Clément Chenevière

Linear Intervals in the Tamari and the Dyck Lattices and in the alt-Tamari Posets

We count the number of linear intervals in the Tamari and the Dyck lattices according to their height, using generating series and Lagrange inversion. Surprisingly, these numbers are the same in both lattices. We define a new family of posets on Dyck paths, which we call alt-Tamari posets. Each alt-Tamari poset depends on the choice of an increment function δ ∈ {0, 1} n . We recover the Tamari and the Dyck lattices as extreme cases with δ = 1 and δ = 0, respectively. We prove that all the alt-Tamari posets have the same number of linear intervals of any given height.

Introduction

In a partially ordered set or poset, when two elements P and Q are comparable, the interval [P, Q] is the subset of all elements R that satisfy P ≤ R ≤ Q. When studying a poset, one may be interested in its intervals and the simplest of them are those which are totally ordered, that we call linear intervals.

It is quite natural to count linear intervals when looking at a finite poset, although this has not been much studied. In this work, we are interested mainly in two classical posets on Catalan objects, namely the Tamari and the Dyck lattices [START_REF] Tamari | The algebra of bracketings and their enumeration[END_REF][START_REF] Stanley | Enumerative combinatorics[END_REF][START_REF] Bernardi | Intervals in Catalan lattices and realizers of triangulations[END_REF]. Surprisingly, these two posets share the same number of linear intervals of any given height, where the height is the cardinality minus one. Furthermore, these numbers are very simple. We prove this result and give an explicit formula for these numbers.

This coincidence was a hint that there was a stronger link between these two posets than what was previously known. In fact, one can define a whole new family of posets that interpolate between the Tamari and the Dyck lattices, and we call them the alt-Tamari posets. We define and study this family in this paper and we also prove that they all have the same number of linear intervals of any fixed height.

There are two other families of posets that generalize the Tamari lattices, namely the Cambrian lattices of type A and the posets of tilting modules. The former were defined by Reading in [START_REF] Reading | Cambrian lattices[END_REF][START_REF] Reading | From the Tamari lattice to Cambrian lattices and beyond[END_REF] as the restriction of the weak order on the subset of c-sortable elements for some standard Coxeter element c. The latter were defined by Happel and Unger, following work of Riedtmann and Schofield [START_REF] Riedtmann | On a simplicial complex associated with tilting modules[END_REF][START_REF] Happel | On a partial order of tilting modules[END_REF] in the context of quiver representations. Interested readers may want to look at the survey article [START_REF] Thomas | The Tamari lattice as it arises in quiver representations[END_REF].

All posets of these two other families also seem to have the same number of linear intervals of any given height as the Tamari lattice.

These three families have even more in common, namely their categories of modules seem to be all derived-equivalent. This has been proven for posets of tilting modules and Cambrian posets by Ladkani in [START_REF] Ladkani | Universal derived equivalences of posets of tilting modules[END_REF][START_REF] Ladkani | Universal derived equivalences of posets of cluster tilting objects[END_REF], but this is only conjectural for the alt-Tamari posets. In particular, these new posets between the Tamari and the Dyck lattices might be useful for proving their expected derived equivalence.

In all posets, intervals of height 0 are the trivial intervals of the form [P, P ] and intervals of height 1 are those reduced to 2 elements, that we call the covering relations and that we denote by P Q.

Nevertheless, some posets do not have any linear interval of height 2 or more, as for instance the third classical poset on Catalan objects, namely the non-crossing partition lattice with coarsening order. The Boolean lattice B n and the set partition lattice P n are two other examples. In both cases, all intervals of height 2 are isomorphic to the Boolean lattice B 2 or to the set partition lattice P 2 , which are not totally ordered. The posets which do not have linear intervals of height 2 are said to be 2-thick, and for lattices of finite length, this is equivalent to be relatively complemented [START_REF] Björner | On complements in lattices of finite length[END_REF].

Let us note that this new family of posets also has close links with the ν-Tamari lattices. On the one hand, there are strong hints that all posets and results would generalize to this larger context. On the other hand, the alt-Tamari posets can be understood as intervals in some ν-Tamari lattices, as understood after discussions with Wenjie Fang.

In further work, we may look at the Cambrian lattices, the posets of tilting modules and other properties than the number of linear intervals. We could also investigate if the techniques in this paper give some results in the case of other posets as for instance the m-Tamari [START_REF] Bergeron | Higher trivariate diagonal harmonics via generalized Tamari posets[END_REF] and the m-Cambrian lattices [START_REF] Stump | Cataland: why the fuss?[END_REF] or the Cambrian lattices and posets of tilting modules in other types. Several other families of posets seem to have nice formulas for their numbers of linear intervals, for instance weak orders on the symmetric groups.

In section 1 we recall classical material on intervals and posets and the Tamari and the Dyck lattices. In sections 2 and 3, we study the structure of linear intervals in the Tamari and the Dyck lattices. In both cases, we first define a notion of left and right intervals, which we prove to be linear. We then show that all linear intervals of height k ≥ 2 are either left or right intervals. Then, we give a bijection between linear intervals of height 1 (resp. k ≥ 2) and a marked Catalan object (resp. with a direction) and one Catalan object (resp. a sequence of k Catalan objects). From this bijection, we deduce equations on the generating series of linear intervals of a given height. Finally, we solve these equations and compute the number of linear intervals.

In section 4, we define the family of alt-Tamari posets which generalizes both the Tamari and the Dyck lattices. We study their linear intervals and show that there are similar notions of left and right intervals and similar decompositions. This gives a bijection between linear intervals of any two alt-Tamari posets that respects the height. It follows that their linear intervals are counted by the same numbers and this is the main result of this paper.

Theorem 0.1. For any increment function of size n, the alt-Tamari poset Tam δ n contains:

• 1 n + 1 2n n linear intervals of height 0, • 2n -1 n -2 linear intervals of height 1, • 2 2n -k n -k -1 linear intervals of height k, for 2 ≤ k < n.
Furthermore, there are no linear interval of height k ≥ n.

We also prove additional results on the family and in particular that there is a boolean structure of refinement of the alt-Tamari posets.

Definitions and first properties 1.Linear intervals

In a poset, if we have two elements P ≤ Q, then a chain of length k from P to Q is a sequence of elements P = P 0 < P 1 < • • • < P k = Q. Recall that the height of an interval [P, Q] is defined as the maximal length of a chain from P to Q and that an interval is linear if it is totally ordered.

In any poset, the intervals of height 0 are those of the form [P, P ] with P some element of the poset. They are clearly linear. We call them trivial intervals and they obviously are in bijection with elements of the poset. Intervals [P, Q] of height 1 are by definition reduced to two elements since otherwise they would contain a chain of length 2 or more. They are linear as well. Such an interval is called a covering relation and is denoted P Q.

A chain is said to be maximal if for all 0 ≤ i < k, P i P i+1 is a covering relation. In a finite poset, an interval [P, Q] is linear if and only if there is a unique maximal chain from P to Q. The length of this maximal chain is exactly the height of the interval.

We can define the Cartesian product of two posets P and Q as a poset P × Q whose elements are couples (p, q) with p ∈ P and q ∈ Q and whose order is (p, q) ≤ (p , q ) component wise. The Cartesian product of two intervals in P and Q gives obviously an interval in P ×Q, but the converse is also true, namely every interval of P × Q comes from the Cartesian product of two intervals.

We can thus keep track of the counting of linear intervals in a multiplicative way in the Cartesian product of two posets. To do so, we have to distinguish trivial intervals and linear intervals which are not trivial. Indeed, the Cartesian product of two intervals is linear if and only if one of the intervals is trivial and the other is linear, and it is trivial if and only if both intervals are trivial.

Let be a formal variable with 2 = 0. Given a poset P , we can define the polynomial L(P ) = T (P ) + U (P ) where T (P ) is the number of trivial intervals of P and U (P ) counts its non trivial linear intervals. Then, we have

L(P × Q) = L(P )L(Q) in Z[ ].
This procedure could be helpful in generalizing some results or in counting linear intervals in posets where intervals are understood through a recursive decomposition.

Trees and Tamari lattices

A planar rooted binary tree is a finite connected acyclic planar graph whose vertices have degree 3 or 1, with one marked vertex of degree 1 called the root. The other vertices of degree 1 are called the leaves and those of degree 3 are the nodes. In what follow all trees will be planar rooted binary trees and we will draw them with leaves at the top and the root at the bottom. The node connected to the root will be called the root node. We will usually not draw the root and have thus only the root node of degree 2 at the bottom. The size of a tree will be its number of nodes. The tree of size 0, reduced to a leaf will be considered as a trivial tree.

For n ≥ 1, let us denote by Y n the set of all trees of size n. The Tamari lattice Tam n of size n is a partial order on Y n , described as the reflexive transitive closure of the left rotations (see fig. 5) . This poset was first defined and studied by Tamari [START_REF] Tamari | The algebra of bracketings and their enumeration[END_REF] and happens to be a lattice, see [START_REF] Huang | Problems of associativity: A simple proof for the lattice property of systems ordered by a semi-associative law[END_REF]. Its covering relations are exactly all left rotations. We will usually not consider the Tamari lattice of size 0 and its unique trivial interval.

We can graft a tree T on a chosen leaf of another tree T . This is done by deleting the root of T and identifying the root node r of the tree T with the chosen leaf of T , producing a tree whose size is the sum of the sizes of T and T , as drawn in fig. 1. Grafting a tree of size 0 does nothing.

We can also plug a tree T into a chosen edge of another tree T , except for the edge between the root and the root node. To do so, we create a new node r Figure 1: Grafting a tree on another one. n on the selected edge of T , and we identify this node with the root of T . If the selected edge was a left son, the tree T will be the right son of n, and vice versa. If T and T are respectively of size m and m , plugging T into an edge of T will produce a tree of size m + m + 1. In fig. 2, we take the same two trees T and T as in the example of grafting, but this time, we plug T into the red dotted edge of T . Note that plugging a tree of size 0 on an edge of a tree T produces a tree that is different from T , with one more vertex. In the following, we will need a special family of trees, namely left and right combs, recursively defined as follows: 1 = r 1 =: Y is the only tree of size 1 and for n ≥ 2, n (resp. r n ) is obtained as the grafting of Y on the leftmost (resp. rightmost) leaf of n-1 (resp. r n-1 ). In fig. 4, the first tree of the first row is the right comb r 4 and the last tree of the second row is the left comb 5 .

There is an involution on trees that exchanges a tree with its mirror image. This can be seen as a vertical symmetry on the drawings. As an example, the right comb r n and the left comb n are exchanged through this involution.

Remark 1.1. This involution on trees is an order-reversing involution on the Tamari lattice.

Proof. Let P Q be a covering relation and P * (resp. Q * ) be the mirror image of P (resp. Q). Then Q * P * is a covering relation in the Tamari lattice.

Structure of intervals in the Tamari lattices

In [START_REF] Chapoton | Sur le nombre d'intervalles dans les treillis de Tamari[END_REF], the author introduces an operadic structure on intervals in the Tamari lattices with the operation of grafting of intervals. See this article for more details on the results of this section.

If I = [P, Q] is an interval in Tam n and I = [P , Q ] is an interval in Tam m , we can define the grafting of I on the k-th leaf of I for some 0 ≤ k ≤ n, producing an interval I ∈ Tam n+m as represented in fig. 3. The bottom element of I is obtained by grafting P on the k-th leaf of P and the top element of I is obtained by grafting Q on the k-th leaf of Q. As an interval, I is isomorphic to the product of the intervals I and I .

r P P' k n-k ≤ r Q Q' k n-k Figure 3: Grafting an interval [P , Q ] on the k-th leaf of the interval [P, Q].
The author also defines new intervals, that cannot be written as an interval grafted on another one. This gives a unique decomposition of intervals into new intervals. Any interval then has the structure of the product of the new intervals in its decomposition.

Remark 1.2. If an interval

I = [P , Q ] is not new, then it decomposes into an interval I = [P , Q ] grafted on the k-th leaf of some interval I = [P, Q] ∈ Tam n for some 0 ≤ k ≤ n.
Thus, there is a common node r in P and Q such that there are k leaves on the left of r and n -k leaves on its right, as show in fig. 3. The node r is precisely the root node of P and Q that has been identified to the selected leaf of P and Q.

In fact, such common nodes between the bottom and top elements of an interval are precisely the root nodes of the new intervals in its decomposition.

Dyck paths and Dyck lattices

A Dyck path of length (or size) n is a path in N 2 consisting of up steps (1, 1) and down steps (1, -1), starting from (0, 0) and ending at (2n, 0). A Dyck path can be seen as a word in letters u (for up steps) and d (for down steps) with each letter appearing n times and such that in any prefix the letter u appears at least as many times as the letter d. Such words are also called Dyck words and we will identify a Dyck path to its Dyck word, and more generally any word in letters u and d with the associated path. The height of a step is defined as the second coordinate of the vertex at which it starts. Equivalently, it is equal to the number of u preceding it minus the number of d preceding it.

In all of the following, a subword B of a word A will be a portion of consecutive letters of A. We will denote B ⊂ A.

There is a natural matching of up and down steps. Given an up step, we can define the excursion starting at this step as the portion of the path that starts at this up step and ends at the first down step that ends at the same height. In other words, it is the smallest subword starting at this letter u that is a Dyck word. Then, the last step of the excursion is the down step that matches the first step of the excursion.

For n ≥ 1, let us denote by Z n the set of all Dyck paths (or words) of size n. A natural order on Z n is the inclusion order, where we say that a path P is included in a path Q if P is always under Q when we draw them together. In other words, for every 1 ≤ k ≤ 2n, the height of the k-th step of P is less than or equal to the height of the k-step of Q. This produces the Dyck lattice Dyck n of size n, sometimes known as the Stanley lattice (see [START_REF] Stanley | Enumerative combinatorics[END_REF][START_REF] Bernardi | Intervals in Catalan lattices and realizers of triangulations[END_REF]).

Given a Dyck path, we define a valley as a down step followed by an up step, and a peak as an up step followed by a down step. The covering relations in Dyck n can be defined as transforming a valley du in a peak ud. Similarly as in the Tamari lattice, the path of size 0 will be considered as a trivial path but we will usually not consider the Dyck lattice of size 0 and its unique trivial interval.

Remark 1.3. We can define another order on the Dyck paths, where the covering relations consist of swapping a down step with the excursion that follows it, if any.

When doing so, we recover a lattice isomorphic to the Tamari lattice that we defined on trees. This alternative description was introduced by Bergeron and Préville-Ratelle in [START_REF] Bergeron | Higher trivariate diagonal harmonics via generalized Tamari posets[END_REF].

Furthermore, the inclusion order refines the Tamari order, in the sense that any interval [P, Q] in the Tamari lattice defined this way will be an interval in Dyck n .

We will prove that both lattices have the same number of linear intervals of any height, however a linear interval in Tam n will usually not be linear in Dyck n .

There is also an involution on Dyck paths that exchanges a path with its mirror image, that can be seen as a vertical symmetry on the drawings. More precisely, it reverses up and down steps and the order of the steps.

Remark 1.4. This involution on Dyck paths is an order-preserving involution on the Dyck lattice. However, it is neither an order-preserving nor an orderreversing involution on the Tamari lattice.

2 Linear intervals in the Tamari lattices

Structure of linear intervals

Let us define two particular intervals L n and R n in Tam n+1 , for n ≥ 2. We will see that they are actually the only linear new intervals in Tam n+1 and also its linear intervals of greatest height, namely n.

The interval R n has the right comb r n+1 as bottom element and its top element is obtained from r n+1 by performing one rotation at each node of the right side, from top to bottom. In other words, the top element of R n is the tree whose root node has a leaf as right son and a right comb r n as left son.

The interval L n is the mirrored version of R n . Its top element is the left comb n+1 and its bottom element is the tree whose root node has a leaf as left son and a left comb n as right son.

Remark 2.1. For n ≥ 2, both intervals L n and R n are linear of height n since there is only one possible way to perform a right rotation to go down in R n at every step (or a left rotation to go up in L n ).

Moreover, they are new since their top and bottom element do not share any common node as explained in remark 1.2. Indeed, all nodes in the bottom element of L n have exactly one leaf strictly on their left whereas all nodes in the top element of L n have no leaves strictly on their left.

We can also define L 1 = R 1 as the interval whose bottom element is r 2 and whose top element is 2 . It is linear of height 1 and new. We now prove that there are four kinds of linear intervals in the Tamari lattice, namely trivial intervals, covering relations and "left" or "right" intervals. Left (resp. right) intervals are defined by first grafting trivial intervals on the leaves of an interval L n (resp. R n ) for some n ≥ 2 and then grafting the result on some leaf of a trivial interval. As an interval, it is the product of trivial intervals and one linear interval, and thus, it is linear. It remains to prove that all linear intervals are of this form. Proof. We prove the result for k = 2 and then by induction. Suppose we have a linear interval of height 2, that is of the form P Q Q . Then we know that P Q is a covering relation at some node s as in fig. 5, for some trees A, B, C, D. More precisely, P (resp. Q) can be described as a comb r 2 (resp. 2 ) grafted on the k-th leaf of a tree A, with the trees B, C, D grafted from left to right on the leaves of r 2 (resp. 2 ).

≤

B C D A 1 C 2 C 3 ≤ B C D A 1 C 2 C 3
Let us study the next possible rotations for the covering relation Q Q . As we can see in fig. 8, if we perform a rotation within A, B, C or D, then we get a square, thus, not a linear interval. More precisely, if the next rotation Q Q is such that Q can be described as a comb 2 grafted on the k-th leaf of a tree A , with the trees B , C , D grafted from left to right on the leaves of 2 , then we can define a tree P as Q where the left comb 2 is replaced by a right comb r 2 . Then we have P P Q and [P, Q ] is not linear. This being excluded, there remains at most three possible nodes for a rotation as in fig. 9, namely the same node s, its left successor t (if any), and its predecessor u (if s is a right son).

If we perform another rotation at the node s, we get a pentagon as in fig. 10, thus not a linear interval.

Otherwise, a rotation at the node t produces a left interval of height 2 as in fig. 6 and a rotation at the node u produces a right interval of height 2 similar to the one in fig. 7, both being linear intervals.

Let now [P, Q] be a linear interval of height k ≥ 3. Let Q be the lower cover of Q in [P, Q], such that [P, Q ] is a linear interval of height k -1 ≥ 2. By induction, it is either a left of a right interval.

Assume [P, Q ] is a left interval. Then, the last rotation in [P, Q ] happens at a node s as in fig. 6. The node s is not a right son, so the only possible rotation to get a linear interval is at its left successor t. Thus, [P, Q] is a left interval of height k.

Assume [P, Q ] is a right interval, and the last rotation happens at a node s, then a rotation to its left son (if possible) would produce a non linear interval as in fig. 10. The only possible rotation is thus at the predecessor of s and [P, Q] is a right interval of height k. We have proven that all linear intervals of height k ≥ 1 decompose as an interval L k or R k , grafted on some tree A with some trees grafted on the leaves of L k or R k . Thus, as long as A or one of the trees we grafted on L k or R k is a non trivial tree, we get an interval which is not new. Consequently, L n and R n are the only linear intervals that are new in Tam n+1 .

A' B' D' C' ≤ A' B' D' C' ≤ ≤ A B D C ≤ A B D C

Combinatorial description of linear intervals

We have described the structure of all linear intervals according to their height. We now deduce a combinatorial description of them in order to count them. A linear interval of height 0 is of the form [P, P ] where P is a tree not reduced to a leaf.

A linear interval of height 1 is a covering relation. As we can see in fig. 5, the part with A, B and D can be understood as a tree with a marked node s, together with another tree C that we plug into the right edge out of s for the bottom element and into the left edge out of s for the top element. Note that A, B, C and D might be just leaves.

Similarly, as in figs. 6 and 7, a linear interval of height k ≥ 2 can be understood as a tree formed of A, B and D, with a marked node s and a direction (left or right), together with a sequence of k trees C 1 , . . . , C k . Those k trees are then plugged into the selected edge going out of s or on a common branch then plugged into the other edge out of s.

This enables us to write down equations on the generating series of linear intervals. Let A(t) be the generating series of trees, including the degenerate case of the tree reduced to a leaf. The standard decomposition of the trees along the left side for instance gives the equation A = 1 + tA 2 , and the number of trees with n nodes is the Catalan number

C n = 1 n + 1 2n n
. Moreover, the generating series of trees with a marked node is equal to tA (t) and there are

nC n = n n + 1 2n n of them.
For k ≥ 0, let S k (t) be the generating series of linear intervals of height k. Since we did not consider the tree reduced to a leaf for the Tamari lattice, we have:

S 0 = A -1.
Then, a covering relation is given by a marked tree and another tree that we connect to it, which creates an additional node, so we can write:

S 1 = (tA )(tA) = t 2 A A.
Similarly, for k ≥ 2, we have the following equation:

S k = (2tA )(tA) k .
Here, the factor 2 is for the choice of the direction, and we plug k trees, creating one node for each of them.

Counting of linear intervals

In this section, we use Lagrange inversion [13, ch. 5] to get the coefficients of S k . Let us introduce the series B = A -1 of non trivial trees. We can then write B = t(B + 1) 2 = tF (B), where F (x) = (x + 1) 2 . Then, it follows that

B = (B + 1) 3 1 -B .
For k ≥ 2, we can write:

S 1 = t 2 B (B + 1) = t 2 φ 1 (B) and S k = 2t k+1 B (B + 1) k = 2t k+1 φ k (B),
where

φ k (x) = (1 + x) k+3 1 -x for k ≥ 1.
In what follows, we use the notation [t n ]S to denote the coefficient of degree n in the series S(t). As φ k (x) is a series in x with a constant term equals to 1 and B(t) is a series in t with no constant term, we have [t 0 ]φ k (B) = 1.

We can write 1 1 -x = k≥0

x k and 1 (1 -x) 2 = k≥0 (k + 1)x k . Let us compute [t n ]φ k (B), using Lagrange inversion, for n ≥ 1:

[t n ]φ k (B) = 1 n [x n-1 ]φ k (x)F (x) n = k + 3 n [x n-1 ] (1 + x) k+2+2n 1 -x + 1 n [x n-1 ] (1 + x) k+3+2n (1 -x) 2 = k + 3 n n-1 j=0 k + 2 + 2n j + 1 n n-1 j=0 k + 3 + 2n j (n -j -1 + 1).
After a few lines of computation, we obtain that [

t n ]φ k (B) = k + 2 + 2n n .
We can notice that this formula is still true for n = 0.

Theorem 2.4. In the Tamari lattice Tam n of size n, there are:

• 1 n + 1 2n n linear intervals of height 0, • 2n -1 n -2 linear intervals of height 1, • 2 2n -k n -k -1 linear intervals of height k, for 2 ≤ k < n.
Furthermore, there are no linear interval of height k ≥ n.

This adds up to

1 n + 1 2n n + 2n -1 n -2 + 2 2n -1 n + 2 linear intervals in Tam n .
Proof. For the intervals of height 0, this is the number of elements in Tam n . As stated in remark 2.3, we already know that any linear interval in Tam n is of height at most n -1. Thus, we can fix n > k > 0 and use the previous results to get the number of intervals of height k in Tam n .

We have [t n ]φ k (B) = k + 2 + 2n n . Thus, we get:

[t n ]t k+1 φ k (B) = 2n -k n -k -1 . Now, for k = 1, we have S 1 = t 2 φ 1 (B) thus there are 2n -1 n -2 intervals of height 1 in Tam n .
For 2 ≤ k < n, we have S k = 2t k+1 φ k (B), and there are 2 2n -k n -k -1 intervals of height k.

Finally, we have

n-1 k=2 2n -k n + 1 = n-3 k=0 n + 1 + k n + 1 = 2n -1 n + 2
. This can be proven combinatorially, as a particular case of the identity 3 Linear intervals in the Dyck lattice

Structure of linear intervals

As in the Tamari lattice, we will first define a family of "left" and "right" intervals that will be linear, then we will show that all linear intervals are either trivial, covering relations, left or right intervals. A left (resp. right) interval is an interval [P, Q] where the Dyck word Q is obtained from P by changing a subword

d k u into ud k (resp du k into u k d) for some k ≥ 2.
Such an interval is indeed linear of height k since at every step, there is only one valley that can be changed into a peak such that the path remains under Q. Note that left and right intervals are exchanged by the mirror involution on Dyck paths. Proof. We prove the result for k = 2 and then by induction.

Suppose we have a linear interval of height 2, that is of the form P Q R. Then the Dyck word Q is obtained from P by transforming a valley du into a peak ud. If the next covering relation uses the last step d of the peak of Q we just produced, then the valley of P was followed by an up step and R is obtained from P by changing a subword duu into uud. Thus, [P, R] is a right interval of height 2.

Similarly, if the next covering relation uses the first step u of the peak of Q we just produced, then the valley of P was preceded by a down step and R is obtained from P by changing a subword ddu into udd. Thus, [P, R] is a left interval of height 2.

If the next covering relation happens at a valley somewhere else in Q, then this valley exists in P and the two covering relations can be performed independently, thus [P, R] would be a square as shown in fig. 12 and not a linear interval. If [P, Q ] is a left interval, then Q is obtained from P by changing a subword d k u into ud k . Furthermore, in the chain from P to Q , the peak created in Q is formed by the two first steps of this subword ud k , which is not followed by an up step in Q . Thus, the covering relation Q Q has to use this first up step u of this subword ud k , as otherwise the interval would not be linear. Then Q is obtained from P by changing a subword d k+1 u into ud k+1 and [P, Q] is a left interval of height k + 1.

d u d u ≤ d u d u ≤ ≤ d u d u ≤ d u d u
If [P, Q ] is a right interval, then Q is obtained from P by changing a subword du k into u k d. Symmetrically, the covering relation Q Q has to use this last down step d and we get that [P, Q] is a right interval of height k + 1.

Combinatorial description of linear intervals

We have described the structure of all linear intervals according to their height. As for the Tamari lattice, we will give a combinatorial description of them in order to write some equations. In fact, we will produce the very same equations as for the Tamari lattice, and thus we will prove that the two lattices have the same number of linear intervals of any fixed height.

For k ≥ 0, let T k (t) be the generating series of linear intervals of height k in the Dyck lattices. Proposition 3.2. For any k ≥ 0, the generating series T k (t) of linear intervals of height k in the Dyck lattices is equal to the generating series S k (t) of linear intervals of height k in the Tamari lattices.

Proof. First, let recall that the Dyck paths of size n are in bijection with the rooted planar binary trees of size n. Thus, the generating series A(t) of binary trees is also the generating series of Dyck paths. Similarly, as every Dyck path of size n has n down (resp. up) steps, the generating series of Dyck paths marked at a down (resp. up) step is equal to tA (t).

As in the Tamari lattice, a linear interval of height 0 is of the form [P, P ] with P a non trivial Dyck path. Thus, we can write:

T 0 = A -1.
Let [P, Q] be a linear interval of height 1, i.e. a covering relation. Then Q is obtained from P by changing a valley du into a peak ud. Let uDd be the excursion whose first step is the up step of this valley of P . Then D as a word is a Dyck word.

In fact, as we can see in fig. 13, any covering relation P Q can be understood as a path P with a marked down step #, before which we insert duD for the bottom element P and udD for the top element Q. We have: Thanks to this, for any k ≥ 2, we finally have:

T 1 = tA tA = t 2 A A.
T k = tA (tA) k + tA (tA) k = 2t k+1 A A k .
We have thus proven that for any k ≥ 0, S k and T k satisfy the same equation, and are indeed equal.

As a corollary we have the following result: Theorem 3.3. For any n ≥ 1 and k ≥ 0, the Tamari lattice Tam n and the Dyck lattice Dyck n have the same number of linear intervals of height k.

Linear intervals in the alt-Tamari posets

One can define a family of posets on the set of Dyck paths Z n for every n ≥ 1, which includes the Tamari and the Dyck lattices that we introduced before. We call them the alt-Tamari posets. More precisely, they depend on an increment function δ : {1, . . . , n} → {0, 1}, and will be denoted Tam δ n . We will have a result of refinement whenever two functions δ and δ are comparable. Moreover, in all these posets, we can define trivial intervals, covering relations, left and right intervals and prove that all linear intervals are of this kind. We can then give a combinatorial description of these intervals. It follows that we have a bijection between linear intervals of any two alt-Tamari posets of the same size which preserves the height. This proves the main result of this part, namely the theorem 0.1 which states that all these posets have the same number of linear intervals, even when distinguished according to their height.

Definition of the alt-Tamari posets

Given a Dyck path of size n, we will number its up steps with integers {1, . . . , n} increasingly from left to right. For example, the path uududdud will be numbered u 1 u 2 du 3 ddu 4 d.

Fix some n ≥ 1. Let δ : {1, . . . , n} → {0, 1} be an increment function. We introduce a notion of δ-altitude, where the function δ encodes that the up step u i increases the δ-altitude by δ(i), while all down steps make the δ-altitude decrease by 1. Thus, we write δ(d) = -1 and δ(u i ) = δ(i).

Let P be a Dyck path of size n and A be a subword of P . We define the δ-elevation δ(A) = s∈A δ(s) as the change of δ-altitude of A.

We define the δ-excursion of an up step u i of P as the smallest subword C i of P starting with u i such that δ(C i ) = 0.

Remark 4.1. The δ-excursion is well defined since the excursion E i as defined in section 1.4 starting at the up step

u i satisfies δ(E i ) ≤ 0. Moreover, C i is always a prefix of E i .
For instance, if δ(i) = 0, the δ-excursion of u i is reduced to u i . If δ(i) = 1 for all i, then the δ-excursion of an up step u i is exactly the excursion starting at u i .

Given an increment function δ and a Dyck path P with a valley du i , we define the δ-rotation of P at the up step u i as the Dyck path Q obtained from P by exchanging the down step that precedes u i with the δ-excursion of u i . In other words, if C i is the δ-excursion of u i , we can write P = AdC i B and Q = AC i dB. Remark 4.2. If Q is the δ-rotation of P at the up step u i , then Q is strictly greater than P in the Dyck lattice.

Proof. Let C i be the δ-excursion of u i in P. We can write P = AdC i B as a word and then we have Q = AC i dB. The Dyck path Q is obtained from P by moving the down step d that precedes u i to the right. This can be achieved as a sequence of covering relations in the Dyck lattice. Indeed, when moving this down step d letter by letter to the right, either it is exchanged with an up step u k and it is a covering relation in the Dyck lattice or it is exchanged with another down step and the Dyck path is unchanged. This proves that there are no cycles of δ-rotations. We can thus define the alt-Tamari poset Tam δ n as the reflexive transitive closure of δ-rotations on the set Z n of Dyck paths of length n. • The first value δ(1) does not matter in the definition of the poset since u 1 is never in a valley.

• The last value δ(n) does not matter either in the definition. Indeed, the δ-excursion of the last step u n can be either u n or u n d and in both cases, a δ-rotation at the last up step will always change du n d into u n dd.

Let us now prove that there is a boolean structure of refinements of the alt-Tamari posets. More precisely, given two increment functions δ and δ of size n, whenever we have δ ≤ δ component-wise, then Tam δ n refines Tam δ n . This means that whenever we have two Dyck paths P and Q such that P ≤ Q in Tam δ n , then P ≤ Q in Tam δ n .

Lemma 4.9. Let P be a Dyck path of size n and δ ≤ δ two increment functions. The δ-excursion E i of u i in P is a prefix of the δ -excursion E i of u i .

Proof. Let w be any strict prefix of E i . As E i is a δ-excursion, we have δ(w) > 0.

Then, as δ ≥ δ, we also have δ (w) > 0. Similarly, δ

(E i ) ≥ δ(E i ) = 0.
As the alt-Tamari lattice Tam δ n is defined as the transitive reflexive closure of δ -rotations, it is sufficient to prove that any δ -rotation P Q defines an interval [P, Q] in Tam δ n .

Proposition 4.10. Let δ ≤ δ be two increment functions of size n. Let P Q be a covering relation in Tam δ n . We have P < Q in Tam δ n .

Proof. Using lemmas 4.4 and 4.9, we can build a chain from P to Q in Tam δ n , just as we did to prove remark 4.2, namely that the Dyck lattice refines all alt-Tamari posets.

We write P = AdC i B and Q = AC i dB with C i the δ -excursion of u i . We will exchange this down step d with down steps of δ-excursions until we reach Q and this will prove the result.

The δ-excursion C i of u i in P is a prefix of C i = C i D. Letting P 1 = AC i dDB, we have P P 1 in Tam δ n . Now, either D starts with a down step and exchanging the two down steps does not change the path or D starts with an up step u j . In this second case, let C j be the δ-excursion of u j in P . Then C j is a prefix of the δ -excursion C j of u j in P . Now, C j and C i are two δ -excursions whose intersection is not empty. Lemma 4.4 ensures that C j is thus a prefix of D and the same holds for C j . Then exchanging d and C j is a δ-rotation.

By induction on the length of D, we build a chain from P to Q in Tam δ n .

Structure of linear intervals

Now we study the linear intervals in the alt-Tamari posets. As in the Tamari lattice and the Dyck lattice, we define left and right intervals, which we prove to be linear. Then we prove that all linear intervals of height k ≥ 2 are either left or right intervals. Lastly, we give a combinatorial description and deduce the same equations as previously.

Lemma 4.11. Let δ be an increment function. Let P Q be a covering relation. We can write P = AdC i B and Q = AC i dB with C i the δ-excursion of u i . There are at most two covering relations Q Q such that [P, Q ] is a linear interval. Precisely, if this second covering relation is not occurring at the valley du i if A ends with a down step d or at the valley du k if B starts with the up step u k , then the interval [P, Q ] is not linear.

Proof. Let Q be the δ-rotation of Q at the up step u j with u j = u i and u j = u k if B starts with the up step u k .

We want to prove that [P, Q ] is not linear. We have one maximal chain P Q Q from P to Q . We will show that there is a different maximal chain from P to Q .

Case 1: Suppose that u j ∈ C i . Then, we can write Q = AC i dB and if P is the δ-rotation of P at u j , we have P = AdC i B, and we have another maximal chain P P Q from P to Q . Thus, [P, Q] is a non linear interval as shown in fig. 16.

d C i ≤ C i d ≤ ≤ C i , d ≤ C i , d
Figure 16: A square in an alt-Tamari poset.

Case 2: Suppose now that u j / ∈ C i . Let C j be the δ-excursion of u j in P . Then, two more cases occur. Either C j is directly followed by C i in P or not. Case 2.1: Suppose that C j = Ed and P = A dEdC i B as in the first case of lemma 4.5. Then, we can write Q = A dEC i dB and Q = A EC i ddB. Set P = A EddC i B and P = A EdC i dB and we have another maximal chain P P P Q from P to Q as we can see in fig. 17. Case 2.2: Suppose that C j is not directly followed by C i in P . Then the δ-excursion of u j in Q is still C j and we can write Q = A C i dB after the δ-rotation at u j . Let P = A dC i B be the δ-rotation of P at u j . We have again another maximal chain P P Q from P to Q and thus, a square similar to the one in fig. 16.

d d C E j Ci ≤ d d E Ci Cj , ≤ Ci d d Cj , E ≤ ≤ d Cj E d Ci ≤ Ci d d E Cj
k and a Dyck path marked at a down (resp. up) step and a sequence of k Dyck paths (possibly trivial). The same decomposition works also for intervals of height k = 1, that is to say, they are in bijection with a pair of Dyck paths, the first being marked at a down step. We start with this case.

In all what follows, we will use the notion of excursions as defined in section 1.4, not to be confused with δ-excursions of an up step as defined in section 4.1, which is always a prefix of the excursion of this up step (remark 4.1). Proposition 4.16. Let δ be an increment function. There is a bijection between covering relations in Tam δ n and pairs (P 0 , P 1 ) of Dyck paths, where P 0 is marked at a down step, and the lengths of P 0 and P 1 add up to n -1.

Proof. Let [P, Q] be a covering relation in Tam δ n . We can write P = AdC i B and Q = AC i dB, where C i is the δ-excursion of the up step u i .

Let E be the excursion of u i as defined in section 1.4. We can write E = C i D with D a prefix of B = DB . Remark that E starts with u i and ends with the down step matching with u i . Thus, we can write E = u i E d with E a possibly empty Dyck path. Now, we can set P 0 = AdB and P 1 = E , so that P 0 is a Dyck path marked at the down step between A and B and P 1 is a possibly empty Dyck path. The lengths of P 0 and P 1 add up to n -1.

Let us prove that it is a bijection. Let P 0 be a Dyck path of length n 0 ≥ 1, marked at a down step, so that we can write P 0 = AdB with d the marked down step of P 0 . Let P 1 be a Dyck path of length n 1 ≥ 0. Let n = n 0 + n 1 + 1 and δ be an increment function of size n.

Suppose that there are j up steps in A and let i = j + 1. We will construct P (resp. Q) by inserting a Dyck path into P 0 between d and B (resp. between A and B ), so that its first up step will become the i-th up step of P and Q.

Let E = u i P 1 d, where the up steps of P 1 are relabelled starting with u i+1 . Obviously, E is the excursion of u i .

Let then C i be the δ-excursion of u i in E, where u i is considered to be the i-th up step. This is possible because a δ-excursion is always a prefix of the excursion. We can write E = C i D and B = DB .

Then, setting P = AdC i B and Q = AC i dB, we have constructed a covering relation P Q in Tam δ n . It is clear that this is the reciprocal of the decomposition described above and thus, this is a bijection. Moreover, we have P = AdEB and this writing does not depend on δ.

Corollary 4.17. Let δ and δ be two increment functions of the same size n. There is a bijection between covering relations in Tam δ n and Tam δ n . Moreover, this bijection preserves the bottom elements of the covering relations.

In particular, there are 2n -1 n -2 covering relations in Tam δ n for any δ. There is a bijection between left intervals of height k in Tam δ n and sequences (P 0 , P 1 , . . . , P k ) of Dyck paths, where P 0 is marked at an up step, and the lengths of P 0 , . . . , P k add up to n -k.

Proof. Let [P, Q] be a left interval of height k ≥ 2 in Tam δ n . By definition of left intervals, we can write P = Ad k C i B and Q = AC i d k B, where C i is the δ-excursion of the up step u i .

Let u i1 , . . . , u i k be the up steps of A matching with the k down steps of P between A and B. Then we can write P = A u i1 P 1 u i2 . . . u i k P k d k C i B, where P 1 , . . . , P k are k possibly trivial Dyck paths. We then set P 0 = A C i B, and it is a Dyck path marked at the first step of C i . It is clear that the total number of up steps in P 0 , . . . , P k is n -k since all up steps of P except u i1 , . . . , u i k are in exactly one of these Dyck paths P 0 , . . . , P k .

Let us prove that this decomposition is bijective by constructing its reciprocal.

Let P 0 be a Dyck path of length n 0 ≥ 1, marked at an up step, so that we can write P 0 = A uB with u the marked up step of P 0 . Let P 1 , . . . , P k be k Dyck paths of respective lengths n 1 , . . . , n k ≥ 0. Let n = k i=0 n i + k and δ be an increment function of size n.

We will build P and Q by inserting these k paths and k additional pairs of up and down steps between A and B . Let i 1 -1 be the number of up steps in A and for 2 ≤ j ≤ k let i j = i j-1 + n j-1 + 1.

Then, we can set A = A u i1 P 1 u i2 . . . u i k P k . We set as well P = Ad k uB and this is by construction a Dyck path of length n. Remark that the construction of P does not depend on δ.

Let now C be the δ-excursion of the up step u preceding B in P . We can write uB = CB so that P = Ad k CB. Setting finally Q = ACd k B, we obtain a left interval [P, Q] of height k. Again, this is clearly the reciprocal of the decomposition of left intervals defined above. There is a bijection between right intervals of height k in Tam δ n and sequences (P 0 , P 1 , . . . , P k ) of Dyck paths, where P 0 is marked at a down step, and the lengths of P 0 , . . . , P k add up to n -k. Let u i1 , . . . , u i k be the steps with which C 1 , . . . , C k start respectively. Let d 1 , . . . , d k be the down steps matching with u i1 , . . . , u i k .

Let D k be the part of the path P starting just after C k and ending just after d k . Then, E k = C k D k starts with u i k and ends with its matching down step d k , so E k is the excursion of u i k . In particular, if E = C k then D k is empty.

For k > j ≥ 1, let D j be the part of P starting just after D j+1 (which may be empty) and ending just after d j . As previously, the excursion of u ij in P is either E j = C j and D j is empty or it is E j = C j E j-1 D j and D j is not empty. Now, we can write P = AdC 1 . . . C k D k . . . D 1 B . Set P 0 = AdB , marked at the down step between A and B and C j D j = u ij P j d j for 1 ≤ j ≤ k and we have the decomposition as stated.

We now prove that this decomposition is bijective by constructing its reciprocal.

  n

Figure 2 :

 2 Figure 2: Plugging a tree into the selected (red dotted) edge of another one.

Figure 4 :

 4 Figure 4: The intervals R 3 with all 4 elements (top) and L 4 (bottom).

Figure 5 :Figure 6 :

 56 Figure 5: A covering relation.

Figure 7 :

 7 Figure 7: A right interval of height 3.

Figure 8 :Figure 9 :

 89 Figure 8: Non linear interval.

Remark 2 . 3 .Figure 10 :

 2310 Figure 10: Two consecutive rotations at the same node produce a non linear interval.

Proposition 3 . 1 .

 31 All linear intervals of height k ≥ 2 in the Dyck lattice are either left or right intervals.

Figure 11 :

 11 Figure 11: A covering relation in the Dyck lattice.

Figure 12 :

 12 Figure 12: A square in the Dyck lattice.

Figure 13 :

 13 Figure 13: Decomposition of a covering relation in the Dyck lattice.

3 DFigure 14 :

 314 Figure 14: Decomposition of a right interval of height 3 in the Dyck lattice.

3 Figure 15 :

 315 Figure 15: The δ-excursion C 3 of u 3 on a Dyck path for δ = (0, 1, 1, 1, 0, 1, 0).

Figure 17 :

 17 Figure 17: A pentagon in an alt-Tamari poset.

Proposition 4 .

 4 18. Let δ be an increment function and k ≥ 2.

Corollary 4 .

 4 19. Let δ and δ be two increment functions of the same size n and k ≥ 2. There is a bijection between left intervals of height k in Tam δ n and Tam δ n . Moreover, this bijection preserves the bottom element of the intervals. In particular, there are 2n -k n + 1 left intervals of height k in Tam δ n for any δ.Proposition 4.20. Let δ be an increment function and k ≥ 2.

  Proof. Let [P, Q] be a right interval of height k ≥ 2 in Tam δ n . By definition of right intervals, we can write P = AdC 1 . . . C k B and Q = AC 1 . . . C k dB with k δ-excursions C 1 , . . . , C k .

• We recover the Tamari lattice when the increment function δ is such that δ(i) = 1 for all i, since the δ-excursion of any up step is always its excursion.

• We recover the Dyck lattice when the increment function δ is such that δ(i) = 0 for all i, since the δ-excursion of any up step is reduced to the up step itself.

Lemma 4.4. Let C i be the δ-excursion of u i in a Dyck path P . Let C j be the δ-excursion of u j in P with i = j.

Either C i and C j are disjoint as subwords and we write C i ∩ C j = ∅ or one is included in the other. Moreover, they do not end at the same step.

Proof. Suppose i < j. Suppose that C i and C j are not disjoint. Then u j is a step of C i .

Write C i = Au j B. Then as C i is the δ-excursion of u i , all its strict prefixes w satisfy δ(w) > 0 and so we have δ(A) > 0.

Moreover, as C j is the δ-excursion of u j , we have δ(W ) ≥ 0 for any prefix of C j . Hence, δ(AW ) = δ(A) + δ(W ) > 0.

It follows that AC j is a prefix of C i , hence C j is included in C i . We can finally notice that C i and C j do not end at the same step, for the same reason.

Let P be a Dyck path of size n and δ an increment function. For 1 ≤ i ≤ n we define h i (P ) as the position of u i in P and h(P ) = (h 1 (P ), . . . , h n (P )). For instance, if P = u 1 u 2 du 3 ddu 4 d then h(P ) = [START_REF] Bergeron | Higher trivariate diagonal harmonics via generalized Tamari posets[END_REF][START_REF] Bernardi | Intervals in Catalan lattices and realizers of triangulations[END_REF][START_REF] Chapoton | Sur le nombre d'intervalles dans les treillis de Tamari[END_REF][START_REF] Huang | Problems of associativity: A simple proof for the lattice property of systems ordered by a semi-associative law[END_REF].

For 1 ≤ i ≤ n, we define i (P ; δ) as the length of the δ-excursion of u i in P and (P ; δ) = ( 1 (P ; δ), . . . , n (P ; δ)). For instance, if P = u 1 u 2 du 3 ddu 4 d and δ(i) = 1 for all i, then (P ; δ) = (6, 2, 2, 2).

On the Dyck path P of fig. 15 we have h(P ) = (1, 2, 5, 6, 8, 9, 12) and (P ; δ) = (1, 2, 7, 2, 1, 2, 1). Lemma 4.5. Let P be a Dyck path with a valley du i and C i be the δ-excursion of u i . Let Q be the result of the δ-rotation of P at u i .

For all j such that u j ∈ C i , we have h j (Q) = h j (P ) -1 and for all other j, h j (Q) = h j (P ).

Furthermore, if there exists some j such that the δ-excursion of u j in P ends with the down step d of the valley du i , then j (Q; δ) = j (P ; δ) + i (P ; δ). For all other j, we have j (Q; δ) = j (P ; δ).

Proof. The statement about the h j (Q) is immediate by definition of δ-rotations.

Suppose that there exists j such that the δ-excursion C j of the up step u j in P ends with the down step of the valley du i . Write C j = Bd, and P = ABdC i D.

The equality C j = Bd implies δ(B) = 1 and for all non empty prefixes w of B, we have δ(w) > 0. Similarly, for all prefixes w of C i , we have δ(w ) ≥ 0. So for all prefixes w of BC i , we have δ(w ) > 0 and δ(BC i d) = δ(BdC i ) = 0.

Thus, BC i d is the δ-excursion of u j in Q and we have j (Q; δ) = j (P ; δ) + i (P ; δ). Suppose that j is such that the δ-excursion C j of the up step u j in P does not end with the down step of the valley du i .

If

If C i ⊆ C j then we write C j = AdC i B in P . The δ-excursion of u j in Q will then be AC i dB and its length does not change either. This proves in particular that for all Q ≥ P in Tam δ n , h(Q) ≤ h(P ) and (Q; δ) ≥ (P ; δ) component-wise.

Remark 4.6. One can wander if we have the converse implication, which would give a characterization of the alt-Tamari order. This converse implication holds for the Dyck and the Tamari lattices.

Lemma 4.7. Let P a Dyck path such that u j is in the δ-excursion of u i .

For all

Proof. It is sufficient to prove this for all covering relations of P . By definition, u i is in the δ-excursion of u j in P , if and only if h i (P ) -h j (P ) ≤ j (P ). Suppose Q is an upper cover of P . Because of lemma 4.5, we have either

In all cases, we have

The last inequality is again guaranteed by lemma 4.5. Finally, the inequality Proof. As the poset is defined as the reflexive transitive closure of δ-rotations, all covering relations are δ-rotations. We have to prove the converse, namely that no δ-rotation can be achieved as a non trivial sequence of δ-rotations.

Suppose Q is the δ-rotation of P at the up step u i , and C i is the δ-excursion of u i in P . We write P = AdC i B and Q = AC i dB.

Then, since A and B are unchanged, no δ-rotation is possible at an up step in A or in B.

Suppose

Then all δ-rotations of the sequence have to happen at steps in C i .

Because h i (Q) = h i (P ) -1 and u i is the only up step of C i that changes h i , then one of those δ-rotations has to happen at u i .

If the first δ-rotation happens at u j = u i , then h j (Q 1 ) = h j (P ) -1 = h j (Q). Thus, h j can not decrease any more. But then, as lemma 4.7 ensures that u j will always remain in the δ-excursion of u i in all the sequence, the δ-rotation at u i would make h j decrease by 1 and this is not possible. This proves that the first δ-rotation has to happen at u i and Q 1 = Q.

Therefore, P ≤ Q is the only chain from P to Q and thus it is a covering relation.

In all these cases, the interval [P, Q ] is not linear. Definition 1. We say that an interval [P, Q] is a left interval if we can write

We say that [P, Q] is a right interval if we have

Remark that in both cases, for k = 1, these would simply describe a covering relation.

Proposition 4.12. A left interval is linear and the k that appears in definition 1 is its height.

Proof. Let [P, Q] be a left interval. We can write

First, we clearly have a maximal chain of length k from P to Q, namely

We want to prove that it is the unique maximal chain from P to Q.

As in the proof of proposition 4.8, for all u j / ∈ C i , h j (Q) = h j (P ) so in any chain from P to Q, only δ-rotations at up steps of C i are possible. Moreover, for all u j ∈ C i , h j (Q) = h j (P ) -k. Remark than only δ-rotations at u i can make h(Q ) decrease by 1 for any Q ∈ [P, Q], so that any maximal chain from P to Q contains k δ-rotations at u i . Furthermore, because of lemma 4.7, for all u j ∈ C i , each of these δ-rotation at u i will make h j (Q ) decrease by 1 for all Q ≥ P . Thus, any maximal chain from P to Q must contain exactly k δ-rotations at u i and no other δ-rotations.

Thus, any left interval is linear. Again, we can write a maximal chain of length k from P to Q, namely P = P 0 P 1 • • • P k = Q, where P j = AC 1 . . . C j dC j+1 . . . C k B. We prove by induction on k that it is the unique maximal chain from P to Q. The case k = 1 is proven in proposition 4.8.

For all up steps u j in one of these k δ-excursions, h j (Q) = h j (P ) -1 and for all other u j , h j (Q) = h j (P ). Furthermore, we also have j (Q; δ) = j (P ; δ).

Let u ij be the first step of C j . Because of lemma 4.7, for all Q ≥ P , any δrotation that moves u ij will also move all up steps of C j . Hence, in any maximal chain from P to Q, the only possible δ-rotations happen at the steps u ij .

Moreover, for all 1 ≤ j ≤ k-1, C j and C j+1 are two consecutive δ-excursions in P so a δ-rotation at u ij+1 would change ij (P ; δ) into ij (P ; δ) + ij+1 (P ; δ), as stated in lemma 4.5. Thus, in any maximal chain from P to Q, the first δ-rotation must happen at u i1 .

Then, [P 1 , Q] is either a covering relation if k = 2 or a right interval with k -1 δ-excursions otherwise. By induction, [P 1 , Q] is a linear interval of height k-1 and thus, [P, Q] is an interval of height k since it contains a unique maximal chain of length k.

It follows that any right interval is linear. Proof. We have two chains P AdEC 2 dB Q and P AC 1 dC 2 dB AC 1 C 2 ddB Q from P to Q. This is exactly the situation of fig. 17.

Proposition 4.15. In the alt-Tamari poset Tam δ n , all linear intervals of height k ≥ 2 are either right or left intervals.

Proof. We already know thanks to lemma 4.11 that there are only two kinds of linear intervals of height 2, and they are precisely left and right intervals. We will prove the result by induction on height.

Let [P, Q] be a linear interval of height k

As the last rotation in [P, Q ] occurs at u i and C i is followed by at least two down steps in Q , the lemma 4.11 assures that the only possible δ-rotation Q Q such that [P, Q] is a linear interval is again at the up step u i . This produces a left interval of height k + 1. The lemma 4.14 shows that a δ-rotation at the first step of C k would produce a non-linear interval. Hence, the only possible δ-rotation Q Q happens at the first step of B and this produces a right interval of height k + 1.

Combinatorial description and counting

We have described the structure of all linear intervals of the alt-Tamari posets, according to their height. We can adapt the combinatorial description that we gave for the Dyck lattice, and produce again a decomposition which generalizes the cases of the Dyck and the Tamari lattices. This proves the main result of this section, namely that all the alt-Tamari posets share the same number of linear intervals of any height.

As all the alt-Tamari posets are defined on Dyck paths, it is immediate that they all have the same number of trivial intervals, or in other words, (linear) intervals of height 0. Furthermore, as a covering relation can be described as a Dyck path with a marked valley, it follows that every Dyck path has the same number of upper covers in every alt-Tamari posets. Thus, the total number of covering relations of all alt-Tamari posets is the same, namely the number of linear intervals of height 1.

In this section, we will prove that for any k ≥ 2, all these posets have the same number of right (resp. left) intervals of height k, namely 2n -k n + 1 . This will be proven through a bijection between a right (resp. left) interval of height Let P 0 be a Dyck path of length n 0 ≥ 1, marked at a down step, so that we can write P 0 = AdB with d the marked up step of P 0 . Let P 1 , . . . , P k be k Dyck paths of respective lengths n 1 , . . . , n k ≥ 0. Let n = k i=0 n i + k and δ be an increment function of size n.

We will build P and Q by inserting these k paths one by one and k additional pairs of up and down steps between A and B .

Let i 1 -1 be the number of up steps in A. Let C 1 be the δ-excursion of the first step of uP 1 d where the steps are relabelled starting with i 1 . For 2 ≤ j ≤ k, let i j -1 be the number of up steps in AC 1 . . . C j-1 and C j be the δ-excursion of uP j d where the steps are relabelled starting with i j . Thus, for all 1 ≤ j ≤ k, we can write uP j d = C j D j . Now, let P = AdC

A Table

Here are the first few coefficients of the series S k . The total sequence can be found on the OEIS [5, A344136].


Stephane Segonds

Guillaume Cohen

Yann Landon

Frederic Monies

Pierre Lagarrigue

Characterising the behaviour of workpieces under the effect of tangential cutting force during NC turning Application to machining of slender workpieces

Keywords: NC turning, Flexure, Slender workpieces, Compensation

Turning slender workpieces usually means a tailstock centre has to be used to in order to achieve geometric tolerances for cylindricality. Indeed, tight tolerances mean the grinding phase can be performed quicker or even rendered superfluous. To limit use of the tailstock centre, that limits tool access when producing slender workpieces using turning, knowledge of the spindle-chuck-workpiece assembly's behaviour is essential. It is proposed to characterise this assembly's behaviour under the action of machining loads in order to introduce a method that in some cases will dispense with the need to use the tailstock centre while keeping the geometric quality of the machined workpiece.

Introduction

Turning slender workpieces generally means a tailstock centre has to be used to limit workpiece flexure during machining. Jig boring has to be implemented first at one end of the workpiece, possibly over a machining oversize that will have to be removed later. Using the tailstock centre thus leads firstly to additional operations to prepare workpieces with the increased production cost this entails. Secondly, its presence limits access to one end of the workpiece and can prevent certain machining operations [START_REF] Anselmetti | Optimisation of a workpiece considering production requirements[END_REF].

For some types of machining, the workpiece-machine link provided by the chuck alone is incapable of sustaining the loads due to cutting and, in this case, using a tailstock centre remains indispensable. However, in many instances of turning, using a tailstock centre mainly allows the workpiece-machine assembly to be rigidified to allow the tolerances imposed by the Design Office to be respected, especially in terms of surface condition and geometric tolerance [START_REF] Anselmetti | Optimisation of a workpiece considering production requirements[END_REF]. Machining a slender workpiece by turning with the workpiece being held by jaws on one side only is thus a delicate matter, both in the domains of strong and * weak cutting conditions. Indeed, machining operations implemented under blank cutting conditions and significant loads lead to workpiece flexures that adversely affect the geometric quality of the workpiece. These geometric defects can be significant enough to prevent a grinding phase being implemented directly. However, machining operations implemented in finishing cutting conditions do not allow surface conditions as determined by predictive models to be obtained, this being a function of the feed rate and the tool corner radius [START_REF] Lin | Modeling the surface roughness and cutting force for turning[END_REF]. Indeed, the workpiece's low rigidity and its assembly often cause vibrations that interfere with machining [START_REF] Lee | Design of a dynamic vibration absorber against vibrations in turning operations[END_REF][START_REF] Anselmetti | Optimisation of a workpiece considering production requirements[END_REF].

In most cases, a setting workpiece is machined in order to determine the optimum values for tool offsets allowing machining defects to be compensated for as best as possible [START_REF] Sandvik | Techniques modernes d'usinage[END_REF]. However, this method involves a major drawback, as it does not allow for due consideration of the fact that the amplitude of the defect generated by the workpiece flexure phenomenon is a function of the co-ordinates of the variable point machined. The correction made to the tool offset is limited to a mean defect correction. One alternative would involve machining with conditions generating significant tangential cutting force to limit vibrations of the workpiece and keep a good surface condition while following a path that compensates for workpiece flexure due to loads. This choice should allow us to respect the geometric specifications of the workpiece to be machined.

The goal of the present study is thus to characterise the behaviour of the workpiece and the workpiece-machine link in the case of turning for slender workpieces with held on one side only so as to be able to compensate for movements and deformations of workpieces during machining without a tailstock centre.

To do so, one can kit the machine up with instruments to measure workpiece flexure and process these data in real time within the control unit. However, this method cannot reasonably be considered in an industrial context [START_REF] Chih-Hao | An application of real-time error compensation on a turning center[END_REF]. The turning workpiece flexure phenomenon is taken into account by inserting offsets in the machining programme in order to best compensate the defect induced by workpiece flexure.

To model behaviour of the workpiece during machining, first a workpiece is machined so as to test validity of the model derived from the literature [START_REF] Dessein | Qualification et optimisation de la précision d'une machineoutil à commande numérique[END_REF][START_REF] Mayer | Prediction of diameter errors in bar turning: a computationally effective model[END_REF][START_REF] Sandvik | Techniques modernes d'usinage[END_REF]. This model is commonly used to determine workpiece flexure during machining. Then, a new modelling is proposed for which the characteristics in relation to machining parameters are determined. To finish off, the validity of the proposed modelling is tested through additional machining operations.

The results derived from this modelling should allow for compensation of the defect due to workpiece flexure to be achieved at a low cost through performing early correction of the tool paths.

Experimental

Experimental procedure

A test workpiece (Fig. 1) was machined to implement measurements of flexure during machining without a tailstock centre. A measurement made during turning meant movement of the axis of seating B could be determined during machining of seating C.

The workpiece was produced on an NC lathe with power on the spindle of 11 kW and equipped with an NUM NC control unit. The chuck was equipped with three hard jaws and clamping pressure set to 5.5 bar for the entire duration of the test (clamping force: 1718 N). Flexure data acquisition was conducted using two Foucault current contact-free distance sensors with a resolution of 6.47 V/mm (measurement range of 3 mm) coupled with a memory multimeter whose maximum frequency of acquisition at full resolution is 40 Hz. The sensor was secured to the machine frame, then calibrated on the test workpiece. The position of the sensors on the Z-axis is noted "K" (Fig. 2) and the direction of measurement of the sensors corresponds to the directions of the lathe Y-and X-axis (directions of cutting load and radial cutting force). The machining loads were measured using an instrumented tool-holder with stress gauges calibrated at the start of the test campaign.

A comparator placed on the spindle allowed us to check that its movement remained within less than 3 m during application of a load of 1500 N. This meant the spindle flexure phenomenon could be neglected during this study [START_REF] Sandvik | Techniques modernes d'usinage[END_REF].

The workpiece was machined in a single phase from a 2017 T4 aluminium blank in accordance with the procedure defined in Fig. 2. The workpiece was held in the jaws throughout the length of seating A (L 1 = 30 mm).

The cutting conditions retained for the different operations are summarised in Fig. 2.

The tool used to produce the workpiece was installed in the tool-holder so as to limit the overhang of the active part of the tool and thus best limit the tool flexure phenomenon [START_REF] Sandvik | Techniques modernes d'usinage[END_REF]. Seating B was machined carefully during operations 1 and 2 (see Fig. 2) and served as a reference to measure workpiece flexure during machining. The sensors allowed the movement dx of the surface of seating B along the X-axis to be measured as well as movement dy of the surface of seating B along the Y-axis during machining of seating C. Movements dx and dy measured during operation 3 when producing the test workpiece are shown in Fig. 3.

Analysis of the distance sensor measurement results provides us with knowledge of the influence exerted by machining loads on flexure at the end of the workpiece. During machining of the test workpiece, the maximum movement of the workpiece exceeded 0.18 mm along the Y-axis. The tangential cutting forces measured during machining were: Tangential cutting force: F y = 1440 N. Longitudinal cutting force: F z = 521 N. Radial cutting force: F x = 388 N.

Analysis of results and application 2.2.1. Determining workpiece flexure from sensor measurements

The abscissa Z = 80 mm (Fig. 3) corresponds to the start of operation 3 of workpiece machining (tool at point I, Fig. 2). Up to the abscissa Z = 1 mm (Fig. 3) corresponding to the end of operation 3 (tool at point J, Fig. 2), workpiece flexure diminished gradually as the tool moved closer to the spindle.

In light of the measurements shown in Fig. 3, it would appear that the movement measured along the X-axis is relatively insignificant compared with the movement measured along the Y-axis. Furthermore, the radial cutting force tends to make the workpiece bend towards the tool. This is due to the positive geometry of the tool used.

During machining, the position sensors meant movements dx and dy along the X-and Y-axis of the reference framework (G 0 , X, Y) could be measured. Using these measurements, flexures f x and f y of the workpiece are determined in planes (X, Z) and (Y, Z) (Fig. 4). In a first approach, the amplitude of flexure of the tool is considered to be negligible in comparison with workpiece flexure.

The surface really generated considering deformations of the workpiece is a circle going through three points P 1 , P 2 and P 3 of co-ordinates (-R 0 , 0), (R 0 + dx, 0) and (0, R 0 + dy) whose centre G 1 of co-ordinates (a, b) in the reference framework (G 0 , X, Y) and radius R 1 verify the system of three equations with three unknowns below (Eq. ( 1)). Eq. ( 1) is a system of equations allowing the real surface machined to be defined. Solving this system means values for a and b, co-ordinates of the centre of the real machined surface and its radius R 1 can be determinated. The solution of the system of equations defining co-ordinates of the centre of the surface and the radius of the circle is presented below:

(-R 0 -a) 2 + b 2 = R 2 1 (R 0 + dx -a) 2 + b 2 = R 2 1 a 2 + (R 0 + dy -b) 2 = R 2 1 (1)
a = dx 2 b = dy 2 -R 0 × dx + 2R 0 × dy 2(R 0 + dy) R 1 = R 0 + dx 2 2 + dy 2 -R 0 × dx + 2R 0 × dy 2(R 0 + dy) 2 (2)
Fig. 3. Change in workpiece/sensor distances during machining of seating C.

According to the relations below (Eq. ( 3)), it is possible to determine the values for loads f x and f y as a function of movements dx and dy measured by the sensors and the radius R 0 of the programmed surface.

R 0 = f x + R 1 + a R 0 = R 1 -f y + b (3)
Fig. 4. Representation of theoretical and real surfaces machined during turning. The relations introduced above (Eq. ( 4)) lead to the expression of loads f x and f y , respectively, along the X-and Y-axis induced by machining loads.

f x = R 0 - R 0 + dx 2 2 + dy 2 -R 0 × dx + 2R 0 × dy 2(R 0 + dy) 2 - dx 2
f y = R 0 + dx 2 2 + dy 2 -R 0 × dx + 2R 0 × dy 2(R 0 + dy) 2 + dy 2 -2R 2 0 -R 0 × dx 2(R 0 + dy) (4)

Theoretical determination of the workpiece diameter

In order to validate the methodology for use of the results and the hypothesis whereby the amplitude of the tool flexure is negligible in comparison with workpiece flexure, it has been verified that the measurements dx and dy allow us to determine the real diameter of the machined workpiece. The theoretical radius R 1 of the machined workpiece is determined as a function of values dx and dy measured during the test (Eq. ( 2)).

Fig. 5 allows for a comparison of the results of diameter measurements made on the test workpiece and the theoretical results determined from values dx and dy measured during machining. The measurements for diameters of the workpiece with a given abscissa shown here are a mean for measurements calculated after a series of five measurements.

In the light of results from comparing the diameter measured on the workpiece and the diameter recalculated in accordance with movement measurements (Fig. 5), it appears that the maximum error committed on determination of the real diameter of the workpiece in accordance with the measurements made during the test does not exceed 3 m. It is thus possible to determine the diameter of the machined workpiece as a function of the abscissa Z from movement measurements during machining. Furthermore, this allows the assumption whereby tool flexure is negligible in comparison with workpiece flexure during machining to be validated.

Comparison with the model derived from the literature

To determine the workpiece flexure, the commonly accepted model [START_REF] Ramesh | Error compensation in machine tools-a review: Part I: geometric, cutting-force induced and fixturedependent errors[END_REF] is derived from the theory of strength of materials shown in Fig. 6. In what follows, this model is called model 1.

The diameter of the beam under consideration is that of the workpiece before the tool cut. Indeed, the cross-section under stress is that located between point O (centre of the workpiece-chuck restraining bond) and the point of application of the load.

This study can be reduced to the case of a recessed beam under stress from deviated flexure. The workpieces for turning revolve around the Z-axis and all the direct orthonormed references containing the Z-axis are thus principal reference frameworks. The problem of flexure shown here can thus be studied separately in the planes (X, Z) and (Y, Z).

The static study performed on the modelling shown in Fig. 6 means that the expression for the deformation f 1 measured at point K can be determined as a function of the abscissa Z of the point of application of the load F and the various parameters characteristic of the beam. Deformations f 1X and f 1Y (Eq. ( 5)) are, respectively, caused by F x and F y along the X-and Y-axis. Expressions of deformations f 1X and f 1Y at point K under the action of loads F x and F y are presented below:

f 1X = 3L -Z 6Z • E • I • F x • Z 3 f 1Y = 3L -Z 6Z • E • I • F y • Z 3 (5)
In Fig. 7, the theoretical loads calculated in accordance with modelling and the experimental loads calculated in accordance with Eq. ( 4) are compared. This comparison in each of the planes (X, Z) and (Y, Z) highlights the inadequacy of the model shown in Fig. 6 with respect to reality. Now, the duration for machining of the test workpiece was less than 5 s. The results of works by Mize and Ziegert [START_REF] Mize | Neural network thermal error compensation of a machining center[END_REF] as well as those of Lee et al. [START_REF] Lee | ICA based thermal source extraction and thermal distortion compensation method for a machine tool[END_REF] allow us to assume that the influence of thermal phenomena remains negligible for the results of the tests. This is confirmed by the fact that the distance between the reference surface on the workpiece and the sensors remains the same before and after machining (without machining load).

The curve shapes for the experimental deformations of the workpiece along the directions X and Y are different from those derived from the modelling shown in Fig. 6. Offsets on the abscissa Z = 1 (Fig. 7) appear between models and experimental measurements. They are noted as 1 x and 1 y .

Oscillations of the curve of flexure along the X-axis could be due to machining vibrations, or to movements of the sensor support. A study is under way to determine the origin of the oscillations measured during the test. Considering the curve shapes for the experimental curves as compared with the curves from the model, it appears that the modelling initially proposed (Fig. 6) needs to be modified.

Modelling the workpiece flexure phenomenon

Proposal for new modelling

The presence of offsets 1 x and 1 y (Fig. 7) can be interpreted as an offset of the centre of the equivalent bond between the workpiece and the spindle in relation to the front side of the jaws. The big gap between the results of calculation for the workpiece deformation during machining from the standard beam model and experimental values led us to propose the modelling of Fig. 8. This model is called model 2.

In order to determine the position of the workpiece-spindle restraining bond, the parameter for offset of the restraining bond centre in relation to the front side of the jaws that is noted as d is introduced.

Deformations f 2X and f 2Y (Eq. ( 6)) are, respectively, induced by F x and F y along the X-and Y-axis.

Expressions of deformations f 2X and f 2Y at point K under the action of loads F x and F y are presented below:

f 2X = 3 × (L + d) -(Z + d) 6 • E • I • F x • (Z + d) 2 f 2Y = 3 × (L + d) -(Z + d) 6 • E • I • F y • (Z + d) 2 (6)
Changes in the workpiece/sensor distances measured during machining of the test workpiece allow the value d of the restraining bond offset in relation to the front side of the jaws to be determined. This value is determined using the Excel software solver in accordance with the criterion for minimisation of the square root transformation of the deviations between experi- The graphs in Fig. 9 show the curves for change in workpiece flexure along the Y and X directions during the test, together with the theoretical trend curves determined from the modelling shown in Fig. 8 for an offset value d of 31 mm.

The curves represented in Fig. 9 show that introducing the parameter for offset of the restraining link centre means you can obtain much improved correlation between experimental measurements and modelling: the mean error is divided by 9 adopting the modelling that integrates an offset of the restraining centre.

The same offset value "d" allows you to obtain good correlation between experimental measurements and modelling results in the planes (X, Z) and (Y, Z). Additional analysis is thus implemented to validate the modelling proposed and allow the value of parameter d to be determined.

Validation of modelling proposed and calculation of the offset parameter 4.2.1. Choice of parameters for the study

In order to validate the modelling proposed in Fig. 8, the influence of certain parameters on the value for offset "d" of the modelling proposed is studied.

The influence of the following parameters is quantified:

Amplitude of loads applied to the workpiece during machining (factor F): In order to check that modelling is independent of the amplitude of machining loads, the influence of the "amplitude of machining loads" factor is quantified. Indeed, modelling must remain identical whatever the loading applied on the workpiece as long as the stresses imposed do not exceed the yield strength stress of the machined material.

Diameter of the machined workpiece (factor D): A change in the diameter of the machined workpiece also modifies the position of the points of contact between the workpiece and the jaws in the reference framework (O, X, Y, Z). This factor could have an influence on the position of the centre of reduction of the workpiece/spindle restraining bond.

Length of workpiece seized in the jaws (factor L): Just like the machined diameter, the length of the workpiece seized modifies the position of the points of contact between the workpiece and the jaws in the reference framework (O, X, Y, Z). This For this first approach, the scope of this study is limited to one machine, one material and one type of jaw (hard jaws). Study of the effect of "machine", "material" and "type of jaw" parameters may be considered within the scope of a broader study, in the light of the results from this first study.

Study of the influence of parameters by the experimental design method 4.2.2.1. Defining the experimental design.

The values for the factors retained for this study are shown in Table 1.

A complete design for four factors on two levels in accordance with Table 2 is implanted, meaning that in principle, no interaction would be neglected. Furthermore, under reserve that the effects of the factors are linear, it is possible to write a raster model allowing for an interpolation of the results obtained over the range of variation for all the parameters. Each of the tests was conducted in the same conditions (tool, material, machine and workpiece seizure) as for machining of the test workpiece (see set-up in Figs. 1 and2). Only the values for the diameters of seatings A, B and C and the length L 1 change as a function of the levels of the factors.

Analysis of results.

Measurements of the movements dx and dy along the X-and Y-axis and the loads F x and F y during machining allow us to determine the optimum value of parameter d. For each of the tests, the value for offset d retained is that which minimises the square root transformation of the offsets between flexure calculated in accordance with measurements of movements and theoretical flexure from model 2 as shown in Fig. 8.

Determination of parameter d for each of the 16 tests allows the graph of effects (Fig. 10) and graphs of interactions for each of the factors (Fig. 11) to be plotted.

Analysis of the graph of effects means that only the "diameter" factor has an influence on parameter d for offset of the restraining link. The fact that the "load" factor has no influence on the value of parameter d means the relevance of modelling 2 proposed in Fig. 8 can be confirmed.

The graphs for interactions of the first order highlight the very low interactions between factors. In order to determine a simple model for characterisation of the value of parameter d, the change in this parameter is modelled solely as a function of the "diameter" factor. Interactions between the "diameter" and "pressure", "diameter" and "length", "diameter" and "load" factors are relatively insignificant and are neglected in this first approach. Considering these assumptions, the raster model by Viguier and Sisson [START_REF] Pillet | Les plans d'expérience par la méthode Taguchi[END_REF] offers an interpolation so that the value of parameter d at intermediate levels from those retained during the tests for the "diameter" factor can be determined. This formulation as a function of the results of tests from the experimental design and indicators for levels (InD1 and InD2) of the "diameter" factor is shown in Eq. ( 7) which presents a raster model for interpolation by Viguier and Sisson.

For each of the workpieces produced, a comparison between the diameter measured and the diameter determined is made using the model introduced previously (Fig. 8). The diameters were measured every 10 mm along seating C (Fig. 2), that is eight measurement points per workpiece. Five series of measurements per workpiece were made. The distribution of errors measured during the five series of measurements on the 16 test workpieces from the experiment design is illustrated in Fig. 12.

It appears that distribution of error on the sample studied can be assimilated to a normal distribution with standard deviation of 0.002 mm. This means the model's uncertainty to ±0.006 mm in relation to reality in 95% of cases.

With the raster model presented in Eq. ( 7), the value of the parameter d can be determined as a function of the diameter of the machined workpiece. This is valid under reserve that the Correlation between the theoretical and experimental results checks the validity of the proposed model (Fig. 8). The assumption based on linearity of parameter "d" needed to write the raster model is still verified (Eq. ( 7)). For this application, modelling thus proves to be independent of the machined material.

Example of application of automatic defect correction

A last workpiece made of 2017T4 aluminium alloy defined identically to the first validation workpiece was produced (see Section 5). Machining was implemented along a path that compensates the flexure due to machining loads in accordance with modelling results (Fig. 8), while path correction was applied according to the mirror method [START_REF] Dessein | Qualification et optimisation de la précision d'une machineoutil à commande numérique[END_REF][START_REF] Larue | Deviation of a machined surface in flank milling[END_REF][START_REF] Seo | Intégration des effets de déformation d'outil en génération de trajectoires d'usinage[END_REF] using the value of parameter d calculated using the raster model defined in Eq. [START_REF] Mayer | Prediction of diameter errors in bar turning: a computationally effective model[END_REF].

Fig. 16 illustrates the diameters measured on the first application workpiece and those measured on the workpiece produced with a corrected path. These values are an average of the measurements calculated in accordance with a series of five measurements performed every 10 mm along the machined seating.

From the appearance of the curves in Fig. 16, it would appear that the error on the diameter of the machined seating is reduced from 22 m to 3 m taking the phenomenon of workpiece flexure during machining into account. To produce this workpiece, the proposed method thus allows for machining error in the diameter of the machined seating to be reduced by more than 85%.

Conclusion

This study's aim is to propose a method for automatic correction of the path to machine slender workpieces.

Having tested the model commonly used to characterise the phenomenon of workpiece flexure during machining, a model for workpiece flexure turning aluminium alloy with a given type of jaws is proposed. This model includes a parameter d for offset from the restraining link centre in relation to the front side of the jaws. Then, it is showed that this parameter is solely a function of the diameter of the machined workpiece. Experimental verification then enables us to validate the model proposed and machining with another material shows how this method can be extrapolated. Finally, machinating with a path compensating for workpiece flexure is implemented. The example studied then shows the adequacy of the resolution method proposed to the problem posed by workpiece flexure during machining.

In the light of this study, that was limited to one type of material and one type of spindle, it appears possible to limit the defect due to workpiece flexure under the effect of loads due to cutting. This can be done by programming a corrected path compensating for workpiece flexure, for example, by carrying out a mirror type correction. This then releases us from the need to use the tailstock centre when the workpiece-spindle link alone is sufficient to sustain loads due to cutting. This method dispenses with the need to use sensors whose data would then have to be processed in real time and would need to be taken into account directly within the NC control unit. It is necessary to study validity of the modelling proposed for machines working on materials other than those to which this first study was limited.

Fig. 1 .

 1 Fig. 1. Test workpiece set-up.

Fig. 2 .

 2 Fig. 2. Procedure.

Fig. 5 .

 5 Fig. 5. Comparison between diameters measured and diameters measured on the test workpiece.

Fig. 6 .

 6 Fig. 6. Initial modelling of the workpiece during machining (model 1).

Fig. 7 .

 7 Fig. 7. Theoretical (f 1 ) and experimental values for deformation along the Xand Y-axis.

Fig. 8 .

 8 Fig. 8. Modified modelling (model 2).

Fig. 9 .

 9 Fig. 9. Theoretical (f 2 ) and experimental values for the deformation along the X-and Y-axis. mental values and theoretical values calculated from the second modelling. The offset value d for modelling relative to the test is: d = 31 mm.The graphs in Fig.9show the curves for change in workpiece flexure along the Y and X directions during the test, together with the theoretical trend curves determined from the modelling shown in Fig.8for an offset value d of 31 mm.The curves represented in Fig.9show that introducing the parameter for offset of the restraining link centre means you can obtain much improved correlation between experimental measurements and modelling: the mean error is divided by 9 adopting the modelling that integrates an offset of the restraining centre.The same offset value "d" allows you to obtain good correlation between experimental measurements and modelling results in the planes (X, Z) and (Y, Z). Additional analysis is thus implemented to validate the modelling proposed and allow the value of parameter d to be determined.

a

  F y = 1440 N and F x = 388 N F y = 720 N and F x = 194 N Length of seizure 25 The values for loads efforts F y = 1440 N and F x = 388 N are obtained by performing a cut of 4 mm in depth at 0.54 mm/rev feed with a cutting speed of 250 m/min. Loads F y = 720 N and F x = 194 N are obtained by making a cut of 3 mm in depth at 0.36 mm/rev feed with a cutting speed of 250 m/min. b A pressure of 3 bar corresponds to a clamping load of 937 N on each of the jaws and a clamping pressure of 5.5 bar corresponds to a clamping load of 1718 N on each of the jaws. parameter could have an influence on the position of the centre of reduction of the workpiece/spindle restraining bond. Jaw clamping pressure (factor P): The jaw clamping pressure is adjustable using a pressure gauge on the test machine. Changing the clamping load of the workpiece can change the rigidity of the workpiece/spindle link and thus modify the parameter d for position of the centre of the equivalent restraining link.

Fig. 10 .

 10 Fig. 10. Graph of effects of factors on parameter d.

Fig. 11 .

 11 Fig. 11. Interaction graphs.

d

  = 59.39 + [-26.73 26.73

Fig. 12 .

 12 Fig. 12. Distribution of error between diameter measured and diameter determined with model 2.

Fig. 13 .

 13 Fig. 13. Defining validation workpiece No. 1.

Fig. 15 .

 15 Fig. 15. Theoretical (f 2 ) and experimental values for the deformation along the X-and Y-axis.

Fig. 16 .

 16 Fig. 16. Comparison of workpiece diameters with and without compensation of workpiece flexures.

Table 1

 1 Summary of factors and values retained for tests

	Factor	Low level	High level
	Diameter	30 mm	50 mm
	Load a

Table 2 Table of

 2of

		tests from the test design
	Test no.	Workpiece no.	Factor 1 (diameter)	Factor 2 (length)	Factor 3 (pressure)	Factor 4 (load)
	1	5	30	25	3	1440
	2	6	30	25	3	720
	3	7	30	25	5.5	1440
	4	8	30	25	5.5	720
	5	1	30	39	3	1440
	6	2	30	39	3	720
	7	3	30	39	5.5	1440
	8	4	30	39	5.5	720
	9	13	50	25	3	1440
	10	14	50	25	3	720
	11	15	50	25	5.5	1440
	12	16	50	25	5.5	720
	13	12	50	39	3	1440
	14	10	50	39	3	720
	15	11	50	39	5.5	1440
	16	9	50	39	5.5

assumption of linearity of parameter d in relation to the diameter of the machined workpiece is satisfied.

Experimental verification of modelling

In order to apply the method for determination of workpiece flexure during machining using the new modelling, a workpiece is produced in conditions different from those retained for tests of the experimental design. This meant the validity of the raster formulation proposed (Eq. ( 7)) could be tested, as well as the linearity assumption for parameter d in relation to the diameter of the machined workpiece.

The machining of a workpiece made of 2017 T4 aluminium alloy whose definition drawing is shown in Fig. 13 is performed. Implementation, scheduling of machining operations, the tools and the machine used remain similar to those retained for the test workpiece and workpieces from the experimental design (Fig. 2).

The cutting parameters adopted to implement operation The graphs in Fig. 14 show the curves for changes in workpiece flexure along the directions Y and X during implementation of the validation workpiece, together with the curves for theoretical changes determined from the modelling shown in Fig. 8 with the value for offset set at 59.39 mm.

Correlation between the theoretical and experimental results checks the validity of the proposed model (Fig. 8). The assumption based on linearity of parameter d as a function of the "diameter" factor, as needed to write the raster model, is also verified (Eq. ( 7)) during this test.

Extrapolation in the case of another material

In order to test the validity of the modelling proposed for machining of different materials, the machining of a C40 steel workpiece with geometry similar to the previous workpiece is performed (see Fig. 13). Implementation, scheduling of machining operations and the machine used remain similar to those retained for the previous workpiece. The tools used involved a PDJNL 20-20 tool-holder and a DCMT 15 06 04-UF insert.

The cutting parameters adopted to implement operation 3 were: