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This question is related to this one (and indeed the goals are similar).

Let $N$ be odd and consider the braided fusion category $\mathcal{C}$ (actually modular) obtained from $U_q\mathfrak{so}_N$ at $q=e^{\pi i/(2N)}$ in the usual way (i.e. take the quotient of the category of tilting modules of Lusztig's integral form by the tensor ideal of negligible morphisms). This is sometimes denoted $SO(N)_2$ by physicists. Let $V$ denote the object in $\mathcal{C}$ analogous to the fundamental spin representation. Then the braid group $\mathcal{B}_n$ acts on the simple $End(V^{\otimes n})$-modules $Hom(W,V^{\otimes n})$ irreducibly, since in this case the image of $\mathcal{B}_n$ generates the centralizer algebras.

Question: how can one explicitly describe the braid group representations, say up to $n=5$? I.e. is there a uniform way of writing down the matrices of the generators of $\mathcal{B}_n$?

What I know:

  • For $q$ generic one can just use the $R$-matrix.
  • For $N=3$ this is a (quotient of) the Temperley-Lieb algebras
  • For $N=5$ this is a (quotient of) BMW-algebras (since $\mathfrak{so}_5$ is $\mathfrak{sp}_4$).
  • For $N=7$ Westbury has a description MR2388243.
  • For $N=k^2$ the category is integral and in fact group-theoretical

So for $N\leq 7$ or a perfect square, one can get a useful description of the braid group action using the irreps of the quotient algebras or by the Drinfeld double construction. It is too much to ask for a description as a quotient of $\mathcal{B}_n$ for $q$ generic, but perhaps there is a way for this particular value of $q$.

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I have a conjecture for this question. My question is: if I give you some matrices and claim they answer this question then how are you going to check my claim? My conjecture is for $n=3$ and $q$ generic. – Bruce Westbury Jun 10 '10 at 17:46
This is the problem, I don't have a way of checking--no generators/relations. I have a conjecture myself for all n and $q$ as above that I can't check either! If substituting the value of $q$ above into your $(N\pm 1)/2$ dimensional irreps gives me reps. factoring over $SL(2,Z_N)$ for $N$ prime then I would be a long way towards believing they are correct... If someone has worked out the 6j-symbols for these values then perhaps one could get somewhere. – Eric Rowell Jun 10 '10 at 19:44
Knowing the $6j$-symbols is equivalent to knowing the representation for $n=3$. There are two bases. In one basis $\sigma_1$ is diagonal and the eigenvalues are known, similarly for $\sigma_2$ in the other basis. The $6j$ symbols give the change of basis. – Bruce Westbury Jun 11 '10 at 6:30

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