Let $\mathfrak{g}$ be a simple lie algebra over $\mathbb{C}$ and let $\hat{\mathfrak{g}}$ be the Kac-Moody algebra obtained as the canonical central extension of the algebraic loop algebra $\mathfrak{g} \otimes \mathbb{C}[t,t^{-1}]$. In a sequence of papers, Kazhdan and Lusztig constructed a braided monoidal structure on (a certain subcategory of) the category of representations of $\hat{\mathfrak{g}}$ of central charge $k – h$ where $k \in \mathbb{C}^* \;\backslash\; \mathbb{Q}_{\geq 0}$ and $h$ is the coxeter number of $\mathfrak{g}$. They then showed that the resulting braided category is equivalent to the braided category of finite dimensional representations of the quantum group $U_q(\mathfrak{g})$ for $q = e^{\frac{\pi i}{k}}$.
My question then is this: is there any conceptual explanation as to why these two braided categories should be equivalent (which does not resort to computing both sides and seeing that they are same)? The representations of $\hat{\mathfrak{g}}$ of various central charges can be considered as twists of the representation theory of the loop algebra $\mathfrak{g} \otimes \mathbb{C}[t,t^{-1}]$. On the other hand, the representation theory of $U_q(\mathfrak{g})$ is a braided deformation (which can be thought of as a form of twisting) of the representation theory of $\mathfrak{g}$ itself. Moreover, the equivalence above only holds for non-trivially deformed/twisted cases. The limiting case of the representations of $\mathfrak{g}$ is recovered by (carefully) taking $q=1$, which corresponds to $k \rightarrow \infty$ and hence does not participate in the game. On the other hand, to obtain central charge $0$ we would need to take $k=h$ which is also excluded (as the proof Kazhdan-Lustig assumes $k \notin \mathbb{Q}_{\geq 0}$). Is there any reason why these two lie algebras would have the same twisted/deformed representations, but not the same representations?
Best Answer
This is not very sophisticated answer, but in a way there isn't much choose from as the representation categories of "deformations of $U(\mathfrak{g})$." For $\mathfrak{g}=\mathfrak{sl}_n$, this can be made precise by [1] as follows. Any semisimple $\mathbb{C}$-linear monoidal category with the fusion rule of $\mathrm{SL}(n)$ has to be a twist of the finite dimensional admissible representations of $U_q(\mathfrak{sl}_n)$ for some $q$, and the twisting data is discrete, given by 3-cohomology of the Pontrjagin dual of the center of $\mathrm{SL}(n)$. If you impose the existence of braiding, the twisting class has to be of order two, so you either have the representation category of $\mathrm{SL}_q(n)$ or "twisting by parity" when $n$ is even. Other cases are probably the same, see for example [2].