††2010 Mathematics Subject Classification: Primary 11K60 ; Secondary 28A80,
37E10.

# On a lower bound of Hausdorff dimension of weighted singular vectors

Taehyeong Kim  Taehyeong Kim. Department of Mathematical Sciences, Seoul
National University<EMAIL_ADDRESS>and  Jaemin Park  Jaemin Park. Department
of Mathematical Sciences, Seoul National University<EMAIL_ADDRESS>
###### Abstract.

Let $w=(w_{1},\dots,w_{d})$ be an ordered $d$-tuple of positive real numbers
such that $\sum_{i}w_{i}=1$ and $w_{1}\geq\cdots\geq w_{d}$. A $d$-dimensional
vector $x=(x_{1},\dots,x_{d})\in\mathbb{R}^{d}$ is said to be $w$-singular if
for every $\epsilon>0$ there exists $T_{0}>1$ such that for all $T>T_{0}$ the
system of inequalities

${\max_{1\leq i\leq
d}|qx_{i}-p_{i}|^{\frac{1}{w_{i}}}<\frac{\epsilon}{T}\quad\text{and}\quad
0<q<T}$

have an integer solution
$(\mathbf{p},q)=(p_{1},\dots,p_{d},q)\in\mathbb{Z}^{d}\times\mathbb{Z}$. In
this paper, we prove that Hausdorff dimension of the set of $w$-singular
vectors in $\mathbb{R}^{d}$ is bounded below by $d-\frac{1}{1+w_{1}}$. This
result extends partially the previous result of Liao et al. [_Hausdorff
dimension of weighted singular vectors in $\mathbb{R}^{2}$_, J. Eur. Math.
Soc. 22 (2020), 833-875].

##  1\. Introduction

In 1937, Khintchine introduced the notion of singularity in the sense of
Diophantine approximation. Recall that a vector
$x=(x_{1},\dots,x_{d})\in\mathbb{R}^{d}$ is said to be singular if, for every
$\epsilon>0$, there exists $T_{0}>1$ such that for all $T>T_{0}$ the system of
inequalities

(1.1) ${\max_{1\leq i\leq
d}|qx_{i}-p_{i}|^{d}<\frac{\epsilon}{T}\quad\text{and}\quad 0<q<T}$

admits an integer solution
$(\mathbf{p},q)=(p_{1},\dots,p_{d},q)\in\mathbb{Z}^{d}\times\mathbb{Z}$.

The name singular is derived from the fact that the set of singular vectors is
a Lebesgue nullset. On the other hand, the computation of the Hausdorff
dimension of the set of singular vectors, or more generally singular matrices,
has been a challenge until the breakthrough [DFSU] using a variational
principle in the parametric geometry of numbers. Historically, the first
breakthrough was made in [Che11] to prove that the Hausdorff dimension of the
set of $2$-dimensional singular vectors is $4/3$, which was extended in [CC16]
to $d$-dimensional singular vectors. They proved that the set of
$d$-dimensional singular vectors has Hausdorff dimension $d^{2}/(d+1)$. For
general $m\times n$ singular matrices, it was proved in [KKLM17] that the
Hausdorff dimension of $m\times n$ singular matrices is at most
$mn(1-\frac{1}{m+n})$ , and finally, it was shown in [DFSU] that the upper
bound is sharp.

In this article, we consider the weighted version of the singularity as
follows: Let $w=(w_{1},\dots,w_{d})\in\mathbb{R}_{>0}^{d}$ be an ordered
$d$-tuple of positive real numbers such that $\sum_{i}w_{i}=1$ and
$w_{1}\geq\cdots\geq w_{d}$. We say that a vector
$x=(x_{1},\dots,x_{d})\in\mathbb{R}^{d}$ is $w$-singular if for every
$\epsilon>0$ there exists $T_{0}>1$ such that for all $T>T_{0}$ the system of
inequalities

(1.2) ${\max_{1\leq i\leq
d}|qx_{i}-p_{i}|^{\frac{1}{w_{i}}}<\frac{\epsilon}{T}\quad\text{and}\quad
0<q<T}$

admits an integer solution
$(\mathbf{p},q)=(p_{1},\dots,p_{d},q)\in\mathbb{Z}^{d}\times\mathbb{Z}$.
Denote by $\operatorname{Sing}(w)$ the set of $w$-singular vectors in
$\mathbb{R}^{d}$. Here and hereafter we always assume that the weight vector
$w$ satisfies the above assumption.

Note that if all weights have the same value, that is, $w_{i}=1/d$ for all
$i=1,\dots,d$, then the weighted system (1.2) gives the unweighted system
(1.1).

In the weighted setting, it was shown in [LSST20] that the set of
$2$-dimensional $w$-singular vectors has Hausdorff dimension
$2-\frac{1}{1+w_{1}}$. The aim of the present article is to extend this
$2$-dimensional result to higher dimensions regarding the lower bound of the
Hausdorff dimension.

######  Theorem 1.1.

For $d\geq 2$, the Hausdorff dimension of $\operatorname{Sing}(w)$ is at least
$d-\frac{1}{1+w_{1}}$.

One of the main ingredients of the proof of Theorem 1.1 is Dani’s
correspondence, which means that $w$-singular vectors correspond to certain
divergent trajectories in the space $\mathcal{L}_{d+1}$ of unimodular lattices
in $\mathbb{R}^{d+1}$. More precisely, let
$a_{t}:=\operatorname{diag}\left(e^{w_{1}t},\dots,e^{w_{d}t},e^{-1}\right)\in\operatorname{SL}_{d+1}(\mathbb{R})$
and let

$h(x):=\left(\begin{matrix}I_{d}&x\\\
0&1\end{matrix}\right)\in\operatorname{SL}_{d+1}(\mathbb{R})\quad\text{ for
}x\in\mathbb{R}^{d},$

where $I_{d}$ is the $d\times d$ identity matrix. Then $x$ is $w$-singular if
and only if the diagonal orbit $\left(a_{t}h(x)\mathbb{Z}^{d+1}\right)_{t\geq
0}$ is divergent.

Our method for the proof of Theorem 1.1 is basically extension of the method
in [LSST20], hence we also have the following result as in [LSST20, Theorem
1.5].

######  Theorem 1.2.

For any $\Lambda\in\mathcal{L}_{d+1}$ and any nonempty open subset $U$ in
$\mathbb{R}^{d}$, the Hausdorff dimension of the set

$\\{x\in U:\left(a_{t}h(x)\Lambda\right)_{t\geq 0}\text{ is divergent}\\}$

is at least $d-\frac{1}{1+w_{1}}$.

Theorem 1.2 implies the following corollary as in [LSST20, Corollary 1.6] (see
also [GS20, AGMS21]).

######  Corollary 1.3.

The Hausdorff dimension of the set

$\\{\Lambda\in\mathcal{L}_{d+1}:\left(a_{t}\Lambda\right)_{t\geq 0}\text{ is
divergent}\\}$

is at least
$\mathrm{dim}\operatorname{SL}_{d+1}(\mathbb{R})-\frac{1}{1+w_{1}}=(d+1)^{2}-1-\frac{1}{1+w_{1}}$.

Recently, Solan [Sol] established a variational principle in the parametric
geometry of numbers for general flows. Following his notations, we consider
the following two subgroups:

$\begin{split}H&=\\{g\in\operatorname{SL}_{d+1}(\mathbb{R}):a_{-t}ga_{t}\to
I_{d+1}\text{ as }t\to\infty\\},\\\
H^{\prime}&=\\{h(x)\in\operatorname{SL}_{d+1}(\mathbb{R}):x\in\mathbb{R}^{d}\\}.\end{split}$

Note that $H$ is the unstable horospherical subgroup of $a_{1}$. In the
unweighted setting ($w_{1}=\dots=w_{d}$), the two subgroups $H$ and
$H^{\prime}$ are the same, but in general, $H$ is bigger than $H^{\prime}$.
One of the applications of the variational principle for general flows in
[Sol] is to give an upper bound of the Hausdorff dimension of the set

$\operatorname{Sing}(H,\Lambda;a_{t})=\\{h\in
H:\left(a_{t}h\Lambda\right)_{t\geq 0}\text{ is divergent}\\}.$

More precisely, [Sol, Corollary 2.34] implies that the Hausdorff dimension of
$\operatorname{Sing}(H,\Lambda;a_{t})$ is at most
$\mathrm{dim}~{}H-\frac{1}{1+w_{1}}$. On the other hand, Theorem 1.2 implies
that the Hausdorff dimension of $\operatorname{Sing}(H,\Lambda;a_{t})$ is at
least $\mathrm{dim}~{}H-\frac{1}{1+w_{1}}$, hence we have the following
corollary.

######  Corollary 1.4.

The Hausdorff dimension of $\operatorname{Sing}(H,\Lambda;a_{t})$ is
$\mathrm{dim}~{}H-\frac{1}{1+w_{1}}$.

The article is organized as follows. In Section 2, we recall fractal
structures and self-affine structures. We estimate the lower bound of
Hausdorff dimension of the associated fractal set. In Section 3, we generalize
the lattice point counting results in [LSST20, Section 3] to higher
dimensional cases. In Section 4, we construct a fractal set contained in the
set of weighted singular vectors and prove Theorem 1.1 by estimating Hausdorff
dimension of the fractal set.

Acknowledgments. We would like to express our gratitude to Seonhee Lim for her
constant help and encouragement. We also would like to thank Frédéric Paulin
for providing helpful comments.

TK and JP were supported by the National Research Foundation of Korea under
Project Number NRF-2020R1A2C1A01011543. TK was supported by the National
Research Foundation of Korea under Project Number NRF-2021R1A6A3A13039948. JP
was supported by Industrial and Mathematical Data Analytics Research Center
under Project Number 2022R1A5A6000840.

##  2\. Fractal sutructure and Hausdorff dimension

###  2.1. Fractal structure

A tree $\mathcal{T}$ is a connected graph without cycles. If we take a vertex
$\tau_{0}$ and fix it (we call it a root), then $\mathcal{T}$ is a rooted
tree. In this paper, we identify $\mathcal{T}$ with the set of vertices of
$\mathcal{T}$. It can be checked directly from the definition of $\mathcal{T}$
that any $\tau\in\mathcal{T}$ can be joined to $\tau_{0}$ by a unique geodesic
edge path. We define the height of $\tau$ as the length of the geodesic edge
path joining $\tau,\tau_{0}$ and denote the set of vertices of height $n$ by
$\mathcal{T}_{n}$. For any $\tau\in\mathcal{T}_{n}$, there exists a unique
$\tau_{n-1}\in\mathcal{T}_{n-1}$ such that $\tau$ and $\tau_{n-1}$ are
adjacent. Then we say $\tau$ is a son of $\tau_{n-1}$ and denote the set of
all sons of $\tau_{n-1}$ by $\mathcal{T}(\tau_{n-1})$. The boundary of
$\mathcal{T}$, denoted by $\partial\mathcal{T}$, is the set of all sequences
$\\{\tau_{n}\\}=\\{\tau_{n}\\}_{n\in\mathbb{N}\cup\\{0\\}}$ where $\tau_{n}$
is a son of $\tau_{n-1}$ for all $n\in\mathbb{N}$.

A fractal structure on $\mathbb{R}^{d}$ is a pair $(\mathcal{T},\beta)$ where
$\mathcal{T}$ is a rooted tree and $\beta$ is a map from $\mathcal{T}$ to the
set of nonempty compact subsets of $\mathbb{R}^{d}$. A fractal associated to
$(\mathcal{T},\beta)$ is a set

${\mathcal{F}(\mathcal{T},\beta)=\bigcup_{\\{\tau_{n}\\}\in\partial\mathcal{T}}\bigcap_{n=0}^{\infty}\beta(\tau_{n}).}$

A fractal structure $(\mathcal{T},\beta)$ is said to be regular if it
satisfies the followings:

  * •

each vertex of $\mathcal{T}$ has at least one son;

  * •

if $\tau$ is a son of $\tau^{\prime}$, then
$\beta(\tau)\subset\beta(\tau^{\prime})$;

  * •

for any $\\{\tau_{n}\\}\in\partial\mathcal{T}$,
$\operatorname{diam}{\beta(\tau_{n})}\to 0$ as $n\to\infty$.

###  2.2. Self-affine structure and lower bound

A self-affine structure on $\mathbb{R}^{d}$ is a fractal structure
$(\mathcal{T},\beta)$ on $\mathbb{R}^{d}$ such that for $\tau\in\mathcal{T}$
the compact subset $\beta(\tau)$ of $\mathbb{R}^{d}$ is given by a
$d$-dimensional rectangle with size $L^{(1)}(\tau)\times\cdots\times
L^{(d)}(\tau)$. A self-affine structure is regular if it is a regular fractal
structure.

The following theorem is a generalization of [LSST20, Theorem 2.1] for
$d$-dimensional self-affine structures.

######  Theorem 2.1.

Let $(\mathcal{T},\beta)$ be a regular self-affine structure on
$\mathbb{R}^{d}$ that associates to sequences
$\\{\rho_{n}\\},\\{C_{n}\\},\\{L_{n}^{(j)}\\}$ for $j=1,\dots,d$ of positive
real numbers indexed by $\mathbb{N}\cup\\{0\\}$ with the following properties:

  1. (1)

The sequence $\\{L_{n}^{(j)}\\}$ is decreasing in $n\in\mathbb{N}\cup\\{0\\}$
for each $j=1,\dots,d$.

  2. (2)

There exists $1\leq\ell<d$ such that

${L_{n}^{(1)}=\cdots=L_{n}^{(\ell)}<L_{n}^{(\ell+1)}\leq\cdots\leq
L_{n}^{(d)}\quad\text{and}\quad L^{(j)}(\tau)=L_{n}^{(j)}}$

for all $n\in\mathbb{N}\cup\\{0\\}$, $j=1,\dots,d$, and
$\tau\in\mathcal{T}_{n}$;

  3. (3)

$C_{0}=1$ and $\\#\mathcal{T}(\tau)\geq C_{n}$ for all $n\in\mathbb{N}$ and
$\tau\in\mathcal{T}_{n-1}$;

  4. (4)

$\rho_{n}\leq 1$ for all $n\in\mathbb{N}$ and

${\textrm{dist}(\beta(\tau),\beta(\kappa))\geq\rho_{n+1}L_{n}^{(1)}}$

for all $\tau_{n}\in\mathcal{T}_{n}$ and distinct
$\tau,\kappa\in\mathcal{T}(\tau_{n})$.

We denote by

$\begin{split}P_{n}&=\prod_{i=0}^{n}C_{i},\\\ D_{n}&=\max\\{i\geq
n:L_{i}^{(d)}\geq L_{n}^{(1)}\\},\\\
s&=\sup\left\\{t>0:\lim_{n\to\infty}\frac{\log(P_{n}(L_{n}^{(1)})^{t}\rho_{n+1}^{t}\cdot\prod_{i=n+1}^{D_{n}}\rho_{i}^{\ell}C_{i})}{\max\\{D_{n}-n,1\\}}=\infty\right\\}.\end{split}$

If $s>d-\ell$, then $\dim_{H}\mathcal{F}(\mathcal{T},\beta)\geq s$.

Using Theorem 2.1, we obtain the following corollary which is a generalization
of [LSST20, Corollary 2.3 and Corollary 2.4] for $d$-dimensional self-affine
structures.

######  Corollary 2.2.

With the notations in Theorem 2.1, suppose that there exists
$k,n_{0}\in\mathbb{N}$ such that for all $n\geq n_{0}$ the followings hold:

  1. (i)

$\frac{L_{kn}^{(d)}}{L_{kn-1}^{(d)}}\leq\frac{L_{n}^{(1)}}{L_{n-1}^{(1)}}\text{
and }L_{kn_{0}-1}^{(d)}<L_{n_{0}-1}^{(1)},$

  2. (ii)

$e^{n/k}\leq C_{n}\leq e^{kn},$

  3. (iii)

$e^{-kn}\leq\rho_{n}\leq e^{-n/k},$

  4. (iv)

$\rho_{n}^{\ell}C_{n}\prod_{j=\ell+1}^{d}L_{n}^{(j)}/L_{n-1}^{(j)}\geq
n^{-k}.$

If the limit

${\lim_{n\to\infty}\frac{\log\left(C_{n}\prod_{j=\ell+1}^{d}L_{n}^{(j)}/L_{n-1}^{(j)}\right)}{-\log\left(L_{n}^{(1)}/L_{n-1}^{(1)}\right)}}$

exists and is equal to $r>0$, then $\dim_{H}\mathcal{F}(\mathcal{T},\beta)\geq
d-\ell+r$.

###### Proof of Corollary 2.2.

By the assumptions (iii) and (iv), since the sequence $\\{L_{n}^{(j)}\\}$ is
decreasing in $n\in\mathbb{N}\cup\\{0\\}$ for each $j=1,\dots,d$, we have

$\log\left(C_{n}\prod_{j=\ell+1}^{d}L_{n}^{(j)}/L_{n-1}^{(j)}\right)=O(n),$

which implies that $-\log\left(L_{n}^{(1)}/L_{n-1}^{(1)}\right)\to\infty$ as
$n\to\infty$. Hence, using

$\frac{\log\left(P_{n}\prod_{j=\ell+1}^{d}L_{n}^{(j)}\right)}{-\log
L_{n}^{(1)}}=\frac{\log\left(C_{0}\prod_{j=\ell+1}^{d}L_{0}^{(j)}\right)+\sum_{i=1}^{n}\log\left(C_{i}\prod_{j=\ell+1}^{d}L_{i}^{(j)}/L_{i-1}^{(j)}\right)}{-\log
L_{0}^{(1)}-\sum_{i=1}^{n}\log\left(L_{i}^{(1)}/L_{i-1}^{(1)}\right)},$

it follows that

$\lim_{n\to\infty}\frac{\log\left(C_{n}\prod_{j=\ell+1}^{d}L_{n}^{(j)}/L_{n-1}^{(j)}\right)}{-\log\left(L_{n}^{(1)}/L_{n-1}^{(1)}\right)}=\lim_{n\to\infty}\frac{\log\left(P_{n}\prod_{j=\ell+1}^{d}L_{n}^{(j)}\right)}{-\log
L_{n}^{(1)}}.$

Let us denote by

$s=d-\ell+r=d-\ell+\lim_{n\to\infty}\frac{\log\left(P_{n}\prod_{j=\ell+1}^{d}L_{n}^{(j)}\right)}{-\log
L_{n}^{(1)}}.$

By the regularity of the given self-affine structure $(\mathcal{T},\beta)$, we
have that $L_{n}^{(1)}\to 0$ as $n\to\infty$, which implies

(2.1)
${s=\sup\left\\{t>0:\lim_{n\to\infty}P_{n}(L_{n}^{(1)})^{t}\prod_{j=\ell+1}^{d}\frac{L_{n}^{(j)}}{L_{n}^{(1)}}=\infty\right\\}.}$

We will show that $\dim_{H}\mathcal{F}(\mathcal{T},\beta)\geq s$ using the
equality (2.1).

Recall that $D_{n}=\max\left\\{i\geq n:L_{i}^{(d)}\geq L_{n}^{(1)}\right\\}$.
Since the sequence $\\{L_{n}^{(j)}\\}$ is decreasing in
$n\in\mathbb{N}\cup\\{0\\}$ for each $j=1,\dots,d$, we have

$\displaystyle L_{kn}^{(d)}$
$\displaystyle=L_{kn_{0}-1}^{(d)}\prod_{i=kn_{0}}^{kn}\frac{L_{i}^{(d)}}{L_{i-1}^{(d)}}\leq
L_{kn_{0}-1}^{(d)}\prod_{i=n_{0}}^{n}\frac{L_{ki}^{(d)}}{L_{ki-1}^{(d)}}$
$\displaystyle\leq
L_{kn_{0}-1}^{(d)}\prod_{i=n_{0}}^{n}\frac{L_{i}^{(1)}}{L_{i-1}^{(1)}}\quad$
by assumption (i)
$\displaystyle=L_{kn_{0}-1}^{(d)}\frac{L_{n}^{(1)}}{L_{n_{0}-1}^{(1)}}$
$\displaystyle\leq L_{n}^{(1)}\quad$ $\displaystyle\text{by assumption
}\ref{FHCorA1}.$

Hence we have $D_{n}\leq kn$.

Given $t>0$, $\epsilon>0$, it follows from the assumptions (ii), (iii), and
$D_{n}\leq kn$ that $\rho_{D_{n}+1}^{\ell}C_{D_{n}+1}\leq e^{k(D_{n}+1)}\leq
e^{k(kn+1)}$. Since $P_{n}^{\epsilon}=(\prod_{i=0}^{n}C_{i})^{\epsilon}\geq
e^{\frac{n(n+1)\epsilon}{2k}}$, we have

(2.2) ${\rho_{D_{n}+1}^{\ell}C_{D_{n}+1}\leq P_{n}^{\epsilon}}$

for all large enough $n\geq 1$. Similarly, it follows from the assumptions
(iii), (iv), and $D_{n}\leq kn$ that

$\begin{split}\rho_{n+1}^{t}\prod_{i=n+1}^{D_{n}+1}\left(\rho_{i}^{\ell}C_{i}\prod_{j=\ell+1}^{d}\frac{L_{i}^{(j)}}{L_{i-1}^{(j)}}\right)&\geq
e^{-tk(n+1)}\prod_{i=n+1}^{D_{n}+1}i^{-k}\geq e^{-tk(n+1)}(kn+1)^{-k(kn-n)}\\\
&\geq e^{-tk(n+1)-k(kn-n)\log(kn+1)}.\end{split}$

The inequality $P_{n}^{-\epsilon}\leq e^{-\frac{n(n+1)\epsilon}{2k}}$ implies
that

(2.3)
${\rho_{n+1}^{t}\prod_{i=n+1}^{D_{n}+1}\left(\rho_{i}^{\ell}C_{i}\prod_{j=\ell+1}^{d}\frac{L_{i}^{(j)}}{L_{i-1}^{(j)}}\right)\geq
P_{n}^{-\epsilon}}$

for all large enough $n\geq 1$.

Fix a real number $t$ with $d-\ell<t<s$ and take sufficiently small $\epsilon$
such that $d-\ell<t/(1-3\epsilon)<s.$ By the equality (2.1), we have

(2.4)
$\lim_{n\to\infty}P_{n}(L_{n}^{(1)})^{t/(1-3\epsilon)}\prod_{j=\ell+1}^{d}\frac{L_{n}^{(j)}}{L_{n}^{(1)}}\geq
1$

for all large enough $n\geq 1$.

For all large enough $n\geq 1$ so that the above inequalities (2.2), (2.3),
and (2.4) hold, we have

$\displaystyle P_{n}$
$\displaystyle(L_{n}^{(1)})^{t}\rho_{n+1}^{t}\prod_{i=n+1}^{D_{n}}\rho_{i}^{\ell}C_{i}\geq
P_{n}^{1-\epsilon}(L_{n}^{(1)})^{t}\rho_{n+1}^{t}\prod_{i=n+1}^{D_{n}+1}\rho_{i}^{\ell}C_{i}$
$\displaystyle\text{by }\eqref{eq211}$
$\displaystyle=P_{n}^{1-\epsilon}(L_{n}^{(1)})^{t}\rho_{n+1}^{t}\left(\prod_{j=\ell+1}^{d}\frac{L_{n}^{(j)}}{L_{D_{n}+1}^{(j)}}\right)\cdot\prod_{i=n+1}^{D_{n}+1}\left(\rho_{i}^{\ell}C_{i}\prod_{j=\ell+1}^{d}\frac{L_{i}^{(j)}}{L_{i-1}^{(j)}}\right)$
$\displaystyle\geq
P_{n}^{1-\epsilon}(L_{n}^{(1)})^{t}\rho_{n+1}^{t}\left(\prod_{j=\ell+1}^{d}\frac{L_{n}^{(j)}}{L_{n}^{(1)}}\right)\cdot\prod_{i=n+1}^{D_{n}+1}\left(\rho_{i}^{\ell}C_{i}\prod_{j=\ell+1}^{d}\frac{L_{i}^{(j)}}{L_{i-1}^{(j)}}\right)$
$\displaystyle\geq
P_{n}^{1-\epsilon}(L_{n}^{(1)})^{t}\left(\prod_{j=\ell+1}^{d}\frac{L_{n}^{(j)}}{L_{n}^{(1)}}\right)P_{n}^{-\epsilon}\quad$
$\displaystyle\text{by }\eqref{eq212}$ $\displaystyle\geq
P_{n}^{1-\epsilon}(L_{n}^{(1)})^{t}\left(\prod_{j=\ell+1}^{d}\frac{L_{n}^{(j)}}{L_{n}^{(1)}}\right)^{1-3\epsilon}P_{n}^{-2\epsilon}P_{n}^{\epsilon}$
$\displaystyle=\left(P_{n}(L_{n}^{(1)})^{t/(1-3\epsilon)}\prod_{j=\ell+1}^{d}\frac{L_{n}^{(j)}}{L_{n}^{(1)}}\right)^{1-3\epsilon}P_{n}^{\epsilon}$
$\displaystyle\geq P_{n}^{\epsilon}\quad$ $\displaystyle\text{by
}\eqref{eq22}.$

It follows that for all large enough $n\geq 1$,

$\log\left(P_{n}(L_{n}^{(1)})^{t}\rho_{n+1}^{t}\prod_{i=n+1}^{D_{n}}\rho_{i}^{\ell}C_{i}\right)\geq\epsilon\log
P_{n}\gg\epsilon n^{2}\geq\epsilon\frac{n}{k-1}(D_{n}-n),$

where the implied constant is independent of $n$. Hence
$\dim_{H}\mathcal{F}(\mathcal{T},\beta)\geq t$ by Theorem 2.1. Since we choose
arbitrary $t$ with $d-\ell<t<s$, it concludes Corollary 2.2. ∎

By elementary squares of $\beta(\tau)$ for $\tau\in\mathcal{T}$, we mean
closed squares contained in $\beta(\tau)$ whose side length is equal to
$L^{(1)}(\tau)$.

######  Lemma 2.3.

For $n\in\mathbb{N}\cup\\{0\\}$ with $D_{n}>n$, let $\kappa\in\mathcal{T}_{n}$
and $\tau\in\mathcal{T}_{i-1}$ where $n+1\leq i\leq D_{n}$. Then for any
elementary square $S$ of $\beta(\kappa)$,

$\\#\\{\tau^{\prime}\in\mathcal{T}(\tau):\beta(\tau^{\prime})\cap
S\neq\varnothing\\}\leq(16d)^{d}\rho_{i}^{-\ell}.$

###### Proof.

Through this proof, we denote the size of a rectange $R$ in $\mathbb{R}^{d}$
by $l_{1}(R)\times\cdots\times l_{d}(R)$.

For a fixed elementary square $S$ of $\beta(\kappa)$, let
$R_{0}=\beta(\tau)\cap S$ and

$\mathcal{S}=\\{\beta(\tau^{\prime})\cap
S:\tau^{\prime}\in\mathcal{T}(\tau),\beta(\tau^{\prime})\cap
S\neq\varnothing\\}.$

If $R_{0}=\varnothing$, then there is nothing to prove since
$\\#\\{\tau^{\prime}\in\mathcal{T}(\tau):\beta(\tau^{\prime})\cap
S\neq\varnothing\\}=0$.

Let $j_{i}\in\\{1,\dots,d-1\\}$ be an integer such that
$L_{i}^{(d)}\geq\cdots\geq L_{i}^{(j_{i}+1)}\geq
L_{n}^{(1)}>L_{i}^{(j_{i})}\geq\cdots\geq L_{i}^{(1)}$. Note that
$j_{i}\geq\ell$. Let $R_{0}^{\prime}$ be the rectangle with the same center as
$R_{0}$ such that

$l_{j}(R_{0}^{\prime})=\begin{cases}4L_{i-1}^{(j)}&\quad\text{for
}j=1,\dots,j_{i}\\\ 4L_{n}^{(1)}&\quad\text{for
}j=j_{i}+1,\dots,d.\end{cases}$

Similarly, for $R\ \in\mathcal{S}$ let $R^{\prime}$ be the rectange with the
same center such that

$l_{j}(R^{\prime})=\begin{cases}L_{i}^{(j)}+\frac{\rho_{i}}{4\sqrt{d}}L_{i-1}^{(1)}&\quad\text{for
}j=1,\dots,j_{i}\\\ L_{n}^{(1)}&\quad\text{for }j=j_{i}+1,\dots,d.\end{cases}$

We denote by $r_{0}$ (resp. $r$) the center of $R_{0}^{\prime}$ (resp.
$R^{\prime}$). Here, we note that $r_{0}$ and $r$ are contained in both
$\beta(\tau)$ and $S$. For $x\in R^{\prime}$ and $j=1,\dots,d$,

$|x_{j}-(r_{0})_{j}|\leq|x_{j}-r_{j}|+|r_{j}-(r_{0})_{j}|\leq\frac{1}{2}l_{j}(R^{\prime})+\min(L_{i-1}^{(j)},L_{n}^{(1)})\leq\frac{1}{2}l_{j}(R_{0}^{\prime}).$

Thus for all $R\in\mathcal{S}$, $R^{\prime}\subset R_{0}^{\prime}$.

For any distinct $R_{1},R_{2}\in\mathcal{S}$, let
$\tau_{1}^{\prime},\tau_{2}^{\prime}\in\mathcal{T}(\tau)$ be such that
$R_{1}=\beta(\tau_{1}^{\prime})\cap S$ and $R_{2}=\beta(\tau_{2}^{\prime})\cap
S$, and let $r_{1},r_{2}$ be the centers of $R_{1}^{\prime},R_{2}^{\prime}$,
respectively. Suppose $\|r_{1}-r_{2}\|_{\infty}=|(r_{1})_{j}-(r_{2})_{j}|>0$
for some $j=1,\dots,j_{i}$. Then for any $x\in R_{1}^{\prime}$ and $y\in
R_{2}^{\prime}$, we have

$\displaystyle|x_{j}-y_{j}|$
$\displaystyle\geq|(r_{1})_{j}-(r_{2})_{j}|-|x_{j}-(r_{1})_{j}|-|y_{j}-(r_{2})_{j}|$
$\displaystyle\geq\frac{1}{\sqrt{d}}\mathrm{dist}(\beta(\tau_{1}^{\prime}),\beta(\tau_{2}^{\prime}))+L_{i}^{(j)}-\frac{1}{2}l_{j}(R_{1}^{\prime})-\frac{1}{2}l_{j}(R_{2}^{\prime})$
$\displaystyle\geq\left(\frac{\rho_{i}}{\sqrt{d}}L_{i-1}^{(1)}+L_{i}^{(j)}\right)-\frac{1}{2}l_{j}(R_{1}^{\prime})-\frac{1}{2}l_{j}(R_{2}^{\prime})$
$\displaystyle=\frac{3\rho_{i}}{4\sqrt{d}}L_{i-1}^{(1)}>0.$

Thus $R_{1}^{\prime}\cap R_{2}^{\prime}=\varnothing$.

Now we suppose $\|r_{1}-r_{2}\|_{\infty}=|(r_{1})_{j}-(r_{2})_{j}|>0$ for some
$j=j_{i}+1,\dots,d$. Observe that

$L_{n}^{(1)}\geq
l_{j}(R_{1})+l_{j}(R_{2})+\frac{1}{\sqrt{d}}\mathrm{dist}(\beta(\tau_{1}^{\prime}),\beta(\tau_{2}^{\prime}))>l_{j}(R_{1})+l_{j}(R_{2}),$

which implies that

$|(r_{1})_{j}-(r_{2})_{j}|=L_{n}^{(1)}-\frac{1}{2}l_{j}(R_{1})-\frac{1}{2}l_{j}(R_{2})>\frac{1}{2}L_{n}^{(1)}.$

Thus, for any fixed $R_{1}\in\mathcal{S}$ and $j=j_{i}+1,\dots,d$,

$\\#\\{R_{2}\in\mathcal{S}\setminus\\{R_{1}\\}:\|r_{1}-r_{2}\|_{\infty}=|(r_{1})_{j}-(r_{2})_{j}|\text{
and }R_{1}^{\prime}\cap R_{2}^{\prime}\neq\varnothing\\}\leq 1.$

Combining above two arguments, we conclude that every points of
$R_{0}^{\prime}$ is covered by at most $d-j_{i}+1$ rectangles of
$\\{R^{\prime}:R\in\mathcal{S}\\}$. It follows that

$\displaystyle\left(\frac{\rho_{i}}{4\sqrt{d}}L_{i-1}^{(1)}\right)^{j_{i}}\left(L_{n}^{(1)}\right)^{d-j_{i}}\\#\mathcal{S}$
$\displaystyle\leq\left(L_{i}^{(j)}+\frac{\rho_{i}}{4\sqrt{d}}L_{i-1}^{(1)}\right)^{j_{i}}\left(L_{n}^{(1)}\right)^{d-j_{i}}\\#\mathcal{S}$
$\displaystyle=\operatorname{vol}(R^{\prime})\\#\mathcal{S}$
$\displaystyle\leq(d-j_{i}+1)\operatorname{vol}(R_{0}^{\prime})$
$\displaystyle\leq
d4^{d}\left(L_{i-1}^{(1)}\right)^{j_{i}}\left(L_{n}^{(1)}\right)^{d-j_{i}},$

hence, using $j_{i}\geq\ell$,

$\\#\mathcal{S}\leq
d^{1+j_{i}/2}4^{d+j_{i}}\rho_{i}^{-j_{i}}\leq(16d)^{d}\rho_{i}^{-\ell}.$

This inequality completes the proof. ∎

Let $\mu$ be the unique probability measure on
$\mathcal{F}(\mathcal{T},\beta)$ satisfying the following property: For all
$y\in\mathcal{F}(\mathcal{T},\beta)$ and $n\in\mathbb{N}$,

(2.5)
${\frac{\mu(\\{x\in\mathcal{F}(\mathcal{T},\beta):\tau_{n}(x)=\tau_{n}(y)\\})}{\mu(\\{x\in\mathcal{F}(\mathcal{T},\beta):\tau_{n-1}(x)=\tau_{n-1}(y)\\})}=\frac{1}{\\#\mathcal{T}(\tau_{n-1}(y))},}$

where $x=\bigcap_{n\geq 0}\beta(\tau_{n}(x))$. We remark that for any
$n\in\mathbb{N}$ and $\kappa\in\mathcal{T}_{n}$, it follows from (2.5) that

(2.6)
${\mu(\beta(\kappa))\leq\frac{\mu(\mathcal{F}(\mathcal{T},\beta))}{C_{0}\dots
C_{n}}=\frac{1}{P_{n}}}$

######  Lemma 2.4.

Let $n\in\mathbb{N}$ and $\kappa\in\mathcal{T}_{n}$. Then for any elementary
square $S$ of $\beta(\kappa)$, one has

$\mu(S)\leq(16d)^{d(D_{n}-n)}P_{n}^{-1}\prod_{i=n+1}^{D_{n}}\rho_{i}^{-\ell}C_{i}^{-1}.$

###### Proof.

If $D_{n}=n$, then it follows from (2.6). Assume $D_{n}>n$. Applying Lemma 2.3
for $i=n+1,\dots,D_{n}$, we have

(2.7) $\\#\\{\tau\in\mathcal{T}_{D_{n}}:\beta(\tau)\cap
S\neq\varnothing\\}\leq(16d)^{d(D_{n}-n)}\prod_{i=n+1}^{D_{n}}\rho_{i}^{-\ell}.$

Since $S\cap\mathcal{F}(\mathcal{T},\beta)$ can be covered by rectangles
$\\{\beta(\tau):\tau\in\mathcal{T}_{D_{n}},\beta(\tau)\cap
S\neq\varnothing\\}$, we have

$\displaystyle\mu(S)$
$\displaystyle\leq\sum_{\begin{subarray}{c}\tau\in\mathcal{T}_{D_{n}}\\\
\beta(\tau)\cap S\neq\varnothing\end{subarray}}\mu(\beta(\tau))$
$\displaystyle\leq\mu(\beta(\kappa))\prod_{i=n+1}^{D_{n}}C_{i}^{-1}\cdot\\#\\{\tau\in\mathcal{T}_{D_{n}}:\beta(\tau)\cap
S\neq\varnothing\\}$
$\displaystyle\leq(16d)^{d(D_{n}-n)}P_{n}^{-1}\prod_{i=n+1}^{D_{n}}\rho_{i}^{-\ell}C_{i}^{-1}.$

In the last inequality, we use (2.6) and (2.7). ∎

Let $U$ be an open subset of $\mathbb{R}^{d}$ with
$U\cap\mathcal{F}(\mathcal{T},\beta)\neq\varnothing$. If
$U\cap\mathcal{F}(\mathcal{T},\beta)$ is a single point set, then we denote by
$n(U)$ the smallest $n\in\mathbb{N}$ such that
$\operatorname{diam}(U)\geq\rho_{n+1}L_{n}^{(1)}$. In that case, there is a
unique $\kappa=\kappa(U)\in\mathcal{T}_{n(U)}$ such that
$U\cap\mathcal{F}(\mathcal{T},\beta)\subset\beta(\kappa)$. If
$U\cap\mathcal{F}(\mathcal{T},\beta)$ contains more than two points, then we
denote by $n(U)$ the largest $n\in\mathbb{N}$ such that
$U\cap\mathcal{F}(\mathcal{T},\beta)\subset\beta(\kappa)$ for some
$\kappa=\kappa(U)\in\mathcal{T}_{n}$. We note that
$\operatorname{diam}(U)\geq\rho_{n(U)+1}L_{n(U)}^{(1)}$ by the assumption (4)
of Theorem 2.1.

######  Lemma 2.5.

Let $U$ be an open subset of $\mathbb{R}^{d}$ with
$U\cap\mathcal{F}(\mathcal{T},\beta)\neq\varnothing$. Let $n=n(U)$ and
$\kappa=\kappa(U)$. Then there is a family $\mathcal{S}$ of elementary squares
of $\beta(\kappa)$ such that

  1. (1)

$\bigcup_{S\in\mathcal{S}}S\supset U\cap\mathcal{F}(\mathcal{T},\beta)$;

  2. (2)

$\left(L_{n}^{(1)}\right)^{t}\cdot\\#\mathcal{S}\leq
2^{d-\ell}\rho_{n+1}^{-t}\operatorname{diam}(U)^{t}$ for all $t\geq d-\ell$.

###### Proof.

If $\operatorname{diam}(U)\leq L_{n}^{(1)}$, then there exists an elementary
square $S$ of $\beta(\kappa)$ such that $S\supset
U\cap\mathcal{F}(\mathcal{T},\beta)$. We set $\mathcal{S}=\\{S\\}$ so that
$\mathcal{S}$ satisfies two conditions.

Now we assume $\operatorname{diam}(U)>L_{n}^{(1)}$. Then
$U\cap\mathcal{F}(\mathcal{T},\beta)$ can be covered by
$\left\lceil\frac{\operatorname{diam}(U)}{L_{n}^{(1)}}\right\rceil^{d-\ell}$
elementary sqaures. Let $\mathcal{S}$ be the family of these elementary
squares. Then

$\displaystyle\left(L_{n}^{(1)}\right)^{t}\cdot\\#\mathcal{S}$
$\displaystyle=\left(L_{n}^{(1)}\right)^{t}\left\lceil\frac{\operatorname{diam}(U)}{L_{n}^{(1)}}\right\rceil^{d-\ell}\leq
2^{d-\ell}\left(\frac{\operatorname{diam}(U)}{L_{n}^{(1)}}\right)^{d-\ell}\left(L_{n}^{(1)}\right)^{t}$
$\displaystyle\leq
2^{d-\ell}\left(\frac{\operatorname{diam}(U)}{L_{n}^{(1)}}\right)^{t}\left(L_{n}^{(1)}\right)^{t}\leq
2^{d-\ell}\rho_{n+1}^{-t}\operatorname{diam}(U)^{t}.$

∎

###### Proof of theorem 2.1.

For a real number $t$ such that $d-\ell\leq t<s$, there exists
$n_{0}=n_{0}(t)$ such that for all $n\geq n_{0}$,

(2.8)
$P_{n}\left(L_{n}^{(1)}\right)^{t}\rho_{n+1}^{t}\prod_{i=n+1}^{D_{n}}\rho_{i}^{\ell}C_{i}\geq(16d)^{d\max\\{D_{n}-n,1\\}}\geq(16d)^{d(D_{n}-n)}.$

Let $\mathcal{U}$ be an open cover of $\mathcal{F}(\mathcal{T},\beta)$. Assume
that for all $U\in\mathcal{U}$, $\operatorname{diam}(U)$ is small enough so
that $n(U)\geq n_{0}$. Since $\mathcal{F}(\mathcal{T},\beta)$ is compact,
there exists a finite subcover $\mathcal{U}_{0}$ such that for all
$U\in\mathcal{U}_{0}$, $U\cap\mathcal{F}(\mathcal{T},\beta)\neq\varnothing$.

For $U\in\mathcal{U}_{0}$, let $\mathcal{S}_{U}$ be a family of elementary
squares given by Lemma 2.5. Let
$\mathcal{Q}=\bigcup_{U\in\mathcal{U}_{0}}\mathcal{S}_{U}$ and $n(S)=n(U)$ for
$S\in\mathcal{S}_{U}$. We note that $S$ may belong to different
$\mathcal{S}_{U}$. However, $n(S)$ is well-difined since a side length of $S$
is $L_{n(U)}^{(1)}$. Then $\mathcal{Q}$ covers
$\mathcal{F}(\mathcal{T},\beta)$ and hence

$\displaystyle\sum_{U\in\mathcal{U}_{0}}\operatorname{diam}(U)^{t}$
$\displaystyle\geq\frac{1}{2^{d-\ell}}\sum_{S\in\mathcal{Q}}\rho_{n(S)+1}^{t}\left(L_{n(S)}^{(1)}\right)^{t}\quad$
by Lemma 2.5
$\displaystyle\geq\frac{1}{2^{d-\ell}}\sum_{S\in\mathcal{Q}}(16d)^{d(D_{n}-n)}P_{n(S)}^{-1}\prod_{i=n+1}^{D_{n}}\rho_{i}^{-\ell}C_{i}^{-1}\quad$
$\displaystyle\text{by }\eqref{eq26}$
$\displaystyle\geq\frac{1}{2^{d-\ell}}\sum_{S\in\mathcal{Q}}\mu(S)\quad$ by
Lemma 2.4 $\displaystyle\geq\frac{1}{2^{d-\ell}}.$

Thus we have $\dim_{H}\mathcal{F}(\mathcal{T},\beta)\geq t$. Since we choose
arbitrary $t$ with $d-\ell\leq t<s$, the proof is completed. ∎

##  3\. Counting lattice points in convex sets

In this section, we will generalize the results in [LSST20, §3.2] for
$\mathbb{R}^{3}$ to the general $\mathbb{R}^{d+1}$. In §3.1, we first recall
the notations and lemmas in [LSST20, §3.1].

###  3.1. Preliminaries for lattice point counting

For a positive integer $D\geq 1$, we write the $D$-dimensional Euclidean space
by $\mathcal{E}_{D}=\mathbb{R}^{D}$ . For a convex body
$K\subset\mathbb{R}^{D}$ and a lattice $\Lambda\subset\mathbb{R}^{D}$, let
$\lambda_{i}(K,\Lambda)$ $(i=1,\dots,D)$ be the $i$-th successive minimum of
$\Lambda$ with respect to $K$, that is, the infimum of those numbers $\lambda$
such that $\lambda K\cap\Lambda$ contains $i$ linearly independent vectors.
Let $\operatorname{vol}(\cdot)$ be the Lebesgue measure on $\mathbb{R}^{D}$
and let $\operatorname{cov}(\Lambda)$ be the covolume of a lattice $\Lambda$,
which is the Lebesgue measure of a fundamental domain of $\Lambda$. Denote by

$\theta(K,\Lambda):=\frac{\operatorname{vol}(K)}{\operatorname{cov}(\Lambda)}.$

For an affine subspace $H$ of $\mathbb{R}^{D}$, let
$\operatorname{vol}_{H}(\cdot)$ be the Lebesgue measure on $H$ with respect to
the subspace Riemannian structure. We write
$\operatorname{vol}_{H}(S)=\operatorname{vol}_{H}(S\cap H)$ for a Borel
measurable subset $S$ of $\mathbb{R}^{D}$ by abuse of notation. We say that a
subspace $H$ of $\mathbb{R}^{D}$ is $\Lambda$-rational if $H\cap\Lambda$ is a
lattice in $H$, and denote by $\operatorname{cov}_{H}(\Lambda)$ the covolume
of the lattice $H\cap\Lambda$ in $H$. We also use the same notations for the
dual vector space $\mathcal{E}_{D}^{*}$ with respect to the standard Euclidean
structure.

We use $\|\cdot\|$ for the Euclidean norms on $\mathbb{R}^{D}$ and
$\mathcal{E}_{D}^{*}$. For a normed vector space $V$, denote by $B_{r}(V)$ (or
$B_{r}$ if $V=\mathbb{R}^{D}$) the ball of radius $r$ centered at $0\in V$. We
use $K$-norms on $\mathbb{R}^{D}$ and $\mathcal{E}_{D}^{*}$ defined by

$\begin{cases}\|\mathbf{v}\|_{K}=\inf\\{r>0:\mathbf{v}\in
rK\\},\quad&\mathbf{v}\in\mathbb{R}^{D},\\\
\|\varphi\|_{K}=\sup_{\mathbf{v}\in
K}|\varphi(\mathbf{v})|,\quad&\varphi\in\mathcal{E}_{D}^{*}.\end{cases}$

Recall that $\mathcal{L}_{D}$ is the space of unimodular lattices in
$\mathbb{R}^{D}$, which can be identified with the homogeneous space
$\operatorname{SL}_{D}(\mathbb{R})/\operatorname{SL}_{D}(\mathbb{Z})$. For
$g\in\operatorname{SL}_{D}(\mathbb{R})$ let $g^{*}$ be the adjoint action on
$\mathcal{E}_{D}^{*}$ defined by $\varphi\mapsto\varphi\circ g$. Then $g^{*}$
can be represented by the transpose of $g$ with respect to the standard basis
$\mathbf{e}_{1},\dots,\mathbf{e}_{D}$ of $\mathbb{R}^{D}$ and the dual basis
$\mathbf{e}_{1}^{*},\dots,\mathbf{e}_{D}^{*}$ of $\mathcal{E}_{D}^{*}$.

The dual lattice of $\Lambda$ in $\mathbb{R}^{D}$ is the lattice in
$\mathcal{E}_{D}^{*}$ defined by

$\Lambda^{*}=\\{\varphi\in\mathcal{E}_{D}^{*}:\varphi(\mathbf{v})\in\mathbb{Z},\
\forall\mathbf{v}\in\Lambda\\}.$

Let us define the following two sets:

$\begin{split}\mathcal{K}_{\epsilon}&=\mathcal{K}_{\epsilon}(D)=\\{\Lambda\in\mathcal{L}_{D}:\|\mathbf{v}\|\geq\epsilon,\
\forall\mathbf{v}\in\Lambda\smallsetminus\\{0\\}\\}=\\{\Lambda\in\mathcal{L}_{D}:\lambda_{1}(B_{1},\Lambda)\geq\epsilon\\};\\\
\mathcal{K}_{\epsilon}^{*}&=\mathcal{K}_{\epsilon}^{*}(D)=\\{\Lambda\in\mathcal{L}_{D}:\|\varphi\|\geq\epsilon,\
\forall\varphi\in\Lambda^{*}\smallsetminus\\{0\\}\\}.\end{split}$

Since $\mathcal{E}_{D}^{*}$ can be naturally identified with
$\bigwedge_{\mathbb{R}}^{D-1}\mathbb{R}^{D}$ with the standard Euclidean
structure, we have $\Lambda^{*}=\bigwedge_{\mathbb{Z}}^{D-1}\Lambda$.

A nonzero vector $\mathbf{v}\in\Lambda$ is said to be primitive if
$(1/n)\mathbf{v}\notin\Lambda$ for all $n\in\mathbb{N}$. The set of primitive
vectors in $\Lambda$ is denoted by $\widehat{\Lambda}$.

We summarize the lemmas in [LSST20, §3.1].

######  Lemma 3.1.

Let $D\geq 2$. For every lattice $\Lambda$ in $\mathbb{R}^{D}$ and every
bounded centrally symmetric convex subset $K$ of $\mathbb{R}^{D}$ with
$\lambda_{d}(K,\Lambda)\leq 1$ we have

${\\#(K\cap\widehat{\Lambda})=\left(\zeta(D)^{-1}+\eta(K,\Lambda)\right)\cdot\theta(K,\Lambda)}$

where $\zeta$ is the Riemann $\zeta$-function and

${|\eta(K,\Lambda)|\ll_{D}\lambda_{D}(K,\Lambda)-\lambda_{D}(K,\Lambda)\log\lambda_{1}(K,\Lambda).}$

######  Lemma 3.2.

Let $D\geq 2$. For every lattice $\Lambda$ in $\mathbb{R}^{D}$ and every
bounded centrally symmetric convex subset $K$ of $\mathbb{R}^{D}$ with
$\lambda_{D}(K,\Lambda)\leq 1$ we have

${\\#(K\cap(\Lambda\smallsetminus\\{0\\}))=(1+\alpha(K,\Lambda))\cdot\theta(K,\Lambda)}$

where $|\alpha(K,\Lambda)|\ll_{D}\lambda_{D}(K,\Lambda)$

######  Lemma 3.3.

Let $K$ and $\Lambda$ be as in Lemma 3.2. Then

${\\#(K\cap\Lambda)\asymp_{D}\theta(K,\Lambda).}$

######  Lemma 3.4.

Let $D\geq 1$. Let $\Lambda$ be a lattice in $\mathbb{R}^{D}$ and $K$ be a
bounded centrally symmetric convex subset of $\mathbb{R}^{D}$ with nonempty
interior. Then

${\\#(K^{\circ}\cap\Lambda)\asymp_{D}\\#(K\cap\Lambda)\asymp_{D}\\#(\overline{K}\cap\Lambda).}$

######  Lemma 3.5.

Let $K$ and $\Lambda$ be as in Lemma 3.2. If $\lambda_{i}(K,\Lambda)\leq s\leq
s^{\prime}\leq\lambda_{j+1}(K,\Lambda)$ where $1\leq i\leq j\leq D$, then

${\left(\frac{s^{\prime}}{s}\right)^{i}\ll_{D}\frac{\\#(s^{\prime}K\cap\Lambda)}{\\#(sK\cap\Lambda)}\ll_{D}\left(\frac{s^{\prime}}{s}\right)^{j}.}$

######  Lemma 3.6.

Let $D\geq 2$. Let $K$ be a bounded centrally symmetric convex subset of
$\mathbb{R}^{D}$ with nonempty interior and let
$\varphi\in\mathcal{E}_{D}^{*}\smallsetminus\\{0\\}$. Then

${\mathrm{vol}_{H_{\varphi}}(K)\asymp_{D}\|\varphi\|\mathrm{vol}(K)/\|\varphi\|_{K}.}$

We need the following auxiliary lemma.

######  Lemma 3.7.

Given $D\geq 2$ and $r>0$, let $\Lambda\in\mathcal{K}_{r}^{*}(D)$, and let
$\mathbf{v},\mathbf{w}\in\Lambda$ be any nonzero linearly independent vectors.
Then there exists a positive constant $c^{\prime}=c^{\prime}(D)>0$ such that
$\|\mathbf{v}\wedge\mathbf{w}\|\geq c^{\prime}r^{D-2}$.

###### Proof.

Let $\Lambda^{\prime}$ be the 2-dimensional sublattice of $\Lambda$ generated
by $\mathbf{v},\mathbf{w}$. By Minkowski’s second theorem, we have

(3.1)
${\|\mathbf{v}\wedge\mathbf{w}\|\geq\text{cov}(\Lambda^{\prime})\gg_{2}\lambda_{1}(B_{1},\Lambda^{\prime})\lambda_{2}(B_{1},\Lambda^{\prime})\geq\lambda_{1}(B_{1},\Lambda)\lambda_{2}(B_{1},\Lambda).}$

Agian by Minkowski’s second theorem, we have

(3.2)
${\begin{split}1\ll_{D}\lambda_{1}(B_{1},\Lambda)\cdots\lambda_{D}(B_{1},\Lambda)&\leq\lambda_{1}(B_{1},\Lambda)\lambda_{2}(B_{1},\Lambda)\lambda_{D}(B_{1},\Lambda)^{D-2}\\\
&\leq\lambda_{1}(B_{1},\Lambda)\lambda_{2}(B_{1},\Lambda)\frac{1}{r^{D-2}}.\end{split}}$

The last inequality comes from $\Lambda\in\mathcal{K}_{r}^{*}(D)$. The result
is following by combining (3.1) and (3.2). ∎

###  3.2. Lattice point counting in $\mathbb{R}^{d+1}$

For $d\geq 2$ and a $(d+1)$-tuple $\mathbf{r}=(r_{1},\dots,r_{d+1})$ of
positive real numbers, we estimate the number of lattice points in the set

${M_{\mathbf{r}}=\\{(x_{1},\dots,x_{d+1})\in\mathbb{R}^{d+1}:|x_{i}|\leq
r_{i},\ \forall i=1,\dots,d+1\\}.}$

Let

${M_{\mathbf{r}}^{*}=\\{\varphi\in\mathcal{E}_{d+1}^{*}:|x_{i}^{\varphi}|\leq
r_{i},\ \forall i=1,\dots,d+1\\},}$

where the element $\varphi\in\mathcal{E}_{d+1}^{*}$ is represented by
$\varphi=\sum_{i=1}^{d+1}x_{i}^{\varphi}\mathbf{e}_{i}^{*}$.

######  Lemma 3.8.

Let $d\geq 2$. For any real number $c_{0}>1$, there exists positive real
number $\tilde{c}<1$ such that for every lattice $\Lambda$ in
$\mathbb{R}^{d+1}$ and every $(d+1)$-tuple $\mathbf{r}$ of positive real
numbers with

${\lambda_{d+1}(M_{\mathbf{r}},\Lambda)\leq\tilde{c}\quad\text{and}\quad-\lambda_{d+1}(M_{\mathbf{r}},\Lambda)\log\lambda_{1}(M_{\mathbf{r}},\Lambda)\leq\tilde{c}}$

one has

${\frac{1}{c_{0}\zeta(d+1)}\theta(M_{\mathbf{r}},\Lambda)\leq\\#(M_{\mathbf{r}}\cap\widehat{\Lambda})\leq\frac{c_{0}}{\zeta(d+1)}\theta(M_{\mathbf{r}},\Lambda).}$

###### Proof.

The proof follows directly from Lemma 3.1. ∎

Now we fix real numbers $s,r_{1},\dots,r_{d+1}$ such that $0<s<1/2$,
$r_{i}\geq 1$ for each $i=1,\dots,d$, and $r_{d+1}=1$. Denote by
$\mathbf{r}=(r_{1},\dots,r_{d+1})$, $r_{M}=\max_{1\leq i\leq d}r_{i}$, and
$r_{m}=\min_{1\leq i\leq d}r_{i}$. Define a norm

${\|\varphi\|_{\mathbf{r}}=\max\left\\{r_{i}|x_{i}^{\varphi}|:i=1,\dots,d+1\right\\}.}$

It follows from the definition that

(3.3)
${\|\varphi\|_{\mathbf{r}}\leq\|\varphi\|_{M_{\mathbf{r}}}\leq(d+1)\|\varphi\|_{\mathbf{r}}.}$

For $q>0$ let

${N_{q}(\mathbf{r},s)=\left\\{\varphi\in\mathcal{E}_{d+1}^{*}:|x_{i}^{\varphi}|\leq
s,\ \forall i=1,\dots,d,\text{ and }\|\varphi\|_{\mathbf{r}}\leq q\right\\}.}$

Note that $N_{q}(\mathbf{r},s)=M^{*}_{\mathbf{r}^{\prime}}$ where
$\mathbf{r}^{\prime}=(r_{1}^{\prime},\dots,r_{d}^{\prime},q)$ with
$r_{i}^{\prime}=\min\\{q/r_{i},s\\}$. For a lattice $\Lambda$ in
$\mathbb{R}^{d+1}$ and $i=1,\dots,d+1$, let $q_{i}(\Lambda,\mathbf{r},s)$ be
the infimum of those positive real number $q$ such that
$N_{q}(\mathbf{r},s)\cap\Lambda$ contains $i$ linearly independent vectors. We
will give an upper bound of the number of

${\mathcal{S}(\Lambda,\mathbf{r},s):=\left\\{\mathbf{v}\in
M_{\mathbf{r}}\cap\widehat{\Lambda}:\varphi(\mathbf{v})=0\text{ for some
}\varphi\in N_{(d+1)sr_{M}}(\mathbf{r},s)\cap\widehat{\Lambda}^{*}\right\\},}$

where $\widehat{\Lambda}^{*}$ is the set of primitive vectors in
$\Lambda^{*}$.

######  Lemma 3.9.

For $d\geq 2$, let $\Lambda$ be a unimodular lattice in $\mathbb{R}^{d+1}$
with $q_{1}(\Lambda,\mathbf{r},s)\geq s^{-2}$. Then

  1. (1)

if $r_{m}=r_{M}$ and $q_{d+1}(\Lambda,\mathbf{r},s)\leq ds^{-1/2}r_{M}$, then

$\\#\mathcal{S}(\Lambda,\mathbf{r},s)\ll
s^{1/2}\cdot\mathrm{vol}(M_{\mathbf{r}});$

  2. (2)

if $r_{m}<r_{M}$ and $q_{d+1}(\Lambda,\mathbf{r},s)\log
q_{d+1}(\Lambda,\mathbf{r},s)\leq sr_{M}$, then

$\\#\mathcal{S}(\Lambda,\mathbf{r},s)\ll
s^{2}\cdot\operatorname{vol}(M_{\mathbf{r}}).$

###### Proof.

For simplicity, we denote by $N_{q}=N_{q}(\mathbf{r},s)$,
$q_{i}=q_{i}(\Lambda,\mathbf{r},s)$ and
$\mathcal{S}=\mathcal{S}(\Lambda,\mathbf{r},s)$. If
$N_{(d+1)sr_{M}}\cap\widehat{\Lambda}^{*}$ is empty then there is nothing to
prove. We assume that $N_{(d+1)sr_{M}}\cap\widehat{\Lambda}^{*}$ is nonempty.
It follows from the definition that

(3.4) ${\\#\mathcal{S}\leq\sum_{\varphi\in
N_{(d+1)sr_{M}}\cap\widehat{\Lambda}^{*}}\\#(H_{\varphi}\cap
M_{\mathbf{r}}\cap\widehat{\Lambda})}$

with the notation $H_{\varphi}=\mathrm{ker}\varphi$.

We first claim that for every $\varphi\in
N_{(d+1)sr_{M}}\cap\widehat{\Lambda}^{*}$,

(3.5) ${\\#(H_{\varphi}\cap
M_{\mathbf{r}}\cap\widehat{\Lambda})\ll\frac{\mathrm{vol}(M_{\mathbf{r}})}{\|\varphi\|_{M_{\mathbf{r}}}}\leq\frac{\operatorname{vol}(M_{\mathbf{r}})}{\|\varphi\|_{\mathbf{r}}}.}$

where the second inequality follows from (3.3). If $\\#(H_{\varphi}\cap
M_{\mathbf{r}}\cap\widehat{\Lambda})<d+1$, then it follows from (3.3) that

${\frac{\mathrm{vol}(M_{\mathbf{r}})}{\|\varphi\|_{M_{\mathbf{r}}}}\geq\frac{2^{(d+1)}r_{1}\dots
r_{d+1}}{(d+1)\|\varphi\|_{\mathbf{r}}}\geq\frac{2^{(d+1)}r_{1}\dots
r_{d+1}}{(d+1)^{2}sr_{M}}\gg\\#(H_{\varphi}\cap
M_{\mathbf{r}}\cap\widehat{\Lambda}).}$

Otherwise, $H_{\varphi}\cap M_{\mathbf{r}}\cap\Lambda$ has $d$ linearly
independent vectors, hence it follows from Lemma 3.3 and Lemma 3.6 that

${\\#(H_{\varphi}\cap
M_{\mathbf{r}}\cap\widehat{\Lambda})\ll\frac{\operatorname{vol}_{H_{\varphi}}(M_{\mathbf{r}})}{\operatorname{cov}_{H_{\varphi}}(\Lambda)}\ll\frac{\|\varphi\|\operatorname{vol}(M_{\mathbf{r}})}{\operatorname{cov}_{H_{\varphi}}(\Lambda)\|\varphi\|_{M_{\mathbf{r}}}}\ll\frac{\operatorname{vol}(M_{\mathbf{r}})}{\|\varphi\|_{M_{\mathbf{r}}}},}$

which concludes the claim.

By (3.4) and (3.5), it suffices to estimate

(3.6) ${\begin{split}\eta:=&\sum_{\varphi\in
N_{(d+1)sr_{M}}\cap\widehat{\Lambda}^{*}}\|\varphi\|_{\mathbf{r}}^{-1}\\\
=&\frac{1}{(d+1)sr_{M}}\\#(N_{(d+1)sr_{M}}\cap\widehat{\Lambda}^{*})+\sum_{\varphi\in
N_{(d+1)sr_{M}}\cap\widehat{\Lambda}^{*}}\int_{\|\varphi\|_{\mathbf{r}}}^{(d+1)sr_{M}}\frac{1}{q^{2}}\mathrm{d}q.\end{split}}$

We denote the first and second terms in the last line by $\eta_{1},\eta_{2}$,
respectively.

Observe that

(3.7) ${\begin{split}\eta_{2}&=\sum_{\varphi\in
N_{(d+1)sr_{M}}\cap\widehat{\Lambda}^{*}}\int_{q_{1}}^{(d+1)sr_{M}}\frac{\mathbbm{1}_{q}(\|\varphi\|_{\mathbf{r}})}{q^{2}}\mathrm{d}q\\\
&=\int_{q_{1}}^{(d+1)sr_{M}}\sum_{\varphi\in
N_{(d+1)sr_{M}}\cap\widehat{\Lambda}^{*}}\frac{\mathbbm{1}_{q}(\|\varphi\|_{\mathbf{r}})}{q^{2}}\mathrm{d}q\\\
&\leq\int_{q_{1}}^{(d+1)sr_{M}}\frac{\\#(N_{q}\cap\widehat{\Lambda}^{*})}{q^{2}}\mathrm{d}q.\end{split}}$

where $\mathbbm{1}_{q}$ denotes the indicator function of the set
$\\{x\in\mathbb{R}:x\leq q\\}$.

For $i=2,\dots,d$, if $q_{i-1}\leq q<q_{i}$ then
$\\#(N_{q}\cap\widehat{\Lambda}^{*})=i\leq d$. Thus

(3.8)
${\int_{q_{1}}^{q_{d}}\frac{\\#(N_{q}\cap\widehat{\Lambda}^{*})}{q^{2}}\mathrm{d}q\leq\int_{q_{1}}^{q_{d}}\frac{d}{q^{2}}\mathrm{d}q\leq\frac{d}{q_{1}}\ll
s^{2}\leq s^{1/2},}$

where the third inequality follows from the assumption $q_{1}\geq s^{-2}$.

###### Proof of the assertion (1).

We claim that $\eta\ll s^{-1/2}$ under the assumption of (1), which concludes
the assertion (1). Assume that $r_{m}=r_{M}$ and $q_{d+1}\leq ds^{-1/2}r_{M}$.
Observe that by definition

(3.9) ${N_{(d+1)s^{-1/2}r_{M}}=M_{(s,\dots,s,(d+1)s^{-1/2}r_{M})}^{*}.}$

We have an upper bound of $\eta_{1}$ as

(3.10)
${\eta_{1}\leq\frac{\\#(N_{(d+1)s^{-1/2}r_{M}}\cap\Lambda^{*})}{(d+1)sr_{M}}\ll\frac{\operatorname{vol}(N_{(d+1)s^{-1/2}r_{M}})}{(d+1)sr_{M}}\ll
s^{d-3/2}\leq s^{1/2}.}$

The first inequality follows from $s<1/2$, the second inequality follows from
Lemma 3.3, and the third inequality follows from (3.9).

For an upper bound of $\eta_{2}$, we first compute

(3.11)
${\begin{split}\int_{sr_{M}}^{(d+1)sr_{M}}\frac{\\#(N_{q}\cap\widehat{\Lambda}^{*})}{q^{2}}\mathrm{d}q&\leq\int_{sr_{M}}^{(d+1)sr_{M}}\frac{\\#(N_{(d+1)sr_{M}}\cap\Lambda^{*})}{q^{2}}\mathrm{d}q\\\
&\leq\frac{\\#(N_{(d+1)sr_{M}}\cap\Lambda^{*})}{sr_{M}}\\\ &\ll
s^{1/2},\end{split}}$

where the last inequality can be shown by the same as (3.10).

If $sr_{M}\leq q_{d}$, then it follows from (3.8) and (3.11) that $\eta_{2}\ll
s^{1/2}$. Now we suppose that $sr_{M}>q_{d}$. For all $q_{d}<q\leq
sr_{M}=sr_{m}$, observe that

$N_{q}=M_{(q/r_{1},\dots,q/r_{d+1})}^{*}=\frac{q}{sr_{M}}N_{sr_{M}}.$

Since
$\lambda_{d}(N_{q},\Lambda)=\lambda_{d}(\frac{q}{sr_{M}}N_{sr_{M}},\Lambda)\leq
1\leq sr_{M}/q$, it follows from Lemma 3.5 that

${\\#(N_{q}\cap\widehat{\Lambda}^{*})\leq\\#\left(\frac{q}{sr_{M}}N_{sr_{M}}\cap\Lambda^{*}\right)\ll\left(\frac{q}{sr_{M}}\right)^{d}\\#(N_{sr_{M}}\cap\Lambda^{*}).}$

By $sr_{M}\leq ds^{-1/2}r_{M}$ and Lemma 3.3, we have

$\displaystyle\\#(N_{q}\cap\widehat{\Lambda}^{*})$
$\displaystyle\ll\left(\frac{q}{sr_{M}}\right)^{d}\\#(N_{ds^{-1/2}r_{M}}\cap\Lambda^{*})$
$\displaystyle\ll\left(\frac{q}{sr_{M}}\right)^{d}\operatorname{vol}(N_{ds^{-1/2}r_{M}})$
$\displaystyle\ll\left(\frac{q}{sr_{M}}\right)^{2}s^{d-1/2}r_{M}\ll\frac{q^{2}s^{-1/2}}{r_{M}}.$

The last line follows from $\frac{q}{sr_{M}}\leq 1$ and $s\leq 1$. Thus we
have

${\int_{q_{d}}^{sr_{M}}\frac{\\#(N_{q}\cap\widehat{\Lambda}^{*})}{q^{2}}\mathrm{d}q\ll\int_{q_{d}}^{sr_{M}}\frac{s^{-1/2}}{r_{M}}\mathrm{d}q\ll
s^{1/2}.}$

It follows that $\eta_{2}\ll s^{1/2}$ under the assumption of (1), which
concludes the assertion (1). ∎

###### Proof of the assertion (2).

We will prove that $\eta\ll s^{2}$ under the assumption of (2). By the
assumption, we have $q_{d+1}\geq q_{1}\geq s^{-2}\geq 4$ so that
$q_{d+1}<sr_{M}<(d+1)sr_{M}$ since $q_{d+1}\log q_{d+1}\leq sr_{M}$. Thus
$N_{(d+1)sr_{M}}\cap\Lambda^{*}$ contains $d+1$ linearly independent vectors.
By Lemma 3.3, we have

(3.12)
${\eta_{1}\leq\frac{\\#(N_{(d+1)sr_{M}}\cap\Lambda^{*})}{(d+1)sr_{M}}\ll\frac{\operatorname{vol}(N_{(d+1)sr_{M}})}{(d+1)sr_{M}}\ll
s^{d}\leq s^{2}.}$

By (3.7), it suffices to show that

${\int_{q_{1}}^{(d+1)sr_{M}}\frac{\\#(N_{q}\cap\widehat{\Lambda}^{*})}{q^{2}}\mathrm{d}q\ll
s^{2}.}$

We split the domain of integration as
$(q_{1},q_{d})\cup(q_{d},q_{d+1})\cup(q_{d+1},sr_{M})\cup(sr_{M},(d+1)sr_{M})$
and estimate upper bounds of the integrals.

For each $q\in(sr_{M},(d+1)sr_{M})$, it follows from Lemma 3.3 that
$\\#(N_{q}\cap\widehat{\Lambda}^{*})\ll\operatorname{vol}(N_{q})\ll s^{d}q$.
Thus we have

(3.13)
${\int_{sr_{M}}^{(d+1)sr_{M}}\frac{\\#(N_{q}\cap\widehat{\Lambda}^{*})}{q^{2}}\mathrm{d}q\ll\int_{sr_{M}}^{(d+1)sr_{M}}\frac{s^{d}}{q}\mathrm{d}q=s^{d}\log(d+1)\ll
s^{2}.}$

For each $q\in(q_{d+1},sr_{M})$, it follows from Lemma 3.3 that
$\\#(N_{q}\cap\widehat{\Lambda}^{*})\ll\operatorname{vol}(N_{q})\ll
s^{d-1}q^{2}/r_{M}$. Thus we have

(3.14)
${\int_{q_{d+1}}^{sr_{M}}\frac{\\#(N_{q}\cap\widehat{\Lambda}^{*})}{q^{2}}\mathrm{d}q\ll\int_{q_{d+1}}^{sr_{M}}\frac{s^{d-1}}{r_{M}}\mathrm{d}q\leq
s^{d}\leq s^{2}.}$

By (3.8), the integral over $(q_{1},q_{d})$ is bounded above by $s^{2}$.

Now it remains to show that the integral over $(q_{d},q_{d+1})$ is bounded
above by $s^{2}$. Let
$H=\operatorname{Span}_{\mathbb{R}}(N_{q_{d}}\cap\Lambda^{*})$. We claim that
for every $q\in(q_{d},q_{d+1})$,

(3.15)
${\operatorname{vol}_{H}(N_{q})\leq\frac{q}{q_{d+1}}\operatorname{vol}_{H}(N_{q_{d+1}}).}$

If $H$ contains $\mathbf{e}_{d+1}^{*}$, then the claim is easily checked from
the definition of $N_{q}$. Otherwise, we let $\operatorname{\mathbf{pr}}^{*}$
be the orthogonal projection onto
$\operatorname{Span}_{\mathbb{R}}\\{\mathbf{e}_{1}^{*},\dots,\mathbf{e}_{d}^{*}\\}$.
Then the volume of $\operatorname{\mathbf{pr}}^{*}(N_{q})$ is at most
$q/q_{d+1}$ times the volume of $\operatorname{\mathbf{pr}}^{*}(N_{q_{d+1}})$
since $q/r_{M}<q_{d+1}/r_{M}<s$. Thus we prove the claim.

For each $q\in(q_{d},q_{d+1})$, we have

$\displaystyle\\#(N_{q}\cap\widehat{\Lambda}^{*})$
$\displaystyle\ll\frac{\operatorname{vol}_{H}(N_{q})}{\operatorname{cov}_{H}(\Lambda^{*})}$
by Lemma 3.3
$\displaystyle\leq\frac{q}{q_{d+1}}\frac{\operatorname{vol}_{H}(N_{q_{d+1}})}{\operatorname{cov}_{H}(\Lambda^{*})}$
by (3.15) $\displaystyle\ll\frac{q}{q_{d+1}}\\#(N_{q_{d+1}}\cap
H\cap\Lambda^{*})$ by Lemma 3.3
$\displaystyle\ll\frac{q}{q_{d+1}}\\#(N_{q_{d+1}}^{\circ}\cap
H\cap\Lambda^{*})$ by Lemma 3.4
$\displaystyle=\frac{q}{q_{d+1}}\\#(N_{q_{d+1}}^{\circ}\cap\Lambda^{*})\leq\frac{q}{q_{d+1}}\\#(N_{q_{d+1}}\cap\Lambda^{*})$
$\displaystyle\ll\frac{q}{q_{d+1}}\operatorname{vol}(N_{q_{d+1}})$ by Lemma
3.3 $\displaystyle\ll s^{d-1}\frac{q_{d+1}q}{r_{M}}.$

Therefore, we have

(3.16)
${\int_{q_{d}}^{q_{d+1}}\frac{\\#(N_{1}\cap\widehat{\Lambda}^{*})}{q^{2}}\mathrm{d}q\ll\int_{q_{d}}^{q_{d+1}}s^{2}\frac{q_{d+1}}{sr_{M}}\frac{1}{q}\mathrm{d}q\leq
s^{2}\frac{q_{d+1}\log q_{d+1}}{sr_{M}}\leq s^{2}.}$

By combining (3.8), (3.13), (3.14), and (3.16), the proof of (2) is completed.
∎

This proves Lemma 3.9. ∎

For a weight vector $w=(w_{1},\dots,w_{d})$ as in the introduction, let
$1\leq\ell\leq d-1$ be the unique integer such that
$w_{1}=\cdots=w_{\ell}>w_{\ell+1}\geq\cdots\geq w_{d}$, and denote by
$\xi=\max(1,\frac{d-\ell}{\ell})$. For a fixed lattice
$\Lambda\subset\mathbb{R}^{d+1}$ and fixed $\mathbf{r},s$, we denote
$q_{i}(\Lambda,\mathbf{r},s)$ by $q_{i}(\Lambda)$ and $N_{q}(\mathbf{r},s)$ by
$N_{q}$ for simplicity. Let us fix a constant $C\geq 1$ which is an implied
constant for the conclusion of Lemma 3.9 (1) and (2).

######  Lemma 3.10.

Let $d\geq 2$, $s=\epsilon^{2}$, $\mathbf{r}=(r_{1},\dots,r_{d+1})=(\epsilon
e^{t},\dots,\epsilon e^{t},1)$,
$\Lambda\in\mathcal{K}_{\epsilon^{2}}^{*}\cap\mathcal{L}_{d+1}^{\prime}$, and
$a_{t}=\mathrm{diag}(e^{w_{1}t},\dots,e^{w_{d}t},e^{-t})$. Then there exists a
positive real number $\tilde{\epsilon}\leq 1$ and $c=c(d)>(d+1)^{1/14}$ such
that for all $\epsilon,t>0$ with
$ce^{-w_{d}t/(2d^{3})}<\epsilon<\tilde{\epsilon}$, one has

$\\#\mathcal{S}(a_{t}\Lambda,\mathbf{r},s)\leq\epsilon^{1/2}\cdot\operatorname{vol}(M_{\mathbf{r}}).$

###### Proof.

We will prove the lemma for $\tilde{\epsilon}<1/C^{2}$ and the constant $c$
will be determined later. By Lemma 3.9 (1), it suffices to show that

(3.17) ${q_{1}(a_{t}\Lambda)\geq s^{-2}\quad\mathrm{and}\quad
q_{d+1}(a_{t}\Lambda)\leq ds^{-1/2}r_{d}.}$

First, note that

$N_{q}\cap(a_{t}\Lambda)^{*}=N_{q}\cap
a_{-t}^{*}\Lambda^{*}=a_{-t}^{*}(a_{t}^{*}N_{q}\cap\Lambda^{*}),$

where $a_{t}^{*}$ denotes the transpose of $a_{t}$. Hence it is enough to show
that $a_{t}^{*}N_{s^{-2}}$ has no nonzero lattice point of $\Lambda^{*}$ for
the first inequality of (3.17). Since $d\geq 2$ and $w_{d}\leq 1/d$, we have

$e^{-\frac{t}{7}}<e^{-\frac{w_{d}t}{2d^{3}}}<ce^{-\frac{w_{d}t}{2d^{3}}}<\epsilon,$

that is, $s^{-2}<r_{1}s$. Thus we have

$N_{s^{-2}}=M^{*}_{(s^{-2}/r_{1},\dots,s^{-2}/r_{1},s^{-2})}=M^{*}_{(\epsilon^{-5}e^{-t},\dots,\epsilon^{-5}e^{-t},\epsilon^{-4})},$

which implies that

$a_{t}^{*}N_{s^{-2}}=M^{*}_{(\epsilon^{-5}e^{(w_{1}-1)t},\dots,\epsilon^{-5}e^{(w_{d}-1)t},\epsilon^{-4}e^{-t})}.$

Since for all $i=1,\dots,d$

$\frac{\epsilon}{(d+1)^{1/14}}>\frac{c}{(d+1)^{1/14}}e^{-\frac{w_{d}t}{2d^{3}}}>e^{-\frac{(d-1)w_{d}t}{7}}\geq
e^{\frac{(w_{i}-1)t}{7}},$

we have $\epsilon^{-5}e^{(w_{i}-1)t}<\frac{\epsilon^{2}}{\sqrt{d+1}}$ for all
$i=1,\dots,d$. It is clear that
$\epsilon^{-4}e^{-t}<\epsilon^{-5}e^{(w_{d}-1)t}<\frac{\epsilon^{2}}{\sqrt{d+1}}$.
Thus $a_{t}^{*}N_{s^{-2}}$ is contained in the interior of
$B_{\epsilon^{2}}(\mathcal{E}_{d+1}^{*})$. Since
$\Lambda\in\mathcal{K}_{\epsilon^{2}}^{*}$, there is no lattice point of
$\Lambda$ in $a_{t}^{*}N_{s^{-2}}$.

To show the second inequality of (3.17), we will construct a basis for
$\Lambda^{*}$ of which vectors are contained in
$N_{ds^{-1/2}r_{d}}=N_{de^{t}}$. Since $de^{t}>r_{d}>r_{d}s$, we have

$a_{t}^{*}N_{de^{t}}=a_{t}^{*}M^{*}_{(s,\dots,s,de^{t})}=M^{*}_{(se^{w_{1}t},\dots
se^{w_{d}t},d)}.$

Let $1/2<r\leq 1$ be such that $r\mathbf{e}_{d+1}\in\widehat{\Lambda}$ from
the assumption $\Lambda\in\mathcal{L}_{d+1}^{\prime}$. Let
$\operatorname{\mathbf{pr}}:\mathbb{R}^{d+1}\to\mathbb{R}^{d}$ be the
orthogonal projection onto
$\operatorname{Span}{(\mathbf{e}_{1},\dots,\mathbf{e}_{d})}$. Note that
$\operatorname{\mathbf{pr}}(\Lambda)$ is a lattice with covolume $1/r$ in
$\mathbb{R}^{d}$. If $\mathbf{v}\in\Lambda$ satisfies
$\|\operatorname{\mathbf{pr}}(\mathbf{v})\|=\lambda_{1}(B_{1},\operatorname{\mathbf{pr}}(\Lambda))$,
then since $\Lambda\in\mathcal{K}_{\epsilon^{2}}^{*}$, it follows from Lemma
3.7 with $D=d+1$ that

(3.18) $\lambda_{1}(B_{1},\operatorname{\mathbf{pr}}(\Lambda))\geq
r\lambda_{1}(B_{1},\operatorname{\mathbf{pr}}(\Lambda))=\|\mathbf{v}\wedge
r\mathbf{e}_{d+1}\|\geq\bar{c}_{1}(\epsilon^{2})^{d-1}$

for some $\bar{c}_{1}=\bar{c}_{1}(d)<1$. Since
$\mathrm{cov}(\operatorname{\mathbf{pr}}(\Lambda))=1/r$, it follows from the
Minkowski’s second theorem and (3.18) that for any $0<c_{1}<\bar{c}_{1}$

$c_{1}^{d-1}(\epsilon^{2})^{(d-1)^{2}}\lambda_{d}(B_{1},\operatorname{\mathbf{pr}}(\Lambda))\leq\lambda_{1}(B_{1},\operatorname{\mathbf{pr}}(\Lambda))\cdots\lambda_{d}(B_{1},\operatorname{\mathbf{pr}}(\Lambda))\ll
1,$

hence there exists $c_{2}=c_{2}(d)>1$ such that

(3.19) ${\lambda_{d}(B_{1},\operatorname{\mathbf{pr}}(\Lambda))\leq
c_{2}(\epsilon^{-2})^{(d-1)^{2}}.}$

Let $\\{v^{(i)}:i=1,\dots,d\\}$ be a Minkowski reduced basis for
$\operatorname{\mathbf{pr}}(\Lambda)$ such that $\|v^{(i)}\|\leq
2^{d}\lambda_{i}(B_{1},\operatorname{\mathbf{pr}}(\Lambda))$. For each
$i=1,\dots,d$, let $\mathbf{v}_{i}\in\Lambda$ be such that
$\operatorname{\mathbf{pr}}(\mathbf{v}_{i})=v^{(i)}$ and
$|\mathbf{e}^{*}_{d+1}(\mathbf{v}_{i})|<1$. Then the vectors
$\mathbf{v}_{1},\dots,\mathbf{v}_{d},\mathbf{v}_{d+1}=r\mathbf{e}_{d+1}$ form
a basis for $\Lambda$. Recall that $\mathcal{E}_{d+1}^{*}$ can be naturally
identified with $\bigwedge_{\mathbb{R}}^{d}\mathbb{R}^{d+1}$ with the standard
Euclidean structure. Under this identification, we have
$\Lambda^{*}=\bigwedge_{\mathbb{Z}}^{d}\Lambda$, hence the vectors
$\bigwedge_{j\neq i}\mathbf{v}_{j}$ for $i=1,\dots,d+1$ forms a basis for
$\bigwedge_{\mathbb{Z}}^{d}\Lambda$. We now claim that the vectors
$\bigwedge_{j\neq i}\mathbf{v}_{j}$ for $i=1,\dots,d+1$ are contained in
$a_{t}^{*}N_{de^{t}}$ via the above identification, which proves that
$q_{d+1}(a_{t}\Lambda)\leq ds^{-1/2}r_{d}$.

For each $i=1,\dots,d+1$, write

$\bigwedge_{j\neq
i}\mathbf{v}_{j}=\sum_{h=1}^{d+1}\left(x_{h}^{(i)}\bigwedge_{k\neq
h}\mathbf{e}_{k}\right).$

Note that $|x_{d+1}^{(d+1)}|=1/r\leq 2\leq d$ and $x_{d+1}^{(i)}=0$ for each
$i=1,\dots,d$ since $\mathbf{v}_{d+1}=r\mathbf{e}_{d+1}$. By the definition of
$\mathbf{v}_{i}$ and (3.19), since $\epsilon<1$, we can choose large enough
$c_{3}=c_{3}(d)>(d+1)^{d^{2}/7}$ for each $i=1,\dots,d$,

$\|\mathbf{v}_{i}\|\leq\sqrt{1+\|v^{(i)}\|^{2}}\leq
2^{d}\sqrt{2}c_{2}(\epsilon^{-2})^{(d-1)^{2}}\leq
c_{3}(\epsilon^{-2})^{(d-1)^{2}}.$

Thus for each $i=1,\dots,d+1$ and $h=1,\dots,d$,

$|x_{h}^{(i)}|\leq\left\|\bigwedge_{j\neq
i}\mathbf{v}_{j}\right\|\leq\prod_{j\neq i}\|\mathbf{v}_{j}\|\leq
c_{3}^{d}(\epsilon^{-2})^{d(d-1)^{2}}.$

From the assumption $ce^{-w_{d}t/(2d^{3})}<\epsilon$, it follows that

$c^{2d^{3}}e^{-w_{d}t}<(\epsilon^{2})^{d^{3}}<(\epsilon^{2})^{d(d-1)^{2}+1}.$

Choosing $c=c_{3}^{1/2d^{2}}>(d+1)^{1/14}$, we have

$|x_{h}^{(i)}|\leq
c_{3}^{d}(\epsilon^{-2})^{d(d-1)^{2}}<\epsilon^{2}e^{w_{d}t}=se^{w_{d}t}\leq
se^{w_{i}t},$

which concludes the claim. ∎

######  Lemma 3.11.

Let $d\geq 2$, $\mathbf{r}=(r_{1},\dots,r_{d+1})$,
$\overline{b}_{t}=\operatorname{diag}\left(\overline{b}_{t,1},\dots,\overline{b}_{t,d+1}\right)$,
and $\Lambda\in\mathcal{K}^{*}_{\epsilon^{2}}$, where

$r_{i}=\begin{cases}\epsilon
e^{\left(\xi-\frac{1}{\ell}(w_{\ell+1}+\cdots+w_{d})\right)t}&\quad\text{if}\quad
1\leq i\leq\ell,\\\ \epsilon e^{(\xi+w_{i})t}&\quad\text{if}\quad\ell+1\leq
i\leq d,\\\ 1&\quad\text{if}\quad i=d+1,\end{cases}$

and

$\overline{b}_{t,i}=\begin{cases}e^{\left(\xi
w_{i}-\frac{1}{\ell}(w_{\ell+1}+\cdots+w_{d})\right)t}&\quad\text{if}\quad
1\leq i\leq\ell,\\\ e^{(1+\xi)w_{i}t}&\quad\text{if}\quad\ell+1\leq i\leq
d,\\\ e^{-\xi t}&\quad\text{if}\quad i=d+1.\end{cases}$

Then there exists a positive real number $\tilde{s}\leq 1$ such that for all
$s,t>0$ with $e^{-\delta t}<\epsilon<s<\tilde{s}$ where
$\delta=\frac{1}{18d^{2}}\min\left(\xi w_{d},\xi
w_{1}-\frac{1}{\ell}(w_{\ell+1}+\cdots+w_{d})\right)$, one has

(3.20) $\\#\mathcal{S}(\overline{b}_{t}\Lambda,\mathbf{r},s)\leq
s\operatorname{vol}(M_{\mathbf{r}}).$

###### Proof.

Note that $r_{m}=r_{1}<r_{M}=r_{\ell+1}$. Take
$t_{0}=t_{0}(w_{1},\dots,w_{d})>0$ such that for any $t>t_{0}$ we have

(3.21) $e^{\frac{w_{d}}{20}t}\geq(\xi+\frac{w_{d}}{2})t.$

Denoting by $c_{4}=e^{-\delta t_{0}}$, then $c_{4}\in(0,1)$ depends only on
the weights $w_{1},\dots,w_{d}$, and the inequality (3.21) holds whenever
$e^{-\delta t}<c_{4}$. Let

${\tilde{s}=\min\left(\frac{1}{C},c_{4},\frac{1}{\sqrt{d+1}},\left(\frac{\operatorname{vol}(B_{1})}{4^{d+1}}\right)^{1/d}\right)\leq
1.}$

By Lemma 3.9 (2), it suffices to show that for $e^{-\delta
t}<\epsilon<s<\tilde{s}$,

$q_{1}(\overline{b}_{t}\Lambda)\geq s^{-2}\quad\mathrm{and}\quad
q_{d+1}(\overline{b}_{t}\Lambda)\log q_{d+1}(\overline{b}_{t}\Lambda)\leq
s\epsilon e^{(\xi+w_{\ell+1})t}.$

Since $e^{-\delta t}<\epsilon<s$, it follows from
$s^{-3}\epsilon^{-1}<\epsilon^{-4}<e^{4\delta t}$ that $s^{-2}/r_{i}<s$ for
all $i=1,\dots,d$, hence

$\begin{split}\overline{b}_{t}^{*}N_{s^{-2}}&=\overline{b}_{t}^{*}M^{*}_{(\frac{s^{-2}}{r_{1}},\dots,\frac{s^{-2}}{r_{d}},s^{-2})}\\\
&=M^{*}_{\left(e^{\xi(w_{1}-1)t}\epsilon^{-1}s^{-2},\dots,e^{\xi(w_{d}-1)t}\epsilon^{-1}s^{-2},e^{-\xi
t}s^{-2}\right)}.\end{split}$

Since $\tilde{s}\leq\frac{1}{\sqrt{d+1}}$, we have for all $i=1,\dots,d$,

$\frac{s^{2}\epsilon^{3}}{\sqrt{d+1}}>\epsilon^{6}>e^{-6\delta t}>e^{-(\xi
w_{1}-\frac{1}{\ell}(w_{\ell+1}+\cdots+w_{d}))t}\geq e^{\xi(w_{i}-1)t},$

and

$\frac{s^{2}\epsilon^{2}}{\sqrt{d+1}}>\epsilon^{5}>e^{-5\delta t}>e^{-\xi
w_{d}t}>e^{-\xi t},$

hence it follows that $\overline{b}_{t}^{*}N_{s^{-2}}$ is contained in the
interior of $B_{\epsilon^{2}}(\mathcal{E}_{d+1}^{*})$. Since
$\Lambda\in\mathcal{K}_{\epsilon^{2}}^{*}$, there is no lattice point of
$\Lambda$ in $\overline{b}_{t}^{*}N_{s^{-2}}$, which concludes
$q_{1}(\overline{b}_{t}\Lambda)\geq s^{-2}$ as in the proof of the first
inequality of (3.17).

Since $\xi=\max(1,\frac{d-\ell}{\ell})<d$, we have

(3.22) ${s\epsilon>\epsilon^{2}>e^{-\frac{1}{9d^{2}}\xi
w_{d}t}>e^{-\frac{1}{9d}w_{d}t},}$

which implies that

$e^{\frac{w_{d}}{2}t}=e^{-\frac{w_{d}}{2}t}e^{w_{d}t}<e^{-\frac{1}{9d}w_{d}t}e^{w_{d}t}<s\epsilon
e^{w_{d}t},$

hence $e^{(\xi+\frac{w_{d}}{2})t}<r_{d}s$. On the other hand, it is clear that
$r_{\ell}s<e^{(\xi+\frac{w_{d}}{2})t}$, hence
$\overline{b}_{t}^{*}N_{e^{(\xi+w_{d}/2)t}}$ is the set of
$\varphi=x_{1}^{\varphi}\mathbf{e}_{1}^{*}+\cdots+x_{d+1}^{\varphi}\mathbf{e}_{d+1}^{*}\in\mathcal{E}_{d+1}^{*}$
such that

$\begin{cases}|x_{i}^{\varphi}|\leq se^{\left(\xi
w_{i}-\frac{1}{\ell}(w_{\ell+1}+\cdots+w_{d})\right)t}\qquad&\text{for }1\leq
i\leq\ell,\\\ |x_{i}^{\varphi}|\leq\epsilon^{-1}e^{(\xi
w_{i}+\frac{w_{d}}{2})t}\qquad&\text{for }\ell+1\leq i\leq d,\\\
|x_{i}^{\varphi}|\leq e^{\frac{1}{2}w_{d}t}\qquad&\text{for
}i=d+1.\end{cases}$

It follows from $\Lambda\in\mathcal{K}_{\epsilon^{2}}^{*}$ that
$\lambda_{1}(B_{1},\Lambda^{*})\geq\epsilon^{2}$. By Minkowski’s second
theorem, we have

$\epsilon^{2d}\lambda_{d+1}(B_{1},\Lambda^{*})\leq\lambda_{1}(B_{1},\Lambda^{*})\cdots\lambda_{d+1}(B_{1},\Lambda^{*})\leq\frac{2^{d+1}}{\operatorname{vol}(B_{1})},$

hence
$\lambda_{d+1}(B_{1},\Lambda^{*})\leq\frac{2^{d+1}}{\operatorname{vol}(B_{1})}\epsilon^{-2d}$.
Thus there exists a Minkowski reduced basis $\varphi_{1},\dots,\varphi_{d+1}$
of $\Lambda^{*}$ such that
$\|\varphi_{i}\|\leq\frac{4^{d+1}}{\operatorname{vol}(B_{1})}\epsilon^{-2d}\leq\epsilon^{-3d}$
for all $i=1,\dots d+1$ since
$\epsilon^{d}<\tilde{s}^{d}\leq\frac{\operatorname{vol}(B_{1})}{4^{d+1}}$.
Recall that $w_{1}=\cdots=w_{\ell}$, hence it can be easily checked that
$\varphi_{i}$’s are contained in $\overline{b}_{t}^{*}N_{e^{(\xi+w_{d}/2)t}}$.
Thus $q_{d+1}(\overline{b}_{t}\Lambda)\leq e^{(\xi+w_{d}/2)t}$ so that

$\displaystyle q_{d+1}(\overline{b}_{t}\Lambda)\log
q_{d+1}(\overline{b}_{t}\Lambda)$ $\displaystyle\leq
e^{(\xi+\frac{w_{d}}{2})t}(\xi+\frac{w_{d}}{2})t$ $\displaystyle\leq
e^{(\xi+\frac{w_{d}}{2})t}e^{\frac{w_{d}t}{20}}\quad$ by (3.21)
$\displaystyle\leq s\epsilon e^{(\xi+w_{d})t}\quad$ by (3.22)
$\displaystyle\leq s\epsilon e^{(\xi+w_{\ell+1})t}.$

∎

##  4\. Lower bound

###  4.1. Construction of the fractal set

For a given weight vector $w=(w_{1},\dots,w_{d})$, recall that $1\leq\ell\leq
d-1$ is the unique integer such that
$w_{1}=\cdots=w_{\ell}>w_{\ell+1}\geq\cdots\geq w_{d}$, and
$\xi=\max(1,\frac{d-\ell}{\ell})$ (see §3.2). Let
$\tilde{c},\tilde{\epsilon},\tilde{s}\leq 1$ be positive real numbers as in
Lemmas 3.8, 3.10, and 3.11, respectively. We fix the constants
$\epsilon,t,r>0$ with the following properties:

  1. (1)

$0<\epsilon<r<\frac{1}{10^{4}4^{d}}\min\\{\tilde{\epsilon},\tilde{s}\\}$;

  2. (2)

$t\geq 1$ will be chosen large enough so that (4.3), (4.8), (4.10), (4.11),
(4.13), (4.14), (4.15), (4.16) hold.

Let $\\{\epsilon_{n}\\}$ and $\\{t_{n}\\}$ be the sequence defined as follows:
for $n\in\mathbb{N}$,

  1. (1)

$\epsilon_{n}=\epsilon/n$;

  2. (2)

$t_{n}-t_{n-1}=\xi nt$ and $t_{0}=1$.

We will construct the tree $\mathcal{T}$ whose vertices are in the set
$\mathbb{Q}^{d}$ of rational vectors and the map $\beta$ from $V\mathcal{T}$
to the set of compact subsets in $\mathbb{R}^{d+1}$, inductively. We first set
the root of $\mathcal{T}$ to be zero, that is, $\tau_{0}=\mathbf{0}$ and
define

${\beta(\tau_{0})=\\{x\in\mathbb{R}^{d}:|(\tau_{0})_{i}-x_{i}|<e^{-w_{i}t_{1}},\forall
i=1,\dots,d\\}.}$

For each $\tau\in\mathcal{T}_{n}$ with $n\geq 1$, let

$\tilde{\beta}(\tau)=\\{x\in\mathbb{R}^{d}:|\tau_{i}-x_{i}|<\epsilon_{n+1}e^{-w_{i}t_{n+1}-t_{n}},\forall
i=1,\dots,d\\}.$

Recall that
$a_{t}=\operatorname{diag}\left(e^{w_{1}t},\dots,e^{w_{d}t},e^{-t}\right)$ and
$h(x)=\left(\begin{matrix}I_{d}&x\\\ 0&1\end{matrix}\right)$ for
$x\in\mathbb{R}^{d}$. Denote by

$b_{n}=\text{diag}(e^{-\frac{1}{\ell}(w_{\ell+1}+\cdots+w_{d})nt},\dots,e^{-\frac{1}{\ell}(w_{\ell+1}+\cdots+w_{d})nt},e^{w_{\ell+1}nt},\dots,e^{w_{d}nt},1).$

Note that the first $\ell$ terms of $b_{n}$ are the same.

For each $\kappa\in\mathcal{T}_{n-1}$, we define $\mathcal{T}(\kappa)$ as the
set of all $\tau\in\tilde{\beta}(\kappa)$ with the following properties:

(4.1)
${\begin{split}a&{}_{t_{n}}h(\tau)\mathbb{Z}^{d+1}\in\mathcal{L}_{d+1}^{\prime},\\\
a&{}_{t_{n}}h(\tau)\mathbb{Z}^{d+1}\in\mathcal{K}_{\epsilon_{n}^{2}}^{*},\\\
b&{}_{n}a_{t_{n}}h(\tau)\mathbb{Z}^{d+1}\in\mathcal{K}_{r}^{*}.\end{split}}$

It follows from the definitions of $\mathcal{T}(\kappa)$ and
$\mathcal{L}_{d+1}^{\prime}$ that $\tau\in\mathbb{Q}^{d}$ and for any
$\tau\in\mathcal{T}(\kappa)$ there exists the unique vector

(4.2) ${\mathbf{v}(\tau)\in\\{r\mathbf{e}_{d+1}:1/2<r\leq 1\\}\cap
a_{t_{n}}h(\tau)\mathbb{Z}^{d+1}.}$

Note that $(d+1)$-th coordinate of $\mathbf{v}(\tau)$ is $qe^{-t_{n}}$ for
some $q\in\mathbb{Z}$ such that $1/2<qe^{-t_{n}}\leq 1$. Since $t_{n}\geq
t_{n-1}+1$, $\mathcal{T}_{n}$ has empty intersection with $\bigcup_{0\leq
i\leq n-1}\mathcal{T}_{i}$, which implies that $\mathcal{T}$ is a rooted tree.

For each $\tau\in\mathcal{T}(\kappa)$ with $\kappa\in\mathcal{T}_{n-1}$,
define

${\beta(\tau)=\\{x\in\mathbb{R}^{d}:|\tau_{i}-x_{i}|<\epsilon_{n}e^{-w_{i}t_{n+1}-t_{n}},\forall
i=1,\dots,d\\}.}$

Note that for each $\tau\in\mathcal{T}(\kappa)$, it follows from the
definitions of $\tilde{\beta}$ and $\beta$ that
$\beta(\tau)\subset\beta(\kappa)$. If follows from Lemma 4.1 below that each
vertex of $\mathcal{T}$ has sons by choosing $t\geq 1$ large enough so that
for any $n\in\mathbb{N}$

(4.3) ${\frac{1}{100}\epsilon_{n}^{d}e^{\xi dnt}\geq 1.}$

Hence the pair $(\mathcal{T},\beta)$ is a regular self-affine structure.

######  Lemma 4.1.

For every $n\in\mathbb{N}$ and $y\in\mathcal{T}_{n-1}$ one has

${\frac{1}{100}\epsilon_{n}^{d}e^{\xi dnt}\leq\\#\mathcal{T}(y)\leq
2^{d+1}\epsilon_{n}^{d}e^{\xi dnt}.}$

For fixed $n\in\mathbb{N}$ and $y\in\mathcal{T}_{n-1}$, we let

$\displaystyle\Lambda=a_{t_{n-1}}h(y)\mathbb{Z}^{d+1}\in\mathcal{L}^{\prime}_{d+1}\cap\mathcal{K}_{\epsilon_{n-1}^{2}}^{*},$
$\displaystyle\Lambda_{1}=a_{t_{n}}h(y)\mathbb{Z}^{d+1}=a_{\xi nt}\Lambda,$
$\displaystyle\Lambda_{2}=b_{n}a_{t_{n}}h(y)\mathbb{Z}^{d+1}=b_{n}a_{\xi
nt}\Lambda,$

and for $x\in\tilde{\beta}(y)$,

$\displaystyle\Lambda_{1}(x)=a_{t_{n}}h(x)\mathbb{Z}^{d+1}=a_{t_{n}}h(x-y)a_{t_{n}}^{-1}\Lambda_{1},$
$\displaystyle\Lambda_{2}(x)=b_{n}a_{t_{n}}h(x)\mathbb{Z}^{d+1}=b_{n}a_{t_{n}}h(x-y)a_{t_{n}}^{-1}b_{n}^{-1}\Lambda_{2}.$

The lattices $\Lambda_{1}(x)$ and $\Lambda_{2}(x)$ satisfy
$\Lambda_{1}(x)\in\mathcal{L}_{d+1}^{\prime}\cap\mathcal{K}_{\epsilon_{n}^{2}}^{*}$
and $\Lambda_{2}(x)\in\mathcal{K}_{r}^{*}$ if and only if
$x\in\mathcal{T}(y)$. Hence Lemma 4.1 follows from the following lemma.

######  Lemma 4.2.

Let $n\in\mathbb{N}$ and $y\in\mathcal{T}_{n-1}$. Then

(4.4) $\displaystyle\frac{1}{10}\epsilon_{n}^{d}e^{\xi
dnt}\leq\\#\\{x\in\tilde{\beta}(y):\Lambda_{1}(x)\in\mathcal{L}_{d+1}^{\prime}\\}\leq
2^{d+1}\epsilon_{n}^{d}e^{\xi dnt},$ (4.5)
$\displaystyle\\#\\{x\in\tilde{\beta}(y):\Lambda_{1}(x)\in\mathcal{L}_{d+1}^{\prime}\smallsetminus\mathcal{K}_{\epsilon_{n}^{2}}^{*}\\}\leq\frac{8}{100}\epsilon_{n}^{d}e^{\xi
dnt},$ (4.6)
$\displaystyle\\#\\{x\in\tilde{\beta}(y):\Lambda_{2}(x)\in\mathcal{L}_{d+1}^{\prime}\smallsetminus\mathcal{K}_{r}^{*}\\}\leq\frac{1}{100}\epsilon_{n}^{d}e^{\xi
dnt}.$

###### Proof.

Let $x\in\tilde{\beta}(y)$ with $\Lambda_{1}(x)\in\mathcal{L}_{d+1}^{\prime}$.
Then there exists $s_{x}$ such that $1/2<s_{x}\leq 1$ and
$\Lambda_{1}(x)\cap\mathbb{R}\mathbf{e}_{d+1}=\\{s_{x}\mathbf{e}_{d+1}\\}$. We
denote $s_{x}\mathbf{e}_{d+1}$ by $\mathbf{v}(x)$.

First, we prove (4.4). It can be checked by a direct calculation that the map
$x\mapsto a_{t_{n}}h(y-x)a_{t_{n}}^{-1}\mathbf{v}(x)$ is a bijection from
$\\{x\in\tilde{\beta}(y):\Lambda_{1}(x)\in\mathcal{L}_{d+1}^{\prime}\\}$ to
$M\cap\widehat{\Lambda}_{1}$ where

$M=\\{(z_{1},\dots,z_{d+1}):\max_{1\leq i\leq d}|z_{i}|\leq\epsilon_{n}e^{\xi
nt}|z_{d+1}|,\ 1/2<|z_{d+1}|\leq 1\\}.$

Thus it suffices to estimate $\\#(M\cap\widehat{\Lambda}_{1})$. Let

$\displaystyle M^{(1)}$ $\displaystyle=\\{(z_{1},\dots,z_{d+1}):\max_{1\leq
i\leq d}|z_{i}|\leq\frac{1}{2}\epsilon_{n}e^{\xi nt},\ |z_{d+1}|\leq 1\\}$
$\displaystyle M^{(2)}$ $\displaystyle=\\{(z_{1},\dots,z_{d+1}):\max_{1\leq
i\leq d}|z_{i}|\leq\frac{1}{2}\epsilon_{n}e^{\xi nt},\
|z_{d+1}|\leq\frac{1}{2}\\}.$

Since $M^{(1)}\smallsetminus M^{(2)}\subset M\subset 2M^{(2)}$, we have

(4.7)
$\\#(M^{(1)}\cap\widehat{\Lambda}_{1})-\\#(M^{(2)}\cap\widehat{\Lambda}_{1})\leq\\#(M\cap\widehat{\Lambda}_{1})\leq\\#(2M^{(2)}\cap\widehat{\Lambda}_{1}).$

We will use Lemma 3.8 to estimate $\\#(M^{(i)}\cap\widehat{\Lambda}_{1})$ for
$i=1,2$. Since
$\Lambda\in\mathcal{K}_{\epsilon_{n-1}^{2}}^{*}\subset\mathcal{K}_{\epsilon_{n}^{2}}^{*}$,
it follows from the natural identification
$\mathcal{E}_{d}^{**}=\mathbb{R}^{d}$ and Minkowski second theorem that there
exist contants $C_{1},C_{2}>0$ depending only on $d$ such that

$\lambda_{1}(B_{1},\Lambda)\geq C_{1}\epsilon_{n}^{2d}\text{ and
}\lambda_{d+1}(B_{1},\Lambda)\leq C_{2}\epsilon_{n}^{-2}.$

Since $\Lambda=a_{\xi nt}^{-1}\Lambda_{1}$, for $i=1,2$, we have

$\begin{split}\lambda_{1}(M^{(i)},\Lambda_{1})&=\lambda_{1}(a_{\xi
nt}^{-1}M^{(i)},\Lambda)\geq\lambda_{1}(a_{\xi nt}^{-1}M^{(1)},\Lambda)\\\
&\geq\lambda_{1}(B_{(d+1)e^{\xi nt}},\Lambda)\geq\frac{C_{1}}{d+1}e^{-\xi
nt}\epsilon_{n}^{2d}\end{split}$

and

$\begin{split}\lambda_{d+1}(M^{(i)},\Lambda_{1})&=\lambda_{d+1}(a_{\xi
nt}^{-1}M^{(i)},\Lambda)\leq\lambda_{d+1}(a_{\xi nt}^{-1}M^{(2)},\Lambda)\\\
&\leq\lambda_{d+1}(B_{\frac{1}{2}\epsilon_{n}e^{(1-w_{1})\xi nt}},\Lambda)\leq
2C_{2}e^{(w_{1}-1)\xi nt}\epsilon_{n}^{-3}.\end{split}$

Thus we can choose $t\geq 1$ large enough so that for all $n\in\mathbb{N}$

(4.8)
${\lambda_{d+1}(M^{(i)},\Lambda_{1})<\widetilde{c}\quad\text{and}\quad-\lambda_{d+1}(M^{(i)},\Lambda_{1})\log\lambda_{1}(M^{(i)},\Lambda_{1})<\widetilde{c}.}$

Using Lemma 3.8 and (4.7), we have

$\left(\frac{2}{c_{0}}-c_{0}\right)\frac{1}{\zeta(d+1)}\epsilon_{n}^{d}e^{\xi
dnt}\leq\\#(M\cap\widehat{\Lambda}_{1})\leq\frac{c_{0}}{\zeta(d+1)}2^{d+1}\epsilon_{n}^{d}e^{\xi
dnt}.$

By the fact that $1<\zeta(d+1)<2$ and choosing $c_{0}$ sufficiently close to
1, we complete the proof of (4.4).

Next, we prove (4.5) and (4.6). Let $s_{1}=\epsilon_{n}^{2}$, $s_{2}=r$,
$a^{(1)}=a_{\xi nt}$, $a^{(2)}=b_{n}a_{\xi nt}$, and

$\mathcal{S}_{j}=\\{x\in\tilde{\beta}(y):\Lambda_{i}(x)\in\mathcal{L}_{d+1}^{\prime}\smallsetminus\mathcal{K}_{s_{j}}^{*}\\}\text{
for }j=1,2.$

Recall that

${\mathcal{S}(\Lambda,\mathbf{r},s)=\left\\{\mathbf{v}\in
M_{\mathbf{r}}\cap\widehat{\Lambda}:\varphi(\mathbf{v})=0\text{ for some
}\varphi\in N_{(d+1)sr_{M}}(\mathbf{r},s)\cap\widehat{\Lambda}^{*}\right\\}.}$

We will show that

(4.9)
$\\#\mathcal{S}_{j}\leq\\#\mathcal{S}(\Lambda_{j},\mathbf{r}_{j},s_{j})\quad(j=1,2)$

for some $\mathbf{r}_{j}$ and apply Lemma 3.10 and 3.11.

Let $z^{(1)}$ and $z^{(2)}$ be vectors in $\mathbb{R}^{d}$ such that

$\displaystyle z^{(1)}_{i}$
$\displaystyle=(y_{i}-x_{i})e^{(w_{i}+1)t_{n}}\quad\text{for }1\leq i\leq d;$
$\displaystyle z^{(2)}_{i}$
$\displaystyle=\begin{cases}(y_{i}-x_{i})e^{-\frac{1}{\ell}(w_{\ell+1}+\cdots+w_{d})nt+(w_{i}+1)t_{n}}\quad&\text{if
}1\leq i\leq\ell,\\\ (y_{i}-x_{i})e^{w_{i}nt+(w_{i}+1)t_{n}}\quad&\text{if
}\ell+1\leq i\leq d.\end{cases}$

Then $h(z^{(j)})=a^{(j)}a_{t_{n-1}}h(y-x)(a^{(j)}a_{t_{n-1}})^{-1}$ for
$j=1,2$ and

$\displaystyle|z^{(1)}_{i}|$ $\displaystyle\leq\epsilon_{n}e^{\xi
nt}=:r^{(1)}_{i}\quad\text{for }1\leq i\leq d;$ $\displaystyle|z^{(2)}_{i}|$
$\displaystyle\leq\begin{cases}\epsilon_{n}e^{\left(\xi-\frac{1}{\ell}(w_{\ell+1}+\cdots+w_{d})\right)nt}=:r^{(2)}_{i}\quad&\text{if
}1\leq i\leq\ell,\\\ \epsilon_{n}e^{(\xi+w_{i})nt}=:r^{(2)}_{i}\quad&\text{if
}\ell+1\leq i\leq d.\end{cases}$

Since $\mathbf{v}(x)\in\Lambda_{1}(x)\cap\Lambda_{2}(x)$, for $j=1,2$,

$\mathbf{w}_{j}(x):=h(z^{(j)})\mathbf{v}(x)\in\Lambda_{j}.$

For $\mathbf{r}_{j}=(r^{(j)}_{1},\dots,r^{(j)}_{d},1)$, the map
$\mathcal{S}_{j}\to M_{\mathbf{r}_{j}}\cap\widehat{\Lambda}_{j}$ given by
$x\mapsto\mathbf{w}_{j}(x)$ is injective. Hence, in order to show (4.9), we
should find $\varphi_{j}\in
N_{(d+1)s_{j}r^{(j)}_{M}}(\mathbf{r}_{j},s_{j})\cap\widehat{\Lambda}^{*}_{i}$
such that $\varphi_{j}(\mathbf{w}_{j}(x))=0$. It follows from the definition
of $\mathcal{S}_{j}$ that for $x\in\mathcal{S}_{j}$,
$a^{(j)}a_{t_{n-1}}h(x)\mathbb{Z}^{d+1}\notin\mathcal{K}_{s_{j}}^{*}$. Then
there exists $\varphi_{j}\in\widehat{\Lambda}^{*}_{j}$ such that
$\|h(z^{(j)})^{*}\varphi_{j}\|<s_{j}$, where $h(z^{(j)})^{*}$ is the adjoint
action defined by $g^{*}\varphi(\mathbf{v})=\varphi(g\mathbf{v})$ for all
$g\in\operatorname{SL}_{d+1}(\mathbb{R})$, $\varphi\in\mathcal{E}_{d+1}^{*}$,
and $\mathbf{v}\in\mathbb{R}^{d+1}$. It follows from direct calculation that

$h(z^{(j)})^{*}\varphi_{j}=\left(\varphi_{j}(\mathbf{e}_{1}),\dots,\varphi_{j}(\mathbf{e}_{d}),\sum_{i=1}^{d}z_{i}^{(j)}\varphi_{j}(\mathbf{e}_{i})+\varphi_{j}(\mathbf{e}_{d+1})\right).$

By choosing $t\geq 1$ large enough so that for all $n\in\mathbb{N}$

(4.10) ${\epsilon e^{\xi nt}\geq 1,}$

it follows from $\|h(z^{(j)})^{*}\varphi_{j}\|<s_{j}$ that

$\displaystyle|\varphi_{j}(\mathbf{e}_{i})|<s_{j}\quad\text{for }1\leq i\leq
d;$
$\displaystyle|\varphi_{j}(\mathbf{e}_{d+1})|<s_{j}+ds_{j}r^{(j)}_{M}<(d+1)s_{j}r^{(j)}_{M}.$

Hence we have $\varphi_{j}\in N_{(d+1)s_{j}r^{(j)}_{M}}$. It follows that

$\displaystyle|\varphi_{j}(\mathbf{w}_{j}(x))|$
$\displaystyle=|h(z^{(j)})^{*}\varphi_{j}(h(-z^{(j)})\mathbf{w}_{j}(x))|=|h(z^{(j)})^{*}\varphi_{j}(\mathbf{v}(x))|$
$\displaystyle\leq|h(z^{(j)})^{*}\varphi_{j}(\mathbf{e}_{d+1})|\leq\|h(z^{(j)})^{*}\varphi_{j}\|<s_{j}<1.$

Since $\varphi_{j}(\mathbf{w}_{j}(x))\in\mathbb{Z}$, it follows that
$\varphi_{j}(\mathbf{w}_{j}(x))=0$. This proves (4.9).

We choose $t\geq 1$ large enough so that for all $n\in\mathbb{N}$

(4.11) ${ce^{-w_{d}\xi nt/(2d^{3})}<\epsilon_{n}\quad\text{and}\quad
e^{-\delta nt}<\epsilon_{n}.}$

Since
$\Lambda\in\mathcal{K}_{\epsilon_{n-1}^{2}}^{*}\subset\mathcal{K}_{\epsilon_{n}^{2}}^{*}$
and (4.11), it follows from Lemma 3.10 and 3.11 that

$\displaystyle\mathcal{S}_{1}\leq\\#\mathcal{S}(\Lambda_{1},\mathbf{r}_{1},s_{1})\leq\sqrt{\epsilon_{n}}\operatorname{vol}(M_{\mathbf{r}_{1}})=2^{d+1}\sqrt{\epsilon_{n}}\epsilon_{n}^{d}e^{\xi
dnt},$
$\displaystyle\mathcal{S}_{2}\leq\\#\mathcal{S}(\Lambda_{2},\mathbf{r}_{2},s_{2})\leq
r\operatorname{vol}(M_{\mathbf{r}_{2}})=2^{d+1}r\epsilon_{n}^{d}e^{\xi dnt}.$

This complete the proof. ∎

The following lemma is $d$-dimensional version of [LSST20, Lemma 4.1].

######  Lemma 4.3.

$\mathcal{F}(\mathcal{T},\beta)\subset\mathrm{Sing}(w)$.

###### Proof.

This lemma directly follows from the same argument in the proof of [LSST20,
Lemma 4.1]. ∎

###  4.2. The lower bound calculation

In this subsection we complete the proof of main results.

######  Proposition 4.4.

Let $w=(w_{1},\dots,w_{d})\in\mathbb{R}_{>0}^{d}$ where
$w_{1}=\cdots=w_{\ell}>w_{\ell+1}\geq\cdots\geq w_{d}>0$ and
$\sum_{i=1}^{d}w_{i}=1$ and let $(\mathcal{T},\beta)$ be the self-affine
strunction on $\mathbb{R}^{d}$ in the previous section. Then

${\dim_{H}\mathcal{F}(\mathcal{T},\beta)\geq d-\frac{1}{1+w_{1}}.}$

We will prove Proposition 4.4 using Corollary 2.2. Let
$C_{n},L_{n}^{(1)},\dots,L_{n}^{(d)}$ be the positive constants defined as
follows:

${C_{n}=\epsilon_{n}^{d}e^{\xi dnt},\quad
L_{n}^{(i)}=2\epsilon_{n}e^{-w_{i}t_{n+1}-t_{n}},\forall i=1,\dots,d.}$

It can be easily checked that a regular self-affine structure
$(\mathcal{T},\beta)$ satisfies assumptions (1), (2), and (3) of Theorem 2.1.
For the assumption (4) of Theorem 2.1, we need the following lemma.

######  Lemma 4.5.

Let $n\in\mathbb{N}$ be large and $\tau\in\mathcal{T}_{n-1}$. Then

${\mathrm{dist}(\beta(x),\beta(y))\geq
L_{n-1}^{(1)}\frac{c^{\prime}r^{d-1}}{4\sqrt{d}\epsilon_{n-1}}e^{\frac{1}{\ell}(w_{\ell+1}+\cdots+w_{d}-\xi\ell)nt},}$

where $x,y\in\mathcal{T}(\tau)$ are distinct and $c^{\prime}$ is the positive
constant in Lemma 3.7.

###### Proof.

By the construction of $\mathcal{T}$ and the definition of $b_{n}$, there are
$1/2\leq s_{x},s_{y}\leq 1$ such that

${s_{x}\mathbf{e}_{d+1}\in b_{n}a_{t_{n}}h(x)\mathbb{Z}^{d+1},\quad
s_{y}\mathbf{e}_{d+1}\in b_{n}a_{t_{n}}h(y)\mathbb{Z}^{d+1}.}$

Let us denote by

${\begin{split}\mathbf{v}&=b_{n}a_{t_{n}}h(y-x)(b_{n}a_{t_{n}})^{-1}s_{x}\mathbf{e}_{d+1}\in
b_{n}a_{t_{n}}h(y)\mathbb{Z}^{d+1},\\\ \mathbf{v}&\wedge
s_{y}\mathbf{e}_{d+1}=s_{x}s_{y}\sum_{i=1}^{d}u_{i}\mathbf{e}_{i}\wedge\mathbf{e}_{d+1}.\end{split}}$

Observe that

$u_{i}=\begin{cases}(y_{i}-x_{i})e^{(w_{i}+1)t_{n}-\frac{1}{\ell}(w_{\ell+1}+\cdots+w_{d})nt}\quad&\text{for
}1\leq i\leq\ell,\\\ (y_{i}-x_{i})e^{(w_{i}+1)t_{n}+w_{i}nt}\quad&\text{for
}\ell+1\leq i\leq d.\end{cases}$

Since $x$ and $y$ are distinct, the vectors $\mathbf{v}$ and
$e_{y}\mathbf{e}_{d+1}$ are linearly independent, hence it follows from Lemma
3.7 that

(4.12) ${\sqrt{d}\|\mathbf{u}\|_{\infty}\geq
s_{x}s_{y}\|\mathbf{u}\|=\|\mathbf{v}\wedge s_{y}\mathbf{e}_{d+1}\|\geq
c^{\prime}r^{d-1},}$

where $\mathbf{u}=(u_{1},\dots,u_{d})\in\mathbb{R}^{d}$ and
$\|\cdot\|_{\infty}$ denotes the max norm.

Let $x^{\prime}\in\beta(x)$ and $y^{\prime}\in\beta(y)$. Suppose that
$\|\mathbf{u}\|_{\infty}=|u_{i}|$ for some $1\leq i\leq\ell$. Then it follows
from (4.12) that

$\displaystyle\|y^{\prime}-x^{\prime}\|$
$\displaystyle\geq|y_{i}^{\prime}-x_{i}^{\prime}|\geq|y_{i}-x_{i}|-|x_{i}-x_{i}^{\prime}|-|y_{i}-y_{i}^{\prime}|$
$\displaystyle\geq
e^{-(w_{i}+1)t_{n}+\frac{1}{\ell}(w_{\ell+1}+\cdots+w_{d})nt}\left(\frac{c^{\prime}r^{d-1}}{\sqrt{d}}-2\epsilon_{n}e^{-\xi
w_{i}(n+1)t-\frac{1}{\ell}(w_{\ell+1}+\cdots+w_{d})nt}\right)$ (4.13)
$\displaystyle\geq
e^{-(w_{i}+1)t_{n}+\frac{1}{\ell}(w_{\ell+1}+\cdots+w_{d})nt}\frac{c^{\prime}r^{d-1}}{2\sqrt{d}}$
$\displaystyle\geq
L_{n-1}^{(i)}\frac{c^{\prime}r^{d-1}}{4\sqrt{d}\epsilon_{n-1}}e^{\frac{1}{\ell}(w_{\ell+1}+\cdots+w_{d}-\xi\ell)nt}$
$\displaystyle\geq
L_{n-1}^{(1)}\frac{c^{\prime}r^{d-1}}{4\sqrt{d}\epsilon_{n-1}}e^{\frac{1}{\ell}(w_{\ell+1}+\cdots+w_{d}-\xi\ell)nt}.$

We choose $t\geq 1$ large enough so that the third line (4.13) holds for all
$n\in\mathbb{N}$.

On the other hand, if $\|\mathbf{u}\|_{\infty}=|u_{i}|$ for some $\ell+1\leq
i\leq d$, then we have

$\displaystyle\|y^{\prime}-x^{\prime}\|$
$\displaystyle\geq|y_{i}^{\prime}-x_{i}^{\prime}|\geq|y_{i}-x_{i}|-|x_{i}-x_{i}^{\prime}|-|y_{i}-y_{i}^{\prime}|$
$\displaystyle\geq
e^{-(w_{i}+1)t_{n}-w_{i}nt}\left(\frac{c^{\prime}r^{d-1}}{\sqrt{d}}-2\epsilon_{n}e^{w_{i}nt-\xi
w_{i}(n+1)t}\right)$ (4.14) $\displaystyle\geq
e^{-(w_{i}+1)t_{n}-w_{i}nt}\frac{c^{\prime}r^{d-1}}{2\sqrt{d}}$
$\displaystyle\geq
L_{n-1}^{(1)}\frac{c^{\prime}r^{d-1}}{4\sqrt{d}\epsilon_{n-1}}e^{(w_{1}-w_{i})t_{n}-(\xi+w_{i})nt}$
$\displaystyle\geq
L_{n-1}^{(1)}\frac{c^{\prime}r^{d-1}}{4\sqrt{d}\epsilon_{n-1}}e^{(w_{1}-w_{\ell+1})t_{n}-(\xi+w_{i})nt}$
(4.15) $\displaystyle\geq
L_{n-1}^{(1)}\frac{c^{\prime}r^{d-1}}{4\sqrt{d}\epsilon_{n-1}}e^{\frac{1}{\ell}(w_{\ell+1}+\cdots+w_{d}-\xi\ell)nt}.$

We choose $t\geq 1$ large enough so that the third line (4.14) and last line
(4.15) hold for all $n\in\mathbb{N}$.

This concludes the proof of the lemma. ∎

We choose $t\geq 1$ large enough so that for all $n\in\mathbb{N}$

(4.16)
${\rho_{n}:=\frac{c^{\prime}r^{d-1}}{4\sqrt{d}\epsilon_{n-1}}e^{\frac{1}{\ell}(w_{\ell+1}+\cdots+w_{d}-\xi\ell)nt}\leq
1}$

since $w_{\ell+1}+\cdots+w_{d}<\xi\ell$. The assumption (4) of Theorem 2.1
follows from Lemma 4.5.

###### Proof of Proposition 4.4.

We prove the proposition applying Corollary 2.2. It can be easily checked that
for $k>4\xi dt$, the assumptions of Corollary 2.2 hold. Then we have

$\displaystyle\frac{\log(C_{n}L_{n}^{(\ell+1)}\cdots
L_{n}^{(d)}/L_{n-1}^{(\ell+1)}\cdots
L_{n-1}^{(d)})}{-\log(L_{n}^{(1)}/L_{n-1}^{(1)})}$ $\displaystyle=\frac{\xi
dnt-\xi(w_{\ell+1}+\cdots+w_{d})(n+1)t-\xi(d-\ell)nt+o(n)}{\xi w_{1}(n+1)t+\xi
nt+o(n)}$
$\displaystyle\to\frac{\ell-(w_{\ell+1}+\cdots+w_{d})}{1+w_{1}}=\ell-\frac{1}{1+w_{1}}\quad\text{
as }n\to\infty$

Hence Corollary 2.2 implies

${\dim_{H}\mathcal{F}(\mathcal{T},\beta)\geq(d-\ell)+\ell-\frac{1}{1+w_{1}}=d-\frac{1}{1+w_{1}}.}$

∎

###### Proof of Theorem 1.1.

If $w_{1}=\cdots=w_{d}$, then the result follows from [CC16, Theorem 1.1]. If
there exists $1\leq\ell\leq d-1$ such that
$w_{1}=\cdots=w_{\ell}>w_{\ell+1}\geq\cdots\geq w_{d}$, then the result
follows from Lemma 4.3 and Proposition 4.4. ∎

###### Proof of Theorem 1.2.

This theorem directly follows from the same argument in the proof of [LSST20,
Theorem 1.5]. ∎

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