Timeline for Existence of orientable finite volume complete cusp hyperbolic 3-manifolds $\mathbb{H}^3 / \Gamma$, where $\Gamma$ has no parabolic generators?
Current License: CC BY-SA 4.0
16 events
| when toggle format | what | by | license | comment | |
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| Oct 14 at 20:36 | vote | accept | T ghosh | ||
| Oct 14 at 20:24 | comment | added | T ghosh | @MoisheKohan now I got your point. So, $\mathbb{Z}\oplus \mathbb{Z}$ can't come as a knot group. Thanks | |
| Oct 14 at 20:19 | comment | added | T ghosh | @MoisheKohan 1st homology group of knot complement is $\mathbb{Z}$ | |
| Oct 14 at 19:48 | comment | added | Moishe Kohan | Did you take an algebraic topology class before trying to learn knot theory and hyperbolic geometry? A standard exercise in algebraic topology is to compute homology of a knot complement. Can you do this? | |
| Oct 14 at 18:14 | comment | added | HJRW | @Tghosh: no commuting lair of loxodromics can generate a subgroup that acts properly on hyperbolic space. You need to understand this basic exercise if you want to understand the answer to your question. | |
| Oct 14 at 18:07 | comment | added | T ghosh | @MoisheKohan Why can't $\mathbb{Z} \oplus \mathbb{Z}$ come as a knot group? If we take $\mathbb{H}^3/\mathbb{Z} \oplus \mathbb{Z}$ then what problem will arise? | |
| Oct 14 at 17:48 | comment | added | Moishe Kohan | Noncommuting generators is automatic, since the only commutative knot groups are $\mathbb Z$, the fundamental group of the unknot. Similarly, if $M$ is a complete hyperbolic 3-manifold of finite volume, then its fundamental group is never commutative. | |
| Oct 14 at 17:31 | comment | added | T ghosh | @MoisheKohan Thank you so much pointing out mistakes, I am new in this area, I very recently started learning cusp hyperbolic 3-manifolds. Noncommutative generators: I mean generators do not commute each other. How I will get always loxodromic generators? | |
| Oct 14 at 17:07 | review | Close votes | |||
| Oct 20 at 3:08 | |||||
| Oct 14 at 17:02 | history | edited | T ghosh | CC BY-SA 4.0 |
added 52 characters in body
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| Oct 14 at 16:56 | history | edited | T ghosh | CC BY-SA 4.0 |
added 52 characters in body
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| Oct 14 at 16:44 | history | edited | Paul Broussous | CC BY-SA 4.0 |
existancy ---> existence
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| Oct 14 at 16:03 | answer | added | Sam Nead | timeline score: 3 | |
| Oct 14 at 15:57 | comment | added | Moishe Kohan | Your writing is unclear. What do you mean by "noncommutative generators"? I know what "noncommuting" means and standard parabolic generators of the figure 8 knot group do not commute. And any nonelementary 2-generated Kleinian group admits a 2-generating set both elements of which are loxodromic. | |
| Oct 14 at 15:55 | comment | added | Lee Mosher | Cross posted on math.stackexchange. | |
| Oct 14 at 15:49 | history | asked | T ghosh | CC BY-SA 4.0 |