$\displaystyle\leq(C^{+})^{n_{0}^{\prime}-2\ell}C^{\ell}L^{C(p-1)\sqrt{\delta}}(\log
L)^{|\mathcal{N}^{rm}|+C(p-1)}\left(\delta^{(2\ell)/4}\int_{(\alpha_{1},\cdots,\alpha_{\ell}):\alpha_{k}\in
S_{k}(1)}\frac{\prod_{k=1}^{\ell}\Phi_{k}(\alpha_{k}-\widetilde{\mu}_{k})}{(\ell-5)!}\right)$
$\displaystyle\leq(C^{+})^{n_{0}^{\prime}-2\ell}C^{\ell}L^{C(p-1)\sqrt{\delta}}(\log
L)^{|\mathcal{N}^{rm}|+C(p-1)}\left(C^{\ell}\delta^{(2\ell)/4}\frac{(\log
L)^{\ell+10}}{(\ell-5)!}\right)$
$\displaystyle\leq(C^{+})^{n_{0}^{\prime}-2\ell}C^{2\ell}L^{C(p-1)\sqrt{\delta}}(\log
L)^{|\mathcal{N}^{rm}|+C(p-1)}\left((\log L)^{15}e^{C\sqrt{\delta}\log
L}\right),$

where $S_{k}(1)$ is a set of measure $\leq L^{10d}$, we denoted by
$\widetilde{\mu}_{k}$ either $0$ or $\mu_{k}$ (depending on whether the $k$-th
module is A or B), and $\Phi_{k}(z)$ is either $\langle z\rangle^{-1}$ or (for
at most $10$ values of $k$) $\langle z\rangle^{-1}\log(2+|z|)$. In the final
step we used the bound $\frac{x^{\ell-5}}{(\ell-5)!}\leq e^{x}$ for any
$\ell$. Note also that in applying the induction hypothesis for
$\mathcal{N}_{new}^{tr}$ we have fixed the value of
$\alpha_{\mathfrak{h}_{1}}=2\theta_{0}$ (using integrability of
$\widehat{\mathcal{K}}$), but it is clear from the proof that fixing the value
of any $\alpha_{\mathfrak{n}}$ will only lead to better estimates than
integrating in $\alpha_{\mathfrak{n}}$. This gives the estimate (10.10) and
finishes the proof. ∎

###  10.3. Proof of Propositions 2.5 and 2.7

We are now ready to prove Propositions 2.5 and 2.7. First we establish the
absolute upper bound for (8.27), which then allows us to control (8.4).

######  Proposition 10.4.

Given one congruence class $\mathscr{F}$ of non-regular marked couples of
scale $n$ as in Definition 8.4, the expression

$\sum_{\mathcal{Q}\in\mathscr{F}}\mathcal{K}_{\mathcal{Q}}(t,s,k)$ (10.15)

can be decomposed into at most $C^{n}$ terms. For each term there is an
integer $1\leq r\leq n$ such that this term is bounded, uniformly in
$(t,s)\in[0,1]^{2}$, by $(C^{+}\delta^{1/4})^{n}\langle k\rangle^{-20d}\cdot
L^{-\nu r}$. Moreover, for each fixed $r$, the number of possibilities of
$\mathcal{Q}$ (or $\mathscr{F}$) that correspond to this $r$ is at most
$C^{n}(Cr)!$.

###### Proof.

As in Section 8 we can reduce to (8.27), and then to (10.2). Note that in
(10.2) the $\mathcal{Q}$ actually means $\mathcal{Q}_{sk}^{\\#}$ by our
notation. Using the decay factors in (8.28) we can gain the power $\langle
k\rangle^{-30d}$, and also restrict to the subset where
$|k_{\mathfrak{l}}-a_{\mathfrak{l}}|\leq 1$ for some fixed parameters
$(a_{\mathfrak{l}})$ (with summability in $(a_{\mathfrak{l}})$ guaranteed).
Using the first inequality in (8.26), we may also fix the value of
$\bm{\lambda}$ (and hence $\mu_{\mathfrak{n}}$).

As in Section 9, by decomposing into at most $C^{n_{0}^{\prime}}$ terms (where
$n_{0}^{\prime}$ is the scale of $\mathcal{Q}_{sk}^{\\#}$), we can add the set
of extra conditions $\mathtt{Ext}$, which also defines the sets
$\widetilde{\mathcal{N}}$ (as in Proposition 10.1), etc., and the value
$r_{1}\geq 1$. Let $r=r_{0}+r_{1}$ as above, then thanks to $\mathtt{Ext}$, we
can use (10.3) to reduce $\mathcal{U}_{\mathcal{Q}_{sk}^{\\#}}$ to
$\mathcal{V}_{\mathcal{Q}_{sk}^{\\#}}$. 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
$k[\mathcal{Q}_{sk}^{\\#}]$ varies (this is because each $k_{\mathfrak{n}}$
belongs to a ball of radius at most $n\leq(\log L)^{3}$ 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 $k[\mathcal{Q}_{sk}^{\\#}]$.

To estimate (10.2) with $\bm{\lambda}$ fixed, we first integrate in
$\bm{\sigma}$. Using (8.28), we can estimate (10.2) using

$\sum_{\mathscr{E}_{sk}^{\\#}}|\epsilon_{\mathscr{E}_{sk}^{\\#}}|\cdot\sup_{\bm{\sigma}}\big{|}\mathcal{V}_{\mathcal{Q}_{sk}^{\\#}}\big{(}t,s,\bm{\sigma},(\delta
L^{2}\zeta_{\mathfrak{n}}\Omega_{\mathfrak{n}}+\lambda_{\mathfrak{n}})_{\mathfrak{n}\in\widetilde{\mathcal{N}}}\big{)}\big{|},$
(10.16)

where $\mathscr{E}_{sk}^{\\#}=k[\mathcal{Q}_{sk}^{\\#}]$ is a $k$-decoration
of $\mathcal{Q}_{sk}^{\\#}$ (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

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

by definition, so if we use (10.4) to sum over $(m_{\mathfrak{n}})$ in the
end, we can further estimate (10.16) using

$\sum_{\mathscr{E}_{sk}^{\\#}}|\epsilon_{\mathscr{E}_{sk}^{\\#}}|\cdot\prod_{\mathfrak{l}}\mathbf{1}_{|k_{\mathfrak{l}}-a_{\mathfrak{l}}|\leq
1}\prod_{\mathfrak{n}}\mathbf{1}_{|\Omega_{\mathfrak{n}}-b_{\mathfrak{n}}|\leq\delta^{-1}L^{-2}},$
(10.17)

where $a_{\mathfrak{l}}$ and $b_{\mathfrak{n}}$ are constants, and we also
include the conditions in $\mathtt{Ext}$. Now (10.17) is almost exactly the
counting problem $\mathfrak{D}(\mathbb{M},\mathtt{Ext})$ stated in Definition
9.8, due to Remark 9.9, 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
$\mu_{\mathfrak{n}}^{\circ}$ as in Lemma 6.6, so by using (6.30), one can
reduce (10.17) to at most $C^{n_{0}^{\prime}}$ counting problems, each of
which having exactly the same form as $\mathfrak{D}(\mathbb{M},\mathtt{Ext})$
in Definition 9.8. Therefore, (10.17) can be bounded using Proposition 9.10
(and using Remark 9.11 if necessary). Collecting all the factors appearing in
the above estimates, we get that

${\langle
k\rangle^{20d}\cdot|(\ref{congsum})|}\leq(C^{+})^{n}\delta^{(n-n_{0}^{\prime})/2}\delta^{3n_{0}^{\prime}/4}L^{-(d-1)n_{0}^{\prime}}\cdot
L^{-2\nu r_{0}}\cdot L^{Cr\sqrt{\delta}}(\log
L)^{Cr}\delta^{-(n_{0}^{\prime}+m)/2}L^{(d-1)n_{0}^{\prime}-2\nu r_{1}},$
(10.18)

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

Finally, suppose we fix $r$, then the base molecule formed by
$\mathcal{Q}_{sk}^{\\#}$ is, up to at most $Cr$ remaining atoms, a union of at
most $Cr$ type II chains with total length at most $n_{0}^{\prime}$. This
clearly has at most $(Cr)!C^{n}$ possibilities. By Proposition 9.6, the number
of choices for $\mathcal{Q}_{sk}^{\\#}$ is also at most $(Cr)!C^{n}$. To form
$\mathcal{Q}_{sk}$ from $\mathcal{Q}_{sk}^{\\#}$ one needs to insert at most
$Cr$ irregular chains with total length at most $n$, which also has at most
$C^{n}$ possibilities. Finally, using Corollary 4.16, we see that
$\mathcal{Q}$ has at most $(Cr)!C^{n}$ choices. The number of choices for
markings, as well as $\mathtt{Ext}$, are also at most $C^{n}$ and can be
accommodated. ∎

###### Proof of Proposition 2.5.

By definition, we have

$\mathbb{E}|(\mathcal{J}_{n})_{k}(t)|^{2}=\sum_{\mathcal{Q}}\mathcal{K}_{\mathcal{Q}}(t,t,k),$

where the sum is taken over all couples
$\mathcal{Q}=(\mathcal{T}^{+},\mathcal{T}^{-})$ such that
$n(\mathcal{T}^{+})=n(\mathcal{T}^{-})=n$. If $\mathcal{Q}$ is regular, then
the number of such $\mathcal{Q}$’s is at most $C^{n}$ by Proposition 4.9, and
for each $\mathcal{Q}$ we have
$|\mathcal{K}_{\mathcal{Q}}(t,t,k)|\lesssim\langle
k\rangle^{-20d}(C^{+}\delta)^{n}$ by Proposition 6.7 and Remark 6.9.
Therefore, the sum over these $\mathcal{Q}$’s is under control.

Now consider non-regular $\mathcal{Q}$. It follows from definition that the
congruence relation (as in Definition 8.4) preserves the scales of both trees
of a couple. Thus, the sum over $\mathcal{Q}$ can be decomposed into sums over
$\mathcal{Q}\in\mathscr{F}$ (i.e. sums of form (10.15)), where $\mathscr{F}$
runs over the (possible) different congruence classes recall that there are
also $\leq C^{n}$ such classes). Applying Proposition 10.4, we can regroup
these terms according to the value of $1\leq r\leq 2n$ (which we call the
index), such that (i) each single term with index $r$ is bounded by
$(C^{+}\delta^{1/4})^{2n}\langle k\rangle^{-20d}\cdot L^{-\nu r}$, and (ii)
the number of terms with index $r$ is at most $(Cr)!C^{2n}$. Hence

$\mathbb{E}|(\mathcal{J}_{n})_{k}(t)|^{2}\lesssim\langle
k\rangle^{-20d}(C^{+}\delta)^{n}+\langle
k\rangle^{-20d}(C^{+}\delta^{1/4})^{2n}\sum_{r=1}^{2n}L^{-\nu
r}(Cr)!C^{2n}\lesssim\langle k\rangle^{-20d}(C^{+}\sqrt{\delta})^{n},$

noticing also that $r\leq 2n\leq 2(\log L)^{3}$. This completes the proof. ∎

###### Proof of Proposition 2.7.

Here we are considering the sum of $\mathcal{K}_{\mathcal{Q}}(t,t,k)$ over all
couples $\mathcal{Q}$ such that $n(\mathcal{Q})=m$ for some fixed value $m$.
If $\mathcal{Q}$ is non-regular, then using the same argument as in the above
proof we can bound the corresponding contribution by $\langle
k\rangle^{-20d}(C^{+}\delta^{1/4})^{m}L^{-\nu}$ since we also have $r\geq 1$.
Therefore we only need to consider regular couples $\mathcal{Q}$. If $m$ is
odd then this sum is zero because the scale of regular couples must be even.
If $m=2n$, we only need to show that

$\bigg{|}\sum_{\begin{subarray}{c}n(\mathcal{Q})=2n\\\
\mathcal{Q}\,\,\mathrm{regular}\end{subarray}}\mathcal{K}_{\mathcal{Q}}(t,t,k)-\mathcal{M}_{n}(t,k)\bigg{|}\lesssim\langle
k\rangle^{-20d}(C^{+}\sqrt{\delta})^{n}L^{-\nu},$

but this is a consequence of Proposition 7.11. This completes the proof. ∎

##  11\. The operator $\mathscr{L}$

In this section we prove Proposition 2.6. The arguments are mostly the same as
in previous sections, so we will only point out the necessary changes in the
proof. First, in order to expand the kernel
$(\mathscr{L}^{n})_{k\ell}^{\zeta}(t,s)$, we need to slightly modify the
definition of trees and couples.

######  Definition 11.1.

A _flower tree_ is a tree $\mathcal{T}$ with one leaf $\mathfrak{f}$
specified, called the _flower_ ; different choices of $\mathfrak{f}$ for the
same tree $\mathcal{T}$ leads to different flower trees. There is a unique
path from the root $\mathfrak{r}$ to the flower $\mathfrak{f}$, which we call
the _stem_. A _flower couple_ is a couple formed by two flower trees, such the
two flowers are paired (in particular they have opposite signs).

The _height_ of a flower tree $\mathcal{T}$ is the number of branching nodes
in the stem of $\mathcal{T}$. Clearly a flower tree of height $n$ is formed by
attaching two sub-trees each time, and repeating $n$ times, starting from a
single node; see Figure 38. We say a flower tree is _admissible_ if all these
sub-trees have scale at most $N$.

Figure 38. A flower tree, as in Definition 11.1. The red leaf $\mathfrak{f}$
is the flower, $\mathfrak{r}$ is the root, and $\mathcal{T}_{j}\,(1\leq j\leq
2n)$ are attached sub-trees, where $n$ is the height.

######  Proposition 11.2.

Given $\zeta\in\\{\pm\\}$, we can make the decomposition (2.27), such that for
each $m$,

$\mathbb{E}|(\mathscr{L}^{n})_{k\ell}^{m,\zeta}(t,s)|^{2}=\sum_{\mathcal{Q}}\widetilde{\mathcal{K}}_{\mathcal{Q}}(t,s,k,\ell),$
(11.1)

where the sum is taken over all _flower_ couples
$\mathcal{Q}=(\mathcal{T}^{+},\mathcal{T}^{-})$, such that both
$\mathcal{T}^{\pm}$ are admissible, have _height_ $n$ and _scale_ $m$, and the
flower of $\mathcal{T}^{\pm}$ has sign $\pm\zeta$. For $t>s$, the quantity
$\widetilde{\mathcal{K}}_{\mathcal{Q}}$ is defined similar to (2.24):

$\widetilde{\mathcal{K}}_{\mathcal{Q}}(t,s,k,\ell):=\bigg{(}\frac{\delta}{2L^{d-1}}\bigg{)}^{2m}\zeta^{*}(\mathcal{Q})\sum_{\mathscr{E}}\epsilon_{\mathscr{E}}\int_{\mathcal{E}}\prod_{\mathfrak{n}\in\mathcal{N}^{*}}e^{\zeta_{\mathfrak{n}}\pi
i\delta
L^{2}\Omega_{\mathfrak{n}}t_{\mathfrak{n}}}\mathrm{d}t_{\mathfrak{n}}\prod\bm{\delta}(t_{\mathfrak{f}^{p}}-s){\prod_{\mathfrak{f}\neq\mathfrak{l}\in\mathcal{L}^{*}}^{(+)}n_{\mathrm{in}}(k_{\mathfrak{l}})}\mathbf{1}_{k_{\mathfrak{f}}=\ell},$
(11.2)

where $\mathscr{E}$ is a $k$-decoration of $\mathcal{Q}$, the other objects
are associated with the couple $\mathcal{Q}$, and the set $\mathcal{E}$ is
defined as in (5.4) but with $s$ replaced by $t$; in the last product we
assume $\mathfrak{l}$ has sign $+$ and is not one of the two flowers
$\mathfrak{f}$ of the flower couple $\mathcal{Q}$.

The differences between (11.2) and (2.24) are the (two) Dirac factors
$\bm{\delta}(t_{\mathfrak{f}^{p}}-s)$, where $\mathfrak{f}^{p}$ is the parent
of $\mathfrak{f}$ for both flowers $\mathfrak{f}$, and the (one) factor
$\mathbf{1}_{k_{\mathfrak{f}}=\ell}$.

###### Proof.

Note that

$\mathscr{L}b=\sum_{n(\mathcal{T}_{1}),n(\mathcal{T}_{2})\leq
N}\big{\\{}\mathcal{I}\mathcal{C}_{+}(b,\mathcal{J}_{\mathcal{T}_{1}},\mathcal{J}_{\mathcal{T}_{2}})+\mathcal{I}\mathcal{C}_{+}(\mathcal{J}_{\mathcal{T}_{1}},\overline{b},\mathcal{J}_{\mathcal{T}_{2}})+\mathcal{I}\mathcal{C}_{+}(\mathcal{J}_{\mathcal{T}_{1}},\mathcal{J}_{\mathcal{T}_{2}},b)\big{\\}},$

where the signs of the trees $\mathcal{T}_{j}$ are determined by the positions
they appear ($+$ for the first and third inputs of $\mathcal{C}_{+}$ and $-$
otherwise). This corresponds to attaching two sub-trees $\mathcal{T}_{1,2}$ to
a single node. Calculating $\mathscr{L}^{n}$ corresponds to repeating this $n$
times (obtaining an admissible flower tree $\mathcal{T}$ of height $n$), and
the linear (or) conjugate linear part of $\mathscr{L}^{n}$ corresponds to the
flower of $\mathcal{T}$ having the same (or opposite) sign as the root. Taking
into account also the time integrations, we get

$(\mathscr{L}^{n})_{k\ell}^{\zeta}(t,s)=\sum_{m}(\mathscr{L}^{n})_{k\ell}^{m,\zeta}(t,s):=\sum_{m}\sum_{\mathcal{T}}\widetilde{\mathcal{J}}_{\mathcal{T}}(t,s,k,\ell),$

where the inner sum in the last expression is taken over all admissible flower
trees $\mathcal{T}$ of height $n$ and scale $m$ such that the
$\zeta_{\mathfrak{r}}=+$ and $\zeta_{\mathfrak{f}}=\zeta$, and

$\widetilde{\mathcal{J}}_{\mathcal{T}}(t,s,k,\ell)=\bigg{(}\frac{\delta}{2L^{d-1}}\bigg{)}^{m}\prod_{\mathfrak{n}\in\mathcal{N}}(i\zeta_{\mathfrak{n}})\sum_{\mathscr{D}}\epsilon_{\mathscr{D}}\int_{\mathcal{D}}\prod_{\mathfrak{n}\in\mathcal{N}^{*}}e^{\zeta_{\mathfrak{n}}\pi
i\delta
L^{2}\Omega_{\mathfrak{n}}t_{\mathfrak{n}}}\mathrm{d}t_{\mathfrak{n}}\cdot\bm{\delta}(t_{\mathfrak{f}^{p}}-s)\prod_{\mathfrak{f}\neq\mathfrak{l}\in\mathcal{L}}\sqrt{n_{\mathrm{in}}(k_{\mathfrak{l}})}\eta_{k_{\mathfrak{l}}}^{\zeta_{\mathfrak{l}}}(\omega)\mathbf{1}_{k_{\mathfrak{f}}=\ell},$
(11.3)

where $\mathscr{D}$ is a $k$-decoration of $\mathcal{T}$, $\mathcal{D}$ is
defined as in (5.2), and the other objects are associated with the tree
$\mathcal{T}$. Note also that if $\mathcal{T}$ is admissible and has height
$n\leq N$ and scale $m$, then $n\leq m\leq(1+2N)n\leq N^{3}$. Then, by
repeating the arguments in Section 2.2.3 using Lemma A.2, we can deduce
(11.2). ∎

###### Proof of Proposition 2.6.

We only need to control the right hand side of (11.1). We will basically
repeat the arguments in Sections 5–10. The main points worth noticing are the
followings.

(1) If $\mathcal{Q}$ is an admissible flower couple, and
$\widetilde{\mathcal{Q}}$ is congruent to $\mathcal{Q}$ in the sense of
Definition 8.4, then $\widetilde{\mathcal{Q}}$ is also an admissible flower
couple, if we choose its flower to be the image of the flower of
$\mathcal{Q}$, and has the same height and scale as $\mathcal{Q}$. This will
enable us to decompose the right hand side of (11.1) into sums of form (10.15)
where $\mathscr{F}$ is a congruence class of marked flower couples (which are
defined similar to Definition 8.4), which then allows for the cancellation
exploited in Section 8.

To prove the above claim, notice that the branching nodes in any irregular
chain in $\mathcal{Q}_{sk}$ are also branching nodes in $\mathcal{Q}$, and
this chain can be divided into two chains such that all branching nodes in the
first one belong to the stem of a tree in $\mathcal{Q}$, and all branching
nodes in the second one contained in one of the $\mathcal{T}_{j}$ sub-trees
that are attached in the process described in Definition 11.1.

We may treat these two chains separately; at the joint of the two chains we
may leave out at most $5$ nodes, but this will be acceptable similar to
Section 8. Similarly we may assume that the first chain avoids the flower, by
shortening it if necessary. For the second chain all branching nodes are
contained in some $\mathcal{T}_{j}$, so modifying it in the sense of
Definition 8.4 has effect only within $\mathcal{T}_{j}$ (and it does not
affect any pairings between $\mathcal{T}_{j}$ and any other
$\mathcal{T}_{j^{\prime}}$), and does not change the scale of
$\mathcal{T}_{j}$. For the first chain all branching nodes belong to the stem,
so modifying it may result in some $\mathcal{T}_{j}$ being replaced by its
conjugate, or being permuted with some other $\mathcal{T}_{j}$, see Figure 22.
Note that the nodes in the chain may not be consecutive nodes on the stem, but
the part of stem between them can be obtained by including a unique path
within each regular tree (represented by a black box in Figure 22). In either
case, this does not change the height or scale of $\mathcal{Q}$, nor the fact
that $n(\mathcal{T}_{j})\leq N$ for each $j$. Therefore, the couple
$\widetilde{\mathcal{Q}}$ is also admissible and has the same height and scale
as $\mathcal{Q}$.

(2) In (11.2) we have the factor $\mathbf{1}_{k_{\mathfrak{f}}=\ell}$ instead
of $n_{\mathrm{in}}(k_{\mathfrak{f}})$. First notice that $k-\ell$ is a linear
combination of the $k_{\mathfrak{l}}$ for
$\mathfrak{f}\neq\mathfrak{l}\in\mathcal{L}^{*}$, so the decay factor $\langle
k-\ell\rangle^{-20d}$ in (2.28) can be obtained from the
$n_{\mathrm{in}}(k_{\mathfrak{l}})$ factors. Moreover, since
$k_{\mathfrak{f}}\in\mathbb{Z}_{L}^{d}$, we can replace
$\mathbf{1}_{k_{\mathfrak{f}}=\ell}$ by $\psi(L(k_{\mathfrak{f}}-\ell))$ for
some suitable cutoff function $\psi$. Using this function in place of
$n_{\mathrm{in}}(k_{\mathfrak{f}})$, we can repeat all the previous arguments,
with at most a $L^{3d}$ loss. For example, in Propositions 6.7 and 6.10 we are
relying on Proposition 6.1, which only requires the norm in (6.2). The norm of
$W$ is bounded by the same norm of the tensor product function
$\prod_{\mathfrak{l}}n_{\mathrm{in}}(k_{\mathfrak{l}})$, as $W$ is obtained
from the latter by a linear change of variables; if one factor in this tensor
product is replaced by $\psi(L(k_{\mathfrak{f}}-\ell))$, then its norm gets
multiplied by a constant power of $L$. Therefore, all the proofs will be the
same, except for a possible loss of at most $L^{3d}$.

(3) In (11.2) we have the Dirac factors $\bm{\delta}(t_{\mathfrak{f}^{p}}-s)$.
This means that in the integral in (11.2) we are omitting the integration in
$t_{\mathfrak{f}^{p}}$ for both flowers $\mathfrak{f}$. However, this
difference will cause at most another $L^{20d}$ loss. This is intuitively
clear as only one node (and one time variable) is affected, and we can
demonstrate it as follows.

Recall the sequence of reductions in Sections 5–10, where we remove from the
couple $\mathcal{Q}$ successively (i) the regular couples and regular trees,
then (ii) the irregular chains, then (iii) the nodes corresponding to atoms in
type II chains of the base molecule. In both steps (ii) and (iii) we can
choose to avoid the two specific nodes $\mathfrak{f}^{p}$, so for each flower
$\mathfrak{f}$, we only need to consider the case where (a) $\mathfrak{f}^{p}$
belongs to a regular couple or a regular chain in step (i), or (b)
$\mathfrak{f}^{p}$ belongs to the rest of $\mathcal{Q}$ after performing steps
(i)–(iii). Let $\mathfrak{m}=\mathfrak{f}^{p}$.

In case (a) we will further reduce the regular couple or regular tree using
Proposition 4.8, and we may assume $\mathfrak{f}^{p}$ belongs to one of the
regular chains in this process (we only consider _Case 2_ in Section 5.1.1;
_Case 1_ is much easier as the expression is much simpler and we can directly
calculate it). The point here is that, if we omit the integration in
$t_{\mathfrak{m}}$, then the resulting expression, which is a function of
$\alpha[\mathcal{N}^{*}]$ as in (5.3), satisfies the same bound as the one
with $t_{\mathfrak{m}}$ integration, but in the _weaker norm_
$L_{\alpha_{\mathfrak{m}}}^{\infty}L_{\alpha[\mathcal{N}^{*}\backslash\\{\mathfrak{m}\\}]}^{1}$
instead of $L_{\alpha[\mathcal{N}^{*}]}^{1}$. To see this, consider

$K(t,\alpha_{1},\cdots,\alpha_{m})=\int_{t>t_{1}>\cdots>t_{2m}>0}e^{\pi
i(\beta_{1}t_{1}+\cdots+\beta_{2m}t_{2m})}\,\mathrm{d}t_{1}\cdots\mathrm{d}t_{2m}$
(11.4)

as in (5.12), where $\beta_{a}\,(1\leq a\leq 2m)$ is a permutation of
$\pm\alpha_{j}\,(1\leq j\leq m)$ associated with a legal partition, as in
Section 5.2; for simplicity we have omitted the $\lambda_{a}$ variables. By
the arguments in Section 5.2, we can bound the
$L_{\alpha_{1},\cdots,\alpha_{m}}^{1}$ norm of $K$ (or we may extract explicit
$\frac{1}{\alpha_{j}}$ factors from $K$ and bound the $L^{1}$ norm in the
other $\alpha_{j}$ variables, see Lemma 5.10; for simplicity we will omit this
case). Now, suppose we insert $\bm{\delta}(t_{2m}-s)$ in (11.4) (note that,
since a child of $\mathfrak{m}$ is a leaf that is paired with a leaf in the
other tree, we must have $t_{\mathfrak{m}}=t_{2m}$ in the regular chain
integration), then we will lose integrability in $\beta_{2m}$; however if we
_fix_ the value of $\beta_{2m}$ then we get the expression

$e^{\pi i\beta_{2m}s}\int_{t>\cdots>t_{2m-1}>s}e^{\pi
i(\beta_{1}t_{1}+\cdots+\beta_{2m-1}t_{2m-1})}\mathrm{d}t_{1}\cdots\mathrm{d}t_{2m-1}$
(11.5)

(note that there is some $a$ such that $\beta_{a}=-\beta_{2m}$ is also fixed).
This has basically the same form as (11.4), except for a harmless class $J$
operator corresponding to integration in $t_{a}$, so we can repeat the proof
in Section 5.2, using the notions of class $J$ and $R$ operators, to obtain
the same bound for this expression in the $L^{1}$ norm in the remaining
$\alpha_{j}$ (i.e. excluding $\beta_{a}$ and $\beta_{2m}$) variables. This
then implies the bound for our expression, for fixed $s$, in the
$L^{\infty}L^{1}$ type weaker norm as desired.

In case (b), the same argument applies, except that we replaced the $L^{1}$
norm by the variant in (10.4). The proof is in fact easier, as the bound
(10.4), after removing the type II molecular chains, follows solely from the
denominators $\langle q_{\mathfrak{n}}\rangle$ occurring in (10.8) in Lemma
10.2. If we omit the integration in $t_{\mathfrak{m}}$, then we are at loss of
only _one_ denominator involving $\alpha_{\mathfrak{m}}$, which does not
affect the presence of all the other denominators. Thus, if
$\alpha_{\mathfrak{m}}$ is fixed, the function can be bounded in the remaining
variables in the norm in (10.4), using the same arguments in Section 10.

In either case, in the end we can obtain the same bound for the modified time
integral, but in weaker norms without integrability in _at most two of the
$\alpha_{\mathfrak{n}}$ variables_. But this bound can easily be transformed
to the $L^{1}$ type bound involving all variables, with at most $L^{10d}$
loss, because each $\alpha_{\mathfrak{m}}$ will be replaced by $\delta
L^{2}\Omega_{\mathfrak{m}}$ in the actual $\mathcal{K}_{\mathcal{Q}}$
expression, which belongs to the union of at most $L^{5d}$ fixed unit
intervals (at least if we restrict $|k_{\mathfrak{l}}|\leq L$ for each
$\mathfrak{l}$; if $\max_{\mathfrak{l}}|k_{\mathfrak{l}}|:=M^{\prime}\geq L$
then we may lose $(M^{\prime})^{10d}$ but this will be covered by the
$(M^{\prime})^{100d}$ gain coming from $n_{\mathrm{in}}$). Therefore
$L_{\alpha_{\mathfrak{m}}}^{\infty}$ bounds imply the corresponding
$L_{\alpha_{\mathfrak{m}}}^{1}$ bounds with a loss of at most $L^{5d}$, once
we insert the suitable cutoff functions adapted to these unit intervals.

In view of the arguments (1)–(3) above, the bound for the right hand side of
(11.1) can be obtained, using the same arguments as in Sections 5–10. This
proves Proposition 2.6. ∎

######  Corollary 11.3.

Fix $M_{0}\geq L$ and $M_{1}\geq L^{(100d)^{3}}$. In (11.2), suppose $|k|\geq
M_{0}^{2}$, and we insert suitable cutoff functions supported in
$|k_{\mathfrak{l}}|\leq M_{0}$ for each
$\mathfrak{f}\neq\mathfrak{l}\in\mathcal{L}^{*}$; moreover, suppose we insert
one (or more) cutoff function supported in $|\Omega_{\mathfrak{n}_{j}}|\geq
M_{1}$ for some $1\leq j\leq n-1$, where $\mathfrak{n}_{j}$ is the $j$-th node
in the stem from top to bottom (in particular
$\mathfrak{n}_{n}=\mathfrak{f}^{p}$), then the resulting expression satisfies
the same bound as (11.2), but with an additional decay factor
$M_{1}^{-1/9}M_{0}^{5d}$. The same holds for the right hand side of (11.1).

###### Proof.

Note that the assumption implies that for any irregular chain in
$\mathcal{Q}_{sk}$ with branching nodes on the stem, the gap $h$ (see
Proposition 8.3) must satisfy $|h|\geq M$(since $|k_{\mathfrak{n}}|\geq
M_{0}^{2}/2$ for any node $\mathfrak{n}$ on the stem, and
$|k_{\mathfrak{n}}|\leq N^{3}M_{0}$ for any node $\mathfrak{n}$ off the stem);
in particular we are in the large gap case (Section 8.3.2) and thus do not
need the cancellation coming from congruence couples obtained by altering this
irregular chain. Thus, in carrying out the arguments in previous sections we
only need to sum over $\mathcal{Q}\in\mathscr{F}^{\prime}$ where
$\mathscr{F}^{\prime}$ (unlike $\mathscr{F}$) is a subset in a fixed
congruence class, formed by altering irregular chains that are completely
contained in some $\mathcal{T}_{j}$. Since altering these chains do not affect
the structure of the stem or any $\Omega_{\mathfrak{n}_{j}}$ factor, we can
bound the resulting expression in the same way as (11.2).

To gain the extra decay in $M_{1}$ using the largeness of
$|\Omega_{\mathfrak{n}_{j}}|$, like in the proof of Proposition 11.2, we may
assume $\mathfrak{n}_{j}$ belongs to either (a) a regular couple or regular
tree, or (b) the rest of the couple after removing all the special structures.
In case (a), if $\mathfrak{n}_{j}$ is not paired (as a branching node) to
$\mathfrak{n}_{n}=\mathfrak{f}^{p}$, we can use (5.10) to gain a power of
$M_{1}$; if $\mathfrak{n}_{j}$ is paired to $\mathfrak{f}^{p}$, then we will
consider _Case 1_ and _Case 2_ (in the sense of Section 5.1.1) separately. By
direct calculation in _Case 1_ and examining (11.5) similar to Section 5.2 in
_Case 2_ , we can also gain a power $M_{1}^{-1/9}$ at a loss of at most
$M_{0}^{5d}$, in either situation. In case (b), the decay comes from the
denominators $q_{\mathfrak{n}}$ in (10.8). If $|\Omega_{\mathfrak{n}_{j}}|\geq
M_{1}$ then one of these denominators, say $\langle\widetilde{q}\rangle$, will
be $\gtrsim M_{1}$; we then sum over this $\widetilde{q}$ to get

$\sum_{\widetilde{q}}\frac{1}{\langle\widetilde{q}\rangle}\lesssim
M_{1}^{-1}M_{0}^{5d}$

since $\widetilde{q}$ belongs to a set of cardinality at most $M_{0}^{5d}$ as
the decoration varies. This provides the needed decay, and the rest of the sum
can be estimated as in Proposition 10.1. ∎

##  12\. The endgame

In this section we prove Theorem 1.1. We will do this in a few steps. Recall
that $A\geq 40d$ is fixed in Section 2.3.1, as is the even integer
$p\gg_{A,\nu}1$ and $\delta\ll_{p,C^{+}}1$.

######  Proposition 12.1.

With probability $\geq 1-L^{-A}$, we have

$|(\mathcal{J}_{n})_{k}(t)|\lesssim\langle
k\rangle^{-9d}(p^{2}C^{+}\sqrt{\delta})^{n/2}L,\quad|\mathcal{R}_{k}(t)|\lesssim\langle
k\rangle^{-9d}(p^{2}C^{+}\sqrt{\delta})^{N/2}L$ (12.1)

for any $k\in\mathbb{Z}_{L}^{d}$, $t\in[0,1]$, and all $0\leq n\leq N^{3}$,
where $\mathcal{R}$ is defined in (2.17). Note in particular that the right
hand side of the second inequality in (12.1) is bounded, due to our choice
$N=\lfloor\log L\rfloor$, by

$(p^{2}C^{+}\sqrt{\delta})^{N/2}L\leq\delta^{N/8}L\leq L^{-(100d)^{3}}.$
(12.2)

###### Proof.

First consider $\mathcal{J}_{n}$. For fixed $k$ and fixed $t$,
$(\mathcal{J}_{n})_{k}(t)$ is a random variable of form (A.4), so using Lemma
A.3 and Proposition 2.5 we get

$\mathbb{E}|\langle k\rangle^{10d}(\mathcal{J}_{n})_{k}(t)|^{p}\lesssim
p^{np}(C^{+}\sqrt{\delta})^{np/2}.$

This being uniform in $t$, we can integrate in $t$ and sum in $k$ to obtain
that

$\mathbb{E}\|\langle
k\rangle^{9d}(\mathcal{J}_{n})_{k}(t)\|_{L_{t,k}^{p}([0,1]\times\mathbb{Z}_{L}^{d})}^{p}\lesssim
p^{np}(C^{+}\sqrt{\delta})^{np/2},$ (12.3)

where $L_{k}^{p}$ is taken with respect to $L^{-d}$ times the counting measure
in $k$. Moreover we also have

$\mathbb{E}|\partial_{t}(\mathcal{J}_{n})_{k}(t)|^{2}\lesssim\langle
k\rangle^{-20d}(C^{+}\sqrt{\delta})^{n}L^{40d},$ (12.4)

which can be proved using the arguments in Section 11, as taking
$\partial_{t}$ derivative just corresponds to omitting the $t_{\mathfrak{r}}$
integration and producing something like (11.2). This then implies that

$\mathbb{E}\|\langle
k\rangle^{9d}\partial_{t}(\mathcal{J}_{n})_{k}(t)\|_{L_{t,k}^{p}([0,1]\times\mathbb{Z}_{L}^{d})}^{p}\lesssim
p^{np}(C^{+}\sqrt{\delta})^{np/2}L^{40dp}.$ (12.5)

By Gagliardo-Nirenberg, we conclude that

$\mathbb{E}\|\langle
k\rangle^{9d}(\mathcal{J}_{n})_{k}(t)\|_{L_{t,k}^{\infty}([0,1]\times\mathbb{Z}_{L}^{d})}^{p}\lesssim
L^{d}p^{np}(C^{+}\sqrt{\delta})^{np/2}L^{40d},$ (12.6)

thus with probability $\geq 1-L^{-p/2}$, we have

$\sup_{t,k}|\langle k\rangle^{9d}(\mathcal{J}_{n})_{k}(t)|\lesssim
p^{n}(C^{+}\sqrt{\delta})^{n/2}L^{41d/p+1/2},$

which implies (12.1). The estimate for $\mathcal{R}$ is the same, with $n$
replaced by $N$; we just need to notice that $\mathcal{R}_{k}(t)$ equals the
sum of $(\mathcal{J}_{\mathcal{T}^{+}})_{k}(t)$ over all trees
$\mathcal{T}^{+}$ of scale $>N$ such that its three sub-trees all have scale
$\leq N$ (in particular the scale of $\mathcal{T}^{+}$ is between $N$ and
$3N$). This property, as well as the similar property for couples, is again
invariant under congruence relations, so the same arguments in the previous
sections apply. ∎

######  Proposition 12.2.

With probability $\geq 1-L^{-A}$, we have

$\|\mathscr{L}^{n}\|_{Z\to Z}\lesssim(p^{2}C^{+}\sqrt{\delta})^{n/2}L^{60d}$
(12.7)

for all $0\leq n\leq N$.

###### Proof.

We only need to show, with probability $\geq 1-L^{-A}$, that

$\sup_{k,\ell}\sup_{0\leq s<t\leq 1}\langle
k-\ell\rangle^{9d}|(\mathscr{L}^{n})_{k\ell}^{m,\zeta}(t,s)|\lesssim(p^{2}C^{+}\sqrt{\delta})^{m/2}L^{55d}$
(12.8)

for any $\zeta\in\\{\pm\\}$ and $n\leq m\leq N^{3}$. The supremum in $(t,s)$
can be treated similar to the proof of Proposition 12.1, so the main point
here is to address the supremum in $(k,\ell)$. This formally has infinitely
many possibilities, but we will use Lemma A.6, which is a variant of Claim 3.7
in [18], to reduce it to finitely many possibilities.

We start by making the decomposition

$1=\sum_{R\geq L}\chi_{R}(k_{\mathfrak{l}}),$ (12.9)

where $\chi_{R}(z)$ is supported in $|z|\lesssim R$ if $R=L$ and in $|z|\sim
R$ if $R>L$, for each $\mathfrak{f}\neq\mathfrak{l}\in\mathcal{L}^{*}$. Let
the maximum of these $R$ for all the $k_{\mathfrak{l}}$ be $M_{0}$. Below we
may assume $M_{0}=L$, since even if $M_{0}>L$, we will lose at most
$M_{0}^{20d}$ in all subsequent arguments (see for example 11.3), which can be
covered by the $M_{0}^{-200d}$ gain from the
$n_{\mathrm{in}}(k_{\mathfrak{l}})$ factor, and clearly the summation over $R$
is not a problem.

Now assume $M_{0}=L$. If $|k|\leq L^{2}$, then the number of possibilities for
$(k,\ell)$ is at most $L^{8d}$. We can replace the $L_{k,\ell}^{\infty}$ norm
in (12.8) by $L_{k,\ell}^{p}$ and argue as in Proposition 12.1, applying
Proposition 2.6 and the corresponding bound for $t$ and $s$ derivatives (which
can be obtained similarly as in Section 11), to get (12.8).

Suppose now $|k|\geq L^{2}$, then we will further make the decomposition
(12.9) for the variables $\Omega_{\mathfrak{n}_{j}}\,(1\leq j\leq n-1)$
defined in Corollary 11.3, but with $L$ replaced by $L^{(100d)^{3}}$. Let the
maximum of these $R$ for all the $\Omega_{\mathfrak{n}_{j}}$ be $M$. For fixed
$M$, let the corresponding contribution to
$(\mathscr{L}^{n})_{k\ell}^{m,\zeta}$ be
$(\mathscr{L}^{n})_{M,k\ell}^{m,\zeta}$, then it suffices to show that

$\mathbb{E}\big{|}\sup_{k,\ell}\sup_{0\leq s<t\leq 1}\langle
k-\ell\rangle^{9d}|(\mathscr{L}^{n})_{M,k\ell}^{m,\zeta}(t,s)|\big{|}^{p}\lesssim
p^{mp}(C^{+}\sqrt{\delta})^{mp/2}L^{50dp}M^{-p/20}$ (12.10)

for $M>L^{(100d)^{3}}$, and the same bound without $M^{-p/20}$ for
$M=L^{(100d)^{3}}$.

Suppose $M$ is fixed. Clearly $|k-\ell|\leq L^{2}$, so we may also fix the
value of $k-\ell$ at a loss of $L^{5d}$. In the formula (11.3) of the terms in
$\mathscr{L}^{n}$, we will fix the values of $k_{\mathfrak{f}^{\prime}}$ and
$k_{\mathfrak{f}^{\prime\prime}}$ in the decoration $\mathscr{D}$, where
$\mathfrak{f}^{\prime}$ and $\mathfrak{f}^{\prime\prime}$ are the two siblings
of $\mathfrak{f}$. Clearly each of them has at most $L^{5d}$ choices, so
fixing them again introduces a factor of at most $L^{15d}$.

Recall that in (11.3), the whole expression depends on
$\Omega_{\mathfrak{f}^{p}}$ _only_ through the integral

$\int e^{\zeta_{\mathfrak{f}^{p}}\pi i\delta
L^{2}\Omega_{\mathfrak{f}^{p}}t_{\mathfrak{f}^{p}}}\bm{\delta}(t_{\mathfrak{f}^{p}}-s)\,\mathrm{d}t_{\mathfrak{f}^{p}}=e^{\zeta_{\mathfrak{f}^{p}}\pi
i\delta L^{2}\Omega_{\mathfrak{f}^{p}}s}.$

Moreover, once $k_{\mathfrak{f}^{\prime}}$ and
$k_{\mathfrak{f}^{\prime\prime}}$ are fixed, and $k_{\mathfrak{f}}=\ell$, then
$\Omega_{\mathfrak{f}^{p}}$ is determined by $\ell$ and no longer depends on
the other parts of the decoration. Therefore one can extract the factor
$e^{\zeta_{\mathfrak{f}^{p}}\pi i\delta L^{2}\Omega_{\mathfrak{f}^{p}}s}$ and
write the contribution currently under consideration in the form of

$e^{\zeta_{\mathfrak{f}^{p}}\pi i\delta
L^{2}\Omega_{\mathfrak{f}^{p}}s}\cdot\bigg{(}\frac{\delta}{2L^{d-1}}\bigg{)}^{m}\prod_{\mathfrak{n}\in\mathcal{N}}(i\zeta_{\mathfrak{n}})\sum_{\mathscr{D}}\epsilon_{\mathscr{D}}\mathcal{A}^{**}(t,s,\delta
L^{2}\cdot\Omega[\mathcal{N}\backslash\\{\mathfrak{f}^{p}\\}])\prod_{\mathfrak{f}\neq\mathfrak{l}\in\mathcal{L}}\sqrt{n_{\mathrm{in}}(k_{\mathfrak{l}})}\eta_{k_{\mathfrak{l}}}^{\zeta_{\mathfrak{l}}}(\omega)\mathbf{1}_{k_{\mathfrak{f}}=\ell},$
(12.11)

where $\mathcal{A}^{**}$ is a function of the _remaining_ variables
$\Omega[\mathcal{N}\backslash\\{\mathfrak{f}^{p}\\}]$ and does not depend on
$\Omega_{\mathfrak{f}^{p}}$. Moreover, when $\Omega_{\mathfrak{f}^{p}}$ is
fixed, the function $\mathcal{A}^{**}$ (or more precisely the functions
generated by $\mathcal{A}^{**}$ that occur in the proofs in the previous
sections) satisfies the bounds described in the proof of Proposition 2.6 in
Section 11, hence one can get the same square moment estimate as in
Proposition 2.6. Moreover if $M>L^{(100d)^{3}}$, then using the same arguments
as in the proof of Corollary 11.3, we can gain an extra $M^{-1/9}$ compared to
Proposition 2.6, with at most $L^{5d}$ loss.

Now, after removing the unimodular factor $e^{\zeta_{\mathfrak{f}^{p}}\pi
i\delta L^{2}\Omega_{\mathfrak{f}^{p}}s}$ in (12.11), we notice that the rest
of (12.11) depends on $k$ _only_ through the resonance factors
$\Omega_{\mathfrak{n}_{j}}$ for $1\leq j\leq n-1$. Moreover we have assumed
that $|\Omega_{\mathfrak{n}_{j}}|\leq M$ for each such $j$. If any
$\mathfrak{n}_{j}$ and $\mathfrak{n}_{j+1}$ have opposite signs, then by
definition and $|\Omega_{\mathfrak{n}_{j}}|\leq M$, we easily see that
$|k|\lesssim M$, so we can replace the $L_{k,\ell}^{\infty}$ norm by
$L_{k,\ell}^{p}$ and close as before, where the loss is at most $M^{C/p}$ and
can either be absorbed by the gain $M^{-1/9}$ coming from Corollary 11.3 if
$M>L^{(100d)^{3}}$, or neglected if $M=L^{(100d)^{3}}$. We may thus assume all
$\mathfrak{n}_{j}$ have the same sign, and in this case we have

$\Omega_{\mathfrak{n}_{j}}=|k+a_{j}|_{\beta}^{2}-|k+b_{j}|_{\beta}^{2}+\Omega_{j}^{\prime}=\langle
k,c_{j}\rangle_{\beta}+\Omega_{j}^{\prime\prime}$

where $(a_{j},b_{j},c_{j})$ are vectors, and
$(\Omega_{j}^{\prime},\Omega_{j}^{\prime\prime})$ are expressions, that do not
depend on $k$ (hence they are bounded by $L^{3}$).

We may then apply Lemma A.6 and assume $k$ is represented by a system
$(r,q,v_{1},\cdots,v_{q},f,y)$. Then, if $\chi_{M}(z)$ is a cutoff function
supported in $|z|\leq M$, and $F$ is an arbitrary function, we have

$\prod_{j=1}^{n-1}\chi_{M}(\Omega_{\mathfrak{n}_{j}})\cdot
F(\Omega_{\mathfrak{n}_{1}},\cdots,\Omega_{\mathfrak{n}_{n-1}})=\prod_{j=1}^{n-1}\mathbf{1}_{S}(c_{j})\chi_{M}(G(f,y,c_{j})+\Omega_{j}^{\prime\prime})\cdot
F((G(f,y,c_{j})+\Omega_{j}^{\prime\prime})_{1\leq j\leq n-1}),$ (12.12)

where $S$ is the set of $z$ whose first $r$ coordinates form a linear
combination of $\\{v_{1},\cdots,v_{q}\\}$ and $G$ is some function. The point
here is that, the right hand side of (12.12) does _not_ explicitly involve
$k$, but only depends on the variables $(v_{1},\cdots,v_{q},f,y)$, which are
bounded in size by $M^{C}$ for a constant $C$ depending only on $d$. This is
in contrast with $k$, which has no a priori upper bound and may have
infinitely many choices.

Therefore, for any $k$, the function $\mathcal{A}^{**}(t,s,\delta
L^{2}\cdot\Omega[\mathcal{N}\backslash\\{\mathfrak{f}^{p}\\}])$, viewed as a
function of $(t,s,k[\mathcal{T}\backslash\\{\mathfrak{r},{\mathfrak{f}}\\}])$,
equals
$\mathcal{G}^{**}(t,s,v_{1},\cdots,v_{q},f,y,k[\mathcal{T}\backslash\\{\mathfrak{r},\mathfrak{f}\\}])$
for some function $\mathcal{G}^{**}$. This means that

$\sup_{k}\big{|}(\cdots[\mathcal{A}^{**}]\cdots)\big{|}=\sup_{(v_{1},\cdots,v_{q},f,y)}\big{|}(\cdots[\mathcal{G}^{**}]\cdots)\big{|},$
(12.13)

where the parenthesis $(\cdots)$ in (12.13) represents (12.11) without the
$e^{\zeta_{\mathfrak{f}^{p}}\pi i\delta L^{2}\Omega_{\mathfrak{f}^{p}}s}$
factor. Now, with the finite volume that $(v_{1},\cdots,v_{q},f,y)$ and
$(t,s)$ occupy, we can bound

$\big{\|}(\cdots[\mathcal{G}^{**}]\cdots)\big{\|}_{L_{t,s}^{\infty}L_{v_{1},\cdots,v_{q},f,y}^{\infty}}\lesssim\big{\|}(\cdots[\mathcal{G}^{**}]\cdots)\big{\|}_{L_{t,s}^{p}L_{v_{1},\cdots,v_{q},f,y}^{p}}^{1-C/p}\big{\|}(\partial_{t,s,y}\cdots[\mathcal{G}^{**}]\cdots)\big{\|}_{L_{t,s}^{p}L_{v_{1},\cdots,v_{q},f,y}^{p}}^{C/p}.$

Then, we take $p$-th power moments and argue as in the proof of Proposition
12.1. Note that by Lemma A.6, for _any_ system $(r,q,v_{1},\cdots,v_{q},f,y)$,
the $\mathcal{G}^{**}$ function is the _limit_ of the $\mathcal{A}^{**}$
functions for some sequence of $k$; the same limit holds for the
$(\partial_{t},\partial_{s})$ derivative, and $\partial_{y}\mathcal{G}^{**}$
is the limit of $\partial_{\Omega_{\mathfrak{n}_{j}}}\mathcal{A}^{**}$
multiplied by some harmless factors (any derivative of $\mathcal{A}^{**}$ can
be treated in the same way as $\mathcal{A}^{**}$ itself), so the $p$-th power
moments for fixed $(v_{1},\cdots,v_{q},f,y)$ can be estimated as in
Proposition 2.6, with the extra decay in $M$ for the term without derivatives
if $M>L^{(100d)^{3}}$.

In summary we get

$\mathbb{E}\big{|}\sup_{k,\ell}\sup_{0\leq s<t\leq 1}\langle
k-\ell\rangle^{9d}|(\mathscr{L}^{n})_{M,k\ell}^{m,\zeta}(t,s)|\big{|}^{p}\lesssim
p^{mp}(C^{+}\sqrt{\delta})^{mp/2}L^{45dp}M^{C-p/18},$

which clearly implies (12.10), if $M>L^{(100d)^{3}}$. Here notice that we gain
the power $M^{-p/18}$ from the $p$-th moment of the
$(\cdots[\mathcal{G}^{**}]\cdots)$ term without derivatives, thanks to
Corollary 11.3, and that all the losses caused by the summation or integration
in $(r,q,v_{1},\cdots,v_{q},f,y)$, or by the
$(\partial_{t},\partial_{s},\partial_{y})$ derivatives, are at most $M^{C}$.
If $M=L^{(100d)^{3}}$ then we do not have the gain $M^{-p/18}$, but the losses
are still at most $M^{C}\leq L^{p}$ which is also acceptable. The proof is now
complete. ∎

######  Proposition 12.3.

With probability $\geq 1-L^{-A}$, the mapping defined by the right hand side
of (2.18) is a contraction mapping from the set $\\{b:\|b\|_{Z}\leq
L^{-500d}\\}$ to itself.

###### Proof.

Suppose we exclude the exceptional set of probability $L^{-A}$ in Propositions
12.1–12.2. Consider the mapping

$b\mapsto(1-\mathscr{L})^{-1}(\mathcal{R}+\mathscr{B}(b,b)+\mathscr{C}(b,b,b)),$

as usual, we just need to prove it maps the given set to itself, and the
contraction property will follow similarly. Suppose $\|b\|_{Z}\leq L^{-500d}$,
note that

$(1-\mathscr{L})^{-1}=(1-\mathscr{L}^{N})^{-1}(1+\mathscr{L}+\cdots+\mathscr{L}^{N-1})$

maps $Z$ to $Z$, where $(1-\mathscr{L}^{N})^{-1}$ can be constructed by
Neumann series; using (12.7) we get that $\|(1-\mathscr{L})^{-1}\|_{Z\to
Z}\lesssim L^{62d}$. Therefore, it suffices to show that

$\|\mathcal{R}\|_{Z}+\|\mathscr{B}(b,b)\|_{Z}+\|\mathscr{C}(b,b,b)\|_{Z}\lesssim
L^{-600d}.$

The bound for $\mathcal{R}$ follows from (12.1) and (12.2), so we only need to
consider $\mathscr{B}$ and $\mathscr{C}$. But this is again easy, using the
loose estimate

$\|\mathcal{I}\mathcal{C}_{+}(u,v,w)\|_{Z}\lesssim\|\mathcal{C}_{+}(u,v,w)\|_{Z}\lesssim
L^{20d}\|u\|_{Z}\|v\|_{Z}\|w\|_{Z},$

together with (12.1) and the assumption $\|b\|_{Z}\leq L^{-500d}$. This
completes the proof. ∎

###### Proof of Theorem 1.1.

By Propositions 12.1–12.3, with probability $\geq 1-L^{-A}$, the solution
$a=a_{k}(t)$ to (2.3)–(2.4) can be written as the ansatz (2.15) for
$t\in[0,1]$, where each $\mathcal{J}_{n}$ satisfies (12.1), and $b$ is
constructed by contraction mapping and satisfies $\|b\|_{Z}\leq L^{-500d}$.
Denote this event by $E$, so that $\mathbb{P}(E)\geq 1-L^{-A}$. Let
$E_{1}\supset E$ be the event that (NLS) has a smooth solution on
$[0,\delta\cdot T_{\mathrm{kin}}]$.

For each $\tau\in[0,\delta]$ we will calculate, with $\widehat{u}$ as in
(1.2), that

$\mathbb{E}(|\widehat{u}(\tau\cdot
T_{\mathrm{kin}},k)|^{2}\mathbf{1}_{E_{1}})=\mathbb{E}(|a_{k}(\delta^{-1}\tau)|^{2}\mathbf{1}_{E_{1}}).$

If we replace $\mathbf{1}_{E_{1}}$ by $\mathbf{1}_{E_{1}\backslash E}$, then
the resulting contribution is bounded by $L^{-A+10d}$, since $|a_{k}(t)|^{2}$
is bounded uniformly in $k$ and $t$ by mass conservation, so we may replace
$\mathbf{1}_{E_{1}}$ by $\mathbf{1}_{E}$. This then reduces to the expression

$\sum_{0\leq n_{1},n_{2}\leq
N}\mathbb{E}((\mathcal{J}_{n_{1}})_{k}(t)\overline{(\mathcal{J}_{n_{2}})_{k}(t)}\mathbf{1}_{E})+2\sum_{n=0}^{N}\mathrm{Re}\,\mathbb{E}((\mathcal{J}_{n})_{k}(t)\overline{b_{k}(t)}\mathbf{1}_{E})+\mathbb{E}(|b_{k}(t)|^{2}\mathbf{1}_{E}),$

where $t=\delta^{-1}\tau$. The terms involving $b$ are obviously bounded by
$L^{-100d}$ using (12.1) and $\|b\|_{Z}\leq L^{-500d}$, so we just need to
consider the correlations between $\mathcal{J}_{n_{1}}$ and
$\mathcal{J}_{n_{2}}$. In these correlations, if we replace $\mathbf{1}_{E}$
by $\mathbf{1}_{E^{c}}$, then the resulting contribution is

$\big{|}\mathbb{E}((\mathcal{J}_{n_{1}})_{k}(t)\overline{(\mathcal{J}_{n_{2}})_{k}(t)}\mathbf{1}_{E^{c}})\big{|}\leq(\mathbb{E}|(\mathcal{J}_{n_{1}})_{k}(t)|^{4})^{1/4}(\mathbb{E}|(\mathcal{J}_{n_{2}})_{k}(t)|^{4})^{1/4}(\mathbb{P}(E^{c}))^{1/2}\lesssim
L^{-A/2+10d}$

using Lemma A.3 and Proposition 2.5, so we may replace $\mathbf{1}_{E}$ by
$1$, thus reducing to the expression

$\sum_{0\leq n_{1},n_{2}\leq
N}\mathbb{E}((\mathcal{J}_{n_{1}})_{k}(t)\overline{(\mathcal{J}_{n_{2}})_{k}(t)})=\sum_{\mathcal{Q}}\mathcal{K}_{\mathcal{Q}}(t,t,k),$

where the last sum is taken over all couples
$\mathcal{Q}=(\mathcal{T}^{+},\mathcal{T}^{-})$ with $n(\mathcal{T}^{\pm})\leq
N$. We may replace this condition by the condition $n(\mathcal{Q})\leq 2N$,
because each term $\mathcal{K}_{\mathcal{Q}}$ in the difference must satisfy
$N\leq n(\mathcal{Q})\leq 2N$, and the set of these $\mathcal{Q}$ is invariant
under congruence, so we can bound the difference by
$(C^{+}\sqrt{\delta})^{N/2}\leq L^{-100d}$. This reduces our target, up to
errors $O(L^{-10d})$, to

$\sum_{n=0}^{2N}\sum_{n(\mathcal{Q})=n}\mathcal{K}_{\mathcal{Q}}(t,t,k)=\sum_{n=0}^{N}\mathcal{M}_{n}(t,k)+O(L^{-\nu})=n(\delta
t,k)+O(L^{-\nu})=n(\tau,k)+O(L^{-\nu}),$

where the last steps are due to Propositions 2.7 and 7.9. In the end we get
that

$\mathbb{E}(|\widehat{u}(\tau\cdot
T_{\mathrm{kin}},k)|^{2}\mathbf{1}_{E_{1}})=n(\tau,k)+O(L^{-\nu}),$

uniformly in $\tau\in[0,\delta]$ and $k\in\mathbb{Z}_{L}^{d}$. This proves
Theorem 1.1. ∎

##  Appendix A Preliminary lemmas

###  A.1. The exceptional set $\mathfrak{Z}$

We will define in Lemma A.1 the Lebesgue null set $\mathfrak{Z}$ used in
Theorem 1.1. Once Lemma A.1 is proved, for the rest of the paper we will fix
one $\beta\in(\mathbb{R}^{+})^{d}\backslash\mathfrak{Z}$.

######  Lemma A.1 (The genericity condition).

There exists a Lebesgue null set $\mathfrak{Z}\subset(\mathbb{R}^{+})^{d}$
such that the followings hold for any
$\beta=(\beta^{1},\cdots,\beta^{d})\in(\mathbb{R}^{+})^{d}\backslash\mathfrak{Z}$.

(1) For any integers $(K^{1},K^{2})\neq(0,0)$, we have

$|\beta^{1}K^{1}+\beta^{2}K^{2}|\gtrsim(1+|K^{1}|+|K^{2}|)^{-1}\log^{-4}(2+|K^{1}|+|K^{2}|);$
(A.1)

(2) The numbers $\beta^{1},\cdots,\beta^{d}$ are algebraically independent
over $\mathbb{Q}$, and for any $R$ we have

$\\#\left\\{(X,Y,Z)\in(\mathbb{Z}^{d})^{3}:|X|,|Y|,|Z|\leq R,\,X\neq
0,\,\max(|\langle X,Y\rangle_{\beta}|,|\langle X,Z\rangle_{\beta}|)\leq
1\right\\}\lesssim R^{3d-4+\frac{1}{6}}.$ (A.2)

###### Proof.

(1) This is standard in Diophantine approximation, see [8].

(2) Without loss of generality we may assume $\beta\in[1,2]^{d}$. If
$X^{j}Y^{j}=X^{j}Z^{j}=0$ for all $j$, since $X\neq 0$, the number of choices
for $(X,Y,Z)$ is clearly at most $R^{2d-1}$ which satisfies (A.2) since $d\geq
3$. If $X^{i}X^{j}(Y^{i}Z^{j}-Y^{j}Z^{i})=0$ for all $(i,j)$, but not all
$X^{j}Y^{j}$ and $X^{j}Z^{j}$ are zero, say $X^{1}Y^{1}\neq 0$, then for fixed
$(X,Y,Z)$, the Lebesgue measure of the set

$E:=\bigg{\\{}\beta\in[1,2]^{d}:\bigg{|}\sum_{\ell=1}^{d}\beta^{\ell}X^{\ell}Y^{\ell}\bigg{|}\leq
1,\,\bigg{|}\sum_{\ell=1}^{d}\beta^{\ell}X^{\ell}Z^{\ell}\bigg{|}\leq
1\bigg{\\}}$

is bounded by $C|X^{1}Y^{1}|^{-1}$. Moreover, once $X^{1}$ and $Y^{1}$ are
fixed, the number of choices for $(X^{j},Y^{j},Z^{j})$ for each $j\geq 2$ is
at most $R^{2}$. This implies that

$\int_{[1,2]^{d}}(\mathrm{left\ hand\ side\ of\
(\ref{generic1.5})})\,\mathrm{d}\beta\leq C\sum_{(X,Y,Z)}|X^{1}Y^{1}|^{-1}\leq
CR^{2d-1+\frac{1}{8}}.$

By Borel-Cantelli lemma, and using that $2d-1\leq 3d-4$, we get (A.2) for any
$R$ and almost all $\beta$.

Now suppose there is $1\leq i<j\leq d$ such that
$X^{i}X^{j}(Y^{i}Z^{j}-Y^{j}Z^{i})\neq 0$, say $(i,j)=(1,2)$. Then for fixed
$(X,Y,Z)$, the Lebesgue measure of $E$ is bounded by
$C|X^{1}X^{2}|^{-1}\cdot|Y^{1}Z^{2}-Y^{2}Z^{1}|^{-1}$, therefore

$\int_{[1,2]^{d}}(\mathrm{left\ hand\ side\ of\
(\ref{generic1.5})})\,\mathrm{d}\beta\leq
C\sum_{(X,Y,Z)}|X^{1}X^{2}|^{-1}\cdot|Y^{1}Z^{2}-Y^{2}Z^{1}|^{-1}.$

The last sum is bounded by $R^{3d-4+\frac{1}{8}}$, by fixing the values of
$X^{1}$, $X^{2}$ and $Y^{1}Z^{2}-Y^{2}Z^{1}$ and using the divisor estimate.
Again by Borel-Cantelli lemma, we obtain (A.2) for almost all $\beta$. ∎

###  A.2. Miscellaneous results

We collect some auxiliary results needed in the main proof.

######  Lemma A.2 (Complex Isserlis’ theorem).

Given $k_{j}\in\mathbb{Z}_{L}^{d}$ (not necessarily distinct) and
$\zeta_{j}\in\\{\pm\\}$ for $1\leq j\leq n$, then

$\mathbb{E}\bigg{[}\prod_{j=1}^{n}\eta_{k_{j}}^{\zeta_{j}}(\omega)\bigg{]}=\sum_{\mathscr{P}}\prod_{\\{j,j^{\prime}\\}\in\mathscr{P}}\mathbf{1}_{k_{j}=k_{j^{\prime}}},$
(A.3)

where the summation is taken over all partitions $\mathscr{P}$ of
$\\{1,2,\cdots,n\\}$ into two-element subsets $\\{j,j^{\prime}\\}$ such that
$\zeta_{j^{\prime}}=-\zeta_{j}$.

###### Proof.

Let all the different vectors in $\\{k_{j}:1\leq j\leq n\\}$ be
$k^{(1)},\cdots,k^{(r)}$. Assume for each $1\leq i\leq r$ and
$\zeta\in\\{\pm\\}$ that the number of $j$’s such that $k_{j}=k^{(i)}$ and
$\zeta_{j}=\zeta$ is $a_{i}^{\zeta}$, then we have

$\mathbb{E}\bigg{[}\prod_{j=1}^{n}\eta_{k_{j}}^{\zeta_{j}}(\omega)\bigg{]}=\mathbb{E}\bigg{[}\prod_{i=1}^{r}(g_{k^{(i)}})^{a_{i}^{+}}(\overline{g_{k^{(i)}}})^{a_{i}^{-}}\bigg{]}=\left\\{\begin{split}(a_{1}^{+})!\cdots(a_{r}^{+})!,&&\mathrm{if\
}a_{i}^{+}=a_{i}^{-}\mathrm{\ for\ all\ }i,\\\
0,&&\mathrm{otherwise}.\end{split}\right.$

On the other hand, for fixed $\mathscr{P}$ the product on the right hand side
of (A.3) is either $0$ or $1$, and equals $1$ if and only if each pair
$\\{j,j^{\prime}\\}\in\mathscr{P}$ is such that $k_{j}=k_{j^{\prime}}=k^{(i)}$
for some $i$, and $\zeta_{j^{\prime}}=-\zeta_{j}$. If $a_{i}^{+}=a_{i}^{-}$
for all $i$, then the number of choices for $\mathscr{P}$ clearly equals
$(a_{1}^{+})!\cdots(a_{r}^{+})!$; otherwise no such $\mathscr{P}$ exists. This
proves (A.3). ∎

######  Lemma A.3 (Gaussian hypercontractivity).

Given $n$ and $\zeta_{j}\in\\{\pm\\}$ for $1\leq j\leq n$, suppose the random
variable $X$ has the form

$X=\sum_{k_{1},\cdots,k_{n}}a_{k_{1}\cdots
k_{n}}\prod_{j=1}^{n}\eta_{k_{j}}^{\zeta_{j}}(\omega),$ (A.4)

where $a_{k_{1}\cdots k_{n}}$ are constants, then for any $q\geq 2$ we have

$\mathbb{E}|X|^{q}\leq(q-1)^{\frac{nq}{2}}\cdot(\mathbb{E}|X|^{2})^{\frac{q}{2}}.$
(A.5)

###### Proof.

This is the standard hypercontractivity estimate for Gaussians, see for
example [52]. ∎

######  Lemma A.4 (A combinatoric inequality).

Given a multi-index $\rho$, we have

$\sum_{\rho^{1}+\cdots+\rho^{9}=\rho}\frac{\rho!}{(\rho^{1})!\cdots(\rho^{9})!}\cdot(2|\rho^{1}|)!\cdots(2|\rho^{9}|)!\leq
C(2|\rho|)!.$ (A.6)

###### Proof.

We first fix $\rho_{*}^{2}:=\rho^{2}+\cdots+\rho^{9}$ and sum over
$(\rho^{2},\cdots,\rho^{9})$, then sum over $(\rho^{1},\rho_{*}^{2})$. The sum
over $(\rho^{2},\cdots,\rho^{9})$ can be bounded by inductively repeating this
process, provided one can bound the sum over $(\rho^{1},\rho_{*}^{2})$. To
bound this latter sum, if $\rho=(a_{1},\cdots,a_{n})$ then it can be written
as

$\sum_{0\leq b_{j}\leq a_{j}}\prod_{j=1}^{n}\binom{a_{j}}{b_{j}}(2A-2B)!(2B)!$

where $A=a_{1}+\cdots+a_{n}$ and $B=b_{1}+\cdots+b_{n}$. If $B$ is fixed, then
the sum over $(b_{1},\cdots,b_{n})$ equals $\binom{A}{B}$ by a simple
application of the binomial theorem (or the Vandermonde identity), so (A.6)
would follow from the inequality

$\sum_{0\leq B\leq A}\binom{A}{B}\binom{2A}{2B}^{-1}\leq C.$ (A.7)

By symmetry, in (A.7) we may assume $B\leq A/2$, so
$\binom{A}{B}/\binom{2A}{2B}\leq 1/\binom{A}{B}\leq 1/\binom{2B}{B}\leq
2^{-B}$, which proves (A.7) and hence (A.6) by induction. ∎

######  Lemma A.5 (A sharp Hua’s lemma).

For $s,r\in\mathbb{R}$ and $h\in\mathbb{Z}$, define the Gauss sums

$G_{h}(s,r,n)=\sum_{p=h}^{h+n}e(sp^{2}+rp),n\in\mathbb{N};\qquad\mathrm{and}\qquad
G_{h}(s,r,x)=G_{h}(s,r,\lfloor x\rfloor),x\in\mathbb{R}_{+},$ (A.8)

where $\lfloor x\rfloor$ is the floor function, and $e(z)=e^{2\pi iz}$. Then
we have

$\|G_{h}(\cdot,r,n)\|_{L^{4}([0,1])}^{4}\lesssim
n^{2}\log(2+n),\quad\|G_{h}(\cdot,r,n)\|_{L^{6}([0,1])}^{6}\lesssim n^{4}$
(A.9)

uniformly in $(r,h)$. The constants involved in $\lesssim$ here are absolute
constants.

###### Proof.

We only need to bound the cardinalities of the sets

$A_{4}=\\{(a,b,c,d)\in[h,h+n]^{4}:a^{2}-b^{2}+c^{2}-d^{2}=0\\},\\\
\mathrm{and}\,\,A_{6}=\\{(a,\cdots,f)\in[h,h+n]^{6}:a^{2}-b^{2}+c^{2}-d^{2}+e^{2}-f^{2}=0\\}.$

By changing variables $(a,b)\mapsto(a+b,a-b)$ etc., we can reduce to the sets

$B_{4}=\\{(a,b,c,d):ab+cd=0\\}\quad\mathrm{and}\quad
B_{6}=\\{(a,\cdots,f):ab+cd+ef=0\\},$

where $|a|,|c|,|e|\leq n$ and $b,d,f\in[2h,2h+2n]$. To count $\\#B_{6}$, we
may assume $a\geq|c|\geq|e|$ (and $a>0$). Note that $f$ belongs to a fixed
residue class modulo $\gcd(a,c)/\gcd(a,c,e)$; once $f$ is fixed, then $d$
belongs to a fixed residue class modulo $a/\gcd(a,c)$. When $f$ and $d$ are
fixed then $b$ is unique. This implies that

$\\#B_{6}\lesssim\sum_{a\geq|c|\geq|e|}\frac{n}{\gcd(a,c)/\gcd(a,c,e)}\cdot\frac{n}{a/\gcd(a,c)}=n^{2}\sum_{a\geq|c|\geq|e|}\frac{\gcd(a,c,e)}{a}.$

For the last sum, let $\gcd(a,c,e)=\Delta$ and $a=a^{\prime}\Delta$ etc., then

$\sum_{a\geq|c|\geq|e|}\frac{\gcd(a,c,e)}{a}\lesssim\sum_{0<\Delta\leq
n}\sum_{0<a^{\prime}\leq n/\Delta}\sum_{|c^{\prime}|,|e^{\prime}|\leq
a^{\prime}}\frac{1}{a^{\prime}}\lesssim\sum_{0<\Delta\leq
n}(n/\Delta)^{2}\lesssim n^{2},$

hence $\\#B_{6}\lesssim n^{4}$. In the same way we can bound $\\#B_{4}\lesssim
n^{2}\log(2+n)$. ∎

######  Lemma A.6.

Fix $M\geq L^{(100d)^{3}}$. Consider
$k=(k^{1},\cdots,k^{d})\in\mathbb{Z}_{L}^{d}$ and a system
$(r,q,v_{1},\cdots,v_{q},f,y)$ where $0\leq q\leq r\leq d$,
$v_{j}\in\mathbb{Z}_{L}^{r}\,(1\leq j\leq q)$ are nonzero orthogonal vectors
and $f=(f^{r+1},\cdots,f^{d})\in\mathbb{Z}_{L}^{d-r}$ and
$y=(y^{1},\cdots,y^{q})\in\mathbb{R}^{q}$, such that $|f|,|v_{j}|,|y|\leq
M^{C_{0}(d)}$ where $C_{0}(d)$ is a fixed large constant depending on $d$, and
that the linear span of $\\{v_{1},\cdots,v_{q}\\}$ does not contain any
coordinate vector in $\mathbb{R}^{r}$. We say that the system
$(r,q,v_{1},\cdots,v_{q},f,y)$ _represents_ $k$, if the followings hold:

  1. (1)

If $z=(z^{1},\cdots,z^{d})\in\mathbb{Z}_{L}^{d}$, $|z|\leq M$ and $|\langle
k,z\rangle_{\beta}|\leq M$, then the vector $(z^{1},\cdots,z^{r})$ is a linear
combination of $\\{v_{1},\cdots,v_{q}\\}$;

  2. (2)

If $(z^{1},\cdots,z^{r})=\gamma^{1}v_{1}+\cdots+\gamma^{q}v_{q}$, then we have
$\langle
k,z\rangle_{\beta}=y^{1}\gamma^{1}+\cdots+y^{q}\gamma^{q}+\beta^{r+1}f^{r+1}z^{r+1}+\cdots+\beta^{d}f^{d}z^{d}$.

Then, each $k\in\mathbb{Z}^{d}$, after possibly permuting the coordinates, is
represented by some system $(r,q,v_{1},\cdots,v_{q},f,y)$. Conversely, for
each system $(r,q,v_{1},\cdots,v_{q},f,y)$ and $\theta>0$, there exists
$k\in\mathbb{Z}^{d}$ represented by a system
$(r,q,v_{1},\cdots,v_{q},f,y^{\prime})$, such that $|y^{\prime}-y|<\theta$.

###### Proof of Lemma A.6.

As $M\geq L^{(100d)^{3}}$, upon multiplying everything by $L$, we may replace
$\mathbb{Z}_{L}$ by $\mathbb{Z}$. As our convention, in the proof $C$ will
denote any large constant depending only on $d$. In the first part, given $k$
we will construct the system $(r,q,v_{1},\cdots,v_{q},f,y)$, which is done by
induction in $d$. The case $d=1$ is obvious, now suppose the result is true
for $d-1$, with constant $C_{0}=C_{0}(d-1)$. We also denote
$k_{\beta}=(k^{1}\beta^{1},\cdots,k^{d}\beta^{d})$ for
$k=(k^{1},\cdots,k^{d})$ and $\beta=(\beta^{1},\cdots,\beta^{d})$, so that
$\langle k,z\rangle_{\beta}=\langle k_{\beta},z\rangle$.

Fix $k\in\mathbb{Z}^{d}$, consider the set $H$ of $z\in\mathbb{Z}^{d}$ such
that $|z|\leq M$ and $|\langle k,z\rangle_{\beta}|\leq M$ (clearly $0\in H$).
Let $q$ be the maximal number of linearly independent vectors in $H$, we may
fix a maximum independent set $\\{w_{1},\cdots,w_{q}\\}\subset H$, and apply
Gram-Schmidt process to get orthogonal vectors $(v_{1},\cdots,v_{q})$. Since
each $w_{j}\in\mathbb{Z}^{d}$ and $|w_{j}|\leq M$, we can easily make
$v_{j}\in\mathbb{Z}^{d}$ and $|v_{j}|\leq M^{C}$. If the linear span of
$\\{v_{1},\cdots,v_{q}\\}$ does not contain any coordinate vector in
$\mathbb{R}^{d}$, then we shall prove the result with $r=d$. In fact, (1) is
already satisfied by definition; since $|\langle k,w_{j}\rangle_{\beta}|\leq
M$ for $1\leq j\leq q$, we also know that $|\langle
k,v_{j}\rangle_{\beta}|\leq M^{C}$ for $1\leq j\leq q$. Let $y^{j}=\langle
k,v_{j}\rangle_{\beta}$, then $|y|\leq M^{C}$ and (2) is also satisfied.

If, instead, the linear span of $\\{v_{1},\cdots,v_{q}\\}$ contains a
coordinate vector in $\mathbb{R}^{d}$, say $e_{d}=(0,\cdots,0,1)$, we shall
apply the induction hypothesis. In this case we have $|\langle
k,v_{j}\rangle_{\beta}|\leq M^{C}$ for $1\leq j\leq q$ and hence $|k^{d}|\leq
M^{C}$. By induction hypothesis (with $M$ replaced by $M^{C}$), the vector
$(k^{1},\cdots,k^{d-1})$ is represented by some system
$(r,q,v_{1},\cdots,v_{q},f,y)$ where $0\leq q\leq r\leq d-1$ and
$|f|,|v_{j}|,|y|\leq M^{CC_{0}}$. Now we claim that $k$ is represented by
$(r,q,v_{1},\cdots,v_{q},f^{\prime},y)$ where $f^{\prime}=(f,k^{d})$; in fact
(2) is satisfied by definition, as for (1), if $|z|\leq M$ and $|\langle
k,z\rangle_{\beta}|\leq M$, then
$|\beta^{1}k^{1}z^{1}+\cdots+\beta^{d-1}k^{d-1}z^{d-1}|\leq M^{C}$ as
$|k^{d}|\leq M^{C}$, so we may apply the induction hypothesis (with $M$
replaced by $M^{C}$) to show that $(z^{1},\cdots,z^{r})$ is a linear
combination of $\\{v_{1},\cdots,v_{q}\\}$. In either case we have constructed
the desired system, with $C_{0}(d)=C\cdot C_{0}(d-1)$.

Now, suppose a system $(r,q,v_{1},\cdots,v_{q},f,y)$, and $\theta>0$, is
fixed. We may choose $k^{j}=f^{j}$ for $r+1\leq j\leq d$, and again notice
that $|z|\leq M$ and $|\langle k,z\rangle_{\beta}|\leq M$ implies that
$|\beta^{1}k^{1}z^{1}+\cdots+\beta^{r}k^{r}z^{r}|\leq M^{C}$. Therefore we
only need to consider $r=d$. Select vectors
$u_{1},\cdots,u_{d-q}\in\mathbb{Z}^{d}$ such that they form an orthogonal
basis with $\\{v_{1},\cdots,v_{q}\\}$, and $|u_{j}|\leq M^{C}$ for $1\leq
j\leq d-q$. Now choose $k^{*}=\rho_{1}u_{1}+\cdots+\rho_{d-q}u_{d-q}$, where
$\rho_{j}$ are large integers, and assume $k$ is chosen such that
$|k_{\beta}-k^{*}|\leq M^{C}$. We will assume $0\leq\rho_{j}\leq B$ and
$B\gg_{M,\theta}1$. Clearly, if $|z|\leq M$ and $|\langle
k,z\rangle_{\beta}|\leq M$, then $|\langle k^{*},z\rangle|\leq M^{C}$.

Suppose $|z|\leq M$ and $|\langle k^{*},z\rangle|\leq M^{C}$. If we decompose
$z=x_{1}u_{1}+\cdots+x_{d-q}u_{d-q}+z^{\prime}$ where $z^{\prime}$ is a linear
combination of $(v_{1},\cdots,v_{q})$, then

$\langle k^{*},z\rangle=\sum_{j=1}^{d-q}\rho_{j}|u_{j}|^{2}x_{j}.$

Each $x_{j}$ is a rational number and $\max_{j}|x_{j}|\geq M^{-C}$ unless all
$x_{j}=0$, and the number of choices for $(x_{1},\cdots,x_{d-q})$ is at most
$M^{C}$ when $z$ varies. For each fixed nonzero $(x_{1},\cdots,x_{d-q})$ the
number of choices for $(\rho_{1},\cdots,\rho_{d-q})$ satisfying $|\langle
k^{*},z\rangle|\leq M^{C}$ is at most $M^{C}B^{d-q-1}$, so for at least
$B^{d-q}-M^{C}B^{d-q-1}$ choices of $(\rho_{1},\cdots,\rho_{d-q})$ (and for
any choice of $k$ satisfying $|k_{\beta}-k^{*}|\leq M^{C}$ given $k^{*}$), we
have that $|z|\leq M$ and $|\langle k,z\rangle_{\beta}|\leq M$ implies that
$z$ is a linear combination of $v_{j}\,(1\leq j\leq q)$.

For such choices we already have (1). Clearly $z$ is then represented by
$(q,v_{1},\cdots,v_{q},y^{\prime})$ (with $r=d$ and $f$ being void) where
$(y^{\prime})^{j}=\langle k_{\beta},v_{j}\rangle=-\langle
k^{*}-k_{\beta},v_{j}\rangle$. Given $(v_{j})$ and $y$, we may fix a vector
$h\in\mathbb{R}^{d}$, where $|h|\leq M^{C}$, such that $\langle
h,v_{j}\rangle=-y^{j}$ for $1\leq j\leq q$; it then suffices to choose $k^{*}$
and $k$ such that $|k^{*}-k_{\beta}-h|\leq M^{-C_{1}}\theta$ with $C_{1}$
larger than all the $C$ appearing above. (note that this also implies
$|k_{\beta}-k^{*}|\leq M^{C}$). Since $k$ can be arbitrarily chosen, it
suffices to have

$\bigg{\\{}\frac{(k^{*})^{j}-h^{j}}{\beta^{j}}\bigg{\\}}<M^{-2C_{1}}\theta\quad\mathrm{for}\quad
1\leq j\leq d,$

where $\\{\cdot\\}$ means the distance to the nearest integer. Clearly

$\frac{(k^{*})^{j}-h^{j}}{\beta^{j}}=\sum_{i=1}^{d-q}\frac{(u_{i})^{j}}{\beta^{j}}\rho_{i}-\frac{h^{j}}{\beta^{j}}.$

Since the linear span of $\\{v_{j}\\}$ does not contain any coordinate vector
in $\mathbb{R}^{d}$, we know that, for each $1\leq j\leq d$, there exists
$1\leq i=i(j)\leq d-q$ such that $(u_{i})^{j}\neq 0$.

We may choose $\rho_{i}$ such that

$\bigg{\\{}\frac{(u_{i})^{j}}{\beta^{j}}\rho_{i}\bigg{\\}}<M^{-3C_{1}}\theta\,\,(i\neq
i(j));\quad\bigg{\\{}\frac{(u_{i})^{j}}{\beta^{j}}\rho_{i}-\frac{h^{j}}{\beta^{j}}\bigg{\\}}<M^{-3C_{1}}\theta\,\,(i=i(j)).$

Given $i$, since all the nonzero numbers in the set
$\\{1,(u_{i})^{j}(\beta^{j})^{-1}:1\leq j\leq d\\}$ are $\mathbb{Q}$-linearly
independent, by Weyl’s equidistribution theorem, we see that the number of
$(\rho_{1},\cdots,\rho_{d-q})$ satisfying all the above conditions is
$\gtrsim_{M,\theta}B^{d-q}$. Therefore, we may choose
$(\rho_{1},\cdots,\rho_{d-q})$, and hence $k^{*}$ and $k$, such that both (1)
and (2) are satisfied. ∎

###  A.3. Lattice point counting bounds

We list the various lattice point counting bounds, which are the main
technical tools used in Section 9.3.

######  Lemma A.7 (Sphere counting).

Uniformly in $a\in\mathbb{Z}_{L}^{d}$ and $\gamma\in\mathbb{R}$, we have the
bound

$\\#\big{\\{}x\in\mathbb{Z}_{L}^{d}:|x-a|\leq
1,\,\,||x|_{\beta}^{2}-\gamma|\leq\delta^{-1}L^{-2}\big{\\}}\lesssim\delta^{-1}L^{d-\frac{4}{3}}.$

###### Proof.

By dividing an interval of length $\delta^{-1}L^{-2}$ into $O(\delta^{-1})$
intervals of length $L^{-2}$ we may assume $\delta=1$. Let
$y=(y^{1},\cdots,y^{d})=(\sqrt{\beta^{1}}x^{1},\cdots,\sqrt{\beta^{d}}x^{d})$,
then $|y|^{2}=\gamma+O(L^{-2})$, where $|y|$ is the usual norm in
$\mathbb{R}^{d}$. If we fix the coordinates $y^{j}\,(3\leq j\leq d)$, noticing
that each $y^{j}\,(3\leq j\leq d)$ has $\lesssim L$ choices, it then suffices
to prove that

$\\#\big{\\{}(u,v)\in(\sqrt{\beta^{1}}\mathbb{Z}_{L})\times(\sqrt{\beta^{2}}\mathbb{Z}_{L}):u^{2}+v^{2}=\gamma+O(L^{-2}),\,\,|u-u_{0}|+|v-v_{0}|\lesssim
1\big{\\}}\lesssim L^{\frac{2}{3}}$ (A.10)

uniformly in $(u_{0},v_{0},\gamma)\in\mathbb{R}^{3}$.

Let $|\gamma|\sim R^{2}$ (we may assume $R\gg L^{-1}$), then $(u,v)$ belongs
to the $O(\varepsilon)$ neighborhood of a circle centered at the origin of
radius $\sim R$, where $\varepsilon=L^{-2}R^{-1}$. Since $(u,v)$ also belongs
to a disc of radius $O(1)$, we know that $(u,v)$ actually belongs to the
$O(\varepsilon)$ neighborhood of an arc of length $O(\min(R,1))$ on the
circle. Let
$\Gamma:=(\sqrt{\beta^{1}}\mathbb{Z}_{L})\times(\sqrt{\beta^{2}}\mathbb{Z}_{L})$
be a fixed lattice, it will suffice to prove that the number of points in
$\Gamma$ that belong to this neighborhood is $\lesssim L^{\frac{2}{3}}$.

Now, we may decompose the above arc of length $O(\min(R,1))$ into at most
$O(L^{\frac{2}{3}})$ sub-arcs, each with length $\ll\rho$, where
$\rho=L^{-\frac{2}{3}}R^{\frac{1}{3}}$, note that $\varepsilon\ll\rho\ll R$.
Thus is suffices to prove that the $O(\varepsilon)$ neighborhood of each sub-
arc contains $O(1)$ points in $\Gamma$. Let this neighborhood be $M$, from
elementary geometry we can calculate that the area of the convex hull of $M$
is

$A\ll\left(\frac{\rho}{R}\right)R\varepsilon+R^{2}\left(\frac{\rho}{R}\right)^{3}=\rho\varepsilon+\frac{\rho^{3}}{R}\lesssim
L^{-2}.$

But any nondegenerate triangle with vertices in $\Gamma$ have area $\gtrsim
L^{-2}$, so the points in $\Gamma\cap M$ must be collinear; however $M$ is
contained in an annulus of width $2\varepsilon$, and any straight line
contains at most two segments in this annulus, each having length at most
$O(\sqrt{\varepsilon R})=O(L^{-1})$, so in any case the number of points in
$\Gamma\cap M$ is at most $O(1)$. ∎

######  Lemma A.8 (Good and bad vectors).

We say a vector $0\neq x\in\mathbb{Z}_{L}^{d}$ is a _bad_ vector, if

$\\#\big{\\{}y\in\mathbb{Z}_{L}^{d}:|y-b|\leq 1,\,|\langle
x,y\rangle_{\beta}-\Gamma|\leq L^{-2}\big{\\}}\geq L^{d-1-\frac{1}{4}}$ (A.11)

for _some_ $b\in\mathbb{Z}_{L}^{d}$ and $\Gamma\in\mathbb{R}$; otherwise we
say $x$ is a _good_ vector. Then for any $a\in\mathbb{Z}_{L}^{d}$, the number
of bad vectors $x$ satisfying $|x-a|\leq 1$ is at most $L^{d-1-\frac{1}{4}}$.

###### Proof.

If (A.11) is true for some $(b,\Gamma)$, then it is actually true for
$b=\Gamma=0$ up to some constant, by fixing $x$ taking the difference between
any two possibilities of $y$. We will show that

$\sum_{x\neq 0,|x-a|\leq 1}\sum_{|y|,|z|\leq 1}\mathbf{1}_{|\langle
x,y\rangle_{\beta}|\leq L^{-2}}\cdot\mathbf{1}_{|\langle
x,z\rangle_{\beta}|\leq L^{-2}}\lesssim L^{3d-4+\frac{1}{6}},$ (A.12)

which implies the desired result, as the left hand side of (A.12) is just the
sum of the square of the left hand side of (A.11) over $|x-a|\leq 1$. However
the left hand side of (A.12) is bounded by the same expression with $a=0$,
again by fixing $(y,z)$ and taking the difference between any two
possibilities of $x$. Moreover, if we set $(x,y,z)=L^{-1}(X,Y,Z)$ with
$(X,Y,Z)\in(\mathbb{Z}^{d})^{3}$, then (A.12) with $a=0$ is just (A.2) which
follows from the definition of $\mathfrak{Z}$. This completes the proof. ∎

######  Lemma A.9 (Atom counting bounds).

For $1\leq j\leq 5$, let $x_{j}\in\mathbb{Z}_{L}^{d}$ be variables,
$a_{j}\in\mathbb{Z}_{L}^{d}$ and $\zeta_{j}\in\\{\pm\\}$ be fixed parameters.
Fix also the parameters $k_{i}\in\mathbb{Z}_{L}^{d}$ and
$\gamma_{i}\in\mathbb{R}$ for $i\in\\{1,2\\}$. We require that
$|x_{j}-a_{j}|\leq 1$ for each $j$. If any of the statements below involves an
equation of form $\sum_{j\in A}\zeta_{j}x_{j}=k_{i}$ with some set $A$, then
we also require that (i) no three of $\zeta_{j}\,(j\in A)$ are the same, and
(ii) if $j,j^{\prime}\in A$ and $\zeta_{j^{\prime}}+\zeta_{j}=0$ then
$x_{j}\neq x_{j^{\prime}}$.

We have the following estimates, where the implicit constants only depend on
$d$ and $\beta$, and do not depend on $(\delta,L)$ or any of the parameters
$(a_{j},\gamma_{i},k_{i})$:

(1) (Two-vector counting) If we require

$\zeta_{1}x_{1}+\zeta_{2}x_{2}=k_{1},\quad\big{|}\zeta_{1}|x_{1}|_{\beta}^{2}+\zeta_{2}|x_{2}|_{\beta}^{2}-\gamma_{1}\big{|}\leq\delta^{-1}L^{-2},$
(A.13)

then the number of choices for $(x_{1},x_{2})$ is
$\lesssim\delta^{-1}L^{d-1}$, and is $\lesssim\delta^{-1}L^{d-1-\frac{1}{3}}$
if $\zeta_{1}=\zeta_{2}$.

(2) (Three-vector counting) If we require

$\zeta_{1}x_{1}+\zeta_{2}x_{2}+\zeta_{3}x_{3}=k_{1},\quad\big{|}\zeta_{1}|x_{1}|_{\beta}^{2}+\zeta_{2}|x_{2}|_{\beta}^{2}+\zeta_{3}|x_{3}|_{\beta}^{2}-\gamma_{1}\big{|}\leq\delta^{-1}L^{-2},$
(A.14)

then the number of choices for $(x_{1},x_{2},x_{3})$ is
$\lesssim\delta^{-1}L^{2d-2}$.

(3) (Four-vector counting 1) If we require

$\left\\{\begin{aligned}
&\zeta_{1}x_{1}+\zeta_{2}x_{2}=k_{1},\quad\big{|}\zeta_{1}|x_{1}|_{\beta}^{2}+\zeta_{2}|x_{2}|_{\beta}^{2}-\gamma_{1}\big{|}\leq\delta^{-1}L^{-2},\\\
&\zeta_{1}x_{1}+\zeta_{3}x_{3}+\zeta_{4}x_{4}=k_{2},\quad\big{|}\zeta_{1}|x_{1}|_{\beta}^{2}+\zeta_{3}|x_{3}|_{\beta}^{2}+\zeta_{4}|x_{4}|_{\beta}^{2}-\gamma_{2}\big{|}\leq\delta^{-1}L^{-2},\end{aligned}\right.$
(A.15)

then the number of choices for $(x_{1},\cdots,x_{4})$ is
$\lesssim\delta^{-2}L^{2d-2-\frac{1}{4}}$.

(4) (Four-vector counting 2) If we require

$\left\\{\begin{aligned}
&\zeta_{1}x_{1}+\zeta_{2}x_{2}+\zeta_{3}x_{3}=k_{1},\quad\big{|}\zeta_{1}|x_{1}|_{\beta}^{2}+\zeta_{2}|x_{2}|_{\beta}^{2}+\zeta_{3}|x_{3}|_{\beta}^{2}-\gamma_{1}\big{|}\leq\delta^{-1}L^{-2},\\\
&\zeta_{1}x_{1}+\zeta_{2}x_{2}+\zeta_{4}x_{4}=k_{2},\quad\big{|}\zeta_{1}|x_{1}|_{\beta}^{2}+\zeta_{2}|x_{2}|_{\beta}^{2}+\zeta_{4}|x_{4}|_{\beta}^{2}-\gamma_{2}\big{|}\leq\delta^{-1}L^{-2},\end{aligned}\right.$
(A.16)

_and assume that $(\zeta_{3},x_{3})\neq(\zeta_{4},x_{4})$_, then the number of
choices for $(x_{1},\cdots,x_{4})$ is
$\lesssim\delta^{-2}L^{2d-2-\frac{1}{4}}$.

(5) (Five-vector counting 1) If we require

$\left\\{\begin{aligned}
&\zeta_{1}x_{1}+\zeta_{2}x_{2}+\zeta_{3}x_{3}=k_{1},\quad\big{|}\zeta_{1}|x_{1}|_{\beta}^{2}+\zeta_{2}|x_{2}|_{\beta}^{2}+\zeta_{3}|x_{3}|_{\beta}^{2}-\gamma_{1}\big{|}\leq\delta^{-1}L^{-2},\\\
&\zeta_{1}x_{1}+\zeta_{4}x_{4}+\zeta_{5}x_{5}=k_{2},\quad\big{|}\zeta_{1}|x_{1}|_{\beta}^{2}+\zeta_{4}|x_{4}|_{\beta}^{2}+\zeta_{5}|x_{5}|_{\beta}^{2}-\gamma_{2}\big{|}\leq\delta^{-1}L^{-2},\end{aligned}\right.$
(A.17)

then the number of choices for $(x_{1},\cdots,x_{5})$ is
$\lesssim\delta^{-2}L^{3d-3-\frac{1}{4}}$.

(6) (Five-vector counting 2) If we require

$\left\\{\begin{aligned}
&\zeta_{1}x_{1}+\zeta_{2}x_{2}+\zeta_{3}x_{3}=k_{1},\quad\big{|}\zeta_{1}|x_{1}|_{\beta}^{2}+\zeta_{2}|x_{2}|_{\beta}^{2}+\zeta_{3}|x_{3}|_{\beta}^{2}-\gamma_{1}\big{|}\leq\delta^{-1}L^{-2},\\\
&\zeta_{1}x_{1}+\zeta_{2}x_{2}+\zeta_{4}x_{4}+\zeta_{5}x_{5}=k_{2},\quad\big{|}\zeta_{1}|x_{1}|_{\beta}^{2}+\zeta_{2}|x_{2}|_{\beta}^{2}+\zeta_{4}|x_{4}|_{\beta}^{2}+\zeta_{5}|x_{5}|_{\beta}^{2}-\gamma_{2}\big{|}\leq\delta^{-1}L^{-2},\end{aligned}\right.$
(A.18)

then the number of choices for $(x_{1},\cdots,x_{5})$ is
$\lesssim\delta^{-2}L^{3d-3-\frac{1}{4}}$.

(7) (Five-vector counting 3) If we require (A.17), and that

$\left\\{\begin{aligned}
&\zeta_{2}x_{2}-\zeta_{4}x_{4}=k_{1}^{*},\quad\big{|}\zeta_{2}|x_{2}|_{\beta}^{2}-\zeta_{4}|x_{4}|_{\beta}^{2}-\gamma_{1}^{*}\big{|}\leq
n\delta^{-1}L^{-2},\\\
&\zeta_{3}x_{3}-\zeta_{5}x_{5}=k_{2}^{*},\quad\big{|}\zeta_{3}|x_{3}|_{\beta}^{2}-\zeta_{5}|x_{5}|_{\beta}^{2}-\gamma_{2}^{*}\big{|}\leq
n\delta^{-1}L^{-2}\end{aligned}\right.$ (A.19)

for some constants $(k_{1}^{*},k_{2}^{*},\gamma_{1}^{*},\gamma_{2}^{*})$ and
$n\geq 1$, _and assume that
$(\zeta_{2},\zeta_{3},x_{2},x_{3})\neq(\zeta_{4},\zeta_{5},x_{4},x_{5})$_,
then the number of choices for $(x_{1},\cdots,x_{5})$ is $\lesssim
n\delta^{-2}L^{2d-2-\frac{1}{4}}$.

Note that, if $n\leq(\log L)^{3}$ then the bound in (7) can be replaced by
$\delta^{-2}L^{2d-2-\frac{1}{6}}$.

###### Proof.

In all the proofs, we may assume $\delta=1$ as above, by dividing an interval
of length $\delta^{-1}L^{-2}$ into $O(\delta^{-1})$ intervals of length
$L^{-2}$.

(1) If $\zeta_{1}=\zeta_{2}$, then $x_{1}+x_{2}=\pm k_{1}$ is fixed. For
$y=x_{1}-x_{2}$, we have that

$|y|_{\beta}^{2}=\pm 2\gamma_{1}-|k_{1}|_{\beta}^{2}+O(L^{-2});$

moreover as $|x_{1}-a_{1}|\leq 1$ we have that $|y-(2a_{1}-k_{1})|\leq 2$.
Lemma A.7 then implies that the number of choices for $y$, and hence for
$(x_{1},x_{2})$, is $\lesssim L^{d-\frac{4}{3}}$.

Otherwise, we may assume $\zeta_{1}=+$ and $\zeta_{2}=-$, then
$x_{1}-x_{2}=k_{1}\neq 0$. Let $y=x_{1}+x_{2}$, then we have that

$\langle k_{1},y\rangle_{\beta}=\gamma_{1}+O(L^{-2});$

moreover as $|x_{1}-a_{1}|\leq 1$ we have that $|y-(2a_{1}-k_{1})|\leq 2$. We
may assume that the first coordinate $k_{1}^{1}$ of $k_{1}$ is nonzero, then
$|k_{1}^{1}|\geq L^{-1}$. Thus, when the coordinates $y^{j}\,(2\leq j\leq d)$
are fixed, $y^{1}$ will belong to an interval of length $\lesssim L^{-1}$ and
will have $\lesssim 1$ choices. As each $y^{j}\,(2\leq j\leq d)$ has $\lesssim
L$ choices, we conclude that the number of choices for $y$, and hence for
$(x_{1},x_{2})$, is $\lesssim L^{d-1}$.

Note that, if in addition we assume $x_{1}-x_{2}$ is a good vector, then by
definition we can bound the number of choices of $(x_{1},x_{2})$ by
$L^{d-1-\frac{1}{4}}$.

(2) By assumption the $\zeta_{j}\,(1\leq j\leq 3)$ are not all equal, so we
may assume $\zeta_{1}=\zeta_{3}=+$ and $\zeta_{2}=-$. Let $y=k_{1}-x_{1}$ and
$z=k_{1}-x_{3}$, then we have that

$\langle
y,z\rangle_{\beta}=\frac{|k_{1}|_{\beta}^{2}-\gamma_{1}}{2}+O(L^{-2});$

moreover as $|x_{1}-a_{1}|\leq 1$ and $|x_{3}-a_{3}|\leq 1$ we have that
$|y-(k_{1}-a_{1})|\leq 1$ and $|z-(k_{1}-a_{3})|\leq 1$. By applying
Proposition 6.1 with $n=1$, where $W$ and $\Psi$ are two fixed (translates of)
nonnegative compactly supported smooth cutoff functions, we know that the
number of choices for $(y,z)$, and hence for $(x_{1},x_{2},x_{3})$, is
$\lesssim L^{2d-2}$. Note that the second inequality in (6.5) is not needed if
one only needs the upper bound instead of asymptotics.

(3) If $\zeta_{3}=\zeta_{4}$, then by (1) we know that $(x_{1},x_{2})$ have at
most $L^{d-1}$ choices, while for $x_{1}$ fixed, $(x_{3},x_{4})$ has at most
$L^{d-1-\frac{1}{3}}$ choices, so the total number of choices for
$(x_{1},\cdots,x_{4})$ is at most $L^{2d-2-\frac{1}{3}}$. The same is true
(with $\frac{1}{4}$ instead of $\frac{1}{3}$) if $\zeta_{3}+\zeta_{4}=0$ and
$x_{3}-x_{4}$ is a good vector. But if $x_{3}-x_{4}$ is a bad vector, then
$x_{1}$ is a fixed translate of a bad vector which belongs to a fixed ball of
radius $1$, so by Lemma A.8, the number of choices for $x_{1}$, and hence
$(x_{1},x_{2})$, is at most $L^{d-1-\frac{1}{4}}$, so we get the same result.

(4) This follows from (3) by taking the difference of the two equations, and
noticing that if $\zeta_{3}=\zeta_{4}$, we must have $x_{3}\neq x_{4}$.

(5) If $\zeta_{4}=\zeta_{5}$, then by (2) we know that $(x_{1},x_{2},x_{3})$
have at most $L^{2d-2}$ choices, while for $x_{1}$ fixed, $(x_{4},x_{5})$ has
at most $L^{d-1-\frac{1}{3}}$ choices, so the total number of choices for
$(x_{1},\cdots,x_{5})$ is at most $L^{3d-3-\frac{1}{3}}$; note that this
estimate is valid even if we allow $\zeta_{3}=\zeta_{4}=\zeta_{5}$. The same
is true (with $\frac{1}{4}$ instead of $\frac{1}{3}$) if
$\zeta_{4}+\zeta_{5}=0$ and $x_{4}-x_{5}$ is a good vector. If $x_{4}-x_{5}$
is a bad vector, then $x_{1}$ has at most $L^{d-1-\frac{1}{4}}$ choices by
Lemma A.8. For $x_{1}$ fixed, the number of $(x_{2},x_{3})$ and
$(x_{4},x_{5})$ can be bounded by $L^{d-1}$ by (1), so we get the same result.

(6) This follows from the first part of (5) by taking the difference of the
two equations, provided $(\zeta_{3},x_{3})\neq(\zeta_{j},x_{j})$ for
$j\in\\{4,5\\}$; now suppose, say, $\zeta_{3}=\zeta_{4}$ and $x_{3}=x_{4}$,
then the value of $x_{5}$ is fixed and the number of choices for
$(x_{1},x_{2},x_{3})$ is at most $L^{2d-2}$ by (2), so the result is still
true.

(7) By subdividing one interval of length $nL^{-2}$ we may assume $n=1$. The
result then follows from (3), since we may assume (for example) either
$\zeta_{2}\neq\zeta_{4}$ or $x_{2}\neq x_{4}$, and simply exploit the first
equation in (A.17) and the first equation in (A.19). ∎

##  Appendix B An example of molecule reduction

Here we provide an example of the molecule reduction algorithm described in
Section 9.4.2. For simplicity we only consider phase two.

Figure 39. The original base molecule “flashlight”. The bonds $\ell_{1}$ to
$\ell_{8}$ will appear in the $\mathtt{Ext}$ condition obtained by the
algorithm.

Suppose the original molecule is a base molecule as in Figure 39. Then,
according to the algorithm, we first treat the two degree 3 atoms (labeled 9
and 10) connected by a single bond. As in (2-b) we claim a checkpoint and
perform either (3S3-1) or (3S3-2G). In either case $\mathbb{M}$ is reduced to
the one in Figure 40.

Figure 40. The molecule obtained after performing (3S3-1) or (3S3-2G). Note
that this is a checkpoint and corresponds to two possible steps (though the
operation on $\mathbb{M}$ is the same and the only difference is
$\Delta\mathtt{Ext}$).

Next, as in (1), we perform (BR) and remove the bridge connecting atoms
labeled 4 and 5. Then $\mathbb{M}$ is reduced to the one in Figure 41.

Figure 41. The molecule obtained after performing (BR).

Next, we treat the two pairs of degree 3 atoms (labeled $(3,4)$ and $(5,7)$)
connected by two single bonds. As in (2-a) we perform (3S3-5G) (Scenario 2)
twice, and reduce $\mathbb{M}$ to the one in Figure 42.

Figure 42. The molecule obtained after performing (3S3-5G) twice.

Next, we treat the two degree 3 atoms (labeled 11 and 12) connected by a
double bond. Since the type II chain continues, as in (3-b) we claim a
checkpoint and perform either (3D3-1) or (3D3-2G). In either case $\mathbb{M}$
is reduced to the one in Figure 43.

Figure 43. The molecule obtained after performing (3D3-1) or (3D3-2G). Again
this is a checkpoint and the only difference between two possible steps is
$\Delta\mathtt{Ext}$.

Next, we treat the two degree 3 atoms (labeled 13 and 14) connected by a
double bond. The type II chain does not continue, so as in (3-c-ii) we perform
(3D3-6G) and reduce $\mathbb{M}$ to the one in Figure 44.

Figure 44. The molecule obtained after performing (3D3-6G).

Finally, we treat the remaining three pairs of degree 2 atoms connected by
three double bonds. As in (7) we perform (2R-5) three times and reduce
$\mathbb{M}$ to the empty graph.

Following the algorithm we have performed at least three good steps ($r\geq
3$). The two checkpoints provide four possible tracks, which correspond to
different possibilities of $\mathtt{Ext}$ in the beginning; for example if we
choose (3S3-1) and (3D3-1) then the $\mathtt{Ext}$ we obtain in the beginning
is

$\\{k_{\ell_{1}}=k_{\ell_{2}},\,\,k_{\ell_{3}}=k_{\ell_{4}},\,\,k_{\ell_{5}}=k_{\ell_{6}},\,\,k_{\ell_{7}}-k_{\ell_{8}}\mathrm{\
is\ a\ good\ vector}\\}.$

In this track (other tracks will have better exponents) we can calculate
$\gamma=18-\frac{1}{2(d-1)}$ at the beginning, so we have, omitting powers of
$\delta$, that

$\sup\\#\mathfrak{D}(\mathbb{M},\mathtt{Ext})\lesssim
L^{18(d-1)-\frac{1}{2}}.$

##  Appendix C Table of notations

Here we list some important notations used in this paper. These are mainly
concerned about trees, couples, molecules and their structures. Table 1
contains the basic notations and the corresponding symbols. Table 2 contains
further notations, including different types of couples. Table 3 contains
notations related to molecules.

Concept | Symbol |  Where defined
---|---|---
Tree | $\mathcal{T}$ |  Definition 2.1
Root, node, leaf | $\mathfrak{r},\mathfrak{n},\mathfrak{l}$ |  Definition 2.1
Leaf set (tree) | $\mathcal{L}$ |  Definition 2.1
Branching node set (tree) | $\mathcal{N}$ |  Definition 2.1
Sign | $\zeta,\zeta_{\mathfrak{n}}$ |  Definition 2.1
Scale (tree) | $n(\mathcal{T})$ |  Definition 2.1
Couple | $\mathcal{Q}$ |  Definition 2.2
Leaf set (couple) | $\mathcal{L}^{*}$ |  Definition 2.2
Branching node set (couple) | $\mathcal{N}^{*}$ |  Definition 2.2
Scale (couple) | $n(\mathcal{Q})$ |  Definition 2.2
Paired tree, saturated paired tree | — |  Definition 2.2
Lone leaf | — |  Definition 2.2
Decoration | $\mathscr{D},\mathscr{E}$ |  Definition 2.4

Table 1. Basic notations about trees and couples.

Concept | Symbol |  Where defined
---|---|---
$(1,1)$-mini couple, mini tree | — |  Definition 4.1
Code (mini couple, mini tree) | — |  Definition 4.1
Regular couple | — |  Definition 4.2
Legal partition, dominant partition | $\mathcal{P}$ |  Definition 4.4
Regular chain, regular double chain | — |  Definition 4.6
Type (regular couple) | — |  Proposition 4.10
Prime couple | — |  Definition 4.12
Skeleton | $\mathcal{Q}_{sk}$ |  Proposition 4.13
Regular tree | — |  Remark 4.15
Dominant couple | — |  Definition 4.17
Special set | $Z$ |  Definition 4.18
Equivalence (dominant couple) | — |  Definition 4.18
Encoded tree, equivalence (encoded tree) | — |  Section 4.5, Definition 4.21
Associated encoded tree | — |  Definition 4.22
Irregular chain | $\mathcal{H},\mathcal{H}^{\circ}$ |  Definition 8.1
Congruence | — |  Definition 8.2, 8.4

Table 2. Further notations about trees and couples.

Concept | Symbol |  Where defined
---|---|---
Molecule, atom, bond | $\mathbb{M},v,\ell$ |  Definition 9.1
Saturated component | — |  Definition 9.1
Base molecule | — |  Proposition 9.2
Molecule associated to a couple | — |  Definition 9.3
Type I and II chains | — |  Definition 9.7
Degenerate atom, tame atom | — |  Definition 9.8
Extra conditions | $\mathtt{Ext}$ |  Definition 9.8
Step, track, checkpoint | — |  Section 9.2
Bridge, special bond | — |  Definition 9.12
Good step, normal step | — |  Beginning of Section 9.3

Table 3. Notations about molecules.

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