1 Introduction

What is a Lie Algebra?

Lie algebra is a vector space with a special operator, bracket operator, which is a non-commutative multiplication.

Definition

Suppose $K$ is a field with $\mathrm{char}K=0$. A Lie algebra over $K$ is a $K$-vector space $g$ together with a bracket operation

$$\begin{array}{rl}\left[,\right]:g\times g& \to g,\\ (X,Y)& \mapsto [X,Y]\end{array}$$

satisfying the following properties:

• $[Y,X]=-[X,Y]$
• (Jacobi identity) $[X,[Y,Z]]+[Y,[Z,X]]+[Z,[X,Y]]=0$.

Conversely, if $\mathcal{A}$ is an associative algebra over a field $K$, then the operation $[X,Y]=XY-YX$ turns $\mathcal{A}$ into a Lie algebra. Note that the bracket is trivial when $\mathcal{A}$ is Abel, so we can say Lie algebra describes where the abelianization fails.

Some Trivial Definitions

homomorphism

For two Lie algebras $g_1,g_2$, a homomorphism $\rho :g_1\to g_2$ is a $k$-linear map satisfying $\rho ([X,Y])=[\rho (X),\rho (Y)]$.

Remark. Note that an algebra does not always have multiplication. So homomorphism does NOT always preserve multiplication. For example, consider the adjoint representation.

ideal

An ideal of a Lie algebra $g$ is a subset $I\subseteq g$ such that $[X,Y]\in I$ for any $X\in I,Y\in g$.

Lie subalgebra

$h\subseteq g$ is a Lie subalgebra if $h$ is closed under the bracket operation.

quotient

In $g/I$, define $[\overline{X},\overline{Y}]:=\overline{[X,Y]}$. Note that $g\mapsto g/I$ is a Lie algebra homomorphism.

representation

A $K$-linear representation of a Lie algebra $g$ is a $K$-vector space $V$ together with a Lie algebra homomorphism:

$$\rho :g\to \mathrm{End}(V).$$

$$\begin{array}{rl}\mathrm{ad}:g& \to \mathrm{End}(g),\\ x& \mapsto \mathrm{ad}X:Y\mapsto [X,Y]\end{array}$$

is a Lie algebra representation.

indecomposable

$\rho$ is called indecomposable if $V$ can not be written as the direct sum of two proper invariant subspaces.

irreducible

$\rho$ is called irreducible if it does not admit non-trivial invariant subspaces.

Abelian, Solvable and Nilpotent

abelian, solvable, nilpotent

Suppose $g$ is a Lie algebra over $K$.

• It is abelian if $[\cdot ,\cdot ]=0$.
• It is solvable if

$$\begin{array}{rl}{g}^{(0)}& =g,g^{(1)}=[g,g]=\{[X,Y]:X,Y\in g\},\cdots ,\\ g^{(n)}& :=[g^{(n-1)},g^{(n-1)}]=0.\end{array}$$

• It is nilpotent if the following series descend to zero $g^1=g,g^2=[g,g],\cdots ,g^n=[g,g^{n-1}]$.

Some Examples

For example,

$$B_n(K)=\left\{\left[\begin{array}{ccc}\ast & \ast & \ast \\ & \ddots & \ast \\ 0& & \ast \end{array}\right]\right\}$$

is solvable, and

$$N_n(K)=\left\{\left[\begin{array}{ccc}0& \ast & \ast \\ & \ddots & \ast \\ 0& & 0\end{array}\right]\right\}$$

is nilpotent.

Semi-simple: Defined by Representations

semi-simple

Suppose $g$ is a finite dimensional Lie algebra over $\mathbb{C}$ and $(\rho ,V)$ is a finite dimensional representation of $g$ over $\mathbb{C}$. Then:

• $\rho$ is called completely reducible if for any $g$-invariant subspace $U$ of $V$, there exists a $g$-invariant subspace $U^{\prime }$ such that $V=U\oplus U^{\prime }$.
• If a finite dimensional Lie algebra over $\mathbb{C}$ with the property that every finite dimensional representation of $g$ over $\mathbb{C}$ is completely reducible, such a Lie algebra is called semi-simple.

Note that there exists non-completely-reducible representation:

Define $L=\left[\begin{array}{ccc}\lambda & 1& 0\\ & \lambda & 1\\ & & \lambda \end{array}\right]$. Then any representation generated by $L$ is not completely reducible.

Remark.

• There is another definition of semi-simple, that is, the direct sum of finitely many non-trivial finite dimensional simple Lie algebra. See here.
• Non-completely-reducible representation roughly comes from the solvable property, as $g$ semi-simple iff $\mathrm{rad}g=0$, where $\mathrm{rad}g$ is the maximal solvable ideal. In addition, representation of solvable parts can be written as upper triangular matrix, and “upper triangular matrix” is roughly non-completely-reducible, as the above example shows.

$\lambda$-weight Space and Generalized $\lambda$-weight Space

weight

Suppose $g$ is a finite dimensional Lie algebra over $\mathbb{C}$ and $\rho :g\to \mathrm{End}(V)$ is a finite dimensional complex representation. For a $1$-dimensional representation $\lambda :g\to \mathbb{C}$, the $\lambda$-weight space is defined to be

$$V_{\lambda }=\{v\in V:\rho (x)v=\lambda (x)v,\forall x\in g\}.$$

When $V_{\lambda }\ne 0$, we call $\lambda$ a weight of $\rho$ and $V_{\lambda }$ the according weight space.

Lemma

Since any representation of a solvable Lie algebra has a weight, we have the following properties:

• One dimensional representations of a finite dimensional Lie algebra $g$ are just linear maps $g\to \mathbb{C}$ that vanish on $[g,g]$. Hence, when $[g,g]⊊g$ is solvable, $g$ has a non-trivial $1$-dimension representation.
• If $g$ is a finite dimensional solvable Lie algebra over $\mathbb{C}$, then every finite dimensional representation of $G$ over $\mathbb{C}$ has a $1$-dim sub-representation.

generalized weight

Then generalized $\lambda$-weight is defined to be

$${\tilde{V}}_{\lambda }=\{v\in V:(\rho (X)-\lambda (X){)}^{n_{x,v}}v=0\}$$

where $n_{x,v}$ depends on $X$ and $V$.