Commuting elements in hom-associative algebras

School of Education, Culture and Communication

Division of Mathematics and Physics

MASTER’S DEGREE PROJECT IN MATHEMATICS

Commuting elements in hom-associative algebras

by

Viktor Klinga

MAA515 — Examensarbete i matematik för masterexamen

DIVISION OF MATHEMATICS AND PHYSICS

MÄLARDALEN UNIVERSITY SE-721 23 VÄSTERÅS, SWEDEN

School of Education, Culture and Communication

Division of Mathematics and Physics

MAA515 — Master’s Degree Project in Mathematics

Date of presentation:

3 June 2021

Project name:

Commuting elements in hom-associative algebras

Author(s):

Viktor Klinga

Version:

5th June 2021

Supervisor(s):

Sergei Silvestrov, German Garcia

Reviewer: Lars Hellström Examiner: Masood Aryapoor Comprising: 30 ECTS credits

Abstract

In this thesis, we consider hom-associative algebras, which is an algebra with multiplication that is not necessarily commutative nor associative, but obeys a twisted version of associativity by a linear homomorphism. We will give some conditions for associativity, which helps us determine commuting elements. Under other conditions, such as different types of unitality conditions, we can also state some results regarding commuting elements in the general, non-associative case.

Contents

1 Introduction 3

1.1 Structure of the report . . . 4

2 Properties of hom-associative algebras 5 2.1 Multiplicative and weakly unital algebras . . . 6

2.2 Unital hom-associative algebras . . . 7

2.3 Hom-Lie algebras . . . 10

2.4 Generalized hom-associative structures . . . 13

3 Commuting elements in hom-associative algebras 19 3.1 Commuting elements in multiplicative and weakly unital algebras . . . 20

3.2 Commuting elements in unital algebras . . . 21

3.3 Examples . . . 24

Chapter 1

Introduction

We will assume that the reader is familiar with basic terminology in abstract algebra, such as fields, linear spaces, homomorphisms, etc.. Furthermore, we will always let 𝐾 denote a field of characteristic 0. Let N denote the set of non-negative integers.

Definition 1.1. Let 𝑉 be a linear space over 𝐾, 𝜇 : 𝑉 × 𝑉 → 𝑉 a bilinear map and 𝛼 : 𝑉 → 𝑉 a

linear space homomorphism. Then a hom-associative algebra is the triple (𝑉 , 𝜇, 𝛼) satisfying 𝜇(𝛼(𝑥), 𝜇(𝑦, 𝑧)) = 𝜇(𝜇(𝑥, 𝑦), 𝛼(𝑧)). (1.1) The name hom-associative algebra arises from the fact that we twist the regular associative law with a homomorphism. There are different ways of defining a hom-associative algebra, but here we use the same definition as they do in [1].

Remark1.2. A bilinear map 𝜇 : 𝑉 × 𝑉 → 𝑉 , where 𝑉 is a linear space over a field 𝐾, is a map satisfying

𝜇(𝑥, 𝑦 + 𝑧) = 𝜇(𝑥, 𝑦) + 𝜇(𝑥, 𝑧), 𝜇(𝑥 + 𝑦, 𝑧) = 𝜇(𝑥, 𝑧) + 𝜇(𝑦, 𝑧),

𝜇(𝑘𝑥, 𝑦) = 𝑘 𝜇(𝑥, 𝑦), 𝜇(𝑥, 𝑘 𝑦) = 𝑘 𝜇(𝑥, 𝑦), for all 𝑥, 𝑦, 𝑧 ∈ 𝑉 , and for all 𝑘 ∈ 𝐾.

Remark 1.3. The notations used in Definition 1.1 will be used frequently throughout the rest of the thesis. Hence, unless stated otherwise, 𝑉 , 𝜇 and 𝛼 will be reserved to be as in the definition. We may also say that an element belongs to the hom-associative algebra, if the element belongs to 𝑉 .

In this thesis, we will study what elements commute in a hom-associative algebra. Finding commuting elements in a non-commutative algebraic structures is a research topic of great interest, since it provides much information on the structure. However, there are results stating conditions for a hom-associative algebra to be associative as well. If we have associativity, we can say quite a bit about the commuting elements in general, even if we do not have hom-associativity. Hence, if we also have associativity, we can say more about commuting elements.

Example 1.4. Obviously, if 𝛼 = id, the identity map, then we have associative multiplication.

In addition to this, if 𝜇 is commutative, then studying commutative elements is trivial, since all elements commute. However, 𝜇 is not in general commutative, e.g. the linear space of 𝑛 × 𝑛 invertible matrices together with 𝜇 being ordinary matrix multiplication, is not commutative.

The hom-Lie algebra structure was first introduced by Hartwig, Larsson and Silvestrov in [2]. This was later extended to the class of quasi-Lie and quasi-hom-Lie in [3, 4]. From this, Makhlouf and Silvestrov introduced the hom-associative algebra in [1]. Especially, it is shown that a hom-associative algebra together with the bracket product defined by the commutator induces a hom-Lie algebra.

In [5], the structure of a multiplicative hom-associative algebra is studied and a special unitality condition is considered. In this thesis, we refer to this as a weakly unital associative algebra. However, several different unitality conditions make sense in a hom-associative algebra, as has been shown in [6].

Different types of algebras which are related to the hom-associative algebras, are studied in [7]. The algebras studied are essentially those with the property

𝜇(𝑥, 𝜇(𝑦, 𝛼(𝑧)) = 𝜇(𝜇(𝛼(𝑥), 𝑦), 𝑧),

or some other property where the linear space homomorphism 𝛼 is not in its usual place. In [8] the authors continue the study of these “alternative” hom-associative algebras, together with the unitality condition, i.e., that there exists an element 1 (a unity) in the linear space, such that

𝜇(1, 𝑥) = 𝜇(𝑥, 1) = 𝑥, for every element 𝑥 in the same space.

An Ore extension, introduced by Ore in [9], is a non-commutative polynomial ring. Bäck, Richter and Silvestrov studied hom-associative Ore-extensions in [10], and especially weak

units. Bäck and Richter prove a more general version for Ore extensions of Hilbert’s basis theorem in [11], which considers both non-associativity and hom-associativity.

1.1

Structure of the report

In chapter 2 we study general properties of hom-associative algebras. The definition, and some properties, of multiplicative and weakly unital algebras is studied in section 2.1. Section 2.2 explores properties when we have a unit element, and we prove that if 𝛼 is injective, then the unital hom-associative algebra is associative. hom-Lie algebras and how it relates to hom-associative algebras is handled in section 2.3. We also look at a more generalized hom-associative algebra in section 2.4, and give a proof of when this algebra is associative.

In chapter 3, we focus on the actual commuting elements of the different cases studied in chapter 2. We look at what elements commute in multiplicative algebras in section 3.1. In section 3.2 we study the commuting elements of unital algebras. We end this chapter with section 3.3, where we present some examples of hom-associative algebras, and try to classify a 3-dimensional hom-associative algebra.

Chapter 2

Properties of hom-associative algebras

Remark2.1. For a bilinear map 𝜇 defined on a linear space 𝑉 over 𝐾, we have that for all 𝑥 ∈ 𝑉 that

𝜇(𝑥, 0) = 𝜇(0, 𝑥) = 0, due to linearity over 𝐾.

Definition 2.2. Given two elements 𝑥 and 𝑦 in a hom-associative algebra, we call

[𝑥, 𝑦] = 𝜇(𝑥, 𝑦) − 𝜇(𝑦, 𝑥) the commutator of 𝑥 and 𝑦.

Remark2.3. Two elements commute if and only if their commutator is equal to 0.

Lemma 2.4. Let 𝛼 be a linear space homomorphism. Then 𝛼 is injective iff Ker(𝛼) = {0}. Proof. Suppose 𝛼 is injective. For any two elements 𝑥 and 𝑦 such that 𝛼(𝑥 − 𝑦) = 0, we have that 𝛼(𝑥) = 𝛼(𝑦). Since 𝛼 is injective, 𝑥 = 𝑦 ⇒ 𝑥 − 𝑦 = 0, i.e. Ker(𝛼) = {0}.

If Ker(𝛼) = {0}, then for any two elements 𝑥 and 𝑦 such that 𝛼(𝑥) = 𝛼(𝑦), we get 𝛼(𝑥 − 𝑦) = 0. But since Ker(𝛼) = {0}, this means that 𝑥 − 𝑦 = 0 ⇒ 𝑥 = 𝑦. Thus 𝛼 is

injective. _

Proposition 2.5. Let (𝑉 , 𝜇, 𝛼) be a hom-associative algebra. Suppose 𝛼(𝑥) = 𝑘𝑥 for all 𝑥 ∈ 𝑉 , where 𝑘 ≠ 0 is some fixed element in 𝐾. Then the associative law holds.

Proof. Let 𝑥, 𝑦, 𝑧 ∈ 𝑉 . Then

𝜇(𝛼(𝑥), 𝜇(𝑦, 𝑧)) = 𝜇(𝜇(𝑥, 𝑦), 𝛼(𝑧)). We get

𝜇(𝑘𝑥, 𝜇(𝑦, 𝑧)) = 𝜇(𝜇(𝑥, 𝑦), 𝑘 𝑧). From bilinearity we obtain

𝑘 𝜇(𝑥, 𝜇(𝑦, 𝑧)) = 𝑘 𝜇(𝜇(𝑥, 𝑦), 𝑧). Since 𝐾 is a field, and 𝑘 ∈ 𝐾, this implies that

𝜇(𝑥, 𝜇(𝑦, 𝑧)) = 𝜇(𝜇(𝑥, 𝑦), 𝑧),

2.1

Multiplicative and weakly unital algebras

Multiplicativehom-associative algebras are a class of hom-associative algebras, which we will present in the next definition.

Definition 2.6. A hom-associative algebra (𝑉 , 𝜇, 𝛼) is multiplicative if

𝛼( 𝜇(𝑥, 𝑦)) = 𝜇(𝛼(𝑥), 𝛼(𝑦)), for all 𝑥, 𝑦 ∈ 𝑉 .

There are different kinds of units in a hom-associative algebra, and next we will define what we call a weak unit.

Definition 2.7. An element 𝑢 ∈ 𝑉 , such that

𝜇(𝑥, 𝑢) = 𝜇(𝑢, 𝑥) = 𝛼(𝑥), (2.1) is a weak unit of the hom-associative algebra (𝑉 , 𝜇, 𝛼). Such an algebra is called weakly unital.

Remark 2.8. Suppose that there exist two elements 𝑢, 𝑣 ∈ 𝑉 , that satisfy the conditions in Definition 2.7. Then

𝛼(𝑣) = 𝜇(𝑣, 𝑢) = 𝛼(𝑢), i.e. this element is unique if 𝛼 is injective.

The next proposition gives a condition when weakly unital hom-associative algebras are multiplicative.

Proposition 2.9. Let (𝑉 , 𝜇, 𝛼) be a weakly unital hom-associative algebra with weak unit 𝑢. If

𝛼(𝑢) = 𝑢, (2.2)

then (𝑉 , 𝜇, 𝛼) is multiplicative.

Proof. For any two elements 𝑥, 𝑦 ∈ 𝑉 , we have that

𝜇(𝛼(𝑥), 𝛼(𝑦)) = 𝜇(𝛼(𝑥), 𝜇(𝑦, 𝑢)) = 𝜇( 𝜇(𝑥, 𝑦), 𝛼(𝑢)),

by using the definition of the weak unit, and hom-associativity. Then using property (2.2) on this previous expression yields

𝜇( 𝜇(𝑥, 𝑦), 𝛼(𝑢)) = 𝜇(𝜇(𝑥, 𝑦), 𝑢) = 𝛼( 𝜇(𝑥, 𝑦)).

Hence 𝜇(𝛼(𝑥), 𝛼(𝑦)) = 𝛼(𝜇(𝑥, 𝑦)), for all 𝑥, 𝑦 ∈ 𝑉 , so (𝑉 , 𝜇, 𝛼) is multiplicative, as desired. 

2.2

Unital hom-associative algebras

We will now study another restriction of the hom-associative algebras, when we assume that we also have a unit element in the linear space, as introduced in [6].

Definition 2.10 ([6]). An element 1 ∈ 𝑉 , such that

𝜇(𝑥, 1) = 𝜇(1, 𝑥) = 𝑥, (2.3) is a unit of the hom-associative algebra (𝑉 , 𝜇, 𝛼). Such a algebra is a called unital.

Remark2.11. Suppose we have two units 1 and 𝑒 in 𝑉 . Then 1 = 𝜇(1, 𝑒) = 𝑒, which implies uniqueness.

From the definition of unital hom-associative algebras, we can prove this next proposition.

Proposition 2.12. In a unital hom-associative algebra, 𝛼(1) = 1 if and only if 𝛼 = id. Proof. If 𝛼 = id, then obviously 𝛼(1) = 1.

Suppose that 𝛼(1) = 1. Then for any 𝑥 ∈ 𝑉 ,

𝑥 = 𝜇(𝑥,1) = 𝜇(𝜇(𝑥, 1), 1) = 𝜇(𝜇(𝑥, 1), 𝛼(1)) = 𝜇(𝛼(𝑥), 𝜇(1, 1)) = 𝛼(𝑥).

Hence 𝛼(𝑥) = 𝑥, and thus, 𝛼 = id. _ We will now present a lemma which will be frequently used when studying other properties and commuting elements.

Lemma 2.13 ([6]). Let 𝑥 and 𝑦 be arbitrary elements in a unital hom-associative algebra. Then

𝜇(𝛼(𝑦), 𝑥) = 𝜇(𝑦, 𝛼(𝑥)), (2.4) 𝛼(𝑦) = 𝜇(𝑦, 𝛼(1)), (2.5) 𝛼( 𝜇(𝑥, 𝑦)) = 𝜇(𝛼(𝑥), 𝑦). (2.6)

Remark2.14. From (2.5), we see that 𝛼(1) is a weak unit.

Proof. We begin with proving equation (2.4). Hence

𝜇(𝛼(𝑦), 𝑥) = 𝜇(𝛼(𝑦), 𝜇(1, 𝑥)) = 𝜇(𝜇(𝑦, 1), 𝛼(𝑥)) = 𝜇(𝑦, 𝛼(𝑥)), by applying first (2.3), then hom-associativity, and then (2.3) again.

We also note that (2.5) follows from (2.4) by letting 𝑥 = 1 in (2.4), and applying (2.3) to the left hand side of (2.5).

Similarly, (2.6) follows from 𝛼( 𝜇(𝑥, 𝑦)) = 𝜇(𝛼(𝜇(𝑥, 𝑦)), 𝜇(1, 1)) = 𝜇( 𝜇( 𝜇(𝑥, 𝑦),1), 𝛼(1)) = 𝜇( 𝜇(𝑥, 𝑦), 𝛼(1)) = 𝜇(𝛼(𝑥), 𝜇(𝑦,1)) = 𝜇(𝛼(𝑥), 𝑦).  Let (𝑉 , 𝜇, 𝛼) be a unital hom-associative algebra. For all 𝑥 and 𝑦 in the algebra, we have 𝛼( 𝜇(𝑥, 𝑦)) = 𝜇(𝛼(𝑥), 𝑦) = 𝜇(𝑥, 𝛼(𝑦)), from Lemma 2.13. If (𝑉 , 𝜇, 𝛼) is multiplicative, we have the new property 𝜇(𝛼(𝑥), 𝑦) = 𝜇(𝛼(𝑥), 𝛼(𝑦)), on which we can apply equation (2.4) on the right hand side to obtain

𝜇(𝛼(𝑥), 𝛼(𝑦)) = 𝜇(𝛼2(𝑥), 𝑦). (2.7) Thus, using the definition of being a unital hom-associative algebra and (2.7), we obtain

𝛼(𝑥) = 𝜇(𝛼(𝑥), 1) = 𝜇(𝛼(𝑥), 𝛼(1)) = 𝜇(𝛼2(𝑥), 1) = 𝛼2(𝑥).

We can iterate this arbitrary many times, and we present the result in the next proposition.

Proposition 2.15. Let 𝑥 be an element in a multiplicative and unital hom-associative algebra

(𝑉 , 𝜇, 𝛼). Then 𝛼𝑘

(𝑥) = 𝛼𝑛

(𝑥), if 𝑘 and 𝑛 are positive integers.

Proof. The process that was presented right before this proposition can be used to get 𝛼𝑘

(𝑥) = 𝛼𝑘+1(𝑥), for any positive integer 𝑘. We can iterate the process arbitrary many times, such that 𝛼𝑘(𝑥) = 𝛼𝑛(𝑥), for any positive integer 𝑛 ≥ 𝑘. 

The next corollary is a direct implication of the previous proposition.

Corollary 2.16. Let (𝑉 , 𝜇, 𝛼) be a multiplicative and unital hom-associative algebra. If 𝛼 is injective, then 𝛼 = id.

Proof. For any 𝑥 in this algebra, we get from Proposition 2.15 that 𝛼2(𝑥) = 𝛼(𝑥) and linearity gives 𝛼(𝛼(𝑥) − 𝑥) = 0, and if 𝛼 is injective, this implies that 𝛼(𝑥) − 𝑥 = 0. Hence 𝛼(𝑥) = 𝑥,

for any 𝑥, as desired. _

Proposition 2.17 ([6]). Let x,y and z be elements in a unital hom-associative algebra. Then the following properties hold:

𝜇(𝛼(𝑥), 𝜇(𝑦, 𝑧)) = 𝜇(𝜇(𝛼(𝑥), 𝑦), 𝑧) (2.8) 𝜇(𝑥, 𝜇(𝛼(𝑦), 𝑧)) = 𝜇(𝜇(𝑥, 𝛼(𝑦)), 𝑧) (2.9) 𝜇(𝑥, 𝜇(𝑦, 𝛼(𝑧))) = 𝜇(𝜇(𝑥, 𝑦), 𝛼(𝑧)) (2.10) 𝛼( 𝜇(𝑥, 𝜇(𝑦, 𝑧))) = 𝛼(𝜇(𝜇(𝑥, 𝑦), 𝑧)) (2.11)

Proof. We begin with proving (2.8). By first using hom-associativity, then equation (2.4), (2.6) and (2.4) again, gives us

𝜇(𝛼(𝑥), 𝜇(𝑦, 𝑧)) = 𝜇(𝜇(𝑥, 𝑦), 𝛼(𝑧)) = 𝜇(𝛼( 𝜇(𝑥, 𝑦)), 𝑧) = 𝜇( 𝜇(𝛼(𝑥), 𝑦), 𝑧), as desired.

Using the properties of Lemma 2.13, together with (2.8) which we just proved, we get 𝜇(𝑥, 𝜇(𝛼(𝑦), 𝑧)) = 𝜇(𝑥, 𝛼(𝜇(𝑦, 𝑧)))

= 𝜇(𝛼(𝑥), 𝜇(𝑦, 𝑧)) = 𝜇( 𝜇(𝛼(𝑥), 𝑦), 𝑧) = 𝜇( 𝜇(𝑥, 𝛼(𝑦)), 𝑧), which is exactly (2.9).

Similarly, we show (2.10) by using Lemma 2.13, as

𝜇(𝑥, 𝜇(𝑦, 𝛼(𝑧))) = 𝜇(𝜇(𝛼(𝑥), 𝑦), 𝑧) = 𝜇(𝛼( 𝜇(𝑥, 𝑦)), 𝑧) = 𝜇( 𝜇(𝑥, 𝑦), 𝛼(𝑧)). Equation (2.11) is shown as expected by Lemma 2.13, and we obtain

𝛼( 𝜇(𝑥, 𝜇(𝑦, 𝑧))) = 𝜇(𝛼(𝑥), 𝜇(𝑦, 𝑧)) = 𝜇( 𝜇(𝑥, 𝑦), 𝛼(𝑧)) = 𝜇(𝛼( 𝜇(𝑥, 𝑦)), 𝑧) = 𝛼( 𝜇( 𝜇(𝑥, 𝑦), 𝑧)).

 We note that all equations in Proposition 2.17 are similar. Studies of what happens if one replaces hom-associativity (still assuming that we have a unit element) with one of the equations (2.8), (2.9), (2.10) and more, have been done in [7] and [8]. Since the properties in Lemma 2.13 arise from the unital condition together with hom-associativity, it may not hold if one uses (2.8), (2.9), (2.10) or (2.11) instead.

The upcoming proposition gives a condition for associativity.

Proposition 2.18. A unital hom-associative algebra (𝑉 , 𝜇, 𝛼) is associative if 𝛼 is injective. Proof. From equation (2.11), it follows that

𝛼( 𝜇(𝑥, 𝜇(𝑦, 𝑧)) − 𝜇( 𝜇(𝑥, 𝑦), 𝑧)) = 0, and since Ker(𝛼) = {0}, this yields

𝜇(𝑥, 𝜇(𝑦, 𝑧)) − 𝜇( 𝜇(𝑥, 𝑦), 𝑧) = 0,

Proposition 2.19 ([6]). If 𝛼 is surjective in a unital hom-associative algebra, then 𝛼 is injective. Proof. Assuming that 𝛼 is surjective, then there exists an element 𝑥 ∈ 𝑉 such that 𝛼(𝑥) = 1. Let 𝛼(𝑦) = 0, for some 𝑦 ∈ 𝑉 . Thus, we can write

0 = 𝜇(𝑥, 0) = 𝜇(𝑥, 𝛼(𝑦)). We get that furthermore

𝜇(𝑥, 𝛼(𝑦)) = 𝜇(𝛼(𝑥), 𝑦), from equation (2.6) in Lemma 2.13.

By the fact that 𝛼(𝑥) = 1, and the unital property, we have 𝜇(𝛼(𝑥), 𝑦) = 𝜇(1, 𝑦) = 𝑦 ⇒ 𝑦 = 0,

hence 𝛼 is injective. _

Remark2.20. In Proposition 2.19, we do not need to assume surjectivity of 𝛼, but only that 1 is in the image of 𝛼, as is shown in the proof.

Corollary 2.21 ([6]). A unital hom-associative algebra (𝑉 , 𝜇, 𝛼) is associative if 𝛼 is surjective. Proof. This follows from Proposition 2.19 and Proposition 2.18. _

Remark 2.22. The proof of Corollary 2.21 can also be done directly. If we take arbitrary elements 𝑥, 𝑦 and 𝑧, we can let 𝑥 = 𝛼(𝑤), for some 𝑤, by surjectivity, and hence get

𝜇(𝑥, 𝜇(𝑦, 𝑧)) = 𝜇(𝛼(𝑤), 𝜇(𝑦, 𝑧)) = 𝛼(𝜇(𝑤, 𝜇(𝑦, 𝑧))) = 𝛼(𝜇(𝜇(𝑤, 𝑦), 𝑧)) = 𝜇(𝛼( 𝜇(𝑤, 𝑦)), 𝑧) = 𝜇( 𝜇(𝛼(𝑤), 𝑦), 𝑧) = 𝜇( 𝜇(𝑥, 𝑦), 𝑧), by Lemma 2.13 and (2.11).

2.3

Hom-Lie algebras

This sections handles the relation between hom-Lie algebras and hom-associative algebras. We will begin with defining a hom-Lie algebra, first presented in [2].

Definition 2.23 ([2]). Let 𝑉 be a linear space, [·, ·] : 𝑉 ×𝑉 → 𝑉 a bilinear map and 𝛼 : 𝑉 → 𝑉

a linear space homomorphism. A hom-Lie algebra is the triple (𝑉 , [·, ·], 𝛼), satisfying,for all 𝑥 , 𝑦, 𝑧 ∈ 𝑉 , the following properties:

[𝑥, 𝑦] = −[𝑦, 𝑥], (skew-symmetry) (2.12) 𝑥 , 𝑦,𝑧 [𝛼(𝑥), [𝑦, 𝑧]] = 0. (Hom-Jacobi identity) (2.13) Here, 𝑥 , 𝑦,𝑧

expresses the cyclic sum over 𝑥, 𝑦, 𝑧, i.e. equation (2.13) means

𝑥 , 𝑦,𝑧

The next proposition states how hom-Lie algebras and hom-associative algebras are related to each other.

Proposition 2.24 ([1]). A hom-associative algebra (𝑉 , 𝜇, 𝛼) defines a hom-Lie algebra (𝑉 , [·, ·], 𝛼), with bracket product defined as the commutator

[𝑥, 𝑦] = 𝜇(𝑥, 𝑦) − 𝜇(𝑦, 𝑥),

for all 𝑥, 𝑦 ∈ 𝑉 .

Proof. We need to prove that this bracket product satisfies skew-symmetry (2.12) and the Hom-Jacobi identity (2.12).

For any two elements 𝑥, 𝑦 ∈ 𝑉 , we have

[𝑥, 𝑦] = 𝜇(𝑥, 𝑦) − 𝜇(𝑦, 𝑥) = − (𝜇(𝑦, 𝑥) − 𝜇(𝑥, 𝑦)) = −[𝑦, 𝑥], hence we have skew-symmetry.

We continue with the Hom-Jacobi identity (2.13). For any three elements 𝑥, 𝑦, 𝑧 ∈ 𝑉 , we have 𝑥 , 𝑦,𝑧 [𝛼(𝑥), [𝑦, 𝑧]] = [𝛼(𝑥), [𝑦, 𝑧]] + [𝛼(𝑦), [𝑧, 𝑥]] + [𝛼(𝑧), [𝑥, 𝑦]] = 𝜇(𝛼(𝑥), [𝑦, 𝑧]) − 𝜇( [𝑦, 𝑧], 𝛼(𝑥)) + 𝜇(𝛼(𝑦), [𝑧, 𝑥]) − 𝜇( [𝑧, 𝑥], 𝛼(𝑦)) + 𝜇(𝛼(𝑧), [𝑥, 𝑦]) − 𝜇( [𝑥, 𝑦], 𝛼(𝑧)) = 𝜇(𝛼(𝑥), 𝜇(𝑦, 𝑧) − 𝜇(𝑧, 𝑦)) − 𝜇( 𝜇(𝑦, 𝑧) − 𝜇(𝑧, 𝑦), 𝛼(𝑥)) + 𝜇(𝛼(𝑦), 𝜇(𝑧, 𝑥) − 𝜇(𝑥, 𝑧)) − 𝜇( 𝜇(𝑧, 𝑥) − 𝜇(𝑥, 𝑧), 𝛼(𝑦)) + 𝜇(𝛼(𝑧), 𝜇(𝑥, 𝑦) − 𝜇(𝑦, 𝑥)) − 𝜇( 𝜇(𝑥, 𝑦) − 𝜇(𝑦, 𝑥), 𝛼(𝑧)). We continue by using bilinearity of 𝜇, and so the previous expression is equal to

𝜇(𝛼(𝑥), 𝜇(𝑦, 𝑧)) − 𝜇(𝛼(𝑥), 𝜇(𝑧, 𝑦)) − 𝜇( 𝜇(𝑦, 𝑧), 𝛼(𝑥)) + 𝜇( 𝜇(𝑧, 𝑦), 𝛼(𝑥)) + 𝜇(𝛼(𝑦), 𝜇(𝑧, 𝑥)) − 𝜇(𝛼(𝑦), 𝜇(𝑥, 𝑧)) − 𝜇( 𝜇(𝑧, 𝑥), 𝛼(𝑦)) + 𝜇( 𝜇(𝑥, 𝑧), 𝛼(𝑦)) + 𝜇(𝛼(𝑧), 𝜇(𝑥, 𝑦)) − 𝜇(𝛼(𝑧), 𝜇(𝑦, 𝑥)) − 𝜇( 𝜇(𝑥, 𝑦), 𝛼(𝑧)) + 𝜇( 𝜇(𝑦, 𝑥), 𝛼(𝑧)) = 𝜇( 𝜇(𝑥, 𝑦), 𝛼(𝑧)) − 𝜇( 𝜇(𝑥, 𝑧), 𝛼(𝑦)) − 𝜇(𝛼(𝑦), 𝜇(𝑧, 𝑥)) + 𝜇(𝛼(𝑧), 𝜇(𝑦, 𝑥)) + 𝜇(𝛼(𝑦), 𝜇(𝑧, 𝑥)) − 𝜇( 𝜇(𝑦, 𝑥), 𝛼(𝑧)) − 𝜇(𝛼(𝑧), 𝜇(𝑥, 𝑦)) + 𝜇( 𝜇(𝑥, 𝑧), 𝛼(𝑦)) + 𝜇(𝛼(𝑧), 𝜇(𝑥, 𝑦)) − 𝜇(𝛼(𝑧), 𝜇(𝑦, 𝑥)) − 𝜇( 𝜇(𝑥, 𝑦), 𝛼(𝑧)) + 𝜇( 𝜇(𝑦, 𝑥), 𝛼(𝑧)) =0.

We get the first equality by applying hom-associativity property to the first, second, third, fourth, sixth and seventh term in the first sum. Then every term in the sum occurs exactly twice with opposite signs. Thus, each term cancels, and the whole expression terminates. Hence,

𝑥 , 𝑦,𝑧

[𝛼(𝑥), [𝑦, 𝑧]] = 0, and the Hom-Jacobi identity is proved. _ Next, we will prove that commuting elements in the hom-associative algebra and the induced hom-Lie algebra are in fact the same elements.

Lemma 2.25. Assume we have a hom-associative algebra (𝑉 , 𝜇, 𝛼), and the induced hom-Lie algebra (𝑉 , [·, ·], 𝛼), with bracket product defined as

[𝑥, 𝑦] = 𝜇(𝑥, 𝑦) − 𝜇(𝑦, 𝑥),

for all 𝑥, 𝑦 ∈ 𝑉 . Then the commuting elements in both algebras are the same.

Proof. Assume 𝑥, 𝑦 ∈ 𝑉 commute in the hom-associative algebra. Then by definition of the commutator, we have that [𝑥, 𝑦] = 0. So in the hom-Lie algebra, we get from the skew-symmetry property that

[𝑥, 𝑦] = 0 = −0 = −[𝑥, 𝑦] = − (−[𝑦, 𝑥]) = [𝑦, 𝑥]. Suppose now that 𝑥, 𝑦 ∈ 𝑉 commute in the hom-Lie algebra. Then

[𝑥, 𝑦] = [𝑦, 𝑥] = −[𝑥, 𝑦],

which we can rewrite as 2[𝑥, 𝑦] = 0, which means that [𝑥, 𝑦] = 0. So the commutator is equal to 0, i.e. 𝑥 and 𝑦 commute in the hom-associative algebra. _ The next proposition gives the form of 𝛼 for a specific hom-Lie algebra. We will use this fact later, when presenting some examples.

Proposition 2.26 ([1]). Let 𝑉 be three dimensional linear space over 𝐾, with (𝐻, 𝐸, 𝐹) as its basis. Suppose that a hom-Lie algebra (𝑉 , [·, ·], 𝛼) is defined such that the bracket satisfies

[𝐻, 𝐸] = 2𝐸, [𝐻, 𝐹] = −2𝐹, [𝐸, 𝐹] = 𝐻.

Then, with respect to the basis (𝐻, 𝐸, 𝐹), the homomorphism 𝛼 is defined by a matrix of the form © ­ « 𝑎 𝑑 𝑐 2𝑐 𝑏 𝑓 2𝑑 𝑒 𝑏 ª ® ¬ where 𝑎, 𝑏, 𝑐, 𝑑, 𝑒, 𝑓 ∈ 𝐾 .

Proof. Since we have a linear space of dimension three with a basis, since 𝛼 is a linear space homomorphism we can represent 𝛼 as a 3 × 3 matrix. Let

𝐴 = © ­ « 𝑎 𝑑 𝑐 𝛾 𝑏 𝑓 𝛿 𝑒 𝛽 ª ® ¬ . We get that 𝛼(𝐻) = © ­ « 𝑎 𝑑 𝑐 𝛾 𝑏 𝑓 𝛿 𝑒 𝛽 ª ® ¬ © ­ « 1 0 0 ª ® ¬ = 𝑎𝐻 + 𝛾𝐸 + 𝛿𝐹,

and similarly that

𝛼(𝐸) = 𝑑𝐻 + 𝑏𝐸 + 𝑒𝐹, 𝛼(𝐹) = 𝑐𝐻 + 𝑓 𝐸 + 𝛽𝐹.

Since we have a hom-Lie algebra, the twisted Jacobi identity holds. By applying the definition of the bracket to the twisted Jacobi identity, and using the expressions we got earlier for 𝛼(𝐻), 𝛼(𝐸) and 𝛼(𝐹), we get

0 = [𝛼(𝐻), [𝐸, 𝐹]] + [𝛼(𝐸), [𝐹, 𝐻]] + [𝛼(𝐹), [𝐻, 𝐸]]

= [𝑎𝐻 + 𝛾𝐸 + 𝛿𝐹, 𝐻] + [𝑑𝐻 + 𝑏𝐸 + 𝑒𝐹,2𝐹] + [𝑐𝐻 + 𝑓 𝐸 + 𝛽𝐹, 2𝐸].

We can simplify this expression by using bilinearity and the property that [𝑥, 𝑥] = 0, for all 𝑥 ∈ 𝑉 . Hence, we get

[𝑎𝐻 + 𝛾𝐸 + 𝛿𝐹, 𝐻] + [𝑑𝐻 + 𝑏𝐸 + 𝑒𝐹, 2𝐹] + [𝑐𝐻 + 𝑓 𝐸 + 𝛽𝐹, 2𝐸] = [𝛾𝐸 + 𝛿𝐹, 𝐻] + [𝑑𝐻 + 𝑏𝐸,2𝐹] + [𝑐𝐻 + 𝛽𝐹, 2𝐸]

= 𝛾 [𝐸, 𝐻] + 𝛿 [𝐹, 𝐻] +2𝑑 [𝐻, 𝐹] + 2𝑏 [𝐸, 𝐹] + 2𝑐[𝐻, 𝐸] + 2𝛽[𝐹, 𝐸] = −2𝛾𝐸 + 2𝛿𝐹 − 4𝑑𝐹 + 2𝑏𝐻 + 4𝑐𝐸 − 2𝛽𝐻.

Since this whole expression should be equal to zero, we have that 0 = −2𝛾𝐸 + 2𝛿𝐹 − 4𝑑𝐹 + 2𝑏𝐻 + 4𝑐𝐸 − 2𝛽𝐻

= (−2𝛾 + 4𝑐)𝐸 + (2𝛿 − 4𝑑)𝐹 + (2𝑏 − 2𝛽)𝐻, which is the same as to say that

−2𝛾 + 4𝑐 = 0, 2𝛿 − 4𝑑 = 0, 2𝑏 − 2𝛽 = 0.

Hence, we see that 𝛾 = 2𝑐, 𝛿 = 2𝑑 and 𝛽 = 𝑏, which is what we set out to prove. _

2.4

Generalized hom-associative structures

In this section we will study the so called generalized hom-associative structure, as introduced in [8], and give a condition for it to be associative.

Definition 2.27 ([8]). A 7-tuple (𝑉 , ★, 𝛼1, 𝛼₂, 𝛼₃, 𝛼₄, 𝛼₅), where 𝑉 is a set, ★ : 𝑉 × 𝑉 → 𝑉 is a binary operation, and maps 𝛼𝑖 : 𝑉 → 𝑉 , 𝑖 = 1, . . . , 5, satisfying

𝛼₅(𝛼₁(𝑥) ★ 𝛼₂(𝛼₃(𝑦) ★ 𝛼₄(𝑧))) = 𝛼₅(𝛼₂(𝛼₄(𝑥) ★ 𝛼₃(𝑦)) ★ 𝛼₁(𝑧)) , (2.14) is called a generalized hom-associative structure.

Remark 2.28. If 𝑉 is a linear space over 𝐾, 𝛼𝑖 = id for 𝑖 = 2, . . . , 5, and 𝛼1 is a linear

space homomorphism, ★ bilinear over 𝐾, then we have the ordinary hom-associative algebra (𝑉 , ★, 𝛼₁).

Remark2.29. We will say that a generalized hom-associative structure, with set 𝑉 , has a unit element, i.e. if there exist an element 𝑢 ∈ 𝑣 , such that

𝑥 ★ 𝑢 = 𝑢 ★ 𝑥 = 𝑥, for all 𝑥 ∈ 𝑉 , similarly as we did for the ordinary case.

Theorem 2.30 ([8]). Suppose a generalized hom-associative structure (𝑉 , ★, 𝛼1, 𝛼₂, 𝛼₃, 𝛼₄, 𝛼₅)

is unital, where 𝛼1, 𝛼₃, 𝛼₄are surjective maps with right-inverses 𝛽₁, 𝛽₃, 𝛽₄respectively, and 𝛼₂, 𝛼₅are injective maps with left-inverses 𝛽₂, 𝛽₅respectively. Then the ★ is associative on 𝑉 .

The proof of this theorem will appear later in this section.

Remark 2.31. We will from now on refer to a generalized unital hom-associative structure (𝑉 , ★, 𝛼₁, 𝛼₂, 𝛼₃, 𝛼₄, 𝛼₅), where 𝛼₁, 𝛼₃, 𝛼₄ are surjective maps and 𝛼₂, 𝛼₅are injective maps, as the condition of Theorem 2.30.

We will present some useful lemmas that will help us with the proof.

Definition 2.32. We say that 𝛽1 : 𝑉 → 𝑉 is the right-inverse of 𝛼 if 𝛼 ◦ 𝛽1 = id., where id

is the identity operator on 𝑉 . In the same way, we say that 𝛽2 : 𝑉 → 𝑉 is the left-inverse of

𝛼: 𝑉 → 𝑉 , if 𝛽₂◦ 𝛼 = id.

Lemma 2.33. There exist a left-inverse of the map 𝛼 : 𝑉 → 𝑉 if and only if 𝛼 is injective. Proof. Assume that 𝛼 has a left-inverse 𝛽 : 𝑉 → 𝑉 . Then for any 𝑥, 𝑦 ∈ 𝑉 , such that 𝛼(𝑥) = 𝛼(𝑦), we have that 𝛽(𝛼(𝑥)) = 𝛽(𝛼(𝑦)), hence 𝑥 = 𝑦. So 𝛼 is injective.

Assume now instead that 𝛼 is injective. Define the map 𝛽 for all 𝑥 ∈ 𝑉 as 𝛽(𝑥) = 𝑦, if there exist a 𝑦 ∈ 𝑉 such that 𝛼(𝑦) = 𝑥. If not such 𝑦 exist, then set 𝛽(𝑥) = 𝑧, for some fixed 𝑧. If we can prove that 𝛽 is well-defined, then it is a left-inverse. In the first case, if 𝛼(𝑦) = 𝑥 and 𝛼( ˜𝑦) = 𝑥, for two elements 𝑦, ˜𝑦 ∈ 𝑉 , this implies 𝑦 = ˜𝑦, since 𝛼 is injective. Hence, 𝛽(𝑥) = 𝑦 maps to a unique element.

In the other case, we have a fixed output 𝑧 ∈ 𝑉 , and thus for any 𝑥 ∈ 𝑉 that is not in the image of 𝛼, we have 𝛽(𝑥) = 𝑧 which is unique. So in both cases 𝛽 is well-defined, which proves that there exist a left-inverse of 𝛼. _

Lemma 2.34. There exist a right-inverse of the map 𝛼 : 𝑉 → 𝑉 if and only if 𝛼 is surjective.

We will skip the proof of Lemma 2.34. The point of Lemma 2.33 and Lemma 2.34, is that under the condition of Theorem 2.30, we can choose right-inverses 𝛽1, 𝛽₃, 𝛽₄ of 𝛼₁, 𝛼₃, 𝛼₄, respectively, since 𝛼1, 𝛼₃, 𝛼₄are surjective. We can also choose 𝛽₂ as the left-inverse of 𝛼₂, since 𝛼2is injective.

Remark 2.35. Note that 𝛼5is also injective, but we do not need to choose a left-inverse of it,

since under that assumption property (2.14) is simplified to

𝛼₁(𝑥) ★ 𝛼₂(𝛼₃(𝑦) ★ 𝛼₄(𝑧)) = 𝛼₂(𝛼₄(𝑥) ★ 𝛼₃(𝑦)) ★ 𝛼₁(𝑧). Using the right-inverse 𝛽3of 𝛼3, we can remove 𝛼3from property (2.14) to get

𝛼₁(𝑥) ★ 𝛼₂(𝑦 ★ 𝛼₄(𝑧)) = 𝛼₂(𝛼₄(𝑥) ★ 𝑦) ★ 𝛼₁(𝑧), (2.15) by replacing 𝑦 by 𝛽3(𝑦).

Note that in the next lemma, we use (2.15) to remove 𝛼3, but we still need the condition

that 𝛼3is surjective, since we need the right-inverse 𝛽3.

Lemma 2.36 ([8]). Under the condition of Theorem 2.30, the following holds

𝛼₁(𝑥) ★ 𝛼₂(𝑦 ★ 𝛼₄( 𝛽₁(𝑧))) = 𝛼₂(𝛼₄(𝑥) ★ 𝑦) ★ 𝑧, (2.16) 𝑥 ★ 𝛼₂(𝛼₄( 𝛽₁(1))) = 𝛼₂(𝛼₄( 𝛽₁(𝑥))), (2.17) 𝛼₄( 𝛽₁(𝛼₁(𝑥))) = 𝛼₄(𝑥), (2.18) 𝛼₂(𝛼₄( 𝛽₁(𝑥)) ★ 𝑦) ★ 𝑧 = 𝑥 ★ 𝛼₂(𝑦 ★ 𝛼₄( 𝛽₁(𝑧))) , (2.19) 𝛼₂(𝛼₄( 𝛽₁(𝑥)) ★ 𝑦) = 𝑥 ★ 𝛼₂(𝑦 ★ 𝛼₄( 𝛽₁(1))) , (2.20) 𝛼₂(𝛼₄( 𝛽₁(1)) ★ 𝑥) = 𝛼₂(𝑥 ★ 𝛼₄( 𝛽₁(1))) , (2.21) 𝛼₄( 𝛽₁(1)) ★ 𝑥 = 𝑥 ★ 𝛼₄( 𝛽₁(1)), (2.22) 𝛼₁( 𝛽₄(𝑥)) ★ 𝛼₂(𝑦) = 𝛼₂(𝑥) ★ (𝛼₁( 𝛽₄(𝑦))), (2.23) 𝛼₂(𝑥) ★ 𝛼₁( 𝛽₄(𝛼₄( 𝛽₁(1)) ★ 𝑦)) = 𝛼₂(𝑥 ★ 𝑦), (2.24) 𝛼₂(𝑥 ★ 𝑦) ★ 𝛼₁(𝑧) = 𝛼₂(𝑥) ★ 𝛼₁( 𝛽₄(𝑦 ★ 𝛼₄(𝑧))), (2.25) 𝛼₂(𝑥 ★ 𝑦) ★ 𝛼₁( 𝛽₄(𝑧)) = 𝛼₂(𝑥) ★ 𝛼₁( 𝛽₄(𝑦 ★ 𝑧)), (2.26) 𝛼₂(𝑥 ★ 𝛼₄( 𝛽₁(1))) ★ 𝑦 = 𝛼₂(𝑥 ★ (𝛼₄( 𝛽₁(𝑦)))), (2.27) (𝑥 ★ 𝛼₄( 𝛽₁(1))) ★ 𝑦 = 𝑥 ★ (𝛼4( 𝛽1(1)) ★ 𝑦), (2.28) for all 𝑥, 𝑦, 𝑧 ∈ 𝑉 .

Proof. We prove (2.16) by substituting 𝛽1(𝑧) in place 𝑧 in equation (2.15) yields the result.

In equation (2.16), let 𝛽1(𝑥) take the place of 𝑥, and also let 𝑦 = 1 and 𝑧 = 1. Then we get

that

𝛼₁( 𝛽₁(𝑥)) ★ 𝛼₂(1 ★ 𝛼₄( 𝛽₁(1))) = 𝛼₂(𝛼₄( 𝛽₁(𝑥)) ★ 1) ★ 1,

and by using the fact that 𝛼1( 𝛽1(𝑥)) = 𝑥, and the definition of the unit 1. Then this previous

equation simplifies to 2.17.

We prove (2.18) by substituting 𝑥 with 𝛼1(𝑥) equation (2.17), and thus

𝛼₂(𝛼₄( 𝛽₁(𝛼₁(𝑥)))) = 𝛼₁(𝑥) ★ 𝛼₂(𝛼₄( 𝛽₁(1))). By equation (2.16), with 𝑦 = 1 and 𝑧 = 1, we get

Since 𝛼2is injective, the proof is done.

Since 𝛼1is surjective, we use its right inverse 𝛽1to get that

𝑥 ★ 𝛼₂(𝑦 ★ 𝛼₄( 𝛽₁(𝑧))) = 𝛼₁( 𝛽₁(𝑥)) ★ 𝛼₂(𝑦 ★ 𝛼₄( 𝛽₁(𝑧))) . Using equation (2.15) on this previous expression, we obtain

𝛼₁( 𝛽₁(𝑥)) ★ 𝛼₂(𝑦 ★ 𝛼₄( 𝛽₁(𝑧))) = 𝛼₂(𝛼₄( 𝛽₁(𝑥)) ★ 𝑦) ★ 𝛼₁( 𝛽₁(𝑧)). Finally, since 𝛼1◦ 𝛽1 =id, this yields

𝛼₂(𝛼₄( 𝛽₁(𝑥)) ★ 𝑦) ★ 𝛼₁( 𝛽₁(𝑧)) = 𝛼₂(𝛼₄( 𝛽₁(𝑥)) ★ 𝑦) ★ 𝑧, which proves (2.19).

To prove equation (2.20), set 𝑧 = 1 in equation (2.19).

Equation (2.21) follows from interchanging 𝑥 and 𝑦 in equation (2.20), and then letting 𝑦=1.

Furthermore, since 𝛼2is injective, we get that equation (2.22) follows from equation (2.21).

We prove (2.23) by using the fact that 𝛼4 has a right inverse 𝛽4, together with the unital

property, and thus

𝛼₁( 𝛽₄(𝑥)) ★ 𝛼₂(𝑦) = 𝛼₁( 𝛽₄(𝑥)) ★ 𝛼₂(1 ★ 𝛼₄( 𝛽₄(𝑦))). Using first (2.15), then the right inverse of 𝛼4again, yields

𝛼₁( 𝛽₄(𝑥)) ★ 𝛼₂(1 ★ 𝛼₄( 𝛽₄(𝑦))) = 𝛼₂(𝛼₄( 𝛽₄(𝑥)) ★ 1) ★ 𝛼₁( 𝛽₄(𝑦)) = 𝛼₂(𝑥) ★ (𝛼₁( 𝛽₄(𝑦))),

which completes the proof.

We move on to prove (2.24) First apply (2.23), in which we have substituted 𝑦 by 𝛼4( 𝛽1(1)★

𝑦,to get

𝛼₂(𝑥) ★ 𝛼₁( 𝛽₄(𝛼₄( 𝛽₁(1)) ★ 𝑦)) = 𝛼₁( 𝛽₄(𝑥)) ★ 𝛼₂(𝛼₄( 𝛽₁(1)) ★ 𝑦). Using equation (2.21), we obtain

𝛼₁( 𝛽₄(𝑥)) ★ 𝛼₂(𝛼₄( 𝛽₁(1)) ★ 𝑦) = 𝛼₁( 𝛽₄(𝑥)) ★ 𝛼₂(𝑦 ★ 𝛼₄( 𝛽₁(1))). Using the identity (2.20) with 𝑥 substituted by 𝛼1( 𝛽4(𝑥)), yields

𝛼₁( 𝛽₄(𝑥)) ★ 𝛼₂(𝑦 ★ 𝛼₄( 𝛽₁(1))) = 𝛼₂(𝛼₄( 𝛽₁(𝛼₁( 𝛽₄(𝑥)))) ★ 𝑦).

Applying equation (2.18), and then using the existence of a right inverse 𝛽4of 𝛼4, gives

𝛼₂(𝛼₄( 𝛽₁(𝛼₁( 𝛽₄(𝑥)))) ★ 𝑦) = 𝛼₂(𝛼₄( 𝛽₄(𝑥)) ★ 𝑦) = 𝛼2(𝑥 ★ 𝑦),

We use that 𝛼4has a right inverse 𝛽4, and then apply (2.15), which yields

𝛼₂(𝑥 ★ 𝑦) ★ 𝛼₁(𝑧) = 𝛼₂(𝛼₄( 𝛽₄(𝑥)) ★ 𝑦) ★ 𝛼₁(𝑧) = 𝛼₁( 𝛽₄(𝑥)) ★ 𝛼₂(𝑦 ★ 𝛼₄(𝑧)). Then we use (2.23) on the previous expression, to get that

𝛼₁( 𝛽₄(𝑥)) ★ 𝛼₂(𝑦 ★ 𝛼₄(𝑧)) = 𝛼₂(𝑥) ★ 𝛼₁( 𝛽₄(𝑦 ★ 𝛼₄(𝑧))), which proves (2.25).

(2.28) is proved by letting 𝑧 in (2.25) be replaced by 𝛽4(𝑧). Then it follows directly from

the fact that 𝛼4has a right inverse 𝛽4. First we have

𝛼₂(𝑥 ★ 𝛼₄( 𝛽₁(1))) ★ 𝑦 = 𝛼₂(𝛼₄( 𝛽₁(1)) ★ 𝑥) ★ 𝑦,

by equation (2.21). By equation (2.16) and by the fact that 𝛼1has a right inverse 𝛽1, this gives

𝛼₂(𝛼₄( 𝛽₁(1)) ★ 𝑥) ★ 𝑦 = 𝛼₁( 𝛽₁(1)) ★ 𝛼₂(𝑥 ★ 𝛼₄( 𝛽₁(𝑦))) =1 ★ 𝛼2(𝑥 ★ 𝛼4( 𝛽1(𝑦)))

= 𝛼₂(𝑥 ★ 𝛼₄( 𝛽₁(𝑦))). Thus, the proof of (2.27) is complete.

Equation (2.24), with 𝑥 replaced by 𝑥 ★ 𝛼4( 𝛽1(1)), yields

𝛼₂( (𝑥 ★ 𝛼₄( 𝛽₁(1))) ★ 𝑦) = 𝛼₂(𝑥 ★ 𝛼₄( 𝛽₁(1))) ★ 𝛼₁( 𝛽₄(𝛼₄( 𝛽₁(1)) ★ 𝑦)). Substituting 𝑦 with 𝛼1( 𝛽4(𝛼4( 𝛽1(1)) ★ 𝑦)) in (2.27), gives us

𝛼₂(𝑥 ★ 𝛼₄( 𝛽₁(1))) ★ 𝛼₁( 𝛽₄(𝛼₄( 𝛽₁(1)) ★ 𝑦)) = 𝛼₂(𝑥 ★ 𝛼₄( 𝛽₁(𝛼₁( 𝛽₄(𝛼₄( 𝛽₁(1)) ★ 𝑦))))). Using (2.18), and that 𝛽4is a right inverse of 𝛼4, we obtain

𝛼₂(𝑥 ★ 𝛼₄( 𝛽₁(𝛼₁( 𝛽₄(𝛼₄( 𝛽₁(1)) ★ 𝑦))))) = 𝛼₂(𝑥 ★ 𝛼₄( 𝛽₄(𝛼₄( 𝛽₁(1)) ★ 𝑦))) = 𝛼₂(𝑥 ★ (𝛼₄( 𝛽₁(1)) ★ 𝑦)),

and since 𝛼2is assumed to be injective, this implies exactly the equation (2.28), as desired.



Proof of Theorem 2.30 [8]. We need to show that the associative law (𝑥 ★ 𝑦) ★ 𝑧 = 𝑥 ★ (𝑦 ★ 𝑧). Since 𝛼2 is injective, we can use the same method as in the proof of equation 2.28. By

substituting 𝑥 in (2.24) by 𝑥 ★ 𝑦, we obtain

𝛼₂( (𝑥 ★ 𝑦) ★ 𝑧) = 𝛼₂(𝑥 ★ 𝑦) ★ 𝛼₁( 𝛽₄(𝛼₄( 𝛽₁(1)) ★ 𝑧)). Then, using (2.25) on this previous expression yields

Furthermore, if we use that 𝛽4is a right inverse of 𝛼4, followed by property (2.28), then

(2.22), and then (2.28) again, we get

𝛼₂(𝑥) ★ 𝛼₁( 𝛽₄(𝑦 ★ 𝛼₄( 𝛽₄(𝛼₄( 𝛽₁(1)) ★ 𝑧)))) = 𝛼₂(𝑥) ★ 𝛼₁( 𝛽₄(𝑦 ★ (𝛼₄( 𝛽₁(1)) ★ 𝑧))) = 𝛼₂(𝑥) ★ 𝛼₁( 𝛽₄( (𝑦 ★ 𝛼₄( 𝛽₁(1))) ★ 𝑧)) = 𝛼₂(𝑥) ★ 𝛼₁( 𝛽₄( (𝛼₄( 𝛽₁(1)) ★ 𝑦) ★ 𝑧)) = 𝛼₂(𝑥) ★ 𝛼₁( 𝛽₄(𝛼₄( 𝛽₁(1))) ★ (𝑦 ★ 𝑧)). Substituting 𝑦 in (2.24) by 𝑦 ★ 𝑧 gives us 𝛼₂(𝑥) ★ 𝛼₁( 𝛽₄(𝛼₄( 𝛽₁(1))) ★ (𝑦 ★ 𝑧)) = 𝛼₂(𝑥 ★ (𝑦 ★ 𝑧)),

and since 𝛼2is injective, these calculations imply (𝑥 ★ 𝑦) ★ 𝑧 = 𝑥 ★ (𝑦 ★ 𝑧), which is what we

wanted to prove.

 Note that from (2.18), that under the conditions of Theorem 2.30, Ker(𝛼1) ⊂ Ker(𝛼4). But

we do not need all the conditions from Theorem 2.30 to prove this fact, as we will see in the next proposition.

Proposition 2.37. Let 𝛼5and 𝛼2be injective, and let 𝛼3be surjective. Also suppose we have a unit 𝑢. ThenKer(𝛼1) ⊂ Ker(𝛼4).

Proof. By surjectivity, we can let 𝛼1(𝑥) = 0, 𝛼1(𝑧) = 𝑢 = 𝛼3(𝑦) for some 𝑦 and 𝑧. Then

0 = 𝛼1(𝑥) ★ 𝑢 = 𝛼1(𝑥) ★ 𝛼2(𝛼3(𝑦) ★ 𝛼4(𝑧)),

and by generalized hom-associativity and some simplifications, we get

𝛼₁(𝑥) ★ 𝛼₂(𝛼₃(𝑦) ★ 𝛼₄(𝑧)) = 𝛼₂(𝛼₄(𝑥) ★ 𝛼₃(𝑦)) ★ 𝛼₁(𝑧) = 𝛼₂(𝛼₄(𝑥) ★ 𝑢) ★ 𝑢 = 𝛼₂(𝛼₄(𝑥)), i.e. 𝛼4(𝑥) = 0 as well, since 𝛼2is injective. This shows that if 𝑥 ∈ Ker(𝛼1), then 𝑥 ∈ Ker(𝛼4),

which is what we wanted to prove. _

Corollary 2.38. Let 𝛼5and 𝛼2be injective, and let 𝛼1and 𝛼3be surjective. Also suppose we have a unit 𝑢, and that 𝛼4is injective. Then 𝛼1is injective.

Proof. Since 𝛼4is injective, we have that Ker(𝛼4) = 0. Proposition 2.37 then forces Ker(𝛼1) =

Chapter 3

Commuting elements in hom-associative

algebras

We will now begin looking at some commuting elements in hom-associative algebras.

Lemma 3.1. Let (𝑉 , 𝜇, 𝛼) be a hom-associative algebra, and 𝑥, 𝑦, 𝑧 ∈ 𝑉 . If [𝑥, 𝑦] = [𝑥, 𝑧] = 0, then [𝑥, 𝑦 + 𝑧] = 0.

Remark3.2. This previous lemma does not actually depend on the hom-associativity, but only from the definition of bilinearity.

Proof. This follows directly from the definition bilinearity of 𝜇, and the assumption that 𝜇(𝑥, 𝑦) = 𝜇(𝑦, 𝑥) and 𝜇(𝑥, 𝑧) = 𝜇(𝑧, 𝑥) as

𝜇(𝑥, 𝑦 + 𝑧) = 𝜇(𝑥, 𝑦) + 𝜇(𝑥, 𝑧) = 𝜇(𝑦, 𝑥) + 𝜇(𝑧, 𝑥) = 𝜇(𝑦 + 𝑧, 𝑥),

which completes the proof. _

Lemma 3.3. Let (𝑉 , 𝜇, 𝛼) be a hom-associative algebra and 𝑥, 𝑦 ∈ 𝑉 . If [𝑥, 𝑦] = 0 and

[𝑦, 𝑧] = 0, then [𝛼(𝑦), 𝜇(𝑥, 𝑧)] = 0.

Proof. This follows from the hom-associativity, as

𝜇(𝛼(𝑦), 𝜇(𝑥, 𝑧)) = 𝜇(𝜇(𝑦, 𝑥), 𝛼(𝑧)) = 𝜇(𝜇(𝑥, 𝑦), 𝛼(𝑧)) = 𝜇(𝛼(𝑥), 𝜇(𝑦, 𝑧)) = 𝜇(𝛼(𝑥), 𝜇(𝑧, 𝑦)) = 𝜇( 𝜇(𝑥, 𝑧), 𝛼(𝑦)),

(3.1) which is what we set out to prove. _

We will let 𝛼𝑛

denote the composition of 𝛼 𝑛 times. We define 𝛼0(𝑥) = 𝑥, for all 𝑥 ∈ 𝑉 .

Proposition 3.4. Let (𝑉 , 𝜇, 𝛼) be a hom-associative algebra and 𝑥, 𝑦, 𝑧 ∈ 𝑉 . If [𝑥, 𝑦] = 0,

[𝑥, 𝑧] = 0 and 𝑛 ∈ N, then [𝛼𝑛 (𝑥), ˜𝜇𝑛(𝑦, 𝑧)] = 0, where ˜ 𝜇1(𝑦, 𝑧) = 𝜇(𝑦, 𝑧), ˜ 𝜇2(𝑦, 𝑧) = 𝜇(𝜇(𝑦, 𝑧), 𝜇(𝑦, 𝑧)), ˜ 𝜇𝑛(𝑦, 𝑧) = 𝜇( ˜𝜇𝑛−1(𝑦, 𝑧), ˜𝜇𝑛−1(𝑦, 𝑧)), 𝑛 ≥ 2.

Proof. The case 𝑛 = 1 follows from Lemma 3.3. Assume that it is true for 𝑛 = 𝑘 − 1 ≥ 1. Then 𝜇(𝛼𝑘(𝑥), ˜𝜇𝑘(𝑦, 𝑧)) = 𝜇(𝛼𝑘(𝑥), 𝜇( ˜𝜇𝑘−1(𝑦, 𝑧), ˜𝜇𝑘−1(𝑦, 𝑧))) = 𝜇( 𝜇(𝛼𝑘−1 (𝑥), ˜𝜇𝑘−1(𝑦, 𝑧)), 𝛼( ˜𝜇𝑘−1(𝑦, 𝑧))) = 𝜇( 𝜇(𝜇˜𝑘−1(𝑦, 𝑧), 𝛼𝑘−1(𝑥)), 𝛼( ˜𝜇𝑘−1(𝑦, 𝑧))) = 𝜇(𝛼(𝜇˜𝑘−1(𝑦, 𝑧)), 𝜇(𝛼𝑘−1(𝑥), ˜𝜇𝑘−1(𝑦, 𝑧)) = 𝜇(𝛼(𝜇˜𝑘−1(𝑦, 𝑧)), 𝜇( ˜𝜇𝑘−1(𝑦, 𝑧), 𝛼𝑘−1(𝑥))) = 𝜇( 𝜇(𝜇˜𝑘−1(𝑦, 𝑧), ˜𝜇𝑘−1(𝑦, 𝑧)), 𝛼𝑘(𝑥)) = 𝜇(𝜇˜𝑘(𝑦, 𝑧), 𝛼𝑘(𝑥)). 

Corollary 3.5. Let 𝑛 ∈ N. If 𝑛 ≥ 1 and 𝑥 ∈ 𝑉, then 𝛼𝑛

(𝑥) and ˜𝜇𝑛(𝑥, 𝑥) commute.

3.1

Commuting elements in multiplicative and weakly unital

algebras

In this section, we will study commuting elements in the multiplicative and weakly unital hom-associative algebras.

We remember that Ker(𝛼) = {0} is equivalent to that 𝛼 is injective, and for the unital hom-associative algebras, this also implies associativity. However, for multiplicative and weakly unital hom-associative algebras, we do not have this condition.

Proposition 3.6. Let 𝑥 and 𝑦 be two elements in a multiplicative hom-associative algebra

(𝑉 , 𝜇, 𝛼), and Ker(𝛼) = {0}. Then 𝑥 and 𝑦 commute if and only if 𝛼(𝑥) and 𝛼(𝑦) commute.

Proof. Assume 𝑥 and 𝑦 commute. By definition

𝜇(𝑥, 𝑦) − 𝜇(𝑦, 𝑥) = 0, which is equivalent to that

𝛼( 𝜇(𝑥, 𝑦) − 𝜇(𝑦, 𝑥)) = 0,

since Ker(𝛼) = {0}. By linearity and the multiplicative property we get 𝛼( 𝜇(𝑥, 𝑦) − 𝜇(𝑦, 𝑥)) = 𝛼(𝜇(𝑥, 𝑦)) − 𝛼(𝜇(𝑦, 𝑥))

= 𝜇(𝛼(𝑥), 𝛼(𝑦)) − 𝜇(𝛼(𝑦), 𝛼(𝑥)) =0, (3.2) i.e. 𝛼(𝑥) and 𝛼(𝑦) commute. If we instead assume that 𝛼(𝑥) and 𝛼(𝑦) commute, then equation (3.2) and Ker(𝛼) = {0} proves that 𝑥 and 𝑦 commute. _

Remark 3.7. If Ker(𝛼) ≠ {0}, then in Proposition 3.6 we still have that if 𝑥 and 𝑦 commute then 𝛼(𝑥) and 𝛼(𝑦) commute, since 𝛼(0) = 0. However, the other direction is not necessarily true. If 𝛼(𝑥) and 𝛼(𝑦) commute, then

This does not imply that 𝜇(𝑥, 𝑦) − 𝜇(𝑦, 𝑥) = 0, if Ker(𝛼) ≠ {0}. Still, 𝜇(𝑥, 𝑦) − 𝜇(𝑦, 𝑥) ∈ Ker(𝛼). This means that if 𝛼 is not injective, the right implication is true but the left is not.

Corollary 3.8. Let Ker(𝛼) = {0} in a hom-associative algebra with weak unit 𝑢, such that it satisfies 𝛼(𝑢) = 𝑢 as in(2.2). Then 𝑥 and 𝑦 commute if and only if 𝛼(𝑥) and 𝛼(𝑦) commute.

Proof. We have a weakly unital hom-associative algebra that satisfies (2.2), and hence by Proposition 2.9 we have multiplicativity. Then the conclusion follows immediately from

Proposition 3.6. _

Corollary 3.9. Let 𝑥 and 𝑦 be two elements in a multiplicative hom-associative algebra

(𝑉 , 𝜇, 𝛼), Ker(𝛼) = {0} and 𝑛 ∈ N. Then 𝑥 and 𝑦 commute if and only if 𝛼𝑛

(𝑥) and 𝛼𝑛

(𝑦)

commute.

Corollary 3.10. Let 𝑥 and 𝑦 be two elements in a multiplicative hom-associative algebra

(𝑉 , 𝜇, 𝛼). If 𝛼(𝑥) and 𝑦 commute, then 𝛼𝑛+1

(𝑥) commutes with 𝛼𝑛

(𝑦), for any 𝑛 ∈ N.

Proof. Let 𝑥 and 𝑦 be two elements in our algebra such that [𝛼(𝑥), 𝑦] = 0. Using multiplic-ativity, together with [𝛼(𝑥), 𝑦] = 0, we obtain

𝜇(𝛼𝑛(𝑦), 𝛼𝑛+1(𝑥)) = 𝛼𝑛(𝑦, 𝛼(𝑥)) = 𝛼𝑛 (𝛼(𝑥), 𝑦) = 𝜇(𝛼𝑛+1 (𝑥), 𝛼𝑛 (𝑦)),

for all 𝑛 ∈ N, as desired. 

Remark 3.11. We note that most results in this section, e.g. Corollary 3.10, are trivial when 𝛼=Id, the identity map.

3.2

Commuting elements in unital algebras

In this section, we will study which elements commute in a unital hom-associative algebra.

Proposition 3.12. If 𝛼(𝑦) and 𝑥 commute in unital hom-associative algebra, then 𝛼(𝑥) and 𝑦 commute as well.

Proof. Suppose that we have two elements 𝑥 and 𝑦, such that [𝛼(𝑦), 𝑥] = 0. From equation 2.4, we get

𝜇(𝛼(𝑦), 𝑥) = 𝜇(𝑦, 𝛼(𝑥)), and

𝜇(𝑥, 𝛼(𝑦)) = 𝜇(𝛼(𝑥), 𝑦).

But since [𝛼(𝑦), 𝑥] = 0, these two previous equations are equal, which means [𝛼(𝑥), 𝑦] = 0,

as desired. _

Lemma 3.13. Let 𝑥 and 𝑦 be two elements in a unital hom-associative algebra, such that

[𝑥, 𝑦] = 0. Then 𝑥 and 𝛼𝑛

(𝑦) commute for any 𝑛 ∈ N.

Proof. We prove this using induction on 𝑛. The case 𝑛 = 0 is trivial because [𝑥, 𝑦] = 0. When 𝑛=1 it follows from first using equation (2.6) and that 𝑥 and 𝑦 commute, since we get

𝜇(𝛼(𝑦), 𝑥) = 𝛼(𝜇(𝑦, 𝑥)) = 𝛼(𝜇(𝑥, 𝑦)). Then we use equation (2.6) and (2.4), to obtain

𝛼( 𝜇(𝑥, 𝑦)) = 𝜇(𝛼(𝑥), 𝑦) = 𝜇(𝑥, 𝛼(𝑦)), which shows that 𝛼(𝑦) and 𝑥 commute.

Assume that it is true for 𝑛 = 𝑘 − 1 ≥ 1. Then by equation (2.6), we have that 𝜇(𝛼𝑘(𝑦), 𝑥) = 𝛼(𝜇(𝛼𝑘−1(𝑦), 𝑥),

and the induction hypothesis gives

𝛼( 𝜇(𝛼𝑘−1(𝑦), 𝑥)) = 𝛼(𝜇(𝑥, 𝛼𝑘−1(𝑦))). Applying (2.6), and then (2.4), gives

𝛼( 𝜇(𝑥, 𝛼𝑘−1(𝑦))) = 𝜇(𝛼(𝑥), 𝛼𝑘−1(𝑦)) = 𝜇(𝑥, 𝛼𝑘(𝑦)).

The results follows. _

Proposition 3.14. Let 𝛼 be injective and let 𝑥 and 𝑦 be two elements in a unital hom-associative algebra. Then [𝑥, 𝑦] =0 if and only if [𝑥, 𝛼𝑛

(𝑦)] = 0, for any 𝑛 ∈ N.

Proof. If [𝑥, 𝑦] = 0, then by Lemma 3.13, [𝑥, 𝛼𝑛

(𝑦)] = 0 for any 𝑛 ∈ N. Now let [𝑥, 𝛼𝑛

(𝑦)] = 0 for some 𝑛 ∈ N. By iteration of equations (2.6) and (2.4), we obtain that

𝛼𝑛( 𝜇(𝑥, 𝑦)) = 𝜇(𝑥, 𝛼𝑛(𝑦)). Then we use that 𝑥 and 𝛼𝑛

(𝑦) commute, and iterate equations (2.6) and (2.4), which yields 𝜇(𝑥, 𝛼𝑛(𝑦)) = 𝜇(𝛼𝑛(𝑦), 𝑥)

= 𝛼𝑛

( 𝜇(𝑦, 𝑥)). Finally, we get that

𝛼𝑛( 𝜇(𝑥, 𝑦) − 𝜇(𝑦, 𝑥)) = 0,

and since 𝛼 is injective, this means that 𝜇(𝑥, 𝑦) − 𝜇(𝑦, 𝑥) = 0 ⇒ [𝑥, 𝑦] = 0, as sought.



Proposition 3.15. Let 𝑥 and 𝑦 be two elements in a unital hom-associative algebra (𝑉 , 𝜇, 𝛼). If 𝑥 and 𝑦 commute, then 𝛼𝑛

(𝑥) and 𝛼𝑚

Proof. Assume that 𝑥 and 𝑦 are two elements in our algebra such that [𝑥, 𝑦] = 0. Fix any 𝑛, 𝑚 ∈ N. Then we iterate (2.4) 𝑛 times, which yields

𝜇(𝛼𝑛(𝑥), 𝛼𝑚(𝑦)) = 𝜇(𝑥, 𝛼𝑚+𝑛(𝑦)).

By the previous Lemma, 𝑥 and 𝛼𝑚+𝑛(𝑦) commute, and using (2.4) 𝑛 times again we get 𝜇(𝛼𝑚+𝑛(𝑦), 𝑥) = 𝜇(𝛼𝑚(𝑦), 𝛼𝑛(𝑥)).

Hence, [𝛼𝑛_{(𝑥), 𝛼}𝑚

(𝑦)] = 0, as desired. _

Proposition 3.16. Let 𝑥, 𝑦 and 𝑧 be three elements in a unital hom-associative algebra (𝑉 , 𝜇, 𝛼). If [𝑥, 𝑦] =0 and [𝑥, 𝑧] = 0, then [𝑥, 𝜇(𝛼𝑘_{(𝑦), 𝛼}𝑛

(𝑧))] = 0, where 𝑛, 𝑘 ∈ N and 𝑛 + 𝑘 ≥ 1.

Proof. If 𝑘 = 0, then we can use the fact that 𝑛 ≥ 1 and 𝜇(𝑦, 𝛼𝑛

(𝑧)) = 𝜇(𝛼𝑛

(𝑦), 𝑧) by Lemma 3.13, enabling us to assume that 𝑘 ≥ 1, without loss of generality. We get from equations (2.6) and (2.4) that

𝜇(𝑥, 𝜇(𝛼𝑘(𝑦), 𝛼𝑛(𝑧))) = 𝜇(𝑥, 𝛼(𝜇(𝛼𝑘−1(𝑦), 𝛼𝑛(𝑧)))) = 𝜇(𝛼(𝑥), 𝜇(𝛼𝑘−1

(𝑦), 𝛼𝑛

(𝑧)). By hom-associativity and using Lemma 3.13, we obtain

𝜇(𝛼(𝑥), 𝜇(𝛼𝑘−1(𝑦), 𝛼𝑛(𝑧))) = 𝜇(𝜇(𝑥, 𝛼𝑘−1(𝑦)), 𝛼𝑛+1(𝑧)) = 𝜇( 𝜇(𝛼𝑘−1 (𝑦), 𝑥), 𝛼𝑛+1 (𝑧)) = 𝜇(𝛼𝑘 (𝑦), 𝜇(𝑥, 𝛼𝑛 (𝑧))) = 𝜇(𝛼𝑘 (𝑦), 𝜇(𝛼𝑛 (𝑧), 𝑥)) = 𝜇( 𝜇(𝛼𝑘−1 (𝑦), 𝛼𝑛 (𝑧)), 𝛼(𝑥)). Using equations (2.4) and (2.6) gives us

𝜇( 𝜇(𝛼𝑘−1(𝑦), 𝛼𝑛(𝑧)), 𝛼(𝑥)) = 𝜇(𝛼(𝜇(𝛼𝑘−1(𝑦), 𝛼𝑛(𝑧))), 𝑥) = 𝜇( 𝜇(𝛼𝑘

(𝑦), 𝛼𝑛

(𝑧)), 𝑥),

as desired. _

Example 3.17. Let 𝑥, 𝑦 and 𝑧 be three elements in a unital hom-associative algebra (𝑉 , 𝜇, 𝛼)

such that [𝑥, 𝑦] = 0 and [𝑥, 𝑧] = 0. By Proposition 3.16, we get [𝑥, 𝜇(𝛼( 𝜇(𝛼(𝑥), 𝑧)), 𝜇(𝛼(𝑧), 𝑦))] = 0, and

[𝑥, 𝜇( 𝜇(𝛼3(𝑥), 𝑧), 𝜇(𝑧, 𝑦))] = 0, by (2.6).

When 𝛼 is injective, as we noted earlier, our unital hom-associative algebra is associative. This in turn does not imply that we have commutative multiplication.

Lemma 3.18. Let 𝛼 be injective, and let 𝑥, 𝑦 and 𝑧 be three elements in a unital hom-associative algebra such that [𝑥, 𝑦] =0 and [𝑦, 𝑧] = 0. Then [𝑥, 𝜇(𝑦, 𝑧)] = 0.

Proof. This follows directly from the associative property. _

Corollary 3.19. Let 𝛼 be injective, and let 𝑥 and 𝑦 be elements in a unital hom-associative algebra such that [𝑥, 𝑦] =0. Then [𝑥, 𝜇(𝑦, 𝑦)] = 0.

Remark 3.20. From the previous corollary, if [𝑥, 𝑦] = 0, then [𝑥, 𝜇(𝜇(𝑦, 𝑦), 𝜇(𝑦, 𝑦)] = 0. From the previous lemma, [𝑥, 𝜇(𝜇(𝑥, 𝑦), (𝜇(𝑦, 𝑥))] = 0, and so on. If 𝑥 and 𝑦 commute and 𝑃(𝑥, 𝑦) and 𝑄 (𝑥, 𝑦) are two elements in the 𝐾-subalgebra generated by 𝑥 and 𝑦, then [𝑃(𝑥, 𝑦), 𝑄 (𝑥, 𝑦)] = 0. We can use associativity to prove relations like:

𝜇(𝑥, 𝜇(𝑥, 𝜇(𝑥, 𝑥))) = 𝜇(𝜇(𝑥, 𝑥), 𝜇(𝑥, 𝑥)),

so we could say that this element is “equal” to 𝑥4. Thus, it does not create any confusion of what element we actually mean, because every product of 𝑛 times multiplied 𝑥, will be equal. Hence, we can always write such an element as 𝑥𝑛

. In non-associative algebras, this is generally not the case.

3.3

Examples

We will now try to classify a 3-dimensional hom-associative algebra, in terms of associativity, multiplicativity and unitality.

Remember that in the introduction, we assumed that the field 𝐾 is of characteristic 0. It is especially important in this section, since some results might not hold if 𝐾 is of characteristic 2.

Later on in this section, we present a example of a non-unital 3-dimensional hom-associative algebra. Assume we have a three dimensional linear space 𝑉 over 𝐾 with basis {𝑥1, 𝑥₂, 𝑥₃}. Define the bilinear map 𝜇, such that it satisfies the following identities:

𝜇(𝑥₁, 𝑥₂) = 2𝑥₂+ 𝜇(𝑥₂, 𝑥₁), (3.3a) 𝜇(𝑥₁, 𝑥₃) = −2𝑥₃+ 𝜇(𝑥₃, 𝑥₁), (3.3b) 𝜇(𝑥₂, 𝑥₃) = 𝑥₁+ 𝜇(𝑥₃, 𝑥₂). (3.3c) Furthermore, assume that (𝑉 , 𝜇, 𝛼) is a hom-associative algebra. Then we get the induced hom-Lie algebra with brackets defined as

[𝑥₁, 𝑥₂] = 2𝑥₂, [𝑥₁, 𝑥₃] = −2𝑥₃, [𝑥₂, 𝑥₃] = 𝑥₁,

which follows from Proposition 2.24. Then from Proposition 2.26, we have that for any such hom-Lie algebra, 𝛼 is given by a matrix on the form

© ­ « 𝑎 𝑑 𝑐 2𝑐 𝑏 𝑓 2𝑑 𝑒 𝑏 ª ® ¬

where 𝑎, 𝑏, 𝑐, 𝑑, 𝑒, 𝑓 ∈ 𝐾 .

Hence, we have a hom-associative algebra with linear space 𝑉 , 𝜇 defined by equations (3.3), and 𝛼 as previous stated. Equations (3.3) yield that this class of hom-associative algebra is not commutative, since no pair of the basis elements commute under 𝜇, however some elements may commute.

This hom-associative algebra is also non-unital. Assume on the contrary, that it is unital. Then there should exist some element 𝑢 ∈ 𝑉 such that 𝜇(𝑢, 𝑥) = 𝜇(𝑥, 𝑢) = 𝑥, for all 𝑥 ∈ 𝑉 . Since {𝑥1, 𝑥₂, 𝑥₃} constitutes a basis for this linear space, we can write 𝑢 as a linear combination of these elements. Hence, let 𝑢 = 𝑟1𝑥₁+ 𝑟₂𝑥₂+ 𝑟₃𝑥₃,where 𝑟₁, 𝑟₂, 𝑟₃∈ 𝐾 . Then

𝜇(𝑥₁, 𝑢) = 𝜇(𝑥₁, 𝑟₁𝑥₁+ 𝑟₂𝑥₂+ 𝑟₃𝑥₃)

= 𝑟₁𝜇(𝑥₁, 𝑥₁) + 𝑟₂𝜇(𝑥₁, 𝑥₂) + 𝑟₃𝜇(𝑥₁, 𝑥₃), 𝜇(𝑢, 𝑥₁) = 𝜇(𝑟₁𝑥₁+ 𝑟₂𝑥₂+ 𝑟₃𝑥₃, 𝑥₁)

= 𝑟1𝜇(𝑥₁, 𝑥₁) + 𝑟₂𝜇(𝑥₂, 𝑥₁) + 𝑟₃𝜇(𝑥₃, 𝑥₁).

If 𝑢 is a unit, then these two expressions should be equal. We note that the first term cancels out, and we are left with

𝑟₂𝜇(𝑥₁, 𝑥₂) + 𝑟₃𝜇(𝑥₁, 𝑥₃) = 𝑟₂𝜇(𝑥₂, 𝑥₁) + 𝑟₃𝜇(𝑥₃, 𝑥₁). This we can rewrite, by subtracting the right hand side, as

0 = 𝑟2𝜇(𝑥₁, 𝑥₂) − 𝑟₂𝜇(𝑥₂, 𝑥₁) + 𝑟₃𝜇(𝑥₁, 𝑥₃) − 𝑟₃𝜇(𝑥₃, 𝑥₁) = 𝑟₂( 𝜇(𝑥₁, 𝑥₂) − 𝜇(𝑥₂, 𝑥₁)) + 𝑟₃( 𝜇(𝑥₁, 𝑥₃) − 𝜇(𝑥₃, 𝑥₁)) = 𝑟₂(2𝑥2) + 𝑟3(−2𝑥3) ,

where in the last step we used the the definition of 𝜇 as in equations (3.3). This implies 𝑟₂𝑥₂ = 𝑟₃𝑥₃, so 𝑟₂ = 𝑟₃ = 0, since {𝑥₁, 𝑥₂, 𝑥₃} is a basis, and hence linearly independent. Furthermore,

𝜇(𝑥₂, 𝑢) = 𝜇(𝑥₂, 𝑟₁𝑥₁+ 𝑟₂𝑥₂+ 𝑟₃𝑥₃)

= 𝑟1𝜇(𝑥₂, 𝑥₁) + 𝑟₂𝜇(𝑥₂, 𝑥₂) + 𝑟₃𝜇(𝑥₂, 𝑥₃), 𝜇(𝑢, 𝑥₂) = 𝜇(𝑟₁𝑥₁+ 𝑟₂𝑥₂+ 𝑟₃𝑥₃, 𝑥₂)

= 𝑟₁𝜇(𝑥₁, 𝑥₂) + 𝑟₂𝜇(𝑥₂, 𝑥₂) + 𝑟₃𝜇(𝑥₃, 𝑥₂). These two equations should be equal, hence

𝑟₁𝜇(𝑥₂, 𝑥₁) − 𝑟₁𝜇(𝑥₁, 𝑥₂) + 𝑟₃𝜇(𝑥₂, 𝑥₃) − 𝑟₃𝜇(𝑥₃, 𝑥₂) = 0.

This simplifies to −2𝑟1𝑥₂ + 𝑟₃𝑥₁ = 0, and since 𝑟₃ = 0, this implies 𝑟₂ = 0. We reach a contradiction, and thus there is no unit element in 𝑉 .

We are now going to study examples of some different choices of the matrix for 𝛼, and examine (hom-)associativity and other properties.

Especially, when we try to disprove associativity in some of these cases, we assume that we actually have associativity in a particular instance, and from that try to reach a contradiction.

Example 3.21. In this algebra, if 𝑎 = 𝑏 ≠ 0, and 𝑐, 𝑑, 𝑒, 𝑓 = 0, then it is associative, since the

matrix is given by the identity matrix multiplied with a constant in 𝐾 . Hence 𝛼(𝑥) = 𝑎𝑥, for every 𝑥 ∈ 𝑉 . Thus the hom-associativity states, that for any 𝑥, 𝑦, 𝑧 ∈ 𝑉 ,

𝜇(𝑎𝑥, 𝜇(𝑦, 𝑧)) = 𝜇(𝜇(𝑥, 𝑦), 𝑎𝑧),

and from bilinearity of 𝜇, we can factor out 𝑎 ≠ 0, and multiply with the inverse of 𝑎. Then the algebra is associative. This is true for a general linear space, that if 𝛼 = 𝑘 · id for some 𝑘 ∈ 𝐾, then associativity follows, as showed in Proposition 2.5.

Let us investigate if it is multiplicative in this case. We have 𝛼( 𝜇(𝑥₁, 𝑥₂)) = 𝛼(2𝑥₂+ 𝜇(𝑥₂, 𝑥₁)) =2𝛼(𝑥2) + 𝛼( 𝜇(𝑥2, 𝑥₁)) =2𝑎𝑥2+ 𝛼( 𝜇(𝑥2, 𝑥₁)) and 𝜇(𝛼(𝑥₁), 𝛼(𝑥₂)) = 𝜇(𝑎𝑥₁, 𝑎𝑥₂) = 𝑎2𝜇(𝑥₁, 𝑥₂) =2𝑎2𝑥₂+ 𝑎2𝜇(𝑥₂, 𝑥₁). If it was multiplicative for the other cases, then we should have

𝛼( 𝜇(𝑥₂, 𝑥₁)) = 𝜇(𝛼(𝑥₂), 𝛼(𝑥₁)) = 𝜇(𝑎𝑥₂, 𝑎𝑥₁) = 𝑎2𝜇(𝑥₂, 𝑥₁).

However, from the previous calculations, this would imply that 2𝑎𝑥2 = 2𝑎2𝑥₂, i.e. it is only true when 𝑎 = 0 or 𝑎 = 1. If 𝑎 = 0, then also 𝑏 = 0, which means that 𝛼 is the zero map. If 𝑎 =1, then 𝑏 = 1, and 𝛼 is the identity operator. In all other cases it is not multiplicative.

Example 3.22. Assume that we have the hom-associative algebra defined in the beginning of

this section. If 𝑎, 𝑏 ≠ 0, 𝑎 ≠ 𝑏 and 𝑐, 𝑑, 𝑒, 𝑓 = 0, then it is not associative. Since 𝛼(𝑥1) = 𝑎𝑥1,

𝛼(𝑥₂) = 𝑏𝑥₂and 𝛼(𝑥₃) = 𝑏𝑥₃, we get from the hom-associativity that 𝜇(𝑎𝑥₁, 𝜇(𝑥₂, 𝑥₃)) = 𝜇(𝜇(𝑥₁, 𝑥₂), 𝑏𝑥₃). Since 𝑎 ≠ 𝑏, and both nonzero, we have

𝜇(𝑥₁, 𝜇(𝑥₂, 𝑥₃)) = 𝑎−1𝑏 𝜇( 𝜇(𝑥₁, 𝑥₂), 𝑥₃),

where 𝑎−1denotes the inverse of 𝑎. Since 𝑎−1𝑏 ≠1, the associativity does not hold. Is it multiplicative? Doing similar calculations as we did for the previous case, we get

𝛼( 𝜇(𝑥₁, 𝑥₂)) = 2𝛼(𝑥₂) + 𝛼( 𝜇(𝑥₂, 𝑥₁)) =2𝑏𝑥2+ 𝛼( 𝜇(𝑥2, 𝑥₁)),

and also

𝜇(𝛼(𝑥₁), 𝛼(𝑥₂)) = 𝜇(𝑎𝑥₁, 𝑏𝑥₂) = 𝑎𝑏𝜇(𝑥₁, 𝑥₂)

=2𝑎𝑏𝑥2+ 𝑎𝑏𝜇(𝑥2, 𝑥₁) =2𝑎𝑏𝑥2+ 𝜇(𝑏𝑥2, 𝑎𝑥₁).

The assumption that 𝛼(𝜇(𝑥2, 𝑥₁)) = 𝜇(𝛼(𝑥₂), 𝛼(𝑥₁)), we have the identity 2𝑏𝑥₂ = 2𝑎𝑏𝑥₂, which forces 𝑎 = 1, since we only treat 𝑏 that are non-zero. Does this mean that we have the multiplicativity for any 𝑏 ∈ 𝐾, as long as 𝑎 = 1? Turns out, this is not true, since 𝛼( 𝜇(𝑥₂, 𝑥₃)) = 𝑎𝑥₁+ 𝛼( 𝜇(𝑥₃, 𝑥₂)), and

𝜇(𝛼(𝑥₂), 𝛼(𝑥₃)) = 𝑏2𝜇(𝑥₂, 𝑥₃)

= 𝑏2𝑥₁+ 𝜇(𝑏𝑥₃, 𝑏𝑥₂).

Assuming that 𝛼(𝜇(𝑥3, 𝑥₂)) = 𝜇(𝛼(𝑥₃), 𝛼(𝑥₂)), previous calculations are equal when 𝑎𝑥₁ = 𝑏2𝑥₂,which forces 𝑎 = 𝑏2 =1. Thus, we could say that 𝑏 is its own inverse, e.g. if 𝐾 = R, 𝑏 could be −1.

However, we still need 𝛼(𝜇(𝑥1, 𝑥₃)) = 𝜇(𝛼(𝑥₁), 𝛼(𝑥₃)). Doing similar calculations as before yields that

𝛼( 𝜇(𝑥₁, 𝑥₃)) = 𝛼(−2𝑥₃+ 𝜇(𝑥₃, 𝑥₁)) = −2𝑏𝑥3+ 𝛼( 𝜇(𝑥3, 𝑥₁)) =2𝑥3+ 𝛼( 𝜇(𝑥3, 𝑥₁)), and 𝜇(𝛼(𝑥₁), 𝛼(𝑥₃)) = 𝜇(𝑎𝑥₁, 𝑏𝑥₃) = −𝜇(𝑥₁, 𝑥₃) = −2𝑥3− 𝜇(𝑥3, 𝑥₁) = −2𝑥3+ 𝜇(𝛼(𝑥3), 𝛼(𝑥1)).

Under the assumption that 𝛼(𝜇(𝑥3, 𝑥₁)) = 𝜇(𝛼(𝑥₃), 𝛼(𝑥₁)), we need that −2𝑥₃ = 2𝑥₃, which forces 𝑥3 =0. So we can not have multiplicativity.

Furthermore, we can not have a unit. Remember, that if we had a unital hom-associative algebra, then 𝜇(𝛼(𝑥), 𝑦) = 𝜇(𝑥, 𝛼(𝑦)). Then e.g. 𝜇(𝛼(𝑥₁), 𝑥₂) = 𝑎𝜇(𝑥₁, 𝑥₂), and also 𝜇(𝑥₁, 𝛼(𝑥₂)) = 𝑏𝜇(𝑥₁, 𝑥₂). This means that 𝑎 = 𝑏, which is a contradiction. Hence, in this case, there is no unit element.

Example 3.23. In the hom-associative algebra defined in the beginning of this section, if 𝑐 = 1,

and 𝑎, 𝑏, 𝑑, 𝑒, 𝑓 = 0, then 𝛼(𝑥1) = 2𝑥2, 𝛼(𝑥2) = 0 and 𝛼(𝑥3) = 𝑥1. This yields

𝜇(𝑥₁, 𝜇(𝑥₂, 𝑥₃)) = 𝜇(𝛼(𝑥₃), 𝜇(𝑥₂, 𝑥₃)) = 𝜇( 𝜇(𝑥₃, 𝑥₂), 𝛼(𝑥₃)) = 𝜇( 𝜇(𝑥₃, 𝑥₂), 𝑥₁)

Hence 𝜇(𝑥1, 𝜇(𝑥₂, 𝑥₃)) − 𝜇( 𝜇(𝑥₂, 𝑥₃), 𝑥₁)) = −𝜇(𝑥₁, 𝑥₁). However, we note 𝜇( 𝜇(𝑥₂, 𝑥₃), 𝑥₁) = 𝜇(𝛼(𝑥₂), 𝜇(𝑥₁, 𝑥₃))

= 𝜇(0, 𝜇(𝑥1, 𝑥₃)) = 0, i.e. we get

𝜇(𝑥₁, 𝜇(𝑥₂, 𝑥₃)) = −𝜇(𝑥₁, 𝑥₁). (3.4) Furthermore, 𝜇(𝑥1, 𝑥₁) = 𝜇(𝑥₁, 𝜇(𝑥₂, 𝑥₃) − 𝜇(𝑥₃, 𝑥₂)), by the definition of 𝜇. Expanding the right hand side of the this expression give us

𝜇(𝑥₁, 𝜇(𝑥₂, 𝑥₃) − 𝜇(𝑥₃, 𝑥₂)) = 𝜇(𝑥₁, 𝜇(𝑥₂, 𝑥₃)) − 𝜇(𝑥₁, 𝜇(𝑥₃, 𝑥₂)), where we note that

𝜇(𝑥₁, 𝜇(𝑥₃, 𝑥₂)) = 𝜇(𝜇(𝑥₃, 𝑥₃), 𝛼(𝑥₂)) = 𝜇(𝜇(𝑥₃, 𝑥₃), 0) = 0.

So, 𝜇(𝑥1, 𝜇(𝑥₃, 𝑥₂)) = 0, and thus 𝜇(𝑥₁, 𝜇(𝑥₂, 𝑥₃)) = 𝜇(𝑥₁, 𝑥₁), but from equation 3.4, we can conclude that this whole expression is equal to zero. And also

𝜇(2𝑥₂, 𝜇(𝑥₃, 𝑥₁)) = 𝜇(𝛼(𝑥₁), 𝜇(𝑥₃, 𝑥₁)) = 𝜇( 𝜇(𝑥₁, 𝑥₃), 𝛼(𝑥₁)) = 𝜇( 𝜇(𝑥1, 𝑥₃), 2𝑥₂) = 𝜇( 𝜇(𝑥₃, 𝑥₁) − 2𝑥₃,2𝑥₂) = 𝜇(𝜇(𝑥₃, 𝑥₁), 2𝑥₂) − 2𝜇(𝑥₃,2𝑥₂). Hence 𝜇(𝑥₂, 𝜇(𝑥₃, 𝑥₁)) − 𝜇( 𝜇(𝑥₃, 𝑥₁), 𝑥₂)) = −2𝜇(𝑥₃, 𝑥₂), Now, note 2𝜇(𝜇(𝑥3, 𝑥₁), 𝑥₂) = 𝜇(𝜇(𝑥₃, 𝑥₁), 𝛼(𝑥₁)) = 𝜇(𝛼(𝑥₃), 𝜇(𝑥₁, 𝑥₁) = 𝜇(𝛼(𝑥₃), 0) = 0, since we know from before that 𝜇(𝑥1, 𝑥₁) = 0. Hence, we have that

𝜇(𝑥₂, 𝜇(𝑥₃, 𝑥₁)) = −2𝜇(𝑥₃, 𝑥₂). (3.5) We expand the left-hand side

𝜇(𝑥₂, 𝜇(𝑥₃, 𝑥₁)) = 1

2𝜇(𝛼(𝑥1), 𝜇(𝑥3, 𝑥1)) = 𝜇( 𝜇(𝑥₁, 𝑥₃), 𝑥₂)

= 𝜇(−2𝑥3+ 𝜇(𝑥3, 𝑥₁), 𝑥₂)

= −2𝜇(𝑥3, 𝑥₂) + 𝜇( 𝜇(𝑥₃, 𝑥₁), 𝑥₂).

Putting these two previous equations together, we obtain 𝜇(𝜇(𝑥3, 𝑥₁), 𝑥₂) = 0. Furthermore 0 = 𝜇(𝜇(𝑥1, 𝑥₁), 𝑥₁) = 𝜇(𝜇(𝑥₁, 𝑥₁), 𝛼(𝑥₃))

= 𝜇(2𝑥2, 𝜇(𝑥₁, 𝑥₃)) = 𝜇(2𝑥2, 𝜇(𝑥₃, 𝑥₁) − 2𝑥₃)