Volume 2, issue 2 (2002)

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Smith equivalence and finite Oliver groups with Laitinen number 0 or 1

Krzysztof Pawałowski and Ronald Solomon

Algebraic & Geometric Topology 2 (2002) 843–895
 arXiv: math.AT/0210373
Abstract

In 1960, Paul A. Smith asked the following question. If a finite group $G$ acts smoothly on a sphere with exactly two fixed points, is it true that the tangent $G$–modules at the two points are always isomorphic? We focus on the case $G$ is an Oliver group and we present a classification of finite Oliver groups $G$ with Laitinen number ${a}_{G}=0$ or $1$. Then we show that the Smith Isomorphism Question has a negative answer and ${a}_{G}\ge 2$ for any finite Oliver group $G$ of odd order, and for any finite Oliver group $G$ with a cyclic quotient of order $pq$ for two distinct odd primes $p$ and $q$. We also show that with just one unknown case, this question has a negative answer for any finite nonsolvable gap group $G$ with ${a}_{G}\ge 2$. Moreover, we deduce that for a finite nonabelian simple group $G$, the answer to the Smith Isomorphism Question is affirmative if and only if ${a}_{G}=0$ or $1$.

Keywords
finite group, Oliver group, Laitinen number, smooth action, sphere, tangent module, Smith equivalence, Laitinen–Smith equivalence
Mathematical Subject Classification 2000
Primary: 57S17, 57S25, 20D05
Secondary: 55M35, 57R65.