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Some of the 5-dimensional manifolds are (co)bordant via oriented cobordism.

  • For example, if I understand correctly, 5-dimensional Dold manifold and Wu manifold are manifolds which are cobordant to each other via 5-dimensional bordism group: $$ \Omega^{SO}_5=\mathbb{Z}_2. $$ In other words, my interpretation is that the Dold and Wu manifolds are both the nontrivial generators of $\Omega^{SO}_5=\mathbb{Z}_2$.

  • For example, consider an oriented bordism, bordisms between $4$-manifolds. The signature $\sigma(X^{4k})$ of an oriented $4k$-manifold $X^{4k}$ is an oriented bordism invariant. Now $\sigma(S^4) = 0$ and $\sigma( C P^2) = 1.$ So $S^4$ and $ C P^2$ are not oriented bordant. In fact, $\Omega_4^{\mathrm{SO}} \cong \Bbb Z$ with generator $[C P^2]$.

Question 1: I suppose that their 5-dimensional analogs, $CP^2 \times S^1$ and $S^4 \times S^1$ are not oriented bordant, either, via $\Omega^{SO}_5=\mathbb{Z}_2$, yes or no? But it looks that both manifolds are not the generators of $\Omega^{SO}_5=\mathbb{Z}_2$, thus could both $CP^2 \times S^1$ and $S^4 \times S^1$ are null bordant?

  • More generally, we can consider the following 5-manifolds,

(1). $CP^2 \times S^1$,

(2). $RP^5$,

(3). $({CP}^2 \times S^1)/\tau$, Dold manifold,

(4). $SU(3)/SO(3)$, Wu manifold,

(5). $({CP}^2 \# \overline{{CP}^2}) \times S^1$,

(6). $RP^5 \# (CP^2 \times S^1)$,

(7). $RP^5 \# (S^4 \times S^1)$,

(8). $S^4 \times S^1$,

(9). $S^5$,

(10). $S^5/\mathbb{Z}_{2m}$ (Lens space), say $2m$ is an even integer.

I think (1),(5) and (8) are mapping tori, and others are not.

Question 2: What are some of which are null bordant, some of which are bordant to each other via $\Omega^{SO}_5 \equiv \Omega^{}_5(BSO) =\mathbb{Z}_2$?

Question 3: We can consider the following new classifying space $BG'$ constructed from the fibrations $$ K(\mathbb{Z}_2,2)\to BG' \to BSO, $$ or equivalently, $$ K(\mathbb{Z}_2,2)\to BG' \to K(SO,1), $$

where $K(\mathbb{Z}_2,2)$ is the Eilenberg–MacLane space. The possible classes of fibrations are classified by Postinikov classes $[\omega]\in H^3(BSO,\mathbb{Z}_2)=\mathbb{Z}_2.$ So there are actually two different fibrational classifying space $BG'$, called either $BG'_1$ and $BG'_2$.

What are some of which are null bordant, some of which are bordant to each other via this new $\Omega^{}_5(BG')$?

Similarly, can you suggest new constructions of $BG'$, such that some of the manifolds $(1)-(10)$ become bordant to each other (even though they may not be bordant through $\Omega^{SO}_5$?

p.s. If you are confident that you can improve my question, please feel free to revise it. I appreciate that.

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    $\begingroup$ For a start, you could compute all of the spaces' Stiefel-Whitney numbers, which determine the unoriented cobordism type. If they are not unoriented cobordant, then they wouldn't be oriented cobordant. $\endgroup$
    – Tobias Shin
    Commented Sep 6, 2018 at 18:13
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    $\begingroup$ More specifically, the Stiefel-Whitney number $w_{2,3} = \langle w_2(M)w_3(M), [M]\rangle$ is a complete invariant for both oriented and unoriented cobordism in dimension 5. $\endgroup$
    – Arun Debray
    Commented Sep 6, 2018 at 18:26
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    $\begingroup$ For some of those manifolds, though, you can directly check they bound without worrying about characteristic numbers: if $M$ is any 4-manifold, $M\times S^1$ bounds $M\times D^2$, which takes care of (1), (5), and (8). (They are indeed mapping tori; (3) is as well.) $\endgroup$
    – Arun Debray
    Commented Sep 6, 2018 at 18:28
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    $\begingroup$ A small point: $\Omega_5^{SO}$ is not the same as $\Omega_5(BSO)$. The latter is cobordism classes of maps of unoriented 5-manifolds to $BSO$, and every element is 2-torsion. Now an oriented manifold comes equipped with a map to $BSO$, as it comes equipped with an oriented vector bundle, so you have a natural map $\Omega_5^{SO} \to \Omega_5^{SO}(BSO)$. But they are still definitely not the same. So I doubt $\Omega_5 BG'$ is actually what you want. Also, $BSO$ is not $K(SO, 1)$. It seems that you get a fibration $B^2(A) \to BH \to BG$ when $G$ is a central extension of $H$. $\endgroup$
    – mme
    Commented Sep 6, 2018 at 21:06
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    $\begingroup$ 1) When $G$ is a topological group (as opposed to a discrete group) $BG$ is not the same as $K(G,1)$. 2) I think that's really odd notation and probably not what you mean. The far right thing usually means "oriented bordism clases of maps from oriented manifolds to $BG$". The thing in the middle usually means "unoriented bordism classes of maps form unoriented manifolds to $BSO \times BG$". $\endgroup$
    – mme
    Commented Sep 22, 2018 at 17:40

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