Timeline for If the universal cover of a manifold is spin, must it admit a finite cover which is spin?
Current License: CC BY-SA 4.0
14 events
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| Apr 6, 2022 at 22:47 | comment | added | Jason DeVito - on hiatus | @R.Rankin: The projection $\pi:\widetilde{M}\rightarrow M$ from the universal cover $\widetilde{M}$ of $M$ to $M$ is a local diffeomorphism, so that, $\pi^\ast(TM) = T\widetilde{M}$. Now compute $w_2$ of both sides and use naturality. | |
| Mar 27, 2022 at 2:41 | comment | added | R. Rankin | What about the other way? If a universal cover of M is not spin does that imply M is not spin? | |
| Sep 15, 2020 at 2:56 | vote | accept | Michael Albanese | ||
| Sep 11, 2020 at 11:50 | comment | added | Johannes Ebert | Of course the universal cover is only stably parallelizable. | |
| Sep 11, 2020 at 11:50 | history | edited | Johannes Ebert | CC BY-SA 4.0 |
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| Sep 10, 2020 at 22:25 | comment | added | Michael Albanese | Thanks for fleshing out the details, I can now see why $w_2(TM) \neq 0$. Is the universal cover of $M$ parallelisable or only stably parallelisable? I can show the latter using the isomorphism $TM\oplus\mathbb{R} \cong \ell^*f^*V\oplus\mathbb{R}$. | |
| Sep 10, 2020 at 7:36 | comment | added | Johannes Ebert | Then by obstruction theory (compute the connectivity of the map $BO(d+n-1)\to BO(d+n)$ and recall that $M$ is $d$-dimensional, you can cancel all of the extra summands except for the last one. Without adding the extra copy, the statement becomes false, as you can see when $d=4$ and you apply this surgery below the middle dimension to the trivial bundle $\mathbb{R}^4 \to \ast$: there does not exist a simply connected $4$-manifold which is parallelizable, by an Euler number argument. | |
| Sep 10, 2020 at 7:33 | comment | added | Johannes Ebert | One only gets an isomorphism $TM \oplus \mathbb{R}\cong \ell^\ast f^\ast V\oplus \mathbb{R}$; I have corrected the statement. To deduce my claim from the result in Wall's book, you pick a complementary bundle $V^\bot$ of large dimension, apply Wall's result to get $TM \oplus \ell^\ast f^\ast V^\bot$ stably trivial, add further trivial bundles to $V^\bot$ to make the sum actually trivial. Add a copy of $\ell^\ast f^\ast V$ to both sides of the equation and get that $TM \oplus \mathbb{R}^n \cong \ell^\ast f^\ast V \oplus \mathbb{R}^n$ for some large $n$. (ctd.) | |
| Sep 10, 2020 at 7:27 | history | edited | Johannes Ebert | CC BY-SA 4.0 |
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| Sep 9, 2020 at 18:57 | comment | added | Michael Albanese | I know this is technique is relatively standard, but it's my first time seeing it. I don't see how the aforementioned theorem allows us to deduce that there is a $2$-connected map $\ell : M \to BG$ with $TM\cong\ell^*f^*V$. Rather, from the definition of normal map on page 19, it seems to me that one obtains a $2$-connected map $\ell : M \to BG$ with $TM\oplus\ell^*f^*V$ stably trivial. What am I missing? | |
| Sep 9, 2020 at 16:01 | comment | added | Johannes Ebert | I don't think you can get it in a more direct way, because you also want to control the tangent/normal bundle along the process. An easier strategy to construct a manifold with fundamental group $G$ which is often mentioned is to embed the 2-skeleton of BG into R^5 (or some larger R^n), to take a regular neighborhood of the image and to take its boundary. The result would, however, be stably parallelizable. | |
| Sep 9, 2020 at 15:16 | comment | added | Michael Albanese | Thanks for your answer. I think I follow the construction up until the penultimate paragraph. I am trying to understand the construction of $M$ via Theorem 1.2 on page 22 here. Is there a more direct result I can apply to deduce the existence of $M$? | |
| Sep 8, 2020 at 22:18 | history | edited | Johannes Ebert | CC BY-SA 4.0 |
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| Sep 8, 2020 at 18:27 | history | answered | Johannes Ebert | CC BY-SA 4.0 |