8
$\begingroup$

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.

$\endgroup$
3
  • $\begingroup$ Any closed manifold has non-zero cup product by Poincare duality. So $S^1\times S^2$ satisfies your criteria. $\endgroup$
    – Ben Webster
    Jun 15, 2015 at 2:15
  • 1
    $\begingroup$ Original math.SE post: math.stackexchange.com/q/1325568/264 $\endgroup$ 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
    Jun 15, 2015 at 2:26

1 Answer 1

11
$\begingroup$

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 $x^2\neq 0$.

$\endgroup$
7
  • $\begingroup$ 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. $\endgroup$ Jun 15, 2015 at 20:04
  • 3
    $\begingroup$ I guess we're just looking for an example in which the involution in question is orientation-preserving. $\endgroup$
    – Dan Ramras
    Jun 16, 2015 at 0:59
  • $\begingroup$ Right. Sorry I didn't make it explicit. $\endgroup$
    – abx
    Jun 16, 2015 at 4:24
  • 3
    $\begingroup$ 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
    Jun 16, 2015 at 16:44
  • 1
    $\begingroup$ $y$ 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
    Jul 5, 2016 at 7:58

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.