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Let $\mathfrak{g}$ be a semisimple Lie algebra, $\mathfrak{g}^L$ be its Langlands dual. Feigin--Frenkel duality says $$ W^k(\mathfrak{g})=W^{k_L}(\mathfrak{g}^L) $$ if $r'(k+h^{'})(k_L+h'_L)=1$, where $r'$ is the maximum number of edges between two vertices in the Dynkin diagram of $\mathfrak{g}$, $h'$ (resp. $h'_L$) is the dual Coxeter number of $\mathfrak{g}$ (resp. $\mathfrak{g}_L$).

A degeneration of this result when $k_L$ tends to infinity was used as local input by Beilinson--Drinfeld to establish some results on geometric Langlands (which was globalized using the theory of chiral algebras). Regarding the non-degenerate result, they say "We do not know if this “doubly quantized” picture can be globalized." Has the non-degenerate result been globalized in some way since then?

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The Feigin-Frenkel isomorphism is globalized by the global quantum geometric Langlands conjecture, proposed by Stoyanovsky, and refined by Gaitsgory and his collaborators. See Gaitsgory's 2016 collection of conjectures, in particular the discussion on page 5.

At irrational level, this appears to be (almost?) a theorem, with a major missing piece worked out earlier this year by Arakawa and E. Frenkel. However, the language in the introduction and section 2 of that paper does not inspire complete confidence. The key sentence is on page 7:

According to Gaitsgory [Gai2], a Ran space version of this statement can be used to prove the existence of the quantum geometric Langlands correspondence $\mathbb{L}_\kappa$ discussed in the Introduction.

and [Gai2] refers to a video of a 2018 lecture rather than a published paper.

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  • $\begingroup$ Any comments why FF is similar to Stoynovsky? Would be helpful... $\endgroup$
    – Alexander Chervov
    Commented Dec 13, 2018 at 14:15
  • $\begingroup$ I think you are misinterpreting the sentence in Arakawa-Frenkel. The result proved there leads to a construction of the "master chiral algebra", which indeed produces a functor from the LHS to the RHS. However, I don't think it is clear that this functor is an equivalence (but maybe this is my ignorance speaking). More generally, my understanding is that global quantum geometric Langlands is still a ways away from being a theorem. $\endgroup$
    – dhy
    Commented Dec 13, 2018 at 17:38
  • $\begingroup$ @dhy That is entirely possible. I was rather surprised by Arakawa's announcement at a conference earlier this year, but I may have misinterpreted the scope of his claims. $\endgroup$
    – S. Carnahan ♦
    Commented Dec 14, 2018 at 5:52

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