Lie algebras, Round Two

More definitions, sure, but also a nontrivial example.

Hi Friends!

Lectures while speaking don’t often translate well to posts, videos or otherwise. The actual first lecture included a third section which we put at the top of today’s release. If you missed the first episode, you can find it here:

Best,

Sean

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The Derived Subalgebra

Any Lie algebra that is a subset of another is called a subalgebra. Subalgebras are strict if their dimension is strictly less than that of the original algebra. They are strict in the same sense that subsets are. An ideal of a Lie algebra g is a subalgebra h where in every bracket which contains at least one member of h maps back to h. We sometimes write this as

\([\mathfrak{h},\mathfrak{g}] \subseteq \mathfrak{h}. \)

More explicitly, we mean that

\([\mathfrak{h},\mathfrak{g}] = \left\{ z \;\Big|\; z = [h,g],\; h \,\mathrm{in} \;\mathfrak{h}\;\mathrm{and}\,g\,\mathrm{in}\;\mathfrak{g} \right\}.\)

While it’s not a particularly informative statement, a Lie algebra can be an ideal of itself. Notice however that [g, g] forms a subalgebra of g. We call this the derived subalgebra.

Simple, Soluble and Nilpotent

A Lie algebra g is called simple if g = [g, g]. In particular, there are no nontrivial subalgebras of a simple Lie algebra.

The dimension of [g, g] relative to that of g will prove useful in studying finite-dimensional Lie algebras. Let us define this number to be the rank¹ of g.

We will also utilize some helpful generalizations of [g, g], to which we now turn.

As above, let g^(n) be defined as

\(\mathfrak{g}^{(n)} = [\mathfrak{g}^{(n-1)},\mathfrak{g}^{(n-1)}],\)

where g^(0) is g itself. A Lie algebra g is called soluble² if g^(n) = 0 for some n. A Lie algebra is called abelian if that n equals 1.

Between the soluble and the simple Lie algebras there is considerable opportunity for structure. As above, let g_(n) be defined as

\(\mathfrak{g}_{(n)} = [\mathfrak{g},\mathfrak{g}_{(n-1)}],\)

where g_(0) is g.

A Lie algebra is called nilpotent if g_(n) = 0 for some n. From abelian to soluble to nilpotent to simple, very little of this makes intuitive sense without explicit examples.

Execises

1. Show that all one-dimensional Lie algebras are abelian.

2. Prove that all abelian Lie algebras are nilpotent and soluble.

Lecture Two

You have shown that one-dimensional Lie algebras are abelian, which is a restriction imposed by the Lie bracket on the vector space. Two-dimensional Lie algebras are similarly restricted, and ofter a very clear demonstration of how the Lie bracket interacts with the vector space structure.

The Case of Two Dimensions

Let g be a two-dimensional vector space and choose a basis with independent vectors x and y. Up to linear transformation there is only one possible bracket. For some scalars a and b, we may write the most general case³ as

\([x,y] = ax + by \)

What follows is the singular most important thing you can prove to follow along with the rest of this chapter.

Proposition 4. Up to Lie algebra isomorphism, there is a unique, two-dimensional Lie algebra such that [x, y] = y.

Proof. Exercise⁴.

In what follows we refer to this two-dimensional Lie algebra as Σ_2.

Notice that Σ_2^(1) = span{y}, and is one-dimensional. So Σ_2 is soluable⁵. But is it nilpotent?

Evidently,

\(\Sigma_{2(1)} = \Sigma_{2}^{(1)},\)

but

\(\Sigma_{2(2)} = [\Sigma_{2},\Sigma_{2(1)}] = \Sigma_{2{(1)}}.\)

Indeed, we can always iterate this so that

\(\Sigma_{2(n)} = \Sigma_{2{(1)}},\)

which never vanishes. Hence Σ_2, while soluble, is not nilpotent.

1. This is a multiplicity of the usage for the term rank in the study of Lie algebras. For simple Lie algebras, Cartan’s classification bestows a slightly different notion of rank. We shall highlight that distinction as needed, later. For now, we are using rank in the linear sense, as in the dimension of the collective image of all the adjoint maps.

2. Many authors prefer the term solvable, which hits my ears like fingernails on a chalkboard. Like I said at the outset, this is an idiosyncratic introduction to Lie algebras.

3. Here’s one way to think of this. Bilinearity restricts the left hand side to this single bracket. The right hand side is simply a statement that the Lie bracket must map to some vector in g.

4. What do you think this says about the relationship between linear structure and Lie algebra structure? Hint: consider a change of basis vectors.

5. Exercise: Why?