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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.

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  • $\begingroup$ Any closed manifold has non-zero cup product by Poincare duality. So $S^1\times S^2$ satisfies your criteria. $\endgroup$
    – Ben Webster ♦
    Commented Jun 15, 2015 at 2:15
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    $\begingroup$ Original math.SE post: math.stackexchange.com/q/1325568/264 $\endgroup$
    – Zev Chonoles
    Commented Jun 15, 2015 at 2:16
  • $\begingroup$ @BenWebster Sorry, I edited the question to clarify that I mean the cup-product $H^1(M;\Bbb Z_2)\times H^1(M;\Bbb Z_2)\to H^2(M;\Bbb Z_2)$. $\endgroup$
    – Irina
    Commented Jun 15, 2015 at 2:26

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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 $x^2\neq 0$.

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  • $\begingroup$ Why is M orientable? If N is $S^4$ and $\sigma$ is the antipodal map, then $M = RP^4$ is an example of the construction you suggest and is not orientable, so there must be an additional condition to get M to be orientable. $\endgroup$
    – Robert Bruner
    Commented Jun 15, 2015 at 20:04
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    $\begingroup$ I guess we're just looking for an example in which the involution in question is orientation-preserving. $\endgroup$
    – Dan Ramras
    Commented Jun 16, 2015 at 0:59
  • $\begingroup$ Right. Sorry I didn't make it explicit. $\endgroup$
    – abx
    Commented Jun 16, 2015 at 4:24
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    $\begingroup$ This is the other way around: all Enriques surfaces have a double covering by a K3 surface, which has therefore an orientation-preserving (and fixed point free) involution. $\endgroup$
    – abx
    Commented Jun 16, 2015 at 16:44
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    $\begingroup$ $y$ is 4-torsion because $2y=\beta x$ is 2-torsion. And I meant what I wrote : if $\pi :N\rightarrow M$ is the quotient map, we have $\pi _*\pi ^*y=2y\neq 0$, hence $\pi ^*y$ is a nonzero torsion class in $H^2(N,\mathbb{Z})$. $\endgroup$
    – abx
    Commented Jul 5, 2016 at 7:58

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