Quantum braid group

Question

Recently I learned the definition of the quantum permutation group $A_s(n)$, which starts from the fundamental representation by permutation matrices and exchanges the entries by noncommuting operators, i.e.:

It is the universal $C^*$-algebra generated by matrices $u = (u_{ij})$ of generators, such that each $u_{ij}$ is a projection and on each row and column of $u$ these projections are orthogonal and sum up to $1$.

From the point of view of generators and relations the braid group is not that far away from the permutation group. Although it is hard to find faithful representations as was done by Bigelow in this paper. An interesting theorem by Artin also shows that the braid group $B_n$ embeds into Aut($F_n$), i.e. the automorphisms of the free group, as those elements that fix $x_1\cdot \dots \cdot x_n$ (see also this question).

Is there a definition of a quantum braid group? Or is this too difficult due to the non-finiteness of $B_n$? Is there a way to define what a free quantum group would be? (to obtain quantum-$B_n$ as a fixed point algebra)

Answer 1

The following answer comprises a collection of remarks.

(1) What, typically, allows a quantum version of a classical group is not relations between elements of the group $G$ (but see (2)), but rather between functions on the group.

For example, where $\mathbf{1}_{j\to i}\in C(S_N)$ is the indicator function for the subset of permutations in $S_N$ that map $j$ to $i$ ($i,j\in\{1,2,\dots,N\}$), we have that these indicator functions are projections in the $\mathrm{C}^*$-algebra $C(S_N)$, and, if we populate a matrix $v\in M_N(C(S_N))$ according to $v_{ij}=\mathbf{1}_{j\to i}$, they also they sum to one along $i=1,\dots,N$ and along $j=1,\dots N$ (note if we are in the $\mathrm{C}^*$-algebraic setting orthogonality for such partitions of unity is automatic (but see here)). Such a matrix is called a magic unitary.

So, the relations between these functions, which generate all of $C(S_N)$, means that the algebra of functions can be given as a universal commutative $\mathrm{C}^*$-algebra: $$C(S_N)=\mathrm{C}_{\operatorname{comm}}(v_{ij}\mid v\text{ a magic unitary}).$$

Now drop commutativity and you get the algebra of continuous functions on the quantum permutation group $S_N^+$. These days the notation $C(S_N^+)$ is used for $A_s(N)$.

This approach allows us to define a quantum orthogonal group, a quantum unitary group, and a quantum automorphism group of a graph. And sitting inside all of these, via the abelianisation, are the classical versions they have "liberated".

Forgetting all about topology, I imagine there are no nice relations on functions on the braid group.

(2) I would say that the typical place that we see generators in the group giving a quantum group is the dual case. So, for example, take whatever presentation you want of $S_N$ and form the group $\mathrm{C}^*$-algebra from this presentation. Then you can give this the structure of the algebra of continuous functions on compact quantum group, the dual of $S_N$: $$C(\widehat{S_N}):=\mathrm{C}^*(S_N).$$

I imagine this can be done for the braid group no problem... but note $\widehat{S_N}$ isn't a quantum version of $S_N$... in fact it is a liberation of $\mathbb{Z}_2$. I think the liberation of $\widehat{B_N}$ should be an abelian group... as a set equal to the set (equivalence classes?) of one dimensional representations of $B_N$.

(3) I have not checked if applicable, but Section 3 of Rollier--Vaes might produce a quantisation of the braid group...

(4) As an addenda, I was just in a seminar talk where it was claimed as a moral that actions of the (possibly infinite) symmetric group in classical probability should be replaced by action of braid groups in quantum probability... see here.