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A Peirce decomposition for (-1,-1)-Freudenthal-Kantor triple systems | OMICS International
ISSN: 1736-4337
Journal of Generalized Lie Theory and Applications
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A Peirce decomposition for (-1,-1)-Freudenthal-Kantor triple systems

Noriaki KAMIYA*

University of Aizu, Aizu-Wakamatsu, 965-8580 Fukushima, Japan

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

Abstract

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

Introduction

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

image

where image In particular, for balanced cases, we have

image

which implies

image

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

image (1.1)

image (1.2)

where

image

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

image

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

image

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

image

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

image

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

An image -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 image -Freudenthal-Kantor triple system, we have the following relation:

image

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 image-Freudenthal-Kantor triple systems, we have the following.

Proposition 1.1 ([7, 12]). Let Uimage be a image-Freudenthal-Kantor triple system. If P is a linear transformation of Uimage such that P < xyz >=< PxPyPz > and P2 = -εδ Id, then (Uimage; [-;-;-]) 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

image

Corollary 1.1. Let Uimage be a image-Freudenthal-Kantor triple system. Then the vector space Timage := Uimage image 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

image

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

image

From these results, it follows that the vector space

image

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

image

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

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

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

a) Uimage is simple,

b) Timage is simple,

c) Limage is simple.

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

image

image

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.

Main results (proof of the Main Theorem)

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

image (2.1)

image (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) = R2(x) - Q2(x) + LR(x) (2.3)

On the other hand,

image (2.4)

Also,

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

Therefore, we obtain

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

and moreover

image

hence,

image (2.6)

On the other hand, from (2.2) we have

image

and so

image

Therefore we have

image (2.7)

Similarly, from (2.5) we have

image (2.8)

Lemma 2.1. There is no vector a ≠ 0 satisfying

image (2.9)

Proof. From (2.3), it follows that

image

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

image

i.e

image (2.10)

From (2.4), it follows that

image (2.11)

From (2.6), it follows that

image (2.12)

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

image (2.13)

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

image

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

image

image

Therefore we have

image (2.14)

This completes the proof.

We use the following notations.

image

Lemma 2.2. The space U is a direct sum of subspaces UR=-L and UR=Id:

image (2.15)

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

image (2.16)

image

Either of (2.16) or (2.17) implies

image (2.18)

We set image 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

image (2.19)

image (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,

image

Using (2.5) and (2.20), we get

image

The other proofs are obtained in the same way.

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

image (2.21)

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

image

Considering the operator image

image

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 Uλ,μ of all vectors a satisfying

image (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 Uμ=-λ the sum of all subspaces with the first possibility and by Uμ=1 with the second one.

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

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

image

By the definition of Uλ,μ we get

image

This implies, according to Lemma 2.3,

image

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 image Then one of the following two possibilities occurs:

image

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

image

that is

image (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

image (2.24)

From (2.4) it follows that

image (2.25)

i.e

image

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

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

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

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

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

image

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),

image (2.27)

and by (2.4)

image (2.28)

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

image (2.29)

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

image (2.30)

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

In the second case, we have

image (2.31)

From (2.29) and (2.31) we have λ = 1. The case of λ = 1 means that both image and image 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;

image(2.32)

where

image

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

image (2.33)

That is, the direct sum

image

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 a1; a2 ≠ 0 such that

image (2.34)

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

In the case of image we have

image (2.35)

It follows from (2.6) that

image (2.36)

image (2.37)

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

image

and so

image

Furthermore, using (2.35),

image

image (2.38)

image

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

image

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 image and by (2.35), R(a2) = -a1. We come to a contradiction with (2.34).

In the second case, image and we have

image (2.39)

Thus it follows that

image (2.40)

image (2.41)

From (2.4), it follows that

image

and so

image

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

image

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

image

Thus we get

image (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 image 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 image and <e/e>= 1 then we have the balanced case image

Examples of (-1,-1)-Freudenthal-Kantor triple systems

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

image

where tx 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) image type's Lie superalgebra and

image

2) image type's Lie superalgebra and

image

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 image 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]:

image

image

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

image

where cx,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 image 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

image

Here, let k ≥ n> l and we set

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

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

Then we have

image

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

image

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

image

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 image

Proof. Indeed, from R2(x) = x, we have

image

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

image

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

image

and

image

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 image we have the following.

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

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

image

image

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

image

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

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

This Lie superalgebra decomposition is image is the copy of image

Also, it holds that image

Acknowledgment

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

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