Enveloping Algebras and The Poincaré-Birkhoff-Witt Theorem

Rahbar Virk

A Lie algebra 𝔤 is not an algebra in the sense that the Lie bracket is not associative. We would like to find an associative algebra U⁢𝔤 such that the modules for 𝔤 are the same as those for U⁢𝔤.

Let 𝔤 be a Lie algebra and let T be the tensor algebra of the vector space 𝔤. Recall that

T=T0⊕T1⊕⋯⊕Tn⊕⋯,

where Tn=𝔤⊗𝔤⊗⋯⁢𝔤 (n times); in particular T0=k⁢.1 and T1=𝔤; the product in T is simply tensor multiplication. Let J be the two sided ideal of T generated by the tensors

x⊗y-x,y,

where x,y∈𝔤. The associative algebra T/J is called the enveloping algebra (or sometimes the universal enveloping algebra) of 𝔤 and is denoted by U⁢𝔤. The composite mapping σ of the canonical mappings 𝔤→T→U⁢𝔤 is termed the canonical mapping of 𝔤 into U⁢𝔤; observe that for all x,y∈𝔤 we have that

σ⁢x⁢σ⁢y-σ⁢y⁢σ⁢x=σ⁢x,y

We denote the canonical image in U⁢𝔤 of T0+T1+⋯+Tq by Uq⁢𝔤.

Lemma 1.1.

Let σ be the canonical mapping of 𝔤 into U⁢𝔤, let A be an algebra with unity, and let τ be a linear mapping of 𝔤 into A such that

τ⁢x⁢τ⁢y-τ⁢y⁢τ⁢x=τ⁢x,y

for all x,y∈𝔤. There exists one and only one homomorphism τ′ of U⁢𝔤 into A such that τ′⁢1=1 and τ′∘σ=τ.

Proof.

Note that U⁢𝔤 is generated by 1 and σ⁢𝔤 so τ′ must be unique. Now let φ be the unique homomorphism of T into A that extends τ and such that φ⁢1=1. For x,y∈𝔤, we have

φ⁢x⊗y-y⊗x-x,y=τ⁢x⁢τ⁢y-τ⁢y⁢τ⁢x-τ⁢x,y=0,

hence φ⁢J=0 and thus φ gives us a homomorphism τ′ of U⁢𝔤 into A such that τ′⁢1=1 and τ′∘σ=τ. ∎

Lemma 1.2.

Let a1,…,ap∈𝔤, σ the canonical mapping of 𝔤 into U⁢𝔤, and π be a permutation of 1,…,p. Then

σ⁢a1⁢⋯⁢σ⁢ap-σ⁢aπ⁢1⁢⋯⁢σ⁢aπ⁢p∈Up-1⁢𝔤.

Proof.

It suffices to prove the statement when π is the transposition of j and j+1. But observe that

σ⁢aj⁢σ⁢aj+1-σ⁢aj+1⁢σ⁢aj=σ⁢aj,aj+1.

The lemma now follows. ∎

We now assume 𝔤 to a be finite dimensional Lie algebra and fix a basis x1,…,xn for 𝔤. We denote the canonical image of xi in U⁢𝔤 by yi. For every finite sequence I=i1,…,ip of integers between 1 and n, we set yI=yi1⁢yi2⁢…⁢yip.

Lemma 1.3.

The yI, for all increasing sequences I of length ≤p, generate the vector space Up⁢𝔤.

Proof.

Clearly the vector space Up⁢𝔤 is generated by yI, for all sequences I of length ≤p. But now the required statement follows by applying the previous lemma. ∎

Let P be the algebra k⁢z1,…,zn of polynomials in n indeterminates z1,…,zn. For every i∈ℤ≥0, let Pi be the set of elements of P of degree ≤i. If I=i1,…,ip is a sequence of integers between 1 and n, we set zI=zi1⁢zi2⁢⋯⁢zip.

Lemma 1.4.

For every integer p≥0, there exists a unique linear mapping fp of the vector space 𝔤⊗Pp into P which satisfies the following conditions:

Ap

fp⁢xi⊗zI=zi⁢zI for i≤I, zI∈Pp;
Bp

fp⁢xi⊗zI-zi⁢zI∈Pq for zI∈Pq, q≤p;
Cp

fp⁢xi⊗fp⁢xj⊗zJ=fp⁢xj⊗fp⁢xi⊗zJ+fp⁢xi,xj⊗zJ for zJ∈Pp-1. [The terms in Cp are meaningful by virtue of (Bp)].

Moreover, the restriction of fp to 𝔤⊗Pp-1 is fp-1.

Proof.

For p=0 the mapping given by f0⁢xi⊗1=zi⊗1, satisfies the conditions A0,B0 and C0, furthemore it is clear that the condition A0 forces this to be our linear mapping. Proceeding by induction, assume the existence and uniqueness of fp-1. If fp exists then fp restricted to 𝔤⊗Pp-1 satisfies Ap-1,Bp-1,Cp-1 and is hence equal to fp-1. Thus, to prove our claim it suffices to show that fp-1 has a unique linear extension fp to 𝔤⊗Pp which satisfies Ap,Bp,Cp. Note that Pp is generated by zI for an increasing sequence I of p elements. Thus, we must define fp⁢xi⊗zI for such a sequence I. If i≤I, then the choice is dictated by Ap. Otherwise we can write I as j,J where j<i and j≤J. Then we must have that

	fp⁢xi⊗zI	=fp⁢xi⊗fp-1⁢zj⊗zJ		Ap-1
		=fp⁢xj⊗fp-1⁢xi⊗zJ+fp-1⁢xi,xj⊗zJ		Cp.

Now fp-1⁢xi⊗zJ=zi⁢zJ+w, with w∈Pp-1 (from Bp-1). Hence

	fp⁢xj⊗fp-1⁢xi⊗zJ	=zj⁢zi⁢zJ+fp-1⁢xj⊗w⁢from ⁢Ap
		=zj⁢zI+fp-1⁢xj⊗w.

The above defines a unique linear extension fp of fp-1 to 𝔤⊗Pp, and by construction this extension satisfies Ap and Bp. All that remains to show is that fp when defined this way satisfies Cp.
Observe that if j≤i and j≤J then Cp is satisfied by construction. Since xj,xi=-xi,xj, it is also satisfied if i<j and i≤J. Since Cp is trivially satisfied if i=j, we see that Cp is satisfied if i≤J or j≤J. Otherwise, J=k,K, where k≤K, k<i and k<j. For the sake of brevity we will write fp⁢x⊗z=x⁢z for x∈𝔤 and z∈Pp. Then from the induction hypothesis we have that

xizJ=xj(xkzK)=xk(xjzK)+[xj,xk]zK(*)

Now xj⁢zK is of the form zj⁢zK+w, where w∈Pp-2. As k≤K and k<j we can apply Cp to xi⁢xk⁢zj⁢zK, and to xi⁢xk⁢w from the induction hypothesis and hence also to xi⁢xk⁢xj⁢zK; using (*) this then gives us that

xi⁢xj⁢zJ=xk⁢xi⁢xj⁢zK+xi,xk⁢xj⁢zK+xj,xk⁢xi⁢zK+xi,xj,xk⁢zk.

Interchanging i and j, this gives us that

	xi⁢xj⁢zJ-xj⁢xi⁢zJ	=xk(xi(xjzK)-xj(xizK)+[xi,[xj,xk]]zK-[xj,[xi,xk]]zK
		=xk⁢xi,xj⁢zK+xi,xj,xk⁢zK+xj,xk,xi⁢zK
		=xi,xj⁢xk⁢zK+xk,xi,xj⁢zK+xi,xj,xk⁢zK+xj,xk,xi⁢zK
		=xi,xj⁢xk⁢zK
		=xi,xj⁢zJ,

as required. ∎

Lemma 1.5.

The yI, for every increasing sequence I, form a basis for the vector space U⁢𝔤.

Proof.

By the previous lemma (whose notation we will also use), there exists a bilinear mapping f of 𝔤 × P into P such that f⁢xi,zI=zi⁢zI for i≤I and

f⁢xi,f⁢xi,zJ=f⁢xj,f⁢xi,zJ+f⁢xi,xj,zJ,

for all i,j,J. Thus, we have a representation ϱ of 𝔤 in P such that ϱ⁢xi⁢zI=zi⁢zI for i≤I. From lemma 1.1 there exists a homomorphism φ of U(𝔤 into End((P) such that φ⁢yi⁢zI=zi⁢zI for i≤I. We thus obtain that, if i1≤i2≤⋯≤ip, then

φ⁢yi1⁢yi2⁢⋯⁢yip⋅1=zi1⁢zi2⁢⋯⁢zip.

Thus, the yI are linearly independent, for I increasing. From lemma 1.3 they generate U⁢𝔤 and hence form a basis. ∎

Proposition 1.6.

The canonical mapping of 𝔤 into U⁢𝔤 is injective.

Proof.

This is immediate from the previous lemma.∎

We thus have that 𝔤 is embedded in U⁢𝔤, so we will identify every element of 𝔤 with its canonical image in U⁢𝔤

Theorem 1.7 (Poincareé-Birkhoff-Witt).

Let x1,…,xn be a basis for the vector space 𝔤. Then the x1λ1⁢x2λ2⁢⋯⁢xnλn, where λ1,…,λn∈ℕ, form a basis for U⁢𝔤.

Proof.

This is again immediate from the previous lemma.∎

Taking the Poincaré-Birkhoff-Witt Theorem into account Lemma 1.1 can be restated as

Lemma 1.8.

Let A be an algebra with unity, τ a linear mapping of 𝔤 into A such that τ⁢x⁢τ⁢y-τ⁢y⁢τ⁢x=τ⁢x,y for all x,y∈𝔤. Then τ can be uniquely extended to a homomorphism of U⁢𝔤 into A which transforms 1 into 1.

Corollary 1.9.

Let V be a vector space, and R and R′ the sets of representations of 𝔤 and U⁢𝔤 in V respectively. For all ϱ∈R, there exists one and only one ϱ′∈R′ which extends ϱ, and the mapping ϱ→ϱ′ is a bijection of R onto R′.