Timeline for Extending a 2-complex embedded in $\mathbb{S}^3$ into a simply connected one
Current License: CC BY-SA 4.0
8 events
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Sep 10, 2021 at 13:29 | comment | added | Agelos | I'm fine with this extended definition, though I think you only used it in the sentence "That is, C′ is a spine for S3", which doesn't affect the rest of the discussion. | |
Sep 10, 2021 at 7:07 | comment | added | Sam Nead | @Agelos - I fixed the typo (about degree zero) you pointed out. Thank you! Also, I suppose that I am working with a more general definition of "spine". Probably I should not do that! But it feels like the correct definition? I mean, why not allow multiple three-balls in the complement of a spine? The extra flexibility is very handy. | |
Sep 10, 2021 at 7:05 | history | edited | Sam Nead | CC BY-SA 4.0 |
fix typo
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Sep 9, 2021 at 20:41 | comment | added | Sam Nead | Yes... I think so. We take the quotient, take a regular neighbourhood $B$ of the orbifold locus, cut it into standard pieces with disks (so, balls about the vertices, solid cylinders about the edges) and add $\partial B$ and the disks to our spine. Then we play the same game as before in $S^3 / H - B$. | |
Sep 9, 2021 at 9:21 | comment | added | Agelos | Suppose now I want $C '$ to be `canonical' in the following sense. I have a finite group $H$ of isometries acting on $S^3$ mapping my original 2-complex $C$ onto intself. I want this action of $H$ on $C$ to extend to $C'$. A possible approach would be to repeat your construction of $C'$ on the quotient orbifold $S^3/H$, and lift the result back to $S^3$. Would this work? | |
Sep 9, 2021 at 9:18 | comment | added | Agelos | Nice, thank you! When you write $\Gamma$ may have vertices of degree zero, I think you mean "degree one". I don't agree that "$C'$ is a spine for $S^3$" (it is if you remove a point from each chamber), but this is irrelevant. | |
Sep 9, 2021 at 9:16 | vote | accept | Agelos | ||
Sep 7, 2021 at 15:08 | history | answered | Sam Nead | CC BY-SA 4.0 |