The Herbrand quotient of a lattice in a representation

Tate's lattice lemma. Let G be a finite cyclic group and V a representation of G over a nontrivial commutative ℚ-algebra K, classically K = ℝ. If two integral representations M and N, free of finite rank over ℤ, are both G-stable lattices in V, that is, if G-equivariant maps M → V and N → V exhibit V as the base change of M and of N to K, then M and N have the same Herbrand quotient.

The Herbrand quotient of a lattice therefore depends only on the representation it spans. This is how the Herbrand quotient of a unit lattice is computed: Dirichlet's logarithmic embedding makes the S-units, together with a copy of ℤ, a lattice in the same real representation as the permutation lattice on the places in S, whose Herbrand quotient is computed in TauCeti.RepresentationTheory.Homological.TateCohomology.Permutation.Basic.

The identity of V, written in the two bases coming from M and from N, is a nonsingular matrix over K intertwining the integer matrices of the actions on M and on N. By Matrix.exists_det_ne_zero_forall_mul_eq_mul_of_algebraMap it may be replaced by a nonsingular rational matrix with the same property, that is, by an equivalence between the rationalizations ℚ ⊗[ℤ] M and ℚ ⊗[ℤ] N. Clearing denominators then gives an injective G-equivariant map M → N with finite cokernel (Representation.Equiv.exists_injective_finite_quotient_range), which preserves the Herbrand quotient by TauCeti.TateCohomology.herbrandQuotient_eq_of_mono_of_finite_cokernel.

Main results

• TauCeti.TateCohomology.herbrandQuotient_eq_of_isBaseChange: two G-stable lattices in one representation over a ℚ-algebra have the same Herbrand quotient.

References

• J. S. Milne, Class Field Theory, Chapter VII, the Herbrand quotient of the unit group.

theorem TauCeti.TateCohomology.herbrandQuotient_eq_of_isBaseChange {G : Type} [Group G] [Fintype G] [IsCyclic G] {K : Type u_1} {V : Type u_2} [CommRing K] [Nontrivial K] [Algebra ℚ K] [AddCommGroup V] [Module K V] (ρ : Representation K G V) {M N : Rep.{0, 0, 0} ℤ G} [Module.Free ℤ ↑M] [Module.Finite ℤ ↑M] [Module.Free ℤ ↑N] {i : ↑M →ₗ[ℤ] V} {j : ↑N →ₗ[ℤ] V} (hi : IsBaseChange K i) (hj : IsBaseChange K j) (hiG : ∀ (g : G) (x : ↑M), i ((M.ρ g) x) = (ρ g) (i x)) (hjG : ∀ (g : G) (y : ↑N), j ((N.ρ g) y) = (ρ g) (j y)) : herbrandQuotient M = herbrandQuotient N

Tate's lattice lemma. Let G be a finite cyclic group, K a nontrivial commutative ℚ-algebra (classically ℝ) and ρ a representation of G on a K-module V. If M and N are integral representations of G, free over ℤ with M of finite rank, and G-equivariant maps i : M → V and j : N → V exhibit V as the base change of M and of N to K, then M and N have the same Herbrand quotient. The rank of N is then finite as well, being that of V.