Restricted geometric Langlands in positive characteristic

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Restricted geometric Langlands in positive characteristic

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Recursos

Take a field, a prime and a reductive group, decide whether the full-support theorem covers them, and explain why no spectral component can then be missing from the restricted Langlands equivalence.

The gap and the theorem

Gaitsgory–Raskin's restricted functor in characteristic p is an equivalence onto Y′, an open-and-closed union of components of the restricted spectral stack Y. A parameter on a component outside Y′ gets no automorphic category, so nothing hands it a Hecke eigensheaf [0:08]. Theorem 1.4 (full support) says Y′ = Y [0:55] in two settings. Both assume the common Hypothesis 1.1:

Regime A: k is the algebraic closure of a finite field, and the curve and a split form of G descend to some F_q. There is no Weyl-group condition.
Regime B: k is any algebraically closed field, p is very good for G and p ∤ |W_G|.

Under either regime, every parameter, semisimple or not, gets a nonzero Hecke eigenobject M_σ.

Check yourself [1:44]: for E8 with p = 7, p is very good, but 7 divides |W(E8)|. Over the closure of F7(t), with no finite-field model, neither regime applies. Give the same curve and group a model over F49 and regime A covers them.

The proof chain

Specialization [2:24]: lift to the Witt vectors. A missing component C acts by zero downstairs. That gives a nonzero compact sheaf F on the generic fiber with ΨF = 0, a "ghost" sheaf.
Test function [3:09]: take a covector outside the nilpotent cone and a function f with that differential. A Levi reduction builds a transverse Hecke modification. This is the only step where regimes A and B enter.
Two Hecke correspondences give the quadratic local model Σ u_i v_i + f = 0, and Ψ vanishes at its vertex.
Projective quadric comparison [3:55]: the triangle of hyperplane classes, the quadric family and a line on f = 0 forces Ψ(F|D) = 0. Incidence extends this to slices whose cotangent planes avoid the nilpotent cone.
Finite slice [4:36]: cut the support of F down to an isolated point u. There Ψ is a finite direct sum of stalks with one nonzero summand, but the slice test says it is zero. That is a contradiction, so Y′ = Y.

Decision rule [5:12]: when an equivalence reaches only some components, check the regime first. Then ask what a missing component would leave behind, and find the test it cannot pass.

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