From: Daniel Ruberman 
Newsgroups: sci.math.research
Subject: Re: Unitary cobordisms
Date: Sat, 3 Apr 2004 16:57:05 -0500
Organization: Brandeis University, Waltham MA
Message-ID: <>

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.