The Bernstein Problem in Dimension 6

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1 JOURNAL OF ALGEBRA 185, ARTICLE NO The Bernstein Problem in Dimension 6 J. Carlos Gutierrez Fernandez Departamento de Matematicas, Uniersidad de Oiedo, CCalo Sotelo, sn, Oiedo, Spain Communicated by E. Kleinfelol Received November, 1995 The solution of the Bernstein problem in the regular and exceptional cases, in all dimensions n, was made by Yu. Lyubich. A. Grishkov proved that there are no nonregular nonexceptional nuclear Bernstein algebras of type Ž 4,. with stochastic realization and therefore the Bernstein problem of type Ž 4,. was completely solved by the present author Ž J. Algebra, to appear.. The aim of this paper is to describe explicitly all simplicial stochastic nonexceptional nonregular Bernstein algebras of type Ž 3, 3.. Since every nonregular nonexceptional Bernstein algebra of dimension 6 is either of type Ž 4,. or of type Ž 3, 3., the Bernstein problem in dimension 6 is completely solved in this paper Academic Press, Inc. 1. INTRODUCTION In works 14, S. N. Bernstein raised and partially solved an important problem concerning mathematical expression of fundamental laws of biological heredity. Let us, following 1, describe the statement of the Bernstein problem. The state of a population in any generation can be described by a stochastic Ž or probabilistic. vector x Ž x. n, so all the x 0 and sž x. i i1 i Ýixi1. The set of all states is the basic simplex n1 n. The vertices Ž e. i i1 n of this simplex are the types of individuals in this population. Let us denote by pij,k the probability that an individual of type ek appears in the next generation from parents whose types are ei and e j,so p 0 and p 1. Ž 1. ij,k Moreover, if the material or paternal origin does not play a role in the production of offsprings types, then p p, i, j, k 1,,..., n $18.00 ij,k ji, k Copyright 1996 by Academic Press, Inc. All rights of reproduction in any form reserved. 40 Ý k ij, k

2 BERNSTEIN PROBLEM 41 Let the population be panmixic, i.e., the mating in it is at random. Then in the absence of selection the state x Ž x. i i1 n in the next generation will be x p x x Ž 1kn.. Ž 3. Ý k ij, k i j i, j These formulas define a mapping V: n1 n1 called the eolutionary operator Ž e.o.. of the given population. A state x is called an equilibrium if Vx x. An evolutionary operator V is called stationary if for every x n1 the corresponding state in the next generation is an equilibrium, or equivalently V V. The Bernstein problem is to describe explicitly all stationary evolutionary operators. The following interpretation of evolutionary operator V, given by the formula Ž. 1, is very useful in the solution of this problem. One can define in the space n a commutative product Ý ee p e Ž 1i, jn.. Ž 4. i j ij, k k k In such a way we obtain a commutative Ž nonassociative. baric algebra A over. The pair Ž A, s. is a simplicial stochastic algebra Žsee 1, p. 150 for a general definition.. This means that the product of any elements x, y n1 n1 belongs to Ž we say stochastic for simplicity.. This commutative algebra A has a nonzero homomorphism s: A. There is a remarkable connection between the stationary properties of an evolutionary operator and some property of the corresponding algebra A. The following theorem was established by Lyubich 1. THEOREM 1.1. The operator V is stationary if and only if the identity Ž x. sž x. x holds. One of most important conclusions that can be drawn from the preceding theorem is that the Bernstein problem is equivalent to finding, for every Bernstein algebra, all stochastic bases with their corresponding multiplication constants. Let us recall some definitions and facts concerning Bernstein algebras Žsee 1, for more information.. A Bernstein algebra Ž A,. over is a commutative baric algebra, that is, : A is a nonzero algebra homomorphism. satisfying the identity Ž x. Ž x. x Ž 5. for every x A. We remark that Bernstein algebras are not necessarily algebras with a stochastic realization; that is, there exist Bernstein algebras

3 4 J. CARLOS GUTIERREZ FERNANDEZ such that for every basis the simplex spanned by this basis is not invariant with respect to the multiplication, i.e., Žsee. 8. A linearly independent set in A is said to be stochastic if. Therefore, a Bernstein algebra is an algebra with a stochastic realization if and only if it has a stochastic basis. A Bernstein algebra has a nonzero idempotent element e. Every such elements yields the Peirce decomposition A e Ue V e, where 4 1 ker Ue V e, Ue x A ex x, Ve xaex 04. We will let N kerž. and denote by IŽ A. the set of idempotent elements. It was proved in 1 that dim U Ž and then dim V. e e does not depend on the choice of the idempotent element, so we can define type A Ž m,., where m 1 dim U and dim V Ž so n dim A m. e e. Prod- ucts between elements of Ue and Ve satisfy the relations Ue V e, UV e eu, e Ve U, e Ž 6. 3 Ž u. u Ž u. u už u. už. Ž u. u 0 Ž 7. for all u Ue and V e. On the other hand, I A euu u U 4 and if e e u u is another idempotent, then e 4 4 Ue u uu u U e, Ve u u V e. 8 Moreover dim U and dimžuv V. e e e e do not depend on the choice of the idempotent element. So we will use the following definitions introduced by Lyubich Žsee 1, 16, 1. : a Bernstein algebra in which UV e e Ž. V 0 is called regular and a Bernstein algebra in which U Ž. e e 0 is said to be exceptional. Every Bernstein algebra of types Ž m,. with m 1 or less than or equal to 1 is regular or exceptional. Further, an algebra A where A A is called nuclear. We now define the basic concept of inariant linear form. This is a linear form, f, on a Bernstein algebra which satisfies the identity fž x. Ž x. fž x. Ž 9. for all x A Žsee 1 for more information.. If we denote by J invariant linear forms of A, then its orthogonal is A the set of J UV V V. Ž 10. A e e e e

4 BERNSTEIN PROBLEM 43 Finally, we will denote by U0 the subset Ue ann U e. This subset is independent of the idempotent considered and is an ideal of A. On the other hand, A AU is a JordanBernstein algebra Žsee 0 5 for general information about U and JordanBernstein algebras. 0. We call a s.e.o. regular, exceptional, orof type Ž m,. if the corresponding stochastic Bernstein algebra is regular, exceptional, or of type Ž m,., respectively. So a s.e.o. is called degenerate if there exists i such that x i 0. Bernstein solved the Bernstein problem for n 3. ŽThe case n is trivial.. For all n 3 the Bernstein problem in the so-called regular case and the exceptional case was completely solved by Yu. Lyubich in a series of papers Žsee also his book 1.. In 7 we solved the problem for type Ž 3,. in the nonregular and nonexceptional case and therefore we solved completely the problem for n 5. Finally, in 10 we solved the Bernstein problem for type Ž n,., for all n, in the nonregular nonexceptional nonnuclear case. Now, we will recall some definitions of Lyubich that are very useful in the solution of this problem Žsee 1, pp. 8, 33 for more information.. Let V be a Bernstein evolutionary operator. Consider an arbitrary face of the basis simplex and let C Int Im V Ž as usual, Int is the interior of the face with respect to its affine hull.. We will call the face essential if C. An essential face is called a k-dimensional essential face if C has dimension k as topological space Ž we say k-essen-. n tial for simplicity. On the other hand, if x Ýi1ie i, i 0 the support of x, denoted suppž x., is the set of basis vector Žor their corresponding indices. for which the coefficient is positive, i.e., i suppž x. if and only if i 0. We can prove the following generalization of Lemma 5.7. of 1 : LEMMA 1.1. Let V be a Bernstein eolutionary operator of type Ž m,.. Then it contains at least m k k-dimensional essential faces, for k 0, 1,..., m 1. Now, we obtain the following LEMMA 1.. Let V be a Bernstein eolutionary operator, and two essential faces of V, and 0-dimensional. Then, implies that. Besides, if is also 0-essential, then either or. Proof. Let x be an element of. Since every essential face is invariant with respect to the evolutionary operator, we have that x Ž.. Consequently, supp x. On the other hand, x C and Ž therefore supp x. is equal to e e 4. i i

5 44 J. CARLOS GUTIERREZ FERNANDEZ We remark that if e,...,e 4 1 n is a basis of a Bernstein algebra with ij,k the corresponding multiplication constants, i.e., eeý i j l ij, le, l then the basis is stochastic if and only if ij,k 0 and Ýlij, l 1 for i, j, k 1,...,n. Thus, let be stochastic; we can consider ij,k as the hereditary coefficients of a stationary evolutionary operator. In the following we will take the elements of a stochastic basis as vectors of the algebra or as types of the s.e.o. indiscriminately. Also, we will call a stochastic basis degenerate if its corresponding s.e.o. is degenerate. A vector ek of the stochastic basis is a disappearing type if x k 0 in the corresponding s.e.o., so ek is disappearing if and only if 0 for i, j 1,...,n. ij,k LEMMA If e,...,e 4 1 n is a stochastic basis of a Bernstein algebra, then Ž e. 1 for i 1,...,n. i Therefore, we obtain the following LEMMA 1.4. If e 4 is a basis of a Bernstein algebra and Ž e. i i 1 for all i, then the basis is stochastic if and only if the coordinates of ej e 1 j with respect to are either greater than or equal to 0 for all j1 and j.. STOCHASTIC BASIS FOR TYPE 3, 3 It can be proved that if A is a nonregular nonexceptional Bernstein algebra of dimension 6, then the type of A is either Ž 4,. or Ž 3, 3.. ŽRemember that the Bernstein problem in the regular and exceptional case was solved by Lyubich.. The type Ž 4,. was solved, in 10, in the nonregular nonexceptional nonnuclear case. Also, A. Grishkov proved that if a Bernstein algebra is of type Ž 4,. nuclear and stochastic, then it is regular. Thus, the Bernstein problem for type Ž 4,. has been completely solved. In this section we will describe all stochastic bases of a nonregular nonexceptional Bernstein algebra of type Ž 3, 3. and therefore, the Bernstein problem will be completely solved in this paper for n 6. For a finite sequence x, x,..., x 4 1 r of elements of a vector space we write ² x, x,..., x : 1 r as the vector subspace spanned by and denote by x, x,..., x the convex hull of. 1 r LEMMA.1 0. Let A be a nonregular nonexceptional Bernstein algebra 3 of type 3, n 3. If e is an idempotent element of A, then U Ž. e 0 and UV e eu 0. Ž. Ž. Ž. Proof. It is enough to prove that U0 0.If Ve 0, then 0 Ve U. In the other case, V Ž. 0 e 0 and as the algebra is nonexceptional UV Ž. e e 0. Therefore, there exist u and such that u 0. If u and u are linearly independent, they are a basis of U and by Ž. 7 it follows that e

6 BERNSTEIN PROBLEM 45 u U 0.If uand u are linearly dependent, we can assume u u, then už u. U. 0 An immediate consequence of the previous lemma is that if A is a nonregular nonexceptional Bernstein -algebra of type Ž 3, n 3., then dim U 1 and type A Ž 3, In the following A will be a Bernstein -algebra of type Ž 3, n 3. and with dim U 1. A basis e, u, u,,..., n3 of A is called standard if e is an idempotent element, u 1, u a basis of U e, 1,...,n3 a basis of V, u U, and u. ŽSee Proposition 4 of e for more information about standard bases of A.. We will denote by j, k and i, k the scalars where j kj, ku and ui k i, ku for j, k 1,...,n3 and i 1,. For every standard basis, we denote by the scalar Ý ž Ý i, k Ý j, k/ kn3 i1, 1jk. PROPOSITION.1 6. Let e, u, u,,..., n3 be a standard basis of A. Then Ž. i a basis of A of the form e, u, u,,..., n3 is standard if and only if there exist, and such that u u u, u 1 1 u and 1 1; ii if e e u u is another idempotent of A, u u1 u, then the ectors u u, u u, ii 1, i, i 1, i u, i,...,n3 form a standard basis of A. LEMMA.. Let e 1, e be two linearly independent elements of A such that with respect to a standard basis e1 e 1u, e e u, and e u e, e. Then the following conditions are equialent: 1 Ž. i The set e, e 1 is 0-essential. Ž ii. 0 and 1, 1 4Ž.. 1,,,, LEMMA.3. Let e 1, e be two linearly independent elements of A such that with respect to a standard basis e1 e u1 1u 1, e e u u, and e u e, e Then the following conditions are equialent: Ž. i The set e, e 1 is 0-essential. Ž ii. 0 and, Ž. 1 1, 1,,,,.

7 46 J. CARLOS GUTIERREZ FERNANDEZ LEMMA.4. Let e 1, e, e3 be three linearly independent elements of A of weight 1 such that e and e, e are 0-essential, e, e, e is 1-essential, and e, e A. Then f e f e f e for all f J U A 0 We now suppose that dim A 6 and e 4 i is a stochastic basis of A. Since dim A 4 the stochastic basis has at most two disappearing types. If has two disappearing types, we can assume that they are e5 and e 6; 4 then e 1, e, e 3, e4 is a stochastic basis of A. If has exactly one disappearing type, we can assume that it is e ; then e,...,e is a stochastic basis of a subalgebra of A that contains A, and therefore is a stochastic basis of a Bernstein subalgebra of type Ž 3,. with dim U0 1. By Lemma.1, A is a regular Bernstein algebra of type Ž 3, 1.. As a reformulation of a result obtained by Lyubich in 13 Žsee also. 8, we have the following THEOREM.1. where 0 and 1. 1 The stochastic bases of A hae the forms e e 1 e e u e eu e eu, By Lemma 1.1 we have the following. LEMMA.5. Eery stochastic basis in A, dim A 5, contains an idempotent element. LEMMA.6. Eery stochastic basis in A, dim A 5 contains two elements in A. Proof. Suppose e 4 i is a stochastic basis in A, with a unique element in A. By Lemma 1.1 for an arrangement of the stochastic basis the faces e, e, e, and e, e are 0-essential, and e, e, e is 1-essential. Suppose now that e is the idempotent in e, e 3. Since the vectors of have weight 1, they are of the form e e z with z kerž. i i i.by Proposition.1 and Lemma.3 there exists A such that e e and e3 e and by Lemma.4 we obtain that e1 e u with u U. Also by Proposition.1 the idempotent of e, e is of the form eu and therefore there exist scalars such that 1 1 e eu u, e eu u

8 BERNSTEIN PROBLEM 47 Since the coordinates of the vectors ee4 and ee5 with respect to the stochastic basis are greater than or equal to 0, we obtain that 1 0. Finally, considering the coordinates of ee, we obtain a contradiction. THEOREM.. Let A be of dimension 5. If e,...,e is a stochastic basis of A, with at most three elements in A, then there exist a standard basis and an arrangement of the stochastic basis, such that the basis has one of the two following forms: 1. The form is e1 e u 1 4 e e e e 3 e eu e eu, where 1 0, 1 0, 1, 1,, 1, 0, and 1, 1.. The form is e1 e e e u e eu e eu e eu, 5 1 Ž. where 1 1 0, the ectors of, Ž 0, and Ž 1, 0., belong to the set Ž,., Ž,., Ž 0,1., and A is regular. 1 1 Proof. If A is nonregular, then the theorem was proved in 7. In the rest of the proof we will suppose that A is regular. Now we have two cases: 4 1. If ei is a stochastic basis with exactly three elements in A, then we can assume that e 1, e, e3 A. There exists a unique element that belongs to A and e, e, which we denote by e. The set e, e, e, e is a stochastic basis of A. By Theorem.1 and Lemmas 1.3 and 1.4, we obtain the desired result. 4. If ei is a stochastic basis with exactly two elements in A, then by Lemma 1.1 after a suitable arrangement we have one of the two following possibilities: Ž. a Faces e, e, e, e are 0-essential. Ž b. Faces e, e, e, e, e are 0-essential

9 48 J. CARLOS GUTIERREZ FERNANDEZ Ž. a If e is the idempotent of e 3, e 4, then there exist A and 0 such that e e and e e Either e1 or e is not in e U0 Ve, for otherwise there exists f, an invariant linear form such that fž e. fž e. i 5 for i 1,, 3, 4, and therefore the vector e5 would be an idempotent, but this is impossible since e5 A. Thus, we can assume that e e u and e e u u with u U and u U. The vector u is different from 0, for otherwise e5 would be an idempotent. Therefore the stochastic basis with respect to the standard basis e, u, u,, 4 is of the form 1 1 e e u, e e u u, e e, e e, e e u u, where 0 and 0. Now, as the multiplication constants of the 1 stochastic basis are greater than or equal to 0, we obtain ee, 0, e 0 and 0 1, 5 1 ee 1, 1 e e 0. 3 Therefore, considering e e 1, we obtain by Proposition.1 that the basis has the desired form. Ž b. For a suitable standard basis e e u, e e u u, and e e for i 3, 4, 5. Multiplying 1 i i 1 i e by e we obtain that 1. Therefore, for e e the stochastic basis 1 1 has the form of the theorem..1. The Degenerate Case with Two Disappearing Types If the algebra A is nonregular and dim A 6, then every stochastic basis of A with two disappearing types has the form e e 1 e e u e eu e eu e e u u e e u u, 6 1 3

10 BERNSTEIN PROBLEM 49 where 0 and the following inequalities hold, 1 0, 1, Ž ; 0, Ž 1. i i 1 1, Ž 13. i, i1 0 1, Ž 14. i 1, i1 j 1, i1 ij 0 Ž i 1, j1 i, j1. 1, Ž 15. j 1, i1 j, i1 i1, j1 for, 4 and i, j 1, 4 i i i i. If e 4 i is a stochastic basis of A with e5 and e6 the disappearing types, then e 1, e, e 3, e4 is a stochastic basis of A. By Lemma 1.4 the inequalities Ž 11. Ž 15. hold... The Degenerate Case with Exactly One Disappearing Type If the algebra A is nonregular and dim A 6, then every stochastic basis of A with exactly one disappearing type has one of the following forms: 1. If there are two integers r, s such that x and x r s are proportional and nondegenerate, then e1 e u eeu111 e3eu11 e e 4 e e 5 e eu u, where 0 and the following inequalities hold, 1, 1,, 0, Ž ; 1 0; 1, Ž 17. 0; 0,3,,3, Ž 18. 1,,,3 1; 0 Ž 1,3 i 1,3. 1, Ž for, and i 1,. 0Ž 1,3,3. 3,3 1, Ž 0.

11 430 J. CARLOS GUTIERREZ FERNANDEZ. If there is no pair of integers r, s such that x and x r s are proportional and nondegenerate, then e1 e e e u e eu e eu e eu 5 1 e eu u, where 0 and the following inequalities hold, 1,, 1,, 0, Ž 1. Ž 1. Ž 1. 0, 1 1 Ž 0,0., Ž 1,0. Ž 1, 1., Ž,., Ž 0,1., Ž 3. 0; 0; 1,3 1, Ž 4. 0Ž 1,3 i 1,3 i,3. 1, Ž 5. 0Ž 1,3,3. 3,3 1, Ž for 0, 1,,, and i 1,, We remark that if e,...,e is a stochastic basis of A with e6 the unique disappearing type, then this basis has at most three elements in A and e,...,e is a stochastic basis of a subalgebra of A that contains A. Therefore, has one of the two forms of Theorem..3. The Nondegenerate Case Stochastic Bases without Disappearing Types If A is nonregular and e, e,...,e is a stochastic basis of A with the corresponding s.e.o. nondegenerate, then there exist a standard basis and an arrangement of the stochastic basis, such that the basis is of one of the following three forms: 1. The form is e1 e u1 1 1 eeu11 e e 3 e e 4 e e u e e u, 6 3

12 BERNSTEIN PROBLEM 431 where 0 and the following inequalities hold, 1, 1, 1,3, 0, Ž 7. ; 4Ž. ; 1,3,3 3,3,3 11,3; 1,3, Ž 8. 1,0; 1 0; 1, Ž 9. 0ijŽ i, i,3., Ž for,, 1, and i, j 1, The form is e e u 1 e e e e 3 e eu 4 1 i e eu e eu, where 0 and the following inequalities hold, 1, 1,3,3 1,3 1,,,3 3,3 0, Ž The form is Ž 1. Ž 1. 0, Ž Ž 0,0., Ž 1,0. Ž 1, 1., Ž,., Ž 0,1., Ž ; 1,, 1, 1. Ž 34. e e u 1 e eu e eu e e 4 3 e e 5 e e, 6 3 where 0 and the following inequalities hold, 1, 1,3 1, 1,3,,3 3,3 0, Ž 35. 1,0; 1 0; 1, Ž 36. 1,, 1. 37,,3,,3

13 43 J. CARLOS GUTIERREZ FERNANDEZ Besides, every family of the above vectors is a stochastic basis of A if it verifies the corresponding inequalities. Now, we will prove that these bases give us a full description of all nondegenerate stochastic bases of A Stochastic Bases without Idempotents Let e 4 i be a stochastic basis of A without idempotent elements. By Lemma 1.1 Ž after a reordering of the stochastic basis. we have the following properties: Ž. i e, e, e, e, e, e are 0-essential faces and Ž ii. e, e, e, e, e, e, e, e are 1-essential faces Now, we will distinguish two cases: 4 1. If e 1, e A, then e 1, e, e 3, e5 is a stochastic basis of A.By Lemmas. and.3 and Theorem.1 we obtain that this basis belongs to Ž. the form 1. The result is analogous if e 3, e4 or e 5, e6 belongs to A.. If e i, ei1 A for i 1, 3, 5, then there exists a standard basis where e1 e 1, e4 e u1 4u 4 1 4, ee, e5e5u5 3, e eu u, e e u, and e u e, e, e u e, e Since ee3 and ee4 are contained in the face e, e, e, e 1 3 4, scalars 1 and are equal to zero. Finally, considering the product e e we obtain a contradiction Stochastic Bases with Exactly One Idempotent and One Element in A Reordering the basis we have the following cases: 1. Faces e, e, e, e, e, e are 0-essential. If e, e, e is a 1-essential face, then there exists a standard basis such that e e u, e e u u, ee, e5eu15u5155 3, e3e, e6eu16u 3, where 0 and e, e, e e u By Lemma 1.1 one of the faces e, e, e, e, e, e, e, e, e is a 1-essential face. Now taking the

14 BERNSTEIN PROBLEM 433 products ee for k 4, 5, 6, e 4, and ee we see that k Therefore, the basis belongs to the form Ž.. If e, e, e 1 3 is not a 1-essential face, then e, e, e, e, e and e, e, e, e are 1-essential faces. We consider a suitable standard basis e1 e u1 1, e4 e 4u1 4u , e e, e e u u, e e, e e u u, where 0 and e e u u. Now we obtain a contradic tion as the following schema shows: ee1 k 4,5,64k 0, eeiž i 4,5,6. 4 0, ee 1 4 1e,e,e and 4 5 0, ee e e, e, e Faces e, e, e, e, e are 0-essential. Ife, e, e is a 1-essential face, then for a suitable standard basis e1 e u, e4 e u1 4u 1 4 3, e e, e eu u, e e, e e u u, where 0, e5 e u1 1, and The set ² e,u : 1,u, e 4, e 5, e6 contains exactly one vector, which we denote by e e u1 u.as ee, ee, ee, and ee belong to e, e, e, e, we conclude that , and the basis belongs to the form Ž.. On the other hand, if e, e, e 1 i i1 is not a 1-essential face for i equal to and 4, then e, e, e j j1 6 is not a 1-essential face for j equal to and 4 and also either e, e, e, e or e, e, e, e is a 1-essential face Žwe will suppose that e, e, e, e is 1-essential, the other case being analogous.. For a suitable standard basis e e u, e e u u, e e, e e u u, e e, e e u,

15 434 J. CARLOS GUTIERREZ FERNANDEZ where 0 6 1, 4 1, 6 0, and e4 e 4u1 u 4 1. The set e, e, e, e, e 4 is stochastic, but by Theorem. this is impossible Nondegenerate Stochastic Bases with One Idempotent and Exactly Two Elements in A Reordering the basis we have the following possibilities: 1. faces e, e, e, e, e, e are 0-essential;. faces e, e, e, e, e are 0-essential; 3. faces e, e, e, e are 0-essential; 4. faces e 1, e 3, e 4, e 5, e6 are 0-essential and e A ; 5. faces e, e, e, e, e, e are 0-essential and e A If our basis verifies case 1, then A is regular, which is impossible.. When the basis verifies case, then with respect to a suitable standard basis e1 e 1u1 u 1 1, e4 e, eeu1 1, e5e 3, e e, e e u u, where e, e, e e 4, 1 1,, 0, , and 6 0. Multiplying e by e we see that 6 0. So the products ee 1 and eeimply 1 11 and Now considering the products of the elements of the stochastic basis we see that A is regular, but this is impossible. 3. If the basis verifies one of the cases 3, 4, or 5, then the set e, e Ž² e, e : A contains exactly one vector, which we denote by e. It is clear that e, e, e, e, e is stochastic. Using Theorem. we can see that the basis is one of the following: Ž. a The basis is e1 e u, e4 e, eeu1 1, e5eu15u515 3, e3e3, e6eu16u61 3, where 0 and e e u. Since ee and ee belong to ,it follows that and so the basis is of the form Ž.. b The basis is e1 e u1 1, e4 e, eeu1 1, e5e5u5 3, e e, e e u,

16 BERNSTEIN PROBLEM 435 where 0 and e e u. It is clear that the basis is of the form 1. Ž. c The basis is Ž. e e, e e u, e eu, e eu, e eu, e eu u, where e e u. Since ee and ee belong to it follows that Also, we can see that the algebra is regular, but this is impossible. Ž d. The basis is e1 e u, e4u141, eeu1 1, e5e5 3, e eu, e e, Ž. where e e. The basis is of the form Stochastic Nondegenerate Bases with One Idempotent and Exactly Three Elements in A We can consider the basis with e1 idempotent element and e and e3 in A. We denote by e the unique element in e 4, e 5, e6 A. It is clear that 4 e, e, e, e is a stochastic basis of A and e, e,...,e e,e,e,e According to Theorem.1 we have two possibilities: 1. The basis is e1 e, e4 e u1 4u, eeu, e5eu15u 3, e eu, e eu u, where 0 and e 6 6 e u1 1. The products ee 1 i for i 4, 5, 6 imply that Considering the products of the vectors of the stochastic basis we can see that the algebra is regular, which is impossible.. The basis is e1 e u, e4 e 4u, eeu1 1, e5e5u 3, e eu, e e u,

17 436 J. CARLOS GUTIERREZ FERNANDEZ where 0 and e 6 6 e. It is easy to prove that , considering the products eei for i 4, 5, 6. Consequently, the basis is of the form Ž THE BERNSTEIN PROBLEM FOR TYPE 3, 3 Now we will formulate our main result about the Bernstein problem. It is a reformulation of the previous results. THEOREM 3.1. Let V be a nonregular nonexceptional nondegenerate stationary eolutionary operator of type Ž 3, 3.. Then it has one of the following three forms: 1. The form is x 1 Ž s p. p 1 1 x Ž p s. p x 3Ž s ps. x 41Ž s ps. x 5 x, 6 1 where s Ý 6 x and p x x are inariant linear forms and i1 i 1 p Ž 1 1. x5ž 1. x 6, q11x3x 4, q1x5x 6, Ž 38. sp Ž p p q q p. q1 qq, 1 Ž ; 01 ; 111, Ž 40. 0; 1 ; 0, Ž i1 j i i j i 1 Ž 1. Ž 1. Ž 1. Ž. 1 Ž i, j1,.. Ž 4.

18 BERNSTEIN PROBLEM 437. The form is x x s Ž x x x Ž x x. sx Ž x x x Ž x x. sx Ž x x x Ž 1. p s p 4 x Ž 1. p s p 5 1 x Ž. sž. p p, where s Ý 6 x and p x x x are inariant linear forms and i1 i Ž 1. 1 Ž 11., Ž ; 11; 1 0; 1, Ž 44. Ž 0,0., Ž 1,0. Ž 1, 1., Ž,., Ž 0,1., Ž 45. 1,, 1; 0; 0. Ž The form is x Ž s p. p 1 1 x Ž p s. p x s Ž. x x x x x Ž x x x. s Ž Ž. x x x. x x Ž x x x. s Ž Ž. x x x. x x6 Ž Ž x4x5x6. s Ž Ž. x4 x6 x6. x 3., 1 where s Ý 6 x and p x x are inariant linear forms and i1 i 1 1,0; 1 0; 1, Ž 47. 1,, 1; 0. Ž 48.

19 438 J. CARLOS GUTIERREZ FERNANDEZ Remark. By Ž 10. and Lemma.1, if V is a nonregular nonexceptional s.e.o. of type Ž 3, 3., then its space of invariant linear forms, J V, has dimension. Consequently if V belongs to one of the above three forms, then J ² s, p :. V It remains to note that from.1 and. also we can describe explicitly all nonregular nonexceptional and degenerate s.e.o. s of type Ž 3, 3.. We conclude that these evolutionary operators are not normal in the sense of 1, p This is a new support for the Lyubich conjecture 1, p. 3. According to the Lyubich conjecture, if V is a normal s.e.o., then the corresponding Bernstein algebra is regular. A nondegenerate s.e.o. V is normal if there is no pair j, j such that x and x 1 j j are proportional 1 and there is no pair j, j such that all x s depend only on x x, x 1 j j1 j i Ž ij, j.. 1 REFERENCES 1. S. N. Bernstein, Mathematical problems in modern biology, Sci. Ukraine 1 Ž 19., S. N. Bernstein, Principle de stationarite et generalisation de la loi de Mendel, C. R. Acad. Sci. Paris 177 Ž 193., S. N. Bernstein, Demonstration mathematique de la loi d heredite de Mendel, C. R. Acad. Sci. Paris 177 Ž 193., S. N. Bernstein, Solution of a mathematical problem related to the theory of inheritance, Uchen. Zap. n.-i. kaf. Ukrainy 1 Ž 194., in Russian 5. S. Gonzalez and C. Martınez, Idempotent elements in a Bernstein algebra, J. London Math. Soc. 4 Ž 1991., S. Gonzalez, J. C. Gutierrez, and C. Martınez, Classification of Bernstein algebras of type Ž 3, n 3,Comm.. Algebra 3, No. 1 Ž 1995., S. Gonzalez, J. C. Gutierrez, and C. Martınez, The Bernstein problem in dimension 5, J. Algebra 177 Ž 1995., S. Gonzalez, J. C. Gutierrez, and C. Martınez, On regular Bernstein algebras, Linear Algebra Appl Ž 1996., A. Grishkov, The Lyubich conjecture for the type Ž 4,., preprint. 10. J. C. Gutierrez Fernandez, The Bernstein problem for the type Ž n,., J. Algebra 181 Ž 1996., P. Holgate, Genetic algebras satisfying Bernstein s stationarity principle, J. London Math. Soc. 9, No. 4 Ž 1975., Yu. I. Lyubich, Basis concepts and theorems of evolutionary genetic for free populations, Russian Math. Sureys 6, No. 5 Ž 1971., Yu. I. Lyubich, Structure of Bernstein populations of the type Ž n 1, 1., Uspekhi Math. Nauk 8, No. 5 Ž 1973., in Russian 14. Yu. I. Lyubich, Two-level Bernstein populations, Math. USSR-Sb. 4, No. 4 Ž 1974., Originally published in Mat. Sb. 95Ž 137., No. 4 Ž 1974., Yu. I. Lyubich, Structure of Bernstein populations of the type Ž, n,uspekhi. Math. Nauk 30, No. 1 Ž 1975., in Russian 16. Yu. I. Lyubich, Quasilinear Bernstein populations, Teor. Funktsil Funktsional. Anal. Appl. 6 Ž 1976., 7984.

20 BERNSTEIN PROBLEM Yu. I. Lyubich, Proper Bernstein populations, Problemy Peredachi Informatsii 13 No. 3 Ž 1977., Transl. Problems Inform. Transmission Ž Jan , 835. in Russian 18. Yu. I. Lyubich, Bernstein algebras, Uspekhi Math. Nauk 3, No. 6 Ž 1977., 616. in Russian 19. Yu. I. Lyubich, Classification of some types of Bernstein algebras, Vestnik Khar ko Uni. Ser. Mech. Mat. 05 Ž 1980., Transl., Selecta Math. Soiet 6, No. 1 Ž 1987., Yu. I. Lyubich, Classification of nonexceptional Bernstein algebras of type Ž 3, n 3,. Vestnik Khar ko Uni. Ser Mech Mat. 54 Ž 1984., 364. Transl. Selecta Math. Soiet 11, No. 1 Ž 199., Yu. I. Lyubich, Mathematical Structures in Population Genetics, Naukova Dumka, Kiev, 1983 in Russian. Transl., Lecture Notes in Biomathematics, Vol., Springer- Verlag, BerlinHeidelberg, A. Worz-Busekros, Algebras in Genetics, Lecture Notes in Biomathematics, Vol. 36, Springer-Verlag, New York, 1980.