From: Daniel RubermanNewsgroups: sci.math.research Subject: Re: Unitary cobordisms Date: Sat, 3 Apr 2004 16:57:05 -0500 Organization: Brandeis University, Waltham MA Message-ID: <Pine.LNX.4.44.0404031628400.26409-100000@groucho.unet.brandeis.edu> On Sat, 3 Apr 2004, John Baez wrote: > I'll assume you know a bit about categories where > the morphisms are cobordisms, as in the study of > topological quantum field theory. > > Suppose I have a morphism in this category, > namely an n-dimensional cobordism > > M: S -> S' > > from some manifold S to some manifold S'. Let > > M*: S' -> S > > be the obvious "reversed" version of M. Suppose > M* is the inverse of M, so > > M M* = 1 > > and > > M* M = 1. > > Can we conclude that M is just a cylinder, i.e. that > M corresponds to some element of the mapping class group of M? > > It's true in low dimensions. It's not true in dimension 4 if you're talking about smooth manifolds, or (I'm pretty sure) in higher dimensions for TOP/PL/DIFF. The phenomena are rather different. The 4-dimensional case relies on the existence of homotopy equivalent (simply connected) 4-manifolds N, N' that are not diffeomorphic. An old theorem of Wall says that they are h-cobordant, ie that there is a 5-manifold M with boundary N,N', that is homotopy equivalent to a product. Let M* be M turned upside down, so that MM* is an h-cobordism from N to itself and M*M is an h-cobordism from N' to itself. Now a nice argument, due in general to D. Barden (Simply connected five-manifolds. Ann. of Math. (2) 82 1965 365--385.) says that an h-cobordism from a simply-connected 4-manifold *to itself* must be a product. (There's actually an interesting technicality here; the map from one end to the other of the h-cobordism gotten by collapsing the h-cobordism to that end must be homotopic to the identity. Without this hypothesis, the theorem fails, as is shown by an example due to Donaldson.) On the other hand, since N and N' are not diffeomorphic, neither M nor M* is a product. In higher dimensions, your statement fails for many simply-connected manifolds. Take your M to be an h-cobordism with non-trivial torsion, say t(M,N). The the torsion of M*, t(M*,N') is gotten (up to signs, related to the parity of the dimension) by a certain involution on the Whitehead group. It follows (I think) that the torsions of both MM* and M*M are trivial, so they are both products. But neither M nor M* will be a product. (A reference for all this, more accurate than my recollection, is Marshall Cohen's beautiful book, A course on simple-homotopy theory.) These observations were used by Stallings to show that N x R and N' x R are diffeomorphic (likewise with R replaced by a circle). If the dimension of M is 4, so the boundaries are 3-manifolds, then I think you'd be hard pressed to prove anything, because no one knows if the h-cobordism theorem is true. I think in fact that the topological s-cobordism theorem is false (Cappell-Shaneson) so there are probably topological counterexamples to your theorem. That leaves the case of 3-dimensional cobordisms, where the result is true (Alexander's theorem when the boundary is a 2-sphere, Waldhausen in general) and 2-dimensional cobordisms, where the result is easy. Danny