# Group Cohomology (Thursday, August 26) > See Cassels-Frohlich, Stein, etc for group cohomology. ## Computing Examples :::{.example} For $G=\ZZ$, take the resolution \[ 0 \to \ZZ[x, x\inv] \mapsvia{x-1} \ZZ[x, x\inv] \to 0 .\] Then $H_*(G; \ZZ)= H^*(G;\ZZ)$ is $\ZZ$ in degrees 0 and 1, and 0 otherwise. For $M\in \mods{G}$, we have \[ H^*(G; M) = H^*(M \mapsvia{x-1} M) = \begin{cases} M^G & *=0 \\ M_G & * = 1 \\ 0 & \text{else}, \end{cases} \\ H_*(G; M) = H_*(M \mapsvia{x-1} M) = \begin{cases} M_G & *=0 \\ M^G & * = 1 \\ 0 & \text{else}. \end{cases} \] ::: :::{.example title="?"} For $G = \ZZ/n$, write $\sigma$ as the generator so that $\ZZ[G] = \ZZ[\sigma] / \gens{\sigma^n - 1}$ We can take a resolution \[ \cdots \to \ZZ[\sigma]/ \gens{\sigma - 1} \mapsvia{\sigma - 1} \ZZ[\sigma]/ \gens{\sigma - 1} \mapsvia{1 + \sigma + \cdots + \sigma^{n-1}} \ZZ[\sigma]/ \gens{\sigma - 1} \mapsvia{\sigma - 1} \ZZ[\sigma]/ \gens{\sigma - 1} \to 0 .\] Now apply $\Hom_{\ZZ[G]}(\wait, \ZZ)$, use that $\Hom_{\ZZ[G]}(\ZZ[G], \ZZ) = \ZZ$, and take homology of the complex \[ \ZZ \mapsvia{\sigma - 1} \ZZ \mapsvia{\sum \sigma^i} \to \ZZ \mapsvia{\sigma - 1} \cdots \ZZ \mapsvia{0} \ZZ \mapsvia{n} \to \ZZ \mapsvia{0} \cdots .\] This yields \[ H^*(G; \ZZ) = \begin{cases} \ZZ & *=0 \\ 0 & * \text{ odd} \\ \ZZ/n & * \text{ even}. \end{cases} \] ::: :::{.remark} For the free abelian group $\ZZ^n$, we get $H^*(\ZZ^n; \ZZ) = \Extalgebra^* (\ZZ^n)$. For the free group $F_n$, we get $H^*(F_n; \ZZ)$ is $\ZZ$ in degree zero (always true for the trivial module, since the invariants are everything) and $\ZZ^n$ in degree 1. ::: :::{.fact} If $X$ is a CW complex with $\pi_0(X) = 0, \pi_1(X) = G, \pi_{>2}(X) = 0$, then $H^*_{\Grp}(G; \ZZ) = H^*_{\Sing}(X; \ZZ)$. Note that $X \mapsvia{\sim} \B G$ in this case, and the proof is easy: take the universal cover, then the simplicial/cellular cohomology resolves $\ZZ$ as a $\ZZ[G]\dash$module. ::: :::{.proposition title="?"} Suppose $G$ is finite and $M\in \mods{G}$, then $H^{>n}(G; M)$ is torsion. 1. It suffices to show this for $*=1$ by using dimension shifting. Choose $M \injects I$ into an injective object to get a SES \[ 0 \to M\to I \to M/I \to 0 \] to get a LES in cohomology, and use that $\Ext$ into injectives vanishes to get $H^*(G; M) \cong H^{*}(G; M/I)[-1]$. 2. We want to show $H^1(G; M) = \Ext_{\ZZ[G]}^1(\ZZ; M)$ is torsion, and it suffices to show $\Ext^1_{\ZZ[G]}(\ZZ; M)\tensor \QQ = 0$, which we can replace with $\Ext_{\ZZ[G]}^1(\QQ, M \tensor \QQ)$. So we consider SESs of the form \[ 0 \to M\tensor Q \to W \to \QQ ,\] which we'd like to split as a SES of $G\dash$representations over $\QQ$. See uniquely divisible groups? This splits by Maschke's theorem: all SESs of irreducible representations of $G$ for $G$ finite over $\ch k = 0$ split. The usual proof over $\CC$ doesn't work for $\QQ$, but one uses a splitting instead of an inner product. ::: ## Functoriality :::{.remark} Given $M\to N \in \mods{G}$ there are maps \[ H^*(G; M) &\to H^*(G; N) \\ H_*(G; M) &\to H_*(G; N) .\] Suppose $\iota: G\to T$ with $M\in \mods{T}$, then there are induced maps \[ \iota^*: H^*(T; M) &\to H^*(G; M) \\ \iota_*: H_*(T; M) &\to H_*(G; M) \] coming from the functoriality of Ext and Tor under change of rings. We'll use the following as a black box: for $G\leq T$ finite index, there is a *trace* map (or *corestriction*) \[ \trace_{G/T}: H^*(G; M) \to H^*(T; M) .\] It's functorial in $M$, and $\trace_{G/T} \circ \iota^*$ is multiplication by $m \da [G: T]$. This yields another proof of the previous element: take $G = 1$ to get $H^*(G; M) = 0$ and check $\trace_{G/T} \circ \iota_*$ is multiplication by $\abs{T}$ and zero, making the group torsion. ::: :::{.remark} Some interpretations: - $H_1(G; \ZZ) = G^\ab = G/[G, G]$ is the abelianization (which can still be torsion). - $H^1(G; \ZZ) = \Hom_{\Grp}(G;\ZZ)$, which is always torsionfree. - $H^2(G; M)$ classifies extensions of $G$ by $M$ in the following sense: $G'$ occurring in a "SES" $\xi: 0\to M\to G' \to G\to 1$ such that the action of $G$ on $M$ by conjugation is the given $G\dash$module structure on $M$. Moreover $\xi = 0$ in $H^2(G; M)$ iff $\xi$ splits, then $G' \cong G \semidirect M$. For $M$ a trivial $G\dash$module, these are *central extensions*. ::: :::{.warnings} Note all SESs yield semidirect products: take $0\to \ZZ \mapsvia{\cdot n} \ZZ \to \ZZ/n \to 0$, which has no sections since $\ZZ$ has no $n\dash$torsion. This in fact represents a generator $H^2(\ZZ/n; \ZZ)$. ::: :::{.definition title="Galois cohomology"} Let $L\slice{k}$ be a finite Galois extension, $M\in \mods{G}$ for $G\da \Gal(L\slice{k})$. Then \[ H^*_\Gal(L\slice{k}; M) \da H^*_\Grp(G; M) .\] If $M$ is a discrete continuous $\Gal(k^s/K)\dash$module, then \[ H^i(k; M) \da \colim_{U \normal \Gal(k^s/k)} H^*( \Gal(k^s/k) / U; M) .\] ::: > The stabilizer of any point is open (and finite index). :::{.definition title="Brauer Groups"} \[ \Br(k) = H^2(K; (k^s )\units) .\] ::: :::{.example title="?"} Consider $\Br(\FF_q)$, then $\Gal(\FF_q^s/\FF_q) = \hat{\ZZ} \gens{\Frob_q}$. Then \[ \Br(\FF_q) &\da H^2\qty{\hat\ZZ \gens{\Frob_q}; \bar \FF_q \units} \\ &=\colim_{U_n \subseteq \hat \ZZ \to \hat \ZZ \to \ZZ/n} H^2\qty{ \ZZ/n; (\bar \FF_q\units)^{U_n}}\\ &=\colim H^2\qty{ \ZZ/n \gens{\Frob_q}; \bar \FF_{q^n}\units } \\ &=\colim H^2( \Gal(\FF_{q^n} / \FF_q); \bar \FF_{q^n}\units ) \\ &= \colim H^2 \qty{ \FF_{q^n}\units \mapsvia{\Frob - 1} \FF_{q^n}\units \mapsvia{\Norm} \FF_{q^n}\units \to \cdots } \\ &= \colim \FF_q\units / \Norm(\FF_{q^n}, \FF_q) \FF_{q^n}\units \\ &= \colim 0 \\ &= 0 .\] Note: we've used that \[ \ker (\Frob - 1: x\mapsto x^{q-1}) = \FF_q\units .\] ::: :::{.exercise title="?"} Show that the norm is surjective. :::