64 lines
1.9 KiB
Text
64 lines
1.9 KiB
Text
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/-
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Copyright (c) 2024 Joseph Tooby-Smith. All rights reserved.
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Released under Apache 2.0 license.
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Authors: Joseph Tooby-Smith
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-/
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import HepLean.AnomalyCancellation.PureU1.Permutations
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/-!
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# The Pure U(1) case with 3 fermion
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We show that S is a solution only if one of its charges is zero.
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We define a surjective map from `LinSols` with a charge equal to zero to `Sols`.
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-/
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universe v u
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open Nat
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open Finset
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namespace PureU1
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variable {n : ℕ}
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namespace Three
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lemma cube_for_linSol' (S : (PureU1 3).LinSols) :
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3 * S.val (0 : Fin 3) * S.val (1 : Fin 3) * S.val (2 : Fin 3) = 0 ↔
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(PureU1 3).cubicACC S.val = 0 := by
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have hL := pureU1_linear S
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simp at hL
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rw [Fin.sum_univ_three] at hL
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change _ ↔ accCube _ _ = _
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rw [accCube_explicit, Fin.sum_univ_three]
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rw [show S.val (0 : Fin 3) = - (S.val (1 : Fin 3) + S.val (2 : Fin 3)) by
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linear_combination hL]
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ring_nf
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lemma cube_for_linSol (S : (PureU1 3).LinSols) :
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(S.val (0 : Fin 3) = 0 ∨ S.val (1 : Fin 3) = 0 ∨ S.val (2 : Fin 3) = 0) ↔
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(PureU1 3).cubicACC S.val = 0 := by
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rw [← cube_for_linSol']
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simp only [Fin.isValue, _root_.mul_eq_zero, OfNat.ofNat_ne_zero, false_or]
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rw [@or_assoc]
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lemma three_sol_zero (S : (PureU1 3).Sols) : S.val (0 : Fin 3) = 0 ∨ S.val (1 : Fin 3) = 0
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∨ S.val (2 : Fin 3) = 0 := (cube_for_linSol S.1.1).mpr S.cubicSol
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/-- Given a `LinSol` with a charge equal to zero a `Sol`.-/
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def solOfLinear (S : (PureU1 3).LinSols)
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(hS : S.val (0 : Fin 3) = 0 ∨ S.val (1 : Fin 3) = 0 ∨ S.val (2 : Fin 3) = 0) :
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(PureU1 3).Sols :=
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⟨⟨S, by intro i; simp at i; exact Fin.elim0 i⟩,
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(cube_for_linSol S).mp hS⟩
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theorem solOfLinear_surjects (S : (PureU1 3).Sols) :
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∃ (T : (PureU1 3).LinSols) (hT : T.val (0 : Fin 3) = 0 ∨ T.val (1 : Fin 3) = 0
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∨ T.val (2 : Fin 3) = 0), solOfLinear T hT = S := by
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use S.1.1
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use (three_sol_zero S)
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rfl
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end Three
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end PureU1
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