On ( non-) freeness of some tridendriform algebras

We present some results on the freeness or non freeness of some tridendriform algebras. In particular, we give a combinatorial proof of the freeness of WQSym, an algebra based on packed words, result already known with an algebraic proof. Then, we prove the non-freeness of an another tridendriform algebra, PQSym, a conjecture remained open. The method of these proofs is generalizable, in particular it has been used to prove the freeness of the dendriform algebra FQSym and the quadrialgebra of 2-permutations. Résumé. Nous présentons des résultats de liberté concernant certaines algèbres tridendriformes. En particulier, nous prouvons par des arguments combinatoires que l’algèbre WQSym est tridendriforme libre, résultat déjà connu, mais obtenu par des méthodes purement algébriques. Puis nous prouvons que PQSym n’est pas une algèbre tridendriforme libre, conjecture restée ouverte jusqu’à présent. Les méthodes utilisées dans les preuves sont généralisable. En particulier, elles ont été utilisées pour prouver la liberté de l’algèbre dendriforme FQSym et de la quadrialgèbre des 2-permutations.


Introduction
For some years now, a lot of algebras arise from combinatorial objects.For examples, permutations, parking functions can be equipped with dendriform products which are half shuffle products ( [LR98], [NT07]).Conversely, some combinatorial objects appear naturally in the theory of operads, which is in part the study of different types of algebras and the relations between them ( [LV12]).For instance, planar binary trees arise naturally from dendriform algebras ( [Lod01]).Indeed, Loday proved that the Hilbert series of the free dendriform algebra over one generator is the series of Catalan numbers ( [Lod01]).Other combinatorial properties have been studied in [HNT05].
Tridendriform algebras were introduced independently in [LR04] and [Cha02].An example is given by WQSym ( [NT06]) and the construction of this algebra comes from products defined on a family of words, the packed words.The freeness of this algebra was proved in [BR10].Another tridendriform algebra coming from a family of words, parking functions, denoted by PQSym was introduced in [NT07], where the freeness of this algebra was conjectured.
The aim of this article is to use the combinatorial interpretation of an algebra in terms of evaluation trees in order to prove the freeness of WQSym and to prove the non-freeness of PQSym.
The paper is structured as follows: we present some background about evaluation trees, tridendriform and and free tridendriform algebras.Then we discuss the freeness of the algebras WQSym and PQSym.

Background
In the sequel we are only interested in graded algebras.

Evaluation trees and algebras
Definition 1.A pair A = (A, P) is called a P-algebra if A is a graded vector space (A = ⊕ n∈N A n ) with A 0 isomorphic to K, P is a finite set of bilinear maps from A to A such that for each B in P, y n in A n and y m in A m , the element B(y n , y m ) is in A n+m , and the element of A 0 identified to 1 K is the neutral element of B. We set Definition 2. A decorated complete binary tree is defined by induction as follows: • the empty set ∅ is a decorated complete binary tree, • the triple (a, ∅, ∅) is a decorated complete binary tree, where a is an element of a certain set, • the triple (a, T 1 , T 2 ) is a decorated complete binary tree if T 1 and T 2 are non-empty complete binary trees, and a is an element.
We denote by CBT (P, B) the set of decorated complete binary trees where the leaves are decorated by elements of B and the internal nodes are decorated by the elements of P and by BT (P, B) the graded vector space freely spanned by CBT (P, B).
Definition 3. Let T = (a, T 1 , T 2 ) and T be two decorated binary trees.We say that T is a prefix of T if T is the empty set; or T = (a, T 1 , T 2 ) and T 1 is a prefix of T 1 and T 2 is a prefix of T 2 .
Definition 4. Let T = (a, T 1 , T 2 ) and T be two decorated binary trees.We say that T contains the pattern T if T is a prefix of T ; or T 1 or T 2 contain the pattern T .If T does not contain the pattern T we say that T avoids the pattern T .
The tree contains the pattern ≺ but avoids .
Definition 5. Let (A, P) be a P-algebra, and B a basis of A + .The vector space of evaluation trees over A denoted by ET (A) is the vector space BT (P, B).Definition 6.Let (A, P) be a P-algebra.The evaluation map Ev is a linear map from ET (A) to A defined on trees by: where internal nodes are decorated by bilinear maps of P and leaves by elements of A.
Example 3. Let us consider the vector space of non-commutative polynomials K A , where A is a totally ordered alphabet.Let us define following three products on non-empty words:

Tridendriform algebras
Definition 7. A tridendriform algebra is a vector space V equipped with three bilinear maps, •, , ≺ such that for each elements a, b, c in A, we have: ( where =≺ + • + . Example 4. The non-commutative polynomials K A equipped with the three laws presented in example 3 is a tridendriform algebra ([NT06]).

WQSym
One of the algebras we are interested in is the algebra WQSym.It was defined in [NT06].
It was proved in a different way by Burgunder and Ronco in [BR10] that is a free tridendriform algebra.We give a proof based on the method presented in [Von15] for the freeness of this algebra in Section 3. Definition 8.A packed word of size n is a word w = w 1 • • • w n over the alphabet N with the following property: if k is a letter of w, then each integer from 1 to k is also a letter of w.
We denote by MT (n) the set of packed word of size n, by MT the set of packed words, by WQSym the graduated vector space freely generated by packed words and by WQSym n the component of degree n generated by MT (n).
Example 5.The word 1132422 is a packed word but 14355 is not.Definition 10.Let us consider an homogeneous basis (M u ) u∈MT of WQSym.We define the three following products in this basis: where max(w) is the greatest letter of w.
Example 7. We have:

PQSym
In order to define the algebra PQSym, we have to define parking functions and an algorithm the parkization ([NT07]).
Definition 11.A parking function w is a word over the alphabet N * such that the non decreasing word w ↑= w 1 w 2 • • • w n associated to w satisfies the following property: Let us denote by PQSym the vector space freely spanned by parking functions and by PQSym n the subspace spanned by parking functions of size n.
Example 8.The word w = 11321 is a parking function since w ↑= 11123 satisfies the property (6) but the word v = 76191128 is not one.Definition 12. Let w = w 1 • • • w n be a word over the alphabet N * .The parkized of the word denoted by Park(w) is obtained by the following process.We set Definition 13.Let (P u ) be an homogeneous basis of PQSym.We define the following three products in this basis:

Free tridendriform algebras
The method presented in this section has been used in [Nov14] and [Von15].
With the evaluation trees formalism, we rewrite the relations (3).For example, one of them becomes: So the free tridendriform algebra over one generator x which we denote by T rid can be seen as the vector space BT ({≺, , •}, {x}) modulo the relations (3).In order to find a basis, we consider the set of trees of CBT ({≺, , •}, {x}) which avoids the following patterns : where ∈ {≺, , •}.We denote this set by TAP ({x}) (for Trees Avoiding Patterns with leaves in {x}).
Proposition 1.The generating series of TAP ({x}) (denoted by G) is the series of the little Schroeder numbers.
Proof.Let us denote by G a the generating series of trees in TAP ({x}) and whose root is decorated by a.So we have the following system: a consequence, the series G satisfies the same equation as the generating series of the little Schroeder numbers.Since this equation has one and only one solution which is a series with positive integers coefficients, we deduce that the coefficients of G are the little Schroeder numbers.
Proof.By induction on the number of leaves n.For n ≤ 2, it is a consequence of the rewriting rules.Let T be a tree with n+1 leaves.If T is equal to ( , T 1 , T 2 ).By applying the induction hypothesis on T 2 , we obtain ( , T 1 , ) where is a linear combination of trees avoiding (11).But one of these trees may have the following shape: ( , T 1 , ( , T , T ")), where T " has a root different to , and avoiding (11).Then we use the following rewriting rule ( ) = ( ) .So we have ( , T 1 , ( , T , T ")) = ( , ( , T 1 , T ) , T ").Thus it is sufficient to apply the induction hypothesis to the left sub-tree.
If the root is decorated by ≺ or •, we first apply the induction hypothesis to the left sub-tree.If a forbidden pattern appears, at the level of the root with the left sub-tree, we apply the associated rewriting rule.Thanks to the induction hypothesis, now the left sub-tree is a leaf.If the root is now decorated by , we already considered that case.Otherwise, it is sufficient to apply the induction hypothesis to the right sub-tree.
Theorem 1.A linear basis of the free tridendriform algebra over one generator x is given by the family TAP ({x}).
Proof.We have seen that family TAP ({x}) spans the free tridendriform algebra over one generator x.It is known that the coefficients of the Hilbert series of this algebra is given by the little Shroeder numbers ( [LR04]).Thanks to Proposition 1, we know that the generating series of TAP ({x}) is also given by the generating series of the little Schroeder numbers.So the family is a basis TAP ({x}).
Theorem 2. Let A be a free tridendriform algebra generated by a free family F. A linear basis of A is given by the evaluations of the family TAP (F) (for Trees Avoiding Patterns with leaves in F).
Proof.Let v ∈ A. Since the family F generates the algebra A, the element v is a linear combination of trees in TAP (F).Conversely, Assume there exists a linear combination of trees in TAP (F) such that: If some of the coefficients are non-zero, we have a relation for the family F.So each of these coefficients are zero.So (Ev(T )) T ) is a basis of A.

Freeness of the tridendriform algebra WQSym
From Theorem 1, in order to prove that WQSym is a free tridendriform algebra, it is sufficient to find a family F of packed words such that the family (Ev(T )) T ∈TAP(F ) is a basis of WQSym.In order to do so, we find three reduced products over packed words satisfying some relations with the three products of WQSym and are compatible with a total order over packed words.Thanks to the reduced products, we construct the family F of indecomposable packed words.Thanks to the order, we deduce that (Ev(T )) T ∈TAP(F ) is uni-triangular and so is a basis.

Definitions and general properties
Definition 14.Let u and v two packed words.We define the following three products: where: • ū is obtained by replacing max(u) by max(u) + max(v), v is obtained by shifting all letters of v by max(u) − 1, Definition 15.Let u be a non-empty packed word.We say that u is indecomposable if : Otherwise, u is decomposable.
we say that u is -connected.
Proposition 3. Let u be a non-empty packed word of size n.The following two propositions are equivalent: 1. the word u is not -connected, 2. there exists 1 < i < n such that all letters u 1 • • • , u i are strictly lesser than any letter u i+1 , • • • , u n .
Proof.Assume that u is not -connected.There exists v and w such that: with the letters of v strictly lesser than any letter of ŵ.So we can take i equals to |v|.Conversely, Assume that there exists 1 < i < n such that all letters u 1 • • • , u i are strictly lesser than any letter u i+1 , • • • , u n .Since u is a packed word, we deduce that u = pack(u Proposition 4. Let u be a non-empty packed word of size n.The following two propositions are equivalent: 1. the word u is decomposable, 2. there exists 1 < i < n such that all letters u 1 • • • , u i (but the maximum of u i if it is one of the u i ) are strictly lesser than any letter u i+1 , • • • , u n .
Proof.Assume that the word u is decomposable.So there exists v, w two packed words and ∈ {≺ , , • } such that u = v w.By taking i = |v|, we deduce that u satisfies the second property.Conversely, if u satisfies the second property, we have:

The factorization Theorem
Theorem 3. Let w be a non-empty packed word.Then w satisfies one and only one of the following properties: In order to prove Theorem 3 we need to prove the following two propositions: Proposition 5. Let w be a non-empty packed word.Assume that there exists u, v, u and v four non-empty packed words such that: where ∈ {≺ , • }.Then u is decomposable.
Proposition 6.Let w be a non-empty packed word.Assume there exists u, v, u and v four non-empty packed words such that: Then u is decomposable.
Proof of Proposition 5. Assume that u v = u v .By construction, we have: Since û and ũ are strict prefixes of w, one is prefix of the other.Since max(w) is not in û but in ũ , it results that û is a strict prefix of ũ .That is to say there exists a non-empty word ω such that : But each letter of û is strictly lesser than all letters of v and so each letter of û is strictly lesser than all letters of ω.Thus, pack( ũ ) = u = û ω .In particular, u is decomposable.
Proof of Proposition 6. Assume that u • v = u ≺ v .By construction, we have: Since v and v are strict suffixes of w, one is suffix of the other.Since max(w) is not in v but in v, it results that v is a strict suffix of v.So u is a strict prefix of ū .That is to say, there exists a non-empty word ω such that ū = uω.
We note that each letter of u but max(w) is strictly lesser than all letters of v and so is strictly lesser than all letters of ω.Thus, the word u is decomposable.
Proof of Theorem 3. Let w be a packed word of size n.Assume that w is decomposable.If w is notconnected, thanks to Proposition 3, there exists u and v two non-empty words such that w = uv, where all letters of u is strictly lesser than all letters of v.By taking u the greatest prefix satisfying this property, we deduce that w = pack(u) pack(v) with pack(v) -connected.Assume that w is decomposable and -connected.Thanks to Proposition 4, there exists u and v two non-empty words such that w = uv, where all letters of u but max(u) is strictly lesser than all letters of v.By taking u the smallest prefix satisfying this property, we deduce that w = pack(u) pack(v) with pack(u) is indecomposable.If max(u) > max(v), we have = and if max(u) = max(v), we have = • .
From Proposition 5 and Proposition 6, we deduce that the four cases are disjoint.Assume that w = u v and w = u v with v and v -connected.If |v| < |v |, we deduce that v is not -connected.By symmetry, we deduce that |v| = |v |.Then u = u and v = v .By similar arguments, it follows that u v = u v with ∈ {• , ≺ } and u, u indecomposable implies that u = u and v = v .
Theorem 4.There is a bijection between packed words and TAP (F) where F is the set of indecomposable packed words.
Proof.By applying Theorem 3 recursively, we obtain the bijection.

Freeness of WQSym
In order to prove that WQSym is a free tridendriform algebra, we need to find an order on packed words which have some compatibility with the products ≺, , • and the products ≺ , , • .
Definition 16.Let u and v be two packed words of same size.We say that u is lesser than v denoted by u ≤ m,lex v if: Example 12.We have: 13123 ≤ m,lex 21122 and 21134 ≤ m,lex 21143.
Proposition 7. Let I be a set of packed words of size m and J be a set of packed words of size n.Let u be the minimum of I and v be the minimum J. Then: Proof.Let w = w 1 w 2 be the minimum of I J, where ∈ {≺, , •} and w 1 be the prefix of size n.Thus w is lesser than u v.By definition of ≤ m,lex , we deduce that max(w) ≥ max (u v).By definition of w, there exists r in I and s in J such that w ∈ r s.Since u is the minimum of I and v in the minimum of J, we deduce that max(r) ≤ max(u) and max(s) ≤ max(v).But max(w) ≤ max(r) + max(s) − 1 if = • and max(w) ≤ max(r) + max(s) in the other cases.It follows that max(r) = max(u) and max(s) = max(v).Let us denote by ũ the prefix of size n of u v and ṽ the corresponding suffix.We have w 1 ≤ lex ũ.But u ≤ lex pack(w 1 ) ≤ lex w 1 ≤ lex ũ.But u and ũ may differ only in maximum value.So pack(w 1 ) = r = u since max(r) is equal to max(u).So we have u ≤ lex w 1 ≤ lex ũ and w 1 and ũ may differ only in maximum value.We have three cases: • if = , we deduce that ũ = u.So w 1 is equal to u.
In all cases, we have w 1 = ũ.Since max(w) = max(u v) and w 1 contains the smallest letters (but the maximum letter), we deduce that w 2 contains the other letters.So w 2 and ṽ have the same letters.By same arguments as previously, we deduce that w 2 = ṽ.
Theorem 5.The algebra WQSym is a free tridendriform algebra.
Proof.Thanks to Theorem 4, we deduce that the family (Ev(T )) T ∈TAP(F ) where F is the set of indecomposable packed words may be a basis.Thanks to the total order, we deduce that the matrix of the family in the basis (M u ) is uni-triangular.Thus the algebra WQSym is a free tridendriform algebra.

The tridendriform algebra PQSym
We prove that PQSym is not a free tridendriform algebra by expliciting relations.Theorem 6.The algebra PQSym is not a free tridendriform algebra.
Proof.We denote by F the tridendriform algebra generated by F. Assume there exists a family F which generates PQSym as a free tridendriform algebra.Let us recall that the valuation of P in PQSym is the first n such that Π n (P ) = 0, where Π n is the projection onto PQSym n .
We consider the elements in F which generate P 1 , P 112 and P 211 .Since they are finite linear combinations of evaluation trees, there is a finite number of leaves.Then we can restrict to a finite free family F n = f 1 , • • • , f n elements of f which generates P 1 , P 112 and P 211 .since one of the f i is valuation one, by linear combinations we can assume without loss of generality that f 1 is the only element which has a component of degree one.Since Π 2 (f 1 • f 1 ), Π 2 (f 1 f 1 and Π 2 (f 1 ≺ f 1 ) span the subspace PQSym 2 we can replace f i by where α i , β i , γ i are respectively the coefficients of P 11 , P 12 , P 21 in f i .From now on, we assume that for i ≥ 2, the valuation of f i is at least 3.
Let us denote by V the vector space spanned by f 1 , • • • , f n and by W the subspace spanned by the family of non empty evaluation trees BT ( , ≺, •, f 1 , • • • , f n ).
By construction, P 1 , P 112 and P 211 are in the vector space V W .By hypothesis, a basis of V is given by f 1 , • • • , f n .By construction, the valuation of elements in W is at least 2. So we have: with w ∈ W and for i ≥ 2 the valuation of f i is at least 3. Since the only element with valuation 1 is f 1 , we deduce that α 1 = 0. Let us write P 112 in the same basis.We have: with w ∈ W .The valuation of w is at least 2 and the valuation of n k=2 β i f i is at least 3. Since the valuation of P 112 is 3, we deduce that β 1 = 0.For i ≥ 2, the valuation of f i is at least 3.So the valuation of w cannot be 2. Thus, the valuation of w is at least 3.By valuation arguments, trees of size greater than 4 and trees whose leaves are different to f 1 have a valuation greater than 4. By computing the other trees, we observe that the coefficients of P 112 and P 113 are equal.So necessarily, the coefficient of one of the f i for i = 1 is different to 0. Without loss of generality, by re-indexing the elements, we can assume that β 2 = 0.
We have:

Definition 9 .
Let w be a word over a totally ordered alphabet A and b 1 < b 2 < • • • < b r the different letters of w.The word pack(w) is the packed word obtained by replacing all occurrences of b i by i. Example 6.If w = 3944577, we have pack(w) = 1522344.
by convention, the minimum of an empty set is ∞.While d(w) ≤ n, we replace w by the word obtained by replacing in w all letters j in w greater than d(w) by j + d(w) − a where a is the smallest letter of w greater than d(w).Example 9.If w = 9541843, we obtain successively d(w) = 2 and w = 8431732, then d(w) = 6 and w = 7431632.Since d(w) = ∞, we deduce that Park(w) = 7431632.
and v is obtained by shifting all letters of v by max(u), • u is obtained by replacing max(u) by max(u)+max(v)−1, and v is obtained by shifting all letters of v by max(u) − 1, Example 11.If u = 2134341 and v = 3123, we have: u v = 21343457567, u ≺ v = 21373716456 and u • v = 21363616456.