Codimension one rational cohomology groups vs Codimension one mod 2 cohomology groups

Question

Let $M$ be an orientable even dimensional manifold of dimension d. I try to find the examples of $M$ such that $H^{d-1}(M;\mathbb{Q})=0$ but $H^{d-1}(M;\mathbb{Z}_{2})$ is not trivial.

Answer 1

There are examples of closed orientable manifolds with the desired property in every dimension greater than two. In dimension two, there are no examples (closed or otherwise).

If $M$ is a closed, orientable $d$-dimensional manifold, then $H^{d-1}(M; R) \cong H_1(M; R) \cong H_1(M;\mathbb{Z})\otimes R$. So $M$ provides an example if and only if $G := H_1(M; \mathbb{Z})$ satisfies $G\otimes\mathbb{Q} = 0$ and $G\otimes\mathbb{Z}_2 \neq 0$. Since $G$ is a finitely generated abelian group, it follows that these two conditions hold if and only if $G$ is finite and contains elements of even order. For example, the groups $G = \mathbb{Z}_2^r$ have this property - note, the first homology of the manifolds mentioned in Aphelli's comment above are of this type.

There are examples of closed, orientable $d$-dimensional manifolds with $H_1(M; \mathbb{Z}) \cong \mathbb{Z}_2$ for every $d \geq 3$. By the above discussion, they have $H^{d-1}(M; \mathbb{Q}) = 0$ and $H^{d-1}(M; \mathbb{Z}_2) \cong \mathbb{Z}_2\otimes\mathbb{Z}_2 \cong \mathbb{Z}_2 \neq 0$.

In the smallest possible dimension, namely three, we can take $M = \mathbb{RP}^3$. For $m \geq 2$, we can also form the manifold $M = \mathbb{RP}^3\times S^m$ where $m \geq 2$. This provides examples in every dimension $m + 3 \geq 5$. In dimension four, consider the manifold $M = (S^2\times S^2)/\sim$ where $(x, y) \sim (-x, -y)$. One can identify $M$ with $\operatorname{Gr}(2,4)$, the grassmannian of unoriented two planes in $\mathbb{R}^4$. Note that $\pi_1(M) \cong \mathbb{Z}_2$ and since the map $(x, y) \to (-x, -y)$ is orientation-preserving, $M$ is orientable.

As for dimension two, note that in the closed case $H_1(M; \mathbb{Z}) \cong \mathbb{Z}^{2g}$, which can be finite ($g = 0$), but never has elements of even order. In the non-closed case, since $H^1(M; R) \cong \operatorname{Hom}(\pi_1(M), R)$ and $\pi_1(M)$ is free, then either $\pi_1(M) \neq 0$, in which case $H^1(M; R) \neq 0$ for every non-trivial $R$, or $\pi_1(M) = 0$, in which case $H^1(M; R) = 0$ for every $R$.