Timeline for $ \mathbb{R}P^n $ bundles over the circle
Current License: CC BY-SA 4.0
11 events
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| Jan 24, 2022 at 0:41 | comment | added | Jason DeVito - on hiatus | @IanGershonTeixeira: The bundle structure is quite explicit. Deleting a ball from $\mathbb{R}P^3$, the resulting space is the total space of the disk bundle in the tautological bundle over $\mathbb{R}P^2$. Taking two copies and gluing, we get a map $\mathbb{R}P^3\sharp \mathbb{R}P^3\rightarrow \mathbb{R}P^2$ with fiber obtained by gluing the two fiber $[0,1]$s to themselves along the boundary, i.e., with fiber $S^1$. (Technically, to make everything glue up nicely, we must use $\mathbb{R}P^3\sharp -\mathbb{R}P^3$, but this is diffeomorphic to $\mathbb{R}P^3\sharp \mathbb{R}P^3$.) | |
| Jan 23, 2022 at 15:16 | comment | added | Igor Belegradek | @IanGershonTeixeira: yes, there is a non-orientable circle bundle over $RP^2$ whose total space is diffeomorphic to $RP^3\# RP^3$. See e.g. p.459 in Scott's "The geometries of 3-manifolds", homepages.warwick.ac.uk/~masgar/Teach/2021_3MFDS/References/…. It is probably not hard to describe the bundle explicitly. | |
| Jan 23, 2022 at 14:56 | comment | added | Ian Gershon Teixeira | Oh goodness! Perhaps the moral of the story here is that topology is not my area and when I talk about it I should be wary of the different notions of equivalence. While I have the attention of you fine gentleman, could I ask your opinion on whether the question "How can I see that $ \mathbb{R}P^3 \# \mathbb{R}P^3 $ is diffeomoprhic to a circle bundle over $ \mathbb{R}P^2 $" is appropriate for MO or if you think I should post it to MSE first? | |
| Jan 23, 2022 at 14:37 | comment | added | Dmitry Vaintrob | @IanGershonTeixeira Just to be clear, the result I wrote only works for Serre fibrations, which are not just the "up to homeomorphism" but also the "up to homotopy" version of bundles (I edited my preamble to make that clearer). I'm not sure whether the "up to homeomorphism" result is true, perhaps someone else can comment on this. | |
| Jan 23, 2022 at 14:35 | comment | added | Igor Belegradek | @IanGershonTeixeira: perhaps your confusion comes from ignoring the structure group. One can classify up to fiber homotopy equivalence (as in Vaintrob's answer), fiber preserving homeomorphism, fiber preserving diffeomorphism (as in Goodwillie's answer), and also one can look at linear $S^n$ bundles over the circle, which have the structure group $O(n+1)$. Since $\pi_1(BO(n+1))\cong \pi_0(O(n+1))\cong\mathbb Z_2$ there are two linear $S^n$ bundles over $S^1$. | |
| Jan 23, 2022 at 14:26 | comment | added | Ian Gershon Teixeira | I am a little confused because Michael Albanese says " there are only two $S^n$-bundles over $S^1$ up to diffeomorphism" in his comment on math.stackexchange.com/questions/4348711/… is that consistent with what you are saying here? | |
| Jan 23, 2022 at 14:22 | comment | added | Ian Gershon Teixeira | I am a little uneducated about exotic smooth structures. Although I am interested in both perspectives, I think the answer by Dmitry Vaintrob about classification up to homeomorphism is more what I am looking for, so I have accepted it. You are right that the MSE question had diffeomorphism in the title. Sorry if that caused any confusion. And thanks for the answer, maybe this will motivate me to look into exotic smooth structures more! | |
| Jan 23, 2022 at 13:55 | history | edited | Tom Goodwillie | CC BY-SA 4.0 |
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| Jan 23, 2022 at 13:51 | comment | added | ThiKu | Or at least it has been proven to equal something else. | |
| Jan 23, 2022 at 13:49 | comment | added | ThiKu | For r=11mod16 the quotient of $\pi_0 Diff^+(RP^r)$ modulo concordance has been computed in Wells:The concordance group of real projective space, jstor.org/stable/1996838?seq=19#metadata_info_tab_contents | |
| Jan 23, 2022 at 13:16 | history | answered | Tom Goodwillie | CC BY-SA 4.0 |