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A vector space is a combination of a set of vectors V and the binary function of the addition of vectors, along with the additive identity and the existence of an additive inverse for each element of V. Furthermore there is a set of scalars belonging to a field such that for any scalar λ and any vector v, λv belongs to V. A Lie Algebra is a vector space L with a binary function [ , ] (called the Lie bracket) which satisfies the following conditions:
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 {B_{i}} a basis for the vector space V of the Lie algebra. Any vector in V including the vectors such as [B_{i},B_{j}] can be represented as a linear combination of the elements of the basis. Thus
The constants C_{i,j}^{k} 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
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
A set S of V is a subalgebra of V if [s_{1},s_{2}] belongs to S for all s_{1} and s_{2} 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.
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 (XY) belongs to S; i.e. XY=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+s_{1}: s_{1} in S} + {Y+s_{2}: s_{2} in S} because for any s_{1} and s_{2} in S, s_{1}+s_{2} is in S.
Now consider the equivalence class of [X,Y]. By the linearity of the Lie bracket
Since S is an ideal subalgebra, [X,s_{2}] belongs to S, [s_{1},Y] belongs to S and [s_{1},s_{}] belongs to S so their sum belongs to S. Thus [X+s_{1},Y+s_{2}] 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.)
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