\section{Introduction}

In 1936, Aleksandr Khintchin showed that there exists a constant $\gamma$ such that the denominators $(q_{n})_{n\geq 0}$
of the convergents of the continued fraction expansions of almost all real
numbers $\theta$ satisfy
\[
\lim_{n\rightarrow\infty}q_{n}^{1/n}=\gamma
\]
(see \cite{Khin1936}). Soon afterward, in
\cite{Levy}, in the footnote page 289, Paul L\'{e}vy gave the explicit value
of the constant,
\[
\gamma=\exp\frac{\pi^{2}}{12\ln 2}.
\]


In 1983, Wieb Bosma, Hendrik Jager and Freek Wiedijk, proved the following
conjecture due to Hendrik Lenstra: for almost all real numbers $\theta,$
\[
\lim_{n\rightarrow\infty}\frac{1}{n}\operatorname*{card}\{0\leq k<n:q_{k}%
\dd(q_{k}\theta,\mathbb{Z})\leq t\}=g(t)
\]
for all $t\in[0,1]$, where 
\[
g(t)=\int_0^t\frac{1}{2\ln 2}\frac{1-\left\vert
1-2s\right\vert}{s}\,ds.
\] 
Later, Jager proved variants of this result in
particular with the quantity $q_{k+1}\dd(q_{k}\theta,\mathbb{Z})$ instead of
$q_{k}\dd(q_{k}\theta,\mathbb{Z})$.

The aim of the paper is to extend to best simultaneous Diophantine
approximations, both L\'{e}vy-Khintchin's result and Bosma, Jager and Wiedijk's result.

Let $d$ and $c$ be two positive integers. Suppose $\mathbb{R}^{d}$ and
$\mathbb{R}^{c}$ are endowed with the standard Euclidean norms $\left\Vert
.\right\Vert _{\mathbb{R}^{d}}$ and $\left\Vert .\right\Vert _{\mathbb{R}^{c}%
}$. We prove

\begin{theorem}
\label{thm:levy-d} There exists a constant $L_{d,c}$ such that for almost all
matrices $\theta\in\M_{d,c}(\mathbb{R})$,
\begin{align*}
\lim_{n\rightarrow\infty}\frac{1}{n}\ln\left\Vert Q_{n}(\theta)\right\Vert
_{\mathbb R^c}  &  =L_{d,c}\,,\\
\lim_{n\rightarrow\infty}\frac{1}{n}\ln(\dd(\theta Q_{n}(\theta),\mathbb{Z}%
^{d}))  &  =-\frac{c}{d}L_{d,c}%
\end{align*}
where $Q_{n}(\theta)\in\mathbb{Z}^{c}$, $n\geq 0$, is the sequence of best
Diophantine approximation denominators of $\theta$ associated with the norms
$\left\Vert .\right\Vert _{\mathbb{R}^{d}}$ and $\left\Vert .\right\Vert
_{\mathbb{R}^{c}}$.
\end{theorem}

(See section 2.1, the definition of best Diophantine approximation denominators).

For a matrix $\theta$ in $\M_{d,c}(\mathbb{R)}$, let denote $\beta_{n}%
(\theta)=\left\Vert Q_{n+1}(\theta)\right\Vert ^{c} _{\mathbb{R}^{c}}\dd(\theta
Q_{n}(\theta),\mathbb{Z}^{d})^{d}$.

\begin{theorem}
\label{thm:jager}1. There exists a probability measure $\nu_{d,c}$ on
$\mathbb{R}$ such that for almost all matrices $\theta\in\M_{d,c}(\mathbb{R}%
)$, $\nu_{d,c}$ is the weak limit of the sequence of probability measures
\[
\frac{1}{n}\sum_{k=0}^{n-1}\delta_{\beta_{k}(\theta)}%
\]

where $\delta_{a}$ is the Dirac measure at $a$. \newline 2. The support of the
measure $\nu_{d,c}$ is included in a  bounded interval, and contains $0$ provided
that $c+d\geq 3$.
\end{theorem}

L\'{e}vy-Khintchin's result has already been extended to multi-dimensional
settings. For instance, for almost all $\theta$ in $\mathbb{R}^{d}$, the
denominators $(J_{n}(\theta))_{n\geq 0}$ of the Jacobi-Perron expansion of
$\theta$ satisfy $\lim_{n\rightarrow\infty}\frac{1}{n}\ln J_{n}(\theta
)=c_{d}$ for some constant $c_{d}$ (see \cite{Broise}). The common proofs of such results use ergodic theory. The one-dimensional
L\'{e}vy-Khintchin's result can be proved with Birkhoff ergodic Theorem, while
the growth rate of the Jacobi-Perron denominators can be derived from Oseledec
multiplicative ergodic Theorem. In both cases, the proof depends on the
existence of an underlying dynamical system: the Gauss map or the
Jacobi-Perron map (see \cite{Schweiger} for many examples of these kinds of
maps). However, no such map associated with best Diophantine approximations is
known when $d+c\geq 3$. One classical way to circumvent this problem is
to use the action of the diagonal flow
\[
g_{t}=\left(
\begin{array}
[c]{cc}%
e^{ct}I_{d} & 0\\
0 & e^{-dt}I_{c}%
\end{array}
\right)  \in\SL (d+c,\mathbb{R)}%
\]
on the space of unimodular lattices $\mathcal{L}_{d+c}=\SL (d+c,\mathbb{R}%
)/\SL (d+c,\mathbb{Z)}$. For instance, in \cite{Chev-Levy} this flow is used
to prove that the sequence of best Diophantine approximation denominators of
almost all $\theta$ in $\M_{d,c}(\mathbb{R})$ satisfies
\[
\lim\sup_{n\rightarrow\infty}\frac{1}{n}\ln\left\Vert Q_{n}(\theta)\right\Vert
_{\mathbb{R}^{c}}\leq K_{d,c}%
\]
for some constant $K_{d,c}$. When $c=1$, it is also possible to derive this inequality
from a Theorem of W. M. Schmidt (see
\cite{Chev-Khin}).

In this work, as in \cite{Chev-Levy}, the flow $(g_{t})$ is the main tool.
Together with the flow, an important ingredient is a surface $S$ of co--dimension $1$ transverse to the flow and the first return map
associated with the flow. Such
transversals have been widely used and we only mention two closely related works.

Firstly, P. Arnoux and A. Nogueira in \cite{Arn-Nog}, have used
transversals to naturally obtain   invariant measures associated with
multidimensional continued fraction algorithms. Furthermore, in the case of
the one dimensional continued fraction algorithm, their approach leads to an
interpretation of the L\'{e}vy's constant as the average return time of the
flow on the transversal.

Secondly, in some cases, the transformation induced on a sub-interval by an interval
exchange transformation $T_{1}$ is an interval exchange transformation $T_{2}$
of the same kind as $T_{1}$. In such situations, the map $T_{1} \rightarrow
T_{2}$ can be seen as the Gauss map of a ``multidimensional continued
fraction algorithm". In \cite{Veech}, W. Veech used a transversal to prove
that this Gauss map admits an unique absolutely continuous invariant measure up
to a scalar multiple.

In our case, the transversal is the set of unimodular lattices the first two  minimums of which are equal (actually, the definition is slightly more restrictive, see
section 3.1 for the exact definition of the transversal). It is crucial to observe
that the visiting times of the transversal are given by a
formula involving best simultaneous Diophantine approximations, see
Lemma \ref{lem:return-best}. Making use of Birkhoff Theorem, this observation leads to a
L\'{e}vy-Khintchin result in the space of lattices and to a formula close to
the Arnoux-Nogueira interpretation of the L\'{e}vy's constant:
\begin{equation}
L_{d,c}=\frac{d}{\mu_{S}(S)}\int_{S}\tau~d\mu_{S}=\frac{d\times\mu
(\mathcal{L}_{d+c})}{\mu_{S}(S)} \label{constant-levy}
\end{equation}
where $\mu$ is the invariant measure in the space of lattices, $\mu_{S}$ the
invariant measure induced by the flow on the transversal $S$ and $\tau$ the
return time to $S$, see Theorem~\ref{Birkhoff} and Corollary \ref{cor:levy-d}.

The second step of the proof of Theorem~\ref{thm:levy-d} consists in converting an
almost all result in the space of lattices $\mathcal{L}_{d+c}$ into an almost all result in
$\M_{d,c}(\mathbb{R})$. To achieve this goal, we prove a general result,
Theorem \ref{thm:reduction}, which might be of independent interest. At
first sight, this result might appear as an easy consequence of the following standard fact: the
set of lattices associated with the matrices $\theta$ in $\M_{d,c}%
(\mathbb{R)}$, is the expanding direction of the flow $g_t$. However, an example
shows that Theorem \ref{thm:reduction} depends on some properties of the transversal, see
section \ref{sect:reduction}.

When $d=1$ or $2$ and $c=1$, the submanifold $S$ and  the measure $\mu_{S}$ can be entirely calculated 
(see section \ref{sec:S}). When $d=c=1$, thanks to  Siegel formula
giving the volume of the modular space $\SL(2,\mathbb R)/\SL(2,\mathbb Z)$,
computing  L\'evy's constant $L_{1,1}=\ln \gamma$ is easy; it is even
possible to determine the first return map to the transversal. It
turns out that this first return map is a $2$-fold extension of the
natural extension of the Gauss map (see subsection \ref{sub:d=1}).
However, when $d=2$ and $c=1$, the calculation of
$L_{2,1}$ leads to a seven-tuple integral and we only succeed in reducing it
to a triple integral that  can be evaluated numerically (see subsection \ref{sub:d=2}).

When $d=c=1$, the double inequality $\frac{1}{2}\leq q_{n+1}\dd(q_{n}%
\theta,\mathbb{Z})\leq 1$ shows that the behaviors of the two sequences
$(\frac{1}{n}\ln q_{n})_n$ and $(\frac{-1}{n}\ln \dd(q_{n}\theta,\mathbb{Z}))_n$ are the
same and each of the limits in Theorem \ref{thm:levy-d} implies the other.
When $d$ is larger or equal than two, no such double inequality exists. Indeed, it
has been proved in $\cite{Chev-Khin}$ that when $c=1$ and $d\geq 2$,
\[
\lim\inf_{n\rightarrow\infty}q_{n+1}\dd(q_{n}\theta,\mathbb{Z}^{d})^{d}=0
\]
for almost all $\theta$ in $\mathbb{R}^{d}$. Observe that  Theorem \ref{thm:jager} implies this latter result; it is an immediate consequence of the
fact that $0$ is in the support of the measure $\nu_{d,c}$. Hopefully for the proof of
Theorem \ref{thm:levy-d}, the weaker inequality
\[
\dd(\theta Q_{n}(\theta),\mathbb{Z}^{d})\geq\frac{1}{\left\Vert Q_{n}%
(\theta)\right\Vert _{\mathbb{R}^{c}}^{c/d}\ln\left\Vert Q_{n}(\theta
)\right\Vert }
\]
which holds almost surely by the convergence part of the Khintchin-Groshev
Theorem, is enough to link both of the limits in Theorem \ref{thm:levy-d}.

In the last section, we extend the aforementioned result of \cite{Chev-Khin}
to best simultaneous approximations of matrices. Our proof leads to the
stronger result
\[
\lim\inf_{n\rightarrow\infty}\left\Vert Q_{n+k}(\theta)\right\Vert
_{\mathbb{R}^{c}}^{c}\dd(\theta Q_{n}(\theta),\mathbb{Z}^{d})^{d}=0
\]
for almost all $\theta$ in $\M_{d,c}(\mathbb{R)}$ and all $k\in\mathbb{N}$.

Obviously, if the matrix $\theta$ is badly approximable, then
\[
\lim\inf_{n\rightarrow\infty}\left\Vert Q_{n+k}(\theta)\right\Vert
_{\mathbb{R}^{c}}^{c}\dd(\theta Q_{n}(\theta),\mathbb{Z}^{d})^{d}>0
\]
because by definition $\lim\inf_{n\rightarrow\infty}\left\Vert Q_{n}(\theta)\right\Vert
_{\mathbb{R}^{c}}^{c}\dd(\theta Q_{n}(\theta
),\mathbb{Z}^{d})^{d}>0$. When $d=2$ and $c=1$, we prove that the set of
$\theta$ with $\lim\inf_{n\rightarrow\infty}q_{n+1}\dd(q_{n}\theta
,\mathbb{Z})^{2}>0$ is not reduced to the set of badly approximable vectors.

Historical Note: L\'{e}vy's proof does not rely on the Ergodic Theorem which
was not known for non-invertible maps at that time. A proof of the Birkhoff
Theorem for non-invertible maps was given by Fr\'{e}d\'{e}ric Riesz in 1945
(see \cite{Riesz}) and then a proof of L\'{e}vy's Theorem using the Ergodic Theorem
was given in \cite{Ryll-Nardzewski}. The authors would like to thank Vitaly
Bergelson for bringing \cite{Ryll-Nardzewski} to their attention.

\section{Notation}
Let $d$ and $c$ be two positive integers.
\subsection{Vectors and distances}
Let $\left\Vert \cdot\right\Vert _{\mathbb{R}^{n}}$ denote the usual
Euclidean norm on $\mathbb{R}^{n}$.

We assume $\mathbb{R}^{d}$ and $\mathbb{R}^{c}$ are equipped with the usual Euclidean norms and $\mathbb{R}^{d+c}$ is equipped with
the norm $\left\Vert (u,h)\right\Vert _{\mathbb{R}^{d+c}}=\max\{\left\Vert
u\right\Vert _{\mathbb{R}^{d}},\left\Vert h\right\Vert _{\mathbb{R}^{c}}\}$. 


For $X=(u,h)$ in $\mathbb{R}^{d+c}$, let  $\left|  X\right|  _{-}=\left\|
h\right\|  _{\mathbb{R}^{c}}$  denote the \textit{height} of the vector $X$ and
$\left|  X\right|  _{+}=\left\|  u\right\|  _{\mathbb{R}^{d}}$ denote the norm of the
projection of $X$ in the horizontal space. We also denote $X_{+}=u$ and
$X_{-}=h$ the vertical and horizontal components of $X$.


For a vector $X$ in $\mathbb{R}^{d+c}$, let  $C(X)$ denote  the closed cylinder
\[
C(X)=B_{\mathbb{R}^{d}}(0,\left\vert X\right\vert _{+}%
)\times B_{\mathbb{R}^{c}}(0,\left\vert X\right\vert _{-}),
\]
and if $Y$ is another vector, let $C(X,Y)$ denote  the closed cylinder
\[
C(X,Y)=B_{\mathbb{R}^{d}}(0,\left\vert X\right\vert _{+}%
)\times B_{\mathbb{R}^{c}}(0,\left\vert Y\right\vert _{-}).
\]


In all situations, let $\dd(x,y)$ denote the distance associated with the underlying norm between the two points $x$ and $y$ and $\dd(x,A)$ the distance between the point $x$ and the set $A$.

\subsection{Matrices}

Let $I_{n}$ denote the identity matrix in $\M_{n}(\mathbb{R)}$. 

 We fix once and
for all a norm on $\M_{d+c}(\mathbb{R})$. All the distances and the balls in
the  space of matrices are associated with this norm. When $E$ is a subset of
$\M_{d+c}(\mathbb{R})$, let $B_{E}(x,r)$ denote the set of matrices in $E$ within a distance from $x$ smaller than $r$. 

Let  $\mathcal{L}=\mathcal{L}_{d+c}$ denote the space
of $(d+c)$-dimensional unimodular lattices in $\mathbb
R^{d+c}$ which we identify  with 
$\SL (d+c,\mathbb{R})/\SL (d+c,\mathbb{Z})$.



For $\theta\in\M_{d,c}(\mathbb{R)}$, let  $M_{\theta}$ denote the matrix
\[
\left(
\begin{array}
[c]{cc}%
I_{d} & -\theta\\
0 & I_{c}%
\end{array}
\right) \in\SL(d+c,\mathbb R)
\]
and $\Lambda_{\theta}=M_{\theta}\mathbb{Z}^{d+c}$ the lattice associated with
$M_{\theta}$. 

Let  $\mathcal{H}_{>0}$ denote the subgroup of all matrices $M_{\theta}$, $\theta\in
\M_{d,c}(\mathbb{R})$,  and let $\mathbb{T}_{d,c}$ denote its image in
$\mathcal{L}_{d+c}$.

In the same manner, let $\mathcal{H}_{<}$ denote the subgroup of $\SL(d+c,\mathbb{R})$ of matrices of
the form
\[
\left(
\begin{array}
[c]{cc}%
I_{d} & 0\\
B & I_{c}%
\end{array}
\right)
\]
and let $\mathcal{H}_{\leq}$ denote the subgroup of $\SL(d+c,\mathbb{R})$ of
matrices of the form
\[
\left(
\begin{array}
[c]{cc}%
A & 0\\
B & C
\end{array}
\right)
\]
where $A\in\GL(d,\mathbb{R)}$, $B\in M_{c,d}(\mathbb{R)}$ and $C\in
\GL(c,\mathbb{R})$.

Let 
\[
g_{t}=\left(
\begin{array}
[c]{cc}%
e^{ct}I_{d} & 0\\
0 & e^{-dt}I_{c}
\end{array}
\right)\in \SL (d+c,\mathbb R),
\]
$t\in\mathbb{R}$, denote the standard diagonal flow,
$E_{-}= \{0\}\times\mathbb R^c $ denote the contracting direction of the flow and
$E_{+}=\mathbb R^d \times \{0\}$ denote the expanding direction of the flow. We also refer to $E_{+}$ as the horizontal subspace and to $E_{-}$ as the vertical subspace.
\subsection{Lattices}
Suppose $\mathbb R^n$ is equipped with a norm $\Vert.\Vert$. For a lattice $\Lambda$ and an integer $i\in\{1,...,n\}$, let $\lambda_i(\Lambda)$ denote the $i$-th minimum of the lattice $\Lambda$ with respect to the norm $\Vert.\Vert$, i.e.,
\[
\lambda_i(\Lambda,\Vert.\Vert)=\min\{\lambda>0:B(0,\lambda)\cap\Lambda \text{ contains at least $i$ independent vectors}\}.
\]
Observe that $\lambda_1(\Lambda)$ is the length of the shortest nonzero vector in $\Lambda$. When there is no ambiguity about the norm we write $\lambda_i(\Lambda)$ instead of $\lambda_i(\Lambda,\Vert.\Vert)$.


\section{Best approximations}

\subsection{Best Diophantine approximations}

Multidimensional extensions of the classical continued fraction expansion cannot conciliate all the properties of the one dimensional expansion. For instance, it is not possible to conciliate the unimodularity and the best approximation property (see \cite{Lag4} and \cite{Mosh}). The ``best simultaneous Diophantine approximations'' is the multidimensional extension based solely on the best approximation property. It has been studied by many Authors, see for instance \cite{Lag1,Lag2,Lag3,Lag4,Lag5,Cheu,Chev-Khin,Chev-Mosc,Mosh}. 

\begin{definition}
Let $\theta\in\M_{d,c}(\mathbb{R)}$. \newline 
1. A nonzero vector
$Q\in\mathbb{Z}^{c}$ is a \textit{best simultaneous Diophantine approximation
denominator} of $\theta$ if for all nonzero $U$ in $\mathbb{Z}^{c}$,
\begin{align*}
\left\Vert U\right\Vert _{\mathbb{R}^{c}}  &  <\left\Vert Q\right\Vert
_{\mathbb{R}^{c}}\Rightarrow \dd(\theta Q,\mathbb{Z}^{d})<\dd(\theta U,\mathbb{Z}^{d})\\
\left\Vert U\right\Vert _{\mathbb{R}^{c}}  &  \leq\left\Vert Q\right\Vert
_{\mathbb{R}^{c}}\Rightarrow \dd(\theta Q,\mathbb{Z}^{d})\leq \dd(\theta U,\mathbb{Z}^{d}).
\end{align*}
2. An element $(P,Q)$ in $\mathbb{Z}^{d}\times\mathbb{Z}^{c}$ is a
\textit{best Diophantine approximation vector} of $\theta$ if $Q$ is a best
simultaneous Diophantine approximation denominator of $\theta$\ and if
\[
\left\Vert \theta Q-P\right\Vert _{\mathbb{R}^{d}}=\dd(\theta Q,\mathbb{Z}%
^{d}).
\]

\end{definition}

If the equation $\theta Q=0 \mod \mathbb Z^{d}$ has no nontrivial solution $Q\in
\mathbb{Z}^{d}$, the set of best Diophantine approximation denominators of
$\theta$ is infinite. Numbering the set of best approximation denominators in
ascending order of the norm $q=\left\Vert Q\right\Vert _{c}$, we obtain two
sequences
\[
q_{0}=q_{0}(\theta)=\lambda_{1}(\mathbb{Z}^{c})<q_{1}=q_{1}(\theta)=\left\Vert
Q_{1}(\theta)\right\Vert _{c}<...<q_{n}=q_{n}(\theta)=\left\Vert Q_{n}%
(\theta)\right\Vert _{c}<....
\]
and 
\[
r_{0}=r_{0}(\theta)=\dd(\theta Q_0,\mathbb{Z}^{d})>r_{1}=r_{1}(\theta)=\dd(\theta Q_1,\mathbb{Z}^{d})>...>r_{n}=r_{n}(\theta)=\dd(\theta Q_n,\mathbb{Z}^{d})>....
\]

When $d=c=1$, by the best approximation property, the integers $q_{0}%
,q_{1},...,q_{n},...$ are the denominators of the ordinary continued fraction
expansion of $\theta$. The only slight difference is that in the ordinary
continued fraction expansion, it can happen that $q_{0}=q_{1}=1$. In this
case, the indices are shifted by one.

\subsection{Minimal vectors in lattices}

The notion of minimal vector goes
back to Vorono\"{\i}. He used minimal vectors to find units in cubic fields (see \cite{Vor} and also \cite{BuchI,BuchII}). It allows to
convert statements about best simultaneous Diophantine approximations of vectors or of
matrices into statements about lattices (see \cite{Cheu,Chev-Mosc}).

\begin{definition}
Let $M\in\SL (d+c,\mathbb{R})$ and let $\Lambda=M\mathbb{Z}^{d+c}\in
\mathcal{L}_{d+c}$. A nonzero vector $X\in\Lambda$ is a minimal vector of
$\Lambda$ (with respect to the norms $\left\Vert \cdot\right\Vert
_{\mathbb{R}^{d}}$ and $\left\Vert \cdot\right\Vert _{\mathbb{R}^{c}}$) if the
only nonzero vectors $Y\in\Lambda$ in the cylinder $C(X)$ are such that
$C(X)=C(Y)$, i.e.,
\[
\left\vert X\right\vert _{+}=\left\vert Y\right\vert _{+},\ \left\vert
X\right\vert _{-}=\left\vert Y\right\vert _{-}.
\]
If two minimal vectors $X$ and $Y$ define the same cylinder we say that they
are equivalent.
\end{definition}

Observe that for each lattice $\Lambda$, there exists a minimal vector $X$ that
is a shortest vector of $\Lambda$ with respect to the norm $\Vert.\Vert_{\mathbb R^{d+c}}$. There might exist other shortest vectors
and even other shortest vector that are minimal. The set of minimal vectors is
generally infinite but might be finite. For $d,c\geq 1$, it is easily shown
that there are at least two linearly independent minimal vectors in any
lattice. This lower bound may be achieved, for instance with $\Lambda
=\mathbb{Z}^{1+1}$.

Given a lattice $\Lambda$ in $\mathcal{L}_{d+c}$, we select one minimal vector
in each equivalent class of minimal vectors. We number these vectors in
ascending order of heights. Such a numbering exists because $0$ is the only
possible limit point of the set of heights of minimal vectors (see the proof of Lemma \ref{lem:lowerrate}) We get a
sequence
\[
...X_{-n}(\Lambda),...,\ X_{-1}\left(  \Lambda\right)  ,\;X_{0}(\Lambda
),\;X_{1}(\Lambda),...
\]
This sequence might be finite, infinite one-sided or two-sided. The sequence
$(\left\vert X_{n}(\Lambda)\right\vert _{+})_{n}$ is decreasing while
the sequence $(\left\vert X_{n}(\Lambda)\right\vert _{-})_{n}$ is 
increasing. The numbering with increasing heights is unique up to a shift on
the indices. Though, this shift is not really relevant, we will fix later a
convenient way of choosing $X_{0}(\Lambda)$ (see the section about return
times).\bigskip

We shall always use the following notations
\[
q_{n}(\Lambda)=\left|  X_{n}(\Lambda)\right|  _{-}\quad\text{ and }\quad
r_{n}(\Lambda)=\left|  X_{n}(\Lambda)\right|  _{+}.
\]


The following Lemma is easy and very important. It shows that for $\theta
\in\M_{d,c}(\mathbb{R})$, the sequences $(q_{n}(\Lambda_{\theta}))_{n}$
and $(q_{n}(\theta))_{n}$ are deduced one another by a shift. Therefore, if one of the two limits
\[
\lim_{n\rightarrow\infty}\frac{1}{n}\ln q_{n}(\theta),\ \text{and }\lim_{n\rightarrow\infty}\frac{1}{n}\ln q_{n}(\Lambda
_{\theta})
\]
exists, then the other exists and have the same value. The same results holds with the sequences $(r_n(\theta))_n$ and $(r_n(\Lambda_{\theta}))_n$.


\begin{lemma}
\label{lem:min-best}
Let $\theta$ be in $\M_{d,c}(\mathbb{R)}$. 
\newline 1. If $X=M_{\theta}Y$ is a
minimal vector of the lattice $\Lambda_{\theta}$ with positive height ,
$\left|  X\right|  _{-}>0$, then $Y$ is best a approximation vector
of $\theta$. 
\newline 2. Conversely if $Y$ in $\mathbb{Z}^{d+c}$ is a best
approximation vector of $\theta$ such that%
\[
\left\|  Y_{+}-\theta Y_{-}\right\|  _{\mathbb{R}^{d}}<\lambda_{1}%
(\mathbb{Z}^{d})
\]
then $X=M_{\theta}Y$ is a minimal vector of $\Lambda_{\theta}.$
\end{lemma}

\begin{proof}
1. Set $Q=Y_-$. Since $X$ is a minimal vector, $\dd(\theta Q,\mathbb{Z}^{d})=\left|  X\right|  _{+}$. Suppose that $U\in\mathbb{Z}^{c}$ and $V\in\mathbb{Z}^{d}$ are such that%
\[
\left\|  \theta U-V\right\|  _{\mathbb{R}^{d}}=\dd(\theta U,\mathbb{Z}^{d})\leq
\dd(\theta Q,\mathbb{Z}^{d})=\left|  X\right|  _{+}%
\]
and%
\[
\left\|  U\right\|  _{\mathbb{R}^{c}}\leq\left\|  Q\right\|  _{\mathbb{R}^{c}%
}=\left|  Y\right|  _{-},
\]
then the vector
\[
M_{\theta}\left(
\begin{array}
[c]{c}%
U\\
V
\end{array}
\right)  =\left(
\begin{array}
[c]{c}%
V-\theta U\\
V
\end{array}
\right)
\]
is in the cylinder $C(X)$, and therefore by definition of minimal vectors, we have%
\[
\left\|  \theta U-V\right\|  _{\mathbb{R}^{d}}=\dd(\theta U,\mathbb{Z}%
^{d})=\dd(\theta Q,\mathbb{Z}^{d})
\]
and%
\[
\left\|  U\right\|  _{\mathbb{R}^{c}}=\left\|  Q\right\|_{\mathbb{R}^{c}}.
\]
It follows that $Q$ is a best approximation denominator.
\newline 2. Conversely suppose
that $Y=(P,Q)$ is a best approximation vector such that%
\[
\left\|  P-\theta Q\right\|  _{\mathbb{R}^{d}}<\lambda_{1}(\mathbb{Z}^{d}).
\]
If a nonzero vector
\[
Z=M_{\theta}\left(
\begin{array}
[c]{c}%
V\\
U
\end{array}
\right)  \in C(X),
\]
then we have%
\[
\left\|  V-\theta U\right\|  _{\mathbb{R}^{d}}\leq\left\|  P-\theta Q\right\|
_{\mathbb{R}^{d}}<\lambda_{1}(\mathbb{Z}^{d}),
\]
hence $U$ is not zero. We also have
\[
\left\|  U\right\|  _{\mathbb{R}^{c}}\leq\left\|  Q\right\|  _{\mathbb{R}^{c}}%
\]
hence by definition of best approximation denominator, we have $\left|
Z\right|  _{-}=\left\|  U \right\|  _{\mathbb{R}^{c}}=\left\|  Q\right\|
_{\mathbb{R}^{c}}=\left|  X\right|  _{-}$ and $\left|  Z\right|  _{+}=\left\|
V-\theta U\right\|  _{\mathbb{R}^{d}}=\left\|  P-\theta Q\right\|
_{\mathbb{R}^{d}}=\left|  X\right|  _{+}$, hence $X$ is minimal.
\end{proof}

 The classical inequality 
 \[
 q_{n+1}r_n\leq 1
 \]
 which holds for the one-dimensional continued fraction expansion of a real number can be extended  to minimal vectors of lattices or to best approximation vectors. This fact is well known but it is worth stating it.

\begin{lemma}
\label{lem:Minkowski}
There is a constant $C_{d,c}$ depending only on $c$ and $d$ such that for all lattice $\Lambda \in \mathcal L_{d+c}$ or all matrices $\theta \in \M_{d,c}$  and all integers $n$, we have 
\begin{align*}
q_{n+1}^c(\Lambda)r_n^d(\Lambda)\leq C_{d,c},\\
q_{n+1}^c(\theta)r_n^d(\theta)\leq C_{d,c}.
\end{align*}
\end{lemma}

\begin{proof}
Just use the first Minkowski Theorem with the cylinder defined by two consecutive minimal vectors or best approximation vectors.
\end{proof}


The classical inequality%
\[
q_{n+2}\geq 2q_{n}%
\]
which holds for the denominators of the one-dimensional continued fraction expansion
of a real number can be extended to minimal vector of lattices. This
inequality has already been extended to best simultaneous Diophantine
approximations, see \cite{Lag3},\ \cite{Lag4} and \cite{Chev-Mosc}. The
extension to minimal vectors of lattices is straightforward.

\begin{lemma}
\label{lem:lowerrate}There is a positive integer constant $A=A(d,c)$ such that for
any $\Lambda$ in $\mathcal{L}_{d+c}$ and any $n\in\mathbb{Z}$, if
$X_{n}(\Lambda),\ X_{n+1}(\Lambda),....,X_{n+A}(\Lambda)$ exist, then%
\begin{align*}
q_{n+A}(\Lambda)  &  \geq 2q_{n}(\Lambda),\\
r_{n+A}(\Lambda)  &  \leq\frac{1}{2}r_{n}(\Lambda)
\end{align*}

\end{lemma}

\begin{proof}
Let $A$ be an integer constant such that if $A$ points $(x_1,y_1),...,(x_A,y_A)$ are in the product  of balls $B_{\mathbb{R}^{d}}(0,r_1)\times B_{\mathbb{R}^{c}}(0,r_2)$ with $r_1,r_2\geq 0$ then
there exist two indices $i\neq j$ such that $\left\Vert x_{i}-x_{j}\right\Vert
_{\mathbb{R}^{d}}\leq\frac{1}{2}r_1$ and $\left\Vert y_{i}-y_{j}\right\Vert
_{\mathbb{R}^{c}}\leq\frac{1}{2}r_2$. With this choice of the constant $A$, if $k\geq A$ is a positive
integer such that $q_{n+k}(\Lambda)\leq 2q_{n}(\Lambda)$, then there are two
integers $0\leq i<j\leq k$ such that the vector $X_{n+j}(\Lambda
)-X_{n+i}(\Lambda)$ satisfies both conditions%
\[
\left\{
\begin{array}
[c]{c}
\left\vert X_{n+j}(\Lambda)-X_{n+i}(\Lambda)\right\vert _{+}\leq\frac{1}%
{2}r_{n}(\Lambda)\\
\left\vert X_{n+j}(\Lambda)-X_{n+i}(\Lambda)\right\vert _{-}\leq \frac{1}{2}2q_{n}%
(\Lambda)
\end{array}
\right.
\]
which contradicts the definition of $X_{n}(\Lambda)$. The same way of
reasoning leads to the other inequality.
\end{proof}

We shall use several times the following very simple Lemma which is a
consequence of the following observation. For any minimal vector $X$ of a
lattice $\Lambda$ in $\mathbb{R}^{d+c}$ and any $t\in\mathbb R$, $g_{t}X$ is a minimal vector of the
lattice $g_{t}\Lambda$. It follows that

\begin{lemma}
\label{lem:flow-minimal}Let $\Lambda$ be in $\mathcal{L}_{d+c}$ and let $t$ be
in $\mathbb{R}$. The  sequence of minimal vectors of the lattice $g_t\Lambda$ is  $(g_{t}
(X_{n}(\Lambda)))_{n}$.
\end{lemma}



\section{The surfaces $S$ and $S^{\prime}$}

We assume that $\mathbb{R}^{d+c}$ is endowed with the norm
\[
\left\|  (x,y)\right\|  _{\mathbb{R}^{d+c}}=\max\{\left\|  x\right\|
_{\mathbb{R}^{d}},\left\|  y\right\|  _{\mathbb{R}^{c}}\}.
\]
The main idea of the proofs of Theorems \ref{thm:levy-d} and \ref{thm:jager}
is to induce the flow $g_{t}$ on the surface
\[
\{\Lambda\in\mathcal{L}_{d+c}:\lambda_{2}(\Lambda)=\lambda_{1}(\Lambda)\}
\]
where $\lambda_{1}(\Lambda)$ and $\lambda_{2}(\Lambda)$ are the first two minima of the lattice $\Lambda$ associated with the above norm. And then, to use Birkhoff ergodic Theorem with the first return map associated with the flow $g_t$. For
technical reason it is better to slightly change the surface. For instance the
above set is not a submanifold of $\mathcal{L}_{d+c}$. It could have some ``branching
points'' while a slightly smaller set is clearly a submanifold, see Lemma
\ref{lem:submanifold}. It will be convenient to use two surfaces $S$ and
$S^{\prime}$ for the proof of Theorem \ref{thm:levy-d}. These two surfaces are
very similar; we state all the results we need for both surfaces but we only perform 
the proofs for the first surface $S$.

\subsection{Definition of $S$}\label{sec:defS}

The surface $S$ is the set of lattices $\Lambda$ in $\mathcal{L}_{d+c}$ such
that there exist two independent vectors $v_0^S(\Lambda)$ and $v_1^S(\Lambda)$ in
$\Lambda$ such that:

\begin{itemize}
\item $\left\vert v_1^S(\Lambda)\right\vert _{+}$ and $\left\vert
v_0^S(\Lambda)\right\vert _{-}$ are $<\left\vert v_1^S(\Lambda)\right\vert
_{-}=\left\vert v_0^S(\Lambda)\right\vert _{+}$,

\item the only nonzero points of $\Lambda$ in the ball $B_{\mathbb{R}^{d+c}%
}(0,\lambda_{1}(\Lambda))$ are $\pm v_0^S(\Lambda)$ and $\pm v_1^S(\Lambda)$.
\end{itemize}

Observe that for $\Lambda$ in $S$, $v_0^S(\Lambda)$ and $v_1^S(\Lambda)$ are
unique up two sign and are consecutive minimal vectors of $\Lambda$.

Since $\pm v_0^S(\Lambda)$ and $\pm v_1^S(\Lambda)$ are the only nonzero
points of $\Lambda$ in the ball $B_{\mathbb{R}^{d+c}}(0,\lambda_{1}(\Lambda
))$, $S$ is included in the set
\[
\{\Lambda\in\mathcal{L}_{d+c}:\lambda_{1}(\Lambda)=\lambda_{2}(\Lambda)\}.
\]


\subsection{Definition of $S^{\prime}$}

The surface $S^{\prime}$ is the set of lattices in $\mathcal{L}_{d+c}$ such
that there exists a vectors $w_{0}^{S'}(\Lambda)$ in $\Lambda$ such that:

\begin{itemize}
\item the only nonzero points of $\Lambda$ in the ball $B_{\mathbb{R}^{d+c}%
}(0,\lambda_{1}(\Lambda))$ are $\pm w_{0}^{S'}(\Lambda)$,

\item the ball $B_{\mathbb{R}^{d+c}}(0,\lambda_{1}(\Lambda))$ is equal to the cylinder $C(w_{0}^{S'}
(\Lambda))$.
\end{itemize}

Observe that $w_{0}^{S'}(\Lambda)$ is unique up to sign and is a minimal vector of
$\Lambda$.

\subsection{Lattices bases and minima}

We shall need the following results about lattices.

\begin{lemma}
Suppose $\mathbb{R}^{n}$ is equipped with any norm $\left\Vert .\right\Vert $.
Let $\Lambda$ be a lattice in $\mathbb{R}^{n}$ and let $v_{1}$, $v_{2}$ be two
independent vectors of $\Lambda$ such that $\left\Vert v_{1}\right\Vert =\lambda
_{1}(\Lambda)$ and $\left\Vert v_{2}\right\Vert =\lambda_{2}(\Lambda)$. Then
$\ \mathbb{Z}v_{1}+\mathbb{Z}v_{2}$ is a primitive sub-lattice of $\Lambda$
unless $\frac{1}{2}(v_{1}+v_{2})\in\Lambda$ and $\left\Vert v_{1}\right\Vert
=\left\Vert v_{2}\right\Vert =\left\Vert \frac{1}{2}(v_{1}+v_{2})\right\Vert $.
\end{lemma}

\begin{proof}
Consider the parallelogram $\mathcal{P}$ defined by the vectors $v_{1}$ and
$v_{2}$. Let $v$ be an element of $\Lambda$ that belongs to the interior of
$\mathcal{P}$. If $v$ is not in the segment joining $v_{1}$ and $v_{2}$ then
the distance from $v$ to either $0$ or $v_{1}+v_{2}$ is of the form
$\left\Vert t_{1}v_{1}+t_{2}v_{2}\right\Vert $ for some positive real numbers
$t_{1}$ and $t_{2}$ with $t_{1}+t_{2}<1$. Hence this distance is
$\leq t_{1}\left\Vert v_{1}\right\Vert +t_{2}\left\Vert v_{2}\right\Vert
<\lambda_{2}(\Lambda)$ which contradicts the definition of $\lambda
_{2}(\Lambda)$. If $v$ is in the the segment joining $v_{1}$ and $v_{2}$ but
is not the point $\frac{1}{2}(v_{1}+v_{2})$ then the distance from $v$ to
either $v_{1}$ or $v_{2}$ is of the form $\left\Vert t(v_{1}-v_{2})\right\Vert
$ with $t<\frac{1}{2}$ which implies that this distance is $<\lambda
_{2}(\Lambda)$, again a contradiction. If $v=\frac{1}{2}(v_{1}+v_{2})$, we
have $\left\Vert v\right\Vert \leq\frac{1}{2}(\left\Vert v_{1}\right\Vert
+\left\Vert v_{2}\right\Vert )$ which is $<\lambda_{2}(\Lambda)$ unless
$\left\Vert v_{1}\right\Vert =\left\Vert v_{2}\right\Vert=\left\Vert v \right\Vert .$
\end{proof}


It follows that when the norm is strictly convex, the sub-lattice $\mathbb{Z}%
v_{1}+\mathbb{Z}v_{2}$ is always primitive. In our setting despite that the
norm is not strictly convex it is possible to use the above Lemma. With our
choice of the norm on $\mathbb{R}^{d+c}$, the triangle inequality is strict
for two vectors one inside the ``top'' of the cylinder $B_{\mathbb{R}^{d+c}%
}(0,r)$ and one inside the lateral side of $B_{\mathbb{R}^{d+c}}(0,r)$. Therefore,

\begin{corollary}
\label{cor:lattice}Let $\Lambda$ be in $S$. Then the vectors $v_0^S(\Lambda)$
and $v_1^S(\Lambda)$ associated with $\Lambda$ are the first two vectors of a
basis of $\Lambda$.
\end{corollary}



\subsection{Geometric properties of $S$ and $S^{\prime}$}

\begin{lemma}
\label{lem:submanifold}$S$ and $S^{\prime}$ are a submanifolds of
$\mathcal{L}_{d+c}$ of dimension $(d+c)^{2}-2$, transverse to the diagonal
flow $g_{t}$.
\end{lemma}

\begin{proof}
Let $\Lambda_{0}$ be in $S$ and call $v_{0}^S(\Lambda_{0})$ and $v_{1}^S(\Lambda_{0})$
the two vectors provided by the definition of $S$. By Corollary
\ref{cor:lattice},  $v_{0}^S(\Lambda_{0})$ and $v_{1}^S(\Lambda_{0})$
are the first two vectors of a basis $(b_{1},...,b_{d+c})$ of $\Lambda_{0}$.
We can find a small enough positive real number $\varepsilon$ such that for
any $(v_{1},...,v_{d+c})$ in the open set
\[
W=B_{\mathbb{R}^{d+c}}(b_{1},\varepsilon)\times...\times B_{\mathbb{R}^{d+c}%
}(b_{d+c},\varepsilon),
\]
\begin{itemize}
\item the matrix $M=M(v_{1},...,v_{d+c})$ the columns of which are the $v_{i}
$, is in $\GL (d+c,\mathbb R)$ and the sets $WP$, $P\in\SL (d+c,\mathbb  Z)$ are disjoint,
\item the vectors $\pm v_{1}$ and $\pm v_{2}$ are the only nonzero vectors of
the lattice $\Lambda=M\mathbb Z^{d+c}$ in the cylinder $C(v_{1},v_{2})$,
\item $|v_1|_{+}>0$,
\item $\left\Vert v\right\Vert >\left\Vert v_{1}\right\Vert $ and $\left\Vert
v_{2}\right\Vert $ for all $v$ in $\Lambda\setminus\{0,\pm v_{1},\pm v_{2}\}$.
\end{itemize}
Consider the map%
\begin{align*}
f  &  :W\rightarrow\mathbb{R}^{2}\\
&  :M=(v_{1},...,v_{d+c})\rightarrow(f_{1}(M)=\det M,f_{2}(M)=\left\vert
v_{1}\right\vert _{+}^{2}-\left\vert v_{2}\right\vert _{-}^{2}).
\end{align*}
Then a lattice $\Lambda=M\mathbb{Z}^{d+1}$ with $M\in W$, is in $S$ iff
$f(M)=(1,0)$. To prove that $S$ is a submanifold, it is enough to show that the differential
$Df(M)$ is onto at every point $M$ in $W$. The differential of $f_{2}$ is
given by
\[
Df_{2}(M).(w_{1},...,w_{d+c})=2v_{1}^{+}.w_{1}^{+}-2v_{2}^{-}.w_{2}^{-}.
\]
The linear map $Df_{2}(M)$ depends only on $w_{1}$ and $w_{2}$ and since
$v_{1}^{+}\neq 0$ for all $M$ in $W$, $Df_{2}(M)$ is never the zero map. The
differential of $f_{1}$ is given by
\[
Df_{1}(M).(w_{1},...,w_{d+c})=\sum_{i,j=1}^{d+c}(-1)^{i+j}\Delta_{i,j}w_{i,j}%
\]
where $w_{j}=(w_{i,j})_{i=1,...,d+c}$, $j=1,...,d+c$ and $\Delta_{i,j}$ is the
$(i,j)$-minor of the matrix $M$. Since $\det M\neq 0$, one at least one of the
minors $\Delta_{i,3}$, $i\leq d+c$, is not zero. Therefore the linear
$Df_{1}(M)$ is not zero and depends on $w_{3}$. It follows that the two linear
maps $Df_{1}(M)$ and $Df_{2}(M)$ are linearly independent for all $M$ in $W$
which implies that $S$ is a submanifold of $\mathcal{L}_{d+c}$.
To show that the flow is transverse to $S$, we have to check that for a matrix
$M=M(v_1,\dots,v_{d+c})$ in $W$ such that $f(M)=0$ we have $Df(M).(w_{1},...,w_{d+c})\neq 0$ when
$w_{i}=(cv_{i}^{+},-dv_{i}^{-})$. Now, for such $w_{i}$, $Df_{2}%
(M).(w_{1},...,w_{d+c})=2c\left\vert v_{1}\right\vert _{+}^{2}$ $+2d\left\vert
v_{2}\right\vert _{-}^{2}>0$, hence $Df(M).(w_{1},...,w_{d+c})$ is not zero.
\end{proof}


\subsection{Negligible sets}
A important ingredient of the proof of Theorem \ref{thm:levy-d} is that, for a given lattice $\Lambda$, the visiting times $t$, i.e. the times $t$ such that $g_t\Lambda\in S$, can be read from the sequence $(X_n(\Lambda))_n$ of minimal vectors.
However, this reading is straightforward only for generic lattices, a small subset of lattices has to be avoided.
\subsubsection{A negligible set $\mathcal N$ in the space lattices }

Let $\mathcal{N}=\mathcal{N}_{d+c}$ be the set of lattices $\Lambda$ in
$\mathcal{L}_{d+c}$ such that either 
\begin{itemize}
\item there exist two vectors $v_{1}$, and $v_2$, such that
$v_{1}\neq\pm v_{2}$ and $\left\vert v_{1}\right\vert _{+}=\left\vert
v_{2}\right\vert _{+}>0$ or $\left\vert v_{1}\right\vert _{-}=\left\vert
v_{2}\right\vert _{-}>0$,
\item or there exists a nonzero vector in $\Lambda$ lying in
the vertical space $\{0\}\times\mathbb{R}^{c}$, 
or in the horizontal subspace
$\mathbb{R}^{d}\times\{0\}$.
\end{itemize}

\begin{remark}
All the lattices $\Lambda_{\theta}$ are in $\mathcal{N}$.
\end{remark}

\begin{lemma}
$\mathcal{N}$ is negligible and $g_{t}$ invariant.
\end{lemma}

\begin{proof}
Clearly $\mathcal{N}$ is $g_{t}$ invariant and the set of lattices with a
nonzero vector in the vertical subspace or in the horizontal subspace is
negligible. So we are reduced to prove that if $X\neq\pm Y$ are two nonzero
vectors in $\mathbb{Z}^{d+c}$ the set of matrices $M$ in $\SL (d+c,\mathbb{R)}%
$ satisfying one of the equations%
\[
\left\vert MX\right\vert _{+}^{2}-\left\vert MY\right\vert _{+}^{2}=0
\]
or
\[
\left\vert MX\right\vert _{-}^{2}-\left\vert MY\right\vert _{-}^{2}=0
\]
is of zero measure. Firstly, by symmetry, it is enough to deal with one of the
equations, say the first. Secondly, by homogeneity it is equivalent to prove
that the set of matrices in $M_{d+c}(\mathbb{R)}$ that satisfy this equation
is of zero measure. Since this is an algebraic equation, it is enough to prove
that there exists at least one matrix $M$ such that $\left\vert MX\right\vert
_{+}^{2}-\left\vert MY\right\vert _{+}^{2}\neq 0$. If $X$ and $Y$ are
proportional just choose a matrix $M$ such that $|MX|_{+}\neq 0$. Otherwise,
first choose a vector $Z$ in the subspace spanned by $X$ and $Y$ that is
orthogonal to $X$. Observe that $Z.Y\neq 0$. Next choose a $d$-dimensional
subspace $V$ of $\mathbb{R}^{d+c}$ containing $Z$ and orthogonal to $X$. A
matrix $M$ the first $d$ rows of which are a basis of $V$, is such that
$\left\vert MX\right\vert _{+}=0$ and $\left\vert MY\right\vert _{+}\neq 0$.
\end{proof}


\begin{remark}
A lattice $\Lambda$ that is not in $\mathcal{N}$ has a
bi-infinite sequence of minimal vectors and is in $S$ iff $\lambda_{1}%
(\Lambda)=\lambda_{2}(\Lambda)$.
\end{remark}

\subsubsection{A negligible set $\mathcal M$ in the space of matrices $\M_{d,c}(\mathbb R)$}

Let $C$ be a positive real constant and let $\mathcal{M}=\mathcal{M}_{d,c}=\mathcal M_{d,c}(C)$ be the set of matrices $\theta \in \M_{d,c}(\mathbb R)$
 such that either 
 \begin{itemize}
 \item there exist two nonzero vectors $X\neq \pm Y$ in $\mathbb Z^{d+c}$ with nonzero heights such that $\left\vert M_{\theta}X\right\vert _{+}=\left\vert M_{\theta}Y\right\vert _{+}$
 \item or there exist infinitely many pairs $X\neq \pm Y$ in $\mathbb Z^{d+c}$  such that $\left\vert X\right\vert _{-}=\left\vert Y\right\vert _{-}\neq 0$ and 
 $\left\vert M_{\theta}X\right\vert _{+},\,\left\vert M_{\theta}Y\right\vert _{+}\leq C\left\vert X \right\vert _{-}^{-\frac{c}{d}}$. 
 \end{itemize}
 The set $\mathcal M$ depends on the constant $C$. Actually, we will only use the value  $C=C_{d,c}$ where $C_{d,c}$ is given by Lemma \ref{lem:Minkowski}.

  \begin{lemma}
  $\mathcal M$ is negligible.
  \end{lemma}

 \begin{proof} We prove that $\mathcal M$ is included in a countable union of negligible sets.
 
Given $X\neq \pm Y$ two nonzero vectors in $\mathbb Z^{d+c}$ with nonzero heights, consider the set $\mathcal M(X,Y)$ of matrices $\theta \in \M_{d,c}(\mathbb R)$  such that
 $\left\vert M_{\theta}X\right\vert _{+}=\left\vert M_{\theta}Y\right\vert _{+}$. In order to show that $\mathcal M(X,Y)$ has zero measure it is enough to show that the polynomial 
\begin{align*}
f(\theta)&=\left\vert M_{\theta}X\right\vert _{+}^2-\left\vert M_{\theta}Y\right\vert _{+}^2\\
&=\left\Vert X_{+}\right\Vert _{\mathbb R^d}^2-\left\Vert  Y_{+}\right\Vert_{\mathbb R^d}^2
-2(X_{+}.\theta X_{-}-Y_{+}.\theta Y_{-})
 +\left\Vert\theta X_{-}\right\Vert _{\mathbb R^d}^2-\left\Vert \theta Y_{-}\right\Vert_{\mathbb R^d}^2
\end{align*}
 is not the zero polynomial.

 
If $X_{-}\neq \pm Y_{-}$, we can choose $\theta_0$ such that $\left\Vert\theta_0 X_{-}\right\Vert _{\mathbb R^d}^2-\left\Vert \theta_0 Y_{-}\right\Vert_{\mathbb R^2} ^2\neq 0$. With this choice, the one variable polynomial $P(t)= f(t\theta_0)$ has a nonzero degree two monomial which implies that the polynomial $f$ is not the zero polynomial. 

If $X_{-}=Y_{-}$ (the case $X_{-}=-Y_{-}$ is similar), $f(\theta)=\left\vert
X\right\vert_{+}^2-\left\vert Y\right\vert_{+}^2-2(X_{+}-Y_{+}).\theta X_{-}.$ Since $X_{-}\neq
0$, the map $\varphi:\theta\in\M_{d,c}(\mathbb R)\rightarrow \theta X_{-}\in \mathbb R^d$ is onto. It follows that we can
choose $\theta$ such that $\theta X_{-}=X_{+}-Y_{+}$. With this value of $\theta$, we obtain $f(\theta)-f(-\theta)=-4\left\vert X_{+}-Y_{+}\right\vert^2 \neq 0$ which implies that the
polynomial $f$ is not the zero polynomial.  It follows that $\mathcal M(X,Y)$ is negligible.\\

Consider now, for a positive integer $n$, the set $\mathcal M_n$ of matrices $\theta \in
\M_{d,c}(\mathbb R)$  such that there a pair of linearly independent vectors $(X, Y)$  in $\mathbb Z^{d+c}\times \mathbb Z^{d+c}$  such 
that $n\leq\left\vert X\right\vert _{-},\,\left\vert Y\right\vert _{-}<n+1$ and 
\[ 
\left\vert M_{\theta}X\right\vert _{+},\,\left\vert M_{\theta}Y\right\vert _{+}\leq C\left\vert
X \right\vert _{-}^{-\frac{c}{d}}.
\]
We want to prove that the set of matrices $\theta$
that are in infinitely many $\mathcal M_n$ is negligible. We can move in the space $\M_{d,c}(\mathbb R/\mathbb Z)$ and consider instead  the set $\mathcal T_n$ of $\theta \in \M_{d,c}(\mathbb R/\mathbb Z)$ such that there exist $q,q' \in \mathbb Z^c$ linearly independent with $n\leq\left\Vert q \right\Vert_{\mathbb R^c},\,\left\Vert q' \right\Vert_{\mathbb R^c}<n+1$ and 
\[
d(\theta q,\mathbb Z^d),\,d(\theta q',\mathbb Z^d)\leq Cn^{-\frac{c}{d}}.
\]
For $q$ fixed, the measure of the set of $\theta \in \M_{d,c}(\mathbb R/\mathbb Z)$ such that
$d(\theta q,\mathbb Z^c)\leq Cn^{-\frac{c}{d}}$ is $a_{d,c} n^{-c}$ where the constant $a_{d,c}$ depends only on $C$ and the dimensions. When the inequality holds simultaneously for two linearly independent integer vectors $q$ and $q'$, the measure is the square of $a_{d,c} n^{-c}$. 
It follows that the measure of $\mathcal T_n$ is bounded above by 
\[
u_n=\text{card}\{(q,q')\in\mathbb Z^c\times \mathbb Z^c:n\leq\left\Vert q \right\Vert_{\mathbb R^c},\,\left\Vert q' \right\Vert_{\mathbb R^c}<n+1\}\times a_{d,c}^{2} n^{-2c}.
\]
By Borel-Cantelli, it is enough to prove that the $\Sigma_n u_n<\infty$. Now $\text{card}\{q\in\mathbb Z^c:n\leq\left\Vert q \right\Vert_{\mathbb R^c}<n+1\}\ll n^{c-1}$, hence
\[
u_n\ll n^{-2} 
\]
and we are done.

\end{proof}



 
\subsection{Visiting and return times}\label{sec:visit}


Let $\Lambda$ be in $\mathcal{L}_{d+c}$. By definition of $S$, when
$g_{t}\Lambda$ is in $S$, $v_{0}^S(g_{t}\Lambda)$ and $v_{1}^S(g_{t}\Lambda)$
are two consecutive minimal vectors of $g_{t}\Lambda$. Therefore, 
\[
\left\{
\begin{array}
[c]{l}
v_{0}^S(g_{t}\Lambda)=g_{t}X_{k}(\Lambda)\\
v_{1}^S(g_{t}\Lambda)=g_{t}X_{k+1}(\Lambda)
\end{array}
\right.  .
\]
for an integer $k$.
Hence $e^{ct}\left\vert X_{k}(\Lambda)\right\vert _{+}=e^{-dt}\left\vert
X_{k+1}(\Lambda)\right\vert _{-}$ which implies
\[
t=\frac{1}{d+c}\ln\frac{q_{k+1}(\Lambda)}{r_{k}(\Lambda)}.
\]
It follows that the set of real numbers $t$ such that $g_{t}\Lambda\in S$ is
included in the set
\[
V_{\Lambda}(S)=\{t_{k}=\frac{1}{d+c}\ln\frac{q_{k+1}(\Lambda)}{r_{k}(\Lambda
)}:\ k\in\mathbb{Z\}}.
\]
It can happen that some values $t_{k}$ are skipped, but in that case, $\Lambda$
must be in $\mathcal{N}$. So, when $\Lambda$ is not in $\mathcal{N}$,
$g_{t}\Lambda\in S$ iff $t\in V_{\Lambda}(S)$. For the surface $S^{\prime}$, the
same results hold with
\[
V_{\Lambda}(S')=\{t_{k}^{\prime}=\frac{1}{d+c}\ln\frac{q_{k}(\Lambda
)}{r_{k}(\Lambda)}:\ k\in\mathbb{Z\}}.
\]
It follows that for almost all $\Lambda$, both the backward trajectory
$(g_{t}\Lambda)_{t\leq 0}$ and the forward trajectory $(g_{t}\Lambda)_{t\geq 0}$
visit the two surfaces  $S$ and $S^{\prime}$  infinitely often. 
Therefore the first return/entrance times in
$S$ and $S^{\prime}$,
\begin{align*}
\tau(\Lambda)  &  =\inf\{t>0:g_{t}(\Lambda)\in S\}\in\mathbb{R}_{>0}%
\cup\{\infty\},\\
\tau^{\prime}(\Lambda)  &  =\inf\{t>0:g_{t}(\Lambda)\in S^{\prime}%
\}\in\mathbb{R}_{>0}\cup\{\infty\},
\end{align*}
are finite almost everywhere and 
the first return/entrance maps 
\begin{align*}
R(\Lambda)  &  =g_{\tau(\Lambda)}\Lambda,\\
R^{\prime}(\Lambda)  &  =g_{\tau^{\prime}(\Lambda)}\Lambda
\end{align*}
are defined for all $\Lambda$ that are not in $\mathcal{N}$.

For an integer $n\geq 1$, denote $\tau_{n}$ the $n$-th return (or entrance)
time in $S$, i.e.%
\[
\tau_{n}(\Lambda)=\sum_{k=0}^{n-1}\tau(R^{k}(\Lambda))
\]
($R^{0}(\Lambda)=\Lambda$ for all $\Lambda$ in $\mathcal{L}_{d+1}$).
It will be convenient to choose the numbering of the sequence of minimal vectors $(X_n(\Lambda))$ in order to have simple formulas for the return time and the return map.
\medskip


\textbf{Numbering convention}: \textit{For a lattice $\Lambda \in \mathcal L_{d+c}$, $n=0$
is the smallest integer $n\in\mathbb{Z}$ such that
\[
\left\vert X_{n+1}(\Lambda)\right\vert _{-}\geq\left\vert X_{n}(\Lambda
)\right\vert _{+}
\] 
when the set of such integers is non empty.}
\medskip

With this numbering convention, for all $\Lambda$ is in $S$, we have 
\[
X_0(\Lambda)=v_0^S(\Lambda)
\]
and for all $\Lambda$ is in $S'$, we have 
\[
X_0(\Lambda)=w_0^{S'}(\Lambda).
\]
Moreover when $\Lambda\notin S$ is not in $\mathcal N$,
\begin{align*}
&\left\vert X_{1}(\Lambda)\right\vert _{-}-\left\vert X_{0}(\Lambda
)\right\vert _{+}> 0,\\
&\tau(\Lambda)=\frac{1}{d+c}\ln\left(\frac{\left\vert X_{1}(\Lambda)\right\vert_{-}}
{\left\vert X_{0}(\Lambda
)\right\vert_{+}}\right),
\end{align*}
and
\[
g_{\tau(\Lambda)}X_{0}(\Lambda)=\pm v_{0}^S(R(\Lambda))=\pm X_{0}(R(\Lambda)).
\]



Let us summarize the above.
\begin{lemma}
\label{lem:return-minimal}
Let $\Lambda$ be a lattice in $\mathcal L_{d+c}$.\newline
1. The set of visiting times in $S$ is included in $V_{\Lambda}(S)$ and the set of visiting times in $S'$ is included in $V_{\Lambda}(S')$. \newline
2. Suppose $\Lambda$ is not in $\mathcal{N}$. The set of visiting times in $S$ is equals to $V_{\Lambda}(S)$ and the set of visiting times in $S'$ is equals to $V_{\Lambda}(S')$.


\end{lemma}

For $\theta$  in $\M_{d,c}(\mathbb R)$, we also need to connect the visiting times of the surface $S$ with the best approximation vectors of $\theta$. 

\begin{lemma}
\label{lem:return-best}
Let $\theta$  be in $\M_{d,c}(\mathbb R)\setminus \mathcal M$. Then for all large enough integers $n$, 

\[
t_{n}(\theta)
=\frac{1}{d+c}\ln\frac{q_{n+1}(\theta)}{r_{n}(\theta)}
\] 
and 
\[
t'_{n}(\theta)
=\frac{1}{d+c}\ln\frac{q_{n}(\theta)}{r_{n}(\theta)}
\]
are visiting times for the surfaces $S$ and $S'$ respectively.

\end{lemma} 

\begin{proof}
Let $\theta$  be in $\M_{d,c}(\mathbb R)\setminus \mathcal M$. Consider the sequence of all best approximation vectors $(Y_n(\theta))_{n\in\mathbb N}$ of $\theta$. By Lemma \ref{lem:min-best}, there are integers $n_1$ and $k$ such that $X_{n+k}(\Lambda_{\theta})=M_{\theta}Y_{n}(\theta)$ for all $n\geq n_1$. Since $\theta$ is not in $\mathcal M$, by Lemma \ref{lem:Minkowski} there is another integer $n_2$ such that for all $n\geq n_2$, the only nonzero vector of $\Lambda_{\theta}$ in the box $\mathcal C(X_{n+k}(\theta),X_{n+k+1}(\theta))$ are $\pm X_{n+k}(\theta)$ and $\pm X_{n+k+1}$. This means that for all $n$ large enough, the times 
\[
t_n(\theta)=\frac{1}{d+c}\ln\frac{q_{n+1}(\theta)}{r_{n}(\theta)}
\text{ and }
t'_{n}(\theta))
=\frac{1}{d+c}\ln\frac{q_{n}(\theta)}{r_{n}(\theta)}
\] 
are  visting times for the surfaces $S$ and $S'$.
\end{proof}

\subsection{Functions defined on $S$}\label{sec:S}

Let $\Lambda$ be in $S$. By definition of $S$, the functions\newline%
\[
\rho,\rho^{\ast}:\Lambda\in S\rightarrow\ln\frac{\left\vert v_1^S(\Lambda)\right\vert _{-}}{\left\vert v_0^S(\Lambda)\right\vert _{-}}%
,\;\ln\frac{\left\vert v_0^S(\Lambda)\right\vert _{+}}{\left\vert
v_1^S(\Lambda)\right\vert _{+}}\in\mathbb{R}_{>0}\cup\{+\infty\}%
\]
are well defined on $S$. The next Lemma is easy, its proof is close to beginning of
the proof of Lemma \ref{lem:submanifold} and is omitted.

\begin{lemma}
The functions $\Lambda\in\mathcal L_{d,c}\rightarrow\, |v_0^S(\Lambda)|_-,\,|v_0^S(\Lambda)|_+,\,|v_1^S(\Lambda)|_-,\,|v_1^S(\Lambda)|_+$ are continuous and thus the functions $\rho$ and $\rho^{\ast}$ are continuous.
\end{lemma}

The following Lemma is important. On the one hand, it will imply that the functions $\rho$ and $\rho^\ast$ are integrable. On the other hand, it will explain the connection between the Lévy's constant
$L_{d,c}$ and the average return times on $S$.

\begin{lemma}
\label{lem:return-rho}Let $\Lambda$ be a lattice in $S\setminus\mathcal{N}$.
Then
\[
\tau(\Lambda)=\frac{1}{d+c}(\rho(R(\Lambda))+\rho^{\ast}(\Lambda)).
\]

\end{lemma}

\begin{proof}
Let $\Lambda$ be in $S\setminus\mathcal{N}$. By definition of $S$,
$q_{1}(\Lambda)=r_{0}(\Lambda)$. Hence, by Lemma \ref{lem:return-minimal},
\begin{align*}
(d+c)\tau(\Lambda)  &  =\ln\frac{q_{2}(\Lambda)}{r_{1}(\Lambda)}\times
\frac{r_{0}(\Lambda)}{q_{1}(\Lambda)}\\
&  =\ln\frac{q_{2}(\Lambda)}{q_{1}(\Lambda)}+\ln\frac{r_{0}(\Lambda)}%
{r_{1}(\Lambda)}\\
&  =\rho(R(\Lambda))+\rho^{\ast}(\Lambda).
\end{align*}
\end{proof}


\section{Finiteness of the induced measure on $S$ and $S^{\prime}$}

Fix a measure $\mu$  on $\mathcal{L}_{d+c}$  invariant by the $\SL
(d+c,\mathbb{R)}$ action. Recall that $\mu$ is unique up to a multiplicative constant. Since $S$ is a submanifold of $\mathcal{L}_{d+c}$
transverse to the flow $(g_{t})_{t\in\mathbb{R}}$ there exists a unique
measure $\mu_{S}$ defined on $S$ by the following property:

\textit{For all }$\Lambda$\textit{ in }$S$,\textit{ there exists a
neighborhood }$W$\textit{ of }$\Lambda$\textit{ in }$S$\textit{ and
}$\varepsilon_{\Lambda}>0$\textit{ such that for all Borel subsets }$V\subset W$\textit{ and
all }$0\leq\varepsilon\leq\varepsilon_{\Lambda}$\textit{,}%
\[
\mu(\cup_{t\in[0,\varepsilon]}g_{t}V)=\varepsilon\mu_{S}(V).
\]
The measure $\mu_S$ is the measure induced by the flow. It is well known that the measure $\mu_{S}$ is $R$-invariant.

The flow induces a measure $\mu_{S'}$ on $S'$ as well. Let us prove that these measures are finite. This is a simple consequence of the next Lemma which will be very important in the proof of Theorem 
\ref{thm:reduction} about the almost sur convergence in $M_{d,c}(\mathbb R)$.

\begin{lemma}
\label{lem:nb-returntimes} 1. There exists an integer constant $A$ such that
$\tau_{A}(\Lambda)\geq 1$ for all $\Lambda$ in $S$
 and  $\tau_{A}^{\prime}(\Lambda)\geq 1$ for all $\Lambda$ in $S'$.
\end{lemma}

\begin{proof}
Let $A$ be the constant given by Lemma \ref{lowerrate} about the growth rate of the sequences $(q_n(\Lambda))_n$ and $(r_n(\Lambda))_n$. For all integers $k$, we have 
\begin{align*}
\frac{1}{d+c}\left(\ln\frac{q_{k+A+1}(\Lambda)}{r_{k+A}(\Lambda)}-
\ln\frac{q_{k+1}(\Lambda)}{r_{k}(\Lambda)}\right)
&\geq 
\frac{1}{d+c}\ln\frac{q_{k+A+1}(\Lambda)}{q_{k+1}(\Lambda)}\\
&\geq\frac{1}{d+c}\ln 2.
\end{align*}
Since by Lemma \ref{lem:return-minimal}, the set of visiting times of $\Lambda$ is included in
\[
V_{\Lambda}(S)=\{t_{k}=\frac{1}{d+c}\ln\frac{q_{k+1}(\Lambda)}{r_{k}(\Lambda
)}:\ k\in\mathbb{Z\}},
\]
$\tau_{A}(\Lambda)\geq \frac{1}{d+c}\ln 2
$. Multiplying $A$ by the smallest integer larger thna $\frac{d+c}{\ln 2}$ we are done.
\end{proof}


\begin{proposition}
$\mu_{S}(S)$ and $\mu_{S^{\prime}}(S^{\prime})$ are finite and nonzero.
\end{proposition}

\begin{proof}
Since $S$ is nonempty and transverse to the flow, $\mu_S(S)>0$.

Since $\mu_{S}$ is $R$-invariant,  for all $k$%
\[
\int_{S}\tau(R^{k}\Lambda)d\mu_{S}(\Lambda)=\int_{S}\tau(\Lambda)d\mu
_{S}(\Lambda),
\]
and by Kac's return time Theorem,
\[
\int_{S}\tau(\Lambda)d\mu
_{S}(\Lambda)=\mu(\mathcal{L}_{d+1}),
\]
therefore,
\[
 \int_{S}\tau_{A}(\Lambda)d\mu_{S}(\Lambda)=\int_{S}\sum_{k=0}^{A-1}\tau
(R^{k}\Lambda)d\mu_{S}(\Lambda)=A\mu(\mathcal{L}_{d+c}).
\]
By the above Lemma, $\tau_A\geq 1$ on $S$, hence $\mu_{S}(S)\leq A\, \mu(\mathcal{L}_{d+c})$ which is finite by Siegel's Theorem.
\end{proof}


\section{Almost sure convergence in the space of lattices}

\subsection{Consequence of the Birkhoff Theorem}

\begin{theorem}\label{thm:Birkhoff}
\label{Birkhoff}There exist two positive constants $L_{d,c}$ and
$L_{d,c}^{\ast}$ such that for almost all lattices $\Lambda$ in $\mathcal{L}_{d+c}$,
\begin{align*}
\lim_{n\rightarrow\infty}\frac{1}{n}\ln q_{n}(\Lambda)  &  =\frac{1}{\mu
_{S}(S)}\int_{S}\rho d\mu_{S}=L_{d,c}>0,\\
\lim_{n\rightarrow\infty}\frac{-1}{n}\ln r_{n}(\Lambda)  &  =\frac{1}{\mu
_{S}(S)}\int_{S}\rho^{\ast}d\mu_{S}=L_{d,c}^{\ast}>0,\\
\lim_{n\rightarrow\infty}\frac{1}{n}\tau_{n}(\Lambda)  &  =\frac{1}%
{d+c}(L_{d,c}+L_{d,c}^{\ast})=\frac{\mu(\mathcal{L}_{d+c})}{\mu_S(S)}.
\end{align*}
Moreover, these two constants do not depend on the particular choice of the
Euclidean norms on $\mathbb{R}^{d}$ and $\mathbb{R}^{c}$.
\end{theorem}

\begin{proof}
Let $\Lambda$ be in $S\setminus\mathcal{N}$. By Lemma \ref{lem:return-rho},
$\tau(\Lambda)=\frac{1}{d+c}(\rho(R(\Lambda))+\rho^{\ast}(\Lambda))$. Because
the spaces of lattices has finite measure, the return time $\tau$ is in
$\mathcal{L}^{1}(S)$ and therefore the non negative functions $\rho\circ R$
and $\rho^{\ast}$ are also in $\mathcal{L}^{1}(S)$. Making use of the
Birkhoff's Theorem with the functions $\rho$ and $\rho^{\ast}$, we obtain the
almost everywhere convergence of the sums%
\[
\frac{1}{N}\sum_{n=0}^{N-1}\rho\circ R^{n},\ \frac{1}{N}\sum_{n=0}^{N-1}%
\rho^{\ast}\circ R^{n}%
\]
on $S$ to $R$-invariant functions. Now the ergodicity of the flow $g_{t}$
implies the ergodicity of the return map $R$. Therefore $\frac{1}{N}\sum_{k=0}^{N-1}\rho\circ
R^{n}$ and $\frac{1}{N}\sum_{n=0}^{N-1}\rho^{\ast}\circ R^{n}$ converge almost
everywhere on $S$ to the constants 
\[L_{d,c}=\frac{1}{\mu_{S}(S)}\int_{S}\rho d\mu_{S} \text{ and }
L_{d,c}^{\ast}=\frac{1}{\mu_{S}(S)}\int_{S}\rho^{\ast}d\mu_{S}.
\]
We would like to see that the Birkhoff sums converge to the same limits almost everywhere in the whole space of lattices. Let $\Lambda$ be in
$\mathcal{L}_{d+c}\setminus(S\cup\mathcal{N})$ and let $n$ be \ a positive integer. By
Lemma \ref{lem:return-minimal} and the numbering convention, for all $n\geq 1$,
\begin{align*}
\rho\circ R^{n}(\Lambda)  &  =\ln\frac{\left\vert X_{1}(R^{n}(\Lambda
))\right\vert _{-}}{\left\vert X_{0}(R^{n}(\Lambda))\right\vert _{-}}\\
&  =\ln\frac{\left\vert g_{\tau_{n}(\Lambda)}(X_{n}(\Lambda))\right\vert
_{-}}{\left\vert g_{\tau_{n}(\Lambda)}(X_{n-1}(\Lambda))\right\vert _{-}}\\
&  =\ln\frac{\left\vert X_{n}(\Lambda)\right\vert _{-}}{\left\vert
X_{n-1}(\Lambda)\right\vert _{-}}%
\end{align*}
and%
\begin{align*}
\rho^{\ast}\circ R^{n}(\Lambda)  &  =\ln\frac{\left\vert X_{0}(R^{n}%
(\Lambda))\right\vert _{+}}{\left\vert X_{1}(R^{n}(\Lambda))\right\vert _{+}%
}\\
&  =\ln\frac{\left\vert X_{n-1}(\Lambda)\right\vert _{+}}{\left\vert
X_{n}(\Lambda)\right\vert _{+}}.
\end{align*}
as well. It follows that if the Birkhoff sums $\frac{1}{N}\sum_{n=1}^{N}%
\rho\circ R^{n}(\Lambda)$ and $\frac{1}{N}\sum_{n=1}^{N}\rho^{\ast}\circ
R^{n}(\Lambda)$ converge to $L_{d,c}$ and
$L_{d,c}^{\ast}$, then
\begin{align*}
\lim_{N\rightarrow\infty}\frac{1}{N}\ln q_{N}(\Lambda)  &  =L_{d,c}\\
\lim_{N\rightarrow\infty}\frac{-1}{N}\ln r_{N}(\Lambda)  & =L_{d,c}^{\ast}.
\end{align*}
Now the image by the map $R:\mathcal{L}_{d+c}\rightarrow S$ of a subset of
nonzero measure in $\mathcal{L}_{d+c}$ is a set of nonzero measure in $S$, therefore the sums $\frac{1}{N}\sum_{n=1}^{N}\rho\circ R^{n}$ and $\frac{1}%
{N}\sum_{n=1}^{N}\rho^{\ast}\circ R^{n}$ converge almost everywhere in $\mathcal{L}_{d+c}$ to
$L_{d,c}$ and $L_{d,c}^{\ast}$.

By Lemma \ref{lem:lowerrate}, we know that the sequences $(q_{n}(\Lambda))_{n}$
and $(r_{n}(\Lambda)^{-1})_{n}$ have at least exponential growth rate;
therefore, the constants $L_{d,c}$ and $L_{d,c}^{\ast}$ are $>0$.


By Lemma \ref{lem:return-rho}, for all $\Lambda$ in $S\setminus\mathcal{N}$
and $k\in\mathbb{N}$,
\[
\tau_{k+1}(\Lambda)-\tau_{k}(\Lambda)=\frac{1}{d+c}(\rho(R^{k+1}%
(\Lambda))+\rho^{\ast}(R^{k}(\Lambda))),
\]
hence
\[
\lim_{n\rightarrow\infty}\frac{1}{n}\tau_{n}(\Lambda)=\frac{1}{d+c}(L_{d,c}+L_{d,c}^{\ast})
\]
almost everywhere.

Finally, let us proof that the constants $L_{d,c}$ and $L_{d,c}^{\ast}$
do not depend on the Euclidean norm on $\mathbb R^d$ and $\mathbb R^c$.  For a matrix $A_d$ in 
$\SL (d,\mathbb R)$ and a matrix $A_c$ in 
$\SL (c,\mathbb R)$,
let denote $A$ the matrix
\[
A=\left(
\begin{array}
[c]{cc}%
A_d & 0\\
0 & A_{c}%
\end{array}
\right) \in\SL(d+c,\mathbb R) .
\]
Since the action of $A$ on
$\mathcal{L}_{d+c}$ is measure preserving,
\begin{align*}
\lim_{n\rightarrow\infty}\frac{1}{n}\ln q_{n}(A\Lambda)  &  =L_{d,c}\\
\lim_{n\rightarrow\infty}\frac{-1}{n}\ln r_{n}(A\Lambda)  &  =L_{d,c}^{\ast}%
\end{align*}
almost everywhere $\mathcal{L}_{d+1}$. Now a vectors $AX$ in the lattice
$A\Lambda$ is  minimal  iff $X$ is a minimal vector of the of the lattice $\Lambda$ with respect to
the new Euclidean norms $\left\Vert .\right\Vert _{A_d,\mathbb{R}^{d}}$  and $\left\Vert .\right\Vert _{A_c,\mathbb{R}^{c}}$ where
\[
\left\Vert u\right\Vert _{A,\mathbb{R}^{d}}=\left\Vert Au\right\Vert
_{\mathbb{R}^{d}}, \text{ and }
\left\Vert v\right\Vert _{A,\mathbb{R}^{c}}=\left\Vert Av\right\Vert
_{\mathbb{R}^{c}}.
\]
Since up to  multiplicative constants, all the Euclidean norms are of the above form, the constants $L_{d,c}$ and $L_{d,c}^{\ast}$ do not depend on the Euclidean norm on $\mathbb R^d$ and $\mathbb R^c$.
\end{proof}


\subsection{A consequence of Borel-Cantelli Lemma}

\begin{proposition}
\label{prop:borel-cantelli}With the notation of Theorem \ref{Birkhoff} we have%
\[
cL_{d,c}=dL_{d,c}^{\ast}.
\]

\end{proposition}

\begin{proof}
The inequality $cL_{d,c}\leq dL_{d,c}^{\ast}$ is easy to prove. By Lemma \ref{lem:Minkowski}, for all lattices $\Lambda$ in $\mathcal{L}_{d+c}$ and all $n$, we have
$q_{n+1}^{c}(\Lambda)r_{n}^{d}(\Lambda)\leq C_{d,c}$. Hence for a lattice $\Lambda$
such that
\begin{align*}
\lim_{n\rightarrow\infty}\frac{1}{n}\ln q_{n}(\Lambda)  &  =L_{d,c}\\
\lim_{n\rightarrow\infty}\frac{-1}{n}\ln r_{n}(\Lambda)  &  =L_{d,c}^{\ast},
\end{align*}
we have,%
\[
cL_{d,c}-dL_{d,c}^{\ast}=c\lim_{n\rightarrow\infty}\frac{\ln q_{n}(\Lambda)}%
{n}+d\lim_{n\rightarrow\infty}\frac{\ln r_{n}(\Lambda)}{n}=\lim_{n\rightarrow
\infty}\frac{\ln q_{n}^{c}(\Lambda)r_{n}^{d}(\Lambda)}{n}\leq 0.
\]
The converse inequality uses Borel-Cantelli Lemma. Let $\varphi:]0,\infty
[\rightarrow]0,\infty[$ be a decreasing function such that
$\sum_{n\geq 1}\varphi(n)<\infty$, for instance $\varphi(t)=\frac{1}{t^{\alpha
}}$ with $\alpha>1$. Since for such a function $\varphi$, $\lim\inf
_{n\rightarrow\infty}\frac{1}{n}\ln\varphi(n)=0$, the inequality
$cL_{d}-dL_{d}^{\ast}\geq 0$ holds provided that for almost all lattices
$\Lambda$, we have $q_{n}^{c}(\Lambda)r_{n}^{d}(\Lambda)\geq\varphi(n)^{d+c}$
for $n$ large enough.
Let $K$ be a constant that will be chosen later. For each integer $n\geq 1$,
consider the set $A_{n}$ of lattices $\Lambda$ in $\mathcal{L}_{d+c}$ such
that
\[
\lambda_{1}(\Lambda)\leq K\varphi(n)
\]
and the set $B_{n}=g_{t_{n}}A_{n}$ where $t_{n}=\frac{1}{d}(\ln\varphi(n)-n)$.
It is well known that the function $\lambda_{1}^{-1}:\mathcal{L}%
_{d+c}\rightarrow\mathbb{R}$ is integrable,  see for instance \cite{Breu} p. 27
(actually, the only important fact is that a positive power of $\lambda
_{1}^{-1}$ is integrable). Making use of the Markov inequality, we obtain
\[
\mu(B_{n})=\mu(A_{n})\leq\frac{\left\Vert \lambda_{1}^{-1}\right\Vert _{1}%
}{\frac{1}{K\varphi(n)}}\ll\varphi(n).
\]
Therefore, by Borel-Cantelli Lemma, the set $\mathcal{B}$ of lattices
$\Lambda$ in $\mathcal{L}_{d+c}$ such that $\Lambda\in B_{n}$ for infinitely
many integers $n$, is negligible.
Suppose now that $\Lambda$ is a lattice such that $q_{n}^{c}(\Lambda)r_{n}%
^{d}(\Lambda)\leq\varphi(n)^{d+c}$ for infinitely many $n$. For each integer
$n\geq 1$, set $k_{n}=k_{n}(\Lambda)=$ $\lfloor\ln q_{n}(\Lambda)\rfloor$. By
Theorem \ref{Birkhoff}, for almost all lattices we have $k_{n}\leq
(L_{d,c}+1)n$ for $n$ large enough. Therefore for almost all lattices, for $n$
large enough, if $q_{n}^{c}(\Lambda)r_{n}^{d}(\Lambda)\leq\varphi(n)^{d+c}$
then the vector $g_{t_{k_{n}}}^{-1}(X_{n}(\Lambda))$ satisfies both%
\begin{align*}
\left\vert g_{t_{k_{n}}}^{-1}(X_{n}(\Lambda))\right\vert _{+}  &
=r_{n}(\Lambda)e^{\frac{c}{d}(k_{n}-\ln\varphi(k_{n}))}\\
&  \leq r_{n}(\Lambda)q_{n}^{\frac{c}{d}}(\Lambda)\varphi(k_{n})^{-\frac{c}%
{d}}\\
&  \leq\varphi(n)^{\frac{d+c}{d}}\varphi(k_{n})^{-\frac{c}{d}}\\
&  =\varphi(k_{n})\times\left(  \frac{\varphi(n)}{\varphi(k_{n})}\right)
^{\frac{d+c}{d}}\\
&  \leq\varphi(k_{n})\times\left(  \frac{\varphi(n)}{\varphi((L_{d,c}%
+1)n)}\right)  ^{\frac{d+c}{d}}\leq K\varphi(k_{n})
\end{align*}
for some constant $K$ depending only on $\varphi$ (we use that $\frac
{\varphi(t)}{\varphi((L_{d,c}+1)t)}$ is bounded above which is obviously true
when $\varphi(t)=\frac{1}{t^{\alpha}}$) and%
\[
\left\vert g_{t_{k_{n}}}^{-1}(X_{n}(\Lambda))\right\vert _{-}=q_{n}%
(\Lambda)e^{-k_{n}+\ln\varphi(k_{n})}\leq e\varphi(k_{n})\leq K\varphi
(k_{n}).
\]
Thus there are infinitely many $n$ such that $\Lambda\in B_{k_{n}}$. Since the
sequence $(k_{n})_{n}$ goes to infinity, $\Lambda\in\mathcal{B}$. It follows
that for almost all lattices $\Lambda$, $q_{n}^{c}(\Lambda)r_{n}^{d}%
(\Lambda)\geq\varphi(n)$ for $n$ large enough and we are done.
\end{proof}


As an immediate consequence of the previous Proposition and of Theorem \ref{thm:Birkhoff} we have:

\begin{corollary}
\label{cor:levy-d}%
\[
L_{d,c}=\frac{d}{\mu_{S}(S)}\int_{S}\tau~d\mu_{S}=\frac{d\times\mu
(\mathcal{L}_{d+1})}{\mu_{S}(S)}.
\]

\end{corollary}



\section{Parametrization of $S$ when $c=1$ \label{sec:S}}


This section is not necessary neither for the proofs of Theorems
\ref{thm:levy-d} and \ref{thm:jager}, nor for sections \ref{sect:reduction} and \ref{sec:qnrn}.

The aim is to show that the computation of the constant $L_{d,c}$ is theoretically feasible in the case $c=1$. However if the case $d=1$ is easy (see below), the case $d=2$ is already difficult. It is possible to give an integral formula for $L_{2,1}$. However, we are not able to compute the integral, only a numerical estimation of the integral has been carried out. 
An exact description of $S$ when $d\geq 3$ seems to be rather difficult.

In this section we assume $c=1$.
\subsection{$rkN$ decomposition }

In this subsection we give a parametrization of a set of lattices that contains $S$.

Let $\Lambda$ be a lattice in $S$ and let $v_{0}^S(\Lambda)=u_{1}$ and
$v_{1}^S(\Lambda)=u_{2}$ be the two vectors associated with $\Lambda$ by the
definition of $S$ (see the definition of $S$ section \ref{sec:defS}). When $d\geq 2$, we suppose these two
vectors have non negative heights and when $d=1$, we only suppose that $u_2$ has a nonnegative height. Since $u_1$ and $u_2$ are independent shortest vectors, by Corollary \ref{cor:lattice}, they
 are the first two vectors of a basis of $\Lambda$. Thus, there is a
matrix $M\in \SL(d+1,\mathbb{R)}$ defining $\Lambda$ the first two columns of
which are the vectors $u_{1}$ and $u_{2}$. 

When $d=1$, using the scaling factor $r=\left\vert u_{1}\right\vert_{+}=\left\vert u_2\right\vert_{-}>0$, we can write
\begin{align*}
M =rN,
\end{align*}
where $N$ is in the set $U_1$ of $2\times 2$ matrices such that
\begin{align*}
\det N &>0,\\
n_{1,1}=n_{2,2}&=1,\\
\left\vert n_{2,1}\right\vert,\,\left\vert n_{1,2}\right\vert&<1.
\end{align*}

When $d\geq 2$, let denote $(e_1,e_2,...,e_{d+1})$ the standard basis of $\mathbb R^{d+1}$. Using the same scaling factor 
$r=\left\vert u_{1}\right\vert_{+}=\left\vert u_2\right\vert_{-}>0$
and an orthogonal matrix $k$ that fixes $e_{d+1}$  and sends $e_{1}$ onto $\frac{1}{r}u_{1,+}$, we can find a matrix
$N=(n_{i,j})_{1\leq i,j\leq d+1}$ such that%
\begin{align*}
& M    =rkN,\\
& \det N    >0,
\end{align*}

\begin{gather}
n_{1,1}=n_{d+1,2}=1>n_{d+1,1}=\left\vert u_1\right\vert_{-}\geq 0,\label{decomposition1}\\
\left\Vert (n_{1,2},...,n_{d,2})\right\Vert _{\mathbb{R}^{d}}<1,
\label{decomposition2}\\
n_{2,1}=...=n_{d,1}=0. \label{decomposition3}
\end{gather}



When $d\geq 2$, $k$ is chosen in the group
\[
K_{d}=\{k\in \SO(d+1):ke_{d+1}=e_{d+1}\}.
\]
and using the decomposition of a $d\times d$ matrix in a product of an orthogonal 
matrix with positive determinant and of an upper triangular matrix we can even suppose that
\begin{equation}
n_{i,j}=0,\text{\ for all }1\leq j<i\leq d. \label{decomposition4}%
\end{equation}


For $d\geq 2$, let denote $U_d$ the set of $(d+1)\times(d+1)$-matrices such that
(\ref{decomposition1}), (\ref{decomposition2}),  and
(\ref{decomposition4}) hold ((\ref{decomposition4} implies (\ref{decomposition3})). 

Since $\det M=1$, the scaling factor $r$ must be equal to $(\det N)^{-\frac{1}{d+1}}$.
Puting $K_1=\{I_2\}$, for all $d\geq 1$, the map
\[
(k,N)\in K_d\times U_d\rightarrow (\det N)^{-\frac{1}{d+1}}kN
\]
provides a natural parametrization of a subset $\Sigma$ in
$\SL(d+1,\mathbb{R)}$ the projection
in $\mathcal{L}_{d+1}$ of which contains $S$. The main problem is now to find
which of these couples $(k,N)$ are such that $rkN\mathbb{Z}^{d+1}\in S$ and
to select a fundamental domain in this set of couples. This problem reduces to
finding the set of matrices $N$ such that

\begin{itemize}
	\item $N\in U_d$,
	
	\item The first two columns $u_{1}$ and $u_{2}$ of $N$ are in the unit ball
	$B_{\mathbb{R}^{d+1}}(0,1)$ and are the only nonzero vectors of the lattice
	$N\mathbb{Z}^{d+1}$ in this ball,
\end{itemize}
then select a fundamental domain in this set of matrices $N$.

This is easy when $d=1$ and doable when $d=2$. When
$d=1$, it is even possible to find the first return map $R$.

Another issue is to find the measure $\mu_{S}$ on $S$ induced by the flow
$g_{t}$ and the invariant measure $\mu$ of $\mathcal{L}_{d+1}$. This comparatively easier
issue can be performed for all $d$ without knowing explicitly $S$.

\subsection{The induced measure $\mu_S$}

Consider the manifold $V_{d}=\mathbb{R}_{>0}
\times\mathbb{R\times}K_{d}\times U_{d}$ and the  submanifold
\[
W=\{(\Delta,t,k,N)\in V_{d}:t=0,\ \Delta=1\}=\{1\}\times\{0\}\times
K_{d}\times U_{d}.
\]%
together with the maps
\begin{align*}
F &  :V_{d}\rightarrow \GL(d+1,\mathbb{R})\\
&  :\left(\Delta,t,k,N\right)\rightarrow\  \left(\frac{\Delta}{\det N}\right)
^{\frac{1}{d+1}}g_tkN\\
\end{align*}
and  $\overline F:V_d\rightarrow\GL(d+1,\mathbb R)/\SL(d+1,\mathbb Z)$ defined by
\[
\overline F(\Delta,t,k,N)=F(\Delta,t,k,N)\mathbb Z^{d+1}.
\]
By the discussion of the previous subsection, $\overline F$ provides a parametrization of $S$:
\[
S\subset \overline F(W).
\]
We would like to compute the measure $\mu_S$ in the coordinates $(1,0,k,N)$. The submanifold $W$ is equipped with the reference measure 
\[
\mu_{K_d}\otimes \lambda_{U_d}
\]
where $\lambda_{U_d}$ is the Lebesgue measure on $U_d$ and  $\mu_{K_d}$ is the invariant measure on $K_d$ associated with the invariant volume form $\gamma$
on $K_{d}$ that is dual to the exterior product of 
the	invariant vector fields generated by the standard skew symmetric matrices
$(A_{i,j}=E_{ij}-E_{ji})_{1\leq j<i\leq d}$.
The induce measure $\mu_S$ can be expressed with the parametrization $\overline{F}$, we give without proof an explicit formula in next Proposition.
\begin{proposition} \label{prop:invariant-measure}
 Assume $d \geq 2$. Suppose that $\mathcal D$ is an open subset of $W$ such that $\overline F(\mathcal D)\subset S$ and the restriction  of $\overline F$ to $\mathcal D$ is one to one. Then the image by $\overline F$ of the measure 
\[
1_{\mathcal{D}}\,(\frac{1}{\det N})^{d+1}(%
	{\displaystyle\prod\limits_{j=1}^{d-1}}
	n_{j,j}^{d-j})\,\mu_{K_d}\otimes \lambda_{U_d}
\]
is the restriction of $\mu_S$ to $\overline F(\mathcal D)$.
\end{proposition}
\begin{remark} \label{rem:d=1measure}
Recall that $n_{1,1}=1$ in the above formula.
\end{remark}
\begin{remark} \label{rem:d=1,2measure}
When $d=1$, the above measure has the density
	\[
	f(N)=f(n_{2,1},n_{1,2})=\frac{1}{(1-n_{2,1}n_{1,2})^2}.
	\]
with respect  to the Lebesgue measure on  
the two dimensional set $U_1$ of matrices $N$.\newline
\end{remark}



\subsection{Determination of the surface $S$, $c=1,\,d=1$ \label{sub:d=1}}
We already have a map 
\begin{align*}
& U_1\rightarrow \mathcal L_2\\
& N\rightarrow (\det N)^{-\frac{1}{2}}N\mathbb Z^2
\end{align*}
that sends $U_1$ onto a set that contains $S$.
By definition of $S$, the image of a matrix
\[
N=\left(
\begin{array}
[c]{cc}
1& n_{1,2} \\
n_{2,1}& 1
\end{array}
\right)\in U_1
\]
is in $S$ iff the only nonzero vectors of the lattice $\Lambda=N\mathbb Z^2$ in the ball $B_{\mathbb R^{1+1}}(0,1)$ are the two columns of $N$ up to sign.  We obtain that  $\Lambda\in S$ iff
\begin{itemize}
	\item $0<\left\vert n_{2,1}\right\vert,\,\left\vert n_{1,2}\right\vert <1$,
	\item the signs of $n_{1,2}$ and $n_{2,1}$ are opposite.
\end{itemize}
So the map $F$ defined on $]0,1[^2\times\{-1,1\}$ defined by 
\[
(x,y,\epsilon)\rightarrow \frac{1}{(1+xy)^{1/2}}\left(
\begin{array}
[c]{cc}
1& -\epsilon x \\
\epsilon y& 1
\end{array}
\right)\mathbb Z^2
\]
provide a parametrization of $S$ and it is easy to see that $F$ is a bijection. By Remark \ref{rem:d=1,2measure}, with these coordinates, the function
\[
f(x,y,\epsilon)=\frac{1}{(1+xy)^2}.
\]  
is density the measure $\mu_S$ with respect to the Lebesgue measure.
Therefore $\mu_S(S)=2\ln 2$.
With the Siegel formula (\cite{Siegel}) and Corollary  \ref{cor:levy-d}, we obtain the L\'evy's constant
\[
L_{1,1}=\frac{\mu(\mathcal{L}_{d+c})}{\mu_{S}(S)}=\frac{\zeta(2)}{2\ln 2}
=\frac{\pi^2}{12\ln 2}.
\] 

\subsubsection{Determination of the first return map, $c=1,\ d=1$}
Let 
\[
\Lambda=F(x,y,\epsilon)=\frac{1}{(1+xy)^{1/2}}\left(
\begin{array}
[c]{cc}
1& -\epsilon x \\
\epsilon y& 1
\end{array}
\right)\mathbb Z^2
\]
be in $S$.   By Lemma \ref{lem:return-minimal}, to find the first return $R(\Lambda)$ in $S$, it is enough to find the minimal vector $X_2(\Lambda)$. Then $R(\Lambda)$ is given by 
$g_{\tau(\lambda)}(\Lambda)$ with 
\[
\tau(\Lambda)=\frac{1}{2}\ln\left(\frac{\left\vert X_{2}(\Lambda)\right\vert_{-}}
{\left\vert X_{1}(\Lambda
	)\right\vert_{+}}\right).
\]
By corollary \ref{cor:lattice}, the first minimal vectors
$X_0(\Lambda)$ and $X_1(\Lambda)$ form a basis of $\Lambda$. The
minimal vector $X_2(\Lambda)$ is the vector of the form
$X=aX_0(\Lambda)+bX_1(\Lambda)$  in the strip $\left\vert X
\right\vert_{+}<\left\vert X_1(\Lambda)\right\vert_{+}=x$
with $a,b\in \mathbb Z$, and with the smallest height. It is not difficult to see that 
\[
X_2(\Lambda)=\epsilon X_0(\Lambda)+\lfloor\frac{1}{x}\rfloor X_1(\Lambda).
\]
So we obtain $R(\Lambda)=F(x',y',\epsilon')$ where
\begin{align*}
\epsilon'&=-\epsilon\\
x'&=\{\frac{1}{x}\}\\
y'&=\frac{1}{y+\lfloor\frac{1}{x}\rfloor}
\end{align*}
and we see that return map $R$ is a two-fold extension of the natural extension of the Gauss map.

\subsection{Value of L\'evy's constant when $d=2$ and $c=1$, \label{sub:d=2}}
An exact description of $S$ is possible when $d=2$ and $c=1$. Together with the expression of the measure $\mu_S$ in Proposition \ref{prop:invariant-measure}, this lead to a closed formula for L\'evy's constant as a seven-tuple integral of an algebraic function over an union of domains the boundaries of which are algebraic surfaces of degree at most two. We are not able to compute this seven-tuple integral. However using Octave, Seraphine Xieu (see \cite{Xieu}) has compute a numerical approximation of Levy's constant
\[
L_{2,1}=1.135256974\dots
\]
This can be compared with the one dimensional Levy's constant
\[
L_{1,1}=1.186569111\dots
\]

\section{Almost sure convergence in M$_{d,c}(\mathbb{R)}\label{sect:reduction}
$}


\subsection{A general result}

Recall that $\mathcal{H}_{\leq}$ is the subgroup of $\SL(d+c,\mathbb{R)}$
defined by%
\[
\mathcal{H}_{\leq}=\{h\in\SL(d+c,\mathbb{R)}:h=\left(
\begin{array}
[c]{cc}%
A & 0\\
B & C
\end{array}
\right)  \}
\]
with $ A\in\GL(d,\mathbb{R)}$, $B\in\M_{c,d}(\mathbb{R)}$ and $C\in
\GL(c,\mathbb{R})$. We say that a function $f:\mathcal{L}_{d+c}\rightarrow
\mathbb{R}$ is \textit{uniformly continuous in the }$\mathcal{H}_{\leq }%
$\textit{-direction} if for all $\varepsilon$ there exists $\beta>0$ such that
for all $\Lambda\in\mathcal{L}_{d+c}$ and all $h\in B_{\mathcal{H}_{\leq }}%
(I_{d+c},\beta)$, $\left\vert f(h\Lambda)-f(\Lambda)\right\vert \leq\varepsilon$.

\begin{theorem}
\label{thm:reduction} 1. Let $\varphi:S\rightarrow\mathbb{R}$ be a function
continuous almost everywhere on $S$. Suppose there exists a non negative function $f:\mathcal{L}_{d+c}\rightarrow\mathbb{R}_{\geq 0}$ that is continuous,
uniformly continuous in the $\mathcal{H}_{\leq}$-direction, integrable and
  such that $\left\vert
\varphi\right\vert \leq f$ on $S$. Then,
\[
\int_{S}fd\mu_{S}<+\infty
\]
and for almost all $\theta$ in $\M_{d,c}(\mathbb{R})$,
\[
\lim_{n\rightarrow\infty}\frac{1}{n}\sum_{k=0}^{n-1}\varphi\circ R^{k}%
(\Lambda_{\theta})=\frac{1}{\mu_{S}(S)}\int_{S}\varphi d\mu_{S}.
\]
2. The same result holds for $S^{\prime}$ instead of $S$
\end{theorem}

We
can formulate Theorem \ref{thm:reduction} for  a general surface $S$. The assumptions about $S$ are:

\begin{itemize}
\item $S$ is a co--dimension one submanifold transverse to the flow,

\item the number of visiting times in a time interval of length $1$ is bounded
above by a universal constant $A$ (Lemma \ref{lem:nb-returntimes}),

\item Lemma \ref{lem:boundary} below holds for $S$.
\end{itemize}

The other assumptions and the conclusion are the same as in Theorem \ref{thm:reduction}.

\medskip

For a compact subset $K$ of the submanifold $S$ and $\delta>0$, let denote
\[
U(K,\delta)=\{g_{t}h\Lambda:t\in[0,1],\ h\in B_{\mathcal{H}_{\leq}}%
(I_{d+c},\delta),\ \Lambda\in S\setminus K\}.
\]


\begin{lemma}
\label{lem:boundary} For all $\varepsilon>0$, there exists a compact subset
$K$ in $S$ and $\delta>0$ such that $\mu(U(K,\delta))\leq\varepsilon$.
\end{lemma}

This Lemma also holds  for $S'$ and is proven below only for $S$. This is the key Lemma because it explains that the
part of $S$ near its \textquotedblleft boundary\textquotedblright\ is not
relevant. 

Next Proposition is an important step toward Theorem \ref{thm:reduction}.
An example shows that without some assumptions about the boundary of
$S$ such as Lemma \ref{lem:boundary}, neither Theorem \ref{thm:reduction} nor Proposition \ref{prop:reduction}  hold. 




\begin{proposition}
\label{prop:reduction} Let $\varphi:S\rightarrow\mathbb{R}$ be a bounded
continuous function. Then for almost all $\theta$ in $\M_{d,c}(\mathbb{R})$,
\[
\lim_{n\rightarrow\infty}\frac{1}{n}\sum_{k=0}^{n-1}\varphi\circ R^{k}%
(\Lambda_{\theta})=\frac{1}{\mu_{S}(S)}\int_{S}\varphi d\mu_{S}.
\]

\end{proposition}

\subsubsection{Auxiliary Lemmas}
We will need three Lemmas the proofs of which are omitted.
The first one only use  that $S$ and $S'$ are transverse to the flow together with the inverse mapping Theorem. 
\begin{lemma}
	\label{lem:K-S} For all compact subset $K$ in $S$ (or in $S'$), there exist $\alpha$ and
	$\eta>0$ such that 
	\newline- the map $(t,\Lambda)\rightarrow g_{t}\Lambda$ is
	one to one on $[-\alpha,\alpha]\times K$,
	\newline- for all $h\in
	B(I_{d+c},\eta)$ and all $\Lambda$ in $K$, there exists an unique
	$t=t(h,\Lambda)\in[-\alpha,\alpha]$ such that $g_{-t}h\Lambda\in S$.
	\newline- the maps $\sigma:(h,\Lambda)\rightarrow t=t(h,\Lambda)$ and $\pi
	:(h,\Lambda)\rightarrow g_{-t}h\Lambda$ are continuous on $B(I_{d+c}%
	,\eta)\times K$ and the values of $\tau$ are in $[-\alpha/4,\alpha/4]$.
\end{lemma}

The second Lemma is a purely  theoretical measure result.

\begin{lemma}
	\label{lem:measure-image} Let $X$ and $Y$ be locally compact second countable
	metric spaces. Let $\mu_{X}$ and $\mu_{Y}$ be two measures on $X$ and $Y$
	finite on compact subsets. Suppose $\psi:X\rightarrow Y$ is a continuous map
	such that every $y$ in $Y$ has at most $N$ preimages and such that for all $x$ in $X$
	there exists a compact neighborhood $\omega_{x}$ of $x$ with the following
	property:
	\newline- $\psi$ is one to one on $\omega_{x}$,
	\newline- the image by
	$\psi$ of the measure $1_{\omega_{x}}\mu_{X}$ is the measure $1_{\psi
		(\omega_{x})}\mu_{Y}$. 
	\newline Then for all nonnegative measurable function
	$f:Y\rightarrow\mathbb{R}$,
	\[
	\int_{X}f\circ\psi\, d\mu_{X}\leq N\int_{Y}f\,d\mu_{Y}.
	\]
	
\end{lemma}


The last Lemma is an easy consequence of the previous Lemma and of  the definition of the induced measure $\mu_S$.

\begin{lemma}
	\label{lem:induce-measure} Let $U$ be an open subset in $\mathcal{L}_{d+c}$
	such that for all $\Lambda$ in $U$, $g_{t}\Lambda\in U$ for all $t$ in a time
	interval $I_{\Lambda}$ of length $1$ containing $0$. Then
	\[
	\mu_{S}(U\cap S)\leq 4A\mu(U)
	\]
	where $A$ is the maximum number of entrance times in $S$ of a flow trajectory
	during a time interval of length $1$ (see Lemma \ref{lem:nb-returntimes}).
\end{lemma}

\begin{remark} The constant $4A$ is certainly not the best one.
\end{remark}
\begin{remark}	
 The assumption $U$ Borel subset should be sufficient. 
\end{remark} 


\subsubsection{An example}

We want to construct a co-dimension one submanifold $V$ in $\mathcal{L}_{d+c}$
transverse to the flow $g_{t}$ together with a bounded continuous function
$\varphi:V\rightarrow\mathbb{R}$ such that for a set of positive measure of
$\theta\in M_{d,c}(\mathbb R)$, \ the sequence $\frac{1}{n}\sum_{k=0}^{n-1}\varphi\circ R_{V}%
^{k}(\Lambda_{\theta})$ does not converge to $\frac{1}{\mu_{V}(V)}\int
_{V}\varphi d\mu_{V}$ where $R_{V}$ is the first return map in $V$ and
$\mu_{V}$ is the invariant measure induced by the flow. The idea is the
following. Take $V$ an open set in $S$. Then $\mu_{V}$ is the restriction of
$\mu_{S}$ to $V$. Suppose that the open set $V$ can be chosen in order that
for all $\theta\in M_{d,c}(\mathbb R)$, and all $k\geq 1$,$\ $%
\[
R_{V}^{k}(\Lambda_{\theta})=R_{S}^{k}(\Lambda_{\theta})(=R^{k}(\Lambda
_{\theta})).
\]
Then if $\varphi:S\rightarrow\mathbb{R}$ is a non negative continuous function
not identically zero with support included in $V$, the sequences
\[
\frac{1}{n}\sum_{k=0}^{n-1}\varphi\circ R_{V}^{k}(\Lambda_{\theta}),\ \frac
{1}{n}\sum_{k=0}^{n-1}\varphi\circ R_{S}^{k}(\Lambda_{\theta})
\]
converge to the same limit which cannot be equal to both $\frac{1}{\mu_{S}%
(S)}\int_{S}\varphi d\mu_{S}$ and $\frac{1}{\mu_{V}(V)}\int_{V}\varphi
d\mu_{V}=\frac{1}{\mu_{S}(V)}\int_{S}\varphi d\mu_{S}$ provided that $\mu
_{S}(V)<\mu_{S}(S)$. So we are reduced to constructing $V$.

Observe that Theorem \ref{thm:reduction} implies that for such a $V$, $\varphi=1_{V}$
is not almost everywhere continuous on $S$ which means that the boundary of
$V$ in $S$ has positive measure. Moreover, it shows that the assumption about the continuity of the function $\varphi$ in Theorem \ref{thm:reduction}, cannot be dropped.

\subsubsection{Construction of $V$}

Consider the set $\mathbb{T}$ of lattices $\Lambda_{\theta}$ such that the
coefficients of $\theta$ are all in $[0,1]$. It is a compact subset in
$\mathcal{L}_{d+c}$ containing all the lattices $\Lambda_{\theta}$. Denote
$W_{\varepsilon}$ the open ball $B(I_{d+c},\varepsilon)$ in $\SL(d+c,\mathbb{R)}$.
We consider the open sets%
\[
U_{n}(\varepsilon)=\bigcup_{t\in[n,n+1]}g_{t}(W_{\varepsilon
}\mathbb{T})
\]
and for a sequence $(\varepsilon_{n})_{n\in\mathbb{N}}$ of positive real
numbers, we consider the open set
\[
U=U((\varepsilon_{n})_{n\in\mathbb{N}})=%
\bigcup_{n\in\mathbb{N}}
U_{n}(\varepsilon_{n}).
\]
Take $V=S\cap U$. For all $t\geq$ and all $\theta$, $g_{t}\Lambda_{\theta
}=g_{t}I_{d+c}\Lambda_{\theta}$ is in $U$, hence for all $k\in\mathbb{N}$, we have
\[
R_{S}^{k}(\Lambda_{\theta})=R_{V}^{k}(\Lambda_{\theta}).
\]
So we are reduce to show that when the sequence $(\varepsilon_{n})_{n}$ is
small enough,
\[
\mu_{S}(V)<\mu_{S}(S).
\]






By definition of $U_{n}(\varepsilon)$, if $\Lambda=g_{t}g\Lambda_{\theta}$ with
$t\in[n,n+1]$, $g\in W_{\varepsilon}$ and $\Lambda_{\theta}\in\mathbb{T},$ then
\[
g_{s}g_{t}g\Lambda_{\theta}\in U_{n}(\varepsilon)
\]
for all $s$ in the interval $[n-t,n+1-t]$. So $U$ satisfies the assumption of the
Lemma \ref{lem:induce-measure} and therefore
\[
\mu_{S}(U\cap S)\leq 4A\sum_{n\in\mathbb{N}}\mu(U_{n}(\varepsilon_{n})).
\]
Using that $g_{t}g\Lambda=g_{n}(g_{t-n}gg_{-(t-n)})g_{t-n}\Lambda$, we see
that
\begin{align*}
U_{n}(\varepsilon)  &  =\{g_{t}\Lambda:t\in[n,n+1],\ \Lambda\in W_{\varepsilon
}\mathbb{T}\}\\
&  \subset g_{n}W_{\varepsilon^{\prime}}\{g_{s}\Lambda:s\in[0,1],\ \Lambda
\in\mathbb{T\}}%
\end{align*}
where $\varepsilon^{\prime}$ is such that $g_{s}W_{\varepsilon}g_{-s}\subset
W_{\varepsilon^{\prime}}$ for all $s$ in $[0,1]$. Furthermore, the compact set
$\{g_{s}\Lambda:s\in[0,1],\ \Lambda\in\mathbb{T\}}$ has zero measure because
it has dimension $cd+1$ which is $<(c+d)^{2}-1$. Therefore
\[
\lim_{\varepsilon^{\prime}\rightarrow 0}\mu(W_{\varepsilon^{\prime}}%
\{g_{s}\Lambda:s\in[0,1],\ \Lambda\in\mathbb{T}\})=0,
\]
which implies
\[
\lim_{\varepsilon\rightarrow 0}\mu(U_{n}(\varepsilon))=0.
\]
So there exists a sequence $(\varepsilon_{n})_{n\in\mathbb{N}}$ such that%
\[
\sum_{n\in\mathbb{N}}\mu(U_{n}(\varepsilon_{n}))<\frac{1}{4A}\mu(S)
\]
and for such a sequence, the sets $U=\cup_{n\in\mathbb{N}}U_{n}(\varepsilon
_{n})$ and $V=U\cap S$ are the ones we are looking for which ends the construction of a counter-example to Theorem \ref{thm:reduction} without assumption about the boundary of $S$.

\subsection{Proof of Proposition \ref{prop:reduction}}

Let $\varphi:S\rightarrow\mathbb{R}$ be a continuous bounded function.

Let $\varepsilon$ be a positive real number, let $K$ and $\delta$ be
associated with $\varepsilon$ by Lemma \ref{lem:boundary}, and $\alpha$ and
$\eta$ associated with $K$ by Lemma \ref{lem:K-S}.

{\it Preliminary observations. }
Let $(a_{n})_{n\in\mathbb N}$ be a decreasing sequence of reals numbers in $]0,\eta[$
tending to zero and set $L_{a_{n}}=B(I_{d+c},a_{n})\times K$. Since
the intersection of all the compact sets $L_{a_{n}}$, $n\in\mathbb N$, is $L_{0}=\{I_{d+c}\}\times
K$ and since the map $\psi(g,\Lambda)=\varphi(\pi(g,\Lambda))-\varphi
(\Lambda)$ is continuous, we have%
\[
\cap_{n\geq 0}\psi(L_{a_{n}})=\psi(\cap_{n\geq 0}L_{a_{n}})=\psi(L_{0})=\{0\}.
\]
Therefore, for $n$ large enough $\psi(L_{a_{n}})\subset]-\varepsilon
,\varepsilon[$ which implies there exists $\beta>0$ such that for all
$\Lambda\in K$ and all $g\in B(I_{d+c},\beta)$,%
\begin{equation}
\left\vert \varphi(\pi(g,\Lambda))-\varphi(\Lambda)\right\vert \leq
\varepsilon. \label{uniform}%
\end{equation}
Finally, let $\gamma>0$ be such that for all $s\geq 0$ and all $h\in
B_{\mathcal{H}_{\leq}}(I_{d+c},\gamma)B_{\mathcal{H}_{\leq}}^{-1}%
(I_{d+c},\gamma)$, 
\[
\dd(g_{s}hg_{-s},I_{d+c})\leq\min(\delta,\beta,\eta).
\]

For $T\geq 0$, $\Lambda$ a lattice, and $E$ a subset of $S,$ denote%
\[
I(T,\Lambda,E)=\{t\in[0,T]:g_{t}\Lambda\in E\}.
\]


For almost all $\theta\in M_{d,c}(\mathbb R)$, we can fix $h_{\theta}\in B_{\mathcal{H}_{\leq}%
}(I_{d+c},\gamma)$ such that the conclusion of Birkhoff Theorem holds for the flot $g_t$ or the first return map in $S$ and the 
lattice $h_{\theta}\Lambda_{\theta}$. Observe that $h_{\theta}=h_{\theta,\varepsilon}$ 
depends on $\varepsilon$.
It is understood that we shall use Birkhoff Theorem in  countably many 
situations. We fix a sequence $\varepsilon_{n}$ going to zero and for each
$\varepsilon=\varepsilon_{n}$ and we use three times Birkhoff Theorem and the ergodicity of the flow: for
almost all $\theta$,
\begin{equation}
\lim_{T\rightarrow\infty}\frac{1}{T}\int_{0}^{T}1_{U(K,\delta)}(g_{t}%
h_{\theta}\Lambda_{\theta})dt
=\frac{1}{\mu(\mathcal{L}_{d+c})}\mu(U(K,\delta))\leq\varepsilon,
\label{birkhoff1}
\end{equation}%
\begin{equation}
\lim_{T\rightarrow\infty}\frac{1}{T}\operatorname{card}I(T,h_{\theta}%
\Lambda_{\theta},S)=\frac{1}{\mu_{S}(S)}\int_{S}\tau_S d\mu_{S}=\frac
{\mu(\mathcal{L}_{d+c})}{\mu_{S}(S)}, \label{birkhoff2}%
\end{equation}
where $\tau_S$ is the first return time in $S$,
and%
\begin{equation}
\lim_{T\rightarrow\infty}\frac{1}{\operatorname*{card}I(T,h_{\theta}%
\Lambda_{\theta},S)}\sum_{t\in I(T,h_{\theta}\Lambda_{\theta},S)}\varphi
(g_{t}h_{\theta}\Lambda_{\theta})=\frac{1}{\mu_{S}(S)}\mu_{S}(\varphi).
\label{birkhoff3}%
\end{equation}
Let $T$ be positive and let $s_{1}<...<s_{m}$ be the elements of $I(T,\Lambda_{\theta},S\setminus K)$,
we have
\[
g_{t}h_{\theta}\Lambda_{\theta}=g_{t-s_{i}}(g_{s_{i}}h_{\theta}g_{-s_{i}%
})g_{s_{i}}\Lambda_{\theta}\in U(K,\delta)
\]
for all $t\in[ s_{i},s_{i}+1]$ and we can extract a subsequence $s_{n_{1}%
},...,s_{n_{p}}$ defined by $n_{1}=1$ and $n_{i+1}=\min\{j:s_{j}\geq s_{n_{i}%
}+1\}$. Now by Lemma \ref{lem:nb-returntimes}, there is an absolute constant $A$ such that there are at most
$A$ elements of $I(T,\Lambda_{\theta},S\setminus K)$ ($\subset I(T,\Lambda
_{\theta},S)$) in an interval of length $1$, hence $Ap\geq m$. Therefore, by
(\ref{birkhoff1})%
\[
\frac{m}{A}\leq p\leq\int_{0}^{T+1}1_{U(K,\delta)}(g_{t}h_{\theta}%
\Lambda_{\theta})dt\leq 2(T+1)\varepsilon
\]
and hence%
\begin{equation}
\operatorname*{card}I(T,\Lambda_{\theta},S\setminus K)\leq 3AT\varepsilon
\label{nbvisitingSK1}%
\end{equation}
for $T$ large enough: $T\geq T(\Lambda_{\theta},\varepsilon)$. We will also
need to bound above the number of elements of $I(T,h_{\theta}\Lambda_{\theta
},S\setminus K)$ and making use of (\ref{birkhoff1}), the same way of
reasoning leads to the same result
\begin{equation}
I(T,h_{\theta}\Lambda_{\theta},S\setminus K)\leq 3AT\varepsilon
\label{nbvisitingSK2}%
\end{equation}
for $T\geq T(\Lambda_{\theta},\varepsilon)$.

{\it Heart of the proof. }We want to compare
\[
\Sigma_{1}=\frac{1}{\operatorname*{card}I(T,\Lambda_{\theta},S)}\sum_{t\in
I(T,\Lambda_{\theta},S)}\varphi(g_{t}\Lambda_{\theta})
\]
with
\[
\Sigma_{2}=\frac{1}{\operatorname*{card}I(T,h_{\theta}\Lambda_{\theta},S)}%
\sum_{t\in I(T,h_{\theta}\Lambda_{\theta},S)}\varphi(g_{t}h_{\theta}%
\Lambda_{\theta})
\]
because by (\ref{birkhoff3}), this latter sum tends to $\frac{1}{\mu_{S}(S)}%
\int_{S}\varphi d\mu_{S}$ when $T$ goes to infinity. We split $\sum_{t\in
I(T,\Lambda_{\theta},S)}$ in two sums $\sum_{t\in I(T,\Lambda_{\theta},K)}$
and $\sum_{t\in I(T,\Lambda_{\theta},S\setminus K)}$. Observe that for $t\in
I(T,\Lambda_{\theta},K)$, $g_{t}h_{\theta}\Lambda_{\theta}=(g_{t}h_{\theta
}g_{-t})g_{t}\Lambda_{\theta}$ is of the form $g\Lambda$ with $g\in
B(I_{d+c},\eta)$ and $\Lambda\in K$, this allows to use Lemma \ref{lem:K-S}.
We use the notation of Lemma \ref{lem:K-S} and for $t$ in $I(T,\Lambda
_{\theta},K)$, we denote $t^{\prime}=\sigma(g_{t}h_{\theta}g_{-t},g_{t}%
\Lambda_{\theta})$. By (\ref{uniform}), we have%
\[
\left\vert \sum_{t\in I(T,\Lambda_{\theta},K)}\varphi(g_{t}\Lambda_{\theta
})-\sum_{t\in I(T,\Lambda_{\theta},K)}\varphi(\pi(g_{t}h_{\theta}g_{-t}%
,g_{t}\Lambda_{\theta}))\right\vert \leq\varepsilon\operatorname{card}%
I(T,\Lambda_{\theta},K).
\]
Now,
\[
\pi(g_{t}h_{\theta}g_{-t},g_{t}\Lambda_{\theta})=g_{-t^{\prime}}g_{t}%
h_{\theta}g_{-t}g_{t}\Lambda_{\theta}=g_{t-t^{\prime}}h_{\theta}%
\Lambda_{\theta}%
\]
hence%
\[
\left\vert \sum_{t\in I(T,\Lambda_{\theta},K)}\varphi(g_{t}\Lambda_{\theta
})-\sum_{t\in I(T,\Lambda_{\theta},K)}\varphi(g_{t-t^{\prime}}h_{\theta
}\Lambda_{\theta})\right\vert \leq\varepsilon\operatorname{card}%
I(T,\Lambda_{\theta},K).
\]
Observe that the map $t\in I(T,\Lambda_{\theta},K)\rightarrow t-t^{\prime}$ is one to one because $t^{\prime}\in
[-\frac{\alpha}{4},\frac{\alpha}{4}]$ and the gap between two visiting times
of $K$ is $\geq\alpha$. Observe also that $t-t'\in
I(T,h_{\theta}\Lambda_{\theta},S)$ except possibly for the first and the last element of $t\in I(T,h_{\theta}\Lambda_{\theta},S)$. On the one hand, it follows that
\begin{multline*}
\left\vert \sum_{t\in I(T,\Lambda_{\theta},S)}\varphi(g_{t}\Lambda_{\theta
})-\sum_{t\in I(T,h_{\theta}\Lambda_{\theta},S)}\varphi(g_{t}h_{\theta}%
\Lambda_{\theta})\right\vert \\
\leq\varepsilon\operatorname{card}I(T,\Lambda_{\theta},K)\\
+\left\Vert \varphi\right\Vert _{\infty}(\operatorname{card}I(T,\Lambda
_{\theta},K\setminus S)+\operatorname*{card}I(T,h_{\theta}\Lambda_{\theta
},S)-\operatorname*{card}I(T,\Lambda_{\theta},K)+2).
\end{multline*}
On the other hand, it follows that%
\[
\operatorname*{card}I(T,h_{\theta}\Lambda_{\theta},S)\geq\operatorname*{card}%
I(T,\Lambda_{\theta},K)-2
\]
and the same way of reasoning leads to
\[
\operatorname*{card}I(T,\Lambda_{\theta},S)\geq\operatorname*{card}%
I(T,h_{\theta}\Lambda_{\theta},K) -2.
\]
Making use of (\ref{nbvisitingSK1}) and (\ref{nbvisitingSK2}), we obtain
\begin{align*}
-2  &  \leq\operatorname*{card}I(T,h_{\theta}\Lambda_{\theta}%
,S)-\operatorname*{card}I(T,\Lambda_{\theta},K)=\\
&  \operatorname*{card}I(T,h_{\theta}\Lambda_{\theta},S)-\operatorname*{card}%
I(T,h_{\theta}\Lambda_{\theta},K)\\
&  +\operatorname*{card}I(T,h_{\theta}\Lambda_{\theta},K)-\operatorname*{card}%
I(T,\Lambda_{\theta},S)\\
&  +\operatorname*{card}I(T,\Lambda_{\theta},S)-\operatorname*{card}%
I(T,\Lambda_{\theta},K)\\
&  \leq 3 AT\varepsilon+2+3AT\varepsilon=6AT\varepsilon+2,
\end{align*}
hence (using (\ref{nbvisitingSK1}) once again)
\[
\left\vert \sum_{t\in I(T,\Lambda_{\theta},S)}\varphi(g_{t}\Lambda_{\theta
})-\sum_{t\in I(T,h_{\theta}\Lambda_{\theta},S)}\varphi(g_{t}h_{\theta}%
\Lambda_{\theta})\right\vert \leq\varepsilon\operatorname{card}I(T,\Lambda
_{\theta},K)+(9A\varepsilon T+2)\left\Vert \varphi\right\Vert _{\infty}%
\]
for $T\geq T(\Lambda_{\theta},\varepsilon)$. We obtain%
\begin{align*}
\left\vert \operatorname*{card}I(T,h_{\theta}\Lambda_{\theta}%
,S)-\operatorname*{card}I(T,\Lambda_{\theta},S)\right\vert  &  \leq\left\vert
\operatorname*{card}I(T,h_{\theta}\Lambda_{\theta},S)-\operatorname*{card}%
I(T,\Lambda_{\theta},K)\right\vert \\
&  +\left\vert \operatorname*{card}I(T,\Lambda_{\theta},S\setminus
K)\right\vert \\
&  \leq 9AT\varepsilon+2
\end{align*}
as well. Relation (\ref{birkhoff2}) implies that $\operatorname{card}%
I(T,h_{\theta}\Lambda_{\theta},S)\geq aT$ for $T\geq T(\Lambda_{\theta
},\varepsilon)$ where $a=\frac{1}{2}\frac{\mu(\mathcal{L)}}{\mu_{S}(S)}$. All
together, for $T\geq T(\Lambda_{\theta},\varepsilon)$, we obtain%
\begin{align*}
\left\vert \Sigma_{2}-\Sigma_{1}\right\vert  &  \leq\left\vert \Sigma
_{2}-\frac{\operatorname{card}I(T,\Lambda_{\theta},S)}{\operatorname*{card}%
I(T,h_{\theta}\Lambda_{\theta},S)}\Sigma_{1}\right\vert +\left\vert
\frac{\operatorname*{card}I(T,\Lambda_{\theta},S)-\operatorname*{card}%
I(T,h_{\theta}\Lambda_{\theta},S)}{\operatorname*{card}I(T,h_{\theta}%
\Lambda_{\theta},S)}\right\vert \left\vert \Sigma_{1}\right\vert \\
&  \leq\frac{1}{\operatorname{card}I(T,h_{\theta}\Lambda_{\theta}%
,S)}\left\vert \sum_{t\in I(T,h_{\theta}\Lambda_{\theta},S)}\varphi(g_{t}%
h_{\theta}\Lambda_{\theta})-\sum_{t\in I(T,\Lambda_{\theta},S)}\varphi(g_{t}\Lambda
_{\theta})\right\vert \\
&  +\left\vert \frac{\operatorname*{card}I(T,\Lambda_{\theta}%
,S)-\operatorname*{card}I(T,h_{\theta}\Lambda_{\theta},S)}%
{\operatorname*{card}I(T,h_{\theta}\Lambda_{\theta},S)}\right\vert \left\Vert
\varphi\right\Vert _{\infty}\\
&  \leq\frac{\varepsilon\operatorname{card}I(T,\Lambda_{\theta}%
,K)+(9AT\varepsilon+2)\left\Vert \varphi\right\Vert _{\infty}}{\operatorname{card}%
I(T,h_{\theta}\Lambda_{\theta},S)}+\frac{(9AT\varepsilon+2)\left\Vert \varphi\right\Vert
_{\infty}}{\operatorname{card}I(T,h_{\theta}\Lambda_{\theta},S)}\\
&  \leq\varepsilon(1+\frac{18AT\left\Vert \varphi\right\Vert _{\infty}}%
{aT})+\frac{4\|\varphi\|_{\infty}}{aT}
\end{align*}
which is $\ll \varepsilon$ when $T$ is large enough.

\subsection{Proof of Theorem \ref{thm:reduction}}

\textbf{Step 1. }Let us show that the restriction of $f$ to $S$ is integrable
with respect to $\mu_{S}$.

We use Lemma \ref{lem:measure-image} with $X=]0,1[\times S$, $Y=\mathcal{L}%
_{d+c}$, the map $\psi:]0,1[\times S\rightarrow\mathcal{L}$ defined by
$\psi(t,\Lambda)=g_{t}\Lambda$, the measures $\mu_{X}=dt\otimes\mu_{S}$ and
$\mu_{Y}=\mu$, and the function $f$. By definition of the induced measure, we
know that the image of the restriction of $dt\otimes\mu_{S}$ to any small enough
open subset $\omega$ is the restriction to $\psi(\omega)$ of the invariant
measure $\mu$ on $\mathcal{L}_{d+c}$. Now, by Lemma \ref{lem:lowerrate} each element of $\mathcal{L}_{d+c}$ has at most
$A+1$ $\psi$-preimages, therefore by Lemma \ref{lem:measure-image}
\[
\int_{0}^{1}\int_{S}f(g_{t}\Lambda)d\mu_{S}dt\leq (A+1)\int_{\mathcal{L}_{d+c}}fd\mu.
\]
Since $f$ is uniformly continuous in the $\mathcal{H}_{\leq}$ direction,
there exists $\Delta>0$ such that for all $\Lambda$, and all $t\in
[0,\Delta],\ f(g_{t}\Lambda)\geq f(\Lambda)-1.$ Therefore%
\[
\int_{0}^{\Delta}\int_{S}(f(\Lambda)-1)d\mu_{S}dt\leq (A+1)\int_{\mathcal{L}_{d+c}}%
fd\mu,
\]
which implies $\int_{S}f(\Lambda)d\mu_{S}\leq\mu_{S}(S)+\frac{A+1}{\Delta}%
\int_{\mathcal{L}_{d+c}}fd\mu<+\infty$.

\textbf{Step 2:} It is enough to prove the Theorem for continuous functions
$\varphi$.

Indeed, since $\varphi$ is continuous almost everywhere and since $\left\vert
\varphi\right\vert \leq f$ with $f$ continuous and in $L^{1}(\mu_{S})$, for all positive
integer $p$, there exist two continuous functions $\varphi_{p}^{-}$ and
$\varphi_{p}^{+}$ such that
\[
-f\leq\varphi_{p}^{-}\leq\varphi\leq\varphi_{p}^{+}\leq f
\]
and
\[
\int_{S}\varphi d\mu_{S}-\frac{1}{p}\leq\int_{S}\varphi_{p}^{-}d\mu_{S}%
\leq\int_{S}\varphi_{p}^{+}d\mu_{S}\leq\int_{S}\varphi d\mu_{S}+\frac{1}{p}.
\]
Therefore, if the convergence holds for almost every $\theta$ for all the functions $\varphi_p^-$ and $\varphi_p^+$, we have%
\begin{align*}
\int_{S}\varphi_{p}^{-}d\mu_{S}  &  =\lim_{n\rightarrow\infty}\frac{1}{n}%
\sum_{k=0}^{n-1}\varphi_{p}^{-}\circ R^{k}(\Lambda_{\theta})\leq\lim
\inf_{n\rightarrow\infty}\frac{1}{n}\sum_{k=0}^{n-1}\varphi\circ R^{k}%
(\Lambda_{\theta})\\
&  \leq\lim\sup_{p\rightarrow\infty}\frac{1}{n}\sum_{k=0}^{n-1}\varphi\circ
R^{k}(\Lambda_{\theta})\leq\lim_{n\rightarrow\infty}\frac{1}{n}\sum
_{k=0}^{n-1}\varphi_{p}^{-}\circ R^{k}(\Lambda_{\theta})=\int_{S}\varphi
_{p}^{+}d\mu_{S}%
\end{align*}
which implies that for almost all $\theta$,
\[
\lim_{n\rightarrow\infty}\frac{1}{n}\sum_{k=0}^{n-1}\varphi\circ R^{k}%
(\Lambda_{\theta})=\int_{S}\varphi d\mu_{S}.
\]
So, we are reduce to prove the Theorem for $\varphi$ continuous.

\textbf{Step 3. }

Writing $\varphi=\varphi^+-\varphi^-$, we can suppose $\varphi\geq 0$. Using Proposition \ref{prop:reduction} with the
minimum of $\varphi$ and of a constant $M$, we obtain for almost all $\theta$,%
\[
\lim_{n\rightarrow\infty}\inf\frac{1}{n}\sum_{k=0}^{n-1}\varphi\circ
R^{k}(\Lambda_{\theta})\geq\lim_{n\rightarrow\infty}\frac{1}{n}\sum
_{k=0}^{n-1}\min(\varphi,M)\circ R^{k}(\Lambda_{\theta})=\frac{1}{\mu_{S}%
(S)}\int_{S}\min(\varphi,M) d\mu_{S},
\]
hence, letting $M$ going to infinity, we obtain
\[
\lim_{n\rightarrow\infty}\inf\frac{1}{n}\sum_{k=0}^{n-1}\varphi\circ
R^{k}(\Lambda_{\theta})\geq\frac{1}{\mu_{S}(S)}\int_{S}\varphi d\mu_{S}.
\]
So we have to bound above the sums $\sum_{k=0}^{n-1}\varphi\circ R^{k}%
(\Lambda_{\theta})$.

Since $f$ is in $L^{1}$ there exists $\varepsilon^{\prime}>0$ such that
for any measurable subset $B$ in $\mathcal{L}_{d+c}$, we have
\[
\mu(B)\leq\varepsilon^{\prime}\Rightarrow\int_{B}fd\mu\leq\varepsilon.
\]
This allows to strengthen Lemma \ref{lem:boundary}:

\begin{lemma}
For all $\varepsilon>0$, there exists a compact subset $K$ in $S$ and
$\delta>0$ such that $\frac{1}{\mu(\mathcal{L}_{d+c})}\int_{U(K,\delta)}fd\mu$
and $\frac{1}{\mu(\mathcal{L}_{d+c})}\mu(U(K,\delta))$ are $\leq\varepsilon$.
\end{lemma}

We keep all the choices and the notations of the proof of Proposition
\ref{prop:reduction}, and we use Birkhoff Theorem with one more function:
\begin{equation}
\lim_{T\rightarrow\infty}\frac{1}{T}\int_{0}^{T}f(g_{t}h_{\theta}%
\Lambda_{\theta})1_{U(K,\delta)}(g_{t}h_{\theta}\Lambda_{\theta})dt=\frac
{1}{\mu(\mathcal{L}_{d+c})}\int_{U(K,\delta)}fd\mu\leq\varepsilon\label{L1}%
\end{equation}
so that (\ref{birkhoff1}), (\ref{birkhoff2}), (\ref{birkhoff3}) and (\ref{L1})
hold for almost all $\theta$.

Since the function $f$ is uniformly continuous in the $\mathcal{H}_{\leq }%
$-direction, there exists $\kappa>0$ such that $f(h\Lambda)\geq f(\Lambda)-\frac
{1}{2}$ for all $\Lambda$ and all $h\in B_{\mathcal{H}_{\leq }}(I_{d+c}%
,\kappa)$. By choosing $\gamma$ small enough we can suppose that
$(g_{s}hg_{-s})\in B_{\mathcal{H}_{\leq}}(I_{d+c},\kappa)$ for all $s\geq 0$
and all $h\in B_{\mathcal{H}_{\leq}}(I_{d+c},\gamma)$. Furthermore, there
exists a positive constant $\Delta=\Delta(\kappa)$ such that $g_{t}\in
B_{\mathcal{H}_{\geq 0}}(I_{d+c},\kappa)$ for all $t\in[0,\Delta]$. Therefore
for all lattices $\Lambda$, all non negative real number $s$, all $h\in
B_{\mathcal{H}_{\leq}}(I_{d+c},\gamma)$ and all $t\in [0,\Delta]$, we have
\[
f(g_{t}(g_{s}hg_{-s})g_{s}\Lambda)\geq f((g_{s}hg_{-s})g_{s}\Lambda)-\frac
{1}{2}\geq f(g_{s}\Lambda)-1
\]
and hence
\begin{equation}
f(g_{t+s}h\Lambda)=f(g_{t}(g_{s}hg_{-s})g_{s}\Lambda)\geq f(g_{s}\Lambda)-1.
\label{uniform-f}%
\end{equation}


As in the proof of Proposition \ref{prop:reduction}, let $s_{1}<...<s_{m}$ be
the elements of $I(T,\Lambda_{\theta},S\setminus K)$. On the one hand $g_{t}h_{\theta
}\Lambda_{\theta}\in U(K,\delta)$ for all $t\in[ s_{i},s_{i}+1]$, and on the other hand, for almost
all $\theta$,  (\ref{nbvisitingSK1}) and (\ref{nbvisitingSK2}) hold for $T\geq
T(\Lambda_{\theta},\varepsilon)$. We can suppose $\Delta<1$ and since there
are at most $A$ elements of $I(T,\Lambda_{\theta},S\setminus K)$ ($\subset
I(T,\Lambda_{\theta},S)$) in an interval of length $1$, by
(\ref{uniform-f}) we obtain
\begin{align*}
\sum_{s\in I(T,\Lambda_{\theta},S\setminus K)}\varphi(g_{s}\Lambda_{\theta})
&  \leq\sum_{s\in I(T,\Lambda_{\theta},S\setminus K)}f(g_{s}\Lambda_{\theta
})\\
&  \leq\sum_{i=1}^{m}\frac{1}{\Delta}\int_{s_{i}}^{s_{i}+\Delta}%
(1+1_{U(K,,\delta)}(g_{t}h_{\theta}\Lambda_{\theta})f(g_{t}h_{\theta}%
\Lambda_{\theta}))dt\\
&  \leq m+\frac{A}{\Delta}\int_{0}^{T+1}1_{U(K,\delta)}(g_{t}h_{\theta
}\Lambda_{\theta})f(g_{t}h_{\theta}\Lambda_{\theta})dt
\end{align*}
and with (\ref{nbvisitingSK1}) and (\ref{L1}), this gives%
\[
\sum_{s\in I(T,\Lambda_{\theta},S\setminus K)}f(g_{s}\Lambda_{\theta}%
)\leq 3AT\varepsilon+\frac{A}{\Delta}3T\varepsilon\leq 6\frac{A}{\Delta
}T\varepsilon
\]
for all $T\geq T(\Lambda_{\theta},\varepsilon)$.

We want to bound above
\[
\Sigma_{1}(T)=\frac{1}{\operatorname*{card}I(T,\Lambda_{\theta},S)}\sum_{t\in
I(T,\Lambda_{\theta},S)}\varphi(g_{t}\Lambda_{\theta})
\]
with
\[
\Sigma_{2}(T)=\frac{1}{\operatorname*{card}I(T,h_{\theta}\Lambda_{\theta},S)}%
\sum_{t\in I(T,h_{\theta}\Lambda_{\theta},S)}\varphi(g_{t}h_{\theta}%
\Lambda_{\theta})
\]
because this last sum tends to $\frac{1}{\mu_{S}(S)}\int_{S}\varphi d\mu_{S}$
when $T$ goes to infinity. We split $\sum_{t\in I(T,\Lambda_{\theta},S)}$ in
two sums $\sum_{t\in I(T,\Lambda_{\theta},K)}$ and $\sum_{t\in I(T,\Lambda
_{\theta},S\setminus K)}$. As in the previous proof for $T$ large enough, we
have%
\[
\left\vert I(T,h_{\theta}\Lambda_{\theta},S)-I(T,\Lambda_{\theta
},S)\right\vert \leq 9AT\varepsilon+2,
\]%
\[
I(T,h_{\theta}\Lambda_{\theta},S)\geq aT,
\]
and%
\[
\left\vert \sum_{t\in I(T,\Lambda_{\theta},K)}\varphi(g_{t}\Lambda_{\theta
})-\sum_{t\in I(T,\Lambda_{\theta},K)}\varphi(g_{t-t^{\prime}}h_{\theta
}\Lambda_{\theta})\right\vert \leq\varepsilon\operatorname{card}%
I(T,\Lambda_{\theta},K)
\]
where $t^{\prime}=\tau(g_{t}h_{\theta}g_{-t},g_{t}\Lambda_{\theta})$ is
defined in Lemma \ref{lem:K-S}. Taking into account of the first element $t_{min}$ and of the last element in $I(T,\Lambda_{\delta},K)$, the latter inequality implies that%
\begin{align*}
\sum_{t\in I(T,\Lambda_{\theta},K)}\varphi(g_{t}\Lambda_{\theta})  &  \leq \varphi(g_{t_{min}}\Lambda_{\theta})+
\sum_{t\in I(T,\Lambda_{\theta},K)\setminus\{t_{min}\}}\varphi(g_{t-t^{\prime}}h_{\theta}%
\Lambda_{\theta})+\varepsilon\operatorname{card}I(T,\Lambda_{\theta},K)\\
&  \leq\varphi(R\Lambda_{\theta})+\sum_{t\in I(T+1,h_{\theta}\Lambda_{\theta},S)}\varphi(g_{t}h_{\theta
}\Lambda_{\theta})+\varepsilon\operatorname{card}I(T,\Lambda_{\theta},K).
\end{align*}
All together, we obtain (recall that $\varphi\geq 0$)%
\begin{align*}
\Sigma_{1}(T)  &  \leq
\frac{1}{T}\varphi(R\Lambda_{\theta})+
\frac{\operatorname{card}I(T+1,h_{\theta}\Lambda_{\theta}%
,S)}{\operatorname*{card}I(T,\Lambda_{\theta},S)}\Sigma_{2}(T+1)+\varepsilon
+\frac{1}{\operatorname*{card}I(T,\Lambda_{\theta},S)}\sum_{s\in
I(T,\Lambda_{\theta},S\setminus K)}f(g_{s}\Lambda_{\theta})\\
&  \leq\frac{1}{T}\varphi(R\Lambda_{\theta})+
\left(  1+\left\vert \frac{\operatorname*{card}I(T+1,h_{\theta}%
\Lambda_{\theta},S)-\operatorname*{card}I(T,\Lambda_{\theta},S)}%
{\operatorname*{card}I(T,\Lambda_{\theta},S)}\right\vert \right)  \Sigma
_{2}(T+1)+\varepsilon\\
&  +\frac{1}{\operatorname*{card}I(T,\Lambda_{\theta},S)}6\frac{A}{\Delta
}T\varepsilon\\
&  \leq
\Sigma_{2}(T+1)+\frac{1}{T}\varphi(R\Lambda_{\theta})+\frac{9AT\varepsilon+2+A}{aT-9AT\varepsilon-2}\Sigma_{2}(T+1)+\left(1+\frac{6A}%
{(aT-9AT\varepsilon-2)\Delta}\right)\varepsilon
\end{align*}
and we are done. $\square\bigskip$

\subsection{Proofs of Lemma \ref{lem:boundary}}

We need an auxiliary Lemma.

\begin{lemma}
Let $E(\lambda,\eta)$ be the set of lattices $\Lambda$ in $\mathcal{L}_{d,c}$ such
that there exist two nonzero vectors $X\neq\pm X^{\prime}$ of $\Lambda$ in the
open ball $B_{\mathbb{R}^{d+c}}(0,\lambda)\ $with $\frac{1}{1+\eta}%
<\frac{\left\vert X\right\vert _{\mathbb{\pm}}}{\left\vert X^{\prime
}\right\vert _{\mathbb{\pm}}}<1+\eta$ or a nonzero vector $X$ in the open ball
$B_{\mathbb{R}^{d+c}}(0,\lambda)$ with $\left\vert X\right\vert _{\pm}<\eta$.
For all $\lambda>0$, we have $\lim_{\eta\rightarrow 0}\mu(E(\lambda,\eta))=0$.
\end{lemma}

\begin{proof}
Since $\lim_{\rho\rightarrow 0}\mu(\{\lambda_{1}(\Lambda)\leq\rho\})=0$, it is
enough to show that for all $\rho>0$, $\mu(E(\lambda,\eta)\cap\{\lambda
_{1}(\Lambda)\geq\rho\})\rightarrow 0$ when $\eta$ goes to $0$.
Choose a Siegel reduction domain $\mathcal{S}\subset \SL(d+1,\mathbb{R)}$. There is a
constant $c=c(\mathcal{S})>0$ such that for all matrices $M$ in $\mathcal{S}$
and all vectors $Y$ in $\mathbb{R}^{d+c}$, we have%
\[
\left\Vert MY\right\Vert _{\mathbb{R}^{d+c}}\geq c\lambda_{1}(\Lambda
)\left\Vert Y\right\Vert _{\mathbb{R}^{d+c}}.
\]
where $\Lambda=M\mathbb{Z}^{d+c}$ (this inequality holds for all norms with a
constant $c$ depending only on the norm, just use the norm equivalence). It
follows that we can find a finite subset $F_{\rho}$ of $\mathbb{Z}^{d+c}$ such
that for all matrices $M$ in $\mathcal{S}$ with $\lambda_{1}(M\mathbb{Z}%
^{d+c})\geq\rho$, the only $Y$ in $\mathbb{Z}^{d+c}$ such that $\left\Vert
MY\right\Vert _{\mathbb{R}^{d+c}}\leq\lambda$, are in $F_{\rho}$. Therefore,
if a matrix $M$ in $\mathcal{S}$ is such that $\Lambda=M\mathbb{Z}^{d+c}$
belongs to $E(\lambda,\eta)\cap\{\lambda_{1}(\Lambda)\geq\rho\}$ then there
exist a nonzero $Y$ in $F_{\rho}$ or two nonzero vectors $Y\neq\pm Y^{\prime}$ in
$F_{\rho}$ with%
\[
\left\vert MY\right\vert _{\pm}\leq\eta
\]
or
\[
\frac{1}{1+\eta}<\frac{\left\vert MY\right\vert _{\mathbb{\pm}}}{\left\vert
MY^{\prime}\right\vert _{\mathbb{\pm}}}<1+\eta.
\]
For a fixed $Y$ or a fixed pair $Y\neq\pm Y^{\prime}$ of nonzero vectors in
$F_{\rho}$, the measure of the set of $M$ in $\mathcal{S}$ for which the above
inequality holds, goes to $0$ as $\eta$ goes to $0$. Since $F_{\rho}$ is
finite and since a Siegel domain contains a fundamental domain we are done.
\end{proof}


\begin{proof}
[Proof of Lemma \ref{lem:boundary} ]Consider the set $V(\lambda,\eta
,\rho)=E(\lambda,\eta)\cup\{\lambda_{1}(\Lambda)<\rho\}$.

\textbf{Step 1: }\textit{ The complementary $V^{C}$ of $V(\lambda,\eta,\rho)$ is a
closed subset of $\mathcal{L}_{d,c}$.}
\newline Let $(\Lambda_{n})_{n\in\mathbb{N}}$
be a sequence of points of $V^{C}$ which converge to $\Gamma$ in $\mathcal{L}_{d,c}%
$. First, since $\lambda_{1}$ is continuous, $\lambda_{1}(\Gamma)=\lim
\lambda_{1}(\Lambda_{n})\geq\rho$. 
There is a sequence of matrices $(M_{n})_{n\in\mathbb{N}}$ such that $\Lambda
_{n}=M_{n}\mathbb{Z}^{d+c}$ for all $n\in\mathbb{N}$ and such that
$(M_{n})_{n\in\mathbb{N}}$ converges to $M$ with $\Gamma=M\mathbb{Z}^{d+c}$.
We have to show
that $\Gamma$ is not in $E(\lambda,\eta)$.
Let $X=MY$ and $X^{\prime}=MY^{\prime}$ be two nonzero vectors in $\Gamma$
with $X\neq\pm X^{\prime}$ and $\left\Vert X\right\Vert _{\mathbb{R}^{d+c}}%
$,\ $\left\Vert X^{\prime}\right\Vert _{\mathbb{R}^{d+c}}<\lambda$. When $n$
is large enough, $X_{n}=M_{n}Y$ and $X_{n}^{\prime}=M_{n}Y^{\prime}$ are in
the open ball $B(0,\lambda)$ and since $\Lambda_{n}=M_{n}\mathbb{Z}^{d+c}$ is
not in $E(\lambda,\eta)$ we have both%
\[
\left\vert M_{n}Y\right\vert _{\pm}\geq\eta
\]
and
\[
\frac{\left\vert M_{n}Y\right\vert _{\pm}}{\left\vert M_{n}Y^{\prime
}\right\vert _{\pm}}\notin]\frac{1}{1+\eta},1+\eta[
\]
and passing through the limit we obtain%
\[
\frac{\left\vert X\right\vert _{\pm}}{\left\vert X^{\prime}\right\vert _{\pm}%
}=\frac{\left\vert MY\right\vert _{\pm}}{\left\vert MY^{\prime}\right\vert
_{\pm}}\notin]\frac{1}{1+\eta},1+\eta[
\]
and
\[
\left\vert X\right\vert _{\pm}\geq\eta.
\]
Therefore $\Gamma$ is not in $E(\lambda,\eta)$.

\textbf{Step 2: }\textit{$F=S\setminus V(\lambda,\eta,\rho)$ is a compact subset of
$S$ when $\lambda\geq 2\max\{\lambda_{1}(\Lambda):\Lambda\in
\mathcal{L}_{d,c}\}$.}
\newline Thanks to Malher compactness Theorem it is enough to prove that
$F$ is a closed subset of $\mathcal{L}_{d,c}$. Let $(\Lambda_{n})_{n\in\mathbb{N}}$
be a sequence of points of $F$ which converges to $\Gamma$ in $\mathcal{L}_{d,c}$. We
want to prove that $\Gamma$ is in $F$. By the first step it is enough to prove
that $\Gamma$ is in $S$. Choose a Siegel domain $\mathcal{S}$. There is a
sequence of matrices $M_{n}\in\mathcal{S}$ such that $\Lambda_{n}%
=M_{n}\mathbb{Z}^{d+c}$ for all $n\in\mathbb{N}$ and such that $(M_{n}%
)_{n\in\mathbb{N}}$ converges to $M\in\mathcal{S}$. For each $n$, there are
two vectors $Y_{1,n}$ and $Y_{2,n}$ in $\mathbb{Z}^{d+c}$ such that
$X_{1,n}=M_{n}Y_{1,n}$ and $X_{2,n}=M_{n}Y_{2,n}$ are the two vectors
associated with $\Lambda_{n}$ by the definition of $S$. Since the matrices
$M_{n}$ are all in the Siegel domain $\mathcal{S}$ and that $\left\Vert
M_{n}Y_{i,n}\right\Vert _{\mathbb{R}^{d+c}}=\lambda_{1}(\Lambda_{n})$,
$i=1,2$, the sequences $(Y_{i,n})_{n\in\mathbb{N}}$, $i=1,2$, are bounded
sequence in $\mathbb{Z}^{d+c}$. Therefore extracting subsequences, we can
suppose that the two sequences $(Y_{i,n})_{n\in\mathbb{N}}$ are constant:
$Y_{i,n}=Y_{i}$ for all $n$, $i=1,2$. It follows that $\left\Vert
MY_{i}\right\Vert _{\mathbb{R}^{d+c}}=\lim_{n\rightarrow\infty}\left\Vert
M_{n}Y_{i}\right\Vert _{\mathbb{R}^{d+c}}=\lim_{n\rightarrow\infty}\lambda
_{1}(\Lambda_{n})=\lambda_{1}(\Gamma)$. Moreover
\[
\left\vert MY_{1}\right\vert _{+}=\lim_{n\rightarrow\infty}\left\vert
M_{n}Y_{1}\right\vert _{+}=\lim_{n\rightarrow\infty}\lambda_{1}(\Lambda
_{n})=\lambda_{1}(\Gamma)
\]
and%
\[
\left\vert MY_{2}\right\vert _{-}=\lim_{n\rightarrow\infty}\left\vert
M_{n}Y_{2}\right\vert _{-}=\lim_{n\rightarrow\infty}\lambda_{1}(\Lambda
_{n})=\lambda_{1}(\Gamma).
\]
Suppose now that $\lambda\geq 2\max\{\lambda_{1}(\Lambda):\Lambda\in
\mathcal{L}_{d,c}\}$, then making use of the first step we conclude that $\Gamma$ is
in $S$.

\textbf{Step 3. }For a neighborhood $W$ of $I_{d+c}$ in $\SL(d+c,\mathbb{R)}$,
set
\[
U_{W}=\{g_{t}h\Lambda:t\in[0,1],\ h\in W,\ \Lambda\in V(\lambda
,\eta,\rho)\}.
\]
Let us show that we can choose $W$ in order that $U_{W}\subset V(2e^{d+c}%
\lambda,5e^{d+c}\eta,2e^{d+c}\rho)$. It will finish the proof of Lemma \ref{lem:boundary}.
Indeed, we first fix $\lambda\geq 2\max\{\lambda_{1}(\Lambda):\Lambda
\in\mathcal{L}_{d,c}\}$, next we take $\eta$ and $\rho$ such that $\mu
(E(2e^{d+c}\lambda,5e^{d+c}\eta))\leq\frac{\varepsilon}{2}$ and $\mu
(\{\lambda_{1}<2e^{d+c}\rho\})\leq\frac{\varepsilon}{2}$, then we take $W$
such that $U_{W}\subset V(2e^{d+c}\lambda,5e^{d+c}\eta,2e^{d+c}\rho)$ and
$\delta$ such that $B(I_{d+c},\delta)\subset W$. Now by the second step
$K=S\setminus V(\lambda,\eta,\rho)$ is compact and since $U(K,\delta)\subset
U_{W}\subset V(2e^{d+c}\lambda,5e^{d+c}\eta,2e^{d+c}\rho)$, we have
$\mu(U(K,\delta))\leq\varepsilon$.

Let $\Lambda$ be in $V(\lambda,\eta,\rho)$, $h$ in $W$ and $t\in[0,1]$.
We explain how to successively reduce $W$ in order to obtain the above inclusion.

\textbf{Case 1. }Suppose $\lambda_{1}(\Lambda)<\rho$. We can choose $W$ small
enough in order that $\left\Vert hX\right\Vert _{\mathbb{R}^{d+c}}%
\leq 2\left\Vert X\right\Vert _{\mathbb{R}^{d+c}}$ for all $h$ in $W$ and all
$X$ in $\mathbb{R}^{d+c}$. This implies that $\lambda_{1}(g_{t}h\Lambda
)\leq 2e^{d+c}\lambda_{1}(\Lambda)<2e^{d+c}\rho$, hence $g_{t}h\Lambda\in
V(2e^{d+c}\lambda,5e^{d+c}\eta,2e^{d+c}\rho)$.

\textbf{Case 2. }Suppose there exist a nonzero vector $X$ in $\Lambda\cap
B(0,\lambda)$ with $\left\vert X\right\vert _{-}<\eta$ (the case $\left\vert
X\right\vert _{+}<\eta$ is easier). Call $p_{\pm}$ the projections on the
subspaces $E_{\pm}$ and $\left\Vert u\right\Vert $ the norm of the linear
operator $u$ associated with the norm $\left\Vert .\right\Vert _{\mathbb{R}%
^{d+c}}$. The vector $g_{t}hX$ is in the open ball $B(0,2\lambda e^{d+c})$ and
we have%
\[
p_{-}g_{t}hX=g_{t}p_{-}hp_{-}X+g_{t}p_{-}hp_{+}X,
\]
hence%
\[
\left\vert g_{t}hX\right\vert _{-}<e^{d+c}\left\Vert p_{-}h\right\Vert
\eta+e^{d+c}\left\Vert p_{-}hp_{+}\right\Vert \lambda.
\]
We can choose $W$ in order that $\left\Vert p_{-}hp_{+}\right\Vert <\frac
{\eta}{\lambda}$ and $\left\Vert p_{-}h\right\Vert \leq 1$. Then we have $\left\vert g_{t}hX\right\vert _{-}%
<3e^{d+c}\eta$ which implies that $g_{t}h\Lambda\in V(2e^{d+c}\lambda
,5e^{d+c}\eta,2e^{d+c}\rho)$.

\textbf{Case 3. }Suppose there exists two distinct nonzero vectors $X$ and
$X^{\prime}$ in $\Lambda\cap B(0,\lambda)$ such that%
\[
\left\vert  X\right\vert  _{-},\left\vert  X^{\prime}\right\vert  _{-}\geq\eta\, \text{ and }\, \frac
{1}{1+\eta}<\frac{\left\vert  X\right\vert  _{-}}{\left\vert  X^{\prime}\right\vert  _{-}%
}<1+\eta.
\]
The case with $\left\vert  .\right\vert  _{+}$ is similar. \newline As above,%
\begin{align*}
\left\vert  hX\right\vert  _{-}  &  <\left\|  p_{-}h\right\|  \left\vert  X\right\vert
_{-}+\left\|  p_{-}hp_{+}\right\|  \lambda\\
&  \leq(\left\|  p_{-}h\right\|  +\left\| p_{-}hp_{+}\right\|  \frac{\lambda
}{\eta})\left\vert  X\right\vert  _{-}.
\end{align*}
We can choose $W$ in order that $\left\|  p_{-}h\right\|  +\left\|
p_{-}hp_{+}\right\|  \frac{\lambda}{\eta}\leq 1+$ $\eta$. We  also have%
\[
\left\vert  hX^{\prime}\right\vert  _{-}\geq\left\|  p_{-}hp_{-}X^{\prime}\right\|
-\left\|  p_{-}hp_{+}\right\|  \left\|  X^{\prime}\right\|  .
\]
We can choose $W$ in order that $\left\|  p_{-}hp_{+}\right\|  \frac{\lambda
}{\eta}\leq\eta$ and 
$\left\Vert p_{-}-p_{-}hp_{-}\right\Vert\leq \eta$. With this choice, we have%
\[
\left\vert  hX^{\prime}\right\vert  _{-}\geq\left\vert  X^{\prime}\right\vert  _{-}%
-\eta\left\vert  X^{\prime}\right\vert  _{-}-\eta\left\vert  X^{\prime}\right\vert  _{-}.
\]
It follows that
\[
\frac{\left\vert  g_{t}hX\right\vert  _{-}}{\left\vert  g_{t}hX^{\prime}\right\vert  _{-}%
}=\frac{\left\vert  hX\right\vert  _{-}}{\left\vert  hX^{\prime}\right\vert  _{-}}%
\leq\frac{\left\vert  X\right\vert  _{-}}{\left\vert  X^{\prime}\right\vert  _{-}}%
\times\frac{1+\eta}{1-2\eta}\leq\frac{(1+\eta)^{2}}{1-2\eta}\leq 1+5\eta
\]
when $\eta$ is small enough. Inverting the role of $X$ and $X^{\prime}$ we get
the inequality $\frac{\left\vert  hX^{\prime}\right\vert  _{-}}{\left\vert  hX\right\vert
_{-}}\leq 1+5\eta$ and we are done.
\end{proof}


\medskip


\subsection{Proofs of Theorems \ref{thm:levy-d} and \ref{thm:jager}.1.}

We begin by the proof of Theorem \ref{thm:levy-d} which is more difficult.
We want to prove that for almost all $\theta$ in $\M_{d,c}(\mathbb{R})$,
\[
\lim_{n\rightarrow\infty}\frac{1}{n}\ln q_{n}(\theta)=L_{d,c}=\frac{1}{\mu
_{S}(S)}\int_{S}\rho(\Lambda)~d\mu_{S}(\Lambda)
\]
and that
\[
\lim_{n\rightarrow\infty}\frac{-1}{n}\ln r_{n}(\theta)=\frac{c}{d}L_{d,c}.
\]
By Khintchin-Groshev Theorem, the convergence almost everywhere of $\frac
{1}{n}\ln q_{n}(\theta)$ to $L_{d,c}=\frac{1}{\mu_{S}(S)}\int_{S}\rho
(\Lambda)~d\mu_{S}(\Lambda)$, implies the convergence almost everywhere of
$\frac{-1}{n}\ln r_{n}(\theta)$ to $\frac{c}{d}L_{d,c}$. Therefore the proof
of Theorem \ref{thm:levy-d} reduces in the first almost everywhere limit. 

As soon as Theorem \ref{thm:levy-d} is proven, 
the formula in the introduction
\[
L_{d,c}=\frac{d}{\mu_{S}(S)}\int_{S}\tau~d\mu_{S}=\frac{d\times\mu
(\mathcal{L}_{d+c})}{\mu_{S}(S)}%
\]
is a consequence of Proposition \ref{prop:borel-cantelli} and Lemma
\ref{lem:return-rho}. Indeed, by Proposition \ref{prop:borel-cantelli} and by
Lemma \ref{lem:return-rho}, $\frac{c}{d}L_{d,c}=L_{d,c}^*=\frac{1}{\mu_{S}(S)}\int
_{S}\rho^{\ast}(\Lambda)~d\mu_{S}(\Lambda)$ and $\tau(\Lambda)=\frac{1}%
{d+c}(\rho(R(\Lambda))+\rho^{\ast}(\Lambda))$ for $\Lambda$ in $S\setminus
\mathcal{N}$, hence%
\begin{align*}
L_{d,c}  &  =\frac{d}{d+c}\times\frac{1}{\mu_{S}(S)}\int_{S}(\rho
(R(\Lambda))+\rho^{\ast}(\Lambda))d\mu_{S}(\Lambda)\\
&  =\frac{d}{\mu_{S}(S)}\int_{S}\tau(\Lambda)d\mu_{S}(\Lambda)\\
&  =\frac{d~\mu(\mathcal{L}_{d+c})}{\mu_{S}(S)}.
\end{align*}


Let us now prove the first almost everywhere limit. We need two Lemmas. The
first one is clear.

\begin{lemma}
For all compact set $K$ in $\mathbb{R}^{d+c}$ and all $\varepsilon>0$ there
exists $\alpha>0$ such that for all $g\in B(I_{d+c},\alpha)$ and all $x\in K$,
$\dd(gx,x)\leq\varepsilon$.
\end{lemma}

\begin{lemma}
Let $\Lambda$ be in $S^{\prime}\setminus\mathcal{N}$. Then return map $R=R_{S}$ is defined on neighborhood of $\Lambda$ and is continuous at $\Lambda$.
\end{lemma}

\begin{proof}
 Consider the minimal
vectors $X_{0}=X_{0}(\Lambda)$ and $X_{1}=X_{1}(\Lambda)$. By definition of
$S^{\prime}$ the only nonzero vector $B(0,\lambda_{1}(\Lambda))=\mathcal{C}%
(X_{0})$ are $\pm X_{0}$. Therefore there exists $\varepsilon>0$ such that all
$X$ in $\Lambda\setminus\{0,\pm X_{0}\}$ are at a distance $\geq\varepsilon$
from $\mathcal{C}(X_{0})$. Since $\Lambda$ is not in $\mathcal{N}$, $\pm
X_{0}$ and $\pm X_{1}$ are the only nonzero vector of $\Lambda$ in the
cylinder $\mathcal{C}(X_{0},X_{1})$. Therefore reducing $\varepsilon$ if
necessary, all $X$ in $\Lambda\setminus\{0,\pm X_{0},\pm X_{1}\}$ are at a
distance $\geq\varepsilon$ from $\mathcal{C}(X_{0},X_{1})$. By the above Lemma
we can choose $\delta>0$ such that
$\forall g\in B(I_{d+c},\delta),\ \forall X\in\mathcal{C}(X_{0},X_{1})+B(0,1), $
\[
 \max(\dd(g^{-1}X,X),\dd(gX,X))\leq\varepsilon/3\text{. }%
\]
It follows that for all $g\in B(I_{d+c},\delta)$, $\pm gX_{0}$ are the only
nonzero vector of $g\Lambda$ in $\mathcal{C}(gX_{0})$ and that $\pm gX_{0}$
and $\pm gX_{1}$ are the only nonzero vectors of $\Lambda$ in $\mathcal{C}%
(gX_{0},gX_{1})$. It follows that if the lattice $\Gamma=g\Lambda$ is in the set of lattices $B(I_{d+c}%
,\delta)\Lambda\cap S^{\prime}$ then $X_{0}(\Gamma)=gX_{0}$ and $X_{1}%
(\Gamma)=gX_{1}$. By definition of $S$ the return times are from $\Lambda$ and $\Gamma$ are well defined we have
\begin{align*}
\tau(\Lambda)  &  =\frac{1}{d+1}\ln\frac{\left|  X_{1}\right|  _{-}}{\left|
X_{0}\right|  _{+}},\\
\tau(\Gamma)  &  =\frac{1}{d+1}\ln\frac{\left|  gX_{1}\right|  _{-}}{\left|
gX_{0}\right|  _{+}},
\end{align*}
hence $R(\Lambda)$ is defined and
\begin{align*}
\left|  \tau(\Lambda)-\tau(\Gamma)\right|   &  =\frac{1}{d+1}\left|  \ln
\frac{\left|  X_{1}\right|  _{-}}{\left|  X_{0}\right|  _{+}}\frac{\left|
gX_{0}\right|  _{+}}{\left|  gX_{1}\right|  _{-}}\right| \\
&  \leq\frac{1}{d+1}(\left|  \ln\frac{\left|  X_{1}\right|  _{-}}{\left|
gX_{1}\right|  _{-}}\right|  +\left|  \ln\frac{\left|  gX_{0}\right|  _{+}%
}{\left|  X_{0}\right|  _{+}}\right|  ).
\end{align*}
which goes to zero $\delta$ goes to zero.
\end{proof}


\begin{proof}[End of proof of Theorem \ref{thm:levy-d}]
We use Theorem \ref{thm:reduction} with $S^{\prime}$ and the function
$\varphi:S^{\prime}\rightarrow\mathbb{R}_{\geq 0}$ defined by $\varphi
(\Lambda)=\rho\circ R(\Lambda)=\ln\frac{q_{1}(\Lambda)}{q_{0}(\Lambda)}$ when $R(\Lambda)$ is defined and by $\varphi(\Lambda)=0$ otherwise.
Since $\rho$ is continuous on $S$ and $R$ is continuous on $S^{\prime
}\setminus\mathcal{N}$, $\varphi$ is almost everywhere continuous on
$S^{\prime}$. We need to find a uniformly continuous function $f:\mathcal{L}%
\rightarrow\mathbb{R}$ such that $\left\vert \varphi\right\vert \leq f$.
Observe that $\varphi$ is nonnegative. By Minkowski convex body Theorem, for
all lattice $\Lambda\in\mathcal{L}$.
\[
q_{1}(\Lambda)^{c}r_{0}(\Lambda)^{d}\leq C=C_{d,c}
\]
where $C_{d,c}$ depends only on $c$ and $d$. It follows that for all $\Lambda$ in
$S^{\prime}$ we have%
\begin{align*}
\varphi(\Lambda) &  =\ln\frac{q_{1}(\Lambda)}{q_{0}(\Lambda)}\\
&  =\ln\frac{q_{1}(\Lambda)r_{0}(\Lambda)^{d/c}}{q_{0}(\Lambda)r_{0}%
(\Lambda)^{d/c}}\\
&  \leq\ln C^{d/c}-\ln q_{0}(\Lambda)r_{0}(\Lambda)^{d/c}.
\end{align*}
For $\Lambda$ is in $S^{\prime}$ we have $q_{0}(\Lambda)=r_{0}(\Lambda)=\lambda_1(\Lambda)$.
Therefore%
\[
\varphi(\Lambda)\leq\ln C^{d/c}-\frac{d+c}{c}\ln\lambda_{1}(\Lambda).
\]
It is well known that the function $\ln\lambda_{1}$ is uniformly continuous
and integrable on $\mathcal{L}_{c+d}$, consequently we can use Theorem
\ref{thm:reduction} with $S^{\prime}$ and $\varphi$. It follows that for
almost all $\theta$ in $\M_{d,c}(\mathbb{R})$ we have%
\[
\lim_{n\rightarrow\infty}\frac{1}{n}\sum_{k=0}^{n-1}\varphi\circ R_{S^{\prime}%
}^{k}(\Lambda_{\theta})=\frac{1}{\mu_{S^{\prime}}(S^{\prime})}\int_{S^{\prime}}%
\varphi~d\mu_{S^{\prime}}%
\]
where $R_{S^{\prime}}$ is the first return map on $S^{\prime}$. Now by Lemmas \ref{lem:min-best} and \ref{lem:return-best}, for almost
all $\theta$, there is an integer $k_0$ such that
\[
\varphi\circ R_{S^{\prime}}^{k}(\Lambda_{\theta})=\ln\frac{q_{k+k_0+1}(\theta)}{q_{k+k_0}(\theta
)}
\]
for all large enough $k$, where $k_0$ depend only on $\theta$. It follows that for almost all $\theta$ 
\[
\lim_{n\rightarrow\infty}\frac{1}{n}\sum_{k=0}^{n-1}\varphi\circ R_{S^{\prime}%
}^{k}(\Lambda_{\theta})=
\lim_{n\rightarrow\infty}\frac{1}{n}\sum_{k=0}^{n-1}\ln\frac{q_{k+k_0+1}(\theta)}{q_{k+k_0}(\theta
)}=
\lim_{n\rightarrow\infty}\frac{1}{n}\ln q_{n}(\theta)
\]
So that the only thing left is the equality
\[
\int_{S}\rho~d\mu_{S}=\int_{S^{\prime}}\varphi~d\mu_{S^{\prime}}.
\]
Now, the image of $\mu_{S^{\prime}}$ by $R$ is $\mu_{S}$, hence
\begin{align*}
\int_{S^{\prime}}\varphi~d\mu_{S^{\prime}}  & =\int_{S^{\prime}}\rho\circ
R\ d\mu_{S^{\prime}}\\
& =\int_{S}\rho\ d\mu_{S}.
\end{align*}
\end{proof}



\begin{proof}[Proof of Theorem \ref{thm:jager}.1]
Consider the map $F:S\rightarrow \mathbb R$ defined by 
\[
F(\Lambda)=q_1^c(\Lambda)r_0^d(\Lambda)=|v_1^S(\Lambda)|_-^c|v_0^S(\Lambda)|^d_+
\]
and call $\nu=\nu_{d,c}$ the image of the measure $\frac{1}{\mu_S(S)}\mu_S$ by $F$.
Let $\varphi:\mathbb R\rightarrow \mathbb R$ be a continuous and bounded function. We want to prove that 
\[
\lim_{n\rightarrow\infty}\frac{1}{n}\sum_{k=0}^{n-1}\varphi(\beta_k(\theta))
=\int_{\mathbb R} \varphi(x) \ d\nu(x)
\] 
for almost all $\theta \in \M_{d,c}(\mathbb R)$. 

Now by Lemmas  \ref{lem:min-best} and \ref{lem:return-best}, for almost all $\theta$, 
\[
F(R^k(\Lambda_{\theta}))=q_{k+k_0+1}^c(\theta)r_{k+k_0}^d(\theta)=\beta_{k+k_0}(\theta)
\]
for all $k$ large enough. Now the function $\varphi\circ F$ is bounded and continuous, thus by Theorem \ref{thm:reduction} (or Proposition \ref{prop:reduction}) we have for almost all $\theta$,
\begin{align*}
\lim_{n\rightarrow\infty}\frac{1}{n}\sum_{k=0}^{n-1}\varphi\circ F(R^k(\Lambda_\theta))
&=\frac{1}{\mu_S(S)}\int_{S}\varphi\circ F(\Lambda)\, d\mu_S(\Lambda)\\
&=\int_{\mathbb R} \varphi(x) \, d\nu(x)
\end{align*}
which  implies that
\[
\lim_{n\rightarrow\infty}\frac{1}{n}\sum_{k=0}^{n-1}\varphi(\beta_k(\theta))
=\lim_{n\rightarrow\infty}\frac{1}{n}\sum_{k=0}^{n-1}\varphi\circ F(R^k(\Lambda_\theta))
=\int_{\mathbb R} \varphi(x) \ d\nu(x)
\]
and finishes the proof of Theorem \ref{thm:jager}.1.
The proof of Theorem \ref{thm:jager}.2. is postponed  at the end of section 9. 
\end{proof}

\section{On $\lim\inf q_{n+k}^{c}r_{n}^{d}$ \label{sec:qnrn}}

For each $\theta$ in $\M_{d,c}(\mathbb{R)}$, we consider the sequence of best
approximation denominators $(Q_{n}(\theta))_{n\in\mathbb{N}}$,  their norms
$q_{n}=\left\Vert Q_{n}(\theta)\right\Vert _{\mathbb{R}^{c}}$, and the sequence
$(r_{n})_{n\geq 0}$ defined by%
\[
r_{n}=d_{\mathbb{R}^{d}}(\theta Q_{n},\mathbb{Z}^{d}).
\]
For a nonnegative integer $k$, call $Bad_{k}$ the subset of $\M_{d,c}(\mathbb{R})$
defined by
\[
Bad_{k}(d,c)=Bad_{k}=\{\theta\in\M_{d,c}(\mathbb{R})\setminus\M_{d,c}(\mathbb{Q}):\inf_{n\in\mathbb N} q_{n+k}^{c}%
r_{n}^{d}>0\}
\]
(if $r_n=0$ for some integer $n$, $\theta$ is not in $Bad_k$). The sequence of sets $(Bad_{k})_{k\geq 0}$ is clearly nondecreasing and the set
$Bad_{0}$ is the usual set of badly approximable matrices. When $d=c=1$, the
classical inequality $q_{n+1}r_{n}\geq\frac{1}{2}$ shows that $Bad_{1}%
=\mathbb{R}\setminus\mathbb{Q}$ while in \cite{Chev-Khin} it has been shown
that for $c=1$ and $d\geq 2$, $Bad_{1}$ is negligible. Our first goal is to
show that $Bad_{1}\setminus Bad_{0}$ is nonempty for $c=1$ and $d=2$. Next we
will prove that the set%
\[
\mathcal{B}(d,c)=\mathcal{B}=\cup_{k\geq 0}Bad_{k}%
\]
is negligible and does not depend on the choice of the norm.

\begin{proposition}
If $c=1$ and $d=2$ then $Bad_{1}\setminus Bad_{0}$ contains uncountably many elements.
\end{proposition}

\begin{remark}The set $\mathbb{Z\theta+Z}^{2}$ is everywhere dense in
$\mathbb{R}^{2}$ for all in $\theta\in Bad_{1}$. Indeed it is known that the
first minimum of the lattice%
\[
\Lambda_{n}=\mathbb{Z}^{2}+\mathbb{Z}\frac{p_{n}}{q_{n}}%
\]
is $\asymp r_{n-1}$ where $p_{n}$ is an integer vector such that $r_{n}%
=\dd(q_{n}\theta,\mathbb{Z}^{2})$. This implies that the second minimum of this
lattice is $\lambda_{2}(\Lambda_{n})\asymp\frac{1}{q_{n}r_{n-1}}$. Now a lower
bound $q_{n}r_{n-1}^{2}\geq\alpha>0$ implies that $\frac{1}{q_{n}r_{n-1}}%
\leq\frac{r_{n-1}}{\alpha}$ which goes to zero when $n\rightarrow\infty$. The
convergence to zero of $\lambda_{2}(\Lambda_{n})$ implies that
$\mathbb{Z\theta+Z}^{2}$ is everywhere dense in $\mathbb{R}^{2}$ (see
\cite{Chev-Mosc} or \cite{Cheu-Chev}).
\end{remark}

\begin{proof}
We assume that $\mathbb{R}^{2}$ is equipped with the standard Euclidean norm.
Set $\theta_{0}=(0,0)$ and $\theta_{1}=(\frac{1}{5},\frac{1}{5})$. We
construct inductively a sequence $(\theta_{n})_{n\geq 0}$ of rational vectors
in $\mathbb{R}^{2}$. For each $n$ in $\mathbb{N}$, let $\Lambda_{n}%
=\mathbb{Z}^{2}+\theta_{n}\mathbb{Z}$ be the lattice associated with
$\theta_{n}$. Observe that the least common denominator $Q_{n}$ of the coordinates of the 
rational vector $\theta_{n}$ is the inverse of the volume of the lattice
$\Lambda_{n}$,\ $\det\Lambda_{n}=\frac{1}{Q_{n}}$ (even for $n=0$). For $1\leq i\leq n$, set%
\begin{align*}
M_{i,n}  &  =\min\{\dd(q\theta_{n},\mathbb{Z}^{2})-\dd(Q_{i-1}\theta
_{n},\mathbb{Z}^{2}):Q_{i-1}<q<Q_{i}\},\\
m_{i,n}  &  =\dd(Q_{i-1}\theta_{n},\mathbb{Z}^{2})-\dd(Q_{i}\theta_{n}%
,\mathbb{Z}^{2})
\end{align*}
The sequence $(\theta_{n})_{n\geq 0}$ is constructed such that the following
properties hold for all $n\geq 1$:
\begin{enumerate}
\item $Q_{0}=1<Q_{1}=5<Q_2<\dots<Q_{n}$ are the best approximations (denominators) of $\theta_{n}$,
\item $Q_{n}>2nQ_{n-1}$ and given $\theta_{0},\theta_{1},\,\dots,\theta_{n-1}$,
there are at least two possible choices of $\theta_{n}$ leading to two
different values of $Q_{n}$ (to ensure that we construct an uncountable set),
\item for all $1\leq i\leq n-1$, $M_{i,n}>0$ (we need to avoid the situation
where $\dd(q\theta_{n},\mathbb{Z}^{2})=\dd(Q_{i-1}\theta_{n},\mathbb{Z}^{2})$ for
some $q$ between $Q_{i-1}$ and $Q_{i}$),
\item $\left\Vert \theta_{n}-\theta_{n-1}\right\Vert \leq\frac{1}{8Q_{n-1}%
}\min\{M_{i,j}:1\leq i<j\leq n-1\},$
\item $\left\Vert \theta_{n}-\theta_{n-1}\right\Vert \leq\frac{1}{8Q_{n-1}%
}\min\{m_{i,j}:1\leq i\leq j\leq n-1\},$
\item $\varepsilon_{n-1}=Q_{n-1}(\theta_{n}-\theta_{n-1})$ is a shortest
vector of $\Lambda_{n}$, i.e. $\lambda_{1}(\Lambda_{n})=\left\Vert
\varepsilon_{n-1}\right\Vert $, and $(-1)^{n-1}\varepsilon_{n-1}$ has positive
coordinates,
\item $ 2\lambda_1(\Lambda_n)\leq\lambda_{2}(\Lambda_{n})\leq 30\lambda_{1}(\Lambda_{n})$.
\end{enumerate}

Observe that, with our choices of $\theta_0$ and $\theta_1$ all these conditions holds for $n=1$ (the conditions 3 and 4 are empty for $n=1$).\\


First, let us show that the above conditions imply that the sequence
$(\theta_{n})_{n\in\mathbb{N}}$ converges to $\theta$ in $Bad_{1}\setminus
Bad_{0}$. By 2 and 4, the sequence $\left\Vert \theta_{n-1}-\theta
_{n}\right\Vert $ converges to $0$ at least at a geometric rate, hence the sequence
$(\theta_{n})_{n\geq 1}$ converge to $\theta\in\mathbb{R}^{2}$. Furthermore, by
4, for all $n\geq 2$,%
\begin{align*}
\left\Vert \theta-\theta_{n}\right\Vert  &  \leq\sum_{p\geq n+1}\left\Vert
\theta_{p}-\theta_{p-1}\right\Vert \\
&  \leq\sum_{p\geq n+1}\frac{1}{8Q_{p-1}}\min\{M_{i,j}:1\leq i<j\leq p-1\}\\
&  \leq\frac{1}{4Q_{n}}\min\{M_{i,n}:1\leq i<n\}.
\end{align*}
Using 5 instead of 4, we obtain%
\[
\left\Vert \theta-\theta_{n}\right\Vert \leq\frac{1}{4Q_{n}}\min
\{m_{i,n}:1\leq i\leq n\}
\]
as well. It follows that for\ all $1\leq i\leq n-1$ and all $Q_{i-1}<q<Q_{i}$,
we have%
\begin{align*}
\dd(q\theta,\mathbb{Z}^{2})  &  \geq \dd(q\theta_{n},\mathbb{Z}^{2})-q\left\Vert
\theta-\theta_{n}\right\Vert \\
&  \geq \dd(Q_{i-1}\theta_{n},\mathbb{Z}^{2})+M_{i,n}-q\left\Vert \theta
-\theta_{n}\right\Vert \\
&  \geq \dd(Q_{i-1}\theta,\mathbb{Z}^{2})-Q_{i-1}\left\Vert \theta-\theta
_{n}\right\Vert +M_{i,n}-q\left\Vert \theta-\theta_{n}\right\Vert \\
&  \geq \dd(Q_{i-1}\theta,\mathbb{Z}^{2})+M_{i,n}-2Q_{i}\left\Vert \theta
-\theta_{n}\right\Vert.
\end{align*}
Since $\left\Vert \theta-\theta_{n}\right\Vert \leq\frac{1}{4Q_{n}}M_{i,n}$,
$\dd(q\theta,\mathbb{Z}^{2})>\dd(Q_{i-1}\theta,\mathbb{Z}^{2})$.
For all $1\leq i\leq n-1$, we also have%
\begin{align*}
\dd(Q_{i}\theta,\mathbb{Z}^{2})  &  \leq \dd(Q_{i}\theta_{n},\mathbb{Z}^{2}%
)+Q_{i}\left\Vert \theta-\theta_{n}\right\Vert \\
&  \leq \dd(Q_{i-1}\theta_{n},\mathbb{Z}^{2})-m_{i,n}+\frac{Q_{i}}{4Q_{n}%
}m_{i,n}\\
&  \leq \dd(Q_{i-1}\theta,\mathbb{Z}^{2})+Q_{i-1}\left\Vert \theta-\theta
_{n}\right\Vert -m_{i,n}+\frac{Q_{i}}{4Q_{n}}m_{i,n}\\
&  \leq \dd(Q_{i-1}\theta,\mathbb{Z}^{2})+\frac{Q_{i-1}}{4Q_{n}}m_{i,n}%
-m_{i,n}+\frac{Q_{i}}{4Q_{n}}m_{i,n}\\
&  <\dd(Q_{i-1}\theta,\mathbb{Z}^{2}).
\end{align*}
It follows that $Q_{0},Q_{1},...,Q_{n-1}$ are the first $n$ best
approximations of $\theta$. Therefore $(Q_{n})_{n\geq 0}$ is the sequence of
best approximations of $\theta$. The standard inequality (see for instance
\cite{Chev-Mosc})%

\[
\lambda_{1}(\Lambda_{n})\asymp \dd(Q_{n-1}\theta,\mathbb{Z}^{2})
\]
together with 7 imply that $\theta\in Bad_{0}\setminus Bad_{1}$.
\newline

Let $n$ be integer $\geq 1$. Let us  explain the construction $\theta_{n+1}$ given that $\theta
_{0},...,\theta_{n}$ are already constructed. First choose a primitive point
$\alpha_{n}=k_{n}\theta_{n}+(a_{n},b_{n})$ of $\Lambda_{n}$ with $0\leq
k_{n}<Q_{n}$ and $(a_n,b_n)\in\mathbb Z^2$, in either $\mathbb{R}_{>0}^{2}$ when $n$ is even or in
$\mathbb{R}_{<0}^{2}$ when $n$ is odd. Just take $\alpha_n$
a point of $\Lambda_n$ in a square $[x,x+1[\times]0,1]$ with
minimal ordinate when $n$ is even and a point of $\Lambda_n$ in
a square $[x,x+1[\times[-1,0[$ with
maximal ordinate when $n$ is odd. Observe that $\left\Vert\alpha_n\right\Vert$ can be made arbitrarily large by choosing$|x|$ large enough. 


Call $L_{n}=\left\Vert \alpha
_{n}\right\Vert $ the length of the segment $[0,\alpha_{n}]$. The (Euclidean) distance
between two consecutive lines of the set $\mathcal{H}_{n}=\Lambda
_{n}+\mathbb{R\alpha}_{n}$ is
\[
d_{n}=\frac{\det\Lambda_{n}}{L_{n}}=\frac{1}{Q_{n}L_{n}}.
\]
We can choose $\alpha_{n}$ such that
\[
L_{n}^{2}\geq n\det\Lambda_{n},
\]
hence $\frac{L_{n}}{d_{n}}=\frac{L_{n}^{2}}{\det\Lambda_{n}}\geq n$. There
are at least two integers $p_{n}\geq 2$ such that%
\[
10\frac{L_{n}}{d_{n}}\leq p_{n}\leq 20\frac{L_{n}}{d_{n}}.
\]
Suppose $p_n$ is one of these and set
\[
\varepsilon_{n}=\frac{1}{p_{n}-\frac{k_{n}}{Q_{n}}}\alpha_{n},
\]%
\[
\theta_{n+1}=\theta_{n}+\frac{\varepsilon_{n}}{Q_{n}},
\]
and
\[
Q_{n+1}=Q_{n}p_{n}-k_{n}.
\]
Since by 1, $Q_{0}=1<Q_{1}=5<...<Q_{j}$ are the best approximations of
$\theta_{j}$, $j=1,...,n$, the real number $\min\{m_{i,j}:1\leq i\leq j\leq
n\}$ is strictly positive. Moreover,%
\[
\left\Vert \varepsilon_{n}\right\Vert \leq\frac{L_{n}}{p_{n}-1}\leq\frac
{L_{n}}{\frac{p_{n}}{2}}\leq 2\frac{L_{n}}{10\frac{L_{n}}{d_{n}}}\leq
\frac{d_{n}}{5}%
\]
and $\alpha_{n}$ can be chosen in order that $d_{n}$ is arbitrarily small,
hence we can choose $\alpha_{n}$ such that $\left\Vert \varepsilon
_{n}\right\Vert <\left\Vert \varepsilon_{n-1}\right\Vert $ and
\[
\left\Vert \theta_{n+1}-\theta_{n}\right\Vert =\frac{1}{Q_{n}}\left\Vert
\varepsilon_{n}\right\Vert \leq\frac{1}{8Q_{n}}\min\{m_{i,j}:1\leq i\leq j\leq
n\}
\]
which is condition 5. Next by 3, $\min\{M_{i,j}:1\leq i<j\leq n\}$ is strictly
positive. As above, it follows that $\alpha_{n}$ can be chosen such that%
\[
\left\Vert \theta_{n+1}-\theta_{n}\right\Vert =\frac{1}{Q_{n}}\left\Vert
\varepsilon_{n}\right\Vert \leq\frac{1}{8Q_{n}}\min\{M_{i,j}:1\leq i<j\leq
n\}
\]
which is condition 4.
Clearly $Q_{n+1}\geq Q_{n}(p_{n}-1)\geq 2nQ_{n}$.
Notice that the lattice $\Lambda_{n+1}=\mathbb{Z\theta}_{n+1}+\mathbb{Z}^{2}$
is included in $\mathcal{H}_{n}$. Next observe that%
\begin{align*}
Q_{n+1}\theta_{n+1}  &  =(Q_{n}p_{n}-k_{n})(\theta_{n}+\frac{\varepsilon_{n}%
}{Q_{n}})\\
&  =(Q_{n}p_{n}-k_{n})(\theta_{n}+\frac{\alpha_{n}}{Q_{n}(p_{n}-\frac{k_{n}%
}{Q_{n}})})\\
&  =Q_{n}p_{n}\theta_{n}-k_{n}\theta_{n}+\alpha_{n}\\
&  =p_{n}Q_{n}\theta_{n}+(a_{n},b_{n})\in\mathbb{Z}^{2}.
\end{align*}
It follows that $Q_{n+1}\det\Lambda_{n+1}=l\in\mathbb{N}$. On the other hand,
consider the one dimensional lattice $\Lambda_{n+1}\cap\mathbb{R\alpha}_{n}$.
Because $Q_{n}\theta_{n}\in\mathbb{Z}^{2}$ and $Q_{n}\theta_{n+1}=Q_{n}%
\theta_{n}+\varepsilon_{n}$, it contains $\varepsilon_{n}$ and is spanned by a
vector $v_{n}=\frac{\varepsilon_{n}}{m}$ where $m$ is an integer. Next observe that $\theta_n\in\Lambda_{n+1}+\mathbb R\alpha_n$, hence $\Lambda_{n+1}+\mathbb R\alpha_n=\mathcal H_n$. It follows that
\begin{align*}
\frac{l}{Q_{n+1}}  &  =\det\Lambda_{n+1}=\left\Vert v_{n}\right\Vert d_{n}\\
&  =\frac{\left\Vert \varepsilon_{n}\right\Vert }{m}d_{n}=\frac{Q_{n}L_{n}%
}{mQ_{n+1}}d_{n}\\
&  =\frac{1}{mQ_{n+1}},
\end{align*}
which implies $m=l=1$. Therefore $\det\Lambda_{n+1}=\frac{1}{Q_{n+1}}$ and
\[
\Lambda_{n+1}=\{0,...,Q_{n}-1\}\theta_{n+1}+\mathbb{Z\varepsilon}%
_{n}+\mathbb{Z}^{2}.
\]
Since $\left\Vert \varepsilon_{n}\right\Vert \leq\frac{d_{n}}{5}$, and
$\Lambda_{n+1}\subset\mathcal{H}_{n}$, $\varepsilon_{n}$ is the shortest
vector of $\Lambda_{n+1}$. The choice of the signs for $\alpha_n$ now implies  that condition 6 holds. Next
\begin{align*}
\lambda_{1}(\Lambda_{n+1})  &  =\left\Vert \varepsilon_{n}\right\Vert ,\\
5\left\Vert \varepsilon_{n}\right\Vert  &  \leq d_{n}\leq\lambda_{2}%
(\Lambda_{n+1})\leq d_{n}+\left\Vert \varepsilon_{n}\right\Vert .
\end{align*}
Since%
\[
\left\Vert \varepsilon_{n}\right\Vert \geq\frac{L_{n}}{p_{n}}\geq\frac{L_{n}%
}{20\frac{L_{n}}{d_{n}}}=\frac{d_{n}}{20},
\]
we obtain%
\[
5\lambda_{1}(\Lambda_{n+1})\leq\lambda_{2}(\Lambda_{n+1})\leq 21\left\Vert
\varepsilon_{n}\right\Vert \leq 30\lambda_{1}(\Lambda_{n+1})
\]
which contains condition 7.
Let us show that $Q_{0},...,Q_{n-1}$ are the first best approximations of
$\theta_{n+1}$.\newline For\ all $1\leq i\leq n-1$ and all $Q_{i-1}<q<Q_{i}$,
we have%
\begin{align*}
\dd(q\theta_{n+1},\mathbb{Z}^{2})  &  \geq \dd(q\theta_{n},\mathbb{Z}%
^{2})-q\left\Vert \theta_{n+1}-\theta_{n}\right\Vert \\
&  \geq \dd(Q_{i-1}\theta_{n},\mathbb{Z}^{2})+M_{i,n}-q\left\Vert \theta
_{n+1}-\theta_{n}\right\Vert \\
&  \geq \dd(Q_{i-1}\theta_{n+1},\mathbb{Z}^{2})-Q_{i-1}\left\Vert \theta
_{n+1}-\theta_{n}\right\Vert +M_{i,n}-q\left\Vert \theta_{n+1}-\theta
_{n}\right\Vert \\
&  \geq \dd(Q_{i-1}\theta_{n+1},\mathbb{Z}^{2})+M_{i,n}-2Q_{i}\left\Vert
\theta_{n+1}-\theta_{n}\right\Vert .
\end{align*}
Since $\left\Vert \theta_{n+1}-\theta_{n}\right\Vert \leq\frac{1}{8Q_{n}%
}M_{i,n}$, $\dd(q\theta_{n+1},\mathbb{Z}^{2})>\dd(Q_{i-1}\theta_{n+1}%
,\mathbb{Z}^{2})$ and hence $M_{i,n+1}>0$. We also have%
\begin{align*}
\dd(Q_{i}\theta_{n+1},\mathbb{Z}^{2})  &  \leq \dd(Q_{i}\theta_{n},\mathbb{Z}%
^{2})+Q_{i}\left\Vert \theta_{n+1}-\theta_{n}\right\Vert \\
&  \leq \dd(Q_{i-1}\theta_{n},\mathbb{Z}^{2})-m_{i,n}+\frac{Q_{i}}{8Q_{n}%
}m_{i,n}\\
&  \leq \dd(Q_{i-1}\theta_{n+1},\mathbb{Z}^{2})+Q_{i-1}\left\Vert \theta
_{n+1}-\theta_{n}\right\Vert -m_{i,n}+\frac{Q_{i}}{8Q_{n}}m_{i,n}\\
&  \leq \dd(Q_{i-1}\theta_{n+1},\mathbb{Z}^{2})+\frac{Q_{i-1}}{8Q_{n}}%
m_{i,n}-m_{i,n}+\frac{Q_{i}}{8Q_{n}}m_{i,n}\\
&  <\dd(Q_{i-1}\theta_{n+1},\mathbb{Z}^{2}).
\end{align*}
It follows that $Q_{0},Q_{1},...,Q_{n-1}$ are the first $n$ best
approximations of $\theta_{n+1}$. The proof will be done once we will have
explained that $Q_{n}$ and $Q_{n+1}$ are the only best approximations that
follow $Q_{n-1}$ and that $M_{n,n+1}>0$. These are the places where the sign
condition 6 plays a role. First observe that $\varepsilon_{n}$ and
$-\varepsilon_{n}$ are the only two shortest vectors of $\Lambda_{n+1}$ and
that%
\[
\varepsilon_{n}=Q_{n}\theta_{n+1}-Q_{n}\theta_{n}%
\]
and%
\begin{align*}
-\varepsilon_{n}&=(Q_{n+1}-Q_{n})\theta_{n+1}-Q_{n+1}\theta_{n+1}+Q_{n}\theta_{n}\\
&=(Q_{n+1}-Q_{n})\theta_{n+1}+\text{ a vector in }\mathbb Z^2
.
\end{align*}
Together with the inequality $Q_{n+1}-Q_{n}>Q_{n}$ this implies that $Q_{n}$ is a best approximation
of $\theta_{n+1}$ and that there is no best approximation of $\theta_{n+1}$
between $Q_{n}$ and $Q_{n+1}$.
Next, denoting by $\equiv$ the equivalence $\operatorname{mod}\mathbb{Z}^{2}$,
we have%
\begin{align*}
Q_{n-1}\theta_{n+1}  &  =Q_{n-1}(\theta_{n}+\frac{\varepsilon_{n}}{Q_{n}%
})=Q_{n-1}(\theta_{n-1}+\frac{\varepsilon_{n-1}}{Q_{n-1}}+\frac{\varepsilon
_{n}}{Q_{n}})\\
&  \equiv\varepsilon_{n-1}+\frac{Q_{n-1}}{Q_{n}}\varepsilon_{n}%
\end{align*}
and%
\begin{align*}
(Q_{n}-Q_{n-1})\theta_{n+1}  &  =(Q_{n}-Q_{n-1})(\theta_{n}+\frac
{\varepsilon_{n}}{Q_{n}})\\
&  \equiv-Q_{n-1}\theta_{n}+(1-\frac{Q_{n-1}}{Q_{n}})\varepsilon_{n}\\
&  =-Q_{n-1}(\theta_{n-1}+\frac{\varepsilon_{n-1}}{Q_{n-1}})+(1-\frac{Q_{n-1}%
}{Q_{n}})\varepsilon_{n}\\
&  \equiv-\varepsilon_{n-1}+(1-\frac{Q_{n-1}}{Q_{n}})\varepsilon_{n},
\end{align*}
by the choice of the signs we obtain that%
\[
\dd(Q_{n-1}\theta_{n+1},\mathbb{Z}^{2})
<\dd((Q_{n}-Q_{n-1})\theta_{n+1},\mathbb{Z}^{2}).
\]
If $q$ is an integer $\neq Q_{n}-Q_{n-1}$ lying in $]Q_{n-1},Q_{n}[$, then
$q\theta_{n}$ cannot be $\equiv\pm\varepsilon_{n-1}$ which are the shortest vectors of $\Lambda_n$. Hence
\[
\dd(q\theta_{n},\mathbb{Z}^{2})\geq\min(2\left\Vert \varepsilon_{n-1}\right\Vert
,\lambda_{2}(\Lambda_{n}))=2\left\Vert \varepsilon_{n-1}\right\Vert .
\]
It follows that%
\begin{align*}
\dd(q\theta_{n+1},\mathbb{Z}^{2})  &  \geq \dd(q\theta_{n},\mathbb{Z}%
^{2})-q\left\Vert \theta_{n+1}-\theta_{n}\right\Vert \\
&  \geq 2\left\Vert \varepsilon_{n-1}\right\Vert -\frac{q}{Q_{n}}\left\Vert
\varepsilon_{n}\right\Vert \\
&  \geq 2\left\Vert \varepsilon_{n-1}\right\Vert -\left\Vert \varepsilon
_{n}\right\Vert >\left\Vert \varepsilon_{n-1}\right\Vert \\
&  >=\|\varepsilon_{n-1}+\frac{Q_{n-1}}{Q_n}\varepsilon_n\|=\dd(Q_{n-1}\theta_{n+1},\mathbb{Z}^{2})
\end{align*}
which implies both that $Q_{n-1}$ and $Q_{n}$ are consecutive best
approximations of $\theta_{n+1}$ and that $M_{n,n+1}>0$.
\end{proof}




It is not clear whether the set $Bad_{1}$ depends on the norm. However, using
an easy result about the relations between best approximation vectors
associated with two norms, we can prove:

\begin{proposition}\label{prop:bad}
The set $\mathcal{B}(d,c)$ does not depend on the norms.
\end{proposition}

\begin{proof}
We give the proof only in the case $c=1$. When $c>1$ one has to extend first,
the following result about best approximations:

\textit{Consider two norms $N$ and
$N^{\prime}$ on $\mathbb{R}^{d}$. For $\theta \in \mathbb R^d$, call $(q_{n})_{n\in\mathbb N}$  the sequence of best approximation denominators
associated with the norm $N$ and $(q_{n}^{\prime})_{n\in\mathbb N}$  the
sequence associated with the norm $N^{\prime}$. Then (see \cite{Chev-Mosc}
)  there exists an integer $k$ depending only on the norms $N$ and
$N^{\prime}$ such that each interval $]q_{n},q_{n+k}]$, contains a best
approximation denominator $q^{\prime}_m$ associated with the norm $N^{\prime}$.}

It is enough to prove that $\mathbb{R}^{d}\setminus Bad_{kp}\subset
\mathbb{R}^{d}\setminus Bad_{p}^{\prime}$ for all $p$. Let $\theta$ be in
$\mathbb{R}^{d}$ and $n\geq k$ be an integer. By the above result, their
exists at least one best approximation denominator in each interval $]q_{n+(j-1)k}%
,q_{n+jk}]$, $j=0,...,p$. Let $q_{n_{j}}^{\prime}$ be the largest best
approximation denominator in each of these intervals $]q_{n+(j-1)k},q_{n+jk}]$. For each
$j$ we have%
\[
r_{n_{j}}^{\prime}\leq Cr_{n+jk}%
\]
where $C$ is the constant involved in the norm equivalence. Making use of the
above inequality with $j=0$, we obtain $q_{n_{0}+p}^{\prime}r_{n_{0}}^{\prime
d}\leq C^{d}q_{n_{0}+p}^{\prime}r_{n}^{d}$. Next $q_{n_{0}+p}^{\prime}\leq
q_{n+kp}$, hence,
\[
q_{n_{0}+p}^{\prime}r_{n_{0}}^{\prime d}\leq C^{d}q_{n+kp}r_{n}^{d}.
\]
It follows that $\lim\inf_{n\rightarrow\infty}q_{n+kp}r_{n}^{d}=0$ implies
$\lim\inf_{n\rightarrow\infty}q_{n+p}^{\prime}r_{n}^{\prime d}=0$.
\end{proof}


\begin{theorem}
\label{thm:badk} The set $\mathcal{B}(d,c)=\cup_{k\geq 0}Bad_{k}$ has zero measure.
\end{theorem}

By the above Proposition $\mathcal{B}(d,c)$ doesn't depend on the norms and we can suppose that $\mathbb{R}^{d}$
and $\mathbb{R}^{c}$ are equipped with the standard Euclidean norms. Let us
show that for each $k$, $Bad_{k}$ has zero measure. We need two lemmas.

\begin{lemma}
\label{lem:perturbation} 1.
Let $a<b$  be two integers, let
$\Lambda=M\mathbb{Z}^{d+c}$ be a lattice in $S$ and let $(Y_{n})_{n=a,...,b}$
be a sequence of vectors in $\mathbb{Z}^{d+c}$. Suppose that for $n=a,...,b$,
\begin{itemize}
\item $X_{n}(\Lambda)=MY_{n}$,
\item the only nonzero points of
$\Lambda$ in the cylinder $C(X_{n}(\Lambda),X_{n+1}(\Lambda))$ are $\pm
X_{n}(\Lambda)$ and $\pm X_{n+1}(\Lambda).$
\end{itemize}
Then there exists a open
neighborhood $W$ of $M$ such that for all lattices $\Lambda^{\prime}%
=M^{\prime}\mathbb{Z}^{d+c}\in \mathcal L_{d+c}$ with
$M^{\prime}$ in $W$, the vectors $Z_n=M'Y_n$ are consecutive
minimal vectors of $\Lambda'$ and
\begin{align*}
\left\vert Z_n\right\vert_+  &  \in[\frac{1}{2}r_{n}(\Lambda),2r_{n}(\Lambda)],\\
\left\vert Z_n\right\vert_- &  \in[\frac{1}{2}q_{n}(\Lambda),2q_{n}(\Lambda)]
\end{align*}
for $n=a,...,b$.\newline
2. Suppose furthermore that $a<0$, $b>1$ and $\Lambda\in S$.
Then for all lattices $\Lambda^{\prime}%
=M^{\prime}\mathbb{Z}^{d+c}\in S$ with $M^{\prime}$ in $W$, we have $X_{n}(\Lambda^{\prime})  =M^{\prime}Y_{n}$ for $n=a,...,b$..
\end{lemma}

\begin{proof}
1. Consider a ball $B_{\mathbb{R}^{d+c}}(0,R)$ that contains all the points $MY_{n}$,
$n=a,...,b$. There is a neighborhood $\omega$ of the identity matrix $I_{d+c}$ such
that for all $A$ in $\omega$ \ and all $X$ in $\mathbb{R}^{d+c}$,
\[
\frac{1}{2}\left\Vert X\right\Vert _{\mathbb{R}^{d+c}}\leq\left\Vert
AX\right\Vert _{\mathbb{R}^{d+c}}\leq 2\left\Vert X\right\Vert _{\mathbb{R}%
^{d+c}},
\]
so that%
\begin{align*}
X  &  \notin B_{\mathbb{R}^{d+c}}(0,8R)\Rightarrow AX\notin B_{\mathbb{R}%
^{d+c}}(0,4R)\\
X  &  \in B_{\mathbb{R}^{d+c}}(0,R)\Rightarrow AX\in B_{\mathbb{R}^{d+c}%
}(0,2R).
\end{align*}
Two vectors $Z_{n}=AMY_{n}$ and $Z_{n+1}=AMY_{n+1}$ are consecutive
minimal vectors of $A\Lambda$ as soon as
\[
\left\vert Z_{n+1}\right\vert _{-}>\left\vert Z_{n}\right\vert _{-}%
,\ \left\vert Z_{n+1}\right\vert _{+}<\left\vert Z_{n}\right\vert _{+}%
\]
and the cylinder $C(Z_{n},Z_{n+1})$ contains no other nonzero vector of
$A\Lambda$ than $\pm Z_{n}$ and $\pm Z_{n+1}$. Since, $\left\vert
X_{n+1}\right\vert _{-}>\left\vert X_{n}\right\vert _{-},\ \left\vert
X_{n+1}\right\vert _{+}<\left\vert X_{n}\right\vert _{+}$, by reducing
$\omega$, we can assume $\left\vert Z_{n+1}\right\vert _{-}>\left\vert
Z_{n}\right\vert _{-}\ $\ and $\left\vert Z_{n+1}\right\vert _{+}<\left\vert
Z_{n}\right\vert _{+}$, $n=a,\dots,b$. Since $C(Z_{n},Z_{n+1})=C(AX_{n},AX_{n+1})\subset
B_{\mathbb{R}^{d+c}}(0,2R)$, the image by $A$ of a vector of $\Lambda$ that is
not in the ball $B_{\mathbb{R}^{d+c}}(0,8R)$, cannot enter in the cylinder
$C(AX_{n},AX_{n+1})$. Therefore, there are only finitely many $X$ in $\Lambda$
such that $AX$ is in $C(AX_{n},AX_{n+1})$. Since by assumption all these
vectors $X$, except $\pm X_n$ and $\pm X_{n+1}$, are at a positive distance from $C(X_{n},X_{n+1})$, we obtain that
$Z_{n}$ and $Z_{n+1}$ are consecutive minimal vectors by reducing once again
$\omega$. It follows that $Z_a,...,Z_b$ are consecutive minimal vectors of the lattice $A\Lambda$. A new reduction of $\omega$ ensures that the two  inequalities of the
Lemma hold.


2. We want to see that there is no shift on the indices. By the numbering 
convention (see section \ref{sec:visit}),
\[
\left\vert X_{0}(\Lambda)\right\vert _{+}=\left\vert X_{1}(\Lambda)\right\vert
_{-},\ \left\vert X_{-1}(\Lambda)\right\vert _{+}>\left\vert X_{0}%
(\Lambda)\right\vert _{-},\ \left\vert X_{1}(\Lambda)\right\vert
_{+}<\left\vert X_{2}(\Lambda)\right\vert _{-}.
\]
By a further reduction of $\omega$, we can assume that the two inequalities
hold for the vectors $Z_{-1}=AMY_{-1}$, $Z_{0}=AMY_{0}$, $Z_{1}=AMY_{1}$ and $Z_{2}=AMY_{2}$.
Therefore if $AM\mathbb{Z}^{d+c}$ is in $S$ we must have%
\[
X_{0}(A\Lambda)=AMY_{0}\text{ and }X_{1}(A\Lambda)=AMY_{1}%
\]
which implies that $X_{n}(A\Lambda)=Z_{n}$ for $n=a,....,b$.
\end{proof}


\begin{lemma}
\label{lem:smallvectors}Assume that $d+c\geq3$. Let $\Gamma$ be a two dimensional lattice in
$\mathcal{L}_{2}\setminus\mathcal{N}_{2}$ which is in $S_{2}$ and let $k$ be a
non negative integer. Then for all positive real number $\delta$, there
exists $\varepsilon<2\delta$ and a lattice $\Lambda_{\varepsilon}$ in
$S\setminus\mathcal{N}$ such that
\begin{align*}
r_{n}(\Lambda_{\varepsilon})  &  \leq\varepsilon r_{n}(\Gamma)\\
q_{n}(\Lambda_{\varepsilon})  &  \leq\varepsilon q_{n}(\Gamma)
\end{align*}
for $n=0,...,k$.
\end{lemma}

\begin{proof}
Let $\Gamma=A\mathbb{Z}^{2}$ be a lattice in $S_{2}\setminus\mathcal{N}_{2}$
where
\[
A=\left(
\begin{array}
[c]{cc}%
a_{11} & a_{12}\\
a_{21} & a_{22}%
\end{array}
\right)  .
\]
Consider the matrix $M_{\delta}\in\SL(d+c,\mathbb R)$ defined by%
\[
M_{\delta}=\left(
\begin{array}
[c]{cccccc}%
\delta a_{11} & 0 & 0 & \ldots & 0 & \delta a_{12}\\
0 & \delta^{-\frac{2}{d+c-2}} & 0 & \ldots & \ldots & 0\\
0 & 0 & \delta^{-\frac{2}{d+c-2}} & 0 & \ldots & 0\\
\vdots & \vdots & \ddots~~ & \ddots~~~~ & \ddots & \vdots\\
0 & 0 & \ldots & 0 & \delta^{-\frac{2}{d+c-2}} & 0\\
\delta a_{21} & 0 & \ldots& \ldots & 0 & \delta a_{22}%
\end{array}
\right)  .
\]
\qquad
Let $(U_{n}=(u_{n,1},u_{n,2}))_{n\in\mathbb{Z}}$ be the sequence of vectors in $\mathbb{Z}^{2}$
such that $(X_{n}(\Gamma)=AU_{n})_{n\in\mathbb Z}$ is the sequence of minimal vectors of
$\Gamma$. For each $n\in\mathbb{Z}$, let $Y_{n}$ be the element of
$\mathbb{Z}^{d+c}$ defined by $y_{1}=u_{n,1},\ y_{2}=...=y_{d+c-1}=0$ and
$y_{d+c}=u_{n,2}$. If $\mathbb{\delta}>0$ is small enough, then for all
$Z\in\mathbb{Z}^{d+c}$ not in the $\mathbb R e_{1}+\mathbb R e_{d+c}$-plane, we have%
\[
\left\Vert M_{\delta}Z\right\Vert _{\mathbb{R}^{d+c}}=\max(\left\vert
M_{\delta}Z\right\vert _{+},\left\vert M_{\delta}Z\right\vert _{-})\geq \delta^{-\frac{2}{d+c-2}}\geq
\max(2\mathbb{\delta}r_{-1}(\Gamma),2\mathbb{\delta}q_{k+1}(\Gamma)).
\]
It follows that none of these vectors $M_{\mathbb{\delta}}Z$ are in one of the
cylinders $C(M_{\mathbb{\delta}}Y_{n},M_{\delta}Y_{n+1})$, $n=-1,...,k$.
Therefore the vectors $X_{n}=M_{\mathbb{\delta}}Y_{n}$, $n=-1,...,k+1$ are all
consecutive minimal vectors of $\Lambda_{\mathbb{\delta}}=M_{\mathbb{\delta}%
}\mathbb{Z}^{d+1}$ and $\Lambda_{\delta}$ is in $S$. With our numbering
convention we have $X_{n}(\Lambda_{\mathbb{\delta}})=X_{n}$ for all
$n=-1,...,k+1$.
Now we fix $\delta$ small enough. By the previous Lemma applied to
$\Lambda_{\delta}$, there is sequence of matrices $(M_{p})_{p}$ in
$S\setminus\mathcal{N}$ which converges to $M_{\mathbb{\delta}}$ such for all
$p$,%
\[
X_{n}(M_{p}\mathbb{Z}^{d+1})=M_{p}Y_{n}%
\]
$n=0,...,k$. When $p$ goes to infinity,
\begin{align*}
r_{n}(M_{p}\mathbb{Z}^{d+1})  &  =\left\vert M_{p}Z_{n}\right\vert
_{+}\rightarrow\left\vert M_{\delta}Z_{n}\right\vert _{+}=\delta r_{n}%
(\Gamma),\\
q_{n}(M_{p}\mathbb{Z}^{d+1})  &  =\left\vert M_{p}Z_{n}\right\vert
_{-}\rightarrow\left\vert M_{\delta}Z_{n}\right\vert _{-}=\delta q_{n}(\Gamma)
\end{align*}
for $n=0,...,k$. So we can take $\Lambda_{\varepsilon}=M_{p}\mathbb{Z}^{d+1}$
for some $p$ large enough.
\end{proof}


\begin{proof}[End of proof of  Theorem \ref{thm:badk}]
Let $k$ and $\eta>0$ be fixed. We want to prove that the set of $\theta$ in
$\M_{d,c}(\mathbb{R})$ such that
\[
\lim\inf_{n\rightarrow\infty}q_{n+k}^{c}(\theta)r_{n}^{d}(\theta)\leq\eta
\]
has full measure.
By Lemma \ref{lem:min-best}, it is enough to show that
\[
\lim\inf_{n\rightarrow\infty}q_{n+k}^{c}(\Lambda_{\theta})r_{n}^{d}(\Lambda
_{\theta})\leq\eta
\]
for almost all $\theta$.
Fix a two-dimensional lattice $\Gamma$ in $S_{2}\setminus \mathcal{N}_{2}$
and let $\delta$ be a positive real number with $\delta\leq\frac{1}{4}%
(\frac{\eta}{q_{k}^{c}(\Gamma)r_{0}^{d}(\Gamma)})^{\frac{1}{d+c}}$. By 
Lemma \ref{lem:smallvectors}, there exist $\varepsilon\leq 2\delta$ and a
lattice $\Lambda_{\varepsilon}$  in $S\setminus\mathcal{N}$ such that
\begin{align*}
r_{n}(\Lambda_{\varepsilon}) &  \leq\varepsilon r_{n}(\Gamma)\\
q_{n}(\Lambda_{\varepsilon}) &  \leq\varepsilon q_{n}(\Gamma)
\end{align*}
for $n=0,...,k$. Hence,
\[
q_{k}^{c}(\Lambda_{\varepsilon})r_{0}^{d}(\Lambda_{\varepsilon})\leq\frac
{\eta}{2^{d+c}}.
\]
By Lemma \ref{lem:perturbation}, there exists an open neighborhood $W$ of
$\Lambda_{\varepsilon}$ such that for all $\Lambda$ in $W$ and some integer $m(\Lambda)$, we have both
\[
r_{m(\Lambda)}(\Lambda)\leq 2\varepsilon r_0(\Gamma)
\]
and 
\[
q_{m(\Lambda)+k}^{c}(\Lambda)r_{m(\Lambda)}^{d}(\Lambda)\leq\eta.
\]


Let us show that if for a given lattice $\Lambda$, there exists a sequence $(t_{n}%
)_{n\in\mathbb{N}}$ going to infinity such that $g_{t_{n}}\Lambda\in$ $W$ for
all $n\in\mathbb{N}$, then
\[
\lim\inf_{n\rightarrow\infty}q_{n+k}^{c}(\Lambda)r_{n}^{d}(\Lambda)\leq\eta.
\]
Indeed, if $g_{t_n}\Lambda \in W$, then for some integer $m(\Lambda,t_n)$ we have 
\[
(e^{-dt_n}q_{m(\Lambda,t_n)+k}(\Lambda))^{c}(e^{ct_n}r_{m(\Lambda,t_n)}(\Lambda))^{d}\leq\eta.
\] 
So, the only thing to see is that $m(\Lambda,t_n)\rightarrow\infty$ when $n\rightarrow\infty$. 
Now $e^{dt_n}r_{m(\Lambda,t_n)}(\Lambda)\leq 2\varepsilon r_0(\Gamma) $,
hence $r_{m(\Lambda,t_n)}(\Lambda)$ goes to zero when $m$ goes
to infinity which implies that $m(\Lambda,t_n)$ goes to infinity.

Making use of Birkhoff Theorem with the flow $g_t$, the proof
would be already finished if our goal were
$\lim\inf_{n\rightarrow\infty}q_{n+k}^{c}(\Lambda)r_{n}^{d}(\Lambda)\leq\eta$
for almost all lattices. However we want an ``almost all" with respect of the Lebesgue measure of $\M_{d,c}(\mathbb R)$. 
 

Let $U$  be a relatively compact nonempty open set in $\mathcal L_{d+c}$ such that
$\overline{U}\subset W$. 
One can find
a neighborhood $V$ of $I_{d+c}$ in $\mathcal{H}_{\leq}$ such
that for all
$\theta\in \M_{d,c}(\mathbb{R)}$, all $t\geq 0$ and all $h\in V$, we have%
\[
g_{t}h\Lambda_{\theta}=(g_{t}hg_{-t})g_{t}\Lambda_{\theta}\in U\Longrightarrow
g_{t}\Lambda_{\theta}\in W.
\]
Call $\mathcal{V}$ the set of $\theta$ such that $g_{t}\Lambda_{\theta}\notin
W$ for all $t$ large enough. By the choices of $U$ and $V$, for all $h\in V$ and all
$\theta\in\mathcal{V}$, $g_{t}h\Lambda_{\theta}\notin U$ for all $t$ large
enough. If the Lebesgue measure of $\mathcal{V}$ were nonzero then the set of
lattices of the form $g_{s}h\Lambda_{\theta}$ with $s\in[0,1]$, $h\in
V$ and $\theta\in\mathcal{V}$, would have a nonzero measure. Now, by Birkhoff Theorem,
for almost all lattices $\Lambda$, there exist a sequence $t_{n}%
\rightarrow\infty$ such that $g_{t_{n}}\Lambda\in U$ for all $n$, therefore
$\mathcal{V}$ has zero measure.
\end{proof}


\begin{proof}[Proof of Theorem \ref{thm:jager}. 2] By the proof of the first part of Theorem \ref{thm:jager}, we know that the measure $\nu_{d,c}$ is the image of the measure $\frac{1}{\mu_S(S)}\mu_S$ by the map $F:S\rightarrow \mathbb R$ defined by 
	\[
	F(\Lambda)=q_1^c(\Lambda)r_0^d(\Lambda)=|v_1^S(\Lambda)|_-^c|v_0^S(\Lambda)|^d_+.
	\]
We want to prove that the support of the measure $\nu_{d,c}$ contains zero, {\it i.e.}, that $\nu_{d,c}([0,\eta])>0$ for all $\eta>0$. By Birkhoff Theorem and by definition of $\nu_{d,c}$, it is enough to prove that 
\[
\lim_{n\rightarrow\infty}\frac{1}{n}\sum_{i=0}^{n-1}1_{[0,\eta]}(F(R^i(\Lambda)))>0
\]
for almost all $\Lambda\in S$. 
By Lemmas \ref{lem:perturbation} and \ref{lem:smallvectors},
there exists a non empty  open set $W$ in $S$ such that 
\[
\left\vert v_1^S(\Lambda)\right\vert^c_-\left\vert v_0^S(\Lambda)\right\vert^d_+\leq\eta
\]
for all $\Lambda\in W$. Hence $1_W\leq 1_{[0,\eta]}\circ F$.
 By Birkhoff Theorem,
 for almost
all $\Lambda$ in $S$
\begin{align*}
\lim_{n\rightarrow\infty}\frac{1}{n}\sum_{i=1}^{n}1_W\circ R^{i}
(\Lambda)=\frac{1}{\mu_{S}(S)}\int_{S}1_W~d\mu_{S}=a>0.
\end{align*}
therefore,
\[
\lim_{n\rightarrow\infty}\frac{1}{n}\sum_{i=0}^{n-1}1_{[0,\eta]}(F(R^i(\Lambda)))\geq a>0.
\]
\end{proof}

\section{Miscellaneous Questions}
\begin{enumerate}
\item 

In Theorems \ref{thm:levy-d} and \ref{thm:jager}, we assume that $\mathbb R^d$ and $\mathbb R^c$ are equipped with the standard Euclidean norms. Do these Theorems hold when $\mathbb R^d$ and $\mathbb R^c$ are equipped with any norms?\\
If Theorem \ref{thm:levy-d} holds for any norms, does the Levy's constant depend on  the norms?
\item Is the measure $\nu_{d,c}$ in Theorem \ref{thm:jager}, absolutely continuous with respect to Lebesgue measure?\\
Is the support of $\nu_{d,c}$ an interval?

\item Suppose $c=1$. 
Consider a flow $(g_t)_{t\in\mathbb R}$ defined by the matrices
\[
g_{t}=\text{Diag}(e^{a_1t},\dots,e^{a_dt},e^{-dt}) \in \SL (d+1,\mathbb R)
\]
where the $a_is$ are positive real numbers with sum $d$.
Best approximation vectors of $\theta\in\mathbb R^d$ with respect to the flow $(g_t)_{t\in\mathbb R}$  can be defined as follow. A nonzero vector $X$ in $\mathbb Z^{d+1}$ is a best approximation vector of $\theta$ if there exists $t\geq0$ such that the interior of the ball $B(g_tM_{\theta}X)\subset \mathbb R^d\times\mathbb R$ contains no nonzero vector  of the lattice $g_tM_\theta\mathbb Z^{d+1}$ (equivalently $\|g_tM_{\theta}X\|_{\mathbb R^{d+1}}=\lambda_1(g_tM_{\theta}\mathbb Z^{d+1}$)). Arranging the set of best approximation vector according to their heights, we obtain a sequence $(X_n(\theta))_{n\in\mathbb N}$ of best approximation vectors associated with $\theta$. Does Theorem \ref{thm:levy-d} hold for these new sequences of best approximation vectors?
\item 
For a fixed $k\geq 1$, does the set
\[
Bad_{k}(d,c)=Bad_{k}=\{\theta\in\M_{d,c}(\mathbb{R})\setminus\M_{d,c}(\mathbb{Q}):\inf_{n\in\mathbb N} q_{n+k}^{c}%
r_{n}^{d}>0\}
\]
depends on the norms used to define best approximations vectors?\\
Observe that by Proposition \ref{prop:bad}, the union $\cup_{k\geq 1}Bad_k$ does not depend on the choice  of the norms.
\end{enumerate}
