Enveloping Algebras and the Poincare-Birkhoff-Witt Theorem
Rabhar Virk
September 20, 2011

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 U𝔤 such that the modules for 𝔤 are the same as those for U𝔤.

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


where Tn=𝔤𝔤𝔤 (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


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


We denote the canonical image in U𝔤 of T0+T1++Tq by Uq𝔤.

Lemma 1.1.

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


for all x,y𝔤. There exists one and only one homomorphism τ of U𝔤 into A such that τ1=1 and τσ=τ.


Note that U𝔤 is generated by 1 and σ𝔤 so τ must be unique. Now let φ be the unique homomorphism of T into A that extends τ and such that φ1=1. For x,y𝔤, we have


hence φJ=0 and thus φ gives us a homomorphism τ of U𝔤 into A such that τ1=1 and τσ=τ. ∎

Lemma 1.2.

Let a1,,ap𝔤, σ the canonical mapping of 𝔤 into U𝔤, and π be a permutation of 1,,p. Then


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


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 U𝔤 by yi. For every finite sequence I=i1,,ip of integers between 1 and n, we set yI=yi1yi2yip.

Lemma 1.3.

The yI, for all increasing sequences I of length p, generate the vector space Up𝔤.


Clearly the vector space Up𝔤 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 kz1,,zn of polynomials in n indeterminates z1,,zn. For every i0, 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=zi1zi2zip.

Lemma 1.4.

For every integer p0, there exists a unique linear mapping fp of the vector space 𝔤Pp into P which satisfies the following conditions:

  1. Ap

    fpxizI=zizI for iI, zIPp;

  2. Bp

    fpxizI-zizIPq for zIPq, qp;

  3. Cp

    fpxifpxjzJ=fpxjfpxizJ+fpxi,xjzJ for zJPp-1. [The terms in Cp are meaningful by virtue of (Bp)].

Moreover, the restriction of fp to 𝔤Pp-1 is fp-1.


For p=0 the mapping given by f0xi1=zi1, 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 𝔤Pp-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 𝔤Pp which satisfies Ap,Bp,Cp. Note that Pp is generated by zI for an increasing sequence I of p elements. Thus, we must define fpxizI for such a sequence I. If iI, then the choice is dictated by Ap. Otherwise we can write I as j,J where j<i and jJ. Then we must have that


Now fp-1xizJ=zizJ+w, with wPp-1 (from Bp-1). Hence

fpxjfp-1xizJ=zjzizJ+fp-1xjwfrom Ap

The above defines a unique linear extension fp of fp-1 to 𝔤Pp, 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 ji and jJ then Cp is satisfied by construction. Since xj,xi=-xi,xj, it is also satisfied if i<j and iJ. Since Cp is trivially satisfied if i=j, we see that Cp is satisfied if iJ or jJ. Otherwise, J=k,K, where kK, k<i and k<j. For the sake of brevity we will write fpxz=xz for x𝔤 and zPp. Then from the induction hypothesis we have that


Now xjzK is of the form zjzK+w, where wPp-2. As kK and k<j we can apply Cp to xixkzjzK, and to xixkw from the induction hypothesis and hence also to xixkxjzK; using (*) this then gives us that


Interchanging i and j, this gives us that


as required. ∎

Lemma 1.5.

The yI, for every increasing sequence I, form a basis for the vector space U𝔤.


By the previous lemma (whose notation we will also use), there exists a bilinear mapping f of 𝔤 × P into P such that fxi,zI=zizI for iI and


for all i,j,J. Thus, we have a representation ϱ of 𝔤 in P such that ϱxizI=zizI for iI. From lemma 1.1 there exists a homomorphism φ of U(𝔤 into End((P) such that φyizI=zizI for iI. We thus obtain that, if i1i2ip, then


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

Proposition 1.6.

The canonical mapping of 𝔤 into U𝔤 is injective.


This is immediate from the previous lemma.∎

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

Theorem 1.7 (Poincareé-Birkhoff-Witt).

Let x1,,xn be a basis for the vector space 𝔤. Then the x1λ1x2λ2xnλn, where λ1,,λn, form a basis for U𝔤.


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 τxτy-τyτx=τx,y for all x,y𝔤. Then τ can be uniquely extended to a homomorphism of U𝔤 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 U𝔤 in V respectively. For all ϱR, there exists one and only one ϱR which extends ϱ, and the mapping ϱϱ is a bijection of R onto R.

  • Updated: 2011-09-20
  • Status: in progress


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