Antichains in the Bruhat Order for the Classes $$\mathcal {A}(n,k)$$

Order Pub Date : 2023-11-28 DOI:10.1007/s11083-023-09654-6
Henrique F. da Cruz
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Abstract

Let \(\varvec{\mathcal {A}(n,k)}\) represent the collection of all \(\varvec{n\times n}\) zero-and-one matrices, with the sum of all rows and columns equalling \(\varvec{k}\). This set can be ordered by an extension of the classical Bruhat order for permutations, seen as permutation matrices. The Bruhat order on \(\varvec{\mathcal {A}(n,k)}\) differs from the Bruhat order on permutations matrices not being, in general, graded, which results in some intriguing issues. In this paper, we focus on the maximum length of antichains in \(\varvec{\mathcal {A}(n,k)}\) with the Bruhat order. The crucial fact that allows us to obtain our main results is that two distinct matrices in \(\varvec{\mathcal {A}(n,k)}\) with an identical number of inversions cannot be compared using the Bruhat order. We construct sets of matrices in \(\varvec{\mathcal {A}(n,k)}\) so that each set consists of matrices with the same number of inversions. These sets are hence antichains in \(\varvec{\mathcal {A}(n,k)}\). We use these sets to deduce lower bounds for the maximum length of antichains in these partially ordered sets.

类的Bruhat序中的反链 $$\mathcal {A}(n,k)$$
设\(\varvec{\mathcal {A}(n,k)}\)表示所有\(\varvec{n\times n}\) 0和1矩阵的集合,所有行和列的和等于\(\varvec{k}\)。这个集合可以通过排列的经典Bruhat顺序的扩展来排序,可以看作是排列矩阵。\(\varvec{\mathcal {A}(n,k)}\)上的Bruhat顺序与排列矩阵上的Bruhat顺序不同,一般来说,排列矩阵不是分级的,这导致了一些有趣的问题。本文主要研究了\(\varvec{\mathcal {A}(n,k)}\)中具有Bruhat阶的反链的最大长度。使我们能够获得主要结果的关键事实是,\(\varvec{\mathcal {A}(n,k)}\)中具有相同逆序数的两个不同矩阵不能使用Bruhat顺序进行比较。我们在\(\varvec{\mathcal {A}(n,k)}\)中构造矩阵集合,使得每个集合由具有相同逆序个数的矩阵组成。这些集合因此是\(\varvec{\mathcal {A}(n,k)}\)中的反链。我们利用这些集合来推导出这些部分有序集合中反链的最大长度的下界。
本文章由计算机程序翻译,如有差异,请以英文原文为准。
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