Cohomology, Extensions and Automorphisms of Skew Braces
aa r X i v : . [ m a t h . G R ] F e b COHOMOLOGY, EXTENSIONS AND AUTOMORPHISMS OF BRACES
NISHANT AND MANOJ K. YADAV
Abstract.
We formalize the definition of low dimensional cohomology of a left brace with coefficientsin a trivial brace with actions, study its connections with extensions of left braces and construct Wells’like exact sequence relating second cohomology group with inducible automorphisms of extensions ofleft braces. Introduction
Classification of set-theoretic solutions of the Yang-Baxter equation, proposed by Drinfeld [11], is awide open problem. In 2007, Rump [21] introduced the notion of braces in connection with non-degenerateset theoretic solutions of the Yang-Baxter equation. The definition of a brace, which we use here andis frequently used in the contemporary literature, is slightly different but equivalent to the one given byRump, and is due to Cedo, Jespers and Onkinski [9]. An algebraic structure ( E, + , ◦ ) is said to be a leftbrace if ( E, +) is an abelian group, ( E, ◦ ) is a group and, for all a, b, c ∈ E , the following compatibilitycondition holds: a ◦ ( b + c ) + a = a ◦ b + a ◦ c. A right brace can be defined analogously. In the present article, we shall consider only left braces.As is well known now (see [9]), a left brace gives rise to a non-degenerate involutive set theoretic solutionof the Yang-Baxter equation (and vice-versa). Not only this, braces have deep connections with manyother algebraic structures; linear cycle sets, radical rings, Hopf-Galois extensions, bijective 1-cocycles, toname some. So each new construction of a left brace, contributes to the solutions of the Yang-Baxterequation and enriches the class of other related structures. Constructing new algebraic structure from theexisting ones is always an important and interesting problem. So is the case for braces. We mention somesuch constructions. Notion of semidirect product of braces was introduced in [22], which was generalisedto asymmetric product of braces in [10]. Asymmetric product of braces was further studied in [3], wherethe authors also studied the wreath product of braces. Matched product of braces was investigated in[1]. Iterated matched product of braces was taken up in [2]. Our interest in this note is the cohomologyand extension theory of braces.Vendramin [15] studied dynamical extensions of finite cycle sets and its relationship with dynamicalco-cycles. The topic was further studied by Castelli, Catino, Miccoli and Pinto [8] for quasi-linear cyclesets. Dynemical extensions give rise to non-degenerate involutive multipermutation solutions of theYang-Baxter equation. Extensions of bijective 1-cocycles were investigated by Ben David and Ginosar[5]. Homology and cohomology theories for solution sets of the Yang-Baxter equations were developed byCarter, Elhambadi and Saito [7]. Different homology theories for various structures related to solutions ofthe Yang-Baxter equations were extensively investigated by Lebel and Vendramin [17]. They also definedsecond cohomology group for a cycle set X with coefficients in an abelian group A and established abijective correspondence between the second cohomology group and equivalence classes of extensions of X by A . Further, a general homology and cohomology theory was developed for linear cycle sets byLebed and Vendramin [16] with trivial actions, where the authors extensively explored close connectionsbetween second cohomology of linear cycle sets and extension theory. Analogous results for left braces Mathematics Subject Classification.
Key words and phrases. left brace, cohomology, extension, automorphism. are also obtained in most of the cases by the authors. J.A. Guccione and J.J. Guccione [12] furtherinvestigated the ideas of [16] and used ‘perturbation lemma’ for computing first and second cohomologygroups and extensions of cyclic linear cycles set acting trivaialy on an abelian group. Bachiller [1]considered extensions of left braces with actions and characterised equivalent extension in the form ofcertain cocycles and relations among them.In the present article, we formalize the ideas developed in [1] in cohomological settings showing that theextension theory in the present case encompasses the extension theory investigated in [16], and present anexact sequence connecting automorphism groups of braces with second cohomology. Let H := ( H, + , ◦ )be a left brace and I := ( I, +) an abelian group viewed as a trivial left brace. Suppose that ( H, ◦ ) actson ( I, +) from left by an action ν and from right by an acton σ . The images of h ∈ H in Aut( I, +),the automorphism group of the group ( I, +), under ν and σ are denoted by ν h and σ h respectively. Ofparticular interest are the pair of actions ( ν, σ ) which satisfy ν h + h ( σ h + h ( y )) + y = ν h ( σ h ( y )) + ν h ( σ h ( y ))for all h , h ∈ H , y ∈ I . Such pairs will be called good pair of actions. For a given good pair of actions,we define H iN ( H, I ), i = 1 ,
2, the ith cohomology group of H with coefficients in I (see Section 3). LetExt( H, I ) denote the set of equivalence classes of extensions of H by I (see Section 2 for the definition).It turns out that Ext( H, I ) = G ( ν,σ ) Ext ν,σ ( H, I ) , where the pairs ( ν, σ ) run over all good pairs of actions of H on I and Ext ν,σ ( H, I ) is the set of allequivalence classes of extensions of H by I whose corresponding good pair of actions is ( ν, σ ) (see Corollary3.5). We can now state our first result Theorem A.
Let H be a left brace, which acts on a trivial brace I by a good pair of actions ( ν, σ ) . Thenthere exists a bijection between H N ( H, I ) and Ext ν,σ ( H, I ) . Let Z N ( H, I ) denote the group of all derivations from H to I as defined in Section 3. For an extension E : 0 → I → E π → H of H by I , where I is viewed as an ideal of E , we denote by Autb I ( E ) the set of all brace automorphismsof E which normalize I . It follows that Autb I ( E ) is a subgroup of Autb( E ), the group of all braceautomorphisms of E . For the extension E , as explained in Section 3, one can associate a unique goodpair of actions ( ν, σ ). LetC ( ν,σ ) := { ( φ, θ ) ∈ Autb( H ) × Autb( I ) | ν h = θ − ν φ ( h ) θ and σ h = θ − σ φ ( h ) θ } . Our next result is an analog for braces of an exact sequence, called the fundamental exact sequenceof Wells [23], relating derivations, automorphisms and cohomology group of groups. For unexplainedsymbols in the following result see Section 5.
Theorem B.
Let E : 0 → I → E π → H be a extension of a left brace H by a trivial left brace I such that [ E ] ∈ Ext ν,σ ( H, I ) . Then we have the following exact sequence of groups → Z N ( H, I ) → Autb I ( E ) ρ ( E ) −→ C ν,σ ω ( E ) −→ H N ( H, I ) , where ω ( E ) is, in general, only a derivation. We remark that the fundamental exact sequence in group theoretical setting has been revisited, refor-mulated and applied by many authors in the past, for example, see [13, 14, 18, 20]. The exposition whichwe present here is analogous to the one in [14]. A unified treatment of the fundamental exact sequenceof Wells with various applications is carried out in all fine details in [19, Chapter 2]. A similar exactsequence for cohomology, extensions and automorphisms of quandles was constructed in [4].We close this section with a quick layout of the article. Section 2 contains basic definitions andobservation on left braces. In Section 3 we formalise the ideas on extension theory developed in [1] in
OHOMOLOGY, EXTENSIONS AND AUTOMORPHISMS OF BRACES 3 cohomological setting and prove Theorem A (as Theorem 3.7). An explicit construction is carried outfor left braces of order 8. In Section 4, generalising some ideas developed in [16], we define generalcohomology theory in a particular case, but with actions. The last section deals with automorphismsof braces, where various actions are exhibited which prepare a road map for the proof of Theorem B(as Theorem 5.7). Finally, a reduction argument for lifting and extension of automorphisms of braces isdiscussed. 2.
Preliminaries
An algebraic structure ( E, + , ◦ ) is said to be a left brace if ( E, +) is an abelian group, ( E, ◦ ) is a groupand, for all a, b, c ∈ E , the following compatibility condition holds: a ◦ ( b + c ) + a = a ◦ b + a ◦ c. (1)Notice that the identity element 0 of ( E, +) coincides with the identity element 1 of ( E, ◦ ).For a left brace E and a ∈ E , define a map λ a : E → E by λ a ( b ) = a ◦ b − a for all b ∈ E . The automorphism group of a group G is denoted by Aut( G ). We have the following resultwhich was proved by Rump [21] in linear cycle set settings. Lemma 2.1.
For each a ∈ E , the map λ a is an automorphism of ( E, +) and the map λ : ( E, ◦ ) → Aut( E, +) given by λ ( a ) = λ a is a group homomorphism. A subbrace I of a left brace E is said to be a left ideal of E if λ a ( y ) ∈ I for all a ∈ E and y ∈ I . Aleft ideal of E is said to be an ideal if ( I, ◦ ) is a normal subgroup of ( E, ◦ ). An ideal I of E is said to be central if y ◦ a = a ◦ y = a + y for all a ∈ A and y ∈ I .The following is an easy but important observation, which will be used several times in what follows. Lemma 2.2.
Let E be a left brace. Then for all a, b ∈ E , the following hold:(i) a + b = a ◦ λ − a ( b ) .(ii) a ◦ b = a + λ a ( b ) . Let E and E be two left braces. A map f : E → E is said to be a brace homomorphism if f ( a + b ) = f ( a ) + f ( b ) and f ( a ◦ b ) = f ( a ) ◦ f ( b ) for all a, b ∈ E . A one-to-one and onto bracehomomorphism from E to itself is called an automorphism of E . The kernel of a homomorphism f : E → E is defined to be the subset { a ∈ E | f ( a ) = 0 } of E . It turns out that Ker( f ), the kernelof f , is an ideal of E . The set of all brace automorphisms of a left brace E , denoted by Autb( E ), is agroup.Let H and I be two left braces. By an extension of H by I , we mean a left brace E with an exactsequence E := 0 → I i → E π → H → , where i and π , respectively, are injective and surjective brace homomorphisms. An extension E is saidto a central extension if the image of I under i is a central ideal of E . A set map s : H → E is called a set-theoretic section of π if π ( s ( h )) = h for all h ∈ H and s (0) = 0. The abbreviation ‘st-section’ will beused for ‘set-theoretic section’ throughout.Let E and E ′ be two extension of H by I , that is, E : 0 → I i → E π → H → E ′ : 0 → I i ′ → E π ′ → H → NISHANT AND MANOJ K. YADAV are exact sequences of braces. The extensions E and E ′ are said to be equivalent if there exists a bracehomomorphism φ : E → E ′ such that the following diagram commutes:0 i −−−−→ I −−−−→ E π −−−−→ H −−−−→ Id y φ y y Id i ′ −−−−→ I −−−−→ E ′ π ′ −−−−→ H −−−−→ . The set of all equivalence classes of extensions of H by I is denoted by Ext( H, I ).Let H be a left brace. An abelian brace I , that is, ( I, ◦ ) is also abelian, is said to be an H -bi-module if there exists a group homomorphism ν : ( H, ◦ ) → Autb( I ) and a group anti-homomorphism σ : ( H, ◦ ) → Autb( I ). This means that H acts on I from left through ν and from right through σ . Foran ideal I of a left brace E , we get the natural exact sequence I ֒ → E π ։ H, where H ∼ = E/I . Suppose that the ideal I is a trivial brace. Let s be an st-section of π . Then we candefine a left action ν : H → Autb( I ) and a right action σ : ( H, ◦ ) → Autb( I ) as defined in Section 3 by(5) and (6) respectively. Thus I becomes an H -bi-module. All H -bi-modules I , in this paper, will betaken trivial braces. More dictinctly, I will always denote a trivial brace throughout.Let H and H ′ be left braces, and I and I ′ be H and H ′ bi-modules with ( ν, σ ) and ( ν ′ , σ ′ ), respectively,as actions. Let α : H ′ → H and ζ : I → I ′ be homomorphisms of braces. The pair ( α, ζ ) is said to be compatible with the pairs of actions ( ν, σ ) and ( ν ′ , σ ′ ) if the following diagram commutes for both leftand right actions: H × I ( ν, σ ) −−−−→ I α x y ζ y ζ H ′ × I ′ ( ν ′ , σ ′ ) −−−−→ I ′ . (2)More precisely, ( α, ζ ) is compatible with the actions, if ζ ( ν α ( h ′ ) ( y )) = ν ′ h ′ ( ζ ( y ))and ζ ( σ α ( h ′ ) ( y )) = σ ′ h ′ ( ζ ( y )) . Let I and I ′ be two H -bi-modules with actions ( ν, σ ) and ( ν ′ , σ ′ ) respectively. A map ζ : I → I ′ is called H -bi-module homomorphism if (Id , ζ ) is compatible with the pairs of actions ( ν, σ ) and ( ν ′ , σ ′ ), whereId : H → H is the identity homomorphism.3. Second Cohomology and Extensions of Braces
Let H := ( H, + , ◦ ) be a brace and I := ( I, +) an abelian group viewed as a trivial brace. Let ν : ( H, ◦ ) → Aut( I, +) and σ : ( H, ◦ ) → Aut( I, +), respectively, be left and right group actions of ( H, ◦ )on ( I, +). We denote the image of h ∈ H under ν and σ by ν h and σ h , respectively. We call a pair ( ν, σ )of such actions a good pair if, for all h , h ∈ H , y ∈ I , the following relation holds: ν h + h ( σ h + h ( y )) + y = ν h ( σ h ( y )) + ν h ( σ h ( y )) . Let Sh r,j − r be the set of all permutations p on the set { , . . . , j } such that p (1) ≤ · · · ≤ p ( r ), p ( r +1) ≤ · · · ≤ p ( j ), where 1 ≤ r ≤ j −
1. For given i ≥ j ≥ r in the range 1 ≤ r ≤ j − h , . . . , h i , h i +1 , . . . , h i + j ) ∈ H i + j , an expression of the form X p ∈ Sh r,j − r ( − p (cid:0) h , . . . , h i , h i + p − (1) , . . . , h i + p − ( j ) (cid:1) OHOMOLOGY, EXTENSIONS AND AUTOMORPHISMS OF BRACES 5 is called a partial shuffle .For i ≥ j ≥
1, let Fun( H i + j , I ) denote the abelian group of all functions from H i + j to I and C ijN := C ijN ( H, I ) denote the subgroup of Fun( H i + j , I ) consisting of all functions whose linearisationvanishes on all partial shuffles defined above and whose values on all degenerate tuples is zero. A tuple( h , . . . , h n ) ∈ H n is said to be degenerate if h i = 0 for at least one i , 1 ≤ i ≤ n . Set C N := I ν = { y ∈ I | ν h ( y ) = y for all h ∈ H } and for i ≥ j ≥
1, set C nN := L i + j = n C i,jN . We are mainly interestedin small values of n , that is, n ≤
3. To be more precise, we take C N = C , N , C N = C , N ⊕ C , N and C N = C , N ⊕ C , N ⊕ C , N . We wish to constitute the following zero-sequence: C N ∂ −→ C N ∂ −→ C N ∂ −→ C N . Define ∂ : C N → C N by ∂ ( y ) = f y , where f y ( h ) = ν h ( σ h ( y )) − y and ∂ : C N → C N by ∂ ( θ ) = ( g, f ), where θ ∈ C N and for h , h ∈ H , g ( h , h ) = θ ( h ) − θ ( h + h ) + θ ( h ) ,f ( h , h ) = ν h ( θ ( h )) − θ ( h ◦ h ) + ν h ◦ h ( σ h ( ν − h ( θ ( h )))) . Next define ∂ : C N → C N by ∂ ( f, g ) = (cid:0) ∂ , v ( g ) , ∂ , h ( g ) − ∂ , v ( f ) , ∂ , h ( f ) (cid:1) , where ( g, f ) ∈ C N and, for h , h , h ∈ H , ∂ , v : C , N → C , N , ∂ , h : C , N → C , N , ∂ , v : C , N → C , N and ∂ , h : C , N → C , N are defined by ∂ , v ( g )( h , h , h ) = g ( h , h ) − g ( h + h , h ) + g ( h , h + h ) − g ( h , h ) ,∂ , h ( g )( h , h , h ) = ν h ( g ( h , h )) − g ( h ◦ h , h ◦ h ) + g ( h , h ◦ ( h + h )) ,∂ , v ( f )( h , h , h ) = f ( h , h ) − f ( h , h + h ) + f ( h , h ) ,∂ , h ( f )( h , h , h ) = ν h ( f ( h , h )) − f ( h ◦ h , h ) + f ( h , h ◦ h ) − ν h ◦ h ◦ h ( σ h ( ν − h ◦ h f ( h , h ))) . We can now prove
Lemma 3.1.
Let ( ν, σ ) be a good pair of actions of H on I . Then the sequence C N ∂ −→ C N ∂ −→ C N ∂ −→ C N is indeed a zero-sequence.Proof. Since ( ν, σ ) is a good pair of actions, it follows that f y ( h + h ) = f y ( h ) + f y ( h ). So, for y ∈ C N ,take ∂ ( y ) = f y and ∂ ( f y ) = ( g, f ). Then for all h , h ∈ H , we get g ( h , h ) = f y ( h ) − f y ( h + h ) + f y ( h ) = 0 . Using the properties of left and right actions ν and σ of H on I , respectively, and the fact that ν actstrivially on C N , we get f ( h , h ) = ν h ( f y ( h )) − f y ( h ◦ h ) + ν h ◦ h ( σ h ( ν − h ( f y ( h ))))= ν h ◦ h ( σ h ( y )) − y − ν h ◦ h ( σ h ◦ h ( y )) + y + ν h ◦ h ( σ h ◦ h ( y )) − ν h ◦ h ( σ h ( y ))= 0 . This shows that ∂ ◦ ∂ = 0. That ∂ ◦ ∂ = 0 can also be easily shown by a direct computation. Wedemonstrate it here only in one case. Let θ ∈ C N and ∂ ( θ ) = ( g, f ). We’ll show that ∂ , h ( g )( h , h , h ) − NISHANT AND MANOJ K. YADAV ∂ , v ( f )( h , h , h ) = 0 for all h , h , h ∈ H . We compute each component separately. ∂ , h ( g )( h , h , h ) = ν h ( g ( h , h )) − g ( h ◦ h , h ◦ h ) + g ( h , h ◦ ( h + h ))= ν h ( θ ( h )) − ν h ( θ ( h + h )) + ν h ( θ ( h )) − θ ( h ◦ h ) + θ ( h + h ◦ ( h + h )) − θ ( h ◦ h )+ θ ( h ) − θ ( h + h ◦ ( h + h )) + θ ( h ◦ ( h + h ))= ν h ( θ ( h )) − ν h ( θ ( h + h )) + ν h ( θ ( h )) − θ ( h ◦ h ) − θ ( h ◦ h ) + θ ( h ) + θ ( h ◦ ( h + h ))and ∂ , v ( f )( h , h , h ) = f ( h , h ) − f ( h , h + h ) + f ( h , h )= ν h ( θ ( h )) − θ ( h ◦ h ) + ν h ◦ h ( σ h ( ν − h ( θ ( h )))) − ν h ( θ ( h + h )) + θ ( h ◦ ( h + h )) − ν h ◦ ( h + h ) ( σ h + h ( ν − h ( θ ( h ))))+ ν h ( θ ( h )) − θ ( h ◦ h ) + ν h ◦ h ( σ h ( ν − h ( θ ( h )))) . Subtracting the second expression from the first and solving on the right hand side, we get ∂ , h ( g )( h , h , h ) − ∂ , v ( f )( h , h , h ) = θ ( h ) + ν h (cid:16) ν h + h ( σ h + h ( ν − h ( θ ( h )))) − ν h ( σ h ( ν − h ( θ ( h )))) − ν h ( σ h ( ν − h ( θ ( h )))) (cid:17) . The pair ( ν, σ ) being good pair, we finally get ∂ , h ( g )( h , h , h ) − ∂ , v ( f )( h , h , h ) = θ ( h ) − ν h ( ν − h ( θ ( h ))) = 0 . The proof is now complete. (cid:3)
Let Z iN ( H, I ) := Ker( ∂ i ) and B iN ( H, I ) := Im( ∂ i − ) for i = 1 ,
2. DefineH iN ( H, I ) := Z iN ( H, I ) /B iN ( H, I ) , the i th cohomology group of H with coefficients in I . Elements of Z iN ( H, I ) and B iN ( H, I ) are called i -cocycles and i -coboundaries respectively. Two 2-cocycles ( β , τ ) and ( β , τ ) are said to be cohomologous if ( β , τ ) − ( β , τ ) ∈ B N ( H, I ); more precsely, if there exists a θ ∈ C N such that( β − β )( h , h ) = θ ( h ) − θ ( h + h ) + θ ( h )and ( τ − τ )( h , h ) = ν h ( θ ( h )) − θ ( h ◦ h ) + ν h ◦ h ( σ h ( ν − h ( θ ( h ))))for all h , h ∈ H . The elements of Z N ( H, I ) are also called derivations .The next result follows from the definitions.
Lemma 3.2.
Let ( g, f ) ∈ C N . Then, for all h , h ∈ H , the following hold:(i) g ( h , h ) = g ( h , h ) .(ii) g ( h ,
0) = g (0 , h ) = 0 .(iii) f ( h ,
0) = f (0 , h ) = 0 . Let ( β, τ ) ∈ C N . Define on H × I , the following operations:( h , y ) + ( h , y ) = (cid:0) h + h , y + y + β ( h , h ) (cid:1) . (3)( h , y ) ◦ ( h , y ) = (cid:0) h ◦ h , ν h ◦ h ( σ h ( ν − h ( y ))) + ν h ( y ) + τ ( h , h ) (cid:1) . (4)Then we have OHOMOLOGY, EXTENSIONS AND AUTOMORPHISMS OF BRACES 7
Theorem 3.3.
Let ( ν, σ ) be a good pair of actions of H on I . Then H × I takes the structure of a leftbrace under the operations defined in the preceding para if and only if ( β, τ ) ∈ Z N ( H, I ) .Proof. If ( β, τ ) ∈ Z N ( H, I ), then, on the lines of the proof of [1, Theorem 3.3], using Lemma 3.2 itfollows that the operations under hypothesis turn H × I into a left brace. Let ( E, + , ◦ ) denote the bracedefined on H × I with the operations under hypothesis. Then, that ( β, τ ) ∈ Z N ( H, I ) follows from theassociativity of ′ + ′ and ′ ◦ ′ and the compatibility condition (1) of the brace E . (cid:3) The left brace structure on H × I (as given in Theorem 3.3) is an extension of H by I , which we denoteby the 6-tuple data ( H, I, ν, σ, β, τ ) with the exact sequence0 → I i → ( H, I, ν, σ, β, τ ) π → H → , where i ( y ) = (0 , y ) and π ( h, y ) = h . Throughout the paper, for such extensions, we’ll always take a fixedst-section s : H → ( H, I, ν, σ, β, τ ) of π given by s ( h ) = ( h, → I → E π → H → H by I . For the ease of notation, we shall view I as anideal of E through the given embedding. Let s : H → E be any st-section of π . Then, for all h ∈ H and y ∈ I , we define ν, σ : ( H, ◦ ) → Aut ( I, +) by ν h ( y ) := s ( h ) ◦ y − s ( h ) = λ s ( h ) ( y ) (5)and σ h ( y ) := s ( h ) − ◦ y ◦ s ( h ) . (6)It is not difficult to see that, for a given st-section s , ν is a homomorphisms and σ is an anti-homomorphismfrom ( H, ◦ ) to Aut ( I, +). Thus ν and σ are, respectively, left and right actions of ( H, ◦ ) on ( I, +).With this setting, we have Proposition 3.4.
Let → I → E π → H → be an extension of a left brace H by a trivial brace I .(1) Then the actions ν and σ are independent of the choice of an st-section. Moreover, the pair ( ν, σ ) is a good pair of actions of H on I .(2) Equivalent extensions have the same pair of actions.Proof. Let s and s be two different st-sections of π . Let ( σ, ν ) and ( σ ′ , ν ′ ) be actions corresponding to s and s respectively. We know that for each h ∈ H , there exists y h ∈ I such that s ( h ) = y h ◦ s ( h ).Thus, using the triviality of the brace I , for all h ∈ H and y ∈ I , we get σ ′ h ( y ) = s ( h ) − ◦ y ◦ s ( h )= ( s ( h )) − ◦ ( y h ) − ◦ y ◦ y h ◦ s ( h )= s ( h ) − ◦ y ◦ s ( h )= σ h ( y )and ν ′ h ( y ) = s ( h ) ◦ y − s ( h )= y h ◦ s ( h ) ◦ y − y h ◦ s ( h )= y h ◦ ( s ( h ) ◦ y − s ( h )) − y h = y h + ( s ( h ) ◦ y − s ( h )) − y h = ν h ( y ) . This shows the independence of the actions on the choice of an st-section. That ( ν, σ ) is a good pair isproved by Bachiller [1, page 1675]. This proves assertion (1).
NISHANT AND MANOJ K. YADAV
For assertion (2), let E and E ′ be two equivalent extensions of H by I . Then there exists a homomor-phism φ : E → E ′ such that the following diagram commutes:0 −−−−→ I −−−−→ E π −−−−→ H −−−−→ Id y φ y y Id −−−−→ I −−−−→ E ′ π ′ −−−−→ H −−−−→ . Let s be an st-section of π . By the commutativity of the diagram, it follows that s ′ : H → E ′ , given by s ′ ( h ) := φ ( s ( h )) for all h ∈ H , is an st-section of π ′ , and φ ( y ) = y for all y ∈ I . Let ( σ, ν ) and ( σ ′ , ν ′ ) beactions corresponding to s and s ′ respectively. Then, using the fact that φ is a homomorphism, we have σ ′ h ( y ) = s ′ ( h ) − ◦ y ◦ s ′ ( h )= φ ( s ( h )) − ◦ y ◦ φ ( s ( h ))= φ (cid:0) s ( h ) − ◦ y ◦ s ( h ) (cid:1) = φ ( σ h ( y ))= σ h ( y )for all h ∈ H and y ∈ I . Thus, σ and σ ′ are the same actions. Similarly, it also follows that the actions ν and ν ′ are same, which completes the proof. (cid:3) Recall that Ext(
H, I ) denotes the set of equivalence classes of all extensions of H by I . Equivalenceclass of an extension E : 0 → I → E → H → E ]. As a consequence of the precedingproposition, it follows that each equivalence class of extension of H by I admits a unique good pair ofactions ( ν, σ ) of H on I , which we call the corresponding pair of actions . Let Ext ν,σ ( H, I ) denote theequivalence classes of those extensions of H by I whose corresponding pair of actions is ( ν, σ ). We caneasily establish Corollary 3.5.
Ext(
H, I ) = F ( ν,σ ) Ext ν,σ ( H, I ) . Next, for the extension 0 → I → E π → H →
0, define maps β, τ : H × H → I by β ( h , h ) := s ( h ) + s ( h ) − s ( h + h ) (7)and τ ( h , h ) := ν h ◦ h (cid:0) s ( h ◦ h ) − ◦ s ( h ) ◦ s ( h ) (cid:1) = s ( h ) ◦ s ( h ) − s ( h ◦ h ) , (8)where ν is defined in (6). Lemma 3.6.
Let → I → E π → H → be an extension of H by I and ν, σ are, respectively, the leftand right actions of H on I defined by (5) and (6) . Then ( β, τ ) ∈ Z N ( H, I ) , where the maps β and τ are defined in (7) and (8) . Moreover, the cohomology class of ( β, τ ) is independent of the choice of anst-section of π .Proof. Let s be an st-section of π . That ∂ , v ( β ) = 0 and ∂ , h ( τ ) = 0 follows from the associativity of ′ + ′ and ′ ◦ ′ in E respectively; more precisely, one achieves this by using the identities: (i) s ( h ) + ( s ( h ) + s ( h )) = ( s ( h ) + s ( h )) + s ( h ) and (ii) s ( h ) ◦ ( s ( h ) ◦ s ( h )) = ( s ( h ) ◦ s ( h )) ◦ s ( h ). We demonstratecomputations in one case. Using the definition of τ , for all h , h , h ∈ H , we have s ( h ) ◦ ( s ( h ) ◦ s ( h )) = s ( h ) ◦ (cid:0) τ ( h , h ) + s ( h ◦ h ) (cid:1) = s ( h ) ◦ τ ( h , h ) − s ( h ) + s ( h ) ◦ s ( h ◦ h )= ν h ( τ ( h , h )) + τ ( h , h ◦ h ) + s ( h ◦ h ◦ h ) . OHOMOLOGY, EXTENSIONS AND AUTOMORPHISMS OF BRACES 9
On the other hand( s ( h ) ◦ s ( h )) ◦ s ( h ) = s ( h ◦ h ) ◦ ν − h ◦ h ( τ ( h , h )) ◦ s ( h )= s ( h ◦ h ) ◦ s ( h ) ◦ σ h ( ν − h ◦ h ( τ ( h , h )))= s ( h ◦ h ◦ h ) ◦ ν − h ◦ h ◦ h ( τ ( h ◦ h , h )) ◦ σ h ( ν − h ◦ h ( τ ( h , h )))= s ( h ◦ h ◦ h ) ◦ (cid:0) ν − h ◦ h ◦ h ( τ ( h ◦ h , h )) + σ h ( ν − h ◦ h ( τ ( h , h ))) (cid:1) = s ( h ◦ h ◦ h ) ◦ ν − h ◦ h ◦ h ( τ ( h ◦ h , h ))+ s ( h ◦ h ◦ h ) ◦ σ h ( ν − h ◦ h ( τ ( h , h ))) − s ( h ◦ h ◦ h )= s ( h ◦ h ◦ h ) ◦ ν − h ◦ h ◦ h ( τ ( h ◦ h , h ))+ ν h ◦ h ◦ h ( σ h ( ν − h ◦ h ( τ ( h , h ))))= τ ( h ◦ h , h ) + s ( h ◦ h ◦ h ) + ν h ◦ h ◦ h ( σ h ( ν − h ◦ h ( τ ( h , h )))) . Comparing the preceding two expressions, we get ν h ( τ ( h , h )) − τ ( h ◦ h , h ) + τ ( h , h ◦ h ) − ν h ◦ h ◦ h ( σ h ( ν − h ◦ h τ ( h , h ))) = 0 . That ∂ , h ( β ) − ∂ , v ( τ ) = 0 follows from the brace condition, that is, s ( h ) ◦ ( s ( h ) + s ( h )) + s ( h ) = s ( h ) ◦ s ( h ) + s ( h ) ◦ s ( h ), by similar computations. We refer the reader to [1, pages 1674-75] for othercomputations. This shows that ( β, τ ) ∈ Z N ( H, I ).For the second assertion, let s and s ′ be two st-sections of π . Let ( β, τ ) and ( β ′ , τ ′ ) be 2-cocyclescorresponding to s and s ′ respectively. Notice that s ( h ) − s ′ ( h ) ∈ I . Define a map θ : H → I by θ ( h ) = s ( h ) − s ′ ( h ) for all h ∈ H . A straightforward computation then shows that the 2-cocycles ( β, τ )and ( β ′ , τ ′ ) differ by ∂ ( θ ). This completes the proof. (cid:3) We are now ready to prove the main result of this section (Theorem A).
Theorem 3.7.
Let H := ( H, + , ◦ ) be a left brace and I := ( I, +) an abelian group viewed as a trivialbrace. Let ( ν, σ ) be a good pair of actions of H on I . Then there is a bijection between H N ( H, I ) and Ext ν,σ ( H, I ) .Proof. Let ( τ, β ) ∈ Z N ( H, I ). Then it follows from Theorem 3.3 that the pair ( τ, β ) gives rise to anextension (
H, I, ν, σ, β, τ ). Let ( β , τ ) , ( β , τ ) ∈ Z N ( H, I ) be cohomologous. Then there exists θ ∈ C N such that ( β − β )( h , h ) = θ ( h ) − θ ( h + h ) + θ ( h ) (9)and ( τ − τ )( h , h ) = ν h ( θ ( h )) − θ ( h ◦ h ) + ν h ◦ h ( σ h ( ν − h ( θ ( h )))) , (10)for all h , h ∈ H . Let ( H, I, ν, σ, β , τ ) and ( H, I, ν, σ, β , τ ) be the corresponding extensions for ( β , τ )and ( β , τ ) respectively. Define a map φ : ( H, I, ν, σ, β , τ ) → ( H, I, ν, σ, β , τ ) by φ ( h, y ) := ( h, y + θ ( h )) . Consider the following diagram:0 −−−−→ I i −−−−→ ( H, I, ν, σ, β , τ ) π −−−−→ H −−−−→ Id y φ y Id y −−−−→ I i ′ −−−−→ ( H, I, ν, σ, β , τ ) π ′ −−−−→ H −−−−→ . Notice that the linearity of φ in ′ + ′ and ′ ◦ ′ follows from the equations (9) and (10) respectively. Itis now obvious that the preceding diagram is commutative. Thus the extensions ( H, I, ν, σ, β , τ ) and( H, I, ν, σ, β , τ ) are equivalent, and therefore the map ψ : H N ( H, I ) → Ext ν,σ ( H, I ) given by ψ ([( β, τ )]) = [( H, I, ν, σ, β, τ )]is well defined.We now show that ψ is surjective. Let E : 0 → I → E π → H → ν, σ ). Let s : H → E be an st-section of π . Then, by Lemma 3.6, we get ( β, τ ) ∈ Z N ( H, I ) as defined in (7) and (8). Applying ψ , we have ψ ([( β, τ )]) = [( H, I, ν, σ, β, τ )]. Now thesurjectivity of ψ is equivalent to showing that E and ( H, I, ν, σ, β, τ ) are equivalent. More precisely, thecohomology class [( β, τ )] of ( β, τ ) will be a pre-image of the extension E . We remark that this mechanismis independent of the choice of an st-section s of π , because whatever s we start with, we finally get anextension equivalent to E . So, for the surjectivity of ψ , we only need to establish the commutativity ofthe diagram 0 −−−−→ I i −−−−→ E π −−−−→ H −−−−→ Id y φ y Id y −−−−→ I i ′ −−−−→ ( H, I, σ, ν, β, τ ) π ′ −−−−→ H −−−−→ , where φ is a brace homomorphism. As mentioned above, π ′ ( h, y ) = h and s ′ , given by s ′ ( h ) := ( h, π ′ , which we’ll use here. Notice that every g ∈ E can be uniquely written as g = s ( h ) + y for some h ∈ H and y ∈ I . Define φ : E → ( H, I, ν, σ, β, τ ) by φ ( s ( h ) + y ) = ( h, y ) . It is easy to see that φ is linear under ′ + ′ . Notice that s ( h ) ◦ y = s ( h ) + ν h ( y ) and s ( h ) + y = s ( h ) ◦ ν − h ( y )for all h ∈ H and y ∈ I . Let s ( h ) + y and s ( h ) + y be two elements of E . Then φ (cid:0) ( s ( h ) + y ) ◦ ( s ( h ) + y ) (cid:1) = φ (cid:0) s ( h ) ◦ ν − h ( y ) ◦ s ( h ) ◦ ν − h ( y ) (cid:1) = φ (cid:0) s ( h ) ◦ s ( h ) ◦ σ h ( ν − h ( y )) ◦ ν − h ( y ) (cid:1) = φ (cid:0) s ( h ◦ h ) ◦ ν − h ◦ h ( τ ( h , h )) ◦ σ h ( ν − h ( y )) ◦ ν − h ( y ) (cid:1) = φ (cid:0) s ( h ◦ h ) + ν h ◦ h (cid:0) ν − h ◦ h ( τ ( h , h )) + σ h ( ν − h ( y )) + ν − h ( y ) (cid:1)(cid:1) = (cid:0) h ◦ h , τ ( h , h ) + ν h ◦ h ( σ h ( ν − h ( y ))) + ν h ( y ) (cid:1) = ( h , y ) ◦ ( h , y )= φ ( s ( h ) + y ) ◦ φ ( s ( h ) + y ) . Hence φ is a brace homomorphism. Also φ ( y ) = (0 , y ) = i ′ ( y ) and π ( s ( h ) + y ) = h = π ′ ( h, y ) = π ′ ( φ ( s ( h ) + y )) , which establishes the commutativity of the preceding diagram.Finally, we proceed to show the injectivity of ψ . Let ( β , τ ) , ( β , τ ) ∈ Z N ( H, I ) such that ψ ([( β , τ )]) = ψ ([( β , τ )]). More precisely, we have the following commutative diagram:0 −−−−→ I −−−−→ ( H, I, ν, σ, β , τ ) π −−−−→ H −−−−→ Id y φ y Id y −−−−→ I −−−−→ ( H, I, ν, σ, β , τ ) π ′ −−−−→ H −−−−→ , where φ is a brace homomorphism. Let s and s ′ be the st-sections of π and π ′ given by s ( h ) = ( h,
0) and s ′ ( h ) = ( h, h ∈ H . It now follows from the commutativity of the diagram that,for each h ∈ H , φ ( h,
0) = ( h, y h ) for some y h ∈ I . Notice that for a given h ∈ H , y h is unique. Define OHOMOLOGY, EXTENSIONS AND AUTOMORPHISMS OF BRACES 11 θ : H → I by θ ( h ) = y h . Since s (0) = (0 , θ ∈ C N . By a regular computation, one canshow that ( β − β )( h , h ) = θ ( h ) − θ ( h + h ) + θ ( h )and ( τ − τ )( h , h ) = ν h ( θ ( h )) − θ ( h ◦ h ) + ν h ◦ h ( σ h ( ν − h ( θ ( h ))))for all h , h ∈ H . This simply means that ( β , τ ) and ( β , τ ) are cohomologous, and the proof iscomplete. (cid:3) Let H be a left brace and I a trivial brace such that H acts trivially on I , that is, both ν h and σ h aretrivial automorphism of ( I, +) for all h ∈ H . It is immediate, in this case, that ( ν, σ ) is a good pair ofactions. Let [( β, τ )] ∈ H N ( H, I ) and E := ( H, I, σ, ν, τ, β ) be the corresponding extension of H by I . Itturns out that the image of I in E is a central ideal of E . On the other hand, if E : 0 → I −→ E π −→ H → E ∈
Ext ν,σ ( H, I ), where ( ν, σ ) is a trivial action of H on I . Let the set of allequivalence classes of central extension of H by I be denoted by CExt( H, I ). Then, as a special case ofTheorem 3.7, we get the following brace theoretic analog of [16, Theorem 5.8].
Corollary 3.8.
Let H := ( H, + , ◦ ) be a left brace which acts trivially on a trivial brace I := ( I, +) . Thenthere is a bijection between H N ( H, I ) and CExt(
H, I ) . We conclude this section with explicit computation of H N ( H, I ), where H = Z / Z and I = Z / Z × Z / Z with two good pairs of actions; one non-trivial and other trivial. It is easy to see that the pair( ν, σ ) with ν h ( a, b ) = ( a + b + h, b ), σ h ( a, b ) = ( a, b + a + h ) is a good pair of actions, where h ∈ H and a, b ∈ I . This pair of actions was considered by Bachiller [1]. Let ( β, τ ) be a 2-cocycle. Notice that both β as well as τ are determined by their values at (1 , ∈ H × H . An easy computation then reveals thatthe following are the only choices for (cid:0) β (1 , , τ (1 , (cid:1) : { (cid:0) (0 , , (0 , (cid:1) , (cid:0) (0 , , (0 , (cid:1) , (cid:0) (1 , , (0 , (cid:1)(cid:0) (1 , , (0 , (cid:1) } . For clarity, let us take these cocycles as ( β , τ ) , ( β , τ ) , ( β , τ ) , ( β , τ ) respectively. If we define θ (0) =(0 ,
0) and θ (1) = (0 , θ ∈ C N ; further ( β , τ ) = ∂ ( θ ), that is, ( β , τ ) is a 2-coboundary. Weclaim that none of ( β , τ ) and ( β , τ ) is a 2-coboundary, but these are cohomologous. Contrarily, assumethat ( β , τ ) is a 2-coboundary. Then there exists θ : C N such that ∂ ( θ ) = ( β , τ ). So θ (0) = (0 ,
0) andlet θ (1) = ( y , y ). Now β (1 ,
1) = θ (1) − θ (0) + θ (1)= ( y , y ) + ( y , y )= (0 , , which is not possible, since β (1 ,
1) = (1 , β , τ ) is not a 2-couboundary. The same computationalso shows that ( β , τ ) too is not a 2-coboundary. Finally, notice that ( β , τ ) − ( β , τ ) = ( β , τ ), whichwe have already shown to be a 2-coboundary. Hence ( β , τ ) and ( β , τ ) are cohomologous, and thereforeH N ( H, I ) = { [( β , τ )] , [( β , τ )] } ∼ = Z / Z . Now we consider the trivial action, that is, ν h = σ h = Id for all h ∈ H . As above, for computing 2-cocycles ( β, τ ) of H , it is enough to know their values on (1 , β i , τ j ),1 ≤ i, j ≤
4, are all possible 2-cocycles of H , where β = (0 ,
0) = τ , β = (0 ,
1) = τ , β = (1 ,
0) = τ and β = (1 ,
1) = τ . Let θ ∈ C N . Notice that θ (0) = 0. Then ∂ ( θ ) = ( g, f ), where g ( h , h ) = f ( h , h ) = θ ( h ) − θ ( h + h ) + θ ( h ) = 0for all h , h ∈ H , since H is a trivial brace and the action is trivial. Hence B N ( H, I ) is the trivial group,and therefore H N ( H, I ) is isomorphic to the elementary abelian group of order 16. Cohomology of Braces : A special case
In this section we generalize the cohomological results of [16, Sections 2, 3] for braces. Let ( H, + , ◦ )be a left brace and ( I, +) an abelian group. Let ν : ( H, ◦ ) → Aut( I, +) and σ : ( H, ◦ ) → Aut( I, +),respectively, again be left and right group actions of ( H, ◦ ) on ( I, +).Set C n ( H ; I ) := Fun( H n , I ) for n ≥
1, where H n denotes the cartesian product of n copies of H .Notice that C n ( H ; I ) is an abelian group for each n . For n ≥
0, define the maps ∂ n : C n → C n +1 by( ∂ n f )( h , . . . , h n +1 ) = ν h ( f ( h , . . . , h n +1 ))+ n X i =1 ( − i f ( h , . . . , h i ◦ h i +1 , . . . , h n +1 )+ ( − n +1 ν h ◦ h ◦···◦ h n +1 ( σ h n +1 ( ν − h ◦ h ◦···◦ h n ( f ( h , . . . , h n )))) . Define RC n ( H ; I ) = { f ∈ F un ( H n , I ) | f is linear in the n th co-ordinate } . Then the restriction of ∂ n , which we still denote by ∂ n , gives the map ∂ n : RC n ( H ; I ) → RC n +1 ( H ; I )for each n ≥
1. Further, let RC nN ( H ; I ) denote the set of all f ∈ RC n ( H ; I ) which vanish on alldegenerate tuples. This becomes a subgroup of RC n ( H ; I ) and the restriction of ∂ n gives a map from RC nN ( H ; I ) → RC n +1 N ( H ; I ). Theorem 4.1.
For n ≥ , ( C n ( H ; I ) , ∂ n ) , ( RC n ( H ; I ) , ∂ n ) and ( RC nN ( H ; I ) , ∂ n ) are cochain complexes.Proof. We shall only indicate a proof of the fact that ( C n ( H ; I ) , ∂ n ) is a cochain complex. Other assertioncan be easily verified from this. We are required to show that ∂ n +1 ∂ n = 0. We can write ∂ n = n +1 X i =0 ( − i ∂ n,i , where ( ∂ n, f )( h , . . . , h n +1 ) = ν h ( f ( h , . . . , h n +1 )) , ( ∂ n,i f )( h , . . . , h n +1 ) = f ( h , . . . , h i ◦ h i +1 , . . . , h n +1 ) , ≤ i ≤ n, ( ∂ n,n +1 f )( h , . . . , h n +1 ) = ν h ◦ h ◦···◦ h n +1 ( σ h n +1 ( ν − h ◦ h ◦···◦ h n ( f ( h , . . . , h n )))) . By a direct computation one can easily show that ∂ n,i ∂ n − ,j = ∂ n,j +1 ∂ n − ,i for 0 ≤ i, j ≤ n , whichimplies ∂ n ∂ n − = 0. To demonstrate, we compute in one case. Let i = 0 and j ≥ ∂ n, ( ∂ n − ,j f )( h . . . h n +1 ) = ν h ( ∂ n − ,j f )( h , . . . , h n +1 )= ν h ( f ( h , . . . , h j +1 ◦ h j +2 , . . . , h n +1 )) . On the other hand ∂ n,j +1 ( ∂ n − , f )( h , . . . h n +1 ) = ( ∂ n − , f )( h , . . . , h j +1 ◦ h j +2 , . . . , h n +1 )= ν h ( f ( h , . . . , h j +1 ◦ h j +2 , . . . , h n +1 )) . Hence ∂ n, ∂ n − ,j = ∂ n,j +1 ∂ n − , . (cid:3) OHOMOLOGY, EXTENSIONS AND AUTOMORPHISMS OF BRACES 13
Let H n ( H, I ), RH n ( H, I ) and RH nN ( H, I ), respectively, denote the n th cohomology groups of thecomplexes ( C n ( H ; I ) , ∂ n ), ( RC n ( H ; I ) , ∂ n ) and ( RC nN ( H ; I ) , ∂ n ).Let STExt ν,σ ( H, I ) be the set of equivalence classes of all extensions of H by I with fixed good pair ofactions ( ν, σ ), and which split as extensions of abelian groups. Let ( β, τ ) ∈ H N ( H, I ). If we take β = 0,then it follows that τ ∈ RH N ( H, I ). More precisely, RH N ( H, I ) = { τ : H × H → I | (0 , τ ) ∈ H N ( H, I ) } .The next result now follows from Theorem 3.7 by taking β = 0. Theorem 4.2.
There exists a one-to-one correspondence between RH N ( H, I ) and STExt ν,σ ( H, I ) . Let ( ν, σ ) be a good pair of actions of H on I . Let H N (( H, ◦ ) , I ) denote the second cohomology groupof ( H, ◦ ) with coefficients in the abelian group I , where ( H, ◦ ) acts on I through σ . For a 2-cocycle f from RC nN ( H ; I ) we can define a map f ′ : ( H, ◦ ) × ( H, ◦ ) → I by setting f ′ ( h , h ) = ν − h ◦ h ( f ( h , h )) (11)for all h , h ∈ ( H, ◦ ). It turns out that f ′ is a 2-cocycle of ( H, ◦ ) with coefficients in I . This gives riseto the following Proposition 4.3.
The cohomology group RH N ( H, I ) of braces embeds in the cohomology group H N (( H, ◦ ) , I ) of groups under the association (11) . Let H and H ′ be left braces, and I and I ′ be H and H ′ bi-modules with ( ν, σ ) and ( ν ′ , σ ′ ), respectively,as actions. Let α : H ′ → H and ζ : I → I ′ be homomorphisms of braces. The pair ( α, ζ ) is said to be compatible with the pairs of actions ( ν, σ ) and ( ν ′ , σ ′ ) if the diagram (2) commutes for both left and rightactions. Recall that, equivalently saying, ( α, ζ ) is compatible with the actions, if ζ ( ν α ( h ′ ) ( y )) = ν ′ h ′ ( ζ ( y ))and ζ ( σ α ( h ′ ) ( y )) = σ ′ h ′ ( ζ ( y )) . Let ( α, ζ ) be compatible with the actions ( ν, σ ) and ( ν ′ , σ ′ ). Let us fix an integer n ≥
1. For f ∈ C n ( H, I ), define f ′ : H ′ n → I ′ by f ′ ( h ′ , . . . , h ′ n ) = ζ (cid:0) f (cid:0) α ( h ) , . . . , α ( h n ) (cid:1)(cid:1) for h ′ i ∈ H ′ . Since ( α, ζ ) is compatible with the actions, it easily follows that f ′ ∈ C n ( H ′ , I ′ ). Define( α, ζ ) n : C n ( H, I ) → C n ( H ′ , I ′ ) by ( α, ζ ) n ( f ) = f ′ for all f ∈ C n ( H, I ). It is not difficult to see that ( α, ζ ) n is a homomorphism and the following diagramcommutes C n ( H, I ) ( α,ζ ) n −−−−→ C n ( H ′ , I ′ ) y ∂ n y ∂ ′ n C n +1 ( H, I ) ( α,ζ ) n +1 −−−−−−→ C n +1 ( H ′ , I ′ ) . Further straightforward computations now give the following
Theorem 4.4.
For each n ≥ , the homomorphism ( α, ζ ) n : C n ( H, I ) → C n ( H ′ , I ′ ) , defined in thepreceding para, induces a homomorphism from H n ( H, I ) to H n ( H ′ , I ′ ) , which further induces homomor-phisms from RH n ( H, I ) to RH n ( H ′ , I ′ ) and RH nN ( H, I ) to RH nN ( H ′ , I ′ ) . Let K be a left ideal of H and α : K → H be the embedding. For a pair of actions ( ν, σ ) of H on I , ( ν ′ , σ ′ ) is a pair of actions of K on I , where ν ′ = ν | K and σ ′ = σ | K . Let Id : I → I be the identityhomomorphism. Obviously ( α, Id) is compatible with these actions. Hence, as a consequence of thepreceding theorem, we have
Corollary 4.5.
There exists a homomorphism from RH nN ( H, I ) to RH nN ( K, I ) . Such a homomorphism is called the restriction homomorphism, denoted by res HK .5. Wells’ type exact sequence for braces
In this section we establish various group actions on the set Ext(
H, I ) and construct an exact sequenceconnecting certain automorphism groups of E with H N ( H, I ) for a given extension E : 0 → I i → E π → H → H ) × Autb( I ) on Ext( H, I ).Let H be any left brace and I a trivial brace. For a pair ( φ, θ ) ∈ Autb( H ) × Autb( I ) of braceautomorphisms and an extension E : 0 → I i → E π → H → H by I , we can define a new extension E ( φ,θ ) : 0 → I iθ −→ E φ − π −→ H → H by I . Let E : 0 → I i → E π → H → E : 0 → I i ′ → E π ′ → H → H by I . Then it is not difficult to show that the extensions E ( φ,θ )1 and E ( φ,θ )2 are also equivalent for any ( φ, θ ) ∈ Autb( H ) × Autb( I ). Thus, for a given ( φ, θ ) ∈ Autb( H ) × Autb( I ),we can define a map from Ext( H, I ) to itself given by[ E ] [ E ( φ,θ ) ] . (12)If φ and θ are identity automorphisms, than obviously E ( φ,θ ) = E . It is also easy to see that[ E ] ( φ ,θ )( φ ,θ ) = (cid:0) [ E ] ( φ ,θ ) (cid:1) ( φ ,θ ) . We conclude that the association (12) gives an action of the group Autb( H ) × Autb( I ) on the set Ext( H, I ).As we know that Ext(
H, I ) = F ( ν,σ ) Ext ν,σ ( H, I ). Let ( ν, σ ) be an arbitrary but fixed good pair ofactions of H on I . Let C ( ν,σ ) denote the stabiliser of Ext ν,σ ( H, I ) in Autb( H ) × Autb( I ); more explicitlyC ( ν,σ ) = { ( φ, θ ) ∈ Autb( H ) × Autb( I ) | ν h = θ − ν φ ( h ) θ and σ h = θ − σ φ ( h ) θ } . Notice that C ( ν,σ ) is a subgroup of Autb( H ) × Autb( I ), and it acts on Ext ν,σ ( H, I ) by the same rule asgiven in (12).Next we consider an action of C ( ν,σ ) on H N ( H, I ). Let ( φ, θ ) ∈ Aut( H ) × Aut( I ) and f ∈ Fun( H n , I ),where n ≥ f ( φ,θ ) : H n → I by setting f ( φ,θ ) ( h , h , . . . , h n ) := θ − (cid:0) f ( φ ( h ) , φ ( h ) , . . . , φ ( h n )) (cid:1) . It is not difficult to see that the group Autb( H ) × Autb( I ) acts on the group Fun( H n , I ) as well as onthe group C i,jN ( H, I ), by automorphisms, given by the association f f ( φ,θ ) . (13)It is also obvious that C ( ν,σ ) acts on both of these sets. We are interested in the action of C ( ν,σ ) onH N ( H, I ). The association (13) induces an action of C ( ν,σ ) on C N = C , N ( H, I ) ⊕ C , N ( H, I ) by setting( β, τ ) (cid:0) β ( φ,θ ) , τ ( φ,θ ) (cid:1) . (14) OHOMOLOGY, EXTENSIONS AND AUTOMORPHISMS OF BRACES 15
Lemma 5.1.
For ( φ, θ ) ∈ C ( ν,σ ) , the following hold:(i) If ( β, τ ) ∈ Z N ( H, I ) , then (cid:0) β ( φ,θ ) , τ ( φ,θ ) (cid:1) ∈ Z N ( H, I ) .(ii) If ( β, τ ) ∈ B N ( H, I ) , then (cid:0) β ( φ,θ ) , τ ( φ,θ ) (cid:1) ∈ B N ( H, I ) .Hence the association (14) gives an action of C ( ν,σ ) on H N ( H, I ) by automorphisms if we define [( β, τ )] ( φ,θ ) = [ (cid:0) β ( φ,θ ) , τ ( φ,θ ) (cid:1) ] . Proof.
That ∂ , v (cid:0) β ( φ,θ ) (cid:1) is a zero function in C N holds trivially. Since ( φ, θ ) ∈ C ( ν,σ ) , we have ν h ( θ − ( τ ( φ ( h ) , φ ( h )))) = θ − (cid:0) ν φ ( h ) ( τ ( φ ( h ) , φ ( h ))) (cid:1) and ν h ◦ h ( σ h ( ν − h ( θ − ( φ ( h ) , φ ( h ))))) = θ − (cid:0) ν φ ( h ◦ h ) ( σ φ ( h ) ( ν − φ ( h ) ( τ ( φ ( h ) , φ ( h ))))) (cid:1) . Using these identities, for all h , h , h ∈ H , it follows that ∂ , h (cid:0) τ ( φ,θ ) (cid:1) ( h , h , h ) = θ − (cid:0) ∂ , h ( τ )( φ ( h ) , φ ( h ) , φ ( h )) (cid:1) = 0 . Similar computations show that (cid:0) ∂ , h (cid:0) β ( φ,θ ) (cid:1) − ∂ , v (cid:0) τ ( φ,θ ) (cid:1)(cid:1) ( h , h , h ) = θ − (cid:0) ( ∂ , h ( β ) − ∂ , v ( τ ))( φ ( h ) , φ ( h ) , φ ( h )) (cid:1) = 0 . This proves assertion (i).Now assume that ( β, τ ) ∈ B N ( H, I ). Then there exists a λ ∈ C N such that ( β, τ ) = ∂ ( λ ). It is nownot difficult to see that (cid:0) β ( φ,θ ) , τ ( φ,θ ) (cid:1) = ∂ (cid:0) λ ( φ,θ ) (cid:1) , which establishes assertion (ii). That the association under consideration gives an action of C ν,σ onH N ( H, I ) by automorphisms, is now straighforward, which completes the proof. (cid:3)
Remark 15.
The action of C ( ν,σ ) on H N ( H, I ), defined in the preceding lemma, can be transferred onExt ν,σ ( H, I ) through the bijection given in Theorem 3.7. Notice that the resulting action of C ( ν,σ ) onExt ν,σ ( H, I ) agrees with the action defined by (12).We now consider the action of H N ( H, I ) onto itself by right translation, which is faithful and transitive.Again using Theorem 3.7, we can transfer this action on Ext ( ν,σ ) ( H, I ) = { [( H, I, ν, σ, β, τ )] | [( β, τ )] ∈ H N ( H, I ) } . More precisely, for [( β , τ )] ∈ H N ( H, I ), the action is given by[(
H, I, ν, σ, β, τ )] [( β ,τ )] = [( H, I, ν, σ, β + β , τ + τ )]for all [( H, I, ν, σ, β, τ )] ∈ Ext ( ν,σ ) ( H, I ). Notice that this action is also faithful and transitive.Consider the semidirect product Γ := C ( ν,σ ) ⋉ H N ( H, I ) under the action defined in Lemma 5.1. Wewish to define an action of Γ on Ext ν,σ ( H, I ). For ( c, h ) ∈ Γ and [ E ] ∈ Ext ν,σ ( H, I ), define[ E ] ( c,h ) = ([ E ] c ) h . (16) Lemma 5.2.
The rule in (16) gives an action of Γ on Ext ν,σ ( H, I ) .Proof. Notice that for ( c , h ) , ( c , h ) ∈ Γ, ( c , h )( c , h ) = ( c c , h c h ). So, it is enough to showthat (cid:0) [ E ] h (cid:1) c = (cid:0) [ E ] c (cid:1) h c for each c ∈ C ( ν,σ ) , h ∈ H N ( H, I ) and [ E ] ∈ Ext ν,σ ( H, I ). We know that[ E ] = [( H, I, ν, σ, β, τ )] for some [( β, τ )] ∈ H N ( H, I ). Then, for h = [( β h , τ h )] ∈ H N ( H, I ), we have (cid:0) [ E ] h (cid:1) c = [( H, I, ν, σ, ( β + β h ) c , ( τ + τ h ) c )]= [( H, I, ν, σ, β c + β ch , τ c + τ ch )]= ([( H, I, ν, σ, β c , τ c )]) h c = (cid:0) [ E ] c (cid:1) h c . The proof is now complete. (cid:3)
Let [ E ] ∈ Ext ν,σ ( H, I ) be a fixed extension. Since the action of H N ( H, I ) on Ext ν,σ ( H, I ) is transitiveand faithful, for each c ∈ C ( ν,σ ) , there exists a unique element (say) h c in H N ( H, I ) such that[ E ] c = [ E ] h c . We thus have a well defined map ω ( E ) : C ( ν,σ ) → H N ( H, I ) given by ω ( E )( c ) = h c (17)for c ∈ C ( ν,σ ) . Lemma 5.3.
The map ω ( E ) : C ( ν,σ ) → H N ( H, I ) given in (17) is a derivation with respect to the actionof C ( ν,σ ) on H N ( H, I ) given in (14) .Proof. Let c , c ∈ C ( ν,σ ) and ω ( E )( c c ) = h c c . Thus, by the definition of ω ( E ), [ E ] c c = [ E ] h c c .Using the fact that (cid:0) [ E ] h (cid:1) c = (cid:0) [ E ] c (cid:1) h c for each c ∈ C ( ν,σ ) , h ∈ H N ( H, I ), we have[ E ] h c c = [ E ] ( c c ) = (cid:0) [ E ] c (cid:1) c = (cid:0) [ E ] h c (cid:1) c = (cid:0) [ E ] c (cid:1) ( h c ) c = (cid:0) [ E ] h c (cid:1) ( h c ) c = [ E ] (cid:0) h c +( h c ) c (cid:1) . Since the action of H N ( H, I ) on Ext ν,σ ( H, I ) is faithful, it follows that h c c = ( h c ) c + h c . This impliesthat ω ( E )( c c ) = (cid:0) ω ( E )( c ) (cid:1) c + ω ( E )( c ); hence ω ( E ) is a derivation. (cid:3) Let E : 0 → I → E π → H be an extension of a left brace H by a trivial brace I such that [ E ] ∈ Ext ν,σ ( H, I ). Let Autb I ( E ) denotethe subgroup of Autb( E ) consisting of all automorphisms of E which normalize I , that is,Autb I ( E ) := { γ ∈ Autb( E ) | γ ( y ) ∈ I for all y ∈ I } . For γ ∈ Autb I ( E ), set γ I := γ | I , the restriction of γ to I , and γ H to be the automorphism of H inducedby γ . More precisely, γ H ( h ) = π ( γ ( s ( h ))) for all h ∈ H , where s is an st-section of π . Notice thatthe definition of γ H is independent of the choice of an st-section. Define a map ρ ( E ) : Autb I ( E ) → Autb( H ) × Autb( I ) by ρ ( E )( γ ) = ( γ H , γ I ) . Although ω ( E ) is not a homomorphism, but we can still talk about its set theoretic kernel, that is,Ker( ω ( E )) = { c ∈ C ( ν,σ ) | [ E ] c = [ E ] } . With this setting we have
Proposition 5.4.
For the extension E , Im( ρ ( E )) ⊆ C ( ν,σ ) and Im( ρ ( E )) = Ker( ω ( E )) . OHOMOLOGY, EXTENSIONS AND AUTOMORPHISMS OF BRACES 17
Proof.
For the first assertion, we are required to show that ν h = γ − I ν γ H ( h ) γ I and σ h = γ − I σ γ H ( h ) γ I forall h ∈ H . Let s be an st-section of π and x ∈ E . Notice that γ − I is the restriction of γ − on I . Alsonotice that for a given x ∈ E , s ( π ( x )) = x ◦ y h for some y h ∈ I . Now for h ∈ H and y ∈ I , we have γ − I ν γ H ( h ) γ I ( y ) = γ − (cid:0) ν π ( γ ( s ( h ))) ( γ ( y )) (cid:1) = γ − (cid:0) s ( π ( γ ( s ( h )))) ◦ γ ( y ) − s ( π ( γ ( s ( h )))) (cid:1) = γ − (cid:0) γ ( s ( h )) ◦ y γ ( s ( h )) ◦ γ ( y ) − γ ( s ( h )) ◦ y γ ( s ( h )) (cid:1) = s ( h ) ◦ γ − ( y γ ( s ( h )) ) ◦ y − s ( h ) ◦ γ − ( y γ ( s ( h )) )= s ( h ) ◦ (cid:0) γ − ( y γ ( s ( h )) ) + y (cid:1) − s ( h ) ◦ γ − ( y γ ( s ( h )) )= s ( h ) ◦ y − s ( h ) (using (1))= ν h ( y ) . Hence ν h = γ − I ν γ H ( h ) γ I . One can similarly show that σ h = γ − I σ γ H ( h ) γ I .Now we prove the second assertion. Let ρ ( E )( γ ) = ( γ H , γ I ) for γ ∈ Autb I ( E ). We know that s ( π ( x )) = x + y x for some y x ∈ I . Thus we have γ − H ( π ( γ ( s ( h )))) = π (cid:0) γ − ( s ( π ( γ ( s ( h ))))) (cid:1) = π (cid:0) γ − (cid:0) γ ( s ( h )) + y γ ( s ( h )) (cid:1)(cid:1) = h, which implies that the diagram0 −−−−→ I −−−−→ E π −−−−→ H −−−−→ Id y γ y Id y −−−−→ I γ I −−−−→ E γ − H π −−−−→ H −−−−→ E )] ( γ H ,γ I ) = [( E )], which shows that Im( ρ ( E )) ⊆ Ker( ω ( E )).Conversely, if ( φ, θ ) ∈ Ker( ω ( E )), then there exists a brace homomorphism γ : E → E such that thediagram 0 −−−−→ I −−−−→ E π −−−−→ H −−−−→ Id y γ y Id y −−−−→ I θ −−−−→ E φ − π −−−−→ H −−−−→ γ ∈ Autb I ( E ), θ = γ I and φ = γ H . Hence ρ ( E )( γ ) = ( φ, θ ), whichcompletes the proof. (cid:3) Continuing with the above setting, set Autb
H,I ( E ) := { γ ∈ Autb I ( E ) | γ I = Id , γ H = Id } . Noticethat Autb H,I ( E ) is precisely the kernel of ρ ( E ). Hence, using Proposition 5.4, we get Theorem 5.5.
Let E : 0 → I → E π → H be a extension of a left brace H by a trivial brace I such that [ E ] ∈ Ext ν,σ ( H, I ) . Then we have the following exact sequence of groups → Autb
H,I ( E ) → Autb I ( E ) ρ ( E ) −→ C ( ν,σ ) ω ( E ) −→ H N ( H, I ) , where ω ( E ) is, in general, only a derivation. We further prove
Proposition 5.6.
Let E : 0 → I → E π → H be an extension of H by I such that [ E ] ∈ Ext ν,σ ( H, I ) .Then Autb
H,I ( E ) ∼ = Z N ( H, I ) . Proof.
We know that every element x ∈ E has a unique expression of the form x = s ( h )+ y = s ( h ) ◦ ν − h ( y )for some h ∈ H and y ∈ I . Let us define a map η : Z N ( H, I ) → Autb
H,I ( E ) by η ( λ )(( s ( h ) + y ) = s ( h ) + λ ( h ) + y, where λ ∈ Z N ( H, I ). Notice that the image of η ( λ ) is independent of the choice of an st-section. Weclaim that η ( λ ) ∈ Autb
H,I ( E ). For h , h ∈ H and y , y ∈ I , we know that τ ( h , h ) = ν h ◦ h (cid:0) s ( h ◦ h ) − ◦ s ( h ) ◦ s ( h ) (cid:1) as defined in (8). Set τ = ν − h ◦ h τ . Also notice that s ( h ) ◦ y = s ( h ) + ν h ( y ). We then have η ( λ ) (cid:0) ( s ( h ) + y ) ◦ ( s ( h ) + y ) (cid:1) = η ( λ ) (cid:0) ( s ( h ) ◦ ν − h ( y )) ◦ ( s ( h ) ◦ ν − h ( y )) (cid:1) = η ( λ ) (cid:0) s ( h ) ◦ s ( h ) ◦ σ h ( ν − h ( y )) ◦ ν − h ( y ) (cid:1) = η ( λ ) (cid:0) s ( h ◦ h ) ◦ (cid:0) τ ( h , h ) + σ h ( ν − h ( y )) + ν − h ( y ) (cid:1)(cid:1) = η ( λ ) (cid:0) s ( h ◦ h ) + ν h ◦ h (cid:0) τ ( h , h ) + σ h ( ν − h ( y )) + ν − h ( y ) (cid:1)(cid:1) = s ( h ◦ h ) + λ ( h ◦ h ) + τ ( h , h ) + ν h ◦ h ( σ h ( ν − h ( y )))+ ν h ( y )= s ( h ◦ h ) + τ ( h , h ) + ν h ◦ h ( σ h ( ν − h ( y + λ ( h ))))+ ν h ( y + λ ( h ))= ( s ( h ) + λ ( h ) + y ) ◦ ( s ( h ) + λ ( h ) + y )= η ( λ )( s ( h ) + y ) ◦ η ( λ )( s ( h ) + y ) . It is easy to see that η ( λ ) (cid:0) ( s ( h ) + y ) + ( s ( h ) + y ) (cid:1) = η ( λ )( s ( h ) + y ) + η ( λ )( s ( h ) + y ) . That η is injective, follows from the injectivity of s and the fact that λ (0) = 0. Surjectivity of η ( λ ) isimmediate. This shows that η ( λ ) is an automorphism of E . We also have η ( λ )( s ( h )) = s ( h ) + λ ( h ) and η ( λ )( y ) = y for all h ∈ H and y ∈ I . Hence η ( λ ) ∈ Autb
H,I ( E ).We’ll now define a map ζ : Autb H,I ( E ) → Z N ( H, I ). For γ ∈ Autb
H,I ( E ) and h ∈ H , there exists aunique element (say) y γh ∈ I such that γ ( s ( h )) = s ( h ) + y γh for some unique y γh ∈ I . Thus, for h ∈ H ,define ζ by ζ ( γ )( h ) = y γh . Notice that this is independent of the choice of an st-section. Since σ − h = σ h − , it follows that γ ( s ( h ) − ) = s ( h ) − ◦ σ h − (cid:0) ν − h (( y γh ) − ) (cid:1) for all h ∈ H . Further, since s ( h ◦ h ) − ◦ s ( h ) ◦ s ( h ) ∈ I for all h , h ∈ H , we have s ( h ◦ h ) − ◦ s ( h ) ◦ s ( h ) = γ (cid:0) s ( h ◦ h ) − ◦ s ( h ) ◦ s ( h ) (cid:1) = γ ( s ( h ◦ h ) − ) ◦ γ ( s ( h )) ◦ γ ( s ( h ))= s ( h ◦ h ) − ◦ σ ( h ◦ h ) − (cid:0) ν − h ◦ h ( y γh ◦ h ) − (cid:1) ◦ s ( h ) ◦ ν − h ( y γh ) ◦ s ( h ) ◦ ν − h ( y γh )= s ( h ◦ h ) − ◦ σ ( h ◦ h ) − (cid:0) ν − h ◦ h ( y γh ◦ h ) − (cid:1) ◦ σ h − ( ν − h ( y γh )) ◦ σ ( h ◦ h ) − (cid:0) ν − h ( y γh ) (cid:1) ◦ s ( h ) ◦ s ( h ) , which, using ( y γh ◦ h ) − = − y γh ◦ h , implies that σ ( h ◦ h ) − (cid:0) ν − h ◦ h ( y γh ◦ h ) (cid:1) = σ h − ( ν − h ( y γh )) + σ ( h ◦ h ) − (cid:0) ν − h ( y γh ) (cid:1) . This, on further simplification, finally gives y γh ◦ h = ν h ◦ h (cid:0) σ h ( ν − h ( y γh )) (cid:1) + ν h ( y γh ) . OHOMOLOGY, EXTENSIONS AND AUTOMORPHISMS OF BRACES 19
For h , h ∈ H , it easily follows that y γh + h = y γh + y γh . We have now proved that ζ ( γ ) is a derivation. That both η and ζ are homomorphisms, and ηζ and ζη are, respectively, the identity elements of Autb H,I ( E ) and Z N ( H, I ) is obvious. Hence Autb
H,I ( E ) ∼ =Z N ( H, I ), and the proof is complete. (cid:3)
We finally get the following Wells’ like exact sequence for braces (Theorem B).
Theorem 5.7.
Let E : 0 → I → E π → H be a extension of a left brace H by a trivial brace I such that [ E ] ∈ Ext ν,σ ( H, I ) . Then we have the following exact sequence of groups → Z N ( H, I ) → Autb I ( E ) ρ ( E ) −→ C ( ν,σ ) ω ( E ) −→ H N ( H, I ) , where ω ( E ) is, in general, only a derivation. A pair of automorphisms ( φ, θ ) ∈ C ( ν,σ ) is said to be inducible if there exists an automorphism γ ∈ Autb( E ) such that ( γ H , γ I ) = ( φ, θ ). Theorem 5.7 tells that a pair ( φ, θ ) ∈ C ( ν,σ ) is inducible if andonly if ( φ, θ ) ∈ Ker( ω ( E )). When H N ( H, I ) = 0, as a consequence of Theorem 5.7, we get the following
Corollary 5.8.
Let E : 0 → I → E π → H be a extension of a left brace H by a trivial brace I such that [ E ] ∈ Ext ν,σ ( H, I ) and H N ( H, I ) = 0 . Then we have the following exact sequence of groups → Z N ( H, I ) → Autb I ( E ) ρ ( E ) −→ C ( ν,σ ) ω ( E ) −→ . Hence every pair ( φ, θ ) ∈ C ( ν,σ ) is inducible. We remark that when ( ν, σ ) is a trivial action, then C ( ν,σ ) = Autb( H ) × Autb( I ). So, if H N ( H, I ) = 0,then every pair ( φ, θ ) ∈ Autb( H ) × Autb( I ) is inducible.We now present a module theoretic interpretation of inducible pairs. Let I and I ′ be two H -bi-moduleswith good pairs of actions ( ν, σ ) and ( ν ′ , σ ′ ) respectively. Recall that a map ζ : I → I ′ is called H -bi-module homomorphism if (Id , ζ ) is compatible (as defined in Section 2) with the pairs of actions ( ν, σ )and ( ν ′ , σ ′ ).Let E : 0 → I → E → H → H by a trivial brace I . Then I can beviewed as an H -bi-module through the corresponding good pair of actions ( ν, σ ) as defined in (5) and(6). An automorphism φ of the brace H defines a new H -bi-module structure on I given by ( νφ, σφ ),which we denote by I φ , where νφ ( h ) = ν φ ( h ) and σφ ( h )( y ) = s ( φ ( h )) − ys ( φ ( h )) for h ∈ H and y ∈ I .It is not difficult to show that the automorphism φ induces an isomorphism φ ∗ of cohomology groups φ ∗ : H N ( H, I ) → H N ( H, I φ ) defined by φ ∗ ([( β, τ )]) = [ (cid:0) β ( φ, Id) , τ ( φ, Id) (cid:1) ] . Further, any H -bi-module isomorphism θ : I → I φ induces an isomorphism θ ∗ of cohomology groups θ ∗ : H N ( H, I ) → H N ( H, I φ ) given by θ ∗ ([( β, τ )]) = [ (cid:0) β (Id ,θ − ) , τ (Id ,θ − ) (cid:1) ] . With this set-up, we have
Theorem 5.9.
A pair of automorphisms ( φ, θ ) ∈ Autb( H ) × Autb( I ) is inducible if and only if thefollowing conditions hold:(i) θ : I → I φ is an isomorphism of H -bi-modules.(ii) θ ∗ ([( β, τ )]) = φ ∗ ([( β, τ )]) . Proof.
Let ( φ, θ ) be inducible. Then there exists an automorphism γ ∈ Autb I ( E ) such that ( φ, θ ) =( γ H , γ I ). For h ∈ H and y ∈ I , we know that γ ( s ( h )) = s ( γ H ( h )) ◦ y h for some y h ∈ I . We have γ I ( ν h ( y )) = γ (cid:0) s ( h ) ◦ y − s ( h ) (cid:1) = γ ( s ( h )) ◦ γ ( y ) − γ ( s ( y ))= s ( γ H ( h )) ◦ y h ◦ γ ( y ) − s ( γ H ( h )) ◦ y h = s ( γ H ( h )) ◦ γ I ( y ) − s ( γ H ( h ))= ν γ H ( h ) ( γ I ( y )) . (18)Similarly it also follows that γ I ( σ h ( y )) = σ γ H ( h ) ( γ I ( y )) . (19)This shows that (Id , γ I ) is compatible with the pairs of actions ( ν, σ ) and ( νγ H , σγ I ); hence condition (i)holds.For each h ∈ H , there exists a unique element (say) λ ( h ) in I such that γ ( s ( h )) = s ( γ H ( h )) + λ ( h ).It turns out that λ : H → I given by h λ ( h ) lies in C N . Given elements x , x ∈ E have uniqueexpressions of the form x = s ( h ) + y and x = s ( h ) + y for some h , h ∈ H and y and y ∈ I . Now γ ( x + x ) = γ (cid:0) s ( h + h ) + β ( h , h ) + y + y (cid:1) = γ ( s ( h + h )) + γ I ( β ( h , h )) + γ I ( y + y )= s (cid:0) γ H ( h ) + γ H ( h ) (cid:1) + λ ( h + h ) + γ I ( β ( h , h )) + γ I ( y + y ) . On the other hand, we have γ ( x ) + γ ( x ) = s ( γ H ( h )) + s ( γ H ( h )) + λ ( h ) + λ ( h ) + γ I ( y + y ) . Since γ ( x + x ) = γ ( x ) + γ ( x ), preceding two equations give γ I ( β ( h , h )) − β ( γ H ( h ) , γ H ( h )) = λ ( h ) − λ ( h + h ) + λ ( h ) . (20)Notice that s ( h ) + λ ( h ) = s ( h ) ◦ ν − h ( λ ( h )) and s ( h ) ◦ λ ( h ) = s ( h ) + ν h ( λ ( h )). By this and using (18)and (19), we have γ ( x ◦ x ) = γ (cid:0) ( s ( h ) + y ) ◦ (( s ( h ) + y ) (cid:1) = γ (cid:0) s ( h ◦ h ) ◦ (cid:0) ν − h ◦ h ( τ ( h , h )) + σ h ( ν − h ( y )) + ν − h ( y ) (cid:1)(cid:1) = γ (cid:0) s ( h ◦ h ) + ν h ◦ h ( σ h ( ν − h ( y ))) + ν h ( y ) + τ ( h , h ) (cid:1) = s ( γ H ( h ◦ h )) + λ ( h ◦ h ) + γ I ( ν h ◦ h ( σ h ( ν − h ( y )))) + γ I ( ν h ( y )) + γ I ( τ ( h , h ))= s ( γ H ( h ◦ h )) + λ ( h ◦ h ) + ν γ H ( h ◦ h ) ( σ γ H ( h ) ( ν − γ H ( h ) ( γ I ( y ))))+ ν γ H ( h ) ( γ I ( y )) + γ I ( τ ( h , h )) . By similar computations, on the other hand, we get γ ( x ) ◦ γ ( x ) = s ( γ H ( h ◦ h )) + ν γ H ( h ◦ h ) ( σ γ H ( h ) ( ν − γ H ( h ) ( γ I ( y ))))+ ν γ H ( h ◦ h ) ( σ γ H ( h ) ( ν − γ H ( h ) ( λ ( h )))) + ν γ H ( h ) ( γ I ( y ))+ ν γ H ( h ) ( λ ( h )) + τ ( γ H ( h ) , γ H ( h )) . Preceding two equations give γ I ( τ ( h , h )) − τ ( γ H ( h ) , γ H ( h )) = ν γ H ( h ◦ h ) ( σ γ H ( h ) ( ν − γ H ( h ) ( λ ( h )))) − λ ( h ◦ h ) + ν γ H ( h ) ( λ ( h )) , which, alongwith (20) proves condition (ii), that is, γ ∗ I ([( β, τ )]) = γ ∗ H ([( β, τ )]).Conversely, let ( φ, θ ) ∈ Autb( H ) × Autb( I ) satisfy conditions (i) and (ii). Condition (ii) guaranteesthe existence of a map λ : H → I in C N such that (cid:0) β ( φ, Id) , τ ( φ, Id) (cid:1) − (cid:0) β (Id ,θ − ) , τ (Id ,θ − ) (cid:1) = ∂ ( λ ) . OHOMOLOGY, EXTENSIONS AND AUTOMORPHISMS OF BRACES 21
Define the map γ : E → E , for all h ∈ H and y ∈ I , by γ ( s ( h ) + y ) = s ( φ ( h )) + λ ( h ) + θ ( y ) . A routine check then shows that γ ∈ Autb I ( E ), γ H = φ and γ I = θ . The proof is now complete. (cid:3) We conclude with a reduction argument on extension and lifting problem for automorphisms of braces.We’ll only deal with brace extensions from STExt ν,σ ( H, I ), with H of finite order, for a given good pairof actions ( ν, σ ) of H on I . Let us start with the following fixed extension of a finite left brace H by atrivial brace I lying in STExt ν,σ ( H, I ): E : 0 → I → E → H → . For this extension E , by Theorem 5.5, we get the exact sequence of groups0 → Autb
H,I ( E ) → Autb I ( E ) ρ ( E ) −→ C ( ν,σ ) ω ( E ) −→ RH N ( H, I ) , (21)where ω ( E ) is a derivation.Let P i denote a Sylow p i -subgroup of ( H, ◦ ), where p i is a prime divisor of the order of H . Noticethat P i is also the Sylow p i -subgroup of ( H, +); hence P i becomes a left ideal of H . Let R i denote thepre-image of P i in E . Notice that ( ν i , σ i ) is a good pair of actions of P i on I , where ν i := ν | P i and σ i := σ | P i , and the extension E i : 0 → I → R i → P i → ν i ,σ i ( H, I ). SetC ( ν i ,σ i ) := { ( φ, θ ) ∈ Autb( P i ) × Autb( I ) | ν ih = θ − ν iφ ( h ) θ and σ ih = θ − σ iφ ( h ) θ } . The extension E i now gives the exact sequence of groups0 → Autb P i ,I ( R i ) → Autb I ( R i ) ρ ( E i ) −→ C ( ν i ,σ i ) ω ( E i ) −→ RH N ( P i , I ) . On the other hand, denote by Autb
I,R i ( E, I ) the subgroup of Autb I ( H, I ) consisting of all automor-phism of E normalising I as well as R i , and set C i ( ν,σ ) := { ( φ, θ ) ∈ C ( ν,σ ) | φ ( P i ) = P i } . Then for theextension E : 0 → I → E → H → , we get the following exact sequence of groups from (21):0 → Autb
H,I ( E ) → Autb
I,R i ( E ) ρ ( E ) −→ C i ( ν,σ ) ω ( E ) −→ RH N ( H, I ) . Let res HP i : RH N ( H, I ) → RH N ( P i , I ) be the restriction homomorphisms as defined in Corollary 4.5.Define r HP i : C i ( ν,σ ) → C ( ν i ,σ i ) by r HP i ( φ, θ ) = ( φ | P i , θ ) . Using the definition of ω ( E ), we now get the following commutative diagram:C i ( ν,σ ) ω ( E ) −−−−→ RH N ( H, I ) y r HPi y res
HPi C ( ν i ,σ i ) ω ( E i ) −−−−→ RH N ( P i , I ) . (22)Recall that H N (( H, ◦ ) , I ) denotes the second cohomology group of the group ( H, ◦ ) with coefficients in I , where the right action of H on I is through σ . Similarly H N (( P i , ◦ ) , I ) denotes the second cohomologygroup of ( P i , ◦ ) with coefficients in I , where the right action of P i on I is through σ i = σ | P i . By Proposition ι : RH N ( H, I ) → H N (( H, ◦ ) , I ) and ι i : RH N ( P i , I ) → H N (( P i , ◦ ) , I ). We nowget the following commutative diagram:RH N ( H, I ) ι −−−−→ H N (( H, ◦ ) , I ) y res HPi y res
HPi RH N ( P i , I ) ι i −−−−→ H N (( P i , ◦ ) , I ) . (23)Let π ( H ) := { p , . . . , p r } be the set of all distinct prime divisors of the order of H . Please note thead-hoc use of π . With this set-up, we finally have Theorem 5.10.
Let ( φ, θ ) ∈ C ( ν,σ ) be such that ( φ | P i , θ ) ∈ C ( ν i ,σ i ) is inducible for some Sylow p i -subgroup of H for each p i ∈ π ( H ) . Then ( φ, θ ) is inducible.Conversely, if ( φ, θ ) ∈ C ( ν,σ ) is inducible, then ( φ | P i , θ ) ∈ C ( ν i ,σ i ) is inducible whenever φ ( P i ) = P i .Proof. Let ( φ, θ ) ∈ C ( ν,σ ) be as in the statement and ω ( E )( φ, θ ) = [ f ]. Then by the hypothesis and thecommutativity of diagram (22), it follows that res HP i ([ f ]) is the identity element of RH N ( P i , I ). Further,by the commutativity of diagram (23), we have res HP i ([ f ′ ]) is the identity element of H N (( P i , ◦ ) , I ). Beingin the cohomology of groups, we can now use corestriction-restriction homomorphism result [6, (9.5)Proposition. (ii)] to deduce that k [ f ′ ] is the zero cohomology class of H N (( H, ◦ ) , I ), where k is the indexof P i in H . Since this is true for at least one Sylow p i -subgroup for each p i ∈ π ( H ), it follows that[ f ′ ] is the identity element of H N (( H, ◦ ) , I ), which, ι being an embedding, implies that [ f ] is the zerocohomology class of H N ( H, I ). Hence ( φ, θ ) is inducible.The converse part is left as an exercise for the reader. (cid:3)
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