Timeline for Cobordant of 5d manifolds, and the generalization of bordisms
Current License: CC BY-SA 4.0
10 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Sep 22, 2018 at 17:40 | comment | added | mme | 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 $BSO \times BG$". | |
| Sep 22, 2018 at 17:37 | comment | added | wonderich | @MikeMiller, feel free let me know my last comment --- is my notation above OK or do you suggest something else? Thank you! | |
| Sep 10, 2018 at 19:46 | comment | added | wonderich | In my notation, I wrote $Ω^{SO}_5(BG)$ is the same as $$Ω_5(B(SO \times G))=Ω_5(BSO \times BG))=Ω^{SO}_5(BG).$$ Does this make sense or should we have a better notation? Thank you! | |
| Sep 6, 2018 at 22:07 | comment | added | wonderich | Yes thanks so much for the comment. This "$B^2(A)→BH→BG$" is what I meant, but I thought it is the same as "$K(A,2)→BH→K(G,1)$," no? | |
| Sep 6, 2018 at 21:06 | comment | added | mme | 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 $H$. | |
| Sep 6, 2018 at 18:57 | comment | added | wonderich | +1 thanks for the excellent comments!!! | |
| Sep 6, 2018 at 18:28 | comment | added | Arun Debray | 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.) | |
| Sep 6, 2018 at 18:26 | comment | added | Arun Debray | 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. | |
| Sep 6, 2018 at 18:13 | comment | added | Tobias Shin | 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. | |
| Sep 6, 2018 at 17:50 | history | asked | wonderich | CC BY-SA 4.0 |