The discrete Weibull model can be adapted to capture different levels of dispersion in the count data. This paper takes into account the direct relationship between explanatory variables and the median of discrete Weibull response variable. Additionally, it provides the Bayesian estimate of the discrete Weibull regression model using the random walk Metropolis algorithm. The prior distributions of the coefficient predictors were carried out based on the uniform non-informative, normal and Laplace distributions. The performance of the Bayes estimators was also compared with the maximum likelihood estimator in terms of the mean square error and the coverage probability through the Monte Carlo simulation study. Meanwhile, a real data set was analyzed to show how the proposed model and the methods work in practice.
Citation: Monthira Duangsaphon, Sukit Sokampang, Kannat Na Bangchang. Bayesian estimation for median discrete Weibull regression model[J]. AIMS Mathematics, 2024, 9(1): 270-288. doi: 10.3934/math.2024016
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The discrete Weibull model can be adapted to capture different levels of dispersion in the count data. This paper takes into account the direct relationship between explanatory variables and the median of discrete Weibull response variable. Additionally, it provides the Bayesian estimate of the discrete Weibull regression model using the random walk Metropolis algorithm. The prior distributions of the coefficient predictors were carried out based on the uniform non-informative, normal and Laplace distributions. The performance of the Bayes estimators was also compared with the maximum likelihood estimator in terms of the mean square error and the coverage probability through the Monte Carlo simulation study. Meanwhile, a real data set was analyzed to show how the proposed model and the methods work in practice.
The theory of cellular algebra was first introduced by Graham and Lehrer [1]. K¨onig and Xi [2] later gave a more structural equivalent definition of cellular theory. Suppose K is a field and A is an associative unital free K-algebra. In the sense of Graham and Lehrer, A is cellular if it has a K-basis {cs,t|λ∈Λ, s,t∈T(λ)}, where (Λ,≥) is a poset (partially ordered set) and T(λ) are finite index sets, such that
(i) The K-linear map ∗:A⟶A defined by cst↦cts for all λ∈Λ, s,t∈T(λ) is an anti-isomorphism of A.
(ii) For any λ∈Λ, t∈T(λ), and a∈A, there exists ratv∈K such that for all s∈T(λ),
csta≡∑v∈T(λ)ratvcsvmodA>λ. |
The basis {cst|λ∈Λ, s,t∈T(λ)} is the so called cellular basis. The existence of a cellular basis implies rich information on representations of A. One of the main uses of a cellular basis is to give the complete set of simple modules of A. According to Graham and Lehrer's theory, the cellular basis determines a cell filtration (a two-sided ideal filtration) A(λ1)⊂A(λ2)⊂⋯⊂A(λk) of A with respect to a total ordering λ1,λ2,…,λk of the poset Λ. As an A-module, each quotient A(λi)/A(λi−1) of the filtration is a direct sum of |T(λi)| copies of cell module C(λi). Moreover, for each λ∈Λ, the cellular basis attaches C(λ) a bilinear form ⟨,⟩λ such that C(λ)/rad⟨,⟩λ is either 0 or an irreducible module. Denote by D(λ) the quotient C(λ)/rad⟨,⟩λ; all the nonzero D(λ)s consist of a complete set of non-isomorphic simple A-modules. For a cellular algebra, it may possess different constructions of cellular bases. By Graham and Lehrer's theory, different cellular bases may determine different parameterizations of simple modules. So the study of the relationship between different parameterizations of simple modules becomes an interesting topic.
In this paper, we fix n as a natural number and ℓ a positive integer. The cyclotomic Hecke algebras of G(ℓ,1,n) was introduced by Ariki, Koike [3] and Brouˊe, Malle [4] independently. Many authors have constructed different cellular bases of cyclotomic Hecke algebras of G(ℓ,1,n). For example, Dipper, James, and Mathas [5] constructed the cellular basis \{m_{{\mathfrak{s}}{\mathfrak{t}}}| {\boldsymbol\lambda}\in\mathcal{P}^\ell_n\ \text{and}\ {\mathfrak{s}}, {\mathfrak{t}}\in\text{Std}({\boldsymbol\lambda})\} with respect to the poset ({\mathcal{P}}^\ell_n, \unrhd) , where {\mathcal{P}}^\ell_n is the set of \ell -partitions of n and \unrhd is the dominance order on {\mathcal{P}}^\ell_n . Through the cellular basis m_{{\mathfrak{s}}{\mathfrak{t}}} , Ariki [6] proved that the simple modules of cyclotomic Hecke algebras of G(\ell, 1, n) are paramaterized by Kleshchev multipartitons. By Brundan–Kleshchev's isomorphism [7], Hu and Mathas [8] constructed the graded cellular basis \{\psi_{{\mathfrak{s}}{\mathfrak{t}}}| {\boldsymbol\lambda}\in{\mathcal{P}}^\ell_n\ \text{and}\ {\mathfrak{s}}, {\mathfrak{t}}\in\text{Std}({\boldsymbol\lambda})\} of cyclotomic Hecke algebras of G(\ell, 1, n) with respect to the poset ({\mathcal{P}}^\ell_n, \unrhd) . Different from m_{{\mathfrak{s}}{\mathfrak{t}}} and \psi_{{\mathfrak{s}}{\mathfrak{t}}} , Bowman [9] constructed an integral graded cellular basis \{c_{{\mathfrak{s}}{\mathfrak{t}}}^\theta| {\boldsymbol\lambda}\in\mathcal{P}^\ell_n\ \text{and}\ {\mathfrak{s}}, {\mathfrak{t}}\in\text{Std}_\theta({\boldsymbol\lambda})\} of cyclotomic Hecke algebras of G(\ell, 1, n) with respect to the poset ({\mathcal{P}}^\ell_n, \unrhd_\theta) , where \unrhd_\theta is the \theta -dominance order on {\mathcal{P}}^\ell_n . Corresponding to Bowman's basis, the simple modules of cyclotomic Hecke algebra of G(\ell, 1, n) are labeled by Uglov multipartitions. We want to study the relationship between these different paramaterizations of simple modules of cyclotomic Hecke algebra of G(\ell, 1, n) . To this aim, it's necessary for us to understand the relationship between dominance order and \theta -dominance order on {\mathcal{P}}_n^\ell .
The content of this paper is organized as follows; In Section 2, we introduce some notations and definitions. In Section 3, we give a combinatorial description of the neighbors with weak \theta -dominance order whenever the loading \theta is strongly separated. In Section 4, we give the main results of this paper: The relationship between weak \theta -dominance order, \theta -dominance order, and dominance order. Throughout this paper, we denote by {\mathbb{N}} the set of natural numbers and {\mathbb{Z}} the set of integers.
A partition of n is a finite non-increasing sequence \lambda = (\lambda_1, \lambda_2, \dots) of non-negative integers with |\lambda| = \sum_i\lambda_i = n . If \lambda is a partition of n , we write \lambda\vdash n . Let {\mathcal{P}}_n be the set of partitions of n . The Young diagram of \lambda is a set
[\lambda] = \{(i,j)|1\leq j\leq\lambda_i,\forall i\geq 1\}. |
The elements of [\lambda] are called the nodes of \lambda . The Young diagram can be identified with a tableau. For example, \lambda = (3, 2, 1) is a partition of 6; its Young diagram
[\lambda] = \{(1,1),(1,2),(1,3),(2,1),(2,2),(3,1)\}, |
it can be identified with the tableau
![]() |
where (i, j) corresponds to the box in the i -th row and j -th column. For a partition \lambda , define its height h(\lambda) = \max\{k\in{\mathbb{N}}|\lambda_k\neq0\} .
A multipartition of n with \ell components is an ordered sequence {\boldsymbol\lambda} = (\lambda^{(1)}, \dots, \lambda^{(\ell)}) of partitions such that |\lambda^{(1)}|+\dots+|\lambda^{(\ell)}| = n . We denote by {\mathcal{P}}_n^\ell the set of multipartitions of n with \ell components. For {\boldsymbol\lambda}\in{\mathcal{P}}_n^\ell , we write {\boldsymbol\lambda}\vdash_\ell n and call {\boldsymbol\lambda} an \ell -partition of n . When \ell = 1 , it is clear {\mathcal{P}}_n^1 = {\mathcal{P}}_n . The Young diagram of {\boldsymbol\lambda} is a set
[ {\boldsymbol\lambda}] = \{(i,j,s)|1\leq s\leq\ell,1\leq j\leq\lambda_i^{(s)},\forall i\geq 1\}. |
The elements of [ {\boldsymbol\lambda}] are called the nodes of {\boldsymbol\lambda} . The Young diagram [ {\boldsymbol\lambda}] can be identified with an ordered sequence of tableaux. For example, {\boldsymbol\lambda} = ((2, 1), \emptyset, (3, 2, 1)) is a 3-partition of 9, the Young diagram
[ {\boldsymbol\lambda}] = \{(1,1,1),(1,2,1),(2,1,1),(1,1,3),(1,2,3),(1,3,3),(2,1,3),(2,2,3),(3,1,3)\}, |
it can be identified with the following ordered sequence of tableaux;
![]() |
where (i, j, s) corresponds to the box in the i -th row and j -th column of the s -th tableau. For simplicity, we identify {\boldsymbol\lambda} with its Young diagram [ {\boldsymbol\lambda}] .
Suppose {\boldsymbol\lambda}\in{\mathcal{P}}_n^\ell . If \alpha\in[ {\boldsymbol\lambda}] and [ {\boldsymbol\lambda}]\setminus\{\alpha\} is a Young diagram of \ell -partition of n-1 , then we call \alpha a removable node of {\boldsymbol\lambda} . If \beta\notin[ {\boldsymbol\lambda}] and [ {\boldsymbol\lambda}]\cup\{\beta\} is a Young diagram of \ell -partition of n+1 , then we call \beta an addable node of {\boldsymbol\lambda} .
Let {\boldsymbol\lambda} = (\lambda^{(1)}, \dots, \lambda^{(\ell)}), {\boldsymbol\mu} = (\mu^{(1)}, \dots, \mu^{(\ell)}) be \ell -partitions of n ; write {\boldsymbol\mu}\unrhd {\boldsymbol\lambda} if
\sum\limits^{s-1}_{a = 1}|\mu^{(a)}|+\sum\limits^{t}_{i = 1}\mu^{(s)}_j\geq\sum\limits^{s-1}_{a = 1}|\lambda^{(a)}|+\sum\limits^{t}_{i = 1}\lambda^{(s)}_j\quad\forall\,1\leq s\leq \ell\ \ \forall\,t\geq1. |
If {\boldsymbol\mu}\unrhd {\boldsymbol\lambda} and {\boldsymbol\mu}\neq {\boldsymbol\lambda} , we write {\boldsymbol\mu}\rhd {\boldsymbol\lambda} . In particular, for \lambda, \mu\in{\mathcal{P}}_n , write \mu\unrhd\lambda if
\sum\limits^{t}_{i = 1}\mu_i\geq\sum\limits^{t}_{i = 1}\lambda_i\quad\forall\,t\geq1. |
We call \unrhd the dominance order.
Let \mathcal{N}^\ell_n = \{(r, c, l)|r, c, l\in{\mathbb{N}}_{\geq1}, r+c\leq2(n+1), 1\leq l\leq\ell\}. The elements of \mathcal{N}^\ell_n are also called nodes and the subsets of \mathcal{N}^\ell_n are called configurations of nodes. By definition, the Young diagrams of \ell -partitions of n are configurations of nodes.
We fix e an element in {\mathbb{N}}_{\geq2}\cup\{\infty\} and I = {\mathbb{Z}}/e{\mathbb{Z}} , where I = {\mathbb{Z}} whenever e = \infty . An e -multicharge is a sequence (\kappa_1, \kappa_2, \dots, \kappa_\ell)\in I^\ell . For \alpha = (r, c, l)\in\mathcal{N}_n^\ell , we define its residue to be \text{res}(\alpha) = c-r+\kappa_l\in I . A loading is a sequence of integers \theta = (\theta_1, \dots, \theta_\ell) such that \theta_i-\theta_j\notin\ell{\mathbb{Z}} for i < j.
Definition 2.1. [9, Definition 1.2] Let \alpha = (r, c, l), \alpha^\prime = (r^\prime, c^\prime, l^\prime)\in\mathcal{N}^\ell_n . We write \alpha^\prime < _{\theta}\alpha if either
(i) \theta_l+\ell(r-c) < \theta_{l^\prime}+\ell(r^\prime-c^\prime) or
(ii) \theta_l+\ell(r-c) = \theta_{l^\prime}+\ell(r^\prime-c^\prime) and r+c < r^\prime+c^\prime .
Moreover, if \text{res}(\alpha) = \text{res}(\alpha^\prime) , then we write \alpha^\prime\lhd_\theta\alpha .
Definition 2.2. [9, Definition 1.2] Let {\boldsymbol\lambda}, {\boldsymbol\mu}\in{\mathcal{P}}^\ell_n , we write {\boldsymbol\mu}\unlhd_\theta {\boldsymbol\lambda} if
|\{\beta\in {\boldsymbol\mu}|\gamma\lhd_{\theta}\beta\}|\leq|\{\beta\in {\boldsymbol\lambda}|\gamma\lhd_{\theta}\beta\}|\quad\forall\gamma\in\mathcal{N}^\ell_n. |
We call \unlhd_\theta the \theta -dominance order.
Deleting the residue condition in the definition of \theta -dominance order, we can get a weak version of it.
Definition 2.3. Let {\boldsymbol\lambda}, {\boldsymbol\mu}\in{\mathcal{P}}^\ell_n , we write {\boldsymbol\mu}\leq_\theta {\boldsymbol\lambda} if
|\{\beta\in {\boldsymbol\mu}|\gamma < _{\theta}\beta\}|\leq|\{\beta\in {\boldsymbol\lambda}|\gamma < _{\theta}\beta\}|\quad \forall\gamma\in\mathcal{N}^\ell_n. |
We call \leq_\theta the weak \theta -dominance order.
Fix a loading \theta = (\theta_1, \dots, \theta_\ell) , if \theta_{i+1}-\theta_i > \ell n for each i = 1, 2, \dots, \ell-1 , then we call \theta a strongly separated loading.
Let {\boldsymbol\lambda} be a configuration of nodes. For i\in{\mathbb{Z}} , we call \{(r, c, l)\in {\boldsymbol\lambda}|\theta_l+\ell(r-c) = i\} the i -diagonal of {\boldsymbol\lambda} and d_{i}^ {\boldsymbol\lambda} = |\{(r, c, l)\in {\boldsymbol\lambda}|\theta_l+\ell(r-c) = i\}| the length of the i -diagonal. Let (r, c, l) be a node in the i -diagonal of {\boldsymbol\lambda} . We call (r, c, l) the terminal node (respectively, head node) in the i -diagonal of {\boldsymbol\lambda} if r^{\prime}+c^{\prime}\leq r+c (respectively, r^{\prime}+c^{\prime}\geq r+c ) for each (r^{\prime}, c^{\prime}, l) in the i -diagonal of {\boldsymbol\lambda} .
We give a rough description of the weak \theta -dominance order by the length of diagonals.
Lemma 3.1. Let {\boldsymbol\lambda}, {\boldsymbol\mu}\in{\mathcal{P}}_{n}^\ell , then {\boldsymbol\mu}\leq_\theta {\boldsymbol\lambda} if and only if
\sum\limits_{i = -\infty}^{t}d_{i}^ {\boldsymbol\lambda}\geq\sum\limits_{i = -\infty}^{t}d_{i}^{ {\boldsymbol\mu}}\quad\forall t\in{\mathbb{Z}}. |
Proof. Firstly, let us prove the necessity. Assume t to be an integer such that
\sum\limits_{i = -\infty}^{t}d_{i}^ {\boldsymbol\lambda} < \sum\limits_{i = -\infty}^{t}d_{i}^{ {\boldsymbol\mu}}. |
Since | {\boldsymbol\lambda}| = | {\boldsymbol\mu}| = n , hence there exists an integer t^\prime > t such that
● the t^\prime -diagonal of {\boldsymbol\lambda} is non-empty, and
● \forall t < t^{\prime\prime} < t^\prime , the t^{\prime\prime} -diagonal of {\boldsymbol\lambda} is empty.
Let \alpha be the head node in the t^\prime -diagonal of {\boldsymbol\lambda} ; then we have
|\{\beta\in {\boldsymbol\lambda}|\alpha < _\theta\beta\}| = \sum\limits_{i = -\infty}^{t}d_{i}^ {\boldsymbol\lambda} < \sum\limits_{i = -\infty}^{t}d_{i}^{ {\boldsymbol\mu}}\leq|\{\beta\in {\boldsymbol\mu}|\alpha < _\theta\beta\}|. |
This contradicts to {\boldsymbol\lambda}\geq_\theta {\boldsymbol\mu} .
Next, let us prove the sufficiency. Suppose \sum_{i = -\infty}^{t}d_{i}^ {\boldsymbol\lambda}\geq\sum_{i = -\infty}^{t}d_{i}^{ {\boldsymbol\mu}} for all t\in{\mathbb{Z}} . Let \gamma = (r, c, l) be a node in the t -diagonal of \mathcal{N}_n^\ell and (r^\prime, c^\prime, l^\prime) be the head node in the t -diagonal of \mathcal{N}_n^\ell . If \gamma\notin {\boldsymbol\lambda}\cup {\boldsymbol\mu} , then
|\{\alpha\in {\boldsymbol\lambda}|\gamma < _\theta\alpha\}| = \sum\limits_{i = -\infty}^{t}d_i^ {\boldsymbol\lambda}\geq\sum\limits_{i = -\infty}^{t}d_{i}^{ {\boldsymbol\mu}} = |\{\beta\in {\boldsymbol\mu}|\gamma < _\theta\beta\}|. |
If \gamma\in {\boldsymbol\lambda}\setminus {\boldsymbol\mu} , then
|\{\alpha\in {\boldsymbol\lambda}|\gamma < _\theta\alpha\}| = r-r^\prime+\sum\limits_{i = -\infty}^{t-1}d_i^ {\boldsymbol\lambda}\geq r-r^\prime+\sum\limits_{i = -\infty}^{t-1}d_{i}^{ {\boldsymbol\mu}}\geq|\{\beta\in {\boldsymbol\mu}|\gamma < _\theta\beta\}|. |
If \gamma\in {\boldsymbol\mu}\setminus {\boldsymbol\lambda} , then
|\{\alpha\in {\boldsymbol\lambda}|\gamma < _\theta\alpha\}| = \sum\limits_{i = -\infty}^{t}d_i^ {\boldsymbol\lambda}\geq \sum\limits_{i = -\infty}^{t}d_{i}^{ {\boldsymbol\mu}} > r-r^\prime+\sum\limits_{i = -\infty}^{t-1}d_{i}^{ {\boldsymbol\mu}} = |\{\beta\in {\boldsymbol\mu}|\gamma < _\theta\beta\}|. |
If \gamma\in {\boldsymbol\lambda}\cap {\boldsymbol\mu} , then
|\{\alpha\in {\boldsymbol\lambda}|\gamma < _\theta\alpha\}| = r-r^\prime+\sum\limits_{i = -\infty}^{t-1}d_i^ {\boldsymbol\lambda}\geq r-r^\prime+\sum\limits_{i = -\infty}^{t-1}d_{i}^{ {\boldsymbol\mu}} = |\{\beta\in {\boldsymbol\mu}|\gamma < _\theta\beta\}|. |
Therefore, {\boldsymbol\lambda}\geq_\theta {\boldsymbol\mu} .
Remark 3.2. Suppose \theta = (\theta_1, \dots, \theta_{\ell}) are strongly separated and {\boldsymbol\lambda} = (\lambda^{(1)}, \dots, \lambda^{(\ell)})\in{\mathcal{P}}_{n}^\ell . Let \alpha = (r, c, s) be a node in the i -diagonal of {\boldsymbol\lambda} and \alpha^\prime = (r^\prime, c^\prime, s+1) be a node in the i^\prime -diagonal of {\boldsymbol\lambda} . Then i^\prime > i . In fact, {\boldsymbol\lambda} is a multipartition of n , hence
i^\prime-i = \theta_{s+1}+\ell(r^\prime-c^\prime)-(\theta_{s}+\ell(r-c)) = (\theta_{s+1}-\theta_{s})+\ell(r^\prime+c-c^\prime-r) > \ell n+\ell(2-n) = 2\ell > 0. |
That is, the s -component \lambda^{(s)} is completely separated from the (s+1) -component \lambda^{(s+1)} .
For {\boldsymbol\lambda}, {\boldsymbol\mu}\in{\mathcal{P}}_n^\ell , we say that {\boldsymbol\lambda} and {\boldsymbol\mu} are neighbors with the weak \theta -dominance order if {\boldsymbol\mu} > _\theta {\boldsymbol\lambda} and there is no {\boldsymbol\gamma}\in{\mathcal{P}}_n^\ell such that {\boldsymbol\mu} > _\theta\boldsymbol\gamma > _\theta {\boldsymbol\lambda} .
In [[10], Theorem 1.4.10], there is a characterization of partitions that are neighbors with the usual dominance order. In the following lemma, let us prove a similar combinatorial description of neighbors with weak \theta -dominance order on partitions. Consequently, it will be clear that the weak \theta -dominance order coincides with the usual dominance order on partitions.
Lemma 3.3. Suppose \lambda, \mu\in{\mathcal{P}}_n and \mu > _\theta\lambda , then \lambda, \mu are neighbors with the weak \theta -dominance order if and only if there exist positive integers r < r^\prime such that one of the following (a) and (b) occurs, where
(a) r^\prime = r+1, \quad\mu_r = \lambda_r+1, \quad\mu_{r+1} = \lambda_{r+1}-1\quad\mathit{\text{and}}\quad\mu_t = \lambda_t\quad\forall t\neq r, r+1 ,
(b) \lambda_r = \lambda_{r^\prime}, \quad\mu_r = \lambda_r+1, \quad \mu_{r^\prime} = \lambda_{r^\prime}-1\quad\mathit{\text{and}}\quad \mu_{t} = \lambda_t\quad\forall t\neq r, r^{\prime} .
Proof. Let \ell = 1 and \theta, \theta^\prime\in{\mathbb{Z}} be different integers. For \alpha, \alpha^\prime\in\mathcal{N}_n^\ell , we have \alpha > _\theta\alpha^\prime if and only if \alpha > _{\theta^\prime}\alpha^\prime . Therefore, for each \lambda, \mu\in{\mathcal{P}}_n , we have \mu > _\theta\lambda if and only if \mu > _{\theta^\prime}\lambda . Hence, for simplicity, we assume \theta = 0 . By this assumption, node (a, b) lies in the (a-b) -diagonal.
First, let us prove the necessity. Assume \mu > _\theta\lambda to be partitions of n and there exist no \gamma\in{\mathcal{P}}_n such that \mu > _\theta\gamma > _\theta\lambda . Define i: = \min\{k\in{\mathbb{Z}}|d^\mu_k\neq d^\lambda_k\} . Then i is well defined since \mu\neq\lambda . Define i^\prime: = \min\Bigl\{k\in{\mathbb{Z}}\big|\sum_{t = -\infty}^kd^\mu_t = \sum_{t = -\infty}^kd^\lambda_t, i < k\Bigr\} . Then i^\prime is well defined since |\lambda| = |\mu| = n . By definition, -n < i < i^\prime < n . Combining with Lemma 3.1, we derive
\begin{align} 0\leq d^\lambda_i < d^\mu_i,\quad d^\mu_{i-1} = d^\lambda_{i-1} \end{align} | (3.4) |
and
\begin{align} d^\lambda_{i^\prime} > d^\mu_{i^\prime}\geq0,\quad d^\mu_{i^\prime+1}\geq d^\lambda_{i^\prime+1}. \end{align} | (3.5) |
We will give the proof of necessity in 3 steps:
Step 1. Let \alpha = (r, c) be the terminal node in the i -diagonal of \mu . Let us prove \lambda_{r-1}\geq\lambda_r+1 and (r, c) is the last node in the r -th row of \mu .
If i\leq 0 , let us prove d_{i-1}^\mu = d_i^\mu-1 . We should prove the i -diagonal and (i-1) -diagonal of \mu is like
![]() |
where \clubsuit and \alpha = (r, c) are the i -diagonal of \mu and \spadesuit are the (i-1) -diagonal of \mu . If d^\mu_{i-1} > d^\mu_{i} , then we derive \eta = (r+1, c+2)\in\mu , hence \gamma = (r+1, c+1)\in\mu . Then the i -diagonal and (i-1) -diagonal of \mu is like
![]() |
where \clubsuit , \alpha = (r, c) and \gamma = (r+1, c+1) are the i -diagonal of \mu , while \spadesuit and \eta = (r+1, c+2) are the (i-1) -diagonal of \mu . This contradicts that \alpha = (r, c) is the terminal node in the i -diagonal of \mu ; therefore, d^\mu_{i-1}\leq d^\mu_i . Similarly, one can prove d_{i-1}^\lambda\leq d_{i}^\lambda . If d^\mu_{i-1} = d^\mu_i , by (3.4), we have d^\lambda_{i-1} = d^\mu_{i-1} = d^\mu_i > d_i^\lambda ; this contradicts d_{i-1}^\lambda\leq d_i^\lambda , hence d^\mu_{i-1} < d^\mu_i . If d_{i-1}^\mu < d_{i}^\mu-1 , then (r-1, c)\notin\mu . Then the i -diagonal and (i-1) -diagonal of \mu are like
![]() |
where \clubsuit and \alpha = (r, c) are the i -diagonal of \mu and \spadesuit are the (i-1) -diagonal of \mu . This contradicts to \alpha = (r, c)\in\mu and \mu_{r-1}\geq\mu_r . Therefore, d_{i-1}^\mu = d_i^\mu-1 . By (3.4), we have d_{i-1}^\lambda = d_{i-1}^\mu = d^\mu_i-1 > d_i^\lambda-1 , hence d_i^\lambda = d_{i-1}^\lambda = d_{i-1}^\mu = d_{i}^\mu-1 . So we derive \alpha = (r, c)\notin\lambda and \delta = (r-1, c)\in\lambda . Hence, the i -diagonal and (i-1) -diagonal of \lambda are like
![]() |
where \clubsuit are the i -diagonal of \lambda and \spadesuit , \delta = (r-1, c) are the (i-1) -diagonal of \lambda . The i -diagonal and (i-1) -diagonal of \mu are like
![]() |
where \clubsuit , \alpha = (r, c) are the i -diagonal of \mu and \spadesuit are the (i-1) -diagonal of \mu . Therefore \lambda_{r-1}\geq\lambda_r+1 . Moreover, (r, c) is the last node in the r -th row of \mu .
For the case when i > 0 , the discussion is tedious and similar to that of i\leq0 , so we don't show it here again.
Step 2. Let \beta = (r^\prime, c^\prime) be the terminal node in the i^\prime -diagonal of \lambda . Let us prove \lambda_{r^\prime}-1\geq\lambda_{r^\prime+1} and r^\prime > r .
If i^\prime\geq 0 , let us prove d_{i^\prime+1}^\lambda = d_{i^\prime}^\lambda-1 . We should prove the i^\prime -diagonal and (i^\prime+1) -diagonal of \lambda are like
![]() |
where \clubsuit and \beta = (r^\prime, c^\prime) are the i^\prime -diagonal of \lambda and \spadesuit are the (i^\prime+1) -diagonal of \lambda . If d_{i^\prime+1}^\lambda > d_{i^\prime}^\lambda , then \eta = (r^\prime+2, c^\prime+1)\in\lambda , hence \gamma = (r^\prime+1, c^\prime+1)\in\lambda . The i^\prime -diagonal and (i^\prime+1) -diagonal of \lambda are like
![]() |
where \clubsuit , \beta = (r^\prime, c^\prime) and \gamma = (r^\prime+1, c^\prime+1) are the i^\prime -diagonal of \lambda and \spadesuit , \eta = (r^\prime+2, c^\prime+1) are the (i^\prime+1) -diagonal of \lambda . This contradicts that \beta = (r^\prime, c^\prime) is the terminal node in the i^\prime -diagonal of \lambda . Therefore, d_{i^\prime+1}^\lambda\leq d_{i^\prime}^\lambda . Similarly, we can prove d^\mu_{i^\prime+1}\leq d_{i^\prime}^\mu . If d_{i^\prime+1}^\lambda = d_{i^\prime}^\lambda , by (3.5), we have d^\mu_{i^\prime+1}\geq d^\lambda_{i^\prime+1} = d_{i^\prime}^\lambda > d^\mu_{i^\prime}. This contradicts to d^\mu_{i^\prime+1}\leq d_{i^\prime}^\mu . If d_{i^\prime+1}^\lambda < d_{i^\prime}^\lambda-1 , then \eta = (r^\prime, c^\prime-1)\notin\lambda. Then the i^\prime -diagonal and (i^\prime+1) -diagonal of \lambda are like
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where \clubsuit and \beta = (r^\prime, c^\prime) are the i^\prime -diagonal of \lambda and \spadesuit are the (i^\prime+1) -diagonal of \lambda . This contradicts to \beta = (r^\prime, c^\prime)\in\lambda . Therefore, d_{i^\prime+1}^\lambda = d^\lambda_{i^\prime}-1 , by (3.5), we have d_{i^\prime+1}^\mu\geq d^\lambda_{i^\prime+1} = d^\lambda_{i^\prime}-1 > d_{i^\prime}^\mu-1. Therefore, we derive d^\mu_{i^\prime} = d^\mu_{i^\prime+1} = d_{i^\prime+1}^\lambda = d_{i^\prime}^\lambda-1. Hence, (r^\prime+1, c^\prime)\notin\lambda, \quad \beta = (r^\prime, c^\prime)\notin\mu. Then the i^\prime -diagonal and (i^\prime+1) -diagonal of \lambda are like
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where \clubsuit and \beta = (r^\prime, c^\prime) are the i^\prime -diagonal of \lambda and \spadesuit are the (i^\prime+1) -diagonal of \lambda . The i^\prime -diagonal and (i^\prime+1) -diagonal of \mu are like
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where \clubsuit are the i^\prime -diagonal of \mu and \spadesuit are the (i^\prime+1) -diagonal of \mu . Therefore, we have \lambda_{r^\prime}-1\geq\lambda_{r^\prime+1} . Next, let us prove r^\prime > r . If r^\prime = r , since (r^\prime, c^\prime)\notin\mu and (r, c) is the last node in the r -th row of \mu , so we have c^\prime > c , hence i^\prime = r-c^\prime < r-c = i , this contradicts to i^\prime > i . If r^\prime < r , since r^\prime-c^\prime = i^\prime > i = r-c , then c^\prime+s < c , where s = r-r^\prime > 0 . Since (r, c) is the last node in the r -th row of \mu , so (r, c^\prime+s)\in\mu . Moreover, since r-(c^\prime+s) = r^\prime-c^\prime = i^\prime , so (r, c^\prime+s) lies in the i^\prime -diagonal of \mu and (r, c^\prime+s) < _\theta(r^\prime, c^\prime) , this contradicts to (r^\prime, c^\prime)\notin\mu . Therefore, we derive r^\prime > r .
For the case when i < 0 , the discussion is also tedious and similar to that of i\geq0 , so we do not show it here again.
Step 3. Now we have proved r^\prime > r , \lambda_{r-1}\geq\lambda_r+1 and \lambda_{r^\prime}-1\geq\lambda_{r^\prime+1} . Hence
\gamma = (\lambda_1,\dots,\lambda_{r-1},\lambda_r+1,\lambda_{r+1},\dots,\lambda_{r^\prime-1},\lambda_{r^\prime}-1,\lambda_{r^\prime+1},\dots) |
is a partition of n . Let \alpha^\prime = (r, \lambda_{r}+1) , \beta^\prime = (r^\prime, \lambda_{r^\prime}) and j = r-(\lambda_r+1) , j^\prime = r^\prime-\lambda_{r^\prime} . Then \alpha^\prime is the terminal node in the j -diagonal of \gamma , and \beta^\prime is the terminal node in the j^\prime -diagonal of \lambda . We can obtain \gamma from \lambda by removing \beta^\prime to \alpha^\prime . Since r^\prime > r , \lambda_{r}+1 > \lambda_{r^\prime} , hence j^\prime = r^\prime-\lambda_{r^\prime} > r-(\lambda_r+1) = j and \beta^\prime = (r^\prime, \lambda_{r^\prime}) < _\theta(r, \lambda_r+1) = \alpha^\prime . So we have \gamma > _{\theta}\lambda . Next, let us prove \mu\geq_{\theta}\gamma . Since \alpha = (r, c)\notin\lambda , so \lambda_r+1\leq c , hence j = r-(\lambda_r+1)\geq r-c = i and \alpha^\prime = (r, \lambda_r+1)\leq_\theta(r, c) = \alpha. Since \beta = (r^\prime, c^\prime)\in\lambda , so \lambda_{r^\prime}\geq c^\prime and j^\prime = r^\prime-\lambda_{r^\prime}\leq r^\prime-c^\prime = i^\prime . Hence \beta = (r^\prime, c^\prime)\leq_\theta(r^\prime, \lambda_{r^\prime}) = \beta^\prime . Therefore,
\alpha = (r,c)\geq_\theta\alpha^\prime = (r,\lambda_r+1) > _\theta\beta^\prime = (r^\prime,\lambda_{r^\prime})\geq_\theta\beta = (r^\prime,c^\prime)\quad\text{and}\quad i\leq j < j^\prime\leq i^\prime. |
Combining with the choice of i, i^\prime and Lemma 3.1, we derive \mu\geq_\theta\gamma . Hence \mu\geq_\theta\gamma > _\theta\lambda . Since \lambda and \mu are neighbors with \geq_\theta , so we have \mu = \gamma .
Finally, let us prove r = r^\prime-1 or \lambda_r = \lambda_{r^\prime} . Otherwise, suppose r\neq r^\prime-1 and \lambda_r\neq\lambda_{r^\prime} , then r < r^\prime-1 and \lambda_r > \lambda_{r^\prime} . Let t = 1+\min\{k|\lambda_{k} > \lambda_{k+1}, r\leq k < r^\prime\} . If t = r^\prime , then \lambda_r = \dots = \lambda_{r^\prime-1} > \lambda_{r^\prime} > 0 , let \nu = (\lambda_1, \dots, \lambda_{r-1}, \lambda_{r}+1, \lambda_{r+1}, \dots, \lambda_{r^\prime-2}, \lambda_{r^\prime-1}-1, \lambda_{r^\prime}, \dots) . Since (r^\prime, \lambda_{r^\prime}) < _\theta(r^\prime-1, \lambda_{r^\prime-1}) < _\theta(r, \lambda_r+1) , hence \mu > _\theta\nu > _\theta\lambda , this contradicts that \mu and \lambda are neighbors with \geq_\theta . If t < r^\prime , then \lambda_r = \dots = \lambda_{t-1} > \lambda_t\geq\dots\geq\lambda_{r^\prime} > 0 , let
\eta: = (\lambda_1,\dots,\lambda_{t-1},\lambda_{t}+1,\dots,\lambda_{r^\prime-1},\lambda_{r^\prime}-1,\lambda_{r^\prime+1},\dots). |
Since (r^\prime, \lambda_{r^\prime}) < _\theta(t, \lambda_{t}+1) < _\theta(r, \lambda_r+1), then \mu > _\theta\eta > _\theta\lambda , this contradicts that \mu and \lambda are neighbors with \geq_\theta . Therefore, r = r^\prime-1 or \lambda_r = \lambda_{r^\prime} . Now we complete the proof of necessity.
Next, let us prove the sufficiency. Suppose \lambda, \mu\in{\mathcal{P}}_n and there exist positive integers r < r^\prime satisfying
(a) r^\prime = r+1, \quad\mu_r = \lambda_r+1, \quad\mu_{r+1} = \lambda_{r+1}-1\quad\text{and}\quad\mu_t = \lambda_t\quad\forall t\neq r, r+1 , or
(b) \lambda_r = \lambda_{r^\prime}, \quad\mu_r = \lambda_r+1, \quad \mu_{r^\prime} = \lambda_{r^\prime}-1\quad\text{and}\quad \mu_{t} = \lambda_t\quad\forall t\neq r, r^{\prime} .
Let \nu\in{\mathcal{P}}_n^\ell such that \mu\geq_\theta\nu > _\theta\lambda and \lambda, \nu are neighbors with \geq_\theta . Let us prove \mu = \nu . Let i = r-(\lambda_r+1) , i^\prime = r^\prime-\lambda_{r^\prime} , it is clear i < i^\prime . By assumption, \mu can be obtained from \lambda by removing (r^\prime, \lambda_{r^\prime}) to (r, \lambda_r+1) . By Lemma 3.1 and the the choice of \nu , we have
\begin{align} d^\mu_t = d^\nu_t = d^\lambda_t,\quad\quad\text{for all}\ t < i\ \text{or}\ t > i^\prime. \end{align} | (3.6) |
By the necessity of this lemma, there exist integers s < s^\prime such that
\nu_s = \lambda_s+1,\quad \nu_{s^\prime} = \lambda_{s^\prime}-1,\quad \nu_t = \lambda_t\quad\forall t\neq s,s^\prime. |
Let j = s-(\lambda_{s}+1), \ j^\prime = s^\prime-\lambda_{s^\prime} , it is clear j < j^\prime . In other words, \nu can be obtained from \lambda by removing (s^\prime, \lambda_{s^\prime}) to (s, \lambda_s+1) . Combining with (3.6), we know i\leq j < j^\prime\leq i^\prime .
If (a) occurs, r = r^\prime-1 , the i -diagonal and i^\prime -diagonal of \lambda are like
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where \spadesuit are the i -diagonal and \clubsuit are the i^\prime -diagonal. The i -diagonal and i^\prime -diagonal of \mu are like
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where \spadesuit are the i -diagonal and \clubsuit are the i^\prime -diagonal. By the above arguments, we have s^\prime-\lambda_{s^\prime} = j^\prime = i^\prime = r^\prime-\lambda_{r^\prime} , s-\lambda_s-1 = j = i = r-\lambda_r-1 . Since the addable node and removable node are unique for the i -diagonal and i^\prime -diagonal of \lambda , respectively. Hence, s = r, \ s^\prime = r^\prime and \mu = \nu .
If (b) occurs, r < r^\prime-1 and \lambda_r = \lambda_{r+1} = \dots = \lambda_{r^\prime} , the i -diagonal and i^\prime -diagonal of \lambda are like
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where \spadesuit are the i -diagonal and nodes \clubsuit are the i^\prime -diagonal. The i -diagonal and i^\prime -diagonal of \lambda are like
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where \spadesuit are the i -diagonal and \clubsuit are the i^\prime -diagonal. By the above arguments, we have s^\prime-\lambda_{s^\prime} = j^\prime = i^\prime = r^\prime-\lambda_{r^\prime}, \quad s-\lambda_s-1 = j = i = r-\lambda_r-1 . Since the addable node and removable node are unique for the i -diagonal and i^\prime -diagonal of \lambda respectively. Hence, s = r, \ s^\prime = r^\prime , and \mu = \nu .
Now we can give a combinatorial description of neighbors with weak \theta -dominance order on multipartitions.
Proposition 3.7. Suppose \theta = (\theta_1, \dots, \theta_\ell) be strongly separated, {\boldsymbol\lambda} = (\lambda^{(1)}, \dots, \lambda^{(\ell)}) and {\boldsymbol\mu} = (\mu^{(1)}, \dots, \mu^{(\ell)}) be \ell -partitions of n with {\boldsymbol\mu} > _\theta {\boldsymbol\lambda} . Then {\boldsymbol\lambda}, {\boldsymbol\mu} are neighbors with the weak \theta -dominance order if and only if one of (a) , (b) , and (c) occurs, where
(a) there exists s < \ell such that
\begin{align*} &\mu^{(r)} = \lambda^{(r)}\quad \forall r\neq s,s+1\\ &\mu^{(s)} = (\lambda_1^{(s)},\dots,\lambda^{(s)}_{k_s},1)\quad\mathit{\text{where }}\quad k_s = h(\lambda^{(s)}) \\ &\mu^{(s+1)}_1 = \lambda^{(s+1)}_1-1\quad\mathit{\text{and}}\quad\mu^{(s+1)}_j = \lambda^{(s+1)}_j\quad \forall j > 1. \end{align*} |
(b) there exist s and i such that
\begin{align*} &\mu^{(r)} = \lambda^{(r)}\quad \forall r\neq s\\ &\mu^{(s)}_j = \lambda_j^{(s)}\quad \forall j\neq i,i+1\\ &\mu^{(s)}_i = \lambda^{(s)}_i+1\quad\mathit{\text{and}}\quad\mu^{(s)}_{i+1} = \lambda^{(s)}_{i+1}-1. \end{align*} |
(c) there exist s and i < i^\prime such that
\begin{align*} &\mu^{(r)} = \lambda^{(r)}\quad \forall r\neq s\\ &\mu^{(s)}_j = \lambda_j^{(s)}\quad \forall j\neq i,i^\prime\\ &\mu^{(s)}_i-1 = \mu^{(s)}_{i^\prime}+1 = \lambda^{(s)}_i = \lambda^{(s)}_{i^\prime}. \end{align*} |
Proof. Let us prove the necessity. We assume {\boldsymbol\lambda} = (\lambda^{(1)}, \dots, \lambda^{(\ell)}) and {\boldsymbol\mu} = (\mu^{(1)}, \dots, \mu^{(\ell)}) are \ell -partitions of n with {\boldsymbol\mu} > _\theta {\boldsymbol\lambda} and there exists no \boldsymbol\gamma = (\gamma^{(1)}, \dots, \gamma^{(\ell)})\in{\mathcal{P}}_n^\ell such that {\boldsymbol\mu} > _\theta\boldsymbol\gamma > _\theta {\boldsymbol\lambda} . Define
s: = \min\{k|\mu^{(k)}\neq\lambda^{(k)}\}, |
then \lambda^{(s)}\neq\mu^{(s)} and \lambda^{(r)} = \mu^{(r)} for all r < s . Since {\boldsymbol\mu} > _\theta {\boldsymbol\lambda} and \theta is strongly separated, combining with Lemma 3.1 and Remark 3.2, we have
\begin{align} \sum\limits_{k = 1}^{t}|\mu^{(k)}|\geq\sum\limits_{k = 1}^{t}|\lambda^{(k)}|\quad\forall 1\leq t\leq\ell. \end{align} | (3.8) |
Hence |\mu^{(s)}|\geq|\lambda^{(s)}| .
Suppose |\mu^{(s)}| > |\lambda^{(s)}| , then s < \ell . Set m_s = h(\mu^{(s)}) , let us prove \mu^{(s)}_{m_s} = 1 . Otherwise, assume \mu^{(s)}_{m_s} > 1 . Let \boldsymbol\gamma = (\gamma^{(1)}, \dots, \gamma^{(s)}, \dots, \gamma^{(\ell)}) , where
\begin{align*} &\gamma^{(r)} = \mu^{(r)}\quad \forall r\neq s,\\ &\gamma^{(s)} = (\mu^{(s)}_1,\dots,\mu^{(s)}_{m_s-1},\mu^{(s)}_{m_s}-1,1). \end{align*} |
Since (m_s, \mu^{(s)}_{m_s}, s) > _\theta(m_s+1, 1, s) , so {\boldsymbol\mu} > _\theta\boldsymbol\gamma . Let us prove \boldsymbol\gamma > _\theta {\boldsymbol\lambda} . Let
u = \theta_s+\ell(1-\mu^{(s)}_1),\ p = \theta_{s}+\ell(m_s-\mu^{(s)}_{m_s}),\ q = \theta_{s}+\ell(m_s+1-1) = \theta_s+\ell m_s. |
Where u\leq p < q . Define
\bigstar: = (1,\mu^{(s)}_1,s)\text{ is the unique node in the } u \text{-diagonal of } {\boldsymbol\mu} |
\clubsuit: = (m_s,\mu^{(s)}_{m_s},s)\text{ is the terminal node in the } p \text{-diagonal of } {\boldsymbol\mu} |
\spadesuit: = (m_s+1,1,s)\text{ is the unique node in the } q \text{-diagonal of } \boldsymbol\gamma . |
\boldsymbol\gamma can be obtained from {\boldsymbol\mu} by removing \clubsuit to \spadesuit . The Young diagrams of \mu^{(s)} and \gamma^{(s)} are like
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By the construction of \boldsymbol\gamma , we have d_{t}^{ {\boldsymbol\mu}} = d_{t}^{\boldsymbol\gamma} for all t\neq p, q . By Remark 3.2 and the choice of s , we have
\sum\limits_{t = -\infty}^{u-1}d_t^{\boldsymbol\gamma} = \sum\limits_{t = -\infty}^{u-1}d_t^{ {\boldsymbol\mu}} = \sum\limits_{t = 1}^{s-1}|\mu^{(t)}| = \sum\limits_{t = 1}^{s-1}|\lambda^{(t)}| = \sum\limits_{t = -\infty}^{u-1}d_t^{ {\boldsymbol\lambda}}. |
By Lemma 3.1, we derive
\begin{align} \sum\limits_{t = u}^{k}d_t^{\boldsymbol\gamma} = \sum\limits_{t = u}^{k}d_t^{ {\boldsymbol\mu}}\geq\sum\limits_{t = u}^{k}d_t^{ {\boldsymbol\lambda}}\quad\forall u\leq k < p \end{align} | (3.9) |
\begin{align} 1+\sum\limits_{t = u}^{k}d_t^{\boldsymbol\gamma} = \sum\limits_{t = u}^{k}d_t^{ {\boldsymbol\mu}}\geq\sum\limits_{t = u}^{k}d_t^{ {\boldsymbol\lambda}}\quad\forall p\leq k < q \end{align} | (3.10) |
and
\sum\limits_{t = -\infty}^{k}d_t^{\boldsymbol\gamma} = \sum\limits_{t = -\infty}^{k}d_t^{ {\boldsymbol\mu}}\geq\sum\limits_{t = -\infty}^{k}d_t^{ {\boldsymbol\lambda}}\quad \forall k\geq q. |
If \boldsymbol\gamma\ngeq_\theta {\boldsymbol\lambda} , then there exists p\leq\epsilon < q such that
\begin{align} \sum\limits_{t = u}^{\epsilon-1}d_{t}^{\boldsymbol\gamma}\geq\sum\limits_{t = u}^{\epsilon-1}d_t^{ {\boldsymbol\lambda}},\qquad\sum\limits_{t = u}^{\epsilon}d_{t}^{\boldsymbol\gamma} < \sum\limits_{t = u}^{\epsilon}d_t^{ {\boldsymbol\lambda}}. \end{align} | (3.11) |
If \epsilon = p , by (3.9)–(3.11), we have
\sum\limits_{t = u}^{p-1}d_t^{\boldsymbol\gamma} = \sum\limits_{t = u}^{p-1}d_t^{ {\boldsymbol\mu}}\geq\sum\limits_{t = u}^{p-1}d_t^{ {\boldsymbol\lambda}}\quad\text{and} \quad 1+\sum\limits_{t = u}^{p}d_t^{\boldsymbol\gamma} = \sum\limits_{t = u}^{p}d_t^{ {\boldsymbol\mu}} = \sum\limits_{t = u}^{p}d_t^{ {\boldsymbol\lambda}}, |
hence d_p^{ {\boldsymbol\lambda}}\geq d_p^{ {\boldsymbol\mu}} . So \clubsuit = (m_s, \mu^{(s)}_{m_s}, s)\in\lambda^{(s)} and hence |\lambda^{(s)}|\geq|\mu^{(s)}| , this contradicts to |\mu^{(s)}| > |\lambda^{(s)}| . If p < \epsilon < q , by (3.9)–(3.11), we have
\sum\limits_{t = u}^{\epsilon-1}d_{t}^{ {\boldsymbol\mu}} > \sum\limits_{t = u}^{\epsilon-1}d_{t}^{\boldsymbol\gamma}\geq\sum\limits_{t = u}^{\epsilon-1}d_t^{ {\boldsymbol\lambda}},\qquad\sum\limits_{t = u}^{\epsilon}d_{t}^{ {\boldsymbol\mu}} = 1+\sum\limits_{t = u}^{\epsilon}d_{t}^{\boldsymbol\gamma} = \sum\limits_{t = u}^{\epsilon}d_t^{ {\boldsymbol\lambda}} |
then d_\epsilon^{ {\boldsymbol\lambda}} > d_\epsilon^{ {\boldsymbol\mu}} and |\lambda^{(s)}| > |\mu^{(s)}| ; this contradicts to |\mu^{(s)}| > |\lambda^{(s)}| . Therefore \boldsymbol\gamma\geq_\theta {\boldsymbol\lambda} . Since |\gamma^{(s)}| = |\mu^{(s)}| > |\lambda^{(s)}| , hence \boldsymbol\gamma\neq {\boldsymbol\lambda} . So we derive
{\boldsymbol\mu} > _\theta\boldsymbol\gamma > _\theta {\boldsymbol\lambda}. |
This contradicts that {\boldsymbol\lambda} and {\boldsymbol\mu} are neighbors with \geq_\theta . So we have \mu^{(s)}_{m_s} = 1 .
Let \boldsymbol\nu: = (\nu^{(1)}, \dots, \nu^{(s)}, \nu^{(s+1)}, \dots, \nu^{(\ell)}) , where
\begin{align*} &\nu^{(r)} = \mu^{(r)}\quad \forall r\neq s,s+1,\\ &\nu^{(s)} = (\mu^{(s)}_1,\dots,\mu^{(s)}_{m_s-1}) \end{align*} |
and \nu^{(s+1)} = (\nu^{(s+1)}_1, \nu^{(s+1)}_2\dots) , where
\nu^{(s+1)}_1 = \mu^{(s+1)}_1+1,\ \nu^{(s+1)}_t = \mu^{(s+1)}_t\quad t > 1. |
Let
\clubsuit: = (m_s,1,s),\quad\spadesuit: = (1,\nu^{(s+1)}_1,s+1) |
and
\begin{align*} u^\prime:& = \theta_s+\ell(m_s-1),\\ q^\prime:& = \theta_{s+1}+\ell(1-\nu_1^{(s+1)}),\\ p^\prime:& = \theta_{s+1}+\ell(1-\lambda^{(s+1)}_1). \end{align*} |
Let r^\prime: = \min\{q^\prime, p^\prime\} , by Remark 3.2, we have u^\prime < r^\prime . Since \mu^{(s)}_{m_s} = 1 , so \clubsuit is the unique node in the u^\prime -diagonal of {\boldsymbol\mu} and \spadesuit is the unique node in the q^\prime -diagonal of \boldsymbol\nu . That is, \boldsymbol\nu can be obtained from {\boldsymbol\mu} by removing the node \clubsuit to \spadesuit . From the point of Young's diagram
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By Lemma 3.1, we have {\boldsymbol\mu} > _\theta\boldsymbol\nu . Next, let us prove \boldsymbol\nu\geq_\theta {\boldsymbol\lambda} . By Lemma 3.1, we have
\sum\limits_{t = -\infty}^{k}d_t^{\boldsymbol\nu} = \sum\limits_{t = -\infty}^{k}d_t^{ {\boldsymbol\mu}}\geq\sum\limits_{t = -\infty}^{k}d_t^ {\boldsymbol\lambda}\quad\text{where}\ k < u^\prime\text{ or }k\geq q^\prime. |
Moreover, by (3.8) and the assumption |\mu^{(s)}| > |\lambda^{(s)}| , we have
\sum\limits_{t = 1}^{s}|\mu^{(t)}| > \sum\limits_{t = 1}^{s}|\lambda^{(t)}|. |
Combining with Remark 3.2 and the definition of u^\prime, r^\prime , we derive
\sum\limits_{t = -\infty}^{k}d_t^{\boldsymbol\nu} = \sum\limits_{t = -\infty}^{k}d_t^{ {\boldsymbol\mu}}-1 = \sum\limits_{t = 1}^{s}|\mu^{(t)}|-1 \geq\sum\limits_{t = 1}^{s}|\lambda^{(t)}|\geq\sum\limits_{t = -\infty}^{k}d_t^{ {\boldsymbol\lambda}} |
where u^\prime\leq k < r^\prime . If r^\prime = q^\prime , then we have proved
\sum\limits_{t = -\infty}^{k}d_t^{\boldsymbol\nu}\geq\sum\limits_{t = -\infty}^{k}d_t^{ {\boldsymbol\lambda}}\quad \forall k\in{\mathbb{Z}}. |
Therefore, \boldsymbol\nu\geq_\theta {\boldsymbol\lambda} . If r^\prime = p^\prime < q^\prime , suppose \boldsymbol\nu\ngeq_\theta {\boldsymbol\lambda} , there exists some r^\prime\leq\epsilon^\prime < q^\prime such that
\sum\limits_{t = -\infty}^{\epsilon^\prime}d_t^{\boldsymbol\nu} < \sum\limits_{t = -\infty}^{\epsilon^\prime}d_t^{ {\boldsymbol\lambda}},\quad \sum\limits_{t = -\infty}^{\epsilon^\prime+1}d_t^{\boldsymbol\nu}\geq\sum\limits_{t = -\infty}^{\epsilon^\prime+1}d_t^{ {\boldsymbol\lambda}}. |
On the other hand, by the definition of p^\prime and q^\prime , we have
\sum\limits_{t = -\infty}^{\epsilon^\prime+1}d_t^{\boldsymbol\nu}\leq 1+\sum\limits_{t = -\infty}^{\epsilon^\prime}d_t^{\boldsymbol\nu} < 1+\sum\limits_{t = -\infty}^{\epsilon^\prime}d_t^{ {\boldsymbol\lambda}} \leq\sum\limits_{t = -\infty}^{\epsilon^\prime+1}d_t^{ {\boldsymbol\lambda}}, |
this contradicts to the choice of \epsilon^\prime .
So we derive \boldsymbol\nu\geq_\theta {\boldsymbol\lambda} , then {\boldsymbol\mu} > _\theta\boldsymbol\nu\geq_\theta {\boldsymbol\lambda} . Since {\boldsymbol\lambda}, {\boldsymbol\mu} are neighbors with \geq_\theta , we derive \boldsymbol\nu = {\boldsymbol\lambda} . That is, {\boldsymbol\lambda} and {\boldsymbol\mu} satisfy (a) .
Suppose |\lambda^{(s)}| = |\mu^{(s)}| . Let \boldsymbol\nu = (\mu^{(1)}, \dots, \mu^{(s)}, \lambda^{(s+1)}, \dots, \lambda^{(\ell)}) , then {\boldsymbol\mu}\geq_\theta\boldsymbol\nu > _\theta {\boldsymbol\lambda} , so \boldsymbol\nu = {\boldsymbol\mu} since {\boldsymbol\lambda} and {\boldsymbol\mu} are neighbors with \geq_\theta . Hence \mu^{(r)} = \lambda^{(r)}, \forall r\neq s . Let m: = |\lambda^{(s)}| = |\mu^{(s)}| , then \mu^{(s)} and \lambda^{(s)} are partitions of m with \mu^{(s)} > _\theta\lambda^{(s)} . If there exist partition \eta of m with \mu^{(s)} > _\theta\eta > _\theta\lambda^{(s)} , then \boldsymbol\eta = (\lambda^{(1)}, \dots, \lambda^{(s-1)}, \eta, \lambda^{(s+1)}, \dots, \lambda^{(\ell)}) , satisfy {\boldsymbol\mu} > _\theta\boldsymbol\eta > _\theta {\boldsymbol\lambda} , this contradicts that {\boldsymbol\lambda} and {\boldsymbol\mu} are neighbors with \geq_\theta . So \mu^{(s)} and \lambda^{(s)} are neighbors with \geq_\theta . Applying the necessity of Lemma 3.3 to \lambda^{(s)} and \mu^{(s)} , we derive that {\boldsymbol\lambda} and {\boldsymbol\mu} satisfy either (b) or (c) .
Next, let us prove the sufficiency. Suppose {\boldsymbol\mu} and {\boldsymbol\lambda} are \ell -partitions of n with {\boldsymbol\mu} > _\theta {\boldsymbol\lambda} and one of (a), (b), (c) holds. Suppose \boldsymbol\nu be a \ell -partition of n with {\boldsymbol\mu}\geq_\theta\boldsymbol\nu > _\theta {\boldsymbol\lambda} and {\boldsymbol\lambda}, \boldsymbol\nu are neighbors with \geq_\theta . Now let us prove {\boldsymbol\mu} = \boldsymbol\nu .
If (a) holds, let p = \theta_s+\ell(k_s+1-1) = \theta_s+\ell k_s and q = \theta_{s+1}+\ell(1-\lambda_1^{(s+1)}) . We have d_p^{ {\boldsymbol\mu}} = d_q^{ {\boldsymbol\lambda}} = 1 . Moreover, by Remark 3.2, Lemma 3.1, and the choice of \boldsymbol\nu , we derive
\begin{align*} d_t^{ {\boldsymbol\mu}}& = d_t^{ {\boldsymbol\lambda}} = d_t^{\boldsymbol\nu}\quad t < p\text{ or }t > q\\ d_{t^\prime}^{ {\boldsymbol\mu}}& = d^{ {\boldsymbol\lambda}}_{t^{\prime\prime}} = 0\quad p < t^\prime\leq q,\ p\leq t^{\prime\prime} < q. \end{align*} |
Therefore,
\begin{align} \sum\limits_{p\leq t\leq q}d_{t}^{\boldsymbol\nu} = \sum\limits_{p\leq t\leq q}d_{t}^{ {\boldsymbol\mu}} = \sum\limits_{p\leq t\leq q}d_{t}^{ {\boldsymbol\lambda}} = 1. \end{align} | (3.12) |
We claim d_t^{\boldsymbol\nu} = 0 for all q < t < p ; otherwise, there must be d_p^{\nu^{(s)}}\neq0 or d_q^{\nu^{(s+1)}}\neq0 , this contradicts (3.12). If d_q^{\boldsymbol\nu} = 1 , then d_q^{\nu^{(s+1)}} = 1 and \boldsymbol\nu = {\boldsymbol\lambda} ; this contradicts {\boldsymbol\lambda}\neq\boldsymbol\nu . Therefore d_{p}^{\boldsymbol\nu} = 1 , hence d_p^{\nu^{(s)}} = 1 , and {\boldsymbol\mu} = \boldsymbol\nu .
If (b) or (c) holds. Combining with the choice of \boldsymbol\nu , we have
\begin{align*} &\mu^{(r)} = \nu^{(r)} = \lambda^{(r)}\quad \text{where}\ r\neq s,\\ &\mu^{(s)}\geq_\theta\nu^{(s)} > _\theta\lambda^{(s)}. \end{align*} |
Apply the sufficiency of Lemma 3.3 to \mu^{(s)} and \lambda^{(s)} ; we derive \mu^{(s)} and \lambda^{(s)} are neighbors with \geq_{\theta} ; hence, \mu^{(s)} = \nu^{(s)} and {\boldsymbol\mu} = \boldsymbol\nu .
Now we can give the relationship between dominance order and weak \theta -dominance order on multipartitions.
Theorem 4.1. Suppose {\boldsymbol\lambda}, {\boldsymbol\mu}\in{\mathcal{P}}^\ell_n and \theta = (\theta_1, \dots, \theta_\ell) are strongly separated. Then {\boldsymbol\mu}\unrhd {\boldsymbol\lambda} if and only if {\boldsymbol\mu}\geq_\theta {\boldsymbol\lambda} .
Proof. The conclusion is clear by Proposition 3.7 and [11, Lemma 6.3].
For {\boldsymbol\lambda}\in{\mathcal{P}}_n^\ell , we define \text{res}({\boldsymbol\lambda}) = \{\text{res}(\alpha)|\alpha\in {\boldsymbol\lambda}\} to be a multi-set. According to Definitions 2.2 and 2.3, by a trivial discussion, one can prove {\boldsymbol\mu}\geq_\theta {\boldsymbol\lambda} and \text{res}({\boldsymbol\mu}) = \text{res}({\boldsymbol\lambda}) whenever {\boldsymbol\mu}\unrhd_\theta {\boldsymbol\lambda} . Finally, as a corollary of Theorem 4.1, we obtain the relationship between dominance order and \theta -dominance order.
Theorem 4.2. Suppose {\boldsymbol\lambda}, {\boldsymbol\mu}\in{\mathcal{P}}_n^\ell and \theta = (\theta_1, \dots, \theta_\ell) be strongly separated. If {\boldsymbol\lambda}\unlhd_\theta {\boldsymbol\mu} , then {\boldsymbol\lambda}\unlhd {\boldsymbol\mu} and \text{res}({\boldsymbol\lambda}) = \text{res}({\boldsymbol\mu}) .
We point out that the inverse of Theorem 4.2 is not true. We can give a counterexample as follows:
Example 4.3. Let \ell = 2 , n = 6 , (\sigma_1, \sigma_2) = (0, 1) , \theta = (0, 25) , \theta is strongly separated. Let {\boldsymbol\lambda} = ((2, 1), (2, 1)) , {\boldsymbol\mu} = ((3), (3)) , the Young diagrams with residue are as follows:
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On one hand, {\boldsymbol\mu}\rhd {\boldsymbol\lambda} and \text{res}({\boldsymbol\lambda}) = \text{res}({\boldsymbol\mu}) . On another hand, let \gamma = (3, 2, 1) ; we have \text{res}(\gamma) = 1 and
|\{\alpha\in {\boldsymbol\lambda}|\gamma\lhd_\theta\alpha\}| = |\{(1,2,1),(2,1,1)\}| > |\{\beta\in {\boldsymbol\mu}|\gamma\lhd_\theta\beta\}| = |\{(1,2,1)\}| |
hence {\boldsymbol\mu}\ntriangleright_\theta {\boldsymbol\lambda} .
In this paper, we prove that the weak \theta -dominance order coincides with the dominance order on multipartitions, whenever the loading \theta is strongly separated. As a corollary, we prove that the \theta -dominance order is stronger than the usual dominance order on multipartitions, whenever the loading \theta is strongly separated.
The author declares he has not used Artificial Intelligence (AI) tools in the creation of this article.
The author was supported by the Natural Science Foundation of Shandong Province of China (No. ZR2023QA093) and the Doctoral Research Start-up Foundation of Shandong Jianzhu University (No. X22021Z). The author appreciates professor Jun Hu and Zhankui Xiao for their helpful discussions. The author also appreciates the reviewers for their helpful comments.
The author declares no conflicts of interest in this paper.
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