# Jan A. Sanders

Curator of ScholarpediaCurator Index: 1

## Articles sponsored or reviewed

Under construction.

## 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. To make things more interesting, the theory is developed for Leibniz algebras, which are not so well known. This approach has the advantage of simplifying things because there is less choice. Of course, at some point, this implies that some creativity is needed to do the necessary generalizations.

## definition of Lie (and Leibniz) algebra

A Leibniz 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:

If, moreover, one has that

then we say that $$\mathfrak{g}$$ is a Lie algebra. Lie algebras have been extensively studied for more than a century, Leibniz algebras are a more recent invention and much less is known about them.

### corollary

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

### 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!).

### 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.

### example of a Leibniz algebra

Let $$\mathcal{A}$$ be an associative algebra. Let $$P$$ be a projector in $$\mathrm{End}(\mathcal{A})$$, that is, $$P^2=P$$. Suppose that $$P(aP(b))=P(a)P(b)$$ and $$P(P(a)b)=P(a)P(b)$$. Denote $$Px$$ by $$\bar{x}$$. Define $[x,y]=\bar{x}y-y\bar{x}$ This provides $$\mathcal{A}$$ with a Leibniz algebra structure: $[[x,y],z]-[x,[y,z]]+[y,[x,z]]=\overline{[x,y]}z-z\overline{[x,y]}-\bar{x}[y,z]+[y,z]\bar{x}+\bar{y}[x,z]-[x,z]\bar{y}$

$=\overline{(\bar{x}y-y\bar{x})}z-z\overline{(\bar{x}y-y\bar{x})} -\bar{x}(\bar{y}z-z\bar{y})+(\bar{y}z-z\bar{y})\bar{x} +\bar{y}(\bar{x}z-z\bar{x})-(\bar{x}z-z\bar{x})\bar{y}$
$=(\bar{x}\bar{y}-\bar{y}\bar{x})z-z(\bar{x}\bar{y}-\bar{y}\bar{x}) -\bar{x}\bar{y}z+\bar{x}z\bar{y}+\bar{y}z\bar{x}-z\bar{y}\bar{x} +\bar{y}\bar{x}z-\bar{y}z\bar{x}-\bar{x}z\bar{y}+z\bar{x}\bar{y}$
$=(\bar{x}\bar{y}-\bar{y}\bar{x})z-z(\bar{x}\bar{y}-\bar{y}\bar{x}) -\bar{x}\bar{y}z-z\bar{y}\bar{x} +\bar{y}\bar{x}z+z\bar{x}\bar{y}$
$=0$

When $$P$$ is the identity on $$\mathcal{A}$$, then one has a Lie algebra.

An example of this is the following. Let $$\mathcal{A}$$ consist of formal power series $$a(z)=\sum_{i\in\mathbb{Z}}a_i z^i$$ and let $$(P a)(z)=a_0$$.

### 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 (Leibniz) algebra morphism.

### 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.

### super remark

A super Leibniz 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 $[[x,y],z]=[x,[y,z]]-(-1)^{|x||y|}[y,[x,z]]$ Observe that $$\mathfrak{g}^0$$ itself is a Leibniz algebra.

Since asymmetry is not assumed in a Leibniz algebra, the order of the elements in an expression cannot be changed around. This makes it a rather trivial exercise to check that the theory to be developed below immediately applies to the super case. For instance, in the corollary above, we just have to keep track of to interchange in the order to obtain $[[x,y]+(-1)^{|x||y|}[y,x],z]=0$ A super Lie algebra is a super Leibniz algebra with $[x,y]=-(-1)^{|x||y|}[y,x],\quad x\in \mathfrak{g}^{|x|},y\in\mathfrak{g}^{|y|}$ Observe that $$\mathfrak{g}^0$$ itself is a Lie algebra.

## 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

### 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 $$\mathrm{ad}(x)y$$.

### 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.

## representations of Leibniz algebras

The definition of a Leibniz algebra representation is best motivated by the construction in the second lecture. The idea is to form a new Leibniz algebra given a Leibniz algebra $$\mathfrak{g}$$ and a module $$\mathfrak{a}$$ as follows. One considers the direct sum (as $$R$$-modules) $$\mathfrak{a}\oplus_R \mathfrak{g}$$ and one requires the Jacobi identity to hold: $[[a_1+x_1,a_2+x_2],a_3+x_3]=[a_1+x_1,[a_2+x_2,a_3+x_3]]-[a_2+x_2,[a_1+x_1,a_3+x_3]],\quad a_i\in\mathfrak{a}, x_i\in\mathfrak{g},\quad i=1,2,3$ One defines $d_+^{(0)}(x)a=[x,a]$ $d_-^{(0)}(x)a=-[a,x]$ Require $$[\mathfrak{a},\mathfrak{a}]=0$$, $$[\mathfrak{a},\mathfrak{g}]\subset\mathfrak{a}$$ and $$[\mathfrak{g},\mathfrak{a}]\subset\mathfrak{a}$$. Then this leads to the following definition.

### definition

If $$d_\pm^{(0)}$$ obeys the following three axioms

then it is called a Leibniz algebra representation Notice that the two conditions in (<ref>Representation1</ref>) give rise to compatibily conditions

### definition

If there is only one representation $$d^{(0)}=d_+^{(0)}=d_-^{(0)}$$, obeying

one speaks of an even representation.

### remark

In the case of a Lie algebra, one simply has $$d_{\pm}^{(0)}=d^{(0)}$$ (One could also take $$d_+^{(0)}=d^{(0)}$$ and $$d_-^{(0)}=0$$; this would however have the later disadvantage that the coboundary operator would not carry antisymmetric forms to antisymmetric forms).$$\quad\square$$

### 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 $$\mathrm{ad}_+(x)y$$ or $$-\mathrm{ad}_-(y)x$$.

### final super remark

In the super case this would have to be be changed to

## 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

<math coboundary0>d^0 a^0 (x)=d_-^{(0)}(x)a^0[/itex].

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

<math coboundary1>d^1 a^1(x,y)=d_+^{(0)}(x)a^1(y)-d_-^{(0)}(y)a^1(x)-a^1([x,y])[/itex].

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

<math coboundary2>d^2 a^2(x,y,z)=d_+^{(0)}(x)a^2(y,z)-d_+^{(0)}(y)a^2(x,z)+d_-^{(0)}(z)a^2(x,y)-a^2([x,y],z)-a^2(y,[x,z])+a^2(x,[y,z])[/itex].

### remark

These definitions are motivated by the central extension problem in the second lecture.

### exercise

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