Active Phase for the Stochastic Sandpile on $\mathbb{Z}$

We start by defining the ‘site-wise’ representation for SSM. We identify configurations of particles as functions $\eta:\operatorname{\mathbb{Z}}\to\operatorname{\mathbb{N}}$, i.e. $\eta(x)$ counts the number of particles at site $x$. To run the dynamics on a finite interval $I\subset\operatorname{\mathbb{Z}}$, every site $x\in I$ is assigned an infinite sequence of iid instructions, $(\xi^{x}_{j}:j\in\operatorname{\mathbb{N}})$ each taking one of the two possible values $\xi_{-,x}$ or $\xi_{+,x}$ with equal probability, where $\xi_{-,x}$ and $\xi_{+,x}$ are operators on the space of particle configurations that act via

and

In the usual discrete-time setup, the system evolves one step by choosing a site $x\in I$ with $\eta(x)\geqslant 2$, and applying the two stack instructions $\xi^{x}_{j},\xi^{x}_{j+1}$ where $j$ is such that all instructions $\xi^{x}_{j^{\prime}}$ for $j^{\prime}<j$ have already been used, but $\xi^{x}_{j}$ has not. In this version, particles always topple in pairs. For an initial particle configuration $\eta$ and a sequence of sites $\alpha=(\alpha_{1},\alpha_{2},\ldots,\alpha_{k}),\alpha_{i}\in I$ to be toppled, let

be the operator obtained by performing all topplings of $\alpha$ in order. The odometer of the pair $(\eta,\alpha)$ is the function which records the total number of times each site was toppled, i.e.

Such a sequence $\alpha$ is called legal for $\eta$ if $\xi_{(\alpha_{1},\ldots,\alpha_{\ell-1})}\eta(\alpha_{\ell})\geqslant 2$ for every $\ell\in\{1,\ldots,k\}.$

The odometer function $m_{I}$ is universal in the sense that if $\alpha,\alpha^{\prime}$ are any two legal toppling sequences for a configuration $\eta$ on interval $I$ such that $\xi_{\alpha}\eta$ and $\xi_{\alpha^{\prime}}\eta$ have no sites with at least two particles, then $\alpha$ is a permutation of $\alpha^{\prime}$ and thus $m_{I}(\eta,\alpha,\cdot)=m_{I}(\eta,\alpha^{\prime},\cdot)$. For our argument, it is only necessary that the half-toppling version of the dynamics gives a lower bound on this odometer; see Lemma 2.2.

We need the following result from [16] which relates the site-wise representation to the continuous-time process $(\eta_{t})_{t\geqslant 0}$ in Section 1. Define

where the supremum is over all finite intervals $I$ and all legal toppling sequences $\alpha$ for the configuration $\eta$ on $I$.

The above lemma says that in order to establish the system staying active a.s., it suffices to show that some site is toppled infinitely many times with positive probability.

To allow more freedom in our toppling procedure (and retain some independence between particles), we work with an equivalent version of the dynamics. Instead of toppling two particles at a time, we topple single particles, i.e. perform  half-topplings. In the usual dynamics, the number of particles toppled at a given site is always even, so there are restrictions on which half-topplings are allowed. Namely, we must keep track of how many times each site has been half-toppled, and only allow another half-toppling if that number is odd, or if that site has at least two particles.

Thus, in the half-toppling scheme, a configuration of particles consists of two functions, $\eta:\mathbb{Z}\to\mathbb{N}$, the number of particles per site, and a parity sequence $\omega:\mathbb{Z}\to\{0,1\}$. A site $x$ is legal for a pair of configurations $(\eta,\omega)$ if either $\eta(x)\geqslant 2$ or $\eta(x)=\omega(x)=1$. At each step of the half-toppling scheme, a legal site $x$ is chosen for the current pair of configurations, and the first unused instruction $\xi_{j}^{x}$ acts upon the pair by sending a single particle at $x$ to a uniform random neighbor and adding 1 (mod 2) to $\omega(x)$. A half-toppling sequence of sites $\alpha=(\alpha_{1},\alpha_{2},\ldots,\alpha_{k})$ is legal for $(\eta,\omega)$ if the chosen site is legal for every step, that is, $\alpha_{\ell}$ is legal for

for every $\ell\in\{1,\ldots,k\}.$

One key fact about this version of the dynamics is that it provides a lower bound on the total odometer. Recall the odometer function $m_{I}$ for the classical dynamics, and for any sequence $\alpha$ of half-topplings, let $m_{I}(\eta,\omega,\alpha,x)$ denote the corresponding half-toppling odometer function, after performing all half-topplings in $\alpha$ started from the configuration $(\eta,\omega)$. We have:

We omit the proof, which follows identically to that of [16, Lemma 6].

Our strategy is to alternate between two types of legal half-toppling sequences, taking place on a sequence of nested intervals in $\mathbb{Z}$.

The first of these is called the ‘carpet/hole toppling procedure’ and is essentially a renormalization scheme of the SSM, detailed in Section 3. The procedure starts from and remains in a very particular set of configurations, and runs until it partially stabilizes the interval – when there are no more legal topplings among a small class of marked (‘free’) particles. We will show that each iteration is well behaved – that many particles reach the boundary – with exponentially high probability. The main technical hurdle in analyzing the carpet/hole procedure is the single block estimate Lemma 4.6, which we deal with in Sections 6, 7 and 8. In Section 4 we use the single block estimate to prove a version of the above statement, via an energy-entropy calculation similar to [9].

The carpet/hole procedure needs to start with relatively nice configurations. Thus when partial stabilization is achieved on an interval, before restarting the carpet/hole procedure on the larger interval, it is necessary to perform the other ‘bootstrap’ toppling procedure to restore the configuration. In Section 5 we analyze the alternation of both procedures and prove Theorem 1.1.

In the remaining sections, we focus on the carpet/hole dynamics inside a single block. In Section 6, to exploit the independence among random walks, we introduce an auxiliary process that only reveals partial information about the parity configuration. In Section 7, we turn to the intricate correlation between particles in our toppling procedure, by carefully controlling the typical number of zeros in the auxiliary process. Finally, in Section 8 we combine these results and prove Lemma 4.6.

Fix integers $a>0$ and $K=a^{4}$. The procedure described in this section is well-defined for any positive integer $a$, although later in Proposition 3.4 and Lemma 4.6 we will need to choose a large enough $a$. The toppling procedure needs to start from relatively nice particle configurations on a finite interval, which we call valid configurations.

The subinterval $[iK,iK+a]$ is called the $i$th block for $i=0,1,\ldots,n-1$, and the complement of the blocks are called transit regions. In other words, a valid configuration has single particles everywhere, except for at most one empty spot per block and possibly some additional particles at the boundary of every block.

Fix a valid configuration $(\eta,\omega)$. In order to perform the carpet/hole toppling procedure starting from $(\eta,\omega)$, we need to identify the special site hole in each block based on $(\eta,\omega)$, as well as classify the particles of $\eta$ into several categories. Note that the definitions given in the current Section 3.1 only apply to the starting configuration $(\eta,\omega)$. In Section 3.2, we will provide rules regarding how the holes move and particles switch types during the procedure.

Every block has exactly one hole, uniquely determined by the pair $(\eta,\omega)$. For each $i=0,\ldots,n-1$, if block $i$ has a position with no particle, then the empty site is the hole of block $i$. If block $i$ has a particle in every position, the leftmost site $x$ in that block with $\omega(x)=1$ is declared the hole. If there is no such $x$, then declare $iK+a$ to be the position of the hole.

With the definition of the holes, we allocate the particles in $\eta$ into several types. By definition, each site in $D_{n}$ (containing both blocks and transit regions) that does not have a hole will have at least one particle. We declare one particle at every such site to be a carpet particle. (If there are multiple particles at a site, we simply pick an arbitrary one and declare it a carpet particle.) All the remaining particles in $D_{n}$ are free particles.

In the configuration $\eta$, free particles are either in a hole or at an endpoint of a block. Free particles are further divided into two groups. For each $i=0,\ldots,n-1$, if block $i$ has $\eta(x)\geqslant 1$ and $\omega(x)=0$ for all $x\in[iK,iK+a]$, then the hole was defined to be at $iK+a$ and we declare an arbitrary (free) particle at $iK+a$ to be a frozen free particle. All the other free particles are thawed. There is always one special thawed particle, the hot particle, that actually gets toppled. We will explain the rule of determining the hot particle in Section 3.2.

We now formally define the carpet/hole toppling procedure inside a given interval $D_{n}=(-K+a,nK)$. Fix a valid configuration $(\eta,\omega)$. Suppose we’ve picked the starting locations of the holes and labelled the particles as ‘carpet’, ‘free’, ‘frozen’ and ‘thawed’ according to Section 3.1. The toppling procedure works as follows:

1. (L1)

   We follow a leftmost priority policy for choosing the hot particle, which implies Lemma 4.4 below. Find the leftmost block $i$ that contains a thawed particle. If no such block exists, the procedure ends and we say the procedure reaches partial stabilization. Among the thawed particles in block $i$, we choose the one inside the hole to be the hot particle if there is one. Otherwise, the thawed particles are at the boundary $iK$ or $iK+a$, and we declare an arbitrary thawed particle the hot one. This particular order of choice is to ensure 2 below. Once we choose the hot particle, we proceed with 2 or 3 depending on whether block $i$ contains a frozen particle.

2. (L2)

   Suppose block $i$ does not contain a frozen particle. If the hot particle is at the hole but the hole is not the leftmost position $y$ with parity value $\omega(y)=1$, then go directly to 2a or 2b. Otherwise, repeatedly topple the hot particle until it returns to the hole of block $i$, or hits either $(i-1)K+a$ or $(i+1)K$, which could be a boundary point of a neighboring block or a boundary point of the interval $D_{n}$. There are three cases:

   1. (L2a)

      Suppose the hot particle returned to the hole of block $i$ and there exists some position in the block $i$ with odd parity. Find the leftmost position $y$ in block $i$ with parity value $\omega(y)=1$. Declare the hot particle at the hole a carpet particle, move the hole in block $i$ to position $y$, and declare the carpet particle at $y$ a thawed and hot particle. Return to 2.

   2. (L2b)

      Suppose the hot particle returned to the hole of block $i$, but there is no site in block $i$ with odd parity. Turn the hot particle at the hole into a carpet particle, move the hole to position $iK+a$, and declare the carpet particle at $iK+a$ a frozen free particle. Return to 1.

   3. (L2c)

      If the hot particle reached $(i-1)K+a$ or $(i+1)K$, declare it no longer hot but still free and thawed. Return to 1.

3. (L3)

   If block $i$ does contain a frozen particle, then every site in block $i$ has a particle which is not the hot particle. Repeatedly topple the hot particle until it reaches $(i-1)K+a$ or $(i+1)K$. Then declare it no longer hot but still free and thawed. If block $i$ now has the leftmost site $y$ with parity value $\omega(y)=1$, then turn the frozen free particle at $iK+a$ into a carpet particle, move the hole to $y$, and declare the carpet particle at $x$ free and thawed. Return to 1.

The block size $a$ is large, so the typical occurrence is 2a: a hot particle makes an excursion inside its block. Our aim is to show that step 2b occurs much less often than 2c.

More broadly, what we will show is that the carpet/hole toppling procedure, as a renormalization scheme of the Stochastic Sandpile Model, ‘converges’ to the Activated Random Walk model (ARW) with a small sleep rate when the block size goes to infinity. Each block and every free particle in our procedure correspond to a single site and a normal particle in the ARW model respectively. Step 2c mimics an ARW particle moving to a neighboring site, whereas the rare 2b step represents an ARW particle becoming sleepy with a small sleep rate. If 2b does happen and a free particle becomes frozen, it can become thawed through 3, just as a sleepy ARW particle gets reactivated by the presence of another particle.

Once we make the ‘convergence’ rigorous in certain sense, the analysis of a 1D ARW-like system with a small sleep rate yields to a classical energy-entropy argument [11]. The overall outcome of these is Proposition 3.4, stated at the end of Section 3.

The following properties hold trivially for the initial configuration $(\eta,\omega)$ and are preserved by the half-toppling sequence, which can be checked by induction.

1. (P1)

   Each block $i$ has exactly one hole which is located at some site $x\in[iK,iK+a]$.

2. (P2)

   Whenever the hot particle is inside the hole of a block without a frozen particle, we always make sure this hole is located at the leftmost site $y$ in that block with parity value $\omega(y)=1$. If there is no site in block $i$ with odd parity, then we make sure the hole is at $iK+a$ and contains a frozen particle.

3. (P3)

   There is exactly one carpet particle at every site in $D_{n}$ except for the holes.

4. (P4)

   All free particles except the hot particle are in a hole, or at a site $iK$ or $iK+a$ for some $i$.

5. (P5)

   We call a block which contains a frozen particle a frozen block. In a frozen block, both the frozen particle and the hole are at $iK+a$. Thus every site in this block has a particle that is not hot.

We also collect some facts that will be useful throughout the paper.

1. (F1)

   All half-topplings performed during this procedure are legal.

2. (F2)

   After the first free particle has been designated hot in block $i$, all free particles in block $i$ except the hot particle (if there is one) are at $iK$ or $iK+a$.

3. (F3)

   A free particle designated hot in a block with a frozen particle reaches a neighboring block before being declared not hot.

4. (F4)

   The number of free particles is conserved during the procedure.

5. (F5)

   When the procedure ends, all free particles are either frozen, with at most one frozen particle per block, or at the boundary points of $D_{n}$. Also, there is at most one particle (either carpet or frozen free particle) at every site inside $D_{n}$.

We prove the following regarding the carpet/hole toppling procedure. Roughly speaking, it says that the dynamics is likely to sustain ‘enough activity’ so that at the end of the procedure, there are few frozen particles left inside every subinterval.

Let $(\eta,\omega)$ be a valid configuration on $D_{n}=(-K+a,nK)$. Run carpet/hole dynamics with initial condition $(\eta,\omega)$ until partial stabilization when there are no more thawed particles inside $D_{n}$. Denote its law by $\mathbb{P}_{(\eta,\omega)}$. For $0\leqslant n_{0}\leqslant n_{1}\leqslant n-1$, define ${\text{Frozen}}(n_{0},n_{1})$ to be the number of frozen free particles remaining in the blocks $n_{0},n_{0}+1,\dots,n_{1}$ at the end of the carpet/hole dynamics on $D_{n}$. Also let $m_{n}(\eta,\omega,x)$ be the (half-toppling) odometer function at site $x\in D_{n}$ resulting from the carpet/hole toppling procedure on $D_{n}$, and define

for some constant $\beta>0$. The purpose of the event $H(n_{0},n_{1})$ is to ensure each block starting from a nice configuration and to facilitate a bootstrap argument later in the proof of Lemma 5.9. Write $H:=H(0,n-1)$ and ${\text{Frozen}}:={\text{Frozen}}(0,n-1)$.

Proposition 3.4 will be proved in the next section. We will rely on Proposition 3.4 to do the analysis in Section 5.

Whenever a free particle is designated as the hot particle in step 1, we designate the block $i$ it is in as the hot block. For each $y\in D_{n}$ in a transit region, we split the stack instructions for $y$ into two independent stacks of independent instructions, $\xi^{y}=(\xi^{y,L},\xi^{y,R})$. The hot particle toppled at site $y$ uses instructions from the $L$ stack if the nearest block to the left of $y$ is the hot block, and from the $R$ stack if the nearest block to the right of $y$ is the hot block. Sites $y$ in a block have just a single instruction stack, $\xi^{y}$. We work with the filtration $\mathcal{F}_{i}$ of all the stack instructions decorated by blocks $j\leqslant i$, i.e.

To index possible ‘trajectories’ of the carpet/hole dynamics, we count the flow of particles between blocks. For each $i\in\{0,\ldots,n-1\}$ and $s\geqslant 0$ let

be the configuration $\eta$ with $s$ additional particles at the right edge of block $i$. We define the following ‘coarse-grained counters’:

Note that $F_{i}^{j}(s),L_{i}^{j}(s)\in\mathcal{F}_{j}$. Our leftmost priority policy for choosing the hot particle guarantees the next lemma.

See the corresponding lemma in [9] for a proof. This allows us to study the number of frozen free particles in block $i$ at the end of the procedure by running the carpet/hole dynamics up to block $i$, plus some random number of extra particles input from the right.

To avoid dependence, we make a uniform argument over all possible values of the input from the right. A vector of integers $\boldsymbol{s}=(s_{0},s_{1},\cdots,s_{n}=0)$ is said to satisfy the mass balance equations if

Note that the random vector $(L_{0},L_{1},\ldots,L_{n}=0)$ satisfies the mass balance equations by definition. In what follows, we will sum over the random set of possible vectors $\boldsymbol{s}$ satisfying these random mass balance equations.

The following Lemma is an upper bound on the number of frozen particles $F_{i}$ in a fixed block. Sections 6, 7 and 8 are devoted to its proof.

Note the lower limit of the summation over $s$ in the first statement. Since we only assume the starting configuration to be valid, we need at least two input particles to ‘fix’ the configuration – see the proof of Lemma 4.6 in Section 8.

In this section we use the single block estimate to prove Proposition 3.4.

To bound Frozen, the number of frozen free particles remaining in $D_{n}$ after the carpet/hole dynamics, we use the fact that $H_{i}\subset\{L_{i}\geqslant 2\}\cup J_{i}^{c}$. Since ${\text{Frozen}}=\sum_{i}F_{i}$,

Recall there are at most $L^{\ast}|D_{n}|$ particles inside $D_{n}$ at the beginning. Let $c_{4}=L^{\ast}(K+1).$ Using Lemma 4.4, we may rewrite the expectation as

where

and

We will inductively show that $\mathcal{S}(k)\leqslant(c_{4}n+1)(c^{\prime}+4e^{2c-c_{2}\wedge c_{3}n})^{k}$ for constants $c_{2},c_{3},c_{4}$. The base case $k=0$ is trivial and the case $k=1$ follows from Lemma 4.6 and the estimate (4.9). Suppose that the inequality is true for $k-2$ and $k-1$. Let

Decomposing the last two sums depending on whether $S_{k-1}\geq 2$ and $S_{k}\geq 2$, we get

By conditioning on $\mathcal{F}_{k-1}$ and using (4.9), the first sum is bounded above by $\mathcal{S}(k-1)\cdot 2e^{c-c_{2}n}$. Similarly, the second sum is at most $\mathcal{S}(k-1)\cdot c^{\prime}$ by conditioning on $\mathcal{F}_{k-2}$ and using the first part of Lemma 4.6. For the last sum, we have

It follows from (4.9) and the second part of Lemma 4.6 that the third sum is upper bounded $\mathcal{S}(k-2)\cdot 2e^{2c-c_{3}n}$ for some $c_{3}>0$. Combining all three bounds finishes the inductive step.

To bound Frozen, we apply Markov’s inequality together with the bound on $\mathcal{S}(n-1)$ to get

By the assumption $\log c^{\prime}<\delta c$, the event in question is exponentially unlikely in $n$. This completes the proof of the case $n_{0}=0$ and $n_{1}=n-1$.

The proof of the general case is almost the same as the above case. The main difference is that when $n_{0}\neq 0$, the counter $L_{n_{0}}$ no longer enjoys the deterministic upper bound as $L_{0}\leqslant c_{4}n$. Instead it suffices to argue that $H_{n_{0}-1}^{c}$ and $L_{n_{0}}>4\beta/(1-\theta)$ occur simultaneously with probability exponentially small in $n$. By the second part of Lemma 4.6, out of every two added particles at $(n_{0}-1)K+a$, with probability at least $1-\theta$ some particle reaches $(n_{0}-2)K+a$ and thus visits $(n_{0}-1)K$. Then a standard concentration bound gives the claim. $\square$

Define a sequence $\{\tilde{M}_{i}\}_{i\geqslant 0}$ with $\tilde{M}_{i+1}=\lfloor(\tilde{M}_{i})(1+\ratio)\rfloor$ for some large $\tilde{M}_{0}$ with $\gamma=.02$. Let $M_{i}=(\tilde{M}_{i}+1)K-a/2$. Consider a sequence of intervals of integers

which is a shifted version of the interval $D_{2\tilde{M}_{i}+1}=(-K+a,(2\tilde{M}_{i}+1)K)$ with $2\tilde{M}_{i}+1$ many blocks. For $i\geqslant 1$, write

and

Also define $S^{+}_{i}:=\sum_{j\in{\text{Interval}_{\text{right}}}}jX(j)$ and $S^{-}_{i}:=\sum_{j\in{\text{Interval}_{\text{left}}}}jX(j)$. Then

Finally, let ${\text{Event}^{\pm}_{4,i}}$ be the event that

The above notation ${\text{Interval}_{i}}$ works for all $i\geqslant 0$. For $i=-1$, we use the convention that ${\text{Interval}_{-1}}:=\{-a/2\}$. Here $-a/2$ is the left endpoint of the center block containing site zero. For $i=0$, we write ${\text{Interval}_{\text{left}}}=\{-M_{0},\dots,-a/2-1\}$ and ${\text{Interval}_{\text{right}}}=\{-a/2+1,\dots,M_{0}\}$.

We will inductively define partial stabilization procedures on the nested sequence of intervals $\{{\text{Interval}_{i}}\}_{i\geqslant 0}$ and the resulting configurations $\{Y_{i}(z)\}_{z\in{\text{Interval}_{i}}}$. Set $Y_{-1}(-a/2)=X(-a/2)$. Suppose $\{Y_{i-1}(z)\}_{z\in{\text{Interval}_{i-1}}}$ is defined, we may extend the definition $Y_{i-1}(z):=X(z)$ for all $z\notin{\text{Interval}_{i-1}}$, and then define $\{Y_{i}(z)\}_{z\in{\text{Interval}_{i}}}$ to be the configuration after the partial stabilization of ${\text{Interval}_{i}}$ with initial configuration $\{Y_{i-1}(z)\}_{z\in{\text{Interval}_{i}}}$ and particles frozen when they get to the boundary of the interval. Let $u_{i}(z)$ be the site odometer at $z$ in the partial stabilization of ${\text{Interval}_{i}}$ from $Y_{i-1}$.

For $i\geqslant 1$, to go from $Y_{i-1}$ to $Y_{i}$ we do the partial stabilization in the following order:

• •

  Run IDLA on the particles in the two intervals

  $${\text{Interval}_{\text{left}}}\setminus\{-M_{i}\}\text{ and }{\text{Interval}_{\text{right}}}\setminus\{M_{i}\}$$

  and freeze particles at the boundaries $-M_{i},-M_{i-1},M_{i-1}$ or $M_{i}$. In other words, keep toppling every particle inside both intervals until it either becomes alone at its site or reaches the boundary.

• •

  Run IDLA on the particles at $-M_{i-1}$ and $M_{i-1}$, freezing particles at $-M_{i}$ and $M_{i}$, until ${\text{Interval}_{\text{left}}}\setminus\{-M_{i}\}$ and ${\text{Interval}_{\text{right}}}\setminus\{M_{i}\}$ are completely filled or we run out of particles at $-M_{i-1}$ and $M_{i-1}$.

• •

  If the configuration inside ${\text{Interval}_{i}}$ becomes valid after the previous two steps, stabilize ${\text{Interval}_{i}}$ according to the carpet/hole dynamics, freezing particles at $-M_{i}$ and $M_{i}$.

For $i=0$, we perform a similar procedure by replacing the interval endpoints $-M_{i-1}$ and $M_{i-1}$ in the first and second steps by $-a/2$.

Let $Z_{i,1}$, $Z_{i,2}$ and $Z_{i,3}=Y_{i}$ the configurations after each of these three steps at stage $i$ respectively. Each of these configurations has at most one particle per site inside ${\text{Interval}_{i}}$ except at $-M_{i-1}$ and $M_{i-1}$ (or $-a/2$ when $i=0$).

Let $L^{\ast}:=\max\{j:\ \mu(j)>0\}\geqslant 2$, and let ${\text{Stage}_{-1}}$ be the event that

• •

  every site in ${\text{Interval}_{0}}$ contains $L^{\ast}$ particles initially,

which occurs with small but positive probability. Suppose ${\text{Stage}_{i-1}}$ happens, we will define the event ${\text{Stage}_{i}}$ inductively. Conditioned on ${\text{Stage}_{i-1}}$, we will observe the following typical behavior during the partial stabilization on ${\text{Interval}_{i}}$:

• •

  After we run IDLA on the particles in

  $${\text{Interval}_{\text{left}}}\setminus\{-M_{i}\}\text{ and }{\text{Interval}_{\text{right}}}\setminus\{M_{i}\},$$

  the density of sites inside each of those intervals that is covered by a particle is between $1-1/(3K)$ and 1. We also expect that ${\text{Event}^{+}_{4,i}}\cap{\text{Event}^{-}_{4,i}}$ occurs when $i\geqslant 1$. Call the intersection of these three events to be ${\text{Event}_{1,i}}$.

• •

  Then we run IDLA on the particles at $-M_{i-1}$ and $M_{i-1}$ (or $-a/2$ when $i=0$) until ${\text{Interval}_{\text{left}}}\setminus\{-M_{i}\}$ and ${\text{Interval}_{\text{right}}}\setminus\{M_{i}\}$ are completely filled. We expect that there are at least $.2\tilde{M}_{i}$ particles left at both $-M_{i-1}$ and $M_{i-1}$ (or $-a/2$ when $i=0$) at the end of this step. Call this event ${\text{Event}_{2,i}}$.

• •

  If the typical events occur up to this point, by definition the configuration inside ${\text{Interval}_{i}}$ will be valid and thus we may carry out the carpet/hole toppling procedure. After we stabilize according to the carpet/hole dynamics, there will be less than $\delta(2\tilde{M}_{i}+1)=.0004(2\tilde{M}_{i}+1)$ blocks without a hole in ${\text{Interval}_{i}}$. At both $-M_{i}$ and $M_{i}$ there will be at least $\tilde{M}_{i}/4$ particles. Moreover, the odometer $u_{i}(z)$ at the left endpoint $z=-a/2+m^{\prime}K$ of every block $m^{\prime}\in\{-\tilde{M}_{i},\dots,\tilde{M}_{i}\}$ during this carpet/hole procedure is at least $\beta(2\tilde{M}_{i}+1)=.0004(2\tilde{M}_{i}+1)$. Call this event ${\text{Event}_{3,i}}$.