An introduction to Lie algebra cohomology/saras/lecture 1

Introduction

Let $G$ be a group with identity $e\ .$

Morphism (Group homomorphism)

Let $G\ ,$ $H$ be two groups; Let $\phi:G\rightarrow H$ be a map. If $\phi(g_1\cdot g_2)=\phi(g_1)\circ\phi(g_2)\ ,$ where $\cdot$ is the operation on G and $\circ$ is the operation on H, then $\phi$ is a group homorphism.

Linear forms

Let $A$ be a module. The space of $n$-linear forms, with arguments in $G$ and values in $A\ ,$ is denoted by $C^n(G,A)\ .$

Representations

Let $A$ be a module or a vector space; $d^{(0)}:G\rightarrow End(A)$ is a representation of $G$ in $A$ if

• $d^{(0)}(g_1 \cdot g_2)=d^{(0)}(g_1),d^{(0)}(g_2),\quad g_1, g_2\in G\ .$

The coboundary operator

We now define the coboundary operators $d^n\ :$ Let $a\in A=C^0(G,A)\ .$ Then define $d^0 a\in C^1(G,A)$ by

• $(d^0 a) (g)=ga-a\ ,$ $g\in G\ .$

Thus $d^0 :C^0(G,A)\rightarrow C^1(G,A)\ .$ Let $b\in C^1(G,A)\ .$ Then define $d^1 b\in C^2(G,A)$ by* $(d^1 b)(g_1,g_2)=g_1 b(g_2)-b(g_1 g_2)+b(g_1)\ .$ Thus $d^1:C^1(G,A)\rightarrow C^2(G,A)\ .$ One checks that $d^1d^0=0\ :$

$$d^1d^0 a(g_1,g_2)=g_1d^0 a(g_2)-d^0 a(g_1g_2)+d^0 a(g_1)=g_1g_2 a- g_1a-g_1g_2 a+a+g_1 a-a=0.$$ Let $c\in C^2(G,A)$ be a two-form. Then define $$\tag{1} d^2 c(g_1,g_2,g_3)=g_1c(g_2,g_3)-c(g_1g_2,g_3)+c(g_1,g_2g_3)-c(g_2,g_3)\ .$$

In general, when one has defined $d^i:C^i(G,A)\rightarrow C^{i+1}(G,A)$ such that $d^{i+1}d^i=0\ ,$ then one calls $d^\cdot$ a coboundary operator.

Exercise

Show that $d^2 d^1=0\ .$

To the second lecture