An Idiosyncratic Introduction to Lie algebras

Of course we have physical applications in mind, but also a sense of narrative.

Howdy!

I still owe you a discussion of the quantum bosonic string. That is absolutely coming, but it’s taking longer than I anticipated. In the meantime, I thought I’d share some notes from a Lie algebra course I put together. This newsletter originally envisioned as an informal place to chat about mathematics and mathematical physics, broadly speaking. So it’s still in keeping with the mission.

The main goal of the course was to revisit and reframe the Bianchi classification of (real) three-dimensional Lie algebras, with an eye to the general case. We assume the reader is pretty familiar with linear algebra, and therefore it should be accessible to advanced undergraduates.

Useful references for this material include John M Lee’s Introduction to Smooth Manifolds and of course Humphreys’ Introduction to Lie algebras and representation theory. We do make implicit use of the somewhat arcane Lie algebras and Lie groups by Serre. You could also pick up some background from our old YouTube channel ( starting here ).

More soon!

Sean

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Lecture 1 : Generalities

Every vector space is complex and finite-dimenstional unless otherwise specified, although the results and ideas are largely field agnostic.

Algebras

A bilinear map on a vector space V is a map

\(\star: V\times V \rightarrow V,\)

which is linear in each argument. A vector space equipped with a bilinear map is called an algebra.

There are a wide variety of algebras whose structure depends on properties of such an operation. For example, the vector space of n-by-n matrices has a natural multiplication that is associative:

\((a\star b)\star c = a\star(b\star c) = a \star b \star c.\)

The exterior product used by the various vector spaces of differential forms is alternating. That is, for each vector a,

\(a \wedge a = 0.\)

A linear map d on an algebra is called a derivation if it obeys the product rule:

\(d : a\star b \mapsto (da)\star b + a \star (db).\)

A Lie bracket is an alternating, bilinear map that satisfies the so-called Jacobi identity:

\([a,[b,c]] + [b,[c,a]] + [c,[a,b]] = 0.\)

Lie Algebras

A vector space equipped with a Lie bracket is called a Lie algebra. For the underinitiated it is sometimes helpful to prove the following.

Claim: Any Lie bracket is antisymmetric in its arguments.

Proof. Exercise.

To each vector z in a Lie algebra g¹ we can associate a linear map called the adjoint:

\(\mathrm{ad}_{z} : \mathfrak{g}\rightarrow \mathfrak{g},\)

where

\(\mathrm{ad}_{z} : x \mapsto [z,x].\)

We denote the space of adjoint maps by ad(g).

Claim: For each Lie algebra g, ad(g) is a vector space.

Proof. Exercise.

Proposition 1: Each adjoint map is a derivation.

Proof. Let x, y and z be elements of a Lie algebra g. The adjoint of the bracket gives

\(\mathrm{ad}_{z}[x,y] = [z,[x,y]].\)

By the Jacobi identity we have

\([z,[x,y]] + [x,[y,z]] + [y,[z,x]] = 0.\)

Owing to the antisymmetry of the Lie bracket, this is implies

\([z,[x,y]] =[[z,x],y] + [x,[z,y]],\)

which is the definition of a Lie algebra derivation. 😎

In some sense, the Jacobi identity is equivalent to demanding that the adjoint map be a Derivation.

Derivations are also fun to consider in their own right.

Proposition 2: Derivations on an arbitrary algebra (A,⋆) form a Lie algebra under commutators.

Proof. Derivations are linear maps and linear maps of course form a vector space in their own right. Let d and D be two derivations on A, and let a and b be two elements of A. The commutator of two derivations is

\([d,D]_{C} = dD - Dd.\)

Here juxtaposition is understood to imply composition. Let us now act with [d,D] on a⋆b and see what happens.

\([d,D]_{C} \,a \star b = dD(a \star b)- Dd(a \star b).\)

Computing,

\([d,D]_{C} \,a \star b = d\left( Da \star b + a \star Db \right)-D\left(da \star b + a\star db\right),\)

\(dD(a \star b) = \left( (dD)a \star b + Da \star db + da \star Db + a \star dDb \right)\)

\(Dd(a \star b) = \left( (Dd)a \star b + da \star Db + Da \star db + a \star Ddb \right),\)

Putting things together we find the diagonal terms cancel so,

\([d,D]_{C} \,a \star b = \left([d,D]_{C}\,a\right)\star b + a\star \left([d,D]_{C}\,b\right).\)

In other words, the commutator of two derivations is itself a derivation, and so the derivations form an algebra under commuators.

It remains to show that the commutator is a Lie bracket. It is clearly alternating. The Jacobi identity can be computed directly from terms like

\([a,[b,c]_{C}]_{C} = [a , {bc - cb} ]_{C} = abc - acb - bca + cba.\)

The other two terms in the identity are

\([b,[c,a]_{C}]_{C} = bca - bac - cab + acb,\)

and

\([c,[a,b]_{C}]_{C} = cab - cba - abc + bac.\)

The first line cancels against terms from the other two. The middle two terms from the second line cancels against the outer terms in the last line. Hence the Jacobi identity is satisfied.😎

Immediately we infer that the adjoint maps of some Lie algebra g forms a Lie algebra under commutators. By definition there is a linear² isomorphism between ad(g) and g itself. We might think there is some conspiracy relating the two. Rightfully so, as it turns out.

Proposition 3. ad(g) is Lie algebra isomorphic to g.
Proof. Let x, y, z, w be elements of a Lie algebra g and suppose [x, y] = w. Consider the commutator of two adjoint maps:

\([\mathrm{ad}_{x},\mathrm{ad}_{y}]_{C}\,z = \mathrm{ad}_x \mathrm{ad}_y z- \mathrm{ad}_y \mathrm{ad}_x z.\)

In terms of the Lie bracket this is

\([\mathrm{ad}_{x},\mathrm{ad}_{y}]_{C}\,z = [x, [y,z]]- [y,[x,z]] = [x,[y,z]] + [y,[z,x]].\)

Using the Jacobi identity, we have

\([\mathrm{ad}_{x},\mathrm{ad}_{y}]_C\,z = -[z,[x,y]] = [w,z] = \mathrm{ad}_{w}z.\)

Given the arbitrary nature of z we can directly infer that

\([\mathrm{ad}_{x},\mathrm{ad}_{y}]_C = \mathrm{ad}_{w}.\)

😎

The algebra ad(g) is sometimes called the adjoint representation of g. We shall learn more about representations in due course.

1. In the discussion of Lie algebras in particular, the mathfrak type face for algebras and subalgebras is standard. We maintain that through the LaTeX code, but will not be able to impose such font adjustments inline. Meanings should be clear from context. Though centuries old, the crew at 99 percent invisible have a good historical discussion of how these typefaces were used more recently for nefarious ends.

2. One theme of these notes will be the interplay of the Lie bracket with the linear structure. Both impose their own restrictions to the classification of finite-dimensional Lie algebras.