Extending Structures for Lie 2-Algebras

Transcription

1 mathematcs Artcle Extendng Structures for Le 2-Algebras Yan Tan * and Zhxang Wu Mathematcs Department, Zhejang Unversty, Hangzhou , Chna; wzx@zju.edu.cn * Correspondence: ytanalg@126.com Receved: 19 May 2019; Accepted: 12 June 2019; Publshed: 18 June 2019 Abstract: The extendng structures problem for strct Le 2-algebras s studed. To provde the theoretcal answer to ths problem, ths paper ntroduces the unfed product of a gven strct Le 2-algebra g and 2-vector space V. The unfed product ncludes some nterestng products such as sem-drect product, crossed product, and bcrossed product. The paper focuses on crossed and bcrossed products, whch gve the answer to the extenson problem and factorzaton problem, respectvely. Keywords: strct Le 2-algebra; extendng structures problem; crossed product; unfed product 1. Introducton Recently, many mathematcans have pad attenton to Le algebra-lke structures. In partcular, they seek category theoretc analogs of them n [1 3]. A knd of algebra, strct Le 2-algebra, has appeared n some parts of the artcles. Le 2-algebras play a part n studyng algebrac structures on Le 2-groups, strng theory, hgher categorcal structures, and multsymplectc structures, Courant algebrods, Drac structures, omn-le 2-algebras, and Hom-Le 2-algebras, and so on [4,5]. For example, Omn-Le 2-algebras are a knd of specal weak Le 2-algebra. Weak Le 2-algebras are a categorfcaton of Le algebras, or an nternal category of Le algebras. Ths paper s gong to study extensons of crossed modules of Le algebras. Meanwhle, crossed modules of Le algebras can be dentfed wth strct Le 2-algebras. The extendng structures problem for some algebra objects such as Le algebras, Hopf algebras, Lebnz algebras, assocatve algebras, left-symmetrc algebras and Le conformal algebras have been studed n [6 11] respectvely. Le 2-algebras were frst ntroduced by J. C. Baez and A. S. Crans n It s a new knd algebra and more sophstcated than the usual Le algebras. By now, many Le 2-algebra theores have not been developed. The extenson theory of Le 2-algebras has been characterzed by cohomologcal groups n [12]. However, n [12], one needs the subalgebras to be abelan. Ths paper wll abandon the commutatve condtons n [12]. More explctly, the paper studes the followng extendng structures problem of strct Le 2-algebras. Problem 1. Let g := ((g 0, [, ] 0 ), (g 1, [, ] 1 ), φ, L) be a strct Le 2-algebra, V := V 1 V 0 a 2-vector space and c : c 1 c 0 a 2-vector space such that g s a 2-vector subspace. Suppose that c = g V as vector spaces for = 0, 1. Descrbe and classfy all strct Le 2-algebra structures on c up to an somorphsm of Le 2-algebras that stablzes g. In fact, ths problem generalzes two mportant algebra problems. One s the extenson problem for strct Le 2-algebras. Problem 2. Gven two Le 2-algebras g := ((g 0, [, ] 0 ), (g 1, [, ] 1 ), φ, L), V := ((V 0, 0 ), (V 1, 1 ),, 2 ). Descrbe and classfy all extensons of V by g whch are strct Le 2-algebras up to an somorphsm of Le 2-algebras that stablzes g. Mathematcs 2019, 7, 556; do: /math

2 Mathematcs 2019, 7, of 18 Here, an extenson of V by g s a Le 2-algebras c whch satsfes short exact sequence 0 g 1 ι 1 c 1 π 1 V 1 0 φ 0 g 0 ι 0 c 0 π 0 V 0 0 where ι, π are lnear maps for = 0, 1,.e., Im(ι ) = Ker(π ), Ker(ι ) = 0 and Im(π ) = V (Ref. [13]). When g s an abelan Le 2-algebra, all extensons of V by g, whch are strct Le 2-algebras up to an somorphsm of Le 2-algebras that stablzes g, can be characterzed by the second cohomology group H 2 (V, φ) defned n [12]. When g s not abelan, all extensons of V by g, whch are strct Le 2-algebras that stablzes g, are exactly the non-abelan extensons of V by g defned n [13]. The other problem s the factorzaton problem for strct Le 2-algebras. Problem 3. Let g := ((g 0, [, ] 0 ), (g 1, [, ] 1 ), φ, L), V := ((V 0, 0 ), (V 1, 1 ),, 2 ) be two strct Le 2-algebras and c : c 1 c 0 a 2-vector space such that g s a 2-vector subspace. Suppose that c = g V as vector spaces for = 0, 1. Descrbe and classfy all strct Le 2-algebra structures on c such that g and V are two sub-le 2-algebras of c up to an somorphsm of Le 2-algebras that stablzes g. Therefore, the study of the extendng structures problem s of sgnfcaton and wll be useful for nvestgatng the structure theory of Le 2-algebras. The paper always assumes that V = 0. Two cohomologcal type objects are constructed by ntroducng the unfed products. Usng ths unfed product, the extenson problem and the factorzaton problem for strct Le 2-algebras are studed n detal. An outlne of ths paper s as follows. In Secton 2, the paper provdes some prelmnares. The unfed product g V of a strct Le 2-algebra g by 2-vector space V assocated wth an extendng datum Ω(g, V) = ( j, j, f j, j, σ, 3, 3 ; j = 0, 1, 2) s ntroduced n Secton 3. Then, the paper presents the suffcent and necessary condtons to ensure that g V s a Le 2-algebra. Next, the paper shows that any strct Le 2-algebra c satsfyng the condton n extendng structures problem s somorphc to a unfed product of g by V. Fnally, the paper constructs two cohomologcal type objects, where one s somorphc to the classfcaton of the extendng structures problem. Some specal cases of unfed products such as crossed product and bcrossed product are ntroduced n Secton 4. Usng the crossed product and bcrossed product, the paper descrbes the extenson problem and factorzaton problem, respectvely. 2. Prelmnares In ths secton, some defntons and results about strct Le 2-algebras are provded. A Le 2-algebra s an object of an nternal category of Le algebras. It has been noted n [12,14,15] and elsewhere that the category of strct Le 2-algebras s equvalent to the category of crossed modules of Le algebras. A crossed module of Le algebras s defned as follows. Defnton 1. A crossed module of Le algebras s a quadruple ((g 0, [, ] 0 ), (g 1, [, ] 1 ), φ, L) where (g, [, ] ) for = 0, 1 are Le algebras, φ : g 1 g 0 s a Le algebra homomorphsm and L : g 0 gl(g 1 ) s a Le algebra acton by dervatons, such that for any x, y g and = 0, 1, φ(l x0 (x 1 )) = [x 0, φ(x 1 )] 0, L φ(x1 )(y 1 ) = [x 1, y 1 ] 1. These equatons are called equvarance and nfntesmal Peffer, respectvely. The homomorphsm of two crossed modules of Le algebras s gven by the followng defnton.

3 Mathematcs 2019, 7, of 18 Defnton 2. Let g := ((g 0, [, ] 0 ), (g 1, [, ] 1 ), φ, L) and g := ((g 0, [, ] 0 ), (g 1, [, ] 1 ), φ, L ) be two strct Le 2-algebras. A Le 2-algebra homomorphsm Φ : g g conssts of lnear maps Φ : g g for = 0, 1, such that the followng equaltes hold for all x, y g, Φ 0 φ = φ Φ 1, Φ [x, y ] = [Φ (x ), Φ (y )], Φ 1 (L x0 x 1 ) = L Φ 0 (x 0 ) Φ 1(x 1 ). If Φ 0, Φ 1 are nvertble, then Φ s an somorphsm. All crossed modules of Le algebras and homomorphsms between them form a category. Gven a 2-vector space V := V 1 V 0. One can construct a strct Le 2-algebra gl() := ((gl() 1, [, ] ), (gl() 0, [, ]),, L ), where gl() 0 = {(F, f ) End(V 1 ) End(V 0 ) : F = f }, gl() 1 = Hom(V 0, V 1 ), (A) = (A, A), [(F 0, f 0 ), (F 1, f 1 )] = (F 0 F 1 F 1 F 0, f 0 f 1 f 1 f 0 ), [A, B] = AB BA and L (F 1, f 1 ) A = F 1 A A f 1 for A, B gl() 1 and (F 0, f 0 ), (F 1, f 1 ) gl() 0 (Ref. [12]). If there s a homomorphsm ρ from a Le 2-algebra g to the Le 2-algebra gl(), then V s called a representaton of the Le 2-algebra g. Suppose that V s a representaton of a Le 2-algebra g. Then V 1 s a representaton of g 1, and V 0 s a representaton of g 0. Moreover, the paper ntroduces the followng concept. Defnton 3. Let g := ((g 0, [, ] 0 ), (g 1, [, ] 1 ), φ, L) be a strct Le 2-algebra, V := V 1 V 0 a 2-vector space and c := c 1 c 0 a 2-vector space such that the dagram 0 g 1 ι 1 c 1 π 1 V 1 0 (1) φ 0 g 0 ι 0 c 0 π 0 V 0 0 commutates, where π : c V are the canoncal projectons of c and ι : g c are the ncluson maps for = 0, 1. For lnear functor ϕ = (ϕ 0, ϕ 1 ) : c c, consder the dagram: g 1 ι 1 g 1 π 1 c 1 ι 1 ϕ 1 c 1 π 1 V 1 V 1 g 0 ι 0 c 0 π 0 V 0 ϕ 0 ι 0 g 0 π 0 c 0 V 0. (2) The paper calls that ϕ stablzes ((g 0, [, ] 0 ), (g 1, [, ] 1 ), φ, L) (resp. co-stablzes V 1 V 0 ) f the left cube (resp. the rght cube) of the dagram (2) s commutatve. Let ([, ] c1, [, ] c0,, L c ) and ([, ] c 1, [, ] c 0,, L c ) be two strct Le 2-algebra structures on c 1 c 0 both contanng ((g 0, [, ] 0 ), (g 1, [, ] 1 ), φ, L) as a sub-le 2-algebra. If there exsts a Le 2-algebra homomorphsm ϕ whch stablzes ((g 1, [, ] 1 ), (g 0, [, ] 0 ), φ, L), then ([, ] c1, [, ] c0,, L c ) and ([, ] c 1, [, ] c 0,, L c ) are called equvalent, whch s denoted by ((c 1, [, ] c1 ), (c 0, [, ] c0 ),, L c ) ((c 1, [, ] c 1 ), (c 0, [, ] c 0 ),, L c ). If there exsts a Le 2-algebra somorphsm ϕ whch stablzes ((g 0, [, ] 0 ), (g 1, [, ] 1 ), φ, L) and co-stablzes V 1 V 0,.e., the dagram (2) commutates, then ([, ] c1, [, ] c0,, L c ) and ([, ] c 1, [, ] c 0,, L c ) are called cohomologous, whch s denoted by ((c 1, [, ] c1 ), (c 0, [, ] c0 ),, L c ) ((c 1, [, ] c 1 ), (c 0, [, ] c 0 ),, L c ).

4 Mathematcs 2019, 7, of 18 It s easy to see that and are equvalence relatons on the set of all strct Le 2-algebra structures on c 1 c 0 contanng ((g 0, [, ] 0 ), (g 1, [, ] 1 ), φ, L) as a sub-le 2-algebra. The set of all equvalence classes va and are denoted by LExd(c, g) and LExd (c, g) respectvely. In addton, t s easy to show that there exsts a canoncal projecton LExd (c, g) LExd(c, g). Proposton 1. Let g := ((g 0, [, ] 0 ), (g 1, [, ] 1 ), φ, L) be a strct Le 2-algebra, V := V 1 V 0 a 2-vector space and c := c 1 c 0 a 2-vector space such that the dagram (1) commutates. Then c s 2-vector space φ+σ+ g 1 + V 1 g 0 + V 0, where σ : V 1 g 0 s a lnear map. Proof. By the defnton of π and ι, c = g + V. Let (x 1 ) = x 0 + v 0 for any x 1 Im(ι 1 ) c 1. Snce the left square of dagram (1) commutates, x 0 + v 0 = (x 1 ) = ι 0 φ(x 1 ) = φ(x 1 ). Thus, (x 1 ) = φ(x 1 ). Smlarly, f (v 1 ) = x 0 + v 0 for v 1 V 1, then v 0 = (v 1 ) as the rght square of the dagram (1) commutates. Defne σ : V 1 g 0 by σ(v) = π (v) for any v V 1, where π : g 0 V 0 g 0 s the canoncal projecton. Then (v 1 ) = φ(v 1 ) + (v 1 ) for any v 1 V 1. Hence (x 1 + v 1 ) = φ(x 1 ) + σ(v 1 ) + (v 1 ) for x 1 + v 1 c 1 and = φ + σ Unfed Products for Le 2-Algebras In ths secton, a unfed product of two Le 2-algebras s ntroduced. Usng ths product, the paper provdes the theoretcal answer to the extendng structure problem. 2-vector space V 1 V 0 and Le 2-algebra ((g 0, [, ] 0 ), (g 1, [, ] 1 ), φ, L) are smply denoted by V and g respectvely n the followng. Defnton 4. Suppose that g s a strct Le 2-algebra and V s a 2-vector space. An extendng datum of g by V s a system Ω(g, V) = ( j, j, f j, j, 3, 3, σ; j = 0, 1, 2) consstng of one lnear map σ : V 1 g 0 and fourteen blnear maps 0 : V 0 g 0 g 0, 0 : V 0 g 0 V 0, f 0 : V 0 V 0 g 0, 0 : V 0 V 0 V 0, 1 : V 1 g 1 g 1, 1 : V 1 g 1 V 1, f 1 : V 1 V 1 g 1, 1 : V 1 V 1 V 1, 2 : V 0 g 1 g 1, 2 : V 0 g 1 V 1, f 2 : V 0 V 1 g 1, 2 : V 0 V 1 V 1, 3 : V 1 g 0 g 1, 3 : V 1 g 0 V 1. Let Ω(g, V) = ( j, j, f j, j, 3, 3, σ; j = 0, 1, 2) be an extendng datum. Defne a new strct Le 2-algebra g Ω(g,V) V as follows. As a 2-vector space, g Ω(g,V) V s equal to g 1 V 1 g 0 V 0, where (y, w) = (φ(y) + σ(w), (w)) for y g 1 and w V 1. The blnear maps {, } : (g V ) (g V ) g V and the lnear map L c : g 0 V 0 gl(g 1 V 1 ) s gven by and {(x, v ), (y, w )} = ([x, y ] + v y w x + f (v, w ), v w + v y w x ) (3) L c (x 0,v 0 ) (x 1, v 1 ) = (L x0 x 1 + v 0 2 x 1 v 1 3 x 0 + f 2 (v 0, v 1 ), v 0 2 v 1 + v 0 2 x 1 v 1 3 x 0 ) (4) respectvely, for = 0, 1 and all x, y g, v, w V. Ths strct Le 2-algebra g Ω(g,V) V s called a unfed product of g and V, Ω(g, V) s called a Le 2-extendng structure of g by V. If only one Le 2-extendng structure Ω(g, V) of g by V s consdered, then g Ω(g,V) V s usually smplfed as g V and the Le 2-extendng structure s smply called extendng datum. The set of all Le 2-extendng structures of g by V s denoted by L(g, V).

5 Mathematcs 2019, 7, of 18 Theorem 1. Suppose that g s a strct Le 2-algebra, V s a 2-vector space. Then Ω(g, V) s an extendng datum of g by V such that g V s a strct Le 2-algebra f and only f the followng condtons hold for any x, y g, v, w, u V and = 0, 1: (L1) f (v, v ) = 0, v v = 0; (L2) (V, ) s a rght g -module; (L3) v [x, y ] = [v x, y ] + [x, v y ] + (v x ) y (v y ) x ; (L4) (v w ) x = v (w x ) + (v x ) w + v (w x ) w (w x ); (L5) (v w ) x = v (w x ) w (v x ) + [x, f (v, w )] + f (v, w x ) + f (v x, w ); (L6) f (v, w u ) + f (w, u v ) + f (u, v w ) + v f (w, u ) + w f (u, v ) + u f (v, w ) = 0; (L7) v (w u ) + w (u v ) + u (v w ) + v f (w, u ) + w f (u, v ) + u f (v, w ) = 0; (L8) (v 1 ) 0 φ(x 1 ) = φ(v 1 1 x 1 ) + σ(v 1 1 x 1 ) + [φ(x 1 ), σ(v 1 )] 0 ; (L9) ((v 1 ) 0 φ(x 1 ) = (v 1 1 x 1 ); (L10) (w 1 ) 0 σ(v 1 ) (v 1 ) 0 σ(w 1 ) = [σ(v 1 ), σ(w 1 )] 0 φ( f 1 (v 1, w 1 )) σ(v 1 1 w 1 ) + f 0 ((v 1 ), (w 1 )); (L11) (w 1 ) 0 σ(v 1 ) (v 1 ) 0 σ(w 1 ) = (v 1 ) 0 (w 1 ) (v 1 1 w 1 ); (L12) L x0 (v 1 1 x 1 ) = (v 1 1 x 1 ) 3 x 0 + v 1 1 (L x0 x 1 ) + [x 1, v 1 3 x 0 ] 1 (v 1 3 x 0 ) 1 x 1 ; (L13) (v 1 3 x 0 ) 1 x 1 = (v 1 1 x 1 ) 3 x 0 + v 1 1 (L x0 x 1 ); (L14) L x0 f 1 (v 1, w 1 ) = (v 1 1 w 1 ) 3 x 0 + w 1 1 (v 1 3 x 0 ) f 1 (v 1 3 x 0, w 1 ) v 1 1 (w 1 3 x 0 ) f 1 (v 1, w 1 3 x 0 ); (L15) (v 1 1 w 1 ) 3 x 0 + w 1 1 (v 1 3 x 0 ) = (v 1 3 x 0 ) 1 w 1 + v 1 1 (w 1 3 x 0 ) + v 1 1 (w 1 3 x 0 ); (L16) v 0 2 [x 1, y 1 ] 1 = [v 0 2 x 1, y 1 ] 1 + (v 0 2 x 1 ) 1 y 1 + [x 1, v 0 2 y 1 ] 1 (v 0 2 y 1 ) 1 x 1 ; (L17) v 0 2 [x 1, y 1 ] 1 = (v 0 2 x 1 ) 1 y 1 (v 0 2 y 1 ) 1 x 1 ; (L18) v 0 2 (v 1 1 x 1 ) + f 2 (v 0, v 1 1 x 1 ) v 1 1 (v 0 2 x 1 ) + f 1 (v 0 2 x 1, v 1 ) + [x 1, f 2 (v 0, v 1 )] 1 (v 0 2 v 1 ) 1 x 1 = 0; (L19) v 0 2 (v 1 1 x 1 ) + v 0 2 (v 1 1 x 1 ) v 1 1 (v 0 2 x 1 ) + (v 0 2 x 1 ) 1 v 1 (v 0 2 v 1 ) 1 x 1 = 0; (L20) v 0 2 f 1 (v 1, w 1 ) + f 2 (v 0, v 1 1 w 1 ) + w 1 1 f 2 (v 0, v 1 ) = f 1 (v 0 2 v 1, w 1 ) + v 1 1 f 2 (v 0, w 1 ) + f 1 (v 1, v 0 2 w 1 ); (L21) v 0 2 f 1 (v 1, w 1 ) + v 0 2 (v 1 1 w 1 ) + w 1 1 f 2 (v 0, v 1 ) = (v 0 2 v 1 ) 1 w 1 + v 1 1 f 2 (v 0, w 1 ) + v 1 1 (v 0 2 w 1 ); (L22) L x0 (v 1 3 y 0 ) = v 1 3 [x 0, y 0 ] 0 + (v 1 3 y 0 ) 3 x 0 + L y0 (v 1 3 x 0 ) (v 1 3 x 0 ) 3 y 0 ; (L23) v 1 3 [x 0, y 0 ] 0 = (v 1 3 x 0 ) 3 y 0 (v 1 3 y 0 ) 3 x 0 ; (L24) L v0 0 x 0 x 1 = (v 0 2 x 1 ) 3 x 0 (v 0 0 x 0 ) 2 x 1 L x0 (v 0 2 x 1 ) + v 0 2 L x0 x 1 ; (L25) (v 0 0 x 0 ) 2 x 1 = (v 0 2 x 1 ) 3 x 0 + v 0 2 L x0 x 1 ; (L26) v 1 3 (v 0 0 x 0 ) v 0 2 (v 1 3 x 0 ) = f 2 (v 0 0 x 0, v 1 ) + L x0 f 2 (v 0, v 1 ) (v 0 2 v 1 ) 3 x 0 + f 2 (v 0, v 1 3 x 0 ); (L27) v 1 3 (v 0 0 x 0 ) v 0 2 (v 1 3 x 0 ) = (v 0 0 x 0 ) 2 v 1 (v 0 2 v 1 ) 3 x 0 + v 0 2 (v 1 3 x 0 ); (L28) L f0 (v 0,w 0 ) x 1 + (v 0 0 w 0 ) 2 x 1 v 0 2 (w 0 2 x 1 ) f 2 (v 0, w 0 2 x 1 ) + w 0 2 (v 0 2 x 1 ) + f 2 (w 0, v 0 2 x 1 ) = 0; (L29) (v 0 0 w 0 ) 2 x 1 v 0 2 (w 0 2 x 1 ) v 0 2 (w 0 2 x 1 ) + w 0 2 (v 0 2 x 1 ) + w 0 2 (v 0 2 x 1 ) = 0; (L30) v 1 3 f 0 (v 0, w 0 ) + f 2 (v 0 0 w 0, v 1 ) v 0 2 f 2 (w 0, v 1 ) f 2 (v 0, w 0 2 v 1 ) + w 0 2 f 2 (v 0, v 1 ) + f 2 (w 0, v 0 2 v 1 ) = 0; (L31) v 1 3 f 0 (v 0, w 0 ) + (v 0 0 w 0 ) 2 v 1 v 0 2 f 2 (w 0, v 1 ) v 0 2 (w 0 2 v 1 ) + w 0 2 f 2 (v 0, v 1 ) + w 0 2 (v 0 2 v 1 ) = 0; (L32) (v 1 ) 0 x 0 = φ(v 1 3 x 0 ) + σ(v 1 3 x 0 ) + [x 0, σ(v 1 )] 0 ; (L33) (v 1 ) 0 x 0 = (v 1 3 x 0 ); (L34) v 0 0 φ(x 1 ) = φ(v 0 2 x 1 ) + σ(v 0 2 x 1 ); (L35) v 0 0 φ(x 1 ) = (v 0 2 x 1 ); (L36) φ( f 2 (v 0, v 1 )) + σ(v 0 2 v 1 ) v 0 0 σ(v 1 ) f 0 (v 0, (v 1 )) = 0; (L37) (v 0 2 v 1 ) v 0 0 σ(v 1 ) v 0 0 (v 1 ) = 0; (L38) v 1 3 φ(x 1 ) = v 1 1 x 1 ; (L39) v 1 3 φ(x 1 ) = v 1 1 x 1 ; (L40) (v 1 ) 2 x 1 = v 1 1 x 1 L σ(v1 ) x 1; (L41) (v 1 ) 2 x 1 = v 1 1 x 1 ; (L42) w 1 3 σ(v 1 ) = f 2 ((v 1 ), w 1 ) f 1 (v 1, w 1 );

6 Mathematcs 2019, 7, of 18 (L43) w 1 3 σ(v 1 ) = (v 1 ) 2 w 1 v 1 1 w 1. Proof. By ([6] Theorem 2.2), g V s a Le algebra f and only f the condtons (L1) (L7) hold. Thus, g V s a strct Le 2-algebra f and only f : g 1 V 1 g 0 V 0 s a Le algebra homomorphsm, L c : g 0 V 0 gl(g 1 V 1 ) s a Le algebra acton by dervatons and satsfyng equvarance and nfntesmal Peffer,.e., {(x 1, v 1 ), (y 1, w 1 )} 1 = {(x 1, v 1 ), (y 1, w 1 )} 0, (5) L c (x 0,v 0 ){(x 1, v 1 ), (y 1, w 1 )} 1 = {L c (x 0,v 0 )(x 1, v 1 ), (y 1, w 1 )} 1 + {(x 1, v 1 ), L c (x 0,v 0 )(y 1, w 1 )} 1, (6) L c {(x 0,v 0 ),(y 0,w 0 )} 0 (x 1, v 1 ) = L c (x 0,v 0 ) Lc (y 0,w 0 )(x 1, v 1 ) L c (y 0,w 0 ) Lc (x 0,v 0 )(x 1, v 1 ), (7) (L c (x 0,v 0 )(x 1, v 1 )) = {(x 0, v 0 ), (x 1, v 1 )} 0, (8) L c (x 1,v 1 )(y 1, w 1 ) = {(x 1, v 1 ), (y 1, w 1 )} 1 (9) for = 0, 1 and all x, y g, v, w V. Snce (x, v ) = (x, 0) + (0, v ) n g V, Equatons (5) (9) hold f and only f they hold for the set {(x, 0) x g } {(0, v ) v V }. Frst, Equaton (5) holds for (x 1, 0), (y 1, 0) as Snce {(x 1, 0), (y 1, 0)} 1 {(x 1, 0), (y 1, 0)} 0 = ([x 1, y 1 ] 1, 0) {(φ(x 1 ), 0), (φ(y 1 ), 0)} 0 = (φ[x 1, y 1 ] 1 [φ(x 1 ), φ(y 1 )] 0, 0) = (0, 0). {(x 1, 0), (0, v 1 )} 1 {(x 1, 0), (0, v 1 )} 0 = ( v 1 1 x 1, v 1 1 x 1 ) {(φ(x 1 ), 0), (σ(v 1 ), (v 1 ))} 0 = ( φ(v 1 1 x 1 ) σ(v 1 1 x 1 ) [φ(x 1 ), σ(v 1 )] 0 + (v 1 ) 0 φ(x 1 ), (v 1 1 x 1 ) + (v 1 ) 0 φ(x 1 )), Equaton (5) holds for (x 1, 0), (0, v 1 ) f and only f (L8) and (L9) hold. Equaton (5) holds for (0, v 1 ), (0, w 1 ) f and only f (L10) and (L11) hold, snce {(0, v 1 ), (0, w 1 )} 1 {(0, v 1 ), (0, w 1 )} 0 = ( f 1 (v 1, w 1 ), v 1 1 w 1 ) {(σ(v 1 ), (v 1 )), (σ(w 1 ), (w 1 ))} 0 = (φ( f 1 (v 1, w 1 )) + σ(v 1 1 w 1 ) [σ(v 1 ), σ(w 1 )] 0 (v 1 ) 0 σ(w 1 ) + (w 1 ) 0 σ(v 1 ) f 0 ((v 1 ), (w 1 )), (v 1 1 w 1 ) (v 1 ) 0 σ(w 1 ) + (w 1 ) 0 σ(v 1 ) (v 1 ) 0 (w 1 )). Then, Equaton (6) holds for (x 0, 0), (x 1, 0), (y 1, 0) as L c (x 0,0){(x 1, 0), (y 1, 0)} 1 {L c (x 0,0)(x 1, 0), (y 1, 0)} 1 {(x 1, 0), L c (x 0,0)(y 1, 0)} 1 = L c (x 0,0)([x 1, y 1 ] 1, 0) {(L x0 x 1, 0), (y 1, 0)} 1 {(x 1, 0), (L x0 y 1, 0)} 1 = (L x0 [x 1, y 1 ] 1 [L x0 x 1, y 1 ] 1 [x 1, L x0 y 1 ] 1, 0) = (0, 0). Snce, for (x 0, 0), (x 1, 0), (0, v 1 ), L c (x 0,0){(x 1, 0), (0, v 1 )} 1 {L c (x 0,0)(x 1, 0), (0, v 1 )} 1 {(x 1, 0), L c (x 0,0)(0, v 1 )} 1 = L c (x 0,0)( v 1 1 x 1, v 1 1 x 1 ) {(L x0 x 1, 0), (0, v 1 )} 1 + {(x 1, 0), (v 1 3 x 0, v 1 3 x 0 )} 1 = ( L x0 (v 1 1 x 1 ) + (v 1 1 x 1 ) 3 x 0 + v 1 1 (L x0 x 1 ) + [x 1, v 1 3 x 0 ] 1 (v 1 3 x 0 ) 1 x 1, (v 1 1 x 1 ) 3 x 0 + v 1 1 (L x0 x 1 ) (v 1 3 x 0 ) 1 x 1 ),

7 Mathematcs 2019, 7, of 18 Equaton (6) holds for (x 0, 0), (x 1, 0), (0, v 1 ) f and only f (L12) and (L13) hold. Equaton (6) holds for (x 0, 0), (0, v 1 ), (0, w 1 ) f and only f (L14) and (L15) hold, snce L c (x 0,0){(0, v 1 ), (0, w 1 )} 1 {L c (x 0,0)(0, v 1 ), (0, w 1 )} 1 {(0, v 1 ), L c (x 0,0)(0, w 1 )} 1 = L c (x 0,0)( f 1 (v 1, w 1 ), v 1 1 w 1 ) + {(v 1 3 x 0, v 1 3 x 0 ), (0, w 1 )} 1 + {(0, v 1 ), (w 1 3 x 0, w 1 3 x 0 )} 1 = (L x0 f 1 (v 1, w 1 ) (v 1 1 w 1 ) 3 x 0 w 1 1 (v 1 3 x 0 ) + f 1 (v 1 3 x 0, w 1 ) + v 1 1 (w 1 3 x 0 ) + f 1 (v 1, w 1 3 x 0 ), (v 1 1 w 1 ) 3 x 0 w 1 1 (v 1 3 x 0 ) + (v 1 3 x 0 ) 1 w 1 + v 1 1 (w 1 3 x 0 ) +v 1 1 (w 1 3 x 0 )). Equaton (6) holds for (0, v 0 ), (x 1, 0), (y 1, 0) f and only f (L16) and (L17) hold, as L c (0,v 0 ){(x 1, 0), (y 1, 0)} 1 {L c (0,v 0 )(x 1, 0), (y 1, 0)} 1 {(x 1, 0), L c (0,v 0 )(y 1, 0)} 1 = L c (0,v 0 )([x 1, y 1 ] 1, 0) {(v 0 2 x 1, v 0 2 x 1 ), (y 1, 0)} 1 {(x 1, 0), (v 0 2 y 1, v 0 2 y 1 )} 1 = (v 0 2 [x 1, y 1 ] 1 [v 0 2 x 1, y 1 ] 1 (v 0 2 x 1 ) 1 y 1 [x 1, v 0 2 y 1 ] 1 + (v 0 2 y 1 ) 1 x 1, v 0 2 [x 1, y 1 ] 1 (v 0 2 x 1 ) 1 y 1 + (v 0 2 y 1 ) 1 x 1 ). Equaton (6) holds for (0, v 0 ), (x 1, 0), (0, v 1 ) f and only f (L18) and (L19) hold. Indeed, L c (0,v 0 ){(x 1, 0), (0, v 1 )} 1 {L c (0,v 0 )(x 1, 0), (0, v 1 )} 1 {(x 1, 0), L c (0,v 0 )(0, v 1 )} 1 = L c (0,v 0 )( v 1 1 x 1, v 1 1 x 1 ) {(v 0 2 x 1, v 0 2 x 1 ), (0, v 1 )} 1 {(x 1, 0), ( f 2 (v 0, v 1 ), v 0 2 v 1 )} 1 = ( v 0 2 (v 1 1 x 1 ) f 2 (v 0, v 1 1 x 1 ) + v 1 1 (v 0 2 x 1 ) f 1 (v 0 2 x 1, v 1 ) [x 1, f 2 (v 0, v 1 )] 1 +(v 0 2 v 1 ) 1 x 1, v 0 2 (v 1 1 x 1 ) v 0 2 (v 1 1 x 1 ) + v 1 1 (v 0 2 x 1 ) (v 0 2 x 1 ) 1 v 1 +(v 0 2 v 1 ) 1 x 1 ) for (0, v 0 ), (x 1, 0), (0, v 1 ). Notce that L c (0,v 0 ){(0, v 1 ), (0, w 1 )} 1 {L c (0,v 0 )(0, v 1 ), (0, w 1 )} 1 {(0, v 1 ), L c (0,v 0 )(0, w 1 )} 1 = L c (0,v 0 )( f 1 (v 1, w 1 ), v 1 1 w 1 ) {( f 2 (v 0, v 1 ), v 0 2 v 1 ), (0, w 1 )} 1 {(0, v 1 ), ( f 2 (v 0, w 1 ), v 0 2 w 1 )} 1 = (v 0 2 f 1 (v 1, w 1 ) + f 2 (v 0, v 1 1 w 1 ) + w 1 1 f 2 (v 0, v 1 ) f 1 (v 0 2 v 1, w 1 ) v 1 1 f 2 (v 0, w 1 ) f 1 (v 1, v 0 2 w 1 ), v 0 2 f 1 (v 1, w 1 ) + v 0 2 (v 1 1 w 1 ) + w 1 1 f 2 (v 0, v 1 ) (v 0 2 v 1 ) 1 w 1 v 1 1 f 2 (v 0, w 1 ) v 1 1 (v 0 2 w 1 )) for (0, v 0 ), (0, v 1 ), (0, w 1 ). Thus, Equaton (6) holds for (0, v 0 ), (0, v 1 ), (0, w 1 ) f and only f (L20) and (L21) hold. Now Equaton (7) holds for (x 0, 0), (y 0, 0), (x 1, 0) as Snce L c {(x 0,0),(y 0,0)} 0 (x 1, 0) L c (x 0,0) Lc (y 0,0)(x 1, 0) + L c (y 0,0) Lc (x 0,0)(x 1, 0) = L c ([x 0,y 0 ] 0,0)(x 1, 0) L c (x 0,0)(L y0 x 1, 0) + L c (y 0,0)(L x0 x 1, 0) = (L [x0,y 0 ] 0 x 1 L x0 L y0 x 1 + L y0 L x0 x 1, 0) = (0, 0). L c {(x 0,0),(y 0,0)} 0 (0, v 1 ) L c (x 0,0) Lc (y 0,0)(0, v 1 ) + L c (y 0,0) Lc (x 0,0)(0, v 1 ) = L c ([x 0,y 0 ] 0,0)(0, v 1 ) + L c (x 0,0)(v 1 3 y 0, v 1 3 y 0 ) L c (y 0,0)(v 1 3 x 0, v 1 3 x 0 ) = ( v 1 3 [x 0, y 0 ] 0 + L x0 (v 1 3 y 0 ) (v 1 3 y 0 ) 3 x 0 L y0 (v 1 3 x 0 ) + (v 1 3 x 0 ) 3 y 0, v 1 3 [x 0, y 0 ] 0 (v 1 3 y 0 ) 3 x 0 + (v 1 3 x 0 ) 3 y 0 )

8 Mathematcs 2019, 7, of 18 for (x 0, 0), (y 0, 0), (0, v 1 ), Equaton (7) holds for (x 0, 0), (y 0, 0), (0, v 1 ) f and only f (L22) and (L23) hold. As L c {(x 0,0),(0,v 0 )} 0 (x 1, 0) L c (x 0,0) Lc (0,v 0 )(x 1, 0) + L c (0,v 0 ) Lc (x 0,0)(x 1, 0) = L c ( v 0 0 x 0, v 0 0 x 0 )(x 1, 0) L c (x 0,0)(v 0 2 x 1, v 0 2 x 1 ) + L c (0,v 0 )(L x0 x 1, 0) = ( L v0 0 x 0 x 1 (v 0 0 x 0 ) 2 x 1 L x0 (v 0 2 x 1 ) + (v 0 2 x 1 ) 3 x 0 + v 0 2 L x0 x 1, (v 0 0 x 0 ) 2 x 1 + (v 0 2 x 1 ) 3 x 0 + v 0 2 L x0 x 1 ), Equaton (7) holds for (x 0, 0), (0, v 0 ), (x 1, 0) f and only f (L24) and (L25) hold. Indeed, L c {(x 0,0),(0,v 0 )} 0 (0, v 1 ) L c (x 0,0) Lc (0,v 0 )(0, v 1 ) + L c (0,v 0 ) Lc (x 0,0)(0, v 1 ) = L c ( v 0 0 x 0, v 0 0 x 0 )(0, v 1 ) L c (x 0,0)( f 2 (v 0, v 1 ), v 0 2 v 1 ) + L c (0,v 0 )( v 1 3 x 0, v 1 3 x 0 ) = (v 1 3 (v 0 0 x 0 ) f 2 (v 0 0 x 0, v 1 ) L x0 f 2 (v 0, v 1 ) + (v 0 2 v 1 ) 3 x 0 v 0 2 (v 1 3 x 0 ) f 2 (v 0, v 1 3 x 0 ), v 1 3 (v 0 0 x 0 ) (v 0 0 x 0 ) 2 v 1 + (v 0 2 v 1 ) 3 x 0 v 0 2 (v 1 3 x 0 ) v 0 2 (v 1 3 x 0 )) for (x 0, 0), (0, v 0 ), (0, v 1 ). Then Equaton (7) holds for (x 0, 0), (0, v 0 ), (0, v 1 ) f and only f (L26) and (L27) hold. Because L c {(0,v 0 ),(0,w 0 )} 0 (x 1, 0) L c (0,v 0 ) Lc (0,w 0 )(x 1, 0) + L c (0,w 0 ) Lc (0,v 0 )(x 1, 0) = L c ( f 0 (v 0,w 0 ),v 0 0 w 0 )(x 1, 0) L c (0,v 0 )(w 0 2 x 1, w 0 2 x 1 ) + L c (0,w 0 )(v 0 2 x 1, v 0 2 x 1 ) = (L f0 (v 0,w 0 ) x 1 + (v 0 0 w 0 ) 2 x 1 v 0 2 (w 0 2 x 1 ) f 2 (v 0, w 0 2 x 1 ) + w 0 2 (v 0 2 x 1 ) + f 2 (w 0, v 0 2 x 1 ), (v 0 0 w 0 ) 2 x 1 v 0 2 (w 0 2 x 1 ) v 0 2 (w 0 2 x 1 ) + w 0 2 (v 0 2 x 1 ) +w 0 2 (v 0 2 x 1 )) for (0, v 0 ), (0, w 0 ), (x 1, 0), Equaton (7) holds for (0, v 0 ), (0, w 0 ), (x 1, 0) f and only f (L28) and (L29) hold. Snce L c {(0,v 0 ),(0,w 0 )} 0 (0, v 1 ) L c (0,v 0 ) Lc (0,w 0 )(0, v 1 ) + L c (0,w 0 ) Lc (0,v 0 )(0, v 1 ) = L c ( f 0 (v 0,w 0 ),v 0 0 w 0 )(0, v 1 ) L c (0,v 0 )( f 2 (w 0, v 1 ), w 0 2 v 1 ) + L c (0,w 0 )( f 2 (v 0, v 1 ), v 0 2 v 1 ) = ( v 1 3 f 0 (v 0, w 0 ) + f 2 (v 0 0 w 0, v 1 ) v 0 2 f 2 (w 0, v 1 ) f 2 (v 0, w 0 2 v 1 ) + w 0 2 f 2 (v 0, v 1 ) + f 2 (w 0, v 0 2 v 1 ), v 1 3 f 0 (v 0, w 0 ) + (v 0 0 w 0 ) 2 v 1 v 0 2 f 2 (w 0, v 1 ) v 0 2 (w 0 2 v 1 ) +w 0 2 f 2 (v 0, v 1 ) + w 0 2 (v 0 2 v 1 )), Equaton (7) holds for (0, v 0 ), (0, w 0 ), (0, v 1 ) f and only f (L30) and (L31) hold. Next Equaton (8) holds for (x 0, 0), (x 1, 0) as (L c (x 0,0)(x 1, 0)) {(x 0, 0), (x 1, 0)} 0 = (L x0 x 1, 0) {(x 0, 0), (φ(x 1 ), 0)} 0 = (φ(l x0 x 1 ) [x 0, φ(x 1 )] 0, 0) = (0, 0). As (L c (x 0,0)(0, v 1 )) {(x 0, 0), (0, v 1 )} 0 = ( v 1 3 x 0, v 1 3 x 0 ) {(x 0, 0), (σ(v 1 ), (v 1 ))} 0 = ( φ(v 1 3 x 0 ) σ(v 1 3 x 0 ) [x 0, σ(v 1 )] 0 + (v 1 ) 0 x 0, (v 1 3 x 0 ) + (v 1 ) 0 x 0 ),

9 Mathematcs 2019, 7, of 18 Equaton (8) holds for (x 0, 0), (0, v 1 ) f and only f (L32) and (L33) hold. Indeed, (L c (0,v 0 )(x 1, 0)) {(0, v 0 ), (x 1, 0)} 0 = (v 0 2 x 1, v 0 2 x 1 ) {(0, v 0 ), (φ(x 1 ), 0)} 0 = (φ(v 0 2 x 1 ) + σ(v 0 2 x 1 ) v 0 0 φ(x 1 ), (v 0 2 x 1 ) v 0 0 φ(x 1 )) for (0, v 0 ), (x 1, 0). Thus, Equaton (8) holds for (0, v 0 ), (x 1, 0) f and only f (L34) and (L35) hold. Equaton (8) holds for (0, v 0 ), (0, v 1 ) f and only f (L36) and (L37) hold, snce (L c (0,v 0 )(0, v 1 )) {(0, v 0 ), (0, v 1 )} 0 = ( f 2 (v 0, v 1 ), v 0 2 v 1 ) {(0, v 0 ), (σ(v 1 ), (v 1 ))} 0 = (φ( f 2 (v 0, v 1 )) + σ(v 0 2 v 1 ) v 0 0 σ(v 1 ) f 0 (v 0, (v 1 )), (v 0 2 v 1 ) v 0 0 σ(v 1 ) v 0 0 (v 1 )). Fnally, Equaton (9) holds for (x 1, 0), (y 1, 0) as L c (x 1,0)(y 1, 0) {(x 1, 0), (y 1, 0)} 1 = L c (φ(x 1 ),0)(y 1, 0) ([x 1, y 1 ] 1, 0) = (L φ(x1 ) y 1 [x 1, y 1 ] 1, 0) = (0, 0). Indeed, L c (x 1,0)(0, v 1 ) {(x 1, 0), (0, v 1 )} 1 = L c (φ(x 1 ),0)(0, v 1 ) + (v 1 1 x 1, v 1 1 x 1 ) = ( v 1 3 φ(x 1 ) + v 1 1 x 1, v 1 3 φ(x 1 ) + v 1 1 x 1 ) for (x 1, 0), (0, v 1 ). Thus, Equaton (9) holds for (x 1, 0), (0, v 1 ) f and only f (L38) and (L39) hold. Snce, for (0, v 1 ), (x 1, 0), L c (0,v 1 )(x 1, 0) {(0, v 1 ), (x 1, 0)} 1 = L c (σ(v 1 ),(v 1 ))(x 1, 0) (v 1 1 x 1, v 1 1 x 1 ) = (L σ(v1 ) x 1 + (v 1 ) 2 x 1 v 1 1 x 1, (v 1 ) 2 x 1 v 1 1 x 1 ), Equaton (9) holds for (0, v 1 ), (x 1, 0) f and only f (L40) and (L41) hold. Equaton (9) holds for (0, v 1 ), (0, w 1 ) f and only f (L42) and (L43) hold, snce L c (0,v 1 )(0, w 1 ) {(0, v 1 ), (0, w 1 )} 1 = L c (σ(v 1 ),(v 1 ))(0, w 1 ) ( f 1 (v 1, w 1 ), v 1 1 w 1 ) = ( w 1 3 σ(v 1 ) + f 2 ((v 1 ), w 1 ) f 1 (v 1, w 1 ), w 1 3 σ(v 1 ) + (v 1 ) 2 w 1 v 1 1 w 1 ). By now the proof s completed. Example 1. Let Ω(g, V) = ( j, j, f j, j, 3, 3, σ; j = 0, 1, 2) be an extendng datum of Le 2-algebra g by 2-vector space V such that j, j, f j, j, 3, 3, σ for j = 0, 1, 2 are trval maps,.e., v x = 0, v x = 0, f (v, w ) = 0, v w = 0, σ(v 1 ) = 0, v 0 2 x 1 = 0, v 0 2 x 1 = 0, f 2 (v 0, v 1 ) = 0, v 0 2 v 1 = 0, v 1 3 x 0 = 0, v 1 3 x 0 = 0 for = 0, 1 and x g, v, w V. Then g V s a Le 2-algebra wth = φ, the brackets and dervaton gven by {(x, v ), (y, w )} = ([x, y ], 0) and L c (x 0,v 0 ) (x 1, v 1 ) = (L x0 x 1, 0) respectvely, for = 0, 1 and all x, y g, v, w V. The paper calls ths Le 2-extendng structure the trval extendng structure of g by V.

10 Mathematcs 2019, 7, of 18 In the sequel, the paper uses the followng conventon: f one of the maps j, j, f j, j, 3, 3, σ for j = 0, 1, 2 of an extendng datum Ω(g, V) = ( j, j, f j, j, 3, 3, σ; j = 0, 1, 2) s trval then the paper wll omt t from ( j, j, f j, j, 3, 3, σ; j = 0, 1, 2). Example 2. Let Ω(g, V) = ( j, j, f j, j, 3, 3, σ; j = 0, 1, 2) be an extendng datum of Le 2-algebra g by 2-vector space V such that j, f j, j, σ, 3 for j = 0, 1, 2 are trval maps. Then Ω(g, V) = ( k ; k = 0, 1, 2, 3) s a Le 2-extendng structure of g by V f and only f ρ : g gl() s a representaton of Le 2-algebra g on 2-vector space V, where ρ 1 (x 1 )v 0 = v 0 2 x 1, ρ 0 0 (x 0)v 0 = v 0 0 x 0 and ρ 1 0 (x 0)v 1 = v 1 3 x 0 for x g, v V. In ths case, the assocatve unfed product g Ω(g,V) V s the sem-drect product ((g 1 ρ 1 0 φ V 1, {, } 1 ), (g 0 ρ 0 0 V 0, {, } 0 ), φ, L c ), where for = 0, 1 and x, y g, v, w V. {(x 0, v 0 ), (y 0, w 0 )} 0 = ([x 0, y 0 ] 0, ρ 0 0 (x 0)w 0 ρ 0 0 (y 0)v 0 ), {(x 1, v 1 ), (y 1, w 1 )} 1 = ([x 1, y 1 ] 1, (ρ 1 0 φ)(x 1)w 1 (ρ 1 0 φ)(y 1)v 1 ), L c (x 0,v 0 ) (x 1, v 1 ) = (L x0 x 1, ρ 1 0 (x 0)v 1 ρ 1 (x 1 )v 0 ) Let Ω(g, V) = ( j, j, f j, j, 3, 3, σ; j = 0, 1, 2) L(g, V) be a Le 2-extendng structure and g V be the assocated unfed product. Then the canoncal ncluson ι = (ι 0, ι 1 ) : g g V, ι (x ) = (x, 0) s an njectve Le 2-algebra homomorphsm. Hence, g can be vewed as a sub-le 2-algebra of g V by the dentfcaton g = ι(g) = g {0}. Conversely, the paper wll prove that any strct Le 2-algebra structure on 2-vector space c contanng g as a sub-le 2-algebra s somorphc to a unfed product. Theorem 2. Let g be a strct Le 2-algebra and c a 2-vector space contanng g as a 2-vector subspace. Suppose that ({, },, L c ; = 0, 1) s a strct Le 2-algebra structure on c such that g s a sub-le 2-algebra n (c, {, },, L c ; = 0, 1) of g by a 2-vector space V. Then there s an somorphsm of Le 2-algebras (c, {, },, L c ; = 0, 1) = g V whch stablzes g and co-stablzes V. Proof. Let p : c g be lnear maps such that p (x ) = x for = 0, 1 and x g. Then V := Ker(p ) s a subspace of c and a complement of g n c. Defne the extendng datum of g by V by the followng formulas: = p : V g g, v x := p ({v, x }), = p : V g V, v x := {v, x } p ({v, x }), f = f p : V V g, f (v, w ) := p ({v, w }), = p : V V V, v w := {v, w } p ({v, w }), 2 : V 0 g 1 g 1, v 0 2 x 1 := p 1 (L c v 0 x 1 ), 2 : V 0 g 1 V 1, v 0 2 x 1 := (L c v 0 x 1 ) p 1 (L c v 0 x 1 ), f 2 : V 0 V 1 g 1, f 2 (v 0, v 1 ) := p 1 (L c v 0 v 1 ), 2 : V 0 V 1 V 1, v 0 2 v 1 := L c v 0 v 1 p 1 (L c v 0 v 1 ), 3 : V 1 g 0 g 1, v 1 3 x 0 := p 1 (L c x 0 v 1 ), 3 : V 1 g 0 V 1, v 1 3 x 0 := L c x 0 v 1 + p 1 (L c x 0 v 1 ) σ : V 1 g 0, σ(v 1 ) := p 0 ((v 1 ))

11 Mathematcs 2019, 7, of 18 for any = 0, 1, x g and v, w V. Frst, the above maps are all well defned. Ths paper shall prove that Ω(g, V) = ( j, j, f j, j, 3, 3, σ; j = 0, 1, 2) s a Le 2-extendng structure of g by V and Ψ = (Ψ 0, Ψ 1 ) : g V c, Ψ (x, v ) := x + v s an somorphsm of Le 2-algebras that stablzes g and co-stablzes V. It s easy to verfy that for = 0, 1 and z c, Ψ 1 = (Ψ0 1, Ψ 1 1 ) : c g V, Ψ 1 (z ) := (p (z ), z p (z )) s an nverse of Ψ : g V c as 2-vector space. Therefore, there s a unque strct Le 2-algebra structure on g V such that Ψ s an somorphsm of strct Le 2-algebras and ths unque Le 2-algebra structure s gven by {(x, v ), (y, w )} := Ψ 1 ({Ψ (x, v ), Ψ (y, w )} ), L c (x 0,v 0 ) (x 1, v 1 ) := Ψ 1 1 (Lc Ψ 0 (x 0,v 0 ) Ψ 1(x 1, v 1 )) for all x, y g and v, w V. Then the proof s suffcent to prove that ths Le 2-algebra structure concdes wth the one defned by (3) and (4) assocated wth the system ( j, j, f j, j, 3, 3, σ; j = 0, 1, 2). Indeed, for any x, y g and v, w V, {(x, v ), (y, w )} = Ψ 1 ({Ψ (x, v ), Ψ (y, w )} ) = Ψ 1 ({x, y } + {x, w } + {v, y } + {v, w } ) = (p ({x, y } ) + p ({x, w } ) + p ({v, y } ) + p ({v, w } ), {x, y } p ({x, y } ) +{x, w } p ({x, w } ) + {v, y } p ({v, y } ) + {v, w } p ({v, w } )) = ([x, y ] w x + v y + f (v, w ), v y + v w w x } ), L c (x 0,v 0 ) (x 1, v 1 ) = Ψ 1 1 (Lc x 0 x 1 + L c x 0 v 1 + L c v 0 x 1 + L c v 0 v 1 ) = (p 1 (L c x 0 x 1 ) + p 1 (L c x 0 v 1 ) + p 1 (L c v 0 x 1 ) + p 1 (L c v 0 v 1 ), L c x 0 x 1 p 1 (L c x 0 x 1 ) +L c x 0 v 1 p 1 (L c x 0 v 1 ) + L c v 0 x 1 p 1 (L c v 0 x 1 ) + L c v 0 v 1 p 1 (L c v 0 v 1 ) = (L x0 x 1 v 1 3 x 0 + v 0 2 x 1 + f 2 (v 0, v 1 ), v 1 3 x 0 + v 0 2 x 1 + v 0 2 v 1 ). Moreover, the followng dagram s commutatve The proof s completed now. g 1 g 1 1 g1 Ψ 1 g 1 V 1 c 1 V 1 V 1 g 0 0 c 0 V 0 Ψ 0 g0 g 0 g 0 V 0 V 0. (10) By Theorem 2, the classfcaton of all strct Le 2-algebra structure on c that contanng g as a sub-le 2-algebra reduces to the classfcaton of all unfed products g V assocated wth all Le 2-extendng structures Ω(g, V) = ( j, j, f j, j, 3, 3, σ; j = 0, 1, 2), for a gven 2-vector space V such that V s a complement of g n c. Lemma 1. Suppose that Ω(g, V) = ( j, j, f j, j, 3, 3, σ; j = 0, 1, 2) and Ω (g, V) = ( j, j, f j, j, σ, 3, 3 ; j = 0, 1, 2) are two Le 2-extendng structures of g by V and g V, g V are the assocated unfed

12 Mathematcs 2019, 7, of 18 products. Then there exsts a bjecton between the set of all morphsms of Le 2-algebras Φ : g V g V whch stablzes g and the set of (r, τ ; = 0, 1), where r : V g and τ : V V are lnear maps satsfyng the followng compatblty condtons for = 0, 1 and any x g, v, w V : (M1) τ (v ) x = τ(v x ); (M2) r (v x ) = {r (v ), x } v x + τ(v ) x ; (M3) τ(v w ) = τ(v ) τ(w ) + τ(v ) r (w ) τ(w ) r (v ); (M4) r (v w ) = {r (v ), r (w )} + τ(v ) r (w ) τ(w ) r (v ) + f (τ(v ), τ(w )) f (v, w ), (M5) τ 1 (v 1 ) = τ 0 (v 1 ), (M6) φ(r 1 (v 1 )) + σ (τ 1 (v 1 )) = σ(v 1 ) + r 0 ((v 1 )), (M7) v 1 3 x 0 + r 1 (v 1 3 x 0 ) = τ 1 (v 1 ) 3 x 0 L x0 r 1 (v 1 ), (M8) τ 1 (v 1 3 x 0 ) = τ 1 (v 1 ) 3 x 0, (M9) v 0 2 x 1 + r 1 (v 0 2 x 1 ) = L r0 (v 0 ) x 1 + τ 0 (v 0 ) 2 x 1, (M10) τ 1 (v 0 2 x 1 ) = τ 0 (v 0 ) 2 x 1, (M11) f 2 (v 0, v 1 ) + r 1 (v 0 2 v 1 ) = L r0 (v 0 ) r 1(v 1 ) + τ 0 (v 0 ) 2 r 1(v 1 ) τ 1 (v 1 ) 3 r 0(v 0 ) + f 2 (τ 0(v 0 ), τ 1 (v 1 )), (M12) τ 1 (v 0 2 v 1 ) = τ 0 (v 0 ) 2 r 1(v 1 ) τ 1 (v 1 ) 3 r 0(v 0 ) + τ 0 (v 0 ) 2 τ 1(v 1 ). Under the above bjecton the morphsm of Le 2-algebras Φ = Φ (r,τ ;=0,1) = (Φ 0, Φ 1 ) : g V g V correspondng to (r, τ ; = 0, 1) s gven by: Φ (x, v ) = (x + r (v ), τ (v )), for any x g, v V and = 0, 1. Moreover, Φ = Φ (r,τ ;=0,1) s an somorphsm f and only f τ : V V s an somorphsm and Φ = Φ (r,τ ;=0,1) co-stablzes V f and only f τ = Id V. Proof. Suppose that Φ = (Φ 0, Φ 1 ) : g V g V s a lnear functor such that g 1 g 1 g1 g 0 g1 Φ 1 g 1 V 1 Φ 0 g 1 V 1 g0 g0 g 0 g 0 V 0 g 1 V 1 (11) commutates. Then t s unquely determned by lnear maps r : V g and τ : V V such that Φ (x, v ) = (x + r (v ), τ (v )) for = 0, 1 and all x g, v V. In fact, let Φ (0, v ) = (r (v ), τ (v )) g V for all v V. Then Φ (x, v ) = (x + r (v ), τ (v )) g V. Next, the paper proves that Φ = Φ (r,τ ;=0,1) s a morphsm of strct Le 2-algebras f and only f the compatblty condtons (M1) (M12) hold. It s suffcent to prove the equatons Φ ({(x, v ), (y, w )} ) = {Φ(x, v ), Φ(y, w )}, (12) (φ + σ + )Φ 1 (x 1, v 1 ) = Φ 0 (φ(x 1 ) + σ(v 1 ), (v 1 )), (13) Φ 1 (L c (x 0,v 0 ) (x 1, v 1 )) = L c Φ0 (x 0,v 0 ) Φ 1(x 1, v 1 ) (14) hold for the set {(x, 0) x g } {(0, v ) v V } and = 0, 1. By ([6], Lemma 2.5), Equaton (12) holds f and only f condtons (M1) (M4) hold. Frst, consder Equaton (13). It s easy to see that Equaton (13) holds for (x 1, 0) g 1 V 1. Equaton (13) holds for (0, v 1 ) g 1 V 1 f and only f (M5) and (M6) hold, snce (φ + σ + )Φ 1 (0, v 1 ) Φ 0 (σ(v 1 ), (v 1 )) = (φ(r 1 (v 1 )) + σ (τ 1 (v 1 )) σ(v 1 ) r 0 ((v 1 ), τ 1 (v 1 ) τ 0 (v 1 )).

13 Mathematcs 2019, 7, of 18 Now consder Equaton (14). It s easy to see that Equaton(14) holds for (x, 0) g V. Snce, for (x 0, 0) g 0 V 0 and (0, v 1 ) g 1 V 1, Φ 1 (L c (x 0,0) (0, v 1)) L c Φ0 (x 0,0) Φ 1(0, v 1 ) = Φ 1 ( v 1 3 x 0, v 1 3 x 0 ) L c (x0,0) (r 1(v 1 ), τ 1 (v 1 )) = ( v 1 3 x 0 r 1 (v 1 3 x 0 ) L x0 r 1 (v 1 ) + τ 1 (v 1 ) 3 x 0, τ 1 (v 1 3 x 0 ) + τ 1 (v 1 ) 3 x 0), Equaton (14) holds for (x 0, 0) g 0 V 0 and (0, v 1 ) g 1 V 1 f and only f (M7) and (M8) hold. Smlarly, t s easy to check that Equaton (14) holds for (0, v 0 ) g 0 V 0 and (x 1, 0) g 1 V 1 f and only f (M9) and (M10) hold; Equaton (14) holds for (0, v 0 ) g 0 V 0 and (0, v 1 ) g 1 V 1 f and only f (M11) and (M12) hold. Assume that τ : V V s bjectve. Then Φ (r,τ ;=0,1) s an somorphsm of Le 2-algebras wth the nverse gven by Φ 1 (r,τ ;=0,1) = (Φ 1 0, Φ 1 1 ), where Φ 1 (x, v ) = (x r (τ 1 (v )), τ 1 (v )) for x g and v V. Conversely, assume that Φ (r,τ ;=0,1) = (Φ 0, Φ 1 ) s somorphc. Then Φ s an somorphsm of Le algebras for = 0, 1. By the proof of ([6], Lemma 2.5), τ s a bjecton for = 0, 1. The last asserton s trval, and the proof s completed now. Defnton 5. Let g be a strct Le 2-algebra and V a 2-vector space. If there exsts lnear maps r : V g and τ : V V for = 0, 1 such that Le 2-algebra extendng structure Ω (g, V) = ( j, j, f j, j, 3, 3, σ ; j = 0, 1, 2) can be yeld from another Le 2-algebra extendng structure Ω(g, V) = ( j, j, f j, j, 3, 3, σ; j = 0, 1, 2) usng (r, τ ; = 0, 1) va: v x = τ (τ 1 (v ) x ) v x = r (τ 1 (v, w ) = f (τ 1 f r (τ 1 +τ 1 v w = τ (τ 1 (v ) x ) + τ 1 (v ), τ 1 (v ) r (τ 1 (w ) r (τ 1 (v ) τ 1 (w )) + r (τ 1 (v ) x + [x, r (τ 1 (v ))] (w ))) τ 1 (v )) (w )) τ (τ 1 (v ) τ 1 (w )) + [r (τ 1 (v ) r (τ 1 (v ) r (τ 1 σ (v 1 ) = σ(τ 1 1 (v 1 )) + r 0 ((τ 1 1 (v 1 ))) φ(r 1 (τ 1 1 (v 1 ))) (w )) + r(τ 1 (v )), r (τ 1 (w ))] (w ) τ 1 (v )) (w ))) + τ (τ 1 (w ) r (τ 1 (v ))) = σ(τ1 1 (v 1 )) + r 0 (τ0 1 ((v 1 ))) φ(r 1 (τ1 1 (v 1 ))) v 1 3 x 0 = r 1 (τ1 1 (v 1 ) 3 x 0 ) + τ1 1 (v 1 ) 3 x 0 + L x0 r 1 (τ1 1 (v 1 )) v 1 3 x 0 = τ 1 (τ1 1 (v 1 ) 3 x 0 ) v 0 2 x 1 = r 1 (τ0 1 (v 0 ) 2 x 1 ) L r0 (τ0 1 (v 0 )) x 1 + τ0 1 (v 0 ) 2 x 1 v 0 2 x 1 = τ 1 (τ0 1 (v 0 ) 2 x 1 ) f 2 (v 0, v 1 ) = r 1 (τ0 1 (v 0 ) 2 τ1 1 (v 1 ) + τ1 1 (v 1 ) 3 r 0 (τ0 1 (v 0 )) τ0 1 (v 0 ) 2 r 1 (τ1 1 (v 1 ))) + f 2 (τ0 1 (v 0 ), τ1 1 (v 1 )) + τ1 1 (v 1 ) 3 r 0 (τ0 1 (v 0 )) τ0 1 (v 0 ) 2 r 1 (τ1 1 (v 1 )) +L r0 (τ0 1 (v 0 )) r 0(τ1 1 (v 1 )) v 0 2 v 1 = τ 1 (τ0 1 (v 0 ) 2 τ1 1 (v 1 ) + τ1 1 (v 1 ) 3 r 0 (τ0 1 (v 0 )) τ0 1 (v 0 ) 2 r 1 (τ1 1 (v 1 ))) for = 0, 1 and any x g, v, w V, then Ω(g, V) and Ω (g, V) are sad to be equvalent, whch s denoted by Ω(g, V) Ω (g, V). In partcular, f τ = Id V for = 0, 1, then Ω(g, V) and Ω (g, V) are called cohomologous, whch s denoted by Ω(g, V) Ω (g, V). The paper concludes ths secton by the followng theorem, whch provdes an answer to the extendng structures problem of strct Le 2-algebras.

14 Mathematcs 2019, 7, of 18 Theorem 3. Suppose that g s a strct Le 2-algebra, V s a 2-vector space and c s a 2-vector space whch contans g as a 2-subspace and V s a complement of g n c for = 0, 1. Then 1. the relaton s an equvalence relaton on the set L(g, V) of all Le 2-extendng structures of g by V, and the map H 2 g(v, g) := L(g, V)/ LExtd(c, g), gven by ( j, j, f j, j, 3, 3, σ; j = 0, 1, 2) (g V, {, }, φ + σ +, L g V ), s bjectve, where ( j, j, f j, j, 3, 3, σ; j = 0, 1, 2) s the equvalence class of ( j, j, f j, j, 3, 3, σ; j = 0, 1, 2) under the equvalent relaton. 2. the relaton s an equvalent relaton on the set L(g, V) of all Le 2-extendng structures of g by V, and the mappng H 2 (V, g) := L(g, V)/ LExtd (c, g) gven by ( j, j, f j, j, 3, 3, σ; j = 0, 1, 2) (g V, {, }, φ + σ +, L g V ) s a bjecton, where ( j, j, f j, j, 3, 3, σ; j = 0, 1, 2) s the equvalence class of ( j, j, f j, j, 3, 3, σ; j = 0, 1, 2) under the equvalent relaton. Proof. It follows from Theorem 1, Theorem 2 and Lemma Specal Cases of Unfed Products In ths secton, two specal cases of unfed products are studed. One corresponds to the extenson problem and the other corresponds to the factorzaton problem Crossed Products and the Extenson Problem Let Ω(g, V) = ( j, j, f j, j, 3, 3, σ; j = 0, 1, 2) be the extendng datum of g by V such that k are trval maps for k = 0, 1, 2, 3. Then Ω(g, V) = ( j, j, f j, j, 3, 3, σ; j = 0, 1, 2) = ( j, f j, j, 3, σ; j = 0, 1, 2) s a Le 2-extendng structure of g by V f and only f (V,, j ; j = 0, 1, 2) s a strct Le 2-algebra and the followng compatbltes hold for = 0, 1 and any x, y g, v, w V : f 0 (v 0, v 0 ) = 0; v 0 0 [x 0, y 0 ] 0 = [v 0 0 x 0, y 0 ] 0 + [x 0, v 0 0 y 0 ] 0 ; (v 0 0 w 0 ) 0 x 0 = v 0 0 (w 0 0 x 0 ) w 0 0 (v 0 0 x 0 ) + [x 0, f 0 (v 0, w 0 )] 0 ; f 0 (v 0, w 0 0 u 0 ) + f 0 (w 0, u 0 0 v 0 ) + f 0 (u 0, v 0 0 w 0 ) + v 0 0 f 0 (w 0, u 0 ) + w 0 0 f 0 (u 0, v 0 ) + u 0 0 f 0 (v 0, w 0 ) = 0; v 0 2 [x 1, y 1 ] 1 = [v 0 2 x 1, y 1 ] 1 + [x 1, v 0 2 y 1 ] 1 ; L x0 (v 1 3 y 0 ) = v 1 3 [x 0, y 0 ] 0 + L y0 (v 1 3 x 0 ); L v0 0 x 0 x 1 = L x0 (v 0 2 x 1 ) + v 0 1 L x0 x 1 ; v 1 3 (v 0 0 x 0 ) v 0 2 (v 1 3 x 0 ) = L x0 f 2 (v 0, v 1 ) (v 0 2 v 1 ) 3 x 0 ; L f0 (v 0,w 0 ) x 1 + (v 0 0 w 0 ) 2 x 1 v 0 2 (w 0 2 x 1 ) + w 0 2 (v 0 2 x 1 ) = 0; v 1 3 f 0 (v 0, w 0 ) + f 2 (v 0 0 w 0, v 1 ) v 0 2 f 2 (w 0, v 1 ) f 2 (v 0, w 0 2 v 1 ) + w 0 2 f 2 (v 0, v 1 ) + f 2 (w 0, v 0 2 v 1 ) = 0; (v 1 ) 0 x 0 = φ(v 1 3 x 0 ) + [x 0, σ(v 1 )] 0 ; v 0 0 φ(x 1 ) = φ(v 0 2 x 1 ); φ( f 2 (v 0, v 1 )) + σ(v 0 2 v 1 ) = v 0 0 σ(v 1 ) + f 0 (v 0, (v 1 )); v 1 3 φ(x 1 ) = v 1 1 x 1 ; (v 1 ) 2 x 1 = v 1 1 x 1 L σ(v1 ) x 1; w 1 3 σ(v 1 ) = f 2 ((v 1 ), w 1 ) f 1 (v 1, w 1 ). In ths case, the assocated unfed product g Ω(g,V) V = g f V s called the crossed product of the Le 2-algebras g and V. A system (g, V, j, f j, 3, σ; j = 0, 1, 2) consstng of two strct Le 2-algebras

15 Mathematcs 2019, 7, of 18 g, V, seven blnear maps : V g g, f : V V g, 2 : V 0 g 1 g 1, 3 : V 1 g 0 g 1, f 2 : V 0 V 1 g 1 for = 0, 1 and one lnear map σ : V 1 g 0 satsfyng the above compatblty condtons wll be called a crossed system of Le 2-algebras. The crossed product assocated wth the crossed system (g, V, j, f j, 3, σ; j = 0, 1, 2) s the strct Le 2-algebra g f V = g V wth the brackets and dervaton gven by: {(x, v ), (y, w )} := ([x, y ] + v y w x + f (v, w ), v w ), L (x0,v 0 )(x 1, v 1 ) := (L x0 x 1 + v 0 2 x 1 v 1 3 x 0 + f 2 (v 0, v 1 ), v 0 2 v 1 ) for = 0, 1 and x, y g, v, w V. Then g = g {0} s an deal of the Le 2-algebra g f V snce {(x, 0), (y, w )} := ([x, y ] w x, 0), L (x0,0)(x 1, v 1 ) := (L x0 x 1 v 1 3 x 0, 0), and L (x0,v 0 )(x 1, 0) := (L x0 x 1 + v 0 2 x 1, 0). Conversely, crossed products descrbe all Le 2-algebra structures on the 2-vector space c such that a gven strct Le 2-algebra g s an deal of c. Corollary 1. Let g be a strct Le 2-algebra, c a 2-vector space contanng g as a 2-vector subspace. Then any strct Le 2-algebra structure on c that contans g as an deal s somorphc to a crossed product of Le 2-algebras g f V. Proof. Let ({, },, L c ; = 0, 1) be a strct Le 2-algebra structure on c such that g s an deal of c. In partcular, g s a 2-vector subalgebra of c. By Theorem 2, the paper obtans the Le 2-extendng structure Ω(g, V) = ( j, j, f j, j, 3, 3, σ; j = 0, 1, 2), where the acton j for j = 0, 1, 2 are trval. Indeed, for any v V and x g, {x, v } g, L x0 v 1 g 1 and L v0 x 1 g 1 and hence p ({x, v } ) = {x, v }, p 1 (L x0 v 1 ) = L x0 v 1 and p 1 (L v0 x 1 ) = L v0 x 1. Thus, the unfed product g Ω(g,V) V = g f V s the crossed product of the Le 2-algebras g and 2-vector space V := Ker(p 1 ) p 1 Ker(p 0 ). Remark 1. By Corollary 1, all crossed products of Le 2-algebras g f V gve the theoretcal answer to Problem 2. In fact, a crossed product of Le 2-algebras g f V also corresponds to the non-abelan extenson structure defned n [13]. Defne 0 (v 0 )(x 0 ) = v 0 0 x 0, 0 (v 0 )(x 1 ) = v 0 2 x 1, 1 (v 1 )(x 0 ) = v 1 3 x 0, ω(v 0, w 0 ) = f 0 (v 0, w 0 ), ν(v 0, v 1 ) = f 2 (v 0, v 1 ), ϕ(v 1 ) = σ(v 1 ) for v, w V, x g and = 0, 1, where, ω, ν, ϕ are lnear maps defned n [13]. The result follows. In partcular, f g s an abelan Le 2-algebra,.e., g s a 2-vector space. Then the set of all extendng structures of the abelan Le 2-algebra g by the 2-vector space V s parameterzed by the set of all ( j, f j, j, 3, σ; j = 0, 1, 2), such that (V,, j ; j = 0, 1, 2) s a strct Le 2-algebra, (g, j, j = 0, 2, 3) s a left V-module and f : V V V, f 2 : V 0 V 1 V 1 are blnear maps such that f (v, v ) = 0 and f 0 (v 0, w 0 0 u 0 ) + f 0 (w 0, u 0 0 v 0 ) + f 0 (u 0, v 0 0 w 0 ) + v 0 0 f 0 (w 0, u 0 ) + w 0 0 f 0 (u 0, v 0 ) + u 0 0 f 0 (v 0, w 0 ) = 0; v 0 2 f 1 (v 1, w 1 ) + f 2 (v 0, v 1 1 w 1 ) + w 1 1 f 2 (v 0, v 1 ) = f 1 (v 0 2 v 1, w 1 ) + v 1 1 f 2 (v 0, w 1 ) + f 1 (v 1, v 0 2 w 1 ); v 1 3 (v 0 0 x 0 ) v 0 2 (v 1 3 x 0 ) = L x0 f 2 (v 0, v 1 ) (v 0 2 v 1 ) 3 x 0 ; v 1 3 f 0 (v 0, w 0 ) = f 2 (v 0 0 w 0, v 1 ) v 0 2 f 2 (w 0, v 1 ) f 2 (v 0, w 0 2 v 1 ) + w 0 2 f 2 (v 0, v 1 ) + f 2 (w 0, v 0 2 v 1 ); φ( f 2 (v 0, v 1 )) + σ(v 0 2 v 1 ) v 0 0 σ(v 1 ) f 0 (v 0, (v 1 )) = 0; (v 1 ) 2 x 1 = v 1 1 x 1 ; w 1 3 σ(v 1 ) = f 2 ((v 1 ), w 1 ) f 1 (v 1, w 1 ) for = 0, 1 and u, v, w V, x g. For such ( j, f j, j, 3, σ; j = 0, 1, 2), the brackets and the dervaton of the extendng structure on c = g V are gven by {(x, v ), (y, w )} := (v y w x + f (v, w ), v w ),

16 Mathematcs 2019, 7, of 18 L (x0,v 0 )(x 1, v 1 ) := (v 0 2 x 1 v 1 3 x 0 + f 2 (v 0, v 1 ), v 0 2 v 1 ) respectvely, where = 0, 1 and x, y g, v, w V. By ([12], Theorem 5.6), the second cohomology group H 2 (V, φ) classfes 2-extensons. To study the relaton between H 2 (V, φ) and Le 2-algebra g V, the paper needs a result of [12]. Lemma 2 ([12], Proposton 5.3). Let ρ be a 2-representaton of strct Le 2-algebra g on V. Gven a trple (ω, α, ϕ) ((g 0 g 0 ) V 0) (g 0 g 1 V 1) (g 1 V 0). Then 0 V 1 ι 1 g 1 ω 1 ρ 1 0 φ V 1 π 1 g 1 0, φ 0 V 0 ι 0 g 0 ω ρ 0 0 V 0 π 0 g 0 0 wth ω 1 (x 1, y 1 ) := ρ 1 (y 1 )ϕ(x 1 ) + α(φ(x 1 ); y 1 ), (x 1, v 1 ) = (φ(x 1 ), (v 1 ) + ϕ(x 1 )), L (x 0,v 0 ) (x 1, v 1 ) = (L x0 x 1, ρ 1 0 (x 0)v 1 ρ 1 (x 1 )v 0 α(x 0 ; x 1 )) s a 2-extensons f and only f the followng equatons are satsfed () ρ 0 0 (x 0)ω(y 0, z 0 ) ω([x 0, y 0 ] 0, z 0 )+ = 0; () ρ 1 (x 1 )ϕ(y 1 ) + α(φ(y 1 ); x 1 )+ = 0; () ρ 1 0 (φ(x 1))(ρ 1 (y 1 )ϕ(z 1 ) + α(φ(z 1 ); y 1 )) ρ 1 (z 1 )ϕ([x 1, y 1 ]) α(([x 1, y 1 ]); z 1 )+ = 0; (v) ω(x 0, φ(x 1 )) = α(x 0 ; x 1 ) + ρ 0 0 (x 0)ϕ(x 1 ) ϕ(l x0 x 1 ); (v) α([x 0, y 0 ]; x 1 ) ρ 1 (x 1 )ω(x 0, x 1 ) = ρ 1 0 (x 0)α(y 0 ; x 1 ) α(y 0 ; L x0 x 1 )+ = 0; (v) For the contracton a x0 := α(x 0 ; ) g 1 V 1 seen as a 1-cocycle wth values n ρ 1 0 φ; for = 0, 1 and x, y, z g, v V. Here, stands for cyclc permutatons. Proposton 2. If g s an abelan Le 2-algebra, then Le 2-algebra g V s the Le 2-algebra defned by a second cohomology group (ω, α, ϕ) H 2 (V, φ) as n Lemma 2. Proof. Gven a Le 2-algebra g V. Let ω = f 0, α = f 2, ϕ = σ. Then the paper obtans a Le 2-algebra determned by the (ω, α, ϕ) H 2 (V, φ). Conversely, suppose that c s a Le 2-algebra determned by a the (ω, α, ϕ) H 2 (V, φ). Let f 0 = ω, f 2 = α, σ = ϕ. Then the paper obtans a Le 2-algebra g V. Thus, the concluson follows. Moreover, ths case Le 2-algebra s also correspondence wth the abelan extenson of V by g whch s defned n [14] Bcrossed Products and the Factorzaton Problem Let Ω(g, V) = ( j, j, f j, j, 3, 3, σ; j = 0, 1, 2) be the extendng datum of g by V such that σ and f j are trval maps for j = 0, 1, 2. Then Ω(g, V) = ( j, j, f j, j, 3, 3, σ; j = 0, 1, 2) = ( j, j, j, 3, 3 ; j = 0, 1, 2) s a Le 2-extendng structure of g by V f and only f (V,, j ; j = 0, 1, 2) s a strct Le 2-algebra, g s a left V-module under (( 0, 2 ), 3 ) : V gl(φ), V s a rght g-module under (( 0, 3 ), 2 ) : g gl() and the followng compatbltes hold for = 0, 1 and any x, y g, v, w V : v [x, y ] = [v x, y ] + [x, v y ] + (v x ) y (v y ) x ; (v w ) x = v (w x ) + (v x ) w + v (w x ) w (w x ); L x0 (v 1 1 x 1 ) = (v 1 1 x 1 ) 3 x 0 + v 1 1 (L x0 x 1 ) + [x 1, v 1 3 x 0 ] 1 (v 1 3 x 0 ) 1 x 1 ;

17 Mathematcs 2019, 7, of 18 (v 1 1 w 1 ) 3 x 0 + w 1 1 (v 1 3 x 0 ) = (v 1 3 x 0 ) 1 w 1 + v 1 1 (w 1 3 x 0 ) + v 1 1 (w 1 3 x 0 ); v 0 2 [x 1, y 1 ] 1 = [v 0 2 x 1, y 1 ] 1 + (v 0 2 x 1 ) 1 y 1 + [x 1, v 0 2 y 1 ] 1 (v 0 2 y 1 ) 1 x 1 ; v 0 2 (v 1 1 x 1 ) + v 0 2 (v 1 1 x 1 ) v 1 1 (v 0 2 x 1 ) + (v 0 2 x 1 ) 1 v 1 (v 0 2 v 1 ) 1 x 1 = 0; L x0 (v 1 3 y 0 ) = v 1 3 [x 0, y 0 ] 0 + (v 1 3 y 0 ) 3 x 0 + L y0 (v 1 3 x 0 ) (v 1 3 x 0 ) 3 y 0 ; L v0 0 x 0 x 1 = (v 0 2 x 1 ) 3 x 0 (v 0 0 x 0 ) 2 x 1 L x0 (v 0 2 x 1 ) + v 0 2 L x0 x 1 ; v 1 3 (v 0 0 x 0 ) v 0 2 (v 1 3 x 0 ) = (v 0 0 x 0 ) 2 v 1 (v 0 2 v 1 ) 3 x 0 + v 0 2 (v 1 3 x 0 ); (v 0 0 w 0 ) 2 x 1 v 0 2 (w 0 2 x 1 ) v 0 2 (w 0 2 x 1 ) + w 0 2 (v 0 2 x 1 ) + w 0 2 (v 0 2 x 1 ) = 0; (v 1 ) 2 x 1 = v 1 1 x 1 ; (v 1 ) 2 x 1 = v 1 1 x 1. In ths case, the assocated unfed product g Ω(g,V) V s called the bcrossed product g V of (g, V, k, k ; k = 0, 1, 2, 3) of the Le 2-algebras g and V. The brackets and dervaton on g V are gven by: {(x, v ), (y, w )} := ([x, y ] + v y w x, v w + v y w x ), L (x0,v 0 )(x 1, v 1 ) := (L x0 x 1 + v 0 2 x 1 v 1 3 x 0, v 0 2 v 1 + v 0 2 x 1 v 1 3 x 0 ) for = 0, 1 and x, y g, v, w V. Theorem 4. Suppose that g := ((g 0, [, ] 0 ), (g 1, [, ] 1 ), φ, L), V := ((V 0, 0 ), (V 1, 1 ),, 2 ) are two Le 2-algebras and c : c 1 c 0 s a 2-vector space such that dagram (1) satsfes. Assume that ({, },, L c ; = 0, 1) s a strct Le 2-algebra structure on c such that g and V are sub-le 2-algebras n (c, {, },, L c ; = 0, 1). Then Le 2-algebra (c, {, },, L c ; = 0, 1) s somorphc to g V of g and V. Proof. It follows from Theorem Conclusons Ths paper contans nformaton on how to construct a strct Le 2-algebra from one strct Le 2-algebra by another strct Le 2-algebra. A Le 2-algebra can be obtaned from a Le 2-group. A strct Le 2-group s usually called a crossed module of Le groups. It s also an nternal object n the category of Le groups. Smlarly, one can study extenson structures of a Le 2-group by another Le 2-group. One natural avenue of further exploraton s to consder the relatons between these two knds of extenson, and, furthermore, the relatons between the cohomologcal groups. Please note that ths paper s lmted to consderaton of extensons of algebras. The paper has neglected the possble topology and geometry of the Le 2-groups, more generalzaton 2-gerbes, of whch the felds are flat connectons classcal, n the sense famlar to physcsts. Another avenue s to dscuss algebrac 2-groups over a feld of characterstc non-zero p. Then, the Frobenus maps gve the correspondng Le algebra [p]-maps. A knd of Le algebra over a feld of characterstc p wth a [p]-map s called restrcted Le algebra. Hence one can study a smlar queston about strct restrcted Le 2-algebras and algebrac 2-groups. A strct Le 2-algebra s smlar to another extenson algebra of Le algebras, namely the two-term L algebra. It s well known that an L algebra s the same thng as a sem-free graded-commutatve dfferental graded algebra. Suppose M s a smooth manfold. Then the algebra Ω(M) of dfferental forms s a graded-commutatve dfferental graded algebra. One can construct a sem-free graded-commutatve dfferental graded algebra CE(L) from any L algebra L, whch s called Chevalley Elenberg algebra. It s a complcated but straghtforward exercse to check that the nlpotence d 2 CE = 0 for Chevalley Elenberg algebra dfferental s exactly equvalent to the homotopy Jacob denttes. Usng ths deal, the authors hope to seek smlar graded-commutatve dfferental graded algebras to reduce a large number of the equatons n Theorem 1. All of these need to be further explored.

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