What os the meaning of a reverse Langlands transform
to which Drinfeld seems to refer?
1 Answer
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My guess is that, traditionally, the Geometric Langlands program seems to be looking for a functor from D_{coh}(Loc,O)\to D_{coh}(^L Bun,D), that is, from the derived category on the space of local systems to the derived category of D-modules on the space of bundles for the Langlands dual. So, by reverse Langlands transformation, he might be asking for a functor in the other direction, to try to establish the equivalence.