Tate cohomology and the Herbrand quotient
TheoremPeriodicity of Tate cohomology for finite cyclic groups#✓fully proved
Let \(R\) be a commutative ring, \(G\) a finite cyclic group and \(A\) a representation of \(G\) over \(R\). Then for every integer \(n\), the Tate cohomology groups \(\hat{H}^n(G, A)\) and \(\hat{H}^{n+2}(G, A)\) are isomorphic as \(R\)-modules.
theorem nonempty_periodicityIso {R G : Type u} [CommRing R] [Group G]
[Fintype G] [IsCyclic G] (A : Rep R G) (n : ℤ) :
Nonempty (tateCohomology A n ≅ tateCohomology A (n + 2))
DefinitionCardinality of Tate cohomology#✓fully proved
Let \(G\) be a finite group, \(R\) a commutative ring, \(M\) an \(R\)-linear representation of \(G\) and \(n\) an integer. This is the cardinality of the Tate cohomology group \(\hat H^n(G, M)\), as a cardinal number (possibly infinite).
noncomputable def tateCohomologyCard (M : Rep R G) (n : ℤ) : Cardinal :=
Cardinal.mk ↥(tateCohomology M n)
DefinitionCardinality of Tate homology#✓fully proved
Let \(G\) be a finite group, \(R\) a commutative ring and \(M\) an \(R\)-linear representation of \(G\). For an integer \(n\), the cardinality \(h_n(M)\) of the degree-\(n\) Tate homology of \(M\) is defined as the cardinal number of the Tate cohomology group \(\hat H^{-n-1}(G, M)\), following the convention \(\hat H_n = \hat H^{-n-1}\).
noncomputable def tateHomologyCard (M : Rep R G) (n : ℤ) : Cardinal :=
tateCohomologyCard M (-n - 1)
noncomputable def tateCohomologyCard (M : Rep R G) (n : ℤ) : Cardinal :=
Cardinal.mk ↥(tateCohomology M n)
MathlibExt/RepresentationTheory/Homological/TateCohomology/HerbrandQuotient.lean:37
Uses 1 · Used by 1
TheoremCardinality of Tate homology in degree zero#✓fully proved
Let \(G\) be a finite group, \(R\) a commutative ring and \(M\) an \(R\)-linear representation of \(G\). The cardinality of the degree-\(0\) Tate homology of \(M\) equals the cardinality of the Tate cohomology group \(\hat H^{-1}(G, M)\).
The cardinality of Tate homology in degree \(n\) is defined as that of \(\hat H^{-n-1}(G, M)\), so this holds by definition.
@[simp] theorem tateHomologyCard_zero (M : Rep R G) :
tateHomologyCard M 0 = tateCohomologyCard M (-1)
noncomputable def tateHomologyCard (M : Rep R G) (n : ℤ) : Cardinal :=
tateCohomologyCard M (-n - 1)
MathlibExt/RepresentationTheory/Homological/TateCohomology/HerbrandQuotient.lean:41
noncomputable def tateCohomologyCard (M : Rep R G) (n : ℤ) : Cardinal :=
Cardinal.mk ↥(tateCohomology M n)
MathlibExt/RepresentationTheory/Homological/TateCohomology/HerbrandQuotient.lean:37
Uses 2
TheoremCardinality of finite Tate cohomology as a natural number#✓fully proved
Let \(G\) be a finite group, \(R\) a commutative ring, \(M\) an \(R\)-linear representation of \(G\) and \(n\) an integer such that the Tate cohomology group \(\hat H^n(G, M)\) is finite. Then the cardinality of \(\hat H^n(G, M)\) equals its number of elements, a natural number viewed as a cardinal number.
theorem tateCohomologyCard_eq_natCast (M : Rep R G) (n : ℤ)
[Finite ↥(tateCohomology M n)] :
tateCohomologyCard M n = Nat.card ↥(tateCohomology M n)
noncomputable def tateCohomologyCard (M : Rep R G) (n : ℤ) : Cardinal :=
Cardinal.mk ↥(tateCohomology M n)
MathlibExt/RepresentationTheory/Homological/TateCohomology/HerbrandQuotient.lean:37
Uses 1
DefinitionFiniteness hypothesis for the Herbrand quotient#✓fully proved
Let \(G\) be a finite group, \(R\) a commutative ring and \(M\) an \(R\)-linear representation of \(G\). The representation \(M\) satisfies the Herbrand finiteness hypothesis when its Tate cohomology groups \(\hat H^0(G, M)\) and \(\hat H^{-1}(G, M)\) are both finite.
def HerbrandFinite (M : Rep R G) : Prop :=
Finite ↥(tateCohomology M 0) ∧ Finite ↥(tateCohomology M (-1))
Used by 6
DefinitionHerbrand quotient#✓fully proved
Let \(G\) be a finite group, \(R\) a commutative ring and \(M\) an \(R\)-linear representation of \(G\) whose Tate cohomology groups \(\hat H^0(G, M)\) and \(\hat H^{-1}(G, M)\) are finite. The Herbrand quotient of \(M\) is the rational number \(h(M) = |\hat H^0(G, M)| / |\hat H^{-1}(G, M)|\), where \(|X|\) is the number of elements of \(X\).
noncomputable def herbrandQuotient (M : Rep R G) (h : HerbrandFinite M) : ℚ :=
haveI := h.1
haveI := h.2
(Nat.card ↥(tateCohomology M 0) : ℚ) / Nat.card ↥(tateCohomology M (-1))
def HerbrandFinite (M : Rep R G) : Prop :=
Finite ↥(tateCohomology M 0) ∧ Finite ↥(tateCohomology M (-1))
MathlibExt/RepresentationTheory/Homological/TateCohomology/HerbrandQuotient.lean:55
TheoremHerbrand quotient as a ratio of orders#✓fully proved
Let \(G\) be a finite group, \(R\) a commutative ring and \(M\) an \(R\)-linear representation of \(G\) whose Tate cohomology groups \(\hat H^0(G, M)\) and \(\hat H^{-1}(G, M)\) are finite. Then the Herbrand quotient of \(M\) equals the rational number \(|\hat H^0(G, M)| / |\hat H^{-1}(G, M)|\), where \(|X|\) is the number of elements of \(X\).
theorem herbrandQuotient_eq (M : Rep R G) (h : HerbrandFinite M) :
herbrandQuotient M h = (Nat.card ↥(tateCohomology M 0) : ℚ) /
Nat.card ↥(tateCohomology M (-1))
def HerbrandFinite (M : Rep R G) : Prop :=
Finite ↥(tateCohomology M 0) ∧ Finite ↥(tateCohomology M (-1))
MathlibExt/RepresentationTheory/Homological/TateCohomology/HerbrandQuotient.lean:55
noncomputable def herbrandQuotient (M : Rep R G) (h : HerbrandFinite M) : ℚ :=
haveI := h.1
haveI := h.2
(Nat.card ↥(tateCohomology M 0) : ℚ) / Nat.card ↥(tateCohomology M (-1))
MathlibExt/RepresentationTheory/Homological/TateCohomology/HerbrandQuotient.lean:59
Uses 2 · Used by 2
TheoremThe denominator of the Herbrand quotient is positive#✓fully proved
Let \(G\) be a finite group, \(R\) a commutative ring and \(M\) an \(R\)-linear representation of \(G\) whose Tate cohomology groups \(\hat H^0(G, M)\) and \(\hat H^{-1}(G, M)\) are finite. Then the number of elements of \(\hat H^{-1}(G, M)\) is positive.
theorem herbrandQuotient_den_pos (M : Rep R G) (h : HerbrandFinite M) :
0 < Nat.card ↥(tateCohomology M (-1))
def HerbrandFinite (M : Rep R G) : Prop :=
Finite ↥(tateCohomology M 0) ∧ Finite ↥(tateCohomology M (-1))
MathlibExt/RepresentationTheory/Homological/TateCohomology/HerbrandQuotient.lean:55
Uses 1 · Used by 1
TheoremThe denominator of the Herbrand quotient is nonzero#✓fully proved
Let \(G\) be a finite group, \(R\) a commutative ring and \(M\) an \(R\)-linear representation of \(G\) whose Tate cohomology groups \(\hat H^0(G, M)\) and \(\hat H^{-1}(G, M)\) are finite. Then the number of elements of \(\hat H^{-1}(G, M)\) is nonzero.
theorem herbrandQuotient_den_ne_zero (M : Rep R G) (h : HerbrandFinite M) :
Nat.card ↥(tateCohomology M (-1)) ≠ 0
def HerbrandFinite (M : Rep R G) : Prop :=
Finite ↥(tateCohomology M 0) ∧ Finite ↥(tateCohomology M (-1))
MathlibExt/RepresentationTheory/Homological/TateCohomology/HerbrandQuotient.lean:55
Uses 2
TheoremPositivity of the Herbrand quotient#✓fully proved
Let \(G\) be a finite group, \(R\) a commutative ring and \(M\) an \(R\)-linear representation of \(G\) whose Tate cohomology groups \(\hat H^0(G, M)\) and \(\hat H^{-1}(G, M)\) are finite. Then the Herbrand quotient of \(M\) is strictly positive.
theorem herbrandQuotient_pos (M : Rep R G) (h : HerbrandFinite M) :
0 < herbrandQuotient M h
def HerbrandFinite (M : Rep R G) : Prop :=
Finite ↥(tateCohomology M 0) ∧ Finite ↥(tateCohomology M (-1))
MathlibExt/RepresentationTheory/Homological/TateCohomology/HerbrandQuotient.lean:55
noncomputable def herbrandQuotient (M : Rep R G) (h : HerbrandFinite M) : ℚ :=
haveI := h.1
haveI := h.2
(Nat.card ↥(tateCohomology M 0) : ℚ) / Nat.card ↥(tateCohomology M (-1))
MathlibExt/RepresentationTheory/Homological/TateCohomology/HerbrandQuotient.lean:59
Uses 3
TheoremHerbrand quotients agree when Tate cohomology in degrees \(0\) and \(-1\) is in bijection#✓fully proved
Let \(G\) be a finite group, \(R\) a commutative ring and \(M\), \(N\) two \(R\)-linear representations of \(G\) whose Tate cohomology groups \(\hat H^0(G, M)\), \(\hat H^{-1}(G, M)\), \(\hat H^0(G, N)\) and \(\hat H^{-1}(G, N)\) are finite. If there are bijections between the underlying sets of \(\hat H^0(G, M)\) and \(\hat H^0(G, N)\) and between those of \(\hat H^{-1}(G, M)\) and \(\hat H^{-1}(G, N)\), then the Herbrand quotients of \(M\) and \(N\) are equal.
theorem herbrandQuotient_congr {M N : Rep R G} {hM : HerbrandFinite M}
{hN : HerbrandFinite N}
(e₀ : ↥(tateCohomology M 0) ≃ ↥(tateCohomology N 0))
(e₁ : ↥(tateCohomology M (-1)) ≃ ↥(tateCohomology N (-1))) :
herbrandQuotient M hM = herbrandQuotient N hN
def HerbrandFinite (M : Rep R G) : Prop :=
Finite ↥(tateCohomology M 0) ∧ Finite ↥(tateCohomology M (-1))
MathlibExt/RepresentationTheory/Homological/TateCohomology/HerbrandQuotient.lean:55
noncomputable def herbrandQuotient (M : Rep R G) (h : HerbrandFinite M) : ℚ :=
haveI := h.1
haveI := h.2
(Nat.card ↥(tateCohomology M 0) : ℚ) / Nat.card ↥(tateCohomology M (-1))
MathlibExt/RepresentationTheory/Homological/TateCohomology/HerbrandQuotient.lean:59
Uses 3
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