组合Alexander对偶——一个简短的初等证明
组合数学
2011-10-25 v3
摘要
设X为以V为基础集的单纯复形。定义其Alexander对偶为单纯复形 X* = {A \subset V: V \setminus A \notin X}。组合Alexander对偶指出，X的第i个约化同调群同构于X*的第(|V|-i-3)个约化上同调群（在给定的交换环R上）。我们给出了一个自洽的证明。
引用
@article{arxiv.0710.1172,
title = {Combinatorial Alexander Duality -- a Short and Elementary Proof},
author = {Anders Björner and Martin Tancer},
journal= {arXiv preprint arXiv:0710.1172},
year = {2011}
}
评论
7 pages, 2 figure; v3: the sign function was simplified