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The KZ equations on the configuration space of $n$ distinct points in $\mathbb C$ give rise to a representation of $B_n$ on $V^{\otimes n}$, where $V$ is any given representation of $SL(2)$ (we'll stick to this case; clearly we could work with other Lie groups as well). The WRT Hilbert space $\mathcal H_n$ associated to the $SL(2)$ character variety of $\mathbb S^2\setminus\{p_0,\ldots,p_n\}$ (where we restrict the monodromy around each $p_i$ to lie in a particular conjugacy class) also carries a natural action of $B_n$.

Now, both of these representations give rise to the Jones polynomial in a more or less straightforward manner, so I would almost like to conjecture that there is a natural isomorphism between the two. Unfortunately, I don't even see a reason why they should have the same dimension (and I'd guess they probably don't!). Thus I have a weaker proposal:

Question: Is there a natural homomorphism $\rho:V^{\otimes n}\to\mathcal H_n$ (or perhaps it should go the other direction) which respects the action of $B_n$?

Remark: There are choices to be made to construct these representations (KZ and WRT). I expect the choice of representation $V$ and Planck constant $\hbar$ in the KZ equations correspond in a straightforward manner with the choice of level $k$ and conjugacy class for the WRT character variety. More specifically, one would most certainly take $\hbar=k^{-1}$ and the conjugacy class to have trace $2\cos(\hbar(\dim V-1))$ [perhaps up to multiplicative constants].

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Can this be phrased in the quantum group setting? Something like comparing the $B_n$ representations on $V^{\ot n}$ where $V$ is a $U_q sl_2$ module and the $B_n$ representations from the mapping of $\mathbb{C}B_n$ in $End(V^{\ot n})$? To get a Hilbert space I think you need the root of unity case (and indeed, if you are fixing a level then this is the case). If so, I can give you an answer. – Eric Rowell Aug 26 '11 at 21:47
Sure! I guess I didn't really need to use the KZ equations here . . . The monodromy of KZ is equivalent to the quantum group construction giving a $B_n$ representation on $V^{\otimes n}$ where $V$ is any representation of $U_q(sl_2)$. The root of unity case for me would suffice. – John Pardon Aug 27 '11 at 2:17
up vote 5 down vote accepted

What I will say is just for quantum $SL(2)$ with $V$ being the 2-dimensional representation (or the analogous object in the representation category). Let us denote by $H$ the quantum group. For $q$ generic you can just use the following: $V^{\otimes n}\cong\bigoplus_i Hom(V_i,V^{\otimes n})\otimes V_i$ as an $H$ module and the image of $\mathbb{C}B_n$ in the centralizer algebra commutes with the action of $H$ so as a $B_n$-representation $V^{\otimes n}\cong \bigoplus_i \dim(V_i)Hom(V_i,V^{\otimes n})$. The WRT representation space is exactly $\mathcal{H}_n=\bigoplus Hom(V_i,V^{\otimes n})$, that is, the minimal faithful module for $End(V^{\otimes n})$. Here $V_i$ are the simple $H$-submodules of $V^{\otimes n}$. So your injection $\rho$ goes the other way.

This proof works for any (topologically) quasi-triangular Hopf algebra. For the root of unity case (fixed level case) there is the problem that the object $V$ no longer has a vector space structure (in the semisimple quotient). This is necessary to consider particularly if you want unitary representations. However, there is something similar one can try, which Wang and I call a "localization." Basically you look for a Yang-Baxter operator $R$ on a vector space $W$ so that $\mathcal{H}_n$ is a $B_n$-subrepresentation of $W^{\otimes n}$. For the $SL(2)$ situation this is only possible at levels $k=1,2,4$ corresponding to roots of unity of degree $3,4$ and $6$.

Some references on the more general question: link text link text

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Thanks! How does one see that $\mathcal H_n=\bigoplus\operatorname{Hom}(V_i,V^{\otimes n})$? (a pointer to a reference would suffice). – John Pardon Aug 27 '11 at 18:19
Probably Turaev's book would have this. Briefly, consider the disk with n punctures each labeled with the object $V$ and the boundary labelled by the object $V_i$. The modular functor of the corresponding TQFT assigns to this surface the vector space $\mathrm{Hom}(V_i,V^{\otimes n})$. From the Jones polynomial/Temperley-Lieb algebra perspective these are the irreducible $TL_n(q)$-modules. So the direct sum of these is the minimal faithful module that one can use to define the Jones polynomial (this by $q$-Schur-Weyl duality). – Eric Rowell Aug 27 '11 at 18:46

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