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
11 events
| when toggle format | what | by | license | comment | |
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| Jan 25, 2021 at 2:03 | comment | added | Michael Albanese | Another reference for the reflection trick is Chapter 11 of Davis' The Geometry and Topology of Coxeter Groups. | |
| Sep 14, 2020 at 22:53 | comment | added | Michael Albanese | I see, $\langle a, b \mid [a, [a, b]] = 1, [b, [a, b]] = 1\rangle$ is a valid presentation. Thanks yet again for your help. I will try to remember this trick. | |
| Sep 14, 2020 at 22:51 | comment | added | Moishe Kohan | @MichaelAlbanese: One common way to eliminate the boundary is to double the manifold along the boundary. However, you typically loose asphericity. Davis trick is a form of doubling which preserves asphericity. | |
| Sep 14, 2020 at 22:50 | comment | added | Moishe Kohan | @MichaelAlbanese: Right. | |
| Sep 14, 2020 at 22:49 | comment | added | Moishe Kohan | @MichaelAlbanese: $t$ is a letter in a presentation: You can eliminate it if you like by writing down relators of the form $[a,[a,b]]=1$. The bottom line is that you need just two generators. | |
| Sep 14, 2020 at 22:49 | comment | added | Michael Albanese | Am I correct in saying that the reason you apply the Davis trick to $Z$ is to obtain a closed aspherical manifold $M$ (as opposed to $Z$ which is an aspherical manifold with boundary)? | |
| Sep 14, 2020 at 22:47 | comment | added | Michael Albanese | [Feel free to ignore this question] If you don't include $t$ as a generator, what does $[a, b] = t$ mean ($t$ is not a word in $a$ and $b$)? | |
| Sep 14, 2020 at 18:54 | comment | added | Moishe Kohan | @MichaelAlbanese If you prefer three, but two suffice, since [𝑎,𝑏]=𝑡. | |
| Sep 14, 2020 at 18:53 | history | edited | Moishe Kohan | CC BY-SA 4.0 |
added 1222 characters in body
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| Sep 14, 2020 at 15:36 | comment | added | Michael Albanese | I think you mean $\pi_1(\partial E)$ has three generators $a$, $b$, and $t$. In the following paragraph, how do you know that $\pi_1(\partial N)$ maps non-trivially to $\pi_1(W)$? Is the inclusion map $\partial N \hookrightarrow N$ followed by the deformation retraction $N \to W$ non-trivial on $\pi_1$? | |
| Sep 9, 2020 at 17:11 | history | answered | Moishe Kohan | CC BY-SA 4.0 |