Decomposition theorems for involutive solutions to the Yang-Baxter equation
aa r X i v : . [ n li n . S I] J a n DECOMPOSITION THEOREMS FOR INVOLUTIVESOLUTIONS TO THE YANG–BAXTER EQUATION
S. RAM´IREZ, L. VENDRAMIN
Abstract.
Motivated by the proof of Rump of a conjecture of Gateva–Ivanova on the decomposability of square-free solutions to the Yang–Baxter equation, we present several other decomposability theoremsbased on the cycle structure of a certain permutation associated withthe solution. Introduction
In [11] Drinfeld considered set-theoretic solutions, i.e. pairs (
X, r ), where X is a non-empty set and r : X × X → X × X is a bijective map such that( r × id)(id × r )( r × id) = (id × r )( r × id)(id × r ) . The first papers about these solutions were written by Etingof, Schedler andSoloviev [12] and Gateva-Ivanova and Van den Bergh [16]. Both paper stud-ied involutive solutions. The theory of such solutions was further developedin several other papers, see for example [4, 6, 7, 13–15, 20, 21].Originally Gateva–Ivanova mainly studied involutive square-free solutions,i.e. solutions (
X, r ) such that r ( x, x ) = ( x, x ) holds for all x ∈ X . Square-free solutions produce algebraic structures with several interesting proper-ties, see for example [13]. Nowadays it is clear that not only square-freesolutions are of interest. In fact, several intriguing connections involvingarbitrary involutive solutions were found recently [22]. These connectionsinclude radical rings [21], homology [18], ordered and diffuse groups [3,9,19]and Garside theory [8, 10].Recent research on the combinatorics of the Yang–Baxter equation ismainly based on the use of powerful algebraic methods. This leads to sev-eral breakthroughs in the area and new families of solutions were found.However, the problem of constructing new classes of solutions is still chal-lenging and a program to classify all solutions is somewhat still out of reach.There are two promising approaches in this context. One is based on in-decomposable solutions and the other one is based on retractable solutions.In both cases, the idea is to understand small solutions suitable to play therole of building blocks for the theory.The starting point of this work is an old conjecture of Gateva–Ivanovaabout the decomposability of certain solutions. In a 1996 talk at the Interna-tional Algebra Conference in Hungary, she conjectured that finite square-free solutions are always decomposable, that is solutions where the underlyingset X that can be decomposed into a disjoint union X = Y ∪ Z of subsets Y and Z of X such that r ( Y × Y ) ⊆ Y × Y and r ( Z × Z ) ⊆ Z × Z . The conjecture was proved by Rump in [20]. In the same paper he alsoshowed that the conjecture cannot be extended to infinite solutions.Based on the work of Etingof, Schedler and Soloviev [12], we considerthe diagonal of a solution (
X, r ), a certain permutation T of the set X .The cycle structure of this permutation turns out to be an invariant of thesolution. Rump’s theorem can then be restated as follows: If the diagonalmap of a solution ( X, r ) fixes all points of X , then ( X, r ) is decomposable.In this work we deal with decomposition theorems similar to that ofRump. The basic tools are the diagonal map of a solution and a recentresult of Ced´o, Jespers and Okni´nski about solutions with a primitive per-mutation group [5]. It seems that Rump’s result is just the tip of the iceberg,as –at least for finite solutions– the bijective map T turns out to encode deepcombinatorial information about the structure of the solution. In particular,the map T could be used to detect (in)decomposability or retractability ofsolutions.The paper is organized as follows. We recall some preliminaries in Sec-tion 2. We recall the classification of involutive solutions with a primenumber of elements found by Etingof, Schedler and Soloviev in [12], Rump’stheorem on decomposability of square-free solutions [20] and the theoremof Ced´o, Jespers and Okni´nski on involutive solutions with primitive per-mutation groups [5]. Our results appear in Section 3. In Theorem 3.2 weprove that indecomposable solutions are always defined by permutations dif-ferent from the identity. We remark that this result might be useful for theproblem of constructing explicitly indecomposable solutions of small size.In Theorem 3.5 we prove that involutive solutions where the diagonal is acycle of maximal length are indecomposable. Theorem 3.10 deals with thedecomposability of those solutions where the diagonal fixes only one point.Theorem 3.13 is devoted to the decomposability of those solutions wherethe diagonal fixes two or three points. We conclude the paper with somequestions and conjectures. 2. Preliminaries A set-theoretic solution to the Yang–Baxter equation is a pair ( X, r ),where X is a non-empty set and r : X × X → X × X is a bijective mapsuch that ( r × id)(id × r )( r × id) = (id × r )( r × id)(id × r ) . By convention, we write r ( x, y ) = ( σ x ( y ) , τ y ( x )) . A solution (
X, r ) is said to be non-degenerate if all the maps σ x : X → X and τ x : X → X are bijective for all x ∈ X , and it is said to be involutive if r = id X × X . Note that for non-degenerate involutive solutions, τ y ( x ) = σ − σ x ( y ) ( x ) and σ x ( y ) = τ − τ y ( x ) ( y ) . By convention, a solution will be a non-degenerate involutive solution tothe Yang–Baxter equation.
ECOMPOSITION THEOREMS 3 A homomorphism between the solutions ( X, r X ) and ( Y, r Y ) is a map f : X → Y such that the diagram X × X f × f (cid:15) (cid:15) r X / / X × X f × f (cid:15) (cid:15) Y × Y r Y / / Y × Y is commutative, i.e. ( f × f ) ◦ r X = r Y ◦ ( f × f ). An isomorphism of solutionsis a bijective homomorphism of solutions. Solutions form a category.A solution ( X, r ) is said to be square-free if r ( x, x ) = ( x, x ) for all x ∈ X .In [12, Proposition 2.2] it is proved that if ( X, r ) is a solution, then themap T : X → X , T ( x ) = τ − x ( x ), is invertible with inverse x σ − x ( x ) and,moreover, τ − x ◦ T = T ◦ σ x for all x ∈ X . Definition 2.1.
The diagonal of a solution (
X, r ) is defined as the permu-tation T : X X , x τ − x ( x ).The permutation group of a solution ( X, r ) is defined as the subgroup G ( X, r ) of Sym X generated by the set { σ x : x ∈ X } . Note that if X is finite,then G ( X, r ) is finite. The group G ( X, r ) naturally acts on X . A solution( X, r ) is said to be indecomposable if the group G ( X, r ) acts transitively on X and decomposable otherwise. This definition of decomposability turnsout to be equivalent to the definition mentioned in the introduction, see forexample the proof of [12, Proposition 2.12]. Using the invertible map T oneproves that G ( X, r ) is isomorphic (as a permutation group on the set X ) tothe group generated by the set { τ x : x ∈ X } . Theorem 2.2 (Etingof–Schedler–Soloviev) . Let p be a prime number. Anindecomposable solution with p elements is a cyclic solution , i.e. a solutionisomorphic to ( Z /p, r ) , where r ( x, y ) = ( y − , x + 1) .Proof. See [12, Theorem 2.13] (cid:3)
We recall Rump’s theorem.
Theorem 2.3 (Rump) . Let ( X, r ) be a finite solution. If the diagonal ofthe solution is the identity, then ( X, r ) is decomposable.Proof. See [20, Theorem 1]. (cid:3)
A solution (
X, r ) is said to be primitive if the group G ( X, r ) acts primi-tively on X . In the Oberwolfach mini-workshop Algebraic Tools for Solvingthe Yang–Baxter Equation , Ballester–Bolinches posed the following conjec-ture [17]: Finite primitive solutions are of size p for some prime number p .The conjecture was proved by Ced´o, Jespers and Okni´nski. The proof usesthe theory of braces and Theorem 2.2. Theorem 2.4 (Ced´o–Jespers–Okni´nski) . If ( X, r ) is a finite primitive so-lution, then | X | is a prime number and ( X, r ) is a cyclic solution.Proof. See [5, Theorem 3.1]. (cid:3)
S. RAM´IREZ, L. VENDRAMIN
In order to prove the conjecture mentioned in the introduction, Rumpintroduced a better way to deal with solutions. A cycle set is a non-emptyset X provided with a binary operation X × X → X , ( x, y ) x · y , suchthat the maps ϕ x : X → X , y x · y , are bijective for all x ∈ X and( x · y ) · ( x · z ) = ( y · x ) · ( y · z )holds for all x, y, z ∈ X .A homomorphism between the cycle sets X and Z is a map f : X → Z such that f ( x · y ) = f ( x ) · f ( y ) for all x, y ∈ X . An isomorphism of cyclesets is a bijective homomorphism of cycle sets. Cycle sets form a category.A cycle set X is said to be non-degenerate if the map X → X , x x · x , isbijective. Rump proved that solutions are in bijective correspondence withnon-degenerate cycle sets. The correspondence is given by r ( x, y ) = (( y ∗ x ) · y, y ∗ x ) , where y ∗ x = z if and only if x = y · z , see [20, Proposition 1]. Moreover,the category of solutions and the category of non-degenerate cycle sets areequivalent.The permutation group G ( X ) of a cycle set X is defined as the permu-tation group of its associated solution. Indecomposable and decomposblecycle sets are then defined in the usual way.The diagonal of a cycle set X is then defined as the diagonal of its asso-ciated solution. Clearly, the cycle structure of the diagonal of a finite cycleset is invariant under isomorphisms. Note that a solution is square-free ifand only if the diagonal of its associated cycle set is the identity.We write Sym X to denote the set of bijective maps X → X . Lemma 2.5.
Let X be a cycle set with diagonal T . If U : X → X is abijective map and U and T are conjugate in Sym X , then there exists anisomorphic cycle set structure on the set X with diagonal U .Proof. Let ( x, y ) x · y be the cycle set operation on X and let γ ∈ Sym X be such that U = γ − ◦ T ◦ γ . A direct calculation shows that the operation y · U z = γ − ( γ ( y ) · γ ( z )) turns X into a cycle set isomorphic to the original X . Moreover, y · U y = γ − ( γ ( y ) · γ ( y )) = γ − ( T ( γ ( y ))) = ( γ − ◦ T ◦ γ )( y ) = U ( y )holds for all y ∈ X . (cid:3) Decomposition theorems
For the first results in this section we use the language of cycle sets. Westart with some easy consequences of Theorem 2.4 and the results of [2].
Theorem 3.1.
Let X be a finite cycle set. If some ϕ x is a cycle of lengthat least | X | / and is coprime with | X | then X is decomposable.Proof. We first note that Theorem 2.4 implies that in none of these cases G ( X ) can act primitively on X . Let y be a fixed point of ϕ x and B a non-trivial block containing y . Since ϕ x fixes y we must have ϕ x ( B ) = B . Ifthis block were to contain an element that is not fixed by ϕ x , and since thispermutation is a cycle, it would have to contain all of them, which cannot ECOMPOSITION THEOREMS 5 happen since the block would then have more than | X | / y was arbitrary this means any block containing a point fixed by ϕ x mustconsist entirely of fixed points. In particular the size of the block mustdivide the number of fixed points. Since it also has to divide | X | and thesenumbers are coprime, there cannot be non-trivial blocks, contradicting theimprimitivity of the solution. (cid:3) Theorem 3.2. If X is a finite cycle set and there exists some x ∈ X suchthat ϕ x = id , then X is decomposable.Proof. Let x ∈ X be such that ϕ x = id. Then y · z = ( x · y ) · ( x · z ) = ( y · x ) · ( y · z )holds for all y, z ∈ X . This implies that ϕ y · x = id for all y ∈ X and hence ϕ g · x = id for all g ∈ G ( X, r ). Let Y = { x ∈ X : ϕ x = id } . If Y = X ,then the cycle set is trivial and hence it is decomposable. If not, then thegroup G ( X ) acts on Y . Thus Y is a non-trivial G ( X )-orbit and hence X isdecomposable. (cid:3) Let us show some direct consequences of the theorem. A cycle set X issaid to be retractable if the equivalence relation on X given by x ∼ y ⇐⇒ ϕ x = ϕ y is non-trivial. This notion corresponds to that of retractable solutions . Corollary 3.3.
Let X be a finite cycle set. If G ( X ) acts regularly on X ,then X is retractable.Proof. If G ( X ) acts regularly on X , then |G ( X ) | = | X | . Since X is indecom-posable, ϕ x = id for all x ∈ X by Theorem 3.2. Then some ϕ x must appearat least twice as a defining permutation of X . (cid:3) The following consequence goes back to Rump, see [20, Proposition 1].
Corollary 3.4. If X is an indecomposable finite cycle set such that G ( X ) is abelian, then X is retractable.Proof. The assumptions imply that G ( X ) acts regularly on X . Thus theclaim follows from Corollary 3.3. (cid:3) Now we present several results similar to that of Rump about decompos-ability of square-free solutions. First we consider the case where the diagonalof the solution is a cycle of maximal length.
Theorem 3.5.
Let X be a cycle set of size n . If T is a cycle of length n ,then X is indecomposable.Proof. Let x, y ∈ X . It is enough to prove that if m ∈ N and T m ( x ) = y ,then x and y belong to the same G ( X )-orbit. This is clearly true if m = 1.Now assume that the result holds for some m >
1, then y = T m +1 ( x ) = T ( T m ( x ))implies that T m ( x ) and y belong to the same G ( X )-orbit. Then the inductivehypothesis implies the claim. (cid:3) S. RAM´IREZ, L. VENDRAMIN
In Theorem 3.5, it is important that the diagonal is a cycle of maximallength. A diagonal that moves all points does not necessarily imply inde-composability.
Example 3.6.
Let X = { , . . . , } and r ( x, y ) = ( σ x ( y ) , τ y ( x )) be given by σ = (123) , σ = (123)(456) , σ = (123)(465) , σ = σ = σ = (456) ,τ = (132) , τ = (132)(465) , τ = (132)(456) , τ = τ = τ = (465) . Then (
X, r ) is decomposable and T = (123)(456) moves all points of X .For the next results it is more convenient to use the language of solutionsinstead of that of cycle sets. We shall need the following lemmas. Lemma 3.7.
Let ( X, r ) be a solution and x, y ∈ X . Then T ( x ) = y if andonly if r ( y, x ) = ( y, x ) .Proof. We first compute r ( T ( x ) , x ) = ( σ T ( x ) ( x ) , τ x T ( x )) = ( σ T ( x ) ( x ) , x ) . Since (
X, r ) is involutive, it follows that y = T ( x ) = σ T ( x ) ( x ). Conversely,if r ( y, x ) = ( y, x ), then τ x ( y ) = x and hence y = T ( x ). (cid:3) Lemma 3.8.
Let ( X, r ) be a solution and x ∈ X be such that T ( x ) = x .Then T ( τ y ( x )) = τ σ x ( y ) ( x ) for all y ∈ X .Proof. The Yang–Baxter equation applied to the tuple ( x, x, y ) yields r ( σ x σ x ( y ) , τ σ x ( y ) ( x ) , τ y ( x )) = ( σ x σ x ( y ) , τ σ x ( y ) ( x ) , τ y ( x )) . Thus r ( τ σ x ( y ) ( x ) , τ y ( x )) = ( τ σ x ( y ) ( x ) , τ y ( x )). Now Lemma 3.7 implies that T ( τ y ( x )) = τ σ x ( y ) ( x ). (cid:3) Applying the lemma to square-free solutions we get the following result.
Proposition 3.9. If ( X, r ) is a square-free solution, then τ x = σ − x for all x ∈ X .Proof. Let x, y ∈ X . We know that all solutions satisfy τ x ( y ) = σ − σ y ( x ) ( y ).Since ( X, r ) is square-free, T ( y ) = y and T ( τ x ( y )) = τ x ( y ). Lemma 3.8implies that τ σ y ( x ) ( y ) = τ x ( y ) = σ − σ y ( x ) ( y ), from where the claim follows. (cid:3) We consider the case where the diagonal fixes one point.
Theorem 3.10.
Let ( X, r ) be a finite solution of size n > . If the diagonalof ( X, r ) is a cycle of length n − , then X is decomposable.Proof. Let x ∈ X be such that T ( x ) = x and Y = X \ { x } . Assumethat ( X, r ) is indecomposable. Let f : Y → X be given by y τ y ( x ).Since ( X, r ) is indecomposable, there exists y ∈ Y such that τ y moves x ,as x = T ( x ) = τ x ( x ) and G ( X, r ) = h τ y : y ∈ Y i . So let y ∈ Y be suchthat f ( y ) ∈ Y . Since T is a cycle of length n −
1, the permutation T moveseach element of Y . This implies that Y = { f ( y ) , T ( f ( y )) , . . . , T n − ( f ( y )) } . Since T ( f ( y )) = T ( τ y ( x )) = τ σ x ( y ) ( x ) = f ( σ x ( y )) ECOMPOSITION THEOREMS 7 by Lemma 3.7, it follows that Y = { f ( y ) , f ( σ x ( y )) , . . . , f ( σ n − x ( y )) } . Then σ x permutes every element of Y cyclically.We claim that G ( X, r ) acts primitively on X . If not, there exists a non-trivial block B . Without loss of generality we may assume that x ∈ B .Then there exists y ∈ Y such that y ∈ B . Since σ x ( x ) = x , it followsthat σ x ( B ) = B . In particular, σ x ( y ) ∈ B and hence Y ⊆ B because σ x is a cyclic permutation on Y , a contradiction.Since G ( X, r ) acts primitively on X , Theorem 2.4 implies that ( X, r ) is acyclic solution. This is a contradiction, as the diagonal of a cyclic solutionhas no fixed points. (cid:3)
By inspection, solutions of size
10 such that the diagonal has only onefixed point are decomposable. This is not necessarily true if the diagonalfixes more than one point.
Example 3.11.
Let X = { , , . . . , } and r ( x, y ) = ( σ x ( y ) , τ y ( x )), where σ = (45) , σ = (36) , σ = (27) , σ = (18) ,σ = (13428657) , σ = (17568243) , σ = (12468753) , σ = (13578642) ,τ = (46) , τ = (35) , τ = (28) , τ = (17) ,τ = (18657243) , τ = (13427568) , τ = (13687542) , τ = (12457863) . Then T = (57)(68) has four fixed points and ( X, r ) is indecomposable.
Example 3.12.
Let X = { , , . . . , } and r ( x, y ) = ( σ x ( y ) , τ y ( x )), where σ = (167925483) , σ = (125983467) , σ = (165923487) ,σ = (158936472) , σ = (136972458) , σ = (156932478) ,σ = (149)(268) , σ = (149)(357) , σ = (268)(357) ,τ = (185639742) , τ = (189632745) , τ = (139642785) ,τ = (198623754) , τ = (128653794) , τ = (194628753) ,τ = (295)(384) , τ = (176)(295) , τ = (176)(384) . Then T = (123)(456) has three fixed points and ( X, r ) is indecomposable.Now we consider the case where the diagonal fixes two or three points.
Theorem 3.13.
Let ( X, r ) be a finite solution of size n . (1) If T is an ( n − -cycle and n is odd, then X is decomposable. (2) If T is an ( n − -cycle and gcd(3 , n ) = 1 , then X is decomposable.Proof. Let m be the length of T , Y = { y ∈ X : T ( y ) = y } and Z = { z ∈ X : T ( z ) = z } . If X = Y ∪ Z is a decomposition of X , there is nothing to prove. If not,there exists z ∈ Z and x ∈ X such that τ x ( z ) ∈ Y . Since T permutes theelements of Y cyclically, the elements τ x ( z ) , T ( τ x ( z )) , . . . , T m − ( τ x ( z )) areall distinct. Moreover, since z is a fixed point, { τ x ( z ) , T ( τ x ( z )) , . . . , T m − ( τ x ( z )) } = { τ x ( z ) , τ σ z ( x ) ( z ) , . . . , τ σ m − z ( x ) ( z ) } . S. RAM´IREZ, L. VENDRAMIN
In particular, x, σ z ( x ) , . . . , σ m − z ( x ) are all distinct elements, so the decom-position of σ z into disjoint cycles contains a cycle of length at least m . Since T is a cycle of length m , T k + m ( τ x ( z )) = T k ( τ x ( z )), i.e. τ σ k + mz ( x )( z ) = τ σ kz ( x ) ( z ) . Moreover, m is the minimal period which implies that the length of thiscycle in σ z has to be divisible by m . By the hypothesis on the m we aretaking this can only happen if the length of the cycle is exactly m .In the case where m = n −
2, since σ z has at least one fixed point z , σ z has to be a cycle of length m . By Theorem 3.1 X must be decomposable.In the case that m = n − σ z is a cycleof length n −
3, and we can conclude the same way as before, or it is aproduct of cycle of length n − z . Like before this blockcannot contain an element of the cycle of length n −
3. If it does not but itcontains an element that belongs to the cycle of length 2 then it also has tocontain the other element of the cycle. This means the block has size 3 andthis is not possible because gcd(3 , n ) = 1. (cid:3)
The assumption on n cannot be removed from the first claim of Theo-rem 3.13. Example 3.14.
Let X = { , , , } and r ( x, y ) = ( σ x ( y ) , τ y ( x )), where σ = (34) , σ = (1324) , σ = (1423) , σ = (12) ,τ = (24) , τ = (1432) , τ = (1234) , τ = (13) . Then T = (12) and ( X, r ) is indecomposable.Indecomposable solutions of small size are known up to size 11 and thereare only 172 such solutions, see Table 3.1 [1].
Table 3.1.
Indecomposable solutions of size n n Question 3.15.
Is it true thatlim n →∞ n n = 1?The number of known indecomposable solutions is quite small for mak-ing reasonable conjectures. However, the following questions seem to beinteresting. Question 3.16.
Let (
X, r ) be a finite solution. Is it true that if some σ x contains a cycle of length coprime with | X | then ( X, r ) is decomposable?
Question 3.17.
Let (
X, r ) be a finite solution. How many fixed points ofthe diagonal of (
X, r ) guarantee the decomposability of (
X, r )?Computer calculations suggest that if the diagonal of a solution (
X, r )fixes more than | X | / X, r ) is decomposable. We do not know,for example, what happens if the diagonal is a transposition or a 3-cycle.
ECOMPOSITION THEOREMS 9
Question 3.18.
Are there indecomposable solutions, where the diagonal isa transposition, different from that of Example 3.14?
Question 3.19.
Are there indecomposable solutions where the diagonal isa 3-cycle?In Table 3.2 we show the cycle structure of the diagonal of small inde-composable solutions.
Table 3.2.
Cycle structure of the diagonal of small inde-composable solutions. n number cycle structure4 3 42 26 6 63 3,31 2,2,28 7 2,225 2,2,2,22 2,2,41 2,614 851 4,49 1 3,36 3,3,39 910 1 2,2,2,2,215 5,520 10 Acknowledgments
Vendramin acknowledges the support of NYU-ECNU Institute of Mathe-matical Sciences at NYU Shanghai. This work is supported by PICT 2016-2481 and UBACyT 20020170100256BA.
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IMAS–CONICET and Depto. de Matem´atica, FCEN, Universidad de BuenosAires, Pabell´on 1, Ciudad Universitaria, C1428EGA, Buenos Aires, Argentina
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