4

One has $\mathbb R^n=\mathbb R^p\times\mathbb R^q$ whenever $p+q=n$. My question is whether or not one can choose different factors in the product.

Do there exist topological spaces $A,B$ so that $A\times B\cong \mathbb R^n$ for some $n$ and $A\not\cong \mathbb R^p$ for any $p$?

Intuitively: no way. But I have no idea how I would tackle such a problem, I can't even tell if it is simple or incredibly difficult.

s.harp
  • 21,879
  • By "cross product", do you mean the "cartesian product"?

    $$A \times B = { (a,b) : a\in A \ \wedge \ b \in B}$$

    The "cross product" works on pairs of vectors in $\mathbb R^3$ and gives something perpendicular.

     – Fly by Night Jan 12 '17 at 16:54
  • @FlybyNight yes, the cartesian product of topological spaces. Sometimes this is called "Kreuzprodukt" in german, it may be incorrect to use the corresponding word in english. – s.harp Jan 12 '17 at 16:56
  • 8
    Yes. Look up Whitehead manifolds. They are contractible three manifolds. If you take the Cartesian product of one of them with $\mathbb{R}$ you get $\mathbb{R}^4$. I bet the Cartesian product of two of them is six space. – Charlie Frohman Jan 12 '17 at 16:56
  • @CharlieFrohman Interesting, thank you. – s.harp Jan 12 '17 at 17:10
  • 1
    There are even worse examples: There is a space $B$ constructed by Bing (the "dogbone space") which is not even a manifold, but $B\times R$ is homeomorphic to $R^4$. – Moishe Kohan Jan 12 '17 at 21:56
  • Related: https://math.stackexchange.com/questions/57375/mathbb-r-x2-as-a-cartesian-product – Watson Nov 12 '18 at 11:26

0 Answers0