Assume $M^n$ and $N^n$ are null bordant, i.e. each can be realized as boundary of an $n+1$ dimensional manifold. Suppose $M^n \times \mathbb R$ is homeomorphic to $N^n\times \mathbb R$. Is there any example shows that $M$ is NOT homeomorphic to $N$?

[**Edit:** I have added some details and a more explicit example by Milnor.]

I will present a couple of examples verifying the conditions required in the question.$\newcommand{\RR}{\mathbb{R}} \newcommand{\TT}{\mathbb{T}} \newcommand{\ZZ}{\mathbb{Z}}$

### $H$-cobordant manifolds

We will make fundamental use of the following fact: if $M$ and $N$ are $h$-cobordant, closed, smooth manifolds of dimension greater than 3, then $M\times\RR$ is diffeomorphic to $N\times\RR$. [*Aside:* A similar statement holds for topological manifolds; simply replace diffeomorphism with homeomorphism.] Here are a couple of quick references for this fact, both making use of the $s$-cobordism theorem:

In a related mathoverflow question, Oscar Randal-Williams provides a fairly direct argument in a comment to one of the answers.

The article "

*Contact manifolds with symplectomorphic symplectizations*" states the result for smooth manifolds as corollary 2.5. [The proof described there applies also to topological manifolds.]

[*Further aside*: Incidentally, the converse also holds — although we will not make use of it. In another mathoverflow question, Igor Belegradek remarks in his answer and the ensuing comments that if $M\times\RR$ is homeomorphic to $N\times\RR$, then $M$ and $N$ are $h$-cobordant.]

So it suffices to find closed smooth manifolds $M$ and $N$ which are $h$-cobordant, yet not homeomorphic. As required by the question, we also want $M$ to be null-cobordant; then $N$ will be null-cobordant as well. As a minor bonus, since $M$ and $N$ are smoothly $h$-cobordant, $M\times\RR$ is actually *diffeomorphic* to $N\times\RR$.

### An example by Farrel and Hsiang

One such example is given by Farrel and Hsiang in their article "*$H$-cobordant manifolds are not necessarily homeomorphic*". There, the manifold $M$ is a product $M=L\times\TT^n$, where $L$ is any 3-dimensional lens space such that $\pi_1(L)\simeq\ZZ/p^2$ for some prime $p$, and $\TT^n=(S^1)^{\times n}$ is a torus of dimension $n\geq 3$. Note that $M$ is null-cobordant since the torus $\TT^n$ is null-cobordant. Unfortunately, the methods used by Farrel and Hsiang do not yield an explicit description of the manifold $N$.

### Milnor's examples

A more explicit example is given in the celebrated article "*Two complexes which are homeomorphic but combinatorially distinct*" by John Milnor. Milnor essentially proves that we can take $M=L(7,1)\times S^{2n}$ and $N=L(7,2)\times S^{2n}$ for any $n>0$, where $L(p,q)$ denotes the lens space of type $(p,q)$. This is *only partially* stated as theorem 4 in that article. By the way, observe that $M$ and $N$ are null-cobordant since spheres are null-cobordant.

I will briefly describe how to conclude from Milnor's article that the preceding $M$ and $N$ meet our requirements. In section 2, Milnor sketches a proof that $M$ and $N$ are $h$-cobordant. At the end of section 4, in the proof of corollary 2, he uses the fact that $L(7,1)$ and $L(7,2)$ have distinct Reidemeister torsions, and concludes that $M$ and $N$ also have distinct Reidemeister torsions, and thus are *not diffeomorphic*. But more is true: since the Reidemeister torsions of $M$ and $N$ differ, $M$ and $N$ are *not homeomorphic*. This follows from the topological invariance (i.e. invariance under homeomorphisms) of Reidemeister torsion, which is itself a simple consequence of the topological invariance of Whitehead torsion — proved by Chapman after the publication of Milnor's article.

In general, the results in sections 2 and 4 of Milnor's article actually prove that we can take $M=K\times S^n$ and $N=L\times S^n$, where:

$K$, $L$ are parallelizable, smooth, closed $k$-manifolds;

$K$, $L$ are homotopy equivalent;

$n$ is even and $n\geq k$;

the Reidemeister torsions of $K$ and $L$ are defined (for some complex-valued representation of their common fundamental group), and have different absolute values.

For instance, we can take $K$ and $L$ to be 3-dimensional lens spaces, and not just $L(7,1)$ and $L(7,2)$. The homotopical classification and Reidemeister torsions of lens spaces are well-known: see http://www.map.mpim-bonn.mpg.de/Lens_spaces.

2more comments