Lie Theory, Universal Enveloping Algebras, and

the Poincare-Birkhoff-Witt Theorem

Lucas Lingle

August 22, 2012

Abstract

We investigate the fundamental ideas behind Lie groups, Lie algebras,and universal enveloping algebras. In particular, we emphasize the use-ful properties of the exponential mapping, which allows us to transitionbetween Lie groups and Lie algebras. From there, we discuss universalenveloping algebras, prove their existence and uniqueness, and after in-troducing the necessary machinery, we prove the Poincare-Birkhoff-WittTheorem.

1 Introduction

In the first section, we introduce Lie groups and prove some basic theoremsabout them. In the second section, we discuss and prove the properties ofthe exponential mapping. In the third section, we introduce Lie algebras andprove some important facts relating Lie groups to Lie algebras. In the fourthsection, we introduce universal enveloping algebras, and prove their existenceand uniqueness. In the fifth and final section, we prove the Poincare-Birkhoff-Witt Theorem and its corollaries.

2 Lie Groups

Definition 2.1. A Lie group G is group which is also a finite-dimensionalsmooth manifold, and in which the group operation and inversion are smoothmaps.

Definition 2.2. The general linear group over the real numbers, denotedGLn(R),is the set of all n × n invertible real matrices, equipped with the operation ofmatrix multiplication.

Similarly, the general linear group over the complex numbers, denotedGLn(C),is the set of all n× n invertible complex matrices, equipped with the operationof matrix multiplication.

1

Since the general linear groups only contain invertible matrices, each matrixin GLn(R) has an inverse in GLn(R), so the general linear groups are closedunder inversion. Since the product AB of any two invertible matrices A and Bis also invertible, and has entries in the same field as A and B, the general lineargroups are closed under the group operation. Lastly, since matrix multiplicationis associative, the elements of GLn(R) associate. Hence, GLn(R) is a group. Theabove logic likewise holds for GLn(C).

More abstractly, the general linear group of a vector space V , written GL(V ),is the automorphism group, whose elements can be written in matrix form butcan also be thought of as operators that form a group under composition.

Definition 2.3. Denote the set of all n× n complex matrices by Mn(C).

Definition 2.4. Let Am be a sequence of complex matrices in Mn(C). We saythat Am converges to a matrix A if each entry of the matrices in the sequenceconverges to the corresponding entry in A. That is, if (Am)kl converges to Aklfor all 1 ≤ k, l ≤ n, we say Am converges to A.

Definition 2.5. A matrix Lie group is any subgroup G of GLn(C) such thatif Am is any sequence of matrices in G converging to some matrix A, theneither A is in G or else A is not invertible.

Thus a matrix Lie group is a set algebraically closed under the inheritedgroup operation from GLn(C), and is also a topologically closed subset ofGLn(C). In other words, a matrix Lie group is a closed subgroup of GLn(C).

Definition 2.6. A matrix Lie group G is said to be compact if the followingtwo conditions are satisfied:

1. If Am is any sequence of matrices in G and Am converges to a matrixA, then A is in G.

2. There is some C ∈ R such that for all matrices A ∈ G, |Aij | ≤ C for all1 ≤ i, j ≤ n.

Definition 2.7. A matrix Lie group G is connected if given any two matricesA,B ∈ G, there exists a continuous path A(t), for a ≤ t ≤ b, so that A(a) = Aand A(b) = B.

Technically, this is what is known as path-connectedness in topology, andgenerally is not the same as connectedness. However, a matrix Lie group isconnected if and only if it is path connected, and so we shall continue to referto matrix Lie groups as connected when they are path-connected.

Definition 2.8. A matrix Lie group G that is not connected can be uniquelydescribed as the union of disjoint sets. Each such disjoint set is called a compo-nent of G.

Proposition 2.9. If G is a matrix Lie group, then the component of G con-taining the identity is a subgroup of G.

2

Proof. Let A and B be two matrices in the component of G containing theidentity. Then there exist two continuous paths A(t) and B(t), with A(0) =B(0) = I, A(1) = A, and B(1) = B. Then A(t)B(t) is a continuous path from Ito AB. But A and B are any two elements of the identity component, and theirproduct AB is also in the identity component, since the continuous path givenby A(t)B(t) goes from I to AB and such a continuous path can only be formedbetween elements of the same component. Let (A(t))−1 denote the inverse ofthe matrix given by A(t), for each t. Then (A(t))−1 goes from I to A−1, and bythe same logic as above, A−1 must be in the identity component as well. Sincethe identity component is closed under the inherited group operation and underinversion, it is a subgroup of G.

Definition 2.10. Let G and H be matrix Lie groups. A map Φ : G → H iscalled a Lie group homomorphism if Φ is continuous and Φ(g1g2) = Φ(g1)Φ(g2)for all g1, g2 ∈ G.

If Φ is a bijective Lie group homomorphism and Φ−1 is continuous, then Φis called a Lie group isomorphism.

3 The Exponential Mapping

Although Lie groups are endowed with some extra structure and thus are aneasier form of manifold to study, they themselves can still be difficult to dealwith. For this reason, we often deal with a more wieldy object, namely the Liealgebra corresponding to the group. In order to transfer information from theLie algebra to the Lie group, we use a function called the exponential mapping.

Definition 3.1. Let X be any matrix. Define the matrix exponential by

eX =

∞∑m=0

Xm

m!.

One might wonder if this even converges. As we will see shortly, the answeris an emphatic yes. First, though, we must introduce a few new concepts.

Definition 3.2. The Hilbert-Schmidt norm of an n× n matrix X is given by

||X|| =( n∑j=1

n∑i=1

|xij |2)1/2

.

It is easy to verify, using the triangle and Cauchy-Schwarz inequalities, thatthe norm obeys the following:

||X + Y || ≤ ||X||+ ||Y ||,

||XY || ≤ ||X|| ||Y ||.

3

Proposition 3.3. For any n × n real or complex matrix X, the series aboveconverges. Furthermore, eX is a continuous function of X.

Proof. Since we are working with matrices having real or complex entries, weknow that there is some entry whose absolute value is the greatest among theentries. Let M denote the maximum, in absolute value, of all entries of thematrix X. Then |(X)ij | ≤M , and since X is a n× n matrix, |(X2)ij | ≤ nM2,and so on. In general, |(Xm)ij | ≤ nm−1Mm. Then

∞∑m=0

nm−1Mm

m!

converges by a simple application of the ratio test. Then since |(Xm)ij | ≤nm−1Mm, we can use the comparison test. Thus, the sum

∞∑m=0

|(Xm)ij |m!

=

∞∑m=0

∣∣∣∣(Xm

m!

)ij

∣∣∣∣converges as well. Then by a basic theorem from analysis, we know that since

∞∑m=0

(Xm

m!

)ij

converges absolutely, it converges in general. By Definition 2.4, we know thesequence (of partial sums) of matrices converges—and hence

∞∑m=0

Xm

m!= eX

converges. It is easy to see that eX is continuous.

Now that we see that the exponential mapping is well-behaved, we can provesome important properties about it.

Proposition 3.4. Let X and Y be arbitrary n×n matrices, and let M∗ denotethe conjugate transpose of a matrix M . Then we have the following:

1. e0 = I,2. (eX)∗ = e(X∗),3. eX is invertible and (eX)−1 = e−X ,4. e(α+β)X = eαXeβX for all α, β ∈ C,5. if XY = Y X, then eX+Y = eXeY = eY eX ,6. if C is invertible, then eCXC

−1

= CeXC−1,7. ||eX || ≤ e||X||.

4

Proof. Point 1 is obvious, and Point 2 follows from taking the conjugate trans-poses term-wise. Points 3 and 4 are special cases of Point 5.

For Point 5, we note that since eZ converges for all Z, eXeY is defined forall X and Y . Furthermore,

eXeY =

(I +X +

X2

2!+ · · ·

)(I + Y +

Y 2

2!+ · · ·

).

Multiplying out, and collecting terms where the power of X plus the power ofY is m, we get

eXeY =

∞∑m=0

m∑k=0

Xk

k!

Y m−k

(m− k)!=

∞∑m=0

1

m!

m∑k=0

m!

k!(m− k)!XkY m−k.

And since X and Y commute,

(X + Y )m =

m∑k=0

m!

k!(m− k)!XkY m−k.

So we get

eXeY =

∞∑m=0

1

m!(X + Y )m = e(X+Y ).

Point 6 follows immediately, since each term of the matrix exponential canbe written as

(CXC−1)m

m!=

(CXC−1)(CXC−1) · · · (CXC−1)(CXC−1)

m!= C

(Xm

m!

)C−1.

For Point 7, notice that for each m ∈ N, by the Cauchy-Schwarz inequality,∣∣∣∣∣∣∣∣Xm

m!

∣∣∣∣∣∣∣∣ =||Xm||m!

≤ ||X||m

m!.

And since ||X|| is a real number,

e||X|| =

∞∑m=0

||X||m

m!

converges. By the comparison test, we know that

∞∑m=0

∣∣∣∣∣∣∣∣Xm

m!

∣∣∣∣∣∣∣∣.converges as well. It follows from the triangle inequality that

SK :=

∣∣∣∣∣∣∣∣ K∑m=0

Xm

m!

∣∣∣∣∣∣∣∣ ≤ K∑m=0

∣∣∣∣∣∣∣∣Xm

m!

∣∣∣∣∣∣∣∣ ≤ K∑m=0

||X||m

m!=: LK .

5

Since the sequence defined by

EK :=

K∑m=0

Xm

m!

converges (to eX), we know that the sequence

SK :=

∣∣∣∣∣∣∣∣ K∑m=0

Xm

m!

∣∣∣∣∣∣∣∣converges as well (to ||eX ||). It follows that

limK→∞

SK =∣∣∣∣eX ∣∣∣∣ ≤ e||X|| = lim

K→∞LK .

Proposition 3.5. Let X be a n × n complex matrix. Then etX is a smoothcurve in GLn(C) and

d

dtetX = XetX = etXX.

In particular,d

dt

∣∣∣∣t=0

etX = X.

Proof. For each i and j, we know (etX)ij is given by an everywhere convergentpower series and so we can find d

dtetX by differentiating the power series for etX

term by term. Everything else follows immediately.

Proposition 3.6. Let X and Y be n× n complex matrices. Then

eX+Y = limm→∞

(e

Xm e

Ym

)m.

Though this result is important, we will not prove it here, as it relies on thematrix logarithm, which we have avoided discussing due to space constraints.A good proof can be found in [1].

Definition 3.7. A function A : R → GLn(C) is a one-parameter subgroup ofGLn(C) if

1. A is continuous,2. A(0) = I,3. A(t+ s) = A(t)A(s) for all t, s ∈ R.

Theorem 3.8. If A is a one-parameter subgroup of GLn(C), then there existsa unique n× n complex matrix X so that A(t) = etX .

Though this is an important result that we will use later, we will not proveit; the proof builds upon the concept of the matrix logarithm. Skeptics shouldconsult [1].

6

4 Lie Algebras

As explained in the previous section, it will be convenient to explore a Lie group’sLie algebra—its tangent space at the identity element. Such inquiry will be quiterewarding, as it will let us discover important and otherwise difficult-to-accessinformation with ease.

Definition 4.1. A finite-dimensional real or complex Lie algebra is a finite-dimensional real or complex vector space g together with a map [·, ·] from g× ginto g with the following properties:

1. [·, ·] is bilinear,2. [X,Y ] = −[Y,X] for all X,Y ∈ g,3. [X, [Y, Z]] + [Y, [X,Z]] + [Z, [X,Y ]] = 0 for all X,Y, Z ∈ g.

This property is called the Jacobi identity.

Definition 4.2. Let G be a matrix Lie group. The Lie algebra of G, denotedg, is the set of all matrices X such that etX ∈ G for all real numbers t, and werefer to g as a matrix Lie algebra.

These definitions may not seem to coincide; after proving the next few propo-sitions, it will become clear that matrix Lie algebras satisfy Definition 4.1.

Proposition 4.3. Let G be a matrix Lie group, and X an element of its Liealgebra. Then eX is an element in the identity component of G.

Proof. By the definition of the Lie algebra for matrix Lie groups, we knowetX ∈ G for all real numbers t. We know A(t) = etX is a continuous functiongoing from I to eX as t goes from 0 to 1, and since I is in the identity componentof G, we know eX is as well.

Proposition 4.4. Let G be a matrix Lie group with Lie algebra g. Let X be anelement of g, and A be an element of G. Then AXA−1 is in g.

Proof. It follows from Proposition 3.4 that etAXA−1

= AetXA−1. By the def-inition of a Lie algebra for a matrix group we know that etX is in G for allreal t, and since G is closed under inversion, we know A−1 is in G; thusAetXA−1 = et(AXA

−1) ∈ G, which implies AXA−1 ∈ g by the definition ofthe Lie algebra for a matrix Lie group.

Now we are well-positioned to prove that matrix Lie algebras are indeed Liealgebras: as this next theorem tells us, they are vector spaces which we canequip with a bilinear antisymmetric operation satisfying the Jacobi identity.

Theorem 4.5. Let G be a matrix Lie group with Lie algebra g, and let X andY be elements of g. Then

1. sX ∈ g for all real numbers s,2. X + Y ∈ g,

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3. XY − Y X ∈ g.

Proof. Part 1. For X ∈ g, and all real t and s, we know e(ts)X ∈ G. Thensince e(ts)X = et(sX), we know that et(sX) ∈ G for all real t and s. Then by thedefinition of a Lie algebra for a matrix Lie group we know sX ∈ g for all real s.

Part 2. If X,Y ∈ g commute, then for all real t, we know et(X+Y ) = etXetY .Clearly etX and etY are in G for all real t, so we know et(X+Y ) is in G as well,which means X + Y ∈ g. In the general case, however, we use Proposition 3.6,the Lie product formula:

et(X+Y ) = limm→∞

(etX/metY/m

)m.

BecauseX,Y ∈ g, we know that etX/m and etY/m are inG and so is (etX/metY/m)m,since G is a group. However, since G is a matrix Lie group, the limit of elementsin G are also in G, so long as the limit is invertible. Since et(X+Y ) is invertible,we know limm→∞(eX/meY/m)m is in G as well. Then limm→∞(etX/metY/m)m

is in G, and hence so is et(X+Y ). This implies that X + Y ∈ g.Part 3. Using Proposition 3.5, we can see that

d

dtetXY

∣∣∣∣t=0

= XY.

Then by the product rule,

d

dtetXY e−tX

∣∣∣∣t=0

= (XY )e0 + (e0Y )(−X) = XY − Y X.

By Parts 1 and 2, we know that g is a vector space. By Proposition 4.4, weknow etYXe−tY ∈ g for all real t. Finally, since γ(t) = etXY e−tX is a smoothcurve through g, we know the derivative of γ with respect to t exists and isalways in g; indeed, the derivative of a curve in a vector space is always in thatvector space. Therefore,

d

dtγ(t)

∣∣∣∣t=0

= XY − Y X

is in g.

Definition 4.6. Given two n× n matrices A and B, the commutator of A andB, denoted [A,B], is defined to be AB −BA.

It is easy to verify that the commutator satisfies all the necessary propertiesthat the bracket of a Lie algebra must have. Furthermore, since the commutatorof two matrices in a matrix Lie algebra is also that matrix Lie algebra, we shalluse the commutator for our bracket operation when dealing with matrix Liealgebras.

It should be noted that [·, ·] is used to denote the Lie bracket of any Liealgebra and need not correspond to the commutator. However, since we tend touse the commutator as our Lie bracket for matrix Lie algebras, it inherits thesomewhat ambiguous bracket notation.

8

Definition 4.7. A subalgebra of a real or complex Lie algebra g is a subspaceh of g such that [H1, H2] ∈ h for all H1, H2 ∈ h. If g is a complex Lie algebraand h is a real subspace of g closed under brackets then we say that h is a realsubalgebra of g.

If g and h are Lie algebras then a linear map φ : g→ h is called a Lie algebrahomomorphism if φ([X,Y ]) = [φ(X), φ(Y )] for all X,Y ∈ g. Furthermore, if φis also bijective, then φ is called a Lie algebra isomorphism. Lastly, a Lie algebraisomorphism with Lie algebra g as both its domain and codomain is called a Liealgebra automorphism.

Theorem 4.8. Let G and H be Lie groups with Lie algebras g and h, respec-tively. Suppose Φ : G→ H is a Lie group homomorphism. Then there exists aunique linear map φ : g→ h such that Φ(eX) = eφ(X), and

1. φ(AXA−1) = Φ(A)φ(X)Φ(A)−1, for all X ∈ g, A ∈ G,2. φ([X,Y ]) = [φ(X), φ(Y )], for all X,Y ∈ g,3. φ(X) = d

dtΦ(etX)|t=0, for all X ∈ g.

Proof. Since Φ is a continuous group homomorphism and etX is also continuous,we know Φ(etX) will be a one-parameter subgroup of H. By Theorem 3.8, weknow there is a unique matrix Z so that Φ(etX) = etZ for all t ∈ R. Furthermore,we know Z ∈ h, since etZ = Φ(etX) ∈ H for all t. Now we simply defineφ(X) = Z and check that the necessary properties are satisfied.

Step 1: Φ(eX) = eφ(X).This follows from a few simple facts: we know etZ = Φ(etX) for all t, and

φ(X) = Z. Thus, etφ(X) = Φ(etX) for all t, and in particular for t = 1, we knoweφ(X) = Φ(eX). Now for linearity!

Step 2: φ(sX) = sφ(X).For all s, t ∈ R, we have etφ(sX) = Φ

(et(sX)

), and et(sφ(X)) = Φ

(et(sX)

).

But for the first equation, we know that φ(sX) is the unique matrix so thatetφ(sX) = Φ(etsX). By the second equation, we know sφ(X) is the uniquematrix so that esφ(X) = Φ(etsX). Hence, sφ(X) = φ(sX).

Step 3: φ(X + Y ) = φ(X) + φ(Y ).By Steps 1 and 2, we know that

etφ(X+Y ) = eφ(t(X+Y )) = Φ(et(X+Y )

).

By the Lie product formula from Proposition 3.6, and the fact that Φ is acontinuous homomorphism, we have

etφ(X+Y ) = Φ

(limm→∞

(etX/metY/m

)m)= limm→∞

(Φ(etX/m)Φ(etY/m)

)m.

But then by the relationship between Φ and φ, and applying the Lie productformula from Proposition 3.6, we know that

limm→∞

(Φ(etX/m)Φ(etY/m)

)m= limm→∞

(etφ(X)/metφ(Y )/m

)m= et(φ(X)+φ(Y )).

9

Thus, etφ(X+Y ) = et(φ(X)+φ(Y )). Using Proposition 3.5, we can differentiateboth of these at t = 0 to get φ(X + Y ) = φ(X) + φ(Y ).

Step 4: φ(AXA−1) = Φ(A)φ(X)Φ(A)−1.By Steps 1 and 2,

etφ(AXA−1) = eφ(tAXA−1) = Φ(etAXA−1

).

By Proposition 3.4 and Step 1, we know that

etφ(AXA−1) = Φ(etAXA−1

) = Φ(AetXA−1) = Φ(A)Φ(etX)Φ(A−1).

And since we know that Φ(A−1) = Φ(A)−1 for any homomorphism Φ, and sinceΦ(etX) = etφ(X), we know

etφ(AXA−1) = Φ(A)etφ(X)Φ(A)−1.

Differentiating at t = 0 we obtain

φ(AXA−1) = Φ(A)φ(X)Φ(A)−1.

Step 5: φ([X,Y ]) = [φ(X), φ(Y )].Recall from the proof of Theorem 4.5, we know that

[X,Y ] =d

dtetXY e−tX

∣∣∣∣t=0

.

Hence,

φ([X,Y ]) = φ

(d

dtetXY e−tX

∣∣∣∣t=0

)=

d

dtφ(etXY e−tX)

∣∣∣∣t=0

,

since a derivative commutes with a linear transformation. Then by Step 1,

φ([X,Y ]) =d

dtΦ(etX)φ(Y )Φ(e−tX)

∣∣∣∣t=0

=d

dtetφ(X)φ(Y )e−tφ(X)

∣∣∣∣t=0

.

Of course, we know from the proof of Theorem 4.5 that the far right side of thisequation is equal to φ(X)φ(Y )− φ(Y )φ(X), so

φ([X,Y ]) = [φ(X), φ(Y )],

and thus φ is a Lie algebra homomorphism.Step 6: φ(X) = d

dtΦ(etX)∣∣t=0

.

To begin with, it is clear that etφ(X) = Φ(etX), so

d

dtΦ(etX)|t=0 =

d

dtetφ(X)

∣∣∣∣t=0

.

10

By Proposition 3.5, we know ddte

tφ(X)

∣∣∣∣t=0

= φ(X), so

φ(X) =d

dtΦ(etX)

∣∣∣∣t=0

.

Step 7: φ is the unique linear map such that Φ(etX) = etφ(X).Suppose ψ is another such linear map. Then,

etψ(X) = eψ(tX) = Φ(etX).

And so by Step 6,

ψ(X) =d

dtΦ(etX)

∣∣∣∣t=0

= φ(X).

Theorem 4.9. Suppose that G, H, and K are matrix Lie groups, with cor-responding Lie algebras g, h, and k. Let Φ : H → K and Ψ : G → H beLie group homomorphisms, and let Λ : G → K be the composition of Φ andΨ, so that Λ(A) = Φ(Ψ(A)) for all A in G. Let φ : h → k, ψ : g → h, andλ : g→ k be the associated Lie algebra homomorphisms such that eφ(X) = Φ(eX),eψ(X) = Ψ(eX), and eλ(X) = Λ(eX). Then for all X ∈ g, λ(X) = φ(ψ(X)).

Proof. For any X ∈ g,

etλ(X) = Λ(etX) = Φ(Ψ(etX)

)= Φ

(etψ(X)

)= etφ(ψ(X)).

Differentiating at t = 0, we know by Proposition 3.5 that

λ(X) =d

dtetλ(X)

∣∣∣∣t=0

=d

dtetφ(ψ(X))

∣∣∣∣t=0

= φ(ψ(X)).

Definition 4.10. Let G be a matrix Lie group with Lie algebra g. Then foreach A ∈ G define a linear map AdA : g→ g by the formula AdA(X) = AXA−1.This map is called the adjoint mapping.

Proposition 4.11. Let G be a matrix Lie group, with Lie algebra g. Let GL(g)denote the group of all invertible linear transformations of g. Then for each A ∈G, AdA is an invertible linear transformation of g with inverse AdA−1 , and themap Ad : A 7→ AdA is a group homomorphism of G into GL(g). Furthermore,for each A ∈ G, AdA satisfies AdA([X,Y ]) = [AdA(X), AdA(Y )] for all X,Y ∈g.

Proof. We can see that for any X ∈ g, and any A ∈ G,

AdA(AdA−1(X)) = A(A−1XA)A−1 = X = A−1(AXA−1)A = AdA−1(AdA(X)).

11

And now

Ad(AB) = AdAB( · ) = AB( · )B−1A−1 = AdA(AdB( · )) = Ad(A)(Ad(B)).

Since multiplication of matrices is the group operation of G and composition oflinear maps is the group operation in GL(g), we know that the Ad operator isa group homomorphism. And lastly, for any X,Y ∈ g and any A ∈ G,

AdA([X,Y ]) = A(XY − Y X)A−1 = AXY A−1 −AYXA−1.

And the far right side of this equality is clearly

AXA−1AY A1 −AY A−1AXA−1 = AdA(X)AdA(Y )−AdA(Y )AdA(X).

Thus AdA([X,Y ]) = [AdA(X), AdA(Y )], and so the map AdA is a Lie algebrahomomorphism for each A ∈ G.

Since g is a real vector space with some dimension k, we can pick a basis for g,and GL(g) can be written as a group of matrices with the group operation beingmatrix multiplication. (This notion is consistent with the traditional meaningof the general linear group of a vector space as the group of automorphisms onthat vector space, having composition as the group operation.) Thus we canregard GL(g) as a matrix Lie group. It is easy to show that Ad : G→ GL(g) iscontinuous and so is a Lie group homomorphism. By Theorem 4.8, there is anassociated real linear map ad taking X to adX from the Lie algebra of G to theLie algebra of GL(g) (that is, from g to gl(g), the space of all endomorphismsof g). Specifically, ad satisfies

eadX = Ad(eX).

Proposition 4.12. Let G be a matrix Lie group with Lie algebra g. Let Ad :G → GL(g) be the Lie group homomorphism defined above. Let ad : g → gl(g)be the associated Lie algebra map. Then for all X,Y ∈ g, adX(Y ) = [X,Y ].

Proof. By Theorem 4.8, we know that ad can be calculated by

adX =d

dtAd(etX)

∣∣∣∣t=0

.

Thus,

adX(Y ) =d

dtAd(etX)(Y )

∣∣∣∣t=0

=d

dtetXY e−tX

∣∣∣∣t=0

= [X,Y ].

Definition 4.13. A representation of a Lie group G on a vector space V is a Liegroup homomorphism ρ : G→ GL(V ), sending elements of G to automorphismson V . Similarly, a representation of a Lie algebra g on a vector space V is a Liealgebra homomorphism ρ : g→ gl(V ), mapping elements of g to endomorphismson V .

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Though we will not prove it here, it turns out that all finite-dimensional Liealgebras can be represented with matrices.

Theorem (Ado) 4.14. Every finite-dimensional real Lie algebra is isomorphicto a real subalgebra of gln(R). Every finite-dimensional complex Lie algebra isisomorphic to a complex subalgebra of gln(C).

5 Universal Enveloping Algebras

Generally speaking, a Lie algebra g does not have any defined notion of asso-ciative multiplication. However, if we consider a representation ρ : g → gl(V )then the product ρ(X)ρ(Y ) is well-defined. (Note that the “product” is actuallycomposition of the operators ρ(X) and ρ(Y ), which are endomorphisms of V .)

With convenience of multiplication and the structure passed on by the com-mutator, we will define the notion of the “universal” associative algebra gener-ated by “products” of operators of the form ρ(X) for X ∈ g. To make thingsboth more formal and more interesting, we will first introduce a few new con-cepts:

Definition 5.1. An associative algebra A is a vector space V over a field Kequipped with an associative, bilinear vector product · : V × V → V . If thereis some element 1 ∈ V such that 1 · a = a = a · 1 for every a ∈ A, then we saythat A is unital, or “has unit.” Often we will describe an associative algebra Aas “having unit over K” to mean that it is a unital algebra with its vector spaceover a field K.

In particular, we will be very interested in an associative algebra called thetensor algebra. But first, we must define the tensor product.

Definition 5.2. Let Vp for 1 ≤ p ≤ k and W be modules over a ring R. Amodule T = TV1,···Vk

together with a multilinear map ⊗ : V1 × · · · × Vk → T iscalled universal for k-multilinear maps on V1 × · · · × Vk if for every multilinearmap µ : V1 × · · ·Vk → W there is a unique linear map µ : T → W such thatµ ⊗ = µ. If such a universal object exists, it will be called a tensor product.

It turns out that the tensor product is unique up to isomorphism.

Proposition 5.3. If (T1,⊗1) and (T2,⊗2) are both universal for k-multilinearmaps on V1 × · · · × Vk, then there is a unique isomorphism Φ : T1 → T2 suchthat Φ ⊗1 = ⊗2.

Proof. By the assumption of universality, we know that there are maps ⊗1 and⊗2 such that Φ ⊗1 = ⊗2, and Φ ⊗2 = ⊗1. Thus we have Φ Φ ⊗1 = ⊗1,and by the uniqueness part of the universality of ⊗1 it follows that Φ Φ = Id.Similarly, Φ Φ ⊗2 = ⊗2, and by the uniqueness part of the universality of ⊗2

it follows that Φ Φ = Id. Thus Φ = Φ−1.

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More concretely, the realization of the tensor product of modules V1, · · ·Vkis, at least roughly, the set of all linear combinations of symbols of the formv1 ⊗ · · · ⊗ vk subject to the multilinear relations

v1 ⊗ · · · ⊗ avi ⊗ · · · ⊗ vk = a(v1 ⊗ · · · ⊗ vi ⊗ · · · ⊗ vk),

and

v1⊗· · ·⊗ (vi+v′i)⊗· · ·⊗vk = (v1⊗· · ·⊗vi⊗· · ·⊗vk)+(v1⊗· · ·⊗v′i⊗· · ·⊗vk).

This space is denoted by V1 ⊗ · · · ⊗ Vk, and is called the tensor product ofthe modules V1, · · · , Vk.

Definition 5.4. The tensor algebra of a vector space V over a field K, denotedT (V ), is the associative algebra of tensors on V , with the tensor product ⊗serving as the associative, bilinear vector product.

For any vector space, we can construct its tensor algebra as follows:Let T 0V = K. For any k ∈ N, define the k-th tensor power of V , denoted

T kV , to be the tensor product of V with itself k times:

T kV = V ⊗ V ⊗ · · · ⊗ V.

From here, we simply take the direct sum of the T kV for k = 0, 1, 2, . . . ,

T (V ) =

∞⊕k=0

T kV = K⊕ V ⊕ (V ⊗ V )⊕ (V ⊗ V ⊗ V ) · · · .

Now we are in a suitable position to discuss what is known as the “universalproperty” of tensor algebras.

Proposition 5.5. Let A be any associative algebra with unit over K, and letf : V → A be a linear map. Then there exists a unique algebra homomorphismf : T (V ) → A such that f = f i, where i : V → T (V ) is the inclusion ofV = T 1V into T (V ).

Proof. Let A be any associative algebra with unit over K. For any linear mapf : V → A, define a linear map f : T (V )→ A by

f(v1 ⊗ · · · ⊗ vk) = f(v1) · · · f(vk).

This will be well-defined for any k ∈ N, and it is easy to see that it is indeed analgebra homomorphism. However, if k = 0, then we must clarify. Fortunately,this is easy: since A is an algebra with unit over K, we can simply let f(`) = `·1,where 1 is the unit element of A and ` ∈ K. The above definition is clearly theonly way to extend f as a homomorphism, since V generates T (V ) as a K-algebra.

Now for one more detail that will be important later:

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Definition 5.6. Let I be the two-sided ideal in T (V ) generated by all theX ⊗Y −Y ⊗X (where X,Y ∈ V ). Define the symmetric algebra of V , denotedSym(V ), by Sym(V ) = T (V )/I.

In other words, the symmetric algebra of V is just the commutative versionof the tensor algebra. With that out of the way, we are ready to move on to thereal topic of this section.

Definition 5.7. Let g be a Lie algebra over a field K. The universal envelopingalgebra of g is a pair (Ug, i), satisfying the following:

1. Ug is an associative algebra with unit over K,2. i : g→ Ug is linear and i(X)i(Y )− i(Y )i(X) = i([X,Y ]), for all X,Y ∈ g,3. for any associative algebra A with unit over K and for any linear mapj : g→ A satisfying j(X)j(Y )− j(Y )j(X) = j([X,Y ]) for each X,Y ∈ g,there exists a unique homomorphism of algebras φ : Ug → A such thatφ i = j.

Notice that the Lie bracket is not necessarily the commutator—after all,there may be no notion of associative multiplication in g—but that applying ito the bracket of any two X,Y ∈ g must give us the commutator of i(X) andi(Y ).

Theorem 5.8. For any Lie algebra g over an arbitrary field K, there exists aunique universal enveloping algebra (Ug, i), up to isomorphism.

Proof. Uniqueness: Suppose that the Lie algebra g has two universal envelopingalgebras (Ug, i) and (Ug′, i′). Then by definition, for each associative K-algebraA there exists a unique homomorphism λA : Ug → A. In particular, sinceUg′ is an associative K-algebra, we have a unique homomorphism of algebrasλ : Ug → Ug′. Switching the roles of Ug and Ug′ and applying the same logic,we know there exists a unique homomorphism of algebras µ : Ug′ → Ug. Thenλ µ = 1Ug, and µ λ = 1Ug′ , which means λ is bijective. But a bijectivehomomorphism of algebras is an isomorphism of algebras. Thus (Ug, i) is uniqueup to isomorphism.

Existence: Let T (g) be the tensor algebra of g, and let J be the two-sidedideal in T (g) generated by all X⊗Y −Y ⊗X−[X,Y ], where X,Y ∈ g. We claimthat Ug = T (g)/J satisfies all the necessary conditions delineated in Definition5.7. Let π : T (g) → Ug be the homomorphism mapping each element of thetensor algebra to its equivalence class in the associative algebra T (g)/J . Clearly,

J ⊂⊕k>0

T kg.

It follows that π maps T 0g = K isomorphically into T (g)/J , and hence Ug atleast contains scalars—great! Now let i : g → Ug be the restriction of π tog ⊂ T (g). Let A be any associative algebra with unit over K, and let j : g→ Abe a linear map satisfying j(X)j(Y ) − j(Y )j(X) = j([X,Y ]) for all X,Y ∈ g.

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The universal property of tensor algebras from Proposition 5.5 gives us a uniquealgebra homomorphism φ′ : T (g)→ A that extends j and sends 1 to 1. It followsfrom the special property of j that X ⊗ Y − Y ⊗ X − [X,Y ] is in Ker(φ′) forall X,Y ∈ g. Thus, since each of the elements in T (g) generated by the termsX⊗Y −Y ⊗X−[X,Y ] gets mapped to zero, we can identify all such elements anda homomorphism shall still exist. In other words, φ′ induces a homomorphismφ : Ug → A such that φ i = j. The uniqueness of φ is evident, since 1 andIm(i) together generate Ug.

Now for a simple example. If g is an abelian Lie algebra (i.e., [X,Y ] = 0 forall X,Y ∈ g), then Ug = T (g)/J , where J is the two-sided ideal generated byall X ⊗ Y − Y ⊗ X − [X,Y ]. But since [X,Y ] is always zero, we simply haveUg = Sym(g).

6 The Poincare-Birkhoff-Witt Theorem

It turns out that g is mapped injectively into Ug, and hence universal envelopingalgebras are indeed “enveloping” in some sense. This becomes quite useful, sincewe can proceed think of each i(X) as simply being the corresponding X ∈ g;the lack of restrictions that characterize the universal enveloping algebra allowus to perform feats that would be cumbersome or impossible in the Lie algebraitself. This important result turns out to be a simple corollary of the muchstronger Poincare-Birkhoff-Witt Theorem.

Definition 6.1. A graded algebra G is an associative algebra that can be de-composed into the direct sum of abelian groups Gk (with the group operationbeing addition of vectors) and is characterized by the fact that if X ∈ Gm andY ∈ Gp then X · Y ∈ Gm+p.

For instance, the tensor algebra T (V ) of any vector space V is a gradedalgebra. Clearly if X ∈ TmV and Y ∈ T pV , then X ⊗ Y ∈ Tm+pV . If we modout by the two-sided ideal I generated by the X ⊗ Y − Y ⊗X, for X,Y ∈ V ,we get Sym(V ) = T (V )/I. This is a graded algebra as well, as we can define agrading SmV = TmV/I.

Definition 6.2. A filtration is an indexed set Qi of subobjects in a givenalgebraic structure Q, where the index i runs over an index set S which istotally ordered, and if i ≤ j then Qi ⊂ Qj .

Definition 6.3. A filtered algebra F is an associative algebra over some fieldK, which has a filtration of linear subspaces, indexed by a set S, satisfying0 ⊂ F0 ⊂ F1 ⊂ · · · ⊂ F, recovering F via the union⋃

s∈SFs = F,

and satisfying the following: if X ∈ Fm and Y ∈ Fp, then X · Y ∈ Fm+p.

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So far, we know very little about Ug, other than the fact that it containsthe scalars which were passed on isomorphically from K. For brevity, we shallwrite T instead if T (g), and Sym instead of Sym(g). Similarly, we shall writeTm instead of Tmg, and Sm instead of Smg. For algebras other than the tensoralgebra itself, we shall also frequently omit the ⊗, choosing a dot or simplyplacing variables next to each other to indicate multiplication.

Define a filtration on T by

Tm := T 0 ⊕ T 1 ⊕ · · · ⊕ Tm.

Recall that π : T → Ug is the quotient map. Let Um = π(Tm), and U−1 = 0.Suppose we have W ∈ Tm, Z ∈ Tp, and define X = π(W ) ∈ Um, and Y =π(Z) ∈ Up. Then W ⊗ Z ∈ Tm+p, which implies that π(W ⊗ Z) ∈ π(Tm+p),and hence

XY = π(W )π(Z) = π(W ⊗ Z) ∈ π(Tm+p) = Um+p.

Thus, for all X ∈ Um, Y ∈ Up, XY ∈ Um+p, so the Um’s form a filtration onUg.

DefineGm := Um/Um−1

(this is just a vector space), and let the multiplication in Ug define a bilinear mapGm ×Gp → Gm+p. This operation is well defined, as we shall see momentarily.Suppose we have two representatives X,X ′ ∈ Up of the same equivalence classin Gp, and two representatives Y, Y ′ ∈ Um of the same equivalence class in Gm.Define W = X −X ′ ∈ Up−1 and Z = Y − Y ′ ∈ Um−1. Then

XY = (X ′ +W )(Y ′ + Z) = X ′Y ′ + (X ′Z +WY ′ +WZ).

But surely WY ′ ∈ Um+p−1, X ′Z ∈ Um+p−1, and WZ ∈ U(m−1)+(p−1) ⊂Um+p−1. Thus, when we mod out by Um+p−1, in accordance with our defi-nition of Gm+p, all the terms in the parentheses vanish. Having gotten that outof the way, we define

G =

∞⊕m=0

Gm.

This gives us a bilinear map G × G → G in accordance with the rules of mul-tiplication for Gm × Gp → Gm+p. It is clear that G is a graded associativealgebra with unit.

Since π maps the last bit Tm of each Tm into Um, it follows that the com-posite linear map φm : Tm → Um → Gm = Um/Um−1 is well defined. And wecan write Tm = Tm−1⊕Tm. Clearly, π maps Tm surjectively onto Um = π(Tm).And surely the mapping from Um to Um/Um−1 is surjective, and so the mapΦm : Tm → Um → Um/Um−1 is surjective as well. And since this map maps ev-erything in Tm−1 to Um−1/Um−1 = 0, we know that the restriction of Φm toTm, which we call φm : Tm → Um → Um/Um−1, hits everything in Um/Um−1,except possibly zero. But we know 0 = 0⊗ · · ·⊗ 0 ∈ Tm, so zero is in the imageof Tm under φm. Hence, φm is surjective onto Gm. The maps φm therefore canbe combined to give us a surjective linear map φ : T → G sending 1 to 1.

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Lemma 6.4. The map φ : T → G is an algebra homomorphism. Moreover, if Iis the two-sided ideal generated by X⊗Y −Y ⊗X for X,Y ∈ g, then I ∈ Ker(φ),and so φ induces a homomorphism ω of Sym = T/I onto G.

Proof. Suppose we have some X ∈ Tm and Y ∈ T p. It follows that φ(X) ∈ Gm,φ(Y ) ∈ Gp, and X ⊗ Y ∈ Tm+p, so φ(X ⊗ Y ) ∈ Gm+p. Then by the definitionof the product in G, it follows that φ(X ⊗ Y ) = φ(X)φ(Y ) for each X,Y ∈ T .Thus, φ is a (surjective) algebra homomorphism.

Let X ⊗ Y − Y ⊗X (for X,Y ∈ g) be a typical generator of the two-sidedideal I described earlier. Then π(X ⊗ Y − Y ⊗ X) ∈ U2, by definition. Onthe other hand, we also know π(X ⊗ Y − Y ⊗ X) = π([X,Y ]) ∈ U1, and soφ(X ⊗ Y − Y ⊗ X) ∈ U1/U1 = 0. Hence I ⊂ Ker(φ), and so if we identifyall the elements of the ideal I with the zero vector, we surely shall still have a(surjective) algebra homomorphism ω : Sym→ G.

It turns out that ω is not only a surjective algebra homomorphism, butis also injective, and hence is an isomorphism of algebras. This fundamentalresult is known as the Poincare-Birkhoff-Witt Theorem, which we shall proveafter introducing the important corollaries it entails.

Theorem (Poincare-Birkhoff-Witt) 6.5. The homomorphism ω : Sym→ Gis an isomorphism of algebras.

Corollary 6.6. Let W be a subspace of Tm. Suppose the canonical map Tm →Sm sends W isomorphically onto Sm. Then π(W ) is a complement to Um−1 inUm.

Proof. Let gm be the quotient map from Tm to Sm, and let hm be the quotientmap from Um to Um/Um−1. By Lemma 6.4, and the definitions, we know thediagram below is commutative:

Tmπ //

gm

Um

hm

Sm

ω// Gm

Since gm sends W ⊂ Tm isomorphically onto Sm by our supposition, andsince ω : Sym → G is an isomorphism by the Poincare-Birkoff-Witt Theorem,we know the map ω gm sends W isomorphically onto Gm. Since W is mappedisomorphically, it is mapped injectively, and hence Ker(hm π) ∩W = 0. Itfollows that Ker(hm)∩ π(W ) = 0 as well. But the kernel of hm is just Um−1,and so Um−1 ∩ π(W ) = 0.

And since W is mapped isomorphically onto Gm, we know that hm is anisomorphism from π(W ) to hm(π(W )) = Um/Um−1 = hm(Um). By the Rank-Nullity Theorem,

Um ∼= Ker(hm)⊕ Im(hm).

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The kernel of hm is just Um−1, and in this context Im(hm) = hm(Um) =hm(π(W )) ∼= π(W ). Hence

Um ∼= Um−1 ⊕ π(W ).

Corollary 6.7. The canonical map i : g→ Ug is injective.

Proof. This is just a special case of Corollary 6.6, with m = 1, and W = T 1 = g.The supposition in Corollary 6.6 tells us that T 1 and S1 are isomorphic holdsbecause we have S1 = gm(T 1) = T 1. Using the same logic as in the proof ofCorollary 6.6, it is clear that W = g must be mapped isomorphically onto U1.Since U1 is in the filtration of Ug we know that g is mapped injectively into Ugitself.

This result is very important—it allows us to identity each X ∈ g withi(X) ∈ Ug, and hence think of Ug as a bigger algebra “enveloping” the Liealgebra g.

Corollary 6.8. Let (x1, x2, x3, · · · ) be any ordered basis of g. Then the elementsxi1 · · ·xim = π(xi1 ⊗ · · · ⊗ xim), where m ∈ N, and i1 ≤ i2 ≤ · · · ≤ im, alongwith 1, form a basis of Ug.

Proof. Let W be the subspace of Tm spanned by all xi1 ⊗ · · · ⊗ xim , wherei1 ≤ i2 ≤ · · · ≤ im. It is evident that W maps isomorphically onto Sm, so weknow by Corollary 6.6 that π(W ) is a complement to Um−1 in Um, and it is easyto see by induction that the union of 1 and the bases of each Um, for m ∈ N,form a basis for Ug.

And now we set out to prove the Poincare-Birkhoff-Witt Theorem itself. Butbefore we do, we shall introduce some new notation.

Fix an ordered basis (xλ : λ ∈ Ω) of g. This choice identifies Sym with thepolynomial algebra in indeterminates zλ, where λ ∈ Ω. For each sequence Σ =(λ1, λ2, · · ·λm) of indices (m is called the length of Σ), let zΣ = zλ1

· · · zλm∈ Sm

and let xΣ = xλ1⊗ · · · ⊗ xλm

∈ Tm. We shall call Σ increasing if λ1 ≤ λ2 ≤· · · ≤ λm in a given ordering of Ω. By fiat, let ∅ be increasing and set z∅ = 1.It follows that the set zΣ |Σ increasing is a basis of Sym. Later on, the factthat Sym is a filtered algebra will be of importance, and so we note it now:associated with the grading of

Sym =

∞⊕k=0

Sk

is a filtrationSk = S0 ⊕ · · · ⊕ Sk.

Lastly, in the following lemmas, we shall write λ ≤ Σ if λ ≤ µ for all µ ∈ Σ.

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Lemma 6.9. For each m ∈ Z+, there exists a unique linear map fm : g⊗Sm →Sym satisfying:

(Am) fm(xλ ⊗ zΣ) = zλzΣ for any λ ≤ Σ, and any zΣ ∈ Sm,(Bm) fm(xλ ⊗ zΣ)− zλzΣ ∈ Sk for any k ≤ m, and any zΣ ∈ Sk,(Cm) fm(xλ ⊗ fm(xµ ⊗ zT )) = fm(xµ ⊗ fm(xλ ⊗ zT )) + fm([xλ, xµ] ⊗ zT )

for all zT ∈ Sm−1.Moreover, the restriction of fm to g ⊗ Sm−1 agrees with fm−1.

Proof. First, note that all of the terms in (Cm) make sense once we have proven(Bm). Note further that the restriction of fm to g⊗Sm−1 must satisfy (Am−1),(Bm−1), and (Cm−1), so this restricted map must be the same as fm−1 due tothe asserted uniqueness. To verify that existence and uniqueness hold for eachfm, we proceed by induction on m. For m = 0, only zΣ = z∅ = 1 occurs; thus,we can let f0(xλ⊗ 1) = zλ and extend linearly to g⊗S0. Evidently, (A0), (B0),and (C0) are satisfied. Furthermore, (A0) shows that our choice of f0 is the onlypossible one.

Assuming the existence of a unique fm−1 satisfying (Am−1), (Bm−1), and(Cm−1), we will show how to extend fm−1 to a map fm. For this purpose, itwill suffice to define fm(xλ ⊗ zΣ) where Σ is an increasing sequence of lengthm.

For the case where λ ≤ Σ, condition (Am) cannot hold unless we definefm(xλ⊗zΣ) = zλzΣ. On the other hand, if λ ≤ Σ fails to hold, then λ is greaterthan some element of Σ. Certainly, then, the first index µ in Σ is strictly lessthan λ, and Σ = (µ, T ), where µ ≤ T and T is of length m − 1. By (Am−1),we know zΣ = zµzT = fm−1(xµ ⊗ zT ). Since µ ≤ T , fm(xµ ⊗ zT ) = zµzT = zΣ

is already defined, so the left side of (Cm) becomes fm(xλ ⊗ zΣ). On the otherhand, (Bm−1), with k = m− 1 implies that

fm(xλ ⊗ zT ) = fm−1(xλ ⊗ zT ) = zλzT + y

for a specific y ∈ Sm−1, since fm−1 is defined uniquely by the induction hypoth-esis. This shows that the right side of (Cm) is already defined:

zµzλzT + fm−1(xµ ⊗ y) + fm−1([xλ, xµ]⊗ zT ),

where y ∈ Sm−1.The preceding remarks show that fm can be defined, and in only one way.

Moreover, (Am) and (Bm) certainly hold, as does (Cm), as long as µ < λ, µ ≤ T .But [xµ, xλ] = −[xλ, xµ], so (Cm) also holds for λ < µ, λ ≤ T . When λ = µ,(Cm) also holds. We now only need to consider the case where neither λ ≤ Tnor µ ≤ T is true. Write T = (v,Ψ), where v ≤ Ψ, v < λ, and v < µ. To keepnotation under control, write fm(x⊗ z) as xz for any x ∈ g and z ∈ Sm.

The induction hypothesis guarantees that xµzT = xv(xµzΨ) + [xµ, xv]zΨ,and we know xµzΨ = zµzΨ + w for some w ∈ Sm−2. Since v ≤ Ψ and v ≤ µ,(Cm) already applies to xλ(xv(zµzΨ)). By induction, we know (Cm) also appliesto xλ(xvw), and thus to xλ(xv(xµzΨ)). Consequently,

xλ(xµzT ) = xv(xλ(xµzΨ)) + [xλ, xv](xµzΨ) + [xµ, xv](xλzΨ) + [xλ, [xµ, xv]]zΨ.

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Recall that λ and µ are interchangeable throughout this argument. If we in-terchange them in the above equation, and subtract the two, we obtain xλ(xµzT )−xµ(xλzT ), which is equivalent to

xv(xλ(xµzΨ))− xv(xµ(xµzT )) + [xλ, [xµ, xv]]zΨ − [xµ, [xλ, xv]]zΨ.

But this is the same as

xv([xλ, xµ]zΨ) + [xλ, [xµ, xv]]zΨ + [xµ, [xv, xλ]]zΨ,

which can be written as

[xλ, xµ](xvzΨ) + ([xv, [xλ, xµ]] + [xλ, [xµ, xv]] + [xµ, [xv, xλ]])zΨ.

And thanks to the Jacobi identity, the terms in parenthesis vanish, and weobtain [xλ, xµ]zΨ. This proves (Cm) and with it the lemma.

Great! Just two more to go!

Lemma 6.10. There exists a representation ρ : g→ gl(Sym) satisfying:

1. ρ(xλ)zΣ = zλzΣ for λ ≤ Σ,2. ρ(xλ)zΣ ≡ zλzΣ (mod Sm), if Σ has length m.

Proof. Lemma 6.9 allows us to define a linear map f : g⊗Sym→ Sym satisfying(Am), (Bm), and (Cm) for all m, since fm restricted to g⊗Sm−1 coincides withfm−1 by the uniqueness part. In other words, Sym becomes a g-module bycondition (Cm), giving us a representation ρ satisfying the conditions 1 and 2listed above, thanks to conditions (Am) and (Bm).

Lemma 6.11. Let t ∈ Tm ∩ J , where J = Ker(π), and π is the quotient mapfrom T to Ug. Then the homogenous component tm of t of degree m lies in I,the kernel of the quotient map taking T to Sym.

Proof. Write tm as a linear combination of basis elements xΣ(i) for 1 ≤ i ≤ r, andwhere each Σ(i) is of length m. The Lie algebra homomorphism ρ : g→ gl(Sym)constructed in Lemma 6.10 extends, by the universal property of Ug = T/J , toan algebra homomorphism ρ′ : T → End(Sym), with J ⊂ Ker(ρ′). So ρ′(t) = 0.Then

ρ′(xΣ(i)) · 1 = ρ′(xΣ(i)1 ⊗ · · · ⊗ xΣ(i)m) · 1 = 0

by the definition of xΣ(i). But this becomes

ρ(xΣ(i)1) · · · ρ(xΣ(i)m) · 1 = 0

due to the fact that ρ′ is an algebra homomorphism and because the restrictionof ρ′ to g is ρ. And then by Lemma 6.10, we obtain zΣ(i). Hence ρ′(t) · 1 isa polynomial whose term of highest degree is the appropriate combination ofthe zΣ(i) (1 ≤ i ≤ r). Therefore this combination of the zΣ(i) is 0 in Sym, andtm ∈ I as required.

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Proof of the Poincare-Birkhoff-Witt Theorem. Let t ∈ Tm, and π : T → Ug bethe quotient map. We must show that π(t) ∈ Um−1 implies t ∈ I. But t ∈ Tmand π(t) ∈ Um−1 together imply that π(t) = π(t′) for some t′ ∈ Tm−1, and sot − t′ ∈ J . Applying Lemma 6.11 to the tensor t − t′ ∈ Tm ∩ J , and using thefact that the homogenous component of degree m is t, we get t ∈ I.

Acknowledgments

First and foremost, I would like to thank my mentor, Jared Bass, for all ofhis help this summer—without his guidance, this paper would not have beenpossible. And of course, I would like to thank Peter May for creating andrunning the REU.

References

[1] Brian Hall. Lie Groups, Lie Algebras, and Representations. Springer-VerlagNew York Inc. 1st Edition. 2003.

[2] Jeffrey Lee. Manifolds and Differential Geometry. American MathematicalSociety. 1st Edition. 2009.

[3] Alexander Kirillov, Jr. An Introduction to Lie Groups and Lie Algebras.Cambridge University Press. 1st Edition. 2008.

[4] James Humphreys. Introduction to Lie Algebras and Representation Theory.Springer-Verlag New York Inc. 3rd Edition. 1980.

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