Timeline for Loop manipulation subgroup of the braid group
Current License: CC BY-SA 4.0
10 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Apr 8 at 14:36 | vote | accept | pgadey | ||
| Apr 7 at 14:19 | comment | added | Jules Lamers | If you'd invert one of the $C_i$'s in equation (4) (which I understand you don't want to do) then the subgroup generated by the $C_i$ and $T_j$ is an affine braid group. More precisely, replacing your $C_i$ by something that's usually called $T_i$, which in addition obeys the quadratic relation $(T_i - q)(T_i + q^{-1})=0$, and writing $Y_i$ for your $T_i$, instead by gives (the Bernstein–Zelevinsky presentation of) the affine Hecke algebra of type $GL_n$. (I'm not aware of an analogue of your $P_i$ in that case.) | |
| Apr 2 at 1:27 | comment | added | Tuomas Laakkonen | I've seen this referred to as the Hilden subgroup of $B_{2n}$. E.g: arXiv:0706.4421 Theorem 5.3 gives a presentation in terms of your generators and 14 relations. | |
| Apr 1 at 21:49 | history | became hot network question | |||
| Apr 1 at 20:10 | comment | added | Jules Lamers | You're thinking of $T_i = \sigma_{2i-1}$ every second braid (twist)? | |
| Apr 1 at 19:08 | answer | added | Sam Nead | timeline score: 8 | |
| Apr 1 at 17:59 | comment | added | pgadey | Thanks. These seem like straightforward generalizations of the usual braid relation. How can I be sure that this list is complete? | |
| Apr 1 at 17:37 | comment | added | Sam Nead | If I've understood things correctly, you need to add the following relations: $P_i C_{i+1} C_i = C_{i+1} C_i P_{i+1}$, and $P_i P_{i+1} C_i = C_{i+1} P_i P_{i+1}$, and $P_i P_{i+1} P_i = P_{i+1} P_i P_{i+1}$. | |
| Apr 1 at 13:59 | history | edited | YCor | CC BY-SA 4.0 |
formatting, added tag
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| Apr 1 at 13:49 | history | asked | pgadey | CC BY-SA 4.0 |