16
$\begingroup$

Bing gave a classical example of spaces $X, Y, Z$ such that $X \times Y = Z$, where $X$ and $Z$ are manifolds but $Y$ isn't. The space $Z$ in his example has dimension four. Is it known if this is best possible? In other words, if $X \times Y =Z$ where $X$ is a manifold and $Z$ is a 3-manifold, then is $Y$ a manifold?

In Bing's example, $Z$ is not compact. Is there a compact example in dimension $4$? In the example, $X$ is the real line, so one can also ask if it possible to get a four dimensional example where $X$ is a surface, or where $X$ is compact.

$\endgroup$
9
$\begingroup$

As asked Dusan Repovs (who is an expert in the theory of topological manifolds), and he sent me the following answer:

This is indeed best possible result, since whenever a product of two spaces is a topological manifold, both factors must be generalized manifolds - which in dimensions below 3 are topological manifolds.

Ref.: A.Cavicchioli, F.Hegenbarth and D.Repovš, Higher-Dimensional Generalized Manifolds: Surgery and Constructions, EMS Series of Lectures in Mathematics, European Mathematical Society, Zurich, 2016.

To the second question: there are also compact examples (in dimensions >3): e.g. take the product of the 3-sphere $S^3$ modulo the Fox-Artin wild arc $A$ and $S^1$. This product is homeomorphic to $S^3\times S^1$.

Ref.: R.J.Daverman, Decompositions of Manifolds, Academic Press, Orlando, 1986.


Added in Edit: Answering a comment of John Samples, Dusan Repovs pointed out that the Chapter 29 of Daverman's book contains the following fact: for any $n,m>2$ there are a generalized $n$-manifold $X$ and a generalized $m$-manifold $Y$ which are not topological manifolds, but their product $X\times Y$ is a topological manifold.

$\endgroup$
  • $\begingroup$ I do not understand how the statement fits together with the example. Is the wild arc a generalized manifold or not? $\endgroup$ – ThiKu Jun 19 '18 at 7:06
  • $\begingroup$ @ThisKu The wild arc is not a generalized manifold (it is a topological arc), but the quotient space of the sphere by the wild acr is a generalized manifold (not being a manifold). $\endgroup$ – Taras Banakh Jun 19 '18 at 7:44
  • $\begingroup$ Now I wonder if there is a pair of (non-manifold) generalized $3$-manifolds whose product gives a manifold, or if one factor will always be a bona fide manifold. $\endgroup$ – John Samples Jun 21 '18 at 22:18
  • $\begingroup$ It should be noted that for some definitions of 'generalized manifold' the properties of separability and metrizability are not assumed. In that setting, there are examples of non-manifold generalized manifolds in dimensions one and two. But factors of separable, metric spaces are separable metric, so since manifolds have these properties everything still works for this factorization problem. I am starting to learn this area, now. $\endgroup$ – John Samples Jun 30 '18 at 2:14

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.