Is there a q-analog to the braid group? - MathOverflow

Is there a q-analog to the braid group?

2010-11-11T06:03:23Z

The braid group $B_n$ on $n$ strands fits into a short exact sequence of groups:

$$ 1 \longrightarrow P_n \longrightarrow B_n \longrightarrow S_n \longrightarrow 1,$$

where $S_n$ is the symmetric group on the strands, and $P_n$ is the normal subgroup of braids that do not permute the strands.

Since symmetric groups are, in some sense, ``general linear groups over the field with one element,'' perhaps there is some corresponding short exact sequence ending with $GL_n(F_q)$ that specializes to the exact sequence above as $q \rightarrow 1$.

In the spirit of the exact sequence above, is there a $q$-analog to the braid group?

Answer by Mark Sapir for Is there a q-analog to the braid group?

2010-11-11T06:21:06Z

You may want to look at these slides of Jon McCammond's talk. I am not sure he wrote a paper about it, but he did introduce the idea similar to what you want and in more general situation (arbitrary Artin group instead of the braid group),

Answer by Romeo for Is there a q-analog to the braid group?

2010-11-11T07:09:15Z

Possibly. 2 ideas in this circle are: (1) The Artin braid groups can be formulated as ``Weyl groups" (corresponding to the $A_n$ Dynkin diagrams), and Weyl groups have q-analogues. Many references exist, but I'm not sure what the best is (hopefully others will know). (2) A second but related perspective can be found in Braids, Q-binomials and Quantum Groups by M Aguilar.

Answer by James Griffin for Is there a q-analog to the braid group?

2010-11-11T11:47:16Z

I suggest you look at the Iwahori-Hecke algebras of type A. These deform the symmetric group algebras with relations that look like

$T_i^2 = q + (1-q)T_i$

for generating elements $T_i$. The braid relations

$T_iT_{i+1}T_i = T_{i+1}T_iT_{i+1}$

still hold though so you get a (surjective) morphism

$kB_n\rightarrow \mathcal{H}_n\rightarrow 0$

(From here on I'm less sure of the details) giving you a `short exact sequence'

$0\rightarrow K_n\rightarrow kB_n \rightarrow \mathcal{H}_n\rightarrow 0$.

To define the kernel you should look at the coinvariants of $kB_n$ w.r.t. the coalgebra map of $\mathcal{H}_n$. This makes $K_n$ an algebra but not necessarily a Hopf algebra (although it may be a braided Hopf algebra in a suitable category).

The Hecke algebra may be the algebra that you want because the q parameter counts the way the Borel double cosets in some $GL_n(k)$ multiply (recall the Bruhat decomposition). When the Borels become trivial then $q$ becomes one.

But notice also that this deformation does not require a deformation of the braid group. I have no idea what the algebra $K_n$ looks like and if indeed it is well defined, it may still turn out to be $kP_n$.

To offer an answer to your final question: there may very well be q-analogues of the braid groups, but they may not be what you should be looking for.

Answer by grok for Is there a q-analog to the braid group?

2010-11-11T13:46:22Z

I don't see why there would be a unique scheme that "converges" to the braid group as $q\to1$. Here are some groups that could reasonably considered as the $q$-analogues of $B_n$:

Consider the non-commutative power series algebra over $\mathbb F_q$ in $n$ variables, usually written $A=\mathbb F_q\langle\langle x_1,\dots,x_n\rangle\rangle$. Its group of automorphisms $B_{n,q}$ is a $q$-analogue of $B_n$; the natural map $A\mapsto A/\langle x_ix_j\rangle$ induces the map $B_{n,q}\to GL(n,q)$.
At least when $q$ is prime, consider the subgroup of $A$ generated by $1+x_1,\dots,1+x_n$. A classical result of Magnus says it's a free group on $n$ generators. Take its closure in $A$ --- that's a free pro-$q$ group. Consider then the subgroup of $B_{n,q}$ that preserves that group. (If $q$ is a prime power, presumably throw in a few more automorphisms to obtain a group mapping onto $GL(n,q)$).

There's a lot of theory on the following filtration of $B_n$: it has a subgroup $P_n$, as the poster mentioned; and $P_n$ has a lower central series converging to $1$, for which the structure of the successive quotients are well understood, as $S_n$-modules. A good $q$-analogue should probably have a filtration by $GL(n,q)$-modules with the same shapes and multiplicities in each degree.

Answer by Sheikraisinrollbank for Is there a q-analog to the braid group?

2010-11-11T15:32:08Z

Here is one strategy that has not been suggested: the braid group is the fundamental group of the space $U/W$, where $U$ is the set of points in the (complexified) reflection representation of $W=S_n$ that have trivial fixer. Now $W=GL_n(\mathbb{F}_q)$ is a reflection group over the finite field $\mathbb{F}_q$; let $U$ be the set of points (over an algebraic closure $F$ of $\mathbb{F}_q$) in the reflection representation of $W$ that have trivial fixer and define the "braid group" to be the etale fundamental group of $U/W$. Essentially by definition it has a surjection onto $W$. It's not clear to me in what sense this q-braid group might converge to the usual one as q goes to 1, but it should definitely play an important role in the study of $GL_n(\mathbb{F}_q)$. For instance, one might be able to define a Hecke algebra deforming $F[GL_n(\mathbb{F}_q)]$ via monodromy representations of this braid group.

Answer by Alexei Oblomkov for Is there a q-analog to the braid group?

2010-11-12T05:58:57Z

There is a nice paper by Etingof and Rains about a non-standard q-deformations of S_n:

http://xxx.lanl.gov/abs/math/0409261

They deform not quadratic relations but the braid relations. It seems to be relevant to the discussion.