Correlations in the continuous multispecies TASEP on a ring

Surjadipta De Sarkar Department of Mathematics, Indian Institute of Science, Bangalore - 560012 [email protected] and Nimisha Pahuja Department of Mathematics, Indian Institute of Science, Bangalore - 560012 [email protected]

Abstract.

In this paper, we prove a conjecture proposed by Aas and Linusson regarding the two-point correlations of adjacent particles in a continuous multispecies TASEP on a ring (AIHPD, 2018). We use the theory of multiline queues as devised by Ferrari and Martin (AOP, 2008) to interpret the conjecture in terms of the placements of numbers in triangular arrays. Further, we use projections to calculate the correlations in the continuous multispecies TASEP using a distribution on these placements.

Key words and phrases:

multispecies, correlation, TASEP, continuous, lumping

1. Introduction

Multispecies exclusion processes and their combinatorial properties have been an intense topic of investigation in recent times [5, 6, 18, 7]. One property which is of great interest is the correlations of adjacent particles in the stationary distribution of the process [1, 3, 9]. The aim of this paper is to prove a conjecture of Aas and Linusson [3] on correlations of two adjacent points in a multispecies totally asymmetric exclusion process (TASEP) on a continuous ring. The definitions will be given in Section 2.

In recent years, a lot of attention has been given to various properties of the multispecies TASEP. Connections with TASEPs to different objects like affine Weyl groups [15, 2], Macdonald polynomials [11], Schubert polynomials [14] and multiline queues [12] have been explored extensively. Multispecies TASEP on a ring has been studied in [8, 4, 10]. Ayyer and Linusson [9] studied multispecies TASEP on a ring where they proved conjectures by Lam [15] on random reduced words in an affine Weyl group and gave results on two-point and three-point correlations.

One of the first instances where the continuous multispecies TASEP on a ring was mentioned is by Aas and Linusson [3]. They studied a distribution of labelled particles on a continuous ring which arises as a certain infinite limit of the stationary distribution of multispecies TASEP on a ring. They also conjectured [3, Conjecture 4.2] a formula for correlations $c_{i,j}$ which is given by the probability that the two particles, labelled $i$ and $j$ are next to each other with $i$ on the left of $j$ in the limit distribution. We prove this conjecture first for the case $i>j$ (Theorem 1.1) in Section 3 and then for the case $i<j$ (Theorem 1.2) in Section 4. The technique we use is similar to and inspired by the work of Ayyer and Linusson [9] where they study correlations in multispecies TASEP on a ring with a finite number of sites.

To carry out the analysis, we use the theory of multiline processes that Ferrari and Martin described in [12]. The multiline process is defined on structures known as multiline queues or MLQs. This process can be projected to the multispecies TASEP using a procedure known as lumping (see [16, Lemma 2.5]). This projection lets us study the stationary distribution of the multiline process to infer results for the stationary distribution of the multispecies TASEP and is defined using an algorithm known as bully path projection which projects a multiline queue to a word. See Section 2 for the precise definitions.

We study the two-point correlations in a continuous TASEP with type $\operatorname{\langle 1^{n}\rangle}=\underbrace{(1,\ldots,1)}_{n}$ . In this case, there are $n$ particles, each with a distinct label from the set $[n]=\{1,\ldots,n\}$ . In this regard, let $c_{i,j}(n)$ denote the probability that the two particles, labelled $i$ and $j$ are next to each other with the label $i$ on the left of label $j$ in the continuous multispecies TASEP with type $\operatorname{\langle 1^{n}\rangle}$ . Aas and Linusson gave an explicit conjecture ([3, Conjecture 4.2]) for calculating $c_{i,j}(n)$ . We prove their conjecture in this paper separately for the two cases.

Theorem 1.1.

For $n\geq 2$ and $i>j$ , we have the following two-point correlations:

c_{i,j}(n)=\begin{cases}\frac{n}{\binom{n+j}{2}}-\frac{n}{\binom{n+i}{2}},&\text{ if }j<i<n,\\ \frac{n(j+1)}{\binom{n+j}{2}}-\frac{n(j-1)}{\binom{n+j-1}{2}}-\frac{n}{\binom{2n}{2}},&\text{ if }j<i=n.\end{cases}

Theorem 1.2.

For $n\geq 2$ and $i<j$ , we have the following two-point correlations:

c_{i,j}(n)=\begin{cases}\frac{n}{\binom{n+j}{2}},&\text{ if }i+1<j\leq n,\\ \frac{n}{\binom{n+j}{2}}+\frac{ni}{\binom{n+i}{2}},&\text{ if }i+1=j\leq n.\end{cases}

2. Preliminaries

A multispecies TASEP is a stochastic process on a graph. We first define multispecies TASEP on a ring before proceeding to study the continuous multispecies TASEP on a ring.

2.1. Multispecies TASEP

A multispecies TASEP is a continuous-time Markov process which can be defined on a ring with $L$ sites. For a tuple $m=(m_{1},\ldots,m_{n}$ ), a multispecies TASEP of type $m$ has $m_{1}+\cdots+m_{n}$ sites occupied with particles. Each particle is assigned a label from the set $[n]$ and there are exactly $m_{k}$ particles with the label $k$ . The unoccupied sites are treated as particles with the highest label $n+1$ . The dynamics of the process are as follows: Each particle carries an exponential clock that rings with rate $1$ . The particle then tries to jump to the site on its left whenever the clock rings. Let this particle be labelled $i$ . The jump is successful only if the site on the left has a label greater than $i$ . In that case, the two particles exchange positions. The states of the multispecies TASEP are words of length $L$ with the letter $k$ occurring $m_{k}$ times for all $k\in[n]$ and $n+1$ occurring $L-\sum_{i}m_{i}$ times.

Now we define the multiline process which can be projected to the multispecies TASEP through lumping. A multiline process is a Markov process defined on a graph which is a collection of disjoint cycle graphs of the same length. We refer to each path graph as a row and the rows are numbered from top to bottom. Each row has the same number of sites and each site may or may not be occupied by a particle. For an $n$ -tuple $m=(m_{1},\ldots,m_{n})$ with $m_{k}\geq 0$ and $L\geq m_{1}+\cdots+m_{n}$ , a multiline queue of type $m$ is a collection of $n$ rows, each having $L$ sites, stacked on top of each other. In the $i^{th}$ row from the top, $S_{i}=m_{1}+\cdots+m_{i}$ of the sites are occupied. See Figure 1 for an example of a multiline queue.

Refer to caption — Figure 1. A multiline queue of type $(2,1,2,2)$ on $13$ sites.

The dynamics of the multiline process are described in detail in [12] via transitions on the multiline queues of a fixed type. The stationary distribution of the process is stated in the following theorem.

Theorem 2.1.

[12, Theorem 3.1] The stationary distribution of the multiline process of type $m$ is uniform.

A multiline queue of type $m$ can be projected to a word of the same type by an algorithm known as the bully path procedure which we define recursively as follows:

(1)

Let $M$ be a multiline queue of type $m$ . We construct bully paths that contain exactly one particle from each row. Start with the first row in $M$ . The bully path starting at any particle in the first row moves downwards and then rightwards along the multiline queue until it runs into a particle in the second row. It again moves downwards and rightwards in the third row till it hits another particle, and so on all the way to the last row. All the particles encountered by this bully path are labelled $1$ . We similarly construct the bully paths starting from other particles in the first row. It turns out that the order in which these paths are constructed starting from the particles in the first row does not matter. At the end of this construction, we have a total of $m_{1}$ bully paths. That is, $m_{1}$ particles of the last row are labelled $1$ . See Figure 2 for the construction of bully paths to the multiline queue in Figure 1.
(2)

Next, we construct bully paths starting with the unlabelled particles in the second row by repeating the same process from Step (1). We label the ends of all such paths as $2$ and they are $m_{2}$ in number. We repeat these steps for all the subsequent rows. Finally, label all the particles that are left unlabelled in the last row as $n$ and all the unoccupied sites as $n+1$ .
(3)

Hence for each $\ell$ , $m_{\ell}$ particles in the last row are labelled as $\ell$ . Let $\omega$ denote the word formed by the labels in the last row. Then, $\omega$ is a configuration of the multispecies TASEP of type $m$ . Let $\mathbf{B}$ denote this projection map. Then, $\omega$ is known as the projected word of $M$ and we write it as $\omega=\mathbf{B}(M)$ . The projected word for the multiline queue in Figure 1 is 3345515525145; see Figure 2.

The connection between the stationary distributions of the multiline process and the multispecies TASEP is established by the following theorem given by Ferrari and Martin [12].

Theorem 2.2.

[12, Theorem 4.1] The process on the last row of the multiline process of type $m$ is the same as the multispecies TASEP of type $m$ .

2.2. Continuous multispecies TASEP

Fix $m=(m_{1},\ldots,m_{n})$ and let $S_{i}=m_{1}+\cdots+m_{i}$ , for all $i\in[n]$ . The continuous multispecies TASEP is a formal limit of the stationary distribution of the multispecies TASEP on a ring with $L$ sites. First, we first consider a multispecies TASEP of type $m$ on a ring with $L$ sites. Let $\Pi_{n}^{L}$ be the stationary distribution of this TASEP. Then, $L$ is taken to infinity keeping the vector $m$ constant, i.e., the number of unoccupied sites tends to infinity. The ring is then scaled to the continuous interval $[0,1)$ . The limit of the stationary distribution $\Pi_{n}^{L}$ gives a distribution $\Pi_{n}$ of labelled particles on a continuous ring. Note that $\Pi_{n}$ is not yet shown to be the stationary distribution of any Markov process yet.

Similar to the multiline queues in [12], we can look at the continuous multiline queues of a given type. For $m=(m_{1},\ldots,m_{n})$ , let $S_{i}=m_{1}+\cdots+m_{i}$ . Then, a continuous multiline queue of type $m$ is a collection of $n$ copies of the continuous ring $[0,1)$ stacked on top of each other with $S_{i}$ particles in $i^{th}$ row from the top.

The location of each particle is considered to be a real number in the continuous interval $[0,1)$ . In the distribution that we will consider, the horizontal position of each particle will almost surely be distinct.

Example 2.3.

See Figure 3 for an example of a continuous multiline queue of type $(1,3,1,2)$ . The rows have $1,4,5$ and $7$ particles respectively. Note that there is no particle directly above or below any other particle.

Consider the labels of the particles in Figure 3. The labels are assigned in the order of the horizontal positions of the particles as seen from left to right. We now refer to an integer representation of a continuous multiline queue which was also used by Aas and Linusson [3].

Definition 2.4.

Let $m=(m_{1},\ldots,m_{n})$ be an $n$ -tuple, $S_{i}=m_{1}+\cdots+m_{i}$ and let $N=\sum_{i=1}^{n}S_{i}$ . A placement of a continuous multiline queue of type $m$ is a triangular array $(p_{i,k})$ with distinct integers from the set $[N]$ such that the integer $p_{i,k}$ stands for the relative horizontal position of $k^{th}$ particle in the $i^{th}$ row of the multiline queue as seen from left to right.

Example 2.5.

The placement corresponding to the continuous multiline queue in Figure 3 is

\begin{array}[]{ccccccc}5&&&&&&\\ 1&3&7&9\\ 8&10&13&15&16\\ 2&4&6&11&12&14&17.\end{array}

Each row of the array is formed by adding the order of the particles in the corresponding row of the continuous multiline queue when their horizontal positions are sorted from left to right.

For our purpose, it is enough to know the relative positions of particles on the rows and hence a continuous multiline queue will be represented by its placement and we will use the two terms interchangeably. The number of different placements of type $m$ is given by $\binom{N}{S_{1}\,,\ldots,\,S_{n}}$ .

The bully path projection is a map from the set of continuous multiline queues of type $m$ to words of type $m$ . Let $M$ be a continuous multiline queue. We define the algorithm recursively which takes $M$ and maps it to a word $\omega$ as follows:

(1)

Consider the placement of $M$ . For each integer $k_{1}$ in the first row, look for the smallest available entry larger than $k_{1}$ in the second row and mark it as $k_{2}$ . If $k_{1}$ is larger than all the available integers in the second row, we mark the smallest available integer in the second row as $k_{2}$ . This is known as wrapping from the first row to the second row. We say that $k_{1}$ “bullies” $k_{2}$ and write it as $k_{1}\rightarrow k_{2}$ ; or $k_{1}\xrightarrow{W}k_{2}$ in the case of wrapping. See Figure 4 for the map applied to Example 2.5.
(2)

Look for the smallest available integer in the third row larger than $k_{2}$ and label it $k_{3}$ and so on. The sequence $k_{1},k_{2},\ldots,k_{n}$ thus obtained is called a bully path starting at $k_{1}$ and these integers are now unavailable for further bullying. Label the endpoints of all such paths as $1$ . There are $m_{1}$ such paths and we call them type $1$ bully paths. In Figure 4, $5\rightarrow 7\rightarrow 8\rightarrow 11$ is a bully path of type $1$ . The order in which we construct the bully paths starting in the first row does not matter.
(3)

Next, we construct bully paths starting with the available integers in the second row by following steps (1) and (2). We label the ends of all such paths with $2$ and they are $m_{2}$ in number. We repeat these steps for all the other rows sequentially. The bully paths of type $n$ are just the integers in the last row that are remaining after the construction of all type $(n-1)$ bully paths.
(4)

Therefore, there are $m_{\ell}$ bully paths of type $\ell$ for each $\ell$ . Let $\omega$ denote the word formed by the labels in the last row. Then, $\omega$ is a configuration for the continuous multispecies TASEP of type $m$ . Let $\mathbf{B}$ denote this projection. Then, $\omega=\mathbf{B}(M)$ is known as the projected word of $M$ . The projected word for the continuous multiline queue in Example 2.5 is 3441222; see Figure 4.

The distribution of the words of type $m$ is the same as the distribution of the last row of continuous multiline queues of type $m$ which can be obtained by taking the limit of the distribution of the last row for discrete multiline queues. Also, the distribution of the last row for discrete multiline queues is $\Pi_{n}^{L}$ by Theorem 2.2. Therefore, $\Pi_{n}$ gives the distribution of the labels on the last row for a uniformly chosen continuous multiline queue. Thus, to study the correlations of two adjacent particles with labels $i$ and $j$ in the continuous multispecies TASEP, it is enough to count the number of placements that project to the words with $i$ and $j$ as adjacent particles. Next, we define an operator on the space of all continuous multiline queues of a fixed type.

Definition 2.6.

Let $M$ be a continuous multiline queue of type $m$ . Let $S$ be an operator such that if $M^{\prime}=S(M)$ , then $M^{\prime}$ is obtained from $M$ by adding $1\mod N$ to each entry of $M$ and then rearranging any row in increasing order if needed. $S$ is known as the shift operator and we will refer to $M^{\prime}$ as the shifted multiline queue.

Example 2.7.

Let $M$ be the multiline queue from Example 2.5, then $S(M)=M^{\prime}$ is given by

M^{\prime}=\begin{array}[]{ccccccc}6&&&&&&\\ 2&4&8&10\\ 9&11&14&16&17\\ 1&3&5&7&12&13&15.\end{array}

The following lemma relates the projected words of a multiline queue and its shifted multiline queue.

Lemma 2.8.

Let $M$ be a continuous multiline queue of type $m$ with $N=\displaystyle\sum_{i=1}^{n}S_{i}$ particles. Shifting $M$ rotates the projected word by one position to the right when the largest entry $N$ is in the last row of the placement and preserves the word otherwise. In other words, if $M^{\prime}=S(M)$ , $\omega=\mathbf{B}(M)$ and $\omega^{\prime}=\mathbf{B}(M^{\prime})$ , then $\omega$ and $\omega^{\prime}$ are related in the following way:

(1)

if $N$ is not in the last row, then $\omega^{\prime}=\omega$ .
(2)

if $N$ is in the last row, then $\omega^{\prime}$ is obtained by rotating $\omega$ one position to the right.

Proof.

Let $N$ be in the $r^{th}$ row of the placement of $M$ . First, let $r<n$ . $M^{\prime}$ is obtained from $M$ by adding $1$ to every integer other than $N$ and by replacing $N$ with $1$ . By the increasing property of the rows, the $r^{th}$ row is now rotated by one position to the right. If $N$ lies on a bully path that starts in some row above $r$ , then all the bully paths remain the same. This is true because $N$ , the largest integer in $M$ , wraps around and bullies the first entry in $(r+1)^{st}$ row available to it. Whereas in $M^{\prime}$ , $1$ being the smallest integer bullies the first entry available to it which is exactly the translation of the entry bullied by $N$ in $M$ . The remaining bully paths are the same since the inequalities among all other elements do not change.

On the other hand, let there be a bully path of type $r$ in $M$ that starts at $N$ . Let $s=r+1$ and the available integers in the $s^{th}$ row in $M$ after the construction of all the bully paths of a type less than $r$ be $s_{1}<s_{2}<\cdots<s_{n_{t}}$ . If $N\xrightarrow{W}s_{x}$ (say) in $M$ , then observe that there exist elements in the $r^{th}$ row, namely $r_{i}\,(1\leq i\leq x-1)$ such that $r_{1}$ bullies $s_{1}$ , $r_{2}$ bullies $s_{2},\ldots,r_{x-1}$ bullies $s_{x-1}$ in $M$ . In $M^{\prime}$ , the bully paths that begin in a row above the $r^{th}$ are the same as those in $M$ . For a type $r$ bully path, $1$ bullies $(s_{1}+1)$ , $(r_{1}+1)$ bullies $(s_{2}+1),\ldots,(r_{x-1}+1)$ bullies $(s_{x}+1)$ . The construction of these type $r$ bully paths in $M^{\prime}$ from here onwards is the same as the construction of those in $M$ . Thus, the projected word remains the same.

Finally, when $r=n$ , i.e., when $N$ is in the last row, the last row rotates by one position to the right when adding $1\mod N$ to each entry in $M$ . The bully paths remain the same and therefore, the projected word which is obtained from the labels of the particles in the last row is rotated by one position to the right. ∎

Let $\operatorname{\langle 1^{n}\rangle}=(1,\ldots,1)$ be an $n$ -tuple. Define $c_{i,j}(n)=\mathbb{P}\{\eta_{a}=i,\eta_{a+1}=j;a\in[n]\}$ . To make the analysis of continuous multispecies TASEP of type $\operatorname{\langle 1^{n}\rangle}$ easy, we use a property known as the projection principle which appears in [9] in a similar context. To obtain $c_{i,j}(n)$ for $i>j$ , it is enough to find the probability that $4$ is followed by a $2$ in the projection of the word of the five-species continuous multiline queue with type $m_{i,j}=(j-1,1,i-j-1,1,n-i).$ This is called the projection principle.

We can further simplify the task by projecting many such continuous multiline queues to a three-species system. Given a continuous multiline queue with type $\operatorname{\langle 1^{n}\rangle}$ , consider its projection to a continuous multiline queue of type $m_{s,t}=(s,t,n-s-t)$ where $j>s$ and $i>s+t$ . Thus, a particle of class $j$ becomes a $2$ and that of class $i$ becomes a $3$ whenever $t,n-s-t>0$ . So, to compute the correlation of $i$ and $j$ in a continuous multispecies TASEP of type $\operatorname{\langle 1^{n}\rangle}$ , we need to look at the correlation of $3$ and $2$ in the projected words of continuous multiline queues of type $m_{s,t}$ . Similarly, to formulate the correlation of $i$ and $j$ for $i<j$ , we need to look at the correlation of $2$ and $3$ in the projected words of continuous multiline queues with type $m_{s,t}$ .

Let $M$ be a continuous multiline queue of type $\operatorname{\langle 1^{n}\rangle}$ and $\eta=\mathbf{B}(M)$ be the word that is projected from $M$ using bully path projection. For $1\leq i,j\leq n$ , let $E^{a}_{i,j}(n)=\mathbb{P}\{\eta_{a}=i,\eta_{a+1}=j\}$ for $a\in[n]$ . Thus,

(2.1)

c_{i,j}=\sum_{a=1}^{n}E^{a}_{i,j}.

Consider a continuous multiline queue $M^{\prime}$ of type $m_{s,t}$ . Let the projected word of $M^{\prime}$ be $\omega$ . If

	$\displaystyle\Theta_{a}^{<}(s,t)=\mathbb{P}\{\omega_{a}=2,\omega_{a+1}=3\},$
	$\displaystyle\text{and }\Theta_{a}^{>}(s,t)=\mathbb{P}\{\omega_{a}=3,\omega_{a+1}=2\}$

for $a\in[n]$ , then by projection principle, we have

	$\displaystyle\Theta_{a}^{<}(s,t)$	$\displaystyle=\sum_{j=s+t+1}^{n}\sum_{i=s+1}^{s+t}E^{a}_{i,j},$
	$\displaystyle\text{ and }\Theta_{a}^{>}(s,t)$	$\displaystyle=\sum_{i=s+t+1}^{n}\sum_{j=s+1}^{s+t}E^{a}_{i,j}.$

Let $\mathscr{T}^{<}$ (respectively $\mathscr{T}^{>}$ ) be the sum of $\Theta_{a}^{<}$ (respectively $\Theta_{a}^{>}$ ) over the index $a$ . We have,

	$\displaystyle\mathscr{T}^{<}(s,t)$	$\displaystyle=\sum_{a=1}^{n}\sum_{j=s+t+1}^{n}\sum_{i=s+1}^{s+t}E^{a}_{i,j}(n)$
		$\displaystyle=\sum_{j=s+t+1}^{n}\sum_{i=s+1}^{s+t}c_{i,j}(n).$

The last equality follows from (2.1). Similarly,

\displaystyle\mathscr{T}^{>}(s,t)

\displaystyle=\sum_{i=s+t+1}^{n}\sum_{j=s+1}^{s+t}c_{i,j}(n).

For $i<j$ , the principle of inclusion-exclusion then gives us

(2.2)

c_{i,j}(n)=\mathscr{T}^{<}(i-1,j-i)-\mathscr{T}^{<}(i,j-i-1)-\mathscr{T}^{<}(i-1,j-i+1)+\mathscr{T}^{<}(i,j-i),

and for $i>j$ , we have

(2.3)

c_{i,j}(n)=\mathscr{T}^{>}(j-1,i-j)-\mathscr{T}^{>}(j,i-j-1)-\mathscr{T}^{>}(j-1,i-j+1)+\mathscr{T}^{>}(j,i-j).

For $a\in[n]$ , if we let

	$\displaystyle T_{a}^{<}(s,t)$	$\displaystyle=\mathbb{P}\{\omega_{a}=2,\omega_{a+1}=3,M^{\prime}_{3,n}=N\},$
	$\displaystyle\text{ and }T_{a}^{>}(s,t)$	$\displaystyle=\mathbb{P}\{\omega_{a}=3,\omega_{a+1}=2,M^{\prime}_{3,n}=N\},$

then the following lemma holds.

Lemma 2.9.

$\mathscr{T}^{<}(s,t)=(n+2s+t)\,T_{a}^{<}(s,t)$ for any $a\in[n]$ .

Proof.

Let $M^{\prime}$ be a continuous multiline queue of type $m_{s,t}$ . Shifting it $N=n+2s+t$ times generates all the rotations of the projected word. By Lemma 2.8, in exactly $n$ out of $N$ shifted continuous multiline queues, the projected word rotates one unit to the right and in the remaining shifts, the projected word remains same. For a fixed $a\in[n]$ , any continuous multiline queue which contributes to $\mathscr{T}^{<}(s,t)$ can be obtained as a rotation of a continuous multiline queue for which

(i)

the projected word has $3$ and $2$ (or $2$ and $3$ ) in positions $a$ and $a+1\mod n$ respectively, and
(ii)

$N$ is in the last row.

Note that if a word has a $3$ followed by a $2$ in $p$ separate positions, it occurs as a rotation of $p$ different words with $\omega_{a}=3,\omega_{a+1}=2$ . Hence, the result. ∎

We also have $\mathscr{T}^{>}(s,t)=(n+2s+t)T_{a}^{>}(s,t)$ for any $a\in[n]$ . Further, Lemma 2.9 holds for $a=1$ in particular. Henceforth, we will write $T_{1}^{<}$ (respectively $T_{1}^{>}$ ) as $T^{<}$ (respectively $T^{>}$ ) for simplicity. Then, (2.2) and (2.3) become

	$\displaystyle c_{i,j}=$	$\displaystyle(n+i+j-2)\,T^{<}(i-1,j-i)-(n+i+j-1)\,T^{<}({i,j-i-1})$
(2.4)			$\displaystyle-(n+i+j-1)\,T^{<}({i-1,j-i+1})+(n+i+j)\,T^{<}({i,j-i}),$

and

	$\displaystyle c_{i,j}=$	$\displaystyle(n+i+j-2)\,T^{>}({j-1,i-j})-(n+i+j-1)\,T^{>}({j,i-j-1})$
(2.5)			$\displaystyle-(n+i+j-1)\,T^{>}({j-1,i-j+1})+(n+i+j)\,T^{>}({j,i-j})$

respectively.

2.3. Standard Young tableaux

Consider a partition $\lambda$ = $(\lambda_{1},\lambda_{2},\ldots,\lambda_{\ell})$ of a positive integer $n$ , where $\lambda_{1}\geq\lambda_{2}\geq\cdots\geq\lambda_{\ell}>0$ and $n=\lambda_{1}+\lambda_{2}+\cdots+\lambda_{\ell}$ . We use the notation $\lambda\vdash n$ or $n=|\lambda|$ to indicate that $\lambda$ is a partition of $n$ . The Young diagram of a partition $\lambda$ is an arrangement of boxes in $\ell$ left-aligned rows, with $\lambda_{i}$ boxes in the $i^{th}$ row. The hook of a box $a$ in the Young diagram is defined as the set of boxes that are directly to the right or directly below $a$ , including $a$ itself. The hook length of a box $a$ , denoted by $h_{a}$ , is the number of boxes in its hook. Below is an example of a Young diagram of shape $(5,3,1)$ with the hook lengths for each box indicated within the box.

\young

(75421,421,1)

A standard Young tableau or an SYT of shape $\lambda$ is a filling of a Young diagram of shape $\lambda$ with the numbers $1,2,\ldots,n$ , where $n=|\lambda|$ , such that the numbers increase strictly down a column and across a row. The following is an example of a standard Young tableau of shape $(5,3,1)$ .

\young

(12569,378,4)

The number of standard Young tableaux of a given shape $\lambda$ can be counted using the following result which is known as the hook-length formula.

Theorem 2.10.

[13] Let $\lambda\vdash n$ be an integer partition. The number of standard Young tableaux of a shape $\lambda$ is given by

(2.6)

f_{\lambda}=\frac{n!}{\Pi_{a\in\lambda}h_{a}},

where the product is over all the boxes in the Young diagram of $\lambda$ and $h_{a}$ is the hook length of a box $a$ .

Example 2.11.

For $\lambda=(a,b)$ , the number of standard Young tableaux of shape $\lambda$ is given by

(2.7)

f_{(a,b)}=\frac{(a+b)!(a-b+1)}{(a+1)!b!}=\frac{a-b+1}{a+1}\binom{a+b}{a}.

Similarly for $\mu=(a,b,c)$ , we have

	$\displaystyle f_{(a,b,c)}$	$\displaystyle=\frac{(a+b+c)!(a-c+2)(a-b+1)(b-c+1)}{(a+2)!(b+1)!c!}$
(2.8)			$\displaystyle=\frac{(a-c+2)(a-b+1)(b-c+1)}{(a+2)(a+1)(b+1)}\binom{a+b+c}{a,b,c}.$

Remark 2.12.

It is straightforward to verify from (2.7), that $f_{(a,b)}$ satisfies an interesting recurrence relation given by:

f_{(a,b)}=f_{(a-1,b)}+f_{(a,b-1)}.

Lemma 2.13.

Let $m=(m_{1},\ldots,m_{n})$ , and let $S_{i}=m_{1}+\cdots+m_{i}$ . The set of continuous multiline queues of type $m$ that have no wrapping under the bully path projection is in bijection with standard Young tableaux of shape $\lambda$ where $\lambda_{i}=S_{n-i+1}=m_{1}+\cdots+m_{n-i+1}$ .

Proof.

Consider a continuous multiline queue of type $m$ , denoted by $M$ , such that the largest integer in $M$ is given by the sum $N=S_{1}+\cdots+S_{n}$ . For $1\leq i\leq n,1\leq j\leq S_{i}$ , let $M_{i,j}$ be distinct integers from $1$ to $N$ . Then, we have

M=\begin{array}[]{cccccccc}M_{1,1}&\ldots&M_{1,S_{1}}\\ M_{2,1}&\ldots&\ldots&M_{2,S_{2}}\\ \vdots\\ M_{n,1}&\ldots&\ldots&\ldots&M_{n,S_{n}}.\end{array}

If we assume that there is no wrapping from the first row to the second row in $M$ , then it follows that $M_{1,S_{1}-k}<M_{2,S_{2}-k}$ for all $0\leq k\leq S_{1}-1$ . Similarly, since there is no wrapping from the second row to the third row in $M$ , $M_{2,S_{2}-k}<M_{3,S_{3}-k}$ for all $0\leq k\leq S_{2}-1$ , and so on. This along with the increasing property of the rows gives us

\begin{array}[]{ccccccccccccc}&&&&M_{1,1}&<&\cdots&<&M_{1,S_{1}-i}&<&\cdots&<&M_{1,S_{1}}\\ &&&&\wedge&&&&\wedge&&\cdots&&\wedge\\ &&M_{2,1}&<&\cdots&<&\cdots&<&M_{2,S_{2}-i}&<&\cdots&<&M_{2,S_{2}}\\ &&\wedge&&&&&&\wedge&&&&\wedge\\ &&&&&&&&&&&&\vdots\\ M_{n,1}&<&\cdots&<&\cdots&<&\cdots&<&M_{n,S_{n}-i}&<&\cdots&<&M_{n,S_{n}}.\end{array}

Hence, $M$ is in bijection with a standard Young tableau $\mathscr{Y}$ of shape $(S_{n},\ldots,S_{1})$ and the bijection is given by $\mathscr{Y}_{i,j}=N-M_{n-i+1,S_{(n-i+1)}-j+1}+1$ . Thus, the number of continuous multiline queues of type $m$ with no wrapping is given by $f_{(S_{n},\ldots,S_{1})}$ . ∎

Definition 2.14.

Given two partitions $\lambda,\mu$ such that $\mu\subseteq\lambda$ (containment order, i.e., $\mu_{i}\leq\lambda_{i}$ for all $i$ ), the skew shape $\lambda/\mu$ is a Young diagram that is obtained by subtracting the Young diagram of shape $\mu$ from that of $\lambda$ .

\ydiagram

2+2,1+2,2,1 \ydiagram2+3,3,2

Figure 5. Skew shapes

(4,3,2,1)/(2,1)

and

(5,3,2)/(2)

respectively

Definition 2.15.

A standard Young tableau of a skew shape $\lambda/\mu$ is a filling of the skew shape by positive integers that are strictly increasing in rows and columns.

Example 2.16.

Following are a few of the many standard Young tableaux of the same shape $(4,3,2,1)/(2,1)$ .

\ytableausetup

notabloids {ytableau} \none& \none 1 4

\none

2 3

5 6

7 \ytableausetupnotabloids {ytableau} \none& \none 1 2

\none

5 7

3 6

4 \ytableausetupnotabloids {ytableau} \none& \none 2 3

\none

1 7

4 6

Let $f_{(\lambda/\mu)}$ denote the number of SYTs of a skew shape $\lambda/\mu$ . This number is counted by the following formula.

Theorem 2.17.

[17, Corollary 7.16.3] Let $|\lambda/\mu|=n$ be the number of boxes in the skew shape $\lambda/\mu$ that has $\ell$ parts. Then,

(2.9)

f_{(\lambda/\mu)}=n!\text{det}\Big{(}\frac{1}{(\lambda_{i}-\mu_{j}-i+j)!}\Big{)}_{i,j=1}^{\ell}

Note that we take $0!=1$ and $k!=0$ for any $k<0$ as a convention.

Example 2.18.

Let $\lambda/\mu=(6,4)/(3)$ . Then, $n=|\lambda/\mu|=7$ and $\ell=2$ . We have,

f_{(\lambda/\mu)}=7!{\Large\begin{vmatrix}\frac{1}{(6-3-1+1)!}&\frac{1}{(6-0-1+2)!}\\ \frac{1}{(4-3-2+1)!}&\frac{1}{(4-0-2+2)!}\end{vmatrix}}=34.

2.4. Two-species continuous TASEP

Let $m=(s,t)$ . In this section, we study continuous TASEP with two kinds of particles and a hole. Because of the simplicity of the structure, it is easy to completely calculate the stationary distribution of the continuous TASEP on two species. Let the correlation $c_{i,j}(n)$ be defined as the probability that the particle labelled $i$ is immediately followed on the ring by a particle labelled $j$ in the limit distribution. Once again, we use the continuous multiline queues of type $m$ to find these correlations.

Let $\Omega_{s,t}$ be the set of continuous multiline queues of type $(s,t)$ and let $n=s+t$ . That is, $M\in\Omega_{s,n-s}$ is a continuous multiline queue that has $s$ integers in the first row and $n$ integers in the second row. For $c\in\{1,2\}$ , let $\delta_{c}(s,n)$ be the number of continuous multiline queues of $\Omega_{s,n-s}$ with the largest integer $n+s$ in the second row such that the projected word has $c$ in the first position. By the well-known property of rotational symmetry of multispecies TASEP (see [9, Proposition 2.1 (iii)]), we have

(2.10)

\delta_{1}(s,n)=\frac{s}{n}\binom{n+s-1}{s},

(2.11)

\delta_{2}(s,n)=\frac{n-s}{n}\binom{n+s-1}{s}.

Remark 2.19.

Note that the number of configurations of $\Omega_{s,n-s}$ with $n+s$ in the second row such that the projected word has $c$ in the $a^{th}$ position is the same for any $a\in[n]$ by rotational symmetry.

Similarly, let $\delta_{c,d}(s,n)$ count the number of continuous multiline queues of type $(s,n-s)$ that have the largest integer $n+s$ in the second row such that the projected word has $c$ in the first place and $d$ in the second place. Using Remark 2.19, we have the following system of independent equations for fixed $s$ and $n$ .

(2.12)		$\displaystyle\delta_{1,1}+\delta_{1,2}=\delta_{1},$
(2.13)		$\displaystyle\delta_{1,1}+\delta_{2,1}=\delta_{1},$
(2.14)		$\displaystyle\delta_{1,2}+\delta_{2,2}=\delta_{2}.$

Therefore, finding $\delta_{c,d}$ for any $(c,d)\in\{1,2\}\times\{1,2\}$ solves the system of equation. In particular, let $c=d=2$ . Consider an arbitrary continuous multiline queue $M^{\prime}$ such that $\pi=\mathbf{B}(M^{\prime})$ with the following configuration:

\begin{array}[]{cccccccc}&a_{1}&a_{2}&\ldots&a_{s}\\ &b_{1}&b_{2}&\ldots&\ldots&b_{n-1}&n+s\\ \hline\cr\pi=&2&2&\ldots&\ldots&\ldots&\ldots\end{array}.

Since $\pi_{1}=\pi_{2}=2$ , $b_{1}$ and $b_{2}$ are not bullied by any $a_{i}$ in $M^{\prime}$ . Therefore, $a_{1}>b_{2}$ . We also have $b_{2}>b_{1}$ by the increasing property of the rows. This forces $b_{1}=1$ and $b_{2}=2$ . Moreover, there is no wrapping from the first row to the second row. This implies that the integers in $M^{\prime}$ other than $b_{1}$ and $b_{2}$ satisfy the following inequalities:

\begin{array}[]{ccccccccccccc}&&&&a_{1}&<&\cdots&<&a_{s-1}&<&a_{s}\\ &&&&\wedge&&\cdots&&\wedge&&\wedge\\ b_{3}&<&\cdots&<&b_{n-s+1}&<&\cdots&<&b_{n-1}&<&n+s.\end{array}

Such configurations are in bijection with standard Young tableaux of shape $(n-2,s)$ by Lemma 2.13. Therefore using (2.7), we get

(2.15)

\delta_{2,2}(s,n)=f_{(n-2,s)}=\frac{n-s-1}{n+s-1}\binom{n+s-1}{s}

Using (2.10)-(2.15), we can solve for all the remaining $\delta_{c,d}$ as follows:

(2.16)		$\displaystyle\delta_{1,2}(s,n)$	$\displaystyle=\delta_{2,1}(s,n)=\frac{n-s+1}{n+s-1}\binom{n+s-1}{s-1},$
(2.17)			$\displaystyle\delta_{1,1}(s,n)=2\binom{n+s-2}{s-2}.$

3. Proof of Theorem 1.1

In this section, we prove Theorem 1.1 using the tools developed in Section 2. Let $m_{s,t}=(s,t,n-s-t)$ and let $\mathscr{S}_{s,t}$ be the set of all continuous multiline queues $M$ of type $m_{s,t}$ that satisfy $\omega_{1}=3$ and $\omega_{2}=2$ where $\omega=\mathbf{B}(M)$ . Recall that $T^{>}(s,t)$ = $\mathbb{P}\{\omega_{1}=3,\omega_{2}=2,M_{3,n}=N\}$ . We first compute $T^{>}(s,t)$ and then substitute it in (2.3) to solve for $c_{i,j}$ for the case $i>j$ . Let $\tau_{s,t}$ denote the cardinality of the set $\mathscr{S}_{s,t}$ . That is, $\tau_{s,t}$ counts the continuous multiline queues which have the following structure where $a_{i},b_{i}$ and $c_{i}$ be distinct integers from the set $[N]$ .

\begin{array}[]{cccccccc}&a_{1}&a_{2}&\ldots&\ldots&a_{s}\\ &b_{1}&b_{2}&\ldots&b_{k}&\ldots&b_{s+t}\\ &c_{1}&c_{2}&\ldots&\ldots&\ldots&\ldots&c_{n}=N.\\ \hline\cr\omega=&3&2&\ldots&\ldots&\ldots&\ldots&\ldots\end{array}

Lemma 3.1.

Let $M$ be a continuous multiline queue of type $m_{s,t}$ . $M\in\mathscr{S}_{s,t}$ if and only if the following conditions hold:

(1)

If $a_{1}\rightarrow b_{k}$ , then $b_{k}>c_{2}$ ,
(2)

$c_{1}=1$ and $b_{1}=2$ ,
(3)

there is no wrapping from any of the two rows.

Proof.

Let $M\in\mathscr{S}_{s,t}$ and suppose that $a_{1}\rightarrow b_{k}$ . If $b_{k}<c_{2}$ , then $b_{k}$ bullies either $c_{1}$ or $c_{2}$ and we get $\omega_{1}=1$ or $\omega_{2}=1$ . Hence, (1) holds. If there is any wrapping from the second row to the third row, then we have $\omega_{1}=1$ or $2$ , which is a contradiction. Further $\omega_{2}=2$ implies that $b_{1}\rightarrow c_{2}$ and $b_{1}$ is not bullied by any $a_{i}$ . Therefore, $c_{1}<b_{1}<c_{2}$ and $b_{1}<a_{1}$ , and hence $c_{1}=1$ and $b_{1}=2$ .

If there is any wrapping from the first row to the second row, then we have $a_{i}\rightarrow b_{1}$ for some $i$ , which is a contradiction, thus proving the remaining half of 3. It is straightforward to verify that the three conditions imply $\omega_{1}=3$ , $\omega_{2}=2$ and $c_{n}=N$ . ∎

Proof of Theorem 1.1.

Let $M\in\mathscr{S}_{s,t}$ . We have $c_{1}=1$ , $b_{1}=2$ and $c_{n}=N$ and we have to find the number of ways the remaining entries of $M$ can be assigned values from the set $\{3,\ldots,N-1\}$ complying with the three conditions of Lemma 3.1.

Any $b_{\ell}$ which is bullied by some $a_{m}$ should be larger than $c_{2}$ . Thus, there are at least $s$ integers in the second row that are larger than $c_{2}$ . Therefore, $c_{2}\in\{3,4,\ldots,s+t+2\}$ as there can be at most $s$ integers in the first row and at most $t$ integers in the second row that are less than $c_{2}$ . Let $c_{2}=i$ and let $a_{1}\rightarrow b_{k}$ . Then, $c_{2}<b_{k}$ and $b_{k-1}<a_{1}<b_{k}$ . So, all the numbers from $3$ to $i-1$ are assigned, in order, to $b_{2}<\cdots<b_{k-1}<a_{1}<\cdots<a_{i-k-1}$ . Therefore, $0\leq i-k-1\leq s$ , i.e., $i-s-1\leq k\leq i-1$ . Also, since $k-1$ is the number of entries in the second row that are smaller than $c_{2}$ , it is bounded by $1$ and $t$ . Hence, $2\leq k\leq t+1$ .

The ways in which the remaining entries can be assigned values are in bijection with standard Young tableaux of shape $\lambda_{i,k}=(n-2,s+t-k+1,s-i+k+1)$ by Lemma 2.13 because there is no wrapping from any of the rows. This gives us

(3.1)

\tau_{s,t}=\sum_{i=3}^{s+t+2}\sum_{k=\max\{2,i-s-1\}}^{\min\{i-1,t+1\}}f_{\lambda_{i,k}},

where $f_{\lambda}$ is the number of standard Young tableaux of shape $\lambda.$ By the hook length formula for a $3$ -row partition (2.11), we have:

$\displaystyle f_{\lambda_{i,k}}$	$\displaystyle=$	$\displaystyle\frac{(n+2s+t-i)!(n-s+i-k-1)(n-s-t+k-2)(t+i-2k+1)}{n!(s+t-k+2)!(s-i+k+1)!}$
	$\displaystyle=$	$\displaystyle\frac{(n-s+i-k-1)(n-s-t+k-2)(t+i-2k+1)}{(n+2s+t-i+3)(n+2s+t-i+2)(n+2s+t-i+1)}$
		$\displaystyle\times\binom{n+2s+t-i+3}{n,s+t-k+2,s+k-i+1}.$

Substituting this formula in (3.1) and changing $k$ to $k^{\prime}=k-2$ and $i$ to $j=i-3$ , we get

(3.2)

\tau_{s,t}=\sum_{j=0}^{s+t-1}\sum_{k^{\prime}=\text{max }\{j-s,0\}}^{\text{ min }\{j,t-1\}}\frac{(n-s+j-k^{\prime})(n-s-t+k^{\prime})(t+j-2k^{\prime})}{(n+2s+t-j)(n+2s+t-j-1)(n+2s+t-j-2)}\\ \times\binom{n+2s+t-j}{n,s+t-k^{\prime},s-j+k^{\prime}}.

Note that when $j\geq s$ and $k^{\prime}\leq j-s$ , the multinomial coefficient in (3.2) becomes 0. Therefore, the index $k^{\prime}$ can equivalently be summed over the range $0$ to $\min\{j,t-1\}$ . Substituting $v$ for $s+t-j-1$ and $u$ for $s-j+k^{\prime}$ in (3.2), we get

	$\displaystyle\tau_{s,t}$	$\displaystyle=\sum_{v=0}^{s+t-1}\sum_{u=v-t+1}^{\min\{s,v\}}\frac{(n-u)(n+u-s-v-1)(s+v+1-2u)}{(n+s+v+1)(n+s+v)(n+s+v-1)}\binom{n+s+v+1}{n,s+v+1-u,u}$
		$\displaystyle=\sum_{v=0}^{s+t-1}\sum_{u=v-t+1}^{\min\{s,v\}}\frac{(n+s+v-2)!}{n!(s+v+1)!}\zeta_{s,t},$

where $\zeta_{s,t}=(n-u)(n+u-s-v-1)(s+v+1-2u)\binom{s+v+1}{u}$ . This sum can be split into two cases for $v<s$ and $v\geq s$ as:

\tau_{s,t}=\sum_{v=0}^{s-1}\frac{(n+s+v-2)!}{n!(s+v+1)!}\sum_{u=v-t+1}^{v}\zeta_{s,t}+\sum_{v=s}^{s+t-1}\frac{(n+s+v-2)!}{n!(s+v+1)!}\sum_{u=v-t+1}^{s}\zeta_{s,t}.

This simplifies to

(3.3)

\tau_{s,t}=\frac{\binom{n+2s+t}{n,s+t,s}}{n+2s+t}\times\frac{nt(s+t)}{(n+s)(n+s+t)}\left\{-1+\frac{s}{n}+\frac{(n+s)(n^{2}+nt-t-s(s+t)-1)}{(n+s-1)(s+t)(n+s+t-1)}\right\}.

By substituting $\tau_{s,t}=\binom{n+2s+t}{n,s+t,s}T^{>}(s,t)$ , we get

(3.4)

(n+2s+t)T^{>}(s,t)=\frac{-nt(s+t)}{(n+s)(n+s+t)}+\frac{st(s+t)}{(n+s)(n+s+t)}\\ +\frac{nt(n^{2}+nt-t-s(s+t)-1)}{(n+s-1)(n+s+t)(n+s+t-1)}.

Let $i<n$ . Denote the terms on the right-hand side of (3.4) by $A(s,t),B(s,t)$ and $C(s,t)$ respectively. We first compute $A(j-1,i-j)-A(j,i-j-1)-A(j-1,i-j+1)+A(j,i-j)=\mathscr{A}$ (say).

	$\displaystyle\mathscr{A}$	$\displaystyle=$	$\displaystyle\frac{n(i-1)}{n+i-1}\left\{\frac{j-i}{n+j-1}+\frac{i-j-1}{n+j}\right\}-\frac{ni}{n+i}\left\{\frac{j-i-1}{n+j-1}+\frac{i-j}{n+j}\right\}$
		$\displaystyle=$	$\displaystyle\frac{n}{2\binom{n+j}{2}}.$

If we define $\mathscr{B}$ and $\mathscr{C}$ similarly as $\mathscr{A}$ using the inclusion-exclusion formula, we get $\mathscr{B}=\frac{n}{2\binom{n+j}{2}}-\frac{n}{2\binom{n+i}{2}}$ and $\mathscr{C}=\frac{-n}{2\binom{n+i}{2}}$ . Since $c_{i,j}=\mathscr{A}+\mathscr{B}+\mathscr{C}$ , we have

c_{i,j}=\frac{n}{\binom{n+j}{2}}-\frac{n}{\binom{n+i}{2}}.

Note that when $s+t=n$ , $T^{>}(s,t)=0$ . When $j<i=n$ , (2.2) becomes

c_{i,j}=(n+i+j-2)\,T^{>}(j-1,i-j)-(n+i+j-1)\,T^{>}(j,i-j-1).

Let $\mathscr{A}^{\prime}=A(j-1,n-j)-A(j,n-j-1)$ and define $\mathscr{B}^{\prime}$ and $\mathscr{C}^{\prime}$ similarly. Simplifying the equations resulting from substituting the expressions for $A(s,t),B(s,t)$ and $C(s,t)$ , we get

$\displaystyle c_{n,j}$	$\displaystyle=$	$\displaystyle\mathscr{A}^{\prime}+\mathscr{B}^{\prime}+\mathscr{C}^{\prime}$
	$\displaystyle=$	$\displaystyle\frac{-1}{2n-1}-\frac{-2n(n-1)}{(n+j-1)(n+j)}+\frac{2n(n-1)}{(n+j-1)(n+j-2)}$
	$\displaystyle=$	$\displaystyle\frac{n(j+1)}{\binom{n+j}{2}}-\frac{n(j-1)}{\binom{n+j-1}{2}}-\frac{n}{\binom{2n}{2}},$

thereby proving Theorem 1.1. ∎

4. Proof of Theorem 1.2

Recall that $m_{s,t}=(s,t,n-s-t)$ and $N=n+2s+t$ . Recall from (2.2) that to find $c_{i,j}(n)$ for $i<j$ , we need to know the probability $T^{<}(s,t)=\mathbb{P}\left\{\omega_{1}=2,\omega_{2}=3,M_{3,n}=N\right\}$ where $\omega$ is the projected word of a continuous multiline queue $M$ of type $m_{s,t}$ .

Let $\mathscr{P}_{s,t}$ be the set of all continuous multiline queues $M$ of type $m_{s,t}$ that satisfy $\omega_{1}=2$ and $\omega_{2}=3$ and $M_{3,n}=N$ where $\omega=\mathbf{B}(M)$ . Let $\theta_{s,t}$ be the cardinality of set $\mathscr{P}_{s,t}$ . First, consider the following continuous multiline queue:

\begin{array}[]{cccccccc}a_{1}&a_{2}&\cdots&a_{s}\\ b_{1}&b_{2}&\cdots&\cdots&b_{s+t}\\ c_{1}&c_{2}&\cdots&\cdots&\cdots&c_{n-1}&N\\ \hline\cr x&y&\cdots&\cdots&\cdots&\cdots&\cdots\end{array}

Define $n_{x,y}(s,t,n)$ as the number of continuous multiline queues of type $m_{s,t}$ with $N$ in the last row that project to a word with $x$ and $y$ in the first and the second place respectively as above, where $(x,y)\in\{1,2,3\}^{2}$ . Note that $\theta_{s,t}=n_{2,3}(s,t,n)$ . Also, let $n_{z}(s,t,n)$ be the number of continuous multiline queues of type $m_{s,t}$ with $N$ in the last row that project to a word with $z$ in the first place. Note that by rotational symmetry of multispecies TASEP in [9, Proposition 2.1 (i)], $n_{z}$ also gives the number of continuous multiline queues that project to a word with $z$ in the second place. Therefore, we have

(4.1)

n_{1,3}+n_{2,3}+n_{3,3}=n_{3},

for fixed $s,t$ and $n$ . Again by rotational symmetry, we have

(4.2)

n_{3}(s,t,n)=\frac{n-s-t}{N}\binom{n+2s+t}{n,s,s+t},

since there are $n-s-t$ particles that are labelled $3$ . We compute $n_{3,3}(s,t,n)$ in the following lemma.

Lemma 4.1.

$n_{3,3}(s,t)=\binom{N-1}{s}f_{(n-2,s+t)}$ .

Proof.

Let $M$ be a continuous multiline queue that is counted by $n_{3,3}(s,t,n)$ , i.e.,

M=\begin{array}[]{ccccccc}a_{1}&a_{2}&\cdots&a_{s}\\ b_{1}&b_{2}&\cdots&\cdots&b_{s+t}\\ c_{1}&c_{2}&\cdots&\cdots&\cdots&c_{n-1}&N\\ \hline\cr 3&3&\cdots&\cdots&\cdots&\cdots&\cdots.\end{array}

Here, $c_{1}$ and $c_{2}$ are not bullied by any entry in the second row. This implies that $b_{1}>c_{2}$ and that there is no wrapping from the second to the third row. It follows that there are no restrictions on $a_{i}$ and hence their values can be chosen from the set $\{1,2,\ldots,N-1\}$ in $\binom{N-1}{s}$ ways. $c_{1}$ and $c_{2}$ take the smallest two integers available after fixing $a_{i}$ ’s. Since there are no wrappings from the second to the third row, the configurations formed by the remaining variables are in bijection with standard Young tableaux of shape $(n-2,s+t)$ by Lemma 2.13. Therefore, they are $f_{(n-2,s+t)}$ in number. ∎

Remark 4.2.

Following the steps in the proof of Lemma 4.1, we can give an alternate proof of (4.2) which is equivalent to saying $n_{3}(s,t,n)=\binom{N-1}{s}f_{(n-1,s+t)}$ .

Thus, given (4.1), it suffices to find $n_{1,3}$ for the sake of our analysis. The values of $n_{1,3}(s,t,n)$ for different $s$ and $t$ for $n=5,6$ are shown in the following tables.

$s$ \ $t$	1	2	3
1	9	14	14
2	126	140	0
3	770	0	0

$s$ \ $t$	1	2	3	4
1	14	28	42	42
2	280	462	504	0
3	2772	3276	0	0
4	15288	0	0	0

Table 1. Data for

n_{1,3}(s,t,n)

for

n=5,6

By studying the values for $n_{1,3}$ for different $s,t$ and $n$ , we formulate the following expression for $n_{1,3}$ , which we prove in Section 5.

Theorem 4.3.

For $s,t\geq 1$ and $n>s+t$ , we have

(4.3)

n_{1,3}(s,t,n)=\binom{N-1}{s-1}f_{(n-1,s+t)}.

Proof of Theorem 1.2.

From straightforward calculations using (4.1), (4.2), Lemma 4.1 and Theorem 4.3, we have

n_{2,3}(s,t,n)=\frac{1}{N}\binom{n+2s+t}{n,s,s+t}\left\{\frac{n+t}{\binom{n+s+t}{s+t}}f_{(n-1,s+t)}-\frac{n+s+t}{\binom{n+s+t}{s+t}}f_{(n-2,s+t)}\right\}.

Since $n_{2,3}(s,t,n)=\theta_{s,t}=\binom{n+2s+t}{n,s,s+t}T^{<}(s,t)$ , we have

(4.4)

(n+2s+t)T^{<}(s,t)=\frac{n+t}{\binom{n+s+t}{s+t}}f_{(n-1,s+t)}-\frac{n+s+t}{\binom{n+s+t}{s+t}}f_{(n-2,s+t)}.

The proof is completed by substituting (4.4) in (2.2). First let $i+1<j$ . Denote the terms of the right-hand side of (4.4) by $C(s,t)$ and $D(s,t)$ respectively. We first compute $C(i-1,j-i)-C(i,j-i-1)-C(i-1,j-i+1)+C(i,j-i)=\mathscr{C}$ (say).

$\displaystyle\mathscr{C}$	$\displaystyle=$	$\displaystyle\frac{f_{(n-1,j-1)}}{\binom{n+j-1}{j-1}}\left\{((n+j-i)-(n+j-i-1)\right\}$
		$\displaystyle-\frac{f_{(n-1,j)}}{\binom{n+j}{j}}\left\{((n+j-i)-(n+j-i+1)\right\}$
	$\displaystyle=$	$\displaystyle\frac{n}{\binom{n+j}{2}}.$

If we define $\mathscr{D}$ similarly as $D(i-1,j-i)-D(i,j-i-1)-D(i-1,j-i+1)+D(i,j-i)$ , we get $\mathscr{D}=0$ . Since, $c_{i,j}(n)=\mathscr{C}+\mathscr{D}$ , we have $c_{i,j}(n)=\frac{n}{\binom{n+j}{2}}$ .

Note that $T^{<}(s,0)=0$ by definition. Therefore when $i+1=j$ , (2.2) becomes

c_{j-1,j}=(n+2j-3)\,T^{<}(j-2,1)-(n+2j-2)\,T^{<}(j-2,2)+(n+2j-1)\,T^{<}(j-1,1).

Let $\mathscr{C}^{\prime}=C(j-2,1)-C(j-2,2)+C(j-1,1)$ and define $\mathscr{D}^{\prime}$ similarly. Then,

	$\displaystyle\mathscr{C}^{\prime}$	$\displaystyle=$	$\displaystyle\frac{n+1}{\binom{n+j-1}{j-1}}f_{(n-1,j-1)}-\frac{n+2}{\binom{n+j}{j}}f_{(n-1,j)}+\frac{n+1}{\binom{n+j}{j}}f_{(n-1,j)}$
	$\displaystyle\mathscr{D}^{\prime}$	$\displaystyle=$	$\displaystyle-\frac{n+j-1}{\binom{n+j-1}{j-1}}f_{(n-2,j-1)}$

Thus, $c_{j-1,j}(n)=\mathscr{C}^{\prime}+\mathscr{D}^{\prime}=\frac{ni}{\binom{n+i}{2}}+\frac{n}{\binom{n+j}{2}},$ completing the proof. ∎

5. Proof of Theorem 4.3

To compute $n_{1,3}(s,t,n)$ , we need to count the number of continuous multiline queues with the following configuration. Here, $N=n+2s+t$ while $a_{i},b_{i}$ and $c_{i}$ are distinct integers from the set $[N]$ .

\begin{array}[]{ccccccccc}&a_{1}&a_{2}&\cdots&a_{s}\\ &b_{1}&b_{2}&\cdots&\cdots&b_{s+t}\\ &c_{1}&c_{2}&\cdots&\cdots&\cdots&c_{n-1}&N\\ \hline\cr\omega=&1&3&\cdots&\cdots&\cdots&\cdots&\cdots\end{array}

Since $\omega_{2}=3$ , $c_{2}$ can not be bullied by any $b_{i}$ . Therefore, there can be at most one wrapping from the second row to the third row. These configurations can be classified into two types:

(1)

there is no wrapping from the second row,
(2)

only $b_{s+t}$ wraps around and bullies $c_{1}$ .

Let us denote the number of continuous multiline queues from the two cases by $\alpha_{1,3}(s,t,n)$ and $\beta_{1,3}(s,t,n)$ respectively. We will enumerate them separately.

Proposition 5.1.

A continuous multiline queue of type $m_{s,t}$ where there is no wrapping from the second row projects to a word $\omega$ with $\omega_{1}=1$ and $\omega_{2}=3$ if and only if

(1)

there exists $1\leq i\leq s$ such that $a_{i}\rightarrow b_{1}\rightarrow c_{1}$ , and
(2)

$b_{2}>c_{2}$ .

Proof.

Let $M$ be a continuous multiline queue satisfying (1) and (2) and let $\omega$ be the projected word of $M$ . The reverse implication is straightforward. We proceed to prove the forward implication. Note that $\omega_{2}=3$ implies $b_{2}>c_{2}$ , otherwise $c_{2}$ is bullied by either $b_{1}$ or $b_{2}$ giving $\omega_{2}<3$ . Since there is no wrapping from the second row to the third row, $\omega_{1}=1$ is only possible when there exist $a_{i},b_{j}$ such that $a_{i}\rightarrow b_{j}\rightarrow c_{1}$ for some $i$ . That is, $b_{j}<c_{1}.$ Further, $b_{j}<c_{1}<c_{2}<b_{2}$ implies $j=1$ . ∎

Theorem 5.2.

Let $\alpha_{1,3}(s,t,n)$ be the number of continuous multiline queues of type $m_{s,t}$ with the largest entry $N$ in the last row and no wrapping from the second row such that the projected word $\omega$ has $\omega_{1}=1$ and $\omega_{2}=3$ . Then,

\alpha_{1,3}(s,t)=\left(\binom{N-2}{s-1}-\binom{N-2}{s-3}\right)f_{(n-1,s+t)}.

Proof.

Let the continuous multiline queues that are counted by $\alpha_{1,3}(s,t,n)$ exhibit the following configuration:

\begin{array}[]{ccccccccc}&a_{1}&a_{2}&\cdots&a_{s}\\ &b_{1}&b_{2}&\cdots&\cdots&b_{s+t}\\ &c_{1}&c_{2}&\cdots&\cdots&\cdots&c_{n-1}&N\\ \hline\cr\omega=&1&3&\cdots&\cdots&\cdots&\cdots&\cdots\end{array}

Let us first assume that $a_{1}<b_{1}$ . Coupled with $b_{1}<c_{1}$ from the proof of Proposition 5.1, we get $a_{1}=1$ , and $a_{1}\rightarrow b_{1}\rightarrow c_{1}$ , which gives $\omega_{1}=1$ . The remaining $a_{i}$ ’s assume increasing integer values between $2$ and $N-1$ in $\binom{N-2}{s-1}$ ways. We also have $c_{1}<c_{2}<b_{2}$ . Thus, $b_{1}\,,c_{1}$ and $c_{2}$ are assigned the three smallest values after eliminating integers selected by the $a_{i}$ ’s. Because there is no wrapping in any of the rows, such configurations are in bijection with standard Young tableaux of shape $\lambda=(n-2,s+t-1)$ (see Lemma 2.13). This gives us $\binom{N-2}{s-1}f_{(n-2,s+t-1)}$ continuous multiline queues that contribute to $\alpha_{1,3}(s,t,n)$ for the case $a_{1}<b_{1}$ .

Now, let us assume that $a_{1}>b_{1}$ . Then, there exists an $a_{i}$ such that $a_{i}\xrightarrow{W}b_{1}$ by Proposition 5.1(1). The inequalities $b_{1}<c_{1}$ and $b_{1}<a_{1}$ imply $b_{1}=1$ . Instead, we first find the number of continuous multiline queues where the only constraints are $b_{1}=1,c_{n}=N$ and $c_{2}<b_{2}$ , with no wrapping from the second row to the third row. For these, $b_{1}\rightarrow c_{1}$ and we get $\omega_{1}\in\{1,2\}$ and $\omega_{2}=3$ . Observe that the $a_{i}$ ’s can take any value other than $1$ and $N$ . $c_{1}$ and $c_{2}$ are then assigned the smallest two integers after eliminating $1$ and the integers selected by the $a_{i}$ ’s. Also, the configurations formed by the remaining $b_{i}$ ’s and $c_{i}$ ’s satisfy the following inequalities

\begin{array}[]{ccccccccccccc}&&b_{2}&<&\ldots&<&b_{s+t-j}&<&\ldots&<&b_{s+t}\\ &&\wedge&&\cdots&&\wedge&&\cdots&&\wedge\\ c_{3}&<&\ldots&<&\ldots&<&c_{n-j}&<&\ldots&<&c_{n},\end{array}

and hence they can be arranged in $f_{(n-2,s+t-1)}$ ways. The required number is given by $\binom{N-2}{s}f_{(n-2,s+t-1)}$ . From this set, in order to eliminate the continuous multiline queues with $\omega_{1}=2$ and $\omega_{2}=3$ , we need to subtract the number of continuous multiline queues where $b_{1}=1$ , $b_{2}<c_{2}$ with no wrapping in any row from the number $\binom{N-2}{s}f_{(n-2,s+t-1)}$ . In this regard, let $c_{2}=k+2$ for some $k\leq s$ . Then, there are $k$ $a_{i}$ ’s that are smaller than $c_{2}$ and there are $k+1$ ways to assign values to $c_{1},a_{1},\cdots a_{k}$ . The remaining entries of the continuous multiline queue satisfy the following inequalities:

\begin{array}[]{ccccccccccccc}&&&&&&a_{k+1}&<&a_{k+2}&<&\cdots&<&a_{s}\\ &&&&&&\wedge&&\wedge&&\cdots&&\wedge\\ &&b_{2}&<&\cdots&<&b_{t-k-1}&<&b_{t-k}&<&\cdots&<&b_{s+t}\\ &&\wedge&&\ldots&&\wedge&&\wedge&&\cdots&&\wedge\\ c_{3}&<&\cdots&&\cdots&&\cdots&&\cdots&&\cdots&<&c_{n}.\end{array}

Such configurations are in bijection with standard Young tableaux of shape $\lambda_{k}=(n-2,s+t-1,s-k)$ which are $f_{(n-2,s+t-1,s-k)}$ in number. Thus, the number of continuous multiline queues contributing to $\alpha_{1,3}(s,t,n)$ where $a_{1}>b_{1}$ is given by

\displaystyle\binom{N-2}{s}f_{(n-2,s+t-1)}-\sum_{k=0}^{s}(k+1)f_{(n-2,s+t-1,s-k)}.

Adding this to $\binom{N-2}{s-1}f_{(n-2,s+t-1)}$ for the case $a_{1}<b_{1}$ , we get

$\displaystyle\alpha_{1,3}(s,t,n)$	$\displaystyle=$	$\displaystyle\binom{N-2}{s-1}f_{(n-2,s+t-1)}+\binom{N-2}{s}f_{(n-2,s+t-1)}$
		$\displaystyle-\frac{f_{(n-2,s+t-1)}}{n(s+t)}\sum_{k=0}^{s}(k+1)(t+k)(n-s+k)\binom{N-k-3}{s-k}$
	$\displaystyle=$	$\displaystyle\frac{(n-s-t)(n+t+2)}{(n+s+t)(n+s+t+1)}\binom{N-1}{n,s-1,s+t}$
	$\displaystyle=$	$\displaystyle\left(\binom{N-2}{s-1}-\binom{N-2}{s-3}\right)f_{(n-1,s+t)}.$

∎

Proposition 5.3.

A continuous multiline queue of type $m_{s,t}$ with $N$ in the last row, such that there is exactly one wrapping from the second row, projects to a word $\omega$ with $\omega_{1}=1$ and $\omega_{2}=3$ if and only if

(1)

there exists $i<s$ such that $a_{i}\rightarrow b_{s+t-1}\rightarrow c_{n}$ and $a_{i+1}\rightarrow b_{s+t}\xrightarrow{W}c_{1}$ , and
(2)

$b_{1}>c_{2}$ .

Proof.

Let $M$ be a continuous multiline queue with $N$ in the last row and exactly one wrapping from the second row such that the projected word $\omega$ has $\omega_{1}=1$ and $\omega_{2}=3$ . If $b_{1}<c_{2}$ in $M$ , then $c_{2}$ is bullied by at least one $b_{i}$ (either by $b_{1}$ or by the wrapping) giving $\omega_{2}<3$ , a contradiction. Therefore, $b_{1}>c_{2}$ .

Then, there exists integers, say $j,v$ such that $a_{j}\rightarrow b_{v}\xrightarrow{W}c_{1}$ to give $\omega_{1}=1$ . If $v<s+t$ , then there exists $w>v$ such that $b_{w}\xrightarrow{W}c_{2}$ , once again giving a contradiction. Therefore, $v=s+t$ . Further, since $a_{j}\rightarrow b_{s+t}\xrightarrow{W}c_{1}$ , there exists $i<j$ and $u<s+t$ such that $a_{i}\rightarrow b_{u}\rightarrow c_{n}$ to give $\omega_{n}=1$ . Otherwise, $b_{s+t}$ bullies $c_{n}=N$ and there is no wrapping from the second row to the third row. Since, $b_{u}\rightarrow c_{n}$ , all of $b_{u+1},b_{u+2},\cdots,b_{s+t}$ wrap around to the third row. As there can be exactly one such wrapping, $u+1=s+t$ . Moreover, since $a_{i}\rightarrow b_{s+t-1}$ , $a_{k}$ wraps around to the second row for all $k>i+1$ , thus proving $j={i+1}$ . The reverse implications are straightforward to verify. ∎

Recall that the number of continuous multiline queues satisfying Proposition 5.3 is counted by $\beta_{1,3}$ . The values of $\beta_{1,3}(s,t,n)$ for different $s$ and $t$ for $n=5,6$ are shown in Table 2.

$s$ \ $t$	0	1	2
2	9	14	14
3	140	154	0
4	924	0	0

$s$ \ $t$	0	1	2	3
2	14	28	42	42
3	280	504	546	0
4	3276	3822	0	0
5	19110	0	0	0

Table 2. Data for

\beta_{1,3}(s,t,n)

for n=5,6

By observing the above data, we formulate the following expression for $\beta_{13}$ .

Theorem 5.4.

$\beta_{1,3}(s,t,n)=\binom{N-1}{s-2}f_{(n-1,s+t)}$ .

Remark 5.5.

It is interesting to see that the techniques we have used to prove the earlier cases do not work here as there is no easy bijection which can be used to prove Theorem 5.4. We use alternative methods to give the proof of Theorem 5.4 in Section 5.1. For now, we independently prove Theorem 4.3 using the properties of continuous multiline queues that are counted by $\beta_{1,3}$ .

We first prove that $\beta_{1,3}(s,t,n)$ satisfies a simple recurrence.

Lemma 5.6.

For $s,t\geq 2$ and $s+t<n$ :

(5.1)

\beta_{1,3}(s,t,n)=\beta_{1,3}(s-1,t+1,n)+\beta_{1,3}(s,t-1,n)+\beta_{1,3}(s,t,n-1).

Proof.

Consider a continuous multiline queue $M$ satisfying the conditions from Proposition 5.3. Let $c_{2}=k+2$ for some $k\leq s$ . Then, $M$ has the following configuration:

\begin{array}[]{ccccccccc}a_{1}&a_{2}&\cdots&a_{k}&\cdots&a_{s}\\ b_{1}&b_{2}&\cdots&b_{k}&\cdots&\cdots&b_{s+t}\\ c_{1}&k+2&\cdots&\cdots&\cdots&\cdots&\cdots&c_{n-1}&N\\ \hline\cr 1&3&\cdots&\cdots&\cdots&\cdots&\cdots&\cdots&1\end{array}

Since $b_{1}>c_{2}$ , there are only $\binom{k+1}{k}=k+1$ ways to assign values to $c_{1},a_{1},\ldots,a_{k}$ from the set $[k+1]$ . Thus, for each $u\leq k,\,a_{u}\rightarrow b_{u}$ . Given that the rows are strictly increasing, exactly one of the following cases is true:

(1)

$a_{k+1}=k+3$ :
Here $a_{k+1}$ bullies $b_{k+1}$ , so $a_{k+1}$ does not bully $b_{s+t-1}$ or $b_{s+t}$ . Deleting $a_{k+1}$ and subtracting $1$ from all the values higher than $k+3$ does not affect the bully paths containing $b_{s+t-1}$ or $b_{s+t}$ . In the projected word $\omega$ , one of the $1$ ’s changes to a $2$ . So, there are $\beta_{1,3}(s-1,t+1,n)$ such continuous multiline queues.
(2)

$b_{1}=k+3$ :
$b_{1}$ bullies $c_{3}$ . Deleting $b_{1}$ and subtracting $1$ from all the values higher than $k+3$ in the continuous multiline queue does not affect the bully paths containing $b_{s+t-1}$ or $b_{s+t}$ . In the projected word $\omega$ , one of the $2$ ’s changes to a $3$ . Thus, there are $\beta_{1,3}(s,t-1,n)$ such continuous multiline queues.
(3)

$c_{3}=k+3$ :
Finally, in this case, $c_{3}$ is not bullied by any $b_{u}$ because $b_{1}>c_{3}$ and there is exactly one wrapping from the second to the third row. Here, deleting $c_{3}$ and subtracting $1$ from all the values higher than $k+3$ does not affect any bully path and the length of the resulting projected is reduced by one. There are $\beta_{1,3}(s,t,n-1)$ such continuous multiline queues.

Therefore, $\beta_{1,3}(s,t,n)$ is obtained by adding the numbers in each of the above cases. ∎

We can verify the equation

\alpha_{1,3}(s,t,n)=\alpha_{1,3}(s-1,t+1,n)+\alpha_{1,3}(s,t-1,n)+\alpha_{1,3}(s,t,n-1),

for $s,t\geq 2$ and $n>s+t$ by plugging the value of $\alpha_{1,3}(s,t,n)$ from Theorem 5.2. This along with Lemma 5.6 gives the recurrence relation

n_{1,3}(s,t,n)=n_{1,3}(s-1,t+1,n)+n_{1,3}(s,t-1,n)+n_{1,3}(s,t,n-1),

for $s,t\geq 2$ and $n>s+t$ . Let $P(s,t,n)$ denote the product $\binom{N-1}{s-1}f_{n-1,s+t}$ from Theorem 4.3. We have

P(s,t,n)=P(s-1,t+1,n)+P(s,t-1,n)+P(s,t,n-1),

using Pascal’s rule and the hook length recurrence relation $f_{(a,b)}=f_{(a-1,b)}+f_{(a,b-1)}$ where $a\geq b>1$ ( see Remark 2.12). As a result, $P(s,t,n)$ satisfies the same recurrence relation as $n_{1,3}(s,t,n)$ . Moreover, $n_{1,3}(s,t,s+t)=0$ as there is no $3$ in the projected word. We also have

\displaystyle n_{1,3}(1,t,n)=\alpha_{1,3}(1,t,n)=f_{(n-1,t+1)},

as $\beta_{1,3}(1,t,n)=0$ for all $t,n$ . This holds because for a continuous multiline queue with exactly one wrapping from the second row to the third row which projects to a word beginning with $(2,3)$ , we need $s$ to be greater than $1$ , by Proposition 5.3. Thus, proving the initial condition $n_{1,3}(s,1,n)=P(s,1,n)$ completes the proof of Theorem 4.3. To that end, we have the following result.

Proposition 5.7.

For $s,n$ such that $1<s+1<n$ , we have

(5.2)

n_{1,3}(s,1,n)=\binom{n+2s}{s-1}f_{(n-1,s+1)}.

Proof.

Recall that $m_{s,t}=(s,t,n-s-t)$ . Also, $n_{x,y}(s,t,n)$ counts the number of continuous multiline queues $M$ of type $m_{s,t}$ with $N$ in the last row, such that $M$ projects to a word $\omega$ where $\omega_{1}=x$ and $\omega_{2}=y$ .

M=\begin{array}[]{ccccccccc}a_{1}&a_{2}&\cdots&a_{s}\\ b_{1}&b_{2}&\cdots&\cdots&b_{s+t}\\ c_{1}&c_{2}&\cdots&\cdots&\cdots&c_{n-1}&N.\\ \hline\cr x&y&\cdots&\cdots&\cdots&\omega_{n-1}&\omega_{n}\end{array}

Let $\mu_{x,y}$ be the probability that a particle labelled $x$ is immediately followed by a particle labelled $y$ in a continuous TASEP of type $m_{s,t}$ , with $x$ followed by $y$ . Then by Lemma 2.9,

\mu_{x,y}=\frac{n+2s+t}{\binom{n+2s+t}{s,s+t,n}}n_{x,y}.

Consider $M^{\prime}$ , a continuous multiline queue of type $m_{u}=(u,n-u)$ such that the largest entry $N^{\prime}=n+u$ is in the last row. Recall from Section 2.4, that $\delta_{c,d}(u,n)$ counts the number of continuous multiline queues $M^{\prime}$ with $N^{\prime}$ in the last row, such that $M^{\prime}$ projects to a word $\omega^{\prime}$ where $\omega^{\prime}_{1}=c$ and $\omega^{\prime}_{2}=d$ .

M^{\prime}=\begin{array}[]{ccccccccc}a_{1}&a_{2}&\cdots&a_{u}\\ b_{1}&b_{2}&\cdots&\cdots&b_{n-1}&N^{\prime}\\ \hline\cr c&d&\cdots&\cdots&\omega^{\prime}_{n-1}&\omega^{\prime}_{n}\end{array}.

From (2.15), we know that $\delta_{2,2}(u,n)=f_{(n-2,u)}$ . Let $\epsilon_{c,d}(n)$ denote the probability that a particle labelled $c$ is immediately followed by one labelled $d$ in a continuous two-species TASEP of type $m_{u}$ . Then, again by Lemma 2.9,

\epsilon_{c,d}=\frac{n+u}{\binom{n+u}{u,n}}\delta_{c,d}.

We can define a lumping (or a colouring) for the continuous three-species TASEP of type $m_{s,t}$ to a continuous two-species TASEP as follows. Let $\Omega_{s,t}$ and $\Omega_{s}$ be the set of labelled words on a ring of type $m_{s,t}$ and $m_{s}=(s,n-s)$ respectively. Let $f:\Omega_{s,t}\rightarrow\Omega_{s}$ be a map defined as follows:

f(\omega_{1},\ldots,\omega_{n})=(f(\omega_{1}),\ldots,f(\omega_{n})),

where

f(i)=\begin{cases}1,&\text{if }i=1,\\ 2,&\text{if }i=2,3.\end{cases}

$x$ \ $y$	1	2	3
1	$\mu_{1,1}$	$\mu_{1,2}$	$\mu_{1,3}$
2	$\mu_{2,1}$	$\mu_{2,2}$	$\mu_{2,3}$
3	$\mu_{3,1}$	$\mu_{3,2}$	$\mu_{3,3}$

Table 3. The table contains the correlations of two adjacent particles in

\Omega_{s,t}

. The table is divided into four parts, and entries of the yellow, green, red and blue sections contribute to the correlations

\epsilon_{1,1},\epsilon_{1,2},\epsilon_{2,1}

and

\epsilon_{2,2}

respectively in

\Omega_{s}

By lumping the Markov process, we have

\epsilon_{2,2}=\{\mu_{2,2}+\mu_{3,2}+\mu_{2,3}+\mu_{3,3}\}.

That is,

(5.3)

\frac{n+s}{\binom{n+s}{s}}\delta_{2,2}=\frac{n+2s+t}{\binom{n+2s+t}{n,s,s+t}}\{n_{2,2}+n_{3,2}+n_{2,3}+n_{3,3}\}.

Let $t=1$ . Then, $n_{2,2}=0$ because there is only one particle with label $2$ in the three-species continuous TASEP. Note that $n_{3,2}(s,t,n)=\tau_{s,t}$ from Section 3. Thus, substituting $t=1$ in (3.3), we obtain

(5.4)

\frac{n+2s+1}{\binom{n+2s+1}{n,s,s+1}}n_{3,2}(s,1,n)=\frac{(n-s-1)(n-s+1)}{(n+s-1)(n+s+1)}.

We also have

(5.5)

\{n_{1,3}+n_{2,3}+n_{3,3}\}(s,1,n)=\frac{n-s-1}{n+2s+1}\binom{n+2s+1}{n,s,s+1},

by substituting $t=1$ in (4.1). Solving (5.5) for $n_{1,3}$ using (5.3) and (5.4) gives

	$\displaystyle n_{1,3}(s,1,n)$	$\displaystyle=\frac{s(n-s-1)}{(n+s+1)(n+2s+1)}\binom{n+2s+1}{n,s,s+1}$
		$\displaystyle=\binom{n+2s}{s-1}f_{(n-1,s+1)},$

proving the result. ∎

5.1. Proof of Theorem 5.4

We can now prove Theorem 5.4 directly using Theorem 4.3 as follows.

Proof of Theorem 5.4.

	$\displaystyle\beta_{1,3}(s,t,n)$	$\displaystyle=n_{1,3}(s,t,n)-\alpha_{1,3}(s,t,n)$
		$\displaystyle=\left(\binom{N-1}{s-1}-\binom{N-2}{s-1}+\binom{N-2}{s-3}\right)f_{(n-1,s+t)}$
		$\displaystyle=\binom{N-1}{s-2}f_{(n-1,s+t)}.$

∎

In addition, we describe the developments made towards a direct proof of Theorem 5.4 using the first principles in this section. Recall the recurrence relation (5.1). The equation holds true for $s,t\geq 2$ and $s+t\leq n$ . It is easy to verify that the product $\binom{N-1}{s-2}f_{(n-1,s+t)}$ satisfies the same recurrence relation as $\beta_{1,3}(s,t,n)$ . Thus, it is sufficient to show that the initial conditions are the same for both quantities. The conditions are:

	$\displaystyle\beta_{1,3}(1,t,n)$	$\displaystyle=0,$
	$\displaystyle\beta_{1,3}(s,t,s+t)$	$\displaystyle=0,$
	$\displaystyle\beta_{1,3}(s,1,n)$	$\displaystyle=\binom{N-1}{s-2}f_{(n-1,s+1)}.$

The first two initial conditions are straightforward. We now provide a formula for $\beta_{1,3}$ for $t=1$ .

Lemma 5.8.

\beta_{1,3}(s,1,n)=\sum_{\ell=2}^{s}\gamma^{\ell}(s,n),

where $\gamma^{\ell}(s,n)$ is the number of continuous multiline queues of type $m_{s,1}$ with $c_{2}=\ell,c_{n}=N$ and $b_{1}>c_{2}$ , such that the projected words have $1$ and $3$ in the first two places respectively.

Proof.

Consider a continuous multiline queue $M$ of type $m_{s,1}$ that is counted by $\beta_{1,3}(s,1,n)$ :

M=\begin{array}[]{cccccccc}a_{1}&a_{2}&\cdots&\cdots&a_{s}\\ b_{1}&b_{2}&\cdots&\cdots&b_{s}&b_{s+1}\\ c_{1}&c_{2}&\cdots&\cdots&\cdots&\cdots&c_{n-1}&N.\\ \hline\cr 1&3&\cdots&\cdots&\cdots&\cdots&\cdots&1\end{array}

Let $c_{2}$ be equal to $\ell$ which is greater than $1$ . According to Proposition 5.3, $c_{2}<b_{1}$ . Therefore, the values $\{1,\ldots,\ell-1\}$ are assigned to $c_{1},a_{1},\ldots,a_{\ell-2}$ . Since there exists $i$ such that $a_{i}\rightarrow b_{s}$ and $a_{i+1}\rightarrow b_{s+1}$ , $\ell$ ranges from $\{2,\ldots,s\}$ . The set of all continuous multiline queues counted by $\beta_{1,3}(s,1,n)$ can be divided into smaller sets depending on the value of $\ell$ . We denote the number of such continuous multiline queues that have $c_{2}=\ell$ as $\gamma^{l}(s,n)$ . By adding over all possible values of $\ell$ , we obtain the required expression. ∎

Next, we prove a formula for $\gamma^{\ell}$ from the first principles. First, consider a skew shape $\lambda/\mu$ , where $\lambda$ and $\mu$ are partitions such that $\mu\subseteq\lambda$ in containment order. Recall from Section 2.3, that the number of standard Young tableaux of a skew shape $\lambda/\mu$ is given by $f_{\lambda/\mu}$ and it can be computed using (2.9).

Theorem 5.9.

(5.6)		$\displaystyle\gamma^{l}(s,n)$	$\displaystyle=$	$\displaystyle(\ell-1)\sum_{i=\ell-1}^{s-1}\,\sum_{j=\ell-2}^{i-2}\,\sum_{k=2}^{n-s+i-1}f_{(n-s+i-3,i-2,j-\ell+2)/(n-s+i-k-1)}\times$
(5.6)				$\displaystyle\left\{f_{(n+i-k-j,s-j,s-j)/(i-j+1,i-j-1)}-f_{(n+i-k-j-1,s-j,s-j)/(i-j,i-j-1)}\right\}.$

Proof.

Let $c_{2}=\ell$ . Then, $\gamma^{l}(s,n)$ counts the number of continuous multiline queues $M$ with the following configuration.

M=\begin{array}[]{ccccccccc}a_{1}&a_{2}&\cdots&a_{s}\\ b_{1}&b_{2}&\cdots&b_{s}&b_{s+1}\\ c_{1}&\ell&\cdots&\cdots&\cdots&N,\\ \hline\cr 1&3&\cdots&\cdots&\cdots&1\end{array}

where $b_{1}>c_{2}$ , such that $c_{1},a_{1},\ldots a_{\ell-2}\in[\ell-1]$ . We can choose $c_{1}$ in $(\ell-1)$ different ways, which determines the values of $a_{1}$ to $a_{\ell-2}$ . These values must satisfy $a_{u}<b_{u}$ for $1\leq u\leq\ell-2$ , implying that $a_{u}\rightarrow b_{u}$ for each $u$ .

Since $M$ is of type $(s,1,n-s-1)$ , there is exactly one $b_{i}$ which is not bullied by any $a_{j}$ in the first iteration of the bully path process. Here, $i$ can range from $\ell-1$ to $s-1$ . For a fixed $i$ , we can split the set of entries of a continuous multiline queue into two sets based on their relation with $b_{i}$ . Let $j$ and $k$ be the largest integers such that $a_{j}<b_{i}$ and $c_{k}<b_{i}$ respectively. The value of $j$ lies between $\ell-2$ and $i-1$ by the choice of $i$ and $j$ . And $n-k>s-i$ ensures that there is no more than one wrapping from the second row to the third row which implies that $k$ lies between $0$ and $n-s+i-1$ . For the continuous multiline queues with at most one wrapping from the second row to the third row, the following inequalities hold for fixed $i,j$ and $k$ according to Proposition 5.3.

\begin{array}[]{ccccccccccccc}&&a_{\ell-1}&<&\ldots&<&a_{j}\\ &&\wedge&&\cdots&&\wedge\\ &b_{1}<&\ldots&<&\ldots&<&b_{i-1}\\ &\wedge&\cdots&&\wedge&&&\\ c_{3}<&\ldots&\ldots&<&c_{k}\end{array}<b_{i}<\begin{array}[]{ccccccccccccc}&&a_{j+1}&<&\ldots&\ldots&<&\ldots&<a_{s}\\ &&\wedge&&\cdots&\wedge&&\wedge\\ &&b_{i+1}&<&\ldots&b_{s}&<&b_{s+1}\\ &&\wedge&&\cdots&\wedge&&\\ c_{k+1}&<&\cdots&<&\cdots&c_{n}.\end{array}

These arrangements are counted by the product of hooks length formulas of appropriate skew shapes, i.e. by,

(5.7)

f_{\lambda_{1}/\mu_{1}}.f_{\lambda_{2}/\mu_{2}},

where $\lambda_{1}/\mu_{1}=(n-s+i-3,i-1,j-\ell)/(n-s+i-k-1)$ , $\lambda_{2}/\mu_{2}=(n+i-k-j,s-j,s-j)/(i-j+1,i-j-1)$ .

To obtain the required number of continuous multiline queues with exactly one wrapping from the second row to the third row, we have to remove the continuous multiline queues with no wrapping from the second row to the third row from the above set. These multiline queues are determined by the inequalities:

\begin{array}[]{ccccccccccccc}&&a_{\ell-1}&<&\cdots&<&a_{j}\\ &&\wedge&&\cdots&&\wedge\\ &b_{1}<&\cdots&<&\cdots&<&b_{i-1}\\ &\wedge&\ldots&&\wedge&&&\\ c_{3}<&\cdots&\cdots&<&c_{k}\end{array}<b_{i}<\begin{array}[]{ccccccccccccc}&&a_{j+1}&<&\cdots&<&\cdots&<a_{s}\\ &&\wedge&&\cdots&&\wedge\\ &&b_{i+1}&<&\cdots&<&b_{s+1}\\ &&\wedge&&\ldots&&\wedge&&\\ c_{k+1}&<&\cdots&<&\cdots&<&c_{n}.\end{array}

The number of these arrangements is

(5.8)

f_{\lambda_{1}/\mu_{1}}.f_{\lambda_{3}/\mu_{3}},

where $\lambda_{3}/\mu_{3}=(n+i-k-j-1,s-j,s-j)/(i-j,i-j-1)$ .

Summing the difference of two products in (5.7) and (5.8) over all possible values of $i,j,k$ , and multiplying the sum with $\ell-1$ for each choice of $c_{1}$ , we prove the result. ∎

Remark 5.10.

Unfortunately, we have not been able to find a closed-form expression of the sum on the right-hand side of (5.6). However, by doing extensive numerical checks, we have a conjecture formulating $\gamma^{\ell}(s,n)$ .

Conjecture 5.11.

We have,

\gamma^{\ell}(s,n)=(\ell-1)\binom{n+2s-\ell}{s-\ell}f_{(n-1,s+1)}.

Assuming Conjecture 5.11, we can immediately prove by summing over $\ell$ that

\beta_{1,3}(s,1,n)=\binom{n+2s}{s-2}f_{(n-1,s+1)},

giving Theorem 5.4. Next, we demonstrate an approach to prove Conjecture 5.11. First, consider the set $\mathscr{H}$ of continuous multiline queues of type $m_{s,1}$ that have $c_{1}=1,\,c_{2}=2$ and $c_{n}=N$ where $N=n+2s+1$ is the largest entry. Let $\rho_{c,d}(s,n)$ denote the number of continuous multiline queues in $\mathscr{H}$ that project to a word starting with $c,d$ . That is, a continuous multiline queue in $\mathscr{H}$ that contributes to $\rho_{c,d}(s,n)$ has the following structure:

\begin{array}[]{cccccccc}a_{1}&a_{2}&\cdots&\cdots&a_{s}\\ b_{1}&b_{2}&\cdots&\cdots&b_{s}&b_{s+1}\\ 1&2&\cdots&\cdots&\cdots&\cdots&N\\ \hline\cr c&d&\cdots&\cdots&\cdots&\cdots&\cdots\end{array}

Similarly, let $\rho_{d}(s,n)$ denote the number of continuous multiline queues in $\mathscr{H}$ that project to a word with $d$ at the second position. Then for $s,n$ we have,

(5.9)

\rho_{1,3}(s,n)+\rho_{2,3}(s,n)+\rho_{3,3}(s,n)=\rho_{3}(s,n).

Note that $\rho_{1,3}(s,n)=\gamma^{2}(s,n)$ . Moreover, $\rho_{3,3}(s,n)=\binom{N-3}{s}f_{(n-2,s+1)}$ following the same lines of arguments as in Lemma 4.1. For $\rho_{3}$ , consider a continuous multiline queue of the form

\begin{array}[]{cccccccc}a_{1}&a_{2}&\cdots&\cdots&a_{s}\\ b_{1}&b_{2}&\cdots&\cdots&b_{s}&b_{s+1}\\ 1&2&\cdots&\cdots&\cdots&\cdots&N\\ \hline\cr\cdot&3&\cdots&\cdots&\cdots&\cdots&\end{array}

To count such configurations, note that there is no restriction on the $a_{i}$ ’s and they can be assigned any values from the set $\{3,\ldots,N-1\}$ . There can at most be one wrapping from the second row to the third row, thus the remaining entries satisfy the following inequalities:

\begin{array}[]{ccccccccccccccc}&&b_{1}&<&\ldots&<&b_{s-1}&<&b_{s}&<&b_{s+1}\\ &&\wedge&&\cdots&&\wedge&&\wedge&\\ c_{3}&<&\ldots&<&\ldots&<&c_{n-1}&<&N.\end{array}

The arrangements are in bijection with standard Young tableaux of skew shape $(n-1,s+1)/(2)$ . Therefore,

(5.10)

\rho_{3}(s,n)=\binom{N-3}{s}f_{(n-1,s+1)/(2)}.

Hence, it is enough to compute $\rho_{2,3}(s,n)$ in order to find $\gamma^{2}(s,n)$ using (5.9), which in turn gives $\beta_{1,3}(s,1,n)$ . The values of $\rho_{2,3}(s,n)$ for different $s$ and $n$ are shown in Table 4. By observing Table 4, we conjecture the following formula for $\rho_{2,3}(s,n)$ .

$s$ \ $n$	3	4	5	6
1	3	4	5	6
2	0	40	70	112
3	0	0	630	1260
4	0	0	0	11088

Table 4.

\rho_{2,3}(s,n)

for different values of

s

and

n

Conjecture 5.12.

We have

\rho_{2,3}(s,n)=\binom{N-3}{n-1}f_{(s,s)}.

Assuming Conjecture 5.12, we have $\rho_{1,3}=\gamma^{2}(s,n)=\binom{N-3}{s-2}f_{(n-1,s+1)}$ , by (5.9) and (5.10).

Since $\binom{N-3}{n-1}=\frac{N-3}{n-1}\binom{N-4}{n-2}$ , proving Conjecture 5.12 is equivalent to proving the following recurrence.

Conjecture 5.13.

We have

(n-1)\rho_{2,3}(s,n)=(n+2s-2)\rho_{2,3}(s,n-1).

We now give a formula for $\rho_{2,3}(s,n)$ using the first principles. The following triple sum formula counts the number of continuous multiline queues with $c_{1}=1,c_{2}=2$ and $c_{n}=N$ that project to words beginning with $(2,3)$ .

Theorem 5.14.

(5.11)

\rho_{2,3}(s,n)=\sum_{i=1}^{s+1}\sum_{j=0}^{i-1}\sum_{k=0}^{i-1}f_{(s-j+k,s-i+k+1,s-i+k)/(k,k)}\times\\ \{f_{(n+i-s-3,i-1,j)/(k)}-f_{(n+s-i-4,i-1,j)/(k-1)}\}

Proof.

There exists a unique $i\in[s+1]$ such that $a_{j}$ does not bully $b_{i}$ for any $j$ . Also $b_{i}\rightarrow c_{1}$ by wrapping around from the second to the third row. We already have $c_{1}=1,c_{2}=2$ and $c_{n}=N$ . We can split the set of remaining entries from each of the three rows into two sets depending on their relation with $b_{i}$ . Let $j$ and $d$ be the largest integers such that $a_{j}<b_{i}$ and $c_{d}<b_{i}$ . Here, $j$ lies between $0$ and $i-1$ . Since there is no more than one wrapping from the second row to the third row, we have $n-d>s+1-i$ which implies that $d<n-s+i-1$ . Further, $d>n-s-1$ as $b_{i}>b_{1}>c_{2}$ . For fixed $i,j,d$ , following inequalities hold:

\begin{array}[]{ccccccccccccc}&&&&a_{1}&<&\cdots&<&a_{j}\\ &&&&\wedge&&\cdots&&\wedge\\ &&b_{1}&<&\cdots&<&\cdots&<&b_{i-1}\\ &&\wedge&&\wedge&&&&&\\ c_{3}&<&\cdots&<&c_{d}\end{array}<b_{i}<\begin{array}[]{ccccccccccccc}&&a_{j+1}&<&\cdots&<&\cdots&<a_{s}\\ &&\wedge&&\cdots&&\wedge\\ &&b_{i+1}&<&\cdots&<&b_{s+1}\\ &&\wedge&&\ldots&&\wedge&&\\ c_{d+1}&<&\cdots&<&\cdots&<&N\end{array}

Let $k$ be the number of extra entries of the third row that are sticking out of the skew shape on the right of $b_{i}$ , i.e., $k=(n-d)-(s-i+1)$ . The range of $k$ as inferred from the range of $d$ is $0\leq k<i$ . These arrangements are counted by the product of hooks length formulas of appropriate skew shapes, i.e. by,

(5.12)

f_{\lambda_{1}/\mu_{1}}.f_{\lambda_{2}/\mu_{2}},

where $\lambda_{1}/\mu_{1}=(n+i-s-3,i-1,j)/k$ and $\lambda_{2}/\mu_{2}=(s-j+k,s-i+k+1,s-i+k)/(k,k)$ .

From these arrangements, for $k>1$ , we have to remove the arrangements which do not result in $b_{i}$ wrapping from the second row to the third row. These are obtained by shifting the bottom row of inequalities in the skew shape on the left of $b_{i}$ by $1$ position towards the right. The resulting inequalities are as follows:

\begin{array}[]{ccccccccccccc}&&&&a_{1}&<&\cdots&<&a_{j}\\ &&&&\wedge&&\cdots&&\wedge\\ &&b_{1}&<&\cdots&<&\cdots&<&b_{i-1}\\ &&\wedge&&\ldots&&\wedge&&&\\ c_{3}&<&\cdots&<&\cdots&<&c_{d}\end{array}<b_{i}<\begin{array}[]{ccccccccccccc}&&a_{j+1}&<&\cdots&<&\cdots&<a_{s}\\ &&\wedge&&\cdots&&\wedge\\ &&b_{i+1}&<&\cdots&<&b_{s+1}\\ &&\wedge&&\ldots&&\wedge&&\\ c_{d+1}&<&\cdots&<&\cdots&<&N\end{array}

The number of these multiline queues is

(5.13)

f_{\lambda_{3}/\mu_{3}}.f_{\lambda_{2}/\mu_{2}},

where $\lambda_{3}/\mu_{3}=(n+i-s-4,i-1,j)/(k-1)$ .

Summing the difference of two products in (5.12) and (5.13) over all possible values of $i,j,k$ , we get Theorem 5.14. ∎

Acknowledgements

We would like to thank our advisor Arvind Ayyer for all the insightful discussions during the preparation of this paper. We are grateful to Svante Linusson for his fruitful suggestions regarding the proof of Proposition 5.7. We thank Christoph Koutschan for helping with an earlier approach towards proof of Theorem 1.2. The second author was supported in part by a SERB grant CRG/2021/001592.

References

[1] Erik Aas, Arvind Ayyer, Svante Linusson, and Samu Potka. The exact phase diagram for a semipermeable TASEP with nonlocal boundary jumps. Journal of Physics A: Mathematical and Theoretical, 52(35):355001, aug 2019.
[2] Erik Aas, Arvind Ayyer, Svante Linusson, and Samu Potka. Limiting directions for random walks in classical affine weyl groups. International Mathematics Research Notices, 12 2021.
[3] Erik Aas and Svante Linusson. Continuous multi-line queues and TASEP. Ann. Inst. Henri Poincaré D, 5(1):127–152, 2018.
[4] Erik Aas and Jonas Sjöstrand. A product formula for the TASEP on a ring. Random Structures & Algorithms, 48(2):247–259, 2016.
[5] Gideon Amir, Omer Angel, and Benedek Valkó. The TASEP speed process. The Annals of Probability, 39(4):1205 – 1242, 2011.
[6] Omer Angel. The stationary measure of a 2-type totally asymmetric exclusion process. Journal of Combinatorial Theory, Series A, 113(4):625–635, 2006.
[7] Chikashi Arita, Atsuo Kuniba, Kazumitsu Sakai, and Tsuyoshi Sawabe. Spectrum of a multi-species asymmetric simple exclusion process on a ring. Journal of Physics A: Mathematical and Theoretical, 42(34):345002, 2009.
[8] Arvind Ayyer and Svante Linusson. An inhomogeneous multispecies TASEP on a ring. Advances in Applied Mathematics, 57, 06 2012.
[9] Arvind Ayyer and Svante Linusson. Correlations in the multispecies TASEP and a conjecture by Lam. Trans. Amer. Math. Soc., 369(2):1097–1125, 2017.
[10] Luigi Cantini. Inhomogenous Multispecies TASEP on a ring with spectral parameters. arXiv e-prints, page arXiv:1602.07921, February 2016.
[11] Sylvie Corteel, Olya Mandelshtam, and Lauren Williams. From multiline queues to Macdonald polynomials via the exclusion process. Amer. J. Math., 144(2):395–436, 2022.
[12] Pablo A. Ferrari and James B. Martin. Stationary distributions of multi-type totally asymmetric exclusion processes. Ann. Probab., 35(3):807–832, 2007.
[13] J Sutherland Frame, G de B Robinson, and Robert M Thrall. The hook graphs of the symmetric group. Canadian Journal of Mathematics, 6:316–324, 1954.
[14] Donghyun Kim and Lauren Williams. Schubert polynomials and the inhomogeneous tasep on a ring. arXiv preprint arXiv:2102.00560, 2021.
[15] Thomas Lam. The shape of a random affine Weyl group element and random core partitions. The Annals of Probability, 43(4):1643 – 1662, 2015.
[16] David A. Levin, Yuval Peres, and Elizabeth L. Wilmer. Markov chains and mixing times. American Mathematical Society, Providence, RI, 2009. With a chapter by James G. Propp and David B. Wilson.
[17] Richard P Stanley. Enumerative Combinatorics: Volume 2, volume 49. Cambridge University Press, 2011.
[18] Masaru Uchiyama and Miki Wadati. Correlation function of asymmetric simple exclusion process with open boundaries. Journal of Nonlinear Mathematical Physics, 12(sup1):676–688, 2005.