2.1 Site percolation: Model and basic definitions

We introduce the model more formally. Given $p \in [0,1]$ , we can choose our probability space to be $(\{0,1\}^{\mathbb Z^d}, \mathcal F, \mathbb P_p)$ , where the $\sigma$ -algebra $\mathcal F$ is generated by the cylinder sets, and $\mathbb P_p = \bigotimes_{x\in\mathbb Z^d} \text{Ber}(p)$ . We call $\omega\in\{0,1\}^{\mathbb Z^d}$ a configuration and say that a site $x\in\mathbb Z^d$ is open or occupied in $\omega$ if $\omega(x)=1$ . If $\omega(x)=0$ , we say that the site x is closed or vacant. We often identify $\omega$ with the set $\{x\in\mathbb Z^d: \omega(x)=1\}$ .

For $k\in\mathbb N$ and a configuration $\omega$ , we call $(v_0,v_1,\dots,v_k)\in(\mathbb Z^d)^{k+1}$ an occupied path of length k from $v_0$ to $v_k$ if $| v_i-v_{i-1}| = 1$ for all $1 \leq i \leq k$ , and $v_i\in\omega$ for $1 \leq i \leq k-1$ . Here, and throughout the paper, we write $|x|= \sum_{i=1}^d|x_i|$ for $x\in\mathbb R^d$ (which is equal to the graph distance in $\mathbb Z^d$ ). For two points $x \neq y\in\mathbb Z^d$ we write $\{x \longleftrightarrow y\}$ (and say that x is connected to y) if there exists an occupied path from x to y of arbitrary length; mind that the this event is irrespective of the occupation status of x and y. We set $\{x \longleftrightarrow x\}=\varnothing$ , that is, x is not connected to itself. Moreover, $|x-y|=1$ implies $\{x \longleftrightarrow y\}=\{0,1\}^{\mathbb Z^d}$ (neighbours are always connected).

We define the cluster of x to be $\mathscr {C}(x) = \{x\} \cup \{y \in \omega: x \longleftrightarrow y\}$ . Note that apart form x itself, points in $\mathscr {C}(x)$ need to be occupied.

The two-point function $\tau_p\colon \mathbb Z^d \to [0,1]$ is defined as $\tau_p(x):=\mathbb P_p(\textbf{0} \longleftrightarrow x)$ , where $\textbf{0}$ denotes the origin in $\mathbb Z^d$ . The percolation probability is defined as $\theta(p) = \mathbb P_p(\textbf{0} \longleftrightarrow \infty) = \mathbb P_p( | \mathscr {C}(\textbf{0})| = \infty)$ . We note that $p \mapsto \theta(p)$ is increasing and define the critical point for $\theta$ as in (1.1). The critical point $p_c$ depends on the underlying graph.

For an absolutely summable function $f\colon\mathbb Z^d \to \mathbb R$ , the discrete Fourier transform is defined as $\widehat f\colon(-\pi,\pi]^d \to \mathbb C$ , where

\begin{equation*} \widehat f(k) = \sum_{x \in \mathbb Z^d} \text{e}^{\text{i} k \cdot x} f(x),\end{equation*}

and $k\cdot x = \sum_{j=1}^{d} k_j x_j$ denotes the scalar product.

2.2 The lace expansion in high dimension

We use this section to state the definitions and results from [Reference Heydenreich and Matzke13] needed in the proof of Theorem 1.1. We note that the below definition uses the notion of disjoint occurrence (denoted $`\circ$ ’) related to the BK inequality (which we will use at a later stage as well). For details on both, see e.g. [Reference Bollobás and Riordan2, Chapter 2] or [Reference Grimmett9, Section 2.3].

Note that we introduce $\Omega=2d$ . For better readability, we stick to using $\Omega$ for the remainder of the paper. We also address the Landau notation $f(\Omega) \leq \mathcal O(g(\Omega))$ that will appear frequently throughout the paper. It is always to be understood in the sense that there exists some $d_0$ and a constant $C(d_0)$ , such that $f(\Omega) \leq C g(\Omega)$ for all $\Omega \geq d_0$ . The constant C may depend on other appearing parameters.

We remark that $\{x \longleftrightarrow y\textrm { in } {\mathbb Z^d}\} = \{x \longleftrightarrow y\} = \{x \longleftrightarrow y\textrm { in } {\omega}\}$ and that $\{u \Longleftrightarrow x\} = \{0,1\}^{\mathbb Z^d}$ for $|u-x| = 1$ . Similarly, $\{u \Longleftrightarrow x\} = \varnothing$ for $u=x$ . We state two elementary observations made in [Reference Heydenreich and Matzke13] involving J that will be important later on.

Observation 2.2 (Convolutions of J, [Reference Heydenreich and Matzke13, Observation 4.4]). Let $m\in\mathbb N$ and $x\in\mathbb Z^d$ with $m \geq |x|$ . Then there is a constant $c=c(m,x)$ with $c \leq m!$ such that

\begin{equation*} J^{\ast m}(x) = c \textrm{1}_{\{m-|x| \text{ is even}\}} \Omega^{(m-|x|)/2 }. \end{equation*}

Observation 2.3 (Elementary bound on $\tau_p^{\ast n}$ , [Reference Heydenreich and Matzke13, Observation 4.5]). Let $m,n \in \mathbb N, p\in[0,1]$ and $x \in\mathbb Z^d$ . Then there is a constant $c=c(m,n)$ such that

\begin{equation*} \tau_p^{\ast n}(x) \leq c \sum_{l=0}^{m-1} p^{l} J^{\ast (l+n)}(x) + c \sum_{j=1}^{n} p^{m+j-n}(J^{\ast m} \ast \tau_p^{\ast j})(x). \end{equation*}

The statement of [Reference Heydenreich and Matzke13, Observation 4.5] is actually slightly stronger than Observation 2.3, but the version stated here suffices for our purpose. The following, more specific definitions are important to define the lace-expansion coefficients:

We remark that $\{u \leftrightarrow{\ {\mathbb Z^d} \ } x\} = \{ u \longleftrightarrow x\}$ . We can now define the lace-expansion coefficients. To this end, let $(\omega_i)_{i\in\mathbb N_0}$ be a sequence of independent site percolation configurations. For an event E taking place on $\omega_i$ , we highlight this by writing $E_i$ . We also stress the dependence of random variables on the particular configuration they depend on. For example, we write $\mathscr {C}(u; \omega_i)$ to denote the cluster of u in configuration i.

Definition 2.5 (Lace-expansion coefficients) Let $n\in\mathbb N_0, x\in\mathbb Z^d$ , and $p \in [0,p_c]$ . We define

\begin{align*}{ \Pi_p^{(0)}(x) := \mathbb P_p(\textbf{0} \Longleftrightarrow x) - J(x), \\ \Pi_p^{(n)}(x) := p^n \sum_{u_0, \ldots, u_{n-1}} \mathbb P_p \Big( \{\textbf{0} \Longleftrightarrow u_0\}_0 \cap \bigcap_{i=1}^{n} E'(u_{i-1},u_i; \mathscr {C}_{i-1})_i \Big), }\end{align*}

where $u_{-1}=\textbf{0}, u_n=x$ and $\mathscr {C}_{i} = \widetilde{\mathscr {C}}^{u_i}(u_{i-1};\omega_i)$ . Let furthermore $\Pi_{p}(x) := \sum_{n=0}^{\infty} (-1)^n \Pi_p^{(n)}(x)$ .

It is proved in [Reference Heydenreich and Matzke13] that the functions $(\Pi_p^{(n)}(x))_{n\in\mathbb N_0}$ are (absolutely) summable for every x and that $\Pi_p$ is thus well defined. We remark that $E'(u_{i-1},u_i; \mathscr {C}_{i-1})_i$ takes place solely on $\omega_i$ only if $\mathscr {C}_{i-1}$ is regarded as a fixed set; otherwise, it takes place on $\omega_{i-1}$ as well as $\omega_i$ . Proposition 2.6 summarises the main results of [Reference Heydenreich and Matzke13] (namely, Theorem 1.1 and Proposition 4.2).

Proposition 2.6 (OZE, infra-red bound and bounds on the lace-expansion coefficients). Let $p \in [0,p_c]$ . Then there is $d_0 \geq 6$ such that, for all $d > d_0$ , $\tau_p$ satisfies the Ornstein–Zernike equation

(2.1) \begin{equation} \tau_p(x) = J(x) + \Pi_p(x) + p\big((J+\Pi_p)\ast\tau_p\big)(x). \end{equation}

Secondly, there is a constant $C=C(d_0)$ such that

(2.2) \begin{equation} p|\widehat\tau_p(k)| \leq \frac{|\widehat D(k)| + C/d}{1-\widehat D(k)}, \end{equation}

where we take the right-hand side to be $\infty$ for $k=0$ . Thirdly, $2dp \leq 1+C/d$ , and lastly, for $n\in\mathbb N_0$ ,

(2.3) \begin{equation} p\sum_{x\in\mathbb Z^d} \Pi_{p}^{(n)}(x) \leq C (C/d)^{n \vee 1}. \end{equation}

As a consequence, we also have $p \sum_{x} \Pi_p(x) \leq C/d$ .

2.3 Diagrammatic bounds

In the proofs to follow, we need another result from [Reference Heydenreich and Matzke13]. We formulate it in terms of a diagrammatic notation, as we are going to make use of this later as well. To this end, we introduce some quantities related to $\tau_p$ .

Definition 2.7 (Modified two-point functions and triangles). Let $x\in\mathbb Z^d$ and define

\begin{equation*} \tau_p^\circ(x) := \delta_{\textbf{0},x} + \tau_p(x), \qquad \tau_p^\bullet(x) = \delta_{\textbf{0},x} + p\tau_p(x).\end{equation*}

Moreover, let $\triangle_p(x) = p^2(\tau_p\ast\tau_p\ast\tau_p)(x)$ , $\triangle_p^{\bullet}(x) = p(\tau_p^\bullet\ast\tau_p\ast\tau_p)(x)$ , $\triangle_p^{\bullet\circ}(x) = p(\tau_p^\bullet\ast\tau_p^\circ\ast\tau_p)(x)$ , and $\triangle_p^{\bullet\bullet\circ}(x) = (\tau_p^\bullet\ast\tau_p^\bullet\ast\tau_p^\circ)(x)$ . We also set

\begin{equation*} \triangle_p = \sup_{x \in\mathbb Z^d} \triangle_p(x), \quad \triangle_p^{\bullet} = \sup_{\textbf{0} \neq x \in\mathbb Z^d} \triangle_p^{\bullet}(x), \quad\triangle_p^{\bullet\circ} = \sup_{\textbf{0} \neq x \in\mathbb Z^d} \triangle_p^{\bullet\circ}(x), \quad \triangle_p^{\bullet\bullet\circ} = \sup_{x\in\mathbb Z^d} \triangle_p^{\bullet\bullet\circ}(x). \end{equation*}

We need the following bounds obtained in [Reference Heydenreich and Matzke13].

Proposition 2.8 (Triangle bounds, [Reference Heydenreich and Matzke13, Lemma 4.7]). Let $p\in[0,p_c]$ . Then there is $d_0\geq 6$ and a constant $C=C(d_0)$ such that, for all $d>d_0$ ,

\begin{equation*} \max\{ \triangle_p, \triangle_p^{\bullet}, \triangle_p^{\bullet\circ}\} \leq C /d, \qquad \max\{\triangle_p^{\bullet}(\textbf{0}), \triangle_p^{\bullet\circ}(\textbf{0}), \triangle_p^{\bullet\bullet\circ}\} \leq C. \end{equation*}

As part of the proof that bounds the functions $\Pi_p^{(i)}$ in [Reference Heydenreich and Matzke13], a first bound is formulated in terms of a long sum over products of the modified two-point functions. In the second step, those are decomposed into products of the modified triangles. We need a formulation of this intermediate bound on $\Pi_p^{(i)}$ for $i\in\{1,2\}$ for Section 5, as well as a pictorial representation. We first state the needed bound on $\Pi_p^{(1)}$ .

Lemma 2.9 (Diagrammatic bound on $\Pi_p^{(1)}$ , [Reference Heydenreich and Matzke13, Lemma 3.10]). Let $p \in [0, p_c]$ . Then

(2.4) \begin{equation} \sum_{x\in\mathbb Z^d} \Pi_p^{(1)}(x) \leq \sum_{\substack{w,u,t,z,x \in\mathbb Z^d: \\ u \neq x, |\{t,z,x\}| \neq 2}} \tau_p^\bullet(w) \tau_p(u)\tau_p(w-u) \tau_p^\circ(z-w) \tau_p^\bullet(t-u) \tau_p^\bullet(z-t) \tau_p^\bullet(x-t) \tau_p^\circ(x-z). \end{equation}

The bounds in [Reference Heydenreich and Matzke13] are formulated only for $p<p_c$ , but as the bounds are increasing in p, a limit argument easily extends them to the critical point. We now show how we represent the bound in (2.4) in terms of pictorial diagrams. As the bound on $\Pi_p^{(2)}$ is even longer to write down, Lemma 2.10 is stated only in terms of these pictorial bounds.

The points w,u,t,z,x summed over are represented as squares, factors of $\tau_p$ are represented as lines and lines with a ‘ $\bullet$ ’ (‘ $\circ$ ’) symbol represent factors of $\tau_p^\bullet$ ( $\tau_p^\circ$ ). For example, the factor $\tau_p(w-u)$ is represented as a line between two squares, which we think of as the points w and u. We interpret the factor $\tau_p(u)$ as a line between u and the origin. We indicate the position of u and x in the below diagrams. After expanding the two cases in (2.4) according to whether $|\{t,z,x\}|=3$ or $|\{t,z,x\}|=1$ , this pictorial representation allows us to rewrite the bound in (2.4) as:

We now formulate the bound on $\Pi_p^{(2)}$ ; more precisely, we are going to insert a case distinguishing indicator, resulting in two bounds.

Lemma 2.10 (Diagrammatic bound on $\Pi_p^{(2)}$ , [Reference Heydenreich and Matzke13, Lemma 3.10]). Let $p\in[0,p_c]$ . Then

(2.5)

and

(2.6)

2.4 Convolution bounds

The last result from [Reference Heydenreich and Matzke13] we need to state is going to be important for the proofs of Section 4.

Lemma 2.11 (Bounds on convolutions of and $\tau_p$ , [Reference Heydenreich and Matzke13, Lemma 4.6]). Let $m,n\in\mathbb N_0$ with $2m+n\geq 2$ . For $p\in[0,p_c]$ and $d>20n/9$ ,

\begin{equation*} \sup_{a\in\mathbb Z^d} p^{2m+n-1} \big(J^{\ast 2m} \ast \tau_p^{\ast n}\big)(a) \leq c \Omega^{1-m} \end{equation*}

for some constant $c=c(m,n)$ .

Lemma 4.6 in [Reference Heydenreich and Matzke13] states only the upper bound $c\Omega^{-1}$ , but an inspection of its proof gives the stronger bound of Lemma 2.11: for $m\ge4$ this is evident from the first bound on page 842 in [Reference Heydenreich and Matzke13], and for $m\le4$ one has to adapt [13, (4.10)] and the subsequent lines accordingly.

Again, Lemma 4.6 in [Reference Heydenreich and Matzke13] is stated only for $p<p_c$ , but the bounds

\begin{equation*} 2dp_c \leq 1 + \mathcal O(\Omega^{-1}) \quad \text{and} \quad \sup_{k\in(-\pi,\pi]^d} \frac{p_c|\widehat\tau_{p_c}(k)|}{\widehat G_1(k)} \leq 1 + \mathcal O(\Omega^{-1}), \end{equation*}

are sufficient for the statement to extend to $p_c$ . While the former bound is a direct consequence of Proposition 2.6, the latter bound (for $k\neq 0$ ) follows from the infra-red bound (2.2) and $|\widehat D(k)| \leq 1$ . The bound for $k=0$ follows from the continuity of the Fourier transform.

5.1 Analysis of $\widehat{\Pi}^{(0)}$

We recall that we write $\widehat\Pi^{(i)} = \widehat\Pi_{p_c}^{(i)}(\textbf{0})$ . We will also abbreviate $\mathbb P=\mathbb P_{p_c}$ and $\tau=\tau_{p_c}$ throughout Section 5. We use (3.3) a lot throughout Section 5, and we recall that it states

\begin{equation*} \Omega p_c= 1+ \mathcal O(\Omega^{-1})\end{equation*}

and follows from Proposition 2.6. Moreover, we will use (4.1) of Lemma 4.2 frequently in the proofs to follow and will not mention every time we do so.

Lemma 5.1 (Finer asymptotics of $\widehat\Pi^{(0)}$ ) As $d\to\infty$ ,

\begin{equation*} \widehat\Pi^{(0)} = \tfrac 12 \Omega^2 p_c^2 + \tfrac 52 \Omega^{-1} + \mathcal O(\Omega^{-2}).\end{equation*}

Proof. Recall that $\widehat{\Pi}^{(0)} = \sum_{x} \big(\mathbb P(\textbf{0} \Longleftrightarrow x) - J(x)\big)$ . This sum only gets contributions from $|x| \geq 2$ . Now,

\begin{align*}{\widehat\Pi^{(0)}= \sum_{|x|\geq 2}\mathbb P(\textbf{0} \Longleftrightarrow x) = \sum_{|x|=2}\mathbb P \big(\textbf{0} \Longleftrightarrow{\,\, ({\leq 4}) \,\,} x\big) + \sum_{|x| \leq 3}\mathbb P\big(\textbf{0} \Longleftrightarrow{\,\, ({= 6}) \,\,} x\big) + \sum_{|x|\geq 2} \mathbb P\big(\textbf{0} \Longleftrightarrow{\,\, ({\geq 8}) \,\,} x\big) \\ = \sum_{|x|=2}\mathbb P \big(\textbf{0} \Longleftrightarrow{\,\, ({\leq 4}) \,\,}x\big) + \sum_{|x| \leq 3}\mathbb P\big(\textbf{0} \Longleftrightarrow{\,\, ({= 6}) \,\,} x\big) + \mathcal O(\Omega^{-2}), }\end{align*}

where the last identity is due to Lemma 4.2. We first consider 4-cycles. The only points x with $|x| \geq 2$ that can form a 4-cycle with the origin are those with $|x|=2, \|x\|_\infty=1$ . There are $\tfrac 12 \Omega(\Omega-2)$ such points. If $x=v_1+v_2$ (with $|v_i|=1$ ) is such a point, then $\{\textbf{0} \Longleftrightarrow{\,\, ({\leq 4}) \,\,} x\}$ holds if and only if $\{v_1, v_2\} \subseteq \omega$ . Therefore,

(5.1) \begin{equation} \sum_{|x| \geq 2} \mathbb P \big(\textbf{0} \Longleftrightarrow{\,\, ({\leq 4}) \,\,} x\big) = \tfrac 12 \Omega(\Omega-2) p_c^2 = \tfrac 12 \Omega^2 p_c^2 - \Omega^{-1} + \mathcal O(\Omega^{-2}). \end{equation}

We are left to consider points $|x| \geq 2$ contained in cycles of length 6 that also contain the origin. Note that this is possible for $|x|\in\{2,3\}$ and $\| x\|_\infty \in\{1,2\}$ . We first claim that $\|x\|_\infty=2$ gives a contribution of order $\mathcal O(\Omega^{-2})$ .

Indeed, there are $\Omega$ points x with $|x| = 2$ and $\|x\|_\infty=2$ , and any such point is contained in at most $c \Omega$ many origin-including cycles of length 6 (where c is some absolute constant). Any given 6-cycle has probability $p_c^4$ of being present, and so the contribution is at most $c \Omega^2 p_c^4 = \mathcal O(\Omega^{-2})$ .

Similarly, there are at most $\Omega(\Omega-2)$ points x with $|x|=3, \|x\|_\infty=2$ , and any such point is contained in exactly one origin-including cycle of length 6. Hence, this contributes at most $\Omega^2 p_c^4 = \mathcal O(\Omega^{-2})$ as well.

Let now $|x|=3, \|x\|_\infty=1$ . There are $\tfrac 16\Omega(\Omega-2)(\Omega-4)$ such points. Such a point spans a (3-dimensional) cube with the origin, in which two internally disjoint paths of respective length 3, making up the sought-after 6-cycle, have to be occupied. There are nine such cycles. By inclusion-exclusion,

(5.2) \begin{equation} \sum_{|x|=3, \|x\|_\infty=1} \mathbb P (\textbf{0} \Longleftrightarrow{\,\, ({= 6}) \,\,} x) \begin{cases} \leq \tfrac 96 \Omega^3 p_c^4 = \tfrac 32 \Omega^{-1} + \mathcal O(\Omega^{-2}), \\ \\[-7pt] \geq \tfrac 16 (\Omega-4)^3 \big[9p_c^4 - \binom{9}{2} p_c^5 \big] = \tfrac 32 \Omega^{-1} + \mathcal O(\Omega^{-2}) \end{cases}, \end{equation}

(for the lower bound, we sum the probabilities for the nine cycles to be occupied and substract the probability that at least two of them are occupied at the same time). Lastly, consider one of the $\tfrac 12\Omega(\Omega-2)$ points $x=v_1+v_2$ with $|x|=2, \|x\|_\infty=1$ , and $|v_i|=1$ . Note that there are precisely two paths of length 2 from $\textbf{0}$ to x, namely the ones using $v_i$ . To produce a relevant contribution to $\{\textbf{0} \Longleftrightarrow{\,\, ({= 6}) \,\,} x\}$ , we claim that exactly one of the two vertices must be vacant and the other occupied. Indeed, if both are occupied, then there is a 4-cycle containing $\textbf{0}$ and x. If both are vacant, then the shortest possible cycle containing $\textbf{0}$ and x is of length 8.

We assume $v_1$ to be occupied and $v_2$ to be vacant (the reverse gives the same contribution by symmetry, and we respect it with a factor of 2). It remains to count the number of paths of length 4 from $\textbf{0}$ to x that avoid $v_1$ and $v_2$ . Avoiding $\pm v_i$ gives $\Omega-4$ options for the first step. There are two options for the second step (namely, to a neighbour of $v_1$ or $v_2$ ). Steps 3 and 4 are now fixed: Out of the two shortest paths to x, one is via $v_i$ , and is not an option. In conclusion, the probability that there is a $\textbf{0}-x$ -path of length 4 traversing some fixed neighbour of $\textbf{0}$ (which is not $\pm v_i$ ) first is $p_c^2 (2p_c-p_c^2)$ . This gives

(5.3) \begin{equation} \sum_{|x|=2, \|x\|_\infty=1} \mathbb P (\textbf{0} \Longleftrightarrow{\,\, ({= 6}) \,\,}x) \begin{cases} \leq \tfrac 12 \Omega^3 4 p_c^4 = 2 \Omega^{-1} + \mathcal O(\Omega^{-2}), \\ \\[-7pt] \geq (\Omega-4)^3 p_c^3 (2p_c - p_c^2) - 4\Omega^2 p_c \binom{\Omega-4}{2} p_c^6 = 2 \Omega^{-1} + \mathcal O(\Omega^{-2}),\end{cases} .\end{equation}

Summing up (5.1), (5.2), and (5.3) finishes the proof.

5.2 Analysis of $\widehat\Pi^{(1)}$

Lemma 5.2 (Finer asymptotics of $\widehat\Pi^{(1)}$ ) As $d\to\infty$ ,

\begin{equation*} \widehat\Pi^{(1)} = \Omega p_c + 2\Omega^2 p_c^2 + 4 \Omega^{-1} + \mathcal O(\Omega^{-2}).\end{equation*}

Proof. Abbreviating $\mathscr {C}_0 = \widetilde{\mathscr {C}}^{u}(\textbf{0};\ \omega_0)$ , we recall that

(5.4) \begin{equation}\widehat\Pi^{(1)} = p_c \sum_{u\in\mathbb Z^d}\sum_{x\in\mathbb Z^d} \mathbb P \big( \{\textbf{0} \Longleftrightarrow u\}_0 \cap E'(u,x; \mathscr {C}_0)_1\big). \end{equation}

While this is a double sum over all points in $\mathbb Z^d$ , we first prove that only small values of u give relevant contributions. To this end, assume that $|u| \geq 3$ . We use the pictorial representation of the bound in Lemma 2.9 and decompose it in terms of modified triangles introduced in Definition 2.7. In the below pictorial diagrams, points over which the supremum is taken (in particular, those points are not summed over) are represented by coloured discs. The indicator that two such points (discs) may not coincide is represented by a disrupted two-sided arrow. Lemma 2.9 together with Proposition 2.8 then gives

(5.5)

where the last identity is due to Lemma 4.2. When we encounter similar diagrams to the ones in (5.5) at later stages of this paper, we decompose them in the same way as performed in (5.5), but in less detail.

We consider the cases of $|u| \in \{1,2\}$ separately. For both, we make further case distinctions according to the value of $|x|$ . The contributions are summarised in the following table:

Contributions of $|u|=1$ . By rotational symmetry, we can drop the sum over u, and rewrite (5.4) as:

(5.6) \begin{align} \ p_c \sum_{|u|=1}\sum_{x\in\mathbb Z^d} \mathbb P \big( \{\textbf{0} \Longleftrightarrow u\}_0 \cap E'(u,x;\ \mathscr {C}_0)_1 \big) \notag \\ = \ p_c \sum_{u,x\in\mathbb Z^d} J(u) \mathbb P \left( E'(u,x;\ \mathscr {C}_0)_1 \right), \end{align}

(5.7) \begin{align} = \Omega p_c \sum_{x\in\mathbb Z^d} \mathbb P \left( E'(u,x;\ \mathscr {C}_0)_1 \right) . \end{align}

In (5.7) and in the following, we take u to be an arbitrary (but fixed) neighbour of the origin. We recall that $\omega_i$ is a sequence of independent percolation configurations and an event with subscript i takes place on $\omega_i$ . Moreover, $E'(u,x;\ \mathscr {C}_0)$ is indexed to take place on configuration 1, which is only accurate if $\mathscr {C}_0$ is regarded as a fixed set; otherwise, the event takes place on $\omega_0$ and $\omega_1$ .

We proceed by splitting the sum over x in (5.7) (respectively, (5.6)) into different cases.

$\underline{\textrm{The case of}\ |u|=1, x=\textbf{0}\ \textrm{contributes}\ \Omega p_c}$ : The event $E'(u,x;\ \mathscr {C}_0)_1$ in (5.7) holds, the sum collapses to 1, and the contribution is $\Omega p_c$ .

$\underline{\textrm{The case of}\ |u|=1=|x|\ \textrm{contributes}\ \Omega^2p_c^2 - 2\Omega^{-1} + \mathcal O(\Omega^{-2})}$ : There are $\Omega-1$ choices for $x\neq u$ . We exclude the special case $x=-u$ first. For other choices of x, we let $v:=x+u$ .

• For $x=-u$ , we have $E'(u,x;\ \mathscr {C}_0)_1 \subseteq \{u \leftrightarrow{\,\, ({4}) \,\,} x\}_1$ by Observation 4.3. Hence, (5.7) is bounded by

  \begin{equation*} \Omega p_c \tau^{(4)}(u-x) = \mathcal O(\Omega^{-2}).\end{equation*}

• Let $x \neq \pm u$ and $v\in \omega_1$ , so that $\textbf{0},u,v,x$ span a ‘square’ in one of the hyperplanes. Note first that there are $\Omega-2$ choices for x, and we can treat them equally by symmetry. Since $x\sim \textbf{0}$ and hence $x\in\langle {\mathscr {C}_0} \rangle$ , we have that on the event $\{v\in\omega_1\}$ , the occurrence of $E'(u,x;\ \mathscr {C}_0)_1$ implies that either v is not pivotal for $\{u\longleftrightarrow x\}$ , or it is pivotal but $v\notin\langle {\mathscr {C}_0} \rangle$ :

  \begin{equation*} E'(u,x;\ \mathscr {C}_0)_1 \cap\{v \in\omega_1\}_1 = \{v \in\omega_1\}_1 \cap \Big( \{v \notin \langle {\mathscr {C}_0} \rangle\}_0 \cup \{v \notin\textsf {Piv}({u,x})\}_1 \Big). \end{equation*}
  Note that all three appearing events on the right are independent of each other. Recalling that $\mathscr {C}_0$ is shorthand for $\tilde{\mathscr {C}}^u(0,\omega_0)$ , we see that $\{0\leftrightarrow v\}_0$ if either $x\in\omega_0$ or if there is an occupied path of length $\ge 4$ in $\omega_0\setminus\{u\}$ :
  \begin{equation*}\mathbb P (v \notin \langle {\mathscr {C}_0} \rangle) =1- \mathbb P\big( x \in\omega_0\big) - \mathbb P\big(\textbf{0} \leftrightarrow{\,\, ({\geq 4}) \,\,} v \text{ in } \omega_0\setminus\{u\}\big) = 1-p_c+\mathcal O(\Omega^{-2}). \end{equation*}
  In order for v to be not pivotal, there must be a ‘second connection’ from u to x, either a short one via $\textbf{0}$ , or via a longer path, that is,
  \begin{equation*} \mathbb P(v \notin\textsf {Piv}({u,x})) = \mathbb P(\textbf{0} \in \omega_1) + \mathbb P\big(u \leftrightarrow{\,\, ({\geq 4}) \,\,} x \text{ in } \omega_1\setminus\{v\}\big) = p_c + \mathcal O(\Omega^{-2}). \end{equation*}
  We can now replace the sum over x in (5.7) by a factor of $(\Omega-2)$ and thus obtain the contribution:
  \begin{align*}{ \Omega p_c (\Omega-2) \mathbb P \big( E'(u,x;\ \mathscr {C}_0)_1 \cap \{v \in\omega_1\}_1 \big) = \Omega(\Omega-2) p_c^2 \Big( 1-p_c+p_c - (1-p_c)p_c \Big) + \mathcal O(\Omega^{-2}) \\ = (\Omega p_c)^2 (1-p_c) - 2\Omega p_c^2 + \mathcal O(\Omega^{-2}) \\ = \Omega^2 p_c^2 - 3 \Omega^{-1} + \mathcal O(\Omega^{-2}). }\end{align*}

• Let $x \neq \pm u, v \notin \omega_1$ and $\textbf{0} \notin \omega_1$ . For $E'(u,x;\ \mathscr {C}_0)_1$ to hold, there needs to be a $\omega_1$ -path between u and x. Its pivotal points cannot lie in $\langle {\mathscr {C}_0} \rangle$ however. First, note that any relevant path between u and x is of length 4, as

  \begin{equation*} \Omega p_c (\Omega-2) \mathbb P\big(E'(u,x;\ \mathscr {C}_0)_1 \cap \{ u \leftrightarrow{\,\, ({\geq 6}) \,\,} x\}_1\big) \leq \Omega^2 p_c \tau^{(6)}(x-u) = \mathcal O(\Omega^{-2}). \end{equation*}
  We now investigate the 4-paths from u to x that avoid $\textbf{0}$ and v—from Lemma 5.1, we already know that there are $2(\Omega-4)$ of them. Let z be one of the $\Omega-4$ unit vectors satisfying $\dim \langle\!\langle {u,x,z} \rangle\!\rangle =3$ , where we let $\langle\!\langle {\cdot} \rangle\!\rangle$ denote the span. We denote by $\gamma_1$ and $\gamma_2$ the two $u-x$ -paths of length 4 that visit $y_1:=u+z$ . W.l.o.g., $\gamma_1$ visits $y_2:=y_1+x$ second and $y_3:=y_2-u$ third, whereas $\gamma_2$ visits z second and $y_3$ third. Let $\{\gamma_i \subseteq \omega_1\}$ denote the event that the three internal vertices of $\gamma_i$ are $\omega_1$ -occupied. See Figure 2a for an illustration.

  We now show that only $\gamma_1$ produces a relevant term. Assume first that $y_2 \notin\omega_1$ , but $\gamma_2\subseteq\omega_1$ . For $E'(u,x;\ \mathscr {C}_0)_1$ to hold, $z\in\langle {\mathscr {C}_0} \rangle$ must not be a pivotal point. Under $\gamma_2 \subseteq \omega_1$ ,

  (5.8) \begin{equation} \{ z \notin \textsf {Piv}({u,x}) \}_1 \subseteq \bigcup_{a\in\{u,y_1\}, b\in\{y_3, x\} } \{a \longleftrightarrow b \text{ in } \omega_1 \setminus \{z\}\}_1. \end{equation}
  Resolving the right-hand side of (5.8) by a union bound gives four connection events. The shortest $\omega_1$ -path from u to x of non-vacant vertices is of length 4. Moreover, the shortest $\omega_1$ -path from $y_1$ to $y_3$ of non-vacant vertices that avoids z is of length 4 as well, and so (5.7) is bounded by:
  \begin{align*}\Omega (\Omega-2) p_c \sum_z \mathbb P\Big(E'(u,x;\ \mathscr {C}_0)_1, \{\textbf{0},v,y_2\}\cap\omega_1 = \varnothing, \gamma_2\subseteq\omega_1 \Big) \\ \leq \Omega^3 p_c^4 \Big( \tau^{(4)}(x-u) + \tau^{(3)}(y_3-u) + \tau^{(3)}(x-y_1) + \tau^{(4)}(y_3-y_1)\Big) = \mathcal O(\Omega^{-2}). \end{align*}
  We now show that $\gamma_1\in\omega_1$ gives a contribution. Note that under $\{\textbf{0},v\notin\omega_1, \gamma_1 \subseteq \omega_1\}$ ,
  (5.9) \begin{equation} E'(u,x;\ \mathscr {C}_0)_1 = \bigcap_{i\in\{1,2,3\}} \Big( \{y_i \notin\textsf {Piv}({u,x})\}_1 \cup \{y_i \notin\langle {\mathscr {C}_0} \rangle\}_0\Big). \end{equation}
  But $\mathbb P(\{y_i \notin\textsf {Piv}({u,x})\}_1 \cup \{y_i \notin\langle {\mathscr {C}_0} \rangle\}_0) \geq 1-\mathbb P(y_i \in \langle {\mathscr {C}_0} \rangle) \geq 1-\tau^{(2)}(y_i) =1-\mathcal O(\Omega^{-1})$ for all i by Lemma 4.2, and so, by inclusion-exclusion,
  \begin{align*}\Omega (\Omega-2) p_c \sum_z \mathbb P\Big(E'(u,x;\ \mathscr {C}_0)_1, \{\textbf{0},v\}\cap\omega_1 = \varnothing, \gamma_1\subseteq\omega_1 \Big) \\ \begin{cases} \leq \Omega(\Omega-2)(\Omega-4) (1-p_c)^2 p_c^4 \big(1-\mathcal O(\Omega^{-1}) \big) = \Omega^{-1} + \mathcal O(\Omega^{-2}), \\ \geq \Omega^3 p_c^4 (1-\mathcal O(\Omega^{-1})) - \Omega^2 \binom{\Omega-4}{2} p_c^7 = \Omega^{-1} + \mathcal O(\Omega^{-2}). \end{cases} \end{align*}

• Let $x \neq \pm u, v\notin\omega_1$ , and $\textbf{0}\in \omega_1$ . By Observation 4.3,

  \begin{equation*} \Omega p_c (\Omega-2) \mathbb P \big( E'(u,x;\ \mathscr {C}_0)_1 \cap \{v \notin\omega_1,\textbf{0}\in\omega_1\} \big) \leq \Omega^2 p_c^2 \tau^{(4)}(x-u) = \mathcal O(\Omega^{-2}). \end{equation*}

Figure 2. An illustration of several appearing cases for $|u|=1$ . In the first two cases, $\textbf{0}$ and v are vacant in $\omega_1$ . In case (a), the black path is $\gamma_1$ , the red and dotted one is $\gamma_2$ . In case (b), the two $\textbf{0}-x$ -paths are marked as black chains of arrows. In case (c), $\{v_1, v_2\} \cap \omega_0 = \{v_1\}$ and the only relevant $u-x$ -path is marked in black:

$\underline{\textrm{The case of}\ |u|=1, |x|=2\ \textrm{contributes}\ {\Omega^2 p_c^2 +\Omega^{-1} + \mathcal O(\Omega^{-2})}}$ : There are $\tfrac 12 \Omega^2$ choices for x. We first consider the $\Omega-1$ choices neighbouring u and, among those, exclude the special case $x=2u$ first. For x a neighbour of u, we set $v:=x-u$ .

• Let $x=2u$ . Since $x \sim u$ , we have $E'(u,x;\ \mathscr {C}_0)_1 = \{x \in \langle {\mathscr {C}_0} \rangle\}_0 \subseteq \{ \textbf{0} \leftrightarrow{\,\, ({4}) \,\,} x \}_0$ , and so the contribution to (5.7) is bounded by $\Omega p_c \tau^{(4)}(x) = \mathcal O(\Omega^{-2})$ .

• Let $2u\neq x\sim u$ and $v\in \omega_0$ . There are $\Omega-2$ choices for x. The event $E'(u,x;\ \mathscr {C}_0)_1$ holds, and so

  \begin{equation*} \Omega p_c \sum_{2u\neq x\sim u} \mathbb E_0\left[ \textrm{1}_{\{v\in\omega_0\}}\mathbb P_1\left( E'(u,x;\ \mathscr {C}_0) \right) \right] = \Omega(\Omega-2) p_c^2 = \Omega^2 p_c^2 - 2\Omega^{-1} + \mathcal O(\Omega^{-2}).\end{equation*}

• Let $2u\neq x\sim u$ and $v\notin \omega_0$ . We partition

  \begin{align*} E'(u,x;\ \mathscr {C}_0)_1 = \Big( E'(u,x;\ \mathscr {C}_0)_1 \cap \{\textbf{0} \leftrightarrow{\,\, ({\leq 4}) \,\,} x \text{ in } \mathbb Z^d\setminus\{u\} \}_0 \Big)\\[3pt]\quad \cup \Big(E'(u,x;\ \mathscr {C}_0)_1 \cap \{\textbf{0} \leftrightarrow{\,\, ({\geq 6}) \,\,} x \text{ in } \mathbb Z^d\setminus\{u\}\}_0\Big) \end{align*}
  and treat the second event by observing
  \begin{equation*} \Omega p_c \sum_{2u\neq x\sim u} \mathbb P \Big( \{v\notin\omega_0\}_0 \cap \{\textbf{0} \leftrightarrow{\,\, ({\geq 6}) \,\,} x \text{ in } \mathbb Z^d\setminus\{u\}\}_0 \cap E'(u,x;\ \mathscr {C}_0)_1 \Big) \leq \Omega^2 p_c \tau^{(6)}(x) = \mathcal O(\Omega^{-2}).\end{equation*}
  As the only 2-paths from $\textbf{0}$ to x go through u and v, respectively, we can focus on paths of length 4 avoiding v and u. Hence, the status of v is independent of such paths. Let z be one of the $\Omega-4$ neighbours of $\textbf{0}$ with $\dim \langle\!\langle {u,v,z} \rangle\!\rangle =3$ . For any such z, there are two $\textbf{0}-x$ -paths of length 4 that first visit z and avoid $\{v,u\}$ . More precisely, these paths are $(\textbf{0},z,u+z,x+z,x)$ and $(\textbf{0},z,v+z,x+z,x)$ . Let $Q_4(z)$ denote the event that at least one of these paths is in $\omega_0$ . See Figure 2b for an illustration. As the events $\{ Q_4(z) \}$ are pairwise independent,
  \begin{align*}{\{v\notin\omega_0\}_0 \cap \{\textbf{0} \leftrightarrow{\,\, ({\leq 4}) \,\,} x \text{ in } \mathbb Z^d\setminus\{u\}\}_0 \cap E'(u,x;\ \mathscr {C}_0)_1 = \{v\notin\omega_0\}_0 \cap \big( \cup_{z} Q_4(z)\big) , \\[3pt] \mathbb P(\cup_z Q_4(z)) = (\Omega-4)\mathbb P( Q_4(z)) +\mathcal O(\Omega^{-4}) = 2(\Omega-4)p_c^3 + \mathcal O(\Omega^{-3}). }\end{align*}
  Consequently,
  \begin{align*}{\Omega p_c \sum_{2u\neq x\sim u} \mathbb P \Big( \{v\notin\omega_0\}_0 \cap \{\textbf{0} \leftrightarrow{\,\, ({\leq 4}) \,\,} x \text{ in } \mathbb Z^d\setminus\{u\}\}_0 \cap E'(u,x;\ \mathscr {C}_0)_1 \Big) \\[3pt] = \Omega p_c (\Omega-2)2(\Omega-4) p_c^3 + \mathcal O(\Omega^{-2}) = 2\Omega^{-1} + \mathcal O(\Omega^{-2}).}\end{align*}

• Let $|u-x|=3$ and $\|x\|_\infty =2$ . There are $\Omega-1$ choices for x. Let $2v = x$ . Note first that

  \begin{align*}{ \Omega(\Omega-1)p_c \mathbb P \Big( \big(\{x \in \langle {\mathscr {C}_0} \rangle\}_0 \cup \{ u \leftrightarrow{\,\, ({5}) \,\,} x\}_1 \big) \cap E'(u,x;\ \mathscr {C}_0)_1 \Big) \\ \leq \Omega^2 p_c \Big( \tau^{(2)}(x) \tau^{(3)}(x-u) + \tau^{(5)}(x-u)\Big) = \mathcal O(\Omega^{-2}). }\end{align*}
  The complementary event is that $x \notin\langle {\mathscr {C}_0} \rangle$ and the presence of a $u-x$ -path of length 3. The former implies $v\notin\omega_0$ . There are at most four potential sites that can make up internal vertices on a $u-x$ -path of length 3, namely $\textbf{0}, v, u+v, u+2v$ . To avoid potential pivotality of $\textbf{0}$ and v and still guarantee a path of length 3, we require $\{v+u, u+2v\} \subseteq \omega_1$ . But both these vertices are of distance at least 2 from the origin, and at least one of them must be in $\langle {\mathscr {C}_0} \rangle$ . In conclusion,
  \begin{align*}{ \Omega(\Omega-1)p_c \mathbb P \Big( \{x \notin \langle {\mathscr {C}_0} \rangle\}_0 \cap \{ u \leftrightarrow{\,\, ({\leq 3}) \,\,} x\}_1 \cap E'(u,x;\ \mathscr {C}_0)_1 \Big) \\ \leq 2 \Omega^2 p_c \tau^{(2)}(u+v) \tau^{(3)}(x-u) = \mathcal O(\Omega^{-2}). }\end{align*}

• Let $|u-x|=3, \| x\|_\infty=1$ and $x\in\langle {\mathscr {C}_0} \rangle$ . Write $x=v_1+v_2$ , where $|v_i|=1$ . We first show that contributions arise when precisely one point in $\{v_1, v_2\}$ is $\omega_0$ -occupied. Note that when both $v_1$ and $v_2$ are vacant in $\omega_0$ , the contribution to (5.7) is bounded by $\Omega^3 p_c \tau^{(4)}(x) \tau^{(3)}(x-u) = \mathcal O(\Omega^{-2})$ . On the other hand, if $\{v_1,v_2\} \subseteq \omega_0$ , then the contribution is bounded by $\Omega^3 p_c^3 \tau^{(3)}(u-x) =\mathcal O(\Omega^{-2})$ . Let now $v_1\in\omega_0$ and $v_2\notin\omega_0$ (the other case is identical and is respected by counting the contribution twice). There are $\tfrac 12 \Omega^2 (1+\mathcal O(\Omega^{-1}))$ choices for x. If $\{u \leftrightarrow{\,\, ({5}) \,\,} x\}_1$ , then the contribution to (5.7) is $\mathcal O(\Omega^{-2})$ . Set $z_1=u+v_2, z_2=u+v_2+v_1$ and set $y=u+v_1$ . We claim that the only $u-x$ -path of length 3 that produces a relevant contribution is $(u,z_1,z_2,x)$ . See Figure 2c for an illustration. First, assume $z_1 \notin \omega_1$ . Note that the only other paths of length 3 from u to x go through either $\textbf{0}$ or y. But $\{0,y\} \subseteq \langle {\mathscr {C}_0} \rangle$ , and so neither $\textbf{0}$ nor y can be a pivotal point. Hence, $E'(u,x;\ \mathscr {C}_0)_1\cap \{z_1\notin \omega_1\}$ enforces $\{\textbf{0},y\} \subseteq \omega_1$ . To get to x and avoid pivotality of any points in $\langle {\mathscr {C}_0} \rangle$ , at least two points in $\{v_1, v_2, z_1\}$ must be occupied, and the contribution to (5.7) is at most

  \begin{equation*} 2 \Omega p_c \big(\tfrac 12 \Omega^2(1+\mathcal O(\Omega^{-1}))\big) p_c^2 \binom{3}{2} p_c^2 = \mathcal O(\Omega^{-2}).\end{equation*}
  If $z_1\in \omega_1$ and $z_2 \notin\omega_1$ , then the only $u-x$ -path of length 3 through $z_1$ visits $v_2\in\langle {\mathscr {C}_0} \rangle$ . This gives a contribution of $\mathcal O(\Omega^{-2})$ by the same bound as above. We may turn to the case $z_i \in\omega_1$ for $i\in\{1,2\}$ . Now, under $\{v_1\in\omega_0, \{z_1, z_2\} \subseteq \omega_1\}$ , we can express $E'(u,x;\ \mathscr {C}_0)_1$ similarly to (5.9), replacing $y_i$ ( $i\in[3]$ ) by $z_i$ ( $i \in[2]$ ). Applying the same bounds, we obtain a contribution to (5.7) of
  \begin{equation*} 2\Omega p_c \big( \tfrac 12 \Omega^2 (1+\mathcal O(\Omega^{-1})) \big) \mathbb P\big(v_1\in \omega_0, \{z_1, z_2\} \subseteq\omega_1 \big) (1-\mathcal O(\Omega^{-1})) = \Omega^{-1} + \mathcal O(\Omega^{-2}).\end{equation*}

• Let $|u-x|=3,\| x\|_\infty=1$ and $x \notin\langle {\mathscr {C}_0} \rangle$ . Let $\gamma$ be a $u-x$ -path in $\omega_1$ . By assumption, there needs to be some $z\in\gamma$ with $z \in\langle {\mathscr {C}_0} \rangle$ . Consequently, z cannot be a pivotal point and so there needs to be another $u-x$ -path $\tilde{\gamma}$ in $\omega_1$ that contains a point $\tilde z \notin \gamma$ with $\tilde z\in\langle {\mathscr {C}_0} \rangle$ . Assume first that both $\gamma, \tilde \gamma$ are paths of length 3. If they are disjoint, then the contribution to (5.7) is at most $9\Omega^3 p_c^5 = \mathcal O(\Omega^{-2})$ . If they share their first vertex, then, in the terminology of Figure 2c, it must be either y or $z_1$ (otherwise $\textbf{0}$ is pivotal). W.l.o.g., $\tilde\gamma$ must then pass through $z_2$ and so $\tilde z=z_2 \in \langle {\mathscr {C}_0} \rangle$ needs to hold, and the contribution to (5.7) is at most $\Omega^3 p_c^4 \tau^{(3)}(z_2) = \mathcal O(\Omega^{-2})$ . Assume next that $\tilde\gamma$ is of length 5. As $\gamma$ and $\tilde\gamma$ share at most one internal vertex (and there are two internal vertices in $\gamma$ ), we count a factor of $p_c$ for the unique vertex of $\gamma$ , and the contribution to (5.7) is at most $18 \Omega^3 p_c^2 \tau^{(5)}(x-u) = \mathcal O(\Omega^{-2})$ . Similarly, when both $\gamma$ and $\tilde\gamma$ are of length at least 5, the contribution is $\mathcal O(\Omega^{-2})$ .

$\underline{\textrm{The case of}\ |u|=1, |x|=3\ \textrm{contributes}\ {2\Omega^{-1} + \mathcal O(\Omega^{-2})}}$ : Note that when $\{u \leftrightarrow{\,\, ({4}) \,\,} x\}_1$ , then the contribution to (5.6) is at most

(5.10) \begin{equation} p_c \sum_{u,x} \triangle^{(8)} (u,x,\textbf{0}) + p_c^2 \sum_{u,t,z,x} \triangleleft \hspace{-3.1 pt} \triangleright^{(9)} (u,t,z,x) = \mathcal O(\Omega^{-2}), \end{equation}

by Lemma 4.2. We can therefore focus on x with $|x-u|=2$ and $\{u \leftrightarrow{\,\, ({=2}) \,\,} x\}_1$ . Moreover, we can assume that there is no $u-x$ -path of length 4. Let $x=u+v_1+v_2$ , where $|v_1|=1=|v_2|$ , and assume first that $\dim\langle\!\langle {u,v_1,v_2} \rangle\!\rangle=3$ . There are $\tfrac 12 (\Omega-2)(\Omega-4)$ choices for x. Let $z_i=u+v_i$ be the two internal vertices of the two shortest $u-x$ -paths—see Figure 2a for an illustration.

We first claim that only $x\in\langle {\mathscr {C}_0} \rangle$ produces a relevant contribution. Indeed, if $x \notin\langle {\mathscr {C}_0} \rangle$ , and as there is no $u-x$ -path of length 4, we must have $z_i \in \omega_1 \cap \langle {\mathscr {C}_0} \rangle$ for $i \in\{1,2\}$ . For $\{\textbf{0} \longleftrightarrow z_i\}_0$ to hold, either $v_i \in\omega_0$ , or $\{ \textbf{0} \leftrightarrow{\,\, ({4}) \,\,} z_i \}_0$ , and so (5.7) is at most

\begin{align*}{ \Omega^3 p_c \mathbb P \Big(\{\{z_1,z_2\} \subseteq \omega_1\} \cap \big(\{\{v_1,v_2\} \subseteq \omega_0\} \cup \{ \textbf{0} \leftrightarrow{\,\, ({4}) \,\,} z_1\}_0 \cup \{\textbf{0} \leftrightarrow{\,\, ({4}) \,\,} z_2\}_0 \big) \Big) \\ = \Omega^3 p_c^3 \big( p_c^2 + \tau^{(4)}(z_1) + \tau^{(4)}(z_2) \big) = \mathcal O(\Omega^{-2}).}\end{align*}

Turning to $x\in \langle {\mathscr {C}_0} \rangle$ , note that when $\{z_1, z_2\} \subseteq \omega_1$ , then (5.7) is at most

\begin{equation*}\Omega^3 p_c \mathbb P \big( \{\textbf{0} \longleftrightarrow x\}_0 \cap \{\{z_1, z_2\} \subseteq \omega_1\}\big) = \Omega^3 p_c^3 \tau^{(3)}(x)=\mathcal O(\Omega^{-2}). \end{equation*}

W.l.o.g., we assume that $z_1\in \omega_1$ (and $z_2 \notin \omega_1$ ) and (by symmetry) count the contribution twice. Now, the contribution to (5.7) is equal to

(5.11) \begin{equation} \Omega(\Omega-2)(\Omega-4) p_c {\mathbb P} \Big(\{x \in \langle {\mathscr {C}_0} \rangle\}_0 \cap \{z_2\notin \omega_1\ni z_1 \}_1 \cap \big( \{z_1 \notin \langle {\mathscr {C}_0} \rangle_0 \cup \{z_1 \notin \textsf {Piv}({u,x})\}_1 \big) \Big). \end{equation}

If $v_1 \in\omega_0$ , then $z_1\in\langle {\mathscr {C}_0} \rangle$ and so $z_1$ cannot be pivotal, which, in turn, forces $\{u \leftrightarrow{\,\, ({4}) \,\,} x\}_1$ . But this was already shown to produce an $\mathcal O(\Omega^{-2})$ contribution. Further, if $\{\textbf{0} \leftrightarrow{\,\, ({5}) \,\,} x\}_0$ , then (5.11) is at most $\Omega^3 p_c^2 \tau^{(5)}(x) = \mathcal O(\Omega^{-2})$ , and so $\textbf{0}$ must be $\omega_0$ -connected to x by a path of length 3.

There are precisely two $\textbf{0}-x$ -paths of length 3 that use neither $v_1$ nor u, namely $\gamma_1=(\textbf{0},v_2,v_1+v_2,x)$ and $\gamma_2=(\textbf{0},v_2,z_2,x)$ . If both are occupied, the contribution is $\mathcal O(\Omega^{-2})$ . Note that

\begin{equation*} \mathbb P ( z_1 \notin \langle {\mathscr {C}_0} \rangle \mid \gamma_i\subseteq \omega_0 ) \geq 1- 3 \tau^{(2)}(z_1) = 1-\mathcal O(\Omega^{-1}), \end{equation*}

and so (5.11) becomes

\begin{equation*} \Omega^3 (1-\mathcal O(\Omega^{-1})) p_c \mathbb P \Big( \big( \cup_{i=1,2} \{ \gamma_i\subseteq \omega_0\}_0 \big), z_1 \in\omega_1 \Big) = 2\Omega^3 p_c^4 (1-\mathcal O(\Omega^{-1})) = 2\Omega^{-1} + \mathcal O(\Omega^{-2}).\end{equation*}

Finally, if $\dim\langle\!\langle {u,v_1,v_2} \rangle\!\rangle \leq 2$ , then the same bounds with at least one factor of $\Omega$ in the choice of x gives a contribution of $\mathcal O(\Omega^{-2})$ .

$\underline{\textrm{The case of}\ |u|=1, |x| \geq 4\ \textrm{contributes}\ \mathcal O(\Omega^{-2})}$ : The bound is the same as in (5.10).

Contributions of $|u|=2$ . If u is one of the $\Omega$ points with $|u|=2=\| u \|_\infty$ , then $\widehat\Pi^{(1)}$ is bounded by $\Omega p_c\sum_x \mathbb P(\textbf{0} \Longleftrightarrow u) \tau(u-x)$ . For fixed $j=|u-x|$ , this is bounded by

\begin{equation*} \Omega^{1+j} p_c \tau^{(2)}(u) \tau^{(4)}(u) \tau^{(j)}(x-u) = \mathcal O(\Omega^{-2}).\end{equation*}

We now show that we can impose some further restrictions on u and x. Recall the bound in (5.5), and observe that if $x\notin\langle {\mathscr {C}_0} \rangle$ , then

Similar considerations enforce that $|x| \leq 3$ and $|x-u| \leq 2$ as well as $\{\textbf{0} \Longleftrightarrow{\,\, ({\leq 4}) \,\,} u\}_0$ . Before going into the different cases, we note that there are $\tfrac 12\Omega(\Omega-2)$ choices for $u=v_1+v_2$ (where $|v_i|=1$ ), and on every choice, $\{v_1,v_2\} \subseteq \omega_0$ need to hold for a relevant contribution to arise. Taking all this into consideration, the contribution to $\widehat\Pi^{(1)}$ becomes

(5.12) \begin{equation} \tfrac 12 \Omega(\Omega-2) p_c^3 \sum_{x\in\mathbb Z^d} \textrm{1}_{\{|x| \leq 3, |u-x|\leq 2\}} \mathbb P \Big( \{x\in\langle {\mathscr {C}_0} \rangle\}_0 \cap E'(u,x;\ \mathscr {C}_0)_1 \mid \{v_1,v_2\}\subseteq \omega_0 \Big), \end{equation}

where $v_1$ and $v_2$ is a pair of arbitrary but fixed independent unit vectors (and $u=v_1+v_2$ ).

$\underline{\textrm{The case of}\ |u|=2, x=\textbf{0}\ \textrm{contributes}\ \mathcal O(\Omega^{-2})}$ : As $|u-x|=2$ , the contribution to (5.12) is at most $\Omega^2 p_c^3 \tau^{(2)}(x-u) = \mathcal O(\Omega^{-2})$ .

$\underline{\textrm{The case of}\ |u|=2, |x|=1\ \textrm{contributes}\ {\Omega^{-1} + \mathcal O(\Omega^{-2})}}$ : Note that we only need to consider $x \in\{v_1,v_2\}$ (otherwise $|u-x| =3$ ). For these choices of x, both $x\in\langle {\mathscr {C}_0} \rangle$ and $E'(u,x;\ \mathscr {C}_0)_1$ hold and the contribution to (5.12) is as claimed.

$\underline{\textrm{The case of}\ |u|=2, |x|=2\ \textrm{contributes}\ \Omega^{-1} + \mathcal O(\Omega^{-2})}$ : By the indicator in (5.12), we only consider $|x-u|=2$ . Let first $\|x\|_\infty=2$ . There are only two such points at distance 2 of u, and so the contribution to (5.12) is at most $\Omega^2 p_c^3 \tau^{(2)}(x-u) = \mathcal O(\Omega^{-2})$ .

Let thus x be one of the $2(\Omega-3)$ points with $\|x\|_\infty=1$ . W.l.o.g., we assume that $x=v_1+v_3$ , where $|v_3|=1$ . If $v_3=-v_2$ , then the contribution is bounded by $\Omega^2 p_c^3 \tau^{(2)}(x-u) = \mathcal O(\Omega^{-2})$ . Let x be one of the remaining $2(\Omega-4)$ points with $\dim\langle\!\langle {v_1,v_2,v_3} \rangle\!\rangle =3$ . As $x \sim v_1$ , the event $x\in\langle {\mathscr {C}_0} \rangle$ holds. We partition $E'(u,x;\ \mathscr {C}_0)_1$ into whether $\{u \leftrightarrow{\,\, ({=2}) \,\,} x\}_1$ or $\{u \leftrightarrow{\,\, ({\geq 4}) \,\,} x\}_1$ and see that in the latter case, the contribution to (5.12) is at most $\Omega^3 p_c^3 \tau^{(4)}(x-u) = \mathcal O(\Omega^{-2})$ .

For the existence of a path of length 2, either $v_1$ or $v_4:=x+v_2$ need to be $\omega_1$ -occupied. As $v_1\in\mathscr {C}_0$ , it cannot be a pivotal point for the $\omega_1$ -connection between u and x and there needs to be another path. The contribution to (5.12) is therefore at most $\Omega^3 p_c^4 \tau^{(2)}(x-u) = \mathcal O(\Omega^{-2})$ . We observe that

\begin{equation*}E'(u,x;\ \mathscr {C}_0)_1 \cap \{v_4\in \omega_1\} = \{v_4\in\omega_1\} \cap \Big(\{v_4\notin\textsf {Piv}({u,x})\}_1 \cup \{\textbf{0} \not \leftrightarrow v_4 \text{ in } \mathbb Z^d\setminus\{u\}\}_0 \Big). \end{equation*}

As previously, $\mathbb P(v_4\notin\textsf {Piv}({u,x})) = \mathcal O(\Omega^{-1})$ and $\mathbb P(\textbf{0} \not \leftrightarrow v_4 \text{ in } \mathbb Z^d\setminus\{u\}) = 1-\mathcal O(\Omega^{-1})$ , and so the contribution to (5.12) is

\begin{equation*}\Omega^3(1-\mathcal O(\Omega^{-1})) p_c^4 (1+\mathcal O(\Omega^{-1})) = \Omega^{-1} + \mathcal O(\Omega^{-2}).\end{equation*}

$\underline{\textrm{The case of}\ |u|=2, |x|=3\ \textrm{contributes}\ {\Omega^{-1} + \mathcal O(\Omega^{-2})}}$ : We only need to consider neighbours of u, otherwise $|u-x|\geq 3$ . Recall that for $|u-x|=1$ , the event $E'(u,x;\ \mathscr {C}_0)_1$ holds precisely when $x \in\langle {\mathscr {C}_0} \rangle$ . Under our conditioning, x must be connected to $\{\textbf{0},v_1,v_2\}$ . Note that there are two choices for x with $\|x\|_\infty=2$ . Since $\mathbb P(x\in\langle {\mathscr {C}_0} \rangle) \leq 3 \max_{y\in\{\textbf{0},v_1,v_2\}} \tau^{(2)}(x-y) = \mathcal O(\Omega^{-1})$ , we may focus on the $\Omega-2$ choices of x with $\|x\|_\infty=1$ .

Let $x=u+v_3$ and set $z_1:=v_1+v_3, z_2:=v_2+v_3$ . If $\{z_1,z_2\} \cap \omega_0 = \varnothing$ , then $\{\textbf{0} \leftrightarrow{\,\, ({5}) \,\,} x\}_0$ holds, and the contribution to (5.12) is at most $\Omega^3p_c^3 \max_{y\in\{\textbf{0},v_1,v_2\}} \tau^{(3)}(x-y) = \mathcal O(\Omega^{-2})$ . If $\{z_1,z_2\} \subset \omega_0$ , then the contribution to (5.12) is at most $\Omega^3 p_c^5 = \mathcal O(\Omega^{-2})$ .

We consider the case where $z_1\notin\omega_0\ni z_2$ and respect the other case with a factor of 2. The contribution to (5.12) is

\begin{equation*}\Omega^3(1+\mathcal O(\Omega^{-1}))\, p_c^4 \,(1+\mathcal O(\Omega^{-1})) = \Omega^{-1} + \mathcal O(\Omega^{-2}).\end{equation*}

This finishes the analysis of $\widehat\Pi^{(1)}$ .

5.3 Analysis of $\widehat\Pi^{(2)}$

Lemma 5.3 (Asymptotics of $\widehat\Pi^{(2)}$ ). As $d\to\infty$ ,

\begin{equation*} \widehat\Pi^{(2)} = 10 \Omega^{-1} + \mathcal O(\Omega^{-2}).\end{equation*}

Proof. For the proof, we recall that

(5.13) \begin{equation} \widehat\Pi^{(2)} = p_c^2 \sum_{u,v,x\in\mathbb Z^d} \mathbb P\Big( \{\textbf{0} \Longleftrightarrow u\}_0 \cap E'(u,v;\ \mathscr {C}_0)_1 \cap E'(v,x;\ \mathscr {C}_1)_2 \Big), \end{equation}

where $\mathscr {C}_0= \widetilde{\mathscr {C}}^{u}(\textbf{0};\ \omega_0)$ and $\mathscr {C}_1=\widetilde{\mathscr {C}}^v(u;\ \omega_1)$ . We first show that when either $v\notin\langle {\mathscr {C}_0} \rangle$ or $x\notin\langle {\mathscr {C}_1} \rangle$ , then the contribution to $\widehat{\Pi}^{(2)}$ is $\mathcal O(\Omega^{-2})$ . Indeed, by Lemma 2.10 and Proposition 2.8,

(5.14)

(5.15)

We expanded the third diagram in (5.14) to get the two diagrams of (5.15). We next show that only $|u|=1$ gives a relevant contribution. Indeed,

We can thus fix u to be an arbitrary neighbour of the origin and need to investigate

(5.16) \begin{equation} \Omega p_c^2 \sum_{v,x\in\mathbb Z^d} \mathbb P \Big( E'(u,v;\ \mathscr {C}_0)_1 \cap E'(v,x;\ \mathscr {C}_1)_2 \cap \{v\in\langle {\mathscr {C}_0} \rangle\}_0 \cap \{x\in\langle {\mathscr {C}_1} \rangle\}_1 \Big). \end{equation}

Before going into specific cases, we exclude some of them right away: When $|x| \vee |u-x| \geq 4$ , then the contribution to (5.16) is

by Lemma 4.2. In the above, a line decorated with a ‘ $\sim$ ’ symbol denotes a direct edge. Similarly, when $|v| \geq 3$ or $|x-v|\geq 3$ , the contribution to (5.16) is at most

We now investigate (5.16) by splitting the double sum over v and x. We organise this by considering the three main cases for $|v|\in\{0,1,2\}$ . An overview of the contributions is given in the following table:

Contributions of $v=\textbf{0}$ . The events $E'(u,v;\ \mathscr {C}_0)_1$ and $\{v\in \langle {\mathscr {C}_0} \rangle\}$ hold.

$\underline{\textrm{The case of}\ |x|=1\ \textrm{contributes}\ 2\Omega^{-1} + \mathcal O(\Omega^{-2})}$ : First, consider the choice of $x=u$ . It is easy to see that the event in (5.16) holds and the contribution is $\Omega p_c^2 = \Omega^{-1} + \mathcal O(\Omega^{-2})$ .

Consider $0\sim x\neq u$ . As $v\sim x$ , we have $E'(v,x;\ \mathscr {C}_1)_2 = \{x \in \langle {\mathscr {C}_1} \rangle\}_1$ . If $x=-u$ , then $\{x \in \langle {\mathscr {C}_1} \rangle\}_1 \subseteq \{u \leftrightarrow{\,\, ({4}) \,\,} x\}_1$ and we receive a contribution of order $\mathcal O(\Omega^{-2})$ . Consider now one of the $\Omega-2$ remaining choices for x and set $z=u+x$ . Then

\begin{equation*} \mathbb P(x \in \langle {\mathscr {C}_1} \rangle) = \mathbb P(z \in\omega_1) + \mathbb P(z \notin\omega_1, x \in \langle {\mathscr {C}_1} \rangle) = p_c + \mathcal O(\tau^{(4)}(x-u)) = p_c + \mathcal O(\Omega^{-2}),\end{equation*}

yielding a contribution to (5.16) of $\Omega(\Omega-2)p_c^3 + \mathcal O(\Omega^{-2}) = \Omega^{-1}+\mathcal O(\Omega^{-2})$ .

$\underline{\textrm{The case of}\ |x|=2\ \textrm{contributes}\ {\Omega^{-1} + \mathcal O(\Omega^{-2})}}$ : If $|u-x|=3$ , then the contribution to (5.16) is bounded by $\Omega^3 p_c^2 \tau^{(2)}(x-v) \tau^{(3)}(u-x) =\mathcal O(\Omega^{-2})$ . Similarly, if $x=2u$ , we obtain a bound of $\Omega p_c^2 \tau^{(2)}(x-v) = \mathcal O(\Omega^{-2})$ . Let therefore x be one of the $\Omega-2$ remaining neighbours of u and note that $\{x\in\langle {\mathscr {C}_1} \rangle\}$ holds.

We set $z=x-u$ . If $z\notin\omega_2$ , then $E'(v,x;\ \mathscr {C}_1)_2 \subseteq \{ v \leftrightarrow{\,\, ({4}) \,\,} x\}_2$ by Observation 4.3, and the contribution to (5.16) is at most $\Omega^2 p_c^2 \tau^{(4)}(x-v) = \mathcal O(\Omega^{-2})$ . If $z\in\omega_2$ , then $E'(v,x;\ \mathscr {C}_1)_2 = \{z \notin\textsf {Piv}({v,x})\}_2 \cup \{z \notin\langle {\mathscr {C}_1} \rangle \}_1$ . By a similar argument to the one below (5.9), the contribution to (5.16) becomes

\begin{equation*} \Omega(\Omega-2) p_c^2 \mathbb P \Big(\{z\in \omega_2 \} \cap \big(\{z \notin\textsf {Piv}({v,x})\}_2 \cup \{z \notin\langle {\mathscr {C}_1} \rangle \}_1\big)\Big) = \Omega^2 p_c^3 (1-\mathcal O(\Omega^{-1})) = \Omega^{-1} + \mathcal O(\Omega^{-2}). \end{equation*}

$\underline{\textrm{The case of}\ |x|=3\ \textrm{contributes}\ {\mathcal O(\Omega^{-2})}}$ : Distinguishing between $|u-x|=4$ (at most $\Omega^3$ choices for x) and $|u-x|=2$ (at most $\Omega^2$ choices), the contribution to (5.16) is at most

\begin{equation*}\Omega p_c^2 \tau^{(3)}(x-v) \big( \Omega^3 \tau^{(4)}(u-x) + \Omega^2 \tau^{(2)}(u-x) \big) = \mathcal O(\Omega^{-2}).\end{equation*}

Contributions of $|v|=1$ . Let us first consider $v=-u$ and show that this case contributes $\mathcal O(\Omega^{-2})$ . Indeed, $E'(u,v;\ \mathscr {C}_0)_1 \subseteq \{u \leftrightarrow{\,\, ({4}) \,\,} v\}_1$ by Observation 4.3. With the further inclusion $E'(v,x;\ \mathscr {C}_1)_2 \cap \{x \in\langle {\mathscr {C}_1} \rangle\} \subseteq \{v \longleftrightarrow x\}_2$ , we have that the contribution to (5.16) is at most

\begin{align*}{ \Omega p_c^2 \tau^{(4)} (u-v) \Big( \sum_{|x|=1} \tau^{(2)}(x-v) + \sum_{x: v\sim x} 1 + \sum_{|x|=2,|x-v|=3} \tau^{(3)}(x-v) \\ \qquad + \sum_{|x|=3,|x-v|=2} \tau^{(2)}(x-v) +\sum_{|x|=3,|x-v|=4} \tau^{(4)}(x-v) \Big) \\ \leq \mathcal O(\Omega^{-3}) \Big(\Omega \mathcal O(\Omega^{-1}) + \Omega + \Omega^2 \mathcal O(\Omega^{-2}) + \Omega^2 \mathcal O(\Omega^{-1}) + \Omega^3 \mathcal O(\Omega^{-3}) \Big) = \mathcal O(\Omega^{-2}).}\end{align*}

We may therefore take $v\neq \pm u$ to be one of the $\Omega-2$ remaining neighbours of the origin. Set $t=v+u$ . We first claim that $t \notin \omega_1$ results in an $\mathcal O(\Omega^{-2})$ contribution. Note that, by Observation 4.3, $E'(u,v;\ \mathscr {C}_0)_1 \cap \{t \notin \omega_1\} \subseteq \{u \leftrightarrow{\,\, ({4}) \,\,} v\}_1$ . As there is only one choice of x such that $u \sim x \sim v$ and at most $\Omega$ choices such that $|x|=3$ and $x \sim v$ , we can bound (5.16) by

\begin{align*}{ \Omega^2 p_c^2 \sum_{x\in\mathbb Z^d} \Big( \tau^{(4)}(v-u) \big(\textrm{1}_{\{x=\textbf{0}\}} + \textrm{1}_{\{|x|=1\}} \tau^{(2)}(x-v) + \textrm{1}_{\{|x|=2, u\sim x \sim v\}} \\ \quad + \textrm{1}_{\{|x|=3, |u-x|=2=|v-x|\}} \tau^{(2)}(x-v) \big) + \textrm{1}_{\{|x|=2, v \sim x\}} \mathbb P \big(E'(u,v;\ \mathscr {C}_0)_1 \cap \{ t \notin\omega_1 \} \cap \{x \in \langle {\mathscr {C}_1} \rangle\} \big) \Big) \\ \leq \mathcal O(\Omega^{-2}) \big( 2 + 2\Omega \tau^{(2)}(x-v)\big) + \mathcal O(1) \sum_{|x|=2, x \sim v} \mathbb P^{(2)} \big(E'(u,v;\ \mathscr {C}_0)_1 \cap \{ t \notin\omega_1 \} \cap \{x \in \langle {\mathscr {C}_1} \rangle\} \big). }\end{align*}

It remains to bound the last probability. There are at most $\Omega$ choices for x. If $\{u \leftrightarrow{\,\, ({5}) \,\,} x\}$ , then the contribution is $\mathcal O(\Omega^{-2})$ . Note that the $u-v$ -path in $\omega_1$ cannot use and is independent of the status of $\textbf{0}$ , as the origin may not be a pivotal point. Hence, if $\textbf{0} \in\omega_1$ , the contribution is at most $\Omega p_c \tau^{(4)}(v-u) = \mathcal O(\Omega^{-2})$ . We therefore assume $\textbf{0}\notin\omega_1$ and aim to bound

(5.17) \begin{equation} \Omega \mathbb P \big(\{u \leftrightarrow{\,\, ({4}) \,\,} v\}_1 \cap \{ \textbf{0},t\notin \omega_1 \} \cap \{ u \leftrightarrow{\,\, ({\leq 3}) \,\,} x \}_1 \big) \end{equation}

When avoiding $\textbf{0}$ and t, there are only two $u-x$ -paths of length 3, namely $\gamma_1=(u,y,z,x)$ and $\gamma_2=(u,y,y-u,x)$ , where $y:=x+u-v$ and $z:=y+v$ . See Figure 3a for an illustration. But now, (5.17) is bounded by:

\begin{align*}{ \Omega \mathbb P \Big( \{\textbf{0},t\notin \omega_1\} \cap \bigcup_{i=1,2} \bigcup_{s \in \gamma_i \setminus\{x\} } \{\gamma_i \subseteq \omega_1\} \circ \{s \longleftrightarrow v\}_1 \Big) \\ \leq 2 \Omega p_c^2 \big( \tau^{(4)}(v-u) + \tau^{(3)}(y-v) + 2\tau^{(2)}(z-v) \big) = \mathcal O(\Omega^{-2}). }\end{align*}

As a consequence, we can focus on $t\in\omega_1$ , and (5.16) reduces to

\begin{equation*} \Omega(\Omega-2) p_c^2 \sum_{x\in\mathbb Z^d} \mathbb P \Big( E'(u,v;\ \mathscr {C}_0)_1 \cap E'(v,x;\ \mathscr {C}_1)_2 \cap \{t\in\omega_1, x\in\langle {\mathscr {C}_1} \rangle\}_1 \Big). \end{equation*}

But under $t \in\omega_1$ , we have $E'(u,v;\ \mathscr {C}_0)_1 = \{t \notin\textsf {Piv}({u,v})\}_1 \cup \{t \notin \langle {\mathscr {C}_0} \rangle\}_0$ . The latter event has probability $1-\mathcal O(\Omega^{-1})$ , and so we can can instead investigate

(5.18) \begin{equation} \Omega^2 p_c^2 (1-\mathcal O(\Omega^{-1})) \sum_{x\in\mathbb Z^d} \mathbb P \Big( E'(v,x;\ \mathscr {C}_1)_2 \cap \{t\in\omega_1, x\in\langle {\mathscr {C}_1} \rangle\}_1 \Big), \end{equation}

where u and v are two arbitrary (but fixed) neighbours of $\textbf{0}$ (satisfying ( $u \neq \pm v$ ).

$\underline{\textrm{The contribution of}\ x=\textbf{0}\ \textrm{is}\ {\Omega^{-1} + \mathcal O(\Omega^{-2})}}$ : Note that $x \in \langle {\mathscr {C}_1} \rangle$ holds, and so does $E'(v,x;\ \mathscr {C}_1)_2$ . Hence, the contribution to (5.18) is $\Omega^{-1} + \mathcal O(\Omega^{-2})$ .

$\underline{\textrm{The contribution of}\ |x|=1\ \textrm{is}\ {\mathcal O(\Omega^{-2})}}$ : If $x \in\{\pm u, -v\}$ , we can bound the contribution to (5.18) by $\Omega^2 p_c^2 \tau^{(2)}(u-v)\tau^{(2)}(x-v) = \mathcal O(\Omega^{-2})$ (as both $\{v \longleftrightarrow x\}_2$ and $\{u\longleftrightarrow v\}_1$ need to hold). Consider thus one of the $\Omega-4$ choices for x satisfying $\dim\langle\!\langle {u,v,x} \rangle\!\rangle=3$ . Conditional on $t \in \omega_1$ , we have $\{x\in\langle {\mathscr {C}_1} \rangle\}_1 \subseteq \{u \leftrightarrow{\,\, ({2}) \,\,} x\}_1 \cup \{t \leftrightarrow{\,\, ({3}) \,\,} x\}_1$ , and so the contribution is at most

\begin{equation*}\Omega^3 p_c^3 \tau_p^{(2)}(x-v) \big( \tau_p^{(2)}(x-u) + \tau_p^{(3)}(x-t) \big) = \mathcal O(\Omega^{-2}).\end{equation*}

$\underline{\textrm{The contribution of}\ |x|=2\ \textrm{is}\ {2\Omega^{-1}+\mathcal O(\Omega^{-2})}}$ : We can restrict to the choices of x where $v\sim x$ by the considerations made in the beginning of the proof.

• Let $x \sim u$ . There is only one choice for x such that $|u-x|=|v-x|=1$ , namely $x=t$ . For this choice, $E'(v,x;\ \mathscr {C}_1)_2$ certainly holds, and also $x\in\langle {\mathscr {C}_1} \rangle$ . We get a contribution of $\Omega^{-1}+\mathcal O(\Omega^{-2})$ .

• Let $x \not\sim u$ . There are $\Omega-2$ choices for x. We first exclude $x=v-u$ . As $\mathbb P(x \in \langle {\mathscr {C}_1} \rangle \mid t\in\omega_1) \leq \tau^{(4)}(x-t) + \tau^{(3)}(x-u) = \mathcal O(\Omega^{-2})$ , the contribution in total is $\mathcal O(\Omega^{-2})$ .

  Let now x be one of the $\Omega-3$ remaining neighbours of v. As $v\sim x$ , we have $E'(v,x;\ \mathscr {C}_1)_2 = \{x \in \langle {\mathscr {C}_1} \rangle\}$ . We set $z=x+u$ (see Figure 3b) and assume first that $z \notin\omega_1$ . Then

  \begin{equation*} \{z\notin\omega_1 \ni t, x\in\langle {\mathscr {C}_1} \rangle\} \subseteq \{z\notin\omega_1 \ni t\} \cap \big(\{u \leftrightarrow{\,\, ({3}) \,\,} x \text{ off } \{t\}\} \cup \{t \leftrightarrow{\,\, ({4}) \,\,} x \} \big) \end{equation*}
  and the contribution to (5.18) is at most $\Omega^2 p_c^3 (1-\mathcal O(\Omega^{-1})) (\tau^{(3)}(x-u) + \tau^{(4)}(x-t)) = \mathcal O(\Omega^{-2})$ . On the other hand, if $z\in\omega_1$ , then $x \in \langle {\mathscr {C}_1} \rangle$ holds and (5.18) becomes
  \begin{equation*} \Omega^2 p_c^2 (1-\mathcal O(\Omega^{-1})) (\Omega-3) \mathbb P(t,z \in \omega_1) = \Omega^{-1} + \mathcal O(\Omega^{-2}).\end{equation*}
  

  Figure 3. An illustration of several appearing cases for $|v|=1$ . In (a), the two paths from u to x of length 3 that avoid $\textbf{0}$ and t are drawn. In (b), the path along t, z which ensures $x \in \langle {\mathscr {C}_1} \rangle$ for a contribution of $\Omega^{-1}$ is drawn. In (c), the scenario $|x-u|=2=|v-x|$ is shown, and the path along z ensuring $\{v \longleftrightarrow x\}_2$ is drawn in black.

$\underline{\textrm{The contribution of}\ |x|=3\ \textrm{is}\ {\Omega^{-1} + \mathcal O(\Omega^{-2})}}$ : There are at most $\Omega^3$ choices for x such that $|u-x|=|v-x|=4$ and there are at most $2\Omega^2$ choices where $|x-u|\neq |v-x|$ . The contribution of those x to (5.18) is therefore bounded by

\begin{equation*}\Omega^2 p_c^2 \Big(\sum_{|x|=3} \tau^{(4)}(x-v) \tau^{(4)}(u-x) + 2 \sum_{|x-v|=2\neq |u-x|} \tau^{(2)}(x-v) \tau^{(4)}(u-x) \Big) = \mathcal O(\Omega^{-2}). \end{equation*}

It remains to investigate those x with $|u-x| = 2 = |v-x|$ . This is only possible when $x\sim t$ . Let first $x = 2u+v$ . By Observation 4.3, $E'(v,x;\ \mathscr {C}_1)_2 \cap \{t\in\omega_1\} \subseteq \{v \leftrightarrow{\,\, ({4}) \,\,} x\}$ , and (5.18) is at most $\Omega^2 p_c^2 \tau^{(4)}(x-v) = \mathcal O(\Omega^{-2})$ .

Let now x be one of the $\Omega-3$ remaining neighbours of t (note that either $\|x\|_\infty=1$ or $x=2v+u$ ). We set $z:=x-u$ and point to Figure 3b for an illustration. As t is occupied in $\omega_1$ , we have $x \in\langle {\mathscr {C}_1} \rangle$ . Assume now $z\notin\omega_2$ . By Observation 4.3, $E'(v,x;\ \mathscr {C}_1)_2 \subseteq \{ v \leftrightarrow{\,\, ({4}) \,\,} x\}_2$ and the contribution to (5.18) is at most $\Omega^2 p_c^3 \tau^{(4)}(x-v) = \mathcal O(\Omega^{-2})$ . On the other hand, if $z\in\omega_2$ , (5.18) becomes

\begin{equation*} (1+\mathcal O(\Omega^{-1})) (\Omega-3) \mathbb P \big( \{t\in\omega_1, z \in\omega_2\} \cap \big(\{z \notin \langle {\mathscr {C}_1} \rangle\}_1 \cup \{z \notin\textsf {Piv}({v,x})\}_2 \big) \big) = \Omega^{-1} + \mathcal O(\Omega^{-2}). \end{equation*}

Again, we have used that $\{z \notin \langle {\mathscr {C}_1} \rangle\}_1$ has probability $1-\mathcal O(\Omega^{-1})$ conditional on $t\in\omega_1$ .

Contributions of $|v|=2$ . We first show that when $|u-v| =3$ , no relevant contributions arise. Indeed, for those v, (5.13) is at most

Moreover, $v=2u$ implies $\{v\in\langle {\mathscr {C}_0} \rangle\}_0 \subseteq \{\textbf{0} \leftrightarrow{\,\, ({4}) \,\,} v\}_0$ . We can thus bound the contribution to (5.16) by $\Omega p_c \triangle_p^{\bullet} \tau^{(4)}(v) = \mathcal O(\Omega^{-2})$ . Let v be one of the $\Omega-2$ remaining neighbours of u, implying $E'(u,v;\ \mathscr {C}_0)=\{v\in\langle {\mathscr {C}_0} \rangle\}_0$ . Let $z=v-u$ . Then for $v \in\langle {\mathscr {C}_0} \rangle$ to hold, either $z \in\omega_0$ or there must be a path of length at least 4. In the latter case, we can bound (5.16) by $p_c^2 \sum_{u,v,t,x} \triangleleft \hspace{-1 pt} \triangleright^{(9)}(u,t,v,x) = \mathcal O(\Omega^{-2})$ . We can therefore restrict to investigating

(5.19) \begin{equation} \Omega(\Omega-2) p_c^3 \sum_{x\in\mathbb Z^d} \mathbb P \Big( E'(v,x;\ \mathscr {C}_1)_2 \cap \{x\in\langle {\mathscr {C}_1} \rangle\}_1 \Big), \end{equation}

where u is an arbitrary (but fixed) neighbour of $\textbf{0}$ and $v \notin\{\textbf{0}, 2u\}$ is some fixed neighbour of u.

$\underline{\textrm{The contribution of}\ x=\textbf{0}\ \textrm{is}\ {\mathcal O(\Omega^{-2})}}$ : As $\{\textbf{0} \longleftrightarrow v\}_2$ needs to hold, we get a bound on (5.19) by $\Omega^2 p_c^3 \tau^{(2)}(v) = \mathcal O(\Omega^{-2})$ .

$\underline{\textrm{The contribution of}\ |x|=1\ \textrm{is}\ {\Omega^{-1}+\mathcal O(\Omega^{-2})}}$ : We only need to consider $|v-x|=1$ , and there are two such choices for x. If $x=v-u$ , then the contribution is bounded by $\Omega^2p_c^3 \tau^{(2)}(u-x) = \mathcal O(\Omega^{-2})$ .

On the other hand, if $x=u$ , both $E'(v,x;\ \mathscr {C}_1)_2$ and $\{x \in\langle {\mathscr {C}_1} \rangle\}_1$ hold and the contribution to (5.19) is $\Omega^{-1} + \mathcal O(\Omega^{-2})$ .

$\underline{\textrm{The contribution of}\ |x|=2\ \textrm{is}\ {\Omega^{-1}+\mathcal O(\Omega^{-2})}}$ : Note that only $|v-x|=2$ may produce relevant contributions. Writing $v=u+z$ , we first consider $x=u-z$ . Again, $E'(v,x;\ \mathscr {C}_1)_2 \subseteq \{v \leftrightarrow{\,\, ({4}) \,\,} x\}_2$ by Observation 4.3, and so the contribution to (5.19) is at most $\Omega^2 p_c^2 \tau^{(4)}(v-x) = \mathcal O(\Omega^{-2})$ . Similarly, If $|u-x|=3$ , the contribution is at most $\Omega^3 p_c^3 \tau^{(2)}(v-x) \tau^{(3)}(u-x) = \mathcal O(\Omega^{-2})$ .

Let now y be one of the $\Omega-4$ unit vectors satisfying $\dim\langle\!\langle {u,z,y} \rangle\!\rangle=3$ . Write $x=u+y$ and set $t=x+z=v+y$ . We claim that we only get a relevant contribution if $t\in\omega_2$ : As $\{t\notin\omega_2\} \subseteq \{v \leftrightarrow{\,\, ({4}) \,\,} x\}_2$ by Observation 4.3, this gives a bound on the contribution to (5.19) by $\Omega^3 p_c^3 \tau^{(4)}(x-v) = \mathcal O(\Omega^{-2})$ . Under $t\in\omega_2$ , (5.19) becomes

(5.20) \begin{align} \Omega^3(1-\mathcal O(\Omega^{-1})) p_c^3 \mathbb P \Big(\{t\in\omega_2\} \cap \big(\{t \notin\textsf {Piv}({v,x})\}_2 \cup \{t \notin\langle {\mathscr {C}_1} \rangle\}_1 \big) \Big) \\ = \Omega^3(1-\mathcal O(\Omega^{-1})) p_c^4 (1-\mathcal O(\Omega^{-1})) = \Omega^{-1} + \mathcal O(\Omega^{-2}). \end{align}

$\underline{\textrm{The contribution of}\ |x|=3\ \textrm{is}\ {\Omega^{-1}+\mathcal O(\Omega^{-2})}}$ : We only need to consider terms where $|v-x|=1$ . Let $x=v+y$ , where $|y|=1$ . If $y=z$ , then $\{x\in\langle {\mathscr {C}_1} \rangle\} \subseteq \{v \leftrightarrow{\,\, ({4}) \,\,} x\}_1$ and the contribution to (5.19) is $\mathcal O(\Omega^{-3})$ . For the other $\Omega-2$ choices for x, we set $t=u+y$ . When $t\notin\omega_2$ , we require $\{x\in\langle {\mathscr {C}_1} \rangle\} \subseteq \{v \leftrightarrow{\,\, ({4}) \,\,} x\}_1$ and the contribution is $\mathcal O(\Omega^{-2})$ . When $t\in\omega_2$ , the contribution is identical to (5.20) and hence $\Omega^{-1} + \mathcal O(\Omega^{-2})$ .