推廣施-董的組合固定點定理至有限分配格

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(1)國立臺灣師範大學數學系博士班博士論文. 指導教授：陳界山博士 (指導教授) 施茂祥博士 (共同指導教授). Generalization of Shih-Dong's combinational fixed point theorem to finite distributive lattices. 研究生：吳樹恆. 中華民國 104 年 7 月.

(2) Generalization of Shih-Dong’s combinational fixed point theorem to finite distributive lattices Shu-Han Wu. Abstract Shih-Dong’s combinational fixed point theorem asserts that if a map from the n-dimensional hypercube into itself satisfies that all the Boolean eigenvalues of the Boolean Jacobian matrix are zero for each element in the hypercube, then it has a unique fixed point. Its equivalent contrapositive form has biological implications. Our goal is to provide an extension of Shih-Dong’s theorem into all finite distributive lattices. Our method of proof is based on Shih-Dong’s “collective effect method” as well as G. Birkhoff’s representation theorem for finite distributive lattices.. Keywords. Discrete dynamical system, Finite distributive lattice, Fixed point, Generalized Boolean Jacobian matrix, Negative circuit, Positive circuit.

(3) Contents 1 Introduction. 1. 2 Definitions and Notations. 3. 3 Fixed points and circuit-free. 8. 4 Fixed points and positive circuits. 17. 5 Fixed points and negative circuits. 19. 6 Concluding remarks. 25. 7 Related open questions. 30. Reference. 32.

(4) 1. Introduction. The Jacobian Conjecture is a long-standing problem in algebraic geometry. It was first stated in 1939 by O. H. Keller [5]. Conjecture 1.1 (The Jacobian Conjecture) Suppose F : Cn → Cn is a polynomial map with the property that the derivative at each point is nonsingular. Then must F be one-to-one. Here F (z) = (f1 (z), · · · , fn (z)), each fi is a polynomial in n variables, and z ∈ Cn . The Jacobian matrix of f at z is denoted by F ′ (z). If F is indeed injective, then it is surjective and has an inverse which is a polynomial map. For an elementary proof of this see [11]. See the excellent survey [1] for the importance, background, and related results. S. Smale lists the Jacobian Conjecture as one of 18 great problems for 21 century [14]. In 1997, Cima, Gasull, and Ma˜ nosas [4] established a fixed point conjecture which is equivalent to the Jacobian Conjecture. It states that if F : Rn → Rn is a polynomial map with ρ(F ′ (x)) < 1 for each x ∈ Rn , then F has a unique fixed point. Here ρ(F ′ (x)) denotes the spectral radius of F ′ (x). In the study of automata networks, Shih and Ho [12] raised the Boolean counterpart conjecture of the fixed point conjecture. It states that if F : {0, 1}n → {0, 1}n is a Boolean map with ρ(F ′ (x)) = 0 for each x ∈ {0, 1}n , then F has a unique fixed point. Here F ′ (x) is the Boolean Jacobian matrix of F at x and ρ(F ′ (x)) is the Boolean spectral radius of F ′ (x), see [10] and [12]. In 2005, Shih and Dong used a collective effect method with an intricate argument to prove Shih-Ho’s conjecture. Hence Shih-Dong’s combinatorial theorem is stated as follows. Theorem 1.2 (Shih-Dong) Let F : {0, 1}n → {0, 1}n . If ρ(F ′ (x)) = 0 for each x ∈ {0, 1}n , then F has a unique fixed point. Let us remark that every function F : {0, 1}n → {0, 1}n , F = (f1 , · · · , fn ), is a Boolean polynomial, that is, each fi is a Boolean polynomial. To see this, let pi : {0, 1}n → {0, 1}n be defined by p1 (t) = t¯, p2 (t) = t, and α1 = 0, α2 = 1. Then pj (αi ) = 1 if i = j, otherwise 0. Thus for i = 1, · · · , n, ∑ fi (x1 , · · · , xn ) = pj1 (x1 ) · · · pjn (xn )fi (αj1 , · · · , αjn ). j1 ,··· ,jn ∈{1,2}. 1.

(5) The equivalent contrapositive form of Theorem 1.2 states that if F : {0, 1}n → {0, 1}n has multiple fixed points or has no fixed point, then there exists a point x ∈ {0, 1}n such that the corresponding network Γ(F ′ (x)) has a circuit. This coincides with the Thomas conjecture which was stated in 1981 by the biologist René Thomas. When studying genetic regulatory networks, biologists often represent the results of their genetic and molecular investigations in terms of finite signed directed graphs. The vertices correspond to the members of the network (e.g., genes, RNA, proteins) and a positive (resp. negative) edge from i to j means that the member i activates (resp. represses) member j. These graphs are so called interaction graphs and biologists often use them as a basis to design dynamical models, using either a differential or a discrete framework [17]. In 1981, the biologist R. Thomas conjectured that: a necessary condition for multistationarity is that the interaction graph has a positive circuit and a necessary condition for sustained oscillations is that the interaction graph has a negative circuit, i.e. the sign of a circuit being defined as the product of the signs of its edges [16]. The multistationarity means that there exist several fixed points in the dynamics. Multistationarity is an important dynamical property since it is related to cell differentiation in biology [15, 16, 17, 18, 19]. Shih-Dong’s theorem has been generalized and studied by Richard [7, 8, 9], Remy, Ruet, Thieffry [6], and C. Soulé [15]. The purpose of this paper is to provide an extension of Shih-Dong’s theorem into all finite distributive lattices.. 2.

(6) 2. Definitions and Notations. This section is to present a conceptual framework for our generalized results. We state some notations and results concerning the spectra of Boolean matrices. The material can be found in the book by Robert [10]. Let {0, 1} be with three operations +, ·,¯defined as follows: 0 + 0 = 0 · 1 = 1 · 0 = 0 · 0 = ¯1 = 0, 1 + 0 = 0 + 1 = 1 + 1 = 1 · 1 = ¯0 = 1. For a, b ∈ {0, 1}, we usually suppress the dot “·” of a · b and simply write ab. For n ∈ N, let {0, 1}n be the set of ordered n-tuples,   x1  .  .  x=  . , xn with components xi ∈ {0, 1} for each i = 1, · · · , n. We also write x = (x1 , · · · , xn ) interchangeably. The zero element of {0, 1}n is the point 0, all of whose coordinates are 0. The order “ ≤ ” in {0, 1} is given by 0 ≤ 0 ≤ 1 ≤ 1. Thus for a, b ∈ {0, 1}, a + b = max{a, b},. ab = min{a, b}.. For x, y ∈ {0, 1}n , x ≤ y is meant that xi ≤ yi for each i = 1, · · · , n. For x, y ∈ {0, 1}n and λ ∈ {0, 1}, define     max{x1 , y1 } min{λ, x1 }     .. ..  , λx =  . x+y = . .     max{xn , yn } min{λ, xn } A Boolean matrix is meant to be a matrix over {0, 1}. Boolean matrix addition and Boolean matrix multiplication are the same as in the case of complex matrices but the concerned sums and products of entries are Boolean. Let F : {0, 1}n → {0, 1}n and let us write F = (f1 , · · · , fn ). According to Robert [10], the incidence matrix of F is the n × n Boolean matrix defined by B(F ) = (bij ), where bij = 0 if fi does not depend on xj , bij = 1 otherwise. Let A be an n × n Boolean matrix. A nonzero element 3.

(7) u ∈ {0, 1}n is called a (Boolean) eigenvector of A if there exists λ ∈ {0, 1} such that Au = λu; λ is called the (Boolean) eigenvalue associated with the eigenvector u. The symbol σ(A) stands for the set of all (Boolean) eigenvalues of A, so that σ(A) ⊂ {0, 1}. The Boolean spectral radius of A, which is denoted by ρ(A), is defined to be the largest (Boolean) eigenvalue of A. Because σ(A) ̸= ∅ (this fact is not a priori obvious, see[10]), ρ(A) = 0 or 1. Also ρ(P t AP ) = ρ(A) for any permutation matrix P . For x ∈ {0, 1}n , let   x1  .   ..      x ˜j =  x ¯j  .  .   .   .  xn The notation x ˜ji is meant that x ˜ji = x ¯j if i = j, x ˜ji = xi if x ̸= j. For x ∈ {0, 1}n and {j1 , · · · , jk } ⊂ {1, · · · , n}, let us define x ˜j1 ,··· ,jk = y by { xi , if i ̸= j1 , · · · , jk , yi = x ¯i , if i = j1 , · · · , jk . The Boolean Jacobian matrix of F at x ∈ {0, 1}n is the (Boolean) n × n matrix defined by F ′ (x) = (fij (x)), where fij (x) = 1 if fi (x) ̸= fi (˜ xj ), fij (x) = 0 otherwise. The discrete metric on {0, 1} is denoted by δ, that is, δ(x, y) = 1 if x ̸= y, δ(x, y) = 0 if x = y. For x, y ∈ {0, 1}n , the Boolean vector distance d(x, y) is defined by   δ(x1 , y1 )   .. . d(x, y) =  .   δ(xn , yn ) Let us recall that the Boolean vector distance d satisfies: (i) d(x, y) = d(y, x) (x, y ∈ {0, 1}n ), (ii) d(x, y) = 0 ⇔ x = y (x, y ∈ {0, 1}n ), (iii) d(x, y) ≤ d(x, z) + d(z, y) (x, y, z ∈ {0, 1}n ), where d(x, z) + d(z, y) is the Boolean sum in {0, 1}n . 4.

(8) The digraph (directed graph) of an n × n Boolean matrix A = (aij ), denoted by Γ(A), is the digraph having the vertex set {1, · · · , n} and a directed edge (j, i) from j to i if aij = 1. A directed path in Γ(A) is a sequence of directed edges i1 i2 , i2 i3 , · · · in Γ(A). A circuit C = [c1 , . . . , ck ] in a directed graph is a sequence of vertices which have edges from ci to ci+1 for each i = 1, . . . , k and c1 = ck . We now state some basic results concerning the spectral theory of Boolean matrices. Theorem 2.1 The following conditions are mutually equivalent: (i) (ii) (iii) (iv). ρ(A) = 1. A contains a principal submatrix which has no zero rows. A contains a principal submatrix which has no zero columns. Γ(A) contains a circuit.. Theorem 2.2 The following conditions are mutually equivalent: (i) ρ(A) = 0. (ii) There exists a permutation matrix P such that P t AP is strictly upper triangular. (iii) There exists a positive integer p ≤ n such that Ap = 0. For a finite lattice L, let us recall that the diagram of L is a graph whose vertices are elements of L and the edges correspond to the covering relation. If x covers y or y covers x then there is no z ∈ L such that y < z < x or x < z < y. We define the metric dp on the diagram of L as follows. We say that P = {p0 , p1 , · · · , pk } is a path of length k which connects p0 and pk if there has an edge between pi and pi+1 in the diagram of L for each i = 0, · · · , k −1. Denote by l(P ) the length of P . In particular, the singleton {x} is a path from x to itself of length 0. For x, y ∈ L, define dp (x, y) = min{l(P ) : P is a path connecting x and y}. Then dp is a metric on L. Note that if dp (x, y) = 1, then x covers y or y covers x. Let us recall that an element a of a lattice is join-irreducible if a = x ∨ y implies that a = x or a = y. We denote the set of join-irreducible elements 5.

(9) of L as J(L) = {a1 , · · · , an }. For each x ∈ L, let η(x) = {ai : ai ≤ x} be the set of all join-irreducible elements of L which are less than or equal to x. Since L is finite, x is the join of η(x); that is, every element of L can be obtained as a (possibly empty) join of a down-set of elements from {a1 , · · · , an }. For x ∈ L and i = 1, · · · , n, the ith switch x ˜i of x is defined by { ∨ (η(x)\ ↑ ai ) if ai ∈ η(x), i x ˜ = ∨ (η(x)∪ ↓ ai ) if ai ∈ / η(x), where ↑ ai = {a ∈ L : a is a join-irreducible element and ai ≤ a}, ↓ ai = {a ∈ L : a is a join-irreducible element and a ≤ ai }. Given a mapping F : L → L and x ∈ L, let F = (f1 , · · · , fn ), where for each i = 1, · · · , n, fi : L → {0L , ai } is given by { ai if ai ∈ η(F (x)), fi (x) = 0L if ai ∈ / η(F (x)). ∨ Then F (x) = {fi (x) : i = 1, · · · , n} for each x ∈ L. Similarly, we denote x = (x1 , · · · , xn ) such that { ai if ai ∈ η(x), xi = 0L if ai ∈ / η(x), ∨ for each i = 1, · · · , n, and x = {xi : i = 1, · · · , n}. We define the generalized Boolean Jacobian matrix of F at x as F ′ (x) = (fij (x)), where { 1 if fi (x) ̸= fi (˜ xj ) and dp (x, x ˜j ) = 1, fij (x) = 0 otherwise. Let Γ(F ′ (x)) be the directed graph associated with F ′ (x) having the vertex set {1, · · · , n} and a directed edge (j, i) from j to i if fij (x) = 1. The directed edge (j, i) is with negative sign if xj = 0L and fi (x) = ai or xj = aj and fi (x) = 0L . The directed edge (j, i) is with positive sign if xj = 0L and fi (x) = 0L or xj = aj and fi (x) = ai . 6.

(10) A circuit C = [c1 , · · · , ck ] in a directed graph is a sequence of vertices with c1 = ck such that there exist an edge from ci to ci+1 for each i = 1, · · · , k. The sign of a circuit C is the product of the signs of its edges. Let IF (x) = {i : fi (x) ̸= xi }. The asynchronous dynamic graph of F , denoted G(F ), is the directed graph whose set of vertices is L and whose set of edges is {(x, y) : x ∈ L, y = x ˜i for some i ∈ IF (x)}. Let us remark that IF (x) = ∅ if and only if x is a fixed point of F . A trap domain of G(F ) is a non-empty subset D ⊂ L such that if any edge (x, y) of G(F ) satisfies x ∈ D, then y ∈ D. An attractor of G(F ) is the smallest trap domain. A cyclic attractor is an attractor of cardinality at least two. For a set U , denote by |U | the cardinality of U . According to the above definitions, Thomas’ conjecture may be stated as follows. Let L ba a finite distributive lattice and F : L → L. ∪ 1. If F has multiple fixed points then x∈L Γ(F ′ (x)) has a positive circuit. ∪ 2. If G(F ) has a cyclic attractor then x∈L Γ(F ′ (x)) has a negative cir∪ cuit.(In particular, if F has no fixed point then x∈L Γ(F ′ (x)) has a negative circuit.). 7.

(11) 3. Fixed points and circuit-free. In this section, we extend Theorem 1.2 from the Boolean case to the finite distributive lattice. We shall establish the following: Theorem 3.1 Let L be a finite distributive lattice with J(L) = {a1 , . . . , ak }. If F : L → L is such that Γ(F ′ (x)) has no circuit for all x ∈ L, then there exists a unique fixed point. To prove the full generality we adopt the thinking of the Chinese proverb,“Many a little makes a mickle.” Thus we need the notion of sublattice [a, b] generated by a, b ∈ L, a < b. Let a, b ∈ L, a < b. We define [a, b] = {x ∈ L : a ≤ x ≤ b}. We call [a, b] a sublattice generated by a, b. Let I = {i : ai < bi }. Then [a, b] = {x ∈ L : xi = ai = bi for all i ̸= I}. Proposition 3.2 Let L be a finite distributive lattice. For a, b ∈ L, a < b, the sublattice [a, b] is a distributive lattice. Proof. For x, y, z ∈ [a, b] ⊂ L, L is distributive, we have x ∧ (y ∨ z) = (x ∧ y) ∨ (x ∧ z). Since [a, b] = {x ∈ L : a ≤ x ≤ b}, we have c ∨ d, c ∧ d ∈ [a, b] for all c, d ∈ [a, b]. This shows [a, b] is a distributive lattice. Let a, b ∈ L, a < b. We say that α ∈ [a, b] is a local fixed point if fi (α) = αi , ∀i ∈ {i : ai < bi }. Theorem 3.3 Let L be a finite distributive lattice with J(L) = {a1 , . . . , ak } and let F : L → L be such that Γ(F ′ (x)) has no circuit for all x ∈ L. Then for every a, b ∈ L with a < b, there exists a unique local fixed point α ∈ [a, b]. To prove Theorem 3.1, we apply Theorem 3.3 by taking a = 0L , b = 1L . Thus we have a generalization of Shih-Dong’s theorem to all finite distributive lattices. 8.

(12) In the case L = {0, 1}n , Theorem 3.3 reduces to a lemma proved by Shih and Dong [13]. The proof of Theorem 3.3 reveals a phenomenon that a global feature of a system emerges from collective behavior of its many components. In terms of the Chinese proverb quoted earlier, Theorem 3.3 is many a little and Theorem 3.1 is the mickle. To establish Theorem 3.3, we need the following lemmas. The proof of Theorem 3.3 is modelled after the proof of Shih-Dong’s theorem. Lemma 3.4 Let L be a finite distributive lattice with J(L) = {a1 , . . . , ak }. Then for each x, y ∈ L with x < y, the set I = {i : xi < yi } is non-empty and y = x ˜i for all i ∈ I when dp (x, y) = 1. Proof. For x < y, if xi = ai , then ai ≤ x < y and hence yi = 1. This shows xi ≤ yi for each i. If I = ∅, then xi = yi for each i. We have x = y, a contradiction. If dp (x, y) = 1, then y covers x. For i ∈ I, since xi = 0L , yi = ai and x ˜ii = ai , we have x < x ˜i ≤ y by the definition of x ˜i . Thus x ˜i = y for each i ∈ I. The collection of down-sets of any partially ordered set forms a lattice in which the lattice’s partial ordering is given by the set inclusion. The join operation corresponds to the set union, and the meet operation corresponds to the set intersection. Since the set union and the set intersection obey the distributive law, the collection of down-sets is a distributive lattice. Birkhoff’s representation theorem asserts that any finite distributive lattice L is isomorphic to the lattice of down-sets of the join-irreducible elements of L [2, 3]. According to Birkhoff’s representation theorem, we can prove the following lemma. Lemma 3.5 Let L be a finite lattice. Then L is distributive if and only if dp (x, y) = |η(x ∨ y)| − |η(x ∧ y)| for all x, y ∈ L. Proof. We first show that the inequality dp (x, y) ≤ |η(x∨y)|−|η(x∧y)| holds for all x, y ∈ L where L is any finite lattice (not necessarily distributive). Let x, y ∈ L. If x ≤ y, then it is readily seen that dp (x, y) ≤ |η(x∨y)|−|η(x∧y)| by the definition of dp (x, y) and Lemma 3.4. 9.

(13) Let J = {i : xi ̸= yi }, A = {i : (x ∧ y)i < xi }, and B = {i : (x ∧ y)i < yi }. It is obvious that A ⊂ J, B ⊂ J, and J ⊂ A ∪ B. If i ∈ A and i ∈ B, then xi = yi = ai . It implies that (x ∧ y)i = ai and i ∈ / A, which is a contradiction. It follows that A ∩ B = ∅ and A ∪ B = J. Thus dp (x, y) ≤ dp (x, x ∧ y) + dp (y, x ∧ y) ≤ |A| + |B| = |J|. Let i ∈ J. Then ai ∈ η(x ∨ y) and ai ∈ / η(x ∧ y). Thus |J| = |η(x ∨ y)| − |η(x ∧ y)|. We have shown that dp (x, y) ≤ |η(x ∨ y)| − |η(x ∧ y)| holds for any x, y in arbitrary finite lattice L. Now we assume that L is distributive. According to Birkhoffs representation theorem, we have the following triangle inequality: |η(x ∨ y)| − |η(x ∧ y)| ≤ |η(x ∨ z)| − |η(x ∧ z)| + |η(y ∨ z)| − |η(y ∧ z)| for all x, y, z ∈ L. Let x, y ∈ L. In case dp (x, y) = 1, we may assume that x < y. Suppose that |η(x ∨ y)| − |η(x ∧ y)| ≥ 2, then there exist ai1 , ai2 ∈ J(L) such that xi = 0L and yi = ai for i = i1 , i2 . If ai1 and ai2 are incomparable, that is ai1 ai2 and ai2 ai1 , then x < x ˜i1 < y by Birkhoff’s representation theorem. It contradicts dp (x, y) = 1. If ai1 < ai2 or ai2 < ai1 , then x < x ˜i1 < y or x < x ˜i2 < y. It also contradicts dp (x, y) = 1 and hence dp (x, y) = 1 = |η(x ∨ y)| − |η(x ∧ y)|. In case dp (x, y) = k, there exists a path {x = p0 , p1 , . . . , pk = y} such that dp (pi , pi+1 ) = 1 = |η(pi ∨ pi+1 )| − |η(pi ∧ pi+1 )| for each i = 0, . . . , k − 1. Thus |η(x ∨ y)| − |η(x ∧ y)| ≤ |η(x ∨ p1 )| − |η(x ∧ p1 )| + |η(p1 ∨ p2 )| − |η(p1 ∧ p2 )| + · · · + |η(pk−1 ∨ y)| − |η(pk−1 ∧ y)| = dp (x, p1 ) + dp (p1 , p2 ) + · · · + dp (pk−1 , y) = k = dp (x, y). It follows that dp (x, y) = |η(x ∨ y)| − |η(x ∧ y)|. Conversely, let dp (x, y) = |η(x ∨ y)| − |η(x ∧ y)| for all x, y ∈ L. If L is not distributive, then L has a sublattice isomorphic to M3 (the diamond) or 10.

(14) 1L. 1L u. a. b. c. w v. 0L. 0L. M3. N5. Figure 1: The diamond lattice M3 (left) and the pentagon lattice N5 (right). N5 (the pentagon), as shown in Figure 1. It is obvious that 2 = dp (0L , 1L ) < |η(1L )| − |η(0L )| = 3 when L is M3 or N5 . If M is a sublattice of L and M is isomorphic to M3 , then there exist a, b, c ∈ M such that 0M < a, b, c < 1M and a ∨ b = b ∨ c = a ∨ c = 1M . ∧ ∨ ∧ ∨ Let L′ = [ M, M ] = {x ∈ L : M ≤ x ≤ M } be a sublattice of L ∧ ∨ and I = {i : ( M )i < ( M )i } = {i1 , i2 , . . . , ik }. Since a ∨ b = b ∨ c = a ∨ c = 1M , there exist ia , ib , ic ∈ I such that if x ∈ M with xi , xj ̸= 0L for ∨ i, j ∈ {ia , ib , ic }, then x = M . ∧ ∧ ∨ ∧ Since dp ( M, M ) = |η( M )| − |η( M )| = |I|, there exists a path ∧ ∨ P = { M = z 0 , z 1 , . . . , z k = M } where dp (z j , z j−1 ) = 1 for each j = 1, . . . , k. By Lemma 3.4, we have z j = z j−1 ∨ aij for some ij ∈ I for each j = 1, . . . , k. However the condition “if x ∈ M with xi , xj ̸= 0L for ∨ i, j ∈ {ia , ib , ic }, then x = M ” implies that z j = z j+1 for some j < k by pigeonhole principle, which is a contradiction. On the other hand, if N is a sublattice of L and N is isomorphic to N5 , then there exist u, v, w ∈ M such that 0N < u, v, w < 1N and u ∨ w = ∧ ∨ ∧ ∨ v ∨ w = 1N . Let L′′ = [ N, N ] = {x ∈ L : N ≤ x ≤ N } be a ∧ ∨ sublattice of L and I = {i : ( N )i < ( N )i } = {i1 , i2 , . . . , ik }. Since u ∨ w = v ∨ w = 1N , there exist iu , iv , iw ∈ I such that if x ∈ M with ∨ ∧ xw , xi ̸= 0L for i ∈ {iu , iv }, then x = N . Let I1 = {i : ( N )i < wi } ∨ and I2 = {i : wi < ( N )i }. Then we have I1 ∩ I2 = ∅ and I1 ∪ I2 = I. In particular, we have iu , iv ∈ I2 . Let l = |I1 |, without loss of generality, we. 11.

(15) may assume that ij ∈ I1 for j = 1 · · · , l and ij ∈ I2 for j = l + 1, . . . , k; this is possible by renumbering the join-irreducible elements ai , i ∈ I. Since ∧ ∧ ∨ ∧ ∧ dp ( N, N ) = |η( N )| − |η( N )| = |I|, there exists a path P = { N = ∨ z 0 , . . . , z l−1 , z l = w, z l+1 , . . . , z k = N } such that dp (z j , z j−1 ) = 1 for each j = 1, . . . , k. By Lemma 3.4, we have z j = z j−1 ∨ aij for some ij ∈ I for each j = 1, . . . , k. However the condition “if xw , xi ̸= 0L for i ∈ {iu , iv }, then ∨ ∨ x = N ” implies that z j = N for some j ∈ I2 and j < k by pigeonhole principle, which is a contradiction. This completes the proof. Lemma 3.6 Let L be a finite distributive lattice and x, y ∈ L with x ̸= y. Then there exists a join-irreducible element ai such that xi ̸= yi and dp (x, x ˜i ) = 1. Proof. Let x, y ∈ L and x ̸= y. Since x ̸= y, there exists j such that xj ̸= yj . We may assume that xj = 0L and yj = aj . If dp (x, x ˜j ) = 1, then we set i = j. If dp (x, x ˜j ) ≥ 2, by the definition of x ˜j and Birkhoff’s representation theorem, then there exists a join-irreducible element ak such that x < x ˜k < x ˜j . We have x ∧ x ˜k = x, x ∧ x ˜j = x, x ∨ x ˜k = x ˜k , and x∨ x ˜j = x ˜j . It implies that dp (x, x ˜k ) < dp (x, x ˜j ) by Lemma 3.5 and xk = 0L , yk = ak (because yj = aj ). Repeating the above process, we get i such that xi ̸= yi and dp (x, x ˜i ) = 1. We proceed now to prove Theorem 3.3. We prove the assertion by induction on dp (a, b). If dp (a, b) = 1, set b=a ˜i , then the hypothesis Γ(F ′ (a)) = 0 implies that fii (a) = 0 and hence (F (a))i = (F (˜ ai ))i = (F (b))i . So exactly one of the statements (F (a))i = ai , (F (b))i = bi is true. Thus the assertion is valid. We now assume that the theorem holds for all a′ , b′ ∈ L such that 1 ≤ dp (a′ , b′ ) < dp (a, b). We first settle the uniqueness question. Suppose, on the contrary, that α and β are two distinct points in the sublattice [a, b] such that (F (α))i = αi and (F (β))i = βi for all i ∈ I = {i : ai < bi }. Case 1. dp (α, β) < dp (a, b). Then α, β ∈ [α ∧ β, α ∨ β] ⊂ [a, b] and (F (α))i = αi and (F (β))i = βi for all i ∈ {i : (α ∧ β)i < (α ∨ β)i }, 12.

(16) in contradiction to the uniqueness assertion of the induction hypothesis. Case 2. dp (α, β) = dp (a, b). It implies that αi ̸= βi for all i ∈ I. Let Iβ = {i ∈ I : dp (β, β˜i ) = 1}. For i ∈ Iβ , dp (α, β˜i ) < dp (α, β) and α, β˜i ∈ [α ∧ β˜i , α ∨ β˜i ]. By the induction hypothesis, there exists j ∈ I \ {i} such that (F (β˜i ))j ̸= β˜ji . Because dp (β, β˜i ) = 1, we have βj = β˜ji and hence β˜ji = (F (β))j . By Lemma 3.6, there exists l ∈ Iβ such that (F (β))l ̸= (F (β˜i ))l . Thus F ′ (β) contains a principal submatrix. i1 .. . im.  i1 ∗  .  ..  ∗∗. · · · im  ··· ∗ . . ..  . .   ··· ∗. of order m which has no zero columns where Iβ = {i1 , . . . , im } and the symbol ∗ denotes the entry possibly 1. We conclude that ρ(F ′ (β)) = 1, contrary to the spectral condition. We now arrive at a contradiction for Cases 1 and 2. This contradiction completes the proof of the uniqueness part. Now we prove the existence part. By Lemma 3.6, there exists l ∈ I such that al = 0, bl = 1 and dp (a, a ˜l ) = 1. Case 1. dp (b, ˜bl ) = 1. According to the induction hypothesis, there exist a unique fixed point x ∈ [a, ˜bl ] and a unique fixed point y ∈ [˜ al , b] such that (F (x))i = xi and (F (y))i = yi for all i ∈ I \ l.. (3.1). We want to use two distinct points x and y to synthesis a point α that satisfies the theorem; it has to split the arguments into two cases. Case 1.1. dp (x, y) < dp (a, b). Then [x ∧ y, x ∨ y] ⊂ [a, b] and d(x ∧ y, x ∨ y) = dp (x, y) < dp (a, b). By the induction hypothesis, there exists a unique z ∈ [x ∧ y, x ∨ y] such that (F (z))i = zi for all i ∈ {i : (x ∧ y)i < (x ∨ y)i }. Since xl = 0 and yl = 1, l ∈ {i : (x ∧ y)i < (x ∨ y)i }. If zl = 0, then x, z ∈ [x ∧ z, x ∨ z] ⊂ [x ∧ y, x ∨ y]. 13. (3.2).

(17) By (3.1), (3.2), and the uniqueness assertion of the induction hypothesis, we have x = z. Thus x satisfies the equations (F (x))i = xi for all i ∈ I. On the other hand, if zl = 1 then y = z. Hence the existence assertion of the theorem follows if we take α = x or α = y. Case 1.2. dp (x, y) = dp (a, b). Then xi ̸= yi for all i ∈ I. By the condition ρ(F ′ (y)) = 0, it follows that there exists an j ̸= l with j ∈ Iy where Iy = {i ∈ I : dp (y, y˜i ) = 1} such that (F (˜ y j ))i = (F (y))i for all i ∈ Iy .. (3.3). If j = l, then y˜ll = 0 and (F (˜ y l ))i = (F (y))i for all i ∈ Iy .. (3.4). By (3.1), (3.4), and Lemma 3.6, we have (F (˜ y l ))i = (F (y))i = yi = y˜il for all i ∈ I \ {l}. Thus x and y˜l would be two distinct points in [a, ˜bl ] such that (F (x))i = xi and (F (˜ y l ))i = y˜il for all i ∈ I \ {l}, contrary to the uniqueness assertion of the induction hypothesis. By (3.3) and Lemma 3.6, we have (F (˜ y j ))i = (F (y))i for all i ∈ I.. (3.5). Since xj = y˜jj , we have dp (x, y˜j ) = dp (x, y) − 1 < dp (x, y). By the induction hypothesis there exists a unique point w ∈ [x ∧ y˜j , x ∨ y˜j ] with (F (w))i = wi for all i ∈ {i : (x ∧ y˜j )i < (x ∨ y˜j )i }.. (3.6). If wl = y˜lj , then w, y˜j ∈ [w ∧ y˜j , w ∨ y˜j ] ⊂ [x ∧ y˜j , x ∨ y˜j ]. By (3.1), (3.5), (3.6), and the uniqueness assertion of the induction hypothesis we have w = y˜j . By (3.3), (F (y))l = (F (˜ y j ))l = y˜lj = yl . 14.

(18) Thus (F (y))i = yi for all i ∈ I. If wl = xl , then w, x ∈ [w ∧ x, w ∨ x] ⊂ [x ∧ y˜j , x ∨ y˜j ]. By (3.1), (3.6), and the uniqueness assertion of the induction hypothesis we have w = x and hence (F (x))l = xl . We have (F (x))i = xi for all i ∈ I. The existence assertion of the theorem follows if we take α = x or α = y. Case 2. dp (b, ˜bl ) ≥ 2 for all l ∈ {i ∈ I : dp (a, a ˜i ) = 1}. Let i1 ∈ I such that dp (a, a ˜i1 ) = 1. By the definition of dp (b, ˜bi1 ) and the definition of ˜bi1 , there exists ai2 ∈ J(L) such that ai1 < ai2 . According to Lemma 3.6, we can assume that dp (b, ˜bi2 ) = 1. According to the induction hypothesis, there exist a unique fixed point x ∈ [a, ˜bi2 ] and a unique fixed point y ∈ [˜ ai1 , b] such that (F (x))i = xi for all i ∈ I \ i2 and (F (y))i = yi for all i ∈ I \ i1 .. (3.7). Notice that xi2 = 0 and yi1 = 1. Similarly, we want to use two points x and y to synthesis a point α that satisfies the theorem; it has to split the arguments into two cases. Case 2.1. xi1 = 0 or yi2 = 1. If (F (x))i2 ̸= xi2 or (F (y))i1 ̸= yi1 , then (F (x))i2 = 1 or (F (y))i1 = 0. Since ai1 < ai2 , we have (F (x))i1 = 1 or (F (y))i2 = 0. It contradicts (3.7). Thus xi1 = 0 or yi2 = 1 implies that (F (x))i2 = xi2 or (F (y))i1 = yi1 and (F (x))i = xi or (F (y))i = yi for all i ∈ I. The existence assertion of the theorem follows if we take α = x or α = y. 15.

(19) Case 2.2. xi1 = 1 and yi2 = 0. Thus xi = yi for i = i1 , i2 , we have dp (x, y) < dp (a, b) and i1 , i2 ∈ / {i : (x ∧ y)i < (x ∨ yi }. By the induction hypothesis there exists a unique point z ∈ [x ∧ y, x ∨ y] such that (F (z))i = zi for all i ∈ {i : (x ∧ y)i < (x ∨ yi )}. According to (3.7) and the induction hypothesis, it implies that x = y and hence (F (x))i = xi for all i ∈ I. Hence the existence assertion of the theorem follows if we take α = x = y. This completes the inductive proof of Theorem 3.3.. 16.

(20) 4. Fixed points and positive circuits. In this section, we prove the first Thomas conjecture. It is stated that if F : L → L has more than one fixed points where L is a finite distributive ∪ lattice, then x∈L Γ(F ′ (x)) has a positive circuit. In fact, we prove the following statement. If F : L → L has more than one fixed points where L is a finite distributive lattice, then there exists a point x ∈ L such that Γ(F ′ (x)) has a positive circuit. We shall establish the following: Theorem 4.1 Let L be a finite distributive lattice and F : L → L. If Γ(F ′ (x)) has no positive circuit for each x ∈ L, then for each a, b ∈ L with a < b, F has at most 1 local fixed point in [a, b]. Proof. We prove the assertion by induction on the distance dp (a, b). The case of dp (a, b) = 1, let b = a ˜i by Lemma 3.6. By hypothesis ρ(F ′ (a)) = 0, it implies that fii (a) = 0 and hence fi (a) = fi (˜ ai ) = fi (b). So exactly one of the statements fi (a) = ai , fi (b) = bi is true. Thus the theorem holds for the case of dp (a, b) = 1. We may assume that the theorem holds for all a′ , b′ ∈ L such that 1 ≤ dp (a′ , b′ ) < dp (a, b). Suppose, on the contrary, that α and β are two distinct points in the sublattice [a, b] such that fi (α) = αi and fi (β) = βi for all i ∈ I = {i : ai < bi }. Notice that [a, b] is a distributive lattice. For x, y, z ∈ [a, b] ⊂ L, L is distributive, we have x ∧ (y ∨ z) = (x ∧ y) ∨ (x ∧ z). Since [a, b] = {x ∈ L : a ≤ x ≤ b}, we have c ∨ d, c ∧ d ∈ [a, b] for all c, d ∈ [a, b]. This shows [a, b] is a distributive lattice. Case 1. dp (α, β) < dp (a, b). We have α, β ∈ [α ∧ β, α ∨ β] ⊂ [a, b]. Since [a, b] is distributive, by Lemma 3.5, dp (α ∨ β, α ∧ β) = |η(α ∨ β)| − |η(α ∧ β)| = dp (α, β) < dp (a, b). However, fi (α) = αi and fi (β) = βi for all i ∈ {i : (α ∧ β)i < (α ∨ β)i }, in contradiction to the induction hypothesis. 17.

(21) Case 2. dp (α, β) = dp (a, b). Let Iα = {i ∈ I : dp (α, α ˜ i ) = 1}. Then Iα ̸= ∅ by Lemma 3.6. For each i ∈ Iα , since dp (a, b) > 1, we have α ˜i ∈ i i i i i [˜ α ∧ β, α ˜ ∨ β] and α ˜ ̸= β. Since [α ˜ ∧ β, α ˜ ∨ β] is distributive, by Lemma 3.5, dp (˜ αi ∨ β, α ˜ i ∧ β) = |η(˜ αi ∨ β)| − |η(˜ αi ∧ β)| = dp (˜ αi , β) = dp (α, β) − 1 < dp (a, b). The induction hypothesis implies that α ˜ i is not a local fixed point in [˜ αi ∧ β, α ˜ i ∨ β]. There exists j ′ ∈ I \ {i} such that fj ′ (˜ αi ) ̸= α ˜ ji ′ = αj ′ . Let z = (a ∨ F (˜ αi )) ∧ b. Then z ∈ [a, b] and zi′ = fi′ (˜ αi ) for each i′ ∈ I and α ̸= z. By Lemma 3.6, there exists j ∈ Iα such that αj ̸= zj = fj (˜ αi ). Thus fij (α) = 1. Let F ′ (α)|Iα = (fij (α)|Iα ) be the principal submatrix of F ′ (α) where fij (α)|Iα = fij (α) if i, j ∈ Iα ; otherwise fij (x)|Iα = 0. Then F ′ (α) contains a principal submatrix which has no zero columns. It follows that F ′ (α)|Iα has a circuit C = [c1 , . . . , ck , ck+1 ] [10, 13]. We have fci+1 (˜ αci ) ̸= fci+1 (α) = αci+1 = σi (αci ) for each i = 1, . . . k, where σi : {0L , aci } → {0L , aci+1 } satisfies σi (0L ) = aci+1 and σi (aci ) = 0L when the edge from ci to ci+1 is negative, or satisfies σi (0L ) = 0L and σi (aci ) = aci+1 when the edge from ci to ci+1 is positive. Since αc1 = αck+1 = (σk ◦ · · · ◦ σ1 )(αc1 ), the sign of the circuit C is positive, which is a contradiction. This completes the proof. By taking a = 0L and b = 1L in Theorem 4.1, we have Theorem 4.2 Let L be a finite distributive lattice and F : L → L. If Γ(F ′ (x)) has no positive circuit for each x ∈ L, then F has at most one fixed point. In the following, we state explicitly the contrapositive form of Theorem 4.2. Theorem 4.3 Let L be a finite distributive lattice. If F : L → L has more than one fixed points, then there exists a point x ∈ L such that Γ(F ′ (x)) has a positive circuit.. 18.

(22) 5. Fixed points and negative circuits. In this section, we prove the second Thomas conjecture. It is stated as follows. Theorem 5.1 Let L be a finite distributive lattice and F : L → L. Let D be ∪ a cyclic attractor of G(F ). Then x∈L Γ(F ′ (x)) contains a negative circuit. As an immediate consequence of Theorem 5.1, we have Theorem 5.2 Let L be a finite distributive lattice. If F : L → L has no ∪ fixed point, then x∈L Γ(F ′ (x)) contains a negative circuit. Proof. If F has no fixed point, then G(F ) contains a cyclic attractor. By ∪ Theorem 5.1, x∈L Γ(F ′ (x)) contains a negative circuit. To prove Theorem 5.1, we need the following lemmas. Lemma 5.3 Let L be a finite lattice and D be an attractor. For x, y ∈ D, there exists a path from x to y in G(F ). Proof. Suppose that there exist x, y ∈ D such that there has no path from x to y in G(F ). Let D′ = {z : there exists a path from x to z in G(F )}. Then D′ is a trap domain and y ∈ / D′ . So D′ D, which contradicts the smallest trap domain D. Lemma 5.4 Let L be a finite distributive lattice. If (x, y) ∈ G(F ), then {i : xi ̸= yi } ⊂ IF (x). Proof. If (x, y) ∈ G(F ), then there exists i ∈ IF (x) such that y = x ˜i . We may assume that x < y. Let j ∈ {i : xi ̸= yi }, we have xj < yj . Case 1. ai < aj . According to Birkhoff’s representation theorem, yj = (˜ xi )j = 0L = xj , which is a contradiction. Case 2. ai and aj are incomparable. According to Birkhoff’s representation theorem, yj = (˜ xi )j = 0L = xj , which is a contradiction. Case 3. aj < ai . Since fi (x) = ai , it implies that ∨ aj < ai ≤ F (x) = {fi (x) : i = 1, . . . , n}. 19.

(23) Thus fj (x) = aj ̸= xj , j ∈ IF (x).. . Let G1 and G2 be two directed graphs with the same set of vertices V and with set of edges A1 and A2 respectively. We say that G1 is a subgraph of G2 if A1 ⊂ A2 . For x ∈ L, we set   if fi (x) > xi ,  1 si (x) = 0 if fi (x) = xi ,   −1 if fi (x) < xi . Lemma 5.5 Let L be a finite distributive lattice and F : L → L. Let {x0 , x1 , . . . , xk } be a path of G(F ) of length k ≥ 1, and let i ∈ IF (xk ). If si (xp ) ̸= si (xk ) for all 0 ≤ p < k, then there exists j ∈ IF (x0 ) such that ∪ ′ 0 k x∈L Γ(F (x)) has a path from j to i with sign sj (x )si (x ). Proof. We prove the assertion by the induction on the length of the path f0 l for some k. In case k = 1, by hypothesis, si (x0 ) ̸= si (x1 ) and x1 = x l ∈ IF (x0 ). Case 1. dp (x0 , x1 ) = 1. We set l = j. Case 1.1. i ̸= j. Then x0i = x1i . We may assume that si (x1 ) = 1. Thus f0 j ) = fi (x1 ) > x1 = x0 ≥ fi (x0 ), it implies that fij (x0 ) = 1. If x0 = 0L , fi (x i. i. j. then sj (x0 ) = 1 and x0j = fi (x0 ) = 0L . It follows that Γ(F ′ (x0 )) has an edge from j to i with positive sign. If x0j = aj , then sj (x0 ) = −1. Since fi (x0 ) = 0L , Γ(F ′ (x0 )) has an edge from j to i with negative sign. Thus Γ(F ′ (x0 )) has an edge from j to i with sign sj (x0 )si (xk ) when i ̸= j. Case 1.2. i = j. Then si (x0 )si (x1 ) = −1. We may assume si (x1 ) = 1. f0 i ) = fi (x1 ) > x1 = fi (x0 ) and x0 ̸= fi (x0 ), it implies that Thus fi (x i. i. fii (x0 ) = 1 and either x0i = 0L or fi (x0 ) = 0L . Thus G(F ′ (x0 )) has an edge from j to i with negative sign. This completes the proof of Case 1. Case 2. dp (x0 , x1 ) > 1. Let {x0 = p0 , p1 , · · · , pm = x1 } be a path which connects x0 and x1 with pr < pr+1 or pr > pr+1 for all r = 0, 1, · · · , m − 1 in the diagram of L. Case 2.1. i ∈ / {s : p0s ̸= pm s }. There exists 0 ≤ q ≤ m − 1 such that q q+1 1 si (p ) ̸= si (p ) = si (x ). By Lemma 3.5 and Lemma 3.6, there exists j j 0 such that peq = pq+1 . By lemma 5.4, j ∈ {s : pqs ̸= pq+1 s } ⊂ {s : ps ̸= q q+1 0 1 0 . We may assume that pm s } = {s : xs ̸= xs } ⊂ IF (x ), and pi = pi 20.

(24) j si (pq+1 ) = 1. Thus fi (peq ) = fi (pq+1 ) > pq+1 = pqi ≥ fi (pq ), it implies i that fij (pq ) = 1. If sj (x0 ) = 1, then fj (x0 ) > x0j = 0L . Thus x0 < x1 and pq < pq+1 , we have shown that Γ(F ′ (pq )) has an edge from j to i with positive sign. If sj (x0 ) = −1, then 0L = fj (x0 ) < x0j . Thus x1 < x0 and pq+1 < pq , we have shown that Γ(F ′ (pq )) has an edge from j to i with negative sign. Thus Γ(F ′ (pq )) has an edge from j to i with sign sj (x0 )si (xk ) when i ∈ / {s : p0s ̸= pm s }. 0 m Case 2.2. i ∈ {s : p0s ̸= pm s }. By lemma 5.4, i ∈ {s : ps ̸= ps } = {s : x0s ̸= x1s } ⊂ IF (x0 ), and si (x0 )si (x1 ) = si (p0 )si (pm ) = −1. We may assume 0 0 that si (pm ) = 1 and si (p0 ) = −1. Since fi (pm ) > pm i and fi (p ) < pi , it r r+1 for all r = 0, 1, · · · , m−1 in the diagram implies that p0i > pm i and p > p of L. There exists 0 ≤ q ≤ m − 1 such that fi (pq ) = 0L and fi (pq+1 ) = ai . j By Lemma 3.5 and Lemma 3.6, there exists j such that peq = pq+1 . It follows that Γ(F ′ (pq )) has an edge from j to i with negative sign. We have shown that Γ(F ′ (pq )) has an edge from j to i with sign sj (x0 )si (xk ) when i ∈ {s : p0s ̸= pm s }. This completes the proof of Case 2. In case k > 1, let i ∈ IF (xk ). By the induction hypothesis, {xk−1 , xk } is a path of G(F ) of length 1. l ] k−1 . Case 1. dp (xk−1 , xk ) = 1. There exists l ∈ IF (xk−1 ) such that xk = x. Thus Γ(F ′ (xk−1 )) has an edge from l to i with sign sl (xk−1 )si (xk ). Case 2. dp (xk−1 , xk ) > 1. There exists a path {xk−1 = p0 , p1 , · · · , pm = xk } which connects xk+1 and xk with pr < pr+1 or pr > pr+1 for all r = 0, 1, · · · , m − 1 in the diagram of L. k−1 ) = s (p0 ) ̸= s (xk ) = Case 2.1. i ∈ / {s : p0s ̸= pm i i s }. Since si (x 0 . It implies that p0 = f (p0 ). (If p0 ̸= f (p0 ), si (pm ), fi (pm ) ̸= pm = p i i i i i i it implies that si (p0 ) = si (pm ), which is a contradiction.) There exists 0 ≤ q ≤ m − 1 such that pqi = fi (pq ) and pq+1 ̸= fi (pq+1 ). By Lemma i l 3.5 and Lemma 3.6, there exists l such that peq = pq+1 . By lemma 5.4, 0 m k−1 ̸= xk } ⊂ I (xk−1 ) = I (p0 ), l ∈ {s : pqs ̸= pq+1 s } ⊂ {s : ps ̸= ps } = {s : xs F F s k m q+1 q+1 and si (x ) = si (p ) = si (p ). We may assume that si (p ) = 1. Thus j fi (peq ) = fi (pq+1 ) > pq+1 = pqi = fi (pq ), it implies that fil (pq ) = 1. If i sl (p0 ) = 1, then fl (p0 ) > p0l = 0L . It implies that pq < pq+1 . Thus Γ(F ′ (pq )) has an edge from l to i with positive sign. If sl (p0 ) = −1, then 0L = fl (p0 ) < p0l . It implies that pq+1 < pq . Thus Γ(F ′ (pq )) has an edge. 21.

(25) from l to i with negative sign. We have shown that Γ(F ′ (pq )) has an edge from l to i with sign sl (xk−1 )si (xk ) when i ∈ / {s : p0s ̸= pm s }. 0 m Case 2.2. i ∈ {s : ps ̸= ps }. By lemma 5.4, i ∈ {s : p0s ̸= pm s } = {s : k−1 k k−1 0 k−1 k 0 xs ̸= xs } ⊂ IF (x ) = IF (p ), and si (x )si (x ) = si (p )si (pm ) = −1. We may assume that si (pm ) = 1 and si (p0 ) = −1. Since fi (pm ) > pm i and 0 0 0 m r r+1 fi (p ) < pi , it implies that pi > pi and p > p for all r = 0, 1, · · · , m − 1 in the diagram of L. There exists 0 ≤ q ≤ m − 1 such that fi (pq ) = 0L and fi (pq+1 ) = ai . By Lemma 3.5 and Lemma 3.6, there exists l such that l peq = pq+1 . It follows that Γ(F ′ (pq )) has an edge from l to i with negative sign. We have shown that Γ(F ′ (pq )) has an edge from l to i with sign sl (xk−1 )si (xk ) when i ∈ {s : p0s ̸= pm s }. This completes the proof of Case 2. According to Cases 1 and 2, there exist l ∈ IF (xk−1 ) and a point x ∈ L such that Γ(F ′ (x)) has an edge from l to i with sign sl (xk−1 )si (xk ). Consider the smallest 0 ≤ p < k such that sl (xp ) = sl (xk−1 ). If p = 0, then l ∈ IF (x0 ) k−1 )) has an edge from l to i with sign s (x0 )s (xk ). Let j = l. and Γ(F ′ (x] i l ∪ Then x∈L Γ(F ′ (x)) contains a path from j to i with sign sj (x0 )si (xk ). If ′ p > 0, then sl (xi ) ̸= sl (xp ) for all 0 ≤ i′ < p. The path {x0 , . . . , xp } satisfies the conditions of the lemma for l ∈ IF (xp ). By the induction hypothesis, ∪ there exists j ∈ IF (x0 ) such that x∈L Γ(F ′ (x)) has a path from j to l with sign sj (x0 )sl (xp ). Since Γ(F ′ (xk−1 )) has an edge from l to i with sign ∪ sl (xk−1 )si (xk ), the graph x∈L Γ(F ′ (x)) contains a path from j to i with sign ∪ sj (x0 )sl (xp )sl (xk−1 )si (xk ). Since sl (xp ) = sl (xk−1 ), x∈L Γ(F ′ (x)) contains a path from j to i with sign sj (x0 )si (xk ). Lemma 5.6 L is a finite distributive lattice and F : L → L. Let D be a cyclic attractor of G(F ). If there exists x ∈ D such that |IF (x)| = 1, then ∪ ′ x∈L Γ(F (x)) contains a negative circuit. Proof. Suppose that there exists x0 ∈ D such that IF (x) = {i}. We have dp (x, x ˜i ) = 1 by Lemma 3.6. (If dp (x, x ˜i ) ≥ 2, then there exists j such that dp (x, x ˜j ) = 1 and xj ̸= x ˜ij . If xi = 0L , we have xj = 0L and aj < ai . Since fi (x) = ai , it implies that fj (x) = aj and j ∈ IF (x), which is a contradiction. Similarly, if xi = ai , then xj = aj and ai < aj . Since fi (x) = 0L , it implies that fj (x) = 0L and j ∈ IF (x), which is a contradiction.). 22.

(26) We need only to prove that the case si (x0 ) = 1, the other case being f0 i . Then G(F ) has an edge from x0 to x1 and x0 < x1 . similar. Let x1 = x i i Since x0 ∈ D, we have x1 ∈ D and there exists a path {x1 , . . . , xk = x0 } from x1 to x0 , and xi ∈ D for all i = 1, . . . , k. If si (xq ) ≥ 0 for all 1 ≤ q < k, then xqi ≤ xq+1 for all 1 ≤ q < k. We have x0i < x1i ≤ xki = x0i , which is a i contradiction. Thus, there exists a smallest 1 ≤ p < k such that fi′ (xp ) = −1. We have {x1 , . . . , xp } is a path in G(F ) and i ∈ IF (xp ), si (xq ) ̸= si (xp ) for all 0 ≤ q < p. According to Lemma 5.5, there exists j ∈ IF (x0 ) such ∪ that x∈L Γ(F ′ (x)) has a path from j to i with sign sj (x0 )si (xp ). Since IF (x0 ) = {i}, we have j = i and sj (x0 )si (xp ) = si (x0 )si (xp ) = −1. Thus ∪ ′ x∈L Γ(F (x)) has a negative circuit from i to i. We proceed now to prove Theorem 5.1. ∪ If there exists x ∈ D such that |IF (x)| = 1, then x∈L Γ(F ′ (x)) contains a negative circuit by Lemma 5.6. We may assume that |IF (x)| ≥ 2 for all ∪ x ∈ D. Let J = x∈D IF (x) and i1 ∈ J be such that ai ≤ ai1 or ai and ai1 are incomparable for all i ∈ J with i ̸= i1 . Let H : L → L be defined by ∨ H(x) = {xi1 , fi (x) : i ̸= i1 }. If x ∈ D, then IH (x) ⊂ IF (x). Let i ∈ IH (x) and hi (x) ̸= xi . If i ∈ / IF (x), i i i then fi (x) = xi . If xi = a , then a ≤ H(x) and hi (x) = a = xi , which is a contradiction. If xi = 0L , then j ∈ / IF (x) for all j such that ai ≤ aj . Thus, ai H(x) and hi (x) = 0L = xi , which is a contradiction. It follows that IH (x) ⊂ IF (x). We may assume that xi1 = 0L . If ii ∈ / IF (x), then H(x) = F (x) and hence IH (x) = IF (x). If ii ∈ IF (x), then i1 ∈ / H(x) by definition of H. If i i ∈ IF (x) \ {i1 } and xi = 0L , then fi (x) = a and hence hi (x) = ai . We have i ∈ IH (x). If i ∈ IF (x) \ {i1 } and xi = ai , then fi (x) = 0L . Thus fj (x) = 0L for all j such that ai ≤ aj . We have ai H(x) and hence hi (x) = 0L ̸= xi and i ∈ IH (x). In particular, if xi1 = 0L , then IH (x) = IF (x) \ {i1 } when ii ∈ IF (x) or IH (x) = IF (x) when ii ∈ / IF (x). We now prove that D is a trap domain of G(H). Let x ∈ D and IH (x) ⊂ IF (x). Since D is a trap domain of G(F ). If i ∈ IH (x) ⊂ IF (x), then x ˜i ∈ D. It follows that D is a trap domain of G(H). By definition, G(H) contains at least one attractor D′ ⊂ D. We may assume that xi1 = 0L for 23.

(27) all x ∈ D′ . Then there exists x ∈ D satisfies xi1 = 0L . Since |IF (x)| ≥ 2 and IH (x) = IF (x) \ {i1 } when ii ∈ IF (x) or IH (x) = IF (x) when ii ∈ / IF (x) for all x ∈ D′ , we have |D′ | ≥ 2. It follows that D′ is a cyclic attractor of G(H). For x ∈ D′ , consider the principle submatrix M (x) = (mij (x)) of H ′ (x) defined by { hij (x) i, j ∈ J, mij (x) = 0 otherwise. Since xi1 = 0L , we have hi (x) = fi (x) for all i ∈ J and M (x) is the principle submatrix of F ′ (x). We have Γ(M (x)) is a subgraph of Γ(F ′ (x)). If there ∪ exists x ∈ D′ such that |IH (x)| = 1, then x∈L Γ(M (x)) has a negative ∪ circuit by Lemma 5.6. Thus x∈L Γ(F ′ (x)) has a negative circuit. ∪ If |IH (x)| ≥ 2 for all x ∈ D′ , consider J ′ = x∈L IH (x) and i2 ∈ J ′ such that ai ≤ ai2 or ai and ai2 are incomparable for all i ∈ J ′ \ {i2 }. Repeating ¯ and a cyclic attractor D ¯ such the above process, we obtain a function H ∪ ′ ¯ ⊂ D and |I ¯ (x)| = 1 for some x ∈ D. ¯ It follows that that D H x∈L Γ(F (x)) has a negative circuit and the proof is completed.. 24.

(28) 6. Concluding remarks. The distributive condition in Theorem 4.3 is required. We have the following example. Example 6.1 Let L be the lattice with 16 elements. The diagram of L is shown in figure 2 which J(L) = {ai ; i = 1, · · · , 11}. Because 1L. q. r. p. a7. 4. a3. a. a9. a5. a10 a11. a6. a2. a8. 1. a. 0L. Figure 2: The diagram of L a9 ∧ (a10 ∨ a11 ) = a9 ∧ r = a9 ̸= a8 = a8 ∨ a8 = (a9 ∧ a10 ) ∨ (a9 ∧ a11 ), L does not satisfy the distributive law. Consider the map F : L → L is the following:. x F (x). 0L 1L. a1 1L. a2 a2. a3 a2. a4 a2. a5 a2. a6 1L. 25. a7 a6. a8 a8. a9 a8. a10 a8. a11 a8. p a6. q a6. r a6. 1L a6.

(29) The generalized Boolean Jacobian  0 0 0   0 0 0   0 0 0   0 0 0    0 0 0  ′ ′ 8 F (0) = F (a )   0 0 0   0 0 0   0 0 0   0 0 0    0 0 0 0 0 0            ′ 1 ′ 2 F (a ) = F (a )           . 0 0 0 0 0 0 0 0 0 0 0. 0 0 1 1 1 1 1 1 1 1 1. 0 0 0 0 0 0 0 0 0 0 0. matrix of F are the following:  0 0 0 0 1 0 0 0  0 0 0 0 1 0 0 0   0 0 0 0 1 0 0 0   0 0 0 0 1 0 0 0    0 0 0 0 1 0 0 0   0 0 0 0 1 0 0 0  ,  0 0 0 0 1 0 0 0   0 0 0 0 0 0 0 0   0 0 0 0 1 0 0 0    0 0 0 0 1 0 0 0  0 0 0 0 1 0 0 0. 0 0 0 0 0 0 0 0 0 0 0. 26. 0 0 0 0 0 0 0 0 0 0 0. 0 0 0 0 0 0 0 0 0 0 0. 0 0 0 0 0 0 0 0 0 0 0. 0 0 0 0 0 0 0 0 0 0 0. 0 0 0 0 0 0 0 0 0 0 0. 0 0 0 0 0 0 0 0 0 0 0. 0 0 0 0 0 0 0 0 0 0 0.            ,          .

(30)            ′ 3 F (a ) =                       ′ 4 F (a ) =                       ′ 5 F (a ) =           . 0 0 0 0 0 0 0 0 0 0 0. 0 0 0 0 0 0 0 0 0 0 0. 0 0 0 0 0 0 0 0 0 0 0. 0 1 0 0 0 1 0 0 0 0 0. 0 1 0 0 0 1 0 0 0 0 0. 0 0 0 0 0 0 0 0 0 0 0. 0 0 0 0 0 0 0 0 0 0 0. 0 0 0 0 0 0 0 0 0 0 0. 0 0 0 0 0 0 0 0 0 0 0. 0 0 0 0 0 0 0 0 0 0 0. 0 0 0 0 0 0 0 0 0 0 0. 0 0 0 0 0 0 0 0 0 0 0. 0 0 0 0 0 0 0 0 0 0 0. 0 1 0 0 0 1 0 0 0 0 0. 0 0 0 0 0 0 0 0 0 0 0. 0 1 0 0 0 1 0 0 0 0 0. 0 0 0 0 0 0 0 0 0 0 0. 0 0 0 0 0 0 0 0 0 0 0. 0 0 0 0 0 0 0 0 0 0 0. 0 0 0 0 0 0 0 0 0 0 0. 0 0 0 0 0 0 0 0 0 0 0. 0 0 0 0 0 0 0 0 0 0 0. 0 0 0 0 0 0 0 0 0 0 0. 0 0 0 0 0 0 0 0 0 0 0. 0 1 0 0 0 1 0 0 0 0 0. 0 1 0 0 0 1 0 0 0 0 0. 0 0 0 0 0 0 0 0 0 0 0. 0 0 0 0 0 0 0 0 0 0 0. 0 0 0 0 0 0 0 0 0 0 0. 0 0 0 0 0 0 0 0 0 0 0. 0 0 0 0 0 0 0 0 0 0 0. 0 0 0 0 0 0 0 0 0 0 0. 0 0 0 0 0 0 0 0 0 0 0. 27.            ,                      ,                      ,          .

(31) . 0 0 0 0 0 0 0 0 0 0 0. 0 0 0 0 0 0 0 0 0 0 0. 0 0 0 0 0 0 0 0 0 0 0. 0 0 0 0 0 0 0 0 0 0 0. 0 0 0 0 0 0 0 0 0 0 0. 0 0 0 0 0 0 0 0 0 0 0. 0 1 1 1 1 0 1 1 1 1 1. 0 0 0 0 0 0 0 0 0 0 0. 0 0 0 0 0 0 0 0 0 0 0. 0 0 0 0 0 0 0 0 0 0 0. 0 0 0 0 0 0 0 0 0 0 0. 0 0 0 0 0 0 0 0 0 0 0. 0 0 0 0 0 0 0 0 0 0 0. 0 0 0 0 0 0 0 0 0 0 0. 0 0 0 0 0 0 0 0 0 0 0. 0 0 0 0 0 0 0 0 0 0 0. 0 0 0 0 0 0 0 0 0 0 0. 0 0 0 0 0 0 0 0 0 0 0. 0 0 0 0 0 0 0 0 0 0 0. 1 0 0 0 0 1 0 1 0 0 0. 1 0 0 0 0 1 0 1 0 0 0. 0 0 0 0 0 0 0 0 0 0 0. 0 0 0 0 0 0 0 0 0 0 0. 0 0 0 0 0 0 0 0 0 0 0. 0 0 0 0 0 0 0 0 0 0 0. 0 0 0 0 0 0 0 0 0 0 0. 0 0 0 0 0 0 0 0 0 0 0. 0 0 0 0 0 0 0 0 0 0 0. 0 0 0 0 0 0 0 0 0 0 0. 1 0 0 0 0 1 0 1 0 0 0. 0 0 0 0 0 0 0 0 0 0 0. 1 0 0 0 0 1 0 1 0 0 0.           ′ 6 ′ 7 F (a ) = F (a )                       ′ 9 F (a ) =                       ′ 10 F (a ) =           . 28. 0 0 0 0 0 0 0 0 0 0 0.            ,          .            ,                      ,          .

(32)            ′ 11 F (a ) =           . 0 0 0 0 0 0 0 0 0 0 0. 0 0 0 0 0 0 0 0 0 0 0. 0 0 0 0 0 0 0 0 0 0 0. 0 0 0 0 0 0 0 0 0 0 0. 0 0 0 0 0 0 0 0 0 0 0 .           ′ ′ ′ ′ F (p) = F (q) = F (r) = F (1L ) =           . 0 0 0 0 0 0 0 0 0 0 0. 0 0 0 0 0 0 0 0 0 0 0. 0 0 0 0 0 0 0 0 0 0 0. 0 0 0 0 0 0 0 0 0 0 0. 0 0 0 0 0 0 0 0 0 0 0. 0 0 0 0 0 0 0 0 0 0 0. 1 0 0 0 0 1 0 1 0 0 0. 0 0 0 0 0 0 0 0 0 0 0. 1 0 0 0 0 1 0 1 0 0 0. 0 0 0 0 0 0 0 0 0 0 0. 0 0 0 0 0 0 0 0 0 0 0. 0 0 0 0 0 0 0 0 0 0 0.            ,          . 0 0 0 0 0 0 0 0 0 0 0. 0 0 0 0 0 0 0 0 0 0 0. 0 0 0 0 0 0 0 0 0 0 0. 0 0 0 0 0 0 0 0 0 0 0. 0 0 0 0 0 0 0 0 0 0 0.            .          . Therefore F has 2 fixed points and ρ(F ′ (x)) = 0 for all x ∈ / {a6 , a7 }, thus the interaction graph Γ(F ′ (x)) has no circuit. The interaction graph of Γ(F ′ (a6 )) = Γ(F ′ (a7 )) is the following directed graph which contains a negative circuit.. 29.

(33) 1 11. 2. 10. 9. - - -8-7 6. 3 4 5. Figure 3: The interaction graph of Γ(F ′ (a6 )) and Γ(F ′ (a7 )). 7. Related open questions. Theorem 3.3 may be extended to finite lattice from finite distributive lattice. It is stated as follows. Conjecture 7.1 Let L be a finite lattice with J(L) = {a1 , . . . , ak } and let F : L → L be such that Γ(F ′ (x)) has no circuit for all x ∈ L. Then for every a, b ∈ L with a < b and I = {i : ai < bi }, there exists a unique point α ∈ [a, b] such that (F (α))i = αi for all i ∈ I. The uniqueness assertion of Conjecture 7.1 is easy to prove by induction on dp (a, b). However, the existence assertion is difficult to prove. Thus Conjecture 7.1 remains open. Conjecture 7.2 Let L be a finite distributive lattice. If F : L → L has no fixed point, then there exists a point x ∈ L such that Γ(F ′ (x)) has a negative circuit. Conjecture 7.2 comes from the equivalent contrapositive form of Theorem 3.1. The equivalent contrapositive form of Theorem 3.1 is stated as follows. Let L be a finite distributive lattice. If F : L → L has multiple fixed points or has no fixed point, then there exists a point x ∈ {0, 1}n such that the corresponding network Γ(F ′ (x)) has a circuit. Theorem 4.3 shows that if F has multiple fixed points, then there exists a point x ∈ {0, 1}n such that the corresponding network Γ(F ′ (x)) has a circuit. On the other hand, Theorem 5.2 states that if F has no fixed point, ∪ then x∈L Γ(F ′ (x)) contains a negative circuit. However, if F has no fixed 30.

(34) point, then there exists a point x ∈ L such that Γ(F ′ (x)) has a circuit. In view of Thomas’ conjecture, the sign of the circuit in Γ(F ′ (x)) should be negative.. 31.

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(36) [12] M.-H. Shih and J.-L. Ho, Solution of the Boolean MarkusVYamabe problem, Advances in Applied Mathematics, 22 (1999), 60–102. [13] M.-H. Shih and J.-L. Dong, A combinatorial analogue of the Jacobian problem in automata networks, Advances in Applied Mathematics, 34 (2005), 30–46. [14] S. Smale, Mathematical problems for the next century, Mathematical Intelligencer, 20 No. 2 (1998), 7–15. [15] C. Soulé, Mathematical approaches to differentiation and gene regulation, C.R. Paris Biologies, 329 (2006), 13–20. [16] R. Thomas, On the relation between the logical structure of systems and their ability to generate multiple steady states or sustained oscillations, Springer Series in Synergetics, 9 (1981), 180–193. [17] R. Thomas and R. d’Ari, Biological Feedback, CRC Press, 1990. [18] R. Thomas, Laws for the dynamics of regulatory networks, Internat. J. Dev. Biol., 42 (1998), 479–485. [19] R. Thomas and M. Kaufman, Multistationarity, the basis of cell differentiation and memory. I. and II., Chaos, 11 (2001) 170-195.. 33.

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