Classification of 5-dimensional restricted Lie algebras over perfect fields, I
aa r X i v : . [ m a t h . R A ] F e b Classification of 5-dimensional restricted Lie algebras overperfect fields, I
Iren Darijani and Hamid Usefi Department of Mathematics and StatisticsMemorial University of NewfoundlandSt. John’s, NL, Canada, A1C 5S7 [email protected] usefi@mun.ca
July 23, 2018 The research was supported in part by NSERC of Canada, grant hapter 1
Introduction
Let L be a Lie algebra over a field F of positive characteristic p . Recall that L is called restricted if L affords a p -map that satisfies the following conditions for all x, y ∈ L and λ ∈ F
1. ( λx ) [ p ] = λ p x [ p ] ;2. ( ad x ) p = ad ( x [ p ] );3. ( x + y ) [ p ] = x [ p ] + y [ p ] + P p − j =1 s j ( x, y ) , where js j ( x, y ) is the coefficient of t j − in ( ad ( tx + y )) p − ( x ), t an indeterminate.Recall that a restricted Lie algebra L is called p -nilpotent, if there exists an integer n such that x [ p ] n = 0 , for all x ∈ L . The purpose of this paper is to classify all p -nilpotent restricted Lie algebrasof dimension 5 over perfect fields of characteristic p > L is finite-dimensional and p -nilpotent then L is nilpotent. Our work builds upon the recent work ofSchneider and Usefi [13] on the classification of p -nilpotent restricted Lie algebras of dimension up to 4over perfect fields of characteristic p . Our method is different than what is used in [13] as we describebelow. The analogous classification for small dimensional nilpotent Lie algebras has a long history.The classification of all nilpotent Lie algebras of dimension up to five over any field has been known fora long time. However, in dimension 6, the characterization depends on the underlying field. In 1958Morozov [9] gave a classification of nilpotent Lie algebras of dimension 6 over a filed of characteristiczero, see also [1, 10, 8] for a classification over other fields. These classifications, however, differ andit was not easy to compare them until recently that de Graaf [5] gave a complete classification overany field of characteristic other than 2. de Graaf’s approach can be verified computationally and waslater revised and extended to characteristic 2 in [2]. The classification in dimensions more than 6 isstill in progress, see for example [14, 11].We now describe the method used in [13] and explain why this method is not applicable in dimen-sion 5. Note that in order to define a p -map on L , it is enough to define it on a basis of L and thenextend it linearly using property (3). Let ϕ , ϕ : L → L be two p -maps on L . Then the restrictedLie algebras ( L, ϕ ) and ( L, ϕ ) are isomorphic if and only if there exists A ∈ Aut( L ) such that A ( ϕ ( x )) = ϕ ( A ( x )) holds for all x ∈ L. Hence, ϕ and ϕ define isomorphic restricted Lie algebras if and only if there exists A ∈ Aut( L )such that Aϕ A − = ϕ ; that is, they are conjugate under the automorphism group of L . In thiscase we say that the p -maps ϕ and ϕ are equivalent . This defines a left action of Aut( L ) on the1 HAPTER 1. INTRODUCTION p -maps and the isomorphism classes of restricted Lie algebras correspond to the Aut( L )-orbitsunder this action. The main task using this approach would be then to find the Aut( L )-orbits. This isexactly what the authors did in [13] to determine all p -nilpotent restricted Lie algebras of dimensionup to 4 over perfect fields. However, this task becomes computationally infeasible to carry out indimension 5.The method we use to classify p -nilpotent restricted Lie algebras of dimension 5 is the analogueof Skjelbred-Sund method [15] for classifying nilpotent Lie algebras. We describe this method below.Let L be a restricted Lie algebra and M a vector space. We view M as a trivial L -module and define Z ( L, M ), the space of 2-cocycles, and the second cohomology group H ( L, M ) of L with coefficientsin M in Section 2.1.1. Let θ = ( φ, ω ) ∈ Z ( L, M ). We set L θ = L ⊕ M as a vector space and definethe Lie bracket and p -map on L θ by:[( x + m ) , ( x + m )] = [ x , x ] + φ ( x , x ) , ( x + m ) [ p ] = x [ p ] + ω ( x ) . We prove in Theorem 2.2.1 that L θ with the given bracket and p -map is a restricted Lie algebra.Now let K be a p -nilpotent restricted Lie algebra. Then its center Z ( K ) is nonzero and there exists x ∈ Z ( K ) such that x [ p ] = 0 . Let M be the one dimensional restricted ideal of K spanned by x ,and set L = K/M . Let π : K → L be the projection map. We have the exact sequence of restrictedLie algebras: 0 → M → K → L → . Choose an injective linear map σ : L → K such that πσ = 1 L . Define φ : L × L → M by φ ( x , x ) = [ σ ( x ) , σ ( x )] − σ ([ x , x ]) and ω : L → M by ω ( x ) = σ ( x ) [ p ] − σ ( x [ p ] ). We show inTheorem 2.2.2 that θ = ( φ, ω ) ∈ Z ( L, M ) and K ∼ = L θ . Therefore, any p -nilpotent restricted Liealgebra K of dimension n can be constructed as a central extensions of a p -nilpotent restricted Liealgebras of dimension n − p ( L ) of restricted automorphisms of L acts on H ( L, M ) in a natural way. Let A ∈ Aut p ( L ) and θ = ( φ, ω ) ∈ Z ( L, M ). We define Aθ = ( Aφ, Aω ), where Aφ ( x, y ) = φ ( A ( x ) , A ( y ))and Aω ( x ) = ω ( A ( x )). Let θ , θ ∈ Z ( L, M ). It follows from Theorem 2.3.5 that if θ and θ are inthe same Aut p ( L )-orbit then exists an isomorphism f : L θ → L θ such that f ( M ) = M . Therefore,we use the action of Aut p ( L ) to reduce the number of isomorphic restricted Lie algebras.There are nine nilpotent Lie algebras of dimension 5 listed in Theorem 2.3.1. Let K be a nilpotentLie algebra of dimension 5. Since the p -maps are p -nilpotent, there exists a central element x ∈ K such that x [ p ] = 0 . Then we let L = K/ h x i and find all 1-dimensional central extensions of L thatlead to K . That is, we choose those θ = ( φ, ω ) ∈ Z ( L, F ) such that L θ is isomorphic as a Lie algebrato K . Then we list all possible p -maps that are obtained via different choices of θ and x . We stillneed to detect and remove the isomorphic algebras from this list. Finally, we shall prove that theremaining algebras in the list are pairwise non-isomorphic.If p > p -map on K is semilinear and it is easier to construct a basis for H ( L, F )as described in Section 2.4. However, for the cases that p = 2 or p = 3 , the p -maps are no longersemilinear and we have to come up with a different way of finding a basis for H ( L, F ). This differencein turn changes the computations and results in possibly different restricted Lie algebra structures on K . We shall treat the cases of p = 2 , ontents p -nilpotent restricted Lie algebras . . . . . . . . . . . . . . . . . . . . . . 72.3 Strategy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 122.4 Finding a basis for H ( L, F ) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19 L , / h x i , trivial p -map) . . . . . . . . . . . . . . . . . . . . . . . . . . 213.2 Extensions of ( L , / h x i , x [ p ]1 = x ) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 213.3 Extensions of ( L , / h x i , x [ p ]1 = x , x [ p ]3 = x ) . . . . . . . . . . . . . . . . . . . . . . . 223.4 Extensions of ( L , / h x i , x [ p ]1 = x , x [ p ]2 = x ) . . . . . . . . . . . . . . . . . . . . . . . 233.5 Extensions of ( L , / h x i , x [ p ]1 = x , x [ p ]2 = x , x [ p ]3 = x ) . . . . . . . . . . . . . . . . . . 23 L , L = L , h x i . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 274.1.1 Extensions of ( L , trivial p -map) . . . . . . . . . . . . . . . . . . . . . . . . . . 284.2 Extensions of L = L , h x i . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 324.2.1 Extensions of ( L , trivial p -map) . . . . . . . . . . . . . . . . . . . . . . . . . . 334.2.2 Extensions of ( L, x [ p ]1 = x ) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 354.2.3 Extensions of ( L, x [ p ]1 = x ) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 374.2.4 Extensions of ( L, x [ p ]1 = x , x [ p ]2 = x ) . . . . . . . . . . . . . . . . . . . . . . . 394.2.5 Extensions of ( L, x [ p ]3 = x ) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 414.2.6 Extensions of ( L, x [ p ]2 = x , x [ p ]3 = x ) . . . . . . . . . . . . . . . . . . . . . . . 424.2.7 Extensions of ( L, x [ p ]4 = x ) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 444.2.8 Extensions of ( L, x [ p ]2 = x , x [ p ]4 = x ) . . . . . . . . . . . . . . . . . . . . . . . 464.3 Detecting Isomorphisms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 48 L , L = L , h x i . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 575.1.1 Extensions of ( L , trivial p -map) . . . . . . . . . . . . . . . . . . . . . . . . . . 595.1.2 Extensions of ( L, x [ p ]1 = x ) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 623 ONTENTS
L, x [ p ]3 = x ) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 655.2 Extensions of L = L , h x i . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 685.2.1 Extensions of ( L , trivial p -map) . . . . . . . . . . . . . . . . . . . . . . . . . . 695.2.2 Extensions of ( L, x [ p ]1 = x ) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 715.2.3 Extensions of ( L, x [ p ]2 = ξx ) . . . . . . . . . . . . . . . . . . . . . . . . . . . . 735.2.4 Extensions of ( L, x [ p ]3 = x ) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 775.2.5 A list of restricted Lie algebra structures on L , . . . . . . . . . . . . . . . . . 815.3 Detecting isomorphisms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 82 L , . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 966.1.1 Extension of L , / h x i via the trivial p -map . . . . . . . . . . . . . . . . . . . 976.2 Restriction maps on L , . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1026.2.1 Extension of L , / h x i via the trivial p -map . . . . . . . . . . . . . . . . . . . 1046.3 Restriction maps on L , . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1076.3.1 Extensions of L , / h x i via the trivial p -map . . . . . . . . . . . . . . . . . . . 1086.4 Restriction maps on L , . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1116.4.1 Extensions of L , / h x i via the trivial p -map . . . . . . . . . . . . . . . . . . . 113 L , L , trivial p -map) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1187.2 Extensions of ( L, x [ p ]1 = x ) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1217.3 Extensions of ( L, x [ p ]3 = x ) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1257.4 Detecting isomorphisms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 128 L , L , trivial p -map) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1338.2 Extensions of ( L, x [ p ]1 = x ) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1378.3 Extensions of ( L, x [ p ]2 = ξx ) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1428.4 Extensions of ( L, x [ p ]3 = x ) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1468.5 Detecting isomorphisms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 149 Bibliography 156 hapter 2
Preliminaries
Throughout this paper F denotes a perfect filed of characteristic p > p > Definition 2.1.1
A restricted Lie algebra of characteristic p > is a Lie algebra L of characteristic p together with a map L → L , denoted by x x [ p ] , that satisfies • ( λx ) [ p ] = λ p x [ p ] , • ( x + y ) [ p ] = x [ p ] + y [ p ] + P p − j =1 s j ( x, y ) where js j ( x, y ) is the coefficient of t j − in ( ad ( tx + y )) p − ( x ) , t an indeterminate, • [ x, y [ p ] ] = [ x, y, . . . , y | {z } p ] , for all x, y ∈ L and all λ ∈ F . The map x x [ p ] is referred to as the p -map. We remark that the second property is equivalent to( x + y ) [ p ] = x [ p ] + y [ p ] + X xj = x or yx x,x y x [ x , x , . . . , x p ] , where x denotes the number of x ’s among the x j . Note that long commutators are left-tapped,that is [ x , . . . , x k , x k +1 ] = [[ x , . . . , x k ] , x k +1 ] . For a subset S of L , we denote by h S i p the restricted subalgebra of L generated by S and by h S i F the subspace spanned by S . Also, we denote by S [ p ] n the restricted subalgebra generated byall x [ p ] n , where x ∈ S . Recall that S is called p -nilpotent if there exists an integer n such that S [ p ] n = 0 . 5 HAPTER 2. PRELIMINARIES L is a Lie algebra over F and M is vector space, a q -dimensional cochain of L with coefficients in M is a skew-symmetric, q -linear map on L taking values in M . We denote the space of q -dimensionalcochains of a Lie algebra L with coefficients in M by C q cl ( L, M ). So, we have C q cl ( L, M ) =
Hom F (Λ q L, M ) . The coboundary map δ q : C q cl ( L, M ) → C q +1cl ( L, M ) is defined by( δ q φ )( ℓ , . . . , ℓ q +1 ) = X ≤ s Let L be a restricted Lie algebra over F and M a vector space. If φ ∈ C cl ( L, M ) and ω : L → M a function, we say ω has the ⋆ -property with respect to φ if for every x, y ∈ L and λ ∈ F , we have1. ω ( λx ) = λ p ω ( x ) ω ( x + y ) = ω ( x ) + ω ( y ) + X xj = x or yx x,x y x φ ([ x , x , . . . , x p − ] , x p ) , where x is the number of x . Now, we define the space of 2-dimensional cochains of a restricted Lie algebra L with coefficientsin M as the subspace spanned by all ( φ, ω ) such that φ is skew-symmetric and ω : L → M has the ⋆ -property with respect to φ . We denote this vector space by C ( L, M ). Evidently if ω and ω ′ havethe ⋆ -property with respect to φ and φ ′ respectively, then ω + ω ′ has the ⋆ -property with respectto φ + φ ′ , and hence C ( L, M ) is a vector space over F by point wise addition in each coordinate.We have adopted Definition 2.1.2 from [7]. However, definition of ⋆ -property in the whole generalitygiven in [7] is ambiguous. Lemma 2.1.3 Let M be a vector space and ψ : L → M a linear map. Then ˜ ψ : L → M defined by ˜ ψ ( x ) = ψ ( x [ p ] ) has the ⋆ -property with respect to δ ψ . Proof. Since ψ [ x , . . . , x p ] = ( δ ψ )([ x , . . . , x p − ] , x p ), we have˜ ψ ( x + y ) = ψ (( x + y ) [ p ] ) = ψ ( x [ p ] ) + ψ ( y [ p ] ) + ψ ( X xj = x or yx x,x y x [ x , x , . . . , x p ])= ψ ( x [ p ] ) + ψ ( y [ p ] ) + X xj = x or yx x,x y x ) ( δ ψ )([ x , . . . , x p − ] , x p ) , for every x, y ∈ L . (cid:4) Let Z ( L, M ) be the set consisting of all ( φ, ω ) ∈ C ( L, M ) such that δ φ = 0 and φ ( x, y [ p ] ) = φ ([ x, y, . . . , y | {z } p − ] , y ) , for all x, y ∈ L . The elements of Z ( L, M ) are called cocycles . Also, let B ( L, M ) be the setconsisting of all ( φ, ω ) ∈ C ( L, M ) such that there exists ψ ∈ C cl ( L, M ) satisfying δ ψ = φ and˜ ψ = ω . The elements of B ( L, M ) are called coboundaries . It is easy to see that Z ( L, M ) and B ( L, M ) are subspaces of C ( L, M ). HAPTER 2. PRELIMINARIES Theorem 2.1.4 B ( L, M ) ⊆ Z ( L, M ) , so that, the quotient H ( L, M ) = Z ( L, M ) /B ( L, M ) is well-defined. Proof. Let ( δ ψ, ˜ ψ ) ∈ B ( L, M ). First, We claim that δ δ ψ = 0 . Indeed, for all x, y, z ∈ L, wehave δ δ ψ ( x, y, z ) = δ ψ ([ x, y ] , z ) + δ ψ ([ y, z ] , x ) + δ ψ ([ z, x ] , y )= ψ ([[ x, y ] , z ]]) + ψ ([[ y, z ] , x ]]) + ψ ([[ z, x ] , y ]])= ψ ([[ x, y ] , z ]] + [[ y, z ] , x ]] + [[ z, x ] , y ]]) , which is equal to zero by jaccobi identity. Next, we claim that( δ ψ )( x, y [ p ] ) = ( δ ψ )([ x, y, ..., y | {z } p − ] , y ) , for all x, y ∈ L . Indeed, for all x, y ∈ L , we have( δ ψ )( x, y [ p ] ) − ( δ ψ )([ x, y, . . . , y | {z } p − ] , y ) = ψ [ x, y [ p ] ] − ψ ([ x, y, . . . , y | {z } p ])= ψ ([ x, y [ p ] ] − [ x, y, . . . , y | {z } p ])=0 . The proof is complete. (cid:4) We call H ( L, M ) the second cohomology group of L with coefficients in M . Let θ = ( φ, ω ) ∈ Z ( L, M ). Then we denote by [ θ ] the image of θ in H ( L, M ). Definition 2.1.5 A restricted Lie algebra M is called strongly abelian if [ M, M ] = 0 and M [ p ] = 0 . p -nilpotent restricted Lie algebras Let L be a restricted Lie algebra, M a vector space and θ = ( φ, ω ) ∈ Z ( L, M ). We construct arestricted extension of L by M as follows. Lemma 2.2.1 Let L θ = L ⊕ M as a vector space and define the Lie bracket and p -map on L θ by: [( x + m ) , ( x + m )] = [ x , x ] + φ ( x , x ) , ( x + m ) [ p ] = x [ p ] + ω ( x ) . Then L θ with the given bracket and p -map is a resticted Lie algebra. Proof. The bracket is clearly bilinear and skew symmetric and it is well known that the jaccobiidentity is equivalent to δ φ = 0 . We claim that L is restricted with the given p -map. Let x , . . . , x k +1 ∈ L , m , . . . m k +1 ∈ M . Note that by induction we have[ x + m , x + m , . . . , x k +1 + m k +1 ] = [ x , . . . , x k +1 ] + φ ([ x , . . . , x k ] , x k +1 ) . (2.1) HAPTER 2. PRELIMINARIES x + m , ( x + m ) [ p ] ] =[ x + m , x [ p ]2 + ω ( x )]=[ x , x [ p ]2 ] + φ ( x , x [ p ]2 ) . On the other hand,[ x + m , x + m , . . . , x + m | {z } p ] = [ x , x , . . . , x | {z } p ] + φ ([ x , x , . . . , x | {z } p − ] , x ) , by equation (2.1). We have φ ( x , x [ p ]2 ) = φ ([ x , x , . . . , x | {z } p − ] , x ) , also we have [ x , x [ p ]2 ] = [ x , x , . . . , x | {z } p ] . Therefore, [ x + m , ( x + m ) [ p ] ] = [ x + m , x + m , . . . , x + m | {z } p ] . Next, we have ( λ ( x + m )) [ p ] = ( λx + λm ) [ p ] = ( λx ) [ p ] + ω ( λx ) = λ p x [ p ] + λ p ω ( x )= λ p ( x [ p ] + ω ( x ))= λ p ( x + m ) [ p ] . Finally, we have(( x + m ) + ( x + m )) [ p ] =(( x + x ) + ( m + m )) [ p ] =( x + x ) [ p ] + ω ( x + x )= x [ p ]1 + x [ p ]2 + X xlj = x x xl x ,xl x x [ x l , x l , . . . , x l p ]+ ω ( x ) + ω ( x ) + X xlj = x x xl x ,xl x x φ ([ x l , x l , . . . , x l p − ] , x l p ) . On the other hand,( x + m ) [ p ] + ( x + m ) [ p ] + X lj =1 or 2 l ,l x + m ) [ x l + m l , x l + m l , . . . , x l p + m l p ]= x [ p ]1 + ω ( x ) + x [ p ]2 + ω ( x ) + X xlj = x x xl x ,xl x x [ x l , x l , . . . , x l p ]+ X xlj = x x xl x ,xl x x φ ([ x l , x l , . . . , x l p − ] , x l p ) , HAPTER 2. PRELIMINARIES x + m ) + ( x + m )) [ p ] = ( x + m ) [ p ] + ( x + m ) [ p ] + X lj =1 or 2 l ,l x + m [ x l + m l , x l + m l , . . . , x l p + m l p ] . The proof is complete. (cid:4) Now let K be a restricted Lie algebra, and suppose that its center Z ( K ) , is nonzero. Let 0 = M ⊆ Z ( K ) such that M [ p ] = 0 , and set L = K/M . Let π : K → L be the projection map. We have theexact sequence 0 → M → K → L → . Choose an injective linear map σ : L → K such that πσ = 1 L . Note that we can easily show that π ([ σ ( x i ) , σ ( x j )] − σ ([ x i , x j ])) = 0 , for every x i , x j ∈ L and π ( σ ( x ) [ p ] − σ ( x [ p ] )) = 0 , for every x ∈ L .Therefore, [ σ ( x i ) , σ ( x j )] − σ ([ x i , x j ]) , σ ( x ) [ p ] − σ ( x [ p ] ) ∈ M, for every x i , x j , x ∈ L . Now, we define φ : L × L → M by φ ( x i , x j ) = [ σ ( x i ) , σ ( x j )] − σ ([ x i , x j ]) and ω : L → M by ω ( x ) = σ ( x ) [ p ] − σ ( x [ p ] ). With these notation, we have: Lemma 2.2.2 Let θ = ( φ, ω ) . Then θ ∈ Z ( L, M ) and K ∼ = L θ . Proof. It is easy to see that φ is a bilinear and skew-symmetric form on L . We claim that ω hasthe ⋆ -property with respect to φ . Indeed: ω ( x + x ) =( σ ( x + x )) [ p ] − σ (( x + x ) [ p ] )=( σ ( x ) + σ ( x )) [ p ] − σ ( x [ p ]1 + x [ p ]2 + X xlj = x x xl x ,xl x x [ x l , x l , . . . , x l p ])= σ ( x ) [ p ] + σ ( x ) [ p ] + X lj =1 or 2 l ,l σ ( x ) [ σ ( x l ) , σ ( x l ) , . . . , σ ( x l p ) ]) − σ ( x [ p ]1 ) − σ ( x [ p ]2 ) − X xlj = x x xl x ,xl x x σ ([ x l , x l , . . . , x l p ])= ω ( x ) + ω ( x ) + X lj =1 or 2 l ,l x [ σ ( x l ) , σ ( x l ) , . . . , σ ( x l p ) ]) − X xlj = x x xl x ,xl x x σ ([ x l , x l , . . . , x l p ]) . HAPTER 2. PRELIMINARIES φ ( x, y ) ∈ M ⊆ Z ( L ), we have σ ([ x l , x l , . . . , x l p ])= [ σ ([ x l , x l , . . . , x l p − ]) , σ ( x l p )] − φ ([ x l , x l , . . . , x l p − ] , x l p )= [[ σ ([ x l , x l , . . . , x l p − ]) , σ ( x l p − )] − φ ([ x l , x l , . . . , x l p − ] , x l p − ) , σ ( x l p )] − φ ([ x l , x l , . . . , x l p − ] , x l p )= [[ σ ([ x l , x l , . . . , x l p − ]) , σ ( x l p − )] , σ ( x l p )] − φ ([ x l , x l , . . . , x l p − ] , x l p ) . If we repeat this procedure, we obtain that σ ([ x l , x l , . . . , x l p ]) = [ σ ( x l ) , σ ( x l ) , . . . , σ ( x l p ) ]) − φ ([ x l , x l , . . . , x l p − ] , x l p ) . Therefore, we have ω ( x + x ) = ω ( x ) + ω ( x ) + X xlj = x x xl x ,xl x x φ ([ x l , x l , . . . , x l p − ] , x l p ) . Next, we claim that δ φ = 0 . Indeed, for all x, y, z ∈ L , we have( δ φ )( x, y, z ) = φ ([ x, y ] , z ) + φ ([ y, z ] , x ) + φ ([ z, x ] , y )=[ σ ([ x, y ]) , σ ( z )] − σ ([[ x, y ] , z ]) + [ σ ([ y, z ]) , σ ( x )] − σ ([[ y, z ] , x ])+[ σ ([ z, x ]) , σ ( y )] − σ ([[ z, x ] , y ])=[ σ ([ x, y ]) , σ ( z )] + [ σ ([ y, z ]) , σ ( x )] + [ σ ([ z, x ]) , σ ( y )] − σ ([[ x, y ] , z ] + [[ y, z ] , x ] + [[ z, x ] , y ]) . By jaccobi identity, [ x, y, z ] + [ y, z, x ] + [ z, x, y ] = 0 . Therefore, δ φ ( x, y, z )= [ σ ([ x, y ]) , σ ( z )] + [ σ ([ y, z ]) , σ ( x )] + [ σ ([ z, x ]) , σ ( y )]= [[ σ ( x ) , σ ( y )] − φ ( x, y ) , σ ( z )] + [[ σ ( y ) , σ ( z )] − φ ( y, z ) , σ ( x )]+ [[ σ ( z ) , σ ( x )] − φ ( z, x ) , σ ( y )]= [[ σ ( x ) , σ ( y )] , σ ( z )] − [ φ ( x, y ) , σ ( z )] + [[ σ ( y ) , σ ( z )] , σ ( x )] − [ φ ( y, z ) , σ ( x )]+ [[ σ ( z ) , σ ( x )] , σ ( y )] − [ φ ( z, x ) , σ ( y )]= [[ σ ( x ) , σ ( y )] , σ ( z )] + [[ σ ( y ) , σ ( z )] , σ ( x )] + [[ σ ( z ) , σ ( x )] , σ ( y )] , which is equal to zero by jaccobi identity.Finally, we claim that φ ( x, y [ p ] ) = φ ([ x, y, . . . , y | {z } p − ] , y ), for all x, y ∈ L . Indeed: φ ( x, y [ p ] ) − φ ([ x, y, . . . , y | {z } p − ] , y )= [ σ ( x ) , σ ( y [ p ] )] − σ ([ x, y [ p ] ]) − [ σ ([ x, y, . . . , y | {z } p − ]) , σ ( y )] + σ ([ x, y, . . . , y | {z } p ])= [ σ ( x ) , σ ( y [ p ] )] − [ σ ([ x, y, . . . , y | {z } p − ]) , σ ( y )] + σ ([ x, y, . . . , y | {z } p ] − [ x, y [ p ] ])= [ σ ( x ) , σ ( y [ p ] )] − [ σ ([ x, y, . . . , y | {z } p − ]) , σ ( y )] . HAPTER 2. PRELIMINARIES σ ([ x, y, . . . , y | {z } p − ]) , σ ( y )] =[[ σ ([ x, y, . . . , y | {z } p − ]) , σ ( y )] − φ ([ x, y, . . . , y | {z } p − ] , y ) , σ ( y )]=[[ σ ([ x, y, . . . , y | {z } p − ]) , σ ( y )] , σ ( y )] . If we repeat this procedure, we obtain that[ σ ([ x, y, . . . , y | {z } p − ]) , σ ( y )] = [ σ ( x ) , σ ( y ) , . . . , σ ( y ) | {z } p ]= [ σ ( x ) , ( σ ( y )) [ p ] ] . Therefore, φ ( x, y [ p ] ) − φ ([ x, y, . . . , y | {z } p − ] , y )= [ σ ( x ) , σ ( y [ p ] )] − [ σ ( x ) , ( σ ( y )) [ p ] ]= [ σ ( x ) , σ ( y ) [ p ] − ω ( x )] − [ σ ( x ) , ( σ ( y )) [ p ] ]= [ σ ( x ) , σ ( y ) [ p ] ] − [ σ ( x ) , ( σ ( y )) [ p ] ]= 0 . Finally, we show that K ∼ = L θ . Let x ∈ K . Then there exist unique y ∈ L and z ∈ M such that x = σ ( y ) + z . Indeed, we can take z = x − σ ( π ( x )) and y = π ( x ). Since, π ( z ) = 0 , we have z ∈ M .Define f : K → L θ by f ( x ) = y + z . Then f is an isomorphism. Indeed: f ( x [ p ] ) = f (( σ ( y ) + z ) [ p ] ) = f (cid:16) σ ( y ) [ p ] + z [ p ] + X g j = σ ( y ) or zg = σ ( y ) , g = z σ ( y )) [ g , . . . , g p ] (cid:17) . Since z ∈ M , we have z [ p ] = 0 , and X g j = σ ( y ) or zg = σ ( y ) , g = z σ ( y )) [ g , . . . , g p ] = 0 . Thus, f ( x [ p ] ) = f ( σ ( y ) [ p ] ) = f ( σ ( y [ p ] ) + ω ( y )) = y [ p ] + ω ( y ) = y [ p ] + σ ( y ) [ p ] − σ ( y [ p ] ) . On the other hand, we have f ( x ) [ p ] = ( y + z ) [ p ] = y [ p ] + ω ( y ) = y [ p ] + σ ( y ) [ p ] − σ ( y [ p ] ) . Thus, f ( x [ p ] ) = f ( x ) [ p ] .Similarly we can show that f ([ x, y ]) = [ f ( x ) , f ( y )]. (cid:4) We conclude that any p -nilpotent restricted Lie algebra K of dimension n can be constructedfrom a restricted Lie algebra of lower dimension. HAPTER 2. PRELIMINARIES Lemma 2.2.3 Let θ = ( φ, ω ) ∈ Z ( L, M ) and η = ( ν, ξ ) ∈ B ( L, M ) . Then we have L θ ∼ = L θ + η . Proof. We have η = ( ν, ξ ) ∈ B ( L, M ). Therefore, there exits ψ ∈ C ( L, M ) such that δ ( ψ ) = ν and ˜ ψ = ξ . Note that f : L θ → L θ + η such that f ( x ) = x + ψ ( x ) for x ∈ L and f ( m ) = m for m ∈ M is an isomorphism. Indeed, let x , x ∈ L and m , m ∈ M . Then we have f ([ x + m , x + m ]) = f ([ x , x ] + φ ( x , x )) = [ x , x ] + ψ ([ x , x ]) + φ ( x , x ) . On the other hand, we have[ f ( x + m ) , f ( x + m )] = [ x + ψ ( x ) + m , x + ψ ( x ) + m ] = [ x , x ] + ( φ + ν )( x , x ) =[ x , x ] + ψ ([ x , x ]) + φ ( x , x ) . Therefore, f ([ x + m , x + m ]) = [ f ( x + m ) , f ( x + m )]. Also, we have f (( x + m ) [ p ] ) = f ( x [ p ]1 + ω ( x )) = x [ p ]1 + ψ ( x [ p ]1 ) + ω ( x ) . On the other hane, we have f ( x + m ) [ p ] = ( x + ψ ( x ) + m ) [ p ] = x [ p ]1 + ( ω + ξ )( x ) = x [ p ]1 + ψ ( x [ p ]1 ) + ω ( x ) . Therefore, f (( x + m ) [ p ] ) = f ( x + m ) [ p ] . (cid:4) In this section we describe our strategy to find a (possibly redundant but complete) list of all possible p -nilpotent p -maps on a given nilpotent Lie algebra K . Since the p -maps are p -nilpotent, thereexists z ∈ Z ( K ) such that z [ p ] = 0 . We let M = h z i p and L = K/M . We know from Theorem2.2.2 that K ∼ = L θ , for some θ ∈ Z ( L, M ). Hence, we can construct every restricted Lie algebra ofdimension n from a restricted Lie algebra of dimension n − p -nilpotent restricted Lie algebras of dimension 4. For reader’s convenience, we include these listsbelow. Theorem 2.3.1 ([5]) The isomorphism class of all nilpotent Lie algebras of dimension 5 over anarbitrary field is as follows: • L , = abelian ; • L , = h x , . . . , x | [ x , x ] = x i ; • L , = h x , . . . , x | [ x , x ] = x , [ x , x ] = x i ; • L , = h x , . . . , x | [ x , x ] = x , [ x , x ] = x i ; • L , = h x , . . . , x | [ x , x ] = x , [ x , x ] = x , [ x , x ] = x i ; • L , = h x , . . . , x | [ x , x ] = x , [ x , x ] = x , [ x , x ] = x , [ x , x ] = x i ; • L , = h x , . . . , x | [ x , x ] = x , [ x , x ] = x , [ x , x ] = x i ; • L , = h x , . . . , x | [ x , x ] = x , [ x , x ] = x i ; HAPTER 2. PRELIMINARIES • L , = h x , . . . , x | [ x , x ] = x , [ x , x ] = x , [ x , x ] = x i . Theorem 2.3.2 ([13]) Let L be a nilpotent Lie algebra of dimension over a perfect field F ofcharacteristic p > . Then the equivalence classes of the [ p ] -maps on L are as follows.(a) If L = h x , x , x , x i , then1. Trivial p -map;2. x [ p ]1 = x ;3. x [ p ]1 = x , x [ p ]3 = x ;4. x [ p ]1 = x , x [ p ]2 = x ;5. x [ p ]1 = x , x [ p ]2 = x , x [ p ]3 = x .(b) If L = h x , x , x , x | [ x , x ] = x i , then1. Trivial p -map;2. x [ p ]1 = x ;3. x [ p ]1 = x ;4. x [ p ]1 = x , x [ p ]2 = x ;5. x [ p ]3 = x ;6. x [ p ]3 = x , x [ p ]2 = x ;7. x [ p ]4 = x ;8. x [ p ]4 = x , x [ p ]2 = x .(c) If L = h x , x , x , x | [ x , x ] = x , [ x , x ] = x i , then1. Trivial p -map;2. x [ p ]1 = x ;3. x [ p ]3 = x ;4. x [ p ]2 = ξx , where ξ ∈ F ∗ and ξ and ξ represent isomorphic algebras if and only if ξ ξ − is a square in F . Next, we need to determine when L θ ∼ = L θ , for θ , θ ∈ Z ( L, M ). We will do this in Theorem2.3.5 which involves an action of Aut p ( L ) on H ( L, M ) that we need to address first. Here, Aut p ( L )denotes the group of restricted automorphisms of L . The group of automorphisms of L where L isjust considered as a Lie algebra is denoted by Aut( L ).So, let L is a restricted Lie algebra and M a strongly abelian restricted Lie algebra which isconsidered as a trivial L-module. Let A ∈ Aut p ( L ) and θ = ( φ, ω ) ∈ Z ( L, M ). We define Aθ =( Aφ, Aω ), where Aφ ( x, y ) = φ ( A ( x ) , A ( y )) and Aω ( x ) = ω ( A ( x )). With these definitions we have: Lemma 2.3.3 Let A ∈ Aut p ( L ) and θ ∈ Z ( L, M ) . Then Aθ ∈ Z ( L, M ) . Furthermore, if θ ∈ B ( L, M ) then Aθ ∈ B ( L, M ) . HAPTER 2. PRELIMINARIES Proof. It is easy to see that Aφ is skew symmetric and bilinear map. Since δ φ = 0 , we have δ Aφ ( x, y ) = δ φ ( A ( x ) , A ( y )) = 0 , which implies that δ Aφ = 0 . Also, Aω satisfies the ⋆ -property with respect to Aφ . To see this, let x, x , x ∈ L and λ ∈ F . Then we have, Aω ( λx ) = ω ( A ( λx )) = ω ( λA ( x )) = λ p ω ( A ( x )) = λ p Aω ( x ) , and Aω ( x + x ) = ω ( A ( x + x )) = ω ( Ax + Ax )= ω ( Ax ) + ω ( Ax ) + X Ax j = Ax orAx Ax ) φ ([ Ax , . . . , Ax p − ] , Ax p )= ω ( Ax ) + ω ( Ax ) + X x j = x orx x ) φ ( A [ x , . . . , x p − ] , Ax p )= ω ( Ax ) + ω ( Ax ) + X x j = x orx x ) Aφ ([ x , . . . , x p − ] , x p )= Aω ( x ) + Aω ( x ) + X x j = x orx x ) Aφ ([ x , . . . , x p − ] , x p ) . We also have Aφ ( x, y [ p ] ) − Aφ ([ x, y, . . . , y | {z } p − ] , y )= φ ( A ( x ) , A ( y [ p ] )) − φ ( A ([ x, y, . . . , y | {z } p − ]) , A ( y ))= φ ( A ( x ) , A ( y ) [ p ] ) − φ (([ A ( x ) , A ( y ) , . . . , A ( y ) | {z } p − ]) , A ( y ))= 0 . Now, we show that Aut p ( L ) preserves B ( L, M ). Let ( δψ, ˜ ψ ) ∈ B ( L, M ), where ψ : L → M islinear. We have A ( δψ, ˜ ψ ) = ( Aδψ, A ˜ ψ ). Then Aδψ ( x, y ) = δψ ( Ax, Ay ) = ψ ([ Ax, Ay ]) = ψ ( A [ x, y ]) = Aψ ([ x, y ]) = δ ( Aψ )( x, y ) , and A ˜ ψ ( x ) = ˜ ψ ( Ax ) = ψ ( A ( x ) [ p ] ) = ψ ( A ( x [ p ] )) = Aψ ( x [ p ] ) = ∼ ( Aψ )( x ) . So A ( δψ, ˜ ψ ) = ( δ ( Aψ ) , ∼ ( Aψ )). (cid:4) It follows from Lemma 2.3.3 that Aut p ( L ) acts on H ( L, M ). Now, let e , . . . , e s be a basis of M and set θ = ( φ, ω ) ∈ Z ( L, M ). Then φ ( x, y ) = s X i =1 φ i ( x, y ) e i , and ω ( x ) = s X i =1 ω i ( x ) e i . HAPTER 2. PRELIMINARIES Lemma 2.3.4 For every i , we have ( φ i , ω i ) ∈ Z ( L, F ) . Proof. Since φ is skew symmetric and bilinear form, so is every φ i . We claim that δ φ i = 0 . Indeed,0 = δ φ ( x, y ) = P si =1 δ φ i ( x, y ) e i , implying that δ φ i ( x, y ) = 0 .Now we show that ω i has ⋆ -property with respect φ i . First note that ω ( λx ) = λ p ω ( x ). Thus, P si =1 ω i ( λx ) e i = λ p P si =1 ω i ( x ) e i and so ω i ( λx ) = λ p ω i ( x ). Furthermore, we have ω ( x + y ) = ω ( x ) + ω ( y ) + X x j = xoryx = x,x = y x ) φ ([ x , . . . , x p − ] , x p ) . Therefore, s X i =1 ω i ( x + y ) e i = s X i =1 ω i ( x ) e i + s X i =1 ω i ( y ) e i + X x j = xoryx = x,x = y x ) s X i =1 φ i ([ x , . . . , x p − ] , x p ) e i , which implies that ω i ( x + y ) = ω i ( x ) + ω i ( y ) + X x j = xoryx = x,x = y x ) φ i ([ x , . . . , x p − ] , x p ) . Finally, we have 0 = φ ( x, y [ p ] ) − φ ([ x, y, . . . , y | {z } p-1 ] , y )= s X i =1 φ i ( x, y [ p ] ) e i − s X i =1 φ i ([ x, y, . . . , y | {z } p-1 ] , y ) e i . Therefore φ i ( x, y [ p ] ) − φ i ([ x, y, ..., y | {z } p-1 ] , y ) = 0 . The proof is complete. (cid:4) Let θ = ( φ, ω ) , θ = ( ψ, η ) ∈ Z ( L, M ) where φ ( x, y ) = s X i =1 φ i ( x, y ) e i ,ψ ( x, y ) = s X i =1 ψ i ( x, y ) e i , and ω ( x ) = P si =1 ω i ( x ) e i and η ( x ) = P si =1 η i ( x ) e i . With these notation we have: Lemma 2.3.5 There exists an isomorphism f : L θ → L θ with f ( M ) = M if and only if there is an A ∈ Aut p ( L ) such that the images of ( Aψ i , Aη i ) ’s in H ( L, F ) span the same subspace of H ( L, F ) as the images of ( φ i , ω i ) ’s. HAPTER 2. PRELIMINARIES Proof. Let f : L θ → L θ be an isomorphism such that f ( M ) = M . Then f induces an isomorphismof L θ /M to L θ /M , that is an automorphism of L . Denote this automorphism by A . Let L bespanned by x , . . . , x n . Then f ( x i ) = A ( x i )+ v i , for some v i ∈ M . Furthermore, f ( e i ) = P sj =1 a ji e j ,and v i = P sl =1 β iℓ e ℓ . Also write [ x i , x j ] L = P nk =1 c kij x k , and x [ p ] i = P nk =1 b ik x k . We claim that ψ l ( A ( x i ) , A ( x j )) = s X k =1 a ℓk φ k ( x i , x j ) + n X k =1 c kij β kℓ . (2.2)To prove the claim, we note that f ([ x i , x j ] L θ ) = [ f ( x i ) , f ( x j )] L θ . Then f ([ x i , x j ] L θ ) = f ([ x i , x j ] L + φ ( x i , x j )) = f ([ x i , x j ] L ) + f ( φ ( x i , x j )) , where f ([ x i , x j ] L ) = f ( n X k =1 c kij x k ) = n X k =1 c kij f ( x k ) = n X k =1 c kij ( A ( x k ) + v k )= n X k =1 c kij A ( x k ) + n X k =1 c kij v k , and f ( φ ( x i , x j )) = f ( s X k =1 φ k ( x i , x j ) e k ) = s X k =1 φ k ( x i , x j ) f ( e k )= s X k =1 φ k ( x i , x j ) s X l =1 a ℓk e ℓ = s X l =1 s X k =1 a ℓk φ k ( x i , x j ) e ℓ . Therefore, we have f ([ x i , x j ] L θ ) = n X k =1 c kij A ( x k ) + n X k =1 c kij v k + s X l =1 s X k =1 a ℓk φ k ( x i , x j ) e ℓ = n X k =1 c kij A ( x k ) + s X l =1 n X k =1 c kij β kℓ e ℓ + s X l =1 s X k =1 a ℓk φ k ( x i , x j ) e ℓ . On the other hand, we have[ f ( x i ) , f ( x j )] L θ = [ A ( x i ) + v i , A ( x j ) + v j ] L θ = [ A ( x i ) , A ( x j )] L + ψ ( A ( x i ) , A ( x j )) , where [ A ( x i ) , A ( x j )] L = A ([ x i , x j ] L ) = A ( n X k =1 c kij x k ) = n X k =1 c kij A ( x k ) , HAPTER 2. PRELIMINARIES ψ ( A ( x i ) , A ( x j )) = s X l =1 ψ ℓ ( A ( x i ) , A ( x j )) e ℓ . Therefore, we have [ f ( x i ) , f ( x j )] L θ = n X k =1 c kij A ( x k ) + s X l =1 ψ ℓ ( A ( x i ) , A ( x j )) e ℓ . As a result, we have s X l =1 ψ ℓ ( A ( x i ) , A ( x j )) e ℓ = s X l =1 n X k =1 c kij β kℓ e ℓ + s X l =1 s X k =1 a ℓk φ k ( x i , x j ) e ℓ , which implies that ψ ℓ ( A ( x i ) , A ( x j )) = n X k =1 c kij β kℓ + s X k =1 a ℓk φ k ( x i , x j ) , for 1 ℓ s . This proves the claim. Next we claim that η ℓ ( A ( x i )) = n X k =1 b ik β kℓ + s X k =1 a ℓk ω k ( x i ) , (2.3)for every 1 ℓ s . To prove the claim, we note that f ( x [ p ] i ) = f ( x i ) [ p ] . So, f ( x [ p ] i ) = f ( ω ( x i ) + x [ p ] i ) = f ( ω ( x i )) + f ( x [ p ] i ) . Now we have f ( ω ( x i )) = f ( s X k =1 ω k ( x i ) e k ) = s X k =1 ω k ( x i ) f ( e k ) = s X k =1 ω k ( x i ) s X ℓ =1 a ℓk e ℓ = s X ℓ =1 s X k =1 a ℓk ω k ( x i ) e ℓ , and f ( x [ p ] i ) = f ( n X k =1 b ik x k ) = n X k =1 b ik f ( x k ) = n X k =1 b ik ( A ( x k ) + v k ) = n X k =1 b ik A ( x k ) + n X k =1 b ik v k . Therefore, we have f ( x [ p ] i ) = n X k =1 b ik A ( x k ) + n X k =1 b ik v k + s X ℓ =1 s X k =1 a ℓk ω k ( x i ) e ℓ = n X k =1 b ik A ( x k ) + s X ℓ =1 n X k =1 b ik β kℓ e ℓ + s X ℓ =1 s X k =1 a ℓk ω k ( x i ) e ℓ . HAPTER 2. PRELIMINARIES f ( x i ) [ p ] = ( A ( x i ) + v i ) [ p ] = η ( A ( x i )) + A ( x i ) [ p ] , where η ( A ( x i )) = s X ℓ =1 η ℓ ( A ( x i )) e ℓ , and A ( x i ) [ p ] = A ( x [ p ] i ) = A ( n X k =1 b ik x k ) = n X k =1 b ik A ( x k ) . Therefore, we have f ( x i ) [ p ] = s X ℓ =1 η ℓ ( A ( x i )) e ℓ + n X k =1 b ik A ( x k ) . As a result, s X ℓ =1 η ℓ ( A ( x i )) e ℓ = s X ℓ =1 n X k =1 b ik β kℓ e ℓ + s X ℓ =1 s X k =1 a ℓk ω k ( x i ) e ℓ , which implies that η ℓ ( A ( x i )) = n X k =1 b ik β kℓ + s X k =1 a ℓk ω k ( x i ) , for 1 ℓ s . This proves the second claim.Now define the linear function f ℓ : L → F by f ℓ ( x k ) = β kℓ . Then( δ f ℓ )( x i , x j ) = f ℓ ([ x i , x j ]) = n X k =1 c kij β kℓ , and˜ f ℓ ( x i ) = f ℓ ( x [ p ] i ) = f ℓ ( n X k =1 b ik x k ) = n X k =1 b ik β kℓ . Hence, ( δ f ℓ , ˜ f ℓ ) ∈ B ( L, F ). We deduce, by Equations (2.2) and (2.3), that the subspace spannedby all the ( Aψ i , Aη i )’s is the same as the subspace spanned by all the ( φ i , ω i )’s modulo B ( L, F ).Conversely, suppose that the image of ( Aψ i , Aη i )’s span the same subspace of H ( L, F ) as thethe image of ( φ i , ω i )’s. Then there are linear functions f ℓ : L → F and a ℓk ∈ F so that Aψ ℓ ( x i , x j ) = s X k =1 a ℓk φ k ( x i , x j ) + f ℓ ([ x i , x j ]) , and Aη ℓ ( x i ) = s X k =1 a ℓk ω k ( x i ) + ˜ f ℓ ( x i ) . If we set β kℓ = f ℓ ( x k ), then we see that Equations (2.2) and (2.3) hold. This means that, if we define f : L θ → L θ by f ( x i ) = A ( x i ) + P ℓ β il e ℓ and f ( e i ) = P j a ji e j , then f is an isomorphism. (cid:4) HAPTER 2. PRELIMINARIES L θ ∼ = L θ , it is hard to applyLemma 2.3.5 in practice. Instead, we consider the following convention: Definition 2.3.6 Let θ ∈ Z ( L, M ) . We define ¯ θ = { Aθ | A ∈ Aut p ( L ) } . We call ¯ θ an Aut p ( L ) -orbit and θ an Aut p ( L ) -orbit representative. We say θ and θ are in the same Aut p ( L ) -orbit ifthere exists A ∈ Aut p ( L ) such that Aθ = θ . Clearly, if θ and θ are in the same Aut p ( L )-orbit then by Theorem 2.3.5, L θ ∼ = L θ . However,the converse is not true, see for instance Lemma 4.2.2. Nevertheless, it is practical to find all Aut p ( L )-orbit representatives and this way we get a list of all possible extensions of L by M . We shall theneliminate all redundancies. H ( L, F ) In this section, we describe how to find a basis for H ( L, F ), where L is a p -nilpotent 4-dimensionalrestricted Lie algebra and p > x , . . . , x be a basis of L and ( φ, ω ) ∈ C ( L, F ). We have φ = P i Let f : L → H be an isomorphism of restricted Lie algebras. Then f induces anisomorphism between L/ h L [ p ] i p and H/ h H [ p ] i p . HAPTER 2. PRELIMINARIES Proof. Let φ : H → H/ h H [ p ] i p be the canonical homomorphism. Then ψ = φf : L → H/ h H [ p ] i p isa surjective homomorphism of restricted Lie algeras with ker ψ = h L [ p ] i p . Therefore, L/ h L [ p ] i p and H/ h H [ p ] i p are isomorphic. (cid:4) hapter 3 Restriction maps on the abelian Liealgebra Let x , . . . , x be a basis of the abelian Lie algebra K = L , of dimension 5. Since we wantto determine p -nilpotent maps on L , , without loss of generality, we assume that x [ p ]5 = 0 . Let L = L , / h x i . Since dim L = 4 , by Theorem 2.3.2, there are five restricted Lie algebra structures on L given by the following p -maps:I.1 Trivial p -map;I.2 x [ p ]1 = x ;I.3 x [ p ]1 = x , x [ p ]3 = x ;I.4 x [ p ]1 = x , x [ p ]2 = x ;I.5 x [ p ]1 = x , x [ p ]2 = x , x [ p ]3 = x . L , / h x i , trivial p -map) We have x [ p ] i ∈ h x i , for all 1 i x [ p ] i = α i x , for some α i ∈ F and for all 1 i α i = 0 , then without loss of generality, we assume α = 0 . Now, we rescale x such that x [ p ]1 = x .Now, if α j = 0 then we rescale x j such that x [ p ] j = x . Finally, if x [ p ] j = x , then we replace x j with x j − x such that x [ p ] j = 0 . Therefore, the possible p -maps are as follows: K = h x , . . . , x i ; K = h x , . . . , x | x [ p ]1 = x i . L , / h x i , x [ p ]1 = x ) In this case we have x [ p ]1 − x ∈ h x i and x [ p ]2 , x [ p ]3 , x [ p ]4 ∈ h x i . Hence, x [ p ]1 = x + α x , x [ p ]2 = α x , x [ p ]3 = α x , and x [ p ]4 = α x , for some α , α , α , α ∈ F . We replace x with x + α x to obtainthat x [ p ]1 = x . We consider two cases: 21 HAPTER 3. RESTRICTION MAPS ON THE ABELIAN LIE ALGEBRA Case 1. α = 0 : First, if α = α = 0 then we have the following p -map: K = h x , . . . , x | x [ p ]1 = x i . Next, if one of α , α is zero and one is nonzero, without loss of generality, we assume that α = 0 .Now, we rescale x so that x [ p ]3 = x . Therefore, we have the following p -map: K = h x , . . . , x | x [ p ]1 = x , x [ p ]3 = x i . Finally, if both α and α are non-zero then we rescale x and x so that x [ p ]3 = x and x [ p ]4 = x .Now, we replace x with x − x so that x [ p ]4 = 0 . Therefore, we obtain the same p -map as theprevious one. Case 2. α = 0 : First, if α = α = 0 then in L , we replace x with α x so that x [ p ]2 = x .Therefore, we have the following p -map: K = h x , . . . , x | x [ p ]1 = x , x [ p ]2 = x i . Next, if one of α , α is zero and one is nonzero, without loss of generality, we assume that α = 0 .Now, in L , we replace x with ( α /α ) /p x and x with α x so that x [ p ]2 = x and x [ p ]3 = x .Then, we replace x with x − x so that x [ p ]3 = 0 . Therefore, we obtain the same p -map as K .Finally, if both α and α are non-zero then in L , we replace x with ( α /α ) /p x , x with( α /α ) /p x and x with α x so that x [ p ]2 = x , x [ p ]3 = x and x [ p ]4 = x . Now, we replace x with x − x and x with x − x so that x [ p ]3 = 0 and x [ p ]4 = 0 . Therefore, we obtain the same p -map as K . L , / h x i , x [ p ]1 = x , x [ p ]3 = x ) In this case we have x [ p ]1 − x ∈ h x i , x [ p ]3 − x ∈ h x i , and x [ p ]2 , x [ p ]4 ∈ h x i . Hence, x [ p ]1 = x + α x , x [ p ]3 = x + α x , x [ p ]2 = α x , and x [ p ]4 = α x for some α , α , α , α ∈ F . We replace x with x + α x so that x [ p ]1 = x and x with x + α x so that x [ p ]3 = x .First, if α = α = 0 then we have the following p -map: K = h x , . . . , x | x [ p ]1 = x , x [ p ]3 = x i . Next, if α = 0 and α = 0 then in L , we replace x with α x so that x [ p ]2 = x . Therefore, wehave the following p -map: K = h x , . . . , x | x [ p ]1 = x , x [ p ]3 = x , x [ p ]2 = x i . Next, if α = 0 and α = 0 then in L , we replace x with α x so that x [ p ]4 = x . Therefore, wehave the following p -map: K = h x , . . . , x | x [ p ]1 = x , x [ p ]3 = x , x [ p ]4 = x i . Finally, if α = 0 and α = 0 then in L , we replace x with (( α /α ) /p ) /p x , x with( α /α ) /p x and x with α x so that x [ p ]2 = x and x [ p ]4 = x . Therefore, we have the following p -map: K = h x , . . . , x | x [ p ]1 = x , x [ p ]2 = x , x [ p ]3 = x , x [ p ]4 = x i . HAPTER 3. RESTRICTION MAPS ON THE ABELIAN LIE ALGEBRA L , / h x i , x [ p ]1 = x , x [ p ]2 = x ) In this case we have x [ p ]1 − x ∈ h x i , x [ p ]2 − x ∈ h x i , and x [ p ]3 , x [ p ]4 ∈ h x i . Hence, x [ p ]1 = x + α x , x [ p ]2 = x + α x , x [ p ]3 = α x , and x [ p ]4 = α x for some α , α , α , α ∈ F . We replace x with x + α x so that x [ p ]1 = x and x with x + α x so that x [ p ]2 = x .First, if α = α = 0 then we have the following p -map: K = h x , . . . , x | x [ p ]1 = x , x [ p ]2 = x i . Next, if α = 0 and α = 0 then in L , we replace x with α x so that x [ p ]3 = x . Therefore, wehave the following p -map: K = h x , . . . , x | x [ p ]1 = x , x [ p ]2 = x , x [ p ]3 = x i . Next, if α = 0 and α = 0 then in L , we replace x with α x so that x [ p ]4 = x . Therefore, wehave the following p -map: K = h x , . . . , x | x [ p ]1 = x , x [ p ]2 = x , x [ p ]4 = x i . Finally, if α = 0 and α = 0 then in L , we replace x with ( α /α ) /p x and x with α x sothat x [ p ]3 = x and x [ p ]4 = x . Now, we replace x with x − x so that x [ p ]4 =0. Therefore, we havethe following p -map: K = h x , . . . , x | x [ p ]1 = x , x [ p ]2 = x , x [ p ]3 = x i . L , / h x i , x [ p ]1 = x , x [ p ]2 = x , x [ p ]3 = x ) In this case we have x [ p ]1 − x ∈ h x i , x [ p ]2 − x ∈ h x i , x [ p ]3 − x ∈ h x i and x [ p ]4 ∈ h x i . Hence, x [ p ]1 = x + α x , x [ p ]2 = x + α x , x [ p ]3 = x + α x , and x [ p ]4 = α x for some α , α , α , α ∈ F .We replace x with x + α x so that x [ p ]1 = x , x with x + α x so that x [ p ]2 = x and x with x + α x so that x [ p ]3 = x . First, if α = 0 then we have the following p -map: K = h x , . . . , x | x [ p ]1 = x , x [ p ]2 = x , x [ p ]3 = x i . Next, if α = 0 then in L , we replace x with α x so that x [ p ]4 = x . Therefore, we have thefollowing p -map: K = h x , . . . , x | x [ p ]1 = x , x [ p ]2 = x , x [ p ]3 = x , x [ p ]4 = x i . HAPTER 3. RESTRICTION MAPS ON THE ABELIAN LIE ALGEBRA L , are as follows: K = h x , . . . , x i ; K = h x , . . . , x | x [ p ]1 = x i ; K = h x , . . . , x | x [ p ]1 = x i ; K = h x , . . . , x | x [ p ]1 = x , x [ p ]3 = x i ; K = h x , . . . , x | x [ p ]1 = x , x [ p ]2 = x i ; K = h x , . . . , x | x [ p ]1 = x , x [ p ]3 = x , x [ p ]2 = x i ; K = h x , . . . , x | x [ p ]1 = x , x [ p ]3 = x i ; K = h x , . . . , x | x [ p ]1 = x , x [ p ]3 = x , x [ p ]4 = x i ; K = h x , . . . , x | x [ p ]1 = x , x [ p ]2 = x , x [ p ]3 = x , x [ p ]4 = x i ; K = h x , . . . , x | x [ p ]1 = x , x [ p ]2 = x i ; K = h x , . . . , x | x [ p ]1 = x , x [ p ]2 = x , x [ p ]3 = x i ; K = h x , . . . , x | x [ p ]1 = x , x [ p ]2 = x , x [ p ]4 = x i ; K = h x , . . . , x | x [ p ]1 = x , x [ p ]2 = x , x [ p ]3 = x i ; K = h x , . . . , x | x [ p ]1 = x , x [ p ]2 = x , x [ p ]3 = x i ; K = h x , . . . , x | x [ p ]1 = x , x [ p ]2 = x , x [ p ]3 = x , x [ p ]4 = x i . Note that K ∼ = K , K ∼ = K , K ∼ = K , K ∼ = K , and K ∼ = K ∼ = K . Furthermore, K is identical to K . Theorem 3.5.1 The list of all restricted Lie algebra structures on L , , up to isomorphism, is asfollows: L , = h x , . . . , x i L , = h x , . . . , x | x [ p ]1 = x i L , = h x , . . . , x | x [ p ]1 = x , x [ p ]3 = x i L , = h x , . . . , x | x [ p ]1 = x , x [ p ]2 = x i L , = h x , . . . , x | x [ p ]1 = x , x [ p ]3 = x , x [ p ]2 = x i L , = h x , . . . , x | x [ p ]1 = x , x [ p ]2 = x , x [ p ]3 = x , x [ p ]4 = x i L , = h x , . . . , x | x [ p ]1 = x , x [ p ]2 = x , x [ p ]3 = x i L , = h x , . . . , x | x [ p ]1 = x , x [ p ]2 = x , x [ p ]3 = x , x [ p ]4 = x i It is clear from the table below that the restricted Lie algebras given in Theorem 3.5.1 are pairwisenon-isomorphic. HAPTER 3. RESTRICTION MAPS ON THE ABELIAN LIE ALGEBRA 25L dim L [ p ] dim L [ p ] dim L [ p ] L , = h x , . . . , x i L , = h x , . . . , x | x [ p ]1 = x i L , = h x , . . . , x | x [ p ]1 = x , x [ p ]3 = x i L , = h x , . . . , x | x [ p ]1 = x , x [ p ]2 = x i L , = h x , . . . , x | x [ p ]1 = x , x [ p ]3 = x , x [ p ]2 = x i L , = h x , . . . , x | x [ p ]1 = x , x [ p ]2 = x , x [ p ]3 = x , x [ p ]4 = x i L , = h x , . . . , x | x [ p ]1 = x , x [ p ]2 = x , x [ p ]3 = x i L , = h x , . . . , x | x [ p ]1 = x , x [ p ]2 = x , x [ p ]3 = x , x [ p ]4 = x i hapter 4 Restriction maps on L , Let K = L , = h x , . . . , x | [ x , x ] = x i F . Note that Z ( L , ) = h x , x , x i F and the group Aut( L , ) consists of invertible matrices of the form a a a a a a r a a a a a a a a a a , where r = a a − a a . There exists an element αx + βx + γx ∈ Z ( L , ) such that ( αx + βx + γx ) [ p ] = 0 , for some α, β, γ ∈ F . If γ = 0 , then consider H = h x ′ , . . . , x ′ | [ x ′ , x ′ ] = x ′ i , where x ′ = x , x ′ = x , x ′ = x , x ′ = x , x ′ = αx + βx + γx . Let φ : K → H such that x x ′ , x x ′ , x x ′ , x x ′ , and x x ′ . It is easy to see that φ is an isomorphism.Therefore, in this case we can suppose that x [ p ]5 = 0 . Next, if γ = 0 and β = 0 then consider H = h x ′ , . . . , x ′ | [ x ′ , x ′ ] = x ′ i , where x ′ = x , x ′ = x , x ′ = x , x ′ = αx + βx , x ′ = x . Let φ : K → H such that x x ′ , x x ′ , x x ′ , x x ′ , and x x ′ . It is easy to see that φ is an isomorphism. Therefore, inthis case we can suppose that x [ p ]4 = 0 . Finally, if γ = β = 0 and α = 0 , then consider H = h x ′ , . . . , x ′ | [ x ′ , x ′ ] = x ′ i , where x ′ = x , x ′ = x , x ′ = αx , x ′ = x , x ′ = x . Let φ : K → H such that x x ′ , x x ′ , x x ′ , x x ′ , and x x ′ . It is easy to see that φ is an isomorphism. Therefore, in this casewe can suppose that x [ p ]3 = 0 . Hence we have three cases:I. x [ p ]3 = 0 ;II. x [ p ]5 = 0 ; 26 HAPTER 4. RESTRICTION MAPS ON L , x [ p ]4 = 0 .Note that cases II and III yield the same restricted Lie algebras because given a restricted Liealgebra structure on L , for which x [ p ]5 = 0 , the automorphism of L , obtained by switching x and x gives rise to a restricted Lie algebra structure on L , for which x [ p ]4 = 0 . L = L , h x i In this section we find all non-isomorphic p -maps on L , such that x [ p ]3 = 0 . We let L = L , h x i ∼ = L , , where L , = h x , x , x , x i . Note that we denote the image of x i in L by x i again. We rename the x i ’s so that y = x , y = x , y = x , y = x , and y = x and at the end we will switch them.Therefore, we have L = L , = h y , y , y , y i . The group Aut( L ) consists of invertible matrices ofthe form a a a a a a a a a a a a a a a a . Lemma 4.1.1 Let K = L , and [ p ] : K → K be a p -map on K such that x [ p ]3 = 0 and let L = KM where M = h x i F . Then K ∼ = L θ where θ = (∆ , ω ) ∈ Z ( L, F ) . Proof. Let π : K → L be the projection map. We have the exact sequence0 → M → K → L → . Let σ : L → K be a linear map given by x i x i , for all i . Then σ is an injective linear map and πσ = 1 L . Now, we define φ : L × L → M by φ ( x i , x j ) = [ σ ( x i ) , σ ( x j )] − σ ([ x i , x j ]), for all i, j and ω : L → M by ω ( x ) = σ ( x ) [ p ] − σ ( x [ p ] ). Note that φ ( x , x ) = [ σ ( x ) , σ ( x )] − σ ([ x , x ]) = [ x , x ] = x ; φ ( x , x ) = [ σ ( x ) , σ ( x )] − σ ([ x , x ]) = [ x , x ] = 0 . Similarly, we can show that φ ( x , x ) = φ ( x , x ) = φ ( x , x ) = φ ( x , x ) = 0 . Therefore, φ = ∆ .Now, by Lemma 2.2.2, we have θ = (∆ , ω ) ∈ Z ( L, F ) and K ∼ = L θ . (cid:4) Note that by Theorem 2.3.2, there are five non-isomorphic restricted Lie algebra structures on L given by the following p -maps:I.1 Trivial p -map;I.2 y [ p ]1 = y ;I.3 y [ p ]1 = y , y [ p ]3 = y ;I.4 y [ p ]1 = y , y [ p ]2 = y ; HAPTER 4. RESTRICTION MAPS ON L , y [ p ]1 = y , y [ p ]2 = y , y [ p ]3 = y .We make L into a restricted Lie algebra by equipping it with each of the above p -maps. Then, ineach case, we find all possible orbit representatives of the form (∆ , ω ) under the action of Aut p ( L )on H ( L, F ). By Lemma 4.1.1, we do get all possible p -maps on K with the property that x [ p ]3 = 0 .Let us consider the case I.2 where L is a restricted Lie algebra with the p -map y [ p ]1 = y . Let[( φ, ω )] ∈ H ( L, F ). Then we must have φ ( x, y [ p ] ) = 0 , for all x, y ∈ L , where φ = a ∆ + b ∆ + c ∆ + d ∆ + e ∆ + f ∆ , for some a, b, c, d, e, f ∈ F . Since L [ p ] = h y i , we get φ ( x, y ) = 0 for all x ∈ L . Therefore, φ ( y , y ) = 0 which implies that a = 0 . Since φ = ∆ gives us L , , we deduceby Lemma 4.1.1 that L , cannot be constructed in this case. Similarly, we can show that in casesI.3, I.4, or I.5 we also get a = 0 . This means that if we equip L with any of the p -maps in the casesI.2, I.3, I.4, or I.5, then L , ≇ L θ , for every θ ∈ H ( L, F ). In the following subsections, we considerthe remaining cases. L , trivial p -map) We have C ( L, F ) = h (∆ , , (∆ , , (∆ , , (∆ , , (∆ , , (∆ , , (0 , f ) , (0 , f ) , (0 , f ) , (0 , f ) i F . First, we find a basis for Z ( L, F ). Let ( φ, ω ) ∈ Z ( L, F ). Then we must have δ φ = 0 and φ ( x, y [ p ] ) = 0 , for all x, y ∈ L . Note that since L is an abelian Lie algebra and the p -map is trivial, δ φ = 0 and φ ( x, y [ p ] ) = 0 , for all x, y ∈ L . Therefore, a basis for Z ( L, F ) is as follows(∆ , , (∆ , , (∆ , , (∆ , , (∆ , , (∆ , , (0 , f ) , (0 , f ) , (0 , f ) , (0 , f ) . Next, we find a basis for B ( L, F ). Let ( φ, ω ) = ( a ∆ + b ∆ + c ∆ + d ∆ + e ∆ + f ∆ , αf + βf + γf + δf ) ∈ B ( L, F ). Then there exists a linear map ψ : L → F such that δ ψ ( x, y ) = φ ( x, y )and ˜ ψ ( x ) = ω ( x ), for all x, y ∈ L . So, we have φ ( y , y ) = δ ψ ( y , y ) = ψ ([ y , y ]) = 0 . But φ ( y , y ) = a . So we deduce that a = 0 . Similarly, we can show that b = c = d = e = f = 0 .Also, we have ω ( y ) = ˜ ψ ( y ) = ψ ( y [ p ]1 ) = 0 . Hence, α = 0 . Similarly, we can show that β = γ = δ = 0 . Therefore, ( φ, ω ) = 0 and hence B ( L, F ) = 0 . We deduce that a basis for H ( L, F ) is as follows[(∆ , , [(∆ , , [(∆ , , [(∆ , , [(∆ , , [(∆ , , [(0 , f )] , [(0 , f )] , [(0 , f )] , [(0 , f )] . Let [( φ, ω )] ∈ H ( L, F ). Then, we have φ = a ∆ + b ∆ + c ∆ + d ∆ + e ∆ + f ∆ , forsome a, b, c, d, e, f ∈ F . Suppose that Aφ = a ′ ∆ + b ′ ∆ + c ′ ∆ + d ′ ∆ + e ′ ∆ + f ′ ∆ , for some a ′ , b ′ , c ′ , d ′ , e ′ , f ′ ∈ F . Then Aφ ( y , y ) = φ ( Ay , Ay )= φ ( a y + a y + a y + a y , a y + a y + a y + a y )= ( a a − a a ) a + ( a a − a a ) b + ( a a − a a ) c + ( a a − a a ) d + ( a a − a a ) e + ( a a − a a ) f, and HAPTER 4. RESTRICTION MAPS ON L , Aφ ( y , y ) = φ ( Ay , Ay )= φ ( a y + a y + a y + a y , a y + a y + a y + a y )= ( a a − a a ) a + ( a a − a a ) b + ( a a − a a ) c + ( a a − a a ) d + ( a a − a a ) e + ( a a − a a ) f, and Aφ ( y , y ) = φ ( Ay , Ay )= φ ( a y + a y + a y + a y , a y + a y + a y + a y )= ( a a − a a ) a + ( a a − a a ) b + ( a a − a a ) c + ( a a − a a ) d + ( a a − a a ) e + ( a a − a a ) f, and Aφ ( y , y ) = φ ( Ay , Ay )= φ ( a y + a y + a y + a y , a y + a y + a y + a y )= ( a a − a a ) a + ( a a − a a ) b + ( a a − a a ) c + ( a a − a a ) d + ( a a − a a ) e + ( a a − a a ) f, and Aφ ( y , y ) = φ ( Ay , Ay )= φ ( a y + a y + a y + a y , a y + a y + a y + a y )= ( a a − a a ) a + ( a a − a a ) b + ( a a − a a ) c + ( a a − a a ) d + ( a a − a a ) e + ( a a − a a ) f, and Aφ ( y , y ) = φ ( Ay , Ay )= φ ( a y + a y + a y + a y , a y + a y + a y + a y )= ( a a − a a ) a + ( a a − a a ) b + ( a a − a a ) c + ( a a − a a ) d + ( a a − a a ) e + ( a a − a a ) f. Therefore, the action of Aut p ( L ) on the set of φ ’s in the matrix form is as follows: a ′ b ′ c ′ d ′ e ′ f ′ = a a − a a a a − a a a a − a a a a − a a a a − a a a a − a a a a − a a a a − a a a a − a a a a − a a a a − a a a a − a a a a − a a a a − a a a a − a a a a − a a a a − a a a a − a a a a − a a a a − a a a a − a a a a − a a a a − a a a a − a a a a − a a a a − a a a a − a a a a − a a a a − a a a a − a a a a − a a a a − a a a a − a a a a − a a a a − a a a a − a a abcdef . The orbit with representative of this action gives us L , .Also, we have ω = αf + βf + γf + δf , for some α, β, γ, δ ∈ F . Suppose that Aω = α ′ f + β ′ f + γ ′ f + δ ′ f , for some α ′ , β ′ , γ ′ , δ ′ ∈ F . Then we can verify that the action of Aut p ( L ) on the HAPTER 4. RESTRICTION MAPS ON L , ω ’s in the matrix form is as follows: α ′ β ′ γ ′ δ ′ = a p a p a p a p a p a p a p a p a p a p a p a p a p a p a p a p αβγδ . Now we find the representatives of the orbits of the action of Aut( L ) on the set of ω ’s such that theorbit represented by is preserved under the action of Aut( L ) on the set of φ ’s.Let ν = αβγδ ∈ F . If ν = , then { ν } is clearly an Aut p ( L )-orbit. Let ν = 0 and supposethat δ = 0 . Then h − β/δ ) /p α/δ ) /p 00 1 ( − γ/δ ) /p α/δ ) /p − γ/δ ) /p ( β/δ ) /p , − α/δ − β/δ − γ/δ ih , αβγδ i = h , δ i , h /δ ) /p /δ ) /p 00 0 0 0 0 (1 /δ ) /p , /δ ih , δ i = h , i . Next, if δ = 0 , but γ = 0 , then h − β/γ ) /p α/γ ) /p − α/γ ) /p − β/γ ) /p , − αγ 00 1 − βγ 00 0 1 00 0 0 1 ih , αβγ i = h , γ i , h /γ ) /p /γ ) /p /γ ) /p , /γ 00 0 0 1 ih , γ i = h , i . HAPTER 4. RESTRICTION MAPS ON L , δ = γ = 0 , but β = 0 , then h − α/β ) /p − α/β ) /p 00 0 0 1 0 00 0 0 0 1 00 0 0 0 0 1 , − α/β ih , αβ i = h , β i , h β /p β /p /β ) /p /β ) /p 00 0 0 0 0 1 , β /β ih , β i = h , i . Finally, if δ = γ = β = 0 , but α = 0 , then h α − /p α − /p α /p α /p 00 0 0 0 0 1 , /α α ih , α i = h , i . Thus the following elements are Aut( L )-orbit representatives: , , , , . Now, we find the restricted Lie algebra structure corresponding to the orbit representative . Wehave = f and hence ω = f y . First, by Lemma 2.2.1 we get y [ p ]1 = y [ p ]1 + ω ( y ) = y ,y [ p ]2 = y [ p ]2 + ω ( y ) = 0 ,y [ p ]3 = y [ p ]3 + ω ( y ) = 0 ,y [ p ]4 = y [ p ]4 + ω ( y ) = 0 . Therefore, the corresponding restricted Lie algebra structure is as follows: K = h y , . . . , y | [ y , y ] = y , y [ p ]1 = y i . HAPTER 4. RESTRICTION MAPS ON L , y = x , y = x , y = x , y = x , and y = x . Hence, we have K = h x , . . . , x | [ x , x ] = x , x [ p ]1 = x i . Similarly, we obtain the restricted Lie algebra structures corresponding to the other orbits. There-fore, the corresponding restricted Lie algebra structures are as follows: K = h x , . . . , x | [ x , x ] = x i ; K = h x , . . . , x | [ x , x ] = x , x [ p ]1 = x i ; K = h x , . . . , x | [ x , x ] = x , x [ p ]2 = x i ; K = h x , . . . , x | [ x , x ] = x , x [ p ]4 = x i ; K = h x , . . . , x | [ x , x ] = x , x [ p ]5 = x i . L = L , h x i In this section we find all non-isomorphic p -maps on L , such that x [ p ]5 = 0 . We let L = L , h x i ∼ = L , , where L , = h x , x , x , x | [ x , x ] = x i . The group Aut( L ) consists of invertible matrices of theform a a a a a a r a a a a , where r = a a − a a = 0 . Lemma 4.2.1 Let K = L , and [ p ] : K → K be a p -map on K such that x [ p ]5 = 0 and let L = KM where M = h x i F . Then K ∼ = L θ where θ = (0 , ω ) ∈ Z ( L, F ) . Proof. Let π : K → L be the projection map. We have the exact sequence0 → M → K → L → . Let σ : L → K such that x i x i , 1 i σ is an injective linear map and πσ = 1 L . Now,we define φ : L × L → M by φ ( x i , x j ) = [ σ ( x i ) , σ ( x j )] − σ ([ x i , x j ]), 1 i, j ω : L → M by ω ( x ) = σ ( x ) [ p ] − σ ( x [ p ] ). Note that φ ( x , x ) = [ σ ( x ) , σ ( x )] − σ ([ x , x ]) = [ x , x ] − σ ( x ) = 0; φ ( x , x ) = [ σ ( x ) , σ ( x )] − σ ([ x , x ]) = 0 . Similarly, we can show that φ ( x , x ) = φ ( x , x ) = φ ( x , x ) = φ ( x , x ) = 0 . Therefore, φ = 0 .Now, by Lemma 2.2.2, we have θ = (0 , ω ) ∈ Z ( L, F ) and K ∼ = L θ . (cid:4) Note that by Theorem 2.3.2, there are eight non-isomorphic restricted Lie algebra structures on L given by the following p -maps: HAPTER 4. RESTRICTION MAPS ON L , p -map;II.2 x [ p ]1 = x ;II.3 x [ p ]1 = x ;II.4 x [ p ]1 = x , x [ p ]2 = x ;II.5 x [ p ]3 = x ;II.6 x [ p ]3 = x , x [ p ]2 = x ;II.7 x [ p ]4 = x ;II.8 x [ p ]4 = x , x [ p ]2 = x . In the following subsections, we make L into a restricted Lie algebra by equipping it with eachof the above p -maps. Then, in each case, we find all possible orbit representatives of the form (0 , ω )under the action of Aut p ( L ) on H ( L, F ). By Lemma 4.2.1, we do get all possible p -maps on K with the property that x [ p ]5 = 0 . L , trivial p -map) First, we find a basis for Z ( L, F ). Let ( φ, ω ) = ( a ∆ + b ∆ + c ∆ + d ∆ + e ∆ + f ∆ , αf + βf + γf + δf ) ∈ Z ( L, F ). Then we must have δ φ ( x, y, z ) = 0 and φ ( x, y [ p ] ) = 0 , for all x, y, z ∈ L .Therefore,0 = ( δ φ )( x , x , x ) = φ ([ x , x ] , x ) + φ ([ x , x ] , x ) + φ ([ x , x ] , x ) = φ ( x , x ) . Thus, we get f = 0 . Since the p -map is trivial, φ ( x, y [ p ] ) = φ ( x, 0) = 0 , for all x, y ∈ L . Therefore,a basis for Z ( L, F ) is as follows:(∆ , , (∆ , , (∆ , , (∆ , , (∆ , , (0 , f ) , (0 , f ) , (0 , f ) , (0 , f ) . Next, we find a basis for B ( L, F ). Let ( φ, ω ) ∈ B ( L, F ). Since B ( L, F ) ⊆ Z ( L, F ), we have( φ, ω ) = ( a ∆ + b ∆ + c ∆ + d ∆ + e ∆ , αf + βf + γf + δf ). So, there exists a linear map ψ : L → F such that δ ψ ( x, y ) = φ ( x, y ) and ˜ ψ ( x ) = ω ( x ), for all x, y ∈ L . So, we have b = φ ( x , x ) = δ ψ ( x , x ) = ψ ([ x , x ]) = 0 . Similarly, we can show that c = d = e = 0 . Also, we have α = ω ( x ) = ˜ ψ ( x ) = ψ ( x [ p ]1 ) = 0 . Similarly, we can show that β = γ = δ = 0 . Therefore, ( φ, ω ) = ( a ∆ , 0) and hence B ( L, F ) = h (∆ , i F . We deduce that a basis for H ( L, F ) is as follows:[(∆ , , [(∆ , , [(∆ , , [(∆ , , [(0 , f )] , [(0 , f )] , [(0 , f )] , [(0 , f )] . Let [ θ ] = [( φ, ω )] ∈ H ( L, F ). Since we want L θ and L , to be isomorphic as Lie algebras, weshould have φ = 0 . Since 0 is preserved under Aut p ( L ), it is enough to find Aut p ( L )-representativesof the ω ’s. HAPTER 4. RESTRICTION MAPS ON L , ω = αf + βf + γf + δf , for some α, β, γ, δ ∈ F . Then, Aω = α ′ f + β ′ f + γ ′ f + δ ′ f , forsome α ′ , β ′ , γ ′ , δ ′ ∈ F . We can verify that the action of Aut p ( L ) on the set of ω ’s in the matrix formis as follows: α ′ β ′ γ ′ δ ′ = a p a p a p a p a p a p a p a p r p 00 0 a p a p αβγδ . Now we find the representatives of the orbits of the action of Aut( L ) on the set of ω ’s. Let ν = αβγδ ∈ F . If ν = , then { ν } is clearly an Aut p ( L )-orbit. Let ν = 0 . Suppose that γ = 0 .Then − α/γ 00 1 − β/γ 00 0 1 00 0 − δ/γ αβγδ = γ , and /γ /γ 00 0 0 1 γ = . Next, if γ = 0 , but δ = 0 , then − α/δ − β/δ αβ δ = δ , and /δ δ = . Next, if γ = δ = 0 , but β = 0 , then − α/β αβ = β , and /β /β 00 0 0 1 β = . Finally, if γ = δ = β = 0 , but α = 0 , then /α /α 00 0 0 1 α = . HAPTER 4. RESTRICTION MAPS ON L , L )-orbit representatives: , , , , . Therefore, the corresponding restricted Lie algebra structures are as follows: K = h x , . . . , x | [ x , x ] = x i ; K = h x , . . . , x | [ x , x ] = x , x [ p ]1 = x i ; K = h x , . . . , x | [ x , x ] = x , x [ p ]2 = x i ; K = h x , . . . , x | [ x , x ] = x , x [ p ]3 = x i ; K = h x , . . . , x | [ x , x ] = x , x [ p ]4 = x i . ( L, x [ p ]1 = x ) First, we find a basis for Z ( L, F ). Let ( φ, ω ) = ( a ∆ + b ∆ + c ∆ + d ∆ + e ∆ + f ∆ , αf + βf + γf + δf ) ∈ Z ( L, F ). Then we must have δ φ ( x, y, z ) = 0 and φ ( x, y [ p ] ) = 0 , for all x, y, z ∈ L .Therefore, we have0 = ( δ φ )( x , x , x ) = φ ([ x , x ] , x ) + φ ([ x , x ] , x ) + φ ([ x , x ] , x ) = φ ( x , x ) = f. Also, we have φ ( x, y [ p ] ) = 0 . Therefore, φ ( x, x ) = 0 , for all x ∈ L and hence φ ( x , x ) = φ ( x , x ) = φ ( x , x ) = 0 which implies that b = d = f = 0 . Therefore, Z ( L, F ) has a basis consisting of:(∆ , , (∆ , , (∆ , , (0 , f ) , (0 , f ) , (0 , f ) , (0 , f ) . Next, we find a basis for B ( L, F ). Let ( φ, ω ) ∈ B ( L, F ). Since B ( L, F ) ⊆ Z ( L, F ), we have( φ, ω ) = ( a ∆ + c ∆ + e ∆ , αf + βf + γf + δf ). Note that there exists a linear map ψ : L → F such that δ ψ ( x, y ) = φ ( x, y ) and ˜ ψ ( x ) = ω ( x ), for all x, y ∈ L . So, we have a = φ ( x , x ) = δ ψ ( x , x ) = ψ ([ x , x ]) = φ ( x , x ) = ψ ( x ) , and c = φ ( x , x ) = δ ψ ( x , x ) = ψ ([ x , x ]) = 0 . Similarly, we can show that e = 0 . Also, we have α = ω ( x ) = ˜ ψ ( x ) = ψ ( x [ p ]1 ) = ψ ( x ) , and β = ω ( x ) = ˜ ψ ( x ) = ψ ( x [ p ]2 ) = 0 . Similarly, we can show that γ = δ = 0 . Note that ψ ( x ) = a = α . Therefore, ( φ, ω ) = ( a ∆ , af )and hence B ( L, F ) = h (∆ , f ) i F . Note that n [(∆ , , [(∆ , , [(∆ , , [(0 , f )] , [(0 , f )] , [(0 , f )] , [(0 , f )] o spans H ( L, F ). Since [(∆ , , f )] = [(∆ , f )] = [0], then [(0 , f )] is an scalar multiple of[(∆ , H ( L, F ). Note that dim H = dim Z − dim B = 6 . Therefore, n [(∆ , , [(∆ , , [(∆ , , [(0 , f )] , [(0 , f )] , [(0 , f )] o HAPTER 4. RESTRICTION MAPS ON L , H ( L, F ).Note that the group Aut p ( L ) consists of invertible matrices of the form a a a a a a r a a a a , where r = a a − a a = 0 , r = a p and a = 0 .Let [ θ ] = [( φ, ω )] ∈ H ( L, F ). As in Section 4.2.1, it is enough to find Aut p ( L )-representatives ofthe ω ’s. We can verify that the action of Aut p ( L ) on the ω ’s in the matrix form is as follows: β ′ γ ′ δ ′ = a p a p a p r p a p a p βγδ . Note that ω = βf + γf + δf . Since ω ( x ) = 0 , we have Aω ( x ) = 0 which implies that a p β + a p γ + a p δ = 0 .Now we find the representatives of the orbits of this action. Note that we need to have a p β + a p γ + a p δ = 0 . Let ν = βγδ ∈ F . If ν = , then { ν } is clearly an Aut p ( L )-orbit. Let ν = 0 .Suppose that γ = 0 . Then − β/γ 00 1 00 − δ/γ βγδ = γ , and (1 /γ ) p − p /γ 00 0 1 γ = . Next, if γ = 0 , but δ = 0 , then − β/δ β δ = δ , and /δ δ = . Next, if γ = δ = 0 , but β = 0 , then we have β . Thus the following elements are Aut p ( L )-orbitrepresentatives: , β , , . HAPTER 4. RESTRICTION MAPS ON L , K = h x , . . . , x | [ x , x ] = x , x [ p ]1 = x i ; K ( β ) = h x , . . . , x | [ x , x ] = x , x [ p ]1 = x , x [ p ]2 = βx i ; K = h x , . . . , x | [ x , x ] = x , x [ p ]1 = x , x [ p ]3 = x i ; K = h x , . . . , x | [ x , x ] = x , x [ p ]1 = x , x [ p ]4 = x i where β ∈ F ∗ . Lemma 4.2.2 We have K ( β ) ∼ = K (1) , for every β ∈ F ∗ . Proof. The following linear map yields the desired automorphism: β /p β ( p − /p β β p − (cid:4) ( L, x [ p ]1 = x ) First, we find a basis for Z ( L, F ). Let ( φ, ω ) = ( a ∆ + b ∆ + c ∆ + d ∆ + e ∆ + f ∆ , αf + βf + γf + δf ) ∈ Z ( L, F ). Then we must have δ φ ( x, y, z ) = 0 and φ ( x, y [ p ] ) = 0 , for all x, y, z ∈ L .Therefore, we have0 = ( δ φ )( x , x , x ) = φ ([ x , x ] , x ) + φ ([ x , x ] , x ) + φ ([ x , x ] , x ) = φ ( x , x ) = f. Also, we have φ ( x, y [ p ] ) = 0 . Therefore, φ ( x, x ) = 0 , for all x ∈ L and hence φ ( x , x ) = φ ( x , x ) = φ ( x , x ) = 0 which implies that c = e = f = 0 . Therefore, Z ( L, F ) has a basis consisting of:(∆ , , (∆ , , (∆ , , (0 , f ) , (0 , f ) , (0 , f ) , (0 , f ) . Next, we find a basis for B ( L, F ). Let ( φ, ω ) ∈ B ( L, F ). Since B ( L, F ) ⊆ Z ( L, F ), we have( φ, ω ) = ( a ∆ + b ∆ + d ∆ , αf + βf + γf + δf ). Note that there exists a linear map ψ : L → F such that δ ψ ( x, y ) = φ ( x, y ) and ˜ ψ ( x ) = ω ( x ), for all x, y ∈ L . So, we have a = φ ( x , x ) = δ ψ ( x , x ) = ψ ([ x , x ]) = ψ ( x ) , and b = φ ( x , x ) = δ ψ ( x , x ) = ψ ([ x , x ]) = 0 . Similarly, we can show that d = 0 . Also, we have α = ω ( x ) = ˜ ψ ( x ) = ψ ( x [ p ]1 ) = ψ ( x ) , and β = ω ( x ) = ˜ ψ ( x ) = ψ ( x [ p ]2 ) = 0 . Similarly, we can show that γ = δ = 0 . Therefore, ( φ, ω ) = ( a ∆ , αf ) and hence B ( L, F ) = h (∆ , , (0 , f ) i F . We deduce that a basis for H ( L, F ) is as follows:[(∆ , , [(∆ , , [(0 , f )] , [(0 , f )] , [(0 , f )] . HAPTER 4. RESTRICTION MAPS ON L , p ( L ) in this case consists of invertible matrices of the form a a a a a a r a a a a , where r = a a − a a = 0 , a = a p and a = a = 0 .Let [ θ ] = [( φ, ω )] ∈ H ( L, F ). As in Section 4.2.1, it is enough to find Aut p ( L )-representatives ofthe ω ’s. We can verify the action of Aut p ( L ) on the ω ’s in the matrix form is as follows: β ′ γ ′ δ ′ = a p a p a p r p 00 0 a p βγδ . Now we find the representatives of the orbits of this action. Let ν = βγδ ∈ F . If ν = ,then { ν } is clearly an Aut p ( L )-orbit. Let ν = 0 . Suppose that δ = 0 . Then − β/δ βγδ = γδ , and (1 /δ ) − p /δ γδ = γ . If γ = 0 , then /γ /γ 00 0 1 γ = . If γ = 0 then we have . Next, if δ = 0 , but γ = 0 , then − β/δ 00 1 00 0 1 βγ = γ , and /γ /γ 00 0 1 γ = . Finally, if δ = γ = 0 then we have /β /β 00 0 1 β = . HAPTER 4. RESTRICTION MAPS ON L , p ( L )-orbit representatives: , , , , . Therefore, the corresponding restricted Lie algebra structures are as follows: K = h x , . . . , x | [ x , x ] = x , x [ p ]1 = x i ; K = h x , . . . , x | [ x , x ] = x , x [ p ]1 = x , x [ p ]2 = x i ; K = h x , . . . , x | [ x , x ] = x , x [ p ]1 = x , x [ p ]3 = x i ; K = h x , . . . , x | [ x , x ] = x , x [ p ]1 = x , x [ p ]4 = x i ; K = h x , . . . , x | [ x , x ] = x , x [ p ]1 = x , x [ p ]3 = x , x [ p ]4 = x i . ( L, x [ p ]1 = x , x [ p ]2 = x ) First, we find a basis for Z ( L, F ). Let ( φ, ω ) = ( a ∆ + b ∆ + c ∆ + d ∆ + e ∆ + f ∆ , αf + βf + γf + δf ) ∈ Z ( L, F ). Then we must have δ φ ( x, y, z ) = 0 and φ ( x, y [ p ] ) = 0 , for all x, y, z ∈ L .Therefore, we have0 = ( δ φ )( x , x , x ) = φ ([ x , x ] , x ) + φ ([ x , x ] , x ) + φ ([ x , x ] , x ) = φ ( x , x ) = f. Also, we have φ ( x, y [ p ] ) = 0 . Therefore, φ ( x, x ) = 0 and φ ( x, x ) = 0 , for all x ∈ L and hence φ ( x , x ) = φ ( x , x ) = φ ( x , x ) = φ ( x , x ) = φ ( x , x ) = 0 which implies that b = c = d = e = f = 0 . Therefore, Z ( L, F ) has a basis consisting of:(∆ , , (0 , f ) , (0 , f ) , (0 , f ) , (0 , f ) . Next, we find a basis for B ( L, F ). Let ( φ, ω ) ∈ B ( L, F ). Since B ( L, F ) ⊆ Z ( L, F ), we have( φ, ω ) = ( a ∆ , αf + βf + γf + δf ). Note that there exists a linear map ψ : L → F such that δ ψ ( x, y ) = φ ( x, y ) and ˜ ψ ( x ) = ω ( x ), for all x, y ∈ L . So, we have a = φ ( x , x ) = δ ψ ( x , x ) = ψ ([ x , x ]) = ψ ( x ) . Also, we have α = ω ( x ) = ˜ ψ ( x ) = ψ ( x [ p ]1 ) = ψ ( x ) , and β = ω ( x ) = ˜ ψ ( x ) = ψ ( x [ p ]2 ) = ψ ( x ) , and γ = ω ( x ) = ˜ ψ ( x ) = ψ ( x [ p ]3 ) = 0 . Similarly, we can show that δ = 0 . Note that ψ ( x ) = a = α . Therefore, ( φ, ω ) = ( a ∆ , af + βf )and hence B ( L, F ) = h (∆ , f ) , (0 , f ) i F . Note that n [(∆ , , [(0 , f )] , [(0 , f )] , [(0 , f )] , [(0 , f )] o spans H ( L, F ). Since [(∆ , , f )] = [(∆ , f )] = [0], then [(0 , f )] is an scalar multiple of[(∆ , H ( L, F ). Note that dim H = dim Z − dim B = 3 . Therefore, n [(∆ , , [(0 , f )] , [(0 , f )] o HAPTER 4. RESTRICTION MAPS ON L , H ( L, F ).Note that the group Aut p ( L ) in this case consists of invertible matrices of the form a a a a a a r a a a a , where r = a a − a a = a p and a = 0 , a = a p , and a = a p .Let [ θ ] = [( φ, ω )] ∈ H ( L, F ). As in Section 4.2.1, it is enough to find Aut p ( L )-representatives ofthe ω ’s. We can verify that the action of Aut p ( L ) on the ω ’s in the matrix form is as follows: (cid:18) γ ′ δ ′ (cid:19) = a p a p a p ! (cid:18) γδ (cid:19) . Now we find the representatives of the orbits of this action. Note that we need to have Aω ( x ) = 0which implies that a p γ + a p δ = 0 . Let ν = (cid:18) γδ (cid:19) ∈ F . If ν = (cid:18) (cid:19) , then { ν } is clearly an Aut p ( L )-orbit. Let ν = 0 . Suppose that γ = 0 . Then (cid:18) − δ/γ (cid:19) (cid:18) γδ (cid:19) = (cid:18) γ (cid:19) , and (cid:18) /γ 00 (1 /γ ) p − (cid:19) (cid:18) γ (cid:19) = (cid:18) (cid:19) . Next, if γ = 0 , but δ = 0 , then we have (cid:18) δ (cid:19) . Thus the following elements are Aut p ( L )-orbitrepresentatives: (cid:18) (cid:19) , (cid:18) (cid:19) , (cid:18) δ (cid:19) . Therefore, the corresponding restricted Lie algebra structures are as follows: K = h x , . . . , x | [ x , x ] = x , x [ p ]1 = x , x [ p ]2 = x i ; K = h x , . . . , x | [ x , x ] = x , x [ p ]1 = x , x [ p ]2 = x , x [ p ]3 = x i ; K ( δ ) = h x , . . . , x | [ x , x ] = x , x [ p ]1 = x , x [ p ]2 = x , x [ p ]4 = δx i where δ ∈ F ∗ . Lemma 4.2.3 We have K ( δ ) ∼ = K (1) , for every δ ∈ F ∗ . Proof. The following linear map yields the desired automorphism: δ /p δ ( p − /p δ /p δ ( p − /p 00 0 0 0 δ p − (cid:4) HAPTER 4. RESTRICTION MAPS ON L , ( L, x [ p ]3 = x ) First, we find a basis for Z ( L, F ). Let ( φ, ω ) = ( a ∆ + b ∆ + c ∆ + d ∆ + e ∆ + f ∆ , αf + βf + γf + δf ) ∈ Z ( L, F ). Then we must have δ φ ( x, y, z ) = 0 and φ ( x, y [ p ] ) = 0 , for all x, y, z ∈ L .Therefore, we have0 = ( δ φ )( x , x , x ) = φ ([ x , x ] , x ) + φ ([ x , x ] , x ) + φ ([ x , x ] , x ) = φ ( x , x ) = f. Also, we have φ ( x, y [ p ] ) = 0 . Therefore, φ ( x, x ) = 0 , for all x ∈ L and hence φ ( x , x ) = φ ( x , x ) = φ ( x , x ) = 0 which implies that c = e = f = 0 . Therefore, Z ( L, F ) has a basis consisting of:(∆ , , (∆ , , (∆ , , (0 , f ) , (0 , f ) , (0 , f ) , (0 , f ) . Next, we find a basis for B ( L, F ). Let ( φ, ω ) ∈ B ( L, F ). Since B ( L, F ) ⊆ Z ( L, F ), we have( φ, ω ) = ( a ∆ + b ∆ + d ∆ , αf + βf + γf + δf ). Note that there exists a linear map ψ : L → F such that δ ψ ( x, y ) = φ ( x, y ) and ˜ ψ ( x ) = ω ( x ), for all x, y ∈ L . So, we have a = φ ( x , x ) = δ ψ ( x , x ) = ψ ([ x , x ]) = ψ ( x ) , and b = φ ( x , x ) = δ ψ ( x , x ) = ψ ([ x , x ]) = 0 . Similarly, we can show that d = 0 . Also, we have γ = ω ( x ) = ˜ ψ ( x ) = ψ ( x [ p ]3 ) = ψ ( x ) , and α = ω ( x ) = ˜ ψ ( x ) = ψ ( x [ p ]1 ) = 0 . Similarly, we can show that β = δ = 0 . Therefore, ( φ, ω ) = ( a ∆ , γf ) and hence B ( L, F ) = h (∆ , , (0 , f ) i F . We deduce that a basis for H ( L, F ) is as follows:[(∆ , , [(∆ , , [(0 , f )] , [(0 , f )] , [(0 , f )] . Note that the group Aut p ( L ) in this case consists of invertible matrices of the form a a a a a a r a a a a , where r = a a − a a and a = a = a = 0 , and a = r p .Let [ θ ] = [( φ, ω )] ∈ H ( L, F ). As in Section 4.2.1, it is enough to find Aut p ( L )-representatives ofthe ω ’s. We can verify that the action of Aut p ( L ) on the ω ’s in the matrix form is as follows: α ′ β ′ δ ′ = a p a p a p a p a p a p r p αβδ . Now we find the representatives of the orbits of this action. Let ν = αβδ ∈ F . If ν = , HAPTER 4. RESTRICTION MAPS ON L , { ν } is clearly an Aut p ( L )-orbit. Let ν = 0 . Suppose that δ = 0 . Then − α/δ − β/δ αβδ = δ , and (1 /δ ) /p /δ δ = . Next, if δ = 0 , but β = 0 , then − α/β 00 1 00 0 1 αβ = β , and /β 00 0 (1 /β ) p β = . Next, if δ = β = 0 , but α = 0 , then /α /α ) p α = . Thus the following elements are Aut p ( L )-orbit representatives: , , , . Therefore, the corresponding restricted Lie algebra structures are as follows: K = h x , . . . , x | [ x , x ] = x , x [ p ]3 = x i ; K = h x , . . . , x | [ x , x ] = x , x [ p ]1 = x , x [ p ]3 = x i ; K = h x , . . . , x | [ x , x ] = x , x [ p ]2 = x , x [ p ]3 = x i ; K = h x , . . . , x | [ x , x ] = x , x [ p ]3 = x , x [ p ]4 = x i . ( L, x [ p ]2 = x , x [ p ]3 = x ) First, we find a basis for Z ( L, F ). Let ( φ, ω ) = ( a ∆ + b ∆ + c ∆ + d ∆ + e ∆ + f ∆ , αf + βf + γf + δf ) ∈ Z ( L, F ). Then we must have δ φ ( x, y, z ) = 0 and φ ( x, y [ p ] ) = 0 , for all x, y, z ∈ L .Therefore, we have0 = ( δ φ )( x , x , x ) = φ ([ x , x ] , x ) + φ ([ x , x ] , x ) + φ ([ x , x ] , x ) = φ ( x , x ) = f. Also, we have φ ( x, y [ p ] ) = 0 . Therefore, φ ( x, x ) = 0 and φ ( x, x ) = 0 , for all x ∈ L and hence φ ( x , x ) = φ ( x , x ) = φ ( x , x ) = φ ( x , x ) = φ ( x , x ) = 0 which implies that b = c = d = e = f = 0 . Therefore, Z ( L, F ) has a basis consisting of:(∆ , , (0 , f ) , (0 , f ) , (0 , f ) , (0 , f ) . HAPTER 4. RESTRICTION MAPS ON L , B ( L, F ). Let ( φ, ω ) ∈ B ( L, F ). Since B ( L, F ) ⊆ Z ( L, F ), we have( φ, ω ) = ( a ∆ , αf + βf + γf + δf ). Note that there exists a linear map ψ : L → F such that δ ψ ( x, y ) = φ ( x, y ) and ˜ ψ ( x ) = ω ( x ), for all x, y ∈ L . So, we have a = φ ( x , x ) = δ ψ ( x , x ) = ψ ([ x , x ]) = ψ ( x ) . Also, we have α = ω ( x ) = ˜ ψ ( x ) = ψ ( x [ p ]1 ) = 0 , and β = ω ( x ) = ˜ ψ ( x ) = ψ ( x [ p ]2 ) = ψ ( x ) , and γ = ω ( x ) = ˜ ψ ( x ) = ψ ( x [ p ]3 ) = ψ ( x ) , and δ = ω ( x ) = ˜ ψ ( x ) = ψ ( x [ p ]4 ) = 0 . Note that ψ ( x ) = a = β . Hence, ( φ, ω ) = ( a ∆ , af + γf ) and B ( L, F ) = h (∆ , f ) , (0 , f ) i F .Note that n [(∆ , , [(0 , f )] , [(0 , f )] , [(0 , f )] o spans H ( L, F ). Since [(∆ , , f )] = [(∆ , f )] = [0], then [(0 , f )] is an scalar multiple of[(∆ , H ( L, F ). Note that dim H = dim Z − dim B = 3 . Therefore, n [(∆ , , [(0 , f )] , [(0 , f )] o forms a basis for H ( L, F ).Note that the group Aut p ( L ) in this case consists of invertible matrices of the form a a a a a a r a a a a , where r = a a − a a = a p and a = a = a = a = 0 , a = r p .Let [ θ ] = [( φ, ω )] ∈ H ( L, F ). As in Section 4.2.1, it is enough to find Aut p ( L )-representatives ofthe ω ’s. We can verify that the action of Aut p ( L ) on the ω ’s in the matrix form is as follows: (cid:18) α ′ δ ′ (cid:19) = (cid:18) a p a p r p (cid:19) (cid:18) αδ (cid:19) . Now we find the representatives of the orbits of this action. Note that we need to have Aω ( x ) = 0which implies that a p α + a p δ = 0 .Let ν = (cid:18) αδ (cid:19) ∈ F . If ν = (cid:18) (cid:19) , then { ν } is clearly an Aut p ( L )-orbit. Let ν = 0 . Suppose that δ = 0 . Then (cid:18) − α/δ (cid:19) (cid:18) αδ (cid:19) = (cid:18) δ (cid:19) , and (1 /δ ) p − p 00 1 /δ ! (cid:18) δ (cid:19) = (cid:18) (cid:19) . HAPTER 4. RESTRICTION MAPS ON L , δ = 0 , but α = 0 , then we have (cid:18) α (cid:19) . Thus the following elements are Aut p ( L )-orbitrepresentatives: (cid:18) (cid:19) , (cid:18) α (cid:19) , (cid:18) (cid:19) . Therefore, the corresponding restricted Lie algebra structures are as follows: K = h x , . . . , x | [ x , x ] = x , x [ p ]2 = x , x [ p ]3 = x i ; K ( α ) = h x , . . . , x | [ x , x ] = x , x [ p ]1 = αx , x [ p ]2 = x , x [ p ]3 = x i ; K = h x , . . . , x | [ x , x ] = x , x [ p ]2 = x , x [ p ]3 = x , x [ p ]4 = x i where α ∈ F ∗ . Lemma 4.2.4 We have K ( α ) ∼ = K (1) , for every α ∈ F ∗ . Proof. The following linear map yields the desired automorphism: α ( p − /p α /p α α p 00 0 0 0 α p − (cid:4) ( L, x [ p ]4 = x ) First, we find a basis for Z ( L, F ). Let ( φ, ω ) = ( a ∆ + b ∆ + c ∆ + d ∆ + e ∆ + f ∆ , αf + βf + γf + δf ) ∈ Z ( L, F ). Then we must have δ φ ( x, y, z ) = 0 and φ ( x, y [ p ] ) = 0 , for all x, y, z ∈ L .Therefore, we have0 = ( δ φ )( x , x , x ) = φ ([ x , x ] , x ) + φ ([ x , x ] , x ) + φ ([ x , x ] , x ) = φ ( x , x ) = f. Also, we have φ ( x, y [ p ] ) = 0 . Therefore, φ ( x, x ) = 0 , for all x ∈ L and hence φ ( x , x ) = φ ( x , x ) = φ ( x , x ) = 0 which implies that b = d = f = 0 . Therefore, Z ( L, F ) has a basis consisting of:(∆ , , (∆ , , (∆ , , (0 , f ) , (0 , f ) , (0 , f ) , (0 , f ) . Next, we find a basis for B ( L, F ). Let ( φ, ω ) ∈ B ( L, F ). Since B ( L, F ) ⊆ Z ( L, F ), we have( φ, ω ) = ( a ∆ + c ∆ + e ∆ , αf + βf + γf + δf ). Note that there exists a linear map ψ : L → F such that δ ψ ( x, y ) = φ ( x, y ) and ˜ ψ ( x ) = ω ( x ), for all x, y ∈ L . So, we have a = φ ( x , x ) = δ ψ ( x , x ) = ψ ([ x , x ]) = ψ ( x ) , and c = φ ( x , x ) = δ ψ ( x , x ) = ψ ([ x , x ]) = 0 . Similarly, we can show that e = 0 . Also, we have δ = ω ( x ) = ˜ ψ ( x ) = ψ ( x [ p ]4 ) = ψ ( x ) , and α = ω ( x ) = ˜ ψ ( x ) = ψ ( x [ p ]1 ) = 0 . HAPTER 4. RESTRICTION MAPS ON L , β = γ = 0 . Note that ψ ( x ) = a = δ . Therefore, ( φ, ω ) = ( a ∆ , af )and hence B ( L, F ) = h (∆ , f ) i F . Note that n [(∆ , , [(∆ , , [(∆ , , [(0 , f )] , [(0 , f )] , [(0 , f )] , [(0 , f )] o spans H ( L, F ). Since [(∆ , , f )] = [(∆ , f )] = [0], then [(0 , f )] is an scalar multiple of[(∆ , H ( L, F ). Note that dim H = dim Z − dim B = 6 . Therefore, n [(∆ , , [(∆ , , [(∆ , , [(0 , f )] , [(0 , f )] , [(0 , f )] o forms a basis for H ( L, F ).Note that the group Aut p ( L ) in this case consists of invertible matrices of the form a a a a a a r a a a a , where r = a a − a a = a p and a = a = 0 .Let [ θ ] = [( φ, ω )] ∈ H ( L, F ). As in Section 4.2.1, it is enough to find Aut p ( L )-representatives ofthe ω ’s. We can verify that the action of Aut p ( L ) on the ω ’s in the matrix form is as follows: α ′ β ′ γ ′ = a p a p a p a p a p a p r p αβγ . Now we find the representatives of the orbits of this action. Note that we need to have Aω ( x ) = 0which implies that a p γ = 0 .Let ν = αβγ ∈ F . If ν = , then { ν } is clearly an Aut p ( L )-orbit. Let ν = 0 . Supposethat γ = 0 . Then − α/γ − β/γ αβγ = γ , and /γ /γ γ = . Next, if γ = 0 , but β = 0 , then − α/β 00 1 00 0 1 αβ = β , and /β 00 0 1 /β β = . HAPTER 4. RESTRICTION MAPS ON L , γ = β = 0 , but α = 0 , then /α /α α = . Thus the following elements are Aut( L )-orbit representatives: , , , . Therefore, the corresponding restricted Lie algebra structures are as follows: K = h x , . . . , x | [ x , x ] = x , x [ p ]4 = x i ; K = h x , . . . , x | [ x , x ] = x , x [ p ]1 = x , x [ p ]4 = x i ; K = h x , . . . , x | [ x , x ] = x , x [ p ]2 = x , x [ p ]4 = x i ; K = h x , . . . , x | [ x , x ] = x , x [ p ]3 = x , x [ p ]4 = x i . ( L, x [ p ]2 = x , x [ p ]4 = x ) First, we find a basis for Z ( L, F ). Let ( φ, ω ) = ( a ∆ + b ∆ + c ∆ + d ∆ + e ∆ + f ∆ , αf + βf + γf + δf ) ∈ Z ( L, F ). Then we must have δ φ ( x, y, z ) = 0 and φ ( x, y [ p ] ) = 0 , for all x, y, z ∈ L .Therefore, we have0 = ( δ φ )( x , x , x ) = φ ([ x , x ] , x ) + φ ([ x , x ] , x ) + φ ([ x , x ] , x ) = φ ( x , x ) = f. Also, we have φ ( x, y [ p ] ) = 0 . Therefore, φ ( x, x ) = 0 and φ ( x, x ) = 0 , for all x ∈ L and hence φ ( x , x ) = φ ( x , x ) = φ ( x , x ) = φ ( x , x ) = φ ( x , x ) = 0 which implies that b = c = d = e = f = 0 . Therefore, Z ( L, F ) has a basis consisting of:(∆ , , (0 , f ) , (0 , f ) , (0 , f ) , (0 , f ) . Next, we find a basis for B ( L, F ). Let ( φ, ω ) ∈ B ( L, F ). Since B ( L, F ) ⊆ Z ( L, F ), we have( φ, ω ) = ( a ∆ , αf + βf + γf + δf ). Note that there exists a linear map ψ : L → F such that δ ψ ( x, y ) = φ ( x, y ) and ˜ ψ ( x ) = ω ( x ), for all x, y ∈ L . So, we have a = φ ( x , x ) = δ ψ ( x , x ) = ψ ([ x , x ]) = ψ ( x ) . Also, we have α = ω ( x ) = ˜ ψ ( x ) = ψ ( x [ p ]1 ) = 0 , and β = ω ( x ) = ˜ ψ ( x ) = ψ ( x [ p ]2 ) = ψ ( x ) , and γ = ω ( x ) = ˜ ψ ( x ) = ψ ( x [ p ]3 ) = 0 , and δ = ω ( x ) = ˜ ψ ( x ) = ψ ( x [ p ]4 ) = ψ ( x ) . Note that ψ ( x ) = a = δ . Hence, ( φ, ω ) = ( a ∆ , βf + af ) and B ( L, F ) = h (∆ , f ) , (0 , f ) i F .Note that n [(∆ , , [(0 , f )] , [(0 , f )] , [(0 , f )] o HAPTER 4. RESTRICTION MAPS ON L , H ( L, F ). Since [(∆ , , f )] = [(∆ , f )] = [0], then [(0 , f )] is an scalar multiple of[(∆ , H ( L, F ). Note that dim H = dim Z − dim B = 3 . Therefore, n [(∆ , , [(0 , f )] , [(0 , f )] o forms a basis for H ( L, F ).Note that the group Aut p ( L ) in this case consists of invertible matrices of the form a a a a a a r a a a a , where r = a a − a a = a p and a = a = 0 , a = a p , and a = a p .Let [ θ ] = [( φ, ω )] ∈ H ( L, F ). As in Section 4.2.1, it is enough to find Aut p ( L )-representatives ofthe ω ’s. We can verify that the action of Aut p ( L ) on the ω ’s in the matrix form is as follows: (cid:18) α ′ γ ′ (cid:19) = (cid:18) a p a p r p (cid:19) (cid:18) αγ (cid:19) . Now we find the representatives of the orbits of this action. Note that we need to have Aω ( x ) = 0which implies that a p γ = 0 .Let ν = (cid:18) αγ (cid:19) ∈ F . If ν = (cid:18) (cid:19) , then { ν } is clearly an Aut p ( L )-orbit. Let ν = 0 . Suppose that γ = 0 . Then (cid:18) − α/γ (cid:19) (cid:18) αγ (cid:19) = (cid:18) γ (cid:19) , and (1 /γ ) p − p 00 1 /γ ! (cid:18) γ (cid:19) = (cid:18) (cid:19) . Next, if γ = 0 , but α = 0 , then we have (cid:18) α (cid:19) . Thus the following elements are Aut p ( L )-orbitrepresentatives: (cid:18) (cid:19) , (cid:18) α (cid:19) , (cid:18) (cid:19) . Therefore, the corresponding restricted Lie algebra structures are as follows: K = h x , . . . , x | [ x , x ] = x , x [ p ]2 = x , x [ p ]4 = x i ; K ( α ) = h x , . . . , x | [ x , x ] = x , x [ p ]1 = αx , x [ p ]2 = x , x [ p ]4 = x i ; K = h x , . . . , x | [ x , x ] = x , x [ p ]2 = x , x [ p ]3 = x , x [ p ]4 = x i where α ∈ F ∗ . Lemma 4.2.5 We have K ( α ) ∼ = K (1) , for every α ∈ F ∗ . HAPTER 4. RESTRICTION MAPS ON L , Proof. The following linear map yields the desired automorphism: α ( p − /p α /p α p α 00 0 0 0 α p − (cid:4) We can easily see that some of the algebras given above are identical. The following is the list of allirredundant restricted Lie algebra structures on L , and yet, as we shall see below, we prove that HAPTER 4. RESTRICTION MAPS ON L , K = h x , . . . , x | [ x , x ] = x i ; K = h x , . . . , x | [ x , x ] = x , x [ p ]1 = x i ; K = h x , . . . , x | [ x , x ] = x , x [ p ]2 = x i ; K = h x , . . . , x | [ x , x ] = x , x [ p ]4 = x i ; K = h x , . . . , x | [ x , x ] = x , x [ p ]5 = x i ; K = h x , . . . , x | [ x , x ] = x , x [ p ]1 = x i ; K = h x , . . . , x | [ x , x ] = x , x [ p ]2 = x i ; K = h x , . . . , x | [ x , x ] = x , x [ p ]3 = x i ; K = h x , . . . , x | [ x , x ] = x , x [ p ]4 = x i ; K = h x , . . . , x | [ x , x ] = x , x [ p ]1 = x , x [ p ]2 = x i ; K = h x , . . . , x | [ x , x ] = x , x [ p ]1 = x , x [ p ]3 = x i ; K = h x , . . . , x | [ x , x ] = x , x [ p ]1 = x , x [ p ]4 = x i ; K = h x , . . . , x | [ x , x ] = x , x [ p ]1 = x i ; K = h x , . . . , x | [ x , x ] = x , x [ p ]1 = x , x [ p ]2 = x i ; K = h x , . . . , x | [ x , x ] = x , x [ p ]1 = x , x [ p ]3 = x i ; K = h x , . . . , x | [ x , x ] = x , x [ p ]1 = x , x [ p ]4 = x i ; K = h x , . . . , x | [ x , x ] = x , x [ p ]1 = x , x [ p ]3 = x , x [ p ]4 = x i ; K = h x , . . . , x | [ x , x ] = x , x [ p ]1 = x , x [ p ]2 = x i ; K = h x , . . . , x | [ x , x ] = x , x [ p ]1 = x , x [ p ]2 = x , x [ p ]3 = x i ; K = h x , . . . , x | [ x , x ] = x , x [ p ]1 = x , x [ p ]2 = x , x [ p ]4 = x i ; K = h x , . . . , x | [ x , x ] = x , x [ p ]3 = x i ; K = h x , . . . , x | [ x , x ] = x , x [ p ]1 = x , x [ p ]3 = x i ; K = h x , . . . , x | [ x , x ] = x , x [ p ]2 = x , x [ p ]3 = x i ; K = h x , . . . , x | [ x , x ] = x , x [ p ]3 = x , x [ p ]4 = x i ; K = h x , . . . , x | [ x , x ] = x , x [ p ]2 = x , x [ p ]3 = x i ; K = h x , . . . , x | [ x , x ] = x , x [ p ]1 = x , x [ p ]2 = x , x [ p ]3 = x i ; K = h x , . . . , x | [ x , x ] = x , x [ p ]2 = x , x [ p ]3 = x , x [ p ]4 = x i ; K = h x , . . . , x | [ x , x ] = x , x [ p ]1 = x , x [ p ]4 = x i ; K = h x , . . . , x | [ x , x ] = x , x [ p ]2 = x , x [ p ]4 = x i ; K = h x , . . . , x | [ x , x ] = x , x [ p ]3 = x , x [ p ]4 = x i ; K = h x , . . . , x | [ x , x ] = x , x [ p ]2 = x , x [ p ]4 = x i ; K = h x , . . . , x | [ x , x ] = x , x [ p ]1 = x , x [ p ]2 = x , x [ p ]4 = x i ; K = h x , . . . , x | [ x , x ] = x , x [ p ]2 = x , x [ p ]3 = x , x [ p ]4 = x i . HAPTER 4. RESTRICTION MAPS ON L , L , ) consists of invertible matrices of the form a a a a a a a a − a a a a a a a a a a a a , where a a − a a = 0 .We have K ∼ = K , K ∼ = K , K ∼ = K by the following automorphism of L , : − − − − 10 0 0 − Consider the following automorphism of L , : − − − − − . Therefore, K ∼ = K , K ∼ = K , K ∼ = K . Finally, using the following automorphism of L , that only switches x and x , we get K ∼ = K , K ∼ = K , K ∼ = K , K ∼ = K . Theorem 4.3.1 The list of all restricted Lie algebra structures on L , , up to isomorphism, is as HAPTER 4. RESTRICTION MAPS ON L , follows: L , = h x , . . . , x | [ x , x ] = x i ; L , = h x , . . . , x | [ x , x ] = x , x [ p ]1 = x i ; L , = h x , . . . , x | [ x , x ] = x , x [ p ]4 = x i ; L , = h x , . . . , x | [ x , x ] = x , x [ p ]1 = x i ; L , = h x , . . . , x | [ x , x ] = x , x [ p ]3 = x i ; L , = h x , . . . , x | [ x , x ] = x , x [ p ]4 = x i ; L , = h x , . . . , x | [ x , x ] = x , x [ p ]1 = x , x [ p ]2 = x i ; L , = h x , . . . , x | [ x , x ] = x , x [ p ]1 = x , x [ p ]3 = x i ; L , = h x , . . . , x | [ x , x ] = x , x [ p ]1 = x , x [ p ]4 = x i ; L , = h x , . . . , x | [ x , x ] = x , x [ p ]1 = x , x [ p ]2 = x i ; L , = h x , . . . , x | [ x , x ] = x , x [ p ]1 = x , x [ p ]3 = x i ; L , = h x , . . . , x | [ x , x ] = x , x [ p ]1 = x , x [ p ]4 = x i ; L , = h x , . . . , x | [ x , x ] = x , x [ p ]1 = x , x [ p ]2 = x , x [ p ]3 = x i ; L , = h x , . . . , x | [ x , x ] = x , x [ p ]1 = x , x [ p ]2 = x , x [ p ]4 = x i ; L , = h x , . . . , x | [ x , x ] = x , x [ p ]3 = x , x [ p ]4 = x i ; L , = h x , . . . , x | [ x , x ] = x , x [ p ]2 = x , x [ p ]3 = x , x [ p ]4 = x i ; L , = h x , . . . , x | [ x , x ] = x , x [ p ]1 = x , x [ p ]4 = x i ; L , = h x , . . . , x | [ x , x ] = x , x [ p ]3 = x , x [ p ]4 = x i ; L , = h x , . . . , x | [ x , x ] = x , x [ p ]2 = x , x [ p ]4 = x i ; L , = h x , . . . , x | [ x , x ] = x , x [ p ]1 = x , x [ p ]2 = x , x [ p ]4 = x i ; L , = h x , . . . , x | [ x , x ] = x , x [ p ]2 = x , x [ p ]3 = x , x [ p ]4 = x i ; L , = h x , . . . , x | [ x , x ] = x , x [ p ]1 = x , x [ p ]3 = x , x [ p ]4 = x i . In the remaining of this section we establish that the algebras given in Theorem 4.3.1 are pairwisenon-isomorphic, thereby completing the proof of Theorem 4.3.1.It is clear that L , is not isomorphic to the other restricted Lie algebras.We claim that L , and L , are not isomorphic. Suppose to the contrary that there exists anisomorphism A : L , → L , . Then A ( x [ p ]4 ) = A ( x ) [ p ] A ( x ) = ( a x + a x + a x ) [ p ] ( a a − a a ) x = 0 . Therefore, a a − a a = 0 which is a contradiction. Note that L , ≇ L , because L , / ( L , ) [ p ] ≇ L , / ( L , ) [ p ] . (4.1) HAPTER 4. RESTRICTION MAPS ON L , L , ≇ L , and L , ≇ L , . It is clear that L , is not isomorphic to the other restrictedLie algebras.Similar argument as in (4.1) shows that L , ≇ L , , L , ≇ L , , and L , ≇ L , . It is clearthat L , is not isomorphic to the other restricted Lie algebras.Next, we claim that L , and L , are not isomorphic. Suppose to the contrary that there existsan isomorphism A : L , → L , . Then A ( x [ p ]3 ) = A ( x ) [ p ] a a − a a ) x ) [ p ] a a − a a ) p x . Therefore, a a − a a = 0 which is a contradiction.Next, we claim that L , and L , are not isomorphic. Suppose to the contrary that there existsan isomorphism A : L , → L , . Then we have A ( x [ p ]1 ) = A ( x ) [ p ] a x + a x + a x + a x + a x ) [ p ] a p x . Therefore, a = 0 . Also we have A ( x [ p ]2 ) = A ( x ) [ p ] a x + a x + a x + a x + a x ) [ p ] a p x . Therefore, a = 0 . Hence, a a − a a = 0 which is a contradiction.It is clear that L , is not isomorphic to the other restricted Lie algebras.Next, we claim that L , and L , are not isomorphic. Suppose to the contrary that there existsan isomorphism A : L , → L , . Then A ( x [ p ]3 ) = A ( x ) [ p ] a a − a a ) x ) [ p ] a a − a a ) p x . Therefore, a a − a a = 0 which is a contradiction.It is clear that L , and L , are not isomorphic to the other restricted Lie algebras.Note that L , and L , are not isomorphic because ( L , ) [ p ] = 0 but ( L , ) [ p ] = 0 . Also, L , and L , are not isomorphic. Suppose to the contrary that there exists an isomorphism A : L , → L , . Then A ( x [ p ]2 ) = A ( x ) [ p ] a x + a x + a x + a x + a x ) [ p ] a p x + a p x . Therefore, a = 0 and a = 0 . Hence, a a − a a = 0 which is a contradiction. HAPTER 4. RESTRICTION MAPS ON L , L , ≇ L , , L , ≇ L , , L , ≇ L , , and L , ≇ L , .Next, we claim that L , and L , are not isomorphic. Suppose to the contrary that there exists anisomorphism A : L , → L , . Then A ( x [ p ]4 ) = A ( x ) [ p ] A ( x ) = ( a x + a x + a x ) [ p ] ( a a − a a ) x = 0 . Therefore, a a − a a = 0 which is a contradiction.Next, we claim that L , and L , are not isomorphic. Suppose to the contrary that there existsan isomorphism A : L , → L , . Then A ( x [ p ]3 ) = A ( x ) [ p ] a a − a a ) x ) [ p ] a a − a a ) p x . Therefore, a a − a a = 0 which is a contradiction.Note that L , is not isomorphic to any of L , nor L , . Because ( L , ) [ p ] = 0 but this is notthe case for L , and L , .It is clear that L , is not isomorphic to the remaining restricted Lie algebras.Note that L , is not isomorphic to any of L , , L , , L , , L , , L , because ( L , ) [ p ] =( L , ) [ p ] = ( L , ) [ p ] = ( L , ) [ p ] = ( L , ) [ p ] = 0 but ( L , ) [ p ] = 0 .Next, we claim that L , and L , are not isomorphic. Suppose to the contrary that there existsan isomorphism A : L , → L , . Then A ( x [ p ]4 ) = A ( x ) [ p ] A ( x ) = ( a x + a x + a x ) [ p ] ( a a − a a ) x = a p x . Therefore, a a − a a = 0 which is a contradiction.Next, we claim that L , and L , are not isomorphic. Suppose to the contrary that there existsan isomorphism A : L , → L , . Then we have A ( x [ p ]1 ) = A ( x ) [ p ] A ( x ) = ( a x + a x + a x + a x + a x ) [ p ] ( a a − a a ) x = a p x + a p x + a p x . Therefore, a a − a a = 0 which is a contradiction.It is clear that L , is not isomorphic to the other restricted Lie algebras.Similar argument as in (4.1) shows that L , is not isomorphic to any of L , , L , , L , , or L , .Next, we claim that L , and L , are not isomorphic. Suppose to the contrary that there existsan isomorphism A : L , → L , . Then A ( x [ p ]4 ) = A ( x ) [ p ] A ( x ) = ( a x + a x + a x ) [ p ] ( a a − a a ) x = a p x . HAPTER 4. RESTRICTION MAPS ON L , a a − a a = 0 which is a contradiction.Note that L , is not isomorphic to any of L , , L , , and L , . Because ( L , ) [ p ] = 0 but thisis not the case for L , , L , , and L , . It is clear that L , is not isomorphic to the other restrictedLie algebras.Next, we claim that L , and L , are not isomorphic. Suppose to the contrary that there existsan isomorphism A : L , → L , . Then A ( x [ p ]3 ) = A ( x ) [ p ] a a − a a ) x ) [ p ] a a − a a ) p x . Therefore, a a − a a = 0 which is a contradiction. Note that L , is not isomorphic to L , , L , , L , , L , nor L , because ( L , ) [ p ] = 0 but ( L , ) [ p ] = 0 , ( L , ) [ p ] = 0 , ( L , ) [ p ] = 0 ,( L , ) [ p ] = 0 and ( L , ) [ p ] = 0 .Next, we claim that L , and L , are not isomorphic. Suppose to the contrary that there existsan isomorphism A : L , → L , . Then A ( x [ p ]4 ) = A ( x ) [ p ] A ( x ) = ( a x + a x + a x ) [ p ] ( a a − a a ) x = 0 . Therefore, a a − a a = 0 which is a contradiction. It is clear that L , is not isomorphic to theother restricted Lie algebras.Note that L , is not isomorphic to any of L , , L , , L , , L , nor L , because ( L , ) [ p ] = 0but ( L , ) [ p ] = 0 , ( L , ) [ p ] = 0 , ( L , ) [ p ] = 0 , ( L , ) [ p ] = 0 and ( L , ) [ p ] = 0 .Next, we claim that L , and L , are not isomorphic. Suppose to the contrary that there existsan isomorphism A : L , → L , . Then A ( x [ p ]4 ) = A ( x ) [ p ] A ( x ) = ( a x + a x + a x ) [ p ] ( a a − a a ) x = a p x . Therefore, a a − a a = 0 which is a contradiction.It is clear that L , is not isomorphic to the other restricted Lie algebras.Next, we claim that L , and L , are not isomorphic. Suppose to the contrary that there existsan isomorphism A : L , → L , . Then A ( x [ p ]3 ) = A ( x ) [ p ] a a − a a ) x ) [ p ] a a − a a ) p x . Therefore, a a − a a = 0 which is a contradiction.Similar argument as in (4.1) shows that L , is not isomorphic to any of to any of L , , L , , or L , .Note that L , is not isomorphic to L , . Because, ( L , ) ′ [ p ] = 0 , but ( L , ) ′ [ p ] = 0 . HAPTER 4. RESTRICTION MAPS ON L , L , is not isomorphic to the other restricted Lie algebras.Next, we claim that L , and L , are not isomorphic. Suppose to the contrary that there existsan isomorphism A : L , → L , . Then A ( x [ p ]3 ) = A ( x ) [ p ] a a − a a ) x ) [ p ] a a − a a ) p x . Therefore, a a − a a = 0 which is a contradiction.Note that L , is not isomorphic to any of L , or L , because ( L , ) [ p ] = 0 but ( L , ) [ p ] = 0and ( L , ) [ p ] = 0 .Next, we claim that L , and L , are not isomorphic. Suppose to the contrary that there existsan isomorphism A : L , → L , . Then A ( x [ p ]3 ) = A ( x ) [ p ] A ( x ) = (( a a − a a ) x ) [ p ] a x + a x + a x = 0 . Therefore, a = 0 which is a contradiction.It is clear that L , is not isomorphic to the other restricted Lie algebras.Note that L , is not isomorphic to any of L , or L , because ( L , ) [ p ] = 0 but ( L , ) [ p ] = 0and ( L , ) [ p ] = 0 .Next, we claim that L , and L , are not isomorphic. Suppose to the contrary that there existsan isomorphism A : L , → L , . Then A ( x [ p ]1 ) = A ( x ) [ p ] A ( x ) = ( a x + a x + a x + a x + a x ) [ p ] ( a a − a a ) x = a p αx + a p x + a p x , which implies that a = 0 and a = 0 . Therefore, a a − a a = 0 which is a contradiction.It is clear that L , is not isomorphic to the other restricted Lie algebras.Similar argument as in (4.1) shows that L , is not isomorphic to any of to any of L , , L , , or L , .Next, we claim that L , and L , are not isomorphic. Suppose to the contrary that there existsan isomorphism A : L , → L , . Then A ( x [ p ]1 ) = A ( x ) [ p ] a x + a x + a x + a x + a x ) [ p ] a p x + a p x + a p x , which implies that a = 0 .Also, we have A ( x [ p ]2 ) = A ( x ) [ p ] a x + a x + a x + a x + a x ) [ p ] a p x + a p x + a p x , HAPTER 4. RESTRICTION MAPS ON L , a = 0 . Therefore, a a − a a = 0 which is a contradiction.It is clear that L , is not isomorphic to the other restricted Lie algebras.Note that L , and L , are not isomorphic because ( L , ) [ p ] = 0 but ( L , ) [ p ] = 0 .Next, we claim that L , and L , are not isomorphic. Suppose to the contrary that there existsan isomorphism A : L , → L , . Then A ( x [ p ]4 ) = A ( x ) [ p ] A ( x ) = ( a x + a x + a x ) [ p ] ( a a − a a ) x = a p x + a p x . Therefore, a a − a a = 0 which is a contradiction. It is clear that L , is not isomorphic to theother restricted Lie algebras.Note that L , is not isomorphic to any of L , , L , nor L , because ( L , ) [ p ] = 0 but( L , ) [ p ] = 0 , ( L , ) [ p ] = 0 , and ( L , ) [ p ] = 0 .It is clear that L , is not isomorphic to the other restricted Lie algebras.Next, we claim that L , and L , are not isomorphic. Suppose to the contrary that there existsan isomorphism A : L , → L , . Then A ( x [ p ]1 ) = A ( x ) [ p ] a x + a x + a x + a x + a x ) [ p ] a p x + a p x , which implies that, a = 0 . Also we have A ( x [ p ]2 ) = A ( x ) [ p ] a x + a x + a x + a x + a x ) [ p ] a p x + a p x , which implies that a = 0 . Therefore, a a − a a = 0 which is a contradiction.Next, we claim that L , and L , are not isomorphic. Suppose to the contrary that there existsan isomorphism A : L , → L , . Then A ( x [ p ]4 ) = A ( x ) [ p ] A ( x ) = ( a x + a x + a x ) [ p ] ( a a − a a ) x = a p x + a p x . Therefore, a a − a a = 0 which is a contradiction.It is clear that L , is not isomorphic to the other restricted Lie algebras.Note that L , is not isomorphic to L , . Because, ( L , ) ′ [ p ] = 0 , but ( L , ) ′ [ p ] = 0 . It is clearthat L , is not isomorphic to the other restricted Lie algebras.Finally, L , and L , are not isomorphic because ( L , ) [ p ] = 0 but ( L , ) [ p ] = 0 . hapter 5 Restriction maps on L , Let K = L , = h x , . . . , x | [ x , x ] = x , [ x , x ] = x i . Note that Z ( L , ) = h x , x i F and the group Aut( L , ) consists of invertible matrices of the form a a a a a a a a a a a a a a a a a . There exists an element αx + βx ∈ Z ( L , ) such that ( αx + βx ) [ p ] = 0 , for some α, β ∈ F . If β = 0 then consider K = h x ′ , . . . , x ′ | [ x ′ , x ′ ] = x ′ , [ x ′ , x ′ ] = x ′ i , where x ′ = x , x ′ = x , x ′ = x , x ′ = x , x ′ = αx + βx . Let φ : K → K given by x i x ′ i , for1 i φ is an isomorphism. Therefore, in this case we can suppose that x [ p ]5 = 0 . If β = 0 then α = 0 and we rescale x so that x [ p ]4 = 0 . Hence we have two cases:I. x [ p ]4 = 0 ;II. x [ p ]5 = 0 . L = L , h x i In this section we find all non-isomorphic p-maps on L , such that x [ p ]4 = 0 . We let L = L , h x i ∼ = L , , where L , = h x , x , x , x | [ x , x ] = x i . Note that we denote the image of x i in L by x i again.We rename x with x and x with x and at the end we will switch them. The group Aut p ( L )57 HAPTER 5. RESTRICTION MAPS ON L , a a a a a a r a a a a , where r = a a − a a = 0 . Lemma 5.1.1 Let K = L , and [ p ] : K → K be a p -map on K such that x [ p ]4 = 0 and let L = KM ,where M = h x i F . Then K ∼ = L θ where θ = (∆ , ω ) ∈ Z ( L, F ) . Proof. Let π : K → L be the projection map. We have the exact sequence0 → M → K → L → . Let σ : L → K such that x i x i , 1 i σ is an injective linear map and πσ = 1 L . Now,we define φ : L × L → M by φ ( x i , x j ) = [ σ ( x i ) , σ ( x j )] − σ ([ x i , x j ]), 1 i, j ω : L → M by ω ( x ) = σ ( x ) [ p ] − σ ( x [ p ] ). Note that φ ( x , x ) = [ σ ( x ) , σ ( x )] − σ ([ x , x ]) = [ x , x ] = x ; φ ( x , x ) = [ σ ( x ) , σ ( x )] − σ ([ x , x ]) = 0 . Similarly, we can show that φ ( x , x ) = φ ( x , x ) = φ ( x , x ) = φ ( x , x ) = 0 . Therefore, φ = ∆ .Now, by Lemma 2.2.2, we have θ = (∆ , ω ) ∈ Z ( L, F ) and K ∼ = L θ . (cid:4) Note that by Theorem 2.3.2, there are eight non-isomorphic restricted Lie algebra structures on L given by the following p -maps:I.1 Trivial p -map;I.2 x [ p ]1 = x ;I.3 x [ p ]1 = x ;I.4 x [ p ]1 = x , x [ p ]2 = x ;I.5 x [ p ]3 = x ;I.6 x [ p ]3 = x , x [ p ]2 = x ;I.7 x [ p ]4 = x ;I.8 x [ p ]4 = x , x [ p ]2 = x .We make L into a restricted Lie algebra by equipping it with each of the above p -maps. Then, ineach case, we find all possible orbit representatives of the form (∆ , ω ) under the action of Aut p ( L )on H ( L, F ). By Lemma 5.1.1, we do get all possible p -maps on K with the property that x [ p ]4 = 0 .Consider case I.2 where the p -map of L is x [ p ]1 = x . Let [( φ, ω )] ∈ H ( L, F ). Then we musthave φ ( x, y [ p ] ) = 0 , for all x, y ∈ L , where φ = a ∆ + b ∆ + c ∆ + d ∆ , for some a, b, c, d ∈ F .Since, L [ p ] = h x i we get φ ( x, x ) = 0 , for all x ∈ L . Therefore, φ ( x , x ) = 0 which implies that a = 0 . Since φ = ∆ gives us L , , we deduce by Lemma 5.1.1 that L , cannot be constructed inthis case. Similarly, we can show that in cases I.4, I.6, I.7, and I.8 we also get a = 0 . In the follwoingsubsections, we consider the remaining cases. HAPTER 5. RESTRICTION MAPS ON L , L , trivial p -map) First, we find a basis for Z ( L, F ). Let ( φ, ω ) = ( a ∆ + b ∆ + c ∆ + d ∆ + e ∆ + f ∆ , αf + βf + γf + δf ) ∈ Z ( L, F ). Then we must have δ φ ( x, y, z ) = 0 and φ ( x, y [ p ] ) = 0 , for all x, y, z ∈ L .Therefore,0 = ( δ φ )( x , x , x ) = φ ([ x , x ] , x ) + φ ([ x , x ] , x ) + φ ([ x , x ] , x ) = φ ( x , x ) . Thus, we get f = 0 . Since the p -map is trivial, φ ( x, y [ p ] ) = φ ( x, 0) = 0 , for all x, y ∈ L . Therefore,a basis for Z ( L, F ) is as follows:(∆ , , (∆ , , (∆ , , (∆ , , (∆ , , (0 , f ) , (0 , f ) , (0 , f ) , (0 , f ) . Next, we find a basis for B ( L, F ). Let ( φ, ω ) ∈ B ( L, F ). Since B ( L, F ) ⊆ Z ( L, F ), we have( φ, ω ) = ( a ∆ + b ∆ + c ∆ + d ∆ + e ∆ , αf + βf + γf + δf ). So, there exists a linear map ψ : L → F such that δ ψ ( x, y ) = φ ( x, y ) and ˜ ψ ( x ) = ω ( x ), for all x, y ∈ L . So, we have b = φ ( x , x ) = δ ψ ( x , x ) = ψ ([ x , x ]) = 0 . Similarly, we can show that c = d = e = 0 . Also, we have α = ω ( x ) = ˜ ψ ( x ) = ψ ( x [ p ]1 ) = 0 . Similarly, we can show that β = γ = δ = 0 . Therefore, ( φ, ω ) = ( a ∆ , 0) and hence B ( L, F ) = h (∆ , i F . We deduce that a basis for H ( L, F ) is as follows:[(∆ , , [(∆ , , [(∆ , , [(∆ , , [(0 , f )] , [(0 , f )] , [(0 , f )] , [(0 , f )] . Let [( φ, ω )] ∈ H ( L, F ). Then, we have φ = a ∆ + b ∆ + c ∆ + d ∆ , for some a, b, c, d ∈ F .Suppose that Aφ = a ′ ∆ + b ′ ∆ + c ′ ∆ + d ′ ∆ , for some a ′ , b ′ , c ′ , d ′ ∈ F . We can verify that theaction of Aut p ( L ) on the set of φ ’s in the matrix form is as follows: a ′ b ′ c ′ d ′ = ra ra a a a a a a a a ra ra a a a a a a a a abcd . (5.1)The orbit with representative of this action gives us L , .Also, we have ω = αf + βf + γf + δf , for some α, β, γ, δ ∈ F . Suppose that Aω = α ′ f + β ′ f + γ ′ f + δ ′ f , for some α ′ , β ′ , γ ′ , δ ′ ∈ F . Then we have Aω ( x ) = ω ( Ax ) = ω ( a x + a x + a x + a x ) = a p α + a p β + a p γ + a p δ ; Aω ( x ) = a p α + a p β + a p γ + a p δ ; Aω ( x ) = r p γ ; Aω ( x ) = a p γ + a p δ. HAPTER 5. RESTRICTION MAPS ON L , α ′ β ′ γ ′ δ ′ = a p a p a p a p a p a p a p a p r p 00 0 a p a p αβγδ . (5.2)Thus, we can write Equations (5.1) and (5.2) together as follows: h ra ra a a a a a a a a ra ra a a a a a a a a , a p a p a p a p a p a p a p a p r p 00 0 a p a p ih abcd , αβγδ i = h a ′ b ′ c ′ d ′ , α ′ β ′ γ ′ δ ′ i . Now we find the representatives of the orbits of the action of Aut( L ) on the set of ω ’s such that theorbit represented by is preserved under the action of Aut( L ) on the set of φ ’s.Let ν = αβγδ ∈ F . If ν = , then { ν } is clearly an Aut p ( L )-orbit. Let ν = 0 . Supposethat γ = 0 . Then h , − α/γ 00 1 − β/γ 00 0 1 00 0 0 1 ih , αβγδ i = h , γδ i , and h γ /p γ − /p 00 0 0 γ − /p , γ γ − /γ 00 0 0 1 ih , γδ i = h , δ i . Next, if δ = 0 , then h /δ ) /p /δ ) /p , /δ ih , δ i = h , i . If δ = 0 , then we have h , i . Next, if γ = 0 , but δ = 0 , then HAPTER 5. RESTRICTION MAPS ON L , h , − α/δ − β/δ ih , αβ δ i = h , δ i , and h /δ ) /p /δ ) /p , /δ ih , δ i = h , i . Next, if γ = δ = 0 , but β = 0 , then h − α/β ) /p 00 1 0 ( − α/β ) /p , − α/β ih , αβ i = h , β i . Finally, if γ = δ = β = 0 , but α = 0 , then h α − /p α /p 00 0 0 α /p , /α α α 00 0 0 1 ih , α i = h , i . Thus the following elements are Aut( L )-orbit representatives: , , , , β , . Therefore, the corresponding restricted Lie algebra structures are as follows: (Note that we needto switch x and x .) K = h x , . . . , x | [ x , x ] = x , [ x , x ] = x i ; K = h x , . . . , x | [ x , x ] = x , [ x , x ] = x , x [ p ]3 = x , x [ p ]5 = x i ; K = h x , . . . , x | [ x , x ] = x , [ x , x ] = x , x [ p ]3 = x i ; K ( β ) = h x , . . . , x | [ x , x ] = x , [ x , x ] = x , x [ p ]2 = βx i ; K = h x , . . . , x | [ x , x ] = x , [ x , x ] = x , x [ p ]1 = x i ; K = h x , . . . , x | [ x , x ] = x , [ x , x ] = x , x [ p ]5 = x i . Lemma 5.1.2 We have K ( β ) ∼ = K ( β ) if and only if β β − ∈ ( F ∗ ) . Proof. Suppose that f = ( a ij ) : K ( β ) → K ( β ) is an isomorphism. Then, we have f ( x [ p ]2 ) = f ( x ) [ p ] which in turn implies that β a a x = β a p x . Hence, β /β = a p − a − ∈ ( F ∗ ) . Toprove the converse, suppose that β /β = ǫ ∈ ( F ∗ ) . Then the following is an isomorphism from HAPTER 5. RESTRICTION MAPS ON L , K ( β ) to K ( β ): ǫ − /p ǫ /p ǫ /p . (cid:4) ( L, x [ p ]1 = x ) First, we find a basis for Z ( L, F ). Let ( φ, ω ) = ( a ∆ + b ∆ + c ∆ + d ∆ + e ∆ + f ∆ , αf + βf + γf + δf ) ∈ Z ( L, F ). Then we must have δ φ ( x, y, z ) = 0 and φ ( x, y [ p ] ) = 0 , for all x, y, z ∈ L .Therefore, we have0 = ( δ φ )( x , x , x ) = φ ([ x , x ] , x ) + φ ([ x , x ] , x ) + φ ([ x , x ] , x ) = φ ( x , x ) = f. Also, we have φ ( x, y [ p ] ) = 0 . Therefore, φ ( x, x ) = 0 , for all x ∈ L and hence φ ( x , x ) = φ ( x , x ) = φ ( x , x ) = 0 which implies that c = e = f = 0 . Therefore, Z ( L, F ) has a basis consisting of:(∆ , , (∆ , , (∆ , , (0 , f ) , (0 , f ) , (0 , f ) , (0 , f ) . Next, we find a basis for B ( L, F ). Let ( φ, ω ) ∈ B ( L, F ). Since B ( L, F ) ⊆ Z ( L, F ), we have( φ, ω ) = ( a ∆ + b ∆ + c ∆ , αf + βf + γf + δf ). Note that there exists a linear map ψ : L → F such that δ ψ ( x, y ) = φ ( x, y ) and ˜ ψ ( x ) = ω ( x ), for all x, y ∈ L . So, we have a = φ ( x , x ) = δ ψ ( x , x ) = ψ ([ x , x ]) = ψ ( x ) , and b = φ ( x , x ) = δ ψ ( x , x ) = ψ ([ x , x ]) = 0 . Similarly, we can show that c = 0 . Also, we have α = ω ( x ) = ˜ ψ ( x ) = ψ ( x [ p ]1 ) = ψ ( x ) , and β = ω ( x ) = ˜ ψ ( x ) = ψ ( x [ p ]2 ) = 0 . Similarly, we can show that γ = δ = 0 . Therefore, ( φ, ω ) = ( a ∆ , αf ) and hence B ( L, F ) = h (∆ , , (0 , f ) i F . We deduce that a basis for H ( L, F ) is as follows:[(∆ , , [(∆ , , [(0 , f )] , [(0 , f )] , [(0 , f )] . Note that the group Aut p ( L ) in this case consists of invertible matrices of the form a a a a a a r a a a a , where r = a a − a a , a = a = 0 , and a = a p . HAPTER 5. RESTRICTION MAPS ON L , φ, ω )] ∈ H ( L, F ). Then, we have φ = a ∆ + c ∆ , for some a, c ∈ F . Suppose that Aφ = a ′ ∆ + c ′ ∆ , for some a ′ , c ′ ∈ F . We determine a ′ , c ′ . Note that Aφ ( x , x ) = φ ( Ax , Ax ) = φ ( a x + a x + a x + a x , rx )= a ra + a rc ; and Aφ ( x , x ) = φ ( Ax , Ax ) = φ ( a x + a x + a x + a x , rx )= a ra + a rc. In the matrix form we can write this as (cid:18) a ′ c ′ (cid:19) = (cid:18) a r a r a r (cid:19) (cid:18) ac (cid:19) . (5.3)The orbit with representative (cid:18) (cid:19) of this action gives us L , .Note that we need to have Aφ ( x , x ) = Aφ ( x , x ) = Aφ ( x , x ) = 0 which implies that a a a + a a c = 0 ,a a a + a a c = 0 . Also, we have ω = βf + γf + δf , for some β, γ, δ ∈ F . Suppose that Aω = β ′ f + γ ′ f + δ ′ f , forsome β ′ , γ ′ , δ ′ ∈ F . We have Aω ( x ) = a p β + a p γ + a p δ ; Aω ( x ) = r p γ ; Aω ( x ) = a p γ + a p δ. In the matrix form we can write this as β ′ γ ′ δ ′ = a p a p a p r p 00 0 a p βγδ . (5.4)Thus, we can write Equations (5.3) and (5.4) together as follows: h r (cid:18) a a a (cid:19) , a p a p a p r p 00 0 a p ih (cid:18) ac (cid:19) , βγδ i = h (cid:18) a ′ c ′ (cid:19) , β ′ γ ′ δ ′ i . Now we find the representatives of the orbits of the action of Aut( L ) on the set of ω ’s such that theorbit represented by (cid:18) (cid:19) is preserved under the action of Aut( L ) on the set of ω ’s and a a a + a a c = 0 ,a a a + a a c = 0 . HAPTER 5. RESTRICTION MAPS ON L , ν = βγδ ∈ F . If ν = , then { ν } is clearly an Aut p ( L )-orbit. Suppose that δ = 0 . Then h (cid:18) (cid:19) , − β/δ ih (cid:18) (cid:19) , βγδ i = h (cid:18) (cid:19) , γδ i , and h δ /p δ − /p δ /p ! , δ /p δ /p 00 0 1 /δ ih (cid:18) (cid:19) , γδ i = h (cid:18) (cid:19) , δ /p γ i . So, if δ = 0 and γ = 0 , then we have γ .If δ = 0 but γ = 0 , then we have h (cid:18) (cid:19) , i . Next, if δ = 0 but γ = 0 , then h (cid:18) (cid:19) , − β/γ 00 1 00 0 1 ih (cid:18) (cid:19) , βγ i = h (cid:18) (cid:19) , γ i , and h (1 /γ ) /p (cid:18) γ /p γ − /p (cid:19) , (1 /γ ) /γ 00 0 γ p ih (cid:18) (cid:19) , γ i = h (cid:18) (cid:19) , i . Next, if δ = γ = 0 , then we have h (cid:18) (cid:19) , β i . Thus the following elements are Aut( L )-orbit representatives: , , , β , γ . Therefore, the corresponding restricted Lie algebra structures are as follows: (Note that we needto switch x and x . ) K = h x , . . . , x | [ x , x ] = x , [ x , x ] = x , x [ p ]1 = x i ; K ( β ) = h x , . . . , x | [ x , x ] = x , [ x , x ] = x , x [ p ]1 = x , x [ p ]2 = βx i ; K = h x , . . . , x | [ x , x ] = x , [ x , x ] = x , x [ p ]1 = x , x [ p ]3 = x i ; K = h x , . . . , x | [ x , x ] = x , [ x , x ] = x , x [ p ]1 = x , x [ p ]5 = x i ; K ( γ ) = h x , . . . , x | [ x , x ] = x , [ x , x ] = x , x [ p ]1 = x , x [ p ]3 = γx , x [ p ]5 = x i Lemma 5.1.3 The restricted Lie algebras K ( β ) and K ( β ) are isomorphic if and only if β β − ∈ ( F ∗ ) . HAPTER 5. RESTRICTION MAPS ON L , Proof. First assume that f = ( a ij ) : K ( β ) → K ( β ) is an isomorphism. Then f ( x [ p ]2 ) = f ( x ) [ p ] which implies that β a a x = β a p x . Hence, β /β = a p − a − ∈ ( F ∗ ) . To prove the converse,suppose that β /β = ǫ ∈ ( F ∗ ) . Then the following is an isomorphism from K ( β ) to K ( β ): ǫ − /p ǫ /p ǫ /p ǫ − . (cid:4) Lemma 5.1.4 The restricted Lie algebras K ( γ ) and K ( γ ) are isomorphic if and only if γ γ − = ǫ p ( p − p − , for some ǫ ∈ F ∗ . Proof. First assume that f = ( a ij ) : K ( γ ) → K ( γ ) is an isomorphism. It follows that γ /γ = a p − a p − , a a = a p , a = a p . The above Equations then imply that γ γ − = a p ( p − p − . To prove the converse, suppose that γ γ − = ǫ p ( p − p − , for some ǫ ∈ F ∗ . Then the following is an isomorphism from K ( γ ) to K ( γ ): ǫ ǫ p − ǫ p − ǫ p 00 0 0 0 ǫ p . (cid:4) ( L, x [ p ]3 = x ) First, we find a basis for Z ( L, F ). Let ( φ, ω ) = ( a ∆ + b ∆ + c ∆ + d ∆ + e ∆ + f ∆ , αf + βf + γf + δf ) ∈ Z ( L, F ). Then we must have δ φ ( x, y, z ) = 0 and φ ( x, y [ p ] ) = 0 , for all x, y, z ∈ L .Therefore, we have0 = ( δ φ )( x , x , x ) = φ ([ x , x ] , x ) + φ ([ x , x ] , x ) + φ ([ x , x ] , x ) = φ ( x , x ) = f. Also, we have φ ( x, y [ p ] ) = 0 . Therefore, φ ( x, x ) = 0 , for all x ∈ L and hence φ ( x , x ) = φ ( x , x ) = φ ( x , x ) = 0 which implies that c = e = f = 0 . Therefore, Z ( L, F ) has a basis consisting of:(∆ , , (∆ , , (∆ , , (0 , f ) , (0 , f ) , (0 , f ) , (0 , f ) . Next, we find a basis for B ( L, F ). Let ( φ, ω ) ∈ B ( L, F ). Since B ( L, F ) ⊆ Z ( L, F ), we have( φ, ω ) = ( a ∆ + b ∆ + c ∆ , αf + βf + γf + δf ). Note that there exists a linear map ψ : L → F such that δ ψ ( x, y ) = φ ( x, y ) and ˜ ψ ( x ) = ω ( x ), for all x, y ∈ L . So, we have a = φ ( x , x ) = δ ψ ( x , x ) = ψ ([ x , x ]) = ψ ( x ) , and b = φ ( x , x ) = δ ψ ( x , x ) = ψ ([ x , x ]) = 0 . HAPTER 5. RESTRICTION MAPS ON L , c = 0 . Also, we have γ = ω ( x ) = ˜ ψ ( x ) = ψ ( x [ p ]3 ) = ψ ( x ) , and α = ω ( x ) = ˜ ψ ( x ) = ψ ( x [ p ]1 ) = 0 . Similarly, we can show that β = δ = 0 . Therefore, ( φ, ω ) = ( a ∆ , γf ) and hence B ( L, F ) = h (∆ , , (0 , f ) i F . We deduce that a basis for H ( L, F ) is as follows:[(∆ , , [(∆ , , [(0 , f )] , [(0 , f )] , [(0 , f )] . Note that the group Aut p ( L ) in this case consists of invertible matrices of the form a a a a a a r a a a a , where r = a a − a a , a = a = a = 0 , and a = r p .Let [ θ ] = [( φ, ω )] ∈ H ( L, F ). Then φ = a ∆ + c ∆ . We can verify that the action of Aut p ( L )on the φ ’s in the matrix form is as follows: (cid:18) a ′ c ′ (cid:19) = (cid:18) a r a ra r a r (cid:19) (cid:18) ac (cid:19) . (5.5)The orbit with representative (cid:18) (cid:19) of this action gives us L , . Note that we need to have Aφ ( x , x ) = Aφ ( x , x ) = Aφ ( x , x ) = 0 which implies that a a a + a a c = 0 ,a a a + a a c = 0 . Let ω = αf + βf + δf , for some α, β, δ ∈ F . Suppose that Aω = α ′ f + β ′ f + δ ′ f , for some α ′ , β ′ , δ ′ ∈ F . We have Aω ( x ) = a p α + a p β + a p δ ; Aω ( x ) = a p α + a p β + a p δ ; Aω ( x ) = a p δ. In the matrix form we can write this as α ′ β ′ δ ′ = a p a p a p a p a p a p r p αβδ . (5.6)Thus, we can write Equations (5.5) and (5.6) together as follows: h r (cid:18) a a a a (cid:19) , a p a p a p a p a p a p r p ih (cid:18) ac (cid:19) αβδ i = h (cid:18) a ′ c ′ (cid:19) , α ′ β ′ δ ′ i . HAPTER 5. RESTRICTION MAPS ON L , p ( L ) on the set of ω ’s such that theorbit represented by (cid:18) (cid:19) is preserved under the action of Aut p ( L ) on the set of φ ’s. Note that wetake a = 0 . Then we have a a a + a a c = 0 ,a a a + a a c = 0 . Let ν = αβδ ∈ F . If ν = , then { ν } is clearly an Aut p ( L )-orbit. Suppose that δ = 0 . Then h (cid:18) (cid:19) , − α/δ − β/δ ih (cid:18) (cid:19) , αβδ i = h (cid:18) (cid:19) , δ i , and h δ − /p δ /p δ − /p ! , δ /p δ − /p 00 0 1 /δ ih (cid:18) (cid:19) , δ i = h (cid:18) (cid:19) , i . Next, if δ = 0 , but β = 0 , then h (cid:18) − α/β ) /p (cid:19) , − α/β 00 1 00 0 1 ih (cid:18) (cid:19) , αβ i = h (cid:18) (cid:19) , β i . Finally, if δ = β = 0 , but α = 0 , then h α /p (cid:18) α − /p α /p (cid:19) , /α α 00 0 α p ih (cid:18) (cid:19) , α i = h (cid:18) (cid:19) , i . Thus the following elements are Aut p ( L )-orbit representatives: , , , β . Therefore, the corresponding restricted Lie algebra structures are as follows: (Note that we needto switch x and x .) K = h x , . . . , x | [ x , x ] = x , [ x , x ] = x , x [ p ]3 = x i ; K = h x , . . . , x | [ x , x ] = x , [ x , x ] = x , x [ p ]1 = x , x [ p ]3 = x i ; K ( β ) = h x , . . . , x | [ x , x ] = x , [ x , x ] = x , x [ p ]2 = βx , x [ p ]3 = x i ; K = h x , . . . , x | [ x , x ] = x , [ x , x ] = x , x [ p ]3 = x , x [ p ]5 = x i where β ∈ F ∗ . Lemma 5.1.5 The restricted Lie algebras K ( β ) and K ( β ) are isomorphic if and only if β β − ∈ ( F ∗ ) . HAPTER 5. RESTRICTION MAPS ON L , Proof. First assume that f = ( a ij ) : K ( β ) → K ( β ) is an isomorphism. Then f ( x [ p ]2 ) = f ( x ) [ p ] which implies that β a a x = β a p x . Hence, β /β = a p − a − ∈ ( F ∗ ) . To prove the converse,suppose that β /β = ǫ ∈ ( F ∗ ) . Then the following is an isomorphism from K ( β ) to K ( β ): ǫ − /p ǫ /p ǫ /p ǫ . (cid:4) L = L , h x i In this section we find all non-isomorphic p-maps on L , such that x [ p ]5 = 0 . We let L = L , h x i ∼ = L , , where L , = h x , x , x , x | [ x , x ] = x , [ x , x ] = x i . The group Aut( L ) consists of invertiblematrices of the form a a a a a r a a a a a r , where r = a a = 0 . Lemma 5.2.1 Let K = L , and [ p ] : K → K be a p -map on K such that x [ p ]5 = 0 and let L = KM where M = h x i F . Then K ∼ = L θ where θ = (0 , ω ) ∈ Z ( L, M ) . Proof. Let π : K → L be the projection map. We have the exact sequence0 → M → K → L → . Let σ : L → K such that x i x i , 1 i σ is an injective linear map and πσ = 1 L . Now,we define φ : L × L → M by φ ( x i , x j ) = [ σ ( x i ) , σ ( x j )] − σ ([ x i , x j ]), 1 i, j ω : L → M by ω ( x ) = σ ( x ) [ p ] − σ ( x [ p ] ). Note that φ ( x , x ) = [ σ ( x ) , σ ( x )] − σ ([ x , x ]) = 0; φ ( x , x ) = [ σ ( x ) , σ ( x )] − σ ([ x , x ]) = 0 . Similarly, we can show that φ ( x , x ) = φ ( x , x ) = φ ( x , x ) = φ ( x , x ) = 0 . Therefore, φ = 0 .Now, by Lemma 2.2.2, we have θ = (0 , ω ) ∈ Z ( L, M ) and K ∼ = L θ . (cid:4) Note that by Theorem 2.3.2, there are four non-isomorphic restricted Lie algebra structures on L given by the following p -maps:II.1 Trivial p -map; HAPTER 5. RESTRICTION MAPS ON L , x [ p ]1 = x ;II.3 x [ p ]2 = ξx ;II.4 x [ p ]3 = x .In the following subsections, we make L into a restricted Lie algebra by equipping it with eachof the above p -maps. Then, in each case, we find all possible orbit representatives of the form (0 , ω )under the action of Aut p ( L ) on H ( L, F ). By Lemma 5.2.1, we do get all possible p -maps on K with the property that x [ p ]5 = 0 . L , trivial p -map) First, we find a basis for Z ( L, F ). Let ( φ, ω ) = ( a ∆ + b ∆ + c ∆ + d ∆ + e ∆ + f ∆ , αf + βf + γf + δf ) ∈ Z ( L, F ). Then we must have δ φ ( x, y, z ) = 0 and φ ( x, y [ p ] ) = 0 , for all x, y, z ∈ L .Therefore,0 = ( δ φ )( x , x , x ) = φ ([ x , x ] , x ) + φ ([ x , x ] , x ) + φ ([ x , x ] , x ) = φ ( x , x ) , and0 = ( δ φ )( x , x , x ) = φ ([ x , x ] , x ) + φ ([ x , x ] , x ) + φ ([ x , x ] , x ) = φ ( x , x ) . Thus, we get f = e = 0 . Since the p -map is trivial, φ ( x, y [ p ] ) = φ ( x, 0) = 0 , for all x, y ∈ L .Therefore, a basis for Z ( L, F ) is as follows:(∆ , , (∆ , , (∆ , , (∆ , , (0 , f ) , (0 , f ) , (0 , f ) , (0 , f ) . Next, we find a basis for B ( L, F ). Let ( φ, ω ) ∈ B ( L, F ). Since B ( L, F ) ⊆ Z ( L, F ), we have( φ, ω ) = ( a ∆ + b ∆ + c ∆ + d ∆ , αf + βf + γf + δf ). So, there exists a linear map ψ : L → F such that δ ψ ( x, y ) = φ ( x, y ) and ˜ ψ ( x ) = ω ( x ), for all x, y ∈ L . So, we have a = φ ( x , x ) = δ ψ ( x , x ) = ψ ([ x , x ]) = ψ ( x ) , and b = φ ( x , x ) = δ ψ ( x , x ) = ψ ([ x , x ]) = ψ ( x ) , and c = φ ( x , x ) = δ ψ ( x , x ) = ψ ([ x , x ]) = 0Similarly, we can show that d = 0 . Also, we have α = ω ( x ) = ˜ ψ ( x ) = ψ ( x [ p ]1 ) = 0 . Similarly, we can show that β = γ = δ = 0 . Therefore, ( φ, ω ) = ( a ∆ + b ∆ , 0) and hence B ( L, F ) = h (∆ , , (∆ , i F . We deduce that a basis for H ( L, F ) is as follows:[(∆ , , [(∆ , , [(0 , f )] , [(0 , f )] , [(0 , f )] , [(0 , f )] . Let [ θ ] = [( φ, ω )] ∈ H ( L, F ). Since we want L θ and L , to be isomorphic as Lie algebras, we shouldhave φ = 0 . Since 0 is preserved under Aut p ( L ), it is enough to find Aut p ( L )-representatives of the ω ’s.Let ω = αf + βf + γf + δf , for some α, β, γ, δ ∈ F . Suppose that Aω = α ′ f + β ′ f + γ ′ f + δ ′ f ,for some α ′ , β ′ , γ ′ , δ ′ ∈ F . We can easily verify that the action of Aut p ( L ) on the ω ’s in the matrixform is as follows: α ′ β ′ γ ′ δ ′ = a p a p a p a p a p a p a p r p a p a p a p r p αβγδ . HAPTER 5. RESTRICTION MAPS ON L , ν = αβγδ ∈ F . If ν = , then { ν } is clearly an Aut p ( L )-orbit. Suppose that δ = 0 . Then − α/δ − β/δ αβγδ = γδ , and − γ/δ γ /δ − γ/δ γδ = δ , and /δ /δ 00 0 0 1 /δ δ = . Next, if δ = 0 , but γ = 0 , then − α/γ 00 1 − β/γ 00 0 1 − β/γ αβγ = γ , and (1 /γ ) γ /γ 00 0 0 (1 /γ ) γ = . Next, if γ = δ = 0 , but β = 0 , then − α/β αβ = β , and (1 /β ) − /β /β ) − 00 0 0 (1 /β ) − β = . Finally, if δ = γ = β = 0 , but α = 0 , then /α /α 00 0 0 (1 /α ) α = . HAPTER 5. RESTRICTION MAPS ON L , p ( L )-orbit representatives: , , , , . Therefore, the corresponding restricted Lie algebra structures are as follows: K = h x , . . . , x | [ x , x ] = x , [ x , x ] = x i ; K = h x , . . . , x | [ x , x ] = x , [ x , x ] = x , x [ p ]1 = x i ; K = h x , . . . , x | [ x , x ] = x , [ x , x ] = x , x [ p ]2 = x i ; K = h x , . . . , x | [ x , x ] = x , [ x , x ] = x , x [ p ]3 = x i ; K = h x , . . . , x | [ x , x ] = x , [ x , x ] = x , x [ p ]4 = x i . ( L, x [ p ]1 = x ) First, we find a basis for Z ( L, F ). Let ( φ, ω ) = ( a ∆ + b ∆ + c ∆ + d ∆ + e ∆ + f ∆ , αf + βf + γf + δf ) ∈ Z ( L, F ). Then we must have δ φ ( x, y, z ) = 0 and φ ( x, y [ p ] ) = 0 , for all x, y, z ∈ L .Therefore, we have0 = ( δ φ )( x , x , x ) = φ ([ x , x ] , x ) + φ ([ x , x ] , x ) + φ ([ x , x ] , x ) = φ ( x , x ) , and0 = ( δ φ )( x , x , x ) = φ ([ x , x ] , x ) + φ ([ x , x ] , x ) + φ ([ x , x ] , x ) = φ ( x , x ) . Thus, we get f = e = 0 . Also, we have φ ( x, y [ p ] ) = 0 . Therefore, φ ( x, x ) = 0 , for all x ∈ L andhence φ ( x , x ) = φ ( x , x ) = φ ( x , x ) = 0 which implies that c = e = f = 0 . Therefore, Z ( L, F )has a basis consisting of:(∆ , , (∆ , , (∆ , , (0 , f ) , (0 , f ) , (0 , f ) , (0 , f ) . Next, we find a basis for B ( L, F ). Let ( φ, ω ) ∈ B ( L, F ). Since B ( L, F ) ⊆ Z ( L, F ), we have( φ, ω ) = ( a ∆ + b ∆ + c ∆ , αf + βf + γf + δf ). Note that there exists a linear map ψ : L → F such that δ ψ ( x, y ) = φ ( x, y ) and ˜ ψ ( x ) = ω ( x ), for all x, y ∈ L . So, we have a = φ ( x , x ) = δ ψ ( x , x ) = ψ ([ x , x ]) = ψ ( x ) , and b = φ ( x , x ) = δ ψ ( x , x ) = ψ ([ x , x ]) = ψ ( x ) , and c = φ ( x , x ) = δ ψ ( x , x ) = ψ ([ x , x ]) = 0 . Also, we have α = ω ( x ) = ˜ ψ ( x ) = ψ ( x [ p ]1 ) = ψ ( x ) , and β = ω ( x ) = ˜ ψ ( x ) = ψ ( x [ p ]2 ) = 0 . Similarly, we can show that γ = δ = 0 . Note that ψ ( x ) = b = α . Therefore, ( φ, ω ) = ( a ∆ + b ∆ , bf ) and hence B ( L, F ) = h (∆ , f ) , (∆ , i F . Note that n [(∆ , , [(∆ , , [(0 , f )] , [(0 , f )] , [(0 , f )] , [(0 , f )] o HAPTER 5. RESTRICTION MAPS ON L , H ( L, F ). Since [(∆ , , f )] = [(∆ , f )] = [0], then [(0 , f )] is an scalar multiple of[(∆ , H ( L, F ). Note that dim H = dim Z − dim B = 5 . Therefore, n [(∆ , , [(∆ , , [(0 , f )] , [(0 , f )] , [(0 , f )] o forms a basis for H ( L, F ).Note that he group Aut p ( L ) in this case consists of invertible matrices of the form a a a a a r a a a a a r , where r = a a = 0 and a r = a p .Let [ θ ] = [( φ, ω )] ∈ H ( L, F ). As in Section 5.2.1, it is enough to find Aut p ( L )-representatives ofthe ω ’s.Let ω = βf + γf + δf , for some β, γ, δ ∈ F . Suppose that Aω = β ′ f + γ ′ f + δ ′ f , for some β ′ , γ ′ , δ ′ ∈ F . We have Aω ( x ) = a p β + a p γ + a p δ ; Aω ( x ) = r p γ + a p a p δ ; Aω ( x ) = a p r p δ. In the matrix form we can write this as β ′ γ ′ δ ′ = a p a p a p r p a p a p a p r p βγδ . Note that we need to have Aω ( x ) = 0 which implies that a p β + a p γ + a p δ = 0 . Now we findthe representatives of the orbits of this action. Note that we take a = a = a = 0 . Therefore, a p β + a p γ + a p δ = 0 . Let ν = βγδ ∈ F . If ν = , then { ν } is clearly an Aut p ( L )-orbit.Suppose that δ = 0 . Then − β/δ βγδ = γδ , and − γ/δ γ /δ − γ/δ γδ = δ , and (1 /δ ) p − p /δ ) p − p 00 0 1 /δ δ = . Next, if δ = 0 , but γ = 0 , then − β/γ 00 1 − β/γ βγ = γ . HAPTER 5. RESTRICTION MAPS ON L , γ = δ = 0 , but β = 0 , then we have β . Thus the following elements are Aut( L )-orbit representatives: , β , γ , . Therefore, the corresponding restricted Lie algebra structures are as follows: K = h x , . . . , x | [ x , x ] = x , [ x , x ] = x , x [ p ]1 = x i ; K ( β ) = h x , . . . , x | [ x , x ] = x , [ x , x ] = x , x [ p ]1 = x , x [ p ]2 = βx i ; K ( γ ) = h x , . . . , x | [ x , x ] = x , [ x , x ] = x , x [ p ]1 = x , x [ p ]3 = γx i ; K = h x , . . . , x | [ x , x ] = x , [ x , x ] = x , x [ p ]1 = x , x [ p ]4 = x i . Lemma 5.2.2 We have K ( β ) ∼ = h x , . . . , x | [ x , x ] = x , [ x , x ] = x , x [ p ]1 = x , x [ p ]2 = x i , forevery β ∈ F ∗ . Proof. The following automorphism of L , gives us the desired isomorphism: /β . (cid:4) Lemma 5.2.3 We have K ( γ ) ∼ = h x , . . . , x | [ x , x ] = x , [ x , x ] = x , x [ p ]1 = x , x [ p ]3 = x i , forfor every γ ∈ F ∗ . Proof. The following automorphism of L , gives us the desired isomorphism: /γ . (cid:4) ( L, x [ p ]2 = ξx ) First, we find a basis for Z ( L, F ). Let ( φ, ω ) = ( a ∆ + b ∆ + c ∆ + d ∆ + e ∆ + f ∆ , αf + βf + γf + δf ) ∈ Z ( L, F ). Then we must have δ φ ( x, y, z ) = 0 and φ ( x, y [ p ] ) = 0 , for all x, y, z ∈ L .Therefore, we have0 = ( δ φ )( x , x , x ) = φ ([ x , x ] , x ) + φ ([ x , x ] , x ) + φ ([ x , x ] , x ) = φ ( x , x ) , and0 = ( δ φ )( x , x , x ) = φ ([ x , x ] , x ) + φ ([ x , x ] , x ) + φ ([ x , x ] , x ) = φ ( x , x ) . HAPTER 5. RESTRICTION MAPS ON L , f = e = 0 . Also, we have φ ( x, y [ p ] ) = 0 . Therefore, φ ( x, x ) = 0 , for all x ∈ L andhence φ ( x , x ) = φ ( x , x ) = φ ( x , x ) = 0 which implies that c = e = f = 0 . Therefore, Z ( L, F )has a basis consisting of:(∆ , , (∆ , , (∆ , , (0 , f ) , (0 , f ) , (0 , f ) , (0 , f ) . Next, we find a basis for B ( L, F ). Let ( φ, ω ) ∈ B ( L, F ). Since B ( L, F ) ⊆ Z ( L, F ), we have( φ, ω ) = ( a ∆ + b ∆ + c ∆ , αf + βf + γf + δf ). Note that there exists a linear map ψ : L → F such that δ ψ ( x, y ) = φ ( x, y ) and ˜ ψ ( x ) = ω ( x ), for all x, y ∈ L . So, we have a = φ ( x , x ) = δ ψ ( x , x ) = ψ ([ x , x ]) = ψ ( x ) , and b = φ ( x , x ) = δ ψ ( x , x ) = ψ ([ x , x ]) = ψ ( x ) , and c = φ ( x , x ) = δ ψ ( x , x ) = ψ ([ x , x ]) = 0 . Also, we have β = ω ( x ) = ˜ ψ ( x ) = ψ ( x [ p ]2 ) = ξψ ( x ) , and α = ω ( x ) = ˜ ψ ( x ) = ψ ( x [ p ]1 ) = 0 . Similarly, we can show that γ = δ = 0 . Note that ψ ( x ) = b = βξ − . Therefore, ( φ, ω ) =( a ∆ + b ∆ , bξf ) and hence B ( L, F ) = h (∆ , ξf ) , (∆ , i F . Note that n [(∆ , , [(∆ , , [(0 , f )] , [(0 , f )] , [(0 , f )] , [(0 , f )] o spans H ( L, F ). Since [(∆ , ξ [(0 , f )] = [(∆ , ξf )] = [0], then [(0 , f )] is an scalar multiple of[(∆ , H ( L, F ). Note that dim H = dim Z − dim B = 5 . Therefore, n [(∆ , , [(∆ , , [(0 , f )] , [(0 , f )] , [(0 , f )] o forms a basis for H ( L, F ).Note that he group Aut p ( L ) in this case consists of invertible matrices of the form a a a a a r a a a a a r , where r = a a = 0 , a = 0 and a r = a p .Let [ θ ] = [( φ, ω )] ∈ H ( L, F ). As in Section 5.2.1, it is enough to find Aut p ( L )-representatives ofthe ω ’s.Let ω = αf + γf + δf , for some α, γ, δ ∈ F . Suppose that Aω = α ′ f + γ ′ f + δ ′ f , for some α ′ , γ ′ , δ ′ ∈ F . We have Aω ( x ) = a p α + a p γ + a p δ ; Aω ( x ) = r p γ + a p a p δ ; Aω ( x ) = a p r p δ. HAPTER 5. RESTRICTION MAPS ON L , α ′ γ ′ δ ′ = a p a p a p r p a p a p a p r p αγδ . Now we find the representatives of the orbits of this action. Note that we need to have Aω ( x ) = 0which implies that a p γ + a p δ = 0 .Let ν = αγδ ∈ F . If ν = , then { ν } is clearly an Aut p ( L )-orbit. Suppose that δ = 0 .Then − α/δ − γ/δ αγδ = δ . Next, if δ = 0 , but γ = 0 , then − α/γ 00 1 00 0 1 αγ = γ . Finally, if γ = δ = 0 , but α = 0 , then we have α . Thus the following elements are Aut( L )-orbitrepresentatives: , α , γ , δ . Therefore, the corresponding restricted Lie algebra structures are as follows: K ( ξ ) = h x , . . . , x | [ x , x ] = x , [ x , x ] = x , x [ p ]2 = ξx i ; K ( α, ξ ) = h x , . . . , x | [ x , x ] = x , [ x , x ] = x , x [ p ]1 = αx , x [ p ]2 = ξx i ; K ( γ, ξ ) = h x , . . . , x | [ x , x ] = x , [ x , x ] = x , x [ p ]2 = ξx , x [ p ]3 = γx i ; K ( δ, ξ ) = h x , . . . , x | [ x , x ] = x , [ x , x ] = x , x [ p ]2 = ξx , x [ p ]4 = δx i where α, γ, δ, ξ ∈ F ∗ . Lemma 5.2.4 We have K ( α, ξ ) ∼ = K (1 , ξ ) , for every α, ξ ∈ F ∗ . Proof. The following automorphism of L , gives us the desired isomorphism: /α . (cid:4) HAPTER 5. RESTRICTION MAPS ON L , Lemma 5.2.5 We have K ( γ, ξ ) ∼ = K (1 , ξ ) , for every γ, ξ ∈ F ∗ . Proof. The following automorphism of L , gives us the desired isomorphism: /γ . (cid:4) Lemma 5.2.6 We have K ( δ, ξ ) ∼ = K (1 , ξ ) , for every δ, ξ ∈ F ∗ . Proof. The following automorphism of L , gives us the desired isomorphism: /δ . (cid:4) Therefore, the corresponding restricted Lie algebra structures are as follows: K ( ξ ) = h x , . . . , x | [ x , x ] = x , [ x , x ] = x , x [ p ]2 = ξx i ; K ( ξ ) = h x , . . . , x | [ x , x ] = x , [ x , x ] = x , x [ p ]1 = x , x [ p ]2 = ξx i ; K ( ξ ) = h x , . . . , x | [ x , x ] = x , [ x , x ] = x , x [ p ]2 = ξx , x [ p ]3 = x i ; K ( ξ ) = h x , . . . , x | [ x , x ] = x , [ x , x ] = x , x [ p ]2 = ξx , x [ p ]4 = x i where ξ ∈ F ∗ . Note that K ( ξ ) is identical to K ( β ), K ( ξ ) is identical to K ( β ), and K ( ξ ) isidentical to K ( β ). Lemma 5.2.7 We have K ( ξ ) ∼ = K ( ξ ) if and only if ξ ξ − ∈ ( F ∗ ) . Proof. If f : K ( ξ ) → K ( ξ ) is an isomorphism, then f ( x [ p ]2 ) = f ( x ) [ p ] which implies that ξ a a x = a p ξ x + a p x . We deduce that ξ ξ − = a a − p ∈ ( F ∗ ) . Conversely, assume that ξ ξ − = ǫ with some ǫ ∈ F ∗ . Then the following automorphism of L , ǫ ǫ ǫ 00 0 0 0 ǫ p is an isomorphism between K ( ξ ) and K ( ξ ). (cid:4) HAPTER 5. RESTRICTION MAPS ON L , ( L, x [ p ]3 = x ) First, we find a basis for Z ( L, F ). Let ( φ, ω ) = ( a ∆ + b ∆ + c ∆ + d ∆ + e ∆ + f ∆ , αf + βf + γf + δf ) ∈ Z ( L, F ). Then we must have δ φ ( x, y, z ) = 0 and φ ( x, y [ p ] ) = 0 , for all x, y, z ∈ L .Therefore, we have0 = ( δ φ )( x , x , x ) = φ ([ x , x ] , x ) + φ ([ x , x ] , x ) + φ ([ x , x ] , x ) = φ ( x , x ) , and0 = ( δ φ )( x , x , x ) = φ ([ x , x ] , x ) + φ ([ x , x ] , x ) + φ ([ x , x ] , x ) = φ ( x , x ) . Thus, we get f = e = 0 . Also, we have φ ( x, y [ p ] ) = 0 . Therefore, φ ( x, x ) = 0 , for all x ∈ L andhence φ ( x , x ) = φ ( x , x ) = φ ( x , x ) = 0 which implies that c = e = f = 0 . Therefore, Z ( L, F )has a basis consisting of:(∆ , , (∆ , , (∆ , , (0 , f ) , (0 , f ) , (0 , f ) , (0 , f ) . Next, we find a basis for B ( L, F ). Let ( φ, ω ) ∈ B ( L, F ). Since B ( L, F ) ⊆ Z ( L, F ), we have( φ, ω ) = ( a ∆ + b ∆ + c ∆ , αf + βf + γf + δf ). Note that there exists a linear map ψ : L → F such that δ ψ ( x, y ) = φ ( x, y ) and ˜ ψ ( x ) = ω ( x ), for all x, y ∈ L . So, we have a = φ ( x , x ) = δ ψ ( x , x ) = ψ ([ x , x ]) = ψ ( x ) , and b = φ ( x , x ) = δ ψ ( x , x ) = ψ ([ x , x ]) = ψ ( x ) , and c = φ ( x , x ) = δ ψ ( x , x ) = ψ ([ x , x ]) = 0 . Also, we have γ = ω ( x ) = ˜ ψ ( x ) = ψ ( x [ p ]3 ) = ψ ( x ) , and α = ω ( x ) = ˜ ψ ( x ) = ψ ( x [ p ]1 ) = 0 . Similarly, we can show that β = δ = 0 . Note that ψ ( x ) = b = γ . Therefore, ( φ, ω ) = ( a ∆ + b ∆ , bf ) and hence B ( L, F ) = h (∆ , f ) , (∆ , i F . Note that n [(∆ , , [(∆ , , [(0 , f )] , [(0 , f )] , [(0 , f )] , [(0 , f )] o spans H ( L, F ). Since [(∆ , , f )] = [(∆ , f )] = [0], then [(0 , f )] is an scalar multiple of[(∆ , H ( L, F ). Note that dim H = dim Z − dim B = 5 . Therefore, n [(∆ , , [(∆ , , [(0 , f )] , [(0 , f )] , [(0 , f )] o forms a basis for H ( L, F ).Note that he group Aut p ( L ) in this case consists of invertible matrices of the form a a a a a r a a a a a r , where r = a a = 0 , a = a = 0 and a r = r p . HAPTER 5. RESTRICTION MAPS ON L , θ ] = [( φ, ω )] ∈ H ( L, F ). As in Section 5.2.1, it is enough to find Aut p ( L )-representatives ofthe ω ’s. Let ω = αf + βf + δf , for some α, β, δ ∈ F . Suppose that Aω = α ′ f + β ′ f + δ ′ f , forsome α ′ , β ′ , δ ′ ∈ F . We have Aω ( x ) = a p α + a p β + a p δ ; Aω ( x ) = a p β + a p δ ; Aω ( x ) = a p r p δ. In the matrix form we can write this as α ′ β ′ δ ′ = a p a p a p a p a p a p r p αβδ . Note that we need to have Aω ( x ) = 0 which implies that a p a p δ = 0 . Now we find the repre-sentatives of the orbits of this action. Note that we take a = 0 . Therefore, a p a p δ = 0 . Let ν = αβδ ∈ F . If ν = , then { ν } is clearly an Aut p ( L )-orbit. Suppose that δ = 0 . Then − α/δ − β/δ αβδ = δ , and (1 /δ ) p − p /δ ) − pp 00 0 1 /δ δ = . Next, if δ = 0 , but β = 0 , then − α/β 00 1 00 0 1 αβ = β . Finally, if β = δ = 0 , but α = 0 , then we have α . Thus the following elements are Aut( L )-orbit representatives: , α , β , . Therefore, the corresponding restricted Lie algebra structures are as follows: K = h x , . . . , x | [ x , x ] = x , [ x , x ] = x , x [ p ]3 = x i ; K ( α ) = h x , . . . , x | [ x , x ] = x , [ x , x ] = x , x [ p ]1 = αx , x [ p ]3 = x i ; K ( β ) = h x , . . . , x | [ x , x ] = x , [ x , x ] = x , x [ p ]2 = βx , x [ p ]3 = x i ; K = h x , . . . , x | [ x , x ] = x , [ x , x ] = x , x [ p ]3 = x , x [ p ]4 = x i . HAPTER 5. RESTRICTION MAPS ON L , Lemma 5.2.8 We have K ( α ) ∼ = h x , . . . , x | [ x , x ] = x , [ x , x ] = x , x [ p ]1 = x , x [ p ]3 = x i , forfor every α ∈ F ∗ . Proof. The following automorphism of L , gives us the desired isomorphism: /α . (cid:4) Lemma 5.2.9 We have K ( β ) ∼ = h x , . . . , x | [ x , x ] = x , [ x , x ] = x , x [ p ]2 = x , x [ p ]3 = x i , forfor every β ∈ F ∗ . Proof. The following automorphism of L , gives us the desired isomorphism: /β . (cid:4) Therefore, the corresponding restricted Lie algebra structures are as follows: K = h x , . . . , x | [ x , x ] = x , [ x , x ] = x , x [ p ]3 = x i ; K = h x , . . . , x | [ x , x ] = x , [ x , x ] = x , x [ p ]1 = x , x [ p ]3 = x i ; K = h x , . . . , x | [ x , x ] = x , [ x , x ] = x , x [ p ]2 = x , x [ p ]3 = x i ; K = h x , . . . , x | [ x , x ] = x , [ x , x ] = x , x [ p ]3 = x , x [ p ]4 = x i . HAPTER 5. RESTRICTION MAPS ON L , HAPTER 5. RESTRICTION MAPS ON L , L , Therefore, the list of all (possibly redundant) restricted Lie algebra structures on L , is as follows: K = h x , . . . , x | [ x , x ] = x , [ x , x ] = x i ; K = h x , . . . , x | [ x , x ] = x , [ x , x ] = x , x [ p ]3 = x , x [ p ]5 = x i ; K = h x , . . . , x | [ x , x ] = x , [ x , x ] = x , x [ p ]3 = x i ; K ( β ) = h x , . . . , x | [ x , x ] = x , [ x , x ] = x , x [ p ]2 = βx i ; K = h x , . . . , x | [ x , x ] = x , [ x , x ] = x , x [ p ]1 = x i ; K = h x , . . . , x | [ x , x ] = x , [ x , x ] = x , x [ p ]5 = x i ; K = h x , . . . , x | [ x , x ] = x , [ x , x ] = x , x [ p ]1 = x i ; K ( β ) = h x , . . . , x | [ x , x ] = x , [ x , x ] = x , x [ p ]1 = x , x [ p ]2 = βx i ; K = h x , . . . , x | [ x , x ] = x , [ x , x ] = x , x [ p ]1 = x , x [ p ]3 = x i ; K = h x , . . . , x | [ x , x ] = x , [ x , x ] = x , x [ p ]1 = x , x [ p ]5 = x i ; K ( γ ) = h x , . . . , x | [ x , x ] = x , [ x , x ] = x , x [ p ]1 = x , x [ p ]3 = γx , x [ p ]5 = x i ; K = h x , . . . , x | [ x , x ] = x , [ x , x ] = x , x [ p ]3 = x i ; K = h x , . . . , x | [ x , x ] = x , [ x , x ] = x , x [ p ]1 = x , x [ p ]3 = x i ; K ( β ) = h x , . . . , x | [ x , x ] = x , [ x , x ] = x , x [ p ]2 = βx , x [ p ]3 = x i ; K = h x , . . . , x | [ x , x ] = x , [ x , x ] = x , x [ p ]3 = x , x [ p ]5 = x i ; K = h x , . . . , x | [ x , x ] = x , [ x , x ] = x i ; K = h x , . . . , x | [ x , x ] = x , [ x , x ] = x , x [ p ]1 = x i ; K = h x , . . . , x | [ x , x ] = x , [ x , x ] = x , x [ p ]2 = x i ; K = h x , . . . , x | [ x , x ] = x , [ x , x ] = x , x [ p ]3 = x i ; K = h x , . . . , x | [ x , x ] = x , [ x , x ] = x , x [ p ]4 = x i ; K = h x , . . . , x | [ x , x ] = x , [ x , x ] = x , x [ p ]1 = x i ; K = h x , . . . , x | [ x , x ] = x , [ x , x ] = x , x [ p ]1 = x , x [ p ]2 = x i ; K = h x , . . . , x | [ x , x ] = x , [ x , x ] = x , x [ p ]1 = x , x [ p ]3 = x i ; K = h x , . . . , x | [ x , x ] = x , [ x , x ] = x , x [ p ]1 = x , x [ p ]4 = x i ; K ( ξ ) = h x , . . . , x | [ x , x ] = x , [ x , x ] = x , x [ p ]2 = ξx i ; K ( ξ ) = h x , . . . , x | [ x , x ] = x , [ x , x ] = x , x [ p ]1 = x , x [ p ]2 = ξx i ; K ( ξ ) = h x , . . . , x | [ x , x ] = x , [ x , x ] = x , x [ p ]2 = ξx , x [ p ]3 = x i ; K ( ξ ) = h x , . . . , x | [ x , x ] = x , [ x , x ] = x , x [ p ]2 = ξx , x [ p ]4 = x i ; K = h x , . . . , x | [ x , x ] = x , [ x , x ] = x , x [ p ]3 = x i ; K = h x , . . . , x | [ x , x ] = x , [ x , x ] = x , x [ p ]1 = x , x [ p ]3 = x i ; K = h x , . . . , x | [ x , x ] = x , [ x , x ] = x , x [ p ]2 = x , x [ p ]3 = x i ; K = h x , . . . , x | [ x , x ] = x , [ x , x ] = x , x [ p ]3 = x , x [ p ]4 = x i . HAPTER 5. RESTRICTION MAPS ON L , We can easily see that some of the algebras above are identical. Theorem 5.3.1 The list of all restricted Lie algebra structures on L , , up to isomorphism, is asfollows: L , = h x , . . . , x | [ x , x ] = x , [ x , x ] = x i ; L , = h x , . . . , x | [ x , x ] = x , [ x , x ] = x , x [ p ]3 = x , x [ p ]5 = x i ; L , = h x , . . . , x | [ x , x ] = x , [ x , x ] = x , x [ p ]3 = x i ; L , ( β ) = h x , . . . , x | [ x , x ] = x , [ x , x ] = x , x [ p ]2 = βx i ; L , = h x , . . . , x | [ x , x ] = x , [ x , x ] = x , x [ p ]1 = x i ; L , = h x , . . . , x | [ x , x ] = x , [ x , x ] = x , x [ p ]5 = x i ; L , = h x , . . . , x | [ x , x ] = x , [ x , x ] = x , x [ p ]1 = x i ; L , ( β ) = h x , . . . , x | [ x , x ] = x , [ x , x ] = x , x [ p ]1 = x , x [ p ]2 = βx i ; L , = h x , . . . , x | [ x , x ] = x , [ x , x ] = x , x [ p ]1 = x , x [ p ]3 = x i ; L , = h x , . . . , x | [ x , x ] = x , [ x , x ] = x , x [ p ]1 = x , x [ p ]5 = x i ; L , = h x , . . . , x | [ x , x ] = x , [ x , x ] = x , x [ p ]3 = x i ; L , = h x , . . . , x | [ x , x ] = x , [ x , x ] = x , x [ p ]1 = x , x [ p ]3 = x i ; L , ( β ) = h x , . . . , x | [ x , x ] = x , [ x , x ] = x , x [ p ]2 = βx , x [ p ]3 = x i ; L , = h x , . . . , x | [ x , x ] = x , [ x , x ] = x , x [ p ]3 = x , x [ p ]5 = x i ; L , = h x , . . . , x | [ x , x ] = x , [ x , x ] = x , x [ p ]2 = x i ; L , = h x , . . . , x | [ x , x ] = x , [ x , x ] = x , x [ p ]4 = x i ; L , = h x , . . . , x | [ x , x ] = x , [ x , x ] = x , x [ p ]1 = x , x [ p ]2 = x i ; L , = h x , . . . , x | [ x , x ] = x , [ x , x ] = x , x [ p ]1 = x , x [ p ]4 = x i ; L , ( ξ ) = h x , . . . , x | [ x , x ] = x , [ x , x ] = x , x [ p ]2 = ξx , x [ p ]4 = x i ; L , = h x , . . . , x | [ x , x ] = x , [ x , x ] = x , x [ p ]2 = x , x [ p ]3 = x i ; L , = h x , . . . , x | [ x , x ] = x , [ x , x ] = x , x [ p ]3 = x , x [ p ]4 = x i ; L , ( γ ) = h x , . . . , x | [ x , x ] = x , [ x , x ] = x , x [ p ]1 = x , x [ p ]3 = γx , x [ p ]5 = x i where β, ξ ∈ T . In the remaining of this section we establish that the algebras given in Theorem 5.3.1 are pairwisenon-isomorphic, thereby completing the proof of Theorem 5.3.1.It is clear that L , is not isomorphic to the other restricted Lie algebras. We claim that L , and L , are not isomorphic. Suppose to the contrary that there exists an isomorphism A : L , → L , . HAPTER 5. RESTRICTION MAPS ON L , A ( x [ p ]5 ) = A ( x ) [ p ] A ( x ) = ( a x + a x ) [ p ] a a x = 0 . Therefore, a = 0 or a = 0 which is a contradiction.Next, we claim that L , and L , ( β ) are not isomorphic. Suppose to the contrary that thereexists an isomorphism A : L , → L , ( β ). Then A ( x [ p ]5 ) = A ( x ) [ p ] A ( x ) = ( a x + a x ) [ p ] a a x = 0 . Therefore, a = 0 or a = 0 which is a contradiction.Next, we claim that L , and L , are not isomorphic. Suppose to the contrary that there existsan isomorphism A : L , → L , . Then A ( x [ p ]5 ) = A ( x ) [ p ] A ( x ) = ( a x + a x ) [ p ] a a x = 0 . Therefore, a = 0 or a = 0 which is a contradiction.Next, we claim that L , and L , are not isomorphic. Suppose to the contrary that there existsan isomorphism A : L , → L , . Then A ( x [ p ]3 ) = A ( x ) [ p ] A ( x ) = ( a a x + a a x ) [ p ] a a x = 0 , which implies that a a = 0 . Therefore, we have a = 0 or a = 0 which is a contradiction.Next, we claim that L , and L , are not isomorphic. Suppose to the contrary that there existsan isomorphism A : L , → L , . Then A ( x [ p ]5 ) = A ( x ) [ p ] A ( x ) = ( a x + a x ) [ p ] a a x = 0 . Therefore, a = 0 or a = 0 which is a contradiction.Next, we claim that L , and L , are not isomorphic. Suppose to the contrary that there existsan isomorphism A : L , → L , . Then A ( x [ p ]5 ) = A ( x ) [ p ] A ( x ) = ( a x + a x ) [ p ] a a x = 0 . HAPTER 5. RESTRICTION MAPS ON L , a = 0 or a = 0 which is a contradiction.Next, we claim that L , and L , are not isomorphic. Suppose to the contrary that there existsan isomorphism A : L , → L , . Then A ( x [ p ]5 ) = A ( x ) [ p ] A ( x ) = ( a x + a x ) [ p ] a a x = 0 . Therefore, a = 0 or a = 0 which is a contradiction.Next, we claim that L , and L , are not isomorphic. Suppose to the contrary that there existsan isomorphism A : L , → L , . Then A ( x [ p ]4 ) = A ( x ) [ p ] a a x ) [ p ] a p a p x . Therefore, a = 0 or a = 0 which is a contradiction.It is clear that L , is not isomorphic to the other restricted Lie algebras.Next, we claim that L , and L , ( β ) are not isomorphic. Suppose to the contrary that thereexists an isomorphism A : L , → L , ( β ). Then A ( x [ p ]2 ) = A ( x ) [ p ] a x + a x + a x + a x ) [ p ] a p βx , which implies that a p β = 0 . Since, β = 0 , we have a = 0 which is a contradiction.Next, we claim that L , and L , are not isomorphic. Suppose to the contrary that there existsan isomorphism A : L , → L , . Then A ( x [ p ]1 ) = A ( x ) [ p ] a x + a x + a x + a x + a x ) [ p ] a p x . Therefore, a = 0 which is a contradiction.Next, we claim that L , and L , are not isomorphic. Suppose to the contrary that there existsan isomorphism A : L , → L , . Then A ( x [ p ]3 ) = A ( x ) [ p ] A ( x ) = ( a a x + a a x ) [ p ] a a x = 0 , which implies that a a = 0 . Therefore, we have a = 0 or a = 0 which is a contradiction. HAPTER 5. RESTRICTION MAPS ON L , L , and L , are not isomorphic. Suppose to the contrary that there existsan isomorphism A : L , → L , . Then A ( x [ p ]1 ) = A ( x ) [ p ] a x + a x + a x + a x + a x ) [ p ] a p x . Therefore, a = 0 which is a contradiction.Next, we claim that L , and L , are not isomorphic. Suppose to the contrary that there existsan isomorphism A : L , → L , . Then A ( x [ p ]3 ) = A ( x ) [ p ] A ( x ) = ( a a x + a a x ) [ p ] a a x = a p a p x , which implies that a a = 0 and a p a p = 0 . Therefore, we have a = 0 or a = 0 which is acontradiction.Next, we claim that L , and L , are not isomorphic. Suppose to the contrary that there existsan isomorphism A : L , → L , . Then A ( x [ p ]2 ) = A ( x ) [ p ] a x + a x + a x + a x ) [ p ] a p x . Therefore, a = 0 which is a contradiction.Next, we claim that L , and L , are not isomorphic. Suppose to the contrary that there existsan isomorphism A : L , → L , . Then A ( x [ p ]4 ) = A ( x ) [ p ] a a x ) [ p ] a p a p x . Therefore, a = 0 or a = 0 which is a contradiction.It is clear that L , is not isomorphic to the other restricted Lie algebras.Next, we claim that L , ( β ) and L , are not isomorphic. Suppose to the contrary that thereexists an isomorphism A : L , ( β ) → L , . Then A ( x [ p ]1 ) = A ( x ) [ p ] a x + a x + a x + a x + a x ) [ p ] a p x . Therefore, a = 0 which is a contradiction. HAPTER 5. RESTRICTION MAPS ON L , L , ( β ) and L , are not isomorphic. Suppose to the contrary that thereexists an isomorphism A : L , → L , ( β ). Then A ( x [ p ]5 ) = A ( x ) [ p ] A ( x ) = ( a x + a x ) [ p ] a a x = 0 . Therefore, a = 0 or a = 0 which is a contradiction.Next, we claim that L , ( β ) and L , are not isomorphic. Suppose to the contrary that thereexists an isomorphism A : L , ( β ) → L , . Then A ( x [ p ]1 ) = A ( x ) [ p ] a x + a x + a x + a x + a x ) [ p ] a p x . Therefore, a = 0 which is a contradiction.Next, we claim that L , ( β ) and L , are not isomorphic. Suppose to the contrary that thereexists an isomorphism A : L , ( β ) → L , . Then A ( x [ p ]3 ) = A ( x ) [ p ] a a x + a a x ) [ p ] a p a p x , which implies that a p a p = 0 . Therefore, we have a = 0 or a = 0 which is a contradiction.Next, we claim that L , ( β ) and L , are not isomorphic. Suppose to the contrary that thereexists an isomorphism A : L , ( β ) → L , . Then A ( x [ p ]2 ) = A ( x ) [ p ] A ( x ) = ( a x + a x + a x + a x ) [ p ] a a x = a p x . Therefore, a = 0 which is a contradiction.Next, we claim that L , ( β ) and L , are not isomorphic. Suppose to the contrary that thereexists an isomorphism A : L , ( β ) → L , . Then A ( x [ p ]4 ) = A ( x ) [ p ] a a x ) [ p ] a p a p x . Therefore, a = 0 or a = 0 which is a contradiction.It is clear that L , ( β ) is not isomorphic to the other restricted Lie algebras.Next, we claim that L , and L , are not isomorphic. Suppose to the contrary that there existsan isomorphism A : L , → L , . Then A ( x [ p ]5 ) = A ( x ) [ p ] A ( x ) = ( a x + a x ) [ p ] a a x = 0 . HAPTER 5. RESTRICTION MAPS ON L , a = 0 or a = 0 which is a contradiction.Next, we claim that L , and L , are not isomorphic. Suppose to the contrary that there existsan isomorphism A : L , → L , . Then A ( x [ p ]1 ) = A ( x ) [ p ] A ( x ) = ( a x + a x + a x + a x + a x ) [ p ] a a x = a p x . Therefore, a = 0 which is a contradiction.Next, we claim that L , and L , are not isomorphic. Suppose to the contrary that there existsan isomorphism A : L , → L , . Then A ( x [ p ]3 ) = A ( x ) [ p ] a a x + a a x ) [ p ] a p a p x , which implies that a p a p = 0 . Therefore, we have a = 0 or a = 0 which is a contradiction.Next, we claim that L , and L , are not isomorphic. Suppose to the contrary that there existsan isomorphism A : L , → L , . Then A ( x [ p ]2 ) = A ( x ) [ p ] a x + a x + a x + a x ) [ p ] a p x . Therefore, a = 0 which is a contradiction.Next, we claim that L , and L , are not isomorphic. Suppose to the contrary that there existsan isomorphism A : L , → L , . Then A ( x [ p ]4 ) = A ( x ) [ p ] a a x ) [ p ] a p a p x . Therefore, a = 0 or a = 0 which is a contradiction.It is clear that L , is not isomorphic to the other restricted Lie algebras.Next, we claim that L , and L , are not isomorphic. Suppose to the contrary that there existsan isomorphism A : L , → L , . Then A ( x [ p ]1 ) = A ( x ) [ p ] a x + a x + a x + a x + a x ) [ p ] a p x . Therefore, a = 0 which is a contradiction. HAPTER 5. RESTRICTION MAPS ON L , L , and L , are not isomorphic. Suppose to the contrary that there existsan isomorphism A : L , → L , . Then A ( x [ p ]3 ) = A ( x ) [ p ] a a x + a a x ) [ p ] a p a p x , which implies that a p a p = 0 . Therefore, we have a = 0 or a = 0 which is a contradiction.Next, we claim that L , and L , are not isomorphic. Suppose to the contrary that there existsan isomorphism A : L , → L , . Then A ( x [ p ]2 ) = A ( x ) [ p ] a x + a x + a x + a x ) [ p ] a p x . Therefore, a = 0 which is a contradiction.Next, we claim that L , and L , are not isomorphic. Suppose to the contrary that there existsan isomorphism A : L , → L , . Then A ( x [ p ]4 ) = A ( x ) [ p ] a a x ) [ p ] a p a p x . Therefore, a = 0 or a = 0 which is a contradiction.It is clear that L , is not isomorphic to the other restricted Lie algebras.Next, we claim that L , and L , are not isomorphic. Suppose to the contrary that there existsan isomorphism A : L , → L , . Then A ( x [ p ]3 ) = A ( x ) [ p ] a a x + a a x ) [ p ] a p a p x , which implies that a p a p = 0 . Therefore, we have a = 0 or a = 0 which is a contradiction.Next, we claim that L , and L , are not isomorphic. Suppose to the contrary that there existsan isomorphism A : L , → L , . Then A ( x [ p ]2 ) = A ( x ) [ p ] a x + a x + a x + a x ) [ p ] a p x . Therefore, a = 0 which is a contradiction. HAPTER 5. RESTRICTION MAPS ON L , L , and L , are not isomorphic. Suppose to the contrary that there existsan isomorphism A : L , → L , . Then A ( x [ p ]4 ) = A ( x ) [ p ] a a x ) [ p ] a p a p x . Therefore, a = 0 or a = 0 which is a contradiction.It is clear that L , is not isomorphic to the other restricted Lie algebras.Next, we claim that L , ( β ) and L , are not isomorphic. Suppose to the contrary that thereexists an isomorphism A : L , ( β ) → L , . Then A ( x [ p ]3 ) = A ( x ) [ p ] a a x + a a x ) [ p ] a p a p x , which implies that a p a p = 0 . Therefore, we have a = 0 or a = 0 which is a contradiction.Note that L , ( β ) is not isomorphic to any of L , , L , , L , , L , ( ξ ), L , , L , ( γ ) because( L , ) [ p ] = 0 but this is not true for those restricted Lie algebras.Next, we claim that L , ( β ) and L , are not isomorphic. Suppose to the contrary that thereexists an isomorphism A : L , ( β ) → L , . Then A ( x [ p ]3 ) = A ( x ) [ p ] a a x + a a x ) [ p ] a p a p x , which implies that a p a p = 0 . Therefore, we have a = 0 or a = 0 which is a contradiction.Next, we claim that L , ( β ) and L , ( β ) are not isomorphic. Suppose to the contrary that thereexists an isomorphism A : L , ( β ) → L , ( β ). Then A ( x [ p ]3 ) = A ( x ) [ p ] a a x + a a x ) [ p ] a p a p x , which implies that a p a p = 0 . Therefore, we have a = 0 or a = 0 which is a contradiction.Next, we claim that L , ( β ) and L , are not isomorphic. Suppose to the contrary that thereexists an isomorphism A : L , ( β ) → L , . Then A ( x [ p ]2 ) = A ( x ) [ p ] A ( βx ) = ( a x + a x + a x + a x ) [ p ] βa a x = a p x . HAPTER 5. RESTRICTION MAPS ON L , a = 0 which is a contradiction.Next, we claim that L , ( β ) and L , are not isomorphic. Suppose to the contrary that thereexists an isomorphism A : L , ( β ) → L , . Then A ( x [ p ]3 ) = A ( x ) [ p ] a a x + a a x ) [ p ] a p a p x , which implies that a p a p = 0 . Therefore, we have a = 0 or a = 0 which is a contradiction.It is clear that L , ( β ) is not isomorphic to the other restricted Lie algebras.Note that L , is not isomorphic to any of L , , L , , L , , L , ( ξ ), L , , L , ( γ ) because( L , ) [ p ] = 0 but this is not true for those restricted Lie algebras.Next, we claim that L , and L , are not isomorphic. Suppose to the contrary that there existsan isomorphism A : L , → L , . Then A ( x [ p ]3 ) = A ( x ) [ p ] A ( x ) = ( a a x + a a x ) [ p ] a a x = a p a p x , which implies that a a = 0 and a p a p = 0 . Therefore, we have a = 0 or a = 0 which is acontradiction.Next, we claim that L , and L , ( β ) are not isomorphic. Suppose to the contrary that thereexists an isomorphism A : L , → L , ( β ). Then A ( x [ p ]2 ) = A ( x ) [ p ] a x + a x + a x + a x ) [ p ] a p βx + a p x , which implies that a p β = 0 . Since, β = 0 we have a = 0 which is a contradiction.Next, we claim that L , and L , are not isomorphic. Suppose to the contrary that there existsan isomorphism A : L , → L , . Then A ( x [ p ]2 ) = A ( x ) [ p ] a x + a x + a x + a x ) [ p ] a p x . Therefore, a = 0 which is a contradiction.Next, we claim that L , and L , are not isomorphic. Suppose to the contrary that there existsan isomorphism A : L , → L , . Then A ( x [ p ]2 ) = A ( x ) [ p ] a x + a x + a x + a x ) [ p ] a p x + a p x . HAPTER 5. RESTRICTION MAPS ON L , a = 0 which is a contradiction.It is clear that L , is not isomorphic to the other restricted Lie algebras.Note that L , is not isomorphic to any of L , , L , ( β ), L , , L , because ( L , ) [ p ] = 0 butthis is not true for those restricted Lie algebras.Next, we claim that L , and L , are not isomorphic. Suppose to the contrary that there existsan isomorphism A : L , → L , . Then A ( x [ p ]3 ) = A ( x ) [ p ] a a x + a a x ) [ p ] a p a p x , which implies that a p a p = 0 . Therefore, we have a = 0 or a = 0 which is a contradiction.Next, we claim that L , and L , are not isomorphic. Suppose to the contrary that there existsan isomorphism A : L , → L , . Then A ( x [ p ]4 ) = A ( x ) [ p ] a a x ) [ p ] a p a p x . Therefore, a = 0 or a = 0 which is a contradiction.Next, we claim that L , and L , ( ξ ) are not isomorphic. Suppose to the contrary that thereexists an isomorphism A : L , → L , ( ξ ). Then A ( x [ p ]4 ) = A ( x ) [ p ] a a x ) [ p ] a p a p x . Therefore, a = 0 or a = 0 which is a contradiction.Next, we claim that L , and L , are not isomorphic. Suppose to the contrary that there existsan isomorphism A : L , → L , . Then A ( x [ p ]3 ) = A ( x ) [ p ] a a x + a a x ) [ p ] a p a p x + a p a p x , which implies that a p a p = 0 . Therefore, we have a = 0 or a = 0 which is a contradiction.Next, we claim that L , and L , ( γ ) are not isomorphic. Suppose to the contrary that thereexists an isomorphism A : L , → L , ( γ ). Then A ( x [ p ]3 ) = A ( x ) [ p ] a a x + a a x ) [ p ] a p a p γx , HAPTER 5. RESTRICTION MAPS ON L , a p a p = 0 . Therefore, we have a = 0 or a = 0 which is a contradiction.It is clear that L , is not isomorphic to the other restricted Lie algebras.Next, we claim that L , and L , are not isomorphic. Suppose to the contrary that there existsan isomorphism A : L , → L , . Then A ( x [ p ]2 ) = A ( x ) [ p ] a x + a x + a x + a x ) [ p ] a p x . Therefore, a = 0 which is a contradiction.Next, we claim that L , and L , are not isomorphic. Suppose to the contrary that there existsan isomorphism A : L , → L , . Then A ( x [ p ]4 ) = A ( x ) [ p ] a a x ) [ p ] a p a p x . Therefore, a = 0 or a = 0 which is a contradiction.It is clear that L , is not isomorphic to the other restricted Lie algebras.Next, we claim that L , and L , ( β ) are not isomorphic. Suppose to the contrary that thereexists an isomorphism A : L , → L , ( β ). Then A ( x [ p ]2 ) = A ( x ) [ p ] a x + a x + a x + a x ) [ p ] a p βx + a p x , which implies that a p β = 0 . Since, β = 0 we have a = 0 which is a contradiction.Note that L , is not isomorphic to any of L , , L , , L , ( ξ ), L , , L , ( γ ) because ( L , ) [ p ] = 0but this is not true for those restricted Lie algebras.Next, we claim that L , and L , are not isomorphic. Suppose to the contrary that there existsan isomorphism A : L , → L , . Then A ( x [ p ]2 ) = A ( x ) [ p ] a x + a x + a x + a x ) [ p ] a p x . Therefore, a = 0 which is a contradiction.Next, we claim that L , and L , are not isomorphic. Suppose to the contrary that there existsan isomorphism A : L , → L , . Then A ( x [ p ]2 ) = A ( x ) [ p ] a x + a x + a x + a x ) [ p ] a p x + a p x . Therefore, a = 0 which is a contradiction. HAPTER 5. RESTRICTION MAPS ON L , L , is not isomorphic to the other restricted Lie algebras.Note that L , is not isomorphic to any of L , , L , , L , ( ξ ), L , , L , ( γ ) because ( L , ) [ p ] = 0but this is not true for those restricted Lie algebras.Next, we claim that L , ( β ) and L , are not isomorphic. Suppose to the contrary that thereexists an isomorphism A : L , ( β ) → L , . Then A ( x [ p ]1 ) = A ( x ) [ p ] a x + a x + a x + a x + a x ) [ p ] a p x + a p x . Therefore, a = 0 which is a contradiction.Next, we claim that L , ( β ) and L , are not isomorphic. Suppose to the contrary that thereexists an isomorphism A : L , ( β ) → L , . Then A ( x [ p ]2 ) = A ( x ) [ p ] A ( βx ) = ( a x + a x + a x + a x ) [ p ] βa a x = a p x + a p x . Therefore, a = 0 which is a contradiction.It is clear that L , ( β ) is not isomorphic to the other restricted Lie algebras.Note that L , is not isomorphic to any of L , , L , because ( L , ) [ p ] = 0 but this is not truefor those restricted Lie algebras.Next, we claim that L , and L , are not isomorphic. Suppose to the contrary that there existsan isomorphism A : L , → L , . Then A ( x [ p ]4 ) = A ( x ) [ p ] a a x ) [ p ] a p a p x . Therefore, a = 0 or a = 0 which is a contradiction.Next, we claim that L , and L , ( ξ ) are not isomorphic. Suppose to the contrary that thereexists an isomorphism A : L , → L , ( ξ ). Then A ( x [ p ]4 ) = A ( x ) [ p ] a a x ) [ p ] a p a p x . Therefore, a = 0 or a = 0 which is a contradiction.Next, we claim that L , and L , are not isomorphic. Suppose to the contrary that there existsan isomorphism A : L , → L , . Then A ( x [ p ]4 ) = A ( x ) [ p ] a a x ) [ p ] a p a p x . HAPTER 5. RESTRICTION MAPS ON L , a = 0 or a = 0 which is a contradiction.Next, we claim that L , and L , ( γ ) are not isomorphic. Suppose to the contrary that thereexists an isomorphism A : L , ( γ ) → L , . Then A ( x [ p ]3 ) = A ( x ) [ p ] A ( γx ) = ( a a x + a a x ) [ p ] a a γx = a p a p x , which implies that a p a p = 0 . Therefore, we have a = 0 or a = 0 which is a contradiction.It is clear that L , is not isomorphic to the other restricted Lie algebras.Next, we claim that L , and L , are not isomorphic. Suppose to the contrary that there existsan isomorphism A : L , → L , . Then A ( x [ p ]4 ) = A ( x ) [ p ] a a x ) [ p ] a p a p x . Therefore, a = 0 or a = 0 which is a contradiction.It is clear that L , and L , are not isomorphic to the other restricted Lie algebras.Note that L , is not isomorphic to any of L , , L , ( ξ ), L , , L , ( γ ) because ( L , ) [ p ] = 0 butthis is not true for those restricted Lie algebras.Next, we claim that L , and L , are not isomorphic. Suppose to the contrary that there existsan isomorphism A : L , → L , . Then A ( x [ p ]3 ) = A ( x ) [ p ] a a x + a a x ) [ p ] a p a p x , which implies that a p a p = 0 . Therefore, we have a = 0 or a = 0 which is a contradiction.Next, we claim that L , and L , ( ξ ) are not isomorphic. Suppose to the contrary that thereexists an isomorphism A : L , → L , ( ξ ). Then A ( x [ p ]2 ) = A ( x ) [ p ] a x + a x + a x + a x ) [ p ] a p x + a p x . Therefore, a = 0 which is a contradiction.Note that L , is not isomorphic to L , because ( L , ) [ p ] = 0 but this is not true for L , .Next, we claim that L , and L , are not isomorphic. Suppose to the contrary that there existsan isomorphism A : L , → L , . Then A ( x [ p ]3 ) = A ( x ) [ p ] a a x + a a x ) [ p ] a p a p x + a p a p x , HAPTER 5. RESTRICTION MAPS ON L , a p a p = 0 . Therefore, we have a = 0 or a = 0 which is a contradiction.Next, we claim that L , and L , ( γ ) are not isomorphic. Suppose to the contrary that thereexists an isomorphism A : L , → L , ( γ ). Then A ( x [ p ]3 ) = A ( x ) [ p ] a a x + a a x ) [ p ] a p a p γx , which implies that a p a p = 0 . Therefore, we have a = 0 or a = 0 which is a contradiction.Note that L , is not isomorphic to L , because ( L , ) [ p ] = 0 but this is not true for L , .Next, we claim that L , ( ξ ) and L , are not isomorphic. Suppose to the contrary that thereexists an isomorphism A : L , ( ξ ) → L , . Then A ( x [ p ]3 ) = A ( x ) [ p ] a a x + a a x ) [ p ] a p a p x + a p a p x , which implies that a p a p = 0 . Therefore, we have a = 0 or a = 0 which is a contradiction.Next, we claim that L , and L , ( γ ) are not isomorphic. Suppose to the contrary that thereexists an isomorphism A : L , → L , ( γ ). Then A ( x [ p ]3 ) = A ( x ) [ p ] a a x + a a x ) [ p ] a p a p γx , which implies that a p a p = 0 . Therefore, we have a = 0 or a = 0 which is a contradiction.Note that L , is not isomorphic to L , and L , ( γ ) because ( L , ) [ p ] = 0 but this is not truefor L , and L , ( γ ).Next, we claim that L , and L , ( γ ) are not isomorphic. Suppose to the contrary that thereexists an isomorphism A : L , ( γ ) → L , . Then A ( x [ p ]4 ) = A ( x ) [ p ] a a x ) [ p ] a p a p x , which implies that a p a p = 0 . Therefore, we have a = 0 or a = 0 which is a contradiction. hapter 6 Restriction maps on Lie algebras of1-dimensional centre L , L , = h x , . . . , x | [ x , x ] = x , [ x , x ] = x i . We have Z ( L , ) = h x i F and so x [ p ]5 = 0 . Let L = L , h x i F ∼ = L , , where L , = h x , x , x , x i . Note that the group Aut( L ) consists of invertible matrices of the form a a a a a a a a a a a a a a a a . Lemma 6.1.1 Let K = L , and [ p ] : K → K be a p -map on K and let L = KM where M = h x i F .Then K ∼ = L θ where θ = (∆ + ∆ , ω ) ∈ Z ( L, F ) . Proof. Let π : K → L be the projection map. We have the exact sequence0 → M → K → L → . Let σ : L → K such that x i x i , 1 i σ is an injective linear map and πσ = 1 L . Now,we define φ : L × L → M by φ ( x i , x j ) = [ σ ( x i ) , σ ( x j )] − σ ([ x i , x j ]), 1 i, j ω : L → M by ω ( x ) = σ ( x ) [ p ] − σ ( x [ p ] ). Note that φ ( x , x ) = [ σ ( x ) , σ ( x )] − σ ([ x , x ]) = [ x , x ] = x ; φ ( x , x ) = [ σ ( x ) , σ ( x )] − σ ([ x , x ]) = x ; φ ( x , x ) = [ σ ( x ) , σ ( x )] − σ ([ x , x ]) = 0;Similarly, we can show that φ ( x , x ) = φ ( x , x ) = φ ( x , x ) = 0 . Therefore, φ = ∆ + ∆ . Now,by Lemma 2.2.2, we have θ = (∆ + ∆ , ω ) ∈ Z ( L, F ) and K ∼ = L θ . (cid:4) HAPTER 6. RESTRICTION MAPS ON LIE ALGEBRAS OF 1-DIMENSIONAL CENTRE L given by the following p -maps:I.1 Trivial p -map;I.2 x [ p ]1 = x ;I.3 x [ p ]1 = x , x [ p ]3 = x ;I.4 x [ p ]1 = x , x [ p ]2 = x ;I.5 x [ p ]1 = x , x [ p ]2 = x , x [ p ]3 = x .We make L into a restricted Lie algebra by equipping it with each of the above p -maps. Then,in each case, we find all possible orbit representatives of the form (∆ + ∆ , ω ) under the actionof Aut p ( L ) on H ( L, F ). By Lemma 6.1.1, we do get all possible p -maps on L , with the propertythat x [ p ]5 = 0 .Consider the case I.2 where the p -map of L is given by x [ p ]1 = x . Let [( φ, ω )] ∈ H ( L, F ). Thenwe must have φ ( x, y [ p ] ) = 0 , for all x, y ∈ L , where φ = a ∆ + b ∆ + c ∆ + d ∆ + e ∆ + f ∆ ,for some a, b, c, d, e, f ∈ F . Since L [ p ] = h x i F , we get φ ( x, x ) = 0 , for all x ∈ L . Therefore, φ ( x , x ) = a ∆ ( x , x ) + b ∆ ( x , x ) + c ∆ ( x , x )+ d ∆ ( x , x ) + a ∆ ( x , x ) + f ∆ ( x , x ) = 0which implies that a = 0 . Since φ = ∆ + ∆ gives us L , , we deduce by Lemma 6.1.1 that L , cannot be constructed in this case. Similarly, in cases I.3, I.4, and I.5 we can show that we cannotconstruct L , . In the following subsections, we consider the remaining cases. L , / h x i via the trivial p -map First, we find a basis for Z ( L, F ). Let ( φ, ω ) = ( a ∆ + b ∆ + c ∆ + d ∆ + e ∆ + f ∆ , αf + βf + γf + δf ) ∈ Z ( L, F ). Then we must have δ φ ( x, y, z ) = 0 and φ ( x, y [ p ] ) = 0 , for all x, y, z ∈ L .Since L is abelian and the p -map is trivial, δ φ ( x, y, z ) = 0 and φ ( x, y [ p ] ) = 0 , for all x, y, z ∈ L .Therefore, a basis for Z ( L, F ) is as follows:(∆ , , (∆ , , (∆ , , (∆ , , (∆ , , (∆ , , (0 , f ) , (0 , f ) , (0 , f ) , (0 , f ) . Next, we find a basis for B ( L, F ). Let ( φ, ω ) ∈ B ( L, F ). Since B ( L, F ) ⊆ Z ( L, F ), we have( φ, ω ) = ( a ∆ + b ∆ + c ∆ + d ∆ + e ∆ + f ∆ , αf + βf + γf + δf ). So, there exists a linearmap ψ : L → F such that δ ψ ( x, y ) = φ ( x, y ) and ˜ ψ ( x ) = ω ( x ), for all x, y ∈ L . So, we have a = φ ( x , x ) = δ ψ ( x , x ) = ψ ([ x , x ]) = 0 . Similarly, we can show that b = c = d = e = f = 0 . Also, we have α = ω ( x ) = ˜ ψ ( x ) = ψ ( x [ p ]1 ) = 0 . Similarly, we can show that β = γ = δ = 0 . Therefore, ( φ, ω ) = (0 , 0) and hence B ( L, F ) = 0 . Wededuce that a basis for H ( L, F ) is as follows:[(∆ , , [(∆ , , [(∆ , , [(∆ , , [(∆ , , [(∆ , , [(0 , f )] , [(0 , f )] , [(0 , f )] , [(0 , f )] . HAPTER 6. RESTRICTION MAPS ON LIE ALGEBRAS OF 1-DIMENSIONAL CENTRE φ, ω )] ∈ H ( L, F ). Then we have φ = a ∆ + b ∆ + c ∆ + d ∆ + e ∆ + f ∆ , for some a, b, c, d, e, f ∈ F . Suppose that Aφ = a ′ ∆ + b ′ ∆ + c ′ ∆ + d ′ ∆ + e ′ ∆ + f ′ ∆ , for some a ′ , b ′ , c ′ , d ′ , e ′ , f ′ ∈ F . Then Aφ ( x , x ) = φ ( Ax , Ax )= φ ( a x + a x + a x + a x , a x + a x + a x + a x )= ( a a − a a ) a + ( a a − a a ) b + ( a a − a a ) c + ( a a − a a ) d + ( a a − a a ) e + ( a a − a a ) f, and Aφ ( x , x ) = φ ( Ax , Ax )= φ ( a x + a x + a x + a x , a x + a x + a x + a x )= ( a a − a a ) a + ( a a − a a ) b + ( a a − a a ) c + ( a a − a a ) d + ( a a − a a ) e + ( a a − a a ) f, and Aφ ( x , x ) = φ ( Ax , Ax )= φ ( a x + a x + a x + a x , a x + a x + a x + a x )= ( a a − a a ) a + ( a a − a a ) b + ( a a − a a ) c + ( a a − a a ) d + ( a a − a a ) e + ( a a − a a ) f, and Aφ ( x , x ) = φ ( Ax , Ax )= φ ( a x + a x + a x + a x , a x + a x + a x + a x )= ( a a − a a ) a + ( a a − a a ) b + ( a a − a a ) c + ( a a − a a ) d + ( a a − a a ) e + ( a a − a a ) f, and Aφ ( x , x ) = φ ( Ax , Ax )= φ ( a x + a x + a x + a x , a x + a x + a x + a x )= ( a a − a a ) a + ( a a − a a ) b + ( a a − a a ) c + ( a a − a a ) d + ( a a − a a ) e + ( a a − a a ) f, and Aφ ( x , x ) = φ ( Ax , Ax )= φ ( a x + a x + a x + a x , a x + a x + a x + a x )= ( a a − a a ) a + ( a a − a a ) b + ( a a − a a ) c + ( a a − a a ) d + ( a a − a a ) e + ( a a − a a ) f. In the matrix form we can write this as a ′ b ′ c ′ d ′ e ′ f ′ = a a − a a a a − a a a a − a a a a − a a a a − a a a a − a a a a − a a a a − a a a a − a a a a − a a a a − a a a a − a a a a − a a a a − a a a a − a a a a − a a a a − a a a a − a a a a − a a a a − a a a a − a a a a − a a a a − a a a a − a a a a − a a a a − a a a a − a a a a − a a a a − a a a a − a a a a − a a a a − a a a a − a a a a − a a a a − a a a a − a a abcdef . HAPTER 6. RESTRICTION MAPS ON LIE ALGEBRAS OF 1-DIMENSIONAL CENTRE of this action gives us L , .Also, we have ω = αf + βf + γf + δf , for some α, β, γ, δ ∈ F . Suppose that Aω = α ′ f + β ′ f + γ ′ f + δ ′ f , for some α ′ , β ′ , γ ′ , δ ′ ∈ F . Then we can verify that the action of Aut( L ) on theset of ω ’s in the matrix form is as follows: α ′ β ′ γ ′ δ ′ = a p a p a p a p a p a p a p a p a p a p a p a p a p a p a p a p αβγδ . Now we find the representatives of the orbits of the action of Aut( L ) on the set of ω ’s such thatthe orbit represented by is preserved under the action of Aut( L ) on the set of φ ’s. Let ν = αβγδ ∈ F . If ν = , then { ν } is clearly an Aut( L )-orbit. Let ν = 0 . Suppose that δ = 0 ,then h − β/δ ) /p α/δ ) /p − α/δ ) /p αβ/δ ) /p − α /δ ) /p ( α/δ ) /p − β/δ ) /p β /δ ) /p − αβ/δ ) /p ( β/δ ) /p β/δ ) /p − α/δ ) /p , − α/δ − β/δβ/δ − α/δ ih , αβγδ i = h , γδ i , and HAPTER 6. RESTRICTION MAPS ON LIE ALGEBRAS OF 1-DIMENSIONAL CENTRE h δ /p ( − γ ) /p /δ ) /p δ /p ( − γ ) /p 00 0 0 0 (1 /δ ) /p 00 0 0 0 0 1 , δ − γ /δ ih , γδ i = h , i . Next, suppose that δ = 0 , but β = 0 , then h γ/β ) /p − γ/β ) /p − α/β ) /p ( − γ /β ) /p ( γ/β ) /p − α/β ) /p 00 0 0 1 0 00 0 0 0 1 00 0 0 0 ( − γ/β ) /p , − α/β − γ/β − γ/β ih , αβγ i = h , β i , and h β /p β /p /β ) /p /β ) /p 00 0 0 0 0 1 , β /β ih , β i = h , i . HAPTER 6. RESTRICTION MAPS ON LIE ALGEBRAS OF 1-DIMENSIONAL CENTRE δ = β = 0 , but α = 0 , then h γ/α ) /p γ/α ) /p γ /α ) /p − γ/α ) /p − γ/α ) /p , γ/α − γ/α ih , α γ i = h , α i , and h /α ) /p /α ) /p α /p α /p 00 0 0 0 0 1 , /α α ih , α i = h , i . Next, if δ = β = α = 0 , but γ = 0 , then h /γ ) /p γ /p /γ ) /p γ /p 00 0 0 0 0 1 , /γ 00 0 0 γ ih , γ i = h , i . Now we claim that the following elements are in the same Aut( L )-orbit representatives. h , i , h , i , h , i , h , i . HAPTER 6. RESTRICTION MAPS ON LIE ALGEBRAS OF 1-DIMENSIONAL CENTRE h − − , − ih , i = h , i , h − − − − , ih , i = h , i , h − − , − ih , i = h , i . Therefore, the following elements are Aut( L )-orbit representatives: h , i , h , i . Theorem 6.1.2 The list of all restricted Lie algebra structures on L , , up to isomorphism, is asfollows: L , = h x , . . . , x | [ x , x ] = x , [ x , x ] = x i ; L , = h x , . . . , x | [ x , x ] = x , [ x , x ] = x , x [ p ]1 = x i . L , L , = h x , . . . , x | [ x , x ] = x , [ x , x ] = x , [ x , x ] = x i . We have Z ( L , ) = h x i F . Let L = L , h x i F ∼ = L , , where L , = h x , x , x , x | [ x , x ] = x i . Note that the group Aut( L ) consists of invertiblematrices of the form a a a a a a r a a a a , where r = a a − a a = 0 . HAPTER 6. RESTRICTION MAPS ON LIE ALGEBRAS OF 1-DIMENSIONAL CENTRE Lemma 6.2.1 Let K = L , and [ p ] : K → K be a p -map on K and let L = KM where M = h x i F .Then K ∼ = L θ where θ = (∆ + ∆ , ω ) ∈ Z ( L, F ) . Proof. Let π : K → L be the projection map. We have the exact sequence0 → M → K → L → . Let σ : L → K such that x i x i , 1 i σ is an injective linear map and πσ = 1 L . Now,we define φ : L × L → M by φ ( x i , x j ) = [ σ ( x i ) , σ ( x j )] − σ ([ x i , x j ]), 1 i, j ω : L → M by ω ( x ) = σ ( x ) [ p ] − σ ( x [ p ] ). Note that φ ( x , x ) = [ σ ( x ) , σ ( x )] − σ ([ x , x ]) = [ x , x ] = x ; φ ( x , x ) = [ σ ( x ) , σ ( x )] − σ ([ x , x ]) = x ; φ ( x , x ) = [ σ ( x ) , σ ( x )] − σ ([ x , x ]) = 0;Similarly, we can show that φ ( x , x ) = φ ( x , x ) = φ ( x , x ) = 0 . Therefore, φ = ∆ + ∆ . Now,by Lemma 2.2.2, we have θ = (∆ + ∆ , ω ) ∈ Z ( L, F ) and K ∼ = L θ . (cid:4) Note that by Theorem 2.3.2, there are eight non-isomorphic restricted Lie algebra structures on L given by the following p -maps:II.1 Trivial p -map;II.2 x [ p ]1 = x ;II.3 x [ p ]1 = x ;II.4 x [ p ]1 = x , x [ p ]2 = x ;II.5 x [ p ]3 = x ;II.6 x [ p ]3 = x , x [ p ]2 = x ;II.7 x [ p ]4 = x ;II.8 x [ p ]4 = x , x [ p ]2 = x . We make L into a restricted Lie algebra by equipping it with each of the above p -maps. Then,in each case, we find all possible orbit representatives of the form (∆ + ∆ , ω ) under the actionof Aut p ( L ) on H ( L, F ). By Lemma 6.2.1, we do get all possible p -maps on L , with the propertythat x [ p ]5 = 0 .Consider the case I.2 where the p -map of L is given by x [ p ]1 = x . Note that L [ p ] = h x i F . Let[( φ, ω )] ∈ H ( L, F ). Then we must have φ ( x, y [ p ] ) = 0 , for all x, y ∈ L , where φ = a ∆ + b ∆ + c ∆ + d ∆ + e ∆ + f ∆ , for some a, b, c, d, e, f ∈ F . Hence, φ ( x , x ) = 0 which implies that b = 0 . Since φ = ∆ +∆ gives us L , , we deduce by Lemma 6.2.1 that L , cannot be constructedin this case. Similarly, in cases II3-II8 we can show that we cannot construct L , . It remains toconsider case II.1. HAPTER 6. RESTRICTION MAPS ON LIE ALGEBRAS OF 1-DIMENSIONAL CENTRE L , / h x i via the trivial p -map First, we find a basis for Z ( L, F ). Let ( φ, ω ) = ( a ∆ + b ∆ + c ∆ + d ∆ + e ∆ + f ∆ , αf + βf + γf + δf ) ∈ Z ( L, F ). Then we must have δ φ ( x, y, z ) = 0 and φ ( x, y [ p ] ) = 0 , for all x, y, z ∈ L .Therefore,0 = ( δ φ )( x , x , x ) = φ ([ x , x ] , x ) + φ ([ x , x ] , x ) + φ ([ x , x ] , x ) = φ ( x , x ) . Thus, we get f = 0 . Since the p -map is trivial, φ ( x, y [ p ] ) = φ ( x, 0) = 0 , for all x, y ∈ L . Therefore,a basis for Z ( L, F ) is as follows:(∆ , , (∆ , , (∆ , , (∆ , , (∆ , , (0 , f ) , (0 , f ) , (0 , f ) , (0 , f ) . Next, we find a basis for B ( L, F ). Let ( φ, ω ) ∈ B ( L, F ). Since B ( L, F ) ⊆ Z ( L, F ), we have( φ, ω ) = ( a ∆ + b ∆ + c ∆ + d ∆ + e ∆ , αf + βf + γf + δf ). So, there exists a linear map ψ : L → F such that δ ψ ( x, y ) = φ ( x, y ) and ˜ ψ ( x ) = ω ( x ), for all x, y ∈ L . So, we have b = φ ( x , x ) = δ ψ ( x , x ) = ψ ([ x , x ]) = 0 . Similarly, we can show that c = d = e = 0 . Also, we have α = ω ( x ) = ˜ ψ ( x ) = ψ ( x [ p ]1 ) = 0 . Similarly, we can show that β = γ = δ = 0 . Therefore, ( φ, ω ) = ( a ∆ , 0) and hence B ( L, F ) = h (∆ , i F . We deduce that a basis for H ( L, F ) is as follows:[(∆ , , [(∆ , , [(∆ , , [(∆ , , [(0 , f )] , [(0 , f )] , [(0 , f )] , [(0 , f )] . Let [( φ, ω )] ∈ H ( L, F ). Then we have φ = a ∆ + b ∆ + c ∆ + d ∆ , for some a, b, c, d ∈ F .Suppose that Aφ = a ′ ∆ + b ′ ∆ + c ′ ∆ + d ′ ∆ , for some a ′ , b ′ , c ′ , d ′ ∈ F . We can verify that theaction of Aut( L ) on the set of φ ’s in the matrix form is as follows: a ′ b ′ c ′ d ′ = ra ra a a a a a a a a ra ra a a a a a a a a abcd . (6.1)The orbit with representative of this action gives us L , .Also, we have ω = αf + βf + γf + δf , for some α, β, γ, δ ∈ F . Suppose that Aω = α ′ f + β ′ f + γ ′ f + δ ′ f , for some α ′ , β ′ , γ ′ , δ ′ ∈ F . We can verify that the action of Aut( L ) on the set of ω ’s in the matrix form is as follows: α ′ β ′ γ ′ δ ′ = a p a p a p a p a p a p a p a p r p 00 0 a p a p αβγδ . (6.2) HAPTER 6. RESTRICTION MAPS ON LIE ALGEBRAS OF 1-DIMENSIONAL CENTRE h ra ra a a a a a a a a ra ra a a a a a a a a , a p a p a p a p a p a p a p a p r p 00 0 a p a p ih abcd , αβγδ i = h a ′ b ′ c ′ d ′ , α ′ β ′ γ ′ δ ′ i . Now we find the representatives of the orbits of the action of Aut( L ) on the set of ω ’s such that theorbit represented by is preserved under the action of Aut( L ) on the set of φ ’s.Let ν = αβγδ ∈ F . If ν = , then { ν } is clearly an Aut( L )-orbit. Let ν = 0 . Suppose that γ = 0 . Then h , − α/γ 00 1 − β/γ 00 0 1 00 0 0 1 ih , αβγδ i = h , γδ i , and h γ /p γ − /p 00 0 0 1 , γ γ − γ − 00 0 0 γ ih , γδ i = h , γ δ i . Furthermore, h δ /p − δ /p − δ /p δ /p − δ /p , δ − δ ih , δ i = h , i . Next, if γ = 0 , but δ = 0 , then h , − α/δ − β/δ ih , αβ δ i = h , δ i . Next, if γ = δ = 0 , but β = 0 , then h − α/β ) /p α/β ) /p − α/β ) /p ( − α/β ) /p α/β ) /p , − α/β α/β ih , αβ i = h , β i . HAPTER 6. RESTRICTION MAPS ON LIE ALGEBRAS OF 1-DIMENSIONAL CENTRE γ = δ = β = 0 , but α = 0 , then h α − /p α /p 00 0 0 1 , α − α α 00 0 0 α − ih , α i = h , i . Thus the following elements are Aut( L )-orbit representatives: , , β , , δ . Theorem 6.2.2 The list of all restricted Lie algebra structures on L , , up to isomorphism, is asfollows: L , = h x , . . . , x | [ x , x ] = x , [ x , x ] = x , [ x , x ] = x i ; L , = h x , . . . , x | [ x , x ] = x , [ x , x ] = x , [ x , x ] = x , x [ p ]1 = x i ; L , ( β ) = h x , . . . , x | [ x , x ] = x , [ x , x ] = x , [ x , x ] = x , x [ p ]2 = βx i ; L , = h x , . . . , x | [ x , x ] = x , [ x , x ] = x , [ x , x ] = x , x [ p ]3 = x i ; L , ( δ ) = h x , . . . , x | [ x , x ] = x , [ x , x ] = x , [ x , x ] = x , x [ p ]4 = δx i where β, γ ∈ F ∗ . The automorphism group of L , consists of a a a a a a a − a a a a a a a d a a a , with d = a a + a a − a a . Lemma 6.2.3 We have L , ( β ) ∼ = L , ( β ) if and only if β β − ∈ ( F ∗ ) . Proof. Suppose that f = ( a ij ) : L , ( β ) → L , ( β ) is an isomorphism. Then we have f ( x [ p ]2 ) = f ( x ) [ p ] which implies that β /β = a − a p − ∈ ( F ∗ ) . To prove the converse, suppose that β /β = ǫ , for some ǫ ∈ F ∗ . It is easy to see that the following is an isomorphism from L , ( β ) to L , ( β ): ǫ − /p ǫ /p ǫ /p ǫ − /p 00 0 0 0 1 (cid:4) HAPTER 6. RESTRICTION MAPS ON LIE ALGEBRAS OF 1-DIMENSIONAL CENTRE Lemma 6.2.4 We have L , ( δ ) ∼ = h x , . . . , x | [ x , x ] = x , [ x , x ] = x , [ x , x ] = x , x [ p ]4 = x i ,for every δ ∈ F ∗ . Proof. δ − δ − δ − . (cid:4) L , L , = h x , . . . , x | [ x , x ] = x , [ x , x ] = x , [ x , x ] = x , [ x , x ] = x i . We have Z ( L , ) = h x i F . Let L = L , h x i F = h x , x , x , x | [ x , x ] = x , [ x , x ] = x i ∼ = L , . The group Aut( L ) consists of invertible matrices of the form a a a a a r a a a a a r , where r = a a = 0 . Lemma 6.3.1 Let K = L , and [ p ] : K → K be a p -map on K and let L = KM where M = h x i F .Then K ∼ = L θ where θ = (∆ + ∆ , ω ) ∈ Z ( L, F ) . Proof. Let π : K → L be the projection map. We have the exact sequence0 → M → K → L → . Let σ : L → K such that x i x i , 1 i σ is an injective linear map and πσ = 1 L . Now,we define φ : L × L → M by φ ( x i , x j ) = [ σ ( x i ) , σ ( x j )] − σ ([ x i , x j ]), 1 i, j ω : L → M by ω ( x ) = σ ( x ) [ p ] − σ ( x [ p ] ). Note that φ ( x , x ) = [ σ ( x ) , σ ( x )] − σ ([ x , x ]) = [ x , x ] = x ; φ ( x , x ) = [ σ ( x ) , σ ( x )] − σ ([ x , x ]) = x ; φ ( x , x ) = [ σ ( x ) , σ ( x )] − σ ([ x , x ]) = 0;Similarly, we can show that φ ( x , x ) = φ ( x , x ) = φ ( x , x ) = 0 . Therefore, φ = ∆ + ∆ . Now,by Lemma 2.2.2, we have θ = (∆ + ∆ , ω ) ∈ Z ( L, F ) and K ∼ = L θ . (cid:4) Note that by Theorem 2.3.2, there are four non-isomorphic restricted Lie algebra structures on L given by the following p -maps: HAPTER 6. RESTRICTION MAPS ON LIE ALGEBRAS OF 1-DIMENSIONAL CENTRE p -map;I.2 x [ p ]1 = x ;I.3 x [ p ]2 = ξx ;I.4 x [ p ]3 = x .We make L into a restricted Lie algebra by equipping it with each of the above p -maps. Then,in each case, we find all possible orbit representatives of the form (∆ + ∆ , ω ) under the actionof Aut p ( L ) on H ( L, F ). By Lemma 6.3.1, we do get all possible p -maps on L , with the propertythat x [ p ]5 = 0 .Consider the case I.2 where the p -map of L is given by x [ p ]1 = x . Note that L [ p ] = h x i F . Let[( φ, ω )] ∈ H ( L, F ). Then we must have φ ( x, y [ p ] ) = 0 , for all x, y ∈ L , where φ = a ∆ + b ∆ + c ∆ + d ∆ + e ∆ + f ∆ , for some a, b, c, d, e, f ∈ F . Hence, φ ( x , x ) = 0 which implies that c = 0 . Since φ = ∆ +∆ gives us L , , we deduce by Lemma 6.3.1 that L , cannot be constructedin this case. Similarly we can show that in cases I.3 and I.4 we also get c = 0 . It remains to considerthe case I.1. L , / h x i via the trivial p -map First, we find a basis for Z ( L, F ). Let ( φ, ω ) = ( a ∆ + b ∆ + c ∆ + d ∆ + e ∆ + f ∆ , αf + βf + γf + δf ) ∈ Z ( L, F ). Then we must have δ φ ( x, y, z ) = 0 and φ ( x, y [ p ] ) = 0 , for all x, y, z ∈ L .Therefore,0 = ( δ φ )( x , x , x ) = φ ([ x , x ] , x ) + φ ([ x , x ] , x ) + φ ([ x , x ] , x ) = φ ( x , x ) , and0 = ( δ φ )( x , x , x ) = φ ([ x , x ] , x ) + φ ([ x , x ] , x ) + φ ([ x , x ] , x ) = φ ( x , x ) . Thus, we get e = f = 0 . Since the p -map is trivial, φ ( x, y [ p ] ) = φ ( x, 0) = 0 , for all x, y ∈ L .Therefore, a basis for Z ( L, F ) is as follows:(∆ , , (∆ , , (∆ , , (∆ , , (0 , f ) , (0 , f ) , (0 , f ) , (0 , f ) . Next, we find a basis for B ( L, F ). Let ( φ, ω ) ∈ B ( L, F ). Since B ( L, F ) ⊆ Z ( L, F ), we have( φ, ω ) = ( a ∆ + b ∆ + c ∆ + d ∆ , αf + βf + γf + δf ). So, there exists a linear map ψ : L → F such that δ ψ ( x, y ) = φ ( x, y ) and ˜ ψ ( x ) = ω ( x ), for all x, y ∈ L . So, we have c = φ ( x , x ) = δ ψ ( x , x ) = ψ ([ x , x ]) = 0 . Similarly, we can show that d = 0 . Also, we have α = ω ( x ) = ˜ ψ ( x ) = ψ ( x [ p ]1 ) = 0 . Similarly, we can show that β = γ = δ = 0 . Therefore, ( φ, ω ) = ( a ∆ + b ∆ , 0) and hence B ( L, F ) = h (∆ , , (∆ , i F . We deduce that a basis for H ( L, F ) is as follows:[(∆ , , [(∆ , , [(0 , f )] , [(0 , f )] , [(0 , f )] , [(0 , f )] . Let [( φ, ω )] ∈ H ( L, F ). Then we have φ = a ∆ + b ∆ , for some a, b ∈ F . Suppose that Aφ = a ′ ∆ + b ′ ∆ , for some a ′ , b ′ ∈ F . We determine a ′ , b ′ . Note that Aφ ( x , x ) = φ ( Ax , Ax ) = φ ( a x + a x + a x + a x , a rx ) = a ra ; Aφ ( x , x ) = φ ( Ax , Ax ) = φ ( a x + a x + a x , rx + a a x ) = a rb. HAPTER 6. RESTRICTION MAPS ON LIE ALGEBRAS OF 1-DIMENSIONAL CENTRE (cid:18) a ′ b ′ (cid:19) = (cid:18) a r a r (cid:19) (cid:18) ab (cid:19) . (6.3)The orbit with representative (cid:18) (cid:19) of this action gives us L , .Also, we have ω = αf + βf + γf + δf , for some α, β, γ, δ ∈ F . Suppose that Aω = α ′ f + β ′ f + γ ′ f + δ ′ f , for some α ′ , β ′ , γ ′ , δ ′ ∈ F . We have Aω ( x ) = ω ( Ax ) = ω ( a x + a x + a x + a x ) = a p α + a p β + a p γ + a p δ ; Aω ( x ) = a p β + a p γ + a p δ ; Aω ( x ) = r p γ + a p a p δ ; Aω ( x ) = a p r p δ. In the matrix form we can write this as α ′ β ′ γ ′ δ ′ = a p a p a p a p a p a p a p r p a p a p a p r p αβγδ . (6.4)Thus, we can write Equations (6.3) and (6.4) together as follows: h r (cid:18) a a (cid:19) , a p a p a p a p a p a p a p r p a p a p a p r p ih (cid:18) ab (cid:19) , αβγδ i = h (cid:18) a ′ b ′ (cid:19) , α ′ β ′ γ ′ δ ′ i . Now we find the representatives of the orbits of the action of Aut( L ) on the set of ω ’s such that theorbit represented by (cid:18) (cid:19) is preserved under the action of Aut( L ) on the set of φ ’s.Let ν = αβγδ ∈ F . If ν = , then { ν } is clearly an Aut( L )-orbit. Let ν = 0 . Suppose that δ = 0 . Then h (cid:18) (cid:19) , − α/δ − β/δ ih (cid:18) (cid:19) , αβγδ i = h (cid:18) (cid:19) , γδ i , and h (cid:18) (cid:19) , − γ/δ γ /δ − γ/δ ih (cid:18) (cid:19) , γδ i = h (cid:18) (cid:19) , δ i . Next, if δ = 0 , but γ = 0 , then h (cid:18) (cid:19) , − α/γ 00 1 − β/γ 00 0 1 − β/γ ih (cid:18) (cid:19) , αβγ i = h (cid:18) (cid:19) , γ i . HAPTER 6. RESTRICTION MAPS ON LIE ALGEBRAS OF 1-DIMENSIONAL CENTRE γ = δ = 0 , but β = 0 , then h (cid:18) (cid:19) , − α/β ih (cid:18) (cid:19) , αβ i = h (cid:18) (cid:19) , β i . Finally, if β = γ = δ = 0 , but α = 0 , then we have h (cid:18) (cid:19) , α i .Thus the following elements are Aut( L )-orbit representatives: , α , β , γ , δ . Theorem 6.3.2 The list of all restricted Lie algebra structures on L , , up to isomorphism, is asfollows: L , = h x , . . . , x | [ x , x ] = x , [ x , x ] = x , [ x , x ] = x , [ x , x ] = x i ; L , ( α ) = h x , . . . , x | [ x , x ] = x , [ x , x ] = x , [ x , x ] = x , [ x , x ] = x , x [ p ]1 = αx i ; L , ( β ) = h x , . . . , x | [ x , x ] = x , [ x , x ] = x , [ x , x ] = x , [ x , x ] = x , x [ p ]2 = βx i ; L , ( γ ) = h x , . . . , x | [ x , x ] = x , [ x , x ] = x , [ x , x ] = x , [ x , x ] = x , x [ p ]3 = γx i ; L , ( δ ) = h x , . . . , x | [ x , x ] = x , [ x , x ] = x , [ x , x ] = x , [ x , x ] = x , x [ p ]4 = δx i where α, γ, δ ∈ T , and β ∈ T , . The automorphism group consists of a a a a a a a a a a a a a u v a , where u = a a + a a − a a , v = a a + a a . Lemma 6.3.3 We have L , ( α ) ∼ = L , ( α ) if and only if α /α = ǫ p − , for some ǫ ∈ F ∗ . Proof. Suppose that f = ( a ij ) : L , ( α ) → L , ( α ) is an isomorphism. Then we have f ( x [ p ]1 ) = f ( x ) [ p ] which implies that α /α = a p − . To prove the converse, suppose that α /α = ǫ p − , forsome ǫ ∈ F ∗ . It is easy to see that the following is an isomorphism from L , ( α ) to L , ( α ): ǫ ǫ ǫ ǫ 00 0 0 0 ǫ (cid:4) HAPTER 6. RESTRICTION MAPS ON LIE ALGEBRAS OF 1-DIMENSIONAL CENTRE Lemma 6.3.4 We have L , ( β ) ∼ = L , ( β ) if and only if β /β = ǫ p − , for some ǫ ∈ F ∗ . Proof. Suppose that f = ( a ij ) : L , ( β ) → L , ( β ) is an isomorphism. Then we have f ( x [ p ]2 ) = f ( x ) [ p ] which implies that α /α = a p − . To prove the converse, suppose that β /β = ǫ p − , forsome ǫ ∈ F ∗ . It is easy to see that the following is an isomorphism from L , ( α ) to L , ( α ): ǫ ǫ ǫ ǫ 00 0 0 0 ǫ (cid:4) Lemma 6.3.5 We have L , ( γ ) ∼ = L , ( γ ) if and only if γ /γ = ǫ p − , for some ǫ ∈ F ∗ . Proof. Suppose that f = ( a ij ) : L , ( γ ) → L , ( γ ) is an isomorphism. Then we have f ( x [ p ]3 ) = f ( x ) [ p ] which implies that γ /γ = a p − . To prove the converse, suppose that γ /γ = a p − , forsome ǫ ∈ F ∗ . It is easy to see that the following is an isomorphism from L , ( γ ) to L , ( γ ): ǫ ǫ ǫ ǫ 00 0 0 0 ǫ (cid:4) Lemma 6.3.6 We have L , ( δ ) ∼ = L , ( δ ) if and only if δ /δ = ǫ p − , for some ǫ ∈ F ∗ . Proof. Suppose that f = ( a ij ) : L , ( δ ) → L , ( δ ) is an isomorphism. Then we have f ( x [ p ]4 ) = f ( x ) [ p ] which implies that δ /δ = a p − . To prove the converse, suppose that δ /δ = ǫ p − , forsome ǫ ∈ F ∗ . It is easy to see that the following is an isomorphism from L , ( δ ) to L , ( δ ): ǫ ǫ ǫ ǫ 00 0 0 0 ǫ (cid:4) L , L , = h x , . . . , x | [ x , x ] = x , [ x , x ] = x , [ x , x ] = x i . We have Z ( L , ) = h x i F . Let L = L , h x i ∼ = L , , HAPTER 6. RESTRICTION MAPS ON LIE ALGEBRAS OF 1-DIMENSIONAL CENTRE L , = h x , . . . , x | [ x , x ] = x , [ x , x ] = x i . Note that the group Aut( L ) consists ofinvertible matrices of the form a a a a a r a a a a a r , where r = a a = 0 . Lemma 6.4.1 Let K = L , and [ p ] : K → K be a p -map on K and let L = KM where M = h x i F .Then K ∼ = L θ where θ = (∆ , ω ) ∈ Z ( L, F ) . Proof. Let π : K → L be the projection map. We have the exact sequence0 → M → K → L → . Let σ : L → K such that x i x i , 1 i σ is an injective linear map and πσ = 1 L . Now,we define φ : L × L → M by φ ( x i , x j ) = [ σ ( x i ) , σ ( x j )] − σ ([ x i , x j ]), 1 i, j ω : L → M by ω ( x ) = σ ( x ) [ p ] − σ ( x [ p ] ). Note that φ ( x , x ) = [ σ ( x ) , σ ( x )] − σ ([ x , x ]) = [ x , x ] = x ; φ ( x , x ) = [ σ ( x ) , σ ( x )] − σ ([ x , x ]) = 0;Similarly, we can show that φ ( x , x ) = φ ( x , x ) = φ ( x , x ) = φ ( x , x ) = 0 . Therefore, φ = ∆ .Now, by Lemma 2.2.2, we have θ = (∆ , ω ) ∈ Z ( L, F ) and K ∼ = L θ . (cid:4) Note that by Theorem 2.3.2, there are four non-isomorphic restricted Lie algebra structures on L , given by the following p -maps:I.1 Trivial p -map;I.2 x [ p ]1 = x ;I.3 x [ p ]2 = ξx ;I.4 x [ p ]3 = x .In the following subsections, we make L into a restricted Lie algebra by equipping it with each ofthe above p -maps. Then, in each case, we find all possible orbit representatives of the form (∆ , ω )under the action of Aut p ( L ) on H ( L, F ). By Lemma 6.4.1, we do get all possible p -maps on L , with the property that x [ p ]5 = 0 .Consider the case I.2 where the p -map of L is given by x [ p ]1 = x . Let [( φ, ω )] ∈ H ( L, F ). Thenwe must have φ ( x, y [ p ] ) = 0 , for all x, y ∈ L , where φ = a ∆ + b ∆ + c ∆ + d ∆ + e ∆ + f ∆ ,for some a, b, c, d, e, f ∈ F . Since L [ p ] = h x i F , we get φ ( x , x ) = 0 which implies that c = 0 . Since φ = ∆ gives us L , , we deduce by Lemma 6.4.1 that L , cannot be constructed in this case.Similarly, we can show in cases I.3 and I.4 we also get c = 0 . It remains to consider the case I.1. HAPTER 6. RESTRICTION MAPS ON LIE ALGEBRAS OF 1-DIMENSIONAL CENTRE L , / h x i via the trivial p -map First, we find a basis for Z ( L, F ). Let ( φ, ω ) = ( a ∆ + b ∆ + c ∆ + d ∆ + e ∆ + f ∆ , αf + βf + γf + δf ) ∈ Z ( L, F ). Then we must have δ φ ( x, y, z ) = 0 and φ ( x, y [ p ] ) = 0 , for all x, y, z ∈ L .Therefore,0 = ( δ φ )( x , x , x ) = φ ([ x , x ] , x ) + φ ([ x , x ] , x ) + φ ([ x , x ] , x ) = φ ( x , x ) , and0 = ( δ φ )( x , x , x ) = φ ([ x , x ] , x ) + φ ([ x , x ] , x ) + φ ([ x , x ] , x ) = φ ( x , x ) . Thus, we get e = f = 0 . Since the p -map is trivial, φ ( x, y [ p ] ) = φ ( x, 0) = 0 , for all x, y ∈ L .Therefore, a basis for Z ( L, F ) is as follows:(∆ , , (∆ , , (∆ , , (∆ , , (0 , f ) , (0 , f ) , (0 , f ) , (0 , f ) . Next, we find a basis for B ( L, F ). Let ( φ, ω ) ∈ B ( L, F ). Since B ( L, F ) ⊆ Z ( L, F ), we have( φ, ω ) = ( a ∆ + b ∆ + c ∆ + d ∆ , αf + βf + γf + δf ). So, there exists a linear map ψ : L → F such that δ ψ ( x, y ) = φ ( x, y ) and ˜ ψ ( x ) = ω ( x ), for all x, y ∈ L . So, we have c = φ ( x , x ) = δ ψ ( x , x ) = ψ ([ x , x ]) = 0 . Similarly, we can show that d = 0 . Also, we have α = ω ( x ) = ˜ ψ ( x ) = ψ ( x [ p ]1 ) = 0 . Similarly, we can show that β = γ = δ = 0 . Therefore, ( φ, ω ) = ( a ∆ + b ∆ , 0) and hence B ( L, F ) = h (∆ , , (∆ , i F . We deduce that a basis for H ( L, F ) is as follows:[(∆ , , [(∆ , , [(0 , f )] , [(0 , f )] , [(0 , f )] , [(0 , f )] . Let [( φ, ω )] ∈ H ( L, F ). Then we have φ = a ∆ + b ∆ , for some a, b ∈ F . Suppose that Aφ = a ′ ∆ + b ′ ∆ , for some a ′ , b ′ ∈ F . We determine a ′ , b ′ . Note that Aφ ( x , x ) = φ ( Ax , Ax ) = φ ( a x + a x + a x + a x , a rx ) = a ra ; Aφ ( x , x ) = φ ( Ax , Ax ) = φ ( a x + a x + a x , rx + a a x ) = a rb. In the matrix form we can write this as (cid:18) a ′ b ′ (cid:19) = (cid:18) a r a r (cid:19) (cid:18) ab (cid:19) . (6.5)The orbit with representative (cid:18) (cid:19) of this action gives us L , .Also, we have ω = αf + βf + γf + δf , for some α, β, γ, δ ∈ F . Suppose that Aω = α ′ f + β ′ f + γ ′ f + δ ′ f , for some α ′ , β ′ , γ ′ , δ ′ ∈ F . We have Aω ( x ) = ω ( Ax ) = ω ( a x + a x + a x + a x ) = a p α + a p β + a p γ + a p δ ; Aω ( x ) = a p β + a p γ + a p δ ; Aω ( x ) = r p γ + a p a p δ ; Aω ( x ) = a p r p δ. HAPTER 6. RESTRICTION MAPS ON LIE ALGEBRAS OF 1-DIMENSIONAL CENTRE α ′ β ′ γ ′ δ ′ = a p a p a p a p a p a p a p r p a p a p a p r p αβγδ . (6.6)Thus, we can write Equations (6.5) and (6.6) together as follows: h r (cid:18) a a (cid:19) , a p a p a p a p a p a p a p r p a p a p a p r p ih (cid:18) ab (cid:19) , αβγδ i = h (cid:18) a ′ b ′ (cid:19) , α ′ β ′ γ ′ δ ′ i . Now we find the representatives of the orbits of the action of Aut( L ) on the set of ω ’s such that theorbit represented by (cid:18) (cid:19) is preserved under the action of Aut( L ) on the set of φ ’s. Let ν = αβγδ ∈ F . If ν = , then { ν } is clearly an Aut( L )-orbit. Let ν = 0 . Suppose that δ = 0 . Then h (cid:18) (cid:19) , − α/δ − β/δ ih (cid:18) (cid:19) , αβγδ i = h (cid:18) (cid:19) , γδ i , and h (cid:18) (cid:19) , − γ/δ γ /δ − γ/δ ih (cid:18) (cid:19) , γδ i = h (cid:18) (cid:19) , δ i , and h δ − /p (cid:18) δ /p δ − /p (cid:19) , δ δ − 00 0 δ − 00 0 0 δ − ih (cid:18) (cid:19) , δ i = h (cid:18) (cid:19) , i . Next, if δ = 0 , but γ = 0 , then h (cid:18) (cid:19) , − α/γ 00 1 − β/γ 00 0 1 − β/γ ih (cid:18) (cid:19) , αβγ i = h (cid:18) (cid:19) , γ i . Next, if γ = δ = 0 , but β = 0 , then h (cid:18) (cid:19) , − α/β ih (cid:18) (cid:19) , αβ i = h (cid:18) (cid:19) , β i . HAPTER 6. RESTRICTION MAPS ON LIE ALGEBRAS OF 1-DIMENSIONAL CENTRE β = γ = δ = 0 , but α = 0 , then we have h α /p (cid:18) α − /p α /p (cid:19) , α − α α 00 0 0 α ih (cid:18) (cid:19) , α i = h (cid:18) (cid:19) , i . Thus the following elements are Aut( L )-orbit representatives: , , β , γ , . Theorem 6.4.2 The list of all restricted Lie algebra structures on L , , up to isomorphism, is asfollows: L , = h x , . . . , x | [ x , x ] = x , [ x , x ] = x , [ x , x ] = x i ; L , = h x , . . . , x | [ x , x ] = x , [ x , x ] = x , [ x , x ] = x , x [ p ]1 = x i ; L , ( β ) = h x , . . . , x | [ x , x ] = x , [ x , x ] = x , [ x , x ] = x , x [ p ]2 = βx i ; L , ( γ ) = h x , . . . , x | [ x , x ] = x , [ x , x ] = x , [ x , x ] = x , x [ p ]3 = γx i ; L , = h x , . . . , x | [ x , x ] = x , [ x , x ] = x , [ x , x ] = x , x [ p ]4 = x i where β, γ ∈ F ∗ . The automorphism group Aut( L , ) consists of a a a a a a a a a a a a a a a a a a a a a . Lemma 6.4.3 We have L , ( β ) ∼ = L , ( β ) if and only if the exists non-zero x, y ∈ F ∗ such that β /β = x y p − . Proof. Let f = ( a ij ) ∈ Aut( L , ). We just need to observe that f is an isomorphism between L , ( β ) and L , ( β ) if and only if f ( x [ p ]2 ) = f ( x ) [ p ] which in turn is equivalent to saying that β /β = a − a p − . (cid:4) Note that the solutions to Equation β /β = x y p − in Lemma 6.4.3 depends on the underlyingfield. For example, over the prime field F p , we have that L , ( β ) ∼ = L , ( β ) if and only if the ratio β /β is a cubic. Lemma 6.4.4 We have L , ( γ ) ∼ = L , ( γ ) if and only if γ /γ ∈ ( F ∗ ) . HAPTER 6. RESTRICTION MAPS ON LIE ALGEBRAS OF 1-DIMENSIONAL CENTRE Proof. Suppose that f = ( a ij ) : L , ( γ ) → L , ( γ ) is an isomorphism. Then we have f ( x [ p ]3 ) = f ( x ) [ p ] which implies that γ /γ = a p − a p − ∈ ( F ∗ ) . To prove the converse, suppose that γ /γ = ǫ , for some ǫ ∈ F ∗ . It is easy to see that the following is an isomorphism from L , ( γ ) to L , ( γ ): ǫ − /p ǫ /p ǫ /p ǫ /p 00 0 0 0 1 . (cid:4) hapter 7 Restriction maps on L , Let K = L , = h x , . . . , x | [ x , x ] = x , [ x , x ] = x i . Then Z ( L , ) = h x , x i F and the group Aut( L , ) consists of invertible matrices of the form a a a a a a a a a a a a a a a a a a a a a . Note that there exists an element αx + βx ∈ Z ( L , ) such that ( αx + βx ) [ p ] = 0 , for some α, β ∈ F . If α = 0 then consider K = h y , . . . , y | [ y , y ] = y , [ y , y ] = y i , where y = x , y = αx + βx , y = x , y = αx + βx , y = x . Let φ : K → K given by x i y i ,for 1 i φ is an isomorphism. Therefore, in this case we can supposethat x [ p ]4 = 0 . If α = 0 then β = 0 and we rescale x so that x [ p ]5 = 0 . Hence we can assume either x [ p ]4 = 0 or x [ p ]5 = 0 . Consider the automorphism of K given by x x , x x , x x , x x and x x . Using this automorphism, we deduce that it is enough to determine all the p -maps on K for which x [ p ]5 = 0 . Lemma 7.0.5 Let K = L , and [ p ] : K → K be a p -map on K such that x [ p ]5 = 0 and let L = KM where M = h x i F . Then K ∼ = L θ where θ = (∆ , ω ) ∈ Z ( L, F ) . Proof. Let π : K → L be the projection map. We have the exact sequence0 → M → K → L → . Let σ : L → K such that x i x i , 1 i σ is an injective linear map and πσ = 1 L . Now,we define φ : L × L → M by φ ( x i , x j ) = [ σ ( x i ) , σ ( x j )] − σ ([ x i , x j ]), 1 i, j ω : L → M by ω ( x ) = σ ( x ) [ p ] − σ ( x [ p ] ). Note that φ ( x , x ) = [ σ ( x ) , σ ( x )] − σ ([ x , x ]) = [ x , x ] = x ; φ ( x , x ) = [ σ ( x ) , σ ( x )] − σ ([ x , x ]) = 0 . HAPTER 7. RESTRICTION MAPS ON L , φ ( x , x ) = φ ( x , x ) = φ ( x , x ) = φ ( x , x ) = 0 . Therefore, φ = ∆ .Now, by Lemma, 2.2.2 we have θ = (∆ , ω ) ∈ Z ( L, F ) and K ∼ = L θ . (cid:4) Let L = L , h x i = h x , x , x , x | [ x , x ] = x i . Then L ∼ = L , and the group Aut( L ) consists of invertible matrices of the form a a a a a a a a a a r , where r = a a − a a = 0 . By Theorem 2.3.2, there are eight non-isomorphic restricted Liealgebra structures on L given by the following p -maps:I.1 Trivial p -map;I.2 x [ p ]1 = x ;I.3 x [ p ]1 = x ;I.4 x [ p ]1 = x , x [ p ]2 = x ;I.5 x [ p ]4 = x ;I.6 x [ p ]4 = x , x [ p ]2 = x ;I.7 x [ p ]3 = x ;I.8 x [ p ]3 = x , x [ p ]2 = x .We make L into a restricted Lie algebra by equipping it with each of the above p -maps. Then, ineach case, we find all possible orbit representatives of the form (∆ , ω ) under the action of Aut p ( L )on H ( L, F ). By Lemma 7.0.5, we do get all possible p -maps on K .Consider the case I.3 where the p -map of L is given by x [ p ]1 = x . Let ( φ, ω ) ∈ Z ( L, F ). Thenwe must have φ ( x, y [ p ] ) = 0 , for all x, y ∈ L , where φ = a ∆ + b ∆ + c ∆ + d ∆ + e ∆ + f ∆ ,for some a, b, c, d, e, f ∈ F . Since L [ p ] = h x i , we get φ ( x , x ) = 0 . So, b = 0 . Since φ = ∆ givesus L , , we deduce by Lemma 7.0.5 that L , cannot be constructed in this case. Similarly, we canshow that in cases I.4, I.5, I.6, and I.8 we also get b = 0 and so we cannot put a p -map on L , . Weconsider the remaining cases in the following sections. L , trivial p -map) First, we find a basis for Z ( L, F ). Let ( φ, ω ) = ( a ∆ + b ∆ + c ∆ + d ∆ + e ∆ + f ∆ , αf + βf + γf + δf ) ∈ Z ( L, F ). Then we must have δ φ ( x, y, z ) = 0 and φ ( x, y [ p ] ) = 0 , for all x, y, z ∈ L .Therefore,0 = ( δ φ )( x , x , x ) = φ ([ x , x ] , x ) + φ ([ x , x ] , x ) + φ ([ x , x ] , x ) = φ ( x , x ) . HAPTER 7. RESTRICTION MAPS ON L , f = 0 . Since the p -map is trivial, φ ( x, y [ p ] ) = φ ( x, 0) = 0 , for all x, y ∈ L . Therefore,a basis for Z ( L, F ) is as follows:(∆ , , (∆ , , (∆ , , (∆ , , (∆ , , (0 , f ) , (0 , f ) , (0 , f ) , (0 , f ) . Next, we find a basis for B ( L, F ). Let ( φ, ω ) ∈ B ( L, F ). Since B ( L, F ) ⊆ Z ( L, F ), we have( φ, ω ) = ( a ∆ + b ∆ + c ∆ + d ∆ + e ∆ , αf + βf + γf + δf ). So, there exists a linear map ψ : L → F such that δ ψ ( x, y ) = φ ( x, y ) and ˜ ψ ( x ) = ω ( x ), for all x, y ∈ L . So, we have b = φ ( x , x ) = δ ψ ( x , x ) = ψ ([ x , x ]) = 0 . Similarly, we can show that c = d = e = 0 . Also, we have α = ω ( x ) = ˜ ψ ( x ) = ψ ( x [ p ]1 ) = 0 . Similarly, we can show that β = γ = δ = 0 . Therefore, ( φ, ω ) = ( a ∆ , 0) and hence B ( L, F ) = h (∆ , i F . We deduce that a basis for H ( L, F ) is as follows:[(∆ , , [(∆ , , [(∆ , , [(∆ , , [(0 , f )] , [(0 , f )] , [(0 , f )] , [(0 , f )] . Let [( φ, ω )] ∈ H ( L, F ). Then we have φ = a ∆ + b ∆ + c ∆ + d ∆ , for some a, b, c, d ∈ F . Supposethat Aφ = a ′ ∆ + b ′ ∆ + c ′ ∆ + d ′ ∆ , for some a ′ , b ′ , c ′ , d ′ ∈ F . We determine a ′ , b ′ , c ′ , d ′ . Notethat Aφ ( x , x ) = a a a + a a b + a a c + a a d ; Aφ ( x , x ) = a rb + a rd ; Aφ ( x , x ) = a a a + a a b + a a c + a a d ; Aφ ( x , x ) = a rb + a rd. In the matrix form we can write this as a ′ b ′ c ′ d ′ = a a a a a a a a a r a ra a a a a a a a a r a r abcd . (7.1)The orbit with representative of this action gives us L , .Also, we have ω = αf + βf + γf + δf , for some α, β, γ, δ ∈ F . Suppose that Aω = α ′ f + β ′ f + γ ′ f + δ ′ f , for some α ′ , β ′ , γ ′ , δ ′ ∈ F . We determine α ′ , β ′ , γ ′ , δ ′ . Note that Aω ( x ) = ω ( Ax ) = ω ( a x + a x + a x + a x ) = a p α + a p β + a p γ + a p δ ; Aω ( x ) = a p α + a p β + a p γ + a p δ ; Aω ( x ) = a p γ + a p δ ; Aω ( x ) = r p δ. In the matrix form we can write this as α ′ β ′ γ ′ δ ′ = a p a p a p a p a p a p a p a p a p a p r p αβγδ . (7.2) HAPTER 7. RESTRICTION MAPS ON L , h a a a a a a a a a r a ra a a a a a a a a r a r , a p a p a p a p a p a p a p a p a p a p r p ih abcd , αβγδ i = h a ′ b ′ c ′ d ′ , α ′ β ′ γ ′ δ ′ i . Now we find the representatives of the orbits of the action of Aut( L ) on the set of ω ’s such that theorbit represented by is preserved under the action of Aut( L ) on the set of φ ’s.Let ν = αβγδ ∈ F . If ν = , then { ν } is clearly an Aut( L )-orbit. Let ν = 0 . Suppose that δ = 0 . Then h − γ/δ − γ/δ , − α/δ − β/δ − γ/δ ih , αβγδ i = h , δ i , and h δ − /p δ /p 00 0 0 δ − /p , /δ δ 00 0 0 1 /δ ih , δ i = h , i . Next, if δ = 0 , but γ = 0 , then h , − α/γ 00 1 − β/γ 00 0 1 00 0 0 1 ih , αβγ i = h , γ i , and h γ /p γ − /p 00 0 0 γ /p , γ /γ 00 0 0 γ ih , γ i = h , i . Next, if δ = γ = 0 , but β = 0 , then h − α/β ) /p 00 1 0 ( − α/β ) /p , − α/β ih , αβ i = h , β i , and h /β ) /p /β ) /p 00 0 0 (1 /β ) /p , /β /β ih , β i = h , i . HAPTER 7. RESTRICTION MAPS ON L , δ = γ = β = 0 , but α = 0 , then h α − /p α /p 00 0 0 α − /p , /α α 00 0 0 1 /α ih , α i = h , i . Thus the following elements are Aut( L )-orbit representatives: , , , , . Therefore, the corresponding restricted Lie algebra structures are as follows: K = h x , . . . , x | [ x , x ] = x , [ x , x ] = x i ; K = h x , . . . , x | [ x , x ] = x , [ x , x ] = x , x [ p ]1 = x i ; K = h x , . . . , x | [ x , x ] = x , [ x , x ] = x , x [ p ]2 = x i ; K = h x , . . . , x | [ x , x ] = x , [ x , x ] = x , x [ p ]3 = x i ; K = h x , . . . , x | [ x , x ] = x , [ x , x ] = x , x [ p ]4 = x i . L, x [ p ]1 = x ) First, we find a basis for Z ( L, F ). Let ( φ, ω ) = ( a ∆ + b ∆ + c ∆ + d ∆ + e ∆ + f ∆ , αf + βf + γf + δf ) ∈ Z ( L, F ). Then we must have δ φ ( x, y, z ) = 0 and φ ( x, y [ p ] ) = 0 , for all x, y, z ∈ L .Therefore, we have0 = ( δ φ )( x , x , x ) = φ ([ x , x ] , x ) + φ ([ x , x ] , x ) + φ ([ x , x ] , x ) = φ ( x , x ) . Thus, we get f = 0 . Also, we have φ ( x, y [ p ] ) = 0 . Therefore, φ ( x, x ) = 0 , for all x ∈ L and hence φ ( x , x ) = φ ( x , x ) = φ ( x , x ) = 0 which implies that c = e = f = 0 . Therefore, Z ( L, F ) has abasis consisting of: (∆ , , (∆ , , (∆ , , (0 , f ) , (0 , f ) , (0 , f ) , (0 , f ) . Next, we find a basis for B ( L, F ). Let ( φ, ω ) ∈ B ( L, F ). Since B ( L, F ) ⊆ Z ( L, F ), we have( φ, ω ) = ( a ∆ + b ∆ + c ∆ , αf + βf + γf + δf ). Note that there exists a linear map ψ : L → F such that δ ψ ( x, y ) = φ ( x, y ) and ˜ ψ ( x ) = ω ( x ), for all x, y ∈ L . So, we have a = φ ( x , x ) = δ ψ ( x , x ) = ψ ([ x , x ]) = ψ ( x ) , and b = φ ( x , x ) = δ ψ ( x , x ) = ψ ([ x , x ]) = 0 . Similarly, we can show that c = 0 . Also, we have α = ω ( x ) = ˜ ψ ( x ) = ψ ( x [ p ]1 ) = ψ ( x ) , and β = ω ( x ) = ˜ ψ ( x ) = ψ ( x [ p ]2 ) = 0 . HAPTER 7. RESTRICTION MAPS ON L , γ = δ = 0 . Note that ψ ( x ) = a = α . Therefore, ( φ, ω ) = ( a ∆ , af )and hence B ( L, F ) = h (∆ , f ) i F . Note that[(∆ , , [(∆ , , [(∆ , , [0 , f ] , [(0 , f )] , [(0 , f )] , [(0 , f )]spans H ( L, F ). Since [(∆ , , f )] = [(∆ , f )] = [0], then [(0 , f )] is an scalar multiple of[(∆ , H ( L, F ). Note that dim H = dim Z − dim B = 6 . Therefore,[(∆ , , [(∆ , , [(∆ , , [(0 , f )] , [(0 , f )] , [(0 , f )] . forms a basis for H ( L, F ).Note that the group Aut( L ) in this case consists of invertible matrices of the form a a a a a a a a a a r , where r = a a − a a = a p and a = 0 .Let [( φ, ω )] ∈ H ( L, F ). Then we have φ = a ∆ + b ∆ + c ∆ , for some a, b, c ∈ F . Supposethat Aφ = a ′ ∆ + b ′ ∆ + c ′ ∆ , for some a ′ , b ′ , c ′ ∈ F . We determine a ′ , b ′ , c ′ . Note that Aφ ( x , x ) = φ ( a x + a x + a x + a x , a x + a x ) = a a a + a a c ; Aφ ( x , x ) = φ ( a x + a x + a x + a x , a x + a x + a x + a x )= ( a a − a a ) a + ( a a − a a ) b + ( a a − a a ) c ; Aφ ( x , x ) = φ ( a x + a x + a x + a x , a x + a x ) = a a a + a a c. In the matrix form we can write this as a ′ b ′ c ′ = a a a a a a a a a a − a a a a abc . (7.3)The orbit with representative of this action gives us L , .Also, we have ω = βf + γf + δf , for some β, γ, δ ∈ F . Suppose that Aω = β ′ f + γ ′ f + δ ′ f ,for some β ′ , γ ′ , δ ′ ∈ F . We have Aω ( x ) = a p β + a p γ + a p δ ; Aω ( x ) = a p γ + a p δ ; Aω ( x ) = r p δ. In the matrix form we can write this as β ′ γ ′ δ ′ = a p a p a p a p a p a p βγδ . (7.4) HAPTER 7. RESTRICTION MAPS ON L , h a a a a a a a a a a − a a a a ! , a p a p a p a p a p a p ih abc ! , βγδ ! i = h a ′ b ′ c ′ ! , β ′ γ ′ δ ′ ! i . Note that Aφ ( x , x ) = Aφ ( x , x ) = Aφ ( x , x ) = 0 and Aω ( x ) = 0 which imply that a p β + a p γ + a p δ = 0 . Now we find the representatives of the orbits of the action of Aut( L ) on the set of ω ’s such that the orbit represented by is preserved under the action of Aut( L ) on the set of φ ’s. Note that we need to have a p β + a p γ + a p δ = 0 .Let ν = βγδ ∈ F . If ν = , then { ν } is clearly an Aut( L )-orbit. Let ν = 0 . Suppose that δ = 0 . Then h , − β/δ − γ/δ ih , βγδ i = h , δ i , and h /δ ) /p 00 0 δ p − p , (1 /δ ) p − p δ /p 00 0 1 /δ ih , δ i = h , i . Next, if δ = 0 , but γ = 0 , then h γ 00 0 γ p − p , γ p − /γ 00 0 γ p ih , βγ i = h , γ p − β i . If β = 0 , then we have . If β = 0 , then we rename γ p − β with β and we have β .Finally, if δ = γ = 0 , but β = 0 , then we have β . Thus the following elements are Aut( L )-orbitrepresentatives: , β , , , β . Lemma 7.2.1 The vectors β and β are in the same Aut( L ) -orbit if and only if β = β . HAPTER 7. RESTRICTION MAPS ON L , Proof. First assume that β and β are in the same Aut( L )-orbit. Then h a a a a a a a a a a − a a a a ! , a p a p a p a p a p a p ih ! , β ! i = h ! , β ! i . From this we obtain that a a = 1 (7.5) a a = 0 (7.6) a p β + a p = β (7.7) a p = 1 . (7.8)Then using Equation (7.6) and (7.7), we get that β β − = a p . Note that we have r = a p which getthat a = a p − and hence β β − = a p = ( a p ) p − . Now using Equation (7.5) and (7.8), we obtainthat a p = 1 which implies that β β − = 1 . Therefore, β = β . The converse is clear. (cid:4) Therefore, the corresponding restricted Lie algebra structures are as follows: K = h x , . . . , x | [ x , x ] = x , [ x , x ] = x , x [ p ]1 = x i ; K ( β ) = h x , . . . , x | [ x , x ] = x , [ x , x ] = x , x [ p ]1 = x , x [ p ]2 = βx i ; K = h x , . . . , x | [ x , x ] = x , [ x , x ] = x , x [ p ]1 = x , x [ p ]3 = x i ; K = h x , . . . , x | [ x , x ] = x , [ x , x ] = x , x [ p ]1 = x , x [ p ]4 = x i ; K ( β ) = h x , . . . , x | [ x , x ] = x , [ x , x ] = x , x [ p ]1 = x , x [ p ]2 = βx , x [ p ]3 = x i where β ∈ F ∗ . Lemma 7.2.2 We have K ( β ) ∼ = h x , . . . , x | [ x , x ] = x , [ x , x ] = x , x [ p ]1 = x , x [ p ]2 = x i , forevery β ∈ F ∗ . Proof. Note that the following automorphism of L , gives us the required result: /β /β . (cid:4) Lemma 7.2.3 We have K ( β ) ∼ = K ( β ) if and only if β /β = ǫ p − , for some ǫ ∈ F ∗ . Proof. Let f = ( a ij ) ∈ Aut( K ). Then f is an isomorphism from K ( β ) to K ( β ) if and onlyif a = a p − , a = a p − , β /β = a p a − a − , HAPTER 7. RESTRICTION MAPS ON L , β /β = a p ( p − . To prove the converse, suppose that β /β = ǫ p − , for some ǫ ∈ F ∗ . Then, it is easy to see that thefollowing is an isomorphism from K ( β ) to K ( β ): ǫ ( p − /p ǫ ( p − /p ǫ /p ǫ ( p − /p 00 0 0 0 ǫ /p . (cid:4) L, x [ p ]3 = x ) First, we find a basis for Z ( L, F ). Let ( φ, ω ) = ( a ∆ + b ∆ + c ∆ + d ∆ + e ∆ + f ∆ , αf + βf + γf + δf ) ∈ Z ( L, F ). Then we must have δ φ ( x, y, z ) = 0 and φ ( x, y [ p ] ) = 0 , for all x, y, z ∈ L .Therefore, we have0 = ( δ φ )( x , x , x ) = φ ([ x , x ] , x ) + φ ([ x , x ] , x ) + φ ([ x , x ] , x ) = φ ( x , x ) . Thus, we get f = 0 . Also, we have φ ( x, y [ p ] ) = 0 . Therefore, φ ( x, x ) = 0 , for all x ∈ L and hence φ ( x , x ) = φ ( x , x ) = φ ( x , x ) = 0 which implies that c = e = f = 0 . Therefore, Z ( L, F ) has abasis consisting of: (∆ , , (∆ , , (∆ , , (0 , f ) , (0 , f ) , (0 , f ) , (0 , f ) . Next, we find a basis for B ( L, F ). Let ( φ, ω ) ∈ B ( L, F ). Since B ( L, F ) ⊆ Z ( L, F ), we have( φ, ω ) = ( a ∆ + b ∆ + c ∆ , αf + βf + γf + δf ). Note that there exists a linear map ψ : L → F such that δ ψ ( x, y ) = φ ( x, y ) and ˜ ψ ( x ) = ω ( x ), for all x, y ∈ L . So, we have a = φ ( x , x ) = δ ψ ( x , x ) = ψ ([ x , x ]) = ψ ( x ) , and b = φ ( x , x ) = δ ψ ( x , x ) = ψ ([ x , x ]) = 0 . Similarly, we can show that c = 0 . Also, we have γ = ω ( x ) = ˜ ψ ( x ) = ψ ( x [ p ]3 ) = ψ ( x ) , and α = ω ( x ) = ˜ ψ ( x ) = ψ ( x [ p ]1 ) = 0 . Similarly, we can show that β = δ = 0 . Note that ψ ( x ) = a = γ . Therefore, ( φ, ω ) = ( a ∆ , af )and hence B ( L, F ) = h (∆ , f ) i F . Note that[(∆ , , [(∆ , , [(∆ , , [0 , f ] , [(0 , f )] , [(0 , f )] , [(0 , f )]spans H ( L, F ). Since [(∆ , , f )] = [(∆ , f )] = [0], then [(0 , f )] is an scalar multiple of[(∆ , H ( L, F ). Note that dim H = dim Z − dim B = 6 . Therefore,[(∆ , , [(∆ , , [(∆ , , [(0 , f )] , [(0 , f )] , [(0 , f )] . HAPTER 7. RESTRICTION MAPS ON L , H ( L, F ).Note that the group Aut( L ) in this case consists of invertible matrices of the form a a a a a a a a a a r , where r = a a − a a = a p and a = a = 0 .Let [( φ, ω )] ∈ H ( L, F ). Then we have φ = a ∆ + b ∆ + c ∆ , for some a, b, c ∈ F . Supposethat Aφ = a ′ ∆ + b ′ ∆ + c ′ ∆ , for some a ′ , b ′ , c ′ ∈ F . We determine a ′ , b ′ , c ′ . Note that Aφ ( x , x ) = φ ( a x + a x + a x + a x , a x + a x ) = a a a + a a c ; Aφ ( x , x ) = φ ( a x + a x + a x + a x , a x + a x + a x + a x )= ( a a − a a ) a + ( a a − a a ) b + ( a a − a a ) c ; Aφ ( x , x ) = φ ( a x + a x + a x + a x , a x + a x ) = a a a + a a c. In the matrix form we can write this as a ′ b ′ c ′ = a a a a a a − a a a a a a abc . (7.9)The orbit with representative of this action gives us L , .Also, we have ω = αf + βf + δf , for some α, β, δ ∈ F . Suppose that Aω = α ′ f + β ′ f + δ ′ f ,for some α ′ , β ′ , δ ′ ∈ F . We have Aω ( x ) = a p α + a p β + a p δ ; Aω ( x ) = a p α + a p β + a p δ ; Aω ( x ) = r p δ. In the matrix form we can write this as α ′ β ′ δ ′ = a p a p a p a p a p a p a p αβδ . (7.10)Thus, we can write Equations (7.9) and (7.10) together as follows: h a a a a a a − a a a a a a , a p a p a p a p a p a p a p ih abc , αβδ i = h a ′ b ′ c ′ , α ′ β ′ δ ′ i . Note that Aφ ( x , x ) = Aφ ( x , x ) = Aφ ( x , x ) = 0 and Aω ( x ) = 0 which imply that a p δ = 0 .Now we find the representatives of the orbits of the action of Aut( L ) on the set of ω ’s such that HAPTER 7. RESTRICTION MAPS ON L , is preserved under the action of Aut( L ) on the set of φ ’s. Note thatwe take a = 0 . Therefore, a p δ = 0 . Let ν = αβδ ∈ F . If ν = , then { ν } is clearly anAut( L )-orbit. Let ν = 0 . Suppose that δ = 0 . Then h , − α/δ − β/δ ih , αβδ i = h , δ i , and h /δ ) /p 00 0 (1 /δ ) p +2 p , δ /p /δ ) p +1 p 00 0 1 /δ ih , δ i = h , i . Next, if δ = 0 , but β = 0 , then h − α/β ) /p , − α/β 00 1 00 0 1 ih , αβ i = h , β i . Finally, if δ = β = 0 , but α = 0 , then h α 00 0 α p +2 p , /α α p +1 00 0 α p ih , α i = h , i . Thus the following elements are Aut( L )-orbit representatives: , , β , . Therefore, the corresponding restricted Lie algebra structures are as follows: K = h x , . . . , x | [ x , x ] = x , [ x , x ] = x , x [ p ]3 = x i ; K = h x , . . . , x | [ x , x ] = x , [ x , x ] = x , x [ p ]1 = x , x [ p ]3 = x i ; K ( β ) = h x , . . . , x | [ x , x ] = x , [ x , x ] = x , x [ p ]2 = βx , x [ p ]3 = x i ; K = h x , . . . , x | [ x , x ] = x , [ x , x ] = x , x [ p ]3 = x , x [ p ]4 = x i where β ∈ F ∗ . Lemma 7.3.1 We have K ( β ) and K ( β ) are isomorphic if and only if β /β = ǫ p +1 , for some ǫ ∈ F ∗ . Proof. Let f = ( a ij ) ∈ Aut( K ). Then f is an isomorphism from K ( β ) to K ( β ) if and onlyif a = a p a − , β /β = a p a − a − , HAPTER 7. RESTRICTION MAPS ON L , β /β = ( a a ) p +1 . To prove the converse, suppose that β /β = ǫ p +1 , for some ǫ ∈ F ∗ . Then, it is easy to see that thefollowing is an isomorphism from K ( β ) to K ( β ): ǫ − /p ǫ /p ǫ − /p . (cid:4) The following is the list of all restricted Lie algebra structures on L , and yet, as we shall see below,we prove that some of them are isomorphic. K = h x , . . . , x | [ x , x ] = x , [ x , x ] = x i ; K = h x , . . . , x | [ x , x ] = x , [ x , x ] = x , x [ p ]1 = x i ; K = h x , . . . , x | [ x , x ] = x , [ x , x ] = x , x [ p ]2 = x i ; K = h x , . . . , x | [ x , x ] = x , [ x , x ] = x , x [ p ]3 = x i ; K = h x , . . . , x | [ x , x ] = x , [ x , x ] = x , x [ p ]4 = x i ; K = h x , . . . , x | [ x , x ] = x , [ x , x ] = x , x [ p ]1 = x i ; K = h x , . . . , x | [ x , x ] = x , [ x , x ] = x , x [ p ]1 = x , x [ p ]2 = x i ; K = h x , . . . , x | [ x , x ] = x , [ x , x ] = x , x [ p ]1 = x , x [ p ]3 = x i ; K = h x , . . . , x | [ x , x ] = x , [ x , x ] = x , x [ p ]1 = x , x [ p ]4 = x i ; K ( β ) = h x , . . . , x | [ x , x ] = x , [ x , x ] = x , x [ p ]1 = x , x [ p ]2 = βx , x [ p ]3 = x i ; K = h x , . . . , x | [ x , x ] = x , [ x , x ] = x , x [ p ]3 = x i ; K = h x , . . . , x | [ x , x ] = x , [ x , x ] = x , x [ p ]1 = x , x [ p ]3 = x i ; K ( β ) = h x , . . . , x | [ x , x ] = x , [ x , x ] = x , x [ p ]2 = βx , x [ p ]3 = x i ; K = h x , . . . , x | [ x , x ] = x , [ x , x ] = x , x [ p ]3 = x , x [ p ]4 = x i . Note that the following automorphism of L , that maps x x , x x , x x , x x and x x HAPTER 7. RESTRICTION MAPS ON L , K ∼ = K , K ∼ = K , K ∼ = K . Moreover, K ( β ) ∼ = K via the automorphism β /p β /p β /p . Theorem 7.4.1 The list of all the restricted Lie algebra structures on L , , up to isomorphism, isas follows: L , = h x , . . . , x | [ x , x ] = x , [ x , x ] = x i ; L , = h x , . . . , x | [ x , x ] = x , [ x , x ] = x , x [ p ]1 = x i ; L , = h x , . . . , x | [ x , x ] = x , [ x , x ] = x , x [ p ]2 = x i ; L , = h x , . . . , x | [ x , x ] = x , [ x , x ] = x , x [ p ]3 = x i ; L , = h x , . . . , x | [ x , x ] = x , [ x , x ] = x , x [ p ]4 = x i ; L , = h x , . . . , x | [ x , x ] = x , [ x , x ] = x , x [ p ]1 = x , x [ p ]2 = x i ; L , = h x , . . . , x | [ x , x ] = x , [ x , x ] = x , x [ p ]1 = x , x [ p ]3 = x i ; L , = h x , . . . , x | [ x , x ] = x , [ x , x ] = x , x [ p ]1 = x , x [ p ]4 = x i ; L , ( β ) = h x , . . . , x | [ x , x ] = x , [ x , x ] = x , x [ p ]2 = βx , x [ p ]3 = x i ; L , = h x , . . . , x | [ x , x ] = x , [ x , x ] = x , x [ p ]3 = x , x [ p ]4 = x i where β ∈ F ∗ . In the remaining of this section we establish that the algebras given in Theorem 7.4.1 are pairwisenon-isomorphic, thereby completing the proof of Theorem 7.4.1.It is clear that L , is not isomorphic to the other restricted Lie algebras.We claim that L , and L , are not isomorphic. Suppose to the contrary that there exists anisomorphism A : L , → L , . Then A ( x [ p ]1 ) = A ( x ) [ p ] a x + a x + a x + a x + a x ) [ p ] a p x . Therefore, a = 0 which is a contradiction.Next, we claim that L , and L , are not isomorphic. Suppose to the contrary that there existsan isomorphism A : L , → L , . Then A ( x [ p ]1 ) = A ( x ) [ p ] a x + a x + a x + a x + a x ) [ p ] a p x . HAPTER 7. RESTRICTION MAPS ON L , a = 0 which is a contradiction.Next, we claim that L , and L , are not isomorphic. Suppose to the contrary that there existsan isomorphism A : L , → L , . Then A ( x [ p ]1 ) = A ( x ) [ p ] a x + a x + a x + a x + a x ) [ p ] a p x . Therefore, a = 0 which is a contradiction.It is clear that L , is not isomorphic to the other restricted Lie algebras.Next, we claim that L , and L , are not isomorphic. Suppose to the contrary that there existsan isomorphism A : L , → L , . Then A ( x [ p ]3 ) = A ( x ) [ p ] A ( x ) = ( a x + a x + a x + a x ) [ p ] a a x + a a x = a p x , which implies that a a = 0 and a a = a p . Therefore, a = 0 or a = 0 . First, if a = 0 ,we have a contradiction. Next, if a = 0 then a a = 0 and a a = 0 which is a contradiction.Next, we claim that L , and L , are not isomorphic. Suppose to the contrary that there existsan isomorphism A : L , → L , . Then A ( x [ p ]4 ) = A ( x ) [ p ] A ( x ) = ( a a x + a a x ) [ p ] a a x + a a x = 0 . Therefore, a a = 0 and a a = 0 which is a contradiction.It is clear that L , is not isomorphic to the other restricted Lie algebras.Next, we claim that L , and L , are not isomorphic. Suppose to the contrary that there existsan isomorphism A : L , → L , . Then A ( x [ p ]4 ) = A ( x ) [ p ] A ( x ) = ( a a x + a a x ) [ p ] a a x + a a x = 0 . Therefore, a a = 0 and a a = 0 which is a contradiction.It is clear that L , and L , are not isomorphic to the other restricted Lie algebras.Next, we claim that L , and L , are not isomorphic. Suppose to the contrary that there existsan isomorphism A : L , → L , . Then A ( x [ p ]3 ) = A ( x ) [ p ] A ( x ) = ( a x + a x + a x + a x ) [ p ] a a x + a a x = a p x , which implies that a a = 0 and a a = a p . Therefore, a = 0 or a = 0 . First, if a = 0 ,we have a contradiction. Next, if a = 0 then a a = 0 and a a = 0 which is a contradiction. HAPTER 7. RESTRICTION MAPS ON L , L , is not isomorphic to any of L , , L , because ( L , ) [ p ] = 0 but ( L , ) [ p ] = 0 ,( L , ) [ p ] = 0 .Next, we claim that L , and L , ( β ) are not isomorphic. Suppose to the contrary that thereexists an isomorphism A : L , ( β ) → L , . Then A ( x [ p ]1 ) = A ( x ) [ p ] a x + a x + a x + a x + a x ) [ p ] a p x + a p x . Therefore, a = 0 which is a contradiction.Note that L , is not isomorphic to any of L , , L , because ( L , ) [ p ] = 0 but ( L , ) [ p ] = 0 ,( L , ) [ p ] = 0 .Next, we claim that L , and L , ( β ) are not isomorphic. Suppose to the contrary that thereexists an isomorphism A : L , ( β ) → L , . Then A ( x [ p ]1 ) = A ( x ) [ p ] a x + a x + a x + a x + a x ) [ p ] a p x + a p x . Therefore, a = 0 which is a contradiction.Note that L , is not isomorphic to L , ( β ), because ( L , ( β )) [ p ] = 0 but ( L , ) [ p ] = 0 .Next, we claim that L , and L , are not isomorphic. Suppose to the contrary that there existsan isomorphism A : L , → L , . Then A ( x [ p ]1 ) = A ( x ) [ p ] a x + a x + a x + a x + a x ) [ p ] a p x + a p x . Therefore, a = 0 which is a contradiction.Note that L , ( β ) is not isomorphic to L , , because ( L , ( β )) [ p ] = 0 but ( L , ) [ p ] = 0 . hapter 8 Restriction maps on L , Let K = L , = h x , . . . , x | [ x , x ] = x , [ x , x ] = x , [ x , x ] = x i . Then Z ( K ) = h x , x i F and the group Aut( L , ) consists of invertible matrices of the form a a a a a a u a a a a − a a a u a ua a a a − a a a u a u , (8.1)where u = a a − a a = 0 . Note that there exists an element αx + βx ∈ Z ( L , ) such that( αx + βx ) [ p ] = 0 , for some α, β ∈ F . If α = 0 then consider K = h x ′ , . . . , x ′ | [ x ′ , x ′ ] = x ′ , [ x ′ , x ′ ] = x ′ , [ x ′ , x ′ ] = x ′ i , where x ′ = αx + βx , x ′ = α − x , x ′ = x , x ′ = αx + βx , x ′ = α − x . Let φ : K → K givenby x i x ′ i , for 1 i φ is an isomorphism. Therefore, in this case wecan suppose that x [ p ]4 = 0 . If α = 0 then β = 0 and we rescale x so that x [ p ]5 = 0 . Hence we canassume either x [ p ]4 = 0 or x [ p ]5 = 0 . Furthermore, we claim that it is enough to consider the case where x [ p ]5 = 0 . Indeed, suppose that there exists a p -map of L , such that x [ p ]4 = 0 . Then the image ofthe x i ’s under the following automorphism of L , yields another basis y i of L , such that y p ]5 = 0 : − − 10 0 0 − . Now we let L = L , h x i F ∼ = L , , HAPTER 8. RESTRICTION MAPS ON L , L , = h x , . . . , x | [ x , x ] = x , [ x , x ] = x i . The group Aut( L ) consists of invertiblematrices of the form a a a a a d a a a a a d , where d = a a . Lemma 8.0.2 Let K = L , and [ p ] : K → K be a p -map on K such that x [ p ]5 = 0 and let L = KM where M = h x i F . Then K ∼ = L θ where θ = (∆ , ω ) ∈ Z ( L, F ) . Proof. Let π : K → L be the projection map. We have the exact sequence0 → M → K → L → . Let σ : L → K such that x i x i , 1 i σ is an injective linear map and πσ = 1 L . Now,we define φ : L × L → M by φ ( x i , x j ) = [ σ ( x i ) , σ ( x j )] − σ ([ x i , x j ]), 1 i, j ω : L → M by ω ( x ) = σ ( x ) [ p ] − σ ( x [ p ] ). Note that φ ( x , x ) = [ σ ( x ) , σ ( x )] − σ ([ x , x ]) = [ x , x ] = x ; φ ( x , x ) = [ σ ( x ) , σ ( x )] − σ ([ x , x ]) = 0 . Similarly, we can show that φ ( x , x ) = φ ( x , x ) = φ ( x , x ) = φ ( x , x ) = 0 . Therefore, φ = ∆ .Now, by Lemma 2.2.2, we have θ = (∆ , ω ) ∈ Z ( L, F ) and K ∼ = L θ . (cid:4) We deduce that any p -map on K can be obtained by an extension of L via θ = (∆ , ω ), for some ω . Note that by Theorem 2.3.2, there are four non-isomorphic restricted Lie algebra structures on L , given by the following p -maps:I.1 Trivial p -map;I.2 x [ p ]1 = x ;I.3 x [ p ]2 = ξx ;I.4 x [ p ]3 = x .In the following subsections we consider each of the above cases and find all possible [ θ ] = [( φ, ω )] ∈ H ( L, F ) and construct L θ . Note that by Lemma 8.0.2, it suffices to assume [ θ ] = [(∆ , ω )] and findall non-isomorphic restricted Lie algebra structures on L , . L , trivial p -map) First, we find a basis for Z ( L, F ). Let ( φ, ω ) = ( a ∆ + b ∆ + c ∆ + d ∆ + e ∆ + f ∆ , αf + βf + γf + δf ) ∈ Z ( L, F ). Then we must have δ φ ( x, y, z ) = 0 and φ ( x, y [ p ] ) = 0 , for all x, y, z ∈ L .Therefore,0 = ( δ φ )( x , x , x ) = φ ([ x , x ] , x ) + φ ([ x , x ] , x ) + φ ([ x , x ] , x ) = φ ( x , x );0 = ( δ φ )( x , x , x ) = φ ([ x , x ] , x ) + φ ([ x , x ] , x ) + φ ([ x , x ] , x ) = φ ( x , x ) . HAPTER 8. RESTRICTION MAPS ON L , e = f = 0 . Since the p -map is trivial, φ ( x, y [ p ] ) = φ ( x, 0) = 0 , for all x, y ∈ L .Therefore, a basis for Z ( L, F ) is as follows:(∆ , , (∆ , , (∆ , , (∆ , , (0 , f ) , (0 , f ) , (0 , f ) , (0 , f ) . Next, we find a basis for B ( L, F ). Let ( φ, ω ) ∈ B ( L, F ). Since B ( L, F ) ⊆ Z ( L, F ), we have( φ, ω ) = ( a ∆ + b ∆ + c ∆ + d ∆ , αf + βf + γf + δf ). So, there exists a linear map ψ : L → F such that δ ψ ( x, y ) = φ ( x, y ) and ˜ ψ ( x ) = ω ( x ), for all x, y ∈ L . So, we have c = φ ( x , x ) = δ ψ ( x , x ) = ψ ([ x , x ]) = 0 . Similarly, we can show that d = 0 . Also, we have α = ω ( x ) = ˜ ψ ( x ) = ψ ( x [ p ]1 ) = 0 . Similarly, we can show that β = γ = δ = 0 . Therefore, ( φ, ω ) = ( a ∆ + b ∆ , 0) and hence B ( L, F ) = h (∆ , , (∆ , i F . We deduce that a basis for H ( L, F ) is as follows:[(∆ , , [(∆ , , [(0 , f )] , [(0 , f )] , [(0 , f )] , [(0 , f )] . Let [( φ, ω )] ∈ H ( L, F ). Then we have φ = a ∆ + b ∆ , for some a, b ∈ F . Suppose that Aφ = a ′ ∆ + b ′ ∆ , for some a ′ , b ′ ∈ F . We determine a ′ , b ′ . Note that Aφ ( x , x ) = φ ( Ax , Ax ) = φ ( a x + a x + a x + a x , a dx ) = a da ; Aφ ( x , x ) = φ ( Ax , Ax ) = φ ( a x + a x + a x , dx + a a x ) = a db. In the matrix form we can write this as (cid:18) a ′ b ′ (cid:19) = (cid:18) a d a d (cid:19) (cid:18) ab (cid:19) . (8.2)The orbit with representative (cid:18) (cid:19) of this action gives us L , .Also we have ω = αf + βf + γf + δf , for some α, β, γ, δ ∈ F . Suppose that Aω = α ′ f + β ′ f + γ ′ f + δ ′ f , for some α ′ , β ′ , γ ′ , δ ′ ∈ F . We have Aω ( x ) = ω ( Ax ) = ω ( a x + a x + a x + a x ) = a p α + a p β + a p γ + a p δ ; Aω ( x ) = a p β + a p γ + a p δ ; Aω ( x ) = d p γ + a p a p δ ; Aω ( x ) = a p d p δ. In the matrix form we can write this as α ′ β ′ γ ′ δ ′ = a p a p a p a p a p a p a p d p a p a p a p d p αβγδ . (8.3)Thus, we can write Equations (8.2) and (8.3) together as follows: h d (cid:18) a a (cid:19) , a p a p a p a p a p a p a p d p a p a p a p d p ih (cid:18) ab (cid:19) , αβγδ i = h (cid:18) a ′ b ′ (cid:19) , α ′ β ′ γ ′ δ ′ i . HAPTER 8. RESTRICTION MAPS ON L , p ( L ) on the set of ω ’s such that theorbit represented by (cid:18) (cid:19) is preserved under the action of Aut p ( L ) on the set of φ ’s.Let ν = αβγδ ∈ F . If ν = , then { ν } is clearly an Aut p ( L )-orbit. Suppose that δ = 0 .Then h (cid:18) (cid:19) , − α/δ − β/δ ih (cid:18) (cid:19) , αβγδ i = h (cid:18) (cid:19) , γδ i , and h (cid:18) (cid:19) , − γ/δ γ /δ − γ/δ ih (cid:18) (cid:19) , γδ i = h (cid:18) (cid:19) , δ i . Hence the set of vectors αβγδ with δ = 0 form an single orbit with orbit representative δ .Next, if δ = 0 , but γ = 0 , then h (cid:18) (cid:19) , − α/γ 00 1 − β/γ 00 0 1 − β/γ ih (cid:18) (cid:19) , αβγ i = h (cid:18) (cid:19) , γ i , and h (1 /γ ) /p (cid:18) (1 /γ ) /p 00 (1 /γ ) − /p (cid:19) , γ − γ /γ 00 0 0 γ − ih (cid:18) (cid:19) , γ i = h (cid:18) (cid:19) , i . Next, if γ = δ = 0 , but β = 0 , then h (cid:18) (cid:19) , − α/β ih (cid:18) (cid:19) , αβ i = h (cid:18) (cid:19) , β i , and h β /p (cid:18) β /p β − /p (cid:19) , β /β β 00 0 0 β ih (cid:18) (cid:19) , β i = h (cid:18) (cid:19) , i . Finally, if β = γ = δ = 0 , but α = 0 , then we have h (cid:18) (cid:19) , α i . HAPTER 8. RESTRICTION MAPS ON L , p ( L )-orbit representatives: , α , , , δ . Lemma 8.1.1 The vectors δ and δ are in the same Aut p ( L ) -orbit if and only if δ δ − ∈ ( F ∗ ) . Proof. First assume that δ and δ are in the same Aut p ( L )-orbit. Then h d (cid:18) a a (cid:19) , a p a p a p a p a p a p a p d p a p a p a p d p ih (cid:18) (cid:19) , δ i = h (cid:18) (cid:19) , δ i . From this we obtain that δ = a p d p δ and that da = 1 which gives that δ δ − = a − p ∈ ( F ∗ ) .Assume next that δ δ − = ǫ , with some ǫ ∈ F ∗ . Then h ǫ /p (cid:18) ( ǫ ) /p 00 ( ǫ ) − /p (cid:19) , ǫ ǫ − ǫ 00 0 0 ǫ ih (cid:18) (cid:19) , δ i = h (cid:18) (cid:19) , δ i , as required. (cid:4) Hence, the corresponding restricted Lie algebra structures are as follows: K = h x , . . . , x | [ x , x ] = x , [ x , x ] = x , [ x , x ] = x i ; K ( α ) = h x , . . . , x | [ x , x ] = x , [ x , x ] = x , [ x , x ] = x , x [ p ]1 = αx i ; K = h x , . . . , x | [ x , x ] = x , [ x , x ] = x , [ x , x ] = x , x [ p ]2 = x i ; K = h x , . . . , x | [ x , x ] = x , [ x , x ] = x , [ x , x ] = x , x [ p ]3 = x i ; K ( δ ) = h x , . . . , x | [ x , x ] = x , [ x , x ] = x , [ x , x ] = x , x [ p ]4 = δx i where α, δ ∈ F ∗ . Lemma 8.1.2 We have K ( α ) ∼ = K ( α ) if and only if α α − ∈ ( F ∗ ) . Proof. Let f = ( a ij ) ∈ Aut( L , ) and suppose that f : K ( α ) → K ( α ) is an isomorphism. Then f ( x [ p ]1 ) = f ( x ) [ p ] which implies that α a a x = α a p x . Hence, α /α = a p − a − ∈ ( F ∗ ) . To HAPTER 8. RESTRICTION MAPS ON L , α /α = ǫ ∈ ( F ∗ ) . Then the following is an isomorphism from K ( α ) to K ( α ): ǫ /p ǫ − /p ǫ /p ǫ /p 00 0 0 0 1 . (cid:4) The conditions on δ and δ such that K ( δ ) ∼ = K ( δ ) reduces to the solutions of an algebraicequation over F as the following lemma shows. However, by Lemma 8.1.1 if δ /δ ∈ ( F ∗ ) then K ( δ ) ∼ = K ( δ ). Lemma 8.1.3 We have K ( δ ) ∼ = K ( δ ) if and only if the equation δ /δ = x p − y p − has asolution in F ∗ . Proof. Let f = ( a ij ) ∈ Aut( L , ) and suppose that f : K ( δ ) → K ( δ ) is an isomorphism. Then f ( x [ p ]4 ) = f ( x ) [ p ] which implies that δ a a x = δ a p a p x . Hence, δ /δ = a p − a p − . (cid:4) Corollary 8.1.4 If δ /δ ∈ F ∗ p then K ( δ ) ∼ = K ( δ ) . L, x [ p ]1 = x ) First, we find a basis for Z ( L, F ). Let ( φ, ω ) = ( a ∆ + b ∆ + c ∆ + d ∆ + e ∆ + f ∆ , αf + βf + γf + δf ) ∈ Z ( L, F ). Then we must have δ φ ( x, y, z ) = 0 and φ ( x, y [ p ] ) = 0 , for all x, y, z ∈ L .Therefore,0 = ( δ φ )( x , x , x ) = φ ([ x , x ] , x ) + φ ([ x , x ] , x ) + φ ([ x , x ] , x ) = φ ( x , x );0 = ( δ φ )( x , x , x ) = φ ([ x , x ] , x ) + φ ([ x , x ] , x ) + φ ([ x , x ] , x ) = φ ( x , x ) . Thus, we get e = f = 0 . Also, we have φ ( x, y [ p ] ) = 0 . Therefore, φ ( x, x ) = 0 , for all x ∈ L andhence φ ( x , x ) = φ ( x , x ) = φ ( x , x ) = 0 which implies that c = e = f = 0 . Therefore, Z ( L, F )has a basis consisting of:(∆ , , (∆ , , (∆ , , (0 , f ) , (0 , f ) , (0 , f ) , (0 , f ) . Next, we find a basis for B ( L, F ). Let ( φ, ω ) ∈ B ( L, F ). Since B ( L, F ) ⊆ Z ( L, F ), we have( φ, ω ) = ( a ∆ + b ∆ + c ∆ , αf + βf + γf + δf ). Note that there exists a linear map ψ : L → F such that δ ψ ( x, y ) = φ ( x, y ) and ˜ ψ ( x ) = ω ( x ), for all x, y ∈ L . So, we have a = φ ( x , x ) = δ ψ ( x , x ) = ψ ([ x , x ]) = ψ ( x ) , and b = φ ( x , x ) = δ ψ ( x , x ) = ψ ([ x , x ]) = ψ ( x ) , and c = φ ( x , x ) = δ ψ ( x , x ) = ψ ([ x , x ]) = 0 . Also, we have α = ω ( x ) = ˜ ψ ( x ) = ψ ( x [ p ]1 ) = ψ ( x ) , and β = ω ( x ) = ˜ ψ ( x ) = ψ ( x [ p ]2 ) = 0 . HAPTER 8. RESTRICTION MAPS ON L , γ = δ = 0 . Note that ψ ( x ) = b = α . Therefore, ( φ, ω ) = ( a ∆ + b ∆ , bf ) and hence B ( L, F ) = h (∆ , , (∆ , f ) i F . Note that[(∆ , , [(∆ , , [0 , f ] , [(0 , f )] , [(0 , f )] , [(0 , f )]spans H ( L, F ). Since [(∆ , , f )] = [(∆ , f )] = [0], then [(0 , f )] is an scalar multiple of[(∆ , H ( L, F ). Note that dim H = dim Z − dim B = 5 . Therefore,[(∆ , , [(∆ , , [(0 , f )] , [(0 , f )] , [(0 , f )]forms a basis for H ( L, F ).Note that the group Aut( L ) in this case consists of invertible matrices of the form a a a a a d a a a a a d , where d = a a and a d = a p .Let [( φ, ω )] ∈ H ( L, F ). Then we have φ = a ∆ + b ∆ , for some a, b ∈ F . Suppose that Aφ = a ′ ∆ + b ′ ∆ for some a ′ , b ′ ∈ F . We determine a, b ′ . Note that Aφ ( x , x ) = φ ( Ax , Ax ) = φ ( a x + a x + a x + a x , dx + a a x ) = a da + a db ; and Aφ ( x , x ) = φ ( Ax , Ax ) = φ ( a x + a x + a x , dx + a a x ) = a db. In the matrix form we can write this as (cid:18) a ′ b ′ (cid:19) = (cid:18) a d a d a d (cid:19) (cid:18) ab (cid:19) . (8.4)The orbit with representative (cid:18) (cid:19) of this action gives us L , .Also, we have ω = βf + γf + δf , for some β, γ, δ ∈ F . Suppose that Aω = β ′ f + γ ′ f + δ ′ f ,for some β ′ , γ ′ , δ ′ ∈ F . We have Aω ( x ) = a p β + a p γ + a p δ ; Aω ( x ) = d p γ + a p a p δ ; Aω ( x ) = a p d p δ. In the matrix form we can write this as β ′ γ ′ δ ′ = a p a p a p d p a p a p a p d p βγδ . (8.5)Thus, we can write Equations (8.4) and (8.5) together as follows: h (cid:18) a d a d a d (cid:19) , a p a p a p d p a p a p a p d p ih (cid:18) ab (cid:19) , βγδ i = h (cid:18) a ′ b ′ (cid:19) , β ′ γ ′ δ ′ i . HAPTER 8. RESTRICTION MAPS ON L , p ( L ) on the set of ω ’s such that theorbit represented by (cid:18) (cid:19) is preserved under the action of Aut p ( L ) on the set of φ ’s. Note that weneed to have Aω ( x ) = 0 which implies that a p β + a p γ + a p δ = 0 .Let ν = βγδ ∈ F . If ν = , then { ν } is clearly an Aut p ( L )-orbit. Suppose that δ = 0 .Then h (cid:18) (cid:19) , − β/δ ih (cid:18) (cid:19) , βγδ i = h (cid:18) (cid:19) , γδ i , and h (cid:18) (cid:19) , − γ/δ γ /δ − γ/δ ih (cid:18) (cid:19) , γδ i = h (cid:18) (cid:19) , δ i . Next, if δ = 0 , but γ = 0 , then h (cid:18) (cid:19) , − β/γ 00 1 − β/γ ih (cid:18) (cid:19) , βγ i = h (cid:18) (cid:19) , γ i . Finally, if γ = δ = 0 , but β = 0 , then we have β . Thus the following elements are Aut p ( L )-orbit representatives: , β , γ , δ . Lemma 8.2.1 The vectors β and β are in the same Aut p ( L ) -orbit if and only if β /β = ǫ, where ǫ p − = 1 . Proof. Let θ = h (cid:18) (cid:19) , β i and θ = h (cid:18) (cid:19) , β i . Then θ and θ are in the same Aut p ( L )-orbit if and only if there exists A = ( a ij ) ∈ Aut p ( L ) such that A · θ = θ . Equivalently, θ and θ are in the same Aut p ( L )-orbit if and only if there exists A = ( a ij ) ∈ Aut p ( L ) such that a d = a p and h (cid:18) a d a d a d (cid:19) , a p a p a p d p a p a p a p d p ih (cid:18) (cid:19) , β i = h (cid:18) (cid:19) , β i . Hence, θ and θ are in the same Aut p ( L )-orbit if and only if β = a p β (8.6) da = 1 ⇔ a = a − (8.7) a d = a p (8.8) a = 0 . (8.9) HAPTER 8. RESTRICTION MAPS ON L , θ and θ are in the same Aut p ( L )-orbit and set ǫ = a p . Then, by Equations(8.6) and (8.8), we have β /β = a p = ǫ ∈ F ∗ . Furthermore, by Equations (8.7) and (8.8), a p − = 1and so ǫ p − = 1 . Conversely, suppose that β /β = ǫ ∈ F ∗ , where ǫ p − = 1 . We let a = ǫ /p , a = a − , and a = 0 . Then we have a d = a p ⇔ a a = a p ⇔ a − = a − p ⇔ a p − = 1 ⇔ ǫ p − = 1 . It remains to verify that Equation (8.6) is satisfied. We have: β /β = ǫ = a p , as required. (cid:4) Lemma 8.2.2 The vectors γ and γ are in the same Aut p ( L ) -orbit if and only if γ /γ = ǫ , where ǫ p − = 1 . Proof. Let θ = h (cid:18) (cid:19) , γ i and θ = h (cid:18) (cid:19) , γ i . Then θ and θ are in the same Aut p ( L )-orbit if and only if there exists A = ( a ij ) ∈ Aut p ( L ) such that A · θ = θ . Equivalently, θ and θ are in the same Aut p ( L )-orbit if and only if there exists A = ( a ij ) ∈ Aut p ( L ) such that h (cid:18) a d a d a d (cid:19) , a p a p a p d p a p a p a p d p ih (cid:18) (cid:19) , γ i = h (cid:18) (cid:19) , γ i . Hence, θ and θ are in the same Aut p ( L )-orbit if and only if γ = d p γ (8.10) da = 1 ⇔ a = a − (8.11) a d = a p (8.12) a = a = 0 . (8.13)First, suppose that θ and θ are in the same Aut p ( L )-orbit and set ǫ = a ( p − / . Then, by Equations(8.10) and (8.12), we have ( γ /γ ) /p = d = a p − = ǫ ∈ ( F ∗ ) . Furthermore, by Equations (8.11)and (8.12), a p − = 1 and so ǫ p − = a ( p − p − / = a (1 − p )(2 p − = 1 . Conversely, suppose that γ /γ = ǫ ∈ ( F ∗ ) . We let a = ǫ − /p , a = a − , and a = a = 0 . Then we have a d = a p ⇔ a a = a p ⇔ a − = a − p ⇔ a p − = 1 ⇔ ǫ p − = 1 . It remains to verify that Equation (8.10) is satisfied. We have: γ /γ = ǫ = a − p = ( a a ) p = d p , as required. (cid:4) Lemma 8.2.3 The vectors δ and δ are in the same Aut p ( L ) -orbit if and only if δ /δ = ǫ , where ǫ p − = 1 . HAPTER 8. RESTRICTION MAPS ON L , Proof. Let θ = h (cid:18) (cid:19) , δ i and θ = h (cid:18) (cid:19) , δ i . Then θ and θ are in the same Aut p ( L )-orbit if and only if there exists A = ( a ij ) ∈ Aut p ( L ) such that A · θ = θ . Equivalently, θ and θ are in the same Aut p ( L )-orbit if and only if there exists A = ( a ij ) ∈ Aut p ( L ) such that h (cid:18) a d a d a d (cid:19) , a p a p a p d p a p a p a p d p ih (cid:18) (cid:19) , δ i = h (cid:18) (cid:19) , δ i . Hence, θ and θ are in the same Aut p ( L )-orbit if and only if δ = a p d p δ (8.14) da = 1 ⇔ a = a − (8.15) a d = a p (8.16) a = a = a = 0 . (8.17)First, suppose that θ and θ are in the same Aut p ( L )-orbit. Then, using Equation (8.15), we getthat a p = a − p which using (8.15) and (8.16) implies that a p − = 1 . Now, by (8.14) and (8.15) ,we have δ /δ = ( a d ) p = ( a a ) p = a − p . To prove the sufficiency, we let a = ǫ − /p , a = a − , and a = a = a = 0 . Then we have a d = a p ⇔ a a = a p ⇔ a − = a − p ⇔ a p − = 1 ⇔ ǫ p − = 1 . It remains to verify that Equation (8.14) is satisfied. We have: δ /δ = ǫ = a − p = ( a d ) p , as required. (cid:4) Hence, the corresponding restricted Lie algebra structurs are as follows: K = h x , . . . , x | [ x , x ] = x , [ x , x ] = x , [ x , x ] = x , x [ p ]1 = x i ; K ( β ) = h x , . . . , x | [ x , x ] = x , [ x , x ] = x , [ x , x ] = x , x [ p ]1 = x , x [ p ]2 = βx i ; K ( γ ) = h x , . . . , x | [ x , x ] = x , [ x , x ] = x , [ x , x ] = x , x [ p ]1 = x , x [ p ]3 = γx i ; K ( δ ) = h x , . . . , x | [ x , x ] = x , [ x , x ] = x , [ x , x ] = x , x [ p ]1 = x , x [ p ]4 = δx i where β, γ, δ ∈ F ∗ .Lemma 8.2.1 provides us with some sufficient conditions as to when K ( β ) ∼ = K ( β ). Over theprime field F p we can say the following: Lemma 8.2.4 Let β , β ∈ F ∗ p . Then K ( β ) ∼ = K ( β ) over F p if and only if β = β . Proof. Let f = ( a ij ) ∈ Aut( L , ), where a ij ∈ F p . Then f : K ( β ) → K ( β ) is an isomorphism ifand only if u = a p − , β /β = a p − u − ∈ ( F ∗ ) . Thus, K ( β ) ∼ = K ( β ) if and only if β = β . (cid:4) HAPTER 8. RESTRICTION MAPS ON L , K ( γ ) ∼ = K ( γ ) and acharacterization is given below: Lemma 8.2.5 We have K ( γ ) ∼ = K ( γ ) if and only if γ /γ = ǫ p − p +3 , for some ǫ ∈ F ∗ . Proof. Let f = ( a ij ) ∈ Aut( L , ). Then f : K ( γ ) → K ( γ ) is an isomorphism if and only if a = a = 0 , u = a p − , γ /γ = u p − a − , which is equivallent to saying that γ /γ = a p − p +311 . (cid:4) Lemma 8.2.3 provides us with some sufficient conditions as to when K ( δ ) ∼ = K ( δ ) and acharacterization is given below: Lemma 8.2.6 We have K ( δ ) ∼ = K ( δ ) if and only if δ /δ = ǫ p − p +3 , for some ǫ ∈ F ∗ . Proof. Let f = ( a ij ) ∈ Aut( L , ). Then f : K ( δ ) → K ( δ ) is an isomorphism if and only if a = a = 0 , u = a p − , δ /δ = ( a u ) p a u , which is equivalent to saying that δ /δ = a p − p +311 . (cid:4) L, x [ p ]2 = ξx ) First, we find a basis for Z ( L, F ). Let ( φ, ω ) = ( a ∆ + b ∆ + c ∆ + d ∆ + e ∆ + f ∆ , αf + βf + γf + δf ) ∈ Z ( L, F ). Then we must have δ φ ( x, y, z ) = 0 and φ ( x, y [ p ] ) = 0 , for all x, y, z ∈ L .Therefore,0 = ( δ φ )( x , x , x ) = φ ([ x , x ] , x ) + φ ([ x , x ] , x ) + φ ([ x , x ] , x ) = φ ( x , x );0 = ( δ φ )( x , x , x ) = φ ([ x , x ] , x ) + φ ([ x , x ] , x ) + φ ([ x , x ] , x ) = φ ( x , x ) . Thus, we get e = f = 0 . Also, we have φ ( x, y [ p ] ) = 0 . Therefore, φ ( x, x ) = 0 , for all x ∈ L andhence φ ( x , x ) = φ ( x , x ) = φ ( x , x ) = 0 which implies that c = e = f = 0 . Therefore, Z ( L, F )has a basis consisting of:(∆ , , (∆ , , (∆ , , (0 , f ) , (0 , f ) , (0 , f ) , (0 , f ) . Next, we find a basis for B ( L, F ). Let ( φ, ω ) ∈ B ( L, F ). Since B ( L, F ) ⊆ Z ( L, F ), we have( φ, ω ) = ( a ∆ + b ∆ + c ∆ , αf + βf + γf + δf ). Note that there exists a linear map ψ : L → F such that δ ψ ( x, y ) = φ ( x, y ) and ˜ ψ ( x ) = ω ( x ), for all x, y ∈ L . So, we have a = φ ( x , x ) = δ ψ ( x , x ) = ψ ([ x , x ]) = ψ ( x ) , and b = φ ( x , x ) = δ ψ ( x , x ) = ψ ([ x , x ]) = ψ ( x ) , and c = φ ( x , x ) = δ ψ ( x , x ) = ψ ([ x , x ]) = 0 . HAPTER 8. RESTRICTION MAPS ON L , β = ω ( x ) = ˜ ψ ( x ) = ψ ( x [ p ]2 ) = ξψ ( x ) , and α = ω ( x ) = ˜ ψ ( x ) = ψ ( x [ p ]1 ) = 0 . Similarly, we can show that γ = δ = 0 . Note that ψ ( x ) = b = βξ − . Therefore, ( φ, ω ) =( a ∆ + b ∆ , bξf ) and hence B ( L, F ) = h (∆ , , (∆ , ξf ) i F . Note that[(∆ , , [(∆ , , [0 , f ] , [(0 , f )] , [(0 , f )] , [(0 , f )]spans H ( L, F ). Since [(∆ , ξ [(0 , f )] = [(∆ , ξf )] = [0], then [(0 , f )] is an scalar multiple of[(∆ , H ( L, F ). Note that dim H = dim Z − dim B = 5 . Therefore,[(∆ , , [(∆ , , [(0 , f )] , [(0 , f )] , [(0 , f )]forms a basis for H ( L, F ).The group Aut( L ) consists of invertible matrices of the form a a a a a d a a a a a d , where d = a a , a = 0 , and a d = a p .Let [( φ, ω )] ∈ H ( L, F ). Then we have φ = a ∆ + b ∆ , for some a, b ∈ F . Suppose that Aφ = a ′ ∆ + b ′ ∆ for some a ′ , b ′ ∈ F . Then we have (cid:18) a ′ b ′ (cid:19) = (cid:18) a d a d a d (cid:19) (cid:18) ab (cid:19) . (8.18)The orbit with representative (cid:18) (cid:19) of this action gives us L , .Also, we have ω = αf + γf + δf , for some α, γ, δ ∈ F . Suppose that Aω = α ′ f + γ ′ f + δ ′ f ,for some α ′ , γ ′ , δ ′ ∈ F . We have Aω ( x ) = a p α + a p γ + a p δ ; Aω ( x ) = d p γ + a p a p δ ; Aω ( x ) = a p d p δ. In the matrix form we can write this as α ′ γ ′ δ ′ = a p a p a p d p a p a p a p d p αγδ . (8.19)Thus, we can write Equations (8.18) and (8.19) together as follows: h (cid:18) a d a d a d (cid:19) , a p a p a p d p a p a p a p d p ih (cid:18) ab (cid:19) , αγδ i = h (cid:18) a ′ b ′ (cid:19) , α ′ γ ′ δ ′ i (8.20) HAPTER 8. RESTRICTION MAPS ON L , L ) on the set of ω ’s such that theorbit represented by (cid:18) (cid:19) is preserved under the action of Aut( L ) on the set of φ ’s. Note that weneed to have Aω ( x ) = 0 which implies that a p γ + a p δ = 0 .Let ν = αγδ ∈ F . If ν = , then { ν } is clearly an Aut( L )-orbit. Suppose that δ = 0 .Then h (cid:18) (cid:19) , − α/δ ih (cid:18) (cid:19) , αγδ i = h (cid:18) (cid:19) , γδ i , and h (cid:18) (cid:19) , − γ/δ ih (cid:18) (cid:19) , γδ i = h (cid:18) (cid:19) , δ i . Next, if δ = 0 , but γ = 0 , then h (cid:18) (cid:19) , − α/γ 00 1 00 0 1 ih (cid:18) (cid:19) , αγ i = h (cid:18) (cid:19) , γ ih (cid:18) (cid:19) , − α/γ 00 1 00 0 1 ih (cid:18) (cid:19) , αγ i = h (cid:18) (cid:19) , γ i Finally, if γ = δ = 0 , but α = 0 , then we get α . Thus the following elements are Aut( L )-orbit representatives: , α , γ , δ . Therefore, the corresponding restricted Lie algebra structures are as follows: K ( ξ ) = h x , . . . , x | [ x , x ] = x , [ x , x ] = x , [ x , x ] = x , x [ p ]2 = ξx i ; K ( ξ, α ) = h x , . . . , x | [ x , x ] = x , [ x , x ] = x , [ x , x ] = x , x [ p ]1 = αx , x [ p ]2 = ξx i ; K ( ξ, γ ) = h x , . . . , x | [ x , x ] = x , [ x , x ] = x , [ x , x ] = x , x [ p ]2 = ξx , x [ p ]3 = γx i ; K ( ξ, δ ) = h x , . . . , x | [ x , x ] = x , [ x , x ] = x , [ x , x ] = x , x [ p ]2 = ξx , x [ p ]4 = δx i , where α, γ, δ, ξ ∈ F ∗ . Lemma 8.3.1 Let a, b ∈ F p and suppose a = 0 . Then the equation x − ay = b has a solution in F p . Proof. Consider the two sets A = { x | x ∈ F p } and B = { ay + b | y ∈ F p } . Since a = 0 , both setshave p +12 elements. Indeed, let { a , . . . , a p − } be the nonzero elements of F p . Then a i ≡ ( p − a i ) mod p , for every1 i p − a , . . . , a p − . HAPTER 8. RESTRICTION MAPS ON L , a i ≡ a j mod p , for1 i, j p − . Then, we have ( a i − a j )( a i + a j ) ≡ p , which implies that either a i − a j ≡ p , or a i + a j ≡ p . Note that if a i + a j ≡ p , then p − a i ≡ a i ≡ − a j mod p ,and hence a i − a j ≡ p ≡ p , which lead us to a i ≡ a j ≡ p , a contradiction. Therefore, a i − a j ≡ p , and so a i ≡ a j mod p . So, the number of elements of A is p +12 . Note thatwe can easily define a bijection from set A to set B and hence the number of elements of B is also p +12 . So, the sets A and B must overlap and hence a solution exists. (cid:4) Lemma 8.3.2 Let ξ , γ , ξ , γ ∈ F p . Then K ( ξ , α ) and K ( ξ , α ) are isomorphic over F p ifand only if ξ ξ α α ∈ ( F ∗ p ) Proof. Let f = ( a ij ) ∈ Aut( L , ), where a ij ∈ F p . Then f : K ( δ ) → K ( δ ) is an isomorphismif and only if a uξ = ξ a ; (8.21) a uα = ξ a ; (8.22) a uξ = α a ; (8.23) a uα = α a . (8.24)Hence, if K ( ξ , α ) ∼ = K ( ξ , α ) then ξ ξ α α = ( a a ) ∈ ( F ∗ p ) . To prove the converse, suppose that ξ ξ α α = ǫ , for some ǫ ∈ F ∗ p . Note that by Lemma 8.3.1, the equation x − ξ α y = α α (8.25)has a solution in F p . Now, we take a = ǫx , a = y , and set a = a ǫ , a = ξ α ǫa . (8.26)Then, u = a a − a a = ǫ α α and we can verify that Equations (8.21)-(8.24) are satisfied. Thus, K ( ξ , α ) ∼ = K ( ξ , α ). The proof is complete. (cid:4) Lemma 8.3.3 Let ξ , γ , ξ , γ ∈ F p . Then, K ( ξ , γ ) and K ( ξ , γ ) are isomorphic over F p ifand only if ξ /ξ = ǫ , for some ǫ ∈ F ∗ p . Proof. Let f = ( a ij ) ∈ Aut( L , ). Then f : K ( ξ , γ ) → K ( ξ , γ ) is an isomorphism if and onlyif a = a = 0 a uξ = ξ a (8.27) a uγ = γ u. (8.28)Hence, ξ /ξ = a and γ /γ = a . The necessity is now clear. For sufficiency, we note that a isdetermined by the ratio ξ /ξ and a is determined from the equation γ /γ = a . (cid:4) Lemma 8.3.4 Let ξ , γ , ξ , γ ∈ F p . Then, K ( ξ , γ ) and K ( ξ , γ ) are isomorphic over F p ifand only if ξ /ξ = ǫ , for some ǫ ∈ F ∗ p . HAPTER 8. RESTRICTION MAPS ON L , Proof. Let f = ( a ij ) ∈ Aut( L , ). Then f : K ( ξ , δ ) → K ( ξ , δ ) is an isomorphism if and onlyif a = a = 0 a uξ = ξ a a uδ = δ a u. Hence, ξ /ξ = a and δ /δ = a a . The necessity is now clear. For sufficiency, we note that a isdetermined by the ratio ξ /ξ and then a is determined from the equation δ /δ = a a . (cid:4) L, x [ p ]3 = x ) First, we find a basis for Z ( L, F ). Let ( φ, ω ) = ( a ∆ + b ∆ + c ∆ + d ∆ + e ∆ + f ∆ , αf + βf + γf + δf ) ∈ Z ( L, F ). Then we must have δ φ ( x, y, z ) = 0 and φ ( x, y [ p ] ) = 0 , for all x, y, z ∈ L .Therefore,0 = ( δ φ )( x , x , x ) = φ ([ x , x ] , x ) + φ ([ x , x ] , x ) + φ ([ x , x ] , x ) = φ ( x , x );0 = ( δ φ )( x , x , x ) = φ ([ x , x ] , x ) + φ ([ x , x ] , x ) + φ ([ x , x ] , x ) = φ ( x , x ) . Thus, we get e = f = 0 . Also, we have φ ( x, y [ p ] ) = 0 . Therefore, φ ( x, x ) = 0 , for all x ∈ L andhence φ ( x , x ) = φ ( x , x ) = φ ( x , x ) = 0 which implies that c = e = f = 0 . Therefore, Z ( L, F )has a basis consisting of:(∆ , , (∆ , , (∆ , , (0 , f ) , (0 , f ) , (0 , f ) , (0 , f ) . Next, we find a basis for B ( L, F ). Let ( φ, ω ) ∈ B ( L, F ). Since B ( L, F ) ⊆ Z ( L, F ), we have( φ, ω ) = ( a ∆ + b ∆ + c ∆ , αf + βf + γf + δf ). Note that there exists a linear map ψ : L → F such that δ ψ ( x, y ) = φ ( x, y ) and ˜ ψ ( x ) = ω ( x ), for all x, y ∈ L . So, we have a = φ ( x , x ) = δ ψ ( x , x ) = ψ ([ x , x ]) = ψ ( x ) , and b = φ ( x , x ) = δ ψ ( x , x ) = ψ ([ x , x ]) = ψ ( x ) , and c = φ ( x , x ) = δ ψ ( x , x ) = ψ ([ x , x ]) = 0 . Also, we have γ = ω ( x ) = ˜ ψ ( x ) = ψ ( x [ p ]3 ) = ψ ( x ) , and α = ω ( x ) = ˜ ψ ( x ) = ψ ( x [ p ]1 ) = 0 . Similarly, we can show that β = δ = 0 . Note that ψ ( x ) = b = γ . Therefore, ( φ, ω ) = ( a ∆ + b ∆ , bf ) and hence B ( L, F ) = h (∆ , , (∆ , f ) i F . Note that[(∆ , , [(∆ , , [0 , f ] , [(0 , f )] , [(0 , f )] , [(0 , f )]spans H ( L, F ). Since [(∆ , , f )] = [(∆ , f )] = [0], then [(0 , f )] is an scalar multiple of[(∆ , H ( L, F ). Note that dim H = dim Z − dim B = 5 . Therefore,[(∆ , , [(∆ , , [(0 , f )] , [(0 , f )] , [(0 , f )] HAPTER 8. RESTRICTION MAPS ON L , H ( L, F ).The group Aut( L ) in this case consists of invertible matrices of the form a a a a a d a a a a a d , where d = a a , a d = d p , and a = a = 0 .Let [( φ, ω )] ∈ H ( L, F ). Then we have φ = a ∆ + b ∆ , for some a, b ∈ F . Suppose that Aφ = a ′ ∆ + b ′ ∆ for some a ′ , b ′ ∈ F . We determine a, b ′ . Note that Aφ ( x , x ) = φ ( Ax , Ax ) = φ ( a x + a x + a x + a x , dx + a a x ) = a da + a db ; and Aφ ( x , x ) = φ ( Ax , Ax ) = φ ( a x + a x + a x , dx + a a x ) = a db. In the matrix form we can write this as (cid:18) a ′ b ′ (cid:19) = (cid:18) a d a d a d (cid:19) (cid:18) ab (cid:19) . (8.29)The orbit with representative (cid:18) (cid:19) of this action gives us L , .Also, we have ω = αf + βf + δf for some α, β, δ ∈ F . Suppose that Aω = α ′ f + β ′ f + δ ′ f ,for some α ′ , β ′ , δ ′ ∈ F . We have Aω ( x ) = a p α + a p β + a p δ ; Aω ( x ) = a p β + a p δ ; Aω ( x ) = a p d p δ. In the matrix form we can write this as α ′ β ′ δ ′ = a p a p a p a p a p a p d p αβδ . (8.30)Thus, we can write Equations (8.29) and (8.30) together as follows: h (cid:18) a d a d a d (cid:19) , a p a p a p a p a p a p d p ih (cid:18) ab (cid:19) , αβδ i = h (cid:18) a ′ b ′ (cid:19) , α ′ β ′ δ ′ i . Now we find the representatives of the orbits of the action of Aut( L ) on the set of ω ’s such that theorbit represented by (cid:18) (cid:19) is preserved under the action of Aut( L ) on the set of φ ’s. Note that weneed to have Aω ( x ) = 0 which implies that a p a p δ = 0 .Let ν = αβδ ∈ F . If ν = , then { ν } is clearly an Aut( L )-orbit. Suppose that δ = 0 . HAPTER 8. RESTRICTION MAPS ON L , h (cid:18) (cid:19) , − α/δ ih (cid:18) (cid:19) , αβδ i = h (cid:18) (cid:19) , βδ i , and h (cid:18) (cid:19) , − β/δ ih (cid:18) (cid:19) , βδ i = h (cid:18) (cid:19) , δ i . Next, if δ = 0 but β = 0 , then we have h (cid:18) (cid:19) , − α/β ih (cid:18) (cid:19) , αβ i = h (cid:18) (cid:19) , α i , Finally, if β = δ = 0 , but α = 0 , then we have α .Thus the following elements are Aut p ( L )-orbit representatives: , α , δ . Therefore, the corresponding restricted Lie algebra structures are as follows: K = h x , . . . , x | [ x , x ] = x , [ x , x ] = x , [ x , x ] = x , x [ p ]3 = x i ; K ( α ) = h x , . . . , x | [ x , x ] = x , [ x , x ] = x , [ x , x ] = x , x [ p ]1 = αx , x [ p ]3 = x i ; K ( δ ) = h x , . . . , x | [ x , x ] = x , [ x , x ] = x , [ x , x ] = x , x [ p ]3 = x , x [ p ]4 = δx i . where α, δ ∈ F ∗ . Lemma 8.4.1 Let α , α ∈ F p . Then, K ( α ) and K ( α ) are isomorphic over F p if and only if α /α = ǫ , for some ǫ ∈ F ∗ p . Proof. Let f = ( a ij ) ∈ Aut( L , ). Then f : K ( α ) → K ( α ) is an isomorphism if and only if a = a = 0 a uα = α a a u = u. Hence, a = 1 and α /α = a − ∈ ( F ∗ p ) . To prove the converse, suppose that α /α = ǫ . We take a = 1 and a = ǫ − . (cid:4) Lemma 8.4.2 Let δ ∈ F p . Then, K ( δ ) ∼ = K (1) . Proof. We can take the following maps that yields the required automorphism K ( δ ) → K (1): /δ /δ /δ 00 0 0 0 1 /δ (cid:4) HAPTER 8. RESTRICTION MAPS ON L , The following is the list of all restricted Lie algebra structures on L , and yet as we shall see belowwe prove that some of them are isomorphic. K = h x , . . . , x | [ x , x ] = x , [ x , x ] = x , [ x , x ] = x i ; K ( α ) = h x , . . . , x | [ x , x ] = x , [ x , x ] = x , [ x , x ] = x , x [ p ]1 = αx i ; K = h x , . . . , x | [ x , x ] = x , [ x , x ] = x , [ x , x ] = x , x [ p ]2 = x i ; K = h x , . . . , x | [ x , x ] = x , [ x , x ] = x , [ x , x ] = x , x [ p ]3 = x i ; K ( δ ) = h x , . . . , x | [ x , x ] = x , [ x , x ] = x , [ x , x ] = x , x [ p ]4 = δx i ; K = h x , . . . , x | [ x , x ] = x , [ x , x ] = x , [ x , x ] = x , x [ p ]1 = x i ; K ( β ) = h x , . . . , x | [ x , x ] = x , [ x , x ] = x , [ x , x ] = x , x [ p ]1 = x , x [ p ]2 = βx i ; K ( γ ) = h x , . . . , x | [ x , x ] = x , [ x , x ] = x , [ x , x ] = x , x [ p ]1 = x , x [ p ]3 = γx i ; K ( δ ) = h x , . . . , x | [ x , x ] = x , [ x , x ] = x , [ x , x ] = x , x [ p ]1 = x , x [ p ]4 = δx i ; K ( ξ ) = h x , . . . , x | [ x , x ] = x , [ x , x ] = x , [ x , x ] = x , x [ p ]2 = ξx i ; K ( α, ξ ) = h x , . . . , x | [ x , x ] = x , [ x , x ] = x , [ x , x ] = x , x [ p ]1 = αx , x [ p ]2 = ξx i ; K ( ξ, γ ) = h x , . . . , x | [ x , x ] = x , [ x , x ] = x , [ x , x ] = x , x [ p ]2 = ξx , x [ p ]3 = γx i ; K ( ξ, δ ) = h x , . . . , x | [ x , x ] = x , [ x , x ] = x , [ x , x ] = x , x [ p ]2 = ξx , x [ p ]4 = δx i ; K = h x , . . . , x | [ x , x ] = x , [ x , x ] = x , [ x , x ] = x , x [ p ]3 = x i ; K ( α ) = h x , . . . , x | [ x , x ] = x , [ x , x ] = x , [ x , x ] = x , x [ p ]1 = αx , x [ p ]3 = x i ; K ( δ ) = h x , . . . , x | [ x , x ] = x , [ x , x ] = x , [ x , x ] = x , x [ p ]3 = x , x [ p ]4 = δx i ;Note that we have the following isomorphisms: − − : K → K , K → K , and − : K ( α ) → K ( α ) , K ( α ) → K ( α, . Theorem 8.5.1 The list of all restricted Lie algebra structures on L , , up to isomorphism, is as HAPTER 8. RESTRICTION MAPS ON L , follows: L , = h x , . . . , x | [ x , x ] = x , [ x , x ] = x , [ x , x ] = x i ; L , ( α ) = h x , . . . , x | [ x , x ] = x , [ x , x ] = x , [ x , x ] = x , x [ p ]1 = αx i ; L , = h x , . . . , x | [ x , x ] = x , [ x , x ] = x , [ x , x ] = x , x [ p ]2 = x i ; L , = h x , . . . , x | [ x , x ] = x , [ x , x ] = x , [ x , x ] = x , x [ p ]3 = x i ; L , ( δ ) = h x , . . . , x | [ x , x ] = x , [ x , x ] = x , [ x , x ] = x , x [ p ]4 = δx i ; L , ( β ) = h x , . . . , x | [ x , x ] = x , [ x , x ] = x , [ x , x ] = x , x [ p ]1 = x , x [ p ]2 = βx i ; L , ( γ ) = h x , . . . , x | [ x , x ] = x , [ x , x ] = x , [ x , x ] = x , x [ p ]1 = x , x [ p ]3 = γx i ; L , ( δ ) = h x , . . . , x | [ x , x ] = x , [ x , x ] = x , [ x , x ] = x , x [ p ]1 = x , x [ p ]4 = δx i ; L , ( ξ, α ) = h x , . . . , x | [ x , x ] = x , [ x , x ] = x , [ x , x ] = x , x [ p ]1 = αx , x [ p ]2 = ξx i ; L , ( ξ, γ ) = h x , . . . , x | [ x , x ] = x , [ x , x ] = x , [ x , x ] = x , x [ p ]2 = ξx , x [ p ]3 = γx i ; L , ( ξ, δ ) = h x , . . . , x | [ x , x ] = x , [ x , x ] = x , [ x , x ] = x , x [ p ]2 = ξx , x [ p ]4 = δx i ; L , ( δ ) = h x , . . . , x | [ x , x ] = x , [ x , x ] = x , [ x , x ] = x , x [ p ]3 = x , x [ p ]4 = δx i . In the remaining of this section we establish that the algebras given in Theorem 8.5.1 are pairwisenon-isomorphic, thereby completing the proof of Theorem 8.5.1.It is clear that L , is not isomorphic to the other restricted Lie algebras. We claim that L , ( α )and L , are not isomorphic. Suppose to the contrary that there exists an isomorphism A : L , ( α ) → L , . Then A ( x [ p ]1 ) = A ( x ) [ p ] . So, A ( αx ) = ( a x + a x + a x + a x + a x ) [ p ] αa ux + αa ux = a p x [ p ]1 + a p x [ p ]2 + a p x [ p ]3 + a p x [ p ]4 + a p x [ p ]5 αa ux + αa ux = a p x , which implies that αa u = 0 . Since u = 0 , α = 0 , we have a = 0 . Thus, u = a a . Also, wehave A ( x [ p ]2 ) = A ( x ) [ p ] a x + a x + a x + a x + a x ) [ p ] a p x , which implies that a = 0 . Therefore u = 0 , which is a contradiction.Next, We claim that L , ( α ) and L , are not isomorphic. Suppose to the contrary that thereexists an isomorphism A : L , ( α ) → L , . Then A ( x [ p ]3 ) = A ( x ) [ p ] ux + ( a a − a a ) x + ( a a − a a x ) [ p ] u p x . HAPTER 8. RESTRICTION MAPS ON L , u = 0 , which is a contradiction.Next, we claim that L , ( α ) and L , ( δ ) are not isomorphic. Suppose to the contrary that thereexists an isomorphism A : L , ( α ) → L , ( δ ). Then A ( x [ p ]4 ) = A ( x ) [ p ] . So,0 =( a ux + a ux ) [ p ] a p u p δx , which implies that α p u p δ = 0 . Since u = 0 , δ = 0 , we have a = 0 . Thus, u = − a a . Also, wehave A ( x [ p ]5 ) = A ( x ) [ p ] a ux + a ux ) [ p ] a p u p δx , which implies that a p u p δ = 0 . Since u = 0 , δ = 0 , we have a = 0 . Therefore u = 0 , which is acontradiction.It is clear that L , ( α ) is not isomorphic to the other restricted Lie algebras.Next, We claim that L , and L , are not isomorphic. Suppose to the contrary that there existsan isomorphism A : L , → L , . Then A ( x [ p ]3 ) = A ( x ) [ p ] ux + ( a a − a a ) x + ( a a − a a x ) [ p ] u p x . Therefore, u = 0 , which is a contradiction.Next, we claim that L , and L , ( δ ) are not isomorphic. Suppose to the contrary that thereexists an isomorphism A : L , → L , ( δ ). Then A ( x [ p ]4 ) = A ( x ) [ p ] . So,0 =( a ux + a ux ) [ p ] a p u p δx , which implies that α p u p δ = 0 . Since u = 0 , δ = 0 , we have a = 0 . Thus, u = − a a . Also, wehave A ( x [ p ]5 ) = A ( x ) [ p ] a ux + a ux ) [ p ] a p u p δx , which implies that a p u p δ = 0 . Since u = 0 , δ = 0 , we have a = 0 . Therefore u = 0 , which is acontradiction. HAPTER 8. RESTRICTION MAPS ON L , L , is not isomorphic to the other restricted Lie algebras.Next, we claim that L , and L , ( δ ) are not isomorphic. Suppose to the contrary that thereexists an isomorphism A : L , → L , ( δ ). Then A ( x [ p ]4 ) = A ( x ) [ p ] . So,0 =( a ux + a ux ) [ p ] a p u p δx , which implies that α p u p δ = 0 . Since u = 0 , δ = 0 , we have a = 0 . Thus, u = − a a . Also, wehave A ( x [ p ]5 ) = A ( x ) [ p ] a ux + a ux ) [ p ] a p u p δx , which implies that a p u p δ = 0 . Since u = 0 , δ = 0 , we have a = 0 . Therefore u = 0 , which is acontradiction.It is clear that L , and L , ( δ ) are not isomorphic to the other restricted Lie algebras.Next, We claim that L , ( β ) and L , ( γ ) are not isomorphic. Suppose to the contrary that thereexists an isomorphism A : L , ( β ) → L , ( γ ). Then A ( x [ p ]3 ) = A ( x ) [ p ] ux + ( a a − a a ) x + ( a a − a a x ) [ p ] u p γx . Therefore, u = 0 , which is a contradiction.Note that L , ( β ) is not isomorphic to any of L , ( δ ), L , ( ξ, δ ), L , ( δ ) because ( L , ( β )) [ p ] = 0but ( L , ( δ )) [ p ] = 0 , ( L , ( ξ, δ )) [ p ] = 0 , ( L , ( δ )) [ p ] = 0 .We shall compare L , ( β ) and L , ( α, ξ ) in Lemma 8.5.2.Next, We claim that L , ( β ) and L , ( ξ, γ ) are not isomorphic. Suppose to the contrary thatthere exists an isomorphism A : L , ( β ) → L , ( ξ, γ ). Then A ( x [ p ]3 ) = A ( x ) [ p ] ux + ( a a − a a ) x + ( a a − a a x ) [ p ] u p γx . Therefore, u = 0 , which is a contradiction.Note that L , ( γ ) is not isomorphic to any of L , ( δ ), L , ( ξ, δ ), L , ( δ ) because ( L , ( γ )) [ p ] = 0but ( L , ( δ )) [ p ] = 0 , ( L , ( ξ, δ )) [ p ] = 0 , ( L , ( δ )) [ p ] = 0 . HAPTER 8. RESTRICTION MAPS ON L , L , ( γ ) and L , ( α, ξ ) are not isomorphic. Suppose to the contrary thatthere exists an isomorphism A : L , ( α, ξ ) → L , ( γ ). Then A ( x [ p ]3 ) = A ( x ) [ p ] ux + ( a a − a a ) x + ( a a − a a x ) [ p ] u p γx . Therefore, u = 0 , which is a contradiction.Next, We claim that L , ( γ ) and L , ( ξ, γ ) are not isomorphic. Suppose to the contrary thatthere exists an isomorphism A : L , ( γ ) → L , ( ξ, γ ). Then A ( x [ p ]2 ) = A ( x ) [ p ] a x + a x + a x + a x + a x ) [ p ] a p ξx + a p γx , which implies that a p ξ = 0 . Since ξ = 0 , we have a = 0 . Thus, u = − a a . Also, we have A ( x [ p ]3 ) = A ( x ) [ p ] A ( γx ) =( ux + ( a a − a a ) x + ( a a − a a x ) [ p ] γa ux + γa ux = u p γx , which imples that a u = 0 . Since u = 0 , we have a = 0 . Therefore, u = 0 , which is a contradiction.Note that L , ( δ ) is not isomorphic to any of L , ( α, ξ ), L , ( ξ, γ ) because ( L , ( α, ξ )) [ p ] =( L , ( ξ, γ )) [ p ] = 0 but ( L , ( δ )) [ p ] = 0 .Next, We claim that L , ( δ ) and L , ( ξ, δ ) are not isomorphic. Suppose to the contrary that thereexists an isomorphism A : L , ( δ ) → L , ( ξ, δ ). Then A ( x [ p ]2 ) = A ( x ) [ p ] a x + a x + a x + a x + a x ) [ p ] a p ξx + a p δx , which implies that a p ξ = 0 . Since ξ = 0 , we have a = 0 . Thus, u = − a a . Also, we have A ( x [ p ]5 ) = A ( x ) [ p ] a ux + a ux ) [ p ] a p u p δx , which implies that a p u p δ = 0 . Since u = 0 , δ = 0 , we have a = 0 . Therefore, u = 0 , which is acontradiction. HAPTER 8. RESTRICTION MAPS ON L , L , ( δ ) and L , ( δ ) are not isomorphic. Suppose to the contrary that thereexists an isomorphism A : L , ( δ ) → L , ( δ ). Then A ( x [ p ]3 ) = A ( x ) [ p ] ux + ( a a − a a ) x + ( a a − a a x ) [ p ] u p x + ( a a − a a ) p δx . Therefore, u = 0 , which is a contradiction.Next, We claim that L , ( α, ξ ) and L , ( ξ, γ ) are not isomorphic. Suppose to the contrary thatthere exists an isomorphism A : L , ( α, ξ ) → L , ( ξ, γ ). Then A ( x [ p ]3 ) = A ( x ) [ p ] ux + ( a a − a a ) x + ( a a − a a x ) [ p ] u p γx . Therefore, u = 0 , which is a contradiction.Note that L , ( α, ξ ) is not isomorphic to any of L , ( ξ, δ ), L , ( δ ) because ( L , ( α, ξ )) [ p ] = 0but ( L , ( ξ, δ )) [ p ] = 0 , ( L , ( δ )) [ p ] = 0 .Note that L , ( ξ, γ ) is not isomorphic to any of L , ( ξ, δ ), L , ( δ ) because ( L , ( ξ, γ )) [ p ] = 0 but( L , ( ξ, δ )) [ p ] = 0 , ( L , ( δ )) [ p ] = 0 .Finally, We claim that L , ( ξ, δ ) and L , ( δ ) are not isomorphic. Suppose to the contrary thatthere exists an isomorphism A : L , ( ξ, δ ) → L , ( δ ). Then A ( x [ p ]3 ) = A ( x ) [ p ] ux + ( a a − a a ) x + ( a a − a a x ) [ p ] u p x + ( a a − a a ) p δx . Therefore, u = 0 , which is a contradiction. Lemma 8.5.2 Let β, α, ξ ∈ F ∗ p . Then L , ( β ) ∼ = L , ( α, ξ ) over F p if and only if β = − and ξ/α ∈ ( F ∗ ) . Proof. Let f = ( a ij ) ∈ Aut( L , ), where a ij ∈ F p . Then f : L , ( β ) → L , ( α, ξ ) is an isomorphismif and only if a u = ξa ; (8.31) βa u = ξa ; (8.32) a u = αa ; (8.33) βa u = αa . (8.34) HAPTER 8. RESTRICTION MAPS ON L , L , ( β ) ∼ = L , ( α, ξ ). Then, Equations (8.31) and (8.32) imply that a a = a a β whereas Equations (8.33) and (8.34) imply that a a β = a a . Hence, β = 1 . Thus,( a a ) = ( a a ) and since a a = a a , we deduce that a a = − a a . So, β = − ξ/α = ( a a ) .To prove the converse, suppose that ξ/α = ǫ and set a = − ξ/ (2 ǫ ) , a = ǫ, a = α/ , a = 1 .Then, we can see that all the Equations (8.31)-(8.34) are satisfied. (cid:4) ibliography [1] R.E. Beck, B. Kolman, Construction of nilpotent Lie algebras over arbitrary fields , Proceedingsof the 1981 ACM Symposium on Symbolic and Algebraic Computation, pages 169–174. ACMNew York, 1981.[2] S. Cical`o, W.A. de Graaf, C. Schneider, Six-dimensional nilpotent Lie algebras , Linear AlgebraAppl. 436 (2012), no. 1, 163-189.[3] H. Cohen, Number Theory: Volume I: Tools and Diophantine Equations , Graduate Texts inMathematics, vol. 239. Springer, New York, 2007.[4] I. Darijani, H. Usefi, The classification of 5-dimensional restricted Lie algebras over perfect fieldsI , Journal of Algebra, 464 (2016) 97–140[5] W.A. de Graaf, Classification of 6-dimensional nilpotent Lie algebras over fields of characteristicnot 2 , J. Algebra (2007), no. 2, 640–653.[6] T.J. Evans, D. Fuchs, A complex for the cohomology of restricted Lie algebras , J. Fixed PointTheory Appl., 3(2008), 159-179.[7] T.J. Evans, Cohomology of Restricted Lie Algebras , PhD thesis, University of California, Califor-nia, U.S., 2000.[8] M.P. Gong, Classification of Nilpotent Lie Algebras of Dimension 7 , PhD thesis, University ofWaterloo, Waterloo, Canada, 1998.[9] V.V. Morozov, Classification of nilpotent Lie algebras of sixth order , Izv. Vysˇs. Uˇcebn. Zaved.Matematika, (4 (5)) (1958) 161–171.[10] O.A. Nielsen, Unitary representations and coadjoint orbits of low-dimensional nilpotent Liegroups , volume 63 of Queen’s Papers in Pure and Applied Mathematics. Queen’s University,Kingston, ON, 1983.[11] M. Romdhani, Classification of Real and Complex Nilpotent Lie Algebras of Dimension 7 , Linearand Multilinear Algebra 24:3 (1989), 167–189.[12] C. Schneider, A computer-based approach to the classification of nilpotent Lie algebras , Experi-ment. Math., 14(2) (2005) 153–160.[13] C. Schneider, H. Usefi, The classification of p -nilpotent restricted Lie algebras of dimension atmost 4 , Forum Mathematicum, 28 (2016), no. 4, 713–727.156 IBLIOGRAPHY , Trans. Amer. Math. Soc. (1993), no. 2,479–496.[15] T. Skjelbred, T. Sund. Sur la classification des alg`ebres de Lie nilpotentes. C. R. Acad. Sci.Paris S´er. A-B , 286(5), 1978.[16] H. Strade, R. Farnsteiner,