Sets with two associative operations

Transcription

1 CEJM 2 (2003) 169{183 Sets with two associative operations Teimuraz Pirashvili A.M. Razmadze Mathematical Inst. Aleksidze str. 1, Tbilisi, , Republic of Georgia Received 7 January 2003; revised 3 March 2003 Abstract: In this paper we consider duplexes, which are sets with two associative binary operations. Dimonoids in the sense of Loday are examples of duplexes. The set of all permutations carries a structure of a duplex. Our main result asserts that it is a free duplex with an explicitly described set of generators. The proof uses a construction of the free duplex with one generator by planary trees. c Central European Science Journals. All rights reserved. Keywords: Trees, permutations, free algebras MSC (2000): 05E99, 20M05, 08B20 1 Introduction Jean-Louis Loday introduced the notion of a dimonoid and a dialgebra. Let us recall, that a dimonoid is a set equipped with two associative operations satisfying 3 more axioms (see [2, 3 or Section 6.2), while a dialgebra is just a linear analog of a dimonoid. In this paper we drop these additional axioms and we consider sets equipped with two associative binary operations. We call such an algebraic structure as a duplex. Thus dimonoids are examples of duplexes. The set of all permutations, gives an example of a duplex which is not a dimonoid (see Section 3). In Section 4 we construct a free duplex generated by a given set via planar trees and then we prove that the set of all permutations forms a free duplex on an explicitly described set of generators. In the last section we consider duplexes coming from planar binary trees and vertices of the cubes as in [5. We prove that these duplexes are free with one generator in appropriate variety of duplexes. 2 Graded sets A graded set is a set X together with a decomposition X = `n2n X n. Here and elsewhere ` denotes the disjoint union of sets. A map of graded sets is a map f : X! Y such

2 170 T. Pirashvili / Central European Journal of Mathematics 2 (2003) 169{183 that f (X n ) Y n, n 0. We let be the category of graded sets. If x 2 X n we write O(x) = n. In this way we obtain a map O : X! N and conversely, if such a map is given then X is graded with X n = O 1 (n): If X and Y are graded sets, then the Cartesian product X Y is also a graded set with O(x; y) := O(x) + O(y); x 2 X ; y 2 Y : The disjoint union of two graded sets X and Y is also graded by (X ` Y ) n = X n ` Yn. A graded set X is called locally nite if X n is nite for any n 2 N. In this case we put X (T) := 1X n=0 (X n )T n 2 Z[[T: An n-ary operation [ on a graded set X is homogeneous if [ : X n graded sets, in other words! X is a map of [(X k1 X kn ) X k1+ +k n : Free semigroups are examples of graded sets. Let us recall that a semigroup is a set equipped with an associative binary operation. For any set S the free semigroup on S is the following graded set (S) = a n 1 n(s); n(s) := S n ; n 0 while the multiplication is given by S n S m! S n+m ; ((x 1 ; ; x n ); (y 1 ; ; y m )) 7! (x 1 ; ; x n ; y 1 ; ; y m ): It is clear that (S)(T) = (S)T 1 (S)T : 3 Two operations on permutations For any n 1 we let n be the set f1; 2; ; ng. Furthermore, we let n be the set of all bijections n! n. An element f 2 n, called a permutation, is speci ed by the sequence (f(1); ; f(n)). The composition law yields the group structure on n. We put 1a := n: n=1 We are going to consider as a graded set with respect of this grading. Thus O(f ) = n means that f : n! n is a bijection. On we introduce two homogeneous associative operations. The rst one is the concatenation, which we denote by. Thus : n m! n+m ;

3 T. Pirashvili / Central European Journal of Mathematics 2 (2003) 169{ is de ned by (f g)(i) = f(i); if 1 µ i µ n; (f g)(i) = n + g(i n); if n + 1 µ i µ n + m: Here O(f) = n and O(g) = m. For example if f 2 3 and g 2 3 are the following permutations then f g 2 6 is given by The second operation \ : n m! n+m ; is de ned by (f\g)(i) = m + f(i); if 1 µ i µ n; (f\g)(i) = g(i n); if n + 1 µ i µ n + m: For example if f 2 3 and g 2 3 are as above, then f\g 2 6 is given by These operations appear also in [5 under the name `over and `under. 3.1 Duplexes One easily checks that both operations ; \ de ned on 1.10 in [5). This suggests the following de nition are associative (see also Lemma De nition 3.1. A duplex is a set D equipped with two associative operations : D D! D and : D D! D. A map f : D! D 0 from a duplex D to another duplex D 0 is a homomorphism, provided f (x y) = f (x) f (y) and f (x y) = f(x) f(y). We let be the category of duplexes. As usual one can introduce the notion of free duplex as follows.

4 172 T. Pirashvili / Central European Journal of Mathematics 2 (2003) 169{183 De nition 3.2. A duplex F is called free if there exists a subset X F, such that for any duplex D and any map f : X! D there exists a unique homomorphism of duplexes g : F! D such that g(x) = f (x) for all x 2 X. If this holds, then we say that F is a free duplex on X. Thus, duplexes generalize the notion of dimonoids introduced by Jean-Louis Loday in [3.Our main result is the following Theorem 3.3. The set equipped with and \ is a free duplex on the set 2. The description of the set 2 and the proof of Theorem 3.3 is given in Section Semigroups ( ; ) and ( ; \) In this section we prove the fact that the semi-group ( ; ) is free. It is easy to see that ( ; ) and ( ; \) are isomorphic semi-groups (see Lemma 3.4) and therefore ( ; \) is free as well. Lemma 3.4. Let! n 2 n be the permutation de ned by! n (i) = n + 1 i; i 2 n and let ¹ n : n! n be the map given by ¹ n(f) =! n f. Then ¹ n ¹ n = Id n and the diagram n m n m n+m n+m n m \ n+m commutes. Therefore the map ¹ = `n ¹ n :! is an isomorphism from the semi-group ( ; ) to the semi-group ( ; \). Proof. Since!! = Id n it follows that ¹ n(¹ n(f )) =! n (! n f ) = f: Furthermore, for any f 2 n, g 2 m and 1 µ i µ n + m, both ¹ n+m(f g)(i) and (¹ n(f )\¹ m(g))(i) are equal to n + m + 1 f (i) or m + 1 g(i n) depending whether i 2 n or not. An element f 2 n is called -decomposable, provided f = gh for some g 2 k ; h 2 m. If such type decomposition is impossible, then f is called -indecomposable. For each f 2 n we put (f) := i fi 2 n j f (i) ig: In particular f (f1; ; (f )g) f1; ; f g. We let f ± 2 on f1; ; (f)g. ±(f ) be the restriction of f

5 T. Pirashvili / Central European Journal of Mathematics 2 (2003) 169{ Lemma 3.5. i) If f = gh, then (f ) = (g) and f ± = g ±. ii) If f 2 then there is a unique f 2 n ±(f) such that n is -decomposable, f = f ± f : iii) An element f 2 n is -indecomposable i f = f ±. Proof. i) It is su±cient to note that if f = gh and O(g) = k, then f (k) f(k). To show ii) and iii) let us assume that f = gh. Then by i) we have (f) = (g) µ O(g) < O(f): Thus f 6= f ±. Conversely, if f 6= f ±, then f = f ± f ; where f 2 k is given by f (i) = f( (f) + i) (f). Here k = O(f) (f) and 1 µ i µ k. The permutation f is well-de ned since f maps the subset f (f ) + 1; ; O(f)g to itself. We let be the set of -indecomposable elements of. We have = `n T n: For example we have n = n 2 = f(2; 1)g; 3 = f(2; 3; 1); (3; 1; 2); (3; 2; 1)g n, where We let u n be the cardinality of the set n. Here are the rst values of u n : 1; 1; 3; 13; 71; 461; 3447; which can be deduced from Corollary 3.7. Jean-Louis Loday informed me that the integer u n is the number of permutations with no global descent [1. Theorem 3.6. The semigroup ( ; ) is free on. Proof. First we show that the set generates. Take f 2. We have to prove that f lies in the subsemigroup generated by. We may assume that f 62. Thus f is -decomposable and we can write f = gh. Since O(g); O(h) < O(f ) we may assume by induction that g and h lie in the subsemigroup generated by. Thus the same is true for f as well. The fact that any f can be written uniquely in the form f = g 1 g k with g i 2 follows from Lemma 3.5. Corollary 3.7. One has ( n ) = n! X p+q=n p!q! + X p+q+r=n Here p; q; r; runs over all strictly positive integers. Proof. We have an isomorphism of graded sets ¹= `n 1 X X n 1 n!t n = X n 1 m 1 p!q!r! mt m n n and therefore

6 174 T. Pirashvili / Central European Journal of Mathematics 2 (2003) 169{183 and the result follows. By transportation of structures we see that ( ; \) is a free semigroup on the set \, which is by de nition the set of all \-indecomposable permutations. Here a permutation f is called \-indecomposable if! n f is -indecomposable, in other words if f (f1; ; ig) 6 fn i + 1; ; ng for all 1 µ i µ n 1. 4 Free duplexes 4.1 Planar trees By tree we mean in this paper a planar rooted tree. Such trees play an important role in the recent work of Jean-Louis Loday and Maria Ronco [6. We let be the set of trees. It is a graded set = a n; n 1 where n is the set of trees with n leaves. The number of elements of n are known as super Catalan numbers and they are denoted by C n. Here are the rst super Catalan numbers: C 1 = 1 = C 2, C 3 = 3, C 4 = 11, C 5 = 45. In general one has the following well-known relation (see for example Section 8.2 of [4). f (T) := 1X C n T n = 1 p T + 1 T2 6T (1) n 1 So 1 has only one element j and similarly 2 has also only one element, while 3 has tree elements,, Any vertex v of a tree t 2 n with n 2, has a level, which is equal to the number of edges in the path connecting v to the root. Thus the root is the unique vertex of level 0. For example, if u 1 =, u 2 = then the tree u 1 has only one vertex (which is of course the root), while u 2 has two vertices the root and a vertex of level one. Let us also recall that on trees there exists an important operation which is called grafting. The grafting de nes a map : n 1 nk! n ; n = n n k : For example, if u 1 and u 2 are as above, then we have: (u 1 ; u 2 ) =

7 T. Pirashvili / Central European Journal of Mathematics 2 (2003) 169{ Let us also note, that (u 1 ; u 2 ) has 4 vertices, two of them of the level one and one of the level two. It is clear that any tree from n, n > 1 can be written uniquely as (t 1 ; ; t k ). Here k is the number of incoming edges at the root. We will say that t is constructed by the grafting of (t 1 ; ; t k ). We need also the following construction on trees. Let t 1 ; ; t k be trees and let I be a subset of k. We consider t = (t 1 ; ; t k ). The tree obtained by contracting the edges of t which connect the root of t with the roots of t i, i 2 I is denoted by I(t 1 ; ; t k ): If I = ;, then I =. For example, if u 1 and u 2 are as above, then we have 1(u 1 ; u 2 ) = 12(u 1 ; u 2 ) = 4.2 The free duplex with one generator : We consider now two di erent copies of the set n for n 2, which are denoted by n and n, n 2. We put := S n 2 n and := S n 2 n: If t 2 :, then we assign to any vertex of t of even level and to any vertex of t of odd level. Similarly, if t 2, then we assign to any vertex of t of even level and to any vertex of t of odd level. So, for example.. * 2 * We call such trees as decorated trees. To be more precise a decorated tree is an element of the set 1[ = (n); n=1 where (1) = 1 and (n) = n [ n ; n 2: We are going to de ne a duplex structure on decorated trees in such a way, that the above tree can be expressed where e =j. More formally, the operations (e e e) (e (e e)); :! ; :! 6 can be de ned as follows. If t 1 ; t then t 1 t 2 := (t 1 ; t 2 ) 2 2; t 1 t 2 := (t 1 ; t 2 ) 2 2 if t 1 ; t 2 2 then t 1 t 2 := 12 (t 1 ; t 2 ) 2 ; t 1 t 2 := (t 1 ; t 2 ) 2

8 176 T. Pirashvili / Central European Journal of Mathematics 2 (2003) 169{183 if t 1 ; t 2 2 then t 1 t 2 := (t 1 ; t 2 ) 2 ; t 1 t 2 := 12 (t 1 ; t 2 ) 2 if t 1 2 and t 2 2, then t 1 t 2 := 2 (t 1 ; t 2 ) 2 ; t 1 t 2 := 1 (t 1 ; t 2 ) 2 if t 1 2 and t 2 2, then t 1 t 2 := 1 (t 1 ; t 2 ) 2 ; t 1 t 2 := 2 (t 1 ; t 2 ) 2 : This concludes the construction of operations on. Let us observe that in all cases t 1 t 2 = I (t 1 ; t 2 ) 2 and t 1 t 2 = J (t 1 ; t 2 ) 2 ; where I f1; 2g consists of i 2 2 such that t i 2 and J f1; 2g consists of i 2 2 such that t i 2 : Lemma 4.1. Both operations and are associative. Proof. One observes that for any decorated trees t 1 ; t 2 ; t 3 in one has (t 1 t 2 ) t 3 = I (t 1 ; t 2 ; t 3 ) = t 1 (t 2 t 3 ) where I f1; 2; 3g consists of the indices i such that t i 2, i = 1; 2; 3. Similarly for. Lemma 4.2. The semigroup ( ; ) is a free semigroup on the set S := 1 [ [ n 1 n and similarly, the semigroup ( ; ) is a free semigroup on the set S := 1 [ [ n 1 n Proof. It su±ces to note that if t 2, but t 62 S, then t can be written uniquely as (t 1 ; ; t k ). If we consider t 1 ; ; t k as elements from S, then we have t = t 1 t k : Similarly for ( ; ). Theorem 4.3. The duplex is a free duplex with one generator j 2 1. Proof. For simplicity we let e be the tree j. Then e generates the duplex. This can be proved by induction. Indeed, (2) = 2 [ 2 has two decorated trees, which are

9 h T. Pirashvili / Central European Journal of Mathematics 2 (2003) 169{ equal respectively to e e and e e. Assume we already proved that any decorated tree from (m) lies in the subduplex generated by a for any m < n and let us prove that the same holds for n = m. Take t 2 (n). Assume t 2 n. Then there exists unique trees t 1 ; ; t k such that t = (t 1 ; ; t k ). We consider t 1 ; ; t k as elements of the set 1 [. Then we have t = t 1 t k and by the induction assumption t can be expressed in the terms of e. Similarly, if t 2. Now we will show that is free on e. We take a duplex D and an element a 2 D. We have to show that there exists a unique homomorphism f :! D such that f (e) = a. We construct such an f by induction. For the tree e we already have f (e) = a. Assume f is de ned for all decorated trees from (m), m < n and take a decorated tree t 2 n. As above we can write t = t 1 t k with unique t 1 ; ; t k and then we put f(t) = f (t 1 ) f(t k ): Similarly, if t 2. Let us prove that f is a homomorphism. The equations f(x y) = f(x) f(y); f (x y) = f (x) f (y) is clear if x; y 2 1. Let us prove only the rst one, because the proof of the second one is completely similar. If x; y 2 then we have x y = (x; y). Thus by de nition f(x y) = f(x) f (y). If x 2 and y 2. Then x y = 1 (x; y) = (x 1 ; ; x k ; y) = x 1 x k y where x = (x 1 ; ; x k ) and x 1 ; ; x k are considered as elements of 1 [. It follows that f(x y) = f(x 1 ) f(x k ) f (y) = f(x) f(y): Similarly if x 2 and y 2. If x; y 2. Then x y = 12 (x; y) = (x 1 ; ; x k ; y 1 ; ; y l ) = x 1 x k y 1 y l where x = (x 1 ; ; x k ), y = (y 1 ; ; y l ) and x 1 ; ; x k ; y 1 ; ; y l are considered as elements of 1 [. It follows that f(x y) = f(x 1 ) f(x k ) f (y 1 ) f (y l ) = f(x) f(y): 4.3 Free duplexes For a set S we consider the set a (S) := n 1 n(s) where n(s) = (n) S n. We de ne maps ; : n(s) m(s)! n+m(s) by (t 1 ; x 1 ; ; x n ) (t 2 ; y 1 ; y m ) := (t 1 t 2 ; x 1 x k ; y 1 ; y m ):

10 178 T. Pirashvili / Central European Journal of Mathematics 2 (2003) 169{183 (t 1 ; x 1 ; ; x n ) (t 2 ; y 1 ; y m ) := (t 1 t 2 ; x 1 x k ; y 1 ; y m ): We leave as an exercise to prove that nite set S one has 5 Proof of Theorem 3.3 (S)(T) = T indecomposable permutations (S) is a free duplex on S. It is clear that for a 1X C n n 2 (S)T n : (2) A permutation f : n! n is called 2-indecomposable if for any i = 1; ; n 1 one has f (i) 6 i; and f(i) 6 fn i + 1; ; ng It is clear that a permutation is 2-indecomposable if it is simultaneously - and \- indecomposable and therefore for the set 2 of 2-indecomposable permutations we have 2 = \ \ : (3) It is clear that = 1 ; 2 = ; = 3 ; 4 = f(2; 4; 1; 3); (3; 1; 4; 2)g 2 2 One checks that 5 has 22 elements. In general for d n = n we have d n = n! 2 X X p!q! + 2 p!q!r! = 2u n n! (4) p+q=n p+q+r=n Here p; q; r; runs over all strictly positive integers. This follows from the formula for u n = n and from Lemma 5.1. Here are the rst values of d n : 1; 0; 0; 2; 22; 202; 1854; Lemma 5.1. If a permutation is -decomposable then it is \-indecomposable. Similarly, if a permutation is \-decomposable then it is -indecomposable. Proof. If not, then there exists a permutation f : n! n and integers 1 µ i µ n 1 µ j µ n 1 with properties 1 and f (i) i; and f(j) 6 fn j + 1; ; ng: Without loss of generality we may assume that j µ i (otherwise we take! n f ). Thus for any k = 1; ; j we have n j < f (k) µ i. Therefore the interval n j; i contains at least j integers. But this is impossible, because i n + j < j.

11 T. Pirashvili / Central European Journal of Mathematics 2 (2003) 169{ Proof of Theorem 3.3 We will prove that is free on 2 as a duplex. First we show that 2 generates. Indeed, if f 62 2 then either f = gh or f = g\h and it is impossible to have both cases. Without loss of generality we may assume that one has the rst possibility. Since ( ; ) is a free semigroup, there exist uniquely de ned f 1 ; ; f k 2 such that f = f 1 f k. If all f i 2 2 we stop, otherwise for some i we have f i 62 \ and therefore one can write f i in unique way as f i = g 1 \ \g m, with g j 2 \. Since O(g i ) < O(f ) this process stops after a few steps. This shows that 2 generates as a duplex. Since in each step there were unique choices we conclude that elements of 2 are free generators. Corollary 5.2. Between the numbers d n and super Catalan numbers C n there is the following relation X X C m d n t n m Furthermore one has n 1 n!t n = X n 1 d n t n + 2 X m 2 Ã 2 + ¹ Ã Ã + ¹ = 0 where Ã(T) = P n 1 n!tn and ¹ (T) = P n 1 d nt n : Proof. The rst part follows from Theorem 3.3 and Equation (2). To get the second part one rewrites the same relation as n 1 Ã(T) = 2f(¹ (T)) ¹ (T) and then use the formula (1) for f(t) = P n 1 C nt n. 6 Duplexes satisfying some additional identities 6.1 Duplex of binary trees We let 1 be the category of duplexes satisfying the identity (a b) c = a (b c) (5) We prove that the free object in 1 is given via binary trees. Let us recall that for us all trees are planar and rooted. A tree is called binary if any vertex is trivalent. We let be the set of all binary trees: := a n 1 n; where n is the set of planar binary trees with n + 1 leaves. We have n n+1. The number of elements of n are known as Catalan numbers and they are denoted by c n. Here are the rst Catalan numbers: c 1 = 1, c 2 = 2, c 3 = 5. In general one has c n = (2n)! n!(n + 1)!

12 180 T. Pirashvili / Central European Journal of Mathematics 2 (2003) 169{183 So 1 has only one element while 2 has two elements, By our de nition the tree j 2 1 is not a binary tree, but it will play also an important role for binary trees as well. We let t be this particular tree. Let us note that any binary tree u can be written in the unique way as gr(u l ; u r ), where u l ; u r 2 [ ftg: The map gr :! is no longer homogeneous, because we changed the grading, but it is of degree one: gr : n m! n+m+1. We introduce two homogeneous operations by ; : n m! n+m u v := gr(u v l ; v r ) u v := gr(u l ; u r v): Here we use induction and the convention u t = t u = u t = t u = u for t = j. These associative operations under the name over and under rst appeared in [5. One observes that (see [5) u v (resp. u v) is the tree obtained by identifying the root of u (resp. v) with the left (resp. right) most leaf of v (resp. u). Theorem 6.1. The duplex Proof. We have and Therefore 1 generated by e = gr(t; t) =. satis es the identity (5). Moreover it is a free object in (u v) l = u v l ; (u v) r = v r ; (u v) l = u l ; (u v) r = u r v: (a b) c = gr(a b l ; b r c) = a (b c): Thus 2 1. Now we prove that is generated by e. First one observes that u e = gr(u; t); e u = gr(t; u): We claim that for any a; b 2 [ ftg one has (a e) b = gr(a; b) Indeed, we can write (a e) b = gr(a; t) b, which is the same as gr(a; t b) thanks to the de nition of the operation. By our convention t b = b and the claim is proved. Take any element u 2. By the claim we have u = (u l e) u r = u l (e u r ) (6)

13 T. Pirashvili / Central European Journal of Mathematics 2 (2003) 169{ Based on this equality one easily proves by induction that e generates the duplex. Let D be any duplex satisfying the equality (5) and take any element a 2 D. We have to show that there exists an unique homomorphism f :! D such that f (e) = a. Since e generates the uniqueness is obvious. We construct recursively f by f(e) = a; and f(u) = (f (u l ) a) f(u r ) and an obvious induction shows that f is indeed a homomorphism. 6.2 Dimonoids We let be the full subcategory of 1 satisfying two more identities: (a b) c = (a b) c (7) a (b c) = a (b c) (8) We refer the reader to [2 for the extensive information on dimonoids and dialgebras which are just linear analogs of dimonoids. 6.3 Duplex of vertices of cubes We let 2 be the full subcategory of 1 satisfying the identity (a b) c = a (b c) (9) Such types of algebras were rst considered in [7 under the name \Doppelalgebren". Free objects in 1 are given via the following construction. For all n 2 we let n be the set of vertices of the n 1 dimensional cube. So n = f 1; 1g n 1 and elements of n are sequences a = (a 1 ; a n 1 ), where a i = 1 or a i = 1. Moreover we let 1 to the singleton feg and := a n: n 1 Following [5 we de ne two homogeneous operations on by e e := ; e e = ; e a := ( 1; a 1 ; ; a n 1 ); a e := (a 1 ; ; a n 1 ; 1); e a := (1; a 1 ; ; a n 1 ); a e := (a 1 ; ; a n 1 ; 1); a b := (a 1 ; ; a n 1 ; 1; b 1 ; ; b m 1 ); a b := (a 1 ; ; a n 1 ; 1; b 1 ; ; b m 1 ): One checks that is a duplex, which satis es both identities (5) and (9).

14 182 T. Pirashvili / Central European Journal of Mathematics 2 (2003) 169{183 Theorem 6.2. [7 The duplex is the free object in 2 generated by e. Proof. We rst show that e generates. For a = (a 1 ; ; a n ) 2 n, we put i = if a i = 1 and i = if a i = 1. Then a = e 1e 2 n 1 e. The fact that this expression does not depend on parentheses follows from the associativity of ; and from the identities (5), (9). Let D be an object of 2 and let x 2 D. One can use the expression of a in terms of e and i to de ne the map f :! D by f(a) = x 1 x 2 n 1 x. Then by induction one shows that f is in fact a homomorphism with f (e) = x. 6.4 Remarks Our results can be used to describe maps between combinatorial objects. It is not di±cult to check that the map à :! constructed in [5 is a homomorphism of duplexes. On the other hand, since is a free duplex with one generator there is a unique homogeneous map :! which is also a homomorphism of duplexes. To specify this map it su±ces to know the image of the generator: (j) = Id 1. Similarly, if one considers as an object of, then there exists a unique surjective homomorphism % :! which takes the generator j of to e = gr(t; t) 2 1. Thus we have % = Ã. Since 2 1, the free object on a set X in 2 is a quotient of the free object on a set X in 1. In particular for X = feg we obtain the canonical quotient homomorphism :!. If one forgets the corresponding algebraic structures this map from binary trees to vertices of cubes coincides with one considered in [5. The composite map % :! from the decorated trees to the vertices has the following description. Take a decorated tree u 2 n and label the leaves of u by the numbers 1; ; n. Then (u) = (a 1 ; ; a n 1 ), where a i is +1 if the sign at the end of the edge coming from the (i + 1)-th leaf is and is 1 if the corresponding sign is. Acknowledgments This paper is in uenced by the work of Jean-Louis Loday, who started to investigate algebras with two associative operations (see [2, 3) and initiated me to the beauty of the combinatorics of trees during several informal discussions. This work was written during my visit at UniversitÄat Bielefeld. I would like to thank Friedhelm Waldhausen for the invitation to Bielefeld. Thanks to my wife Tamar Kandaurishvili for various helpful discussions on the subject and encouraging me to write this paper. The author was partially supported by the grant INTAS and TMR Network ERB FMRX CT References [1 M. Aguiar and F. Sottile: Structures of the Malvenuto-Reutenauer Hopf algebra of permutations, Preprint (2002),

15 T. Pirashvili / Central European Journal of Mathematics 2 (2003) 169{ [2 J.-L. Loday: \Alg ebras ayant deux op erations associatives (dig ebres)", C. R. Acad. Sci. Paris S er. I Math., Vol. 321, (1995), pp. 141{146. [3 J.-L. Loday: \Dialgebras", In: Dialgebras and related operads, Lecture Notes in Math. 1763, Springer, Berlin, 2001, pp. 7{66. [4 J.-L. Loday: \Arithmetree", J. of Algebra, Vol. 258, (2002), pp. 275{309. [5 J.-L. Loday and M. O. Ronco: \Order structure on the algebra of permutations and of planar binary trees", J. Algebraic Combin., Vol. 15, (2002), pp. 253{270 [6 J.-L. Loday and M. O. Ronco: Trialgebras and families of polytopes, Preprint (2002), [7 B. Richter: Dialgebren, Doppelalgebren und ihre Homologie, Diplomarbeit, UniversitÄat Bonn, 1997,