corresponding to $v_{1}$ and the $\beta$-th child of the branching node
corresponding to $v_{2}$ are paired, if and only if $v_{1}$ and $v_{2}$ are
connected by a bond $\ell$ such that the codes of $(v_{1},\ell)$ and
$(v_{2},\ell)$ are $\alpha$ and $\beta$ respectively.

Finally, note that a node $\mathfrak{n}$ is a root if and only if it is not a
child of any other node, so we can uniquely identify the roots of the trees.
Permuting these $2R$ roots leads to at most $(CR)!$ choices, and once a
permutation is fixed, the garden $\mathcal{G}$ will also be fixed as the
structure of each tree, as well as the leaf pairing structure, has been fixed
as above. This gives at most $(CR)!C^{m}$ possible choices for $\mathcal{G}$.
Note that, if one of the trees in $\mathcal{G}$ is trivial, then the
reconstruction will be slightly different, but this does affect the result. ∎

######  Definition 6.5 (Definition 9.7 in [12]).

We define the _type I_ and _type II (molecular) chains_ in a molecule
$\mathbb{M}$, as in Figure 7. Note that type I chains are formed by double
bonds, and type II chains are formed by double bonds and pairs of single
bonds. For type I chains, we require that the two bonds in any double bond
have opposite directions. For type II chains, we require that any pair of
single bonds have opposite directions, see Figure 7.

Figure 7. The two types of molecular chains. For type II, the single bonds of
the same color are paired single bonds, and must have opposite directions.

Given a molecule $\mathbb{M}$, the main subject of this section is the
following counting problem associated with $\mathbb{M}$, similar to [12].

######  Definition 6.6 (Definition 9.8 in [12]).

Given a molecule $\mathbb{M}$ and a set $S$ of atoms. Suppose we fix (i)
$a_{\ell}\in\mathbb{Z}_{L}^{d}$ for each bond $\ell\in\mathbb{M}$, (ii)
$c_{v}\in\mathbb{Z}_{L}^{d}$ for each (non-isolated, same below) atom
$v\in\mathbb{M}$, assuming $c_{v}=0$ if $v$ has degree 4, (iii)
$\Gamma_{v}\in\mathbb{R}$ for each atom $v$, and (iv)
$f_{v}\in\mathbb{Z}_{L}^{d}$ for each $v\in S$ with $d(v)<4$. Define
$\mathfrak{D}(\mathbb{M})$ to be the set of vectors
$k[\mathbb{M}]:=(k_{\ell})_{\ell\in\mathbb{M}}$, such that each
$k_{\ell}\in\mathbb{Z}_{L}^{d}$ and $|k_{\ell}-a_{\ell}|\leq 1$, and

$\sum_{\ell\sim v}\zeta_{v,\ell}k_{\ell}=c_{v},\quad\bigg{|}\sum_{\ell\sim
v}\zeta_{v,\ell}|k_{\ell}|_{\beta}^{2}-\Gamma_{v}\bigg{|}\leq\delta^{-1}L^{-2}$
(6.1)

for each atom $v$. Here in (6.1) the sum is taken over all bonds $\ell\sim v$,
and $\zeta_{v,\ell}$ equals $1$ if $\ell$ is outgoing from $v$, and equals
$-1$ otherwise. We also require that (a) the values of $k_{\ell}$ for
different $\ell\sim v$ are all equal given each $v\in S$, and this value
equals $f_{v}$ if also $d(v)<4$, and (b) for any $v\not\in S$ and any bonds
$\ell_{1},\ell_{2}\sim v$ of opposite directions (viewing from $v$), we have
$k_{\ell_{1}}\neq k_{\ell_{2}}$. Note that this actually makes $\mathfrak{D}$
depending on $S$, but we will omit this dependence for simplicity. We say an
atom $v$ is _degenerate_ if $v\in S$, and is _tame_ if moreover $d(v)<4$.

In addition, we may add some extra conditions to the definition of
$\mathfrak{D}(\mathbb{M})$. These conditions are independent of the
parameters, and have the form of (combinations of)
$(k_{\ell_{1}}-k_{\ell_{2}}\in E)$ for some bonds
$\ell_{1},\ell_{2}\in\mathbb{M}$ and fixed subsets
$E\subset\mathbb{Z}_{L}^{d}$. Let $\mathtt{Ext}$ be the set of these extra
conditions, and denote the corresponding set of vectors $k[\mathbb{M}]$ be
$\mathfrak{D}(\mathbb{M},\mathtt{Ext})$. We are interested in the quantities
$\sup\\#\mathfrak{D}(\mathbb{M},\mathtt{Ext})$, where the supremum is taken
over all possible choices of parameters $(a_{\ell},c_{v},\Gamma_{v},f_{v})$.

######  Remark 6.7.

The vectors $k[\mathbb{M}]$ will come from decorations of the garden
$\mathcal{G}$ from which $\mathbb{M}$ is obtained. In fact, if
$k[\mathcal{G}]$ is a $(k_{1},\cdots,k_{2R})$-decoration of $\mathcal{G}$,
then it uniquely corresponds to a vector $k[\mathbb{M}]$. Let $v\in\mathbb{M}$
be an atom corresponding to a branching node $\mathfrak{n}\in\mathcal{G}$.
Then $d(v)=4$ unless $\mathfrak{n}$ is the root of some $\mathcal{T}_{j}$, or
some other $\mathcal{T}_{i}$ is a trivial tree paired with a child of
$\mathfrak{n}$ (there may be more than one such $i$).

It is easy to check, using Definitions 3.3 and 6.6, that the followings hold.
If $d(v)=4$ then $\sum_{\ell\sim v}\zeta_{v,\ell}k_{\ell}=0$, and
$\sum_{\ell\sim
v}\zeta_{v,\ell}|k_{\ell}|_{\beta}^{2}=-\zeta_{\mathfrak{n}}\Omega_{\mathfrak{n}}$.
If $d(v)<4$, then the right hand sides of the above equations should be
corrected by suitable algebraic sums of $k_{j}$ and (or) $k_{i}$, and
$|k_{j}|_{\beta}^{2}$ and (or) $|k_{i}|_{\beta}^{2}$, where $j$ and $i$ are
associated with $\mathfrak{n}$ as stated above. Note that all these $k_{j}$
and $k_{i}$ are fixed when considering the decoration $k[\mathcal{G}]$.
Moreover if
$(k_{\mathfrak{n}_{1}},k_{\mathfrak{n}_{2}},k_{\mathfrak{n}_{3}})\in\mathfrak{S}$,
then either the values of $k_{\ell}$ for different $\ell\sim v$ are all equal
(and this value equals $k_{j}$ if $d(v)<4$ where $j$ is as above), or for any
bonds $\ell_{1},\ell_{2}\sim v$ of opposite directions we have
$k_{\ell_{1}}\neq k_{\ell_{2}}$. Note that a degenerate atom corresponds
exactly to a branching node $\mathfrak{n}$ for which
$\epsilon_{k_{\mathfrak{n}_{1}}k_{\mathfrak{n}_{2}}k_{\mathfrak{n}_{3}}}=-1$.

######  Proposition 6.8.

Let $\mathbb{M}$ be the molecule associated with a mixed garden $\mathcal{G}$
of width $2R$ and scale $m$, where $R,m\leq(\log L)^{20}$. Suppose also that
$\mathbb{M}$ _does not contain any triple bond_. Then,
$\mathfrak{D}(\mathbb{M})$ is the union of at most $C^{m}$ subsets. Each
subset has the form $\mathfrak{D}(\mathbb{M},\mathtt{Ext})$, and there exists
$0\leq r\leq m$, and a collection of at most $C(r+R)$ molecular chains of
either type I or type II in $\mathbb{M}$, such that (i) the number of atoms
not in one of these chains is at most $C(r+R)$, and (ii) for any type II chain
in the collection and any two paired single bonds $(\ell_{1},\ell_{2})$ in
this chain (see Figure 7), the set $\mathtt{Ext}$ includes the condition
$(k_{\ell_{1}}=k_{\ell_{2}})$. Moreover we have the estimate that

$\sup\\#\mathfrak{D}(\mathbb{M},\mathtt{Ext})\leq(C^{+})^{m}\delta^{-\frac{m+m_{1}}{2}}L^{(d-1)(m-R/2)-2\nu
r},$ (6.2)

where $m_{1}$ is the number of atoms in the union of type I chains.

######  Remark 6.9.

In view of Remark 5.5, in Definition 6.6 we may also fix some set $S^{*}$ of
atoms such that neither (a) nor (b) is required for $v\in S^{*}$, but we are
allowed to multiply the left hand side of (6.2) by $L^{-40d\cdot|S^{*}|}$. In
this way we can restate Proposition 6.8 appropriately, and the new result can
be easily proved with little difference in the arguments, due to the large
power gains. For simplicity we will not include this in the proof below.

###### Proof of Proposition 6.8.

The proof is basically the same as the proof of Proposition 9.10 in [12]. We
define the same steps as in Section 9.3 of [12], including the good and normal
steps, and apply the same algorithm as in Section 9.4 of [12]. Let the total
number of good steps in the process be $r\geq 0$ (we may assume $r\leq m$ up
to a constant because the total number of steps is at most $O(m)$), then we
may repeat the proofs in Section 9.5 of [12]. The only difference here is the
initial state of the molecule (as $\mathbb{M}$ is obtained from a mixed garden
rather than a couple), but in the current case we still have $V_{<4}+F=O(R)$,
where $V_{<4}$ and $F$ are the number of atoms with degree $<4$ and the number
of connected components and the constant in $O$ depends only on $d$.

Note that in the proof of of Proposition 9.10 in [12], the quantities that are
monitored include $V$, $E$, $F$, $V_{j}$ for $0\leq j\leq 4$ (which is the
number of atoms with degree $j$), $V_{2}^{*}$ (which is the number of degree
$2$ atoms with two single bonds), and $\xi$ (which is the number of “special
bonds” connecting two degree $3$ atoms that have a special form, see
Definition 9.12 of [12]). Since $V_{<4}+F=O(R)$, it is clear that in the
beginning, the value of each of these quantities in the current case is the
same as in [12], up to errors of size $O(R)$. Thus, the same proof as in [12]
yields that $\mathbb{M}$ contains at most $C(R+r)$ type I or II molecular
chains, such that the number of atoms not in one of these chains is at most
$C(R+r)$. Moreover

$\sup\\#\mathfrak{D}(\mathbb{M},\mathtt{Ext})\leq(C^{+})^{m}\delta^{-\kappa}L^{(d-1)\gamma},$

where $\kappa$ and $\gamma$ are calculated retrospectively from the algorithm,
as described in Section 9.2 of [12]. The calculation for $\kappa$ is the same
up to $O(r)$ errors, so we have $\kappa=\frac{m+m_{1}}{2}$ up to errors
$O(R+r)$ which are acceptable. To calculate $\gamma$, note that in [12] we are
actually calculating $\gamma-\chi$, and the same proof yields that
$(d-1)(\gamma-\chi)\leq-2\nu r$ for the initial molecule. Now by Proposition
6.3 we know $\chi\leq m-R/2$, hence

$\sup\\#\mathfrak{D}(\mathbb{M},\mathtt{Ext})\leq(C^{+})^{m}\delta^{-\frac{m+m_{1}}{2}}L^{(d-1)(m-R/2)-2\nu
r},$

as desired. ∎

##  7\. $L^{1}$ coefficient bounds

We now return to the study of the expression (5.13). Let
$\mathcal{G}_{sk}^{\\#}$ and $(r_{0},r_{\mathrm{irr}})$ be as in Section 5.4.
For simplicity, in this section we will write $\mathcal{G}_{sk}^{\\#}$ simply
as $\mathcal{G}$, and the associated sets $(\mathcal{N}_{sk}^{\\#})^{*}$ as
$\mathcal{N}^{*}$ etc. Recall, by (5.15), that the total length of the
irregular chains in $\mathcal{G}$ is at most $C(r_{0}+r_{\mathrm{irr}})$. Let
$\Xi$ be a subset of $\mathcal{N}^{*}$, we may define, as in (5.13), the
function

$\mathcal{U}_{\mathcal{G}}(t,s,\bm{\sigma},\alpha[\mathcal{N}^{*}])=\int_{\widetilde{\mathcal{I}}}\prod_{\mathfrak{n}\in\mathcal{N}^{*}}e^{\pi
i\alpha_{\mathfrak{n}}t_{\mathfrak{n}}}\,\mathrm{d}t_{\mathfrak{n}},$ (7.1)

where $\bm{\sigma}=\sigma[\Xi]\in[0,1]^{\Xi}$, and the domain
$\widetilde{\mathcal{I}}$ is defined as in (4.2), but with the extra
conditions $t_{\mathfrak{n}^{p}}>t_{\mathfrak{n}}+\sigma_{\mathfrak{n}}$ for
$\mathfrak{n}\in\Xi$, where $\mathfrak{n}^{p}$ is the parent of
$\mathfrak{n}$. Then, let $m_{0}^{\prime}$ be the scale of $\mathcal{G}$, we
can write

$(\ref{irrechainsum3})=(C^{+}\delta)^{\frac{m-m_{0}^{\prime}}{2}}\bigg{(}\frac{\delta}{2L^{d-1}}\bigg{)}^{m_{0}^{\prime}}\zeta^{*}(\mathcal{G}){\int_{\mathbb{R}^{\mathcal{N}^{*}}\times\mathbb{R}^{2}}}G(\bm{\lambda})\,\mathrm{d}\bm{\lambda}\int_{[0,1]^{\Xi}}\mathrm{d}\bm{\sigma}\\\
\times\sum_{\mathscr{I}}\epsilon_{\mathscr{I}}\mathcal{U}_{\mathcal{G}}\big{(}t,s,\bm{\sigma},(\delta
L^{2}\zeta_{\mathfrak{n}}\Omega_{\mathfrak{n}}+\lambda_{\mathfrak{n}})_{\mathfrak{n}\in\mathcal{N}^{*}}\big{)}\cdot\mathcal{X}_{\mathrm{tot}}(\bm{\lambda},\bm{\sigma},k[\mathcal{G}]).$
(7.2)

Let $\mathbb{M}$ be the molecule associated with $\mathcal{G}$ as in
Definition 6.2. It is easy to see that $\mathbb{M}$ contains no triple bond,
as triple bonds in $\mathbb{M}$ can only come from $(1,1)$-mini couples and
mini trees (as in Definition 3.4) in $\mathcal{G}$. By the proofs in Section
6, we can introduce at most $C^{m_{0}^{\prime}}$ sets of extra conditions
$\mathtt{Ext}$, such that the summation in $\mathscr{I}=k[\mathcal{G}]$ in
(7.2) can be decomposed into the summations with each of these sets of extra
conditions imposed on $k[\mathcal{G}]$. Moreover, for each choice of
$\mathtt{Ext}$ there is $1\leq r_{1}\leq n_{0}^{\prime}$ such that the
conclusion of Proposition 6.8, including (6.2), holds true (with $r$ replaced
by $r_{1}$).

Notice that a type I chain in $\mathbb{M}$ can _only_ be obtained from either
one irregular chain, or the union of two irregular chains in $\mathcal{G}$;
for couples this can be proved in the same way as in Section 10.1.2 of [12]
(which deals with type II chains), and the same proof works also for gardens.
Therefore, the total length $p$ of type I chains in $\mathbb{M}$ is bounded by
the total length of irregular chains in $\mathcal{G}$, which is at most
$C(r_{0}+r_{\mathrm{irr}})$. However, each irregular chain in $\mathcal{G}$
also corresponds to a type I chain in the base molecule, so
$r_{\mathrm{irr}}\leq C(r_{1}+R)$, hence $p\leq C(r+R)$, where
$r=r_{0}+r_{1}$. This means the number of atoms in $\mathbb{M}$ that are not
in one of the (at most $C(r+R)$) type II chains is at most $C(r+R)$.

Now, suppose $\mathfrak{n}$ and $\mathfrak{n}^{\prime}$ are two branching
nodes in $\mathcal{G}$ which correspond to two atoms in $\mathbb{M}$ _that are
connected by a double bond in a type II chain_ , then we must have
$\zeta_{\mathfrak{n}^{\prime}}\Omega_{\mathfrak{n}^{\prime}}=-\zeta_{\mathfrak{n}}\Omega_{\mathfrak{n}}$
under the extra conditions in $\mathtt{Ext}$, see Remark 6.7. In fact we will
restrict $\\{\mathfrak{n},\mathfrak{n}^{\prime}\\}$ to the _interior_ of this
type II chain by omitting $5$ pairs of atoms at both ends of the chain, in the
same way as in Definition 5.4. Then, we make such
$\\{\mathfrak{n},\mathfrak{n}^{\prime}\\}$ a pair, and choose one node from
each such pair to form a set $\widetilde{\mathcal{N}}^{ch}$. If it happens
that one of $\\{\mathfrak{n},\mathfrak{n}^{\prime}\\}$ is a parent of the
other, we assume the parent belongs to $\widetilde{\mathcal{N}}^{ch}$. Let
$\mathcal{N}^{rm}$ be the set of branching nodes not in these pairs, and
define
$\widetilde{\mathcal{N}}=\widetilde{\mathcal{N}}^{ch}\cup\mathcal{N}^{rm}$.

We will be interested in estimates on the function $\mathcal{U}_{\mathcal{G}}$
in (7.1) where $\alpha_{\mathfrak{n}}=\delta
L^{2}\zeta_{\mathfrak{n}}\Omega_{\mathfrak{n}}+\lambda_{\mathfrak{n}}$, which
means that
$\alpha_{\mathfrak{n}}+\alpha_{\mathfrak{n}^{\prime}}=\mu_{\mathfrak{n}}$ for
each $\mathfrak{n}\in\widetilde{\mathcal{N}}^{ch}$, where
$\mathfrak{n}^{\prime}$ is the node paired to $\mathfrak{n}$ and
$\mu_{\mathfrak{n}}=\lambda_{\mathfrak{n}}+\lambda_{\mathfrak{n}^{\prime}}$ is
a parameter depending on $\bm{\lambda}$. Under this assumption on
$\alpha_{\mathfrak{n}}$, we can write

$\mathcal{U}_{\mathcal{G}}(t,s,\bm{\sigma},\alpha[\mathcal{N}^{*}])=\mathcal{V}_{\mathcal{G}}(t,s,\bm{\sigma},\alpha[\widetilde{\mathcal{N}}])$
(7.3)

for some function $\mathcal{V}_{\mathcal{G}}$. This function actually depends
also on the parameters $\mu_{\mathfrak{n}}$ for
$\mathfrak{n}\in\widetilde{\mathcal{N}}^{ch}$, but we will omit this for
notational convenience. We then have prove the following:

######  Proposition 7.1.

Suppose $\mathcal{G}$ has scale $m_{0}^{\prime}$. For each
$\mathfrak{n}\in\widetilde{\mathcal{N}}$, suppose
$S_{\mathfrak{n}}\subset\mathbb{Z}$ and $\\#S_{\mathfrak{n}}\leq L^{10d}$.
Then, uniformly in $(t,s)$, in the choices of
$(S_{\mathfrak{n}})_{\mathfrak{n}\in\widetilde{\mathcal{N}}}$, and in the
parameters
$(\mu_{\mathfrak{n}})_{\mathfrak{n}\in\widetilde{\mathcal{N}}^{ch}}$, we have

$\delta^{m_{0}^{\prime}/4}\cdot\sum_{(m_{\mathfrak{n}}):m_{\mathfrak{n}}\in
S_{\mathfrak{n}}}\sup_{(\alpha_{\mathfrak{n}}):|\alpha_{\mathfrak{n}}-m_{\mathfrak{n}}|\leq
1}\sup_{\bm{\sigma}}\big{|}\mathcal{V}_{\mathcal{G}}(t,s,\bm{\sigma},\alpha[\widetilde{\mathcal{N}}])\big{|}\leq(C^{+})^{m_{0}^{\prime}}L^{C(r+R)\sqrt{\delta}}(\log
L)^{C(r+R)},$ (7.4)

where $r=r_{0}+r_{1}$.

###### Proof.

The proof is exactly the same as the proof of Proposition 10.1 in [12], except
that all the “couples” should be replaced by “gardens”. The reason is that,
the proof in [12] goes by induction; moreover in each inductive step we remove
either two branching nodes or one chain containing modules A and B (see
Sections 10.1–10.2 of [12]). In either case this step involves at most two
trees and the other trees are not affected, so the proof is the same for
couples and for general gardens.

In the proof in [12] we have also introduced the simpler structures (for the
purpose of induction) of unsigned couples and double-trees, which are
naturally replaced by unsigned gardens and multi-trees (the collection of $2R$
trees with some branching nodes paired, compared to two trees in [12]). The
rest of the proof is exactly the same. Note also that the exponents $Cr$ in
Proposition 10.1 of [12] are replaced by $C(r+R)$ as the number of type II
chains in $\mathbb{M}$, as well as the number of atoms not in one of these
chains, is now $C(r+R)$ instead of $Cr$ due to Proposition 6.8. ∎

##  8\. Proof of Theorem 1.3

In this section we prove Propositions 4.7 and 4.8, which completes the proof
of Theorem 1.3.

###  8.1. Proof of Proposition 4.7

Note that, if $\mathcal{G}$ and $\mathcal{G}^{\prime}$ are congruent in the
sense of Definition 5.4, then they have the same width, same signature, and
the same scale for each of their component trees; moreover
$\mathcal{G}^{\prime}$ is mixed if and only if $\mathcal{G}$ is mixed. Thus,
the sum in (4.4) can be decomposed into different terms, where each term has
the form (5.4) for one congruence class $\mathscr{F}$.

For any fixed $\mathscr{F}$, consider (5.4), which then equals (5.13) and
(7.2). Note that in (7.2) the $\mathcal{G}$ actually means
$\mathcal{G}_{sk}^{\\#}$ by our notation. Using the decay factors in (5.14) we
can restrict to the subset where $|k_{\mathfrak{l}}-a_{\mathfrak{l}}|\leq
1\,(\forall\mathfrak{l}\in(\mathcal{L}_{sk}^{\\#})^{*})$ for some fixed
parameters $(a_{\mathfrak{l}})$, with summability in $(a_{\mathfrak{l}})$
guaranteed. Using the first inequality in (5.12), we may also fix the value of
$\bm{\lambda}$ (and hence $\mu_{\mathfrak{n}}$).

As in Section 6, by decomposing into at most $C^{m_{0}^{\prime}}$ terms (where
$m_{0}^{\prime}$ is the scale of $\mathcal{G}_{sk}^{\\#}$), we can add the set
of extra conditions $\mathtt{Ext}$, which also defines the sets
$\widetilde{\mathcal{N}}$ (as in Section 7), etc., and the value $r_{1}\geq
1$. Let $r=r_{0}+r_{1}$ as above, then thanks to $\mathtt{Ext}$, we can use
(7.3) to reduce $\mathcal{U}_{\mathcal{G}_{sk}^{\\#}}$ to
$\mathcal{V}_{\mathcal{G}_{sk}^{\\#}}$. Moreover, for each
$\mathfrak{n}\in\widetilde{\mathcal{N}}$, the value $\delta
L^{2}\zeta_{\mathfrak{n}}\Omega_{\mathfrak{n}}+\lambda_{\mathfrak{n}}$ belongs
to some subset of $\mathbb{R}$ of cardinality at most $L^{3d}$, as
$k[\mathcal{G}_{sk}^{\\#}]$ varies (this is because each $k_{\mathfrak{n}}$
belongs to a ball of radius at most $n\leq(\log L)^{20}$ under our
assumptions). In particular the value $m_{\mathfrak{n}}=\lfloor\delta
L^{2}\zeta_{\mathfrak{n}}\Omega_{\mathfrak{n}}+\lambda_{\mathfrak{n}}\rfloor$
belongs to a set $S_{\mathfrak{n}}\subset\mathbb{Z}$ with cardinality at most
$L^{3d}$, for all possible choices of $k[\mathcal{G}_{sk}^{\\#}]$.

To estimate (7.2) with $\bm{\lambda}$ fixed, we first integrate in
$\bm{\sigma}$. Using (5.14), we can estimate (7.2) using

$\sum_{\mathscr{I}_{sk}^{\\#}}|\epsilon_{\mathscr{I}_{sk}^{\\#}}|\cdot\sup_{\bm{\sigma}}\big{|}\mathcal{V}_{\mathcal{G}_{sk}^{\\#}}\big{(}t,s,\bm{\sigma},(\delta
L^{2}\zeta_{\mathfrak{n}}\Omega_{\mathfrak{n}}+\lambda_{\mathfrak{n}})_{\mathfrak{n}\in\widetilde{\mathcal{N}}}\big{)}\big{|},$
(8.1)

where $\mathscr{I}_{sk}^{\\#}=k[\mathcal{G}_{sk}^{\\#}]$ is a $k$-decoration
of $\mathcal{G}_{sk}^{\\#}$ (we also have additional factors that will be
collected at the end). We next fix the values of $m_{\mathfrak{n}}\in
S_{\mathfrak{n}}$ for each $\mathfrak{n}$; note that then

$\sup_{\bm{\sigma}}\big{|}\mathcal{V}_{\mathcal{G}_{sk}^{\\#}}\big{(}t,s,\bm{\sigma},(\delta
L^{2}\zeta_{\mathfrak{n}}\Omega_{\mathfrak{n}}+\lambda_{\mathfrak{n}})_{\mathfrak{n}\in\widetilde{\mathcal{N}}}\big{)}\big{|}\leq\sup_{(\alpha_{\mathfrak{n}}):|\alpha_{\mathfrak{n}}-m_{\mathfrak{n}}|\leq
1}\sup_{\bm{\sigma}}\big{|}\mathcal{V}_{\mathcal{G}_{sk}^{\\#}}(t,s,\bm{\sigma},\alpha[\widetilde{\mathcal{N}}])\big{|}$

by definition, so if we use (7.4) to sum over $(m_{\mathfrak{n}})$ in the end,
we can further estimate (8.1) using

$\sum_{\mathscr{I}_{sk}^{\\#}}|\epsilon_{\mathscr{I}_{sk}^{\\#}}|\cdot\prod_{\mathfrak{l}}\mathbf{1}_{|k_{\mathfrak{l}}-a_{\mathfrak{l}}|\leq
1}\prod_{\mathfrak{n}}\mathbf{1}_{|\Omega_{\mathfrak{n}}-b_{\mathfrak{n}}|\leq\delta^{-1}L^{-2}},$
(8.2)

where $a_{\mathfrak{l}}$ and $b_{\mathfrak{n}}$ are constants, and we also
include the conditions in $\mathtt{Ext}$. Now (8.2) is almost exactly the
counting problem $\mathfrak{D}(\mathbb{M},\mathtt{Ext})$ stated in Definition
6.6, due to Remark 6.7, except that we only assume
$|k_{\mathfrak{l}}-a_{\mathfrak{l}}|\leq 1$ for _leaves_ $\mathfrak{l}$.
However, for any branching node $\mathfrak{n}$ there exists a child
$\mathfrak{n}^{\prime}$ of $\mathfrak{n}$ such that $k_{\mathfrak{n}}\pm
k_{\mathfrak{n}^{\prime}}$ belongs to a fixed ball of radius
$\mu_{\mathfrak{n}}^{\circ}$ (with $\mu_{\mathfrak{n}}^{\circ}$ defined Lemma
3.6), so by using (3.3), one can reduce (8.2) to at most $C^{m_{0}^{\prime}}$
counting problems, each of which having exactly the same form as
$\mathfrak{D}(\mathbb{M},\mathtt{Ext})$ in Definition 6.6. Therefore, (8.2)
can be bounded using Proposition 6.8 (and using Remark 6.9 if necessary).
Collecting all the factors appearing in the above estimates, we get that

$|(\ref{irrechainsum})|\leq(C^{+})^{m}\delta^{(m-m_{0}^{\prime})/2}\delta^{3m_{0}^{\prime}/4}L^{-(d-1)m_{0}^{\prime}}L^{-2\nu
r_{0}}\\\ \times L^{C(r+R)\sqrt{\delta}}(\log
L)^{C(r+R)}\delta^{-(m_{0}^{\prime}+q)/2}L^{(d-1)(m_{0}^{\prime}-R/2)-2\nu
r_{1}},$ (8.3)

which is then bounded by
$(C^{+}\delta^{1/4})^{m}L^{-3\nu(r+R)/2}\delta^{-q/2}$, where $q$ is the total
length of type I chains in the molecule obtained from
$\mathcal{Q}_{sk}^{\\#}$. We know $q\leq C(r+R)$ so $\delta^{-q/2}\leq
L^{\nu(r+R)/2}$, which implies that

$|(\ref{irrechainsum})|\leq(C^{+}\delta^{1/4})^{m}L^{-\nu(r+R)}.$ (8.4)

Finally, suppose we fix $r$, then the molecule associated with
$\mathcal{G}_{sk}^{\\#}$ (see Definition 6.2) is, up to at most $C(r+R)$
remaining atoms, a union of at most $C(r+R)$ type II chains with total length
at most $m_{0}^{\prime}$. This clearly has at most $(C(r+R))!C^{m}$
possibilities. By Proposition 6.4, the number of choices for
$\mathcal{G}_{sk}^{\\#}$ is also at most $(C(r+R))!C^{m}$. To form
$\mathcal{G}_{sk}$ from $\mathcal{G}_{sk}^{\\#}$ one needs to insert at most
$C(r+R)$ irregular chains with total length at most $m$, which also has at
most $C^{m}$ possibilities. Finally, using Corollary 4.6, we see that
$\mathcal{G}$ has at most $(C(r+R))!C^{m}$ choices. The number of choices for
markings, as well as $\mathtt{Ext}$, are also at most $C^{m}$ and can be
accommodated. This means that, if we decompose (4.4) into terms of form (5.4),
then further decompose by markings and/or $\mathtt{Ext}$ etc., then each of
the resulting term has an index $r\geq 0$, such that each term of index $r$ is
bounded by $(C^{+}\delta^{1/4})^{m}L^{-\nu(r+R)}$, see (8.4), and that the
number of terms with index $r$ is at most $(C(r+R))!C^{m}$. Therefore

$(\ref{mixedsum})\leq\sum_{r=0}^{m}(C^{+}\delta^{1/4})^{m}L^{-\nu(r+R)}\cdot(C(r+R))!C^{m}\leq(C^{+}\delta^{1/4})^{m}L^{-\nu
R},$ (8.5)

because in any case $r+R$ is bounded by a power of $\log L$, which is $\ll
L^{\nu}$. This completes the proof of Proposition 4.7.

###  8.2. Proof of Proposition 4.8

The proof is almost identical with the corresponding proofs in [12], which we
briefly present here.

First, by Chebyshev’s inequality, to prove (4.6) it suffices to show that

$\mathbb{E}\big{|}\sup_{k,t}\langle
k\rangle^{9d}(\mathcal{J}_{n})_{k}(t)\big{|}^{2}\leq(C^{+}\sqrt{\delta})^{n}L^{100d}$
(8.6)

and

$\mathbb{E}\big{|}\sup_{k,t}\langle
k\rangle^{9d}\mathcal{R}_{k}(t)\big{|}^{2}\leq(C^{+}\sqrt{\delta})^{N}L^{100d}.$
(8.7)

Note that due to our choice $N=\lfloor(\log L)^{4}\rfloor$ instead of
$N=\lfloor\log L\rfloor$, the proof of (4.6) is conceptually easier than [12]
as we do not need the hypercontractivity property (Lemma A.3 of [12]) or the
higher moment estimates.

Now, to prove (8.6), we argue as in the proof of Proposition 12.1 of [12] (but
with $p$ replaced by $2$), and apply Gagliardo-Nirenberg to bound the left
hand side of (8.6), up to a multiple $L^{10d}$, by

$\sup_{k,t}\langle
k\rangle^{20d}\big{(}\mathbb{E}|(\mathcal{J}_{n})_{k}(t)|^{2}+\mathbb{E}|\partial_{t}(\mathcal{J}_{n})_{k}(t)|^{2}\big{)},$

which is then bounded by $(C^{+}\sqrt{\delta})^{n}L^{40d}$ in the same way as
[12]. In fact, the bound for $\mathbb{E}|(\mathcal{J}_{n})_{k}(t)|^{2}$ is as
in Proposition 2.5 of [12] (again our choice $N=\lfloor(\log L)^{4}\rfloor$
here does not affect the proof), while the bound for
$\mathbb{E}|\partial_{t}(\mathcal{J}_{n})_{k}(t)|^{2}$ is as in (12.4) of
[12], which is proved by similar arguments. This settles (8.6). The proof of
(8.7) is the same, except that $\mathcal{J}_{n}$ is replaced by $\mathcal{R}$
and $n$ is replaced by $N$.

Finally, to prove (4.7), again by Chebyshev’s inequality, it suffices to show
that the kernel $(\mathscr{L}^{n})_{k\ell}^{\zeta}(t,s)$ of the
$\mathbb{R}$-linear operator $\mathscr{L}^{n}$ (with $\zeta\in\\{\pm\\}$
indicating the linear and conjugate linear parts) can be decomposed as

$(\mathscr{L}^{n})_{k\ell}^{\zeta}(t,s)=\sum_{n\leq m\leq
N^{3}}(\mathscr{L}^{n})_{k\ell}^{m,\zeta}(t,s)$

and that for each $n\leq m\leq N^{3}$, the kernel
$(\mathscr{L}^{n})_{k\ell}^{m,\zeta}(t,s)$ satisfies that

$\mathbb{E}\big{|}\sup_{k,\ell}\sup_{0\leq s<t\leq 1}\langle
k-\ell\rangle^{9d}(\mathscr{L}^{n})_{k\ell}^{m,\zeta}(t,s)\big{|}^{2}\leq(C^{+}\sqrt{\delta})^{m}L^{100d}.$
(8.8)

Now the decomposition is provided as in Proposition 11.2 of [12], and (8.8) is
proved as in the proof of Proposition 12.2 of [12] (in particular this proof
does require the hypercontractivity property). Note that in that proof, we
actually further decompose $(\mathscr{L}^{n})_{k\ell}^{m,\zeta}(t,s)$ into
$(\mathscr{L}^{n})_{M,k\ell}^{m,\zeta}(t,s)$ for dyadic $M$, and proves (8.8)
for $(\mathscr{L}^{n})_{M,k\ell}^{m,\zeta}$ with the right hand side involving
a negative power of $M$; see (12.10) of [12]. Both proofs carry over to the
current case with our choice $N=\lfloor(\log L)^{4}\rfloor$ with out any
change, which then proves (4.7) and completes the proof of Proposition 4.8.

##  9\. The non-Gaussian case

In this section we briefly discuss the non-Gaussian case, i.e. Theorem 1.4,
which we prove in Sections 9.1–9.2 (Theorem 1.5 basically follows from it and
is proved separately in Section 9.3). Since much of the proof will be
identical with Theorem 1.3, we will only elaborate on the parts where the
proofs are different.

First, in the Gaussian case our proof yields uniform estimates as long as
$R\leq\log L$ (or $R\leq 2\log L$); here we will make slightly stronger
assumptions $R\leq\log L/(\log\log L)^{2}$. Again we may consider $2R$ at some
places, but it does not affect the result.

Next, using the expansion (3.6), we can reduce the proof of Theorem 1.4 to
analyzing the correlations

$\mathbb{E}\bigg{(}\prod_{j=1}^{2R}(\mathcal{J}_{\mathcal{T}_{j}})_{k_{j}}^{\zeta_{j}}(t)\bigg{)}.$
(9.1)

The rest of the proof, including the treatment of the remainder term $b$, can
be done in the same way as with these correlations, see Section 8.2.
Specifically, the Gaussian hypercontractivity inequality, which is used in
Section 8.2, can be substituted by similar inequalities for the current
density function thanks to the assumed bound on $\mu_{r}\leq(Cr)!$; an
instance of such argument can be found in Lemma 3.1 of [11] which treats the
particular case of the uniform distribution on the unit circle, but the
general case can be treated in the same manner. Therefore, below we will focus
on the study of (9.1).

###  9.1. A substitute for Isserlis’ theorem

The obvious difference in the study of (9.1) in the non-Gaussian case is that
the Isserlis’ theorem is not available. Instead we have the following
substitute:

######  Lemma 9.1.

Recall all the random variables $\eta_{k}$ are i.i.d. with radial law, and
$\mathbb{E}|\eta_{k}|^{2r}=\mu_{r}$ with $\mu_{1}=1$ and $\mu_{r}\leq(Cr)!$
for a positive integer $C$. Then, for any $k_{j}\in\mathbb{Z}_{L}^{d}\,(1\leq
j\leq n)$ and $\zeta_{j}\in\\{\pm\\}\,(1\leq j\leq n)$ we have

$\mathbb{E}\bigg{(}\prod_{j=1}^{n}\eta_{k_{j}}^{\zeta_{j}}\bigg{)}=\sum_{\mathcal{O}}\lambda(\mathcal{O})\cdot\sum_{\mathscr{O}}\prod_{A\in\mathscr{O}}\prod_{j,j^{\prime}\in
A}\mathbf{1}_{k_{j}=k_{j^{\prime}}}.$ (9.2)

Here in (9.2), $\mathcal{O}$ runs over all partitions of $n$ into _even_
positive integers (in particular if $n$ is odd then the right hand side of
(9.2) is zero). For fixed $\mathcal{O}$, the $\mathscr{O}$ runs over all
_over-pairings_ of the set $\\{1,\cdots,n\\}$ subordinate to $\mathcal{O}$ (we
use the notation $\mathscr{O}\models\mathcal{O}$), which are partitions of
$\\{1,\cdots,n\\}$ such that the cardinalities of the subsets exactly form the
partition $\mathcal{O}$, and for each subset $A$ exactly half of the signs
$\zeta_{j}\,(j\in A)$ are $+$ and half are $-$.

The coefficients $\lambda(\mathcal{O})$ are constants depending only on
$\mathcal{O}$ (and $n$), and $\lambda(2,\cdots,2)=1$; in general we have
$\lambda(2,\cdots,2,2a_{1},\cdots,2a_{r})=\lambda(2a_{1},\cdots,2a_{r})$.
Moreover, let $q$ be the sum of the elements in $\mathcal{O}$ that are _at
least $4$_, then we have $|\lambda(\mathcal{O})|\leq C_{1}^{n}n^{C_{1}q}$ for
some constant $C_{1}\gg C$.

###### Proof.

We may assume $n$ is even and half the signs $\zeta_{j}\,(1\leq j\leq n)$ are
$+$ and half are $-$, since otherwise both sides of (9.2) are zero. Denote by
$|\mathcal{O}|$ the number of elements in $\mathcal{O}$ (counted with
multiplicity). For two partitions $\mathcal{O}$ and $\mathcal{O}^{\prime}$, we
write $\mathcal{O}^{\prime}\preceq\mathcal{O}$, if $\mathcal{O}^{\prime}$ can
be formed by further partitioning some elements in $\mathcal{O}$ into even
integers (also define $\succeq$ and $\prec$ etc. accordingly). Similarly for
set partitions $\mathscr{O}$ and $\mathscr{O}^{\prime}$, we write
$\mathscr{O}^{\prime}\preceq\mathscr{O}$ if $\mathscr{O}^{\prime}$ can be
formed by further partitioning some subsets in $\mathscr{O}$ (still keeping
half of the signs $+$ and half $-$ in each subset).

Now, for $\mathcal{O}^{\prime}\preceq\mathcal{O}$, we define
$\xi_{\mathcal{O},\mathcal{O}^{\prime}}$ as follows: given a partition
$\mathscr{O}\models\mathcal{O}$, consider the number of partitions
$\mathscr{O}^{\prime}\preceq\mathscr{O}$ such that
$\mathscr{O}^{\prime}\models\mathcal{O}^{\prime}$. The number of choices for
$\mathscr{O}^{\prime}$ is independent of the choice of $\mathscr{O}$, and we
define it to be $\xi_{\mathcal{O},\mathcal{O}^{\prime}}$. Obviously
$\xi_{\mathcal{O},\mathcal{O}}=1$. We define the coefficients
$\lambda(\mathcal{O})$ for each $\mathcal{O}$, such that they satisfy the
following recurrence relation: first $\lambda(2,\cdots,2)=1$, and for each
$\mathcal{O}$, we have

$\prod_{2b\in\mathcal{O}}\mu_{b}=\sum_{\mathcal{O}^{\prime}\preceq\mathcal{O}}\xi_{\mathcal{O},\mathcal{O}^{\prime}}\cdot\lambda(\mathcal{O}^{\prime}).$
(9.3)

Here the product is taken over all elements $2b$ appearing in $\mathcal{O}$,
counted with multiplicity. Clearly the values of of $\lambda(\mathcal{O})$ for
each $\mathcal{O}$ are uniquely determined by (9.3). To prove
$\lambda(2,\cdots,2,2a_{1},\cdots,2a_{r})=\lambda(2a_{1},\cdots,2a_{r})$, we
simply notice $\mu_{1}=1$ and that if $\mathcal{O}$ contains a certain number
of terms $2$, then any $\mathcal{O}^{\prime}\preceq\mathcal{O}$ must contain
at least the same number of $2$’s. Then we may proceed inductively using
(9.3).

Next we prove (9.2) with $\lambda(\mathcal{O})$ defined by (9.3). Assume all
different values of these $k_{j}$ are $m_{i}\,(1\leq i\leq r)$, where for each
$j$ there are $a_{i}$ copies of $m_{i}$ with corresponding $\zeta_{j}=+$, and
$b_{j}$ copies with $\zeta_{j}=-$. We may assume $a_{i}=b_{i}$ (otherwise it
is easy to check that both sides of (9.2) are zero), hence the left hand side
of (9.2) is reduced to

$\mathbb{E}\bigg{(}\prod_{i=1}^{r}|\eta_{m_{i}}|^{2a_{i}}\bigg{)}=\prod_{j=1}^{r}\mu_{a_{i}}.$

Note also that $\mathcal{O}_{*}=(2a_{1},\cdots,2a_{r})$ is a partition of $n$
and the sets $\\{j:k_{j}=m_{i}\\}$ form a partition
$\mathscr{O}_{*}\models\mathcal{O}_{*}$. Moreover, on the right hand side of
(9.2), in order for the product $\prod_{A\in\mathscr{O}}\prod_{j,j^{\prime}\in
A}\mathbf{1}_{k_{j}=k_{j^{\prime}}}$ to be nonzero, one must have
$\mathcal{O}\preceq\mathcal{O}_{*}$ and $\mathscr{O}\preceq\mathscr{O}_{*}$,
and in this case this product equals $1$. Thus the right hand side of (9.2)
equals

$\sum_{\mathcal{O}\preceq\mathcal{O}_{*}}\lambda(\mathcal{O})\sum_{\mathscr{O}\preceq\mathscr{O}_{*},\mathscr{O}\models\mathcal{O}}1=\sum_{\mathcal{O}\preceq\mathcal{O}_{*}}\xi_{\mathcal{O}_{*},\mathcal{O}}\cdot\lambda(\mathcal{O})=\prod_{i=1}^{r}\mu_{a_{i}}$

using the definition of $\xi_{\mathcal{O}_{*},\mathcal{O}}$ and (9.3), as
desired.

Next we prove that

$|\lambda(\mathcal{O})|\leq\frac{(n/2)!}{(|\mathcal{O}|)!}\prod_{2b\in\mathcal{O}}(C_{0}b)!$
(9.4)

for $C_{0}=C+40$. The base case $\mathcal{O}=(2,\cdots,2)$ is clear. By
induction, and using that $\mu_{r}\leq(Cr)!$, we only need to prove that for
any $\mathcal{O}$, we have

$\sum_{\mathcal{O}^{\prime}\prec\mathcal{O}}\xi_{\mathcal{O},\mathcal{O}^{\prime}}\frac{(|\mathcal{O}|)!}{(|\mathcal{O}^{\prime}|)!}\frac{\prod_{2b\in\mathcal{O}^{\prime}}(C_{0}b)!}{\prod_{2b\in\mathcal{O}}(C_{0}b)!}\leq\frac{1}{2}.$
(9.5)

Now, fix a partition $\mathscr{O}$ of $\\{1,\cdots,n\\}$ subordinate to
$\mathcal{O}$. To construct $\mathscr{O}^{\prime}$, we first fix a partition
of each element of $\mathcal{O}$ into even positive integers, such that the
terms in these partitions exactly constitute $\mathcal{O}^{\prime}$. Let the
number of choices for these partitions be
$\eta_{\mathcal{O},\mathcal{O}^{\prime}}$. Once these partitions are fixed, we
have that

$\mathrm{Number\ of\ choices\ for\ }\mathscr{O}^{\prime}\mathrm{\ is\
}\leq\bigg{(}\frac{\prod_{2b\in\mathcal{O}}b!}{\prod_{2b\in\mathcal{O}^{\prime}}b!}\bigg{)}^{2}\quad\Rightarrow\quad\xi_{\mathcal{O},\mathcal{O}^{\prime}}\leq\bigg{(}\frac{\prod_{2b\in\mathcal{O}}b!}{\prod_{2b\in\mathcal{O}^{\prime}}b!}\bigg{)}^{2}\cdot\eta_{\mathcal{O},\mathcal{O}^{\prime}}.$
(9.6)

In fact, consider any subset $A\in\mathscr{O}$, say $|A|=2a$, which is
partitioned $a=b_{1}+\cdots+b_{q}$ as described above. To divide $A$ into
subsets of cardinalities $2b_{j}\,(1\leq j\leq q)$ to form part of
$\mathscr{O}^{\prime}$ (we may call this part $\mathscr{O}_{A}$), we need to
divide the set of elements with $+$ sign and the set of elements with $-$ sign
separately, leading to at most $(a!)^{2}/((b_{1})!\cdots(b_{q})!)^{2}$
choices, considering also that there may be repetitions due to symmetry.
Applying this for each $A$, we get the upper bound (9.6).

We write $\mathcal{O}=(2a_{1},\cdots,2a_{r})$, and define
$G(a)=(C_{0}a)!/(a!)^{2}$. Using (9.6), we can bound the left hand side of
(9.5) by

$\sum_{\mathcal{O}^{\prime}\prec\mathcal{O}}\eta_{\mathcal{O},\mathcal{O}^{\prime}}\frac{r!}{(|\mathcal{O}^{\prime}|)!}\frac{\prod_{2b\in\mathcal{O}^{\prime}}G(b)}{G(a_{1})\cdots
G(a_{r})}.$

Using the definition of $\eta_{\mathcal{O},\mathcal{O}^{\prime}}$, we can
further reduce this to

$\sum_{(\mathcal{P}_{1},\cdots,\mathcal{P}_{r})}\frac{r!}{(|\mathcal{P}_{1}|+\cdots+|\mathcal{P}_{r}|)!}\prod_{i=1}^{r}\frac{1}{G(a_{i})}\prod_{2b\in\mathcal{P}_{i}}G(b)$

where each $\mathcal{P}_{i}$ is a partition of $2a_{i}$ into even positive
integers, and at least one $\mathcal{P}_{i}$ is nontrivial (i.e. contains at
least two elements). Suppose $q$ of these $r$ partitions are nontrivial, then
$|\mathcal{P}_{1}|+\cdots+|\mathcal{P}_{r}|\geq r+q$. Thus we get an upper
bound

$\sum_{q=1}^{r}\binom{r}{q}\frac{r!}{(r+q)!}\bigg{(}\sup_{a}\sum_{\mathcal{P}}\frac{1}{G(a)}\prod_{2b\in\mathcal{P}}G(b)\bigg{)}^{q},$
(9.7)

where $\mathcal{P}$ runs over all nontrivial even partitions of $2a$. As
$\binom{r}{q}\frac{r!}{(r+q)!}\leq 1/q!$, it suffices to show that

$\sum_{\mathcal{P}}\frac{1}{G(a)}\prod_{2b\in\mathcal{P}}G(b)\leq\frac{1}{4},$
(9.8)

which would then imply that $(\ref{upperbd2})\leq 1/2$ and thus complete the
induction.

The proof of (9.8) is easy. Note that $\log G$ is convex, so if
$|\mathcal{P}|=s$, then

$\frac{1}{G(a)}\prod_{2b\in\mathcal{P}}G(b)\leq\frac{((C_{0})!)^{s-1}G(a-s+1)}{G(a)}\leq\min\bigg{(}\frac{(C_{0}!)^{s-1}}{(C_{0}(s-1))!},\frac{(C_{0}!)^{s-1}}{(C_{0}(a-s+1))^{C_{0}(s-1)}}\bigg{)}a^{2(s-1)}.$

Using also that $(m/3)^{m}\leq m!\leq m^{m}$, we can further bound this by

$\frac{1}{G(a)}\prod_{2b\in\mathcal{P}}G(b)\leq\frac{C_{0}^{C_{0}(s-1)}}{(C_{0}\max((s-1)/3,a-s+1))^{C_{0}(s-1)}}\leq(a/4)^{-C_{0}(s-1)}a^{2(s-1)}.$

The number of choices of $\mathcal{P}$ is at most $a^{s-1}$, so the left hand
side of (9.8) is bounded by

$\sum_{s=2}^{a}a^{s-1}(a/4)^{-C_{0}(s-1)}a^{2(s-1)}\leq\frac{1}{4}$

provided $C_{0}$ is big enough (we may assume $a\geq 5$, since the cases
$a\leq 4$ are easily verified). This proves (9.8), and finishes the inductive
proof of (9.4).

Finally, suppose the sum of elements in $\mathcal{O}$ that are at least $4$ is
$q$, then $|\mathcal{O}|\geq(n-q)/2$. Thus by (9.4) we have

$|\lambda(\mathcal{O})|\leq\frac{(n/2)!}{((n-q)/2)!}(C_{0}!)^{(n-q)/2}(C_{0}q/2)!\leq(C_{0}!)^{n/2}(n/2)^{q/2}(C_{0}q/2)^{C_{0}q/2}\leq
C_{1}^{n}n^{C_{1}q},$

which completes the proof. ∎

######  Remark 9.2.

Note that $\xi_{\mathcal{O},(2,\cdots,2)}=\prod_{2b\in\mathcal{O}}b!$. In the
Gaussian case, where $\mu_{b}=b!$, (9.3) implies that $\lambda(\mathcal{O})=0$
for _any_ $\mathcal{O}\neq(2,\cdots,2)$, hence we reduce to the Isserlis’
theorem.

###  9.2. Over-gardens

Using Lemma 9.2 instead of Isserlis’ theorem, we can replace (4.3) by

$\mathbb{E}\bigg{(}\prod_{j=1}^{2R}(\mathcal{J}_{\mathcal{T}_{j}})_{k_{j}}^{\zeta_{j}}(t)\bigg{)}=\sum_{\mathscr{O}}\lambda(\mathcal{O})\mathcal{M}_{\mathcal{O}\mathcal{G}}(t,k_{1},\cdots,k_{2R}).$
(9.9)

Here $\mathscr{O}$ is a set of _over-pairings_ of the leaves of the trees
$\mathcal{T}_{j}\,(1\leq j\leq 2R)$, which is a partition of these leaves into
subsets, such that the number of leaves with sign $+$ in each subset is equal
to the number of leaves with sign $-$. The total number of these leaves is
$2(m+R)$ where $m$ is the sum of scales of $\mathcal{T}_{j}\,(1\leq j\leq
2R)$, and $\mathscr{O}$ induces an even partition of $2(m+R)$ which we denote
by $\mathcal{O}$, such that $\mathscr{O}\models\mathcal{O}$ in the sense of
Lemma 9.1. The coefficient $\lambda(\mathcal{O})$ is as in Lemma 9.1, and
$\mathcal{O}\mathcal{G}$ is the set of these $2R$ trees together with the set
of over-pairing $\mathscr{O}$, which we refer to as an _over-garden_. Note
that we may also write $\mathcal{O}\mathcal{G}\models\mathcal{O}$ instead of
$\mathscr{O}\models\mathcal{O}$. Finally, like (4.1) we have

$\mathcal{M}_{\mathcal{O}\mathcal{G}}(t,k_{1},\cdots,k_{2R})=\bigg{(}\frac{\delta}{2L^{d-1}}\bigg{)}^{m}\zeta^{*}(\mathcal{O}\mathcal{G})\sum_{\mathscr{I}}\epsilon_{\mathscr{I}}\int_{\mathcal{I}}\prod_{\mathfrak{n}\in\mathcal{N}^{*}}e^{\pi
i\zeta_{\mathfrak{n}}\cdot\delta
L^{2}\Omega_{\mathfrak{n}}t_{\mathfrak{n}}}\,\mathrm{d}t_{\mathfrak{n}}\cdot\prod_{\mathfrak{l}\in\mathcal{L}^{*}}^{(+)}n_{\mathrm{in}}(k_{\mathfrak{l}}),$
(9.10)

where all the objects are defined as in (4.1), except that for the decoration
$\mathscr{I}$ we require that $k_{\mathfrak{n}}\,(\mathfrak{n}\in A)$ are all
equal for each $A\in\mathscr{O}$.

Clearly an over-garden $\mathcal{O}\mathcal{G}$ can be turned into a garden
$\mathcal{G}$ by dividing each over-pairing $A\in\mathscr{O}$ into leaf pairs;
below we will write $\mathcal{O}\mathcal{G}\sim\mathcal{G}$ for this. In this
case $\mathcal{M}_{\mathcal{O}\mathcal{G}}$ is the same as
$\mathcal{M}_{\mathcal{G}}$, except for the finitely many additional
conditions of form $k_{\mathfrak{l}}=k_{\mathfrak{l}^{\prime}}$ associated
with the over-pairings structure of $\mathcal{O}\mathcal{G}$, which are added
to the decoration $\mathscr{I}$ in the summation (9.10). Now by (9.9) we have

$\mathbb{E}\bigg{(}\prod_{j=1}^{2R}(\mathcal{J}_{m_{j}})_{k_{j}}^{\zeta_{j}}(t)\bigg{)}=\sum_{\mathcal{O}}\lambda(\mathcal{O})\sum_{\mathcal{O}\mathcal{G}\models\mathcal{O}}\mathcal{M}_{\mathcal{O}\mathcal{G}}(t,k_{1},\cdots,k_{2R}),$
(9.11)

where $\mathcal{O}$ is an even partition of $2(m+R)$ with
$m=m_{1}+\cdots+m_{2R}$, and
$\mathcal{O}\mathcal{G}=(\mathcal{T}_{1},\cdots,\mathcal{T}_{2R})\models\mathcal{O}$
is the over-garden of width $2R$ and signature $(\zeta_{1},\cdots,\zeta_{2R})$
such that the scale of $\mathcal{T}_{j}$ is $m_{j}$ for $1\leq j\leq 2R$. Note
that for fixed $\mathcal{G}$, the number of $\mathcal{O}\mathcal{G}$ such that
$\mathcal{O}\mathcal{G}\models\mathcal{O}$ and
$\mathcal{O}\mathcal{G}\sim\mathcal{G}$ depends only on $\mathcal{O}$;
similarly, for fixed $\mathcal{O}\mathcal{G}\models\mathcal{O}$, the number of
$\mathcal{G}$ such that $\mathcal{O}\mathcal{G}\sim\mathcal{G}$ also depends
only on $\mathcal{O}$. We denote them by $\sigma_{1}(\mathcal{O})$ and
$\sigma_{2}(\mathcal{O})$ respectively. Thus (9.11) can be rewritten as

$\mathbb{E}\bigg{(}\prod_{j=1}^{2R}(\mathcal{J}_{m_{j}})_{k_{j}}^{\zeta_{j}}(t)\bigg{)}=\sum_{\mathcal{O}}\frac{\lambda(\mathcal{O})}{\sigma_{2}(\mathcal{O})}\sum_{\mathcal{G}}\sum_{\mathcal{O}\mathcal{G}\sim\mathcal{G},\mathcal{O}\mathcal{G}\models\mathcal{O}}\mathcal{M}_{\mathcal{O}\mathcal{G}}(t,k_{1},\cdots,k_{2R}),$
(9.12)

where $\mathcal{G}$ satisfies the same condition as $\mathcal{O}\mathcal{G}$
but is a runs over gardens instead of over-gardens.

Now the study of (9.1) reduces to the study of the quantities
$\mathcal{M}_{\mathcal{O}\mathcal{G}}$ for over-gardens
$\mathcal{O}\mathcal{G}$. To this end we introduce the notion of regular over-
gardens, and one simple linear algebra lemma.

######  Definition 9.3.

Define an over-garden $\mathcal{O}\mathcal{G}$ to be a _regular_ over-garden,
if there exists $\mathcal{G}$ such that
$\mathcal{O}\mathcal{G}\sim\mathcal{G}$, and (i) $\mathcal{G}$ is a regular
multi-couple (Definition 4.1), and (ii) for each leaf $\mathfrak{l}$ in each
over-pairing $A\in\mathscr{O}$ with $|A|\geq 4$, the tree $\mathcal{T}_{j}$
containing $\mathfrak{l}$ _must_ be a regular tree and $\mathfrak{l}$ must be
its lone leaf.

######  Lemma 9.4.

Let $A\subset B$ be two sets of affine linear equations posed on
$\mathbb{R}^{n}$, in terms of the coordinates $(x_{1},\cdots,x_{n})$. Let
$\mathcal{A}\supset\mathcal{B}$ be the affine submanifold of $\mathbb{R}^{n}$
determined by equations in $A$ and $B$ respectively, assume
$\mathcal{B}\neq\varnothing$, and denote
$p=\mathrm{codim}_{\mathcal{A}}(\mathcal{B})$. Let $1\leq r\leq n-1$.

For any fixed $x=(x_{1},\cdots,x_{r})\in\mathbb{R}^{r}$, consider the affine
submanifold $\mathcal{A}_{x}\supset\mathcal{B}_{x}$ of
$(x_{r+1},\cdots,x_{n})\in\mathbb{R}^{n-r}$ determined by equations in $A$ and
$B$ for $(x,x_{r+1},\cdots,x_{n})$ respectively. Let
$\mathcal{A}^{\circ}\supset\mathcal{B}^{\circ}$ be the affine manifold for $x$
such that $\mathcal{A}_{x}\neq\varnothing$ and
$\mathcal{B}_{x}\neq\varnothing$ respectively. Then, if
$\mathrm{codim}_{\mathcal{A}^{\circ}}(\mathcal{B}^{\circ})=p^{\circ}$, then
for any $x\in\mathcal{B}^{\circ}$ we have
$\mathrm{codim}_{\mathcal{A}_{x}}(\mathcal{B}_{x})=p-p^{\circ}$.

###### Proof.

We omit the proof as it is elementary. ∎

Now we can state the main estimate for the non-Gaussian case.

######  Proposition 9.5.

Fix $R$ and $(\zeta_{1},\cdots,\zeta_{2R})$ and $(k_{1},\cdots,k_{2R})$ and
$(m_{1},\cdots,m_{2R})$ as in Proposition 4.7. Assume $R\leq 2\log L/(\log\log
L)^{2}$ and $m_{j}\leq N\,(1\leq j\leq 2R)$, and set $m:=m_{1}+\cdots+m_{2R}$.
Consider the sum $\mathscr{I}^{\prime}$ as in (9.12), but we restrict to _non-
regular over-gardens_ $\mathcal{O}\mathcal{G}$ in the summation. Then we have

$|\mathscr{I}^{\prime}|\leq(C^{+}\delta^{1/4})^{m}L^{-\nu}$ (9.13)

uniformly in $t$ and $(k_{1},\cdots,k_{2R})$.

###### Proof.

Let $\mathcal{O}\mathcal{G}$ be an over-garden and
$\mathcal{O}\mathcal{G}\sim\mathcal{G}$. Since the expression
$\mathcal{M}_{\mathcal{O}\mathcal{G}}$ in (9.10) is just the expression
$\mathcal{M}_{\mathcal{G}}$ in (4.1) with finitely many extra requirements of
form $k_{\mathfrak{l}}=k_{\mathfrak{l}^{\prime}}$ in the summation, it is
clear that $\mathcal{M}_{\mathcal{O}\mathcal{G}}$ can at least be estimated in
the same way as $\mathcal{M}_{\mathcal{G}}$ with no power loss. Also the
number of $\mathcal{O}$ is at most $C^{m}$, and for fixed $\mathcal{G}$ and
$\mathcal{O}$, the third sum in (9.12) contains $\sigma_{1}(\mathcal{O})$
terms, where $\sigma_{1}(\mathcal{O})$ has the same upper bound as
$\lambda(\mathcal{O})$ in Lemma 9.2.

Therefore, compared to the bounds for $\mathcal{M}_{\mathcal{G}}$ that we
already know, we only have two tasks in proving the desired result: first,
gain the extra power $L^{-\nu}$, and second, gain enough extra powers to
cancel the factor $|\lambda(\mathcal{O})|$ in case the latter is too large.
Note also that if $\mathcal{O}$ is given and $\mathcal{G}$ and
$\mathcal{G}^{\prime}$ are congruent in the sense of Definition 5.4, then the
over-gardens
$\mathcal{O}\mathcal{G}\models\mathcal{O},\,\mathcal{O}\mathcal{G}\sim\mathcal{G}$
are in one-to-one correspondence with the over-gardens
$\mathcal{O}\mathcal{G}^{\prime}\models\mathcal{O},\,\mathcal{O}\mathcal{G}^{\prime}\sim\mathcal{G}^{\prime}$,
and cancellations between the terms $\mathcal{M}_{\mathcal{O}\mathcal{G}}$ and
$\mathcal{M}_{\mathcal{O}\mathcal{G}^{\prime}}$ are the same as the
cancellations between $\mathcal{M}_{\mathcal{G}}$ and
$\mathcal{M}_{\mathcal{G}^{\prime}}$ in Section 5, up to minor modifications.
As such, we can exploit the same cancellation for irregular chains in
$\mathcal{G}$ as in Section 5 and [12].

Now let us go over the process of studying $\mathcal{M}_{\mathcal{G}}$, and
see what the extra conditions $k_{\mathfrak{l}}=k_{\mathfrak{l}^{\prime}}$ may
do at each step of this process. Below let $q_{0}$ be the number of
independent extra equations $k_{\mathfrak{l}}=k_{\mathfrak{l}^{\prime}}$, then
$q_{0}\sim q$, where $q$ is the sum of elements in $\mathcal{O}$ that are at
least $4$ as in Lemma 9.1. In fact, we have

$q=\sum_{4\leq a\in\mathcal{O}}a,\quad q_{0}=\sum_{4\leq
a\in\mathcal{O}}\big{(}\frac{a}{2}-1\big{)},$

hence $2q_{0}\leq q\leq 4q_{0}$. We will keep track of the codimension
introduced by these $q_{0}$ extra equations to the affine manifold of all
possible decorations $(k_{\mathfrak{l}})$ using Lemma 9.4.

(1) Assume that $\mathcal{G}$ is a regular multi-couple. In this case, we
shall estimate the summation (together with the integration) in (9.10) using
the method in Section 6 of [12] (note that here we have to treat all the
regular couples in $\mathcal{G}$ together—instead of one at a time in Section
6 of [12]—because of the extra conditions linking different regular couples
together, but this will cause minor changes to the proof). In particular, we
define the variables $x_{\mathfrak{n}}$ and $y_{\mathfrak{n}}$ as in the proof
of Proposition 6.7 of [12]. Note that there are $2R$ linear equations that any
decoration of the leaves of $\mathcal{G}$ must satisfy (and such decoration of
leaves uniquely determines the full decoration of $\mathcal{G}$); moreover the
set of decorations satisfying these $2R$ equations is in affine bijection with
the set of free variables $(x_{\mathfrak{n}},y_{\mathfrak{n}})$, see the proof
of Proposition 6.7 of [12].

Now, with the extra conditions, the dimension of the affine manifold of all
possible decorations $(k_{\mathfrak{l}})$ gets strictly lower, and the
codimension $r$ introduced satisfies $r\gtrsim\max(1,q_{0}-O(R))$. In fact we
have $r>0$ because at least one extra condition must take the form
$k_{\mathfrak{l}}=k_{\mathfrak{l}^{\prime}}$ where $\mathfrak{l}$ is not the
lone leaf of a regular tree by Definition 9.3, and this equation will be
independent of the $2R$ original equations stated above (since the only way
for this extra condition to be dependent is for the two trees containing
$\mathfrak{l}$ and $\mathfrak{l}^{\prime}$ to be distinct and coupled, which
would easily imply that they are two regular trees with lone leaves
$\mathfrak{l}$ and $\mathfrak{l}^{\prime}$). The lower bound $q_{0}-O(R)$ is
because the number of independent extra equations is $q_{0}$, and we subtract
$O(R)$ because some extra equations combined may imply some of the $2R$
original equations.

Using the affine bijection, we know that the
$(x_{\mathfrak{n}},y_{\mathfrak{n}})$ variables must satisfy $r$ independent
linear equations. Then, we proceed as in the proof of Proposition 6.1 in [12],
and sum over the $(x_{\mathfrak{n}},y_{\mathfrak{n}})$ variables one by one.
At each step, suppose we are summing over the pair $(x_{j},y_{j})$, depending
on the extra equations satisfied by these variables, we have one of three
possibilities: (a) there is no restriction on $(x_{j},y_{j})$ and we are
summing over all choices of $(x_{j},y_{j})$; (b) we are summing over
$(x_{j},y_{j})$ that satisfies one linear equation (such as
$x_{j}=\mathrm{const}$ or $y_{j}=\mathrm{const}$ or $x_{j}\pm
y_{j}=\mathrm{const}$); (c) we are summing over $(x_{j},y_{j})$ that satisfies
two linear equations, i.e. over only one point $(x_{j}^{*},y_{j}^{*})$. In
either case the summation can be performed in the same way as in [12], and in
cases (b) and (c) we are gaining a power of $L$ in this summation, compared to
the factor $L^{2d-2}$ in [12]. Also, by repeating Lemma 9.4, we know that case
(b) or (c) must occur at least $\gtrsim r$ times during this process.

Therefore, putting altogether, with these extra conditions we can gain power
$L^{-cr}$ for some small constant $c$ with $r\gtrsim\max(1,q-O(R))$, in the
estimate of $\mathcal{M}_{\mathcal{O}\mathcal{G}}$ compared to
$\mathcal{M}_{\mathcal{G}}$. This already provides the needed $L^{-\nu}$ gain.
It also covers any possible loss due to $\lambda(\mathcal{O})$ because
$|\lambda(\mathcal{O})|\leq C^{m}m^{Cq}\leq C^{m}(\log L)^{Cq_{0}}$ by Lemma
9.1, hence

$|\lambda(\mathcal{O})|\leq C^{m}(\log L)^{O(R)}\cdot(\log
L)^{C\max(1,q_{0}-O(R))}\leq C^{m}L^{\nu/2}\cdot(\log L)^{O(r)}$

by our choice $R\leq\log L/(\log\log L)^{2}$, which can then be covered by the
$L^{-\nu}$ gain; also the various loss of $C^{m}$ is unimportant as they can
be absorbed into $(C^{+})^{m}$ in (9.13).

(2) Now we assume $\mathcal{G}$ is _not_ a regular multi-couple. Then the sum
of $\mathcal{M}_{\mathcal{G}}$ already gains the power $L^{-\nu}$ in view of
Proposition 4.7 and also Proposition 10.4 in [12]. It then suffices to cover
the possible loss due to $\lambda(\mathcal{O})$. We perform the reduction
steps as in previous sections, and analyze the extra conditions appearing in
each step. As in (1), the total codimension introduced by the $q_{0}$ extra
equations is $r\gtrsim\max(1,q_{0}-O(R))$; we may assume $q_{0}\gg R$ because
otherwise the loss $|\lambda(\mathcal{O})|\leq C^{m}m^{Cq_{0}}\leq
C^{m}L^{\nu/2}$ can already be covered by the guaranteed $L^{-\nu}$ gain.
Therefore we now have $r\gtrsim q_{0}$.

_Step 1_. We first remove the regular couples and regular trees to reduce
$\mathcal{G}$ to its skeleton $\mathcal{G}_{sk}$ as in Proposition 4.4. In
this process we are fixing all the remaining $k_{\mathfrak{n}}$ variables
(which are determined by the variables $k_{\mathfrak{l}_{1}}$ for leaves
$\mathfrak{l}_{1}$ of $\mathcal{G}_{sk}$) and sum over the variables
$k_{\mathfrak{l}_{2}}$, where $\mathfrak{l}_{2}$ are leaves of these regular
couple and regular trees, similar to (1) above. By Lemma 9.4, there exists
$r_{1}+r_{2}=r$, such that for any fixed $(k_{\mathfrak{l}_{1}})$, the
codimension of the submanifold formed by the $(k_{\mathfrak{l}_{2}})$
variables is $r_{2}$, and the codimension of the submanifold formed by the
$(k_{\mathfrak{l}_{1}})$ variables is $r_{1}$. By repeating the argument in
(1) above, we can gain a power $L^{-cr_{2}}$ in summing over the
$(k_{\mathfrak{l}_{2}})$ variables. Note that some extra equations satisfied
by the $(k_{\mathfrak{l}_{2}})$ variables may be of form
$k_{\mathfrak{l}_{2}}=\mathrm{const}$ instead of
$k_{\mathfrak{l}_{2}}=k_{\mathfrak{l}_{2}^{\prime}}$ as in (1), but this does
not affect the proof.

_Step 2_. We further remove the irregular chains from the skeleton
$\mathcal{G}_{sk}$ and exploit the cancellation as in Section 5. Note that if
$\mathcal{G}$ is not a regular multi-couple, then _any_
$\mathcal{O}\mathcal{G}\sim\mathcal{G}$ must be non-regular in the sense of
Definition 9.3, so for fixed $\mathcal{G}$, the summation in
$\mathcal{O}\mathcal{G}$ we are studying here is still the same as the one in
(9.12) even though we have made the restriction that $\mathcal{O}\mathcal{G}$
is non-regular. Thus, as said above, the cancellation for irregular chains
works the same way in the current situation as in Section 5. The extra
conditions again lead to gain of powers in $L$. Like in _Step 1_ , we can
write $r_{1}=r_{3}+r_{4}$, such that we can gain a power $L^{-cr_{3}}$ in the
current step, and after removing the irregular chains, the remaining
decoration (of the remaining garden $\mathcal{G}_{sk}^{\\#}$, see Section 5.4)
still satisfies $r_{4}$ extra linear equations.

_Step 3_. Finally we reduce the estimate of the remaining expression to the
counting problem associated with the molecule formed from
$\mathcal{G}_{sk}^{\\#}$. Here we only need to show that, if, in addition to
the equations in the counting problem, the variables in question also satisfy
$r_{4}$ additional independent linear equations, then we can improve the upper
bound for the counting problem by a power $L^{-cr_{4}}$ with a small constant
$c$.

To see this, we follow the procedure described in Section 6, and in particular
apply the algorithm introduced in Section 9.4 of [12]. In this process, where
we fix some of the variables in each step, we keep track of the codimension,
or the number $p$ of independent equations satisfied by the remaining
variables. Initially we have $p\gtrsim r_{4}$, while in the end we have $p=0$.
Therefore, there must be at least $\gtrsim r_{4}$ steps where $p$ strictly
decreases. If this step is a good step in the sense of Section 9.3 of [12],
then we are gaining a constant power $L^{-\nu}$ here; even if it is a normal
step, since $\Delta p<0$, by Lemma 9.4, in doing the counting estimate for
this step, we can take into account an additional independent linear equation
satisfied by the variables in consideration. For example, if we perform the
step (3R-1) defined in Section 9.3.8 of [12], then the corresponding counting
problem we solve is (say)

$\left\\{\begin{aligned} &a-b+c=\mathrm{const},\\\
&|a|_{\beta}^{2}-|b|_{\beta}^{2}+|c|_{\beta}^{2}=\mathrm{const}+O(L^{-2}),\\\
&a,b,c\in\mathbb{Z}_{L}^{d},\,|a|,|b|,|c|\lesssim 1,\end{aligned}\right.$

which has $O(L^{2d-2})$ solutions. However, if we add to this system another
independent linear equation $\alpha a+\beta b+\gamma c=\mathrm{const}$, where
$(\alpha,\beta,\gamma)$ is not a multiple of $(1,-1,1)$, then the number of
solutions will be at most $L^{d}$ with $d<2d-2$. This leads to a power gain in
each such step, so in total we can gain a power $L^{-cr_{4}}$ for some
constant $c$.

After the above three steps, the total power we gain would be
$L^{-c(r_{2}+r_{3}+r_{4})}=L^{-cr}$, which is enough to cover the loss
$C^{m}m^{Cq}$ from $\lambda(\mathcal{O})$ because $r\gtrsim q_{0}\gtrsim q$.
Therefore in any case we can cover the possible loss with an extra gain of
$L^{-\nu}$, hence (9.13) holds. This completes the proof. ∎

With Proposition 9.5 we can now prove Theorem 1.4.

###### Proof of Theorem 1.4.

We use (3.6) to expand
$\mathbb{E}\big{(}\prod_{j=1}^{2R}a_{k_{j}}^{\zeta_{j}}(t)\big{)}$. The
estimate for the remainder term $b$ can be done using arguments similar to
Section 8.2, which we shall omit. Then, using also (9.12), we can write

$\mathbb{E}\bigg{(}\prod_{j=1}^{2R}a_{k_{j}}^{\zeta_{j}}(t)\bigg{)}=\sum_{\mathcal{G},\mathcal{O}}\frac{\lambda(\mathcal{O})}{\sigma_{2}(\mathcal{O})}\sum_{\mathcal{O}\mathcal{G}\sim\mathcal{G},\mathcal{O}\mathcal{G}\models\mathcal{O}}\mathcal{M}_{\mathcal{O}\mathcal{G}}(t,k_{1},\cdots,k_{2R})+O(L^{-\nu}),$
(9.14)

where $\mathcal{G}$ runs over all gardens of width $2R$ such that the scale of
each tree is at most $N$, and $\mathcal{O}$ runs over all even partitions of
$2(m+R)$ where $m$ is the scale of $\mathcal{G}$. By Proposition 9.5 and
summing over all possible $m_{j}$ as in the proof of Theorem 1.3 above, we see
that with $R$ fixed and $L\to\infty$, the contribution of _non-regular_ over-
gardens $\mathcal{O}\mathcal{G}$ decays like $L^{-\nu}$ in the limit. Thus, we
only need to consider _regular_ over-gardens $\mathcal{O}\mathcal{G}$. Suppose
$\mathcal{O}\mathcal{G}\sim\mathcal{G}$, then $\mathcal{G}$ is a regular
multi-couple. Therefore, unless we can divide $\\{1,\cdots,2R\\}$ into pairs
such that for each pair $\\{i,j\\}$ we have $k_{i}=k_{j}$ and
$\zeta_{i}=-\zeta_{j}$, the contribution of regular over-gardens must vanish,
in particular (1.3) is true. Now we only need to prove (1.7).

For any regular couple $\mathcal{Q}=(\mathcal{T}_{1},\mathcal{T}_{2})$, the
tree $\mathcal{T}_{1}$ is a regular tree if and only if $\mathcal{T}_{2}$ is a
regular tree (and hence the two lone leafs are paired). In this case we say
$\mathcal{Q}$ is _tangential_ (since the two trees only have one leaf-pair in
common), otherwise say $\mathcal{Q}$ is _non-tangential_. Note that by the
proof of Theorem 1.3 in [12] we have

$\sum_{\mathcal{Q}}\mathcal{M}_{\mathcal{Q}}(t,t,k)=n(\delta
t,k)+O(L^{-\nu}),$ (9.15)

where $\mathcal{Q}$ runs over all regular couples with both trees having scale
at most $N$. If we restrict to _tangential_ couples only, then the sum should
be approximated by $n_{0}(\delta t,k)$ where $n_{0}$ is defined in (WKE-0).
The easiest way to see this is that, the expression
$\mathcal{M}_{\mathcal{Q}}(t,t,k)$ contains a factor $n_{\mathrm{in}}(k)$ if
and only if $\mathcal{Q}$ is tangential, because for any regular tree
$\mathcal{T}$ with root $\mathfrak{r}$ and lone leaf $\mathfrak{l}$ we must
have $k_{\mathfrak{l}}=k_{\mathfrak{r}}=k$. Thus, since the sum (9.15) over
all couples exactly matches the Taylor expansion of $n(\delta t,k)$ (as shown
in [12]), we know that the sum over tangential couples will exactly match the
terms in the Taylor expansion _that contain the factor $n_{\mathrm{in}}(k)$_.
Due to the form of (1.1), it is easy to see that the sum of these terms is
exactly $n_{0}(t,k)$, hence the result. Therefore we have

$\sum_{\mathcal{Q}\mathrm{\
\\!tangential}}\mathcal{M}_{\mathcal{Q}}(t,t,k)=n_{0}(\delta
t,k)+O(L^{-\nu}),$ (9.16) $\sum_{\mathcal{Q}\textrm{\ \\!non-
tangential}}\mathcal{M}_{\mathcal{Q}}(t,t,k)=n_{+}(\delta t,k)+O(L^{-\nu}).$
(9.17)

We now return to the sum (9.14) over regular over-gardens
$\mathcal{O}\mathcal{G}$. For (1.7), we may rename $(k_{1},\cdots,k_{2R})$
such that there are $2a_{j}$ copies of $k_{i}^{*}$ for $1\leq i\leq r$ (with
half of them having sign $+$ and half having sign $-$) where the $k_{i}^{*}$
are all different and $a_{1}+\cdots+a_{r}=R$. For simplicity we will write
$k_{i}$ instead of $k_{i}^{*}$ below. Clearly the $2a_{i}$ trees corresponding
to the input variable $k_{i}$ must form $a_{i}$ couples in $\mathcal{G}$;
assume $b_{i}$ of these $a_{i}$ couples are tangential and the rest are non-
tangential, where $0\leq b_{i}\leq a_{i}$. Note also that for any
$\mathcal{O}\mathcal{G}\sim\mathcal{G}$ we have
$\mathcal{M}_{\mathcal{O}\mathcal{G}}=\mathcal{M}_{\mathcal{G}}$ because over-
pairings can only happen at lone leaves of regular trees. Therefore, for fixed
$(b_{1},\cdots,b_{r})$, we can calculate the contribution of regular over-
gardens as

$\mathscr{I}_{b_{1},\cdots,b_{r}}=\prod_{i=1}^{r}\binom{a_{i}}{b_{i}}^{2}(a_{i}-b_{i})!\bigg{(}\sum_{\mathcal{Q}:\mathrm{non-
tangential}}\mathcal{M}_{\mathcal{Q}}(t,t,k_{i})\bigg{)}^{a_{i}-b_{i}}\\\
\times\sum_{\mathcal{G}^{\prime},\mathcal{O}^{\prime}}\frac{\lambda(\mathcal{O}^{\prime})\sigma_{3}(\mathcal{O}^{\prime})}{\sigma_{2}(\mathcal{O}^{\prime})}\mathcal{M}_{\mathcal{G}^{\prime}}(t,k_{1},\cdots,k_{1},\cdots,k_{r},\cdots,k_{r}).$
(9.18)

Here $\mathcal{G}^{\prime}$ runs over all gardens _of width
$2(b_{1}+\cdots+b_{r})$_ which are multi-couples formed by _tangential
couples_ (and otherwise same as $\mathcal{G}$), $\mathcal{O}^{\prime}$ runs
over all _even partitions of $2(b_{1}+\cdots+b_{r})$_ such that
$\mathcal{O}^{\prime}\preceq(2b_{1},\cdots,2b_{r})$, and
$\sigma_{3}(\mathcal{O}^{\prime})$ is the number of regular over-gardens
$\mathcal{O}\mathcal{G}$ such that $\mathcal{O}\mathcal{G}\sim\mathcal{G}$ and
$\mathcal{O}\mathcal{G}\models\mathcal{O}$ where
$\mathcal{O}=(\mathcal{O}^{\prime},2,\cdots,2)$. We can further reduce the
inner sum in (9.18) to

$\sum_{\mathcal{G}^{\prime},\mathcal{O}^{\prime}}\frac{\lambda(\mathcal{O}^{\prime})\sigma_{3}(\mathcal{O}^{\prime})}{\sigma_{2}(\mathcal{O}^{\prime})}\mathcal{M}_{\mathcal{G}^{\prime}}(t,k_{1},\cdots,k_{1},\cdots,k_{r},\cdots,k_{r})=\prod_{i=1}^{r}\bigg{(}\sum_{\mathcal{Q}\mathrm{\
\\!tangential}}\mathcal{M}_{\mathcal{Q}}(t,t,k_{i})\bigg{)}^{b_{i}}\\\
\times\sum_{\mathcal{O}^{\prime}}\frac{\lambda(\mathcal{O}^{\prime})\sigma_{3}(\mathcal{O}^{\prime})}{\sigma_{2}(\mathcal{O}^{\prime})}(b_{1})!\cdots(b_{r})!.$
(9.19)

Finally, in the above summation $\mathcal{O}^{\prime}$ must be
$\preceq(2b_{1},\cdots,2b_{r})$ otherwise
$\sigma_{3}(\mathcal{O}^{\prime})=0$, and by definitions and a double counting
argument we can show that under this assumption we have

$\xi_{(2b_{1},\cdots,2b_{r}),\mathcal{O}^{\prime}}=\sigma_{3}(\mathcal{O}^{\prime})\frac{(b_{1})!\cdots(b_{r})!}{\sigma_{2}(\mathcal{O}^{\prime})}.$

In fact, fix $\mathscr{O}\models(2b_{1},\cdots,2b_{r})$. To construct
$\mathscr{O}^{\prime}$ such that $\mathscr{O}^{\prime}\preceq\mathscr{O}$ and
$\mathscr{O}^{\prime}\models\mathcal{O}^{\prime}$, we first divide each subset
in $\mathscr{O}$ into pairs (which has $(b_{1})!\cdots(b_{r})!$ choices), thus
obtaining a garden $\mathcal{G}$, then construct
$\mathcal{O}\mathcal{G}\sim\mathcal{G}$ and
$\mathcal{O}\mathcal{G}\models\mathcal{O}$ (which has
$\sigma_{3}(\mathcal{O}^{\prime})$ choices) and form $\mathscr{O}^{\prime}$
accordingly. Note that each $\mathscr{O}^{\prime}$ is counted exactly
$\sigma_{2}(\mathcal{O}^{\prime})$ times (which is just the number of choices
of $\mathcal{G}$ with fixed $\mathcal{O}\mathcal{G}$), hence the result.

Now we can reduce the last sum in (9.18) to

$\sum_{\mathcal{O}^{\prime}}\frac{\lambda(\mathcal{O}^{\prime})\sigma_{3}(\mathcal{O}^{\prime})}{\sigma_{2}(\mathcal{O}^{\prime})}(b_{1})!\cdots(b_{r})!=\sum_{\mathcal{O}^{\prime}\preceq(2b_{1},\cdots,2b_{r})}\lambda(\mathcal{O}^{\prime})\xi_{(2b_{1},\cdots,2b_{r}),\mathcal{O}^{\prime}}=\prod_{i=1}^{r}\mu_{b_{i}}$

using (9.3). Putting altogether, using (9.16) and (9.17), and then summing
over $b_{i}\,(1\leq i\leq r)$, we get

$\mathbb{E}\bigg{(}\prod_{j=1}^{2R}a_{k_{j}}^{\zeta_{j}}(t)\bigg{)}=\sum_{0\leq
b_{i}\leq a_{i}\,(1\leq i\leq
r)}\prod_{i=1}^{r}\binom{a_{i}}{b_{i}}^{2}(a_{i}-b_{i})!\mu_{b_{i}}(n_{0}(\delta
t,k))^{b_{i}}(n_{+}(\delta t,k))^{a_{i}-b_{i}}+O(L^{-\nu})$

which is just (1.7) given (1.9). This completes the proof. ∎

###  9.3. Evolution of density

Finally we prove Theorem 1.5. Note that if $\mu_{r}\leq C^{r}(2r)!$, then by
(1.9) for any $t$ we also have $\mu_{r}(t,k)\leq C^{r}(2r)!$ perhaps for some
different $C$. Thus, convergence in law will be a consequence of the following
lemma:

######  Lemma 9.6.

Suppose $\\{X_{n}\\}$ are $\mathbb{R}^{d}$ valued random variables, such that
for any multi-index $\mu$ the limit

$A_{\mu}:=\lim_{n\to\infty}\mathbb{E}(X_{n}^{\mu})$

exists and $|A_{\mu}|\leq C^{|\mu|}(|\mu|)!$. Then $\\{X_{n}\\}$ converges in
law to a random variable $X$ satisfying $\mathbb{E}(X^{\mu})=A_{\mu}$ for any
multi-index $\mu$. Moreover, the law with these given moments is unique.

###### Proof.

First the assumption implies that $\mathbb{E}|X_{n}|^{2}$ is uniformly bounded
in $n$, thus the sequence of laws of $X_{n}$ is tight. For any subsequence
$X_{n_{k}}$ we then have a subsequence $X_{n_{k_{\ell}}}$ that converges in
law to (say) some random variable $X$. For any $\mu$, since
$\mathbb{E}|X_{n}|^{2|\mu|}$ are bounded in $n$ it is easy to see that
$\mathbb{E}(X^{\mu})=\lim\mathbb{E}(X_{n}^{\mu})=A_{\mu}$. Therefore, it
suffices to prove that the law of the random variables $X$ with
$\mathbb{E}(X^{\mu})=A_{\mu}$ is unique. This is true because, since
$|A_{\mu}|\leq C^{|\mu|}(|\mu|)!$, we then have
$\mathbb{E}(e^{\varepsilon|X|})<\infty$ for small enough $\varepsilon$, hence
$f(\xi)=\mathbb{E}(e^{i\xi\cdot X})$ is well-defined and holomorphic in the
region $|\mathrm{Im}\xi|<\varepsilon$. The moments $\mathbb{E}(X^{\mu})$
uniquely determines the Taylor expansion of $f(\xi)$ at $\xi=0$, hence
uniquely determines the value of $f(\xi)$ in a neighborhood of $0$—and
consequently in the whole region $|\mathrm{Im}\xi|<\varepsilon$ by
analyticity. In particular the moments uniquely determine the value of
$f(\xi)$ on $\mathbb{R}$, which is the characteristic function of $X$. Thus
the law of $X$ is unique, as desired. ∎

Now, for any $t\in[0,\delta]$ and $k\in\mathbb{Z}_{L}^{d}$, consider the
unique radial density $\rho=\rho_{k}(t,v)$ (where $v\in\mathbb{C}$ also viewed
as an $\mathbb{R}^{2}$ vector) such that

$\int_{\mathbb{C}}(v_{1}^{2}+v_{2}^{2})^{r}\rho_{k}(t,v)\,\mathrm{d}v_{1}\mathrm{d}v_{2}=\mu_{r}(t,k)=\sum_{p=0}^{r}\binom{r}{p}^{2}(r-p)!\mu_{p}\cdot(n_{0}(t,k))^{p}(n_{+}(t,k))^{r-p},$
(9.20)

then by Theorem 1.4 and Lemma 9.6 we have that convergence in law described by
(1.13) and (1.14) is true. Thus to prove Theorem 1.5 it suffices to show that
$\rho_{k}(t,v)$ solves the equation (1.12). The initial data $\rho_{k}(0)$ is
verified by definition since the right hand side of (9.20) equals
$\mu_{r}n_{\mathrm{in}}(k)^{r}$ when $t=0$. Now, by approximating from below,
we may assume $\\{\mu_{p}\\}$ is bounded, and consider

$L_{k}(t,\xi)=\int_{\mathbb{C}}e^{i\xi(v_{1}^{2}+v_{2}^{2})}\rho_{k}(t,v)\,\mathrm{d}v_{1}\mathrm{d}v_{2}=\sum_{0\leq
p\leq
r}\frac{(i\xi)^{r}}{p!}\binom{r}{p}\mu_{p}\cdot(n_{0}(t,k))^{p}(n_{+}(t,k))^{r-p}.$

Taking time derivative and calculating using (WKE), (1.1), (WKE-0) and (1.8)
yields

$\partial_{t}L_{k}=\sum_{0\leq p\leq
r}\frac{(i\xi)^{r}}{p!}\binom{r}{p}\mu_{p}\cdot\big{\\{}p(n_{0}(t,k))^{p}(n_{+}(t,k))^{r-p}\gamma_{k}(t)\\\
+(r-p)(n_{0}(t,k))^{p}(n_{+}(t,k))^{r-p-1}(\sigma_{k}(t)+\gamma_{k}(t)n_{+}(t,k))\big{\\}}$
(9.21)

where $\sigma_{k}(t)$ and $\gamma_{k}(t)$ are defined in (1.10) and (1.11).
This simplifies to

$\partial_{t}L_{k}=\gamma_{k}(t)\cdot\sum_{0\leq p\leq
r}r\cdot\frac{(i\xi)^{r}}{p!}\binom{r}{p}\mu_{p}\cdot(n_{0}(t,k))^{p}(n_{+}(t,k))^{r-p}\\\
+\sigma_{k}(t)\cdot\sum_{0\leq p\leq
r}(r-p)\cdot\frac{(i\xi)^{r}}{p!}\binom{r}{p}\mu_{p}\cdot(n_{0}(t,k))^{p}(n_{+}(t,k))^{r-p-1}.$
(9.22)

The first sum on the right hand side equals $\xi\partial_{\xi}L_{k}$, and the
second sum equals

$\sum_{0\leq p\leq
r}(r+1)\cdot\frac{(i\xi)^{r+1}}{p!}\binom{r}{p}\mu_{p}\cdot(n_{0}(t,k))^{p}(n_{+}(t,k))^{r-p}$

by changing $r$ to $r+1$, which then equals
$(i\xi+i\xi^{2}\partial_{\xi})L_{k}$. Therefore we have

$\partial_{t}L_{k}=\gamma_{k}(t)\cdot\xi\partial_{\xi}L_{k}+\sigma_{k}(t)\cdot(i\xi+i\xi^{2}\partial_{\xi})L_{k}.$

Finally, by taking the inverse Fourier transform and switching between
Cartesian and polar coordinates, we can verify that the density
$\rho_{k}(t,v)$ solves (1.12). This completes the proof of Theorem 1.5.

##  10\. Proof of Theorem 1.6

In this last section we prove Theorem 1.6. The proof is of “soft” nature and
has a different flavor from the other parts of this paper. The main idea,
already hinted at in Section 1.4, is to represent $n_{r}$ as an average of
tensor products for which one can apply Theorem 1.3. This is demonstrated in
the following lemma, which is a variant of the classical Hewitt-Savage
theorem.

######  Lemma 10.1 (a variant of Hewitt-Savage).

Suppose $(n_{r})_{\mathrm{in}}$ is as in the statement of Theorem 1.6,
satisfies (1.16), and is admissible in the sense of (1.17). Recall the set
$\mathcal{A}$ defined in (1.18). Then, there exists a unique probability
measure $\zeta$ on $\mathcal{A}$ such that for any $r$ and any distinct
$(k_{1},\cdots,k_{r})$ we have

$(n_{r})_{\mathrm{in}}(k_{1},\cdots,k_{r})=\int_{\mathcal{A}}m(k_{1})\cdots
m(k_{r})\,\mathrm{d}\zeta(m).$ (10.1)

###### Proof.

For simplicity we will write $(n_{r})_{\mathrm{in}}$ just as $n_{r}$. We may
assume $\mathfrak{X}=1$ (otherwise simply replace $n_{r}$ by
$\mathfrak{X}^{-r}n_{r}$ and $C_{1}$ by $\mathfrak{X}^{-1}C_{1}$). Then, each
$n_{r}$ is the density of the joint distribution of some $r$ random variables
valued in $\mathbb{R}^{d}$, and $n_{r-1}$ is a marginal of $n_{r}$. By
Kolmogorov extension theorem, we can find an infinite sequence of
$\mathbb{R}^{d}$-valued random variables $(X_{1},X_{2},\cdots)$ such that
$n_{r}$ is the density of the joint distribution of $(X_{1},\cdots,X_{r})$. By
symmetry of $n_{r}$, these random variables $X_{j}$ are exchangeable, so by
Hewitt-Savage theorem [23], there exists a unique probability measure $\zeta$
on the set $\mathcal{M}$ of _all probability measures_ $m$ on $\mathbb{R}^{d}$
such that (10.1) is true with $\mathcal{A}$ replaced by $\mathcal{M}$.

Now it remains to prove that $\zeta$ is supported on $\mathcal{A}$. We will
apply the beautiful argument of Rosenzweig-Staffilani in Section 5 of [30],
which goes back to the work of Spohn [31]. Fix a Schwarz function $\varphi$ on
$\mathbb{R}^{d}$ and indices $\alpha,\beta$ with $|\alpha|,|\beta|\leq 40d$;
for any _even_ integer $r$, we may test both sides of (10.1) by the tensor
product $(\partial_{k}^{\beta}k^{\alpha}\varphi)^{\otimes r}$ to get

$\int_{\mathcal{M}}(m,\partial_{k}^{\beta}k^{\alpha}\varphi)^{r}\,\mathrm{d}\zeta(m)=\int_{(\mathbb{R}^{d})^{r}}n_{r}(k_{1},\cdots,k_{r})\prod_{j=1}^{r}k_{j}^{\alpha}\partial_{k_{j}}^{\beta}\varphi(k_{j})\,\mathrm{d}k_{1}\cdots\mathrm{d}k_{r},$

where $(m,\partial_{k}^{\beta}k^{\alpha}\varphi)$ is the integral of
$\partial_{k}^{\beta}k^{\alpha}\varphi$ against the measure $m$, in the usual
distributional sense. Recall that $\|n_{r}\|_{\mathfrak{S}^{40d;r}}\leq
C_{1}^{r}$ from (1.16), so the right hand side is bounded by

$\int_{(\mathbb{R}^{d})^{r}}n_{r}(k_{1},\cdots,k_{r})\prod_{j=1}^{r}\partial_{k_{j}}^{\beta}k_{j}^{\alpha}\varphi(k_{j})\,\mathrm{d}k_{1}\cdots\mathrm{d}k_{r}\leq\|n_{r}\|_{\mathfrak{S}^{40d;r}}\cdot\|\varphi\|_{L^{2}}^{r}\leq(C_{1}\|\varphi\|_{L^{2}})^{r}.$

For any $C_{2}>C_{1}$, by Chebyshev, this implies that

$\zeta\big{(}\big{\\{}m:|(m,\partial_{k}^{\beta}k^{\alpha}\varphi)|\geq
C_{2}\|\varphi\|_{L^{2}}\big{\\}}\big{)}\leq(C_{2}\|\varphi\|_{L^{2}})^{-r}\int_{\mathcal{M}}(m,\partial_{k}^{\beta}k^{\alpha}\varphi)^{r}\,\mathrm{d}\zeta(m)\leq(C_{1}/C_{2})^{r}.$

Since $r$ is arbitrary, we conclude that

$|(m,\partial_{k}^{\beta}k^{\alpha}\varphi)|\leq C_{2}\|\varphi\|_{L^{2}}$
(10.2)

holds $\zeta$-almost surely for any fixed $\varphi$, fixed $(\alpha,\beta)$,
and fixed $C_{2}>C_{1}$. By choosing $\varphi$ in a countable dense subset of
Schwartz space, enumerating finitely many possible $(\alpha,\beta)$, and
choosing a sequence $C_{2}=(1+\varepsilon)C_{1}$, we know that $\zeta$-almost
surely in $m$, (10.2) actually holds for _all_ $(\varphi,\alpha,\beta)$, and
with $C_{2}$ replaced by $C_{1}$. By Riesz representation theorem, this
implies that

$\|k^{\alpha}\partial_{k}^{\beta}m\|_{L^{2}}\leq
C_{1},\quad\forall|\alpha|,|\beta|\leq 40d$

holds $\zeta$-almost surely. By definition this means that $m\in\mathcal{A}$
because $m$ is a probability measure and $\mathfrak{X}=1$. Therefore we get
$\zeta(\mathcal{M}\backslash\mathcal{A})=0$, and hence we can define the
measure $\zeta$ on $\mathcal{A}$ such that (10.1) is true. ∎

We are now ready to prove Theorem 1.6.

###### Proof of Theorem 1.6.

First, if $(n_{r})_{\mathrm{in}}$ is hybrid then it must be admissible. In
fact, by the definition of hybrid initial data we have

$\int_{\mathcal{A}}m(k_{1})\cdots
m(k_{r})\,\mathrm{d}\zeta(m)=(n_{r})_{\mathrm{in}}(k_{1},\cdots,k_{r})$ (10.3)

for any distinct $k_{j}\in\mathbb{Z}_{L}^{d}\,(1\leq j\leq r)$. Since $m$ and
$(n_{r})_{\mathrm{in}}$ are all continuous functions, by taking suitable
limits and letting $L\to\infty$ we know that (10.3) is actually true for all
$k_{j}\in\mathbb{R}^{d}\,(1\leq j\leq r)$. Then we simply integrate (10.3) in
$k_{r}$, using the integral condition in the definition of $\mathcal{A}$, to
get (1.17).

From now on we shall assume $(n_{r})_{\mathrm{in}}$ is admissible, then by
Lemma 10.1, we can find a unique measure $\zeta$ such that (10.3) holds.
Consider the hybrid data $u_{\mathrm{in}}$ described in the statement of
Theorem 1.6. We can view it as obtained by first randomly selecting
$m\in\mathcal{A}$ with law given by $\zeta$, then working in the same setting
as in Theorems 1.1 and 1.3–1.5 with the particular choice $n_{\mathrm{in}}=m$.
Since $m\in\mathcal{A}$, by Theorem 1.1 we know that the conditional
probability $\mathbb{P}(E|m)\geq 1-L^{-A}$ for any $m\in\mathcal{A}$, where
$E$ is the event that (NLS) has a smooth solution up to time $T$. This implies
that

$\mathbb{P}(E)=\int_{\mathcal{A}}\mathbb{P}(E|m)\,\mathrm{d}\zeta(m)\geq
1-L^{-A}.$

Finally we prove (1.20). As is clear from the proof of Theorem 1.4, the
remainder in (1.7) is in fact $O(L^{-\nu})$ for some absolute constant
$\nu>0$, where the implicit constant in $O(\cdot)$ may depend on $r$, but is
uniform in $(t,k_{j})$ and $n_{\mathrm{in}}=m$, as long as $m\in\mathcal{A}$.
Thus, by Theorem 1.4, for the hybrid data $u_{\mathrm{in}}$, for $t\in[0,T]$
we have that

$\mathbb{E}\bigg{(}\prod_{j=1}^{r}|\widehat{u}(t,k_{j})|^{2}\bigg{)}=\int_{\mathcal{A}}\mathbb{E}\bigg{(}\prod_{j=1}^{r}|\widehat{u}(t,k_{j})|^{2}\,\bigg{|}\,m\bigg{)}\,\mathrm{d}\zeta(m)=\int_{\mathcal{A}}\prod_{j=1}^{r}\widetilde{m}\bigg{(}\frac{t}{T_{\mathrm{kin}}},k_{j}\bigg{)}\,\mathrm{d}\zeta(m)+O(L^{-\nu})$

where $\widetilde{m}(t,k)$ is the solution to (WKE) with initial data
$\widetilde{m}(0,k)=m(k)$. It remains to show that

$n_{r}(t,k_{1},\cdots,k_{r})=\int_{\mathcal{A}}\prod_{j=1}^{r}\widetilde{m}(t,k_{j})\,\mathrm{d}\zeta(m).$

In fact, for any $m\in\mathcal{A}$, by definition $\widetilde{m}$ is the
solution to (WKE) with initial data $m$, so by direct calculation we see that
$(\widetilde{m})^{\otimes r}$ is a solution to (WKH) with initial data
$m^{\otimes r}$. Since (WKH) is linear, we know that

$\int_{\mathcal{A}}(\widetilde{m})^{\otimes r}\,\mathrm{d}\zeta(m)$ (10.4)

is a solution to (WKH) with initial data

$\int_{\mathcal{A}}m^{\otimes r}\,\mathrm{d}\zeta(m)=(n_{r})_{\mathrm{in}},$

due to (10.3). Moreover, since $m\in\mathcal{A}$, for short time
$t\in[0,\delta]$, the solution (10.4) clearly belongs to the class
$C_{t}^{0}\mathfrak{L}_{s,\epsilon}^{\infty}$ defined in [30]; since short
time uniqueness is proved in [30] for solutions in this class, we know that
(10.4) has to equal $n_{r}$, which is the unique solution to (WKH) in this
class with initial data $(n_{r})_{\mathrm{in}}$. The admissibility condition
(1.17) follows from the fact that $n_{r}$ equals (10.4) which is an average of
factorized solutions for which admissibility is clearly true (Here we use that
the (WKE) conserves the total mass $\int n(t,\xi)\,d\xi$). This completes the
proof. ∎

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