Random surface growth with a wall and Plancherel measures for O(infinity)

Introduction

The principal object of study in this paper is a one-parameter family of probability measures on certain interlacing two-dimensional particle systems that can be defined in at least three different ways.

Random lozenge tilings. Consider the domain pictured on the left in Figure 1 drawn on the regular triangular lattice, and consider all possible tilings of this domain by lozengesA lozenge consists of two neighboring elementary triangles glued together..

An example of lozenge tiling can be seen in the middle of Figure 1. To each tiling we assign a weight equal to $\frac{1}{2}$ raised to the number of vertical lozenges on the left border of the domain (three such lozenges are highlighted on the figure). Let us normalize the weights so that the total weight of all tilings is 1; then we obtain a probability distribution with three parameters $a,b,c$ that represent side lengths of our domain.

Let us further consider the limit $a,b,c\to\infty$ so that $2ab/c\to t>0$, and focus on the part of the tiling that is of finite distance to the bottom-left corner of the domain. One can show that in this limit our probability distributions weakly converge to a probability measure $\mathcal{M}_{t}$ on lozenge tilings of the quarter-plane, and it is the limiting measure that we are interested in.

Lozenge tilings are also commonly viewed as stepped surfaces (when three types of lozenges are interpreted as three faces of $1\times 1\times 1$ cubes in a three-dimensional space), as nonintersecting paths (see the right-most part of Figure 1), and as dimers on the hexagonal lattice (see Figure 5 in Section 2.3 below). Theory of dimer models is a rapidly developing subject, see for a recent review and references.

In terms of nonintersecting paths, the initial $(a,b,c)$-measures give an extra factor of 2 every time the left-most path passes by the wall. Thus, it is natural to say that this path reflects off the wall.

Random surface growth. Any lozenge tiling is uniquely determined by locations of lozenges of a single type. Let us introduce coordinates on the plane as shown in Figures 1 and 5, and mark the midpoints of all vertical lozenges; call them particles. Denote the horizontal coordinates of all particles with vertical coordinate $m$ by $y^{m}_{1}>y^{m}_{2}>\dots$. Then $\mathcal{M}_{t}$ is a probability measure on particle configurations

that satisfy the interlacing conditions $y_{k+1}^{m+1}<y_{k}^{m}<y_{k}^{m+1}$ for all meaningful values of $k$ and $m$.

We show that $\mathcal{M}_{t}$ is the time $t$ distribution of a continuous time Markov chain defined as follows.

The initial condition is a single particle configuration when all the particles are as much to the left as possible, i.e. $y^{m}_{k}=m-2k+1$ for all $k,m$. Now let us describe the evolution.

We say that a particle $y^{m}_{k}$ is blocked on the right if $y^{m}_{k}+1=y^{m-1}_{k-1}$, and it is blocked on the left if $y^{m}_{k}-1=y^{m-1}_{k}$ (if the corresponding particle $y^{m-1}_{k-1}$ or $y^{m-1}_{k}$ does not exist, then $y^{m}_{k}$ is not blocked).

Each particle has two exponential clocks of rate $\frac{1}{2}$; all clocks are independent. One clock is responsible for the right jumps, while the other is responsible for the left jumps. When the clock rings, the particle tries to jump by 1 in the corresponding direction. If the particle is blocked, then it stays still. If the particle is against the wall (i.e. $y^{m}_{[\frac{m+1}{2}]}=0$) and the left jump clock rings, the particle is reflected, and it tries to jump to the right instead.

In other words, the particles with smaller upper indices can be thought of as heavier than those with larger upper indices, and the heavier particles block and push the lighter ones so that the interlacing conditions are preserved.

Figure 2 depicts three possible first jumps: Left clock of $y_{1}^{1}$ rings first (it gets reflected by the wall), then right clock of $y_{1}^{5}$ rings, and then left clock of $y_{1}^{1}$ again.

In terms of the underlying stepped surface, the evolution can be described by saying that we add possible “sticks” with base $1\times 1$ and arbitrary length of a fixed orientation with rate 1/2, remove possible “sticks” with base $1\times 1$ and a different orientation with rate 1/2, and the rate of removing sticks that touch the left border is doubled.This phrase is based on the convention that is a figure of a $1\times 1\times 1$ cube. If one uses the dual convention that this is a cube-shaped hole then the orientations of the sticks to be added and removed have to be interchanged, and the tiling representations of the sticks change as well.

A computer simulation of this dynamics can be found at $\mathtt{http://www.math.caltech.edu/papers/Orth\_Planch.html}$.

Similar Markov chains have been previously studied in without the wall, and in with a different (“symplectic”) interaction with the wall.

Representation Theory. Let $O(N)$ be the group of $N\times N$ orthogonal matrices with real entries. The group $O(N)$ is embedded in $O(N+1)$ as a subgroup of matrices fixing the $(N+1)$st basis vector. Let $O(\infty)=\bigcup_{N=1}^{\infty}O(N)$ be the infinite-dimensional orthogonal group.

The measures $\mathcal{M}_{t}$ are the Fourier transforms of the distinguished one-parameter family of indecomposable characters of $O(\infty)$ (the indecomposable characters of $O(\infty)$ were classified in as a part of a solution of a much more general problem). It is natural to call them the Plancherel measures. Details can be found in Section 2.

Similarly defined Plancherel measures for the infinite symmetric group $S(\infty)$ and the infinite-dimensional unitary group $U(\infty)$ have been thoroughly studied, see for $S(\infty)$ and for $U(\infty)$.

Results. We first prove, see Theorem 3.12 below, that representation theoretic and Markov chain descriptions of $\mathcal{M}_{t}$ given above are equivalent (the lozenge tiling description of $\mathcal{M}_{t}$ is a simple corollary of the representation theoretic one and Theorem 1.4 of ). This equivalence is far from being obvious, and we employ the general formalism of to give a proof.

Our second result (Theorem 4.1) shows that $\mathcal{M}_{t}$, viewed as a measure on particle configurations $\{y^{m}_{k}\}$, is a determinantal random point process (see Appendix A for basic definitions), and it also provides an explicit formula for the correlation kernel. In fact, we prove such a result for random point processes associated with arbitrary indecomposable characters of $O(\infty)$.

We then focus on the asymptotics of $\mathcal{M}_{t}$ as $t\to\infty$. Note that, at first reading, one could look at the asymptotic results without the construction in Sections 2 and 3.

As one might anticipate from previous results on dimer models and Plancherel measures, cf. , as $t\to\infty$ the quarter-plane should split into “frozen” parts and a “liquid” part. In each frozen part the tiling asymptotically consists of lozenges of only one type, while in the liquid part the random tiling locally (i.e. on the lattice scale) converges to the unique (thanks to ) translation invariant Gibbs measure of a certain slope; the slope depends on the location in the liquid region. The underlying random surface should also converge, in a suitable metric, to the deterministic smooth limit surface, and the slopes of the Gibbs measures are the slopes of the tangent planes to this limit shape.

In Theorem 5.2 we prove the statements about local convergence. The frozen and liquid phases can be clearly seen in Figure 4. More exactly, we prove the convergence of our correlation kernel to the incomplete beta-kernel first obtained in , see for a detailed discussion of the Gibbs properties of the corresponding determinantal process. In Section 5.2, we also provide a formula for the hypothetical limit shape, although we do not address the concentration of measure phenomenon.

From previously known results it is also natural to expect that near the boundaries between frozen and liquid regions away from the wall, our determinantal process converges in an appropriate scaling to the so-called Airy process, see e.g. Section 4.5 of for an analogous results in the case of Plancherel measures for $U(\infty)$. This is indeed correct, and since the result and the method of proving it are well known by now, we did not include them in this paper.

The main novel feature of the model analyzed in this paper is the wall, and we focus on the corresponding scaling limits.

The simplest case is the neighborhood of the origin (the corner of the quarter-plane). Taking $t\to\infty$ asymptotics in Theorem 2.8, one can easily show (although we do not do this in the paper) that as one scales the horizontal coordinate by $\sqrt{t}$ and keeps the vertical coordinate finite, $\mathcal{M}_{t}$ converges to the antisymmetric GUE minor process (aGUEM) of , see also . Note that the way this process was obtained in from lozenge tilings of a half-hexagon is also similar to what we are doing. The aGUEM process can also be obtained from the evolution of interacting Brownian motions with a reflecting wall, which can be seen as a limit of the Markov chain described above; see for details.

We call it the symmetric Pearcey kernel because of the similarity of the above expression to the Pearcey kernel that has previously appeared in . Using the nonintersecting paths interpretation mentioned above, it seems plausible that the symmetric Pearcey kernel should also appear in the model treated in at the critical location when the paths touch the wall. Indeed, we were informed by the authors of that this is indeed the case, cf. .

Acknowledgements. The authors are very grateful to Grigori Olshanski for a number of valuable remarks. The first named author (A. B.) was partially supported by the NSF grant DMS-0707163.

Measures on partitions

Let $O(N)$ denote the group of all real-valued $N\times N$ orthogonal matrices. For each $N$, $O(N)$ is naturally embedded in $O(N+1)$ as the subgroup fixing the $(N+1)$-st basis vector. Equivalently, $O\in O(N)$ can be thought of as an $(N+1)\times(N+1)$ matrix by setting $O_{i,N+1}=O_{N+1,j}=0$ for $1\leq i,j\leq N$ and $O_{N+1,N+1}=1$. The union $\bigcup_{N=1}^{\infty}O(N)$ is denoted by $O(\infty)$.

Let us review some basic results from the representation theory of finite- and infinite-dimensional orthogonal groups, see e.g. .

The special orthogonal group, denoted by $SO(N)$, is the subgroup of $O(N)$ consisting of matrices with determinant $1$. Let $O\in SO(N)$. If $N=2m$ is even, then the spectrum of any $O$ is of the form $\{z_{1},z_{1}^{-1},\ldots,z_{m},z_{m}^{-1}\}$, while if $N=2m+1$ is odd, then the spectrum of any $O\in SO(N)$ is of the form $\{z_{1},z_{1}^{-1},\ldots,z_{m},z_{m}^{-1},1\}$, where in both cases $z_{i}$ are complex numbers having absolute value $1$. For this paper, if $\chi$ is a character of $O(\infty),SO(2m)$ or $SO(2m+1)$, then $\chi(O)$ is written interchangably with $\chi(z_{1},\ldots,z_{m})$. Using this notation, any $\omega\in\Omega$ defines a function on $SO(\infty)=\bigcup_{N=1}^{\infty}SO(N)$ by (see Theorem 1.4 of )

Note that the infinite product converges because $\sum(\alpha_{i}+\beta_{i})$ is finite. As $\omega$ ranges over $\Omega$, the functions $\chi^{\omega}$ range over all the extreme characters of $O(\infty)$ (Theorem 5.2 of ).

where $\chi^{\lambda}$ is the character of $SO(2N+1)$ or $SO(2N)$ parameterized by $\lambda$. Evaluating both sides of the equation at the identity of the group shows that the sum of the weights is one. Furthermore, the weights are nonnegative because the characters are positive definite, so we obtain probability measures. Note that if the parameters of $\omega$ are $\alpha=0,\beta=0,\gamma=0$, then $P_{N}^{\omega}$ is the delta measure supported at the partition $(0,0,\ldots,0)$.

There is a useful explicit formula for $\chi^{\lambda}$. Let $J_{k}^{(a,b)}(x)$ denote the $k$-th Jacobi polynomial with parameters $a,b$ see e.g. . Define the constant $c_{k}$ to be

and let $\mathsf{J}_{k}^{(a,b)}(x)=J_{k}^{(a,b)}(x)/c_{k}$. The character of $O\in SO(2N)$ or $SO(2N+1)$ is

Expressions (4) and (5) can be simplified using

Let $h_{k}^{(a,b)}$ denote the squared norm of $J_{k}^{(a,b)}$,

For proofs of these equations, see §1 of , Chapter 4 of , and Chapter 24 of .

In Section 2.5, a formula for $P^{\omega}_{N,a}$ will be proved.

2 Central Measures

Note that $\lambda\prec\mu$ is equivalent to the following relation from representation theory. Let $V_{\mu}$ be the representation of $SO(M)$ corresponding to $\mu$ and let $V_{\lambda}$ be the representation of $SO(M-1)$ corresponding to $\lambda$. With this notation, $V_{\lambda}$ is a subrepresentation of $V_{\mu}|_{SO(M-1)}$ iff $\lambda\prec\mu$. See .

The definition of $\varkappa(\lambda,\mu)$ is motivated by the branching rules

Also for a finite path, define the weight $w_{u}$ to be

The brancing rules imply that if we sum $w_{u}$ over all finite paths $u$ that end at $\lambda$, we get $\dim\lambda=\dim V_{\lambda}$. Note that $\dim V_{\lambda}=\dim V_{\lambda^{*}}$.

for any two finite paths $u,v$ that end at the same partition. For a more general definition of central measures, see section 6 of .

Furthermore, set $d(n_{0},a_{0};n_{1},a_{1})=|2(n_{1}-n_{0})+a_{1}-a_{0}|$. In other words, $d(n_{0},a_{0};n_{1},a_{1})$ is the distance between the levels $(n_{0},a_{0})$ and $(n_{1},a_{1})$.

We identify $\mathfrak{X}$ and $\mathfrak{Y}$ via the bijection

Note that $\iota(\mathcal{L}_{\mathfrak{X}}(\boldsymbol{\lambda}))=\mathcal{L}_{\mathfrak{Y}}(\boldsymbol{\lambda})$ for any $\boldsymbol{\lambda}$.

In Figure 5, black dots mark the elements of $\mathcal{L}_{\mathfrak{Y}}(\boldsymbol{\lambda})$ for $\boldsymbol{\lambda}=((1)\prec(1)\prec(1,0)\prec(2,1)\prec(2,1,1)\prec(3,1,1))$.

4 Preliminary Lemmas

Before continuing, a couple of lemmas will be needed. Since they will be used several times throughout this paper, it is convenient to gather them in this section. Lemma 2.1 is a variant of the Cauchy-Binet formula.

The proof is almost identical to the proof of Theorem 1.2.1 of . ∎

For $a=\pm 1/2,-1\leq\zeta\leq 1$, and a test function $T\in C^{1}$,

Plugging in $a=-1/2$ and setting $x=\cos\phi$, $\zeta=\cos\theta$, (4.1.7) of and (11) give

Since $T$ is $C^{1}$, the Fourier series of $T$ converges to $T$ (see Chapter 3, Section 6 of ):

In the case when $a=1/2$, (4.1.8) of and (11) tell us

and the rest of the argument is similar. ∎

The previous two lemmas also imply the next one.

Let $-1\leq\zeta_{1},\ldots,\zeta_{N}\leq 1$, and suppose $\varphi\in C^{1}$. Let $l_{1},\ldots,l_{N}$ be nonnegative integers. Set

The normalized Jacobi polynomials $\mathsf{J}^{(\pm 1/2,-1/2)}_{s}$ satisfy the following properties:

(a) $\displaystyle\sum_{r=0}^{s}W^{(-1/2,-1/2)}(r)\mathsf{J}_{r}^{(-1/2,-1/2)}(x)=\mathsf{J}_{s}^{(1/2,-1/2)}(x)$,

(b) $\displaystyle\sum_{r=0}^{s-1}\mathsf{J}_{r}^{(1/2,-1/2)}(x)=\frac{\mathsf{J}_{s}^{(-1/2,-1/2)}(x)-1}{x-1}$,

(c) $\displaystyle\frac{1}{\pi}\int_{-1}^{1}\mathsf{J}_{s}^{(1/2,-1/2)}(x)(1-x)^{-1/2}(1+x)^{-1/2}dx=1.$

(a),(b) Let $z$ be on the unit circle such that $(z+z^{-1})/2=x$. Using (6) and (7), the sum becomes a geometric series, which can be evaluated explicitly.

which equals $1$ by the orthogonality relations. ∎

Let $T\in C^{1}$. The following identities hold:

The first equality follows from (11) and $\mathsf{J}_{r}^{(-1/2,-1/2)}(1)=1$, and the second equality follows from Lemma 2.2.

Subtracting this from the sum in (a) proves (b). ∎

Since the series in Lemma 2.5(a) converges,

Taking $s\rightarrow\infty$ and using Corollary 2.6,

Subtracting these two sums proves the lemma. ∎

5 Explicit Formula for the Measures

In this section, we prove the following statement:

Theorem 2.8 follows from (1) and the following statement with $E=E^{\omega}$. It is an orthogonal group analog of Lemma 6.5 of .

Let $E(x)\in C^{1}$. Fix complex numbers $z_{1},\ldots,z_{N}$ on the unit circle and set $\zeta_{k}=(z_{k}+z_{k}^{-1})/2$. Then

where $f_{j}^{(N,a)}(k)$ is defined by (14), with $E(x)$ in place of $E^{\omega}(x)$.

Let $E_{i}(x)=x^{N-i}E(x)$ for $1\leq i\leq N$. Using equation (11) and Lemma 2.2 shows that for $1\leq i\leq N$,

where $\lambda_{i}=k_{i}+i-N$. Equation (15) can be rewritten as

Set $\zeta_{k}=(z_{k}+z_{k}^{-1})/2$. First consider the case when $a=1/2$. Using (4), equation (16) becomes

Now consider the case when $a=-1/2$. Using (5), equation (16) becomes

Stochastic Dynamics

where $f_{j}^{(N,a)}$ is defined by (14), with $E$ instead of $E^{\omega}$.

Multiply both sides by $\mathsf{J}_{l}^{(a,-1/2)}(x)$ and integrate over $$. Then the above definition is equivalent to

Let $A$, $B$, $C$ be the following matrices:

Then $C=AB$, so by the Cauchy-Binet formula,

For any $\varphi\in C^{1}$, the rows of $T_{N,a}^{\varphi}$ sum to $\varphi(1)^{N}$. In particular, if $\varphi(1)=1$, then the rows of $T_{N,a}^{\varphi}$ sum to $1$.

Let $z_{1},\ldots,z_{N}$ be complex numbers on the unit circle and set $\zeta_{i}=(z_{i}+z_{i}^{-1})/2$. Using the notation and statement of Lemma 2.3,

Taking $z_{1},\ldots,z_{N}=1$ shows that the rows of $T_{N,a}^{\varphi}$ sum to $\varphi(1)^{N}$. ∎

If $\varphi(x)=p_{0}+p_{1}x$ with $p_{0}>p_{1}\geq 0$, then each entry of $T_{N,a}^{\varphi}$ is nonnegative. The same holds if $\varphi(x)=e^{t(x-1)}$, where $t\geq 0$. Additionally, the diagonal entries of $T_{r,\pm 1/2}^{p_{0}+p_{1}x}$ are bounded below by

First consider the situation when $\varphi(x)=p_{0}+p_{1}x$. Note that by (8)–(10), $I_{a}^{\varphi}(k,l)=0$ if $|k-l|>1$.

If $\mu_{i}<\lambda_{i}-1$ for some $i$ then $\mu_{k}<\lambda_{l}-1$ for $k\geq i$ and $l\leq i$, which implies that $I_{a}^{\varphi}(\mu_{k},\lambda_{l})=0$ for such $k,l$, and thus the determinant in question is . If $\mu_{i}>\lambda_{i}+1$ then $\mu_{k}>\lambda_{l}+1$ for $k\leq i$ and $l\geq i$, which means $I_{a}(\mu_{k},\lambda_{l})=0$, and the determinant is again. So $\det[I_{a}^{\varphi}]$ is zero if $|\lambda_{i}-\mu_{i}|>1$ for some $i$. Hence, it remains to consider the case when $|\lambda_{i}-\mu_{i}|\leq 1$ for all $1\leq i\leq N$.

Split $\{\mu_{i}-i+N\}_{i=1}^{N}$ into blocks of neighbouring integers wth distance between blocks being at least $2$. Then it is easy to see that $\det[I_{a}(\mu_{i}-i+N,\lambda_{j}-j+N)]$ splits into the product of determinants corresponding to blocks. It suffices to show that the determinant corresponding to each block is nonnegative, so assume without loss of generality that $\{\mu_{i}-i+N\}_{i=1}^{N}$ is one such block. In other words, assume that all $\mu_{i}$ are equal. Then there exist $m$ and $n$, $1\leq m\leq n\leq N$ such that $\mu_{i}=\lambda_{i}-1$ for $1\leq i<m$, and $\mu_{i}=\lambda_{i}$ for $m\leq i<n$, and $\mu_{i}=\lambda_{i}+1$ for $n\leq i\leq N$. The determinant is the product of determinants of three matrices. We shall examine each of these matrices.

Because $I_{a}^{\varphi}(k,k\pm 1)\geq 0$, the matrix parametrized by $1\leq i,j<m$ is triangular with nonnegative diagonal entries, so has nonnegative determinant. Similarly, the matrix parametrized by $n\leq i,j\leq N$ also is triangular with nonnegative diagonal entries. It remains to consider the matrix parametrized by $m\leq i,j<n$.

For now, assume $\lambda_{n-1}\neq 0.$ Then this matrix is tridiagonal, with $p_{0}$ in the diagonal entries and $p_{1}/2$ in the subdiagonal and superdiagonal entries. If this matrix has size $r\times r$, let $D_{r}$ denote its determinant. Then $D_{r}$ satisfies the recurrence relation

If $\lambda_{n-1}=0$ and $a=1/2$, then the entry in the $r$th row and $r$th column is $p_{0}-p_{1}/2$ instead of $p_{0}$. The other entries are $p_{0},p_{1}/2$, and , as before. In this case, the determinant is

Note that $A_{r}$ is positive on the interval $(p_{1}/2,\infty)$, which contains $(p_{0}+\sqrt{p_{0}^{2}-p_{1}^{2}})/2$, and is negative on the interval $(0,p_{1}/2)$, which contains $(p_{0}-\sqrt{p_{0}^{2}-p_{1}^{2}})/2$. Therefore the above expression is nonnegative.

If $\lambda_{n-1}=0$ and $a=-1/2$, then the only modified entry is in the $r$th row and $(r-1)$st column. It equals $p_{1}$ instead of $p_{1}/2$. In this case, the determinant is

We have already shown that $D_{r}\geq(p_{1}/2)D_{r-1}$, so therefore $D_{r}\geq(p_{1}^{2}/4)D_{r-2}$. So the above expression is also nonnegative.

For the last claim in the proposition, note that $D_{r}\geq D_{r}-(p_{1}/2)^{2}D_{r-2}\geq D_{r}-(p_{1}/2)D_{r-1}=\eqref{SmallestDet}$.

Now let $\varphi(x)=e^{t(x-1)}$. For $n>2t$, let $\varphi_{n}(x)=\left(1+\frac{t(x-1)}{n}\right)^{n}$. By Lemma 3.4 below, $T_{N,a}^{\varphi_{n}}=T_{N,a}^{1+t(x-1)/n}\ldots T_{N,a}^{1+t(x-1)/n}$. We just showed that all entries of each $T_{N,a}^{1+t(x-1)/n}$ are nonnegative. Therefore all entries of $T_{N,a}^{\varphi_{n}}$ are nonnegative. By Lemma 3.5, $T_{N,a}^{\varphi}(\mu,\lambda)$ equals $\lim_{n\rightarrow\infty}T_{N,a}^{\varphi_{n}}(\mu,\lambda)$, so all entries of $T_{N,a}^{\varphi}$ are nonnegative. ∎

If $\varphi_{1},\varphi_{2}\in C^{1}$, then $T_{N,a}^{\varphi_{1}\varphi_{2}}=T_{N,a}^{\varphi_{1}}T_{N,a}^{\varphi_{2}}$.

This is a straightforward computation using the Lemmas 2.1 and 2.2. ∎

Since $\varphi_{n}$ converges uniformly to $\varphi$ and $\varphi$ is bounded, the dominated convergence theorem implies that each $I_{a}^{\varphi_{n}}(i,j)$ converges to $I_{a}^{\varphi}(i,j)$. Since $T_{N,a}^{\varphi_{n}}(\mu,\lambda)$ is continuous in the variables $I_{a}^{\varphi_{n}}(\mu_{i}-i+N,\lambda_{j}-j+N)$, it must converge to $T_{N,a}^{\varphi}(\mu,\lambda)$. ∎

2 Generalities on Multivariate Markov Chains

Let $\mathcal{S}_{1},\ldots,\mathcal{S}_{n}$ be discrete sets. For $1\leq k\leq n$, let $T_{k}$ be a stochastic matrix with rows and columns indexed by $\mathcal{S}_{k}$. For $2\leq k\leq n$, let $\Lambda^{k}_{k-1}$ be a stochastic matrix with rows indexed by $\mathcal{S}_{k}$ and columns indexed by $\mathcal{S}_{k-1}$. Assume these matrices commute:

The state space for the multivariate Markov Chain is

Write $X_{n}=(x_{1},\ldots,x_{n}),Y_{n}=(y_{1},\ldots,y_{n})\in\mathcal{S}_{\Lambda}^{(n)}$. The probability of a transition from $X_{n}$ to $Y_{n}$ is

Let $\mathbf{T}$ denote this matrix of transition probabilities. One could think of $\mathbf{T}$ as follows.

Starting from $X=(x_{1},\ldots,x_{n})$, first choose $y_{1}$ according to the transition matrix $T_{1}(x_{1},y_{1})$, then choose $y_{2}$ using $\frac{T_{2}(x_{2},y_{2})\Lambda^{2}_{1}(y_{2},y_{1})}{\Delta^{2}_{1}(x_{2},y_{1})}$ , which is the conditional distribution of the middle point in the successive application of $T_{2}$ and $\Lambda^{2}_{1}$ provided that we start at $x_{2}$ and finish at $y_{1}$. Then choose $y_{3}$ using the conditional distribution of the middle point in the successive application of $T_{3}$ and $\Lambda^{3}_{2}$ provided that we start at $x_{3}$ and finish at $y_{2}$, and so on. Thus, one could say that $Y$ is obtained by the sequential update .

Let $m_{n}(x_{n})$ be a probability measure on $\mathcal{S}_{n}$. Consider the evolution of the measure $m_{n}(x_{n})\Lambda_{n-1}^{n}(x_{n},x_{n-1})\ldots\Lambda^{2}_{1}(x_{2},x_{1})$ on $\mathcal{S}_{\Lambda}^{(n)}$ under the Markov chain $\mathbf{T}$, and denote by $(x_{1}(j),\ldots,x_{n}(j))$ the result after $j=0,1,2,\ldots$ steps. Then for any $k_{1}\geq k_{2}\geq\ldots\geq k_{n}\geq 0$, the joint distribution of

coincides with the stochastic evolution of $m_{n}$ under transition matrices

Let $L$ be the linear subspace of $l^{1}(\mathcal{S}^{(n)}_{\Lambda})$ spanned by elements of the form $m_{n}(x_{n})\Lambda^{n}_{n-1}(x_{n},x_{n-1})\ldots\Lambda^{2}_{1}(x_{2},x_{1})$, where $m_{n}$ is a summable function on $\mathcal{S}_{n}$. Then $\mathbf{T}$ can be thought of as a bounded linear operator of $L$. Similarly, $T_{n}$ is a bounded linear operator on $l^{1}(\mathcal{S}_{n})$.

If $f\in L$, then $f$ must have the form $m_{n}(x_{n})\Lambda^{n}_{n-1}(x_{n},x_{n-1})\ldots\Lambda^{2}_{1}(x_{2},x_{1})$ for some $m_{n}\in l^{1}(\mathcal{S}_{n})$. Since the matrices $\Lambda_{k}^{k+1}$ are all stochastic, $\|f\|_{L}=\|m_{n}\|_{l^{1}(\mathcal{S}_{n})}$. By Proposition 3.6, $\mathbf{T}f=(T_{n}m_{n})(x_{n})\Lambda^{n}_{n-1}(x_{n},x_{n-1})\ldots\Lambda^{2}_{1}(x_{2},x_{1})$, which implies $\|\mathbf{T}f-f\|_{L}=\|T_{n}m_{n}-m_{n}\|_{l^{1}(\mathcal{S}_{n})}$. Thus, the lemma holds. ∎

3 Markov Chain on Multiple Levels

Recall the definition of $\varkappa$ in Section 2.2.

The argument is standard, see e.g. Proposition 3.4 of . The proof of the second formula is exactly the same. ∎

The matrices $T^{N,+}_{N,-}$ and $T_{N-1,+}^{N,-}$ are stochastic.

First let us show that $T^{N,+}_{N,-}$ is stochastic. Taking dimensions of both sides of (12) yields

By Lemma 3.8, $T^{N,+}_{N,-}$ has nonnegative entries, so it is stochastic.

Now we will show that $T_{N-1,+}^{N,-}$ is stochastic. Taking dimensions in (13) yields

The nonnegativity also follows from Lemma 3.8. ∎

Assume $\varphi(1)=1$. For any $N\geq 1$, we have the following commutation relations:

We start by proving the first relation. By Lemma 2.1,

By applying Lemma 2.7 to the right hand side and Lemma 2.4(a) to the left hand side, one sees that both sides are equal to

Now we prove the second relation. Expanding $\det[\phi]_{1}^{N}$ along the $N$th column and using Lemma 2.1, we obtain

where it is agreed that all matrix elements in the $N$th column ($j=N$) of 22 are equal to $1$. Similarly,

By Lemma 2.5(a), the $N$th column of (23) equals $\varphi(1)\mathsf{J}_{\mu_{i}-i+N}^{(-1/2,-1/2)}(1)=1$. Therefore the $N$th columns of (22) and (23) are equal.

By Lemma 2.4(b), for $j\neq N$, the $(i,j)$-entry of (22) equals

By Lemma 2.5(b), for $j\leq N$, the $(i,j)$-entry of (23) equals

Their difference only depends on $j$, so for $j\neq N$,

so the matrix in (23) is obtained from the matrix in (22) by elementary column operations. This means that their determinants are equal. ∎

By the construction in section 3.2, $A_{N,\pm 1/2}^{\varphi}$ are stochastic for $\varphi(x)=1-p_{1}+p_{1}x$, $0\leq p_{1}\leq 1/2$, and $\varphi(x)=e^{t(x-1)},t\geq 0$.

For any $E(x)\in C^{1}$ such that $E(1)\neq 0$, let $P^{(N),a}$ be the (possibly signed) measure

where $\widetilde{P}_{N,a}$ is defined from the function $\widetilde{E}(x)=\varphi(x)E(x)$.

4 A Continuous-time Markov Chain on Multiple Levels

Case 1. This occurs when there exist $(n_{0},a_{0})\trianglelefteq(n_{1},a_{1})$ and $k\leq n_{0}$ such that the numbers $\mu_{k}^{(n^{*}),a^{*}}-1,\lambda_{k}^{(n^{*}),a^{*}}$ are all equal for $(n_{0},a_{0})\trianglelefteq(n^{*},a^{*})\trianglelefteq(n_{1},a_{1})$. Furthermore, $\mu_{l}^{(\bar{n}),\bar{a}}=\lambda_{l}^{(\bar{n}),\bar{a}}$ for all other $\bar{n},\bar{a},l$.

Case 1a. When case 1 is satisfied and $a_{0}=-1/2$ and $\lambda_{k}^{(n_{0}),a_{0}}=0$.

Case 1b. When case 1 is satisfied and case 1a is not satisfied.

Case 2. This occurs when there exist $(n_{0},a_{0})\trianglelefteq(n_{1},a_{1})$ and $k\leq n_{0}$ such that the numbers $\lambda_{k+d(n^{*},a^{*};n_{0},a_{0})}^{(n^{*}),a^{*}},\mu_{k+d(n^{*},a^{*};n_{0},a_{0})}^{(n^{*}),a^{*}}+1$ are all equal for $(n_{0},a_{0})\trianglelefteq(n^{*},a^{*})\trianglelefteq(n_{1},a_{1})$. Recall that $d(n_{1},a_{1};n_{0},a_{0})=|2n_{1}+a_{1}-2n_{0}-a_{0}|$. Furthermore, $\mu_{l}^{(\bar{n}),\bar{a}}=\lambda_{l}^{(\bar{n}),\bar{a}}$ for all other $\bar{n},\bar{a},l$.

Case 3. This occurs when the two paths $\lambda$ and $\mu$ are not equal and neither case 1 nor case 2 is satisfied.

When case 1b or case 2 occurs, the corresponding element of $Q_{N,a}$ is $1/2$. When case 1a occurs, the corresponding element is $1$. When case 3 occurs, the corresponding element is . The diagonal entries are defined so that the rows of $Q_{N,a}$ sum to .

Under the map $\mathcal{L}_{\mathfrak{Y}}$, the cases can be described more easily. Let $\{y_{k}^{m}\}=\mathcal{L}_{\mathfrak{Y}}(\boldsymbol{\lambda})$ and $\{z_{k}^{m}\}=\mathcal{L}_{\mathfrak{Y}}(\boldsymbol{\mu})$. Case 1 occurs when there exist $m_{0}\leq m_{1}$ and $k\leq[\frac{m_{0}}{2}]$ such that

Furthermore, $z_{l}^{\bar{m}}=y_{l}^{\bar{m}}$ for all other $l,\bar{m}$.

Case 1a occurs when case 1 is satisfied and $y_{k}^{m_{0}}=0$.

Case 1b occurs when case 1 is satisfied and case 1a is not satisfied.

Case 2 occurs when there exist $m_{0}\leq m_{1}$ and $k\leq[\frac{m_{0}}{2}]$ such that

Furthermore, $z_{l}^{\bar{m}}=y_{l}^{\bar{m}}$ for all other $l,\bar{m}$.

Case 3 occurs when $\{y_{k}^{m}\}\neq\{z_{k}^{m}\}$ and neither case 1 nor case 2 is satisfied.

It is not hard to see that $Q_{N,a}$ is the generator of the continuous-time Markov Chain defined in Section 1. In general, if $Q$ is a matrix with countably many rows and columns such that its rows add up to , its off-diagonal entries are nonnegative, and its diagonal entries are uniformly bounded, then there is a unique continuous-time Markov chain with $Q$ as its generator (see e.g. Proposition 2.10 of ). In words, this Markov chain satisfies

In state $i$, a jump takes place after exponential waiting time with parameter $-Q_{ii}$.

The system makes a jump to state $j$ with probability $-Q_{ij}/Q_{ii}$.

This theorem relies on the following proposition. It can be found as Theorem 9.6.1 in .

Let $\{A(t):t>0\}$ be bounded linear operators on a Banach space $B$ such that $A(s+t)=A(s)A(t)$ for all $s,t>0$. If $\displaystyle\lim_{t\rightarrow 0^{+}}\|A(t)-I\|=0$, then there exists a bounded linear operator $Q$ on $B$ such that $A(t)=e^{tQ}$ for $t\geq 0$.

Since $T_{N,a}^{\varphi_{t}}$ is stochastic, it is equivalent to show that

We prove that $T_{r,\pm}^{\varphi_{t}}(\lambda^{(r),\pm},\lambda^{(r),\pm})\geq\exp(-t(r+1/2))$. Let $\varphi_{t}^{(n)}=(1+t(x-1)/n)^{n}$. By Lemmas 3.5, 3.4, and Proposition 3.3, respectively,

Finally, notice that as $n\rightarrow\infty$,

We have just shown that $A^{\varphi_{t}}_{N,a}\cdot P^{(N),a}=e^{tQ}\cdot P^{(N),a}$ for some $Q$. To finish the proof, we show that $Q=\tfrac{d}{dt}A^{\varphi_{t}}_{N,a}|_{t=0}=Q_{N,a}$. Since we only need to calculate $Q$ up to terms of order $O(t^{2})$, we can replace $\varphi_{t}(x)=e^{t(x-1)}$ with $1-t+tx$.

The problem now is to calculate $A_{N,a}^{\varphi_{t}}$ up to terms of order $O(t^{2})$. There are three cases to consider: when all the particles on the $m$th level stay still, when one of the particles on the $m$th level is pushed by a particle on a lower level, and when one of the particles on the $m$th level moves by itself. As an example, consider particles on the $(m,1/2)$ level when one of them is pushed.

The expression that needs to be calculated is

Assume that $\mu^{(m),-1/2}\nprec\lambda^{(m),1/2}$. Since $\mu^{(m),-1/2}\prec\mu^{(m),1/2}$, this implies that $\lambda^{(m),1/2}\neq\mu^{(m),1/2}$. Similarly, $\lambda^{(m),-1/2}\neq\mu^{(m),-1/2}$, which means one of the particles on the $(m,-1/2)$ level is pushing a particle on the $(m,1/2)$ level. Conversely, if the $k$th particle on the $(m,-1/2)$ level is pushing a particle on the $(m,1/2)$ level, then $\mu^{(m),-1/2}_{k}>\lambda^{(m),1/2}_{k}$, so $\mu^{(m),-1/2}\not\prec\lambda^{(m),1/2}$.

The transition probability on the $(m,1/2)$ level is (because of (8))

Similarly, when all the particles on a level stay still, the contribution is $1+O(t)$. When a particle against the wall moves, the contribution is $t+O(t^{2})$. When a particle not against the wall moves without being pushed by a particle on a lower level, the contribution is $t/2+O(t^{2})$. ∎

The Correlation Kernel

For any $\omega\in\Omega$ with parameter $\beta_{1}<1$, the point process $\mathcal{P}^{\omega}_{\mathfrak{X}}$ is determinantal. Denote its correlation kernel by $K^{\omega}(n_{1},a_{1},s_{1};n_{2},a_{2},s_{2})$. If $(n_{1},a_{1})\trianglerighteq(n_{2},a_{2})$, then $K^{\omega}(n_{1},a_{1},s_{1};n_{2},a_{2},s_{2})$ equals

If $(n_{1},a_{1})\triangleleft(n_{2},a_{2})$, then $K^{\omega}(n_{1},a_{1},s_{2};n_{2},a_{2},s_{2})$ equals

The $u$-contour $C$ is a positively oriented simple loop that encircles the interval $$but does not encircle any zeroes of$E^{\omega}$. Recall that the functions$E^{\omega},\mathsf{J}_{s}$and$W$ were defined in Section 2.1.

Remark. The case $\beta_{1}=1$ can be obtained by the limiting transition $\beta_{1}\rightarrow 1$ from Theorem 4.1.

With the definition of the particle-hole involution $\Delta$ given in Appendix A, if $(n_{2},a_{2})\triangleleft(n_{1},a_{1})$, then $K^{\omega}_{\Delta}(n_{1},a_{1},s_{1};n_{2},a_{2},s_{2})$ equals

If $(n_{1},a_{1})\trianglelefteq(n_{2},a_{2})$, then $K^{\omega}_{\Delta}(n_{1},a_{1},s_{2};n_{2},a_{2},s_{2})$ equals

This result follows from the orthogonality relations

for $a=\pm 1/2$. Note that in Theorem 4.1, the two cases are $(n_{2},a_{2})\trianglelefteq(n_{1},a_{1})$ and $(n_{1},a_{1})\triangleleft(n_{2},a_{2})$. Here, the two cases are $(n_{2},a_{2})\triangleleft(n_{1},a_{1})$ and $(n_{1},a_{1})\trianglelefteq(n_{2},a_{2})$. ∎

This proof uses Theorem 4.2 from , which we will describe in the next subsection. We will alter the notation to make it more convenient later.

Theorem 2.8 and Lemma 3.8 imply that the measure of a finite path $\boldsymbol{\lambda}=(\lambda^{(1),-1/2}\prec\lambda^{(1),1/2}\prec\lambda^{(2),-1/2}\prec\ldots\prec\lambda^{(N),a})$ is

where $\phi$ and $\mathcal{T}$ were defined in Section 3.3, and $f_{j}^{N,a}$ was defined in the statement of Theorem 2.8. If $a=-1/2$, then the final determinant with the $\mathcal{T}$ does not occur.

Recall that we have set $\lambda_{n}^{(n-1),1/2}$ to be equal to zero, so $\lambda_{n}^{(n-1),1/2}-n+n-1=-1$. We will refer to $-1$ as a “virtual variable,” or “virt.”

Observe that Span$\{f_{1},\ldots,f_{N}\}=$ Span$\{\Psi_{1},\ldots,\Psi_{N}\}$. Thus, if we replace $f_{l}$ in (27) with $\Psi_{N-l}$, the measure is not going to change.

Let $*$ denote convolution. More explicitly,

For $(n_{1},a_{1})\triangleleft(n_{2},a_{2})$, set

For $(n_{1},a_{1})\trianglerighteq(n_{2},a_{2})$, set $\phi^{(n_{1},a_{1}),(n_{2},a_{2})}=0$.

Let $M$ be the $N\times N$ matrix with entries

For $k\leq N$, define $\Psi_{n-k}^{n,a_{1}}=\Psi_{N-k}^{N,a}*\phi^{(n,a_{1}),(N,a)}$.

Theorem 4.2 from also says that if $M$ is upper triangular and invertible, then there exist functions $\Phi_{n-k}^{n,a_{2}}(s)$ such that

$\{\Phi^{n,a_{2}}_{n-k}(s)\}_{k=1,\ldots,n}$ is a basis of the linear span of

The formula for the correlation kernel is given by

In section 4.2, we prove that $M$ is upper triangular. In section 4.3, we calculate $\Psi_{n_{1}-k}^{n_{1},a_{1}}$. In section 4.4, we calculate $\displaystyle\sum_{k=1}^{n_{2}}\Psi_{n_{1}-k}^{n_{1},a_{1}}(s_{1})\Phi_{n_{2}-k}^{n_{2},a_{2}}(s_{2})$. In section 4.5, we calculate $\phi^{(n_{1},a_{1}),(n_{2},a_{2})}(s_{2},s_{1})$. Finally in section 4.6, we add all these expressions together to get the expression in Theorem 4.1.