**INTRODUCTION**

Associated with any algebra Θ over a field of char. ≠ 2 are two algebras denoted by Θ^{-} and Θ^{+}. There algebras have the same underlying vector space as Θ but are given the products [x, y] = x * y-y * x and x○y = ½(x * y+y * x), respectively, where * is the multiplication in Θ.

Albert (1948), proposed the problem of classifying all power-associative flexible Lie admissible algebras, whose commutator algebras are semi simple Lie algebras. Without assuming flexibility, Benkart (1984) classified all third power-associative Lie admissible algebras whose commutator algebras are semisimple Lie algebras. Myung (1985) classified all third power-associative Lie admissible algebras associated with the Virasoro algebra and the Witt albebra. Jeong *et al*. (1997) determined all third power-associative Lie admissible algebras whose commentator algebras are Kac-Moody algebras. In this study, we will determine all third power associative (-1, 1) algebras whose commutator algebras are simple generalized Kac-Moody algebras. Thedy (1975) collected information on two natural concepts in a right alternative algebra R, the submodule M generated by all alternators (x, x, y) and a new nucleus called alternative nucleus. He also studied the properties of (-1, 1) rings. Hentzel (1974), characterized the properties of (-1, 1) rings and showed that all simple (-1, 1) rings are associative. He developed several identities for (-1, 1) rings. Benkart (1982), classified all power-associative products * that can be defined on the algebra A of n/sup x/n matrices over a 2, 3-torsion free field F satisfying the condition x*y-y*x = xy-yx for all x, y ∈ A and stated that such product * are automatically Lie-admissible and they are Jordan-admissible too. Myung (1996) studied the properties of Lie-admissible algebras along with Virasoro algebra and determined third power associative Lie-admissible algebras whose commutator algebras are Virasoro algebras. Humphreys (1972), developed the basic general theory of Lie algebras to give a first insight into basics of the structure theory and representation theory of semisimple Lie algebras. It is noted that a 2, 3-torsion free right alternative algebra is a (-1, 1) algebra if it satisfies the identity (x, y, z) + (y, z, x) + (z, x, y) = 0.

**MATERIALS AND METHODS**

**Preliminaries:** We begin with some of the basic facts about (-1, 1) algebras. Let (Θ, *) be a (nonassociative) algebra over the complex field with multiplication * and let (Θ‾, [,]) be its commutator algebra, where the underlying space is the same as Θ and the bracket operation [,]:Θ×Θ→Θ is defined by [x, y] = x * y-y * x for all x, y ∈ Θ.

An algebra Θ is called a (-1, 1) algebra if it satisfies the identity:

where, (x, y, z) = (xy)z - x(yz) denotes the associator of elements x, y and z. Identity (1) is meant to imply that Θ is right alternative and (2) implies that Θ is Lie-admissible, that is, the commutator algebra Θ‾ with a product [x, y] = xy - yx is a Lie algebra.

For a (-1, 1) ring (Θ, *), the multiplication * can be written as

Equation 3 is equivalent to

for all x ∈ R which can be linearized to

for all x, y ∈ R. Again linearizing (6) we obtain

for all x, y, z ∈R

The structure of generalized Kac-Moody algebras has been considered by many authors (Kac, 1990). A generalized Kac-Moody algebra is a Lie algebra that is similar to a Kac-Moody algebra, except that it is allowed to have imaginary simple roots.

Generalized Kac-Moody algebras are also sometimes called GKM algebras. The best known example is the monster Lie algebra. Most properties of GKM algebras are straight forward extensions of the usual properties of Kac-Moody algebras. A GKM algebra has an invariant symmetric bilinear form such that (e_{i}, f_{i}) = 1. There is a character formula for highest weight modules, similar to the Weyl-Kac character formula for Kac-Moody algebras except that it has correction terms for the imaginary simple roots.

A symmetrized Cartan matrix is a (Possibly infinite) square matrix with entries x_{ij} such that:

The universal generalized Kac-Moody algebra with given symmetrized cartan matrix is defined by generator e_{i}, f_{i} and h_{i} and relations:

These differ from the relations of a (symmetrizable) Kac-Moody algebra mainly by allowing the diagonal entries of a Cartan matrix to the nonpositive. In other words we allow simple roots to be imaginary, whereas in a Kac-Moody algebra simple roots are always real.

A generalized Kac-Moody algebra is obtained from a universal one by changing the certain matrix by the operations of killing something in the center or taking a central extension or adding outer derivations.

We will use the identities (3), (4), (5), (6) and (7) frequently in determining the third power-associative (-1, 1) multiplications on generalized Kac-Moody algebras and our approach is based on the techniques developed by Benkart (1984), Jeong *et al*. (1997) and Myung (1996).

Let A = (a_{ij})_{i, j∈I} be a generalized Cartan matrix where I is the finite index and g = g (A) be the generalized Kac-Moody algebra associated with A. We will determine all the third power associative (-1, 1) structures on the generalized Kac-Moody algebra g = g (A). When A is of finite type, this problem was settled by Benkart (1984). Hence we will assume that the generalized Cartan matrix is not of finite type.

We start with two technical Lemmas, which support the rest of the results and the proof of these Lemmas can be found in (1997).

**Main section**

**Lemma 1:** The algebra of linear functional on K is an integral domain. In particular, if x ∈ Φ is a root of g and f: K → is a linear functional on K satisfying f(h)x(hʹ) + f(hʹ)x(h) = 0 for all h, hʹ ∈ K then f = 0.

**Lemma 2:** Let A = (a_{ij})_{i, j ∈ I }be a generalized cartan matrix which is not of finite type. If a_{ij}≠0 for some indices i, j ∈ I, there exists a root β = Σ_{xεI} c_{k} x_{k} ∈ Φ of g such that c_{i}>0, c_{j}>0 and c_{i}≠c_{j}.

**Remark:** Lemma 2 holds for all generalized cartan matrices that are not of type A_{n} see (1972), (1990).

Let B be an abelian group, K be the vector space over F. Denote by FB the group algebra of B over F. The elements t^{c}, c ∈ B from a basis of this algebra and the multiplication is defined by t^{c}t^{d} = t^{c + d}. We shall write 1 instead of t^{0}. The tensor product M = FB ⊗_{F} K is a free left FB- module. Denote an arbitrary element of K by d. For the sake of simplicity we write t^{c}δ instead of c^{c}⊗δ. We fix a pairing Φ: K× B → F which is F-linear in the first variable and additive in the second one. For convenience we use the following notations:

for arbitrary δ ∈ k and c ∈ R.

The following multiplications:

for arbitrary c, d ∈ R and δ_{1},δ_{2} ∈ K make g into a Lie algebra, called a generalized Kac-Moody algebra.

Let g_{c} = t^{c} k for x ∈ R, in particular g_{o} = K then [g_{c}, g_{d}] ⊂ g_{c+d} holds for all c, d ∈ B. This means that g is a B-graded Lie algebra. It is clear that [δ, t^{c}δ_{1}] = δ(x)+t^{c}δ_{1}, hence δ is semisimple. Consequently K is a torus.

Let B_{o} = {x ∈ B/<δ, x> = 0, ∀ δ∈ K} and K_{o} = {δ ∈ K/<δ, x> = 0, ∀ δ ∈ B}. Φ is said nondegenerate if B_{o} = 0 and K_{o} = 0.

**Lemma 3: **

(i) |
The Cartan sub algebra K is closed under ο. |

(ii) |
For each x ∈ B/{0}, there exists a linear functional f_{x}: K→C and a by linear map u_{x}: K×K→K such that: |

for all δ ∈ K, t^{x}δ_{x} ∈ g_{x} and the f_{x} satisfies:

for all δ, δʹ ∈ K.

**Proof:**

(i) |
By bilinearity and commutativity of ο it suffices to show that δοδ ∈ K. write: |

It follows from [δ, δοδ] = 0 that

Hence, if δ(x) = 0, then δ_{1}(x)t^{x}δ_{x} = 0 for all δ_{1} ∈ K which implies δ_{x} = 0. Therefore δοδ = δʹ ∈ T.

(ii) |
Write δοt^{x}δ_{x} = δʹ + By (6), we have |

Similarly, for δ_{1} ∈ K, we have:

By (7), we have [δ, δ_{1}οt^{x}δ_{x}] + [δ_{1}, t^{x}δ_{x} δ] + [t^{x}δ_{x}, δοδ_{1}] = 0 which yields

By (10), one can easily find that

Denote δοt^{x}δ_{x} = u_{x}(δ, δ_{x}) + t^{x}δ_{(δ, δx) }where u_{x}(δ, δ_{x}) and δ_{(δ, δx) }∈ K. By_{:}

We can find:

Fixing δʹ ∈ K so that δʹ(x) = 1, we see that if δ(x) = 0, δ_{(δ, δx)} = (δοδʹ)(x)t^{x}δ_{x} and 2 δ_{(δʹ, δx)} = (δʹοδʹ) (x)δ_{x}

Define for all δ ∈ Ker x: = {δ ∈ K/δ(x) = 0} and .

Since, K = Ker_{x} ⊗ Fδʹ, f_{x} is well defined. Then δοt^{x}δ_{x} = f_{x}(δ)t^{x}δ_{x} + u_{x}(δ, δ_{x}) since, δ_{(δ, δx)} = f_{x}(δ)δ_{x}. It follows from (13) that (11) is true. It is clear that u_{x}: K× K → K is a bilinear map.

From the following Lemma we can see that the linear functional f_{x} are the same for all x ∈ B/{0}.

**Lemma 4**:

(i) |
For all x, y ∈ B/{0}, f_{x} = f_{y}. We denote it by f |

(ii) |
For all δ, δʹ ∈ K, δοδʹ = f(δʹ)δ + f(δ)δʹ |

**Proof:** If dim K = 1, then K = Fδ and for any x ∈ B/{0}, δ(x) ≠ 0 since, B_{0} = 0. Hence for δ_{1}, δ_{2} ∈ K, we have δ_{1 }= δ_{2} if δ_{1}(x) = δ_{2}(x) for some x ∈ B/{0}.

from (9), we have 2f_{x}(δ)δ(x) = (δοδ)(x).

So this means that f_{x} = f_{y} for any y ∈ B/{0}.

Now we suppose that dim K>1. If T_{x} ≠ Ty, then K = T_{x} + T_{y} and there exists δ_{1}, δ_{2} ∈ K such that δ_{1}(x) = 1, δ_{1}(y) = 0, δ_{2}(x) = 0, δ_{2}(y) = 1 and K = Fδ_{1} ⊗ Fδ_{2}(y) ⊗ (T_{x} Ç T_{y}). Let δ_{3 }∈ T_{x} ∩ T_{y}. For any δ, δʹ ∈ K, it follows from (9) that

In (12), let δ = δʹ = δ_{1} we have f_{x+y}(δ_{1}) = f_{x}(δ_{1})

Let δ = δ_{1}, δʹ = δ_{3} we have f_{x+y}(δ_{3}) = f_{x}(δ_{3}).

Similarly, we have f_{x+y}(δ_{2}) = f_{y}(δ_{2}), f_{x+y}(δ_{3}) = f_{y}(δ_{3})

Let δ = δ_{1}, δʹ = δ_{2}, we have:

Now let z = 2x+y in (11) we have

In (16), let δ = δʹ = δ_{1}, we have f_{z}(δ_{1}) = f_{x}(δ_{1})

Similarly we have f_{z}(δ_{2}) = f_{y}(δ_{2})

Let δ = δ_{1}, δʹ = δ_{2}, we show

From 13 and 15 we know that f_{y}(δ_{2}) = f_{x}(δ_{2}), f_{y}(δ_{1}) = f_{x}(δ_{1}). Hence, f_{x} = f_{y}. If T_{x} = T_{y}, we can choose z ∈ B/{0} such that T_{z}≠T_{x}. Then f_{x} = f_{z} = f_{y}. (ii) follows from first part (i) and 9

**Lemma 5: **For each x ∈ B/{0}, there exists a linear functional λ_{x}: K → F and a symmetric bilinear map σ_{x}: K×K → K such that t^{x}δ_{1}οt^{x}δ_{2 }= λ_{x}(δ_{2}) t^{x}δ_{1} + λ_{x} (δ_{1}) t^{x}δ_{2 }+ σ_{x}(δ_{1}, δ_{2}) for all δ_{1}, δ_{2} ∈ K

**Proof:** For x, y ∈ B/{0}, by (7) and Lemma 3 and Lemma 4 we have

For all d ∈ K. It follows that t^{x}δ_{1}οt^{y}δ_{2} ∈ K + g_{x }+ g_{y}.

Let x = y, we have t^{x}δ_{1}οt^{x}δ_{2} ∈ K+g_{x}. Then t^{x}δ_{1}οt^{x}δ_{2 }= ^{tx}δʹ+σ_{x}(δ_{1}, δ_{2}) for some symmetric bilinear map σ_{x}: K×K → K and δʹ ∈ K. For any d ∈ K, we have [δ, t^{x}δʹ_{1}οt^{x}δ_{2 }] = δ(x)t^{x}δʹ. Using 16 we have δ(x)δʹ = u_{x}(δ, δ_{1})(x)δ_{2} + u_{x}(δ, δ_{2})(x)δ_{1}. Fixing δ_{0} ∈ K such that

for all δ_{1} ∈ K. It is clear that λ_{x} (δ_{1}) does not depend on the choice of δ_{0} with δ_{0}(x) ≠ 0. So t^{x}δ_{1}οt^{y}δ_{2} = λ_{x}(δ_{1}) t^{x}δ_{2} + λ_{2}(δ_{2}) t^{x}δ_{1 }+ σ_{x}(δ_{1}, δ_{2}).

**Lemma 6:**

(i) |
For any x, y ∈ B/{0}, there exists a bilinear s_{xy}: K´ K → K such that: |

(ii) |
For any δ, δ_{1} ∈ K, x ∈ B/{0}, δοt^{x}δ_{1} = f(δ)t^{x}δ_{1} + λ_{x}(δ_{1})δ |

**Proof:**

(i) |
Let x, y ∈ B/{0} and x ≠ y. For any δ_{1}, δ_{2} ∈ K we know that t^{x}δ_{1}οt^{y}δ_{2} = t^{x}δʹ_{1}+t^{y}δʹ_{2} + σ_{xy}(δ_{1}, δ_{2}) for some δʹ_{1}, δʹ_{2}, σ_{xy}(δ_{1}, δ_{2}) ∈ K. It is clear that σ_{xy} is a bilinear map from K×K to K. For any δ ∈ K, by [δ, t^{x}δ_{1}οt^{y}δ_{2}] = δ(x)t^{x}δʹ_{1} + δ(y)t^{y}δʹ_{2} and (16) we have δ(x)δ = u_{y}(δ, δ_{2})(x)δ_{1} and δ(y)= u_{x}(δ, δ_{1})(y)δ_{2}. Choosing δ_{0}, δʹ ∈ K such that δ_{0}(x) = 1, δʹο(y) = 1, we can define λ^{y}_{x}(δ_{2}) = u_{y}(δ_{0}, δ_{2})(x) for all δ_{2} ∈ K and λ^{x}_{y} (δ_{1}) = u_{x}(δʹο, δ_{1})(y) for all δ_{1} ∈ K. Then t^{x}δ_{1}οt^{y}δ_{2} = λ^{x}_{y}(δ_{2}) t^{x}δ_{1} + λ^{x}_{y}(δ_{1})t^{y}δ_{2 }+ σ_{xy}(δ_{1}, δ_{2}). By [δ, t^{x}δ_{1}οt^{y}δ_{2}] + [t^{x}δ_{1}, t^{y}δ_{2}οδ] + [t^{y}δ_{1}, δʹο t^{x}δ_{1}] = 0, |

we have λ^{x}_{y}(δ_{2}) δ(x)t^{x}δ_{1 }+ λ^{x}_{y}(δ_{1}) δ(y)t^{y}δ_{2}-u_{y}(δ, δ_{2})(x)t^{x}δ_{1}-u_{x}(δ, δ_{1})(y)t^{y}δ_{2 }= 0. Hence

for all δ_{1}, δ ∈ K. Denote λ^{x}_{x} = λ_{x}, we have that 17 holds for all y ∈ B/{0}

Now we shall prove that λ^{y}_{x} = λ^{z}_{x} for all x, y, z ∈ B/{0}.

If Ker y = Ker z, since λ^{x}_{y}(δ_{1})δ-u_{x}(δ, δ_{1})) (*y*) = 0, we know λ^{x}_{y}(δ_{1})δ - u_{x}(δ, δ_{1}) ∈ Ker y = Ker z and hence (λ^{x}_{y}(δ_{1})δ-u_{x}(δ, δ_{1}))(z) = 0. Then λ^{x}_{y}(δ_{1})δ(z) = u_{x}(δ, δ_{1})(z) = λ^{z}_{x}(δ_{1})δ(z) for all δ, δ_{1} ∈ K . Hence λ^{y}_{x} = λ^{z}_{x}.

If Ker y ≠ Ker z, we can choose δʹ, δʹʹ ∈ K such that δʹ(y) = 1, δʹ(z) = 0, δʹʹ(y) = 0, δʹʹ (z) = 1. By (17), we can show

Let δ = δʹ and δʹʹ, we show

For all d_{1} ∈ K . Hence, we have λ^{y}_{x} = λ^{x}_{x} = λ_{x} for all x, y ∈ B/{0}.

(ii) |
For any z ∈ B/{0} we have u_{x}(δ, δ_{1})(z) = λ^{z}_{x}(δ_{1})δ(*z*) = λ_{x}(δ_{1})δ(z) |

Hence, u_{x}(δ, δ_{1}) = λ_{x}(δ_{1}). From Lemma 3 we know (ii) is true.

**Lemma 7:** For any x, y ∈ B/{0}, x ≠ y and δ_{1}, δ_{2} ∈ K, we have.

• |
σ_{x}(δ_{1}, δ_{2}) = 0, where σ_{x}: K×K → K is the symmetric bilinear map defined in Lemma 5 |

• |
σ_{xy}(δ_{1}, δ_{2}) = 0, where σ_{xy}: K×K → K is the bilinear map defined in Lemma 6 |

**Proof: **Let x ∈ B/{0}. From [t^{x}δ_{1}, t^{x}δ_{1}οt^{x}δ_{1}] = 0 for all δ_{1} ∈ K, we have σ_{x}(δ_{1}, δ_{1})(x) = 0. Since, σ_{x} is a symmetric linear map we have σ_{x}(δ_{1}, δ_{2})(x) = 0 for all δ_{1}, δ_{2} ∈ K.

For δ_{1}, δ_{2} ∈ K, let y ∈ B/{0, x} and δ_{3} ∈ K. Since, [t^{y}δ_{3}, t^{x}δ_{1}οt^{x}δ_{2}] = [t^{y}δ_{3}οt^{x}δ_{2}, t^{x}δ_{1}] + [t^{y}δ_{3}οt^{x}δ_{1}, t^{x}δ_{2}], we have σ_{x}(δ_{1}, δ_{2})(y) t^{y}δ_{3} = σ_{yx}(δ_{3}, δ_{2})(x) t^{x}δ_{1} + σ_{yx}(δ_{3}, δ_{1})(x) t^{x}δ_{2} which implies σ_{x}(δ_{1}, δ_{2})(y) = 0. So σ_{x}(δ_{1}, δ_{2}) ∈ K_{0 }= 0. Hence σ_{x}(δ_{1}, δ_{2}) = 0 for all δ_{1}, δ_{2} ∈ K. Similarly, we can prove (ii).

Now we can give the following main theorem of this paper.

**Theorem 1: **If * is a third power-associative (-1, 1) multiplication on the simple generalized Kac-Moody algebra g, then there exist a linear functional τ: g → such that

for all u, v ∈ g.

Conversely, if * is a multiplication on g defined by (18), then * is a third power-associative (-1, 1) multiplication on g

**Proof: **Define a linear functional τ: g → by:

It is clear that (18) is true.

From (1) we have (x * y) * y = x * (y * y) for all x, y ∈ B which is equivalent to [x, y]οy = [x, yοy]. Using theorem 1 we can determine all (-1, 1) multiplication on simple generalized Kac-Moody algebras.