This question arose during my Differential Geometry course. Possibly there is an obvious answer, but I do not see it, and I could not find it in the literature. The same question was asked yesterday on MSE, but it did not get much attention.

Question. Does it exist a closed, non-orientable smooth manifold that can be written as the union of exactly two simply-connected charts? If so, what is a reference?

Of course, the intersection of the two charts must be disconnected. As noted in the linked MSE question, the open Möbius band and the closed Möbius band give examples in the open case and in the non-empty boundary case, respectively. Moreover, the Klein bottle is covered by two charts, both homeomorphic to cylinders, but they are not simply-connected.


Take a non-orientable $S^n$ bundle over $S^1$ with $n \geq 2$ (*), (sometimes called generalised Klein bottles) then covering $S^1$ by two intervals and taking preimages should work.

(*) Let $\tau : S^n \rightarrow S^{n}$ be a non-orientable diffeomorphism of $S^n$ given by $(x_{1},\ldots,x_{n+1}) \mapsto (x_1,\ldots,-x_{n+1}) $. Then such a bundle is given by taking the quotient of $[0,1] \times S^{n}$ by the equivelance closure of the relation $$(0,p) \sim (1,\tau(p)). $$

Such bundles are discussed in the following paper https://academic.oup.com/plms/article-abstract/s3-4/1/196/1497906?redirectedFrom=PDF

The non-orientable $S^2$-bundle over $S^1$ appear's in Hatchers 3-manifold notes, since it plays a role in the prime decomposition theorem for non orientable $3$-manifolds. See page 8 of https://pi.math.cornell.edu/~hatcher/3M/3Mdoublepage.pdf

  • 1
    $\begingroup$ Thank you for the nice answer. I am not an expert in the field, could you please suggest a reference for these manifolds? $\endgroup$ – Francesco Polizzi Apr 21 at 10:46
  • 3
    $\begingroup$ Ok, I added a definition of the bundles and a couple of references. $\endgroup$ – Nick L Apr 21 at 14:27
  • 1
    $\begingroup$ If a manifold is covered by two contractible charts, it has the homotopy type of the suspension of the intersection by the strong van Kampen theorem. This puts quite a strong constraint on its homotopy type - I would be skeptical it's possible to do it.. $\endgroup$ – Denis Nardin Apr 22 at 6:10
  • 1
    $\begingroup$ @FrancescoPolizzi The (unreduced) suspension of two points is not contractible. To be more explicit the theorem I'm referring to is that $U\cup V$ is the homotopy pushout of $U←U\cap V→V$, and it is true without hypotheses. In this case $U,V$ are contractible so the homotopy pushout is the unreduced suspension of $U\cap V$. In your example $U\cap V$ is homotopy equivalent to two points and $U\cup V$ is homotopy equivalent to $S^1$. $\endgroup$ – Denis Nardin Apr 22 at 7:28
  • 1
    $\begingroup$ @FrancescoPolizzi: It's not possible with contractible charts. The resulting manifold would have (reduced) Lusternik-Schnirelmann category equal to one, so all cup products in its mod 2 cohomology would vanish. But the first Stiefel-Whitney class, which is non-zero if the manifold is orientable, pairs with something by mod 2 Poincaré duality. $\endgroup$ – Mark Grant Apr 22 at 12:17

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.