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Let $$a \in H^1(\mathbb{RP}^5,\mathbb{Z}_2)=\mathbb{Z}_2,$$

  • When $a' \in H^1(\mathbb{RP}^5,\mathbb{Z}_2)=\mathbb{Z}_2$ is a nontrivial generator, the Poincaré dual (4-manifold generator) PD$(a')$ of the nontrivial class $H^1(\mathbb{RP}^5,\mathbb{Z}_2)$ is $\mathbb{RP}^4$ in $\mathbb{RP}^5$ (Yes?).
  • When $a \in H^1(\mathbb{RP}^5,\mathbb{Z}_2)=\mathbb{Z}_2$, what is the Poincaré dual (4-manifold generator) PD$(a)$ of the trivial class $H^1(\mathbb{RP}^5,\mathbb{Z}_2)$ in $\mathbb{RP}^5$? Can it be $\mathbb{CP}^2$?
wonderich
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    Arun Debray’s comment to your question here https://mathoverflow.net/questions/309604/non-spin-5-manifold-and-22-bockstein-homomorphism answers your (first) question. – Aleksandar Milivojević Sep 03 '18 at 16:39
  • "Poincaré dual for 0∈H1(M;ℤ/2) is any embedded closed 4-manifold which bounds" - it means that $\mathbb{CP}^2$ is also OK, but it looks too general to be true to me -- that is why I asked. – wonderich Sep 03 '18 at 16:48
  • @wonderich: $\mathbb{CP}^2$ does not bound. – Michael Albanese Sep 03 '18 at 18:07
  • If I am correct, $\mathbb{CP}^2=S^{5}/S^1$, why $\mathbb{CP}^2$ does not bound? (You mean $\mathbb{CP}^2$ does not bound a point or a line or what?) Thanks. – wonderich Sep 03 '18 at 20:51
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    $\mathbb{CP}^2$ cannot be the boundary of a compact five-dimensional manifold with boundary; i.e. $\mathbb{CP}^2$ is not nullcobordant. One way to see this is that it has odd Euler characteristic, whereas every manifold which bounds (i.e. is nullcobordant) has even Euler characteristic. See this answer for more details. – Michael Albanese Sep 03 '18 at 21:37

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