Why should affine lie algebras and quantum groups have equivalent representation theories?

Question

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?

Answer 1

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].

1. David Kazhdan and Hans Wenzl, Reconstructing monoidal categories, I. M. Gel′fand Seminar, Adv. Soviet Math., vol. 16, Amer. Math. Soc., Providence, RI, 1993, pp. 111–136. MR 1237835 (95e:18007)
2. Imre Tuba and Hans Wenzl, On braided tensor categories of type $BCD$, J. Reine Angew. Math. 581 (2005), 31--69.

Answer 2

(Written on my phone - apologies for any typos.)

A few comments:

a) First, as to the source of the braided monoidal structure on the Kazhdan-Lusztig category. The category of integrable affine Lie algebra reps is naturally a factorization category, which is close morally to an E2/braided monoidal category, see DBZ's answer to mathoverflow.net/questions/53988/what-is-the-motivation-for-a-vertex-algebra/54008#54008. In some cases this can be made precise, and this gives you your braided monoidal category here.

b) Now as to why one might expect these two braided monoidal categories to agree, I think the key is exactly the $\kappa\rightarrow\infty$ case you mention (corresponding to $q=1.$) A general "limiting to infinity" procedure is described in https://arxiv.org/abs/1708.05108, but here I will proceed in a more ad hoc way. Before taking this limit, let me restate the category we are interested in. The Kazhdan-Lusztig category is the category of finitely-generated $U_{\kappa}(g((t)))$-modules equipped with an action of $G$, with the conditions that the two induced $g$ actions agree and that elements of $tg[[t]]$ act nilpotently.

For our $\kappa\rightarrow\infty$ limit, we need to figure out how to degenerate $U_{\kappa}(g((t)))$. Writing $\kappa=c\kappa'$ for a fixed nondegenerate $\kappa'$, we can describe $U_{\kappa}(g((t)))$ as the free algebra on $g((t))$, mod the relations $[s,t]=[s,t]_0+c\kappa'(s,t).$ We can't directly limit $c$ to infinity, but note that we can rescale the generators and rewrite the relations as $[s,t]=\frac{1}{c}[s,t]_0+\kappa'(s,t),$ which we can limit to $[s,t]=\kappa'(s,t).$ So we can reasonably set $U_{\infty}(g((t)))$ to be the tensor product of $\operatorname{Sym} g$ and a Weyl algebra on $tg[[t]]\oplus (tg[[t]])^*$ (recall $\kappa'$ induces a perfect pairing between $g[[t]]$ and $t^{-1}g[t^{-1}]$.)

Now our original integrability condition limits to the conditions that $tg[[t]]$ acts nilpotently, that $g$ acts by zero (not that it agrees with the action of $G$ - recall our rescaling!), and that the $G$ actions on our representation and on the Weyl algebra are compatible. The only representations satisfying these conditions are of the form $V\otimes W$, where $V$ is a finite dimensional representation of $G$ and $W$ is the standard rep of the Weyl algebra. So our final category is indeed the category of $G$-reps.

c) A more highbrow way to limit $\kappa$ to infinity is via geometric Langlands. GL switches level infinity with critical level, and the Kazhdan-Lusztig category with the Whittaker category on the affine Grassmannian for the Langlands dual group (since here we are working with abelian categories and critical level, you can think instead of the category of spherical D-modules.) Now geometric Satake tells you this gives you back $Rep(G)$.

This is kind of perverse, in that you're using Langlands duality twice. The point here though is that this suggests one can prove the Kazhdan-Lusztig equivalence via similar methods to geometric Satake, by describing both sides via root datum. I believe Dennis Gaitsgory has a new proof following these lines.

Answer 3

I don't have the references to hand but as no-one else has offered an explanation, here is how I understand it. The representation category of the Kac-Moody algebra is the fusion category of a rational conformal field theory. The category associated to the quantum group needs to be defined a bit more carefully than in your post. However this is the category associated to a 2+1 topological field theory. There is a relation, I hesitate to say correspondence, between rational conformal field theories and 2+1 TQFT. Then the claim is that these two examples correspond.