2 Group Cohomology
2.1 Introduction
The aim of this chapter is to define the basic concepts of group cohomology and to prove the following theorem of Tate:
Let \(M\) be a representation of a finite group \(G\) over the ring \({\mathbb Z}\), and suppose that for all subgroups \(S\) of \(G\) the following two conditions hold:
\(H^1(S,M) \cong 0\),
\(H^2(S,M) \cong {\mathbb Z}/ |S| \cdot {\mathbb Z}\).
Then there is an isomorphism
If \(\sigma \) is a 2-cocycle representing a generator of \(H^2(G,M)\) then the inverse of the isomorphism is given by
In later chapters we shall see that the hypotheses of the theorem are satisfied in the following cases.
If \(l/k\) is a finite Galois extension of local fields and \(G = \mathrm{Gal}(l/k)\) then we may regard the group \(l^\times \) as a representation of \(G\). This representation satisfies the hypotheses above so we have an isomorphism:
\[ \mathrm{reciprocity}: k^\times / N(l^\times ) \cong \mathrm{Gal}(l/k)^{\mathrm{ab}}. \]If \(l/k\) is a finite Galois extension of number fields and \(G = \mathrm{Gal}(l/k)\) then we may regard the idele class group \(\mathrm{Cl}_l = {\mathbb A}^\times / l^\times \) as a representation of \(G\). This representation satisfies the hypotheses above so we have an isomorphism:
\[ \mathrm{reciprocity}: \mathrm{Cl}_k / N(Cl_l) \cong \mathrm{Gal}(l/k)^{\mathrm{ab}} . \]
2.2 Group cohomology and homology
Let \(G\) be a group and \(R\) a commutative ring. By a representation of \(G\) over \(R\), we shall mean an \(R\)-module \(M\) with an action of \(G\) by \(R\)-linear maps. We shall use the notation \(g \bullet m\) for the action of an element \(g \in G\) on an element \(m \in M\). We’ll call \(M\) a trivial representation if for all \(g \in G\) and all \(m \in M\) we have \(g \bullet m = m\). We’ll write \(\mathbf{Mod}(R)\) for the category of \(R\)-modules and \(\mathbf{Rep}(R,G)\) for the category of such representations. In Mathlib these are called ModuleCat R and Rep R G.
If \(S\) is a subgroup of \(G\) and \(M\) is a representation of \(G\) then \(M \downarrow S\) will mean the same \(R\)-module \(M\), but regarded only as a representation of \(S\). We shall write \(M^S\) for the \(R\)-submodule of \(S\)-invariant vectors in \(M\). If the subgroup \(S\) is normal then \(M^S\) is a representation of the quotient group \(G/S\).
The map \(M \mapsto M \downarrow S\) defines a functor \(\mathbf{res}: \mathbf{Rep}(R,G) \to \mathbf{Rep}(R,S)\). If \(S\) is a normal subgroup of \(G\) then then map \(M \mapsto M^S\) defines a functor \(\mathbf{invar}: \mathbf{Rep}(R,G) \to \mathbf{Rep}(R, G/S)\). These functors are called restriction and inflation respectively. These are defined in Mathlib as Rep.res and Rep.quotientToInvariantsFunctor.
Given a representation \(M\) of \(G\), there is a cochain complex of \(R\)-modules
where each term \(C^n(G,M)\) is the space of functions \(G^n \to M\) with some appropriately defined linear maps \(d^i\) linking them. The zeroth module \(C^0(G,M)\) should be interpreted as just \(M\). The first few of these linear maps \(d^i\) are
The cochain complex \(C^\bullet (G,M)\) is functorial in \(M\) and is defined in Mathlib as groupCohomology.cochainsFunctor. The cohomology groups of \(C^\bullet (G,M)\) are called the cohomology groups of the \(G\)-module \(M\), and are written \(H^n(G,M)\). These are defined in Mathlib as groupCohomology M n, or (groupCohomology.functor n).obj M.
For example \(H^0(G,M)\) is the kernel of the map \(d^0 : M \to (G \to M)\). Since \((d^0m)(g) = g \bullet m - m\), an element \(m\) is in this kernel if \(m\) is in \(M^G\), so we have \(H^0(G,M) \cong M^G\).
Suppose \(M\) is a trivial representation of \(G\). Then the map \(d^0\) is zero, so \(H^1(G,M)\) is the kernel of the map \((d^1f)(x,y) = f(x) + f(y) - f(xy)\). A function \(f: G \to M\) is in this kernel if it is a group homomorphism, so we have \(H^1(G,M) \cong \mathrm{Hom}(G,M)\).
If \(G\) is the trivial group then for all \(n{\gt}0\), \(H^n(G,M)\cong 0\).
Each of the modules \(C^n(G,M)\) may be identified with \(M\), and the coboundary maps reduce to alternating sums of the form \(d^n(m) = \sum _{i=0}^{n+1} (-1)^i m\). Hence \(d^n\) is the identity map for odd \(n\), and is the zero map for even \(m\). It follows that the image of \(d^n\) is equal to the kernel of \(d^{n+1}\).
The functions \(d^i : (G^i \to M) \to (G^{i+1} \to M)\) do not depend on the ring \(R\). Consequently, the additive groups \(H^n(G,M)\) do not depend on \(R\). More precisely there are forgetful functors \(F_1 : \mathbf{Rep}(R,G) \to \mathbf{Rep}({\mathbb Z},G)\) and \(F_2 : \mathbf{Mod}(R) \to \mathbf{Mod}({\mathbb Z})\), and there are isomorphisms of functors
That said, it is very little extra work to prove the cohomological results discussed here in the generality of \(\mathbf{Rep}(R,G)\) rather than \(\mathbf{Rep}({\mathbb Z},G)\).
At this point it’s worth stressing one particular aspect of this theory. Many of the proofs involve showing that certain diagrams commute. When writing proofs on paper, some mathematicians don’t give too much thought to the reasons why a particular diagram commutes but when formalizing theorems we need to be more careful than this. For that reason it is important to define all of our objects in as functorial a way as possible. In the example from the previous remark, saying that there is an isomorphism \(H^n(G,F_1(M)) \cong F_2(H^n(G,M))\) is not as good as saying that there is an isomorphism of functors \(H^n \circ F_1 \cong F_2 \circ H^n\). The isomorphism of functors implies that for every map \(f : A \to B\) in \(\mathbf{Rep}(R,G)\) we have a commuting square in \(\mathbf{Mod}({\mathbb Z})\), in which the vertical arrows are the isomorphisms of the objects
This commuting square might be crucial in some other proof. This is also the reason why we work with the categories Rep R G and ModuleCat R, rather than with Representation R G M and Module R M which could be more familiar.
If \(S\) is a subgroup of \(G\), then we write \(H^n(S,M)\) for the cohomology groups of the restricted representation \(M \downarrow S\). If \(f : G^n \to M\) is an element of \(C^n(G,M)\), then we may restrict \(f\) to a function \(S^n \to M\). Restricting functions in this way defines a map of cochain complexes \(C^\bullet (G,M) \to C^\bullet (S,M)\), and hence a map of cohomology groups
This map is called the restriction map, and is a morphism of functors from \(H^n(G,-)\) to \(H^n(S, -) \circ \mathbf{res}\).
If \(S\) is a normal subgroup of \(G\), then we write \(H^n(G/S,M^S)\) for the cohomology groups of the representation \(M^S\) of \(G/S\). If \(f : (G/S)^n \to M^S\) is an element of \(C^n(G/S,M^S)\), then we may “inflate” \(f\) to a function \(G^n \to M\). This inflation process defines a map of cochain complexes \(C^\bullet (G/S,M^S) \to C^\bullet (G,M)\), and hence a map of cohomology groups, called the inflation map:
More precisely, the inflation map is a morphism of functors from \(H^n (G/S, -)\circ \mathbf{invar}\) to \(H^n(G,-)\).
The functor taking \(M\) to \(C^\bullet (G,M)\) is exact. I.e. if \(0 \to A \to B \to C \to 0\) is a short exact sequence in \(\mathbf{Rep}(R,G)\). Then the corresponding sequence of cochain complexes is exact:
This is already in Mathlib.
As a consequence of this, we have the following (which is in Mathlib):
Given a short exact sequence \(0 \to A \stackrel{f}\to B \stackrel{g}\to C \to 0\) in \(\mathbf{Rep}(R,G)\), the corresponding sequence of cochain complexes is exact: \(0 \to C^n(G,A) \to C^n(G,B) \to C^n(G,C) \to 0\). This implies that there exist “connecting homomorphisms” \(\delta : H^n(G,C) \to H^{n+1}(G,A)\), such that the following is a long exact sequence:
Let \(S\) be a subgroup of \(G\) and suppose we have a short exact sequence \(0 \to A \to B \to C \to 0\) in \(\mathbf{Rep}(R,G)\). Then the sequence the sequence \(0 \to A \downarrow S \to B \downarrow S \to C \downarrow S \to 0\) is exact in \(\mathbf{Rep}(R,S)\). The following diagram commutes, where the rows are the long exact sequences for \(0 \to A \to B \to C \to 0\) and for its restriction to \(S\), and the vertical maps are restriction.
Suppose now that \(S\) is a normal subgroup of \(G\). Then we have for every map \(f : A \to B\) in \(\mathbf{Rep}(R,G)\) a commuting square in which the vertical maps are inflation.
If \(0 \to A \to B \to C \to 0\) is exact in \(\mathbf{Rep}(R,G)\) and its inflation \(0 \to A^S \to B^S \to C^S \to 0\) is also exact in \(Rep(R,G/S)\), then we have a commutative diagram in which the rows are the corresponding long exact sequences and the vertical maps are inflation:
Most of the commuting squares have already been proved if inflation and restriction are defined as morphisms of functors. The remaining two squares are:
Both of these can be deduced from the following statement in Mathlib:
HomologicalComplex.HomologySequence.\(\delta \)_naturality
There is also a chain complex of \(R\)-modules:
whose \(n\)-th term is the space of finitely supported functions \(f : G^n \to _0 M\), with appropriately defined boundary maps \(d_n\). In the literature \(C_n(G,M)\) is often defined as \(R[G]^{\otimes n} \otimes _R M\), to which it is canonically isomorphic. In the case \(n=0\) this is interpreted as meaning \(C_0(G,M) = M\). The homology groups of \(C_n(G,M)\) are called the homology groups of \(M\) and are written \(H_n(G,M)\).
We’ll sometimes write \(\mathrm{single}(g,m)\) for the function with value \(m\) at \(g\) and value zero elsewhere. The \(R\)-module \(C_1(G,M)\) is spanned by the elements \(\mathrm{single}(g,m)\) for \(g \in G\) and \(m \in M\) (Finsupp.single in Mathlib). For such elements we have
We shall write \(I_G M\) for the submodule of \(M\) spanned by elements of the form \(g \bullet m - m\). The quotient \(M / I_GM\) is commonly called the coinvariants of \(M\) and is written \(M_G\); this is the largest quotient module on which \(G\) acts trivially. It follows that \(H_0(G,M) \cong M_G\).
In the case that \(G=1\) is the trivial group then for all \(n {\gt} 0\) we have \(H_n(1,M)\). This follows in a similar way to 5
If \(M\) is a trivial representation of \(G\) then \(H_1(G,M) \cong G^{\mathrm{ab}} \otimes {\mathbb Z}M\). The isomorphism takes a 1-cycle \(\mathrm{single}(x,m)\) to \(xG' \otimes m\) for \(x \in G\) and \(m \in M\). (This result is already in Mathlib).
Since \(M\) is trivial, the map \(d_0 : C_1(G,M) \to C_0(G,M)\) is zero, so every \(1\)-chain is a 1-cycle. The map \(d_1 : C_1(G,M) \to C_0(G,M)\) is given by
Therefore \(H_1(G,M)\) is the quotient of \(G \to _0 M\) by the relations
If we fix an \(m \in M\), then these relations imply that the map \(G \to H_1(G,M)\) defined by \(x \mapsto \mathrm{single}(x,m)\) is a group homomorphism. In particular this map descends to a homomorphism \(G^{\mathrm{ab}} \to H_1(G,M)\). If we fix \(x \in G\) and allow \(m\) to vary, then \(\mathrm{single}(x,m)\) is \({\mathbb Z}\)-linear in \(m\). It is therefore a \({\mathbb Z}\)-bilinear map \(G^\mathrm{ab}\times M \to H_1(G,M)\). Such a map lifts to a linear map \(\Phi : G^{\mathrm{ab}} \otimes _{{\mathbb Z}} M \to H_1(G,M)\) given by
To construct an inverse of \(\Phi \), we note that all of the relations are in the kernel of the linear map \((G\to _0 M) \to (G^\mathrm{ab}\otimes _{\mathbb Z}M)\) defined by \(\mathrm{single}(x,m) \mapsto xG' \otimes m\). Therefore this map descends to a linear map \(\Psi : H_1(G,M) \to (G^\mathrm{ab}\otimes _{\mathbb Z}M)\). It is trivial to check that \(\Phi \) and \(\Psi \) are inverses.
2.3 Tate Cohomology
Throughout this section, the group \(G\) is assumed to be finite. Under this assumption, we show that the homology and cohomology groups may both be regarded as part of a bigger cohomology theory, which is called Tate cohomology.
Let \(G\) be a finite group and \(M\) a representation of \(G\) over a commutative ring \(R\). There is a canonical linear map \(N_G : M \to M\) called the norm, defined by
We shall also regard the norm as a linear map from \(C_0(G,M)\) to \(C^0(G,M)\), both of which may be identified with \(M\).
(We’ll see in the next lemma that \(N_G\) commutes with the action of \(G\), so is a morphism in \(\mathbf{Rep}(R,G)\). However, we shall only regard it as a morphism in \(\mathbf{Mod}(R)\). The reason is that the chain and cochain modules \(C^0(G,M)\) and \(C_0(G,M)\) are regarded as \(R\)-modules rather than representations of \(G\).)
For any \(g \in G\) and \(m \in M\) we have \(g \bullet N_G (m) = N_G (m)\) and \(N_G (g \bullet m) = N_G (m)\).
These equalities follow by reindexing the sums defining \(N_G (m)\):
The composition \(d^0 \circ N_G\) is zero.
The map \(d^0 : M \to (G \to M)\) is given by \((d^0 m)(g) = m - g\bullet m\). Using this formula, we obtain (by Lemma 16) \(d^0 (N_G m) (g) = g \bullet N_G m - N_G m = 0\).
The composition \(N_G \circ d_0\) is zero.
Since the elements \(\mathrm{single}(g,m)\) span \(C_1(G,M)\), it’s sufficient to check that these are all mapped to \(0\). We have by 16
For every map \(f : A \to B\) in \(\mathbf{Rep}(R,G)\) we have a commuting square:
Equivalently, \(N_G\) is an endomorphism of the forgetful functor \(\mathbf{Rep}(R,G) \to \mathbf{Mod}(R)\).
For \(m \in M\) we have
Recall that we have a cochain complex \(C^n(G,M)\), indexed by \(n \in {\mathbb N}\), whose zeroth term may be identified with \(M\). We also have a chain complex \(C_n(G,M)\) whose zeroth term may be identified with \(M\). By 17 and 18, we may glue these together with the map \(N_G : M \to M\) to obtain a cochain complex indexed by \({\mathbb Z}\):
We shall write \(C^n_{\mathrm{Tate}}(G,M)\) for this cochain complex, and we normalize the indices so that for natural numbers \(n\) we have \(C^n_{\mathrm{Tate}}(G,M) = C^n(G,M)\). This implies \(C^{-n-1}_{\mathrm{Tate}}(G,M) = C_n(G,M)\). It follows from 19 that \(C^\bullet _{\mathrm{Tate}}(G,M)\) is functorial in \(M\).
For an integer \(n\), we shall write \(H^n_{\mathrm{Tate}}(G,M)\) for the \(n\)-th cohomology of the complex \(C^n_{\mathrm{Tate}}(G,M)\); this is called the \(n\)-th Tate cohomology of \(M\), and is often written \(\hat H^n(G,M)\) or (confusingly) just \(H^n(G,M)\) in the literature. We stress that Tate cohomology exists only in the case that \(G\) is a finite group.
Let \(G\) be a finite group and \(M\) a representation of \(G\).
The zeroth Tate cohomology \(H^0_{\mathrm{Tate}}(G,M)\) is isomorphic to \(M^G / N_G(M)\). In particular if \(M\) is a trivial representation of \(G\) then \(H^0_{\mathrm{Tate}}(G,M) \cong M / |G|M\).
For \(n {\gt}0 \) we have (an isomorphism of functors in the variable \(M\))
\[ H^{n}_{\mathrm{Tate}}(G,M) \cong H^{n}(G,M). \]There is an isomorphism
\[ H^{-1}_{\mathrm{Tate}}(G,M) \cong \ker (N_G : M \to M ) / I_G M, \]Where \(I_GM\) is the submodule of \(M\) generated by elements of the form \(g \bullet m - m\). In particular if \(M\) is a trivial representation of \(G\) then \(H^{-1}_{\mathrm{Tate}}(G,M)\) is isomorphic to the \(|G|\)-torsion in \(M\).
For \(n {\lt} -1\) we have (an isomorphism of functors in the variable \(M\))
\[ H^{n}_{\mathrm{Tate}}(G,M) \cong H_{-1-n} (G,M). \]
This result is clear from the definition for \(n {\gt} 0\) and \(n {\lt} -1\). We’ll discuss the two remaining cases.
The \(0\)-cocycle submodule is the kernel of the map \(d^0 : C^0(G,M) \to C^1(G,M)\). This is the same as \(H^0(G,M)\), which is isomorphic to \(M^G\). On the other hand \(B^0_{\mathrm{Tate}}(G,M)\) is by definition the image of \(N_G : M \to M\).
Similarly, \(H_0(G,M)\) is the quotient of \(M\) by \(I_GM\), and \(H^{-1}_{\mathrm{Tate}}(G,M)\) is by definition the quotient of \(\ker (N_G : M \to M)\) by the same submodule.
Since the functors \(C^\bullet (G,-)\) and \(C_\bullet (G,-)\) are both exact, it follows that \(C^\bullet _{\mathrm{Tate}}(G,-)\) is an exact functor. Hence, given any short exact sequence in \(\mathbf{Rep}(R,G)\):
we obtain a short exact sequence of Tate complexes and therefore connecting homomorphisms \(\delta : H^n_{\mathrm{Tate}}(G,C) \to H^{n+1}_{\mathrm{Tate}}(G,A)\) such that the following is a long exact sequence (for \(n \in {\mathbb Z}\)):
The exactness statements are in Mathlib in the namespace HomologicalComplex.HomologySequence. The connecting maps \(\delta : H^n_{\mathrm{Tate}}(G,C) \to H^n_{\mathrm{Tate}}(G,A)\) coincide with those for cohomology for \(n \ge 1\) and with those for homology for \(n \le -3\).
2.4 Coinduction and induction
Let \(M\) be a representation of \(G\) over a ring \(R\).
\(M\) is said to have trivial cohomology if for every subgroup \(S \le G\) and every \(n {\gt} 0\), \(H^n(S,M) \cong 0\).
\(M\) is said to have trivial homology if for every subgroup \(S \le G\) and every \(n {\gt} 0\), \(H_n(S,M) \cong 0\).
Suppose the group \(G\) is finite. Then \(M\) is said to have trivial Tate cohomology if for every subgroup \(S \le G\) and every \(n \in {\mathbb Z}\), \(H^n_{\mathrm{Tate}}(S,M) \cong 0\).
(We will later see that for a finite group \(G\), the three concepts are equivalent.)
In this section we shall describe certain representations with trivial homology and cohomology.
Let \(\phi : S \to G\) be a group homomorphism and let \(M\) be a representation of \(S\). We define a representation \(\mathrm{coind}_\phi (M)\) of \(G\) as follows. As an \(R\)-module we have
This space can be regarded as the \((G \to M)^S\), where the action of \(S\) is \((s\bullet f)(x)=s \bullet f(\phi (s^{-1})\bullet x)\). The action of an element \(g \in G\) on \(f \in \mathrm{coind}_\phi (M)\) is defined by
The map \(M \mapsto \mathrm{coind}_\phi (M)\) is a functor from \(\mathbf{Rep}(R,S)\) to \(\mathbf{Rep}(R,G)\), and is implemented in Mathlib as Rep.coindFunctor.
There is also a functor \(\mathrm{ind}_\phi : \mathbf{Rep}(R,S) \to \mathbf{Rep}(R,G)\) which is defined as a quotient of \(G \to _0 M\) rather than a subspace of \(G \to M\):
The action of an element \(g \in G\) on \(G \to _0 M\) is defined by \((g \bullet f)(x) = f(g^{-1} x)\); this action descends to \(\mathrm{ind}_\phi (M)\). The functor \(\mathrm{ind}_\phi : \mathbf{Rep}(R,S) \to \mathbf{Rep}(R,G)\) is implemented in Mathlib as Rep.ind.
The functors \(\mathrm{ind}_\phi \) and \(\mathrm{coind}_\phi \) are particularly important in the case that \(S\) is a subgroup of \(G\), and \(\phi : S \to G\) is the inclusion map. This is because of the following result:
Let \(S\) be a subgroup of \(G\). Then there are isomorphisms for all \(n \ge 0\):
More precisely these are isomorphisms of functors.
This is already in Mathlib.
We shall be interested in the special case in which the group \(S\) is the trivial group \(1\). In this case the \(S\)-representation \(M\) may be regarded as simply an \(R\)-module. This case is described again below.
Let \(G\) be a group, \(R\) a commutative ring and \(A\) an \(R\)-module.
There is a representation of \(G\) over \(R\) on the space of all functions \(f : G \to A\). The action of an element \(g \in G\) on \(f\) is defined by
\[ (g \bullet f) (x) = f(xg). \]This representation is called the coinduced representation and is denoted \(\mathrm{coind}_1(G,A)\).
There is a representation of \(G\) over \(R\) on the space of all finitely supported functions \(f : G \to _0 A\). The action of an element \(g \in G\) on \(f\) is defined by
\[ (g \bullet f) (x) = f(xg), \qquad i.e.\; g \bullet \mathrm{single}(x,m) = \mathrm{single}(xg^{-1},m). \]This representation is called the induced representation and is denoted \(\mathrm{ind}_1(G,A)\).
More precisely, \(\mathrm{coind}_1(G,-)\) and \(\mathrm{ind}_1(G,-)\) are both functors from \(\mathbf{Mod}(R)\) to \(\mathbf{Rep}(R,G)\).
The representation \(\mathrm{coind}_1(G,A)\) has trivial cohomology.
Let \(S\) be a subgroup of \(G\) and let \(R\) be a set of representatives for the cosets \(rS\) of \(S\) in \(G\). There is an isomorphism of \(S\)-representations
This isomorphism takes a function \(f : G \to M\) to function \(S \to (R \to M)\) defined by \(s \mapsto (r \mapsto f(rs))\).
The isomorphism, together with 24 and 5 implies for \(n {\gt}0\):
Let \(S\) be a normal subgroup of \(G\). Then \(\mathrm{coind}_1(G,A)^S\) is isomorphic to \(\mathrm{coind}_1(G/S,A)\). In particular \(\mathrm{coind}_1(G,A)^S\) is has trivial cohomology as a representation of \(G/S\).
Let \(f : G \to A\). Then \(f\) is in the subspace \(\mathrm{coind}_1(G,A)^S\) if \(f\) is constant on cosets of \(S\), i.e. it descends to a function \(G/S \to A\). This gives a linear bijection \(\mathrm{coind}_1(G,A)^S \cong \mathrm{coind}_1(G/S,A)\), and it’s trivial to check that this map is compatible with the action of \(G / S\).
The representation \(\mathrm{ind}_1(G,A)\) has trivial homology.
There is a morphism of representations \(\mathrm{ind}_1(G,A) \to \mathrm{coind}_1(G,A)\), which takes a finitely supported function \(f : G \to _0 A\) to the function \(f\). If the group \(G\) is finite then this map is an isomorphism. More precisely, this is an isomorphism of functors \(\mathrm{ind}_1(G,-) \cong \mathrm{coind}_1(G,-)\).
If the group \(G\) is finite then \(\mathrm{ind}_1(G,A)\) and \(\mathrm{coind}_1(G,A)\) have trivial Tate cohomology.
These representations are isomorphic, so it’s sufficient to prove that \(\mathrm{ind}_1(G,A)\) has trivial Tate cohomology (this is the more convenient case to prove). We already know that \(\mathrm{ind}_1(G,A)\) has trivial homology. Also, since it is isomorphic to \(\mathrm{coind}_1(G,A)\), it must have trivial cohomology. It only remains to prove that \(H^0_{\mathrm{Tate}}(S,\mathrm{ind}_1(M))\) and \(H^{-1}_{\mathrm{Tate}}(S,\mathrm{ind}_1(M))\) are zero for all subgroups \(S\) of \(G\).
Recall that \(\mathrm{ind}_1(G,M)\) is the space of functions \(G \to _0 M\), and the action of \(G\) is by right-translation:
To prove that \(H^0_{\mathrm{Tate}}(S,\mathrm{ind}_1(G,A))=0\), we use the isomorphism 21:
A function \(f : G \to _0 M\) is \(S\)-invariant if \(f\) is constant on cosets \(gS\) of \(S\). If we let \(\{ g_i\} \) be a set of coset representatives, then we have \(f = N_S (\sum _i \mathrm{single}(g_i, f(g_i)))\). Therefore \(H^0_{\mathrm{Tate}}(S,\mathrm{ind}_1(G,A))=0\).
For the \(n=-1\) case we use the isomorphism 21:
where \(I_G \mathrm{ind}_1(M)\) is generated by elements of the form \(s \bullet f - f\) for \(s \in S\) and \(f : G \to M\). Suppose \(f:G \to _0 M\) is in the kernel of \(N_S\). This implies that the sum of the values of \(f\) over each coset of \(S\) is zero. We can then write \(f\) in the form
Therefore \(f \in I_S \mathrm{ind}_1(M)\). This shows that \(H^{-1}(S,\mathrm{ind}_1(M)) = 0\).
2.5 Dimension-shifting
2.5.1 Shifting \(\mathrm{up}\)
Let \(G\) be a group and \(M\) a representation of \(G\) over a commutative ring \(R\). There is a representation \(\mathrm{coind}_1' (M)\) on the \(R\)-module of functions \(G \to M\). The action of an element \(g \in G\) on a function \(f : G \to M\) is given by
The representations \(\mathrm{coind}_1'(M)\) and \(\mathrm{coind}_1(G,M)\) of \(G\) are isomorphic. More precisely there is an isomorphism of functors \(\mathrm{coind}_1' \cong \mathrm{coind}_1(G,-) \circ \mathbf{forget}\), where \(\mathbf{forget}: \mathbf{Rep}(R,G) \to \mathbf{Mod}(R)\) is the forgetful functor.
The map \(f \mapsto (x \mapsto x \bullet f(x))\) is an isomorphism from \(\mathrm{coind}_1'(M)\) to \(\mathrm{coind}_1(G,M)\).
The representation \(\mathrm{coind}_1'(M)\) has trivial cohomology.
Let \(S\) be a normal subgroup of \(G\). Then \(\mathrm{coind}_1'(M)^S\) has trivial cohomology as a representation of \(G/S\).
There is an injective morphism \(M \to \mathrm{coind}_1'(M)\) which takes a vector \(m \in M\) to the constant function on \(G\) with value \(m\). We define a representation \(\mathrm{up}(M)\) to be the cokernel of this map, so that we have a short exact sequence
This construction is functorial in \(M\); in particular for every \(f : M_1 \to M_2\) in \(\mathbf{Rep}(R,G)\), there is a commutative diagram
Let \(S\) be any subgroup of \(G\) and let \(n \ge 1\). Then the connecting map from the long exact sequence \(H^{n}(S,\mathrm{up}(M)) \to H^{n+1}(S,M)\) is an isomorphism. The corresponding map \(H^{0}(S,\mathrm{up}(M)) \to H^{1}(S,M)\) is surjective.
The isomorphism \(H^{n}(S,\mathrm{up}(-)) \cong H^{n+1}(S,-)\) is an isomorphism of functors. This means that for every morphism \(f : M \to N\) of representations, the following square commutes:
We have already shown in Corollary 33 that \(\mathrm{coind}_1'(M)\) has trivial cohomology, so \(H^{r}(S,\mathrm{coind}_1'(M))=0\) for all \(r{\gt}0\). This implies that the connecting maps are isomorphisms.
The commuting square follows from HomologicalComplex.HomologySequence.\(\delta \)_naturality because the short exact sequence \(0 \to M \to \mathrm{coind}_1'(M) \to \mathrm{up}(M) \to 0\) is functorial in \(M\).
2.5.2 Shifting \(\mathrm{down}\)
Let \(G\) be a group and \(M\) a representation of \(G\) over a commutative ring \(R\).
There is a representation \(\mathrm{ind}_1' (M)\) on the \(R\)-module of finitely supported functions \(G \to _0 M\). The action of an element \(g \in G\) on a function \(f : G \to _0 M\) is given by
The map \(\mathrm{ind}_1'\) is functorial in \(M\).
The representations \(\mathrm{ind}_1'(M)\) and \(\mathrm{ind}_1(G,M)\) are isomorphic; more precisely the functors \(\mathrm{ind}_1'\) and \(\mathrm{ind}_1(G,-) \circ \mathbf{forget}\) are isomorphic.
The data of the isomorphism is contained in the lean file; the isomorphism takes \(f : G \to _0 M\) to the finitely supported function
It remains to check linearity and naturality.
The representation \(\mathrm{ind}_1'(M)\) has trivial homology.
We’ve shown that \(\mathrm{ind}_1'(M)\) is isomorphic to \(\mathrm{ind}_1(M)\), which is already known to have trivial homology by 28.
For any representation \(M\), there is a surjective morphism \(\mathrm{ind}_1'(M) \to M\), which takes a finitely supported function \(f : G \to _0 M\) to the sum \(\sum _{x \in G} f (x)\). We define \(\mathrm{down}(M)\) to be the kernel of this map. There is therefore a short exact sequence
Both \(\mathrm{down}(M)\) and the short exact sequence are functors of \(M\); in particular for every map \(f : M \to N\) in \(\mathbf{Rep}(R,G)\), we have a commutative diagram:
If \(M\) is a representation of a finite group \(G\) then the representations \(\mathrm{ind}_1'(M)\) and \(\mathrm{coind}_1'(M)\) have trivial Tate cohomology.
If the group \(G\) is finite then for every subgroup \(S\) of \(G\) and every \(n \in {\mathbb Z}\) we have isomorphisms
These are the connecting homomorphisms from the short exact sequences linking \(\mathrm{up}(M)\) and \(\mathrm{down}(M)\) to \(M\). They are isomorphisms because \(\mathrm{coind}_1'(M)\) and \(\mathrm{ind}_1'(M)\) have trivial Tate cohomology.
As an example we consider the case of the trivial representation \(R\). The induced representation is then the group ring \(RG\), which is referred to Mathlib as Rep.leftRegular R G; this is a free \(R\)-module with basis \(\mathrm{single}(g,1)\) for \(g \in G\). For simplicity we shall write \([g]\) for the basis vector \(\mathrm{single}(g,1)\). The map \(\mathrm{ind}_1'(R) \to R\) takes \(\sum _{g\in G} x_g [g]\) to \(\sum _{g \in G} x_g\). This map is commonly called the augmentation, and its kernel \(\mathrm{down}(R)\) the augmentation module. We shall write \(\mathrm{aug}(R,G)\) for this kernel. The kernel \(\mathrm{aug}(R,G)\) is spanned by the elements \([g]-[1]\) for \(g \in G\).
Let \(S\) be a subgroup of a finite group \(G\). Then there is an isomorphism of \(R\)-modules
which takes an element \(s \otimes 1\) for \(s \in S\) to the coset of \([s]-[1]\) in \(H^{-1}(S,\mathrm{aug}(R,G)) \cong \mathrm{aug}(R,G) / I_G \mathrm{aug}(R,G)\). In particular, taking \(R = {\mathbb Z}\) we have an isomorphism
Recall that by 14, 21 we already have an isomorphism \(S^\mathrm{ab}\otimes R \cong H_1(S,R) \cong H^{-2}_{\mathrm{Tate}}(S,R)\) which takes \(s \otimes 1\) to (the coset of) \(\mathrm{single}(s,1)\). Furthermore by 42 there is an isomorphism \(H^{-2}_{\mathrm{Tate}}(S,R) \cong H^{-1}_{\mathrm{Tate}}(S,\mathrm{aug}(G,R))\). It remains to check that the image of \(s \otimes 1\) in \(H^{-1}_{\mathrm{Tate}}(S,\mathrm{aug}(G,R))\) is \([s]-[1]\).
In the following diagram the rows are short exact sequences and the vertical maps are the differentials in the Tate complex.
By 14, the image of \(s \otimes 1\) in \(H^{-2}_{\mathrm{Tate}}(S,R)\) is represented by the element \(\mathrm{single}(s,1)\) in \(C^{-2}_{\mathrm{Tate}}(S,R)\). To calculate the image in \(H^{-1}_{\mathrm{Tate}}(S,\mathrm{aug}(R,G))\), we take a pre-image of \(\mathrm{single}(s,1)\) in \(C^{-2}_{\mathrm{Tate}}(S,RG)\); map that preimage into \(C^{-1}_{\mathrm{Tate}}(S,RG)\), and then take its preimage in \(C^{-1}_{\mathrm{Tate}}(S,\mathrm{aug}(R,G))\).
An obvious pre-image of \(\mathrm{single}(s,1)\) in \(C^{-2}_{\mathrm{Tate}}(S,RG)\) is \(\mathrm{single}(s,[1])\). The image of this in \(C^{-1}_{\mathrm{Tate}}(S,RG) \cong RG\) is \([s]-[1]\). Therefore the image of \(s \otimes 1\) in \(H^{-1}_{\mathrm{Tate}}(S,\mathrm{aug}(R,G))\) is \([s]-[1]\).
2.6 The inflation-restriction sequence
Let \(S\) be a normal subgroup of a group \(G\) and let \(n\) be a positive integer. Assume that for all natural numbers \(0 {\lt}i {\lt} n\) we have \(H^{i}(S,M) \cong 0\). Then the following sequence is exact:
where the first map is inflation and the second is restriction.
This is already in Mathlib for \(n=1\). Assume the result is true for some \(n\ge 1\); we will prove it for \(n+1\) by dimension-shifting.
Let \(M\) be a representation such that \(H^i(S,M)\cong 0\) for all \(0 {\lt} i {\lt} n+1\). This implies \(H^i(S,\mathrm{up}(M)) \cong 0\) for all \(0 {\lt} i {\lt} n\). Hence by the inductive hypothesis the following sequence is exact:
Recall that we have a short exact sequence of representations of \(G\):
Since \(0 {\lt} 1 {\lt} n+1\), our condition on \(M\) implies \(H^1(S,M) \cong 0\), so by taking \(S\)-invariants we obtain a short exact sequence of \(G/S\)-modules:
By 34, \(\mathrm{coind}_1'(M)^S\) has trivial cohomology, so we have an isomorphism
We now have a diagram where the horizontal maps are inflation and restriction maps and the vertical maps are dimension-shifting isomorphisms.
The diagram commutes by 10. The first row is exact by the inductive hypothesis. Therefore the second row is also exact.
2.7 Corestriction
Let \(S\) be a subgroup of \(G\). We have already discussed the restriction map \(H^\bullet (G,M) \to H^\bullet (S,M)\). In the case that \(S\) has finite index in \(G\) there is also a “corestriction map” which goes in the other direction, i.e. \(\mathrm{cor}: H^\bullet (S,-)\) to \(H^\bullet (G,-)\). We define this now, and point out some easy consequences of the definition.
Let \(S\) be a subgroup of finite index in \(G\) and let \(\{ r_i\} \) be a set of representatives for the cosets \(r_i S\). For any representation \(M\) of \(G\) there is a linear map \(N_{G/S} : M^S \to M^G\) defined by
This map does not depend on the choince of coset representatives. Also, the map \(N_{G/S}\) is a morphism of functors, i.e. for every map \(f : A \to B\) in \(\mathbf{Rep}(R,G)\) we have a commuting square in \(\mathbf{Mod}(R)\):
where the horizontal maps are induced by \(f\) and the vertical maps are \(N_{G/S}\).
The corestriction maps \(\mathrm{cor}^n : H^n(S,M) \to H^n(G,M)\) are defined recursively as follows:
The map \(\mathrm{cor}^0 : H^0(S,M) \to H^0(G,M)\) is defined to be \(N_{G/S}\).
Assume that we have defined \(\mathrm{cor}^n\). For any \(M\) we have a commutative diagram in which the rows are exact and the vertical arrows are \(\mathrm{cor}^n\):
\[ \begin{matrix} H^n(S,\mathrm{coind}_1’(M)) & \to & H^n(S,\mathrm{up}(M)) & \to & H^{n+1}(S,M) & \to & 0 \\ \downarrow & & \downarrow \\ H^n(S,\mathrm{coind}_1’(M)) & \to & H^n(S,\mathrm{up}(M)) & \to & H^{n+1}(S,M) & \to & 0 \end{matrix}. \]It follows that there is a unique linear map \(\mathrm{cor}^{n+1} : H^{n+1}(S,M) \to H^{n+1}(G,M)\) such that the following square commutes:
\[ \begin{matrix} H^n(S,\mathrm{up}(M)) & \to & H^{n+1}(S,M) \\ \downarrow & & \downarrow \\ H^n(S,\mathrm{up}(M)) & \to & H^{n+1}(S,M) \end{matrix}. \]
The map \(\mathrm{cor}^{n} : H^n(G , -) \to H^n(S,-)\) is a morphism of functors.
For all \(\sigma \in H^n(G,M)\) we have \(\mathrm{cor}(\mathrm{rest}(\sigma )) = [G:S] \cdot \sigma \).
We’ll prove the result by induction on \(n\). In the case \(n = 0\), this follows from the relation for all \(m \in M^G\):
The identity above holds because each term in the sum defining \(N_{G/S}(m)\) is equal to \(m\).
Let’s assume that the lemma is true for some \(n\). We then have a diagram in which the vertical maps are the dimension shifting maps, which are all surjective:
By the inductive hypothesis, the composition of the maps on the top row is multiplication by \([G:S]\). The square on the left commutes by 10, and the square on the right commutes by definition of the corestriction map. Therefore the composition on the bottom row is multiplication by \([G:S]\).
If \(M\) is a representation of a finite group \(G\) then for all \(n \in {\mathbb Z}\) and all \(\sigma \in H^n_{\mathrm{Tate}}(G,M)\) we have \(|G| \cdot \sigma = 0\).
By dimension-shifting (42) it’s enough to prove the result for \(n {\gt} 0\), in which case by 21 Tate cohomology is isomorphic to cohomology. Take \(S\) to be the trivial subgroup \(1\) of \(G\). By 47 it’s sufficient to prove that \(\mathrm{cor}(\mathrm{rest}(\sigma ))= 0\). This follows because \(\mathrm{rest}(\sigma ) \in H^n(1,M) \cong 0\) by 5.
Let \(M\) be a representation of a finite group \(G\) and let \(S_p\) be a Sylow \(p\)-subgroup of \(G\) for some prime number \(p\). Then for any \(n \in {\mathbb Z}\), \(H^n_{\mathrm{Tate}}(G,M)[p^\infty ]\) is isomorphic to an \(R\)-submodule of \(H^n_{\mathrm{Tate}}(S_p,M)\).
By dimension-shifting it’s enough to prove the result for \(n {\gt} 0\), in which case Tate cohomology is isomorphic to cohomology. It follows from 47 that the composition \(\mathrm{cor}\circ \mathrm{rest}\) is injective on \(H^n(G,M)[p^\infty ]\). Therefore the restriction map is an injective map from \(H^n(G,M)[p^\infty ]\) to \(H^n(S_p,M)\).
2.8 Periodicity for finite cyclic groups
In this section we shall assume that \(G\) is a finite cyclic group of order \(n\). We shall write \(\mathrm{gen}\) for a fixed generator of \(G\).
Given any representation \(M\) of \(G\), there is a map \(\mathrm{map}_1 : \mathrm{coind}_1'(M) \to \mathrm{coind}_1'(M)\) which takes a function \(f : G \to M\) to the function
The kernel of \(\mathrm{map}_1\) consists of the constant functions \(G \to M\), i.e. the image of the map \(M \to \mathrm{coind}_1'(M)\). Hence by the first isomorphism theorem, the image of \(\mathrm{map}_1\) is isomorphic to \(\mathrm{up}(M)\).
Since \(G\) is finite, the representations \(\mathrm{coind}_1'(M)\) and \(\mathrm{ind}_1'(M)\) are isomorphic, and we define \(\mathrm{map}_2\) to be the corresponding map \(\mathrm{ind}_1'(M) \to \mathrm{ind}_1'(M)\). This is given by
The image of \(\mathrm{map}_2 : \mathrm{ind}_1'(M) \to \mathrm{ind}_1'(M)\) is precisely the set of functions \(G \to _0 M\) which sum to zero. This is the kernel of the map \(\mathrm{ind}_1'(M) \to M\), which we are calling \(\mathrm{down}(M)\).
It’s clear that the values of \(\mathrm{map}_2(f)\) sum to \(0\), so the image of \(\mathrm{map}_2\) is contained in the kernel. Conversely suppose \(h : G \to _0 M\) satisifies \(\sum _{i=0}^{n-1} h(\mathrm{gen}^i) = 0\). Then we have
Furthermore each of the terms \(\mathrm{single}(\mathrm{gen}^i, m) - \mathrm{single}(1, m)\) is in the image of \(\mathrm{map}_2\):
We have a commutative square with vertical isomorphisms:
It follows that \(\mathrm{im}(\mathrm{map}_1) \cong \mathrm{im}(\mathrm{map}_2)\), i.e.
This is an isomorphism of functors; i.e. for each map \(f : M \to N\) in \(\mathbf{Rep}(R,G)\) we have a commuting square:
Let \(G\) be a finite cyclic group. For all \(n {\gt} 0\) and all representations \(M\) we have an isomorphism \(H^{n}(G,M) \cong H^{n+2}(G,M)\). Similarly for all integers \(n\) we have isomorphisms \(H^{n}_{\mathrm{Tate}}(G,M) \cong H^{n+2}_{\mathrm{Tate}}(G,M)\).
By the dimension-shifting isomorphisms we have \(H^{n}(G,M) \cong H^{n+1}(G,\mathrm{down}(M)) \cong H^{n+1}(G,\mathrm{up}(M)) \cong H^{n+2}(G,M)\), and similarly for Tate cohomology.
A very important example for us is the trivial representation of \(G\) on \({\mathbb Z}\), which we describe very precisely here:
Let \(G\) be a finite cyclic group of order \(n\). Then \(H^1(G,{\mathbb Z}) \cong 0\) and \(H^2(G,{\mathbb Z}) \cong {\mathbb Z}/n{\mathbb Z}\).
Since the module \({\mathbb Z}\) is trivial, we have \(H^1(G,{\mathbb Z})\cong \mathrm{Hom}(G,{\mathbb Z}) \cong 0\).
It follows from 21 that \(H^0_{\mathrm{Tate}}(G,{\mathbb Z}) \cong {\mathbb Z}/n{\mathbb Z}\). Hence by periodicity there is an isomorphism \(H^2(G,{\mathbb Z}) \cong {\mathbb Z}/ n{\mathbb Z}\).
Let \(G\) be a finite cyclic group of order \(n\) generated by an element \(\mathrm{gen}\). Then the map
is called the local invariant of \(G\).
Let \(G\) be a finite cyclic group of order \(n\) generated by an element \(\mathrm{gen}\). The local invariant \(\mathrm{inv}_G : H^2(G,{\mathbb Z}) \to {\mathbb Z}/n{\mathbb Z}\) is an isomorphism. The pre-image of \(1 \in {\mathbb Z}/n{\mathbb Z}\) is the cohomology class of the cocycle
It is easy to check that the formula for \(\mathrm{inv}_{\mathbb Z}\) defines a homomorphism \(H^2(G,{\mathbb Z}) \to {\mathbb Z}/ n {\mathbb Z}\) (i.e. the coboundaries are in the kernel).
\(\sigma _G\) is a 2-cocycle since one can write \(\sigma _G(\mathrm{gen}^i, \mathrm{gen}^j)\) as \(\tilde i + \tilde j - \tilde(i + j)\) for \(\tilde: {\mathbb Z}/n{\mathbb Z}\to {\mathbb Z}\) a section
It follows from the definitions that all the terms in the sum defining the local invariant are zero apart from the term \(i=n-1\), so we have
It follows that \(\mathrm{inv}_G : H^2(G,{\mathbb Z}) \to {\mathbb Z}/n{\mathbb Z}\) is surjective. By 53, \(|H^2(G,{\mathbb Z})| = |{\mathbb Z}/n{\mathbb Z}|\), therefore \(\mathrm{inv}_G\) is in fact an isomorphism.
2.9 Herbrand quotients
Let \(G\) be a finite cyclic group and \(M\) a representation of \(G\). Recall that there are isomorphisms \(H^n_{\mathrm{Tate}}(G,M) \cong H^{n+2}_{\mathrm{Tate}}(G,M)\). We define the Herbrand quotient of \(M\) to be
If either of the two cohomology groups are infinite then \(h(G,M)\) defaults to \(0\).
If \(G\) is a cyclic group and \({\mathbb Z}\) has the trivial action of \(G\) then \(h(G,{\mathbb Z}) = |G|\). This follow immediately from 53.
If \(M\) is finite then \(h(G,M)=1\).
Let \(\mathrm{gen}\) be a generator of \(G\). Recall that \(H^0_{\mathrm{Tate}}(G,M) \cong M^G / N_GM\). Also, we can write \(M^G\) as \(\ker (1-\mathrm{gen}: M \to M)\). Similarly \(H^{-1}_{\mathrm{Tate}}(G,M)\) is isomorphic to \(\ker (N_G : M \to M) / \mathrm{im}(1-\mathrm{gen}: M \to M)\). The result follows because
Suppose we have a short exact sequence of representations of a finite cyclic group \(G\):
If two of the representations \(A\), \(B\), \(C\) have non-zero Herbrand quotient then so does the third, and \(h(G,B) = h(G,A) \cdot h(G,C)\).
It follows from the long exact sequence that if two of the representations \(A,B,C\) have finite cohomology groups then so does the third. Also, by periodicity, the long exact sequence reduces to an exact hexagon:
The result follows because the alternating product of the finite group orders in the hexagon is \(1\).
2.10 The Triviality Criterion
Recall that a representation \(M\) of a group \(G\) has trivial cohomology if for all subgroups \(S\) of \(G\) and all \(n {\gt} 0\), the cohomology groups \(H^{n}(S,M)\) are zero.
Let \(M\) be a representation of a finite solvable group \(G\). Suppose we have positive natural numbers \(e\) and \(o\) with \(e\) even and \(o\) odd, such that for all subgroups \(S\) of \(G\) we have
Then \(M\) has trivial cohomology.
We must prove that \(H^{n}(S,M) = 0\) for all \(S\) and all \(n {\gt} 0\). We’ll prove this by induction on \(S\). The result is true for the trivial subgroup of \(G\). Assume that the result is true for \(S\), and assume that \(S' / S\) is cyclic. The inductive hypothesis implies that (for all \(n\)) the inflation restriction sequence is exact:
We therefore have isomorphisms \(H^{n} (S'/S, M^S) \cong H^{n}(S' , M)\). In particular we have \(H^{e} (S'/S, M^S) \cong 0\) and \(H^{o} (S'/S, M^S) \cong 0\). Using periodicity of the cohomology of a cyclic group, we have \(H^{n}(S'/S,M^S) \cong 0\) for all \(n{\gt}0\).
Let \(M\) be a representation of a finite group \(G\) (no longer assumed to be solvable). Suppose we have positive natural numbers \(e\) and \(o\) with \(e\) even and \(o\) odd, such that for all subgroups \(S\) of \(G\) we have
Then \(M\) has trivial cohomology.
Let \(S\) be a subgroup of \(G\). Fix a prime number \(p\) and let \(S_p\) be a Sylow \(p\)-subgroup of \(S\). Since \(S_p\) is solvable, 60 implies that \(H^n(S_p,M) \cong 0\) for all \(n {\gt} 0\). By 49, it follows that \(H^n(S,M)[p^\infty ] \cong 0\). Since this holds for all primes \(p\), 48 implies that \(H^n(S,M) \cong 0\).
If \(M\) is a representation of a finite group \(G\) and \(M\) has trivial cohomology then \(\mathrm{up}(M)\) and \(\mathrm{down}(M)\) have trivial cohomology.
For each subgroup \(S\) of \(G\) we have
By 61 \(\mathrm{up}(M)\) has trivial cohomology. Similarly
so \(\mathrm{down}(M)\) has trivial cohomology.
Let \(M\) be a representation of a finite group \(G\), and assume that \(M\) has trivial cohomology. Then \(M\) has trivial Tate cohomology.
Fix an integer \(n\) and choose a natural number \(m\) such that \(m + n {\gt} 0\). By 62, \(\mathrm{down}^m(M)\) has trivial cohomology. Therefore
2.11 The splitting module of a 2-cocycle
Let \(\sigma \in H^2(G,M)\). In this section we describe a representation of \(G\) called the splitting module of \(\sigma \).
We shall write \(\sigma '\) for an inhomogeneous 2-cocycle representing \(\sigma \). This means \(\sigma ' : G \times G \to M\) satisfies the following 2-cocycle relation for all \(x,y,z\in G\)
The splitting module of \(\sigma '\) is the \(R\)-module \(M \times \mathrm{aug}(R,G)\), with the action of an element \(g \in G\) given by
We’ll write \(\mathrm{split}(\sigma ')\) for the splitting module. Although we don’t need this fact, it’s worth knowing that up to isomorphism, the splitting module depends only on the cohomology class of \(\sigma '\).
There is evidently a short exact sequence of representations of \(G\).
The image of \(\sigma '\) in \(H^2(G,\mathrm{split}(\sigma ))\) is zero.
We can check that the cocycle \(\sigma '\) is the coboundary of the \(1\)-cochain \(\tau : G \to \mathrm{split}(\sigma )\) defined by
(Here we are using the notation \([x]\) to mean the function with value \(1\) at \(x\) and \(0\) elsewhere).
By definition we have
It remains to prove that \(\sigma '(x,1) = x \bullet \sigma '(1,1)\). This follows from the \(2\)-cocycle relation 3 in the case \(y=z=1\).
2.12 The Reciprocity Isomorphism
In this section \(G\) is a finite group; \(M\) is a representation of \(G\) over a commutative ring \(R\). We shall call \((R,G,M)M\) is a finite class formation if:
The ring \(R\) has no additive torsion. This implies for all subgroups \(S\) of \(G\):
\[ H^2(S,\mathrm{aug}(R,G)) \cong H^1(S,R) \cong \mathrm{Hom}(S,R) = 0. \]For all subgroups \(S \le G\) we have \(H^1(S,M) \cong 0\).
For all subgroups \(S \le G\), \(H^2(S,M)\) is isomorphic as an \(R\)-module to \(R / |S| \cdot R\).
If \(M\) is a finite class formation then a generator \(\sigma \) for \(H^2(G,M)\) is called a fundamental class.
In this section we’ll show that if \((R,G,M)\) is a finite class formation then there is an isomorphism (called the reciprocity isomorphism)
The reciprocity isomorphism depends on the choice of a fundamental class.
Let \(I\) be an ideal of a commutative ring \(R\) and let \(f:R/I \to R/I\) be a surjective \(R\)-linear map. Then \(f\) is injective.
(This lemmma is already in Mathlib.) Without loss of generality \(I=0\), since such a map is also \(R/I\)-linear. Let \(c\in R\) be a preimage of \(1\) and let \(d = f(1)\). We have \(cd = cf(1) = f(c) = 1\). The map \(f\) is given by \(f(x) = dx\) and the map \(x \mapsto cx\) is an inverse.
Let \(\sigma \in H^2(G,M)\) be a fundamental class. Then the restriction of \(\sigma \) to any subgroup \(S\) of \(G\) is a generator for \(H^2(S,M)\).
The restriction and corestriction maps are \(R\)-linear maps
We need to prove that the restriction map is surjective. By the conditions on \(M\), we can think of these maps as
Since \(R\) has no additive torsion, the image of \(R/|S|\) in \(R/|G|\) is contained in the \(S\)-torsion, which is \([G:S] R / |G|\). Furthermore, since the composition is \([G:S]\), it follows that the image of \(\mathrm{cor}\) contains \([G:S] R / |G|\). Therefore the image of \(\mathrm{cor}\) is precisely \([G:S] R / |G|\), which is isomorphic to \(R/|S|\). Hence \(\mathrm{cor}\) may be regarded as a surjective linear map \(R/|S| \to R/|S|\). By 68, \(\mathrm{cor}\) is injective with image \([G:S] H^2(G,M)\).
Now let \(b \in H^2(S,M)\). We have \(\mathrm{cor}(b) = [G:S]c\) for some \(c \in H^2(G,M)\). This implies \(\mathrm{cor}(b) = \mathrm{cor}(\mathrm{rest}( c))\), and since \(\mathrm{cor}\) is injective we have \(b = \mathrm{rest}(c)\).
Let \(\sigma '\) be a 2-cocycle representing a fundamental class in \(H^2(G,M)\). Then \(\mathrm{split}(\sigma ')\) has trivial cohomology.
By 61, it’s enough to prove for every subgroup \(S\) of \(G\) that \(H^1(S,\mathrm{split}(\sigma ')) \cong 0\) and \(H^2(S,\mathrm{split}(\sigma ')) \cong 0\). We have a long exact sequence with the following terms:
By 69 and 65, the map \(H^2(S,M) \to H^2(S,\mathrm{split}(\sigma '))\) is zero. In particular \(H^2(S,\mathrm{split}(\sigma ')) \cong 0\). The \(R\)-modules \(H^1(S,\mathrm{aug}(R,G))\) and \(H^2(S,M)\) are both isomorphic to \(R / |S|R\), and the map from one to the other is surjective. By 68, the map from \(H^1(S,\mathrm{aug}(R,G))\) to \(H^2(S,M)\) is also injective. Therefore \(H^1(S,\mathrm{split}(\sigma ')) \cong 0\).
The theorem implies that we have isomorphisms for all \(n\in {\mathbb Z}\) (which depend of \(\sigma \)):
In particular in the case \(n = -1\) we have the reciprocity isomorphism
Here \(\delta \) is the connecting map for the short exact sequence 4.
We next prove a formula for the reciprocity isomorphism in terms of a cocycle \(\sigma '\) representing the fundamental class \(\sigma \).
The reciprocity isomorphism for a fundamental class \(\sigma \in H^2(G,M)\) is given by
This depends only on the cohomology class \(\sigma \) rather than the cocycle \(\sigma '\).
We have a diagram with exact rows. Note that \(C^0(G,M)\) and \(C^{-1}(G,M)\) are both \(M\) and the vertical maps are Tate coboundary maps, which are all \(N_G\).
Choose an element \(g \in G\). By 44, the element \(gG' \otimes 1 \in G^{\mathrm{ab}} \otimes R\) maps to the cocycle \([g]-[1] \in C^{-1}(G,\mathrm{aug}({\mathbb Z},G))\). An obvious pre-image of this element in \(C^{-1}(G,\mathrm{split}(\sigma ))\) is the element \((0,[g]-[1])\). The image of \((0,[g]-[1])\) in \(C^{0}(G,\mathrm{split}(\sigma ))\) is
By the cocycle relation we have \(\sigma '(x,1) = x \bullet \sigma '(1,1)\) for all \(x\in G\). Hence the reciprocity map is given by
If we modify \(\sigma '\) by a coboundary \(d\tau \):
then we will add the folowing term to \(\mathrm{reciprocity}(g)\):
The second and third terms canel each other out after reindexing, so we are left with \(N_G(\tau (g))\), which is zero in \(H^0_{\mathrm{Tate}}(G,M)\).
2.13 Compatibility in towers
Let \((R,G,M)\) be a finite class formation with a fundamental class \(\sigma _G\) and let \(S\) be a subgroup of \(G\). Then \((R, S, M \downarrow S)\) is a finite class formation. The restriction \(\sigma _S\) of \(\sigma _G\) to \(S\) is a fundamental class in \(H^2(S,M)\). Furthermore there is a commuting square
The horizontal maps are the reciprocity isomorphisms defined by the fundamental classes \(\sigma _G\) and \(\sigma _S\); the left hand vertical map is \(sS' \mapsto sG'\) and the right hand map is induced by \(N_{G/S} :M^S \to M^G\).
The fact that \((R, S, M \downarrow S)\) is a finite class formation is a tautology. The fact that \(\sigma _S\) is a fundamental is 69. It remains to prove that the diagram commutes. We’ll write \(\theta _G\) and \(\theta _S\) for the horizontal maps in the diagram. For an element \(s \in S\) we have by 72:
Let \(R\) be a set of representatives for the cosets \(rS\) of \(S\) in \(G\). Then we have
Using the cocycle relation we have
The first and last terms cancel after reindexing, and we are left with
The right hand side is \(N_{G/S} \theta _S(s)\) by 72.
For a finite subgroup \(S\) of \(G\), we shall call \(N_{G/S}M^S\) the norm submodule corresponding to \(S\). This is a submodule of \(M^G\) containing \(N_G M\). By the commutative diagram in 73, the image of \(SG'/G' \otimes R\) under the reciprocity map is equal to \(N_{G/S}(M^S) / N_G M\).
Let \(({\mathbb Z},G,M)\) be a finite class formation and let \(S_1\) and \(S_2\) be two subgroups of \(G\). Then \(S_1 G' \subseteq S_2 G'\) if and only if \(N_{G/S_1}(M^{S_1}) \subseteq N_{G/S_2}(M^{S_2})\).
Let \((R,G,M)\) be a finite class formation. Then
Suppose \(({\mathbb Z},G,M)\) is a finite class formation with \(G\) abelian. Let \(S_1\) and \(S_2\) be subgroups of \(G\). Then \(S_1 \subseteq S_2\) if and only if \(N_{G/S_1}(M^{S_1}) \subseteq N_{G/S_2}(M^{S_2})\)