# Construction of subspaces with known Diophantine exponents for the last
angle

Gaétan GUILLOT

###### Abstract

Schmidt generalized in 1967 the theory of classical Diophantine approximation
to subspaces of $\mathbb{R}^{n}$. We consider Diophantine exponents for linear
subspaces of $\mathbb{R}^{n}$ which generalize the irrationality measure for
real numbers. Using geometry of numbers, we construct subspaces of
$\mathbb{R}^{n}$ for which we are able to compute the associated exponents for
the last angle.

##  1 Introduction

Classical Diophantine approximation deals with how closely real numbers (or
points in $\mathbb{R}^{n}$) can be approximated by rational numbers (or
rational points). In 1967, Schmidt [9] proposed a broader version of this
problem, focusing on the approximation of subspaces of $\mathbb{R}^{n}$ by
rational subspaces. In this context, we will briefly outline the key concepts
needed for this study, according to the definitions and notation from [9],
[4], [5], and [6]. The result presented in this article comes from the Ph.D
thesis of the author, see [2, Chapter 9].

A subspace of $\mathbb{R}^{n}$ is rational if it admits a basis of vectors in
$\mathbb{Q}^{n}$. The set of all rational subspaces of dimension $e$ in
$\mathbb{R}^{n}$ is denoted by $\mathcal{R}_{n}(e)$. To such a rational
subspace $B$, we can associate a point
$\eta=(\eta_{1},\ldots,\eta_{N})\in\mathbb{P}^{N}(\mathbb{R})$ with
$N=\binom{n}{e}$ known as the Grassmann (or Plücker) coordinates of $B$. We
can select a representative vector $\eta$ with coprime integer coordinates and
we define $H(B)=\|\eta\|$ where $\|\cdot\|$ denotes the Euclidean norm on
$\mathbb{R}^{N}$. Note that if $X_{1},\ldots,X_{e}$ is a $\mathbb{Z}$-basis of
$B\cap\mathbb{Z}^{n}$, then $H(B)=\|X_{1}\wedge\ldots\wedge X_{e}\|$. Further
details on the height can be found in [9] and [10]. Let us fix
$n\in\mathbb{N}\setminus\\{0\\}$. For $d,e\in\llbracket 1,n\rrbracket^{2}$
such that $d+e\leq n$ and $j\in\llbracket 1,\min(d,e)\rrbracket$ we say that a
subspace $A$ of dimension $d$ of $\mathbb{R}^{n}$ is $(e,j)\text{-}$irrational
if $\forall\,B\in\mathcal{R}_{n}(e),\quad\dim(A\cap B)<j.$ We denote by
$\mathcal{I}_{n}(d,e)_{j}$ the set of all $(e,j)\text{-}$irrational subspaces
$A$ of dimension $d$ of $\mathbb{R}^{n}$.

We now introduce the concept of proximity between two subspaces. Let
$X,Y\in\mathbb{R}^{n}\setminus\\{0\\}$, and define

$\omega(X,Y)=\frac{\|X\wedge Y\|}{\|X\|\cdot\|Y\|},$

where $X\wedge Y$ represents the exterior product of $X$ and $Y$.
Geometrically, $\omega(X,Y)$ denotes the absolute value of the sine of the
angle between $X$ and $Y$. Consider two subspaces $A$ and $B$ of
$\mathbb{R}^{n}$ with dimensions $d$ and $e$ respectively. As in [9], we
construct by induction $t=\min(d,e)$ angles between $A$ and $B$. Let us define

$\psi_{1}(A,B)=\min\limits_{\begin{subarray}{c}X\in A\setminus\\{0\\}\\\ Y\in
B\setminus\\{0\\}\end{subarray}}\omega(X,Y),$

and choose $(X_{1},Y_{1})\in A\times B$ be such that
$\omega(X_{1},Y_{1})=\psi_{1}(A,B)$. Assume that
$\psi_{1}(A,B),\ldots,\psi_{j}(A,B)$ and $(X_{1},Y_{1}),\ldots,(X_{j},Y_{j})$
have been constructed for $j\in\llbracket 1,t-1\rrbracket$. Let $A_{j}$ and
$B_{j}$ be respectively the orthogonal complements of
$\text{Span}(X_{1},\ldots,X_{j})$ in $A$ and $\text{Span}(Y_{1},\ldots,Y_{j})$
in $B$.

We define

$\psi_{j+1}(A,B)=\min\limits_{\begin{subarray}{c}X\in A_{j}\setminus\\{0\\}\\\
Y\in B_{j}\setminus\\{0\\}\end{subarray}}\omega(X,Y),$

and let $(X_{j+1},Y_{j+1})\in A\times B$ such that
$\omega(X_{j+1},Y_{j+1})=\psi_{j+1}(A,B)$.

We now have all the tools to define the Diophantine exponents studied in this
paper.

######  Definition 1.1.

Let $(d,e)\in\llbracket 1,n-1\rrbracket^{2}$ be such that $d+e\leq n$ and
$j\in\llbracket 1,\min(d,e)\rrbracket$, and $A\in\mathcal{I}_{n}(d,e)_{j}$. We
define $\mu_{n}(A|e)_{j}$ as the supremum of the set of all $\mu>0$ such that
there exist infinitely many $B\in\mathcal{R}_{n}(e)$ such that

$\psi_{j}(A,B)\leq H(B)^{-\mu}.$

The angle corresponding with $j=\min(d,e)$ is the most natural to study.
Indeed, $\psi_{\min(d,e)}$ is "almost" a distance in the sense:

$\displaystyle\psi_{\min(d,e)}(A,B)=0\Longleftrightarrow A\subset B\text{ or
}B\subset A.$

In particular, if $d=e$, then $\psi_{d}$ is a distance on the Grassmannian of
$d$-dimensional vector subspaces of $\mathbb{R}^{n}$. In [8], de Saxcé
describes the image of $\mathcal{I}_{n}(d,e)_{\min(d,e)}$ by
$\mu_{n}(\cdot|e)_{\min(d,e)}$. In this article we study the joint spectrum
associated to the last angle, that is to say the image of the tuple of
functions
$(\mu_{n}(\cdot|1)_{\min(d,1)},\ldots,\mu_{n}(\cdot|n-d)_{\min(d,n-d)})$ on
$\bigcap\limits_{e=1}^{n-d}\mathcal{I}_{n}(d,e)_{\min(d,e)}$. We prove here
the following theorem.

######  Theorem 1.2.

Let $(d,q)\in(\mathbb{N}\setminus\\{0\\})^{2}$. We set $n=(q+1)d$. There
exists an explicit constant $C_{d}$ such that for any $\alpha\geq C_{d}$,
there exists a subspace $A$ of $\mathbb{R}^{n}$ of dimension $d$ such that
$A\in\bigcap\limits_{e=1}^{n-d}\mathcal{I}_{n}(d,e)_{\min(d,e)}$ and

$\displaystyle\forall\,e\in\llbracket d,n-d\rrbracket,\quad$
$\displaystyle\mu_{n}(A|e)_{d}=\frac{\alpha^{q_{e}+1}}{r_{e}+(d-r_{e})\alpha},$
$\displaystyle\forall\,e\in\llbracket 1,d-1\rrbracket,\quad$
$\displaystyle\mu_{n}(A|e)_{e}=\frac{\alpha}{r_{e}}=\frac{\alpha}{e},$

where $q_{e}$ and $r_{e}$ are the quotient and the remainder of the Euclidean
division of $e$ by $d$.

In the case $d=1$, the image of
$(\mu_{n}(\cdot|1)_{1},\ldots,\mu_{n}(\cdot|n-1)_{1})$ has been determined by
Roy [7]. In the case $d>1$, the determination of the joint spectrum remains an
open problem. In [3], assuming that $d$ divides $n$, the author provides
examples of points in this image (see Theorem 1.8) and, in particular,
identifies a non-empty open subset within it (see the proof of Theorem 1.5).
While in [3], one explicitly identifies the rational subspaces of best
approximations of a subspace $A$ to compute $\mu_{n}(A|e)_{\min(d,e)}$, here
we employ tools of geometry of numbers (mainly Minkowski’s theorem on convex
bodies) to find specific vectors with integer coordinates in the best
approximations. Theorem 1.2 provides us with values taken by the tuple of
functions
$(\mu_{n}(\cdot|1)_{\min(d,1)},\ldots,\mu_{n}(\cdot|n-d)_{\min(d,n-d)})$ which
we are able to compute using the method developed here.

The definition of $C_{d}$ and the construction of the subspace $A$ are carried
out in Section 2. Section 3 is devoted to the study of rational vectors and
rational subspaces that shall give good approximations of $A$. To prove
Theorem 1.2, we consider two cases depending on whether $e<d$ or $e\geq d$.
In both cases, the lower bound of the exponent $\mu_{n}(A|e)_{\min(d,e)}$ is
shown by presenting a family of rational subspaces that approximate $A$ well
(see Lemmas 4.4 and 5.2). For $e\geq d$, we get that the "best" subspaces $C$
approximating $A$ contain a certain rational subspace $B_{N+1,q_{e}}$ (Lemma
4.6). We can then bound the height of the subspace $C$ from below (Lemma 4.5),
and we conclude that $\mu_{n}(A|e)_{d}$ cannot be too large, thanks to Lemma
4.7. For $e<d$, we show that the "best" subspaces $C$ approximating $A$
intersect non-trivially a certain rational subspace $D_{N,d}$ (Lemma 5.3). We
achieve the upper bound of $\mu_{n}(A|e)_{e}$ in Lemma 5.4 by bounding
$\psi_{1}(C\cap D_{N,d},A)$ from below and thus, a fortiori, $\psi_{e}(C,A)$.

Lemma 4.7 and Lemma 5.3 represent the most challenging aspects of the proof of
Theorem 1.2, and they crucially involve the geometry of numbers.

##  2 Construction of the subspace $A$

We define the constant $C_{d}$ as the smallest real number such that for any
$\alpha\geq C_{d}$, we have:

$\displaystyle-\alpha^{2}+\alpha(2d+2)-d$ $\displaystyle\leq 0$ (2.1)
$\displaystyle-\frac{\alpha}{2}+d(d-1)+1$ $\displaystyle\leq 0$ (2.2)
$\displaystyle-\alpha^{2}+(1+2d)\alpha-d$ $\displaystyle\leq 0,$ (2.3)

and for any $e\in\llbracket d,qd\rrbracket:$

$\displaystyle dr_{e}-(d-r_{e})\alpha\leq 0$ (2.4)
$\displaystyle\frac{\alpha^{q_{e}}}{d-r_{e}+\frac{1}{2}}-\frac{\alpha^{q_{e}+1}}{r_{e}+(d-r_{e})\alpha}+1\leq
0$ (2.5)

where $q_{e}$ and $r_{e}$ are the quotient and the remainder of the Euclidean
division of $e$ by $d$. Elementary computations show that $2<C_{d}\leq
3d(d+4)$, see [2, Lemma 9.3].

We fix $\alpha\geq C_{d}$. Recall that $n=(q+1)d$ and therefore $qd=n-d$.
Here, we construct the subspace $A$ from Theorem 1.2. Let $\theta$ be a prime
number greater than or equal to $5$. For $j\in\llbracket 1,d\rrbracket$,
define $\phi_{j}:\mathbb{N}\to\llbracket 0,qd-1\rrbracket$ by:

$\displaystyle\phi_{j}(k)=k+(j-1)q\text{ mod }(qd)$

where $x\text{ mod }(qd)$ is the remainder of the Euclidean division of $x$ by
$qd$.
Throughout this paper, we denote
$\sigma_{i,j}=\sum\limits_{k=0}^{+\infty}\frac{u^{(i,j)}_{k}}{\theta^{\lfloor\alpha^{k}\rfloor}}$
for $i\in\llbracket 0,qd-1\rrbracket$ and $j\in\llbracket 1,d\rrbracket$, with
sequences $u^{(i,j)}$ that we will chose using the following lemma. Using the
fact that $\alpha\geq 2$, Roth’s theoreme implies that $\sigma_{i,j}$ is
transcendental, see [1] for further details on the number constructed here.

######  Lemma 2.1.

There exist sequences
$u^{(0,1)},\ldots,u^{(qd-1,1)},\ldots,u^{(0,d)},\ldots,u^{(qd-1,d)}$
satisfying $\forall\,i\in\llbracket
0,qd-1\rrbracket,\quad\forall\,j\in\llbracket
1,d\rrbracket,\quad\forall\,k\in\mathbb{N}$:

$\displaystyle u^{(i,j)}_{k}\left\\{\begin{array}[]{ll}\in\\{2,3\\}&\text{if
}i=k+(j-1)q\text{ mod }(qd)\\\ =0&\text{otherwise}\end{array}\right.$ (2.8)

and such that the family
$(\sigma_{0,1},\ldots,\sigma_{qd-1,1},\ldots,\sigma_{0,d},\ldots,\sigma_{qd-1,d})$
is algebraically independent over $\mathbb{Q}$.

###### Proof.

Let
$\sigma_{0,1},\ldots,\sigma_{qd-1,1},\ldots,\sigma_{0,d},\ldots,\sigma_{qd-1,d}$
be arbitrarily ordered and denoted by $\sigma_{1},\ldots,\sigma_{d(n-d)}$, and
let us reason by induction on $t\in\llbracket 1,d(n-d)\rrbracket$. We denote
the sequences $u_{k}^{(i,j)}$ associated with $\sigma_{i,j}$ by
$u_{k}^{1},\ldots,u_{k}^{d(n-d)}$.

The set of algebraic numbers over $\mathbb{Q}$ is countable and the set of
sequences $(u_{k}^{1})$ satisfying (2.8) is uncountable. Therefore, we choose
a sequence such that $\sigma_{1}$ is transcendental over $\mathbb{Q}$.
Now, suppose that we have constructed $\sigma_{1},\ldots,\sigma_{t}$ as an
algebraically independent family over $\mathbb{Q}$ with $t\in\llbracket
1,d(n-d)-1\rrbracket$. The set of algebraic numbers over
$\mathbb{Q}(\mathcal{R},\sigma_{1},\ldots,\sigma_{t})$ is countable, but the
set of sequences $(u_{k}^{t+1})_{k\in\mathbb{N}}$ satisfying (2.8) is
uncountable. Therefore, we can choose a sequence such that $\sigma_{t+1}$ is
transcendental over $\mathbb{Q}(\sigma_{0},\ldots,\sigma_{t})$, completing the
induction.

∎

From now on, we assume that the sequences $(u_{k}^{(i,j)})_{k\in\mathbb{N}}$
and $\sigma_{i,j}$ for $i\in\llbracket 0,qd-1\rrbracket$ and $j\in\llbracket
1,d\rrbracket$ satisfy the conclusion of Lemma 2.1.

######  Remark 2.2.

For fixed $k$ and $j$, the integer $i=k+(j-1)q\text{ mod }(qd)$ is the unique
integer in $\llbracket 0,qd-1\rrbracket$ such that $u_{k}^{(i,j)}\neq 0$.

Before defining the subspace $A$, we state a lemma in order to study more
precisely the sequences $(u_{k}^{(i,j)})_{k\in\mathbb{N}}$.

######  Lemma 2.3.

Let $i\in\llbracket 0,qd-1\rrbracket$ and $N\in\mathbb{N}$. There exists a
unique pair $(k,j)\in\llbracket 0,q-1\rrbracket\times\llbracket 1,d\rrbracket$
such that $u_{N+k}^{(i,j)}\neq 0$.

###### Proof.

• Uniqueness: Suppose there exist $\ell_{1},\ell_{2}\in\llbracket
0,q-1\rrbracket$ and $j_{1},j_{2}\in\llbracket 1,d\rrbracket$ such that
$u_{N+\ell_{1}}^{(i,j_{1})}\neq 0\text{ and }u_{N+\ell_{2}}^{(i,j_{2})}\neq
0.$ By the definition of $u_{k}^{(i,j)}$, we have:

$\displaystyle N+\ell_{1}+(j_{1}-1)q\equiv N+\ell_{2}+(j_{2}-1)q\text{ mod
}{(qd)}.$

By the uniqueness of Euclidean division by $q$, since
$\ell_{1},\ell_{2}\in\llbracket 0,q-1\rrbracket$, we have $\ell_{1}=\ell_{2}$.
Thus, $(j_{1}-1)q\equiv(j_{2}-1)q\text{ mod }{(qd)}$ and hence
$(j_{1}-j_{2})q\equiv 0\text{ mod }{(qd)}$. Since $j_{1},j_{2}\in\llbracket
1,d\rrbracket$, we have $j_{1}=j_{2}$.

• Existence: We write the Euclidean divisions of $i$ and $N$ by $q$,
$i=qu+v\text{ and }N=qu^{\prime}+v^{\prime}$ with $v,v^{\prime}\in\llbracket
0,q-1\rrbracket$.
If $v\geq v^{\prime}$, we set $k=v-v^{\prime}\in\llbracket 0,q-1\rrbracket$
and $j=(u-u^{\prime}\text{ mod }d)+1\in\llbracket 1,d\rrbracket$. We then
verify that $i=N+k+(j-1)q\text{ mod }{(qd)}$:

$\displaystyle N+k+(j-1)q\text{ mod }{(qd)}$
$\displaystyle=qu^{\prime}+v^{\prime}+v-v^{\prime}+q(u-u^{\prime})\text{ mod
}{(qd)}=qu+v\text{ mod }{(qd)}=i.$

If $v<v^{\prime}$, we set $k=v-v^{\prime}+q\in\llbracket 0,q-1\rrbracket$ and
$j=(u-u^{\prime}-1\text{ mod }d)+1\in\llbracket 1,d\rrbracket$. We then verify
that $i=N+k+(j-1)q\text{ mod }{(qd)}$:

$\displaystyle N+k+(j-1)q\text{ mod }{(qd)}$
$\displaystyle=qu^{\prime}+v^{\prime}+v-v^{\prime}+q+q(u-u^{\prime}-1)\text{
mod }{(qd)}=qu+v\text{ mod }{(qd)}=i.$

∎

We define for $j\in\llbracket 1,d\rrbracket$, the vector $Y_{j}$ in
$\mathbb{R}^{n}$ as:

$\displaystyle
Y_{j}=\begin{pmatrix}0&\cdots&0&1&0&\cdots&0&\sigma_{0,j}&\cdots&\sigma_{qd-1,j}\end{pmatrix}^{\intercal}$

where the $j-$th coordinate is equal to $1$, the last $n-d$ are
$\sigma_{0,j},\ldots,\sigma_{qd-1,j}$ and the others are zero. We then define
the subspace $A$ from Theorem 1.2 as $A=\text{Span}(Y_{1},\ldots,Y_{d})$. By
considering the first $d$ coefficients of the vectors $Y_{j}$, it is clear
that $\dim(A)=d$.

######  Lemma 2.4.

Let $e\in\llbracket 1,n-d\rrbracket$, then the subspace $A$ is
$(n-d,\min(d,e))\text{-}$irrational.

###### Proof.

Let $B$ be a rational subspace of dimension $e$. Suppose by contradiction that
$\dim(A\cap B)\geq\min(d,e)$. Let us distinguish between two cases, depending
on whether $d\geq e$ or $d\leq e$.
• If $d\geq e$ then $A\cap B=B$. So there exists $X\in
A\cap\mathbb{Q}^{n}\setminus\\{0\\}$ of the form
$X=\lambda_{1}Y_{1}+\ldots+\lambda_{d}Y_{d}$. By looking at the first $d$
coordinates of $X$ we have
$(\lambda_{1},\ldots,\lambda_{d})\in\mathbb{Q}^{n}$. For $i\in\llbracket
0,qd-1\rrbracket$, we have a $\mathbb{Q}$-linear relation between
$1,\sigma_{i,1},\ldots,\sigma_{i,d}$ by looking at the $(d+1+i)-$th coordinate
of $X$ which leads to a contradiction.
• If $d\leq e$ then $A\cap B=A$. In particular, we then have $Y_{1}\in B$.
Denoting by $Z_{1},\ldots,Z_{n-d}$ a rational basis of $B$, we have
$Y_{1}\wedge Z_{1}\wedge\ldots\wedge Z_{n-d}=0$. This equality implies the
nullity of any minor of size $n-d+1$ of the matrix
$\begin{pmatrix}Y_{1}&Z_{1}&\cdots&Z_{n-d}\end{pmatrix}\in\mathcal{M}_{n,n-d+1}(\mathbb{R}).$
Every such minor is polynomial with rational coefficients in the coefficients
of $Y_{1}$, as the $Z_{i}$ are rational. We can also view each determinant as
a polynomial in $\mathbb{Q}[X_{0},\ldots,X_{n-d-1}]$ evaluated at the
$\sigma_{i,1}$. Since these coefficients form an algebraically independent
family over $\mathbb{Q}$, then each polynomial is identically zero. Thus, we
can replace the coefficients $\sigma_{i,1}$ of $Y_{1}$ by any real family, and
the determinant will be zero. Using this, we shall show that any minor of size
$n-d$ of the matrix $Q=\begin{pmatrix}Z_{1}&\cdots&Z_{n-d}\end{pmatrix}$
vanishes. Let $\Delta$ be a submatrix of size $(n-d)\times(n-d)$ of $Q$, we
denote by $\text{Ind}(\Delta)$ the set of indices $1\leq
i_{1}<\ldots<i_{n-d}\leq n$ of the rows of $Q$ from which $\Delta$ is
extracted. We distinguish between two cases.

  * $\diamond$

If $1\notin\text{Ind}(\Delta)$, then we set
$\sigma_{0,j}=\ldots=\sigma_{qd-1,0}=0$ and we compute the minor of size
$n-d+1$ of the matrix$\begin{pmatrix}Y_{1}&Z_{1}&\cdots&Z_{n-d}\end{pmatrix}$
corresponding to the rows $1$ and $\text{Ind}(\Delta)$. This minor is equal to
$\pm\det(\Delta)$ and is zero, so $\det(\Delta)$ vanishes.

  * $\diamond$

If $1\in\text{Ind}(\Delta)$, let us fix $i\in\llbracket 0,qd-1\rrbracket$ such
that $d+i+1\notin\text{Ind}(\Delta)$. We set $\sigma_{i,1}=1$ and
$\sigma_{k,1}=0$ for $k\in\llbracket 0,qd-1\rrbracket\setminus\\{i\\}$. We
compute the minor of size $n-d+1$ of the
matrix$\begin{pmatrix}Y_{1}&Z_{1}&\cdots&Z_{n-d}\end{pmatrix}$ corresponding
to the rows $d+i+1$ and $\text{Ind}(\Delta)$; it is equal to:

$\displaystyle\pm\det(\Delta)+\det(\Delta^{\prime})$

where $\Delta^{\prime}$ is a submatrix of $(n-d)\times(n-d)$ of $Q$ with
$1\notin\text{Ind}(\Delta^{\prime})=\text{Ind}(\Delta)\setminus\\{1\\}$. Using
the first case, we have $\det(\Delta^{\prime})=0$ and so $\det(\Delta)=0$.

We have thus shown that every minor of size $n-d$ of $Q$ vanishes. In
particular, $\text{rank}(Q)<n-d$ which is contradictory since
$Z_{1},\ldots,Z_{n-d}$ form a basis of $B$.

∎

######  Remark 2.5.

We can actually prove that
$A\in\mathcal{I}_{n}(d,n-d)_{1}=\bigcap\limits_{e=1}^{n-d}\mathcal{I}_{n}(d,e)_{1}$,
see [2, Lemma 9.6].

##  3 Rational subspaces of good approximation

In this section we define rational subspaces. We will show later that theses
subspaces achieve good approximations of $A$.

For $i\in\llbracket 0,qd-1\rrbracket$, $j\in\llbracket 1,d\rrbracket$, and
$N\in\mathbb{N}\setminus\\{0\\}$, we define the truncated sum:
$\sigma_{i,j,N}=\sum\limits_{k=0}^{N}\frac{u_{k}^{(i,j)}}{\theta^{\lfloor\alpha^{k}\rfloor}}\in\frac{1}{\theta^{\lfloor\alpha^{N}\rfloor}}\mathbb{Z}$.
Now, for $j\in\llbracket 1,d\rrbracket$, we define the vector in
$\mathbb{Z}^{n}$:

$\displaystyle
X_{N}^{j}=\theta^{\lfloor\alpha^{N}\rfloor}\begin{pmatrix}0&\cdots&0&1&0&\cdots&0&\sigma_{0,j,N}&\cdots&\sigma_{qd-1,j,N}\end{pmatrix}^{\intercal}$

where the $j-$th coordinate is equal to $1$, the last $n-d$ are
$\sigma_{0,j,N},\ldots,\sigma_{qd-1,j,N}$ and the others are zero.

Next, we denote for $N$ and $v$ two positive integers:

$\displaystyle
B_{N,v}=\text{Span}(X_{N}^{1},X_{N+1}^{1},\ldots,X_{N+v-1}^{1},X_{N}^{2},\ldots,X_{N+v-1}^{2},\ldots,X_{N}^{d},\ldots,X_{N+v-1}^{d})$
(3.1)

which is a rational subspace by definition. For $j\in\llbracket
1,d\rrbracket$, we note that:

$\displaystyle
X_{N+1}^{j}=\theta^{\lfloor\alpha^{N+1}\rfloor-\lfloor\alpha^{N}\rfloor}X_{N}^{j}+U_{N+1}^{j}\text{
with
}U_{N+1}^{j}=\begin{pmatrix}0&\cdots&0&u_{N+1}^{(0,j)}&\cdots&u_{N+1}^{(qd-1,j)}\end{pmatrix}^{\intercal}.$
(3.2)

We set $V_{N}^{j}=\frac{U_{N}^{j}}{\|U_{N}^{j}\|}\in\mathbb{Z}^{n}$ because
the vectors $U_{N}^{j}$ have a unique non-zero coordinate according to the
construction in (2.8). The vectors $V_{N}^{j}$ are thus vectors of the
canonical basis of $\mathbb{R}^{n}$. We also introduce the vectors:

$\displaystyle Z_{N}^{j}=\frac{1}{\theta^{\lfloor\alpha^{N}\rfloor}}X_{N}^{j}$
(3.3)

and we note that $Z_{N}^{j}\underset{N\to+\infty}{\longrightarrow}Y_{j}$ so we
deduce that there exist constants $c_{1}$ and $c_{2}$ independent of $N$ such
that for any $N\in\mathbb{N}$:

$\displaystyle
c_{\ref{8cons_minor_norme_XN}}\theta^{\alpha^{N}}\leq\|X_{N}^{j}\|\leq
c_{\ref{8cons_major_norme_XN}}\theta^{\alpha^{N}}.$ (3.4)

More precisely for $j\in\llbracket 1,d\rrbracket$, we have:

$\displaystyle\psi_{1}(\text{Span}(Y_{j}),\text{Span}(X_{N}^{j}))=\omega(Y_{j},Z_{N}^{j})\leq\frac{\|Y_{j}-Z_{N}^{j}\|}{\|Y_{j}\|}\leq
c_{\ref{8cons_major_angle_Yj_XNj}}\theta^{-\alpha^{N+1}}$ (3.5)

with $c_{3}>0$ independent of $N$.

######  Lemma 3.1.

Let $v\in\llbracket 1,q\rrbracket$. Then the subspace $B_{N,v}$ has dimension
$dv$. Moreover, the vectors $(X_{N}^{j})_{j\in\llbracket
1,d\rrbracket}\cup(V_{k}^{j})_{j\in\llbracket 1,d\rrbracket,k\in\llbracket
N+1,N+v-1\rrbracket}$ form a $\mathbb{Z}$-basis of
$B_{N,v}\cap\mathbb{Z}^{n}$.

###### Proof.

By induction on $v$ and using (3.2), we have:

$\displaystyle
B_{N,v}=\text{Span}(X_{N}^{1},V_{N+1}^{1},\ldots,V_{N+v-1}^{1},X_{N}^{2},V_{N+1}^{2},\ldots,V_{N+v-1}^{2},\ldots,X_{N}^{d},V_{N+1}^{d},\ldots,V_{N+v-1}^{d}).$

Remark 2.2 allows us to assert that the $V_{k}^{j}$ considered here are all
different. Furthermore, we recall that these are vectors of the canonical
basis. We deduce in particular that $(X_{N}^{j})_{j\in\llbracket
1,d\rrbracket}\cup(V_{k}^{j})_{j\in\llbracket 1,d\rrbracket,k\in\llbracket
N+1,N+v-1\rrbracket}$ form a free family since the $j-$th coefficient of
$X_{N}^{j}$ is $\theta^{\left\lfloor\alpha^{N}\right\rfloor}$ while the $j-$th
coefficient of the other vectors of the family is zero for any $j\in\llbracket
1,d\rrbracket$.
We first prove the lemma for $v=q$. Let $(a_{j,k})_{{j\in\llbracket
1,d\rrbracket,k\in\llbracket 0,q-1\rrbracket}}\in\left[0,1\right]^{qd}$ be
such that:

$\displaystyle
U=\sum\limits_{j=1}^{d}a_{j,0}X_{N}^{j}+\sum\limits_{k=1}^{q-1}\sum\limits_{j=1}^{d}a_{j,k}V_{N+k}^{j}\in\mathbb{Z}^{n}$
(3.6)

By examining the first $d$ coordinates of $U$, we find that
$a_{j,0}\theta^{\lfloor\alpha^{N}\rfloor}\in\mathbb{Z}$ for all
$j\in\llbracket 1,d\rrbracket$. We can write $a_{j,0}=\frac{x_{j}}{y_{j}}$
with $y_{j}\mid\theta^{\lfloor\alpha^{N}\rfloor}$ and $\gcd(x_{j},\theta)=1$
for all $j\in\llbracket 1,d\rrbracket$. Let $j_{\max}\in\llbracket
1,d\rrbracket$ be such that $\max\limits_{j=1}^{d}y_{j}=y_{j_{\max}}$.
Now consider the integer $i=N+q(j_{\max}-1)\text{ mod }(qd)\in\llbracket
0,qd-1\rrbracket$. By the definition of $u_{k}^{(i,j)}$ in (2.8), we have
$u^{(i,j_{\max})}_{N}\neq 0$ and therefore $\forall\,k\in\llbracket
1,q-1\rrbracket,\quad\forall\,j\in\llbracket 1,d\rrbracket,\quad
u_{N+k}^{(i,j)}=0$ according to Lemma 2.3. By the definition of $U_{N+k}$ in
(3.2) and $V_{N+k}=\frac{U_{N+k}}{\|U_{N+k}\|}$, the $(d+i)$-th coordinate of
the vector $\sum\limits_{k=1}^{q-1}\sum\limits_{j=1}^{d}a_{j,k}V_{N+k}^{j}$ is
therefore zero. Hence, according to (3.6)
$\sum\limits_{j=1}^{d}a_{j,0}\theta^{\lfloor\alpha^{N}\rfloor}\sigma_{i,j,N}=\sum\limits_{j=1}^{d}\frac{x_{j}}{y_{j}}\theta^{\lfloor\alpha^{N}\rfloor}\sigma_{i,j,N}\in\mathbb{Z}.$
Since the $y_{j}$ are powers of $\theta$ and
$\max\limits_{j=1}^{d}y_{j}=y_{j_{\max}}$, we have
$\frac{y_{j_{\max}}}{y_{j}}\in\mathbb{Z}$ and:

$\displaystyle\sum\limits_{j=1}^{d}x_{j}\frac{y_{j_{\max}}}{y_{j}}\theta^{\lfloor\alpha^{N}\rfloor}\sigma_{i,j,N}\in
y_{j_{\max}}\mathbb{Z}.$ (3.7)

Now we notice that for $j\neq j_{\max}$, we have $i\not\equiv
N+1+q(j-1)\pmod{qd}$ and thus $u_{N}^{(i,j)}=0$ in this case. Since
$\lfloor\alpha^{N-1}\rfloor<\lfloor\alpha^{N}\rfloor$, we have for all $j\neq
j_{\max}$,
$\theta\mid\theta^{\lfloor\alpha^{N}\rfloor}\sigma_{i,j,N}=\theta^{\lfloor\alpha^{N}\rfloor}\sum\limits_{k=0}^{N}\frac{u_{k}^{(i,j)}}{\theta^{\lfloor\alpha^{k}\rfloor}}.$
If $\theta\mid y_{j_{\max}}$, then using (3.7) $\theta\mid
x_{j_{\max}}\theta^{\lfloor\alpha^{N}\rfloor}\sigma_{i,j_{\max},N}.$ Since
$u_{n}^{(i,j_{\max})}$ is non-zero and coprime with $\theta$, we have
$\gcd(\theta^{\lfloor\alpha^{N}\rfloor}\sigma_{i,j_{\max},N},\theta)=1$.
We finally deduce that $\theta\mid x_{j_{\max}}$, which is contradictory to
$\gcd(x_{j_{\max}},\theta)=1$. All $y_{j}$ are therefore equal to 1, and thus
all $a_{j,0}$ are integers. Returning to (3.6), we find
$\sum\limits_{k=1}^{q-1}\sum\limits_{j=1}^{d}a_{j,k}V_{N+k}^{j}\in\mathbb{Z}^{n}.$
Since all the vectors $V_{N+k}^{j}$ are distinct and come from the canonical
basis, $a_{j,k}$ is an integer for all $j\in\llbracket 1,d\rrbracket$ and
$k\in\llbracket 0,q-1\rrbracket$. This shows that $(X_{N}^{j})_{j\in\llbracket
1,d\rrbracket}\cup(V_{k}^{j})_{j\in\llbracket 1,d\rrbracket,k\in\llbracket
N+1,N+q-1\rrbracket}$ forms a $\mathbb{Z}$-basis of
$B_{N,q}\cap\mathbb{Z}^{n}$, and in particular that $\dim(B_{N,q})=qd$.

Now let $v\in\llbracket 1,q-1\rrbracket$. We have $(X_{N}^{j})_{j\in\llbracket
1,d\rrbracket}\cup(V_{k}^{j})_{j\in\llbracket 1,d\rrbracket,k\in\llbracket
N+1,N+v-1\rrbracket}\subset(X_{N}^{j})_{j\in\llbracket
1,d\rrbracket}\cup(V_{k}^{j})_{j\in\llbracket 1,d\rrbracket,k\in\llbracket
N+1,N+q-1\rrbracket}$. Since it is contained in a $\mathbb{Z}-$basis, the
family $(X_{N}^{j})_{j\in\llbracket
1,d\rrbracket}\cup(V_{k}^{j})_{j\in\llbracket 1,d\rrbracket,k\in\llbracket
N+1,N+v-1\rrbracket}$ forms a $\mathbb{Z}$-basis of the $\mathbb{Z}$-module it
generates. This $\mathbb{Z}$-module is $B_{N,v}\cap\mathbb{Z}^{n}$ and the
lemma is proved.

∎

Using this lemma, we can compute the height of some specific rational
subspaces in Lemmas 4.2, 4.5, 4.7 and 5.1.

##  4 Computation of the exponent in the case $e\geq d$

In this section, we consider $e\in\llbracket d,qd\rrbracket$. Recall that
$e=q_{e}d+r_{e}$ is the Euclidean division of $e$ by $d$. In particular, we
have $1\leq q_{e}\leq q$. The goal of this section is to compute
$\mu_{n}(A|e)_{d}$.

###  4.1 Lower bound on the exponent

In this section, we introduce a sequence of rational subspaces of dimension
$e$ that approximate $A$ well, which allows us to bound $\mu_{n}(A|e)_{d}$
from below. Let $N\in\mathbb{N}$. We define the subspace $C_{N,e}$ by:

$\displaystyle C_{N,e}$
$\displaystyle=\text{Span}(X_{N+1}^{1},\ldots,X_{N+q_{e}}^{1},\ldots,X_{N+1}^{d},\ldots,X_{N+q_{e}}^{d})\bigoplus\text{Span}(X_{N}^{1},\ldots,X_{N}^{r_{e}})$
(4.1)
$\displaystyle=B_{N+1,q_{e}}\bigoplus\text{Span}(X_{N}^{1},\ldots,X_{N}^{r_{e}})$

which is a rational subspace. Using (3.2) and reasoning by induction for each
$j\in\llbracket 1,d\rrbracket$, we have:

$\displaystyle C_{N,e}=\,$
$\displaystyle\text{Span}(X_{N}^{1},V_{N+1}^{1},\ldots,V_{N+q_{e}}^{1},\ldots,X_{N}^{r_{e}},V_{N+1}^{r_{e}},\ldots,V_{N+q_{e}}^{r_{e}})$
$\displaystyle\bigoplus\text{Span}(X_{N+1}^{r_{e}+1},V_{N+2}^{r_{e}+1},\ldots,V_{N+q_{e}}^{r_{e}+1}\ldots,X_{N+1}^{d},V_{N+2}^{d},\ldots,V_{N+q_{e}}^{d}).$
(4.2)

######  Remark 4.1.

Note that in the case where $r_{e}=0$, by the definition in (4.1), we have
$C_{N,e}=B_{N+1,q_{e}}$.
In all cases, we have $B_{N+1,q_{e}}\subset C_{N,e}\subset B_{N,q_{e}+1}$.

######  Lemma 4.2.

We have $\dim(C_{N,e})=e$. Moreover, there exist constants $c_{4}>0$ and
$c_{5}>0$ independent of $N$ such that

$\displaystyle
c_{\ref{8cons_haut_CN_minor}}\theta^{r_{e}{\alpha^{N}}+(d-r_{e}){\alpha^{N+1}}}\leq
H(C_{N,e})\leq
c_{\ref{8cons_haut_CN_major}}\theta^{r_{e}{\alpha^{N}}+(d-r_{e}){\alpha^{N+1}}}.$

###### Proof.

If $q_{e}=q$, then $C_{N,e}=B_{N+1,q}$ because $e=qd$ and $r_{e}=0$; otherwise
$q_{e}<q$ and in this case $C_{N,e}\subset B_{N,q}$. In each case, Lemma 3.1
states that the vectors considered in (4.1) come from a $\mathbb{Z}$-basis of
$B_{N+1,q}\cap\mathbb{Z}^{n}$ or $B_{N,q}\cap\mathbb{Z}^{n}$ respectively. The
relation (4.1) gives directly
$\dim(C_{N,e})=(q_{e}+1)r_{e}+(d-r_{e})q_{e}=q_{e}d+r_{e}=e$.

These vectors thus form a $\mathbb{Z}-$basis of $C_{N,e}\cap\mathbb{Z}^{n}$.
Taking up the notation
$Z_{N}^{j}=\frac{1}{\theta^{\left\lfloor\alpha^{N}\right\rfloor}}X_{N}^{j}$,
we have:

$\displaystyle H(C_{N,e})=$
$\displaystyle\theta^{r_{e}\left\lfloor\alpha^{N}\right\rfloor+(d-r_{e})\left\lfloor\alpha^{N+1}\right\rfloor}\|H_{N}\|\leq\theta^{r_{e}{\alpha^{N}}+(d-r_{e}){\alpha^{N+1}}}\|H_{N}\|$
(4.3)

where $H_{N}$ is the exterior product of the vectors

$\displaystyle(Z_{N}^{j})_{j\in\llbracket
1,r_{e}\rrbracket}\cup(Z_{N+1}^{j})_{j\in\llbracket
r_{e}+1,d\rrbracket}\cup(V_{N+k}^{j})_{j\in\llbracket
1,r_{e}\rrbracket,k\in\llbracket
1,q_{e}\rrbracket}\cup(V_{N+k}^{j})_{j\in\llbracket
r_{e}+1,d\rrbracket,k\in\llbracket 2,q_{e}\rrbracket}.$ (4.4)

We can bound this norm from above by
$\|H_{N}\|\leq\|Z_{N}^{1}\wedge\ldots\wedge Z_{N}^{r_{e}}\wedge
Z_{N+1}^{r_{e}+1}\wedge\ldots\wedge Z_{N+1}^{d}\|$ since the norms of the
vectors $V_{k}^{j}$ are equal to $1$. Now the quantity
$\|Z_{N}^{1}\wedge\ldots\wedge Z_{N}^{r_{e}}\wedge
Z_{N+1}^{r_{e}+1}\wedge\ldots\wedge Z_{N+1}^{d}\|$ converges to
$\|Y_{1}\wedge\ldots\wedge Y_{d}\|$ as $N$ tends to infinity and is therefore
bounded independently of $N$. There exists $c_{\ref{8cons_haut_CN_major}}>0$,
independent of $N$, such that:

$\displaystyle\|H_{N}\|\leq c_{\ref{8cons_haut_CN_major}}.$ (4.5)

Furthermore, let us define the matrix $M$ whose column vectors are the vectors
of $(\ref{8vecteurs_dans_haut_produit})$. Then, by taking up the construction
of the vectors $V_{N}^{j}$ in $(\ref{8rec_Xn_Un})$, $M$ takes the form
$M=\begin{pmatrix}\begin{matrix}I_{d}\\\
\Sigma_{N}\end{matrix}&\begin{matrix}0\\\ V_{N}\end{matrix}\end{pmatrix}$
where $\Sigma_{N}$ is a matrix whose coefficients are $\sigma_{i,j,N}$ or
$\sigma_{i,j,N+1}$, and $V_{N}$ is a matrix of
$\mathcal{M}_{qd,e-d}(\mathbb{Z})$ of rank $e-d$ since its columns are $e-d$
distinct vectors from the canonical basis.

Let $\Delta$ be a non-zero minor of $V_{N}$ of size $e-d$. We can then extract
a square matrix $M^{\prime}$ of size $e$ from $M$ by selecting the first $d$
rows and $e-d$ among the last ones, corresponding to the minor $\Delta$. The
determinant of $M^{\prime}$ is an integer because it is the product of
$\det(I_{d})=1$ and $\Delta$ which is a minor of a matrix in
$M_{e-d}(\mathbb{Z})$. Hence, we have $|\det(M^{\prime})|\geq 1$. Now
$\det(M^{\prime})$ is a minor of size $e$ of $M$, so we have
$\|H_{N}\|\geq|\det(M^{\prime})|\geq 1$. By combining this with (4.3) and
(4.5), we obtain:

$\displaystyle
c_{\ref{8cons_haut_CN_minor}}\theta^{r_{e}{\alpha^{N}}+(d-r_{e}){\alpha^{N+1}}}\leq\theta^{r_{e}\left\lfloor\alpha^{N}\right\rfloor+(d-r_{e})\left\lfloor\alpha^{N+1}\right\rfloor}\leq
H(C_{N,e})\leq
c_{\ref{8cons_haut_CN_major}}\theta^{r_{e}{\alpha^{N}}+(d-r_{e}){\alpha^{N+1}}}$

by setting $c_{\ref{8cons_haut_CN_minor}}=\theta^{-d}$.

∎

We now focus on the angle $\psi_{d}(A,C_{N,e})$. To do this, we first study
the angle $\psi_{1}(\text{Span}(Y_{1}),C_{N,e})$.

######  Lemma 4.3.

There exists a constant $c_{6}>0$ independent of $N$ such that:

$\displaystyle\psi_{1}(\text{Span}(Y_{1}),C_{N,e})\geq
c_{\ref{8cons_minor_angle_y1_CNe}}\theta^{-\alpha^{N+q_{e}+1}}.$

###### Proof.

Let $X\in C_{N,e}\setminus\\{0\\}$ such that
$\psi_{1}(\text{Span}(Y_{1}),C_{N,e})=\omega(Y_{1},X)$. We use the basis of
$C_{N,e}$ as explicitly detailed in (4.1), and we denote $(a_{k,j})$ a family
of real numbers such that $X$ can be written as:

$\displaystyle\sum\limits_{j=1}^{r_{e}}a_{0,j}\theta^{-\left\lfloor\alpha^{N}\right\rfloor}X_{N}^{j}+\sum\limits_{j=r_{e}+1}^{d}a_{0,j}\theta^{-\left\lfloor\alpha^{N+1}\right\rfloor}X_{N+1}^{j}+\sum\limits_{j=1}^{r_{e}}\sum\limits_{k=1}^{q_{e}}a_{k,j}V_{N+k}^{j}+\sum\limits_{j=r_{e}+1}^{d}\sum\limits_{k=1}^{q_{e}-1}a_{k,j}V_{N+k+1}^{j}.$

Without loss of generality, we assume that:

$\displaystyle\sum\limits_{j=1}^{r_{e}}a_{0,j}^{2}+\sum\limits_{j=r_{e}+1}^{d}a_{0,j}^{2}+\sum\limits_{j=1}^{r_{e}}\sum\limits_{k=1}^{q_{e}}a_{k,j}^{2}+\sum\limits_{j=r_{e}+1}^{d}\sum\limits_{k=1}^{q_{e}-1}a_{k,j}^{2}=1.$
(4.6)

The norms of the vectors
$\theta^{-\left\lfloor\alpha^{N}\right\rfloor}X_{N}^{j}$ and $V_{N+k}^{j}$ are
bounded by a constant independent of $N$, and we have
$\omega(Y_{1},X)=\frac{\|Y_{1}\wedge X\|}{\|Y_{1}\|\cdot\|X\|}$. Therefore, it
suffices to show that if $N$ is large enough:

$\displaystyle\|Y_{1}\wedge X\|\geq
c_{\ref{8cons_minor_angle_y1_CNe}}\theta^{-\left\lfloor\alpha^{N+q_{e}+1}\right\rfloor}\geq
c_{\ref{8cons_minor_angle_y1_CNe}}\theta^{-{\alpha^{N+q_{e}+1}}}.$ (4.7)

Recall that
$Y_{1}=\begin{pmatrix}1&0&\cdots&0&\sigma_{0,1}&\cdots&\sigma_{qd-1,1}\end{pmatrix}^{\intercal}$
and let us explicit $X$ as:

$\displaystyle\begin{pmatrix}a_{0,1}\\\ \vdots\\\ a_{0,d}\\\
\sum\limits_{j=1}^{r_{e}}a_{0,j}\sigma_{0,j,N}+\sum\limits_{j=r_{e}+1}^{d}a_{0,j}\sigma_{0,j,N+1}+\sum\limits_{j=1}^{r_{e}}\sum\limits_{k=1}^{q_{e}}a_{k,j}v_{N+k,0}^{j}+\sum\limits_{j=r_{e}+1}^{d}\sum\limits_{k=1}^{q_{e}-1}a_{k,j}v_{N+k+1,0}^{j}\\\
\vdots\\\
\sum\limits_{j=1}^{r_{e}}a_{0,j}\sigma_{qd-1,j,N}+\sum\limits_{j=r_{e}+1}^{d}a_{0,j}\sigma_{qd-1,j,N+1}+\sum\limits_{j=1}^{r_{e}}\sum\limits_{k=1}^{q_{e}}a_{k,j}v_{N+k,qd-1}^{j}+\sum\limits_{j=r_{e}+1}^{d}\sum\limits_{k=1}^{q_{e}-1}a_{k,j}v_{N+k+1,qd-1}^{j}\end{pmatrix}$

with
$V_{N+k}^{j}=\begin{pmatrix}0&\cdots&0&v_{N+k,0}^{j}&\cdots&v_{N+k,qd-1}^{j}\end{pmatrix}^{\intercal}$.
We then prove (4.7) by considering different cases. Let $\sigma\geq 1$ be an
upper bound on the $\sigma_{i,j}$, especially for all $N\in\mathbb{N}$,
$\sigma\geq\sigma_{i,j,N}$.

• First case: If there exists $j\in\llbracket 2,d\rrbracket$ such that
$|a_{0,j}|\geq\frac{\theta^{-\left\lfloor\alpha^{N+q_{e}+1}\right\rfloor}}{(\sigma
qd)^{2}}$ then, by bounding $\|X\wedge Y_{1}\|$ from below by the minor of
$(Y_{1}|X)$ corresponding to the first row and the $j$-th row, we find
$\|X\wedge Y_{1}\|\geq\left|\det\begin{pmatrix}1&a_{0,1}\\\
0&a_{0,j}\end{pmatrix}\right|\geq|a_{0,j}|\geq\frac{\theta^{-\left\lfloor\alpha^{N+q_{e}+1}\right\rfloor}}{(\sigma
qd)^{2}}$ which yields $(\ref{8minor_angle_y1_CNe_proof})$.
• Second case: Otherwise $\forall\,j\in\llbracket
2,d\rrbracket,\quad|a_{0,j}|<\frac{\theta^{-\left\lfloor\alpha^{N+q_{e}+1}\right\rfloor}}{(\sigma
qd)^{2}}.$ According to $(\ref{8proof_aik_carre1})$, we have:

$\displaystyle
a_{0,1}^{2}+\sum\limits_{j=1}^{r_{e}}\sum\limits_{k=1}^{q_{e}}a_{k,j}^{2}+\sum\limits_{j=r_{e}+1}^{d}\sum\limits_{k=1}^{q_{e}-1}a_{k,j}^{2}\geq
1-(d-1)\left(\frac{\theta^{-\left\lfloor\alpha^{N+q_{e}+1}\right\rfloor}}{(\sigma
qd)^{2}}\right)^{2}.$

In particular, if $N$ is large enough, there exists $(j^{\prime},k^{\prime})$
such that $|a_{k^{\prime},j^{\prime}}|\geq\frac{1}{qd}$ with
$k^{\prime}>0\text{ or }(k^{\prime}=0\text{ and }j^{\prime}=1).$ First, let us
consider the case $k^{\prime}>0$. We then set
$i=N+k^{\prime}+(j^{\prime}-1)q\text{ mod }(qd)$. By definition of
$u_{N+k}^{(i,j)}$ (and thus of $v_{N+k,i}^{j}$) in
$(\ref{8construc_suite_u})$, we have:

$\displaystyle v_{N+k^{\prime},i}^{j^{\prime}}=1\text{ and
}\forall\,(j,k)\neq(j^{\prime},k^{\prime}),\quad v_{N+k,i}^{j}=0.$

By bounding $\|X\wedge Y_{1}\|$ from below by the minor of $(Y_{1}|X)$
corresponding to the first row and the $(i+1+d)$-th row, we find:

$\displaystyle\|X\wedge Y_{1}\|$
$\displaystyle\geq\left|\det\begin{pmatrix}1&a_{0,1}\\\
\sigma_{i,1}&\sum\limits_{j=1}^{r_{e}}a_{0,j}\sigma_{i,j,N}+\sum\limits_{j=r_{e}+1}^{d}a_{0,j}\sigma_{i,j,N+1}+a_{k^{\prime},j^{\prime}}v_{N+k^{\prime},i}^{j^{\prime}}\end{pmatrix}\right|$
$\displaystyle=\left|a_{0,1}(\sigma_{i,1,N}-\sigma_{i,1})+\sum\limits_{j=2}^{r_{e}}a_{0,j}\sigma_{i,j,N}+\sum\limits_{j=r_{e}+1}^{d}a_{0,j}\sigma_{i,j,N+1}+a_{k^{\prime},j^{\prime}}v_{N+k^{\prime},i}^{j^{\prime}}\right|.$

Now,
$|\sigma_{i,1,N}-\sigma_{i,1}|=\sum\limits_{k=N+1}^{+\infty}\frac{u_{k}^{(i,1)}}{\theta^{\left\lfloor\alpha^{k}\right\rfloor}}\leq
4\theta^{-\left\lfloor\alpha^{N+1}\right\rfloor}$ if $N$ is large enough.
Thus,

$\displaystyle\|X\wedge Y_{1}\|$
$\displaystyle\geq|a_{j^{\prime},k^{\prime}}v_{N+k^{\prime},i}^{j^{\prime}}|-|a_{0,1}(\sigma_{i,1,N}-\sigma_{i,1})|-\left|\sum\limits_{j=2}^{r_{e}}a_{0,j}\sigma_{i,j,N}\right|-\left|\sum\limits_{j=r_{e}+1}^{d}a_{0,j}\sigma_{i,j,N+1}\right|$
$\displaystyle\geq\frac{1}{qd}-4\theta^{-\left\lfloor\alpha^{N+1}\right\rfloor}-d\sigma\left(\frac{\theta^{-\left\lfloor\alpha^{N+q_{e}+1}\right\rfloor}}{(\sigma
qd)^{2}}\right)$
$\displaystyle\geq\theta^{-\left\lfloor\alpha^{N+q_{e}+1}\right\rfloor}$

if $N$ is large enough, which proves $(\ref{8minor_angle_y1_CNe_proof})$. Now,
let us suppose that $k^{\prime}=0$ and $j^{\prime}=1$, so
$|a_{0,1}|\geq\frac{1}{qd}$. We set $i=N+q_{e}+1\text{ mod }(qd)$. By
definition of $u_{N+k}^{(i,j)}$ in $(\ref{8construc_suite_u})$, we have
$\forall\,k\in\llbracket 1,q_{e}\rrbracket,\quad\forall\,j\in\llbracket
2,d\rrbracket,\quad v_{N+k,i}^{j}=0$ and
$u_{N+1}^{(i,1)}=\ldots=u_{N+q_{e}}^{(i,1)}=0$ and
$u_{N+q_{e}+1}^{(i,1)}\in\\{2,3\\}$. In particular,
$\sigma_{i,1,N}=\sum\limits_{k=0}^{N}\frac{u_{k}^{(i,1)}}{\theta^{\left\lfloor\alpha^{k}\right\rfloor}}=\sum\limits_{k=0}^{N+q_{e}}\frac{u_{k}^{(i,1)}}{\theta^{\left\lfloor\alpha^{k}\right\rfloor}}=\sigma_{i,1,N+q_{e}}$
and
$\sum\limits_{j=1}^{r_{e}}\sum\limits_{k=1}^{q_{e}}a_{k,j}v_{N+k,i}^{j}+\sum\limits_{j=r_{e}+1}^{d}\sum\limits_{k=1}^{q_{e}-1}a_{k,j}v_{N+k+1,i}^{j}=0.$
By bounding $\|X\wedge Y_{1}\|$ from below by the minor of $(X|Y_{1})$
corresponding to the first row and the $(i+1+d)$-th row, we find:

$\displaystyle\|X\wedge Y_{1}\|$
$\displaystyle\geq\left|\det\begin{pmatrix}1&a_{0,1}\\\
\sigma_{i,1}&\sum\limits_{j=1}^{r_{e}}a_{0,j}\sigma_{i,j,N}+\sum\limits_{j=r_{e}+1}^{d}a_{0,j}\sigma_{i,j,N+1}\end{pmatrix}\right|$
$\displaystyle=\left|a_{0,1}(\sigma_{i,1,N}-\sigma_{i,1})+\sum\limits_{j=2}^{r_{e}}a_{0,j}\sigma_{i,j,N}+\sum\limits_{j=r_{e}+1}^{d}a_{0,j}\sigma_{i,j,N+1}\right|.$

Now,
$|\sigma_{i,1,N}-\sigma_{i,1}|=|\sigma_{i,1,N+q_{e}}-\sigma_{i,1}|=\sum\limits_{k=N+q_{e}+1}^{+\infty}\frac{u_{k}^{(i,1)}}{\theta^{\left\lfloor\alpha^{k}\right\rfloor}}\geq\frac{u_{N+q_{e}+1}^{(i,1)}}{\theta^{\left\lfloor\alpha^{N+q_{e}+1}\right\rfloor}}\geq
2\theta^{-\left\lfloor\alpha^{N+q_{e}+1}\right\rfloor}$. Therefore,

$\displaystyle\|X\wedge Y_{1}\|$
$\displaystyle\geq|a_{0,1}(\sigma_{i,1,N}-\sigma_{i,1})|-\left|\sum\limits_{j=2}^{r_{e}}a_{0,j}\sigma_{i,j,N}\right|-\left|\sum\limits_{j=r_{e}+1}^{d}a_{0,j}\sigma_{i,j,N+1}\right|$
$\displaystyle\geq\frac{2\theta^{-\left\lfloor\alpha^{N+q_{e}+1}\right\rfloor}}{qd}-d\sigma\left(\frac{\theta^{-\left\lfloor\alpha^{N+q_{e}+1}\right\rfloor}}{(\sigma
qd)^{2}}\right)$
$\displaystyle\geq\frac{\theta^{-\left\lfloor\alpha^{N+q_{e}+1}\right\rfloor}}{qd}$

which proves $(\ref{8minor_angle_y1_CNe_proof})$. Therefore,
$(\ref{8minor_angle_y1_CNe_proof})$ holds in all cases, which concludes the
proof of the lemma.

∎

We can now estimate the last angle (corresponding to $j=d)$ between $A$ and
$C_{N,e}$.

######  Lemma 4.4.

There exist constants $c_{7}>0$ and $c_{8}>0$ independent of $N$ such that

$\displaystyle
c_{\ref{8cons_minor_angle_psid}}H(C_{N,e})^{\frac{-\alpha^{q_{e}+1}}{r_{e}+(d-r_{e})\alpha}}\leq\psi_{d}(A,C_{N,e})\leq
c_{\ref{8cons_major_angle_psid}}H(C_{N,e})^{\frac{-\alpha^{q_{e}+1}}{r_{e}+(d-r_{e})\alpha}}.$

###### Proof.

Recall that for $j\in\llbracket 1,d\rrbracket$ and
$N\in\mathbb{N},\psi_{1}(\text{Span}(Y_{j}),\text{Span}(X_{N+q_{e}}^{j}))\leq
c_{\ref{8cons_major_angle_Yj_XNj}}\theta^{-\alpha^{N+q_{e}+1}}$ according to
$(\ref{8major_Yj_XNj})$. By construction of $C_{N,e}$, we have
$\text{Span}(X_{N+q_{e}}^{1},\ldots,X_{N+q_{e}}^{d})\subset C_{N,e}$ and thus

$\psi_{d}(A,C_{N,e})\leq\psi_{d}(A,\text{Span}(X_{N+q_{e}}^{1},\ldots,X_{N+q_{e}}^{d}))$

by the corollary of Lemma 12 of [9]. According to Theorem $1.2$ of [6], we
have:

$\displaystyle\psi_{d}(A,\text{Span}(X_{N+q_{e}}^{1},\ldots,X_{N+q_{e}}^{d}))$
$\displaystyle\leq
c_{\ref{8cons_maj_prop45_appliquee}}\sum\limits_{j=1}^{d}\psi_{1}(\text{Span}(Y_{j}),\text{Span}(X_{N+q_{e}}^{j}))$

where $c_{9}$ depends on $Y_{1},\ldots,Y_{d}$ and $n$. Hence,

$\displaystyle\psi_{d}(A,\text{Span}(X_{N+q_{e}}^{1},\ldots,X_{N+q_{e}}^{d}))$
$\displaystyle\leq
c_{\ref{8cons_maj_prop45_appliquee}}c_{\ref{8cons_major_angle_Yj_XNj}}d\theta^{-{\alpha^{N+q_{e}+1}}}$
$\displaystyle\leq
c_{\ref{8cons_maj_prop45_appliquee}}c_{\ref{8cons_major_angle_Yj_XNj}}dc_{\ref{8cons_haut_CN_major}}^{\frac{{\alpha^{N+q_{e}+1}}}{r_{e}{\alpha^{N}}+(d-r_{e}){\alpha^{N+1}}}}H(C_{N,e})^{\frac{-{\alpha^{N+q_{e}+1}}}{r_{e}{\alpha^{N}}+(d-r_{e}){\alpha^{N+1}}}}$
$\displaystyle=c_{\ref{8cons_major_angle_psid}}H(C_{N,e})^{\frac{-{\alpha^{q_{e}+1}}}{r_{e}+(d-r_{e})\alpha}}$

since $H(C_{N,e})\leq
c_{\ref{8cons_haut_CN_major}}\theta^{r_{e}{\alpha^{N}}+(d-r_{e}){\alpha^{N+1}}}$
by Lemma 4.2. This proves the upper bound of the lemma with
$c_{\ref{8cons_major_angle_psid}}=c_{\ref{8cons_maj_prop45_appliquee}}c_{\ref{8cons_major_angle_Yj_XNj}}dc_{\ref{8cons_haut_CN_major}}^{\frac{{\alpha^{q_{e}+1}}}{r_{e}+(d-r_{e}){\alpha}}}$.

To establish the lower bound, we use the fact that
$\psi_{d}(A,C_{N,e})\geq\psi_{1}(\text{Span}(Y_{1}),C_{N,e})$ according to
Lemma 2.3 of [6]. Lemma 4.3 then gives $\psi_{d}(A,C_{N,e})\geq
c_{\ref{8cons_minor_angle_y1_CNe}}\theta^{-\alpha^{N+q_{e}+1}}.$ Since
$H(C_{N,e})\geq
c_{\ref{8cons_haut_CN_minor}}\theta^{r_{e}{\alpha^{N}}+(d-r_{e}){\alpha^{N+1}}}$
according to Lemma 4.2, we have the lower bound

$\displaystyle\psi_{d}(A,C_{N,e})\geq
c_{\ref{8cons_minor_angle_y1_CNe}}c_{\ref{8cons_haut_CN_minor}}^{\frac{{\alpha^{N+q_{e}+1}}}{r_{e}{\alpha^{N}}+(d-r_{e}){\alpha^{N+1}}}}H(C_{N,e})^{\frac{{-\alpha^{N+q_{e}+1}}}{r_{e}{\alpha^{N}}+(d-r_{e}){\alpha^{N+1}}}}=c_{\ref{8cons_minor_angle_psid}}H(C_{N,e})^{\frac{-{\alpha^{q_{e}+1}}}{r_{e}+(d-r_{e})\alpha}}$

where
$c_{\ref{8cons_minor_angle_psid}}=c_{\ref{8cons_minor_angle_y1_CNe}}c_{\ref{8cons_haut_CN_minor}}^{\frac{{\alpha^{q_{e}+1}}}{r_{e}+(d-r_{e}){\alpha}}}$,
proving the lower bound of the lemma.

∎

We have thus constructed an infinite family of rational subspaces $C_{N,e}$ of
dimension $e$ such that $\psi_{d}(A,C_{N,e})\leq
c_{\ref{8cons_major_angle_psid}}H(C_{N,e})^{\frac{-{\alpha^{q_{e}+1}}}{r_{e}+(d-r_{e})\alpha}}$
which implies in particular that
$\mu_{n}(A|e)_{d}\geq\frac{{\alpha^{q_{e}+1}}}{r_{e}+(d-r_{e})\alpha}.$

###  4.2 Upper bound on the exponent

We shall now show that the lower bound found in the previous section is
optimal, meaning that we will bound $\mu_{n}(A|e)_{d}$ from above by
$\frac{{\alpha^{q_{e}+1}}}{r_{e}+(d-r_{e})\alpha}$. We state a first technical
lemma which will be useful in the proof of the upper bound. This lemma
actually generalizes the lower bound of Lemma 4.2.
Recall that for $N$ and $v$ two positive integers, we have defined:

$\displaystyle
B_{N,v}=\text{Span}(X_{N}^{1},X_{N+1}^{1},\ldots,X_{N+v-1}^{1},X_{N}^{2}\ldots,X_{N+v-1}^{2},\ldots,X_{N}^{d},\ldots,X_{N+v-1}^{d}).$

######  Lemma 4.5.

Let $N\in\mathbb{N}$, $v\in\llbracket 1,q-1\rrbracket$, and $r\in\llbracket
0,d-1\rrbracket$. For $W$ a rational subspace of
$\text{Span}(X_{N}^{1},\ldots,X_{N}^{d})$ of dimension $r$, we have:

$\displaystyle H(B_{N+1,v}\oplus W)\geq
c_{\ref{8cons_minoration_somme_espace}}\theta^{r{\alpha^{N}}+(d-r){\alpha^{N+1}}}$

where $c_{10}>0$ is independent of $N$ and $W$.

###### Proof.

We fix $U_{1},\ldots,U_{r}$ be a $\mathbb{Z}$-basis of $W\cap\mathbb{Z}^{n}$.
Since $W\subset\text{Span}(X_{N}^{1},\ldots,X_{N}^{d})$ and these vectors form
a $\mathbb{Z}$-basis of
$\text{Span}(X_{N}^{1},\ldots,X_{N}^{d})\cap\mathbb{Z}^{n}$, we can write for
$i\in\llbracket 1,r\rrbracket$:

$\displaystyle U_{i}=\sum\limits_{j=1}^{d}a_{i,j}X_{N}^{j}$ (4.8)

with $a_{i,j}\in\mathbb{Z}$. We note that $B_{N+1,v}\oplus W$ is a direct sum
and has dimension $\dim(B_{N+1,v})+r=vd+r$ according to Lemma 3.1. The same
lemma states also that the vectors $(X_{N+1}^{j})_{j\in\llbracket
1,d\rrbracket}\cup(V_{k}^{j})_{j\in\llbracket 1,d\rrbracket,k\in\llbracket
N+2,N+v\rrbracket}$ form a $\mathbb{Z}$-basis of
$B_{N+1,v}\cap\mathbb{Z}^{n}$. By concatenating this base of $B_{N+1,v}$ with
the chosen basis $(U_{1},\ldots,U_{r})$ of $W$, we form a basis of the real
vector subspace $B_{N+1,v}\oplus W$ (but not necessarily of the
$\mathbb{Z}$-module $(B_{N+1,v}\oplus W)\cap\mathbb{Z}^{n}$). Moreover, since
the vectors of this basis have integers coordinates, the formula $(7)$ stated
in [9, section 3] gives the height of $B_{N+1,v}\oplus W$:

$\displaystyle H(B_{N+1,v}\oplus
W)=\frac{\left\|(\bigwedge_{j=1}^{d}X_{N+1}^{j})\wedge\left(\bigwedge_{k=N+2}^{N+v}(V_{k}^{1}\wedge\ldots\wedge
V_{k}^{d})\right)\wedge U_{1}\wedge\ldots\wedge U_{r}\right\|}{N(I)}$ (4.9)

where $N(I)$ is the norm of the ideal $I$ generated by the Grassmannian
coordinates associated with this basis, which, as a reminder, are the minors
of size $vd+r$ of the matrix associated with the vectors
$(X_{N+1}^{j})_{j\in\llbracket 1,d\rrbracket}\cup(V_{k}^{j})_{j\in\llbracket
1,d\rrbracket,k\in\llbracket N+2,N+v\rrbracket}\cup(U_{i})_{i\in\llbracket
1,r\rrbracket}$. According to (4.8) we have for $i\in\llbracket
1,r\rrbracket$:

$\displaystyle
U_{i}=\sum\limits_{j=1}^{d}a_{i,j}X_{N}^{j}=\sum\limits_{j=1}^{d}a_{i,j}\frac{X_{N+1}^{j}-U_{N+1}^{j}}{\theta^{\lfloor\alpha^{N+1}\rfloor-\lfloor\alpha^{N}\rfloor}}$

using also the formula (3.2). Thus:

$\displaystyle\|(\bigwedge_{j=1}^{d}X_{N+1}^{j})\wedge\left(\bigwedge_{k=N+2}^{N+v}(V_{k}^{1}\wedge\ldots\wedge
V_{k}^{d})\right)\wedge U_{1}\wedge\ldots\wedge U_{r}\|$
$\displaystyle=\left(\frac{1}{\theta^{\lfloor\alpha^{N+1}\rfloor-\lfloor\alpha^{N}\rfloor}}\right)^{r}\|(\bigwedge_{j=1}^{d}X_{N+1}^{j})\wedge\left(\bigwedge_{k=N+2}^{N+v}(V_{k}^{1}\wedge\ldots\wedge
V_{k}^{d})\right)\wedge\sum\limits_{j=1}^{d}a_{1,j}U_{N+1}^{j}\wedge\ldots\wedge\sum\limits_{j=1}^{d}a_{r,j}U_{N+1}^{j}\|$
$\displaystyle=\theta^{r\lfloor\alpha^{N}\rfloor+(d-r)\lfloor\alpha^{N+1}\rfloor}\|(\bigwedge_{j=1}^{d}Z_{N+1}^{j})\wedge\left(\bigwedge_{k=N+2}^{N+v}(V_{k}^{1}\wedge\ldots\wedge
V_{k}^{d})\right)\wedge\sum\limits_{j=1}^{d}a_{1,j}U_{N+1}^{j}\wedge\ldots\wedge\sum\limits_{j=1}^{d}a_{r,j}U_{N+1}^{j}\|$
(4.10)

with $Z_{N+1}^{j}=\frac{1}{\theta^{\lfloor\alpha^{N+1}\rfloor}}X_{N+1}^{j}$.
Now the norm of the exterior product that appears in (4.10) can be bounded
below by the absolute value of any minor of size $dv+r$ of the matrix $M$
whose columns are the vectors $(Z_{N+1}^{j})_{j\in\llbracket
1,d\rrbracket}\cup(V_{k}^{j})_{j\in\llbracket 1,d\rrbracket,k\in\llbracket
N+2,N+v\rrbracket}\cup(\sum\limits_{j=1}^{d}a_{i,j}U_{N+1}^{j})_{i\in\llbracket
1,r\rrbracket}$. We have $M=\begin{pmatrix}\begin{matrix}I_{d}\\\
\Sigma_{N+1}\end{matrix}&\begin{matrix}A\end{matrix}\end{pmatrix}\in\text{M}_{n,dv+r}(\mathbb{R})$
where $\Sigma_{N+1}=(\sigma_{i,j}^{N+1})_{i\in\llbracket
0,qd-1\rrbracket,j\in\llbracket 1,d\rrbracket}$ and
$A\in\text{M}_{n,d(v-1)+r}(\mathbb{R})$ whose columns are the vectors
$(V_{k}^{j})_{j\in\llbracket 1,d\rrbracket,k\in\llbracket
N+2,N+v\rrbracket}\cup(\sum\limits_{j=1}^{d}a_{i,j}U_{N+1}^{j})_{i\in\llbracket
1,r\rrbracket}$. According to the construction of the vectors $V_{k}^{j}$ and
$U_{N+1}^{j}$ in (3.2) and since $a_{i,j}\in\mathbb{Z}$, there exists
$A^{\prime}\in\text{M}_{n-d,d(v-1)+r}(\mathbb{Z})$ a matrix with integer
coefficients such that:

$\displaystyle M=\begin{pmatrix}\begin{matrix}I_{d}\\\
\Sigma_{N+1}\end{matrix}&\begin{matrix}A\end{matrix}\end{pmatrix}=\begin{pmatrix}\begin{matrix}I_{d}\\\
\Sigma_{N+1}\end{matrix}&\begin{matrix}0\\\
A^{\prime}\end{matrix}\end{pmatrix}.$

Moreover, the matrix $A^{\prime}$ has rank $d(v-1)+r$ because
$\operatorname{rank}(M)=dv+r$ since the columns of $M$ are vectors from a
basis. Thus, we can extract a minor from
$\begin{pmatrix}\begin{matrix}I_{d}\\\
\Sigma_{N+1}\end{matrix}&\begin{matrix}0\\\
A^{\prime}\end{matrix}\end{pmatrix}$ that is non-zero and integer. In
particular, it is bounded below in absolute value by 1, and by referring to
(4.10) we have:

$\displaystyle\|(\bigwedge\limits_{j=1}^{d}X_{N+1}^{j})\wedge\left(\bigwedge\limits_{k=N+2}^{N+v}(V_{k}^{1}\wedge\ldots\wedge
V_{k}^{d})\right)\wedge U_{1}\wedge\ldots\wedge
U_{r}\|\geq\theta^{r\lfloor\alpha^{N}\rfloor+(d-r)\lfloor\alpha^{N+1}\rfloor}.$
(4.11)

According to (4.9), it remains to show that $N(I)$ is bounded by a constant
depending only on $A$. For this, we show that the family
$(X_{N+1}^{j})_{j\in\llbracket 1,d\rrbracket}\cup(V_{k}^{j})_{j\in\llbracket
1,d\rrbracket,k\in\llbracket N+2,N+v\rrbracket}\cup(U_{i})_{i\in\llbracket
1,r\rrbracket}$ forms "almost" a $\mathbb{Z}$-basis of $(B_{N+1,v}\oplus
W)\cap\mathbb{Z}^{n}$. With this aim in mind, let $(u_{j})_{j\in\llbracket
1,d\rrbracket}\cup(w_{j,k})_{j\in\llbracket 1,d\rrbracket,k\in\llbracket
N+2,N+v\rrbracket}\cup(v_{i})_{i\in\llbracket 1,r\rrbracket}\in[0,1]^{r+vd}$
such that:

$\displaystyle
X=\sum\limits_{j=1}^{d}u_{j}X_{N+1}^{j}+\sum\limits_{\underset{j\in\llbracket
1,d\rrbracket}{k\in\llbracket
N+2,N+v\rrbracket}}w_{j,k}V_{k}^{j}+\sum\limits_{i=1}^{r}v_{i}U_{i}\in\mathbb{Z}^{n}.$
(4.12)

We decompose $X$ in the $\mathbb{Z}$-basis of $B_{N,v+1}\cap\mathbb{Z}^{n}$
explicitly described in Lemma 3.1. We have
$U_{i}=\sum\limits_{j=1}^{d}a_{i,j}X_{N}^{j}$ according to (4.8), and
$X_{N+1}^{j}=\theta^{\lfloor\alpha^{N+1}\rfloor-\lfloor\alpha^{N}\rfloor}X_{N}^{j}+\|U_{N+1}^{j}\|V_{N+1}^{j}$
using the formula (3.2). For $k\in\llbracket N+2,N+v\rrbracket$, the
$V_{k}^{j}$ do not appear in the decomposition of $U_{i}$ and $X_{N+1}^{j}$.
By definition of a $\mathbb{Z}$-basis we then have $\forall\,j\in\llbracket
1,d\rrbracket,\quad\forall\,k\in\llbracket N+2,N+v\rrbracket,\quad
w_{j,k}\in\mathbb{Z}.$ In particular, $\sum\limits_{\underset{j\in\llbracket
1,d\rrbracket}{k\in\llbracket
N+2,N+v\rrbracket}}w_{j,k}V_{k}^{j}\in\mathbb{Z}^{n}$ and the relation (4.12)
then gives
$\sum\limits_{j=1}^{d}u_{j}\left(\theta^{\lfloor\alpha^{N+1}\rfloor-\lfloor\alpha^{N}\rfloor}X_{N}^{j}+\|U_{N+1}^{j}\|V_{N+1}^{j}\right)+\sum\limits_{j=1}^{d}\left(\sum\limits_{i=1}^{r}v_{i}a_{i,j}\right)X_{N}^{j}\in\mathbb{Z}^{n}.$
According to Lemma 3.1, the $V_{N+1}^{j}$ and $X_{N}^{j}$ form a
$\mathbb{Z}$-basis of $V\cap\mathbb{Z}^{n}$, where $V$ is the real vector
subspace generated by these vectors. So we have for all $j\in\llbracket
1,d\rrbracket$:

$\displaystyle u_{j}\|U_{N+1}^{j}\|\in\mathbb{Z},$ (4.13) $\displaystyle
u_{j}\theta^{\lfloor\alpha^{N+1}\rfloor-\lfloor\alpha^{N}\rfloor}+\sum\limits_{i=1}^{r}v_{i}a_{i,j}\in\mathbb{Z}.$
(4.14)

Since $\|U_{N+1}^{j}\|=2$ or $3$, the relation (4.13) gives
$6u_{j}\in\mathbb{Z}$ for all $j\in\llbracket 1,d\rrbracket$. By the second
relation (4.14), we then have
$\sum\limits_{i=1}^{r}6v_{i}a_{i,j}\in\mathbb{Z}$ for all $j\in\llbracket
1,d\rrbracket$. Finally:
$\sum\limits_{i=1}^{r}6v_{i}U_{i}=\sum\limits_{j=1}^{d}\left(\sum\limits_{i=1}^{r}6v_{i}a_{i,j}\right)X_{N}^{j}\in\mathbb{Z}^{n}$
and as $U_{1},\ldots,U_{r}$ form a $\mathbb{Z}$-basis of $W\cap\mathbb{Z}^{n}$
we find $6v_{i}\in\mathbb{Z}$ for all $i\in\llbracket 1,r\rrbracket$. So we
finally have:

$\displaystyle w_{j,k}\in\mathbb{Z}$ $\displaystyle\text{ for
}(j,k)\in\llbracket 1,d\rrbracket\times\llbracket N+2,N+v\rrbracket,$
$\displaystyle 6u_{j}\in\mathbb{Z}$ $\displaystyle\text{ for }j\in\llbracket
1,d\rrbracket,$ $\displaystyle 6v_{i}\in\mathbb{Z}$ $\displaystyle\text{ for
}i\in\llbracket 1,r\rrbracket.$

In particular, this gives $N(I)\leq 6^{d+r}\leq 6^{2d}$. By combining this
with (4.11) in (4.9) we find $H(B_{N+1,v}\oplus W)\geq
6^{-2d}\theta^{r\lfloor\alpha^{N}\rfloor+(d-r)\lfloor\alpha^{N+1}\rfloor}\geq
c_{\ref{8cons_minoration_somme_espace}}\theta^{r\alpha^{N}+(d-r)\alpha^{N+1}}.$
This is the expected result with
$c_{\ref{8cons_minoration_somme_espace}}=6^{-2d}\theta^{-d}$.

∎

We can now bound from below the $d$-th angle that $A$ forms with any rational
subspace of dimension $e$. We first show that any good approximation of $A$
contains a subspace $B_{N+1,q_{e}}$ for a specific $N\in\mathbb{N}$.

######  Lemma 4.6.

Let $\varepsilon>0$ and $C$ be a rational subspace of dimension $e$ such that
$\psi_{d}(A,C)\leq
H(C)^{-\frac{{\alpha^{q_{e}+1}}}{r_{e}+(d-r_{e}){\alpha}}-\varepsilon}.$
Assume $H(C)$ is sufficiently large, and let $N\in\mathbb{N}$ be such that:

$\displaystyle\theta^{\alpha^{N+q_{e}}}\leq
H(C)^{\frac{\alpha^{q_{e}+1}}{r_{e}+(d-r_{e})\alpha}+\frac{\varepsilon}{2}-1}<\theta^{{\alpha^{N+q_{e}+1}}}.$
(4.15)

Then $B_{N+1,q_{e}}\subset C$.

###### Proof.

Let $N$ be the unique integer satisfying $(\ref{8choixN})$. Let
$Z_{1},\ldots,Z_{e}$ be a $\mathbb{Z}$-basis of $C\cap\mathbb{Z}^{n}$. For
$j\in\llbracket 1,d\rrbracket$, we study
$\mathcal{D}_{j,k}=\|X_{N+k}^{j}\wedge Z_{1}\wedge\ldots\wedge Z_{e}\|$ for
$k\in\llbracket 1,q_{e}\rrbracket$. Given a vector $x\in\mathbb{R}^{n}$ and a
subspace $V\subset\mathbb{R}^{n}$, we denote by $p_{V}(x)$ the orthogonal
projection of $x$ onto $V$. One has
$\mathcal{D}_{j,k}=\|p_{B^{\perp}}(X_{N+k}^{j})\|\|Z_{1}\ldots\wedge
Z_{e}\|=\psi_{1}(\text{Span}(X_{N+k}^{j}),C)\|X_{N+k}^{j}\|H(C).$ We then have
$\mathcal{D}_{j,k}\leq
c_{\ref{8cons_major_norme_XN}}\theta^{\alpha^{N+k}}H(C)\left(\omega(X_{N+k}^{j},Y_{j})+\psi_{1}(\text{Span}(Y_{j}),C)\right)$
because $\|X_{N+k}^{j}\|\leq
c_{\ref{8cons_major_norme_XN}}\theta^{\alpha^{N+k}}$ from $(\ref{8norme_XN})$,
and
$\psi_{1}(\text{Span}(X_{N+k}^{j}),C)\leq\omega(X_{N+k}^{j},Y_{j})+\psi_{1}(\text{Span}(Y_{j}),C)$
by the triangle inequality (see [9, section 8]). Now
$\psi_{1}(Y_{j},C)\leq\psi_{d}(A,C)$ by Lemma 2.3 of [6] since $e\geq d$, and
$\omega(X_{N+k}^{j},Y_{j})\leq
c_{\ref{8cons_major_angle_Yj_XNj}}\theta^{-{\alpha^{N+k+1}}}$ from
$(\ref{8major_Yj_XNj})$, thus:

$\displaystyle\mathcal{D}_{j,k}$ $\displaystyle\leq
c_{\ref{8cons_maj_1Djk}}\theta^{{\alpha^{N+k}}}H(C)\left(\theta^{-{\alpha^{N+k+1}}}+H(C)^{\frac{-\alpha^{q_{e}+1}}{r_{e}+(d-r_{e})\alpha}-\varepsilon}\right)$

where $c_{11}>0$ is independent of $N$. Furthermore,
$\theta^{{\alpha^{N+k}}}\leq\theta^{{\alpha^{N+q_{e}}}}\leq
H(C)^{\frac{\alpha^{q_{e}+1}}{r_{e}+(d-r_{e})\alpha}+\frac{\varepsilon}{2}-1}$
from the choice of $N$ in $(\ref{8choixN})$, hence $\mathcal{D}_{j,k}\leq
c_{\ref{8cons_maj_1Djk}}\theta^{{\alpha^{N+k}}-{\alpha^{N+k+1}}}H(C)+c_{\ref{8cons_maj_1Djk}}H(C)^{-\frac{\varepsilon}{2}}.$
Moreover,
$H(C)^{\frac{\alpha^{q_{e}+1}}{r_{e}+(d-r_{e})\alpha}+\frac{\varepsilon}{2}-1}\leq\theta^{{\alpha^{N+q_{e}+1}}}$
from $(\ref{8choixN})$, hence $\theta\geq
H(C)^{\frac{\frac{\alpha^{q_{e}+1}}{r_{e}+(d-r_{e})\alpha}+\frac{\varepsilon}{2}-1}{{\alpha^{N+q_{e}+1}}}}.$
Thus, we obtain the upper bound:

$\displaystyle\mathcal{D}_{j,k}$ $\displaystyle\leq
c_{\ref{8cons_maj_1Djk}}H(C)^{1+\frac{({\alpha^{N+k}}-{\alpha^{N+k+1}})(\frac{{\alpha^{q_{e}+1}}}{r_{e}+(d-r_{e})\alpha}+\frac{\varepsilon}{2}-1)}{{\alpha^{N+q_{e}+1}}}}+c_{\ref{8cons_maj_1Djk}}H(C)^{-\frac{\varepsilon}{2}}.$
(4.16)

We examine the exponent, and since $k\geq 1$:

$\displaystyle
1+\frac{({\alpha^{N+k}}-{\alpha^{N+k+1}})(\frac{{\alpha^{q_{e}+1}}}{r_{e}+(d-r_{e})\alpha}+\frac{\varepsilon}{2}-1)}{{\alpha^{N+q_{e}+1}}}$
$\displaystyle\leq
1+\frac{({\alpha^{N+1}}-{\alpha^{N+2}})(\frac{{\alpha^{q_{e}+1}}}{r_{e}+(d-r_{e})\alpha}+\frac{\varepsilon}{2}-1)}{{\alpha^{N+q_{e}+1}}}$
$\displaystyle=1+\frac{(1-{\alpha})(\frac{{\alpha^{q_{e}+1}}}{r_{e}+(d-r_{e})\alpha}+\frac{\varepsilon}{2}-1)}{{\alpha^{q_{e}}}}$
$\displaystyle\leq\frac{\alpha^{q_{e}}+(1-{\alpha})(\frac{{\alpha^{q_{e}}}}{d}+\frac{\varepsilon}{2}-1)}{{\alpha^{q_{e}}}}$

because
$\frac{{\alpha^{q_{e}+1}}}{r_{e}+(d-r_{e})\alpha}\geq\frac{{\alpha^{q_{e}}}}{d}\geq
1$ and $1-\alpha<-1$. Moreover,

$\displaystyle\alpha^{q_{e}}+(1-{\alpha})(\frac{{\alpha^{q_{e}}}}{d}-1)\leq\frac{1}{d}\left(-\alpha^{q_{e}+1}+\alpha^{q_{e}}(d+1)+d(\alpha-1)\right)\leq\frac{1}{d}\left(-\alpha^{2}+\alpha(2d+2)-d\right)$

since $e\geq d$, hence $q_{e}\geq 1$. By inequality
$(\ref{8lem_inegal_alpha1})$, we have $-\alpha^{2}+\alpha(2d+2)-d\leq 0$.
Then,
$\frac{\alpha^{q_{e}}+(1-{\alpha})(\frac{{\alpha^{q_{e}}}}{d}+\frac{\varepsilon}{2}-1)}{{\alpha^{q_{e}}}}\leq\frac{(1-\alpha)\varepsilon}{2\alpha^{q_{e}}}\leq\frac{-\varepsilon}{2\alpha^{q}}$
because $1-\alpha<-1$ and $q_{e}\leq q$. Thus, by revisiting (4.16):

$\displaystyle\mathcal{D}_{j,k}$ $\displaystyle\leq
c_{\ref{8cons_maj_1Djk}}H(C)^{-\frac{\varepsilon}{2\alpha^{q}}}+c_{\ref{8cons_maj_1Djk}}H(C)^{-\frac{\varepsilon}{2}}\leq
2c_{\ref{8cons_maj_1Djk}}H(C)^{-\frac{\varepsilon}{2\alpha^{q}}}.$

For $H(C)$ sufficiently large depending on $\varepsilon,\alpha,n$, and
$c_{\ref{8cons_maj_1Djk}}$, we have $\mathcal{D}_{j,k}<1$ for all
$j\in\llbracket 1,d\rrbracket$ and $k\in\llbracket 1,q_{e}\rrbracket$. It
implies that for such $(j,k)$ the infinite norm $\|X_{N+k}^{j}\wedge
Z_{1}\wedge\ldots\wedge Z_{e}\|_{\infty}$ vanishes since it is an integer
bounded from above by $\mathcal{D}_{j,k}$. Thus, we have a linear relation
between the vectors $X_{N+k}^{j},Z_{1},\ldots,Z_{e}$ that gives
$\forall\,j\in\llbracket 1,d\rrbracket,\quad\forall\,k\in\llbracket
1,q_{e}\rrbracket,\quad X_{N+k}^{j}\in C.$ These vectors generate
$B_{N+1,q_{e}}$, hence $B_{N+1,q_{e}}\subset C$.

∎

######  Lemma 4.7.

Let $\varepsilon>0$. For all but a finite number of rational subspaces $C$ of
dimension $e$, we have

$\psi_{d}(A,C)>H(C)^{-\frac{{\alpha^{q_{e}+1}}}{r_{e}+(d-r_{e}){\alpha}}-\varepsilon}.$

###### Proof.

Assume the contrary. Then, for some $\varepsilon>0$, there exist infinitely
many rational subspaces $C$ of dimension $e$ satisfying:

$\displaystyle\psi_{d}(A,C)\leq
H(C)^{\frac{-\alpha^{q_{e}+1}}{r_{e}+(d-r_{e}){\alpha}}-\varepsilon}.$ (4.17)

According to Lemma 4.6, if such a $C$ has sufficiently large height, which we
can assume this there infinitely many such $C$, then $B_{N+1,q_{e}}\subset C$
with $N$ satisfying (4.15). We divide the proof into two cases depending on
the value of $r_{e}$.

• First case $r_{e}=0$: In this case, we have $e=q_{e}d$ and
$\frac{{\alpha^{q_{e}+1}}}{r_{e}+(d-r_{e}){\alpha}}=\frac{\alpha^{q_{e}}}{d}$.
We already have $B_{N+1,q_{e}}\subset C$; by equality of dimensions and since
$r_{e}=0$, we have $C=B_{N+1,q_{e}}=C_{N,e}$ by Remark 4.1. Thus,
$H(C)=H(C_{N,e})$ and by Lemma 4.4 we have
$c_{\ref{8cons_minor_angle_psid}}H(C)^{\frac{-{\alpha^{q_{e}+1}}}{r_{e}+(d-r_{e}){\alpha}}}\leq\psi_{d}(A,C_{N,e})=\psi_{d}(A,C)\leq
H(C)^{-\frac{{\alpha^{q_{e}+1}}}{r_{e}+(d-r_{e}){\alpha}}-\varepsilon}.$ This
implies
$\frac{{\alpha^{q_{e}+1}}}{r_{e}+(d-r_{e}){\alpha}}\geq\frac{{\alpha^{q_{e}+1}}}{r_{e}+(d-r_{e}){\alpha}}+\frac{\varepsilon}{2}$
since we can make $H(C)$ tend to $+\infty$. Thus, we obtain
$\frac{\varepsilon}{2}\leq 0$, which is a contradiction.

• Second case $r_{e}\neq 0$: In this case, $e=q_{e}d+r_{e}$ and in particular
$1\leq q_{e}\leq q-1$. Let us prove that
$\text{Span}(X_{N}^{1},\ldots,X_{N}^{d})\cap C$ has dimension greater than or
equal to $r_{e}$ by showing by induction on $r\in\llbracket
0,r_{e}-1\rrbracket$ that there exist $r+1$ linearly independent integer
vectors in $\text{Span}(X_{N}^{1},\ldots,X_{N}^{d})\cap C$. Let
$r\in\llbracket 0,r_{e}-1\rrbracket$. Suppose there exist $U_{1},\ldots,U_{r}$
linearly independent integer vectors in
$\text{Span}(X_{N}^{1},\ldots,X_{N}^{d})\cap C$; if $r=0$, this assumption is
vacuous and thus true. Let us show that there exists
$U_{r+1}\in\text{Span}(X_{N}^{1},\ldots,X_{N}^{d})\cap C\cap\mathbb{Z}^{n}$
such that $U_{1},\ldots,U_{r+1}$ are linearly independent.

As $q_{e}\leq q-1$ and the family of vectors involved in the definition of
$B_{N+1,q_{e}}$ in (3.1) is free, we have
$B_{N+1,q_{e}}\cap\text{Span}(X_{N}^{1},\ldots,X_{N}^{d})=\\{0\\}$, and
therefore, according to Lemma 3.1,
$\dim(B_{N+1,q_{e}}\oplus\text{Span}(U_{1},\ldots,U_{r}))=dq_{e}+r.$ We denote
$G_{r}=B_{N+1,q_{e}}\oplus\text{Span}(U_{1},\ldots,U_{r})$ and
$D_{r}=G_{r}^{\perp}\cap C$ the orthogonal complement of $G_{r}$ in $C$. We
have $\dim(D_{r})=e-dq_{e}-r=r_{e}-r\geq 1$. Let $\pi_{r}:C\to D_{r}$ be the
orthogonal projection onto $D_{r}$. We define
$\Delta_{r}=\pi_{r}(C\cap\mathbb{Z}^{n})$. Then $\Delta_{r}$ is a Euclidean
lattice of $D_{r}$ with determinant $d(\Delta_{r})=\frac{H(C)}{H(G_{r})}$ ;
this result can be found in the proof of Theorem 2 of [9]. According to
Minkowski’s theorem (see [10, Lemma $4B$]), there exists
$X_{r}^{\prime}\in\Delta_{r}\setminus\\{0\\}\subset D_{r}\cap\mathbb{Q}^{n}$
such that:

$\displaystyle\|X_{r}^{\prime}\|\leq
c_{\ref{8cons_minko_1}}d(\Delta_{r})^{\frac{1}{\dim(D_{r})}}\leq
c_{\ref{8cons_minko_1}}\left(\frac{H(C)}{H(G_{r})}\right)^{\frac{1}{r_{e}-r}}$
(4.18)

with $c_{12}>0$ a constant depending only on $e$. Since
$X_{r}^{\prime}\in\Delta_{r}$, there exists $X_{r}\in C\cap\mathbb{Z}^{n}$
such that $\pi_{r}(X_{r})=X_{r}^{\prime}$; we have $X_{r}\notin G_{r}$ so that
$X_{r}\notin B_{N+1,q_{e}}.$ We define $E_{r}$ as the exterior product of the
vectors $(X_{N+k}^{j})_{j\in\llbracket 1,d\rrbracket,k\in\llbracket
0,q_{e}\rrbracket}$ and $X_{r}$. We aim to show that $E_{r}=0$ and for this,
we examine its norm. We have:

$\displaystyle\|E_{r}\|$
$\displaystyle=\|X_{r}\wedge\bigwedge\limits_{k=0}^{q_{e}}(X_{N+k}^{1}\wedge\ldots\wedge
X_{N+k}^{d})\|=\|\pi_{r}(X_{r})\wedge\bigwedge\limits_{k=0}^{q_{e}}(X_{N+k}^{1}\wedge\ldots\wedge
X_{N+k}^{d})\|$

since $X_{r}-\pi_{r}(X_{r})\in G_{r}\subset B_{N,q_{e}+1}$. Now
$\|\pi_{r}(X_{r})\|=\|X_{r}^{\prime}\|\leq
c_{\ref{8cons_minko_1}}\left(\frac{H(C)}{H(G_{r})}\right)^{\frac{1}{r-r_{e}}}$
according to (4.18) and thus $\|E_{r}\|$ is bounded from above by

$\displaystyle\|\pi_{r}(X_{r})\|\cdot\|\bigwedge\limits_{k=0}^{q_{e}}(X_{N+k}^{1}\wedge\ldots\wedge
X_{N+k}^{d})\|\leq
c_{\ref{8cons_minko_1}}\left(\frac{H(C)}{H(G_{r})}\right)^{\frac{1}{r-r_{e}}}\|(X_{N}^{1}\wedge\ldots\wedge
X_{N}^{d})\wedge\bigwedge\limits_{k=1}^{q_{e}}(U_{N+k}^{1}\wedge\ldots\wedge
U_{N+k}^{d})\|$

using the formula (3.2) on the $X_{N}^{j}$ with
$U_{N+k}^{j}=\begin{pmatrix}0&\cdots&0&u_{N+k}^{(0,j)}&\cdots&u_{N+k}^{(qd-1,j)}\end{pmatrix}^{\intercal}$.
According to the construction of the $u_{k}^{(i,j)}$ in (2.8) we have
$\|U_{k}^{j}\|\leq 3$ for all $k$ and $j$. This yields:

$\displaystyle\|E_{r}\|$ $\displaystyle\leq
c_{\ref{8cons_minko_1}}\left(\frac{H(C)}{H(G_{r})}\right)^{\frac{1}{r_{e}-r}}\|X_{N}^{1}\wedge\ldots\wedge
X_{N}^{d}\|\prod\limits_{k=1}^{q_{e}}(\|U_{N+k}^{1}\|\cdots\|U_{N+k}^{d}\|)$
$\displaystyle\leq
3^{dq}c_{\ref{8cons_minko_1}}\left(\frac{H(C)}{H(G_{r})}\right)^{\frac{1}{r-r_{e}}}\|X_{N}^{1}\wedge\ldots\wedge
X_{N}^{d}\|.$

Now $\text{Span}(X_{N}^{1},\ldots,X_{N}^{d})=B_{N,1}$ according to the
construction in (3.1). According to Lemma 3.1 $X_{N}^{1},\ldots,X_{N}^{d}$
form a $\mathbb{Z}$-basis of $B_{N,1}\cap\mathbb{Z}^{n}$ and thus
$H(B_{N,1})=\|X_{N}^{1}\wedge\ldots\wedge X_{N}^{d}\|$. Now since
$B_{N,v}=C_{N-1,dv}$ for $v\in\llbracket 1,q\rrbracket$, Lemma 4.2 gives
$H(B_{N,1})\leq c_{\ref{8cons_haut_CN_major}}\theta^{d{\alpha^{N}}}$ and thus
$\|X_{N}^{1}\wedge\ldots\wedge X_{N}^{d}\|\leq
c_{\ref{8cons_haut_CN_major}}\theta^{d{\alpha^{N}}}$. On the other hand, Lemma
4.5 yields $H(G_{r})=H(B_{N+1,q_{e}}\oplus\text{Span}(U_{1},\ldots,U_{r}))\geq
c_{\ref{8cons_minoration_somme_espace}}\theta^{r{\alpha^{N}}+(d-r){\alpha^{N+1}}}$.
Thus we have:

$\displaystyle\|E_{r}\|\leq
3^{dq}c_{\ref{8cons_minko_1}}c_{\ref{8cons_minoration_somme_espace}}^{\frac{-1}{r_{e}-r}}c_{\ref{8cons_haut_CN_major}}\theta^{d{\alpha^{N}}-\frac{r{\alpha^{N}}+(d-r){\alpha^{N+1}}}{r_{e}-r}}H(C)^{\frac{1}{r_{e}-r}}=3^{dq}c_{\ref{8cons_minko_1}}c_{\ref{8cons_minoration_somme_espace}}^{\frac{-1}{r_{e}-r}}c_{\ref{8cons_haut_CN_major}}\theta^{\frac{\alpha^{N}(d(r_{e}-r)-r-(d-r){\alpha})}{r_{e}-r}}H(C)^{\frac{1}{r_{e}-r}}.$

Now $d(r_{e}-r)-r-(d-r){\alpha}\leq dr_{e}-(d-r_{e})\alpha\leq 0$ according to
the inequality $(\ref{8lem_inegal_alpha4})$, and the choice of $N$ in (4.15)
gives
$H(C)^{\frac{{\alpha^{q_{e}}}}{\alpha^{N+q_{e}+1}(d-r_{e}+\frac{1}{2})}}\leq
H(C)^{\frac{\frac{\alpha^{q_{e}+1}}{r_{e}+(d-r_{e})\alpha}+\frac{\varepsilon}{2}-1}{\alpha^{N+q_{e}+1}}}\leq\theta$
because
$\frac{\alpha^{q_{e}}}{d-r_{e}+\frac{1}{2}}\leq\frac{\alpha^{q_{e}+1}}{r_{e}+(d-r_{e})\alpha}-1+\frac{\varepsilon}{2}$
according to the inequality $(\ref{8lem_inegal_alpha5})$. The previous
inequality then becomes:

$\displaystyle\|E_{r}\|\leq
3^{dq}c_{\ref{8cons_minko_1}}c_{\ref{8cons_minoration_somme_espace}}^{\frac{-1}{r_{e}-r}}c_{\ref{8cons_haut_CN_major}}H(C)^{\frac{\alpha^{N}\alpha^{q_{e}}(d(r_{e}-r)-r-(d-r){\alpha})}{\alpha^{N+q_{e}+1}(r_{e}-r)(d-r_{e}+\frac{1}{2})}+\frac{1}{r_{e}-r}}=3^{dq}c_{\ref{8cons_minko_1}}c_{\ref{8cons_minoration_somme_espace}}^{\frac{-1}{r_{e}-r}}c_{\ref{8cons_haut_CN_major}}H(C)^{\frac{(d(r_{e}-r)-r-(d-r){\alpha})+\alpha(d-r_{e}+\frac{1}{2})}{\alpha(r_{e}-r)(d-r_{e}+\frac{1}{2})}}.$

We study the exponent denoted by
$\delta=\frac{(d(r_{e}-r)-r-(d-r){\alpha})+\alpha(d-r_{e}+\frac{1}{2})}{\alpha(r_{e}-r)(d-r_{e}+\frac{1}{2})}$,
and we can then bound it from above as follows:

$\displaystyle\delta=\frac{1}{\alpha(r_{e}-r)(d-r_{e}+\frac{1}{2})}(-\alpha(r_{e}-r-\frac{1}{2})+d(r_{e}-r)-r)\leq\frac{1}{\alpha(r_{e}-r)(d-r_{e}+\frac{1}{2})}(-\frac{\alpha}{2}+d(d-1))$

because $0\leq r\leq r_{e}-1$ and $1\leq r_{e}-r\leq d-1$. Finally, according
to inequality $(\ref{8lem_inegal_alpha2})$, we have
$-\frac{\alpha}{2}+d(d-1)\leq-1$, thus
$\delta\leq\frac{-1}{\alpha(r_{e}-r)(d-r_{e}+\frac{1}{2})}$ and $\|E_{r}\|\leq
3^{dq}c_{\ref{8cons_minko_1}}c_{\ref{8cons_minoration_somme_espace}}^{\frac{-1}{r_{e}-r}}c_{\ref{8cons_haut_CN_major}}H(C)^{\frac{-1}{\alpha(r_{e}-r)(d-r_{e}+\frac{1}{2})}}$.
In particular, if $H(C)$ is large enough, we have $\|E_{r}\|<1$. Now, since
the vectors considered in the exterior product
$E_{r}=X_{r}\wedge\bigwedge\limits_{k=0}^{q_{e}}(X_{N+k}^{1}\wedge\ldots\wedge
X_{N+k}^{d})$ belong to $\mathbb{Z}^{n}$, this exterior product vanishes.

This implies the existence of
$U_{r+1}\in(B_{N+1,q_{e}}\oplus\text{Span}(X_{r}))\cap\text{Span}(X_{N}^{1},\ldots,X_{N}^{d})\setminus\\{0\\}$.
Recall that $X_{r}\notin B_{N+1,q_{e}}$ and since the subspaces under
consideration are rational, we can take $U_{r+1}\in\mathbb{Z}^{n}$. As
$(B_{N+1,q_{e}}\oplus\text{Span}(X_{r}))\subset C$, we also have $U_{r+1}\in
C$.

Let us now show that the vectors $U_{1},\ldots,U_{r+1}$ are linearly
independent over $\mathbb{R}$. Suppose
$\sum\limits_{k=1}^{r+1}\lambda_{k}U_{k}=0$ is a linear dependency relation.
We apply $\pi_{r}$, the orthogonal projection onto
$D_{r}=(B_{N+1,q_{e}}\oplus\text{Span}(U_{1},\ldots,U_{r}))^{\perp}\cap C$,
yielding:

$\displaystyle 0=\pi_{r}\left(\sum\limits_{k=1}^{r+1}\lambda_{k}U_{k}\right)$
$\displaystyle=\sum\limits_{k=1}^{r+1}\lambda_{k}\pi_{r}(U_{k})=\lambda_{r+1}\pi_{r}(U_{r+1}).$
(4.19)

Now, $U_{r+1}=Z+\mu X_{r}$ with $Z\in B_{N+1,q_{e}}$ and $\mu\in\mathbb{R}$.
Also, since $U_{r+1}\in\text{Span}(X_{N}^{1},\ldots,X_{N}^{d})$ and
$\text{Span}(X_{N}^{1},\ldots,X_{N}^{d})\cap B_{N+1,q_{e}}=\\{0\\}$, we have
$\mu\neq 0$. Hence $\pi_{r}(U_{r+1})=\mu X_{r}^{\prime}\neq 0$, implying
$\lambda_{r+1}=0$ using (4.19). Finally, we find
$\lambda_{1}=\ldots=\lambda_{r}=0$ using the induction hypothesis that
$U_{1},\ldots,U_{r}$ are linearly independent, which concludes the induction.

Thus, we have shown that there exists a rational subspace
$W\subset\text{Span}(X_{N}^{1},\ldots,X_{N}^{d})\cap C$ of dimension $r_{e}$.
Since $B_{N+1,q_{e}}\subset C$ and
$\text{Span}(X_{N}^{1},\ldots,X_{N}^{d})\cap B_{N+1,q_{e}}=\\{0\\}$, we have
$C=B_{N+1,q_{e}}\oplus W$ by equality of dimensions. Lemma 4.5 then gives:

$\displaystyle H(C)\geq
c_{\ref{8cons_minoration_somme_espace}}\theta^{r_{e}{\alpha^{N}}+(d-r_{e}){\alpha^{N+1}}}.$
(4.20)

Furthermore, $C=B_{N+1,q_{e}}\oplus W\subset C_{N-1,(q_{e}+1)d}$, and applying
Lemma 4.3 with $N^{\prime}=N-1$ and $e^{\prime}=q_{e^{\prime}}d$ where
$q_{e^{\prime}}=q_{e}+1$, we get:

$\displaystyle\psi_{1}(\text{Span}(Y_{1}),C_{N-1,(q_{e}+1)d})\geq
c_{\ref{8cons_minor_angle_y1_CNe}}\theta^{-\alpha^{N^{\prime}+q_{e^{\prime}}}}=c_{\ref{8cons_minor_angle_y1_CNe}}\theta^{-\alpha^{N+q_{e}+1}}$
(4.21)

Now, using Lemma 2.3 of [6], since $Y_{1}\in A\setminus\\{0\\}$ and
$\dim(A)=d$, we have:

$\displaystyle\psi_{d}(A,C)\geq\psi_{d}(A,C_{N-1,(q_{e}+1)d})\geq\psi_{1}(\text{Span}(Y_{1}),C_{N-1,(q_{e}+1)d}).$
(4.22)

Combining inequalities $(\ref{8minor_hautC_rappel})$,
$(\ref{8minor_rappel_y1_C})$, and $(\ref{8minor_angle_C})$, we get the
existence of a constant $c_{13}>0$ independent of $C$, such that
$\psi_{d}(A,C)\geq
c_{\ref{8cons_proof_infinit_derniere_cons}}H(C)^{\frac{-\alpha^{N+q_{e}+1}}{r_{e}{\alpha^{N}}+(d-r_{e}){\alpha^{N+1}}}}=c_{\ref{8cons_proof_infinit_derniere_cons}}H(C)^{\frac{-\alpha^{q_{e}+1}}{r_{e}+(d-r_{e}){\alpha}}}.$
Recalling the assumption made in $(\ref{8hyp_absurde_proof_lem14})$, we have
$c_{\ref{8cons_proof_infinit_derniere_cons}}H(C)^{\frac{-\alpha^{q_{e}+1}}{r_{e}+(d-r_{e}){\alpha}}}\leq
H(C)^{\frac{-\alpha^{q_{e}+1}}{r_{e}+(d-r_{e}){\alpha}}-\varepsilon}.$ In
particular, for all $C$ with sufficiently large $H(C)$, we have
$c_{\ref{8cons_proof_infinit_derniere_cons}}\leq H(C)^{-\varepsilon}$, and
thus $c_{\ref{8cons_proof_infinit_derniere_cons}}=0$. This is contradictory
and concludes the proof of the lemma.

∎

Lemma 4.7 yields
$\mu_{n}(A|C)_{d}\leq\frac{\alpha^{q_{e}+1}}{r_{e}+(d-r_{e}){\alpha}}.$ Since
the other inequality has been proven in section 4.1, this completes the proof
of theorem 1.2 in the case $d\leq e$.

##  5 Computation of the exponent in the case $e<d$

In this section, we show that $\mu_{n}(A|e)_{e}=\frac{\alpha}{e}$ for
$e\in\llbracket 1,d-1\rrbracket$. It is worth noting that in the case where
$e=d$, this equality also holds; indeed, we proved Theorem 1.2 in this case.
Since this case also appears in the proofs of subsequent sections, we will
reprove that $\mu_{n}(A|d)_{d}=\frac{\alpha}{d}$. Throughout the following, we
fix $e\in\llbracket 1,d\rrbracket$.

###  5.1 Lower bound on the exponent

The lower bound on the exponent $\mu_{n}(A|e)_{e}$ follows the same ideas as
in the case $e\geq d$. Here, we introduce a sequence of subspaces that
approximate $A$ very well. For $N\in\mathbb{N}$, we define:

$\displaystyle D_{N,e}=\text{Span}(X_{N}^{1},\ldots,X_{N}^{e})$ (5.1)

which is a rational subspace of dimension $e$ since
$X_{N}^{i}\in\mathbb{Z}^{n}$ for all $i$.

######  Lemma 5.1.

The vectors $X_{N}^{1},\ldots,X_{N}^{e}$ form a $\mathbb{Z}$-basis of
$D_{N,e}\cap\mathbb{Z}^{n}$ and

$\displaystyle c_{\ref{8cons_minor_haut_DN}}\theta^{e{\alpha^{N}}}\leq
H(D_{N,e})\leq c_{\ref{8cons_major_haut_DN}}\theta^{e{\alpha^{N}}}$

with $c_{14}>0$ and $c_{15}>0$ independent of $N$.

###### Proof.

By Lemma 3.1, the vectors $X_{N}^{1},\ldots,X_{N}^{e}$ are vectors of a
$\mathbb{Z}$-basis of $B_{N,v}\cap\mathbb{Z}^{n}$ introduced in (3.1). They
thus form in particular a $\mathbb{Z}$-basis of $W\cap\mathbb{Z}^{n}$ where
$W$ is the subspace they generate, namely $D_{N,e}$. By definition of the
height, we have $H(D_{N,e})=\|X_{N}^{1}\wedge\ldots\wedge X_{N}^{e}\|$.
We recall the notation
$Z_{N}^{j}=\frac{1}{\theta^{\lfloor\alpha^{N}\rfloor}}X_{N}^{j}$ for
$i\in\llbracket 1,e\rrbracket$ and we have
$Z_{N}^{j}\underset{N\to+\infty}{\longrightarrow}Y_{j}$. In particular, we
have $\|Z_{N}^{1}\wedge\ldots\wedge
Z_{N}^{e}\|\underset{N\to+\infty}{\longrightarrow}\|Y_{1}\wedge\ldots\wedge
Y_{e}\|\neq 0.$ This implies

$\displaystyle\theta^{-e\lfloor\alpha^{N}\rfloor}\|X_{N}^{1}\wedge\ldots\wedge
X_{N}^{e}\|=\|Z_{N}^{1}\wedge\ldots\wedge
Z_{N}^{e}\|\underset{N\to+\infty}{\longrightarrow}\|Y_{1}\wedge\ldots\wedge
Y_{e}\|\neq 0.$

Thus, there exist $c_{16}>0$ and $c_{17}>0$ such that for all
$N\in\mathbb{N}$,
$c_{\ref{8cons_proof_haut_DN1}}\theta^{e\lfloor\alpha^{N}\rfloor}\leq
H(D_{N,e})\leq
c_{\ref{8cons_proof_haut_DN2}}\theta^{e\lfloor\alpha^{N}\rfloor}$ which allows
us to conclude since
$\theta^{e{\alpha^{N}}-e}\leq\theta^{e\lfloor\alpha^{N}\rfloor}\leq\theta^{e\alpha^{N}}$.

∎

######  Lemma 5.2.

There exists a constant $c_{18}>0$ independent of $N$ such that:

$\displaystyle\psi_{e}(A,D_{N,e})\leq
c_{\ref{8cons_major_psie_A_DNe}}H(D_{N,e})^{-\alpha/e}$

for any $N$ sufficiently large.

###### Proof.

We recall that for $j\in\llbracket 1,d\rrbracket$ and $N\in\mathbb{N}$, we
have $\psi_{1}(\text{Span}(Y_{j}),\text{Span}(X_{N}^{j}))\leq
c_{\ref{8cons_major_angle_Yj_XNj}}\theta^{-\alpha^{N+1}}$ according to
$(\ref{8major_Yj_XNj})$. Since $\text{Span}(Y_{1},\ldots,Y_{e})\subset A$, it
follows that
$\psi_{e}(A,D_{N,e})\leq\psi_{e}(\text{Span}(Y_{1},\ldots,Y_{e}),D_{N,e})$ and
according to Lemma 6.1 of [5] we have
$\psi_{e}(\text{Span}(Y_{1},\ldots,Y_{e}),D_{N,e})\leq
c_{\ref{8cons_maj_prop45}}\sum\limits_{j=1}^{e}\psi_{1}(\text{Span}(Y_{j}),\text{Span}(X_{N}^{j}))$
with $c_{19}>0$ depending only on $Y_{1},\ldots,Y_{e}$ and $n$. Thus,

$\displaystyle\psi_{e}(A,D_{N,e})\leq
c_{\ref{8cons_maj_prop45}}c_{\ref{8cons_major_angle_Yj_XNj}}e\theta^{-{\alpha^{N+1}}}\leq
c_{\ref{8cons_maj_prop45}}c_{\ref{8cons_major_angle_Yj_XNj}}dc_{\ref{8cons_major_haut_DN}}^{\frac{{\alpha^{N+1}}}{e{\alpha^{N}}}}H(D_{N,e})^{\frac{-{\alpha^{N+1}}}{e{\alpha^{N}}}}=c_{\ref{8cons_major_psie_A_DNe}}H(D_{N,e})^{-{\alpha}/{e}}$

where
$c_{\ref{8cons_major_psie_A_DNe}}=c_{\ref{8cons_maj_prop45}}c_{\ref{8cons_major_angle_Yj_XNj}}dc_{\ref{8cons_major_haut_DN}}^{{\alpha}/{e}}$
and because $H(D_{N,e})\leq
c_{\ref{8cons_major_haut_DN}}\theta^{e{\alpha^{N}}}$ according to Lemma 5.1.

∎

Therefore, we have constructed an infinite sequence of rational subspaces
$D_{N,e}$ of dimension $e$ such that $\psi_{e}(A,D_{N,e})\leq
c_{\ref{8cons_major_psie_A_DNe}}H(D_{N,e})^{-\alpha/e}$ ; this implies in
particular $\mu_{n}(A|e)_{e}\geq\frac{\alpha}{e}.$

###  5.2 Upper bound of the exponent

The aim of this section is to establish an upper bound on $\mu_{n}(A|e)_{e}$.
We first give a necessary condition on the subspaces of best approximation of
$A$.

######  Lemma 5.3.

Let $\varepsilon>0$ and $C$ be a rational subspace of dimension $e$ such that
$\psi_{e}(A,C)\leq H(C)^{-\frac{\alpha}{e}-\varepsilon}.$ Then, if $H(C)$ is
large enough, there exist $N\in\mathbb{N}$ and
$Z\in\mathbb{Z}^{n}\setminus\\{0\\}$ such that $Z\in C\cap D_{N,d}$ and

$\displaystyle\|Z\|\leq c_{\ref{8cons_enonce_lem_minko}}H(C)^{{1}/{e}}$

with $c_{20}>0$ independent of $Z$ and $N$.

###### Proof.

According to Minkowski’s theorem (see [10, Lemma $4B$]), there exists $Z\in
C\cap\mathbb{Z}^{n}\setminus\\{0\\}$ such that:

$\displaystyle\|Z\|\leq c_{\ref{8cons_minko2}}H(C)^{{1}/{e}}$ (5.2)

with $c_{21}>0$ a constant independent of $Z$. It remains to show that there
exists some $N$ such that $Z\in
D_{N,d}=\text{Span}(X_{N}^{1},\ldots,X_{N}^{d})$. Let $Z^{A}$ be the
orthogonal projection of $Z$ onto $A$. We introduce
$a_{1},\ldots,a_{d}\in\mathbb{R}$ such that
$Z^{A}=\sum\limits_{j=1}^{d}a_{j}Y_{j}.$ We seek to show that there exists $N$
such that $\|Z\wedge X_{N}^{1}\wedge\ldots\wedge X_{N}^{d}\|$ vanishes. Since
$X_{N}^{1},\ldots,X_{N}^{d}$ form a $\mathbb{Z}$-basis of
$D_{N,d}\cap\mathbb{Z}^{n}$ by Lemma 5.1, we have, for $N\in\mathbb{N}$:

$\displaystyle\|Z\wedge X_{N}^{1}\wedge\ldots\wedge X_{N}^{d}\|$
$\displaystyle=\psi_{1}(\text{Span}(Z),D_{N,d})H(D_{N,d})\|Z\|$
$\displaystyle\leq\omega(Z,\sum\limits_{j=1}^{d}a_{j}X_{N}^{j})H(D_{N,d})\|Z\|$
$\displaystyle\leq\left(\omega(Z,Z^{A})+\omega(Z^{A},\sum\limits_{j=1}^{d}a_{j}X_{N}^{j})\right)H(D_{N,d})\|Z\|.$
(5.3)

Recalling the notation
$Z_{N}^{j}=\theta^{-\left\lfloor\alpha^{N}\right\rfloor}X_{N}^{j}$, we have:

$\displaystyle\omega(Z^{A},\sum\limits_{j=1}^{d}a_{j}X_{N}^{j})=\omega(Z^{A},\sum\limits_{j=1}^{d}a_{j}Z_{N}^{j})\leq\frac{\left\|Z^{A}-\sum\limits_{j=1}^{d}a_{j}Z_{N}^{j}\right\|}{\|Z^{A}\|}\leq\left\|\sum\limits_{j=1}^{d}a_{j}Y_{j}\right\|^{-1}\sum\limits_{j=1}^{d}|a_{j}|\left\|Y_{j}-Z_{N}^{j}\right\|.$

By construction of the $Y_{j}$, we have
$\left\|\sum\limits_{j=1}^{d}a_{j}Y_{j}\right\|\geq\sqrt{\sum\limits_{j=1}^{d}|a_{j}|^{2}}\geq\sum\limits_{j=1}^{d}\frac{|a_{j}|}{d}$.
Indeed, for $i\in\llbracket 1,d\rrbracket$, the $i$-th coordinate of $Y_{j}$
is equal to $1$ if $i=j$ and $0$ otherwise. Furthermore, we have
$\|Y_{j}-Z_{N}^{j}\|\leq\|Y_{j}\|c_{\ref{8cons_major_angle_Yj_XNj}}\theta^{-{\alpha^{N+1}}}$
by $(\ref{8major_Yj_XNj})$. Therefore, we have:

$\displaystyle\omega(Z^{A},\sum\limits_{j=1}^{d}a_{j}X_{N}^{j})\leq
c_{\ref{8cons_angle_ZA_DN}}\theta^{-{\alpha^{N+1}}}$ (5.4)

with
$c_{22}={c_{\ref{8cons_major_angle_Yj_XNj}}}{d}\max\limits_{j\in\llbracket
1,d\rrbracket}\|Y_{j}\|>0$. On the other hand, using Lemma 2.3 of [6] since
$Y_{1}\in A\setminus\\{0\\}$ and $\dim(C)=e$, we have
$\omega(Z,Z^{A})=\psi_{1}(\text{Span}(Z),A)\leq\psi_{e}(C,A).$ The hypothesis
of the lemma then gives:

$\displaystyle\omega(Z,Z^{A})\leq H(C)^{-\frac{\alpha}{e}-\varepsilon}.$ (5.5)

Combining (5.4) and (5.5) with (5.3), we get: $\|Z\wedge
X_{N}^{1}\wedge\ldots\wedge
X_{N}^{d}\|\leq\left(H(C)^{-\frac{\alpha}{e}-\varepsilon}+c_{\ref{8cons_angle_ZA_DN}}\theta^{-\alpha^{N+1}}\right)H(D_{N,d})\|Z\|.$
Additionally, by Lemma 5.1, we know that $H(D_{N,d})\leq
c_{\ref{8cons_major_haut_DN}}\theta^{d\alpha^{N}}$ and by (5.2), we have
$\|Z\|\leq c_{\ref{8cons_minko2}}H(C)^{\frac{1}{e}}$. Thus, for
$N\in\mathbb{N}$, we get:

$\displaystyle\|Z\wedge X_{N}^{1}\wedge\ldots\wedge
X_{N}^{d}\|\leq\left(H(C)^{-\frac{\alpha}{e}-\varepsilon}+c_{\ref{8cons_angle_ZA_DN}}\theta^{-\alpha^{N+1}}\right)c_{\ref{8cons_major_haut_DN}}\theta^{d\alpha^{N}}c_{\ref{8cons_minko2}}H(C)^{\frac{1}{e}}\leq
c_{\ref{8cons_major_final_ext_preuve}}\left(H(C)^{-\frac{\alpha-1}{e}-\varepsilon}\theta^{d\alpha^{N}}+\theta^{\alpha^{N}(d-\alpha)}H(C)^{\frac{1}{e}}\right)$

with
$c_{23}=c_{\ref{8cons_major_haut_DN}}c_{\ref{8cons_minko2}}\max(1,c_{\ref{8cons_angle_ZA_DN}})>0$
independent of $C$ and $N$. We now choose $N$ as the integer such that:

$\displaystyle\theta^{d\alpha^{N}}\leq
H(C)^{\frac{\alpha-1}{e}+\frac{\varepsilon}{2}}<\theta^{d\alpha^{N+1}}.$ (5.6)

This gives $\theta^{d\alpha^{N}}\leq
H(C)^{\frac{\alpha-1}{e}+\frac{\varepsilon}{2}}$ and
$\theta>H(C)^{\frac{\frac{\alpha-1}{e}+\frac{\varepsilon}{2}}{d\alpha^{N+1}}}$
hence, since $\alpha>d$:

$\displaystyle\|Z\wedge X_{N}^{1}\wedge\ldots\wedge X_{N}^{d}\|$
$\displaystyle\leq
c_{\ref{8cons_major_final_ext_preuve}}\left(H(C)^{-\frac{\varepsilon}{2}}+H(C)^{\alpha^{N}(d-\alpha)\left(\frac{\frac{\alpha-1}{e}+\frac{\varepsilon}{2}}{d\alpha^{N+1}}\right)+\frac{1}{e}}\right)$
$\displaystyle\leq
c_{\ref{8cons_major_final_ext_preuve}}\left(H(C)^{-\frac{\varepsilon}{2}}+H(C)^{\frac{(d-\alpha)(2\alpha-2+e\varepsilon)+2d\alpha}{2ed\alpha}}\right).$
(5.7)

We now study the exponent of the second term:
$\frac{(d-\alpha)(2\alpha-2+e\varepsilon)+2d\alpha}{2ed\alpha}=\frac{1}{ed\alpha}\left(-\alpha^{2}+(1+2d)\alpha-d\right)-\frac{\alpha-d}{2d\alpha}\varepsilon.$
According to inequality (2.3), we have $-\alpha^{2}+(1+2d)\alpha-d\leq 0$, and
thus $\frac{(d-\alpha)(2\alpha-2+e\varepsilon)+2d\alpha}{2ed\alpha}\leq-
c_{\ref{8cons_major_epsilon}}\varepsilon$ with
$c_{24}=\frac{\alpha-d}{2d\alpha}>0$ since $\alpha>d$. Thus, inequality (5.7)
becomes $\|Z\wedge X_{N}^{1}\wedge\ldots\wedge X_{N}^{d}\|\leq
c_{\ref{8cons_major_final_ext_preuve}}\left(H(C)^{-\frac{\varepsilon}{2}}+H(C)^{-c_{\ref{8cons_major_epsilon}}\varepsilon}\right)$
with $c_{\ref{8cons_major_final_ext_preuve}}$ and
$c_{\ref{8cons_major_epsilon}}$ independent of $C$. For sufficiently large
$H(C)$, we have $\|Z\wedge X_{N}^{1}\wedge\ldots\wedge
X_{N}^{d}\|_{\infty}\leq\|Z\wedge X_{N}^{1}\wedge\ldots\wedge X_{N}^{d}\|<1$
and then

$\displaystyle Z\in\text{Span}(X_{N}^{1},\ldots,X_{N}^{d})=D_{N,d}$

for $N$ satisfying (5.6), which completes the proof.

∎

We can now prove the result that allows us to provide a lower bound for the
exponent.

######  Lemma 5.4.

Let $\varepsilon>0$. For all but a finite number of rational subspaces $C$ of
dimension $e$, we have:

$\displaystyle\psi_{e}(A,C)>H(C)^{-\frac{\alpha}{e}-\varepsilon}.$

###### Proof.

We argue by contradiction and assume that there exists infinitely many
rational subspaces $C$ of dimension $e$ such that:

$\displaystyle\psi_{e}(A,C)\leq H(C)^{-\frac{\alpha}{e}-\varepsilon}.$ (5.8)

Let $C$ be such a subspace. Lemma 5.3 then gives $N\in\mathbb{N}$ and
$Z\in\mathbb{Z}^{n}\setminus\\{0\\}$ such that $Z\in C\cap D_{N,d}$ with
$\|Z\|\leq c_{\ref{8cons_enonce_lem_minko}}H(C)^{\frac{1}{e}}.$ Since $Z\in
D_{N,d}\cap\mathbb{Z}^{n}$, we can write
$Z=\sum\limits_{j=1}^{d}v_{j}X_{N}^{j}$ with $v_{j}\in\mathbb{Z}$ because the
$X_{N}^{j}$ form a $\mathbb{Z}$-basis of $D_{N,d}\cap\mathbb{Z}^{n}$ according
to Lemma 5.1. For $j\in\llbracket 1,d\rrbracket$, let
$z_{j}=\theta^{\left\lfloor\alpha^{N}\right\rfloor}v_{j}$, which gives
$Z=\sum\limits_{j=1}^{d}z_{j}Z_{N}^{j}$. We now bound $\psi_{1}(Z,A)$ from
below. Recall the notation $Z^{A}$ for the orthogonal projection of $Z$ onto
$A$. We introduce $a_{1},\ldots,a_{d}\in\mathbb{R}$ such that
$Z^{A}=\sum\limits_{j=1}^{d}a_{j}Y_{j}.$ We also define
$\Delta=Z^{A}-\sum\limits_{j=1}^{d}z_{j}Y_{j},$ and $\omega=\|Z^{A}-Z\|.$
Since the $Z_{N}^{j}$ and $Y_{j}$ have their $i$-th coordinate equal to $1$ if
$i=j$ and $0$ otherwise, this gives:

$\displaystyle
Z^{A}-Z=\begin{pmatrix}a_{1}-z_{1}&\cdots&a_{d}-z_{d}&\star&\ldots&\star\end{pmatrix}^{\intercal}.$

We then have $\omega^{2}\geq\sum\limits_{j=1}^{d}(a_{j}-z_{j})^{2},$ and thus
for all $j\in\llbracket 1,d\rrbracket$, $|a_{j}-z_{j}|\leq\omega$. This gives
us in particular:

$\displaystyle\|\Delta\|=\left\|\sum\limits_{j=1}^{d}(a_{j}-z_{j})Y_{j}\right\|\leq
c_{\ref{8cons_maj_Delta_par_omega}}\omega$ (5.9)

with $c_{25}=d\max\limits_{j\in\llbracket 1,d\rrbracket}\|Y_{j}\|>0$. We can
compute $\|Z\wedge
Z^{A}\|=\|Z\wedge\left(\sum\limits_{j=1}^{d}z_{j}Y_{j}+Z^{A}-\sum\limits_{j=1}^{d}z_{j}Y_{j}\right)\|$
and then bound it from below:

$\displaystyle\|Z\wedge
Z^{A}\|=\|Z\wedge\sum\limits_{j=1}^{d}z_{j}Y_{j}+Z\wedge\Delta\|\geq\|Z\wedge\sum\limits_{j=1}^{d}z_{j}Y_{j}\|-\|Z\wedge\Delta\|.$
(5.10)

On the one hand, for all $j_{0}\in\llbracket 1,d\rrbracket$ and
$i\in\llbracket 0,qd-1\rrbracket$, recalling the definitions of $Y_{j}$ and
$\sigma_{i,j}$ given in section 2 and considering the $j_{0}$-th and
$(d+i)$-th rows of the matrix
$\begin{pmatrix}Z&\left|\>\sum\limits_{j=1}^{d}z_{j}Y_{j}\right.\end{pmatrix}$,
we have
$\|Z\wedge\sum\limits_{j=1}^{d}z_{j}Y_{j}\|=\|\sum\limits_{j=1}^{d}z_{j}Z_{N}^{j}\wedge\sum\limits_{j=1}^{d}z_{j}Y_{j}\|\geq\left|\det\begin{pmatrix}z_{j_{0}}&z_{j_{0}}\\\
\sum\limits_{j=1}^{d}z_{j}\sigma_{i,j}^{N}&\sum\limits_{j=1}^{d}z_{j}\sigma_{i,j}\end{pmatrix}\right|$
and thus :

$\displaystyle\|Z\wedge\sum\limits_{j=1}^{d}z_{j}Y_{j}\|\geq|z_{j_{0}}|\left|\sum\limits_{j=1}^{d}z_{j}(\sigma_{i,j}-\sigma_{i,j}^{N})\right|=|z_{j_{0}}|\left|\sum\limits_{j=1}^{d}z_{j}\sum\limits_{k=N+1}^{+\infty}\frac{u_{k}^{(i,j)}}{\theta^{\left\lfloor\alpha^{k}\right\rfloor}}\right|.$

We choose $j_{0}$ such that $|z_{j_{0}}|=\max\limits_{j=1}^{d}{|z_{j}|}\neq
0$; such a $j_{0}$ exists because $Z\neq 0$. In particular, we have
$|z_{j_{0}}|=\theta^{\left\lfloor\alpha^{N}\right\rfloor}|v_{j_{0}}|\geq\theta^{\left\lfloor\alpha^{N}\right\rfloor}$
since $v_{j_{0}}\in\mathbb{Z}$. We choose $i$ such that
$u_{N+1}^{(i,j_{0})}\neq 0$, that is $i=\phi_{j_{0}}(N+1)$ using the notations
from section 2. Then, we have $u_{N+1}^{(i,j_{0})}\geq 2$ and
$u_{N+1}^{(i,j)}=0$ for all $j\neq j_{0}$, hence:

$\displaystyle\|Z\wedge\sum\limits_{j=1}^{d}z_{j}Y_{j}\|$
$\displaystyle\geq|z_{j_{0}}|\left|\sum\limits_{j=1}^{d}z_{j}\sum\limits_{k=N+1}^{+\infty}\frac{u_{k}^{(i,j)}}{\theta^{\left\lfloor\alpha^{k}\right\rfloor}}\right|$
$\displaystyle\geq|z_{j_{0}}|\left|z_{j_{0}}\frac{u_{N+1}^{(i,j_{0})}}{\theta^{\left\lfloor\alpha^{N+1}\right\rfloor}}+\sum\limits_{j=1}^{d}z_{j}\sum\limits_{k=N+2}^{+\infty}\frac{u_{k}^{(i,j)}}{\theta^{\left\lfloor\alpha^{k}\right\rfloor}}\right|$
$\displaystyle\geq|z_{j_{0}}|\left(|z_{j_{0}}|\frac{2}{\theta^{\left\lfloor\alpha^{N+1}\right\rfloor}}-d|z_{j_{0}}|\sum\limits_{k=N+2}^{+\infty}\frac{u_{k}^{(i,j)}}{\theta^{\left\lfloor\alpha^{k}\right\rfloor}}\right)$
$\displaystyle\geq|z_{j_{0}}|^{2}\frac{1}{\theta^{\left\lfloor\alpha^{N+1}\right\rfloor}}$

since if $N$ is large enough, we have
$d\sum\limits_{k=N+2}^{+\infty}\frac{u_{k}^{(i,j)}}{\theta^{\left\lfloor\alpha^{k}\right\rfloor}}\leq\frac{1}{\theta^{\left\lfloor\alpha^{N+1}\right\rfloor}}$.
We deduce that:

$\displaystyle\|Z\wedge\sum\limits_{j=1}^{d}z_{j}Y_{j}\|$
$\displaystyle\geq\frac{\theta^{2\left\lfloor\alpha^{N}\right\rfloor}}{\theta^{\left\lfloor\alpha^{N+1}\right\rfloor}}.$
(5.11)

On the other hand, according to (5.9), we have:

$\displaystyle\|Z\wedge\Delta\|\leq\|Z\|\|\Delta\|\leq
c_{\ref{8cons_maj_Delta_par_omega}}\omega\|Z\|.$ (5.12)

Combining (5.10) with (5.11) and (5.12), we find $\|Z\wedge
Z^{A}\|\geq\frac{\theta^{2\left\lfloor\alpha^{N}\right\rfloor}}{\theta^{\left\lfloor\alpha^{N+1}\right\rfloor}}-c_{\ref{8cons_maj_Delta_par_omega}}\omega\|Z\|.$
We have $\omega=\|Z^{A}-Z\|=\|Z\|\omega(Z^{A},Z)$ and thus, since
$\|Z^{A}\|\leq\|Z\|$, we get:

$\displaystyle\omega=\|Z\|\frac{\|Z^{A}\wedge
Z\|}{\|Z\|\cdot\|Z^{A}\|}\geq\frac{\|Z^{A}\wedge
Z\|}{\|Z\|}\geq\frac{\theta^{2\left\lfloor\alpha^{N}\right\rfloor}}{\|Z\|\theta^{\left\lfloor\alpha^{N+1}\right\rfloor}}-c_{\ref{8cons_maj_Delta_par_omega}}\omega.$

Finally, we have
$\omega\geq\frac{c_{\ref{8cons_minor_omega}}}{\|Z\|\theta^{\left\lfloor\alpha^{N+1}\right\rfloor-2\left\lfloor\alpha^{N}\right\rfloor}}$
with $c_{26}=(1+c_{\ref{8cons_maj_Delta_par_omega}})^{-1}>0$. We also have
$\omega(Z,Z^{A})=\frac{\omega}{\|Z\|}\geq\frac{c_{\ref{8cons_minor_omega}}}{\|Z\|^{2}\theta^{\left\lfloor\alpha^{N+1}\right\rfloor-2\left\lfloor\alpha^{N}\right\rfloor}},$
where $c_{\ref{8cons_minor_omega}}>0$ is independent of $N$ and $Z$. Recalling
that $Z\in C$, and thus
$\omega(Z,Z^{A})=\psi_{1}(\text{Span}(Z),A)\leq\psi_{e}(C,A)$, by (5.8) we
have:

$\displaystyle\frac{c_{\ref{8cons_minor_omega}}}{\|Z\|^{2}\theta^{\left\lfloor\alpha^{N+1}\right\rfloor-2\left\lfloor\alpha^{N}\right\rfloor}}\leq
H(C)^{-\frac{\alpha}{e}-\varepsilon}.$ (5.13)

Moreover, we have chosen $Z$ such that $\|Z\|\leq
c_{\ref{8cons_enonce_lem_minko}}H(C)^{{1}/{e}}$. Hence, inequality (5.13)
becomes
$\frac{c_{\ref{8cons_minor_omega}}}{c_{\ref{8cons_enonce_lem_minko}}^{\alpha+e\varepsilon}\|Z\|^{2}\theta^{\left\lfloor\alpha^{N+1}\right\rfloor-2\left\lfloor\alpha^{N}\right\rfloor}}\leq\|Z\|^{-{\alpha}-e\varepsilon}.$
In particular, we have:

$\displaystyle\frac{c_{\ref{8cons_derniere_contradiction}}}{\|Z\|^{2}\theta^{\left\lfloor\alpha^{N+1}\right\rfloor-2\left\lfloor\alpha^{N}\right\rfloor}}\leq\|Z\|^{-{\alpha}-e\varepsilon}$
(5.14)

with
$c_{27}=c_{\ref{8cons_minor_omega}}c_{\ref{8cons_enonce_lem_minko}}^{-\alpha-e\varepsilon}>0$.
On the other hand, by the construction of $Z_{N}^{j}$, we have:
$\forall\,j\in\llbracket 1,d\rrbracket,\quad\|Z\|\geq|z_{j}|.$ Indeed, for
$i\in\llbracket 1,d\rrbracket$, the $i\text{-}$th coordinate of $Z_{N}^{j}$ is
equal to $1$ if $i=j$ and $0$ otherwise. In particular, we have
$\|Z\|\geq|z_{j_{0}}|\geq\theta^{\left\lfloor\alpha^{N}\right\rfloor}$ since
$z_{j_{0}}\neq 0$. Combining this with (5.14), we find
$c_{\ref{8cons_derniere_contradiction}}\leq\theta^{-\left\lfloor\alpha^{N}\right\rfloor(\alpha+e\varepsilon-2)+\left\lfloor\alpha^{N+1}\right\rfloor-2\left\lfloor\alpha^{N}\right\rfloor}$
with, as a reminder, $c_{\ref{8cons_derniere_contradiction}}>0$ being a
constant independent of $N$. We have:

$\displaystyle-\left\lfloor\alpha^{N}\right\rfloor(\alpha+e\varepsilon-2)+\left\lfloor\alpha^{N+1}\right\rfloor-2\left\lfloor\alpha^{N}\right\rfloor=\left\lfloor\alpha^{N+1}\right\rfloor-\left\lfloor\alpha^{N}\right\rfloor(\alpha+e\varepsilon)\underset{N\to+\infty}{\longrightarrow}-\infty.$

If $H(C)$ is large, then $N$ is also large by (4.15). Therefore, as $H(C)$
tends to $+\infty$, we find $c_{\ref{8cons_derniere_contradiction}}=0$, which
is contradictory and completes the proof of the lemma.

∎

Lemma 5.4 then yields for $e\in\llbracket 1,d-1\rrbracket$,
$\mu_{n}(A|C)_{e}\leq\frac{\alpha}{e}.$ Thus, we have proven Theorem 1.2 in
the case where $e<d$.

Acknowledgements: I warmly thank Stéphane Fischler for his reading and
comments of this paper.

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Gaétan Guillot, Université Paris-Saclay, CNRS, Laboratoire de mathématiques
d’Orsay, 91405 Orsay, France.
Email adress<EMAIL_ADDRESS>
MSC: 11J13, 11H25