5
$\begingroup$

For a simple Lie algebra $\mathfrak{g}$, we have the W-algebra of level $k$, denoted by $\mathcal{W}^k(\mathfrak{g})$. Using Wakimoto free field realization and screening operators, Feigin and Frenkel established the celebrated isomorphism $$\mathcal{W}^k(\mathfrak{g})\simeq\mathcal{W}^{^Lk}(^L\mathfrak{g}).$$ Here $^L\mathfrak{g}$ is the Langlands dual of $\mathfrak{g}$, $^Lk\in\mathbb{C}$ satisfies the equation $$r^\vee(k+h^\vee)({^Lk}+{^Lh^\vee})=1,$$ where $r^\vee$ is the lacing number of $\mathfrak{g}$ and $h^\vee$ (resp. $^Lh^\vee$) is the dual Coxeter number of $\mathfrak{g}$ (resp. $^L\mathfrak{g}$).

Qustion: is there a functorial isomorphism here? Namely, can we construct a functor $$F\colon \mathcal{W}^k(\mathfrak{g})\text{-}\mathsf{Mod}\to\mathcal{W}^{^Lk}(^L\mathfrak{g})\text{-}\mathsf{Mod}$$ that is an equivalence? Moreover, since the quantum Drinfeld–Sokolov reduction gives a functor from the category of restricted $\hat{\mathfrak{g}}$-modules of level $k$ to $\mathcal{W}^k(\mathfrak{g})\text{-}\mathsf{Mod}$, is it possible to lift the functor $F$ to a functor between certain $\hat{\mathfrak{g}}$-modules and $\widehat{^L\mathfrak{g}}$-modules?

$\endgroup$
3
  • $\begingroup$ For your first question, the FF theorem is an isomorphism of vertex algebras (or topological associative algebras if you pass to enveloping algebras) so tautologically gives an equivalence of their categories of modules $\endgroup$ Commented May 5, 2023 at 17:41
  • $\begingroup$ For the second question, yes, this is an affine version due to Sam Raskin of the much easier Skryabin theorem for Whittaker modules for a reductive Lie algebra— arxiv.org/abs/1611.04937 $\endgroup$ Commented May 5, 2023 at 17:43
  • $\begingroup$ @DavidBen-Zvi Yes, I understand that there is an equivalence of categories of modules, but I want to know if there is an explicit way to describe it. $\endgroup$
    – Estwald
    Commented May 6, 2023 at 5:04

0

You must log in to answer this question.