††2020 Mathematics Subject Classification: Primary 11J13, 11K55; Secondary
37A17††The first named author is supported by the ERC grant ”Dynamics on
Homogeneous Spaces” (no. 754475).††The second named author is supported by the
ERC grant HomDyn, ID 833423.

# Hausdorff Dimension of Singular Vectors in Function Fields

Noy Soffer Aranov  Noy Soffer Aranov. Department of Mathematics, Technion,
Haifa, Israel<EMAIL_ADDRESS>and  Taehyeong Kim  Taehyeong Kim.
The Einstein Institute of Mathematics, Edmond J. Safra Campus, Givat Ram
The Hebrew University of Jerusalem, Jerusalem, 91904, Israel
<EMAIL_ADDRESS>
###### Abstract.

We compute the Hausdorff dimension of the set of singular vectors in function
fields and bound the Hausdorff dimension of the set of $\varepsilon$-Dirichlet
improvable vectors in this setting. This is a function field analogue of the
results of Cheung and Chevallier [Duke Math. J. 165 (2016), 2273–2329].

##  1\. Introduction

A fundamental theorem in Diophantine approximation is Dirichlet’s Theorem
which says that for all
$\boldsymbol{\theta}=(\theta_{1},\dots,\theta_{d})\in\mathbb{R}^{d}$ and $T>1$
there exists
$(\mathbf{a},b)=(a_{1},\dots,a_{d},b)\in\mathbb{Z}^{d}\times\mathbb{Z}$ such
that

$\max_{1\leq i\leq
d}|b\theta_{i}-a_{i}|<\frac{1}{T^{\frac{1}{d}}}\qquad\text{and}\qquad
0<|b|\leq T.$

It is natural to ask whether Dirichlet’s Theorem can be improved, and this
question leads to the following definition. Given $\varepsilon>0$, a vector
$\boldsymbol{\theta}\in\mathbb{R}^{d}$ is said to be $\varepsilon$-Dirichlet
improvable if there is $T_{0}>1$ such that for all $T>T_{0}$ there exists
$(\mathbf{a},b)\in\mathbb{Z}^{d}\times\mathbb{Z}$ such that

$\max_{1\leq i\leq
d}|b\theta_{i}-a_{i}|<\frac{\varepsilon}{T^{\frac{1}{d}}}\qquad\text{and}\qquad
0<|b|\leq T.$

We say that a vector $\boldsymbol{\theta}\in\mathbb{R}^{d}$ is singular if it
is $\varepsilon$-Dirichlet improvable for every $\varepsilon>0$. Denote by
$\mathbf{DI}_{d}(\varepsilon)$ and $\mathbf{Sing}_{d}$ the set of
$\varepsilon$-Dirichlet improvable vectors and singular vectors in
$\mathbb{R}^{d}$, respectively.

Khintchine [Khi26] originally introduced singular vectors and showed that
$\mathbf{Sing}_{d}$ has Lebesgue measure zero. Davenport and Schmidt [DS70]
proved that $\mathbf{DI}_{d}(\varepsilon)$ has Lebesgue measure zero for any
$0<\varepsilon<1$. It is worth noting that $\mathbf{Sing}_{1}=\mathbb{Q}$, but
if $d\geq 2$, the singular set is nontrivial. The computation of the Hausdorff
dimension of $\mathbf{Sing}_{d}$ with $d\geq 2$ has been a challenge until the
first breakthrough by Cheung [Che11], who proved that the Hausdorff dimension
of $\mathbf{Sing}_{2}$ is $4/3$ and obtained a certain bound for the Hausdorff
dimension of $\mathbf{DI}_{2}(\varepsilon)$. This was extended by Cheung and
Chevallier [CC16] to higher dimensions as follows:

######  Theorem 1.1.

[CC16] Let $d\geq 2$. Then

  1. (1)

the Hausdorff dimension of $\mathbf{Sing}_{d}$ is $\frac{d^{2}}{d+1}$;

  2. (2)

for any $t>d$, there is a constant $C>0$ such that for all small enough
$\varepsilon>0$,

$\frac{d^{2}}{d+1}+\varepsilon^{t}\leq\mathrm{dim}_{H}\mathbf{DI}_{d}(\varepsilon)\leq\frac{d^{2}}{d+1}+C\varepsilon^{\frac{d}{2}}.$

Here and hereafter, $\mathrm{dim}_{H}$ stands for the Hausdorff dimension. We
remark that this result improves the bounds in [Bak77, Bak92, Ryn90] and refer
to [KKLM17, DFSU24] for the Hausdorff dimension of singular matrices.

In this paper, we shall prove a function field analogue of Theorem 1.1. We
first introduce the function field setting. Let $p$ be a prime, $q$ be a power
of $p$, and let $\mathcal{R}=\mathbb{F}_{q}[x]$ be the ring of polynomials
over $\mathbb{F}_{q}$. Let $\mathcal{K}=\mathbb{F}_{q}(x)$ be the field of
rational functions over $\mathbb{F}_{q}$. We define an absolute value on
$\mathcal{R}$ by $|f|=q^{\mathrm{deg}(f)}$ and extend it to an absolute value
on $\mathcal{K}$ by
$\left|\frac{f}{g}\right|=q^{\mathrm{deg}(f)-\mathrm{deg}(g)}$. Then the
topological completion of $\mathcal{K}$ with respect to the metric
$d(f,g)=|f-g|$ is the field of Laurent series $\widetilde{\mathcal{K}}$
defined by

$\widetilde{\mathcal{K}}\>=\mathbb{F}_{q}\left(\left(x^{-1}\right)\right)=\Bigg{\\{}\sum_{n=-N}^{\infty}a_{n}x^{-n}:a_{n}\in\mathbb{F}_{q},N\in\mathbb{Z}\Bigg{\\}}.$

We equip $\widetilde{\mathcal{K}}^{d}$ with the supremum norm
$\|\mathbf{v}\|=\max_{i=1,\dots,d}|v_{i}|$ for
$\mathbf{v}=(v_{1},\dots,v_{d})\in\widetilde{\mathcal{K}}$ and consider the
Hausdorff dimension of subsets of $\widetilde{\mathcal{K}}^{d}$ with respect
to this supremum norm.

We remark that the ring of polynomials $\mathcal{R}$ is discrete in
$\widetilde{\mathcal{K}}$ and plays a similar role to the ring of integers
$\mathbb{Z}$ in $\mathbb{R}$. In this point of view, there have been numerous
developments on Diophantine approximation in the function field setting, see
for instance [dM70, Las00]. In particular, Dirichlet’s Theorem also holds (see
e.g. [dM70, Chapter IV]). Hence, it makes sense to consider the following
definition as in the real case:

Given $\varepsilon>0$, we say that
$\boldsymbol{\theta}\in\widetilde{\mathcal{K}}^{d}$ is $\varepsilon$-Dirichlet
improvable if there is $T_{0}>1$ such that for all $T>T_{0}$ there exists
$(\mathbf{a},b)\in\mathcal{R}^{d}\times\mathcal{R}$ such that

$\|b\boldsymbol{\theta}-\mathbf{a}\|<\frac{\varepsilon}{T^{\frac{1}{d}}}\qquad\text{and}\qquad
0<|b|\leq T.$

We say that a vector $\boldsymbol{\theta}\in\widetilde{\mathcal{K}}^{d}$ is
singular if it is $\varepsilon$-Dirichlet improvable for every
$\varepsilon>0$. Denote by $\mathbf{DI}_{d}(\varepsilon)$ and
$\mathbf{Sing}_{d}$ the set of $\varepsilon$-Dirichlet improvable vectors and
singular vectors in $\widetilde{\mathcal{K}}^{d}$, respectively.

Compared to intensive studies in the real setting as mentioned above, there
has been less attention to singular sets over function fields. In particular,
it was proved in [GG17, Theorem 3.6] that $\mathbf{DI}_{d}(\varepsilon)$ has
zero Haar measure on $\widetilde{\mathcal{K}}^{d}$ for every $\varepsilon$
small enough. Hence, $\mathbf{Sing}_{d}$ has also zero Haar measure. Recently,
it has been proved by Bang, the second named author, and Lim [BKL24] that the
Hausdorff dimension of $\mathbf{Sing}_{d}$ is at most $\frac{d^{2}}{d+1}$. We
prove that their upper bound is sharp, and moreover we bound the Hausdorff
dimension of $\mathbf{DI}_{d}(\varepsilon)$.

######  Theorem 1.2 (Corollaries 5.5+6.13+6.14).

Let $d\geq 2$. Then

  1. (1)

the Hausdorff dimension of $\mathbf{Sing}_{d}$ is $\frac{d^{2}}{d+1}$;

  2. (2)

for any $\varepsilon>0$, we have

$\begin{split}&\frac{d^{2}}{d+1}+\frac{d}{d+1}\log_{q}\left(1+\frac{(q-1)^{2}}{q^{2d+\frac{d^{2}}{d-1}}}\left(\frac{q-1}{q}-\varepsilon^{d-1}\right)\varepsilon^{d}\right)\\\
&\leq\mathrm{dim}_{H}\mathbf{DI}_{d}(\varepsilon)\leq\frac{d^{2}}{d+1}+\frac{d}{d+1}\log_{q}\left(1+\sqrt{(q-1)^{2}(q+1)^{d}q^{2d}\varepsilon^{d}}\right).\end{split}$

######  Remark 1.3.

We remark several things about the second argument in Theorem 1.2.

  1. (1)

Since $\log_{q}(1+y)=y/\log q+o(y)$ as $y\to 0$, Theorem 1.2 has the same
order of magnitude as in Theorem 1.1 for all small enough $\varepsilon$.

  2. (2)

Note that the lower bound is increasing with
$0<\varepsilon\leq\left(\frac{d(q-1)}{(2d-1)q}\right)^{\frac{1}{d-1}}$ and the
upper bound is nontrivial for $0<\varepsilon\leq(q+1)^{-1}q^{-2}$, otherwise
the upper bound is greater than $d$. Therefore, the nontrivial regions of
$\varepsilon$ are
$\left(0,\left(\frac{d(q-1)}{(2d-1)q}\right)^{\frac{1}{d-1}}\right]$ and
$(0,(q+1)^{-1}q^{-2}]$ for the lower bound and the upper bound, respectively.

  3. (3)

While Theorem 1.1 only works for $\varepsilon$ small enough, our result works
for relatively large $\varepsilon$. We are able to obtain this stronger result
by the ultrametric nature which will be explained at the end of the
introduction. In both real and function field cases, it is totally open to
determine/bound reasonably the Hausdorff dimension of
$\mathbf{DI}_{d}(\varepsilon)$ for any $\varepsilon>0$. See [KM23, KSY22] for
$\varepsilon<1$ sufficiently near $1$.

We will now discuss a dynamical interpretation of our main theorem. Let
$\operatorname{SL}_{d+1}(\widetilde{\mathcal{K}})$ be the group of invertible
$(d+1)\times(d+1)$ matrices over $\widetilde{\mathcal{K}}$ with $|\det(g)|=1$
and let $\operatorname{SL}_{d+1}(\mathcal{R})$ be the subgroup of
$\operatorname{SL}_{d+1}(\widetilde{\mathcal{K}})$ with entries in
$\mathcal{R}$. Then, the homogeneous space
$\mathcal{L}_{d+1}=\operatorname{SL}_{d+1}(\widetilde{\mathcal{K}})/\operatorname{SL}_{d+1}(\mathcal{R})$
is identified with the space of unimodular lattices in
$\widetilde{\mathcal{K}}^{d+1}$ under the identification

$g\operatorname{SL}_{d+1}(\mathcal{R})\mapsto g\mathcal{R}^{d+1}.$

There is a well-known dynamical interpretation of singular vectors. For
$\boldsymbol{\theta}\in\widetilde{\mathcal{K}}^{d}$, consider the lattice
$\Delta_{\boldsymbol{\theta}}=\begin{pmatrix}\mathbf{I}_{d}&\boldsymbol{\theta}\\\
0&1\end{pmatrix}\mathcal{R}^{d+1}$ and the diagonal flow
$g_{t}=\operatorname{diag}\\{x^{t},\dots,x^{t},x^{-dt}\\}$. Then,
$\boldsymbol{\theta}$ is singular if and only if the orbit
$\\{g_{t}\Delta_{\boldsymbol{\theta}}:t\in\mathbb{N}\\}$ is divergent, that is
eventually leaves every compact set in $\mathcal{L}_{d+1}$. See for instance
[BKL24, Lemma 2.1].

This relationship was originally studied by Dani [Dan85] in the real setting.
Divergent trajectories for diagonal flows have been well studied in the real
setting (see for example [KW05, Wei04, Wei06, Tam19, ST23, DFSU24, KKLM17]).

Theorem 1.2 implies the following dynamical result as in [CC16, Corollary
1.2].

######  Corollary 1.4.

For $d\geq 2$, the Hausdorff dimension of the set

$\\{\Delta\in\mathcal{L}_{d+1}:\text{ the orbit
}\\{g_{t}\Delta:t\in\mathbb{N}\\}\text{ is divergent}\\}$

is
$\mathrm{dim}_{H}\operatorname{SL}_{d+1}(\widetilde{\mathcal{K}})-\frac{d}{d+1}$.

###### Proof.

The same proof verbatim of [CC16, Corollary 1.2] works in our setting. See
also the proof of [BKL24, Proposition 6.5]. ∎

Finally, we discuss how we are able to obtain stronger results than Theorem
1.1. We basically follow the proof of [CC16], but the ultrametric nature
enables us to obtain better results. By the ultrametric nature, we mean the
following lemma.

######  Lemma 1.5 (Ultrametric inequality).

For every $\mathbf{v},\mathbf{w}\in\widetilde{\mathcal{K}}^{d}$, we have

$\|\mathbf{v}+\mathbf{w}\|\leq\max\\{\|\mathbf{v}\|,\|\mathbf{w}\|\\}.$ (1.1)

Moreover, the inequality in (1.1) is an equality if
$\|\mathbf{v}\|\neq\|\mathbf{w}\|$.

There are two main reasons for the difference between Theorems 1.1 and 1.2.
The first reason is about the characterization of $\varepsilon$-Dirichlet
improvability in terms of the length of the shortest vector in the Farey
lattice given in Definition 4.9 with respect to best approximations. This
corresponds to [CC16, Corollary 4.4]. In the real case, the characterization
works upto $1/2$ and $2$ multiplications, but the ultrametric nature allows us
to obtain the exact equivalence as in Corollary 4.7. The second reason comes
from Minkowski’s second theorem (Theorem 2.4). While Minkowski’s second
theorem on $\mathbb{R}^{d}$ works with $2^{d}/d!$ and $2^{d}$ multiplications,
we have the equality over $\widetilde{\mathcal{K}}^{d}$ as in Theorem 2.4.
These two reasons ultimately ensure that the fractal structures we will build
work well for all $0<\varepsilon<1$, unlike [CC16].

The article is organized as follows. In section 2, we recall some basic
estimates in number theory and geometry of numbers which will be used later.
In section 3, we recall the definition of self-similar structures and some
related dimension results given in [CC16]. In section 4, we characterize
$\varepsilon$-Dirichlet improvability in terms of best approximations, define
the Farey lattice, and connect it with best approximations. In section 5 and
section 6, we define the fractal structures using the Farey lattices and
compute the upper and lower bounds of the Hausdorff dimensions of
$\mathbf{DI}_{d}(\varepsilon)$ and $\mathbf{Sing}_{d}$, respectively.

##  2\. Preliminaries

###  2.1. Basic number theory

In this subsection, we gather some notation and facts about basic number
theory in function fields following [Ros13].

Recall that $\mathbb{F}_{q}$ is a finite field with $q$ elements for a prime
power $q$ and $\mathcal{R}=\mathbb{F}_{q}[x]$ is the ring of polynomials over
$\mathbb{F}_{q}$. A non-zero polynomial $f\in\mathcal{R}$ is called a monic
polynomial if the leading coefficient is equal to 1. Monic polynomials play
the role of positive integers. For $f,g\in\mathcal{R}$, the greatest common
divisor of $f$ and $g$, denoted $\mathrm{gcd}(f,g)$, is the monic polynomial
of largest degree which divides both $f$ and $g$. Define the Euler totient
function $\phi(f)$ of $f\in\mathcal{R}$ to be the number of non-zero
polynomials of degree less than $\mathrm{deg}(f)$ and relatively prime to $f$,
that is,

$\phi(f)=\\#\\{g\in\mathcal{R}\setminus\\{0\\}:\mathrm{deg}(g)<\mathrm{deg}(f)\
\text{ and }\mathrm{gcd}(f,g)=1\\}.$

Consider the divisor function $D_{1}(f)$ of $f\in\mathcal{R}$ defined by
$D_{1}(f)=\sum_{g|f}|g|$, where $\sum_{g|f}$ stands for the summation over all
monic divisors of $f$. We will need the following estimates:

######  Lemma 2.1.

[Ros13, Proposition 2.7] For every integer $\ell\geq 1$, we have

$\sum_{\mathrm{deg}(f)=\ell}\phi(f)=\frac{(q-1)^{2}}{q}q^{2\ell}.$

###### Proof.

It follows from [Ros13, Proposition 2.7] that

$\sum_{\begin{subarray}{c}\mathrm{deg}(f)=\ell\\\ f\
\text{monic}\end{subarray}}\phi(f)=\frac{q-1}{q}q^{2\ell}.$

Given monic polynomial $f\in\mathcal{R}$, if $g\in\mathbb{F}_{q}^{*}f$, then
$\phi(g)=\phi(f)$. Hence the lemma follows. ∎

######  Lemma 2.2.

For every integer $\ell\geq 1$, we have

$\sum_{\mathrm{deg}(f)=\ell}\phi(f)D_{1}(f)\leq(q-1)q^{3\ell}.$

###### Proof.

Given a non-zero polynomial $f\in\mathcal{R}$, let $f=\alpha
P_{1}^{e_{1}}P_{2}^{e_{2}}\cdots P_{t}^{e_{t}}$ be its prime decomposition,
where, $\alpha\in\mathbb{F}_{q}^{*}$, $P_{i}$ are distinct monic irreducible
polynomials, and $e_{i}$ are non-negative integers. It follows from [Ros13,
Propositions 1.7 and 2.4] that

$\phi(f)=\prod_{i=1}^{t}\left(|P_{i}|^{e_{i}}-|P_{i}|^{e_{i}-1}\right)\quad\text{and}\quad
D_{1}(f)=\prod_{i=1}^{t}\frac{|P_{i}|^{e_{i}+1}-1}{|P_{i}|-1}.$

Hence, we have

$\phi(f)D_{1}(f)=|f|\prod_{i=1}^{t}\frac{|P_{i}|^{e_{i}+1}-1}{|P_{i}|}=|f|^{2}\prod_{i=1}^{t}\left(1-\frac{1}{|P_{i}|^{e_{i}+1}}\right)\leq|f|^{2}.$

Therefore,

$\sum_{\mathrm{deg}(f)=\ell}\phi(f)D_{1}(f)\leq\sum_{\mathrm{deg}(f)=\ell}q^{2\ell}=(q-1)q^{3\ell}.$

∎

###  2.2. Geometry of numbers

An important concept in the geometry of numbers is successive minima.
Minkowski’s second theorem connects between the successive minima and the
determinant/covolume of a lattice. The function field version of this theorem
was proved by Mahler [Mah41]. In order to state it, we define the determinant
of a lattice.

######  Definition 2.3.

For a lattice $\Lambda=g\mathcal{R}^{d}$ in $\widetilde{\mathcal{K}}^{d}$ with
$g\in\mathrm{GL}_{d}(\widetilde{\mathcal{K}})$, we define its determinant by

$\det(\Lambda)=|\det(g)|.$

######  Theorem 2.4 (Minkowski’s second theorem).

For any lattice $\Lambda\subseteq\widetilde{\mathcal{K}}^{d}$ and $i=1,...d$,
let

$\lambda_{i}(\Lambda)=\min\\{r>0:\text{ there exist }i\text{ linearly
independent vectors in }\Lambda\text{ of norm }\leq r\\}.$

Then,

$\prod_{i=1}^{d}\lambda_{i}(\Lambda)=\det(\Lambda).$

Throughout this paper, $B(\mathbf{x},r)$ stands for the closed ball of radius
$r>0$ centered at $\mathbf{x}\in\widetilde{\mathcal{K}}^{d}$, that is,
$B(\mathbf{x},r)=\\{\mathbf{y}\in\widetilde{\mathcal{K}}^{d}:\|\mathbf{x}-\mathbf{y}\|\leq
r\\}$. The following lattice point counting result will be used for the upper
bound of $\mathrm{dim}_{H}\mathbf{DI}_{d}(\varepsilon)$.

######  Proposition 2.5.

[BK23, Lemma 6.2] Let $\Lambda\subseteq\widetilde{\mathcal{K}}^{d}$ be a
lattice and $B(0,r)\subseteq\widetilde{\mathcal{K}}^{d}$ be the closed ball
with radius $r>0$. Then

$\\#(\Lambda\cap
B(0,r))=\prod_{i=1}^{d}\left\lceil\frac{qr}{\lambda_{i}(\Lambda)}\right\rceil.$

We shall prove positive characteristic analogues of the results of [CC16,
Section 2]. Firstly, given $X\subseteq\widetilde{\mathcal{K}}^{d}$, we denote
the radius of the largest ball disjoint from $X$ by

$e(X)=\sup\\{r>0:\exists\alpha:B(\alpha,r)\cap X=\emptyset\\}$

Let $\mathcal{O}$ be the maximal compact subgroup of
$\widetilde{\mathcal{K}}$, that is

$\mathcal{O}=\mathbb{F}_{q}\left[\left[x^{-1}\right]\right]=\\{f\in\widetilde{\mathcal{K}}:|f|\leq
1\\}.$

Then, the group of isometries for the
$\operatorname{SL}_{d}(\widetilde{\mathcal{K}})$ action on $\mathcal{L}_{d}$
is $\operatorname{SL}_{d}(\mathcal{O})$.

######  Lemma 2.6.

The function $e$ is $\operatorname{SL}_{d}(\mathcal{O})$ invariant.

###### Proof.

Denote $r:=e(X)$ and let $u\in\operatorname{SL}_{d}(\mathcal{O})$. Assume
towards a contradiction that for every $\alpha\in\widetilde{\mathcal{K}}^{d}$,
$B(\alpha,r)\cap uX\neq\emptyset$. Let $\mathbf{v}\in B(\alpha,r)\cap uX$.
Then, $u^{-1}\mathbf{v}\in X$ and

$\|u^{-1}\alpha-u^{-1}\mathbf{v}\|=\|\alpha-\mathbf{v}\|\leq r.$

Hence, for every $\alpha\in\widetilde{\mathcal{K}}^{d}$,
$B(u^{-1}\alpha,r)\cap X\neq\emptyset$, which is a contradiction to the
definition of $r$. Thus, $e(uX)\geq e(X)$, and by replacing $u$ with $u^{-1}$
and $X$ with $uX$, we obtain that $e(X)=e(uX)$. ∎

######  Lemma 2.7.

For any lattice $\Lambda\subseteq\widetilde{\mathcal{K}}^{d}$,

$q^{2}e(\Lambda)=\lambda_{d}(\Lambda)$

###### Proof.

Since $e$ is $\operatorname{SL}_{d}(\mathcal{O})$ invariant, by [Ara23,
Theorem 1.15], it suffices to assume that
$\Lambda=\mathcal{R}x^{\log_{q}\lambda_{1}(\Lambda)}\mathbf{e}_{1}\oplus\dots\oplus\mathcal{R}x^{\log_{q}\lambda_{d}(\Lambda)}\mathbf{e}_{d}$.
Define
$\mathbf{v}=\sum_{i=1}^{d}x^{\log_{q}\lambda_{i}(\Lambda)-1}\mathbf{e}_{i}$.
We claim that

$B\left(\mathbf{v},\frac{1}{q^{2}}\lambda_{d}(\Lambda)\right)\cap\Lambda=\emptyset.$

Assume towards a contradiction that
$\mathbf{u}=\sum_{i=1}^{d}a_{i}x^{\log_{q}\lambda_{i}(\Lambda)}\mathbf{e}_{i}\in
B\left(\mathbf{v},\frac{1}{q^{2}}\lambda_{d}(\Lambda)\right)\cap\Lambda$,
where $a_{1}\dots a_{d}\in\mathcal{R}$. Then,

$\|\mathbf{u}-\mathbf{v}\|=\max_{i=1,\dots,d}\lambda_{i}(\Lambda)\left|a_{i}-\frac{1}{x}\right|\leq
q^{-2}\lambda_{d}(\Lambda).$ (2.1)

In particular, (2.1) holds for $i=d$, so that
$\left|a_{i}-\frac{1}{x}\right|\leq q^{-2}$, but this cannot hold for
$a_{i}\in\mathcal{R}$. Hence,
$B\left(\mathbf{v},\frac{1}{q^{2}}\lambda_{d}(\Lambda)\right)\cap\Lambda\neq\emptyset$,
so that $e(\Lambda)\geq q^{-2}\lambda_{d}(\Lambda)$.

On the other hand, by [Ara23, Theorem 1.13],

$\Lambda+B\left(0,\frac{1}{q}\lambda_{d}(\Lambda)\right)=\widetilde{\mathcal{K}}^{d}$

Thus, every ball of radius $\frac{1}{q}\lambda_{d}(\Lambda)$ intersects
$\Lambda$. Hence, $e(\Lambda)\leq q^{-2}\lambda_{d}(\Lambda)$. ∎

######  Lemma 2.8.

Let $L\subseteq\widetilde{\mathcal{K}}^{d}$, let $L^{\prime}\subset L$ be a
sublattice of dimension $d-1$ with minimal volume and let
$H^{\prime}=\operatorname{span}_{\widetilde{\mathcal{K}}}(L^{\prime})$. Then,
$q^{2}e(L+H^{\prime})=\lambda_{d}(L)$.

###### Proof.

Since $e$ is $\operatorname{SL}_{d}(\mathcal{O})$ invariant, then, by [Ara23,
Theorem 1.15] , it suffices to assume that
$L=\mathcal{R}x^{\log_{q}\lambda_{1}(L)}\mathbf{e}_{1}\oplus\dots\oplus\mathcal{R}x^{\log_{q}\lambda_{d}(L)}\mathbf{e}_{d}$.
In this case, Theorem 2.4 implies that

$H^{\prime}=\operatorname{span}_{\widetilde{\mathcal{K}}}\big{\\{}x^{\log_{q}\lambda_{1}(L)}\mathbf{e}_{1},\dots,x^{\log_{q}\lambda_{d-1}(L)}\mathbf{e}_{d-1}\big{\\}}.$

Thus,

$L+H^{\prime}=x^{\log_{q}\lambda_{d}(L)}\mathbf{e}_{d}+H^{\prime}.$

If $\mathbf{v}=\sum_{i=1}^{d}a_{i}x^{\log_{q}\lambda_{d}(L)}\mathbf{e}_{i}\in
B\left(x^{\log_{q}\lambda_{d}(L)-1},\frac{1}{q^{2}}\lambda_{d}(L)\right)\cap(L+H^{\prime})$,
where $a_{d}\in\mathcal{R}$, then,

$\lambda_{d}(L)\left|a_{d}-x^{-1}\right|=\left|a_{d}x^{\log_{q}\lambda_{d}(L)}-x^{\log_{q}\lambda_{d}(L)-1}\right|\leq\frac{1}{q^{2}}\lambda_{d}(L).$

Thus, $\left|a_{d}-x^{-1}\right|\leq q^{-2}$, but this cannot hold for
$a_{d}\in\mathcal{R}$. Hence,

$B\left(x^{\log_{q}\lambda_{d}(L)}\mathbf{e}_{d},\frac{1}{q^{2}}\lambda_{d}(L)\right)\cap(L+H^{\prime})=\emptyset,$

so that $e(L+H^{\prime})\geq\frac{1}{q^{2}}\lambda_{d}(L)$. On the other hand,
since $L\subseteq L+H^{\prime}$, then, $e(L+H^{\prime})\leq
e(L)=\frac{1}{q^{2}}\lambda_{d}(L)$. Hence,
$e(L+H^{\prime})=\frac{1}{q^{2}}\lambda_{d}(L)$. ∎

##  3\. Self-similar coverings

In this section, we recall the definition of self-similar structures and some
related dimension results following [Che07, CC16].

######  Definition 3.1.

Let $X$ be a metric space with a metric $d_{X}$. A self-similar structure on
$X$ is a triple $(J,\sigma,B)$, where $J$ is countable, $\sigma\subseteq
J\times J$, and $B$ is a map from $J$ into the set of bounded subsets of $X$.
For $\alpha\in J$, we denote

$\sigma(\alpha)=\\{\beta\in J:(\alpha,\beta)\in\sigma\\}.$

We say that a sequence $\\{\alpha_{k}\\}_{k\in\mathbb{N}}\subseteq J$ is
$\sigma$-admissible if for every $k\in\mathbb{N}$,
$\alpha_{k+1}\in\sigma(\alpha_{k})$. For a bounded subset $E$ of $X$, we
denote $\operatorname{diam}(E)=\sup\\{d_{X}(y,z):y,z\in E\\}$.

For a subset $S$ of $X$, a self-similar covering of $S$ is a self-similar
structure $(J,\sigma,B)$ such that for every $\theta\in S$, there exists a
$\sigma$-admissible sequence $\\{\alpha_{k}\\}_{k\in\mathbb{N}}$ in $J$
satisfying

  * •

$\lim_{k\to\infty}\operatorname{diam}B(\alpha_{k})=0$;

  * •

$\bigcap_{k\in\mathbb{N}}B(\alpha_{k})=\\{\theta\\}$.

In this case, we say that the self-similar structure $(J,\sigma,B)$ covers the
subset $S$.

We call a triple $(J,\sigma,B)$ a strictly nested self-similar structure on
$X$ if it is a self-similar structure on $X$ satisfying

  * •

for all $\alpha\in J$, $\sigma(\alpha)$ is finite, $B(\alpha)$ is a nonempty
compact subset of $X$, and for all $\beta\in\sigma(\alpha)$ we have
$B(\beta)\subset B(\alpha)$;

  * •

for all $\sigma$-admissible sequence $(\alpha_{k})_{k\in\mathbb{N}}$,
$\lim_{k\to\infty}\operatorname{diam}B(\alpha_{k})=0$;

  * •

for all $\alpha\in J$ and each $\beta\in\sigma(\alpha)$,
$\operatorname{diam}B(\beta)<\operatorname{diam}B(\alpha)$.

######  Theorem 3.2.

[Che07, Theorem 5.3] Let $X$ be a metric space, $S$ be a subset of $X$,
$(J,\sigma,B)$ be a self-similar covering of $S$, and $s>0$. If for every
$\alpha\in J$,

$\sum_{\beta\in\sigma(\alpha)}\operatorname{diam}(B(\beta))^{s}\leq\operatorname{diam}(B(\alpha))^{s},$

then $\mathrm{dim}_{H}(S)\leq s$.

######  Theorem 3.3.

[CC16, Theorem 3.4] Let $X$ be a metric space and $S$ be a subset of $X$.
Suppose that there is a strictly nested self-similar structure $(J,\sigma,B)$
that covers a subset of $S$ and constants $c,s\geq 0$ and $\rho\in(0,1)$ such
that

  1. (1)

for all $\alpha\in J$ and all $\beta\in\sigma(\alpha)$, there are at most $c$
points $\gamma$ in $\sigma(\alpha)\setminus\\{\beta\\}$ such that

$d_{X}\left(B(\beta),B(\gamma)\right)\leq\rho\operatorname{diam}B(\alpha);$

  2. (2)

for all $\alpha\in J,$

$\sum_{\beta\in\sigma(\alpha)}\operatorname{diam}B(\beta)^{s}\geq(c+1)\operatorname{diam}B(\alpha)^{s}.$

Then $S$ contains a subset of positive $s$-dimensional Hausdorff measure.

######  Theorem 3.4.

[CC16, Theorem 3.6] Let $X$ be a metric space and $S$ be a subset of $X$.
Suppose that there is a strictly nested self-similar structure $(J,\sigma,B)$
that covers a subset of $S$, a subset $J_{0}\subset J$ that contains a tail of
any $\sigma$-admissible sequence, a function $\rho:J\to(0,1)$, and two
constants $c,s\geq 0$ such that

  1. (1)

for all $\alpha\in J_{0}$ and all $\beta\in\sigma(\alpha)$, there are at most
$c$ points $\gamma$ in $\sigma(\alpha)\setminus\\{\beta\\}$ such that

$d_{X}(B(\beta),B(\gamma))\leq\rho(\alpha)\operatorname{diam}B(\alpha);$

  2. (2)

for all $\alpha\in J_{0}$,

$\sum_{\beta\in\sigma(\alpha)}\left(\rho(\beta)\operatorname{diam}B(\beta)\right)^{s}\geq(c+1)\left(\rho(\alpha)\operatorname{diam}B(\alpha)\right)^{s};$

  3. (3)

for all $\alpha\in J_{0}$ and all $\beta\in\sigma(\alpha)$,
$\rho(\beta)\operatorname{diam}B(\beta)<\rho(\alpha)\operatorname{diam}B(\alpha)$.

Then $S$ contains a subset of positive $s$-dimensional Hausdorff measure.

##  4\. Best approximation vectors, orthogonality, and Farey lattices

In this section, we recall the definition of best approximation vectors and
prove several properties of best approximations. Then, we introduce
orthogonality and Farey lattices over function fields and prove some related
properties. In particular, we are able to obtain stronger properties than
those in the real case due to Lemma 1.5.

###  4.1. Best approximation vectors

We denote

$Q=\\{(a_{1},\dots,a_{d},b)\in\mathcal{R}^{d+1}:\mathrm{gcd}(a_{1},\dots,a_{d},b)=1,b\neq
0\\}.$

For $\mathbf{u}=(a,b)=(a_{1},\dots,a_{d},b)\in Q$, we denote
$\hat{\mathbf{u}}=\frac{a}{b}$ and $|\mathbf{u}|=|b|$. Given
$\boldsymbol{\theta}\in\widetilde{\mathcal{K}}^{d}$ and $\mathbf{u}\in Q$, we
denote $A(\boldsymbol{\theta},\mathbf{u})=\|b\boldsymbol{\theta}-a\|$.

######  Definition 4.1.

We say that $\mathbf{u}=(a,b)\in Q$ is a best approximation of
$\boldsymbol{\theta}\in\widetilde{\mathcal{K}}^{d}$ if the following
conditions hold.

  1. (1)

For every $\mathbf{v}\in Q$ with $|\mathbf{v}|<|\mathbf{u}|$,
$A(\boldsymbol{\theta},\mathbf{u})<A(\boldsymbol{\theta},\mathbf{v})$;

  2. (2)

For every $\mathbf{v}\in Q$ with $|\mathbf{v}|\leq|\mathbf{u}|$,
$A(\boldsymbol{\theta},\mathbf{u})\leq A(\boldsymbol{\theta},\mathbf{v})$.

######  Remark 4.2.

See [BZ19, KLP23] for the existence of a best approximation sequence of
$\boldsymbol{\theta}\in\widetilde{\mathcal{K}}^{d}$. If $\mathbf{u}=(a,b)\in
Q$ is a best approximation of
$\boldsymbol{\theta}\in\widetilde{\mathcal{K}}^{d}$, then every element of
$\mathbb{F}_{q}^{*}\mathbf{u}$ is also a best approximation. Moreover, $b$ is
unique up to $\mathbb{F}_{q}^{*}$-multiples. To see this, if
$\mathbf{v}=(a^{\prime},b^{\prime})\in Q$ is a best approximation of
$\boldsymbol{\theta}$ with $b^{\prime}\notin\mathbb{F}_{q}^{*}b$ and
$|b^{\prime}|=|b|$, we may assume that $0<|b-b^{\prime}|<|b|$. Since
$A(\boldsymbol{\theta},\mathbf{u})=A(\boldsymbol{\theta},\mathbf{v})$, we have

$\|(b-b^{\prime})\boldsymbol{\theta}-(a-a^{\prime})\|\leq\max\\{A(\boldsymbol{\theta},\mathbf{u}),A(\boldsymbol{\theta},\mathbf{v})\\}=A(\boldsymbol{\theta},\mathbf{u}).$

This is a contradiction to the definition of a best approximation.

######  Lemma 4.3.

Assume that $\mathbf{u}\in Q$ is a best approximation of
$\boldsymbol{\theta}\in\widetilde{\mathcal{K}}^{d}$. Then, for every
$\mathbf{v}\in Q$ such that $|\mathbf{v}|\leq|\mathbf{u}|$ and
$\mathbf{v}\notin\mathbb{F}_{q}^{*}\mathbf{u}$, we have

$A(\boldsymbol{\theta},\mathbf{v})=A(\hat{\mathbf{u}},\mathbf{v}).$

###### Proof.

Write $\mathbf{u}=(a,b)$ and $\mathbf{v}=(a^{\prime},b^{\prime})$. Since
$|b^{\prime}|\leq|b|$, by Lemma 1.5 it follows that

$\begin{split}A(\boldsymbol{\theta},\mathbf{v})&=\|b^{\prime}\boldsymbol{\theta}-a^{\prime}\|=|b^{\prime}|\Bigg{\|}\boldsymbol{\theta}-\frac{a^{\prime}}{b^{\prime}}\Bigg{\|}\leq|b^{\prime}|\cdot\max\Bigg{\\{}\bigg{\|}\boldsymbol{\theta}-\frac{a}{b}\bigg{\|},\bigg{\|}\frac{a^{\prime}}{b^{\prime}}-\frac{a}{b}\bigg{\|}\Bigg{\\}}\\\
&\leq\max\Bigg{\\{}\|b\boldsymbol{\theta}-a\|,\bigg{\|}b^{\prime}\frac{a}{b}-a^{\prime}\bigg{\|}\Bigg{\\}}=\max\\{A(\boldsymbol{\theta},\mathbf{u}),A(\hat{\mathbf{u}},\mathbf{v})\\}.\end{split}$
(4.1)

By Remark 4.2, since $b$ is unique up to $\mathbb{F}_{q}^{*}$-multiples, if
$b^{\prime}\notin\mathbb{F}_{q}^{*}b$, then we have
$A(\boldsymbol{\theta},\mathbf{u})<A(\boldsymbol{\theta},\mathbf{v})$. Thus,
the right hand side of (4.1) is equal to $A(\hat{\mathbf{u}},\mathbf{v})$,
that is, $A(\boldsymbol{\theta},\mathbf{v})\leq
A(\hat{\mathbf{u}},\mathbf{v})$. Suppose that
$b^{\prime}\in\mathbb{F}_{q}^{*}b$ and we may assume that $b^{\prime}=b$ and
$A(\boldsymbol{\theta},\mathbf{u})=A(\boldsymbol{\theta},\mathbf{v})$. Since
$\mathbf{v}\notin\mathbb{F}_{q}^{*}\mathbf{u}$ and
$A(\boldsymbol{\theta},\mathbf{u})\leq 1$, it follows that
$A(\hat{\mathbf{u}},\mathbf{v})=\|a-a^{\prime}\|\geq 1\geq
A(\boldsymbol{\theta},\mathbf{v})$.

On the other hand, by Lemma 1.5

$\begin{split}A(\hat{\mathbf{u}},\mathbf{v})&=\Bigg{\|}b^{\prime}\frac{a}{b}-a^{\prime}\Bigg{\|}\leq\max\Bigg{\\{}A(\boldsymbol{\theta},\mathbf{v}),\left|\frac{b^{\prime}}{b}\right|\|b\boldsymbol{\theta}-a\|\Bigg{\\}}\\\
&\leq\max\\{A(\boldsymbol{\theta},\mathbf{v}),A(\boldsymbol{\theta},\mathbf{u})\\}=A(\boldsymbol{\theta},\mathbf{v}).\end{split}$

Therefore, we have
$A(\hat{\mathbf{u}},\mathbf{v})=A(\boldsymbol{\theta},\mathbf{v})$. ∎

######  Definition 4.4.

For $\mathbf{u}\in Q$ and $r>0$, denote

$\mathcal{B}(\mathbf{u},r)=\\{\boldsymbol{\theta}\in\widetilde{\mathcal{K}}^{d}:A(\boldsymbol{\theta},\mathbf{u})<r\\}\quad\text{and}\quad\bar{\mathcal{B}}(\mathbf{u},r)=\\{\boldsymbol{\theta}\in\widetilde{\mathcal{K}}^{d}:A(\boldsymbol{\theta},\mathbf{u})\leq
r\\}.$

For $\mathbf{v}\in Q$, define

$\Delta_{\mathbf{v}}(\mathbf{u})=\\{\boldsymbol{\theta}\in\widetilde{\mathcal{K}}^{d}:A(\boldsymbol{\theta},\mathbf{u})<A(\boldsymbol{\theta},\mathbf{v})\\}\quad\text{and}\quad\bar{\Delta}_{\mathbf{v}}(\mathbf{u})=\\{\boldsymbol{\theta}\in\widetilde{\mathcal{K}}^{d}:A(\boldsymbol{\theta},\mathbf{u})\leq
A(\boldsymbol{\theta},\mathbf{v})\\}.$

Then, $\mathbf{u}$ is a best approximation vector for $\boldsymbol{\theta}$ if
and only if $\boldsymbol{\theta}\in\Delta(\mathbf{u})$, where

$\Delta(\mathbf{u})=\left(\bigcap_{|\mathbf{v}|<|\mathbf{u}|}\Delta_{\mathbf{v}}(\mathbf{u})\right)\cap\left(\bigcap_{|\mathbf{v}|=|\mathbf{u}|}\bar{\Delta}_{\mathbf{v}}(\mathbf{u})\right).$

In a view of Lemma 4.3, we define

$r(\mathbf{u})=\min_{\begin{subarray}{c}\mathbf{v}\in
Q:|\mathbf{v}|\leq|\mathbf{u}|,\\\
\mathbf{v}\notin\mathbb{F}_{q}^{*}\mathbf{u}\end{subarray}}A(\hat{\mathbf{u}},\mathbf{v}).$

######  Lemma 4.5.

For $\mathbf{u}\in Q$, we have

$\mathcal{B}(\mathbf{u},r(\mathbf{u}))\subseteq\Delta(\mathbf{u})\subseteq\bar{\mathcal{B}}(\mathbf{u},r(\mathbf{u})).$

###### Proof.

If $\boldsymbol{\theta}\in\Delta(\mathbf{u})$, that is, $\mathbf{u}$ is a best
approximation vector for $\boldsymbol{\theta}$, then it follows from Lemma 4.3
that for every $\mathbf{v}\in Q$ with $|\mathbf{v}|\leq|\mathbf{u}|$ and
$\mathbf{v}\notin\mathbb{F}_{q}^{*}\mathbf{u}$, we have

$A(\boldsymbol{\theta},\mathbf{u})\leq
A(\boldsymbol{\theta},\mathbf{v})=A(\hat{\mathbf{u}},\mathbf{v}).$

Hence,
$\Delta(\mathbf{u})\subseteq\bar{\mathcal{B}}(\mathbf{u},r(\mathbf{u}))$.

If $\boldsymbol{\theta}\in\mathcal{B}(\mathbf{u},r(\mathbf{u}))$, then we have
$A(\boldsymbol{\theta},\mathbf{u})<A(\hat{\mathbf{u}},\mathbf{v})$ for every
$\mathbf{v}\in Q$ with $|\mathbf{v}|\leq|\mathbf{u}|$ and
$\mathbf{v}\notin\mathbb{F}_{q}^{*}\mathbf{u}$. It follows that for every such
$\mathbf{v}$,

$A(\boldsymbol{\theta},\mathbf{u})<A(\hat{\mathbf{u}},\mathbf{v})=|\mathbf{v}|\|\hat{\mathbf{u}}-\hat{\mathbf{v}}\|\leq|\mathbf{v}|\max\left\\{\|\boldsymbol{\theta}-\hat{\mathbf{u}}\|,\|\boldsymbol{\theta}-\hat{\mathbf{v}}\|\right\\}\leq\max\\{A(\boldsymbol{\theta},\mathbf{u}),A(\boldsymbol{\theta},\mathbf{v})\\}.$

Hence, $A(\boldsymbol{\theta},\mathbf{u})<A(\boldsymbol{\theta},\mathbf{v})$,
that is, $\boldsymbol{\theta}\in\Delta_{\mathbf{v}}(\mathbf{u})$ for every
such $\mathbf{v}$. Since
$A(\boldsymbol{\theta},\mathbf{u})=A(\boldsymbol{\theta},\mathbf{v})$ for
every $\mathbf{v}\in\mathbb{F}_{q}^{*}\mathbf{u}$, we finally have
$\boldsymbol{\theta}\in\Delta(\mathbf{u})$. ∎

######  Lemma 4.6.

Let
$\boldsymbol{\theta}\in\widetilde{\mathcal{K}}^{d}\setminus\mathcal{K}^{d}$
and let $\\{\mathbf{u}_{n}\\}_{n\in\mathbb{N}}$ be the sequence of best
approximation vectors for $\boldsymbol{\theta}$. Then, for every
$i,j\in\mathbb{N}$, we have

$A(\hat{\mathbf{u}}_{i+j},\mathbf{u}_{i})=A(\boldsymbol{\theta},\mathbf{u}_{i})=r(\mathbf{u}_{i+1}).$

###### Proof.

For the first equality, observe that by Lemma 4.3,
$A(\boldsymbol{\theta},\mathbf{v})=A(\hat{\mathbf{u}}_{i+j},\mathbf{v})$ for
every $\mathbf{v}\in Q$ with $|\mathbf{v}|\leq|\mathbf{u}_{i+j}|$ and
$\mathbf{v}\notin\mathbb{F}_{q}^{*}\mathbf{u}_{i+j}$. In particular, this
holds for $\mathbf{v}=\mathbf{u}_{i}$, which gives rise to the first equality.

For the second part, by Lemma 4.3, we have

$r(\mathbf{u}_{i+1})=\min_{\begin{subarray}{c}\mathbf{v}\in
Q:|\mathbf{v}|\leq|\mathbf{u}_{i+1}|,\\\
\mathbf{v}\notin\mathbb{F}_{q}^{*}\mathbf{u}_{i+1}\end{subarray}}A(\hat{\mathbf{u}}_{i+1},\mathbf{v})=\min_{\begin{subarray}{c}\mathbf{v}\in
Q:|\mathbf{v}|\leq|\mathbf{u}_{i+1}|,\\\
\mathbf{v}\notin\mathbb{F}_{q}^{*}\mathbf{u}_{i+1}\\\
\end{subarray}}A(\boldsymbol{\theta},\mathbf{v}).$

Then it is clear that $r(\mathbf{u}_{i+1})\leq
A(\boldsymbol{\theta},\mathbf{u}_{i})$. Denoting $\mathbf{u}_{i+1}=(a,b)$ and
using Remark 4.2, it follows from the definition of best approximations that
$A(\boldsymbol{\theta},\mathbf{v})\geq A(\boldsymbol{\theta},\mathbf{u}_{i})$
for every $\mathbf{v}=(a^{\prime},b^{\prime})\in Q$ with
$|\mathbf{v}|\leq|\mathbf{u}_{i+1}|$ and
$b^{\prime}\notin\mathbb{F}_{q}^{*}b$. For every
$\mathbf{v}=(a^{\prime},b^{\prime})\in Q$ with
$b^{\prime}\in\mathbb{F}_{q}^{*}b$ but
$\mathbf{v}\notin\mathbb{F}_{q}^{*}\mathbf{u}_{i+1}$, we have
$A(\hat{\mathbf{u}}_{i+1},\mathbf{v})\geq 1$. Hence, we have the second
equality of the lemma. ∎

######  Corollary 4.7.

Let
$\boldsymbol{\theta}\in\widetilde{\mathcal{K}}^{d}\setminus\mathcal{K}^{d}$
and let $\\{\mathbf{u}_{n}\\}_{n\in\mathbb{N}}$ be the sequence of best
approximation vectors for $\boldsymbol{\theta}$. Then
$\boldsymbol{\theta}\in\mathbf{DI}_{d}(\varepsilon)$ if and only if
$|\mathbf{u}_{n}|^{1/d}r(\mathbf{u}_{n})<\varepsilon$ for all large enough
$n\geq 1$. In particular, $\boldsymbol{\theta}\in\mathbf{Sing}_{d}$ if and
only if $|\mathbf{u}_{n}|^{1/d}r(\mathbf{u}_{n})\to 0$ as $n\to\infty$.

###### Proof.

Write $\mathbf{u}_{n}=(a_{n},b_{n})$. It is clear that
$\boldsymbol{\theta}\in\mathbf{DI}_{d}(\varepsilon)$ if and only if
$|b_{n}|^{1/d}\|b_{n-1}\boldsymbol{\theta}-a_{n-1}\|<\varepsilon$ for all
large enough $n\geq 2$. Hence, the corollary follows from Lemma 4.6. ∎

###  4.2. Orthogonality and Farey lattice

One of the main differences from the real case is orthogonality. We use the
following definition of orthogonality from [KST16, Lemma 3.3].

######  Definition 4.8.

A subset $\\{\mathbf{x}_{1},\dots,\mathbf{x}_{k}\\}$ of
$\widetilde{\mathcal{K}}^{d}$ is said to be orthogonal if

$\|r_{1}\mathbf{x}_{1}+\cdots+r_{k}\mathbf{x}_{k}\|=\max\\{|r_{1}|\|\mathbf{x}_{1}\|,\dots,|r_{k}|\|\mathbf{x}_{k}\|\\}$

for any $r_{1},\dots,r_{k}\in\widetilde{\mathcal{K}}$.

By the Hadamard inequality [RW17, Lemma 2.4] and Theorem 2.4, any lattice
$\Lambda$ in $\widetilde{\mathcal{K}}^{d}$ admits an orthogonal basis
$\xi_{1},\dots,\xi_{d}$ of the lattice $\Lambda$ with
$\|\xi_{i}\|=\lambda_{i}(\Lambda)$ for each $i=1,\dots,d$. This corresponds to
a “Minkowski reduced basis” in the real case.

######  Definition 4.9.

For $\mathbf{u}\in Q$, we define the Farey lattice $\Lambda_{\mathbf{u}}$ in
$\widetilde{\mathcal{K}}^{d}$ by

$\Lambda_{\mathbf{u}}=\operatorname{span}_{\mathcal{R}}\\{\mathbf{e}_{1},\dots,\mathbf{e}_{d},\hat{\mathbf{u}}\\}=\pi_{\mathbf{u}}(\mathcal{R}^{d+1}),$

where $\pi_{\mathbf{u}}(a,b)=b\hat{\mathbf{u}}-a$, where
$(a,b)\in\widetilde{\mathcal{K}}^{d}\times\widetilde{\mathcal{K}}$. Note that
$\det(\Lambda_{\mathbf{u}})=|\mathbf{u}|^{-1}$. We define

$\hat{\lambda}_{i}(\Lambda_{\mathbf{u}})=|\mathbf{u}|^{\frac{1}{d}}\lambda_{i}(\Lambda_{\mathbf{u}})\quad\text{for
every }i=1,\dots,d,$

and we say that $\hat{\lambda}_{i}$ are the normalized successive minima of
$\Lambda_{\mathbf{u}}$.

For $\mathbf{u}\in Q$, let $\Lambda_{\mathbf{u}}^{\prime}$ be the sublattice
of $\Lambda_{\mathbf{u}}$ of minimal covolume, and denote
$H_{\mathbf{u}}^{\prime}=\operatorname{span}_{\widetilde{\mathcal{K}}}(\Lambda_{\mathbf{u}}^{\prime})$
and $H_{\mathbf{u}}=\pi_{\mathbf{u}}^{-1}(H_{\mathbf{u}}^{\prime})$. Let
$\xi_{1},\dots,\xi_{d}$ be an orthogonal basis of $\Lambda_{\mathbf{u}}$ with
$\|\xi_{i}\|=\lambda_{i}(\Lambda_{\mathbf{u}})$ for each $i=1,\dots,d$. It can
be easily checked that
$\Lambda_{\mathbf{u}}^{\prime}=\mathrm{span}_{\mathcal{R}}\\{\xi_{1},\dots,\xi_{d-1}\\}$
and
$H_{\mathbf{u}}^{\prime}=\mathrm{span}_{\widetilde{\mathcal{K}}}\\{\xi_{1},\dots,\xi_{d-1}\\}$.

One of the reasons we consider the Farey lattice is because
$\lambda_{1}(\Lambda_{\mathbf{u}})$ is indeed equal to $r(\mathbf{u})$ in
Definition 4.4.

######  Lemma 4.10.

For $\mathbf{u}\in Q$, we have
$\lambda_{1}(\Lambda_{\mathbf{u}})=r(\mathbf{u}).$

###### Proof.

First, observe that

$\lambda_{1}(\Lambda_{\mathbf{u}})=\min_{\begin{subarray}{c}(a,b)\in\mathcal{R}^{d}\times\mathcal{R}\\\
b\hat{\mathbf{u}}-a\neq
0\end{subarray}}\|b\hat{\mathbf{u}}-a\|=\min_{\mathbf{v}\in
Q,\mathbf{v}\notin\mathbb{F}_{q}^{*}\mathbf{u}}A(\hat{\mathbf{u}},\mathbf{v})\leq
r(\mathbf{u}).$

If $\lambda_{1}(\Lambda_{\mathbf{u}})=A(\hat{\mathbf{u}},\mathbf{v})$ for some
$\mathbf{v}\in Q$ with $|\mathbf{v}|>|\mathbf{u}|$, then writing
$\mathbf{u}=(a,b),\mathbf{v}=(a^{\prime},b^{\prime})$ and using Euclidean
division of polynomials, we have $b^{\prime}=kb+r$ for some
$k,r\in\mathcal{R}$ with $0<|r|<|b|$. Since
$A(\hat{\mathbf{u}},\mathbf{v})=A(\hat{\mathbf{u}},\mathbf{v}-k\mathbf{u})$
and $0<|\mathbf{v}-k\mathbf{u}|<|\mathbf{u}|$, we have
$\lambda_{1}(\Lambda_{\mathbf{u}})\geq r(\mathbf{u})$. ∎

The following two lemmas are a function field analogue of the results of
[CC16, Section 5.2] for the geometry associated with the subspace
$H_{\mathbf{u}}$.

######  Lemma 4.11.

Let $\mathbf{u},\mathbf{v}\in Q$. Then, $\mathbf{v}\in H_{\mathbf{u}}$ if and
only if $\hat{\mathbf{v}}\in\hat{\mathbf{u}}+H_{\mathbf{u}}^{\prime}$.

###### Proof.

Notice that $\mathbf{v}\in H_{\mathbf{u}}$ if and only if
$\pi_{\mathbf{u}}(\mathbf{v})\in H_{\mathbf{u}}^{\prime}$. Hence, if we write
$\mathbf{v}=(a,b)$, then it is equivalent to $b\hat{\mathbf{u}}-a\in
H_{\mathbf{u}}^{\prime}$. Hence,
$b^{-1}\pi_{\mathbf{u}}(\mathbf{v})=\hat{\mathbf{u}}-\hat{\mathbf{v}}\in
H_{\mathbf{u}}^{\prime}$. Thus, it is equivalent to
$\hat{\mathbf{v}}\in\hat{\mathbf{u}}+H_{\mathbf{u}}^{\prime}$. ∎

######  Lemma 4.12.

Let $\mathbf{u},\mathbf{v}\in Q$ be such that $\mathbf{v}\in H_{\mathbf{u}}$
and
$\|\hat{\mathbf{u}}-\hat{\mathbf{v}}\|\leq\frac{\lambda_{1}(\Lambda_{\mathbf{u}})}{|\mathbf{u}|}$,
and let $\alpha\in\Lambda_{\mathbf{v}}\setminus H_{\mathbf{u}}^{\prime}$. Then
we have $\|\alpha\|\geq\lambda_{d}(\Lambda_{\mathbf{u}})\geq
q^{-2}\lambda_{d}(\Lambda_{\mathbf{v}})$.

###### Proof.

Since $\mathbf{v}\in H_{\mathbf{u}}$, by Lemma 4.11, we have
$\hat{\mathbf{v}}-\hat{\mathbf{u}}\in H_{\mathbf{u}}^{\prime}$. Hence,
$\Lambda_{\mathbf{v}}\subseteq\Lambda_{\mathbf{u}}+\widetilde{\mathcal{K}}(\hat{\mathbf{v}}-\hat{\mathbf{u}})\subseteq\Lambda_{\mathbf{u}}+H_{\mathbf{u}}^{\prime}$.
Let $\xi_{1},\dots,\xi_{d}$ be an orthogonal basis of $\Lambda_{\mathbf{u}}$
with $\|\xi_{i}\|=\lambda_{i}(\Lambda_{\mathbf{u}})$ for each $i=1,\dots,d$.
Then we have $\alpha\in(\Lambda_{\mathbf{u}}+H_{\mathbf{u}}^{\prime})\setminus
H_{\mathbf{u}}^{\prime}=(\mathcal{R}\setminus\\{0\\})\xi_{d}+H_{\mathbf{u}}^{\prime}$.
Since
$H_{\mathbf{u}}^{\prime}=\mathrm{span}_{\widetilde{\mathcal{K}}}\\{\xi_{1},\dots,\xi_{d-1}\\}$,
it follows from the orthogonality of $\\{\xi_{1},\dots,\xi_{d}\\}$ that
$\|\alpha\|\geq\lambda_{d}(\Lambda_{\mathbf{u}})$.

We now claim that $e(\Lambda_{\mathbf{v}})\leq q^{2}e(\Lambda_{\mathbf{u}})$.
Then, by Lemma 2.7, we have

$\|\alpha\|\geq\lambda_{d}(\Lambda_{\mathbf{u}})=q^{2}e(\Lambda_{\mathbf{u}})\geq
e(\Lambda_{\mathbf{v}})=q^{-2}\lambda_{d}(\Lambda_{\mathbf{v}}),$

which completes the proof.

Observe that by definition of $e(\Lambda_{\mathbf{u}})$, for any
$\beta\in\widetilde{\mathcal{K}}^{d}$,
$B(\beta,q^{2}e(\Lambda_{\mathbf{u}}))\cap\Lambda_{\mathbf{u}}\neq\emptyset$.
Thus, there exist $(a,b)\in\mathcal{R}^{d}\times\mathcal{R}$ with
$|b|\leq|\mathbf{u}|$, such that $\|\beta-(b\hat{\mathbf{u}}+a)\|\leq
q^{2}e(\Lambda_{\mathbf{u}})$. Since
$\|\hat{\mathbf{u}}-\hat{\mathbf{v}}\|\leq\frac{\lambda_{1}(\Lambda_{\mathbf{u}})}{|\mathbf{u}|}$
and
$q^{2}e(\Lambda_{\mathbf{u}})=\lambda_{d}(\Lambda_{\mathbf{u}})\geq\lambda_{1}(\Lambda_{\mathbf{u}})$,
we have

$\|\beta-(b\hat{\mathbf{v}}+a)\|\leq\max\\{\|\beta-(b\hat{\mathbf{u}}+a)\|,\|b(\hat{\mathbf{u}}-\hat{\mathbf{v}})\|\\}\leq\max\\{q^{2}e(\Lambda_{\mathbf{u}}),\lambda_{1}(\Lambda_{\mathbf{u}})\\}=q^{2}e(\Lambda_{\mathbf{u}}).$

Hence, $e(\Lambda_{\mathbf{v}})\leq q^{2}e(\Lambda_{\mathbf{u}})$, which
completes the proof. ∎

We will often consider inverse images of the map $\pi_{\mathbf{u}}$ of lattice
vectors in $\Lambda_{\mathbf{u}}$. Therefore, we will need the following
lemma.

######  Lemma 4.13.

Let $\mathbf{u}\in Q$, $\alpha\in\Lambda_{\mathbf{u}}$, and
$k\in\mathbb{Z}_{\geq 0}$. Then we have

$\\#\\{\mathbf{v}\in\pi^{-1}_{\mathbf{u}}(\alpha)\cap
Q:|\mathbf{v}|=q^{k}|\mathbf{u}|\\}\leq(q-1)q^{k}.$

Moreover, if $\alpha\in\Lambda_{\mathbf{u}}$ is primitive, then

$\\#\\{\mathbf{v}\in\pi^{-1}_{\mathbf{u}}(\alpha)\cap
Q:|\mathbf{v}|=q^{k}|\mathbf{u}|\\}=(q-1)q^{k}.$

###### Proof.

Denote $\mathbf{u}=(a_{0},b_{0})$. Using Euclidean division of polynomials, we
can choose
$(a_{1},b_{1})\in\pi_{\mathbf{u}}^{-1}(\alpha)\cap\mathcal{R}^{d+1}$ with
$|b_{1}|<|b_{0}|$. We first claim that

$\\{\mathbf{v}\in\pi_{\mathbf{u}}^{-1}(\alpha)\cap\mathcal{R}^{d+1}:|\mathbf{v}|=q^{k}|\mathbf{u}|\\}=\big{\\{}(a_{1}+a_{0}s,b_{1}+b_{0}s):s\in\mathcal{R}\text{
with }|s|=q^{k}\big{\\}}.$

It is clear that the right hand side is contained in the left hand side. For
$\mathbf{v}=(a,b)\in\mathcal{R}^{d+1}$ with
$\pi_{\mathbf{u}}(\mathbf{v})=\alpha$ and $|\mathbf{v}|=q^{k}|\mathbf{u}|$, we
have $\alpha=b\frac{a_{0}}{b_{0}}-a=b_{1}\frac{a_{0}}{b_{0}}-a_{1}$. Thus,
$(b-b_{1})\frac{a_{0}}{b_{0}}=a-a_{1}$. Since $\mathrm{gcd}(a_{0},b_{0})=1$,
it follows that $b_{0}$ divides $b-b_{1}$. Hence, there exists some
$s\in\mathcal{R}$ such that $b=b_{1}+b_{0}s$. Thus, $a=a_{1}+a_{0}s$. Since
$|b_{1}|<|b_{0}|$, we have

$q^{k}|\mathbf{u}|=|\mathbf{v}|=|b|=|b-b_{1}|=|b_{0}s|=|s||\mathbf{u}|,$

which completes the proof of the claim.

Since $\\#\\{s\in\mathcal{R}:|s|=q^{k}\\}=(q-1)q^{k}$, the first argument of
the lemma follows from this claim. For the second argument, it is enough to
show that if $\alpha\in\Lambda_{\mathbf{u}}$ is primitive, then
$(a_{1}+a_{0}s,b_{1}+b_{0}s)\in Q$ for any $s\in\mathcal{R}$ with $|s|=q^{k}$.
Indeed, if not, $(a_{1}+a_{0}s,b_{1}+b_{0}s)=s_{0}\mathbf{w}$ for some
$s_{0}\in\mathcal{R}$ with $|s_{0}|>1$ and $\mathbf{w}\in Q$. Since
$\pi_{\mathbf{u}}$ is linear, we have

$\alpha=\pi_{\mathbf{u}}(a_{1}+a_{0}s,b_{1}+b_{0}s)=s_{0}\pi_{\mathbf{u}}(\mathbf{w}).$

This is a contradiction to the primitivity of $\alpha$. Hence, the second
argument follows.

∎

##  5\. Upper Bound

In order to estimate the upper bound of the Hausdorff dimension of
$\mathbf{DI}_{d}(\varepsilon)$, we will define the following self-similar
structure for $\mathbf{DI}_{d}(\varepsilon)$.

######  Definition 5.1.

For $\mathbf{u}\in Q$, we define

$D(\mathbf{u})=\bigg{\\{}\mathbf{v}\in Q:\ |\mathbf{u}|\leq|\mathbf{v}|,\
\mathbf{v}\in H_{\mathbf{u}},\ \text{and
}\|\hat{\mathbf{u}}-\hat{\mathbf{v}}\|\leq\frac{\lambda_{1}(\Lambda_{\mathbf{u}})}{|\mathbf{u}|}\bigg{\\}}.$

For $\mathbf{u}\in Q$ and $\mathbf{v}\in D(\mathbf{u})$, we define

$E(\mathbf{u},\mathbf{v},\varepsilon)=\Bigg{\\{}\mathbf{w}\in Q:\
|\mathbf{v}|<|\mathbf{w}|,\ \mathbf{w}\notin H_{\mathbf{u}},\ \text{and
}\|\hat{\mathbf{v}}-\hat{\mathbf{w}}\|<\frac{\varepsilon}{|\mathbf{v}||\mathbf{w}|^{1/d}}\Bigg{\\}},$

and

$\sigma_{\varepsilon}(\mathbf{u})=\bigcup_{\mathbf{v}\in
D(\mathbf{u})}E(\mathbf{u},\mathbf{v},\varepsilon),\quad
Q_{\varepsilon}=\\{\mathbf{u}\in
Q:\hat{\lambda}_{1}(\Lambda_{\mathbf{u}})<\varepsilon\\},\quad
B(\mathbf{u})=B\left(\hat{\mathbf{u}},\frac{1}{|\mathbf{u}|^{1+1/d}}\right).$

For given $\varepsilon>0$, it follows from Corollary 4.7 and Lemma 4.10 that

$\mathbf{DI}_{d}(\varepsilon)\setminus\mathcal{K}^{d}=\\{\boldsymbol{\theta}\in\widetilde{\mathcal{K}}^{d}\setminus\mathcal{K}^{d}:\mathbf{u}_{n}\in
Q_{\varepsilon}\text{ for all large enough }n\geq 1\\},$

where $\\{\mathbf{u}_{n}\\}_{n\geq 1}$ is the sequence of best approximation
vectors for $\boldsymbol{\theta}$. We denote by
$\widetilde{\mathbf{DI}}_{d}(\varepsilon)$ the above set in the equation.
Since the Hausdorff dimension of the rationals $\mathcal{K}^{d}$ is zero, we
can work with $\widetilde{\mathbf{DI}}_{d}(\varepsilon)$ instead of
$\mathbf{DI}_{d}(\varepsilon)$. We also define

$\widetilde{\mathbf{DI}}_{d}^{*}(\varepsilon)=\\{\boldsymbol{\theta}=(\theta_{1},\dots,\theta_{d})\in\widetilde{\mathbf{DI}}_{d}(\varepsilon):1,\theta_{1},\dots,\theta_{d}\text{
are linearly independent over }\mathcal{K}\\}.$

######  Proposition 5.2.

The triple $(Q_{\varepsilon},\sigma_{\varepsilon},B)$ is a self-similar
covering of $\widetilde{\mathbf{DI}}_{d}^{*}(\varepsilon)$.

###### Proof.

Fix $\boldsymbol{\theta}\in\widetilde{\mathbf{DI}}_{d}^{*}(\varepsilon)$. We
first claim that the sequence of best approximations
$\\{\mathbf{u}_{n}\\}_{n\in\mathbb{N}}$ of $\boldsymbol{\theta}$ contains a
$\sigma_{\varepsilon}$-admissible sequence. By definition, there exists some
$n_{0}$ such that $\mathbf{u}_{n}\in Q_{\varepsilon}$ for every $n>n_{0}$.
Choose $n_{1}=n_{0}+1$ and assume that we have already constructed a
$\sigma_{\varepsilon}$-admissible sequence
$\mathbf{u}_{n_{1}},\dots,\mathbf{u}_{n_{i}}$. We let $n_{i+1}$ be the
smallest $m>n_{i}$ such that $\mathbf{u}_{m}\notin H_{\mathbf{u}_{n_{i}}}$.
Since $1,\theta_{1}\dots\theta_{d}$ are linearly independent over
$\mathcal{K}$, such $n_{i+1}$ indeed exists. Now we let
$\mathbf{u}=\mathbf{u}_{n_{i}}$, $\mathbf{v}=\mathbf{u}_{n_{i+1}-1}$, and
$\mathbf{w}=\mathbf{u}_{n_{i+1}}$. Then,
$|\mathbf{u}|\leq|\mathbf{v}|<|\mathbf{w}|$, $\mathbf{v}\in H_{\mathbf{u}}$,
and $\mathbf{w}\notin H_{\mathbf{u}}$. Moreover, it follows from Lemmas 4.6
and 4.10 that

$\|\hat{\mathbf{u}}-\hat{\mathbf{v}}\|=\frac{A(\hat{\mathbf{v}},\mathbf{u})}{|\mathbf{u}|}<\frac{\lambda_{1}(\Lambda_{\mathbf{u}})}{|\mathbf{u}|}$

and

$\|\hat{\mathbf{v}}-\hat{\mathbf{w}}\|=\frac{A(\hat{\mathbf{w}},\mathbf{v})}{|\mathbf{v}|}=\frac{\lambda_{1}(\Lambda_{\mathbf{w}})}{|\mathbf{v}|}=\frac{\hat{\lambda}_{1}(\Lambda_{\mathbf{w}})}{|\mathbf{v}|\cdot|\mathbf{w}|^{1/d}}<\frac{\varepsilon}{|\mathbf{v}|\cdot|\mathbf{w}|^{1/d}}.$

Thus, $\mathbf{u}_{n_{i+1}}\in\sigma_{\varepsilon}(\mathbf{u}_{n_{i}})$ so
that $\\{\mathbf{u}_{n_{i}}\\}_{i\in\mathbb{N}}$ is
$\sigma_{\varepsilon}$-admissible.

It is obvious that $\mathrm{diam}\ B(\mathbf{u}_{n_{i}})\to 0$ as
$i\to\infty$. We shall now prove that
$\bigcap_{i\in\mathbb{N}}B(\mathbf{u}_{n_{i}})=\\{\boldsymbol{\theta}\\}$.
Note that Theorem 2.4 implies
$\lambda_{1}(\Lambda_{\mathbf{u}})\leq|\mathbf{u}|^{-1/d}$ for any
$\mathbf{u}\in Q$. By Lemmas 4.5 and 4.10, for any best approximation vector
$\mathbf{u}$ of $\boldsymbol{\theta}$, we have

$\Delta(\mathbf{u})\subseteq\bar{\mathcal{B}}(\mathbf{u},r(\mathbf{u}))=B\left(\hat{\mathbf{u}},\frac{\lambda_{1}(\Lambda_{\mathbf{u}})}{|\mathbf{u}|}\right)\subseteq
B\left(\hat{\mathbf{u}},|\mathbf{u}|^{-(1+1/d)}\right)=B(\mathbf{u}).$

This implies that $\boldsymbol{\theta}\in B(\mathbf{u})$ for every best
approximation vector $\mathbf{u}$ of $\boldsymbol{\theta}$. Therefore, we have
$\bigcap_{i\in\mathbb{N}}B(\mathbf{u}_{n_{i}})=\\{\boldsymbol{\theta}\\}$. ∎

######  Proposition 5.3.

Let $t>d$ and $\mathbf{u}\in Q$. Then,

$\sum_{\mathbf{v}\in
D(\mathbf{u})}\left(\frac{|\mathbf{u}|}{|\mathbf{v}|}\right)^{t}\leq\frac{(q-1)q^{t-1}}{q^{t-d}-1}.$

###### Proof.

For $k\geq 0$, denote

$D_{k}(\mathbf{u})=\\{\mathbf{v}\in
D(\mathbf{u}):|\mathbf{v}|=q^{k}|\mathbf{u}|\\}.$

Then,

$\sum_{\mathbf{v}\in
D(\mathbf{u})}\left(\frac{|\mathbf{u}|}{|\mathbf{v}|}\right)^{t}=\sum_{k=0}^{\infty}\sum_{\mathbf{v}\in
D_{k}(\mathbf{u})}\frac{1}{q^{kt}}=\sum_{k=0}^{\infty}\frac{\\#D_{k}(\mathbf{u})}{q^{kt}}.$
(5.1)

To estimate $\\#D_{k}(\mathbf{u})$, consider the map $\pi_{\mathbf{u}}$
restricted to $D_{k}(\mathbf{u})$. By definition, we have

$\pi_{\mathbf{u}}(D_{k}(\mathbf{u}))\subseteq\Lambda_{\mathbf{u}}^{\prime}=\Lambda_{\mathbf{u}}\cap\pi_{\mathbf{u}}(H_{\mathbf{u}})\quad\text{and}\quad\|\pi_{\mathbf{u}}(\mathbf{v})\|=|\mathbf{v}|\|\hat{\mathbf{u}}-\hat{\mathbf{v}}\|\leq
q^{k}\lambda_{1}(\Lambda_{\mathbf{u}})$

for any $\mathbf{v}\in D_{k}(\mathbf{u})$. Hence, it follows that
$\pi_{\mathbf{u}}(D_{k}(\mathbf{u}))\subseteq
B(0,q^{k}\lambda_{1}(\Lambda_{\mathbf{u}}))\cap\Lambda_{\mathbf{u}}^{\prime}$.
Thus, by Lemma 4.13, Proposition 2.5, and the fact that
$\lambda_{1}(\Lambda_{\mathbf{u}})\leq\lambda_{i}(\Lambda_{\mathbf{u}}^{\prime})$
for every $i=1,\dots,d-1$, we obtain

$\begin{split}\\#D_{k}(\mathbf{u})\leq(q-1)q^{k}\\#\left(\Lambda_{\mathbf{u}}^{\prime}\cap
B(0,q^{k}\lambda_{1}(\Lambda_{\mathbf{u}}))\right)&=(q-1)q^{k}\prod_{i=1}^{d-1}\Bigg{\lceil}\frac{q^{k+1}\lambda_{1}(\Lambda_{\mathbf{u}})}{\lambda_{i}(\Lambda_{\mathbf{u}}^{\prime})}\Bigg{\rceil}\\\
&\leq(q-1)q^{d-1}q^{kd}.\end{split}$ (5.2)

Hence, by plugging in (5.2) into (5.1), we obtain that

$\sum_{\mathbf{v}\in
D(\mathbf{u})}\left(\frac{|\mathbf{u}|}{|\mathbf{v}|}\right)^{t}\leq(q-1)q^{d-1}\sum_{k=0}^{\infty}q^{-k(t-d)}=\frac{(q-1)q^{d-1}}{1-q^{-(t-d)}}=\frac{(q-1)q^{t-1}}{q^{t-d}-1}.$

∎

######  Proposition 5.4.

For every $t>d$, we have

$\sum_{\mathbf{w}\in
E(\mathbf{u},\mathbf{v},\varepsilon)}\left(\frac{|\mathbf{v}|}{|\mathbf{w}|}\right)^{t}\leq\frac{(q-1)(q+q^{2})^{d}\varepsilon^{d}}{q^{t-d}-1}.$

###### Proof.

For $k\geq 1$, denote

$E_{k}(\mathbf{u},\mathbf{v},\varepsilon)=\\{\mathbf{w}\in
E(\mathbf{u},\mathbf{v},\varepsilon):|\mathbf{w}|=q^{k}|\mathbf{v}|\\}.$

Then,

$\sum_{\mathbf{w}\in
E(\mathbf{u},\mathbf{v},\varepsilon)}\left(\frac{|\mathbf{v}|}{|\mathbf{w}|}\right)^{t}=\sum_{k=1}^{\infty}\sum_{\mathbf{w}\in
E_{k}(\mathbf{u},\mathbf{v},\varepsilon)}\frac{1}{q^{kt}}=\sum_{k=1}^{\infty}\frac{\\#E_{k}(\mathbf{u},\mathbf{v},\varepsilon)}{q^{kt}}.$
(5.3)

Observe that for any $\mathbf{w}\in E_{k}(\mathbf{u},\mathbf{v},\varepsilon)$,

$\|\pi_{\mathbf{v}}(\mathbf{w})\|=|\mathbf{w}|\|\hat{\mathbf{v}}-\hat{\mathbf{w}}\|<|\mathbf{w}|\frac{\varepsilon}{|\mathbf{v}||\mathbf{w}|^{1/d}}=\varepsilon
q^{k\frac{d-1}{d}}|\mathbf{v}|^{-1/d}.$

Since $\mathbf{v}\in H_{\mathbf{u}}$ and $\mathbf{w}\notin H_{\mathbf{u}}$, it
follows from Lemma 4.11 that $\pi_{\mathbf{v}}(\mathbf{w})\notin
H_{\mathbf{u}}^{\prime}$. Hence, we have

$\pi_{\mathbf{v}}\left(E_{k}(\mathbf{u},\mathbf{v},\varepsilon)\right)\subseteq
B\left(0,\varepsilon
q^{k\frac{d-1}{d}}|\mathbf{v}|^{-1/d}\right)\cap\Lambda_{\mathbf{v}}\setminus
H_{\mathbf{u}}^{\prime}.$

If the set $B\left(0,\varepsilon
q^{k\frac{d-1}{d}}|\mathbf{v}|^{-1/d}\right)\cap\Lambda_{\mathbf{v}}\setminus
H_{\mathbf{u}}^{\prime}$ is non-empty, then it follows from Lemma 4.12 that

$\lambda_{d}(\Lambda_{\mathbf{v}})\leq q^{2}\varepsilon
q^{k\frac{d-1}{d}}|\mathbf{v}|^{-1/d}.$ (5.4)

It follows from Theorem 2.4, Proposition 2.5, Lemma 4.13, and (5.4) that

$\begin{split}\\#E_{k}(\mathbf{u},\mathbf{v},\varepsilon)&\leq(q-1)q^{k}\\#\left(B(0,q^{\varepsilon
k\frac{d-1}{d}}|\mathbf{v}|^{-\frac{1}{d}})\cap\Lambda_{\mathbf{v}}\setminus
H_{\mathbf{u}}^{\prime}\right)\\\
&=(q-1)q^{k}\prod_{i=1}^{d}\Bigg{\lceil}\frac{q\varepsilon
q^{k\frac{d-1}{d}}|\mathbf{v}|^{-1/d}}{\lambda_{i}(\Lambda_{\mathbf{v}})}\Bigg{\rceil}\\\
&\leq(q-1)q^{k}\prod_{i=1}^{d}\left(\frac{q\varepsilon
q^{k\frac{d-1}{d}}|\mathbf{v}|^{-1/d}+\lambda_{d}(\Lambda_{\mathbf{v}})}{\lambda_{i}(\Lambda_{\mathbf{v}})}\right)\\\
&\leq(q-1)q^{k}\prod_{i=1}^{d}\frac{(q+q^{2})\varepsilon
q^{k\frac{d-1}{d}}|\mathbf{v}|^{-1/d}}{\lambda_{i}(\Lambda_{\mathbf{v}})}\\\
&=(q-1)q^{k}(q+q^{2})^{d}\varepsilon^{d}q^{k(d-1)}.\end{split}$ (5.5)

Hence, by plugging in (5.5) into (5.3), we obtain that

$\begin{split}\sum_{\mathbf{w}\in
E(\mathbf{u},\mathbf{v},\varepsilon)}\left(\frac{|\mathbf{v}|}{|\mathbf{w}|}\right)^{t}\leq\sum_{k=1}^{\infty}\frac{(q-1)(q+q^{2})^{d}\varepsilon^{d}q^{kd}}{q^{kt}}&=(q-1)(q+q^{2})^{d}\varepsilon^{d}\sum_{k=1}^{\infty}q^{-k(t-d)}\\\
&=\frac{(q-1)(q+q^{2})^{d}\varepsilon^{d}}{q^{t-d}-1}.\end{split}$

∎

######  Corollary 5.5.

For every $\varepsilon>0$, we have

$\mathrm{dim}_{H}\mathbf{DI}_{d}(\varepsilon)\leq\frac{d^{2}}{d+1}+\frac{d}{d+1}\log_{q}\left(1+\sqrt{(q-1)^{2}(q+1)^{d}q^{2d}\varepsilon^{d}}\right).$

In particular, we have
$\mathrm{dim}_{H}\mathbf{Sing}_{d}\leq\frac{d^{2}}{d+1}$.

###### Proof.

For any $\frac{d^{2}}{d+1}<s\leq d$, it follows from Propositions 5.3 and 5.4
with $t=\frac{d+1}{d}s$ that

$\begin{split}\sum_{\mathbf{w}\in\sigma_{\varepsilon}(\mathbf{u})}\frac{\operatorname{diam}(B(\mathbf{w}))^{s}}{\operatorname{diam}(B(\mathbf{u}))^{s}}&=\sum_{\mathbf{w}\in\sigma_{\varepsilon}(\mathbf{u})}\left(\frac{|\mathbf{u}|}{|\mathbf{w}|}\right)^{(1+1/d)s}\\\
&\leq\sum_{\mathbf{v}\in
D(\mathbf{u})}\left(\frac{|\mathbf{u}|}{|\mathbf{v}|}\right)^{(1+1/d)s}\sum_{\mathbf{w}\in
E(\mathbf{u},\mathbf{v},\varepsilon)}\left(\frac{|\mathbf{v}|}{|\mathbf{w}|}\right)^{(1+1/d)s}\\\
&\leq\frac{(q-1)^{2}(q+q^{2})^{d}\varepsilon^{d}q^{\frac{d+1}{d}s-1}}{\left(q^{\frac{d+1}{d}s-d}-1\right)^{2}}\leq\frac{(q-1)^{2}(q+1)^{d}q^{2d}\varepsilon^{d}}{\left(q^{\frac{d+1}{d}s-d}-1\right)^{2}}.\end{split}$
(5.6)

If
$s\geq\frac{d^{2}}{d+1}+\frac{d}{d+1}\log_{q}\left(1+\sqrt{(q-1)^{2}(q+1)^{d}q^{2d}\varepsilon^{d}}\right)$,
then the right hand side of (5.6) is $\leq 1$. Thus, by Theorem 3.2 and
Proposition 5.2,

$\mathrm{dim}_{H}\widetilde{\mathbf{DI}}_{d}^{*}(\varepsilon)\leq\frac{d^{2}}{d+1}+\frac{d}{d+1}\log_{q}\left(1+\sqrt{(q-1)^{2}(q+1)^{d}q^{2d}\varepsilon^{d}}\right).$
(5.7)

Since the set of $\boldsymbol{\theta}\in\widetilde{\mathcal{K}}^{d}$ such that
$1,\theta_{1},\cdots,\theta_{d}$ are linearly dependent over $\mathcal{K}$ has
Hausdorff dimension $d-1$, the corollary follows from (5.7). ∎

##  6\. Lower bound

In this section, we will estimate the lower bound of the Hausdorff dimension
of $\mathbf{DI}_{d}(\varepsilon)$. For simplicity, we will use the following
notation throughout this section: for $\mathbf{u}\in Q$ and $i=1,\dots,d$,

$\lambda_{i}(\mathbf{u})=\lambda_{i}(\Lambda_{\mathbf{u}})\qquad\text{and}\qquad\hat{\lambda}_{i}(\mathbf{u})=\hat{\lambda}_{i}(\Lambda_{\mathbf{u}}).$

###  6.1. Self-similar structure

While a certain orthogonal projection was used in [CC16], we will utilize the
orthogonal basis given in Definition 4.9. The basic idea is to directly use
the lattice given in [CC16, Lemma 8.7]. For this, we need the following
definition.

Given $\mathbf{u}\in Q$, we fix an orthogonal basis $\xi_{1},\dots,\xi_{d}$ of
the lattice $\Lambda_{\mathbf{u}}$ with $\|\xi_{i}\|=\lambda_{i}(\mathbf{u})$
for each $i=1,\dots,d$ as in Definition 4.9. Note that by Theorem 2.4,
$\det(\Lambda_{\mathbf{u}})=\lambda_{1}(\mathbf{u})\cdots\lambda_{d}(\mathbf{u})=1/|\mathbf{u}|$.

######  Definition 6.1.

Consider $\alpha\in\Lambda_{\mathbf{u}}$ of the form
$\alpha=\sum_{i<d}m_{i}\xi_{i}+n\xi_{d}$ with $n\neq 0$.

  1. (1)

Define the lattice

$\Lambda_{\alpha}=\mathcal{R}\xi_{1}+\cdots+\mathcal{R}\xi_{d-1}+\mathcal{R}\sum_{i<d}\frac{m_{i}}{n}\xi_{i}.$

Recall that
$H_{\mathbf{u}}^{\prime}=\mathrm{span}_{\widetilde{\mathcal{K}}}\\{\xi_{1},\dots,\xi_{d-1}\\}$.
Define the linear map $\pi_{\alpha}:\widetilde{\mathcal{K}}^{d}\to
H_{\mathbf{u}}^{\prime}$ by

$\pi_{\alpha}(\xi)=c_{1}\xi_{1}+\cdots+c_{d-1}\xi_{d-1}-c_{d}\sum_{i<d}\frac{m_{i}}{n}\xi_{i},$

where $\xi=\sum_{i\leq d}c_{i}\xi_{i}$. One can easily check that
$\pi_{\alpha}(\Lambda_{\mathbf{u}})=\Lambda_{\alpha}$ and
$\mathrm{Ker}(\pi_{\alpha})=\widetilde{\mathcal{K}}\alpha$.

  2. (2)

Let $\mathbf{v}\in Q$ be such that $\pi_{\mathbf{u}}(\mathbf{v})=\alpha$, and
assume that $\alpha$ is primitive and $\|\alpha\|=|n|\lambda_{d}(\mathbf{u})$.
By Lemma 6.2, we have

$\det(\Lambda_{\alpha})=\frac{1}{|\mathbf{u}|\|\alpha\|}=\frac{1}{|\mathbf{u}\wedge\mathbf{v}|},$

where
$|\mathbf{u}\wedge\mathbf{v}|=|\mathbf{u}||\mathbf{v}|\|\hat{\mathbf{u}}-\hat{\mathbf{v}}\|$.
For each $i=1\dots,d-1$, denote
$\lambda_{i}(\alpha)=\lambda_{i}(\Lambda_{\alpha})$ and define
$\hat{\lambda}_{i}(\alpha)=\hat{\lambda}_{i}(\Lambda_{\alpha})=|\mathbf{u}\wedge\mathbf{v}|^{\frac{1}{d-1}}\lambda_{i}(\Lambda_{\alpha})$.

We first calculate the covolume of the lattice $\Lambda_{\alpha}$ when
$\alpha$ is primitive.

######  Lemma 6.2.

Let $\mathbf{u}\in Q$ and let
$\alpha=\sum_{i<d}m_{i}\xi_{i}+n\xi_{d}\in\Lambda_{\mathbf{u}}$ be primitive
with $n\neq 0$. Then

$\det(\Lambda_{\alpha})=\frac{\lambda_{1}(\mathbf{u})\cdots\lambda_{d-1}(\mathbf{u})}{|n|}=\frac{1}{|\mathbf{u}||n|\lambda_{d}(\mathbf{u})}.$

###### Proof.

Note that
$\Lambda_{\mathbf{u}}^{\prime}=\mathrm{span}_{\mathcal{R}}\\{\xi_{1},\dots,\xi_{d-1}\\}$
and by the orthogonality of $\\{\xi_{1},\dots,\xi_{d-1}\\}$ we have
$\det(\Lambda_{\mathbf{u}}^{\prime})=\|\xi_{1}\|\cdots\|\xi_{d-1}\|=\lambda_{1}(\mathbf{u})\cdots\lambda_{d-1}(\mathbf{u})$.
Since $\Lambda_{\mathbf{u}}^{\prime}$ is a sublattice of $\Lambda_{\alpha}$ of
full rank,

$\det(\Lambda_{\mathbf{u}}^{\prime})=[\Lambda_{\alpha}:\Lambda_{\mathbf{u}}^{\prime}]\det(\Lambda_{\alpha}),$

where $[\Lambda_{\alpha}:\Lambda_{\mathbf{u}}^{\prime}]$ denotes the index of
the subgroup $\Lambda_{\mathbf{u}}^{\prime}$ of $\Lambda_{\alpha}$. Consider
the group homomorphsim $\varphi:\Lambda_{\alpha}\to\mathcal{R}/n\mathcal{R}$
given by

$\varphi\left(\sum_{i<d}c_{i}\xi_{i}+c_{d}\sum_{i<d}\frac{m_{i}}{n}\xi_{i}\right)=c_{d}+n\mathcal{R}.$

Since $\alpha$ is primitive, we have $\mathrm{gcd}(m_{1},\dots,m_{d-1},n)=1$.
Hence, $\varphi$ is well-defined. Since $\varphi$ is surjective and
$\mathrm{Ker}(\varphi)=\Lambda_{\mathrm{u}}^{\prime}$, we have
$[\Lambda_{\alpha}:\Lambda_{\mathbf{u}}^{\prime}]=\\#\mathcal{R}/n\mathcal{R}=|n|$.
∎

######  Lemma 6.3.

Let $\mathbf{u}\in Q$,
$\alpha=\sum_{i<d}m_{i}\xi_{i}+n\xi_{d}\in\Lambda_{\mathbf{u}}$ be primitive
with $n\neq 0$, and $\mathbf{v}\in Q$ be such that
$\pi_{\mathbf{u}}(\mathbf{v})=\alpha$. Assume that
$\|\alpha\|=|n|\lambda_{d}(\mathbf{u})$. Then
$\frac{|\mathbf{u}\wedge\mathbf{v}|}{|\mathbf{v}|}\leq\lambda_{1}(\alpha)$
implies
$\lambda_{1}(\mathbf{v})=\frac{|\mathbf{u}\wedge\mathbf{v}|}{|\mathbf{v}|}$.

###### Proof.

We first claim that for any $\xi\in\widetilde{\mathcal{K}}^{d}$,
$\|\xi\|\geq\|\pi_{\alpha}(\xi)\|$. Indeed, writing $\xi=\sum_{i\leq
d}c_{i}\xi_{i}$, we have

$\|\pi_{\alpha}(\xi)\|\leq\max_{i=1,\dots,d-1}\left|c_{i}-c_{d}\frac{m_{i}}{n}\right|\|\xi_{i}\|\leq\max_{i=1,\dots,d-1}\max\Big{\\{}|c_{i}|\lVert\xi_{i}\|,\left|c_{d}\frac{m_{i}}{n}\right|\|\xi_{i}\|\Big{\\}}.$

Since $\|\alpha\|=|n|\lambda_{d}(\mathbf{u})$, we have
$|\frac{m_{i}}{n}|\|\xi_{i}\|\leq\|\xi_{d}\|$. Hence by the orthogonality of
$\\{\xi_{1},\dots,\xi_{d}\\}$,
$\|\pi_{\alpha}(\xi)\|\leq\max_{i=1,\dots,d}|c_{i}|\|\xi_{i}\|=\|\xi\|$.

Observe that since
$\mathrm{Ker}(\pi_{\alpha})=\widetilde{\mathcal{K}}\alpha=\widetilde{\mathcal{K}}(\widehat{\mathbf{u}}-\widehat{\mathbf{v}})$,
it follows that
$\pi_{\alpha}(b\widehat{\mathbf{u}}-a)=\pi_{\alpha}(b\widehat{\mathbf{v}}-a)$
for any $(a,b)\in\mathcal{R}^{d}\times\mathcal{R}$. Hence
$\pi_{\alpha}(\Lambda_{\mathbf{u}})=\pi_{\alpha}(\Lambda_{\mathbf{v}})=\Lambda_{\alpha}$.

Now, let $\beta=\pi_{\mathbf{v}}(\mathbf{u})$. For any
$\gamma\in\Lambda_{\mathbf{v}}\setminus\widetilde{\mathcal{K}}\beta$, since
$\mathrm{Ker}(\pi_{\alpha})=\widetilde{\mathcal{K}}\beta$, it follows from the
claim and the observation that
$\|\gamma\|\geq\|\pi_{\alpha}(\gamma)\|\geq\lambda_{1}(\alpha)\geq\frac{|\mathbf{u}\wedge\mathbf{v}|}{|\mathbf{v}|}$.
Since $\alpha$ is primitive, $\beta$ is also primitive. Hence, if
$\gamma\in\Lambda_{\mathbf{v}}\cap\widetilde{\mathcal{K}}\beta\setminus\\{0\\}$,
then
$\|\gamma\|\geq\|\beta\|=\frac{|\mathbf{u}\wedge\mathbf{v}|}{|\mathbf{v}|}$.
Therefore, we have
$\lambda_{1}(\mathbf{v})=\frac{|\mathbf{u}\wedge\mathbf{v}|}{|\mathbf{v}|}$. ∎

We shall now define the fractal structure for the lower bound of the Hausdorff
dimension of $\mathbf{DI}_{d}(\varepsilon)$.

######  Definition 6.4.

Given $\mathbf{u}\in Q$ and $\varepsilon>0$, define

$\Lambda_{\mathbf{u}}(\varepsilon)=\\{\alpha\in\Lambda_{\mathbf{u}}:\alpha\text{
is primitive and }\hat{\lambda}_{1}(\alpha)>\varepsilon\\}.$

Given $n\in\mathcal{R}$ and $N\in\mathbb{N}$, define

$C_{n}^{\prime}({\mathbf{u}})=\Bigg{\\{}\mathbf{v}=\sum_{i=1}^{d-1}\alpha_{i}\xi_{i}+n\xi_{d}:\alpha_{i}\in\widetilde{\mathcal{K}},|\alpha_{i}|<\frac{|n|\lambda_{d}(\mathbf{u})}{\lambda_{i}(\mathbf{u})}\Bigg{\\}};\quad
C_{N}(\mathbf{u})=\bigcup_{0<|n|\leq q^{N}}C_{n}^{\prime}(\mathbf{u}),$

and define

$F_{N}(\mathbf{u},\varepsilon)=\bigcup_{\alpha\in\Lambda_{\mathbf{u}}(\varepsilon)\cap
C_{N}(\mathbf{u})}\zeta(\mathbf{u},\alpha,\varepsilon);\qquad
F(\mathbf{u},\varepsilon)=\bigcup_{N\in\mathbb{N}}F_{N}(\mathbf{u},\varepsilon),$

where

$\zeta(\mathbf{u},\alpha,\varepsilon)=\Bigg{\\{}\mathbf{v}\in
Q:\pi_{\mathbf{u}}(\mathbf{v})=\alpha,\left(\frac{|\mathbf{u}\wedge\mathbf{v}|}{\varepsilon}\right)^{\frac{d}{d-1}}\leq|\mathbf{v}|\leq\left(\frac{q|\mathbf{u}\wedge\mathbf{v}|}{\varepsilon}\right)^{\frac{d}{d-1}}\Bigg{\\}}.$

Finally, we define

$\sigma_{\varepsilon,N}(\mathbf{u})=F_{N}(\mathbf{u},\varepsilon),\quad
Q_{\varepsilon,N}=\bigcup_{\mathbf{u}\in
Q}\sigma_{\varepsilon,N}(\mathbf{u}),\quad
B(\mathbf{u})=B\left(\hat{\mathbf{u}},\frac{\lambda_{1}(\mathbf{u})}{q|\mathbf{u}|}\right).$

######  Lemma 6.5.

Let $0<\varepsilon<1$ and $\mathbf{v}\in F(\mathbf{u},\varepsilon)$. Then
$\lambda_{1}(\mathbf{v})=\frac{|\mathbf{u}\wedge\mathbf{v}|}{|\mathbf{v}|}$,
$\frac{1}{q}\varepsilon\leq\hat{\lambda}_{1}(\mathbf{v})\leq\varepsilon$, and
$|\mathbf{v}|>|\mathbf{u}|$.

###### Proof.

Let $\alpha=\pi_{\mathbf{u}}(\mathbf{v})$. Since

$\frac{\varepsilon}{q|\mathbf{u}\wedge\mathbf{v}|}\leq\frac{1}{|\mathbf{v}|^{\frac{d-1}{d}}}\leq\frac{\varepsilon}{|\mathbf{u}\wedge\mathbf{v}|}$

and
$\hat{\lambda}_{1}(\alpha)=|\mathbf{u}\wedge\mathbf{v}|^{\frac{1}{d-1}}\lambda_{1}(\alpha)>\varepsilon$,
it follows from $\varepsilon<1$ that

$\frac{|\mathbf{u}\wedge\mathbf{v}|}{|\mathbf{v}|}\leq\frac{\varepsilon^{\frac{d}{d-1}}}{|\mathbf{u}\wedge\mathbf{v}|^{\frac{1}{d-1}}}<\varepsilon^{\frac{1}{d-1}}\lambda_{1}(\alpha)<\lambda_{1}(\alpha).$

Since $\alpha$ is primitive, by Lemma 6.3, we have
$\lambda_{1}(\mathbf{v})=\frac{|\mathbf{u}\wedge\mathbf{v}|}{|\mathbf{v}|}$.
Thus,

$\frac{\varepsilon}{q}\leq\hat{\lambda}_{1}(\mathbf{v})=\frac{|\mathbf{u}\wedge\mathbf{v}|}{|\mathbf{v}|^{\frac{d-1}{d}}}\leq\varepsilon.$

Since $\alpha\in\Lambda_{\mathbf{u}}\setminus H_{\mathbf{u}}^{\prime}$, by
Lemma 4.12, we have $\|\alpha\|\geq\lambda_{d}(\mathbf{u})$. Note that by
Theorem 2.4,
$\hat{\lambda}_{1}(\mathbf{u})\cdots\hat{\lambda}_{d}(\mathbf{u})=1$. Hence,
$\hat{\lambda}_{d}(\mathbf{u})\geq 1$. Therefore,

$1\leq\hat{\lambda}_{d}(\mathbf{u})=|\mathbf{u}|^{1/d}\lambda_{d}(\mathbf{u})\leq|\mathbf{u}|^{1/d}\|\alpha\|.$
(6.1)

It follows from (6.1) that

$\frac{|\mathbf{u}|}{|\mathbf{v}|}\leq\frac{\varepsilon^{\frac{d}{d-1}}|\mathbf{u}|}{|\mathbf{u}\wedge\mathbf{v}|^{\frac{d}{d-1}}}=\frac{\varepsilon^{\frac{d}{d-1}}|\mathbf{u}|}{\|\alpha\|^{\frac{d}{d-1}}|\mathbf{u}|^{\frac{d}{d-1}}}=\frac{\varepsilon^{\frac{d}{d-1}}}{(|\mathbf{u}|^{\frac{1}{d}}\|\alpha\|)^{\frac{d}{d-1}}}\leq\varepsilon^{\frac{d}{d-1}}.$

Since $\varepsilon<1$, we have $|\mathbf{v}|>|\mathbf{u}|$. ∎

######  Lemma 6.6.

Let $0<\varepsilon<1$ and $\mathbf{v}\in F(\mathbf{u},\varepsilon)$. Then,
$B\left(\hat{\mathbf{v}},\frac{\lambda_{1}(\mathbf{v})}{|\mathbf{v}|}\right)\subset
B\left(\hat{\mathbf{u}},\frac{\lambda_{1}(\mathbf{u})}{q|\mathbf{u}|}\right)$.

###### Proof.

We first bound $\|\hat{\mathbf{u}}-\hat{\mathbf{v}}\|$ and
$\frac{\lambda_{1}(\mathbf{v})}{|\mathbf{v}|}$. Let
$\alpha=\pi_{\mathbf{u}}(\mathbf{v})$. Then it follows from
$\xi_{1}\in\Lambda_{\alpha}$ that
$\lambda_{1}(\alpha)\leq\|\xi_{1}\|=\lambda_{1}(\mathbf{u})$. Since
$\varepsilon<1$, we have

$\|\hat{\mathbf{u}}-\hat{\mathbf{v}}\|=\frac{|\mathbf{u}\wedge\mathbf{v}|}{|\mathbf{u}||\mathbf{v}|}\leq\frac{\varepsilon^{\frac{d}{d-1}}}{|\mathbf{u}\wedge\mathbf{v}|^{\frac{1}{d-1}}|\mathbf{u}|}<\frac{\varepsilon^{\frac{1}{d-1}}\hat{\lambda}_{1}(\alpha)}{|\mathbf{u}\wedge\mathbf{v}|^{\frac{1}{d-1}}|\mathbf{u}|}\leq\varepsilon^{\frac{1}{d-1}}\frac{\lambda_{1}(\mathbf{u})}{|\mathbf{u}|}<\frac{\lambda_{1}(\mathbf{u})}{|\mathbf{u}|}.$
(6.2)

By Lemma 6.5 and (6.2), we have

$\frac{\lambda_{1}(\mathbf{v})}{|\mathbf{v}|}=\frac{|\mathbf{u}\wedge\mathbf{v}|}{|\mathbf{v}|^{2}}=\frac{|\mathbf{u}|}{|\mathbf{v}|}\|\hat{\mathbf{u}}-\hat{\mathbf{v}}\|<\frac{\lambda_{1}(\mathbf{u})}{|\mathbf{v}|}<\frac{\lambda_{1}(\mathbf{u})}{|\mathbf{u}|}.$
(6.3)

Now let $\boldsymbol{\theta}\in
B\left(\hat{\mathbf{v}},\frac{\lambda_{1}(\mathbf{v})}{|\mathbf{v}|}\right)$.
By Lemma 1.5, (6.2), and (6.3),

$\|\boldsymbol{\theta}-\hat{\mathbf{u}}\|\leq\max\\{\|\hat{\mathbf{u}}-\hat{\mathbf{v}}\|,\|\boldsymbol{\theta}-\hat{\mathbf{v}}\|\\}<\frac{\lambda_{1}(\mathbf{u})}{|\mathbf{u}|}.$

Since the image of the norm $\|\cdot\|$ is in $q^{\mathbb{Z}}\cup\\{0\\}$, we
have
$\|\boldsymbol{\theta}-\hat{\mathbf{u}}\|\leq\frac{\lambda_{1}(\mathbf{u})}{q|\mathbf{u}|}$,
that is, $\boldsymbol{\theta}\in
B\left(\hat{\mathbf{u}},\frac{\lambda_{1}(\mathbf{u})}{q|\mathbf{u}|}\right)$.
∎

######  Proposition 6.7.

For $0<\varepsilon<1$, the triple
$(Q_{\varepsilon,N},\sigma_{\varepsilon,N},B)$ is a strictly nested self-
similar structure covering a subset of $\mathbf{DI}_{d}(\varepsilon)$.

###### Proof.

We need to verify that for every $\sigma_{\varepsilon,N}$ admissible sequence
$\\{\mathbf{u}_{k}\\}_{k\in\mathbb{N}}$,

  1. (1)

for all $k$, $B(\mathbf{u}_{k+1})\subset B(\mathbf{u}_{k})$ and
$\operatorname{diam}(B(\mathbf{u}_{k+1}))<\operatorname{diam}(B(\mathbf{u}_{k}))$;

  2. (2)

$\lim_{k\rightarrow\infty}\operatorname{diam}(B(\mathbf{u}_{k}))=0$ and
$\bigcap_{k=1}^{\infty}B(\mathbf{u}_{k})$ is a point contained in
$\mathbf{DI}_{d}(\varepsilon)$.

By Lemma 6.6, it is clear that (1) holds. Moreover, due to the definition of
$\sigma_{\varepsilon,N}$ and Lemma 6.5, $|\mathbf{u}_{k}|\rightarrow\infty$.
Therefore,
$\lim_{k\rightarrow\infty}\operatorname{diam}(B(\mathbf{u}_{k}))=0$. Hence,
$\bigcap_{k=1}^{\infty}B(\mathbf{u}_{k})$ is a single point
$\boldsymbol{\theta}$.

In order to prove that $\boldsymbol{\theta}\in\mathbf{DI}_{d}(\varepsilon)$,
we observe that $\boldsymbol{\theta}\in B(\mathbf{u}_{k})$ for every $k\geq
1$. Thus, for every $k\geq 1$, we have
$\|\boldsymbol{\theta}-\hat{\mathbf{u}}_{k}\|\leq\frac{\lambda_{1}(\mathbf{u}_{k})}{q|\mathbf{u}_{k}|}$.
Thus, it follows that

$A(\boldsymbol{\theta},\mathbf{u}_{k})=\|\boldsymbol{\theta}-\hat{\mathbf{u}}_{k}\||\mathbf{u}_{k}|<\lambda_{1}(\mathbf{u}_{k}).$

By Lemma 4.5, $\boldsymbol{\theta}\in\Delta(\mathbf{u}_{k})$, so that for
every $k\geq 1$, $\mathbf{u}_{k}$ is a best approximation of
$\boldsymbol{\theta}$.

Fix any large enough $T\geq 1$ so that there is $k\geq 1$ such that
$|\mathbf{u}_{k}|\leq T<|\mathbf{u}_{k+1}|$. Since $\mathbf{u}_{k}$’s are best
approximations of $\boldsymbol{\theta}$, it follows from Lemma 4.6 and
$|\mathbf{u}_{k+1}|\geq\left(\frac{|\mathbf{u}_{k}\wedge\mathbf{u}_{k+1}|}{\varepsilon}\right)^{\frac{d}{d-1}}$
that

$A(\boldsymbol{\theta},\mathbf{u}_{k})=A(\hat{\mathbf{u}}_{k+1},\mathbf{u}_{k})=\frac{|\mathbf{u}_{k}\wedge\mathbf{u}_{k+1}|}{|\mathbf{u}_{k+1}|}\leq\varepsilon|\mathbf{u}_{k+1}|^{-\frac{1}{d}}<\varepsilon
T^{-\frac{1}{d}}.$

This proves that $\boldsymbol{\theta}\in\mathbf{DI}_{d}(\varepsilon)$. ∎

###  6.2. Well-separatedness

We will need the following claim regarding the distance between two balls in
$\widetilde{\mathcal{K}}^{d}$.

######  Lemma 6.8.

Let $\mathbf{u},\mathbf{u}^{\prime}\in\widetilde{\mathcal{K}}^{d}$ and let
$r,r^{\prime}>0$. Then,

$d\left(B(\mathbf{u},r),B(\mathbf{u}^{\prime},r^{\prime})\right)=\begin{cases}0&\|\mathbf{u}-\mathbf{u}^{\prime}\|\leq\max\\{r,r^{\prime}\\}\\\
\|\mathbf{u}-\mathbf{u}^{\prime}\|&\text{else}.\end{cases}$

###### Proof.

Notice that an element in $B(\mathbf{u},r)$ will be of the form
$\mathbf{u}+\mathbf{v}$ where $\|\mathbf{v}\|\leq r$. Hence, if
$\|\mathbf{u}-\mathbf{u}^{\prime}\|\leq\max\\{r,r^{\prime}\\}$, then,
$-(\mathbf{u}-\mathbf{u}^{\prime})\in B(0,\max\\{r,r^{\prime}\\})$. Thus,

$d\left(B(\mathbf{u},r),B(\mathbf{u}^{\prime},r^{\prime})\right)=\inf_{\|\mathbf{v}\|\leq
r,\|\mathbf{v}^{\prime}\|\leq
r^{\prime}}\|\mathbf{u}-\mathbf{u}^{\prime}+\mathbf{v}-\mathbf{v}^{\prime}\|\leq\|\mathbf{u}-\mathbf{u}^{\prime}-(\mathbf{u}-\mathbf{u}^{\prime})\|=0.$

On the other hand, assume that
$\|\mathbf{u}-\mathbf{u}^{\prime}\|>\max\\{r,r^{\prime}\\}$. Then, for every
$\|\mathbf{v}\|\leq r$ and $\|\mathbf{v}^{\prime}\|\leq r^{\prime}$, we have
that

$\|\mathbf{v}-\mathbf{v}^{\prime}\|\leq\max\\{\|\mathbf{v}\|,\|\mathbf{v}^{\prime}\|\\}\leq\max\\{r,r^{\prime}\\}<\|\mathbf{u}-\mathbf{u}^{\prime}\|.$

Thus, by Lemma 1.5,

$d\left(B(\mathbf{u},r),B(\mathbf{u}^{\prime},r^{\prime})\right)=\inf_{\|\mathbf{v}\|\leq
r,\|\mathbf{v}^{\prime}\|\leq
r^{\prime}}\|\mathbf{u}-\mathbf{u}^{\prime}+\mathbf{v}-\mathbf{v}^{\prime}\|=\|\mathbf{u}-\mathbf{u}^{\prime}\|.$

∎

######  Lemma 6.9.

Let $\mathbf{v},\mathbf{w}\in F(\mathbf{u},\varepsilon)$ be distinct. Then,
$d(B(\mathbf{v}),B(\mathbf{w}))=\|\hat{\mathbf{v}}-\hat{\mathbf{w}}\|$.

###### Proof.

By Lemma 6.8, it suffices to prove that
$\|\hat{\mathbf{v}}-\hat{\mathbf{w}}\|>\max\Bigg{\\{}\frac{\lambda_{1}(\mathbf{w})}{q|\mathbf{w}|},\frac{\lambda_{1}(\mathbf{v})}{q|\mathbf{v}|}\Bigg{\\}}$.
Assume towards a contradiction that
$\|\hat{\mathbf{w}}-\hat{\mathbf{v}}\|\leq\max\Bigg{\\{}\frac{\lambda_{1}(\mathbf{w})}{q|\mathbf{w}|},\frac{\lambda_{1}(\mathbf{v})}{q|\mathbf{v}|}\Bigg{\\}}$,
and assume without loss of generality that
$\frac{\lambda_{1}(\mathbf{v})}{|\mathbf{v}|}\leq\frac{\lambda_{1}(\mathbf{w})}{|\mathbf{w}|}$.
It follows that

$\frac{\lambda_{1}(\mathbf{v})}{|\mathbf{w}|}\leq\frac{\|\pi_{\mathbf{v}}(\mathbf{w})\|}{|\mathbf{w}|}=\|\hat{\mathbf{w}}-\hat{\mathbf{v}}\|\leq\frac{\lambda_{1}(\mathbf{w})}{q|\mathbf{w}|}.$

Thus,
$\lambda_{1}(\mathbf{v})\leq\frac{\lambda_{1}(\mathbf{w})}{q}<\lambda_{1}(\mathbf{w})$.
It follows from Lemma 6.5 that
$\lambda_{1}(\mathbf{v})=\frac{|\mathbf{u}\wedge\mathbf{v}|}{|\mathbf{v}|}=|\mathbf{u}|\|\hat{\mathbf{u}}-\hat{\mathbf{v}}\|$,
$\lambda_{1}(\mathbf{w})=\frac{|\mathbf{u}\wedge\mathbf{w}|}{|\mathbf{w}|}=|\mathbf{u}|\|\hat{\mathbf{u}}-\hat{\mathbf{w}}\|$,
and

$\|\hat{\mathbf{u}}-\hat{\mathbf{w}}\|=\frac{\lambda_{1}(\mathbf{w})}{|\mathbf{u}|}>\frac{\lambda_{1}(\mathbf{w})}{|\mathbf{w}|}>\|\hat{\mathbf{w}}-\hat{\mathbf{v}}\|.$

By Lemma 1.5 we have

$\|\hat{\mathbf{w}}-\hat{\mathbf{v}}\|<\|\hat{\mathbf{u}}-\hat{\mathbf{w}}\|\leq\max\\{\|\hat{\mathbf{w}}-\hat{\mathbf{v}}\|,\|\hat{\mathbf{u}}-\hat{\mathbf{v}}\|\\}.$

It follows that
$\|\hat{\mathbf{w}}-\hat{\mathbf{v}}\|<\|\hat{\mathbf{u}}-\hat{\mathbf{v}}\|$,
so that by Lemma 1.5,
$\|\hat{\mathbf{u}}-\hat{\mathbf{w}}\|=\|\hat{\mathbf{u}}-\hat{\mathbf{v}}\|$.
Thus, we have

$\frac{\lambda_{1}(\mathbf{w})}{|\mathbf{u}|}=\|\hat{\mathbf{u}}-\hat{\mathbf{w}}\|=\|\hat{\mathbf{u}}-\hat{\mathbf{v}}\|=\frac{\lambda_{1}(\mathbf{v})}{|\mathbf{u}|}.$

This implies that $\lambda_{1}(\mathbf{w})=\lambda_{1}(\mathbf{v})$, which is
a contradiction. Hence
$d\left(B(\mathbf{w}),B(\mathbf{v})\right)=\|\hat{\mathbf{w}}-\hat{\mathbf{v}}\|$.
∎

######  Lemma 6.10.

Let $\mathbf{v},\mathbf{w}\in F_{N}(\mathbf{u},\varepsilon)$ be distinct.
Then,

$d\left(B(\mathbf{v}),B(\mathbf{w})\right)\geq\left(\frac{\varepsilon}{q^{N+1}\hat{\lambda}_{d}(\mathbf{u})}\right)^{\frac{2d}{d-1}}\frac{\lambda_{1}(\mathbf{u})}{|\mathbf{u}|}.$

###### Proof.

By Lemma 6.5, it follows that

$\|\hat{\mathbf{v}}-\hat{\mathbf{w}}\|=\frac{\|\pi_{\mathbf{v}}(\mathbf{w})\|}{|\mathbf{w}|}\geq\frac{\lambda_{1}(\mathbf{v})}{|\mathbf{w}|}=\frac{|\mathbf{u}\wedge\mathbf{v}|}{|\mathbf{v}||\mathbf{w}|}=\frac{\|\pi_{\mathbf{u}}(\mathbf{v})\||\mathbf{u}|}{|\mathbf{v}||\mathbf{w}|}\geq\frac{\lambda_{1}(\mathbf{u})}{|\mathbf{u}|}\left(\frac{|\mathbf{u}|}{|\mathbf{v}|}\right)\left(\frac{|\mathbf{u}|}{|\mathbf{w}|}\right).$
(6.4)

By definition of $F_{N}(\mathbf{u},\varepsilon)$, we have
$|\mathbf{v}|\leq\left(\frac{q|\mathbf{u}\wedge\mathbf{v}|}{\varepsilon}\right)^{\frac{d}{d-1}}$
and $\|\pi_{\mathbf{u}}(\mathbf{v})\|\leq q^{N}\lambda_{d}(\mathbf{u})$. Thus,

$\frac{|\mathbf{u}|}{|\mathbf{v}|}\geq\frac{|\mathbf{u}|\varepsilon^{\frac{d}{d-1}}}{(q|\mathbf{u}\wedge\mathbf{v}|)^{\frac{d}{d-1}}}=\frac{\varepsilon^{\frac{d}{d-1}}}{(q|\mathbf{u}|^{\frac{1}{d}}\|\pi_{\mathbf{u}}(\mathbf{v})\|)^{\frac{d}{d-1}}}\geq\left(\frac{\varepsilon}{q^{N+1}\hat{\lambda}_{d}(\mathbf{u})}\right)^{\frac{d}{d-1}}.$
(6.5)

By the same argument with $\mathbf{w}$ instead of $\mathbf{v}$, we have (6.5)
for $\mathbf{w}$. Combining with Lemma 6.9, (6.4), and (6.5) for $\mathbf{v}$
and $\mathbf{w}$, it follows that

$d\left(B(\mathbf{v}),B(\mathbf{w})\right)=\|\hat{\mathbf{v}}-\hat{\mathbf{w}}\|\geq\left(\frac{\varepsilon}{q^{N+1}\hat{\lambda}_{d}(\mathbf{u})}\right)^{\frac{2d}{d-1}}\frac{\lambda_{1}(\mathbf{u})}{|\mathbf{u}|}.$

∎

###  6.3. Lower bound calculations

In this subsection, we will estimate lower bounds of Hausdorff dimensions for
the sets $\mathbf{DI}_{d}(\varepsilon)$ and $\mathbf{Sing}_{d}$ assuming the
following proposition.

######  Proposition 6.11.

Let $0<\varepsilon<1$, $N\in\mathbb{N}$, and $\mathbf{u}\in
Q_{\varepsilon,N}$. For any $s>0$, we have

$\sum_{\mathbf{v}\in
F_{N}(\mathbf{u},\varepsilon)}\left(\frac{\lambda_{1}(\mathbf{v})}{|\mathbf{v}|}\right)^{s}\left(\frac{|\mathbf{u}|}{\lambda_{1}(\mathbf{u})}\right)^{s}\geq\left(\frac{q-1}{q}-\varepsilon^{d-1}\right)\frac{\varepsilon^{d}(q-1)^{2}}{q^{d+s}}\sum_{T_{0}\leq
k\leq T_{N}}q^{-(\frac{d+1}{d}s-d)k},$

where

$T_{i}=\frac{d}{d-1}\left(i+1+\log_{q}\frac{\hat{\lambda}_{d}(\mathbf{u})}{\varepsilon}\right).$

This proposition will be proved in section 6.4.

######  Remark 6.12.

Note that $\hat{\lambda}_{d}(\mathbf{u})\geq 1$ for any $\mathbf{u}\in Q$. On
the other hand, if $\mathbf{u}\in Q_{\varepsilon,N}$, then it follows from the
definition of $Q_{\varepsilon,N}$ and Lemma 6.5 that
$\hat{\lambda}_{1}(\mathbf{u})\geq\frac{1}{q}\varepsilon$. Hence, by
Minkowski’s second theorem (Theorem 2.4), we have
$\hat{\lambda}_{d}(\mathbf{u})\leq\hat{\lambda}_{1}^{-(d-1)}(\mathbf{u})\leq\frac{q^{d-1}}{\varepsilon^{d-1}}$.
Thus,

$T_{0}\leq\frac{d}{d-1}\left(d+\log_{q}\frac{1}{\varepsilon^{d}}\right)=\log_{q}\left(\frac{q}{\varepsilon}\right)^{\frac{d^{2}}{d-1}}\quad\text{and}\quad
T_{N}\geq\frac{d}{d-1}\left(N+1+\log_{q}\frac{1}{\varepsilon}\right).$

######  Corollary 6.13.

Let $d\geq 2$. For every
$0<\varepsilon<\left(\frac{q-1}{q}\right)^{\frac{1}{d-1}}$, we have

$\mathrm{dim}_{H}\mathbf{DI}_{d}(\varepsilon)\geq\frac{d^{2}}{d+1}+\frac{d}{d+1}\log_{q}\left(1+\frac{(q-1)^{2}}{q^{2d+\frac{d^{2}}{d-1}}}\left(\frac{q-1}{q}-\varepsilon^{d-1}\right)\varepsilon^{d}\right).$

###### Proof.

Let $(Q_{\varepsilon,N},\sigma_{\varepsilon,N},B)$ be the triple given in
Definition 6.4, which is a strictly nested self-similar structure covering a
subset of $\mathbf{DI}_{d}(\varepsilon)$ by Proposition 6.7. In order to use
Theorem 3.3, we first verify the condition (1) of Theorem 3.3. Indeed, since
$\hat{\lambda}_{d}(\mathbf{u})\geq 1$, it follows from Lemma 6.10 that (1) of
Theorem 3.3 holds with $c=0$ and
$\rho=q\left(\frac{\varepsilon}{q^{N+1}}\right)^{\frac{2d}{d-1}}$.

Fix small $\delta>0$ and let

$f(\varepsilon)=\log_{q}\left(1+\frac{(1-\delta)(q-1)^{2}}{q^{2d+\frac{d^{2}}{d-1}}}\left(\frac{q-1}{q}-\varepsilon^{d-1}\right)\varepsilon^{d}\right).$

We claim that
$\mathrm{dim}_{H}\mathbf{DI}_{d}(\varepsilon)\geq\frac{d^{2}}{d+1}+\frac{d}{d+1}f(\varepsilon)$.
Since $\delta>0$ is arbitrary, this claim implies the corollary. By
Proposition 6.11 and Remark 6.12 with
$s=\frac{d^{2}}{d+1}+\frac{d}{d+1}f(\varepsilon)$, it follows that for large
enough $N$ and $\mathbf{u}\in Q_{\varepsilon,N}$,

$\begin{split}\sum_{\mathbf{v}\in\sigma_{\varepsilon,N}(\mathbf{u})}&\frac{\operatorname{diam}(B(\mathbf{v}))^{s}}{\operatorname{diam}(B(\mathbf{u}))^{s}}=\sum_{\mathbf{v}\in
F_{N}(\mathbf{u},\varepsilon)}\left(\frac{\lambda_{1}(\mathbf{v})}{|\mathbf{v}|}\right)^{s}\left(\frac{|\mathbf{u}|}{\lambda_{1}(\mathbf{u})}\right)^{s}\\\
&\geq\left(\frac{q-1}{q}-\varepsilon^{d-1}\right)\frac{\varepsilon^{d}(q-1)^{2}}{q^{d+s}}\sum_{T_{0}\leq
k\leq T_{N}}q^{-f(\varepsilon)k}\\\
&\geq\left(\frac{q-1}{q}-\varepsilon^{d-1}\right)\frac{\varepsilon^{d}(q-1)^{2}}{q^{2d}}\frac{q^{-\lceil
T_{0}\rceil f(\varepsilon)}(1-q^{-(\lfloor T_{N}\rfloor-\lceil
T_{0}\rceil)f(\varepsilon)})}{1-q^{-f(\varepsilon)}}\\\
&\geq\left(\frac{q-1}{q}-\varepsilon^{d-1}\right)\frac{\varepsilon^{d}(q-1)^{2}}{q^{2d}}\left(\frac{\varepsilon}{q}\right)^{\frac{d^{2}}{d-1}f(\varepsilon)}\frac{1-\delta}{q^{f(\varepsilon)}-1}.\end{split}$

Now, we claim that
$\left(\frac{\varepsilon}{q}\right)^{\frac{d^{2}}{d-1}f(\varepsilon)}\geq
q^{-\frac{d^{2}}{d-1}}$. Indeed, it is enough to show that
$f(\varepsilon)\log_{q}\frac{q}{\varepsilon}\leq 1$ and since $\log(1+y)\leq
y$ and $\log y\leq y$ for any $y>0$, we have

$f(\varepsilon)\log_{q}\frac{q}{\varepsilon}\leq\frac{(1-\delta)(q-1)^{2}}{(\log
q)^{2}q^{2d-1+\frac{d^{2}}{d-1}}}\left(\frac{q-1}{q}-\varepsilon^{d-1}\right)\varepsilon^{d-1}\leq\frac{(1-\delta)(q-1)^{2}}{(\log
q)^{2}q^{2d-1+\frac{d^{2}}{d-1}}}\left(\frac{q-1}{2q}\right)^{2}\leq 1.$

Therefore, we have

$\sum_{\mathbf{v}\in\sigma_{\varepsilon,N}(\mathbf{u})}\frac{\operatorname{diam}(B(\mathbf{v}))^{s}}{\operatorname{diam}(B(\mathbf{u}))^{s}}\geq\left(\frac{q-1}{q}-\varepsilon^{d-1}\right)\frac{\varepsilon^{d}(q-1)^{2}(1-\delta)}{q^{2d+\frac{d^{2}}{d-1}}}\frac{1}{q^{f(\varepsilon)}-1}=1.$

Since the condition (2) of Theorem 3.3 with $c=0$ holds, we have
$\mathrm{dim}_{H}\mathbf{DI}_{d}(\varepsilon)\geq\frac{d^{2}}{d+1}+\frac{d}{d+1}f(\varepsilon)$.

∎

######  Corollary 6.14.

For $d\geq 2$, $\mathrm{dim}_{H}\mathbf{Sing}_{d}\geq\frac{d^{2}}{d+1}$.

###### Proof.

Consider two real sequences $(\varepsilon_{i})_{i\geq 0}$ and $(N_{i})_{i\geq
0}$ given by

$\varepsilon_{i}=\frac{1}{\log(i+1)}\quad\text{and}\quad N_{i}=i+1.$

Let $\mathbf{u}_{-1}\in Q$ and let $\mathbf{u}_{0}\in
F_{N_{0}}(\mathbf{u}_{-1},\varepsilon_{0})$. Define
$Q_{0}=\\{\mathbf{u}_{0}\\}$ and $Q_{i+1}=\bigcup_{\mathbf{u}\in
Q_{i}}F_{N_{i}}(\mathbf{u},\varepsilon_{i})$ for each $i\geq 0$. For each
$\mathbf{u}\in Q_{i}$, define
$\sigma^{\prime}(\mathbf{u})=F_{N_{i}}(\mathbf{u},\varepsilon_{i})$ and
$B(\mathbf{u})=B\left(\hat{\mathbf{u}},\frac{\lambda_{1}(\mathbf{u})}{q|\mathbf{u}|}\right)$.
Finally, let $Q^{\prime}=\bigcup_{i\geq 0}Q_{i}$ be a disjoint union. By the
same arguments of the proof of Proposition 6.7, the triple
$(Q^{\prime},\sigma^{\prime},B)$ is a strictly nested self-similar structure
covering a subset of $\mathbf{Sing}_{d}$.

By Lemma 6.10, for $\mathbf{u}\in Q_{i}$ and any distinct
$\mathbf{v},\mathbf{w}\in F_{N_{i}}(\mathbf{u},\varepsilon_{i})$ we have

$d\left(B(\mathbf{v}),B(\mathbf{w})\right)\geq\left(\frac{\varepsilon_{i}}{q^{N_{i}+1}\hat{\lambda}_{d}(\mathbf{u})}\right)^{\frac{2d}{d-1}}\frac{\lambda_{1}(\mathbf{u})}{|\mathbf{u}|}\geq
q\left(\frac{\varepsilon_{i}}{q^{N_{i}+1}}\right)^{\frac{2d}{d-1}}\frac{\lambda_{1}(\mathbf{u})}{q|\mathbf{u}|}.$

Hence, we define the map $\rho:Q^{\prime}\to(0,1)$ by

$\rho(\mathbf{u})=q\left(\frac{\varepsilon_{i}}{q^{N_{i}+1}}\right)^{\frac{2d}{d-1}}\quad\text{for
each }\mathbf{u}\in Q_{i}.$

Then the condition (1) of Theorem 3.4 holds with $c=0$ and the above $\rho$.
Moreover, since the map $\rho$ decreases with $i$, the condition (3) of
Theorem 3.4 follows from (6.3).

By applying Proposition 6.11 with $s=\frac{d^{2}}{d+1}$, we obtain that for
each $\mathbf{u}\in Q_{i}$

$\begin{split}\sum_{\mathbf{v}\in\sigma^{\prime}(\mathbf{u})}\frac{(\rho(\mathbf{v})\operatorname{diam}B(\mathbf{v}))^{s}}{(\rho(\mathbf{u})\operatorname{diam}B(\mathbf{u}))^{s}}&=\left(\frac{\varepsilon_{i+1}}{\varepsilon_{i}q^{N_{i+1}-N_{i}}}\right)^{\frac{2d}{d-1}s}\sum_{\mathbf{v}\in
F_{N_{i}}(\mathbf{u},\varepsilon_{i})}\left(\frac{\lambda_{1}(\mathbf{v})}{|\mathbf{v}|}\right)^{s}\left(\frac{|\mathbf{u}|}{\lambda_{1}(\mathbf{u})}\right)^{s}\\\
&\geq\left(\frac{\varepsilon_{i+1}}{\varepsilon_{i}q}\right)^{\frac{2d}{d-1}s}\left(\frac{q-1}{q}-\varepsilon_{i}^{d-1}\right)\frac{\varepsilon_{i}^{d}(q-1)^{2}}{q^{d+s}}\left(\frac{d}{d-1}N_{i}-2\right).\end{split}$

Since $\frac{\varepsilon_{i+1}}{\varepsilon_{i}}\to 1$ and
$\varepsilon_{i}^{d}N_{i}\rightarrow\infty$ as $i\to\infty$, the last line of
the above equation diverges to infinity as $i\to\infty$. Hence condition (2)
of Theorem 3.4 holds for all large enough $i$. Therefore, by Theorem 3.4, we
have $\mathrm{dim}_{H}\mathbf{Sing}_{d}\geq\frac{d^{2}}{d+1}$. ∎

###  6.4. Proof of Proposition 6.11

By Lemma 6.5, for every $\mathbf{v}\in F_{N}(\mathbf{u},\varepsilon)$,
$\frac{1}{q}\varepsilon\leq\hat{\lambda}_{1}(\mathbf{v})\leq\varepsilon$, that
is,
$\frac{1}{q}\varepsilon|\mathbf{v}|^{-\frac{1}{d}}\leq\lambda_{1}(\mathbf{v})\leq\varepsilon|\mathbf{v}|^{-\frac{1}{d}}$.
Similarly, since $\mathbf{u}\in Q_{\varepsilon,N}$, we have
$\frac{1}{q}\varepsilon|\mathbf{u}|^{-\frac{1}{d}}\leq\lambda_{1}(\mathbf{u})\leq\varepsilon|\mathbf{u}|^{-\frac{1}{d}}$.
Hence, for any $s>0$, we have

$\sum_{\mathbf{v}\in
F_{N}(\mathbf{u},\varepsilon)}\left(\frac{\lambda_{1}(\mathbf{v})}{|\mathbf{v}|}\right)^{s}\left(\frac{|\mathbf{u}|}{\lambda_{1}(\mathbf{u})}\right)^{s}\geq\frac{1}{q^{s}}\sum_{\mathbf{v}\in
F_{N}(\mathbf{u},\varepsilon)}\left(\frac{|\mathbf{u}|}{|\mathbf{v}|}\right)^{\frac{d+1}{d}s}.$
(6.6)

As in the proofs of Propositions 5.3 and 5.4, we will divide
$F_{N}(\mathbf{u},\varepsilon)$ by the following subsets

$F_{N,k}(\mathbf{u},\varepsilon)=\\{\mathbf{v}\in
F_{N}(\mathbf{u},\varepsilon):|\mathbf{v}|=q^{k}|\mathbf{u}|\\}$

for each $k\geq 1$. It is possible since $|\mathbf{v}|>|\mathbf{u}|$ for any
$\mathbf{v}\in F_{N}(\mathbf{u},\varepsilon)$ by Lemma 6.5. By definition, it
follows that

$F_{N,k}(\mathbf{u},\varepsilon)=\bigcup_{\alpha\in\Lambda_{\mathbf{u}}(\varepsilon)\cap
C_{N}(\mathbf{u})}\zeta_{k}(\mathbf{u},\alpha,\varepsilon),$

where

$\zeta_{k}(\mathbf{u},\alpha,\varepsilon)=\Bigg{\\{}\mathbf{v}\in
Q:|\mathbf{v}|=q^{k}|\mathbf{u}|,\ \pi_{\mathbf{u}}(\mathbf{v})=\alpha,\
\left(\frac{|\mathbf{u}\wedge\mathbf{v}|}{\varepsilon}\right)^{\frac{d}{d-1}}\leq|\mathbf{v}|\leq\left(\frac{q|\mathbf{u}\wedge\mathbf{v}|}{\varepsilon}\right)^{\frac{d}{d-1}}\Bigg{\\}}.$

If $\mathbf{v}\in\zeta_{k}(\mathbf{u},\alpha,\varepsilon)$, then

$\left(\frac{\|\alpha\||\mathbf{u}|}{\varepsilon}\right)^{\frac{d}{d-1}}\leq|\mathbf{v}|=q^{k}|\mathbf{u}|\leq\left(\frac{q\|\alpha\||\mathbf{u}|}{\varepsilon}\right)^{\frac{d}{d-1}},$

which is equivalent to

$\varepsilon
q^{k\frac{d-1}{d}-1}|\mathbf{u}|^{-\frac{1}{d}}\leq\|\alpha\|\leq\varepsilon
q^{k\frac{d-1}{d}}|\mathbf{u}|^{-\frac{1}{d}}.$

Hence, we have

$F_{N,k}(\mathbf{u},\varepsilon)=\bigcup_{\begin{subarray}{c}\alpha\in\Lambda_{\mathbf{u}}(\varepsilon)\cap
C_{N}(\mathbf{u}),\\\ \varepsilon
q^{k\frac{d-1}{d}-1}|\mathbf{u}|^{-\frac{1}{d}}\leq\|\alpha\|\leq\varepsilon
q^{k\frac{d-1}{d}}|\mathbf{u}|^{-\frac{1}{d}}\end{subarray}}\left\\{\mathbf{v}\in\pi_{\mathbf{u}}^{-1}(\alpha)\cap
Q:|\mathbf{v}|=q^{k}|\mathbf{u}|\right\\}.$ (6.7)

Since $\|\alpha\|\leq q^{N}\lambda_{d}(\mathbf{u})$, if
$k>T_{N}=\frac{d}{d-1}\left(N+1+\log_{q}\frac{\hat{\lambda}_{d}(\mathbf{u})}{\varepsilon}\right)$,
then there is no such $\alpha$ as in (6.7). Since
$\alpha\in\Lambda_{\mathbf{u}}(\varepsilon)$ is primitive, by Lemma 4.13,
$\\#\\{\mathbf{v}\in\pi_{\mathbf{u}}^{-1}(\alpha)\cap
Q:|\mathbf{v}|=q^{k}|\mathbf{u}|\\}=(q-1)q^{k}$. Therefore, we have

$\sum_{\mathbf{v}\in
F_{N}(\mathbf{u},\varepsilon)}\left(\frac{|\mathbf{u}|}{|\mathbf{v}|}\right)^{\frac{d+1}{d}s}=\sum_{1\leq
k\leq
T_{N}}\frac{q-1}{q^{k\left(\frac{d+1}{d}s-1\right)}}\\#\left\\{\alpha\in\Lambda_{\mathbf{u}}(\varepsilon)\cap
C_{N}(\mathbf{u}):\frac{\varepsilon
q^{k\frac{d-1}{d}-1}}{|\mathbf{u}|^{\frac{1}{d}}}\leq\|\alpha\|\leq\frac{\varepsilon
q^{k\frac{d-1}{d}}}{|\mathbf{u}|^{\frac{1}{d}}}\right\\}$ (6.8)

We will prove Proposition 6.11 assuming the following proposition.

######  Proposition 6.15.

For $k\geq
T_{0}=\frac{d}{d-1}\left(1+\log_{q}\frac{\hat{\lambda}_{d}(\mathbf{u})}{\varepsilon}\right)$,
we have

$\\#\Bigg{\\{}\alpha\in\Lambda_{\mathbf{u}}(\varepsilon)\cap
C_{N}(\mathbf{u}):\frac{\varepsilon
q^{k\frac{d-1}{d}-1}}{|\mathbf{u}|^{\frac{1}{d}}}\leq\|\alpha\|\leq\frac{\varepsilon
q^{\frac{d-1}{d}k}}{|\mathbf{u}|^{\frac{1}{d}}}\Bigg{\\}}\geq\left(\frac{q-1}{q}-\varepsilon^{d-1}\right)\frac{\varepsilon^{d}(q-1)}{q^{d}}q^{(d-1)k}.$

###### Proof of Proposition 6.11 assuming Proposition 6.15.

Combining with (6.6), (6.8), and Proposition 6.15, we have

$\sum_{\mathbf{v}\in
F_{N}(\mathbf{u},\varepsilon)}\left(\frac{\lambda_{1}(\mathbf{v})}{|\mathbf{v}|}\right)^{s}\left(\frac{|\mathbf{u}|}{\lambda_{1}(\mathbf{u})}\right)^{s}\geq\left(\frac{q-1}{q}-\varepsilon^{d-1}\right)\frac{\varepsilon^{d}(q-1)^{2}}{q^{d+s}}\sum_{T_{0}\leq
k\leq T_{N}}q^{-(\frac{d+1}{d}s-d)k}.$

∎

In order to prove Proposition 6.15, we follow the strategy of [CC16, Section
8]. Let $\mathbf{u}\in Q$ and $\varepsilon>0$ be given. For any nonzero
$n\in\mathcal{R}$, we will first give an lower bound of
$\\#(\Lambda_{\mathbf{u}}(\varepsilon)\cap C_{n}^{\prime}(\mathbf{u}))$. By
definition, we have

$\Lambda_{\mathbf{u}}(\varepsilon)\cap
C_{n}^{\prime}(\mathbf{u})=\left\\{\alpha=\sum_{i<d}m_{i}\xi_{i}+n\xi_{d}:\begin{matrix}\forall
i,\ m_{i}\in\mathcal{R}\text{ and
}|m_{i}|\leq\frac{|n|\lambda_{d}(\mathbf{u})}{\lambda_{i}(\mathbf{u})},\\\
\alpha\text{ is primitive},\ \text{and
}\hat{\lambda}_{1}(\alpha)>\varepsilon\end{matrix}\right\\}.$

Since
$\det(\Lambda_{\alpha})=\frac{1}{|\mathbf{u}||n|\lambda_{d}(\mathbf{u})}$ by
Lemma 6.2, observe that

$\hat{\lambda}_{1}(\alpha)>\varepsilon\iff\lambda_{1}(\alpha)>\varepsilon(|\mathbf{u}||n|\lambda_{d}(\mathbf{u}))^{-\frac{1}{d-1}}\iff\lambda_{1}(n\Lambda_{\alpha})>\varepsilon|n|(|\mathbf{u}||n|\lambda_{d}(\mathbf{u}))^{-\frac{1}{d-1}}.$

Hence we consider the following lattice: for given
$\mathbf{m}=(m_{1},\dots,m_{d-1})\in\mathcal{R}^{d-1}$ and
$n\in\mathcal{R}\setminus\\{0\\}$, we denote

$L(\mathbf{m},n)=\mathcal{R}n\xi_{1}+\cdots+\mathcal{R}n\xi_{d-1}+\mathcal{R}\sum_{i<d}m_{i}\xi_{i}.$

Since $\alpha=\sum_{i<d}m_{i}\xi_{i}+n\xi_{d}$ is primitive if and only if
$\mathrm{gcd}(\mathbf{m},n)=1$, we have

$\Lambda_{\mathbf{u}}(\varepsilon)\cap
C_{n}^{\prime}(\mathbf{u})=\left\\{\sum_{i<d}m_{i}\xi_{i}+n\xi_{d}:\begin{matrix}\forall
i,\ m_{i}\in\mathcal{R}\text{ and
}|m_{i}|\leq\frac{|n|\lambda_{d}(\mathbf{u})}{\lambda_{i}(\mathbf{u})},\
\mathrm{gcd}(\mathbf{m},n)=1,\\\
\lambda_{1}(L(\mathbf{m},n))>\varepsilon|n|(|\mathbf{u}||n|\lambda_{d}(\mathbf{u}))^{-\frac{1}{d-1}}\end{matrix}\right\\}.$
(6.9)

Therefore, we can estimate $\\#(\Lambda_{\mathbf{u}}(\varepsilon)\cap
C_{n}^{\prime}(\mathbf{u}))$ by counting $\mathbf{m}\in\mathcal{R}^{d-1}$
satisfying the above conditions in (6.9). Consider the set $X_{n}$ given by

$X_{n}=\left\\{\mathbf{m}\in\mathcal{R}^{d-1}:\begin{matrix}\|\mathbf{m}\|<|n|,\
\mathrm{gcd}(\mathbf{m},n)=1,\\\
\lambda_{1}(L(\mathbf{m},n))>\varepsilon|n|(|\mathbf{u}||n|\lambda_{d}(\mathbf{u}))^{-\frac{1}{d-1}}\end{matrix}\right\\}.$

For any $\mathbf{m}\in X_{n}$ and $\mathbf{k}\in\mathcal{R}^{d-1}$, observe
that $\mathrm{gcd}(\mathbf{m}+n\mathbf{k},n)=1$,
$L(\mathbf{m}+n\mathbf{k},n)=L(\mathbf{m},n)$, and
$|m_{i}+nk_{i}|\leq\frac{|n|\lambda_{d}(\mathbf{u})}{\lambda_{i}(\mathbf{u})}$
if and only if
$|k_{i}|\leq\frac{\lambda_{d}(\mathbf{u})}{\lambda_{i}(\mathbf{u})}$ for any
$i$. Since all translates of the form $X_{n}+n\mathbf{k}$ with
$\mathbf{k}\in\mathcal{R}^{d-1}$ are disjoint, it follows from (6.9) that

$\begin{split}\\#(\Lambda_{\mathbf{u}}(\varepsilon)\cap
C_{n}^{\prime}(\mathbf{u}))&\geq\\#X_{n}\times\\#\Bigg{\\{}\mathbf{k}\in\mathcal{R}^{d-1}:\forall
i,\
|k_{i}|\leq\frac{\lambda_{d}(\mathbf{u})}{\lambda_{i}(\mathbf{u})}\Bigg{\\}}\\\
&=\\#X_{n}\times\frac{\lambda_{d}^{d-1}(\mathbf{u})}{\lambda_{1}(\mathbf{u})\cdots\lambda_{d-1}(\mathbf{u})}\end{split}$
(6.10)

Hence, it suffices to compute $\\#X_{n}$. For this, we will follow the proof
of [CC16, Lemma 8.4]. Recall the functions $\phi$ and $D_{1}$ defined in
section 2.1.

######  Proposition 6.16.

For any nonzero $n\in\mathcal{R}$, we have

$\\#X_{n}\geq\phi(n)\left(|n|^{d-2}-\varepsilon^{d-1}D_{1}(n)|n|^{d-3}\right).$

###### Proof.

For simplicity, let us denote
$\eta=\varepsilon|n|\left(|\mathbf{u}||n|\lambda_{d}(\mathbf{u})\right)^{-\frac{1}{d-1}}$.
First, note that

$\lambda_{1}(L(\mathbf{m},n))\leq\eta\iff\widetilde{L}(\mathbf{m},n)\cap\prod_{i=1}^{d-1}B\left(0,\frac{\eta}{\lambda_{i}(\mathbf{u})}\right)\neq\\{0\\},$

where
$\widetilde{L}(\mathbf{m},n)=\mathcal{R}n\mathbf{e}_{1}+\cdots+\mathcal{R}n\mathbf{e}_{d-1}+\mathcal{R}\sum_{i<d}m_{i}\mathbf{e}_{i}$.
Consider the following sets:

$\begin{split}Y_{n}&=\\{\mathbf{m}\in\mathcal{R}^{d-1}:\|\mathbf{m}\|<|n|\text{
and }\mathrm{gcd}(m_{1},n)=1\\};\\\ Z_{n}&=\left\\{\mathbf{m}\in
Y_{n}:\widetilde{L}(\mathbf{m},n)\cap\prod_{i=1}^{d-1}B\left(0,\frac{\eta}{\lambda_{i}(\mathbf{u})}\right)\neq\\{0\\}\right\\}.\end{split}$

Since $X_{n}\supset Y_{n}\setminus Z_{n}$ and $\\#Y_{n}=\phi(n)|n|^{d-2}$, it
is enough to show that
$\\#Z_{n}\leq\varepsilon^{d-1}\phi(n)D_{1}(n)|n|^{d-3}$.

Note that the lattice $\widetilde{L}(\mathbf{m},n)$ depends only on
$m_{1},\dots,m_{d-1}$ modulo $n$ and $\\{r\in\mathcal{R}:|r|<|n|\\}$ is a
complete set of representatives for $\mathcal{R}/n\mathcal{R}$. Hence, by
considering
$B=\prod_{i=1}^{d-1}B\left(0,\frac{\eta}{\lambda_{i}(\mathbf{u})}\right)$ as
the box in $(\mathcal{R}/n\mathcal{R})^{d-1}$, we have

$Z_{n}=\\{\mathbf{m}\in(\mathcal{R}/n\mathcal{R})^{d-1}:\mathrm{gcd}(m_{1},n)=1,\mathcal{R}\mathbf{m}\cap
B\neq\\{0\\}\\}.$

For $m\in\mathcal{R}/n\mathcal{R}$ with $\mathrm{gcd}(m,n)=1$, let

$Z_{n}(m)=\\{\mathbf{r}\in(\mathcal{R}/n\mathcal{R})^{d-2}:\mathcal{R}(m,\mathbf{r})\cap
B\neq\\{0\\}\\}.$

It follows from $\mathrm{gcd}(m,n)=1$, or equivalently,
$m\in(\mathcal{R}/n\mathcal{R})^{*}$ that $Z_{n}(m)=mZ_{n}(1)$. Therefore,
$\\#Z_{n}=\phi(n)\\#Z_{n}(1)$, and we need to show that
$\\#Z_{n}(1)\leq\varepsilon^{d-1}D_{1}(n)|n|^{d-3}$.

Since $\mathbf{r}\in Z_{n}(1)$ if and only if there exists $(a,\mathbf{b})\in
B\setminus\\{0\\}$ such that $\mathbf{b}=a\mathbf{r}$, we have

$\\#Z_{n}(1)\leq\sum_{(a,\mathbf{b})\in B\setminus\\{0\\}}\\#E(a,\mathbf{b}),$
(6.11)

where
$E(a,\mathbf{b})=\\{\mathbf{r}\in(\mathcal{R}/n\mathcal{R})^{d-2}:\mathbf{b}=a\mathbf{r}\\}$.
For $(a,\mathbf{b})\in B\setminus\\{0\\}$, if $a=0$ and $\mathbf{b}\neq 0$,
then $E(a,\mathbf{b})$ is empty. So, we may assume that $a\neq 0$. If
$\mathbf{b}\in(\mathcal{R}a)^{d-2}$, then there is
$\mathbf{r}_{0}\in\mathcal{R}^{d-2}$ such that $\mathbf{b}=a\mathbf{r}_{0}$,
and the general solutions to the equation $\mathbf{b}=a\mathbf{r}$ is given by

$\mathbf{r}=\frac{n}{\mathrm{gcd}(a,n)}\mathbf{w}+\mathbf{r_{0}}\quad\text{for
}\mathbf{w}\in\mathcal{R}^{d-2}.$

Therefore, the number of solutions to the equation $\mathbf{b}=a\mathbf{r}$ in
$(\mathcal{R}/n\mathcal{R})^{d-2}$ is $|\mathrm{gcd}(a,n)|^{d-2}$ if
$\mathbf{b}\in(\mathcal{R}a)^{d-2}$, and $0$ otherwise. It follows from (6.11)
that

$\begin{split}\\#Z_{n}(1)&\leq\sum_{\begin{subarray}{c}(a,\mathbf{b})\in
B\setminus\\{0\\},\\\
\mathbf{b}\in(\mathcal{R}a)^{d-2}\end{subarray}}|\mathrm{gcd}(a,n)|^{d-2}\leq\sum_{g|n}|g|^{d-2}\sum_{\begin{subarray}{c}a\in
B(0,\frac{\eta}{\lambda_{1}(\mathbf{u})})\\\
g|a\end{subarray}}\\#\left(\prod_{i=2}^{d-1}B(0,\frac{\eta}{\lambda_{i}(\mathbf{u})})\cap(\mathcal{R}a)^{d-2}\right)\\\
&\leq\sum_{g|n}|g|^{d-2}\sum_{\begin{subarray}{c}a\in
B(0,\frac{\eta}{\lambda_{1}(\mathbf{u})})\\\
g|a\end{subarray}}\prod_{i=2}^{d-1}\frac{\eta}{|g|\lambda_{i}(\mathbf{u})}=\sum_{g|n}|g|^{d-2}\prod_{i=1}^{d-1}\frac{\eta}{|g|\lambda_{i}(\mathbf{u})}\\\
&=\varepsilon^{d-1}|n|^{d-2}\sum_{g|n}\frac{1}{|g|}=\varepsilon^{d-1}|n|^{d-2}\sum_{g|n}|g/n|=\varepsilon^{d-1}D_{1}(n)|n|^{d-3}.\end{split}$

This completes the proof. ∎

Combining with Lemmas 2.1, 2.2, and Proposition 6.16, we have

######  Lemma 6.17.

For every $\ell\in\mathbb{N}$, we have

$\sum_{\mathrm{deg}(n)=\ell}\\#X_{n}\geq q^{\ell
d}(q-1)\left(\frac{q-1}{q}-\varepsilon^{d-1}\right).$

###### Proof.

It follows from Lemmas 2.1, 2.2, and Proposition 6.16 that

$\begin{split}\sum_{\mathrm{deg}(n)=\ell}\\#X_{n}&\geq\sum_{\mathrm{deg}(n)=\ell}\phi(n)\left(|n|^{d-2}-\varepsilon^{d-1}D_{1}(n)|n|^{d-3}\right)\\\
&=q^{\ell d}(q-1)\left(\frac{q-1}{q}-\varepsilon^{d-1}\right).\end{split}$

∎

Now we are ready to prove Proposition 6.15.

###### Proof of Proposition 6.15.

For $k\geq 1$, we have

$\displaystyle\\#\left\\{\alpha\in\Lambda_{\mathbf{u}}(\varepsilon)\cap
C_{N}(\mathbf{u}):\frac{\varepsilon
q^{k\frac{d-1}{d}-1}}{|\mathbf{u}|^{\frac{1}{d}}}\leq\|\alpha\|\leq\frac{\varepsilon
q^{k\frac{d-1}{d}}}{|\mathbf{u}|^{\frac{1}{d}}}\right\\}\quad\quad$
$\displaystyle=\sum_{\frac{\varepsilon
q^{k\frac{d-1}{d}-1}}{|\mathbf{u}|^{\frac{1}{d}}\lambda_{d}(\mathbf{u})}\leq|n|\leq\frac{\varepsilon
q^{k\frac{d-1}{d}}}{|\mathbf{u}|^{\frac{1}{d}}\lambda_{d}(\mathbf{u})}}\\#(\Lambda_{\mathbf{u}}(\varepsilon)\cap
C_{n}^{\prime}(\mathbf{u}))\quad\quad$ by Definition 6.4
$\displaystyle\geq\frac{\lambda_{d}^{d-1}(\mathbf{u})}{\lambda_{1}(\mathbf{u})\cdots\lambda_{d-1}(\mathbf{u})}(q-1)\left(\frac{q-1}{q}-\varepsilon^{d-1}\right)\sum_{\ell\in
I_{k}\cap\mathbb{Z}_{\geq 0}}q^{\ell d}\quad\quad$ $\displaystyle\text{by
\eqref{Eq_counting_trans} and Lemma \ref{lem_XnSum}},$

where $I_{k}$ is the closed interval given by

$I_{k}=\left[k\frac{d-1}{d}-1+\log_{q}\frac{\varepsilon}{\hat{\lambda}_{d}(\mathbf{u})},k\frac{d-1}{d}+\log_{q}\frac{\varepsilon}{\hat{\lambda}_{d}(\mathbf{u})}\right].$

Note that for each $k\geq 1$ the interval $I_{k}$ is of length $1$. Hence,
$I_{k}$ must contain at least one integer point. If
$k\geq\frac{d}{d-1}\left(1+\log_{q}\frac{\hat{\lambda}_{d}(\mathbf{u})}{\varepsilon}\right)$,
then $I_{k}\cap\mathbb{Z}_{\geq 0}$ is non-empty. Hence, we have

$\sum_{\ell\in I_{k}\cap\mathbb{Z}_{\geq 0}}q^{\ell d}\geq
q^{(d-1)k-d}\left(\frac{\varepsilon}{\hat{\lambda}_{d}(\mathbf{u})}\right)^{d}=q^{(d-1)k}\frac{\varepsilon^{d}}{q^{d}}\frac{\lambda_{1}(\mathbf{u})\cdots\lambda_{d-1}(\mathbf{u})}{\lambda_{d}^{d-1}(\mathbf{u})}.$

This completes the proof.

∎

## References

  * [Ara23] Noy Soffer Aranov, _Geometry of numbers in positive characteristic_ , 2023.
  * [Bak77] R. C. Baker, _Singular $n$-tuples and hausdorff dimension i_, Math. Proc. Cambridge Philos.Soc. 81 (1977), 377–385.
  * [Bak92] R.C. Baker, _Singular n-tuples and hausdorff dimension, ii_ , Math. Proc. Cambridge Philos. Soc. 111 (1992), 577–584.
  * [BK23] Christian Bagshaw and Bryce Kerr, _Lattices in Function Fields and Applications_ , 2023.
  * [BKL24] Gukyeong Bang, Taehyeong Kim, and Seonhee Lim, _Singular linear forms over global function fields_ , 2024.
  * [BZ19] Yann Bugeaud and Zhenliang Zhang, _On homogeneous and inhomogeneous diophantine approximation over the fields of formal power series_ , Pacific Journal of Mathematics 302(2) (2019), 453–480.
  * [CC16] Yitwah Cheung and Nicolas Chevallier, _Hausdorff dimension of singular vectors_ , Duke Mathematical Journal 165 (2016), 2273–2329.
  * [Che07] Yitwah Cheung, _Hausdorff dimension of the set of points on divergent trajectories of a homogeneous flow on a product space_ , Ergodic Theory and Dynamical Systems 27(1) (2007), 65–85.
  * [Che11] by same author, _Hausdorff dimension of the set of singular pairs_ , Annals of Mathematics 173 (2011), no. 1, 127–167.
  * [Dan85] S. G. Dani, _Divergent trajectories of flows on homogeneous spaces and diophantine approximations_ , J. Reine Angew. Math. 359 (1985), 55–89.
  * [DFSU24] Tushar Das, Lior Fishman, David Simmons, and Mariusz Urbański, _A variational principle in the parametric geometry of numbers_ , Advances in Mathematics 437 (2024), 109435.
  * [dM70] Bernard de Mathan, _Approximations diophantiennes dans un corps local_ , Mémoires de la Société Mathématique de France 21 (1970).
  * [DS70] Harold Davenport and Wolfgang Schmidt, _Dirichlet’s theorem on diophantine approximation_ , Symposia Mathematica IV (1970), 113–132.
  * [GG17] Arijit Ganguly and Anish Ghosh, _Dirichlet’s Theorem in Function Fields_ , Canadian Journal of Mathematics 69 (2017), no. 3, 532–547.
  * [Khi26] Aleksandr Khintchine, _Über eine klasse linearer diophantische approximationen_ , Rend. Circ.Mat.Palermo 50 (1926), 170–195.
  * [KKLM17] Shirali Kadyrov, Dmitry Kleinbock, Elon Lindenstrauss, and Gregory Margulis, _Singular systems of linear forms and non-escape of mass in the space of lattices_ , J. Anal. Math. 133 (2017), 253–277.
  * [KLP23] Taehyeong Kim, Seonhee Lim, and Frédéric Paulin, _On hausdorff dimension in inhomogeneous diophantine approximation over global function fields_ , Journal of Number Theory 251 (2023), 102–146.
  * [KM23] Dmitry Kleinbock and Shahriar Mirzadeh, _On the dimension drop conjecture for diagonal flows on the space of lattices_ , Advances in Mathematics 425 (2023), 109058.
  * [KST16] Dmitry Kleinbock, Ronggang Shi, and Georges Tomanov, _S-adic Version of Minkowski’s Geometry of Numbers and Mahler’s Compactness Criterion_ , Journal of Number Theory 174 (2016), 150–163.
  * [KSY22] Dmitry Kleinbock, Andreas Strömbergsson, and Shucheng Yu, _A measure estimate in geometry of numbers and improvements to dirichlet’s theorem_ , Proceedings of the London Mathematical Society 125 (2022), no. 4, 778–824.
  * [KW05] Dmitry Kleinbock and Barak Weiss, _Friendly measures, homogeneous flows, and singular vectors in algebraic and topological dynamics_ , Contemp. Math. 385 (2005), 281–292.
  * [Las00] A. Lasjaunias, _A survey of diophantine approximation in fields of power series_ , Monatshefte fur Mathematik 130 (2000), no. 3, 211 – 229, Cited by: 67.
  * [Mah41] Kurt Mahler, _An Analogue to Minkowski’s Geometry of Numbers in a Field of Series_ , Annals of Mathematics 42 (1941), no. 2, 488–522.
  * [Ros13] Michael Rosen, _Number theroy in function fields_ , Springer Verlag, 2013.
  * [RW17] Damien Roy and Michel Waldschmidt, _Parametric Geometry of Numbers in Function Fields_ , arXiv: Number Theory (2017), 1–20.
  * [Ryn90] B. P. Rynne, _A lower bound for the hausdorff dimension of sets of singular $n$-tuples_, Math. Proc. Cambridge Philos. Soc. 107 (1990), 387–394.
  * [ST23] Omri N. Solan and Nattalie Tamam, _On topologically big divergent trajectories_ , Duke Math. J. 172(18) (2023), 3429 – 3474.
  * [Tam19] Nattalie Tamam, _Existence of non-obvious divergent trajectories in homogeneous spaces_ , Israel Journal of Mathematics 247 (2019), 459–478.
  * [Wei04] Barak Weiss, _Divergent trajectories on noncompact parameter spaces_ , Geom. Funct. Anal. 14 (2004), 94–149.
  * [Wei06] by same author, _Divergent trajectories and q-rank_ , Israel J. Math. 152 (2006), 221–227.