# Asymptotic Height Distribution in High-Dimensional Sandpiles

Open Access
Article

## Abstract

We give an asymptotic formula for the single-site height distribution of Abelian sandpiles on $$\mathbb {Z}^d$$ as $$d \rightarrow \infty$$, in terms of $$\mathsf {Poisson}(1)$$ probabilities. We provide error estimates.

## Keywords

Abelian sandpile Uniform spanning forest Wilson’s method Loop-erased random walk

## Mathematics Subject Classification (2010)

Primary 60K35 Secondary 82C20

## 1 Introduction

We consider the Abelian sandpile model on the nearest neighbour lattice $$\mathbb {Z}^d$$; see Sect. 1.1 for definitions and background. Let $$\mathbf {P}$$ denote the weak limit of the stationary distributions $$\mathbf {P}_L$$ in finite boxes $$[-L,L]^d \cap \mathbb {Z}^d$$. Let $$\eta$$ denote a sample configuration from the measure $$\mathbf {P}$$. Let $$p_d(i) = \mathbf {P}[\eta (o) = i]$$, $$i = 0, \dots , 2d-1$$, denote the height probabilities at the origin in d dimensions. The following theorem is our main result that states the asymptotic form of these probabilities as $$d \rightarrow \infty$$.

### Theorem 1.1

1. (i)
For $$0 \le i \le d^{1/2}$$, we have
\begin{aligned} p_d(i) = \sum _{j=0}^i \frac{e^{-1} \frac{1}{j!}}{2d-j} + O\Big (\frac{i}{d^2}\Big ) = \frac{1}{2d} \sum _{j=0}^i e^{-1} \frac{1}{j!} + O\Big (\frac{i}{d^2}\Big ). \end{aligned}
(1.1)

2. (ii)
If $$d^{1/2} < i \le 2d-1$$, we have
\begin{aligned} p_d(i) = p_d(d^{1/2}) + O(d^{-3/2}). \end{aligned}
In particular, $$p_d(i) \sim (2d)^{-1}$$, if $$i,d \rightarrow \infty$$.

The appearance of the $$\mathsf {Poisson}(1)$$ distribution in the above formula is closely related to the result of Aldous  that the degree distribution of the origin in the uniform spanning forest in $$\mathbb {Z}^d$$ tends to 1 plus a $$\mathsf {Poisson}(1)$$ random variable as $$d \rightarrow \infty$$. Indeed our proof of (1.1) is achieved by showing that in the uniform spanning forest of $$\mathbb {Z}^d$$, the number of neighbours w of the origin o, such that the unique path from w to infinity passes through o is asymptotically the same as the degree of o minus 1, that is, $$\mathsf {Poisson}(1)$$.

In , we compared the formula (1.1) to numerical simulations in $$d = 32$$ on a finite box with $$L = 128$$, and there is excellent agreement with the asymptotics already for these values.

Other graphs where information on the height distribution is available are as follows. Dhar and Majumdar  studied the Abelian sandpile model on the Bethe lattice, and the exact expressions for various distribution functions including the height distribution at a vertex were obtained using combinatorial methods. For the single-site height distribution they obtained (see [7, Eqn. (8.2)])
\begin{aligned} p_{\mathrm {Bethe},d}(i) = \frac{1}{(d^2 - 1) \, d^d} \sum _{j = 0}^i \left( {\begin{array}{c}d+1\\ j\end{array}}\right) (d-1)^{d-j+1}. \end{aligned}
If one lets the degree $$d \rightarrow \infty$$ in this formula, one obtains the form in the right hand side of (1.1) for any fixed i (with 2d replaced by d).

Exact expressions for the distribution of height probabilities were derived by Papoyan and Shcherbakov  on the Husimi lattice of triangles with an arbitrary coordination number q. However, on d-dimensional cubic lattices of $$d\ge 2$$, exact results for the height probability are only known for $$d = 2$$; see [13, 14, 18, 21, 22].

### 1.1 Definitions and Background

Sandpiles are a lattice model of self-organized criticality, introduced by Bak, Tang and Wiesenfeld  and have been studied in both physics and mathematics. See the surveys [6, 9, 10, 15, 23]. Although the model can easily be defined on an arbitrary finite connected graph, in this paper we will restrict to subsets of $$\mathbb {Z}^d$$.

Let $$V_L = [-L,L]^d \cap \mathbb {Z}^d$$ be a box of radius L, where $$L \ge 1$$. For simplicity, we suppress the d-dependence in our notation. We let $$G_L = (V_L \cup \{s \},E_L)$$ denote the graph obtained from $$\mathbb {Z}^d$$ by identifying all vertices in $$\mathbb {Z}^d {\setminus } V_L$$ that becomes s, and removing loop-edges at s. We call s the sink. A sandpile$$\eta$$ is a collection of indistinguishable particles on $$V_L$$, specified by a map $$\eta : V_L \rightarrow \{ 0, 1, 2, \dots \}$$.

We say that $$\eta$$ is stable at $$x\in V_L$$, if $$\eta (x) < 2d$$. We say that $$\eta$$ is stable, if $$\eta (x) < 2d$$, for all $$x\in V_L$$. If $$\eta$$ is unstable (i.e. $$\eta (x) \ge 2d$$ for some $$x\in V_L$$), x is allowed to topple which means that x passes one particle along each edge to its neighbours. When the vertex x topples, the particles are re-distributed as follows:
\begin{aligned} \begin{aligned}&\eta (x) \rightarrow \eta (x) - 2d; \\&\eta (y) \rightarrow \eta (y) + 1, \quad y \in V_L, y \sim x. \end{aligned} \end{aligned}
Particles arriving at s are lost, so we do not keep track of them. Toppling a vertex may generate further unstable vertices. Given a sandpile $$\xi$$ on $$V_L$$, we define its stabilization
\begin{aligned} \xi ^\circ \in \Omega _L := \{\hbox {all stable sandpiles on}\; V_L\} = \{0,1,\dots ,2d-1\}^{V_L} \end{aligned}
by carrying out all possible topplings, in any order, until a stable sandpile is reached. It was shown by Dhar  that the map $$\xi \mapsto \xi ^\circ$$ is well-defined, that is, the order of topplings does not matter.

We now define the sandpile Markov chain. The state space is the set of stable sandpiles $$\Omega _L$$. Fix a positive probability distribution p on $$V_L$$, i.e. $$\sum _{x\in V_L} p(x) = 1$$ and $$p(x) > 0$$ for all $$x\in V_L$$. Given the current state $$\eta \in \Omega _L$$, choose a random vertex $$X \in V$$ according to p, add one particle at X and stabilize. The one-step transition of the Markov chain moves from $$\eta$$ to $$(\eta + \mathbf {1}_X)^{\circ }$$. Considering the sandpile Markov chain on $$G_L$$, there is only one recurrent class . We denote the set of recurrent sandpiles by $$\mathcal {R}_L$$. It is known  that the invariant distribution $$\mathbf {P}_{L}$$ of the Markov chain is uniformly distributed on $$\mathcal {R}_L$$.

Majumdar and Dhar  gave a bijection between $$\mathcal {R}_L$$ and spanning trees of $$G_L$$. This maps the uniform measure $$\mathbf {P}_L$$ on $$\mathcal {R}_L$$ to the uniform spanning tree measure $$\mathsf {UST}_L$$. A variant of this bijection was introduced by Priezzhev  and is described in more generality in [8, 12]. The latter bijection enjoys the following property that we will exploit in this paper. Orient the spanning tree towards s, and let $$\pi _L(x)$$ denote the oriented path from a vertex x to s. Let
\begin{aligned} W_L = \{ x \in V_L : o \in \pi _L(x) \}. \end{aligned}
Then, we have that
\begin{aligned}&\text {conditional on } \deg _{W_L}(o) = i, \text { the height } \eta (o)\text { is uniformly}\nonumber \\&\text {distributed over the values } i, i+1, \dots , 2d-1. \end{aligned}
(1.2)
This has the following consequence for the height probabilities. Let $$q^L(i) = \mathsf {UST}_L [ \deg _{W_L}(o) = i ]$$, $$i = 0, \dots , 2d-1$$. Then,
\begin{aligned} p^L(i) := \mathbf {P}_L [ \eta (o) = i ] = \sum _{j=0}^{i} \frac{q^L(j)}{2d - j}. \end{aligned}
The measures $$\mathbf {P}_L$$ have a weak limit $$\mathbf {P}= \lim _{L \rightarrow \infty } \mathbf {P}_L$$ , and hence, $$p(i) = \lim _{L \rightarrow \infty } p^L(i)$$ exist, $$i = 0, \dots , 2d-1$$. Although the $$q^L(i)$$ depend on the non-local variable $$W_L$$, one also has that $$q(i) = \lim _{L \rightarrow \infty } q^L(i)$$ exist, $$i = 0, \dots , 2d-1$$; see . In fact, q(i) is given by the following natural analogue of its finite volume definition. Consider the uniform spanning forest measure $$\mathsf {USF}$$ on $$\mathbb {Z}^d$$; defined as the weak limit of $$\mathsf {UST}_L$$; see [16, Chapter 10]. Let $$\pi (x)$$ denote the unique infinite self-avoiding path in the spanning forest starting at x, and let
\begin{aligned} W = \{ x \in \mathbb {Z}^d : o \in \pi (x) \}. \end{aligned}
Then, $$q(i) = \mathsf {USF}[ \deg _{W}(o) = i ]$$, $$i = 0, \dots , 2d-1$$.
Therefore, we have
\begin{aligned} p(i) := \mathbf {P}[ \eta (o) = i ] = \sum _{j=0}^{i} \frac{q(j)}{2d - j}. \end{aligned}
(1.3)

### 1.2 Wilson’s Method

Given a finite path $$\gamma = [s_0, s_1, \ldots , s_k]$$ in $$\mathbb {Z}^d$$, we erase loops from $$\gamma$$ chronologically, as they are created. We trace $$\gamma$$ until the first time t, if any, when $$s_t \in \{s_0, s_1, \ldots , s_{t-1}\}$$, i.e. there is a loop. We suppose $$s_t = s_i$$, for some $$i \in \{0,1,\ldots ,t-1\}$$ and remove the loop $$[s_i,s_{i+1},\ldots ,s_t=s_i]$$. Then, we continue tracing $$\gamma$$ and follow the same procedure to remove loops until there are no more loops to remove. This gives the loop-erasure $$\pi = LE(\gamma )$$ of $$\gamma$$, which is a self-avoiding path . If $$\gamma$$ is generated from a random walk process, the loop-erasure of $$\gamma$$ is called the loop-erased random walk (LERW).

When $$d \ge 3$$, the $$\mathsf {USF}$$ on $$\mathbb {Z}^d$$ can be sampled via Wilson’s method rooted at infinity [4, 16, Section 10], that is described as follows. Let $$s_1, s_2,\dots$$ be an arbitrary enumeration of the vertices and let $$\mathcal {T}_0$$ be the empty forest with no vertices. We start a simple random walk $$\gamma _n$$ at $$s_n$$ and $$\gamma _n$$ stops when $$\mathcal {T}_{n-1}$$ is hit, otherwise we let it run indefinitely. $$LE(\gamma _n)$$ is attached to $$\mathcal {T}_{n-1}$$, and the resulting forest is denoted by $$\mathcal {T}_n$$. We continue the same procedure until all the vertices are visited. The above gives a random sequence of forests $$\mathcal {T}_1 \subset \mathcal {T}_2 \subset \dots$$, where $$\mathcal {T}= \cup _{n} \mathcal {T}_n$$ is a spanning forest of $$\mathbb {Z}^d$$. The extension of Wilson’s theorem  to transient infinite graphs proved in  implies that $$\mathcal {T}$$ is distributed as the $$\mathsf {USF}$$.

## 2 Proof of the Main Theorem

Let $$(S_n^x)_{n\ge 0}$$ be a simple random walk started at x (independent between x’s on $$\mathbb {Z}^d$$) and let $$\pi (x)$$ be the path in the USF from x to infinity. We introduce the events:
\begin{aligned} \begin{aligned}&E_i = \Big \{ |\{ w \sim o : \pi (w)\text { passes through } o\}| = i \Big \}, \quad i = 0, \dots , 2d-1;\\&E_i(x_1,x_2,\dots ,x_i) = \Big \{ \{ w \sim o : \pi (w) \text { passes through } o\} = \{x_1, x_2,\dots ,x_i\} \Big \}. \end{aligned} \end{aligned}
Then, recall that
\begin{aligned} q_d(i) = \mathbf {P}[\deg _W(o) = i] = \mathbf {P}[E_i] = \sum _{\begin{array}{c} x_1,\dots ,x_i \sim o \\ \text {distinct} \end{array}} \mathbf {P}[E_i(x_1,\dots ,x_i)]. \end{aligned}
(2.1)

### Lemma 2.1

We have $$\mathbf {P}[S^o_n = o\text { for some } n \ge 2] = O(1/d)$$ and $$\mathbf {P}[S^o_n = o\text { for some } n \ge 4] = O(1/d^2)$$, as $$d \rightarrow \infty$$.

### Proof

Let $${\hat{D}}(k)$$ = $$\frac{1}{d} \sum _{j=1}^{d} \cos (k_j)$$, $$k \in [-\pi ,\pi ]^d$$ be the Fourier transform in d dimensions of the one-step distribution of RW. Lemma A.3 in  states that for all non-negative integers n and all $$d\ge 1$$, we have
\begin{aligned} \Vert {\hat{D}}^n\Vert _1 = (2\pi )^{-d}\int _{[-\pi ,\pi ]^d}|{\hat{D}}(k)^n|d^dk \le \left( \frac{\pi d}{4n}\right) ^{d/2}. \end{aligned}
Based on above, we have
\begin{aligned} \begin{aligned} \mathbf {P}[S^o_n = o\text { for some } n \ge 4]&\le \frac{1}{(2\pi )^d} \sum _{n=4}^{\infty } \int {\hat{D}}^n(k)dk \\&\le \frac{1}{(2\pi )^d} \sum _{n=4}^{d-1} \int {\hat{D}}^n(k)dk + \sum _{n=d}^{\infty } \Big (\frac{\pi d}{4n}\Big )^{d/2}. \end{aligned} \end{aligned}
(2.2)
Since $$\int {\hat{D}}^4(k) dk$$ and $$\int {\hat{D}}^6(k)dk$$ state the probability that $$S^o$$ returns to o in 4 and 6 steps each, by counting the number of ways to return, they are bounded by dimension-independent multiples of $$1/d^2$$ and $$1/d^3$$, respectively. We have $$\int {\hat{D}}^n(k) dk = 0$$ with odd n, and for $$6 < n \le d-1$$ and n even, we have $$\int {\hat{D}}^n(k)dk \le \int {\hat{D}}^6(k)dk$$. Hence,
\begin{aligned} \frac{1}{(2\pi )^d} \int {\hat{D}}^n(k)dk = O\Big (\frac{1}{d^3}\Big ), \quad 6 \le n \le d-1. \end{aligned}
The last sum in (2.2) can be bounded as:
\begin{aligned} \begin{aligned} \Big (\frac{\pi d}{4}\Big )^{d/2} \sum _{n=d}^{\infty } n^{-d/2}&\le \Big (\frac{\pi d}{4}\Big )^{d/2} \int _{d-1}^\infty x^{-d/2} dx = \Big (\frac{\pi d}{4}\Big )^{d/2} \frac{(d-1)^{1-\frac{d}{2}}}{d/2-1}\\&= \Big (\frac{d-1}{d/2-1}\Big )\Big (\frac{d}{d-1}\Big )^{\frac{d}{2}} \Big (\frac{\pi }{4}\Big )^{\frac{d}{2}} \le Ce^{-cd}, \end{aligned} \end{aligned}
since we can take $$d > 4$$ and $$\frac{\pi }{4} < 1$$.
Hence, we have the required results
\begin{aligned} \begin{aligned} \mathbf {P}[S^o_n = o\text { for some } n \ge 4]&\le \int {\hat{D}}^4(k) dk + d \int {\hat{D}}^6(k)dk + C e^{-cd}\\&= O\Big (\frac{1}{d^2}\Big ) + d\times O\Big (\frac{1}{d^3}\Big ) = O\Big (\frac{1}{d^2}\Big ),\\ \mathbf {P}[S^o_n = o\text { for some } n \ge 2]&\le \Big (\frac{1}{2d}\Big ) + \mathbf {P}[S^o_n = o\text { for some } n \ge 4] = O\Big (\frac{1}{d}\Big ). \end{aligned} \end{aligned}
$$\square$$

### 2.2 Lower Bounds

Let us fix the vertices $$x_1, \dots , x_i \sim o$$. Let
\begin{aligned} A_0 = \Big \{ S_1^o \not \in \{ x_1, \dots , x_i \},\, S_n^o \not \in \mathcal {N} \text { for } n \ge 2 \Big \}, \end{aligned}
where $$\mathcal {N} = \{ y \in \mathbb {Z}^d : |y| \le 1\}$$.

### Lemma 2.2

We have $$\mathbf {P}[A_0] \ge 1- O(i/d).$$

### Proof

\begin{aligned} \mathbf {P}[A_0] = \mathbf {P}[S_1^o \ne x_1,\dots ,x_i] \mathbf {P}[S_n^o \not \in \mathcal {N}\text { for } n \ge 2 | S_1^o \ne x_1,\dots ,x_i]. \end{aligned}
We have $$\mathbf {P}[S_1^o \ne x_1, \dots x_i] = 1-O(i/d)$$ and the probability for the remaining steps is at least $$1-O(1/d)$$, shown as follows. The probabilities $$\mathbf {P}[S_2^o \ne o | S_1^o \ne x_1,\dots ,x_i]$$ and $$\mathbf {P}[S_3^o \not \in \mathcal {N} | S_2^o \ne o, S_1^o \ne x_1,\dots ,x_i]$$ are both equal to $$1-O(1/d)$$. Considering the s.r.w starting at the position $$S_3^o$$, it hits at most three neighbours of o in two further steps, the remaining neighbours will need at least 4 steps to hit, so, by Lemma 2.1, we have
\begin{aligned} \begin{aligned} \sum _{\text {at most } 3\text { neighbours } x_j} \sum _{k\ge 1} P_{2k}(S_3^o,x_j)&\le O\left( \frac{1}{d}\right) , \\ \sum _{\text {the remaining neighbours } x_{j'}} \sum _{k\ge 2} P_{2k}(S_3^o,x_{j'})&\le O(d)O\left( \frac{1}{d^2}\right) = O\left( \frac{1}{d}\right) , \end{aligned} \end{aligned}
since $$P_{2k}(x,y) \le P_{2k}(o,o)$$ for all xy. Therefore, combining above results together, we get $$\mathbf {P}[S_n^o \not \in \mathcal {N}\text { for } n \ge 2 | S_1^o \ne x_1,\dots ,x_i] \ge 1 - O(1/d)$$ as required. $$\square$$

Let us label the neighbours of o different from $$x_1, \dots , x_i$$ as $$x_{i+1}, \dots , x_{2d}$$, in any order. On the event $$A_0$$, the first step of $$\pi (o)$$ is to a neighbour of o in $$\{x_{i+1},\dots ,x_{2d}\}$$ and we could assume $$x_{2d}$$ to be the first step of $$\pi (o)$$. Then, $$\pi (o)$$ does not visit other vertices in $$\mathcal {N} \backslash \{o\}$$. Define $$A_j =\{S_1^{x_j} = o\}$$ for $$j = 1,2,\dots ,i$$ and then $$\mathbf {P}[A_j] = 1/2d$$.

Using Wilson’s algorithm, consider random walks first started at $$o, x_1, .., x_i$$ and then started at $$x_{i+1},\dots ,x_{2d-1}$$. We obtain the following:
\begin{aligned} \begin{aligned} \mathbf {P}[E_i(x_1,\dots ,x_i)]&\ge \mathbf {P}[A_0] \times \prod _{j=1}^i \mathbf {P}[A_j] \times \mathbf {P}[E_i(x_1,..,x_i)|A_0 \cap A_1 \cap \dots \cap A_i] \\&\ge \Big (1-O\Big (\frac{i}{d}\Big )\Big )\Big (\frac{1}{2d}\Big )^i \mathbf {P}[E_i(x_1,..,x_i)|A_0 \cap A_1 \cap \dots \cap A_i]. \end{aligned}\nonumber \\ \end{aligned}
(2.3)
Define $$B_k = \{S_1^{x_k} \ne o, S_n^{x_k} \not \in \{x_1,\dots ,x_i\} \text { for } n\ge 2\}$$ for $$k = i+1,\dots ,2d-1$$.

### Lemma 2.3

$$\mathbf {P}[B_k] \ge 1 - 1/2d - O(i/d^2)$$, where $$i+1 \le k \le 2d-1$$.

### Proof

We have $$\mathbf {P}[S_1^{x_k} \ne o] = 1 - 1/2d$$. If the first step is not to o, the first step could be in one of the $$e_1,\dots ,e_i$$ directions, say $$e_j$$, with probability i / 2d. Then, the probability to hit $$x_j$$ is $$1/2d + O(1/d^2)$$. Hence, the probability that $$S^{x_k}$$ hits $$\{x_1,\dots ,x_i\}$$ is $$O(i/d^2)$$. $$\square$$

### Lemma 2.4

$$q_d(i) \ge e^{-1} \frac{1}{i!}\left( 1+O\left( \frac{i^2}{d}\right) \right) .$$

### Proof

By (2.3), we have
\begin{aligned} \mathbf {P}[E_i(x_1,\dots ,x_i)] \ge \Big (1- O\Big (\frac{i}{d}\Big )\Big )\Big (\frac{1}{2d}\Big )^i \Big (1-\frac{1}{2d} + O\Big (\frac{i}{d^2}\Big )\Big )^{2d-1-i}. \end{aligned}
Then, by (2.1),
\begin{aligned} \begin{aligned} q_d(i)&\ge {2d \atopwithdelims ()i}\Big (1- O\Big (\frac{i}{d}\Big )\Big )\Big (\frac{1}{2d}\Big )^i \Big (1-\frac{1}{2d}+O\Big (\frac{i}{d^2}\Big )\Big )^{2d-1-i}\\&=\frac{2d(2d-1)\dots (2d-i+1)}{i!(2d)^i}\Big (1-O\Big (\frac{i}{d}\Big )\Big )\\&\quad \Big (1-\frac{1}{2d}+O\Big (\frac{i}{d^2}\Big )\Big )^{2d} \Big (1+O\Big (\frac{i}{d}\Big )\Big ), \end{aligned} \end{aligned}
where
\begin{aligned} \begin{aligned} \Big (1 - \frac{1}{2d} + O\Big (\frac{i}{d^2}\Big )\Big )^{2d}&= \exp \Big (2d\times \log \Big (1 - \frac{1}{2d} + O\Big (\frac{i}{d^2}\Big )\Big )\\&= \exp \Big (2d\Big (- \frac{1}{2d} + O\Big (\frac{i}{d^2}\Big )\Big ) \\&= \exp \Big (-1 + O\Big (\frac{i}{d}\Big )\Big ) = e^{-1}\Big (1+O\Big (\frac{i}{d}\Big )\Big ), \end{aligned} \end{aligned}
and
\begin{aligned} \begin{aligned} \frac{2d(2d-1)\dots (2d-i+1)}{(2d)^i}&= 1\Big (1-\frac{1}{2d}\Big )\Big (1-\frac{2}{2d}\Big )\dots \Big (1-\frac{i}{2d}+\frac{1}{2d}\Big )\\&= \Big (1 + O\Big (\frac{i^2}{d}\Big )\Big ). \end{aligned} \end{aligned}
Then, the result follows
\begin{aligned} q_d(i)\ge & {} e^{-1} \frac{1}{i!}\Big (1+O\Big (\frac{i}{d}\Big )\Big ) \Big (1+O\Big (\frac{i^2}{d}\Big )\Big ) \Big (1+O\Big (\frac{i}{d}\Big )\Big ) \Big (1-O\Big (\frac{i}{d}\Big )\Big ) \\= & {} e^{-1} \frac{1}{i!}\Big (1+O\Big (\frac{i^2}{d}\Big )\Big ). \end{aligned}
$$\square$$

The above lemma gives a lower bound for $$q_d$$, and we now prove an upper bound.

### 2.3 Upper Bounds

Recall that $$\pi (o)$$ denotes the unique infinite self-avoiding path in the spanning forest starting at o and let $${\bar{A}}_o = \{\pi (o)\text { visits only one neighbour of } o\}$$.

### Lemma 2.5

$$\mathbf {P}[\pi (o) \text { visits more than one neighbour of } o] = P[{\bar{A}}_o^c] = O(1/d).$$

### Proof

The first step of $$\pi (o)$$ must visit a neighbour of o, denoted by w, then $$P[{\bar{A}}_o^c]$$
\begin{aligned} \begin{aligned}&= \mathbf {P}[\text {The second step of } \pi (o)\text { visits } x\ne 2w, \text { the third step visits } w' \sim o, w'\ne w] \\&\quad + O\Big (\frac{1}{d^2}\Big )\\&=\Big (\frac{1}{2d}\Big )\Big (\frac{2d-1}{2d}\Big ) + O\Big (\frac{1}{d^2}\Big ) = O\Big (\frac{1}{d}\Big ).\\ \end{aligned} \end{aligned}
$$\square$$

Let $${\bar{A}}_\mathrm{all} = \{ \forall w \sim o{:} \text { either } \pi (w) \text { does not visit } o \text { or } \pi (w) \text { visits } o~\mathrm{at~the~first~step} \}$$.

### Lemma 2.6

$$\mathbf {P}[\exists w\sim o: \pi (w)\text { visits } o \text { but not at the first step}] = \mathbf {P}[{\bar{A}}_\mathrm{all}^c] = O(1/d).$$

### Proof

For a given w, $$w \sim o$$, use Wilson’s algorithm with a walk started at w. Consider that if $$S_1^w \ne o$$, or $$S_1^w = o$$, but $$S^w$$ returns to w subsequently and then this loop starting from w in $$S^w$$ is erased, $$\pi (w)$$ does not visit o at the first step. Hence, we have the inequality:
\begin{aligned} \begin{aligned}&\mathbf {P}[\pi (w)\text { visits } o\text { but not at the first step}] \\&\le \mathbf {P}[S^w\text { visits } o \text { but not at the first step}] + \mathbf {P}[S_1^w = o, S_n^w = w \text { for some } n\ge 2]. \end{aligned}\nonumber \\ \end{aligned}
(2.4)
We bound the two terms as follows. For the first term, let us append a step from o to w at the beginning of the walk and analyse it as if the walk started at o. Since $$S_1^o \in \mathcal {N} \backslash \{o\}$$, by symmetry, we may assume $$S_1^o = w$$. Then, if $$S_2^o \ne o$$, $$S^o$$ will need at least 2 more steps to return to o.
For the second term in the right hand side of (2.4), we first note that we have $$\mathbf {P}[S_1^w = o, S_2^w = w] = 1/(2d)^2$$. If $$S^w$$ does not return to w in the first two steps, $$S^w$$ will need at least 4 steps to return to w. Then, we have that the right hand side of (2.4) is
\begin{aligned} \begin{aligned}&\le \mathbf {P}[S^o\text { returns to } o \text { in at least } 4 \text { steps}] + \frac{1}{(2d)^2} + \mathbf {P}[S^w\text { returns to } w \text { in at least } 4 \text { steps}]\\&= 2\times \mathbf {P}[S^o\text { returns to } o \text { in at least } 4 \text { steps}] + O\Big (\frac{1}{d^2}\Big ). \end{aligned} \end{aligned}
Therefore, by Lemma 2.1, we have the required result
\begin{aligned} \begin{aligned}&\mathbf {P}[\exists w\sim o: pi(w)\text { visits } o \text { but not at the first step }]\\&= 2d\times \mathbf {P}[\pi (w)\text { visits } o \text { but not at the first step for a fixed } w \sim o] =O\Big (\frac{1}{d}\Big ). \end{aligned} \end{aligned}
$$\square$$
Due to Lemmas 2.5 and 2.6, we have
\begin{aligned} q_d(i)\le & {} O\Big (\frac{1}{d}\Big ) + \mathbf {P}[{\bar{A}}_o \cap {\bar{A}}_\mathrm{all} \cap E_i] \\= & {} O\Big (\frac{1}{d}\Big ) + \sum _{\begin{array}{c} x_1,\dots ,x_i \sim o \\ \text {distinct} \end{array}} \mathbf {P}[{\bar{A}}_o \cap {\bar{A}}_\mathrm{all} \cap E_i(x_1,\dots ,x_i)]. \end{aligned}
Here,
\begin{aligned} \begin{aligned}&{\bar{A}}_o \cap {\bar{A}}_\mathrm{all} \cap E_i(x_1,\dots ,x_i) \\&\subset {\bar{A}}_o \cap {\bar{A}}_\mathrm{all} \cap \{\text {the first step of } \pi (x_j) \text { is to } o, j = 1,\dots ,i\} \cap F_i(x_1,\dots ,x_i), \end{aligned}\nonumber \\ \end{aligned}
(2.5)
where
\begin{aligned} F_i(x_1,\dots ,x_i) = \{\pi (x_j)\text { does not go through } o, j = i+1, \dots ,2d\}. \end{aligned}
The right hand side of (2.5) is contained in the event
\begin{aligned} {\bar{A}}_o \cap \{\pi (o) \text { does not visit } x_1,\dots ,x_i\} \cap {\bar{A}}_\mathrm{rest} \cap \bigcap _{1\le j \le i} H_j \cap F_i(x_1, \dots , x_i), \end{aligned}
where
\begin{aligned} {\bar{A}}_\mathrm{rest}= & {} \{\pi (x_j)\text { goes through at most one }\\ x_{j'}, j= & {} i+1,\dots ,2d, i+1 \le j' \le 2d, j' \ne j \} \end{aligned}
and $$H_j = \{\text {the first step of } \pi (x_j) \text { is to } o\}$$ for $$j = 1,\dots ,i$$.
We denote $${\bar{A}}_o\cap \{\pi (o)\text { does not visit } x_1,\dots ,x_i\}$$ by $${\bar{A}}_{o,x_1,\dots ,x_i}$$. Then,
\begin{aligned} \begin{aligned}&\mathbf {P}\Big [{\bar{A}}_{o,x_1,\dots ,x_i} \cap {\bar{A}}_\mathrm{rest} \cap \bigcap _{1\le j \le i} H_j \cap F_i(x_1, \dots , x_i)\Big ] \\&\quad =\mathbf {P}[{\bar{A}}_{o,x_1,\dots ,x_i}] \prod _{j=1}^i\mathbf {P}\Big [H_j\Big |\bigcap _{1\le j' < j} H_{j'} \cap {\bar{A}}_{o,x_1,\dots ,x_i}\Big ] \\&\qquad \times \mathbf {P}\Big [F_i(x_1,\dots ,x_i)\cap {\bar{A}}_\mathrm{rest} \Big |{\bar{A}}_{o,x_1,\dots ,x_i}\cap \bigcap _{1\le j \le i}H_j\Big ]. \end{aligned} \end{aligned}
Therefore, we have
\begin{aligned} \begin{aligned} q_d(i)&\le O\Big (\frac{1}{d}\Big ) +\sum _{\begin{array}{c} x_1,\dots ,x_i\sim o \\ \mathrm {distinct} \end{array}} \Big (\prod _{j=1}^i\mathbf {P}\Big [H_j \Big |\bigcap _{1\le j' < j}H_{j'}\cap {\bar{A}}_{o,x_1,\dots ,x_i}\Big ]\Big ) \\&\quad \times \mathbf {P}\Big [F_i(x_1,\dots ,x_i)\cap {\bar{A}}_\mathrm{rest} \Big |{\bar{A}}_{o,x_1,\dots ,x_i}\cap \bigcap _{1\le j \le i}H_j\Big ]. \end{aligned} \end{aligned}
(2.6)

### Lemma 2.7

$$\mathbf {P}[H_j | {\bar{A}}_{o,x_1,\dots ,x_i} \cap \bigcap _{1\le j' < j}H_{j'}] = 1/2d + O(1/d^2)$$, where $$j = 1,\dots ,i$$.

### Proof

Given that $$\pi (o)$$ visits only one neighbour of o which is not in $$\{x_1,\dots ,x_i\}$$ and the first steps of $$\pi (x_1),\dots ,\pi (x_{j-1})$$ are all to o, the probability that $$H_j$$ happens is $$\mathbf {P}[S_1^{x_j} = o] = 1/2d$$ with the error term of $$O(1/d^2)$$ due to the loop-erasure. $$\square$$

### Lemma 2.8

\begin{aligned}&\mathbf {P}\Big [F_i(x_1,\dots ,x_i)\cap {\bar{A}}_\mathrm{rest} \Big |{\bar{A}}_{o,x_1,\dots ,x_i}\cap \bigcap _{1\le j \le i}H_j\Big ]\nonumber \\&\quad \le \mathbf {E}\Big [\Big (1-\frac{1}{2d}+O\Big (\frac{1}{d^2}\Big )\Big )^{2d-i-1-N} \mathbf {1}_{{\bar{A}}_\mathrm{rest}}\Big ], \end{aligned}
(2.7)
where $$N=|\{i+1\le j\le 2d-1:\exists i+1\le j'<j$$ s.t. $$\pi (x_{j'})$$ goes through $$x_j\}|$$.

### Proof

Consider Wilson’s algorithm with random walks started at the remaining neighbours $$x_{i+1}, \dots ,x_{2d}$$. Assume $$x_{2d}$$ to be the neighbour of o that $$\pi (o)$$ goes through. The probability that $$\pi (x_k)$$ does not go through o is $$1 - 1/2d + O(1/d^2)$$ for $$k\in \{i+1,\dots ,2d-1\}$$.

If $$\pi (x_k)$$ visits $$x_{k'}$$, where $$k < k' \le 2d-1$$, the probability that $$\pi (x_{k'})$$ does not go through o is 1 instead of $$1-1/2d+O(1/d^2)$$, since the LERW from $$x_{k'}$$ stops immediately and $$\pi (x_{k'}) \subset \pi (x_k)$$, which does not go through o. $$\square$$

### Lemma 2.9

On the event $${\bar{A}}_\mathrm{rest}$$, $$N \le B$$, where $$B \sim \mathsf {Binom}(2d-i-1, p)$$, $$p = 1/2d + O(1/d^2)$$.

### Proof

Since we have $$(2d-i-1)$$ trials with probability at most $$1/2d+O(1/d^2)$$. $$\square$$

Due to Lemma 2.9, we have that the right hand side of (2.7) is
\begin{aligned} \le \Big (1-\frac{1}{2d}+O\Big (\frac{1}{d^2}\Big )\Big )^{2d} \Big (1+O\Big (\frac{i}{d}\Big )\Big ) \mathbf {E}\Big [\frac{1}{(1-\frac{1}{2d}+O(\frac{1}{d^2}))^B}\Big ], \end{aligned}
(2.8)
where $$\mathbf {E}[z^B] = \sum _{j=0}^{2d-i-1} z^j {2d-i-1 \atopwithdelims ()j}p^j(1-p)^{2d-i-1-j} = (1-p-zp)^{2d-i-1}$$.
Hence, (2.8) is
\begin{aligned} \begin{aligned}&\le e^{-1}\Big (1+O\Big (\frac{1}{d}\Big )\Big )\Big (1+O\Big (\frac{i}{d}\Big )\Big ) \Big (1-\frac{1}{2d} + O\Big (\frac{1}{d^2}\Big ) + \frac{\frac{1}{2d} + O(\frac{1}{d^2})}{1-\frac{1}{2d} + O(\frac{1}{d^2})}\Big )^{2d-i-1}\\&= e^{-1}\Big (1+O\Big (\frac{1}{d}\Big )\Big ) \Big (1+O\Big (\frac{i}{d}\Big )\Big )\Big (1+O\Big (\frac{1}{d^2}\Big )\Big )^{2d-i-1} = e^{-1}\Big (1+O\Big (\frac{i}{d}\Big )\Big ). \end{aligned} \end{aligned}
(2.9)

### Lemma 2.10

$$q_d(i) \le O(\frac{1}{d}) + e^{-1}\frac{1}{i!}(1+O(\frac{i}{d})).$$

### Proof

Due to Lemma 2.7, (2.6) and (2.9), we have
\begin{aligned} \begin{aligned} q_d(i)&\le O\Big (\frac{1}{d}\Big ) + {2d\atopwithdelims ()i}\Big (\frac{1}{2d}+O\Big (\frac{1}{d^2}\Big )\Big )^i e^{-1}\Big (1+O\Big (\frac{i}{d}\Big )\Big )\\&= O\Big (\frac{1}{d}\Big ) + e^{-1}\frac{2d(2d-1)\dots (2d-i+1)}{i!} \Big (\frac{1}{2d}\Big )^i\Big (1+O\Big (\frac{1}{d}\Big )\Big )^i\\&\quad \Big (1+O\Big (\frac{i}{d}\Big )\Big )\\&\le O\Big (\frac{1}{d}\Big ) + e^{-1}\frac{1}{i!}\Big (1+O\Big (\frac{i}{d}\Big )\Big ). \end{aligned} \end{aligned}
$$\square$$

### Lemma 2.11

For $$k = 1, \dots , 3$$ and distinct $$w_1, \dots , w_k \sim o$$, we have
\begin{aligned} \mathbf {P}[ \pi (w_i)\text { passes through } o \text { for } i = 1,\dots ,k ] = \left( \frac{1}{2d} \right) ^k + O (d^{-k-1}). \end{aligned}

This lemma can be proved using ideas used to prove Lemma 2.7.

### Proof of Theorem 1.1

We first prove part (i). By Wilson’s algorithm,
\begin{aligned} p_d(i) = \sum _{j=0}^{i} \frac{q_d(j)}{2d-j}. \end{aligned}
Due to Lemmas 2.4 and 2.10 , we have
\begin{aligned} p_d(i) \ge \sum _{j=0}^{i} \frac{e^{-1}\frac{1}{j!}\left( 1 + O\left( \frac{j^2}{d}\right) \right) }{2d-j} =\sum _{j=0}^{i}\frac{e^{-1}\frac{1}{j!}}{2d-j} +\sum _{j=0}^{i} \frac{\frac{1}{j!} O\left( \frac{j^2}{d}\right) }{2d-j}, \end{aligned}
(2.10)
and
\begin{aligned} p_d(i)\le & {} \sum _{j=0}^{i}\frac{O(\frac{1}{d}) + e^{-1}\frac{1}{j!}\left( 1+O\left( \frac{j}{d}\right) \right) }{2d-j} =\sum _{j=0}^{i}\frac{e^{-1}\frac{1}{j!}}{2d-j}\nonumber \\&+\sum _{j=0}^{i}\frac{O\left( \frac{1}{d}\right) +\frac{1}{j!}O\left( \frac{j}{d}\right) }{2d-j}. \end{aligned}
(2.11)
Here, using that $$0\le j\le d^{1/2}$$, we have
\begin{aligned} \sum _{j=0}^{i} \frac{\frac{1}{j!} O\left( \frac{j^2}{d}\right) }{2d-j} \le \frac{1}{2d-d^{1/2}}O\left( \frac{1}{d}\right) \sum _{j=0}^{i}\frac{j^2}{j!} =O\left( \frac{1}{d^2}\right) . \end{aligned}
Similarly,
\begin{aligned} \begin{aligned} \sum _{j=0}^{i} \frac{O\left( \frac{1}{d}\right) + \frac{1}{j!} O\left( \frac{j}{d}\right) }{2d-j} \le \sum _{j=0}^i O(d^{-2}) + \sum _{j=0}^i \frac{j}{j!} O(d^{-2}) = O( i/d^2 ). \end{aligned} \end{aligned}
Putting these error bounds together with (2.10) and (2.11), we prove statement (i) of the theorem.
Let us now use that
\begin{aligned} \frac{1}{2d}e^{-1}\sum _{j=0}^{i} \frac{1}{j!} \le \sum _{j=0}^{i} \frac{e^{-1} \frac{1}{j!}}{2d-j} \le \frac{1}{2d-i}e^{-1}\sum _{j=0}^{i}\frac{1}{j!}. \end{aligned}
When $$i \le d^{1/2}$$, and $$i, d \rightarrow \infty$$, we have $$\frac{1}{2d-i} \sim \frac{1}{2d}$$ and $$\sum _{j=0}^{i}\frac{1}{j!} \rightarrow e$$. Hence,
\begin{aligned} \sum _{j=0}^{i} \frac{e^{-1} \frac{1}{j!}}{2d-j} \sim \frac{1}{2d}, \quad \text {as } i, d \rightarrow \infty . \end{aligned}
We are left to prove statement (ii). The uniform distribution for $$d^{1/2} \le i \le 2d-1$$ can be obtained from the monotonicity:
\begin{aligned} p_d(d^{1/2}) \le p_d(i) \le p_d(2d-1), \quad d^{1/2} \le i \le 2d-1, \end{aligned}
if we show that $$p_d(2d-1) = p_d(d^{1/2}) + O(d^{-3/2})$$.
We write
\begin{aligned} p_d(2d-1)= & {} \sum _{j=0}^{2d-1} \frac{q_d(j)}{2d-j} = p_d(d^{1/2}) + \sum _{j = d^{1/2}}^{2d-1} \frac{q_d(j)}{2d-j} \le p_d(d^{1/2}) \nonumber \\&+ \sum _{j = d^{1/2}}^{2d-1} q_d(j). \end{aligned}
(2.12)
Introducing the random variable
\begin{aligned} X:= |\left\{ w \sim o : o \in \pi (w) \right\} |, \end{aligned}
the last expression in (2.12) equals
\begin{aligned} p_d(d^{1/2}) + \mathbf {P}[ X \ge d^{1/2} ] \le p_d(d^{1/2}) + \mathbf {P}[ X^3 \ge d^{3/2} ] \le p_d(d^{1/2}) + \frac{\mathbf {E}[ X^3 ]}{d^{3/2}}. \end{aligned}
Therefore, it remains to show that $$\mathbf {E}[ X^3 ] = O(1)$$. This follows from Lemma 2.11, by summing over $$w_1, \dots , w_3$$ (not necessarily distinct). The cases $$k=1,2$$ of the lemma are used to sum the contributions where one or more of the $$w_i$$’s coincide. $$\square$$

## References

1. 1.
Aldous, D.J.: The random walk construction of uniform spanning trees and uniform labelled trees. SIAM J. Discrete Math. 3(4), 450–465 (1990)
2. 2.
Athreya, S.R., Járai, A.A.: Infinite volume limit for the stationary distribution of abelian sandpile models. Commun. Math. Phys. 249(1), 197–213 (2004)
3. 3.
Bak, P., Tang, C., Wiesenfeld, K.: Self-organized criticality: an explanation of the 1/ f noise. Phys. Rev. Lett. 59(4), 381–384 (1987)
4. 4.
Benjamini, I., Lyons, R., Peres, Y., Schramm, O.: Uniform spanning forests. Ann. Probab. 29(1), 1–65 (2001)
5. 5.
Dhar, D.: Self-organized critical state of sandpile automaton models. Phys. Rev. Lett. 64(14), 1613–1616 (1990)
6. 6.
Dhar, D.: Theoretical studies of self-organized criticality. Physica A 369(1), 29–70 (2006)
7. 7.
Dhar, D., Majumdar, S.N.: Abelian sandpile model on the Bethe lattice. J. Phys. A: Math. Gen. 23(19), 4333–4350 (1990)
8. 8.
Gamlin, S.L., Járai, A.A.: Anchored burning bijections on finite and infinite graphs. Electron. J. Probab. 19(117), 1–23 (2014)
9. 9.
Holroyd, A.E., Levine, L., Mészáros, K., Peres, Y., Propp, J., Wilson, D.B.: Chip-firing and rotor-routing on directed graphs. In: Sidoravicius, V., Vares, M.E. (eds.) In and Out of Equilibrium 2, pp. 331–364. Birkhäuser, Basel (2008)
10. 10.
Járai, A.A.: Sandpile models. Probab. Surv. 15, 243–306 (2018)
11. 11.
Járai, A.A., Sun, M.: Toppling and height probabilities in sandpiles. J. Stat. Mech. Theory Exp. 2019(11), 113204 (2019).
12. 12.
Járai, A.A., Werning, N.: Minimal configurations and sandpile measures. J. Theor. Probab. 27(1), 153–167 (2014)
13. 13.
Jeng, M., Piroux, G., Ruelle, P.: Height variables in the abelian sandpile model: scaling fields and correlations. J. Stat. Mech. Theory Exp. 2006(10), P10015 (2006)
14. 14.
Kenyon, R.W., Wilson, D.B.: Spanning trees of graphs on surfaces and the intensity of loop-erased random walk on planar graphs. J. Am. Math. Soc. 28(4), 985–1030 (2015)
15. 15.
Levine, L., Peres, Y.: Laplacian growth, sandpiles, and scaling limits. Bull. Am. Math. Soc. 54(3), 355–382 (2017)
16. 16.
Lyons, R., Peres, Y.: Probability on Trees and Networks. Cambridge University Press, Cambridge (2016)
17. 17.
18. 18.
Majumdar, S.N., Dhar, D.: Height correlations in the abelian sandpile model. J. Phys. A: Math. Gen. 24(7), L357–L362 (1991)
19. 19.
Majumdar, S.N., Dhar, D.: Equivalence between the abelian sandpile model and the $$q \rightarrow 0$$ limit of the potts model. Physica A 185(1), 129–145 (1992)
20. 20.
Papoyan, V.V., Shcherbakov, R.R.: Distribution of heights in the abelian sandpile model on the Husimi lattice. Fractals 4(1), 105–110 (1996)
21. 21.
Poghosyan, V.S., Priezzhev, V.B., Ruelle, P.: Return probability for the loop-erased random walk and mean height in the abelian sandpile model: a proof. J. Stat. Mech. Theory Exp. 2011(10), P10004 (2011)
22. 22.
Priezzhev, V.B.: Structure of two-dimensional sandpile. I. Height probabilities. J. Stat. Phys. 74(5), 955–979 (1994)
23. 23.
Redig, F.: Mathematical aspects of the abelian sandpile model. In: Bovier, A. (ed.) Mathematical Statistical Physics: Lecture Notes of the Les Houches Summer School 2005, pp. 657–729. Elsevier, Amsterdam (2006)
24. 24.
Wilson, D.B.: Generating random spanning trees more quickly than the cover time. In: Proceedings of the Twenty-Eighth Annual ACM Symposium on Theory of Computing, STOC ’96. ACM, New York, pp. 296–303 (1996)Google Scholar