This is a follow up of my previous MO question "Non orientable, closed manifold covered by two simply-connected charts." Nick L's nice answer shows that such manifolds actually exist, examples being provided by some non-orientable $S^n$-bundles over $S^1$, with $n \geq 2$.

In these examples, the two charts have the homotopy type of $S^n$. So, let me ask the following

Question. Does it exist a closed, non-orientable smooth manifold that can be written as the union of exactly two contractible charts?

Note. A comment by Denis Nardin to the aforementioned question shows that, by the strong form of Seifert-van Kampen theorem, a manifold covered by two contractible charts $U$, $V$ has the same homotopy type of the suspension of $U \cap V$.

  • 2
    $\begingroup$ A closed manifold covered by two contractible charts has LS category 1, and hence is homeomorphic to a sphere. I don't know a reference but this fact is mentioned on the top of p.2 of arxiv.org/pdf/0706.1625.pdf. LS category is defined on p.7 (definition 3.1). $\endgroup$ – Igor Belegradek Apr 22 at 12:58

No. Recall that the Lusternik-Schnirelmann category of a space $X$, denoted $\operatorname{cat}(X)$, is the minumum $k$ such that $X$ may be covered by open sets $U_0,U_1,\ldots, U_k$ such that each inclusion is null-homotopic. The standard lower bound for LS-category is the cup-length of reduced cohomology: if $R$ is a commutative ring, and $x_1,\ldots, x_k\in \tilde{H}^*(X;R)$ are cohomology classes whose cup product $x_1\cdot \cdots \cdot x_k\in \tilde{H}^*(X;R)$ is non-zero, then $\operatorname{cat}(X)\geq k$. The proof of this is a nice exercise in the naturality of relative cup products.

Now suppose $N$ is a closed non-orientable $n$-manifold which is covered by two contractible charts $U_0$ and $U_1$. It follows that $\operatorname{cat}(N)=1$ (it can't be $0$, because closed manifolds are never contractible), and all cup products in $\tilde{H}^*(N;\mathbb{Z}/2)$ are trivial.

Since $N$ is non-orientable, its first Stiefel-Whitney class $w_1(N)\in H^1(N;\mathbb{Z}/2)$ is non-zero. Poincaré duality gives a non-singular pairing $H^1(N;\mathbb{Z}/2)\times H^{n-1}(N;\mathbb{Z}/2)\to H^n(N;\mathbb{Z}/2)$. In particular $w_1(N)\cdot y\neq 0$ for some $y\in H^{n-1}(N;\mathbb{Z}/2)$. Contradiction.

  • 3
    $\begingroup$ Another way of seeing that the cup product is trivial is to notice that the manifold is homotopy equivalent to the suspension of $U_0\cap U_1$ by the van Kampen theorem. $\endgroup$ – Denis Nardin Apr 22 at 13:14

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.