a|_{h+1} − (|a − 1|_{j} − |a − 1|_{h+1})
≥ (r − q) − (|a − 1|_{j} − |a − 1|_{h+1}).

But

|a − 1|_{j} − |a − 1|_{h+1} = |{l ∈ {l_{1}, …, l_{r}} | St(l) ∈ [j, h]}|
≤ |{l ∈ {l_{1}, …, l_{r}} | St(l) ∈ [j, j_{q})}|
≤ |{l_{q+1}, …, l_{r}}| = r − q.

Corollary 5.8. Runner a of λ contains a degenerate good bead if and only if

∑_{i=0}^{h} d_{i}(a) > 0 for any h ∈ [0, z].

Proof. This follows immediately from Corollary 5.7 and Proposition 4.4

6. Neighbour runner sizes for λ and m(λ).

The purpose of this section is to prove Corollary 6.7, which expresses |A_{0} + ε_{0} − a|^{j} −|A_{0} +ε_{0} −a−1|^{j} in terms of the |a|_{j} −|a−1|_{j}. (Recall that we use superscripts to represent quantities associated with m(λ), and subscripts for λ.)

Let γ be a p-regular partition and let Γ be an abacus for γ. Throughout this section we use elements of the residue system [0, p) to represent residue classes modulo p, when discussing runner numbers.

Notation. Denote by Γ̃ the abacus obtained from Γ by moving all beads up along their runners as far as they can go.

Example. If

Γ = \begin{matrix}· & k & · & · & l \\ · & m & · & n & · \\ q & · & r & s & · \\ · & t & u & · & v \end{matrix}

then

Γ̃ = \begin{matrix}q & k & r & n & l \\ · & m & u & s & v \\ · & t \end{matrix}

We need to recall some notions concerning p-cores of Young diagrams. Details can be found in [11].

A rim p-hook of a Young diagram is a connected part of its rim, having length p, and such that after its removal, what remains is again a Young diagram. A partition is called a p-core if and only if its Young diagram does not have any rim p-hooks. For any partition γ one can remove rim p-hooks from its Young diagram one after another (in any order) until a diagram of a p-core is obtained. The latter partition (which is well defined, cf. [11]) is called the p-core of γ. We denote it by core(γ).

We make use of the following results.

Proposition 6.1. [9] Let Γ be an abacus for γ.

(i) There is a 1-1 correspondence between rim p-hooks of γ and beads m of Γ such that the position immediately above m (i.e. position P_{Γ}(m) − p) is unoccupied in Γ. Moreover, the abacus obtained from Γ by moving such a bead m up one position is an abacus for the partition obtained from γ by removing the corresponding rim p-hook.

(ii) γ is a p-core if and only if whenever position f of Γ is occupied, all positions

of the form f − vp are occupied (i.e. Γ = Γ̃). (iii) Γ̃ is an abacus for core(γ).

Proposition 6.2. [21] If γ is a p-regular partition then core(m(λ)) = core(λ)′ (recall that “′” means conjugation).

Notation. (i) For any abacus Γ we denote by {a}_{Γ} the number of all (not just

proper) beads in runner a of Γ. (ii) Let Γ be an abacus for a p-core, cf. Proposition 6.1(ii), and let

d = max_{a∈[0,p)}{a}_{Γ}.

We define an abacus Γ′ via the following:

(1) Γ′ is an abacus for a p-core;

(2) {a}_{Γ′} = d − {p − 1 − a}_{Γ} for all a ∈ [0, p).

Lemma 6.3. Let Γ be an abacus for a p-core partition γ. Then Γ′ is an abacus for γ′.

Proof. Let Γ have T beads. Then Γ is constructed from the sequence s_{T} (γ) = {s_{0}, s_{1}, …, s_{v}, … } of spaces and beads as explained in Section 1.

If U is such that s_{v} = “·” for any v > U, we define a new sequence s_{T,U}(γ) = {t_{0}, t_{1}, …, t_{v}, … } by setting

t_{v} = \begin{cases}“·” & if v > U or v ≤ U, s_{U−v} = “×” \\ “×” & otherwise. \end{cases}

Example. If γ = (4, 2^{2}, 1^{3}), T = 7, U = 15 then

s_{T} (γ) = × · × × × · × × · · ×,
s_{T,U}(γ) = × × × × · × × · · × ⋯ ×.

It follows from the definitions that s_{T,U}(γ) = s_{U−T} (γ′). Now it remains to notice that if U is minimal with respect to s_{v} = “·” for any v > U and U ≡ 0, then the abacus constructed from s_{T,U}(γ) is Γ′.

Lemma 6.4. Let Γ be an abacus with T beads, V of which are proper. Then

|a|_{Γ} − |a − 1|_{Γ} = \begin{cases}\{a\}_{Γ} − \{a − 1\}_{Γ} − 1, & if T − V ≢ 0, a ≡ 0; \\ \{a\}_{Γ} − \{a − 1\}_{Γ} + 1, & if T − V ≢ 0, a ≡ T − V; \\ \{a\}_{Γ} − \{a − 1\}_{Γ}, & otherwise. \end{cases}

Proof. The improper beads of Γ occupy positions from 0 to T − V − 1 in Γ. So if we write T − V = Qp + R with Q, R ∈ ℤ, 0 ≤ R < p, then

|a|_{Γ} = \begin{cases}\{a\}_{Γ} − Q, & if a ≡ x with 0 ≤ x < p, x ≥ R; \\ \{a\}_{Γ} − Q − 1, & if a ≡ x with 0 ≤ x < p, x < R. \end{cases}

The lemma follows.

Lemma 6.5. Let Γ_{1}, Γ_{2} be two abaci for γ with T_{1} and T_{2} beads, respectively. Then |a|_{Γ_{1}} = |a + T_{2} − T_{1}|_{Γ_{2}} for any a ∈ [0, p).

Proof. Assume without loss of generality that T_{2} ≥ T_{1}. Let s_{T_{1}}(γ) = {s_{0}, s_{1}, …, s_{v}, … }, s_{T_{2}}(γ) = {t_{0}, t_{1}, …, t_{v}, … }. Then t_{v} = “×” for v ∈ [0, T_{2} − T_{1}) and t_{v+T_{2}−T_{1}} = s_{v} for v ≥ 0. The lemma follows.

Proposition 6.6. Let Γ be the abacus with R beads for a p-regular partition γ and let Ω be the abacus with S beads for m(γ). Assume that r beads of Γ are proper and s beads of Ω are proper. For any a ∈ [0, p), put â ≡ R + S − a. Then

|â|_{Ω} − |â − 1|_{Ω} = \begin{cases}|a|_{Γ} − |a − 1|_{Γ} − 1, & if a ≡ R − r, r + s ≢ 0; \\ |a|_{Γ} − |a − 1|_{Γ} + 1, & if a ≡ R + s, r + s ≢ 0; \\ |a|_{Γ} − |a − 1|_{Γ}, & otherwise. \end{cases}

Proof. For any abacus Λ we let δ_{Λ}|a| = |a|_{Λ} − |a − 1|_{Λ} and δ_{Λ}{a} = {a}_{Λ} − {a − 1}_{Λ}.

Note that it suffices to prove that δ_{Ω}|â|−δ_{Γ}|a| ∈ {0, ±1} and that δ_{Ω}|â| = δ_{Γ}|a|+1 is equivalent to a ≡ R + s, r + s ≢ 0. Indeed, if this is proved then applying it to Γ and Ω interchanged we have that δ_{Ω}|â| = δ_{Γ}|a|−1 is equivalent to â ≡ S+r, r+s ≢ 0, which is in turn equivalent to a ≡ R − r, r + s ≢ 0 (since â ≡ R + S − a).

By definition we have

{a}_{Γ} = {a}_{Γ} = d − {p − 1 − a}_{Γ′}.

Hence

δ_{Γ′}{−a} = δ_{Γ}{a}. (3)

By Proposition 6.1(iii), Γ̃ is an abacus for core(γ). So Proposition 6.2 and Lemma 6.3 imply that Γ̃′ is an abacus for core(m(γ)). Let Γ̃′ have U beads. Then U ≡ −R since Γ̃ has R beads.

Let Ω_{1} be an abacus for m(γ) having V ≡ −R beads. Then by Proposition 6.1(iii), Ω̃_{1} is an abacus for core(m(γ)) having V beads. Since V ≡ U, we have, for any b ∈ [0, p):

δ_{Ω_{1}}{b} = δ_{Γ′}{b}. (4)

Since {b}_{Ω_{1}} = {b}_{Ω_{1}}, equations (3) and (4) give us

δ_{Ω_{1}}{−a} = δ_{Γ}{a}. (5)

Now we use Lemma 6.4. We have

(6)  δ_{Ω_{1}}|−a| = \begin{cases}δ_{Ω_{1}}\{−a\} − 1, & if −R − s ≢ 0, −a ≡ 0; \\ δ_{Ω_{1}}\{−a\} + 1, & if −R − s ≢ 0, −a ≡ −R − s; \\ δ_{Ω_{1}}\{−a\}, & otherwise; \end{cases}

and

(7)  δ_{Γ}\{a\} = \begin{cases}δ_{Γ}|a| + 1, & if R − r ≢ 0, a ≡ 0; \\ δ_{Γ}|a| − 1, & if R − r ≢ 0, a ≡ R − r; \\ δ_{Γ}|a|, & otherwise. \end{cases}

It follows from (5)–(7) that δ_{Ω_{1}}|−a| − δ_{Γ}|a| ∈ {0, ±1, ±2}. Moreover, δ_{Ω_{1}}|−a| = δ_{Γ}|a| + 2 if and only if −R − s ≢ 0, −a ≡ −R − s and R − r ≢ 0, a ≡ 0, which is impossible. Similarly δ_{Ω_{1}}| − a| = δ_{Γ}|a| − 2 is impossible.

Let us prove that δ_{Ω_{1}}| − a| = δ_{Γ}|a| + 1 if and only if a ≡ R + s, r + s ≢ 0. Then the proposition will follow since δ_{Ω}|−a + S − V | = δ_{Ω_{1}}|−a| in view of Lemma 6.5, and −a + S − V ≡ −a + S + R ≡ â.

It follows from (5)–(7) that δ_{Ω_{1}}| − a| = δ_{Γ}|a| + 1 if and only if one of the following conditions holds.

(1) −R − s ≢ 0, −a ≡ −R − s, R − r ≡ 0;

(2) −R − s ≢ 0, −a ≡ −R − s, R − r ≢ 0, a ≢ 0, a ≢ R − r;

(3) −R − s ≡ 0, R − r ≢ 0, a ≡ 0;

(4) −R − s ≢ 0, −a ≢ 0, −a ≢ −R − s, R − r ≢ 0, a ≡ 0. Now it suffices to remark that

(1) is equivalent to a ≡ R + s, r + s ≢ 0, a ≡ r + s;

(2) is equivalent to a ≡ R + s, r + s ≢ 0, a ≢ 0, a ≢ r + s;

(3) is equivalent to a ≡ R + s, r + s ≢ 0, a ≡ 0;

(4) is impossible.

Corollary 6.7. Put â = A_{0} + ε_{0} − a. Then

|â|^{j} − |â − 1|^{j} = \begin{cases}|a|_{j} − |a − 1|_{j} − 1, & if a ≡ R_{0} − R_{j}, A_{j} ≢ 0, A_{j} ≢ −1; \\ |a|_{j} − |a − 1|_{j} + 1, & if a ≡ R_{0} + S_{j}, A_{j} ≢ 0, A_{j} ≢ −1; \\ |a|_{j} − |a − 1|_{j}, & otherwise. \end{cases}

Proof. Recall that R_{j} + S_{j} = A_{j} + ε_{j}. Moreover, A_{j} + ε_{j} ≢ 0 if and only if A_{j} ≢ 0, A_{j} ≢ −1. Now it suffices to recall that λ^{(j)} (resp. m(λ)^{(j)}) is the abacus with R_{0} (resp. S_{0}) beads, R_{j} (resp. S_{j}) of which are proper, and apply Proposition 6.6.

7. The occurrence of degenerate beads for λ and m(λ).

The main result of this section says that runner a of λ has a degenerate good bead if and only if runner â of m(λ) does. Throughout, j ∈ [0, z] and â ≡ A_{0} + ε_{0} − a.

Lemma 7.1. Let d_{j}(a) > 0. Then d^{j}(â) ≥ d_{j}(a) − 1, and if d^{j}(â) = d_{j}(a) − 1 then

j < z, a ≡ R_{0} + S_{j+1}, A_{j+1} ≢ 0, and A_{j+1} ≢ −1.

Proof. If a ≡ R_{0} − R_{j} and A_{j} ≢ 0, then it follows from Lemmas 2.5, 2.6, and 2.7 that d_{j}(a) ≤ 0, giving a contradiction. So Corollary 6.7 implies

|â|^{j} − |â − 1|^{j} = |a|_{j} − |a − 1|_{j} + x

where x = 1 if a ≡ R_{0} + S_{j}, A_{j} ≢ 0, A_{j} ≢ −1, and x = 0 otherwise. Again by 6.7 we have

|â|^{j+1} − |â − 1|^{j+1} − (|a|_{j+1} − |a − 1|_{j+1}) ∈ {0, ±1}.

Summarizing, we have d^{j}(â) ≥ d_{j}(a) − 1; and d^{j}(â) = d_{j}(a) − 1 implies

|â|^{j+1} − |â − 1|^{j+1} = |a|_{j+1} − |a − 1|_{j+1} + 1.

Therefore j+1 ≤ z and, in view of 6.7, a ≡ R_{0}+S_{j+1}, A_{j+1} ≢ 0, and A_{j+1} ≢ −1.

Lemma 7.2. Let d^{j}(â) < 0. Then d_{j}(a) ≤ 0, and if d_{j}(a) > d^{j}(â) then either

a ≡ R_{0} − R_{j}, A_{j} ≢ 0, and A_{j} ≢ −1

or

j < z, a ≡ R_{0} + S_{j+1}, A_{j+1} ≢ 0, and A_{j} ≢ −1.

Moreover, if d_{j}(a) = d^{j}(â) + 2 then both of the above conditions hold.

Proof. If d_{j}(a) > 0 then d^{j}(â) ≥ 0 by the previous lemma. So d_{j}(a) ≤ 0. The remainder of the lemma follows from Corollary 6.7.

Lemma 7.3. If

a ≡ R_{0} + S_{j}, A_{j} ≢ 0, A_{j} ≢ −1

then d_{j}(a) < 0 and either

d^{j}(â) > d_{j}(a)

or

d^{j}(â) = d_{j}(a), j < z, a ≡ R_{0} + S_{j+1}, A_{j+1} ≢ 0, A_{j+1} ≢ −1.

Proof. We have a ≡ R_{0} + S_{j} = R_{0} + A_{j} + ε_{j} − R_{j} = R_{0} − R_{j} + A_{j} + 1, with A_{j} ≢ 0. So by Lemmas 2.8(iii) and 2.9, we have d_{j}(a) < 0. The remainder follows from Corollary 6.7.

Lemma 7.4. Let j > 0, a ≡ R_{0} − R_{j}, A_{j} ≢ 0, and A_{j} ≢ −1. Then one of the following occurs.

(i) d_{j−1}(a) < 0 and either

d^{j−1}(â) > d_{j−1}(a)

or

d^{j−1}(â) = d_{j−1}(a), a ≡ R_{0} − R_{j−1}, A_{j−1} ≢ 0, and A_{j−1} ≢ −1.

(ii) d_{j−1}(a) ≥ 0 and d^{j−1}(â) > d_{j−1}(a).

Proof. By Corollary 6.7, we have d^{j−1}(â) = d_{j−1}(a) + x where x = 0 if a ≡ R_{0} − R_{j−1}, A_{j−1} ≢ 0, A_{j−1} ≢ −1 and x > 0 otherwise. If d_{j−1}(a) < 0 we get (i). Let d_{j−1}(a) ≥ 0. We have to show that x > 0.

If on the contrary x = 0, then a ≡ R_{0} −R_{j−1}, A_{j−1} ≢ 0, and A_{j−1} ≢ −1. Since by assumption a ≡ R_{0} − R_{j}, we get R_{j−1} − R_{j} ≡ 0. Thus A_{j−1} ≢ 0, A_{j−1} ≢ R_{j−1} − R_{j}, A_{j−1} ≢ R_{j−1} − R_{j} − 1, and a ≡ R_{0} − R_{j}. It follows from 2.8(ii) that d_{j−1}(a) = −1, giving a contradiction.

Lemma 7.5. Let

j > 0, a ≡ R_{0} + S_{j−1}, A_{j−1} ≢ 0, A_{j−1} ≢ −1

and

a ≡ R_{0} − R_{j}, A_{j} ≢ 0, A_{j} ≢ −1.

Then d_{j−1}(a) = −2 and d^{j−1}(â) = 0.

Proof. It follows from Corollary 6.7 that d^{j−1}(â) = d_{j−1}(a) + 2. If d_{j−1}(a) > −2 then d^{j−1}(â) > 0. But by Lemma 7.1 (applied to m(λ) in place of λ), d^{j−1}(â) > 0 implies d_{j−1}(a) ≥ d^{j−1}(â) − 1, giving a contradiction.

In the following two propositions we write d_{j} for d_{j}(a) and d^{j} for d^{j}(â).

Proposition 7.6. Let d_{0} > 0 and

{j_{1} < ⋯ < j_{f}} = {j ∈ [0, z] | d^{j} < d_{j}}.

For every t ∈ [1, f], there exist either x_{t} or y_{t} or both such that

(i) 0 ≤ x_{t} < j_{t}, d^{x_{t}} > d_{x_{t}}, d^{v} = d_{v} for all v ∈ (x_{t}, j_{t}); (ii) j_{t} < y_{t} ≤ z, d^{y_{t}} > d_{y_{t}}, d^{v} = d_{v} for all v ∈ (j_{t}, y_{t}), and d_{v} < 0 for all

v ∈ (j_{t}, y_{t}]; (iii) If d^{j_{t}} = d_{j_{t}} − 2 then both x_{t} and y_{t} exist; (iv) If y_{t} exists and t < f then y_{t} < j_{t+1}. If x_{t} exists and t > 0 then x_{t} > j_{t−1}.

(v) If y_{t} and x_{t+1} both exist then y_{t} ≤ x_{t+1}, and y_{t} = x_{t+1} implies d_{y_{t}} = −2, d^{y_{t}} = 0.

Proof. Let t ∈ [1, f]. If d_{j_{t}} > 0 then by Lemma 7.1,

d^{j_{t}} = d_{j_{t}} − 1, j_{t} < z, a ≡ R_{0} + S_{j_{t}+1}, A_{j_{t}+1} ≢ 0, and A_{j_{t}+1} ≢ −1.

So Lemma 7.3 implies that d_{j_{t}+1} < 0 and either

d^{j_{t}+1} > d_{j_{t}+1},

in which case we put y_{t} = j_{t} + 1, or

d^{j_{t}+1} = d_{j_{t}+1}, j_{t} + 1 < z, a ≡ R_{0} + S_{j_{t}+2}, A_{j_{t}+2} ≢ 0, and A_{j_{t}+2} ≢ −1.

Applying Lemma 7.3 repeatedly we find an integer y_{t} such that

j_{t} < y_{t} ≤ z, a ≡ R_{0} + S_{v}, A_{v} ≢ 0, A_{v} ≢ −1, d_{v} < 0

for all v ∈ (j_{t}, y_{t}], d^{v} = d_{v} for all v ∈ (j_{t}, y_{t}), and d^{y_{t}} > d_{y_{t}}. (8)

Now assume d_{j_{t}} ≤ 0. Then j_{t} > 0 (since d_{0} > 0 by assumption), and d^{j_{t}} < 0. By Lemma 7.2, either

a ≡ R_{0} − R_{j_{t}}, A_{j_{t}} ≢ 0, A_{j_{t}} ≢ −1

or

j_{t} < z, a ≡ R_{0} + S_{j_{t}+1}, A_{j_{t}+1} ≢ 0, A_{j_{t}+1} ≢ −1,

and both of these conditions hold if d^{j_{t}} = d_{j_{t}} − 2.

If the latter condition holds then applying Lemma 7.3 repeatedly as above we show that (8) holds.

If the former condition is true then by Lemma 7.4 we have either

d^{j_{t}−1} > d_{j_{t}−1},

in which case we put x_{t} = j_{t} − 1, or

d^{j_{t}−1} = d_{j_{t}−1} < 0, a ≡ R_{0} − R_{j_{t}−1}, A_{j_{t}−1} ≢ 0, and A_{j_{t}−1} ≢ −1.

The assumption d_{0} > 0 now implies j_{t}−1 > 0. Applying these arguments repeatedly we find an integer x_{t} ≥ 0 such that

a ≡ R_{0} − R_{v}, A_{v} ≢ 0, A_{v} ≢ −1 for all v ∈ (x_{t}, j_{t}],

d^{v} = d_{v} for all v ∈ (x_{t}, j_{t}), and d^{x_{t}} > d_{x_{t}}. (9)

Assume that y_{t} exists for some t < f. Since d^{v} = d_{v} for any v ∈ (j_{t}, y_{t}) and d^{y_{t}} > d_{y_{t}}, we get y_{t} < j_{t+1}. Similarly x_{t} > j_{t−1}.

Assume that both y_{t} and x_{t+1} exist. Then d^{v} = d_{v} for any v ∈ (j_{t}, y_{t})∪(x_{t+1}, j_{t+1}), d^{y_{t}} > d_{y_{t}}, and d^{x_{t}} > d_{x_{t}} imply y_{t} ≤ x_{t+1}.

Finally notice that if y_{t} = x_{t+1} =: u then it follows from (8) and (9) that a ≡ R_{0} + S_{u}, A_{u} ≢ 0, A_{u} ≢ −1 and a ≡ R_{0} − R_{u+1}, A_{u+1} ≢ 0, A_{u+1} ≢ −1. So it suffices to apply Lemma 7.5 to complete the proof of the proposition.

Proposition 7.7. Assume that S_{h} := d_{0} +d_{1} +⋯+d_{h} > 0 for any h ∈ [0, z]. Then S^{h} := d^{0} + d^{1} + ⋯ + d^{h} > 0 for any h ∈ [0, z].

Proof. Since d_{0} = S_{0} > 0 we may apply Proposition 7.6. Let

{j_{1} < ⋯ < j_{f}} = {j ∈ [0, z] | d^{j} < d_{j}}

and x_{t}, y_{t} be as in 7.6.

First, we prove by induction on h the following intermediate fact.

(∗) S^{h} ≥ S_{h} − 1, and if S^{h} = S_{h} − 1 then h ∈ [j_{t}, y_{t}) for some t such that y_{t} exists.

If h = 0 and d^{0} = S^{0} < S_{0} = d_{0} then Lemma 7.1 implies S^{0} = S_{0} − 1. Moreover, we have j_{1} = 0 and, by Proposition 7.6, y_{1} exists.

Let h > 0 and assume that S^{h} < S_{h}.

If h ∉ {j_{1}, …, j_{f}}, i.e. d^{h} ≥ d_{h}, then S^{h−1} < S_{h−1}. By the inductive hypothesis we have S^{h−1} = S_{h−1} − 1, and h − 1 ∈ [j_{t}, y_{t}) for some t such that y_{t} exists. If h − 1 < y_{t} − 1 then h ∈ (j_{t}, y_{t}), hence d^{h} = d_{h} and S^{h} = S_{h} − 1, h ∈ [j_{t}, y_{t}), as desired. If h − 1 = y_{t} − 1 then h = y_{t}, d^{h} > d_{h}, and S^{h} ≥ S_{h}, contrary to our assumption that S^{h} < S_{h}.

Now assume that h = j_{t} for some t ∈ [1, f].

We first show that if x_{t} exists then S^{h−1} > S_{h−1}. Indeed, we prove first that S^{x_{t}} > S_{x_{t}}. By Proposition 7.6, we have that one of the following holds:

• y_{t−1} does not exist;

• y_{t−1} < x_{t}; or

• y_{t−1} = x_{t}, d^{x_{t}} = 0, d_{x_{t}} = −2. In the first two cases x_{t−1} ∉ [j_{v}, y_{v}) for any v in view of 7.6. So by the inductive

hypothesis we get S^{x_{t}−1} = S_{x_{t}−1} or x_{t} = 0. So d^{x_{t}} > d_{x_{t}} implies S^{x_{t}} > S_{x_{t}}. In the third case by the inductive hypothesis we get S^{x_{t}−1} ≥ S_{x_{t}−1} − 1 or x_{t} = 0. So d^{x_{t}} = d_{x_{t}} + 2 implies S^{x_{t}} > S_{x_{t}} again. Now, since d^{v} = d_{v} for any v ∈ (x_{t}, j_{t}) we conclude that S^{h−1} > S_{h−1}.

If d^{j_{t}} = d_{j_{t}} − 2 then by Proposition 7.6 both x_{t} and y_{t} exist and it remains to notice that S^{h} = S^{h−1} + d^{j_{t}} ≥ S_{h−1} + 1 + d_{j_{t}} − 2 = S_{h} − 1.

Let d^{j_{t}} ≥ d_{j_{t}} − 1. If x_{t} exists then it follows from S^{h−1} > S_{h−1} that S^{h} ≥ S_{h}. Assume that x_{t} does not exist. Then by Proposition 7.6 y_{t} exists, and we have only to prove that S^{h} ≥ S_{h} −1. Note that S^{h−1} ≥ S_{h−1} by the inductive hypothesis since h−1 ∉ [j_{v}, y_{v}) for any v such that y_{v} exists. Indeed, if on the contrary h−1 ∈ [j_{v}, y_{v}) then j_{t} = h ∈ (j_{v}, y_{v}] whence d^{j_{t}} ≥ d_{j_{t}}, giving a contradiction. Now S^{h} ≥ S_{h} − 1 follows.

Thus (∗) is completely proved.

Now assume that there exists some h with S^{h} ≤ 0. If S^{h} < 0 then S_{h} ≤ S^{h} + 1 implies S_{h} ≤ 0, giving a contradiction. So S^{h} = 0, S_{h} = 1, and h ∈ [j_{t}, y_{t}) for some t such that y_{t} exists. Then by Proposition 7.6(ii), h + 1 ≤ z, d_{h+1} < 0, whence S_{h+1} ≤ 0, again giving a contradiction.

Corollary 7.8. Runner a of λ contains a degenerate good bead if and only if runner â of m(λ) contains a degenerate good bead.

Proof. It follows from Proposition 7.7 that ∑_{i=0}^{h} d_{i}(a) > 0 for any h ∈ [0, z] if and only if ∑_{i=0}^{h} d^{i}(â) > 0 for any h ∈ [0, z]. Now it suffices to apply Corollary 5.8.

8. Proof of the main theorem

In this section we first translate some of the results obtained above from the language of abaci to the language of Young diagrams, and then prove the main theorem.

We recall some notions from [16]:

For a partition γ = (l_{1} ≥ l_{2} ≥ ⋯ ≥ l_{m} > 0) we consider its Young diagram as the subset

γ = {(i, j) ∈ ℤ × ℤ | 1 ≤ i, 1 ≤ j ≤ l_{i}} ⊆ ℤ × ℤ,

where we imagine the i-axis increasing downward and the j-axis increasing to the right.

All elements (i, j) ∈ ℤ × ℤ are called nodes, while the elements (i, j) ∈ γ are called nodes of the diagram γ.

If A = (i, j) is a node we denote the residue class (j − i) (mod p) by res A and call it the residue of A.

If A = (i, j) and B = (i′, j′) are nodes we say that B is above A (or that A is below B) if and only if i′ < i. In this case we write A ↗ B.

Young diagrams μ and ν are said to have the same residue content if and only if for any residue α modulo p the number of nodes of residue α in μ is equal to the number of nodes of residue α in ν.

If l_{i} > l_{i+1} then the node (i, l_{i}) is called a removable node for γ (l_{m+1} is interpreted as 0).

We call (i, j) an indent node for γ if i = 1 and j = l_{1} + 1, or if i > 1, l_{i} < l_{i−1}, and j = l_{i} + 1.

If A = (i, l_{i}) is a removable node for γ we denote by γ_{A} the partition (l_{1}, …, l_{i−1}, l_{i}− 1, l_{i+1}, …) of n − 1 whose Young diagram of is γ \ {A}. T_{A}(λ)

Definition 8.1. Let A be a removable node for γ. Set

M_{A} = {B | B is an indent node for γ, A ↗ B, res B = res A}.

We call A normal if and only if for any B ∈ M_{A} there exists a removable node C(B) such that the following three conditions hold:

(1) A ↗ C(B) ↗ B;

(2) res C(B) = res A;

(3) |{C(B) | B ∈ M_{A}}| = |M_{A}|.

We call the node A good if it is normal and A ↗ D for any normal node D with res D = res A.

We call a good node A for γ degenerate if and only if (λ_{A})^{(1)} = λ^{(1)}. Otherwise we call it non-degenerate.

Lemma 8.2. Let γ be a p-regular partition and let Γ be an abacus for γ with R beads. Assume B_{1} ↗ … ↗ B_{q} are all the removable nodes for γ and C_{1} ↗ … ↗ C_{q} are the top q indent nodes for γ (there are q + 1 indent nodes for γ). Then there exist q ends and q beginnings for Γ.

Moreover, let k_{1}, …, k_{q} and l_{1}, …, l_{q} be all the beginnings and ends, respectively, for Γ, with P_{Γ}(k_{i}) < P_{Γ}(k_{i+1}) and P_{Γ}(l_{i}) < P_{Γ}(l_{i+1}) for i ∈ [1, q). Then for any i ∈ [1, q] the following hold.

(i) Γ(k_{i}) is an abacus for γ_{B_{i}}; (ii) k_{i} ∈ (a)_{Γ} if and only if res B_{i} = a − R; (iii) l_{i} ∈ (a − 1)_{Γ} if and only if res C_{i} = a − R; (iv) k_{i} is a normal bead for Γ if and only if B_{i} is a normal node for γ;

(v) k_{i} is a good bead for Γ if and only if B_{i} is a good node for γ; (vi) k_{i} is a degenerate good bead for Γ if and only if B_{i} is a degenerate good node

for γ.

Proof. If we represent γ in the canonical form (1) from the introduction, then k = q and we can write

γ = (ν_{1}^{α_{1}}, …, ν_{q}^{α_{q}}), ν_{1} > ⋯ > ν_{q} > 0, α_{i} > 0.

The numbers of beginnings and ends for Γ are both equal to the number of strings of proper beads in Γ which in turn equals q.

Part (i) follows immediately from the definitions.

Note that

B_{q+1−i} = (α_{1} + ⋯ + α_{i}, ν_{i})

and

P_{Γ}(k_{q+1−i}) = (R − (α_{1} + ⋯ + α_{q}) − 1) + ν_{i} + (α_{i+1} + ⋯ + α_{q}) + 1
= R + ν_{i} − (α_{1} + ⋯ + α_{i}).

This implies (ii), and (iii) is obtained similarly. The remaining parts of the lemma follow from (i)–(iii) and the definitions.

Let λ be a p-regular partition of n. Recall the Mullineux symbols

G_{p}(λ) = (\begin{matrix}A_{0} & A_{1} & … & A_{z} \\ R_{0} & R_{1} & … & R_{z} \end{matrix}) and G_{p}(m_{n}(λ)) = (\begin{matrix}A_{0} & A_{1} & … & A_{z} \\ S_{0} & S_{1} & … & S_{z} \end{matrix}) .

Lemma 8.3. (i) If A is a non-degenerate good node for λ of residue α then there exists a good node C for λ^{(1)}, with res C = α and (λ_{A})^{(1)} = (λ^{(1)})_{C}. Moreover, in this case the first column of the Mullineux symbol G_{p}(λ_{A}) is

(\begin{matrix}A_{0} \\ R_{0} \end{matrix}).

(ii) There exists a degenerate good node of residue α for λ if and only if there

exists a degenerate good node of residue −α for m(λ). (iii) If A is a degenerate good node for λ then the first column of the Mullineux

symbol G_{p}(λ_{A}) is equal to (\begin{matrix}A_{0} − 1 \\ R_{0} − ξ \end{matrix}) where ξ = 0 or ξ = 1.

(iv) Let A be a degenerate good node for λ and ξ be defined as in (iii). Then

A_{0} ≡ 1 implies ξ = 1, and A_{0} ≡ 0 implies ξ = 0.

(v) If there exists a good node of residue α for λ^{(1)} then there exists a good node of residue α for λ.

Proof. According to Lemma 8.2, there exists a good node of residue α for λ (resp., λ^{(1)}) if and only if there exists a good bead in runner a = α + R_{0} of the canonical abacus for λ (resp., abacus with R_{0} beads for λ^{(1)}). By the same lemma, this good node is degenerate if and only if the corresponding good bead is so.

Similarly, to a (degenerate) good node of residue −α for m(λ) there corresponds a (degenerate) good bead in runner −α + S_{0} of the canonical abacus for m(λ).

Notice that â ≡ R_{0} + S_{0} − a ≡ −α + S_{0}. Now part (i) follows from Lemma 4.3(i) and (iii); part (ii), from Corollary 7.8; part (iii), from Lemma 4.3(iii); part (iv), from Lemma 4.3(iv); and, finally, part (v) follows from Corollary 3.3.

In the proof of 8.7 we shall need the following three results.

Lemma 8.4 ([11, 2.7.41]). Suppose that ν and μ are partitions of the same integer. Then core(ν) = core(μ) if and only if the Young diagrams of ν and μ have the same residue content.

Lemma 8.5. Let ν and μ be p-regular partitions with

G_{p}(ν) = (\begin{matrix}A_{0} & A_{1} & … & A_{z} \\ R_{0} & R_{1} & … & R_{z} \end{matrix}) , G_{p}(μ) = (\begin{matrix}A_{0} & A_{1} & … & A_{z} \\ R_{0} − 1 & R_{1} & … & R_{z} \end{matrix}) .

If A_{0} ≢ 0 then core(ν) ≠ core(μ).

Proof. We have ν^{(1)} = μ^{(1)}. Since A_{0} ≢ 0 we get from 1.1 that R_{0} − 1 − R_{1} > 0, or R_{0} ≥ R_{1} + 2. Let A_{0} = pd + r, with d, r ∈ ℤ, r ∈ (0, p).

Any p-segment of length p contains exactly one node of every residue. Moreover, the bottom p-segments of both ν and μ contain r nodes. Since R_{0} ≥ R_{1} + 2, the bottom p-segment of ν contains nodes of residues −(R_{0} −1)+x, x ∈ [0, r), while the bottom p-segment of μ contains nodes of residues −(R_{0} − 2) + x, x ∈ [0, r). Since 0 < r < p, we conclude that ν and μ have distinct residue contents. An application of Lemma 8.4 completes the proof.

Lemma 8.6. Let γ be a p-regular partition, and α be a residue modulo p. Then the number of nodes in γ of residue α is equal to the number of the nodes in m(γ) of residue −α.

Proof. By Proposition 6.2, core(m(γ)) = (core(γ))′. So the number of the nodes of residue −α in core(m(γ)) is equal to the number of the nodes of residue α in core(γ). Now it remains to notice that any rim p-hook contains exactly one node of every residue.

Notation. If (\begin{matrix}B_{0} & B_{1} & … & B_{z} \\ T_{0} & T_{1} & … & T_{z} \end{matrix}) is the Mullineux symbol of a p-regular partition

γ, then for any i ∈ [0, z] we write:

ε(B_{i}) = \begin{cases}0, & if p|B_{i}, \\ 1, & otherwise; \end{cases}

f(B_{i}, T_{i}) = B_{i} + ε(B_{i}) − T_{i}.

Then by definition

(\begin{matrix}B_{0} & B_{1} & … & B_{z} \\ f(B_{0}, T_{0}) & f(B_{1}, T_{1}) & … & f(B_{z}, T_{z}) \end{matrix})

is the Mullineux symbol for m(γ).

Theorem 8.7. Suppose A is a good node for a p-regular partition λ of n, with res A = α. Then there exists a good node B for m(λ) with res B = −α, and m(λ_{A}) = m(λ)_{B}.

Proof. We prove the theorem by induction on n. If n = 1, it is trivial. Assume n > 1 and the theorem has been proved for all n′ < n. Put

G_{p}(λ) = (\begin{matrix}A_{0} & A_{1} & … & A_{z} \\ R_{0} & R_{1} & … & R_{z} \end{matrix}) , G_{p}(λ_{A}) = (\begin{matrix}B_{0} & B_{1} & … & B_{x} \\ T_{0} & T_{1} & … & T_{x} \end{matrix}) .

Then

G_{p}(m(λ)) = (\begin{matrix}A_{0} & A_{1} & … & A_{z} \\ S_{0} & S_{1} & … & S_{z} \end{matrix}) , G_{p}(m(λ_{A})) = (\begin{matrix}B_{0} & B_{1} & … & B_{x} \\ U_{0} & U_{1} & … & U_{x} \end{matrix})

where S_{i} = f(A_{i}, R_{i}) for i ∈ [0, z], and U_{j} = f(B_{j}, T_{j}) for j ∈ [0, x].

Suppose first that A is non-degenerate. Then, in view of Lemma 8.3(i), there is a good node C for λ^{(1)}, with res C = α and (λ_{A})^{(1)} = (λ^{(1)})_{C}. By the inductive hypothesis there is a good node D for m(λ^{(1)}) = m(λ)^{(1)}, with res D = −α and m((λ^{(1)})_{C}) = m(λ^{(1)})_{D}. According to Lemma 8.3(v), there is a good node B for m(λ) with res B = −α. In view of 8.3(ii), B is non-degenerate. Therefore (m(λ)_{B})^{(1)} = (m(λ)^{(1)})_{D}. Let

G_{p}(m(λ)_{B}) = (\begin{matrix}C_{0} & C_{1} & … & C_{y} \\ V_{0} & V_{1} & … & V_{y} \end{matrix}) .

Since

m((λ_{A})^{(1)}) = m((λ^{(1)})_{C}) = m(λ^{(1)})_{D} = (m(λ)_{B})^{(1)},

we have x = y and B_{i} = C_{i}, U_{i} = V_{i} for i ∈ [1, x]. By Lemma 8.3(i), B_{0} = A_{0}, T_{0} = R_{0} and C_{0} = A_{0}, V_{0} = S_{0}. So B_{0} = C_{0}. It suffices to use the equalities U_{0} = f(B_{0}, T_{0}) and S_{0} = f(A_{0}, R_{0}) to prove that U_{0} = V_{0}. Thus G_{p}(m(λ)_{B}) = G_{p}(m(λ_{A})), hence m(λ)_{B} = m(λ_{A}).

Assume that A is degenerate. By Lemma 8.3(ii), there is a degenerate good node B for m(λ) with res B = −α.

By definition, (λ_{A})^{(1)} = (λ)^{(1)}. So x = z, B_{i} = A_{i}, and T_{i} = R_{i} for i ∈ [1, z]. Moreover, it follows from Lemma 8.3, parts (iii) and (iv), that B_{0} = A_{0} − 1, T_{0} = R_{0} − ξ where ξ ∈ {0, 1}, and A_{0} ≡ 1 implies ξ = 1, A_{0} ≡ 0 implies ξ = 0. Thus,

G_{p}(m(λ_{A})) = (\begin{matrix}A_{0} − 1 & A_{1} & … & A_{z} \\ f(A_{0} − 1, R_{0} − ξ) & S_{1} & … & S_{z} \end{matrix}) .

Similarly

G_{p}(m(λ)_{B}) = (\begin{matrix}A_{0} − 1 & A_{1} & … & A_{z} \\ S_{0} − η & S_{1} & … & S_{z} \end{matrix})

where η ∈ {0, 1}, and A_{0} ≡ 1 implies η = 1, A_{0} ≡ 0 implies η = 0.

We consider three cases.

(a) A_{0} ≡ 0. Then

ξ = 0, η = 0;
S_{0} = f(A_{0}, R_{0}) = A_{0} + ε(A_{0}) − R_{0} = A_{0} − R_{0};
f(A_{0} − 1, R_{0} − ξ) = f(A_{0} − 1, R_{0}) = A_{0} − 1 + ε(A_{0} − 1) − R_{0}
= A_{0} − 1 + 1 − R_{0} = A_{0} − R_{0} = S_{0} − η.
Thus G_{p}(m(λ)_{B}) = G_{p}(m(λ_{A})), i.e. m(λ)_{B} = m(λ_{A}).
(b) A_{0} ≡ 1. Then
ξ = 1, η = 1;
S_{0} = f(A_{0}, R_{0}) = A_{0} + ε(A_{0}) − R_{0} = A_{0} + 1 − R_{0};
f(A_{0} − 1, R_{0} − ξ) = f(A_{0} − 1, R_{0} − 1) = A_{0} − 1 + ε(A_{0} − 1) − (R_{0} − 1)
= A_{0} − 1 − R_{0} + 1 = A_{0} − R_{0} = S_{0} − η.
Thus G_{p}(m(λ)_{B}) = G_{p}(m(λ_{A})), i.e. m(λ)_{B} = m(λ_{A}).

(c) A_{0} ≢ 0, A_{0} ≢ 1. Then

S_{0} = f(A_{0}, R_{0}) = A_{0} + ε(A_{0}) − R_{0} = A_{0} + 1 − R_{0};

f(A_{0} − 1, R_{0} − ξ) = f(A_{0} − 1, R_{0} − ξ) = A_{0} − 1 + ε(A_{0} − 1) − (R_{0} − ξ)
= A_{0} − 1 + 1 − R_{0} + ξ = A_{0} − R_{0} + ξ.

Thus

S_{0} − η = f(A_{0} − 1, R_{0} − ξ) + 1 − η − ξ,

which gives

f(A_{0} − 1, R_{0} − ξ) − 1 ≤ S_{0} − η ≤ f(A_{0} − 1, R_{0} − ξ) + 1.

We have core(m(λ_{A})) = core(m(λ)_{B}) in view of Lemmas 8.4 and 8.6. Now Lemma 8.5 implies S_{0} −η = f(A_{0} −1, R_{0} −ξ), i.e. G_{p}(m(λ)_{B}) = G_{p}(m(λ_{A})), or m(λ)_{B} = m(λ_{A}).

Thus we have proved Conjecture 3. Now the Main Theorem follows from Theorem 4 (cf. Introduction).

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Department of Mathematics GN-50, University of Washington, Seattle, Washington 98195, U.S.A.

E-mail address: ford@math.washington.edu

Department of Mathematics, University of Oregon, Eugene OR 97403, U.S.A. E-mail address: klesh@math.uoregon.edu