# 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 $p:X\rightarrow X/G$ 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 $n\geq 0$. 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 $BG^{*}$, the universal over of $BG$. Hence the cellular resolution for $BG^{*}$, denoted, $C_{*}(BG^{*})$, is a sequence of free $ZH$- modules and $ZH$- linear maps. Taking coefficients in some $ZH$- module $A$, we have

 $H^{n}(G;A)=H^{n}(C_{*}(BG^{*});A)\,\,\rm{and}\,\,H_{n}(G;A)=H_{n}(C_{*}(BG^{*}% );A)$

In particular, when $G$ is discrete, $p$ must be the covering map associated to a universal cover. Hence $X=BG^{*}$ and $C_{*}(BG^{*})$ 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,

 $H^{n}(G;A)=Ext^{n}_{ZG}(Z,A)\,\,\rm{and}\,\,H_{n}(G;A)=Tor^{n}_{ZG}(Z,A)$

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 $G=SU_{1}$. Note topologically, $SU_{1}=S^{1}=k(Z,1)$. As $\pi_{n}(G)=\pi_{n+1}(BG)$ for $n\geq 0$, we know that $BSU_{1}=k(Z,2)=CP^{\infty}$.

More explicitly, we may identify $SU_{1}$ with the unit complex numbers. This acts freely on the infinite complex sphere (which is contractible) leaving a quotient of $CP^{\infty}$.

Hence $H^{n}(SU_{1},Z)=Z$ if $2$ divides $n$ and $0$ otherwise.

Similiarly $BC_{2}=RP^{\infty}$ and $BSU_{2}=HP^{\infty}$, as $C_{2}$ and $SU_{2}$ are isomorphic to U(R) and U(H) respectively. So $H^{n}(C_{2},Z_{2})=Z_{2}$ for all $n$ and $H^{n}(SU_{2},Z)=Z$ if $4$ divides $n$ and $0$ otherwise.

Title group cohomology (topological definition) GroupCohomologytopologicalDefinition 2013-03-22 14:32:24 2013-03-22 14:32:24 whm22 (2009) whm22 (2009) 18 whm22 (2009) Definition msc 55N25 CohomologyGroupTheorem group cohomology classifying spaces