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The $2^2$-Bockstein is $\beta_4$ is associated to $$0\to\mathbb{Z}/2\to\mathbb{Z}/{8}\to\mathbb{Z}/{4}\to 0,$$

(The $2^n$-Bockstein homomorphism $$\beta_{2^n}:H^*(-,\mathbb{Z}/{2^n})\to H^{*+1}(-,\mathbb{Z}/2)$$ is associated to the short exact sequence $$0\to\mathbb{Z}/2\to\mathbb{Z}/{2^{n+1}}\to\mathbb{Z}/{2^n}\to 0.$$ Note $\beta_2=Sq^1$ is the Steenrod square.)

Question: What are some closed 5-dimensional manifold $M$ satisfy all the criteria below:

1) $M$ is a non-spin manifold.

2) $\beta_{4}$ is nonzero. $$\beta_{4}:H^1(M,\mathbb{Z}/{4})\to H^{2}(M,\mathbb{Z}/2).$$

3) There exists a non-zero generator $a \in H^1(M,\mathbb{Z}/2)$, such that its Poincare dual PD$(a)$ is an orientable 4-manifold.

If so, what is this sub-manfiold generator $H^1(M,\mathbb{Z}/2)$ and what is the this orientable 4-manifold PD$(a)$? What is this $M$?

Note that the $\mathbb{RP}^5$ satisfies 1) and 2), but it does not satisfy 3), because the PD$(a)$ for $\mathbb{RP}^5$ is a non-orientable 4-manifold $\mathbb{RP}^4$.

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  • $\begingroup$ Are there any restrictions on $a$? If not, you could let $a = 0$. A Poincaré dual for $0\in H^1(M;\mathbb Z/2)$ is any embedded closed 4-manifold which bounds, such as a small $S^4$ around a point, and this is orientable. $\endgroup$
    – Arun Debray
    Commented Sep 1, 2018 at 15:38
  • $\begingroup$ @Arun, thanks, but $a$ needs to be nonzero. $\endgroup$
    – wonderich
    Commented Sep 1, 2018 at 15:50
  • $\begingroup$ Do you also want $M$ to be connected? $\endgroup$
    – Arun Debray
    Commented Sep 1, 2018 at 15:56
  • $\begingroup$ Not necessarily connected, disjoint union is OK. But the connected one is of course better. $\endgroup$
    – wonderich
    Commented Sep 1, 2018 at 15:59
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    $\begingroup$ If you are okay with taking connected sums, you can just take $\mathbb{R}\mathbb{P}^5 \# (S^1\times S^4)$. Conditions 1 and 2 are satisfied because of the $\mathbb{R}\mathbb{P}^5$ summand, and for condition 3 you can take $a$ to be the generator of $H^1(S^1)$ in the second summand. Its dual is then the generator of $H_4(S^4)$, and $PD(a)$ is represented by the $S^4$ factor in the second summand. $\endgroup$
    – Aleksandar Milivojević
    Commented Sep 1, 2018 at 16:58

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