The Killing forms and decomposition theorems of Lie supertriple systems

Acta Mathematica Scientia 2009,29B(2):360–370 http://actams.wipm.ac.cn

THE KILLING FORMS AND DECOMPOSITION THEOREMS OF LIE SUPERTRIPLE SYSTEMS∗

)

Zhang Zhixue (

Institute of Mathematics and Computers, Hebei University, Baoding 071002, China

Jia Peipei (


)

Fundamental Department, Hebei College of Finance, Baoding 071000, China

Abstract In this article, the Killing form of a Lie supertriple system (LSTS) and that of its imbedding Lie superalgebra (LSA) are investigated, and a unique decomposition theorem for a quasiclassical LSTS with trivial center is established by means of the parallel decomposition theorem for a quasiclassical LSA. Key words Killing form, Lie supertriple system, quasiclassical 2000 MR Subject Classification

17A40, 17B05

Recently, the theory of Lie and anti-Lie (super)triple systems admitting nondegenerate bilinear forms attracted considerable attention due to its applications in the areas of mathematics and physics (see, for example, [1]–[6]). In this article, we investigate the properties of the Killing form for a Lie supertriple system and the relationship between a LSTS and its standard imbedding LSA, and we qlso prove two decomposition theorems for LSTS and for quasiclassical LSTS. In this article algebras and triples are assumed to be finite dimensional over a field F of characteristic zero. For basic concepts not specified in this article, the reader is be referred to [1].

1

Lie Supertriple Systems

In this section, we will briefly sketch the notion of LSTSs as well as LSAs, and then introduce the standard imbedding LSA of a LSTS. Let V = V0 ⊕ V1 be a Z2 -graded space over F, where V0 and V1 are called bosonic and fermionic space, respectively, in physics literature. We denote the degree by ⎧ ⎨ 0, if x ∈ V , 0 d(x) = ⎩ 1, if x ∈ V1 . ∗ Received

(A2005000088)

November 30, 2006. Supported by the Natural Science Foundation of Hebei Province of China

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and write (−1)d(x)d(y) = (−1)xy . Any element considered in this article is always assumed to be homogeneous, that is, either x ∈ V0 or x ∈ V1 . Notice that the associative algebra EndV is the superalgebra EndV = End0 V ⊕ End1 V , where Endα V = {a ∈ EndV | aVs ⊆ Vs+α , s = 0, 1}, α = 0, 1. Furthermore, it becomes a LSA under the bracket [a, b] = ab − (−1)ab ba. For later use we recall the following fact on the supertrace of the operators in EndV . Lemma 1.1 ([7], Proposition 1.1.2) str([a, b]) = 0 for any a, b ∈ EndV . A LSA is a Z2 -graded space L = L0 ⊕ L1 over F with a bracket satisfying the following conditions: (1) d([x, y]) = (d(x) + d(y)) (mod 2). (1.1a) (2) [y, x] = −(−1)xy [x, y]. (1.1b) (3) [x, [y, z]] = [[x, y], z] + (−1)xy [y, [x, z]]. (1.1c) Suppose that in addition L possesses a bilinear nondegenerate form B satisfying conditions: (1) B(x, y) = 0, unless d(x) = d(y). (consistence) xy (2) B(y, x) = (−1) B(x, y). (supersymmetry) (3) B([x, y], z) = B(x, [y, z]). (invariance) Then, (L, B) is called quasiclassical (or quadratic, See [8], [9]). A LSTS is a Z2 -graded space T = T0 ⊕ T1 over F with a ternary composition [ , , ], which is trilinear and satisfies the following conditions: (1) d([x, y, z]) = (d(x) + d(y) + d(z)) (mod 2). (1.2a) (2) [y, x, z] = −(−1)xy [x, y, z]. (1.2b) (3) (−1)xz [x, y, z] + (−1)yx [y, z, x] + (−1)zy [z, x, y] = 0. (1.2c) (4) [u, v, [x, y, z]] = [[u, v, x], y, z] + (−1)(u+v)x [x, [u, v, y], z] + (−1)(u+v)(x+y) [x, y, [u, v, z]]. (1.2d) Obviously, the even part T0 itself is a Lie triple system. A subspace I of an LSTS T is called an ideal of T if [I, T, T ] ⊆ I. Obviously, the center C(T ) = {x ∈ T |[x, y, z] = 0, ∀y, z ∈ T } is an ideal of T . Example 1.1 Let L be an LSA. If we introduce a ternary composition [x, y, z] in L by [x, y, z] := [[x, y], z], then L becomes an LSTS. The definitions of solvability, simplicity, and semisimplicity for Lie supertriple systems are the same as those for Lie triple systems. Define the left and right multiplication operators on T by L(x, y)z = [x, y, z], R(x, y)z = (−1)(x+y)z [z, x, y]. Definition 1.1 An automorphism of an LSTS T is an endomorphism ϕ ∈ End0 T satisfying (1) ϕ is an isomorphism of vector space T , and

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(2) ϕ[x, y, z] = [ϕ(x), ϕ(y), ϕ(z)], x, y, z ∈ T . We denote by Aut(T ) the automorphism group of T . Clearly, d(ϕ(x)) = d(x), for x ∈ T , ϕ ∈ Aut(T ). Definition 1.2 A derivation of degree s, s ∈ Z2 , of an LSTS T is an endomorphism D ∈ Ends T satisfying D[x, y, z] = [D(x), y, z] + (−1)xs [x, D(y), z] + (−1)(x+y)s [x, y, D(z)]. Lemma 1.2 Let Der(T ) be the set of all derivations of T . Then, Der(T ) is an LSA under the bracket [D1 , D2 ] = D1 D2 − (−1)D1 D2 D2 D1 , D1 , D2 ∈ Der(T ). Proof A routine verification. Der(T ) is called the derivation LSA of T . Eq.(1.2d) shows that each left multiplication L(x, y) is a derivation of T , and evidently d(L(x, y)) = (d(x) + d(y)) mod 2. Moreover, it follows that from the eqs.(1.2), L(y, x) = −(−1)xy L(x, y),

(1.3a)

[L(u, v), L(x, y)] = L([u, v, x], y) + (−1)(u+v)x L(x, [u, v, y]).

(1.3b)

We refer to an element D ∈ Der(T ) as an inner derivation, if it can be written as a sum of some left multiplications. Eqs.(1.3) implies that the set of all inner derivations of T becomes a subalgebra of Der(T ), which is called inner derivation LSA of T , denoted by H or L(T, T ). Clearly,  H = L(T, T ) =  L(xi , yj )| xi , yj ∈ T . Furthermore, H is an ideal of Der(T ), because [D, L(x, y)] = L(D(x), y) + (−1)xD L(x, D(y)), for D ∈ Der(T ), x, y ∈ T . As an LSA, H has the Z2 -grading expression H = H0 ⊕ H1 , where Hα = {L(x, y)| x, y ∈ T, d(x) + d(y) = α mod 2}, and α = 0, 1. We are now in a position to construct an important LSA larger than H. Consider the direct sum of vector spaces L(T ) = T ⊕ H, and define a bracket in L(T ) by [x, y] = L(x, y), x, y ∈ T, [L(x, y), z] = (−1)(x+y)z [z, L(x, y)] = [x, y, z], x, y, z ∈ T, and in addition to [L(u, v), L(x, y)] given by eq.(1.3b), L(T ) can be readily verified to be an LSA, for grading of d(L(x, y)) = (d(x) + d(y)) mod 2. L(T ) is called the standard imbedding LSA of T. Evidently, if we write L(T ) as a Z2 -graded form: L(T ) = L0 (T ) ⊕ L1 (T ), then L0 (T ) = T0 ⊕ H0

and L1 (T ) = T1 ⊕ H1 .

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Define a map θ : L(T ) → L(T ) by a + h → −a + h, a ∈ T , and h ∈ H. It is easy to verify that θ is an automorphism of L(T ) with θ2 = 1. The above observation can be summarized to be the following. Theorem 1.1 Let T be an LSTS and L(T ) its standard imbedding LSA, then (1) θ is an involution automorphism of L(T ), T and H are (∓)-eigenspaces of θ, respectively; (2) [T, T ] = H; (3) [x, y, z] = [[x, y], z], for x, y, z ∈ T . The following theorem can be seen as the inverse in some sense of Theorem 1.1. Theorem 1.2 Let σ be an automorphism of an LSA L, T and H be the (∓)-eigenspaces respectively, then (1) T is an LSTS together with the ternary composition [x, y, z] = [[x, y], z], x, y, z ∈ T , and H is a subalgebra of L; (2) If L is simple, then T is simple, and H = [T, T ] ∼ = L(T, T ), L ∼ = L(T ), where L(T, T ) and L(T ) are the inner derivation LSA and the standard imbedding LSA of T , respectively. Proof (1) A straightforward verification. (2) Obviously, L has the decomposition L = T ⊕H, and [T, T ] ⊆ H, [T, H] ⊆ T . Moreover, T ⊕ [T, T ] is clearly an ideal of L, so L = T ⊕ [T, T ] by the simplicity of L, and hence [T, T ] = H. Suppose I = 0 is an ideal of T , then I ⊕ [I, T ] is an ideal of L, hence I ⊕ [I, T ] = T ⊕ [T, T ], which forces I = T , that is, T is simple. Now, we have two LSA L = T ⊕ [T, T ] and L(T ) = T ⊕ L(T, T ). Define ϕ : L → L(T ) by x + [y, z] → x + L(y, z), then it is easy to prove that ϕ is an isomorphism of LSA, which completes the proof. We conclude this section with the definition and some examples of the QLSTSs Let T be an LSTS. In addition, suppose that T possesses a bilinear nondegenerate form b satisfying conditions: (1) b(x, y) = 0 unless d(x) = d(y). (consisitence) (1.4a) xy (2) b(x, y) = (−1) b(y, x). (supersymmetry) (1.4b) (3) b([x, y, u], v) = −(−1)(x+y)u b(u, [x, y, v]). (invariance) (1.4c) Then (T, b) is called quasiclassical (or quadratic). Theorem 1.3 ([1]) The following conditions are equivalent: (1) b([x, y, u], v) = −(−1)(x+y)u b(u, [x, y, v]). (2) b([x, y, u], v) = −(−1)(u+v)y b(x, [u, v, y]). (3) b(x, [y, u, v]) = (−1)xy+uv b(y, [x, v, u]). Example 1.2 Let (L, B) be a quasiclassical LSA (QLSA). Define [x, y, z] := [[x, y], z], ∀x, y, z ∈ L, then (L, B) becomes a QLSTS. Notice that we then have B([x, y, u], v) = B([x, y], [u, v]), from which the validity of equation (1.4) can be verified. Example 1.3 Let T = T0 ⊕T1 be a Z2 -graded vector space with a nondegenerate bilinear form b satisfying (1) b(x, y) = 0 unless d(x) = d(y); (2) b(y, x) = (−1)xy b(x, y).

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Then, the ternary composition [x, y, z] := b(y, z)x − (−1)xy b(x, z)y, defines a QLSTS.

2

The Killing Forms Let L be an LSA. The Killing form on L is defined by K(x, y) = str(adxady), x, y ∈ L,

where str stands for the supertrace. Theorem 2.1 The Killing form K on an LSA L = L0 ⊕ L1 has the following properties: (1) K(x, y) = 0, for x ∈ L0 , y ∈ L1 . (consistence) (2) K(x, y) = (−1)xy K(y, x). (supersymmetry) (3) K([x, y], z) = K(x, [y, z]). (invariance) (4) K(Ax, Ay) = K(x, y), for A ∈ AutL. (5) K(Dx, y) + (−1)Dx K(x, Dy) = 0, for D ∈ Der(L), where AutL and Der(L) are the automorphism group and the derivation Lie superalgebra of L, respectively. Proof (1)–(3) are the content of Proposition 2.3.1 in [7]. (4) ∀A ∈ AutL, we have A[x, y] = [Ax, Ay], x, y ∈ L, then adAx = AadxA−1 , so K(Ax, Ay) = str(adAx · adAy) = str(AadxadyA−1 ) = str(adxady) = K(x, y). (5) ∀D ∈ DerL, we have [D, adx] = adDx, x ∈ L. Then, K(Dx, y) + (−1)Dx K(x, Dy) = str([D, adx] · ady) + (−1)Dx str(adx[D, ady]) = str(D · adx · ady − (−1)Dx adx · D · ady) + (−1)Dx str(adx · D · ady − (−1)Dy adx · ady · D) = (−1)Dx str(adx · D · ady − adx · D · ady) + str(D · adx · ady − (−1)D(x+y) adx · ady · D) = str([D, adxady]) = 0, by Lemma 1.1, which complete the proof. Remark 2.1 Clearly, in Theorem 2.1, (5) implies (3). Remark 2.2 If the Killing form K of an LSA L is nondegenerate, then L is quasiclassical. Let T be an LSTS, and L(T ), H as in Section 1. We know that the Killing form of T is defined to be the bilinear form ρ(x, y) = str(R(x, y) + (−1)xy R(y, x)), x, y ∈ T. Let ρ, K, and KH be the Killing form of T , L(T ), and H, respectively, the following theorem describes the relationship among them. Theorem 2.2 ρ, K, and KH are as above. Then, (1) K(T, H) = 0. (2) K(h, h ) = KH (h, h ) + str(h · h ), for h, h ∈ H. (3) K(a, a ) = ρ(a, a ), for a, a ∈ T , that is, ρ = K|T ×T . Proof (1) We know that the map θ : L(T ) → L(T ) by a + h → −a + h, a ∈ T , and h ∈ H, is an automorphism of LSA L(T ), then −K(h, a) = K(h, −a) = K(h, θa) = K(θh, a) = K(h, a).

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So, we have K(h, a) = 0, which proves (1). (2) Since H and T are invariant under the operator adh · adh , we obtain strL(T ) (adh · adh ) = strH (adh · adh ) + strT (adh · adh ). But adh · adh (a) = [h, [h , a]] = (h · h )(a), for a ∈ T, therefore, strT (adh · adh ) = str(h · h ). Hence, (2) holds. (3) Write T = T0 ⊕ T1 , H = H0 ⊕ H1 , and suppose {ai }, {bi }, {ui }, and {vi } are the bases of T0 , T1 , H0 , and H1 , respectively. Since R(ai , bj )T0 ⊆ T1 , and R(ai , bj )T1 ⊆ T0 , we have strR(ai , bj ) = 0. Hence, to see (3), it suffice to verify K(ai , aj ) = ρ(ai , aj ), and K(bi , bj ) = ρ(bi , bj ). Assume that the operators adai are expressed relative to these bases as follows:  p  p [ai , aj ] ∈ H0 , [ai , aj ] = Sij up , [ai , bj ] ∈ H1 , [ai , bj ] = Kij vp , [ai , uj ] ∈ T0 , [ai , uj ] =

p

p





p

p Xij ap ,

Then,

[ai , vj ] ∈ T1 , [ai , vj ] =

p



[ai , [aj , ak ]] = [ai ,

p Sjk up ] =

p



[ai , [aj , bk ]] = [ai ,

p Kjk vp ] =

p

[ai , [aj , uk ]] = [ai ,

 p

[ai , [aj , vk ]] = [ai ,

p Xjk ap ] =

 p

Therefore, strT (adai · adaj ) =

p Yjk bp ] =

 k,p

strH (adai · adaj ) =

 k,p

 l,p

 l,p

 l,p

 l,p

p k Sjk Xip − p k Xjk Sip −

Yijp bp .

p l Sjk Xip al , p Kjk Yipl bl , p l Xjk Sip ul , p l Yjk Kip vl .

 k,p

 k,p

p Kjk Yipk , p k Yjk Kip .

Clearly, strH (adai · adaj ) = strT (adaj · adai ).

(2.1)

strH (adbi · adbj ) = −strT (adbj · adbi ).

(2.2)

R(ai , aj ) = (adaj · adai )|T ,

(2.3)

Similarly, we can prove

It is easy to see that

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R(bi , bj ) = −(adbj · adbi )|T .

(2.4)

For example, we check (2.4) as follows: R(bi , bj )ak = [ak , bi , bj ] = [[ak , bi ], bj ] = [bj , [ak , bi ]] = −[bj , [bi , ak ]] = −adbj adbi (ak ), R(bi , bj )bk = [bk , bi , bj ] = [[bk , bi ], bj ] = −[bj , [bk , bi ]] = −[bj , [bi , bk ]] = −adbj adbi (bk ). Hence, by (2.1)–(2.4), we have K(ai , aj ) = strH (adai · adaj ) + strT (adai · adaj ) = strT (adaj · adai ) + strT (adai · adaj ) = strR(ai , aj ) + strR(aj , ai ) = ρ(ai , aj ), K(bi , bj ) = strH (adbi · adbj ) + strT (adbi · adbj ) = −strT (adbj · adbi ) + strT (adbi · adbj ) = strR(bi , bj ) − strR(bj , bi ) = ρ(bi , bj ). Theorem 2.3 Let T = T0 ⊕ T1 be an LSTS, and ρ(x, y) its Killing form. Then (1) ρ(T0 , T1 ) = 0, (2) ρ(x, y) = (−1)xy ρ(y, x), (3) ρ(Ax, Ay) = ρ(x, y), A ∈ Aut(T ), (4) ρ is invariant, that is, ρ(L(u, v)x, y) = −(−1)(u+v)x ρ(x, L(u, v)y), ρ(R(u, v)x, y) = (−1)(u+v)x+uv ρ(x, R(v, u)y), (5) If ρ is nondegenerate, then ρ(x, y) = 2strR(x, y). Proof (1) follows from the two inclusion relations R(T0 , T1 )T1 ⊆ T0 and R(T0 , T1 )T0 ⊆ T1 , while (2) follows from the definition of ρ(x, y). (3) ∀A ∈ Aut(T ), A[x, y, z] = [Ax, Ay, Az]. Since d(x) = d(Ax), we have AR(y, z)x = R(Ay, Az)Ax, AR(y, z) = R(Ay, Az)A, R(Ay, Az) = AR(y, z)A−1 , then, ρ(Ax, Ay) = str(R(Ax, Ay) + (−1)xy R(Ay, Ax)) = str(A(R(x, y) + R(y, x))A−1 ) = str(R(x, y) + R(y, x)) = ρ(x, y) (4) Using Theorem 2.1(2), (3), and Theorem 2.2(3), we obtain that, for u, v, x, y ∈ T , ρ(L(u, v)x, y) = K(L(u, v)x, y) = K([u, v], [x, y]) = −(−1)xy K([u, v], [y, x]) = −(−1)xy K([u, v, y], x) = −(−1)xy+(u+v+y)x K(x, [u, v, y]) = −(−1)(u+v)x ρ(x, L(u, v)y).

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Using the same means, we have, for u, v, x, y ∈ T , ρ(R(u, v)x, y) = (−1)(u+v)x ρ([x, u, v], y) = (−1)(u+v)x K([x, u], [v, y]) = (−1)(u+v)x K(x, [u, [v, y]]) = −(−1)(u+v)x+vy K(x, [u, [y, v]]) = (−1)(u+v)x+vy+(y+v)u K(x, [[y, v], u]) = (−1)(u+v)x+uv K(x, R(v, u)y). Notice that the above identity can be easily rewritten in the following symmetric form: ρ(x, [y, u, v]) = (−1)xy+uv ρ(y, [x, v, u]). (5) If ρ is nondegenerate on T , then the restrictions of ρ on T0 and on T1 are also nondegenerate, because ρ(T0 , T1 ) = 0. Let {ai } (resp. {bj }) be a basis of T0 (resp. T1 ): T0 = a1 , a2 , · · ·,

T1 = b1 , b2 , · · ·.

If u, v ∈ T0 , then we have, by (4), ρ(R(u, v)ai , aj ) = ρ(ai , R(v, u)aj ),

ρ(R(u, v)bi , bj ) = ρ(bi , R(v, u)bj ).

If u, v ∈ T1 , then we have ρ(R(u, v)ai , aj ) = −ρ(ai , R(v, u)aj ),

ρ(R(u, v)bi , bj ) = −ρ(bi , R(v, u)bj ).

Let R∗ (u, v) denote the adjoint endomorphism in T of R(u, v) relating to ρ, then R∗ (u, v) = R(v, u), u, v ∈ T0 ,

R∗ (u, v) = −R(v, u), u, v ∈ T1 .

Since strR(u, v) = strR∗ (u, v), we get ρ(u, v) = str(R(u, v) + R(v, u)) = 2strR(u, v), u, v ∈ T0 , ρ(u, v) = str(R(u, v) − R(v, u)) = 2strR(u, v), u, v ∈ T1 . Thus, for all a, a ∈ T , we have ρ(a, a ) = 2strR(a, a ). Remark 2.3 Theorem 2.3 shows that, if the Killing form ρ of an LSTS T is nondegenerate, then (T, ρ) is quasiclassical. Moreover, in this case, the standard imbedding LSA (L(T ), K) is also quasiclassical by Theorem 2.1 and Theorem 2.2. We can also prove a more general result as follows. Theorem 2.4 [1] Let (T, b) be a QLSTS. Then, (L(T ), B) is a QLSA, where B is the unique extension of b to L(T ) such that B(T, H) = 0. Proof Define the bilinear form B on L(T ) by B(L(x, y), L(u, v)) = b([x, y, u], v), B(L(x, y), z) = B(z, L(x, y)) = 0, for x, y, u, v ∈ T .

B(x, y) = b(x, y),

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The Decomposition Theorems

In this section, we shall investigate the decomposability properties of a finite-dimensional LSTS, and present two decomposition theorems. Theorem 3.1 Let T be an LSTS with trivial center. Then, (1) T can be decomposed into a direct sum of its indecomposable ideals. (2) The decomposition of T is unique, up to permutations, that is, if T = T1 ⊕ · · · ⊕ Tm ,

T = T1 ⊕ · · · ⊕ Ts

are two decompositions of T , where Ti , Tj , 1 ≤ i ≤ m, and 1 ≤ j ≤ s, are indecomposable ideals of T , then m = s and by changing the subscripts if necessary, Ti = Ti , i = 1, · · · , m. Proof This is a standard result on an algebraic structure with an n-ary (n ≥ 2) composition. The proof of Theorem 3.1 is similar to the case of n-Lie algebra (n ≥ 2) (see [10] or [11]). Definition 3.1 Suppose that (T, b) is quasiclassical. A graded ideal I of T is called nondegenerate, if the restriction of b to I is nondegenerate. A QLSTS (T, b) is called b-irreducible if T contains no nontrivial nondegenerate graded ideals. The above definitions are perfactly suitable for QLSA. Before working with the decomposition theorem of a QLSTS, we will recall the following decomposition theorem of QLSA, which can be found in [9]. r  Theorem 3.2 Let (L, B) be a QLSA with trivial center C(L). Then, L = Li such i=1

that for all 1 ≤ i ≤ r, (1) Li is a nondegenerate irreducible graded ideal of L. (2) For i = j, Li and Lj are orthogonal relative to B. (3) The above decomposition is unique, up to permutations. Our aim in this section is to prove the following analog of Theorem 3.2 for the case of QLSTS using Theorem 2.4. n  Theorem 3.3 Let (T, b) be a QLSTS with trivial center C(T ). Then, T = Ti , where i=1

for all 1 ≤ i ≤ n, (1) Ti is a nondegenerate irreducible graded ideal of T . (2) For i = j, Ti and Tj are orthogonal relative to b. (3) The above decomposition is unique, up to permutations. Let us first establish two Lemmas before the proof of Theorem 3.3. Lemma 3.1 Let T be an LSTS, and L(T ) the standard imbedding LSA of T . Then, C(T ) = C(L(T )). Proof C(L(T )) is invariant under θ, so C(L(T )) = C(L(T )) ∩ T ⊕ C(L(T )) ∩ H. Now, [T, C(L(T )) ∩ H] = 0, so C(L(T )) ∩ H consists of endomorphisms of T which act trivially on T , hence C(L(T )) ∩ H = 0 and C(L(T )) ⊆ T . Thus, if x ∈ C(L(T )) and y, z ∈ T , then by (1.1c), L(x, y) = [x, y] ∈ C(L(T )) ∩ H, so L(x, y) = 0, and [x, y, z] = L(x, y)z = 0. Hence x ∈ C(T ), so C(L(T )) ⊆ C(T ). Since [C(T ), H] = [C(T ), T, T ] = 0, C(T ) ⊆ C(L(T )), which completes the proof. Lemma 3.2 Let (T, b) be a QLSTS and (L(T ), B) its standard imbedding QLSA as in Theorem 2.4. Then,

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(1) [T, T, T ]⊥ = C(T ). (2) If I is a graded ideal of T , then I ⊥ is a graded ideal of T . Furthermore, if I is nondegenerate, then I ⊥ is also nondegenerate and T = I ⊕ I ⊥ . Proof (1) Clearly, x ∈ [T, T, T ]⊥ ⇔ b(x, [y, u, v]) = 0, ∀y, u, v ∈ T, ⇔ (−1)xy+uv b(y, [x, v, u]) = 0, (by Theorem 1.3) ⇔ [x, v, u] = 0, ∀v, u ∈ T, ⇔ x ∈ C(T ). (2) Let x ∈ I ⊥ , y, u ∈ T , and v ∈ I, then by Theorem 1.3, b([x, y, u], v) = −(−1)(u+v)y b(x, [u, v, y]) = 0. Hence [x, y, u] ∈ I ⊥ , and I ⊥ is an ideal of T . If I is nondegenerate and x ∈ I ∩ I ⊥ , then b(x, I) = 0 which forces x = 0 and I ∩ I ⊥ = 0. Note that dim T = dim I ⊕ dim I ⊥ . So we have T = I ⊕ I ⊥ , which complete the proof. Proof of Theorem 3.3 Let (L, B) be the standard imbedding QLSA of (T, b) as stated in Theorem 2.4, where L := L(T ) = T ⊕ H. Note that T is the (−1)-eigenspace and H is the (+1)-eigenspace of θ ∈ Aut(L(T )), defined by θ(t + h) = −t + h, t ∈ T , h ∈ H. By Lemma r  3.1, C(L) = 0, then (L, B) has a unique decomposition L = Li as described in Theorem 3.2. i=1

Since θ2 = 1, by renumbering the components of L, we obtain

L = L1 ⊕ θL1 ⊕ · · · ⊕ Lp ⊕ θLp ⊕ Lp+1 ⊕ · · · ⊕ Ln , (n + p = r), where Li = θLi , for i > p. Let Mi = Li ⊕ θLi for i ≤ p; Mj = Lj for i > p. Then, each Mi n  is an invariant ideal of L under θ. If x ∈ T , x = xi , and xi ∈ Mi , then θxi = −xi , hence i=1

T = T1 ⊕ · · · ⊕ Tn , where Ti = T ∩ Mi . Similarly, H = H1 ⊕ · · · ⊕ Hn , where Hi = H ∩ Mi , and so Mi = Ti ⊕ Hi . Since [Ti , Ti ] ⊆ Hi and H = [T, T ], we have Hi = [Ti , Ti ]. Since [Ti , Tj ] = 0, for i = j, it follows that [Ti , T, T ] = [Ti , Ti , Ti ] ⊆ Ti , and Ti is an ideal of T . By Theorem 3.2, each Li is nondegenerate and B(Li , Lj ) = 0, for i = j. Then, each Mi is nondegenerate. Let Bi = B|Mi , bi = b|Ti . Since Mi = Ti ⊕ Hi , it follows that (Mi , Bi ) is the standard imbedding QLSA of the QLSTS (Ti , bi ), and therefore each Ti is nondegenerate because Bi (Ti , Hi ) = 0 and Bi |Ti = bi . Now, we verify that each Ti is irreducible. Suppose that I = 0 is a nondegenerate ideal of Ti . Clearly, I ⊕ [I, Ti ] is an ideal in Mi . Moreover, I ⊕ [I, Ti ] is nondegenerate, because I and Ti are nondegenerate and both bi and Bi are invariant. Note that I ⊕ [I, Ti ] is invariant under θ and Mi has no invariant nondegenerate ideals except itself and 0, so I ⊕ [I, Ti ] = Mi , and hence I = Ti , because Mi = Ti ⊕ [Ti , Ti ]. This completes the proof of (1). The assertion (2) is obvious. n  It remains only to prove the uniqueness of the decomposition. Let T = Ti and T = m  j=1

i=1

Tj

⊥

be two such decompositions. By Lemma 3.2, [T, T, T ] = 0 and so T = [T, T, T ], which

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forces Tj = [Tj , Tj , Tj ], j = 1, 2, · · · , m. Clearly, n 

(Ti ∩

i=1

hence Tj =

n 

Tj )

⊆

Tj

=

[T, T, Tj]

=

n 

[Ti , Ti , Tj ]

i=1

(Ti ∩ Tj ).

i=1

Suppose that Tk ∩Tj = 0 for some k, then b(Tk ∩Tj ,

n  ⊆ (Ti ∩ Tj ), i=1



(Ti ∩Tj )) = 0 because b(Tk , Ti ) = 0,

i=k Tj is

nondegenerate. The irreducibility k = i. Then, Tk ∩Tj is a nondegenerate ideal of Tj , since      of Tj implies Tk ∩ Tj = Tj , so Tj ⊆ Tk , which means Tj = Tk . Similarly, for each Ti , 1 ≤ i ≤ n, we have Ti = Tl , for some l, 1 ≤ l ≤ m. Thus, m = n and assertion (3) holds. The proof of Theorem 3.3 is completed. References [1] Okubo S, Kamiya N. Quasiclassical Lie superalgebras and Lie supertriple systems. Comm Alg, 2002, 30(8): 3825–3850 [2] Okubo S, Kamiya N. Jordan-Lie superalgebra and Jordan-Lie triple system. J of Alg, 1997, 198: 388–411 [3] Okubo S. Triple products and Yang-Baxter equation(II):orthogonal and sysplectic ternary systems. J Math Phys, 1993, 34: 3293–3315 [4] Okubo S. Parastatistic as Lie supertriple systems. J Math Phys, 1994, 35(6): 2785–2803 [5] Zhang Zhixue, Shi Yiqian, Zhao Lina. Invariant symmetric bilinear forms on Lie triple systems. Comm Alg, 2002, 30(11): 5563–5573 [6] Zhang Zhixue, Li Huajun, Dong Lei. Invariant bilinear forms on anti-Lie triple systems. Chinese Journal of Contemporary Mathematics, 2004, 25(3): 237–244 [7] Kac V G. Lie superalgebras. Adv Math, 1977, 26: 8–96 [8] Benamor H, Benayadi S. Double extension of quadratic Lie superalgebras. Comm Alg, 1999, 27(1): 67–88 [9] Zhu Linsheng, Meng Daoji. The unique decomposition theorems of Lie superalgebras and quadratic Lie superalgebras. Science in China (Ser A), 2002, 32(3): 226–231 (in Chinese) [10] Bai Ruipu, Meng Daoji. The decomposition of n-Lie algebras and uniqueness. Chinese Journal of contemporary mathematics, 2004, 25(2): 117–124 [11] Meng Daoji. Some results on complete Lie algebras. Comm Alg, 1994, 22(13): 5457–5507 [12] Bai Ruipu, Jia Peipei. The real compact n-Lie algebras and invariant bilinear forms. Acta Math Sci, 2007, 27A(6): 1074–1081 [13] Chen Liangyun, Meng Daoji. Some result of module of Lie superalgebras. Acta Math Sci, 2006, 26B(3): 401–409 [14] Chen Liangyun, Zhang Yongzheng, Meng Daoji. On the Decomposition of Restricted Lie Superalgebras. Acta Math Sci, 2007, 27A(4): 577–583