SMS scnews item created by Haotian Wu at Wed 4 Apr 2018 0929
Type: Seminar
Distribution: World
Expiry: 4 Jul 2018
Calendar1: 12 Apr 2018 1200-1300
CalLoc1: Carslaw 351
CalTitle1: Geometry & Topology Seminar: Hillman -- Quotients of S^2xS^2 by groups of order 4
Auth: haotianw@60-242-94-203.static.tpgi.com.au (hawu2800) in SMS-WASM

# Quotients of S^2xS^2 by groups of order 4

### Jonathan Hillman (Sydney)

Thursday 12 April 12:00–13:00 in Carslaw 351.

Please join us for lunch after the talk!

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].

If you are registered you may mark the scnews item as read.
School members may try to .