与拉丁方自拓扑有关的0/1-多面体
0/1-Polytopes related to Latin squares autotopisms
摘要：
$n$阶的拉丁方的集合LS（n）可以在$\mathbb{R}^{n^3}$中表示为$(n-1)^3$ -维0/1-多面体。给定一个自同构$Θ=(α,β,γ)\in\mathfrak{A}_n$，本文研究了LS（n）的自同构群中含有$Θ$的子集所对应的0/1-多面体.具体地说，我们证明了这种多面体结构是由一个多面体在$\mathbb{R}^{((\mathbf{n}_α-\mathbf{l}_α^1)\cdot n^2 + \mathbf{l}_α^1\cdot \mathbf{n}_β\cdot n)-(\mathbf{l}_α^1\cdot \mathbf{l}_β^1\cdot (n -\mathbf{l}_γ^1) + \mathbf{l}_α^1\cdot \mathbf{l}_γ^1\cdot (\mathbf{n}_β -\mathbf{l}_β^1) + \mathbf{l}_β^1\cdot \mathbf{l}_γ^1\cdot (\mathbf{n}_α -\mathbf{l}_α^1))}$，其中$\mathbf{n}_α$和$\mathbf{n}_β$是$α$和$β$的循环数，分别和$\mathbf{l}_δ^1$是$δ$的不动点的数量，对于所有$δ\in \{α,β,γ\}$。此外，我们研究了这两个多面体的拉丁方的阶数为9。
英文摘要：
The set LS(n) of Latin squares of order $n$ can be represented in $\mathbb{R}^{n^3}$ as a $(n-1)^3$-dimensional 0/1-polytope. Given an autotopism $Θ=(α,β,γ)\in\mathfrak{A}_n$, we study in this paper the 0/1-polytope related to the subset of LS(n) having $Θ$ in their autotopism group. Specifically, we prove that this polyhedral structure is generated by a polytope in $\mathbb{R}^{((\mathbf{n}_α-\mathbf{l}_α^1)\cdot n^2 + \mathbf{l}_α^1\cdot \mathbf{n}_β\cdot n)-(\mathbf{l}_α^1\cdot \mathbf{l}_β^1\cdot (n -\mathbf{l}_γ^1) + \mathbf{l}_α^1\cdot \mathbf{l}_γ^1\cdot (\mathbf{n}_β -\mathbf{l}_β^1) + \mathbf{l}_β^1\cdot \mathbf{l}_γ^1\cdot (\mathbf{n}_α -\mathbf{l}_α^1))}$, where $\mathbf{n}_α$ and $\mathbf{n}_β$ are the number of cycles of $α$ and $β$, respectively, and $\mathbf{l}_δ^1$ is the number of fixed points of $δ$, for all $δ\in \{α,β,γ\}$. Moreover, we study the dimension of these two polytopes for Latin squares of order up to 9.
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