lecture 1

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< An introduction to Lie algebra cohomology

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Author: Dr. Jan A. Sanders, Vrije Universiteit Amsterdam
Author: Dr. Sara Lombardo, Vrije Universiteit Amsterdam

Under construction.

1 introduction
2 definition of Lie algebra
3 representations of Lie algebras
- 3.1 example of a representation
- 3.2 representation of
4 the coboundary operator
- 4.1 remark
- 4.2 exercise

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introduction

The plan is to give an introduction to Lie algebra cohomology that can be followed on different levels. The development of the cohomological theory will require nothing beyond the basic rules for Lie algebras and representations. The treatment is not quite standard, since the forms will not necessarily be antisymmetric.

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definition of Lie algebra

A Lie algebra $\mathfrak{g}$ is a module or vector space over a ring or a field R (think of $\mathbb{R}$ or $\mathbb{C}$ ) with a bilinear operation $[\cdot,\cdot]$ obeying the following rule:

(1)

$[[x,y],z]=[x,[y,z]]-[y,[x,z]],\quad x,y,z\in\mathfrak{g}$

and such that

(2)

$[x,y]+[y,x]=0,\quad x,y\in\mathfrak{g}$ .

Lie algebras have been extensively studied for more than a century.

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example class of a Lie algebra

Let $\mathcal{A}$ be an associative algebra, that is, $(xy)z=x(yz)$ for all $x,y,z\in\mathcal{A}$ (in other words, one can forget the brackets around the multiplication). Then define a bracket by

$[x,y]=xy-yx$

This defines a Lie algebra structure on $\mathcal{A}$ (Check!).

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the Lie algebra $\mathfrak{sl}_2$

Consider the triple $\langle M, N, H \rangle$ with commutation relations

$H=[M,N]\quad , [H,M]=2M,\quad [H,N]=-2N$

Checking the Jacobi identity is a lot of trivial work, which can be avoided by realizing the Lie algebra as an associative algebra.

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morphism

Let $\phi:\mathfrak{a}\rightarrow\mathfrak{b}$ be a linear map. If $\phi([x,y]_{\mathfrak{a}})=[\phi(x),\phi(y)]_{\mathfrak{b}}$ then $\phi$ is a Lie algebra morphism.

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linear forms

The space of $n$ -linear (linear in the $R$ -module structure) forms, with arguments in $\mathfrak{g}$ and values in $\mathfrak{a}$ , is denoted by $C^n(\mathfrak{g},\mathfrak{a})$ . Notice that these are not required to be antisymmetric, contrary to the common Lie algebra cohomology convention.

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super remark

A super Lie algebra is a module $\mathfrak{g}=\mathfrak{g}^0\oplus\mathfrak{g}^1$ and a bracket such that

$[\mathfrak{g}^i,\mathfrak{g}^j]\subset\mathfrak{g}^{i+j \mathrm{mod} 2}$

obeing, with $x\in\mathfrak{g}^{|x|}$ and $y\in\mathfrak{g}^{|y|}$ (where $|\cdot|:\mathfrak{g}^i\mapsto i$ ) and $z\in\mathfrak{g}$ , the super Jacobi identity

$[[x,y],z]=[x,[y,z]]-(-1)^{|x||y|}[y,[x,z]]$

and

$[x,y]=-(-1)^{|x||y|}[y,x],\quad x\in \mathfrak{g}^{|x|},y\in\mathfrak{g}^{|y|}$

Observe that $\mathfrak{g}^0$ itself is a Lie algebra. The abstract theory of super Lie algebras is follows the ordinary theory, with some extra administration. If one is careful not to change the order of the elements too much, one can always insert the necessary factors at the end of the computation. At some point, when the reasoning depends on the antisymmetry of the Lie bracket, one has to be careful again. In these lectures the super signs are not put in, and to do so is left to the reader.

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representations of Lie algebras

Let $\mathfrak{g}$ be a Lie algebra and $\mathfrak{a}$ be a module or a vector space. Then we say that $d^{(0)}:\mathfrak{g}\rightarrow End(\mathfrak{a})$ is a representation of $\mathfrak{g}$ in $\mathfrak{a}$ if

(3)

$d^{(0)}([x,y])=[d^{(0)}(x),d^{(0)}(y)]=d^{(0)}(x)d^{(0)}(y)-d^{(0)}(y)d^{(0)}(x),\quad x,y\in\mathfrak{g}$ .

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example of a representation

Take $\mathfrak{a}=\mathfrak{g}$ and $d^{(0)}(x)y=[x,y]$ . This is called the adjoint representation and written as $\mathrm{ad}(x)y$ .

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representation of $\mathfrak{sl}_2$

Let $\mathfrak{a}=\R^2$ . Take

$d^{(0)}(M)=\begin{bmatrix} 0&1\\0&0\end{bmatrix}, \quad d^{(0)}(N)=\begin{bmatrix} 0&0\\1&0\end{bmatrix} ,\quad d^{(0)}(H)=\begin{bmatrix} 1&0\\0&-1\end{bmatrix}$

Then $d^{(0)}([H,M])=[d^{(0)}(H),d^{(0)}(M)]$ , etc, that is, $d^{(0)}$ is a representation of $\mathfrak{sl}_2$ in $\R^2$ . Since $d^{(0)}(x_1 N + x_2 M +x_3 H)=0$ implies $x_1=x_2=x_3=0$ , one can now easily check the Jacobi identity for $\mathfrak{sl}_2$ , since it follows from the Jacobi identity in the case of an associative algebra.

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the coboundary operator

We now define the first instance of the coboundary operator $d^0$ : Let $a^0\in\mathfrak{a}=C^0(\mathfrak{g},\mathfrak{a})$ . Then define $d^0 a^0\in C^1(\mathfrak{g},\mathfrak{a})$ by

(4)

$d^0 a^0 (x)=d^{(0)}(x)a^0$ .

Thus $d^0 :C^0(\mathfrak{g},\mathfrak{a})\rightarrow C^1(\mathfrak{g},\mathfrak{a})$ . By itself, the zeroth order coboundary operator is not much fun. But there is more. Let $a^1\in C^1(\mathfrak{g},\mathfrak{a})$ . Then define $d^1 a^1\in C^2(\mathfrak{g},\mathfrak{a})$ by

(5)

$d^1 a^1(x,y)=d^{(0)}(x)a^1(y)-d^{(0)}(y)a^1(x)-a^1([x,y])$ .

Thus $d^1:C^1(\mathfrak{g},\mathfrak{a})\rightarrow C^2(\mathfrak{g},\mathfrak{a})$ . One checks that $d^1d^0=0$ :

$d^1d^0 a^0(x,y)=d^{(0)}(x)d^0a^0(y)-d^{(0)}(y)d^0a^0(x)-d^0a^0([x,y])= d^{(0)}(x)d^{(0)}(y)a^0-d^{(0)}(y)d^{(0)}(x)a^0-d^{(0)}([x,y])a^0= 0.$

In general, when one has defined $d^i:C^i(\mathfrak{g},\mathfrak{a})\rightarrow C^{i+1}(\mathfrak{g},\mathfrak{a})$ such that $d^{i+1}d^i=0$ , then one calls $d^\cdot$ a coboundary operator. To treat the example of central extensions one needs one more coboundary operator. Let $a^2\in C^2(\mathfrak{g},\mathfrak{a})$ be a two-form. Then define