While working on something apparently unrelated I encountered a "braid-like" group, which is a relatively geometric subgroup of a braid group and seems to be itself an Artin group. It seems like something that might be important for other things or might have been encountered in a different situation, or might have a better description than what I'm about to give... But I don't know much about braids groups (and even less about Artin groups), so I'm hoping someone here might recognize it...
I'm curious to hear anything that one might tell me about this group. In particular:
- Has it been studied or encountered somewhere else?
- Is there a better (ideally more geometric) way to describe it?
- Is there a nice way to characterize its elements directly as braids? (instead of specifying generators).
- Is the group presentation I give at the end correct?
Ok, so the group I'm interested in is a subgroup of the braid group $B_n$ generated by two subgroups $B_p$ and $B_q$ with $n=p+q$ (but they don't commute, so it's not $B_p \times B_q$).
I'm fixing a partition $P$ of $\{1,\dots,n\}$ in two subsets, that I'll call "red" and "blue" with $p$ red elements and $q$ blue elements. The subgroup I'm going to construct is formed of "coloured braids", that is each strand connects two points of the same colour (which I call the colour of the strand).
So:
- I'm identifying $B_p$ with the subgroup of $B_n$ of coloured braids whose blue strands are vertical and just go over all the red strands (which can form any braid on $p$ strands).
- I'm identifying $B_q$ with the subgroup of $B_n$ of coloured braids whose red strands are vertical and just go over all the blue strands (which can form any braid on $p$ strands).
The group I'm interested in is the subgroup $T_P \subset B_n$ generated by these two subgroups $B_q$ and $B_p$.
Of course, if I replace one of the two "over" in the description above with "under" then the two groups would commute and I would simply obtain something isomorphic to $B_p \times B_q$, but here some non-trivial braiding between the red and blue strands can appear. Similarly, in the special case where the partition $P$ has all red elements on one side and all the blue elements on the other, we have $T_P \simeq B_p \times B_q$.
But if I chose the partition of $\{1,2,3,4\}$ as "red-blue-red-blue", then $T_P$ is the group generated by the braids $\sigma_1^{-1} \sigma_2 \sigma_1 = \sigma_2 \sigma_1 \sigma_2^{-1}$ (crossing the first and third strands under the second) and the braid $\sigma_2^{-1} \sigma_3 \sigma_2 = \sigma_3 \sigma_2 \sigma_3^{-1}$ (crossing the second and fourth strands under the third), which I believe generates a free group on two generators.
(and I apologize for the lack of pictures - I haven't figured out a nice way to draw braids on MO yet, but if I can figure it out, I'll add some pictures)
Conjectured presentation:
I believe the group I'm talking about admits a presentation as an Artin-Tits group as follows. One takes as generators the usual generators of the $B_p$ and $B_q$, which I'll denote by $\rho_{ij}$ and $\beta_{uv}$ where $\rho_{ij}$ exchange the two red strands $i$ and $j$ that are only separated by blue strands, and the $\beta_{uv}$ exchange the two blue strands $u$ and $v$ separated only by red strands.
The $p-1$ elements $\rho_{ij}$ together satisfy the usual braids relations to make them the generators of the $B_p$ described above, and the same things for the $q-1$ elements $\beta_{uv}$.
The relation between $\rho_{ij}$ and $\beta_{uv}$ are as follows:
- If they are "completely disjoint" (that is if $i<j<u<v$ of $u<v<i<j$) or if one is "included" in the other (that is if $i < u <v < j$ or $u<i<j<v$) then they commute.
- If they are "interlaced" (that is if $i < u < j <v$ or $u<i<v<j$) then they satisfy no relations at all.
The elements I've introduced are clearly generators, and the relation mentioned above are fairly clearly satisfied. I see other relations between these elements in general but I'm sure how to prove that this is actually a presentation.
