The Margulis–Platonov conjecture over number fields

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The Margulis–Platonov conjecture over number fields

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Clase 43 de 100

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Recursos

Take an absolutely almost simple, simply connected group G over a number field. Find its compact anisotropic places, read off what its normal subgroups must be, and see how the proof cuts the problem down to one finite obstruction that it then kills.

Predict the answer from the compact places

Let A be the finite set of nonarchimedean places v where G has local rank zero. Each G(k_v) for v in A is compact. The theorem says every noncentral normal subgroup N of G(k) is N = δ_A⁻¹(W) for some open normal subgroup W of H_A = ∏_{v∈A} G(k_v) [0:57].

Worked case [1:14]: G = SL₁(D), where D is a quaternion algebra over ℚ ramified at 2 and 3. Then A = {2, 3}, and N is cut out by an open normal subgroup of SL₁(D₂) × SL₁(D₃).
Types D4 and E6 [1:50]: a group that is anisotropic at a nonarchimedean place has type A, so A = ∅ and N = G(k). That is equality, which is stronger than "dense" or "finite index".

Reduce to the finite defect

N has finite index, so pick an exponent e that kills G(k)/N, and let R = ⟨g^e⟩. The closure of δ(R) is an open subgroup P₀, and V₀ = δ⁻¹(P₀). Approximation only sees V₀, so the gap is the finite group V = V₀/R [2:39]. If V = 1, then G(k)/R ≅ H_A/P₀, and that gives N as a preimage.

Kill cosets with three tools

Power-factor approximation [3:03]: Φ(a) = x∏ k_j η(a_j)^e k_j⁻¹ never leaves the coset xR. Because η(la) = η(a), it can be pushed into prescribed local open sets.
Norm-torus Hasse principle [3:41]: Ш² is detected on cyclic subgroups and vanishes for S_m × C₂. So a torsor with local points everywhere has a rational point.
Restriction lemma [4:05]: a piece that lies in a subgroup where (MP) is known, and is an e-th power at that subgroup's compact places, is killed.

Outer type A, D4 and E6 combine these tools in different ways [4:22]. Check from the class: in E6, write g = d·h with d in a simply connected D4 subgroup. Then q(d) = 1 with no local condition at all, because A(D4) = ∅ [5:06].

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