Unique factorization of manifolds?

Question

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?

Answer 1

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.

Answer 2

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.