An introduction to Lie algebra cohomology/saras/lecture 4

The third lecture

Lifting the representation to the forms

Let $A$ be a $G$-module. In order to give a general definition of a coboundary operator $d^n \ ,$ one defines first an induced representation on $C^n (G,A)$ as follows.

definition

Let $\mathfrak{a}$ be an $\mathfrak{l}$-module. In order to give a general definition of a coboundary operator $d^n \ ,$ one defines first an induced representation on $C^n (G,A)$ as follows. Let, for $h\in G\ ,$

$$(d^{(n)}(h)a)(g_1,\cdots,g_n)=a(g_1,\cdots,g_n h)-a(g_1,\cdots,g_{n-1}, h)$$ This is indeed a representation. Let $h,k\in G\ .$ Then $$d^{(n)}(h)d^{(k)}(z)a(g_1,\cdots,g_n)=$$ $$=d^{(n)}(k)a(g_1,\cdots,g_n k)-d^{(n)}(k)a(g_1,\cdots,g_{n-1}, k)=$$ $$=a(g_1,\cdots,g_n kh)-a(g_1,\cdots,g_{n-1},h)-a(g_1,\cdots,g_{n-1} kh)+a(g_1,\cdots,g_{n-1},h)$$ $$=a(g_1,\cdots,g_n kh)-a(g_1,\cdots,g_{n-1} kh)=(d^{(n)}(hk)a)(g_1,\cdots,g_n).$$ By definition $$d^{(0)}(h)a(g_1,\cdots,g_n)=ha(g_1,\cdots,g_n).$$ We can then write the coboundary operators as follows $$(d^{1}b)(g_1,g_2)=gd^{(k)}(z)a(g,h)=ga(h)-d^{(1)}(h)a(g)-a(h)+a(g)=$$ $$=(ga(h)-a(h))-(d^{(1)}(h)a(g)-a(g))=$$ $$=d^{(0)}(g)a(h)-(d^{(1)}(h)a(g)-a(g))=d^{(0)}(g)a(h)-\Lambda^{1} (h)a(g),$$ where for convenience we define $$\Lambda^{n} (h)$$

example

Trivial module. Every abelian group $A$ can be viewed as a $G$-module defining the (trivial) action $ga=a\ ,$ $\forall g\in G\ .$ Such a $G$-module is called trivial $G$-module.

example

Let $A=\mathbb{Z}$ and let $G=\mathbb{Z}_2=\{e,g\}\ ,$ $g^2=e\ ,$ then defining the map $gn=-n$ $A$ becomes a $G$-module (nontrivial).

remark

Any abelian group is a $\mathbb{Z}$-module.

remark

Any $G$-module is a $\mathbb{Z}$-module as well. More examples:

example

Let $A=\{e,a,b,ab\}$ be the Klein's (Viergruppe) group, let $G=\mathbb{Z}_2$ and let $\phi(g)$ be the homomorphism from $G$ to $Aut(A)$ which exchanges $a$ and $b$ and keeps $ab$ fixed. This makes $A$ into a $G$-module.

example

Let $G$ be a finite group. Let $\mathcal{A}=\mathbb{C}(\lambda)$ be the field of rational functions in $\lambda\ .$ Suppose we have a fixed representation $\sigma: G \rightarrow PSL(2,\mathbb{C})\ .$ Then this induces a representation of $G$ in $\mathcal{A}$ by identifying $PSL(2,\mathbb{C})$ with the fractional linear transformations of $\mathbb{C}\ .$ This makes $\mathcal{A}$ into a $G$-module. Let, for instance, $G=\mathbb{Z}_2=\{e,g\}\ ,$ $g^2=e\ ,$ and identify the group with the group of fractional-linear transformation in the complex $\lambda$-plane generated by $\sigma^g(\lambda)=\lambda^{-1}\ .$ This makes $A$ into a $G$-module; indeed $(1)$ and $(2)$ are obvious while $(3)$ can be verified as follows

$$\sigma^g(f_1 (\lambda)+f_2 (\lambda))=f_1 (\lambda^{-1})+f_2 (\lambda^{-1})= \sigma^g(f_1 (\lambda))+ \sigma^g(f_2 (\lambda^{-1}))\,.$$ In what follows we will consider $G$ to be a finite group of order $|G|=N\ .$

definition

The group Ring $\mathbb{Z}[G]$ is the ring whose additive group is the abelian group of all formal sums

$$\left\{ \sum_{g\in G}n_{g}\,g\,\,n_{g}\in \mathbb{Z},\,g\in G\right\}\,,$$ and whose multiplication operation is defined by the multiplication in $G\ ,$ extended $\mathbb{Z}$-linearly to $\mathbb{Z}[G]\ .$ Two such formal sums $\sum_{g\in G}n_{g}\,g$ and $\sum_{g\in G}m_{g}\,g$ are equal iff $n_{g}=m_{g}\ .$ Addition in $\mathbb{Z}[G]$ is componentwise, while, according to the definition, multiplication is determined in the natural way by the multiplication in $G\ ;$ more in detail $$\sum_{g\in G}n_{g}\,g+\sum_{g\in G}m_{g}\,g=\sum_{g\in G}(n_{g}+m_{g})\,g\,,$$ $$\left(\sum_{g\in G}n_{g}\,g\right)\left(\sum_{g\in G}m_{g}\,g\right)=\sum_{g\in G}\left(\sum_{\tau \rho\in G}n_{\tau}m_{\rho}\right)\,g=\sum_{g\in G}\left(\sum_{\tau\in G}n_{\tau}m_{\tau^{-1}g}\right)\,g\,.$$ Sometimes the group Ring $\mathbb{Z}[G]$ is identified with its additive group, that is the abelian group of formal integer linear combinations of elements of $G\ .$ We may view $G$ as imbedded in $\mathbb{Z}[G]$ under the identification of $g\in G$ with $1g\in \mathbb{Z}[G]\ .$

remark

A $G$-module structure on $A$ is equivalent to a $\mathbb{Z}[G]$-module structure via

$$\mathbb{Z}[G]\rightarrow Aut(A)$$ $$\left(\sum_{g\in G}n_{g}\,g\right)a=\sum_{g\in G}n_{g}\,ga,\,\,$$ $$\forall a\in A.$$ As abelian group, $\mathbb{Z}[G]$ is itself a $G$-module.

The $G$-module $X_q$

For each $q\ge 1$ we define a $q$-cell, this is a set

$$\{g_{1},g_{2},\ldots,g_{q}\,\,|\,\,g_{i}\in G\}\,.$$ $g_{1},g_{2},\ldots,g_{q}$ are elements of the group $G\ .$ We use the $q$-cells as free generators of our $G$-modules $X_{q}\ ,$ we namely define $$X_{q}$$ $$=\sum_{g_{1},\ldots,g_{q}\in G}\oplus\mathbb{Z}[G](g_{1},g_{2},\ldots,g_{q}).$$ For $q=0$ we set $$X_{0}=$$ $$\mathbb{Z}[G],$$ considering the $0$-cell generated by $e\in \mathbb{Z}[G]\ .$ By construction the modules $$X_{0},\,\,X_{1},X_{-2},\ldots$$ are $G$-free, where

definition

A $G$-module $A$ is $G$-free (or $\mathbb{Z}[G]$-free) if it is the direct sum of $G$-modules isomorphic to $\mathbb{Z}[G]$

$$A=\sum_{i}\bigoplus\Gamma_{i}, \quad\textrm{with}\,\,\,\Gamma_{i}\simeq \mathbb{Z}[G].$$

example

Let $G=\mathbb{Z}_{2}=\{e,g\}\ ;$ then

$$X_{1}=X_{-2}=\sum_{g\in G}\oplus\mathbb{Z}[G](g)=\mathbb{Z}[G]\oplus\mathbb{Z}[G](g).$$ $$X_{2}=X_{-3}=\sum_{g_{1},g_{2}\in G}\oplus\mathbb{Z}[G](g_{1},g_{2})=$$ $$\mathbb{Z}[G](e,e)\oplus\mathbb{Z}[G](e,g)\oplus\mathbb{Z}[G](g,e)\oplus\mathbb{Z}[G](g,g).$$ Let us now define the $G$-homomorphisms $d_{q}\,:\;X_{q+1}\to X_{q}\ ;$ since $Z_q$ are $G$-free, to define the $G$-homomorphisms $d_{q}$ it is sufficient to specify the value they take on the generators $(g_{1},g_{2},\ldots,g_{q})\ .$ We set $$q=0$$ $d_{0}\, e=N_{G}=\sum_{g\in G}g;$ $$q=1$$ $d_{1}\,(g_{1})=g_{1}-e;$ $$q>1$$ $d_{q}\,(g_{1},g_{2},\ldots,g_{q}) = g_{1}(g_{2},\ldots,g_{q})+\sum_{i=1}^{q-1}(-1)^{i}(g_{1},\ldots,g_{i-1},g_{i}g_{i+1},g_{i+2},\ldots,g_{q})+\ :$

$$+(-1)^{q}(g_{1},g_{2},\ldots,g_{q-1});$$

Let $A$ be a $G$-module, we define

$$A_{q}=Hom_{G}(X_{q},A).$$

definition

The elements of $A_{q}$ are called \textbf{$q$-cochains} of A. They are the $G$-homomorphisms

$$x\,\,:\,\,\,X_{q}\to A$$ Recall that the cochains group $A_{q}=A_{-q-1}=Hom_{G}(X_{q},A)\ ,$ $q\ge 1$ is the group of all $G$-homomorphisms $x\,\,:\,\,X_{q}\to A\ .$ Recall also that, by definition, $X_{q}$ is free generated by the $q$-cells $(g_{1},\ldots,g_{q})\ ,$ $g_{i}\in G\ .$ So one can uniquely specify the $G$-homomorphism $x$ through its value on the $q$-cells $(g_{1},\ldots,g_{q})\ .$ Each cochain can be interpret as a function $$x\,\,:\,\,\underbrace{G\times G\times\ldots\times G}_{q-times}\,\to\, A$$ so that we can identify $$A_{q}=$$ $$\{x\,\,:\,\,\underbrace{G\times G\times\ldots\times G}_{q-times}\,\to\,A\},\quad q\ge 1$$ with $$A_{0}=Hom_{G}(\mathbb{Z}[G],A)\simeq A.$$ From the definition of $G$-homomorphisms $d_{q}$ in the standard complex we obtain the following for $\partial_{q}$

itemize

$$q=0$$ $(\partial_{0} x)e=N_{G}x=\sum_{g\in G}g x,\,\,\,\textrm{for}\,\,\, x\in A_{-1}=A;$ $$q=1$$ $(\partial_{1}x)(g_{1})=g_{1}x-ex,\,\,\,\textrm{for}\,\,\, x\in A_{0}=A;$ $$q\ge 1$$ $(\partial_{q}x)\,(g_{1},g_{2},\ldots,g_{q})=g_{1}x(g_{2},\ldots,g_{q})+\sum_{i=1}^{q-1}(-1)^{i}x(g_{1},\ldots,g_{i-1},g_{i}g_{i+1},g_{i+2},\ldots,g_{q})+\ :$

$$+(-1)^{q}x(g_{1},g_{2},\ldots,g_{q-1}),\,\,\,\textrm{for}\,\,\,x\in A_{q-1}$$

In this setting, we can define cocycles as maps

$$x\,\,:\,\,\underbrace{G\times G\times\ldots\times G}_{q-times}\,\to\, A$$ $$\,\,\,\,\textrm{s.t.}\,\,\,\,\partial_{q+1}x=0.$$ Coboundaries are those cocycles for which $$\exists\,\, y\in A_{q-1}\,\,\,\,\textrm{s.t.}\,\,\,\,x=\partial_{q} y.$$

The fourth lecture