$\mathrm{Red}=\\{u_{1},\dots,u_{\mathcal{R}}\\}$, which are all measurable
w.r.t $\mathcal{F}_{t_{1}}$. Let the (random) distribution
$\mathfrak{L}(x,j)$, for $x\in\mathcal{X}$, $0\leq j\leq\mathcal{R}$, be
defined by

$\mathfrak{L}(x,j)=\nu_{\mathrm{Blue}\cup\\{u_{1},\dots,u_{j}\\}}\otimes\delta_{X^{x}_{t_{1}}(\mathrm{Red}\setminus\\{u_{1},\dots,u_{j}\\})}\otimes\delta_{X_{t_{1}}(\mathrm{Green})},$

which means that $\mathfrak{L}(x,j)$ is the law of the random configuration
obtained by resampling the spins on the sites in $\\{u_{1},\dots,u_{j}\\}$ of
$X^{x}_{t_{1}}$ according to i.i.d. $\mathfrak{R}_{\bar{\rho}}$ independent of
$X^{x}_{t_{1}}$. By Lemma 28, almost surely, $\mathfrak{L}(x,0)$ is exactly
the conditional distribution of $X^{x}_{t_{1}}$ given $\mathcal{F}_{t_{1}}$,
and

$\mathfrak{L}(x,\mathcal{R})=\nu_{\mathrm{Blue}\cup\mathrm{Red}}\otimes\delta_{X_{t_{1}}(\mathrm{Green})},$

which does not depend on $x$. So we can write $\mathfrak{L}(\mathcal{R})$ for
$\mathfrak{L}(x,\mathcal{R})$. The point here is that the sequence
$(\mathfrak{L}(x,j))_{j=0,\dots,\mathcal{R}}$ forms a “path” from
$\mathfrak{L}(x,0)$ to $\mathfrak{L}(\mathcal{R})$. We see that, for any
$x_{1},x_{2}\in\mathcal{X}$,

$\displaystyle\textrm{d}_{\textsc{tv}}\left(\mathbb{P}_{x_{1}}\left[X_{t}\in\cdot\right],\mathbb{P}_{x_{2}}\left[X_{t}\in\cdot\right]\right)$
$\displaystyle\leq\mathbb{E}\left[\left\|\mathbb{P}_{x_{1}}\left[X_{t}\in\cdot|\mathcal{F}_{t_{1}}\right]-\mathbb{P}_{x_{2}}\left[X_{t}\in\cdot|\mathcal{F}_{t_{1}}\right]\right\|_{TV}\right]$
$\displaystyle=\mathbb{E}\left[\left\|\mathbb{P}_{\mathfrak{L}(x_{1},0)}\left[X_{1}\in\cdot\right]-\mathbb{P}_{\mathfrak{L}(x_{2},0)}\left[X_{1}\in\cdot\right]\right\|_{TV}\right]$
$\displaystyle\leq
2\max_{x\in\mathcal{X}}\mathbb{E}\left[\left\|\mathbb{P}_{\mathfrak{L}(x,0)}\left[X_{1}\in\cdot\right]-\mathbb{P}_{\mathfrak{L}(\mathcal{R})}\left[X_{1}\in\cdot\right]\right\|_{TV}\right].$
(69)

By the triangle inequality, for any $x\in\mathcal{X}$,

$\displaystyle\left\|\mathbb{P}_{\mathfrak{L}(x,0)}\left[X_{1}\in\cdot\right]-\mathbb{P}_{\mathfrak{L}(\mathcal{R})}\left[X_{1}\in\cdot\right]\right\|_{TV}\leq\sum_{j=0}^{\mathcal{R}-1}\left\|\mathbb{P}_{\mathfrak{L}(x,j)}\left[X_{1}\in\cdot\right]-\mathbb{P}_{\mathfrak{L}(x,j+1)}\left[X_{1}\in\cdot\right]\right\|_{TV}.$
(70)

By Proposition 31, there exists a constant $\kappa$ such that

$\displaystyle\left\|\mathbb{P}_{\mathfrak{L}(x,j)}\left[X_{1}\in\cdot\right]-\mathbb{P}_{\mathfrak{L}(x,j+1)}\left[X_{1}\in\cdot\right]\right\|_{TV}$
$\displaystyle\leq\kappa\left(\dfrac{\log
L}{\sqrt{L}}+\sqrt{\mathbb{P}_{\mathrm{Blue}\cup\\{u_{1},\dots,u_{j+1}\\}}\left[\exists
s\in[0,1]:Z_{s}\in\mathrm{BAD}\right]}\right)$
$\displaystyle\leq\kappa\left(\dfrac{\log
L}{\sqrt{L}}+\sqrt{\mathbb{P}_{\mathrm{Blue}}\left[\exists
s\in[0,1]:Z_{s}\in\mathrm{BAD}\right]}\right).$ (71)

We have used the attractiveness of $Z$ and the fact that $\mathrm{BAD}$ is a
decreasing subset of $\Lambda_{L}$ in the last inequality. The equations
(6.1), (70), (6.1) together imply

$\displaystyle\textrm{d}_{\textsc{tv}}\left(\mathbb{P}_{x_{1}}\left[X_{t}\in\cdot\right],\mathbb{P}_{x_{2}}\left[X_{t}\in\cdot\right]\right)$
$\displaystyle\leq 2\mathbb{E}\left[\mathcal{R}\,\kappa\left(\dfrac{\log
L}{\sqrt{L}}+\sqrt{\mathbb{P}_{\mathrm{Blue}}\left[\exists
s\in[0,1]:Z_{s}\in\mathrm{BAD}\right]}\right)\right]$ $\displaystyle\leq
2\dfrac{\kappa\log
L}{\sqrt{L}}\mathbb{E}\left[\mathcal{R}\right]+2\kappa\mathbb{E}\left[\mathcal{R}\sqrt{\mathbb{P}_{\mathrm{Blue}}\left[\exists
s\in[0,1]:Z_{s}\in\mathrm{BAD}\right]}\right].$ (72)

The latter term in the last expression above can be bounded by the Cauchy-
Schwarz inequality:

$\displaystyle\mathbb{E}\left[\mathcal{R}\sqrt{\mathbb{P}_{\mathrm{Blue}}\left[\exists
s\in[0,1]:Z_{s}\in\mathrm{BAD}\right]}\right]$
$\displaystyle\leq\sqrt{\mathbb{E}\left[\mathcal{R}^{2}\right]\mathbb{E}\left[\mathbb{P}_{\mathrm{Blue}}\left[\exists
s\in[0,1]:Z_{s}\in\mathrm{BAD}\right]\right]}$
$\displaystyle\leq\sqrt{L^{2}\mathbb{P}_{\O}\left[\exists
s\in[t_{1},t_{1}+1]:Z_{s}\in\mathrm{BAD}\right]}.$

Here, we have used the facts that $\mathcal{R}\leq L$,
$\mathrm{Blue}=Z^{\O}_{t_{1}}$ (by definition) and the Markov property of the
process $Z^{\O}$ at time $t_{1}$. Then, by Lemma 30, provided that
$t_{1}\geq\beta_{3}$,

$\displaystyle\mathbb{E}\left[\mathcal{R}\sqrt{\mathbb{P}_{\mathrm{Blue}}\left[\exists
s\in[0,1]:Z_{s}\in\mathrm{BAD}\right]}\right]$
$\displaystyle\leq\sqrt{L^{2}\mathcal{O}\left(1/L^{6}\right)}$
$\displaystyle=\mathcal{O}\left(1/L^{2}\right).$

It remains to estimate the first term in (72). Thanks to Lemma 29 and Lemma
18, we can take $t_{1}=\dfrac{\log
L}{2|R^{\prime}(\rho_{*})|}+\dfrac{2\log\log L}{|R^{\prime}(\rho_{*})|}$ to
make $\dfrac{\log
L}{\sqrt{L}}\mathbb{E}\left[\mathcal{R}\right]=\mathcal{O}\left(1/\log
L\right)$. Clearly, $t_{1}\geq\beta_{3}$ when $L$ is large enough. This
implies that

$\displaystyle\textrm{d}_{\textsc{tv}}\left(\mathbb{P}_{x_{1}}\left[X_{t}\in\cdot\right],\mathbb{P}_{x_{2}}\left[X_{t}\in\cdot\right]\right)=\mathcal{O}\left(1/\log
L\right)+\mathcal{O}\left(1/L^{2}\right)=\mathcal{O}\left(1/\log L\right),$

for any $x_{1},x_{2}\in\mathcal{X}$. This finishes our proof.

∎

The rest of the paper is devoted to proving Lemma 30 and Proposition 31. In
Subsection 6.2, we prove Lemma 30, and finally, in Subsection 6.3 and
Subsection 6.4, we prove Proposition 31.

###  6.2 Proof of Lemma 30: the bad set is rarely visited

First, we need the following lemmas.

######  Lemma 32 (the bad set has small mass).

There exist positive constants $\beta_{2}$ and $\beta_{3}$ such that for any
$t\geq\beta_{3}$, and for any deterministic subset $E$ of $\Lambda_{L}$ of
cardinality at least $\left\lceil\beta_{2}\log L\right\rceil$,

$\max_{z\in\mathcal{Z}}\mathbb{P}_{z}\left[Z_{t}\cap
E=\O\right]=\mathcal{O}\left(1/L^{10}\right).$ (73)

######  Lemma 33 (Inertia in a short time).

Let $\tau_{out}$ be the time at which the process $Z$ jumps out of the initial
state $Z_{0}$:

$\tau_{out}=\inf\left\\{t\geq 0:Z_{t}\neq Z_{0}\right\\}.$ (74)

Then there is a constant $\beta>0$ such that,

$\min_{z\in\mathcal{Z}}\mathbb{P}_{z}\left[\tau_{out}>\dfrac{1}{L^{3}}\right]\geq\beta.$
(75)

Now we can prove Lemma 30.

###### Proof of Lemma 30.

Let $z$ be an element of $\mathcal{Z}$. Let $\beta_{2},\beta_{3}$ be as in
Lemma 32. We write $\mathrm{BAD}$ for $\mathrm{BAD}_{\beta_{2}}$ to lighten
the notation. We apply the union bound in Lemma 32 for the subsets
$E_{j}=\left\llbracket j\left\lceil\beta_{2}\log
L\right\rceil,(j+1)\left\lceil\beta_{2}\log
L\right\rceil-1\right\rrbracket,\,j\in\left\llbracket
0,\left\lceil\dfrac{L}{\left\lceil\beta_{2}\log
L\right\rceil}\right\rceil\right\rrbracket$, to obtain:

$\forall
t\geq\beta_{3},\;\mathbb{P}_{z}\left[Z_{t}\in\mathrm{BAD}\right]=\mathcal{O}\left(1/L^{9}\right).$

This implies that for any $t\geq\beta_{3}$,

$\mathbb{E}_{z}\left[\int_{t}^{t+1+1/L^{3}}\mathds{1}_{\left\\{Z_{s}\in\mathrm{BAD}\right\\}}\mathrm{d}s\right]=\int_{t}^{t+1+1/L^{3}}\mathbb{P}_{z}\left[Z_{s}\in\mathrm{BAD}\right]\mathrm{d}s=\mathcal{O}\left(1/L^{9}\right).$
(76)

Let

$\displaystyle\tau_{\mathrm{BAD}}:=\inf\left\\{t\geq
0:Z_{t}\in\mathrm{BAD}\right\\}.$

By the Markov property,

$\displaystyle\mathbb{E}_{z}\left[\int_{t}^{t+1+1/L^{3}}\mathds{1}_{\left\\{Z_{s}\in\mathrm{BAD}\right\\}}\mathrm{d}s\right]$
$\displaystyle=\mathbb{E}_{z}\left[\mathbb{E}_{Z_{t}}\left[\int_{0}^{1+1/L^{3}}\mathds{1}_{\left\\{Z_{s}\in\mathrm{BAD}\right\\}}\mathrm{d}s\right]\right]$
$\displaystyle\geq\mathbb{E}_{z}\left[\mathbb{E}_{Z_{t}}\left[\mathds{1}_{\left\\{\tau_{\mathrm{BAD}}\leq
1\right\\}}\int_{0}^{1+1/L^{3}}\mathds{1}_{\left\\{Z_{s}\in\mathrm{BAD}\right\\}}\mathrm{d}s\right]\right]$
$\displaystyle\geq\mathbb{E}_{z}\left[\mathbb{E}_{Z_{t}}\left[\mathds{1}_{\left\\{\tau_{\mathrm{BAD}}\leq
1\right\\}}\mathbb{E}_{Z_{\tau_{\mathrm{BAD}}}}\left[\int_{0}^{1+1/L^{3}-\tau_{\mathrm{BAD}}}\mathds{1}_{\left\\{Z_{s}\in\mathrm{BAD}\right\\}}\mathrm{d}s\right]\right]\right].$
(77)

Note that if $\tau_{\mathrm{BAD}}\leq 1$, then
$\int_{0}^{1+1/L^{3}-\tau_{\mathrm{BAD}}}\mathds{1}_{\left\\{Z_{s}\in\mathrm{BAD}\right\\}}\mathrm{d}s\geq\int_{0}^{1/L^{3}}\mathds{1}_{\left\\{Z_{s}\in\mathrm{BAD}\right\\}}\mathrm{d}s$.
On the other hand, with $\beta$ as in Lemma 33, we have

$\mathbb{E}_{Z_{\tau_{\mathrm{BAD}}}}\left[\int_{0}^{1/L^{3}}\mathds{1}_{\left\\{Z_{s}\in\mathrm{BAD}\right\\}}\mathrm{d}s\right]\geq\mathbb{E}_{Z_{\tau_{\mathrm{BAD}}}}\left[\dfrac{1}{L^{3}}\mathds{1}_{\left\\{\tau_{out}>\dfrac{1}{L^{3}}\right\\}}\right]\geq\min_{z\in\mathcal{Z}}\dfrac{1}{L^{3}}\mathbb{P}_{z}\left[\tau_{out}>\dfrac{1}{L^{3}}\right]\geq\dfrac{\beta}{L^{3}}.$

This and (6.2) together imply

$\displaystyle\mathbb{E}_{z}\left[\int_{t}^{t+1+1/L^{3}}\mathds{1}_{\left\\{Z_{s}\in\mathrm{BAD}\right\\}}\mathrm{d}s\right]\geq\dfrac{\beta}{L^{3}}\mathbb{E}_{z}\left[\mathbb{E}_{Z_{t}}\left[\mathds{1}_{\left\\{\tau_{\mathrm{BAD}}\leq
1\right\\}}\right]\right].$ (78)

This and (76) together imply

$\mathbb{E}_{z}\left[\mathbb{P}_{Z_{t}}\left[\tau_{\mathrm{BAD}}\leq
1\right]\right]=\mathcal{O}\left(1/L^{6}\right).$ (79)

Note that the left-hand side of the equation above is exactly
$\mathbb{P}_{z}\left[\exists s\in[t,t+1]:Z_{s}\in\mathrm{BAD}\right]$. So this
leads to what we want. ∎

We now prove Lemma 33 and Lemma 32.

###### Proof of Lemma 33.

$\tau_{out}$ is at least the time that the first point of $\Xi$ appears, which
has distribution $\exp(L^{2}\times L+\lambda L)$, where
$\lambda=\sum_{i=1}^{q}\lambda_{i}$. Hence

$\displaystyle\mathbb{P}_{z}\left[\tau_{out}>\dfrac{1}{L^{3}}\right]$
$\displaystyle\geq\mathbb{P}\left[\exp(L^{3}+\lambda
L)>\dfrac{1}{L^{3}}\right]$ $\displaystyle=\exp\left(-\dfrac{L^{3}+\lambda
L}{L^{3}}\right)$ $\displaystyle=\exp(-1-\dfrac{\lambda}{L^{2}}).$

We can choose, for example, $\beta=e^{-1-\lambda}$. ∎

To finish this subsection, we prove Lemma 32.

###### Proof of Lemma 32.

Recall that the process $Z$ is attractive. Note also that for any
$E\subset\Lambda_{L}$, the set $\\{z\subset\Lambda_{L}|z\cap E=\O\\}$ is
decreasing. So, we only need to prove the statement for the initial condition
$z:=\O$. To know whether $E\cap Z^{\O}_{t}=\O$, a natural way, again, is to
trace the history of the set $E$ backward in time. We do it by placing at each
site in $E$ at time $t$ a particle labeled by the corresponding element in
$\mathcal{A}$ and trace them back to time $0$, given the collection of marks
generated by $\Xi$.

  * •

When a particle meets an exclusion mark, it jumps to the other endpoint of the
corresponding edge.

  * •

When a particle, say $\mathsf{w}$, meet a correlation mark, it is removed, and
$Z^{\O}_{t}(\mathsf{w}):=0$.

  * •

When a particle, say $\mathsf{w}$, meet a refresh mark, it is removed, and
$Z^{\O}_{t}(\mathsf{w}):=1$.

  * •

If a particle $\mathsf{w}$ reaches time $0$, then $Z^{\O}_{t}(\mathsf{w}):=0$.

With those rules, we can construct $Z^{\O}_{t}(E)$.

It is more convenient to generate the history in the forward direction. Let
$\Xi^{*}=(\Xi^{\text{exclusion,*}},\Xi^{\text{Glauber,*}},\Xi^{\text{refresh,*}})$
be an independent copy of $\Xi$. In another copy of the space-time slab, we
construct the process $(E_{s})_{0\leq s\leq t}$ takings values in
$\mathcal{Z}$ and the collection of marks $\mathcal{C}^{*}$ as follows.

  * •

$E_{0}:=E$.

  * •

Whenever $\Xi^{\text{exclusion,*}}_{u}$ jumps, say at time $s$, we perform the
following operations:

    1. 1.

Put an exclusion mark on the edge $(u,u+1)$,

    2. 2.

$E_{s}:=E_{s-}^{u\leftrightarrow u+1}$.

  * •

Whenever $\Xi^{\text{refresh,*}}_{u}$ jumps, $0\leq u\leq L-1$, say at time
$s$,

    * –

If $u\leq\left|E_{s-}\right|$, put a refresh mark on the $u-$th site in
$E_{s-}$ , say site $u^{\prime}$. Then
$E_{s}:=E_{s-}\setminus\\{u^{\prime}\\}$.

    * –

If $u>\left|E_{s-}\right|$, put a refresh mark on the
$(u-\left|E_{s-}\right|)$-th site in $\Lambda_{L}\setminus E_{s-}$. Then
$E_{s}:=E_{s-}$.

  * •

Whenever $\Xi^{\text{Glauber,*}}_{i,u}$ jumps, say at time $s$,

    * –

if $u\leq\left|B(E_{s-},m)\right|$, put a Glauber mark on the $u-$th site in
$B(E_{s-},m)$, say site $u^{\prime}$, and put correlation marks on all the
sites in $B(u^{\prime},m)$. Then $E_{s}:=E_{s-}\setminus B(u^{\prime},m)$.

    * –

if $u>\left|B(E_{s-},m)\right|$, put a Glauber mark on the
$(u-\left|B(E_{s-},m)\right|)-$th site in $\Lambda_{L}\setminus B(E_{s-},m)$,
say site $u^{\prime}$, and put correlation marks on all the sites in
$B(u^{\prime},m)$. Then $E_{s}:=E_{s-}$.

By the homogeneity of the Poisson processes $\Xi$ and $\Xi^{*}$, the
collection of marks $\mathcal{C}^{*}|_{[0,t]}$ has the same distribution as
$\mathcal{C}|_{[0,t]}$ viewed backward in time. In particular, the process
$(E_{s})_{0\leq s\leq t}$ has the same distribution as the set of particles
alive when we track the history of the set $E$ backward from time $t$. In
particular, $\mathbb{P}\left[Z^{\O}_{t}\cap E=\O\right]$ is the same as the
probability that no site is removed from $(E_{s})_{0\leq s\leq t}$ due to the
appearance of a refresh mark.

Here, we have rearranged the Poisson clocks so that the first clocks in
$\Xi^{\text{refresh,*}}$ indicate the refresh of the sites in $(E_{s})_{0\leq
s\leq t}$, and the first clocks in $\Xi^{\text{Glauber,*}}$ indicate the
Glauber updates of the sites in the neighbors of the process $(E_{s})_{0\leq
s\leq t}$. This makes our computation easier. More precisely, let
$\mathcal{G}_{s}$ be the $\sigma$-algebra generated by $\Xi^{*}$ up to time
$s$. Let $\mathcal{T}_{0}=0$ and $(\tau_{i},\mathcal{T}_{i})_{i\geq 1}$ be
defined recursively as follows.

$\displaystyle\mathcal{T}_{i+1}$
$\displaystyle=\inf\left\\{t>\mathcal{T}_{i}:\,\text{a clock among
$(\Xi^{\text{refresh,*}}_{u})_{1\leq
u\leq\left|E_{\mathcal{T}_{i}}\right|},(\Xi^{\text{Glauber,*}}_{u,i})_{3\leq
i\leq q,1\leq u\leq(2m+1)\left|E_{\mathcal{T}_{i}}\right|}$ rings at time
$t$}\right\\}$ $\displaystyle\tau_{i+1}$
$\displaystyle=\mathcal{T}_{i+1}-\mathcal{T}_{i}.$

Then we have the following:

  1. 1.

Conditionally on $\mathcal{G}_{\mathcal{T}_{i}}$,
$\tau_{i+1}\sim\exp\left((\lambda_{1}+\lambda_{2})\left|E_{\mathcal{T}_{i}}\right|+(\lambda-\lambda_{1}-\lambda_{2})(2m+1)\left|E_{\mathcal{T}_{i}}\right|\right)$.

  2. 2.

$\left|E_{s}\right|=\left|E_{\mathcal{T}_{i}}\right|,\,\forall
s\in[\mathcal{T}_{i},\mathcal{T}_{i+1});\;\left|E_{\mathcal{T}_{i+1}}\right|\geq\left|E_{\mathcal{T}_{i}}\right|-(2m+1),$

and hence

$\left|E_{\mathcal{T}_{i}}\right|\geq|E|-i(2m+1).$

This is because at most $(2m+1)$ sites are removed each time a Glauber mark
appears.

  3. 3.

$\displaystyle\mathbb{P}\left[\text{the ring at $\mathcal{T}_{i+1}$ is among
$\Xi^{\text{refresh,*}}_{1},\dots,\Xi^{\text{refresh,*}}_{\left|E_{\mathcal{T}_{i}}\right|}$}\big{|}\mathcal{G}_{\mathcal{T}_{i}}\right]$
$\displaystyle=\dfrac{\lambda_{1}+\lambda_{2}}{(\lambda_{1}+\lambda_{2})+(\lambda-\lambda_{1}-\lambda_{2})(2m+1)}$
$\displaystyle=:1-\alpha.$ (80)

Based on the arguments above,

$\displaystyle\mathbb{P}\left[Z^{\O}_{t}\cap
E=\O\right]=\mathbb{P}\left[\text{For any $\mathcal{T}_{i}\leq t$, the ring at
$\mathcal{T}_{i}$ is among $\Xi^{\text{Glauber,*}}$}\right].$

Note that, for any $l\in\mathbb{Z}_{+}$,

$\displaystyle\mathbb{P}\left[\text{For any $\mathcal{T}_{i}\leq t$, the ring
at $\mathcal{T}_{i}$ is among $\Xi^{\text{Glauber,*}}$}\right]$
$\displaystyle\leq\mathbb{P}\left[\mathcal{T}_{l}>t\right]+\mathbb{P}\left[\mathcal{T}_{l}\leq
t,\text{the ring at $\mathcal{T}_{1},\dots,\mathcal{T}_{l}$ are among
$\Xi^{\text{Glauber,*}}$}\right].$ (81)

By equation (3),

$\mathbb{P}\left[\mathcal{T}_{l}\leq t,\text{the ring at
$\mathcal{T}_{1},\dots,\mathcal{T}_{l}$ are among
$\Xi^{\text{Glauber,*}}$}\right]\leq\alpha^{l}.$ (82)

We can choose $l=10\left\lceil\dfrac{\log L}{-\log\alpha}\right\rceil$ to make
this term smaller than $\dfrac{1}{L^{10}}$. The bound for the first term on
the right side of (81) is a consequence of the concentration of the sum
$\tau_{1}+\dots+\tau_{l}$. Let
$\beta:=(\lambda_{1}+\lambda_{2})+(\lambda-\lambda_{1}-\lambda_{2})(2m+1)$. By
Claims 1 and 2, conditionally on $\mathcal{G}_{\mathcal{T}_{i}}$, $\tau_{i+1}$
is stochastically dominated by
$\exp\left(\beta(\left|E\right|-i(2m+1)\right)$. Set $\theta:=(2m+1)l$, then
provided that $|E|>\theta+\dfrac{2\theta}{\beta}$,

$\dfrac{\beta(|E|-(i-1)(2m+1))}{\beta(|E|-(i-1)(2m+1))-\theta}\leq 2,\,\forall
1\leq i\leq l.$

Therefore,

$\displaystyle\mathbb{E}\left[e^{\theta\mathcal{T}_{l}}\right]$
$\displaystyle=\mathbb{E}\left[\mathbb{E}\left[e^{\theta(\mathcal{T}_{l-1}+\tau_{l})}|\mathcal{G}_{\mathcal{T}_{l-1}}\right]\right]$
$\displaystyle=\mathbb{E}\left[e^{\theta\mathcal{T}_{l-1}}\mathbb{E}\left[e^{\tau_{l}}|\mathcal{G}_{\mathcal{T}_{l-1}}\right]\right]$
$\displaystyle\leq\mathbb{E}\left[e^{\theta\mathcal{T}_{l-1}}\times\dfrac{\beta(|E|-(l-1)(2m+1))}{\beta(|E|-(l-1)(2m+1))-\theta}\right]$
$\displaystyle\leq 2\mathbb{E}\left[e^{\theta\mathcal{T}_{l-1}}\right].$

By induction, we deduce that

$\displaystyle\mathbb{E}\left[e^{\theta\mathcal{T}_{l}}\right]\leq 2^{l}.$

Hence, by Chernoff’s bound,

$\mathbb{P}\left[\mathcal{T}_{l}>t\right]\leq e^{-\theta
t}2^{l}=(2e^{-(2m+1)t})^{l}.$

With $l$ chosen as above, there exists $\beta_{3}$ such that when
$t\geq\beta_{3}$, the last expression is smaller than $1/L^{10}$. We finish
the proof by choosing $\beta_{2}$ such that $\beta_{2}\log
L\geq\theta+\theta/\beta$, which clearly exists when
$\theta=(2m+1)l,\;l=10\left\lceil\dfrac{\log L}{-\log q}\right\rceil$. ∎

###  6.3 First step of the proof of Proposition 31: reformulation

The rest of the paper is devoted to proving Proposition 31. This is done in
two steps. In this subsection, we do the first step: reduce our task to
estimating the local time that a certain Markov process spends in a certain
subset of its state space. In the next subsection, we use techniques from
excursion theory to estimate this local time.

Without loss of generality, we can suppose $u=0$. Throughout this subsection,
let $z,\pi_{1},\pi_{2}$ be as in the statement of Proposition 31. We first
recall a result about perturbations of a product measure.

######  Lemma 34 (Perturbation of a product measure).

Let $\Omega=\\{-1,1\\}^{n}$. For each subset $S\subset[n]$, let $\varphi_{S}$
be a distribution on $\\{-1,1\\}^{S}$. Let $\rho\in(-1,1)$, and let $\nu$ be
the product measure $\mathfrak{R}_{\rho}^{\otimes n}$ on $\Omega$. Let $\mu$
be the measure on $\Omega$ obtained by first sampling a subset $S\subset[n]$
via some measure $\tilde{\mu}$, and then, conditionally on $S$, generating
independently the values on $S$ via $\varphi_{S}$ and the values on
$[n]\setminus S$ via $\mathfrak{R}_{p}^{\otimes[n]\setminus S}$. Then

$4\textrm{d}_{\textsc{tv}}\left(\mu,\nu\right)^{2}\leq\left\|\dfrac{\mu}{\nu}-1\right\|^{2}_{L^{2}(\nu)}\leq\mathbb{E}\left[\theta^{\left|S\cap
S^{\prime}\right|}\right]-1,$

where $S,S^{\prime}$ are i.i.d. with law $\tilde{\mu}$, and
$\theta=\max\left\\{\dfrac{2}{1+\rho},\dfrac{2}{1-\rho}\right\\}$.

Originally introduced by Miller and Peres in [49], this result has been
successfully used to prove cutoff for the Glauber dynamics by Lubetzky and Sly
in [41, 42, 43]. It has recently been used in conjunction with negative
dependence to prove cutoff for SSEP with reservoirs in [55, 63]. For the proof
of Lemma 34, see Lemma $1$ in [55].

We will use the following graphical construction. We will construct a coupling
$(X,Z)$ where $X$ starts from $\pi_{1}$ or $\pi_{2}$, and $Z$ starts from
$z\cup\\{0\\}$. Recall that a site in $Z$ is viewed as colored blue. From now
on, we will color all sites not in $Z$ black. The content of a site $u$ at
time $t$ is its spin $X_{t}(u)\in\\{-1,1\\}$ and its color
$Z_{t}(u)\in\\{0,1\\}$. From now on, we allow a collection of marks to contain
new types of marks: blue and black marks, which can appear on the edge of
$\Lambda_{L}$, and whose effects are explained later.

#### Graphical Construction 3.

Let $\Xi^{\text{refresh}}$ and $\Xi^{\text{Glauber}}$ be defined as in
Graphical Construction $2$. They generate the refresh and Glauber marks as
before. However, we replace the effect of the exclusion marks with that of the
blue and black marks, as follows. Let $(\Xi^{\text{blue}}_{u})_{0\leq u\leq
L-1}$ and $(\Xi^{\text{black}}_{u})_{0\leq u\leq L-1}$ be independent of
$\Xi^{\text{refresh}},\Xi^{\text{Glauber}}$ and defined as follows.

  * •

$(\Xi^{\text{black}}_{u})_{0\leq u\leq L-1}$ are independent Poisson clocks of
intensity $L^{2}$ on the edges. When $\Xi^{\text{black}}_{u}$ jumps, we put a
black mark on the edge $(u,u+1)$.

  * •

$(\Xi^{\text{blue}}_{u})_{0\leq u\leq L-1}$ are independent Poisson clocks of
intensity $2L^{2}$ on the edges. When $\Xi^{\text{blue}}_{u}$ jumps, we put a
blue mark on the edge $(u,u+1)$.

From now on, let
$\Xi:=(\Xi^{\text{refresh}},\Xi^{\text{Glauber}},\Xi^{\text{black}},\Xi^{\text{blue}})$.
Each realization of $\Xi$ defines a collection of marks consisting of refresh,
Glauber, black, and blue marks. We construct the process $X$ from this
collection of marks as follows.

  * •

The effect of the refresh marks and Glauber marks are as before.

  * •

Each time we see a black mark, say on the edge $(u,u+1)$, if at least one of
the endpoints of the edge $(u,u+1)$ is black, we exchange the contents of
those endpoints: $x\mapsto x^{u\leftrightarrow u+1},z\mapsto
z^{u\leftrightarrow u+1}$.

  * •

Each time we see a blue mark, say on the edge $(u,u+1)$, if both endpoints are
blue, we exchange their spins with probability $1/2$ (we decide by sampling an
independent Bernoulli variable with parameter $1/2$).

With this construction, $X$ is still a Glauber-Exclusion process, and $Z$
still tracks the independent sites of $X$ if $X$ starts from $\pi_{2}$. The
idea is to reveal $\Xi$ but not the Rademacher variables
$\mathfrak{R}_{\bar{\rho}}$ used to refresh the sites at the refresh marks and
the Bernoulli variables used to make decisions at the blue marks.

#### The conditional distribution of $X$.

Conditionally on $\Xi([0,1])$, let
$(u_{1},\tau_{1}),(u_{2},\tau_{2}),\dots,(u_{k},\tau_{k})$ be the positions of
the correlation marks that appear on the blue sites, i.e. $\forall
i,\;(u_{i},\tau_{i}-)\in Z_{\tau_{i}-}$. In other words, they are the
correlation marks that turn some blue sites black. We claim that,
conditionally on $\Xi$, almost surely, there is a function $\Phi$ independent
of the initial configuration of $X$ such that

$X_{1}=\Phi(y,X_{\tau_{1}-}(u_{1}),\dots,X_{\tau_{k}-}(u_{k}),X_{1}(Z_{1})).$
(83)

The explanation for the formula above is as follows. At time $1$, almost
surely, each site is either black or blue. The spins of the blue sites are
given by $X_{1}(Z_{1})$. The spins of the black sites can be constructed from

  1. 1.

The spins of the black sites at the beginning, i.e. $y$,

  2. 2.

The trajectories of the black sites given by $\Xi^{\text{black}}$ and the
update functions given by $\Xi^{\text{Glauber}}$. All that information is
encoded in the function $\Phi$,

  3. 3.

The spins of the blue sites used in the update functions at the Glauber marks.
Those spins are exactly $X_{\tau_{1}-}(u_{1}),\dots,X_{\tau_{k}-}(u_{k})$.

Let $\mathfrak{L}_{1}$ (resp. $\mathfrak{L}_{2}$) be the distribution of
$(X_{\tau_{1}-}(u_{1}),\dots,X_{\tau_{k}-}(u_{k}),X_{1}(Z_{1}))$ when the
Glauber-Exclusion process starts from $\pi_{1}$ (resp. $\pi_{2}$). We remark
here that if we start with the distribution $\pi_{2}$, then
$\mathfrak{L}_{2}=\mathfrak{R}_{\bar{\rho}}^{\otimes
k}\otimes\mathfrak{R}_{\bar{\rho}}^{\otimes Z_{1}}$, and $\mathfrak{L}_{1}$ is
a perturbation of $\mathfrak{L}_{2}$. Now, we can take advantage of the
information-percolation framework in [42]. Thanks to partially revealing the
randomness of the graphical construction, we transform the problem into
comparing a product law with its perturbation, which can be done using Lemma
34. More formally, we have the following lemma.

######  Lemma 35 (Projection).

$\displaystyle\left\|\mathbb{P}_{\pi_{1}}\left[X_{1}\in\cdot\right]-\mathbb{P}_{\pi_{2}}\left[X_{1}\in\cdot\right]\right\|_{TV}\leq\mathbb{E}\left[\left\|\mathfrak{L}_{1}-\mathfrak{L}_{2}\right\|_{TV}\right].$
(84)

###### Proof.

We have

$\displaystyle\left\|\mathbb{P}_{\pi_{1}}\left[X_{1}\in\cdot\right]-\mathbb{P}_{\pi_{2}}\left[X_{1}\in\cdot\right]\right\|_{TV}$
$\displaystyle\leq\mathbb{E}\left[\left\|\mathbb{P}_{\pi_{1}}\left[X_{1}\in\cdot\big{|}\Xi([0,1])\right]-\mathbb{P}_{\pi_{2}}\left[X_{1}\in\cdot\big{|}\Xi([0,1])\right]\right\|_{TV}\right]$
$\displaystyle\leq\mathbb{E}\left[\left\|\mathfrak{L}_{1}-\mathfrak{L}_{2}\right\|_{TV}\right],$

where in the last inequality, we have use (83) and the definitions of
$\mathfrak{L}_{1},\mathfrak{L}_{2}$. ∎

It remains to estimate the right-hand side in equation (84). First, we
introduce a random walk used to track the position of the perturbation.

#### The perturbation position.

Let the random walk $(U_{1}(t))_{t\geq 0}$ taking values in
$\Lambda_{L}\cup\\{{\dagger}\\}$ be constructed as follows.

  * •

$U_{1}(0)=0$ (the walk $U_{1}$ starts at site $0$).

  * •

Whenever a refresh mark or a correlation mark appears on the site occupied by
$U_{1}$, the walk $U_{1}$ is killed (it becomes the cemetery state
${\dagger}$).

  * •

Whenever a black mark appears, if $U_{1}$ is at one of the two endpoints of
the corresponding edge and the other endpoint is black, then $U_{1}$ takes the
edge ($U_{1}$ jumps to the other endpoint).

  * •

Whenever a blue mark appears, if $U_{1}$ is at one of the endpoints of the
mark, with probability $1/2$ (decided by an independent Bernoulli variable
with parameter $1/2$), it attempts to take the edge, and it succeeds if the
other endpoint is blue and fails if the other endpoint is black.

Let $U_{2}$ be an independent copy of $U_{1}$ conditionally on $\Xi$, i.e. the
decisions of the two walks each time a blue mark appears are independent. The
law of $U_{1}$ (or $U_{2}$) conditionally on $\Xi$ is the averaging process,
introduced by Aldous in [3], defined as follows.

#### The averaging process.

Let $H=(H_{t})_{t\geq 0}$ be the piece-wise constant process taking values in
the space of function from $\Lambda_{L}$ to $[0,1]$, constructed as follows:

  * •

$H_{0}(0)=1;\;H_{0}(u)=0,\,\forall u\in\Lambda_{L}\setminus\\{0\\}$.

  * •

When we see a refresh mark, say at site $u$, we make the transition $h\mapsto
h^{u,0}$, where $h^{u,0}$ is obtained from $h$ by replacing the values at $u$
by $0$, i.e.

$\forall
u^{\prime}\in\Lambda_{L},\;h^{u,0}(u^{\prime})=h(u^{\prime})\times\mathds{1}_{\left\\{u^{\prime}\neq
u\right\\}}.$

  * •

When we see a Glauber mark, say at site $u$, we make the transition $h\mapsto
h^{B(u,m),0}$, where $h^{B(u,m),0}$ is obtained from $h$ by replacing the
values in $B(u,m)$ by $0$, i.e.

$\forall
u^{\prime}\in\Lambda_{L},\;h^{B(u,m),0}(u^{\prime})=h(u^{\prime})\times\mathds{1}_{\left\\{u^{\prime}\notin
B(0,m)\right\\}}.$

  * •

When a black mark appears, say on the edge $(u,u+1)$, if the process $Z$ makes
the transition $z\mapsto z^{u\leftrightarrow u+1}$, we make the transition
$h\mapsto h^{u\leftrightarrow u+1}$, where $h^{u\leftrightarrow u+1}$ is
obtained from $h$ by exchanging the values at sites $u$ and $u+1$, i.e.

$h^{u\leftrightarrow
u+1}(u^{\prime})=h(u^{\prime})\times\mathds{1}_{\left\\{u^{\prime}\notin\\{u,u+1\\}\right\\}}+h(u)\times\mathds{1}_{\left\\{u^{\prime}=u+1\right\\}}+h(u+1)\times\mathds{1}_{\left\\{u^{\prime}=u\right\\}}.$

  * •

When a blue mark appears, say at site $u$, if both sites $u$ and $u+1$ are
blue, we replace $h(u)$ and $h(u+1)$ by their average
$\dfrac{h(u)+h(u+1)}{2}$, i.e. $h\mapsto\dfrac{h+h^{u\leftrightarrow
u+1}}{2}$.

The averaging process is an interesting Markov process by itself. We refer the
readers to [54, 7] for recent development. By construction,
$(H_{s})_{s\in[0,t]}$ is measurable w.r.t $\Xi([0,t])$. Moreover,

$H_{t}(u)=\mathbb{P}\left[U_{1}(t)=u|\Xi\right],\;\forall
i\in\Lambda_{L},\forall t\geq 0,$

and in a particular

$H_{t}(u)^{2}=\mathbb{P}\left[U_{1}(t)=U_{2}(t)=u|\Xi\right].$ (85)

Recall that $\mathsf{dist}\left(\cdot,\cdot\right)$ denotes the shortest-path
distance on $\Lambda_{L}$. By convention, we define

$\displaystyle\mathsf{dist}\left(u,{\dagger}\right):=\mathsf{dist}\left({\dagger},u\right):=\infty,\;\forall
u\in\Lambda_{L}\cup\\{{\dagger}\\}.$

The purpose of this subsection is to prove the following result.

######  Lemma 36 (Comparision with local time).

There exists a constant $\beta$ such that

$\displaystyle\left\|\mathbb{P}_{\pi_{1}}\left[X_{1}\in\cdot\right]-\mathbb{P}_{\pi_{2}}\left[X_{1}\in\cdot\right]\right\|_{TV}^{2}\leq\beta\mathbb{E}\left[\int_{0}^{1}\mathds{1}_{\left\\{\mathsf{dist}\left(U_{1}(s),U_{2}(s)\right)=0\right\\}}\mathrm{d}s\right].$
(86)

Thanks to Lemma 35, we only need to show the following results to prove Lemma
36.

######  Lemma 37 (Conditional upper bound by the Averaging process).

$\left\|\mathfrak{L}_{1}-\mathfrak{L}_{2}\right\|_{TV}^{2}\leq\left(\max\left\\{\dfrac{2}{1+\bar{\rho}},\dfrac{2}{1-\bar{\rho}}\right\\}-1\right)\left(\sum_{j=1}^{k}H_{\tau_{j}-}^{2}(i_{j})+\left\|H_{1}\right\|_{2}^{2}\right).$
(87)

######  Lemma 38.

There exist a constant $\beta$ such that

$\mathbb{E}\left[\sum_{j=1}^{k}H_{\tau_{j}-}^{2}(i_{j})+\left\|H_{1}\right\|_{2}^{2}\right]\leq\beta\mathbb{E}\left[\int_{0}^{1}\mathds{1}_{\left\\{\mathsf{dist}\left(U_{1}(s),U_{2}(s)\right)=0\right\\}}\mathrm{d}s\right].$
(88)

It remains to show Lemma 37 and Lemma 38. We first show Lemma 37.

###### Proof of Lemma 37.

By construction, $\mathfrak{L}_{2}$ is a product of
$\mathfrak{R}_{\bar{\rho}}$, and $\mathfrak{L}_{1}$ is a perturbed version of
$\mathfrak{L}_{2}$, and the walk $U_{1}$ tracks the perturbation position. Let

$\displaystyle S_{1}$
$\displaystyle:=\\{(u_{j},\tau_{j}-)|j\in[k],\,U_{1}(\tau_{j}-)=u_{j}\\}\cup\\{(u,1)|u\in
Z_{1},\,U_{1}(1)=u\\}.$

In words, $S_{1}$ contains places perturbed in the vector
$(X_{\tau_{1}-}(u_{1}),\dots,X_{\tau_{k}-}(u_{k}),X_{1}(Z_{1}))$. Note that
the walk $U_{1}$ is killed when it meets a correlation mark or a refresh mark.
Hence $|S_{1}|\leq 1$. Let $S_{2}$ be defined as $S_{1}$ when we replace
$U_{1}$ by $U_{2}$. Then, by Lemma 34,

$\displaystyle\left\|\mathfrak{L}_{1}-\mathfrak{L}_{2}\right\|_{TV}^{2}\leq\mathbb{E}\left[\theta^{\left|S_{1}\cap
S_{2}\right|}-1|\Xi\right],$ (89)

where
$\theta=\max\left\\{\dfrac{2}{1+\bar{\rho}},\dfrac{2}{1-\bar{\rho}}\right\\}$.
Note that, as $|S_{1}\cap S_{2}|\leq 1$,

$\displaystyle\mathbb{E}\left[\theta^{\left|S_{1}\cap
S_{2}\right|}-1\big{|}\Xi\right]$ (90)
$\displaystyle=\mathbb{E}\left[(\theta-1)\left|S_{1}\cap
S_{2}\right|\bigg{|}\Xi\right]$ (91)
$\displaystyle=(\theta-1)\left(\left(\sum_{j=1}^{k}\mathbb{P}\left[U_{1}(\tau_{j}-)=U_{2}(\tau_{j}-)=u_{j}\big{|}\Xi\right]\right)+\left(\sum_{u\in\Lambda_{L}}\mathbb{P}\left[U_{1}(1)=U_{2}(1)=u|\Xi\right]\right)\right)$
(92)
$\displaystyle=(\theta-1)\left(\sum_{j=1}^{k}H_{\tau_{j}-}^{2}(u_{j})\times+\left\|H_{1}\right\|_{2}^{2}\right),$
(93)

which is what we want. ∎

Now we show Lemma 38.

###### Proof of Lemma 38.

Recall that $(u_{1},\tau_{1}),\dots,(u_{k},\tau_{k})$ are the correlation
marks that turn some blue sites green up to time $1$. Let
$(u_{i},\tau_{i})_{i\geq k+1}$ be the correlation marks that turn some blue
sites green from time $1$ onwards. Let us define

$F_{t}:=\sum_{j=1}^{\infty}H^{2}_{\tau_{j}-}(u_{j})\mathds{1}_{\left\\{\tau_{j}\leq
t\right\\}}.$ (94)

Then $(Z_{t},H_{t},F_{t})_{t\geq 0}$ is a Markov process. In particular, let
$\mathcal{L}$ be the generator corresponding to this Markov process. Then for
any function $\varphi$ on the state space of this process,

$\displaystyle\mathcal{L}\varphi(z,h,f)=$
$\displaystyle\sum_{u=0}^{L-1}L^{2}\left(\varphi(z^{u\leftrightarrow
u+1},h^{u\leftrightarrow
u+1},f)-\varphi(z,h,f)\right)\left(\mathds{1}_{\left\\{z(u)=0\right\\}}\vee\mathds{1}_{\left\\{z(u+1)=0\right\\}}\right)$
$\displaystyle+\sum_{u=0}^{L-1}2L^{2}\left(\varphi\left(z,\dfrac{h+h^{u\leftrightarrow
u+1}}{2},f\right)-\varphi(z,h,f)\right)\mathds{1}_{\left\\{z(u)=z(u+1)=1\right\\}}$
$\displaystyle+\sum_{u=0}^{L-1}(\lambda_{1}+\lambda_{2})\left(\varphi(z^{u,1},h^{i,0},f)-\varphi(z,h,f)\right)$
$\displaystyle+\sum_{i=3}^{q}\sum_{u=0}^{L-1}\lambda_{i}\left(\varphi\left(z\setminus
B(u,m),h^{B(u,m),0},f+\sum_{u^{\prime}\in
B(0,m)}h^{2}(u^{\prime})\mathds{1}_{\left\\{z(u^{\prime})=1\right\\}}\right)-\varphi(z,h,f)\right).$

Let $\varphi$ be the projection onto the third coordinate. Then

$\displaystyle\mathcal{L}\varphi(z,h,f)$
$\displaystyle=\sum_{i=3}^{q}\sum_{u=0}^{L-1}\lambda_{i}\sum_{u^{\prime}\in
B(u,m)}h^{2}(u^{\prime})\mathds{1}_{\left\\{z(u^{\prime})=1\right\\}}$
$\displaystyle=(\lambda-\lambda_{1}-\lambda_{2})\sum_{u=0}^{L-1}\sum_{u^{\prime}=0}^{L-1}h^{2}(u^{\prime})\mathds{1}_{\left\\{z(u^{\prime})=1\right\\}}\mathds{1}_{\left\\{u^{\prime}\in
B(u,m)\right\\}}$
$\displaystyle=(\lambda-\lambda_{1}-\lambda_{2})\sum_{u^{\prime}=0}^{L-1}\sum_{u=0}^{L-1}h^{2}(u^{\prime})\mathds{1}_{\left\\{z(u^{\prime})=1\right\\}}\mathds{1}_{\left\\{u\in
B(u^{\prime},m)\right\\}}$
$\displaystyle\leq(\lambda-\lambda_{1}-\lambda_{2})(2m+1)\sum_{u^{\prime}=0}^{L-1}h^{2}(u^{\prime})\mathds{1}_{\left\\{z(u^{\prime})=1\right\\}}$
$\displaystyle=(\lambda-\lambda_{1}-\lambda_{2})(2m+1)\left\|h\right\|_{2}^{2}.$

A classical result (see e.g. [19]) says that

$F_{t}-F_{0}-\int_{0}^{t}\mathcal{L}\varphi(Z_{s},H_{s},F_{s})ds$

is a zero-mean martingale, and hence

$\mathbb{E}\left[F_{t}\right]=\mathbb{E}\left[F_{0}\right]+\mathbb{E}\left[\int_{0}^{t}\mathcal{L}\varphi(Z_{s},H_{s},F_{s})ds\right]\leq(\lambda-\lambda_{1}-\lambda_{2})(2m+1)\mathbb{E}\left[\int_{0}^{t}\left\|H_{s}\right\|_{2}^{2}ds\right].$
(95)

Moreover, $\left\|H_{t}\right\|_{2}^{2}$ almost surely decreases, hence

$\left\|H_{1}\right\|_{2}^{2}\leq\int_{0}^{1}\left\|H_{s}\right\|_{2}^{2}\mathrm{d}s.$
(96)

Let

$\displaystyle\beta:=(\lambda-\lambda_{1}-\lambda_{2})(2m+1)+1.$

Two equations (95), (96), and Fubini’s theorem together imply that

$\displaystyle\mathbb{E}\left[\sum_{j=1}^{k}H_{\tau_{j}-}^{2}(i_{j})+\left\|H_{1}\right\|_{2}^{2}\right]\leq\beta\int_{0}^{1}\mathbb{E}\left[\left\|H_{s}\right\|_{2}^{2}\right]\mathrm{d}s.$
(97)

Note that for any $s\in[0,1]$,

$\displaystyle\mathbb{E}\left[\left\|H_{s}\right\|_{2}^{2}\right]=\mathbb{E}\left[\sum_{u\in\Lambda_{L}}\mathbb{P}\left[U_{1}(s)=U_{2}(s)=u|\Xi\right]\right]=\mathbb{E}\left[\mathds{1}_{\left\\{\mathsf{dist}\left(U_{1}(s),U_{2}(s)\right)=0\right\\}}\right].$

Therefore

$\displaystyle\int_{0}^{1}\mathbb{E}\left[\left\|H_{s}\right\|_{2}^{2}\right]\mathrm{d}s=\int_{0}^{1}\mathbb{E}\left[\mathds{1}_{\left\\{\mathsf{dist}\left(U_{1}(s),U_{2}(s)\right)=0\right\\}}\right]\mathrm{d}s=\mathbb{E}\left[\int_{0}^{1}\mathds{1}_{\left\\{\mathsf{dist}\left(U_{1}(s),U_{2}(s)\right)=0\right\\}}\mathrm{d}s\right]$
(98)

Two equations (97), (98) lead to what we want. ∎

###  6.4 Second step of the proof of Proposition 31: the excursion coupling

This subsection is devoted to estimating the right-hand side of (88). It
should be more convenient to consider a modified version
$(\bar{Z},\bar{U}_{1},\bar{U}_{2})$ of $(Z,U_{1},U_{2})$, where we do not kill
the walks when we see refresh marks or correlation marks, and we always color
the sites occupied by $\bar{U}_{1},\bar{U}_{2}$ blue. Note that before $U_{1}$
or $U_{2}$ are killed, $(\bar{Z},\bar{U}_{1},\bar{U}_{2})=(Z,U_{1},U_{2})$.
Therefore, almost surely,

$\displaystyle\int_{0}^{1}\mathds{1}_{\left\\{\mathsf{dist}\left(U_{1}(s),U_{2}(s)\right)=0\right\\}}\mathrm{d}s\leq\int_{0}^{1}\mathds{1}_{\left\\{\mathsf{dist}\left(\bar{U}_{1}(s),\bar{U}_{2}(s)\right)=0\right\\}}\mathrm{d}s.$
(99)

Besides, not killing the walks only increases the blue sites, i.e. , almost
surely,

$\displaystyle\bar{Z}_{s}\geq Z_{s},\,\forall s\in\mathbb{R}_{+}.$ (100)

An important fact here is the observation of Aldous in [3]:
$(\bar{Z},\bar{U}_{1},\bar{U}_{2})$ itself is a Markov process (without
conditional probability), constructed as follows. We still use the collection
of marks given in Graphical Construction 3. Suppose that we are at state
$(z,u_{1},u_{2})$.

  * •

When a refresh mark appears, say at site $u$, make the transition
$(z,u_{1},u_{2})\mapsto(z^{u,1},u_{1},u_{2})$.

  * •

When a Glauber mark appears, say at site $u$, make the transition
$(z,u_{1},u_{2})\mapsto((z\setminus
b(u,m))\cup\\{u_{1},u_{2}\\},u_{1},u_{2})$.

  * •

When a black mark appears, say on the edge $(u,u+1)$, if $z(u)=0$ or
$z(u+1)=0$, make the transition $(z,u_{1},u_{2})\mapsto(z^{u\leftrightarrow
u+1},u_{1}^{u\leftrightarrow u+1},u_{2}^{u\leftrightarrow u+1})$, where
$u_{1}^{u\leftrightarrow u+1}$ is given by

$u_{1}^{u\leftrightarrow u+1}=\begin{cases}u+1&\text{if $u_{1}=u$},\\\
u&\text{if $u_{1}=u+1$},\\\ u_{1}&\text{if
$u_{1}\notin\\{u,u+1\\}$},\end{cases}$

and $u_{2}^{u\leftrightarrow u+1}$ is defined similarly.

  * •

When a blue mark appears, say on edge $(u,u+1)$, if $z(u)=z(u+1)=1$,

$\displaystyle(z,u_{1},u_{2})$ $\displaystyle\mapsto(z,u_{1},u_{2})$
$\displaystyle\text{with probability $1/4$},$ $\displaystyle(z,u_{1},u_{2})$
$\displaystyle\mapsto(z,u_{1}^{u\leftrightarrow u+1},u_{2})$
$\displaystyle\text{with probability $1/4$},$ $\displaystyle(z,u_{1},u_{2})$
$\displaystyle\mapsto(z,u_{1},u_{2}^{u\leftrightarrow u+1})$
$\displaystyle\text{with probability $1/4$},$ $\displaystyle(z,u_{1},u_{2})$
$\displaystyle\mapsto(z,u_{1}^{u\leftrightarrow u+1},u_{2}^{u\leftrightarrow
u+1})$ $\displaystyle\text{with probability $1/4$}.$

By abuse of notation, we write
$\mathbb{P}_{z,u_{1},u_{2}}\left[\cdot\right],\mathbb{E}_{z,u_{1},u_{2}}\left[\cdot\right]$
for the probability and expectation taken with respect to the process
$(\bar{Z},\bar{U}_{1},\bar{U}_{2})$ starting from $(z,u_{1},u_{2})$. We will
compare $(\bar{U}_{1},\bar{U}_{2})$ with its idealized version
$(\widetilde{U}_{1},\widetilde{U}_{2})$, defined§ as follows.

#### Idealized version of $(\bar{U}_{1},\bar{U}_{2})$.

Let two random walks $(\widetilde{U}_{1},\widetilde{U}_{2})$ on $\Lambda_{L}$
be constructed as follows. Each edge of the lattice is associated with a
Poisson clock of intensity $2L^{2}$. Whenever a clock rings, if
$\widetilde{U}_{1}$ (or $\widetilde{U}_{2}$) is at one endpoint of the edge,
it decides to take the edge with probability $1/2$. If both can take the edge,
their decisions are made independently. Informally,
$(\widetilde{U}_{1},\widetilde{U}_{2})$ is the version of
$(\bar{U}_{1},\bar{U}_{2})$ where we assume that all the sites are always
blue.

Figure 4: Transition rates of
$\mathsf{dist}\left(\widetilde{U}_{1},\widetilde{U}_{2}\right)$ when $L$ is
even.

#### The excursion.

Let $\tau^{0\to 1}_{i}$ be the time for $i$-th passage from $0$ to $1$, and
let $\tau^{1\to 0}_{i}$ be the time for $i$-th passage from $1$ to $0$ of
$\left(\mathsf{dist}\left(\bar{U}_{1}(t),\bar{U}_{2}(t)\right)\right)_{t\geq
0}$. More precisely, let $\tau^{0\to 1}_{i},\tau^{1\to 0}_{i}$, and
$\mathcal{T}_{i}$ be defined recursively as follows.

$\displaystyle\mathcal{T}_{j}$ $\displaystyle=\sum_{i=1}^{j-1}(\tau^{0\to
1}_{i}+\tau^{1\to 0}_{i}),$ (101) $\displaystyle\tau^{0\to 1}_{i}$
$\displaystyle=\inf\left\\{t\geq
0:\mathsf{dist}\left(\bar{U}_{1}(\mathcal{T}_{i}+t),\bar{U}_{2}(\mathcal{T}_{i}+t)\right)=1\right\\},$
(102) $\displaystyle\tau^{1\to 0}_{i}$ $\displaystyle=\inf\left\\{t\geq
0:\mathsf{dist}\left(\bar{U}_{1}(\mathcal{T}_{i}+\tau^{0\to
1}_{i}+t),\bar{U}_{2}(\mathcal{T}_{i}+\tau^{0\to 1}_{i}+t)\right)=0\right\\}.$
(103)

The sequence $(\tilde{\tau}^{0\to 1}_{i},\tilde{\tau}^{1\to
0}_{i},\tilde{\mathcal{T}}_{i})_{i\geq 1}$ is defined similarly by replacing
$(\bar{U}_{1},\bar{U}_{2})$ with $(\widetilde{U}_{1},\widetilde{U}_{2})$. The
idea is to express the local time in terms of the excursions:

$\displaystyle\int_{0}^{1}\mathds{1}_{\left\\{\mathsf{dist}\left(\bar{U}_{1}(s),\bar{U}_{2}(s)\right)=0\right\\}}\mathrm{d}s$
$\displaystyle=\sum_{i\geq
1}\mathds{1}_{\left\\{\mathcal{T}_{i}<1\right\\}}\times(\tau^{0\to
1}_{i}\wedge[1-\mathcal{T}_{i}]),$ (104)
$\displaystyle\int_{0}^{1}\mathds{1}_{\left\\{\mathsf{dist}\left(\widetilde{U}_{1}(s),\widetilde{U}_{2}(s)\right)=0\right\\}}\mathrm{d}s$
$\displaystyle=\sum_{i\geq
1}\mathds{1}_{\left\\{\tilde{\mathcal{T}}_{i}<1\right\\}}\times(\tilde{\tau}^{0\to
1}_{i}\wedge[1-\tilde{\mathcal{T}}_{i}]).$ (105)

Let us define $\Upsilon$ by:

$\Upsilon=\sum_{i\geq
1}\mathds{1}_{\left\\{\tilde{\mathcal{T}}_{i}<1\right\\}}\times\tilde{\tau}^{0\to
1}_{i}.$ (106)

From now on, we fix a choice of $\beta_{2},\beta_{3}$ as in Lemma 32, and we
alway write $\mathrm{BAD}$ for $\mathrm{BAD}_{\beta_{2}}$. We have the
following results.

######  Lemma 39.

$\mathsf{dist}\left(\widetilde{U}_{1},\widetilde{U}_{2}\right)$ is a Markov
process with transition rates given in Figure 4.

######  Proposition 40 (Excursion coupling).

There is a coupling of $(\bar{U}_{1},\bar{U}_{2})$ and
$(\widetilde{U}_{1},\widetilde{U}_{2})$ where

$\mathsf{dist}\left(\bar{U}_{1}(0),\bar{U}_{2}(0)\right)=\mathsf{dist}\left(\widetilde{U}_{1}(0),\widetilde{U}_{2}(0)\right)=0$

such that, almost surely, for all $i\geq 1$,

$\displaystyle\tau^{0\to 1}_{i}$ $\displaystyle\geq\tilde{\tau}^{0\to 1}_{i},$
(107) $\displaystyle\tau^{1\to 0}_{i}$ $\displaystyle=\tilde{\tau}^{1\to
0}_{i}.$ (108)

######  Lemma 41 (Fast passage to $1$ from good environment).

There exists a constant $\beta>0$ such that

$\max_{t\geq 0}\max_{z\notin\mathrm{BAD},u\in
z}\mathbb{E}_{z,u,u}\left[(\tau^{0\to 1}_{1}\wedge
t)\times\mathds{1}_{\left\\{\bar{Z}_{s}\notin\mathrm{BAD},\,\forall
s\in[0,t]\right\\}}\right]\leq\beta\dfrac{\log^{2}L}{L^{2}}.$

######  Lemma 42 (Anti-concentration, integral form).

$\mathbb{E}\left[\Upsilon\right]=\mathcal{O}\left(1/L\right).$ (109)

######  Remark 11.

In dimension $2$, the bound on the right-hand side of (109) is replaced by
$\mathcal{O}\left(\dfrac{\log L}{L^{2}}\right)$.

Now we can prove Proposition 31.

###### Proof of Proposition 31..

Thanks to Lemma 36 and equation (99), we now only have to estimate
$\mathbb{E}\left[\int_{0}^{1}\mathds{1}_{\left\\{\mathsf{dist}\left(\bar{U}_{1}(s),\bar{U}_{2}(s)\right)=0\right\\}}\mathrm{d}s\right]$.
We have

$\displaystyle\mathbb{E}\left[\int_{0}^{1}\mathds{1}_{\left\\{\mathsf{dist}\left(\bar{U}_{1}(s),\bar{U}_{2}(s)\right)=0\right\\}}\mathrm{d}s\right]$
$\displaystyle=\mathbb{E}\left[\int_{0}^{1}\mathds{1}_{\left\\{\mathsf{dist}\left(\bar{U}_{1}(s),\bar{U}_{2}(s)\right)=0\right\\}}\mathrm{d}s\left(\mathds{1}_{\left\\{\exists
s\in[0,1]:\bar{Z}_{s}\in\mathrm{BAD}\right\\}}+\mathds{1}_{\left\\{\forall
s\in[0,1]:\bar{Z}_{s}\notin\mathrm{BAD}\right\\}}\right)\right].$ (110)

For the first term, we write

$\displaystyle\mathbb{E}\left[\int_{0}^{1}\mathds{1}_{\left\\{\mathsf{dist}\left(\bar{U}_{1}(s),\bar{U}_{2}(s)\right)=0\right\\}}\mathrm{d}s\mathds{1}_{\left\\{\exists
s\in[0,1]:\bar{Z}_{s}\in\mathrm{BAD}\right\\}}\right]$
$\displaystyle\leq\mathbb{P}\left[\exists
s\in[0,1]:\bar{Z}_{s}\in\mathrm{BAD}\right]$
$\displaystyle\leq\mathbb{P}\left[\exists
s\in[0,1]:Z_{s}\in\mathrm{BAD}\right],$ (111)

where we have used (100) in the last inequality. Now, we estimate the second
term. We will prove that

$\displaystyle\mathbb{E}\left[\int_{0}^{1}\mathds{1}_{\left\\{\mathsf{dist}\left(\bar{U}_{1}(s),\bar{U}_{2}(s)\right)=0\right\\}}\mathrm{d}s\mathds{1}_{\left\\{\forall
s\in[0,1]:\bar{Z}_{s}\notin\mathrm{BAD}\right\\}}\right]=\mathcal{O}\left(\log^{2}L\mathbb{E}\left[\Upsilon\right]\right).$
(112)

Thanks to (104), (105), and Lemma 42, we only need to prove that there exists
a constant $\beta$ such that

$\mathbb{E}\left[\mathds{1}_{\left\\{\forall
s\in[0,1]:\bar{Z}_{s}\notin\mathrm{BAD}\right\\}}\mathds{1}_{\left\\{\mathcal{T}_{i}<1\right\\}}\times(\tau^{0\to
1}_{i}\wedge[1-\mathcal{T}_{i}])\right]\leq\beta\log^{2}L\times\mathbb{E}\left[\mathds{1}_{\left\\{\tilde{\mathcal{T}}_{i}<1\right\\}}\times\tilde{\tau}^{0\to
1}_{i}\right].$ (113)

Let $\beta_{1}$ be the constant in Lemma 41. By the Markov property at time
$\mathcal{T}_{i}$,

$\displaystyle\mathbb{E}\left[\mathds{1}_{\left\\{\forall
s\in[0,1]:\bar{Z}_{s}\notin\mathrm{BAD}\right\\}}\mathds{1}_{\left\\{\mathcal{T}_{i}<1\right\\}}\times(\tau^{0\to
1}_{i}\wedge[1-\mathcal{T}_{i}])\right]$
$\displaystyle=\mathbb{E}\left[\mathds{1}_{\left\\{\forall
s\in[0,\mathcal{T}_{i}]:\bar{Z}_{s}\notin\mathrm{BAD}\right\\}}\mathds{1}_{\left\\{\mathcal{T}_{i}<1\right\\}}\times\mathbb{E}_{\bar{Z}(\mathcal{T}_{i}),\bar{U}_{1}(\mathcal{T}_{i}),\bar{U}_{2}(\mathcal{T}_{i})}\left[\mathds{1}_{\left\\{\forall
s\in[0,1-\mathcal{T}_{i}]:\bar{Z}_{s}\notin\mathrm{BAD}\right\\}}\times\left(\tau^{0\to
1}_{1}\wedge(1-\mathcal{T}_{i})\right)\right]\right]$
$\displaystyle\leq\mathbb{E}\left[\mathds{1}_{\left\\{\forall
s\in[0,\mathcal{T}_{i}]:\bar{Z}_{s}\notin\mathrm{BAD}\right\\}}\mathds{1}_{\left\\{\mathcal{T}_{i}<1\right\\}}\dfrac{\beta_{1}\log^{2}L}{L^{2}}\right]$
$\displaystyle\leq\dfrac{\beta_{1}\log^{2}L}{L^{2}}\mathbb{P}\left[\mathcal{T}_{i}<1\right]$
$\displaystyle\leq\dfrac{\beta_{1}\log^{2}L}{L^{2}}\mathbb{P}\left[\tilde{\mathcal{T}}_{i}<1\right].$
(114)

Here we have used Lemma 41 in the first inequality and the fact that
$\mathcal{T}_{i}\geq\tilde{\mathcal{T}}_{i}$ (due to Proposition 40) in the
last inequality. On the other hand,

$\displaystyle\mathbb{E}\left[\mathds{1}_{\left\\{\tilde{\mathcal{T}}_{i}<1\right\\}}\times\tilde{\tau}^{0\to
1}_{i}\right]$
$\displaystyle=\mathbb{E}\left[\mathds{1}_{\left\\{\tilde{\mathcal{T}}_{i}<1\right\\}}\mathbb{E}_{\widetilde{U}_{1}(\tilde{\mathcal{T}}_{i}),\widetilde{U}_{1}(\tilde{\mathcal{T}}_{i})}\left[\tilde{\tau}^{0\to
1}_{1}\right]\right]$
$\displaystyle=\mathbb{E}\left[\mathds{1}_{\left\\{\tilde{\mathcal{T}}_{i}<1\right\\}}\dfrac{1}{2L^{2}}\right].$
(115)

Here we have used the fact that
$\mathsf{dist}\left(\widetilde{U}_{1}(\tilde{\mathcal{T}}_{i}),\widetilde{U}_{2}(\tilde{\mathcal{T}}_{i})\right)=0$
by construction, and that the expected time needed for
$\mathsf{dist}\left(\widetilde{U}_{1},\widetilde{U}_{2}\right)$ to reach $1$
from this time onwards is $\dfrac{1}{2L^{2}}$, due to Lemma 39. So (114) and
(6.4) together imply (113). The equations (86), (99), (6.4), (112), (111)
together imply that

$\left\|\mathbb{P}_{\pi_{1}}\left[X_{t}\in\cdot\right]-\mathbb{P}_{\pi_{2}}\left[X_{t}\in\cdot\right]\right\|_{TV}^{2}=\mathcal{O}\left(\dfrac{\log^{2}L}{L}+\mathbb{P}_{z\cup\\{0\\}}\left[\exists
s\in[0,1]:Z_{s}\in\mathrm{BAD}\right]\right).$

Together with the inequality $\sqrt{a+b}\leq\sqrt{a}+\sqrt{b},\,a,b>0$, the
inequality above gives us what we want. ∎

It remains to prove Lemma 39, Proposition 40, Lemma 41, and Lemma 42. We will
need the following result to prove Lemma 39 and Proposition 40.

######  Lemma 43 (Strong lumpability of Markov process).

Let $\mathcal{X}$ and $\mathcal{Y}$ be two finite sets, and let
$f:\mathcal{X}\to\mathcal{Y}$ be a surjective function. Let $(X_{t})_{t\geq
0}$ be a Markov process with a generator $\mathcal{L}$ on $\mathcal{X}$. Then
$(f(X_{t}))_{t\geq 0}$ is a Markov process for any initial condition of
$(X_{t})_{t\geq 0}$ if and only if

$\displaystyle\forall y_{1},y_{2}\in\mathcal{Y},y_{1}\neq y_{2},\forall
x_{1},x^{\prime}_{1}\in f^{-1}(y_{1}),\;\sum_{x\in
f^{-1}(y_{2})}\mathcal{L}(x_{1},x)=\sum_{x\in
f^{-1}(y_{2})}\mathcal{L}(x^{\prime}_{1},x).$ (116)

Moreover, if (116) is satisfied, then the generator $\mathcal{L}^{\prime}$ of
$(f(X_{t}))_{t\geq 0}$ is determined by

$\displaystyle\forall y_{1},y_{2}\in\mathcal{Y},y_{1}\neq y_{2},\forall
x_{1}\in f^{-1}(y_{1}),\mathcal{L}^{\prime}(y_{1},y_{2})=\sum_{x\in
f^{-1}(y_{2})}\mathcal{L}(x_{1},x).$ (117)

Lemma 43 is a classical result. For its proof, the readers can see Theorem 2.4
in [4], see also Lemma 2.5 in [37] for the analogous result for discrete-time
Markov chains. Now we prove Lemma 39 and Proposition 40.

###### Proof of Lemma 39.

This is just a direct application of Lemma 43 for the process
$(\widetilde{U}_{1},\widetilde{U}_{2})$ and the function
$\mathsf{dist}\left(\cdot,\cdot\right)$. ∎

###### Proof of Proposition 40.

The intuition is that $(\bar{U}_{1},\bar{U}_{2})$ evolves exactly like
$(\widetilde{U}_{1},\widetilde{U}_{2})$ when $\bar{U}_{1}$ and $\bar{U}_{2}$
are not on a same site. This explains (108). The only difference is when
$\bar{U}_{1}$ and $\bar{U}_{2}$ are on a same site since they might be forced
to move together when their neighbors are not blue. This explains (107).

More precisely, for any initial configuration of
$(\bar{Z},\bar{U}_{1},\bar{U}_{2})$, consider the modified version
$(\bar{Z}^{mod},\bar{U}^{mod}_{1},\bar{U}^{mod}_{2})$ of
$(\bar{Z},\bar{U}_{1},\bar{U}_{2})$ which is stopped at
$E:=\\{(z,u,u)|z\in\mathcal{Z},u\in\Lambda_{L}\\}$. Then
$(\bar{Z}^{mod},\bar{U}^{mod}_{1},\bar{U}^{mod}_{2})$ is also a Markov
process, and up to the time that $(\bar{Z},\bar{U}_{1},\bar{U}_{2})$ reaches
$E$, those two processes are the same. The modified version
$(\widetilde{U}^{mod}_{1},\widetilde{U}^{mod}_{2})$ of
$(\widetilde{U}_{1},\widetilde{U}_{2})$ is defined similarly.

We apply Lemma 43 for the function

$\displaystyle f:(z,u_{1},u_{2})\mapsto\mathsf{dist}\left(u_{1},u_{2}\right)$

and the chain $(\bar{Z}^{mod},\bar{U}^{mod}_{1},\bar{U}^{mod}_{2})$ to see
that $f(\bar{Z}^{mod},\bar{U}^{mod}_{1},\bar{U}^{mod}_{2})$ is also a Markov
process. Let $\mathcal{L}$ be the generator associated with this process. By
direct computation, the process with generator $\mathcal{L}$ can be described
as in Figure 4, except that the transition from $0$ to $1$ is suppressed.

Similarly, we apply Lemma 43 for the function

$\displaystyle\tilde{f}:(u_{1},u_{2})\mapsto\mathsf{dist}\left(u_{1},u_{2}\right)$

and the process $(\widetilde{U}^{mod}_{1},\widetilde{U}^{mod}_{2})$ to see
that $\tilde{f}(\widetilde{U}^{mod}_{1},\widetilde{U}^{mod}_{2})$ is also a
Markov process. By direct computation, we can see that its generator is also
$\mathcal{L}$.

Our coupling is as follows. Let $\mathcal{G}_{t}$ be the $\sigma$-algebra
generated by $\Xi([0,t])$. Recall the definition of $(\mathcal{T}_{i})$ in
(101).

  * •

Conditionally on $\mathcal{G}_{\mathcal{T}_{i}}$, let $\tau_{i}$ be the first
time counted from $\mathcal{T}_{i}$ onwards that $\bar{U}_{1}$ meets a blue
mark and the two walks $\bar{U}_{1},\bar{U}_{2}$ take different decisions.
Then,

$\displaystyle\tau_{i}$ $\displaystyle\leq\tau^{0\to 1}_{i},$

where the inequality is strict if the other endpoint of the blue mark is a
black site.

  * •

Conditionally on $\mathcal{G}_{\mathcal{T}_{i}+\tau^{0\to 1}_{i}}$,
$\mathsf{dist}\left(\bar{U}_{1},\bar{U}_{2}\right)$ evolves as a Markov
process with generator $\mathcal{L}$ until it reaches $0$. This gives us the
excursion $\tau^{1\to 0}_{i}$.

Note that, by construction,

  * •

$\tau_{i}$ is independent of $\mathcal{G}_{\mathcal{T}_{i}}$, and hence of
$(\tau_{j},\tau^{1\to 0}_{j})_{1\leq j\leq i-1}$.

  * •

$\tau^{1\to 0}_{i}$ is independent of $\mathcal{G}_{\mathcal{T}_{i}+\tau^{0\to
1}_{i}}$ and hence of $(\tau_{j},\tau^{1\to 0}_{j})_{1\leq j\leq i-1}$.

  * •

For any $i$, $\tau_{i}\sim\exp(2L^{2})$.

  * •

For any $i$, $\tau^{1\to 0}_{i}$ is the time of an excursion of a Markov
process with generator $\mathcal{L}$ starting at state $1$ until reaching
state $0$.

Hence $(\tau_{i},\tau^{1\to 0}_{i})_{i\geq 1}$ are i.i.d., and they have the
same distribution as the excursions $(\tilde{\tau}^{0\to
1}_{i},\tilde{\tau}^{1\to 0}_{i})_{i\geq 1}$. This finishes our proof. ∎

Now we prove Lemma 41. We will need the following result.

######  Lemma 44 (Fast transition in good environment).

There exist constants $\beta,\beta^{\prime}>0$ such that

$\min_{z\notin\mathrm{BAD},u\in z}\mathbb{P}_{z,u,u}\left[\tau^{0\to
1}_{1}<\dfrac{\beta^{\prime}\log^{2}L}{L^{2}}\right]\geq\beta.$

###### Proof of Lemma 41.

Let $z\in\mathrm{BAD}$, $u\in z$ arbitrary. Let $\beta,\beta^{\prime}$ be as
in Lemma 44. Then

$\displaystyle\mathbb{E}_{z,u,u}\left[(\tau^{0\to 1}_{1}\wedge
t)\times\mathds{1}_{\left\\{\bar{Z}_{s}\notin\mathrm{BAD},\,\forall
s\in[0,t]\right\\}}\right]$ $\displaystyle\leq\sum_{i=0}^{\left\lfloor
tL^{2}/(\beta^{\prime}\log^{2}L)\right\rfloor}\mathbb{E}_{z,u,u}\left[\tau^{0\to
1}_{1}\times\mathds{1}_{\left\\{\bar{Z}_{s}\notin\mathrm{BAD},\,\forall
s\in[0,t]\right\\}}\mathds{1}_{\left\\{\dfrac{i\beta^{\prime}\log^{2}L}{L^{2}}\leq\tau^{0\to
1}_{1}<\dfrac{(i+1)\beta^{\prime}\log^{2}L}{L^{2}}\right\\}}\right]$
$\displaystyle\;\;\;\;+\mathbb{E}_{z,u,u}\left[t\times\mathds{1}_{\left\\{\bar{Z}_{s}\notin\mathrm{BAD},\,\forall
s\in[0,t]\right\\}}\mathds{1}_{\left\\{\tau^{0\to 1}_{1}>t\right\\}}\right]$
$\displaystyle\leq\left(\sum_{i=0}^{\left\lfloor
tL^{2}/(\beta^{\prime}\log^{2}L)\right\rfloor}\dfrac{(i+1)\beta^{\prime}\log^{2}L}{L^{2}}\mathbb{E}_{z,u,u}\left[\mathds{1}_{\left\\{\bar{Z}_{s}\notin\mathrm{BAD},\,\forall
s\in[0,t]\right\\}}\mathds{1}_{\left\\{\dfrac{i\beta^{\prime}\log^{2}L}{L^{2}}\leq\tau^{0\to
1}_{1}\right\\}}\right]\right)$
$\displaystyle\;\;\;\;+t\mathbb{E}_{z,u,u}\left[\mathds{1}_{\left\\{\bar{Z}_{s}\notin\mathrm{BAD},\,\forall
s\in[0,t]\right\\}}\mathds{1}_{\left\\{\tau^{0\to 1}_{1}>\left\lfloor
tL^{2}/(\beta^{\prime}\log^{2}L)\right\rfloor(\beta^{\prime}\log^{2}L)/(L^{2})\right\\}}\right]$
$\displaystyle\leq\left(\sum_{i=0}^{\left\lfloor
tL^{2}/(\beta^{\prime}\log^{2}L)\right\rfloor}\dfrac{(i+1)\beta^{\prime}\log^{2}L}{L^{2}}(1-\beta)^{i}\right)+t(1-\beta)^{\left\lfloor
tL^{2}/(\beta^{\prime}\log^{2}L)\right\rfloor}$
$\displaystyle=\mathcal{O}\left(\log^{2}L/L^{2}\right).$

∎

Now we prove Lemma 44. We recall the following classical result about the
hitting time of a simple random walk.

######  Lemma 45 (Hitting time of a symmetric simple random walk).

Let $U$ be a simple random walk on $\mathbb{Z}$ where edges have conductance
$1$. Let $\tau:=\inf\left\\{t\geq 0:U(t)=0\right\\}$ be the hitting time of
the site $0$. Then there exists a constant $\beta>0$ such that

$\displaystyle\max_{u\in\mathbb{Z}}\mathbb{P}_{u}\left[\tau\leq
2|u|^{2}\right]>\beta.$ (118)

The conclusion holds also for a walk in $\mathbb{Z}^{2}$.

###### Proof of Lemma 44.

First, consider the case where one of the neighbors of $u$, say $u+1$, is
blue. Then there is a probability bounded away from $0$ that a blue mark
appears on edge $(u,u+1)$ before time $1/L^{2}$ and before any blue or black
mark appears on edges $(u+1,u+2)$ and $(u-1,u)$. When this happens, with
probability $1/2$, the two walks $\bar{U}_{1},\bar{U}_{2}$ can take different
decisions. Therefore, $\tau^{0\to 1}_{1}<1/L^{2}$ with a probability greater
than some constant $\beta_{4}>0$.

Coming back to the general case, since $z\notin\mathrm{BAD}$, there exists a
blue site $u^{\prime}$ at distance at most $2\beta_{2}\log L$ from $u$. Let
$U_{3}$ be the walk starting at $u^{\prime}$, which behaves like $U_{1}$ (see
paragraph “the perturbation position”) when it meets the marks.

Let $\beta_{1}$ be a constant that we will choose later. The number of Glauber
marks appearing in the interval
$\left[0,\dfrac{\beta_{1}\log^{2}L}{L^{2}}\right]$ is a Poisson variable with
average $\dfrac{\lambda\beta_{1}\log^{2}L}{L}$. Hence, for $L$ large enough,

$\displaystyle\mathbb{P}\left[\text{no Glauber mark appears in
$\left[0,\dfrac{\beta_{1}\log^{2}L}{L^{2}}\right]$}\right]\geq 1/2.$

Conditionally on this event, $U_{3}$ and $\bar{U}_{1}$ evolve as two
independent random walks on $\Lambda_{L}$ in
$\left[0,\dfrac{\beta_{1}\log^{2}L}{L^{2}}\right]$, until they are at distance
$1$ from each other. Let

$\displaystyle\tau:=\inf\left\\{t\geq
0|\mathsf{dist}\left(U_{3}(t),\bar{U}_{1}(t)\right)=1\right\\}.$

Note that $U_{3}-\bar{U}_{1}$ is a simple random walk on $\Lambda_{L}$ where
edges have conductance $2L^{2}$ until it reaches $1$ or $-1$. So for $\beta$
as in Lemma 45, there exists a constant $\beta_{1}$ such that

$\displaystyle\mathbb{P}\left[\tau<\dfrac{\beta_{1}\log^{2}L}{L^{2}}\Bigg{|}\text{no
Glauber mark appears in
$\left[0,\dfrac{\beta_{1}\log^{2}L}{L^{2}}\right]$}\right]\geq\beta.$

Hence,

$\displaystyle\mathbb{P}\left[\tau<\dfrac{\beta_{1}\log^{2}L}{L^{2}},\text{no
Glauber mark appears in
$\left[0,\dfrac{\beta_{1}\log^{2}L}{L^{2}}\right]$}\right]\geq\beta/2.$

By the strong Markov property at time $\tau$ and the first case, we can deduce
that

$\displaystyle\mathbb{P}\left[\tau^{0\to
1}_{1}\leq\tau+1/L^{2},\tau<\dfrac{\beta_{1}\log^{2}L}{L^{2}},\text{no Glauber
mark appears in
$\left[0,\dfrac{\beta_{1}\log^{2}L}{L^{2}}\right]$}\right]\geq\beta_{4}\beta/2.$

This finishes our proof. ∎

######  Remark 12.

The proof in dimension $d=2$ is similar. Lemma 45 is still valid for this
dimension. $U_{3}$ and $\bar{U}_{1}$ still evolve as 2 independent walks until
they are at distance $1$ from each others or until $U_{3}$ is killed when a
Glauber mark appears. The number of Glauber marks appearing in the time
interval $\left[0,\dfrac{\beta_{1}\log^{2}L}{L^{2}}\right]$ is
$\mathcal{O}\left(\log^{2}L\right)$. Each time there is a ring among the
Glauber clocks, the chance that the mark appearing kills $U_{3}$ is
$\mathcal{O}\left(1/L^{2}\right)$. So the chance that $U_{3}$ is killed in the
time interval $\left[0,\dfrac{\beta_{1}\log^{2}L}{L^{2}}\right]$ is
$\mathcal{O}\left(\dfrac{\log^{2}L}{L^{2}}\right)=o(1)$. So $U_{3}$ can still
meet $\bar{U}_{1}$ in time interval
$\left[0,\dfrac{\beta_{1}\log^{2}L}{L^{2}}\right]$ with a chance bounded away
from $0$, and the proof is adapted accordingly.

It remains to prove Lemma 42. It is not hard to prove the result if we replace
$(\widetilde{U}_{1},\widetilde{U}_{2})$ with two independent SRWs. We will
prove the results for two independent SRWs and use a comparison argument to
transfer the result back to the pair $(\widetilde{U}_{1},\widetilde{U}_{2})$.

Let $\check{\mathcal{U}}_{1}$ be a SRW starting from $0$ on $\Lambda_{L}$,
where every edge has conductance $L^{2}$. Let $\check{\mathcal{U}}_{2}$ be an
independent copy of $\check{\mathcal{U}}_{1}$. Let $(\check{\tau}^{0\to
1}_{i},\check{\tau}^{1\to 0}_{i},\check{\mathcal{T}}_{i})_{i\geq 1}$ be
defined as in (101), (102), (103) when we replace $(\bar{U}_{1},\bar{U}_{2})$
by $(\check{\mathcal{U}}_{1},\check{\mathcal{U}}_{2})$. For two real random
variables $\zeta_{1},\zeta_{2}$, we write
$\zeta_{1}\overset{d}{\leq}\zeta_{2}$ if $\zeta_{2}$ dominates stochastically
$\zeta_{1}$. We will need the following results.

Figure 5: Transition rate of
$\mathsf{dist}\left(\check{\mathcal{U}}_{1},\check{\mathcal{U}}_{2}\right)$
when $L$ is even.

######  Lemma 46.

$\displaystyle\check{\tau}^{1\to 0}_{1}$
$\displaystyle\overset{d}{\leq}\tilde{\tau}^{1\to 0}_{1},$ (119)
$\displaystyle\check{\tau}^{0\to 1}_{1}$
$\displaystyle\overset{d}{\leq}\tilde{\tau}^{0\to 1}_{1}.$ (120)

######  Lemma 47.

$\displaystyle\sum_{i=1}^{\infty}\mathbb{P}\left[\check{\mathcal{T}}_{i}<1\right]=\mathcal{O}\left(L\right).$
(121)

Now we prove Lemma 42.

###### Proof of Lemma 42.

In this proof, $\mathbb{E}_{0}\left[\cdot\right]$ means the expectation taken
with respect to the law of the process
$\mathsf{dist}\left(\widetilde{U}_{1},\widetilde{U}_{2}\right)$ starting from
$0$. We have

$\displaystyle\mathbb{E}\left[\Upsilon\right]$ $\displaystyle=\sum_{i\geq
1}\mathbb{E}\left[\mathds{1}_{\left\\{\tilde{\mathcal{T}}_{i}<1\right\\}}\times\check{\tau}^{0\to
1}_{i}\right]$ $\displaystyle=\sum_{i\geq
1}\mathbb{E}\left[\mathds{1}_{\left\\{\tilde{\mathcal{T}}_{i}<1\right\\}}\times\mathbb{E}_{0}\left[\tilde{\tau}^{0\to
1}_{1}\right]\right]$ $\displaystyle=\sum_{i\geq
1}\dfrac{1}{2L^{2}}\mathbb{P}\left[\tilde{\mathcal{T}}_{i}<1\right],$

where we have used the Markov property for the process
$\mathsf{dist}\left(\widetilde{U}_{1},\widetilde{U}_{2}\right)$ in the second
equality, and the formula of the generator of
$\mathsf{dist}\left(\widetilde{U}_{1},\widetilde{U}_{2}\right)$ in the third
equality. Note that, by Lemma 46,
$\check{\mathcal{T}}_{i}\overset{d}{\leq}\tilde{\mathcal{T}}_{i},\,\forall
i\geq 1$. Therefore,

$\displaystyle\sum_{i\geq
1}\dfrac{1}{2L^{2}}\mathbb{P}\left[\tilde{\mathcal{T}}_{i}<1\right]$
$\displaystyle\leq\dfrac{1}{2L^{2}}\sum_{i\geq
1}\mathbb{P}\left[\check{\mathcal{T}}_{i}<1\right]$
$\displaystyle=\dfrac{1}{2L^{2}}\mathcal{O}\left(L\right)$
$\displaystyle=\mathcal{O}\left(L\right).$

This finishes our proof. ∎

It remains to prove Lemma 46 and Lemma 47. We first prove Lemma 46.

###### Proof of Lemma 46.

See Figure 4 and 5 for intuition. A direct application of Lemma 43 shows that
$\mathsf{dist}\left(\check{\mathcal{U}}_{1},\check{\mathcal{U}}_{2}\right)$ is
a Markov process with transition rates as in Figure 5. Then we see that

$\displaystyle\tilde{\tau}^{0\to 1}_{1}$ $\displaystyle\sim\exp(2L^{2}),$
$\displaystyle\check{\tau}^{0\to 1}_{1}$ $\displaystyle\sim\exp(4L^{2}).$

This proves (120). Now let $\tau^{2\to 1}$ be the (random) time it takes for
$\mathsf{dist}\left(\widetilde{U}_{1},\widetilde{U}_{2}\right)$ to reach state
$1$ from state $2$. By Figure 4 and Figure 5, it is also the time it takes for
$\mathsf{dist}\left(\check{\mathcal{U}}_{1},\check{\mathcal{U}}_{2}\right)$ to
reach state $1$ from state $2$. Let $(\zeta_{i})_{i\geq 1},(\tau^{2\to
1}_{i})_{i\geq 1}$, and $(\xi_{i})_{i\geq 1}$ be three independent sequences
of i.i.d. random variables such that

  * •

$(\zeta_{i})_{i\geq 1}$ is a sequence of i.i.d. $\exp(L^{2})$.

  * •

$(\tau^{2\to 1}_{i})_{i\geq 1}$ is a sequence of independent copies of
$\tau^{2\to 1}$.

  * •

$(\xi_{i})_{i\geq 1}$ is a sequence of i.i.d. Bernoulli random variables with
parameter $2/3$.

Then we see that

$\displaystyle\tilde{\tau}^{1\to 0}_{1}$
$\displaystyle\overset{d}{=}\dfrac{\zeta_{1}}{3}+\xi_{1}\tau^{2\to
1}_{1}+\xi_{1}\dfrac{\zeta_{2}}{3}+\xi_{1}\xi_{2}\tau^{2\to 1}_{2}+\dots$
$\displaystyle=\sum_{i=1}^{\infty}\left[\left(\dfrac{\zeta_{i}}{3}\prod_{j=1}^{i-1}\xi_{i}\right)+\left(\tau^{2\to
1}_{i}\prod_{j=1}^{i}\xi_{i}\right)\right].$

This is because to go to state $0$ from state $1$, first, we need to jump out
of state $1$, which takes a time $\exp(3L^{2})$. Then, with probability $1/3$,
we succeed to jump to state $0$, and with probability $2/3$, we jump to state
$2$. Then we have to come back to state $1$ and try again. Similarly, let
$(\xi^{\prime}_{i})_{i\geq 1}$ be a sequence of i.i.d. Bernoulli random
variables with parameter $1/2$, independent of $(\zeta_{i})_{i\geq 1}$ and
$(\tau^{2\to 1}_{i})_{i\geq 1}$. Then

$\displaystyle\check{\tau}^{1\to
0}_{1}\overset{d}{=}\sum_{i=1}^{\infty}\left[\left(\dfrac{\zeta_{i}}{4}\prod_{j=1}^{i-1}\xi^{\prime}_{i}\right)+\left(\tau^{2\to
1}_{i}\prod_{j=1}^{i}\xi^{\prime}_{i}\right)\right].$

Since $\xi^{\prime}_{i}\overset{d}{\leq}\xi_{i}$, we conclude that
$\check{\tau}^{1\to 0}_{1}\overset{d}{\leq}\tilde{\tau}^{1\to 0}_{1}$, which
finishes our proof. ∎

Finally, we show Lemma 47. We will need the following classical result about
anticoncentration of SRW on the lattice.

######  Lemma 48 (Local limit theorem).

Let $\check{\mathcal{U}}$ be a SRW on $\Lambda_{L}$ starting from $0$, where
every edge has conductance $L^{2}$. Then,

$\displaystyle\max_{u\in\Lambda_{L}}\mathbb{P}_{0}\left[\check{\mathcal{U}}(t)=u\right]=\mathcal{O}\left(\dfrac{1}{L}\left(\dfrac{1}{\sqrt{t}}+1\right)\right).$

###### Proof of Lemma 47.

Let us define

$\displaystyle\check{\Upsilon}:=\sum_{i\geq
1}\mathds{1}_{\left\\{\check{\mathcal{T}}_{i}<1\right\\}}\times\check{\tau}^{0\to
1}_{i}.$

We see that

$\displaystyle\check{\Upsilon}-\int_{0}^{1}\mathds{1}_{\left\\{\mathsf{dist}\left(\check{\mathcal{U}}_{1}(s),\check{\mathcal{U}}_{2}(s)\right)=0\right\\}}\mathrm{d}s=\mathds{1}_{\left\\{\mathsf{dist}\left(\check{\mathcal{U}}_{1}(1),\check{\mathcal{U}}_{2}(1)\right)=0\right\\}}\times\left(\inf\left\\{t\geq
1|\mathsf{dist}\left(\check{\mathcal{U}}_{1}(t),\check{\mathcal{U}}_{2}(t)\right)=1\right\\}-1\right)$

Taking the expectation and using the Markov property at time $1$, we get

$\displaystyle\mathbb{E}\left[\check{\Upsilon}-\int_{0}^{1}\mathds{1}_{\left\\{\mathsf{dist}\left(\check{\mathcal{U}}_{1}(s),\check{\mathcal{U}}_{2}(s)\right)=0\right\\}}\mathrm{d}s\right]$
$\displaystyle=\mathbb{P}\left[\mathsf{dist}\left(\check{\mathcal{U}}_{1}(1),\check{\mathcal{U}}_{2}(1)\right)=0\right]\mathbb{E}\left[\check{\tau}^{0\to
1}_{1}\right]$ $\displaystyle\leq 1\times\dfrac{1}{4L^{2}}$
$\displaystyle=\dfrac{1}{4L^{2}}.$

Note that
$\mathds{1}_{\left\\{\mathsf{dist}\left(\check{\mathcal{U}}_{1}(s),\check{\mathcal{U}}_{2}(s)\right)=0\right\\}}=\mathds{1}_{\left\\{\check{\mathcal{U}}_{1}(s)-\check{\mathcal{U}}_{2}(s)=0\right\\}}$,
and note also that
$(\check{\mathcal{U}}_{1}(s)-\check{\mathcal{U}}_{2}(s))_{s\geq 0}$ is a SRW
on the lattice, where edges have conductance $2L^{2}$. Hence by Lemma 48,

$\displaystyle\mathbb{P}\left[\check{\mathcal{U}}_{1}(s)-\check{\mathcal{U}}_{2}(s)=0\right]=\mathcal{O}\left(\dfrac{1}{L}\left(\dfrac{1}{\sqrt{s}}+1\right)\right).$

Therefore,

$\displaystyle\mathbb{E}\left[\check{\Upsilon}\right]$
$\displaystyle\leq\dfrac{1}{4L^{2}}+\mathbb{E}\left[\int_{0}^{1}\mathds{1}_{\left\\{\mathsf{dist}\left(\check{\mathcal{U}}_{1}(s),\check{\mathcal{U}}_{2}(s)\right)=0\right\\}}\mathrm{d}s\right]$
$\displaystyle=\dfrac{1}{4L^{2}}+\int_{0}^{1}\mathbb{P}\left[\check{\mathcal{U}}_{1}(s)-\check{\mathcal{U}}_{2}(s)=0\right]\mathrm{d}s$
$\displaystyle=\dfrac{1}{4L^{2}}+\mathcal{O}\left(\int_{0}^{1}\dfrac{1}{L}\left(\dfrac{1}{\sqrt{s}}+1\right)\mathrm{d}s\right)$
$\displaystyle=\mathcal{O}\left(\dfrac{1}{L}\right).$

On the other hand,

$\displaystyle\mathbb{E}\left[\check{\Upsilon}\right]=\sum_{i\geq
1}\mathbb{P}\left[\check{\mathcal{T}}_{i}<1\right]\times\dfrac{1}{4L^{2}}.$

The two equations above imply that

$\displaystyle\sum_{i\geq
1}\mathbb{P}\left[\check{\mathcal{T}}_{i}<1\right]=\mathcal{O}\left(L\right),$

which finishes our proof. ∎

## Appendix A

In this appendix, we prove Lemma 21. We will need the following result.

######  Lemma 49 (Anticoncentration of SRW, integral form).

Let $\check{\mathcal{U}}_{1},\check{\mathcal{U}}_{2}$ be two i.i.d. SRWs on
$\Lambda^{d}_{L}$ where edges have conductance $L^{2}$. Let $\theta$ be a
strictly positive number and $\zeta\sim\exp(\theta)$, independent of
$(\check{\mathcal{U}}_{1},\check{\mathcal{U}}_{2})$. Then, for any
$k\in\mathbb{Z}_{+}$,

$\max_{u_{1},u_{2}\in\Lambda^{d}_{L}}\mathbb{P}_{u_{1},u_{2}}\left[\mathsf{dist}\left(\check{\mathcal{U}}_{1}(\zeta),\check{\mathcal{U}}_{2}(\zeta)\right)\leq
k\right]=\begin{cases}\mathcal{O}_{\theta,k}(1/L)&\text{if $d=1$},\\\
\mathcal{O}_{\theta,k}(\log L/L^{2})&\text{if $d=2$},\\\
\mathcal{O}_{\theta,k}(1/L^{2})&\text{if $d\geq 3$}.\end{cases}$

For $u_{1},u_{2}\in\Lambda_{L}$, we write $u_{1}\preceq u_{2}$ if
$\mathsf{dist}\left(u_{1},0\right)\leq\mathsf{dist}\left(u_{2},0\right)$. In
higher dimensions, for $u_{1},u_{2}\in\Lambda^{d}_{L}$, we write $u_{1}\preceq
u_{2}$ if there exists a permutation $\sigma$ of $[d]$ such that $\forall
i\in[d],u_{1}(i)\preceq u_{2}(\sigma(i))$. We will also need the following
result.

######  Lemma 50 (Distance between $2$ walks in an $IP(2)$ is bigger than that
between 2 independent SRWs).

Consider the lattice $\Lambda^{d}_{L}$ where every edge has conductance
$L^{2}$. Let $(\mathcal{U}_{1},\mathcal{U}_{2})$ be an IP(2) and
$(\check{\mathcal{U}}_{1},\check{\mathcal{U}}_{2})$ be a pair of independent
SRWs on $\Lambda^{d}_{L}$. There is a Markovian coupling of
$(\mathcal{U}_{1},\mathcal{U}_{2})$ and
$(\check{\mathcal{U}}_{1},\check{\mathcal{U}}_{2})$ such that if the initial
condition $(u_{1},u_{2},\check{u}_{1},\check{u}_{2})$ satisfies
$(\check{u}_{2}-\check{u}_{1})\preceq(u_{2}-u_{1})$, then almost surely,

$\displaystyle\forall t\geq
0,\,(\check{\mathcal{U}}_{2}(t)-\check{\mathcal{U}}_{1}(t))\preceq(\mathcal{U}_{2}(t)-\mathcal{U}_{1}(t)),$

and in particular,

$\displaystyle\mathsf{dist}\left(\check{\mathcal{U}}_{2}(t),\check{\mathcal{U}}_{1}(t)\right)\leq\mathsf{dist}\left(\mathcal{U}_{2}(t),\mathcal{U}_{1}(t)\right).$

Clearly, Lemma 49 and Lemma 50 together imply Lemma 21.

###### Proof of Lemma 21.

Let $u_{1},u_{2}\in\Lambda_{L}$. Consider the coupling of the IP(2)
$(\mathcal{U}_{1},\mathcal{U}_{2})$ and a pair of independent SRWs
$(\check{\mathcal{U}}_{1},\check{\mathcal{U}}_{2})$, both starting from
$u_{1},u_{2}$, independent of $\zeta$, that satisfies Lemma 50. Then, by Lemma
50,

$\displaystyle\mathbb{P}\left[\mathsf{dist}\left(\mathcal{U}_{1}(\zeta),\mathcal{U}_{2}(\zeta)\right)\leq
k\right]\leq\mathbb{P}\left[\mathsf{dist}\left(\check{\mathcal{U}}_{1}(\zeta),\check{\mathcal{U}}_{2}(\zeta)\right)\leq
k\right].$

This and Lemma 49 lead to what we want. ∎

Lemma 49 is just the integral form of Lemma 48.

###### Proof of Lemma 49.

We denote by $\check{\mathcal{U}}_{1}(t,i)$ the $i$-th coordinate of
$\check{\mathcal{U}}_{1}(t)$, and similarly for $\check{\mathcal{U}}_{2}$. We
see that
$(\check{\mathcal{U}}_{1}(\cdot,i)-\check{\mathcal{U}}_{2}(\cdot,i))_{1\leq
i\leq d}$ are $d$ independent SRWs on $\Lambda_{L}$, where every edge has
conductance $2L^{2}$. Hence, for any $t\geq 0$,

$\displaystyle\mathbb{P}\left[\mathsf{dist}\left(\mathcal{U}_{1}(t),\mathcal{U}_{2}(t)\right)\leq
k\right]$
$\displaystyle\leq\prod_{i=1}^{d}\mathbb{P}\left[\left(\check{\mathcal{U}}_{1}(t,i)-\check{\mathcal{U}}_{2}(t,i)\right)\in\\{-k,\dots,k\\}\right]$
$\displaystyle\leq\prod_{i=1}^{d}\left((2k+1)\max_{u\in\Lambda_{L}}\mathbb{P}\left[\check{\mathcal{U}}_{1}(t,i)-\check{\mathcal{U}}_{2}(t,i)=u\right]\right)$
$\displaystyle=\mathcal{O}_{k}\left(\dfrac{1}{L^{d}}\left(\dfrac{1}{t^{d/2}}+1\right)\right).$

Note that

$\displaystyle\mathbb{P}\left[\mathsf{dist}\left(\check{\mathcal{U}}_{1}(\zeta),\check{\mathcal{U}}_{2}(\zeta)\right)\leq
k\right]=\int_{0}^{\infty}\mathbb{P}\left[\mathsf{dist}\left(\check{\mathcal{U}}_{1}(t),\check{\mathcal{U}}_{2}(t)\right)\leq
k\right]\theta e^{-\theta t}\mathrm{d}t.$

We split the integral above into three intervals
$[0,1/L^{2}],\,[1/L^{2},1],\,[1,\infty)$. First, we estimate the integral on
$[0,1/L^{2}]$:

$\displaystyle\int_{0}^{1/L^{2}}\mathbb{P}\left[\mathsf{dist}\left(\check{\mathcal{U}}_{1}(t),\check{\mathcal{U}}_{2}(t)\right)\leq
k\right]\theta e^{-\theta t}\mathrm{d}t$
$\displaystyle\leq\int_{0}^{1/L^{2}}\theta e^{-\theta t}\mathrm{d}t$
$\displaystyle\leq\theta/L^{2}.$

Now we estimate the integral on $[1,\infty)$:

$\displaystyle\int_{1}^{\infty}\mathbb{P}\left[\mathsf{dist}\left(\check{\mathcal{U}}_{1}(t),\check{\mathcal{U}}_{2}(t)\right)\leq
k\right]\theta e^{-\theta t}\mathrm{d}t$
$\displaystyle=\mathcal{O}_{k}\left(\int_{1}^{\infty}\dfrac{1}{L^{d}}\left(\dfrac{1}{t^{d/2}}+1\right)\theta
e^{-\theta t}\mathrm{d}t\right)$
$\displaystyle=\mathcal{O}_{k}\left(\dfrac{1}{L^{d}}\int_{1}^{\infty}\theta
e^{-\theta t}\mathrm{d}t\right)$
$\displaystyle=\mathcal{O}_{k}\left(\dfrac{1}{L^{d}}\right).$

Finally, we estimate the integral on $[1/L^{2},1]$:

$\displaystyle\int_{1/L^{2}}^{1}\mathbb{P}\left[\mathsf{dist}\left(\check{\mathcal{U}}_{1}(t),\check{\mathcal{U}}_{2}(t)\right)\leq
k\right]\theta e^{-\theta t}\mathrm{d}t$
$\displaystyle=\mathcal{O}_{k}\left(\int_{1/L^{2}}^{1}\dfrac{1}{L^{d}}\left(\dfrac{1}{t^{d/2}}+1\right)\theta
e^{-\theta t}\mathrm{d}t\right)$
$\displaystyle=\mathcal{O}_{k}\left(\dfrac{\theta}{L^{d}}\int_{1/L^{2}}^{1}\dfrac{1}{t^{d/2}}\mathrm{d}t\right)$
$\displaystyle=\begin{cases}\mathcal{O}_{k}\left(\dfrac{\theta}{L}\left(2\sqrt{t}\Big{|}^{1}_{1/L^{2}}\right)\right)=\mathcal{O}_{k}\left(\dfrac{\theta}{L}\right)&\text{if
$d=1$},\\\ \mathcal{O}_{k}\left(\dfrac{\theta}{L^{2}}\left(\log
t\Big{|}^{1}_{1/L^{2}}\right)\right)=\mathcal{O}_{k}\left(\dfrac{\theta\log
L}{L^{2}}\right)&\text{if $d=2$},\\\
\mathcal{O}_{k}\left(\dfrac{\theta}{L^{d}}\left(\dfrac{t^{1-d/2}}{1-d/2}\Bigg{|}^{1}_{1/L^{2}}\right)\right)=\mathcal{O}_{k}\left(\dfrac{\theta}{L^{2}}\right)&\text{if
$d\geq 3$}.\end{cases}$

The estimates above lead to what we want. ∎

Now we prove Lemma 50.

###### Proof of Lemma 50.

The intuition is simple: an IP(2) evolves exactly like $2$ independent SRWs
when the two particles are not next to each other. The distance between 2
walks in IP(2) is forbidden to jump to $0$, while the distance between two
independent SRWs can. This makes the distance between 2 particles in $IP(2)$
always bigger than that of $2$ independent SRWs in distribution.

More formally, note that $Y:=(\mathcal{U}_{1}-\mathcal{U}_{2})$ and
$\tilde{Y}:=(\check{\mathcal{U}}_{1}-\check{\mathcal{U}}_{2})$ are also Markov
processes. We only have to construct a coupling of $Y$ and $\tilde{Y}$.

Let $\mathcal{L}$ (resp. $\tilde{\mathcal{L}}$) be the generator of $Y$ (resp
$\tilde{Y}$). Let $\mathfrak{e}_{1},\dots,\mathfrak{e}_{d}$ be the unit
vectors in $\Lambda^{d}_{L}$. Then

$\displaystyle\tilde{\mathcal{L}}(u,u^{\prime})=\begin{cases}2L^{2}&\text{ if
$u^{\prime}=u\pm\mathfrak{e}_{i}$ for some $i\in[d]$},\\\ 0&\text{ if
not},\end{cases}$

and

$\displaystyle\mathcal{L}(u,u^{\prime})=\begin{cases}2L^{2}&\text{ if
$u^{\prime}=u\pm\mathfrak{e}_{i}$ for some $i\in[d]$ and $u^{\prime}\neq
0$},\\\ L^{2}&\text{ if
$(u,u^{\prime})\in\\{(\pm\mathfrak{e}_{i},\mp\mathfrak{e}_{i})|i\in[d]\\}$},\\\
0&\text{ if not},\end{cases}$

To construct the coupling kernel $\mathcal{L}^{coupling}$, we only need to
construct the jump rates from each configuration $(u,\check{u})$ such that
$u\preceq\check{u}$. Without loss of generality, we can suppose that
$\mathsf{dist}\left(u(i),0\right)\leq\mathsf{dist}\left(\check{u}(i),0\right),\,\forall
i\in[d]$. The kernel is as follows. For any coordinate $i$,

  * •

If
$\mathsf{dist}\left(u(i),0\right)<\mathsf{dist}\left(\check{u}(i),0\right)$,

$\displaystyle\mathcal{L}^{coupling}((u,\check{u}),(u\pm\mathfrak{e}_{i},\check{u}))=\mathcal{L}^{coupling}((u,\check{u}),(u,\check{u}\pm\mathfrak{e}_{i}))=2L^{2}.$

  * •

If
$\mathsf{dist}\left(u(i),0\right)=\mathsf{dist}\left(\check{u}(i),0\right)$,
then $u(i)=\pm\check{u}(i)$, and therefore there exists $\xi\in\\{-1,1\\}$
such that $\mathsf{dist}\left(u(i)\pm
1,0\right)=\mathsf{dist}\left(\check{u}(i)\pm\xi,0\right)$. Then

$\displaystyle\mathcal{L}^{coupling}((u,\check{u}),(u\pm\mathfrak{e}_{i},\check{u}\pm\xi\mathfrak{e}_{i}))=2L^{2}\mathds{1}_{\left\\{u\pm
e_{i}\neq 0\right\\}}.$

If $u\in\\{\pm\mathfrak{e}_{i}\\}$, then we necessarily have
$\check{u}\in\\{\pm\mathfrak{e}_{i}\\}$, and in this case,

$\displaystyle\mathcal{L}^{coupling}((u,\check{u}),(-u,\check{u}))$
$\displaystyle=L^{2},$
$\displaystyle\mathcal{L}^{coupling}((u,\check{u}),(u,0))$
$\displaystyle=2L^{2}.$

For any $(u^{\prime},\check{u}^{\prime})\neq(u,\check{u})$ not listed in the
cases above,

$\displaystyle\mathcal{L}^{coupling}((u,\check{u}),(u^{\prime},\check{u}^{\prime}))=0.$

We can verify that this gives us a coupling kernel for $(Y,\tilde{Y})$ that
preserves the relation $\preceq$. This finishes our proof. ∎

## References

  * [1] M. Aizenman and R. Holley.  Rapid convergence to equilibrium of stochastic Ising models in the Dobrushin Shlosman regime.  In Percolation theory and ergodic theory of infinite particle systems (Minneapolis, Minn., 1984–1985), volume 8 of IMA Vol. Math. Appl., pages 1–11. Springer, New York, 1987.
  * [2] D. Aldous and P. Diaconis.  Shuffling cards and stopping times.  American Mathematical Monthly, 93(5):333–348, 1986.
  * [3] D. Aldous and D. Lanoue.  A lecture on the averaging process.  Probability Surveys, 9:90–102, 2012.
  * [4] F. Ball and G. F. Yeo.  Lumpability and marginalisability for continuous-time Markov chains.  Journal of Applied Probability, 30(3):518–528, 1993.
  * [5] L. Bertini, A. De Sole, D. Gabrielli, G. Jona-Lasinio, and C. Landim.  Large deviations for the boundary driven symmetric simple exclusion process.  Mathematical Physics, Analysis and Geometry. An International Journal Devoted to the Theory and Applications of Analysis and Geometry to Physics, 6(3):231–267, 2003.
  * [6] A. Bovier and F. den Hollander.  Metastability, volume 351 of Grundlehren der mathematischen Wissenschaften [Fundamental Principles of Mathematical Sciences].  Springer, Cham, 2015.  A potential-theoretic approach.
  * [7] P. Caputo, M. Quattropani, and F. Sau.  Cutoff for the averaging process on the hypercube and complete bipartite graphs.  Electronic Journal of Probability, 28:Paper No. 100, 31, 2023.
  * [8] J. Chayes, L. Chayes, and R. Schonmann.  Exponential decay of connectivities in the two-dimensional ising model.  Journal of Statistical Physics, 1987.
  * [9] A. De Masi, P. A. Ferrari, and J. L. Lebowitz.  Rigorous derivation of reaction-diffusion equations with fluctuations.  Physical Review Letters, 55(19):1947–1949, 1985.
  * [10] A. De Masi, P. A. Ferrari, and J. L. Lebowitz.  Reaction-diffusion equations for interacting particle systems.  Journal of Statistical Physics, 44(3-4):589–644, 1986.
  * [11] A. De Masi, T. Funaki, E. Presutti, and M. E. Vares.  Fast-reaction limit for Glauber-Kawasaki dynamics with two components.  ALEA. Latin American Journal of Probability and Mathematical Statistics, 16(2):957–976, 2019.
  * [12] P. Diaconis.  The cutoff phenomenon in finite Markov chains.  Proceedings of the National Academy of Sciences of the United States of America, 93(4):1659–1664, 1996.
  * [13] P. Diaconis and M. Shahshahani.  Generating a random permutation with random transpositions.  Zeitschrift für Wahrscheinlichkeitstheorie und Verwandte Gebiete, 57(2):159–179, 1981.
  * [14] P. Diaconis and M. Shahshahani.  Time to reach stationarity in the Bernoulli-Laplace diffusion model.  SIAM Journal on Mathematical Analysis, 18(1):208–218, 1987.
  * [15] J. Ding, E. Lubetzky, and Y. Peres.  The mixing time evolution of Glauber dynamics for the mean-field Ising model.  Communications in Mathematical Physics, 289(2):725–764, 2009.
  * [16] J. Ding, E. Lubetzky, and Y. Peres.  Mixing time of critical Ising model on trees is polynomial in the height.  Communications in Mathematical Physics, 295(1):161–207, 2010.
  * [17] R. L. Dobrushin and S. B. Shlosman.  Completely analytical interactions: constructive description.  Journal of Statistical Physics, 46(5-6):983–1014, 1987.
  * [18] D. Elboim and A. Sly.  Infinite cycles in the interchange process in five dimensions, 2023.
  * [19] S. N. Ethier and T. G. Kurtz.  Markov processes – characterization and convergence.  Wiley Series in Probability and Mathematical Statistics: Probability and Mathematical Statistics. John Wiley & Sons Inc., New York, 1986.
  * [20] D. Gabrielli, G. Jona-Lasinio, C. Landim, and M. E. Vares.  Microscopic reversibility and thermodynamic fluctuations.  In Boltzmann’s legacy 150 years after his birth (Rome, 1994), volume 131 of Atti Convegni Lincei, pages 79–87. Accad. Naz. Lincei, Rome, 1997.
  * [21] S. Ganguly and I. Seo.  Information percolation and cutoff for the random-cluster model.  Random Structures & Algorithms, 57(3):770–822, 2020.
  * [22] P. Gonçalves, M. Jara, O. Menezes, and A. Neumann.  Non-equilibrium and stationary fluctuations for the SSEP with slow boundary.  Stochastic Processes and their Applications, 130(7):4326–4357, 2020.
  * [23] P. Gonçalves, M. Jara, R. Marinho, and O. Menezes.  Sharp convergence to equilibrium for the ssep with reservoirs.  Oct. 2021.
  * [24] P. Hohenberg and B. Halperin.  Theory of dynamic critical phenomena.  Reviews of Modern Physics, 1977.
  * [25] R. Holley.  On the asymptotics of the spin-spin autocorrelation function in stochastic Ising models near the critical temperature.  In Spatial stochastic processes, volume 19 of Progr. Probab., pages 89–104. Birkhäuser Boston, Boston, MA, 1991.
  * [26] R. Holley and D. Stroock.  Logarithmic Sobolev inequalities and stochastic Ising models.  Journal of Statistical Physics, 46(5-6):1159–1194, 1987.
  * [27] R. A. Holley and D. W. Stroock.  Uniform and $L^{2}$ convergence in one-dimensional stochastic Ising models.  Communications in Mathematical Physics, 123(1):85–93, 1989.
  * [28] D. Ioffe.  Exact large deviation bounds up to $T_{c}$ for the Ising model in two dimensions.  Probability Theory and Related Fields, 102(3):313–330, 1995.
  * [29] C. Kipnis and C. Landim.  Scaling limits of interacting particle systems, volume 320 of Grundlehren der mathematischen Wissenschaften [Fundamental Principles of Mathematical Sciences].  Springer-Verlag, Berlin, 1999.
  * [30] C. Kipnis, S. Olla, and S. R. S. Varadhan.  Hydrodynamics and large deviation for simple exclusion processes.  Communications on Pure and Applied Mathematics, 42(2):115–137, 1989.
  * [31] H. Lacoin.  The cutoff profile for the simple exclusion process on the circle.  The Annals of Probability, 44(5):3399–3430, 2016.
  * [32] H. Lacoin.  Mixing time and cutoff for the adjacent transposition shuffle and the simple exclusion.  The Annals of Probability, 44(2):1426–1487, 2016.
  * [33] H. Lacoin.  The simple exclusion process on the circle has a diffusive cutoff window.  Annales de l’Institut Henri Poincaré Probabilités et Statistiques, 53(3):1402–1437, 2017.
  * [34] H. Lacoin and R. Leblond.  Cutoff phenomenon for the simple exclusion process on the complete graph.  ALEA. Latin American Journal of Probability and Mathematical Statistics, 8:285–301, 2011.
  * [35] K. B. Lauritsen and H. C. Fogedby.  Critical exponents from power spectra.  Journal of Statistical Physics, 72(1-2):189–205, 1993.
  * [36] T.-Y. Lee and H.-T. Yau.  Logarithmic Sobolev inequality for some models of random walks.  The Annals of Probability, 26(4):1855–1873, 1998.
  * [37] D. A. Levin and Y. Peres.  Markov chains and mixing times.  American Mathematical Society, Providence, RI, 2017.  Second edition of [ MR2466937], With contributions by Elizabeth L. Wilmer, With a chapter on “Coupling from the past” by James G. Propp and David B. Wilson.
  * [38] T. M. Liggett.  Continuous time Markov processes, volume 113 of Graduate Studies in Mathematics.  American Mathematical Society, Providence, RI, 2010.  An introduction.
  * [39] S. L. Lu and H.-T. Yau.  Spectral gap and logarithmic Sobolev inequality for Kawasaki and Glauber dynamics.  Communications in Mathematical Physics, 156(2):399–433, 1993.
  * [40] E. Lubetzky and A. Sly.  Critical Ising on the square lattice mixes in polynomial time.  Communications in Mathematical Physics, 313(3):815–836, 2012.
  * [41] E. Lubetzky and A. Sly.  An exposition to information percolation for the Ising model.  Annales de la Faculté des Sciences de Toulouse. Mathématiques. Série 6, 24(4):745–761, 2015.
  * [42] E. Lubetzky and A. Sly.  Information percolation and cutoff for the stochastic Ising model.  Journal of the American Mathematical Society, 29(3):729–774, 2016.
  * [43] E. Lubetzky and A. Sly.  Universality of cutoff for the Ising model.  The Annals of Probability, 45(6A):3664–3696, 2017.
  * [44] E. Lubetzky and A. Sly.  Fast initial conditions for Glauber dynamics.  Probability Theory and Related Fields, 181(1-3):647–667, 2021.
  * [45] F. Martinelli.  On the two-dimensional dynamical Ising model in the phase coexistence region.  Journal of Statistical Physics, 76(5-6):1179–1246, 1994.
  * [46] F. Martinelli.  Lectures on Glauber dynamics for discrete spin models.  In Lectures on probability theory and statistics (Saint-Flour, 1997), volume 1717 of Lecture Notes in Math., pages 93–191. Springer, Berlin, 1999.
  * [47] F. Martinelli and E. Olivieri.  Approach to equilibrium of Glauber dynamics in the one phase region. I. The attractive case.  Communications in Mathematical Physics, 161(3):447–486, 1994.
  * [48] F. Martinelli and E. Olivieri.  Approach to equilibrium of Glauber dynamics in the one phase region. II. The general case.  Communications in Mathematical Physics, 161(3):487–514, 1994.
  * [49] J. Miller and Y. Peres.  Uniformity of the uncovered set of random walk and cutoff for lamplighter chains.  The Annals of Probability, 40(2):535–577, 2012.
  * [50] B. Morris.  The mixing time for simple exclusion.  The Annals of Applied Probability, 16(2):615–635, 2006.
  * [51] R. I. Oliveira.  Mixing of the symmetric exclusion processes in terms of the corresponding single-particle random walk.  The Annals of Probability, 41(2):871–913, 2013.
  * [52] L. Onsager.  Crystal statistics. I. A two-dimensional model with an order-disorder transition.  Physical Review. Series II, 65:117–149, 1944.
  * [53] J. G. Propp and D. B. Wilson.  Exact sampling with coupled Markov chains and applications to statistical mechanics.  In Proceedings of the Seventh International Conference on Random Structures and Algorithms (Atlanta, GA, 1995), volume 9, pages 223–252, 1996.
  * [54] M. Quattropani and F. Sau.  Mixing of the averaging process and its discrete dual on finite-dimensional geometries.  The Annals of Applied Probability, 33(2):936–971, 2023.
  * [55] J. Salez.  Universality of cutoff for exclusion with reservoirs.  The Annals of Probability, 51(2):478–494, 2023.
  * [56] R. Schonmann.  Second order large deviation estimates for ferromagnetic systems in the phase coexistence region.  Communications in Mathematical Physics, 1987.
  * [57] D. W. Stroock and B. a. Zegarliński.  The equivalence of the logarithmic Sobolev inequality and the Dobrushin-Shlosman mixing condition.  Communications in Mathematical Physics, 144(2):303–323, 1992.
  * [58] D. W. Stroock and B. a. Zegarliński.  The logarithmic Sobolev inequality for continuous spin systems on a lattice.  Journal of Functional Analysis, 104(2):299–326, 1992.
  * [59] D. W. Stroock and B. a. Zegarliński.  The logarithmic Sobolev inequality for discrete spin systems on a lattice.  Communications in Mathematical Physics, 149(1):175–193, 1992.
  * [60] R. Tanaka and K. Tsunoda.  Glauber-exclusion dynamics: rapid mixing regime.  Electronic Journal of Probability, 27:Paper No. 141, 26, 2022.
  * [61] L. Thomas.  Bound on the mass gap for finite volume stochastic ising models at low temperature.  Communications in Mathematical Physics, 1989.
  * [62] B. Tóth.  Improved lower bound on the thermodynamic pressure of the spin $1/2$ Heisenberg ferromagnet.  Letters in Mathematical Physics, 28(1):75–84, 1993.
  * [63] H.-Q. Tran.  Cutoff for the non reversible ssep with reservoirs, 2022.
  * [64] K. Tsunoda.  Exponentially slow mixing and hitting times of rare events for a reaction-diffusion model.  ALEA. Latin American Journal of Probability and Mathematical Statistics, 19(2):1161–1184, 2022.
  * [65] F.-G. Wang and C.-K. Hu.  Universality in dynamic critical phenomena.  Journal of Physics A: Mathematical and General, 56(2):2310–2313, 1997.
  * [66] D. B. Wilson.  Mixing times of Lozenge tiling and card shuffling Markov chains.  The Annals of Applied Probability, 14(1):274–325, 2004.
  * [67] B. a. Zegarliński.  On log-Sobolev inequalities for infinite lattice systems.  Letters in Mathematical Physics, 20(3):173–182, 1990.
  * [68] B. a. Zegarliński.  Dobrushin uniqueness theorem and logarithmic Sobolev inequalities.  Journal of Functional Analysis, 105(1):77–111, 1992.