group cohomology (topological definition)

Let $G$ be a topological group. Suppose some contractible space $X$ admits a fixed point free action of $G$, so that the quotient map $p:X\to X/G$ is a fibre map. Then $X/G$, denoted $BG$ is called the classifying space of $G$. Classifying spaces always exist and are unique up to homotopy. Further, if $G$ has the structure of a CW- complex, we can choose $BG$ to have one too.

The group (co)homology of $G$ is defined to be the (co)homology of $BG$. From the long-exact sequence associated to the fibre map, $p$, we know that ${\pi }_n(G)={\pi }_{n+1}(BG)$ for $n\ge 0$. In particular the fundamental group of $BG$ is ${\pi }_0(G)$, which inherits a group structure as a quotient of $G$. Let $H$ denote ${\pi }_0(G)$. Then $H$ acts freely on the cells of $B{G}^{*}$, the universal over of $BG$. Hence the cellular resolution for $B{G}^{*}$, denoted, $C_{*}(B{G}^{*})$, is a sequence of free $ZH$- modules and $ZH$- linear maps. Taking coefficients in some $ZH$- module $A$, we have

$$H^n(G;A)=H^n(C_{*}(B{G}^{*});A)\mathrm{and}{\mathrm{H}}_{\mathrm{n}}(\mathrm{G};\mathrm{A})={\mathrm{H}}_{\mathrm{n}}({\mathrm{C}}_{*}({\mathrm{BG}}^{*});\mathrm{A})$$

In particular, when $G$ is discrete, $p$ must be the covering map associated to a universal cover. Hence $X=B{G}^{*}$ and $C_{*}(B{G}^{*})$ is exact, as $X$ is contractible and hence has trivial homology. Note in this case $H=G$. So for a discrete group $G$, we have,

$$H^n(G;A)=Ex{t}_{ZG}^n(Z,A)\mathrm{and}{\mathrm{H}}_{\mathrm{n}}(\mathrm{G};\mathrm{A})={\mathrm{Tor}}_{\mathrm{ZG}}^{\mathrm{n}}(\mathrm{Z},\mathrm{A})$$

Also, as passing to the universal cover preserves ${\pi }_n$ for $n>1$, we know that ${\pi }_n(BG)=0$ for $n>1$. $BG$ is always connected and for a discrete group ${\pi }_0(G)=G$ so we have $BG=k(G,1)$, the Eilenberg - Maclane space.

As an example take $G=S{U}_1$. Note topologically, $S{U}_1=S^1=k(Z,1)$. As ${\pi }_n(G)={\pi }_{n+1}(BG)$ for $n\ge 0$, we know that $BS{U}_1=k(Z,2)=C{P}^{\mathrm{\infty }}$.

More explicitly, we may identify $S{U}_1$ with the unit complex numbers. This acts freely on the infinite complex sphere (which is contractible) leaving a quotient of $C{P}^{\mathrm{\infty }}$.

Hence $H^n(S{U}_1,Z)=Z$ if $2$ divides $n$ and $0$ otherwise.

Similiarly $B{C}_2=R{P}^{\mathrm{\infty }}$ and $BS{U}_2=H{P}^{\mathrm{\infty }}$, as $C_2$ and $S{U}_2$ are isomorphic to U(R) and U(H) respectively. So $H^n(C_2,Z_2)=Z_2$ for all $n$ and $H^n(S{U}_2,Z)=Z$ if $4$ divides $n$ and $0$ otherwise.