Appalachian State University

Department of Mathematics

John Hall

Free Leibniz Algebras

c© 2018

A Directed Research Paper

in Partial Fulfillment of the Requirements

for the Degree of Master of Arts

May 2018

Approved:

Dr. William J. Cook

Member

Dr. Vicky W. Klima

Member

Contents

1 Introduction 2

2 A Survey of Free Objects 32.1 Free Monoids . . . . . . . . . . . . . . . . . . . . . . . . . . . 52.2 Free Groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . 72.3 Vector Spaces Are Free . . . . . . . . . . . . . . . . . . . . . . 10

2.3.1 The Tensor Product . . . . . . . . . . . . . . . . . . . 122.4 Free Algebras . . . . . . . . . . . . . . . . . . . . . . . . . . . 15

2.4.1 Algebras . . . . . . . . . . . . . . . . . . . . . . . . . . 162.4.2 The Tensor Algebra . . . . . . . . . . . . . . . . . . . . 172.4.3 Free Non-associative Algebras . . . . . . . . . . . . . . 19

2.5 Free Rings and Free Fields . . . . . . . . . . . . . . . . . . . . 202.5.1 Free Rings . . . . . . . . . . . . . . . . . . . . . . . . . 202.5.2 Free Fields . . . . . . . . . . . . . . . . . . . . . . . . . 21

3 Lie Algebras 223.1 Basic Properties . . . . . . . . . . . . . . . . . . . . . . . . . . 223.2 The Universal Enveloping Algebra . . . . . . . . . . . . . . . . 273.3 Free Lie Algebras . . . . . . . . . . . . . . . . . . . . . . . . . 30

4 Leibniz Algebras 324.1 Basic Properties . . . . . . . . . . . . . . . . . . . . . . . . . . 324.2 Free Leibniz Algebras . . . . . . . . . . . . . . . . . . . . . . . 35

References 37

Abstract

Leibniz algebras are a generalization of Lie algebras. While Liealgebras are well-known, Leibniz algebras are still in development.Our main goal is to examine the free Lie and free Leibniz algebras. Indoing so, we also discuss free objects in general. The paper concludesby examining a fairly new result for free Leibniz algebras discoveredby [MI].

1

1 Introduction

Lie algebras were introduced by S. Lie in the 1800s. He studied Liealgebras to gain insight into group theory. In particular, Lie algebras arerelated to a group, called the Lie group. Other mathematicians, such as E.Cartan and W. Killing, began to study Lie algebras, and their efforts helpedbring about a complete classification of finite dimensional simple Lie alge-bras [CA]. Leibniz algebras, on the other hand, are a fairly new algebra.They were introduced by J. Loday in the 1990s [LO]. Leibniz algebras are ageneralization of Lie algebras, and it was recently discovered that there areanalogs of Lie’s Theorem, Engel’s Theorem, Cartan’s criterion, and Levi’sTheorem for Leibniz algebras [DE].

The goal of this paper is to discuss free Lie algebras and free Leibniz al-gebra. Essentially, we start from the ground up by examining free objects ingeneral. We define free objects in terms of their universal mapping propertieswhile refraining from the categorical definition of a free object as being leftadjoint to the forgetful functor. Therefore, a working knowledge of under-graduate modern algebra and linear algebra should be all that is necessaryto follow along in the paper. We conclude by discussing the basic proper-ties of Lie and Leibniz algebras, and emphasize certain results pertaining tothe universal enveloping algebra for free Lie algebras, and the basis for freeLeibniz algebras.

2

2 A Survey of Free Objects

Let C be at category of algebraic objects and morphisms. We may think of Cas containing groups with group homomorphisms or real vector spaces withlinear maps. Definition 2.1 states the requirements for an algebraic object inC to be free.

Definition 2.1. Let X be an arbitrary set. Given an object F (X) and afunction i : X → F (X), we say that F (X) is free on X if given any object Aand any function ϕ : X → A, there exists a unique morphism ϕ : F (X)→ Asuch that ϕ ◦ i = ϕ.

X A

F (X)

i

ϕ

ϕ

Figure 1: The commutative diagram for F (X).

Figure 1 represents the theorem diagrammatically; the function ϕ is rep-resented using a dotted line to indicate the existence of ϕ must be provedand is therefore not part of the initial conditions of the definition. Becauseof the universal quantifiers on ϕ and A in definition 2.1, we say that the freeobject F (X) on X satisfies the universal mapping property, which wehereafter refer to as the UMP. We list several properties of free objects thatare all consequences of the UMP.

Theorem 2.1. Let X and Y be arbitrary sets and C be a category thatcontains free objects. If F (X) is the free object on X, and F (Y ) is the freeobject on Y , then

1. F (X) is unique up to isomorphism;

2. if |X| = |Y |, then F (X) ∼= F (Y );

3. every object in C is a homomorphic image of a free object in C.

3

Proof of Theorem 2.1.

1. Assume there are two free objects on X, say F1(X) and F2(X) withi1 : X → F1(X) and i2 : X → F2(X). The universal mapping propertyof F1(X) implies there is a unique morphism ϕ1 from F1(X)→ F1(X)such that ϕ1 ◦ i1 = i1. But, since the identity map IdF1(X) is also amorphism from F1(X) to F1(X) such that IdF1(X) ◦ i1 = i1, it followsfrom the uniqueness of ϕ1 that

ϕ1 ◦ i1 = IdF1(X) ◦ i1 = i1. (1)

That is, the identity map on F1(X) is the only morphism that cansatisfy equation 1. Similarly, the identity map on F2(X) is the onlymorphism such that

IdF2(X) ◦ i2 = i2. (2)

Next, the universal mapping property of F1(X) implies there exists aunique morphism ϕ1 : F1(X)→ F2(X) such that

ϕ1 ◦ i1 = i2.

The universal mapping property of F2(X) implies there exists a uniquemorphism ϕ2 : F2(X)→ F1(X) such that

ϕ2 ◦ i2 = i1.

Consequently,ϕ1 ◦ i1 = ϕ1 ◦ ϕ2 ◦ i2 = i2.

Equation 2 implies that ϕ1◦ ϕ2 = IdF2(X). In a similar fashion, we have

ϕ2 ◦ i2 = ϕ2 ◦ ϕ1 ◦ i1 = i1.

Equation 1 implies that ϕ2◦ϕ1 = IdF1(X). Hence, ϕ1 and ϕ2 are inversesof each other, and F1(X) is isomorphic to F2(X).

2. Since free objects are unique up to isomorphism, we show that F (X)also satisfies the universal mapping property of Y . There are bijectionsf : X → Y and f−1 : Y → X. Let A be an object and ϕ : Y → A amap. Because F (X) is free on X, there is a map i : X → F (X) and aunique morphism ϕ : F (X)→ A such that

ϕ ◦ i = ϕ ◦ f.

4

This implies thatϕ ◦ i ◦ f−1 = ϕ.

Hence, F (X), when paired with the map i ◦ f−1 : Y → F (X), is freeon Y , so F (X) ∼= F (Y ).

3. Let A be an object and F (A) the free object on A (note that A can bethought of as an generating set.) Then there is a map i : A → F (A)and a unique morphism ϕ : F (A)→ A such that

ϕ ◦ i = IdA,

where IdA is the identity map on A. This implies that ϕ is surjectiveas desired.

The existence clause of the UMP (that is, the existence of ϕ) impliesthere are no nontrivial relations in F (X), while the uniqueness of ϕ impliesthe elements of X generate F (X). When we say there no nontrivial relationsin F (X), we mean that every relation that exists in F (X) can be impliedsolely from the axioms that define the object. For example, if F (X) is thefree group on X, then for a collection of elements gi ∈ F (X), if we have therelation gi1 ·gi2 · ... ·gin = gj1 ·gj2 · ... ·gjm , then the axioms that define a groupare all that suffice to prove this relation exists. One may therefore think offree objects as having no imposed relations.

The elements of X generate F (X) in the sense that every element in F (X)can be expressed as a combination of elements in X. For example, if F (X)is the free vector space on X, then for all y ∈ F (X), y ∈ span(x1, x2, ..., xn),where each xi ∈ X. We now proceed to explore examples of free objects forspecific categories. The properties of F (X) in theorem 2.1 apply to all freeobjects – assuming they exist in the first place. Indeed, some categories donot have free objects, such as fields, which we discuss in section 2.5.

2.1 Free Monoids

The free monoid is a monoid that satisfies the UMP of definition 2.1. Let Xbe a set, and define M(X) to be the set of all associative words on X. Forexample, if X = {a, b, c}, then

a, ab, and ccabac

5

are examples of associative words on X. Associative words on X are justfinite sequences of elements of X. We call elements of M(X) associativewords to indicate that there is no need to parenthesize the words. Later on,when we study objects that can be non-associative, such as algebras, the needto distinguish between associative and non-associative words will be moreclear. Equip M(X) with the binary operation ∗ : M(X) ×M(X) → M(X)given by w1 ∗ w2 = w1w2. The operation is just the concatenation of words,which is associative. Notice that ∗ is not commutative because we are treatingelements like ab and ba as distinct words. M(X) is a monoid, where the emptyword, denoted ε, is the unit. Notice that elements of X are also words onX – we call them singleton words. We are therefore justified in defining theinclusion map

i : X →M(X),

which is given by i(x) = x. We claim that M(X) is the free monoid on theset X.

Theorem 2.2. M(X) is the free monoid on the set X.

Proof. Let A be a monoid and ϕ : X → A be a map. We must show thereexists a unique monoid homomorphism ϕ fromM(X) to A such that ϕ◦i = ϕ.For a nonempty word w = x1x2...xn ∈M(X), define ϕ : M(X)→ A by

ϕ(w) = ϕ(x1x2...xn) = ϕ(x1) · ϕ(x2) · ... · ϕ(xn),

where · denotes the operation of A. If w is the empty word, then defineϕ(ε) = eA, where eA is the unit of A. Then ϕ is a homomorphism such thatϕ ◦ i = ϕ. It remains to show that ϕ is unique. Suppose that ϕ is anotherhomomorphism from M(X) to A such that ϕ◦i = ϕ. Then for any nonemptyword w = x1x2...xn ∈ X, we have

ϕ(w) = ϕ(x1 ∗ x2 ∗ ... ∗ xn).

= ϕ(x1) · ϕ(x2) · ... · ϕ(xn).

= ϕ ◦ i(x1) · ϕ ◦ i(x2) · ... · ϕ ◦ i(xn).

= ϕ(x1) · ϕ(x2) · ... · ϕ(xn).

= ϕ ◦ i(x1) · ϕ ◦ i(x2) · ... · ϕ ◦ i(xn).

= ϕ(x1) · ϕ(x2) · ... · ϕ(xn).

= ϕ(x1 ∗ x2 ∗ ... ∗ xn).

= ϕ(w).

6

Also, ϕ(ε) = ϕ(ε). Thus, ϕ is unique, and M(X) is the free monoid onX.

Example 2.1. Let X = {1}. Then the free monoid on X is just the set of allfinite sequences of ones and the empty word: M(X) = {ε, 1, 11, 111, 1111, ...}.Let Z≥0 be the monoid of nonnegative integers under addition. ThenM(X) ∼=Z≥0.

Example 2.2. Let X = ∅ be the empty set. For any monoid A, the maps j :X →M(X) and ϕ : X → A are the canonical empty functions. Here, M(X)is the set containing the empty word, and ϕ is the trivial homomorphism thatsends the empty word to the identity element in A. Note the requirement thatϕ◦j = ϕ is vacuously true. Also, the requirement that ϕ is unique stems fromthe fact that our only element is the identity and a homomorphism must sendthe identity to the identity. Thus there is only one possible homomorphismfrom the trivial monoid.

In the next section, we explore free groups. The construction of the freemonoid was fairly simply, and although the constructions of other free objectsare similar in some ways, the reader may notice that as the complexity ofthe algebraic object increases, the tools needed to construct the free objectbecome more advanced.

2.2 Free Groups

The construction of the free group on a set X is similar to that of the freemonoid. For an arbitrary set X = {x1, x2, ...}, define X−1 = {x−11 , x−12 , ...}.That is, X−1 is the set obtained from X by inserting the superscript, −1, oneach of the elements of X. Let G(X) be the set of all associative words onX ∪X−1. This time, we denote the empty word by eG. We allow for words inG(X) to be concatenated, but concatenation alone is not enough to ensurethat G(X) is a group since we are treating elements like xx−1 and x−1x asdistinct words. We deal with inverses by defining an equivalence relation.

Definition 2.2. Let X be a set. Let G(X) denote the set of all associativewords obtained from X ∪X−1. For w, u ∈ G(X), if u can be obtained fromw by a finite sequence of insertions or deletions of words of the form xx−1

or x−1x, where x, x−1 ∈ X ∪X−1, then we say w is related to u and writew ∼ u.

7

For example, if X = {a, b, c, d}, then

a ∼ a.

a ∼ aaa−1.

abc−1 ∼ b−1babdd−1c−1aa−1.

eG ∼ d−1db−1bcc−1a−1a.

ba 6∼ cbc−1a.

Hopefully, the examples above are enough to convince the reader that ∼is an equivalence relation. For w ∈ G(X), let w = {u ∈ G(X) | w ∼ u}.That is, w is the equivalence class of the word w. Let G(X) denote the setof all equivalence classes of words in G(X). Then G(X) is a group under theoperation w ·u = wu, where wu is the word w and u concatenated. Althoughwe do not prove it, the group operation is well-defined – see [RO]. We needto establish some notation. For a ∈ X−1, we have that a = b−1 for someb ∈ X. So, we let a−1 = (b−1)−1 = b. That is, (b−1)−1 = b for all b ∈ X.Suppose w is a word, say w = x1x2...xn, where xi ∈ X ∪X−1, then we definew−1 = x−1n ...x−12 x−11 . It follows that the inverse of w is w−1, and the identityelement is eG. The check for associativity is rather tedious; for a completeproof that G(X) satisfies all of the group axioms, see [RO].

Theorem 2.3. G(X) is the free group on X.

Proof. Let H be a group and ϕ : X → H be a map. First, we need a mapfrom X to G(X). To this end, we can define the inclusion map

i : X → G(X)

by letting i(x) = x. Next, we need a unique group homomorphism ϕ fromG(X) to H such that ϕ ◦ i = ϕ. We do this rather carefully by first noticingthat every nonempty word of G(X) can be expressed as

wε11 wε22 ...w

εnn ,

where each wi ∈ X and εi ∈ {−1, 1}. For all eG 6= w ∈ G(X), let

ϕ : G(X)→ H

be given by

ϕ(w) = ϕ(wε11 wε22 ...w

εnn ) = ϕ(w1)

ε1 ∗ ϕ(w2)ε2 ∗ · · · ∗ ϕ(wn)εn ,

8

where ∗ is the operation for H. Let ϕ(eG) = eH, where eH is the identityelement of H. It may not be obvious that ϕ is well-defined. So, first observethat for x, x−1 ∈ X ∪X−1,

ϕ(xx−1) = ϕ(x) ∗ ϕ(x)−1 = eH = ϕ(x)−1 ∗ ϕ(x) = ϕ(x−1x).

Hence, ϕ(xx−1) = ϕ(x−1x) = eH. With that in mind, if w = y ∈ G(X), thenw and y must differ by the insertion or deletion of elements of the form xx−1

or x−1x. Since the image of the equivalence classes of such elements under ϕare the identity element of H, it follows that ϕ(w) = ϕ(y). Hence, ϕ is well-defined. It should be clear that ϕ is a homomorphism such that ϕ ◦ i = ϕ. Itremains to show that ϕ is unique, so let ϕ′ be another homomorphism suchthat ϕ′ ◦ i = ϕ. Then ϕ ◦ i = ϕ′ ◦ i, and for eG 6= w ∈ G(X), we have

ϕ(w) = ϕ(wε1wε2...w

εn).

= ϕ(w1)ε1 ∗ ϕ(w2)

ε2 ∗ · · · ∗ ϕ(wn)εn .

= (ϕ ◦ i(w1))ε1 ∗ (ϕ ◦ i(w2))

ε2 ∗ · · · ∗ (ϕ ◦ i(wn))εn .

= (ϕ′ ◦ i(w1))ε1 ∗ (ϕ′ ◦ i(w2))

ε2 ∗ · · · ∗ (ϕ′ ◦ i(wn))εn .

= ϕ′(w1)ε1 ∗ ϕ′(w2)

ε2 ∗ · · · ∗ ϕ′(wn)εn .

= ϕ′(wε11 ) ∗ ϕ′(wε22 ) ∗ · · · ∗ ϕ′(wεnn ).

= ϕ′(wε11 wε22 ...w

εnn ).

= ϕ′(w).

Also, ϕ(eG) = eH = ϕ′(eG). Hence, ϕ is unique, and G(X) is the freegroup on X.

Example 2.3. The integers under addition is a free group. To see why, letX = {1}. For notational convenience, let a = 1 and b = 1−1. Then

G(X) ={eG, a, b, aa, bb, ...

}is the free group on X. It is isomorphic to the group of integers of underaddition.

Example 2.4. We show that the free group generated by more than oneelement is non-abelian. Let X be an arbitrary set with |X| ≥ 2. We maywrite X = {a, b} ∪ Y , where Y ∩ {a, b} = ∅. Consider the symmetric groupon three elements: S3 = {(1), (12), (13), (23), (123), (132)}. We can define amap ϕ : X → S3 by letting

9

ϕ(x) =

(12) x = a

(13) x = b

(1) x ∈ Y

We have a unique homomorphism ϕ : G(X)→ S3, but notice that

ϕ(ab) = ϕ(a)ϕ(b) = ϕ(a)ϕ(b) = (12)(13) = (132),

andϕ(ba) = ϕ(b)ϕ(a) = ϕ(b)ϕ(a) = (13)(12) = (123).

Since ϕ(ab) 6= ϕ(ab), we have ab 6= ba, which means G(X) is not abelian.

As with free monoids, the free group on the empty set is the trivial group.Free groups are interesting in their own right, and have useful applicationsto combinatorial group theory. For this paper, the purpose of discussingfree monoids and free groups is to merely give concrete examples of how toconstruct free objects. Our goal is discuss free Lie algebras and free Leibnizalgebras, and sections 2.3 and 2.4 provide the tools that we need to studythose objects.

2.3 Vector Spaces Are Free

In this section, our main goal is to introduce the tensor product of vectorspaces. To do so, we must discuss vector spaces. We work over arbitraryfields, and the results that we discuss are valid for both finite and infinitedimensional vector spaces.

Let X be a set and F a field. Let V (X) be the set of functions from Xinto F such that only finitely many elements of x are mapped to zero. Thatis, V (X) = {f : X → F | f(x) 6= 0 for at most finitely many x ∈ X}. Forf, g ∈ V (X), let f+g be given by (f+g)(x) = f(x)+g(x). Notice that f+gis indeed in V (X). For c ∈ F, let cf be given by (cf)(x) = cf(x). Again, itshould be clear that cf ∈ V (X). With these operations, V (X) is a vectorspace. We omit the tedious proof that V (X) satisfies all of the axioms for avector space. We proceed by finding a basis for V (X).

First, we define a characteristic function. For x ∈ X, let δx : X → F begiven by δx(x) = 1 and δx(y) = 0 for all y 6= x. There is only one element

10

whose image under δx is non-zero; therefore, δx ∈ V (X). We show thatβ = {δx | x ∈ X} is a basis for V (X).

Let f ∈ V (X). Then there are only a finite number of elements in Xwhose image under f is non-zero. Let x1, ..., xn be those elements. Note thatf(x1), ..., f(xn) ∈ F. Define g(x) =

∑ni=1 f(xi)δxi(x). Then we have

g(xj) =n∑i=1

f(xi)δxi(xj) = 0 + ...+ f(xj) + ...+ 0 = f(xj).

This shows that f and g agree on x1, ..., xn. For y 6= xi, we have

g(y) =n∑i=1

f(xi)δxi(y) = 0 = f(y).

Hence, f = g, which proves that β spans V (X).As for linear independence, assume that

∑ni=1 ciδxi = 0, where ci ∈ F and

xi ∈ X. Then for j ≤ n, we have

0 =n∑i=1

ciδxi(xj) = cj.

Hence, β is linearly independent, and so β is a basis for V (X). The proofthat V (X) is free on X follows quickly.

Theorem 2.4. V (X) is the free vector space on X.

Proof. Let i : X → V (X) be defined by i(x) = δx, where δx is the character-istic function from earlier. Let W be a vector space and ϕ : X → W a map.For any f =

∑ni=1 ciδxi ∈ V (X), let ϕ : V (X)→ W be given by

ϕ(f) =n∑i=1

ciϕ(xi).

It should be apparent that ϕ is a linear map such that ϕ ◦ i = ϕ. Hence,V (X) is free on X.

In the proof of theorem 2.4, we see that every vector space is free on itsbasis. Since every vector space has a basis, all vector spaces are therefore

11

free. Indeed, if V is a vector space and β = {b1, b2, ...} a basis for V , then forany map ϕ from β into a vector space W , there exists a unique linear mapfrom V to W given by

ϕ(v) =n∑i=1

ciϕ(bi).

We say that any map defined on a basis may be extended by linearity toa unique linear map defined on the vector space spanned by the basis. So,from this point on, we bypass saying that V (X) is the free vector space onX, and instead simply say that V (X) is the vector space with basis X.

2.3.1 The Tensor Product

In this section, we define the tensor product of vector spaces and sketch aformal construction. Before doing so, we give the definition of a multilinearmap.

Definition 2.3. Let V1, V2, ..., Vn, and W be vector spaces over a field F.We say that γ : V1 × V2 × · · · × Vn → W is a multilinear map if for allv1 ∈ V1, ..., vi, v′i ∈ Vi, ..., vn ∈ Vn and c ∈ F,

1. γ(v1, ..., vi + v′i, ..., vn) = γ(v1, ..., vi, ..., vn) + γ(v1, ..., v′i, ..., vn);

2. γ(v1, ..., cvi, ..., vn) = cγ(v1, ..., vi, ..., vn).

We see that a multilinear map is map defined on a cartesian product ofvector spaces that is linear in each of is arguments when all other argumentsare fixed. A multilinear map defined a two-fold cartesian product is calledbilinear. Similar to linear maps, if β1, ..., βn are bases for V1, ..., Vn respec-tively, then for any vector space W and any map f : β1 × · · · × βn → W ,there exists a unique multilinear map γ : V1×· · ·×Vn → W such that γ andf agree on β1 × · · · × βn.

Definition 2.4. Let V1, ..., Vn be vector spaces over a field F. Let V be avector space over F and T : V1 × · · · × Vn → V a multilinear map. Wesay that the vector space V equipped with the multilinear map T is a tensorproduct of V1, ..., Vn if, given a vector space W over F and a multilinear mapγ : V1 × · · · × Vn → W , there exists a unique linear map γ : V → W suchthat γ ◦ T = γ.

12

V1 × · · · × Vn W

V

T

γ

γ

Figure 2: The commutative diagram for V .

Note that a tensor product consists of a vector space equipped with aparticular multilinear map. We may therefore refer to a tensor product asthe pair (V , T ). To prove the existence of the tensor product, one begins withthe vector space V (X) with basis X = V1 × · · · × Vn. Let S be the subspaceof V (X) spanned by elements of the form

(v1, ..., vi + v′i, ..., vn)− (v1, ..., vi, ..., vn)− (v1, ..., v′i, ..., vn)

and(v1, ..., cvi, ..., vn)− c(v1, ..., vi, ..., vn),

where vi, v′i ∈ Vi and ci ∈ F. Then the quotient space, V (X)/S, when

equipped with the multilinear map T : V1 × · · · × Vn → V (X)/S given byT (v1, ..., vn) = (v1, ..., vn) + S, is a vector space that satisfies the universalmapping property of definition 2.4 and is therefore the tensor product ofV1, ..., Vn. For a complete proof that V (X)/S satisfies the universal mappingproperty, see [CO]. It is customary to use different notation for tensor prod-ucts. In particular, we let V (X)/S = V1⊗· · ·⊗Vn denote the tensor productof V1, ..., Vn. Elements of tensor products are cosets, but again for notationalpurposes, we set T (v1, ..., vn) = (v1, ..., vn) + S = v1 ⊗ · · · ⊗ vn. We call theelements of a tensor product tensors. There are two properties of tensorsthat are inherit from the multilinearity of T . For each vi, v

′i ∈ Vi and c ∈ F,

we have the following.

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

2. v1 ⊗ · · · ⊗ cvi ⊗ · · · ⊗ vn = c(v1 ⊗ · · · ⊗ vi ⊗ · · · ⊗ vn).

Moreover, one may show that tensor products are unique up to isomor-phism.

13

Lemma 2.5. Suppose that (V , T ) and (V ′, T ′) are tensor products of thevector spaces V1, ..., Vn over F. Then there exists an isomorphism φ : V → V ′such that φ ◦ T = T ′.

Proof. The universal mapping property of V implies there exists a uniquelinear map φ : V → V ′ such that φ ◦ T = T ′. Similarly, there is a uniquelinear map φ′ : V ′ → V such that φ′◦T ′ = T . This implies that φ′◦φ◦T = T ,yet the uniqueness clause of the universal mapping property of V impliesthat φ′ ◦ φ = IdV . Similarly, φ ◦ φ′ = IdV ′ , which shows that V and V ′ areisomorphic.

We use the lemma to show how to find a basis for the tensor product.

Theorem 2.6. Let V1, ..., Vn be vector spaces over a field F with bases β1, ..., βnrespectively. Then the set B = {b1 ⊗ · · · ⊗ bn | bi ∈ βi} is a basis forV1 ⊗ · · · ⊗ Vn.

Proof. Set X = V1 × · · · × Vn and X ′ = β1 × · · · × βn. Referencing ourprevious notation, we have V1 ⊗ · · · ⊗ Vn = V (X)/S, where V (X) is thevector space with basis X, and S is the subspace of V (X) defined earlier.We also have the corresponding multilinear map T : X → V (X)/S definedearlier. Consider the vector space V (X ′) with X ′ as basis. Then we havethe inclusion maps i1 : X ′ → X and i2 : X ′ → V (X ′). The map i2 can beextended to a unique multilinear map T ′ : X → V (X ′) such that T ′ ◦ i1 = i2.We claim that V (X ′) equipped with T ′ is also a tensor product of V1, ..., Vn.

To that end, let γ be a multilinear map from X into some vector spaceW . Let γ′ be the restriction of γ to X ′. Then because X ′ is a basis forV (X ′), γ′ can be extended to a unique linear map γ′ from V (X ′) to W suchthat γ′ ◦ i2 = γ′. We will be done if we can show that γ′ ◦ T ′ = γ. Notethat having γ′ ◦ i2 = γ′ implies that γ′ and γ agree on X ′. Since γ′ and γare multilinear, they must also agree on X. Recall that T ′ ◦ i1 = i2. Thisimplies that γ′ ◦ i2 = γ′ ◦T ′ ◦ i1 = γ′, but since γ′ and γ agree on X, we havethat γ′ ◦ T ′ = γ. Therefore, V (X ′) equipped with T ′ is a tensor product forV1, ..., Vn.

Ultimately, we know from lemma 2.5 that there is an isomorphism φ :V (X ′)→ V (X)/S such that φ ◦ T ′ = T . Since X ′ is a basis for V (X ′), and

14

φ is an isomorphism, φ(X ′) must be a basis for V (X)/S. It follows that

b1 ⊗ · · · ⊗ bn = T (b1, ..., bn).

= φ ◦ T ′(b1, ..., bn).

= φ ◦ T ′ ◦ i1(b1, ..., bn).

= φ ◦ i2(b1, ..., bn).

= φ(b1, ..., bn).

Therefore, φ(X ′) = {b1 ⊗ · · · ⊗ bn | bi ∈ βi} is a basis for V1 ⊗ · · · ⊗ Vn.

Example 2.5. Take V1 = R and V2 = R2. Then a basis for V1 ⊗ V2 is givenby {1⊗ (0, 1), 1⊗ (1, 0)}.

Example 2.6. Take V1 = R2, and V2 = P≤1[x], where P≤1[x] is the set ofall polynomials with real coefficients of degree less than or equal to 1 andindeterminate x. Then {(0, 1)⊗ 1, (0, 1)⊗ x, (1, 0)⊗ 1, (1, 0)⊗ x} is a basisfor V1 ⊗ V2.

If V1, ..., Vm are finite dimensional vector spaces with dimensions n1, ..., nmrespectively, then V1 ⊗ · · · ⊗ Vn has dimension n1 · · ·nm. We also have thefollowing theorems whose proofs can be found in [CO].

Theorem 2.7. Let V1, ..., Vn and W1, ...,Wm be vector spaces over a field F.Then (V1 ⊗ · · · ⊗ Vn)⊗ (W1 ⊗ · · · ⊗Wm) ∼= V1 ⊗ · · · ⊗ Vn ⊗W1 ⊗ · · · ⊗Wm.

Theorem 2.8. Let V1, ..., Vn be vector spaces over a field F and π a permua-tion of {1, 2, ..., n}. Then V1⊗ · · · ⊗ Vn ∼= Vπ(1)⊗ · · · ⊗ Vπ(n) via a linear mapthat takes v1 ⊗ · · · ⊗ vm to vπ(1) ⊗ · · · ⊗ vπ(n).

2.4 Free Algebras

After discussing algebras and some of their basic properties, we constructthe free associative algebra. We rely on the free associative algebra in theconstruction of the free Lie algebra in section 3.3. Then, we construct the freenon-associative algebra, which we use to construct the free Leibniz algebrain section 4.2.

15

2.4.1 Algebras

Vector spaces have two operations – vector addition and scalar multiplica-tion. When equipped with a bilinear multiplication operator, a vector spacebecomes an algebra, and a plethora of new avenues to explore opens up as aconsequence.

Definition 2.5. An algebra A is a vector space over a field F equipped witha bilinear map, called multiplication, from A× A to A. That is, there existsa map (, ) : A× A→ A such that

1. (x+ y, z) = (x, z) + (y, z);

2. (x, y + z) = (x, y) + (x, z);

3. (cx, y) = (x, cy) = c(x, y) for all x, y, z ∈ A and c ∈ F.

The three conditions above follow directly from definition 2.3. The prop-erties of a bilinear map agree with our natural inclination for how multiplica-tion should work. For example, we have a property analogous to the classicFOIL method. For x, y, w, z ∈ A, we see that

(x+ y, w + z) = (x+ y, w) + (x+ y, z).

= (x,w) + (y, w) + (x, z) + (y, z).

We say that A is an associative algebra if ((x, y), z) = (x, (y, z)) for allx, y, z ∈ A. For associative algebras, it is often convenient to denote themultiplication by juxtaposition. If (x, y) = (y, x) for all x, y ∈ A, then A is acommutative algebra. If there is a vector 1 ∈ A such that (1, x) = (x, 1) = xfor all x ∈ A, then we say A is a unital algebra with unit 1.

Example 2.7. A field itself is an associative, commutative, and unital alge-bra.

Example 2.8. The vector space of all n × n matrices over a field withthe usual operations of scalar multiplication, matrix addition, and matrixmultiplication is an associative, unital algebra.

Example 2.9. When equipped with the cross-product, the three-dimensionalvector space A = R3 becomes an algebra. Let i, j, and k denote the standardunit vectors. Recall that the cross-product may be defined on the standard

16

unit vectors by −(j, i) = (i, j) = k,−(k, j) = (j,k) = i, −(i,k) = (k, i) = j,and (i, i) = (j, j) = (k,k) = 0. This algebra is not associative, commutative,nor is it unital. It should be mentioned that it suffices to define an algebra’smultiplication on the basis vectors. This is because bilinear maps (multilin-ear maps) are uniquely determined by their action on pairs (tuples) of basisvectors. Consequently, since {i, j,k} is a basis for A, one may verify that Ais an algebra by checking that the vector cross-product rule is bilinear for thebasis vectors. One can then extend the algebra’s multiplication by bilinearityso that it is applied to all vectors in A. So, in general, knowing how an alge-bra’s multiplication behaves on the basis determines how the multiplicationbehaves on the entire algebra.

Algebra homomorphisms behave as expected.

Definition 2.6. Let A1 and A2 be algebras over a field F. Then a linear mapϕ : A1 → A2 is an algebra homomorphism if for all x, y ∈ A1,

ϕ(x · y) = ϕ(x) ∗ ϕ(y),

where · is the multiplication of A1 and ∗ is the multiplication of A2.

An associative algebra homomorphism is simply a homomorphism be-tween associative algebras. Next, we discuss free associative algebras over afield.

2.4.2 The Tensor Algebra

For a vector space V over a field F, let T (V )0 = F, T (V )1 = V, T (V )2 = V ⊗V,and in general

T (V )n = V ⊗ · · · ⊗ V (n factors )

We say that T (V )n is the nth tensor power of V . Consider the vector spaceof the direct sum of tensor powers of V given by

T (V ) =∞⊕n=0

T (V )n = T (V )0 ⊕ T (V )1 ⊕ T (V )2 ⊕ · · ·

Elements of T (V ) are finite sums:

v0 + v1 + · · ·+ vn,

17

where each vk ∈ T (V )k take on the form v1 ⊗ · · · ⊗ vk for v1, ..., vk ∈ V .Suppose v = v0 + v1 + · · ·+ vn and w = w0 + w1 + · · ·+ wm are elements ofT (V ). Without loss of generality, suppose that n ≤ m. For the sum v + w,we define addition component-wise. In the case where n is strictly less thanm, we can insert a finite number of 0′s, where each 0 ∈ T (V )k for somek > n, so that v and w have the same number of components. We definescalar multiplication component-wise as well.

If β is a basis for V , then a basis for T (V ) is given by

B = {b1 ⊗ · · · ⊗ bn | n ≥ 0, bi ∈ β}.

For n = 0, we denote the empty tensor product as 1. In particular, b0 = 1is the multiplicative identity of T (V )0 = F. Note that elements of B aremembers of T (V )j for some j. With that in mind, if we define a bilinear mapfrom T (V )n × T (V )m to T (V )n+m satisfying

(v1 ⊗ · · · ⊗ vn) · (w1 ⊗ · · · ⊗ wm) = v1 ⊗ · · · ⊗ vn ⊗ w1 ⊗ · · · ⊗ wm, (3)

where vi, wj ∈ V , then we can extend this map by linearity to a multiplicationmap from

T (V )× T (V )→ T (V ).

With this multiplication map, T (V ) becomes an associative, unital algebracalled the tensor algebra of V . The fact that T (V ) is an associative algebrafollows from theorem 2.7. The unit of T (V ) is the element b0 = 1 ∈ T (V )0.Note the fact that (3) is bilinear follows directly from the properties of tensorsgiven in section 2.3.1. The tensor algebra is the free associative algebra onany basis for V . We use the tensor algebra to construct the free associativealgebra on any set X.

Theorem 2.9. The free associative algebra over a field on a set X is thetensor algebra of the vector space with basis X.

Proof. Let V be the vector space with basisX and T (V ) be the tensor algebraof V . We have the inclusion map i : X → T (V ). Let A be an associativealgebra, and ϕ : X → A a map. Since X is basis for V , the map ϕ can beextended to a unique linear map ϕ′ : V → A such that ϕ′ ◦ i = ϕ. Since thebasis vectors of V generate T (V ), the map ϕ′ can be extended to a uniqueassociative algebra homomorphism ϕ : T (V )→ A such that ϕ ◦ i = ϕ.

18

2.4.3 Free Non-associative Algebras

Let X be a set. Similar to how we handle free monoids and free groups, weconsider the set F (X) of all non-associative words on X. Non-associativewords on X are finite sequences of elements of X that are parenthesized. Wedefine the length of a word on X to be the number of elements of X thatare used to construct the word. For instance, if X = {x, y}, then (x) is anon-associative word of length 1, (x, y) is a non-associative word of length 2,and ((x, (x, y)), y) is a non-associative word of length 4. We define the emptyword ε to have length 0. Let A be the vector space with basis F (X). Thatis, A is the vector space whose basis is the set of all non-associative wordson X. We define a bilinear multiplication by concatenation. For example,if w1 = (x1, x2) and w2 = (y1, y2) are non-associative words on X, wherex1, x2, y1, y2 ∈ X, then w1 · w2 = (w1, w2) = ((x1, x2), (y1, y2)). With theseoperations, A becomes a non-associative unital algebra. The unit of A isthe empty word ε. We show that A satisfies the universal property of freeobjects.

Theorem 2.10. A is the free non-associative algebra on X.

Proof. We have the inclusion map i : X → A. Let B be a unital, non-associative algebra with multiplication denoted by ∗ and ϕ : X → B a map.We need to construct a unique algebra homomorphism ϕ : A → B such thatϕ ◦ i = ϕ. Since the set of all non-associative words on X forms a basis forA, defining how ϕ acts on non-associative words uniquely determines how ϕbehaves on A. We define ϕ to send the empty word to the unit of B. Weproceed by induction on the length of the word. For x ∈ X, consider thenon-associative word w = (x), which is of length 1. We define ϕ(w) = ϕ(x).Note that for any non-associative word w on X of length k, we can expressw as the product of non-associative words w1 and w2 whose lengths are lessthan or equal to k. Continuing in that vein, we can uniquely express anynon-empty, non-associative word w on X as a product of elements of X(singleton words of length 1). With that in mind, assume that ϕ is definedfor non-associative words of length less than n, where n > 1. Then for a non-associative word w′ of length n, there exists non-associative words w′1 and w′2of length less than n such that w′ = w′1 ·w′2. The inductive hypothesis impliesthat ϕ(w′1) ∗ ϕ(w′2) = ϕ(w′1 ·w′2) = ϕ(w′) is defined. Therefore, by induction,ϕ is defined for all non-associative words on X. Note that the base case ofthe induction proof implies that ϕ ◦ i = ϕ, the inductive hypothesis implies

19

that ϕ is a homomorphism, and the unique representation of any non-empty,non-associative word w on X as a product of elements of X ensures that ϕis well-defined.

2.5 Free Rings and Free Fields

In this supplementary section, we discuss free rings, and we show why freefields do not exist. Our analysis of free Lie and free Leibniz algebras donot depend on a working knowledge of free rings or free fields, so the readermay skip this section without jeopardizing his or her understanding of thesubsequent sections.

2.5.1 Free Rings

In this section we work in the category of unital rings (i.e., rings with mul-tiplicative identity). Recall that in this category ring homomorphisms mustsend the multiplicative identity of the domain to the multiplicative identityof the codomain.

LetX be a set. LetX∗ denote the set of all associative words onX. Recallthat elements of X∗ are finite sequences of elements in X. Consider the setZ[X∗] = {n1w1 + · · ·+nlwl | wi ∈ X∗, ni ∈ Z}. Let v = n1v1 + · · ·+npvp, u =m1u1 + · · · + mkuk ∈ Z[X∗]. Consider the set {v1, . . . , vp, u1, . . . , uk} andrelabel elements of this set as {w1, . . . , wl}, renumber coefficients and pad outwith 0’s as necessary and get: v = n1w1+· · ·+nlwl and u = m1w1+· · ·+mlwl.Define addition so that Z[X∗] by

(n1w1 + · · ·+nlwl) + (m1w1 + · · ·+mlwl) = (n1 +m1)w1 + · · ·+ (nl +ml)wl.

We are defining nx = x+ x+ · · ·+ x︸ ︷︷ ︸n−times

. Define multiplication on Z[X∗] by

( l∑i=1

niwi

)( k∑j=1

mjuj

)=

l∑i=1

k∑j=1

nimjwiuj,

where wiuj is the concatenation of the words wi and uj. Notice that theempty word ε is the unit of Z[X∗]. These operations meet the requirementsfor Z[X∗] to become a unital ring. We claim that Z[X∗] is the free unitalring on X.

20

Theorem 2.11. Z[X∗] is the free unital ring on X.

Proof. Let R be a ring with unit 1 and ϕ : X → R a map. We have theinclusion map i : X → Z[X∗]. Note that if w is a word on X, then we maywrite w = x1 · · ·xn, where xi ∈ X. Let ϕ : Z[X∗]→ R be given by

ϕ

( l∑i=1

niwi

)=

l∑i=1

niϕ(xi1) · · ·ϕ(ximi),

where wi = xi1 · · ·ximiwith xik ∈ X. In particular, ϕ(ε) = 1. It should be

apparent that ϕ is a homomorphism such that ϕ ◦ i = ϕ. As for uniqueness,assume there is another such homomorphism ϕ′ such that ϕ′ ◦ i = ϕ. Thenwe have that ϕ ◦ i(x) = ϕ(x) = ϕ′ ◦ i(x) for all x ∈ X. Thus ϕ and ϕ′ agreeon a set of generators for Z[X∗] (since X generates this ring), so we deducethat ϕ = ϕ′. Hence, ϕ is unique, and Z(X∗) is the free ring on X.

Example 2.10. When X = ∅, Z[X∗] = {nε | n ∈ Z} ∼= Z. Let X = {x}.Then Z[X∗] = {n0ε + n1x + n2xx + · · · + nlxx · · ·x | l ≥ 0;n0, . . . , nl ∈ Z},which is isomorphic to Z[x], where Z[x] is the polynomial ring with indeter-minate x and integer coefficients. In general, if X = {x1, ..., xn}, then Z[X∗]is isomorphic to the polynomial ring with noncommuting indeterminantesx1, ..., xn and integer coefficients (often denoted Z〈x1, . . . , xn〉).

2.5.2 Free Fields

Unfortunately, not every category has free objects. In this section, we showthat free fields do not exist. Let X be a set. Assume that the free field onX exists. Call it F(X), and let i denote the corresponding map from X intoF(X). Then for any map ϕ from X into a field K, there exists a unique fieldhomomorphism ϕ : F(X) → K such that ϕ ◦ i = ϕ. There are two cases toconsider.

If X is nonempty, let F(X) have characteristic m (m = 0 or m is prime).Let K be any field of any characteristic other than m. Then the only homo-morphism from F(X) to K is the zero map. Forcing ϕ to be the zero mapimplies that ϕ ◦ i = ϕ sends everything to zero as well. But we are free to letϕ send elements of X to anything in K that we want. In particular, ϕ doesnot have to send everything to zero (contradiction).

If X is empty, then i and ϕ are the canonical empty functions. So theuniversal mapping property just asserts that for each field K there exists

21

a unique homomorphism ϕ : F (∅) → K (no compatibility criterion needsto be met). But there are two homomorphisms from F(∅) to itself, namelythe zero morphism and the identity map. This contradicts the uniquenessrequirement. Therefore, free fields cannot exist.

If we require that homomorphisms send the unit of the domain to theunit of the codomain, then we can modify this argument to notice that F(X)has no morphisms into a field of characteristic not m. Thus no free fields canexist.

3 Lie Algebras

In this section, we define the Lie algebra and examine its basic properties.We then move on to study the universal enveloping algebra in section 3.2,and the free Lie algebra in section 3.3.

3.1 Basic Properties

Definition 3.1. A Lie algebra L over a field F is an algebra such thatits multiplication is alternating and obeys the Jacobi identity. That is, werequire that

1. [x, x] = 0 (alternating), and

2. [x, [y, z]] + [y, [z, x]] + [z, [x, y]] = 0 for all x, y, z ∈ L (Jacobi identity).

Recall from section 2.4.1 that an algebra’s multiplication is by definitionbilinear. The bilinear and alternating multiplication imply that for all x andy in L, [x, y] = [−y, x]. Indeed, we have

0 = [x+ y, x+ y].

= [x, x] + [x, y] + [y, x] + [y, y].

= [x, y] + [y, x].

Hence, [x, y] = −[y, x]. We call this property skew-symmetry. If the charac-teristic of our field is not two, then we can show that skew-symmetry impliesalternating. If skew-symmetry holds, then

[x, x] = −[x, x]. (1)

[x, x] + [x, x] = 0. (2)

(1 + 1)[x, x] = 0. (3)

22

If the characteristic of the field is not two, then we know that 1 + 1 6= 0,and so equation 3 implies that [x, x] = 0. If the characteristic of the field is2, then 1 + 1 = 0, and so we can not conclude that [x, x] = 0.

We say that a Lie algebra L is abelian if [x, y] = 0. Note that becauseof skew-symmetry, we have that [x, y] = 0 if and only if [x, y] = [y, x]. Notethat all one dimensional Lie algebras are abelian.

Using skew-symmetry and bilinearity, we can rewrite the Jacobi identity.

[x, [y, z]] + [y, [z, x]] + [z, [x, y]] = 0.

[x, [y, z]] = −[z, [x, y]]− [y, [z, x]].

= [[x, y], z] + [y,−[z, x]].

= [[x, y], z] + [y, [x, z]].

The equation [x, [y, z]] = [[x, y], z] + [y, [x, z]] is a more useful form of theJacobi identity, for it reveals a structure similar to that of the product rule forderivatives. To see this, first define the left multiplication operator, Lx(y) :L → L, to be given by Lx(y) = [x, y]. The operator simply multiplies avector on the left by x. We rewrite the more useful form of the Jacobi interms of the left multiplication operator.

[x, [y, z]] = [[x, y], z] + [y, [x, z]].

Lx([y, z]) = [Lx(y), z] + [y, Lx(z)].

The left multiplication operator on the product [y, z] behaves like differen-tiation on the product of two functions. We formalize this observation withthe following definition.

Definition 3.2. Let A be an algebra over a field F. If ∂ : A→ A is a linearmap such that

∂(x · y) = ∂(x) · y + x · ∂(y),

then ∂ is a derivation of A.

The Jacobi identity implies that the left multiplication operator Lx is aderivation of a Lie algebra.

Example 3.1. Let L = R3. Let i, j, and k denote the standard unit vectors.Define the multiplication by the vector cross-product rule: [i, j] = k, [j,k] = i,and [k, i] = j. This makes L into a Lie algebra.

23

Example 3.2. Let A be any associative algebra. Then A is a Lie algebraunder the commutator bracket, [, ] : A × A → A, which is given by[x, y] = x · y − y · x, where · denotes the associative multiplication. Thecommutator bracket is a very useful way of turning any associative algebrainto a Lie algebra. For example, the set of all n×n matrices is a Lie algebraunder the commutator bracket since matrix multiplication is associative. Weadopt the notation that if A is an associative algebra, then [A] denotes theLie algebra obtained from A by equipping it with the commutator bracket.

Next, we define subalgebras and ideals.

Definition 3.3. A subset S of a Lie algebra L is a subalgebra of L if

1. S is a subspace of L, and

2. [S, S] ⊆ S. That is, for all x, y ∈ S, [x, y] ∈ S.

Definition 3.4. A subset I of a Lie algebra L is an ideal of L if

1. I is a subspace of L, and

2. [I, L] ⊆ I. That is, for all x ∈ I and y ∈ L, [x, y] ∈ I.

Because of skew-symmetry, we have [I, L] ⊆ I if and only if [L, I] ⊆ I.This makes it easier to check that a subspace of a Lie algebra is an ideal. Thisfact also applies to subalgebras. It follows from the definitions that all idealsare subalgebras, but not all subalgebras are ideals. Also, as with subspaces,the intersection of two ideals is an ideal, yet the union of two ideals is notan ideal unless one of the ideals contains the other.

Example 3.3. Let gl(2,R) denote the vector space of all 2 × 2 matricesover R. From example 2, because matrix multiplication is associative, wecan turn gl(2,R) into a Lie algebra by equipping it with the commutatorbracket: [x, y] = xy− yx, where xy is the usual product of matrices x and y.If b(2,R) is the set of all 2 × 2 upper triangular matrices, then b(2,R) is asubalgebra of g(2,R). To check that b(2,R) is a subalgebra, one would needto verify that it is indeed a subspace of g(2,R) and that the multiplicationis closed. It suffices to check closure on any basis. So, if β = {e11, e12, I},where I is the 2 × 2 identity matrix and eij is the matrix with a one in theith row and jth column and zeros elsewhere, then observe that

[e11, e12] = e11e12 − e12e11 = e12 ∈ b(2,R);

24

[e11, I] = e11I − Ie11 = 0;

[e12, I] = e12I − Ie12 = 0.

We know all other multiplications are closed because of alternating and skew-symmetry. Note that b(2,R) is not an ideal of gl(2,R). To check this, weobserve that although

y =

[0 11 0

]∈ gl(2,R),

the product

[e11, y] =

[1 00 0

] [0 11 0

]−[0 11 0

] [1 00 0

]=

[0 1−1 0

]/∈ b(2,R).

Example 3.4. Let L be a Lie algebra over a field F. We define the center ofL to be the set Z(L) = {x ∈ L | [x, y] = 0 for all y ∈ L}. We show that thecenter is an ideal of L. Observe that if x1, x2 ∈ Z(L), y ∈ L, and c ∈ F, then[x1 + cx2, y] = [x1, y] + c[x2, y] = 0. Hence, Z(L) is a subspace of L. Next,we must show that if x ∈ Z(L) and z ∈ L, then [x, z] ∈ Z(L). The Jacobiidentity implies that [[x, z], y] = [x, [z, y]]− [z, [x, y]] = 0. Hence, Z(L) is anideal of L.

Next, we define the homomorphism, kernel, and quotient space for Liealgebras.

Definition 3.5. Let L1 and L2 be Lie algebras over a field F. A linear mapϕ : L1 → L2 is a homomorphism if it preserves the multiplication. Thatis,

ϕ[x, y] = [ϕ(x), ϕ(y)] for all x, y ∈ L1.

Definition 3.6. Let L1 and L2 be Lie algebras over a field F. If ϕ is ahomomorphism from L1 into L2, then the kernel of ϕ, denoted ker(ϕ), isthe set of all elements in L1 whose image under ϕ is zero. That is,

ker(ϕ) = {x ∈ L1 | ϕ(x) = 0}.

Note that the kernel of a Lie algebra homomorphism is an ideal.

Definition 3.7. Let L be a Lie algebra over a field F and I an ideal of L.The quotient space is defined as

L/I = {x+ I | x ∈ L}.

25

Addition and scalar multiplication are defined by

(x+ I) + (y + I) = (x+ y) + I,

andc(x+ I) = (cx) + I

for all x, y ∈ L and c ∈ F. Multiplication on L/I is given by

[x+ I, y + I] = [x, y] + I.

Note that the multiplication on L/I bilinear since the multiplication on Lis bilinear. To check that the multiplication is well-defined, let x+ I = x′+ Iand y+ I = y′+ I for x, x′, y, y′ ∈ L. It follows that x = x′+u and y = y′+vfor some u, v ∈ I. Therefore,

[x, y]− [x′, y′] = [x′ + u, y′ + v]− [x′, y′].

= [x′, y′] + [x′, v] + [u, y′] + [u, v]− [x′, y′].

= [x′, v] + [u, y′] + [u, v] ∈ I

because I is an ideal. Thus, [x, y] + I = [x′, y′] + I, which implies that

[x+ I, y + I] = [x, y] + I = [x′, y′] + I = [x′ + I, y′ + I].

This shows that the multiplication is well-defined. We also have the canonicalhomomorphism ϕ : L→ L/I given by

ϕ(x) = x+ I

for all x ∈ L. We can readily verify that ϕ is a homomorphism since

ϕ(cx+ y) = (cx+ y) + I = c(x+ I) + (y + I) = cϕ(x) + ϕ(y),

andϕ([x, y]) = [x, y] + I = [x+ I, y + I] = [ϕ(x), ϕ(y)]

for all x, y ∈ L and c ∈ F. In particular, this canonical homomorphism isactually an epimorphism.

The usual isomorphism theorems hold for Lie algebras. We make use ofthe first isomorphism theorem, so we restate it here.

Theorem 3.1. Let L1 and L2 be Lie algebras over a field F and ϕ : L1 → L2

be a homomorphism. Then L1/ker(ϕ) ∼= ϕ(L1). If ϕ is surjective, thenL1/ker(ϕ) ∼= L2.

Now that we have established some of the basic properties of Lie algebras,we move our attention to universal enveloping algebras of Lie algebras.

26

3.2 The Universal Enveloping Algebra

Let L be a Lie algebra over a field F. Let

T =∞⊕n=0

T (L)n = T (L)0 ⊕ T (L)1 ⊕ T (L)2 ⊕ · · ·

denote the tensor algebra of L. Recall from section 2.4.2 that T (L)0 = F,T (L)1 = L, T (L)2 = L⊗ L and so on. Let J be the ideal of T generated byall elements of the form

x⊗ y − y ⊗ x− [x, y],

where x, y ∈ L. The quotient space U(L) = T/J is an associative algebraover F called the universal enveloping algebra of L.

Example 3.5. Let L be an n dimensional abelian Lie algebra over F. Conse-quently, if β = {x1, ..., xn} is a basis for L, then [xi, xj] = 0 for all xi, xj ∈ β.The ideal J is therefore generated by all elements of the form x⊗ y − y ⊗ xfor x, y ∈ L. This implies that U(L) = T/J is a commutative algebra. Inparticular, U(L) is generated by β, and the identity 1 ∈ F, which impliesU(L) is also isomorphic to the polynomial algebra F[x1, ..., xn].

In general, we have the inclusion map i : L → T , and the canonicalprojection map π : T → U(L). Let σ : L → U(L) be the composition of πwith i. Note that σ is linear. Recall from example 3.2 that for an associativealgebra A, [A] denotes the Lie algebra obtained from A by equipping itwith the commutator bracket. We proceed by showing that U(L) satisfies auniversal mapping property.

Theorem 3.2. Let L be a Lie algebra over F and A be an associative algebrawith unit 1 over F, and [A] the corresponding Lie algebra. Then given anyLie algebra homomorphism θ : L → [A], there exists a unique associativealgebra homomorphism φ : U(L)→ A such that φ ◦ σ = θ.

27

L A

T/J

σ

θ

φ

Figure 3: The commutative diagram for U(L) = T/J .

Proof. First, note that if β = {b1, b2, ...} is a basis for L, then recall thatB = {bi1 ⊗ · · · ⊗ bin | bik ∈ β} is a basis for T . Consider the map θ : B→ Agiven by

θ(bi1 ⊗ · · · ⊗ bin) = θ(bi1) · · · θ(bin).

Because T is free on B, this map can be extended to a unique associativealgebra homomorphism θ′ from T to A such that θ′ and θ agree on B. Notethat since β ⊂ B, θ and θ′ must also agree on β, which means they agree onL. Let x, y ∈ L. Then

θ′(x⊗ y − x⊗ y − [x, y]) = θ′(x⊗ y)− θ′(y ⊗ x)− θ′([x, y]).

= θ(x⊗ y)− θ(y ⊗ x)− θ([x, y]).

= θ(x)θ(y)− θ(y)θ(x)− θ([x, y]).

= 0

where the last equality holds because θ : L→ [A] is a Lie algebra homomor-phism. Thus, all the generators of the ideal J are in the kernel of θ′. Sincethe kernel is also an ideal, J must lie in the kernel of θ′. Therefore, there isan induced homomorphism φ : T/J → A such that φ ◦ π = θ′. If we restrictthe domain to L, then we get that φ ◦ σ = θ.

As for the uniqueness of φ, suppose that φ′ : U(L) → A is another suchassociative algebra homomorphism. Then φ ◦ σ = φ′ ◦ σ. This implies thatφ and φ′ agree on σ(L). Since L generates T , it follows that σ(L) generatesU(L) = T/J . Because φ and φ′ agree on the generators of U(L), they mustalso agree on U(L), and so they are the same.

28

Like many of the objects in previous sections, the universal mappingproperty of universal enveloping algebras ensures that they are unique up toisomorphism. The next theorem provides us with a basis for U(L).

Theorem 3.3 (Poincare-Birkhoff-Witt). Let L be a Lie algebra with basis{xi ; i ∈ I}. Let < be a total order on the index set I. Let σ : L → U(L)be the natural linear map from L into its enveloping algebra. Let σ(xi) = yi.Then the elements

yr1i1 ⊗ · · · ⊗ yrnin

for all n ≥ 0, all ri ≥ 0, and all i1, ..., in ∈ I with i1 < · · · < in form a basisfor U(L).

For a proof of the Poincare-Birkhoff-Witt theorem, see [CA]. We give anexample of how to use the Poincare-Birkhoff-Witt theorem to find a basis forthe universal enveloping algebra of sl2(F).

Example 3.6. Consider the Lie algebra sl2(F) of all 2 × 2 matrices over F

with trace 0. It has a basis given by {e, f, h}, where e =

[0 10 0

], f =

[0 01 0

],

and h =

[1 00 −1

]. It follows that a basis for the universal enveloping algebra

of sl2(F) is given by

{1, e, f, h, e⊗ e, e⊗ f, e⊗ h, f ⊗ f, f ⊗ h, h⊗ h, ...}.

Note the ordering of the basis elements. The Poincare-Birkhoff-Witt theoremrequires that we establish an ordering of the Lie algebra basis vectors, so inthis example, we choose to order the basis for sl2(F) so that e < f < h. Notethat although sl2(F) is finite-dimensional, its universal enveloping algebra isinfinite dimensional. This is true in general for all nontrivial Lie algebras.That is, the universal enveloping algebra of a Lie algebra is always infinite-dimensional, so long as the Lie algebra is not the zero algebra. Also, itis convenient to use juxtaposition as opposed to tensors when multiplyingelements in the universal enveloping algebra (we are justified in doing thissince tensors are associative.) We can therefore rewrite the basis as

{1, e, f, h, e2, ef, eh, f 2, fh, h2, ...}.

We prove two corollaries of the Poincare-Birkhoff-Witt theorem.

29

Corollary 3.3.1. The map σ : L→ U(L) is injective.

Proof. Let {xi ; i ∈ I} be a basis for L. Suppose that v ∈ ker(σ). Thenσ(v) = 0 and we have

σ(v) = σ

( n∑i=1

cixi

)for some scalars c1, ..., cn.

=n∑i=1

ciσ(xi).

=n∑i=1

ciyi = 0.

Because each of the yi are linearly independent, it must be the case thatc1 = · · · = cn = 0, which implies that v = 0. So, the kernel of σ is zero.

Corollary 3.3.2. The subspace σ(L) is a Lie subalgebra of [U(L)] isomorphicto L. Thus, σ identifies L with a Lie subalgebra of [U(L)].

Proof. By corollary 3.3.1, we know that σ : L → σ(L) is bijective, and soL ∼= σ(L). We know that σ(L) is a subspace of U(L) since it is the range ofa linear transformation. To check closure under the bracket, note that theelements yi, i ∈ I, form a basis of σ(L) and that

[yi, yj] = [σ(xi), σ(xj)] = σ([xi, xj]).

Hence, [yi, yj] ∈ σ(L) and so σ(L) is a Lie subalgebra of [U(L)].

The universal enveloping algebra has many uses in the theory of Lie al-gebras. For our purposes, the universal enveloping algebra of a Lie algebrahelps study free Lie algebras.

3.3 Free Lie Algebras

Let X be a set, and V (X) the vector space with basis X. Consider the tensoralgebra F (X) of V (X). Recall that F (X) is the free associative algebra onX. Since F (X) is associative, let [F (X)] be the corresponding Lie algebraobtained from F (X) by equipping it with the commutator bracket. Notethat X is a subset of [F (X)]. Let FL(X) be the intersection of all Liesubalgebras of [F (X)] that contain X. That is, FL(X) is the subalgebra of[F (X)] generated by the X. We show that FL(X) is the free Lie algebra onX.

30

Theorem 3.4. FL(X) is the free Lie algebra on the set X.

Proof. We have the inclusion map i : X → FL(X). Let L be a Lie algebraand θ : X → L a map. We have the composition θ′ = σ ◦ θ : X → U(L),where σ is the natural linear map from the Lie algebra L into its universalenveloping algebra U(L). SinceX is basis for the vector space V (X), θ′ can beextended to a unique linear map φ : V (X)→ U(L) such that φ and θ′ agreeon X. Because V (X) is a basis for the tensor algebra F (X) of V (X), φ can beextended to a unique associative algebra homomorphism φ′ : F (X) → U(L)such that φ′ and φ agree on V (X). This same map gives a Lie algebrahomomorphism from [F (X)] to [U(L)]. We also see that φ′ and θ′ agree onX and so φ′(X) = θ′(X) = σ ◦ θ(X). This implies that φ′(X) ⊂ σ(L). Weknow from corollary 3.3.2 that σ(L) is a Lie subalgebra of [U(L)] isomorphicto L. The set S = {x ∈ [F (X)] | φ′(x) ∈ σ(L)} is also a subalgebra of[F (X)]. In particular, S contains X, which means it also contains FL(X).Therefore, by restricting the domain, we obtain φ′ : FL(X) → σ(L). Sincethe map σ : L→ σ(L) is bijective, we are justified in defining ϕ : FL(X)→ Lby ϕ = σ−1 ◦ φ′. We check that ϕ and θ agree on X by observing that forx ∈ X, we have

ϕ ◦ i(x) = σ−1 ◦ φ′ ◦ i(x).

= σ−1 ◦ φ(x).

= σ−1 ◦ θ′(x).

= σ−1 ◦ σ ◦ θ(x).

= θ(x).

Thus, we have a homomorphism of the required type. To show that ϕ isunique, let ϕ′ be another such homomorphism. Let S ′ be the set of elementsin FL(X) for which ϕ and ϕ′ agree. Because ϕ◦ i = ϕ′ ◦ i, we see that ϕ andϕ′ agree on X. Moreover, S ′ is the kernel of the homomorphism (ϕ−ϕ′), soit is necessarily a subalgebra of FL(X). Since X generates FL(X) as a Liesubalgebra, we conclude that ϕ and ϕ′ agree on FL(X).

Example 3.7. Not surprisingly, if X = ∅, then FL(∅) = {0}. If X isa singleton set, say X = {x}, then the free Lie algebra on X is given byFL(X) = span{x}, which is a one dimensional abelian Lie algebra (there isonly one such algebra up to isomorphism). If |X| ≥ 2, then FL(X) is infinitedimensional.

31

The next theorem illustrates that we may identify the free associativealgebra on a set X with the universal enveloping algebra of the free Liealgebra on X.

Theorem 3.5. The universal enveloping algebra U(FL(X)) is isomorphic toF (X).

Proof. We show that F (X) is also a universal enveloping algebra for the freeLie algebra FL(X). We do this by showing that F (X) satisfies the universalmapping property given in theorem 3.2. So, let A be an associative algebrawith unit 1 and θ : FL(X)→ [A] be a Lie algebra homomorphism. We havethe inclusion map σ : FL(X) → F (X). Restricting the domain of σ, weobtain a map from X to F (X). Observe that since F (X) is the free associa-tive algebra on X, there exists a unique associative algebra homomorphismφ : F (X) → A such that φ and θ agree on X. This same map gives a Liealgebra homomorphism from [F (X)] to [A]. Since X generates FL(X) asa Lie subalgebra, and FL(X) ⊂ F (X), we deduce that φ and θ must alsoagree on FL(X). As for uniqueness, recall that φ is the unique homomor-phism from F (X) to A such that φ and θ agree on X. Since X ⊂ FL(X), itmust also be that φ is the unique homomorphism from F (X) to A such thatφ and θ agree on FL(X).

One may use what are called Lyndon words to construct a basis for thefree Lie algebra, called the Lyndon basis. Lyndon bases are not easily acces-sible without a working knowledge of combinatorics. Therefore, we do notdiscuss them in this paper. We proceed by discussing Leibniz algebras.

4 Leibniz Algebras

Leibniz algebras are a generalization of Lie algebras. In this section, we beginwith the basis properties of Leibniz algebras and conclude with free Leibnizalgebras.

4.1 Basic Properties

Definition 4.1. A left Leibniz algebra A is an algebra that obeys theLeibniz identity. That is, we require that

[x, [y, z]] = [[x, y], z] + [y, [x, z]] for all x, y, z ∈ A. (Leibniz Identity)

32

Note that the Leibniz identity is actually just the more useful form ofthe Jacobi identity that we derived in section 3.1. It follows that all Liealgebras are Leibniz algebras. What separates the two is that a Leibnizalgebra’s multiplication is not necessarily alternating. Consequently, Leibnizalgebras do not necessarily have skew-symmetry either. This requires that wedistinguish between left and right Leibniz algebras. We make this distinctionby noticing that the Leibniz identity in definition 4.1 shows that the leftmultiplication operator Lx, which is given by Lx(y) = [x, y], is actually aderivation of A. Therefore, a left Leibniz algebra is an algebra whose leftmultiplication operator is a derivation. Similarily, a right Leibniz algebra isan algebra whose right multiplication operator Rx, which is given by Rx(y) =[y, x], is a derivation of A. Like Lie algebras, we say that a Leibniz algebrais abelian if [x, y] = 0 for all x, y ∈ A.

Example 4.1. Let A be a two dimensional algebra over a field F with mul-tiplications given by

[x, y] = [x, x] = y and [y, x] = [y, y] = 0.

Then A is a non-Lie, left Leibniz algebra because the left multiplicationoperator is a derivation. However, note that

Ry([x, x]) = [[x, x], y] = 0,

yet[Ry(x), x] + [x,Ry(x)] = [[x, y], x] + [x, [x, y]] = y 6= 0.

Thus, A is not a right Leibniz algebra because the right multiplication oper-ator is not a derivation.

Example 4.2. Let A be a two dimensional algebra over a field F with mul-tiplications given by

[x, x] = [y, x] = y and [x, y] = [y, y] = 0.

Then A is a non-Lie, right Leibniz algebra. However, note that

Lx([x, x]) = [x, [x, x]] = 0,

yet[Lx(x), x] + [x, Lx(x)] = [[x, x], x] + [x, [x, x]] = y 6= 0.

Thus, A is not a left Leibniz algebra.

33

Let A be a left (right) Leibniz algebra and a ∈ A. Then we define an

inductively by letting a1 = a and an+1 = [a, an] (an+1 = [an, a]). Usinginduction, one can show that if k ≥ 2, then Lak = 0 (Rak = 0), where thebase case is a direct consequence of the Leibniz identity. That is, for leftLeibniz algebras, left multiplication by a an element of power greater than 1yields zero, and for right Leibniz algebras, right multiplication by an elementof power greater than 1 yields zero. Next, we give an example of a Leibnizalgebra that both a left and a right Leibniz algebra.

Example 4.3. Let A be a two dimensional algebra over a field F generatedby a single element. That is, A = span{x, x2}. Let the multiplications begiven by

[x2, x] = [x, x2] = 0.

Then A is both a left and a right Leibniz algebra. In particular, we say thatA is a cyclic, nilpotent Leibniz algebra.

Since left Leibniz algebras are analogs of right Leibniz algebras, for theremainder of this paper, when we refer to a Leibniz algebra, we will assume itis a left Leibniz algebra. Subalgebras for Leibniz algebras are defined in thesame way they are for Lie algebras. However, since we are not guaranteedskew-symmetry, we must distinguish between left and right ideals of Leibnizalgebras.

Definition 4.2. Let A be a Leibniz algebra over F and I a subspace of A.Then I is a left ideal if [A, I] ⊆ I, and I is a right ideal if [I, A] ⊆ I. IfI is both a left and a right ideal, then we say I is an ideal of A.

Example 4.4. Let A be the two-dimensional Leibniz algebra over F with[y, x] = x and all other multiplications zero. Then I = span{x} is an idealof A, for it is both a left and a right ideal. Although J = span{y} is a leftideal, it is not a right ideal since [y, x] = x /∈ J .

Example 4.5. Let A be a Leibniz algebra over a field F. We define theLeib(A) = span{[a, a] | a ∈ A}. The Leib(A) is an ideal of A. To see why,let a, b ∈ A and note that [[b, b], a] = 0, which implies that Leib(A) is a rightideal. Keeping in mind that left multiplication by a power greater than 1 iszero, observe that

[a, [b, b]] = [a, a] + [a, [b, b]] + [[b, b], a] + [[b, b], [b, b]]− [a, a].

= [a+ [b, b], a+ [b, b]]− [a, a].

∈ Leib(A).

34

Hence, Leib(A) is both a left and a right ideal, so we conclude that Leib(A)is an ideal of A. The significance of the Leib(A) is that it is the smallest idealsuch that A/Leib(A) is a Lie algebra. To see why, suppose that I is anotherideal of A such that A/I is Lie. Then for all y ∈ A, I = [y+I, y+I] = [y, y]+I,which implies that [y, y] ∈ I. Hence, Leib(A) ⊆ I.

Now that we have established some of the basic properties of Leibnizalgebras, we move our focus to free Leibniz algebras.

4.2 Free Leibniz Algebras

As usual, we begin with a set X. Recall from section 2.4.3 that we have thefree non-associative algebra A on X. Let J be the ideal of A generated byelements of the form

[x, [y, z]]− [[x,y], z]− [y, [x, z]],

where x,y, and z are non-associative words on X. Then the quotient spaceA(X) = A/J is a Leibniz algebra, called the free Leibniz algebra on X. Weshow that A(X) satisfies the universal mapping property for free objects.

Theorem 4.1. The Leibniz algebra A(X) is the free Leibniz algebra on X.

Proof. Let A1 be a Leibniz algebra and ϕ : X → A1 a map. We have theinclusion map i : X → A, and the projection map π : A → A/J . Letσ = π ◦ i : X → A/J . We need a unique homomorphism ϕ : A/J → A1 suchthat ϕ ◦ σ = ϕ. Because A is the free non-associative algebra on X, there isa unique algebra homomorphism ϕ′ : A → A1 such that ϕ′ and ϕ agree onX. Observe that for all x,y, z ∈ A, we have that

ϕ′[[x, [y, z]− [[x,y], z]− [y, [x, z]]

]= [ϕ′(x), [ϕ′(y), ϕ′(z)]]− [[ϕ′(x), ϕ′(y)], ϕ′(z)]− [ϕ′(y), [ϕ′(x), ϕ′(z)]].

= 0.

The last equality holds because A1 is a Leibniz algebra. Thus, we see thatthe generators of J lie in the kernel of ϕ′. Since the kernel is an ideal, we wededuce that J ⊆ ker(ϕ′). This implies there is an induced homomorphismϕ : A/J → A1. Since ϕ and ϕ agree on X, we deduce that ϕ ◦ σ = ϕ.Because X generates A, we know that the homomorphism is unique.

35

Our ultimate result shows how to find a basis for the free Leibniz algebra.Although certain combinatorial methods are required to give an extensivetreatment of free Leibniz algebras, we can, unlike free Lie algebras, at leastunderstand what the free Leibniz algebra basis looks like without a workingknowledge of combinatorics.

Theorem 4.2. Let A(X) be the free Leibniz algebra on X. Let β be the setcontaining elements of the form

[x1, [x2, ..., [xn−1, xn]...]],

where x1, ..., xn ∈ X. Then β forms a basis for A(X).

We prove that the basis spans the free Leibniz algebra. Showing linearindependence requires combinatorial techniques that are outside the scope ofthis paper. For a complete proof, see [MI].

Proof (spanning set.) Recall from section 2.4.3 that we may define the lengthof a non-associative word on X as the number of elements of X used toconstruct the word. For example, the length of x is 1, the length of [[x, x], z]is 3, and so on. We only consider non-empty words since the theorem triviallyholds for empty words. Any non-associative word of length 1 or 2 is inspan(β). As for words of length 3, the Leibniz identity, when re-writtenas [[x, y], z] = [x, [y, z]] − [y, [x, z]], ensures that all words of length 3 areelements of the span of β. We proceed by induction on the lengths of thewords. Assume that any word of length less than n is an element of thespan of β. Let [A,B] represents a word of length n. We must show that[A,B] ∈ span(β).

Suppose the length of A is 1. Then let A = x1. The length of B must beless than n, so the inductive hypothesis implies B = [x2, [x3, ...[xn−1, xn]]...].Therefore, [A,B] = [x1, [x2, ...[xn−1, xn]]...].

We make a second inductive argument by assuming all elements of theform [C,D] are in the span of β, where the length of [C,D] = n, and thelength of C is less than k (which is also necessarily less than n). Suppose Ahas length k. Then k must be less than n, so by our first inductive hypothesisA ∈ span(β). Let

A =m∑i=1

[xi1 , [xi2 , [...[xik−1, xik ]]...].

36

From bilinearity, we have

[A,B] =m∑i=1

[[xi1 , [xi2 , [...[xik−1, xik ]]...], B].

For notational purposes, let C = [xi2 , [...[xik−1, xik ]]...]. Then we can write

[A,B] = [[xi1 , C], B] = [xi1 , [C,B]] − [C, [xi1 , B]]. Notice that the length of[C,B] < n. So, the first inductive hypothesis implies [xi1 , [C,B]] ∈ span(β).As for [C, [xi1 , B]], note the length of C equals the length of A−1 = k−1 < k.Thus, the second inductive hypothesis implies [C, [xi1 , B]] ∈ span(β). Hence,[A,B] ∈ span(β).

Example 4.6. Let X = {x}. The set A(X) = span{x, x2, x3, ...} is the freeLeibniz algebra on the set X. We see that the free Leibniz algebra generatedby one element is an infinite dimensional cyclic algebra. In general, all freeLeibniz algebras are infinite dimensional with the only exception being thecase where the generating set X is empty (in such a case A(∅) = {0}).

References

[CA] R. Carter, Lie Algebras of Finite and Affine Type, Cambridge Uni-versity Press, 2005, p. 155-159.

[CO] B. Cooperstein, Advanced Linear Algebra, Taylor and FrancisGroup, 2010, p. 300 - 302.

[DE] I. Demir, K.C. Misra, E. Stitzinger, “On Some Structures of LeibnizAlgebras,”Contemporary Mathematics, 623:41-54.

[LO] J.-L., Loday, “Une version non commutative des algebres de Lie:les algebres de Leibniz,” Enseign. Math. (2), 1993, 39(3-4):269-293.

[MI] A.A. Mikhalev, V. Shpilrain, J. Yu, Combinatorial Methods: FreeGroups, Polynomials, and Free Algebras, Springer, 2000, p. 275-276.

[RO] J.J. Rotman, An Introduction to the Theory of Groups, 4th ed.,Springer, 1995, p. 344-345

37