Timeline for 2-generator subgroups of an Artin group of small type
Current License: CC BY-SA 3.0
12 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Feb 17, 2018 at 19:52 | comment | added | Harry Reed | @HJRW Thanks! This is an extremely useful paper for what I'm thinking about. | |
| Feb 16, 2018 at 15:57 | comment | added | Robert Bell | This (web.math.ucsb.edu/~mccammon/papers/mysterious-geometry.pdf) article of Jon McCammond describes the frontier of what we can say about Artin-Tits groups in general and those of small type in particular. | |
| Feb 16, 2018 at 9:23 | comment | added | HJRW | You might be interested in this paper of Leininger—Margalit, which addresses this question for pure braid groups. arxiv.org/abs/0812.1783v1 | |
| Feb 16, 2018 at 8:30 | comment | added | Harry Reed | @DerekHolt You are correct. It's only known in very few select cases. We don't even know if the Artin group of small type defined by the complete bipartite graph $K_{5,1}$, a 5-pointed star, has solvable word problem. | |
| Feb 16, 2018 at 8:26 | comment | added | Derek Holt | Am I correct in thinking that it Is unknown whether the word problem is solvable in Artin groups of small type? | |
| Feb 15, 2018 at 19:18 | comment | added | Robert Bell | This is also true for RAAGs. If $x, y, z$ are generators of a RAAG $G$, then the subgroup generated by $x$ and $z^{-1}yz$ is also a RAAG. In fact, it is a parabolic subgroup of the RAAG which is the kernel of the map $G \to \mathbb{Z}/2$ which maps $z$ to $1$ and every other generator to $0$. (This kernel is isomorphic to the RAAG defined by doubling the original defining graph along the star of $z$; new vertices correspond to conjugates of the original vertices.) It then follows by induction on the length of an element $g$ that the subgroup generated by $x$ and $g^{-1}yg$ is a RAAG. | |
| S Feb 14, 2018 at 19:39 | history | edited | Misha | CC BY-SA 3.0 |
Add possibility of the group being cyclic
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| S Feb 14, 2018 at 19:39 | history | suggested | Michal Buran | CC BY-SA 3.0 |
Add possibility of the group being cyclic
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| Feb 14, 2018 at 19:18 | comment | added | Misha | In the Coxeter setting a similar statement indeed holds and is easy to prove. | |
| Feb 14, 2018 at 18:33 | comment | added | user6976 | Is the similar statement for all Coxeter groups (the subgroup $\langle g, h^x\rangle$ is either free or isomorphic to a visible subgroup of $G$) true? How about RACGs? | |
| Feb 14, 2018 at 17:43 | review | Suggested edits | |||
| S Feb 14, 2018 at 19:39 | |||||
| Feb 14, 2018 at 17:17 | history | asked | Harry Reed | CC BY-SA 3.0 |