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

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?

• For generic $q$ and generic level $k$, the categories of representations are all equivalent as abelian categories, with irreducible objects parametrized by the usual highest-weight theory for $\mathfrak{g}$. The subtlety comes in comparing the braided monoidal structures. Jul 31 '14 at 22:33
• That's true, although the theorem applies also to the case where $k$ is a negative rational number, in which case $q$ is a root of unity and hence the representation theory of $U_q(\mathfrak{g})$ is not the same as that of $\mathfrak{g}$, even on the level of the abelian category. This just reinforces the question of understanding this equivalence conceptually as a whole. Aug 1 '14 at 7:57
• Just to correct a small inaccuracy, looking again in Kazhdan-Lusztig I noticed that they actually require $k \in \mathbb{C} \;\backslash\; \mathbb{R}_{\geq -r}$ for a certain rational $r \geq 0$ depending on $\mathfrak{g}$, such that $r=0$ for $\mathfrak{g}$ of type $A_n$ or $D_n$ (see "tensor structures arising from affine lie algebras IV", page 421). But this doesn't affect the question too much. Aug 1 '14 at 8:58

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.
• Kuperberg proved a result like this for G2. But the larger exceptionals are beyond the range of current techniques. Mar 4 '16 at 14:29
• Thanks a lot! I found your slides from 2010 and a recent paper with Scott and Emily. For this purpose (to use the fusion rule only) don't we need these rather than Kuperberg's paper from 1994? Mar 7 '16 at 1:12
• Yeah, you're right, I was oversimplifying a little. Kuperberg's result gets you most of the way there, but you need a bit more to turn it into a full Wenzl-style recognition theorem. Our trivalent paper gets further towards that, but one needs a bit more as discussed in those slides. There's still a few cases from those slides that haven't made it into papers yet, but should soon. At any rate, Kuperberg's paper is enough to suggest that these techniques should be enough for G2 even though he didn't prove precisely this. Mar 7 '16 at 2:45

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

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.

• Is there a quick explanation/reference as to why the category of integrable affine Lie algebra representations is a factorization category? Jun 22 '19 at 11:29

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.

• A possible source of references: ncatlab.org/nlab/show/AdS3-CFT2+and+CS-WZW+correspondence. This feels like restating the mysterious thing rather than explaining it, though. Aug 2 '14 at 6:02
• @QiaochuYuan Thanks; and yes, I agree.
– BWW
Aug 2 '14 at 8:13
• Thanks, but there is something I still don't understand. In order to obtain a RCFT from representations of the Kac-Moody algebra you actually need to take $k$ to be a positive integer bigger than $h$ and restrict to a certain finitely generated subcategory of representations (the integrable ones), yielding a modular tensor category. However, the theorem of Kazhdan-Lusztig doesn't cover this case at all: first they assume $k \notin \mathbb{Q}_{\geq 0}$ (and this is necessary), and second, they deal with a bigger category of representations, which has infinitely many irreducible objects. Aug 3 '14 at 12:11
• Related: mathoverflow.net/questions/178113/… The "natural" relation indeed involves $k$ positive integer and integrable representations but the proof uses the theorem of Kazhdan-Lusztig via a relation between level $k-h$ and $-k-h$ due to Finkelberg (-k<0 so one can applies Kazhdan-Lusztig). Dec 30 '15 at 20:03