18.745 Introduction to Lie Algebras Fall 2004

Lecture 1 — September 7, 2004

Prof. Victor Kač Scribe: Patrick Lam

Definition 1 (a) An algebra is a vector space over a field 𝔽, endowed with a multiplication ab, which is bilinear:

a(λb + μc) = λab + μac
(λb + μc)a = λba + μca

An altebra is associative if (ab)c = a(bc).

(b) A Lie algebra is an algebra 𝔤 with product [a, b], called the bracket of a and b, subject to two axioms:

• skew commutativity: [a, a] = 0

• Jacobi identity: [a, [b, c]] + [b, [c, a]] + [c, [a, b]] = 0.

c  a  b

Remark. In a Lie algebra, one has [b, a] = −[a, b].

Proof. 0 = [a + b, a + b] = [a, b] + [b, a] + [figure: [a, a]] + [figure: [b, b]] □

Examples.

1. 𝔤 a vector space with bracket [a, b] = 0. This is called an abelian Lie algebra.

2. ℝ^{3} with vector multiplication × (cross product).

3. If A is an associative algebra, then [a, b] = ab−ba satisfies the two identities. This Lie algebra is denoted by A_{−}.

Exercise 1.1. Check the Jacobi identity on [a, b] = ab − ba. Moreover, this is true if A is only quasi-associative, i.e. (ab)c − a(bc) is symmetric in a, b (= (ba)c − b(ac)).

Solution. First, we show that associativity implies quasi-associativity. Let (ab)c = a(bc). Then (ab)c − a(bc) = 0 = (ba)c − b(ac). Hence we only need to show that [a, b] satisfies the Jacobi identity if A is quasi-associative.

Here are some consequences of quasi-associativity.

(ab)c − a(bc) − (ba)c + b(ac) = 0
(cb)a − c(ba) − (bc)a + b(ca) = 0
(ac)b − a(cb) − (ca)b + c(ab) = 0
(ba)c − b(ac) − (ab)c + a(bc) = 0

We expand the Jacobi identity, group terms, and apply quasi-associativity:

[a, [b, c]] + [b, [c, a]] + [c, [a, b]]
= a(bc − cb) − (bc − cb)a + b(ca − ac) − (ca − ac)b + c(ab − ba) − (ab − ba)c
= [(cb)a − c(ba) − (bc)a + b(ca)]
+ [(ac)b − a(cb) − (ca)b + c(ab)]
+ [(ba)c − b(ac) − (ab)c + a(bc)]
= 0 (since all these terms = 0 by quasi-associativity)

□

A special case is A = End V, then A_{−} = gl_{V} is called the general linear Lie algebra. In particular, A = Mat_{n}𝔽, then A_{−} = gl_{n}(𝔽).

4. Any subalgebra of a Lie algebra is a Lie algebra.

Notation: for subsets M, N of 𝔤 we denote [M, N] the span of all commutators [m, n], where m ∈ M and n ∈ N. For example, subspace 𝔥 is a subalgebra of 𝔤 if [𝔥, 𝔥′] ∈ 𝔥.

Example. sl_{n}(𝔽) = {a ∈ gl_{n}(𝔽) | tr a = 0}

Exercise 1.2. Show that tr[a, b] = 0 when a, b ∈ gl_{n}(𝔽). Also show that if f : gl_{n}(𝔽) → 𝔽 is a linear function such that f([a, b]) = 0, a, b ∈ gl_{n}(𝔽), then f(k) = c · tr k.

Solution. We have

tr [a, b] = tr ab − tr ba = ∑_{i} ∑_{j} a_{ji}b_{ij} − ∑_{i} ∑_{j} b_{ji}a_{ij} = 0

Now, any matrix e_{ij}, i ≠ j can be expressed as a commutator [a, b] where a, b ∈ gl_{n}(𝔽), because e_{ij}e_{jj} − e_{jj}e_{ij} = e_{ij}. Hence f(e_{ij}) = 0 for all such matrices.

But f(e_{ii}) = f(e_{jj}) = c for all i, j because f(e_{ii})−f(e_{jj}) = f(e_{ii})−f(e_{jj}) = f(e_{ij}e_{ji}−e_{ji}e_{ij}) = 0.

Any x ∈ gl_{n}(𝔽) can be split into x′ + x′′ where x′ is the sum of e_{ij}, i ≠ j, and x′′ is of the

form ∑_{i} λ_{i}e_{ii}. By linearity,

f(x) = 0 + f(x′′) = c ∑_{i} λ_{i} = c · tr.

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Definition 2 An ideal 𝔪 ⊂ 𝔤 is a subspace such that [𝔪, 𝔤] ⊂ 𝔪.

Example. sl_{n}(𝔽) is an ideal of gl_{n}(𝔽) by Exercise 1.2.

5. Factor algebras: If 𝔤 is a Lie algebra and 𝔪 is an ideal, then 𝔤/𝔪 is a Lie algebra with bracket [a + 𝔪, b + 𝔪] = [a, b] + 𝔪.

6. Direct sum of two (Lie) algebras 𝔤_{1} ⊕ 𝔤_{2}: [(a, b), (a_{1}, b_{1})] = ([a, a_{1}], [b, b_{1}]) where a, a_{1} ∈ 𝔤_{1}, b, b_{1} ∈ 𝔤_{2}.

More examples of subalgebras of gl_{V}: Let B be a bilinear (𝔽-valued) form on a vector space V over 𝔽, define o_{V,B} = {a ∈ gl_{V} | B(a(u), v) + B(u, a(v)) = 0}.

Exercise 1.3. Let B be a bilinear 𝔽-valued form on a vector space V over 𝔽. Define

o_{V,B} = {a ∈ gl_{V} | B(a(u), v) + B(u, a(v)) = 0

Check that this is a subalgebra of the Lie algebra gl_{V}.

Solution. To show that o_{V,B} is a subalgebra (it is clearly a subspace), we need only show that o_{V,B} is closed under the bracket. Let x, y ∈ o_{V,B}.

Consider

B(x(y(u)), v) = −B(y(u), x(v)) = B(u, y(x(v)))

But since B is bilinear and yx = [y, x] − xy, B(u, y(x(v))) = B(u, [x, y] − xy) = −B(u, x(y(v))) + B(u, [x, y]) where B(u, [x, y]) is 0 from above, implying that B(x(y(u)), v) = −B(u, x(y(v))), giving closure:

B(x(y(u)), v) + B(u, x(y(v))) = 0

□

Important special cases: dim V < ∞, B is non-degenerate (i.e. det of the matrix of B in some basis is non-zero).

• case 1: B is symmetric. B(a, b) = B(b, a), then o_{V,B} is called the orthogonal Lie algebra, notation so_{V,B}.

• case 2: B is skew-symmetric. B(a, b) = −B(b, a), then o_{V,B} is called the symplectic Lie algebra, notation sp_{V,B}.

Exercise 1.4. Suppose dim V = n, choose a basis of V, let so_{V,B} and sp_{V,B} ⊂ gl_{n}. Let B be the matrix of the bilinear form. Show

so_{V,B} = {a ∈ gl_{n}(𝔽) | a^{T} B + Ba = 0}
sp_{V,B} = {a ∈ gl_{n}(𝔽) | a^{T} B + Ba = 0}

Solution. Recall that gl_{n} is associative, and that so_{V,B} is the set of a ∈ A such that B(u, a(v)) + B(a(u), v) = 0 and B symmetric; similarly, sp_{V,B} is the corresponding set when B is skewsymmetric. For so_{V,B}, we expand the definition of the bilinear form to get

u^{T} Ba(v) + (a(u))^{T} Bv = 0

and expressing a as matrix multiplication,

u^{T} (Ba)v + (au)^{T} Bv = u^{T} (Ba)v + u^{T} a^{T} Bv = 0

which is equivalent to the matrix condition

a^{T} B + Ba = 0

when B is a symmetric matrix; similarly, for a skew-symmetric B, we expand to get

u^{T} Ba(v) + (a(u))^{T} Bv = 0

and expressing a as matrix multiplication,

u^{T} (Ba)v + (au)^{T} Bv = u^{T} (Ba)v + u^{T} a^{T} Bv = 0

which is equivalent to the matrix condition

a^{T} B + Ba = 0.

□

Definition 3 The derived algebra of 𝔤 is [𝔤, 𝔤]. Obviously, this is an ideal, and hence a subalgebra.

We now classify Lie algebras in dimensions 1 and 2.

dim 1. 𝔤 = 𝔽a, [a, a] = 0. Only the abelian one.

dim 2. 𝔤 = 𝔽a + 𝔽b, [𝔤, 𝔤] = 𝔽[a, b]. case 1. [a, b] = 0. Then 𝔤 abelian. case 2. [a, b] = c ≠ 0. So 𝔤′ = 𝔽c. Take d ∉ 𝔤′ such that d ≠ 0. Then [d, c] = αc, since 𝔽c is an ideal, and α ≠ 0, since 𝔤 nonabelian. Replacing d by \frac{1}{α}d, we get [d, c] = c. Thus we have a unique non-abelian Lie algebra, [a, b] = b.

The two most important ways to construct Lie algebras:

1. as a subalgebra (of gl_{n});

2. by structure constants: choose a basis e_{1}, …, e_{n} of 𝔤; then [e_{i}, e_{j}] = ∑_{k=1}^{n} c_{ij}^{k}e_{k}. The scalars c_{ij}^{k} are called structure constants. Of course, c_{ii}^{k} = 0 and c_{ij}^{k} = −c_{ij}^{k} by skew-commutativity and a quadratic equation which is the Jacobi identity.

Remark 1 The non-abelian 2-dimensional Lie algebra is

\{( \begin{matrix}α & β \\ 0 & 0 \end{matrix} )\} ⊂ gl_{2}(𝔽)

If a = ( \begin{matrix}1 & 0 \\ 0 & 0 \end{matrix} ) and b = ( \begin{matrix}0 & 1 \\ 0 & 0 \end{matrix} ), then [a, b] = b.

Example 1 The Heisenberg Lie algebra ℌ_{n} has basis p_{i}, q_{i}(i = 1, …, n), c. (dim ℌ_{n} = 2n + 1) where [p_{i}, q_{j}] = δ_{ij}c, [c, p_{i}] = 0, [c, q_{i}] = 0, [p_{i}, p_{j}] = 0, [q_{i}, q_{j}] = 0. Jacobi trivially holds. Realization by operators: p_{i} = \frac{∂}{∂x}, q_{i} = x_{i}, c = 1 on ℂ[x_{1}, …, x_{n}].

The first important meaning of the Jacobi identity: Rewrite it as follows:

[a, [b, c]] = [[a, b], c] + [b, [a, c]].  (1)

Definition 4 For any algebra A, an endomorphism D is called a derivation if the Leibniz rule holds, i.e.

D(ab) = (Da)b + a(Db).

Definition 5 Given an element a ∈ 𝔤, define the operator ad a (adjoint) on 𝔤 by:

(ad a)b = [a, b].

Equation 1 means that ad a is a derivation of the Lie algebra 𝔤. It is called an inner derivation.

Notation: Given an algebra A, denote by Der A(⊂ End A) the space of derivations of A.

Exercise 1.5. a) Der A is closed under the bracket in End A i.e. bracket of two derivations is a derivation, or Der A is a subalgebra of gl_{A}. (b) If A = 𝔤 is a Lie algebra, then [D, ad a] = ad (D(a)) for any derivation F ∈ Der 𝔤 and a ∈ 𝔤. Hence inner derivations form an ideal of the Lie algebra Der 𝔤.

Solution. a) Let D_{1}, D_{2} be derivations. Consider (by parts) [D_{1}, D_{2}]:

D_{1}D_{2}(ab) = D_{1}((D_{2}a)b + a(D_{2}b))
= D_{1}((D_{2}a)b) + D_{1}(a(D_{2}b))
= (D_{1}(D_{2}a))b + (D_{2}a)(D_{1}b) + (D_{1}a)(D_{2}b) + a(D_{1}D_{2}b)
D_{2}D_{1}(ab) = (D_{2}(D_{1}a))b + (D_{1}a)(D_{2}b) + (D_{2}a)(D_{1}b) + a(D_{2}D_{1}b)
D_{1}D_{2}(ab) − D_{2}D_{1}(ab) = (D_{1}D_{2}a − D_{2}D_{1}a)b + a(D_{1}D_{2}b − D_{2}D_{1}b)
= ((D_{1}D_{2} − D_{2}D_{1})a)b + a((D_{1}D_{2} − D_{2}D_{1})b)

showing that the bracket is a derivation.

b) Consider

[D, ad a] = (D(ad a))b − ((ad a)D)b
= D[a, b] − [a, D(b)]
= D(ab) − D(ba) − aD(b) + D(b)a
= D(a)b + [figure: aD(b)] − [figure: D(b)a] − bD(a) − [figure: aD(b)] + [figure: D(b)a]
= [D(a), b]
= ad D(a)

which shows that inner derivations form an ideal of the Lie algebra Der 𝔤. □