# Unique factorization of manifolds?

I wonder if there is a result on the unique factorization of manifolds.

Call a topological manifold to be indecomposable if it is not homeomorphic to a product of manifolds of positive dimension. Is every manifold a unique (up to order) product of indecomposable ones?

I couldn't find any statements on this simple question. Are there any results on this? Any result in different categories (smooth, complex, Riemannian or whatever) or with extra conditions is fine.

 The answer seems to be No in most cases. Can we impose strong conditions so that the answer is positive?

• Regarding your edit: you should ask a new question, linking back to this question. Because 1/ your edit means that the existing answers are now incomplete, 2/ you are now asking two different things ("Is there a unique factorization" and "What conditions can we impose to enforce unique factorization"). Dec 14, 2015 at 8:34
• I understood. I was afraid of spamming question and had no other intensions. Dec 14, 2015 at 9:14
• There are some limits on the number of question you can ask (6 per day, 30 per month I think), but as long as your questions are interesting, well-motivated etc, don't be afraid of asking them! This is what this website is for, after all. (By the way you can include links in both directions: link the new question here, and link this old question in your new question) Dec 14, 2015 at 9:15

Nope.

Consider lens spaces $L(p,q)$. They are all indecomposable, by investigation of the fundamental group. Then the main result of this paper is that $L(p,q) \times L(p,q)$ is a manifold $X_p$ which depends only on $p$! So, for instance, $L(p,1) \times L(p,1) \cong L(p,2) \times L(p,2)$, even though $L(p,1) \not\cong L(p,2)$; there are older examples of non-homeomorphic manifolds with diffeomorphic squares, too.

I can't really think of a way to talk about unique factorization that this example doesn't break.

• Thanks. I hope I can see some positive answers for this even if under strong conditions. Lens spaces are not complex manifolds at least, so maybe still a slight chance? I will wait a few days and select your answer. Dec 14, 2015 at 8:06
• @Hwang: I don't have an example off the top of my head, but you should get trouble already from products of tori.
– user98602
Dec 14, 2015 at 20:08
• This is a great example. +1. Dec 17, 2015 at 16:11

Generally the answer is no. For example, $TS^2$ is indecomposible. But $TS^2 \times \mathbb R \simeq S^2 \times \mathbb R^3$, so $S^2 \times \mathbb R^3$ splits as a product of indecomposibles in several different ways.

You could use $\mathbb C$ instead of $\mathbb R$ if you want complex manifolds.

You get similar things happening for Riemann manifolds as well.

• Thanks. Now I see such property is almost hopeless. Dec 14, 2015 at 9:15
• Why is $TS^2$ indecomposable? Apr 21, 2017 at 2:08
• @user251222: it follows from the Poincare-Hopf index theorem, for example. Apr 21, 2017 at 2:43