$+1$). Hence (and since $F$ is supported on the open interval $(a,b)$) this
term can be written $d_{a,b,m-1}\int
F(\vec{y},\vec{\sigma};u)\,\mathrm{d}\mu^{a,b,m-1}_{\rho}(\vec{y},\vec{\sigma})$.
If $\sigma_{k+\ell}=-1$, a similar thing happens, except that now we integrate
against $\mu_{\rho}^{a,b,m+1}$ (and we get a prefactor of $d_{a,b,m+1}$). This
gives the first terms in each of the integrals on the right hand side of
(4.14). For the other term on the right hand side of (4.19), which involves a
$\delta_{y_{1}=a}$ factor, we proceed similarly and get terms which involve
the walk going from $\pm 1$ to $m$. In order to express them in terms of
integration against $\mu_{\rho}^{a,b,m\mp 1}$ we need to shift the path
associated to $u$ and $(\vec{y},\vec{\sigma})$ to paths going from $0$ to
$m\mp 1$, and to this end we need to adjust the absolute height parameter $u$
to $u\pm 1$, yielding the other two terms on the right hand side of (4.14). ∎

###### Proof of Prop. 4.2.

First of all, because of the explicit construction described at the beginning
of this section, the martingale problem for $\mathscr{L}_{L}$ is well-posed
(Thms. 4.1 and 2.6 of [EK86]) and therefore by Echeverria’s theorem (Prop. 9.2
of [EK86]), a measure is invariant for the process if for all $f$ in a core
for $(\mathscr{L}_{L},\mathscr{D}(\mathscr{L}_{L}))$,
$\int\mathscr{L}_{L}f\mathrm{d}\mu=0$. This is the second statement of Prop.
4.3, so we know that each $\mu_{L,\rho}$ is invariant for the process on
$[-L,L)$ with periodic boundary conditions. Now start the full line
$(\vec{y},\vec{\sigma})$ process with measure $\mu_{\rho}$ and let $F$ be any
local function and $t>0$. We run the periodic process on $[-L,L)$ using the
same noise as the full line process and using the same initial data, all
restricted to $[-L,L)$. If $L$ is chosen large enough depending on the range
of $F$ and $t$, then the configuration of the two processes at time $t$
coincides within the range of $F$, because of the finite speed of propagation.
But we have shown that the distribution of the periodic process is
$\mu_{L,\rho}$. Therefore for the full line process
$\mathbb{E}[F(\vec{y}(t),\vec{\sigma}(t))]=\mathbb{E}_{\mu_{\rho}}[F]$. Since
local functions $F$ are enough to determine the measure $\mu_{\rho}$, we have
proven that $\mu_{\rho}$ is invariant. ∎

###  4.4. Adjoint generator

Our next goal is to compute the adjoint of $\mathscr{L}$ with respect to the
$L^{2}(E^{a,b},\mu_{\rho})$ inner product (for some fixed $\rho>0$). Recall
that we are interpreting elements of $E^{a,b}$ as encoding the jumps of an
upper semi-continuous function, and endowing it with the topology that it
inherits from $\operatorname{\mathscr{U\\!C}}$. Let us now interpret instead
the elements of $E^{a,b}$ as the jumps of a lower semi-continuous function
$g\in\operatorname{\mathscr{L\\!C}}\coloneqq\\{f\\!:-f\in\operatorname{\mathscr{U\\!C}}\\}$,
with the topology of local Hausdorff convergence of epigraphs. A function
$G\\!:E^{a,b}\longrightarrow\mathbb{R}$ is therefore continuous if its
restrictions to $E^{a,b}_{n}$ satisfy the _dual matching conditions_

$\mathop{\mathop{\text{{lim}}}}\displaylimits_{y_{i}-y_{i-1}\searrow
0}G_{n}(\ldots,(y_{i-1},+1),(y_{i},-1),\ldots)=G_{n-2}(\ldots,(y_{i-2},\sigma_{i-2}),(y_{i+1},\sigma_{i+1}),\ldots).$
(4.23)

Define

$\mathscr{L}^{*}_{\operatorname{\text{{cr}}}}G_{n}(\vec{x},\vec{\sigma})=2\int_{-\infty}^{\infty}\bigl{(}G_{n+2}(C^{*}_{x}(\vec{y},\vec{\sigma}))-G_{n}(\vec{y},\vec{\sigma})\bigr{)}\mathrm{d}\vec{y},$
(4.24)

where the configuration $C^{*}_{z}(\vec{y},\vec{\sigma})$ is obtained by
inserting $(z,-1)$ and $(z,+1)$ into $(\vec{y},\vec{\sigma})$. We will also
use $\nu_{\rho}$ to denote the product measure of counting measure on the
height $h(0)\in\mathbb{Z}$ at the origin with the invariant measure
$\mu_{\rho}$, $\rho>0$.

######  Proposition 4.4.

(Skew time-reversal invariance)

  1. 1.

In the above setting, the adjoint $\mathscr{L}^{*}$ of $\mathscr{L}$ on
$L^{2}(\mu_{\rho})$ is given by

$\mathscr{L}^{*}=-\mathscr{L}_{\operatorname{\text{{det}}}}+\mathscr{L}^{*}_{\operatorname{\text{{cr}}}}.$
(4.25)

  2. 2.

Let $\hat{\mathscr{L}}^{*}$ be the adjoint of $\hat{\mathscr{L}}$ on
$L^{2}(\nu_{\rho})$. Then $\hat{\mathscr{L}}^{*}$ is the generator of
$-h(t,x)$ or, what is the same, $h(-t,x)$, where $h(t,x)$ is the PNG process.

###### Proof.

The second statement is direct, so we focus on the first one. Assume $F$ and
$G$ have common support $(a,b)$. Integration by parts shows that
$\int(\mathscr{L}_{\operatorname{\text{{det}}}}F)G\,\mathrm{d}\mu_{\rho}+\int
F(\mathscr{L}_{\operatorname{\text{{det}}}}G)\,\mathrm{d}\mu_{\rho}$ equals

$\textstyle\sum_{k,\ell=0}^{\infty}\sum_{\vec{\sigma}\in\\{-1,1\\}^{k+\ell}}\upsilon_{k,\ell}\left[\sum_{i=2}^{k+\ell}\sigma_{i}\int_{y_{i-1}=y_{i}}F_{n}G_{n}-\sum_{i=1}^{k+\ell-1}\sigma_{i}\int_{y_{i}=y_{i+1}}F_{k+\ell}G_{k+\ell}\right]\\\
\textstyle=\sum_{k,\ell=0}^{\infty}\sum_{\vec{\sigma}\in\\{-1,1\\}^{k+\ell}}\upsilon_{k,\ell}\sum_{i=1}^{k+\ell-1}(\sigma_{i+1}-\sigma_{i})\int_{y_{i}=y_{i+1}}F_{k+\ell}G_{k+\ell}.$
(4.26)

The terms on the right hand side with $\sigma_{i}=\sigma_{i+1}$ vanish, while
those with $\sigma_{i}\neq\sigma_{i+1}$ correspond to a
$(\sigma_{i},-\sigma_{i})$ pair at $y_{i}$, and thus using (4.10) and (4.23)
we get that the above equals

$\textstyle\sum_{k,\ell=1}^{\infty}\sum_{\vec{\sigma}\in\\{-1,1\\}^{k+\ell}}\upsilon_{k,\ell}\sum_{i=1}^{k+\ell-1}\\!\int_{y_{i}=y_{i+1}}\\!\big{(}F^{\wedge
i}_{k+\ell-2}G_{k+\ell}\hskip
0.50003pt\mathbf{1}_{\sigma_{i+1}=-\sigma_{i}=1}-F_{k+\ell}G^{\wedge
i}_{k+\ell-2}\hskip 0.50003pt\mathbf{1}_{\sigma_{i}=-\sigma_{i+1}=1}\big{)},$
(4.27)

where we are using the notation from the proof of Prop. 4.3. We have
restricted the sum in $k$ and $\ell$ to start from $1$, because the indicator
functions inside the integral imply that there has to be at least one up jump
and at least one down jump. Now we may reinterpret the integral
$\sum_{i=1}^{k+\ell-1}\int_{y_{i}=y_{i+1}}F^{\wedge
i}_{k+\ell-2}G_{k+\ell}\mathbf{1}_{\sigma_{i+1}=-\sigma_{i}=1}$ as
$\int_{a}^{b}\left(\int
F_{k+\ell-2}G_{k+\ell}(C_{z}(\vec{y},\vec{\sigma}))\right)\mathrm{d}z$, where
the inner integral is over $k+\ell-2$ variables. The analogous thing can be
done for the other term in the integral and then, after changing
$(k,\ell)\longmapsto(k-1,\ell-1)$ and using that the change of measure
$r^{2}p(1-p)=1$, we get that the above expression equals

$\textstyle
2\sum_{k,\ell=0}^{\infty}\sum_{\vec{\sigma}\in\\{-1,1\\}^{k+\ell}}\upsilon_{k,\ell}\int\left(\int\big{(}F_{k+\ell}G_{k+\ell+2}(C_{z}(\vec{y},\vec{\sigma}))-F_{k+\ell+2}(C_{z}(\vec{y},\vec{\sigma}))G_{k+\ell})\right)\mathrm{d}z.$

From (4.13) and (4.24), the right hand side is exactly
$-\int\big{(}(\mathscr{L}_{\operatorname{\text{{cr}}}}F)G-F(\mathscr{L}_{\operatorname{\text{{cr}}}}^{*}G)\big{)}\mathrm{d}\mu_{\rho}$.
∎

###  4.5. Backward equation

It would be convenient if we could conclude from (4.8) that our transition
probabilities $F(t,h)$, defined by

$F(t,h;\vec{x},\vec{r})=\mathbb{P}_{h}(h(t,x_{i})\leq
r_{i},i=1,\ldots,n)=\mathscr{P}_{t}f(h),\quad f=\mathbf{1}_{h(x_{i})\leq
r_{i},i=1,\ldots,n},$ (4.28)

satisfy the backward equation $\partial_{t}F=\hat{\mathscr{L}}F$.
Unfortunately, this function $f$ is _not_ continuous on
$\operatorname{\mathscr{U\\!C}}$, and it does not follow. In fact, even for
$t>0$, $F(t,h)\not\in\mathscr{D}(\hat{\mathscr{L}})$ and the backward equation
cannot hold in the classical sense. However, we are forced to deal with them
as it is only this particular class of functions for which we have exact
formulas. We now show that the backward equation is satisfied in a weak sense.

It is not too hard to see that if $h$ corresponds to $(y_{i},\sigma_{i})$ then
the transition probabilities (4.28), for given $x_{1},\dotsc,x_{n}$, can have
step discontinuities on the _shock set_ ,

$\mathcal{S}(\vec{x})=\\{(\vec{y},\vec{\sigma},t)\\!:\,y_{i}-\sigma_{i}t=x_{j}\text{{
for some }}i,j\\}.$ (4.29)

Note that this set depends on $\vec{x}$. It is essentially just a finite
collection of lines of slope $\pm 1$ and its complement is open. We will refer
to any such object below, for some finite $\vec{x}$, as “a shock set”. One can
also think of it in terms of the height function $h$. It is the collection of
$h$ with an up jump at some point $x_{k}+t$ or a down jump at some point
$x_{k}-t$, for some $k$. At such points, $\partial_{t}F$ and
$\partial_{y_{i}}F$ are ill-defined for particular values of the $r_{i}$’s and
it is impossible for $\partial_{t}F=\hat{\mathscr{L}}F$ to hold in the
classical sense. On the other hand, it is always true at such points that the
Dirac delta functions in $(\partial_{t}-\partial_{y_{i}})F$ cancel, leaving at
most a bounded singularity. In other words, $F$ satisfies
$(\partial_{t}-\hat{\mathscr{L}})F=0$ off a co-dimension $1$ set
$\mathcal{S}(\vec{x})$ where it is still well behaved, and this can be used to
give meaning to the backward equation for $F$. For our purposes, the following
weak version of the backward equation is the most convenient:

######  Proposition 4.5.

$F$ given by (4.28) is a weak solution of the backward equation in the
following sense: for any $\varphi=\varphi(s,h)$ satisfying 1\. $\varphi$ is
continuously differentiable off a shock set, 2\. $\varphi(s,h)=0$ for all
$s\in[0,t]$ if
$\mathop{\mathop{\text{{max}}}}\displaylimits_{i}r_{0}(t,x_{i})\leq-K$ for
some $K>0$, 3\. $|\varphi(s,h)|$ and
$|(\partial_{s}+\hat{\mathscr{L}}^{*})\varphi(s,h)|$ are bounded uniformly in
$s\in[0,t]$ and $h$ in the support of $\nu_{\rho}$, we have

$\int_{0}^{t}\int(-\partial_{s}-\hat{\mathscr{L}}^{*})\varphi F\hskip
1.00006pt\mathrm{d}\nu_{\rho}\hskip
1.00006pt\mathrm{d}s=-\int\varphi(t)F(t)\hskip
1.00006pt\mathrm{d}\nu_{\rho}+\int\varphi(0)F(0)\hskip
1.00006pt\mathrm{d}\nu_{\rho}.$ (4.30)

###### Proof.

Fix $\vec{x},\vec{r}$ and recall $\nu_{\rho}$ is the product of counting
measure on $h(0)\in\mathbb{Z}$ with the invariant measure $\mu_{\rho}$. The
function
$f_{\varepsilon}(h)=(2\varepsilon)^{-n}\int_{|z_{i}-x_{i}|<\varepsilon,\hskip
0.35002pti=1,\dotsc,n}\mathbf{1}_{h(z_{i})\leq r_{i},\hskip
0.35002pti=1,\ldots,n}\hskip 1.00006pt\mathrm{d}\vec{z}$ is in
$\mathscr{C}_{0}$ and so, by the general theory described at the beginning of
Sec. 4, $\mathscr{P}_{t}f_{\varepsilon}=\mathbb{E}_{h}[f_{\varepsilon}(h(t))]$
does satisfy $(\partial_{t}-\hat{\mathscr{L}})\mathscr{P}_{t}f=0$. Integrating
by parts, (4.30) holds with $F$ replaced by $\mathscr{P}_{t}f_{\varepsilon}$.
Furthermore, from the explicit description of the process at the beginning of
Sec. 4, it is clear that $\mathscr{P}_{t}f_{\varepsilon}(h)\longrightarrow
F(t,h)=F(t,h;\vec{x},\vec{r})$ as $\varepsilon\searrow 0$ uniformly in $h$ and
bounded sets of $t$. Therefore, taking $\varepsilon\searrow 0$ in the (weak)
backward equation (4.30) for $\mathscr{P}_{t}f_{\varepsilon}$, we deduce that
the same equation holds for $F$ provided that we can pass the limit through
the integration.

In order to justify this, the only real issue is the sum over
$h(0)\in\mathbb{Z}$ implicit in $\nu_{\rho}$. Let $I\subseteq\mathbb{R}$ be
the smallest (finite) interval containing the origin, each $x_{i}$, and every
point at distance at most $t+1$ from any of these points. Let $\overline{h}$
and $\underline{h}$ be the supremum and infimum of $h$ over this interval. The
event $A_{k}=\\{\overline{h}-\underline{h}=k\\}$ satisfies
$\nu_{\rho}(A_{k})\leq\tilde{C}^{k}/k!$ with a $\tilde{C}$ depending on
$\rho$, $t$ and the $x_{i}$’s. On the other hand, on $A_{k}$, $\varphi(s,h)=0$
for all $s\in[0,t]$ and $h$ with $h(0)+k\leq-K$, since
$\overline{h}\geq\mathop{\mathop{\text{{max}}}}\displaylimits_{i}r_{0}(t,x_{i})$.
Similarly, on $A_{k}$ we have $r_{0}(s,x_{i})$ is bounded below by $h(0)-k$
for each $i$ and each $s\in[0,t]$, and hence that the height function at time
$s\in[0,t]$ is also bounded below by $h(0)-k$, so that (by definition)
$F(s,h)$ vanishes for
$h(0)-k>\overline{r}\coloneqq\mathop{\mathop{\text{{max}}}}\displaylimits\\{r_{1},\dotsc,r_{n}\\}$.
This last fact is also valid for $\mathscr{P}_{t}f_{\varepsilon}(h)$ if
$\varepsilon<1$ in view of our choice of dependent region $I$. Now we consider
the left hand side of (4.30) with $\mathscr{P}_{t}f_{\varepsilon}(h)$.
Splitting the $\nu_{\rho}$ integral according to the value of $k$, using these
facts to truncate the $h(0)$ sum, bounding $F$ by $1$ in the remaining region,
and using the assumption to bound
$|(\partial_{s}+\hat{\mathscr{L}}^{*})\varphi|$ uniformly, the absolute value
of the integrand can be bounded uniformly in $\varepsilon\in(0,1)$ in a way
which makes the whole integral be bounded by

$\textstyle C\hskip 0.50003ptt\sum_{k\geq
0}\sum_{h(0)=-K-k}^{\underline{r}+k}\frac{1}{k!}\tilde{C}^{k}<\infty.$

The same holds for the right hand side of (4.30), and this shows that we may
pass the $\varepsilon\searrow 0$ limit through the integration in (4.30) as
needed. ∎

The next theorem says that (a truncated version of) the Fredholm determinant
satisfies the backward equation in the same sense.

######  Theorem 4.6.

Fix $r_{1},\dotsc,r_{n}\in\mathbb{Z}$, $x_{1},\dotsc,x_{n}\in\mathbb{R}$, and
$h\in\operatorname{\mathscr{U\\!C}}$, and let $K_{r}$ be the kernel defined in
(2.13) using these fixed parameters and $h_{0}=h$. For $t>0$ let

$\hat{F}(t,h)=\mathop{\mathop{\text{{det}}}}\nolimits(I-K_{r})_{\oplus_{n}\ell^{2}(\mathbb{Z}_{>0})}\mathbf{1}_{r_{i}\geq
r_{0}(t,x_{i}),\hskip 0.70004pti=1,\dotsc,n}$ (4.31)

with $r_{0}$ defined using the given $h$. Then:

  1. (i)

$\mathop{\mathop{\text{{lim}}}}\displaylimits_{t\searrow
0}\hat{F}(t,h)=\mathbf{1}_{r_{1}\geq h(x_{1}),\dotsc,r_{n}\geq h(x_{n})}$.

  2. (ii)

$\hat{F}$ satisfies 1\. $(\partial_{t}-\hat{\mathscr{L}})\hat{F}=0$ off
${\mathcal{S}}(\vec{x})$; 2\. $\hat{F}(s,h)=0$ for all $s\in[0,t]$ in a finite
interval if
$\mathop{\mathop{\text{{inf}}}}\displaylimits_{i=1,\dotsc,n}\mathop{\mathop{\text{{inf}}}}\displaylimits_{|z-x_{i}|\leq
t}h(z)$ is sufficiently large; 3\. $|\hat{F}(s,h)|$ and
$|(\partial_{s}-\hat{\mathscr{L}})\hat{F}(t,h)|$ are bounded uniformly in
$s\in[0,t]$ and $h$ in the support of $\nu_{\rho}$.

  3. (iii)

For any _admissible_ $\varphi$ (meaning $\varphi$ satisfies 1,2,3 of Prop.
4.5),

$\int_{0}^{t}\int(-\partial_{s}-\hat{\mathscr{L}}^{*})\varphi\hat{F}\hskip
1.00006pt\mathrm{d}\nu_{\rho}\hskip
0.50003pt\mathrm{d}s=-\int\varphi(t)\hat{F}(t)\hskip
1.00006pt\mathrm{d}\nu_{\rho}+\int\varphi(0)\hat{F}(0)\hskip
1.00006pt\mathrm{d}\nu_{\rho}.$ (4.32)

Note that we have set $\hat{F}(t,h)=0$ in (4.31) when $r_{i}<r_{0}(t,x_{i})$
for some $i$. That $\hat{F}$ should be zero in this region is clear
physically: our ultimate goal is to show that $\hat{F}(t,h)$ corresponds to
the transition probabilities for the PNG height function (with initial data
$h$), and these probabilities are $0$ whenever $r_{i}<r_{0}(t,x_{i})$ for some
$i$. We need to introduce this explicit truncation because some of the
computations on $K_{r}(u,v)$ which we will perform to prove the backward
equation become singular for $u$ or $v$ near or below one of the
$r_{0}(t,x_{i})$’s. We conjecture, however, that the Fredholm determinant
$\mathop{\mathop{\text{{det}}}}\nolimits(I-K_{r})_{\oplus_{n}\ell^{2}(\mathbb{Z}_{>0})}$
is already $0$ if $r_{i}<r_{0}(t,x_{i})$ for some $i$. Proving such a
statement in general for a Fredholm determinant is hard, and we do not attempt
it here. Alternatively, one could attempt to prove this by taking a scaling
limit of formulas for parallel TASEP, which we will do in an upcoming paper.

On the other hand, it is easy to see that the Fredholm determinant
$\mathop{\mathop{\text{{det}}}}\nolimits(I-K_{r})_{\oplus_{n}\ell^{2}(\mathbb{Z}_{>0})}$
as $t\searrow 0$, satisfies the initial condition. From (2.11), we have
$e^{-x_{i}\Delta}\mathscr{T}^{h}_{x_{i},x_{j}}e^{x_{j}\Delta}=P^{\operatorname{\text{{hit}}}(h)}_{x_{i},x_{j}}$
(recall here that, by convention, the right hand side is zero if
$x_{i}>x_{j}$), and using this in (2.12) yields

$K^{\text{{ext}}}\big{|}_{t=0}(x_{i},\cdot;x_{j},\cdot)=-P^{\operatorname{\text{{no
hit}}}(h)}_{x_{i},x_{j}}\mathbf{1}_{x_{i}<x_{j}}+\bar{\chi}_{h(x_{i})}\mathbf{1}_{x_{i}=x_{j}}.$
(4.33)

By Prop. B.2, $K_{r}(\varepsilon,h)\longrightarrow K_{r}(0,h)$ as
$\varepsilon\to 0$ in trace norm in $\oplus_{n}\ell^{2}(\mathbb{Z}_{>0})$,
after conjugation, while from (2.13) and (4.33) we have that $K(0,h)$ is upper
triangular with the $i$-th entry in the diagonal equal to
$\bar{\chi}_{h(x_{i})-r_{i}}$. Hence, since the Fredholm determinant is
continuous in trace norm, after conjugating as in Appdx. B.2 we deduce that

$\textstyle\mathop{\mathop{\text{{det}}}}\nolimits(I-K(\varepsilon,h))\xrightarrow[\varepsilon\to
0]{}\mathop{\mathop{\text{{det}}}}\nolimits(I-K(0,h))=\prod_{i=1}^{n}\mathop{\mathop{\text{{det}}}}\nolimits(I-\bar{\chi}_{h(x_{i})-r_{i}})_{\ell^{2}(\mathbb{Z}>0)}=\prod_{i=1}^{n}\mathbf{1}_{r_{i}\leq
h(x_{i})}.$ (4.34)

This yields the initial condition (i) for $\hat{F}$: if some $r_{i}<h(x_{i})$
then $r_{i}<r_{0}(t,x_{i})$ for all $t>0$ and the condition holds trivially,
while if $r_{i}\geq h(x_{i})$ for all $i$ then by upper semi-continuity of $h$
we have $r_{i}\geq r_{0}(\varepsilon,x_{i})$ for small enough $\varepsilon$,
and hence for such $\varepsilon$ we have
$\hat{F}(\varepsilon,h)=\mathop{\mathop{\text{{det}}}}\nolimits(I-K_{r}(\varepsilon,h))$
and the initial condition is provided by (4.34).

Before we turn to the proof of (4.32) in the next subsection, let us see how
it proves Thm. 2.3.

######  Proposition 4.7.

For all finite $\vec{x}$, $\vec{r}$ and all $t>0$, $F(t,h)=\hat{F}(t,h)$
$\nu_{\rho}$-a.e.

###### Proof.

From (4.30) and (4.32), for any $\varphi$ which is admissible (i.e. satisfying
1,2,3 of Prop. 4.5),

$\textstyle\int_{0}^{t}\int(-\partial_{s}-\hat{\mathscr{L}}^{*})\varphi(F-\hat{F})\mathrm{d}\nu_{\rho}\hskip
0.50003pt\mathrm{d}s=-\int\varphi(t)(F(t)-\hat{F}(t))\mathrm{d}\nu_{\rho}.$
(4.35)

Let $m>n$. For any $y_{1},\ldots,y_{m}$ containing $x_{1},\ldots,x_{n}$, and
$q_{1},\ldots,q_{m}$ we can construct an admissible $\varphi$ with
$(\partial_{s}+\hat{\mathscr{L}}^{*})\varphi=0$ off its shock set and
$\varphi(t,h)=\mathbf{1}_{h(y_{i})=q_{i},i=1,\ldots,m}$. This is done by
solving the model backwards in time using Thm. 4.6 and skew time-reversal
invariance (Prop. 4.4) to produce a Fredholm determinant solution of
$(\partial_{s}+\hat{\mathscr{L}}^{*})\tilde{\varphi}=0$ off the shock set with
final condition $\mathbf{1}_{a_{1}\leq h(y_{1}),\dotsc,a_{m}\leq h(y_{m})}$
for any $a_{1},\dotsc,a_{m}$. Thm. 4.6 ensures that these $\tilde{\varphi}$’s
are admissible, and since the class of admissible $\varphi$ is closed under
finite sums and differences, we can use them to produce such a $\varphi$ with
final data of the form $\mathbf{1}_{h(y_{i})=q_{i},i=1,\ldots,m}$. The reason
to include $x_{1},\ldots,x_{n}$ among the $y_{i}$’s is to ensure that
condition 2 of Prop. 4.5 holds. For this choice of $\varphi$ (4.35) becomes

$\textstyle\int_{h(y_{i})=q_{i},i=1,\ldots,m}\big{(}F(t,h)-\hat{F}(t,h)\big{)}\mathrm{d}\nu_{\rho}=0,$
(4.36)

and now the fact that this holds for any $m>n$ and any $y_{1},\dotsc,y_{m}$
(including the $x_{i}$’s) and $q_{1},\dotsc,q_{m}$ implies the theorem. ∎

###### Proof of Thm. 2.3.

The previous lemma shows that (2.28) holds whenever $h$ is in the support of
$\nu_{\rho}$. To go to general $h\in\operatorname{\mathscr{U\\!C}}$ we will
approximate by functions in this class. The probabilities (4.28) are not
continuous with respect to the $\operatorname{\mathscr{U\\!C}}$ topology, but
$\operatorname{\mathscr{U\\!C}}$ has a partial order, $h_{1}\prec h_{2}$ if
$h_{1}(x)\leq h_{2}(x)$, and both sides of (2.28) are continuous with respect
to local hypograph convergence with decreasing limits (for the Fredholm
determinant on the right hand side of (2.28) this is proved in Prop. B.2).
Since every $\operatorname{\mathscr{U\\!C}}$ function is the decreasing limit
of functions in the support of $\nu_{\rho}$, the result follows. ∎

###  4.6. Why hit kernels solve the Kolmogorov equation

The key to the proof of Thm. 4.6 is the following result, which is a
consequence of time reversibility and preservation of order together with our
invariant measure computations from Sec. 4.3:

######  Proposition 4.8.

Fix $h\in\hat{\mathcal{X}}\subseteq\operatorname{\mathscr{U\\!C}}$. For any
$a\leq b$ and any $u>h(a)+1$, $v>h(b)+1$,

$\hat{\mathscr{L}}P^{\operatorname{\text{{hit}}}(h)}_{a,b}(u,v)=2[P^{\operatorname{\text{{hit}}}(h)}_{a,b},\nabla](u,v),$
(4.37)

where the bracket denotes the commutator, i.e. $[A,B]=AB-BA$.

Recall that $\hat{\mathcal{X}}$, defined in (4.9), is the restricted state
space on which we are working. The condition $u>h(a)+1$, $v>h(b)+1$ in the
result is necessary: when there is a jump of $h$ at $a$, (4.37) fails for
$u\leq h(a)+1$, and similarly at the other edge of the interval. This is
already apparent in Lem. 3.4, and will play a role later in the proof of Thm.
4.6.

Here is the idea of how the proposition is used. From the structure of $K$
(see (2.11) and (2.12), and also Lem. 3.4) differentiating in $t$ produces a
commutator with $2\nabla$, so (4.37) shows
$(\partial_{t}-\hat{\mathscr{L}})K=0$. Modulo an unfortunate number of
technicalities, we will show
$(\partial_{t}-\hat{\mathscr{L}})\hat{F}=\hat{F}\operatorname{\text{{tr}}}((I-K)^{-1}(\partial_{t}-\hat{\mathscr{L}})K)$,
which therefore vanishes.

Before we start the proof we introduce some notation and make some comments.
Let $\Phi(g)$ be the indicator of the event that a path $g$ ever goes below
$0$,

$\Phi(g)=1-\mathbf{1}_{g>0}.$ (4.38)

Consider a random path $g$ in $[a,b]$ which has the same law as $\mathsf{N}$
conditioned on $\mathsf{N}(a)=u$ and $\mathsf{N}(b)=v$. We choose $g$ to be
lower semi-continuous, and extend it as $\infty$ outside the interval $[a,b]$.
Then

$P_{a,b}^{\operatorname{\text{{hit}}}(h)}(u,v)=\mathbf{E}_{a,u;b,v}(\Phi(g-h))e^{(b-a)\Delta}(u,v),$
(4.39)

where $\mathbf{E}_{a,u;b,v}$ denotes the expectation with respect to the law
of the random path $g$.

The key idea of the proof is to attempt to use the backward PNG dynamics to
rewrite the left hand side of (4.37) as
$\mathbf{E}_{a,u;b,v}(\hat{\mathscr{L}}^{*}\Phi(g-h))e^{(b-a)\Delta}(u,v)$
where $\hat{\mathscr{L}}^{*}$ acts on $g$. If the expecatation were over a
full invariant measure, this would vanish ($\hat{\mathscr{L}}^{*}$ has the
same invariant measures as $\hat{\mathscr{L}}$.) But since the expectation is
over the conditioned measure on a finite box, we pick up boundary terms, given
by the right hand side of (4.37). That these boundary terms are exactly what
one gets from the full kernel (2.12) when differentiating in $t$ is the
mechanism which makes the backward equation work.

There is a technical issue that $\Phi(\cdot-h)$ is not in the domain of
$\hat{\mathscr{L}}^{*}$ (it is not even continuous.) The solution is to
perform a small average in $h$, to regularize, then remove the regularization.
Such a regularization has to be done carefully, and in fact will not work for
all $u,v$.

###### Proof of Prop. 4.8.

We divide the proof in several steps.

_Step 1._ For $u$ and $v$ as in the proposition,
$P^{\operatorname{\text{{hit}}}(h)}_{a,b}(u,v)\in\mathscr{D}(\hat{\mathscr{L}})$.
Now define $h_{\alpha}$ to be the $\operatorname{\mathscr{U\\!C}}$ function
obtained from $h$ by shifting each $y_{i}$ of $h$ in a slighly larger box,
say, $[a-1,b+1]$ by $\alpha_{i}$, and let $A_{\varepsilon}$ be the set of such
$\alpha$ where $|\alpha_{i}|\leq\varepsilon$ for each $i$. Then we claim that
for all $u>h(a)$ and $v>h(b)$,

$\text{{Av}}_{\alpha\in
A_{\varepsilon}}P^{\operatorname{\text{{hit}}}(h_{\alpha})}_{a,b}(u,v)\xrightarrow[\varepsilon\to
0]{}P^{\operatorname{\text{{hit}}}(h)}_{a,b}(u,v),~{}\quad\hat{\mathscr{L}}\text{{Av}}_{\alpha\in
A_{\varepsilon}}P^{\operatorname{\text{{hit}}}(h_{\alpha})}_{a,b}(u,v)\xrightarrow[\varepsilon\to
0]{}\hat{\mathscr{L}}P^{\operatorname{\text{{hit}}}(h)}_{a,b}(u,v),$ (4.40)

where $\text{{Av}}_{\alpha\in A_{\varepsilon}}$ denotes average over
$\alpha\in A_{\varepsilon}$.

Consider the first limit in (4.40) and assume first that there is no $y_{i}$
at the boundary of $[a,b]$. Then we may choose $\varepsilon$ to be small
enough so that this property is conserved as we perform the average, and then
it is not hard to see that we can control $\big{|}\text{{Av}}_{\alpha\in
A_{\varepsilon}}P^{\operatorname{\text{{hit}}}(h_{\alpha})}_{a,b}(u,v)-P^{\operatorname{\text{{hit}}}(h)}_{a,b}(u,v)\big{|}$
by the probability that $\mathsf{N}$ has a jump in
$[y_{i}-\varepsilon,y_{i}+\varepsilon]$ for some $y_{i}\in(a,b)$ (for if the
walk stays constant on each $[y_{i}-\varepsilon,y_{i}+\varepsilon]$ interval,
then it will hit $\operatorname{\text{{hypo}}}(h)$ if and only if it hits
$\operatorname{\text{{hypo}}}(h_{\alpha})$). But $h\in\hat{\mathcal{X}}$ so
there is a fixed, finite number $M$ of $y_{i}$’s inside $(a,b)$ in the
$(\vec{y},\vec{\sigma})$ configuration associated to $h$, and thus the
probability is of order $\varepsilon$. If there is a jump at $a$ or at $b$
then the above argument does not work in general, because a fixed random walk
path may or may not hit $\operatorname{\text{{hypo}}}(h_{\alpha})$ as a down
jump at the left edge or an up jump at the right edge enters or leaves the
interval. But since we are taking $u>h(a)$, $v>h(b)$, it is easy to see that
the relevant event also has a probability of order at most $\varepsilon$ by
upper semi-continuity of $h$.

Now consider the second limit. We claim first that
$\hat{\mathscr{L}}P^{\operatorname{\text{{hit}}}(h)}_{a,b}(u,v)=\mathscr{L}P^{\operatorname{\text{{hit}}}(h)}_{a,b}(u,v)$.
The reason is that $P^{\operatorname{\text{{hit}}}(h)}_{a,b}(u,v)$ only
depends on $h$ through its value at $a$ and the configuration of jumps inside,
say, $[a-1,b+1]$. Therefore, when computing the action of the generator on
$P^{\operatorname{\text{{hit}}}(h)}_{a,b}(u,v)$, the kernel can be regarded as
a function only of the jump configuration, the value of $h$ at $a$ being
fixed, and thus the effect of $\hat{\mathscr{L}}$ and of $\mathscr{L}$ on
$\hat{\mathscr{L}}P^{\operatorname{\text{{hit}}}(h)}_{a,b}(u,v)$ is the same.
Essentially the same argument shows that
$\hat{\mathscr{L}}\text{{Av}}_{\alpha\in
A_{\varepsilon}}P^{\operatorname{\text{{hit}}}(h_{\alpha})}_{a,b}(u,v)=\mathscr{L}\text{{Av}}_{\alpha\in
A_{\varepsilon}}P^{\operatorname{\text{{hit}}}(h_{\alpha})}_{a,b}(u,v)$.
Therefore it is enough to prove the second limit in (4.40) with
$\hat{\mathscr{L}}$ replaced by
$\mathscr{L}=\mathscr{L}_{\operatorname{\text{{det}}}}+\mathscr{L}_{\operatorname{\text{{cr}}}}$.

Focus first on the creation part. It is easy to see that the integral coming
from the definition (4.13) of $\mathscr{L}_{\operatorname{\text{{cr}}}}$ can
be interchanged with the average, and then essentially the same argument used
above proves the convergence. In that argument we used the assumption
$u>h(a)$, $v>h(b)$, which could be destroyed by a creation event at the
boundary of $[a,b]$, but creations at a specific, given point can be thrown
out from the integral in (4.13). For
$\mathscr{L}_{\operatorname{\text{{det}}}}$ one can similarly interchange
derivatives and the average, and the convergence follows again from Lem. 3.6,
the assumptions on $u$ and $v$ and the above argument.

_Step 2._ Our next goal is to prove that, for $u>h(a)$ and $v>h(b)$,

$\hat{\mathscr{L}}\text{{Av}}_{\alpha\in
A_{\varepsilon}}P^{\operatorname{\text{{hit}}}(h_{\alpha})}_{a,b}(u,v)=2[\text{{Av}}_{\alpha\in
A_{\varepsilon}}P^{\operatorname{\text{{hit}}}(h_{\alpha})}_{a,b},\nabla](u,v).$
(4.41)

For the proposition we only need to consider $u>h(a)+1$, $v>h(b)+1$, but the
extension to $u=h(a)+1$ and $v=h(b)+1$ will be useful later in this section.
From the Kolmogorov backward equation,

$\hat{\mathscr{L}}\text{{Av}}_{\alpha\in
A_{\varepsilon}}P^{\operatorname{\text{{hit}}}(h_{\alpha})}_{a,b}(u,v)=\mathop{\mathop{\text{{lim}}}}\displaylimits_{\delta\searrow
0}\frac{1}{\delta}\mathbb{E}\Big{(}\text{{Av}}_{\alpha\in
A_{\varepsilon}}P^{\operatorname{\text{{hit}}}(h(\delta,\cdot;h_{\alpha}))}_{a,b}(u,v)-\text{{Av}}_{\alpha\in
A_{\varepsilon}}P^{\operatorname{\text{{hit}}}(h_{\alpha})}_{a,b}(u,v)\Big{)},$
(4.42)

where $\mathbb{E}$ denotes the expectation with respect to the PNG dynamics,
and the PNG height function $h(\delta,\cdot;h_{\alpha})$ is taken to start at
$h(0,\cdot)=h_{\alpha}$. From (4.39),

$P_{a,b}^{\operatorname{\text{{hit}}}(h(\delta,\cdot;h_{\alpha}))}(u,v)=\mathbf{E}_{a,u;b,v}(\Phi(g(0,\cdot)-h(\delta,\cdot;h_{\alpha})))e^{(b-a)\Delta}(u,v)$
(4.43)

where $\mathbf{E}_{a,u;b,v}$ denotes the expectation with respect to the law
of the random path $g(0,\cdot)$.

Run the PNG dynamics backwards in time, starting at time $\delta$ and down to
time $0$, using the same space-time Poisson process used to evolve
$h(\delta,\cdot)$ from $h(0,\cdot)$, and apply it to each of $h(\delta,\cdot)$
and $g(0,\cdot)$ separately. The first of the two functions becomes
$h(0,\cdot)=h_{\alpha}$, while the second one yields a new curve, call it
$g(-\delta,\cdot)$. The PNG dynamics preserves order, so almost surely we have
$h_{\alpha}(x)\leq g(-\delta,x)$ for some $x$ if and only if $h(\delta,x)\leq
g(0,x)$ for some $x$. Hence

$\mathbb{E}\big{(}\mathbf{E}_{a,u;b,v}(\Phi(g(0,\cdot)-h(\delta,\cdot;h_{\alpha})))\big{)}=\mathbb{E}\big{(}\mathbf{E}_{a,u;b,v}(\Phi(g(-\delta,\cdot)-h_{\alpha}))\big{)}.$
(4.44)

Thus (4.42) can be rewritten as

$\mathbf{E}_{a,u;b,v}\Big{(}\mathop{\mathop{\text{{lim}}}}\displaylimits_{\delta\searrow
0}\frac{1}{\delta}\Big{(}\mathbb{E}\big{(}\Phi_{\varepsilon}(g(-\delta,\cdot))\big{)}-\Phi_{\varepsilon}(g(0,\cdot))\Big{)}\Big{)}e^{(b-a)\Delta}(u,v),$
(4.45)

where $\Phi_{\varepsilon}(g)=1-\text{{Av}}_{\alpha\in
A_{\varepsilon}}\mathbf{1}_{g-h_{\alpha}>0}$. Here we first used Fubini to
change the order of expectations, justified by the definite sign of the
integrands. Then the $\delta$ limit was taken inside the expectation over the
path $g$, justified by the average over $A_{\varepsilon}$, which keeps all
relevant derivatives of the function $\Phi_{\varepsilon}$ bounded by
$C\varepsilon^{-1}$, and the dominated convergence theorem. Inside the
$\mathbb{E}$ expectation the PNG dynamics is run, backwards, on $g$, and thus
from the Kolmogorov backward equation and Prop. 4.4 we get that the above
limit equals

$\mathbf{E}_{a,u;b,v}\Big{(}\hat{\mathscr{L}}^{*}\Phi_{\varepsilon}(g(0,\cdot))\Big{)}e^{(b-a)\Delta}(u,v),$
(4.46)

where $\hat{\mathscr{L}}^{*}$ is applied on $g=g(0)$. This is justified by the
fact that $\Phi_{\varepsilon}\in\mathscr{D}(\hat{\mathscr{L}}^{*})$. The law
of $\mathsf{N}$ is invariant for the backwards PNG dynamics, but in (4.46) (in
our choice of $g(0,\cdot)$) we are conditioning on the values at the
endpoints, so we are in the setting of Prop. 4.3, albeit with
$\mathscr{L}^{*}$ instead of $\mathscr{L}$. Slightly abusing notation, and
writing $\Phi_{\varepsilon}(\vec{y},\vec{\sigma};u)$ for
$\Phi_{\varepsilon}(g(0,\cdot))$, Prop. 4.3 tells us that (4.46) is given by

$-d_{a,b,v-u-1}\int(\Phi_{\varepsilon}(\vec{y},\vec{\sigma};u)+\Phi_{\varepsilon}(\vec{y},\vec{\sigma};u+1))\,\mathrm{d}\mu_{\rho}^{a,b,v-u-1}\\\
\quad+d_{a,b,v-u+1}\int(\Phi_{\varepsilon}(\vec{y},\vec{\sigma};u)+\Phi_{\varepsilon}(\vec{y},\vec{\sigma};u-1))\,\mathrm{d}\mu_{\rho}^{a,b,v-u+1}$
(4.47)

(the sign difference coming from the fact that we are dealing with
$\mathscr{L}^{*}=-\mathscr{L}_{\operatorname{\text{{det}}}}+\mathscr{L}^{*}_{\operatorname{\text{{cr}}}}$
now). In view of our choice of $\Phi_{\varepsilon}$, and since
$\mu_{\rho}^{a,v,v-u}$ corresponds to a conditioned measure while
$P^{\operatorname{\text{{hit}}}(h)}_{a,b}(u,v)$ is unconditioned, the above
expression equals

$\text{{Av}}_{\alpha\in
A_{\varepsilon}}\Big{[}-P^{\operatorname{\text{{hit}}}(h_{\alpha})}_{a,b}(u,v-1)-P^{\operatorname{\text{{hit}}}(h_{\alpha})}_{a,b}(u+1,v)+P^{\operatorname{\text{{hit}}}(h_{\alpha})}_{a,b}(u,v+1)+P^{\operatorname{\text{{hit}}}(h_{\alpha})}_{a,b}(u-1,v)\Big{]},$

which is exactly the right hand side of (4.41).

_Step 3._ We finish the proof by taking $\varepsilon\to 0$ in (4.41). The left
hand side converges to
$\hat{\mathscr{L}}P^{\operatorname{\text{{hit}}}(h_{\alpha})}_{a,b}(u,v)$ by
(4.40). For the right hand side, the commutator involves evaluating
$\text{{Av}}_{\alpha\in
A_{\varepsilon}}P^{\operatorname{\text{{hit}}}(h_{\alpha})}_{a,b}$ at $(u\pm
1,v)$ and $(u,v\pm 1)$, and in all cases (4.40) ensures the necessary
convergence, so the limit of the the right hand side is
$2[P^{\operatorname{\text{{hit}}}(h)}_{a,b},\nabla](u,v)$ as needed. ∎

Before turning to the proof of the backward equation we need three additional
results. The first one will be useful in handling the creation part of the
generator:

######  Lemma 4.9.

Let $h\in\operatorname{\mathscr{U\\!C}}$, $t\geq 0$, and $x_{1}<\dotsm<x_{n}$,
and let $K(h)$ denote the PNG kernel (2.12) defined using the given $x_{i}$’s,
$t$, and $h_{0}=h$. Then for any fixed $z\in\mathbb{R}$,
$K(h+\mathbf{1}_{z})-K(h)$ is rank one.

###### Proof.

In order to compare the two kernels we need to look at the hit probabilities
involving $h$ and $h+\mathbf{1}_{z}$. We have

$P^{\operatorname{\text{{hit}}}(h+\mathbf{1}_{z})}_{x_{i}-t,x_{j}+t}=P^{\operatorname{\text{{hit}}}(h)}_{x_{i}-t,x_{j}+t}+P^{\operatorname{\text{{no
hit}}}(h)}_{x_{i}-t,z}\delta_{h(z)+1}P^{\operatorname{\text{{no
hit}}}(h)}_{z,x_{j}+t},$

where for convenience we set $P^{\operatorname{\text{{no hit}}}(h)}_{a,b}=0$
if $a>b$ (so that this covers both the case when $z$ is in $[x_{i}-t,x_{j}+t]$
and when it is not). Then
$K(h+\mathbf{1}_{z})(x_{i},\cdot;x_{j},\cdot)-K(h)(x_{i},\cdot;x_{j},\cdot)$
equals

$\displaystyle
e^{-2t\nabla-x\Delta}\big{(}\mathscr{T}^{h+\mathbf{1}_{z}}_{x_{i}-t,x_{j}+t}-\mathscr{T}^{h}_{x_{i}-t,x_{j}+t}\big{)}e^{2t\nabla+x^{\prime}\Delta}=e^{-2t\nabla-t\Delta}\big{(}P^{\operatorname{\text{{hit}}}(h+\mathbf{1}_{z})}_{x_{i}-t,x_{j}+t}-P^{\operatorname{\text{{hit}}}(h)}_{x_{i}-t,x_{j}+t}\big{)}e^{2t\nabla-t\Delta}$
(4.48) $\displaystyle\hskip
115.63243pt=e^{-2t\nabla-t\Delta}P^{\operatorname{\text{{no
hit}}}(h)}_{x_{i}-t,z}\delta_{h(z)+1}P^{\operatorname{\text{{no
hit}}}(h)}_{z,x_{j}+t}e^{2t\nabla-t\Delta}=f_{i}\otimes g_{j}$ (4.49)

with $f_{i}(u)=e^{-2t\nabla-t\Delta}P^{\operatorname{\text{{no
hit}}}(h)}_{x_{i}-t,z}(u,h(z)+1)$ and $g_{j}(v)=P^{\operatorname{\text{{no
hit}}}(h)}_{z,x_{j}+t}e^{2t\nabla-t\Delta}(h(z)+1,v)$. ∎

The second result we need is an extension of (4.40) up to the curve $h$. We
state it only for evaluation at $h$ at the right edge of the interval $[a,b]$,
but of course an analog result holds at the left edge.

######  Lemma 4.10.

For all $a\leq b$, $h\in\hat{\mathcal{X}}$ and $u>h(a)+1$ we have

$\mathop{\mathop{\text{{lim}}}}\displaylimits_{\varepsilon\to
0}\text{{Av}}_{\alpha\in
A_{\varepsilon}}P^{\operatorname{\text{{hit}}}(h_{\alpha})}_{a,b}(u,h(b))=P^{\operatorname{\text{{hit}}}(h)}_{a,b}(u,h(b))-\tfrac{1}{2}P^{\operatorname{\text{{no
hit}}}(h)}_{a,b^{-}}(u,h(b)).$ (4.50)

Note that the second term on the right hand side of (4.50) vanishes unless $h$
has an up jump at $b$. This extra boundary term will be important in the proof
of Thm. 4.6 below, as it will match another boundary term which will arise
from (3.48).

###### Proof.

From the argument used to prove (4.40) in step 1 of the proof of Prop. 4.8 we
see that the limit equals $P^{\operatorname{\text{{hit}}}(h)}_{a,b}(u,h(b))$
unless $h$ has an up jump at $b$. Now suppose there is such a jump, and take
$\varepsilon$ to be small enough so that there are no other jumps of $h$ in
$[b-\varepsilon,b+\varepsilon]$. The average over shifting the jumps of $h$
other than the one at $b$ has the same effect as in the proof of Prop. 4.8.
Now write $P^{\operatorname{\text{{hit}}}(h)}_{a,b}(u,h(b))\big{|}_{b\hskip
0.70004pt\mapsto s}$ for value of the kernel obtained when the jump of $h$ at
$b$ is moved to $s$. We have
$P^{\operatorname{\text{{hit}}}(h)}_{a,b}(u,h(b))\big{|}_{b\hskip
0.70004pt\mapsto
s}=P^{\operatorname{\text{{hit}}}(h)}_{a,b}(u,h(b))+\mathcal{O}(\varepsilon)$
for $s\in[b-\varepsilon,b)$ (since the difference between the two hit
probabilities involves a jump of $\mathsf{N}$ inside the interval
$[b-\varepsilon,b]$), but

$P^{\operatorname{\text{{hit}}}(h)}_{a,b}(u,h(b))\big{|}_{b\hskip
0.70004pt\mapsto s}=P^{\operatorname{\text{{hit}}}(h)}_{a,b^{-}}(u,h(b))$

for $s\in(b,b+\varepsilon]$. Using this it is not hard to see that
$\mathop{\mathop{\text{{lim}}}}\displaylimits_{\varepsilon\to
0}\text{{Av}}_{\alpha\in
A_{\varepsilon}}P^{\operatorname{\text{{hit}}}(h_{\alpha})}_{a,b}(u,h(b))$
equals

$\displaystyle\tfrac{1}{2}P^{\operatorname{\text{{hit}}}(h)}_{a,b}(u,h(b))+\tfrac{1}{2}P^{\operatorname{\text{{hit}}}(h)}_{a,b^{-}}(u,h(b))$
$\displaystyle=P^{\operatorname{\text{{hit}}}(h)}_{a,b}(u,h(b))+\tfrac{1}{2}\big{(}P^{\operatorname{\text{{hit}}}(h)}_{a,b^{-}}(u,h(b))-P^{\operatorname{\text{{hit}}}(h)}_{a,b}(u,h(b))\big{)}$
(4.51)
$\displaystyle=P^{\operatorname{\text{{hit}}}(h)}_{a,b}(u,h(b))-\tfrac{1}{2}P^{\operatorname{\text{{no
hit}}}(h)}_{a,b^{-}}(u,h(b)).\qed$ (4.52)

The last thing we need before proving Thm. 4.6 is an abstract result about
Fredholm determinants. To motivate it, consider the case $r_{i}>r_{0}(t,x)$
for each $i$ and suppose for a moment that we knew that the kernel
$K=\chi_{r}K^{{\text{{ext}}}}\chi_{r}$ is such that $I-K$ is invertible; this
fact will actually follow from Thm. 2.3, because its Fredholm determinant
equals $F(t,x_{1},\dotsc,x_{n},r_{1},\dotsc,r_{n})$, which is positive under
our assumption on the $r_{i}$’s (see the argument in Sec. 3.2), but of course
our proof of Thm. 2.3 depends on Thm. 4.6. Assuming that $I-K$ is invertible
we have the standard identity

$\partial_{t}\mathop{\mathop{\text{{det}}}}\nolimits(I-K)=-\mathop{\mathop{\text{{det}}}}\nolimits(I-K)\operatorname{\text{{tr}}}((I-K)^{-1}\partial_{t}K),$

provided $\partial_{t}K$ exists in trace class (in our setting this fails at
the shock set, but we ignore this issue in this explanation). Similarly
$\mathscr{L}_{\operatorname{\text{{det}}}}\mathop{\mathop{\text{{det}}}}\nolimits(I-K)=-\mathop{\mathop{\text{{det}}}}\nolimits(I-K)\operatorname{\text{{tr}}}((I-K)^{-1}\mathscr{L}_{\operatorname{\text{{det}}}}K)$,
because $\mathscr{L}_{\operatorname{\text{{det}}}}$ is a first order
differential operator. The key is that the same holds for the creation part
thanks to the last lemma and another standard identity: if $A$ is rank one and
$I-K$ is invertible, then

$\mathop{\mathop{\text{{det}}}}\nolimits(I-K+A)-\mathop{\mathop{\text{{det}}}}\nolimits(I-K)=-\mathop{\mathop{\text{{det}}}}\nolimits(I-K)\operatorname{\text{{tr}}}((I-K)^{-1}A).$
(4.53)

What is crucial here is that $\partial_{t}$,
$\mathscr{L}_{\operatorname{\text{{det}}}}$ and
$\mathscr{L}_{\operatorname{\text{{cr}}}}$ all give rise to expressions of the
same form, leading to

$(\partial_{t}-\mathscr{L})\mathop{\mathop{\text{{det}}}}\nolimits(I-K)=-\mathop{\mathop{\text{{det}}}}\nolimits(I-K)\operatorname{\text{{tr}}}((I-K)^{-1}(\partial_{t}-\mathscr{L})K),$

which has the advantage of reducing the proof of $(\partial_{t}-\mathscr{L})F$
to the kernel identity $(\partial_{t}-\mathscr{L})K=0$.

In order to achieve something similar in our situation, where we do not known
a priori that $I-K$ is invertible, we proceed as follows. Let $\mathcal{H}$ be
a separable Hilbert space and let $\mathcal{B}_{1}$ be the space of trace
class operators on $\mathcal{H}$. Define
$\mathsf{D}\\!:\mathcal{B}_{1}\longrightarrow{\mathbb{C}}$ through

$\mathsf{D}(K)=\mathop{\mathop{\text{{det}}}}\nolimits(I-K),$ (4.54)

where the Fredholm determinant is computed on $\mathcal{H}$.

######  Lemma 4.11.

$\mathsf{D}$ is Fréchet differentiable with Fréchet derivative
$\mathsf{D}^{\prime}(K)$. In particular, if $K_{a}\in\mathcal{B}_{1}$ depends
on a parameter $a\in\mathbb{R}$ and the derivative $\partial_{a}K_{a}$ exists
in trace norm, then

$\partial_{a}\mathop{\mathop{\text{{det}}}}\nolimits(K_{a})=\mathsf{D}^{\prime}(K_{a})(\partial_{a}K_{a}).$
(4.55)

On the other hand, if $K\in\mathcal{B}_{1}$ and $A$ is a rank one operator
acting on $\mathcal{H}$, then

$\mathop{\mathop{\text{{det}}}}\nolimits(I-K-A)-\mathop{\mathop{\text{{det}}}}\nolimits(I-K)=\mathsf{D}^{\prime}(K)(A).$
(4.56)

The main point of the result (which is simple, but which we did not find
stated in the literature) is that (4.56) holds without any assumption on $K$.

###### Proof.

That $\mathsf{D}$ is Frèchet differentiable is standard, see e.g. [Sim05, Cor.
5.2], while (4.55) follows from that and the chain rule. The same result in
[Sim05] shows that, in the special case when $I-K$ is invertible, the Fréchet
derivative of $\mathsf{D}$ has the following form: for each
$L\in\mathcal{B}_{1}$,

$\mathsf{D}^{\prime}(K)(L)=-\mathsf{D}(K)\operatorname{\text{{tr}}}((I-K)^{-1}L).$
(4.57)

Now fix $K,A\in\mathcal{B}_{1}$ with $A$ rank one and, for
$\mu\in{\mathbb{C}}$, define

$\mathsf{R}(\mu)=\mathsf{D}(\mu K+A)-\mathsf{D}(\mu K)-\mathsf{D}^{\prime}(\mu
K)(A).$

The three objects on the right hand side are analytic in $\mu$ (for
$\mathsf{D}(\mu K)$ this is standard, see e.g. [Sim05, Lem. 3.3], while for
the other two it is implicit in Cor. 5.2 of the same book), so
$\mu\longmapsto\mathsf{R}(\mu)$ is an analytic function. On the other hand, if
$\mu^{-1}$ is not in the spectrum of $K$ then, by (4.53) and (4.57),

$\mathsf{D}(\mu K+A)-\mathsf{D}(\mu K)=-\mathsf{D}(\mu
K)\operatorname{\text{{tr}}}((I-\mu K)^{-1}A)=\mathsf{D^{\prime}}(\mu K)(A).$

Hence $\mathsf{R}(\mu)=0$ whenever $\mu^{-1}$ is not in the spectrum of $K$.
$K$ is compact, so its spectrum is discrete. Hence we deduce that
$\mathsf{R}(\mu)=0$ for all $\mu\in{\mathbb{C}}$ and thus, in particular, for
$\mu=1$. ∎

###### Proof of Thm. 4.6.

The initial condition (i) was already verified. The second point of (ii) is
straightforward from the definition (4.31) of $\hat{F}$, while (iii) follows
directly from 1 in (ii) and integration by parts. To get the estimate on
$|\hat{F}(s,h)|$ in the third point of (ii) we use Prop. B.2 together with the
bound $|\mathop{\mathop{\text{{det}}}}\nolimits(I-A)|\leq e^{\|A\|_{1}}$ for
any trace class operator $A$ on given (separable) Hilbert space, with
$\|A\|_{1}$ its trace norm [Sim05, Lem. 3.3]. They give $|\hat{F}(s,h)|\leq
C(1+e^{c\hskip
0.35002pt\mathop{\mathop{\text{{max}}}}\displaylimits_{i}(r_{0}(s,x_{i})-r_{i})})\mathbf{1}_{r_{i}\geq
r_{0}(s,x_{i}),\hskip 0.70004pti=1,\dotsc,n}$ with constants $c,C>0$ valid for
all $s\in[0,t]$, which is clearly bounded uniformly as needed (note the key
role played here by the truncation). The estimate on
$|(\partial_{s}-\hat{\mathscr{L}})\hat{F}|(s,h)$ follows from 1 of (ii) off
the shock set, and will be proved in Case 3b of this proof on
$\mathcal{S}(\vec{x})$.

The rest of the proof will be devoted to showing the two remaining conditions:
that $(\partial_{t}-\hat{\mathscr{L}})\hat{F}=0$ off $\mathcal{S}(\vec{x})$
and that $|(\partial_{t}-\hat{\mathscr{L}})\hat{F}(t,h)|$ is bounded uniformly
on $\mathcal{S}(\vec{x})$ for $t$ in a compact interval and
$h\in\operatorname{\mathscr{U\\!C}}$. We divide the proof in several cases,
according to the position of the parameters $r_{i}$ and whether we are not at
a shock. Throughout the proof all determinants and traces are computed on
$\ell^{2}(\\{x_{1},\dots,x_{n}\\}\times\mathbb{Z})$.

_Case 1_ : Suppose $r_{k}>r_{0}(t,x_{k})$ for each $k$ and write

$K=\chi_{r}K^{\text{{ext}}}\chi_{r},\qquad\text{{so
that}}\quad\hat{F}=\mathop{\mathop{\text{{det}}}}\nolimits(I-K).$

$K$ only depends on the jumps of $h$ inside
$[\mathop{\mathop{\text{{min}}}}\displaylimits\\{x_{1},\dotsc,x_{n}\\}-t,\mathop{\mathop{\text{{max}}}}\displaylimits\\{x_{1},\dotsc,x_{n}\\}+t]$,
so $\mathscr{L}_{\operatorname{\text{{det}}}}$ acts as a sum of finitely many
derivatives. $\partial_{t}-\mathscr{L}_{\operatorname{\text{{det}}}}$ thus
acts as a first order differential operator, and from the arguments in Appdx.
B.3 it follows that (the limit which defines)
$(\partial_{t}-\mathscr{L}_{\operatorname{\text{{det}}}})K$ exists in trace
norm. Then by (4.55) we have

$(\partial_{t}-\mathscr{L}_{\operatorname{\text{{det}}}})\hat{F}=\mathsf{D}^{\prime}(K)((\partial_{t}-\mathscr{L}_{\operatorname{\text{{det}}}})K).$
(4.58)

Now we consider ${\mathscr{L}}_{\operatorname{\text{{cr}}}}$. Using (4.13) and
the notation from Lem. 4.9 we have

$\textstyle{\mathscr{L}}_{\operatorname{\text{{cr}}}}\mathop{\mathop{\text{{det}}}}\nolimits(I-K(h))=\int_{-\infty}^{\infty}\big{(}\mathop{\mathop{\text{{det}}}}\nolimits(I-K(h+\mathbf{1}_{z}))-\mathop{\mathop{\text{{det}}}}\nolimits(I-K(h))\big{)}\hskip
0.50003pt\mathrm{d}z,$

and since $K(h+\mathbf{1}_{z})$ is a rank one perturbation of $K(h)$ by the
same lemma, Lem. 4.11 now shows that the difference in the integrand on the
right hand side equals $\mathsf{D}^{\prime}(K)(K(h+\mathbf{1}_{z})-K(h))$.
This gives

$\textstyle{\mathscr{L}}_{\operatorname{\text{{cr}}}}\mathop{\mathop{\text{{det}}}}\nolimits(I-K)=\mathsf{D}^{\prime}(K)\\!\left(\int_{-\infty}^{\infty}(K(h+\mathbf{1}_{z})-K(h))\,\mathrm{d}z\right)=\mathsf{D}^{\prime}(K)(\mathscr{L}_{\operatorname{\text{{cr}}}}K).$
(4.59)

From this and (4.58) we get
$(\partial_{t}-\mathscr{L}_{\operatorname{\text{{det}}}}-\mathscr{L}_{\operatorname{\text{{cr}}}})\hat{F}=\mathsf{D}^{\prime}(K)((\partial_{t}-\mathscr{L}_{\operatorname{\text{{det}}}}-\mathscr{L}_{\operatorname{\text{{cr}}}})K)$
and then arguing as in step 2 of the proof of Prop. 4.8 (using
$(\hat{\mathscr{L}}-\mathscr{L})K=0$) we conclude that

$(\partial_{t}-\hat{\mathscr{L}})\hat{F}=\mathsf{D}^{\prime}(K)((\partial_{t}-\hat{\mathscr{L}})K).$
(4.60)

Therefore (recalling $K$ includes the projections $\chi_{r}$)
$(\partial_{t}-\hat{\mathscr{L}})\hat{F}=0$ will follow in this case if we
prove that, for $K_{ij}=K(x_{i},\cdot;x_{j},\cdot)$,
$(\partial_{t}-\hat{\mathscr{L}})K_{ij}(u,v)=0$ for all $u>r_{i}$, $v>r_{j}$
or slightly more generally, due to the assumption $r_{i}>r_{0}(t,x_{i})$,
$r_{j}>r_{0}(t,x_{j})$, that

$(\partial_{t}-\hat{\mathscr{L}})K_{ij}(u,v)=0\qquad\forall\,u>h(x_{i}-t)+1,~{}v>h(x_{j}+t)+1.$
(4.61)

Note that in this case we are getting
$(\partial_{t}-\hat{\mathscr{L}})\hat{F}=0$ irrespective of whether we are or
not at a shock.

By definition of $K_{ij}$ and of $\mathscr{T}^{h}_{a,b}$, we clearly have

$\hat{\mathscr{L}}K_{ij}=e^{-2t\nabla-
x_{i}\Delta}\big{(}\hat{\mathscr{L}}\mathscr{T}^{h}_{x_{i}-t,x_{j}+t}\big{)}e^{2t\nabla+x_{j}\Delta}=e^{-2t\nabla-t\Delta}\big{(}\hat{\mathscr{L}}P^{\operatorname{\text{{hit}}}(h)}_{x_{i}-t,x_{j}+t}\big{)}e^{2t\nabla-t\Delta}.$
(4.62)

On the other hand, using (2.12) we have

$\partial_{t}K_{ij}(u,v)=e^{-2t\nabla-
x_{i}\Delta}\Bigl{(}2[\mathscr{T}^{h}_{x_{i}-t,x_{j}+t},\nabla]+\partial_{t}\mathscr{T}^{h}_{x_{i}-t,x_{j}+t}\Bigr{)}e^{2t\nabla+x_{j}\Delta}(u,v)$
(4.63)

(here we have used the fact that $\nabla$ and $\Delta$, and hence also their
exponentials, commute). Now take $u>h(x-t_{i})+1$, $v>h(x_{j}+t)+1$ as in
(4.61). Lem. 3.4 shows that the second term inside the parenthesis vanishes,
and thus using the definition of $\mathscr{T}^{h}_{a,b}$ we get

$\partial_{t}K_{ij}(u,v)=e^{-2t\nabla-t\Delta}\big{(}2[P^{\operatorname{\text{{hit}}}(h)}_{x_{i}-t,x_{j}+t},\nabla]\big{)}e^{2t\nabla-t\Delta}(u,v).$

Therefore, using this and (4.62), in order to prove (4.61) we need to show
that

$e^{-2t\nabla-t\Delta}\big{(}2[P^{\operatorname{\text{{hit}}}(h)}_{x_{i}-t,x_{j}+t},\nabla]-\hat{\mathscr{L}}P^{\operatorname{\text{{hit}}}(h)}_{x_{i}-t,x_{j}+t}\big{)}e^{2t\nabla-t\Delta}(u,v)=0$

for $u>h(x_{i}-t)+1$, $v>h(x_{j}+t)+1$. Since, by Lem. 3.7 and the fact that
$e^{-2t\nabla-t\Delta}=(e^{2t\nabla-t\Delta})^{*}$, $e^{2t\nabla-t\Delta}$ is
upper triangular and $e^{-2t\nabla-t\Delta}$ is lower triangular, in the last
expression we only need to consider
$\hat{\mathscr{L}}P^{\operatorname{\text{{hit}}}(h)}_{x_{i}-t,x_{j}+t}(y,z)$
with $y>h(x_{i}-t)+1$, $z>h(x_{j}+t)+1$. But, for such $(y,z)$, Prop. 4.8
yields
$\hat{\mathscr{L}}P^{\operatorname{\text{{hit}}}(h)}_{x_{i}-t,x_{j}+t}(y,z)=2[P^{\operatorname{\text{{hit}}}(h)}_{x_{i}-t,x_{j}+t},\nabla](y,z)$,
which shows that the desired identity holds.

_Case 2_ : Suppose now that $r_{k}<r_{0}(t,x_{k})$ for some $k$ and that we
are not at a shock in the $x_{k}$ variable (i.e., that $h$ does not have a
jump at $x_{k}+t$ or $x_{k}-t$). Then $r_{k}<r_{0}(s,x_{k})$ for all $s$ in a
neighborhood of $t$, and for all such $s$ we then have $\hat{F}(s,h)=0$, so
$\partial_{t}\hat{F}(t,h)=0$. Now consider $\hat{\mathscr{L}}\hat{F}$. The
deterministic part of the generator only moves the positions of the jumps of
$h$, which lie away from $x_{k}\pm t$, while the creation part only moves $h$
up, which means that the inequality $r_{k}<r_{0}(t,x_{k})$ (and hence the fact
that $\hat{F}=0$) is preserved through the action of the generator. Hence
$\hat{\mathscr{L}}\hat{F}=0$, which gives us what we want.

_Case 3_ : Next suppose that $r_{k}<r_{0}(t,x_{k})$ for some $k$ and we are at
a shock in the $x_{k}$ variable. We split in three subcases.

_Case 3a_ : Suppose $r_{j}<r_{0}(t,x_{j})$ for some $j\neq k$ for which we are
not at a shock in the $x_{j}$ variable. Then the argument of case 2 applies,
and we get again $(\partial_{t}-\hat{\mathscr{L}})\hat{F}=0$.

_Case 3b_ : Suppose now that $r_{j}\geq r_{0}(t,x_{j})$ for all $j\neq k$. We
will only consider the case when the $x_{k}$ shock comes from a jump of $h$ at
$x_{k}+t$; a jump at the other edge, or at both edges, are handled similarly.
Additionally, we assume first for simplicity that there is only one jump at
$x_{k}+t$, and explain at the end the extension to the case when there is a
$(+1,-1)$ pair at that location. Let $(\vec{y},\vec{\sigma})$ be the
configuration of jump locations and signs associated to $h$. By relabeling, we
may assume that the (unique) jump at $x_{k}+t$ has label $0$, i.e.
$y_{0}=x_{k}+t$. If $\sigma_{0}=-1$, i.e. there is a down jump at $x_{k}+t$,
then by upper semi-continuity of $h$ we have that $r_{0}(s,z)$ is constant for
$s\in[t-\varepsilon,t+\varepsilon]$,
$z\in[x_{k}-\varepsilon,x_{k}+\varepsilon]$ and small $\varepsilon$, so
$\partial_{t}F=0$ as above, and for the same reason
$\hat{\mathscr{L}}\hat{F}=0$ (note here we need to consider the movement of
the $y_{0}$ jump, but it also does not affect $r_{0}(t,x_{k})$).

We turn now to the case $\sigma_{0}=1$. In particular, in this case we have
$r_{k}<r_{0}(t+\varepsilon,x_{k})$, and hence $\hat{F}(t+\varepsilon,h)=0$,
for small $\varepsilon>0$. Note that, since we are assuming that the jump of
$h$ at $x_{k}+t$ is up by one,
$\mathop{\mathop{\text{{lim}}}}\displaylimits_{s\nearrow
t}r_{0}(s,x_{k})=r_{k}(t,x_{k})-1$. If $r_{k}<r_{0}(t,x_{k})-1$ then we also
have $r_{k}<r_{0}(t-\varepsilon,x_{k})$ for small $\varepsilon>0$, and as in
the previous cases we have $\partial_{t}\hat{F}=\hat{\mathscr{L}}\hat{F}=0$.

The interesting situation is thus when $\sigma_{0}=1$ and
$r_{k}=r_{0}(t,x_{k})-1$. This is the case for which
$(\partial_{t}-\hat{\mathscr{L}})\hat{F}$ is not $0$; we will show that is is
bounded as needed. In this case we have as above $\hat{F}(t+\varepsilon,h)=0$,
so

$\hat{F}(t+\varepsilon,h)-\hat{F}(t-\varepsilon,h)=-\hat{F}(t-\varepsilon,h).$

Now we study $\hat{\mathscr{L}}\hat{F}$. Once again, since the creation part
of the generator only moves $h$ up (and we are assuming
$r_{k}<r_{0}(t,x_{k})$), we have
$\mathscr{L}_{\operatorname{\text{{cr}}}}\hat{F}=0$. On the other hand, it is
clear that the deterministic part of the generator restricted to the jumps of
$h$ other than the one at $y_{0}=x_{k}+t$ cannot change the inequality
$r_{k}<r_{0}(t,x)$, so that part of the action of the generator on $F$
vanishes. This tells us that the only non-zero contribution to
$\hat{\mathscr{L}}F$ comes from the limit as $\varepsilon\to 0$ of
$(2\varepsilon)^{-1}$ times

$\hat{F}(t,h)|_{y_{0}=x_{k}+t-\varepsilon}-\hat{F}(t,h)|_{y_{0}=x_{k}+t+\varepsilon},$

i.e. the difference between computing $\hat{F}$ with the $y_{0}$ jump moved to
$x_{k}+t-\varepsilon$ and to $x_{k}+t+\varepsilon$ (recall that up steps move
to the left). We may assume that $\varepsilon$ is small enough so that the
only jump of $h$ in $[x_{k}+t-\varepsilon,x_{k}+t+\varepsilon]$ is the one at
$x_{k}+t$. When the $y_{0}$ jump is moved to $x_{k}+t-\varepsilon$ the
inequality $r_{k}<r_{0}(t,x_{k})$ remains (since $r_{0}(t,x_{k})$ does not
change) but, under our assumptions, when it is moved to $x_{k}+t+\varepsilon$,
$r_{0}(t,x_{k})$ changes to $r_{0}(t,x_{k})-1=r_{k}$, which means that the
above difference equals $-\hat{F}(t,h)|_{y_{0}=x_{k}+t+\varepsilon}$. The
conclusion from all this is that, in the current situation,

$(\partial_{t}-\hat{\mathscr{L}})\hat{F}=\mathop{\mathop{\text{{lim}}}}\displaylimits_{\varepsilon\to
0}(2\varepsilon)^{-1}\big{(}-\hat{F}(t-\varepsilon,h)+\hat{F}(t,h)|_{y_{0}=x_{k}+t+\varepsilon}\big{)}.$

Under our assumptions, when evaluating $\hat{F}$ at $t-\varepsilon$ with
$y_{0}=x_{k}+t$ or at $t$ with $y_{0}=x_{k}+t+\varepsilon$, we are in the
regime where $r_{i}\geq r_{0}(t,x_{i})$ for each $i$, so that the $\hat{F}$’s
on the right hand side are given by the corresponding Fredholm determinants,
i.e.

$(\partial_{t}-\hat{\mathscr{L}})\hat{F}=\mathop{\mathop{\text{{lim}}}}\displaylimits_{\varepsilon\to
0}(2\varepsilon)^{-1}\big{(}\mathop{\mathop{\text{{det}}}}\nolimits(I-K(t,h)|_{y_{0}=x_{k}+t+\varepsilon})-\mathop{\mathop{\text{{det}}}}\nolimits(I-K(t-\varepsilon,h))\big{)},$
(4.64)

where we are including the dependence of $K$ on $t$, $h$ and $y_{0}$ in the
same way as for $F$. Now we claim that there is a constant $C<\infty$ which is
independent of $\varepsilon$ and $h$ in the support of $\nu_{\rho}$ and
bounded uniformly for $t$ in a compact interval such that

$\big{|}\mathop{\mathop{\text{{det}}}}\nolimits(I-K(t,h)|_{y_{0}=x_{k}+t+\varepsilon})-\mathop{\mathop{\text{{det}}}}\nolimits(I-K(t-\varepsilon,h))\big{|}\leq
C\hskip 0.50003pt\varepsilon.$ (4.65)

We prove this in Appdx. B.4. It shows that
$|(\partial_{t}-\hat{\mathscr{L}})\hat{F}|\leq C$ uniformly in $t$ in a finite
set and $h\in\operatorname{\mathscr{U\\!C}}$, which is what we wanted.

The only situation left to consider in this case is when there is a $(+1,-1)$
pair at $x_{k}+t$, but it is easy to see that the above argument still works
in this case: when the generator is applied to $\hat{F}$, the up jump moves
left while the down jump moves right, but all the necessary inequalities as
well as the estimates in Appdx. B.4 remain valid.

_Case 3c_ : The last possibility is that there are several indices
$j_{1},\dotsc,j_{\ell}$ (including $k$) for which
$r_{j_{i}}<r_{0}(t,x_{j_{i}})$ and we are at a shock in each of those
variables. The proof in this case is basically the same. When computing the
limit defining $(\partial_{t}-\hat{\mathscr{L}})\hat{F}$ there will be terms
as above coming from the $t$ derivative for each of the $j_{i}$’s, and
matching terms coming from $\hat{\mathscr{L}}\hat{F}$ coming from the
deterministic movement of the jumps of $h$ at the corresponding points. Most
of these terms will vanish as above, and those which do not will lead a sum of
at most $\ell$ differences as in (4.64), which can be estimated separately.

_Case 4_ : All that is left is to consider the case when
$r_{k}=r_{0}(t,x_{k})$ for some $k$. As in case 3, we will focus on the case
where this happens for a single index $k$; the extension to several indices
can be done in the same way. Also, if $r_{j}<r_{0}(t,x_{j})$ for some $j\neq
k$ then we fall back to the situations covered in cases 2 and 3, so we assume
$r_{j}>r_{0}(t,x_{j})$ for all $j\neq k$. In this case we will show that
$(\partial_{t}-\hat{\mathscr{L}})\hat{F}=0$

We are again in the situation of case 1: with the same notation as in that
case, $\hat{F}=\mathop{\mathop{\text{{det}}}}\nolimits(I-K)$ and the argument
leading to (4.60) applies in the same way, so that what we need is to show
that

$(\partial_{t}-\hat{\mathscr{L}})K_{ij}(u,v)=0\qquad\forall\,u>r_{i},~{}v>r_{j}$
(4.66)

for each $i,j$. We begin by noting that, due to our assumption on the
$r_{i}$’s, (4.66) is covered by (4.61) in all cases except when $i$ or $j$
equal $k$, and we evaluate at $r_{k}+1$. So all we need to show is that

$(\partial_{t}-\hat{\mathscr{L}})K_{ik}(u,r_{k}+1)=0\quad\forall\,u>r_{i},\qquad(\partial_{t}-\hat{\mathscr{L}})K_{kj}(r_{k}+1,v)=0\quad\forall\,v>r_{j}.$
(4.67)

We will only prove the first identity; the second one can be proved in the
same way (and, in fact, it follows from the first one by symmetry). Note that
the identity is also covered by (4.61) if $r_{0}(t,x_{k})>h(x_{k}+t)$, so we
may assume $r_{k}=r_{0}(t,x_{k})=h(x_{k}+t)$.

Consider the case $i\neq k$; the case $i=k$ is similar, and we comment on it
at the end. In (4.67) we have $u>r_{i}>r_{0}(t,x_{i})$, so we may assume
$u>h(x_{i}-t)+1$ throughout the computation. In view of (4.63) and (3.48), we
have

$\partial_{t}K_{ik}(u,r_{k}+1)=2[K_{ik},\nabla](u,r_{k}+1)-\tfrac{1}{2}e^{2t-2t\nabla-t\Delta}P^{\operatorname{\text{{no
hit}}}(h)}_{x_{i}-t,(x_{k}+t)^{-}}(u,h(x_{k}+t)).$ (4.68)

The extra term on the right hand side is simply the boundary term at
$v=h_{0}(x^{\prime}+t)+1$ appearing in Lem. 3.4, and we need to show that the
same extra term appears in the application of $\hat{\mathscr{L}}$ in this
case.

We have (as in (4.62))

$\hat{\mathscr{L}}K_{ik}(u,r_{k}+1)=e^{-2t\nabla-t\Delta}\big{(}\hat{\mathscr{L}}P^{\operatorname{\text{{hit}}}(h)}_{x_{i}-t,x_{k}+t}\big{)}e^{2t\nabla-t\Delta}(u,r_{k}+1).$
(4.69)

As in previous proofs, Lem. 3.7 implies that in the above expression we only
need to worry about
$\hat{\mathscr{L}}P^{\operatorname{\text{{hit}}}(h)}_{x_{i}-t,x_{k}+t}(\eta,\xi)$
with $\eta\geq u$, $\xi\geq r_{k}+1$. In other words, and since
$u>h(x_{i}-t)+1$ and $r_{k}=r_{0}(t,x_{k})\geq h(x_{k}+t)$, it is enough to
compute $\hat{\mathscr{L}}P^{\operatorname{\text{{hit}}}(h)}_{a,b}(\eta,\xi)$
for $a\leq b$, $\eta>h(a)+1$ and $\xi>h(b)$. Prop. 4.8 shows that this kernel
equals $2[P^{\operatorname{\text{{hit}}}(h)}_{a,b},\nabla](\eta,\xi)$ when
$\xi>h(b)+1$, so all that it left to compute is
$\hat{\mathscr{L}}P^{\operatorname{\text{{hit}}}(h)}_{a,b}(\eta,h(b)+1)$.
While the proposition does not apply in this case, the regularization argument
which we used in its proof can be reused in conjunction with Lem. 4.10 to do
the computation. To this end we use (4.41) (note it was derived in a setting
which includes our present choices), which gives

$\hat{\mathscr{L}}\text{{Av}}_{\alpha\in
A_{\varepsilon}}P^{\operatorname{\text{{hit}}}(h_{\alpha})}_{a,b}(\eta,h(b)+1)=2[\text{{Av}}_{\alpha\in
A_{\varepsilon}}P^{\operatorname{\text{{hit}}}(h_{\alpha})}_{a,b},\nabla](\eta,h(b)+1).$
(4.70)

The left hand side converges to
$\hat{\mathscr{L}}P^{\operatorname{\text{{hit}}}(h)}_{a,b}(\eta,h(b)+1)$ by
(4.40), so the problem reduces to taking $\varepsilon\to 0$ on the right hand
side, which equals
$2[B_{\varepsilon},\nabla](\eta,\xi)=B_{\varepsilon}(\eta,\xi-1)-B_{\varepsilon}(\eta,\xi+1)-B_{\varepsilon}(\eta+1,\xi)+B_{\varepsilon}(\eta-1,\xi)$
with $B_{\varepsilon}=\text{{Av}}_{\alpha\in
A_{\varepsilon}}P^{\operatorname{\text{{hit}}}(h_{\alpha})}_{a,b}$. (4.40)
implies again that each term converges to the corresponding one with
$B_{\varepsilon}$ replaced by
$P^{\operatorname{\text{{hit}}}(h)}_{a,b}(\eta,\xi)$ except for
$B_{\varepsilon}(\eta,\xi-1)$ in the case $\xi=h(b)+1$, which yields an extra
term coming from Lem. 4.10. From this we deduce that the right hand side of
(4.70) converges to
$2[P^{\operatorname{\text{{hit}}}(h_{\alpha})}_{a,b},\nabla](\eta,h(b)+1)-\tfrac{1}{2}P^{\operatorname{\text{{no
hit}}}(h)}_{a,b^{-}}(\eta,h(b))$ as $\varepsilon\to 0$, which is thus the
value of
$\hat{\mathscr{L}}P^{\operatorname{\text{{hit}}}(h)}_{a,b}(\eta,h(b)+1)$. We
have shown that

$\hat{\mathscr{L}}P^{\operatorname{\text{{hit}}}(h)}_{a,b}(\eta,\xi)=2[P^{\operatorname{\text{{hit}}}(h)}_{a,b},\nabla](\eta,\xi)-\tfrac{1}{2}P^{\operatorname{\text{{no
hit}}}(h)}_{a,b^{-}}(\eta,h(b))\mathbf{1}_{\xi=h(b)+1}$ (4.71)

for $\eta>h(a)+1$, $\xi>h(b)$. Using this in (4.69) we get

$\displaystyle\hat{\mathscr{L}}K_{ik}(u,r_{k}+1)$
$\displaystyle=2e^{-2t\nabla-t\Delta}[P^{\operatorname{\text{{hit}}}(h)}_{a,b},\nabla]e^{2t\nabla-t\Delta}(u,r_{k}+1)$
(4.72) $\displaystyle\hskip
32.52127pt-\tfrac{1}{2}e^{-2t\nabla-t\Delta}P^{\operatorname{\text{{no
hit}}}(h)}_{x_{i}-t,(x_{k}+t)^{-}}(u,h(x_{k}+t))e^{-2t\nabla-t\Delta}(h(x_{k}+t)+1,r_{k}+1)$
(4.73)
$\displaystyle=2[K_{ik},\nabla](u,r_{k}+1)-\tfrac{1}{2}e^{2t-2t\nabla-t\Delta}P^{\operatorname{\text{{no
hit}}}(h)}_{x_{i}-t,(x_{k}+t)^{-}}(u,h(x_{k}+t)),$ (4.74)

where the last equality follows as in previous computations. Comparing with
(4.68), this proves that (4.67) holds as needed.

Finally consider the case $i=k$. The only new kernel entry which we need to
consider is $K_{kk}(r_{k}+1,r_{k}+1)$. We can compute the $t$ derivative of
this quantity as above, but now there will be boundary terms coming from both
identities in (3.47); we get

$\partial_{t}K_{kk}(r_{k}+1,r_{k}+1)=2[K_{kk},\nabla](r_{k}+1,r_{k}+1)-\tfrac{1}{2}P^{\operatorname{\text{{no
hit}}}(h)}_{(x_{k}-t)^{+},x_{k}+t}e^{2t-2t\nabla+t\Delta}(r_{k},r_{k}+1)\\\
-\tfrac{1}{2}e^{2t-2t\nabla-t\Delta}P^{\operatorname{\text{{no
hit}}}(h)}_{x_{k}-t,(x_{k}+t)^{-}}(r_{k}+1,r_{k}).$ (4.75)

On the other hand, repeating the above argument but now using both Lem. 4.10
and its analog at the left edge of $[a,b]$, we see that in order to compute
$\hat{\mathscr{L}}K_{kk}(r_{k}+1,r_{k}+1)$ in the current situation we need to
account for boundary terms coming from the left and right edges of
$[x_{k}-t,x_{k}+t]$, and following the computation through one obtains exactly
the two boundary terms, from which
$(\partial_{t}-\hat{\mathscr{L}})K_{kk}(r_{k}+1,r_{k}+1)=0$ follows. This
finishes case 4, and the proof. ∎

section

##  Appendix A Initial data for non-Abelian Toda

In this appendix we compute the initial condition for the non-Abelian 2D Toda
equation for PNG. We focus on the equation (2.22) in terms of $U_{r}$ and
$V_{r}$, and give expressions for these two matrices at $t=0$. We stress that
this is far from providing the whole boundary data for the equation, which we
do not attempt here. In everything that follows we work under the assumption
$r_{i}\geq r_{0}(t,x_{i})$, which for $t=0$ means we take, for each $i$,

$r_{i}\geq h_{0}(x_{i}).$ (A.1)

The formula (4.33) for $K^{\text{{ext}}}\big{|}_{t=0}$ has the same structure
as the (slightly formal, in that case) formula derived in [QR22], in the
setting of the KPZ fixed point, for the Brownian scattering operator at $t=0$.
Thus the same basic argument applies in our case, and it will lead to a
formula for $Q_{r}\big{|}_{t=0}$ in (A.8) below, providing a discrete analog
of the formula derived in [QR22]. However, in our setting we need to calculate
$Q^{-1}_{r}\big{|}_{t=0}$ and $\partial_{\eta}Q_{r}\big{|}_{t=0}$ as well.
This will require an additional computation, leading to more complicated
formulas. To express them we will need some:

######  Notation.

For $i<j$ and $i\leq k\leq m\leq j$, let

$P_{i,j}^{\geq
h_{0},\leq\check{\mathfrak{d}}_{k,m}}(u,v)=\mathbb{P}_{\mathsf{N}(x_{i})=u}\big{(}\mathsf{N}(y)\geq
h_{0}(y)~{}\forall\,y\in[x_{i},x_{j}],\\\ \mathsf{N}(x_{\ell})\leq
r_{\ell}\text{{ for each
}}x_{\ell}\in[x_{k},x_{m}]\cap(x_{i},x_{j}),\,\mathsf{N}(x_{j})=v\big{)}.$
(A.2)

Note that in the probability we ask $\mathsf{N}$ to stay above $h_{0}$ inside
the whole interval $[x_{i},x_{j}]$, and to stay below the heights $r_{\ell}$
only at those points $x_{\ell}$ among the $x_{1}<\dotsm<x_{n}$ with indices
between $k$ and $m$ but excluding, if necessary, the two endpoints $x_{i}$ and
$x_{j}$. The notation $\check{\mathfrak{d}}_{k,m}$ represents an upside-down,
multiple narrow wedge, defined as $\check{\mathfrak{d}}_{k,m}(x)=r_{\ell}$ if
$x=x_{\ell}$ for some $x_{\ell}\in\\{x_{k},\dotsc,x_{m}\\}$ and $\infty$
otherwise (so the event with upper bound inside the probability is the same as
asking for $\mathsf{N}(y)\leq\check{\mathfrak{d}}_{k,m}(y)$ for
$y\in[x_{i},x_{j}]$). We extend this notation in several ways:

(i) If $k=m$ then we write $\check{\mathfrak{d}}_{k}$ instead of
$\check{\mathfrak{d}}_{k,k}$. (ii) If we change, remove or add inequalities in
the superscript on the left hand side, then all we do is change, remove or add
the corresponding inequalities in the probability on the right hand side.
(iii) When $u=r_{i}$ and $v=r_{j}$, we simply write

$\bar{P}_{i,j}^{\geq h_{0},\leq\check{\mathfrak{d}}_{i,j}}=P_{i,j}^{\geq
h_{0},\leq\check{\mathfrak{d}}_{i,j}}(r_{i},r_{j}),$ (A.3)

and similarly if any of the inequalities are modified. (iv) We turn the last
definition into an $n\times n$ matrix $\bar{P}^{\geq
h_{0},<\check{\mathfrak{d}}}_{\vec{r}}$ defined through (here for convenience
we make the dependence on $\vec{r}=(r_{1},\dots,r_{n})$ explicit)

$(\bar{P}_{\vec{r}}^{\geq
h_{0},<\check{\mathfrak{d}}})_{ij}=\mathbf{1}_{i<j}\bar{P}^{\geq
h_{0},<\check{\mathfrak{d}}_{i,j}}_{i,j}$ (A.4)

for each $i,j$ (note that this is using implicitly the convention that each
$\bar{P}_{ij}=0$ for $i\geq j$). (v) We let $\text{\hbox
to0.0pt{\raisebox{1.0pt}{$\hskip 1.00006pt\cdot$}\hss}$\geq$}h_{0}$ denote the
event that $\mathsf{N}$ stays $\geq h_{0}$ but hits exactly $h_{0}$ along the
way, i.e.

$P^{\text{\hbox to0.0pt{\raisebox{1.0pt}{$\hskip
0.70004pt\cdot$}\hss}$\geq$}h_{0}}_{i,j}=P^{\geq
h_{0}}_{i,j}-P^{>h_{0}}_{i,j}.$ (A.5)

We extend this notation as before if we need to add further restrictions in
the paths of $\mathsf{N}$, and also to the case $\text{\hbox
to0.0pt{\raisebox{1.0pt}{$\,\,\hskip
1.00006pt\cdot$}\hss}$\leq$}\check{\mathfrak{d}}$. (vi) Finally, we use the
notation $\text{\hbox to0.0pt{\raisebox{1.0pt}{$\hskip
1.00006pt\cdot$}\hss}$\geq$}h_{0}\hskip
0.50003pt\scalebox{0.65}{$\curvearrowright$}\hskip 0.50003pt\text{\hbox
to0.0pt{\raisebox{1.0pt}{$\,\,\hskip
1.00006pt\cdot$}\hss}$\leq$}\check{\mathfrak{d}}_{i,j}$ if we require the
$\check{\mathfrak{d}}_{i,j}$ touching to occur after the first $h_{0}$
touching, and we also write

$\big{(}\bar{P}^{\text{\hbox to0.0pt{\raisebox{1.0pt}{$\hskip
0.70004pt\cdot$}\hss}$\geq$}h_{0},\leq\check{\mathfrak{d}}}_{+,\vec{r}}\big{)}_{ij}=P^{\text{\hbox
to0.0pt{\raisebox{1.0pt}{$\hskip
0.70004pt\cdot$}\hss}$\geq$}h_{0},\leq\check{\mathfrak{d}}_{i,j}}_{i,j}(r_{i}+1,r_{j})\qquad\text{and}\qquad\big{(}\bar{P}^{\text{\hbox
to0.0pt{\raisebox{1.0pt}{$\hskip 0.70004pt\cdot$}\hss}$\geq$}h_{0}\hskip
0.35002pt\scalebox{0.65}{$\curvearrowright$}\hskip 0.35002pt\text{\hbox
to0.0pt{\raisebox{1.0pt}{$\,\,\hskip
0.70004pt\cdot$}\hss}$\leq$}\check{\mathfrak{d}}}_{+,\vec{r}}\big{)}_{ij}=P^{\text{\hbox
to0.0pt{\raisebox{1.0pt}{$\hskip 0.70004pt\cdot$}\hss}$\geq$}h_{0}\hskip
0.35002pt\scalebox{0.65}{$\curvearrowright$}\hskip 0.35002pt\text{\hbox
to0.0pt{\raisebox{1.0pt}{$\,\,\hskip
0.70004pt\cdot$}\hss}$\leq$}\check{\mathfrak{d}}_{i,j}}_{i,j}(r_{i}+1,r_{j}).$
(A.6)

Additionally, we introduce the shift operators

$\theta_{r}f(u)=f(u+r),\qquad\theta^{*}_{r}f(u)=f(u-r).$

The action of $\theta_{r}$ is meant as taking place in all of
$\ell^{2}(\mathbb{Z})$ (which is why we are distinguishing $\theta_{1}$ and
$\sigma$). In terms of our formulas, where all (scalar) kernel products take
place in $\ell^{2}(\mathbb{Z}_{>0})$, it is better to think of
$\theta_{r}$/$\theta^{*}_{r}$ simply as a convenient notation, which modifies
the kernel to its right/left, rather than as the action of an operator. The
point is that for a given kernel $A$ we have
$\theta_{r}A\theta^{*}_{r}(u,v)=A(u+r,v+r)$ so we can, for instance, rewrite
(4.33) (using also (A.1)) as

$(K_{r})_{ij}=-\mathbf{1}_{i<j}\theta_{r_{i}}\chi_{r_{i}}P^{>h_{0}}_{i,j}\chi_{r_{j}}\theta^{*}_{r_{j}}=-\mathbf{1}_{i<j}\theta_{r_{i}}P^{>h_{0}}_{i,j}\theta^{*}_{r_{j}},$
(A.7)

since $\theta_{r}\chi_{r}=\chi_{0}\theta_{r}$,
$\chi_{r}\theta_{r}^{*}=\theta_{r}^{*}\chi_{0}$ and our kernels act on
$\ell^{2}(\mathbb{Z}_{>0})$ so the $\chi_{0}$ factors may be omitted.

We turn now to the computation. We begin with a formula for
$Q_{r}\big{|}_{t=0}$:

######  Proposition A.1.

Under the condition (A.1) we have

$Q_{r}\big{|}_{t=0}=I-\bar{P}_{\vec{r}}^{\geq
h_{0},<\check{\mathfrak{d}}}\qquad\text{and}\qquad
Q^{-1}_{r}\big{|}_{t=0}=I+\bar{P}_{\vec{r}}^{\geq
h_{0},\leq\check{\mathfrak{d}}}.$ (A.8)

###### Proof.

In view of (A.7), $K_{r}$ is strictly upper triangular, so
$(R_{r})_{ij}=\sum_{m=0}^{j-i}(K^{m})_{ij}$ and we have $(R_{r})_{ii}=I$ and
$(R_{r})_{ij}=0$ for $i>j$. For $i<j$, on the other hand, we decompose
$(K^{m}_{r})_{ij}$ as
$\sum_{\pi}\prod_{\ell=1}^{m}(K_{r})_{\pi(\ell)\pi(\ell+1)}$, where the sum is
over strictly increasing paths $\pi=(\pi(1),\dotsc,\pi(m+1))$ going from $i$
to $j$ along indices $\\{1,\dotsc,n\\}$. Applying the inclusion-exclusion
principle and using the fact that we are acting on $\ell^{2}(\mathbb{Z}_{>0})$
and $\theta^{*}_{r}\chi_{0}\theta_{r}=\chi_{r}$ in $\ell^{2}(\mathbb{Z})$, we
deduce that

$(R_{r})_{ij}=\mathbf{1}_{i=j}-\mathbf{1}_{i<j}\theta_{r_{i}}P^{>h_{0},\leq\check{\mathfrak{d}}_{i,j}}_{i,j}\theta^{*}_{r_{j}}$
(A.9)

(see [QR22, Eqn. (A.6)] for a similar argument), and then evaluating at
$(1,1)$ gives

$Q_{ij}=\mathbf{1}_{i=j}I-\mathbf{1}_{i<j}P^{>h_{0},\leq\check{\mathfrak{d}}_{i,j}}(r_{i}+1,r_{j}+1)=\mathbf{1}_{i=j}I-\mathbf{1}_{i<j}P^{\geq
h_{0},<\check{\mathfrak{d}}_{i,j}}(r_{i},r_{j}),$

where in the second equality we used the translation invariance of
$\mathsf{N}$. This is the first identity in (A.8).

For $Q^{-1}_{r}\big{|}_{t=0}$ it is now enough to check that
$(I-\bar{P}_{\vec{r}}^{\geq
h_{0},<\check{\mathfrak{d}}})(I+\bar{P}_{\vec{r}}^{\geq
h_{0},\leq\check{\mathfrak{d}}})=I$, which is equivalent to
$\bar{P}_{\vec{r}}^{\geq
h_{0},\leq\check{\mathfrak{d}}}-\bar{P}_{\vec{r}}^{\geq
h_{0},<\check{\mathfrak{d}}}=\bar{P}_{\vec{r}}^{\geq
h_{0},\leq\check{\mathfrak{d}}}\bar{P}_{\vec{r}}^{\geq
h_{0},<\check{\mathfrak{d}}}$. The left hand side equals
$\bar{P}_{\vec{r}}^{\geq h_{0},\text{\hbox
to0.0pt{\raisebox{1.0pt}{$\,\,\hskip
0.70004pt\cdot$}\hss}$\leq$}\check{\mathfrak{d}}}$, using the notation (v)
introduced above. The right hand side is trivially equal to this at or below
the diagonal, while a simple computation using the upper triangularity of each
factor shows that, above the diagonal, the product provides a decomposition of
$\bar{P}_{\vec{r}}^{\geq h_{0},\text{\hbox
to0.0pt{\raisebox{1.0pt}{$\,\,\hskip
0.70004pt\cdot$}\hss}$\leq$}\check{\mathfrak{d}}}$ according to the last time
the walk hits $\check{\mathfrak{d}}$. ∎

Next we need to compute $\partial_{\eta}Q_{r}|_{t=0}$.

######  Proposition A.2.

With the notation introduced above, and under condition (A.1), we have

$\partial_{\eta}Q_{r}\big{|}_{t=0}=\bar{P}^{\text{\hbox
to0.0pt{\raisebox{1.0pt}{$\hskip
0.70004pt\cdot$}\hss}$\geq$}h_{0},\leq\check{\mathfrak{d}}}_{+,\vec{r}}-\bar{P}^{\text{\hbox
to0.0pt{\raisebox{1.0pt}{$\hskip 0.70004pt\cdot$}\hss}$\geq$}h_{0}\hskip
0.35002pt\scalebox{0.65}{$\curvearrowright$}\hskip 0.35002pt\text{\hbox
to0.0pt{\raisebox{1.0pt}{$\,\,\hskip
0.70004pt\cdot$}\hss}$\leq$}\check{\mathfrak{d}}}_{+,\vec{r}}.$ (A.10)

###### Proof.

In all that follows we set $t=0$. We have

$\partial_{\eta}Q_{r}=\partial_{\eta}R_{r}(1,1)=R_{r}\partial_{\eta}K_{r}R_{r}(1,1).$
(A.11)

$R_{r}$ is upper triangular while $\partial_{\eta}K_{r}$ is strictly upper
triangular, so the same holds for $\partial_{\eta}K_{r}R_{r}$ and
$R_{r}\partial_{\eta}K_{r}R_{r}$. In view of this, in the computation that
follows it is enough to consider $i<j$.

Recall $\partial_{\eta}K_{r}(u,v)=-K_{r}(u+1,v)+K_{r}(u,v-1)$, so using (A.7)
we have

$(\partial_{\eta}K_{r})_{ij}=\theta_{r_{i}+1}P^{>h_{0}}_{i,j}\theta^{*}_{r_{j}}-\theta_{r_{i}}P^{>h_{0}}_{i,j}\theta^{*}_{r_{j}-1}=\theta_{r_{i}}P^{\geq
h_{0}}_{i,j}\theta^{*}_{r_{j}-1}-\theta_{r_{i}}P^{>h_{0}}_{i,j}\theta^{*}_{r_{j}-1}=\theta_{r_{i}}P^{\text{\hbox
to0.0pt{\raisebox{1.0pt}{$\hskip
0.70004pt\cdot$}\hss}$\geq$}h_{0}}_{i,j}\theta^{*}_{r_{j}-1},$

where we used translation invariance and the notation introduced in (A.5).
Using now (A.9) we get that

$\textstyle(R_{r}\partial_{\eta}K_{r})_{ij}=\theta_{r_{i}}P^{\text{\hbox
to0.0pt{\raisebox{1.0pt}{$\hskip
0.70004pt\cdot$}\hss}$\geq$}h_{0}}_{i,j}\theta^{*}_{r_{j}-1}-\sum_{i<k<j}\theta_{r_{i}}P^{>h_{0},\leq\check{\mathfrak{d}}_{i+1,k-1},>\check{\mathfrak{d}}_{k},\text{\hbox
to0.0pt{\raisebox{1.0pt}{$\hskip
0.70004pt\cdot$}\hss}$\geq$}h_{0}}_{i,j}\theta^{*}_{r_{j}-1},$ (A.12)

where in the second $P_{i,j}$ operator the sequence of inequalities means in
particular that $\mathsf{N}>h_{0}$ in $[x_{i},x_{k}]$ and
$\mathsf{N}\;\text{\hbox to0.0pt{\raisebox{1.0pt}{$\hskip
1.00006pt\cdot$}\hss}$\geq$}\;h_{0}$ in $[x_{k}+1,x_{j}]$ (there is no
ambiguity at $x_{k}$ since, under (A.1), the condition
$>\check{\mathfrak{d}}_{k}$ implies $\mathsf{N}(x_{k})>h_{0}(x_{k})$ anyway),
and we used again that $\theta^{*}_{r}\chi_{0}\theta_{r}=\chi_{r}$ in
$\ell^{2}(\mathbb{Z})$.

Next we compute $(R_{r}\partial_{\eta}K_{r}R_{r})_{ij}$. Using the last
formula and (A.9), it is given by

$\textstyle\theta_{r_{i}}P^{\text{\hbox to0.0pt{\raisebox{1.0pt}{$\hskip
0.70004pt\cdot$}\hss}$\geq$}h_{0}}_{i,j}\theta^{*}_{r_{j}-1}-\sum_{i<k<j}\theta_{r_{i}}P^{>h_{0},\leq\check{\mathfrak{d}}_{i+1,k-1},>\check{\mathfrak{d}}_{k},\text{\hbox
to0.0pt{\raisebox{1.0pt}{$\hskip
0.70004pt\cdot$}\hss}$\geq$}h_{0}}_{i,j}\theta^{*}_{r_{j}-1}-\sum_{i<k<j}\theta_{r_{i}}P^{\text{\hbox
to0.0pt{\raisebox{1.0pt}{$\hskip
0.70004pt\cdot$}\hss}$\geq$}h_{0}}_{i,k}\theta^{*}_{r_{k}-1}\theta_{r_{k}}P^{>h_{0},\leq\check{\mathfrak{d}}_{k,j}}_{k,j}\theta^{*}_{r_{j}}\\\
\textstyle+\sum_{i<k<\ell<j}\theta_{r_{i}}P^{>h_{0},\leq\check{\mathfrak{d}}_{i+1,k-1},>\check{\mathfrak{d}}_{k},\text{\hbox
to0.0pt{\raisebox{1.0pt}{$\hskip
0.70004pt\cdot$}\hss}$\geq$}h_{0}}_{i,\ell}\theta^{*}_{r_{\ell}-1}\theta_{r_{\ell}}P^{>h_{0},\leq\check{\mathfrak{d}}_{\ell,j}}_{\ell,j}\theta^{*}_{r_{j}}.$
(A.13)

Consider the factor
$\theta^{*}_{r_{k}-1}\theta_{r_{k}}P^{>h_{0},\leq\check{\mathfrak{d}}_{k,j}}_{k,j}\theta^{*}_{r_{j}}$.
If we make explicit the fact that everything acts on
$\ell^{2}(\mathbb{Z}_{>0})$ except for the $\theta_{r}$’s to their right and
the $\theta^{*}_{r}$’s to their left, we can write this as

$\theta^{*}_{r_{k}-1}\chi_{0}\theta_{r_{k}}P^{>h_{0},\leq\check{\mathfrak{d}}_{k,j}}_{k,j}\theta^{*}_{r_{j}}=\chi_{r_{k}-1}\theta_{1}P^{>h_{0},\leq\check{\mathfrak{d}}_{k,j}}_{k,j}\theta^{*}_{r_{j}}=\chi_{r_{k}-1}P^{\geq
h_{0},<\check{\mathfrak{d}}_{k,j}}_{k,j}\theta^{*}_{r_{j}-1},$ (A.14)

where once again we used translation invariance. Let us for a moment replace
the projection $\chi_{r_{k}-1}$ on the left by $\chi_{r_{k}}$, and use this in
the last two terms of the above formula. We get

$\textstyle\theta_{r_{i}}P^{\text{\hbox to0.0pt{\raisebox{1.0pt}{$\hskip
0.70004pt\cdot$}\hss}$\geq$}h_{0}}_{i,j}\theta^{*}_{r_{j}-1}-\sum_{i<k<j}\theta_{r_{i}}P^{>h_{0},\leq\check{\mathfrak{d}}_{i+1,k-1},>\check{\mathfrak{d}}_{k},\text{\hbox
to0.0pt{\raisebox{1.0pt}{$\hskip
0.70004pt\cdot$}\hss}$\geq$}h_{0}}_{i,j}\theta^{*}_{r_{j}-1}-\sum_{i<k<j}\theta_{r_{i}}P^{\text{\hbox
to0.0pt{\raisebox{1.0pt}{$\hskip
0.70004pt\cdot$}\hss}$\geq$}h_{0},>\check{\mathfrak{d}}_{k},<\check{\mathfrak{d}}_{k+1,j-1},\geq
h_{0}}_{i,j}\theta^{*}_{r_{j}-1}\\\
\textstyle+\sum_{i<k<\ell<j}\theta_{r_{i}}P^{>h_{0},\leq\check{\mathfrak{d}}_{i+1,k-1},>\check{\mathfrak{d}}_{k},\text{\hbox
to0.0pt{\raisebox{1.0pt}{$\hskip
0.70004pt\cdot$}\hss}$\geq$}h_{0},>\check{\mathfrak{d}}_{\ell},<\check{\mathfrak{d}}_{\ell-1,j},\geq
h_{0}}_{i,j}\theta^{*}_{r_{j}-1}.$ (A.15)

The second and third terms correspond to $\mathsf{N}$ paths which stay $\geq
h_{0}$ and which at some point touch $h_{0}$ and at some other point go
strictly above one of the $r_{k}$’s; the second sum corresponds to a
decomposition of the event that the path goes above some $r_{k}$ before it
first touches $h_{0}$, while the third sum corresponds to the event that the
path goes above some $r_{k}$ after it last touches $h_{0}$. Now we note that
the sum in the last term is precisely a decomposition of the probability of
the intersection of these two events. Thus the last three terms above
correspond to minus the probability that $\mathsf{N}\;\text{\hbox
to0.0pt{\raisebox{1.0pt}{$\hskip 1.00006pt\cdot$}\hss}$\geq$}\;h_{0}$ and it
goes strictly above one of the $r_{k}$’s. Using this in the last expression
and evaluating at $(1,1)$ yields $P^{\text{\hbox
to0.0pt{\raisebox{1.0pt}{$\hskip
0.70004pt\cdot$}\hss}$\geq$}h_{0},\leq\check{\mathfrak{d}}_{i,j}}_{i,j}(r_{i}+1,r_{j})$.

What remains is to account for the difference between $\chi_{r_{k}-1}$ and
$\chi_{r_{k}}$ in (A.14). This difference is just $\mathbf{1}_{\\{r_{k}\\}}$,
which corresponds to evaluation at $r_{k}$. So, upon evaluating at $(1,1)$,
the remaining terms we get from (A.13)/(A.14) amount to

$\textstyle-\sum_{i<k<j}P^{\text{\hbox to0.0pt{\raisebox{1.0pt}{$\hskip
0.70004pt\cdot$}\hss}$\geq$}h_{0}}_{i,k}(r_{i}+1,r_{k})P^{\geq
h_{0},<\check{\mathfrak{d}}_{k,j}}_{k,j}(r_{k},r_{j})\\\
\textstyle+\sum_{i<k<\ell<j}P_{i,\ell}^{>h_{0},\leq\check{\mathfrak{d}}_{i+1,k-1},>\check{\mathfrak{d}}_{k},\text{\hbox
to0.0pt{\raisebox{1.0pt}{$\hskip
0.70004pt\cdot$}\hss}$\geq$}h_{0}}(r_{i}+1,r_{\ell})P_{\ell,j}^{\geq
h_{0},<\check{\mathfrak{d}}_{\ell,j}}(r_{\ell},r_{j}).$ (A.16)

The first sum is a decomposition of the probability that
$\mathsf{N}\;\text{\hbox to0.0pt{\raisebox{1.0pt}{$\hskip
1.00006pt\cdot$}\hss}$\geq$}\;h_{0}$ and it touches some of the $r_{k}$’s at
some point after it first touches $h_{0}$, remaining below
$\check{\mathfrak{d}}$ after that. The second sum corresponds to something
similar, except that we also ask for $\mathsf{N}$ to go strictly above some of
the $r_{k}$’s, excluding $k=i$, before touching $h_{0}$. The difference of the
two sums then corresponds to the event that $\mathsf{N}\;\text{\hbox
to0.0pt{\raisebox{1.0pt}{$\hskip 1.00006pt\cdot$}\hss}$\geq$}\;h_{0}$,
$\mathsf{N}\;\text{\hbox to0.0pt{\raisebox{1.0pt}{$\,\,\hskip
1.00006pt\cdot$}\hss}$\leq$}\;\check{\mathfrak{d}}$ (with the obvious
meaning), and that there is a $\check{\mathfrak{d}}$ touching after the first
$h_{0}$ touching. In the notation introduced above, we have proved that
$(\partial_{\eta}Q_{r}|_{t=0})_{ij}=P^{\text{\hbox
to0.0pt{\raisebox{1.0pt}{$\hskip
0.70004pt\cdot$}\hss}$\geq$}h_{0},\leq\check{\mathfrak{d}}_{i,j}}_{i,j}(r_{i}+1,r_{j})-P^{\text{\hbox
to0.0pt{\raisebox{1.0pt}{$\hskip 0.70004pt\cdot$}\hss}$\geq$}h_{0}\hskip
0.35002pt\scalebox{0.65}{$\curvearrowright$}\hskip 0.35002pt\text{\hbox
to0.0pt{\raisebox{1.0pt}{$\,\,\hskip
0.70004pt\cdot$}\hss}$\leq$}\check{\mathfrak{d}}_{i,j}}_{i,j}(r_{i}+1,r_{j})$.
This yields (A.10). ∎

Finally we can write the initial data for the non-Abelian Toda equations
(2.22). Like the Rosetta stone, all notations can be deciphered from the
preceding explanations.

######  Proposition A.3.

Assuming $r_{i}\geq h_{0}(x_{i})$ for each $i$,

$U_{r}\big{|}_{t=0}=\big{(}I-\bar{P}_{\vec{r}}^{\geq
h_{0},<\check{\mathfrak{d}}}\big{)}\\!\big{(}I+\bar{P}_{\vec{r}}^{>h_{0},\leq\check{\mathfrak{d}}}\big{)},\qquad
V_{r}\big{|}_{t=0}=\big{(}\bar{P}^{\text{\hbox
to0.0pt{\raisebox{1.0pt}{$\hskip 0.70004pt\cdot$}\hss}$\geq$}h_{0}\hskip
0.35002pt\scalebox{0.65}{$\curvearrowright$}\hskip 0.35002pt\text{\hbox
to0.0pt{\raisebox{1.0pt}{$\,\,\hskip
0.70004pt\cdot$}\hss}$\leq$}\check{\mathfrak{d}}}_{+,\vec{r}}-\bar{P}^{\text{\hbox
to0.0pt{\raisebox{1.0pt}{$\hskip
0.70004pt\cdot$}\hss}$\geq$}h_{0},\leq\check{\mathfrak{d}}}_{+,\vec{r}}\big{)}\\!\big{(}I+\bar{P}_{\vec{r}}^{\geq
h_{0},<\check{\mathfrak{d}}}\big{)}.$ (A.17)

###### Proof.

This follows directly from (A.8) and (A.10), together with the identity
$Q^{-1}_{r-1}\big{|}_{t=0}=I+\bar{P}_{\vec{r}}^{>h_{0},\leq\check{\mathfrak{d}}}$,
which in turn follows from (A.8) and translation invariance. ∎

##  Appendix B Trace class estimates

###  B.1. Continuity

The PNG kernel and transition probabilities are not continuous in the initial
data $h_{0}$ in $\operatorname{\mathscr{U\\!C}}$. Consider, for simplicity,
the one-point case. The issue arises at the shock set $\mathcal{S}(x)$. As an
example, take $h_{\varepsilon}(z)=\mathbf{1}_{z\geq t+\varepsilon}$,
converging to $h_{0}(z)=\mathbf{1}_{z\geq t}$. But
$P^{\operatorname{\text{{hit}}}(h_{0})}_{-t,t}=e^{2t\Delta}\bar{\chi}_{0}$
while $P^{\operatorname{\text{{hit}}}(h_{0}^{\varepsilon})}_{-t,t}=0$. Writing
$K_{h}$ for $K^{\text{{ext}}}_{h}(0,\cdot;0,\cdot)$ we get
$K_{h_{0}}=e^{-2t\nabla+t\Delta}\bar{\chi}_{0}e^{2t\nabla-t\Delta}$ and
$K_{h_{\varepsilon}}=0$. Moreover $\mathbb{P}_{h_{0}}(h(t,0)\leq 0)=0$ while
$\mathbb{P}_{h_{\varepsilon}}(h(t,0)\leq 0)=1$.

In the above example, $h_{\varepsilon}\leq h_{0}$. We will see that if we
restrict instead to approximations of $h_{0}$ from above, continuity does
hold. To express this more precisely, we say that _$h_{n}\longrightarrow h$
from above_ in $\operatorname{\mathscr{U\\!C}}$ if $h_{n}\longrightarrow h$ in
$\operatorname{\mathscr{U\\!C}}$ and $h_{n}(x)\geq h(x)$ for all
$x\in\mathbb{R}$ and all $n\in\mathbb{N}$. A function
$F\\!:\operatorname{\mathscr{U\\!C}}\longrightarrow\mathbb{R}$, is _continuous
from above_ if $F(h_{n})\longrightarrow F(h)$ whenever $h_{n}\longrightarrow
h$ from above.

######  Proposition B.1.

  1. (i)

For each $a<b$ in $\mathbb{R}$ and $u,v$ in $\mathbb{Z}$,
$P_{a,b}^{\operatorname{\text{{hit}}}(h)}(u,v)$ is continuous from above as a
function of $h\in\operatorname{\mathscr{U\\!C}}$.

  2. (ii)

For each $h\in\operatorname{\mathscr{U\\!C}}$, $x\in\mathbb{R}$,
$u,v\in\mathbb{Z}$, $P^{\operatorname{\text{{hit}}}(h)}_{x-t,x+t}(u,v)$ is
right continuous in $t\geq 0$.

###### Proof.

(i) It is not hard to see that if $h_{n}$ are converging to $h$ from above in
$\operatorname{\mathscr{U\\!C}}$, then the Lebesgue measure of the subset
$A_{n}$ of $[a,b]$ on which $h$ is strictly less than $h_{n}$ is going to
zero. Since
$P_{a,b}^{\operatorname{\text{{hit}}}(h_{n})}(u,v)-P_{a,b}^{\operatorname{\text{{hit}}}(h)}(u,v)\geq
0$ and it is bounded above by the probability that the walk $\mathsf{N}$ has a
downward jump in $A_{n}$, which goes to $0$, we conclude that the desired
property holds. (ii) Let $t_{\varepsilon}\searrow t$. Since
$h\in\operatorname{\mathscr{U\\!C}}$, for $\varepsilon$ sufficiently small
$h(x+t_{\varepsilon})\leq h(x+t)$ and $h(x-t_{\varepsilon})\leq h(x-t)$. So it
is not hard to see that the difference between
$P^{\operatorname{\text{{hit}}}(h)}_{x-t_{\varepsilon},x+t_{\varepsilon}}(u,v)$
and $P^{\operatorname{\text{{hit}}}(h)}_{x-t,x+t}(u,v)$ is controlled by the
probability of a jump in $[x-t_{\varepsilon},x-t]\cup[x+t,x+t_{\varepsilon}]$,
which is of order $t_{\varepsilon}-t$. ∎

###  B.2. The kernel

For fixed $x_{1}<\dotsc<x_{n}$ we introduce the multiplication operators

$\vartheta
f(x_{i},u)=\vartheta_{i}f(x_{i},\cdot)(u)\coloneqq\vartheta_{i}(u)f(x_{i},u)\qquad\text{{with}}\qquad\vartheta_{i}(u)=(1+u^{2})^{i}.$
(B.1)

######  Proposition B.2.

For fixed $t>0$, $x_{1}<\dotsc<x_{n}$, $r_{1},\dotsc,r_{n}\in\mathbb{Z}$ and
$h_{0}\in\operatorname{\mathscr{U\\!C}}$, the conjugated kernel
$\vartheta\chi_{r}K^{\text{{ext}}}_{h_{0}}\chi_{r}\vartheta^{-1}$ on
$\ell^{2}(\\{x_{1},\dotsc,x_{n}\\}\times\mathbb{Z})$ is:

  1. (i)

Trace class, with a bound on the trace norm of the form
$C(1+e^{c\mathop{\mathop{\text{{max}}}}\displaylimits_{i}(r_{0}(t,x_{i})-r_{i})})$,
with $r_{0}$ as in (2.14), for some $c,C<\infty$ which can be chosen uniformly
in bounded sets of $t$.

  2. (ii)

Continuous from above on $\operatorname{\mathscr{U\\!C}}$ in trace norm.

  3. (iii)

Right continuous in $t$ in trace norm.

It is important in (i) to get a bound on the trace norm depending on $h_{0}$
only through the $r_{0}(t,x_{i})$’s. The reason is that in our proof of the
backward equation in Sec. 4.5 we need a bound on
$\hat{F}(t,h_{0})=\mathop{\mathop{\text{{det}}}}\nolimits(I-K_{r})_{\oplus_{n}\ell^{2}(\mathbb{Z}_{>0})}\mathbf{1}_{r_{i}\geq
r_{0}(t,x_{i}),\hskip 0.70004pti=1,\dotsc,n}$ in terms of the supremum of
$h_{0}$ on the region of dependence. The abstract bound available for the
absolute value of the Fredholm determinant is $e^{\|K_{r}\|_{1}}$. So thanks
to the truncation coming from the indicator function in the definition of
$\hat{F}$, the bound on the trace norm provided by the proposition in fact
yields a _uniform_ bound on $|\hat{F}|$.

###### Proof.

(i) It is enough to show that each entry
$\chi_{r_{i}}\vartheta_{i}K^{\text{{ext}}}(x_{i},\cdot;x_{j},\cdot)\vartheta^{-1}_{j}\chi_{r_{j}}$
of the kernel is trace class on $\ell^{2}(\mathbb{Z})$. The main estimate we
will use is the following very simple bound, valid for any $t,x\in\mathbb{R}$:

$\big{|}e^{2t\nabla+x\Delta}(u,v)\big{|}\leq C\hskip
0.50003pte^{c(|t|+|x|)}\lambda^{|u-v|}$ (B.2)

for any fixed $\lambda\in(0,1)$ and constants $c,C<\infty$ which depend only
on $\lambda$. This estimate follows directly from (2.32) after choosing
$\gamma_{0}$ as a circle of radius $\lambda$ when $u>v$ and $1/\lambda$
otherwise.

Consider the first term on the right hand side of (2.12), which (after
conjugating and projecting) equals

$\chi_{r_{i}}\vartheta_{i}e^{(x_{j}-x_{i})\Delta}\vartheta^{-1}_{j}\chi_{r_{j}}=\big{(}\chi_{r_{i}}\vartheta_{i}e^{\frac{1}{2}(x_{j}-x_{i})\Delta}\vartheta^{-1/2}_{i+j}\big{)}\big{(}\vartheta_{i+j}^{1/2}e^{\frac{1}{2}(x_{j}-x_{i})\Delta}\vartheta^{-1}_{j}\chi_{r_{j}}\big{)}$
(B.3)

for fixed $i<j$, where the operators $\vartheta^{\pm 1/2}_{k}$ correspond to
multiplication by $\vartheta_{k}(u)^{\pm 1/2}$. Using the bound
$\|AB\|_{1}\leq\|A\|_{2}\|B\|_{2}$ is enough to show that each of the two
factors is Hilbert-Schmidt. By (B.2) we have

$\textstyle\|\chi_{r_{i}}\vartheta_{i}e^{\frac{1}{2}(x_{j}-x_{i})\Delta}\vartheta^{-1/2}_{i+j}\|_{2}^{2}\leq
C^{2}\hskip
0.50003pte^{c(x_{j}-x_{i})}\sum_{u>r_{i}}(1+u^{2})^{2i}\sum_{v\in\mathbb{Z}}\lambda^{2|u-v|}(1+v^{2})^{-i-j}.$

The sum in $v$ restricted to $|v|\leq|u|/2$ is bounded by
$2\kappa|u|\lambda^{|u|}$ while the sum over $|v|>|u|/2$ is bounded by
$C\hskip 0.50003pt2^{2(i+j)}(1+u^{2})^{-i-j}$ for some $C>0$ (depending on
$\lambda)$, so

$\textstyle\|\chi_{r_{i}}\vartheta_{i}e^{\frac{1}{2}(x_{j}-x_{i})\Delta}\vartheta^{-1/2}_{i+j}\|_{2}^{2}\leq
C\hskip
0.50003pte^{c(x_{j}-x_{i})}\sum_{u>r_{i}}((1+u^{2})^{2i}|u|\lambda^{2(1-\kappa)|u|}+(1+u^{2})^{-(j-i)})<\infty,$

since $j-i\geq 1$. The second factor on the right hand side of (B.3) can be
estimated in the same way.

Now we turn to the second term in (2.12). Fix $i,j$ and write
$\underline{x}=x_{i}-t$, $\overline{x}=x_{j}+t$. The kernel we are interested
in equals
$\chi_{r_{i}}\vartheta_{i}e^{-2t\nabla-t\Delta}P^{\operatorname{\text{{hit}}}(h_{0})}_{\underline{x},\overline{x}}e^{2t\nabla-t\Delta}\vartheta^{-1}_{j}\chi_{r_{j}}$.
If $\underline{x}>\overline{x}$ then
$P^{\operatorname{\text{{hit}}}(h_{0})}_{\underline{x},\overline{x}}$ vanishes
and there is nothing to prove, so assume that $\underline{x}\leq\overline{x}$.
$h_{0}$ is upper semi-continuous so we may define

$\textstyle
m_{0}=\mathop{\mathop{\text{{max}}}}\displaylimits_{z\in[\underline{x},\mathop{\mathop{\text{{min}}}}\displaylimits\\{x_{i}+t,\overline{x}\\}]}h_{0}(z),\qquad
x_{0}=\operatorname{\text{{argmax}}}_{z\in[\underline{x},\mathop{\mathop{\text{{min}}}}\displaylimits\\{x_{i}+t,\overline{x}\\}]}h_{0}(z),$

where in the argmax we take, say, the rightmost point in the interval where
the maximum is attained. Write
$e^{-2t\nabla-t\Delta}P^{\operatorname{\text{{hit}}}(h_{0})}_{\underline{x},\overline{x}}e^{2t\nabla-t\Delta}$,
using the inclusion-exclusion principle, as

$e^{-2t\nabla-t\Delta}\big{(}P^{\operatorname{\text{{hit}}}(h_{0})}_{\underline{x},x_{0}}e^{(\overline{x}-x_{0})\Delta}+e^{(x_{0}-\underline{x})\Delta}P^{\operatorname{\text{{hit}}}(h_{0})}_{x_{0},\overline{x}}-P^{\operatorname{\text{{hit}}}(h_{0})}_{\underline{x},x_{0}}P^{\operatorname{\text{{hit}}}(h_{0})}_{x_{0},\overline{x}}\big{)}e^{2t\nabla-t\Delta}.$
(B.4)

The reason we choose this decomposition, splitting the hit probabilities at
$x_{0}$, is that it will allow us to get a bound on the trace norm that only
depends on $h_{0}$ through $r_{0}(t,x_{i})$, its maximum on the interval
$[x_{i}-t,x_{i}+t]$ (which is contained in $[\underline{x},x_{0}]$). This
dependence will come from estimates on the factors involving
$P^{\operatorname{\text{{hit}}}(h_{0})}_{\underline{x},x_{0}}$ in this
decomposition. As we will see, the factors
$P^{\operatorname{\text{{hit}}}(h_{0})}_{x_{0},\overline{x}}$ are a bit
simpler to handle, basically because in the conjugation they end up next to
decaying factor $\vartheta_{j}^{-1}$ (as opposed to the factor $\vartheta_{i}$
on the left, which is blowing up). In fact, we will only estimate the trace
norm of the third term coming from the above expression after conjugating and
projecting; the bound for the other two is slightly simpler, and can be
adapted easily from the one we present here.

Note that if $m_{0}=-\infty$ (as it may be) then
$P^{\operatorname{\text{{hit}}}(h_{0})}_{\underline{x},x_{0}}=0$ and there is
nothing to prove. So we assume $m_{0}>-\infty$. Our goal is to estimate the
trace norm of
$\chi_{r_{i}}\vartheta_{i}e^{-2t\nabla-t\Delta}P^{\operatorname{\text{{hit}}}(h_{0})}_{\underline{x},x_{0}}P^{\operatorname{\text{{hit}}}(h_{0})}_{x_{0},\overline{x}}e^{2t\nabla-t\Delta}\vartheta_{j}^{-1}\chi_{r_{j}}$.
We can write the kernel of this operator as an average of rank one kernels
$A_{s_{1}}(u,\eta_{1})B_{s_{2}}(\eta_{2},v)$,

$\textstyle\int_{\underline{x}}^{x_{0}}\int_{x_{0}}^{\overline{x}}\sum_{\eta_{1},\eta_{2}\in\mathbb{Z}}\alpha(\eta_{1},\mathrm{d}s_{1},\eta_{2},\mathrm{d}s_{2})A_{s_{1}}(u,\eta_{1})B_{s_{2}}(\eta_{2},v)$
(B.5)

where $A_{s}=\chi_{r_{i}}\vartheta_{i}e^{-2t\nabla+(x_{0}-s-t)\Delta}$,
$B_{s}=e^{2t\nabla+(\overline{x}-s_{2}-t)\Delta}\vartheta_{j}^{-1}\chi_{r_{j}}$
and $\alpha(\eta_{1},\mathrm{d}s_{1},\eta_{2},\mathrm{d}s_{2})$ is given by
$\sum_{\xi\in\mathbb{Z}}\hat{p}_{\underline{x},\xi}(\eta_{1},\mathrm{d}s_{1})p_{x_{0},\xi}(\eta_{2},\mathrm{d}s_{2})$,
with
$p_{x,\xi}(\eta,\mathrm{d}s)=\mathbb{P}_{\mathsf{N}(x)=\xi}(\tau\in\mathrm{d}s,\,\mathsf{N}(\tau)=\eta)$
and
$\hat{p}_{x,\xi}(\eta,\mathrm{d}s)=\mathbb{P}_{\mathsf{N}(x)=\xi}(\hat{\tau}\in\mathrm{d}s,\,\mathsf{N}(\hat{\tau})=\eta)$.
Here $\tau$ and $\hat{\tau}$ are the hitting times by $\mathsf{N}$ (restricted
to $[x_{0},\overline{x}]$ and to $[\underline{x},x_{0}]$) of
$\operatorname{\text{{hypo}}}(h_{0})$ and
$\operatorname{\text{{hypo}}}(\hat{h}_{0})$, with $\hat{h}_{0}$ the reversed
barrier given by $h_{0}(s)=h_{0}(\underline{x}+x_{0}-s)$; the first one is a
straightforward decomposition using the hitting time $\tau$ while the second
one is similar after using the reversibility of the walk $\mathsf{N}$. We note
that in the case $x_{0}=\overline{x}$, the above is interpreted as forcing
$\tau=x_{0}$ and $\eta=y$ provided $y\leq h_{0}(x_{0})$, and $0$ otherwise,
and similarly in the case $x_{0}=\underline{x}$.

The trace norm of the average can be bounded by the average of the trace norms
of the rank one operators. But the square of the trace norm of a rank one
operator is just the product of $\ell^{2}$ norms. Each $\ell^{2}$ norm can be
estimated by (B.2). Thus the trace norm of
$\big{\|}\chi_{r_{i}}\vartheta_{i}e^{-2t\nabla-t\Delta}P^{\operatorname{\text{{hit}}}(h_{0})}_{\underline{x},x_{0}}P^{\operatorname{\text{{hit}}}(h_{0})}_{x_{0},\overline{x}}e^{2t\nabla-t\Delta}\vartheta_{j}^{-1}\chi_{r_{j}}\big{\|}_{1}$
is bounded by a constant depending on $\lambda$, $t$, $\underline{x}$ and
$\overline{x}$ times

$\textstyle\sum_{\eta_{1},\eta_{2},\xi}\mathbb{P}_{\mathsf{N}(\underline{x})=\xi}(\hat{\tau}\in[\underline{x},x_{0}],\mathsf{N}(\hat{\tau})=\eta_{1})\mathbb{P}_{\mathsf{N}(x_{0})=\xi}(\tau\in[x_{0},\overline{x}],\mathsf{N}(\tau)=\eta_{2})\\\
\times(\vartheta_{i}(\eta_{1})^{2}+\vartheta_{i}(r_{i})^{2})^{1/2}\lambda^{(r_{i}-\eta_{1})\vee
0}\lambda^{(r_{j}-\eta_{2})\vee 0}.$ (B.6)

If $\xi\leq m_{0}$, then $\hat{\tau}=\underline{x}$ and
$\mathsf{N}(\tau)=\xi$, which forces $\eta_{1}=\xi$. Thus the part of the sum
over $\xi\leq m_{0}$ is bounded by

$\textstyle\sum_{\xi\leq
m_{0},\eta_{2}\in\mathbb{Z}}\mathbb{P}_{\mathsf{N}(x_{0})=\xi}(\tau\in[x_{0},\overline{x}],\mathsf{N}(\tau)=\eta_{2})(\vartheta_{i}(\xi)^{2}+\vartheta_{i}(r_{i})^{2})^{1/2}\lambda^{(r_{i}-\xi)\vee
0}.$ (B.7)

We can sum over $\eta_{2}$ in the probabilities, then drop the resulting
$\mathbb{P}_{\mathsf{N}(x_{0})=\xi}(\tau\in[x_{0},\overline{x}])$. The
resulting expression is summable, with a result which can be bounded by
$C(1+e^{c(m_{0}-r_{i})})$. On the other hand, if $\xi>m_{0}$, we use the fact
that $\mathsf{N}(\hat{\tau})\leq m_{0}$ if
$\hat{\tau}\in[\underline{x},x_{0}]$, so we may restrict the $\eta_{1}$ sum to
$\eta_{1}\leq m_{0}$. So this part of the sum is less than or equal to

$\textstyle\sum_{\eta_{1}\leq
m_{0},\xi>m_{0}}\mathbb{P}_{\mathsf{N}(\underline{x})=\xi}(\hat{\tau}\in[\underline{x},x_{0}],\mathsf{N}(\hat{\tau})=\eta_{1})(\vartheta_{i}(\eta_{1})^{2}+\vartheta_{i}(r_{i})^{2})^{1/2}\lambda^{(r_{i}-\eta_{1})\vee
0}.$ (B.8)

Summing in $\eta_{1}$ gives an expression which is bounded by
$C(1+e^{c(m_{0}-r_{i})})$ again, and it only remains to estimate
$\sum_{\xi>m_{0}}\mathbb{P}_{\mathsf{N}(\underline{x})=\xi}(\hat{\tau}\in[\underline{x},x_{0}])$.
Now
$\mathbb{P}_{\mathsf{N}(\underline{x})=\xi}(\tau\in[x_{0},\overline{x}])\leq\mathbb{P}_{\mathsf{N}(x_{0})=\xi}\\!\left(\mathop{\mathop{\text{{inf}}}}\displaylimits_{y\in[\underline{x},x_{0}]}\mathsf{N}(x)\leq
m_{0}\right)$ since $m_{0}$ maximizes $\hat{h}_{0}$ inside
$[\underline{x},x_{0}]$. We can compute the latter by the reflection
principle, and bound it by $C\hskip 0.50003pt\lambda^{\xi-m_{0}}$, and summing
over $\xi>m_{0}$ yields a new constant. The conclusion is that (B.6) is
bounded by $C(1+e^{c(m_{0}-r_{i})})$ for some $c,C<\infty$, as desired. This
proves (i).

To prove (ii) and (iii), first of all if $h_{\varepsilon}\to h$ in
$\operatorname{\mathscr{U\\!C}}$ then there is a global constant
$\bar{h}<\infty$ such that $h_{\varepsilon}(x)\leq\bar{h}$ for $x$ in the
finite region of support $[x_{1}-t,x_{n}+t]$ up to time $t$, for small
$\varepsilon>0$. The kernel
$\vartheta\chi_{r}K^{\text{{ext}}}_{h_{0}}\chi_{r}\vartheta^{-1}$ is a
composition of kernels each of which is continuous from above in
$\operatorname{\mathscr{U\\!C}}$ or right continuous in $t$, pointwise, by
Prop. B.1 in the case of the hit kernel, or from (2.32) in the case of the
conjugating operators $e^{\pm 2t\nabla\pm x\Delta}$. If we look at one of the
differences for which we have to control the trace norm, say
$\chi_{r}\vartheta
K^{\text{{ext}}}_{h_{\varepsilon}}\chi_{r}\vartheta^{-1}-\vartheta\chi_{r}K^{\text{{ext}}}_{h_{0}}\vartheta^{-1}\chi_{r}$
(the other one is similar), and consider its $(i,j)$ entry, we may factor it
as
$\chi_{r_{i}}\vartheta_{i}e^{-2t\nabla-t\Delta}(P^{\operatorname{\text{{hit}}}(h_{\varepsilon})}_{x_{i}-t,x_{j}+t}-P^{\operatorname{\text{{hit}}}(h)}_{x_{i}-t,x_{j}+t})e^{2t\nabla-t\Delta}\vartheta^{-1}_{j}\chi_{r_{j}}$
We follow the estimation procedure from the proof of (i), keeping the absolute
value of this difference. Replacing the absolute value of the difference by
the sum, and the $h_{\varepsilon}$ by the constant function $\bar{h}$ (which
can only increase the hit probability), we obtain a term that by the
estimation procedure of (i) is bounded, and serves as a dominating function
for the difference. Hence we can use the dominated convergence theorem to
conclude the trace norm convergence from the pointwise convergence of the
individual kernels. ∎

###  B.3. Differentiability

Now we turn to checking that the (conjugated and projected) PNG kernel
satisfies (2) and (3) from Thm. 3.1 with the derivatives holding in trace
class. The arguments are very similar to the ones in Sec. B.2, and we only
write them in the case of the $\eta$ derivative, the $\zeta$ derivative being
completely analogous. As above, it is enough to check this separately for each
entry $K^{{\text{{ext}}}}(x_{i},\cdot;x_{j},\cdot)$, after conjugating and
adding the projections. We fix $i,j$ and let $K_{\eta,\zeta}$ denote this
entry, where we have made the dependence on $\eta$ and $\zeta$ explicit. Then
we need to prove that

$\big{\|}\chi_{r_{i}}\vartheta_{i}\Big{(}\delta^{-1}\big{(}K_{\eta+\delta,\zeta}-K_{\eta,\zeta}\big{)}-\partial_{\eta}K_{\eta,\zeta}\Big{)}\vartheta_{j}^{-1}P_{r_{j}}\big{\|}_{1}\xrightarrow[\delta\to
0]{}0$

with
$\partial_{\eta}K_{\eta,\zeta}(u,v)=K_{\eta,\zeta}(u,v-1)-K_{\eta,\zeta}(u+1,v)$
under the assumption $r_{i}>r_{0}(t,x_{i})$, $r_{j}>r_{0}(t,x_{j})$. Using
(2.12) we can write
$K_{\eta,\zeta}=-\mathbf{1}_{x_{i}<x_{j}}e^{(x_{j}-x_{i})\Delta}+e^{-2(\eta_{i}-\zeta_{i})\nabla-(\eta_{i}+\zeta_{i})\Delta}\mathscr{T}^{h_{0}}_{2\eta_{i},2\zeta_{j}}e^{2(\eta_{j}-\zeta_{j})\nabla+(\eta_{j}+\zeta_{j})\Delta}$,
with $\eta_{k}=\frac{1}{2}(x_{k}+t)$, $\zeta_{k}=\frac{1}{2}(x_{k}-t)$. Then

$K_{\eta+\delta,\zeta}-K_{\eta,\zeta}=(e^{-2\delta\nabla-\delta\Delta}-I)K_{\eta,\zeta}+e^{-2\delta\nabla-\delta\Delta}K_{\eta,\zeta}(e^{2\delta\nabla+\delta\Delta}-I)\\\
+e^{-2(\eta_{i}+\delta-\zeta_{i})\nabla-(\eta_{i}+\delta+\zeta_{i})\Delta}\big{(}\mathscr{T}^{h_{0}}_{2\eta_{i}+2\delta,2\zeta_{j}}-\mathscr{T}^{h_{0}}_{2\eta_{i},2\zeta_{j}}\big{)}e^{2(\eta_{j}+\delta-\zeta_{j})\nabla+(\eta_{j}+\delta+\zeta_{j})\Delta}.$
(B.9)

Focus on the first term on the right hand side. After dividing by $2\delta$,
the factor $e^{-2\delta\nabla-\delta\Delta}-I$ converges to
$-(\nabla+\frac{1}{2}\Delta)$, so we expect that the whole kernel, divided by
$\delta$ and evaluated at $(u,v)$, should converge to
$K_{\eta,\zeta}(u,v)-K_{\eta,\zeta}(u+1,v)$. This convergence has to be proved
in trace norm, after properly conjugating and adding the projections.
Following the arguments of Sec. B.2 we see that this essentially depends on
estimating the $\ell^{2}$ norm of

$\chi_{r_{i}}\vartheta_{i}(u)\big{(}(2\delta)^{-1}\big{(}e^{-2(t+\delta)\nabla-(x_{i}+\delta)\Delta}(u,\eta)-e^{2t\nabla+x_{i}\Delta}(u,\eta)\big{)}-e^{2t\nabla+x_{i}\Delta}(u,\eta)+e^{2t\nabla+x_{i}\Delta}(u+1,\eta)\big{)}$

for fixed $\eta\in\mathbb{Z}$, as a function of $u$. Now (2.32) shows that
$\tfrac{1}{2}\big{(}e^{-2(t+\delta)\nabla-(x+\delta)\Delta}(u,\eta)-e^{2t\nabla+x\Delta}(u,\eta)\big{)}=\frac{1}{2\pi{\rm
i}}\oint_{\gamma}\mathrm{d}z\,\frac{e^{t(z^{-1}-z)-x(z+z^{-1})+2x}}{z^{\eta-u+1}}\frac{e^{-2\delta
z+2\delta}-1}{2}=\frac{\delta}{2\pi{\rm
i}}\oint_{\gamma}\mathrm{d}z\,\frac{e^{t(z^{-1}-z)+x(z+z^{-1})+2x}}{z^{\eta-u+1}}(1-z+\mathcal{O}(\delta))$,
where the $\mathcal{O}(\delta)$ term can be bounded uniformly in $z$ and
$\eta$. Using this in the above expression we see that the desired $\ell^{2}$
norm is bounded by $C\hskip
0.50003pt\delta\|\chi_{r_{i}}\vartheta_{i}e^{-2t\nabla-t\Delta}(\cdot,\eta)\|_{2}$,
and using this in the estimates of Sec. B.2 (see e.g. (B.5) and (B.6)) shows
the desired convergence of
$(2\delta^{-1})(e^{-2\delta\nabla-\delta\Delta}-I)K^{\eta,\zeta}$.

Going back to (B.9), the second term converges in essentially the same way as
above, after dividing by $2\delta$, to
$K_{\eta,\zeta}(u,v-1)-K_{\eta,\zeta}(u,v)$. The last term in (B.9) involves
what in Sec. 3.2 was defined as $W^{(\eta)}_{t,x_{i},x_{j}}$ (here there is an
extra conjugation by $e^{-2t\nabla-t\Delta}$, which is of no consequence). In
Lem. 3.4 we showed that $W^{(\eta)}_{t,x_{i},x_{j}}(u,v)$ vanishes for
$u>r_{0}(t,x_{i})+1$, $v>r_{0}(t,x_{j})+1$. This is precisely our setting,
because we are surrounding by projections $\chi_{r_{i}}$ and $\chi_{r_{j}}$
and assuming $r_{i}>r_{0}(t,x_{i})$, $r_{j}>r_{0}(t,x_{j})$. What is left is
to upgrade this pointwise computation to a convergence in trace norm, but this
can be done without too much difficulty by using the arguments of this section
in the proof of Lem. 3.4; we leave the details to the reader. This finishes
the proof of the desired trace norm convergence of the $\zeta$ derivative.

###  B.4. Proof of (4.65)

Recall the setting of Case 3 in the proof of Thm. 4.6. We are given a
$\operatorname{\mathscr{U\\!C}}$ function $h$ with locally finite $\pm 1$
jumps and parameters $t$, $x_{1},\dotsc,x_{n}$ and $r_{1},\dotsc,r_{n}$ such
that $h$ has an up jump of size one at $x_{k}+t$, no jumps at $x_{k}-t$ nor at
$x_{j}\pm t$ for each $j\neq k$, and $r_{j}\geq r_{0}(t,x_{j})$ for each
$j\neq k$ while $r_{0}(t^{-},x_{k})=r_{k}=r_{0}(t,x_{k})-1$.
$(\vec{y},\vec{\sigma})$ is the configuration of jump locations and signs
associated to $h$, and as a convention we assume that the jump at $x_{k}+t$
has label $0$, i.e. $y_{0}=x_{k}+t$ (and $\sigma_{0}=1$). Letting $K(t,h)$
denote the PNG extended kernel with these choices, which we will abbreviate as
$K_{t}^{0}$, and $K(t,h)|_{y_{0}=x_{k}+t+\varepsilon}$ denote the same kernel
with the jump at $y_{0}=x_{k}+t$ moved to $x_{k}+t+\varepsilon$, which we will
abbreviate as $K_{t}^{\varepsilon}$, we want to show that

$\big{|}\mathop{\mathop{\text{{det}}}}\nolimits(I-\chi_{r}K_{t}^{\varepsilon}\chi_{r})-\mathop{\mathop{\text{{det}}}}\nolimits(I-\chi_{r}K_{t-\varepsilon}^{0}\chi_{r})\big{|}\leq
C$ (B.10)

for some $C>$ which does not depend on $\varepsilon$ and $h$ and which is
bounded uniformly in choices of $t$ on a compact set. We will put tildes on
top of kernels when they are conjugated by $\vartheta$ and multiplied by
projections $\chi_{r}$ on both sides as in the previous subsections. Prop. B.2
shows that both $\tilde{K}_{t-\varepsilon}^{0}$ and
$\tilde{K}_{t}^{\varepsilon}$ are trace class, with trace norms which are
bounded uniformly in $t$ in a compact set and
$h\in\operatorname{\mathscr{U\\!C}}$, and in fact from the argument it is not
hard to see that these are also bounded uniformly in $\varepsilon\in(0,1)$.
Then, using the bound
$|\mathop{\mathop{\text{{det}}}}\nolimits(I-A)-\mathop{\mathop{\text{{det}}}}\nolimits(I-B)|\leq\|A-B\|_{1}e^{1+\|A\|_{1}+\|B\|_{1}}$
for trace class operators $A$ and $B$ we deduce that, for some suitably
uniformly chosen constant $C$,

$\big{|}\mathop{\mathop{\text{{det}}}}\nolimits(I-\tilde{K}_{t}^{\varepsilon})-\mathop{\mathop{\text{{det}}}}\nolimits(I-\tilde{K}_{t-\varepsilon}^{0})\big{|}\leq
C\big{\|}\tilde{K}_{t-\varepsilon}^{0}-\tilde{K}_{t}^{\varepsilon}\big{\|}_{1}.$
(B.11)

In order to estimate the norm on the right hand side of (B.11) it is enough to
do the same for each of the $i,j$ entries $\Delta_{ij}^{\varepsilon}\coloneqq
K_{t-\varepsilon}^{0}(x_{i},\cdot;x_{j},\cdot)-K_{t}^{\varepsilon}(x_{i},\cdot;x_{j},\cdot)$,
which we decompose as

$\Delta_{ij}^{\varepsilon}=\big{(}e^{-2(t-\varepsilon)\nabla-(t-\varepsilon)\Delta}P^{\operatorname{\text{{hit}}}(h)}_{x_{i}-t,x_{j}+t}e^{2(t-\varepsilon)\nabla-(t-\varepsilon)\Delta}-e^{-2t\nabla-t\Delta}P^{\operatorname{\text{{hit}}}(h)}_{x_{i}-t,x_{j}+t}e^{2t\nabla-t\Delta}\big{)}\\\
+e^{-2t\nabla-t\Delta}\big{(}P^{\operatorname{\text{{hit}}}(h)}_{x_{i}-t,x_{j}+t}-P^{\operatorname{\text{{hit}}}(h^{\varepsilon})}_{x_{i}-t,x_{j}+t}\big{)}e^{2t\nabla-t\Delta},$
(B.12)

where $h^{\varepsilon}$ denotes $h$ with the jump at $y_{0}=x_{k}+t$ moved to
$x_{k}+t+\varepsilon$. The difference on the right hand side of the first line
can be estimated in trace norm, after conjugating and projecting, using the
arguments of Sec. B.2, and it is not hard to obtain a bound of the form
$C\hskip 0.50003pt\varepsilon$; we omit the details. Hence, in order to finish
the proof we need to show that the same holds for the second line of (B.12),
i.e. that

$\big{\|}\vartheta_{i}\chi_{r_{i}}e^{-2t\nabla-t\Delta}\big{(}P^{\operatorname{\text{{hit}}}(h)}_{x_{i}-t,x_{j}+t}-P^{\operatorname{\text{{hit}}}(h^{\varepsilon})}_{x_{i}-t,x_{j}+t}\big{)}e^{2t\nabla-t\Delta}\vartheta_{j}^{-1}\chi_{r_{j}}\big{\|}_{1}\leq
C\hskip 0.50003pt\varepsilon.$ (B.13)

To prove (B.13) we need to separate cases according to where $x_{k}+t$ lies
with respect to $x_{i}-t$ and $x_{j}+t$. The simplest situation is
$x_{k}+t\notin[x_{i}-t,x_{j}+t]$; in this case $h=h^{\varepsilon}$ inside
$[x_{i}-t,x_{j}+t]$ (for small enough $\varepsilon$) so the left hand side of
(B.13) vanishes. If $k=j$ then
$\big{(}P^{\operatorname{\text{{hit}}}(h)}_{x_{i}-t,x_{j}+t}-P^{\operatorname{\text{{hit}}}(h^{\varepsilon})}_{x_{i}-t,x_{j}+t}\big{)}(u,\xi)=P^{\operatorname{\text{{hit}}}(h)}_{x_{i}-t,x_{j}+t}(u,h(x_{k}+t)-1)\mathbf{1}_{\xi=h(x_{k}+t)-1}$,
which gives
$\big{(}P^{\operatorname{\text{{hit}}}(h)}_{x_{i}-t,x_{j}+t}-P^{\operatorname{\text{{hit}}}(h^{\varepsilon})}_{x_{i}-t,x_{j}+t}\big{)}\vartheta_{j}^{-1}\chi_{r_{j}}(u,v)=P^{\operatorname{\text{{hit}}}(h)}_{x_{i}-t,x_{j}+t}(u,h(x_{k}+t)-1)e^{2t\nabla-t\Delta}(h(x_{k}+t)-1,v)\vartheta_{j}(v)\mathbf{1}_{v>r_{j}}$,
but in this case $r_{j}=r_{k}=r_{0}(t,x_{k})-1\geq h(x_{k}+t)-1$, so by Lem.
3.7 the kernel vanishes again.

Finally we consider (B.13) when $x_{k}+t\in[x_{i}-t,x_{j}+t)$. In particular,
this implies that the interval is non-empty, and we may assume that it also
includes $x_{k}+t+\varepsilon$. Note that the event inside the probabilities
$P^{\operatorname{\text{{hit}}}(h)}_{x_{i}-t,x_{j}+t}$ and
$P^{\operatorname{\text{{hit}}}(h^{\varepsilon})}_{x_{i}-t,x_{j}+t}$ is the
same if we restrict to paths where the random walk $\mathsf{N}$ stays put
between time $x_{k}+t$ and $x_{k}+t+\varepsilon$, and event whose complement
has probability $1-e^{-2\varepsilon}\leq 2\varepsilon$. The same argument as
in the proof of Prop. B.2 can now be employed to turn this into the desired
bound (B.13) (note that, $h$ and $h^{\varepsilon}$ are, in fact, at Hausdorff
distance $\varepsilon$).

Acknowledgements. KM was partially supported by NSF grant DMS-1953859. JQ was
supported by the Natural Sciences and Engineering Research Council of Canada.
DR was supported by Centro de Modelamiento Matemático Basal Funds FB210005
from ANID-Chile and by Fondecyt Grant 1201914.

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