﻿ Fundamentals of Lie Algebras
San José State University

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 Fundamentals of Lie Algebras

A Lie Algebra is a vector space V with a binary function [ , ] (called the Lie bracket) which satisfies the following conditions:

• For all X and Y belonging to V, {X,Y] belongs to V
• For all X, Y and Z in V, [X, Y+Z] = [X,Y] + [X,Z]
• For all λ belonging to field F over which the vector space V is defined, [X,λY] = λ[X,Y]
• For all X and Y belonging to V, [Y,X] = -[X,Y].
• For all X,Y,Z belonging to V,
[X,[Y,Z]] + [Y,[Z,X] + [Z,[X,Y]]
(Jacobi Identity)

An example of a Lie algebra is the vector space of 3D vectors over the field of real numbers (3D vectors with real number components) with the Lie bracket [ , ] being the vector cross product.

For any Lie bracket there is a set of structure constants defined as follows. Let {Bi} a basis for the vector space V of the Lie algebra. Any vector in V including the vectors such as [Bi,Bj] can be represented as a linear combination of the elements of the basis. Thus

#### [Bi,Bj] = ΣkCi,jkBk

The constants Ci,jk are the structure constants of the algebra. They provide an alternate way to define the Lie bracket.

The structure constants have to satisfy certain conditions. Since the Lie bracket is antisymmetric

#### Cj,ik = -Ci,jk

The Jacobi identity requires a more complex equation that the structure constants have to satisfy.

For the example of the Lie algebra of 3D real vectors with the vector cross product as the bracket the structure constants for the usual basis are

## Subalgebras

A set S of V is a subalgebra of V if [s1,s2] belongs to S for all s1 and s2 in S. S has to be a vector space in its own right which means it is closed with respect to the field operations of the vector space V. A subalgebra T of V is an ideal subalgebra if [t,v] belongs to T for all t in T and v in V. In other words, T is an ideal algebra if it is closed under right bracket operations on it by all the elements of V.

## Quotient Algebras With Respect to Ideal Subalgebras

Let S is be an ideal subalgebra of V. An element X of V is said to be equivalent to an element Y of V modulo S if (X-Y) belongs to S; i.e. X-Y=s for some s in S. This is usually denoted as X=Y(mod S). Equivalence modulo S is an equivalence relation because it is closed, symmetric and transitive as a result of S being a subalgebra. Under this equivalence relation V is partitioned into equivalence classes.

The equivalence class that an element X belongs to can be generated as {X+s: s in S}. The equivalence class of two elements of V, X and Y, is the equivalence class of their sum X+Y; i.e., {X+Y+s: s in S} = {X+s1: s1 in S} + {Y+s2: s2 in S} because for any s1 and s2 in S, s1+s2 is in S.

Now consider the equivalence class of [X,Y]. By the linearity of the Lie bracket

#### [X+s1,Y+s2] = [X,Y] + [X,s2] + [s1,Y] + [s1,s]

Since S is an ideal subalgebra, [X,s2] belongs to S, [s1,Y] belongs to S and [s1,s] belongs to S so their sum belongs to S. Thus [X+s1,Y+s2] is in the same equivalence class as [X,Y] so a Lie bracket for the equivalence classes is well defined and thus the equivalence classes form a Lie algebra. This Lie algebra of the equivalence classes of V modulo an ideal subalgebra S is denoted as V/S.

(To be continued.)