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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.

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

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  • $\begingroup$ 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"). $\endgroup$ Dec 14, 2015 at 8:34
  • $\begingroup$ I understood. I was afraid of spamming question and had no other intensions. $\endgroup$
    – Hwang
    Dec 14, 2015 at 9:14
  • $\begingroup$ 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) $\endgroup$ Dec 14, 2015 at 9:15

2 Answers 2

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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.

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  • $\begingroup$ 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. $\endgroup$
    – Hwang
    Dec 14, 2015 at 8:06
  • $\begingroup$ @Hwang: I don't have an example off the top of my head, but you should get trouble already from products of tori. $\endgroup$
    – user98602
    Dec 14, 2015 at 20:08
  • $\begingroup$ This is a great example. +1. $\endgroup$ Dec 17, 2015 at 16:11
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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.

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  • $\begingroup$ Thanks. Now I see such property is almost hopeless. $\endgroup$
    – Hwang
    Dec 14, 2015 at 9:15
  • $\begingroup$ Why is $TS^2$ indecomposable? $\endgroup$
    – 54321user
    Apr 21, 2017 at 2:08
  • $\begingroup$ @user251222: it follows from the Poincare-Hopf index theorem, for example. $\endgroup$ Apr 21, 2017 at 2:43

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