Relationship Between Mullineux Involution and the Generalized Regularization

TL;DR

This paper explores the relationship between the Mullineux involution and generalized regularization on partitions, establishing conditions for their equivalence and connecting these concepts to algebraic structures like Iwahori-Hecke algebras.

Contribution

It generalizes previous work by characterizing when the Mullineux transpose map coincides with the generalized column regularization, linking combinatorics to algebraic representations.

Findings

Identifies conditions where Mullineux transpose equals generalized column regularization.

Connects combinatorial maps to Iwahori-Hecke algebra and crystal basis theory.

Proposes conjectures on q-decomposition numbers and further generalizations.

Abstract

The Mullineux involution is an important map on $p$-regular partitions that originates from the modular representation theory of $\mathcal{S}_n$. In this paper we study the Mullineux transpose map and the generalized column regularization and prove a condition under which the two maps are exactly the same. Our results generalize the work of Bessenrodt, Olsson and Xu, and the combinatorial constructions is related to the Iwahori-Hecke algebra and the global crystal basis of the basic $U_q(\widehat{\mathfrak{sl}}_b)$-module. In the conclusion, we provide several conjectures regarding the $q$-decomposition numbers and generalizations of results due to Fayers.

Equations

$${\rho }_{{\lambda }^{M_p}}={\rho }_{\lambda }\otimes \mathrm{sgn}.$$

$$\left(\frac{b}{a}-1\right)l_{i,j}<a_{i,j}+1,$$

$$\{(i,j)\in \mathbb{N}\times \mathbb{N}:1\le j,1\le i\le {\lambda }_j\}.$$

$$\left(\frac{b}{a}-1\right)l_{i,j}<a_{i,j}+1.$$

$$\left(\frac{b}{a}-1\right)a_{i,j}<l_{i,j}+1.$$

$${\lambda }^{J_b}:=({\mu }_1+1,...,{\mu }_{k-1}+1,{\mu }_k+\delta ),$$

$$\delta =\{\begin{array}{l}0\text{ if }b\nmid \varphi (\lambda )\\ 1\text{ if }b\mid \varphi (\lambda )\end{array}.$$

$$j_i=|{\lambda }^{J_b^{i-1}}|-|{\lambda }^{J_b^i}|.$$

$${\lambda }^{M_{b}T}=(|\lambda |-|{\lambda }^{J_b}|)\oplus {\lambda }^{J_b{M}_{b}T}.$$

$${\ell }_A={\ell }_{i_0,j_0}:=\{(i,j)\in {\mathbb{N}}_{>0}^2:\exists t\in \mathbb{Z},i=i_0-ta,j=j_0+t(b-a)\}.$$

$${\hat{\ell }}_A={\hat{\ell }}_{i_0,j_0}:=\{(i,j)\in {\mathbb{N}}_{>0}^2:\exists t\in \mathbb{Z},i=i_0-t(b-a),j=j_0+ta\}.$$

$${\lambda }^{{\mathrm{Colreg}}_{a,b}}=(|\lambda |-|{\lambda }^{{\mathrm{Colseg}}_{a,b}}|)\oplus {\lambda }^{{\mathrm{Colseg}}_{a,b}{\mathrm{Colreg}}_{a,b}}.$$

$$(13,10,9,7,5,2,2,1{)}^{{\mathrm{Colseg}}_{2,5}}=(10,10,7,6,2,2,1).$$

$${\mathrm{Core}}_b({\lambda }^{{\mathrm{Reg}}_{a,b}})={\mathrm{Core}}_b(\lambda )(\text{resp. }{\mathrm{Core}}_b({\lambda }^{{\mathrm{Colreg}}_{a,b}})={\mathrm{Core}}_b(\lambda ))$$

$$\mathcal{S}=$$

$$\mathcal{E}=$$

$$x_0=i_0+a<x_1<\dots <x_k=i_c+1$$

$$\psi (j)\ge \psi (j-1)\ge x_{j-1}.$$

$$H_{i,j}=tb,l_{i,j}=ta\text{ for some }t\in {\mathbb{N}}_{>0}.$$

$$r_{\omega (\lambda )}^x=a,r_{\omega (\lambda )}^y=b-a+1,$$

$${\lambda }^{{\mathrm{Colseg}}_{a,b}}={\lambda }_{[1,\psi (\lambda )]}^{J_b}\oplus {\lambda }_{[\psi (\lambda )+2,l(\lambda )]}.$$

$${\lambda }_{[\psi (\lambda )+1,l({\lambda }^{{\mathrm{Colseg}}_{a,b}})]}^{{\mathrm{Colseg}}_{a,b}}={\lambda }_{[\psi (\lambda )+2,l(\lambda )]}.$$

$${\lambda }_{1+\alpha a}\le {\lambda }_{1+(\alpha -1)a}-(b-a),\alpha =1,\dots \omega (\lambda ).$$

$${\delta }_{\alpha }:={\lambda }_{1+(\alpha -1)a}-(b-a)-{\lambda }_{1+\alpha a},\alpha =1,\dots \omega (\lambda ).$$

$$G_{\alpha }(\lambda )=\{(a\alpha +1,j):a\in [1,\omega (\lambda )],(a\alpha -1,j-1)\in \text{rim of }\lambda \setminus b\text{-rim of }\lambda \}.$$

$$({\lambda }^{\mathrm{Colreg}}{)}_1=|\lambda |-|{\lambda }^{{\mathrm{Colseg}}_{a,b}}|=|\lambda |-|{\lambda }^{J_b}|=({\lambda }^{M_{b}T}{)}_1=|\{y:1\le y\le {\lambda }_1,{\ell }_{1,y}\subset \lambda \}|.$$

$$\left(\frac{b}{a}-1\right)l_{i,j}<a_{i,j}+1.$$

$$\text{either}\left(\frac{b}{a}-1\right)l_{i,j}<a_{i,j}+1\text{or}\left(\frac{b}{a}-1\right)a_{i,j}<l_{i,j}+1,$$

$$\{G(\mu ):\mu \text{ is }b\text{-regular}\},$$

$$G(\mu )=\sum _{\lambda }{d}_{\lambda ,\mu }(q)\lambda .$$

$$d_{{\lambda }^T,{\lambda }^{M_b}}(q)=d_{{\lambda }^T,{\lambda }^{T{\mathrm{Reg}}_{a,b}}}(q).$$

Full text

1. 1 Introduction

2. 2 Preliminaries

3. 2.1 Mullineux Transpose

4. 2.2 Regularization and Column Regularization

5. 2.3 Cores and Quotients

6. 3 Rectangular Decomposition and Proof to the Main Theorem

7. 4 Iwahori-Hecke Algebras and Conjectures

1. Introduction

The Mullineux involution appears in the study of modular representations of symmetric groups, and the difficult combinatorial construction of the Mullineux involution is known to admit a conceptual description in terms of modular representations and crystals. Since the irreducible $p\textrm{-}$modular representations of the symmetric group $\mathcal{S}_{n}$ are labeled by the $p\textrm{-}$regular partitions of $n$, our study of the Mullineux involution as an algebraic operator is related to the underlying combinatorics of partitions. Given an irreducible representation $\rho_{\lambda}$, the new representation obtained by taking the tensor product with the one-dimensional sign representation $\operatorname{sgn}$ corresponds to the Mullineux involution of $\lambda^{\operatorname{M}_{p}}$. This notion defines $\lambda^{\operatorname{M}_{p}}$ via the equation:

[TABLE]

There are a few combinatorial definitions of $\operatorname{M}_{p}$ in [Kle96, FK97] where $p$ is not necessarily prime, and all of them are highly non-trivial. In this paper, we focus on the combinatorics of $\operatorname{M}_{p}$, so we use the parameter $b$ instead of $p$ since our parameters are generally positive integers.

Walker, Bessenrodt, Olsson, Xu, and Fayers studied the combinatorial properties of Mullineux involution by relating it to another operation named (resp. column) regularization $\operatorname{Reg}_{b}$ (resp. $\operatorname{Colreg}_{b}$). These two operations are naturally related by the Iwahori-Hecke algebra $\mathcal{H}=\mathcal{H}_{\mathbb{F},q}(S_{n})$ of the symmetric group where $\mathbb{F}$ is a field and $q\in\mathbb{F}\setminus\{0\}$. Let $b$ be the minimal integer satisfying $1+q+\cdots+q^{b-1}=0$ and the decomposition numbers $d_{\lambda,\mu}=\left[S^{\lambda}:D^{\mu}\right]$ be the multiplicity of the simple module $D^{\mu}$ ($\mu$ is $b$-regular) inside the Specht module $S^{\lambda}$ of $\mathcal{H}$. Previous works show that the identities for these decomposition numbers involve certain combinatorics of partitions. James proved that $d_{\lambda,\mu}=0$ unless $\mu\unlhd\lambda^{\operatorname{Reg}_{1,b}}$ and $d_{\lambda,\lambda^{\operatorname{Reg}_{b}}}=1$ [Jam76]. And later, Lascoux, Leclerc and Thibon proved the identity $d_{\lambda,\mu}=d_{\lambda^{\operatorname{T}},\mu^{\operatorname{M}_{b}}}$ [LLT96]. Therefore, it is natural to study the relationship between the Mullineux involution and the (resp. column) regularization map. In combinatorics, Walker proved that when the partition $\lambda$ is horizontal or row-stable, it satisfies $\lambda^{\operatorname{M}_{b}\operatorname{T}}=\lambda^{\operatorname{Colreg}_{b}}$ [Wal94, Wal96]. Later, Bessenrodt, Olsson, and Xu showed in [BOX99] that Walker’s conditions can be broadened to the short-legged (or shallow) partitions, namely for every hook in the partition divisible by $b$, the length of the corresponding arm is at least $(b-1)$ times that of the leg. Even better, Bessenrodt, Olsson, and Xu proved that those partitions are the only ones satisfying $\lambda^{\operatorname{M}_{b}\operatorname{T}}=\lambda^{\operatorname{Colreg}_{b}}$. Fayers generalized Bessenrodt, Olsson, and Xu’s result by considering partitions that are not necessarily $b$-regular. The Specht module $S^{\lambda}$ is reducible when $\lambda^{\operatorname{Reg}_{b}\operatorname{M}_{b}}\neq\lambda^{\operatorname{T}\operatorname{Reg}_{b}}$ and Fayers proved in [Fay08] that the identity $\lambda^{\operatorname{Reg}_{b}\operatorname{M}_{b}}=\lambda^{\operatorname{T}\operatorname{Reg}_{b}}$ holds if and only if hooks divisible by $b$ must be either shallow or steep. However, in all these previous works, the operators $\operatorname{Reg}_{b}$ and $\operatorname{Colreg}_{b}$ only involved a single parameter $b$.

A second parameter was added by Dimakis and the second author in [DY18] where the parameters of (resp. column) regularization $\operatorname{Reg}_{a,b}$ (resp. $\operatorname{Colreg}_{a,b}$) were extended to any positive rational number $\frac{a}{b}$ in the unit interval. In [DY18], the composition of a certain series of column regularizations and Mullineux transposes were shown to be same when applied to the one-row partition, giving a series of monotonically decreasing partitions. On one hand, this result gives a special situation where the simpler operation, column regularization, can be used to understand the Mullineux map, whose combinatorial definition is more convoluted; on the other hand, it proves a special case of Bezrukavnikov’s conjecture stated in the appendix of [DY18]. In this paper, we extend the idea of choosing a suitable parameter $a$ and finding the condition under which Mullineux transpose is identical to the generalized column regularization, generalizing the result of Bessenrodt, Olsson, and Xu in [BOX99], meanwhile shedding light on other cases of Bezrukavnikov’s conjecture. The main theorem is as follows:

Theorem 1.1**.**

Given positive integers $a<b$, let $\lambda$ be a partition satisfying $\lambda^{\operatorname{Colreg}_{a,b}}\in\mathcal{P}$ such that all hooks $H_{i,j}$ in $\lambda$ with $b\mid H_{i,j}$ satisfy:

[TABLE]

then $\lambda$ is $b\textrm{-}$regular and $\lambda^{\operatorname{M}_{b}\operatorname{T}}=\lambda^{\operatorname{Colreg}_{a,b}}$.

The theorem is proved combinatorially in Section 3 by analyzing the Young diagrams. The proof follows after completely characterizing the shape of the partitions satisfying the inequalities in Equation (1).

The rest of the paper is organized as follows. Section 2 is an overview of combinatorics involved. Section 3 presents a detailed description of partitions satisfying the conditions in Theorem 1.1 and concludes with the proof Theorem 1.1. Section 4 first recalls the basic facts about Iwahori-Hecke algebras that motivates the representation-theoretic significance of the main result. The section concludes with conjectures regarding the converse of Theorem 1.1 and a generalization of Fayers’s main theorem from [Fay08].

Acknowledgements. The authors thank Roman Bezrukavnikov and Richard Stanley for useful conversations and comments. They would also like to thank Panagiotis Dimakis for discussions and proofreadings of the manuscript. The first author is grateful to the MIT-PRIMES program for facilitating a part of this research.

2. Preliminaries

In this section, we introduce the fundamental vocabulary and notation needed for the remainder of the paper.

A partition $\lambda=(\lambda_{1},\ldots,\lambda_{k})$ of $n\in\mathbb{N}$ is a tuple of non-increasing positive integers, i.e. $\lambda_{1}\geq\cdots\geq\lambda_{k}>0$, and $|\lambda|=\sum_{i=1}^{k}\lambda_{i}=n$ is called the size of $\lambda$. When appropriate, we also append infinite zeros at the end of $\lambda$, i.e., $\lambda_{k+1}=\lambda_{k+2}=\cdots=0$. We denote $\mathcal{P}_{n}$ to be the set of all partitions of $n$ and $\mathcal{P}=\bigcup_{n\geq 0}\mathcal{P}_{n}$. Denote $l(\lambda)=k$ to be the number of nonzero parts of $\lambda$. Given two positive integers $i\leq j$, denote $\lambda_{[i,j]}$ to be the subpartition $(\lambda_{i},\ldots,\lambda_{j})$. Given a positive integer $b$, we say $\lambda$ is $b$-regular if there is no index $i$ such that $\lambda_{i}=\cdots=\lambda_{i+b-1}>0$. And for simplicity, we define the concatenation of two finite positive integer sequences $\lambda$ and $\mu$ as the tuple $\lambda\oplus\mu=(\lambda_{1},\ldots,\lambda_{l(\lambda)},\mu_{1},\ldots,\mu_{l(\mu)})$.

We identify each partition with its corresponding Young diagram. In this paper, except Section 2.3, we adopt the English convention for the Young diagram: rotate the plane to orient the positive $x$-axis pointing south and the positive $y$-axis pointing east. Recall that the Young diagram consists of boxes, which are unit squares parallel to the axis with vertices in integer coordinates. For the sake of notation, we identify a box and its southeast vertex by the same name $(i,j)$ where a box $(i,j)\in\lambda$ iff $j\leq\lambda_{i}$.

The transpose (or conjugate) $\lambda^{\operatorname{T}}$ of a Young diagram $\lambda$ is given by:

[TABLE]

A box $A=(i,j)\in\lambda$ is called a removable box of $\lambda$ if $\lambda\setminus A\in\mathcal{P}_{n-1}$. A box $B\notin\lambda$ is called an addable box of $\lambda$, if $\lambda\cup B\in\mathcal{P}_{n+1}$. Fix a positive integer $b$, the residue of $A$ with respect to $b$, denoted by $\operatorname{res}_{b}A$, is defined to be the residue class $(j-i)$ mod $b$.

Given a box $(i,j)\in\lambda$, the corresponding arm $a_{i,j}=a_{i,j}(\lambda)$ is the set of boxes $(i,j^{\prime})\in\lambda$ with $j<j^{\prime}$. We use $a_{i,j}$ to denote either this set of boxes or the number of elements of the above set interchangeably. Similarly, the leg $l_{i,j}=l_{i,j}(\lambda)$ is the set of boxes $(i^{\prime},j)\in\lambda$ with $i<i^{\prime}$. We use $l_{i,j}$ to denote either the above set or the number of elements of the above set interchangeably as well. Finally the hook $H_{i,j}=H_{i,j}(\lambda)$ is the union of sets $\{(i,j)\}\cup a_{i,j}\cup l_{i,j}$. Again, the number of elements of the hook is also denoted by $H_{i,j}$ and is equal to $1+a_{i,j}+l_{i,j}$. The northeast-most (resp. southwest-most) box in $H_{i,j}$ is called the hand (resp. foot) box associated to $(i,j)$, denoted by $\mathfrak{h}_{i,j}=\mathfrak{h}_{i,j}(\lambda)=(i,j+a_{i,j}$) (resp. $\mathfrak{f}_{i,j}=\mathfrak{f}_{i,j}(\lambda)=(i+l_{i,j},j)$).

There are two special classes of hooks of particular interest:

Definition 2.1**.**

Given two positive integers $a<b$ and a partition $\lambda$. A hook $H_{i,j}$ in $\lambda$ is $(a,b)$-shallow if it satisfies:

[TABLE]

Dually, a hook $H_{i,j}$ is $(a,b)$-steep if it satisfies:

[TABLE]

Remark 2.2*.*

When $b\mid H_{i,j}$, $H_{i,j}$ is $(a,b)$-shallow iff $l_{i,j}\leq ta-1$ and $a_{i,j}\geq t(b-a)$ for some $t\in\mathbb{N}_{>0}$; and similarly, $H_{i,j}$ is $(a,b)$-steep iff $a_{i,j}\leq ta-1$ and $l_{i,j}\geq t(b-a)$ for some $t\in\mathbb{N}_{>0}$. In particular, the above inequalities hold when $t=H_{i,j}/b.$

Lemma 2.3**.**

A hook divisible by $b$ can be both $(a,b)$-shallow and $(a,b)$-steep only if $b<2a$.

Proof.

By the above remark, by taking $t=H_{i,j}/b$, we know $t(b-a)\leq a_{i,j}\leq ta-1$ and $t(b-a)\leq l_{i,j}\leq ta-1$ for some $t\in\mathbb{N}_{>0}$. Hence $t(b-a)\leq ta-1$, and we conclude $b<2a$. ∎

Finally, for two partitions $\lambda$ and $\mu$ of the same size, the dominance order is defined as $\lambda\unlhd\mu$ if $\sum_{i=1}^{k}\lambda_{i}\leq\sum_{i=1}^{k}\mu_{i}$ is satisfied for all $k$. Note that $\lambda\unlhd\mu$ iff $\mu^{\operatorname{T}}\unlhd\lambda^{\operatorname{T}}$.

2.1. Mullineux Transpose

We abbreviate the composition of Mullineux involution and transpose as Mullineux transpose. There are multiple definitions for Mullineux transpose, but here we will follow the approach of Bessenrodt, Olsson and Xu from [BOX99].

Definition 2.4**.**

The rim of a partition $\lambda$ is the set of boxes $\{(i,j)\in\lambda\mid(i+1,j+1)\notin\lambda\}$. If $\lambda$ is $b$-regular, we define its $b$-rim to be the a subset of its rim obtained through the following procedure:

The $b\textrm{-}$rim consists of the several pieces where each piece, except possibly the last one, contains $b$ boxes. We choose the first $b$ boxes from the rim, beginning with the east-most box of the first row and moving contiguously southwestwards in the rim of $\lambda$. If the last box of this piece is chosen from the $i_{0}\textrm{-}$th row of $\lambda$, then we choose the second piece of $b$ boxes beginning with the right-most box of the next row ${i_{0}+1}$. Continue this procedure until we reach the last piece ending in the last row.

We call a maximal set of contiguous boxes of the $b$-rim a segment. When the number of boxes in this segment is a multiple $b$, it is a $b$-segment. Otherwise, it is called a $b^{\prime}$-segment. By construction, every $\lambda$ has at most one $b^{\prime}$-segment.

Finally, denote $\lambda^{\operatorname{I}_{b}}$ to be the partition obtained by removing the $b$-rim from $\lambda$.

Definition 2.5**.**

Given $\lambda=(\lambda_{1},...,\lambda_{k})$ and $\lambda^{\operatorname{I}_{b}}=(\mu_{1},...,\mu_{k})$, where $k=l(\lambda)$ and some of the $\mu_{i}$’s at the end of $\lambda^{\operatorname{I}_{b}}$ are allowed to be zero, define

[TABLE]

where $\phi(\lambda)=|\lambda|-|\lambda^{\operatorname{I}_{b}}|$ and

[TABLE]

The set of boxes $\lambda\setminus\lambda^{\operatorname{J}_{b}}$ is called the truncated $b$-rim of $\lambda$.

Definition 2.6**.**

For a $b\textrm{-}$regular partition $\lambda$, the operator $\operatorname{X}_{b}$ is defined as $\lambda^{\operatorname{X}_{b}}\coloneqq(j_{1},j_{2},\ldots)$, where

[TABLE]

The Mullineux map $\operatorname{M}_{b}$ for a $b$-regular partition is defined to be $\lambda^{\operatorname{M}_{b}}=\lambda^{\operatorname{X}_{b}\operatorname{T}}$.

Recursively, we can write

[TABLE]

Remark 2.7*.*

In case $b$ is a prime number, Definition 2.6 of the Mullineux map is a combinatorial algorithm of calculating the tensor product of an irreducible representation of the symmetric group with the sign representation in characteristic $b$ given in the Introduction. This result was conjectured by Mullineux in 1979 and later proved by Ford and Kleshchev in [FK97].

There are two other equivalent combinatorial ways to define the Mullineux map, see [FK97, DY18] for details. It is easy to see Mullineux map is an involution from the representation-theoretic definition, but not so obvious from Definition 2.6.

Example 2.8**.**

The $3$-rims of $(7,5,1,1)$ and $(7,2,1)$ are labelled by integers $1,2,3$ in Figure 1. The partition $(7,5,1,1)$ has two $3$-segments and one $3^{\prime}$-segment, and $(7,2,1)$ has two $3$-segments and no $3^{\prime}$-segment. Their truncated $3$-rims are shaded respectively in Figure 1. Thus, $(7,5,1,1)^{\operatorname{J}_{3}}=(5,3,1)$ and $(7,2,1)^{\operatorname{J}_{3}}=(5,1,1)$.

In the proof of the main theorem, we need to fully characterize the shape of the $b$-rim of certain partitions. The concept of rectangular decomposition allows for a simple description.

Definition 2.9**.**

Given a partition $\lambda$, we label the boxes in its $b$-rim with positive integers from 1 to $N$ in order from northeast to southwest ($N$ is the number of boxes in the $b$-rim). Suppose $N=bk+r$ where $k\in\mathbb{N}$ and $0\leq r<b$. The $b$-rectangular decomposition associated to the $b$-rim is the sequence of rectangles (of boxes) $r_{1},\ldots,r_{k}$ such that each $r_{i}$ is the smallest rectangle containing the consecutive boxes labelled by $b(i-1)+1,\ldots,\min\{bi,bk+r\}$.

We let $r_{i}^{x}$ (resp. $r_{i}^{y}$) be the dimension of the rectangle $r_{i}$ in the north-south (resp. west-east) direction. Informally, we may refer to $r_{i}^{x}$ as the width of the rectangle and $r_{i}^{y}$ as the length of the rectangle.

Remark 2.10*.*

More specifically, the “smallest” rectangle containing a set $S$ of boxes is the intersection of all rectangles parallel to the axes that contain $S$. So the smallest rectangle that contains $b$-consecutive boxes on the $b$-rim necessarily satisfies $r_{i}^{x}+r_{i}^{y}=b+1.$ In the context of Definition 2.9, for every $i$ except possibly the final one (when $\lambda$ contains a $b^{\prime}$-segment), we have $r_{i}^{x}+r_{i}^{y}=b+1$.

Example 2.11**.**

The $5$-rectangular decomposition of $(12,9,9,7,5,2,2,1)$ is the sequence of rectangles $r_{1}$, $r_{2}$, $r_{3}$, $r_{4}$ as shown in Figure 2. Note that $r_{i}^{x}=2$ for $i=1,2,3,4$, and $r_{j}^{y}=4$ for $j=1,2,3$ and $r_{4}^{y}=2$. Moreover, boxes from 1 to 10 form a $b$-segment, and boxes from 11 to 18 form a $b^{\prime}$-segment. Also note that by construction, the rectangles may have at most one overlap in the columns, but they occupy all rows of $\lambda$ without overlapping.

2.2. Regularization and Column Regularization

At the beginning of Section 2, we embed the Young diagram of a partition into the coordinate plane and refer to a box and its southeast vertex by the same name $(i,j)$. From now on, any unit square with vertices in $\mathbb{Z}^{2}$ will also be called a box and identified with its southeast vertex, i.e., when we refer to a box $(i,j)\in\mathbb{Z}^{2}$, we mean the unit square whose southeast vertex has coordinates $(i,j)$.

Definition 2.12**.**

Given two nonnegative integers $a<b$ and a partition $\lambda$, we define the following:

1. (1)

For any box $A=(i_{0},j_{0})\in\mathbb{Z}^{2}$, the ladder passing through $A$ is defined to be the finite collection of boxes:

[TABLE] 2. (2)

We define the column regularization $\lambda^{\operatorname{Colreg}_{a,b}}$ of $\lambda$ through the following procedure. For each ladder $\ell_{A}:A\in\mathbb{Z}^{2}$, if $\lambda\cap\ell_{A}\neq\emptyset$, slide the boxes in the intersection southwards along the ladder $\ell_{A}$ to the $|\lambda\cap\ell_{A}|$ south-most boxes of $\ell_{A}$. The resulting set of boxes is $\lambda^{\operatorname{Colreg}_{a,b}}$, which is not necessarily a partition. 3. (3)

For any box $A=(i_{0},j_{0})\in\mathbb{Z}^{2}$, the dual ladder passing through $A$ is defined to be:

[TABLE] 4. (4)

The regularization $\lambda^{\operatorname{Reg}_{a,b}}$ of $\lambda$ is defined to be $\lambda^{\operatorname{T}\operatorname{Colreg}_{a,b}\operatorname{T}}$. Equivalently, $\lambda^{\operatorname{Reg}_{a,b}}$ is the resulting partition after sliding boxes in each nonempty $\lambda\cap\hat{\ell}_{A}:A\in\mathbb{Z}^{2}$ northwards along the dual ladders $\hat{\ell}_{A}$ to the $|\lambda\cap\hat{\ell}_{A}|$ north-most boxes of $\hat{\ell_{A}}$.

We say that $\lambda$ is $\operatorname{Colreg}_{a,b}$-valid (resp. $\operatorname{Reg}_{a,b}$-valid) if $\lambda^{\operatorname{Colreg}_{a,b}}\in\mathcal{P}$ (resp. $\lambda^{\operatorname{Reg}_{a,b}}\in\mathcal{P}$).

Remark 2.13*.*

1. (1)

Note that though $A=(i_{0},j_{0})$ might not be in $\lambda$ or even not in the first quadrant, $\ell_{A}$ (resp. $\hat{\ell}_{A}$) only denotes some of the boxes in the first quadrant on the “underlying line”: $y-j_{0}+\frac{b-a}{a}(x-i_{0})=0$ (resp. $y-j_{0}+\frac{a}{b-a}(x-i_{0})=0$). 2. (2)

Different subscripts can be used to indicate the same (resp. dual) ladder, for example when $a=2,b=5$, $\ell_{1,9}=\ell_{-1,12}=\ell_{3,6}$. But for the sake of clarity, in all the figures in this paper, we will only draw out the “underlying line” when referring to a (resp. dual) ladder. 3. (3)

Only when $a$ and $b$ are co-prime, the (resp. dual) ladder coincide with the set of all positive integer points (boxes) on its “underlying line.” 4. (4)

All the boxes on a (resp. dual) ladder have the same $b$-residue, we say this residue to be the residue of the (resp. dual) ladder, denoted by $\operatorname{res}_{b}\ell$ (resp. $\operatorname{res}_{b}\hat{\ell}$).

Example 2.14**.**

The partition $(3,2,2,1)$ is $\operatorname{Colreg}_{2,3}$-valid and $(3,2,2,1)^{\operatorname{Colreg}_{2,3}}=(2,2,2,1,1)$, but $\operatorname{Colreg}_{2,3}$ maps $(3,2,2)$ to $(2,1,2,1,1)\notin\mathcal{P}$, hence $(3,2,2)$ is not $\operatorname{Colreg}_{2,3}$-valid, which is shown in Figure 3.

By definition, $\lambda$ is $\operatorname{Reg}_{a,b}$-valid exactly when $\lambda^{\operatorname{T}}$ is $\operatorname{Colreg}_{a,b}$-valid, so we only specify the criteria for $\lambda$ being $\operatorname{Colreg}_{a,b}$-valid.

Lemma 2.15**.**

Given $\lambda\in\mathcal{P}$ and positive integers $a<b$, $\lambda$ is $\operatorname{Colreg}_{a,b}$-valid if and only if for all $y\in[1,\lambda_{1}]$, either $\ell_{1,y}\subset\lambda$, or there exists a box $(i,j)\in\lambda\cap\ell_{0,y}$ such that $(i+1,j)\notin\lambda$. In the first case where $\ell_{1,y}\subset\lambda$, we say that the ladder $\ell_{1,y}$ is “full”.

Proof.

$\lambda^{\operatorname{Colreg}_{a,b}}\in\mathcal{P}$ is equivalent to the following:

For every $(i,j)\in\lambda^{\operatorname{Colreg}_{a,b}}$,

1. (1)

if $j\geq 2$, $(i,j-1)\in\lambda^{\operatorname{Colreg}_{a,b}}$; 2. (2)

if $i\geq 2$, $(i-1,j)\in\lambda^{\operatorname{Colreg}_{a,b}}$.

(1) is always true for any $\lambda$. In fact for every $(x,y)\in\lambda$, $y\geq 2$, consider the two ladders $\ell_{x,y}$ and $\ell_{x,y-1}$. If the ladder $\ell_{x,y}$ contains a box of the form $(i,1)$, then $|\ell_{x,y-1}|=|\ell_{x,y}|-1$ and $|(\ell_{x,y-1}\cap\lambda)|\geq|(\ell_{x,y}\cap\lambda)|-1$ since (1) is satisfied by $\lambda$. If $\ell_{x,y}$ does not contain a box of the form $(i,1)$, then $|\ell_{x,y-1}|=|\ell_{x,y}|$ and $|(\ell_{x,y-1}\cap\lambda)|\geq|(\ell_{x,y}\cap\lambda)|$. Because $\operatorname{Colreg}_{a,b}$ fixes the number of boxes on each ladder, and moves the boxes on each ladder to the south-most positions, for any $(i,j)\in\lambda^{\operatorname{Colreg}_{a,b}}$ and $j\geq 2$, $(i,j-1)\in\lambda^{\operatorname{Colreg}_{a,b}}$.

To satisfy condition (2), for every $(x,y)\in\lambda$, $x\geq 2$, we consider $\ell_{x,y}$ and $\ell_{x-1,y}$. Similar as case (1), if $\ell_{x,y}$ contains a box of the form $(1,j)$, then $|\ell_{x-1,y}|=|\ell_{x,y}|-1$ and $|(\ell_{x-1,y}\cap\lambda)|\geq|(\ell_{x,y}\cap\lambda)|-1$ since when $i\geq 2$ for each $(i,j)\in\lambda$, $(i-1,j)\in\lambda$. If $\ell_{x,y}$ does not contain a box of the form $(1,j)$, then $|\ell_{x-1,y}|=|\ell_{x,y}|$ and $|(\ell_{x-1,y}\cap\lambda)|\geq|(\ell_{x,y}\cap\lambda)|$. Hence $\lambda^{\operatorname{Colreg}_{a,b}}\notin\mathcal{P}$ occurs only in the former case where $|(\ell_{x-1,y}\cap\lambda)|=|(\ell_{x,y}\cap\lambda)|-1$, $\ell_{x,y}$ contains through a box of the form $(1,j)$, and $\ell_{x,y}\not\subset\lambda$. In this case, without loss of generality, we suppose $x=1$ and $y\in[1,\lambda_{1}]$. Then the only situation when $\lambda^{\operatorname{Colreg}_{a,b}}\notin\mathcal{P}$ is when $\ell_{1,y}\not\subset\lambda$, and pairs of boxes $(i,j)\in\ell_{1,y}$, $i\geq 2$ and $(i-1,y)\in\ell_{0,y}$ are either both present in $\lambda$ or both not in $\lambda$ simultaneously. Under these conditions, the box $(1,y)\in\ell_{1,y}$ slides to a position where the box immediately north to it is not in $\lambda^{\operatorname{Colreg}_{a,b}}$. ∎

Remark 2.16*.*

Referring to Example 2.14, the box $(3,2)\in\ell_{0,3}\cap(3,2,2)^{\operatorname{Colreg}_{2,3}}$ while $(2,2)\notin(3,2,2)^{\operatorname{Colreg}_{2,3}}.$ In fact, in $(3,2,2)$, there is no box $(i,j)\in\ell_{0,3}\cap\lambda$ where $(i+1,j)\notin\lambda,$ so it is not $\operatorname{Colreg}_{a,b}$-valid.

It is important to note the following proposition, which in part illustrates the subtleties when a second parameter is added. In previous works, the notions of $\operatorname{Reg}$-valid and $\operatorname{Colreg}$-valid were irrelevant since only one parameter was used. In fact, for any value of $b$ and any partition $\lambda$, the maps $\operatorname{Reg}_{b}$ and $\operatorname{Colreg}_{b}$ would be well-defined operators on $\lambda.$

Proposition 2.17**.**

Every partition $\lambda$ is $\operatorname{Reg}_{1,b}$-valid and $\operatorname{Colreg}_{1,b}$-valid for all $b$.

Proof.

To show that $\lambda$ is $\operatorname{Colreg}_{1,b}$-valid, we directly apply Lemma 2.15. For all $y\in[1,\lambda_{1}]$, either $\ell_{1,y}\subset\lambda$ or there exists a box $(i+1,j)$ furthest north along $\ell_{1,y}$ such that $(i+1,j)\notin\lambda$. Then the box $(i,j+b-1)$, which is the box immediately northeast of $(i+1,j)$ along the ladder $\ell_{1,y}$, must be in $\lambda$. Therefore, $(i,j)\in\lambda\cap\ell_{0,y}$ while $(i+1,j)\notin\lambda$, so the condition in Lemma 2.15 is satisfied.

We apply the same argument to all ladders $\ell_{1,y}$ for $y\in[1,\lambda_{1}]$ and conclude that $\lambda$ is $\operatorname{Colreg}_{1,b}$-valid. By applying the proof to $\lambda^{\operatorname{T}},$ we see that $\lambda$ is $\operatorname{Reg}_{1,b}$-valid as well. ∎

Definition 2.18**.**

Given positive integers $a<b$, a partition $\lambda$ is $(a,b)$-regular if $\lambda^{\operatorname{Reg}_{a,b}}=\lambda.$

Remark 2.19*.*

$(1,b)$-regular is equivalent to the notion of $b$-regular. Also $(a,b)$-regular implies $b$-regular for any positive integer $a$ with $a<b$.

For positive integers $a<b$, we wish to define an intermediate operator called the column semi-regularization $\operatorname{Colseg}_{a,b}$ to compute $\operatorname{Colreg}_{a,b}$ in a row-by-row fashion (parallel to that of the Mullineux transpose as in Equation (4)): $\lambda^{\operatorname{Colreg}_{a,b}}_{i}=|\lambda^{\operatorname{Colseg}_{a,b}^{i-1}}|-|\lambda^{\operatorname{Colseg}_{a,b}^{i}}|$, or recursively,

[TABLE]

The only way to do this is the following:

Definition 2.20**.**

Given a $\operatorname{Colreg}_{a,b}$-valid partition $\lambda$, the column semi-regularization $\lambda^{\operatorname{Colseg}_{a,b}}$ of $\lambda$ is defined by the following procedure. For all $(1,y)\in\lambda$, if $\ell_{1,y}\not\subset\lambda$, slide $(1,y)$ to the north-most position $(i,j)$ in $\ell_{1,y}\setminus\lambda$ such that $(i-1,j)\in\lambda$. Then, we remove what remains of the first row, and the resulting partition is defined to be $\lambda^{\operatorname{Colseg}_{a,b}}$.

It is clear that $\operatorname{Colseg}_{a,b}$ is well-defined by Lemma 2.15, and $\lambda^{\operatorname{Colseg}_{a,b}}$ is $\operatorname{Colreg}_{a,b}$-valid if $\lambda$ is. Note that we only perform $\operatorname{Colseg}_{a,b}$ to $\operatorname{Colreg}_{a,b}$-valid partitions.

Example 2.21**.**

In the diagram shown in Figure 4, the boxes $(1,12)$ and $(1,13)$ shift under $\operatorname{Colseg}_{2,5}$ to $(3,10)$ and $(5,6)$ respectively. We have

[TABLE]

2.3. Cores and Quotients

Cores and quotients are crucial concepts of partitions, and many attempts have been made to understand the Mullineux map in terms of cores and quotients. We first recall the definition of cores and refer to [Mac98] for that of quotients and their basic properties.

Definition 2.22**.**

The $b\textrm{-}$core $\operatorname{Core}_{b}(\lambda)$ of any partition $\lambda$ is the partition that remains after removing as many $b\textrm{-}$ribbons in succession as possible. It is well-known that the result is independent of the choice of removals. If $\lambda=\operatorname{Core}_{b}(\lambda)$, then $\lambda$ itself is called a $b$-core.

Definition 2.23**.**

The $b$-content of the partition $\lambda$ is a tuple $(c_{0},\ldots,c_{b-1})$ where $c_{i}$ is the number of boxes in $\lambda$ with residue $i$.

Proposition 2.24** ([Mac98]).**

The $b$-content determines the $b$-core of a partition.

Since the residue of boxes on each ladder is the same, the $b$-content is fixed when boxes move within the same ladders. Thus, $b$-core is an invariant under the (column) regularization process, and we have the following result:

Lemma 2.25**.**

For a $\operatorname{Reg}_{a,b}$-valid (resp. $\operatorname{Colreg}_{a,b}$-valid) partition $\lambda$, we have

[TABLE]

In particular, we have $|\lambda^{\operatorname{Reg}_{a,b}}|_{b}=|\lambda|_{b}$ (resp. $|\lambda^{\operatorname{Colreg}_{a,b}}|_{b}=|\lambda|_{b}$).

Remark 2.26*.*

Mullineux transpose also fixes the core of a $b$-regular partition, as shown in [FK97], which is compatible with Theorem 1.1.

3. Rectangular Decomposition and Proof to the Main Theorem

In this section we will completely characterize the shape of $\lambda$ satisfying the conditions restraining hook shape from Equations (1). Using this description, we then conclude with a proof of Theorem 1.1. It happens that the conditions from Equations (1) give a strong result on the dimensions of any rectangle that contains $b$ consecutive boxes on the rim of $\lambda$. Although we will often apply the lemma below to rectangles in the rectangular decomposition of $\lambda$, it is proven in greater generality here. The following result gives a strong restriction on the dimensions $r^{x}$ and $r^{y}$ (see Definition 2.9) of the rectangles.

Lemma 3.1**.**

Given a partition $\lambda$, suppose $r_{\alpha}$ and $r_{\beta}$ are two smallest rectangles each containing $b$ consecutive boxes of the rim of $\lambda$, i.e. $r_{\alpha}^{x}+r_{\alpha}^{y}=r_{\beta}^{x}+r_{\beta}^{y}=b+1$. Let the box $B_{0}=(i_{0},j_{0})$ be the box furthest northeast in $r_{\alpha}$. Suppose the following conditions are satisfied:

1. (1)

$(i_{0},j_{0}+1)\notin\lambda$; 2. (2)

$r_{\alpha}^{x}>a\geq r_{\beta}^{x}$; 3. (3)

$r_{\beta}$* contains a box strictly southwest of $r_{\alpha}$.*

Then $\lambda$ contains a hook with leg length $a$ and arm length $b-a-1.$

Note that $r_{\alpha}$ and $r_{\beta}$ are not necessarily rectangles from the rectangular decomposition of $\lambda$; the indices $\alpha$ and $\beta$ are irrelevant to the statement of the Lemma but will be referenced later in the section when we apply Lemma 3.1 to rectangles in the rectangular decomposition.

Example 3.2**.**

The example in Figure 5 illustrates the statement of Lemma 3.1. The rectangles $r_{\alpha}$ and $r_{\beta}$ are colored yellow and the corresponding boxes in the rim are outlined in red. We see that $B_{0}=(i_{0},j_{0})=(1,8)$ and $(1,9)\notin\lambda$, thus the first condition is satisfied. We have $r_{\alpha}^{x}=3$, $r_{\alpha}^{y}=3$, $r_{\beta}^{x}=2$, and $r_{\beta}^{y}=4$, thus the second condition is satisfied. Finally, from the diagram, the third condition is satisfied as well. Lemma 3.1 implies that there is a hook of leg length $2$ and arm length $2$, which is outlined in blue.

Proof of Lemma 3.1.

Let $B_{c}=(i_{c},j_{c})$ be the box furthest southwest in $r_{\beta}$. We must have $j_{c}<j_{0}$ and $i_{c}>i_{0}$ since $r_{\beta}$ contains a box strictly southwest of $r_{\alpha}$. Denote by $\mathcal{B}$ the set of consecutive boxes in the rim of $\lambda$ starting at $B_{0}$ and ending at $B_{c}$. We define

[TABLE]

For an example of this definition, see Figure 8.

We first show that $\mathcal{S}\cap\mathcal{E}\neq\emptyset$, then we use a box in this intersection to find a corresponding hook of leg length $a$ and arm length $b-a-1$. We prove that $\mathcal{S}\cap\mathcal{E}\neq\emptyset$ by examining the sequence of boxes in $\mathcal{S}$ and $\mathcal{E}$ when written from the northeast to southwest direction. Our approach is to show that since $\mathcal{S}$ begins south of $\mathcal{E}$ and ends north of $\mathcal{E}$, an “intermediate value” argument implies that the two sets must intersect.

Since there is at least one box in the rim for each $y$-coordinate and $x$-coordinate, the set $\mathcal{S}$ contains exactly one box for every $y$-coordinate in $[j_{c}+1,j_{0}]$, and $\mathcal{E}$ contains one box for every $x$-coordinate in $[i_{0}+a,i_{c}+a].$ We sort the boxes in $\mathcal{S}$ and $\mathcal{E}$ in order from northeast to southwest and consider the two sequences. Since $B_{0}=(i_{0},j_{0})\in\lambda$ and $(i_{0},j_{0}+1)\notin\lambda$ by the first condition, the box $(i_{0}+a,j_{0}-(b-a-1))$ must lie in $\mathcal{E},$ and we denote this box, the first box of $\mathcal{E}$, by $E_{i_{0}+a}=(i_{0}+a,j_{0}-(b-a-1))$. Denote the southwest-most box of $r_{\alpha}$ by $B_{b-1}=(i_{b-1},j_{b-1})=(i_{0}+r_{\alpha}^{x}-1,j_{0}-r_{\alpha}^{y}+1)$. We know that $E_{i_{0}+a}$ is northwest to $B_{b}$ (not necessarily strict) and these two boxes determine a line of slope $1$. Since $j_{c}<j_{0}-(b-a-1)\leq j_{b-1}=j_{0}-r_{\alpha}^{y}+1$, the unique box $(i^{\prime},j_{0}-(b-a-1))\in\mathcal{S}$ in the same column of $E_{i_{0}+a}$ is not north of $E_{i_{0}+a}$, i.e. $i^{\prime}\geq i_{0}+a$.

Next, we consider the unique box $E_{i_{c}}=(i_{c},j^{\prime})$ in $\mathcal{E}$, which has $x$-coordinate $i_{c}$. Since $r_{\beta}^{y}\geq b-a+1$, we find that $j^{\prime}>j_{c}$. Next, we find the unique box $(i^{\prime\prime},j^{\prime})\in\mathcal{S}$ in the same column of $E_{i_{c}}$. Then $i^{\prime\prime}\leq i_{c}$ since $(i^{\prime\prime},j^{\prime})\in\mathcal{S}\subset\lambda$. This argument is illustrated by Figure 6.

We now consider boxes in $\mathcal{S}$ and $\mathcal{E}$ inside the big rectangle (with sides parallel to the axis) whose northeast-most box is $E_{i_{0}+a}$ and southwest-most box is $E_{i_{c}}$. Boxes in $\mathcal{E}$ divide the rows into:

[TABLE]

such that the boxes in $\mathcal{E}$ on rows $[x_{j-1},x_{j}-1]$ lie in the same column $\varphi(j)$ and $\varphi(j)<\varphi(j-1)$ for all $j=1,\ldots,k$ (see Figure 7). Next, we define the function $\psi$ such that $(\psi(j),\varphi(j))$ is the unique box in $\mathcal{S}$ and column $\varphi(j)$.

Suppose on the contrary that $\mathcal{S}\cap\mathcal{E}=\emptyset$. The above argument about $\mathcal{S}$ and $\mathcal{E}$ gives the conditions $\psi(1)\geq x_{1}$ and $\psi(k)<x_{k}$. Also note that $\psi(j)\geq\psi(j-1)$ because $\varphi(j)<\varphi(j-1)$ and we are considering the boxes of $\mathcal{S}$ from northeast to southwest. We show by induction that $\psi(j)\geq x_{j}$ for all $j$; this will give the necessary contradiction since we have already shown $\psi(k)<x$.

The base case, where $j=1$, is true from our reasoning above. Suppose that $\psi(j)\geq x_{j}$ holds for indices strictly less than a fixed $j$. So

[TABLE]

We know that $(x,\varphi(j))\in\mathcal{E}$ for $x\in[x_{j-1},x_{j}-1]$. Since $\mathcal{S}\cap\mathcal{E}=\emptyset$ by assumption, $\psi(j)\geq x_{j}$, which completes the induction. However, this implies that $\psi(k)\geq x_{k}=i_{c}+1$, which is a contradiction, so we must have $\mathcal{S}\cap\mathcal{E}\neq\emptyset.$ Figure 7 illustrates this part of the proof.

Pick $(i,j)\in\mathcal{S}\cap\mathcal{E}.$ Then $(i,j)\in\lambda$, and $(i+1,j)\notin\lambda$. Also, $(i-a,j+(b-a-1))\in\lambda$, and $(i-a,j+b-a)\notin\lambda$. Hence, $H_{i-a,j}$ has corresponding leg length $a$ and arm length $b-a-1$, as desired. See Figure 8 for an illustration of this argument. ∎

At this point, we note that there is a special type of hook of the shape

[TABLE]

Note that this type of hook is not $(a,b)$-shallow, and in the case of $b>2a$, it is not $(a,b)$-steep either.

Lemma 3.3**.**

Consider a $\operatorname{Colreg}_{a,b}$-valid partition $\lambda$ containing no hook of shape (6). For a ladder $\ell_{1,y}:y\in[1,\lambda_{1}]$ that is not full (i.e. $\ell_{1,y}\not\subset\lambda$), let $(i,j)$ be the box furthest north in $\ell_{1,y}\setminus\lambda$. Then, $(i-1,j)\in\lambda$.

Example 3.4**.**

We illustrate the statement of Lemma 3.3 in Figure 9 in the case of $a=2$ and $b=5$. On the left side, in $(8,5,4,1)$, the boxes $(5,1)$ and $(3,5)$ are the north-most boxes on ladders $\ell_{1,7}$ and $\ell_{1,8}$ respectively that are not contained in $\lambda$. The boxes immediately north of $(5,1)$ and $(3,5)$ are in $\lambda$ as Lemma 3.3 states.

On the right hand side, however, in $(8,4,4,2)$, the box $(3,5)$ is the north-most box on $\ell_{1,8}$ not contained in $\lambda$. In this case $(2,5)\notin\lambda$, which contradicts Lemma 3.3. Then Lemma 2.15 implies that $(4,2)\in\lambda$ and $(5,2)\notin\lambda$. This gives hook $H_{2,2}$ of shape (6), which is not allowed by the statement of the lemma.

Proof of Lemma 3.3.

Let $\ell_{1,y_{1}}$ be the westmost ladder that is not full, passing through a box in the first row. Then all $\ell_{1,y}$ east of $\ell_{1,y_{1}}$ are not full either. Thus, we denote by $\ell_{1,y_{1}},\ell_{1,y_{1}+1},\ldots,\ell_{1,\lambda_{1}}$ to be all the ladders passing through a box in the first row of $\lambda$ that are not fully contained in $\lambda$ (if there are any). We proceed by induction to show that the lemma holds for all of these ladders.

In the base case, we consider the ladder $\ell_{1,y_{1}}$. Let $(i,j)$ be the box furthest north on $\ell_{1,y_{1}}$ that is not in $\lambda$. Note that $i>1$ because $\ell_{1,y_{1}}$ contains a box in the first row. Suppose for the sake of contradiction that $(i-1,j)\notin\lambda$. Since $\lambda$ is $\operatorname{Colreg}_{a,b}$-valid, Lemma 2.15 implies $\exists\ t\in\mathbb{N}$ such that $(i+ta,j-t(b-a))\notin\lambda$ and $(i+ta-1,j-t(b-a))\in\lambda.$ By construction $\ell_{1,y_{1}-1}$ is completely contained in $\lambda$, so $(i,j-1)\in\ell_{1,y_{1}-1}\subset\lambda$, and thus $(i-1,j-1)\in\lambda$ as well. Then $H_{i-1,j-t(b-a)}$ satisfies $l_{i-1,j-t(b-a)}=ta$ and $a_{i-1,j-t(b-a)}=t(b-a)-1$, which contradicts the hypothesis that $\lambda$ contains no hook of shape (6). Hence, our initial assumption was false, and $(i-1,j)\in\lambda$, which completes the base case. Figure 10 illustrates this part of the proof.

For the inductive step, we reason similarly as we did in the base case. Assume that the lemma holds for all $\ell_{1,y}$: $y\in[y_{1},y^{\prime}]$ for a fixed $y^{\prime}<\lambda_{1}$. Let $(i,j)$ be the box furthest north on $\ell_{1,y^{\prime}}$ that is not in $\lambda$. Once again, note that $i>1$ because $\ell_{1,y^{\prime}}$ contains a box in the first row. Suppose for the sake of contradiction that $(i-1,j)\notin\lambda$. Since all boxes in $\ell_{1,y^{\prime}}$ north of $(i,j)$ are in $\lambda$, all boxes in $\ell_{1,y^{\prime}-1}$ north of $(i,j)$ are also in $\lambda$.

Consider the box $(i,j-1)$ on ladder $\ell_{1,y^{\prime}}$. If $(i,j-1)\in\lambda$, then $(i-1,j-1)\in\lambda$ since it lies immediately north of $(i,j-1)$; if $(i,j-1)\notin\lambda$, then $(i,j-1)$ is the north-most box in $\ell_{1,y^{\prime}}\setminus\lambda$. In this case, the inductive hypothesis also guarantees $(i-1,j-1)\in\lambda$. Thus, we conclude $(i-1,j-1)\in\lambda$ and $(i-1,j)\notin\lambda$ in all cases.

Using the exact same reasoning from the proof of the base case, we can use Lemma 2.15 to find a box $(i+ta-1,j-t(b-a))\in\lambda$ such that $(i+ta,j-t(b-a))\notin\lambda$. This gives a hook $H_{i-1,j-t(b-a)}=tb$ with leg length $ta$ and contradicts the hypothesis that $\lambda$ contains no hook of shape (6). Thus, we must have $(i-1,j)\in\lambda$, which completes the induction. ∎

Consider the $b$-rectangular decomposition of $\lambda$ in Definition 2.9. Define $\omega(\lambda)$ to be largest index such that

[TABLE]

and define $\psi(\lambda)=\sum_{i=1}^{\omega(\lambda)}r_{i}^{x}.$ When no rectangle of such shape exists, we let $\omega(\lambda)=\psi(\lambda)=0$.

Example 3.5**.**

When $a=2$ and $b=5$, $\omega((2,1,1))=\psi((2,1,1))=0$; and $\omega((8,4,1))=2,\psi((8,4,1))=3$, which is shown in Figure 11.

We first give a description of the general shape of all $\lambda$ that $\operatorname{Colreg}_{a,b}$-valid and contain no hooks given by (6). Note that these conditions are strictly weaker than the conditions of Theorem 1.1. Indeed, using only these two restrictions, we can already find a very specific description for the rectangular decomposition of $\lambda.$

Proposition 3.6**.**

Given a $\operatorname{Colreg}_{a,b}$-valid partition $\lambda$ containing no hooks of shape (6), denote by $r_{1},\ldots,r_{k}$ to be the rectangles in its $b$-rectangular decomposition, defined by Definition 2.9, with $r_{\alpha}^{x}+r_{\alpha}^{y}=b+1:\alpha\in[1,k]$. (Note that these are all the rectangles in the $b$-rectangular decomposition except for possibly the last one.) Then for all $\alpha\in[1,\omega(\lambda)]$, the widths of the rectangles satisfy $r^{x}_{\alpha}=a$, and for all $\alpha\in[\omega(\lambda)+1,k]$ the widths satisfy $r^{x}_{\alpha}>a$.

Example 3.7**.**

We illustrate the statement of Proposition 3.6 on $(11,8,7,5,3,2,2,2)$ in the case of $a=2$ and $b=5$. In Figure 12, the rectangles in the $5$-rectangular decomposition are $r_{1},r_{2},$ and $r_{3}$. Since $\omega(\lambda)=2$, $r_{1}^{x}=r_{2}^{x}=2=a$ and $r_{3}^{x}=4>a$.

Proof of Proposition 3.6.

We first show that $r_{1}^{x}\geq a$. Suppose for the sake of contradiction that $r_{1}^{x}\leq a-1$ and $r_{1}^{y}\geq b-a+2$. Consider the ladder $\ell_{1,\lambda_{1}}$ and $\ell_{0,\lambda_{1}}$. Due to the dimensions of $r_{1}$, the box $(a+1,\lambda_{1}-(b-a))\in\ell_{1,\lambda_{1}}\setminus\lambda$ and $(a,\lambda_{1}-(b-a))\in\ell_{0,\lambda_{1}}\setminus\lambda$. However, this contradicts Lemma 3.3 since $(a+1,\lambda_{1}-(b-a))$ is the north-most box of $\ell_{1,\lambda_{1}}$ not contained in $\lambda$ and $(a,\lambda_{1}-(b-a))\notin\lambda$ either. Thus, we must have $r_{1}^{x}\geq a$. We now consider the cases where $r_{1}^{x}>a$ and $r_{1}^{x}=a$ separately.

If $r_{1}^{x}>a$, Lemma 3.1 implies that for all $\alpha>1$ we have $r_{\alpha}^{x}>a$. In this case, $\omega(\lambda)=0$, and both parts of the proposition are satisfied.

If $r_{1}^{x}=a$, then $\omega(\lambda)\geq 1$. To prove the first part of the proposition, that for all $\alpha\in[1,\omega(\lambda)]$ the widths of the rectangles satisfy $r^{x}_{\alpha}=a$, we proceed by induction on $\alpha$.

Since $r_{1}^{x}=a$, the base case has already been done. Now suppose that $r^{x}_{\alpha}=a$ for $\alpha\in[1,\alpha^{\prime}]$ for a fixed $\alpha^{\prime}<\omega(\lambda)$ and consider $r_{\alpha^{\prime}+1}$. Denote the northeast-most box in $r_{\alpha^{\prime}+1}$ by $(i,j)$. By the inductive hypothesis, all rectangles in the $b$-rectangular decomposition north of $r_{\alpha^{\prime}+1}$ have width $r^{x}=a$, so the ladder $\ell_{i,j}$ must pass through the first row of $\lambda$, and $(i,j)$ is the north-most box in $\ell_{i,j}$ not contained in $\lambda$. Then proceeding as we did in the base case, if $r^{x}_{\alpha^{\prime}+1}<a$, the two boxes $(i+a,j-(b-a))$ and $(i+a-1,j-(b-a))$ are both not in $\lambda$, which contradicts Lemma 3.3 (see Figure 13). Thus, $r^{x}_{\alpha^{\prime}}\geq a$. If $\alpha^{\prime}<\omega(\lambda)$, then applying Lemma 3.1 on $r_{\alpha^{\prime}}$ and $r_{\omega(\lambda)}$ forces $r^{x}_{\alpha^{\prime}}\leq r^{x}_{\omega(\lambda)}=a$. In particular, $r^{x}_{\alpha^{\prime}}=a$ exactly, which completes the inductive step.

Finally, we show the second part of the proposition, that for all $\alpha\in[\omega(\lambda)+1,k]$ the widths of the rectangles satisfy $r^{x}_{\alpha}>a$. For any $\alpha$ in this interval, Lemma 3.1 implies that $r^{x}_{\alpha}\geq a$, and the definition of $\omega(\lambda)$ being the maximal index such that $r^{x}_{\alpha}=a$ forces $r^{x}_{\alpha}>a$. ∎

Lemma 3.8**.**

Given a $\operatorname{Colreg}_{a,b}$-valid partition $\lambda$ containing no hook of shape (6), for $i>\psi(\lambda)$, any hook $H_{i,j}$ divisible by $b$ is not $(a,b)$-shallow.

Proof.

Let $H_{i,j}=tb$ for some $t$ and $i>\psi(\lambda)$. We assume for the sake of contradiction that $H_{i,j}$ is $(a,b)$-shallow. By Proposition 3.6, $r_{\omega(\lambda)+1}^{x}>a$. We show that among the $tb$ consecutive boxes on the rim that determine $H_{i,j}$, we can always find $b$ consecutive boxes which determine a rectangle $r_{\beta}$ such that $r_{\beta}^{x}\leq a$.

Number the $tb$ boxes in the rim corresponding to $H_{i,j}$ from northeast to southwest, and denote $s_{\beta}:\beta\in[1,t]$ to be the smallest rectangles each containing the $[(\beta-1)b+1,\beta b]$ boxes in the set of $tb$ boxes. Since $H_{i,j}$ is $(a,b)$-shallow, $l_{i,j}\leq at-1$. The sum of the widths of all the rectangles cannot exceed $l_{i,j}+1$, so we have $\sum_{\beta=1}^{t}(s_{\beta}^{x}-1)\leq at$. Hence, there exists a $\beta$, such that $s_{\beta}^{x}\leq a$. Applying Lemma 3.1 to $r_{\omega(\lambda)+1}$ and $r_{\beta}$ gives a hook of shape (6), which is a contradiction. Therefore, there cannot be any $(a,b)$-shallow hooks strictly south of $x=\psi(\lambda).$ ∎

Lemma 3.9**.**

Given a $\operatorname{Colreg}_{a,b}$-valid and $b$-regular partition $\lambda$ containing no hook of shape (6), we have the identity

[TABLE]

Example 3.10**.**

We illustrate Lemma 3.9 on $\lambda=(13,10,9,7,5,2,2,1)$ with $a=2$ and $b=5$. In the left half of Figure 14, we are applying $\operatorname{Colseg}_{2,5}$. The rectangles of the $5$-rectangular decomposition are colored yellow, and the arrows indicate where boxes in the first row shift to.

On the right half, we are applying $\operatorname{J}_{5}$. The $5$-rim is labeled, and the boxes that are shaded are removed. For the sake of illustration, we remove the south-most box in each column of $\lambda_{[\psi(\lambda)+1,l(\lambda)]}$, which is equivalent to removing $\lambda_{\psi(\lambda)+1}$. Visually, it is easy to see that both operations gives the partition $(10,10,7,6,2,2,1)$.

Proof of Lemma 3.9.

We first claim that $\ell_{\psi(\lambda)+1,\lambda_{\psi(\lambda)+1}}\subset\lambda$, i.e. the ladder passing through the box $(\psi(\lambda)+1,\lambda_{\psi(\lambda)+1})$ is completely contained in $\lambda$. (if $\psi(\lambda)+1>l(\lambda)$, the argument below still holds by taking $\lambda_{\psi(\lambda)+1}=0$).

By Proposition 3.6, all boxes on $\ell_{\psi(\lambda)+1,\lambda_{\psi(\lambda)+1}}$ that lie strictly north of $(\psi(\lambda)+1,\lambda_{\psi(\lambda)+1})$ are contained in $\lambda$. If $\lambda_{\psi(\lambda)+1}=0$, then the claim is immediate. Otherwise suppose on the contrary that there are boxes south of $(\psi(\lambda)+1,\lambda_{\psi(\lambda)+1})$ that are not contained in $\lambda$. Then consider the north-most box $(i,j)\in\ell_{\psi(\lambda)+1,\lambda_{\psi(\lambda)+1}}\setminus\lambda$. Since $r_{1}^{x}=\ldots=r_{\omega(\lambda)}^{x}=a,$ $\ell_{\psi(\lambda)+1,\lambda_{\psi(\lambda)+1}}$ passes through a box in the first row. By Lemma 3.3, the box $(i-1,j)\in\lambda$. Then the hook $H_{\psi(\lambda)+1,j}=tb$, for some $t\in\mathbb{N}_{>0}$, must have $l_{\psi(\lambda)+1,j}=ta-1$, hence it is $(a,b)$-shallow, which contradicts Lemma 3.8. Thus, our initial assumption was false, and we conclude that $\ell_{\psi(\lambda)+1,\lambda_{\psi(\lambda)+1}}\subset\lambda$.

In the case where $\omega(\lambda)=\psi(\lambda)=0$, the ladder $\ell_{1,\lambda_{1}}$ is completely full so no boxes move when applying $\operatorname{Colseg}_{a,b}$. Hence $\lambda^{\operatorname{Colseg}_{a,b}}=\lambda_{[2,l(\lambda)]}$, which is the desired identity.

In the case where $\omega(\lambda)\geq 1$, and $\psi(\lambda)\geq a$, we know from above that no changes occur south of row $\psi(\lambda)+1$ since the ladder $\ell_{\psi(\lambda)+1,\lambda_{\psi(\lambda)+1}}$ passes through the first row and has all its boxes contained in $\lambda$. So we must have

[TABLE]

Consider the description of the $b$-rectangular decomposition stated in Proposition 3.6. For $\alpha=1,\ldots,\omega(\lambda)$, the northeast-most box in $r_{\alpha}$ (which is contained in $\lambda$), has coordinates $(1+(\alpha-1)a,\lambda_{1+(\alpha-1)a})$. From the shape of the $r_{\alpha}$’s given by Proposition 3.6, we know that

[TABLE]

Denote

[TABLE]

We define the $b$-gaps of $\lambda$ as a collection of boxes in the first $\psi(\lambda)+1$ rows of $\lambda$ and immediately southeast to a box in $\{\text{rim of }\lambda\}\setminus\{b\text{-rim of }\lambda\}$ (see Figure 15). More specifically, we let

[TABLE]

By Proposition 3.6, ladders passing through boxes in the first row also pass through some box in the first row of $r_{\alpha}:\alpha\in[1,\omega(\lambda)]$ and the row $\psi(\lambda)+1$. Hence, for $\alpha\in[1,\omega(\lambda)]$, $G_{\alpha}(\lambda)$ are exactly the $\delta_{\alpha}$ boxes $(1+\alpha a,\lambda_{1+\alpha a}+k_{\alpha}):k_{\alpha}\in[1,\delta_{\alpha}]$ that become occupied under $\operatorname{Colseg}_{a,b}$. Applying $\operatorname{Colseg}_{a,b}$ to $\lambda$ precisely slides boxes $(1,\lambda_{1}),\ldots,(1,\lambda_{1}-\sum_{\beta=1}^{\omega(\lambda)}\delta_{\beta})$ in order to the boxes in $G_{1}(\lambda)\sqcup\ldots\sqcup G_{\alpha}$ when sorted in order from northeast to southwest, and $\operatorname{Colseg}_{a,b}$ finishes by removing what remains in the first row.

Using the definition of $\operatorname{J}_{b}$, we know that $\lambda^{\operatorname{J}_{b}}_{i}=\lambda_{i+1}$ for all $i\in\{1,\ldots,\psi(\lambda)\}:i\neq\beta a$ for some $\beta\in[1,\omega(\lambda)]$. And for all $\beta\in[1,\omega(\lambda)]$, we have $\lambda^{\operatorname{J}_{b}}_{\beta a}=\lambda_{\beta a+1}+\delta_{\beta}$. Thus, we conclude that $\lambda^{\operatorname{Colseg}_{a,b}}_{[1,\psi(\lambda)]}=\lambda^{\operatorname{J}_{b}}_{[1,\psi(\lambda)]}.$

Since they agree in every part, $\lambda^{\operatorname{Colseg}_{a,b}}=\lambda_{[1,\psi(\lambda)]}^{\operatorname{J}_{b}}\oplus\lambda_{[\psi(\lambda)+2,l(\lambda)]},$ as desired. ∎

Remark 3.11*.*

Before proving Theorem 1.1, we again emphasize that in all previous results, we have only used the conditions that $\lambda$ is $\operatorname{Colreg}_{a,b}$-valid and $\lambda$ contains no hooks of shape (6). In the next section, we will claim that these results can be used towards proving a generalization of Theorem 1.1.

To finally prove Theorem 1.1, we now for the first time use the full extent of the condition that $\lambda$ contains no $(a,b)$-shallow hooks. The theorem is restated below for reference:

See 1.1

Proof of Theorem 1.1.

First, we note that $\lambda$ is $b$-regular. If on the contrary, there exists $i$ such that $\lambda_{i}=\ldots=\lambda_{i+b-1}>\lambda_{b}$, then the hook $H_{i,\lambda_{i}}$ is divisible by $b$ and has leg $b-1$ and arm [math], which is not $(a,b)$-shallow. Thus, $\lambda$ must be $b$-regular.

We proceed by induction on $|\lambda|.$ In the base case, we have $(1)^{\operatorname{M}_{b}\operatorname{T}}=(1)^{\operatorname{Colreg}_{a,b}}=(1)$ for all $a,b$.

Consider a partition $\lambda$ that is $\operatorname{Colreg}_{a,b}$-valid with all hooks divisible by $b$ being $(a,b)$-shallow. We assume that $\mu^{\operatorname{M}_{b}\operatorname{T}}=\mu^{\operatorname{Colreg}_{a,b}}$ holds for all $\mu$ with the same constraints and $|\mu|<|\lambda|$.

We claim that the $b$-rim strictly south of row $\psi(\lambda)$ forms a single $b^{\prime}$-segment. If there is a $b$-segment strictly south of $x=\psi(\lambda)$, it must be $(a,b)$-shallow by the conditions of the theorem. However, Lemma 3.8 implies that such an $(a,b)$-shallow hook cannot exist. Thus, the $b$-rim of $\lambda$ strictly south of $x=\psi(\lambda)$ is a single $b^{\prime}$-segment. Hence $\lambda^{\operatorname{J}_{b}}_{[\psi(\lambda)+1,l(\lambda^{\operatorname{J}_{b}})]}=\lambda_{[\psi(\lambda)+2,l(\lambda)]}$. Lemma 3.9 then implies $\lambda^{\operatorname{J}_{b}}=\lambda^{\operatorname{Colseg}_{a,b}}$, so

[TABLE]

In order to apply the inductive hypothesis, it remains to show that $\lambda^{\operatorname{J}_{b}}=\lambda^{\operatorname{Colseg}_{a,b}}$ is $\operatorname{Colreg}_{a,b}$-valid and has all hooks divisible by $b$ being $(a,b)$-shallow. The $\operatorname{Colreg}_{a,b}$-validity is clear since $\operatorname{Colreg}_{a,b}$ is defined recursively by $\operatorname{Colseg}_{a,b}$ via equation (5) and $\lambda$ is $\operatorname{Colreg}_{a,b}$-valid to begin with.

It remains to show that all hooks divisible by $b$ in $\lambda^{\operatorname{Colseg}_{a,b}}$ are $(a,b)$-shallow. Consider all hooks $H_{i,j}({\lambda^{\operatorname{Colseg}_{a,b}}})=tb$ for some $t\in\mathbb{N}_{>0}$ and compare them with $H_{i+1,j}({\lambda})$ (recall that when performing $\operatorname{Colseg}_{a,b}$, the first row is removed, so there is index shifting). If these two hooks have the same dimension, then $H_{i,j}({\lambda^{\operatorname{Colseg}_{a,b}}})$ is $(a,b)$-shallow because all hooks divisible by $b$ in $\lambda$ are $(a,b)$-shallow. They can only be different in the following three situations:

1. (1)

The foot box $\mathfrak{f}_{i,j}(\lambda^{\operatorname{Colseg}_{a,b}})=(i^{\prime},j)$ comes from a box in the first row of $\lambda$, but the hand box does not.

Suppose on the contrary that $H_{i,j}({\lambda^{\operatorname{Colseg}_{a,b}}})$ is not $(a,b)$-shallow, the corresponding leg $l_{i,j}({\lambda^{\operatorname{Colseg}_{a,b}}})\geq at$. Then $(i^{\prime}-ta,j+t(b-a))$, which is on the ladder $\ell_{i^{\prime},j}$, is not contained in $\lambda^{\operatorname{Colseg}_{a,b}}$ and north of $(i^{\prime},j)$, contradicting the definition of $\operatorname{Colseg}_{a,b}$. Thus, $H_{i,j}({\lambda^{\operatorname{Colseg}_{a,b}}})$ must be $(a,b)$-shallow. 2. (2)

The hand box $\mathfrak{h}_{i,j}(\lambda^{\operatorname{Colseg}_{a,b}})=(i,j^{\prime})$ comes from a box in the first row of $\lambda$, but the foot box does not.

Since boxes only slide to rows that are multiples of $a$ under $\operatorname{Colseg}_{a,b}$, in this case $i=\alpha a-1$ for $\alpha\geq 1$. Suppose on the contrary $H_{i,j}({\lambda^{\operatorname{Colseg}_{a,b}}})$ is not $(a,b)$-shallow, then we claim that $H_{i-a+1,j}(\lambda)$ is also divisible by $b$ and not $(a,b)$-shallow either.

$H_{i-a+1,j}(\lambda)$ is also divisible by $b$ is clear because the rectangle in the $b$-rectangular decomposition of $\lambda$ occupying rows $i-a+1$ to $i$ has $r^{x}=a$ and $r^{y}=b-a+1$ as shown in Figure 16. Then $H_{i-a+1,j}(\lambda)=(t+1)b$.

Then, $l_{i-a+1,j}(\lambda)=l_{i,j}(\lambda^{\operatorname{Colseg}_{a,b}})+a\geq(t+1)a$, so $H_{i-a+1,j}(\lambda)=(t+1)b$ is not $(a,b)$-shallow, a contradiction. Thus, we conclude $H_{i,j}(\lambda^{\operatorname{Colseg}_{a,b}})$ must be $(a,b)$-shallow. 3. (3)

Both the foot box $\mathfrak{f}_{i,j}(\lambda^{\operatorname{Colseg}_{a,b}})=(i^{\prime},j)$ and the hand box $\mathfrak{h}_{i,j}(\lambda^{\operatorname{Colseg}_{a,b}})=(i,j^{\prime})$ come from boxes in the first row of $\lambda$.

Proposition 3.6 implies that in this case both $i$ and $i^{\prime}$ are multiples of $a$, so we let $l_{i,j}(\lambda^{\operatorname{Colseg}_{a,b}})=i^{\prime}-i=a\alpha$. But we also have $a_{i,j}(\lambda^{\operatorname{Colseg}_{a,b}})=j^{\prime}-j\geq(b-a)\alpha+1$ since there are $\alpha$ rectangles between $y=j$ and $y=j^{\prime}$. In particular, these inequalities imply that $H_{i,j}(\lambda^{\operatorname{Colseg}_{a,b}})$ must be $(a,b)$-shallow.

Now we can apply the inductive hypothesis: $\lambda^{\operatorname{J}_{b}\operatorname{M}_{b}\operatorname{T}}=\lambda^{\operatorname{Colseg}_{a,b}\operatorname{Colreg}_{a,b}}.$ By Equation (5) and Equation (4), $\lambda^{\operatorname{Colreg}_{a,b}}=(|\lambda|-|\lambda^{\operatorname{Colseg}_{a,b}}|)\oplus\lambda^{\operatorname{Colseg}_{a,b}\operatorname{Colreg}_{a,b}}$ and $\lambda^{\operatorname{M}_{b}\operatorname{T}}=(|\lambda|-|\lambda^{\operatorname{J}_{b}}|)\oplus\lambda^{\operatorname{J}_{b}\operatorname{M}_{b}\operatorname{T}},$ so we find that $\lambda^{\operatorname{M}_{b}\operatorname{T}}=\lambda^{\operatorname{Colreg}_{a,b}}$. ∎

Remark 3.12*.*

Note that the proofs in this section do not require $a$ and $b$ to be co-prime.

4. Iwahori-Hecke Algebras and Conjectures

In this section we state the following two conjectures, which have been proven in the case of $a=1$ by Bessenrodt, Olsson, and Xu in [BOX99] and Fayers in [Fay08] respectively:

Conjecture 4.1**.**

Given positive co-prime integers $a<b$ and a $b$-regular, $\operatorname{Colreg}_{a,b}$-valid partition $\lambda$, if $\lambda^{\operatorname{M}_{b}\operatorname{T}}=\lambda^{\operatorname{Colreg}_{a,b}}$, then all hooks $H_{i,j}$ in $\lambda$ with $b\mid H_{i,j}$ must satisfy Equation (1):

[TABLE]

Conjecture 4.2**.**

Given positive integers $a$, $b$ with $2a<b$ and a partition $\lambda$ that is both $\operatorname{Reg}_{a,b}$-valid and $\operatorname{Colreg}_{a,b}$-valid, we have the following:

1. (1)

If all hooks $H_{i,j}$ in $\lambda$ with $b\mid H_{i,j}$ satisfy:

[TABLE]

then $\lambda^{\operatorname{Reg}_{a,b}\operatorname{M}_{b}}=\lambda^{\operatorname{T}\operatorname{Reg}_{a,b}}$ (i.e. $\lambda^{\operatorname{Reg}_{a,b}\operatorname{M}_{b}\operatorname{T}}=\lambda^{\operatorname{Colreg}_{a,b}})$. 2. (2)

If $a$ and $b$ are co-prime, and $\lambda$ satisfies $\lambda^{\operatorname{Reg}_{a,b}\operatorname{M}_{b}}=\lambda^{\operatorname{T}\operatorname{Reg}_{a,b}}$, then all hooks divisible by $b$ satisfy Equation (7).

Conjecture 4.1 and part (2) of Conjecture 4.2 are closely related to the $q$-decomposition numbers and we first recall the main facts about Iwahori-Hecke algebras that motivates the representation-theoretic relationship of Mullineux involution and column regularization.

Let $\mathbb{F}$ be a field and $q\in\mathbb{F}\setminus\{0\}$ and denote $\mathcal{H}=\mathcal{H}_{\mathbb{F},q}(S_{n})$ to be the Iwahori-Hecke algebra of $S_{n}$. We refer to [Mat99] for definitions and representation theory of $\mathcal{H}$. Let $b$ be the minimal integer satisfying $1+q+\cdots+q^{b-1}=0$ with $b\geq 2.$ For every partition $\lambda$ of $n$, there is a Specht module $S^{\lambda}$ of $\mathcal{H}$, and if $\lambda$ is $b$-regular, there is a simple module $D^{\lambda}$ which is the co-socle of $S^{\lambda}$. The decomposition numbers $d_{\lambda,\mu}=\left[S^{\lambda}:D^{\mu}\right]$ satisfy the following important properties:

Proposition 4.3** ([Jam76, LLT96]).**

The decomposition numbers $d_{\lambda,\mu}$ satisfies the following:

1. (1)

$d_{\lambda,\mu}=0$* unless $\lambda$ and $\mu$ have the same $b$-core;* 2. (2)

$d_{\lambda,\mu}=0$* unless $\mu\unlhd\lambda^{\operatorname{Reg}_{1,b}}$ and $d_{\lambda,\lambda^{\operatorname{Reg}_{1,b}}}=1$;* 3. (3)

If $\mu$ is $b$-regular, $d_{\lambda,\mu}=d_{\lambda^{\operatorname{T}},\mu^{\operatorname{M}_{b}}}$.

The $q$-analogue of decomposition numbers $d_{\lambda,\mu}(q)$ is defined by considering the graded multiplicity of $D^{\mu}$ inside $S^{\lambda}$ and we refer to [Kle10] for details regarding its definitions and properties. Lascoux, Leclerc and Thibon conjectured in [LLT96] that when $\mathbb{F}=\mathbb{C}$, the $d_{\lambda,\mu}(q)$’s are the same coefficients in the expansion of the lower global crystal basis $\{G(\mu)\mid\mu\in\mathcal{P}\text{ is }b\textrm{-}\text{regular}\}$ explained as follows. This result was proved by Ariki in [Ari96].

The readers may refer to [LLT96] for details on the quantized affine Lie algebra $U_{q}(\widehat{\mathfrak{sl}}_{b})$ and its action on the Fock space $\mathcal{F}=\bigoplus_{\lambda\in\mathcal{P}}\mathbb{Q}(q)\lambda$, which was originally developed by Misra-Miwa [MM90] using the work of Hayashi [Hay90]. In our discussion, we state without proof Kashiwara’s existence and uniqueness of crystal bases of the integrable highest weight modules of affine quantum algebra:

Theorem 4.4** ([K*+*91]).**

Denote $M(\Lambda_{0})_{\mathbb{Q}}=U_{\mathbb{Q}}^{-}\emptyset$ where $U^{-}_{\mathbb{Q}}\subset U_{q}(\widehat{\mathfrak{sl}}_{b})$ is the $\mathbb{Q}[q,q^{-1}]$-sub-algebra generated by the Chevalley generators $f_{i}^{(m)}$ of $U_{q}(\widehat{\mathfrak{sl}}_{b})$. Then there exists a unique $\mathbb{Q}[q,q^{-1}]$-basis (which is called the lower crystal basis) of $M(\Lambda_{0})_{\mathbb{Q}}$:

[TABLE]

such that the following holds:

1. (1)

$G(\mu)\equiv\mu\text{ mod }qL$; 2. (2)

$\overline{G(\mu)}=G(\mu)$.

We expand $G(\mu)$ using the standard basis $\mathcal{P}$:

[TABLE]

The $q$-decomposition numbers $d_{\lambda,\mu}(q)$ satisfies the following properties:

Theorem 4.5** ([LLT96, VV*+*99, Sch00]).**

The $q$-decomposition numbers satisfy the following:

1. (1)

$d_{\lambda,\mu}(q)\in\mathbb{N}[q]$; 2. (2)

$d_{\mu,\mu}(q)=1$* and $d_{\lambda,\mu}(q)=0$ unless $\lambda\unlhd\mu$, $|\lambda|=|\mu|$, and $\operatorname{Core}_{b}(\lambda)=\operatorname{Core}_{b}(\mu)$;* 3. (3)

$d_{\lambda^{\operatorname{T}},\mu^{\operatorname{M}_{b}}}(q)=q^{|\mu|_{b}}d_{\lambda,\mu}(q^{-1})$, where $|\mu|_{b}$ is the $b$-weight of $\mu$, which is the number of hooks in $\mu$ that is divisible by $b$.

The third property in the above theorem is important in relating the Mullineux involution with the $q$-decomposition numbers.

Conjecture 4.6**.**

Suppose that $\lambda$ and $\mu$ are partitions of the same size and $\lambda$ is $\operatorname{Reg}_{a,b}$-valid and $\mu$ is $(a,b)$-regular. Then $d_{\lambda,\lambda^{\operatorname{Reg}_{a,b}}}(q)=q^{s(\lambda)}$, where $s(\lambda)$ is the number of $(a,b)$-steep hooks in $\lambda$ that are divisible by $b$.

Conjecture 4.6 generalizes a part of Fayers’s main theorem [Fay07, Theorem 2.2] in terms of the generalized (resp. column) regularization. In fact, this is the only ingredient needed for Conjecture 4.1 and part (2) of Conjecture 4.2. Assume that Conjecture 4.6 holds, then we have:

Proof of Conjecture 4.1.

Suppose a $\operatorname{Colreg}_{a,b}$-valid partition $\lambda$ satisfies $\lambda^{\operatorname{M}_{b}\operatorname{T}}=\lambda^{\operatorname{Colreg}_{a,b}}$, then

[TABLE]

By Conjecture 4.6,

[TABLE]

where $s(\lambda^{\operatorname{T}})$ is also the number of $(a,b)$-shallow hooks in $\lambda.$

From Theorem 4.5, we know that

[TABLE]

Therefore, $|\lambda|_{b}=s(\lambda^{\operatorname{T}}),$ so all hooks in $\lambda$ divisible by $b$ must be $(a,b)$-shallow. ∎

Proof of part (2) of Conjecture 4.2.

Suppose we have a both $\operatorname{Colreg}_{a,b}$-valid and $\operatorname{Reg}_{a,b}$-valid partition $\lambda$ satisfying $\lambda^{\operatorname{Reg}_{a,b}\operatorname{M}_{b}}=\lambda^{\operatorname{T}\operatorname{Reg}_{a,b}}$. Then we have

[TABLE]

On the one hand, by Conjecture 4.6,

[TABLE]

On the other hand,

[TABLE]

where the first equality follows from Theorem 4.5, the second from Lemma 2.25 and the last one from Conjecture 4.6. Therefore, we find that $s(\lambda)+s(\lambda^{\operatorname{T}})=|\lambda|_{b}$. Lemma 2.3 indicates no hooks can be both $(a,b)$-shallow and $(a,b)$-steep in case of $2a<b$, which means all hooks divisible by $b$ must be either shallow or steep. ∎