I am looking for an example of a closed orientable 4-manifold $M$ with $H^1(M;\Bbb Z\_2)=\Bbb Z\_2$ and non-zero cup product $H^1(M;\Bbb Z\_2)\times H^1(M;\Bbb Z\_2)\to H^2(M;\Bbb Z\_2)$.


A non-orientable example is $\Bbb RP^4$. An orientable example of dimension 3 is $\Bbb RP^3$.


I have asked at math stackexchange and the question was upvoted but no answers have been given.



Closed orientable 4-manifold with $H^1(M;\Bbb Z_2)=\Bbb Z_2$ and non-zero cup product $H^1\times H^1\to H^2$
Take a closed oriented simply-connected fourfold $N$ with a fixed point free involution $\sigma $, and put $M=N/\sigma $ (for a typical example, take for $M$ an Enriques surface). Then $H^1(M,\mathbb{Z}\_2)=$ $\mathrm{Hom}(\pi \_1(M),\mathbb{Z}\_2)=\mathbb{Z}\_2$. Let $x$ be the nonzero element of $H^1(M,\mathbb{Z}\_2)$; the square $x^2$ is the reduction (mod. 2) of $\beta x$, where $\beta :H^1(M,\mathbb{Z}\_2)\rightarrow H^2(M,\mathbb{Z})$ is the Bockstein homomorphism. If $x^2=0$, one has $\beta x=2y$ for some class $y$ in $H^2(M,\mathbb{Z})$. But $\beta x\neq 0$ because $H^1(M,\mathbb{Z})=0$, hence $y$ is a 4-torsion class, which is impossible since $H^2(N,\mathbb{Z})$ is torsion free. Hence $x^2\neq 0$.


