Do there exist cancellable manifolds?

Question

I do not know whether there exists a terminology for that property, but let us say that a closed manifold $C$ is cancellable if for every closed manifolds $M_1$ and $M_2$, $C \times M_1$ and $C \times M_2$ are homeomorphic iff $M_1$ and $M_2$ so are. My question is:

Does there exist a closed cancellable $n$-manifold for any $n \geq 1$?

In fact, I even do not know whether there exists such a manifold in one dimension.

Motivation: It seems to be a common belief that the classification of closed (ie. connected, compact, boundaryless) manifolds (up to homeomorphism) becomes harder as the dimension increases. However, I did not find justifications for such an affirmation; moreover, it seems to be harder to solve Poincaré's conjecture in dimension three.

But if there existed a cancellable $n$-manifold $C$, then classifying $(n+k)$-manifolds would be at least as hard as classifying $k$-manifolds.

Answer 1

There was a post here on math stackexchange answering in the negative for dimension 1. i.e. the circle is not a cancelable manifold. They establish that there are manifolds with different homotopy types whose products with the circle are diffeomorphic.

In the same post they reference this paper which establishes that the odd dimensional spheres are not cancelable. (I haven't done more than briefly scan the paper).

This appears in general to be a pretty tricky question to answer. There are results about unique factorization of simply connected Riemannian manifolds, but I haven't been able to find any results relating to any other category of manifolds.

Answer 2

If a manifold $C$ has Euler characteristic $0$, then it is not cancelable, so for instance no odd dimensional manifold is cancelable. I will sketch the argument, which makes use of the basic properties of Whitehead torsion, including the s-cobordism theorem. The question asked about homeomorphism; the argument would work (and indeed has fewer technical prerequisites) in the smooth or PL setting, where the theory of Whitehead torsion and the s-cobordism theorem is simpler. It takes a fair amount of technical work to establish those theorems in the topological setting.

Start with a pair of manifolds of dimension at least $5$, say $N_1$ and $N_2$ that are h-cobordant but not homeomorphic. These aren't so easy to come by, but you can find a discussion of some examples of this eg in this Mathoverflow post. Let $W$ be an h-cobordism between $N_1$ and $N_2$; this means that the boundary of $W$ is $N_1 \cup N_2$ and the inclusion maps $N_i \to W$ are homotopy equivalences. Let $\tau(W,N_1)$ be the Whitehead torsion of this h-cobordism; this is an element of the Whitehead group of the fundamental group of $W$ (and $N_1$ and $N_2$ for that matter.) The key property is that if $\tau(W,N_1) = 0$ then $W$ is homeomorphic to a product.

The product theorem for torsion (Kwun and Szczarba, Product and sum theorems for Whitehead torsion, Ann. of Math. 82, 183-190) says that the torsion $(W \times C, N_1 \times C)$ is $\tau$ times the Euler characteristic of $C$, and so by assumption must vanish. By the s-cobordism theorem, $W \times C$ is homeomorphic to $N_1 \times C \times I$, so passing to the boundary gets that $N_1 \times C$ is homeomorphic to $N_2 \times C$.

I would bet that a similar argument proves that there is no cancelable manifold at all. One approach would be to find, for every natural number $k$, non-homeomorphic manifolds $N_1(k)$ and $N_2(k)$ as above, where the torsion of an h-cobordism $W(k)$ between them is an element of order $k$ in the Whitehead group. (So the torsion is torsion, so to speak!) Then for $\chi{(}C{)} = k$, you would multiply by $W(k)$ to get an example as above. But I have no idea how to construct such pairs $N_i(k)$.