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TauCeti.LinearAlgebra.Matrix.GeneralLinearGroup.Symplectic.Perfect

Perfectness of the standard symplectic group #

The standard symplectic group Sp₂ₘ over a field is perfect except in three small cases: Sp₂(𝔽₂) ≅ S₃, Sp₂(𝔽₃) = SL₂(𝔽₃) and Sp₄(𝔽₂) ≅ S₆. In each case below, every root element x_α(c) is shown to lie in the commutator subgroup; the root subgroups generate the symplectic group over a field, so the commutator subgroup is everything.

Together these give the classical statement for rank at least two: Sp₂ₘ(K) is perfect unless m = 2 and K = 𝔽₂. In particular every finite symplectic group Sp₂ᵣ(𝔽_q) with r ≥ 3 is perfect, including those over 𝔽₂ and 𝔽₃.

Main results #

References #

The torus argument follows the special-linear perfectness proof Matrix.SpecialLinearGroup.commutator_eq_top, with the torus conjugation formula TauCeti.GLSymplecticFin.diagonal_mul_rootSubgroup_mul_inv in place of diagonal conjugation of transvections and TauCeti.GLSymplecticFin.eq_top_of_root_subgroups for generation. The commutator-relation arguments use the relations of TauCeti.LinearAlgebra.Matrix.GeneralLinearGroup.Symplectic.ChevalleyRelations and the root generation TauCeti.GLSymplecticFin.RootSubgroupIndex.hom_apply_mem_of_difference_of_long.

The commutator of a diagonal torus element with a root element is the root element whose parameter is scaled by the root character minus one.

theorem TauCeti.GLSymplecticFin.RootSubgroupIndex.exists_character_ne_one {m : ℕ} {R : Type u} [CommMonoid R] (root : RootSubgroupIndex m) {u : Rˣ} (hu : u * u ≠ 1) :
∃ (t : Fin m → Rˣ), root.character t ≠ 1

Root characters are nontrivial. If the coefficient monoid has a unit whose square is not one, then every root character takes a value different from one on the diagonal torus.

theorem TauCeti.GLSymplecticFin.rootSubgroup_mem_commutator {m : ℕ} {K : Type u} [Field K] {a : K} (ha : a ≠ 0) (hasq : a ^ 2 ≠ 1) (root : RootSubgroupIndex m) (c : Multiplicative K) :

Every root element of the standard symplectic group is in the commutator subgroup if the field contains a nonzero element whose square is not one.

theorem TauCeti.GLSymplecticFin.commutator_eq_top {m : ℕ} {K : Type u} [Field K] {a : K} (ha : a ≠ 0) (hasq : a ^ 2 ≠ 1) :

The standard symplectic group is perfect over a field with a nonzero element whose square is not one, in every rank, including rank zero.

theorem TauCeti.GLSymplecticFin.differenceShortRootUnit_mem_commutator {m : ℕ} {R : Type u} [CommRing R] {i j k : Fin m} (hij : i ≠ j) (hik : i ≠ k) (hkj : k ≠ j) (c : R) :

Difference-root elements are commutators when a third index is available: for pairwise distinct i, j, k, the structure-constant-one Chevalley relation gives x_{eᵢ-eⱼ}(c) = ⁅x_{eᵢ-eₖ}(c), x_{eₖ-eⱼ}(1)⁆.

theorem TauCeti.GLSymplecticFin.rootSubgroup_mem_commutator_of_difference {m : ℕ} {R : Type u} [CommRing R] (hm : 2 ≤ m) (hdiff : ∀ {i j : Fin m} (hij : i ≠ j) (c : R), differenceShortRootUnit hij c ∈ commutator ↥(GLSymplecticFin m R)) (root : RootSubgroupIndex m) (c : Multiplicative R) :

In rank at least two, the difference roots control the commutator subgroup. If the commutator subgroup of the standard symplectic group contains every difference-root element, then it contains every root element: the long-root strings make the long-root elements commutators, and the difference and long roots generate every root subgroup.

In rank at least three, every root element of the standard symplectic group is in the commutator subgroup, over every commutative ring.

In rank at least two, if 2 is a unit, every root element of the standard symplectic group is in the commutator subgroup, over every commutative ring.

The standard symplectic group is perfect in rank at least three, over every field, including 𝔽₂ and 𝔽₃.

The standard symplectic group is perfect in rank at least two over a field of characteristic other than two, including 𝔽₃.

theorem TauCeti.GLSymplecticFin.commutator_eq_top_of_two_le {m : ℕ} {K : Type u} [Field K] (hm : 2 ≤ m) (hK : 3 ≤ m ∨ ∃ (a : K), a ≠ 0 ∧ a ≠ 1) :

The standard symplectic group Sp₂ₘ(K) with m ≥ 2 is perfect unless m = 2 and K = 𝔽₂. The second hypothesis says that the rank is at least three or the field has an element other than 0 and 1. The exception is genuine: Sp₄(𝔽₂) ≅ S₆ is not perfect.