In the following, if we write $H^ i(G, M)$ we will mean that $G$ is a topological group and $M$ a discrete $G$-module with continuous $G$-action and $H^ i(G, -)$ is the $i$th right derived functor on the category $\text{Mod}_ G$ of such $G$-modules, see Definitions 59.57.1 and 59.57.2. This includes the case of an abstract group $G$, which simply means that $G$ is viewed as a topological group with the discrete topology.
When the module has a nondiscrete topology, we will use the notation $H^ i_{cont}(G, M)$ to indicate the continuous cohomology groups introduced in [Tate], see Section 59.58.
Definition 59.57.1. Let $G$ be a topological group.
A $G$-module, sometimes called a discrete $G$-module, is an abelian group $M$ endowed with a left action $a : G \times M \to M$ by group homomorphisms such that $a$ is continuous when $M$ is given the discrete topology.
A morphism of $G$-modules $f : M \to N$ is a $G$-equivariant homomorphism from $M$ to $N$.
The category of $G$-modules is denoted $\text{Mod}_ G$.
Let $R$ be a ring.
An $R\text{-}G$-module is an $R$-module $M$ endowed with a left action $a : G \times M \to M$ by $R$-linear maps such that $a$ is continuous when $M$ is given the discrete topology.
A morphism of $R\text{-}G$-modules $f : M \to N$ is a $G$-equivariant $R$-module map from $M$ to $N$.
The category of $R\text{-}G$-modules is denoted $\text{Mod}_{R, G}$.
The condition that $a : G \times M \to M$ is continuous is equivalent with the condition that the stabilizer of any $x \in M$ is open in $G$. If $G$ is an abstract group then this corresponds to the notion of an abelian group endowed with a $G$-action provided we endow $G$ with the discrete topology. Observe that $\text{Mod}_{\mathbf{Z}, G} = \text{Mod}_ G$.
The category $\text{Mod}_ G$ has enough injectives, see Injectives, Lemma 19.3.1. Consider the left exact functor
We sometimes denote $M^ G = H^0(G, M)$ and sometimes we write $M^ G = \Gamma _ G(M)$. This functor has a total right derived functor $R\Gamma _ G(M)$ and $i$th right derived functor $R^ i\Gamma _ G(M) = H^ i(G, M)$ for any $i \geq 0$.
The same construction works for $H^0(G, -) : \text{Mod}_{R, G} \to \text{Mod}_ R$. We will see in Lemma 59.57.3 that this agrees with the cohomology of the underlying $G$-module.
Definition 59.57.2. Let $G$ be a topological group. Let $M$ be a discrete $G$-module with continuous $G$-action. In other words, $M$ is an object of the category $\text{Mod}_ G$ introduced in Definition 59.57.1.
The right derived functors $H^ i(G, M)$ of $H^0(G, M)$ on the category $\text{Mod}_ G$ are called the continuous group cohomology groups of $M$.
If $G$ is an abstract group endowed with the discrete topology then the $H^ i(G, M)$ are called the group cohomology groups of $M$.
If $G$ is a Galois group, then the groups $H^ i(G, M)$ are called the Galois cohomology groups of $M$.
If $G$ is the absolute Galois group of a field $K$, then the groups $H^ i(G, M)$ are sometimes called the Galois cohomology groups of $K$ with coefficients in $M$. In this case we sometimes write $H^ i(K, M)$ instead of $H^ i(G, M)$.
Lemma 59.57.3. Let $G$ be a topological group. Let $R$ be a ring. For every $i \geq 0$ the diagram whose vertical arrows are the forgetful functors is commutative.
\[ \xymatrix{ \text{Mod}_{R, G} \ar[rr]_{H^ i(G, -)} \ar[d] & & \text{Mod}_ R \ar[d] \\ \text{Mod}_ G \ar[rr]^{H^ i(G, -)} & & \textit{Ab} } \]
Proof. Let us denote the forgetful functor $F : \text{Mod}_{R, G} \to \text{Mod}_ G$. Then $F$ has a left adjoint $H : \text{Mod}_ G \to \text{Mod}_{R, G}$ given by $H(M) = M \otimes _\mathbf {Z} R$. Observe that every object of $\text{Mod}_ G$ is a quotient of a direct sum of modules of the form $\mathbf{Z}[G/U]$ where $U \subset G$ is an open subgroup. Here $\mathbf{Z}[G/U]$ denotes the $G$-modules of finite $\mathbf{Z}$-linear combinations of right $U$ congruence classes in $G$ endowed with left $G$-action. Thus every bounded above complex in $\text{Mod}_ G$ is quasi-isomorphic to a bounded above complex in $\text{Mod}_ G$ whose underlying terms are flat $\mathbf{Z}$-modules (Derived Categories, Lemma 13.15.4). Thus it is clear that $LH$ exists on $D^-(\text{Mod}_ G)$ and is computed by evaluating $H$ on any complex whose terms are flat $\mathbf{Z}$-modules; this follows from Derived Categories, Lemma 13.15.7 and Proposition 13.16.8. We conclude from Derived Categories, Lemma 13.30.2 that
for $M$ in $\text{Mod}_{R, G}$. Observe that $H^0(G, -) = \mathop{\mathrm{Hom}}\nolimits (\mathbf{Z}, -)$ on $\text{Mod}_ G$ where $\mathbf{Z}$ denotes the $G$-module with trivial action. Hence $H^ i(G, -) = \text{Ext}^ i(\mathbf{Z}, -)$ on $\text{Mod}_ G$. Similarly we have $H^ i(G, -) = \text{Ext}^ i(R, -)$ on $\text{Mod}_{R, G}$. Combining everything we see that the lemma is true. $\square$
Lemma 59.57.4. Let $G$ be a topological group. Let $R$ be a ring. Let $M$, $N$ be $R\text{-}G$-modules. If $M$ is finite projective as an $R$-module, then $\text{Ext}^ i(M, N) = H^ i(G, M^\vee \otimes _ R N)$ (for notation see proof).
Proof. The module $M^\vee = \mathop{\mathrm{Hom}}\nolimits _ R(M, R)$ endowed with the contragredient action of $G$. Namely $(g \cdot \lambda )(m) = \lambda (g^{-1} \cdot m)$ for $g \in G$, $\lambda \in M^\vee $, $m \in M$. The action of $G$ on $M^\vee \otimes _ R N$ is the diagonal one, i.e., given by $g \cdot (\lambda \otimes n) = g \cdot \lambda \otimes g \cdot n$. Note that for a third $R\text{-}G$-module $E$ we have $\mathop{\mathrm{Hom}}\nolimits (E, M^\vee \otimes _ R N) = \mathop{\mathrm{Hom}}\nolimits (M \otimes _ R E, N)$. Namely, this is true on the level of $R$-modules by Algebra, Lemmas 10.12.8 and 10.78.9 and the definitions of $G$-actions are chosen such that it remains true for $R\text{-}G$-modules. It follows that $M^\vee \otimes _ R N$ is an injective $R\text{-}G$-module if $N$ is an injective $R\text{-}G$-module. Hence if $N \to N^\bullet $ is an injective resolution, then $M^\vee \otimes _ R N \to M^\vee \otimes _ R N^\bullet $ is an injective resolution. Then
Since the left hand side computes $\text{Ext}^ i(M, N)$ and the right hand side computes $H^ i(G, M^\vee \otimes _ R N)$ the proof is complete. $\square$
Lemma 59.57.5. Let $G$ be a topological group. Let $k$ be a field. Let $V$ be a $k\text{-}G$-module. If $G$ is topologically finitely generated and $\dim _ k(V) < \infty $, then $\dim _ k H^1(G, V) < \infty $.
Proof. Let $g_1, \ldots , g_ r \in G$ be elements which topologically generate $G$, i.e., this means that the subgroup generated by $g_1, \ldots , g_ r$ is dense. By Lemma 59.57.4 we see that $H^1(G, V)$ is the $k$-vector space of extensions
of $k\text{-}G$-modules. Choose $e \in E$ mapping to $1 \in k$. Write
for some $v_ i \in V$. This is possible because $g_ i \cdot 1 = 1$. We claim that the list of elements $v_1, \ldots , v_ r \in V$ determine the isomorphism class of the extension $E$. Once we prove this the lemma follows as this means that our Ext vector space is isomorphic to a subquotient of the $k$-vector space $V^{\oplus r}$; some details omitted. Since $E$ is an object of the category defined in Definition 59.57.1 we know there is an open subgroup $U$ such that $u \cdot e = e$ for all $u \in U$. Now pick any $g \in G$. Then $gU$ contains a word $w$ in the elements $g_1, \ldots , g_ r$. Say $gu = w$. Since the element $w \cdot e$ is determined by $v_1, \ldots , v_ r$, we see that $g \cdot e = (gu) \cdot e = w \cdot e$ is too. $\square$
Lemma 59.57.6. Let $G$ be a profinite topological group. Then
$H^ i(G, M)$ is torsion for $i > 0$ and any $G$-module $M$, and
$H^ i(G, M) = 0$ if $M$ is a $\mathbf{Q}$-vector space.
Proof. Proof of (1). By dimension shifting we see that it suffices to show that $H^1(G, M)$ is torsion for every $G$-module $M$. Choose an exact sequence $0 \to M \to I \to N \to 0$ with $I$ an injective object of the category of $G$-modules. Then any element of $H^1(G, M)$ is the image of an element $y \in N^ G$. Choose $x \in I$ mapping to $y$. The stabilizer $U \subset G$ of $x$ is open, hence has finite index $r$. Let $g_1, \ldots , g_ r \in G$ be a system of representatives for $G/U$. Then $\sum g_ i(x)$ is an invariant element of $I$ which maps to $ry$. Thus $r$ kills the element of $H^1(G, M)$ we started with. Part (2) follows as then $H^ i(G, M)$ is both a $\mathbf{Q}$-vector space and torsion. $\square$
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