About a question of Gateva-Ivanova and Cameron on square-free set-theoretic solutions of the Yang-Baxter equation
aa r X i v : . [ m a t h . QA ] A p r About a question of Gateva-Ivanova and Cameron onsquare-free set-theoretic solutions of the Yang-Baxterequation
Marco Castelli, Francesco Catino ∗ , Giuseppina Pinto Dipartimento di Matematica e Fisica ”Ennio De Giorgi”, Universit`a del Salento, ViaProvinciale Lecce-Arnesano, 73100 Lecce, Italy
Abstract
In this paper, we introduce a new sequence ¯ N m to find a new estimation of thecardinality N m of the minimal involutive square-free solution of level m . Asan application, using the first values of ¯ N m , we improve the estimations of N m obtained by Gateva-Ivanova and Cameron in [11] and by Lebed and Vendraminin [16]. Following the approach of the first part, in the last section we constructseveral new counterexamples to the Gateva-Ivanova’s Conjecture. Keywords:
Cycle set, set-theoretic solution, Yang-Baxter equation
1. Introduction
The Yang-Baxter equation is one of the basic equations in mathematical physics.Finding all the solutions of this equation is still an open problem, for this reasonDrinfeld [8] posed the question of finding a particular subclass of these solutions,the so-called set-theoretic solution , i.e. the pair (
X, r ), where X is a non-emptyset and r : X × X → X × X a bijective map, satisfying r r r = r r r , where r := r × id X and r := id X × r . Recall that, if λ x : X → X and ρ y : X → X are maps such that r ( x, y ) = ( λ x ( y ) , ρ y ( x )) ∗ Corresponding author
Email addresses: [email protected] (Marco Castelli), [email protected] (Francesco Catino), [email protected] (Giuseppina Pinto) or all x, y ∈ X , a set-theoretic solution of the Yang-Baxter equation ( X, r ) issaid to be a left [ right ] non-degenerate if λ x ∈ Sym ( X ) [ ρ x ∈ Sym ( X ) ] forevery x ∈ X and non-degenerate if it is left and right non-degenerate. Moreovera solution is called involutive if r = id X × X and square-free if r ( x, x ) = ( x, x )for every x ∈ X . The involutive square-free solutions have received a lot ofattention since the work due to Gateva-Ivanova and Van der Bergh [13] wherethey showed that there exist several relations between square-free set-theoreticsolutions of the Yang-Baxter equation and semigroups of I-type, semigroups ofskew-polynomial type and Bieberbach groups. In particular, in [10, 12] mul-tipermutational square-free solutions of finite order were considered. Recallthat if ( X, r ) is an involutive non-degenerate solution, it is possible to consideran equivalence relation ∼ on X which induces in a classical way an involutivesolution Ret ( X, r ) := ( X/ ∼ , ¯ r ), the so-called retraction of ( X, r ) (for moredetails see [9]). An involutive non-degenerate solution is said to be retractable if | Ret ( X ) | < | X | , otherwise it is called irretractable . Moreover, an involu-tive solution is called multipermutational of level m if | Ret m − ( X, r ) | > | Ret m ( X, r ) | = 1, where Ret k ( X, r ) is defined inductively as
Ret k ( X, r ) :=
Ret ( Ret k − ( X, r )) for every natural number k greater than 1.In this context, in 2004 Gateva-Ivanova [10, Question 2.28] conjectured thatevery finite square-free involtive non-degenerate solution ( X, r ) is multipermu-tational. Ced´o, Jespers and Okni´nski [6] proved that the conjecture is true if theassociated permutation group G ( X, r ) is abelian, while some year later Gateva-Ivanova [11] and Ced´o, Jespers and Okni´nski [7], by using different teqniques,showed that if G ( X, r ) is abelian, the conjecture is also true without the finite-ness of X . In 2016, Vendramin [19] proved that in full generality the conjectureis false costructing an involutive square-free irretractable solution of cardinality8. Some years later, in [1, 3], other counterexamples were constructed: even ifseveral examples of irretractable involutive square-free solutions were obtained,the construction of further irretractable square-free solutions still to be veryhard. Recently, Vendramin [20, Problem 19] formally posed the question offinding other irretractable involutive square-free solutions, emphasizing the re-search among those having cardinality 9.On the other hand, because of their links with other algebraic structures, mul-tipermutational square-free involutive solutions have been considered in severalpapers [11, 19, 6, 7]. In that regard, several methods to construct multipermuta-tional solutions were developed: for example, in [7] Ced´o, Jespers and Okni´nskiconstructed the first family of square-free involutive solutions X m of level m andabelian associated permutation group. In 2011, Gateva-Ivanova and Cameron[11] posed the following question: Question. [11, Open Question 6.13] For each positive integer m denote by N m the minimal integer so that there exists a square-free involutive multipermu-2ational solution ( X m , r m ) of order | X m | = N m , and with mpl ( X m , r m ) = m .How does N m depend on m ?They showed that N m ≤ m − + 1 and they noted that for m ∈ { , , } theequality holds. For this reason they conjectured the equality for every m ∈ N .In 2016, Vendramin [19, Example 3.2] answered in negative sense constructingan involutive square-free solution of cardinality 6 and multipermutational level4. The next year Lebed and Vendramin [16] inspected the involutive finite so-lutions of small size and they showed that N = 6 and N = 8. Moreover, theyconsidered the relation between two consecutive terms of the succession N m and they showed that N m +1 ≤ N m : in this way, since N = 8 they indirectlyobtained that N m ≤ m − , for every m > N m , introducing a newsequence ¯ N k , defined as the cardinality of the minimal square-free involutivesolution X of multipermutational level k , having an automorphism α such that σ [ k − ( x ) = σ [ k − ( α ( x )) for an x ∈ X , where σ [ n ] is the epimorphism from X to Ret n ( X ) defined inductively as σ [0] ( x ) := x and σ [ n ] ( x ) := σ σ [ n − ( x ) . In themain result of the paper we will show that N m ≤ ¯ N k · m − k − + 1 (1)for every k < m . As an application, working on the first values of the sequence¯ N k , we will improve the estimations of N m due by Gateva-Ivanova and Cameron[11] and Lebed and Vendramin [16].The main tool of the paper is the algebraic structure of left cycle sets , introducedby Rump in [17] and also considered in several papers (see for example [18, 16,19, 3, 4, 2, 14]). Recall that a non-empty set X with a binary operation · is a left cycle set if each left multiplication σ x : X −→ X, y x · y is invertibleand ( x · y ) · ( x · z ) = ( y · x ) · ( y · z )for all x, y, z ∈ X . Moreover, a left cycle set ( X, · ) is non-degenerate if thesquaring map q : X X, x x · x is bijective. Rump proved [17] that if( X, · ) is a non-degenerate left cycle set, the map r : X × X X × X , definedby r ( x, y ) = ( λ x ( y ) , ρ y ( x )), where λ x ( y ) := σ − x ( y ) and ρ y ( x ) := λ x ( y ) · x , is anon-degenerate involutive solution of the Yang-Baxter equation. Conversely, if( X, r ) is a non-degenerate involutive solution and · the binary operation givenby x · y := λ − x ( y ) for all x, y ∈ X , then ( X, · ) is a non-degenerate left cycleset. The existence of this bijective correspondence allows to move the studyof involutive non-degenerate solutions to non-degenerate left cycle sets. In thiscontext, we prove the inequality (1) by a mixture of two well-known extension-tools of left cycle sets: the one-sided extension of left cycle sets, developed interms of set-theoretic solutions [9] by Etingof, Schedler and Soloviev, and the dynamical extension of left cycle sets developed in [19] by Vendramin.3n the last section we will see that the same approach is useful to constructfurther interesting examples of left cycle sets: referring to [20, Problem 19], weprovide several counterexamples to the Gateva-Ivanova’s Conjecture, in additionto those obtained in [19, 1, 3].
2. Some preliminary results
A non-empty set X with a binary operation · is a left cycle set if the leftmultiplication σ x : X −→ X, y x · y is invertible and( x · y ) · ( x · z ) = ( y · x ) · ( y · z ) (2)for all x, y, z ∈ X . A left cycle set ( X, · ) is non-degenerate if the squaring map q : X X, x x · x is bijective. Rump proved [17] that if ( X, · ) is a leftcycle set, the map r : X × X → X × X given by r ( x, y ) = ( λ x ( y ) , ρ y ( x )),where λ x ( y ) := σ − x ( y ) and ρ y ( x ) := λ x ( y ) · x is a non-degenerate involutivesolution of the Yang-Baxter equation. Conversly, if ( X, r ) is a non-degeneratesolution, the binary operation · defined by x · y := λ − x ( y ) for all x, y ∈ X makes X into a left cycle set. The existence of this correspondence allows to movethe study of involutive non degenerate solutions to left cycle sets, as recentlydone in [19, 3, 4, 16, 15, 18, 5, 2], and clearly to translate in terms of left cycleset the classical concepts related to the non-degenerate involutive set-theoreticsolutions.Therefore, a left cycle set is said to be square-free if the squaring-map q is theidentity on X . The image σ ( X ) of the map σ : X −→ Sym ( X ) , x σ x can beendowed with an induced binary operation σ x · σ y := σ x · y which satisfies (2). Rump [17] showed that ( σ ( X ) , · ) is a non-degenerate leftcycle set if and only if ( X, · ) is non-degenerate. The left cycle set σ ( X ) is calledthe retraction of ( X, · ).The left cycle set ( X, · ) is said to be irretractable if ( σ ( X ) , · ) is isomorphic to( X, · ), otherwise it is called retractable .A non-degenerate left cycle set ( X, · ) is said to be multipermutational of level m , if m is the minimal non-negative integer such that σ m ( X ) has cardinalityone, where σ ( X ) := X and σ n ( X ) := σ ( σ n − ( X )) , for n ≥ . In this case we write mpl ( X ) = m . Obviously, a multipermutational left cycleset is retractable but the converse is not necessarly true.From now on, by a left cycle set we mean a non-degenerate left cycle set. The4ermutation group G ( X ) of X is the subgroup of Sym ( X ) generated by theimage σ ( X ) of σ .In order to construct new examples of left cycle sets, Vendramin [19] introducedthe concept of dynamical cocycle . If I is a left cycle set and S a non-empty set,then α : I × I × S −→ Sym ( S ), ( i, j, s ) α i,j ( s, − ) is a dynamical cocycle if α i · j,i · k ( α i,j ( r, s ) , α i,k ( r, t )) = α j · i,j · k ( α j,i ( s, r ) , α j,k ( s, t )) (3)for all i, j, k ∈ I , r, s, t ∈ S . Moreover, if α is a dynamical cocycle, then the leftcycle set S × α I := ( S × I, · ), where( s, i ) · ( t, j ) := ( α ij ( s, t ) , i · j ) (4)for all i, j ∈ I , s, t ∈ S , is called dynamical extension of I by α .A dynamical cocycle α : I × I × S −→ Sym ( S ) is said to be constant [19] if α ( i,j ) ( r, − ) = α ( i,j ) ( s, − ) for all i, j ∈ I , r, s ∈ S . An example of a constantdynamical cocycle, extensively used in [16] because of its simplicity to compute,is the following: Example 1.
Let S be a finite abelian group, X a left cycle set and f a functionfrom X × X to S such that f ( i, k ) + f ( i · j, i · k ) = f ( j, k ) + f ( j · i, j · k )for all i, j, k ∈ X . Then, α : X × X × S −→ Sym ( S ) given by α ( i,j ) ( s, t ) := t + f ( i, j ),for all i, j ∈ X and s, t ∈ S , is a constant dynamical cocycle. Example 2.
Let X be a left cycle set, k a natural number and S := Z /k Z . Let α : X × X × S −→ Sym ( S ) be the function given by α ( i,j ) ( s, t ) := ( t if i = jt + 1 if i = j Then, α is a constant dynamical cocycle and so S × α X is a left cycle set.An important family of dynamical extensions was obtained by Bachiller, Ced´o,Jespers and Okni´nski. Proposition 1 ([1], Section 2) . Let A and B be non-trivial abelian groupsand let I be a set with | I | > . Let ϕ : A −→ B be a function such that ϕ ( − a ) = ϕ ( a ) for every a ∈ A and let ϕ : B −→ A be a homomorphism. On X ( A, B, I ) := A × B × I we define the following operation ( a, b, i ) · ( c, d, j ) := ( ( c, d − ϕ ( a − c ) , j ) , if i = j ( c − ϕ ( b ) , d, j ) , if i = j or all a, c ∈ A , b, d ∈ B and i, j ∈ I . Then ( X ( A, B, I ) , · ) is a non-degenerateleft cycle set. Recently, we constructed a large family of dynamical extensions that includesthe one obtained by Bachiller et al.
Proposition 2 ([3], Theorem 2) . Let
A, B be a non-empty sets, I a non-degenerate left cycle set and β : A × A × I −→ Sym ( B ) , γ : B −→ Sym ( A ) . Put β ( a,c,i ) := β ( a, c, i ) , γ b := γ ( b ) for a, c ∈ A and b ∈ B . Assume that1) γ b γ d = γ d γ b ,2) β ( a,c,i ) = β ( γ b ( a ) ,γ b ( c ) ,j · i ) ,3) γ β ( a,c,i ) ( d ) γ b = γ β ( c,a,i ) ( b ) γ d ,4) β ( a,c,i · i ) β ( a ′ ,c,i ) = β ( a ′ ,c,i · i ) β ( a,c,i ) hold for all a, a ′ , c ∈ A , b, d ∈ B and i, j ∈ I , i = j .Let · be the operation on A × B × I defined by: ( a, b, i ) · ( c, d, j ) := ( ( c, β ( a,c,i ) ( d ) , i · j ) , if i = j ( γ b ( c ) , d, i · j ) , if i = j . (5) Then X ( A, B, I, β, γ ) := ( A × B × I, · ) is a non-degenerate left cycle set.
3. Left cycle sets and automorphisms
Before being able to prove our main result in the next section, some prelim-inary results are requested. At first, one of the needed tool is the concept ofautomorphism of left cycle sets. If X is a left cycle set, an element α ∈ Sym ( X )is an automorphism of X if α ( x · y ) = α ( x ) · α ( y ) for all x, y ∈ X .In this section, we want to show the importance of the automorphisms groupin the study of the left cycle sets. At this purpose we will see that the auto-morphisms of a left cycle set are useful to construct further examples of leftcycle sets and to understand the structure of particular families of left cyclesets. Moreover it is natural ask which is the automorphism group Aut ( X ) of agiven left cycle set X . For example, if X is the left cycle set given by x · y := y for all x, y ∈ X , then Aut ( X ) = Sym ( X ). At this stage of studies, facing thisproblem in the general case seems to be very hard. However, the following twopropositions are useful to find some automorphisms of particular left cycle sets. Proposition 3.
Let X be a left cycle set and α ∈ Aut ( X ) . Consider theretraction σ ( X ) and let ¯ α : σ ( X ) −→ σ ( X ) be the function given by ¯ α ( σ x ) := σ α ( x ) for every x ∈ X . Then ¯ α ∈ Aut ( σ ( X )) . roof. If x, y ∈ X , then σ x = σ y ⇔ ∀ z ∈ X x · z = y · z ⇔ ∀ z ∈ X α ( x ) · α ( z ) = α ( y ) · α ( z ) ⇔ ∀ z ∈ X σ α ( x ) = σ α ( y ) hence ¯ α is well-defined and injective. With a straightforward calculation, it ispossible to show that ¯ α is an epimorphism. Proposition 4.
Let X be a left cycle set, f ∈ Aut ( X ) and set inductively X := X and X m the left cycle set on X m − × Z / Z given by ( x, s ) · ( y, t ) := ( ( x · y, t ) if x = y ( x · y, t + 1) if x = y for all m ∈ N . Then, the function f m : X m −→ X m given by f m ( x, s , ..., s m ) :=( f ( x ) , s , ..., s m ) for every ( x, s , ..., s m ) ∈ X m is an automorphism of X m forevery m ∈ N .Proof. We prove the thesis by induction on m . If m = 1 we have f (( x, s ) · ( y, t )) = f ( x · y, t + 1 − δ x,y )= ( f ( x ) · f ( y ) , t + 1 − δ f ( x ) ,f ( y ) )= ( f ( x ) , s ) · ( f ( y ) , t )= f ( x, s ) · f ( y, t )for all ( x, s ) , ( y, t ) ∈ X × Z / Z = X . Now, let us assume the thesis true for anatural number m . Then, f m +1 (( x, s , ..., s m +1 ) · ( y, t , ..., t m +1 ))= f m +1 (( x, s , ..., s m ) · ( y, t , ..., t m ) , t m +1 + 1 − δ ( x,s ,...,s m ) , ( y,t ,...,t m ) )= f m +1 ( y, t , ..., t m + 1 − δ ( x,s ,...,s m − ) , ( y,t ,...,t m − ) , t m +1 + 1 − δ ( x,s ,...,s m ) , ( y,t ,...,t m ) )= ( f ( y ) , t , ..., t m + 1 − δ ( x,s ,...,s m − ) , ( y,t ,...,t m − ) , t m +1 + 1 − δ ( x,s ,...,s m ) , ( y,t ,...,t m ) )= ( f m (( x, s , ..., s m ) · ( y, t , ..., t m )) , t m +1 + 1 − δ f m ( x,s ,...,s m ) ,f m ( y,t ,...,t m ) )= ( f m ( x, s , ..., s m ) · f m ( y, t , ..., t m ) , t m +1 + 1 − δ f m ( x,s ,...,s m ) ,f m ( y,t ,...,t m ) )= ( f m ( x, s , ..., s m ) , s m +1 ) · ( f m ( y, t , ..., t m ) , t m +1 )= ( f ( x ) , s , ..., s m , s m +1 ) · ( f ( y ) , t , ..., t m , t m +1 )= f m +1 ( x, s , ..., s m +1 ) · f m +1 ( y, t , ..., t m +1 )hence f m +1 ∈ Aut ( X m +1 ). 7ateva-Ivanova and Cameron [11] and Etingof et al. [9] showed that automor-phisms of left cycle sets allows to construct other examples of left cycle sets. Weprove that, if X is a left cycle set, α ∈ Aut ( X ) and z / ∈ X , then, under suitablehypothesis, the retraction σ ( X ∪ { z } ) is isomorphic to the left cycle set havingthe disjoint union σ ( X ) ∪ { α } as underlying set. Proposition 5.
Let X be a left cycle set, α ∈ Aut ( X ) , z / ∈ X and ( X ∪ { z } , ◦ ) the algebraic structure given by x ◦ y := x · y if x, y ∈ Xy if y = zα ( y ) if y ∈ X, x = z. Then the pair ( X ∪ { z } , ◦ ) is a left cycle set.Moreover, suppose that α = σ x for all x ∈ X . Then the retraction σ ( X ∪ { z } ) is isomorphic to the left cycle set ( σ ( X ) ∪ { σ z } , ◦ ) given by σ x ◦ σ y := σ x · σ y if σ x , σ y ∈ Xσ y if y = zσ α ( y ) if y ∈ X, x = z. Proof.
By [9, Section 2], ( X ∪ { z } , ◦ ) is a left cycle set. Now, since α = σ x ,for every x ∈ X , there is a natural bijection φ from σ ( X ∪ { z } ) to the disjointunion σ ( X ) ∪ { σ z } given by φ ( σ x ) := σ ′ x , for every σ x ∈ σ ( X ∪ { z } ), where σ ′ x and σ x are the left multiplications in σ ( X ) ∪ { σ z } and σ ( X ∪ { z } ) respectively.It is easy to see that φ is a homomorphism, so the thesis follows.
4. A new estimation of N m The goal of this section is to provide an estimation of N m depending on anothersequence, which we denote by ¯ N k , that will allow us to improve the estimationobtained by Lebed and Vendramin [16].Following Gateva-Ivanova and Cameron [11], if X is a left cycle set and n anatural number, we indicate by σ [ n ] the epimorphism from X to σ n ( X ) definedinductively by σ [0] ( x ) := x σ [ n ] ( x ) := σ σ [ n − ( x ) for all n ∈ N and x ∈ X . The following Lemma, due to Gateva-Ivanova andCameron, involves the function σ [ n ] and it will be useful in our work. Lemma 6 ([11], Proposition 7.8(3)) . Let X be a finite square-free left cycle setof multipermutational level k . Then, the sets σ − k − ( x ) are G ( X ) -invariant.
8e indicate by ¯ N k the cardinality of the minimal square-free left cycle set X of level k having an automorphism α such that there exists x ∈ X with σ [ k − ( x ) = σ [ k − ( α ( x )). For example, let X := { a, b } be the left cycle oflevel 1 given by σ a = σ b := id X and put α := ( a b ). Then, α ∈ Aut ( X ) and σ [0] ( a ) = σ [0] ( α ( a )), therefore ¯ N = 2.In order to prove the main result of the paper, we need some preliminaryresults. Lemma 7.
Let X be a square-free left cycle set of level k , { z } a set with asingle element such that z / ∈ X and α ∈ Aut ( X ) such that there exists x ∈ X with σ [ k − ( x ) = σ [ k − ( α ( x )) . Then, the left cycle set ( X ∪ { z } , ◦ ) given by x ◦ y := x · y if x, y ∈ Xy if y = zα ( y ) if y ∈ X, x = z. has level k + 1 .Proof. We prove the thesis by induction on k . If k = 1, then x · y = y for all x, y ∈ X and there exist x, y ∈ X such that x = y and α ( x ) = y . This impliesthat for the left cycle set ( X ∪ { z } , ◦ ) we have that σ x = id X ∪{ z } for every x ∈ X and σ z = α = id X ∪{ z } , hence | σ ( X ∪ { z } ) | = 2 and mpl ( X ∪ { z } ) = 2.Now, if X has level k , then mpl ( X ∪ { z } , ◦ ) = 1 + mpl ( σ ( X ∪ { z } )) and, byLemma 6, σ z = σ x for every x ∈ X . Hence, by Proposition 5, σ ( X ∪ { z } ) isisomorphic to the left cycle set ( σ ( X ) ∪ { σ z } , ◦ ) given by σ x ◦ σ y := σ x · σ y if σ x , σ y ∈ Xσ y if y = zσ α ( y ) if y ∈ X, x = z. Moreover, by Proposition 3, we have that σ σ z is an element ot Aut ( σ ( X )). If x and y are elements of X such that σ [ k − ( x ) = σ [ k − ( y ) and α ( x ) = y , it followsthat σ σ z ( σ x ) = σ z · x = σ α ( x ) = σ y and σ [ k − ( σ x ) = σ [ k − ( x ) = σ [ k − ( y ) = σ [ k − ( σ y )hence we can apply the inductive hypothesis. Therefore, mpl ( X ∪ { z } , ◦ ) = 1 + mpl ( σ ( X ) ∪ { σ z } ) = 1 + ( k − k + 1and the thesis follows. 9 emma 8. Let X be a square-free left cycle set of level k , { z } a set with asingle element z / ∈ X and α ∈ Aut ( X ) such that there exists x ∈ X with σ [ k − ( x ) = σ [ k − ( α ( x )) . Moreover, set inductively X = Z := X , X m the leftcycle set on X m − × Z / Z given by ( x, s ) · ( y, t ) := ( ( x · y, t ) if x = y ( x · y, t + 1) if x = y for all x, y ∈ X m − and s, t ∈ Z / Z and ( Z m , ◦ ) the algebraic structure givenby Z m := X m − ∪ { z } and x ◦ y := x · y if x, y ∈ X m − y if y = z ( α ( y ) , ..., y m − ) if y = ( y , ..., y m − ) ∈ X m − , x = z. for every m ∈ N . Then Z m is a square-free left cycle set of level k + m .Proof. By Propositions 4 and 5, the pair ( Z m , ◦ ) is a square-free left cycle set.By induction on m we prove that ( Z m , ◦ ) has level k + m . If m = 1, the thesisfollows by the previous proposition. Now, let us suppose the thesis true for anatural number m . Since mpl ( Z m +1 ) = 1 + mpl ( σ ( Z m +1 )) = 1 + mpl ( σ ( X m ∪ { z } ))and, by Propositions 4 and 5, σ ( X m ∪ { z } ) is isomorphic to the left cycle set( σ ( X m ) ∪ { σ z } , ◦ ) given by σ x ◦ σ y := σ x · σ y if σ x , σ y ∈ X m σ y if y = zσ α ( y ) if y ∈ X m , x = z. it follows that mpl ( Z m +1 ) = 1 + mpl ( σ ( X m ) ∪ { σ z } ). Finally, by [16, Theorem10.6 and Corollary 10.7], σ ( X m ) is isomorphic to X m − , and so we obtain that σ ( X m ) ∪ { σ z } is isomorphic to Z m . By the inductive hypothesis, we have that mpl ( Z m +1 ) = 1 + mpl ( Z m ) = 1 + k + m, hence the thesis. Theorem 9.
Let N m be the cardinality of the minimal square-free left cycle setof level m and ¯ N k the cardinality of the minimal square-free left cycle set X oflevel k such that there exists an automorphism α and an element x ∈ X with σ [ k − ( x ) = σ [ k − ( α ( x )) . Then, the inequality N m ≤ ¯ N k · m − k − + 1 (6) holds for every k < m . roof. If r is a natural number and X is a left cycle set of level k and cardinality¯ N k having an automorphism α such that there exists x ∈ X with σ [ k − ( x ) = σ [ k − ( α ( x )), then the left cycle set Z r , constructed as in the previous Lemma,is a square-free left cycle set of level k + r and cardinality ¯ N k · r − + 1, hence N r + k ≤ ¯ N k · r − + 1. Setting m := r + k , we obtain N m ≤ ¯ N k · m − k − + 1hence the thesis.
5. Some examples and comments
The goal of this section is to calculate the first values of the sequence ¯ N m and to use these to improve the estimations of the sequence N m obtained byGateva-Ivanova and Cameron in [11] and by Lebed and Vendramin in [16].In the following examples we calculate the numbers ¯ N k for k ∈ { , , , } . Examples 1.
1) If X := { , } is the left cycle set given by x · y := y forall x, y ∈ X , then the permutation α := (1 2) is an automorphism of X .Moreover, σ [0] (1) = 1 = 2 = σ [0] (2), hence ¯ N = N = 2.2) The unique square-free left cycle set of size 3 and level 2 is given by σ = σ := id X and σ := (1 2) and the group of automorphism isgenerated by σ . Since σ [1] (1) = σ [1] (2), necessarily ¯ N >
3. Now, let X := { , , , } be the left cycle set given by σ = σ := (3 4) σ = σ := (1 2) . Then, X has level 2, α := (1 3)(2 4) is an automorphism of X and σ [1] (1) = σ [1] (3), hence ¯ N = 4.3) We know that N = 5 and, by calculation, there are two left cycle setsof level 3 and size 5. The first one is given by X := { , , , , } , σ = σ := (3 4), σ = σ := (1 2) and σ := (1 3 2 4). So, the fibers of σ [2] are σ − (1) = { , , , } and σ − (5) = { } , but an automorphism thatmaps 5 to an other element of X can not exists because there are notleft multiplications of X conjugate to σ . Using the same argument forthe other left cycle set of level 3 and size 5, we prove that ¯ N >
5. Let X := { , , , , , } be the left cycle set given by σ = σ := id X σ := (5 6) σ := (1 2)(5 6) σ := (3 4) σ := (1 2)(3 4) . Then X has level 3, σ [2] (3) = σ [2] (5) and α := (3 5)(4 6) ∈ Aut ( X ), hence¯ N = 6. 11) By calculation, we know that ¯ N must be greater than 7. Let X := { , , , , , , , } be the left cycle set given by σ = σ := (7 8) σ := (5 6) σ := (1 7)(2 8)(5 6) σ := (3 4) σ := (1 7)(2 8)(5 6) σ = σ := (1 2) . Then X has level 4, σ [3] (3) = σ [3] (5) and α := (3 5)(4 6) is an automor-phism of X , hence ¯ N = 8.Unfortunately, for k > N k .However, the following example allow us to give an estimation of ¯ N . Example 3.
Let X := { , , , , , , , , , } be the left cycle set given by σ = σ := id X σ := (5 6) σ := (1 2)(5 6) σ := (3 4) σ := (1 2)(3 4) σ := (9 10) σ := (3 5)(4 6)(9 10) σ := (7 8) σ := (3 5)(4 6)(7 8) . Then X has level 5, σ [4] (7) = σ [4] (9) and α := (7 9)(8 10) ∈ Aut ( X ), hence¯ N ≤ N ≤
10, it follows that N m ≤ m − − · m − + 1 (7)for every m >
5. This improve the estimation N m ≤ m − + 1, for every m ∈ N ,due to Gateva-Ivanova and Cameron [11], and the estimation N m ≤ m − (8)for every m >
4, implicitly obtained by Lebed and Vendramin [16]. For example,by (8) we have that N ≤
16 while using (7) we obtain that N ≤
11. If wewant to estimate N , even if we use the better estimation of N obtained inthe previous step and even if we use instead of (8) the inequality N m +1 ≤ N m we obtain N ≤
22 while, by (7), we have that N ≤
21. Using a similarargument one can see that the estimation (7) is actually the best possible forevery m > N k for some k >
6. Irretractable square-free left cycle sets
The multipermutational left cycle sets constructed in the previous sections areobtained by a mixture between some dynamical extensions [19] and a particularcase of an extension-tool developed by Etingof, Schedler and Soloviev called one-sided extension (for more details see [9, Section 2]).12sing the same approach, in this section we find new examples of irretractableleft cycle sets: in particular, according to [20, Problem 19], we obtain severalcounterexamples to the Gateva-Ivanova’s Conjecture that are different fromthose obtained in [3, 1].First of all, we recall the constructions of the irretractable left cycle sets ob-tained by Bachiller, Ced´o, Jespers and Okni´nski [1] and by the authors [3].
Proposition 10 ([1], Theorem 3.3(a)-(b)) . Let
A, B, I, ϕ , ϕ and X ( A, B, I ) be as in Proposition 1.1) X ( A, B, I ) is square-free if ad only if ϕ (0) = 0 ;2) If ϕ − (0) = { } and ϕ is injective, then X ( A, B, I ) is irretractable. Proposition 11 ([3], Proposition 3) . Let
A, B, I, β, γ be as in Proposition 2, | A | , | B | , | I | > . Then if: A × A × { i } ∩ β − ( β ( a,a ) ( i, − )) ⊆ { ( k, k, i ) | k ∈ A } for all i ∈ I and a ∈ A , γ is injective,the non-degenerate left cycle set X ( A, B, I, β, γ ) is irretractable. In a similar way for the left cycle sets considered in Section 3, we can easilyobtain some automorphisms of the left cycle sets constructed by Bachiller, Ced´o,Jespers and Okni´nski.
Lemma 12.
Let X ( A, B, I ) be the left cycle set of the Proposition 1, m ∈ Sym ( I ) and let ψ m the function given by ψ m ( a, b, i ) := ( a, b, m ( i )) for every ( a, b, i ) ∈ X ( A, B, I ) . Then, ψ m is an automorphism and the set G := { ψ m | m ∈ Sym ( I ) } is a subgroup of Aut ( X ( A, B, I )) isomorphic to Sym ( I ) .Proof. Clearly ψ m is bijective. Moreover, since i = j if and only if m ( i ) = m ( j ),we have ψ m ( a, b, i ) · ψ m ( c, d, j ) = ( a, b, m ( i )) · ( c, d, m ( j ))= ( c + (1 − δ m ( i ) ,m ( j ) ) ϕ ( b ) , d + δ m ( i ) ,m ( j ) ϕ ( a − c ) , m ( j ))= ψ m (( c + (1 − δ i,j ) ϕ ( b ) , d + δ i,j ϕ ( a − c ) , j ))= ψ m (( a, b, i ) · ( c, d, j ))for all ( a, b, i ) , ( c, d, j ) ∈ A × B × I , where δ k,l = 1 if k = l and δ k,l = 0 otherwise.The rest of the proof is a straightforward calculation.From now on, for the left cycle set X ( A, B, I ), we will indicate by ψ m theautomorphism associated as in the previous Lemma, for every m ∈ Sym ( I ). Proposition 13.
Let X ( A, B, I ) be the irretractable left cycle set of the Propo-sition 10, ( Y, · ) a left cycle set and α : Y −→ Aut ( X ( A, B, I )) an injective unction such that α ( Y ) ⊆ G , where G is the subgroup of Aut ( X ( A, B, I )) ofthe previous Lemma. Moreover, suppose that α ( a · b ) α ( a ) = α ( b · a ) α ( b ) for all a, b ∈ Y and let ( X ( A, B, I ) ∪ Y, ◦ ) be the algebraic structure given by x ◦ y := x · y if x, y ∈ X ( A, B, I ) y if y = zα ( y ) if y ∈ X ( A, B, I ) , x = z. Then, the pair ( X ( A, B, I ) ∪ Y ) ◦ := ( X ( A, B, I ) ∪ Y, ◦ ) is an irretractable leftcycle set.Proof. By [9, Section 2], we have that ( X ( A, B, I ) ∪ Y ) ◦ is a left cycle set, so it issufficient to show that σ x = σ y for every x = y . If x, y ∈ X ( A, B, I ) or x, y ∈ Y then clearly σ x = σ y . Now, suppose that x := ( a, b, i ) ∈ X ( A, B, I ) and y ∈ Y .If α ( y ) = id I , since σ x ( A × B × { j } ) = A × B × { j } and σ y ( A × B × { j } ) = A × B ×{ α ( y )( j ) } for every j ∈ I , then necessarily σ x = σ y . Finally, if α ( y ) = id I and σ x = σ y , then σ x ( c, d, j ) = ( c, d, j ) for all c ∈ A, d ∈ B, j ∈ I and this implies σ ( a,b,i ) = σ ( c,b,i ) for every c ∈ A , in contraddiction with the irretractability of X ( A, B, I ). Example 4.
Let X ( A, B, I ) be the irretractable square-free left cycle set hav-ing 8 element, m := id I , Y := { m } the left cycle set having 1 element and α : Y −→ Aut ( X ( A, B, I )) the function given by α ( m ) := ψ m . By the previousProposition, ( X ( A, B, I ) ∪ Y ) ◦ is an irretractable left cycle set; moreover, since X ( A, B, I ) is square-free, we have that ( X ( A, B, I ) ∪ Y ) ◦ is square-free. There-fore, ( X ( A, B, I ) ∪ Y ) ◦ is a counterexample of the Gateva-Ivanova’s Conjectureof cardinality 9. Example 5.
Let X ( A, B, I ) be the irretractable square-free left cycle set hav-ing 8 element, m := (1 2), Y := { m } the left cycle set having 1 element and α : Y −→ Aut ( X ( A, B, I )) the function given by α ( m ) := ψ m . By Propo-sition 13, ( X ( A, B, I ) ∪ Y ) ◦ is an irretractable left cycle set; moreover, since X ( A, B, I ) is square-free, we have that ( X ( A, B, I ) ∪ Y ) ◦ is square-free. There-fore, ( X ( A, B, I ) ∪ Y ) ◦ is a counterexample of the Gateva-Ivanova’s Conjectureof cardinality 9. Since σ x = id X ( A,B,I ) ∪ Y for every x ∈ X ( A, B, I ) ∪ Y , this leftcycle set can not be isomorphic to the previous one.Finding automorphism of the square-free left cycle set having 8 element differ-ent from those obtained in Lemma 12 can be useful to find other counterexam-ples to the Gateva-Ivanova’s Conjecture of size 9, as we can see in the followingexample. Example 6.
Let X ( A, B, I ) be the irretractable square-free left cycle set having8 element and f : X ( A, B, I ) −→ X ( A, B, I ) the function given by f ( a, b, i ) :=14 a + 1 , b, i ) for every ( a, b, i ) ∈ A × B × I . Then, f ∈ Aut ( X ( A, B, I )): indeed f is clearly bijective and f ( a, b, i ) · f ( c, d, j ) = ( a + 1 , b, i ) · ( c + 1 , d, j )= ( c + 1 + (1 − δ i,j ) ϕ ( b ) , d + δ i,j ϕ ( a − c ) , j )= f ( c + (1 − δ i,j ) ϕ ( b ) , d + δ i,j ϕ ( a − c ) , j )= f (( a, b, i ) · ( c, d, j ))for all ( a, b, i ) , ( c, d, j ) ∈ A × B × I . Therefore, ( X ( A, B, I ) ∪ Y ) ◦ is a coun-terexample of the Gateva-Ivanova’s conjecture of cardinality 9, where Y := { w } is a left cycle set of size 1 and α ( w ) := f . Since σ x = id X ( A,B,I ) ∪ Y for every x ∈ X ( A, B, I ) ∪ Y , this left cycle set can not be isomorphic to the one inExample 4. Moreover, this left cycle set can not be isomorphic to the previ-ous one: indeed, the previous example has two orbits respect to the associatedpermutation group, while this left cycle set has three orbits.However, Proposition 13 allow us to obtain many other irretractable left cyclesets arbitrarily large, as we can see in the following examples. Example 7.
Let p be an odd prime and k ∈ N such that p = 4 k +1. Moreover,let A = B := Z / Z , I := { , ..., k } , ϕ = ϕ = id A and consider the irretractableleft cycle set X ( A, B, I ) as in Proposition 10. Moreover, let m ∈ Sym ( I ), Y := { m } the left cycle set of 1 element and α : Y −→ Aut ( X ( A, B, I )) givenby α ( m ) := ψ m . Then, ( X ( A, B, I ) ∪ Y ) ◦ is an irretractable square-free leftcycle set having p elements. Example 8.
Let p be a prime number such that p ≡ mod
4) and k ∈ N suchthat 3 p = 4 k + 3. Moreover, let A = B := Z / Z , I := { , ..., k } , ϕ = ϕ = id A and consider the irretractable left cycle set X ( A, B, I ) as in Proposition 10.Moreover, let m := id I , m := id I and m := (1 2) and Y := { m , m , m } theleft cycle set given by σ m = σ m := id Y and σ m := ( m m ). Set α ( m ) = α ( m ) := id X ( A,B,I ) and α ( m ) := ψ m . Then, ( X ( A, B, I ) ∪ Y, ◦ ), where ◦ isthe binary operation defined as in Proposition 13, is a square-free left cycle sethaving 3 p elements. It is not irretractable since σ m = σ m = id X ( A,B,I ) ∪ Y .If we consider the left cycle set X ( A, B, I, β, γ ) of Proposition 2, we can easilygeneralize the argument of Proposition 13.
Lemma 14.
Let X ( A, B, I, β, γ ) be the left cycle set of Theorem 2, m ∈ Aut ( I ) such that β ( a,b ) ( i, − ) = β ( a,b ) ( m ( i ) , − ) for every ( a, b, i ) ∈ A × B × I and let ψ m be the function given by ψ m ( a, b, i ) := ( a, b, m ( i )) for every ( a, b, i ) ∈ X ( A, B, I ) .Then, ψ m is an automorphism of X ( A, B, I, β, γ ) and the set G := { ψ m | ψ m ∈ Aut ( X ( A, B, I, β, γ )) } is a subgroup of Aut ( X ( A, B, I, β, γ )) isomorphic to asubgroup of Aut ( I ) . roposition 15. Let X ( A, B, I, β, γ ) be the irretractable left cycle set of Propo-sition 11, ( Y, · ) a left cycle set and α : Y −→ Aut ( X ( A, B, I, β, γ )) an injectivefunction such that α ( Y ) ⊆ G , where G is the subgroup of Aut ( X ( A, B, I, β, γ )) of the previous Lemma. Moreover, suppose that α ( a · b ) α ( a ) = α ( b · a ) α ( b ) forall a, b ∈ Y and let ( X ( A, B, I, β, γ ) ∪ Y, ◦ ) be the algebraic structure given by x ◦ y := x · y if x, y ∈ X ( A, B, I, β, γ ) y if y = zα ( y ) if y ∈ X ( A, B, I, β, γ ) , x = z. Then, the pair ( X ( A, B, I, β, γ ) ∪ Y, ◦ ) is an irretractable left cycle set. We leave the proofs of the previous results to the reader because they are sim-ilar to the ones of Lemma 12 and Proposition 13.
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