Timeline for Invertible 2-knots in $S^4$
Current License: CC BY-SA 4.0
10 events
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| Jun 23, 2021 at 0:59 | history | edited | Danny Ruberman | CC BY-SA 4.0 |
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| Jun 22, 2021 at 13:17 | comment | added | Danny Ruberman | @HJRW Yes this is the case; the edge group injects into the $G_i$ since the inclusions are isomorphisms on $H_1$. This is implicit in Victor's comment above about the infinite cyclic cover. | |
| Jun 22, 2021 at 13:00 | comment | added | HJRW | Regarding the "I think" bit, as long as $G_1 *_\mathbb{Z} G_2$ means an actual amalgamated free product (i.e. the edge group $\mathbb{Z}$ injects into the vertex groups $G_i$), then this is true by Britton's lemma, which asserts that the vertex groups also embed into the total group. (See this MO question for further discussion of Britton's lemma: mathoverflow.net/q/343958/1463 .) | |
| Jun 22, 2021 at 8:41 | vote | accept | Victor | ||
| Jun 22, 2021 at 0:59 | history | edited | Danny Ruberman | CC BY-SA 4.0 |
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| Jun 22, 2021 at 0:22 | comment | added | Victor | Sorry, I realized I meant a slightly different question. I changed it. It is quite obvious that $\tilde C_{K_1\cdot K_2}\simeq\tilde C_{K_1}\vee\tilde C_{K_2}$, where $\tilde C_K$ is the infinite cyclic cover of a knot complement. That's the standard argument that works in any dimension. | |
| Jun 21, 2021 at 21:59 | comment | added | Danny Ruberman | Yes, that is true. It follows pretty readily from duality in the infinite cyclic cover (= the universal cover). | |
| Jun 21, 2021 at 20:42 | comment | added | Victor | Do you mean that the fundamental group being $\mathbb Z$ implies that the complement is homotopy equivalent to $S^1$? I know it is not true in higher dimensions. | |
| Jun 21, 2021 at 20:42 | vote | accept | Victor | ||
| Jun 22, 2021 at 0:10 | |||||
| Jun 21, 2021 at 19:21 | history | answered | Danny Ruberman | CC BY-SA 4.0 |