Cohomology group for trivial group action

Definition

Let $G$ be a group and $A$ be an abelian group.

The cohomology groups for trivial group action $\!H^{n}(G,A)$, also denoted $\!H^{n}(G;A)$ ($n=0,1,2,3,\dots$) are abelian groups defined in the following equivalent ways.

Definition in terms of classifying space

$\!H^{n}(G,A)$ can be defined as the cohomology group $H^{n}(BG,A)$, where $BG$ is the classifying space of $G$ and the cohomology group is understood to be in the topological sense (singular cohomology or cellular cohomology, or any of the equivalent cohomology theories satisfying the axioms).

Definitions as cohomology group for an action taken as the trivial action

The cohomology groups for trivial group action $H^{n}(G,A)$ are defined as the cohomology groups $H_{\varphi }^{n}(G,A)$ where $\varphi :G\to \operatorname {Aut} (A)$ is the trivial map. In other words, we treat $A$ as a $G$-module with trivial action of $G$ on $A$ (i.e., every element of $G$ fixes every element of $A$. We thus also treat $A$ as a trivial $\mathbb {Z} G$-module, where $\mathbb {Z} G$ is a group ring of $G$ over the ring of integers $\mathbb {Z}$.

The definitions below are basically adaptations of the general definitions of cohomology group to the case where the action is trivial.

No.	Shorthand	Detailed description of $H^{n}(G,A)$, the $n^{th}$ cohomology group
1	Explicit, using the bar resolution	$H^{n}(G,A)$ is defined as the quotient $Z^{n}(G,A)/B^{n}(G,A)$ where $Z^{n}(G,A)$ is the group of cocycles for the trivial group action for the action and $B^{n}(G,A)$ is the group of coboundaries for the trivial action.
1'	Explicit, using the normalized bar resolution	Same as definition (1), but we use normalized cocycles and normalized coboundaries instead of arbitrary cocycles and coboundaries.
2	Complex based on arbitrary projective resolution	Let ${\mathcal {F}}$ be a projective resolution for $\mathbb {Z}$ as a $\mathbb {Z} G$-module with the trivial action. Let ${\mathcal {C}}$ be the complex $\operatorname {Hom} _{\mathbb {Z} G}({\mathcal {F}},A)$. The cohomology group $H^{n}(G,A)$ is defined as the $n^{th}$ cohomology group for this complex.
3	Complex based on arbitrary injective resolution (works if category of $\mathbb {Z} G$-modules has enough injectives!)	Let ${\mathcal {I}}$ be an injective resolution for $A$ as a $\mathbb {Z} G$-module with the trivial group action. Let ${\mathcal {D}}$ be the complex $\operatorname {Hom} _{\mathbb {Z} G}(\mathbb {Z} ,{\mathcal {I}})$ where $\mathbb {Z}$ has the structure of a trivial action $\mathbb {Z} G$-module. The cohomology group $H_{\varphi }^{n}(G,A)$ is defined as the $n^{th}$ cohomology group for this complex.
4	As an $\operatorname {Ext}$ functor	$\operatorname {Ext} _{\mathbb {Z} G}^{n}(\mathbb {Z} ,A)$ where $\mathbb {Z}$ and $A$ are both treated as trivial $\mathbb {Z} G$-modules.
5	As a right derived functor	$H_{\varphi }^{n}(G,A)=R^{n}(-^{G})(A)$, i.e., it is the $n^{th}$ right derived functor of the invariants functor for $G$ (denoted $-^{G}$) evaluated at $A$ viewed as a trivial $\mathbb {Z} G$-module. The invariants functor sends a $\mathbb {Z} G$-module to its submodule of elements fixed by all elements of $G$.

Equivalence of definitions

To show the equivalence of definitions between the topological and algebraic definitions, we can proceed as follows:

• Prove the equivalence of definitions of homology groups with coefficients in the integers.
• Note that the dual universal coefficients theorem holds in both cases, and can be used to get natural isomorphisms.