Deformation of the Hopf algebra of plane posets

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1 Deformation of the Hopf algebra of plane posets Loïc Foissy Laboratoire de Mathématiques, Université de Reims Moulin de la Housse - BP REMS Cedex, France loic.foissy@univ-reims.fr ABSTRACT. We describe and study a four parameters deformation of the two products and the coproduct of the Hopf algebra of plane posets. We obtain a family of braided Hopf algebras, which are generically self-dual. We also prove that in a particular case (when the second parameter goes to zero and the first and third parameters are equal), this deformation is isomorphic, as a self-dual braided Hopf algebra, to a deformation of the Hopf algebra of free quasi-symmetric functions FQSym. KEYWORDS. Plane posets; Deformation; Braided Hopf algebras; Self- duality. AMS CLASSFCATON. 06A11, 16W30, 16S80. Contents 1 Backgrounds and notations Double and plane posets Algebraic structures on plane posets Pairings Morphism to free quasi-symmetric functions Deformation of the products 8.1 Construction Particular cases Subalgebras and quotients Dual coproducts Constructions Particular cases Compatibilities with the products Self-duality results A first pairing on H q Properties of the pairing Comparison of pairings with colinear parameters Non-degeneracy of the pairing Morphism to free quasi-symmetric functions A second Hopf pairing on H (q1,0,q, 1,q 4 ) Quantization of the Hopf algebra of free quasi-symmetric functions

2 ntroduction A double poset is a finite set with two partial orders. As explained in [1], the space generated by the double posets inherits two products and one coproduct, here denoted by, and, making it both a Hopf and an infinitesimal Hopf algebra [10]. Moreover, this Hopf algebra is self dual. When the second order is total, we obtain the notion of special posets, also called labelled posets [13] or shapes [1]. A double poset is plane if its two partial orders satisfy a (in)compatibility condition, see Definition 1. The subspace H PP generated by plane posets is stable under the two products and the coproduct, and is self-dual as a Hopf algebra [3, 4]: in particular, two Hopf pairings are defined on it, using the notion of picture [6, 8, 9, 14]. Moreover, as proved in [4], it is isomorphic to the Hopf algebra of free quasi-symmetric functions FQSym, also known as the Malvenuto-Reutenauer Hopf algebra of permutations. An explicit isomorphism Θ is given by the linear extensions of plane posets, see Definition 11. We define in this text a four parameters deformation of the products and the coproduct of H PP, together with a deformation of the two pairings and of the morphism from H PP to FQSym. f q (q 1, q, q 3, q 4 ) K 4, the product m q (P Q) of two plane posets P and Q is a linear span of plane posets R such that R P Q as a set, P and Q being plane subposets of R. The coefficients are defined with the help of the two partial orders of R, see Theorem 14, and are polynomials in q. n particular: m (1,0,0,0), m (0,1,0,0) op, m (0,0,1,0), m (0,0,0,1) op. We also obtain the product dual to the coproduct (considering the basis of double posets as orthonormal) as m (1,0,1,1), and its opposite given by m (0,1,1,1). Dually, we define a family of coassociative coproducts q. For any plane poset P, (P ) is a linear span of terms (P \ ), running over the plane subposets of P, the coefficients being polynomials in q. n the particular cases where at least one of the components of q is zero, only h-ideals, r-ideals or biideals can appear in this sum (Definition 7 and Proposition ). We study the compatibility of q with both products and on H PP (Proposition 3). n particular, (H PP,, q ) satisfies the axiom q (x y) q x q y q q y q 4 (x q y q) (x q y q ). 3 q x P Q f q 3 1, it is a braided Hopf algebra, with the braiding given by c q (P Q) q 4 Q P ; if q 3 1 and q 4 1, this is a Hopf algebra, and if q 3 1 and q 4 0 this is an infinitesimal Hopf algebra. f q 4 1, this is the coopposite (or the opposite) of a braided Hopf algebra. Similar results hold if we consider the second product, permuting the roles of (q 3, q 4 ) and (q 1, q ). We define a symmetric pairing, q such that: x y, q (z) q x y, z q for all x, y, z H PP. f q (1, 0, 1, 1), we recover the first "classical" pairing of H PP. We prove that in the case q 0, this pairing is nondegenerate if, and only if, q 1 0 (Corollary 36). Consequently, this pairing is generically nondegenerate. The coproduct of FQSym is finally deformed, in such a way that the algebra morphism Θ from H PP to FQSym becomes compatible with q, if q has the form q (q 1, 0, q 1, q 4 ). Deforming the second pairing, of H PP and the usual Hopf pairing of FQSym, the map Θ becomes also an isometry (Theorem 40). Consequently, the deformation, q is nondegenerate if, and only if, q 1 q 4 0.

3 This text is organized as follows. The first section contains reminders on the Hopf algebra of plane posets H PP, its two products, its coproducts and its two Hopf pairings, and on the isomorphism Θ from H PP to FQSym. The deformation of the products is defined in section, and we also consider the compatibility of these products with several bijections on PP and the stability of certain families of plane posets under these products. We proceed to the dual construction in the next section, where we also study the compatibility with the two (undeformed) products. The deformation of the first pairing is described in section 4. The compatibilities with the bijections on PP or with the second product are also given, and the nondegeneracy is proved for q (q 1, 0, q 3, q 4 ) if q 1 0. The last section is devoted to the deformation of the second pairing and of the morphism to FQSym. 1 Backgrounds and notations 1.1 Double and plane posets Definition 1 1. [1] A double poset is a triple (P, 1, ), where P is a finite set and 1, are two partial orders on P.. A plane poset is a double poset (P, h, r ) such that for all x, y P, such that x y, x and y are comparable for h if, and only if, x and y are not comparable for r. The set of (isoclasses of) plane posets will be denoted by PP. For all n N, the set of (isoclasses of) plane posets of cardinality n will be denoted by PP(n). Examples. Here are the plane posets of cardinal 4. They are given by the Hasse graph of h ; if x and y are two vertices of this graph which are not comparable for h, then x r y if y is more on the right than x. PP(0) {1}, PP(1) { }, PP() {, }, PP(3) {,,,,, PP(4) },,,,,,,,,,,,,,,,,,,,,,,,. The following proposition is proved in [3] (proposition 11): Proposition Let P PP. We define a relation on P by: Then is a total order on P. (x y) if (x h y or x r y). As a consequence, substituing to r, plane posets are also special posets [1], that is to say double posets such that the second order is total. For any plane poset P PP(n), we shall assume that P {1,..., n} as a totally ordered set. The following theorem is proved in [4]: Theorem 3 Let σ be a permutation in the nth symmetric group S n. We define a plane poset P σ in the following way: P σ {1,..., n} as a set. 3

4 f i, j P σ, i h j if i j and σ(i) σ(j). f i, j P σ, i r j if i j and σ(i) σ(j). Note that the total order on {1,..., n} induced by this plane poset structure is the usual one. Then for all n 0, the following map is a bijection: { Sn PP(n) Ψ n : σ P σ. Examples. We here represent the permutation σ S n by the word (σ(1)... σ(n)). n particular, the identity element of S n is represented by the word (1... n). Ψ ((1)), Ψ ((1)), Ψ 3 (13)), Ψ 3 ((13)), Ψ 3 (13)), Ψ 3 ((31)), Ψ 3 (31)), Ψ 3 ((31)). We define several bijections on PP: Definition 4 Let P (P, h, r ) PP. We put: Remarks. 1. Graphically: ι(p ) (P, r, h ), α(p ) (P, h, r ), β(p ) (P, h, r ), γ(p ) (P, h, r ). A Hasse graph of α(p ) is obtained from a Hasse graph of P by a horizontal symmetry. A Hasse graph of β(p ) is obtained from a Hasse graph of P by a vertical symmetry. A Hasse graph of γ(p ) is obtained from a Hasse graph of P by a rotation of angle π.. These bijections generate a group of permutations of PP of cardinality 8. t is described by the following array: This is a dihedral group D 4. α β γ ι ι α ι β ι γ α d γ β ι β ι γ ι ι α β γ d α ι α ι ι γ ι β γ β α d ι γ ι β ι α ι ι ι α ι β ι γ d α β γ ι α ι ι γ ι β β γ d α ι β ι γ ι ι α α d γ β ι γ ι β ι α ι γ β α d 1. Algebraic structures on plane posets Two products are defined on PP. The first is called composition in [1] and denoted by in [3]. Definition 5 Let P, Q PP. 1. The double poset P Q is defined as follows: 4

5 P Q P Q as a set, and P, Q are plane subposets of P Q. if x P and y Q, then x r y in P Q.. The double poset P Q is defined as follows: P Q P Q as a set, and P, Q are plane subposets of P Q. if x P and y Q, then x h y in P Q. Examples. 1. The Hasse graph of P Q is the concatenation of the Hasse graphs of P and Q.. Here are examples for :,,,. The vector space generated by PP is denoted by H PP. These two products are linearly extended to H PP ; then (H PP, ) and (H PP, ) are two associative, unitary algebras, sharing the same unit 1, which is the empty plane poset. Moreover, they are both graded by the cardinality of plane posets. They are free algebras, as implied that the following theorem, proved in [3]: Theorem 6 1. (a) Let P be a nonempty plane poset. We shall say that P is h-irreducible if for all Q, PP, P Q R implies that Q 1 or R 1. (b) Any plane poset P can be uniquely written as P P 1... P k, where P 1,..., P k are h-irreducible. We shall say that P 1,..., P k are the h-irreducible components of P.. (a) Let P be a nonempty plane poset. We shall say that P is r-irreducible if for all Q, PP, P QR implies that Q 1 or R 1. (b) Any plane poset P can be uniquely written as P P 1... P k, where P 1,..., P k are r-irreducible. We shall say that P 1,..., P k are the r-irreducible components of P. Remark. The Hasse graphs of the h-irreducible components of H are the connected components of the Hasse graph of (P, h ), whereas the Hasse graphs of the r-irreducible components of H are the connected components of the Hasse graph of (P, r ). Definition 7 Let P (P, h, r ) be a plane poset, and let P. 1. We shall say that is a h-ideal of P, if, for all x, y P : (x, x h y) (y ).. We shall say that is a r-ideal of P, if, for all x, y P : (x, x r y) (y ). 3. We shall say that is a biideal of P if it both an h-ideal and a r-ideal. Equivalently, is a biideal of P if, for all x, y P : (x, x y) (y ). The following proposition is proved in [3] (proposition 9): 5

6 Proposition 8 H PP is given a coassociative, counitary coproduct in the following way: for any plane poset P, (P ) (P \ ). h-ideal of P Moreover, (H PP,, ) is a Hopf algebra, and (H PP,, ) is an infinitesimal Hopf algebra [10], both graded by the cardinality of the plane posets. n other words, using Sweedler s notations (x) x (1) x (), for all x, y H PP : (x y) x (1) y (1) x () y (), (xy) xy (1) y () + x (1) x () y x y. Remarks. The following compatibilities are satisfied: 1. For all P, Q PP: ι(p Q) ι(p )ι(q), ι(p Q) ι(p ) ι(q), α(p Q) α(p ) α(q), α(p Q) α(q)α(p ), β(p Q) β(q) β(p ), β(p Q) β(p )β(q), γ(p Q) γ(q) γ(p ), γ(p Q) γ(q)γ(p ).. Moreover, α (α α) op, β (β β), and γ (γ γ) op. 1.3 Pairings We also defined two pairings on H PP, using the notion of pictures [6, 8, 9, 14]: Definition 9 Let P, Q be two elements of PP. 1. We denote by S(P, Q) the set of bijections σ : P Q such that, for all i, j P : (i h j in P ) (σ(i) r σ(j) in Q). (σ(i) h σ(j) in Q) (i r j in P ).. We denote by S (P, Q) the set of bijections σ : P Q such that, for all i, j P : (i h j in P ) (σ(i) σ(j) in Q). (σ(i) h σ(j) in Q) (i j in P ). The following theorem is proved in [3, 4, 1]: Theorem 10 We define two pairings:, :, : { HPP H PP K P Q P, Q Card(S(P, Q)), { HPP H PP K P Q P, Q Card(S (P, Q)). They are both homogeneous, symmetric, nondegenerate Hopf pairings on the Hopf algebra H PP (H PP,, ). 6

7 1.4 Morphism to free quasi-symmetric functions We here briefly recall the construction of the Hopf algebra FQSym of free quasi-symmetric functions, also called the Malvenuto-Reutenauer Hopf algebra [, 11]. As a vector space, a basis of FQSym is given by the disjoint union of the symmetric groups S n, for all n 0. By convention, the unique element of S 0 is denoted by 1. The product of FQSym is given, for σ S k, τ S l, by: στ ɛ Sh(k,l) (σ τ) ɛ, where Sh(k, l) is the set of (k, l)-shuffles, that is to say permutations ɛ S k+l such that ɛ 1 (1) <... < ɛ 1 (k) and ɛ 1 (k +1) <... < ɛ 1 (k +l). n other words, the product of σ and τ is given by shifting the letters of the word representing τ by k, and then summing all the possible shufflings of this word and of the word representing σ. For example: (13)(1) (1354) + (1534) + (1534) + (5134) + (1543) +(1543) + (5143) + (1543) + (5143) + (5413). ( ) Let σ Σ n. For all 0 k n, there exists a unique triple σ (k) 1, σ(k), ζ k S k S n k ( ) Sh(k, n k) such that σ ζ 1 k σ (k) 1 σ (k). The coproduct of FQSym is then defined by: For example: (σ) n k0 σ (k) 1 σ (k). ((4315)) 1 (4315) + (1) (314) + (1) (13) +(31) (1) + (431) (1) + (4315) 1. Note that σ (k) 1 and σ (k) are obtained by cutting the word representing σ between the k-th and the (k + 1)-th letter, and then standardizing the two obtained words, that is to say applying to their letters the unique increasing bijection to {1,..., k} or {1,..., n k}. Moreover, FQSym has a nondegenerate, homogeneous, Hopf pairing defined by σ, τ δ σ,τ 1 for all permutations σ and τ. Definition 11 [5, 1] 1. Let P (P, h, r ) a plane poset. Let x 1 <... < x n be the elements of P, which is totally ordered. A linear extension of P is a permutation σ S n such that, for all i, j {1,..., n}: (x i h x j ) (σ 1 (i) < σ 1 (j)). The set of linear extensions of P will be denoted by S P.. The following map is an isomorphism of Hopf algebras: H PP FQSym Θ : P PP σ. σ S P moreover, for all x, y H PP, Θ(x), Θ(y) x, y. 7

8 Deformation of the products.1 Construction Definition 1 Let P PP and X, Y P. We put: 1. h Y X {(x, y) X Y / x h y in P }.. rx Y {(x, y) X Y / x r y in P }. Lemma 13 Let X and Y be disjoint parts of a plane poset P. Then: h Y X + h X Y + rx Y + ry X X Y. Proof. ndeed, h Y X + hx Y + ry X + rx Y {(x, y) X Y x < y or x > y} X Y. Remark. f it more generally possible to prove that for any parts X and Y of a plane poset: h Y X + h X Y + r Y X + r X Y 3 X Y + X Y. Theorem 14 Let q (q 1, q, q 3, q 4 ) K 4. We consider the following map: H PP H PP H PP m q : P Q q h R\ 1 q hr\ q r R\ 3 q rr\ 4 R, (R,) PP R, R\P, Q where P, Q PP. Then (H PP, m q ) is an associative algebra, and its unit is the empty plane poset 1. Proof. Let P, Q, R PP. We put: { P1 (m q d) m q (P Q R), P (d m q ) m q (P Q R). Then: P 1 P h 1 R1 +h \ q 1 R \ 1 q h R 1 \ 1 +h R \ 1 q r 1 +r R 1 \ 1 R \ 3 q r R 1 \ 1 +r R \ 1 (R 1, 1,R, ) E 1 h 1 R1 +h \ q 1 R \ 1 q h R 1 \ 1 +h R \ 1 q r 1 +r R 1 \ 1 R \ 3 q r R 1 \ 1 +r R \ 1 (R 1, 1,R, ) E 4 R, 4 R, With: E 1 {(R 1, 1, R, ) PP 4 / 1 R 1, 1 Q, R 1 \ 1 P, R, R, R \ R 1 }, E {(R 1, 1, R, ) PP 4 / 1 R 1, 1 R, R 1 \ 1 Q, R, R 1, R \ P }. We shall also consider: E {(R, J 1, J, J 3 ) PP 4 / R J 1 J J 3, J 1 P, J Q, J 3 R}. First step. Let us consider the following maps: { E φ : 1 E (R 1, 1, R, ) (R, R 1 \ 1, 1, ), { φ E E : 1 (R, J 1, J, J 3 ) (J 1 J, J, R, J 3 ). 8

9 By definition of E and E 1, these maps are well-defined, and an easy computation shows that φ φ d E and φ φ d E1, so φ is a bijection. Let (R 1, 1, R, ) E 1. We put φ(r 1, 1, R, ) (R, J 1, J, J 3 ). Then: Similar computations finally give: P 1 h 1 R 1 \ 1 + h R \ h J J 1 + h J 3 J 1 J h J J 1 + h J 3 J 1 + h J 3 J. (R,J 1,J,J 3 ) E q hj J +h J 3 1 J +h J 3 1 J 1 q h J 1 J +h J 1 J +h J 3 J 3 q r J J +r J 3 1 J +r J 3 1 J 3 q r J 1 J +r J 1 J 3 +r J J 3 4 R. Second step. Let us consider the following maps: { E ψ : E (R 1, 1, R, ) (R, R 1 \ R 1, R 1 \ 1, 1 ), { ψ E E : (R, J 1, J, J 3 ) (J J 3, J 3, R, J J 3 ). By definition of E and E, these maps are well-defined, and a simple computation shows that ψ ψ d E and ψ ψ d E, so ψ is a bijection. Let (R 1, 1, R, ) E. We put ψ(r 1, 1, R, ) (R, J 1, J, J 3 ). Then: Similar computations finally give: P So m q is associative. h 1 R 1 \ 1 + h R \ h J 3 J + h J J 3 J 1 h J J 1 + h J 3 J 1 + h J 3 J. (R,J 1,J,J 3 ) E q hj J +h J 3 1 J +h J 3 1 J 1 q h J 1 J +h J 1 J +h J 3 J 3 q r J J +r J 3 1 J +r J 3 1 J 3 q r J 1 J +r J 1 J 3 +r J J 3 4 R. Last step. Let P PP. Then: P.1 (R,) PP R, R\P, 1 Similarly, for all Q PP, 1.Q Q. q h R 1 q hr\ q r R\ 3 q rr\ 4 R q h1 P 1 qhp 1 qr1 P 3 q rp 1 4 P P. Examples. m q ( ) q 3 q 4 + q 1 q, m q ( ) q3 + q4 + q (q 3 + q 4 ) + q 1 (q 3 + q 4 ) + (q1 + q 1 q + q), m q ( ) q1 + q + (q 1 + q )q 4 + (q 1 + q )q 3 + (q3 + q 3 q 4 + q4), m q ( ) q3 + q4 + q 1 (q 3 + q 4 ) + q (q 3 + q 4 ) + (q1 + q 1 q + q), m q ( ) q + q1 + (q 1 + q )q 4 + (q 1 + q )q 3 + (q3 + q 3 q 4 + q4). Proposition 15 Let (q 1, q, q 3, q 4 ) K 4. Then: m op (q 1,q,q 3,q 4 ) m (q,q 1,q 4,q 3 ), Proof. mmediate. m (q1,q,q 3,q 4 ) (ι ι) ι m (q3,q 4,q 1,q ), m (q1,q,q 3,q 4 ) (α α) α m (q,q 1,q 3,q 4 ), m (q1,q,q 3,q 4 ) (β β) β m (q1,q,q 4,q 3 ), m (q1,q,q 3,q 4 ) (γ γ) γ m (q,q 1,q 4,q 3 ). 9

10 . Particular cases Lemma 16 Let P PP and X P. 1. X is a h-ideal of P if, and only if, h P \X X 0.. X is a r-ideal of P if, and only if, r P \X X The h-irreducible components of P are the h-irreducible components of X and P \ X if, and only if, h X \X P \X hp X The r-irreducible components of P are the r-irreducible components of X and P \ X if, and only if, rp X \X \X rp X P X (P \ X) if, and only if, h X P \X hp \X X r X P \X P X(P \ X) if, and only if, r X P \X rp \X X h X P \X 0. Proof. We give the proofs of points 1, 3 and 5. The others are similar. 1.. Let x X, y P, such that x h y. As h P \X X 0, y / P \ X, so y X: X is a h-ideal.. Then, for all x X, y P \ X, x h y is not possible. So h P \X X Let us put P P 1... P k, where P 1,..., P k are the h-irreducible components of P. By hypothesis, X is the disjoint union of certain P i s, and P \X is the disjoint union of the other P i s. So h X \X P \X hp X 0.. Let be a h-irreducible component of P. f X and (P \X), then for any x X and any y (P \ X), x and y are not comparable for h : contradiction. So is included in or in P \ X, so is a h-irreducible component of X or P \ X. 5.. f x X and y P \ X, then x < r y, so h X \X P \X hp X rp X \X 0.. f x X and y P \ X, by hypothesis we do not have x < h y, x > h y nor x > r y, so x < r y. Hence, P X (P \ X). Proposition Let us assume that q 3 q 4 0. Let P P 1... P k and P P k+1... P k+l be two plane posets, decomposed into their r-connected components. Then: m q (P P ) Q i,j (σ) P i P j P σ(1)... P σ(k+l), σ Sh(k,l) 1 i k<j k+l where Q i,j (σ) q 1 if σ 1 (i) < σ 1 (j) and q if σ 1 (i) > σ 1 (j).. Let us assume that q 1 q 0. Let P P 1... P k and P P k+1... P k+l be two plane posets, decomposed into their r-connected components. Then: m q (P P ) Q i,j(σ) P i P j P σ(1)... P σ(k+l), σ Sh(k,l) 1 i k<j k+l where Q i,j (σ) q 3 if σ 1 (i) < σ 1 (j) and q 4 if σ 1 (i) > σ 1 (j). Proof. Let us prove the first point; the proof of the second point is similar. Let us consider a plane poset R such that the coefficient of R in m q (P P ) is not zero. So there exists R such that R \ P and P. Moreover, as q 3 q 4 0, rr\ rr\ 0. By Lemma 16-4, the r-connected components of R are the r-components of R \ and. As a consequence, 10

11 there exists a (k, l)-shuffle σ, such that R P σ(1)... P σ(k+l). Then h R\ is the sum of P i P j, where 1 i k < j k + l, such that σ 1 (i) < σ 1 (j); h R\ is the sum of P i P j, where 1 i k < j k + l, such that σ 1 (i) > σ 1 (j). This immediately implies the announced result. Remarks. 1. The first point implies that: m (q1,0,0,0)(p P ) q P P 1 P P. n particular, m (1,0,0,0). m (0,q,0,0)(P P ) q P P P P. n particular, m (0,1,0,0) op. m (0,0,q3,0)(P P ) q P P 3 P P. n particular, m (1,0,0,0). m (0,0,0,q4 )(P P ) q P P 4 P P. n particular, m (1,0,0,0) op.. t is possible to define m q on the space of double posets. The same arguments prove that it is still associative. However, m (1,0,0,0) is not equal to on H PP and m (0,0,1,0) is not equal to ; for example, if we denote by the double poset with two elements x, y, x and y being not comparable for h and r :.3 Subalgebras and quotients m (1,0,0,0) ( ) +, m (0,0,1,0) ( ) +. These two particular families of plane posets are used in [3, 4]: Definition 18 Let P PP. 1. We shall say that P is a plane forest if it does not contain set of plane forests is denoted by PF. as a plane subposet. The. We shall say that P is WN ("without N") if it does not contain nor. The set of WN posets is denoted by WN P. Examples. A plane poset is a plane forest if, and only if, its Hasse graph is a rooted forest. PF(0) {1}, PF(1) { }, PF() {, }, PF(3),,,, }, { PF(4),,,,,,,,,,,, },, WN P(0) {1}, WN P(1) { }, WN P() {, }, WN P(3) {,,,,, WN P(4) Definition 19 We denote by: },,,,,,,,,,,,,,,,,,,,,,. 11

12 H WN P the subspace of H PP generated by WN plane posets. H PF the subspace of H PP generated by plane forests. WN P the subspace of H PP generated by plane posets which are not WN. PF the subspace of H PP generated by plane posets which are not plane forests. Note that H WN P and H PF are naturally identified with H PP / WN P and H PP / PF. Proposition 0 Let q (q 1, q, q 3, q 4 ) K H WN P is a subalgebra of (H PP, m q ) if and only if, q 1 q 0 or q 3 q H PF is a subalgebra of (H PP, m q ) if and only if, q 1 q WN P and PF are ideals of (H PP, m q ). Proof. 1.. We use the notations of Proposition f q 3 q 4 0, let us consider two WN posets P and P. then the P i s are also WN, so for any σ S k+l, P σ 1 (1)... P σ 1 (k+l) is WN. As a conclusion, m q (P P ) H WN P. The proof is similar if q 1 q 0, using Proposition Let us consider the coefficients of and in certain products. We obtain: m q ( ) q 1 q3 q 1 q4 m q ( ) q q4 q q3. f H WN P is a subalgebra of (H PP, m q ), then these four coefficients are zero, so, from the first row, q 1 0 or q 3 q 4 0 and from the second row, q 0 or q 3 q 4 0. As a conclusion, q 1 q 0 or q 3 q We use the notations of Proposition 17-. f q 1 q 0, let us consider two plane forests P and P. Then the P i s are plane trees, so for any σ S k+l, P σ 1 (1)... P σ 1 (k+l) is a plane forest. As a conclusion, m q (P P ) H PF... Let us consider the coefficients of in certain products. We obtain: m q ( ) q m q ( ) q1. f H PF is a subalgebra of (H PP, m q ), then q 1 q Let P and P be two plane posets such that P or P is not WN. Let us consider a plane poset R such that the coefficient of R in m q (P P ) is not zero. There exists R, such that R \ P and P. As P or P is not WN, or R \ contains or, so R contains or : R is not WN. So m q (P P ) WN P. The proof is similar for PF, using instead of and. 1

13 3 Dual coproducts 3.1 Constructions Dually, we give H PP a family of coproducts q, for q K 4, defined for all P PP by: q (P ) P q h P \ 1 q hp \ q r P \ 3 q rp \ 4 (P \ ). These coproducts are coassociative; their common counit is given by: { H ε : PP K P PP δ 1,P. Examples. We put, for all P PP, nonempty, q (P ) (P ) P 1 1 P. q ( ) (q 1 + q ), q ( ) (q 3 + q 4 ), ( ) (q1 + q 1 q + q) + (q1 + q 1 q + q), ( ) q (q 3 + q 4 ) + q 1 (q 3 + q 4 ) + q1 + q, ( ) q 1 (q 3 + q 4 ) + q (q 3 + q 4 ) + q + q 1, ( ) q4 + q3 + (q 1 + q )q 3 + (q 1 + q )q 4, ( ) q 3 + q 4 + (q 1 + q )q 4 + (q 1 + q )q 3, ( ) (q 3 + q 3 q 4 + q 4) + (q 3 + q 3 q 4 + q 4). Dualizing Proposition 15: Proposition 1 Let (q 1, q, q 3, q 4 ) K 4. Then: op (q 1,q,q 3,q 4 ) (q,q 1,q 4,q 3 ), 3. Particular cases Proposition Let P PP. Then: 1. (q,q,q,q) (P ) P (ι ι) (q1,q,q 3,q 4 ) (q3,q 4,q 1,q ) ι, (α α) (q1,q,q 3,q 4 ) (q,q 1,q 3,q 4 ) α, (β β) (q1,q,q 3,q 4 ) (q1,q,q 4,q 3 ) β, (γ γ) (q1,q,q 3,q 4 ) (q,q 1,q 4,q 3 ) γ. q P \ (P \ ).. 3. (0,q,q 3,q 4 )(P ) (q1,0,q 3,q 4 )(P ) (q1,q,0,q 4 )(P ) (q1,q,q 3,0)(P ) q h P \ q rp \ 3 q r P \ h-ideal of P q h P \ 1 q rp \ 4 q r P \ h-ideal of P r-ideal of P r-ideal of P 4 (P \ ), 3 (P \ ). \ hp q1 q h P \ q r P \ 4 (P \ ), \ hp q q h P \ 1 q r P \ 3 (P \ ). 13

14 4. (0,q,0,q 4 )(P ) (q1,0,q 3,0)(P ) biideal of P biideal of P q h P \ q r P \ 4 (P \ ), q h P \ 1 q r P \ 3 (P \ ). 5. f P P 1 P k, where the P i s are h-connected, (0,0,q3,q 4 )(P ) {1,,k} q α P () 3 q α P ({1,...,k}\) 4 P P {1,,k}\, with, for all J {j 1,, j l }, 1 j 1 < < j l k, P J P j1 P jl, and α P (J) P i P j. i J,j / J,i<j 6. f P P 1 P k, where the P i s are r-connected, (q1,q,0,0)(p ) {1,,k} q β P () 1 q β P ({1,...,k}\) P P {1,,k}\, 7. with, for all J {j 1,, j l }, 1 j 1 < < j l k, P J P j1 P jl, and β P (J) P i P j. i J,j / J,i<j (q,0,0,0) (P ) (0,q,0,0) (P ) (0,0,q,0) (P ) (0,0,0,q) (P ) q P 1 P P 1 P, P 1 P P P 1 P P P 1 P P P 1 P P q P 1 P P P 1, q P 1 P P 1 P, q P 1 P P P (0,0,0,0) (P ) P P if P 1. Proof. We only prove points 1,, 5, 7 and 8; the others are proved in the same way. 1. mmediate, as h \ P\ + hp + rp\ \ + rp P \ by Lemma 13.. By Lemma 16: (q1,0,q 3,q 4 )(P ) q h P \ 1 q r P \ P h P \ 0 3 q rp \ 4 (R \ ) P \ h-ideal of P As (0,q,q 3,q 4 ) op (0,q,q 4,q 3 ), we obtain also the first assertion. 5. By Lemma 16-3: (0,0,q3,q 4 )(P ) q r P \ 3 q rp \ P h \ hp P \ 0 4 (R \ ) {1,,k} q h P \ 1 q r P \ q rp P \P 3 q r P \P P and it is immediate that r P P \P α P () and r P \P P α P ({1,..., k} ). 3 q rp \ 4 (P \ ). 4 P P {1,,k}\, 14

15 7. By Lemma 16: (0,0,q,0) (P ) P h \ hp P \ r P \ 0 q r P \ (P \ ) P 1 P P q rp P 1 P 1 P P 1 P P q P 1 P P 1 P. 8. Let P, such that h P \ \ hp rp \ \ rp 0. Then: so 1 or P. P \ h P \ \ + hp + rp \ \ + rp 0, Remark. n particular, the coproduct defined in section 1 is (1,0,1,1). The coproduct of deconcatenation, dual of m, is (0,0,1,0) and the coproduct of deconcatenation, dual of, is (1,0,0,0). 3.3 Compatibilities with the products Proposition 3 Let x, y H PP. We put q (x) x q x q and q (y) y q y q, with the x q s, x q s, y q s, y q s homogeneous. Then: 1. q (x y) q x q y q. q (xy) q x q y q 3 q x q y q 1 q x q y q 4 (x q y q) (x q y q ). (x qy q) (x qy q ). Proof. We only prove the first point; the proof of the second point is similar. For the sake of simplicity, we simply denote in this proof the composition of two plane posets P et Q by P Q, as in [1]. Let P, Q PP. Then: q (P Q) P, J Q P, J Q Moreover, for all P, J Q: q hj (P Q)\(J) 1 q h(p Q)\(J) J q rj (P Q)\(J) 3 q r(p Q)\(J) J q hj (P Q)\(J) 1 q hp Q\J J q rj P Q\J 3 q rp Q\J J 4 ((P Q) \ (J)) (J) 4 ((P \ )(Q \ J)) (J). For all x P \, y J, x < r y in P Q, so h J P \ 0. Similarly, h Q\J 0. Hence: h J P Q\J h P \ + hj Q\J + hj P \ + h Q\J h P \ + hj Q\J. n the same way, h P Q\J J h P \ + h Q\J J + h P \ J + h Q\J h P \ + h Q\J J. For all x P \, y J, x < r y in P Q, so rp J \ P \ J. For all x Q \ J, y, x > r y, so rq\j 0. Hence: rp J Q\J r P \ + rj Q\J + rj P \ + r Q\J h P \ + hj Q\J + P \ J. So: n the same way, r P Q\J J r P \ + r Q\J J + r P \ J + r Q\J r P \ + r Q\J J + Q \ J. which is the announced formula. q (P Q) q P q Q q P q Q 3 q q 4 (P qq q) (P q Q q), Examples. 15

16 1. f q 3 1, then (H PP,, q ) is a braided Hopf algebra. The braiding is given, for all P, Q PP, by: P Q c q4 (P Q) q 4 Q P. n particular, if q 4 1, it is a Hopf algebra; if q 4 0, it is an infinitesimal Hopf algebra.. f q 4 0, the compatibility becomes the following: q (x y) q x y q 3 (x y q) y q + q x q y 3 x q (x q y) q x y 3 x y. n particular, if q 3 1, then it is an infinitesimal Hopf algebra. 3. f q q 4 0, then for all x, y H PP : q (x y) (x y) (x y) + ε(x)( q (y) y 1 1 y) +ε(y)( q (x) x 1 1 x) + ε(x)ε(y)1 1. n other terms, for all x, y H PP, such that ε(x) ε(y) 0, x y is primitive. 4 Self-duality results 4.1 A first pairing on H q Notations. f P, Q PP, we denote by Bij(P, Q) the set of bijections from P to Q. Definition 4 Let P, Q be two double posets and let σ Bij(P, Q). We put: φ 1 (σ) {(x, y) P x < h y and σ(x) < h σ(y)} + {(x, y) P x < h y and σ(x) > h σ(y)} + {(x, y) P x < h y and σ(x) < r σ(y)} + {(x, y) P x < r y and σ(x) < h σ(y)}, φ (σ) {(x, y) P x < h y and σ(x) < h σ(y)} + {(x, y) P x < h y and σ(x) > h σ(y)} + {(x, y) P x < h y and σ(x) > r σ(y)} + {(x, y) P x < r y and σ(x) > h σ(y)}, φ 3 (σ) {(x, y) P x < r y and σ(x) < r σ(y)}, φ 4 (σ) {(x, y) P x < r y and σ(x) > r σ(y)}. Lemma 5 Let P, Q be two double posets and let σ Bij(P, Q). For all i {1,..., 4}, φ i (σ 1 ) φ i (σ). Proof. ndeed, as σ is a bijection: φ 1 (σ 1 ) {(x, y) Q x < h y and σ 1 (x) < h σ 1 (y)} + {(x, y) P x < h y and σ 1 (x) > h σ 1 (y)} + {(x, y) P x < h y and σ 1 (x) < r σ 1 (y)} + {(x, y) P x < r y and σ 1 (x) < h σ 1 (y)} {(x, y) P x < h y and σ(x) < h σ(y)} + {(x, y) P x > h y and σ(x) < h σ(y)} + {(x, y) P x < r y and σ(x) < h σ(y)} + {(x, y) P x < h y and σ(x) < r σ(y)} φ 1 (σ). The other equalities are proved similarly. 16

17 Lemma 6 Let P 1, P, Q be double posets. There is a bijection: Bij(P 1 P, Q) Bij(P 1, R \ ) R(P, ) Q σ (σ P1, σ P ), with σ(p ). Let σ Bij(P 1 P, Q) and let (σ 1, σ ) be its image by this bijection. Then: φ 1 (σ) φ 1 (σ 1 ) + φ 1 (σ ) + h Q\, φ (σ) φ (σ 1 ) + φ (σ ) + h Q\, φ 3 (σ) φ 3 (σ 1 ) + φ 3 (σ ) + r Q\, φ 4 (σ) φ 4 (σ 1 ) + φ 4 (σ ) + r Q\. Proof. We put P P 1 P. if x P 1 and y P, then x < r y, so: φ 3 (σ) {(x, y) P 1 x < r y and σ(x) < r σ(y)} + {(x, y) P x < r y and σ(x) < r σ(y)} + {(x, y) P 1 P x < r y and σ(x) < r σ(y)} + {(x, y) P P 1 x < r y and σ(x) < r σ(y)} φ 3 (σ) + φ 3 (σ ) + {(x, y) P 1 P σ(x) < r σ(y)} + 0 φ 3 (σ) + φ 3 (σ ) + {(x, y) (Q \ ) x < r y} φ 3 (σ 1 ) + φ 3 (σ ) + r Q\. The other equalities are proved similarly. Theorem 7 Let q (q 1, q, q 3, q 4 ) K 4. We define a pairing on H PP by: P, Q q q φ 1(σ) 1 q φ (σ) q φ 3(σ) 3 q φ 4(σ) 3. σ Bij(P,Q) This pairing is symmetric and for all x, y, z H PP, x y, z q x y, q (z) q. Proof. Let P, Q be two double posets. Then, by Lemma 5: P, Q q q φ 1(σ) 1 q φ (σ) q φ 3(σ) 3 q φ 4(σ) 3 σ Bij(P,Q) σ Bij(Q,P ) σ Bij(Q,P ) Q, P q. q φ 1(σ 1 ) 1 q φ (σ 1 ) q φ 3(σ 1 ) 3 q φ 4(σ 1 ) 3 q φ 1(σ) 1 q +φ (σ) q φ 3(σ) 3 q φ 4(σ) 3 So this pairing is symmetric. Let P 1, P, Q be three double posets. By Lemma 6: P 1 P, Q q q φ 1(σ) 1 q φ (σ) q φ 3(σ) 3 q φ 4(σ) 3 Q σ Bij(P 1 P,Q) q φ 1(σ 1 )+φ 1 (σ )+h Q\ 1 σ 1 Bij(P 1,Q\) σ Bij(P,) q φ (σ 1 )+φ (σ )+h Q\ q φ 3(σ 1 )+φ 3 (σ )+rq\ 1 q hr\ 3 q rr\ 3 q φ 4(σ 1 )+φ 4 (σ )+r Q\ 4 q h R\ q r R\ 4 P 1, Q \ q P, q Q P 1 P, q (Q) q. 17

18 So, q is a Hopf pairing. Remark. More generally, adapting this proof, it is possible to show that for all x, y, z H PP : m (0,0,q,0) (x y), (z) (q1,q,q 3,q 4 ) x y, (qq1,qq,qq 3,qq 4 )(z) (q1,q,q 3,q 4 ). Examples., q 1. The pairing of plane posets of degree is given by the following array: q 1 q q 1 + q q 1 + q q 3 + q 4 Proposition 8 For all double posets P, Q: P, Q (q1,0,q 3,q 4 ) σ S(P,Q) q φ 1(σ) 1 q φ 3(σ) 3 q φ 4(σ) 4. Consequently,, (1,0,1,1) is the pairing, described in section 1. Proof. Let σ Bij(P, Q), such that the contribution of σ in the sum defining P, Q (q1,0,q 3,q 4 ) is non zero. As q 0, necessarily φ (σ) 0. Hence, if x < h y in P, then σ(x) < r σ(y) in Q; if σ(x) < h σ(y) in Q, then x < r y in P. So σ S(P, Q). 4. Properties of the pairing Lemma 9 Let P 1, P, Q be double posets. There is a bijection: Bij(P 1 P, Q) Bij(P 1, Q \ ) Bij(P, ) Q σ (σ P1, σ P ), with σ(p ). Let σ Bij(P 1 P, Q) and let (σ 1, σ ) be its image by this bijection. Then: φ 1 (σ) φ 1 (σ 1 ) + φ 1 (σ ) + h Q\ + hq\ + rq\, φ (σ) φ (σ 1 ) + φ (σ ) + h Q\ + hq\ ψ (σ) ψ (σ 1 ) + ψ (σ ) + h Q\ + hq\, φ 3 (σ) φ 3 (σ 1 ) + φ 3 (σ ), φ 4 (σ) φ 4 (σ 1 ) + φ 4 (σ ). + r Q\, Proof. We put P P 1 P. Let x, y P. f x < h y, then x, y P 1, or x, y P, or x P 1, y P. f x < r y, then x, y P 1, or x, y P. Moreover, if x P 1 and x P, then x < h y. So: φ 1 (σ) φ 1 (σ 1 ) + φ 1 (σ ) + {(x, y) P 1 P (x < h y) and σ(x) < h σ(y))} + {(x, y) P 1 P (x < h y) and σ(x) > h σ(y))} + {(x, y) P 1 P (x < h y) and σ(x) < r σ(y))} φ 1 (σ 1 ) + φ 1 (σ ) + {(x, y) P 1 P σ(x) < h σ(y))} + {(x, y) P 1 P σ(x) > h σ(y))} + {(x, y) P 1 P σ(x) < r σ(y))} The other equalities are proved similarly. φ 1 (σ 1 ) + φ 1 (σ ) + h Q\ + hq\ + r Q\. 18

19 Proposition 30 For all x, y, z H PP, xy, z q x y, (q1 q,q 1 q,q 1,q )(z) q. Proof. Let P 1, P, Q be three double posets. By Lemma 9: P 1 P, Q q Q σ Bij(P 1 P,Q) q φ 1(σ) 1 q φ (σ) q φ 3(σ) 3 q φ 4(σ) 3 q φ 1(σ 1 )+φ 1 (σ )+h Q\ +hq\ +rq\ 1 σ 1 Bij(P 1,Q\) σ Bij(P,) q φ (σ 1 )+φ (σ )+h Q\ +hq\ +r Q\ q φ 3(σ 1 )+φ 3 (σ ) 3 q φ 4(σ 1 )+φ 4 (σ ) 4 (q 1 q ) h R\ (q 1 q ) hr\ q r R\ P 1, Q \ q P, q Q 1 q rr\ P 1 P, (q1 q,q 1 q,q 1,q )(Q) q, which is the announced formula. Remarks. 1. n particular, if q (1, 0, 1, 1), for all P 1, P, Q PP: P 1 P, Q P 1 P, (0,0,1,0) (Q) P 1 P, Q 1 Q. QQ 1 Q This formula was already proved in [3].. More generally, it is possible to show that for all x, y, z H PP : m (q,0,0,0) (x y), (z) (q1,q,q 3,q 4 ) x y, (qq1 q,qq 1 q,qq 1,qq )(z) (q1,q,q 3,q 4 ). Proposition 31 For all x, y HPP: x, β(y) (q1,q,q 3,q 4 ) x, y (q1,q,q 4,q 3 ), x, γ(y) (q1,q,q 3,q 4 ) x, y (q,q 1,q 4,q 3 ), α(x), α(y) (q1,q,q 3,q 4 ) x, y (q,q 1,q 3,q 4 ), α(x), β(y) (q1,q,q 3,q 4 ) x, y (q1,q,q 4,q 3 ), α(x), γ(y) (q1,q,q 3,q 4 ) x, y (q,q 1,q 4,q 3 ), β(x), β(y) (q1,q,q 3,q 4 ) x, y (q,q 1,q 3,q 4 ), γ(x), γ(y) (q1,q,q 3,q 4 ) x, y (q1,q,q 3,q 4 ). Proof. Let P, Q be two double posets. We put P α(p ) and Q α(q). Note that Bij(P, Q) Bij(P, Q ). Let σ Bij(P, Q). We denote it by σ is we consider it as an element of Bij(P, Q ). By definition of P and Q, it is clear that φ 1 (σ ) φ (σ), φ (σ ) φ (σ), φ 3 (σ ) φ 3 (σ), and φ 4 (σ ) φ 4 (σ). Hence: P, Q q1,q,q 3,q 4 ) σ Bij(P,Q ) σ Bij(P,Q) P, Q (q,q 1,q 3,q 4 ). q φ 1(σ ) 1 q φ (σ ) q φ 3(σ ) 3 q φ 4(σ ) 4 q φ (σ) 1 q φ 1(σ) q φ 3(σ) 3 q φ 4(σ) 4 The other assertions are proved in the same way. Remark. n general, the pairing defined by x y x, α(y) (q1,q,q 3,q 4 ) is not a pairing, q. However, when q 1 q, it is possible to prove that for all x, y H PP, x, α(y) (q1,q 1,q 3,q 4 ) x, y (q1,q 1,q 3,q 4 ). Similarly, β(x), γ(y) (q1,q 1,q 3,q 4 ) x, y (q1,q 1,q 4,q 3 ). 19

20 4.3 Comparison of pairings with colinear parameters Proposition 3 Let q K. We define the following map: { H υ q : PP H PP P PP q h(p ) P, where ( h(p ) {(x, y) ) P ( x < h y}. Then υ q ) is an algebra and coalgebra morphism from HPP,, (qq1,qq,q 3,q 4 ) to HPP,, (q1,q,q 3,q 4 ). Moreover, for all x, y HPP : υ q (x), υ q (y) (q1,q,q 3,q 4 ) x, y (qq1,qq,q 3,q 4 ). Proof. Let P 1, P PP. Then h(p 1 P ) h(p 1 ) + h(p ), so υ q (P 1 P ) υ q (P 1 ) υ q (P ), and υ q is an algebra morphism. Let P PP. For any P, as P (P \ ) ( (P \ )) ((P \ ) ): Consequently: h(p ) h() + h(p \ ) + h P \ + hp \. (q1,q,q 3,q 4 ) υ q (P ) (qq 1 ) h \ P \ hp (qq ) q r P \ 4 q h(p \) (P \ ) q h() P (υ q υ q ) (qq1,qq,q 3,q 4 )(P ). 3 q rp \ So υ q is a coalgebra morphism. Let P, Q PP and let σ Bij(P, Q). We define: a 1 {(x, y) P x < h y, σ(x) < h σ(y)} a {(x, y) P x < h y, σ(x) > h σ(y)} a 3 {(x, y) P x < h y, σ(x) < r σ(y)} a 4 {(x, y) P x < h y, σ(x) > r σ(y)} a 5 {(x, y) P x < r y, σ(x) < h σ(y)} a 6 {(x, y) P x < r y, σ(x) > h σ(y)} a 7 {(x, y) P x < r y, σ(x) < r σ(y)} a 8 {(x, y) P x < r y, σ(x) > r σ(y)}. n order to sum up the notations, we give a following array: {(x, y) P... x < h y,... x > h y,... x < r y,... x > r y,... σ(x) < h σ(y)} a 1 a a 5 a 6 σ(x) > h σ(y)} a a 1 a 6 a 5 σ(x) < r σ(y)} a 3 a 4 a 7 a 8 σ(x) > r σ(y)} a 4 a 3 a 8 a 7 So φ 1 (σ) a 1 + a + a 3 + a 5, φ (σ) a 1 + a + a 4 + a 6, φ 3 (σ) a 7 and φ 4 (σ) a 8 ; h(p ) a 1 + a + a 3 + a 4 and h(q) a 1 + a + a 5 + a 6. Then: q h(p ) q (h(q) q φ 1(σ) 1 q φ (σ) q φ 3(σ) 3 q φ 4(σ) 4 (q q 1 q ) a 1 (q q 1 q ) a (qq 1 ) a 3 (qq ) a 4 (qq 1 ) a 5 (qq ) a 6 q a 7 3 qa 8 4 (qq 1 ) φ1(σ) (qq ) φ(σ) q φ 3(σ) 3 q φ 4(σ) 4. Finally: υ q (P ), υ q (Q) (q1,q,q 3,q 4 ) σ Bij(P,Q) σ Bij(P,Q) P, Q (qq1,qq,q 3,q 4 ). 0 q h(p ) q (h(q) q φ 1(σ) 1 q φ (σ) q φ 3(σ) 3 q φ 4(σ) 4 (qq 1 ) φ 1(σ) (qq ) φ (σ) q φ 3(σ) 3 q φ 4(σ) 4

21 So υ q is an isometry. Remark. t is easy to prove that for P, Q PP, υ q (P Q) υ q (P )υ q (Q). Similarly, one can prove: Proposition 33 Let q K. We define the following map: { υ q H : PP H PP P PP q r(p ) P, where ( r(p ) {(x, y) ) P ( x < r y}. Then υ ) q is an algebra and coalgebra morphism from HPP,, (q1,q,qq 3,qq 4 ) to HPP,, (q1,q,q 3,q 4 ). Moreover, for all x, y HPP : 4.4 Non-degeneracy of the pairing υ q(x), υ q(y) (q1,q,q 3,q 4 ) x, y (q1,q,qq 3,qq 4 ). Lemma 34 For all P PP(n), S(P, ι(p )) is reduced to a single element and: P, ι(p ) (q1,0,q 3,q 4 ) q n(n 1) 1. Proof. Clearly, d P : P P belongs to S(P, ι(p )). Let σ : P P be a bijection. Then σ S(P, ι(p )) if, and only if, for all i, j P : (i h j in P ) (σ(i) h σ(j) in P ). (σ(i) r σ(j)) in P (i r j in P ). t is clear S(P, ι(p )) is a submonoid of S P. As the group S P is finite, S(P, ι(p )) is a subgroup of S P. So if σ S(P, ι(p )), then σ 1 S(P, ι(p )). Hence, if σ S(P, ι(p )): (i h j in P ) (σ(i) h σ(j) in P ). (i r j in P ) (σ(i) r σ(j) in P ). So σ is the unique increasing bijection from P to P, that is to say d P. Moreover: φ 1 (d P ) {(x, y) P x < h y and x < r y} + {(x, y) P x < h y and x > r y} + {(x, y) P x < h y and x < h y} + {(x, y) P x < r y and x < r y} {(x, y) P x < h y} + {(x, y) P x < r y} {(x, y) P x < y} n(n 1), φ 3 (d P ) {(x, y) P (x < r y) and (x < h y)} 0, φ 4 (d P ) {(x, y) P (x > r y) and (x > h y)} 0. So P, ι(p ) (q1,0,q,q 3 ) q n(n 1) 1. For all n N, we totally order S n lexicographically. For example, in S 3 : (13) (13) (13) (31) (31) (31). 1

22 We then define a total order on PP(n) by P Q if, and only if, Ψ n (P ) Ψ n (Q) in S n. For example, if n 3:. For all P PP(n), we put: Lemma 35 For all P PP, m(p ) ι(p ). m(p ) min{q PP(n) / S(P, Q) }. Proof. By Lemma 34, m(p ) ι(p ). Let Q PP, such that S(P, Q). Let us prove that ι(p ) Q. We denote σ Ψ n (P ) and τ Ψ n (Q); we can suppose that P Φ n (σ) and Q Φ n (τ). Moreover, it is not difficult to prove that Ψ n (ι(p )) (n 1) σ. First step. Let us prove that τ(1) n σ(1) + 1. n P Ψ n (σ), 1 h j if, and only if, σ(1) σ(j). So there are exactly n σ(1) + 1 elements of P which satisfy 1 h j in P. Let α S(P, Q) (which is non-empty by hypothesis). Then if 1 h j, α(1) r α(j): there are at least n σ(1)+1 elements of Q which satisfy α(1) r j. Let us put α 1 (1) i. Then α(i) 1 α(1) in Q, that is to say 1 h α(1) or 1 r α(1) in Q. f 1 α(i) h α(1) in Q, then i r 1 in P, so i 1: in both cases, 1 r α(1) in P. So there are at least n σ(1) + 1 elements of Q which satisfy 1 r j. n Q, 1 r j if, and only if, τ(j) τ(1), so there are exactly τ(1) such elements. As a consequence, τ(1) n σ(1) + 1. Moreover, if there is equality, necessarily 1 α(1) for all α S(P, Q). Second step. We consider the assertion H i : if τ(1) n σ(1)+1,, τ(i 1) n σ(i 1)+1, then τ(i) n σ(i) + 1 and, if there is equality, then α(1) 1,, α(i) i for all α S(P, Q). We proved H 0 in the first step. Let us prove H i by induction on i. Let us assume H i 1, 1 i n, and let us prove H i. Let α S(P, Q) (non empty by hypothesis). By H i 1, as the equality is satisfied, α(1) 1,, α(i 1) i 1. n P, the number N i of elements j such that i h j is the cardinality of σ 1 ({σ(i),, n}) {i,, n}, so N i n σ(i) + 1 {k i / σ(k) > σ(i)}. Using α, there exists at least N i elements j of Q such that α(i) r j in Q. Let us put α 1 (i) j. As α(1) 1,, α(i 1) i 1, j i and i α(i). f i h α(i) in Q, then j r i in P, so j i: we always have i r α(i) in Q, and at least N i elements k of Q satisfy i r j in Q. Hence: N i τ 1 ({1,, τ(i)}) {i,, n} n σ(i) + 1 {k i / σ(k) > σ(i)} τ(i) {k i / τ(k) < τ(i)} τ(i) {k i / n σ(k) + 1 < n σ(i) + 1} τ(i) {k i / σ(k) > σ(i)}, so τ(i) n σ(i) + 1. f there is equality, then necessarily i α(i) for all α S(P, Q). Conclusion. The hypothesis H i is true for all 0 i n. So τ (n 1) σ in S n, so ι(p ) Q in PP(n). As a conclusion, ι(p ) m(p ). Corollary 36 Let us assume that q 0. Then the pairing, q is non-degenerate if, and only if, q 1 0. Proof. Let us fix an integer n N. We consider the basis PP(n) of H(n), totally ordered by. n this basis, the matrix of, q restricted to H(n) has the following form, coming from

23 Lemmas 34 and 35: 0 0 q n(n 1) q n(n 1) q n(n 1) q n(n 1) 1 so its determinant is ±q n!n(n 1) 1. Hence,, q is non-degenerate if, and only if, q 1 0. Remark. With the help of the isometry α (Proposition 31), it is possible to prove that if q 1 0,, q is non-degenerate if, and only if, q 0., Corollary 37 f q 1, q, q 3, q 4 are algebraically independent over Z, then the pairing, q is non-degenerate. Proof. For all n, let us consider the matrix of the pairing restricted to H PP (n) in the basis formed by the plane posets of degree n. ts determinant D n is clearly an element of Z[q 1, q, q 3, q 4 ]. Moreover, D n (1, 0, q 3, q 4 ) 0 by Corollary 36, so D n is a non-zero polynomial. As a consequence, if q 1, q, q 3, q 4 are algebraically independent over Z, D n (q 1, q, q 3, q 4 ) 0. 5 Morphism to free quasi-symmetric functions 5.1 A second Hopf pairing on H (q1,0,q, 1,q 4 ) We here assume that q 0 and q 1 q 3. For all double poset P, (q1,0,q 1,q 4 )(P ) q h R\ +r R\ 1 q rp \ h-ideal of P 4 (P \ ). For any h-ideal of P, h P \ 0, so: h R\ + r R\ {(x, y) (P \ ) x < y}, r P \ r P \ + h P \ {(x, y) (P \ ) x > y}. Hence: (q1,0,q 1,q 4 )(P ) {(x,y) (P \) x<y} {(x,y) (P \) x<y} q1 q h-ideal of P 4 (P \ ). Notations. Let P, Q P(n) and let σ Bij(P, Q). As totally ordered sets, P Q {1,..., n}, so σ can be seen as an element of the symmetric group S n. ts length l(σ) is then its length in the Coxeter group S n [7]. Theorem 38 We define a pairing, q : H q H q K by: P, Q q σ S (P,Q) q n(n 1) l(σ) 1 q l(σ) 4, for all P, Q PP, with n deg(p ). Then, q is an homogeneous symmetric Hopf pairing on the braided Hopf algebra H q (H,, q ). 3

24 Proof. This pairing is clearly homogeneous. For any double posets P and Q, the map from S (P, Q) to S (Q, P ) sending σ to its inverse is a bijection and conserves the length: this implies that the P, Q q Q, P q. Let P, Q, R be three double posets. There is a bijection: Bij(P Q, R) Bij(P, R \ ) Bij(Q, ) Θ : R σ (σ P, σ Q ), with σ(q). Let us first prove that σ S (P Q, R) if, and only if, is a h-ideal of R and (σ 1, σ ) S (P, R \ ) S (Q, ).. Let x and let y R such that x h y. We put x σ(x) and y σ(y). As σ S (P Q, R), x y. As x Q, y Q, so y : is a h-ideal of R. By restriction, σ P S (P, R \ ) and σ Q S (Q, ).. Let us assume that x h y in P Q. Then x, y P or x, y Q. As σ P S (P, R \ ) and σ Q S (Q, ), σ(x) σ(q) in R. Let us assume that σ(x) h σ(y) in R. As is a h-ideal, there are two possibilities: σ(x), σ(y) R \ or σ(x), σ(y). Hence, x, y P or x, y Q. As σ P S (P, R \ ) and σ Q S (Q, ), x y in P Q. σ(x) R \ and σ(y). Hence, x P and y Q, so x y in P Q. Finally, we obtain a bijection: S (P Q, R) S (P, R \ ) S (Q, ) Θ : h-ideal of R σ (σ P, σ Q ), Let σ S (P Q, R) and we put Θ(σ) (σ 1, σ ). Let R \ {i 1,..., i k } and {j 1,..., j l }. Let ζ S k+l defined by ζ 1 (1) i 1..., ζ 1 (k) i k, ζ 1 (k + 1) j 1,..., ζ 1 (k + l) j l. Then ζ is a (k, l)-shuffle and σ ζ 1 (σ 1 σ ), so l(σ) l(ζ) + l(σ 1 ) + l(σ ). Moreover: l(ζ) {(p, q) j q < i p } {(i, j) (R \ ) j < i} r R\ + h R\ r R\, as is a h-ideal of R. Finally, l(σ) l(σ 1 ) + l(σ ) + r R\. Moreover, if R n and k, by Lemma 13: so: h R\ + r R\ + rr\ k(n k) + h R\ + r R\ + hr\ + r R\ h R\ + k(k 1) (n k)(n k 1) r R\ + + k(k 1) (n k)(n k 1) + n(n 1) r R\. 4 k(n k), r R\

25 Finally: P Q, R q σ S (P Q,R) q n(n 1) l(σ) 1 q l(σ) 4 q n(n 1) l(σ 1 ) l(σ ) r R\ 1 q l(σ 1)+l(σ )+r R\ 4 h-ideal of R σ 1 S (P,R\),σ S (Q,) q h R\ +r R\ + k(k 1) + (n k)(n k 1) 1 q l(σ 1)+l(σ )+r R\ 4 h-ideal of R σ 1 S (P,R\),σ S (Q,) q h R\ +r R\ 1 q rr\ h-ideal of R P Q, q (R) q. 4 P, R \ q Q, q So this pairing is a Hopf pairing on H q. Remark. n particular,, (1,0,1,1) is the pairing, of the section 1. Examples. Here are the matrices of the pairing, q restricted to H q (n), for n 1,, 3. 1 q 1 q 1 q 1 q 1 + q 4 q1 3 q1 3 q1 3 q1 3 q1 3 q1 3 q1 3 q1 (q 1 + q 4 ) q1 3 q1 (q 1 + q 4 ) q1 3 q1 (q 1 + q 4 ) q1 3 q1 3 q1 (q 1 + q 4 ) q1 3 q1 (q 1 + q 4 ) q1 (q 1 + q 4 ) q 3 1 q1 (q 1 + q 4 ) q1 3 q1 (q 1 + q 4 ) q 1 (q1 + q4 1 ) q 1(q1 + q 1q 4 + q4 ) q1 3 q1 3 q1 (q 1 + q 4 ) q 1 (q1 + q4 1 ) q 1 (q 1 + q 4 ) q 1 (q1 + q 1q 4 + q4 ) q1 3 q1 (q 1 + q 4 ) q1 (q 1 + q 4 ) q 1 (q1 + q 1q 4 + q4 ) q 1(q1 + q 1q 4 + q4 ) (q 1 + q 4 )(q1 + q 1q 4 + q4 ) Proposition 39 For all x, y, z H q, xy, z q x y, (q1,0,q 1,0)(z) q. Proof. Let P, Q, R PP. We consider the following map: Bij(P Q, R) Bij(P, R \ ) Bij(Q, ) Θ : R σ (σ P, σ Q ), with σ(q). Let us first prove that σ S (P Q, R) if, and only if, is a biideal of R and (σ 1, σ ) S (P, R \ ) S (Q, ).. Obviously, (σ 1, σ ) S (P, R \ ) S (Q, ). Let x, y R, such that x y. We put x σ(x) and y σ(y). f y / Q, then y P and x Q, so y < h x. As σ S (P Q, R), y < x : contradiction. So y Q and y.. Let x, y P Q, such that x h y. Two cases are possible. f x, y P or x, y Q, as (σ 1, σ ) S (P, R \ ) S (Q, ), σ(x) σ(y) in R. f x P and y Q, as σ(q) is a biideal, it is not possible to have y x, so x y. 5

26 Let x, y P Q, such that σ(x) h σ(y) in R. possible. As σ(q) is a biideal, two cases are f x, y P or x, y Q, as (σ 1, σ ) S (P, R \ ) S (Q, ), x y in P Q. f x P and y Q, then x y in P Q. Moreover, if σ(q) is a biideal of R, then {k + 1,..., k + l} R, where k P and l Q. Then σ σ 1 σ, so l(σ) l(σ 1 ) + l(σ ). So, with n R : P Q, R q biideal of R σ 1 S (P,R\),σ S (Q,) biideal of R biideal of R biideal of R biideal of R q n(n 1) k(k 1) (n k)(n k 1) P Q, (q1,0,q 1,0)(R) q, q n(n 1) l(σ 1 ) l(σ ) 1 q l(σ 1)+l(σ ) 4 1 P Q, (R \ ) q q (n k)k 1 P Q, (R \ ) q q R\. 1 P Q, (R \ ) q q h R\ +r R\ 1 P Q, (R \ ) q with the observation that, as is a biideal, h R\ r R\ 0, so R \. h R\ + r R\. Remark. n particular, if q (1, 0, 1, 1), for all P, Q, R PP: P Q, R P Q, (1,0,1,0) (R) This formula was already proved in [4]. biideal of R P Q, (R \ ). 5. Quantization of the Hopf algebra of free quasi-symmetric functions Theorem We define a coproduct on FQSym in the following way: for all σ S n, q (σ) n q k(n k) l(σ)+l(σ(1) k )+l(σ() k ) 1 q l(σ) l(σ(1) k ) l(σ() k ) 4 σ (1) k0 k σ () For all q K, FQSym q (FQSym, m, q ) is a graded braided Hopf algebra.. We define a Hopf pairing on FQSym q by σ, τ q n(n 1) l(σ) 1 q l(σ) 4 δ σ,τ The following map is an isomorphism of braided Hopf algebras: H q FQSym q Θ : P PP σ. σ S P Moreover, for all x, y H q, Θ(x), Θ(y) q x, y q. Remarks. k. 6

27 1. Let σ S n. f 0 k n, we put σ({1,..., k}) {i 1,..., i k } and σ({k + 1,..., n}) {j 1,..., j l }. Let ζ be the (k, l)-shuffle defined by ζ 1 (1) i 1..., ζ 1 (k) i k, ζ 1 (k + 1) j 1,..., ζ 1 (k + l) j l. Then σ ζ 1 (σ (1) k σ () k ), and l(σ) l(ζ) + l(σ(1) k ) + l(σ() k ). Moreover, 0 l(ζ) k(n k), so: 0 l(σ) l(σ (1) k ) l(σ() k ) k(n k). As a consequence, the coproduct on FQSym q is well-defined, even if q 1 0 or q t is clear that the pairing, q defined on FQSym q is non degenerate if, and only if, q 1 0 and q 0. Proof. From [1], we already know that Θ is an algebra isomorphism. Moreover, it is also proved there (Lemma 3.1) that S (P, Q) S(P ) S(Q) 1 for any P, Q P P, so: So Θ is an isometry. Θ(P ), Θ(Q) q σ S(P ),τ S(Q) σ S(P ),τ S(Q) σ, τ q q n(n 1) l(σ) 1 q l(σ) 4 δ σ,τ 1 n(n 1) l(σ) q1 q l(σ) 4 σ S(P ) S(Q) 1 σ S (P,Q) P, Q q. Consequently, for all x H q, y, z FQSym q : q n(n 1) l(σ) 1 q l(σ) 4 Θ(x), y z q Θ(x), yz q x, Θ 1 (yz) q x, Θ 1 (y)θ 1 (z) q q (x), Θ 1 (y) Θ 1 (z) q (Θ Θ) q (x), y z q. Let us now assume that q 1, q 4 0. Then the pairing, q on FQSym q is non-degenerate, so Θ (Θ Θ). Hence, Θ is an isometric isomorphism of braided Hopf algebras. This proves the three points immediately. Up to an extension, we can assume that K is infinite. t is not difficult to show that the set A of elements (q 1, q 4 ) K such that the three points are satisfied is given by polynomial equations with coefficients in Z, As it contains (K {0}) from the preceding observations, which is dense in K, it is is equal to K, so the result also holds for any (q 1, q 4 ). Corollary 41 The pairing, q is non-degenerate if, and only if, q 1 0 and q 4 0. Proof. This result is implied via the isomorphism Θ from the fact that the pairing, q on FQSym q is non-degenerate if, and only if, q 1 0 and q

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