Basis-flip criterion for vanishing coinvariants
ProvedLegacyAlgebra.basisFlipCoinvariantsLet ρ be a linear representation of a group G on a vector space V over a field k in which 2 is nonzero. Suppose V has a basis b such that, for every basis vector b_i, some element of G sends b_i to −b_i. Then the span of all differences ρ(g)v − v is V, so the coinvariant space is zero. The basis and group may be infinite.
import Mathlib set_option autoImplicit false
theorem LegacyAlgebra.basisFlipCoinvariants
(k G V ι : Type*) [Field k] [Group G]
[AddCommGroup V] [Module k V]
(ρ : Representation k G V) (b : Module.Basis ι k V)
(hTwo : (2 : k) ≠ 0)
(hFlip : ∀ i : ι, ∃ g : G, ρ g (b i) = -(b i)) :
Representation.Coinvariants.ker ρ = ⊤ := by
sorry
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What the Lean code literally says, in plain math · Codex independent blind auditor; exact model identifier unavailable
For every field , every group , every additive commutative group equipped with a -module structure, every type , every representation , and every -basis of , assume that in and that for each there exists a group element such that . Then the -linear span of all vectors , for and , equals the whole space . The group element that negates a basis vector may depend on its index. No finiteness is required of , , or the dimension of ; an empty basis is allowed, in which case the basis condition forces to be the zero space and the negation hypothesis is vacuous.
Confirmed by the mission captain (proposal self-audit).