An introduction to Lie algebra cohomology/saras/lecture 3
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G-modules
Let \(G\) be a group with identity \(e\ ;\)
definition
A \(G\)-module \(A\) is an abelian group \(A\ ,\) equipped with an action by \(G\) (a left action), i.e. a map \[ G\times A\to A \] \[ (g,a)\mapsto ga\] with the following properties: \[ea=a, \] \[(gh)a=g(ha), \] \[g(a+b)=ga+gb\,, \quad\forall h\,,\quad g\in G\,, \quad\forall a,\,\,b\in A\,.\] Recall that giving an additive group action is the same as giving a group homomorphism \[G\to Aut(A)\,\] where \(Aut(A)\) is the group of automorphisms of \(A\ .\)
\(G\)-Homomorphisms
definition
Let \(A\) and \(B\) be two \(G\)-modules. The homomorphism \[f:\,\,\,A\to B\] is called \(G\)-homomorphism if \[f(ga)=g\,f(a),\quad \forall g\in G\,,\,\,a\in A.\] A \(G\)-homomorphism is a homomorphism of abelian groups commuting with the action of \(G\ .\)
remark
If we regard \(A\) and \(B\) simply as abelian groups, that is as \(\mathbb{Z}\)-modules, then we refer to \(f\) as to \(\mathbb{Z}\)-modules and \(\mathbb{Z}\)-homomorphisms to distinguish them from \(G\)-modules and \(G\)-homomorphisms. Given two \(G\)-modules \(A\) and \(B\) we can construct two more modules: \[\tag{1} \mathrm{ the\ }G-\mathrm{module\ } Hom(A,B)\]
and
\[\tag{2} \mathrm{the\ }G-\mathrm{module\ }A\otimes B\ .\]
The \(G\)-module \(Hom(A,B)\)
exercise
Verify that \(Hom(A,B)\) is indeed a \(G\)-module.
remark
The group of all \(G\)-homomorphisms from \(A\) to \(B\ ,\) \(Hom_{G}(A,B)\ ,\) is a subgroup of \(Hom(A,B)\ .\)
lemma
\[Hom_{G}(A,B)=Hom(A,B)^{G}\,.\] Proof. Let \(f\in Hom_{G}(A,B)\ ;\) that is \(g(f)(a)=gf(g^{-1}a)\) and \(f(ga)=gf(a)\) (by definition of \(G\)-homomorphism), so we have \[g(f)(a)=gg^{-1}f(a)=f(a)\] i.e. \(f\in Hom(A,B)^{G}\ .\) On the other hand, suppose that \(f\in Hom(A,B)^{G}\ ;\) then \(g(f)(a)=f(a)\) for all \(g\in G\ .\) Or \(g(f)(a)=gf(g^{-1}a)=f(a)\ .\) This implies \(gf(g^{-1}a)=f(a)\ ,\) that is \(g^{-1}f(a)=f(g^{-1}a)\) and so \(f\) is a \(G\)-homomorphism.
exercise
Show that
- [1] \(Hom_{G}(\mathbb{Z}[G],A)\simeq A\) (as groups);
- [2] \(Hom(\mathbb{Z},A)\simeq A\) (as \(G\)-modules).
Let \(A\) and \(B\) be \(G\)-modules and let \[l\;\;:\;\; A\to A'\,,\quad m\;\;:\;\; B\to B'\] be \(G\)-homomorphisms. Then they induce the \(G\)-homomorphisms \[Hom(A',B)\to Hom(A,B)\,,\quad f\mapsto f\circ l\] and \[Hom(A,B)\to Hom(A,B')\,,\quad f\mapsto m\circ f\,.\] For this reason \(Hom\) is a \emph{contravariant} functor in the first entry and a \emph{covariant }functor in the second one. Obvuiously \[Hom(A,B)\to Hom(A',B')\,,\quad f\mapsto m\circ f\circ l\,. \]
