发表机构
Institute of Mathematics, Nanjing Normal University（南京师范大学数学研究所）
机构由 AI 辅助整理，请以论文原文为准。AI 中文总结
该研究确定了局部紧阿贝尔群上相容局部拟凸拓扑偏序集的大小，证明非预紧群的偏序集包含幂集序同构副本，回答了相关公开问题并给出精确基数公式。
AI 中文摘要
对于Hausdorff局部拟凸阿贝尔群$G$，令$\boldsymbol{\frak{C}}(G)$为其基础群上所有与$G$具有相同连续特征的Hausdorff局部拟凸群拓扑构成的偏序集。我们证明，当$G$非预紧时，$\boldsymbol{\frak{C}}(G)$包含$(\text{Pow}(\boldsymbol{\frak{c}}),\boldsymbol{\frak{c}})$的一个序同构副本，其中$\boldsymbol{\frak{c}}=2^{\boldsymbol{\frak{\text{aleph}}}_0}$。该嵌入的取值介于Bohr拓扑与原拓扑之间。由此，每个无限离散阿贝尔群$D$满足$|\boldsymbol{\frak{C}}(D)|=\text{width}\boldsymbol{\frak{C}}(D)=2^{2^{|D|}}$。对局部紧阿贝尔群的离散约化可得到所有这类群的精确基数与宽度公式；特别地，对每个非紧$\boldsymbol{\frak{\text{sigma}}}$紧局部紧阿贝尔群，这两个不变量均等于$2^{\boldsymbol{\frak{c}}}$。这些结果回答了L. Außenhofer与D. Dikranjan在文献[\text{AD20}]中提出的问题6.1--6.3、6.5--6.7，以及问题6.4的局部紧情形。该嵌入也适用于非紧完备可度量化局部拟凸群。一个具有唯一相容拓扑的非紧预紧核群表明，仅非紧性在核群情形下并不充分。最后，$\boldsymbol{\frak{R}}^{\boldsymbol{\frak{N}}}$的相容偏序集与任何离散阿贝尔群的相容偏序集都不是序同构的。
英文摘要
For a Hausdorff locally quasi-convex abelian group $G$, let $\C(G)$ be the poset of all Hausdorff locally quasi-convex group topologies on its underlying group having the same continuous characters as $G$. We prove that, whenever $G$ is non-precompact, $\C(G)$ contains an order-isomorphic copy of $(\Pow(\cont),\subseteq)$, where $\cont=2^{\aleph_0}$. The embedding takes values between the Bohr topology and the original topology. Consequently, every infinite discrete abelian group $D$ satisfies $|\C(D)|=\width\C(D)=2^{2^{|D|}}$. The discrete reduction for locally compact abelian groups then yields exact cardinality and width formulas for all such groups; in particular, both invariants equal $2^{\cont}$ for every non-compact $σ$-compact locally compact abelian group. These results answer Questions 6.1--6.3 and 6.5--6.7, and the locally compact case of Question 6.4, posed by L.~Außenhofer and D.~Dikranjan in \cite{AD20}. The embedding also applies to non-compact complete metrizable locally quasi-convex groups. A non-compact precompact nuclear group with a unique compatible topology shows that non-compactness alone does not suffice in the nuclear setting. Finally, the compatible poset of $\R^{\N}$ is not order-isomorphic to that of any discrete abelian group.
