MHB What Are the Basic Properties of This Lie Algebra?

[topsquark]

Warning: This is going to be a bit long.

(Apparently my post was too long so it wouldn't render at all. I've split this into two threads.)

I worked out some basic Algebraic properties of a Lie Algebra. This is similar to my previous thread about SU(2) but as I don't know this example I'm going ask someone to look it over for me. I found the example in my text but it doesn't list a name for the Algebra so I couldn't look it up on the web. Some of this I'm confident of and some I have questions. For ease of referencing I'm going to highlight the areas where I have questions by putting a number (1) on it. In what follows I'm going to refer to a Lie Algebra simply as Algebra.

I have a vector space with the basis I, U, V, W. There isn't much to say about the vector space because I have little knowledge about what U, V, and W are. I'm simply going to say that the most general member of the vector space can be written as [math]g = aU + bV + cW[/math]. (I'm ignoring the identity as it doesn't really affect anything about the discussion.) a, b, and c belong to some field, which we might as well call the real numbers. I don't know of any property discussed that would require a specific field to be named.

The Algebra is defined by the following Lie brackets: [U, W] = [V, W] = 0, [U, V] = W. To make things easier for me I'm going to use g as both an expression of the most general vector in the space and to refer to the Lie Algebra. It should be clear from the context which I mean.

The Jacobi identity can easily be proved. All we need to know is the Lie Brackets. Two of the terms are trivial to work with and the expression [W,[U,V]] = [W, W] = 0.

Subalgebras:
There are two proper Lie subalgebras that can be formed. {U, W} and {V, W}. They are both Abelian and thus the Jacobi identity is trivial.
(1) I can also form each of U, V, W to be Lie subalgebras. Should they be considered or are they trivial?

An ideal of an Algebra g is a subalgebra h such that [math][h, g] \subseteq h[/math] for all g.

I'll run through the highlights of the first one.
Let h = {U, W}.

[math][h, g] = [aU + bW, pU + qV+rW] [/math]

[math]= aq [U, V] + ar [U, W] + bp[W, U] + bq[W, V] = aqW \subset h[/math].

The other ideal is {V, W}.

Derived Algebra:

The derived series is defined by [math]g' = [g, g] \text{, } g^{i} = [ g^{i - 1}, g^{i - 1} ][/math].

The series is simple enough. Take the most general element of the Algebra and take the commutator. To save some typing I will simply give the results:
[math]g' = [g, g] = [aU + bV + cW, pU + qV + rW] \propto W[/math]

[math]g'' = [g', g'] = 0[/math]

By definition this means that g is solvable and since {U, W} and {V, W} are subalgebras of g thus they are also solvable Algebras.

[math]g_{rad}[/math] is the maximal solvable ideal, which in this case is [math]\{ U,W \} \cup \{ V, W \} = g[/math] so [math]g_{rad} = g[/math].

Lower Central Series:

The lower central series is defined as [math]g_1 = g' \text{, } g_i = [g, g_{i - 1}] [/math].

We start with
[math]g_1 = g' = W[/math]. Then [math]g_2 = [g, g'] = [g, cW] = 0[/math]

All Lie brackets for the Lower Central Series is 0 so g is nilpotent.

Center, Centralizer, and Normalizer:

The center of an Algebra is defined as [math]Z(g) = \{ x \in g | [x, y] = 0 \}[/math].

(2) It seems to me that all we need to do is to find [math][U, g] \propto W[/math], [math] [V, g] \propto W[/math], and [math] [W, g] = 0 [/math]. Thus [math]Z(g) = W[/math].

The Centralizer of an Algebra g is the subset k of g such that [math]C_g(k) = \{ x \in g | [x, k] = 0 \}[/math].

(3) The text seems to be implying subsets k of g. But aren't we really looking for a set of subsets of g that have this property?

I'm going to spare you the work here, but as two examples:
k = pU + qV: [math] [aU + bV + cW, pU + qV] \propto W \neq 0[/math]

and
k = rW: [math] [aU + bV + cW, rW] = 0[/math]

The only element of the Centralizer is W. Thus [math]C_g{k} = W[/math].

-Dan

 

[topsquark]

Testing, uno, beth, 3.

-Dan