×

Cancellation of lattices and finite two-complexes. (English) Zbl 0779.57002

Certain cancellation problems arise naturally in algebra and topology. For example, if \(M\), \(M'\), \(N\) are modules with \(M\oplus N\cong M'\oplus N\), is \(M\cong M'\)? If \(K\), \(K'\) are finite two-complexes with \(K\vee rS^ 2\simeq K'\vee rS^ 2\), is \(K\simeq K'\)? In this paper we consider these questions for modules over orders (e.g. integral group rings \(Z\pi\), \(\pi\) a finite group) and two-complexes with finite fundamental group.
The following is our main result about finite two-complexes. The analogous result for “homotopy type” instead of “simple homotopy type” was proved by W. Browning.
Theorem B: Let \(K\) and \(K'\) be finite 2-complexes with the same Euler characteristic and finite fundamental group. Let \(\alpha:\pi_ 1(K,x_ 0)\to\pi_ 1(K',x_ 0')\) be a given isomorphism and suppose that \(K\simeq K_ 0\vee S^ 2\). Then there is a simple homotopy equivalence \(f:K\to K'\) inducing \(\alpha\) on the fundamental groups.
This is the best possible result in general, but for special fundamental groups it can sometimes be improved.
Theorem 2.1: Let \(\pi\) be a finite subgroup of SO(3). If \(K\) and \(K'\) are finite 2-complexes with fundamental group \(\pi\) and the same Euler characteristic, and \(\alpha:\pi_ 1(K,x_ 0)\to\pi_ 1(K',x_ 0')\) a given isomorphism, then there is a simple homotopy equivalence \(f:K\to K'\) inducing \(\alpha\) on the fundamental groups.
For \(\pi\) cyclic or \(\pi=\mathbb{Z}/2\times\mathbb{Z}/2\), this was proved by M. Dyer with A. Sieradski and W. Metzler respectively. The result for \(\pi=D(4n)\), the dihedral group of order \(4n\), has recently been obtained by P. Latiolais. Our methods give a new proof in these cases.

MSC:

57M20 Two-dimensional complexes (manifolds) (MSC2010)
57Q10 Simple homotopy type, Whitehead torsion, Reidemeister-Franz torsion, etc.
PDFBibTeX XMLCite
Full Text: Crelle EuDML