eurm10 msam10

# Inertial torque on a squirmer

F. Candelier1 J. Qiu2 L. Zhao2 G. Voth3 B. Mehlig4 1Aix Marseille Univ, CNRS,
IUSTI, Marseille, France
2AML, Department of Engineering Mechanics, Tsinghua University, 100084
Beijing, China
3Department of Physics, Wesleyan University, Middletown, CT 06459, USA
4Department of Physics, Gothenburg University, 41296 Gothenburg, Sweden

###### Abstract

A small spheroid settling in a quiescent fluid experiences an inertial torque
that aligns it so that it settles with its broad side first. Here we show that
an active particle experiences such a torque too, as it settles in a fluid at
rest. For a spherical squirmer, the torque is
$\boldsymbol{T}^{\prime}=-{\tfrac{9}{8}}m_{f}(\boldsymbol{v}_{s}^{(0)}\wedge\boldsymbol{v}_{g}^{(0)})$
where $\boldsymbol{v}_{s}^{(0)}$ is the swimming velocity,
$\boldsymbol{v}_{g}^{(0)}$ is the settling velocity in the Stokes
approximation, and $m_{f}$ is the equivalent fluid mass. This torque aligns
the swimming direction against gravity: swimming up is stable, swimming down
is unstable.

##  1 Introduction

The motion of small plankton in the turbulent ocean is overdamped (Visser,
2011). Accelerations play no role, and hydrodynamic forces and torques can be
computed in the Stokes approximation. Turbulence rotates these small
organisms, yet they manage to navigate upwards towards the ocean surface.
Gyrotactic organisms make use of gravity to achieve this. These bottom-heavy
swimmers experience a gravity torque that tends to align against the direction
of gravity, so that they swim upwards (Kessler, 1985; Durham et al., 2013;
Gustavsson et al., 2016). Also density or shape asymmetries give rise to
torques in the Stokes approximation that can change the swimming direction
(Roberts, 1970; Jonsson, 1989; Roberts & Deacon, 2002; Candelier & Mehlig,
2016; Roy et al., 2019).

Larger organisms accelerate the surrounding fluid as they move, and this
changes the hydrodynamic force the swimmer experiences (Wang & Ardekani, 2012;
Khair & Chisholm, 2014; Chisholm et al., 2016; Redaelli et al., 2022b, a).
Three different mechanisms cause such fluid-inertia effects, a non-zero slip
velocity (Oseen problem with non-dimensional parameter Rep, the particle
Reynolds number), velocity gradients of the disturbance flow (Saffman problem,
shear Reynolds number Res), and unsteady fluid inertia (with parameter ${\rm
Re}_{p}{\rm Sl}$, where ${\rm Sl}$ is the Strouhal number).

Fluid inertia gives rise to hydrodynamic torques. For a passive spheroid in
spatially inhomogeneous flow, there are ${\rm Re}_{s}$-corrections to
Jeffery’s torque (Subramanian & Koch, 2005; Einarsson et al., 2015; Rosén et
al., 2015). A passive spheroid settling in a quiescent fluid experiences an
inertial torque, a Rep-effect. This Khayat-Cox torque tends to align the
particle so that it settles with its broad side down (Brenner, 1961; Cox,
1965; Khayat & Cox, 1989; Klett, 1995; Dabade et al., 2015; Menon et al.,
2017; Lopez & Guazzelli, 2017; Kramel, 2017; Gustavsson et al., 2019; Jiang et
al., 2021; Cabrera et al., 2022). For a passive sphere, spherical symmetry
ensures that the Khayat-Cox torque vanishes.

\begin{overpic}[width=284.52756pt,clip]{Fig1.pdf} \end{overpic} Figure 1:  (a)
Squirmer with swimming velocity $\boldsymbol{v}_{s}$ and settling velocity
$\boldsymbol{v}_{g}$, see Section 2. Gravity points in the negative $\hat{\bf
e}_{2}$-direction. (b,c,d) Disturbance flow created by a squirmer with
$B_{2}=0$ (schematic). Shown are the flow lines in the frame that translates
with the body. The centre-of-mass velocity $\dot{\boldsymbol{x}}$ is shown in
green.

In this paper, we show that a small spherical squirmer experiences an inertial
torque analogous to the Khayat & Cox torque when it settles in a quiescent
fluid. Using asymptotic matching, we calculate the torque to leading order in
the particle Reynolds number

$\displaystyle{\rm Re}_{p}={au_{\rm c}}/{\nu}\,,$ (1)

where $u_{\rm c}$ is a velocity scale, $a$ is the radius of the squirmer, and
$\nu$ is the kinematic viscosity of the fluid. The calculation shows that the
inertial torque does not vanish for a spherical swimmer because swimming
breaks rotational symmetry. We describe how the torque aligns the squirmer,
and compare its effect with gyrotactic torques, and with the Khayat-Cox torque
for a nonspherical passive particle.

##  2 Model

We consider a steady spherical squirmer, an idealised model for a motile
microorganism developed by Lighthill (1952) and Blake (1971). In this model,
one imposes an active axisymmetric tangential surface-velocity field of the
form

$\displaystyle(B_{1}\sin\theta+B_{2}\sin\theta\cos\theta)\hat{\bf
e}_{\theta}\,,$ (2)

with parameters $B_{1}$ and $B_{2}$, and where $\theta$ is the angle between
the swimming direction (unit vector $\boldsymbol{n}$) and the vector
$\boldsymbol{r}$ from the particle centre to a point on its surface. The
tangential unit vector at this point is denoted by $\hat{\bf e}_{\theta}$. One
distinguishes two types of squirmers depending on the parameter
$\beta=B_{2}/B_{1}$ (Lauga & Powers, 2009): ‘pushers’ ($\beta<0$) and
‘pullers’ with $\beta>0$. In the Stokes limit, a squirmer moving with velocity
$\dot{\boldsymbol{x}}$ in a fluid at rest experiences the hydrodynamic force

$\displaystyle{\boldsymbol{F}^{\prime}}^{(0)}=6\pi\varrho_{f}\nu
a\big{(}\tfrac{2}{3}B_{1}\boldsymbol{n}-\dot{\boldsymbol{x}}\big{)}\,.$ (3)

Here the superscript denotes the Stokes approximation, and $\varrho_{f}$ is
the mass density of the fluid. Following Candelier et al. (2019), we use a
prime to indicate that this is the hydrodynamic force on the squirmer, due to
the disturbance it creates.

Plankton tends to be slightly heavier than the fluid. Therefore we allow the
squirmer to settle subject to the buoyancy force

$\displaystyle\boldsymbol{F}_{g}=\tfrac{4\pi}{3}a^{3}(\varrho_{s}-\varrho_{f})\boldsymbol{g}\,,$
(4)

where $\varrho_{s}$ is the mass density of the squirmer, and $\boldsymbol{g}$
is the gravitational acceleration. In the overdamped limit, the steady centre-
of-mass velocity of the squirmer is determined by the zero-force condition
${\boldsymbol{F}^{\prime}}^{(0)}+\boldsymbol{F}_{g}=\boldsymbol{0}$. This
yields $\dot{\boldsymbol{x}}
=\tfrac{2}{3}B_{1}\boldsymbol{n}+\tfrac{2}{9}\big{(}\tfrac{\varrho_{s}}{\varrho_{f}}-1\big{)}\tfrac{a^{2}}{\nu}\boldsymbol{g}\equiv\boldsymbol{v}_{s}^{(0)}+\boldsymbol{v}_{g}^{(0)}$.
Again, the superscript denotes the Stokes limit. In this limit, the squirmer
experiences no torque in a fluid at rest,
${\boldsymbol{T}^{\prime}}^{(0)}=\boldsymbol{0}$.

##  3 Inertial torque

Assume that the squirmer swims with swimming velocity $\boldsymbol{v}_{s}$ and
settles with settling velocity $\boldsymbol{v}_{g}$. The angle between
$\boldsymbol{v}_{s}$ and $\boldsymbol{v}_{g}$ is denoted by $\alpha$, as shown
in Fig. 1(a). Symmetry dictates the form of the inertial torque
$\boldsymbol{T}^{\prime}$. It has the units mass $\times$ velocity2. Since the
torque is an axial vector, it must be proportional to the vector product
between the two velocities. The torque can therefore be written as

${\boldsymbol{T}^{\prime}}^{(1)}=Cm_{f}\,(\boldsymbol{v}_{s}\wedge\boldsymbol{v}_{g})\,,$
(5)

where $m_{f}=\tfrac{4\pi}{3}a^{3}\varrho_{f}$ is the equivalent fluid mass,
$C$ is a non-dimensional constant, and the superscript indicates that this is
the first inertial correction to the torque. Eq. (5) says that torque vanishes
when the swimmer swims against gravity [$\alpha=\pi$ in Fig. 1(a)], and when
it swims in the direction of gravity ($\alpha=0$). Bifurcation theory implies
that one of these fixed points is stable, the other one unstable. The sign of
the coefficient $C$ determines which of the two is the stable fixed point.

Inertial torques can be understood as a consequence of advection of fluid
momentum. In the frame translating with the squirmer, far-field momentum is
advected by the transverse disturbance flow generated by the squirmer. At non-
zero ${\rm Re}_{p}$, the head of the squirmer – the north pole of the axial
velocity field (2) – experiences more drag than its rear, because some of the
momentum imparted to the fluid by the head is advected to the trailing end, in
the direction transverse to gravity. So when $\boldsymbol{v}_{s}$ is not co-
linear with $\boldsymbol{v}_{g}$ there is an inertial torque which rotates the
swimmer so that $\boldsymbol{v}_{s}$ becomes closer to anti-parallel with
$\boldsymbol{v}_{g}$. Comparing with Eq. (5), this means that the coefficient
$C$ must be negative. Note that the mechanism described above is the same that
creates Khayat-Cox torques on non-spherical passive particles sedimenting in
quiescent fluid. For a fibre, for example, the far-field momentum is advected
by the transverse flow along the fibre, leading to a torque that aligns the
fibre perpendicular to gravity (Khayat & Cox, 1989).

##  4 Perturbation theory for the coefficient $C$

The inertial torque is computed from

$\displaystyle{{\boldsymbol{T}}^{\prime}}^{(1)}=\int_{\mathscr{S}}\\!\\!\boldsymbol{r}\wedge(\bbsigma^{(1)}{\rm
d}\boldsymbol{s})\,,$ (6)

where $\sigma_{mn}^{(1)} =-p^{(1)}\delta_{mn}+2\mu S_{mn}^{(1)}$ are the
elements of the stress tensor $\bbsigma^{(1)}$ with pressure $p^{(1)}$,
$S_{mn}^{(1)}$ are the elements of the strain-rate tensor of the disturbance
flow, and $\mu=\varrho_{f}\nu$ is the dynamic viscosity. The integral goes
over the particle surface $\mathscr{S}$, $\boldsymbol{r}$ is the vector from
the particle centre to a point on the particle surface, and ${\rm
d}\boldsymbol{s}$ is the outward surface normal at this point. In the Stokes
approximation the torque vanishes,
${{\boldsymbol{T}}^{\prime}}^{(0)}=\boldsymbol{0}$, as mentioned above.

The disturbance stress tensor is determined by solving the steady Navier-
Stokes equations for the incompressible disturbance flow $\boldsymbol{w}$,

$-{\rm Re}_{p}\dot{\boldsymbol{x}}\cdot\boldsymbol{\nabla}\boldsymbol{w}+{\rm
Re}_{\rm{p}}\>\boldsymbol{w}\cdot\boldsymbol{\nabla}\boldsymbol{w}=-\boldsymbol{\nabla}p+\boldsymbol{\triangle}\boldsymbol{w}\,,$
(7)

with boundary conditions
$\boldsymbol{w}=\dot{\boldsymbol{x}}+(B_{1}\sin\theta+B_{2}\sin\theta\cos\theta)\hat{\bf
e}_{\theta}$ for $|\boldsymbol{r}|=1$, and $\boldsymbol{w}\to\boldsymbol{0}$
as $|\boldsymbol{r}|\to\infty$. Here we assumed that the squirmer has no
angular velocity. We non-dimensionalised Eq. (7) using the radius $a$ of the
squirmer as a length scale, and with the velocity scale $u_{\rm
c}=v_{g}^{(0)}$. Forces are non-dimensionalised by $\mu au_{\rm c}$, and
torques by $\mu a^{2}u_{\rm c}$. The acceleration terms on the l.h.s. of Eq.
(7) are singular perturbations of the r.h.s., the Stokes part (Hinch, 1995).
We use matched asymptotic expansions in ${\rm Re}_{p}$ to determine the
solution for small Rep. Near the squirmer, one expands

${\boldsymbol{w}}_{\mbox{\scriptsize in}}={\boldsymbol{w}}_{\mbox{\scriptsize
in}}^{(0)}+{\rm Re}_{p}\>{\boldsymbol{w}}_{\mbox{\scriptsize
in}}^{(1)}+\ldots\quad\mbox{and}\quad{p}_{\mbox{\scriptsize
in}}={p}_{\mbox{\scriptsize in}}^{(0)}+{\rm Re}_{p}\>{p}_{\mbox{\scriptsize
in}}^{(1)}+\ldots\>.$ (8)

This inner expansion is matched, term by term, to an outer expansion

$\hat{{\boldsymbol{w}}}_{\mbox{\scriptsize
out}}=\hat{\mathcal{T}}_{\mbox{\scriptsize reg}}^{(0)}+{\rm
Re}_{p}(\hat{\mathcal{T}}_{\mbox{\scriptsize
reg}}^{(1)}+\hat{\mathcal{T}}_{\mbox{\scriptsize sing}}^{(1)})+\ldots\>.$ (9)

Here $\hat{\mathcal{T}}_{\mbox{\scriptsize reg}}^{(0,1)}$ are regular terms in
the outer expansion, while $\hat{\mathcal{T}}_{\mbox{\scriptsize sing}}^{(1)}$
is singular in $\boldsymbol{k}$-space, proportional to
$\delta(\boldsymbol{k})$ (Meibohm et al., 2016). The outer solution is
obtained by replacing the boundary condition on the surface of the squirmer by
a singular source term in Eq. (7), a Dirac $\delta$-function (Schwartz, 1966)
with amplitude
$\boldsymbol{F}^{(0)}=-6\pi(\tfrac{2}{3}B_{1}\boldsymbol{n}-\dot{\boldsymbol{x}})$.
Since the non-linear term (quadratic in $\boldsymbol{w}$) is negligible far
from the particle, the resulting equation can be solved by Fourier transform,
yielding explicit expressions for $\hat{\mathcal{T}}_{\mbox{\scriptsize
reg}}^{(0,1)}$ and $\hat{\mathcal{T}}_{\mbox{\scriptsize sing}}^{(1)}$ which
serve as boundary conditions for the inner problems.

The inner problem to order ${\rm Re}_{p}^{0}$ is the homogeneous Stokes
problem

$\displaystyle-\boldsymbol{\nabla}{p}_{\mbox{\scriptsize
in}}^{(0)}+\boldsymbol{\triangle}{\boldsymbol{w}}_{\mbox{\scriptsize
in}}^{(0)}=\boldsymbol{0}\>,\>\>\boldsymbol{\nabla}\cdot{\boldsymbol{w}}_{\mbox{\scriptsize
in}}^{(0)}=\boldsymbol{0}\,,$ (10a) with boundary conditions
$\displaystyle{\boldsymbol{w}}_{\mbox{\scriptsize
in}}^{(0)}=\dot{\boldsymbol{x}}\\!+\\!(B_{1}\sin\theta\\!+\\!B_{2}\sin\theta\cos\theta)\hat{\bf
e}_{\theta}\>\>\mbox{for}\>\>{|\boldsymbol{r}|=1}\,,\>\>{\boldsymbol{w}}_{\mbox{\scriptsize
in}}^{(0)}\sim\boldsymbol{\mathcal{T}}_{\mathrm{reg}}^{(0)}\>\>\mbox{as}\>\>|\boldsymbol{r}|\to\infty\>.$
(10b)

This problem is solved in the standard fashion using Lamb’s solution (Happel &
Brenner, 1965). The ${\rm Re}_{p}^{1}$-order inner problem is inhomogeneous:

$\displaystyle-\boldsymbol{\nabla}{p}_{\mbox{\scriptsize
in}}^{(1)}+\boldsymbol{\triangle}{\boldsymbol{w}}_{\mbox{\scriptsize
in}}^{(1)}=-{\rm
Re}_{p}\dot{\boldsymbol{x}}\cdot\boldsymbol{\nabla}{\boldsymbol{w}}_{\mbox{\scriptsize
in}}^{(0)}+{\rm Re}_{p}{\boldsymbol{w}}_{\mbox{\scriptsize
in}}^{(0)}\cdot\boldsymbol{\nabla}{\boldsymbol{w}}_{\mbox{\scriptsize
in}}^{(0)}\,,\quad\boldsymbol{\nabla}\cdot{\boldsymbol{w}}_{\mbox{\scriptsize
in}}^{(1)}=0\,,$ (11a) $\displaystyle{\boldsymbol{w}}_{\mbox{\scriptsize
in}}^{(1)}=\boldsymbol{0}\>\>\mbox{for}\>\>{|\boldsymbol{r}|=1}\quad\mbox{and}\quad\quad{\boldsymbol{w}}_{\mbox{\scriptsize
in}}^{(1)}\sim\boldsymbol{\mathcal{T}}_{\mathrm{reg}}^{(1)}+\boldsymbol{\mathcal{T}}_{\mathrm{sing}}^{(1)}\>\>\mbox{for}\>\>|\boldsymbol{r}|\to\infty\>.$
(11b)

To solve Eqs. (11), we make the ansatz ${\boldsymbol{w}}_{\mbox{\scriptsize
in}}^{(1)}=(\boldsymbol{w}_{\mbox{\scriptsize
p}}+\boldsymbol{\mathcal{T}}_{\mathrm{sing}}^{(1)})+\boldsymbol{w}_{\mbox{\scriptsize
h}}$, where $\boldsymbol{w}_{\mbox{\scriptsize p}}$ is a particular solution,
and $\boldsymbol{w}_{\mbox{\scriptsize h}}$ is the homogeneous solution of
Eqs. (11). For the pressure we write $p_{\rm in}^{(1)}=p^{(1)}_{\rm
p}+p^{(1)}_{\rm h}$. We first determine the particular solution
$\boldsymbol{w}_{\mbox{\scriptsize p}}$ and $p^{(1)}_{\rm p}$ using Fourier
transform. Then $\boldsymbol{w}_{\mbox{\scriptsize h}}$ and $p^{(1)}_{\rm h}$
are determined using Lamb’s solution. The boundary condition for
$\boldsymbol{w}_{\mbox{\scriptsize h}}$ is $\boldsymbol{w}_{\mbox{\scriptsize
h}}=-\boldsymbol{w}_{\mbox{\scriptsize
p}}-\boldsymbol{\mathcal{T}}_{\mathrm{sing}}^{(1)}$ on the particle surface.
Having obtained ${\boldsymbol{w}}_{\mbox{\scriptsize in}}^{(1)}$, we compute
the torque from Eq. (6). The torque comes from the particular solution of the
first-order inner problem. For a passive spherical particle, spherical
symmetry ensures that the particular solution does not contribute to the
torque. Swimming breaks spherical symmetry, and this is the reason that torque
does not vanish. Given $p^{(1)}$ and $\boldsymbol{w}_{\rm in}^{(1)}$, we can
determine the inertial correction to the stress tensor, $\bbsigma^{(1)}$.
Integrating it over the particle surface as specified in Eq. (6), we find to
leading order in ${\rm Re}_{p}$

$\displaystyle{\boldsymbol{T}^{\prime}}^{(1)}=-\tfrac{3\pi}{2}{\rm
Re}_{p}(\boldsymbol{v}_{s}^{(0)}\wedge\boldsymbol{v}_{g}^{(0)})\,.$ (12)

In dimensional units, this corresponds to
${\boldsymbol{T}^{\prime}}^{(1)}=-{\tfrac{9}{8}}m_{f}(\boldsymbol{v}_{s}^{(0)}\wedge\boldsymbol{v}_{g}^{(0)})$.
The coefficient $C=-{\tfrac{9}{8}}$ is negative, as predicted by the argument
summarised in Section 3. So a spherical organism swimming downwards
experiences a torque that tends to turn it upwards, causing the organism to
swim against gravity.

##  5 Direct numerical simulations

\begin{overpic}[width=341.43306pt,clip]{Fig2.pdf} \end{overpic} Figure 2: (a)
Non-dimensional inertial torque
${\boldsymbol{T}^{\prime}}^{(1)}={T^{\prime}_{3}}^{(1)}\hat{\bf e}_{3}$ on a
spherical squirmer. Shown is the theory, Eq. (12) (solid line), in comparison
with direct numerical-simulation results (Section 5, $\circ$). The torque was
non-dimensionalised with the factor $\mu a^{2}u_{\rm c}$, where $u_{\rm
c}=v_{g}^{(0)}$. The angle $\alpha$ is defined in Fig. 1(a). Parameters:
$B_{1}=1$, $B_{2}=0$, $v_{s}^{(0)}=2/3$, $v_{g}^{(0)}=1$, and Re${}_{p}=1$.
(b) Non-dimensional torque for $\alpha=90$ as a function of ${\rm Re}_{p}$,
other parameters same as in panel (a). Also shown is the theory, Eq. (12). (c)
Non-dimensional torque for $\alpha=90$ as a function of $\beta=B_{2}/B_{1}$,
other parameters the same as in panel (a). (d) Non-dimensional swimming speed
($\diamond$) and settling speed ($\Box$) from direct numerical simulations for
the same parameters as in panel (a). Also shown are the Stokes estimates for
$v_{s}^{(0)}$ (dashed line) and $v_{g}^{(0)}$ (solid line).

We solved the three-dimensional Navier-Stokes equations for the incompressible
flow using an immersed-boundary method (Peskin, 2002). The interaction between
squirmer and fluid was implemented by the direct-force method (Uhlmann, 2005):
in order to satisfy the boundary condition (2), the algorithm calculates the
predicted fluid velocity on the surface of the squirmer. Based on the mismatch
between the predicted velocity and Eq. (2), an appropriate immersed-boundary
force is applied to the fluid phase to maintain the boundary conditions (2) on
the surface of the squirmer. We implemented the improved algorithm described
in (Kempe & Fröhlich, 2012; Breugem, 2012; Lambert et al., 2013), because it
is more precise for nearly-neutrally buoyant particles. We used a cubic
computational domain of side length $L=20a$ with periodic boundary conditions.
This is large enough to account for convective inertia for Re${}_{p}>1$. The
computational domain was discretised using a cubic mesh with resolution
$\Delta x$. The Navier-Stokes equations were integrated using a second-order
Crank-Nicholson scheme for the time-integration (Kim et al., 2002) with time
step $\Delta t$, while the motion of the squirmer was integrated using a
second-order Adams-Bashforth method (Hairer et al., 2000).

The numerical simulation of solid-body motion in a fluid is challenging at
small Rep. The mesh resolution $\Delta x$ must be fine enough to resolve the
shape of the body, so that the viscous stresses near its surface are
accurately represented (Andersson & Jiang, 2019). In addition, the time step
$\Delta t$ must be small enough to resolve the viscous diffusion of the
disturbance, $\Delta t<\Delta x^{2}/\nu$. In Appendix A we briefly describe
our convergence checks. We found that our algorithm fails to converge for Rep
smaller than unity. In the following we discuss our numerical results for
Re${}_{p}\geq 1$.

To determine the torque, we froze the orientation of the squirmer at a given
angle $\alpha$, but allowed the squirmer to translate. It was initially at
rest. We measured the centre-of-mass velocity and the torque after the
transient, when the disturbance flow was fully established. Figure 2(a) shows
the numerical results for the inertial torque on a spherical squirmer for
Re${}_{p}=1$, in comparison with the theory (12). The remaining parameter
values used in the simulations are quoted in the Figure caption. Although the
theory is valid for Re${}_{p}\ll 1$, it nevertheless agrees qualitatively with
the numerical results for Re${}_{p}=1$. This is encouraging, because it allows
as to draw qualitative conclusions about the effects of the torque on small
plankton (Section 6). Panel (a) shows that the theory overestimates the
amplitude of the torque by a factor of two, but that its angular dependence is
roughly the same. We note, however, that the numerical results exhibit an
asymmetry in their dependence on $\alpha$. Since the small-Rep theory yields a
symmetric angular dependence of the torque, we attribute the asymmetry to
higher-order Rep-corrections. Panel (b) confirms that the difference between
theory and numerical simulation increases for larger Rep, as expected.

The small-Rep theory (12) says that the torque is independent of $\beta$.
Panel (c) shows that this is not the case for the numerical results at
Re${}_{p}=1$. This suggests, again, that higher-Rep corrections matter at
Re${}_{p}=1$, and that they exhibit a $\beta$-dependence. Another indication
that higher-order ${\rm Re}_{p}$-corrections may be important comes from
measuring settling and swimming speeds in the numerical simulations. We
extracted the swimming speed using $\dot{\boldsymbol{x}}
=v_{s}\boldsymbol{n}-v_{g}\hat{\bf e}_{2}$. Solving for $v_{s}$ gives
$v_{s}=\dot{\boldsymbol{x}}\cdot\hat{\bf e}_{1}/(\boldsymbol{n}\cdot\hat{\bf
e}_{1})$. Fig. 2(d) compares the measured swimming and settling speeds. The
settling speed is substantially smaller than the Stokes estimate, consistent
with a significant ${\rm Re}_{p}$-correction. The swimming speed is much
closer to the Stokes estimate. This is because the data shown is for
$\beta=0$, and the known ${\rm Re}_{p}$-corrections to the swimming speed
(Khair & Chisholm, 2014),

$\boldsymbol{v}_{s}=\tfrac{2}{3}B_{1}\boldsymbol{n}[1-\tfrac{3\beta}{20}{\rm
Re}_{p}+(\tfrac{\beta}{8}+\tfrac{11987}{470400}\beta^{2}){\rm
Re}_{p}^{2}+\ldots]\,,$ (13)

vanish for $\beta=0$.

##  6 Conclusions

We showed that a spherical squirmer settling in a fluid at rest experiences an
inertial torque, and computed the torque using matched asymptotic expansions.
The calculation is similar to that of Cox (1965) for the inertial torque on a
nearly spherical, passive particle settling in a quiescent fluid. This torque
vanishes for a passive sphere, a consequence of spherical symmetry. A
spherical swimmer experiences an inertial torque because swimming breaks this
symmetry. The torque causes the squirmer to align with gravity so that it
swims upwards. In other words, this torque acts just like Kessler’s gyrotactic
torque for bottom-heavy organisms.

For plankton, the effect of the inertial torque is much smaller than the
gyrotactic torque, at least for spherical shapes. We can see this by comparing
the corresponding reorientation times. This time scale is defined as
$\tau_{I}=\tfrac{1}{2}(8\pi\mu a^{3})/T_{\rm max}$, where $8\pi\mu a^{3}$ is
the rotational resistance coefficient for a sphere (Kim & Karrila, 2013), and
$T_{\rm max}$ is the maximal magnitude of the torque. For the inertial torque,
one obtains $\tau_{I}={8}\nu/({3}v_{s}^{(0)}v_{g}^{(0)})$ (this and all
following expressions are quoted in dimensional units). The reorientation time
for the gyrotactic torque is $\tau_{G}=3\varrho_{s}\nu/(\varrho_{f}gh)$
(Pedley & Kessler, 1987), where $h$ is the offset between the centre-of-mass
and the geometrical centre of the squirmer, and $g=|\boldsymbol{g}|$. The
ratio of these time scales is $\tau_{I}/\tau_{G}\sim
gh/v_{s}^{(0)}v_{g}^{(0)}$, assuming $\varrho_{s}\approx\varrho_{f}$. Taking
$h\sim 10^{-7}$ m [Table 1 in (Kessler, 1986)], we see that swimming and
settling speeds need to be of the order mm/s for the reorientation times to be
comparable. For small plankton, typical speeds tend to be much smaller
(Kessler, 1986). For larger organisms, however, the inertial torque can be
significant. With typical values for a small copepod (Titelman & Kiørboe,
2003), $v_{s}=1\,\rm{mm/s}$, $v_{g}=0.2\,\rm{mm/s}$, as well as
$\nu=10^{-6}$m2/s, one finds an inertial reorientation time of the order of
$\tau_{I}\sim 10\,$s. Kolmogorov times for ocean turbulence range from
$\tau_{\rm K}=\sqrt{\nu/\mathscr{E}}=100$ s for dissipation rate per unit mass
$\mathscr{E}=10^{-6}$ cm2/s3 to $\tau_{\rm K}=1$ s for $\mathscr{E}=10^{-2}$
cm2/s3. So the non-dimensional reorientation parameter
$\Psi=\tau_{I}/\tau_{\rm K}$ (Durham et al., 2013) ranges from $0.1$ for weak
turbulence to $10$ for strong turbulence. The Reynolds number is of order
Re${}_{p}\sim 1$ for speeds of the order of $1$ mm. This means that the Rep-
perturbation theory does not strictly apply, but we can nevertheless conclude
that for weak turbulence, the inertial torque can have a significant effect on
the angular dynamics of the organism.

Some motile microorganisms are non-spherical (Berland et al., 1995; Faust &
Gulledge, 2002; Smayda, 2010). It has been suggested that this can give
additional contributions to the torque (Qiu et al., 2022). Since the boundary
conditions are different, and since swimming breaks fore-aft symmetry, these
additional contributions may be different from the Khayat-Cox torque for
passive particles. However, we expect that the torque is still determined by
the same physical mechanism, advection of fluid-momentum transverse to
gravity. This may give rise to terms proportional to $\sin(2\alpha)$, whereas
the torque is proportional to $\sin(\alpha$) for the spherical squirmer. To
make these speculations definite, one could compute the inertial torque for a
nearly spherical squirmer in perturbation theory. A second open question is to
determine the inertial torque for bottom-heavy, non-spherical organisms, the
analogue of the inertial torque on passive particles with mass-density
asymmetries (Roy et al., 2019).

More generally, although the small-Rep perturbation theory may become
quantitatively inaccurate for Reynolds numbers of order unity – where the
torque begins to make a significant difference – the results tell us which
non-dimensional parameters matter, and how to reason about the effect of
boundary conditions, and the symmetries of the problem. The calculation
illustrates the conceptual insight that the inertial torque comes from fluid
motion transverse to the direction of gravity. Fluid momentum in this
direction is advected along the swimmer by the transverse fluid velocity,
resulting in a torque. In our case, the boundary conditions are different from
those for passive particle, and so is the symmetry of the problem, because
swimming breaks fore-aft symmetry. Nevertheless, the fundamental mechanism
generating the torque is the same.

Acknowledgements. BM was supported by Vetenskapsrådet (grant no. 2021-4452)
and by the Knut-and-Alice Wallenberg Foundation (grant no. 2019.0079). LZ was
supported by the National Natural Science Foundation of China (grant nos.
11911530141 and 91752205). This research was also supported in part by a
collaboration grant from the joint China-Sweden mobility programme [National
Natural Science Foundation of China (NSFC)-Swedish Foundation for
International Cooperation in Research and Higher Education (STINT)], grant
nos. 11911530141 (NSFC) and CH2018-7737 (STINT).

##  Appendix A Details regarding the direct numerical simulations

\begin{overpic}[width=256.0748pt,clip]{Fig3.pdf} \end{overpic} Figure 3:
Convergence tests changing mesh resolution $\Delta x$ (left column) and
changing integration-time step $\Delta t$ (right column). Settling speed of a
passive sphere (first row), compared with the experimental data of Maxworthy,
extracted from Fig. 4 of (Vesey II & Goldenfeld, 2007). Swimming speed of
neutrally buoyant spherical squirmer with $\beta=0$, compared with Eq. (13),
(Khair & Chisholm, 2014), second row. The third row shows the inertial torque.

To find out the required mesh and time resolution, we considered two test
cases at $\rm{Re}_{p}=1$: a passive sphere settling under gravity, and a
neutrally buoyant squirmer with $\beta=0$. To check convergence as the mesh
resolution increases, we changed $2a/\Delta x$ ranging from 16 to 48, keeping
$\nu\Delta t/\Delta x^{2}=0.58$ constant. Settling and swimming speeds reached
a plateau when we increased the mesh resolution $2a/\Delta x$. The settling
speed varied about 0.1% and swimming speed varied about 1% when we changed
$2a/\Delta x$ from 36 to 48.

We then checked for convergence as the step size $\Delta t$ was reduced, for
fixed $2a/\Delta x=36$. Again, both settling and swimming speeds reached
plateaus as $\nu\Delta t/\Delta x^{2}$ decreased. When we halved $\nu\Delta
t/\Delta x^{2}$ from 0.58 to 0.29, the settling and swimming speeds varied
about 1% and 2%, respectively. Therefore, we used $2a/\Delta x=36$ and $\Delta
t=0.58\Delta x^{2}/\nu$ for our numerical simulations at $\rm{Re}_{p}=1.0$
that are shown in the main text. At these parameter values, the simulated
settling speed of a passively settling particle is about 3% larger than the
measurement of Maxworthy, taken from Fig. 4 of (Vesey II & Goldenfeld, 2007).
The swimming speed of the squirmer is about half a percent larger than the
theoretical value $v_{s}^{(0)}=2B_{1}/3$. The convergence for the inertial
torque was slightly worse. The torque varied about 0.5% when we changed
$2a/\Delta x$ from 36 to 48 for $\nu\Delta t/\Delta x^{2}=0.58$, and it varied
about 4% when we halved $\nu\Delta t/\Delta x^{2}$ from 0.58 to 0.29 for
$2a/\Delta x=36$. We also performed convergence checks for Re${}_{p}=0.32$,
reducing the particle Reynolds number by increasing $\nu$. We found that the
values of $\Delta x$ and $\Delta t$ quoted above are too large for the
numerical scheme to converge at this Reynolds number. Therefore we only show
results for Re${}_{p}\geq 1$ in the main text.

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can apply the dominated convergence theorem to the expectation $\mathbf{E}$ in
(5.48). For this purpose, we remark that since $f\in D(H)$, both $\|f\|_{2}$
and $\|f^{\prime}\|_{2}$ are finite; hence $f$ is bounded (see, e.g., [2, Fact
3.1] and [9, Lemma 3.1]). If we combine this with the facts that
$Q_{i,j}(x)\leq\eta(x)$ with
$\eta(x)=\kappa\big{(}\log(1+|x|)\big{)}^{\mathfrak{b}}+\nu$ for some
$\kappa,\nu>0$ in Cases 1 and 2 (Assumption 2.8-(3)), that there exists some
$c>0$ such that $S(i,x)\geq-c$ for all $1\leq i\leq r$ (Assumption 2.8-(1)),
then we can bound

(5.49) $\sup_{t\in(0,1)}\big{|}\mathfrak{Y}_{i,j}(t;x)\big{|}\\\
\leq\begin{cases}\eta\left(\sup_{t\in(0,1)}|B^{x}(t)|\right)\mathrm{e}^{c}\|f\|_{\infty}&\text{Case
1,}\vspace{5pt}\\\
\eta\left(\sup_{t\in(0,1)}X^{x}(t)\right)\mathrm{e}^{c+\overline{\alpha}_{\mathrm{m}}\mathfrak{L}^{0}_{1}(X^{x})}\|f\|_{\infty}&\text{Case
2,}\vspace{5pt}\\\ \left(\sup_{1\leq i,j\leq
r~{}y\in(0,b)}|Q_{i,j}(y)|\right)\mathrm{e}^{c+\overline{\alpha}_{\mathrm{m}}\mathfrak{L}^{0}_{1}(Y^{x})+\overline{\beta}_{\mathrm{m}}\mathfrak{L}^{b}_{1}(Y^{x})}\|f\|_{\infty}&\text{Case
3.}\end{cases}$

By (5.24) and the fact that Brownian motion suprema have Gaussian tails, this
has a finite expectation for any $x$; thus (5.46) follows by dominated
convergence.

We now conclude the proof of (5.6) by showing that (5.46) can be improved to
convergence in $L^{2}(\mathcal{A},\mathbb{F})$. For this purpose, we once
again use the Vitali convergence theorem. That is, we must prove for every
$\varepsilon>0$, there exists a $\delta>0$ such that

(5.50)
$\displaystyle\sup_{t\in(0,1)}\int_{\mathcal{K}}\big{|}\mathfrak{G}_{1}\big{(}t;(i,x)\big{)}\big{|}^{2}~{}\mathrm{d}\mu(a)<\varepsilon\qquad\text{for
all }\mathcal{K}\subset\mathcal{A}\text{ with }\mu(\mathcal{K})<\delta,$

and there also exists a $\mathcal{K}\subset\mathcal{A}$ with
$\mu(\mathcal{K})<\infty$ and

(5.51)
$\displaystyle\sup_{t\in(0,1)}\int_{\mathcal{A}\setminus\mathcal{K}}\big{|}\mathfrak{G}_{1}\big{(}t;(i,x)\big{)}\big{|}^{2}~{}\mathrm{d}\mu(a)<\varepsilon.$

For this purpose, we note that by Jensen’s inequality,

$\big{|}\mathfrak{G}_{1}\big{(}t;(i,x)\big{)}\big{|}^{2}\leq(r-1)\sum_{j\neq
i}\mathbf{E}\big{[}\big{|}\mathfrak{Y}_{i,j}(t;x)\big{|}\big{]}^{2};$

then the same argument used to obtain (5.49) (except that we do not bound $f$
by its infinity norm) yields that for every $t\in(0,1)$,

(5.52)
$\big{|}\mathfrak{G}_{1}\big{(}t;(i,x)\big{)}\big{|}^{2}\leq(r-1)\mathrm{e}^{2c}\sum_{j\neq
i}\cdots\\\
\begin{cases}\mathbf{E}\left[\eta\left(\sup_{t\in(0,1)}|B^{x}(t)|\right)\big{|}f\big{(}j,B^{x}(t)\big{)}\big{|}\right]^{2}&\text{Case
1,}\vspace{5pt}\\\
\mathbf{E}\left[\eta\left(\sup_{t\in(0,1)}X^{x}(t)\right)\mathrm{e}^{\overline{\alpha}_{\mathrm{m}}\mathfrak{L}^{0}_{1}(X^{x})}\big{|}f\big{(}j,X^{x}(t)\big{)}\big{|}\right]^{2}&\text{Case
2,}\vspace{5pt}\\\ \left(\sup_{1\leq i,j\leq
r~{}y\in(0,b)}|Q_{i,j}(y)|\right)^{2}\mathbf{E}\left[\mathrm{e}^{\overline{\alpha}_{\mathrm{m}}\mathfrak{L}^{0}_{1}(Y^{x})+\overline{\beta}_{\mathrm{m}}\mathfrak{L}^{b}_{1}(Y^{x})}\big{|}f\big{(}j,Z^{x}(t)\big{)}\big{|}\right]^{2}&\text{Case
3.}\end{cases}$

By Hölder’s inequality, the second line of (5.52) is bounded above by

$\begin{cases}\mathbf{E}\left[\eta\left(\sup_{t\in(0,1)}|B^{x}(t)|\right)^{2}\right]\mathbf{E}\big{[}\big{|}f\big{(}j,B^{x}(t)\big{)}\big{|}^{2}\big{]}&\text{Case
1,}\vspace{5pt}\\\
\mathbf{E}\left[\eta\left(\sup_{t\in(0,1)}X^{x}(t)\right)^{2p}\right]^{1/p}\mathbf{E}\left[\mathrm{e}^{2q\overline{\alpha}_{\mathrm{m}}\mathfrak{L}^{0}_{1}(X^{x})}\right]^{1/q}\mathbf{E}\big{[}\big{|}f\big{(}j,X^{x}(t)\big{)}\big{|}^{2}\big{]}&\text{Case
2,}\vspace{5pt}\\\
C\mathbf{E}\left[\mathrm{e}^{2\overline{\alpha}_{\mathrm{m}}\mathfrak{L}^{0}_{1}(Y^{x})+2\overline{\beta}_{\mathrm{m}}\mathfrak{L}^{b}_{1}(Y^{x})}\right]\mathbf{E}\big{[}\big{|}f\big{(}j,Z^{x}(t)\big{)}\big{|}^{2}\big{]}&\text{Case
3,}\end{cases}$

for any $p,q>1$ such that $1/p+1/q=1$, where $C$ is a constant independent of
$t$.

By (5.24), this means that in Case 3 the uniform integrability follows from
(5.41) and (5.42). We now deal with Cases 1 and 2: Recall that
$\eta(x)=\kappa\big{(}\log(1+|x|)\big{)}^{\mathfrak{b}}+\nu$. If we combine
the coupling (5.12) with the triangle and Jensen’s inequalities, and
$\log(1+|x|+|y|)^{\mathfrak{b}}\leq
C\log(1+|x|)^{\mathfrak{b}}+C|y|^{\mathfrak{b}}$ for some constant $C>0$, we
can find some $\overline{\kappa},\overline{\nu}>0$ large enough so that

$\textstyle\mathbf{E}\left[\eta\left(\sup_{t\in(0,1)}|B^{x}(t)|\right)^{2}\right],\mathbf{E}\left[\eta\left(\sup_{t\in(0,1)}X^{x}(t)\right)^{2p}\right]^{1/p}\leq\overline{\kappa}\big{(}\log(1+|x|)\big{)}^{2\mathfrak{b}}+\overline{\nu}.$

Here, we used the fact that

$\textstyle\mathbf{E}\left[\sup_{t\in(0,1)}|W^{0}(s)|^{\mathfrak{c}}\right]<\infty\qquad\text{for
all }\mathfrak{c}>0,$

and thus this expectation can be absorbed in the constant $\overline{\nu}$. By
(5.24),

$\mathbf{E}\left[\mathrm{e}^{2q\overline{\alpha}_{\mathrm{m}}\mathfrak{L}^{0}_{1}(X^{x})}\right]^{1/q}<\infty.$

Thus, by (5.41) and (5.42), the uniform integrability in (5.50) and (5.51) in
Cases 1 and 2 follows from this: For every $\varepsilon>0$, there is some
$\delta>0$ such that

(5.53)
$\displaystyle\sup_{t\in(0,1)}\int_{\mathcal{K}}\big{(}\log(1+|x|)\big{)}^{2\mathfrak{b}}\mathbf{E}\left[\big{|}f\big{(}i,Z^{x}(t)\big{)}\big{|}^{2}\right]~{}\mathrm{d}x<\varepsilon\qquad\text{for
all }\mathcal{K}\subset I\text{ with }|\mathcal{K}|<\delta,$

and there also exists a $\mathcal{K}\subset I$ with $|\mathcal{K}|<\infty$ and

(5.54)
$\displaystyle\sup_{t\in(0,1)}\int_{I\setminus\mathcal{K}}\big{(}\log(1+|x|)\big{)}^{2\mathfrak{b}}\mathbf{E}\left[\big{|}f\big{(}i,Z^{x}(t)\big{)}\big{|}^{2}\right]~{}\mathrm{d}x<\varepsilon.$

Toward this end, we note that for any set $\mathcal{K}\subset I$, one has

$\displaystyle\int_{\mathcal{K}}\big{(}\log(1+|x|)\big{)}^{2\mathfrak{b}}\mathbf{E}\left[\big{|}f\big{(}i,Z^{x}(t)\big{)}\big{|}^{2}\right]~{}\mathrm{d}x$
$\displaystyle=\int_{I\times
I}\big{(}\log(1+|x|)\big{)}^{2\mathfrak{b}}\mathbf{1}_{\\{x\in\mathcal{K}\\}}\Pi_{Z}(t;x,y)|f(i,y)|^{2}~{}\mathrm{d}x\mathrm{d}y$
$\displaystyle=\int_{I}\mathbf{E}\left[\big{(}\log(1+|y+Z^{0}(t)|)\big{)}^{2\mathfrak{b}}\mathbf{1}_{\\{y+Z^{0}(t)\in\mathcal{K}\\}}\right]|f(i,y)|^{2}\mathrm{d}y$
$\displaystyle\leq
C\int_{I}\mathbf{E}\left[\big{(}\log(1+|y|\big{)}^{2\mathfrak{b}}+|Z^{0}(t)|^{2\mathfrak{b}}\big{)}\mathbf{1}_{\\{y+Z^{0}(t)\in\mathcal{K}\\}}\right]|f(i,y)|^{2}\mathrm{d}y$

for some $C>0$ independent of $t$. Then, an application of Hölder’s inequality
yields

(5.55)
$\int_{\mathcal{K}}\big{(}\log(1+|x|)\big{)}^{2\mathfrak{b}}\mathbf{E}\left[\big{|}f\big{(}i,Z^{x}(t)\big{)}\big{|}^{2}\right]~{}\mathrm{d}x\leq
C\int_{I}\mathbf{P}\big{[}Z^{y}(t)\in\mathcal{K}\big{]}\log(1+|y|\big{)}^{2\mathfrak{b}}|f(i,y)|^{2}\mathrm{d}y\\\
+C\int_{I}\mathbf{E}\left[|Z^{0}(t)|^{4\mathfrak{b}}\right]^{1/2}\mathbf{P}\big{[}Z^{y}(t)\in\mathcal{K}\big{]}^{1/2}|f(i,y)|^{2}\mathrm{d}y$

for some $C>0$ independent of $t$. Regarding the second line in (5.55), we
note that

$\sup_{t\in(0,1)}\mathbf{E}\left[|Z^{0}(t)|^{4\mathfrak{b}}\right]^{1/2}<\infty,$

and then, by Jensen’s inequality and the fact that
$\int|f(i,y)|^{2}~{}\mathrm{d}y<\infty$, there exists some $C>0$ independent
of $t$ such that

$\int_{I}\mathbf{P}\big{[}Z^{y}(t)\in\mathcal{K}\big{]}^{1/2}|f(i,y)|^{2}\mathrm{d}y\leq
C\left(\int_{I}\mathbf{P}\big{[}Z^{y}(t)\in\mathcal{K}\big{]}|f(i,y)|^{2}\mathrm{d}y\right)^{1/2}.$

Thus, (5.55) becomes

$\displaystyle\int_{\mathcal{K}}\big{(}\log(1+|x|)\big{)}^{2\mathfrak{b}}\mathbf{E}\left[\big{|}f\big{(}i,Z^{x}(t)\big{)}\big{|}^{2}\right]~{}\mathrm{d}x$
$\displaystyle\leq
C\int_{I}\mathbf{P}\big{[}Z^{y}(t)\in\mathcal{K}\big{]}\log(1+|y|\big{)}^{2\mathfrak{b}}|f(i,y)|^{2}\mathrm{d}y+C\left(\int_{I}\mathbf{P}\big{[}Z^{y}(t)\in\mathcal{K}\big{]}|f(i,y)|^{2}\mathrm{d}y\right)^{1/2}$
$\displaystyle=C\int_{\mathcal{K}}\mathbf{E}\left[\log(1+|Z^{x}(t)|\big{)}^{2\mathfrak{b}}\big{|}f\big{(}i,Z^{x}(t)\big{)}\big{|}^{2}\right]~{}\mathrm{d}x+C\left(\int_{\mathcal{K}}\mathbf{E}\left[\big{|}f\big{(}i,Z^{x}(t)\big{)}\big{|}^{2}\right]~{}\mathrm{d}x\right)^{1/2}$

for some $C>0$ independent of $t$. At this point, if we show that
$\log(1+|y|\big{)}^{2\mathfrak{b}}|f(i,y)|^{2}$ is both integrable and
bounded, then (5.53) and (5.54) follows from the same arguments used in
(5.43)–(5.45) (replace $|f(i,y)|^{2}$ therein by
$\log(1+|y|\big{)}^{2\mathfrak{b}}|f(i,y)|^{2}$).

Toward this end, since $f\in D(H)$ we know that $\|S_{+}^{1/2}f\|_{2}<\infty$
(see Proposition 4.3 and the definition of $D(\mathcal{E}_{i})$ therein).
Thus, by Assumption 2.8-(3), this implies that

(5.56)
$\displaystyle\int_{I}\log(1+|y|\big{)}^{\mathfrak{c}}|f(i,y)|^{2}~{}\mathrm{d}y<\infty\qquad\text{for
every }\mathfrak{c}>0;$

in particular for $\mathfrak{c}=2\mathfrak{b}$. Regarding boundedness, up to
expanding $f(i,y)$ into a sum of its real and imaginary parts, we may assume
without loss of generality that $f(i,y)$ is real-valued. Under this
assumption, by the fundamental theorem of calculus,

$\displaystyle\log(1+|x|\big{)}^{2\mathfrak{b}}f(i,x)^{2}\leq\left|\int_{x}^{\infty}\tfrac{\mathrm{d}}{\mathrm{d}y}\big{(}\log(1+|y|\big{)}^{2\mathfrak{b}}f(i,y)^{2}\big{)}~{}\mathrm{d}y\right|,\qquad
x\in I.$

Using the product rule and Jensen’s inequality, this implies

$\log(1+|x|\big{)}^{2\mathfrak{b}}f(i,x)^{2}\leq
2\int_{I}\big{(}\log(1+|y|)\big{)}^{2\mathfrak{b}}\big{|}f(i,y)f^{\prime}(i,y)\big{|}+\frac{\mathfrak{b}\log(1+|y|)^{2\mathfrak{b}-1}f(i,y)^{2}}{1+|y|}~{}\mathrm{d}y;$

thus it only remains to check that the above integral is finite. On the one
hand, since $\frac{1}{1+|y|}\leq 1$, we get that

$\int_{I}\frac{\mathfrak{b}\log(1+|y|)^{2\mathfrak{b}-1}f(i,y)^{2}}{1+|y|}~{}\mathrm{d}y<\infty$

by (5.56). On the other hand, if we use the inequality $|xy|\leq x^{2}+y^{2}$,
we get

$\int_{I}\big{(}\log(1+|y|)\big{)}^{2\mathfrak{b}}\big{|}f(i,y)f^{\prime}(i,y)\big{|}~{}\mathrm{d}y\leq\int_{I}\big{(}\log(1+|y|)\big{)}^{4\mathfrak{b}}f(i,y)^{2}+f^{\prime}(i,y)^{2}~{}\mathrm{d}y;$

this is finite by a combination of (5.56) and $\|f^{\prime}\|_{2}<\infty$.
With this in hand, the proof of (5.6) is now complete.

####  5.3.3. Proof of Lemma 5.2 Part 3

We now finish the proof of Lemma 5.2—and therefore of Theorem 2.10—by
establishing (5.7). Recalling the definition of $\mathfrak{G}_{2}$ in (5.4),
by Jensen’s and Hölder’s inequalities,

(5.57) $\big{|}\mathfrak{G}_{2}\big{(}t;(i,x)\big{)}\big{|}^{2}\leq
t^{-2}\mathbf{E}\left[|\mathfrak{n}_{t}(Q,A^{(i,x)})|^{4}\mathrm{e}^{-4\int_{0}^{t}S(A^{(i,x)}(s))~{}\mathrm{d}s-4\mathfrak{B}_{t}(A^{(i,x)})}\right]^{1/2}\\\
\cdot\mathbf{P}\big{[}N(t)\geq
2\big{]}^{1/2}\mathbf{E}\Big{[}\big{|}f\big{(}A^{(i,x)}(t)\big{)}\big{|}^{2}\mathbf{1}_{\\{N(t)\geq
2\\}}\Big{]}.$

Then, by (5.28) (but replacing $f$ by $|f|^{2}$), we get that

(5.58)
$\displaystyle\mathbf{E}\Big{[}\big{|}f\big{(}A^{(i,x)}(t)\big{)}\big{|}^{2}\mathbf{1}_{\\{N(t)\geq
2\\}}\Big{]}\leq r\,\mathbf{P}\big{[}N(t)\geq
2\big{]}\mathbf{E}\Big{[}\big{|}f\big{(}u,Z^{x}(t)\big{)}\big{|}^{2}\Big{]},$

where $u$ is uniform on $\\{1,\ldots,r\\}$ and independent of $Z$. If we now
combine Proposition 5.6 (in the case $p=4$) with (5.23), (5.24), and the fact
that $\mathrm{e}^{-t\kappa|x|^{\overline{\mathfrak{a}}}}\leq 1$, then we get

$\sup_{t\in(0,1),~{}(i,x)\in\mathcal{A}}\mathbf{E}\left[|\mathfrak{n}_{t}(Q,A^{(i,x)})|^{4}\mathrm{e}^{-4\int_{0}^{t}S(A^{(i,x)}(s))~{}\mathrm{d}s-4\mathfrak{B}_{t}(A^{(i,x)})}\right]^{1/2}<\infty.$

If we combine this with (5.57) and (5.58), given that $\mathbf{P}[N(t)\geq
2]=O(t^{2})$ as $t\to 0$, then we conclude that there exists a constant $C>0$
independent of $t$ such that

(5.59) $\displaystyle\|\mathfrak{G}_{2}(t;\cdot)\|_{2}^{2}\leq
Ct\int_{\mathcal{A}}\mathbf{E}\Big{[}\big{|}f\big{(}u,Z^{x}(t)\big{)}\big{|}^{2}\Big{]}~{}\mathrm{d}\mu(i,x)=0.$

We then obtain (5.7) from (5.32).

##  6\. Proof of Corollary 2.19

We proceed in two steps. First, we verify that the potential
$V=S+Q=(S_{0}+S_{1})+Q$ in Assumption 2.15 satisfies Assumption 2.8 almost
surely; this immediately implies that the conclusions of Proposition 2.9 and
Theorem 2.10 hold. Then, we verify that (2.17) holds by direct calculation.

###  6.1. Step 1. Verifying Assumption 2.8

In Case 3, the fact that $V$ satisfies Assumption 2.8 is trivial: In that
case, we only need to check items (1) and (2) in Assumption 2.8; item (2)
follows directly by the hypothesis that the $Q_{i,j}$ are continuous in
Assumption 2.15, and item (1) follows from the fact that $S_{0}$ is already
assumed to be bounded below and locally integrable, and since $S_{1}$ is
continuous, the same is true for $S=S_{0}+S_{1}$.

Consider then Cases 1 and 2. First, we note that by a Borel-Cantelli argument
(e.g., [9, Corollary B.2]), the fact that $S_{1}(i,\cdot)$ and $Q_{i,j}$ are
continuous and stationary implies that there exists a finite random variable
$\nu>0$ such that, almost surely,

$|S(i,x)|,|Q_{i,j}(x)|\leq\nu\sqrt{\log(2+|x|)},\qquad x\in I.$

Since $S_{0}(i,\cdot)$ is already assumed to grow faster than $\log$, this
immediately implies Assumption 2.8-(3) (with $\mathfrak{b}=1/2$).

###  6.2. Step 2. Moment Formula

By Theorem 2.10,

$\prod_{k=1}^{n}K(t_{k};a_{k},b_{k})=\prod_{k=1}^{n}\Pi_{A}(t_{k};a_{k},b_{k})\mathbf{E}_{A}\left[\mathfrak{n}_{t}(Q,A^{a_{k},b_{k}}_{t_{k}})\,\mathrm{e}^{-\int_{0}^{t}S(A^{a_{k},b_{k}}_{t_{k}}(s))~{}\mathrm{d}s+\mathfrak{B}_{t}(A^{a_{k},b_{k}}_{t_{k}})}\right];$

here we note that the randomness in $V$ is independent of $A$ in the above
expectations. In order to calculate the expectation of this integral, we aim
to apply Fubini’s theorem twice: Once to combine the product of expectations
$\mathbf{E}_{A}$ into a single expectation with respect to a concatenated path
$A^{\boldsymbol{a},\boldsymbol{b}}_{\boldsymbol{t}}$, and once more to bring
the expectation with respect to $V$ inside the expectation with respect to
$A^{\boldsymbol{a},\boldsymbol{b}}_{\boldsymbol{t}}$. In order to be able to
apply Fubini’s theorem, it suffices to verify that

(6.1)
$\displaystyle\mathbf{E}_{V}\left[\prod_{k=1}^{n}\mathbf{E}_{A}\left[|\mathfrak{n}_{t}(Q,A^{a_{k},b_{k}}_{t_{k}})|\mathrm{e}^{-\int_{0}^{t}S(A^{a_{k},b_{k}}_{t_{k}}(s))~{}\mathrm{d}s+\mathfrak{B}_{t}(A^{a_{k},b_{k}}_{t_{k}})}\right]\right]<\infty.$

By successive applications of Tonelli’s theorem, we can write (6.1) as

(6.2)
$\mathbf{E}_{V}\left[\mathbf{E}_{A}\left[|\mathfrak{n}_{|\boldsymbol{t}|}(Q,A^{\boldsymbol{a},\boldsymbol{b}}_{\boldsymbol{t}})|\mathrm{e}^{-\int_{0}^{|\boldsymbol{t}|}S(A^{\boldsymbol{a},\boldsymbol{b}}_{\boldsymbol{t}}(s))~{}\mathrm{d}s+\mathfrak{B}_{|\boldsymbol{t}|}(A^{\boldsymbol{a},\boldsymbol{b}}_{\boldsymbol{t}})}\right]\right]\\\
=\mathbf{E}_{A}\left[\mathbf{E}_{V}\left[|\mathfrak{n}_{|\boldsymbol{t}|}(Q,A^{\boldsymbol{a},\boldsymbol{b}}_{\boldsymbol{t}})|\mathrm{e}^{-\int_{0}^{|\boldsymbol{t}|}S(A^{\boldsymbol{a},\boldsymbol{b}}_{\boldsymbol{t}}(s))~{}\mathrm{d}s+\mathfrak{B}_{|\boldsymbol{t}|}(A^{\boldsymbol{a},\boldsymbol{b}}_{\boldsymbol{t}})}\right]\right],$

Then, since $S=S_{0}+S_{1}$, and $S_{1}$ and $Q$ are independent (by
Assumption 2.15-(4)), we can write (6.1) as

(6.3)
$\displaystyle\mathbf{E}_{A}\left[\mathbf{E}_{Q}\big{[}|\mathfrak{n}_{|\boldsymbol{t}|}(Q,A^{\boldsymbol{a},\boldsymbol{b}}_{\boldsymbol{t}})|\big{]}\mathbf{E}_{S_{1}}\left[\mathrm{e}^{-\int_{0}^{|\boldsymbol{t}|}S_{1}(A^{\boldsymbol{a},\boldsymbol{b}}_{\boldsymbol{t}}(s))~{}\mathrm{d}s}\right]\mathrm{e}^{-\int_{0}^{|\boldsymbol{t}|}S_{0}(A^{\boldsymbol{a},\boldsymbol{b}}_{\boldsymbol{t}}(s))~{}\mathrm{d}s+\mathfrak{B}_{|\boldsymbol{t}|}(A^{\boldsymbol{a},\boldsymbol{b}}_{\boldsymbol{t}})}\right].$

On the one hand, since $S_{1}(i,\cdot)$ are i.i.d. Gaussian processes with
mean zero and covariance $\gamma$, if we condition on the process $A$, then
the integral $\int_{0}^{|\boldsymbol{t}|}S_{1}(A(s))~{}\mathrm{d}s$ is a
Gaussian variable with mean zero and variance

$\mathbf{E}_{S_{1}}\left[\left(\int_{0}^{|\boldsymbol{t}|}S_{1}(A(s))~{}\mathrm{d}s\right)^{2}\right]=\int_{[0,{|\boldsymbol{t}|}]^{2}}\mathbf{1}_{\\{U(s_{1})=U(s_{2})\\}}\gamma\big{(}Z(s_{1})-Z(s_{2})\big{)}~{}\mathrm{d}\boldsymbol{s}=\sigma^{\gamma}_{|\boldsymbol{t}|}(A).$

Thus, by a Gaussian moment generating function calculation,

(6.4)
$\displaystyle\mathbf{E}_{S_{1}}\left[\mathrm{e}^{-\int_{0}^{|\boldsymbol{t}|}S_{1}(A(s))~{}\mathrm{d}s}\right]=\mathrm{e}^{\frac{1}{2}\sigma^{\gamma}_{|\boldsymbol{t}|}(A)}.$

On the other hand, by definition of $\mathfrak{n}_{t}$ in (2.8) and Hölder’s
inequality,

(6.5)
$\displaystyle\mathbf{E}_{Q}\big{[}|\mathfrak{n}_{|\boldsymbol{t}|}(Q,A)|\big{]}\leq\mathrm{e}^{(r-1)|\boldsymbol{t}|}\prod_{k=1}^{N(|\boldsymbol{t}|)}\mathbf{E}_{Q}\Big{[}\big{|}Q_{J_{k}}\big{(}Z(\tau_{k})\big{)}\big{|}^{N(|\boldsymbol{t}|)}\Big{]}^{1/N(|\boldsymbol{t}|)}.$

By Assumption 2.15, the components of $Q_{i,j}$ (or $Q_{i,j}$ itself if
$\mathbb{F}=\mathbb{R}$) are i.i.d. centered Gaussian processes with
stationary covariance $\varrho$. Therefore, there exists some deterministic
constant $\nu>0$ (which depends only on $|\boldsymbol{t}|$ and $\mathbb{F}$)
such that

$\mathbf{E}_{Q}\big{[}|\mathfrak{n}_{|\boldsymbol{t}|}(Q,A)|\big{]}\leq\mathrm{e}^{(r-1)|\boldsymbol{t}|}\prod_{k=1}^{N(|\boldsymbol{t}|)}\mathbf{E}_{Q}\Big{[}\big{|}Q_{J_{k}}(0)\big{|}^{N(|\boldsymbol{t}|)}\Big{]}^{1/N(|\boldsymbol{t}|)}\leq\nu\big{(}2\varrho(0)\big{)}^{N(|\boldsymbol{t}|)/2}\Gamma\left(\frac{N(|\boldsymbol{t}|)+1}{2}\right),$

where $\Gamma$ is the Gamma function.

With these last two calculations in hand, we obtain (6.1) from (6.3) by noting
that

  1. (1)

$\mathbf{E}_{N}\left[\big{(}2\varrho(0)\big{)}^{N(|\boldsymbol{t}|)/2}\Gamma\left(\frac{N(|\boldsymbol{t}|)+1}{2}\right)\right]<\infty$
since $N(|\boldsymbol{t}|)$ is Poisson distributed,

  2. (2)

since $\gamma$ is positive definite,
$\sigma^{\gamma}_{|\boldsymbol{t}|}(A)\leq|\boldsymbol{t}|^{2}\gamma(0)$,

  3. (3)

since $S_{0}\geq-\nu$ for some deterministic $\nu>0$,
$\mathrm{e}^{-\int_{0}^{|\boldsymbol{t}|}S_{0}(A(s))~{}\mathrm{d}s}\leq\mathrm{e}^{\nu|\boldsymbol{t}|}$,
and

  4. (4)

$\mathbf{E}\big{[}\mathrm{e}^{\mathfrak{B}_{t}(A^{x,y}_{t})}\big{]}<\infty$
for all $t>0$ and $x,y\in I$ (this is a straightforward consequence of the
finiteness and continuity of the Feynman-Kac kernels in one dimension (2.10)).

With (6.1) established, we may now apply Fubini’s theorem; by replicating the
steps in (6.2) and (6.3), but replacing $|\mathfrak{n}_{t}|$ by
$\mathfrak{n}_{t}$, we are led to

$\mathbf{E}\big{[}K(t_{1};a_{1},b_{1})\cdots
K(t_{n};a_{n},b_{n})\big{]}=\left(\prod_{k=1}^{n}\Pi_{A}(t_{k};a_{k},b_{k})\right)\\\
\cdot\mathbf{E}_{A}\left[\mathbf{E}_{Q}\big{[}\mathfrak{n}_{|\boldsymbol{t}|}(Q,A^{\boldsymbol{a},\boldsymbol{b}}_{\boldsymbol{t}})\big{]}\mathbf{E}_{S_{1}}\left[\mathrm{e}^{-\int_{0}^{|\boldsymbol{t}|}S_{1}(A^{\boldsymbol{a},\boldsymbol{b}}_{\boldsymbol{t}}(s))~{}\mathrm{d}s}\right]\mathrm{e}^{-\int_{0}^{|\boldsymbol{t}|}S_{0}(A^{\boldsymbol{a},\boldsymbol{b}}_{\boldsymbol{t}}(s))~{}\mathrm{d}s+\mathfrak{B}_{|\boldsymbol{t}|}(A^{\boldsymbol{a},\boldsymbol{b}}_{\boldsymbol{t}})}\right].$

In this expression, we can reuse the calculation in (6.4). Thus, to prove
Corollary 2.19, it only remains to establish that for every $t>0$, one has

(6.6)
$\displaystyle\mathbf{E}_{Q}\big{[}\mathfrak{n}_{t}(Q,A)\big{]}=\mathfrak{m}_{t}^{\varrho}(A).$

When $\mathbb{F}=\mathbb{R},\mathbb{C}$, this is a straightforward consequence
of Isserlis’ theorem [21]. The argument is somewhat similar for
$\mathbb{F}=\mathbb{H}$, but rather more involved since the product in
$\mathbb{H}$ is not commutative. We thus finish the proof of Corollary 2.19
with a case-by-case proof of (6.6):

####  6.2.1. Proof of (6.6) for $\mathbb{F}=\mathbb{R},\mathbb{C}$

For $\mathbb{F}=\mathbb{R}$ or $\mathbb{C}$, we invoke the statement of
Isserlis’ theorem from [24, (1.7) and (1.8)], whereby

$\mathbf{E}_{\mathrm{d}Q}\left[\prod_{k=1}^{N(t)}Q_{J_{k}}\big{(}X(\tau_{k})\big{)}\right]=\mathbf{1}_{\\{N(t)\text{
is even}\\}}\sum_{p\in\mathcal{P}_{N(t)}}\prod_{\\{\ell_{1},\ell_{2}\\}\in
p}\mathbf{E}_{\mathrm{d}Q}\left[Q_{J_{\ell_{1}}}\big{(}X(\tau_{\ell_{1}})\big{)}Q_{J_{\ell_{2}}}\big{(}X(\tau_{\ell_{2}})\big{)}\right]$

(recall the definition of $\mathcal{P}_{n}$ in Definition 2.16). By
independence,

$\mathbf{E}\big{[}Q_{i,j}(x)Q_{k,\ell}(y)\big{]}=0$

whenever $(i,j)\neq(k,\ell)$ and $(i,j)\neq(\ell,k)$. Otherwise, the analysis
depends on $\mathbb{F}$:

If $\mathbb{F}=\mathbb{R}$, then

$\mathbf{E}\big{[}Q_{i,j}(x)Q_{i,j}(y)\big{]}=\mathbf{E}\big{[}Q_{i,j}(x)Q_{j,i}(y)\big{]}=\varrho(x-y).$

(6.6) for $\mathbb{R}$ then follows by the definition of
$\mathfrak{C}_{t}(p,U)$ in Definition 2.17-(1).

If $\mathbb{F}=\mathbb{C}$, then (recall the definition of $a(x)$ and $b(x)$
in (2.13))

$\mathbf{E}\big{[}Q_{i,j}(x)Q_{i,j}(y)\big{]}=\tfrac{1}{2}\mathbf{E}\big{[}a(x)a(y)+\big{(}a(y)b(x)+a(x)b(y)\big{)}\mathsf{i}-b(x)b(y)\big{]}\\\
=\tfrac{1}{2}\varrho(x-y)+0-\tfrac{1}{2}\varrho(x-y)=0.$

and similarly

$\mathbf{E}\big{[}Q_{i,j}(x)Q_{j,i}(y)\big{]}=\mathbf{E}\big{[}Q_{i,j}(x)Q_{i,j}(y)^{*}\big{]}\\\
=\tfrac{1}{2}\mathbf{E}\big{[}a(x)a(y)+\big{(}a(y)b(x)-a(x)b(y)\big{)}\mathsf{i}+b(x)b(y)\big{]}=\varrho(x-y).$

(6.6) for $\mathbb{C}$ then follows by the definition of
$\mathfrak{C}_{t}(p,U)$ in Definition 2.17-(2).

####  6.2.2. Proof of (6.6) for $\mathbb{F}=\mathbb{H}$

In the case of quaternions, Isserlis’ theorem does not apply directly due to
the lack of commutativity. To get around this, we proceed as follows: Let
$q=a+b\mathsf{i}+c\mathsf{j}+d\mathsf{k}$ be a quaternion. We recall that
there exists an injective homomorphism from $\mathbb{F}$ to the space of
$2\times 2$ complex matrices via the representation

(6.11) $\displaystyle
q=\left[\begin{array}[]{cc}a+b\mathsf{i}&c+d\mathsf{i}\\\
-(c-d\mathsf{i})&a-b\mathsf{i}\end{array}\right]=\left[\begin{array}[]{cc}q^{(0,0)}&q^{(0,1)}\\\
q^{(1,0)}&q^{(1,1)}\end{array}\right].$

(Note that we index the entries $q^{(h,l)}$ of $q$ using binary pairs
$(h,l)\in\mathcal{B}_{1}$, recalling Definition 2.16.) Moreover, if $q$ is
random, then
$\mathbf{E}[q]=\mathbf{E}[a]+\mathbf{E}[b]\mathsf{i}+\mathbf{E}[c]\mathsf{j}+\mathbf{E}[d]\mathsf{k}$
implies by linearity of expectation that

$\mathbf{E}[q]=\left[\begin{array}[]{cc}\mathbf{E}[a]+\mathbf{E}[b]\mathsf{i}&\mathbf{E}[c]+\mathbf{E}[d]\mathsf{i}\\\
-\mathbf{E}[c]+\mathbf{E}[d]\mathsf{i}&\mathbf{E}[a]-\mathbf{E}[b]\mathsf{i}\end{array}\right]=\left[\begin{array}[]{cc}\mathbf{E}[a+b\mathsf{i}]&\mathbf{E}[c+d\mathsf{i}]\\\
\mathbf{E}[-c+d\mathsf{i}]&\mathbf{E}[a-b\mathsf{i}]\end{array}\right];$

in other words, $\mathbf{E}[q]^{(h,l)}=\mathbf{E}[q^{(h,l)}]$ for all $0\leq
h,l\leq 1$. Therefore, by a matrix product,

$\mathbf{E}_{\mathrm{d}Q}\left[\prod_{k=1}^{N(t)}Q_{J_{k}}\big{(}X(\tau_{k})\big{)}\right]^{(h,l)}=\sum_{\begin{subarray}{c}m\in\mathcal{B}^{h,l}_{N(t)}\end{subarray}}\mathbf{E}\left[\prod_{k=1}^{N(t)}Q_{J_{k}}\big{(}X(\tau_{k})\big{)}^{(m_{k-1},m_{k})}\right],\qquad
0\leq h,l\leq 1,$

where we recall the definition of $\mathcal{B}^{h,l}_{n}$ in Definition 2.16.
Since the entries of the matrices (6.11) are complex-valued, we may now apply
the classical statement of Isserlis’ theorem (e.g., [24, (1.8)]), whereby

(6.12)
$\mathbf{E}_{\mathrm{d}Q}\left[\prod_{k=1}^{N(t)}Q_{J_{k}}\big{(}X(\tau_{k})\big{)}\right]^{(h,l)}=\mathbf{1}_{\\{N(t)\text{
is
even}\\}}\sum_{m\in\mathcal{B}^{h,l}_{N(t)}}\sum_{p\in\mathcal{P}_{N(t)}}\Bigg{\\{}\\\
\prod_{\\{\ell_{1},\ell_{2}\\}\in
p}\mathbf{E}\left[Q_{J_{\ell_{1}}}\big{(}X(\tau_{\ell_{1}})\big{)}^{(m_{\ell_{1}-1},m_{\ell_{1}})}Q_{J_{\ell_{2}}}\big{(}X(\tau_{\ell_{2}})\big{)}^{(m_{\ell_{2}-1},m_{\ell_{2}})}\right]\Bigg{\\}}.$

In order to calculate the above expectations we use the following:

######  Lemma 6.1.

Let $x,y\in I$ be arbitrary. On the one hand,

(6.13)
$\displaystyle\mathbf{E}\left[Q_{J_{\ell_{1}}}(x)^{(m_{\ell_{1}-1},m_{\ell_{1}})}Q_{J_{\ell_{2}}}(y)^{(m_{\ell_{2}-1},m_{\ell_{2}})}\right]=0$

whenever $J_{\ell_{1}}\not\in\\{J_{\ell_{2}},J_{\ell_{2}}^{*}\\}$. On the
other hand:

  1. (1)

If $J_{\ell_{1}}=J_{\ell_{2}}$, then

$\mathbf{E}\left[Q_{J_{\ell_{1}}}(x)^{(m_{\ell_{1}-1},m_{\ell_{1}})}Q_{J_{\ell_{2}}}(y)^{(m_{\ell_{2}-1},m_{\ell_{2}})}\right]\\\
=\begin{cases}\frac{\varrho(x-y)}{2}&\text{if
$\big{(}(m_{\ell_{1}-1},m_{\ell_{1}}),(m_{\ell_{2}-1},m_{\ell_{2}})\big{)}=\big{(}(0,0),(1,1)\big{)}$
or $\big{(}(1,1),(0,0)\big{)}$},\\\ -\frac{\varrho(x-y)}{2}&\text{if
$\big{(}(m_{\ell_{1}-1},m_{\ell_{1}}),(m_{\ell_{2}-1},m_{\ell_{2}})\big{)}=\big{(}(0,1),(1,0)\big{)}$
or $\big{(}(1,0),(0,1)\big{)}$},\\\ 0&\text{otherwise}.\end{cases}$

  2. (2)

If $J_{\ell_{1}}=J_{\ell_{2}}^{*}$, then

$\mathbf{E}\left[Q_{J_{\ell_{1}}}(x)^{(m_{\ell_{1}-1},m_{\ell_{1}})}Q_{J_{\ell_{2}}}(y)^{(m_{\ell_{2}-1},m_{\ell_{2}})}\right]\\\
=\begin{cases}\frac{\varrho(x-y)}{2}&\text{if
\footnotesize{$\big{(}(m_{\ell_{1}-1},m_{\ell_{1}}),(m_{\ell_{2}-1},m_{\ell_{2}})\big{)}=\big{(}(0,0),(0,0)\big{)}$,
$\big{(}(1,1),(1,1)\big{)}$, $\big{(}(0,1),(1,0)\big{)}$, or
$\big{(}(1,0),(0,1)\big{)}$}},\\\ 0&\text{otherwise}.\end{cases}$

  * Proof.

We can write the processes

$Q_{i,j}(x)=\tfrac{1}{2}\big{(}a_{i,j}(x)+b_{i,j}(x)\mathsf{i}+c_{i,j}(x)\mathsf{j}+d_{i,j}(x)\mathsf{k}\big{)},\qquad
1\leq i<j\leq d,~{}x\in I$

where $a_{i,j},b_{i,j},c_{i,j},$ and $d_{i,j}$ are i.i.d. copies of a centered
Gaussian process with covariance $\varrho(x-y)$ (see (2.13)). Therefore, for
any $1\leq i<j\leq r$ and $x\in I$, one has

$Q_{i,j}(x)=\frac{1}{2}\left[\begin{array}[]{cc}a_{i,j}(x)+b_{i,j}(x)\mathsf{i}&c_{i,j}(x)+d_{i,j}(x)\mathsf{i}\\\
-\big{(}c_{i,j}(x)-d_{i,j}(x)\mathsf{i}\big{)}&a_{i,j}(x)-b_{i,j}(x)\mathsf{i}\end{array}\right]$

and

$Q_{j,i}(x)=Q_{i,j}(x)^{*}=\frac{1}{2}\left[\begin{array}[]{cc}a_{i,j}(x)-b_{i,j}(x)\mathsf{i}&-\big{(}c_{i,j}(x)+d_{i,j}(x)\mathsf{i}\big{)}\\\
c_{i,j}(x)-d_{i,j}(x)\mathsf{i}&a_{i,j}(x)+b_{i,j}(x)\mathsf{i}\end{array}\right].$

With this in mind, (6.13) follows immediately by independence. As for the
calculations stated as items (1) and (2) of this lemma, we use a simple matrix
product along with the remarks that

$\mathbf{E}\big{[}(a_{i,j}(x)\pm b_{i,j}(x)\mathsf{i})(c_{i,j}(y)\pm
d_{i,j}(y))\big{]}=0,$
$\mathbf{E}\big{[}\big{(}a_{i,j}(x)+b_{i,j}(x)\mathsf{i}\big{)}\big{(}a_{i,j}(y)+b_{i,j}(y)\mathsf{i}\big{)}\big{]}=\mathbf{E}\big{[}\big{(}c_{i,j}(x)+d_{i,j}(x)\mathsf{i}\big{)}\big{(}c_{i,j}(y)+d_{i,j}(y)\mathsf{i}\big{)}\big{]}=0,$

and

$\tfrac{1}{4}\mathbf{E}\big{[}\big{(}a_{i,j}(x)+b_{i,j}(x)\mathsf{i}\big{)}\big{(}a_{i,j}(y)-b_{i,j}(y)\mathsf{i}\big{)}\big{]}\\\
=\tfrac{1}{4}\mathbf{E}\big{[}\big{(}c_{i,j}(x)+d_{i,j}(x)\mathsf{i}\big{)}\big{(}c_{i,j}(y)-d_{i,j}(y)\mathsf{i}\big{)}\big{]}=\frac{\varrho(x-y)}{2},$

all of which are straightforward from the definitions of
$a_{i,j},b_{i,j},c_{i,j},$ and $d_{i,j}$. ∎

We may now wrap up the proof of (6.6): Firstly, we claim that

(6.14)
$\displaystyle\mathbf{E}_{\mathrm{d}Q}\left[\prod_{k=1}^{N(t)}Q_{J_{k}}\big{(}X(\tau_{k})\big{)}\right]^{(0,1)}=\mathbf{E}_{\mathrm{d}Q}\left[\prod_{k=1}^{N(t)}Q_{J_{k}}\big{(}X(\tau_{k})\big{)}\right]^{(1,0)}=0.$

To see this, by Lemma 6.1, the only binary sequences
$m\in\mathcal{B}_{N(t)}^{h,l}$ that can contribute to (6.12) are those made up
of some permutation of pairs of steps in the following set:

$\Big{\\{}\big{\\{}(0,0),(1,1)\big{\\}},\big{\\{}(0,1),(1,0)\big{\\}},\big{\\{}(0,0),(0,0)\big{\\}},\big{\\{}(1,1),(1,1)\big{\\}}\Big{\\}}.$

In particular, there is only one type of pair that induces steps that go from
$0$ to $1$ or vice versa— namely, $\big{\\{}(0,1),(1,0)\big{\\}}$—and each
such pair induces one $(0,1)$ step and one $(1,0)$ step. Therefore, any
$m\in\mathcal{B}_{N(t)}^{h,l}$ that contributes to (6.12) must have an equal
number of $(1,0)$ and $(0,1)$ steps, meaning that $h=m_{0}=m_{N(t)}=l$. Hence
(6.14) holds.

Secondly, we note that, by (6.11),

$\mathbf{E}_{\mathrm{d}Q}\left[\prod_{k=1}^{N(t)}Q_{J_{k}}\big{(}X(\tau_{k})\big{)}\right]^{(0,0)}=\left(\mathbf{E}_{\mathrm{d}Q}\left[\prod_{k=1}^{N(t)}Q_{J_{k}}\big{(}X(\tau_{k})\big{)}\right]^{(1,1)}\right)^{*}.$

By combining (6.12) and Lemma 6.1, it is clear that the above entries have no
complex part; hence we conclude that

$\mathbf{E}_{\mathrm{d}Q}\left[\prod_{k=1}^{N(t)}Q_{J_{k}}\big{(}X(\tau_{k})\big{)}\right]=\mathbf{E}_{\mathrm{d}Q}\left[\prod_{k=1}^{N(t)}Q_{J_{k}}\big{(}X(\tau_{k})\big{)}\right]^{(0,0)},$

noting furthermore that the above is just a real number. With this in hand, we
obtain (6.6) in the case $\mathbb{F}=\mathbb{H}$ by combining (6.12) with the
fact that the constant $\mathfrak{D}_{t}(p,U)$ in Definition 2.17-(3) is
constructed by a direct application of Lemma 6.1. More specifically, the
binary sequences that respect $(p,J)$ are exactly those that give a
nonvanishing contribution to (6.12) (as per the covariance calculations in
Lemma 6.1-(1) and -(2)), and the flips correspond to the terms in Lemma
6.1-(1) that have a negative covariance of $-\frac{\varrho(x-y)}{2}$.

##  7\. Justification for Conjecture 2.25

If we formally apply Corollary 2.19 with $\gamma=c_{1}\delta_{0}$ and
$\varrho=c_{2}\delta_{0}$ for some $c_{1},c_{2}>0$, then we get the moment
formula

(7.1) $\mathbf{E}\big{[}K(t_{1};a_{1},b_{1})\cdots
K(t_{n};a_{n},b_{n})\big{]}=\Pi_{A}(t_{1};a_{1},b_{1})\cdots\Pi_{A}(t_{n};a_{n},b_{n})\\\
\cdot\mathbf{E}\left[\mathfrak{m}^{c_{2}\delta_{0}}_{|\boldsymbol{t}|}(Q,A^{\boldsymbol{a},\boldsymbol{b}}_{\boldsymbol{t}})\mathrm{e}^{-\int_{0}^{|\boldsymbol{t}|}S_{0}(A^{\boldsymbol{a},\boldsymbol{b}}_{\boldsymbol{t}}(s))~{}\mathrm{d}s+\frac{1}{2}\sigma^{c_{1}\delta_{0}}_{|\boldsymbol{t}|}(A^{\boldsymbol{a},\boldsymbol{b}}_{\boldsymbol{t}})+\mathfrak{B}_{|\boldsymbol{t}|}(A^{\boldsymbol{a},\boldsymbol{b}}_{\boldsymbol{t}})}\right].$

If we recall the definition of $A$’s local time in (2.18), then we get the
formal equalities

$\displaystyle\sigma^{c_{1}\delta_{0}}_{t}(A)$
$\displaystyle=c_{1}\int_{[0,t]^{2}}\mathbf{1}_{\\{U(s_{1})=U(s_{2})\\}}\delta_{0}\big{(}Z(s_{1})-Z(s_{2})\big{)}~{}\mathrm{d}\boldsymbol{s}$
$\displaystyle=c_{1}\int_{\mathcal{A}^{2}}L_{t}^{(i,x)}(A)\mathbf{1}_{\\{i=j\\}}\delta_{0}(x-y)L_{t}^{(i,x)}(A)~{}\mathrm{d}\mu(i,x)\mathrm{d}\mu(j,x)$
$\displaystyle=c_{1}\|L_{t}(A)\|_{2}^{2}.$

This explains the appearance of the term
$\frac{c_{1}}{2}\|L_{t}(\widehat{U},Z)\|$ inside (2.21). It now remains to
explain the presence of $\frac{(r-1)^{2}}{2}\|L_{t}(Z)\|_{2}^{2}$ in (2.21),
as well as the coupled process $(\widehat{U},Z)$ instead of $A=(U,Z)$.

For this purpose, our aim is to write (7.1) as nested sums/integrals involving
all the possible realizations of $U$ (i.e., the jumps and jump times). That
said, the fact that we are conditioning on the endpoints of $U$ in
$A^{\boldsymbol{a},\boldsymbol{b}}_{\boldsymbol{t}}$ changes the distribution
of $N$, which complicates the analysis. In order to get around this, we recast
the conditioning on the endpoints of $U$’s path by writing (7.1) as follows:

(7.2)
$\Pi_{Z}(t_{1};x_{1},y_{1})\cdots\Pi_{Z}(t_{n};x_{n},y_{n})\mathbf{E}\Bigg{[}\mathfrak{m}^{c_{2}\delta_{0}}_{|\boldsymbol{t}|}\big{(}Q,(U^{\boldsymbol{i}}_{\boldsymbol{t}},Z^{\boldsymbol{x},\boldsymbol{y}}_{\boldsymbol{t}})\big{)}\\\
\cdot\mathrm{e}^{-\int_{0}^{|\boldsymbol{t}|}S_{0}(U^{\boldsymbol{i}}_{\boldsymbol{t}},Z^{\boldsymbol{x},\boldsymbol{y}}_{\boldsymbol{t}}(s))~{}\mathrm{d}s+\frac{c_{1}}{2}\|L_{t}(U^{\boldsymbol{i}}_{\boldsymbol{t}},Z^{\boldsymbol{x},\boldsymbol{y}}_{\boldsymbol{t}})\|_{2}^{2}+\mathfrak{B}_{|\boldsymbol{t}|}(U^{\boldsymbol{i}}_{\boldsymbol{t}},Z^{\boldsymbol{x},\boldsymbol{y}}_{\boldsymbol{t}})}\mathbf{1}_{\\{\forall
k:~{}U^{\boldsymbol{i}}_{\boldsymbol{t}}(t_{0}+\cdots+t_{k-1}+s)\to
j_{k}\text{ as }s\to t_{k}^{-}\\}}\Bigg{]},$

where we recall the notations of $\boldsymbol{i}=(i_{1},\ldots,i_{n})$,
$\boldsymbol{j}=(j_{1},\ldots,j_{n})$, $\boldsymbol{x}=(x_{1},\ldots,x_{n})$,
and $\boldsymbol{y}=(y_{1},\ldots,y_{n})$ in the statement of Conjecture 2.25.
With this in hand, we now point out that drawing a realization of
$U^{\boldsymbol{i}}_{\boldsymbol{t}}$ requires the following steps:

  1. $\bullet$

Sample $N(|\boldsymbol{t}|)$ according to a Poisson variable with parameter
$(r-1)|\boldsymbol{t}|$:

$\mathbf{P}\big{[}N(|\boldsymbol{t}|)=m\big{]}=\frac{\big{(}(r-1)|\boldsymbol{t}|\big{)}^{m}}{m!}\mathrm{e}^{-(r-1)|\boldsymbol{t}|},\qquad
m=0,1,2,3,\ldots$

  2. $\bullet$

Given $N(|\boldsymbol{t}|)=m$, sample $(\tau_{1},\ldots,\tau_{m})$ by ordering
i.i.d. uniform variables on $[0,|\boldsymbol{t}|]$. That is, if we recall the
notation of $\overrightarrow{s_{i}}$ from Theorem 1.2-(2),

$\mathbf{P}\big{[}(\tau_{1},\ldots,\tau_{m})\in\mathrm{d}(\overrightarrow{s_{1}},\ldots,\overrightarrow{s_{m}})\big{]}=\frac{1}{|\boldsymbol{t}|^{m}}\mathrm{d}\boldsymbol{s},\qquad\boldsymbol{s}=(s_{1},\ldots,s_{m})\in[0,|\boldsymbol{t}|]^{m}.$

  3. $\bullet$

Given $N(|\boldsymbol{t}|)=m$ and $\tau_{i}=\overrightarrow{s_{i}}$ for
$\boldsymbol{s}=[0,|\boldsymbol{t}|]^{m}$, we let
$\boldsymbol{z}(|\boldsymbol{t}|,m,\boldsymbol{s})=(z_{1},\ldots,z_{n})$
denote the partition of the integers $\\{1,\ldots,m\\}$ such that

$z_{k}=\big{|}\big{\\{}1\leq\ell\leq
m:\overrightarrow{s_{\ell}}\in[t_{0}+\cdots+t_{k-1},t_{k})\big{\\}}\big{|},\qquad
t_{0}=0,~{}1\leq k\leq n.$

That is, $z_{k}$ is the number of jumps
$\tau_{\ell}=\overrightarrow{s_{\ell}}$ that land inside the time interval of
the $k^{\mathrm{th}}$ segment of the concatenated path
$U^{\boldsymbol{i}}_{\boldsymbol{t}}$. Finally, let
$\mathcal{W}^{\boldsymbol{i}}_{\boldsymbol{z}(|\boldsymbol{t}|,m,\boldsymbol{s})}$
denote the set of $m$-tuples of jumps

$\boldsymbol{u}=(u^{(1)}_{1},\ldots,u^{(1)}_{z_{1}},u^{(2)}_{1},\ldots,u^{(2)}_{z_{2}},\ldots,u^{(n)}_{1},\ldots,u^{(n)}_{z_{n}}),$

where for every $1\leq k\leq n$, the sequence of jumps

$u^{(k)}_{1},\ldots,u^{(k)}_{z_{k}}$

form a random walk path on the complete graph with vertices $\\{1,\ldots,r\\}$
(without loops) that starts at $i_{k}$. We note that

$|\mathcal{W}^{\boldsymbol{i}}_{\boldsymbol{z}(|\boldsymbol{t}|,m,\boldsymbol{s})}|=\prod_{k=1}^{n}(r-1)^{z_{k}}=(r-1)^{m},$

and that given $N(|\boldsymbol{t}|)=m$ and $\tau_{i}=\overrightarrow{s_{i}}$,
the jumps $(J_{k}:1\leq k\leq m)$ in the concatenated path
$U^{\boldsymbol{i}}_{\boldsymbol{t}}$ are sampled uniformly at random within
$\mathcal{W}^{\boldsymbol{i}}_{\boldsymbol{z}(|\boldsymbol{t}|,m,\boldsymbol{s})}$.

Based on all of this, given some $m\geq 0$,
$\boldsymbol{s}\in[0,|\boldsymbol{t}|]^{m}$, and
$\boldsymbol{u}\in\mathcal{W}^{\boldsymbol{i}}_{\boldsymbol{z}(|\boldsymbol{t}|,m,\boldsymbol{s})}$,
we let $\mathcal{U}_{(\boldsymbol{s},\boldsymbol{u})}$ denote the realization
of $U^{\boldsymbol{i}}_{\boldsymbol{t}}$ with $m$ jumps, whose jump times are
given by $\overrightarrow{s_{i}}$, and whose jumps are given by
$\boldsymbol{u}$. Moreover, to alleviate notation, we henceforth denote

$F(U,Z)=\mathrm{e}^{-\int_{0}^{|\boldsymbol{t}|}S_{0}(U,Z(s))~{}\mathrm{d}s+\frac{c_{1}}{2}\|L_{t}(U,Z)\|_{2}^{2}+\mathfrak{B}_{|\boldsymbol{t}|}(U,Z)}\mathbf{1}_{\\{\forall
k:~{}U(t_{0}+\cdots+t_{k-1}+s)\to j_{k}\text{ as }s\to t_{k}^{-}\\}}.$

By an application of the law of total expectation (conditioning with respect
to $N(|\boldsymbol{t}|)$, $\tau_{k}$, and $J_{k}$) and the fact that $U$ and
$Z$ are independent, we may write the expectation in (7.2) as follows:

(7.3)
$\mathbf{E}\Bigg{[}\sum_{m=0}^{\infty}\Bigg{(}\frac{1}{|\boldsymbol{t}|^{m}}\int_{[0,|\boldsymbol{t}|]^{m}}\frac{1}{|\mathcal{W}^{\boldsymbol{i}}_{\boldsymbol{z}(|\boldsymbol{t}|,m,\boldsymbol{s})}|}\\\
\cdot\sum_{\boldsymbol{u}\in\mathcal{W}^{\boldsymbol{i}}_{\boldsymbol{z}(|\boldsymbol{t}|,m,\boldsymbol{s})}}\mathfrak{m}^{c_{2}\delta_{0}}_{|\boldsymbol{t}|}\big{(}\mathcal{U}_{(\boldsymbol{s},\boldsymbol{u})},Z^{\boldsymbol{x},\boldsymbol{y}}_{\boldsymbol{t}}\big{)}F\big{(}\mathcal{U}_{(\boldsymbol{s},\boldsymbol{u})},Z^{\boldsymbol{x},\boldsymbol{y}}_{\boldsymbol{t}}\big{)}~{}\mathrm{d}\boldsymbol{s}\Bigg{)}\frac{\mathrm{e}^{-(r-1)|\boldsymbol{t}|}\big{(}(r-1)|\boldsymbol{t}|\big{)}^{m}}{m!}\Bigg{]}.$

Recalling the definition $\mathfrak{m}_{|\boldsymbol{t}|}^{\varrho}$ in
(2.15), a formal writing of this with $\varrho=c_{2}\delta_{0}$ yields

(7.4)
$\mathfrak{m}^{c_{2}\delta_{0}}_{|\boldsymbol{t}|}\big{(}\mathcal{U}_{(\boldsymbol{s},\boldsymbol{u})},Z\big{)}=\\\
\mathrm{e}^{(r-1)|\boldsymbol{t}|}\mathbf{1}_{\\{m\text{ is
even}\\}}c_{2}^{m/2}\sum_{p\in\mathcal{P}_{m}}\mathfrak{C}_{|\boldsymbol{t}|}(p,\mathcal{U}_{(\boldsymbol{s},\boldsymbol{u})})\prod_{\\{\ell_{1},\ell_{2}\\}\in
p}\delta_{0}\big{(}Z(\overrightarrow{s_{\ell_{1}}})-Z(\overrightarrow{s_{\ell_{2}}})\big{)}.$

As explained in Section 2.3.3, we aim to interpret the product of Dirac masses
in (7.4) as density functions of the self-intersection probability measures.
That said, the fact that we have the ordered points
$\overrightarrow{s_{\ell_{i}}}$ in (7.4) instead of $s_{\ell_{i}}$ is
problematic for this purpose. We thus introduce a change of variables in
(7.4): Given any permutation $\pi$ of the set $\\{1,\ldots,m\\}$ and
$p\in\mathcal{P}_{m}$, we let $\pi(p)\in\mathcal{P}_{m}$ denote the perfect
matching such that $\\{\ell_{1},\ell_{2}\\}\in p$ if and only if
$\\{\pi(\ell_{1}),\pi(\ell_{2})\\}\in\pi(p)$. Then, given
$\boldsymbol{s}=(s_{1},\ldots,s_{m})$, we let $\pi_{\boldsymbol{s}}$ denote
the specific permutation of $\\{1,\ldots,m\\}$ such that for any
$1\leq\ell\leq m$, one has
$\overrightarrow{s_{\pi_{\boldsymbol{s}}(\ell)}}=s_{\ell}$, or equivalently
$\overrightarrow{s_{\ell}}=s_{\pi^{-1}_{\boldsymbol{s}}(\ell)}$. Then, by the
change of variables $q=\pi_{\boldsymbol{s}}^{-1}(p)$, we can rewrite the
second line in (7.4) as

$\mathrm{e}^{(r-1)|\boldsymbol{t}|}\mathbf{1}_{\\{m\text{ is
even}\\}}c_{2}^{m/2}\sum_{q\in\mathcal{P}_{m}}\mathfrak{C}_{|\boldsymbol{t}|}\big{(}\pi_{\boldsymbol{s}}(q),\mathcal{U}_{(\boldsymbol{s},\boldsymbol{u})}\big{)}\prod_{\\{\ell_{1},\ell_{2}\\}\in
q}\delta_{0}\big{(}Z(s_{\ell_{1}})-Z(s_{\ell_{2}})\big{)}F\big{(}\mathcal{U}_{(\boldsymbol{s},\boldsymbol{u})},Z\big{)}.$

With this self-intersection connection in hand, we may now formally rewrite
(7.3) by replacing the uniform measure on $[0,|\boldsymbol{t}|]^{m}$ by the
self-intersection measures, and at the same time recast the sum over perfect
matchings as a random perfect matching. Before making this more precise, we
make a few remarks:

  1. $\bullet$

In (7.3), the two appearances of the term $|\boldsymbol{t}|^{m}$ cancel.

  2. $\bullet$

The term $\mathrm{e}^{-(r-1)|\boldsymbol{t}|}$ in (7.3) cancels with
$\mathrm{e}^{(r-1)|\boldsymbol{t}|}$ in (7.4).

  3. $\bullet$

Thanks to the indicator $\mathbf{1}_{\\{m\text{ is even}\\}}$, we can
substitute $m$ by $2m$ in (7.3) and (7.4), and then remove the indicator.

  4. $\bullet$

$|\mathcal{P}_{2m}|=(2m-1)!!$ and $\frac{(2m-1)!!}{(2m)!}=\frac{1}{2^{m}m!}$.

Combining all of these simplifications together, we can then rewrite (7.3) as

(7.5)
$\mathbf{E}\Bigg{[}\sum_{m=0}^{\infty}\Bigg{(}\frac{1}{\|L_{|\boldsymbol{t}|}(Z^{\boldsymbol{x},\boldsymbol{y}}_{\boldsymbol{t}})\|_{2}^{2m}}\int_{[0,|\boldsymbol{t}|]^{2m}}\frac{1}{|\mathcal{W}^{\boldsymbol{i}}_{\boldsymbol{z}(|\boldsymbol{t}|,2m,\boldsymbol{s})}|}\\\
\cdot\sum_{\boldsymbol{u}\in\mathcal{W}^{\boldsymbol{i}}_{\boldsymbol{z}(|\boldsymbol{t}|,m,\boldsymbol{s})}}\Bigg{\\{}\frac{c_{2}^{m}}{(2m-1)!!}\sum_{q\in\mathcal{P}_{2m}}\mathfrak{C}_{|\boldsymbol{t}|}(\pi_{\boldsymbol{s}}(q),\mathcal{U}_{(\boldsymbol{s},\boldsymbol{u})})\prod_{\\{\ell_{1},\ell_{2}\\}\in
q}\delta_{0}\big{(}Z^{\boldsymbol{x},\boldsymbol{y}}_{\boldsymbol{t}}(s_{\ell_{1}})-Z^{\boldsymbol{x},\boldsymbol{y}}_{\boldsymbol{t}}(s_{\ell_{2}})\big{)}\\\
\cdot
F\big{(}\mathcal{U}_{(\boldsymbol{s},\boldsymbol{u})},Z^{\boldsymbol{x},\boldsymbol{y}}_{\boldsymbol{t}}\big{)}~{}\mathrm{d}\boldsymbol{s}\Bigg{\\}}\Bigg{)}\frac{\big{(}(r-1)\|L_{|\boldsymbol{t}|}(Z^{\boldsymbol{x},\boldsymbol{y}}_{\boldsymbol{t}})\|_{2}\big{)}^{2m}}{2^{m}m!}\Bigg{]},$

At this point, we obtain the informal statement of Conjecture 2.25 by noting
that

$\sum_{m=0}^{\infty}\frac{\big{(}(r-1)\|L_{|\boldsymbol{t}|}(Z)\|_{2}\big{)}^{2m}}{2^{m}m!}=\mathrm{e}^{\frac{(r-1)^{2}}{2}\|L_{|\boldsymbol{t}|}(Z)\|_{2}^{2}}$

(thus explaining the presence of that term in (2.21)) and thus that the
sums/integrals (7.5) can be interpreted as an expectation with respect to the
process $\widehat{U}$, which was expanded using the law of total expectation.

##  8\. Proof of Proposition 1.9

For ease of notation, let us write

$\mathfrak{O}_{i,j}(t)=\mathbf{E}\left[\left(\mathrm{e}^{\theta
t}\prod_{k=1}^{N(t)}\frac{F_{J_{k}}(\tau_{k})}{G_{J_{k}}}\right)\mathrm{e}^{\int_{0}^{t}F_{M^{i}(s),M^{i}(s)}(s)~{}\mathrm{d}s}\mathbf{1}_{\\{M^{i}(t)=j\\}}\right].$

To prove (1.16), we use the matrix ODE definition of the OE in (1.1), namely,

(8.1) $\displaystyle\lim_{h\to
0}\frac{\mathfrak{O}_{i,j}(t+h)-\mathfrak{O}_{i,j}(t)}{h}=\sum_{\ell=1}^{r}\mathfrak{O}_{i,\ell}(t)F_{\ell,j}(t).$

For this purpose, we write $\mathfrak{O}_{i,j}(t+h)$ as the sum of the
following three terms:

(8.2)
$\displaystyle\mathbf{E}\left[\left(\mathrm{e}^{\theta(t+h)}\prod_{k=1}^{N(t+h)}\frac{F_{J_{k}}(\tau_{k})}{G_{J_{k}}}\right)\mathrm{e}^{\int_{0}^{t+h}F_{M^{i}(s),M^{i}(s)}(s)~{}\mathrm{d}s}\mathbf{1}_{\\{M^{i}(t+h)=j\\}}\mathbf{1}_{\\{N(t+h)-N(t)=0\\}}\right]$
(8.3)
$\displaystyle\mathbf{E}\left[\left(\mathrm{e}^{\theta(t+h)}\prod_{k=1}^{N(t+h)}\frac{F_{J_{k}}(\tau_{k})}{G_{J_{k}}}\right)\mathrm{e}^{\int_{0}^{t+h}F_{M^{i}(s),M^{i}(s)}(s)~{}\mathrm{d}s}\mathbf{1}_{\\{M^{i}(t+h)=j\\}}\mathbf{1}_{\\{N(t+h)-N(t)=1\\}}\right]$
(8.4)
$\displaystyle\mathbf{E}\left[\left(\mathrm{e}^{\theta(t+h)}\prod_{k=1}^{N(t+h)}\frac{F_{J_{k}}(\tau_{k})}{G_{J_{k}}}\right)\mathrm{e}^{\int_{0}^{t+h}F_{M^{i}(s),M^{i}(s)}(s)~{}\mathrm{d}s}\mathbf{1}_{\\{M^{i}(t+h)=j\\}}\mathbf{1}_{\\{N(t+h)-N(t)\geq
2\\}}\right].$

We analyze the behavior of each of these terms as $h\to 0$ individually.

Consider first (8.2). We can rewrite this term as

$\displaystyle\mathrm{e}^{\theta
h+\int_{t}^{t+h}F_{j,j}(s)~{}\mathrm{d}s}\mathbf{E}\left[\left(\mathrm{e}^{\theta
t}\prod_{k=1}^{N(t)}\frac{F_{J_{k}}(\tau_{k})}{G_{J_{k}}}\right)\mathrm{e}^{\int_{0}^{t}F_{M^{i}(s),M^{i}(s)}(s)~{}\mathrm{d}s}\mathbf{1}_{\\{M^{i}(t)=j\\}}\mathbf{1}_{\\{N(t+h)-N(t)=0\\}}\right].$

Since $\big{(}M^{i}(s):0\leq s\leq t\big{)}$ is independent of $N(t+h)-N(t)$,
we can rewrite this as

(8.5) $\displaystyle\mathrm{e}^{\theta
h+\int_{t}^{t+h}F_{j,j}(s)~{}\mathrm{d}s}\mathbf{P}\big{[}N(t+h)-N(t)=0\big{]}\mathfrak{O}_{i,j}(t).$

Since $\mathbf{P}\big{[}N(t+h)-N(t)=0\big{]}=\mathrm{e}^{-\theta h}$ and

$\lim_{h\to
0}\frac{\mathrm{e}^{\int_{t}^{t+h}F_{j,j}(s)~{}\mathrm{d}s}-1}{h}=F_{j,j}(t),$

This means that the contribution of (8.2) to the derivative (8.1) is
$\mathfrak{O}_{i,j}(t)F_{j,j}(t).$

We now consider the term (8.3). We can write this term as

(8.6) $\sum_{\ell\neq
j}\mathbf{E}\Bigg{[}\left(\mathrm{e}^{\theta(t+h)}\prod_{k=1}^{N(t+h)}\frac{F_{J_{k}}(\tau_{k})}{G_{J_{k}}}\right)\mathrm{e}^{\int_{0}^{t+h}F_{M^{i}(s),M^{i}(s)}(s)~{}\mathrm{d}s}\\\
\cdot\mathbf{1}_{\\{M^{i}(t)=\ell\\}}\mathbf{1}_{\\{M^{i}(t+h)=j\\}}\mathbf{1}_{\\{N(t+h)-N(t)=1\\}}\Bigg{]}.$

By the Markov property we can write (8.6) as

$\sum_{\ell\neq j}\mathfrak{O}_{i,\ell}(t)\mathbf{E}\left[\mathrm{e}^{\theta
h}\frac{F_{\ell,j}(t+\tau_{1})}{G_{\ell,j}}\mathrm{e}^{\int_{0}^{h}F_{M^{\ell}(s),M^{\ell}(s)}(s)~{}\mathrm{d}s}\mathbf{1}_{\\{M^{\ell}(h)=j\\}}\mathbf{1}_{\\{N(h)=1\\}}\right].$

Therefore, if we can prove that

(8.7) $\displaystyle\lim_{h\to
0}\frac{1}{h}\mathbf{E}\left[\mathrm{e}^{\int_{0}^{h}F_{M^{\ell}(s),M^{\ell}(s)}(s)~{}\mathrm{d}s+\theta
h}\frac{F_{\ell,j}(t+\tau_{1})}{G_{\ell,j}}\mathbf{1}_{\\{M^{\ell}(h)=j\\}}\mathbf{1}_{\\{N(h)=1\\}}\right]=F_{\ell,j}(t),$

then we can conclude that the contribution of (8.3) to the derivative (8.1) is

$\sum_{\ell\neq j}\mathfrak{O}^{F}_{i,\ell}(t)F_{\ell,j}(t).$

Given that $\mathbf{P}\big{[}N(h)=1\big{]}=\theta h\mathrm{e}^{-\theta h}$, we
can write the limit in (8.7) as

$\lim_{h\to
0}\mathbf{E}\left[\mathrm{e}^{\int_{0}^{h}F_{M^{\ell}(s),M^{\ell}(s)}(s)~{}\mathrm{d}s}\frac{F_{\ell,j}(t+\tau_{1})}{P_{\ell,j}}\mathbf{1}_{\\{M^{\ell}(h)=j\\}}~{}\Bigg{|}~{}N(h)=1\right],$

where we recall that $G_{\ell,j}=\theta P_{\ell,j}$ for $\ell\neq j$.
Conditional on the event $\\{N(h)=1\\}$, the jump time $\tau_{1}$ is uniform
on $[0,h]$. Thus, since the jumps in $M$ are independent of the jump times and

$\mathbf{P}[M(h)=j|N(h)=1,~{}M(0)=\ell]=P_{\ell,j},$

we can write the above limit as

$\lim_{h\to
0}\frac{1}{h}\int_{0}^{h}\mathrm{e}^{\int_{0}^{T}F_{\ell,\ell}(s)~{}\mathrm{d}s+\int_{T}^{h}F_{j,j}(s)~{}\mathrm{d}s}F_{\ell,j}(t+T)~{}\mathrm{d}T.$

The fact that this limit equals $F_{\ell,t}(t)$ can now be seen from

$\lim_{h\to
0}\sup_{T\in[0,h]}\left|\mathrm{e}^{\int_{0}^{T}F_{\ell,\ell}(s)~{}\mathrm{d}s+\int_{T}^{h}F_{j,j}(s)~{}\mathrm{d}s}F_{\ell,j}(t+T)-F_{\ell,t}(t)\right|=0,$

which itself is a consequence of the continuity of $F$’s entries.

It now only remains to ensure that (8.4) is of order $o(h)$ as $h\to 0$. By
Hölder’s inequality and the Markov property, we can bound the modulus of (8.4)
by

$\mathrm{e}^{\theta(t+h)}\mathbf{E}^{i}\left[\prod_{k=1}^{N(t+h)}\left|\frac{F_{J_{k}}(\tau_{k})}{G_{J_{k}}}\right|^{2p}\right]^{1/2p}\mathbf{E}\left[\mathrm{e}^{2p\int_{0}^{t+h}F_{M^{i}(s),M^{i}(s)}(s)~{}\mathrm{d}s}\right]^{1/2p}\mathbf{P}[N(h)\geq
2]^{1/q}$

for any $p,q>1$ such that $1/p+1/q=1$. By continuity of $F$, it is clear that
the two expectations above can be bounded uniformly in $h\in(0,1)$ for any
choice of $p>1$. Thus, the result follows by taking $q<2$, in which case

$\mathbf{P}[N(h)\geq 2]^{1/q}=O(h^{2/q})=o(h),\qquad h\to 0.$

##  9\. Proof of Corollary 3.4

(3.1) is an immediate consequence of Proposition 1.9 due to the continuity of
$q(t,\cdot)$. As for (3.2), we begin by writing the squared modulus using
(3.1), which yields

(9.1)
$\left|\overleftarrow{\mathcal{T}}\left\\{\mathrm{e}^{-\mathsf{i}\int_{0}^{t}\mathcal{H}(s)~{}\mathrm{d}s}\right\\}_{x,y}\right|^{2}=\overleftarrow{\mathcal{T}}\left\\{\mathrm{e}^{-\mathsf{i}\int_{0}^{t}\mathcal{H}(s)~{}\mathrm{d}s}\right\\}_{x,y}\overleftarrow{\mathcal{T}}\left\\{\mathrm{e}^{-\mathsf{i}\int_{0}^{t}\mathcal{H}(s)~{}\mathrm{d}s}\right\\}_{x,y}^{*}\\\
=\mathbf{E}_{w}\left[\mathrm{e}^{2dt(1-\mathsf{i})}\mathsf{i}^{N(t)}\mathrm{e}^{-\mathsf{i}\int_{0}^{t}q(s,w^{x}(s))~{}\mathrm{d}s}\mathbf{1}_{\\{w^{x}(t)=y\\}}\right]\mathbf{E}_{w}\left[\mathrm{e}^{2dt(1+\mathsf{i})}\mathsf{i}^{-N(t)}\mathrm{e}^{\mathsf{i}\int_{0}^{t}q(s,w^{x}(s))~{}\mathrm{d}s}\mathbf{1}_{\\{w^{x}(t)=y\\}}\right].$

Since the random variables inside these expectations are bounded, we can apply
Fubini’s theorem. In particular, we can combine the two expectations above
into a single one that involves two independent copies $w^{x}$ and
$\overline{w}^{x}$:

(9.2)
$\displaystyle\mathbf{E}\left[\mathrm{e}^{4dt}\mathsf{i}^{N(t)-\overline{N}(t)}\mathrm{e}^{-\mathsf{i}\int_{0}^{t}q(s,w^{x}(s))-q(s,\overline{w}^{x}(s))~{}\mathrm{d}s}\mathbf{1}_{\\{w^{x}(t)=y=\overline{w}^{x}(t)\\}}\right].$

Since $q$ is a centered Gaussian process, if we condition on a realization of
$w$,

$\int_{0}^{t}q\big{(}s,w(s)\big{)}-q\big{(}s,\overline{w}(s)\big{)}~{}\mathrm{d}s$

is Gaussian with mean zero and variance

$\mathbf{E}\left[\left(\int_{0}^{t}q\big{(}s,w(s)\big{)}-q\big{(}s,\overline{w}(s)\big{)}~{}\mathrm{d}s\right)^{2}\right]=\int_{[0,t]^{2}}\mathbf{E}\Big{[}q\big{(}s_{1},w(s_{1})\big{)}q\big{(}s_{2},w(s_{2})\big{)}\Big{]}\\\
+\mathbf{E}\Big{[}q\big{(}s_{1},\overline{w}(s_{1})\big{)}q\big{(}s_{2},\overline{w}(s_{2})\big{)}\Big{]}-2\mathbf{E}\Big{[}q\big{(}s_{1},w(s_{1})\big{)}q\big{(}s_{2},\overline{w}(s_{2})\big{)}\Big{]}~{}\mathrm{d}s\mathrm{d}\overline{s}=\chi_{t}^{\gamma,\varrho}(w,\overline{w}),$

where the last equality follows from the covariance structure of $q(t,x)$ in
Assumption 3.1. Thus, by Fubini’s theorem and (9.1) and (9.2), we can write

(9.3)
$\displaystyle\mathbf{E}\left[\left|\overleftarrow{\mathcal{T}}\left\\{\mathrm{e}^{-\mathsf{i}\int_{0}^{t}\mathcal{H}(s)~{}\mathrm{d}s}\right\\}_{x,y}\right|^{2}\right]=\mathbf{E}\left[\mathrm{e}^{4dt-\frac{1}{2}\chi_{t}^{\gamma,\varrho}(w,\overline{w})}\mathsf{i}^{N(t)-\overline{N}(t)}\mathbf{1}_{\\{w^{x}(t)=y=\overline{w}^{x}(t)\\}}\right].$

By the law of total expectation,

(9.4)
$\displaystyle\mathbf{E}\left[\mathrm{e}^{4dt-\frac{1}{2}\chi_{t}^{\gamma,\varrho}(w,\overline{w})}\mathsf{i}^{N(t)-\overline{N}(t)}\mathbf{1}_{\\{w^{x}(t)=y=\overline{w}^{x}(t)\\}}\right]=\sum_{m,\overline{m}=0}^{\infty}\mathsf{i}^{m-\overline{m}}F(m,\overline{m}),$

where

$F(m,\overline{m})=\mathbf{E}\left[\mathrm{e}^{4dt-\frac{1}{2}\chi_{t}^{\gamma,\varrho}(w,\overline{w})}\mathbf{1}_{\\{w^{x}(t)=y=\overline{w}^{x}(t)\\}}\bigg{|}N(t)=m,~{}\overline{N}(t)=\overline{m}\right].$

Note that $F(m,\overline{m})\in\mathbb{R}$ and
$F(m,\overline{m})=F(\overline{m},m)$. Therefore,
$\sum_{m,\overline{m}=0}^{\infty}\mathsf{i}^{m-\overline{m}}F(m,\overline{m})$
is equal to its conjugate
$\sum_{m,\overline{m}=0}^{\infty}\mathsf{i}^{\overline{m}-m}F(m,\overline{m})=\sum_{m,\overline{m}=0}^{\infty}\mathsf{i}^{\overline{m}-m}F(\overline{m},m)$,
hence real. Since

$\Re\big{(}\mathsf{i}^{m-\overline{m}}F(m,\overline{m})\big{)}=\begin{cases}F(m,\overline{m})&\text{if
}m-\overline{m}\equiv 0\text{ mod }4,\\\ -F(m,\overline{m})&\text{if
}m-\overline{m}\equiv 2\text{ mod }4,\\\ 0&\text{otherwise},\end{cases}$

we therefore obtain (3.2) from (9.3) and (9.4).

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##  Appendix A Illustrations

Figure 1. Contour plot of $|\Psi(t,x)|^{2}$ for a free particle (i.e.,
$\mathcal{H}=-\Delta$) on the 1-D periodic lattice $\\{1,\ldots,40\\}$ with
$\Psi_{0}(x)=\mathbf{1}_{\\{x=20\\}}$. (Yellow$=$large; blue$=$small.) This
clearly shows a diffusion effect in the free particle’s probability
amplitudes. Figure 2. Same as Figure 1, except that we replaced the
Hamiltonian by $\mathcal{H}=-\Delta+q(x)$, with $q(x)$ i.i.d. centered
Gaussians. Here we notice the distinct lack of diffusion; hence localization
occurs. Figure 3. Same as Figure 2, except that we replaced the Hamiltonian by
$\mathcal{H}(t)=-\Delta+q(t,x)$, where $q(t,x)$ is a time-dependent Gaussian
potential ($x\mapsto q(t,x)$ are i.i.d. Gaussians for all fixed $t$, and
$t\mapsto q(t,x)$ is a continuous Gaussian process with a smooth and compactly
supported covariance function
$\mathbf{E}\big{[}q(t,x)q(t,x)\big{]}=\varrho(t-s)$ for all fixed $x$). Once
again the particle appears to be lozalized. Figure 4. This is a realization
$A=(U,Z)$. The colors represent the value of the process $U$ (the jump times
therein are represented by black circles). The fact that $Z$’s path has a
discontinuity comes from the fact that this particular figure is a realization
of the concatenated path conditioning
$A^{\boldsymbol{a},\boldsymbol{b}}_{\boldsymbol{t}}$ with
$\boldsymbol{a}=\big{(}(1,1),(3,0)\big{)}$,
$\boldsymbol{b}=\big{(}(1,1),(3,0)\big{)}$, and $\boldsymbol{t}=(1,1)$ (i.e.,
the first path goes from $(i,x)=(1,1)$ to itself in time $1$, and the second
path goes from $(i,x)=(3,0)$ to itself in time $1$). The result is thus a
colored path on the time inverval $[0,2]$ with a jump in the middle. Figure 5.
In this figure, the arrows represent the first six jumps of $U$ by time $t$,
namely, $(1,2),(2,3),(3,1),(1,2),(2,3),(3,1)$. The partition above the arrows
serve to highlight which jumps were paired by the perfect matching
$p=\big{\\{}\\{1,4\\},\\{2,5\\},\\{3,6\\}\big{\\}}$. With this in hand, we
conclude that this particular combination of $(p,U)$ is such that
$\mathfrak{C}_{t}(p,U)\neq 0$ when $\mathbb{F}=\mathbb{R},\mathbb{H}$, since
$J_{1}=J_{4}$, $J_{2}=J_{5}$, and $J_{3}=J_{6}$. However,
$\mathfrak{C}_{t}(p,U)=0$ when $\mathbb{F}=\mathbb{C}$ because paired jumps
are not reversed. Figure 6. In this figure, the six jumps by time $t$ are
$(1,2),(2,1),(1,3),(3,1),(1,2),(2,1)$, and
$p=\big{\\{}\\{1,6\\},\\{2,5\\},\\{3,4\\}\big{\\}}$. In this case,
$\mathfrak{C}_{t}(p,U)\neq 0$ for all
$\mathbb{F}=\mathbb{R},\mathbb{C},\mathbb{H}$, since every paired jumps are
reversed (i.e., $J_{1}=J_{6}^{*}$, $J_{2}=J_{5}^{*}$, and $J_{3}=J_{4}^{*}$).
Figure 7. In this figure, the black arrows are the steps $J_{k}$ until $N(t)$,
the blue arrows are the steps of a binary sequence
$m\in\mathcal{B}^{0,0}_{N(t)}$, and the partition above all that represents
the pairs given by $p$. In order to help decipher whether this choice of $m$
contributes to $\mathfrak{D}_{t}(p,U)$, the matched pairs of steps such that
$J_{\ell_{1}}=J_{\ell_{2}}$ are colored in green, and the matched pairs such
that $J_{\ell_{1}}=J_{\ell_{2}}^{*}$ are colored in red. In this case, $m$
does not respect $(p,J)$, and thus it does not contribute to
$\mathfrak{D}_{t}(p,U)$. (E.g., $J_{4}$ and $J_{5}$ are matched and in
opposite directions, but the corresponding steps in the binary sequence,
$\big{(}(0,0),(0,1)\big{)}$, are not among those listed in Definition
2.17-(3.2)) Figure 8. In this figure, $m$ contributes to
$\mathfrak{D}_{t}(p,U)$, since every matched pairs of jumps
$J_{\ell_{1}},J_{\ell_{2}}$ are accompanied by steps in the binary sequence
that satisfy Definition 2.17-(3.1) and -(3.2). Moreover, there is one flip
that occurs in this sequence: $J_{1}=J_{6}$ and
$\big{(}(m_{0},m_{1}),(m_{5},m_{6})\big{)}=\big{(}(0,1),(1,0)\big{)}$; hence
$f(m,p,J)=1$. Figure 9. This is another example where $m$ contributes to
$\mathfrak{D}_{t}(p,U)$. Moreover, in this case we have $f(m,p,J)=2$; the two
flips occur at the pairs $J_{1},J_{5}$ and $J_{2},J_{6}$. Figure 10. In this
figure, the colored path represents a realization of $(\widehat{U},Z)$ with
$Z(0)=1=Z(1)$. The circles along the path represent the jump times
$\widehat{\tau}_{k}$, and the different colors represent the values of the
process $\widehat{U}(t)$. The jumps $J_{k}$ and the random matching
$\widehat{\pi}(P)$ (the latter of which is represented by the partition above
$Z$’s path) are the same as $J_{k}$ and $p$ in Figure 6. In this illustration,
we observe that $Z(\widehat{\tau}_{\ell_{1}})=Z(\widehat{\tau}_{\ell_{2}})$
for every matched pair $\\{\ell_{1},\ell_{2}\\}\in\widehat{\pi}(P)$.