Quotients of \(S^2\times{S^2}\) by groups of order 4

Jonathan Hillman (Sydney)

Abstract

We consider closed 4-manifolds \(M\) with universal cover \(\widetilde{M}\cong{S^2\times{S^2}}\) and \(\chi(M)=1\). The group \(\pi=\pi_1(M)\) then has order 4, and \(M\) is non-orientable. All such manifolds with \(\pi\cong\mathbb{Z}/4\mathbb{Z}\) are homotopy equivalent, but there are four homeomorphism classes. When \(\pi\cong(\mathbb{Z}/2\mathbb{Z})^2\) there are three homotopy types, each with between two and eight homeomorphism classes. In each case the homeomorphism classes occur in pairs \(M,*M\) with opposite \(KS\) stable smoothing invariants. Establishing this involves some ad hoc algebraic topology and quoting results from surgery theory. We shall sketch this reduction and concentrate on the constructive aspects. In particular, we give a smooth quotient with \(\pi\cong\mathbb{Z}/4\mathbb{Z}\) which may not be homeomorphic (or diffeomorphic?) to the geometric quotient \(S^2\times{S^2}/\langle\sigma\rangle\), where \(\sigma(s,t)=(t,-s)\) for \(s,t\in{S^2}\). We also give an example with \(\pi\cong(\mathbb{Z}/2\mathbb{Z})^2\) which is not homotopy equivalent to either \(RP^2\times{RP^2}\) or the nontrivial bundle space \(RP^2\tilde\times{RP^2}\).

This is joint work with Ian Hambleton. See arXiv:1712.04572 [math.GT].