Algebra with ternary cyclic relations, representations and quark model

We propose a unital associative algebra, which is motivated by a generalization of the Pauli exclusion principle proposed within the framework of the quark model. The generators of this algebra satisfy the following relations: The sum of squares of all generators is equal to zero (binary relation) and the sum of cyclic permutations of factors in any triple product of generators is equal to zero (ternary relations). We study the structure of this algebra and calculate the dimensions of spaces spanned by homogeneous monomials. It is shown how the algebra we propose is related to irreducible representations of the rotation group. Particularly we show that the 10-dimensional space spanned by triple monomials is the space of a double irreducible unitary representation of the rotation group. We use ternary q - and ¯ q -commutators, where q, ¯ q are primitive 3rd order roots of unity, to split the 10-dimensional space spanned by triple monomials into a direct sum of two 5-dimensional subspaces. We endow these subspaces with a Hermitian scalar product by means of an orthonormal basis of triple monomials. In each subspace there is an irreducible unitary representation so (3) → su (5) . We calculate the matrix of this representation and the structure of matrix indicates a possible connection between our algebra and the Georgi-Glashow model.


Introduction
Algebras with an n-ary law of multiplication find more and more applications in theoretical physics.An n-Lie algebra, where n ≥ 3, is an n-ary extension of the concept of Lie algebra, introduced and studied by V.T. Filippov [8].n-Lie algebras with a totally skew-symmetric n-ary Lie bracket, which satisfies a Filippov-Jacobi identity, were used in the theory of M2 and M5 branes to construct a generalization of the Nahm's equation [4], [5], [6].Closely related to generalization of Lie algebra, proposed by V.T. Filippov, is an approach initiated by Y. Nambu, who proposed an n-ary generalization of the Poisson bracket [15], which is now called a Nambu-Poisson bracket.Y. Nambu proposed his n-ary generalization of the Poisson bracket independently of V.T. Filippov, but later it turned out that Nambu-Poisson bracket satisfies the Filippov-Jacobi identity, that is, it defines the structure of an n-Lie algebra.It is worth to mention that originally one of the motivations for Y. Nambu to introduce an n-ary generalization of the Poisson bracket was the quark model.
The quark model poses questions to theoretical physics that still have no clear answers.Undoubtedly, one of the most important problems of the quark model is confinement.However, no less intriguing is the question of why quarks have three colors and whether the coincidence of the number of colors with the dimension of our space is accidental.Since the quark model still contains unsolved problems, it serves as a source of new ideas and approaches for theoretical physicists.R. Kerner proposed a generalization of the Pauli exclusion principle within the framework of the quark model [13], [14].This generalization can be formulated as follows: Three quarks having identical quantum characteristics cannot form a stable configuration perceived as a strongly interacting particle.Note that this generalization of the Pauli exclusion principle allows for a coexistence of two quarks with the same isospin value.R. Kerner proposes to call the generalization of the Pauli exclusion principle formulated above the ternary generalization of the Pauli principle.Consider a wave function ψ(u, u, u) that represents the tensor product of three identical quantum states | u >.According to the ternary generalization of the Pauli exclusion principle, this function must vanish ψ(u, u, u) = 0. Suppose now that we have a superposition of three different quantum states According to the ternary generalization of the Pauli exclusion principle, we have ψ(v, v, v) = 0. Let indices a, b, c run the values 1,2,3.Using the linearity of a wave function ψ and taking into account the arbitrariness of complex numbers λ, µ, ν, we get that ψ(v, v, v) is equal to zero if and only if for any pair of integers a = b, ψ(u a , u b , u b ) + ψ(u b , u b , u a ) + ψ(u b , u a , u b ) = 0, (1) and for any triple of integers a = b = c ψ(u a , u b , u c )+ψ(u b , u c , u a ) + ψ(u c , u a , u b ) + ψ(u c , u b , u a ) + ψ(u b , u a , u c ) + ψ(u a , u c , u b ) = 0. (2) Obviously, the above conditions are satisfied if we assume that a wave function ψ is skew-symmetric in all arguments.In this case, rearranging any two arguments in a wave function ψ(u a , u b , u c ) and taking the sum with ψ(u a , u b , u c ) , we will get zero, that is, ψ(u a , u b , u c ) + ψ(u b , u a , u c ) = 0, ψ(u a , u b , u c ) + ψ(u a , u c , u b ) = 0.
However, this corresponds to the classical Pauli exclusion principle, which excludes the coexistence of two quarks with the same isospin, i.e. in the case of skew-symmetry of a wave function each term in ( 1) is separately equal to zero However, there is another possibility, so to speak, of solving conditions (1), (2), different from the skew-symmetry of a wave function ψ and consistent with the ternary generalization of Pauli exclusion principle.Indeed, if we assume that for any integers a, b, c (we admit a possibility of equal values) a wave function ψ has the following cyclic property then, as a consequence of this assumption, we obtain (1), (2).
Mathematically the classical Pauli principle can be expressed by means of the skew symmetry of a wave function of fermions, which in turn leads to a Grassmann algebra, that is, an algebra generated by anticommuting generators.Arguing in a similar way, we conclude that an algebra corresponding to the ternary generalization of Pauli principle must be an algebra with generators, and the equation (3) suggests that these generators must obey the ternary relations where θ a are generators of an algebra.
It should be noted that in the papers [1], [2], [9], [11], [12] it was proposed to construct an algebra corresponding to a ternary generalization of the Pauli principle using generators ϑ a that obey the ternary relations where q, q are primitive third order roots of unity.In this paper, we propose an algebra whose ternary relations (4) are more general than (5).Indeed it is easy to see that the relations (4) follow from the relations (5), but not vice versa.However, there are no binary relations in algebra (5), but in an algebra proposed in this paper, in addition to ternary relations 4, there is also one binary relation.
The above reasoning serves as a motivation for the introduction of an algebra R with ternary cyclic relations.R is an associative unital algebra over complex numbers C generated by θ 1 , θ 2 , . . ., θ N which obey the following relations where a, b, c is any triple of integers 1, 2, . . ., N .Here it would be appropriate to explain why we add one more binary relation (6) to the ternary cyclic relations (4).The point is that the binary relation ( 6), together with ternary relations (7), leads to the fact that there is a double irreducible representation so(3) → su (5) in the space spanned by triple monomials.And this, in turn, indicates a connection with the 5×5-matrix used in the Georgi-Glashow model in the case when there are only quarks and no electron and electron neutrino.
In this paper, we study the structure of an algebra R. We show that R is a finite-dimensional algebra and find the dimensions of subspaces spanned by homogeneous monomials.Let R p ⊂ R be a subspace spanned by homogeneous monomials of degree p.We show in the case of three generators how the algebra R is related to irreducible representations of the rotation group.In particular, we show that the 10-dimensional space R 3 spanned by monomials of degree 3 is the space of a double irreducible representation of weight 2 of the rotation group.We construct a basis for this 10-dimensional space by means of monomials of degree 3. Making use of the eigenvalues q, q of substitution operator, where q is a primitive third order root of unity, we split 10-dimensional space R 3 into the direct sum of two 5-dimensional subspaces R 3 q , R 3 q .In each subspace we have an irreducible weight 2 representation SO(3) → SU (5).We show that the eigenvectors corresponding to the eigenvalue q can be constructed by means of a ternary q-commutator [θ a , θ b , θ c ] q = θ a θ b θ c + q θ b θ c θ a + q θ c θ a θ b , and the eigenvectors corresponding to the eigenvalue q by means of a ternary q-commutator We introduce a Hermitian inner product into the 5-dimensional spaces R 3 q , R 3 q by constructing an orthonormal basis and find the 5 × 5-matrix of the operator induced by an infinitesimal rotation in space R 1 .Due to the unitarity of a representation of the rotation group, this matrix is skew-Hermitian and its structure is similar to the structure of the matrix for quarks in the Georgi-Glashow model [7].

Algebra with ternary cyclic relations
The aim of this section is to introduce and study an algebra described in Introduction.In this section we study the symmetries of algebraic relations of this algebra and find dimensions of subspaces spanned by homogeneous monomials.
First of all we define an algebra motivated by the ternary generalization of the Pauli exclusion principle as follows: Definition 1 An algebra R is a unital associative algebra over C generated by θ 1 , θ 2 , . . ., θ N which obey the binary relation and for any triple of integers a, b, c a ternary one The identity element of an algebra R will be denoted by 1. Assuming equal superscripts a, b, c in a ternary relation (9) we immediately get that the cube of every generator of R is zero, that is, (θ a ) 3 = 0 for any a = 1, 2, . . ., N .On the other hand, in contrast to a Grassmann algebra, it does not follow from the relations (8), ( 9) of an algebra R that the square of a generator is equal to zero, i.e. for any a we have (θ a ) 2 = 0.
In order to make forthcoming formulas more compact it is useful to introduce the following expressions of generators It is also useful to denote the left-hand sides of the ternary relations (9) In analogy with a binary case we will call {θ a , θ b , θ c } the ternary cyclic anticommutator of generators θ a , θ b , θ c .Now, using the introduced notations, the relations of algebra R can be written in a compact form Recall that the subspace of algebra R spanned by homogeneous monomials of degree p is denoted by R p .Thus the subspace R 1 is spanned by the generators θ 1 , θ 2 , . . ., θ N .The group of non-degenerate matrices GL(N, C) acts on this subspace according to the formula θa = A a b θ b , where A = (A a b ) ∈ GL(N, C), that can be considered as a transition to set of new generators θa of the algebra R. It is easy to see that the new generators θa satisfy the same ternary relations (9), that is, the ternary relations of algebra R are invariant under the action of the group GL(N, C).Indeed we have where the last equality follows from the fact that in the algebra R the ternary cyclic anticommutator of any three generators θ i , θ j , θ k is equal to zero.Obviously, if we wish to find a group under which all relations of the algebra R are invariant, including the quadratic one (8), we must pass from the group GL(N, C) to its subgroup of orthogonal matrices O(N, C).
and from Ω = 0 it immediately follows that Ω = 0. Note that when the quadratic relation ( 8) is multiplied either on the left or on the right by generator θ a , it generates the set of independent ternary relations These relations are also invariant under the action of orthogonal group.Indeed we have θa , and similarly we prove the invariance of ternary relations obtained with the help of multiplication by θ a from the right.Hence we proved that the relations of the algebra R are invariant under the action of the orthogonal group O(N, C).Particularly the relations of the algebra R are invariant under the action of SO(N, C), O(N, R) and SO(N, R).
Now we assume that N = 3, that is, we assume that the algebra R is generated by θ 1 , θ 2 , θ 3 .Then it is easy to see that the dimension of the subspace R 0 spanned by the identity element 1 is 1.The dimension of the subspace R 2 spanned by binary products of generators θ a θ b is 8, because we have nine possible products θ a θ b and only one binary relation (8).
The dimension of the subspace R 3 spanned by triple monomials θ a θ b θ c is 10.Indeed, in total we have twenty seven possible triple monomials θ a θ b θ c .From this number we must subtract the number of independent ternary relations of algebra R. Previously we showed that the ternary relations {θ a , θ b , θ c } = 0, θ a Ω = 0, Ω θ a = 0, form the set of all independent ternary relations of the algebra R. Let us first consider the ternary relations with the ternary cyclic anticommutator on the left-hand side.It is easy to see that the ternary cyclic anticommutator is invariant under cyclic permutations of the superscripts a, b, c, that is, Thus, if all three superscripts a, b, c are different, for example their values are a = 1, b = 2, c = 3, we have two independent relations If two of the three are equal but different from the third, we have only one independent relation.Thus, in this case, we have a total of six independent relations Finally, equal superscripts a = b = c give us three more independent relations (θ 1 ) 3 = 0, (θ 2 ) 3 = 0, (θ 3 ) 3 = 0. Summing up all numbers of independent ternary relations with the ternary cyclic anticommutator on the left-hand side, we get 11.
The second part of the ternary relations θ a Ω = 0, Ω θ a = 0 give us six more independent ternary relations and these ternary relations can be written as follows Finally, we get that totally there are 11+6=17 independent ternary relations in the algebra R. Hence the dimension of the space R 3 spanned by triple monomials is dim R 3 = 27 − 17 = 10.We can construct a basis for the space R 3 as follows.For example we consider the combination a = 1, b = 2, c = 2.It determines the following ternary relation By choosing monomials θ 1 θ 2 θ 2 , θ 2 θ 2 θ 1 as elements of a basis, we see that the monomial θ 2 θ 1 θ 2 can be expressed in terms of two chosen monomials.Relations (11) show that the monomials θ 1 θ 3 θ 3 , θ 3 θ 3 θ 1 can also be expressed in terms of the chosen monomials θ 1 θ 2 θ 2 , θ 2 θ 2 θ 1 .By virtue of the relation the monomial θ 3 θ 1 θ 3 can be expressed in terms of θ 1 θ 3 θ 3 , θ 3 θ 3 θ 1 , which means that it can also be expressed in terms of θ 1 θ 2 θ 2 , θ 2 θ 2 θ 1 .Similar reasoning leads to the conclusion that each pair of integers (1, 2), (2, 3), (3,1) gives two linearly independent monomials.We get four more monomials for the case when all indices are different, that is, a = 1, b = 2, c = 3 and a = 3, b = 2, c = 1.
Thus, as the basis for the 10-dimensional space R 3 , we can take the following monomials: For the numbering of the elements of this basis, we will use the capital letters from the beginning of the Latin alphabet, assuming that their values are integers from 1 to 5. Then the above basis can be written in the form {f A , f A+5 }, where A = 1, 2, ..., 5.The dimensions of spaces of higher degree monomials were found by means of the methods of computer algebra.The dimension of the space R 4 spanned by monomials of the 4th degree is 7.In the case of monomials of the 5th degree, the number of all possible monomials 243 exactly coincides with the number of independent relations of the 5th order of the algebra R, and thus the dimension of the space R 5 is 0. Thus, the dimension of the vector space of the whole algebra R is 29.

Irreducible representations of rotation group
The purpose of this section is to establish a connection between the algebra R (generated by three generators θ 1 , θ 2 , θ 3 ) introduced in the previous section and the irreducible tensor representations of the rotation group in a real 3dimensional space.The rotation group will be denoted by SO(3) (omitting an indication to real numbers R).More precisely, in the previous section we showed that the binary relation Ω = 0 of algebra R, as well as its ternary relations {θ a , θ b , θ c } = 0, are invariant under the action of the group O(3, C).Therefore, they are invariant under the action of the rotation group SO(3).We will show that the 10-dimensional space R 3 of the algebra R spanned by triple monomials is the space of a double irreducible unitary representation of weight 2 of the rotation group SO(3).In the next section, we propose a method for splitting this 10-dimensional space into two 5-dimensional subspaces and in each of these two subspaces there is an irreducible representation SO(3) → SU (5).
Let g = (g a b ) ∈ SO(3) be a 3rd order special orthogonal matrix.Then an action Π g : R 1 → R 1 of the rotation group SO(3) in the subspace R 1 ⊂ R spanned by the generators θ 1 , θ 2 , θ 3 is given by Π g (θ a ) = g a b θ b .We extend this action to the whole algebra R by Π g (θ a1 θ a2 . . .θ ap ) = g a1 b1 g a2 b2 . . .g In this section, we are primarily interested in a representation of the rotation group in the subspace R 3 spanned by the triple monomials of algebra R. Thus, we consider the restriction of the representation g ∈ SO(3) → Π g ∈ GL(R) to the subspace R 3 , i.e. we consider g ∈ SO(3) → Π g ∈ GL(R 3 ).Our aim is to prove that this representation is a double irreducible representation of weight 2 of the rotation group.To prove this statement, we will use the infinitesimal version of representation g ∈ SO(3) → Π g ∈ GL(R 3 ), that is, the representation of the Lie algebra so(3) of the rotation group induced by it.This representation will be denoted by L ∈ so(3) → π L ∈ gl(R 3 ), where L = (L a b ) is a skew-symmetric 3rd order matrix.Obviously π L : R p → R p is a derivation of the algebra R and its action on a homogeneous monomial of degree p is given by Our proof is based on the fact [10] that the triple monomials θ a θ b θ c transforming according to an irreducible representation of weight 2 of the rotation group satisfy the equation where C π is the Casimir operator, that is, In addition, it follows from general considerations of representation theory that the subspace of solutions of equation ( 17) is the space of a twofold irreducible representation of the rotation group.
Our goal is to show that the subspace of solutions of equation ( 17) coincides with the subspace R 3 determined by the ternary relations of algebra R. First of all, for further calculations it is convenient to introduce the 3rd order matrices ρ ab = ǫ abc L c , where ǫ abc is totally skew-symmetric tensor.If (ρ ab ) c d are the entries of a matrix ρ ab and we define Taking into account that π ab acts on a triple monomial as an algebra derivation we obtain Making use of (18), we can calculate the square of an operator π dh and, acting by the square of this operator on a triple monomial θ a θ b θ c , we get the expression Now we calculate the Casimir operator C π = (π 23 ) 2 + (π 31 ) 2 + (π 12 ) 2 .From formula (18) it follows that π aa = 0, π ab = −π ba .Hence (π ab ) 2 = (π ba ) 2 .If we now take the sum of the squares of operators π dh , where the superscripts d, h run independently of each other the values 1,2,3, then we get the doubled Casimir operator 2 C π .This means that in expression (20) for the square of an operator π dh we take sum over the superscripts d, h and the resulting expression will be equal to 2 C π .We get where Substituting the right-hand side of (21) into the basic equation ( 17) we obtain Recall that the space of solutions of this equation is the space of a twofold irreducible unitary representation of the rotation group.Now we can prove that the subspace of solutions of equation ( 23) coincides with the subspace R 3 spanned by triple monomials of the algebra R.
The subspace R 3 is defined by the following set of ternary relations It is easy to show that in the subspace R 3 in addition to the above ternary relations we also have the ternary relations Ω a = 0. Indeed, we can write Consequently Ω a = δ bc {θ b , θ c , θ a } − Ω θ a − θ a Ω, and the relations Ω a = 0, where a = 1, 2, 3, follow immediately from the ternary relations (24).It is easy to see that equation (23) vanishes identically on subspace R 3 .It is easy to prove the opposite, that is, if we have a solution to equation (23), it lies in the subspace R 3 .Let us assume that we have a solution of the equation ( 23).In the left-hand side of this equation take the sum over a = b.This yields Ωθ a = 0. Analogously summation over equal pairs b, c and a, c yield θ a Ω = 0, Ω a = 0. Hence a solution should also satisfy {θ a , θ b , θ c } = 0 and this shows that a solution of equation ( 23) belongs to the subspace R 3 of the algebra R.

Substitution operator and its eigenvectors
Now our task is to split the 10-dimensional subspace R 3 of the algebra R into a direct sum of two 5-dimensional subspaces in a way invariant with respect to the representation of the rotation group.Then, according to the theory of representations [10], in each 5-dimensional subspace we will an irreducible unitary representation of the rotation group SO(3) → SU (5).This decomposition of R 3 into two 5-dimensional subspaces in a way invariant with respect to a representation of the rotation group can be made with the help of eigenvectors of a substitution operator.Let σ be a cyclic substitution such that σ(1) = 2, σ(2) = 3, σ(3) = 1.Then we can define a substitution operator S σ acting on triple monomials as follows S σ (θ a1 θ a2 θ a3 ) = θ a σ(1) θ a σ(2) θ a σ(3) = θ a2 θ a3 θ a1 .
Then one can easily verify that the substitution operator maps the subspace R 3 of the algebra R into the same subspace, that is, S σ : R 3 → R 3 .This follows from the fact that the system of ternary relations defining subspace R 3 is invariant with respect to the substitution operator S σ , that is, the substitution operator S σ transforms each relation of the system again into a relation of the same system or their linear combination.The system of relations defining R 3 has the form (24). Obviously, each of the relations in (24) with a ternary Z 3 -anticommutator on the left-hand side is invariant under the substitution operator S σ .Indeed, a cyclic permutation of arguments in a ternary Z 3 -anticommutator does not change its structure.Hence As it is mentioned before, the relations θ a Θ = Θθ a = 0, where c = 1, 2, 3, can be arranged in the table (11), ( 12), ( 13).If we apply the substitution operator S σ to the first (from the left) relation in (11), we get the second relation in that row of the table.Indeed we have If we apply the substitution operator S σ to the second (from the left) relation in (11), we get a new relation that is not in the table.Indeed we have It is easy to show that the obtained relation does indeed hold, that is, it follows from the algebraic relations ( 8), ( 9) of the algebra R and is not independent.That is why the relations is not included in the table ( 11) - (13).
By virtue of the property σ 3 = id (id is the identical substitution) of the substitution σ, we have S 3 σ = I, where I is the identity operator in the vector space R 3 .Thus it follows that the eigenvalues of the substitution operator S σ are cubic roots of unity 1, q, q, where q = exp(2π i/3).This means that we can decompose the space R 3 spanned by triple monomials into the direct sum of the subspaces spanned by the eigenvectors corresponding to the eigenvalues 1, q, q.Let us denote these subspaces by R 3  1 , R 3 q , R 3 q .Then R 3 = R 3 1 ⊕R 3 q ⊕R 3 q .First of all, note that the ternary Z 3 -anticommutator can be written using the substitution operator S σ in the form Due to the invariance of the ternary Z 3 -anticommutator with respect to the substitution operator S σ , which we noted above (26), it is easy to see that the ternary Z 3 -anticommutator of generators θ a of the algebra R is an eigenvector of the substitution operator with eigenvalue 1.Since in our algebra the sum of cyclic permutations for any triple product of generators equals zero (9), we obtain that the subspace of eigenvectors with eigenvalue 1 degenerates into the trivial subspace consisting of the zero vector.Below we will show that the ternary Z 3 -anticommutators ( 27) give all eigenvectors of the substitution operator with eigenvalue 1.Thus R 3 1 = {0} and we obtain R 3 = R 3 q ⊕ R 3 q .The structure of an eigenvector of the substitution operator S σ with eigenvalue 1, that is, the ternary Z 3 -anticommutator, suggests how to construct the eigenvectors of the substitution operator with eigenvalues q, q.To do this, consider the following linear combinations of triple products of generators, which are constructed by analogy with the ternary Z 3 -anticommutator It is easy to verify that (28), (29) are the eigenvectors of S σ .Indeed we have and we see that the 3rd order polynomial [θ a , θ b , θ c ] q is an eigenvector of the substitution operator with the eigenvalue q.Analogously one can verify that [θ a , θ b , θ c ] q is an eigenvector of the substitution operator with the eigenvalue q.
5 Cyclic Z 3 -extension of a Lie algebra At the end of the previous section, we used expressions ( 28) and (29) to construct the eigenvectors of the substitution operator S σ .We have used square brackets to denote these expressions.This is due to the fact that we interpret these expressions as a ternary analogue of a Lie bracket (or commutator) based on the cyclic group Z 3 .In what follows we will call the expressions [θ a , θ b , θ c ] q , [θ a , θ b , θ c ] q ternary cyclic q-commutator and ternary cyclic q-commutator of generators, respectively.Here we need to clarify why we use the square brackets and the term "commutator".We propose to use the following analogy, which makes it possible to extend the concept of a commutator of two noncommuting variables to a larger number of variables.Suppose we are considering a subgroup G of a permutation group of p integers {1, 2, . . ., p} and this subgroup has a representation ρ : G → C in the field of complex numbers.Suppose that u 1 , u 2 , . . ., u p are elements of an associative but non-commutative algebra over the field of complex numbers.We can form the sum of products of these elements as follows.Taking u 1 • u 2 • . . .• u p as the initial product (dot stands for a multiplication in our algebra), we choose those permutations of the factors of this product, which are determined by the substitutions σ of a subgroup G. Now we take the sum of the obtained products, and each product in this sum is taken with a scalar coefficient ρ(σ), σ ∈ G determined by a representation ρ of a subgroup G. Hence we get the expression If G = S p is a symmetric group and ρ : S p → {−1, +1} is the representation which assigns +1 to even permutations and -1 to odd permutations then the expression (30) yields the well known totally antisymmetric p-ary commutator used in p-Lie algebras and Nambu-Poisson algebras.Note that in the case p = 2 we have Let us take as the next subgroup G the subgroup of cyclic permutations Z 3 of the permutation group of three elements, that is, G = Z 3 .In this case we have two faithful representations of the third order cyclic group in the field of complex numbers ρ : {id, σ, σ 2 } → {1, q, q} ρ : {id, σ, σ 2 } → {1, q, q}.
Accordingly, the formula (30) give us two different (conjugate) expressions for a ternary commutator and Thus, we consider the expressions on the right-hand side of equalities ( 28), (29) as an Z 3 -extension of the notion of a binary commutator to the case of three non-commuting variables.
We can consider the ternary cyclic q-commutator (31) (and the ternary cyclic q-commutator (32) which leads to a similar structure) from a more general point of view with the help of the following definition.
Definition 2 Let g be a Lie algebra and ρ : a ∈ g → ρa ∈ gl(V ) be its representation in a vector space V .A trilinear mapping is called a ternary Z 3 -bracket compatible with a representation ρ of a Lie algebra and for any a ∈ g the endomorphism ρa of a vector space V is a derivation of a mapping (33), that is, A vector space V equipped with a ternary Z 3 -bracket compatible with a representation ρ of a Lie algebra g will be called a ternary Z 3 -extension of a Lie algebra g.In particular, if a ternary Z 3 -bracket (33) has the following symmetries with respect to cyclic permutations of its arguments where q is a primitive 3rd order root of unity, then it will be referred to as a ternary cyclic q-bracket (or q-bracket) compatible with a representation ρ of a Lie algebra g.
Let A be a unital associative algebra over the field of complex numbers.The identity element of this algebra will be denoted by e.We can consider algebra A as a Lie algebra by equipping it with the commutator [u, v] = u v−v u, where u, v ∈ A. To emphasize the fact that we are considering A as a Lie algebra, we will denote it by A L .As a representation of this Lie algebra, we use the adjoint representation ad : u ∈ A L → ad u ∈ gl(A), where ad u (x) = [u, x].

Theorem 1 Define
[[x, y, z]] = x y z + q y z x + q z x y, x, y, z ∈ A. (37) Then (37) is a ternary cyclic q-bracket compatible with the adjoint representation of the Lie algebra A L and A equipped with (37) is a ternary Z 3 -extension of the algebra Lie A L .
For any elements x, y, z, v of an algebra A, there is an identity that can be verified by direct calculations, and this proves the assertion of the theorem.We think that it would be appropriate here to compare the proposed structure with a 3-Lie algebra proposed by Filippov [8].Indeed, identity (38) resembles the Filippov-Jacobi identity, which shows that a Filippov ternary bracket, applied twice, is a derivation of an inner bracket.However, the fundamental difference between our proposed structure and Filippov ternary bracket lies in properties (34) (or (36)) of our ternary bracket.Recall that a ternary Filippov bracket is totally skew-symmetric and, consequently, the sum of cyclic permutations of its elements is not equal to zero.Thus, a ternary bracket we propose is based on a different symmetry, not skew symmetry, but cyclic permutations Z 3 .
The second remark concerning the proved theorem is that the ternary bracket (37) is not associative.Recall that in the case of a ternary operation there are two types of associativity.These are associativity of the first kind where x, y, z, u, v ∈ A. In the case of the ternary bracket (37) neither the associativity of the first kind nor the associativity of the second kind takes place.This is natural, since in the binary case the commutator on an associative algebra defines a Lie algebra, which is a non-associative algebra.
The final remark is that the identity element e of an algebra A makes it possible to reduce the ternary bracket (37) to the usual commutator of two Thus S 2 σ (f A ) = −f A − S σ (f A ). Substituting this expression into the left-hand side of the equation (43), we get We get the system of linear equations Substituting −λx A instead of x A+5 in the second equation, we get Taking the first eigenvalue λ = 1, we see that this equation, and hence the system of equations ( 45), has only solution x A = x A+5 = 0.This proves that the subspace of eigenvectors with eigenvalue 1 trivial, as stated above.Since q, q are the roots of the equation 1 + λ + λ 2 = 0, in the case of λ = q or λ = q the system (45) has non-trivial solutions, which can be written as Thus if X is a q-eigenvector of the substitution operator S σ then it can be written in the form If X is a q-eigenvector of the substitution operator S σ then it can be written in the form It is easily verified that the vectors f A − q S σ (f A ) and f A − q S σ (f A ) are linearly independent q-eigenvectors and q-eigenvectors of the substitution operator S σ and they form the bases for the subspaces R 3 q and R 3 q respectively.Formulae (46), (47) show that linearly independent q-eigenvectors and q-eigenvectors form complete systems of vectors for the subspaces R 3 q and R 3 q respectively.Thus they are the bases for these subspaces.
The transition from the bases f A −q S σ (f A ) and f A − q S σ (f A ) to the systems of polynomials (40) and (41) respectively is carried out by multiplying each polynomial f A − q S σ (f A ) (f A − q S σ (f A )) by the number 1 − q (1 − q).Indeed for the first polynomial of the basis f A − q S σ (f A ), i.e.A = 1, we have Now we introduce a Hermitian scalar product in the 5-dimensional complex space R 3 q by means of an orthonormal basis e A in such a way that the derivation operator D L is a skew-Hermitian, that is, for any subscripts A, B a Hermitian scalar product should satisfy Simple calculations show that, assuming the vectors of the initial basis f ′ A to be orthogonal, i.e. < f ′ we will obtain the solution of equation ( 48), except for those cases when one vector e A belongs to the set {f ′ 1 , f ′ 2 , f ′ 3 }, and the other e B belongs to {f ′ 4 , f ′ 5 }.In these cases, the condition (48) is satisfied if This indicates that we have to renormalize the vectors f ′ 1 , f ′ 2 , f ′ 3 .Hence the Hermitian scalar product in the 5-dimensional complex space R 3 q is determined by the orthonormal basis e 4 = [θ 1 , θ 2 , θ 3 ] q = (1 − q) (f 4 − qS σ (f 4 ), e 5 = [θ 3 , θ 2 , θ 1 ] q = (1 − q) (f 5 − qS σ (f 5 ), and the matrix of the operator of derivation D L in this basis is skew-Hermitian.Similar constructions in the 5-dimensional complex space R 3 q lead to the following orthonormal basis Calculating the matrix of the operator of derivation D L in the basis e A , we get the matrix and the similar matrix (with the replacement of q by q and vice versa) in the basis ēA .

Conclusion
We have introduced and studied the structure of an algebra generated by θ a , which are subjected to one binary relation and the set of ternary relations.This algebra is motivated by a ternary generalization of the Pauli exclusion principle in the framework of the quark model.We have shown that the 10-dimensional space spanned by triple monomials of our algebra is a representation space of double irreducible representation of the rotation group.We have proposed a ternary Z 3 -generalization of the notion of commutator based on primitive 3rd order roots of unity and using these ternary commutators we split the 10dimensional representation space into two 5-dimensional subspaces and proved that each subspace is a representation space for irreducible representation of the rotation group.We have constructed an orthonormal basis in each of the subspaces and computed the matrix of a derivation operator.The standard model for quarks and leptons fit nicely into representations of SU (5).Two matrices [7] used in Georgi-Glashow model have the form .
Comparison of matrix 10 with the matrix of a derivation operator D L of algebra R shows the similarity of the structures of these matrices.If we limit ourselves to the quark part of the matrix 10 (without an electron e c ), the similarity in structure becomes even greater.We think that this is another evidence in favor of the fact that a ternary generalization of the Pauli exclusion principle and the algebra constructed on its basis are an adequate description of the quark model.