A strongly rigid binary relation

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1 A strongly rigid binary relation Anne Fearnley 8 November 1994 Abstract A binary relation ρ on a set U is strongly rigid if every universal algebra on U such that ρ is a subuniverse of its square is trivial. Rosenberg (1973) found a strongly rigid relation on every universe U of at least 3 elements. We exhibit a new strongly rigid relation for every finite U with U 3. We also show that for U = 3 there are only 2 strongly rigid relations up to isomorphism. AMS classification: 08A40 1 Introduction Let U be a non-empty universe and let n be a positive integer. A function f : U n U is an n-ary function on U. We denote by O (n) U the set of all n-ary functions on U and set O U := 0<n<ω O(n) U. For a positive integer h a subset ρ of U h is an h-ary relation on U (we shall often display h-tuples as column vectors). A function f O (n) U preserves ρ U h if whenever (a 1i a 2i... a hi ) T ρ for all i = 1... n we have that (f(a a 1n ) f(a a 2n )... f(a h1... a hn )) T ρ i.e. ρ is a subuniverse of U; f h. For 1 i n the i-th projection e n i is defined by setting en i (x 1... x n ) x i (the symbol indicates an identity on U i.e. both sides are equal for all x 1... x n U). Clearly every projection preserves ρ. We denote by Polρ the set of all functions on U preserving ρ. It is well known and easy to verify that Polρ is a clone on U i.e. it is a subset of O U which is closed under composition (or substitution) and contains all projections. A relation ρ is strongly rigid if every function on U preserving ρ is a projection [2] i.e. Polρ is the smallest clone; the clone of all projections. From now on relations are always binary U is finite and U 3. Without loss of generality U = n + 1 = {0... n} where n 2. 1

2 It is shown in [2] that the relation {( ) ( ) ( ) ( ) n σ n :=... ( 0 ) 1( ) 2( ) n n n 2 3 ( 1 n ( n 2 n 1 is strongly rigid. In part 2 we show that the relation {( ) ( ) ( ) 1 2 n 1 ρ n :=... ( 2 ) ( 3 ) ( ) ( n ) ( n ) ( ) 2... )} n (1) ) ( n 0 ) } (2) = n-1 n is strongly rigid. In part 3 we show that on three elements (i.e. n = 2) every strongly rigid binary relation is isomorphic to either σ 2 or ρ 2. 2 A new strongly rigid relation Theorem 1 Let n 2. Then the binary relation ρ n defined in (2) is strongly rigid. Proof. We write ρ instead of ρ n. We write a b if (a b) ρ and a b if a b and b a. Recall that e 1 1(x) x. Let U := n 1. Set C := Polρ and C (i) := Polρ O (i) U for i = Claim 1 C (1) = {e 1 1}. Proof. Let h C (1) and a i := h(i) for i = 1... n. From 0 i we have a 0 a i for all i {1... n} likewise n implies that a 1 a 2... a n. Notice that a 1 a 2 therefore a 0 = 0. Now a 1... a n {1... n} satisfy a 1 a 2... a n which is only possible when a i = i for all i {1... n}. Claim 2 Let a {1... n} A := {0 a} m > 1 and f C (m). Then f(b 1... b m ) A whenever b 1... b m A (i.e. A is a subuniverse of U; Polρ ). Proof. By Claim 1 we have f(x... x) x (i.e. f is idempotent). Let b 1... b m A. We distinguish two cases: 2

3 1. If a < n then b i a+1 for all i {1... m}; hence b := f(b 1... b m ) f(a a + 1) = a + 1 proving that b {0 a} = A. 2. If a = n then n 1 b i for all i {1... m}; hence n 1 = f(n 1... n 1) f(b 1... b m ) =: b proving that b {0 n} = A. For a {1... n} and x A = {0 a} we define x a by setting 0 a := a and a a := 0. Notice that the self-map x 1 is the negation from the propositional calculus of logic. An m-ary function f on U is a-selfdual if f(x a 1... x a m) f(x 1... x m ) a for any x 1... x m {0 a}. Claim 3 Let a {1... n}. Then every f C is a-selfdual. Proof. Let f C (m) and b 1... b m A := {0 a}. Set b := f(b 1... b m ) and b := f(b a 1... b a m). Since 0 a we have b i b a i for all i {1... m} therefore b b. But by Claim 2 both b b A which implies that b = b a ; and Claim 3 is verified. Now consider the case a = 1. Let f C (m) and denote by g the restriction of f to {0 1}. By Claim 2 we have g : {0 1} m {0 1} so g is a Boolean function. We say that g is monotone (or order-preserving) if x 1 y 1... x m y m imply that g(x 1... x m ) g(y 1... y m ). Claim 4 The restriction of every f C to {0 1} is a monotone Boolean function. Proof. Let f C (m). Suppose that the restriction of f to {0 1} is not monotone. That means that 1 = f(a) > f(b) = 0 for some a = (a 1... a m ) {0 1} m and b = (b 1... b m ) {0 1} m such that a 1 b 1... a m b m. Set { 1 ai if a c i := i = b i 2 otherwise and c := (c 1... c m ). We must show that b i c i a i for all i {1... m}. Indeed if a i = b i then c i = 1 a i and b i = a i 1 a i (= c i ) a i. Otherwise a i < b i therefore a i = 0 b i = 1 and c i = 2 and thus b i = 1 2(= c i ) 0 = a i. Therefore 0 = f(b) f(c) f(a) = 1 which is a contradiction. Claim 5 The restriction of every f C to {0 1} is a projection. Proof. Let f C. Suppose that g the restriction of f to {0 1} is not a projection. From the Claims 3 and 4 we know that the Boolean function g is monotone and self-dual. It is known [1] that the clone D 2 of Boolean functions which are both monotone and self-dual is a minimal clone (i.e. a clone having the clone of 3

4 projections as its only proper subclone) generated by the majority function m (where m(x y z) := 0 if x + y + z 1 and m(x y z) := 1 otherwise). Since g is not a projection and D 2 is minimal it follows that m is a composition of g (i.e. m can be expressed by a formula made up entirely of the symbol g and the three variables x y z). the same composition with f replacing g everywhere gives us a new function k such that the restriction of k to {0 1} is the majority function m. In other words the clone C contains a function k such that k {01} = m. Since 0 2 and 1 0 we have 1 = k(0 1 1) k(2 0 0) therefore from Claim 2 we obtain k(2 0 0) = 0. Symmetrically k(0 0 2) = 0 and k(0 2 0) = 0. By Claim 3 we have k(0 2 2) = k(2 2 0) = k(2 0 2) = 2. Set a := k(0 1 2). We have 0 = k(2 0 0) k(0 1 2) = a therefore a {1... n}. Now k(1 0 1) k(0 1 2) k(2 2 0) i.e. 1 a 2 which is impossible for a {1... n}. Claim 6 Every f C is a projection. Proof. Let f C (m). Then g := f {01} is a projection by Claim 5 i.e. there exists 1 p m such that f(a 1... a m ) = a p for all a 1... a m {0 1}. By induction on k = 2... n we show that f(b 1... b m ) = b p for all b 1... b m k := {0... k 1}. The statement is true for k = 2. Suppose that the statement is true for some 2 k < n. Let b 1... b m k + 1 and b := f(b 1... b m ). For i = 1... m set { { 1 if bi = 0 1 if bi = 0 d i := and e 0 otherwise i := b i 1 otherwise Then e i b i d i {0 1} for all i = 1... m and so f(e 1... e m ) f(b 1... b m ) f(d 1... d m ). Moreover by the induction hypothesis f(d 1... d m ) = d p and f(e 1... e m ) = e p since d 1... d m e 1... e m k. Therefore e p b d p {0 1}. Now if d p = 1 then b = 0 = b p and we are done. Thus let d p = 0. Then b p 0 and from e p b 0 we obtain b 0. If b p > 1 then b p 1 = e p b therefore b = b p since b 0. Finally let b p = 1. Now suppose to the contrary that b > 1. For i = 1... m set { 2 if bi 1 c i := 0 otherwise and set c := f(c 1... c m ). We have b i c i for i = 1... m; consequently b c. Since c 1... c m {0 2} we know by Claim 2 that c {0 2}. Now b > 1 and b c {0 2} therefore c = 0. Observe that c p = 2 since b p = 1. Set { { 2 if ci = 0 0 if ci = 0 g i := and h 0 if c i = 2 i := 1 if c i = 2 Note that here g p = 0 and h p = 1. By Claim 3 f(g 1... g m ) = f(2 c c m ) = 2 f(c 1... c m ) = 2 c = 2. Since h 1... h m {0 1} we have f(h 1... h m ) = h p = 1 (by Claim 5). Now g i h i for all i {1... m} therefore 2 = g(g 1... g m ) f(h 1... h m ) = 1 a contradiction. This concludes the induction step. For k = n we have the statement of the claim. 4

5 3 Strongly rigid relations on a three-element universe We now show that on {0 1 2} the 6 binary relations isomorphic to σ 2 and the 6 binary relations isomorphic to ρ 2 are the only strongly rigid binary relations. σ 2 : ρ 2 : Here the relation σ 2 is the strongly rigid relation from [2] and the relation ρ 2 is the strongly rigid relation from Section 2. It remains to be shown that all other binary relations on {0 1 2} are not strongly rigid. This is done by finding for each relation a non-trivial (i.e. distinct from a projection) function which preserves it. Let ρ be a binary relation on 3. Every function preserves so ρ. If there exists an a in 3 such that (a a) ρ then the constant function f(x) a preserves ρ. Therefore let ρ be an irreflexive relation. Then ρ 6. If ρ 2 then every minority function preserves ρ (f : 3 3 is a minority function if f(x x y) f(x y x) f(y x x) y). Since Polρ = Polρ 1 (where ρ 1 := {(x y) : (y x) ρ}) if is sufficient to consider only one of the relations ρ and ρ 1. In Table 1 we run through all the irreflexive binary relations ρ with 3 ρ 6 such that ρ is not equivalent to either ρ 2 or σ 2 (the column unary function lists f(0) f(1) f(2)). We have the following possibilities 1. Let ρ = 3. Then either a) ρ consists of a (full) edge and an arc so is of type 1 or b) ρ consists of three arcs and is thus of type 2 or Let ρ = 4. Then either a) ρ consists of two edges and so is of type 4 or b) ρ consists of an edge and two arcs and is thus either isomorphic to σ 2 or of type Let ρ = 5. Then ρ consists of two edges and an arc and is thus isomorphic to ρ Let ρ = 6. Then ρ is of type 6. References [1] E. L. Post. The two-valued iterative systems of mathematical logic. Number 5 in Annals of Math. Studies. Princeton Univ. Press [2] I. Rosenberg. Strongly rigid relations. Rocky Mountain Journal of Mathematics 3:

6 Type ρ ρ Name Function # of rels unary binary ρ strict chain max cycle Table 1: The irreflexive binary relations of cardinality greater than 2 which are not strongly rigid Anne Fearnley Département de Mathématiques et de Statistique Université de Montréal Montreal QC Canada fearnley@dms.umontreal.ca 6