# Non orientable, closed manifold covered by two contractible charts

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 $$U \cap V$$.

• 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). Apr 22, 2021 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.
• 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. Apr 22, 2021 at 13:14