arXivDaily arXiv每日学术速递 周一至周五更新
arXiv周末暂无论文更新，休息一下吧，周末愉快~~
arXiv 2510.04471math.CO
收藏

高维树距离矩阵的Smith标准形

The Smith normal form of distance matrices of high dimensional trees

Carlos A. Alfaro, Jesús Uriel Medrano, Iván Téllez Téllez

阅读 PDF
arXiv 页面 ↗HTML 原文 ↗

AI总结：

该研究将普通树距离矩阵的Smith标准形结论推广到高维k-树，证明固定k和n时任意n顶点k-树的k-距离矩阵SNF一致，给出具体形式并拓展了已有经典结果。

摘要：

Graham-Lovász-Pollak \cite{GL,GP} 得出了著名公式$$\det({\sf D}(T_{n+1}))=(-1)^nn2^{n-1},$$用于计算任意具有$n+1$个顶点的树$T_{n+1}$的距离矩阵${\sf D}(T_{n+1})$的行列式。随后，Hou和Woo \cite{HW} 将该公式推广到Smith标准形（SNF），证明对于任意具有$n+1$个顶点的树$T_{n+1}$，有$\SNF({\sf D}(T_{n+1}))={\sf I}_2\oplus 2{\sf I}_{n-2}\oplus [2n]$。\n $k$-树的定义为：要么是具有$k$个顶点的完全图，要么是在一个更小的$k$-树基础上，添加一个新顶点并将其通过$k$条边连接到一个$k$-团所得到的图。若$τ$和$τ'$是$k$-树$T$中的$d$-团，则$τ$与$τ'$之间的$d$-游走是有限序列$τ_1σ_1τ_2σ_2\cdotsτ_l$，其中$τ_1=τ$，$τ_l=τ'$，且$d$-团$τ_i$与$τ_{i+1}$都关联于同一个$(d+1)$-团$σ_i$。对于$d\in\{1,\dots,k\}$，$d$-团$τ$与$τ'$之间的$d$-距离是指从$τ$到$τ'$的最短$d$-游走中$(d+1)$-团的数量，记为$\dist^d(τ,τ')$。设$c_d$表示$k$-树$T$中$d$-团的数量，则$k$-树$T$的$d$-距离矩阵${\sf D}^d(T)$是$c_d\times c_d$矩阵，以$T$的$d$-团为索引，当$i=j$时$(i,j)$元为0，否则为$\dist^d(τ_i,τ_j)$。本文证明，当$k$和$n$固定时，任意具有$n$个顶点的$k$-树的$k$-距离矩阵的SNF均相同。具体而言，对于任意具有$n$个顶点且满足$n\geq k+2$的$k$-树$T_n$，${\sf D}^{k}(T_{n})$的Smith标准形为$${\sf I}_{(k-1)(n-k)+2}\oplus (k+1){\sf I}_{n-k-2}\oplus [k(k+1)(n-k)],$$这一结果推广了Graham-Lovász-Pollak和Hou-Woo的结论。

英文摘要：

Graham-Lovász-Pollak \cite{GL,GP} obtained the celebrated formula $$\det({\sf D}(T_{n+1}))=(-1)^nn2^{n-1},$$ for the determinant of the distance matrix ${\sf D}(T_{n+1})$ for any tree $T_{n+1}$ with $n+1$ vertices. Later, Hou and Woo \cite{HW} extended this formula to the Smith normal form (SNF) obtaining that $\SNF({\sf D}(T_{n+1}))={\sf I}_2\oplus 2{\sf I}_{n-2}\oplus [2n]$, for any tree $T_{n+1}$ with $n+1$ vertices. A $k$-{\it tree} is either a complete graph on $k$ vertices or a graph obtained from a smaller $k$-tree by adjoining a new vertex together with $k$ edges connecting it to a $k$-clique. If $τ$ and $τ'$ are $d$-cliques in a $k$-tree $T$, a $d$-{\it walk} between $τ$ and $τ'$ is a finite sequence $τ_1σ_1τ_2σ_2\cdotsτ_l$, where $τ_1=τ$, $τ_l=τ'$, and the $d$-cliques $τ_i$ and $τ_{i+1}$ are incident to the same $(d+1)$-clique $σ_i$. For $d\in\{1,\dots,k\}$, the $d$-{\it distance} from the $d$-cliques $τ$ and $τ'$ is the number of $(d+1)$-cliques in a minimum $d$-walk from $τ$ and $τ'$, and is denoted by $\dist^d(τ,τ')$. Let $c_d$ denote the number of $d$-cliques in the $k$-tree $T$. Then the $d$-distance matrix ${\sf D}^d(T)$ of the $k$-tree $T$ is the $c_d\times c_d$ matrix, indexed by the $d$-cliques of $T$, such that the $(i,j)$-entry is $0$ if $i=j$, and $\dist^d(τ_i,τ_j)$ otherwise. Here, we show that, for $k$ and $n$ fixed, the SNF of the $k$-distance matrix is the same for any $k$-tree with $n$ vertices. Specifically, for any $k$-tree $T_{n}$ with $n$ vertices such that $n\geq k+2$, the Smith normal form of ${\sf D}^{k}(T_{n})$ is $${\sf I}_{(k-1)(n-k)+2}\oplus (k+1){\sf I}_{n-k-2}\oplus [k(k+1)(n-k)],$$ which extends Graham-Lovász-Pollak and Hou-Woo results.

机构、总结、中文摘要均由AI分析而来，可能存在错误，请以论文正文为准。

↑