The structure of alternative tableaux.

# The structure of alternative tableaux.

Philippe Nadeau Faculty of Mathematics, University of Vienna, Nordbergstraße 15, 1090 Vienna, Austria.
###### Abstract.

In this paper we study alternative tableaux introduced by Viennot [17]. These tableaux are in simple bijection with permutation tableaux, defined previously by Postnikov [12]. We exhibit a simple recursive structure for alternative tableaux. From this decomposition, we can easily deduce a number of enumerative results. We also give bijections between these tableaux and certain classes of labeled trees. Finally, we exhibit a bijection with permutations, and relate it to some other bijections that already appeared in the literature.

## Introduction

Alternative tableaux are certain fillings of Ferrers diagrams introduced by Xavier Viennot [17], in simple bijection with permutation tableaux introduced by Postnikov [12] in his study of the totally positive part of the grassmannian. These permutation tableaux have then been considered by several authors.

Alternative tableaux are related to the stationary distribution of a certain Markov process from statistical physics, the Asymmetric Simple Exclusion Process (ASEP). They are more precisely connected to a Matrix Ansatz describing a general solution for this stationary distribution (see Proposition 1.5). Formulas for this distribution have been first computed by Sasamoto, Uchiyama and Wadati [15], and involve the famous Askey-Wilson orthogonal polynomials. An understanding of alternative tableaux thus seem to be a key in order to get a better grasp of these polynomials, for which no combinatorial interpretation is known. The articles [5, 6, 7] develop such connections between tableaux and some special cases of the ASEP model.

Another line of research is related to permutations; the starting point here is that permutation tableaux of size are counted simply by . They were first studied from a combinatorial point of view in [14], in which a bijection between these tableaux and permutations of was studied in detail. Since then, other bijections have been described [2, 4, 16]. These tableaux also showed up in the work of Lam and Williams [11] where permutation tableaux were shown to fit naturally into the type A case of the classification of Coxeter systems.

In this article, we show that alternative tableaux admit a natural recursive structure, which is best expressed when considering alternative tableaux as labeled combinatorial objects. The central part of this work is thus Section 2, in which we exhibit this recursive decomposition. These structural results are then applied in the following sections, first by giving enumerative results in a very straightforward manner, and then by encoding the recursive decomposition by certain classes of labeled trees, which are then themselves in bijection with permutations.

Let us give a more precise outline of the paper. We introduce some elementary definitions and properties concerning alternative and permutation tableaux in Section 1. We then give our main results relative to the structure of alternative tableaux in Section 2, the recursive structure being a consequence of Proposition 2.5 and Theorem 2.14 in particular. Using this decomposition, we prove several enumeration results in Section 3: this gives in particular elementary proofs of certain results of [4, 7, 11], as well as some new results. We then describe how the recursive decomposition is naturally associated with certain labeled trees in Section 4. Finally we exhibit in Section 5 a bijection from alternative tableaux to permutations, and stress its connection to bijections which have already appeared in the literature.

## 1. Tableaux

### 1.1. Shapes and tableaux

We call shape a staircase diagram (also called Ferrers shape) with possible empty rows or columns, cf. Figure 1. The length of a shape is the number of rows plus the number of columns of the shape. Note that a shape is determined by its south east border, which is the path from the top right corner of the shape to its bottom left corner; it is the path labeled by the integers on the left of Figure 1. There are thus shapes of length , since one can choose to go down or left at each step.

Rows and columns of shapes will be labeled by integers in the following manner: let be a shape of length , and a set of integers. Then is labeled by if the numbers are attached to the rows and columns, in increasing order following the south east border of from top right to bottom left. If , then we say that has the standard labeling; this is the case of the shape of Figure 1. Note that although the labeling is always defined with respect to the south east border, we will actually write the labels on the top and left side for an improved readability, as shown on the right of Figure 1.

Given a labeled shape, we will sometimes say row or column when we actually refer to the row with the label or the column with the label . Then the cell lying at the intersection of row and column is denoted by , where we have necessarily by definition of the labelings.

We now define two possible ways to fill these shapes, called permutation tableaux and alternative tableaux: the two are in bijection by Viennot’s Theorem 1.3. What we will show in this paper is that it is better to study alternative tableaux when it comes to discover the intrinsic structure of these combinatorial objects.

###### Definition 1.1 (Permutation tableau).

A permutation tableau is a shape with a filling of each of its cells by or such that the following properties hold: (i) each column contains at least one , and (ii) there is no cell filled by a which has simultaneously a above it in the same column and a to its left in the same row.

Note that permutation tableaux cannot have any empty columns because of the first condition. A in a permutation tableau is restricted if there is a above it in the same column; it is rightmost restricted if it is the rightmost such in its row. A row is unrestricted if it does not contain a restricted . A is superfluous if, somewhere above it in the same column, there is another cell containing a . A permutation tableau is represented on the left of Figure 2; lines and are unrestricted, the cells in the top row filled by appear in columns and , and there are four superfluous ones in cells and .

###### Definition 1.2 (Alternative tableau).

An alternative tableau is a shape with a partial filling of the cells with left arrows and up arrows , such that all cells left of a left arrow, or above an up arrow are empty. In other words, all cells pointed by an arrow must be empty.

In an alternative tableau, a free row is a row with no left arrow, and a free column is a column with no up arrow. Thus rows (respectively columns) that are not free are in bijection with left (resp. up) arrows. A free cell is a cell which is not filled, and such that there exists no left arrow to its right and no up arrow under it; in other words, the cell is empty and no arrow points toward it. We will let and denote the number of free rows, free columns and free cells of a given tableau. For the tableau which is represented on the right of Figure 2, the free rows are and while the free columns are and . There are four free cells, namely and . Thus we have and .

We can now state the fundamental result of Xavier Viennot, showing that alternative tableaux are actually a new simple encoding of permutation tableaux:

###### Theorem 1.3 ([17]).

There is a bijection between permutation tableaux of length and alternative tableaux of length . If is labeled by , and is the set minus its smallest element, then we label by , and we have:

• columns of with a in their top row correspond to free columns of ;

• unrestricted rows of (the top one excepted) correspond to free rows of ,

• and cells of filled with superfluous correspond to free cells of .

Proof: We just give a description of the bijection and its inverse, and refer to [17] for more details about the proof. Given a permutation tableau , transform all non superfluous to up arrows, and all rightmost restricted to left arrows; then erase all the remaining and , and finally remove the first row from ; the result is . An illustration is given on Figure 2.

For the inverse bijection, given an alternative tableau , add a new top row on top of it, and fill by all cells of this row that lie above a free column of ; then change all up arrows and free cells to , and all remaining cells to . The resulting tableau is .

### 1.2. Alternative tableaux and the ASEP

An important application of permutation tableaux, due to Corteel and Williams in a series of papers [5, 6, 7], is related to a certain model of statistical mechanics, the ASEP: we will briefly talk about some of the connections. The ASEP model(Asymmetric Simple Exclusion Process) is a model that can be described as the following Markov chain (see [8]). Let be real numbers in , and a nonnegative integer. The states of the Markov chain are the words of length on the symbols and . Positions in the words represent sites, which can be either empty () or occupied by a particle (). The transition probabilities between two states and model the way particles can jump from site to site, enter or exit the system:

• If and , then and .

• If and , then and .

• If and , then and .

• If do not correspond to any of these cases, then we set .

• Finally, we have naturally .

The model is simple because there can be at most one particle in each site. We illustrate schematically this model in Figure 3.

It was shown by Derrida et al.[10] that this model has a unique stationary distribution, and that moreover this distribution could be computed through the following Matrix Ansatz: suppose that we can find two matrices , a column vector and a row vector such that the following relations hold:

 (1.1) ⎧⎨⎩DE=qED+D+E(βD−δE)V=VW(αE−γD)=W

Now let be a state of the ASEP, and let be the word of length in and obtained through the substitutions , : for instance, the state is associated to the word . Then Derrida et al show :

###### Proposition 1.4.

If satisfy the relations (1.1), and words in are interpreted as matrix products, the probability to be in state in is given by

 (1.2) Pn(s)=Ws(E,D)VZn   with   Zn=W(D+E)nV

If we have a word in the letters and , we can create a shape by reading the word from left to right and interpreting each as a south step and each E as an east step, thus defining the south east boundary of a shape ; for instance the shape of Figure 1 is associated to the word ; if is a state of the ASEP, we will write simply for the shape . Now we can state the connection with alternative tableaux, first noticed by Corteel and Williams in a series of papers [5, 6] and expressed in terms of permutation tableaux, and later reformulated by Viennot [17] in the following way:

###### Proposition 1.5.

If are matrices that verify the first relation in (1.1), and is any word in , then we have the following identity:

 w=∑Tqfcell(T)Efcol(T)Dfrow(T)

where the sum is over all alternative tableaux of shape .

In [10] it was in fact shown that in the case there exists matrices verifying (1.1). But note that in this particular case, the vectors and become respectively left and right eigenvectors for and . So from Propositions 1.4 and 1.5 we obtain the following:

###### Corollary 1.6.

When we have:

 Pn(s)=∑T~{}of shape~{}λ(s)qfcell(T)α−fcol(T)β−frow(T)∑T~{}of size~{}nqfcell(T)α−fcol(T)β−frow(T)

In the ASEP model where are general but , Corteel and Williams found a similar expression for the stationary probabilities in terms of certain enriched alternative tableaux, see Corollary 4.2 in [7], by slightly generalizing the Matrix Ansatz.

## 2. The structure of alternative tableaux

We define different operations on tableaux, and use them to exhibit a natural recursive structure on alternative tableaux.

### 2.1. First properties of alternative tableaux

We denote by the set of alternative tableaux of length , and those with free rows and free columns. We also denote by and the tableaux having free rows and free columns respectively.

#### 2.1.1. Transposition

We let be the operation of transposing a tableau, which is the reflection across the main diagonal, i.e. the line going south east from the top left corner; in this reflection, we naturally exchange up and left arrows. We have the following immediate result, which we state as a proposition for future reference:

###### Proposition 2.1.

Transposition is an involution on alternative tableaux. For all ,it exchanges and .

In fact it is easily checked that the transposition operation coincides with the involution defined in Section 7 of [6] for permutation tableaux: more precisely, if is a permutation tableau, then we have , where is the bijection between permutation tableaux and alternative tableaux of Theorem 1.3. Note then that the trivial result on alternative tableaux from Proposition 2.1 demanded a much greater effort in [6] where the authors worked with permutation tableaux.

#### 2.1.2. Packed tableaux

As already noticed, the arrows of a tableau are in bijection with its non free rows and columns. This implies immediately that the tableaux in have exactly arrows. The maximum of arrows, i.e. the case , cannot actually be attained if : indeed, if for instance a tableau has no free row, then the leftmost column of this tableau cannot contain any up arrow and thus is free. A total of arrows in a tableau can actually be reached, and in fact constitutes a fundamental class of tableaux as we will see:

###### Definition 2.2 (Packed tableau).

A packed tableau of length is a tableau with arrows. Equivalently, it is a member of either or .

In fact, the following proposition shows that the unique free column of a tableau in is necessarily the leftmost one:

###### Proposition 2.3.

If and is a tableau in , the top left cell of contains a left arrow.

Proof: First note that the tableau has at least one cell (so that is well defined) since tableaux with no cells have at least one free row or two free columns (note that we chose ). has no free rows, so there is a left arrow in the top row in particular. The column where this arrow lies is necessary free, because any up arrow in it would violate the alternative tableau property. But the leftmost column of is also free, because the presence of any up arrow in it would force the row where this arrow lies to be free, which is excluded. As has just one free column, this implies that contains indeed a left arrow.

By transposition, the top left cell of tableaux in is filled by an up arrow. But there is a simpler way to go bijectively from to when : simply change the filling of the cell in the top left corner from to . This also explains why we decided to call these two sets of tableaux with the same name: up to this arrow and the case , they are identical.

#### 2.1.3. Cutting rows and columns

###### Definition 2.4 (cutc and cutr).

For a nonempty alternative tableau with at least one column and no empty rows, we let be the operation of deleting its leftmost column (so that all row lengths decrease by one); we define similarly for deleting the topmost row.

When the tableau from which we start is labeled, we obtain naturally a labeled tableau as a result by simply keeping the labels of the remaining rows and columns in each case. This is illustrated on Figure 4.

Given a tableau in , add a new top row, and fill by up arrows all cells in this row that lie above the free columns of ; this construction will be called to reflect the fact that no free column remains after its application. We define symmetrically. Then we have the following properties:

###### Proposition 2.5.

For all , the operation is a bijection between and ; its inverse is . For all , the operation is a bijection between and ; its inverse is .

Proof: We just prove the claim concerning , the one for being equivalent after transposition. Tableaux in have no empty columns (because such columns are free), and their first row is free, since a left arrow in a cell from this row would force the corresponding column to be free. This shows that the restrictions of and are well defined. It is immediate that is the identity on . Now given a tableau in , we noticed that it has no left arrows in its top row, and that the up arrows in this row occur exactly in columns that are free in . This implies that is the identity on , and the proposition is proved.

If we do the union of the sets above for all and for all respectively, we get bijections between and , and between and . The special cases and are also of interest, and we get the following corollary:

###### Corollary 2.6.

For all , the operation is a bijection between and , and between and ; is a bijection between and , and between and . Therefore for all , we have

 (2.1) A(n)=A0,∗(n+1)=A∗,0(n+1)=A0,1(n+2)=A1,0(n+2)

### 2.2. Splitting a tableau

Now we exhibit a more complicated decomposition, which can be traced back to the last part of Burstein’s work [2]. Nevertheless in his work Burstein did not exhibit a complete recursive decomposition, mainly because he was working with permutation tableaux which are less easy to manipulate than alternative tableaux.

Let be an alternative tableau of size , labeled by , and be the label of one of its free rows (we suppose there is such a row). We compute iteratively a set of labels in the following way: first we set . Then we add to all columns such that there is an arrow in the cell . Afterwards, we add to all rows such that there is an arrow in for one of the columns in added at a previous stage. And so on, we keep adding row and column labels alternatively until there are no new rows or columns to add. The procedure is finite since the set is finite, and in the end we set .

Example: consider the free row labeled in the alternative tableau on the right of Figure 2; so contains . Now the cell is the only cell on row containing an up arrow, so we add to the set . In this column there are two left arrow in cells and , so contains also the row labels and . There is no other arrow on row , and there is one on row in the cell , so also belongs to . Since there is no other arrow in column , we have finally .

Another equivalent characterization of is the following, which we take as a definition:

###### Definition 2.7 (the label subset T(i0)).

Given a tableau labeled by and a free row or column , is the smallest set (wrt. inclusion) which contains , and is such that, for every cell filled by an arrow, belongs to if and only if belongs to .

If stands for the set of labels of free rows and columns of , then we have a collection of subsets of given by .

###### Lemma 2.8.

Let , and . Then the elements of other than label non free rows and columns of .

Proof: By the iterative definition of , a row label belongs to if there exists a column label and a left arrow in the cell . In particular, row is not free in . The proof is similar for column labels.

###### Lemma 2.9.

Let be a tableau labeled by . Then the sets form a partition of .

Proof: Suppose first that there exists an integer in that does not belong to any subset . We assume without loss of generality that is the label of a row, and choose the minimal such . First, cannot label a free row (since it would belong to ) so there exists such that contains a left arrow . Now is not the label of a free column, since otherwise would belong to ; so there exists a row label such that contains an up arrow . We have because otherwise the up arrow in would point towards the up arrow in . But then belongs to a set by minimality of , which entails that and also belong to this set, which contradicts the hypothesis that belongs to no such set.

We have thus shown that the sets cover , we now have to prove that they are disjoint. Let belong to a set ; we will show that we can uniquely determine from . If is free, then because there is only one free row or column in by Lemma 2.8. Now suppose is not free, and let us assume that labels a row: there exists a (necessarily unique) column with containing a left arrow. Now if is free, we know that and we are done. Otherwise we have an up arrow in for a unique . If is free, then and we are done, otherwise we continue this process, and stop until we hit upon a free row or column, and we know this is the index . To conclude, we just need to be sure that the process will end: this is indeed the case because the row labels that we encounter are strictly decreasing (and the column labels strictly increasing).

Given a tableau with label set , and any subset , one can form a new tableau by selecting in only the rows and columns with labels in :

###### Definition 2.10 (T[a] and T[k]).

Let be a tableau labeled by , and . The tableau is defined as the tableau labeled by the subset , where labels a row (respectively a column) in if it labels a row (resp. a column) in , and such that the cell has the same filling as the cell in . We write for simplicity if is a free row or a free column.

Then from Lemma 2.8 we deduce immediately:

###### Proposition 2.11.

Let be a labeled tableau and . The tableau is a packed tableau in which labels the only free row or column.

### 2.3. Merging tableaux

Since we described a way to split a tableau into smaller tableaux, it is natural to try to reconstruct the original tableau, so we need to define a way to merge tableaux together.

###### Definition 2.12 (the function merge).

Let and be two alternative tableaux labeled on disjoint integer sets and . Then is a labeled tableau defined as follows: its label set is , where labels a row in if and only if it labels a row in either or . Then the cell is filled with a left arrow if one of the following two cases occur: either and is a left arrow in , or and is a left arrow in . Up arrows in are defined similarly, and the other cells are left empty.

Note that the empty cells of correspond either to empty cells in or , or to cells for which one of belongs to and the other to . An example of merging is given on Figure 5.

We make a slight abuse of notation in writing , since the operation of merging depends crucially on the labels and not merely on the tableaux. This will not cause any problem in the rest of the paper, since we will always use it when the labels of the tableaux are clear from the context. We now record some immediate properties of the merging procedure in the following proposition:

###### Proposition 2.13.

Given as above, then is an alternative tableau. If , then either and , or and .

Moreover, the application is symmetric, i.e. ; it is also associative, in the sense that if are tableaux labeled by pairwise disjoint sets, then .

The last two properties allow us to extend the domain of definition of : given a finite collection of tableaux with pairwise disjoint label sets, we can merge all tableaux in by defining

 merge(C)=merge(Ti1,merge(Ti2,…,merge(Tit−1,Tit))),

where is any ordering of the index set ; this is well defined thanks to the properties of symmetry and associativity.

### 2.4. Decomposition of tableaux

The following theorem is the main structural result of this paper; together with Corollary 2.6, it describes a recursive structure that completely characterizes alternative tableaux. This will be applied in the remaining sections, first to easily obtain old and new enumerative results, and then to give bijections between alternative tableaux, certain classes of trees, and permutations of integers.

###### Theorem 2.14.

Let be nonnegative integers, and be a label set. The function is a bijection between:

1. Tableaux in labeled by , and

2. Sets of packed tableaux, with of them in and in , all labeled in such a way that their label sets form a partition of .

The inverse bijection is the operation .

Proof: First, the fact that is well defined is a consequence of Lemma 2.9 and Proposition 2.11 . We have also is equal to the identity function thanks to Proposition 2.13. What remains to be proven is that is the identity on : that is, we need to show that given a tableau , merging the labeled tableaux gives back the tableau . Let us then denote by the tableau , and show that we have . We note immediately that the (labeled) shapes of and coincide, so we have to show that the contents of all cells are identical. Let then (respectively ) be the content of a cell in (resp. in ). If and are labels of the same tableau (for a certain ), then is the content of in by the definition of ; but by definition of , this is also equal to . Otherwise, and belong respectively to tableaux and with , and in this case is necessarily empty by Lemma 2.9; and by the definition of again, is also empty. Thus and the result is proved.

An immediate corollary of the theorem is the following, which gives a different way of decomposing tableaux:

###### Corollary 2.15.

There is a bijection between tableaux in labeled by a set , and pairs of tableaux labeled by sets and such that is a partition of .

Proof: Let be a tableau in labeled by a set . First use the bijection of the previous theorem, and, among the tableaux obtained, separate the ones in and the ones in ; merge separately each of these two collections to obtain the tableaux and of the theorem.

There is a more direct way to obtain the same bijection: consider the subsets of labels and , where (respectively ) goes through the labels of the free rows of (resp. the free columns). Then define simply and .

## 3. Enumeration

We will show that, using the structure of alternative tableaux discovered in Section 2, it is easy to prove various enumeration results in a simple way, starting with the plain enumeration of alternative tableaux according to their size.

### 3.1. Labeled combinatorial classes

From the decompositions of Theorem 2.14 and Corollary 2.15, one can easily write down equations for the combinatorial class of alternative tableaux, in the manner of Flajolet and Sedgewick [9]. Indeed, Theorem 2.14 says that the number of tableaux labeled on a set is the same as the number of ways to partition and then choose, for each block of this partition, a tableau labeled on belonging to either or ; in the language of [9], this is written:

 (3.1) A=SET(A0,1+A1,0).

Similarly, a consequence of Corollary 2.15 is

 (3.2) A=A0,∗⋆A∗,0.

This means that an alternative tableau labeled on is obtained by choosing two alternative tableaux in and , labeled respectively by and which are disjoint and whose union is equal to . The advantage of describing our theorems in this way is that there is an automatic way to write down equations for the corresponding exponential generating functions, with the added possibility of taking into account certain parameters; this is what we will do in the rest of this Section.

As a matter of fact, the natural framework for the study of alternative tableaux is arguably the theory of species on a totally ordered set, cf. [1, Chapter 5]. This is not needed for the results in this work and therefore we will not develop this approach.

### 3.2. The number of alternative tableaux.

We will give first a simple proof of the well known fact that alternative tableaux of size are enumerated by ; not surprisingly, that is how the original permutation tableaux got their name. Let , and be the exponential generating functions of tableaux in , and according to their length, that is:

 A(z)=∑n≥0A(n)znn!, B(z)=∑n≥0A0,∗(n)znn!,~{}and~{}C(z)=∑n≥0A0,1(n)znn!.

On the one hand, Corollary 2.5 implies the following relations on generating functions:

 (3.3) B′(z)=A(z)~{}~{}and~{}~{}C′′(z)=A(z).

On the other hand, note that and have respectively the generating functions and : this is immediate by transposition (cf. Proposition 2.1). So we can use the combinatorial equations (3.2) and (3.1) to obtain the functional equations and , by an application of the principles found in [9, Chapter II]. Together with (3.3), we get the differential equations

 B′(z)=B(z)2andC′′(z)=exp(2C(z)).

With the obvious initial conditions , the solutions to these are respectively and . Taking coefficients, we obtain and , which both give us by Corollary 2.6. To sum up we have:

###### Proposition 3.1.

We have the following expressions:

 A(z)=1(1−z)2,  B(z)=11−z,~{}and~{}C(z)=−log(1−z)

### 3.3. Refined enumeration

In fact we can do much better by introducing some statistics. Let be the number of tableaux in with rows, where we allow or . We define the corresponding generating functions and . We have then the following refined enumeration:

###### Theorem 3.2.
 (3.4) A(z,u,x,y)=exp(zy(1−u)+(x+y)ln(1−u1−uexp(z(1−u))))

Proof: By Theorem 2.14, we know that the number of free rows (respectively free columns) of a tableau is equal to the number of tableaux in (resp. ) under the bijection . We can then use Equation (3.1) and insert parameters and in it (cf. [9, Chapter III]) and this gives the equation:

 (3.5) A(z,u,x,y)=exp(xA1,0(z,u))exp(yA0,1(z,u))

Now we have if , by the remark following Proposition 2.3. Taking into account , we obtain on the level of generating functions ; plugging into Equation (3.5) gives

 (3.6) A(z,u,x,y)=exp((x+y)A1,0(z,u)+zy(1−u)).

Using the bijections and , we get the following refinements of Corollary 2.6

 A(n,k)=A0,∗(n+1,k)=A∗,0(n+1,k+1)=A1,0(n+2,k+1),

for . This translates into the following equations for the generating functions, where all derivatives here are taken with respect to the variable :

 (3.7) A′0,∗(z,u) =A(z,u); (3.8) A∗,0(z,u) =uA0,∗(z,u)+(1−u); (3.9) A′1,0(z,u) =uA0,∗(z,u).

Since the number of rows of is the sum of the number of rows of and , we have the equation by Equation (3.2). Using Equations (3.7) and (3.8) we get . Taking into account the initial condition , this differential equation is easily solved and gives us

 (3.10) A0,∗(z,u)=(1−u)exp(z(u−1))−u.

Now we use Equation (3.9), and by immediate integration of (3.10) we obtain

 A1,0(z,u)=ln(1−u1−uexp(z(1−u))).

Now it suffices to replace in (3.6) and the result follows.

From this theorem, we have the following corollary, first proved in [4] by a complicated recurrence:

###### Corollary 3.3.

Define the polynomial ; then we have the following expression:

 (3.11) An(x,y)=n−1∏i=0(x+y+i)

Proof: It is easily seen that for the expression inside the logarithm in (3.4) boils down to , so

 A(z,1,x,y) =exp(−(x+y)log(1−z))=(1−z)−(x+y) =∑n≥0(x+y)(x+y+1)⋯(x+y+n−1)znn!.

It suffices to take the coefficient of on both sides to obtain the result. Note that in fact we just need Equation (3.5) from the proof of Theorem 3.2, and then the expression of follows from the fact that both and are equal to by Proposition 3.1.

### 3.4. Decorated tableaux

In their study of the ASEP model in the case , Corteel and Williams [7] managed to express the stationary distribution in terms of alternative tableaux with certain weights. In particular, the so called partition function can be expressed combinatorially. Following [7], let us call decorated alternative tableau an alternative tableau where each arrow can be in two states, marked and unmarked: a usual alternative tableau with arrows thus gives rise to different decorated alternative tableaux.

###### Theorem 3.4 ([7]).

The number of decorated alternative tableaux of length is equal to .

We give a simple proof of this fact based on the recursive structure of tableaux:

Proof: Let (respectively ) be the exponential generating function of decorated tableaux (resp. decorated tableaux such that the underlying alternative tableau belongs to ). Note that, by transposition, can equivalently be defined by replacing by . Remember that arrows correspond to non free rows and columns: so this is an additive parameter of tableaux with respect to the decomposition . Thus from Equation (3.1) we get immediately

 (3.12) ˜A(z)=exp(2˜C(z)).

But since tableaux in (for ) have exactly non free rows and columns, each of them gives rise to decorated tableaux. In terms of generating functions this means that . Now we know that by Proposition 3.1, so after substituting in Equation (3.12) we get

 ˜A(z)=11−2z

and the result follows by taking the coefficient of on both sides.

The proof in [7] is more involved, but has the nice feature of being bijective. It turns out that the proof above can be easily “bijectivized”:

###### Proposition 3.5.

There is a bijection between decorated tableaux of length , and tableaux in such that all rows and columns can be marked.

Since we will give in Section 5 a bijection between tableaux of and permutations on elements, this will indeed give a fully bijective proof of Theorem 3.4.

Proof: Let be a tableau of length , with standard labeling, and let be the tableaux respectively in and obtained by the procedure of Corollary 2.15, together with their label sets and : we also naturally let rows and columns of and be marked whenever they were originally marked in . Now define a marked labeled tableau as follows: the underlying tableau is and the labels are given by . For the marks, note that transposition exchanges (free) rows and (free) columns. We keep the marks of in for all non free rows and columns. Now has no free columns (so that has no free rows), and all its free rows are unmarked by the definition of decorated tableaux: the corresponding free columns in are defined to be all marked.

and are two marked, labeled tableaux in , therefore is a marked, labeled tableau in , and we claim that is the desired bijection. Indeed, let us describe the inverse bijection. Given a marked tableau in , let be the labels of the marked free columns, and the labels of the unmarked free columns. Define then (respectively ) as the tableau where is the subset of labels (resp. ); these are both labeled, marked tableaux in . Now transpose the tableau , keeping all marks except the ones corresponding to the original labels which are deleted. Merge the resulting tableau with , and let be the resulting labeled, marked tableau: it is clear that it has no marks on free rows and columns, and is thus a decorated tableau. It is then easy to see that is the wanted inverse bijection.

### 3.5. Symmetric tableaux

We call a tableau symmetric if it is fixed by the operation of transposition defined in Section 1. Clearly symmetric tableaux have even length since they have the same number of rows and columns. We have then the following enumeration

###### Proposition 3.6.

The number of symmetric tableaux of size is .

Proof: Let be a symmetric tableau of size with standard labeling. If labels a row, then labels a free column. In fact, even more is true: the tableau labeled by is the transpose of the tableau labeled by , and the labels verify . By the bijection of Corollary 2.15, symmetric tableaux are thus in one to one correspondence with pairs of labeled tableaux , where and the labels verify as well as .

Thus all symmetric tableaux are obtained in the following manner: pick an alternative tableau in , and for each pair pick one of the two integers; let be the set of the chosen integers and the complement of in . Then merge labeled by and labeled by : this is a symmetric tableau. Since has elements and there are clearly choices for the labels , the result follows. We illustrate the correspondence on Figure 8.

So symmetric tableaux of size are equinumerous with signed permutations of , which are permutations on in which letters can be barred; we actually describe a bijection at the end of Section 5.2.

In the work of Lam and Williams [11], permutation tableaux for the type are defined, also enumerated by . It is actually possible to show that their tableaux are in bijection with symmetric alternative tableaux, by adapting suitably the bijection from Theorem 1.3 to the symmetric case.

We note that these permutation tableaux for the type were defined as a certain subclass of diagrams that appeared naturally in the context of Coxeter groups of type . It is quite surprising that when one starts with alternative tableaux (originally related to the ASEP), and then considers those that are symmetric, one obtains configurations that are in simple bijection with permutation tableaux of type . This raises the following question: given a finite Coxeter system , is there a natural way to associate to each of the elements of a certain generalized alternative tableau ?

## 4. Alternative trees and forests

In this section we will give bijections from alternative tableaux to various families of planes and forests, bijections which are based on the decompositions of Section 2.

All trees considered are rooted and plane, by which we mean as usual that the children of every vertex are linearly ordered. Furthermore, we will consider labeled trees and forests, where the labels will be pairwise different integers attached to the vertices; these integers form the label set of the tree or forest.

###### Definition 4.1 (Minimal and Maximal vertices).

Given a vertex in a labeled tree, we say that is minimal (respectively maximal) if its label is smaller (resp. bigger) than all its descendants.

### 4.1. Plane alternative trees and forests

###### Definition 4.2 (Plane alternative tree).

A plane alternative tree is a labeled rooted plane tree with black and white vertices, such that:

• each white vertex is minimal, its children are black and have decreasing labels from left to right;

• each black vertex is maximal, its children are white and have increasing labels from left to right.

A plane alternative forest is a set of plane alternative trees.

We represent a plane tree on Figure 9.

In this subsection we show that these trees are the natural objects encoding the recursive structure described in Section 2.

First we define the function which goes from labeled packed tableaux to alternative trees. Let be labeled; then is an element of for a certain , by Corollary 2.6. Let be the labeled tableaux of given by (cf. Theorem 2.14), and be the label of the top row of . Symmetrically, if , then we let the be the tableaux of obtained by applying in succession and , and be the label of the leftmost column.

###### Definition 4.3 (Tree and Forest).

We define recursively to be the tree whose root is white (respectively black) and labeled by , and whose subtrees attached to the roots trees are , arranged from left to right in increasing (respectively decreasing) order of the labels of their roots if (resp. ).

If is any labeled alternative tableau, and are the labeled packed tableaux given by from Theorem 2.14, we define as the labeled forest consisting of the trees .

The forest for the alternative tableau of Figure 2 is represented on Figure 10.

###### Theorem 4.4.

is a bijection from packed labeled alternative tableaux of length to plane alternative trees with vertices (with the same label set). is a bijection from labeled alternative tableaux of length to plane alternative forests with vertices (with the same label set).

Proof: Note first that the claim about is an immediate corollary of the result for , thanks to Theorem 2.14.

The proof is mostly straightforward, and consists simply of noticing that the recursive structure of tableaux given by Theorem 2.14 and Lemma 2.5 is naturally encoded by alternative trees. The only point that needs to be checked is the minimality of white vertices (the maximality of black vertices being clearly proved symmetrically): when a white vertex is added in a tree, its label is the first row of a tableau , and the labels of its descendants are the labels of the other rows and columns of , which by definition of the labeling of tableaux are indeed larger than .

So is well defined, and the inverse (recursive) construction is clear: given a tree with root labeled , construct (recursively) the labeled packed tableau for each root subtree , then merge all these tableaux to get a tableau , and finish by applying or (according to the root color) to , labeling the new row or column by : the result is .

### 4.2. Arc diagrams

We now introduce alternative arc diagrams, that turn out to be a nice representation of plane alternative forests. We will call arc diagram the data of points aligned horizontally, labeled increasingly from left to right by integers and of arcs where are two of the labels. It is thus a particular representation of a labeled (simple, loopless) graph where the vertices are ordered according to their value. Given an arc diagram, we say that an arc is topmost on its right side if there is no arc with , and that it is topmost on its left side if there is no arc with .

###### Definition 4.5 (alternative arc diagram).

Let be a label set of size at least , with minimal and maximal elements and respectively. An arc diagram with points labeled by is called alternative if the following three conditions are verified:

1. at each vertex , there are no two arcs and for some integers .

2. as an abstract graph, it is a tree;

3. each arc is topmost on exactly one of its sides.

An example is shown on Figure 11. Every arc has been oriented from its topmost side for clarity; moreover, a vertex is colored white when all arcs adjacent to it are of the form with ; it is colored black otherwise. Let be a plane alternative forest labeled on , and consider points aligned horizontally with labels from left to right. Add an arc between points and for each edge of the forest, an arc for each black root and an arc for each white root . Finally put an arc between and , and let the resulting arc diagram be . For instance, the diagram of Figure 11 corresponds through to the forest of Figure 10.

###### Proposition 4.6.

The procedure is a bijection from alternative forests to alternative arc diagrams .

Proof: We first check the three conditions in Definition 4.5, then show the bijectivity. So let us be given a set of arcs on points coming from a forest , and show that it is an arc diagram. Condition is trivial for points and ; every other point is the label of a vertex in . If this vertex is black, then all its descendants have smaller labels, and its father also; therefore in the diagram, all arcs go to the left of . A similar proof shows that all white vertices become points from which all arcs go right. Condition is clear, because is a forest by hypothesis, and the arcs , and make it into a tree.

We finally want to check condition ; let an arc , , be given. If or the result is immediate; now suppose is topmost in ; we’ll show that it’s not topmost in , and by symmetry we’ll have that if is topmost in then it’s not topmost in , which will conclude the proof. But topmost in means that is the father of in ; by minimality, the father of will necessarily be less than , so that is not topmost in . And if has no father in , then it’s a black root, thus there is an arc so that is not topmost in in this case either.

Consider now the following construction: given an arc diagram, color in white (respectively black) all points () at which arcs go to the left (resp. to the right). Then destroy all arcs and : the corresponding vertices and are then roots of certain trees, which form a forest. It is immediate that this is precisely the inverse of .

### 4.3. Crossings in alternative arc diagrams.

There is a very elementary way to describe the composition of with the bijection ; let us call this bijection from labeled tableaux to diagrams.

Given a tableau of length with standard labeling, and points labeled from to . Then draw an arc for all cells filled with an arrow (up or left). Draw also an arc for each free column , an arc for each free row , and finally an arc ; the result is the alternative arc diagram . We have then the result

###### Proposition 4.7.

The construction is a bijection from alternative tableaux of length to alternative arc diagrams on the labels , and coincides with the composition .

In an alternative arc diagram, we call crossing a pairs of arcs with