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

Pawałowski, Krzysztof¹ ; Solomon, Ronald²

¹Faculty of Mathematics and Computer Scienc, Adam Mickiewicz University, ul. Umultowska 87, 61-614 Poznań, Poland
²Department of Mathematics, The Ohio State University, 231 West 18th Avenue, Columbus, OH 43210–1174, USA

Algebraic and Geometric Topology, Tome 2 (2002) no. 2, pp. 843-895

Résumé

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 aG = 0 or 1. Then we show that the Smith Isomorphism Question has a negative answer and aG ≥ 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 aG ≥ 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 aG = 0 or 1.

Détail

DOI : 10.2140/agt.2002.2.843

Keywords: finite group, Oliver group, Laitinen number, smooth action, sphere, tangent module, Smith equivalence, Laitinen–Smith equivalence

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