How to obtain the Klein bottle as a product of manifolds?

Question

I know the Klein bottle $K$ is a fiber bundle over $S^1$, but my question is: is it possible to find a manifold $M$ such that $K = S^1 \times M$ without the need to take an equivalence relation afterwards? My thoughts are, maybe something like taking $M$ as the image of $S^1$ under $z \to z^2$ or something in that direction. Do you think I will be able to get somewhere? Thanks in advance.

Answer 1

The Klein bottle cannot be expressed as a product of one dimensional manifolds. There are a few ways of showing this. One useful starting point is the result that there are only two connected one dimensional manifolds (without boundary, up to homeomorphism), $S^1$ and $\mathbb{R}$. Products of non-connected manifolds are non-connected, so if the Klein bottle is a product of two one dimensional manifolds, they must each be either $S^1$ or $\mathbb{R}$. However, none of the products, $S^1\times S^1$, $\mathbb{R}^2$ or $S^1\times\mathbb{R}$ are homoemorphic to the Klein bottle (e.g. because all three are orientable, while the Klein bottle is not).

In general, being a nontrivial product of lower dimensional manifolds is a very strict condition, and excludes a lot of interesting cases, such as the Klein bottle. It's for precisely this reason that constructions like fiber bundles exist, allowing for more general decompositions of manifolds into simpler pieces.