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Does there exist a smooth, closed, non-orientable $6$-manifold $M$ such that $H_4(M;\mathbb{Z})=\mathbb{Z}/2$?

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    $\begingroup$ How about $S^2\times RP^2\times RP^2?$ $\endgroup$
    – Gabriel C. Drummond-Cole
    Commented May 23, 2015 at 17:08
  • $\begingroup$ I am curious to know if the existence of some manifold like this is meant to be an obstruction for existence of some specific class in bordism groups of immersions, and if so then what is it?!? $\endgroup$
    – user51223
    Commented May 23, 2015 at 19:09
  • $\begingroup$ @GabrielC.Drummond-Cole: Thanks, I think it works! $\endgroup$
    – Mark Grant
    Commented May 24, 2015 at 9:25
  • $\begingroup$ @user51223: This question came up when studying mod 2 cohomology classes realizable by immersions/embeddings. $\endgroup$
    – Mark Grant
    Commented May 24, 2015 at 9:27

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$M=S^2\times \mathbb{RP}^2\times\mathbb{RP}^2$.

Since $Tor(\mathbb{Z}/2,\mathbb{Z}/2)=\mathbb{Z}/2$, the Künneth formula tells you that the homology is:

  • $H_0(M,\mathbb{Z})=\mathbb{Z}$
  • $H_1(M,\mathbb{Z})=\mathbb{Z}/2 \oplus \mathbb{Z}/2$
  • $H_2(M,\mathbb{Z})=\mathbb{Z}/2\oplus \mathbb{Z}$
  • $H_3(M,\mathbb{Z})=\mathbb{Z}/2\oplus\mathbb{Z}/2\oplus \mathbb{Z}/2$
  • $H_4(M,\mathbb{Z})=\mathbb{Z}/2$
  • $H_5(M,\mathbb{Z})=\mathbb{Z}/2$
  • $H_6(M,\mathbb{Z})=0$.
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