Bol. Soc. Mat. Mex. DOI 10.1007/s40590-016-0133-7 ORIGINAL ARTICLE

The Lie bialgebra structure of the vector space of cyclic words Ana González1

Received: 2 April 2015 / Accepted: 28 April 2016 © Sociedad Matemática Mexicana 2016

Abstract This paper presents a combinatorial proof of the existence of a Lie bialgebra structure over the vector space of reduced cyclic words. Any surface with non-empty boundary has an associated vector space determined by the corresponding surface symbol, this space is known as the space of reduced cyclic words. The Lie bialgebra structure over this space was introduced by Chas in the article Combinatorial Lie bialgebras of curves on surfaces, where a proof of the existence of this structure is given. This proof is based on the construction of an isomorphism between the space of reduced cyclic words and the space of curves on a surface. Keywords Lie bialgebra · Space of cyclic words · Linked pair Mathematical Subject Classification 17B62 · 05E15 · 57M99

1 Introduction The goal of this work is to give a combinatorial proof of the existence of a Lie bialgebra structure over the space of reduced cyclic words. We denote this space by V. In [1] M. Chas gave a proof of this result by means of an isomorphism of Lie bialgebras between V and the space of curves on a surface. In [2] Goldman proved that the space of curves on a surface admits a Lie algebra structure. Subsequently Turaev, in [3], defined the Lie coalgebra structure on V ending with the compatibility between the algebra and the coalgebra structures.

B 1

Ana González [email protected] IMERL-Facultad de Ingeniería, Universidad de la República, Montevideo, Uruguay

A. González

The main contribution of this paper is to give a new proof of the existence of a Lie bialgebra structure over the vector space of reduced cyclic words, but through a combinatorial method. Apparently, this proof seems more complex than the proof presented in [1], however the techniques herein used are elementary and purely combinatorial, which it makes approachable to a wider audience. In Sect. 2, we develop the concept of Lie bialgebras. In Sect. 3, we construct the space of cyclic words V, and we set up the terminology and main definitions that allow us to define the Lie algebra and coalgebra associated to this space. In the Sect. 4, we prove that V admits a Lie colagebra structure. The final two sections are dedicated to provide the Lie algebra structure of V and to prove the compatibility between these two structures. Finally, in the Appendix we introduce the concepts of linked pair and their signs developed in [1].

2 Lie bialgebras Lie bialgebras are so-called quasi-classical limits of quantum groups (more precisely, quantum universal enveloping algebras). The structure of a Lie bialgebra also arises on the tangent space to the unit element of a Poisson–Lie group. If A is a k-vector space we can consider two linear maps, s : A ⊗ A → A ⊗ A given by s x ⊗y = y⊗x and ε : A⊗ A⊗ A → A⊗ A⊗ A defined by ε x ⊗y⊗z = z⊗x ⊗y.     Definition 1 The triple A, [ , ],  is a Lie bialgebra if A, [ , ] is a Lie algebra         [ , ] ◦ s = −[ , ] and [ , ] ◦ id ⊗ [ , ] ◦ id + ε + ε2 = 0 , A,  is       a Lie coalgebra s ◦  = − and id + ε + ε2 ◦ id ⊗  ◦  = 0 , and the compatibility  [x, y] = x · (y) − y · (x) holds for every x, y ∈ A,  equation  where x · y ⊗ z = [x, y] ⊗ z + y ⊗ [x, z].     Definition 2 The triple A, [ , ],  is an involutive Lie bialgebra if A, [ , ],  is a Lie bialgebra and [ , ] ◦  = 0 on A.

3 The vector space of cyclic words In this section we construct, from a finite set of symbols, the vector space of cyclic words V. For detail see [1]. Definition 3 For each non-negative integer n, the n-alphabet or alphabet is the set of  n symbols An = a1 , . . . , an . If An is an alphabet, then An is a set disjoint from An , for which there exists a bijection An → An , denoted by x → x.   Definition 4 A word on An is a sequence W = (x1 , x2 , . . .), where xi ∈ An ∪An ∪ 1 for all i, and there is an integer n ∈ N∗ such that xk = 1 for all k > n. We denote the linear word W as W = x1 x2 . . . xn . In particular, the constant sequence (1, 1, 1, . . .) is a word, called the empty word, and it is also denoted by 1. The length of the empty word is defined as 0; the length of W = x1 x2 . . . xn is defined as l(W ) = n.

The Lie bialgebra structure of the vector...

a6

a1

a7

a8 a2

a2

a5

a1

Fig. 1 A cyclic word in the letters of A8

Definition 5 If W = x1 x2 . . . xn is a word, then its inverse is the word W = xn . . . x2 x1 . A subword of W = x1 x2 . . . xn is either the empty word or a word of the form V = xi xi+1 . . . x j , where 1 ≤ i ≤ j ≤ n. Definition 6 A word W is reduced if either W is empty or W = x1 x2 . . . xn , where: for all x ∈ An , x and x arenever adjacent,  i.e., if xi = x then xi+1 = x and if xi = x then xi+1 = x, for all i ∈ 1, 2, . . . , n . The multiplication of reduced words W and V is the reduced word obtained from W V after cancelations. More precisely, there is a (possibly empty) subword U of W with W = W  U such that U is a subword of V with V = U V  and such that W  V  is reduced. Definition 7 Let F be the group of reduced words presented in the previous paragraphs. In F we consider the following equivalence relation: if V , W ∈ F we say that V and W are conjugated, V ∼ W , if there exist U ∈ F such that W = U V U . A cyclic word V is the equivalence class of V in F := F/ ∼. A good representation of cyclic words was introduced by Chas [1], the idea is to consider a cyclic word as an oriented circle. This representation clearly illustrates the definition of cyclic word, since all the words that are obtained by cutting in some letter, respecting the orientation of the circle, are equivalent. In Fig. 1, we present the cyclic word W = c(a1 a2 a7 a1 a5 a2 a8 a6 ) as a circle. Definition 8 A linear word W is a linear representative of a cyclic word W if W can be obtained from W by making a cut between two consecutive letters of W. In such a case, we write W = c(W ). If W is a cyclic word, W is a linear representative of W, n and n is a positive integer, we define W n as c(W n ), W as c(W ), and W −n as W . Definition 9 A cyclic word is reduced if it is non-empty and all its linear representatives are reduced. A reduced cyclic word is primitive if it cannot be written as W r for some r ≥ 2 and some reduced cyclic word W. We denote by V the vector space generated by non-empty reduced cyclic words with letters in An . To define the Lie bracket and the Lie cobracket it is necessary to introduce the concept of linked pairs. This concept, developed in [1], is presented in the Appendix.

A. González

4 The Lie coalgebra structure The Lie cobracket on the space V was introduced in [1] by Chas. However, the tools necessary to give a combinatorial proof that this map gives a Lie coalgebra structure on V are developed in this section. The idea is to split the cyclic word, on all its linked pairs, in two subwords such that the gluing of these generates the original word. To prove that this cobracket satisfies the condition of co-Jacobi we show that in the final sum, after applying the corresponding map of this property to a cyclical word, each of the addends is canceled with another term. Definition 10 We define  : V → V⊗V as the linear map such that for every reduced cyclic word W, (W) =



sg(P, Q)δ1 (P, Q) ⊗ δ2 (P, Q),

(P,Q)∈LP1 (W )

where LP1 (W) is the set of linked pairs (P, Q) of W and the cyclic words δ1 (P, Q) and δ2 (P, Q) are defined as δ1 (P, Q) = c(W1 ) and δ2 (P, Q) = c(W2 ) where, if (P, Q) is the type (1) or (2) we make two cuts on W, one immediately before p2 and the other immediately before q2 . Then W1 is the linear word starting at p2 , and W2 is the linear word starting at q2 ; if (P, Q) is the type (3) let W1 be the linear subword of W starting at p2 and ending at q1 , and let W2 be the linear subword of W starting at q2 and ending at p1 . Remark 1 This sum is finite as a consequence that the set LP1 (W) is finite, see Proposition 2.9 of [1].   Theorem 1 The pair V,  is a Lie coalgebra. To give a combinatorial proof of this result, we need a series of technical lemmas. Lemma 1 Let W be a reduced cyclic word. If (P, Q) ∈ LP1 (W), then δ1 (P, Q) = δ2 (Q, P) and δ2 (P, Q) = δ1 (Q, P). Proof We only need to prove the first identification because the second is a consequence of the fact that (Q, P) is a linked pair. (i) If (P, Q) is a linked pair of type Definition 12(1), then we have that W1 = p2 Bq1 and W2 = q2 Ap1 . Consequently, δ1 (P, Q) = c( p2 Bq1 ). On the other hand, applying Remark 4, we obtain (Q, P) ∈ LP1 (W) and δ2 (Q, P) = c(V2 ) where V2 = p2 Bq1 . Finally, δ2 (Q, P) = c( p2 Bq1 ) = δ1 (P, Q). (ii) If (P, Q) is a linked pair of type Definition 12(2), then W1 = p2 Bq1 Y and W2 = q2 Ap1 Y . Therefore, δ1 (P, Q) = c( p2 Bq1 Y ). As (Q, P) ∈ LP1 (W) we have that δ2 (Q, P) = c(V2 ), where V2 = p2 Bq1 Y . Then δ2 (Q, P) = c( p2 Bq1 Y ) = δ1 (P, Q). (iii) Finally, if (P, Q) is a linked pair of type Definition 12(3), then we have W1 = p2 Bq1 , W2 = q2 Ap1 and δ1 (P, Q) = c( p2 Bq1 ). As before (Q, P) ∈ LP1 (W) and δ2 (Q, P) = c(V2 ), where V2 = p2 Bq1 . Then δ2 (Q, P) = c( p2 Bq1 ) =   δ1 (P, Q).

The Lie bialgebra structure of the vector...

As a consequence we have the following result. Proposition 1 For W a reduced cyclic word, the cobracket  satisfies the following identity 

(W) =

  sg(P, Q) δ1 (P, Q) ⊗ δ2 (P, Q) − δ2 (P, Q) ⊗ δ1 (P, Q) .

{(P,Q),(Q,P)}⊂LP1 (W )

The following result is central for the demonstration of the condition of co-Jacobi in Theorem 1. The reason is the following, if we consider W a reduced cyclic word and apply the map (id + ε + ε2 ) ◦ (id ⊗ ) ◦  to it we get a sum of tensor products of cyclic subwords of W. The Proposition 2 ensures that every term in the expression added that annuls it. (id + ε + ε2 ) ◦ (id ⊗ ) ◦ (W) has an equal and opposite   For (P, Q) ∈ LP1 (W), if we take (R, S) ∈ LP1 δi (P, Q) , with i ∈ {1, 2}, we can cut δi (P, Q) by (R, S) and construct the cyclic words δ j (R, S)δi (P,Q) , where j ∈ {1, 2}. We will denote by δ ji (R, S) these new cyclic words. Proposition 2 Let W be a cyclic reduced word. For each (P, Q) ∈ LP1 (W) and (R, S) ∈ LP1 (δi (P, Q)), i = 1, 2, there exist linked pairs (R  , S  ) ∈ LP1 (W) and (P  , Q  ) ∈ LP1 (δ j (R  , S  )), for some j ∈ {1, 2}, such that sg(P  , Q  ) = sg(P, Q), sg(R  , S  ) = sg(R, S) and the following conditions are satisfied For i = 1     If j = 1 then : δ11 P  , Q = δ11 (R, S), if j = 2 then : δ1 R  , S  = δ11 (R, S)    δ12  P , Q   = δ21 (R, S) δ2 R , S = δ21 (R, S)   δ22 P  , Q  = δ2 (P, Q). δ21 P , Q = δ2 (P, Q)

For i = 2     If j = 1 then : δ11 P  , Q = δ12 (R, S), if j = 2 then : δ1 R  , S  = δ12 (R, S) δ12  P  , Q   = δ1 (P, Q) δ2 R  , S  = δ1 (P, Q)   δ22 P  , Q  = δ22 (R, S). δ21 P , Q = δ22 (R, S)

The conditions that determine in which component is the new linked pair (P  , Q  ), what kind of linked pairs are they and their relative positions, are the signs of (P, Q) and (R, S). Proof We present the proof when (P, Q) and (R, S) are linked pairs of type (2) and i = 1. The other cases are analogous. In order to build the new linked pairs we study the relative position of the pairs (P, Q) and (R, S) within the cyclic word W. We use segments to represent words and intervals to represent linked pairs within them to illustrate each case.   Let be (P, Q) ∈ LP1 (W) and (R, S) ∈ LP1 (δ1 (P, Q)), with P = p1 X p2 , Q = q1 Xq2 and R = r1 Y r2 , S = s1 Y s2 . Then δ1 (P, Q) = c( p2 Bq1 X ) and δ2 (P, Q) = c(q2 Ap1 X ) with W = c( p2 Bq1 Xq2 Ap1 X ). If (R, S) appear in δ1 (P, Q) as δ1 (P, Q) = c( p2 B1r1 Y r2 Cs1 Y s2 B2 q1 X ), we can represent the word δ1 (P, Q) and a linear representation W of W as follows:

A. González

X

B

δ1(P,Q) :

p2

=

q1

B1 p

and W:

B1 p

2

Y r1

r2

C

s1

X

s2

s1

r2

B2

Y

C

Y r1

2

Y

s2

q1 B2

X

q2

q1

A

p1

X

Then, we conclude that (R, S) ∈ LP1 (W) with δ1 (R, S) = c(r2 Cs1 Y ) and δ2 (R, S) = c(s2 B2 q1 Xq2 Ap1 X p2 B1r1 Y ), where (P, Q) ∈ LP1 (δ2 (R, S)) and   c q2 Ap1 X = δ22 (P, Q), δ2 (P, Q) = δ1 (R, S) =  c r2 Cs1 Y =  δ11 (R, S), δ12 (P, Q) = c p2 B1r1 Y s2 B2 q1 X = δ21 (R, S). In this case it is sufficient to take (R  , S  ) = (R, S) and (P  , Q  ) = (P, Q). Now, suppose that the situation is a little different, δ1 (P, Q) = c( p2 Bq1 X ) = c( p2 Y2 s2 B2 r1 Y r2 B1 s1 Y1 q1 X ), where Y = Y1 q1 X p2 Y2 .

B q1

p2

X

=

δ1(P,Q) :

p

Y2

2

s2

B2 r1

Y

B1 r2

s1

Y1

X q1

Then   δ11 (R, S) = c r2 B1 s1 Y1 q1 X p2 Y2 ,     δ21 (R, S) = c s2 B2 r1 Y = c s2 B2 r1 Y1 q1 X p2 Y2 and W = p2 Y2 s2 B2 r1 Y r2 B1 s1 Y1 q1 Xq2 Ap1 X. We consider the subwords R  = r1 Y1 q1 X p2 and S  = s1 Y1 q1 Xq2 of W . If sg(P, Q) = sg(R, S), then (R  , S  ) ∈ LP1 (W) and   δ1 R  , S  =    δ2 R , S =

  c p2 Y2 r2 B1 s1 Y1 q1 X = δ11 (R, S),   c q2 Ap1 X p2 Y2 s2 B2 r1 Y1 q1 X ,

therefore, (P, Q) ∈ LP1 (δ2 (R  , S  )) and   δ12 (P, Q) = c p2 Y2 s2 B2 r1 Y1 q1 X = δ21 (R, S),   δ22 (P, Q) = c q2 Ap1 X = δ2 (P, Q). Finally, we resolve the case sg(P, Q) = sg(R, S). As before W = p2 Y2 s2 B2 r1 Y r2 B1 s1 Y1 q1 Xq2 Ap1 X . Note that R  = q1 X p2 Y2 r2 , S  = p1 X p2 Y2 s2 are subwords of

The Lie bialgebra structure of the vector...

W and (R  , S  ) ∈ LP1 (W) with     δ1 R  , S  = c r2 B1 s1 Y1 q1 Xq2 Ap1 X p2 Y2 ,     δ2 R  , S  = c s2 B2 r1 Y1 q1 X p2 Y2 = δ21 (R, S).    Then (P, Q) ∈ LP1 δ1 R  , S  and   δ11 (P, Q) = c p2 Y2 r2 B1 s1 Y1 q1 X = δ11 (R, S),   δ21 (P, Q) = c q2 Ap1 X = δ2 (P, Q). Example 1 Let O = c(abab) be the surface symbol and the linear word W = baaabaabaaaba. We consider the linked pair (P, Q) ∈ LP1 (W), where P = baaa and Q = aaab. In this case we have that     δ1 (P, Q) = c abaabaaa , δ2 (P, Q) = c babaa . Let R = aaaa and S = baab a linked pair of δ1 (P, Q), then     δ11 (R, S) = c abaa , δ21 (R, S) = c baaa . The next step is to construct the linked pairs (R  , S  ) ∈ LP(W) and (P  , Q  ) ∈ LP(δ2 (R  , S  )) satisfying the Proposition 2. The candidates are R  = babaaa and S  = aabaab. First, we note that sg(R  , S  ) = sg(R, S) and (R  , S  ) ∈ LP1 (W), where         δ1 R  , S  = c abaa , δ2 R  , S  = c baaababaa . We take (P  , Q  ) ∈ LP1 (δ2 (R  , S  )), where P  = baab and Q  = aaab. Note that sg(P  , Q  ) = sg(P, Q) and         δ12 P  , Q  = c baaa , δ22 P  , Q  = c babaa . Finally, we have that       δ2 (P, Q) = δ22 P  , Q  , δ1 R  , S  = δ11 (R, S), δ12 P  , Q  = δ21 (R, S).

A. González

Proof (Theorem 1) (1) Coskew symmetry: s ◦  = −, ⎛

⎞



s ◦ (W) = s ⎝

sg(P, Q)δ1 (P, Q) ⊗ δ2 (P, Q)⎠

(P,Q)∈LP1 (W )



=

sg(P, Q)δ2 (P, Q) ⊗ δ1 (P, Q)

(P,Q)∈LP1 (W )



=

−sg(Q, P)δ2 (P, Q) ⊗ δ1 (P, Q)

(Q,P)∈LP1 (W )



=−

sg(Q, P)δ1 (Q, P) ⊗ δ2 (Q, P)

(Q,P)∈LP1 (W )

= −(W). (2) Co-Jacobi identity: (id + ε + ε2 ) ◦ (id ⊗ ) ◦  = 0. The Lie cobracket in W, applying the Proposition 1, is given by (W ) =



  sg(P, Q) δ1 (P, Q) ⊗ δ2 (P, Q) − δ2 (P, Q) ⊗ δ1 (P, Q) .

{(P,Q),(Q,P)}⊂LP1 (W )

Then (id ⊗ )((W)) =





{(P,Q),(Q,P)}⊂LP1 (W ) {(R,S),(S,R)}⊂LP1 (δ2 (P,Q))

sg(P, Q)sg(R, S)δ1 (P, Q) ⊗ δ12 (R, S) ⊗ δ22 (R, S) − sg(P, Q)sg(R, S)δ1 (P, Q) ⊗ δ22 (R, S) ⊗ δ12 (R, S)  −



{(P,Q),(Q,P)}⊂LP1 (W ) {(R,S),(S,R)}⊂LP1 (δ1 (P,Q))

sg(P, Q)sg(R, S)δ2 (P, Q) ⊗ δ11 (R, S) ⊗ δ21 (R, S) − sg(P, Q)sg(R, S)δ2 (P, Q) ⊗ δ21 (R, S) ⊗ δ11 (R, S). For (P, Q) ∈ LP1 (W) and (R, S) ∈ LP1 (δ2 (P, Q)) the corresponding term in (id ⊗ )((W)) is    id ⊗ (R,S) (P,Q) (W) = sg(P, Q)sg(R, S)δ1 (P, Q) ⊗ δ12 (R, S) ⊗ δ22 (R, S) − sg(P, Q)sg(R, S)δ1 (P, Q) ⊗ δ22 (R, S) ⊗ δ12 (R, S).

The Lie bialgebra structure of the vector...

Note that, by Proposition 2, there are pairs (P  , Q  ) and (R  , S  ) such that (P  , Q  ) ∈ LP1 (δ2 (R  , S  )) or (P  , Q  ) ∈ LP1 (δ1 (R  , S  )). If (P  , Q  ) ∈ LP1 (δ2 (R  , S  )) the corresponding term in (id ⊗ )((W)) is    id ⊗ (P  ,Q  ) (R  ,S  ) (W) = sg(P  , Q  )sg(R  , S  )δ1 (R  , S  ) ⊗ δ12 (P  , Q  ) ⊗ δ22 (P  , Q  ) − sg(P  , Q  )sg(R  , S  )δ1 (R  , S  ) ⊗ δ22 (P  , Q  ) ⊗ δ12 (P  , Q  ). Again applying the Proposition 2 this term can be written as    id ⊗ (P  ,Q  ) (R  ,S  ) (W) = sg(P, Q)sg(R, S)δ12 (R, S) ⊗ δ1 (P, Q) ⊗ δ22 (P, Q) − sg(P, Q)sg(R, S)δ12 (R, S) ⊗ δ22 (R, S) ⊗ δ1 (P, Q). Finally, applying the map id + ε + ε2 we have that         id + ε + ε 2 id ⊗ (R,S) (P,Q) (W ) + id + ε + ε 2 id ⊗ (P  ,Q  ) (R  ,S  ) (W ) = 0.

If (P  , Q  ) ∈ LP1 (δ1 (R  , S  )) the corresponding term in (id ⊗ )((W)) is   id ⊗ (P  ,Q  ) (R  ,S  ) (W) = sg(R  , S  )sg(P  , Q  )δ2 (R  , S  )

⊗ δ21 (P  , Q  ) ⊗ δ11 (P  , Q  ) − sg(R  , S  )sg(P  , Q  )δ2 (R  , S  ) ⊗ δ11 (P  , Q  ) ⊗ δ21 (P  , Q  ).

As before, applying the Proposition 2 this term can be written as   id ⊗ (P  ,Q  ) (R  ,S  ) (W) = sg(R, S)sg(P, Q)δ1 (P, Q) ⊗ δ22 (R, S) ⊗ δ12 (R, S) − sg(R, S)sg(P, Q)δ1 (P, Q) ⊗ δ12 (R, S) ⊗ δ22 (R, S). Finally, applying the map id + ε + ε2 we have that         id + ε + ε 2 id ⊗ (R,S) (P,Q) (W ) + (id + ε + ε 2 ) id ⊗ (P  ,Q  ) (R  ,S  ) W = 0.

 

A. González

Example 2 If we return to the Example 1 we see that:  The corresponding co-Jacobi term of (P, Q), (R, S) is:    id ⊗ (R,S) (P,Q) (W) = sg(P, Q)sg(R, S)δ2 (P, Q) ⊗ δ21 (R, S) ⊗ δ11 (R, S) − sg(P, Q)sg(R, S)δ2 (P, Q) ⊗ δ11 (R, S) ⊗ δ21 (R, S).   And the corresponding co-Jacobi term of (R  , S  ), (P  , Q  ) is:    id ⊗ (P  ,Q  ) (R  ,S  ) (W) = sg(P  , Q  )sg(R  , S  )δ1 (R  , S  )

⊗ δ12 (P  , Q  ) ⊗ δ22 (P  , Q  ) − sg(P  , Q  )sg(R  , S  )δ1 (R  , S  )

⊗ δ22 (P  , Q  ) ⊗ δ12 (P  , Q  ) = sg(P, Q)sg(R, S)δ11 (R, S) ⊗ δ21 (R, S) ⊗ δ2 (P, Q) − sg(P, Q)sg(R, S)δ11 (R, S) ⊗ δ2 (P, Q) ⊗ δ21 (R, S). Then     id + ε + ε2 id ⊗ (R,S) (P,Q) (W)     + id + ε + ε2 id ⊗ (P  ,Q  ) (R  ,S  ) (W) = 0.

5 The Lie algebra structure In this section, we develop the tools necessary to prove that V supports a structure of Lie algebra. In the next paragraph we give a short definition of the Lie bracket, for more details see the section 2.3 of [1]. The Lie bracket [ , ] : V ⊗ V → V is defined as the linear map such that for each pair of cyclic words, W and Z,  W, Z =



sg(P, Q)γ (P, Q),

(P,Q)∈LP2 (W ,Z )

where LP2 (V, W) is the set of all linked pairs (P, Q) for which there exist positive integers j and k such that P is an occurrence of a subword of V j and Q is an occurrence of a subword of W k . The cyclic word γ (P, Q) is defined as c(W1 Z 1 ), where W1 and Z 1 are defined as follows: if (P, Q) a linked pair of type (1) or (2), see Definition 12, then W1 is the representative of W obtained by cutting W immediately before p2 and Z 1 is the representative of Z obtained by cutting Z immediately before q2 ; if (P, Q) a linked pair of type (3), W1 is the linear subword of W that starts right after the end

The Lie bialgebra structure of the vector...

of Y and ends right before the first letter of Y , and Z 1 is the subword of Z that starts right after the last letter of Y and ends right before the beginning of Y . Remark 2 By Proposition 2.9 of [1], LP2 (W, Z) is a finite set and as a consequence the bracket is well defined. If V and W are reduced cyclic words and (P, Q) ∈ LP2 (V, W), then γ (P, Q) = γ (Q, P). Remark 3 Let V and W be reduced cyclic words and (P, Q) ∈ LP2 (V, W). Then we have the following consequences: (i) If l(V) = l(W), then P ⊂ V and Q ⊂ W. (ii) If l(V) < l(W), then W1 is not a power of V1 . The aim of this section is to prove Theorem 2, where the skew-symmetry condition is a consequence of Remark 2, but the proof of Jacobi condition is more complicate. The following results are the key to this demonstration. As in the proof of Theorem 1, we build cyclic words that cancel each addend in the condition of Jacobi. We give the proof of these results for linked pairs of type 2, the remaining cases are shown in the same way. Proposition 3 For (P, Q) ∈ LP2 (V, W), with l(W) ≥ l(V), we have that Q ⊂ W j for j ≤ 2. Moreover, if j = 2 and W1 = q2 B1 q1 B2 , then W 2 = c(q2 B1 q1 X ). Proof If l(W) = l(V), by Remark 3, we have that P ⊂ V and Q ⊂ W. Then j = 1. l(V) If l(W) > l(V), by Proposition 2.13 of [1], we conclude that j ≤ 2 + < 3. l(W) Suppose now that j = 2. Since (P, Q) is a linked pair of type Definition 12(2), then we have linear representations of V and W as V1 = p2 A1 p1 A2 and W1 = q2 B1 q1 B2 , respectively. We know that Q ⊂ W 2 and Q ⊂ W, then Q = q1 B2 q2 B1 q1 B2 q2 ,   X = B2 q2 B1 q1 B2 and W12 = q2 B1 q1 q X . Proposition 4 For (P, Q) ∈ LP2 (V, W), there exist (P1 , Q 1 ) ∈ LP2 (γ (P, Q), W) and (P2 , Q 2 ) ∈ LP2 (V, γ (P, Q)) such that P1 = p1 X W1 p2 , Q 1 = q1 X W1 q2 , P2 = p1 X V1 p2 and Q 2 = q1 X V1 q2 . Proof For l(V) = l(W), we have that Q ⊂ W, P ⊂ V, V1 = p2 Ap1 X and W1 = q2 Bq1 X . Therefore, γ (P, Q) = c( p2 Ap1 Xq2 Bq1 B) and P1 = p1 X W1 p2 , Q 2 = q1 X V1 q2 are linear subwords of γ (P, Q)2 . Note that W 2 = c(q1 X W1 q2 X ) since Q 1 is a linear subword of W 2 . On the other hand V 2 = c( p1 X V1 p2 A) hence P2 is a linear subword of V 2 . For l(V) < l(W), we are going to prove only the case j = 2, the other case, for j = 1, is similar. Let P = p1 X p2 , Q = q1 Xq2 and V1 = p2 A1 p1 A2 , W1 = q2 B1 q1 B2 be linear representatives of V and W, respectively. Applying Proposition 3 we have that W12 = q2 B1 q1 X , where X = B2 q2 B1 q1 B2 . On the other hand, since P ∈ V i , then we conclude that V1i = p2 A1 p1 X and X = A2 V1i−1 , where i ≥ 2. This is because l(W) > l(V) and j = 2.

A. González

W1 X q2

q1

B1

p2

A1

B2

A2 C p1 p 2

q2

D

p2

q1

E p2

B2

F

p1

W12

V1 i

Fig. 2 Representation of the power of W and V as segments and subwords as interval

In Fig. 2, we give a pictorial representation of these words and the relative position of (P, Q) in each one of them.   Now, since l(V) < l(W) and l(X ) < l(V)+l(W), then we have that l B2 < l(V), W1 = q2 DV1l = q2 DV1l−1 p2 Eq1 B2 and V1 = p2 Eq1 B2 = p2 Cq2 D (see Fig. 2). Moreover, B2 = A2 p2 C = F p1 A2 . Since (P, Q) is a linked pair, then the pairs (P1 , Q 1 ) and (P2 , Q 2 ) are linked pairs too. To complete the proof we need to prove that the pairs lie in the corresponding words. 1. Q 1 is a subword of some power of W: this is because W1 = q2 B1 q1 B1 and X = B2 W1 . Hence W13 = q2 B1 q1 X W1 . In particular Q 1 ⊂ c(W13 ). 2. P1 is a subword of some power of γ (P, Q): first, note that W1 V1 = q2 Dp2 C W1 , where V1 = p2 Cq2 D. Then γ (P, Q)2 = c(W1 V1 )2 = c(q2 Dp2 C W1 W1 V1 ) = c(q2 Ap1 A2 p2 C W1 W1 V1 ) = c(q2 Ap1 X W1 V1 ). Therefore, P1 ⊂ γ (P, Q)2 . 3. P2 is a subword of some power of V: similar as in the first case we have that V1i = p2 A1 p1 X , hence V1i+1 = p2 A1 p1 X V1 and consequently P2 ⊂ c(V1i+1 ). 4. Q 2 is a subword of some power of γ (P, Q): first, note that V1l = p2 Eq1 B2 . Hence       γ (P, Q)2 = c p2 Eq1 B2 W1 V1 W1 = c p2 Eq1 X V1 W1 = c p2 Eq1 X V1 q2 B1 q1 B2

and for that reason we have Q 2 ⊂ γ (P, Q)2 .

 

Corollary 1 For (P, Q) ∈ LP2 (V, W), with P = p1 X p2 and Q = q1 Xq2 , we have that X ⊂ γ (P, Q)2 . The following technical results are needed to demonstrate the Jacobi  condition. (V, W) and (R, S) ∈ LP They guarantee that given (P, Q) ∈ LP 2 2 γ (P, Q), Z      there exist P1 , Q 1 and R1 , S1 such that the term [V, W](P,Q) , Z (R,S) is canceled   by a term in [W, Z], V or [Z, V], W related with the linked pairs in the sum      [ , ] ◦ (id ⊗ [ , ]) ◦ id + ε + ε2 V ⊗ W ⊗ Z . The following lemma is crucial to prove the Jacobi condition, but his proof is tedious since to guarantee the existence of new linked pairs it is necessary to do an exhaustive

The Lie bialgebra structure of the vector...

q2

W1

s1

p2

V1

s1

q2

(3)

W1 s1

(2)

p2 s1

V1

s2

2

(W1V1)

(1)

Fig. 3 Possible position of the point s1 ∈ γ (P, Q)2

study of all the possible relative positions of the pairs (P, Q) and (R, S) in addition to differentiate between the signs of them. Lemma 2 For V, W and Z cyclic words such that l(Z) ≤ l(V) ≤ l(W) and (P, Q) ∈ LP2 (V, W), (R, S) ∈ LP2 (Z, γ (P, Q)), there exists a linked pair (T, U ) ∈ LP2 (W, Z) or (T, U ) ∈ LP2 (V, Z). Proof We are considering P = p1 X p2 , Q = q1 Xq2 , R = ri Y r2 and S = s1 Y s2 . Applying the Lemma 2.12 of [1] we have that l(Y ) < l(V) + l(W) + l(Z). This is because Y is a subword of a power of Z and γ (P, Q). / V1 the procedure is analogous). Note We can suppose that s2 ∈ V1 (for s2 ∈ that since l(γ (P, Q)) > l(Z), then S ⊂ γ (P, Q)2 . Figure 3 represents the possible positions of the point s1 in the linear word (W1 V1 )2 . Next we study at each of these possible positions that s1 may have. 1. For s1 ∈ V1 , we have that (R, S) ∈ LP2 (Z, V). Then we define T = S and U = R. 2. For s1 ∈ W1 , note that X W1 ⊂ Y . In this case Y intersects W1 . (a) If sg(P, Q) = sg(R, S), then we can construct a linked pair (T, U ) ∈ LP2 (W, Z). Since s1 ∈ W1 , we can find Z ⊂ W1 such that W1 = Z 1 s1 Z . For this reason we define T = s1 Zq2 and U = r1 Z p2 . (b) If sg(P, Q) = sg(R, S), we have two possible cases. If s1 ∈ X V1 , then / X V1 , then it is possible to construct (T, U ) ∈ (R, S) ∈ LP2 (V, Z) but if s1 ∈ LP2 (V, Z). Since s2 ∈ V1 , then we can decompose V1 = p2 Bs2 C and we can define T = p1 X p2 Bs2 and U = q1 X p2 Br2 . 3. For s1 ∈ V1 or s1 ∈ W1 , similarly as before, the following subcases divide: (a) suppose X W1 ⊂ Y . In this case (P1 , Q 1 ) ∈ LP2 (γ (P, Q), W) is a possible solution. This is because P1 = p1 X W1 p2 ⊂ γ (P, Q)2 and X W1 ⊂ Y . Then T = Q 1 , U = P1 and (T, U ) ∈ LP2 (W, Z); (b) suppose that X W1 ⊂ Y . If s1 ∈ V1 and sg(P, Q) = sg(R, S), then we can define T = s1 X 1 W1 q2 and U = r1 X 1 W1 p2 , where V1 = X 2 s1 X 2 and (T, U ) ∈ LP2 (W, Z). / If sg(P, Q) = sg(R, S) and s1 ∈ X V1 , then (R, S) ∈ LP2 (V, Z). But if s1 ∈ X V1 , then (T, U ) ∈ LP2 (V, Z), where T = p1 X p2 Bs2 and U = q1 X p2 Br2 . Finally, if s1 ∈ W1 and sg(P, Q) = sg(R, S), then (T, U ) ∈ LP2 (V, Z),   where T = s1 X p2 and U = r1 Xq2 . Lemma 3 For V, W and Z cyclic words such that l(Z) ≤ l(V) ≤ l(W) and (P, Q) ∈ LP2 (V, W), (R, S) ∈ LP2 (Z, γ (P, Q)), then we have the following consequences:

A. González

(a) for (T, U ) ∈ LP2 (W, Z) there exists (T  , U  ) ∈ LP2 (V, γ (T, U )) such that   γ(T,U ) T  , U  = γ(P,Q) (R, S) and   V, [W, Z](T,U ) T  ,U   = − Z, V, W](P,Q) (R,S) ; and (b) for (T, U ) ∈ LP2 (V, Z) there exists (T  , U  ) ∈ LP2 (W, γ (T, U )) such that   γ(T,U ) T  , U  = γ(P,Q) (R, S) and   W, [Z, V](U,T ) T  ,U   = − Z, [V, W](P,Q) (R,S) . Proof For (P, Q) ∈ LP2 (V, W) and (R, S) ∈ LP2 (Z, γ (P, Q)), we need to compute  Z, [V, W](P,Q) (R,S) = sg(P, Q)sg(R, S)γ(P,Q) (R, S). By definition γ (P, Q) = c(V1 W1 ), where V1 and W1 are linear representatives of V and W, respectively. On the other hand, as (R, S) ∈ LP2 (Z, γ (P, Q)) then γ(P,Q) (R, S) = c(Z 2 (V1 W1 )2 ), where Z 2 is a linear representative of Z starting in r2 and (V1 W1 )2 is a linear representative of γ (P, Q) starting in s2 . We suppose that s2 ∈ V1 (the other case is an exercise for the reader). As a consequence of the assumption we have that V1 = p2 Bs2 A. In this case we have two possibilities with respect to Z: (i) the  Z 2 can be expressed as Z 2 = r2 C p2 B. In this case we have that  word V1 W1 2 = s2 AW1 p2 B and       γ(P,Q) (R, S) = c r2 C p2 Bs2 AW1 p2 B = c p2 Bs2 AW1 p2 Br2 C = c V1 W1 Z 1 , (ii) the word p2 B can be expressed as p2 B = Z 1i p2 D and Z 1 = p2 Dr2 C, Z 2 = r2 C p2 D. As a consequence         c Z 2 V1 W1 2 = r2 C p2 Ds2 AW1 p2 B = r2 C p2 Ds2 AW1 Z 1i p2 D       i+1 = c Z 1 p2 Ds2 AW1 = c Z 1 p2 Bs2 AW1 = c V1 W1 Z 1 . Finally, we have that γ(P,Q) (R, S) = c(V1 W1 Z 1 ). We prove the result for some cases. (1) For X W1 ⊂ Y , there is (T, U ) ∈ LP2 (W, Z), where T = q1 X 1 q2 and U = p1 X 1 p2 (X 1 = X W1 ). By Proposition 2 we have that there exists (U2 , T2 ) ∈ LP2 (Z, γ (T, U )), where U2 = p1 X 1 Z 1 p2 and T2 = q1 X 1 Z 1 q2 . As T2 is a subword of some power of γ (T, U ) and Q ⊂ T2 is Q a subword of some power of γ (T, U ), then (P, Q) ∈ LP2 (V, γ (T, U )). In this case we define T  = P and U  = Q. Moreover,         γ(T,U ) T  , U  = c V1 Z 1 W1 1 = c V1 W1 Z 1 = γ(P,Q) (R, S).

The Lie bialgebra structure of the vector...

 Now, we determine the term V, [W, Z](T,U ) T  ,U   .      V, [W, Z](T,U ) T  ,U   = sg T  , U  sg(T, U )γ(T,U ) T  , U  =

sg(P, Q)sg(Q, P)γ(P,Q) (R, S) = −γ(P,Q) (R, S).

 Note that sg(P, Q) = sg(R, S), hence V, [W, Z](T,U ) T  ,U   =  − Z, [V, W](P,Q) (R,S) . But, if sg(P, Q) = sg(R, S), then this is not true. (2) For X W1 ⊂ Y and sg(P, Q) = sg(R, S), we consider T = p1 X 2 Bs2 and U = q1 X p2 Br2 , with (T, U ) ∈ LP2 (V, Z) and T  = P, U  = Q, where it is easy to prove that (P, Q) ∈ LP2 (γ (T, U ), W). We can write V2 = s2 Ap2 B and Z 2 = r2 C p2 B, then V2 Z 2 = s2 Ap2 Br2 C p2 B, and (V2 Z 2 )1 = p2 Br2 C p2 Bs2 A = Z 1 V1 . Moreover, γ(T,U ) (T  , U  ) = c((V2 Z 2 )1 W1 ) = c(Z 1 V1 W1 ) = c(V1 W1 Z 1 ). Consequently,      W, [Z, V](U,T ) U  ,T   = sg U  , T  sg(U, T )γ(U,T ) U  , T    = sg(Q, P)sg(Q, P)γ(U,T ) U  , T   = − Z, [V, W](P,Q) (R,S) . (3) Let now X W1 ⊂ Y and sg(P, Q) = sg(R, S), by Lemma 2, there is a (T, U ) ∈ LP2 (W, Z), with T = r1 X 2 q2 and U = s1 X 2 p2 . We need to guaranty the existence of another linked pair (T  , U  ) ∈ LP2 (V, γ (T, U )). / X V1 . In this situation, we can construct, by Proposition 2, Suppose the case s1 ∈ the linked pair (R2 , S2 ) ∈ LP2 (W, γ (T, U )), where R2 = r1 X 2 W1 q2 and S2 = s1 X 2 W 1 p 2 . Note that Q ⊂ S2 , hence (P, Q) ∈ LP2 (V, γ (R1 , S1 )). Therefore, we can define T  = P and U  = Q. As (T  , U  ) ∈ LP2 (V, γ (T, U )) we have that γ(T,U ) (T  , U  ) = c(V1 (W1 Z 1 )1 )   = c(V1 W1 Z 1 ). Then ) (T , U ). Finally, we can calculate  γ(T,U  γ(P,Q) (R, S) =  the following term V, [W, Z](T,U ) T  ,U  .      V, [W, Z](T,U ) T  ,U   = sg T  , U  sg(T, U )γ(T,U ) T  , U    = sg(P, Q)sg(R, S)γ(T,U ) T  , U   = − Z, [V, W](P,Q) (R,S) .   Theorem 2 The pair (V, [ , ]) is a Lie algebra.

A. González

Proof (1) Skew-symmetry condition: [ , ] ◦ s = −[ , ]. 

   [ , ] ◦ s V ⊗ W = W, V =

sg(P, Q)γ (P, Q)

(P,Q)∈LP2 (W ,V )

 sg(Q, P)γ (Q, P) = − W, V .

=−

(Q,P)∈LP2 (V ,W )

    (2) Jacobi identity: [ , ] ◦ id ⊗ [ , ] ◦ id + ε + ε2 = 0. 

    [ , ] ◦ (id ⊗ [ , ]) ◦ id + ε + ε2 V ⊗ W ⊗ Z = [V, [W, Z]] + [W, [Z, V]] + [Z, [V, W]],

and 



 Z, V, W =

 sg(P, Q) Z, γ (P, Q)

(P,Q)∈LP2 (V ,W )



=

 

(P,Q)∈LP2 (V ,W ) (R,S)∈LP2 Z ,γ (P,Q)



 Notation: Z, V, W (P,Q)

(R,S)



sg(P, Q)sg(R, S)γ (R, S)γ (P,Q) .

= sg(P, Q)sg(R, S)γ (R, S)γ (P,Q) .

We can suppose, without any lost of generality, that l(Z) ≤ l(V) ≤ l(W). For (P, Q) ∈ LP2 (V, W) and (R, S) ∈ LP2 (Z, γ (P, Q)), by Lemmas 2 and 3, there is (T, U ) ∈ LP2 (V, Z) and (T  , U  ) ∈ LP2 (W, γ (T, U )) such that   W, [Z, V](U,T ) T  ,U   = − Z, [V, W](P,Q) (R,S)     or there is (T, U ) ∈ LP2 (W, Z) and T  , U  ∈ LP2 V, γ (T, U ) such that   V, [W, Z](T,U ) T  ,U   = − Z, [V, W](P,Q) (R,S) . Therefore, for each   pair (P, Q) ∈ LP2 (V, W) and for each pair (R, S) ∈ Z, γ (P, Q) LP , there exist another pair of linked pairs such that  2 Z, [V, W](P,Q) (R,S) is canceled by the term corresponding to these new pairs.    This implies that Z, [V, W] + W, [Z, V] + V, [W, Z] = 0.  

6 Compatibility Finally, in this section we prove the compatibility between the two Lie structures that supports the space V studied in the previous sections.

The Lie bialgebra structure of the vector...

  Lemma  4 For (P, Q) ∈ LP1 (V) and (R, S)∈ LP2 δ1(P, Q),  δ2 (P,  Q) , there exist  R1 , S1 ∈ LP1 (V) such that (Q, P) ∈ LP2 δ1 R1 , S1 , δ2 R1 , S1 with   sg R1 , S1 = sg(R, S) and γ(P,Q) (R, S) = γ R

1 ,S1

 (Q, P).

Proof We construct the linked pair for some cases. With this list of cases the reader can easily complete the study of cases. We can suppose that l(δ2 (P, Q)) ≥ l(δ1 (P, Q)). 1. For l(δ2 (P, Q)) = l(δ1 (P, Q)), since we have that R ⊂ δ1 (P, Q) and S ⊂ δ2 (P, Q), then (R, S) ∈ LP1 (V). Remember that V1 is the linear representative of δ1 (P, Q) starting at p2 . As a consequence, we can suppose that V1 = p2 B1r1 Y1 q1 Y2 r2 X 2 , where Y = Y1 q1 Y2 and X = Y2 r2 X 2 . In a similar way we can consider that V2 = q2 A1 s1 Y s2 A2 p1 X. Therefore, V1 V2 = p2 B1r1 Y1 q1 Y2 r2 X 2 q2 A1 s1 Y s2 A2 p1 X and     δ1 (R, S) = cr2 X 2 q2 A1 s1 Y1 q1 Y2 = c A1 s1 Y1 q1 Xq2  Q, δ2 (R, S) = c s2 A2 p1 X p2 B1r1 Y  P.   Hence (Q, P) ∈ LP2 δ1 (R, S), δ2 (R, S) . Finally, we compute γ(P,Q) (R, S) and γ(R,S) (Q, P). Therefore,   γ(P,Q) (R, S) = c r2 X 2 p2 B1r1 Y s2 A2 p1 Xq2 A1 s1 Y   = c p2 B1r1 Y s2 A2 p1 Xq2 A1 s1 Y r2 X 2   = c p2 B1r1 Y s2 A2 p1 Xq2 A1 s1 Y1 q1 X , and   γ(R,S) (Q, P) = c q2 A1 s1 Y1 q1 X p2 B1r1 Y s2 A2 p1 X   = c p2 B1r1 Y s2 A2 p1 Xq2 A1 s1 Y1 q1 X , and we conclude that γ(P,Q) (R, S) = γ(R,S) (Q, P). 2. For l(δ2 (P, Q)) > l(δ1 (P, Q)) we consider the case j = 1 and sg(P, Q) = sg(R, S). Since j = 1 we have S ⊂ δ2 (P, Q), and as a consequence V2 = q2 A1 s1 Y s2 A2 p1 X and V1 = p2 B1r1 C2 r2 B2 q1 X , where Y = C2 r2 B2 q1 X p2 B1 r1 C2 . In this case we have V1 V2 = p2 B1r1 C2 r2 B2 q1 Xq2 A1 s1 C2 r2 B2 q1 X p2 B1r1 C2 s2 A2 p1 X

A. González

and R1 = r1 C2 r2 B2 q1 Xq2 , S1 = s1 C2 r2 B2 q1 X p2 , is a linked pair of V and     δ1 R1 , S1 = c q2 A1 s1 C2 r2 B2 q1 X  Q.     δ2 R1 , S1 = c p2 B1r1 C2 s2 A2 p1 X p2 B1r1 C2 r2 B2 q1 X  P. Finally, we compute γ(P,Q) (R, S) and γ(R1 ,S1 ) (Q, P).   γ(P,Q) (R, S) = c r2 B2 q1 X p2 B1r1 C2 s2 A2 p1 Xq2 A1 s1 Y ,   γ R ,S  (Q, P) = c q2 A1 s1 Y r2 B2 q1 X p2 B1r1 C2 s2 A2 p1 X , 1

1

and we conclude that γ(P,Q) (R, S) = γ R ,S  (Q, P). 1 1     3. For l δ2 (P, Q) > l δ1 (P, Q) we consider the case j = 2 and sg(P, Q) = sg(R, S). This case is a little different, but with some care it is possible to construct linear representatives of δ1 (P, Q) and δ2 (P, Q) as follows: V1 = p2 B4 s1 D2 r1 C2 r2 D1 s2 B3 q1 X and V2 = q2 A1 s1 E 2 s2 A2 p1 X, where E 2 = D2 r1 C2 = C2 r2 D1 . Then V1 V2 = p2 B4 s1 E 2 r2 D1 s2 B3 q1 Xq2 A1 s1 E 2 s2 A2 p1 X, with A1 = H2 r2 D4 = H2 r2 D1 s2 B3 q1 X p2 B4 . Therefore, we can find a new linked pair in V: R1 = p1 X p2 B4 s1 E 2 r2 , S1 = q1 X p2 B4 s1 E 2 s2 , and

  δ1 R1 , S1 =   δ2 R1 , S1 =

  c r2 D1 s2 B3 q1 Xq2 H2 r2 D1 s2 B3 q1 X p2 B4 s1 E 2  Q,   c s2 A2 p1 X p2 B4 s1 E 2  P.

As before, it rests to determine γ(P,Q) (R, S) and γ R

1 ,S1

 (Q, P):

 (Q, P) = cq2 H2 r2 D4 s1 E 2 r2 D4 s1 E 2 s2 A2 p1 X    = c q 2 A 1 s1 E 2 r 2 D 4 s1 E 2 s2 A 2 p 1 X ,   γ(P,Q) (R, S) = c r2 D4 s1 D2 r1 C2 s2 A2 p1 Xq2 A1 s1 E 2   = c q 2 A 1 s1 E 2 r 2 D 4 s1 E 2 s2 A 2 p 1 X .

γ R

1 ,S1

 

The Lie bialgebra structure of the vector...

Corollary 2 With the same conditions of the last lemma we have that:    [ , ](R,S) δ(P,Q) (V) = −[ , ](Q,P) δ R

1 ,S1

  (V) .

  Theorem 3 The triple V, [ , ],  is an involutive Lie bialgebra. Proof (1) Involutive: [ , ] ◦  = 0. 

[ , ] ◦ (V) =

 sg(P, Q) δ1 (P, Q), δ2 (P, Q) .

(P,Q)∈LP1 (V )





δ1 (P, Q), δ2 (P, Q) =



(R,S)∈LP2 δ1 (P,Q),δ2 (P,Q)



sg(P, Q)sg(R, S)γ (R, S).

Consequently, we have that 

  [ , ] (V) =

 

(P,Q)∈LP1 (V ) (R,S)∈LP δ (P,Q),δ (P,Q) 2 1 2



sg(P, Q)sg(R, S)γ (R, S).

To prove this result, we need to guarantee that the term sg(P, Q)sg(R, S)γ (R, S) is canceled by another term in the sum. This statement is a consequence of the Lemma 4 and the Corollary 2. (2) Compatibility condition: ([V, W]) = V · (W) − W · (V), where the action is given by Z · (V ⊗ W) = [Z, V] ⊗ W + V ⊗ [Z, W].     V, W = 



 sg(P, Q)γ (P, Q)

(P,Q)∈LP2 (V ,W )

=





sg(P, Q)sg(R, S)δ1 (R, S) ⊗ δ2 (R, S).

(P,Q)∈LP2 (V ,W ) (R,S)∈LP1 (γ (P,Q))

V · (W) =





(P,Q)∈LP1 (W ) (R,S)∈LP2 (V ,δ1 (P,Q))

sg(P, Q)sg(R, S)γ (R, S) ⊗ δ2 (P, Q)   + (P,Q)∈LP1 (W ) (R,S)∈LP2 (V ,δ2 (P,Q))

W · (V) =

sg(P, Q)sg(R, S)δ1 (P, Q) ⊗ γ (R, S).   (P,Q)∈LP1 (V ) (R,S)∈LP2 (W ,δ1 (P,Q))

sg(P, Q)sg(R, S)γ (R, S) ⊗ δ2 (P, Q)

A. González



+



(P,Q)∈LP1 (V ) (R,S)∈LP2 (W ,δ2 (P,Q))

sg(P, Q)sg(R, S)δ1 (P, Q) ⊗ γ (R, S). First, we note that 

([V, W]) =

(P,Q)∈LP2 (V ,W ),(R,S)∈LP1 (γ (P,Q))

 sg(P, Q)sg(R, S) δ1 (R, S) ⊗ δ2 (R, S)  − δ2 (R, S) ⊗ δ1 (R, S) .

(1)

If we have (R, S) ∈ LP1 (γ (P, Q)), then there are three different possible situations: (R, S) ∈ LP1 (V), (R, S) ∈ LP1 (W) or (R, S) ∈ LP2 (V, W)\(P, Q). Now we illustrate the proof of the compatibility condition. We suppose that l(V) = l(W), consequently P ⊂ V and Q ⊂ W. • For (R, S) ∈ LP1 (V) we suppose that V1 = p2 A1r1 Y r2 A2 s1 A3 p1 X 1 s2 X 2 and W1 = q2 Bq1 X , where X = X 1 s2 X 2 and Y = A3 p1 X 1 . First, we determine the following:     (R,S) [V , W ](P,Q) = sg(P, Q)sg(R, S) δ1 (R, S) ⊗ δ2 (R, S) − δ2 (R, S) ⊗ δ1 (R, S) .

Since γ (P, Q) = c( p2 A1r1 Y r2 A2 s1 A3 p1 X 1 s2 X 2 q2 Bq1 X ), then δ1 (R, S) = c(r2 A2 s1 A3 p1 X 1 ) and δ2 (R, S) = c(s2 X 2 q2 Bq1 X p2 A1r1 Y ). Subsequently by (R, S) ∈ LP1 (V), we have δ1 (R, S)V = c(r2 A2 s1 A3 p1 X 1 ) and δ2 (R, S)V = c(s2 X 2 p2 A1r1 Y ) = c(s2 X 2 p2 A1 r1 A3 p1 X 1 ) = c(A1r1 A3 p1 X 1 s2 X 2 p2 ) = c(A1r1 A3 p1 X p2 )  P. We conclude that (Q, P) ∈ LP2 (W, δ2 (R, S)V ), and as a result we have associated to these pairs the term W ·(Q,P) (V)(R,S) .  W ·(Q,P) (V)(R,S) = sg(Q, P)sg(R, S) δ1 (R, S) ⊗ γ (Q, P) − γ (Q, P)  ⊗ δ1 (R, S)    = −sg(P, Q)sg(R, S) c r2 A2 s1 A3 p1 X 1   ⊗ c q2 Bq1 X p2 A1r1 A3 p1 X     + c q2 Bq1 X p2 A1r1 A3 p1 X ⊗ c r2 A2 s1 A3 p1 X 1    = −sg(P, Q)sg(R, S) c r2 A2 s1 A3 p1 X 1   ⊗ c q2 Bq1 X p2 A1r1 Y s2 X 2     + c q2 Bq1 X p2 A1r1 Y s2 X 2 ⊗ c r2 A2 s1 A3 p1 X 1 . Therefore, (R,S) ([V, W](P,Q) ) = −W ·(Q,P) (V)(R,S) . • For (R, S) ∈ LP1 (W) we suppose that W1 = q2 B1r1 Y r2 B2 s1 B3 q1 X 1 s2 X 2 and V1 = p2 Ap1 X , where X = X 1 s2 X 2 and Y = B3 q1 X 1 .

The Lie bialgebra structure of the vector...

In this case we have that δ1 (R, S)W = c(r2 B2 s1 B3 q1 X 1 ) and δ2 (R, S)W = c(s2 X 2 q2 B1r1 Y ) = c(s2 X 2 q2 B1r1 B3 q1 X 1 ) = c(B1r1 B3 q1 X 1 s2 X 2 q2 ) = (B1r1 B3 q1 Xq2 )  Q. Hence, in this case we have that (P, Q) ∈ LP2 (V, 2 (R, S)W ), and we have associated the term V ·(P,Q) (W)(R,S) to these pairs. Note that, in this case    (R,S) [V, W](P,Q) = sg(P, Q)sg(R, S) δ1 (R, S)

 ⊗ δ2 (R, S) − δ2 (R, S) ⊗ δ1 (R, S)    = sg(P, Q)sg(R, S) c r2 B2 s1 B3 q1 X 1   ⊗ c s2 X 2 p2 Ap1 Xq2 B1r1 Y   − c s2 X 2 p2 Ap1 Xq2 B1r1 Y   ⊗ c r2 B2 s1 B3 q1 X 1 .

On the other hand  V ·(P,Q) (W)(R,S) = sg(P, Q)sg(R, S) δ1 (R, S)  ⊗ γ (P, Q) − γ (P, Q) ⊗ δ1 (R, S)    = sg(P, Q)sg(R, S) c r2 B2 s1 B3 q1 X 1   ⊗ c p2 Ap1 Xq2 B1r1 B3 q1 X     − c p2 Ap1 Xq2 B1r1 B3 q1 X ⊗ c r2 B2 s1 B3 q1 X 1    = sg(P, Q)sg(R, S) c r2 B2 s1 B3 q1 X 1   ⊗ c p2 Ap1 Xq2 B1r1 Y s2 X 2     − c p2 Ap1 Xq2 B1r1 Y s2 X 2 ⊗ c r2 B2 s1 B3 q1 X 1 .   Hence (R,S) [V, W](P,Q) = V ·(P,Q) (W)(R,S) . • Finally, we suppose that R ⊂ V and S ⊂ W. In this case we consider V1 = p2 A1r1 Y r2 A2 p1 X and W1 = q2 B1 s1 Y1 q1 Y2 s2 X 1 , where Y = Y1 q1 Y2 and X = Y2 s2 X 1 . Consequently,   γ (P, Q) = c p2 A1r1 Y r2 A2 p1 Xq2 B1 s1 Y1 q1 Y2 s2 X 1 and    (R,S) [V, W](P,Q) = sg(P, Q)sg(R, S) δ1 (R, S)  ⊗ δ2 (R, S) − δ2 (R, S) ⊗ δ1 (R, S)    = sg(P, Q)sg(R, S) c r2 A2 p1 Xq2 B1 s1 Y   ⊗ c s2 X 1 p 2 A 1 r 1 Y     − c s2 X 1 p2 A1r1 Y ⊗ c r2 A2 p1 Xq2 B1 s1 Y

A. González

   = sg(P, Q)sg(R, S) c r2 A2 p1 Xq2 B1 s1 Y   ⊗ c p2 A1r1 Y1 q1 X     − c p2 A1r1 Y1 q1 X ⊗ c r2 A2 p1 Xq2 B1 s1 Y .

Since R ⊂ V and S ⊂ W we have that V and W have linear representatives starting at r2 and s2 , respectively, as follows: V2 = r2 A2 p1 Y2 s2 X 1 p2 A1r1 Y1 q1 Y2 and W2 = s2 X 1 q2 B1 s1 Y1 q1 Y2 . As a consequence   γ (R, S) = c r2 A2 p1 Y2 s2 X 1 p2 A1r1 Y1 q1 Y2 s2 X 1 q2 B1 s1 Y1 q1 Y2   = c r2 A2 p1 X p2 A1r1 Y1 q1 Xq2 B1 s1 Y .   Therefore, (P,  Q) ∈ LP1 γ (R, S) . In this situation we can calculate (P,Q) [V, W](R,S) as follows:    (P,Q) [V, W](R,S) = sg(P, Q)sg(R, S) δ1 (P, Q)  ⊗ δ2 (P, Q) − δ2 (P,   Q) ⊗ δ1 (P, Q) = sg(P,  Q)sg(R, S) c p2A1r1 Y1 q1 X ⊗ cq2 B1 s1 Y r2 A2 p1 X    − c q2 B1 s1 Y r2 A2 p1 X ⊗ c p2 A1r1 Y1 q1 X .     Note that (R,S) [V, W](P,Q) = −(P,Q) [V, W](R,S) , then these terms kill each other in the expression (1). As a consequence, we have that the compatibility is satisfied.  

Appendix 7 Linked pairs I consider it important to describe in this section the concepts of linked pair and their signs, which were introduced by M. Chas in Section 2 of [1]. Let An be a fixed alphabet and O (see Definition 3) be a reduced cyclic word such that every letter of An ∪ An appears exactly once, this word is called a surface symbol.   Definition 11 To each cyclic word W, we associate a number, o(W) ∈ −1, 0, 1 as follows: (i) If W is reduced and there exists an injective orientation preserving map, from the letters of W to the letters of O then o(W) = 1. (ii) If W is reduced and there exists an injective orientation reversing map, from the letters of W to the letters of O then o(W) = −1.

The Lie bialgebra structure of the vector...

(iii) In all other cases (that is, if W is not reduced or if there is no such orientation preserving or reversing map) o(W) = 0. Definition 12 Let P and Q be two linear words. The ordered pair (P, Q) is O-linked if P and Q are reduced words of length at least two and one of the following conditions holds: (1) P = p1 p2 , Q = q1 q2 and o(c( p 1 q 1 p2 q2 )) = 0; (2) P = p1 Y p2 , Q = q1 Y q2 , p1 = q1 , p2 = q2 and Y is a linear word of length at least one and if Y = x1 X x2 , then o(c( p 1 q 1 x1 )) = o(c( p2 q2 x 2 )); (3) P = p1 Y p2 , Q = q1 Y q2 , p1 = q 2 , p2 = q 1 and Y is a linear word of length at least one and if Y = x1 X x2 , then o(c(q2 p 1 x1 )) = o(c(q 1 p2 x 2 )). Definition 13 To each linked pair (P, Q) one associates a sign as follows: (i) If (P, Q) is a linked pair of type Definition 12(1), then sg(P, Q) := o(c( p1 q1 p2 q2 )); (ii) if (P, Q) is a linked pair of type Definition 12(2), then sg(P, Q) := o(c( p1 q1 x1 )); (iii) if (P, Q) is a linked pair of type Definition 12(3), then sg(P, Q) := o(c(q2 p1 x1 )). Remark 4 (a) For every linked pair (P, Q), sg(P, Q) = 1 or sg(P, Q) = −1. (b) If (P, Q) is a linked pair, then (Q, P) is also a linked pair and sg(P, Q) = −sg(Q, P).

References 1. Chas, M.: Combinatorial Lie bialgebras of curves on surfaces. Topology 43, 543–568 (2004) 2. William, M.: Goldman, Invariant functions on Lie groups and Hamiltonian flows of surface group representations. Invent. Math. 85, 263–302 (1986) 3. Turaev, V.G.: Skein quanization of poisson algebras of loops on surfaces. Ann. Sci. Éc. Norm. Sup. 24, 635–704 (1991)