Braided Crossed Modules and Loday-Pirashvili category

TESE DE DOUTORAMENTO Braided Crossed Modules and Loday-Pirashvili category ALEJANDRO FERNÁNDEZ-FARIÑA ESCOLA DE DOUTORAMENTO INTERNACIONAL PROGRAMA DE DOUTORAMENTO EN MATEMÁTICAS SANTIAGO DE COMPOSTELA 2021 DECLARACIÓN DO AUTOR DA TESE Braided Crossed Modules and Loday-Pirashvili category D. Alejandro Fernández Fariña Presentoamiña tese, seguindooprocedementoaxeitado aoRegulamento, e declaro que: 1) A tese abarca os resultados da elaboración do meu traballo. 2) De selo caso, na tese faise referencia ás colaboracións que tivo este traballo. 3) A tese é a versión definitiva presentada para a súa defensa e coincide coa versión enviada en formato electrónico. 4) Confirmo que a tese non incorre en ningún tipo de plaxio doutros autores nin de traballos presentados por min para a obtención doutros títulos. En Santiago de Compostela, 11 de abril de 2021 Asdo.: Alejandro Fernández Fariña AUTORIZACIÓN DO DIRECTOR / TITOR DA TESE Braided Crossed Modules and Loday-Pirashvili category D. Manuel Ladra González INFORMA: Que a presente tese, correspóndese co traballo realizado por D. Alejandro Fernández Fariña, baixo amiñadirección, e autorizoa súapresentación, considerando que reúne os requisitos esixidos no Regulamento de Estudos de Doutoramento da USC, e que como director desta non incorre nas causas de abstención establecidas na Lei 40/2015. De acordo co indicado no Regulamento de Estudos de Doutoramento, declara tamén que a presente tese de doutoramento é idónea para ser defendida en base á modalidade de Monográfica con reproducción de publicaciones, nos que a participación do/a doutorando/a foi decisiva para a súa elaboración e as publicacións se axustan ao Plan de Investigación. En Santiago de Compostela, 11 de abril de 2021 Asdo.: Manuel Ladra González Braided Crossed Modules and Loday-Pirashvili category by ALEJANDRO FERNÁNDEZ-FARIÑA DISSERTATION Submitted for the degree of DOCTOR EN MATEMÁTICAS UNIVERSIDAD DE SANTIAGO DE COMPOSTELA Santiago de Compostela, 2021 The results presented in this thesis were obtained thanks to a grant from the Xunta de Galicia (Spain), with reference ED481A-2017/064, the project MTM2016-79661- P from the Ministerio de Ciencia e Innovación and Agencia Estatal de Investigación (European FEDER support included, UE), and by the Consellería de Cultura, Educación e Universidade da Xunta de Galicia, through the Competitive Reference Groups (GRC), ED431C 2019/10 (European FEDER support included, UE). MTM2016-79661-P Unión Europea Fondo Europeo de Desarrollo Regional Una manera de hacer EUROPA ED481A-2017/064 Xunta de Galicia ED431C 2019/10 categories. Keeping in mind what is done for groups, in this thesis we will give definitions of braidings for the aforementioned internal categories and crossed modules. The case of associative algebras is not complex because the associativity allows us to work in a natural way with braidings on semigroupal categories [17]. The notion of braiding for Lie algebras was already given by Ulualan [50]. On the other hand, Ellis [20] defined the notion of 2-crossed module of Lie algebras, also studied by Martins and Picken [46]. We will use a slightly different definition for braiding for crossed modules of Lie algebras than the one given by Ulualan [50], since we want a parallelism between the examples of braided crossed modules of groups and braided crossed modules of Lie algebras, and we also require braided crossed modules to be a particular case of 2-crossed modules, as it happens in the case of groups. Leibniz algebras appear in mathematics as a “non-antisymmetric” case of Lie algebras. Bearing this in mind, in this thesis, we will show how to extend the idea of braiding for crossed modules and internal categories of Lie algebras to the Leibniz setting. After introducing these notions, we will prove the equivalence between braided crossed modules of Leibniz algebras and braided categorical Leibniz algebras, and we will show the parallelism between its examples and the ones given for groups, associative algebras and Lie algebras. Although Lie algebras are a subvariety of the variety of Leibniz algebras, Loday and Pirashvili found in [44] that Leibniz algebras can be seen as a full coreflective subcategory of a specific type of Lie objects. They introduced a new tensor product in the category of linear maps of vector spaces and internalised the concept of Lie algebra in a (braided) symmetric monoidal category. This realisation proved to be very handful studying different problems in Leibniz algebras, as Lie theory is much better developed, see [12,23,49] for example. We will use this category to extend the concept of braiding from the Lie case to the Leibniz case. The concept of central extension of groups or Lie algebras is highly relevant in mathematics, and it plays a fundamental role in several areas of physics as well. This notion was extended to crossed modules of groups or Lie algebras. The study of central extensions in the categories of crossed modules was initiated in [48] for groups and in [13] for Lie algebras, and it remains a current research topic, as shown by the different literature tackling this issue. Since crossed modules of groups and Lie algebras are a generalisation of groups and Lie algebras, it is essential to search, in the category of crossed modules of groups or Lie algebras, extensions of classical results in the theory of groups or Lie algebras. In [26], Fukushi gave a braided version of the results on universal central extenxvi sions of crossed modules of groups provided by Norrie in [48]. He found a natural braiding on the universal central extension of a crossed module of groups which behaves well with one braided crossed module. However, it is not the archetype of universal central extension in the category of braided crossed modules since, in this category, it is necessary to add additional restrictions including the braiding on the notions of centre and commutator. In this work, we will devise a braided version of the results given by Casas and Ladra in [13] for braided crossed modules of Lie 𝐾-algebras; more precisely, we will study universal central extensions in the category of braided Lie crossed modules BX(LieAlg𝐾). For that purpose, we will need the definition of centre and commutator given by Huq in [35] in the braided context. Note that the framework of Chapter 3 is different from that given in [14], since the category X(LieAlg𝐾)is not a Birkhoff subcategory of BX(LieAlg𝐾). The study of the internalisation of Lie algebras is also a very handful tool, as we will see in Section 2.4. It also allows proving different properties in several kinds of categories at the same time, such as Lie superalgebras, ℤ-graded Lie algebras, differential graded Lie algebras or regular Hom-Lie algebras [33]. For example, two important properties that characterise the variety of Lie algebras amongst all the varieties of non-associative algebras, the existence of algebraic exponents [29,30] or the representability of actions [28], hold also in the categories of Lie objects over certain types of monoidal categories [27,34]. We want to generalise Loday and Pirashvili construction out of the linear maps category, defining a new tensor product in certain kinds of categories with operations, with the least amount of properties needed to do so, to obtain the Loday-Pirashvili category. Then, we will prove that the Leibniz objects (the internalisation of Leibniz algebras), can be seen as a particular case of Lie objects in the Loday-Pirashvili category. In the particular case of vector spaces, this construction generalises the one given in [44]. Throughout this text, we will suppose that 𝐾is a field. Structure of the thesis This manuscript is organized as follows. In the preliminaries (Chapter 1), we will recall some basic definitions, and we will give the notion of braiding for semigroupal categories. In Chapter 2, we will study the braidings for crossed modules and internal categories. We will start showing the first case of braiding, the case of groups (Secxvii tion 2.1), and we will take it as a base to introduce the notions of braided categorical associative algebra and braided crossed module of associative algebras (Section 2.2). We will show the equivalence of the associative case. Then, in Section 2.3, we will motivate the definition given by Ulualan [50] for braided crossed modules of Lie algebras using our definition of braiding for crossed modules of associative algebras, and we give a simpler definition when char(𝐾)≠2. We will also discuss a different definition of braided crossed module of Lie algebras showing its relationship with the associative case. From there, in Section 2.4 we will study the Leibniz algebras case. We show the internalization of a crossed module’s notion with a left Lie action of Lie objects in an arbitrary category. We will also define braidings for crossed modules of Lie objects and categorical Lie objects. Then we apply this definition to the Loday- Pirashvili category 𝐾, and we will obtain the concepts of braiding for crossed modules of Leibniz algebras and categorical Leibniz algebras. With the new definition of braiding, we will prove the equivalence between braided categories in the Leibniz algebras case, and finally, in Section 2.5, we will see the non-abelian tensor product of groups as an example of a braided crossed module of groups. Furthermore, with our definition of braiding for crossed modules of Lie algebras, we obtain similarly an example of braiding using the non-abelian tensor product of Lie algebras. The same is true for our definition of braiding for crossed modules of Leibniz algebras. In Chapter 3, we will study two ideas of universal central extensions for braiding crossed modules of Lie algebras. In Section 3.1, we provide the definitions for central extensions in the category of Lie crossed modules X(LieAlg𝐾)and B-central extensions in BX(LieAlg𝐾), necessary for developing the chapter. In Section 3.2, we construct the universal B-central extension for a B-perfect braided Lie crossed module and prove that a braided Lie crossed module admits a universal B-central extension if and only if it is B-perfect. In Section 3.3, we construct the universal 𝔘-central extension for braided crossed modules, which are perfect as Lie crossed modules, where 𝔘∶BX(LieAlg𝐾)⟶X(LieAlg𝐾)is the forgetful functor. In Section 3.4, we study the relation between the universal B-central extension and the universal 𝔘-central extension of a braided Lie crossed module. Finally, we prove that both universal extensions exist and coincide for a B-perfect braided Lie crossed module. In Chapter 4, we will define the LP category for different tensor categories, and then we study the Lie objects in some kind of LP categories and their relationship with the Leibniz objects in the base category. In Section 4.1, we will study the different tensor categories: categories with operations, (braided) semigroupal categories and (braided) monoidal categories, and we will construct their Loday-Pirashvili category. In Section 4.2, we will talk about additive categories and we will show that, with xviii some assumptions, we can recover many properties in the maps between the tensor product and the “+” operation. The last section (Section 4.3) is devoted to study the internalization of a Leibniz object and Lie object in a category , showing that the Liesation functor exists between these categories. Then we will provide a better understanding of Lie objects in the Loday-Pirashvili category of . To conclude, we will prove that the category of Leibniz objects in is a full coreflective subcategory of the Lie objects in the Loday-Pirashvili category of . xix xx ⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄ Objectives and hypotheses ⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄ This thesis has the following hypothesis and objectives: Hyp. 1 Strict monoidal categories can be seen as categorical monoids in Set. Similarly, we can think that a strict semigroupal category over an internal category in Vect𝐾is really an internal associative 𝐾-algebra. Obj. 1 We want to use the idea of braiding in a semigroupal category to make a braiding for categorical associative 𝐾-algebras using that the same idea as the fact that the category of crossed modules of groups has a natural idea of braiding utilising the idea of monoidal category. Hyp. 2 In the category of crossed modules of groups, we can define the concept of braiding. With that construction, we have en equivalence between braided crossed modules of groups and braided categorical groups. Obj. 2 We want to construct a braiding for crossed modules of associative algebras such that this new category is equivalent to braiding categorical associative 𝐾- algebras. Hyp. 3 The category of associative 𝐾-algebras and Lie 𝐾-algebras are related with a functor (−)∶AssAlg𝐾←←→ LieAlg𝐾, which takes an associative algebra 𝐴and gives back a Lie 𝐾-algebra 𝐴. This Lie algebra has the same 𝐾-vector space as underlying structure and has as multiplication [𝑥, 𝑦] ∶= 𝑥𝑦 −𝑦𝑥. Obj. 3 Use the idea of functor (−)to make a new functor from crossed modules of associative 𝐾-algebras to crossed modules of Lie 𝐾-algebras. We will also define a functor from its braided versions, showing the naturalness when one defines the braiding for the Lie case. We will also show an equivalent (in the 1 2 Objectives and hypotheses sense of categories) definition for braiding crossed modules, which will give us a good example using the non-abelian tensor product. Hyp. 4 TheLoday-Pirashvili category providesuswith a way toseeLeibniz𝐾-algebras as a particular case of Lie objects when it is well known that Lie 𝐾-algebras are a particular case of Leibniz 𝐾-algebras. Obj. 4 Using the Loday-Pirashvili category and internalization, we want to use the definition of braiding for the Lie case to define braiding for crossed modules of Leibniz 𝐾-algebras and categorical Leibniz 𝐾-algebras. Once done, we will show that these new structures have the Lie case as a particular example. Also, we will have an excellent example of braiding crossed modules of Leibniz 𝐾- algebras using the non-abelian tensor product. Hyp. 5 The braiding for the Lie case gives equivalent categories. Obj. 5 We want to show that the braidings for the Leibniz case give equivalent categories. Hyp. 6 The braiding crossed modules of groups have a universal 𝔘-central extension. Obj. 6 We want to define the 𝔘-central extension for Lie algebras category since many results are true in the group case are true in the Lie case. We also describe the B-central extensions using the idea of centre give by Huq [35]. In general, the 𝔘-central extensions and B-central extension do not coincide, but we want to show the relationship between the universal ones. Hyp. 7 The construction of the LP-category given by Loday and Pirashvili can be defined in categories with a small set of properties. Obj. 7 We want to define the Loday-Pirashvili category using the least properties that are possible. We will do that for categories with operations, (braided) semigroupal categories and (braided) monoidal categories, showing that the tensor product in the Loday-Pirashvili category gives a tensor category of the same type. Hyp. 8 There is a Liesation functor from Leibniz 𝐾-algebras to Lie 𝐾-algebras that is a left adjoint to the forgetful functor. Obj. 8 We want to construct a Liesation functor for Leibniz objects to Lie objects for any category with a small set of properties. Objectives and hypotheses 3 Obj. 9 We will prove that the category of Leibniz objects in is a full coreflective subcategory of the Lie objects in the Loday-Pirashvili category of . 4 Objectives and hypotheses CHAPTER 1 ⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄ Preliminaries ⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄ In this chapter, we will give the basic concepts to delve into the rest of the chapters. 1.1 Internal categories Definition 1.1.1. Let Cbe a category with pullbacks. An internal category in Cconsist of two objects 𝐶1(morphisms object) and 𝐶0 (objects object) of C, together with the four following morphisms 𝑠, 𝑡, 𝑒, 𝑘: 𝐶0𝑒//𝐶1 𝑠 oo 𝑡 oo𝐶1×𝐶0𝐶1, 𝑘 oo where 𝐶1×𝐶0𝐶1is the pullback of 𝑡and 𝑠. 𝑠is called source morphism,𝑡is called target morphism,𝑒is called identity mapping morphism and 𝑘is called composition morphism. In addition, the morphisms must satisfy commutative diagrams that express the usual category laws (see [6]): (I1) 𝐶0𝐶1 𝐶0. 𝑒 Id𝐶0 𝑠(I2) 𝐶0𝐶1 𝐶0. 𝑒 Id𝐶0 𝑡 5 12 1 Preliminaries Let us begin with the definition of 𝔄on objects. Let (𝑀𝜕 ←←←←←←→ 𝑁, ∗) be a crossed module in AssAlg𝐾. We will take the semidirect product 𝑀⋊𝑁with ∗. Consider the following diagram: (𝑀⋊𝑁) ×𝑁(𝑀⋊𝑁)𝑀⋊𝑁 𝑁  𝑘  𝑡 𝑠 𝑒 with the following maps: 𝑠((𝑚, 𝑛)) = 𝑏, 𝑡((𝑚, 𝑛)) = 𝜕(𝑚) + 𝑛,𝑒(𝑛) = (0, 𝑛)and  𝑘(((𝑚, 𝑛),(𝑚′, 𝜕(𝑚) + 𝑛))) = (𝑚+𝑚′, 𝑛)for all 𝑚, 𝑚′∈𝑀,𝑛∈𝑁. Note that if ((𝑛, 𝑚),(𝑛′, 𝑚′)) ∈ (𝑀⋊𝑁) ×𝑁(𝑀⋊𝑁),𝑛′=𝑠((𝑚′, 𝑛′)) =  𝑡((𝑚, 𝑛)) = 𝜕(𝑚) + 𝑛, so the definition of  𝑘makes sense. It is necessary to prove that 𝑠, 𝑡,𝑒 and  𝑘are morphisms in AssAlg𝐾, that is, they preserve all the operations. Since it is obvious that 𝑠 and 𝑒 preserve the operations, we will focus on sketching how to prove that  𝑡and  𝑘preserve the sum and the product. Calculations are quite long, so we will not include them, although we will point out the crucial ideas required to complete them. Regarding  𝑡, it preserves the sum directly from the fact that 𝜕preserve it. Furthermore, the fact that 𝜕is an 𝑁-equivariant associative morphism is the key to prove that  𝑡preserves the product. Concerning  𝑘, note that the elements in (𝑀⋊𝑁) ×𝑁(𝑀⋊𝑁)are of the form ((𝑚, 𝑛),(𝑚′, 𝜕(𝑚)+𝑛)), with 𝑚, 𝑚′∈𝑀,𝑛∈𝑁. Immediately below we will show the calculationsrequiredtoprovethat  𝑘preservesthe sum. Let ((𝑚𝑖, 𝑛𝑖),(𝑚′ 𝑖, 𝜕(𝑚𝑖)+𝑛𝑖)) ∈ (𝑀⋊𝑁) ×𝑁(𝑀⋊𝑁)for 𝑖= 1,2. On one hand we have that  𝑘(((𝑚1, 𝑛1),(𝑚′ 1, 𝜕(𝑚1) + 𝑛1)) + ((𝑚2, 𝑛2),(𝑚′ 2, 𝜕(𝑚2) + 𝑛2))) = 𝑘((𝑚1, 𝑛1)+(𝑚2, 𝑛2),(𝑚′ 1, 𝜕(𝑚1) + 𝑛1)+(𝑚′ 2, 𝜕(𝑚2) + 𝑛2)) = ((𝑚1+𝑚2, 𝑛1+𝑛2),(𝑚′ 1+𝑚′ 2, 𝜕(𝑚1) + 𝑛1+𝜕(𝑚2) + 𝑛2)) = (𝑚1+𝑚2+𝑚′ 1+𝑚′ 2, 𝑛1+𝑛2), On the other hand,  𝑘(((𝑚1, 𝑛1),(𝑚′ 1, 𝜕(𝑚1) + 𝑛1))) +  𝑘(((𝑚2, 𝑛2),(𝑚′ 2, 𝜕(𝑚2) + 𝑛2))) 1.3.2 Crossed modules of associative algebras 13 = (𝑚1+𝑚′ 1, 𝑛1)+(𝑚2+𝑚′ 2, 𝑛2) = (𝑚1+𝑚′ 1+𝑚2+𝑚′ 2, 𝑛1+𝑛2), by making use of the definition of  𝑘and the addition in 𝑀⋊𝑁. Hence,  𝑘preserves the sum. Calculations for the product are similar, but involving distributivity and the Peiffer identity. Commutativity of the diagrams of the internal categories is easy. Defining 𝔄on morphisms is quite obvious. Given a morphism of crossed modules (𝑓1, 𝑓2)between (𝑀𝜕 ←←←←←←→ 𝑁, ∗) and (𝑀′𝜕′ ←←←←←←←←→ 𝑁′,∗′), its corresponding internal functor is given by 𝑓1×𝑓2∶𝑀⋊𝑁→𝑀′⋊𝑁′and 𝑓2∶𝑁→𝑁′, where 𝑓1×𝑓2((𝑎, 𝑏)) = (𝑓1(𝑎), 𝑓2(𝑏)). Commutativity of the diagrams for internal functors follows from the definitions of 𝑠, 𝑠′, 𝑡, 𝑡′,𝑒, 𝑒′, 𝑘and  𝑘′, along with the equality 𝑓2◦𝜕=𝜕′◦𝑓1. 𝔄is clearly a functor with the previous assignments for objects and morphisms. Now let us define the functor 𝔄. Let 𝐶= (𝐶1, 𝐶0, 𝑠, 𝑡, 𝑒, 𝑘)be an internal category in AssAlg𝐾. Consider ker(𝑠)and the morphism 𝑡|ker(𝑠)∶ ker(𝑠)→𝐶0. We will write 𝜕𝑡in order to ease notation. We define an associative action (𝑒∗,∗𝑒)with 𝑎𝑒∗𝑥=𝑒(𝑎)𝑥and 𝑥∗𝑒𝑎=𝑥𝑒(𝑎)with 𝑎∈𝐶0, 𝑥 ∈ ker(𝑠). It is easy that the maps are well defined. It only remains to prove that (ker(𝑠)𝜕𝑡 ←←←←←←←←→ 𝐶0,(𝑒∗,∗𝑒)) satisfies is a crossed module. Given 𝑎∈𝐶0and 𝑥∈ ker(𝑠), 𝜕𝑡(𝑎𝑒∗𝑥) = 𝜕𝑡(𝑒(𝑎)𝑥) = 𝑡(𝑒(𝑎)𝑥) = 𝑡(𝑒(𝑎))𝑡(𝑥) = 𝑎𝜕𝑡(𝑥), 𝜕𝑡(𝑥∗𝑒𝑎) = 𝜕𝑡(𝑥𝑒(𝑎)) = 𝑡(𝑥𝑒(𝑎)) = 𝑡(𝑥)𝑡(𝑒(𝑎)) = 𝜕𝑡(𝑥)𝑎. Note that we use that in an internal category 𝑡◦𝑒= Id𝐶0. To prove the Peiffer identity, let 𝑥1, 𝑥2∈ ker(𝑠). 𝜕𝑡(𝑥1)𝑒∗𝑥2=𝑒(𝑡(𝑥1))𝑥2. We need to show that 𝑒(𝑡(𝑥1))𝑥2=𝑥1𝑥2. We will take 𝑧= (𝑒(𝑡(𝑥1)) − 𝑥1). 𝑡(𝑧) = 𝑡(𝑒(𝑡(𝑥1))) − 𝑡(𝑥1) = 𝑡(𝑥1) − 𝑡(𝑥1)=0so we can compose with 𝑒(0) since. 𝑡(𝑧) = 0 = 𝑠(0) = 𝑠(𝑒(0)). Since 𝑥2∈ ker(𝑠), we can take 𝑘((𝑒(0), 𝑥2)), 14 1 Preliminaries because 𝑠(𝑦) = 0 = 𝑡(0) = 𝑡(𝑒(0)). In addition we have that in an internal category 𝑘((𝑧, 𝑒(0))) = 𝑧and 𝑘((𝑒(0), 𝑥2)) = 𝑥2. 0 = 𝑘((0,0)) = 𝑘((𝑧𝑒(0), 𝑒(0)𝑥2)) =𝑘((𝑧, 𝑒(0))(𝑒(0), 𝑥2)) = 𝑘((𝑧, 𝑒(0)))𝑘((𝑒(0), 𝑥2)) = 𝑧𝑥2. Finally, we have: 0 = 𝑧𝑥2= (𝑒(𝑡(𝑥1)) − 𝑥1)𝑥2=𝑒(𝑡(𝑥1))𝑥2−𝑥1𝑥2, which establishes one of the Peiffer identities. Similar arguments apply to the other Peiffer identity. Defining 𝔄on morphisms is also quite obvious. Let 𝐶= (𝐶1, 𝐶0, 𝑠, 𝑡, 𝑒, 𝑘)and 𝐶′= (𝐶′ 1, 𝐶′ 0, 𝑠′, 𝑡′, 𝑒′, 𝑘′)be two internal categories in AssAlg𝐾and 𝐹∶𝐶→𝐶′ an internal functor, with 𝐹1∶𝐶1→𝐶′ 1and 𝐹0∶𝐶0→𝐶′ 0. Its corresponding morphism of crossed modules is given by (𝐹𝑠 1, 𝐹0), with 𝐹𝑠 1(𝑥) = 𝐹1(𝑥)for 𝑥∈ ker(𝑠), which follows from the diagrams of internal functors. It is easy to check that, with the previous assignments, 𝔄is indeed a functor. 𝔄and 𝔄establish an equivalence between the categories where the natural isomorphisms IdX(AssAlg𝐾) 𝛼𝔄 ≅𝔄◦𝔄and IdICat(AssAlg𝐾) 𝛽𝔄 ≅𝔄◦𝔄are given by: ∙if = (𝑀𝜕 ←←←←←←→ 𝑁, (∗1,∗2)) is a crossed module of associative 𝐾-algebras, then 𝛼𝔄 = (𝛼𝔄 𝑀,Id𝑁), with 𝛼𝔄 𝑀∶𝑀←←→ (𝑀, 0) defined as 𝛼𝑀(𝑚) = (𝑚, 0); ∙if = (𝐶1, 𝐶0, 𝑠, 𝑡, 𝑒, 𝑘)is a categorical associative 𝐾-algebra, then 𝛽𝔄 = (𝛽𝔄 𝑠,Id𝐶0), with 𝛽𝔄 𝐶1∶𝐶1←←→ ker(𝑠)⋊𝐶0is defined as 𝛽𝔄 𝐶1(𝑥)=(𝑥−𝑒(𝑠(𝑥)), 𝑠(𝑥)). 1.3.3 Crossed modules of Lie algebras We have an analogous definition for the case of Lie 𝐾-algebras. Crossed modules of Lie 𝐾-algebras were introduced by Kassel and Loday in [40]. Definition 1.3.12. Let 𝑀and 𝑁two Lie 𝐾-algebras. A Lie (left-)action of 𝑁on 𝑀 is a 𝐾-bilinear map ⋅∶𝑁×𝑀⟶𝑀,(𝑛, 𝑚)⟼𝑛⋅𝑚, satisfying: [𝑛, 𝑛′]⋅𝑚=𝑛⋅(𝑛′⋅𝑚) − 𝑛′⋅(𝑛⋅𝑚), 1.3.3 Crossed modules of Lie algebras 15 𝑛⋅[𝑚, 𝑚′] = [𝑛⋅𝑚, 𝑚′]+[𝑚, 𝑛 ⋅𝑚′], 𝑛, 𝑛′∈𝑁, 𝑚, 𝑚′∈𝑀. If we denote ⋅= [−,−], the two identities are the two possible rewrites of the Jacobi identity by taking two elements in 𝑁or two in 𝑀. In particular, if 𝑀is a Lie 𝐾-algebra and 𝑥∈𝑀, we have that the adjoint map ad(𝑥)∶ 𝑀←←→ 𝑀,ad(𝑥)(𝑦) = [𝑥, 𝑦], is a Lie action of 𝑀on itself. Definition 1.3.13. Acrossed module of Lie 𝐾-algebras is a pair (𝑀𝜕 ←←←←←←→ 𝑁, ⋅)where 𝑀and 𝑁are Lie 𝐾-algebras, ⋅is a Lie action of 𝑁on 𝑀, and 𝑀𝜕 ←←←←←←→ 𝑁is a Lie 𝐾-homomorphism satisfying: -𝜕is an 𝑁-equivariant Lie 𝐾-homomorphism (we suppose the adjoint action of 𝑁 on itself), i.e. 𝜕(𝑛⋅𝑚) = ad(𝑛)(𝜕(𝑚)) = [𝑛, 𝜕(𝑚)], 𝑛 ∈𝑁, 𝑚 ∈𝑀, -Peiffer identity: 𝜕(𝑚)⋅𝑚′= ad(𝑚)(𝑚′) = [𝑚, 𝑚′], 𝑚, 𝑚′∈𝑀. Example 1.3.14. 1. As in the previous cases we have the example of crossed module of Lie 𝐾- algebras (𝑀Id𝑀 ←←←←←←←←←←←←←←←→ 𝑀, [−,−]), with the adjoint action, 𝑚⋅𝑚′= [𝑚, 𝑚′], where 𝑀is a Lie 𝐾-algebra. 2. Any central extension of Lie algebras 𝑀𝜕 ←←←←←←→→ 𝑁is a crossed module, with the action 𝜕(𝑚)⋅𝑚′= [𝑚, 𝑚′]. Conversely, a simply connected crossed module (i.e. 𝜕is surjective) is a central extension. In particular, 𝑀ad ←←←←←←←←←→→ IDer(𝑀),𝑚↦ad(𝑚), with the action, ad(𝑚)⋅𝑚′= [𝑚, 𝑚′], is a Lie crossed module, where IDer(𝑀)are the inner derivations of a Lie algebra 𝑀. Definition 1.3.15. Ahomomorphism of crossed modules of Lie 𝐾-algebras between (𝑀𝜕 ←←←←←←→ 𝑁, ⋅)and (𝑀′𝜕 ←←←←←←→ 𝑁′,∗) is a pair of Lie 𝐾-homomorphisms, 𝑓1∶𝑀←←→ 𝑀′ and 𝑓2∶𝑁←←→ 𝑁′such that: 𝑓1(𝑛⋅𝑚) = 𝑓2(𝑛) ∗ 𝑓1(𝑚),(XLieH1) 16 1 Preliminaries 𝜕′◦𝑓1=𝑓2◦𝜕, (XLieH1) 𝑚∈𝑀, 𝑛 ∈𝑁. There is a natural way to correlate the crossed modules of associative 𝐾-algebras with the crossed modules of Lie 𝐾-algebras. The following results that relate both can be seen in [21]. Lemma 1.3.16. Let 𝑀and 𝑁be two associative 𝐾-algebras. We denote by 𝐴the Lie 𝐾-algebra associated to an associative 𝐾-algebra 𝐴, i.e. the Lie 𝐾-algebra with the operation [𝑎, 𝑎′] = 𝑎𝑎′−𝑎′𝑎. (i) If ∗= (∗1,∗2)is an associative action of 𝑁on 𝑀, then we have that the map [−,−]∗∶𝑁×𝑀←←→ 𝑀, defined as [𝑛, 𝑚]∗=𝑛∗1𝑚−𝑚∗2𝑛, is a Lie action of 𝑁on 𝑀. (ii) If (𝑀𝜕 ←←←←←←→ 𝑁, ∗) is a crossed module of associative 𝐾-algebras, then we have that (𝑀𝜕 ←←←←←←→ 𝑁,[−,−]∗)is a crossed module of Lie 𝐾-algebras. Remark 1.3.17.With the previous property we can see that the examples given for the associative algebras case, (𝑀Id𝑀 ←←←←←←←←←←←←←←←→ 𝑀, (∗,∗)), and for the Lie case for the associate Lie algebra 𝑀,(𝑀Id𝑀 ←←←←←←←←←←←←←←←←←←←←→, 𝑀,[−,−]∗), are related. We denote by X(LieAlg𝐾)the category of crossed modules of Lie 𝐾-algebras and their homomorphisms. Remark 1.3.18.The previous lemma gives us a functor (−) ∶X(AssAlg𝐾)←←←←←←←←←←←←→ X(LieAlg𝐾). We have the next proposition which relates the categorical associative 𝐾-algebras with the categorical Lie 𝐾-algebras. Proposition 1.3.19. If (𝐶1, 𝐶0, 𝑠, 𝑡, 𝑒, 𝑘)is a categorical associative 𝐾-algebra, then (𝐶 1, 𝐶 0, 𝑠, 𝑡, 𝑒, 𝑘)is a categorical Lie 𝐾-algebra. Proof. Immediate since (𝐶1×𝐶0𝐶1)=𝐶 1×𝐶 0𝐶 1. They are the same underlying vector space and have the same operation. 1.3.4 Crossed modules of Leibniz algebras 17 Remark 1.3.20.The previous proposition gives us a functor (−) ∶ICat(AssAlg𝐾)←←←←←←←←←←←←→ ICat(LieAlg𝐾). Remark 1.3.21.As in the case of groups and associative 𝐾-algebras, the categories ICat(LieAlg𝐾)and X(LieAlg𝐾)are equivalent (see [5,22,25]). It is easy to check that the equivalence functors commute with the functors (−)  and (−) . We only need to show that (−)preserves the semidirect product. Definition 1.3.22. Let 𝑀and 𝑁be two Lie 𝐾-algebras and ⋅a Lie action of 𝑁on 𝑀. We define its semidirect product, denoted by 𝑀⋊𝑁, as the 𝐾-vector space 𝑀×𝑁 with the following bracket: [(𝑚, 𝑛),(𝑚′, 𝑛′)] = ([𝑚, 𝑚′] + 𝑛⋅𝑚′−𝑛′⋅𝑚, [𝑛, 𝑛′]). Proposition 1.3.23. Let 𝑀and 𝑁be associative 𝐾-algebras. If ∗is an associative action of 𝑁on 𝑀(then [−,−]∗is a Lie action of 𝑁in 𝑀), then we have that (𝑀⋊𝑁)=𝑀⋊𝑁. Proof. Since the underlying vector space is the same, we only need to prove that the bracket is the same. (𝑚, 𝑛)(𝑚′, 𝑛′)−(𝑚′, 𝑛′)(𝑚, 𝑛) = (𝑚𝑚′+𝑛∗1𝑚′+𝑚∗2𝑛′, 𝑛𝑛′)−(𝑚′𝑚+𝑛′∗1𝑚+𝑚′∗2𝑛, 𝑛′𝑛) = (𝑚𝑚′+𝑛∗1𝑚′+𝑚∗2𝑛′−𝑚′𝑚−𝑛′∗1𝑚−𝑚′∗2𝑛, 𝑛𝑛′−𝑛′𝑛) = ([𝑚, 𝑚′]+[𝑛, 𝑚′]∗− [𝑛′, 𝑚]∗,[𝑛, 𝑛′]), where (𝑚, 𝑛),(𝑚′, 𝑛′) ∈ 𝑀×𝑁. 1.3.4 Crossed modules of Leibniz algebras The definition of crossed modules of Leibniz 𝐾-algebras, “non-antisymmetric” case of Lie 𝐾-algebras, was introduced by Loday and Pirashvili in [43]. 18 1 Preliminaries Definition 1.3.24. Let 𝑁and 𝑀be two Leibniz 𝐾-algebras. A Leibniz action of 𝑁 on 𝑀is a pair ⋅= (⋅1,⋅2)where ⋅1∶𝑁×𝑀←←→ 𝑀and ⋅2∶𝑀×𝑁←←→ 𝑀are 𝐾-bilinear maps and the following properties are satisfied 𝑛⋅1[𝑚, 𝑚′]=[𝑛⋅1𝑚, 𝑚′]−[𝑛⋅1𝑚′, 𝑚],(ALeib1) [𝑚, 𝑛 ⋅1𝑚′]=[𝑚⋅2𝑛, 𝑚′]−[𝑚, 𝑚′]⋅2𝑛, (ALeib2) [𝑚, 𝑚′⋅2𝑛]=[𝑚, 𝑚′]⋅2𝑛− [𝑚⋅2𝑛, 𝑚′],(ALeib3) 𝑚⋅2[𝑛, 𝑛′] = (𝑚⋅2𝑛)⋅2𝑛′− (𝑚⋅2𝑛′)⋅2𝑛, (ALeib4) 𝑛⋅1(𝑚⋅2𝑛′) = (𝑛⋅1𝑚)⋅2𝑛′− [𝑛, 𝑛′]⋅1𝑚, (ALeib5) 𝑛⋅1(𝑛′⋅1𝑚) = [𝑛, 𝑛′]⋅1𝑚− (𝑛⋅1𝑚)⋅2𝑛′.(ALeib6) 𝑚, 𝑚′∈𝑀, 𝑛, 𝑛′∈𝑁. Remark 1.3.25.If we change the notation of ⋅1and ⋅2by [−,−] in both cases, the axioms of the Leibniz actions are all possible rewritings of the Leibniz identity when we choose two elements in 𝑀and one in 𝑁(the first three) or one in 𝑀and two 𝑁 (the last three). In particular, we have that the pair ([−,−],[−,−]) where [−,−] is the Leibniz bracket of the Leibniz 𝐾-algebra 𝑀is a Leibniz action of 𝑀on itself. Definition 1.3.26. A crossed module of Leibniz 𝐾-algebras is a pair (𝑀𝜕 ←←←←←←→ 𝑁, ⋅) where 𝑀and 𝑁are Leibniz 𝐾-algebras, ⋅= (⋅1,⋅2)is a Leibniz action of 𝑁on 𝑀, 𝜕∶𝑀←←→ 𝑁is a Leibniz 𝐾-homomorphism, and the following properties are satisfied: -𝜕is an 𝑁-equivariant Leibniz 𝐾-homomorphism (we suppose that the bracket gives the action in 𝑁), i.e. 𝜕(𝑛⋅1𝑚) = [𝑛, 𝜕(𝑚)] and 𝜕(𝑚⋅2𝑛)=[𝜕(𝑚), 𝑛], 𝑛 ∈𝑁, 𝑚 ∈𝑀, -Peiffer identity: 𝜕(𝑚)⋅1𝑚′= [𝑚, 𝑚′] = 𝑚⋅2𝜕(𝑚′)𝑚, 𝑚′∈𝑀, 𝑛 ∈𝑁. Example 1.3.27. As for the previous cases, we have that if 𝑀is a Leibniz 𝐾-algebra then (𝑀Id𝑀 ←←←←←←←←←←←←←←←→ 𝑀, ([−,−],[−,−])) is a crossed module of Leibniz 𝐾-algebras. 1.3.4 Crossed modules of Leibniz algebras 19 The next immediate propositions give a relation between crossed modules of Lie and Leibniz 𝐾-algebras. Proposition 1.3.28. Let 𝑀and 𝑁be two Lie 𝐾-algebras. Then, ⋅is a Lie action of 𝑁on 𝑀if and only if (⋅,⋅−)is a Leibniz action of 𝑁on 𝑀, where ⋅−∶𝑀×𝑁←←→ 𝑀 is defined by 𝑚⋅−𝑛∶= −𝑛⋅𝑚. That is, the Lie action is a particular case of a Leibniz action when the action is “anticommutative”. Proposition 1.3.29. Let 𝑀and 𝑁be Lie K-algebras. Then, (𝑀𝜕 ←←←←←←→ 𝑁, ⋅)is a crossed module of Lie 𝐾-algebras if and only if (𝑀𝜕 ←←←←←←→ 𝑁, (⋅,⋅−)) is a crossed module of Leibniz 𝐾-algebras. Remark 1.3.30.With the previous property we can see that the examples given for the Lie algebra case, (𝑀Id𝑀 ←←←←←←←←←←←←←←←→ 𝑀, [−,−]), and for the Leibniz algebra case, taking a Lie algebra 𝑀,(𝑀Id𝑀 ←←←←←←←←←←←←←←←→ 𝑀, ([−,−],[−,−]), are related, since using anticommutativity [−,−]−= [−,−]. Definition 1.3.31. Let (𝑀𝜕 ←←←←←←→ 𝑁, ⋅)and (𝑀′𝜕′ ←←←←←←←←→ 𝑁′,∗) be crossed modules of Leibniz 𝐾-algebras. A homomorphism is a pair of Leibniz 𝐾-homomorphisms, 𝑓1∶𝑀←←→ 𝑀′ and 𝑓2∶𝑁←←→ 𝑁′such that 𝑓1(𝑛⋅1𝑚) = 𝑓2(𝑛) ∗1𝑓1(𝑚), 𝑓1(𝑚⋅2𝑛) = 𝑓1(𝑚) ∗2𝑓2(𝑛), 𝑛 ∈𝑁, 𝑚 ∈𝑀, and 𝜕′◦𝑓1=𝑓2◦𝜕. We will denote by X(LeibAlg𝐾)the category of crossed modules of Leibniz 𝐾- algebras and its homomorphisms. Remark 1.3.32.As in the case of groups and Lie 𝐾-algebras, we have an equivalence between the categories X(LeibAlg𝐾)and ICat(LeibAlg𝐾). A proof of this can be found in [22]. X(LieAlg𝐾)can be seen as a full subcategory of the category X(LeibAlg𝐾)using Proposition 1.3.29 (we actually have a functorial isomorphism between a full subcategory of X(LeibAlg𝐾)and X(LieAlg𝐾)). 20 1 Preliminaries Since the pullbacks in LieAlg𝐾and LeibAlg𝐾are the same, it is immediate to show that ICat(LieAlg𝐾)is a full subcategory of ICat(LeibAlg𝐾). The equivalence in the Leibniz case generalizes the equivalence in the Lie case since the bracket gives the action in the functors (which was presented in [22]), and then, it is anticommutative when we have Lie 𝐾-algebras. We only have to check that the Leibniz semidirect product generalizes the Lie semidirect product, but this is immediate from definition (since 𝑚⋅2𝑛′= −𝑛′⋅1𝑚is the Lie case). Definition 1.3.33. Let 𝑀and 𝑁be two Leibniz 𝐾-algebras and ⋅a Leibniz action of 𝑁on 𝑀. The semidirect product, denoted by 𝑀⋊𝑁, is the 𝐾-vector space 𝑀×𝑁 with the bracket [(𝑚, 𝑛),(𝑚′, 𝑛′)] ∶= ([𝑚, 𝑚′] + 𝑛⋅1𝑚′+𝑚⋅2𝑛′,[𝑛, 𝑛′]), 𝑚, 𝑚′∈𝑀, 𝑛, 𝑛′∈𝑁. 1.4 Braided semigroupal Category A bifunctor is a functor whose source category is a product category. Let 𝐹∶C×D←←→ Ebe a bifunctor. For 𝐴∈ Ob(C)and 𝐵∈ Ob(D), we denote by 𝐴𝐹and 𝐹𝐵the functors: 𝐴𝐹∶D←←→ E,𝐴𝐹(𝐷𝑓 ←←←←←←←→ 𝐷′) = 𝐹(𝐴, 𝐷)𝐹(Id𝐴,𝑓) ←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←→ 𝐹(𝐴, 𝐷′), 𝐹𝐵∶C←←→ E, 𝐹𝐵(𝐶𝑔 ←←←←←←→ 𝐶′) = 𝐹(𝐶, 𝐵)𝐹(𝑔,Id𝐵) ←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←→ 𝐹(𝐶′, 𝐵). Definition 1.4.1. Given the categories, 𝐂,𝐃and 𝐄, we have the functor 𝐴𝐂,𝐃,𝐄∶ (𝐂×𝐃) × 𝐄←←→ 𝐂× (𝐃×𝐄) defined as 𝐴𝐂,𝐃,𝐄(((𝐴, 𝐵), 𝐶)((𝑓,𝑔),ℎ) ←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←→ ((𝐴′, 𝐵′), 𝐶′))=(𝐴, (𝐵, 𝐶))(𝑓,(𝑔,ℎ)) ←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←→ (𝐴′,(𝐵′, 𝐶′)), called associator functor for the categorical product of the given categories. It is always a functor isomorphism with the obvious inverse. 1.4 Braided semigroupal Category 21 Crane and Yetter defined in [17] the notion of semigroupal category. Definition 1.4.2. Asemigroupal category is a triple = (𝐂, ⊗, 𝑎)where 𝐂is a category, ⊗∶𝐂×𝐂→𝐂is a bifunctor, and 𝑎∶⊗◦(⊗×Id𝐂)←←→ ⊗◦(Id𝐂×⊗)◦𝐴𝐂,𝐂,𝐂 is a natural isomorphism called the associator, such that for all 𝑋, 𝑌 , 𝑍, 𝑊 ∈ Ob(𝐂) the following associative coherence diagram (pentagon axiom) holds: ((𝑋 ⊗ 𝑌 )⊗ 𝑍)⊗ 𝑊 (𝑋 ⊗ (𝑌 ⊗ 𝑍)) ⊗ 𝑊 𝑋 ⊗ ((𝑌 ⊗ 𝑍)⊗ 𝑊 )𝑋 ⊗ (𝑌 ⊗ (𝑍 ⊗ 𝑊 )) (𝑋 ⊗ 𝑌 )⊗(𝑍 ⊗ 𝑊 ) 𝑎𝑋,𝑌 ,𝑍 ⊗Id𝑊 𝑎𝑋,𝑌 ⊗𝑍,𝑊 Id𝑋⊗𝑎𝑌 ,𝑍,𝑊 𝑎𝑋,𝑌 ,𝑍⊗𝑊 𝑎𝑋⊗𝑌 ,𝑍,𝑊 We say that a semigroupal category is strict if the isomorphism 𝑎is the identity morphism. In this case we have that (𝑋 ⊗ 𝑌 )⊗ 𝑍 =𝑋 ⊗ (𝑌 ⊗ 𝑍). It is known that the coherence diagram implies that any diagram made in the same way to more tensor products will be commutative. The definition of monoidal category was given in [7,45]. Definition1.4.3. Amonoidal category is a6-tuple= (𝐂, ⊗, 𝑎, 𝐼, 𝑙, 𝑟)where(𝐂, ⊗, 𝑎) is a semigroupal category, 𝐼is an object of 𝐂(called the tensor unit), and the pair 𝑙∶ (𝐼 ⊗ −) ←←→ Id𝐂,𝑟∶ (− ⊗ 𝐼)←←→ Id𝐂are natural isomorphisms (called the left and right unitors, respectively), such that for all 𝑋, 𝑌 ∈ Ob(𝐂)the unit coherence diagram (triangle equation) holds: (𝑋 ⊗ 𝐼)⊗ 𝑌 𝑎𝑋,𝐼,𝑌 // 𝑟𝑋⊗Id𝑌'' 𝑋 ⊗ (𝐼 ⊗ 𝑌 ) Id𝑋⊗𝑙𝑌 ww 𝑋 ⊗ 𝑌 We say that a monoidal category is strict if the isomorphisms 𝑎,𝑙and 𝑟are the identity morphisms. In this case (𝑋 ⊗ 𝑌 )⊗ 𝑍 =𝑋 ⊗ (𝑌 ⊗ 𝑍),𝑋 ⊗ 𝐼 =𝑋=𝐼 ⊗ 𝑋. 28 2 Braidings for crossed modules and internal objects Definition 2.2.3. Abraided internal functor between two braided categorical associative 𝐾-algebras is an internal functor (𝐹1, 𝐹0)such that 𝐹1(𝜏𝑎,𝑏) = 𝜏′ 𝐹0(𝑎),𝐹0(𝑏), where 𝜏and 𝜏′are the braidings and 𝑎, 𝑏 ∈𝐶0. We denote by BICat(AssAlg𝐾)the category of braided categorical associative 𝐾-algebras and braided internal functors between them. We will introduce the notion of braiding for crossed modules of associative algebras looking for an equivalence between braided crossed modules and braided internal categories of associative algebras, as it happens in the case of groups. Definition 2.2.4. Let (𝑀𝜕 ←←←←←←→ 𝑁, ∗= (∗1,∗2)) be a crossed module of associative 𝐾- algebras. A braiding (or Peiffer lifting) is a 𝐾-bilinear map {−,−}∶ 𝑁×𝑁←←→ 𝑀 satisfying: 𝜕{𝑛, 𝑛′}=[𝑛, 𝑛′],(BXAs1) {𝜕𝑚, 𝜕𝑚′}=[𝑚, 𝑚′],(BXAs2) {𝜕𝑚, 𝑛} = −[𝑛, 𝑚]∗,(BXAs3) {𝑛, 𝜕𝑚}=[𝑛, 𝑚]∗,(BXAs4) {𝑛, 𝑛′𝑛′′} = 𝑛′∗1{𝑛, 𝑛′′}+{𝑛, 𝑛′} ∗2𝑛′′,(BXAs5) {𝑛𝑛′, 𝑛′′} = 𝑛∗1{𝑛′, 𝑛′′}+{𝑛, 𝑛′′} ∗2𝑛′,(BXAs6) with 𝑚, 𝑚′∈𝑀,𝑛, 𝑛′, 𝑛′′ ∈𝑁. Here, [𝑛, 𝑚]∗=𝑛∗1𝑚−𝑚∗2𝑛and [𝑥, 𝑦] = 𝑥𝑦 −𝑦𝑥. (𝑀𝜕 ←←←←←←→ 𝑁, ∗,{−,−}) is a braided crossed module of associative 𝐾-algebras. Example 2.2.5. The commutator map [−,−] is a braiding on the crossed module (𝑀𝜕 ←←←←←←→ 𝑀, (∗,∗)). Definition 2.2.6. Ahomomorphism of braided crossed modules of associative 𝐾- algebras (𝑓1, 𝑓2)∶ (𝑀𝜕 ←←←←←←→ 𝑁, ⋅,{−,−}) ←←←←←←←→ (𝑀′𝜕′ ←←←←←←←←→ 𝑁′,∗,{−,−}′)is a homomorphism of crossed modules of associative 𝐾-algebras such that 𝑓1({𝑛, 𝑛′}) = {𝑓2(𝑛), 𝑓2(𝑛′)}′, 𝑛, 𝑛′∈𝑁. 2.2 Braiding for the Associative Case 29 We denote by BX(AssAlg𝐾)the category of braided crossed modules of associative 𝐾-algebras and their homomorphisms. Proposition 2.2.7. Let = (𝑀𝜕 ←←←←←←→ 𝑁, (∗1,∗2),{−,−}) be a braided crossed module of associative 𝐾-algebras. Then ∶= (𝑀⋊𝑁, 𝑁, 𝑠,  𝑡, 𝑒,  𝑘, 𝜏)is a braided categorical associative 𝐾- algebra where 𝑠,  𝑡, 𝑒,  𝑘are defined in Proposition 1.3.11 and the braiding is: 𝜏 ∶𝑁×𝑁←←→ 𝑀⋊𝑁, 𝜏𝑛,𝑛′= (−{𝑛, 𝑛′}, 𝑛𝑛′). Proof. We only need to check the braiding axioms for this internal category since (𝑀⋊𝑁, 𝑁, 𝑠,  𝑡, 𝑒,  𝑘)is a categorical associative 𝐾-algebra by Proposition 1.3.11. We will start with AsB1. Let 𝑛, 𝑛′∈𝑁. 𝑠(𝜏𝑛, 𝑛′) = 𝑠((−{𝑛, 𝑛′}, 𝑛𝑛′)) = 𝑛𝑛′,  𝑡(𝜏𝑛,𝑛′) =  𝑡((−{𝑛, 𝑛′}, 𝑛𝑛′)) = −𝜕{𝑛, 𝑛′} + 𝑛𝑛′= −[𝑛, 𝑛′] + 𝑛𝑛′=𝑛′𝑛, where we use (BXAs1). We will prove now AsB2. Let 𝑥= (𝑚, 𝑛), 𝑦 = (𝑚′, 𝑛′) ∈ 𝑀⋊𝑁. We need to show that 𝜏𝑡(𝑥),𝑡(𝑦)◦𝑥𝑦 =𝑦𝑥◦𝜏𝑠(𝑥),𝑠(𝑦). 𝜏𝑡(𝑥),𝑡(𝑦)◦𝑥𝑦 = 𝑘(((𝑚, 𝑛)(𝑚′, 𝑛′),(−{ 𝑡((𝑚, 𝑛)), 𝑡((𝑚′, 𝑛′))}, 𝑡((𝑚, 𝑛)) 𝑡((𝑚′, 𝑛′))))) = 𝑘(((𝑚, 𝑛)(𝑚′, 𝑛′),(−{𝜕𝑚 +𝑛, 𝜕𝑚′+𝑛′},(𝜕𝑚 +𝑛)(𝜕𝑚′+𝑛′)))) = 𝑘(((𝑚𝑚′+𝑛∗1𝑚′+𝑚∗2𝑛′, 𝑛𝑛′),(−{𝜕𝑚 +𝑛, 𝜕𝑚′+𝑛′}, (𝜕𝑚 +𝑛)(𝜕𝑚′+𝑛′)))) = (𝑚𝑚′+𝑛∗1𝑚′+𝑚∗2𝑛′− {𝜕𝑚 +𝑛, 𝜕𝑚′+𝑛′}, 𝑛𝑛′) = (𝑚𝑚′+𝑛∗1𝑚′+𝑚∗2𝑛′− {𝜕𝑚, 𝜕𝑚′}−{𝜕𝑚, 𝑛′}−{𝑛, 𝜕𝑚′}−{𝑛, 𝑛′}, 𝑛𝑛′) = (𝑚𝑚′+𝑛∗1𝑚′+𝑚∗2𝑛′− [𝑚, 𝑚′]+[𝑛′, 𝑚]∗− [𝑛, 𝑚′]∗− {𝑛, 𝑛′}, 𝑛𝑛′) = (𝑚′𝑚+𝑚′∗2𝑛+𝑛′∗1𝑚− {𝑛, 𝑛′}, 𝑛𝑛′), 30 2 Braidings for crossed modules and internal objects where we use (BXAs2), (BXAs3) and (BXAs4) in the sixth equality. On the other hand, 𝑦𝑥◦𝜏𝑠(𝑥),𝑠(𝑦) = 𝑘(((−{𝑠((𝑚, 𝑛)), 𝑠((𝑚, 𝑛′))}, 𝑠((𝑚, 𝑛))𝑠((𝑚′, 𝑛′))),(𝑚′, 𝑛′)(𝑚, 𝑛))) = 𝑘(((−{𝑛, 𝑛′}, 𝑛𝑛′),(𝑚′, 𝑛′)(𝑚, 𝑛))) = 𝑘(((−{𝑛, 𝑛′}, 𝑛𝑛′),(𝑚′𝑚+𝑛′∗1𝑚+𝑚′∗2𝑛, 𝑛′𝑛))) = (−{𝑛, 𝑛′} + 𝑚′𝑚+𝑛′∗1𝑚+𝑚′∗2𝑛, 𝑛𝑛′). We will verify AsB3. If 𝑛, 𝑛′, 𝑛′′ ∈𝑁, then (𝜏𝑛,𝑛′′ 𝑒(𝑛′))◦(𝑒(𝑛)𝜏𝑛′,𝑛′′ ) =  𝑘(( 𝑒(𝑛)𝜏𝑛′,𝑛′′ , 𝜏𝑛,𝑛′′ 𝑒(𝑛′))) = 𝑘((0, 𝑛)(−{𝑛′, 𝑛′′}, 𝑛′𝑛′′),(−{𝑛, 𝑛′′}, 𝑛𝑛′′)(0, 𝑛′)) = 𝑘(−𝑛∗1{𝑛′, 𝑛′′}, 𝑛(𝑛′𝑛′′),(−{𝑛, 𝑛′′} ∗2𝑛′,(𝑛𝑛′′)𝑛′)) = (−𝑛∗1{𝑛′, 𝑛′′}−{𝑛, 𝑛′′} ∗2𝑛′, 𝑛(𝑛′𝑛′′)) = (−{𝑛𝑛′, 𝑛′′},(𝑛𝑛′)𝑛′′) = 𝜏𝑛𝑛′,𝑛′′ , where we have used (BXAs6) and associativity. Finally, we will show that AsB4 is satisfied. If 𝑛, 𝑛′, 𝑛′′ ∈𝑁, then (𝑒(𝑛′)𝜏𝑛,𝑛′′ )◦(𝜏𝑛,𝑛′𝑒(𝑛′′)) =  𝑘(𝜏𝑛,𝑛′𝑒(𝑛′′), 𝑒(𝑛′)𝜏𝑛,𝑛′′ ) = 𝑘((−{𝑛, 𝑛′}, 𝑛𝑛′)(0, 𝑛′′),(0, 𝑛′)(−{𝑛, 𝑛′′}, 𝑛𝑛′′)) = 𝑘((−{𝑛, 𝑛′} ∗2𝑛′′,(𝑛𝑛′)𝑛′′),(−𝑛′∗1{𝑛, 𝑛′′}, 𝑛′(𝑛𝑛′′))) = (−𝑛′∗1{𝑛, 𝑛′′}−{𝑛, 𝑛′} ∗2𝑛′′,(𝑛𝑛′)𝑛′′) = (−{𝑛, 𝑛′𝑛′′}, 𝑛(𝑛′𝑛′′)) = 𝜏𝑛,𝑛′𝑛′′ , where we use (BXAs5) along with the associativity in the second equality. Proposition 2.2.8. We have a functor 𝔄∶BX(AssAlg𝐾)←←→ BICat(AssAlg𝐾)defined by 𝔄((𝑓1,𝑓2) ←←←←←←←←←←←←←←←←←←←←←←←←→ ′) =  (𝑓1×𝑓2,𝑓2) ←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←→ ′ where is described in the previous proposition. 2.2 Braiding for the Associative Case 31 Proof. We know that the pair (𝑓1×𝑓2, 𝑓2)is an internal functor between the respective internal categories since what we are doing is extending an existing functor (see Proposition 1.3.11) to the braided case. In the same way, as it is an extension, we only need to show that it is well defined, since it satisfies the properties of functor because the composition and identity are the same as in the categories without braiding. So, to conclude the proof, it is enough to see that (𝑓1×𝑓2, 𝑓2)is a braided internal functor of braided categorical associative 𝐾-algebras. (𝑓1×𝑓2)( 𝜏𝑛,𝑛′) = (𝑓1×𝑓2)((−{𝑛, 𝑛′}, 𝑛𝑛′)) = (−𝑓1({𝑛, 𝑛′}), 𝑓2(𝑛𝑛′)) = (−{𝑓2(𝑛), 𝑓2(𝑛′)}′, 𝑓2(𝑛)𝑓2(𝑛′)) = 𝜏′ 𝑓2(𝑛),𝑓2(𝑛′), where we use that (𝑓1, 𝑓2)is a homomorphism of braided crossed modules of associative algebras. Proposition 2.2.9. Let = (𝐶1, 𝐶0, 𝑠, 𝑡, 𝑒, 𝑘, 𝜏)be a braided categorical associative 𝐾-algebra. Then ∶= (ker(𝑠)𝜕𝑡 ←←←←←←←←→ 𝐶0,(𝑒∗,∗𝑒),{−,−}𝜏)is a braided crossed module of associative 𝐾-algebras, where (𝑒∗,∗𝑒), 𝜕𝑡are defined in Proposition 1.3.11 and the braiding is: {−,−}𝜏∶𝐶0×𝐶0←←→ ker(𝑠),{𝑎, 𝑏}𝜏∶=𝑒(𝑎𝑏) − 𝜏𝑎,𝑏. Proof. We only need to show that {−,−}𝜏is a braiding on a crossed module, since under the above assumptions, (ker(𝑠)𝜕𝑡 ←←←←←←←←→ 𝐶0,(∗𝑒,𝑒∗)) is a crossed module of associative 𝐾-algebras by Proposition 1.3.11. First, we see that it is well defined, i.e. {𝑎, 𝑏}𝜏∈ ker(𝑠)for 𝑎, 𝑏 ∈𝐶0. 𝑠({𝑎, 𝑏}𝜏) = 𝑠(𝑒(𝑎𝑏) − 𝜏𝑎,𝑏) = 𝑎𝑏 −𝑎𝑏 = 0, where we use AsB1. Now, we will check (BXAs1). If 𝑎, 𝑏 ∈𝐶0, then 𝜕𝑡{𝑎, 𝑏}𝜏=𝑡(𝑒(𝑎𝑏) − 𝜏𝑎,𝑏) = 𝑎𝑏 −𝑏𝑎 = [𝑎, 𝑏], 32 2 Braidings for crossed modules and internal objects where we have used (AsB1). We proceed to check (BXAs2). If 𝑥, 𝑦 ∈ ker(𝑠), then {𝜕𝑡𝑥, 𝜕𝑡𝑦}𝜏=𝑒(𝜕𝑡𝑥𝜕𝑡𝑦) − 𝜏𝜕𝑡𝑥,𝜕𝑡𝑦=𝑒(𝑡(𝑥)𝑡(𝑦)) − 𝜏𝑡(𝑥),𝑡(𝑦). We need to show that 𝑒(𝑡(𝑥)𝑡(𝑦)) − 𝜏𝑡(𝑥),𝑡(𝑦)= [𝑥, 𝑦]. By axiom AsB2 we know the equality 𝑘((𝑥𝑦, 𝜏𝑡(𝑥),𝑡(𝑦))) = 𝑘((𝜏𝑠(𝑥),𝑠(𝑦), 𝑦𝑥)). As 𝑥∈ ker(𝑠), we have that 𝑠(𝑥)=0(in the same way for 𝑦), and 𝜏𝑠(𝑥),𝑠(𝑦)= 0 by 𝐾-bilinearity. We have then that 𝑘((𝜏𝑠(𝑥),𝑠(𝑦), 𝑦𝑥)) = 𝑘((0, 𝑦𝑥)), and therefore the equality 𝑘((𝑥𝑦, 𝜏𝑡(𝑥),𝑡(𝑦))) = 𝑘((0, 𝑦𝑥)). Using now the 𝐾-linearity of 𝑘in the previous expression, we obtain 0 = 𝑘((𝑥𝑦, 𝜏𝑡(𝑥),𝑡(𝑦)−𝑦𝑥)). Since 𝑡(𝜏𝑡(𝑥),𝑡(𝑦)−𝑦𝑥) = 𝑡(𝑦)𝑡(𝑥)−𝑡(𝑦)𝑡(𝑥) = 0 = 𝑠(𝑒(0)) we can write 𝑘((𝜏𝑡(𝑥),𝑡(𝑦)− 𝑦𝑥, 𝑒(0))). Further 𝑘((𝜏𝑡(𝑥),𝑡(𝑦)−𝑦𝑥, 𝑒(0))) = 𝜏𝑡(𝑥),𝑡(𝑦)−𝑦𝑥 by the internal category axioms. Adding both equalities and using the 𝐾-linearity of 𝑘, we get 𝑘((𝑥𝑦 +𝜏𝑡(𝑥),𝑡(𝑦)−𝑦𝑥, 𝜏𝑡(𝑥),𝑡(𝑦)−𝑦𝑥)) = 𝜏𝑡(𝑥),𝑡(𝑦)−𝑦𝑥. Therefore, by grouping, we have 𝑘(([𝑥, 𝑦] + 𝜏𝑡(𝑥),𝑡(𝑦), 𝜏𝑡(𝑥),𝑡(𝑦)−𝑦𝑥)) = 𝜏𝑡(𝑥),𝑡(𝑦)−𝑦𝑥. As 𝑠(𝜏𝑡(𝑥),𝑡(𝑦)−𝑦𝑥) = 𝑡(𝑥)𝑡(𝑦) + 0 = 𝑡(𝑥)𝑡(𝑦)(we use that 𝑥or 𝑦are in ker(𝑠)) it makes sense to talk about the composition 𝑘((𝑒(𝑡(𝑥)𝑡(𝑦)), 𝜏𝑡(𝑥),𝑡(𝑦)−𝑦𝑥)), which is equal to 𝜏𝑡(𝑥),𝑡(𝑦)−𝑦𝑥. 2.2 Braiding for the Associative Case 33 Subtracting both equalities and using the 𝐾-linearity of 𝑘, we obtain 𝑘(([𝑥, 𝑦] + 𝜏𝑡(𝑥),𝑡(𝑦)−𝑒(𝑡(𝑥)𝑡(𝑦)),0)) = 0. Again, using the properties for internal categories, we have 0 = 𝑘(([𝑥, 𝑦] + 𝜏𝑡(𝑥),𝑡(𝑦)−𝑒(𝑡(𝑥)𝑡(𝑦)),0)) =𝑘(([𝑥, 𝑦] + 𝜏𝑡(𝑥),𝑡(𝑦)−𝑒(𝑡(𝑥)𝑡(𝑦)), 𝑒(0))) = [𝑥, 𝑦] + 𝜏𝑡(𝑥),𝑡(𝑦)−𝑒(𝑡(𝑥)𝑡(𝑦)), which gives us the required equality. As an observation to the above, in the part of the proof that is related to 𝑥, 𝑦 ∈ ker(𝑠), it is sufficient that one of the two is in that kernel. By using this fact we have the following equalities for 𝑥∈ ker(𝑠)and 𝑦∈𝐶1: 𝑒(𝑡(𝑥)𝑡(𝑦)) − 𝜏𝑡(𝑥),𝑡(𝑦)= [𝑥, 𝑦], 𝑒(𝑡(𝑦)𝑡(𝑥)) − 𝜏𝑡(𝑦),𝑡(𝑥)= [𝑦, 𝑥]. Now with these equalities, we will prove (BXAs3) and (BXAs4). Let 𝑎∈𝐶0and 𝑥∈ ker(𝑠). Then {𝜕𝑡𝑥, 𝑎}𝜏=𝑒(𝑡(𝑥)𝑡(𝑒(𝑎))) − 𝜏𝑡(𝑥),𝑡(𝑒(𝑎)) = [𝑥, 𝑒(𝑎)] = 𝑥𝑒(𝑎) − 𝑒(𝑎)𝑥=𝑥∗𝑒𝑎−𝑎𝑒∗𝑥, {𝑎, 𝜕𝑡𝑥}𝜏=𝑒(𝑡(𝑒(𝑎))𝑡(𝑥)) − 𝜏𝑡(𝑒(𝑎)),𝑡(𝑥)= [𝑒(𝑎), 𝑥] = 𝑒(𝑎)𝑥−𝑥𝑒(𝑎) = 𝑎𝑒∗𝑥−𝑥∗𝑒𝑎. We will see now the last conditions, starting with (BXAs5). Let 𝑎, 𝑏, 𝑐 ∈𝐶0. {𝑎, 𝑏𝑐}𝜏=𝑒(𝑎(𝑏𝑐)) − 𝜏𝑎,𝑏𝑐 =𝑒(𝑎(𝑏𝑐)) − ((𝑒(𝑏)𝜏𝑎,𝑐)◦(𝜏𝑎,𝑏𝑒(𝑐))) =𝑒(𝑎(𝑏𝑐)) − 𝑒(𝑏)𝜏𝑎,𝑐 −𝜏𝑎,𝑏𝑒(𝑐) + 𝑒(𝑡(𝜏𝑎,𝑏𝑒(𝑐))) =𝑒((𝑎𝑏)𝑐) − 𝑒(𝑏)𝜏𝑎,𝑐 −𝜏𝑎,𝑏𝑒(𝑐) + 𝑒((𝑏𝑎)𝑐) =𝑒(𝑏)𝑒(𝑎𝑐) − 𝑒(𝑏)𝜏𝑎,𝑐 +𝑒(𝑎𝑏)𝑒(𝑐) − 𝜏𝑎,𝑏𝑒(𝑐) =𝑒(𝑏){𝑎, 𝑐}𝜏+ {𝑎, 𝑏}𝜏𝑒(𝑐) = 𝑏𝑒∗ {𝑎, 𝑐}𝜏+ {𝑎, 𝑏}𝜏∗𝑒𝑐, where we have used (AsB4), Lemma 2.1.1 and the associativity. To conclude we will check (BXAs6). {𝑎𝑏, 𝑐}𝜏=𝑒((𝑎𝑏)𝑐) − 𝜏𝑎𝑏,𝑐 =𝑒((𝑎𝑏)𝑐) − ((𝜏𝑎,𝑐𝑒(𝑏))◦(𝑒(𝑎)𝜏𝑏,𝑐)) 34 2 Braidings for crossed modules and internal objects =𝑒((𝑎𝑏)𝑐) − 𝜏𝑎,𝑐𝑒(𝑏) − 𝑒(𝑎)𝜏𝑏,𝑐 +𝑒(𝑡(𝑒(𝑎)𝜏𝑏,𝑐)) =𝑒(𝑎(𝑏𝑐)) − 𝜏𝑎,𝑐𝑒(𝑏) − 𝑒(𝑎)𝜏𝑏,𝑐 +𝑒(𝑎(𝑐𝑏)) =𝑒(𝑎)𝑒(𝑏𝑐) − 𝑒(𝑎)𝜏𝑏,𝑐 +𝑒(𝑎𝑐)𝑒(𝑏) − 𝜏𝑎,𝑐𝑒(𝑏) =𝑒(𝑎){𝑏, 𝑐}𝜏+ {𝑎, 𝑐}𝜏𝑒(𝑏) = 𝑎𝑒∗ {𝑏, 𝑐}𝜏+ {𝑎, 𝑐}𝜏∗𝑒𝑏, where we have used (AsB3), Lemma 2.1.1 and associativity. Proposition 2.2.10. We have a functor 𝔄∶BICat(AssAlg𝐾)←←→ BX(AssAlg𝐾)defined by 𝔄((𝐹1,𝐹0) ←←←←←←←←←←←←←←←←←←←←←←←←←→ ′) =  (𝐹𝑠 1,𝐹0) ←←←←←←←←←←←←←←←←←←←←←←←←←←→ ′, where is described in the previous proposition and 𝐹𝑠 1∶ ker(𝑠)←←→ ker(𝑠′)is determined by 𝐹𝑠 1(𝑥) = 𝐹1(𝑥), with 𝑥∈ ker(𝑠). Proof. We only need to show that 𝔄can be extended to the braided case since is a functor between the categories without braiding (see Proposition 1.3.11). For this, we have to satisfy the axioms of the homomorphisms of braided crossed modules of associative 𝐾-algebras. 𝐹𝑠 1({𝑎, 𝑏}𝜏) = 𝐹1(𝑒(𝑎𝑏) − 𝜏𝑎,𝑏) = 𝐹1(𝑒(𝑎, 𝑏)) − 𝐹1(𝜏𝑎,𝑏) =𝑒′(𝐹0(𝑎𝑏)) − 𝜏′ 𝐹0(𝑎),𝐹0(𝑏)=𝑒′(𝐹0(𝑎)𝐹0(𝑏)) − 𝜏′ 𝐹0(𝑎),𝐹0(𝑏) = {𝐹0(𝑎), 𝐹0(𝑏)}𝜏′. Remark 2.2.11.Note that, if (𝑀𝜕 ←←←←←←→ 𝑁, (∗1,∗2),{−,−}) is a braided crossed module of associative 𝐾-algebras, then ker(𝑠) = {(𝑚, 0) ∈ 𝑀⋊𝑁∣𝑚∈𝑀} =∶ (𝑀, 0), where 𝑠 is defined for the functor 𝔄. Proposition 2.2.12. The categories BX(AssAlg𝐾)and BICat(AssAlg𝐾)are equivalent categories. Further, the functors 𝔄and 𝔄are inverse equivalences, where the natural isomorphisms IdBX(AssAlg𝐾) 𝛼𝔄 ≅𝔄◦𝔄and IdBICat(AssAlg𝐾) 𝛽𝔄 ≅𝔄◦𝔄are given by: ∙if = (𝑀𝜕 ←←←←←←→ 𝑁, (∗1,∗2),{−,−}) is a braided crossed module of associative 𝐾- algebras, then 𝛼𝔄 = (𝛼𝔄 𝑀,Id𝑁), with 𝛼𝔄 𝑀∶𝑀←←→ (𝑀, 0) defined as 𝛼𝑀(𝑚) = (𝑚, 0); 2.3 Braiding for the Lie Case 35 ∙if = (𝐶1, 𝐶0, 𝑠, 𝑡, 𝑒, 𝑘, 𝜏)is a braided categorical associative 𝐾-algebra, then 𝛽𝔄 = (𝛽𝔄 𝑠,Id𝐶0), with 𝛽𝔄 𝐶1∶𝐶1←←→ ker(𝑠)⋊𝐶0defined as 𝛽𝔄 𝐶1(𝑥)=(𝑥−𝑒(𝑠(𝑥)), 𝑠(𝑥)). Proof. We only need to show that 𝛼𝔄 and 𝛽𝔄 are isomorphisms between braided objects since that they are well-defined maps, isomorphisms in the categories without braiding, as well as are natural isomorphisms (see [22]). So, it is sufficient to prove that 𝛼𝔄 and 𝛽𝔄 satisfy the braided axioms, since the bijective morphisms are isomorphisms in both categories. Let = (𝑀𝜕 ←←←←←←→ 𝑁, (⋅1,⋅2),{−,−}) a braided crossed module of associative 𝐾- algebras. We will check that 𝛼𝔄 = (𝛼𝔄 𝑀,Id𝑁)is a homomorphism. Id𝑁({𝑛, 𝑛′}𝜏) = {𝑛, 𝑛′}𝜏 =𝑒(𝑛𝑛′) − 𝜏𝑛,𝑛′= (0, 𝑛𝑛′) − (−{𝑛, 𝑛′}, 𝑛𝑛′) = ({𝑛, 𝑛′},0) = 𝛼𝔄 𝑀({𝑛, 𝑛′}),where 𝑛, 𝑛′∈𝑁. Let = (𝐶1, 𝐶0, 𝑠, 𝑡, 𝑒, 𝑘, 𝜏)be a braided categorical associative 𝐾-algebra. We will check that 𝛽𝔄 = (𝛽𝔄 𝑠,Id𝐶0)is a morphism. If 𝑎, 𝑏 ∈𝐶0, we have Id𝐶0(𝜏𝑎,𝑏) = 𝜏𝑎,𝑏 = (−{𝑎, 𝑏}𝜏, 𝑎𝑏)=(𝜏𝑎,𝑏 −𝑒(𝑎𝑏), 𝑎𝑏) = (𝜏𝑎,𝑏 −𝑒(𝑠(𝜏𝑎,𝑏)), 𝑠(𝜏𝑎,𝑏)) = 𝛽𝔄 𝐶1(𝜏𝑎,𝑏). Therefore, the equivalence of categories is obtained since they are morphisms, and we know that they are natural isomorphisms. 2.3 Braiding for categorical Lie algebras and crossed modules of Lie algebras In this section, we will show that the definition given by Ulualan in [50] for braided categorical Lie 𝐾-algebras appears naturally from the previous one, using the fact that we can transform an associative 𝐾-algebra 𝑀in a Lie 𝐾-algebra 𝑀with bracket [𝑥, 𝑦] = 𝑥𝑦 −𝑦𝑥. Now, we will suppose that 𝐾is a field of char(𝐾)≠2to change a little the definition of braiding. Doing this we will obtain the definition given in [24], where 36 2 Braidings for crossed modules and internal objects the equivalence is proven with the category of braided crossed modules of Lie 𝐾- algebras when char(𝐾)≠2. The notion of braiding for categorical Lie 𝐾-algebras was introduced by Ulualan in [50]. Definition 2.3.1 ( [50]).Let = (𝐶1, 𝐶0, 𝑠, 𝑡, 𝑒, 𝑘)be a categorical Lie 𝐾-algebra. Abraiding on is a 𝐾-bilinear map 𝜏∶𝐶0×𝐶0←←→ 𝐶1,(𝑎, 𝑏)↦𝜏𝑎,𝑏, satisfying: 𝜏𝑎,𝑏 ∶ [𝑎, 𝑏]←←→ [𝑏, 𝑎],(LieT1) [𝑠(𝑥), 𝑠(𝑦)] [𝑡(𝑥), 𝑡(𝑦)] [𝑠(𝑦), 𝑠(𝑥)] [𝑡(𝑦), 𝑡(𝑥)], 𝜏𝑠(𝑥),𝑠(𝑦) [𝑥,𝑦] 𝜏𝑡(𝑥),𝑡(𝑦) [𝑦,𝑥] ,(LieT2) 𝜏[𝑎,𝑏],𝑐 = [𝜏𝑎,𝑐, 𝑒(𝑏)] + [𝑒(𝑎), 𝜏𝑏,𝑐],(LieB3) 𝜏𝑎,[𝑏,𝑐]= [𝑒(𝑏), 𝜏𝑎,𝑐]+[𝜏𝑎,𝑏, 𝑒(𝑐)],(LieB4) for 𝑎, 𝑏, 𝑐 ∈𝐶0,𝑥, 𝑦 ∈𝐶1. We say that (𝐶0, 𝐶1, 𝑠, 𝑡, 𝑒, 𝑘, 𝜏)is a braided categorical Lie 𝐾-algebra. Remark 2.3.2.The lack of associativity of the Lie bracket motivates the use of the addition in LieB3 and LieB4 instead of the composition. This choice makes sense since the source and the target are the same using the Jacobi identity. We want to show that the definition of braiding for associative 𝐾-algebras is well related with the definition of braiding for Lie 𝐾-algebras. Proposition 2.3.3. Let (𝐶1, 𝐶0, 𝑠, 𝑡, 𝑒, 𝑘, 𝜏)be a braided categorical associative 𝐾- algebra, then (𝐶 1, 𝐶 0, 𝑠, 𝑡, 𝑒, 𝑘, 𝜏𝐿𝑖𝑒)is a braided categorical Lie 𝐾-algebra, where 𝜏𝐿𝑖𝑒 ∶𝐶 0×𝐶 0←←→ 𝐶 1, 𝜏𝐿𝑖𝑒 𝑎,𝑏 ∶=𝜏𝑎,𝑏 −𝜏𝑏,𝑎. Proof. It is easy to see that AsB1 implies LieT1 and AsB2 implies LieT2. By using Lemma 2.1.1 we obtain LieB3 and LieB4 from AsB3 and AsB4, respectively. 2.3 Braiding for the Lie Case 37 Another definition for braided internal category of Lie 𝐾-algebras was given in [24] to make the equivalence with the braided crossed modules of Lie 𝐾-algebras. The equivalence was proven for a field with char(𝐾)≠2, so we will show that the two definitions are equivalent. The definition given in [24] is the following one. Definition 2.3.4 ( [24]).Let = (𝐶1, 𝐶0, 𝑠, 𝑡, 𝑒, 𝑘)be a categorical Lie 𝐾-algebra. Abraiding on is a 𝐾-bilinear map 𝜏∶𝐶0×𝐶0←←→ 𝐶1,(𝑎, 𝑏)↦𝜏𝑎,𝑏, satisfying LieT1, LieT2 and the following equalities: 𝜏[𝑎,𝑏],𝑐 =𝜏𝑎,[𝑏,𝑐]−𝜏𝑏,[𝑎,𝑐],(LieT3) 𝜏𝑎,[𝑏,𝑐]=𝜏[𝑎,𝑏],𝑐 −𝜏[𝑎,𝑐],𝑏,(LieT4) for 𝑎, 𝑏, 𝑐 ∈𝐶0. In the following proposition we show that the two definitions are equivalent when char(𝐾)≠2. Proposition 2.3.5. Let 𝐾be a field of char(𝐾)≠2and (𝐶1, 𝐶0, 𝑠, 𝑡, 𝑒, 𝑘)a categorical Lie 𝐾-algebra. If 𝜏∶𝐶0×𝐶0←←→ 𝐶1is a 𝐾-bilinear map satisfying LieT1 and LieT2, then 𝜏𝑎,[𝑏,𝑐]= [𝑒(𝑎), 𝜏𝑏,𝑐]and 𝜏[𝑏,𝑐],𝑎 = [𝜏𝑏,𝑐, 𝑒(𝑎)]. In particular, by the anticommutativity, we have that 𝜏𝑎,[𝑏,𝑐]= −𝜏[𝑏,𝑐],𝑎. Proof. Using LieT1 and LieT2, we have the following commutative diagram: [𝑎, [𝑏, 𝑐]] [𝑎, [𝑐, 𝑏]] [[𝑏, 𝑐], 𝑎] [[𝑐, 𝑏], 𝑎]. 𝜏𝑎,[𝑏,𝑐] [𝑒(𝑎),𝜏𝑏,𝑐] 𝜏𝑎,[𝑐,𝑏] [𝜏𝑏,𝑐,𝑒(𝑎)] That is, we have the equality 𝑘(([𝑒(𝑎), 𝜏𝑏,𝑐], 𝜏𝑎,[𝑐,𝑏])) = 𝑘((𝜏𝑎,[𝑏,𝑐],[𝜏𝑏,𝑐, 𝑒(𝑎)])), 44 2 Braidings for crossed modules and internal objects Example 2.4.6. If we take the as the category of vector spaces with the usual tensor product, the previous definition recovers the definition of Lie Algebra if the characteristic of the prefixed field is not 2. Since the generalization is only true for char(𝐾)≠2, we will assume it for the rest of the section. We want to explain what are an object and a morphism in Lie(𝐾). Definition 2.4.7. Let 𝑉be a 𝐾-vector space and 𝑀be a Lie 𝐾-algebra. We say that (𝑉 , ⋅)is a right 𝑀-module if ⋅∶𝑉×𝑀←←→ 𝑉is a 𝐾-bilinear map (𝑣, 𝑚)↦𝑣⋅𝑚such that: 𝑣⋅[𝑚1, 𝑚2] = (𝑣⋅𝑚1)⋅𝑚2− (𝑣⋅𝑚2)⋅𝑚1, for 𝑣∈𝑉,𝑚1, 𝑚2∈𝑀. We say that (𝑉 , ⋅)is a left 𝑀-module if ⋅∶𝑀×𝑉←←→ 𝑉is a 𝐾-bilinear map (𝑚, 𝑣)↦𝑚⋅𝑣such that: [𝑚1, 𝑚2]⋅𝑣=𝑚1⋅(𝑚2⋅𝑣) − 𝑚2⋅(𝑚1⋅𝑣), for 𝑣∈𝑉,𝑚1, 𝑚2∈𝑀. Definition 2.4.8. Let 𝛼∶𝑀←←→ 𝑁be a Lie 𝐾-homomorphism. Let (𝑉 , ⋅)be a right (resp. left) 𝑀-module and (𝑊 , ∗) a right (resp. left) 𝑁-module. A 𝐾-linear map 𝑉𝑓 ←←←←←←←→ 𝑊is (𝛼∶𝑀←←→ 𝑁, ⋅,∗)-equivariant if we have that 𝑓(𝑣⋅𝑚) = 𝑓(𝑣) ∗ 𝛼(𝑚) (resp. 𝑓(𝑚⋅𝑣) = 𝛼(𝑚) ∗ 𝑓(𝑣)),for 𝑣∈𝑉 , 𝑚 ∈𝑀. When 𝑁=𝑀and 𝛼= Id𝑀we said that 𝑓is (𝑀, ⋅,∗)-equivariant. Let (𝑉 , ⋅)be a left 𝑀-module and (𝑊 , ∗) a right 𝑁-module. A 𝐾-linear map 𝑉𝑓 ←←←←←←←→ 𝑊is (𝛼∶𝑀←←→ 𝑁, ⋅,∗)-equivariant if we have that 𝑓(𝑚⋅𝑣) = −𝑓(𝑣) ∗ 𝛼(𝑚),for 𝑣∈𝑉 , 𝑚 ∈𝑀. When 𝑁=𝑀and 𝛼= Id𝑀we said that 𝑓is (𝑀, ⋅,∗)-equivariant. 2.4 Braiding for the Leibniz Case 45 Remark 2.4.9.It is easy to check that if 𝑀is a Lie 𝐾-algebra, then it is a right and left 𝑀-module. Moreover, if ⋅is a Lie action of 𝑁in 𝑀, we have that (𝑀, ⋅)is a left 𝑁-module. Using this, we can see in [44] that a Lie object in 𝐾is the following data: Definition 2.4.10. A Lie object in 𝐾is a triple ( 𝑀 𝑁 𝑓,∗𝑀 𝑁,[−,−]𝑁) where •(𝑁, [−,−]𝑁)is a Lie 𝐾-algebra. •∗𝑀 𝑁∶𝑀×𝑁←←→ 𝑀is such that (𝑀, ∗𝑀 𝑁)is an (𝑁, [−,−]𝑁)-module. •𝑓is ((𝑁, [−,−]𝑁),∗𝑀 𝑁,[−,−]𝑁)-equivariant. As in the case of Lie 𝐾-algebras, we will denote a Lie object in 𝐾using the 𝐾-linear map on which it is defined when there is no confusion. Remark 2.4.11.The “anticommutative” property of Lie object for 𝐾allows to recover the Lie product 𝜇= (𝜇1, 𝜇2)for 𝑀 𝑁 𝑓with the maps 𝜇2= [−,−]𝑁and 𝜇1∶ (𝑀 ⊗𝑁)⊕(𝑁 ⊗𝑀)←←→ 𝑀, with 𝜇1((𝑚⊗𝑛)+(𝑛′⊗𝑚′)) = 𝑚∗𝑀 𝑁𝑛−𝑚′∗𝑀 𝑁𝑛′. Definition 2.4.12. Let 𝑀 𝑁 𝑓and 𝐿 𝐻 𝑔be Lie objects. A Lie morphism in 𝐾between them is an 𝐾morphism (𝛼1, 𝛼2)such that: •𝛼2∶𝑁←←→ 𝐻is a Lie 𝐾-homomorphism. •𝛼1∶𝑀←←→ 𝐿is an (𝛼2∶𝑁←←→ 𝐻, ∗𝑀 𝑁,∗𝐿 𝐻)-equivariant map. In [44] is shown a way to see the Leibniz 𝐾-algebras as a particular case of Lie objects in 𝐾. We show it in the next example. Example 2.4.13. Let 𝑀be a Leibniz 𝐾-algebra. We denote for 𝐼𝑀the ideal generated by elements of the form [𝑥, 𝑥]with 𝑥∈𝑀. It is evident that the quotient Leibniz 𝐾-algebra is a Lie 𝐾-algebra. We will denote its Lie bracket as [−,−], and the elements of the quotient as 𝑚with 𝑚∈𝑀. 46 2 Braidings for crossed modules and internal objects Lie(𝑀)∶=𝑀 𝐼𝑀 is known as Liesation (note that if 𝑀is a Lie 𝐾-algebra, then Lie(𝑀)is trivially naturally isomorphic to 𝑀), and it is functorial. We consider the following Lie object in 𝐾: We take 𝑀 Lie(𝑀) 𝜋𝑀where 𝜋(𝑚) = 𝑚is the natural map. It is a Lie object in 𝐾 with the following data: •𝑚∗𝑀 Lie(𝑀)𝑚′= [𝑚, 𝑚′], •[𝑚, 𝑚′]Lie(𝑀)= [𝑚, 𝑚′]∶= [𝑚, 𝑚′]. It is evident that 𝜋is (Lie(𝑀),∗𝑀 Lie(𝑀),[−,−]Lie(𝑀))-equivariant. So, we have a functor Φ∶ LeibAlg𝐾←←→ Lie(𝐾), that is trivially full. This functor is also injective on objects and morphisms, because there is a functor Ψ∶ Lie(𝐾)←←→ LeibAlg𝐾such that Ψ◦Φ = IdLeibAlg𝐾(see [44]). The functor Ψon objects is described in the following proposition. Proposition 2.4.14 ( [44]).Let 𝑀 𝑁 𝑓be a Lie object in 𝐾. Then (𝑀, [−,−]), where [𝑚, 𝑚′]∶=𝑚∗𝑀 𝑁𝑓(𝑚′), is a Leibniz 𝐾-algebra. In [23], we can see that the previous construction can be extended to crossed modules of Lie algebras in 𝐾. They did a crossed module with a right action. In this paper, we will define which is a crossed module with a left action, or simply a crossed module of Lie objects. Definition 2.4.15. Let = (C, ⊗, 𝑎, )be a braided semigroupal category where C is an additive category. If (𝐴, 𝜇𝐴)and (𝐵, 𝜇𝐵)are Lie objects, then a (left) Lie action of (𝐵, 𝜇𝐵)on (𝐴, 𝜇𝐴) is a morphism 𝑝∶𝐵 ⊗ 𝐴 ←←→ 𝐴such that 𝑝◦(𝜇𝐵⊗Id𝐴) = 𝑝◦(Id𝐵⊗𝑝)◦𝑎𝐵,𝐵,𝐴◦(Id(𝐵⊗𝐵)⊗𝐴 −(𝜏𝐵,𝐵 ⊗Id𝐴)), 𝑝◦(Id𝐵⊗𝜇𝐴)◦𝑎𝐵,𝐴,𝐴 =𝜇𝐴◦(𝑝 ⊗ Id𝐴)◦(Id(𝐵⊗𝐵)⊗𝐴 −(𝑎−1 𝐵,𝐴,𝐴◦(Id𝐵⊗𝜏𝐴,𝐴)◦𝑎𝐵,𝐴,𝐴)). 2.4 Braiding for the Leibniz Case 47 We said that ((𝐴, 𝜇𝐴)𝜕 ←←←←←←→ (𝐵, 𝜇𝐵), 𝑝)is a crossed module of Lie objects if 𝑝is a Lie action of (𝐵, 𝜇𝐵)on (𝐴, 𝜇𝐴)and 𝜕∶ (𝐴, 𝜇𝐴)←←→ (𝐵, 𝜇𝐵)is a Lie morphism such that 𝜕◦𝑝=𝜇𝐵◦(Id𝐵⊗𝜕), 𝜇𝐴=𝑝◦(𝜕 ⊗ Id𝐴). A morphism between two crossed modules of Lie objects ((𝐴, 𝜇𝐴)𝜕 ←←←←←←→ (𝐵, 𝜇𝐵), 𝑝, ) and ((𝐶, 𝜇𝐶)𝛿 ←←←←←←→ (𝐷, 𝜇𝐷), 𝑞)is a pair of Lie morphisms (𝛼, 𝛽),𝛼∶ (𝐴, 𝜇𝐴)←←→ (𝐶, 𝜇𝐶) and 𝛽∶ (𝐵, 𝜇𝐵)←←→ (𝐷, 𝜇𝐷), which satisfies the following diagrams: 𝐵 ⊗ 𝐴 𝐴 𝐷 ⊗ 𝐶 𝐶, 𝛽⊗𝛼 𝑝 𝛼 𝑞 𝐴 𝐵 𝐶 𝐷. 𝛼 𝜕 𝛽 𝛿 We have the category XLie()with the usual composition in C×Cfor pairs of morphisms of Lie morphisms. Example 2.4.16. We have that XLie(Vect𝐾)and X(LieAlg𝐾)are isomorphic categories with the usual tensor product in Vect𝐾(we assume char(𝐾)≠2). Now, we describe the category XLie(𝐾). Definition 2.4.17. Let 𝑀 𝑁 𝑓and 𝐿 𝐻 𝑔be Lie objects in 𝐾. A (left) Lie action of 𝐿 𝐻 𝑔 on 𝑀 𝑁 𝑓in 𝐾is a triple ⋅= (⋅1,⋅2, 𝜉⋅)where •⋅1∶𝐻×𝑀←←→ 𝑀is a 𝐾-bilinear map such that (𝑀, ⋅1)is a left 𝐻-module; •⋅2∶𝐻×𝑁←←→ 𝑁is a Lie action of 𝐻on 𝑁; •𝜉⋅∶𝐿×𝑁←←→ 𝑀is a 𝐾-bilinear map; such that the following properties are satisfied: 48 2 Braidings for crossed modules and internal objects •⋅1and ⋅2are compatible actions with ∗𝑀 𝑁. That is, for ℎ∈𝐻,𝑛∈𝑁,𝑚∈𝑀, we have ℎ⋅1(𝑚∗𝑀 𝑁𝑛)=(ℎ⋅1𝑚) ∗𝑀 𝑁𝑛+𝑚∗𝑀 𝑁(ℎ⋅2𝑛); •𝑓is an (𝐻, ⋅1,⋅2)-equivariant map; •𝜉⋅satisfies, for 𝑙∈𝐿,𝑛, 𝑛′∈𝑁,ℎ∈𝐻, the following equalities 𝑓(𝜉⋅(𝑙, 𝑛)) = 𝑔(𝑙)⋅2𝑛, 𝜉⋅(𝑙∗𝐿 𝐻ℎ, 𝑛) = 𝜉⋅(𝑙, ℎ ⋅2𝑛) − ℎ⋅1𝜉⋅(𝑙, 𝑛), 𝜉⋅(𝑙, [𝑛, 𝑛′]𝑁) = 𝜉⋅(𝑙, 𝑛) ∗𝑀 𝑁𝑛′−𝜉⋅(𝑙, 𝑛′) ∗𝑀 𝑁𝑛. Remark 2.4.18.An action is, in fact, a pair ⋅= (⋅1, ⋅2), with the two maps ⋅1∶ (𝐿 ⊗ 𝑁)⊕(𝐻 ⊗ 𝑁)←←→ 𝑀and ⋅2∶𝐻 ⊗ 𝑁 ←←→ 𝑁 satisfying the general properties, but we can easily obtain the previous definition taking ⋅2∶=⋅2and recovering ⋅1((𝑙 ⊗ 𝑛)+(ℎ⊗𝑚)) =∶ 𝜉⋅(𝑙, 𝑛) + ℎ⋅1𝑚. Definition 2.4.19. A crossed module of Lie objects in 𝐾is a pair ( 𝑀 𝑁 𝑓 𝜕 ←←←←←←→ 𝐿 𝐻 𝑔, ⋅) where 𝑀 𝑁 𝑓and 𝐿 𝐻 𝑔are Lie objects in 𝐾,⋅is a Lie action of 𝐿 𝐻 𝑔on 𝑀 𝑁 𝑓, and 𝜕= (𝜕1, 𝜕2)∶ 𝑀 𝑁 𝑓←←→ 𝐿 𝐻 𝑔is a Lie morphism in 𝐾such that •(𝑁, 𝐻, ⋅2, 𝜕2)is a crossed module of Lie 𝐾-algebras; •𝜕1is an (𝐻, ⋅1,∗𝐿 𝐻)-equivariant map; •𝜕1(𝜉⋅(𝑙, 𝑛)) = 𝑙∗𝐿 𝑁𝜕2(ℎ)and 𝜉⋅(𝜕1(𝑚), 𝑛) = 𝑚∗𝑀 𝑁𝑛= −𝜕2(𝑛)⋅1𝑚,ℎ∈𝐻, 𝑙∈𝐿,𝑚∈𝑀,𝑛∈𝑁. 2.4 Braiding for the Leibniz Case 49 Definition 2.4.20. Let ( 𝑀 𝑁 𝑓 𝜕 ←←←←←←→ 𝐿 𝐻 𝑔, ⋅) and ( 𝑋 𝑌 𝑘 𝛿 ←←←←←←→ 𝑉 𝑊 ℎ, ⋆) be crossed modules of Lie objects in 𝐾. A morphism of crossed modules of Lie objects in 𝐾is a pair (𝛼, 𝛽) of Lie morphisms 𝛼= (𝛼1, 𝛼2)∶ 𝑀 𝑁 𝑓←←→ 𝑋 𝑌 𝑘and 𝛽= (𝛽1, 𝛽2)∶ 𝐿 𝐻 𝑔←←→ 𝑉 𝑊 ℎsuch that •(𝛼2, 𝛽2)∶ (𝑁, 𝐻, ⋅2, 𝜕2)←←→ (𝑌 , 𝑊 , ⋆2, 𝛿2)is an homomorphism of crossed modules of Lie 𝐾-algebras; •𝛼1(𝜉⋅(𝑙, 𝑛)) = 𝜉⋆(𝛽1(𝑙), 𝛼2(𝑛)), for 𝑙∈𝐿,𝑛∈𝑁; •𝛼1is an (𝐻𝛽2 ←←←←←←←←←→ 𝑊 , ⋅1, ⋆1)-equivariant map; •𝛽1◦𝜕1=𝛿1◦𝛼1. As in the case of Leibniz 𝐾-algebras we want to have a pair of functors between the categories XLie(𝐾)and XLeibAlg𝐾. For this purpose, we give the following propositions of which we omit their proofs because they are immediate. The first is symmetrical to the construction we can see in [23] for crossed modules with right actions. Proposition2.4.21. Let (𝑀𝜕 ←←←←←←→ 𝑁, (⋅1,⋅2)) be acrossedmoduleof Leibniz 𝐾-algebras. Then ( 𝑀 𝑀 [𝑀,𝑁]𝑥 𝜋𝑀,𝑁 Lie(𝑁) 𝜋𝑁, ⋅, 𝜕)is a crossed module of Lie objects in 𝐾, where •𝑀 [𝑀,𝑁]𝑥 is the Lie 𝐾-algebra quotient of 𝑀by the ideal [𝑀, 𝑁]𝑥whose generators are [𝑚, 𝑚]for 𝑚∈𝑀and 𝑛⋅1𝑚+𝑚⋅2𝑛for 𝑛∈𝑁,𝑚∈𝑀; we denote the natural map by 𝜋𝑀∶𝑀←←→ 𝑀 [𝑀,𝑁]𝑥 , and the elements of 𝑀 [𝑀,𝑁]𝑥 by 𝑚, •⋅1∶ Lie(𝑁) × 𝑀←←→ 𝑀,(𝑛, 𝑚)↦−𝑚⋅2𝑛, •⋅2∶ Lie(𝑁) × 𝑀 [𝑀,𝑁]𝑥 ←←→ 𝑀 [𝑀,𝑁]𝑥 ,(𝑛, 𝑚)↦𝑛⋅1𝑚= −𝑚⋅2𝑛, •𝜉⋅∶𝑁×𝑀 [𝑀,𝑁]𝑥 ←←→ 𝑀,(𝑛, 𝑚)←←→ 𝑛⋅1𝑚, •𝜕1∶𝑀←←→ 𝑁,𝑚↦𝜕(𝑚), 50 2 Braidings for crossed modules and internal objects •𝜕2∶𝑀 [𝑀,𝑁]𝑥 ←←→ Lie(𝑁),𝑚↦𝜕𝑚. Remark 2.4.22.We will say that the bottom part (𝑀 [𝑀,𝑁]𝑥 𝜕2 ←←←←←←←←←→ Lie(𝑁), ⋅2)is the Liesation of the crossed module of Leibniz 𝐾-algebras. In this way we found a similar relation with the Leibniz and Lie object case. This Liesation satisfies again that applied on a crossed module of Lie 𝐾-algebras, thought as a crossed module of Leibniz 𝐾-algebras with the action (⋅,⋅−), is naturally isomorphic to itself. That occurs because, in the quotient, the second generators are null too: 𝑛⋅1𝑚+𝑚⋅2𝑛=𝑛⋅𝑚+𝑚⋅−𝑛=𝑛⋅𝑚−𝑛⋅𝑚= 0. Proposition 2.4.23. Let (𝑀 𝑁 𝑓 𝜕 ←←←←←←→ 𝐿 𝐻 𝑔, ⋅)be a crossed module of Lie objects in 𝐾, then (𝑀𝜕1 ←←←←←←←←←→ 𝐿, (⋅1,⋅2)) is a crossed module of Leibniz 𝐾-algebras, where •The Leibniz brackets are given by: [𝑚, 𝑚′] = 𝑚∗𝑀 𝑁𝑓(𝑚′), for 𝑚, 𝑚′∈𝑀and [𝑙, 𝑙′] = 𝑙∗𝐿 𝐻𝑔(𝑙′), for 𝑙, 𝑙′∈𝑀; •⋅1∶𝐿×𝑁←←→ 𝑀is defined by 𝑙⋅1𝑚=𝜉⋅(𝑙, 𝑓(𝑚)) for 𝑙∈𝐿,𝑚∈𝑀; •⋅2∶𝑀×𝐿←←→ 𝑀is defined by 𝑚 ⋅2𝑙= −𝑔(𝑙)⋅1𝑚for 𝑙∈𝐿,𝑚∈𝑀. We have the functors X(LeibAlg𝐾) 𝑋Φ//XLie(𝐾) 𝑋Ψ oosatisfying 𝑋Ψ◦𝑋Φ = IdX(LeibAlg𝐾), and so, the functor 𝑋Φis a full inclusion functor. 2.4.1 Braiding for crossed modules of Lie objects in 𝐾and crossed modules of Leibniz algebras We want to define the notion of braiding for crossed modules of Leibniz algebras. We will use the idea that the braiding for crossed module of Leibniz 𝐾-algebras must be a particular case of braiding for Lie objects in 𝐾, satisfying symmetrical properties to the previous ones. 2.4.1 Braiding for crossed modules of Lie objects in 𝐾51 Definition 2.4.24. Let = (C, ⊗, 𝑎, )be a braided semigroupal category where C is an additive category. Let = ((𝐴, 𝜇𝐴)𝜕 ←←←←←←→ (𝐵, 𝜇𝐵), 𝑝)be a crossed module of Lie objects in . A braiding (or Peiffer lifting) on is a morphism 𝔗∶𝐵 ⊗ 𝐵 ←←→ 𝐴satisfying: 𝜕◦𝔗=𝜇𝐵, 𝔗◦(𝜕 ⊗ 𝜕) = 𝜇𝐴, −𝔗◦(𝜕 ⊗ Id𝐵) = 𝑝◦𝐴,𝐵, 𝔗◦(Id𝐵⊗𝜕) = 𝑝, 𝔗◦(Id𝐵⊗𝜇𝐵)⊗ 𝑎𝐵,𝐵,𝐵 =𝔗◦(𝜇𝐵⊗Id𝐵)◦(Id(𝐵⊗𝐵)⊗𝐵 −(𝑎−1 𝐵,𝐵,𝐵◦(Id𝐵⊗𝐵,𝐵)◦𝑎𝐵,𝐵,𝐵)), 𝔗◦(𝜇𝐵⊗Id𝐵) = 𝔗◦(Id𝐵⊗𝜇𝐵)◦𝑎𝐵,𝐵,𝐵◦(Id(𝐵⊗𝐵)⊗𝐵 −(𝐵,𝐵 ⊗Id𝐵)). ((𝐴, 𝜇𝐴)𝜕 ←←←←←←→ (𝐵, 𝜇𝐵), 𝑝, 𝔗)will be called a braided crossed module of Lie objects in . A morphism (𝛼, 𝛽)∶ ((𝐴, 𝜇𝐴)𝜕 ←←←←←←→ (𝐵, 𝜇𝐵), 𝑝, 𝔗)→((𝐶, 𝜇𝐶)𝛿 ←←←←←←→ (𝐷, 𝜇𝐷), 𝑞, 𝔜) of braided crossed modules of Lie objects is a morphism of crossed modules of Lie objects in the category satisfying the following commutative diagram 𝐵 ⊗ 𝐵 𝐴 𝐷 ⊗ 𝐷 𝐵. 𝛽⊗𝛽 𝔗 𝛼 𝔜 We denote this new category as BXLie(). Example 2.4.25. As inthe previouscases, we have thatBXLie(Vect𝐾)andBX(LieAlg𝐾) are isomorphic, taking in Vect𝐾the usual tensor product. BXLie(𝐾)is described in the following definitions. Definition 2.4.26. Let =(𝑀 𝑁 𝑓 𝜕 ←←←←←←→ 𝐿 𝐻 𝑔, ⋅) be a crossed module of Lie objects in 𝐾. A braiding (or Peiffer lifting) for is given by a triple of maps 𝑇{−,−} = ({−,−}𝐿𝐻 ,{−,−}𝐻𝐿,{−,−}2) where 52 2 Braidings for crossed modules and internal objects •{−,−}2∶𝐻×𝐻←←→ 𝑁is a 𝐾-bilinear map such that (𝑁, 𝐻, ⋅2, 𝜕2,{−,−}2)is a braided crossed module of Lie 𝐾-algebras. •{−,−}𝐿𝐻 ∶𝐿×𝐻←←→ 𝑀and {−,−}𝐻𝐿 ∶𝐻×𝐿←←→ 𝑀are 𝐾-bilinear maps, which with {−,−}2satisfy the following properties for 𝑙∈𝐿,ℎ, ℎ′∈𝐻, 𝑚∈𝑀,𝑛∈𝑁: 𝑓({𝑙, ℎ}𝐿𝐻 ) = {𝑔(𝑙), ℎ}2, 𝑓({ℎ, 𝑙}𝐻𝐿) = {ℎ, 𝑔(𝑙)}2, 𝜕1{𝑙, ℎ}𝐿𝐻 =𝑙∗𝐿 𝐻ℎ, 𝜕1{ℎ, 𝑙}𝐻𝐿 = −𝑙∗𝐿 𝐻ℎ, {𝜕1(𝑚), 𝜕2(𝑛)}𝐿𝐻 =𝑚∗𝑀 𝑁𝑛, {𝜕2(𝑛), 𝜕1(𝑚)}𝐻𝐿 = −𝑚∗𝑀 𝑁𝑛, {𝜕1(𝑚), ℎ}𝐿𝐻 = −ℎ⋅1𝑚, {𝜕2(𝑛), 𝑙}𝐻𝐿 = −𝜉⋅(𝑙, 𝑛), {𝑙, 𝜕2(𝑛)} = 𝜉⋅(𝑙, 𝑛),{ℎ, 𝜕1(𝑚)} = ℎ⋅1𝑚, {𝑙, [ℎ, ℎ′]𝐻}𝐿𝐻 = {𝑙∗𝐿 𝐻ℎ, ℎ′}𝐿𝐻 − {𝑙∗𝐿 𝐻ℎ′, ℎ}𝐿𝐻 , {[ℎ, ℎ′]𝐻, 𝑙}𝐻𝐿 = −{ℎ, 𝑙 ∗𝐿 𝐻ℎ′}𝐻𝐿 − {𝑙∗𝐿 𝐻ℎ, ℎ′}𝐿𝐻 , {𝑙, [ℎ, ℎ′]𝐻}𝐿𝐻 = {𝑙∗𝐿 𝐻ℎ, ℎ′}𝐿𝐻 + {ℎ, 𝑙 ∗𝐿 𝐻ℎ′}𝐻𝐿, {[ℎ, ℎ′]𝐻, 𝑙}𝐻𝐿 = −{ℎ, 𝑙 ∗𝐿 𝐻ℎ′}𝐻𝐿 + {ℎ′, 𝑙 ∗𝐿 𝐻ℎ}𝐻𝐿. We will say that ( 𝑀 𝑁 𝑓 𝜕 ←←←←←←→ 𝐿 𝐻 𝑔, ⋅, 𝑇{−,−}) is a braided crossed module of Lie objects in 𝐾. Remark 2.4.27.A braiding is a pair 𝑇{−,−} = (𝑇1 {−,−}, 𝑇 2 {−,−}), but for simplicity we denote 𝑇1 {−,−} ∶ (𝐿⊗𝐻)⊕(𝐻 ⊗𝐿)←←→ 𝑀with 𝑇1 {−,−}((𝑙⊗ℎ)+(ℎ′⊗𝑙′)) = {𝑙, ℎ}𝐿𝐻 + {ℎ′, 𝑙′}𝐻𝐿 and 𝑇2 {−,−}(ℎ, ℎ′) = {ℎ, ℎ′}2. Definition 2.4.28. Let ( 𝑀 𝑁 𝑓 𝜕 ←←←←←←→ 𝐿 𝐻 𝑔, ⋅, 𝑇{−,−}) and ( 𝑋 𝑌 𝑘 𝛿 ←←←←←←→ 𝑉 𝑊 ℎ, ⋆, 𝑇{−,−}′) be two braided crossed modules of Lie objects in 𝐾. A morphism of braided crossed modules of Lie objects in 𝐾is a morphism (𝛼, 𝛽)of crossed modules of Lie objects in 𝐾 satisfying: •(𝛼2, 𝛽2)∶ (𝑁, 𝐻, ⋅2, 𝜕2,{−,−}2)←←→ (𝑌 , 𝑊 , ⋆2, 𝛿2,{−,−}′ 2)is an morphism of braided crossed modules of Lie 𝐾-algebras, 2.4.1 Braiding for crossed modules of Lie objects in 𝐾53 •𝛼1({𝑙, ℎ}𝐿𝐻 ) = {𝛽1(𝑙), 𝛽2(ℎ)}′ 𝑉 𝑊 , for 𝑙∈𝐿,ℎ∈𝐻, •𝛼1({ℎ, 𝑙}𝐻𝐿)={𝛽2(ℎ), 𝛽1(𝑙)}′ 𝑊 𝑉 , for 𝑙∈𝐿,ℎ∈𝐻. We want to use the concept of braiding on crossed modules of Lie objects in 𝐾to obtain a definition for crossed modules of Leibniz 𝐾-algebras. For that, we will take a braiding on ( 𝑀 𝑀 [𝑀,𝑁]𝑥 𝜋𝑀 𝜕 ←←←←←←→ 𝑁 Lie(𝑁) 𝜋𝑁, ⋅). If we try to take one 𝐾-bilinear map {−,−} we would find problems with the way of defining the corresponding maps because we have that the first properties add one more quotient that we would like to be trivial for Lie 𝐾-algebras, or if we take it to be trivial, the rest of properties prevent it from being made for the general case of Leibniz 𝐾-algebras (if we take {𝑛, 𝑛′}𝑁Lie(𝑁)= {𝑛, 𝑛′}={𝑛, 𝑛′}Lie(𝑁)𝑁for example, the third and fourth property leads us to prove that 𝑀must be Lie 𝐾-algebra). For this, as in the case of the two actions, we will take for braiding two 𝐾- bilinear maps {−,−},⟨−,−⟩∶𝑁×𝑁←←→ 𝑀, and define {𝑛, 𝑛′}𝑁Lie(𝑁)= {𝑛, 𝑛′}, {𝑛, 𝑛′}Lie(𝑁)𝑁= −⟨𝑛′, 𝑛⟩and {𝑛, 𝑛′}2= {𝑛, 𝑛′} = −⟨𝑛′, 𝑛⟩, where we can see that we introduce a new quotient in 𝑀. Definition 2.4.29. Let = (𝑀𝜕 ←←←←←←→ 𝑁, (⋅1,⋅2)) be a crossed module of Leibniz 𝐾- algebras. A braiding (or Peiffer lifting) on is a pair ({−,−},⟨−,−⟩)of 𝐾-bilinear maps {−,−},⟨−,−⟩∶𝑁×𝑁←←→ 𝑀,(𝑛, 𝑛′)↦{𝑛, 𝑛′}and (𝑛, 𝑛′)↦⟨𝑛, 𝑛′⟩, satisfying: 𝜕{𝑛, 𝑛′} = [𝑛, 𝑛′] = 𝜕⟨𝑛, 𝑛′⟩,(BXLeib1) {𝜕𝑚, 𝜕𝑚′} = [𝑚, 𝑚′] = ⟨𝜕𝑚, 𝜕𝑚′⟩,(BXLeib2) {𝜕𝑚, 𝑛} = 𝑚⋅2𝑛=⟨𝜕𝑚, 𝑛⟩,(BXLeib3) {𝑛, 𝜕𝑚} = 𝑛⋅1𝑚=⟨𝑛, 𝜕𝑚⟩,(BXLeib4) {𝑛, [𝑛′, 𝑛′′]} = {[𝑛, 𝑛′], 𝑛′′}−{[𝑛, 𝑛′′], 𝑛′},(BXLeib5) ⟨𝑛, [𝑛′, 𝑛′′]⟩= {[𝑛, 𝑛′], 𝑛′′}−⟨[𝑛, 𝑛′′], 𝑛′⟩,(BXLeib6) {𝑛, [𝑛′, 𝑛′′]} = {[𝑛, 𝑛′], 𝑛′′}−⟨[𝑛, 𝑛′′], 𝑛′⟩,(BXLeib7) ⟨𝑛, [𝑛′, 𝑛′′]⟩=⟨[𝑛, 𝑛′], 𝑛′′⟩−⟨[𝑛, 𝑛′′], 𝑛′⟩,(BXLeib8) 60 2 Braidings for crossed modules and internal objects (𝐴×𝐶𝐵)⊗(𝐴×𝐶𝐵)𝐴 ⊗ 𝐴 𝐴×𝐶𝐵 𝐴 𝐵 ⊗ 𝐵 𝐵 𝐶. 𝜇𝐴×𝐶𝐵 𝜋𝐵⊗𝜋𝐵 𝜋𝐴⊗𝜋𝐴 𝜇𝐴 𝜋𝐴 𝜋𝐵𝑓 𝜇𝐵𝑔 It is straightforward to see that 𝜇𝐴×𝐶𝐵is well defined. Now, we will prove that (𝐴×𝐶𝐵, 𝜇𝐴×𝐶𝐵)is a Lie object checking the first axiom. For simplicity of notation, we will denote 𝐷∶=𝐴×𝐶𝐵. Let 𝑋∈ {𝐴, 𝐵}. Using universal properties we have: 𝜋𝑋◦(−𝜇𝐷◦𝐷,𝐷) = −𝜋𝑋◦𝜇𝐷◦𝐷,𝐷 = −𝜇𝑋◦(𝜋𝑋⊗ 𝜋𝑋)◦𝐷,𝐷. Since is a natural isomorphism and that (𝑋, 𝜇𝑋)is a Lie object, we get 𝜋𝑋◦(−𝜇𝐷◦𝐷,𝐷) = −𝜇𝑋◦𝑋,𝑋◦(𝜋𝑋⊗ 𝜋𝑋) = 𝜇𝑋◦(𝜋𝑋⊗ 𝜋𝑋). We conclude that 𝜇𝐷= −𝜇𝐷◦𝐷,𝐷 because 𝜇𝐷is the unique morphism that satisfies the previous equality for 𝑋∈ {𝐴, 𝐵}. Now, we will check the second axiom of Lie object. For the first summand, we have: 𝜋𝑋◦𝜇𝐷◦(Id𝐷⊗𝜇𝐷)◦𝑎𝐷,𝐷,𝐷 =𝜇𝑋◦(𝜋𝑋⊗ 𝜋𝑋)◦(Id𝐷⊗𝜇𝐷)◦𝑎𝐷,𝐷,𝐷 =𝜇𝑋◦(𝜋𝑋⊗(𝜋𝑋◦𝜇𝐷))◦𝑎𝐷,𝐷,𝐷 =𝜇𝑋◦(𝜋𝑋⊗(𝜇𝑋◦(𝜋𝑋⊗ 𝜋𝑋)))◦𝑎𝐷,𝐷,𝐷 =𝜇𝑋◦(Id𝑋⊗𝜇𝑋)◦(𝜋𝑋⊗(𝜋𝑋⊗ 𝜋𝑋))◦𝑎𝐷,𝐷,𝐷. Using that 𝑎is a natural isomorphism, we have 𝜋𝑋◦𝜇𝐷◦(Id𝐷⊗𝜇𝐷)◦𝑎𝐷,𝐷,𝐷 =𝜇𝑋◦(Id𝑋⊗𝜇𝑋)◦𝑎𝑋,𝑋,𝑋◦((𝜋𝑋⊗ 𝜋𝑋)⊗ 𝜋𝑋). Doing the same for the second and third summands (the naturalness of 𝑎gives the same naturalness to 𝑎−1), we have that: 𝜋𝑋◦𝜇𝐷◦(𝜇𝐷⊗Id𝐷)◦𝑎−1 𝐷,𝐷,𝐷◦(Id𝐷⊗𝐷,𝐷)◦𝑎𝐷,𝐷,𝐷 =𝜇𝑋◦(𝜇𝑋⊗Id𝑋)◦𝑎−1 𝑋,𝑋,𝑋◦(Id𝑋⊗𝑋,𝑋)◦𝑎𝑋,𝑋,𝑋◦((𝜋𝑋⊗ 𝜋𝑋)⊗ 𝜋𝑋), 2.4.2 Braiding for categorical Lie objects in 𝐾61 𝜋𝑋◦(−𝜇𝐷◦(𝜇𝐷⊗Id𝐷)) = −𝜇𝑋◦(𝜇𝑋⊗Id𝑋)◦((𝜋𝑋⊗ 𝜋𝑋)⊗ 𝜋𝑋). Adding the three last equalities and using the distributivity of the composition, we have 𝜋𝑋◦𝐷=𝑋◦((𝜋𝑋⊗ 𝜋𝑋)⊗ 𝜋𝑋), where denote by 𝑌the morphism that is in the first term of the equality of the second axiom for a Lie object (𝑌 , 𝜇𝑌). Since (𝑋, 𝜇𝑋)is a Lie object, we get 𝑋=(𝑋⊗𝑋)⊗𝑋0𝑋, and so 𝜋𝑋◦𝐷=(𝐷⊗𝐷)⊗𝐷0𝑋. Now, by the universal property, we have that 𝐷=(𝐷⊗𝐷)⊗𝐷0𝐷and therefore (𝐷, 𝜇𝐷)is a Lie object. To conclude the proof it is enough to check that the morphism given by the pullback in Cis a Lie morphism, but this is a routine verification. Definition 2.4.42. Let = (C, ⊗, 𝑎, )be a braided semigroupal category where C is an additive category with pullbacks. Let ℭ= ((𝐶1, 𝜇𝐶1),(𝐶0, 𝜇𝐶0), 𝑠, 𝑡, 𝑒, 𝑘)be a categorical Lie object in Lie(). A braiding on ℭis a morphism 𝜏∶𝐶0⊗ 𝐶0←←→ 𝐶1satisfying: •𝑠◦𝜏=𝜇𝐶0and 𝑡◦𝜏=𝜇𝐶0◦𝐶0,𝐶0, •We define 𝐶0⊗𝐶0 𝜇𝐶1×𝐶0(𝜏◦(𝑡⊗𝑡)),(𝜏◦(𝑠⊗𝑠))×𝐶0(𝜇𝐶1◦) ←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←→ 𝐶1×𝐶0𝐶1as the two unique morphisms which satisfy the universal property, respectively, in the following diagrams: 𝐶1⊗ 𝐶1 𝐶1×𝐶0𝐶1𝐶1 𝐶1𝐶0 𝜇𝐶1 𝜏◦(𝑡⊗𝑡) 𝜋1 𝜋2𝑡 𝑠 𝐶1⊗ 𝐶1 𝐶1×𝐶0𝐶1𝐶1 𝐶1𝐶0 𝜏◦(𝑠⊗𝑠) 𝜇𝐶1◦𝐶1,𝐶1 𝜋1 𝜋2𝑡 𝑠 , and the equality 𝑘◦(𝜇𝐶1×𝐶0(𝜏◦(𝑡 ⊗ 𝑡))) = 𝑘◦((𝜏◦(𝑠⊗𝑠)) ×𝐶0(𝜇𝐶1◦)). 62 2 Braidings for crossed modules and internal objects •It must satisfy 𝜏◦(Id𝐶0⊗𝜇𝐶0)⊗ 𝑎𝐶0=𝜏◦(𝜇𝐶0⊗Id𝐶0)◦(Id(𝐶0⊗𝐶0)⊗𝐶0−(𝑎−1 𝐶0 ◦(Id𝐶0⊗𝐶0)◦𝑎𝐶0)), 𝜏◦(𝜇𝐶0⊗Id𝐶0) = 𝜏◦(Id𝐶0⊗𝜇𝐶0)◦𝑎𝐶0◦(Id(𝐶0⊗𝐶0)⊗𝐶0−(𝐶0⊗Id𝐶0)). We denote 𝑎𝐶0,𝐶0,𝐶0=∶ 𝑎𝐶0and 𝐶0,𝐶0=∶ 𝐶0. We will say that ((𝐶1, 𝜇𝐶1),(𝐶0, 𝜇𝐶0), 𝑠, 𝑡, 𝑒, 𝑘, 𝜏)is a braided categorical Lie object in . An internal functor ((𝐶1, 𝜇𝐶1),(𝐶0, 𝜇𝐶0), 𝑠, 𝑡, 𝑒, 𝑘, 𝜏)(𝐹1,𝐹0) ←←←←←←←←←←←←←←←←←←←←←←←←←→ ((𝐶′ 1, 𝜇𝐶′ 1),(𝐶′ 0, 𝜇𝐶′ 0), 𝑠′, 𝑡′, 𝑒′, 𝑘′, 𝜏′) is said to be a braided internal functor of braided categorical Lie objects in if it satisfies the following diagram: 𝐶0⊗ 𝐶0𝐶1 𝐶′ 0⊗ 𝐶′ 0𝐶1. 𝐹0⊗𝐹0 𝜏 𝐹1 𝜏′ We denote this new category as BICat(Lie()). Example 2.4.43. Wehavethat thecategoriesBICat(Lie(Vect𝐾)) and BICat(LieAlg𝐾) are isomorphic, taking in Vect𝐾the usual tensor product (we assume char(𝐾)≠2). Definition 2.4.44. Let =( 𝐶1 𝐷1 𝑓1, 𝐶0 𝐷0 𝑓0, 𝑠, 𝑡, 𝑒, 𝑘) be a categorical Lie object in 𝐾. A braiding on is a triple 𝜏= (𝜏𝐶0,𝐷0, 𝜏𝐷0,𝐶0, 𝜏2)where •𝜏2∶𝐷0×𝐷0←←→ 𝐷1is a 𝐾-bilinear map such that (𝐷1, 𝐷0, 𝑠, 𝑡, 𝑒, 𝑘, 𝜏2)is a braided crossed module of Lie 𝐾-algebras, •𝜏𝐷0,𝐶0∶𝐷0×𝐶0←←→ 𝐶1and 𝜏𝐶0,𝐷0∶𝐶0×𝐷0←←→ 𝐶1are 𝐾-bilinear maps which, with 𝜏2, satisfy the following properties for 𝑐∈𝐶0,𝑑, 𝑑′∈𝐷0,𝑥∈𝐶1, 𝑦∈𝐷1: 𝑓1(𝜏𝐶0,𝐷0 𝑐,𝑑 ) = 𝜏2 𝑓0(𝑐),𝑑 and 𝑓1(𝜏𝐷0,𝐶0 𝑑,𝑐 ) = 𝜏2 𝑑,𝑓0(𝑐), 2.4.2 Braiding for categorical Lie objects in 𝐾63 𝜏𝐶0,𝐷0 𝑐,𝑑 ∶𝑐∗𝐶0 𝐷0𝑑←←→ −𝑐∗𝐶0 𝐷0𝑑and 𝜏𝐷0,𝐶0 𝑑,𝑐 ∶ − 𝑐∗𝐶0 𝐷0𝑑←←→ 𝑐∗𝐶0 𝐷0𝑑. The following diagrams are satisfied in the internal category: 𝑠1(𝑥) ∗𝐶0 𝐷0𝑠2(𝑦)𝑡1(𝑥) ∗𝐶0 𝐷0𝑡2(𝑦) −𝑠1(𝑥) ∗𝐶0 𝐷0𝑠2(𝑦) −𝑡1(𝑥) ∗𝐶0 𝐷0𝑡2(𝑦) 𝜏𝐶0,𝐷0 𝑠1(𝑥),𝑠2(𝑦) 𝑥∗𝐶1 𝐶0𝑦 𝜏𝐶0,𝐷0 𝑡1(𝑥),𝑡2(𝑦) −𝑥∗𝐶1 𝐷1𝑦 , −𝑠1(𝑥) ∗𝐶0 𝐷0𝑠2(𝑦) −𝑡1(𝑥) ∗𝐶0 𝐷0𝑡2(𝑦) 𝑠1(𝑥) ∗𝐶0 𝐷0𝑠2(𝑦)𝑡1(𝑥) ∗𝐶0 𝐷0𝑡2(𝑦) 𝜏𝐷0,𝐶0 𝑠2(𝑦),𝑠1(𝑥) −𝑥∗𝐶1 𝐶0𝑦 𝜏𝐷0,𝐶0 𝑡2(𝑦),𝑡1(𝑥) 𝑥∗𝐶1 𝐷1𝑦 . Moreover, we have the following properties: 𝜏𝐶0,𝐷0 𝑐,[𝑑,𝑑′]𝐷0 =𝜏𝐶0,𝐷0 𝑐∗𝐶0 𝐷0𝑑,𝑑′−𝜏𝐶0,𝐷0 𝑐∗𝐶0 𝐷0𝑑′,𝑑, 𝜏𝐷0,𝐶0 [𝑑,𝑑′]𝐷0,𝑐 = −𝜏𝐷0,𝐶0 𝑑,𝑐∗𝐶0 𝐷0𝑑′−𝜏𝐶0,𝐷0 𝑐∗𝐶0 𝐷0𝑑,𝑑′, 𝜏𝐶0,𝐷0 𝑐,[𝑑,𝑑′]𝐷0 =𝜏𝐶0,𝐷0 𝑐∗𝐶0 𝐷0𝑑,𝑑′+𝜏𝐷0,𝐶0 𝑑,𝑐∗𝐶0 𝐷0𝑑′, 𝜏𝐷0,𝐶0 [𝑑,𝑑′]𝐷0,𝑐 = −𝜏𝐷0,𝐶0 𝑑,𝑐∗𝐶0 𝐷0𝑑′+𝜏𝐷0,𝐶0 𝑑′,𝑐∗𝐶0 𝐷0𝑑. We will say that ( 𝐶1 𝐷1 𝑓1, 𝐶0 𝐷0 𝑓0, 𝑠, 𝑡, 𝑒, 𝑘, 𝜏) is a braided categorical Lie object in 𝐾. Remark 2.4.45.A braiding is a pair 𝜏= (𝜏1, 𝜏2)but, for simplicity, we take for the definition 𝜏2(𝑑, 𝑑′) = 𝜏2 𝑑,𝑑′and 𝜏1∶ (𝐶0⊗ 𝐷0)⊕(𝐷0⊗ 𝐶0)←←→ 𝐶1by the expression 𝜏1((𝑐 ⊗ 𝑑)+(𝑑′⊗ 𝑐′)) = 𝜏𝐶0,𝐷0 𝑐,𝑑 +𝜏𝐷0,𝐶0 𝑑′,𝑐′. Definition2.4.46. Let ( 𝐶1 𝐷1 𝑓1, 𝐶0 𝐷0 𝑓0, 𝑠, 𝑡, 𝑒, 𝑘, 𝜏) and ( 𝐶′ 1 𝐷′ 1 𝑔1, 𝐶′ 0 𝐷′ 0 𝑔0, 𝑠′, 𝑡′, 𝑒′, 𝑘′, 𝜓) be braided categorical Lie objects in 𝐾. A braided internal functor between categorical Lie 64 2 Braidings for crossed modules and internal objects objects in 𝐾is an internal functor ((𝐹1 1, 𝐹0 1),(𝐹1 0, 𝐹0 0)) between the respective categorical Lie objects which satisfies: •(𝐹0 1, 𝐹0 0)∶ (𝐷1, 𝐷0, 𝑠2, 𝑡2, 𝑒2, 𝑘2, 𝜏2)←←→ (𝐷′ 1, 𝐷′ 0, 𝑠′ 2, 𝑡′ 2, 𝑒′ 2, 𝑘′ 2, 𝜓2)is a braided internal functor between categorical Lie 𝐾-algebras. •𝐹1 1(𝜏𝐶0,𝐷0 𝑐,𝑑 ) = 𝜓𝐶′ 0,𝐷′ 0 𝐹1 0(𝑐),𝐹0 0(𝑑)for 𝑐∈𝐶0,𝑑∈𝐷0. •𝐹1 1(𝜏𝐷0,𝐶0 𝑑,𝑐 ) = 𝜓𝐷′ 0,𝐶′ 0 𝐹0 0(𝑑),𝐹1 0(𝑐)for 𝑐∈𝐶0,𝑑∈𝐷0. To introduce a braiding for the categorical Leibniz 𝐾-algebras with the previous scheme, we will use two 𝐾-bilinear maps 𝜏, 𝜓 ∶𝐶0×𝐶0←←→ 𝐶1, as in the case of a braiding of crossed modules of Leibniz 𝐾-algebras. Consider for the inclusion Lie object in 𝐾the braiding 𝜏 defined by 𝜏𝐶0,Lie(𝐶0) 𝑎,𝑏 =𝜏𝑎,𝑏,𝜏Lie(𝐶0),𝐶0 𝑎,𝑏 = −𝜓𝑏,𝑎 and 𝜏2 𝑎,𝑏 =𝜏𝑎,𝑏 = −𝜓𝑏,𝑎, where we introduce a quotient in 𝐶1whose elements we will denote as 𝑥. Definition 2.4.47. A braiding for the categorical Leibniz 𝐾-algebra (𝐶1, 𝐶0, 𝑠, 𝑡, 𝑒, 𝑘) is a pair (𝜏, 𝜓)of 𝐾-bilinear maps 𝜏, 𝜓 ∶𝐶0×𝐶0←←→ 𝐶1,(𝑎, 𝑏)↦𝜏𝑎,𝑏 and (𝑎, 𝑏)↦𝜓𝑎,𝑏, satisfying: 𝜏𝑎,𝑏 ∶ [𝑎, 𝑏]←←→ −[𝑎, 𝑏]and 𝜓𝑎,𝑏 ∶ [𝑎, 𝑏]←←→ −[𝑎, 𝑏],(LeibT1) [𝑠(𝑥), 𝑠(𝑦)] [𝑡(𝑥), 𝑡(𝑦)] −[𝑠(𝑥), 𝑠(𝑦)] −[𝑡(𝑥), 𝑡(𝑦)], 𝜏𝑠(𝑥),𝑠(𝑦) [𝑥,𝑦] 𝜏𝑡(𝑥),𝑡(𝑦) −[𝑥,𝑦] [𝑠(𝑥), 𝑠(𝑦)] [𝑡(𝑥), 𝑡(𝑦)] −[𝑠(𝑥), 𝑠(𝑦)] −[𝑡(𝑥), 𝑡(𝑦)], 𝜓𝑠(𝑥),𝑠(𝑦) [𝑥,𝑦] 𝜓𝑡(𝑥),𝑡(𝑦) −[𝑥,𝑦] (LeibT2) 𝜏𝑎,[𝑏,𝑐]=𝜏[𝑎,𝑏],𝑐 −𝜏[𝑎,𝑐],𝑏,(LeibT3) 𝜓𝑎,[𝑏,𝑐]=𝜏[𝑎,𝑏],𝑐 −𝜓[𝑎,𝑐],𝑏,(LeibT4) 𝜏𝑎,[𝑏,𝑐]=𝜏[𝑎,𝑏],𝑐 −𝜓[𝑎,𝑐],𝑏,(LeibT5) 𝜓𝑎,[𝑏,𝑐]=𝜓[𝑎,𝑏],𝑐 −𝜓[𝑎,𝑐],𝑏, 𝑎, 𝑏, 𝑐 ∈𝐶0, 𝑥, 𝑦 ∈𝐶1.(LeibT6) We will saythat (𝐶1, 𝐶0, 𝑠, 𝑡, 𝑒, 𝑘, (𝜏, 𝜓)) is a braided categoricalLeibniz𝐾-algebra. 2.4.2 Braiding for categorical Lie objects in 𝐾65 Definition 2.4.48. Let (𝐶1, 𝐶0, 𝑠, 𝑡, 𝑒, 𝑘, (𝜏, 𝜓)) and (𝐶′ 1, 𝐶′ 0, 𝑠′, 𝑡′, 𝑒′, 𝑘′,(𝜏′, 𝜓′)) be two braided categorical Leibniz 𝐾-algebras. An internal functor (𝐶1, 𝐶0, 𝑠, 𝑡, 𝑒, 𝑘)(𝐹1,𝐹0) ←←←←←←←←←←←←←←←←←←←←←←←←←→ (𝐶′ 1, 𝐶′ 0, 𝑠′, 𝑡′, 𝑒′, 𝑘′)is said to be a braided internal functor between two braided categorical Leibniz 𝐾-algebras if it satisfies: 𝐹1(𝜏𝑎,𝑏) = 𝜏′ 𝐹0(𝑎),𝐹0(𝑏),(LeibHT1) 𝐹1(𝜓𝑎,𝑏) = 𝜓′ 𝐹0(𝑎),𝐹0(𝑏), 𝑎, 𝑏 ∈𝐶0.(LeibHT2) We denote the category of braided categorical Leibniz 𝐾-algebras and braided internal functors between them as BICat(LeibAlg𝐾). We want to see the braided categorical Lie 𝐾-algebras as a particular case of braided categorical Leibniz 𝐾-algebras. Proposition 2.4.49. Let 𝐶1and 𝐶0be Lie 𝐾-algebras. Then, (𝐶1, 𝐶0, 𝑠, 𝑡, 𝑒, 𝑘, 𝜏)is a braided categorical Lie 𝐾-algebra if and only if (𝐶1, 𝐶0, 𝑠, 𝑡, 𝑒, 𝑘, (𝜏, 𝜏−)) is a braided categorical Leibniz 𝐾-algebra. 𝜏−∶𝐶0×𝐶0←←→ 𝐶1is defined as 𝜏− 𝑎,𝑏 = −𝜏𝑏,𝑎. Proof. (LeibT1) and (LeibT2) can be rewritten as (LieT1) and LieT2, respectively, using the anticommutativity. Moreover, it is clear that LeibT3 and LieT4 are identical, and that (LeibT6) is equivalent to LieT3. To see the last equivalences, (LeibT4) with (LieT3), and (LeibT5) with (BXLie4), we must prove 𝜏[𝑎,𝑏],𝑐 = −𝜏𝑐,[𝑎,𝑏], for 𝑎, 𝑏, 𝑐 ∈𝐶0. •In the Lie case, it is true using Proposition 2.3.5. •In the Leibniz case it is not true in general, because we need 𝜏𝑎,𝑏 = −𝜓𝑏,𝑎; but using (LeibT4) and (LeibT5) we can observe that 𝜏𝑎,[𝑏,𝑐]=𝜓𝑎,[𝑏,𝑐]=𝜏− 𝑎,[𝑏,𝑐]= −𝜏[𝑏,𝑐],𝑎. Proposition 2.4.50. Let (𝐶1, 𝐶0, 𝑠, 𝑡, 𝑒, 𝑘, (𝜏, 𝜓)) be a braided categorical Leibniz 𝐾- algebra. Then 66 2 Braidings for crossed modules and internal objects ( 𝐶1 𝐶1 [𝜏𝐶0,𝐶0] 𝜋𝐶1, 𝐶0 Lie(𝐶0) 𝜋𝐶0,(𝑠, 𝑠),(𝑡, 𝑡),(𝑒, 𝑒),(𝑘, 𝑘), 𝜏)is a braided categorical Lie object in 𝐾, where 𝐶1 [𝜏𝐶0,𝐶0]is the Lie 𝐾-algebra which is a Leibniz quotient of 𝐶1by the ideal generated by elements of the form [𝑥, 𝑥]and 𝜏𝑎,𝑏 +𝜓𝑏,𝑎,𝑥∈𝐶1,𝑎, 𝑏 ∈𝐶0; and the maps are the following ones: •𝑠∶𝐶1 [𝜏𝐶0,𝐶0]←←→ Lie(𝐶0)defined as 𝑠(𝑥) = 𝑠(𝑥)for 𝑥∈𝐶1 [𝜏𝐶0,𝐶0]; •𝑡∶𝐶1 [𝜏𝐶0,𝐶0]←←→ Lie(𝐶0)defined as 𝑡(𝑥) = 𝑡(𝑥)for 𝑥∈𝐶1 [𝜏𝐶0,𝐶0]; •𝑒∶ Lie(𝐶0)←←→ 𝐶1 [𝜏𝐶0,𝐶0]defined as 𝑒(𝑎) = 𝑒(𝑎)for 𝑎∈ Lie(𝐶0); • 𝑘∶𝐶1 [𝜏𝐶0,𝐶0]×Lie(𝐶0) 𝐶1 [𝜏𝐶0,𝐶0]←←→ 𝐶1 [𝜏𝐶0,𝐶0]defined as 𝑘((𝑥, 𝑦)) =  𝑘(𝑥, 𝑦)for (𝑥, 𝑦) ∈ 𝐶1 [𝜏𝐶0,𝐶0]×Lie(𝐶0) 𝐶1 [𝜏𝐶0,𝐶0], where  𝑘is again the extension to the product  𝑘(𝑥, 𝑦) = 𝑥+𝑦−𝑒(𝑠(𝑦)) (we can take  𝑘′(𝑥, 𝑦) = 𝑥+𝑦−𝑒(𝑡(𝑥)) too, because in the quotient it will not change anything); •𝜏𝐶0,Lie(𝐶0)∶𝐶0× Lie(𝐶0)←←→ 𝐶1defined as 𝜏𝐶0,Lie(𝐶0) 𝑎,𝑏 =𝜏𝑎,𝑏 for 𝑎∈𝐶0,𝑏∈ Lie(𝐶0); •𝜏Lie(𝐶0),𝐶0∶ Lie(𝐶0) × 𝐶0←←→ 𝐶1defined as 𝜏Lie(𝐶0),𝐶0 𝑎,𝑏 = −𝜓𝑏,𝑎 for 𝑎∈ Lie(𝐶0), 𝑏∈𝐶0; •𝜏2∶ Lie(𝐶0) × Lie(𝐶0)←←→ 𝐶1 [𝜏𝐶0,𝐶0]defined as 𝜏2 𝑎,𝑏 =𝜏𝑎,𝑏 = −𝜓𝑏,𝑎 for 𝑎, 𝑏 ∈ Lie(𝐶0). Remark 2.4.51.The bottom part (𝐶1 [𝜏𝐶0,𝐶0],Lie(𝐶0), 𝑠, 𝑡, 𝑒, 𝑘, 𝜏2)will be called Liesation, and it is again functorial. If we apply this Liesation on a braided categorical Lie 𝐾-algebra, thought as a crossed module of Leibniz 𝐾-algebras with the action with the braiding (𝜏, 𝜏−), the new generator is null 𝜏𝑎,𝑏 +𝜓𝑏,𝑎 =𝜏𝑎,𝑏 +𝜏− 𝑏,𝑎 =𝜏𝑎,𝑏 −𝜏𝑎,𝑏 = 0. 2.4.3 The equivalence between the categories of braided crossed modules and braided internal categories in the case of Leibniz algebras67 Proposition 2.4.52. Let ( 𝐶1 𝐷1 𝑓1, 𝐶0 𝐷0 𝑓0, 𝑠, 𝑡, 𝑒, 𝑘, 𝜏)be a braided categorical Lie object in 𝐾. Then (𝐶1, 𝐶0, 𝑠1, 𝑡1, 𝑒1, 𝑘1,(𝜏𝜏, 𝜓𝜏)) is a braided categorical Leibniz 𝐾-algebra, where [𝑥, 𝑦]𝐶1=𝑥∗𝐶1 𝐷1𝑦and [𝑎, 𝑏]𝐶0=𝑎∗𝐶0 𝐷0𝑏for 𝑥, 𝑦 ∈𝐶1, 𝑎, 𝑏 ∈𝐶0, and 𝜏𝜏 𝑎,𝑏 =𝜏𝐶0,𝐷0 𝑎,𝑓0(𝑏),𝜓𝜏 𝑎,𝑏 = −𝜏𝐷0,𝐶0 𝑓(𝑏),𝑎 for 𝑎, 𝑏 ∈𝐶0. We have again the pair of functors BICat(LeibAlg𝐾) 𝐵𝐼Φ//BICat(Lie(𝐾)) 𝐵𝐼Ψ oo satisfying 𝐵𝐼Ψ◦𝐵𝐼Φ = IdBICat(LeibAlg𝐾), and so, the functor 𝐵𝐼Φis a full inclusion functor. 2.4.3 The equivalence between the categories of braided crossed modules and braided internal categories in the case of Leibniz algebras First, we will prove that BICat(LeibAlg𝐾)and BX(LeibAlg𝐾)are equivalent, as in the case of groups and Lie 𝐾-algebras. Moreover, the equivalence must generalize the Lie 𝐾-algebras case (i.e. the braidings of the Leibniz 𝐾-algebras must satisfy {𝑛, 𝑛′} = −⟨𝑛′, 𝑛⟩and 𝜏𝑎,𝑏 = −𝜓𝑏,𝑎 and the functors for the Lie case would be recovered) and must be an extension of the one given to the non-braiding case. Proposition 2.4.53. Let = (𝑀𝜕 ←←←←←←→ 𝑁, (⋅1,⋅2)({−,−},⟨−,−⟩)) be a braided crossed module of Leibniz 𝐾-algebras. Then ∶= (𝑀⋊𝑁, 𝑁, 𝑠,  𝑡, 𝑒,  𝑘, (𝜏, 𝜓)) is a braided categorical Leibniz 𝐾- algebra where •𝑠∶𝑀⋊𝑁←←→ 𝑁,𝑠((𝑚, 𝑛)) = 𝑛, • 𝑡∶𝑀⋊𝑁←←→ 𝑁, 𝑡((𝑚, 𝑛)) = 𝜕𝑚 +𝑛, •𝑒∶𝑁←←→ 𝑀⋊𝑁,𝑒(𝑛) = (0, 𝑛), • 𝑘∶ (𝑀⋊𝑁) ×𝑁(𝑀⋊𝑁)←←→ 𝑀⋊𝑁, where the source is the pullback of  𝑡 with 𝑠, defined as 𝑘(((𝑚, 𝑛),(𝑚′, 𝜕𝑚 +𝑛))) = (𝑚+𝑚′, 𝑛), 68 2 Braidings for crossed modules and internal objects •𝜏 ∶𝑁×𝑁←←→ 𝑀⋊𝑁,𝜏𝑛,𝑛′= (−2{𝑛, 𝑛′},[𝑛, 𝑛′]), •𝜓 ∶𝑁×𝑁←←→ 𝑀⋊𝑁,𝜓𝑛,𝑛′= (−2⟨𝑛, 𝑛′⟩,[𝑛, 𝑛′]). Proof. We only need to check the braiding axioms, since (𝑀⋊𝑁, 𝑁, 𝑠,  𝑡, 𝑒,  𝑘)is a categorical Leibniz 𝐾-algebra (see [22]). We will start with (LeibT1). Let 𝑛, 𝑛′∈𝑁. 𝑠(𝜏𝑛,𝑛′) = 𝑠((−2{𝑛, 𝑛′},[𝑛, 𝑛′])) = [𝑛, 𝑛′],  𝑡(𝜏𝑛,𝑛′) =  𝑡((−2{𝑛, 𝑛′},[𝑛, 𝑛′])) = −2𝜕{𝑛, 𝑛′}+[𝑛, 𝑛′] = −2[𝑛, 𝑛′]+[𝑛, 𝑛′] = −[𝑛, 𝑛′], where we use (BXLeib1). In the same way we can prove this property of 𝜓 by the symmetry of the construction. We will prove now (LeibT2). Again, we will only check this for 𝜏. Let 𝑥= (𝑚, 𝑛), 𝑦 = (𝑚′, 𝑛′) ∈ 𝑀⋊𝑁. We need to show that 𝜏𝑡(𝑥),𝑡(𝑦)◦[𝑥, 𝑦] = −[𝑥, 𝑦]◦𝜏𝑠(𝑥),𝑠(𝑦). Now, we will write the equalities in function of the data given by the braided crossed module. 𝜏𝑡(𝑥),𝑡(𝑦)◦[𝑥, 𝑦] = 𝑘(([(𝑚, 𝑛),(𝑚′, 𝑛′)],(−2{ 𝑡((𝑚, 𝑛)), 𝑡((𝑚′, 𝑛′))},[ 𝑡((𝑚, 𝑛)), 𝑡((𝑚′, 𝑛′))]))) = 𝑘(([(𝑚, 𝑛),(𝑚′, 𝑛′)],(−2{𝜕𝑚 +𝑛, 𝜕𝑚′+𝑛′},[𝜕𝑚 +𝑛, 𝜕𝑚′+𝑛′]))) = 𝑘((([𝑚, 𝑚′] + 𝑛⋅1𝑚′+𝑚⋅2𝑛′,[𝑛, 𝑛′]),(−2{𝜕𝑚 +𝑛, 𝜕𝑚′+𝑛′},[𝜕𝑚 +𝑛, 𝜕𝑚′+𝑛′]))) = ([𝑚, 𝑚′] + 𝑛⋅1𝑚′+𝑚⋅2𝑛′− 2{𝜕𝑚 +𝑛, 𝜕𝑚′+𝑛′},[𝑛, 𝑛′]) = ([𝑚, 𝑚′] + 𝑛⋅1𝑚′+𝑚⋅2𝑛′− 2{𝜕𝑚, 𝜕𝑚′} − 2{𝜕𝑚, 𝑛′} − 2{𝑛, 𝜕𝑚′} − 2{𝑛, 𝑛′},[𝑛, 𝑛′]) = ([𝑚, 𝑚′] + 𝑛⋅1𝑚′+𝑚⋅2𝑛′− 2[𝑚, 𝑚′] − 2(𝑚⋅2𝑛′) − 2(𝑛⋅1𝑚′) − 2{𝑛, 𝑛′},[𝑛, 𝑛′]) = (−[𝑚, 𝑚′] − 𝑛⋅1𝑚′−𝑚⋅2𝑛′− 2{𝑛, 𝑛′},[𝑛, 𝑛′]), where we use (BXLeib2), (BXLeib3) and (BXLeib4) in the sixth equality. In the other way, − [𝑥, 𝑦]◦𝜏𝑠(𝑥),𝑠(𝑦) = 𝑘(((−2{𝑠((𝑚, 𝑛)), 𝑠((𝑚, 𝑛′))},[𝑠((𝑚, 𝑛)), 𝑠((𝑚′, 𝑛′))]),−[(𝑚, 𝑛),(𝑚′, 𝑛′)])) 2.4.3 Equivalence between categories 69 = 𝑘(((−2{𝑛, 𝑛′},[𝑛, 𝑛′]),−[(𝑚, 𝑛),(𝑚′, 𝑛′)])) = 𝑘(((−2{𝑛, 𝑛′},[𝑛, 𝑛′]),(−[𝑚, 𝑚′] − 𝑛⋅1𝑚′−𝑚⋅2𝑛′,−[𝑛, 𝑛′]))) = (−2{𝑛, 𝑛′}−[𝑚, 𝑚′] − 𝑛⋅1𝑚′−𝑚⋅2𝑛′,[𝑛, 𝑛′]). We will verify (LeibT3) below. Let 𝑛, 𝑛′, 𝑛′′ ∈𝑁. Then 𝜏𝑛,[𝑛′,𝑛′′]= (−2{𝑛, [𝑛′, 𝑛′′]},[𝑛, [𝑛′, 𝑛′′]]) = (−2({[𝑛, 𝑛′], 𝑛′′} − {[𝑛, 𝑛′′], 𝑛′}),[[𝑛, 𝑛′], 𝑛′′] − [[𝑛, 𝑛′′], 𝑛′]) = (−2{[𝑛, 𝑛′], 𝑛′′},[[𝑛, 𝑛′], 𝑛′′]) − (−2{[𝑛, 𝑛′′], 𝑛′},[[𝑛, 𝑛′′], 𝑛′]) =𝜏[𝑛,𝑛′],𝑛′′ −𝜏[𝑛,𝑛′′],𝑛′, where we use (BXLeib5) and the Leibniz identity in the second equality. The same argument is valid for (LeibT6), using (BXLeib8) and by the symmetry of the properties. Finally, we will show that (LeibT4) and (LeibT5) are satisfied. 𝜓𝑛,[𝑛′,𝑛′′]= (−2⟨𝑛, [𝑛′, 𝑛′′]⟩,[𝑛, [𝑛′, 𝑛′′]]) = (−2({[𝑛, 𝑛′], 𝑛′′} − ⟨[𝑛, 𝑛′′], 𝑛′⟩),[[𝑛, 𝑛′], 𝑛′′] − [[𝑛, 𝑛′′], 𝑛′]) = (−2{[𝑛, 𝑛′], 𝑛′′},[[𝑛, 𝑛′], 𝑛′′]) − (−2⟨[𝑛, 𝑛′′], 𝑛′⟩,[[𝑛, 𝑛′′], 𝑛′]) =𝜏[𝑛,𝑛′],𝑛′′ −𝜓[𝑛,𝑛′′],𝑛′ = (−2({[𝑛, 𝑛′], 𝑛′′} − ⟨[𝑛, 𝑛′′], 𝑛′⟩),[[𝑛, 𝑛′], 𝑛′′] − [[𝑛, 𝑛′′], 𝑛′]) = (−2{𝑛, [𝑛′, 𝑛′′]},[𝑛, [𝑛′, 𝑛′′]]) = 𝜏𝑛,[𝑛′,𝑛′′], where we use (BXLeib6) along with the Leibniz identity in the second equality; and (BXLeib7) with the Leibniz identity in the penultimate equality. Remark 2.4.54.Note that if is a braided crossed module of Lie 𝐾-algebras, then 𝜏𝑛,𝑛′= (−2{𝑛, 𝑛′},[𝑛, 𝑛′]) = −(−2⟨𝑛′, 𝑛⟩,[𝑛′, 𝑛]) = − 𝜓𝑛′,𝑛 and we recover the construction for the Lie case (see [24]). 76 2 Braidings for crossed modules and internal objects The following proposition is given for a general case in [10, 11], using actions which are denominated compatible actions for make the tensor product. Proposition 2.5.2 ( [10,11]).Let 𝐺be a group. Then (𝐺 ⊗ 𝐺 𝜕 ←←←←←←→ 𝐺, ⋅)is a crossed module of groups where 𝐺 ⊗ 𝐺 is the non-abelian tensor product of 𝐺with itself using the conjugation action. The action ⋅∶𝐺× (𝐺 ⊗ 𝐺)←←→ (𝐺 ⊗ 𝐺)and the map 𝜕∶𝐺 ⊗ 𝐺 ←←→ 𝐺are defined on generators as 𝑔⋅(𝑔1⊗ 𝑔2) = 𝑔𝑔1𝑔−1 ⊗ 𝑔𝑔2𝑔−1 and 𝜕(𝑔1⊗ 𝑔2)=[𝑔1, 𝑔2]. The next example shows that this crossed module can be associated with a natural braiding (see [26]). Example 2.5.3. Let 𝐺be a group. The map {−,−}∶ 𝐺×𝐺←←→ 𝐺 ⊗ 𝐺 defined as {𝑔1, 𝑔2} = 𝑔1⊗ 𝑔2is a braiding on (𝐺 ⊗ 𝐺 𝜕 ←←←←←←→ 𝐺, ⋅). Using the properties of the non-abelian tensor product of groups (see [47, Proposition 1.2.3]) and the definition, the result follows easily. Once given the example in groups, we look for its analogue in Lie 𝐾-algebras. For this we need the concept of non-abelian tensor product of Lie 𝐾-algebras, introduced by Ellis in [19]. Definition 2.5.4. Let 𝑀and 𝑁be two Lie 𝐾-algebras such that 𝑀acts in 𝑁by ⋅ and 𝑁acts in 𝑀with ∗. The non-abelian tensor product, denoted by 𝑀 ⊗ 𝑁, is the Lie 𝐾-algebra generated by the symbols 𝑚⊗𝑛, where 𝑚∈𝑀,𝑛∈𝑁, with the relations 𝜆(𝑚 ⊗ 𝑛) = 𝜆𝑚 ⊗ 𝑛 =𝑚 ⊗ 𝜆𝑛, (T1) (𝑚+𝑚′)⊗ 𝑛 =𝑚⊗𝑛+𝑚′⊗ 𝑛, (T2) 𝑚 ⊗ (𝑛+𝑛′) = 𝑚⊗𝑛+𝑚⊗𝑛′, [𝑚, 𝑚′]⊗ 𝑛 =𝑚 ⊗ (𝑚′⋅𝑛) − 𝑚′⊗(𝑚⋅𝑛),(T3) 𝑚 ⊗ [𝑛, 𝑛′]=(𝑛′∗𝑚)⊗ 𝑛 − (𝑛∗𝑚)⊗ 𝑛′, [(𝑚 ⊗ 𝑛),(𝑚′⊗ 𝑛′)] = −(𝑛∗𝑚)⊗(𝑚′⋅𝑛′),(T4) where 𝑚, 𝑚′∈𝑀,𝑛, 𝑛′∈𝑁,𝜆∈𝐾. 2.5 The non-abelian tensor product as example of braiding 77 The next proposition, following the pattern of the case of groups, was proved more generally in [19], but we restrict ourselves to the case that interests us. Proposition 2.5.5 ( [19]).Let 𝑀be a Lie 𝐾-algebra. Then (𝑀 ⊗ 𝑀 𝜕 ←←←←←←→ 𝑀, ⋅)is a crossed module of Lie 𝐾-algebras, where 𝑀 ⊗ 𝑀 is the non-abelian tensor product of 𝑀with itself using the adjoint action. The action ⋅∶𝑀× (𝑀 ⊗ 𝑀)←←→ (𝑀 ⊗ 𝑀)and the map 𝜕∶𝑀 ⊗ 𝑀 ←←→ 𝑀 are defined on generators as 𝑚⋅(𝑚1⊗ 𝑚2)=[𝑚, 𝑚1]⊗ 𝑚2+𝑚1⊗[𝑚, 𝑚2]and 𝜕(𝑚1⊗ 𝑚2) = [𝑚1, 𝑚2], where [−,−] is the bracket of 𝑀. Remark 2.5.6.We will rewrite, for clarity, the relations (T3) and (T4) for the case of 𝑀 ⊗ 𝑀 with the adjoint action of 𝑀on itself. (T3) [𝑚1, 𝑚2]⊗ 𝑚3=𝑚1⊗[𝑚2, 𝑚3] − 𝑚2⊗[𝑚1, 𝑚3], 𝑚1⊗[𝑚2, 𝑚3] = [𝑚3, 𝑚1]⊗ 𝑚2− [𝑚2, 𝑚1]⊗ 𝑚3, (T4) [(𝑚1⊗ 𝑚2),(𝑚3⊗ 𝑚4)] = [𝑚1, 𝑚2]⊗[𝑚3, 𝑚4], where 𝑚1, 𝑚2, 𝑚3, 𝑚4∈𝑀. For the last relation we use the anticommutativity. Now, we show an example for the case of Lie 𝐾-algebras analogous to the case of groups. Example 2.5.7. Let 𝑀be a Lie 𝐾-algebra. The 𝐾-bilinear map {−,−}∶ 𝑀×𝑀←←→ 𝑀 ⊗ 𝑀 defined by {𝑚1, 𝑚2} = 𝑚1⊗ 𝑚2is a braiding on the crossed module of Lie 𝐾-algebras (𝑀 ⊗ 𝑀 𝜕 ←←←←←←→ 𝑀, ⋅). We will check (BXLie1). If 𝑚, 𝑚′∈𝑀, then 𝜕{𝑚, 𝑚′} = 𝜕(𝑚⊗𝑚′) = [𝑚, 𝑚′]. To check (BXLie2), we willworkon generators by the 𝐾-linearity and 𝐾-bilinearity, since the general case is only a sum of them. If 𝑚1⊗ 𝑚2and 𝑚3⊗ 𝑚4are generators of 𝑀 ⊗ 𝑀, then {𝜕(𝑚1⊗ 𝑚2), 𝜕(𝑚3⊗ 𝑚4)} = {[𝑚1, 𝑚2],[𝑚3, 𝑚4]} = [𝑚1, 𝑚2]⊗[𝑚3, 𝑚4] = [(𝑚1⊗ 𝑚2),(𝑚3⊗ 𝑚4)], 78 2 Braidings for crossed modules and internal objects where the last equality is given by (T4). For the following properties we need a previous result. We will use (T3) to prove 𝑚1⊗[𝑚2, 𝑚3] = −[𝑚2, 𝑚3]⊗ 𝑚1. [𝑚1, 𝑚2]⊗ 𝑚3=𝑚1⊗[𝑚2, 𝑚3] − 𝑚2⊗[𝑚1, 𝑚3] =𝑚1⊗[𝑚2, 𝑚3]−[𝑚3, 𝑚2]⊗ 𝑚1+ [𝑚1, 𝑚2]⊗ 𝑚3. Simplifying we have 0 = 𝑚1⊗[𝑚2, 𝑚3]+[𝑚2, 𝑚3]⊗ 𝑚1. Now, we will show (BXLie3). Let 𝑚∈𝑀and 𝑚1⊗ 𝑚2∈𝑀 ⊗ 𝑀. {𝜕(𝑚1⊗ 𝑚2), 𝑚} = {[𝑚1, 𝑚2], 𝑚} = [𝑚1, 𝑚2]⊗ 𝑚 =𝑚1⊗[𝑚2, 𝑚] − 𝑚2⊗[𝑚1, 𝑚] = −𝑚1⊗[𝑚, 𝑚2] + 𝑚2⊗[𝑚, 𝑚1] = −𝑚1⊗[𝑚, 𝑚2]−[𝑚, 𝑚1]⊗ 𝑚2= −𝑚⋅(𝑚1⊗ 𝑚2), where we use (T3) together with the previous result. Now, we will verify (BXLie4). {𝑚, 𝜕(𝑚1⊗ 𝑚2)} = 𝑚 ⊗ [𝑚1, 𝑚2] = −[𝑚1, 𝑚2]⊗ 𝑚 = −{𝜕(𝑚1⊗ 𝑚2), 𝑚} = −(−𝑚⋅(𝑚1⊗ 𝑚2)) = 𝑚⋅(𝑚1⊗ 𝑚2), where we use (BXLie3) and 𝑚 ⊗ [𝑚1, 𝑚2] = −[𝑚1, 𝑚2]⊗ 𝑚. Now, we will verify (BXLie5) and (BXLie6). Let 𝑚, 𝑚′, 𝑚′′ ∈𝑀. {𝑚, [𝑚′, 𝑚′′]} = 𝑚 ⊗ [𝑚′, 𝑚′′] = [𝑚′′, 𝑚]⊗ 𝑚′− [𝑚′, 𝑚]⊗ 𝑚′′ = [𝑚, 𝑚′]⊗ 𝑚′′ − [𝑚, 𝑚′′]⊗ 𝑚′= {[𝑚, 𝑚′], 𝑚′′} − {[𝑚, 𝑚′′], 𝑚′}, {[𝑚, 𝑚′], 𝑚′′} = [𝑚, 𝑚′]⊗ 𝑚′′ =𝑚 ⊗ [𝑚′, 𝑚′′] − 𝑚′⊗[𝑚, 𝑚′′] = {𝑚, [𝑚′, 𝑚′′]} − {𝑚′,[𝑚, 𝑚′′]}. We use (T3) in the second equality of both chains of equalities. So, we have shown that {𝑚, 𝑚′} = 𝑚⊗𝑚′is a braiding. Remark 2.5.8.Note that the action given in the previous example is actually given by 𝑚⋅(𝑚1⊗ 𝑚2) = 𝑚 ⊗ [𝑚1, 𝑚2]. 2.5 The non-abelian tensor product as example of braiding 79 The non-abelian tensor product of Leibniz 𝐾-algebras was introduced by Gnedbaye in [32], where the tensor product is denoted as 𝑀 ⋆ 𝑁, and its generators as 𝑚∗𝑛and 𝑛∗𝑚. In the general case it does not give rise to confusion, but in the case 𝑀=𝑁these generators would be denoted in the same way, giving rise to confusion. To avoid this, we change the nomenclature, meaning 𝑚∗𝑛as 𝑚 ⊗ 𝑛 and 𝑛∗𝑚as 𝑛⊛𝑚. Definition 2.5.9. Let 𝑀and 𝑁two Leibniz 𝐾-algebras together with two Leibniz actions ⋅= (⋅1,⋅2)of 𝑀on 𝑁and ∗= (∗1,∗2)of 𝑁on 𝑀. The non-abelian tensor product of 𝑀and 𝑁, denoted by 𝑀 ⋆ 𝑁, is the Leibniz 𝐾-algebra generated by the symbols 𝑚⊗𝑛and 𝑛⊛𝑚with 𝑚∈𝑀,𝑛∈𝑁, together with the relations: 𝜆(𝑚 ⊗ 𝑛) = 𝜆𝑚 ⊗ 𝑛 =𝑚 ⊗ 𝜆𝑛, (RTLeib1) 𝜆(𝑛 ⊛ 𝑚) = 𝜆𝑛 ⊛ 𝑚 =𝑛 ⊛ 𝜆𝑚, (𝑚+𝑚′)⊗ 𝑛 =𝑚⊗𝑛+𝑚′⊗ 𝑛, (RTLeib2) 𝑚 ⊗ (𝑛+𝑛′) = 𝑚⊗𝑛+𝑚⊗𝑛′, (𝑛+𝑛′)⊛ 𝑚 =𝑛⊛𝑚+𝑛′⊛ 𝑚, 𝑛 ⊛ (𝑚+𝑚′) = 𝑛⊛𝑚+𝑛⊛𝑚′, 𝑚 ⊗ [𝑛, 𝑛′] = (𝑚∗2𝑛)⊗ 𝑛′− (𝑚∗2𝑛′)⊗ 𝑛, (RTLeib3) 𝑛 ⊛ [𝑚, 𝑚′] = (𝑛⋅2𝑚)⊛ 𝑚′− (𝑛⋅2𝑚′)⊛ 𝑚, [𝑚, 𝑚′]⊗ 𝑛 = (𝑚⋅1𝑛)⊛ 𝑚′−𝑚 ⊗ (𝑛⋅2𝑚′), [𝑛, 𝑛′]⊛ 𝑚 = (𝑛∗1𝑚)⊗ 𝑛′−𝑛 ⊛ (𝑚∗2𝑛′), 𝑚 ⊗ (𝑚′⋅1𝑛)=−𝑚 ⊗ (𝑛⋅2𝑚′),(RTLeib4) 𝑛 ⊛ (𝑛′∗1𝑚) = −𝑛 ⊛ (𝑚∗2𝑛′), (𝑚∗2𝑛)⊗(𝑚′⋅1𝑛′) = [𝑚 ⊗ 𝑛, 𝑚′⊗ 𝑛′] = (𝑚⋅1𝑛)⊛(𝑚′∗2𝑛′),(RTLeib5) 80 2 Braidings for crossed modules and internal objects (𝑚∗2𝑛)⊗(𝑛′⋅2𝑚′) = [𝑚 ⊗ 𝑛, 𝑛′⊛ 𝑚′] = (𝑚⋅1𝑛)⊛(𝑛′∗1𝑚′), (𝑛∗1𝑚)⊗(𝑛′⋅2𝑚′) = [𝑛 ⊛ 𝑚, 𝑛′⊛ 𝑚′] = (𝑛⋅2𝑚)⊛(𝑛′∗1𝑚′), (𝑛∗1𝑚)⊗(𝑚′⋅1𝑛′) = [𝑛 ⊛ 𝑚, 𝑚′⊗ 𝑛′] = (𝑛⋅2𝑚)⊛(𝑚′∗2𝑛′), 𝑚, 𝑚′∈𝑀, 𝑛, 𝑛′∈𝑁. Proposition 2.5.10 ( [32]).Let 𝑀be a Leibniz 𝐾-algebra. Then (𝑀 ⋆ 𝑀 𝜕 ←←←←←←→ 𝑀, (⋅1,⋅2)) is a crossed module of Leibniz 𝐾-algebras, where 𝑀 ⋆ 𝑀 is the non-abelian tensor product of 𝑀with itself using the actions given by the Leibniz bracket, where •the left action on generators is given by 𝑚⋅1(𝑚1⊗ 𝑚2) = [𝑚, 𝑚1]⊗ 𝑚2− [𝑚, 𝑚2]⊛ 𝑚1,𝑚⋅1(𝑚1⊛ 𝑚2) = [𝑚, 𝑚1]⊛ 𝑚2− [𝑚, 𝑚2]⊗ 𝑚1; •the right action on generators is given by (𝑚1⊗ 𝑚2)⋅2𝑚= [𝑚1, 𝑚]⊗ 𝑚2+ 𝑚1⊗[𝑚2, 𝑚],(𝑚1⊛ 𝑚2)⋅2𝑚= [𝑚1, 𝑚]⊛ 𝑚2+𝑚1⊛[𝑚2, 𝑚]; •the map 𝜕is defined on generators as 𝜕(𝑚1⊗ 𝑚2) = [𝑚1, 𝑚2] = 𝜕(𝑚1⊛ 𝑚2). Remark 2.5.11.We will show how are the relations (RTLeib3)–(RTLeib5) for the non-abelian tensor product 𝑀 ⋆ 𝑀 with the action ([−,−],[−,−]) on itself: 𝑚1⊗[𝑚2, 𝑚3] = [𝑚1, 𝑚2]⊗ 𝑚3− [𝑚1, 𝑚3]⊗ 𝑚2,(RTLeib3) 𝑚1⊛[𝑚2, 𝑚3] = [𝑚1, 𝑚2]⊛ 𝑚3− [𝑚1, 𝑚3]⊛ 𝑚2, [𝑚1, 𝑚2]⊗ 𝑚3= [𝑚1, 𝑚3]⊛ 𝑚2−𝑚1⊗[𝑚3, 𝑚2], [𝑚1, 𝑚2]⊛ 𝑚3= [𝑚1, 𝑚3]⊗ 𝑚2−𝑚1⊛[𝑚3, 𝑚2], 𝑚1⊗[𝑚2, 𝑚3] = −𝑚1⊗[𝑚3, 𝑚2],(RTLeib4) 𝑚1⊛[𝑚2, 𝑚3] = −𝑚1⊛[𝑚3, 𝑚2], [𝑚1, 𝑚2]⊗[𝑚3, 𝑚4] = [𝑚1⊗ 𝑚2, 𝑚3⊗ 𝑚4] = [𝑚1, 𝑚2]⊛[𝑚3, 𝑚4],(RTLeib5) [𝑚1, 𝑚2]⊗[𝑚3, 𝑚4] = [𝑚1⊗ 𝑚2, 𝑚3⊛ 𝑚4] = [𝑚1, 𝑚2]⊛[𝑚3, 𝑚4], [𝑚1, 𝑚2]⊗[𝑚3, 𝑚4] = [𝑚1⊛ 𝑚2, 𝑚3⊛ 𝑚4] = [𝑚1, 𝑚2]⊛[𝑚3, 𝑚4], [𝑚1, 𝑚2]⊗[𝑚3, 𝑚4] = [𝑚1⊛ 𝑚2, 𝑚3⊗ 𝑚4] = [𝑚1, 𝑚2]⊛[𝑚3, 𝑚4], 𝑚1, 𝑚2, 𝑚3, 𝑚4∈𝑀. 2.5 The non-abelian tensor product as example of braiding 81 The following example shows the necessity of a pair of braidings for the Leibniz 𝐾-algebras case since they will be different. Example 2.5.12. Let 𝑀be a Leibniz 𝐾-algebra. The pair of 𝐾-bilinear maps {−,−},⟨−,−⟩∶𝑀×𝑀←←→ 𝑀 ⋆ 𝑀 defined as {𝑚1, 𝑚2} = 𝑚1⊗ 𝑚2and ⟨𝑚1, 𝑚2⟩=𝑚1⊛ 𝑚2is a braiding on the crossed module of Leibniz 𝐾-algebras (𝑀 ⋆ 𝑀 𝜕 ←←←←←←→ 𝑀, (⋅1,⋅2)). First, will check (BXLeib1). 𝜕{𝑚1, 𝑚2} = 𝜕(𝑚1⊗ 𝑚2) = [𝑚1, 𝑚2] = 𝜕(𝑚1⊛ 𝑚2) = 𝜕⟨𝑚1, 𝑚2⟩, 𝑚1, 𝑚2∈𝑀. Now, we will prove (BXLeib2). {𝜕(𝑚1⊗ 𝑚2), 𝜕(𝑚3⊗ 𝑚4)} = [𝑚1, 𝑚2]⊗[𝑚3, 𝑚4] = [𝑚1⊗ 𝑚2, 𝑚3⊗ 𝑚4], {𝜕(𝑚1⊗ 𝑚2), 𝜕(𝑚3⊛ 𝑚4)} = [𝑚1, 𝑚2]⊗[𝑚3, 𝑚4]=[𝑚1⊗ 𝑚2, 𝑚3⊛ 𝑚4], {𝜕(𝑚1⊛ 𝑚2), 𝜕(𝑚3⊗ 𝑚4)} = [𝑚1, 𝑚2]⊗[𝑚3, 𝑚4] = [𝑚1⊛ 𝑚2, 𝑚3⊗ 𝑚4], {𝜕(𝑚1⊛ 𝑚2), 𝜕(𝑚3⊛ 𝑚4)} = [𝑚1, 𝑚2]⊗[𝑚3, 𝑚4] = [𝑚1⊛ 𝑚2, 𝑚3⊛ 𝑚4], ⟨𝜕(𝑚1⊗ 𝑚2), 𝜕(𝑚3⊗ 𝑚4)⟩= [𝑚1, 𝑚2]⊛[𝑚3, 𝑚4] = [𝑚1⊗ 𝑚2, 𝑚3⊗ 𝑚4], ⟨𝜕(𝑚1⊗ 𝑚2), 𝜕(𝑚3⊛ 𝑚4)⟩= [𝑚1, 𝑚2]⊛[𝑚3, 𝑚4] = [𝑚1⊗ 𝑚2, 𝑚3⊛ 𝑚4], ⟨𝜕(𝑚1⊛ 𝑚2), 𝜕(𝑚3⊗ 𝑚4)⟩= [𝑚1, 𝑚2]⊛[𝑚3, 𝑚4] = [𝑚1⊛ 𝑚2, 𝑚3⊗ 𝑚4], ⟨𝜕(𝑚1⊛ 𝑚2), 𝜕(𝑚3⊛ 𝑚4)⟩= [𝑚1, 𝑚2]⊛[𝑚3, 𝑚4] = [𝑚1⊛ 𝑚2, 𝑚3⊛ 𝑚4], where in all the cases we used (RTLeib5). Before to check the following axioms, we need to check a property that can be proven using (RTLeib3) and (RTLeib4). Using (RTLeib4) in the last equality of relation (RTLeib3) and rewriting that equality and the second one, we get 𝑚1⊛[𝑚2, 𝑚3] = [𝑚1, 𝑚2]⊛ 𝑚3− [𝑚1, 𝑚3]⊛ 𝑚2, 𝑚1⊛[𝑚2, 𝑚3] = [𝑚1, 𝑚2]⊛ 𝑚3− [𝑚1, 𝑚3]⊗ 𝑚2. Subtracting, we obtain the equality [𝑚1, 𝑚3]⊗ 𝑚2= [𝑚1, 𝑚3]⊛ 𝑚2. Using this last equality and the first and second equality of (RTLeib3), we obtain 𝑚1⊗[𝑚2, 𝑚3] = [𝑚1, 𝑚2]⊗ 𝑚3− [𝑚1, 𝑚3]⊗ 𝑚2 82 2 Braidings for crossed modules and internal objects = [𝑚1, 𝑚2]⊛ 𝑚3− [𝑚1, 𝑚3]⊛ 𝑚2=𝑚1⊛[𝑚2, 𝑚3]. Let us verify now the first equality of (BXLeib3) with 𝑚, 𝑚1, 𝑚2∈𝑀, {𝜕(𝑚1⊗ 𝑚2), 𝑚} = [𝑚1, 𝑚2]⊗ 𝑚 = [𝑚1, 𝑚]⊛ 𝑚2−𝑚1⊗[𝑚, 𝑚2] = [𝑚1, 𝑚]⊛ 𝑚2+𝑚1⊗[𝑚2, 𝑚] = [𝑚1, 𝑚]⊗ 𝑚2+𝑚1⊗[𝑚2, 𝑚] = (𝑚1⊗ 𝑚2)⋅2𝑚, where we use (RTLeib3) and (RTLeib4). The second equality is analogous: {𝜕(𝑚1⊛ 𝑚2), 𝑚} = [𝑚1, 𝑚2]⊗ 𝑚 = [𝑚1, 𝑚]⊛ 𝑚2+𝑚1⊗[𝑚2, 𝑚] = [𝑚1, 𝑚]⊛ 𝑚2+𝑚1⊛[𝑚2, 𝑚]=(𝑚1⊛ 𝑚2)⋅2𝑚. Using the exchange properties between ⊗and ⊛again, we will see the remaining equalities: ⟨𝜕(𝑚1⊗ 𝑚2), 𝑚⟩= [𝑚1, 𝑚2]⊛ 𝑚 = [𝑚1, 𝑚2]⊗ 𝑚 = (𝑚1⊗ 𝑚2)⋅2𝑚, ⟨𝜕(𝑚1⊛ 𝑚2), 𝑚⟩= [𝑚1, 𝑚2]⊛ 𝑚 = [𝑚1, 𝑚2]⊗ 𝑚 = (𝑚1⊛ 𝑚2)⋅2𝑚. Now we will check the next axiom, (BXLeib4), where we will use again that we can exchange the symbols if in one side is the bracket. Starting with the first equality, we have {𝑚, 𝜕(𝑚1⊗ 𝑚2)} = 𝑚 ⊗ [𝑚1, 𝑚2] = [𝑚, 𝑚1]⊗ 𝑚2− [𝑚, 𝑚2]⊗ 𝑚1 = [𝑚, 𝑚1]⊗ 𝑚2− [𝑚, 𝑚2]⊛ 𝑚1=𝑚⋅1(𝑚1⊗ 𝑚2), where we use (RTLeib3). Analogously we obtain the second equality: {𝑚, 𝜕(𝑚1⊛ 𝑚2)} = 𝑚 ⊗ [𝑚1, 𝑚2] = [𝑚, 𝑚1]⊗ 𝑚2− [𝑚, 𝑚2]⊗ 𝑚1 = [𝑚, 𝑚1]⊛ 𝑚2− [𝑚, 𝑚2]⊗ 𝑚1=𝑚⋅1(𝑚1⊛ 𝑚2). So, the following properties are immediate: ⟨𝑚, 𝜕(𝑚1⊗ 𝑚2)⟩=𝑚 ⊛ [𝑚1, 𝑚2] = 𝑚 ⊗ [𝑚1, 𝑚2] = 𝑚⋅1(𝑚1⊗ 𝑚2), 2.5 The non-abelian tensor product as example of braiding 83 ⟨𝑚, 𝜕(𝑚1⊛ 𝑚2)⟩=𝑚 ⊛ [𝑚1, 𝑚2] = 𝑚 ⊗ [𝑚1, 𝑚2] = 𝑚⋅1(𝑚1⊛ 𝑚2). Tofinalize, we willprove(BXLeib5), because, ifitis satisfied, equalities(BXLeib6)– (BXLeib8) will be fulfilled using the following properties: {𝑚, [𝑚′, 𝑚′′]} = 𝑚 ⊗ [𝑚′, 𝑚′′] = 𝑚 ⊛ [𝑚′, 𝑚′′] = ⟨𝑚, [𝑚′, 𝑚′′]⟩, {[𝑚, 𝑚′], 𝑚′′} = [𝑚, 𝑚′]⊗ 𝑚′′ = [𝑚, 𝑚′]⊛ 𝑚′′ =⟨[𝑚, 𝑚′], 𝑚′′⟩. By using (RTLeib3), we have (BXLeib5): {𝑚, [𝑚′, 𝑚′′]} = 𝑚 ⊗ [𝑚′, 𝑚′′] = [𝑚, 𝑚′]⊗ 𝑚′′ − [𝑚, 𝑚′′]⊗ 𝑚′ = {[𝑚, 𝑚′], 𝑚′′} − {[𝑚, 𝑚′′], 𝑚′}, Remark 2.5.13.Note that the actions can be written with a simpler notation, given by 𝑚⋅1(𝑚1⊗ 𝑚2) = 𝑚⋅1(𝑚1⊛ 𝑚2) = 𝑚 ⊗ [𝑚1, 𝑚2] = 𝑚 ⊛ [𝑚1, 𝑚2], (𝑚1⊗ 𝑚2)⋅2𝑚= (𝑚1⊛ 𝑚2)⋅2𝑚= [𝑚1, 𝑚2]⊗ 𝑚 = [𝑚1, 𝑚2]⊛ 𝑚. Remark 2.5.14.Example 2.5.12 generalizes the Lie example, since if we have that 𝑚1⊛𝑚2= −𝑚2⊗ 𝑚1as a new relation, we obtain the Lie non-abelian tensor product of 𝑀with itself using the adjoint action. 84 2 Braidings for crossed modules and internal objects CHAPTER 3 ⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄ Universal central extension of braided crossed modules of Lie algebras ⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄ In this chapter, we will study two notions of centre and commutator for the braided crossed modules and Lie algebras. With that, we will define its respective central extensions, and we will show the relationship between them. 3.1 Centre and commutator objects The category of Lie crossed modules X(LieAlg𝐾)is a semi-abelian category in the sense of [37]. Thenotionof thecentreof anobjectwasdefinedin [35], inacategorywith specific properties. This construction only needs that the category has finite products and zero object. The category X(LieAlg𝐾)has centres in the sense of Huq [35], and they were constructed in [13]. Definition 3.1.1. The centre of a Lie crossed module = (𝑀𝜕 ←←←←←←→ 𝑁, ⋅)is the crossed submodule 𝑍()=(𝑀𝑁𝜕|𝑀𝑁 ←←←←←←←←←←←←←←←←←←←←←→ st𝑁(𝑀) ∩ 𝑍(𝑁),⋅𝑍), where: •𝑀𝑁= {𝑚∈𝑀∣𝑛⋅𝑚= 0, 𝑛 ∈𝑁}, •𝑍(𝑁) = {𝑛∈𝑁∣ [𝑛, 𝑛′] = 0, 𝑛′∈𝑁}is the centre of the Lie 𝐾-algebra 𝑁, •st𝑁(𝑀) = {𝑛∈𝑁∣𝑛⋅𝑚= 0, 𝑚 ∈𝑀}, 85 92 3 Universal central extension of braided crossed modules of Lie algebras where we have used (BXLie5). For Φ2is true using a similar argument together with the Jacobi identity in both equalities. The proof of (T4) for Φ2follows since both equalities are [[𝑛1, 𝑛2],[𝑛3, 𝑛4]] after applying Φ2. For Φ1we have the following equalities: Φ1([𝑛1⊗ 𝑛2, 𝑛3⊗ 𝑛4]) = [Φ1(𝑛1⊗ 𝑛2),Φ1(𝑛3⊗ 𝑛4)] = [{𝑛1, 𝑛2},{𝑛3, 𝑛4}] = {𝜕{𝑛1, 𝑛2}, 𝜕{𝑛3, 𝑛4}} = {[𝑛1, 𝑛2],[𝑛3, 𝑛4]} = Φ1([𝑛1, 𝑛2]⊗[𝑛3, 𝑛4]), where we have used (BXLie2) and (BXLie1). So, Φ1and Φ2are well defined and are Lie 𝐾-homomorphisms. For the second part, we have that Im Φ1=𝐵𝑁(𝑀)and Im Φ2= [𝑁, 𝑁]. Therefore, Φ1and Φ2are simultaneously surjective if and only if the braided Lie crossed module is B-perfect. Lemma 3.2.2. Let (𝑀𝜕 ←←←←←←→ 𝑁, ⋅,{−,−}) be a braided Lie crossed module, and consider the braided Lie crossed module (𝑁 ⊗ 𝑁 Id𝑁⊗𝑁 ←←←←←←←←←←←←←←←←←←←←←←←←←→ 𝑁 ⊗ 𝑁, [−,−],[−,−]) (see Example 2.3.8 (1)). Then (Φ1,Φ2)∶ (𝑁 ⊗𝑁 Id𝑁⊗𝑁 ←←←←←←←←←←←←←←←←←←←←←←←←←→ 𝑁 ⊗𝑁, [−,−],[−,−]) ⟶(𝑀𝜕 ←←←←←←→ 𝑁, ⋅,{−,−}) is a morphism in BXLie, with Φ1and Φ2defined in Lemma 3.2.1,. Besides, ker(Φ1)⊂(𝑁 ⊗ 𝑁)(𝑁⊗𝑁)and ker(Φ2)⊂Z𝐵(𝑁 ⊗ 𝑁). Proof. For the proof, we will denote the action [−,−] of 𝑁 ⊗ 𝑁 Id𝑁⊗𝑁 ←←←←←←←←←←←←←←←←←←←←←←←←←→ 𝑁 ⊗ 𝑁 as ∗, and its braiding as ⟦−,−⟧. First, we will show (XLieH1). Let 𝑛⊗𝑛′, 𝑛′′ ⊗ 𝑛′′′ ∈𝑁 ⊗ 𝑁. Φ1((𝑛 ⊗ 𝑛′) ∗ (𝑛′′ ⊗ 𝑛′′′)) = Φ1([𝑛 ⊗ 𝑛′, 𝑛′′ ⊗ 𝑛′′′]) = Φ1([𝑛, 𝑛′]⊗[𝑛′′, 𝑛′′′]) = {[𝑛, 𝑛′],[𝑛′′, 𝑛′′′]} = {[𝑛, 𝑛′], 𝜕{𝑛′′, 𝑛′′′}} = [𝑛, 𝑛′]⋅{𝑛′′, 𝑛′′′} = Φ2(𝑛⊗𝑛′)⋅Φ1(𝑛′′ ⊗ 𝑛′′′), 3.2 The universal B-central extension 93 where we have used (BXLie1) and (BXLie4). Now, we will show (XLieH1). 𝜕◦Φ1(𝑛⊗𝑛′) = 𝜕{𝑛, 𝑛′} = [𝑛, 𝑛′] = Φ2(Id𝑁⊗𝑁 (𝑛⊗𝑛′)), where we have used (BXLie2). Now, we will prove (BXLieH3). Φ1(⟦𝑛⊗𝑛′, 𝑛′′ ⊗ 𝑛′′′⟧) = Φ1([𝑛 ⊗ 𝑛′, 𝑛′′ ⊗ 𝑛′′′]) = Φ1([𝑛, 𝑛′]⊗[𝑛′′, 𝑛′′′]) = {[𝑛, 𝑛′],[𝑛′′, 𝑛′′′]} = {Φ2(𝑛⊗𝑛′),Φ2(𝑛′′ ⊗ 𝑛′′′)}. So, (Φ1,Φ2)is a morphism in BXLie. We will now prove that the inclusions hold. If 𝑛⊗𝑛′∈ ker(Φ1)then {𝑛, 𝑛′} = 0. Using (BXLie1) we have that 0 = 𝜕{𝑛, 𝑛′} = [𝑛, 𝑛′]. Since (𝑁 ⊗𝑁)(𝑁⊗𝑁)= {𝑥∈𝑁 ⊗𝑁 ∣ (𝑛′′ ⊗𝑛′′′) ∗ 𝑥= 0, 𝑛′′ ⊗𝑛′′′ ∈𝑁 ⊗𝑁} (it is enough to work on generators), we have (𝑛′′ ⊗ 𝑛′′′) ∗ (𝑛⊗𝑛′) = [𝑛′′ ⊗ 𝑛′′′, 𝑛 ⊗ 𝑛′] = [𝑛′′, 𝑛′′′]⊗[𝑛, 𝑛′] = [𝑛′′, 𝑛′′′]⊗0 = 0. Therefore, we have that 𝑛⊗𝑛′∈ (𝑁 ⊗ 𝑁)(𝑁⊗𝑁)and ker(Φ1)⊂(𝑁 ⊗ 𝑁)(𝑁⊗𝑁). For the second inclusion, we take 𝑛⊗𝑛′∈ ker(Φ2), i.e. [𝑛, 𝑛′] = 0. Since it is enough to work on generators, we have that Z𝐵(𝑁⊗𝑁) = {𝑥∈𝑁⊗𝑁 ∣⟦𝑥, 𝑛′′⊗𝑛′′′⟧= 0 = ⟦𝑛′′⊗𝑛′′′, 𝑥⟧, 𝑛′′⊗𝑛′′′ ∈𝑁⊗𝑁}. Taking into account that ⟦𝑛′′ ⊗ 𝑛′′′, 𝑛 ⊗ 𝑛′⟧= [𝑛′′ ⊗ 𝑛′′′, 𝑛 ⊗ 𝑛′] = [𝑛′′, 𝑛′′′]⊗[𝑛, 𝑛′] = [𝑛′′, 𝑛′′′]⊗0 = 0, ⟦𝑛⊗𝑛′, 𝑛′′ ⊗ 𝑛′′′⟧= [𝑛 ⊗ 𝑛′, 𝑛′′ ⊗ 𝑛′′′] = [𝑛, 𝑛′]⊗[𝑛′′, 𝑛′′′] = 0 ⊗[𝑛′′, 𝑛′′′] = 0, we deduce 𝑛⊗𝑛′∈𝑍𝐵(𝑁 ⊗ 𝑁), which proves that ker(Φ2)⊂ 𝑍𝐵(𝑁 ⊗ 𝑁). Corollary 3.2.3. The morphism given in Lemma 3.2.2 is a B-central extension if and only if (𝑀𝜕 ←←←←←←→ 𝑁, ⋅,{−,−}) is a B-perfect braided Lie crossed module. 94 3 Universal central extension of braided crossed modules of Lie algebras Proof. It will be a B-central extension if and only if (Φ1,Φ2)is an extension, since Lemma 3.2.2 establishes the two inclusions and they have the restricted operations as a braided Lie crossed module. Moreover, (Φ1,Φ2)is an extension if and only if Φ1and Φ2are simultaneously surjective, and by Lemma 3.2.1 that it happens if and only if the braided Lie crossed module (𝑀𝜕 ←←←←←←→ 𝑁, ⋅,{−,−}) is B-perfect. Proposition 3.2.4. If (𝑋1 𝛿 ←←←←←←→ 𝑋2,∗,⦅−,−⦆)𝑓=(𝑓1,𝑓2) ←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←→→ (𝑀𝜕 ←←←←←←→ 𝑁, ⋅,{−,−}) is a B- central extension, then we have a morphism in BX(LieAlg𝐾), ℎ∶ (𝑁 ⊗ 𝑁 Id𝑁⊗𝑁 ←←←←←←←←←←←←←←←←←←←←←←←←←→ 𝑁 ⊗ 𝑁, [−,−],[−,−]) ←←←←←←←→ (𝑋1 𝛿 ←←←←←←→ 𝑋2,∗,⦅−,−⦆), defined by: •ℎ1∶𝑁 ⊗ 𝑁 ←←→ 𝑋1,𝑛 ⊗ 𝜂 ↦⦅𝑛, 𝜂⦆, where 𝑛, 𝜂 ∈𝑋2are elements such that 𝑓2(𝑛) = 𝑛and 𝑓2(𝜂) = 𝜂; •ℎ2∶𝑁 ⊗ 𝑁 ←←→ 𝑋2,𝑛 ⊗ 𝜂 ↦[𝑛, 𝜂], where 𝑛, 𝜂 ∈𝑋2are elements such that 𝑓2(𝑛) = 𝑛and 𝑓2(𝜂) = 𝜂. Besides, 𝑓◦ℎ= Φ, i.e. ℎis a morphism between the extensions. Proof. We need to prove that ℎ1and ℎ2are well defined. We will start with ℎ1. We will take 𝑛, 𝑛, 𝜂, 𝜂 ∈𝑋2such that 𝑓2(𝑛) = 𝑓2(𝑛) = 𝑛 and 𝑓2(𝜂) = 𝑓2(𝜂) = 𝜂and prove that ⦅𝑛, 𝜂⦆=⦅𝑛, 𝜂⦆. Since 𝑓2(𝑛) = 𝑓2(𝑛)and 𝑓= (𝑓1, 𝑓2)is a B-central extension, we have that 𝑛−𝑛 ∈ ker(𝑓2)⊂Z𝐵(𝑋2). By the definition of Z𝐵(𝑋2)we get that ⦅𝑛−𝑛, 𝜂⦆= 0 and so ⦅𝑛, 𝜂⦆=⦅𝑛, 𝜂⦆. Using an analogue reasoning, we have that 𝜂−𝜂 ∈ ker(𝑓2)⊂Z𝐵(𝑋2), and, ⦅𝑛, 𝜂 −𝜂⦆= 0. So ⦅𝑛, 𝜂⦆=⦅𝑛, 𝜂⦆. With both equalities, we have that ⦅𝑛, 𝜂⦆=⦅𝑛, 𝜂⦆=⦅𝑛, 𝜂⦆, and ℎ1is independent of the choice. Since Z𝐵(𝑋2)⊂Z(𝑋2)we can change the proof for ℎ1taking the equalities for [−,−] instead of ⦅−,−⦆which proves that ℎ2is independent of the choice. 3.2 The universal B-central extension 95 We can use an analogue argument as in Lemma 3.2.1 to prove that ℎ1and ℎ2are well defined, i.e. they preserve the relations. So, they are Lie 𝐾-homomorphisms since they are determined on generators. To prove that ℎ= (ℎ1, ℎ2)is a morphism of braided Lie crossed modules, we also use similar reasoning as the one done in Lemma 3.2.2, since we can make the changes in the choice inside the braidings and brackets. To finish, if 𝑛⊗𝜂∈𝑁 ⊗ 𝑁, then 𝑓1◦ℎ1(𝑛⊗𝜂) = 𝑓1(⦅𝑛, 𝜂⦆) = {𝑓2(𝑛), 𝑓2(𝜂)} = {𝑛, 𝜂}=Φ1(𝑛⊗𝜂), 𝑓2◦ℎ2(𝑛⊗𝜂) = 𝑓2([𝑛, 𝜂]) = [𝑓2(𝑛), 𝑓2(𝜂)] = [𝑛, 𝜂]=Φ2(𝑛⊗𝜂). Therefore, 𝑓◦ℎ= Φ. Lemma 3.2.5. If 𝑁is a perfect Lie 𝐾-algebra, i.e. 𝑁= [𝑁, 𝑁], then (𝑁 ⊗ 𝑁 Id𝑁⊗𝑁 ←←←←←←←←←←←←←←←←←←←←←←←←←→ 𝑁 ⊗ 𝑁, [−,−],[−,−]) is a B-perfect braided Lie crossed module. In particular, if (𝑀𝜕 ←←←←←←→ 𝑁, ⋅,{−,−}) is a B-perfect braided Lie crossed module, then (𝑁⊗𝑁 Id𝑁⊗𝑁 ←←←←←←←←←←←←←←←←←←←←←←←←←→ 𝑁⊗𝑁, [−,−],[−,−]) is a B-perfect braided Lie crossed module. Proof. Since the braiding in (𝑁 ⊗ 𝑁 Id𝑁⊗𝑁 ←←←←←←←←←←←←←←←←←←←←←←←←←→ 𝑁 ⊗ 𝑁, [−,−],[−,−]) is the bracket, we have that [𝑁 ⊗ 𝑁, 𝑁 ⊗ 𝑁] = 𝐵𝑁⊗𝑁 (𝑁 ⊗ 𝑁), and so it is enough to prove that [𝑁 ⊗ 𝑁, 𝑁 ⊗ 𝑁] = 𝑁 ⊗ 𝑁. Moreover, it is enough to prove that the generators [𝑛1, 𝑛2]⊗[𝑛3, 𝑛4]are inside [𝑁 ⊗ 𝑁, 𝑁 ⊗ 𝑁]since 𝑁= [𝑁, 𝑁]. Using (T4) we have that [𝑛1, 𝑛2]⊗[𝑛3, 𝑛4] = [𝑛1⊗ 𝑛2, 𝑛3⊗ 𝑛4] ∈ [𝑁 ⊗ 𝑁, 𝑁 ⊗ 𝑁]. For the second part, if (𝑀𝜕 ←←←←←←→ 𝑁, ⋅,{−,−}) is B-perfect, then 𝑁= [𝑁, 𝑁], and we conclude using the first part. Proposition 3.2.6. Let (𝑌1 𝜚 ←←←←←←→ 𝑌2, ⋆, ⟦−,−⟧)Ψ ←←←←←←←←→ (𝑀𝜕 ←←←←←←→ 𝑁, ⋅,{−,−}) be a morphism of braided Lie crossed modules such that (𝑌1 𝜚 ←←←←←←→ 𝑌2, ⋆, ⟦−,−⟧)is B-perfect. 96 3 Universal central extension of braided crossed modules of Lie algebras If (𝑋1 𝜌 ←←←←←←→ 𝑋2,∗,⦅−,−⦆)𝑓 ←←←←←←←→→ (𝑀𝜕 ←←←←←←→ 𝑁, ⋅,{−,−}) is a B-central extension and exists (𝑌1 𝜚 ←←←←←←→ 𝑌2, ⋆, ⟦−,−⟧)ℎ ←←←←←←→ (𝑋1 𝜌 ←←←←←←→ 𝑋2,∗,⦅−,−⦆)such that Ψ = 𝑓◦ℎ, then h is the unique that satisfies the equality. Proof. Suppose that there are 𝑔, ℎ∶ (𝑌1 𝜚 ←←←←←←→ 𝑌2, ⋆, ⟦−,−⟧)←←←←←←←→ (𝑋1 𝜌 ←←←←←←→ 𝑋2,∗,⦅−,−⦆) such that Ψ = 𝑓◦ℎ=𝑓◦𝑔, i.e. Ψ1=𝑓1◦ℎ1=𝑓1◦𝑔1and Ψ2=𝑓2◦ℎ2=𝑓2◦𝑔2. If 𝑦∈𝑌2then 𝑓2◦ℎ2(𝑦) = 𝑓2◦𝑔2(𝑦), i.e. ℎ2(𝑦) − 𝑔2(𝑦) ∈ ker(𝑓2). Then there is 𝑘𝑦∈ ker(𝑓2)such that ℎ2(𝑦) = 𝑔2(𝑦) + 𝑘𝑦. Since 𝑓is a B-central extension we have that ker(𝑓2)⊂Z𝐵(𝑋2)⊂Z(𝑋2). If we take 𝑦, 𝑧 ∈𝑌2, and since 𝑘𝑦, 𝑘𝑧∈ Z(𝑋2), we have [𝑘𝑦, 𝑔2(𝑧)] = [𝑘𝑦, 𝑘𝑧] = [𝑔2(𝑦), 𝑘𝑧] = 0. Using this fact, we have: ℎ2([𝑦, 𝑧]) = [ℎ2(𝑦), ℎ2(𝑧)] = [𝑔2(𝑦) + 𝑘𝑦, 𝑔2(𝑧) + 𝑘𝑧] = [𝑔2(𝑦), 𝑔2(𝑧)] + [𝑘𝑦, 𝑔2(𝑧)] + [𝑘𝑦, 𝑘𝑧]+[𝑔2(𝑦), 𝑘𝑧] = [𝑔2(𝑦), 𝑔2(𝑧)] = 𝑔2([𝑦, 𝑧]). So, 𝑔2=ℎ2since (𝑌1 𝜚 ←←←←←←→ 𝑌2, ⋆, ⟦−,−⟧)is B-perfect. Besides, since ker(𝑓2)⊂Z𝐵(𝑋2), for 𝑦, 𝑧 ∈𝑌2, we have that: ℎ1(⟦𝑦, 𝑧⟧) = ⦅ℎ2(𝑦), ℎ2(𝑧)⦆=⦅𝑔2(𝑦) + 𝑘𝑦, 𝑔2(𝑧) + 𝑘𝑧⦆ =⦅𝑔2(𝑦), 𝑔2(𝑧)⦆+⦅𝑘𝑦, 𝑔2(𝑧)⦆+⦅𝑘𝑦, 𝑘𝑧⦆+⦅𝑔2(𝑦), 𝑘𝑧⦆ =⦅𝑔2(𝑦), 𝑔2(𝑧)⦆=𝑔1(⟦𝑦, 𝑧⟧), where we have used that 𝑘𝑦, 𝑘𝑧∈ Z𝐵(𝑋2). Therefore, 𝑔1=ℎ1because (𝑌1 𝜚 ←←←←←←→ 𝑌2, ⋆, ⟦−,−⟧)is B-perfect, i.e. 𝑌1=𝐵𝑌2(𝑌1) is generated by the images of the braiding. Corollary 3.2.7. If = (𝑀𝜕 ←←←←←←→ 𝑁, ⋅,{−,−}) is a B-perfect Lie braided crossed module, then = (𝑁 ⊗ 𝑁 Id𝑁⊗𝑁 ←←←←←←←←←←←←←←←←←←←←←←←←←→ 𝑁 ⊗ 𝑁, [−,−],[−,−]) Φ=(Φ1,Φ2) ←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←→→ = (𝑀𝜕 ←←←←←←→ 𝑁, ⋅,{−,−}) (UBCE) 3.2 The universal B-central extension 97 is the universal B-central extension of , where Φ1,Φ2were defined in Lemma 3.2.1. Proof. Since is B-perfect, Corollary 3.2.3 states that the morphism Φ ←←←←←←←←→→ is aB-central extension. We need to prove that it is universal. If we have another B-central extension 𝑓 ←←←←←←←→→ then by Proposition 3.2.4 there is ℎsuch that Φ = 𝑓◦ℎ. The uniqueness of this morphism is given by Proposition 3.2.6. We can use the previous proposition since is B-perfect by Lemma 3.2.5 and the fact that is B-perfect. Let us see the converse of Corollary 3.2.7. Proposition 3.2.8. Let (𝑌1 𝜚 ←←←←←←→ 𝑌2, ⋆, ⟦−,−⟧)Ψ ←←←←←←←←←←←←←→→ (𝑀𝜕 ←←←←←←→ 𝑁, ⋅,{−,−}) be an extension in BXLie such that (𝑌1 𝜚 ←←←←←←→ 𝑌2, ⋆, ⟦−,−⟧)is B-perfect. Then (𝑀𝜕 ←←←←←←→ 𝑁, ⋅,{−,−}) is B-perfect. Proof. Ψ1and Ψ2are surjective maps since Ψis an extension, and 𝑌1=𝐵𝑌2(𝑌1)and 𝑌2= [𝑌2, 𝑌2]because (𝑌1 𝜚 ←←←←←←→ 𝑌2, ⋆, ⟦−,−⟧)is B-perfect. Since the elements ⟦𝑦, 𝑧⟧, with 𝑦, 𝑧 ∈𝑌2are the generators of 𝑌1, we have that Ψ1(⟦𝑦, 𝑧⟧)are the generators of Im Ψ1=𝑀. Since Ψ1(⟦𝑦, 𝑧⟧) = {Φ2(𝑦),Φ2(𝑧)}, we get that the generators of 𝑀are braided elements and 𝑀=𝐵𝑁(𝑀). We know that the elements [𝑦, 𝑧], with 𝑦, 𝑧 ∈𝑌2, are the generators of 𝑌2. Therefore, Φ2([𝑦, 𝑧]) = [Φ2(𝑦),Φ2(𝑧)] are the generators of Im Φ2=𝑁, and then 𝑁= [𝑁, 𝑁]. So, (𝑀𝜕 ←←←←←←→ 𝑁⋅,{−,−}) is B-perfect. Lemma 3.2.9. Let = (𝑌1 𝜚 ←←←←←←→ 𝑌2, ⋆, ⟦−,−⟧)Ψ ←←←←←←←←←←←←←→→ = (𝑀𝜕 ←←←←←←→ 𝑁, ⋅,{−,−}) be a B- central extension in BXLie such that is not B-perfect. Then exists another extension 𝑓 ←←←←←←←←←←←←→→ and two different morphisms ℎ, 𝑔 ∶←←→ such that Ψ = 𝑓◦ℎ=𝑓◦𝑔. Proof. Let (𝐵𝑌2(𝑌1) 𝜚|𝐵𝑌2(𝑌1) ←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←→ [𝑌2, 𝑌2], ⋆𝐶,⟦−,−⟧𝐶)𝑖=(𝑖1,𝑖2) ←←←←←←←←←←←←←←←←←←←←←←←←←←←←→ (𝑌1 𝜚 ←←←←←←→ 𝑌2, ⋆, ⟦−,−⟧)be the inclusion morphism of the B-commutator braided crossed submodule. 98 3 Universal central extension of braided crossed modules of Lie algebras Taking the cokernel of 𝑖we have the Lie crossed module = ( 𝑌1 𝐵𝑌2(𝑌1) 𝜚 ←←←←←←→ 𝑌2 [𝑌2, 𝑌2], ⋆, ⟦−,−⟧). We will denote in the same way, by abuse of notation, the braidings in and in its quotient . We will represent the elements in 𝑌1 𝐵𝑌2(𝑌1)as 𝑥,𝑥∈𝑌1, and the ones in 𝑌2 [𝑌2,𝑌2]as 𝑦,𝑦∈𝑌2. We take now the product in the category BX(LieAlg𝐾)and we construct ×. We denote as 𝜋1the first projection morphism. Since 𝜋1 1and 𝜋1 2are surjective maps, we have that ×𝜋1 ←←←←←←←←←←→→ is an extension. We will denote the braiding in the product as ⦃−,−⦄. We will prove that it is a B-central extension, i.e. we need to prove the inclusions ker(𝜋1 1)⊂(𝑀×𝑌1 𝐵𝑌2(𝑌2))(𝑁×𝑌2 [𝑌2,𝑌2])and ker(𝜋1 2)⊂Z𝐵(𝑁×𝑌2 [𝑌2,𝑌2]). If 𝑎∈ ker(𝜋1 1)then 𝑎= (0, 𝑥)with 𝑥∈𝑌1. If we take (𝑛, 𝑦) ∈ 𝑁×𝑌2 [𝑌2,𝑌2]then: (𝑛, 𝑦)(⋅×⋆)(0, 𝑥) = (𝑛⋅0, 𝑦 ⋆ 𝑥) = (0, 𝑦 ⋆ 𝑥). But 𝑦⋆𝑥= 0 since 𝑦⋆𝑥∈𝐷𝑌2(𝑌1)⊂ 𝐵𝑌2(𝑌1). So ker(𝜋1 1)⊂(𝑀×𝑌1 𝐵𝑌2(𝑌2))(𝑁×𝑌2 [𝑌2,𝑌2]). If 𝑎∈ ker(𝜋1 2)then 𝑎= (0, 𝑦)with 𝑦∈𝑌2. If we take (𝑛, 𝑦1) ∈ 𝑁×𝑌2 [𝑌2,𝑌2]then: ⦃(0, 𝑦),(𝑛, 𝑦1)⦄= ({0, 𝑛},⟦𝑦, 𝑦1⟧) = (0,⟦𝑦, 𝑦1⟧), ⦃(𝑛, 𝑦1),(0, 𝑦)⦄= ({𝑛, 0},⟦𝑦1, 𝑦⟧) = (0,⟦𝑦1, 𝑦⟧). Moreover, ⟦𝑦, 𝑦1⟧=⟦𝑦1, 𝑦⟧= 0 since ⟦𝑦1, 𝑦⟧,⟦𝑦, 𝑦1⟧∈𝐵𝑌2(𝑌1). Therefore ker(𝜋1 2)⊂Z𝐵(𝑁×𝑌2 [𝑌2,𝑌2]), and so 𝜋1is a B-central extension. If 𝑖𝑐∶←←←←←←←→→ is the cokernel of 𝑖, then we have two morphisms, induced by the product, with domain and ×as codomain. They are ℎ= (Ψ,0) and 3.3 Braiding on a universal extension of Lie crossed modules 99 𝑔= (Ψ, 𝑖𝑐). Since they are induced by the universal property of the product, we have that Ψ = 𝜋1◦ℎ=𝜋1◦𝑔. To finish the proof, we only must prove that they are different. Since the braided Lie crossed module is not B-perfect and 𝑖𝑐 1and 𝑖𝑐 2are surjective we know that 𝑖𝑐 1≠0 or 𝑖𝑐 2≠0(if both were the zero morphisms, then would be B-perfect), and so ℎ≠𝑔. Corollary 3.2.10. If is a braided Lie crossed module, then its universal B-central extension, if it exists, is B-perfect. Proof. If the universal extension is not B-perfect, then using Lemma 3.2.9 we have another B-central extension ←←←←←←←→→ for which there exist two different morphisms from the universal B-central extension to ←←←←←←←→→ , which contradicts the universality. Theorem 3.2.11. A braided Lie crossed module admits a universal B-central extension if and only if it is B-perfect. Proof. It is a consequence of Corollary 3.2.7, Corollary 3.2.10 and Proposition 3.2.8. 3.3 Braiding on a universal extension of Lie crossed modules Universal central extensions of braided crossed modules of groups are not studied in [26]. However, the author constructed a canonical braiding on the universal central extension of a crossed module of groups [48], when the given crossed module is braided as well, and showed that it was universal in a sense that we will explain in this section. In this part of the paper, we will consider braided Lie crossed modules extensions, but unlike the previous section, we will construct a braiding on the universal central extension of a braided Lie crossed module though as Lie crossed module and with the centre in X(LieAlg𝐾), which we have called 𝔘-central extension. In this sense, we 100 3 Universal central extension of braided crossed modules of Lie algebras will obtain similar results given by Fukushi in [26] for crossed modules of groups in the category BX(LieAlg𝐾). Casas and Ladra in [13] proved that the universal central extension of a perfect Lie crossed module (𝑀←←→ 𝑁, ⋅)in X(LieAlg𝐾)is given by: (𝑁 ⊗ 𝑀 Id𝑁⊗𝜕 ←←←←←←←←←←←←←←←←←←←←←←←←←←→ 𝑁 ⊗ 𝑁, ∗) 𝑐=(𝑐1,𝑐2) ←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←→→ (𝑀𝜕 ←←←←←←→ 𝑁, ⋅),(UCE) where 𝑁 ⊗𝑀 is given by the actions ⋅of 𝑁on 𝑀and 𝑚⋆𝑛 = [𝜕(𝑚), 𝑛]of 𝑀on 𝑁; the action of 𝑁 ⊗ 𝑁 on 𝑁 ⊗ 𝑀 is given by (𝑛 ⊗ 𝑛′) ∗ (𝑛′′ ⊗ 𝑚) = [[𝑛, 𝑛′], 𝑛′′]⊗ 𝑚 + 𝑛′′ ⊗[𝑛, 𝑛′]⋅𝑚for 𝑛, 𝑛′, 𝑛′′ ∈𝑁, 𝑚 ∈𝑀; and the morphisms are 𝑐1(𝑛 ⊗ 𝑚) = 𝑛⋅𝑚 and 𝑐2(𝑛⊗𝑛′) = [𝑛, 𝑛′]. Proposition 3.3.1. If (𝑀𝜕 ←←←←←←←←→ 𝑁, ⋅,{−,−}) is a braided Lie crossed module then ⦃−,−⦄∶ (𝑁⊗𝑁)×(𝑁⊗𝑁)←←→ 𝑁⊗𝑀, defined on generators by ⦃𝑛⊗𝑛′, 𝑛′′⊗𝑛′′′⦄= [𝑛, 𝑛′]⊗{𝑛′′, 𝑛′′′}, is a braiding for the Lie crossed module (𝑁⊗𝑀 Id𝑁⊗𝜕 ←←←←←←←←←←←←←←←←←←←←←←←←←←→ 𝑁⊗𝑁, ∗). Proof. The braiding ⦃−,−⦄is well defined since it preserves the relations (T1) and (T2) using the 𝐾-bilinearity of [−,−] and {−,−}, and (T3) and (T4) are fulfilled too since {−,−} and [−,−] satisfy it. It is sufficient to prove the axioms of braidings. Let 𝑛, 𝑛, 𝑛′, 𝑛′′ ∈𝑁,𝑚, 𝑚′∈𝑀. Then (Id𝑁⊗𝜕)(⦃𝑛⊗𝑛′, 𝑛′′ ⊗ 𝑛′′′⦄) = (Id𝑁⊗𝜕)([𝑛, 𝑛′]⊗{𝑛′′, 𝑛′′′}) = [𝑛, 𝑛′]⊗ 𝜕{𝑛′′, 𝑛′′′} = [𝑛, 𝑛′]⊗[𝑛′′, 𝑛′′′] = [𝑛⊗𝑛′, 𝑛′′ ⊗ 𝑛′′′](BXLie1), ⦃(Id𝑁⊗𝜕)(𝑛 ⊗ 𝑚),(Id𝑁⊗𝜕)(𝑛′⊗ 𝑚′)⦄ =⦃𝑛⊗𝜕(𝑚), 𝑛′⊗ 𝜕(𝑚′)⦄= [𝑛, 𝜕(𝑚)] ⊗{𝑛′, 𝜕(𝑚′)} = −(𝑚⋆𝑛)⊗(𝑛′⋅𝑚′) = [𝑛 ⊗ 𝑚, 𝑛′⊗ 𝑚′](BXLie2), ⦃(Id𝑁⊗𝜕)(𝑛 ⊗ 𝑚), 𝑛′⊗ 𝑛′′⦄=⦃𝑛⊗𝜕(𝑚), 𝑛′⊗ 𝑛′′⦄= [𝑛, 𝜕(𝑚)] ⊗{𝑛′, 𝑛′′} = −(𝑚⋆𝑛)⊗{𝑛′, 𝑛′′} = 𝑛 ⊗ [𝑚, {𝑛′, 𝑛′′}] − {𝑛′, 𝑛′′}⋆ 𝑛 ⊗ 𝑚 =𝑛 ⊗ {𝜕(𝑚), 𝜕({𝑛′, 𝑛′′})} − [𝜕({𝑛′, 𝑛′′}), 𝑛]⊗ 𝑚 3.3 Braiding on a universal extension of Lie crossed modules 101 = −𝑛 ⊗ [𝑛′, 𝑛′′]⋅𝑚− [[𝑛′, 𝑛′′], 𝑛]⊗ 𝑚 = −(𝑛′⊗ 𝑛′′) ∗ (𝑛⊗𝑚)(BXLie3), where we have used the second relation of (T3) in the third equality. ⦃𝑛′⊗ 𝑛′′,(Id𝑁⊗𝜕)(𝑛 ⊗ 𝑚)⦄=⦃𝑛′⊗ 𝑛′′, 𝑛 ⊗ 𝜕(𝑚)⦄= [𝑛′, 𝑛′′]⊗{𝑛, 𝜕(𝑚)} = [𝑛′, 𝑛′′]⊗(𝑛⋅𝑚) = 𝑛 ⊗ [𝑛, 𝑛′]⋅𝑚+ [[𝑛′, 𝑛′′], 𝑛]⊗ 𝑚 = (𝑛′⊗ 𝑛′′) ∗ (𝑛⊗𝑚)(BXLie4), where we have used the first relation of (T3) in the third equality. ⦃𝑛1⊗ 𝑛′ 1,[𝑛2⊗ 𝑛′ 2, 𝑛3⊗ 𝑛′ 3]⦄=⦃𝑛1⊗ 𝑛′ 1,[𝑛2⊗ 𝑛′ 2]⊗[𝑛3⊗ 𝑛′ 3]⦄ = [𝑛1, 𝑛′ 1]⊗{[𝑛2, 𝑛′ 2],[𝑛3, 𝑛′ 3]} = [𝑛1, 𝑛′ 1]⊗[{𝑛2, 𝑛′ 2},{𝑛3, 𝑛′ 3}] = ({𝑛3, 𝑛′ 3}⋆[𝑛1, 𝑛′ 1]) ⊗{𝑛2, 𝑛′ 2} − ({𝑛2, 𝑛′ 2}⋆[𝑛1, 𝑛′ 1]) ⊗{𝑛3, 𝑛′ 3} = [𝜕({𝑛3, 𝑛′ 3}),[𝑛1, 𝑛′ 1]] ⊗{𝑛2, 𝑛′ 2}−[𝜕({𝑛2, 𝑛′ 2}),[𝑛1, 𝑛′ 1]] ⊗{𝑛3, 𝑛′ 3} = [[𝑛3, 𝑛′ 3],[𝑛1, 𝑛′ 1]] ⊗{𝑛2, 𝑛′ 2} − [[𝑛2, 𝑛′ 2],[𝑛1, 𝑛′ 1]] ⊗{𝑛3, 𝑛′ 3} = −[[𝑛1, 𝑛′ 1],[𝑛3, 𝑛′ 3]] ⊗{𝑛2, 𝑛′ 2} + [[𝑛1, 𝑛′ 1],[𝑛2, 𝑛′ 2]] ⊗{𝑛3, 𝑛′ 3} = −⦃[𝑛1, 𝑛′ 1]⊗[𝑛3, 𝑛′ 3], 𝑛2⊗ 𝑛′ 2⦄+⦃[𝑛1, 𝑛′ 1]⊗[𝑛2, 𝑛′ 2], 𝑛3⊗ 𝑛′ 3⦄ =⦃[𝑛1⊗ 𝑛′ 1, 𝑛2⊗ 𝑛′ 2], 𝑛3⊗ 𝑛′ 3⦄−⦃[𝑛1⊗ 𝑛′ 1, 𝑛3⊗ 𝑛′ 3], 𝑛2⊗ 𝑛′ 2⦄(BXLie5), ⦃[𝑛1⊗ 𝑛′ 1, 𝑛2⊗ 𝑛′ 2], 𝑛3⊗ 𝑛′ 3⦄=⦃[𝑛1, 𝑛′ 1]⊗[𝑛2, 𝑛′ 2], 𝑛3⊗ 𝑛′ 3⦄ = [[𝑛1, 𝑛′ 1],[𝑛2, 𝑛′ 2]] ⊗{𝑛3, 𝑛′ 3} = [𝑛1, 𝑛′ 1]⊗[𝑛2, 𝑛′ 2]⋅{𝑛3, 𝑛′ 3}−[𝑛2, 𝑛′ 2]⊗[𝑛1, 𝑛′ 1]⋅{𝑛3, 𝑛′ 3} = [𝑛1, 𝑛′ 1]⊗{[𝑛2, 𝑛′ 2], 𝜕({𝑛3, 𝑛′ 3})} − [𝑛2, 𝑛′ 2]⊗{[𝑛1, 𝑛′ 1], 𝜕({𝑛3, 𝑛′ 3})} = [𝑛1, 𝑛′ 1]⊗{[𝑛2, 𝑛′ 2],[𝑛3, 𝑛′ 3]} − [𝑛2, 𝑛′ 2]⊗{[𝑛1, 𝑛′ 1],[𝑛3, 𝑛′ 3]} =⦃𝑛1⊗ 𝑛′ 1,[𝑛2, 𝑛′ 2]⊗[𝑛3, 𝑛′ 3]⦄−⦃𝑛2⊗ 𝑛′ 2,[𝑛1, 𝑛′ 1]⊗[𝑛3, 𝑛′ 3]⦄ =⦃𝑛1⊗ 𝑛′ 1,[𝑛2⊗ 𝑛′ 2, 𝑛3⊗ 𝑛′ 3]⦄−⦃𝑛2⊗ 𝑛′ 2,[𝑛1⊗ 𝑛′ 1, 𝑛3⊗ 𝑛′ 3]⦄(BXLie6). In all equalities, we have used the properties of {−,−} and relations of the tensor product. 108 3 Universal central extension of braided crossed modules of Lie algebras CHAPTER 4 ⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄ On the Loday-Pirashvili Category ⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄ In this chapter, we generalize the Loday-Pirashvili category to different kinds of tensor categories. Then we use it to prove relationships between internal Lie objects and internal Leibniz objects. 4.1 Tensor Categories In this section, we will generalize the Loday-Pirashvili category ( [44]) to different kinds of tensor categories. The definitions of (braided) semigroupal categories and (braided) monoidal categories are already shown in Section 1.4. 4.1.1 Categories with operations Definition 4.1.1. Let 𝐂be a category and ⊗∶×←←→ a bifunctor. We will say that the pair = (𝐂, ⊗)is a category with an operation. Definition 4.1.2. Let 𝐂be a category. We will denote as Hom(𝐂)the category given by: •Ob (Hom(𝐂))= Arw(𝐂) •Arw (Hom(𝐂))⊂Arw(𝐂)×Arw(𝐂)with 𝐴 𝐵 𝑓 𝛼=(𝛼1,𝛼2) ←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←→ 𝐶 𝐷 𝑔⇔ 𝐴 𝐶 𝐵 𝐷. 𝑓 𝛼1 𝑔 𝛼2 109 110 4 On the Loday-Pirashvili Category •Id𝑓= (Id𝐴,Id𝐵)and 𝛽◦𝛼= (𝛽1◦𝛼1, 𝛽2◦𝛼2). Since Hom(𝐂)is defined by pairs of arrows is immediate that if 𝐂has finite coproducts, then 𝐴 𝐵 𝑓⊕ 𝐶 𝐷 𝑔∶= 𝐴⊕𝐶 𝐵 ⊕ 𝐷, 𝑓⊕𝑔 and Hom(𝐂)has finite coproducts. In the same way, if 0is the zero object for 𝐂, then Id0is the zero object for 𝐂(same for initial object and final object). Proposition 4.1.3. Let = (𝐂, ⊗)be a category with an operation such that 𝐂has finite coproducts. The correspondence  ⊗∶ Hom(𝐂) × Hom(𝐂)←←→ Hom(𝐂)defined: •in objects as 𝐴 𝐵 𝑓 ⊗ 𝐶 𝐷 𝑔∶= (𝐴⊗𝐷)⊕(𝐵 ⊗ 𝐶) 𝐵 ⊗ 𝐷 [(𝑓⊗Id𝐷),(Id𝐵⊗𝑔)] •in arrows, if 𝑓𝛼 ←←←←←←→ 𝑓′and 𝑔𝛽 ←←←←←←→ 𝑔′,𝛼 ⊗𝛽 =((𝛼1⊗ 𝛽2)⊕(𝛼2⊗ 𝛽1), 𝛼2⊗ 𝛽2) is a functor. Therefore, (Hom(𝐂), ⊗)is a category with an operation. Proof. Since ⊗and ⊕are functors, it is just a matter of checking that 𝛼 ⊗𝛽 is a morphism in the category. To do this, we take 𝐴′𝑓′ ←←←←←←←←←→ 𝐵′and 𝐶′𝑔′ ←←←←←←←←→ 𝐷′, and we easily see that the following diagram is commutative: (𝐴⊗𝐷)⊕(𝐵 ⊗ 𝐶)(𝛼1⊗𝛽2)⊕(𝛼2⊗𝛽1)// [(𝑓⊗Id𝐷),(Id𝐵⊗𝑔)]  (𝐴′⊗ 𝐷′)⊕(𝐵′⊗ 𝐶′) [(𝑓′⊗Id𝐷′),(Id𝐵′⊗𝑔′)]  𝐵 ⊗ 𝐷 𝛼2⊗𝛽2 //𝐵′⊗ 𝐷′ Definition 4.1.4. Let = (𝐂, ⊗)be a category with an operation such that 𝐂has finite coproducts. The category with an operation (Hom(𝐂), ⊗)defined in Proposition 4.1.3 will be denoted as LP()and it will be called the Loday-Pirashvili category with an operation of . 4.1.2 Semigroupal categories 111 4.1.2 Semigroupal categories Definition 4.1.5. Let = (𝐂, ⊗)be a category with an operation and consider three objects 𝐴, 𝐵, 𝐶 in such that there exists the coproduct of 𝐴with 𝐵and 𝐴 ⊗ 𝐶 with 𝐵 ⊗ 𝐶. We define the R-⊗-distributor on 𝐴, 𝐵, 𝐶 as the morphism 𝜀𝐴,𝐵,𝐶 given by 𝐴⊗𝐶 𝜄1// 𝜄1⊗Id𝐶(( (𝐴⊗𝐶)⊕(𝐵 ⊗ 𝐶) 𝜀𝐴,𝐵,𝐶  (𝐵 ⊗ 𝐶) 𝜄2 oo 𝜄2⊗Id𝐶 vv (𝐴⊕𝐵)⊗ 𝐶 Similarly, if there exists the coproduct of 𝐴with 𝐵and 𝐶 ⊗ 𝐴 with 𝐶 ⊗ 𝐵 we define the L-⊗-distributor on 𝐴, 𝐵, 𝐶 as the morphism 𝛾𝐴,𝐵,𝐶 given by 𝐶 ⊗ 𝐴 𝜄1// Id𝐶⊗𝜄1(( (𝐶 ⊗ 𝐴)⊕(𝐶 ⊗ 𝐵) 𝛾𝐴,𝐵,𝐶  (𝐶 ⊗ 𝐵) 𝜄2 oo Id𝐶⊗𝜄2 vv 𝐶 ⊗ (𝐴⊕𝐵) Proposition 4.1.6. The correspondences 𝜀and 𝛾defined above are functorial. Furthermore, they are natural transformations. Proof. To check that 𝜀is a natural transformation we need to see that the following diagram is commutative, (𝐴⊗𝐶)⊕(𝐵 ⊗ 𝐶)𝜀𝐴,𝐵,𝐶 // (𝑓⊗ℎ)⊕(𝑔⊗ℎ)  (𝐴⊕𝐵)⊗ 𝐶 (𝑓⊕𝑔)⊗ℎ  (𝐴′⊗ 𝐶′)⊕(𝐵′⊗ 𝐶′)𝜀𝐴′,𝐵′,𝐶′//(𝐴′⊕ 𝐵′)⊗ 𝐶′ where 𝐴𝑓 ←←←←←←←→ 𝐴′, 𝐵 𝑔 ←←←←←←→ 𝐵′, 𝐶 ℎ ←←←←←←→ 𝐶′∈ Arw(𝐂). Since the domain is a coproduct, it easily follows by checking that they coincide when composed with the natural injections. A similar argument works for 𝛾. Definition 4.1.7. Let = (𝐂, ⊗)be a category with an operation where 𝐂has finite coproducts. Take 𝜀and 𝛾from Definition 4.1.5. 112 4 On the Loday-Pirashvili Category •We say that 𝐂is Right Multiplication distributive with ⊗or 𝑅-⊗-distributive if 𝜀is a natural isomorphism. In this case, we will say that is 𝑅-distributive. •We say that 𝐂is Left Multiplication distributive with ⊗or 𝐿-⊗-distributive if 𝛾is a natural isomorphism. In this case, we will say that is 𝐿-distributive. •We say that 𝐂is distributive with ⊗or ⊗-distributive if 𝜀and 𝛾are natural isomorphisms. In this case, we will say that is distributive. Theorem 4.1.8. Let (𝐂, ⊗, 𝑎)be a semigroupal category such that = (𝐂, ⊗)is ⊗- distributive. For 𝐴 𝐵 𝑓, 𝐶 𝐷 𝑔, 𝐸 𝐹 ℎ∈ Ob(Hom(𝐂)) we define using the universal property of coproducts the morphism ((𝐴⊗ 𝐷)⊕(𝐵 ⊗ 𝐶)) ⊗ 𝐹 (((𝐴 ⊗ 𝐷)⊕(𝐵 ⊗ 𝐶)) ⊗ 𝐹 )⊕((𝐵 ⊗ 𝐷)⊗ 𝐸) (𝐵 ⊗ 𝐷)⊗ 𝐸 ((𝐴⊗ 𝐷)⊗ 𝐹 )⊕((𝐵 ⊗ 𝐶)⊗ 𝐹)𝐵 ⊗ (𝐷 ⊗ 𝐸) (𝐴⊗(𝐷 ⊗ 𝐹 )) ⊕(𝐵 ⊗ (𝐶 ⊗ 𝐹 )) 𝐵 ⊗ ((𝐶 ⊗ 𝐹 )⊕(𝐷 ⊗ 𝐸)) (𝐴⊗(𝐷 ⊗ 𝐹 )) ⊕(𝐵 ⊗ ((𝐶 ⊗ 𝐹 )⊕(𝐷 ⊗ 𝐸))) 𝜄1 𝜀−1 𝛼𝑓,𝑔,ℎ 𝜄2 𝑎 𝑎⊕𝑎 Id ⊗𝜄2 Id ⊕(Id ⊗𝜄1)𝜄2 Then 𝑎𝑓,𝑔,ℎ = (𝛼𝑓,𝑔,ℎ, 𝑎𝐵,𝐷,𝐹 )gives an associator for (Hom(𝐂), ⊗). The inverse of this natural isomorphism is given by (𝜔𝑓,𝑔,ℎ, 𝑎−1 𝐵,𝐷,𝐹 )where 𝜔𝑓,𝑔,ℎ is defined by the diagram 𝐴⊗(𝐷 ⊗ 𝐹 ) (𝐴 ⊗ (𝐷 ⊗ 𝐹 )) ⊕(𝐵 ⊗ ((𝐶 ⊗ 𝐹 )⊕(𝐷 ⊗ 𝐸))) 𝐵 ⊗ ((𝐶 ⊗ 𝐹 )⊕(𝐷 ⊗ 𝐸)) (𝐴⊗ 𝐷)⊗ 𝐹 (𝐵 ⊗ (𝐶 ⊗ 𝐹 )) ⊕(𝐵 ⊗ (𝐷 ⊗ 𝐸)) ((𝐴⊗ 𝐷)⊕(𝐵 ⊗ 𝐶)) ⊗ 𝐹 ((𝐵 ⊗ 𝐶)⊗ 𝐹)⊕((𝐵 ⊗ 𝐷)⊗ 𝐸) (((𝐴⊗ 𝐷)⊕(𝐵 ⊗ 𝐶)) ⊗ 𝐹 )⊕((𝐵 ⊗ 𝐷)⊗ 𝐸) 𝑎−1 𝜄1 𝜔𝑓,𝑔,ℎ 𝜄2 𝛾−1 𝜄1⊗Id𝐹𝑎−1⊕𝑎−1 𝜄1(𝜄2⊗Id)⊕Id 4.1.2 Semigroupal categories 113 Proof. First, we will prove that 𝑎𝑓,𝑔,ℎ = (𝛼𝑓,𝑔,ℎ, 𝑎𝐵,𝐷,𝐹 )∶ (𝑓 ⊗𝑔) ⊗ℎ ←←→ 𝑓 ⊗(𝑔 ⊗ℎ), that means, we have the diagram (((𝐴 ⊗ 𝐷)⊕(𝐵 ⊗ 𝐶)) ⊗ 𝐹 )⊕((𝐵 ⊗ 𝐷)⊗ 𝐸) (𝐴 ⊗ (𝐷 ⊗ 𝐹)) ⊕(𝐵 ⊗ ((𝐶 ⊗ 𝐹)⊕(𝐷 ⊗ 𝐸))) (𝐵⊗ 𝐷)⊗ 𝐹 𝐵 ⊗ (𝐷 ⊗ 𝐹 ). 𝛼𝑓,𝑔,ℎ [([(𝑓⊗Id),(Id ⊗𝑔)]⊗Id),(Id ⊗ℎ)] [(𝑓⊗Id),(Id ⊗[(𝑔⊗Id),(Id ⊗ℎ)])] 𝑎 Note that we use the notation [−,−] for the natural morphism to the coproduct. The domain is a coproduct so that we can study each part separately. The first one is ((𝐴 ⊗ 𝐶)⊕(𝐵 ⊗ 𝐷)) ⊗ 𝐹 and the first morphism going down is 𝜀−1, so we will prove that the diagram is commutative when precomposed with 𝜀, since it is an isomorphism. Then the domain is again a coproduct so that we can analyse both parts separately again. To see the diagram’s commutativity is just a matter of resolving the coproduct injections, as we can see in the following diagrams. The second part and the other ones can be checked similarly. The outer diagram is commutative since the inner diagrams are easily commutative. (((𝐴⊗ 𝐷)⊕(𝐵 ⊗ 𝐶)) ⊗ 𝐹 )⊕((𝐵 ⊗ 𝐷)⊗ 𝐸) ((𝐴 ⊗ 𝐷)⊕(𝐵 ⊗ 𝐶)) ⊗ 𝐹 ((𝐴 ⊗ 𝐷)⊗ 𝐹 )⊕((𝐵 ⊗ 𝐶)⊗ 𝐹 ) (𝐴⊗ 𝐷)⊗ 𝐹 (𝐵⊗ 𝐷)⊗ 𝐹 𝐵⊗(𝐷 ⊗ 𝐹 )𝐴 ⊗ (𝐷 ⊗ 𝐹) [([(𝑓⊗Id) ,(Id ⊗𝑔) ] ⊗Id) ,(Id ⊗ℎ) ] 𝜄1 [(𝑓⊗Id) ,(Id ⊗𝑔) ] ⊗Id 𝜀 𝜄1⊗Id 𝜄1 𝑎 (𝑓⊗Id)⊗Id 𝑎 𝑓⊗Id 114 4 On the Loday-Pirashvili Category (((𝐴 ⊗ 𝐷)⊕(𝐵 ⊗ 𝐶)) ⊗ 𝐹 )⊕((𝐵 ⊗ 𝐷)⊗ 𝐸) ((𝐴 ⊗ 𝐷)⊕(𝐵 ⊗ 𝐶)) ⊗ 𝐹 ((𝐴 ⊗ 𝐷)⊗ 𝐹 )⊕((𝐵 ⊗ 𝐶)⊗ 𝐹 ) ((𝐴 ⊗ 𝐷)⊗ 𝐹 )⊕((𝐵 ⊗ 𝐶)⊗ 𝐹 ) (𝐴 ⊗ 𝐷)⊗ 𝐹 (𝐴⊗(𝐷 ⊗ 𝐹 )) ⊕(𝐵 ⊗ (𝐶 ⊗ 𝐹 )) 𝐴 ⊗ (𝐷 ⊗ 𝐹 ) (𝐴 ⊗ (𝐷 ⊗ 𝐹 )) ⊕(𝐵 ⊗ ((𝐶 ⊗ 𝐹 )⊕(𝐷 ⊗ 𝐸))) 𝐵 ⊗ (𝐷 ⊗ 𝐹 ) 𝛼 𝜄1 𝜀−1 𝜀 𝑎⊕𝑎 𝜄1 𝑎 Id ⊕(Id ⊗𝜄1) 𝜄1 𝜄1 𝑓⊗Id [(𝑓⊗Id) ,(Id ⊗[(𝑔⊗Id) ,(Id ⊗ℎ) ]) Note that the rightmost part is the same for these two diagrams, so we conclude that the leftmost part is the same when precomposed with the injections and 𝜀. Therefore, we have that 𝑎𝑓,𝑔,ℎ is a morphism. Let us now see that it is a natural transformation. Let us consider the following objects 𝐴 𝐵 𝑓 𝜆 ←←←←←←←←←←←→ 𝐴′ 𝐵′ 𝑓′, 𝐶 𝐷 𝑔 𝛽 ←←←←←←←←←←←→ 𝐶′ 𝐷′ 𝑔′, 𝐸 𝐹 ℎ 𝜎 ←←←←←←←←←←←←→ 𝐸′ 𝐹′ ℎ′. The lower part of the naturalness is satisfied, since there the operation is the standard one. We will focus in the upper part. (((𝐴⊗ 𝐷)⊕(𝐵 ⊗ 𝐶)) ⊗ 𝐹)⊕((𝐵 ⊗ 𝐷)⊗ 𝐸) (𝐴 ⊗ (𝐷 ⊗ 𝐹 )) ⊕(𝐵 ⊗ ((𝐶 ⊗ 𝐹 )⊕(𝐷 ⊗ 𝐸))) (((𝐴′⊗𝐷′)⊕(𝐵′⊗ 𝐶′)) ⊗ 𝐹 ′)⊕((𝐵′⊗ 𝐷′)⊗ 𝐸′) (𝐴′⊗(𝐷′⊗ 𝐹 ′)) ⊕(𝐵′⊗((𝐶′⊗ 𝐹 ′)⊕(𝐷′⊗ 𝐸′))). 𝛼𝑓,𝑔,ℎ (((𝜆1⊗𝛽2)⊕(𝜆2⊗𝛽1))⊗𝜎2)⊕((𝜆2⊗𝛽2)⊗𝜎1) (𝜆1⊗(𝛽2⊗𝜎2))⊕(𝜆2⊗((𝛽1⊗𝜎2)⊕(𝛽2⊗𝜎1))) 𝛼𝑓′,𝑔′,ℎ′ Again, the domain is a coproduct so that we can study each part separately. It is the same coproduct as the proof of well defined, so we need to use the same steps to prove it, using the injections and 𝜀, as shown in the following diagrams. The other cases can be checked similarly. 4.1.2 Semigroupal categories 115 (((𝐴⊗ 𝐷)⊕(𝐵 ⊗ 𝐶)) ⊗ 𝐹)⊕((𝐵 ⊗ 𝐷)⊗ 𝐸) ((𝐴 ⊗ 𝐷)⊕(𝐵 ⊗ 𝐶)) ⊗ 𝐹 ((𝐴 ⊗ 𝐷)⊗ 𝐹 )⊕((𝐵 ⊗ 𝐶)⊗ 𝐹 ) ((𝐴⊗ 𝐷)⊗ 𝐹 )⊕((𝐵 ⊗ 𝐶)⊗ 𝐹 ) (𝐴 ⊗ 𝐷)⊗ 𝐹 (𝐴⊗(𝐷 ⊗ 𝐹 )) ⊕(𝐵 ⊗ (𝐶 ⊗ 𝐹 )) 𝐴 ⊗ (𝐷 ⊗ 𝐹 ) (𝐴⊗(𝐷 ⊗ 𝐹 )) ⊕(𝐵 ⊗ ((𝐶 ⊗ 𝐹 )⊕(𝐷 ⊗ 𝐸))) (𝐴′⊗(𝐷′⊗𝐹′)) ⊕(𝐵′⊗((𝐶′⊗ 𝐹′)⊕(𝐷′⊗ 𝐸′))) 𝐴′⊗(𝐷′⊗ 𝐹′) (𝐴′⊗ 𝐷′)⊗ 𝐹 ′, 𝛼 𝜄1 𝜀−1 𝜀 𝑎⊕𝑎 𝜄1 𝑎 (𝜆1⊗𝛽2)⊗𝜎2 Id ⊕(Id ⊗𝜄1) 𝜄1 𝜄1 𝜆1⊗(𝛽2⊗𝜆2) (𝜆1⊗(𝛽2⊗𝜎2))⊕(𝜆2⊗((𝛽1⊗𝜎2)⊕(𝛽2⊗𝜎1))) 𝜄1𝑎 (((𝐴⊗ 𝐷)⊕(𝐵 ⊗ 𝐶)) ⊗ 𝐹)⊕((𝐵 ⊗ 𝐷)⊗ 𝐸) ((𝐴 ⊗ 𝐷)⊕(𝐵 ⊗ 𝐶)) ⊗ 𝐹 ((𝐴 ⊗ 𝐷)⊗ 𝐹 )⊕((𝐵 ⊗ 𝐶)⊗ 𝐹 ) ((𝐴′⊗𝐷′)⊕(𝐵′⊗ 𝐶′)) ⊗ 𝐹 ′(𝐴 ⊗ 𝐷)⊗ 𝐹 ((𝐴′⊗𝐷′)⊗ 𝐹′)⊕((𝐵′⊗ 𝐶′)⊗ 𝐹 ′) (𝐴′⊗ 𝐷′)⊗ 𝐹′ (((𝐴′⊗𝐷′)⊕(𝐵′⊗ 𝐶′)) ⊗ 𝐹 ′)⊕((𝐵′⊗ 𝐷′)⊗ 𝐸′) (𝐴′⊗(𝐷′⊗ 𝐹 ′)) ⊕(𝐵′⊗(𝐶′⊗ 𝐹 ′)) (𝐴′⊗(𝐷′⊗𝐹′)) ⊕(𝐵′⊗((𝐶′⊗ 𝐹′)⊕(𝐷′⊗ 𝐸′))) 𝐴′⊗(𝐷′⊗ 𝐹′). (((𝜆1⊗𝛽2)⊕(𝜆2⊗𝛽1))⊗𝜎2)⊕((𝜆2⊗𝛽2)⊗𝜎1) 𝜄1 ((𝜆1⊗𝛽2)⊕(𝜆2⊗𝛽2))⊗𝜎2 𝜀 𝜄1 𝜀−1 (𝜆1⊗𝛽1)⊗𝜎2 𝜄1⊗Id 𝜄1 𝑎⊕𝑎 𝜄1⊗Id 𝑎 𝜄1 𝛼Id ⊕(Id ⊗𝜄1) 𝜄1 𝜄1 Therefore, 𝑎∶ ⊗◦( ⊗× IdHom(𝐂))⇒ ⊗◦(IdHom(𝐂)× ⊗)◦𝐴Hom(𝐂),Hom(𝐂),Hom(𝐂)is a natural transformation. We will shownow thatitisa natural isomorphism. We need toshowthat for 𝑓, 𝑔, ℎ ∈ Hom(𝐂), the morphism 𝑎𝑓,𝑔,ℎ is an isomorphism with inverse (𝜔𝑓,𝑔,ℎ, 𝑎−1 𝐵,𝐷,𝐹 ). Since 𝑎𝐵,𝐷,𝐹 is an isomorphism with inverse 𝑎−1 𝐵,𝐷,𝐹 we only need to prove that 𝛼𝑓,𝑔,ℎ is an isomorphism with inverse 𝜔𝑓,𝑔,ℎ. We will prove 𝜔𝑓,𝑔,ℎ◦𝛼𝑓,𝑔,ℎ = Id, the other composition is analogous. As in the 116 4 On the Loday-Pirashvili Category other proofs, the domain is a coproduct. In fact, it is the same coproduct as in the other proofs, so we need to use the same steps to prove it, using the injections and 𝜀, as can see in the following diagrams. The other cases are checked similarly. (((𝐴 ⊗ 𝐷)⊕(𝐵 ⊗ 𝐶)) ⊗ 𝐹 )⊕((𝐵 ⊗ 𝐷)⊗ 𝐸) ((𝐴 ⊗ 𝐷)⊕(𝐵 ⊗ 𝐶)) ⊗ 𝐹 ((𝐴 ⊗ 𝐷)⊗ 𝐹 )⊕((𝐵 ⊗ 𝐶)⊗ 𝐹 ) ((𝐴 ⊗ 𝐷)⊗ 𝐹 )⊕((𝐵 ⊗ 𝐶)⊗ 𝐹 ) (𝐴 ⊗ 𝐷)⊗ 𝐹 (𝐴⊗(𝐷 ⊗ 𝐹 )) ⊕(𝐵 ⊗ (𝐶 ⊗ 𝐹 )) 𝐴 ⊗ (𝐷 ⊗ 𝐹 ) (𝐴 ⊗ (𝐷 ⊗ 𝐹 )) ⊕(𝐵 ⊗ ((𝐶 ⊗ 𝐹 )⊕(𝐷 ⊗ 𝐸))) (𝐴 ⊗ 𝐷)⊗ 𝐹 (((𝐴⊗ 𝐷)⊕(𝐵 ⊗ 𝐶)) ⊗ 𝐹 )⊕((𝐵 ⊗ 𝐷)⊗ 𝐸) (𝐴 ⊗ 𝐷)⊕(𝐵 ⊗ 𝐶)⊗ 𝐹 ((𝐴 ⊗ 𝐷)⊗ 𝐹 )⊕((𝐵 ⊗ 𝐶)⊗ 𝐹 ) 𝛼 𝜄1 𝜀−1 𝜀 𝑎⊕𝑎 𝜄1 𝑎 Id ⊕(Id ⊗𝜄1) 𝜄1 𝜄1 𝑎−1 𝜔𝜄1⊗Id 𝜄1 𝜄1 𝜀 For completion we will show another diagram, composing first with 𝜄1and then with 𝜄2, to show when the morphism 𝛾from the L-⊗-distributivity appears. (((𝐴 ⊗ 𝐷)⊕(𝐵 ⊗ 𝐶)) ⊗ 𝐹 )⊕((𝐵 ⊗ 𝐷)⊗ 𝐸) ((𝐴 ⊗ 𝐷)⊕(𝐵 ⊗ 𝐶)) ⊗ 𝐹 ((𝐴 ⊗ 𝐷)⊗ 𝐹)⊕((𝐵 ⊗ 𝐶)⊗ 𝐹 ) ((𝐴⊗ 𝐷)⊗ 𝐹 )⊕((𝐵 ⊗ 𝐶)⊗ 𝐹 ) (𝐵 ⊗ 𝐶)⊗ 𝐹 (𝐴 ⊗ (𝐷 ⊗ 𝐹 )) ⊕(𝐵 ⊗ (𝐶 ⊗ 𝐹 )) 𝐵 ⊗ (𝐶 ⊗ 𝐹) (𝐴⊗(𝐷 ⊗ 𝐹 )) ⊕(𝐵 ⊗ ((𝐶 ⊗ 𝐹 )⊕(𝐷 ⊗ 𝐸))) 𝐵 ⊗ ((𝐶 ⊗ 𝐹)⊕(𝐷 ⊗ 𝐸)) (𝐵 ⊗ (𝐶 ⊗ 𝐹 )) ⊕(𝐵 ⊗ (𝐷 ⊗ 𝐸)) ((𝐵 ⊗ 𝐶)⊗ 𝐹 )⊕((𝐵 ⊗ 𝐷)⊗ 𝐸) (𝐵 ⊗ 𝐶)⊗ 𝐹 (((𝐴⊗ 𝐷)⊕(𝐵 ⊗ 𝐶)) ⊗ 𝐹 )⊕((𝐵 ⊗ 𝐷)⊗ 𝐸) ((𝐴 ⊗ 𝐷)⊕(𝐵 ⊗ 𝐶)) ⊗ 𝐹 ((𝐴 ⊗ 𝐷)⊗ 𝐹)⊕((𝐵 ⊗ 𝐶)⊗ 𝐹 ) 𝛼 𝜄1 𝜀−1 𝜀 𝑎⊕𝑎 𝜄2 𝑎 Id ⊕(Id ⊗𝜄1) 𝜄2 Id ⊗𝜄1 𝑎−1 𝜄1 𝜔 𝜄2 𝛾−1 𝑎−1⊕𝑎−1 (𝜄2⊗Id)⊕Id 𝜄1 𝜄2 𝜄2⊗Id 𝜄1𝜀 4.1.2 Semigroupal categories 117 To prove that the natural isomorphism 𝑎 is an associator, we need to prove the associativity coherence diagram. ((𝑓 ⊗𝑔) ⊗ℎ) ⊗𝑘 (𝑓 ⊗𝑔) ⊗(ℎ ⊗𝑘) (𝑓 ⊗(𝑔 ⊗ℎ))  ⊗𝑘 𝑓 ⊗((𝑔 ⊗ℎ) ⊗𝑘)𝑓 ⊗(𝑔 ⊗(ℎ ⊗𝑘)). 𝑎𝑓 ⊗𝑔,ℎ,𝑘 𝑎𝑓,𝑔,ℎ  ⊗Id𝑘 𝑎𝑓,𝑔,ℎ ⊗𝑘 𝑎𝑓,𝑔  ⊗ℎ,𝑘 Id𝑓 ⊗𝑎𝑔,ℎ,𝑘 That means that we need to obtain two commutative diagrams given by the domain and codomain of the tensor product  ⊗. The lower part is immediate since it is the coherence diagram for the associator 𝑎. The upper part is the following one: 𝔄 𝔈 𝔅 ℭ 𝔇, 𝛼𝑓 ⊗𝑔,ℎ,𝑘 (𝛼𝑓,𝑔,ℎ⊗Id)⊕(𝑎𝐵,𝐷,𝐸 ⊗Id) 𝛼𝑓,𝑔,ℎ  ⊗𝑘 𝛼𝑓,𝑔  ⊗ℎ,𝑘 (Id ⊗𝑎𝐷,𝐹 ,𝐻 )⊕(Id ⊗𝛼𝑔,ℎ,𝑘) Where, for clarity, we will denote: 𝔄∶= (((((𝐴⊗𝐷)⊕(𝐵 ⊗ 𝐶)) ⊗ 𝐹 )⊕((𝐵 ⊗ 𝐷)⊗ 𝐸)) ⊗ 𝐻)⊕(((𝐵 ⊗ 𝐷)⊗ 𝐹 )⊗ 𝐺), 𝔅∶= (((𝐴 ⊗ (𝐷 ⊗ 𝐹 )) ⊕(𝐵 ⊗ ((𝐶 ⊗ 𝐹)⊕(𝐷 ⊗ 𝐸)))) ⊗ 𝐻)⊕((𝐵 ⊗ (𝐷 ⊗ 𝐹)) ⊗ 𝐺), ℭ∶= (𝐴 ⊗ ((𝐷 ⊗ 𝐹 )⊗ 𝐻)) ⊕(𝐵 ⊗ ((((𝐶 ⊗ 𝐹 )⊕(𝐷 ⊗ 𝐸)) ⊗ 𝐻)⊕((𝐷 ⊗ 𝐹)⊗ 𝐺))), 𝔇∶= (𝐴 ⊗ (𝐷 ⊗ (𝐹 ⊗ 𝐻))) ⊕(𝐵 ⊗ ((𝐶 ⊗ (𝐹 ⊗ 𝐻)) ⊕(𝐷 ⊗ ((𝐸 ⊗ 𝐻)⊕(𝐹 ⊗ 𝐺))))), 𝔈∶= (((𝐴 ⊗ 𝐷)⊕(𝐵 ⊗ 𝐶)) ⊗(𝐹 ⊗ 𝐻)) ⊕((𝐵 ⊗ 𝐷)⊗((𝐸 ⊗ 𝐻)⊕(𝐹 ⊗ 𝐺))). Following other cases, this can be proved by resolving injections to the coproduct, as we can see in the following diagrams taking the first injection. 124 4 On the Loday-Pirashvili Category (((𝐴⊗ 𝐷)⊕(𝐵 ⊗ 𝐶)) ⊗ 𝐹 )⊕((𝐵 ⊗ 𝐷)⊗ 𝐸) ((𝐴 ⊗ 𝐷)⊕(𝐵 ⊗ 𝐶)) ⊗ 𝐹 ((𝐴 ⊗ 𝐷)⊗ 𝐹 )⊕((𝐵 ⊗ 𝐶)⊗ 𝐹) 𝐹 ⊗ ((𝐴 ⊗ 𝐷)⊕(𝐵 ⊗ 𝐶)) (𝐴 ⊗ 𝐷)⊗ 𝐹 (𝐹 ⊗ ((𝐴 ⊗ 𝐷)⊕(𝐵 ⊗ 𝐶))) ⊕(𝐸 ⊗ (𝐵 ⊗ 𝐷)) 𝐹 ⊗ (𝐴⊗𝐷) (𝐸⊗(𝐵 ⊗ 𝐷)) ⊕(𝐹 ⊗ ((𝐴 ⊗ 𝐷)⊕(𝐵 ⊗ 𝐶))) 𝐹 ⊗ ((𝐴 ⊗ 𝐷)⊕(𝐵 ⊗ 𝐶)) 𝜏⊕𝜏 𝜏1 𝑓 ⊗𝑔,ℎ 𝜄1 𝜏 𝜀 𝜄1 𝜄1 𝜏 𝜄1⊗Id 𝜏⊕Id ⊗𝜄1 Id ⊗𝜄1 𝜄2 𝜄1 4.1.4 Monoidal categories 125 Definition 4.1.11. Let = (𝐂, ⊗, 𝑎, 𝜏)be a braided semigroupal category such that 𝐂is ⊗-distributive. The braided semigroupal category (Hom(𝐂), ⊗, 𝑎, 𝜏)defined in Theorem 4.1.10, will be denoted as LP()and will be called the Loday-Pirashvili braided semigroupal category of . Definition 4.1.12. Let = (𝐂, ⊗, 𝑎, 𝜏)be a braided semigroupal category. We say that is symmetric if for all objects 𝐴, 𝐵 of 𝐂is satisfied that 𝜏−1 𝐴,𝐵 =𝜏𝐵,𝐴. Proposition 4.1.13. Let = (𝐂, ⊗, 𝑎, 𝜏)be a braided semigroupal category such that 𝐂is ⊗-distributive. Then LP()is symmetric if and only if is symmetric. Proof. Let us assume first that LP()is symmetric. For any 𝐴, 𝐵 ∈ Ob(𝐂)we know that (𝜏Id𝐴,Id𝐵)−1 =𝜏Id𝐵,Id𝐴. Then, by taking the second component of the morphisms we have 𝜏−1 𝐴,𝐵 =𝜏𝐵,𝐴 Assume now that 𝐂is symmetric. For any 𝐴 𝐵 𝑓, 𝐶 𝐷 𝑔∈ Ob (Hom(𝐂))we have to prove that (𝜏𝑓,𝑔)−1 =𝜏𝑔,𝑓 . That means that we need to prove that (𝜏1 𝑓,𝑔, 𝜏𝐴,𝐵)−1 = (𝜏1 𝑔,𝑓 , 𝜏𝐵,𝐴), i.e. (𝜏1 𝑓,𝑔)−1 =𝜏1 𝑔,𝑓 and 𝜏−1 𝐴,𝐵◦𝜏𝐵,𝐴. The second one is true by hypothesis. We know that 𝜏1 𝑓,𝑔 is an isomorphism in 𝐂, since it is the component of an isomorphism, so it is just enough to prove 𝜏1 𝑔,𝑓 ◦𝜏1 𝑓,𝑔 = Id(𝐴⊗𝐷)⊕(𝐵⊗𝐶). (𝐴⊗𝐷)⊕(𝐵 ⊗ 𝐶) 𝜏1 𝑓,𝑔 //(𝐶 ⊗ 𝐵)⊕(𝐷 ⊗ 𝐴) 𝜏1 𝑔,𝑓 //(𝐴⊗𝐷)⊕(𝐵 ⊗ 𝐶) But this is obviously true, since what 𝜏1 𝑓,𝑔 does is insert each part of the coproduct into its correspondent one and then twist it. But these twists are inverse to each other by assumption. 4.1.4 Monoidal categories Definition 4.1.14. Let (𝐂, ⊗)be a category with an operation such that 𝐂has an initial object Λ. Then: 126 4 On the Loday-Pirashvili Category •𝐂is said to be left-⊗-annihilated (right-⊗-annihilated) if the unique morphism Λ𝐴⊗Λ∶ Λ →𝐴 ⊗ Λ(respectively, ΛΛ⊗𝐴 ∶ Λ →Λ⊗ 𝐴) is an isomorphism. That means 𝐴 ⊗ Λ(Λ⊗ 𝐴) is an initial object. •𝐂is ⊗-annihilated if it is both left-⊗-annihilated and right-⊗-annihilated. Remark 4.1.15.The definition of left-⊗-annihilated does not depend on the initial object. Assume that Λ′is also another initial object, we have ΛΛ′and Λ′ Λare inverse to each other, so Λ′ 𝐴⊗Λ′= (Id𝐴⊗ΛΛ′)◦Λ𝐴⊗Λ◦Λ′ Λ and Λ′ 𝐴⊗Λ′is an isomorphism by composition, since ⊗preserves isomorphisms. Theorem 4.1.16. Let (𝐂, ⊗, 𝑎, 𝐼, 𝑙, 𝑟)be a monoidal category such that (𝐂, ⊗)is distributive and annihilated. Let = (𝐂, ⊗, 𝑎)and consider its Loday-Pirashvili semigroupal category LP() = (Hom(𝐂), ⊗, 𝑎). We will denote  𝐼∶= Λ 𝐼 Λ𝐼, and take 𝐴 𝐵 𝑓∈ Ob(Hom(𝐂)). Let  𝑙𝑓∶= ( 𝑙1 𝑓, 𝑙𝐵)and 𝑟𝑓∶= (𝑟1 𝑓, 𝑟𝐵), where  𝑙1 𝑓is the following composition: (Λ ⊗ 𝐵)⊕(𝐼 ⊗ 𝐴) Λ−1 Λ⊗𝐵⊕Id𝐼⊗𝐴 //Λ⊕(𝐼 ⊗ 𝐴)(𝜄2)−1 //𝐼 ⊗ 𝐴 𝑙𝐴//𝐴 , and 𝑟1 𝑓is the following composition: (𝐴⊗𝐼)⊕(𝐵 ⊗ Λ) Id𝐴⊗𝐼 ⊕Λ−1 𝐵⊗Λ//(𝐴⊗𝐼)⊕Λ(𝜄1)−1 //𝐴⊗𝐼 𝑟𝐴//𝐴 . Then (Hom(𝐂), ⊗, 𝑎,  𝐼, 𝑙,𝑟)is a monoidal category. Proof. We will first prove that 𝑟𝑓is well defined. This is the same as showing that the following diagram is commutative: (𝐴⊗𝐼)⊕(𝐵 ⊗ Λ) [(𝑓⊗Id𝐼),(Id𝐵⊗Λ𝐼)]  𝑟1 𝑓//𝐴 𝑓  𝐵 ⊗ 𝐼 𝑟𝐵//𝐵 4.1.4 Monoidal categories 127 The composition with the second coproduct injection is trivial since 𝐵⊗Λis an initial object. The first one follows by the commutative diagram: (𝐴⊗𝐼)⊕(𝐵 ⊗ Λ) [(𝑓⊗Id),(Id ⊗Λ)]  𝑟1 𝑓// Id ⊗Λ−1 )) 𝐴 𝑓  (𝐴⊗𝐼)⊕Λ(𝜄1)−1 //𝐴⊗𝐼 𝑟<< 𝐴⊗𝐼 Id 88 𝑓⊗Id uu 𝜄1 cc 𝐵 ⊗ 𝐼 𝑟//𝐵 The fact that 𝑟 is a natural isomorphism is immediate by being composition of natural isomorphisms. The same arguments work for  𝑙. So we have to prove the triangle equation. For that, we will take 𝐶 𝐷 𝑔. The lower part is immediate since it is the triangle equation for the original monoidal category. We have to prove the triangle equation for the upper part, i.e. we need to prove the following diagram: (((𝐴 ⊗ 𝐼)⊕(𝐵 ⊗ Λ)) ⊗ 𝐷)⊕((𝐵 ⊗ 𝐼)⊗ 𝐶) (𝐴 ⊗ (𝐼 ⊗ 𝐷)) ⊕(𝐵 ⊗ ((Λ ⊗ 𝐷)⊕(𝐼 ⊗ 𝐶))) (𝐴 ⊗ 𝐷)⊕(𝐵 ⊗ 𝐶) 𝛼𝑓, 𝐼,𝑔 (𝑟1 𝑓⊗Id𝐷)⊕(𝑟𝐵⊗Id𝐶)(Id𝐴⊗𝑙𝐷)⊕(Id𝐵⊗ 𝑙1 𝑔) As in the previous theorem, we will use that the domain is a coproduct to study each part separately. The first one is ((𝐴 ⊗ 𝐼)⊕(𝐵 ⊗ Λ)) ⊗ 𝐷 and the first morphism from there is 𝜀, so we can precompose with that isomorphism. Now we have again a coproduct that we can study separately, but the second part is immediate since its domain, (𝐵 ⊕ Λ) ⊕ 𝐷, is an initial object. This is true because, since (𝐂, ⊗)is annihilated. Left-annihilation say that (𝐵 ⊗ Λ) is an initial object, and since annihilation does not depend on the initial object selected, right-annihilation say that (𝐵 ⊗ Λ) ⊗𝐶 128 4 On the Loday-Pirashvili Category is an initial object. So, we will prove the first part. The second part of the original coproduct is established similarly. (((𝐴⊗ 𝐼)⊕(𝐵 ⊗ Λ)) ⊗ 𝐷)⊕((𝐵 ⊗ 𝐼)⊗ 𝐶) ((𝐴 ⊗ 𝐼)⊕(𝐵 ⊗ Λ)) ⊗ 𝐷 ((𝐴 ⊗ 𝐼)⊗ 𝐷)⊕((𝐵 ⊗ Λ) ⊗ 𝐷) ((𝐴⊗ 𝐼)⊗ 𝐷)⊕((𝐵 ⊗ Λ) ⊗ 𝐷) (𝐴 ⊗ 𝐼)⊗ 𝐷 (𝐴⊗(𝐼 ⊗ 𝐷)) ⊕(𝐵 ⊗ ((Λ ⊗ 𝐷)⊕(𝐼 ⊗ 𝐶))) (𝐴 ⊗ (𝐼 ⊗ 𝐷)) ⊕(𝐵 ⊗ (Λ ⊗ 𝐷)) 𝐴 ⊗ (𝐼 ⊗ 𝐷) (𝐴⊗ 𝐷)⊕(𝐵 ⊗ 𝐶)𝐴 ⊗ 𝐷 𝛼𝑓, 𝐼,𝑔 𝜄1 𝜀−1 𝜀 𝑎⊕𝑎 𝜄1 𝑟⊗Id 𝑎 (Id ⊗𝑙)⊕(Id ⊗ 𝑙1 𝑔) Id ⊕(Id ⊗𝜄1)𝜄1 𝜄1 Id ⊗𝑙 𝜄1 (((𝐴 ⊗ 𝐼)⊕(𝐵 ⊗ Λ)) ⊗ 𝐷)⊕((𝐵 ⊗ 𝐼)⊗ 𝐶) ((𝐴 ⊗ 𝐼)⊕(𝐵 ⊗ Λ)) ⊗ 𝐷 ((𝐴 ⊗ 𝐼)⊗ 𝐷)⊕((𝐵 ⊗ Λ) ⊗ 𝐷) (𝐴⊗ 𝐷)⊕(𝐵 ⊗ 𝐶)𝐴 ⊗ 𝐷 (𝐴 ⊗ 𝐼)⊗ 𝐷 (𝑟1 𝑓⊗Id)⊕(𝑟⊗Id) 𝜄1 𝑟1 𝑓⊗Id 𝜀 𝜄1 𝜄1 𝑟⊗Id 𝜄1⊗Id With this we know that the triangular equation holds and (Hom(𝐂), ⊗, 𝑎,  𝐼, 𝑙,𝑟)is a monoidal category. Definition 4.1.17. Let = (𝐂, ⊗, 𝑎, 𝐼, 𝑙, 𝑟)be a monoidal category such that 𝐂is ⊗- distributive and ⊗-annihilated. The monoidal category (Hom(𝐂), ⊗, 𝑎,  𝐼, 𝑙,𝑟)defined in Theorem 4.1.16 will be denoted as LP()and will be called the Loday-Pirashvili monoidal category of . 4.1.5 Braided monoidal categories The notion of braided monoidal category was introduced by Joyal and Street in [38]. Definition 4.1.18. Abraided symmetric monoidal category is a braided monoidal category = (𝐂, ⊗, 𝑎, 𝐼, 𝑙, 𝑟, 𝜏)where (𝐂, ⊗, 𝑎, 𝜏)is a braided symmetric semigroupal category. 4.2 Additive Categories with operations 129 Definition 4.1.19. Let = (𝐂, ⊗, 𝑎, 𝐼, 𝑙, 𝑟, 𝜏)be a braided (symmetric) monoidal category such that 𝐂is ⊗-distributive and ⊗-annihilated. The braided (symmetric) monoidal category (Hom(𝐂), ⊗, 𝑎,  𝐼, 𝑙,𝑟, 𝜏)defined between Theorem 4.1.10 and Theorem 4.1.16, will be denoted as LP()and will be called the Loday-Pirashvili braided monoidal category of . Lemma 4.1.20. Let = (𝐂, ⊗, 𝑎, 𝐼, 𝑙, 𝑟, 𝜏)be a braided monoidal category such that 𝐂is ⊗-distributive and ⊗-annihilated. Then LP()is symmetric if and only if  is symmetric. 4.2 Additive Categories with operations In this section, we will try to give properties to the sum of two morphisms in an additive category with an operation in that category. Lemma 4.2.1. Let (𝐂, ⊗)be a category with an operation, where 𝐂has finite coproducts. Then the following diagrams are commutative: (𝐴⊗𝐵)⊕(𝐴⊗𝐵) 𝛾𝐴,𝐵,𝐵  ▽𝐴⊗𝐵 (( (𝐴⊗𝐵)⊕(𝐴⊗𝐵) 𝜀𝐴,𝐴,𝐵  ▽𝐴⊗𝐵 (( 𝐴 ⊗ 𝐵 𝐴 ⊗ 𝐴 𝐴 ⊗ (𝐵 ⊕ 𝐵) Id𝐴⊗▽𝐵 66 (𝐴⊕𝐴)⊗ 𝐵 ▽𝐴⊗Id𝐵 66 where 𝜀and 𝛾are defined in Definition 4.1.5. Proof. Since the domain of both are coproducts we can use the universal property. Take 𝑘∈ {1,2}. (Id𝐴⊗▽𝐵)◦𝛾𝐴,𝐵,𝐵◦𝜄𝑘= (Id𝐴⊗▽𝐵)◦(Id𝐴⊗𝜄𝑘) = Id𝐴⊗(▽𝐵◦𝜄𝑘) = Id𝐴⊗Id𝐵= Id𝐴⊗𝐵 =▽𝐴⊗𝐵◦𝜄𝑘 The second one follows analogously. 130 4 On the Loday-Pirashvili Category Theorem 4.2.2. Let (𝐂, ⊗)a category with an operation, distributive and annihilated, where 𝐂has biproducts. Then, (i) (𝑓+𝑔)⊗ ℎ = (𝑓 ⊗ ℎ)+(𝑔 ⊗ ℎ)for 𝐴𝑓,𝑔 ←←←←←←←←←←←←←→ 𝐵,𝐶ℎ ←←←←←←→ 𝐷, (ii) 𝑓 ⊗ (𝑔+ℎ) = (𝑓 ⊗ 𝑔)+(𝑓 ⊗ ℎ)for 𝐴𝑓 ←←←←←←←→ 𝐵,𝐶𝑔,ℎ ←←←←←←←←←←←←→ 𝐷. Proof. We will show part (i) since the second one is completely analogue. We need to prove that the following diagram is commutative: (𝐴 ⊕ 𝐴)⊗ 𝐶 (𝐵 ⊕ 𝐵)⊗ 𝐷 𝐴⊗ 𝐶 𝐵 ⊗ 𝐷 (𝐴⊗ 𝐶)⊕(𝐴⊗𝐶) (𝐵 ⊗ 𝐷)⊕(𝐵 ⊗ 𝐷) (𝑓⊕𝑔)⊗ℎ ▽⊗Id △⊗Id △ (𝑓⊗ℎ)⊕(𝑔⊗ℎ) ▽ Besides, to check that the previous diagram is commutative is equivalent to prove that the following subdiagrams are commutative: (𝐴 ⊕ 𝐴)⊗ 𝐶 (𝐵 ⊕ 𝐵)⊗ 𝐷 𝐴⊗ 𝐶 𝐵 ⊗ 𝐷 (𝐴⊗ 𝐶)⊕(𝐴⊗𝐶) (𝐵 ⊗ 𝐷)⊕(𝐵 ⊗ 𝐷) (𝑓⊕𝑔)⊗ℎ ▽⊗Id △⊗Id𝐶 △ 𝜀 (𝑓⊗ℎ)⊕(𝑔⊗ℎ) 𝜀 ▽ The rightmost subdiagram has already appeared in Lemma 4.2.1, whereas the middle subdiagram is naturalness from Proposition 4.1.6. Let us prove that the left subdiagram is commutative. For doing this, we will define 𝛿𝑋⊕𝑌 𝑖,𝑗 ∶= 𝜄𝑋⊕𝑌 𝑖◦𝜋𝑋⊕𝑌 𝑗, i.e. 𝛿𝑖,𝑗 = Id if 𝑖=𝑗and 𝛿𝑖,𝑗 = 0 if 𝑖≠𝑗. 4.2 Additive Categories with operations 131 Given the following diagram (𝐴⊕𝐴)⊗ 𝐶 𝜀−1  𝜋1⊗Id  𝐴⊗𝐶 △⊗Id 66 △(( 𝐴⊗𝐶 𝜄𝑖⊗Id hh 𝜄𝑖 vv𝜄𝑖◦𝜋𝑗%% (𝐴⊗𝐶)⊕(𝐴⊗𝐶) 𝜀 OO 𝜋𝑗//𝐴⊗𝐶 for 𝑖, 𝑗 ∈ {1,2}. Note that 𝜄𝑖◦𝜋𝑗= Id𝑋⊗𝑌 if 𝑖=𝑗, and 𝜄𝑖◦𝜋𝑗= 0, if 𝑖≠𝑗. Then, it is straightforward that each subdiagram of the right is commutative: if 𝑖=𝑗, it is immediate, and if 𝑖≠𝑗, the topmost triangle comes from left-⊗-annihilation (0⊗ Id𝐶= 0). Therefore, using that the object in the bottom is a coproduct, the outer right triangle is commutative: (𝐴⊕𝐴)⊗ 𝐶 𝜀−1  𝜋1⊗Id  𝐴⊗𝐶 △⊗Id 66 △(( (𝐴⊗𝐶)⊕(𝐴⊗𝐶) 𝜀 OO 𝜋𝑗//𝐴⊗𝐶 Since the object of the bottom is also a product and we have the out square is commutative, we conclude that the left triangle is also commutative. Corollary 4.2.3. Let (𝐂, ⊗)a category with an operation distributive and annihilated, where 𝐂is additive. Then, for 𝐴𝑓 ←←←←←←←→ 𝐵,𝐶𝑔 ←←←←←←→ 𝐷morphisms and 𝐸0 ←←←←←←→ 𝐹, we have: (i) 𝑓 ⊗ 0∶ 𝐴⊗𝐸→𝐵 ⊗ 𝐹 is the zero morphism. (ii) 0⊗ 𝑓 ∶𝐸 ⊗ 𝐴 →𝐹 ⊗ 𝐵 is the zero morphism (iii) 𝑓 ⊗ (−𝑔) = (−𝑓)⊗ 𝑔 = −(𝑓 ⊗ 𝑔) 132 4 On the Loday-Pirashvili Category Proof. It follows using the previous theorem together with the Abelian group properties of the homomorphisms. Remark 4.2.4.Since the definition of additive category resides in natural morphisms, it is immediate that if 𝐂has biproducts, then the category Hom(𝐂)also has biproducts. In fact, it can be easily proved that the operation in this category is (𝑓, 𝑔)+(ℎ, 𝑘) = (𝑓+ℎ, 𝑔+𝑘). Thus, if 𝐂is additive, then Hom(𝐂)is additive and −(𝑓, 𝑔) = (−𝑓, −𝑔). 4.3 Lie and Leibniz objects in LP 4.3.1 Lie objects and Leibniz Objects In this subsection, we will study the Lie objects and the Leibniz objects. The definitions of that concepts can be seen in Definition 2.4.3 and Definition 2.4.5. Lemma 4.3.1. Let = (𝐂, ⊗, 𝑎, 𝜏)be a symmetric semigroupal category where 𝐂is an additive category. Let (𝐿, 𝜇)be a Leibniz object. Then, 𝜇◦(Id𝐿⊗𝜇) = −𝜇◦(Id𝐿⊗𝜇)◦(Id𝐿⊗𝜏𝐿,𝐿). Proof. By symmetry, if we compose the Leibniz identity with 𝑎−1◦(Id𝐿⊗𝜇)◦𝑎, we get 𝜇◦(𝜇 ⊗ Id𝐿)◦𝑎−1 𝐿◦(Id𝐿⊗𝜏𝐿,𝐿)◦𝑎𝐿=𝜇◦(𝜇 ⊗ Id𝐿) + 𝜇◦(Id𝐿⊗𝜇)◦(Id𝐿⊗𝜏𝐿,𝐿)◦𝑎𝐿. Substituting it in the Leibniz identity, we obtain 𝜇◦(𝜇 ⊗ Id𝐿) = 𝜇◦(Id𝐿⊗𝜇)◦𝑎𝐿+𝜇◦(𝜇 ⊗ Id𝐿) +𝜇◦(Id𝐿⊗𝜇)◦(Id𝐿⊗𝜏𝐿,𝐿)◦𝑎𝐿 Finally, subtracting 𝜇◦(𝜇 ⊗ Id𝐿) + 𝜇◦(Id𝐿⊗𝜇)◦(Id𝐿⊗𝜏𝐿,𝐿)◦𝑎𝐿in both sides of the identity, we get −𝜇◦(Id𝐿⊗𝜇)◦(Id𝐿⊗𝜏𝐿,𝐿)◦𝑎𝐿=𝜇◦(Id𝐿⊗𝜇)◦𝑎𝐿, which composed with 𝑎−1 𝐿gives us the desired identity. 4.3.1 Lie objects and Leibniz Objects 133 Proposition 4.3.2. Let = (𝐂, ⊗, 𝑎, 𝜏)be a braided symmetrical semigroupal category where 𝐂is an additive ⊗-distributive ⊗-annihilated category. The following identity is called the Jacobi identity: 𝜇◦(Id𝐿⊗𝜇)◦(Id𝐿⊗(𝐿⊗𝐿)+𝑎𝐿◦𝜏𝐿,𝐿⊗𝐿 +𝜏𝐿⊗𝐿,𝐿◦𝑎−1 𝐿) = 0.(Jac) Then (𝐿, 𝜇)is a Lie object if and only if it satisfies (AC) and (Jac). Proof. Let (𝐿, 𝜇)be a Lie object and let us prove that (Jac) holds. We can rewrite the Leibniz identity as 0 = 𝜇◦(Id𝐿⊗𝜇) + 𝜇◦(𝜇 ⊗ Id𝐿)◦𝑎−1 𝐿◦(Id𝐿⊗𝜏𝐿,𝐿) − 𝜇◦(𝜇 ⊗ Id𝐿)◦𝑎−1 𝐿. The first and third summands are equal to the first and third summands of (Jac), respectively, since 𝜏𝐿,𝐿◦(𝜇 ⊗ Id𝐿) = (Id𝐿⊗𝜇)◦𝜏𝐿⊗𝐿,𝐿. To see the second one, we will use (AC), the naturalness and symmetry of 𝜏, the second hexagon equation and Lemma 4.3.1: 𝜇◦(𝜇 ⊗ Id𝐿)◦𝑎−1 𝐿◦(Id𝐿⊗𝜏𝐿,𝐿) = −𝜇◦𝜏𝐿,𝐿◦(𝜇 ⊗ Id𝐿)◦𝑎−1 𝐿◦(Id𝐿⊗𝜏𝐿,𝐿) = −𝜇◦(Id𝐿⊗𝜇)◦𝜏𝐿⊗𝐿,𝐿◦𝑎−1 𝐿◦(Id𝐿⊗𝜏𝐿,𝐿) = −𝜇◦(Id𝐿⊗𝜇)◦𝜏−1 𝐿,𝐿⊗𝐿◦𝑎−1 𝐿◦(Id𝐿⊗𝜏𝐿,𝐿) = −𝜇◦(Id𝐿⊗𝜇)◦𝑎𝐿◦(𝜏𝐿,𝐿 ⊗Id𝐿)−1◦𝑎−1 𝐿 = −𝜇◦(Id𝐿⊗𝜇)◦𝑎𝐿◦(𝜏𝐿,𝐿 ⊗Id𝐿)◦𝑎−1 𝐿 =𝜇◦(Id𝐿⊗𝜇)◦(Id𝐿⊗𝜏𝐿,𝐿)◦𝑎𝐿◦(𝜏𝐿,𝐿 ⊗Id𝐿)◦𝑎−1 𝐿 =𝜇◦(Id𝐿⊗𝜇)◦𝑎𝐿◦𝜏𝐿,𝐿⊗𝐿. Conversely, using Theorem 4.2.2 and Corollary 4.2.3, the identity (AC) gives us the same equality given in Lemma 4.3.1. −𝜇◦Id𝐿⊗𝜇◦Id𝐿⊗𝜏𝐿,𝐿 =𝜇◦Id𝐿⊗(−𝜇)◦Id𝐿⊗𝜏𝐿,𝐿 =𝜇◦Id𝐿⊗(−𝜇◦𝜏𝐿,𝐿) = 𝜇◦Id𝐿⊗𝜇. Then, the identity (Lb) automatically follows.