A family of solutions of the Yang-Baxter equation
aa r X i v : . [ m a t h . R A ] D ec A family of solutions of the Yang-Baxter equation
David Bachiller Ferran Ced´o
Abstract
A new method to construct involutive non-degenerate set-theoreticsolutions ( X n , r ( n ) ) of the Yang-Baxter equation from an initial solution( X, r ) is given. Furthermore, the permutation group G ( X n , r ( n ) ) associ-ated to the solution ( X n , r ( n ) ) is isomorphic to a subgroup of G ( X, r ), andin many cases G ( X n , r ( n ) ) ∼ = G ( X, r ). Keywords:
Yang-Baxter equation, involutive non-degenerate solutions, bra-ce, IYB group
MSC:
The quantum Yang-Baxter equation is one of the basic equations in mathe-matical physics named after the authors of the two first works in which theequation arose: the solution of the delta function Fermi gas by C. N. Yang[15], and the solution of the 8-vertex model by R. J. Baxter [1]. It also lies atthe foundation of the theory of quantum groups. One of the important openproblems related to this equation is compute all its solutions. Those are linearmaps R : V ⊗ V → V ⊗ V , with V a vector space, that satisfy R ◦ R ◦ R = R ◦ R ◦ R , where R ij denotes the map R ij : V ⊗ V ⊗ V → V ⊗ V ⊗ V acting as R on the( i, j ) tensor factors and as the identity on the remaining factor.Finding all the solutions of the Yang-Baxter equation is a difficult task farfrom being solved. Nevertheless, many solutions have been found during the last20 years and the related algebraic structures (Hopf algebras) have been studied.In [5], Drinfeld suggested the study of a simpler case: solutions induced bya linear extension of a mapping R : X × X → X × X , where X is a basis for V .In this case, one says that R is a set-theoretic solution of the quantum Yang-Baxter equation. It is not difficult to see that, if τ : X → X is the map definedby τ ( x, y ) = ( y, x ), then the map R : X → X is a set-theoretic solution ofthe quantum Yang-Baxter equation if and only if the mapping r = τ ◦ R is asolution of the equation r ◦ r ◦ r = r ◦ r ◦ r , r ij is the map from X to X that acts as r on the ( i, j ) components andas the identity on the remaining component. In the sequel, we will always workwith this last equivalent equation.We study solutions with some additional conditions: involutively and non-degeneracy. A map r : X × X −→ X × X ( x, y ) ( σ x ( y ) , γ y ( x ))is said to be involutive if r ◦ r = id X . Moreover, it is said to be left (resp. right)non-degenerate if each map σ x (respectively, γ y ) is bijective, and it is said to benon-degenerate if it is left and right non-degenerate. If r is involutive and leftnon-degenerate, it can be checked that it satisfies r ◦ r ◦ r = r ◦ r ◦ r ifand only if it satisfies σ x ◦ σ σ − x ( y ) = σ y ◦ σ σ − y ( x ) for all x, y ∈ X (see the proofof [12, Theorem 9.3.10]).In what follows, by a solution of the YBE we will mean a non-degenerateinvolutive set-theoretic solution of the Yang-Baxter equation.In the last years, solutions of the YBE have received a lot of attention[3, 4, 6, 7, 8, 9, 10, 11, 13, 14]. In this case each solution ( X, r ) of the YBE hasan associated structure group, denoted by G ( X, r ), and defined by G ( X, r ) := h x ∈ X | xy = zt if and only if r ( x, y ) = ( z, t ) i . When X is finite, groups isomorphic to some G ( X, r ) are called groups of I -type.There is another important group associated to every solution of the YBE, itspermutation group G ( X, r ), which is the subgroup of Sym X generated by thebijections σ x , for all x ∈ X . It can be proved that G ( X, r ) is a homomorphicimage of G ( X, r ). When X is finite, groups isomorphic to some G ( X, r ) arecalled IYB groups.In [3], in order to characterize the groups of I -type, it is suggested to followtwo steps: Step 1:
Determine the finite groups that are IYB groups.
Step 2:
Given an IYB group G , find all the solutions of the YBE ( Y, s ) with Y finite such that G ( Y, s ) ∼ = G .Nowadays, these two problems remain unsolved. In the recent Ph.D. thesis ofNir Ben David [2], it is claimed that the result corresponding to [2, CorollaryD] solves Step 2 in homological terms. In fact, this result reduces Step 2 to thefollowing problem. Problem.
Let G be an IYB group. Let π : G −→ A be a bijective -cocycle overa G -module A . Let n be a positive integer. Find all the extensions of G -modules −→ Z n −→ E −→ A −→ , where Z n is a trivial G -module, E ∼ = Z n as abelian groups and there is a basis Y of E , as free abelian group, which is invariant by the action of G on E . X, r ) of theYBE, we state and prove a procedure to define, for each n ∈ N , a solution( X n , r ( n ) ) such that G ( X n , r ( n ) ) is isomorphic to a subgroup of G ( X, r ), andthen we provide sufficient conditions for G ( X n , r ( n ) ) to be isomorphic to G ( X, r ).One of the key steps of the proof of this result is the use of the properties of thebrace structure.
We only present here very briefly the results about braces that we will need later.For a much more detailed account, see [4]. We begin recalling the definition ofleft brace.
Definition 2.1 A left brace is a set G with two operations + and · such that( G, +) is an abelian group, ( G, · ) is a group, and every a, b, c ∈ G satisfy a ( b + c ) + a = ab + ac. We will refer to this property as the brace property. We call ( G, +) the additivegroup, and ( G, · ) the multiplicative group of the left brace. Right braces aredefined similarly, changing the brace property by ( b + c ) a + a = ba + ca .For any a ∈ G , we define a map λ a : G → G by λ a ( b ) = ab − a . In thestudy of braces, these maps play an important role; here is a list of some of theirproperties. Lemma 2.2
Let G be a left brace. The following properties hold:(i) λ a is bijective and λ − a = λ a − .(ii) λ a ( x + y ) = λ a ( x ) + λ a ( y ) ; that is, λ a is an automorphism of the abeliangroup ( G, +) .(iii) λ a λ b = λ ab ; that is, the map λ : ( G, · ) → Aut( G, +) , defined by λ ( a ) = λ a is a homomorphism of groups. iv) a + b = a · λ − a ( b ) . (v) a · λ − a ( b ) = b · λ − b ( a ) . (vi) λ a λ λ − a ( b ) = λ b λ λ − b ( a ) .(vii) The map r : G × G → G × G defined by r ( x, y ) = (cid:16) λ x ( y ) , λ − λ x ( y ) ( x ) (cid:17) is asolution of the YBE. It is called solution associated to the left brace G . Proof.
See [4, Lemmas 2.9 and 4.1].So any left brace gives us a solution of the YBE. There are other relationsbetween braces and solutions of the YBE, like the two next results, which are acharacterization of groups of I-type and IYB groups through braces.
Proposition 2.3
A group is isomorphic to G ( X, r ) for some solution of theYBE ( X, r ) if and only if it is the multiplicative group of a left brace.In particular, a finite group G is an IYB group if and only if it is the mul-tiplicative group of a finite left brace. Proof.
See [4, Corollary 4.6].
Proposition 2.4
A group G is of I -type if and only if it is isomorphic to themultiplicative group of a left brace B such that the additive group of B is a freeabelian group with a finite basis X such that λ x ( y ) ∈ X , for all x, y ∈ X . Proof.
See [4, Proposition 5.2].The next theorem is an essential tool for the proof of the main result of thispaper. It allows us to embed any solution of the YBE (
X, s ) inside a left brace,and then we can use all the additional algebraic properties of this structure.
Theorem 2.5
Let ( X, s ) be a solution of the YBE. Then G ( X, s ) is isomorphicto the multiplicative group of a left brace H such that, if ( H, r ) is the solutionassociated to it, then there exists a subset Y of H such that ( Y, r ′ ) , where r ′ is the restriction of r to Y , is a solution of the YBE isomorphic to ( X, r ) .Furthermore, λ h ( y ) ∈ Y , for all y ∈ Y and all h ∈ H . Proof.
See [4, Theorem 4.4] and its proof.4
Solutions with fixed permutation group
In this section, we focus on the problem of the construction of solutions of theYBE with a fixed permutation group.There is a simple way to produce solutions with the same permutation group.Note that, for any set X , the map r ( x, y ) = ( y, x ), for all x, y ∈ X , is a solutionof the YBE with trivial permutation group; we call ( X, r ) the trivial solution on X . Note also that, if we have solutions ( X i , r i ) of the YBE with permutationgroup G i = G ( X i , r i ), for all i = 1 , . . . , n , and X = ·∪ ni =1 X i is the disjoint unionof the sets X i , then ( X, r ) is a solution of the YBE, where r ( x, y ) = (cid:26) r i ( x, y ) if there exists an i such that x, y ∈ X i , ( y, x ) otherwisewith permutation group isomorphic to G ×· · ·× G n . Hence, since { id }× G ∼ = G ,we can associate to each solution ( X, r ) of the YBE infinitely many solutionsof the YBE with the same permutation group. However, this construction onlyincreases the size of our set adding points that behave as the trivial solution, sowe want to find “less trivial” constructions.We will generalize [3, Lemma 5.2], which gives a non-obvious constructionof an infinite family of solutions of the YBE associated to a fixed permutationgroup. This result is stated in [3] in terms of cycle sets. Translated to thelanguage of solutions of the YBE equation, the result could be stated as follows.
Proposition 3.1
Let ( X, r ) be a solution of the YBE, with r ( x, y ) = ( σ x ( y ) , γ y ( x )) .Then, for x , x ∈ X , the map f ( x ,x ) : X −→ X defined by f ( x ,x ) ( y , y ) = ( σ x σ x ( y ) , σ − σ x σ x ( y ) σ x σ x σ y ( y )) , for y , y ∈ X , is bijective and ( X , r (2) ) , where r (2) : X × X −→ X × X isthe map defined by r (2) (( x , x ) , ( y , y )) = ( f ( x ,x ) ( y , y ) , f − f ( x ,x ( y ,y ) ( x , x )) , is a solution of the YBE. Moreover, if σ z = id for some z ∈ X , then G ( X , r (2) ) ∼ = G ( X, r ) . Remark 3.2
The assumption that σ x = id, for some x ∈ X , is not true ingeneral. However, increasing the size of our set, we can always construct asolution with this property preserving the same permutation group. Namely,consider the solution ( X ·∪{ z } , r ′ ) for the disjoint union of X with a set of onepoint, defined as r ′ ( x, y ) = r ( x, y ) if x and y belong to X and r ′ ( x, y ) = ( y, x )if either x = z or y = z (note that it is the same trivial construction given atthe beginning of this section).Thus, beginning with an initial solution ( X, r ) of the YBE and applyingProposition 3.1 to X , X , ( X ) . . . we can construct infinitely many solutionsof the YBE with the same associated permutation group G ( X, r ).5n order to generalize this result we will need the following lemma, which isa generalization of [3, Lemma 5.1]. We use the notation x = ( x , . . . , x n ) forthe elements of X n . Lemma 3.3
Let X be a set and let σ : X −→ Sym X be a map. Consider themap ψ : Sym X −→ Sym X n defined by ψ ( τ )( y ) = ( t τ ( y ) , . . . , t τn ( y )) , for all τ ∈ Sym X and all y ∈ X n , where t τ ( y ) = τ ( y ) ,t τj +1 ( y ) = σ ( t τj ( y )) − · · · σ ( t τ ( y )) − τ σ ( y ) · · · σ ( y j )( y j +1 ) , for j > . Then ψ is a monomorphism. Proof.
Let τ ∈ Sym X . It is easy to check that ψ ( τ ) is a bijective map from X n to X n with inverse( y , . . . , y n ) ( T τ ( y ) , . . . , T τn ( y )) , where T τ ( y ) = τ − ( y ) ,T τj +1 ( y ) = σ ( T τj ( y )) − · · · σ ( T τ ( y )) − τ − σ ( y ) · · · σ ( y j )( y j +1 ) . Thus ψ is well-defined.To prove that ψ is a morphism, we have to check ψ ( τ ◦ ξ ) = ψ ( τ ) ◦ ψ ( ξ ) forall τ, ξ ∈ Sym X . Component by component, this is equivalent to verify t τ ◦ ξj ( y , . . . , y n ) = t τj ( t ξ ( y ) , . . . , t ξn ( y )) , ∀ j = 1 , . . . , n. This is done by induction. The first component is almost immediate; we write t ξ ( y ) = ( t ξ ( y ) , . . . , t ξn ( y )) for short: t τ ◦ ξ ( y ) = τ ( ξ ( y )) = τ ( t ξ ( y )) = t τ ( t ξ ( y )) . Now, assume that we have checked it up to the j -th component, and we wantto prove it for the ( j + 1)-th component: t τ ◦ ξj +1 ( y ) = σ ( t τξj ( y )) − · · · σ ( t τξ ( y )) − τ ξσ ( y ) · · · σ ( y j )( y j +1 )= σ ( t τj ( t ξ ( y ))) − · · · σ ( t τ ( t ξ ( y ))) − τ ξσ ( y ) · · · σ ( y j )( y j +1 )= σ ( t τj ( t ξ ( y ))) − · · · σ ( t τ ( t ξ ( y ))) − τ · σ ( t ξ ( y )) · · · σ ( t ξj ( y )) σ ( t ξj ( y )) − · · · σ ( t ξ ( y )) − ξσ ( y ) · · · σ ( y j )( y j +1 )= σ ( t τj ( t ξ ( y ))) − · · · σ ( t τ ( t ξ ( y ))) − τ · σ ( t ξ ( y )) · · · σ ( t ξj ( y )) t ξj +1 ( y ) = t τj +1 ( t ξ ( y )) . The second equality comes from the induction hypothesis, and at the end weuse the definition of t ξj +1 ( y ).On the other hand, to prove that ψ is injective, suppose that ψ ( τ ) = ψ ( ξ )for some τ, ξ ∈ Sym X . Then, ψ ( τ )( y ) = ψ ( ξ )( y ) for all y ∈ X n . Looking at thefirst component, we get τ ( y ) = ξ ( y ) for all y ∈ X , so τ = ξ .The next two results give the announced generalization of [3, Lemma 5.2].6 heorem 3.4 Let ( X, r ) be a solution of the YBE, with r ( x, y ) = ( σ x ( y ) , γ y ( x )) .Let n be an integer greater that . For x ∈ X n , consider the map f x : X n −→ X n defined by f x ( y ) = ( h ( x, y ) , h ( x, y ) , . . . , h n ( x, y )) , for y ∈ X n , where the h j is defined recursively by h ( x, y ) = σ x · · · σ x n ( y ) , and h j ( x, y ) = σ − h j − ( x,y ) · · · σ − h ( x,y ) σ x · · · σ x n σ y · · · σ y j − ( y j ) , for j = 2 , . . . , n . Then f x is bijective and ( X n , r ( n ) ) , where r ( n ) : X n × X n −→ X n × X n is the map defined by r ( n ) ( x, y ) = ( f x ( y ) , f − f x ( y ) ( x )) , is a solution of the YBE. Proof.
Let σ : X −→ Sym X be the map defined by σ ( x ) = σ x , for all x ∈ X .Consider the map ψ : Sym X −→ Sym X n defined as in Lemma 3.3. It is clearthat f x = ψ ( σ x · · · σ x n ). Hence f x is bijective.By Theorem 2.5, we may assume that X is a subset of a left brace H andthat σ x is the restriction of λ x to X , for all x ∈ X . Recall that λ a ( b ) = ab − a ,for all a, b ∈ H . Therefore, by Lemma 2.2(iii), h ( x, y ) = σ x · · · σ x n ( y ) = λ x ··· x n ( y ) . We claim that λ x ··· x n ( y · · · y j ) = h ( x, y ) · · · h j ( x, y ) , (1)for all 1 ≤ j ≤ n and x , . . . x n , y , . . . , y n ∈ X .We prove this claim by induction on j . We know that it is true for j = 1.Suppose that j > j −
1. We have λ x ··· x n ( y · · · y j ) = λ x ··· x n ( y · · · y j − + λ y ··· y j − ( y j ))= λ x ··· x n ( y · · · y j − ) + λ x ··· x n λ y ··· y j − ( y j )(by Lemma 2.2(ii))= h ( x, y ) · · · h j − ( x, y ) + λ x ··· x n λ y ··· y j − ( y j )(by induction hypothesis)= h ( x, y ) · · · h j − ( x, y ) λ − h ( x,y ) ··· h j − ( x,y ) λ x ··· x n λ y ··· y j − ( y j )(by Lemma 2.2(iv))By Lemma 2.2(iii), λ − h ( x,y ) ··· h j − ( x,y ) λ x ··· x n λ y ··· y j − ( y j ) = h j ( x, y ) . h j ( x, y ) = λ x ··· x n ( y · · · y j − ) − λ x ··· x n ( y · · · y j ) , (2)for 2 ≤ j ≤ n . Let H ( x, y ) = λ − x ··· x n ( y ) and H j ( x, y ) = λ − x ··· x n ( y · · · y j − ) − λ − x ··· x n ( y · · · y j ) , for 2 ≤ j ≤ n . It is easy to check that the map X n −→ X n defined by y ( H ( x, y ) , . . . , H n ( x, y )) is f − x .By the definition of r ( n ) , it is straightforward to check that r ( n ) ◦ r ( n ) = id.Thus, in order to prove that ( X n , r ( n ) ) is an involutive non-degenerate set-theoretic solution of the Yang-Baxter equation, we should show that(a) f x f f − x ( y ) ( z ) = f y f f − y ( x ) ( z ) , for all x, y, z ∈ X n , and(b) the map γ y : X n −→ X n defined by γ y ( x ) = f − f x ( y ) ( x ) is bijective.By (1) and the definition of H j ( x, y ), the first component of f x f f − x ( y ) ( z ) = f x f ( H ( x,y ) ,...,H n ( x,y )) ( z ) is λ x ··· x n λ H ( x,y ) ··· H n ( x,y ) ( z ) = λ x ··· x n λ λ − x ··· xn ( y ··· y n ) ( z )= λ y ··· y n λ λ − y ··· yn ( x ··· x n ) ( z ) , where the last equality follows from Lemma 2.2(vi). For j >
1, the j -th compo-nent of f x f f − x ( y ) ( z ) = f x f ( H ( x,y ) ,...,H n ( x,y )) ( z ) is( λ x ··· x n λ H ( x,y ) ··· H n ( x,y ) ( z · · · z j − )) − · ( λ x ··· x n λ H ( x,y ) ··· H n ( x,y ) ( z · · · z j ))= ( λ x ··· x n λ λ − x ··· xn ( y ··· y n ) ( z · · · z j − )) − · ( λ x ··· x n λ λ − x ··· xn ( y ··· y n ) ( z · · · z j ))and, by Lemma 2.2(vi), we can interchange the x ’s and the y ’s, and (a) follows.To prove (b), first we shall see that γ y is injective. Let x, z ∈ X n beelements such that γ y ( x ) = γ y ( z ). Hence H j (( h ( x, y ) , . . . , h n ( x, y )) , x ) = H j (( h ( z, y ) , . . . , h n ( z, y )) , z ), for all j = 1 , . . . , n . That is λ − h ( x,y ) ··· h n ( x,y ) ( x ) = λ − h ( z,y ) ··· h n ( z,y ) ( z )and λ − h ( x,y ) ··· h n ( x,y ) ( x · · · x j − ) − λ − h ( x,y ) ··· h n ( x,y ) ( x · · · x j )= λ − h ( z,y ) ··· h n ( z,y ) ( z · · · z j − ) − λ − h ( z,y ) ··· h n ( z,y ) ( z · · · z j ) , for all j = 2 , . . . , n . Therefore λ − h ( x,y ) ··· h n ( x,y ) ( x · · · x j ) = λ − h ( z,y ) ··· h n ( z,y ) ( z · · · z j ) , for all j = 1 , . . . , n . By (1), λ − λ x ··· xn ( y ··· y n ) ( x · · · x j ) = λ − λ z ··· zn ( y ··· y n ) ( z · · · z j ) , (3)8or all j = 1 , . . . , n . By Lemma 2.2(vii), since λ − λ x ··· xn ( y ··· y n ) ( x · · · x n ) = λ − λ z ··· zn ( y ··· y n ) ( z · · · z n ) , we have that x · · · x n = z · · · z n . Thus, by (3), x · · · x j = z · · · z j , for all j = 1 , . . . , n . Hence x j = z j , for all j = 1 , . . . , n . Therefore γ y is injective.We shall see that γ y is surjective. Let z = ( z , . . . , z n ) ∈ X n . By Lemma 2.2(vii),there exists a n ∈ H such that λ − λ an ( y ··· y n ) ( a n ) = z · · · z n . Let a j = λ λ an ( y ··· y n ) ( z . . . z j ) , for all j = 1 , . . . , n −
1. By Theorem 2.5, a ∈ X . Let x = a and x i = a − i − a i ,for 1 < i ≤ n . We shall prove that x i ∈ X , for all i , and γ y ( x ) = z . Supposethat i > x , . . . , x i − ∈ X . We have that a i = λ λ an ( y ··· y n ) ( z . . . z i ) = λ λ an ( y ··· y n ) ( z . . . z i − + λ z ...z i − ( z i ))= λ λ an ( y ··· y n ) ( z . . . z i − ) + λ λ an ( y ··· y n ) ( λ z ...z i − ( z i )) (by Lemma 2.2(ii))= a i − + λ λ an ( y ··· y n ) z ...z i − ( z i ) (by Lemma 2.2)( iii )= a i − λ − a i − ( λ λ an ( y ··· y n ) z ...z i − ( z i )) (by Lemma 2.2)( iv ) . Hence x i = a − i − a i = λ − a i − ( λ λ an ( y ··· y n ) z ...z i − ( z i )) ∈ X , by Theorem 2.5. Thus,by induction, x , . . . , x n ∈ X . Now we have γ y ( x ) = ( H (( h ( x, y ) , . . . , h n ( x, y )) , x ) , . . . , H n (( h ( x, y ) , . . . , h n ( x, y )) , x )) . By (1) and the definition of H j , γ y ( x ) = ( λ − λ x ··· xn ( y ··· y n ) ( x ) , λ − λ x ··· xn ( y ··· y n ) ( x ) − λ − λ x ··· xn ( y ··· y n ) ( x x ) , . . . ,λ − λ x ··· xn ( y ··· y n ) ( x · · · x n − ) − λ − λ x ··· xn ( y ··· y n ) ( x · · · x n ))= ( λ − λ an ( y ··· y n ) ( a ) , λ − λ an ( y ··· y n ) ( a ) − λ − λ an ( y ··· y n ) ( a ) , . . . ,λ − λ an ( y ··· y n ) ( a n − ) − λ − λ an ( y ··· y n ) ( a n )) = ( z , z , . . . , z n ) . Therefore γ y is bijective and (b) follows. This finishes the proof.The next proposition describes the permutation group of the solutions ( X n , r ( n ) )as a certain subgroup of the permutation group of ( X, r ). After that, we givetwo cases in which G ( X n , r ( n ) ) ∼ = G ( X, r ). Proposition 3.5
With the above notation, G ( X n , r ( n ) ) is isomorphic to thesubgroup of G ( X, r ) generated by all the permutations σ x · · · σ x n , for all x i ∈ X .In particular,1. If σ z = id for some z ∈ X , then G ( X n , r ( n ) ) ∼ = G ( X, r ) for all n .2. If X is a finite set and |G ( X, r ) | = m < + ∞ , then, for all n such that gcd( m, n ) = 1 , we have G ( X n , r ( n ) ) ∼ = G ( X, r ) . roof. Recall that r ( x, y ) = ( σ x ( y ) , γ y ( x )). Let σ : X −→ Sym X be the mapdefined by σ ( x ) = σ x , for all x ∈ X . Consider the map ψ : Sym X −→ Sym X n defined as in Lemma 3.3. We have that G ( X n , r ( n ) ) = h f x : x ∈ X n i = h ψ ( σ x · · · σ x n ) : x i ∈ X i = ψ ( h σ x · · · σ x n : x i ∈ X i ) ∼ = h σ x · · · σ x n : x i ∈ X i , using in the last isomorphism that ψ is a monomorphism.In particular,1. If σ z = id for some z ∈ X , then any σ x can be written as a product of n permutations using σ x = σ n − z σ x , so h σ x · · · σ x n : x i ∈ X i = h σ x : x ∈ X i ,and G ( X n , r ( n ) ) ∼ = h σ x · · · σ x n : x i ∈ X i = h σ x : x ∈ X i = G ( X, r ) .
2. Suppose that X is finite, and that n is a positive integer coprime with m = |G ( X, r ) | . We can find a positive integer k such that nk ≡ m ).Then, σ x = σ nkx for all x ∈ X , which implies h σ x · · · σ x n : x i ∈ X i = h σ x : x ∈ X i , and the conclusion follows as in the previous case. Acknowledgments
Research partially supported by DGI MINECO MTM2011-28992-C02-01, byFEDER UNAB10-4E-378 ”Una manera de hacer Europa”, and by the Comis-sionat per Universitats i Recerca de la Generalitat de Catalunya.