18.745 Lecture Notes - Lecture 2 Introduction and Basic Definitions - Part II

Alan M. Dunn^{∗}

(Dated: November 7, 2004)

A homomorphism of Lie algebras 𝔤 is a linear map ϕ : 𝔤 → 𝔤_{1} which preserves the bracket, that is, ϕ([a, b]) = [ϕ(a), ϕ(b)]. The kernel of a homomorphism ϕ, denoted Ker ϕ, is a subset of 𝔤 such that Ker ϕ = {a ∈ 𝔤|ϕ(a) = 0}. The image of a such a homomorphism, Im ϕ ≡ {ϕ(a)|a ∈ 𝔤_{1}} is a subset of 𝔤_{1}. A homomorphism ϕ is called an isomorphism if it is bijective, i.e. if Ker ϕ = 0, Im ϕ = 𝔤_{1}.

Exercise 2.1: Prove the following claims:

a) Ker ϕ is an ideal in 𝔤.

b) Im ϕ is a subalgebra of 𝔤.

c) The Lie algebra Im ϕ is isomorphic to the Lie algebra 𝔤/Ker ϕ.

Proof.

a) Given a ∈ Ker ϕ, b ∈ 𝔤. ϕ([a, b]) = [ϕ(a), ϕ(b)] = [0, ϕ(b)] = 0 → [a, b] ∈ Ker ϕ → [Ker ϕ, 𝔤] ⊂ Ker ϕ → Ker ϕ is an ideal.

b) a, b ∈ Im ϕ → a = ϕ(a′), b = ϕ(b′) → ϕ(λa′ +νb′) = λa+νb (by the linearity of ϕ) → λa+νb ∈ Im ϕ → Im ϕ a subspace. [a, b] = [ϕ(a′), ϕ(b′)] = ϕ([a′, b′]) → [a, b] ∈ Im ϕ → Im ϕ a subalgebra of 𝔤

c) Define ψ : 𝔤/ Ker ϕ → Im ϕ, ψ(a + Ker ϕ) = ϕ(a). ψ(𝔤/ Ker ϕ) is evidently equal to Im ϕ, which implies ψ is surjective. ψ is well defined, since choosing any representatives c, d ∈ a + Ker ϕ will both yield ϕ(a) under ψ. ψ(a + Ker ϕ) = ψ(b + Ker ϕ) → ϕ(a) = ϕ(b) → a = b + c, c ∈ Ker ϕ → a + Ker ϕ = b + Ker ϕ → ψ is bijective, which implies that ψ is an isomorphism.

The center of a Lie algebra 𝔤, denoted Z(𝔤), is the set of elements commuting with 𝔤 i.e. Z(𝔤) = {a ∈ 𝔤|[a, 𝔤] = 0}. This is obviously an ideal.

Exercise 2.2: If 𝔤 is a non-abelian Lie algebra, then dim Z(𝔤) ≤ dim(𝔤) − 2.

Proof. 𝔤 non-abelian implies that there are at least two linearly independent elements a, b ∈ 𝔤 such that [a, b] = c, c ≠ 0 → a, b ∉ Z(𝔤) → dim Z(𝔤) ≤ dim(𝔤) − 2 since the subspace spanned by a and b is not in Z(𝔤): [Aa+Bb, a] = −Bc, and [Aa+Bb, b] = Ac for A, B ∈ 𝔽, which are only 0 for B and A equal to 0 respectively.

Exercise 2.3: If 𝔤 is a Lie algebra of dimension n ≥ 3 with a 1 dimensional derived subalgebra, then 𝔤, up to isomorphism, is one of the following:

• b_{2} ⊕ ab_{n−2}

• H_{k} ⊕ ab_{n−2k−1}

where b_{2} is the unique two-dimensional non-abelian Lie algebra, ab_{k} is a k-dimensional abelian Lie algebra, and H_{k} is the k-th Heisenberg algebra of dimension 2k + 1.

Proof. [𝔤, 𝔤] 1 dimensional implies that [𝔤, 𝔤] = 𝔽a, a ∈ 𝔤, a ≠ 0.

∗Electronic address: amdunn@mit.edu

Case 1 ∃b ∈ 𝔤|[b, a] ≠ 0 → [b, a] = λa (λ ≠ 0). Send b → \frac{1}{λ}b, that is, instead choose b′ = \frac{1}{λ}b and take that as a basis element. Then [b, a] = a. We can now construct a basis for the remaining n − 2 ≥ 1 dimensions of the algebra: Take any element that is linearly independent from a and b, call it e_{1}. [a, e_{1}] = λ_{1}a → [a, e_{1} + λ_{1}b] = 0, so send e_{1} → e_{1}+λ_{1}b. [b, e_{1}] = λ_{2}a → [b, e_{1}−λ_{2}a] = 0, and also [a, e_{1}−λ_{2}a] = 0 → sending e_{1} → e_{1}−λ_{2}a creates e_{1} that commutes with both a and b, while leaving e_{1} linearly independent of a and b. Now we can construct a mutually commuting basis {e_{1}, …, e_{n−2}} that commutes with a and b by induction: Given {e_{1}, …, e_{i}} mutually commuting and commuting with a and b, select e_{i+1} linearly independent of {a, b, e_{1}, …, e_{i}}, which can be done if i ≤ n + 2. e_{i+1} can be made to commute with a and b by the same process as before. Furthermore, given any e_{a}, [e_{i+1}, e_{a}] = λa, and thus by the Jacobi identity λa = [b, [e_{i+1}, e_{a}]] = [[b, e_{i+1}], e_{a}] + [e_{i+1}, [b, e_{a}]] = 0 and thus e_{i+1} will commute with all the e_{a} for a ≤ i + 1. Thus we have {a, b} composing b_{2}, the unique two dimensional non-abelian Lie algebra, and {e_{1}, ⋯, e_{n−2}} which span an n − 2 dimensional abelian Lie algebra, and thus 𝔤 = b_{2} ⊕ ab_{n−2} Case 2 ∄b ∈ 𝔤|[b, a] ≠ 0, 𝔤 non-abelian since [𝔤, 𝔤] ≠ 0 and thus ∃p_{1}, q_{1} ∈ 𝔤|[p_{1}, q_{1}] ≠ 0 → [p_{1}, q_{1}] = λa, sending p_{1} → \frac{1}{λ}p_{1} → [p_{1}, q_{1}] = a. Now if 𝔤 is three dimensional we have 𝔤 = H_{1}, otherwise we can construct a basis for the remaining n − 3 ≥ 1 dimensional subspace. Choose d linearly independent from p_{1}, q_{1}, and a. [d, a] = 0, [d, p_{1}] = λ_{1}a → send d to d + λ_{1}q_{1} → [d, p_{1}] = 0, [d, q_{1}] = λ_{2}a → send d to d − λ_{2}p_{1} → [d, q_{1}] = 0 while keeping [d, p_{1}] = 0. Thus d commutes with the vectors p_{1}, q_{1}, a. While we do not have enough basis vectors we can use the previous algorithm to construct more. Now that we have {v_{1}, …, v_{n−3}} such that these all commute with {a, p_{1}, q_{1}}. Let H = {p_{1}, q_{1}}. We can put the basis into an appropriate form as follows:

1. Consider the complement subspace H^{c} to H. Since a is in this subspace, the subspace is a Lie algebra.

2. If [H^{c}, H^{c}] = 0, all the vectors in H^{c} commute and we are done.

3. If [H^{c}, H^{c}] ≠ 0, ∃p, q ∈ H^{c}|[p, q] = λa ≠ 0, sending p → \frac{1}{λ}p will make [p, q] = a. We can also make p, q commute with all the vectors in H by taking the vectors in H pairwise and performing the same process as before. Move the new vectors we have created from H^{c} to H and go to the first step (recalculate H^{c}).

Since the last step decreases the dimension of H^{c} and dim(H^{c}) begins at a finite number, the algorithm must terminate, and then H ∪ {a} forms the Heisenberg algebra H_{dim(H)/2} since H contains pairs of vectors {p, q} such that [p, q] = a, [p, t] = 0 for t not proportional to q and [q, t] = 0 for t not proportional to p. The vectors still in H^{c} will all commute with each other and form an abelian subalgebra of dimension n − dim(H) − 1. So 𝔤 = H_{dim(H)/2} ⊕ ab_{n−dim(H)−1}.

A digression on Lie algebras formed from algebraic groups:

An algebraic group over a field 𝔽 is a collection of polynomials {p_{α}|α ∈ J} for some indexing set J on the space of n by n matrices Mat_{n}(F) such that for any commutative associative unital algebra A over F, the set G(A) ≡ {g ∈ Mat_{n}(A)|g invertible and p_{α}(g) = 0 for all α ∈ I} is a group with respect to matrix multiplication.

Examples:

a) If we set {p_{α}} = ϕ then the corresponding algebraic group is called the general linear algebraic group and is denoted GL_{n}. GL_{n}(A) = { invertible matrices a ∈ Mat_{n}(A)}.

b) Since the determinant of a matrix is a polynomial in the entries of the matrix, we can set {p_{α}} = {det−1}. The corresponding algebraic group is known as the special algebraic group, denoted SL_{n}. SL_{n}(A) = {a ∈ Mat_{n}(A)|det a = 1}.

c) Let B ∈ Mat_{n}(𝔽). Define O_{n,B}(A) ≡ {a ∈ GL_{n}(A)|a^{T} Ba = B}.

Exercise 2.4: Prove that O_{n,B}(A) is a subgroup of GL_{n}(A).

Proof. O_{n,B} = {a ∈ GL_{n}(A)|a^{T} Ba = B}.

Let a, b ∈ O_{n,B}. a^{T} Ba = B, b^{T} Bb = B → (ab)^{T} B(ab) = b^{T} a^{T} Bab = b^{T} Bb = B → ab ∈ O_{n,B}. a^{T} Ba = B → (a^{−1})^{T} Ba^{−1} = B → a^{−1} ∈ O_{n,B}.

If B is non-degenerate and symmetric (respectively, skew-symmetric) then O_{n,B}(A) is called the orthogonal (respectively, symplectic) algebraic group.

Let D = 𝔽[ϵ]/(ϵ^{2}) = {a + bϵ|a, b ∈ 𝔽, ϵ^{2} = 0}. D is called the algebra of dual numbers. Multiplication in D is carried out as follows:

(a + bϵ)(c + dϵ) = ac + (bc + ad)ϵ

The Lie algebra of an algebraic group G, denoted Lie G, is a subalgebra {X ∈ 𝔤𝔩_{n}(𝔽)|I + ϵX ∈ G(D)}.

Theorem 2.5: Lie G is a subalgebra of the Lie algebra 𝔤𝔩_{n}(𝔽).

Proof. X ∈ Lie G ⟺ p_{α}(I + ϵX) = 0 for all α and I + ϵX is invertible in Mat_{n}(D).

(I + ϵX)^{−1} = I − ϵX since ϵ^{2} = 0 → I + ϵX invertible. Also, p_{α}(I) = 0 since G(D) is a group. We can Taylor expand the constraints for elements of the algebraic group as follows:

0 = p_{α}(I + ϵX) = p_{α}(I) + ∑_{i,j} \frac{∂p_{α}}{∂X_{ij}} (I)ϵX_{ij}(ϵ^{2} = 0)

Hence X ∈ Lie G if and only if

∑_{i,j} \frac{∂p_{α}}{∂X_{ij}} (I)X_{ij} = 0, α ∈ J.

Thus Lie G is a subspace of Mat_{n}(𝔽). If X, Y ∈ Lie G, consider two elements in G(D): 1 + ϵX ∈ G(𝔽[ϵ]/(ϵ^{2})) and 1 + ϵ′Y ∈ G(𝔽[ϵ′]/(ϵ^{′2})). Both of these are in G(𝔽[ϵ, ϵ′]/(ϵ^{2}, ϵ^{′2})) and we can consider an equation in this group:

(1 + ϵX)(1 + ϵ′Y)(1 + ϵX)^{−1}(1 + ϵ′Y)^{−1} =  (1 + ϵX)(1 + ϵ′Y)(1 − ϵX)(1 − ϵY)
=  1 + ϵϵ′(XY − Y X) ∈ G(𝔽[ϵϵ′]/(ϵϵ′)^{2})

which is isomorphic again to the algebra of dual numbers. So Lie G is a subalgebra and we are finished.

Examples:

a) If G = GL_{n}(𝔽), then Lie G = 𝔤𝔩_{n}(𝔽).

b) If G = SL_{n}(𝔽), then Lie G = {X|det(I + ϵX) = 1}. By writing out the determinant we can see that:

| \begin{matrix}1 + ϵx_{11} & ϵx_{12} & ⋯ & ϵx_{1n} \\ ϵx_{21} & 1 + ϵx_{12} & ⋯ & ϵx_{2n} \\ ⋮ & & ... \\ ϵx_{n1} & ϵx_{n2} & ⋯ & 1 + ϵx_{nn} \end{matrix} | = 0

= 1 + ϵ(x_{11} + x_{22} + ⋯ + x_{nn}) + O(ϵ^{2}) = 1 + ϵtrX → trX = 0.
So, Lie SL_{n}(𝔽) = 𝔰𝔩_{n}(𝔽).

Exercise 2.6: Lie O_{n,B} = 𝔬_{n,B}(𝔽).

Proof. Let G = O_{n,B}. X ∈ Lie G → (1+ϵX)^{T} B(1+ϵX) = B → B +ϵ(X^{T} B +BX) = B → X^{T} B +BX = 0.

Next we prepare to prove Engel’s theorem by establishing the following definition and lemma:

Definition: An operator A ∈ End V is nilpotent if A^{N} = 0 for some positive integer N.

Lemma: If A is nilpotent (on V), then ad A is nilpotent on gl_{V}.

Proof. Note ad A = L_{A} − R_{A}, where L_{A} and R_{A} are the operations of left and right multiplication by A (in the space of endomorphisms of V) respectively. Since A is nilpotent, L_{A} and R_{A} raised to some power N are each zero for large enough N. Moreover, L_{A} and R_{B} commute since matrix composition in the space of endomorphisms of V is associative. (ad A)^{2N} = ∑_{i=0}^{2N} \binom{2N}{i}L_{A}^{2N−i}R_{A}^{i}, and for all i, 2N − i ≥ N or i ≥ N → L_{A}^{2N−i} = 0 or R_{A}^{i} = 0 → (ad A)^{2N} = 0.