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- Corresponding Author:
- Noriaki KAMIYA

University of Aizu, Aizu-Wakamatsu

965-8580 Fukushima, Japan

**E-mail:**[email protected]

**Received Date:** October 17, 2007; **Revised Date:** March 02, 2008

**Visit for more related articles at** Journal of Generalized Lie Theory and Applications

In this paper, we study a Peirce decomposition for (-1,-1)-Freudenthal-Kantor triple sys- tems and give several examples.

Our aim is to give a characterization of many mathematical and physical fields by means of concept of triple systems (here, triple systems mean a vector space equipped with a triple product <xyz >). It seems that such concept is useful to an application of nonassociative algebras as well as the characterization of Yang-Baxter equations, and the construction of Lie (super)algebras and Jordan (super)algebras ([3]-[10], [12], [13], [15]).

In this article, first we will consider a Peirce decomposition of (-1,-1)-Freudenthal-Kantor triple systems, and give an example of a Peirce decomposition of simple (-1,-1)- Freudenthal- Kantor triple systems. Second, we will study the decomposition of simple Lie superalgebras associated with such triple systems.

Throughout this paper, we shall be concerned with algebras and triple systems over a field Φ that is characteristic not 2 and do not assume that our algebras and triple systems are finite dimensional, unless otherwise specified.

Summarizing briely this article we will mainly discuss the following result.

**Main Theorem.** Let U be a (-1,-1)-Freudenthal-Kantor triple system with a tripotent element
e, i.e (eee)=e. Then we have the Peirce decomposition

where In particular, for balanced cases, we have

which implies

where R(x) = (xee).

In the rest of this section, I shall give the definition and some results for a certain triple system in order to make this paper as self-contained as possible.

For and, a vector space over Φ with the triple product <-;-;-> is called a (ε, δ)-Freudenthal-Kantor triple system if

(1.1)

(1.2)

where

A triple system is said to be a generalized Jordan triple system, if ε = -1 and only the identity (1.1) is satisfied.

The triple products are generally denoted by

Example 1.1. Let V be a vector space equipped with a bilinear form Then V is a (ε, ε)-Freudenthal-Kantor triple system with respect to the product

Example 1.2. Let V be a Jordan triple system. Then this triple system is a special case of the (-1,1)-Freudenthal-Kantor triple system, because the identity K(a; b)c≡0 (identically zero) implies that <acb>=<bca>, and the identity (1.1) implies that

If its product satisfies <abc>= - <cba> and

then this triple system is called an anti-Jordan triple system.

An -Freudenthal-Kantor triple system over Φ is said to be balanced if there exists a bilinear form <·;·> such that K(x;y) =<x/y> Id, where <x/y>ε Φ*.

Remark 1.1. For a balanced -Freudenthal-Kantor triple system, we have the following relation:

For convenience (in the section 3 of this paper), the notation <x/y> will be used by means of 2 <x/y>. That is, the notation <aab>=<aba>=<a/a> b imply the balanced property.

For the δ-Lie triple systems associated with -Freudenthal-Kantor triple systems, we have the following.

**Proposition 1.1** ([7, 12]). Let U be a -Freudenthal-Kantor triple system. If P is a
linear transformation of U such that P < xyz >=< PxPyPz > and P2 = -εδ Id, then
(U; [-;-;-]) is a Lie triple system for the case of δ = 1 and an anti-Lie triple system for
the case of δ = -1 with respect to the product

Corollary 1.1. Let U be a -Freudenthal-Kantor triple system. Then the vector space T := U becomes a Lie triple system for the case of δ = 1 and an anti-Lie triple system for the case of δ = -1 with respect to the triple product defined by

**Proposition 1.2.** Let V be an anti-Jordan triple system (that is, it satisfies the condition (L1) becomes an anti-Lie triple system with
respect to the product defined by

From these results, it follows that the vector space

where T is a δ Lie triple system and Inn Der makes a Lie algebra (δ = 1) or Lie superalgebra (δ = -1) by

We denote by L the Lie algebras or Lie superalgebras obtained from these constructions associated with U) and call these algebras a standard embedding.

A -Freudenthal-Kantor triple system U) is said to be unitary if the linear span k of the set contains the identity endomorphism Id.

**Proposition 1.3** ([6, 7]). For a unitary -Freudenthal-Kantor triple system U over Φ,
let T be the Lie or anti-Lie triple system and L be the standard embedding Lie algebra
or superalgebra associated with U. The following are equivalent:

a) U is simple,

b) T is simple,

c) L is simple.

For these standard embedding Lie algebras or superalgebras L, we have the following 5 grading subspaces:

This is one reason why we study about a characterization of triple systems, as properties of Lie superalgebras are closely related to those of the triple systems as well as Lie algebras.

From now, we will only consider a (-1;-1){Freudenthal-Kantor triple system, unless otherwise specified, because the case of δ = 1 had considered in other papers ([3]{[7], [11]), that is, one deal with the relations

(2.1)

(2.2)

where K(a,b)x = (axb) + (bxa) and L(a,b)x = (abx).

Remark 2.1. We note that the relations (2.1) and (2.2) coincide with (1.1) and (1.2) with the case of ε = -1, and δ = -1.

Let e be a tripotent, i.e (eee) = e and denote.

L(x) = (eex); R(x) = (xee); Q(x) = (exe)

**Remark 2.2. **We note for the notation that

K(e,e)e = (eee) + (eee) = 2e

K(x,e)e = K(e,x)e = (xee) + (eex)

K(e,e)x = 2Q(x) = 2(exe)

From (2.1), we have

(xe(eee)) = ((xee)ee) - (e(exe)e) + (ee(xee))

and so

(xee) = ((xee)ee) - (e(exe)e) + (ee(xee))

Hence we get

R(x) = R^{2}(x) - Q^{2}(x) + LR(x) (2.3)

On the other hand,

(2.4)

Also,

(ee(xee)) = ((eex)ee) - (x(eee)e) + (xe(eee))

Therefore, we obtain

LR(x) = RL(x) (2.5)

and moreover

hence,

(2.6)

On the other hand, from (2.2) we have

and so

Therefore we have

(2.7)

Similarly, from (2.5) we have

(2.8)

**Lemma 2.1. **There is no vector a ≠ 0 satisfying

(2.9)

**Proof.** From (2.3), it follows that

and if there is an element a satisfying (2.9), then we have

i.e

(2.10)

From (2.4), it follows that

(2.11)

From (2.6), it follows that

(2.12)

We set Q(a) = b, from (2.10) we have

(2.13)

By (2.12), L(b) = 3b and by (2.11),

Hence we have R(b) = -b. From (2.7),

Therefore we have

(2.14)

This completes the proof.

We use the following notations.

Lemma 2.2. The space U is a direct sum of subspaces U_{R=-L} and U_{R=Id}:

(2.15)

**Proof.** It follows from (2:7) and (2:8) that

(2.16)

Either of (2.16) or (2.17) implies

(2.18)

We set To prove that P = 0, assume that P ≠ 0. Then we get R = -L = Id on P. Hence we come to a contradiction with Lemma 2.1.

**Remark 2.3.** We note that

(2.19)

(2.20)

**Lemma 2.3. **The subspaces UR=-L and UR=Id are invariant with respect to operators L and R.

**Proof.** Let us prove, for example,

Using (2.5) and (2.20), we get

The other proofs are obtained in the same way.

**Corollary 2.1.** There is no vector a ≠ 0 such that

(2.21)

**Proof.** If there is an element Then it follows from Lemma 2.3 that

Considering the operator

Now, as one comes to a contradiction of Lemma 2.1, hence the Corollary holds.

An eigensubspace of the tripotent e with respect to the eigenvalues λ;μ is called a subspace
U_{λ,μ} of all vectors a satisfying

(2.22)

Now we prove that there are only two possibilities, i.e 1) μ = -λ, 2) μ = 1. To show this, we
will denote by U_{μ=-λ} the sum of all subspaces with the first possibility and by U_{μ=1} with the
second one.

**Lemma 2.4.** Let Then either of the following holds:

**Proof.** Suppose By Lemma 2.2, we can see that

By the definition of U_{λ,μ} we get

This implies, according to Lemma 2.3,

Therefore, both a1 and a2 are common eigenvectors of L and R. This implies the fact that μ = -λ and μ = 1, and hence we get λ = -1. This contradicts the Corollary of Lemma 2.1.

**Lemma 2.5.** Let Then one of the following two possibilities occurs:

**Proof.** First, suppose that Q(a) = 0: Then (2.3) is equivalent to

that is

(2.23)

Thus λ = 0 and μ = 0; and one come to the case (a).

Let Q(a) ≠ 0: From (2.6) it follows that

(2.24)

From (2.4) it follows that

(2.25)

i.e

Subtracting (2:25) from (2:24), we get

RQ(a) = -Q(a) (2.26)

Thus this implies and from Lemma 2.4 follows In the first case, we have which implies that both and Thus in this case, we come to the case b).

In the second case, Then it follows from (2.26) that This case does not appear. This completes the proof.

**Lemma 2.6.** Let Then one of the following two possibilities occurs:

**Proof.** First suppose that Q(a) = 0. Then (2.3) is equivalent to L(a) = 0. This means that
λ = 0. Thus one come to the case a’).

Suppose that Q(a) ≠ 0. Then by (2.6),

(2.27)

and by (2.4)

(2.28)

Subtracting (2:28) from (2:27), we get

(2.29)

Thus and from Lemma 2.4 follows In the first case, we have

(2.30)

Thus from (2.29) and (2.30) we get λ = 0 which implies because Thus this case does not appear.

In the second case, we have

(2.31)

From (2.29) and (2.31) we have λ = 1. The case of λ = 1 means that both and This comes to the case b’). This completes the proof.

Combining Lemma 2.5 and Lemma 2.6, we have the following.

**Theorem 2.1.** Let the three linear operators L;R;Q be defined on a (-1;-1)-Freudenthal-
Kantor triple system. Then the space U is a direct sum of four subspaces;

(2.32)

where

Proof. To prove (2.32), it is enough to prove

(2.33)

That is, the direct sum

follows from Lemma 2.5 and Lemma 2.6. The equalities (2.33) mean that the operator L has
no Jordan blocks of second degree, i.e there are no vectors a_{1}; a_{2} ≠ 0 such that

(2.34)

Indeed, according to Lemma 2.2 and Lemma 2.3, we can consider the two cases: and separately.

In the case of we have

(2.35)

It follows from (2.6) that

(2.36)

(2.37)

Using (2:36) and (2:37), we obtain from (2.4)

and so

Furthermore, using (2.35),

(2.38)

Now, applying (2.7) to Q(a2) and using the above relations, we obtain

From Lemma 2.5 and Lemma 2.6, the cases of λ = 1 and Q(a1) = 0 are impossible. Hence one comes to Q(a1) = 0 and Q(a2) = 0. Therefore, by Lemma 2.5, we have and by (2.35), R(a2) = -a1. We come to a contradiction with (2.34).

In the second case, and we have

(2.39)

Thus it follows that

(2.40)

(2.41)

From (2.4), it follows that

and so

Then using (2.39), (2.40) and (2:41) we get

Similarly to the first case, by applying (2.7) to Q(a2) and using the above relations, we obtain

Thus we get

(2.42)

Hence, from the same method as for the first case, we have Q(a1) = 0 and Q(a2) = 0. Therefore by Lemma 2.6, it follows that This implies a contradiction of the assumption on (2.34). This completes the proof.

Remark 2.4. We note that the case of more than one tripotent elements will be discussed elsewhere. In particular, if there exists a tripotent element e such that and <e/e>= 1 then we have the balanced case

In this section, we will consider the standard embedding Lie superalgebras of the B(m; n) and D(m; n) types associated with the anti-Lie triple system and the (-1,-1)-Freudenthal-Kantor triple system. Furthermore, we will study a Peirce decomposition of such types.

**Theorem 3.1** ([10]). Let U be a vector space of Mat(k; n; Φ). Then the space U is a unitary
(-1,-1)-Freudenthal-Kantor triple system with respect to the triple product

where t_{x} denotes the transpose matrix of x.

**Remark 3.1.** For this triple system, by straightforward calculations, using the results in Sec.
1, we have the following:

1) type's Lie superalgebra and

2) type's Lie superalgebra and

That is, summarizing these, we have the following.

**Theorem 3.2.** Let U be the same triple system as that described in Theorem 3.1 and L(U) be
the standard embedding Lie superalgebras associated with are Lie
superalgebras of type D(m,n) or B(m,n) if k=2m or k= 2m+1, respectively.

For a bilinear trace form of a (-1,-1)-Freudenthal-Kantor triple system, we have the formula as follows [7]:

Thus for U = Mat(k; n; Φ) of the above Theorem 3.1, by straightforward calculations, we obtain the identity

where c_{x,y} is a constant element in Φ with dependent x,y ε U.

This implies that the trace form γ (x; y) is degenerate if m = n + 1 (the case of k = 2m). Thus this fact is related to the degenerate property of the Killing form of the Lie superalgebra D(n+1; n): For the correspondence between the trace form of the anti-Lie triple system and the trace form (Killing form) of the standard embedding Lie superalgebra L(U) associated with (-1,-1)-FKTS U we refer to the author's previous paper [7] and do not go into detail here.

**Remark 3.2.** For the construction of balanced types of Lie algebras and superalgebras, that
is, in the case of dim k = dimL-2 = dimL2 = 1, we refer to [1, 4, 13].

**Example 3.1.** For the triple system given in Theorem 3.1, we have the following decomposition.
Let U = Mat(k; n; Φ) be the triple system defined by

Here, let k ≥ n> l and we set

is (k; n) matrix; El is (l; l) identity matrix

A : (l; l); B : (l; n - l); C : (k - l; l); D : (k - l; n - l) matrix

Then we have

In particular, for this triple system, let k ≥ n = l, and we set

Then we have eex = x, for all x ε U and Thus by straightforward calculations, we can obtain the decomposition

As the final topic in this section, we shall give several simple examples of Peirce decomposi- tions of triple systems.

In a generalized Jordan triple system equipped with xxy = xyx =<x/x> y and <x/y>=< y/x >, that is, this means the balanced property defined in section one, by straightforward calculations, we have the following.

**Proposition 3.1. **Let U be a balanced generalized Jordan triple system. Then the decomposition
is given by

Proof. Indeed, from R^{2}(x) = x, we have

and the proof is verified.

**Remark 3.3** ([13]). For the standard embedding Lie superalgebra L(U) associated with a simple
balanced (-1,-1)-FKTS U, we have the following decomposition. For convenience, we set

where eee = e,<eje>= 1. In this Lie superalgebra, and we have

and

For the examples of balanced (-1,-1)-Freudenthal-Kantor triple systems U and the standard embedding Lie superalgebras L(U), we refer to [1] and [13].

On the other hand, for a quadratic triple system [14], that is, in a triple system equipped with we have the following.

**Proposition 3.2. **Let U be a quadratic triple system. Then the decomposition is given by

**Example 3.2.** Let U be a triple system satisfying Then this triple system U is a (-1;-1)-Freudenthal-Kantor triple system, but it is not balanced.
Furthermore, we have

**Example 3.3** ([8], anti-Jordan triple system). Let U be a vector space with an anti-symmetric
bilinear form such that Jordan triple system with
respect to the triple product

This product means an anti-J.T.S.

Furthermore, by means of an element e such that eee = 0, we have the decomposition

This Lie superalgebra decomposition is is the copy of

Also, it holds that

The author would like to thank Dr D. Mondoc for several suggestions in preparation of this paper.

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