# Decorrelation estimates for translated measures under diagonal flows

Michael Björklund  ,  Reynold Fregoli  and  Alexander Gorodnik

###### Abstract.

A profound link between Homogeneous Dynamics and Diophantine Approximation is
based on an observation that Diophantine properties of a real matrix $B$ are
encoded by the corresponding lattice $\Lambda_{B}$ translated by a multi-
parameter semigroup $a(t)$. We establish quantitative decorrelation estimates
for measures supported on leaves $a(t)\Lambda_{B}$ with the error terms
depending only on the minimum of the pairwise distances between the
parameters. The proof involves a careful analysis of the translated measures
in the products of the spaces of unimodular lattices and establishes
quantitative equidistributions to measures supported on various intermediate
homogeneous subspaces.

###### Contents

  1. 1 Introduction
  2. 2 Basic notation
  3. 3 Mixing and equidistribution in intermediate homogeneous spaces
  4. 4 Decorrelation estimates
  5. 5 A proposition about weights
  6. 6 Multiple equidistribution
  7. 7 Quasi-independence for separated tuples
  8. 8 Proof of main theorem

##  1\. Introduction

When studying Diophantine properties of a real $(m\times n)$-matrix $B$, it is
often natural to consider the corresponding unimodular lattice

$\Lambda_{B}:=\\{(p+Bq,q):\,\,p\in\mathbb{Z}^{m},\,q\in\mathbb{Z}^{n}\\}\subset\mathbb{R}^{m+n},$

and view it as a point in the space $X$ of all $(m+n)$-dimensional unimodular
lattices. It turns out that many Diophantine properties of $B$ can be
equivalently expressed in terms of the asymptotic behaviour of the orbit of
$\Lambda_{B}$ under the _diagonal flow_

$a(t):=\hbox{diag}\big{(}e^{t^{(1)}},\ldots,e^{t^{(m)}},e^{-t^{(m+1)}},\ldots,e^{-t^{(m+n)}}\big{)}\in\operatorname{SL}_{m+n}(\mathbb{R}),$

where $t=(t^{(1)},\ldots,t^{(m+n)})$ belongs to the additive sub-semigroup
$A^{+}\subset\mathbb{R}_{+}^{m+n}$ defined by

$A^{+}:=\big{\\{}t\in\mathbb{R}_{+}^{m+n}:\,t^{(1)}+\cdots+t^{(m)}=t^{(m+1)}+\cdots+t^{(m+n)}\big{\\}}.$

For instance, Kleinbock and Margulis [13] observed that Khinchin’s classical
theorem in Diophantine approximate is intimately related to the visits to
shrinking cuspidal neighborhoods in $X$ of the orbit of $\Lambda_{B}$ under
the sub-flow $g_{t}=a(t/m,\ldots,t/m,t/n,\ldots,t/n)$ for $t\geq 0$. In
particular, the crucial Borel-Cantelli property needed in their proof was
derived from the asymptotic ”independence” of the lattices
$g_{t_{1}}\Lambda_{B}$ and $g_{t_{2}}\Lambda_{B}$ as $t_{1},t_{2}$ and
$|t_{1}-t_{2}|$ all tend to infinity. More generally, the orbit of
$\Lambda_{B}$ under larger sub-flows of $a(A^{+})$ encode _multiplicative_
Diophantine approximation properties of the matrix $B$ (see e.g. [3]).

Systematic implementation of the idea that various objects in Number Theory
are asymptotically independent goes back to the monograph [15] of W. Philipp,
who developed a general framework for proving probabilistic limit theorems for
weakly independent processes having arithmetic applications in mind. In
particular, this led to profound limit theorems for continued fractions, where
the asymptotic independence was derived from properties of the Gauss map. In
the higher-dimensional setting, the asymptotic independence property of
translated lattices $a(t)\Lambda_{B}$ provides a promising approach to
generalize these ideas and hopefully to establish new limit theorems in
Diophantine Approximation and Geometry of Numbers. For instance, we refer to
recent works [1, 4, 6, 7, 10, 11], where this approach was successfully
applied.

Our paper is devoted to proving a general asymptotic independence result for
lattices of the form $a(t)\Lambda_{B}$ with a random $(m\times n)$-matrix $B$.
To make this more precise, let us introduce some notation. Fix integers
$m,n\geq 1$ once and for all. Let $X$ denote the space of all unimodular
lattices in $\mathbb{R}^{m+n}$ and let $\mu$ denote the unique
$\operatorname{SL}_{m+n}(\mathbb{R})$-invariant probability measure on $X$.
Define the sub-manifold

$Y:=\big{\\{}\Lambda_{B}:\,B\in\hbox{Mat}_{m\times
n}(\mathbb{R}/\mathbb{Z})\big{\\}}\subset X,$

which can be identified with the standard $(m+n)$-torus
$\mathbb{R}^{m+n}/\mathbb{Z}^{m+n}$, and equip $Y$ with the Lebesgue measure
$\nu$ coming from this torus. Our aim is to show that, given any smooth
functions $\varphi_{1},\ldots,\varphi_{r}$ on $X$, and $t_{1},\ldots,t_{r}\in
A^{+}$, we have

$\int_{Y}\left(\prod_{s=1}^{r}\varphi_{s}\circ
a(t_{s})\right)\,d\nu\approx\prod_{s=1}^{r}\int_{Y}\varphi_{s}\circ
a(t_{s})\,d\nu$ (1.1)

with an explicit error term which _only_ depends on the function

$\Delta(t_{1},\ldots,t_{r}):=\min_{i\neq j}\|t_{i}-t_{j}\|,$

where $\|\cdot\|$ denotes the maximum norm on $\mathbb{R}^{m+n}$. Note that
(1.1) only involves correlations with respect to $\nu$, and not $\mu$. We
refer to (1.1) as (higher order) _decorrelation_ , as opposed to (higher
order) _equidistribution_ when one instead proves (1.1) with $\mu$ on the
right-hand side, but often with a significantly worse error term (see e.g. [3,
4, 5]), which is really only useful when all coordinates in $t$ tend to
infinity. More precisely, the error term in this case depend on the minimum of
$\Delta(t_{1},\ldots,t_{r})$ and the function

$\min(\lfloor t_{1}\rfloor,\ldots,\lfloor t_{r}\rfloor),$

where $\lfloor t\rfloor=\min t^{(s)}$ for $t=(t^{(1)},\ldots,t^{(r)})\in
A^{+}$. To prove (1.1) we need to (quantitatively) understand _all_ limits of
translated measures $a(t)\nu$ as $t$ tends to infinity in $A^{+}$ (allowing
for some coordinates in $t$ to stay bounded).

The problem of limiting distribution of measures $a(t)_{*}\nu$ in $X$ can be
studied using Ratner‘s measure classification [16] for measures invariant
under unipotent flows. In particular, the case of measures supported on
horospherical orbits was handled by Dani [8, 9] and generalized by Shah [18].
More generally, this approach can be potentially used to study the
distribution of translated measures embedded diagonally in
$X\times\cdots\times X$, but this method does not provide quantitative bounds.
Regarding quantitative estimates, when $a_{t}$ is a _one-parameter_ partially
hyperbolic flow and $\nu$ is a smooth measure supported on an unstable leaf of
$a_{t}$, the quantitative asymptotic behaviour of the measures
$(a_{t})_{*}\nu$ as $t\to\infty$ was investigated by Kleinbock and Margulis in
[12]. The main difficulty of our present setting is that $Y$ is not an
unstable leaf for the family of transformations $a(t)$. The first quantitative
result in this setting was established by Kleinbock and Margulis in [14]:

######  Theorem 1.1 (Kleinbock, Margulis).

There exist $\ell\geq 1$ and $\delta>0$ such that for every $f\in
C^{\infty}(Y)$, $\varphi\in C_{c}^{\infty}(X),$ and $t\in A^{+}$,

$\int_{Y}f\cdot(\varphi\circ
a(t))\,d\nu=\left(\int_{Y}f\,d\nu\right)\left(\int_{X}\varphi\,d\mu\right)+O\Big{(}e^{-\delta\lfloor
t\rfloor}\,\|f\|_{C^{\ell}}\,\mathcal{N}_{\ell}(\varphi)\Big{)},$

where
$\mathcal{N}_{\ell}(\varphi):=\max\left(\|\varphi\|_{C^{0}},\|\varphi\|_{Lip},\|\varphi\|_{L^{2}_{\ell}}\right)$.
Here $\|\cdot\|_{Lip}$ and $\|\cdot\|_{L^{2}_{\ell}}$ refers to the Lipschitz
norm and the $L^{2}$-Sobolev norm of order $\ell$ respectively.

Shi [19] established the following extension to higher-order correlations:

######  Theorem 1.2 (Shi).

There exist $\ell\geq 1$ and $\delta^{\prime}>0$ such that for every $f\in
C^{\infty}(Y)$, $\varphi_{1},\ldots,\varphi_{r}\in C_{c}^{\infty}(X),$ and
$t_{1},\ldots,t_{r}\in A^{+}$,

$\displaystyle\int_{Y}f\left(\prod_{s=1}^{r}\varphi_{s}\circ
a(t_{s})\right)\,d\nu=$
$\displaystyle\int_{Y}f\,d\nu\prod_{s=1}^{r}\int_{X}\varphi_{s}\,d\mu$
$\displaystyle\quad+O\left(e^{-\delta^{\prime}\min\big{(}\lfloor
t_{1}\rfloor,\lfloor t_{2}-t_{1}\rfloor,\ldots,\lfloor
t_{r}-t_{r-1}\rfloor\big{)}}\|f\|_{C^{\ell}}\prod_{s=1}^{r}\mathcal{N}_{\ell}(\varphi_{s})\right).$

Later, Björklund and Gorodnik [5] established a similar extension but with a
different error term:

######  Theorem 1.3 (Björklund, Gorodnik).

There exist $d_{r}\geq 1$ and $\delta_{r}>0$ such that for every $f\in
C^{\infty}(Y)$, $\varphi_{1},\ldots,\varphi_{r}\in C_{c}^{\infty}(X),$ and
$t_{1},\ldots,t_{r}\in A^{+}$,

$\displaystyle\int_{Y}f\left(\prod_{s=1}^{r}\varphi_{s}\circ
a(t_{s})\right)\,d\nu=$
$\displaystyle\int_{Y}f\,d\nu\prod_{s=1}^{r}\int_{X}\varphi_{s}\,d\mu$
$\displaystyle\quad+O\left(e^{-\delta_{r}\min\big{(}\lfloor
t_{1}\rfloor,\ldots,\lfloor
t_{r}\rfloor,\Delta(t_{1},\ldots,t_{r})\big{)}}\|f\|_{W}\prod_{s=1}^{r}\mathcal{N}_{\ell,d_{r}}(\varphi_{s})\right),$

where
$\mathcal{N}_{\ell,d}(\varphi):=\max\left(\|\varphi\|_{C^{0}},\|\varphi\|_{Lip},\|\varphi\|_{L^{2^{d}}_{\ell}}\right)$.
Here $\|\cdot\|_{W}$ refers to the Wiener norm of functions on $Y$, defined by
(2.3) below.

The main result of this paper is:

######  Theorem 1.4.

There exist $\ell\geq 1$ and $\eta_{r}>0$ such that for every
$\varphi_{1},\ldots,\varphi_{r}\in C_{c}^{\infty}(X),$ and
$t_{1},\ldots,t_{r}\in A^{+}$,

$\displaystyle\int_{Y}\left(\prod_{s=1}^{r}\varphi_{s}\circ
a(t_{s})\right)\,d\nu=\prod_{s=1}^{r}\int_{Y}\varphi_{s}\circ
a(t_{s})\,d\nu+O\left(e^{-\eta_{r}\Delta(t_{1},\ldots,t_{r})}\prod_{s=1}^{r}\|\varphi_{s}\|_{C^{\ell}}\right).$

We again stress that the novelty here is that the error term depends only on
the distances $\|t_{i}-t_{j}\|$ with $i\neq j$, while the previous results are
only applicable when $\min\big{(}\lfloor t_{1}\rfloor,\ldots,\lfloor
t_{r}\rfloor\big{)}\to\infty$. We also note that the measures $a(t)_{*}\nu$,
as $t\rightarrow\infty$ in $A^{+}$, can have many different limits that need
to be analysed in order to prove Theorem 1.4. We illustrate this by the
following example.

######  Example 1.1.

Let $m=2$ and $n=1$. According to Theorem 1.1, for $\varphi\in C_{c}(X)$,

$\int_{Y}\varphi\circ a(s,t)\,d\nu\to\int_{X}\varphi\,d\mu\quad\hbox{as
$\min(s,t)\to\infty$.}$

However, the family of measures $a(s,t)_{*}\nu$ also exhibit other limiting
behaviors: one can show, in particular, that

$\int_{Y}\varphi\circ
a(s,0)\,d\nu\to\int_{X_{1}}\varphi\,d\mu_{1}\quad\hbox{as $s\to\infty$},$

and

$\int_{Y}\varphi\circ
a(0,t)\,d\nu\to\int_{X_{2}}\varphi\,d\mu_{2}\quad\hbox{as $t\to\infty$},$

where

$X_{1}:=\left(\begin{tabular}[]{ccc}$*$&$*$&$*$\\\ 0&$*$&0\\\
$*$&$*$&$*$\end{tabular}\right)\Gamma\quad\quad\hbox{and}\quad\quad
X_{2}:=\left(\begin{tabular}[]{ccc}$*$&$*$&$*$\\\ 0&$*$&$*$\\\
0&$*$&$*$\end{tabular}\right)\Gamma$

are the homogeneous subspaces of $X$ equipped with invariant probability
measures $\mu_{1}$ and $\mu_{2}$. More generally, let us embed $\nu$ in
$X\times X$ diagonally and consider the family of measures
$\big{(}a(s_{1},t_{1}),a(s_{2},t_{2}\big{)})_{*}\nu$ with
$(s_{1},t_{1}),(s_{2},t_{2})\in A^{+}$. Here one has 16 possible limiting
behaviors. For example, one can show that for $\varphi_{1},\varphi_{2}\in
C_{c}(X)$,

$\int_{Y}\big{(}\varphi_{1}\circ a(s,0)\big{)}\cdot\big{(}\varphi_{2}\circ
a(0,t)\big{)}\,d\nu\to\left(\int_{X_{1}}\varphi_{1}\,d\mu_{1}\right)\left(\int_{X_{2}}\varphi_{2}\,d\mu_{2}\right)\quad\hbox{as
$s,t\to\infty$.}$ (1.2)

In order to prove Theorem 1.4, one needs to analyze all such possibilities.
Ultimately, a general version of (1.2), which describes various limiting
behaviors, is given by Theorem 6.1 below.

Organization of the paper. In Section 2, we introduce basic notation and, in
particular, a family of homogeneous subspaces $X_{I}$ that will play a crucial
role for the rest of the paper. In Section 3, we study mixing and
equidistribution properties on those spaces. While for certain one-parameter
subgroups a required de-correlation estimate is given by Corollary 3.2, this
fails for other one-parameter subgroups. This issue in addressed in Section 4,
where additional decorrelation estimates are established. In Section 5, we
prove an elementary result about weights that subsequently will allow us to
run our inductive argument using the decorrelation estimates from Sections
3–4. In Section 6, we prove a general multiple equidistribution result
(Theorem 6.1) generalizing Theorem 1.3. Finally, Theorem 1.4 is proved in
Sections 7–8 with a help of Theorem 6.1.

##  2\. Basic notation

Throughout the paper, $X$ denotes the space of unimodular lattice in
$\mathbb{R}^{m+n}$:

$X\simeq G/\Gamma,\quad\hbox{with $G:=\hbox{SL}_{m+n}(\mathbb{R})$ and
$\Gamma:=\hbox{SL}_{m+n}(\mathbb{Z})$},$

equipped with the unique $G$-invariant probability measure $\mu$. We consider
the orbit

$Y:=U\Gamma\subset X$

of the group

$U:=\left(\begin{tabular}[]{cc}$I_{m}$&$*$\\\
$0$&$I_{n}$\end{tabular}\right),$

which can be identified with $(m\times n)$-dimensional torus. We equip $Y$
with the normalized Lebesgue measure $\nu$ (which is the unique $U$-invariant
probability measure on the $U$-orbit $Y$).

Let us now introduce a family of intermediate homogeneous subspaces $X_{I}$ of
$X$ that will play a crucial role in our arguments. We call a subset
$I\subset\\{1,\ldots,m+n\\}$ _admissible_ if

$I\cap\\{1,\ldots,m\\}\neq\emptyset\quad\hbox{and}\quad
I\cap\\{m+1,\ldots,m+n\\}\neq\emptyset.$ (2.1)

Let us first suppose for simplicity that

$I=I_{0}:=\\{m-k_{1}+1,\ldots,m+k_{2}\\}$ (2.2)

for some $k_{1},k_{2}\geq 1$. In this case, we define

$G_{I_{0}}:=\left(\begin{tabular}[]{ccc}$I_{m-k_{1}}$&$*$&$*$\\\
$0$&$\hbox{SL}_{k_{1}+k_{2}}(\mathbb{R})$&$*$\\\
$0$&$0$&$I_{n-k_{1}}$\end{tabular}\right).$

We note that $G_{I_{0}}=S_{I_{0}}\ltimes U_{I_{0}}$, where

$S_{I_{0}}:=\left(\begin{tabular}[]{ccc}$I_{m-k_{1}}$&$0$&$0$\\\
$0$&$\hbox{SL}_{k_{1}+k_{2}}(\mathbb{R})$&$0$\\\
$0$&$0$&$I_{m-k_{1}}$\end{tabular}\right)\quad\hbox{and}\quad
U_{I_{0}}=\left(\begin{tabular}[]{ccc}$I_{m-k_{1}}$&$*$&$*$\\\
$0$&$I_{k_{1}+k_{2}}$&$*$\\\ $0$&$0$&$I_{m-k_{1}}$\end{tabular}\right).$

For general admissible $I$, we take a permutation $\sigma_{I}$ such that

$\displaystyle\sigma_{I}(\\{1,\ldots,m\\}\backslash I)$
$\displaystyle=\\{1,\ldots,m-k_{1}\\},$
$\displaystyle\sigma_{I}(I\cap\\{1,\ldots,m\\})$
$\displaystyle=\\{m-k_{1}+1,\ldots,m\\},$
$\displaystyle\sigma_{I}(I\cap\\{m+1,\ldots,m+n\\})$
$\displaystyle=\\{m+1,\ldots,m+k_{2}\\},$
$\displaystyle\sigma_{I}(\\{m+1,\ldots,m+n\\}\backslash I)$
$\displaystyle=\\{m+k_{2}+1,\ldots,m+n\\}.$

Let $w_{I}\in G$ be the corresponding permutation matrix and define

$G_{I}:=w_{I}G_{I_{0}}w_{I}^{-1},\quad S_{I}:=w_{I}S_{I_{0}}w_{I}^{-1},\quad
U_{I}:=w_{I}U_{I_{0}}w_{I}^{-1},$

and $G_{I}=S_{I}\ltimes U_{I}$. The orbits

$X_{I}:=G_{I}\Gamma\subset X$

are finite volume homogeneous spaces. We denote by $\mu_{I}$ the unique
$G_{I}$-invariant probability measure on $X_{I}$. For instance, when
$I=\\{1,\ldots,m+n\\}$, we have $X_{I}=X$.

We recall that the space

$Y=U\Gamma\simeq\mathbb{R}^{mn}/\mathbb{Z}^{mn}$

is (canonically) isomorphic to a torus, and under this isomorphism the measure
$\nu$ corresponds to the Lebesgue probability measure. Let $\Xi_{Y}$ denote
the set of continuous characters of $Y$ (viewed as a torus). For $f\in
L^{1}(Y)$, we write

$\widehat{f}(\xi):=\int_{Y}f\cdot\bar{\xi}\,d\nu.$

The _Wiener norm_ of $f$ is defined by

$\|f\|_{W}:=\sum_{\xi\in\Xi_{Y}}|\widehat{f}(\xi)|.$ (2.3)

When $\|f\|_{W}<\infty$, one has a uniformly convergent expansion

$f=\sum_{\xi\in\Xi_{Y}}\widehat{f}(\xi)\xi.$ (2.4)

We recall that for $f\in C^{\infty}(Y)$,

$\|f\|_{C^{0}}\leq\|f\|_{W}\leq\|f\|_{C^{1}},$ (2.5)

and for $f_{1},f_{2}\in C^{\infty}(Y)$,

$\|f\|_{W}\leq\|f_{1}\|_{W}\cdot\|f_{2}\|_{W}.$

In particular, it follows that for $\phi\in C_{c}^{\infty}(X)$,

$\big{\|}\phi|_{Y}\big{\|}_{W}\leq\|\phi\|_{C^{1}}.$

##  3\. Mixing and equidistribution in intermediate homogeneous spaces

Let $I\subset\\{1,\ldots,m+n\\}$ be an admissible subset. If $s\in G$, we
write $\|s\|$ the operator norm of $s$ with respect to the
$\ell^{\infty}$-norm of $\mathbb{R}^{m+n}$.

######  Theorem 3.1.

The action of $S_{I}$ on $X_{I}$ is exponentially mixing, that is, there exist
$\ell\geq 1$ and $\tau>0$ such that for every $\varphi,\psi\in
C_{c}^{\infty}(X_{I})$ and $s\in S_{I}$,

$\int_{X_{I}}(\varphi\circ
s)\cdot\psi\,d\mu_{I}=\left(\int_{X_{I}}\varphi\,d\mu_{I}\right)\left(\int_{X_{I}}\psi\,d\mu_{I}\right)+O\Big{(}\max\big{(}1,\|s\|\big{)}^{-\tau}\,\|\varphi\|_{L^{2}_{\ell}}\|\psi\|_{L^{2}_{\ell}}\Big{)}.$

###### Proof.

Without loss of generality, we may assume that $I$ is given as in (2.2). In
particular, $S_{I}\simeq\hbox{SL}_{k_{1}+k_{2}}(\mathbb{R})$. The exponential
mixing property will follow once we show that the action of $S_{I}$ on $X_{I}$
has the spectral gap property, namely, that the unitary representation of
$S_{I}$ on $L_{0}^{2}(X_{I})$, the space of $L^{2}$-integrable functions with
integral zero, is isolated from the trivial representation (see, for instance,
[12, Sec. 2.4]).

We observe that the space of $U_{I}$-invariant functions
$L^{2}(X_{I})^{U_{I}}$ is invariant under $G_{I}$ and

$L_{0}^{2}(X_{I})^{U_{I}}\simeq
L_{0}^{2}\big{(}\hbox{SL}_{k_{1}+k_{2}}(\mathbb{R})/\hbox{SL}_{k_{1}+k_{2}}(\mathbb{Z})\big{)}$

as $S_{I}$-spaces. The spectral gap property on the latter space is well-
known. Therefore, it is sufficient to verify the spectral gap property on the
orthogonal complement $\mathcal{H}_{I}$ of $L_{0}^{2}(X_{I})^{U_{I}}$. We
observe that

$U_{I}=U^{(1)}_{I}U^{(2)}_{I}U^{(3)}_{I},$

where

$\displaystyle U^{(1)}_{I}$
$\displaystyle:=\left(\begin{tabular}[]{ccc}$I_{m-k_{1}}$&$*$&$0$\\\
$0$&$I_{k_{1}+k_{2}}$&$0$\\\ $0$&$0$&$I_{n-k_{2}}$\end{tabular}\right),$
$\displaystyle U^{(2)}_{I}$
$\displaystyle:=\left(\begin{tabular}[]{ccc}$I_{m-k_{1}}$&$0$&$0$\\\
$0$&$I_{k_{1}+k_{2}}$&$*$\\\ $0$&$0$&$I_{n-k_{2}}$\end{tabular}\right),$
$\displaystyle U^{(3)}_{I}$
$\displaystyle:=\left(\begin{tabular}[]{ccc}$I_{m-k_{1}}$&$0$&$*$\\\
$0$&$I_{k_{1}+k_{2}}$&$0$\\\ $0$&$0$&$I_{n-k_{2}}$\end{tabular}\right).$

We note that $U_{I}$ is generated by $U^{(1)}_{I}$ and $U^{(2)}_{I}$.

Let $V_{i}$, $i=1\ldots,m-k_{1}$, denote the subgroups $U^{(1)}_{I}$
corresponding to the $i$-th row in
$\hbox{M}_{m-k_{1},k_{1}+k_{2}}(\mathbb{R})$ and $W_{j}$, $j=m+k_{2}+1,m+n$,
the subgroup of $U^{(2)}_{I}$ corresponding to the $j$-th column in
$\hbox{M}_{k_{1}+k_{2},n-k_{2}}(\mathbb{R})$. We note that

$S_{I}V_{i}\simeq\hbox{SL}_{k_{1}+k_{2}}(\mathbb{R})\ltimes\mathbb{R}^{k_{1}+k_{2}}\quad\hbox{and}\quad
S_{I}W_{j}\simeq\hbox{SL}_{k_{1}+k_{2}}(\mathbb{R})\ltimes\mathbb{R}^{k_{1}+k_{2}}.$

When $k_{1}+k_{2}\geq 3$, the group
$S_{I}\simeq\hbox{SL}_{k_{1}+k_{2}}(\mathbb{R})$ has Kazhdan property (T), so
it is sufficient to check that $\mathcal{H}_{I}$ contains no non-zero
$S_{I}$-invariant vectors. Suppose that such a vector exists. We apply Mautner
Lemma (see, for instance, [2, Lemma 1.4.8]) to the representations of
$S_{I}V_{i}$ and $S_{I}W_{j}$, and deduce that this vector is also invariant
under all $V_{i}$’s and $W_{j}$’s. Hence, it is invariant under $U^{(1)}_{I}$
and $U^{(2)}_{I}$, so that it is $U_{I}$-invariant and hence is zero because
$\mathcal{H}_{I}$ is the orthogonal complement of $L_{0}^{2}(X_{I})^{U_{I}}$.
This verifies that $\mathcal{H}_{I}$ contains no non-zero $S_{I}$-invariant
vectors and implies the spectral gap property.

Now suppose that $k_{1}+k_{2}=2$, that is,
$S_{I}\simeq\hbox{SL}_{2}(\mathbb{R})$. In this case, we use that for any
unitary representation of $\hbox{SL}_{2}(\mathbb{R})\ltimes\mathbb{R}^{2}$
without $\mathbb{R}^{2}$-invariant vectors, its restriction to
$\hbox{SL}_{2}(\mathbb{R})$ has the spectral gap property (see [2, Corollary
1.4.13]). In particular, it follows that the action of $S_{I}$ on
$(\mathcal{H}^{V_{i}})^{\perp}$ and on $(\mathcal{H}^{W_{j}})^{\perp}$ has the
spectral gap property, so that the action of $S_{I}$ on

$\mathcal{H}^{0}_{I}:=\big{(}\oplus_{i}(\mathcal{H}_{I}^{V_{i}})^{\perp}\big{)}\oplus\big{(}\oplus_{j}(\mathcal{H}_{I}^{W_{j}})^{\perp}\big{)}$

has the spectral gap property. Moreover,

$(\mathcal{H}^{0}_{I})^{\perp}=\big{(}\cap_{i}\mathcal{H}_{I}^{V_{i}}\big{)}\cap\big{(}\cap_{j}\mathcal{H}_{I}^{W_{j}}\big{)}=\mathcal{H}_{I}^{U_{I}^{(1)}}\cap\mathcal{H}_{I}^{U_{I}^{(2)}}=\mathcal{H}_{I}^{U_{I}}=0.$

This completes the proof. ∎

Using the above results, and following the proof in [5, Lemma 4.3], we get the
following corollary:

######  Corollary 3.2.

There exist $\ell\geq 1$ and $\tau>0$ such that for every $\phi,\psi\in
C_{c}^{\infty}(X_{I})$, $u_{1},u_{2}\in\mathbb{R}$, and $w\in\hbox{\rm
Lie}(U\cap S_{I})$,

$\displaystyle\int_{X_{I}}(\phi\circ\exp(u_{1}w))\cdot(\psi\circ\exp(u_{2}w))\,d\mu_{I}=$
$\displaystyle\left(\int_{X_{I}}\phi\,d\mu_{I}\right)\left(\int_{X_{I}}\psi\,d\mu_{I}\right)$
$\displaystyle+O\left(\max\big{(}1,|u_{1}-u_{2}|\|w\|\big{)}^{-\tau}\|\phi\|_{L^{2}_{\ell}}\|\psi\|_{L^{2}_{\ell}}\right).$

### General version of Theorem 1.1

Given an admissible $I\subset\\{1,\ldots,m+n\\}$, we set

$A^{+}_{I}=\big{\\{}t\in\mathbb{R}^{d}_{+}\,:\,t_{j}=0,\enskip\textrm{for all
$j\notin I$}\big{\\}}.$

For $I\subset\\{1,\ldots,m+n\\}$ and $t\in A^{+}$, we define

$\lfloor t\rfloor_{I}:=\min_{i\in I}t_{i}.$

The following theorem is a straightforward generalization of Theorem 1.1.

######  Theorem 3.3.

Let $I\subset\\{1,\ldots,m+n\\}$ be admissible subset. Then there exist
$\ell\geq 1$ and $\delta>0$ such that for every $f\in C^{\infty}(Y)$,
$\varphi\in C_{c}^{\infty}(X_{I})$, and $t\in A_{I}^{+}$,

$\int_{Y}f\cdot(\varphi\circ
a(t))\,d\nu=\left(\int_{Y}f\,d\nu\right)\left(\int_{X_{I}}\varphi\,d\mu_{I}\right)+O\Big{(}e^{-\delta\lfloor
t\rfloor_{I}}\|f\|_{C^{\ell}}\mathcal{N}_{\ell}(\varphi)\Big{)},$

where
$\mathcal{N}_{\ell}(\varphi):=\max\big{(}\|\varphi\|_{C^{0}},\|\varphi\|_{Lip},\|\varphi\|_{L^{2}_{\ell}}\big{)}.$

###### Proof.

The proof can be completed as in [14] (explicit dependence of the functions in
this argument was worked out in [4, Sec. 2]). Without loss of generality, we
may assume that $I$ is given as in (2.2). We also note that the space $X_{I}$
is a bundle

$\pi_{I}:X_{I}\to S_{I}/(S_{I}\cap\Gamma)$

with the base

$S_{I}/(S_{I}\cap\Gamma)\simeq\hbox{SL}_{k_{1}+k_{2}}(\mathbb{R})/\hbox{SL}_{k_{1}+k_{2}}(\mathbb{Z})$

and compact fibers isomorphic to $U_{I}/s(U_{I}\cap\Gamma)s^{-1}$, $s\in
S_{I}$. We view the base as the space of unimodular lattices in
$\mathbb{R}^{k_{1}+k_{2}}$ and define an exhaustion of $X_{I}$ by compact
subsets

$K_{\varepsilon}:=\\{x\in X_{I}:\,\|v\|\geq\varepsilon\quad\hbox{for all non-
zero $v\in\pi_{I}(x)$}\\}.$

Then one has the non-divergence estimate as [14, Corollary 3.4] for the
translates $a(t)Ux$ with $t\in A^{+}_{I}$ when $\lfloor t\rfloor_{I}$ is
sufficiently large. It is also clear that for $x\in K_{\varepsilon}$, the
injectivity radius $\iota(x)$ of $x$ can be estimated as in [14, Proposition
3.5]. Let us also consider a one-parameter sub-semigroup of $A^{+}_{I}$

$g_{t}:=\hbox{diag}\big{(}1,\ldots,1,e^{t/k_{1}},\ldots,e^{t/k_{1}},e^{-t/k_{2}},\ldots,e^{-t/k_{2}},1,\ldots,1\big{)}.$

We observe that

$\hbox{Lie}(G_{I})=L\oplus\hbox{Lie}(U_{I}),$

where $L$ is the subspace spanned by matrices

$\displaystyle E_{ij},\quad 1\leq i\leq m+k_{2},\;m-k_{1}+1\leq j\leq m,$
$\displaystyle E_{ij},\quad m+1\leq i\leq m+k_{2},\;m+1\leq j\leq m+n.$

One easily verifies that the adjoint action
$\hbox{Ad}(g_{t})X=g_{t}Xg_{t}^{-1}$ is non-expanding on $L$. Moreover,
according to Theorem 3.1, there exist $\ell\geq 1$ and $\delta^{\prime}>0$
such that for every $\varphi,\psi\in C_{c}^{\infty}(X_{I}),$ and $t\geq 0$,

$\int_{X_{I}}(\varphi\circ
g_{t})\cdot\psi\,d\mu_{I}=\left(\int_{X_{I}}\varphi\,d\mu_{I}\right)\left(\int_{X_{I}}\psi\,d\mu_{I}\right)+O\Big{(}e^{-\delta^{\prime}t}\,\|\varphi\|_{L^{2}_{\ell}}\|\psi\|_{L^{2}_{\ell}}\Big{)}.$

Now we can apply the argument as in [14, Theorem 2.3] to derive quantitative
equidistribution of the leaves $g_{t}Ux$: there exists $c,\gamma>0$ such that
for every $F\in C_{c}^{\infty}(U_{I})$ with
$\hbox{dist}(\operatorname{supp}(F),e)\leq\rho$, $x\in X_{I}$ with
$\iota(x)>2\rho$, $\varphi\in C_{c}^{\infty}(X_{I}),$ and $t\geq 0$,

$\displaystyle\int_{U_{I}}F(u)\varphi(g_{t}ux)\,du=$
$\displaystyle\left(\int_{U_{I}}F(u)\,du\right)\left(\int_{X_{I}}\varphi\,d\mu_{I}\right)$
$\displaystyle\quad\quad\quad+O\Big{(}\rho\|F\|_{L^{1}}\|\varphi\|_{Lip}+\rho^{-c}e^{-\gamma
t}\|F\|_{C^{\ell}}\|\varphi\|_{L^{2}_{\ell}}\Big{)}.$

With those ingredients the proof can be completed as in [14]. ∎

##  4\. Decorrelation estimates

One of the challenges of our present argument is that the decorrelation
estimate in Corollary 3.2 only holds for one-parameter subgroups in directions
$w\in\hbox{Lie}(U\cap S_{I})$. Nonetheless, we establish that certain
”partial” decorrelation estimates also hold for other directions. A basic
prototype of this phenomenon is the following elementary result for circle
rotations:

######  Lemma 4.1.

For every $\phi,\psi\in L^{2}(\mathbb{R}/\mathbb{Z})$ and $L\geq 1$,

$\displaystyle\frac{1}{L^{2}}\int_{0}^{L}\int_{0}^{L}\int_{\mathbb{R}/\mathbb{Z}}\phi(x+s_{1})\overline{\psi(x+s_{2})}\,dx\,ds_{1}ds_{2}=$
$\displaystyle\left(\int_{\mathbb{R}/\mathbb{Z}}\phi(x)\,dx\right)\left(\int_{\mathbb{R}/\mathbb{Z}}\overline{\psi(x)}\,dx\right)$
$\displaystyle\quad\quad+O\Big{(}L^{-2}\|\phi\|_{L^{2}}\|\psi\|_{L^{2}}\Big{)}.$

###### Proof.

We have the Fourier expansions:

$\phi(x)=\sum_{n\in\mathbb{Z}}\widehat{\phi}(n)e^{2\pi
inx}\quad\hbox{and}\quad\psi(x)=\sum_{n\in\mathbb{Z}}\widehat{\psi}(n)e^{2\pi
inx},$

so that

$\int_{\mathbb{R}/\mathbb{Z}}\phi(x+s_{1})\overline{\psi(x+s_{2})}\,dx=\sum_{n\in\mathbb{Z}}\widehat{\phi}(n)\overline{\widehat{\psi}(n)}e^{2\pi
i(s_{1}-s_{2})n}.$

Since for $n\neq 0$,

$\frac{1}{L^{2}}\int_{0}^{L}\int_{0}^{L}e^{2\pi
i(s_{1}-s_{2})n}\,ds_{1}ds_{2}\ll n^{-2}L^{-2},$

it follows that from the Plancherel Formula that

$\displaystyle\frac{1}{L^{2}}\int_{0}^{L}\int_{0}^{L}\int_{\mathbb{R}/\mathbb{Z}}\phi(x+s_{1})\overline{\psi(x+s_{2})}\,dx\,ds_{1}ds_{2}=$
$\displaystyle\widehat{\phi}(0)\overline{\widehat{\psi}(0)}+O(L^{-2})\|\phi\|_{L^{2}}\|\psi\|_{L^{2}},$

which proves the lemma. ∎

###  4.1. Mean decorrelation for flows - special case

We will need a version of Lemma 4.1 for flows on the space of affine lattices.
Let

$\mathcal{L}:=H/\Delta\quad\quad\hbox{with
$H:=\hbox{SL}_{d}(\mathbb{R})\ltimes\mathbb{R}^{d}$ and
$\Delta:=\hbox{SL}_{d}(\mathbb{Z})\ltimes\mathbb{Z}^{d}$}$ (4.1)

with $d\geq 2$. We observe that $\mathcal{L}$ can be identified with the space
of affine unimodular lattices in $\mathbb{R}^{d}$:

$\mathcal{L}\simeq\big{\\{}\Lambda+x:\,\hbox{$\Lambda$ -- unimodular lattice
in $\mathbb{R}^{d}$},\;x\in\mathbb{R}^{d}/\Lambda\big{\\}}.$

We denote by

$\mathcal{L}^{\star}:=\hbox{SL}_{d}(\mathbb{R})/\hbox{SL}_{d}(\mathbb{Z})$

the space of unimodular lattices, and we have the natural projection map

$\pi:\mathcal{L}\to\mathcal{L}^{\star}:\Lambda+x\mapsto\Lambda,$

whose fibers are tori $\pi^{-1}(\Lambda)\simeq\mathbb{R}^{d}/\Lambda$. For
$z\in\mathcal{L}^{\star}$, we denote $\tau_{z}$ the probability invariant
measure supported on the torus $\pi^{-1}(z)$. Then the probability invariant
measure on $\mathcal{L}$ can be written as

$\mu=\int_{\mathcal{L}^{\star}}\tau_{z}\,d\mu^{\star}(z),$

where $\mu^{\star}$ denotes the invariant probability measure on
$\mathcal{L}^{\star}$. For $\phi\in C_{c}(\mathcal{L})$, we define
$\phi^{\star}\in C_{c}(\mathcal{L}^{\star})$ by

$\phi^{\star}(\Lambda):=\int_{\mathbb{R}^{d}/\Lambda}\phi(y+\Lambda)\,d\tau_{\Lambda}(y+\Lambda).$

We think of $\phi^{\star}$ as a function on $\mathcal{L}$ which is constant on
the fibers. Finally, given a non-zero $w\in\mathbb{R}^{d}$, we get a one-
parameter flow $\sigma_{w}:\mathbb{R}\rightarrow H$ by

$\sigma_{w}(u)=(0,uw)\in H,\quad u\in\mathbb{R}.$

######  Theorem 4.2.

For every $w\in\mathbb{R}^{d}\setminus\\{0\\}$, and for all $\phi,\psi\in
C_{c}^{\infty}(\mathcal{L})$ and $L>0$,

$\displaystyle\frac{1}{L^{2}}\int_{0}^{L}\int_{0}^{L}\int_{\mathcal{L}}(\phi\circ\sigma_{w}(u_{1}))\cdot$
$\displaystyle(\bar{\psi}\circ\sigma_{w}(u_{2}))\,d\mu\,du_{1}du_{2}$
$\displaystyle=$
$\displaystyle\int_{\mathcal{L}}(\phi^{\star}\cdot{\bar{\psi}^{\star}})\,d\mu+O\Big{(}\max\big{(}1,L\|w\|\big{)}^{-2/3}\|\phi\|_{C^{\ell}}\|\psi\|_{C^{\ell}}\Big{)}$

for sufficiently large $\ell$.

###### Proof.

Without loss of generality, we may assume that $\|w\|=1$. For a lattice
$\Lambda$ in $\mathcal{L}^{\star}$, we write

$\phi^{\Lambda}(x):=\phi(\Lambda+x)\quad\hbox{for
$x\in\mathbb{R}^{d}/\Lambda$}.$

Then

$(\phi\circ\sigma_{w}(u))^{\Lambda}(x)=\phi^{\Lambda}(x+uw)\quad\hbox{and}\quad(\psi\circ\sigma_{w}(u))^{\Lambda}(x)=\psi^{\Lambda}(x+uw).$

The functions $\phi^{\Lambda}$ and $\psi^{\Lambda}$ have the Fourier
expansions:

$\phi^{\Lambda}(x)=\sum_{\lambda\in\Lambda^{\perp}}\widehat{\phi}^{\Lambda}(\lambda)e^{2\pi
i\left<x,\lambda\right>}\quad\hbox{and}\quad\psi^{\Lambda}(x)=\sum_{\lambda\in\Lambda^{\perp}}\widehat{\psi}^{\Lambda}(\lambda)e^{2\pi
i\left<x,\lambda\right>},$

where $\Lambda^{\perp}$ denotes the dual lattice for $\Lambda$. Since $\phi$
and $\psi$ are smooth, the series converge uniformly, and thus

$\displaystyle\int_{\mathcal{L}}\phi(\sigma(u_{1})z){\bar{\psi}(\sigma(u_{2})z)}\,d\mu(z)=\int_{\mathcal{L}^{\star}}\int_{\mathbb{R}^{d}/\Lambda}\phi^{\Lambda}(x+u_{1}w){\bar{\psi}^{\Lambda}(x+u_{2}w)}dx\,d\mu^{\star}(\Lambda),$

and

$\int_{\mathbb{R}^{d}/\Lambda}\phi^{\Lambda}(x+u_{1}w){\bar{\psi}^{\Lambda}(x+u_{2}w)}dx=\sum_{\lambda\in\Lambda^{\perp}}\widehat{\phi}^{\Lambda}(\lambda)\overline{\widehat{\psi}^{\Lambda}(\lambda)}e^{2\pi
i(u_{1}-u_{2})\left<w,\lambda\right>}.$

Since

$\frac{1}{L^{2}}\int_{0}^{L}\int_{0}^{L}e^{2\pi
i(u_{1}-u_{2})\left<w,\lambda\right>}\,du_{1}du_{2}=\omega\big{(}\pi
L\left<w,\lambda\right>\big{)},$

where $\omega(t)=\frac{\sin^{2}(t)}{t^{2}}$ for $t\neq 0$ and $\omega(0)=1$,
we conclude that

$\displaystyle\frac{1}{L^{2}}\int_{0}^{L}\int_{0}^{L}\int_{\mathbb{R}^{d}/\Lambda}\phi^{\Lambda}(x+u_{1}w){\bar{\psi}^{\Lambda}(x+u_{2}w)}dx\,du_{1}du_{2}$
$\displaystyle=$
$\displaystyle\sum_{\lambda\in\Lambda^{\perp}}\widehat{\phi}^{\Lambda}(\lambda)\overline{\widehat{\psi}^{\Lambda}(\lambda)}\omega\big{(}\pi
L\left<w,\lambda\right>\big{)}$ $\displaystyle=$
$\displaystyle\phi^{\star}(\Lambda)\bar{\psi}^{\star}(\Lambda)+\sum_{\lambda\in\Lambda^{\perp}\backslash\\{0\\}}\widehat{\phi}^{\Lambda}(\lambda)\overline{\widehat{\psi}^{\Lambda}(\lambda)}\omega\big{(}\pi
L\left<w,\lambda\right>\big{)}.$

Now it remains to estimate the integral of the last sum. Since for
$\phi,\psi\in C_{c}^{\infty}(\mathcal{L})$,

$\big{|}\widehat{\phi}^{\Lambda}(\lambda)\big{|}\ll_{\ell}\max\big{(}1,\|\lambda\|\big{)}^{-\ell}\,\|\phi\|_{C^{\ell}}\quad\hbox{and}\quad\big{|}\widehat{\psi}^{\Lambda}(\lambda)\big{|}\ll_{\ell}\max\big{(}1,\|\lambda\|\big{)}^{-\ell}\,\|\psi\|_{C^{\ell}}$
(4.2)

for every $\ell\geq 1$. This leads to the estimate

$\displaystyle\frac{1}{L^{2}}\int_{0}^{L}\int_{0}^{L}\int_{\mathcal{L}}\phi(\sigma(u_{1})z){\bar{\psi}(\sigma(u_{2})z)}\,d\mu(z)\,du_{1}du_{2}=$
$\displaystyle\int_{\mathcal{L}^{\star}}(\phi^{\star}\cdot\bar{\psi^{\star}})\,d\mu^{\star}$
$\displaystyle\quad\quad+O\Big{(}\Omega(L)\|\phi\|_{C^{\ell}}\|\psi\|_{C^{\ell}}\Big{)},$

where

$\Omega(L):=\int_{\mathcal{L}^{\star}}\left(\sum_{\lambda\in\Lambda^{\perp}\backslash\\{0\\}}\max\big{(}1,\|\lambda\|\big{)}^{-2\ell}\omega\big{(}\pi
L\left<w,\lambda\right>\big{)}\right)\,d\mu^{\star}(\Lambda).$

Since $\phi^{\star}\cdot\bar{\psi^{\star}}$ is constant on the fibers,

$\int_{\mathcal{L}^{\star}}(\phi^{\star}\cdot\bar{\psi^{\star}})\,d\mu^{\star}=\int_{\mathcal{L}}(\phi^{\star}\cdot\bar{\psi^{\star}})\,d\mu.$

By Siegel’s Formula [17],

$\Omega(L)=\int_{\mathbb{R}^{d}}\max\big{(}1,\|x\|\big{)}^{-2\ell}\omega\big{(}\pi
L\left<w,x\right>\big{)}\,dx.$

Applying an orthogonal change of variables, we may assume that $w=e_{1}$. For
$C\geq 1$, we obtain

$\displaystyle\int_{\mathbb{R}^{d}}\max\big{(}1,\|x\|\big{)}^{-2\ell}\omega\big{(}\pi
Lx_{1}\big{)}\,dx$ $\displaystyle\leq\int_{|x_{1}|\leq
C^{-1}}\max\big{(}1,\|x\|\big{)}^{-2\ell}\,dx$
$\displaystyle\quad\quad+\frac{C^{2}}{\pi^{2}L^{2}}\int_{|x_{1}|\geq
C^{-1}}\max\big{(}1,\|x\|\big{)}^{-2\ell}\,dx$
$\displaystyle\ll_{\ell}\,C^{-1}+L^{-2}C^{2},$

provided that $\ell$ is sufficiently large. Taking $C=L^{2/3}$, we obtain the
stated estimate. ∎

###  4.2. Mean decorrelation for flows - general version

Let us now extend the previous result to more general homogeneous spaces. Let
us fix $d\geq 2$ and $d^{\prime}\geq 1$ and set

$H:=\left(\begin{tabular}[]{ccc}1&$\hbox{Mat}_{1,d}(\mathbb{R})$&$\hbox{Mat}_{1,d^{\prime}}(\mathbb{R})$\\\
0&$\hbox{SL}_{d}(\mathbb{R})$&$\hbox{Mat}_{d,d^{\prime}}(\mathbb{R})$\\\
0&0&$I_{d^{\prime}}$\end{tabular}\right)\quad\hbox{and}\quad\Delta:=H\cap\hbox{SL}_{d+d^{\prime}+1}(\mathbb{Z}).$
(4.3)

We consider the homogeneous space

$Z:=H/\Delta.$ (4.4)

We note that the space $\mathcal{L}$ of affine unimodular lattices, which we
introduced above, can be considered as a degenerate case of this definition
with $d^{\prime}=0$.

As above we have the natural projection map

$\pi:Z\to Z^{\star},$

where $Z^{\star}$ denotes the homogeneous space of $(d+d^{\prime})$-matrices
obtained by removing the first row and the first column. The space $Z^{\star}$
can be viewed as the spaces of tuples

$(\Lambda,v_{1},\ldots,v_{d^{\prime}}),$

where $\Lambda$ is a unimodular lattice in $\mathbb{R}^{d}$, and
$v=(v_{1},\ldots,v_{d^{\prime}})\in(\mathbb{R}^{d}/\Lambda)^{d^{\prime}}$. The
fibers of this map are tori given by

$\pi^{-1}(\Lambda,v)\simeq\mathbb{R}^{d+d^{\prime}}/\Lambda_{v},$

where

$\Lambda_{v}:=\big{\\{}(\delta,m_{1}-\delta\cdot
v_{1},\ldots,m_{d^{\prime}}-\delta\cdot
v_{d^{\prime}}):\,\delta\in\Lambda^{\perp},m\in\mathbb{Z}^{d^{\prime}}\big{\\}}.$

We note that the dual lattice is given by

$\Lambda_{v}^{\perp}=\big{\\{}(\delta+m_{1}v_{1}+\cdots+m_{d^{\prime}}v_{d^{\prime}},m):\,\delta\in\Lambda,m\in\mathbb{Z}^{d^{\prime}}\big{\\}}.$

For $z\in Z^{\star}$, we denote $\tau_{z}$ the probability invariant measure
supported on the torus $\pi^{-1}(z)$. Then the probability invariant measure
on $Z$ is

$\mu=\int_{Z}\tau_{z}\,d\mu^{\star}(z),$

where $\mu^{\star}$ denotes the invariant probability measure on $Z^{\star}$.
For $\phi\in C_{c}(Z)$, we define $\phi^{\star}\in C_{c}(Z^{\star})$ by

$\phi^{\star}(z):=\int_{Z}\phi\,d\tau_{z}.$

We think of $\phi^{\star}$ as a function on $Z$ which is constant on the
fibers. Finally, given a non-zero $w\in\textrm{Mat}_{1,d}(\mathbb{R})$, we get
a one-parameter flow $\sigma_{w}:\mathbb{R}\rightarrow H$ by

$\sigma_{w}(u)=\left(\begin{tabular}[]{ccc}1&$uw$&0\\\ 0&$I_{d}$&$0$\\\
0&0&$I_{d^{\prime}}$\end{tabular}\right),\quad\textrm{for $u\in\mathbb{R}$}.$

######  Theorem 4.3.

For every $w\in\textrm{Mat}_{1,d}(\mathbb{R})\setminus\\{0\\}$ and for all
$\phi,\psi\in C_{c}^{\infty}(Z)$ and $L>0$,

$\displaystyle\frac{1}{L^{2}}\int_{0}^{L}\int_{0}^{L}\int_{Z}(\phi\circ\sigma_{w}(u_{1}))\cdot$
$\displaystyle(\bar{\psi}\circ\sigma_{w}(u_{2}))\,d\mu\,du_{1}du_{2}$
$\displaystyle=$
$\displaystyle\int_{Z}(\phi^{\star}\cdot{\bar{\psi}^{\star}})\,d\mu+O\Big{(}\max\big{(}1,L\|w\|\big{)}^{-2/3}\|\phi\|_{C^{\ell}}\|\psi\|_{C^{\ell}}\Big{)}$

for sufficiently large $\ell$.

###### Proof.

For a tuple $(\Lambda,v)$ in $Z^{\star}$, we write

$\phi^{(\Lambda,v)}(x,y)=\phi|_{\pi^{-1}(\Lambda,v)}(x,y)\quad\hbox{for
$(x,y)\in(\mathbb{R}^{d}\times\mathbb{R}^{d^{\prime}})/\Lambda_{v}$}.$

Then

$(\phi\circ\sigma_{w}(u))^{(\Lambda,v)}(x,y)=\phi^{(\Lambda,v)}(x+uw,y)\quad\hbox{and}\quad(\psi\circ\sigma_{w}(u))^{(\Lambda,v)}(x,y)=\psi^{(\Lambda,v)}(x+uw,y).$

The function $\phi^{(\Lambda,v)}$ and $\psi^{(\Lambda,v)}$ have the Fourier
expansions:

$\phi^{(\Lambda,v)}(x,y)=\sum_{\lambda\in\Lambda_{v}^{\perp}}\widehat{\phi}^{(\Lambda,v)}(\lambda)e^{2\pi
i\left<(x,y),\lambda\right>}\quad\hbox{and}\quad\psi^{(\Lambda,v)}(x,y)=\sum_{\lambda\in\Lambda_{v}^{\perp}}\widehat{\psi}^{(\Lambda,v)}(\lambda)e^{2\pi
i\left<(x,y),\lambda\right>}.$

Since $\phi$ and $\psi$ are smooth, the series converge uniformly. So we
obtain

$\displaystyle\int_{Z}\phi(\sigma_{w}(u_{1})z){\bar{\psi}(\sigma_{w}(u_{2})z)}\,d\mu(z)$
$\displaystyle=$
$\displaystyle\int_{Z^{\star}}\int_{(\mathbb{R}^{d}\times\mathbb{R}^{d^{\prime}})/\Lambda_{v}}\phi^{(\Lambda,v)}(x+u_{1}w,y){\bar{\psi}^{(\Lambda,v)}(x+u_{2}w,y)}dxdy\,d\mu^{\star}(\Lambda,v),$

and

$\displaystyle\int_{(\mathbb{R}^{d}\times\mathbb{R}^{d^{\prime}})/\Lambda_{v}}\phi^{(\Lambda,v)}(x+u_{1}w){\bar{\psi}^{(\Lambda,v)}(x+u_{2}w)}\,$
$\displaystyle dxdy$ $\displaystyle=$
$\displaystyle\sum_{\lambda\in\Lambda_{v}^{\perp}}\widehat{\phi}^{(\Lambda,v)}(\lambda)\overline{\widehat{\psi}^{(\Lambda,v)}(\lambda)}e^{2\pi
i(u_{1}-u_{2})\left<(w,0),\lambda\right>}.$

Since

$\frac{1}{L^{2}}\int_{0}^{L}\int_{0}^{L}e^{2\pi
i(u_{1}-u_{2})\left<(w,0),\lambda\right>}\,du_{1}du_{2}=\omega\big{(}\pi
L\left<(w,0),\lambda\right>\big{)},$

we conclude that

$\displaystyle\frac{1}{L^{2}}\int_{0}^{L}\int_{0}^{L}\int_{(\mathbb{R}^{d}\times\mathbb{R}^{d^{\prime}})/\Lambda_{v}}\phi^{(\Lambda,v)}(x+u_{1}w,y){\bar{\psi}^{(\Lambda,v)}(x+u_{2}w,y)}dxdy\,du_{1}du_{2}$
$\displaystyle=$
$\displaystyle\sum_{\lambda\in\Lambda_{v}^{\perp}}\widehat{\phi}^{(\Lambda,v)}(\lambda)\overline{\widehat{\psi}^{(\Lambda,v)}(\lambda)}\omega\big{(}\pi
L\left<(w,0),\lambda\right>\big{)}$ $\displaystyle=$
$\displaystyle\phi^{\star}(\Lambda,v)\bar{\psi}^{\star}(\Lambda,v)+\sum_{\lambda\in\Lambda_{v}^{\perp}\backslash\\{0\\}}\widehat{\phi}^{(\Lambda,v)}(\lambda)\overline{\widehat{\psi}^{(\Lambda,v)}(\lambda)}\omega\big{(}\pi
L\left<(w,0),\lambda\right>\big{)}.$

Now it remains to estimate the integral of the last sum. Since the
$\phi,\psi\in C_{c}^{\infty}(Z)$,

$\big{|}\widehat{\phi}^{(\Lambda,v)}(\lambda)\big{|}\ll_{\ell}\max\big{(}1,\|\lambda\|\big{)}^{-\ell}\,\|\phi\|_{C^{\ell}}\quad\hbox{and}\quad\big{|}\widehat{\psi}^{(\Lambda,v)}(\lambda)\big{|}\ll_{\ell}\max\big{(}1,\|\lambda\|\big{)}^{-\ell}\,\|\psi\|_{C^{\ell}}$
(4.5)

for every $\ell\geq 1$, this leads to the estimate

$\displaystyle\frac{1}{L^{2}}\int_{0}^{L}\int_{0}^{L}\int_{Z}\phi(\sigma_{w}(u_{1})z){\bar{\psi}(\sigma_{w}(u_{2})z)}\,$
$\displaystyle d\mu(z)\,du_{1}du_{2}$ $\displaystyle=$
$\displaystyle\int_{Z^{\star}}(\phi^{\star}\cdot\bar{\psi^{\star}})\,d\mu^{\star}+O\Big{(}\Omega(L)\|\phi\|_{C^{\ell}}\|\psi\|_{C^{\ell}}\Big{)},$

where

$\displaystyle\Omega(L):=$
$\displaystyle\int_{Z^{\star}}\left(\sum_{\lambda\in\Lambda_{v}^{\perp}\backslash\\{0\\}}\max\big{(}1,\|\lambda\|\big{)}^{-2\ell}\omega\big{(}\pi
L\left<(w,0),\lambda\right>\big{)}\right)\,d\mu^{\star}(\Lambda,v)$
$\displaystyle=$
$\displaystyle\int_{Z^{\star}}\left(\sum_{(\delta,m)\in(\Lambda\times\mathbb{Z}^{d^{\prime}})\backslash\\{0\\}}\max\Big{(}1,\Big{\|}(\delta+{\sum}_{i}m_{i}v_{i},m)\Big{\|}\Big{)}^{-2\ell}\right.$
$\displaystyle\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\times\left.\omega\big{(}\pi
L\big{<}w,\delta+{\sum}_{i}m_{i}v_{i}\big{>}\big{)}\right)\,d\mu^{\star}(\Lambda,v).$

Since the part of the sum with $m_{1}=\cdots=m_{d^{\prime}}=0$ has been
handled in the proof of the previous theorem, it remains to estimate

$\displaystyle\Omega_{1}(L)=$
$\displaystyle\int_{Z^{\star}}\left(\sum_{\delta\in\Lambda,m\in\mathbb{Z}^{d^{\prime}}\backslash\\{0\\}}\max\Big{(}1,\Big{\|}(\delta+{\sum}_{i}m_{i}v_{i},m)\Big{\|}\Big{)}^{-2\ell}\right.$
$\displaystyle\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\times\left.\omega\big{(}\pi
L\big{<}w,\delta+{\sum}_{i}m_{i}v_{i}\big{>}\big{)}\right)\,d\mu^{\star}(\Lambda,v).$

Let us suppose, for example, that $m_{1}\neq 0$. Then we consider the change
of variables

$(\mathbb{R}^{d}/\Lambda)^{d^{\prime}}\to(\mathbb{R}^{d}/\Lambda)^{d^{\prime}}:\tilde{v}_{1}=\sum_{i=1}^{d^{\prime}}m_{i}v_{i},\tilde{v}_{2}=v_{2},\ldots,\tilde{v}_{d^{\prime}}=v_{d^{\prime}}.$

We observe that this transformation is measure-preserving because it is a
composition of the maps

$v_{1}\mapsto m_{1}v_{1},v_{2}\mapsto v_{2},\ldots,v_{d^{\prime}}\mapsto
v_{d^{\prime}}\quad\hbox{and}\quad v_{1}\mapsto v_{1}+m_{i}v_{i},v_{2}\mapsto
v_{2},\ldots,v_{d^{\prime}}\mapsto v_{d^{\prime}}.$

Using this the integration reduces to the space $\mathcal{L}$ of affine
lattices in $\mathbb{R}^{d}$. Namely, we obtain

$\displaystyle\Omega_{1}(L)=$
$\displaystyle\int_{\mathcal{L}^{\star}}\left(\sum_{\delta\in\Lambda,m\in\mathbb{Z}^{d^{\prime}}\backslash\\{0\\}}\max\big{(}1,\|(\delta+\tilde{v}_{1},m)\|\big{)}^{-2\ell}\omega\big{(}\pi
L\left<w,\delta+\tilde{v}_{1}\right>\big{)}\right)\,d\mu_{\mathcal{L}}(\Lambda,\tilde{v}_{1}),$

where $\mu_{\mathcal{L}}$ denotes the invariant measure on $\mathcal{L}$.
Applying Siegel’s Formula for the space of affine lattices, we obtain

$\displaystyle\Omega_{1}(L)$ $\displaystyle=\int_{\mathbb{R}^{d}}\sum_{m\neq
0}\max\big{(}1,\|(x,m)\|\big{)}^{-2\ell}\omega\big{(}\pi
L\left<w,x\right>\big{)}\,dx$
$\displaystyle\ll\int_{\mathbb{R}^{d}}\sum_{m\neq
0}\big{(}\|x\|+|m|\big{)}^{-2\ell}\omega\big{(}\pi
L\left<w,x\right>\big{)}\,dx$ $\displaystyle\ll\left(\sum_{m\neq
0}m^{-\ell}\right)\int_{\mathbb{R}^{d}}\max\big{(}1,\|x\|\big{)}^{-\ell}\omega\big{(}\pi
L\left<w,x\right>\big{)}\,dx,$

where we used the Cauchy–Schwarz inequality. Now the estimate can be completed
as in the previous proof. ∎

More generally, we derive a version of the previous result for products of
spaces as above. Let $d_{1},\ldots,d_{k}\in\mathbb{N}$ and
$d^{\prime}_{1},\ldots,d^{\prime}_{k}\in\mathbb{N}\cup\\{0\\}$. When
$d_{s}\geq 2$ and $d_{s}^{\prime}=0$, we define $H_{s}$, $\Delta_{s}$, and
$Z_{s}$ as in (4.3)–(4.4). When $d_{s}\geq 2$ and $d_{s}^{\prime}=0$, we
define $H_{s}$, $\Delta_{s}$, and $Z_{s}$ as in (4.1). When $d_{s}=1$, we set

$H_{s}:=\mathbb{R},\quad\Delta_{s}:=\mathbb{Z},\quad\hbox{and}\quad
Z_{s}:=H_{s}/\Gamma_{s}.$

We recall that when $d_{s}\geq 2$ the spaces $Z_{s}$ are equipped with the
invariant probability measures $\mu_{s}$, and they fiber over the spaces
$Z_{s}^{\star}$ which is equipped with the invariant probability measure
$\mu_{s}^{\star}$. When $d_{s}\geq 2$ and $d^{\prime}_{s}=0$, $Z_{s}^{\star}$
is the space of unimodular lattices $\Lambda_{s}$ in $\mathbb{R}^{d_{s}}$.
When $d_{s}\geq 2$ and $d^{\prime}_{s}\geq 1$, $Z_{s}^{\star}$ is the spaces
of tuples $(\Lambda_{s},v_{s})$, where $\Lambda_{s}$ is a unimodular lattice
in $\mathbb{R}^{d_{s}}$ and
$v_{s}=(v_{s,1},\ldots,v_{s,d_{s}^{\prime}})\in(\mathbb{R}^{d_{s}}/\Lambda)^{d_{s}^{\prime}}$.
To have consistent notation, we write the elements of $Z_{s}^{\star}$ as
$(\Lambda_{s},v_{s})$ in both cases thinking that $v_{s}=0$ when
$d_{s}^{\prime}=0$. When $d_{s}=1$, we view $Z_{s}^{\star}$ as a one-point
space equipped with the Dirac measure $\mu_{s}^{\star}$.

Let

$w_{s}\in\hbox{Mat}_{1,d_{s}}(\mathbb{R})\subset H_{s}$, $s=1,\ldots,k$, with
$\|w_{1}\|\geq\cdots\geq\|w_{k}\|$

and $\sigma_{s}(u)=uw_{s}$ be the corresponding one-parameter subgroups.

######  Theorem 4.4.

There exists $\delta>0$ such that for every $\phi_{s},\psi_{s}\in
C_{c}^{\infty}(Z_{s})$, $s=1,\ldots,k$, and $L>0$,

$\displaystyle\frac{1}{L^{2}}\int_{0}^{L}\int_{0}^{L}$
$\displaystyle\prod_{s=1}^{k}\int_{Z_{s}}(\phi_{s}\circ\sigma_{s}(u_{1}))\cdot(\bar{\psi}_{s}\circ\sigma_{s}(u_{2}))\,d\mu_{s}\,du_{1}du_{2}$
$\displaystyle=$
$\displaystyle\prod_{s=1}^{k}\int_{Z_{s}}(\phi_{s}^{\star}\cdot{\bar{\psi}_{s}^{\star}})\,d\mu_{s}+O\Big{(}\max\big{(}1,L\|w_{k}\|\big{)}^{-\frac{2}{k(k+2)}}{\prod}_{s=1}^{k}\|\phi_{s}\|_{C^{\ell}}\|\psi_{s}\|_{C^{\ell}}\Big{)}$

for sufficiently large $\ell$.

###### Proof.

As in the previous proof, we obtain that

$\displaystyle\int_{(\mathbb{R}^{d_{s}}\times\mathbb{R}^{d_{s}^{\prime}})/\Lambda_{s,v_{s}}}\phi^{(\Lambda_{s},v_{s})}(x_{s}+u_{1}w_{s},y_{s})$
$\displaystyle{\bar{\psi}^{(\Lambda_{s},v_{s})}(x_{s}+u_{2}w_{s},y_{s})}\,dx_{s}dy_{s}$
$\displaystyle=\sum_{\lambda_{s}\in\Lambda_{s,v_{s}}^{\perp}}\widehat{\phi}^{(\Lambda_{s},v_{s})}(\lambda_{s})\overline{\widehat{\psi}^{(\Lambda_{s},v_{s})}(\lambda_{s})}e^{2\pi
i(u_{1}-u_{2})\left<(w_{s},0),\lambda_{s}\right>}.$

Then

$\displaystyle\frac{1}{L^{2}}\int_{0}^{L}\int_{0}^{L}\prod_{s=1}^{k}\int_{(\mathbb{R}^{d_{s}}\times\mathbb{R}^{d_{s}^{\prime}})/\Lambda_{s,v_{s}}}\phi^{(\Lambda_{s},v_{s})}(x_{s}+u_{1}w_{s},y_{s}){\bar{\psi}^{(\Lambda_{s},v_{s})}(x_{s}+u_{2}w_{s},y_{s})}dx_{s}dy_{s}\,du_{1}du_{2}$
$\displaystyle=\sum_{\lambda_{1},\ldots,\lambda_{s}}\prod_{s=1}^{k}\widehat{\phi}^{(\Lambda_{s},v_{s})}(\lambda_{s})\overline{\widehat{\psi}^{(\Lambda_{s},v_{s})}(\lambda_{s})}\,\omega\Big{(}\pi
L{\sum}_{j=1}^{k}\left<(w_{j},0),\lambda_{j}\right>\Big{)},$

where the sum is taken over $\lambda_{1}\in\Lambda_{1,v_{1}}^{\perp}$, …,
$\lambda_{k}\in\Lambda_{k,v_{k}}^{\perp}$. Since the function $\omega$ is
uniformly bounded, in view of estimates (4.5), this sum converges uniformly.
We note that the term with $\lambda_{1}=\cdots=\lambda_{k}=0$ gives

$\prod_{s=1}^{k}\int_{Z_{s}^{\star}}\widehat{\phi}^{\Lambda_{s}}(0)\overline{\widehat{\psi}^{\Lambda_{s}}(0)}\,d\mu_{s}^{\star}(\Lambda_{s},v_{s})=\prod_{s=1}^{k}\int_{Z^{\star}_{s}}(\phi_{s}^{\star}\cdot{\bar{\psi}_{s}^{\star}})\,d\mu_{s}^{\star}=\prod_{s=1}^{k}\int_{Z_{s}}(\phi_{s}^{\star}\cdot{\bar{\psi}_{s}^{\star}})\,d\mu_{s},$

which is the main term in the theorem. Hence, it remains to estimate the
integrals of remaining summands over
$(\lambda_{1},\ldots,\lambda_{k})\neq(0,\ldots,0)$. In fact, we only discuss
the sum over $\lambda_{1},\ldots,\lambda_{k}\neq 0$ because the remaining sums
reduces to similar estimates with fewer number of factors. Now we need to
estimate the expression

$\displaystyle\int_{Z_{1}^{\star}\times\cdots\times
Z_{k}^{\star}}{\sum_{\lambda_{1},\ldots,\lambda_{k}}}^{\\!\\!\\!\\!\\!*}\,\prod_{s=1}^{k}\max$
$\displaystyle\big{(}1,\|\lambda_{s}\|\big{)}^{-2\ell}$
$\displaystyle\times\omega\Big{(}\pi
L{\sum}_{j=1}^{k}\left<(w_{j},0),\lambda_{j}\right>\Big{)}\,d\mu_{1}^{\star}(\Lambda_{1},v_{1})\cdots
d\mu_{k}^{\star}(\Lambda_{k},v_{k}),$

where the sum is taken
$\lambda_{1}\in\Lambda_{1,v_{1}}^{\perp}\backslash\\{0\\},\ldots,\lambda_{k}\in\Lambda_{k,v_{k}}^{\perp}\backslash\\{0\\}$.
More explicitly, when $d_{s}^{\prime}\geq 1$, the parameters $\lambda_{s}$ run
over

$\lambda_{s}=(\delta_{s}+m_{s,1}v_{s,1}+\cdots+m_{s,d^{\prime}}v_{s,d^{\prime}},m_{s})\quad\hbox{with
$\delta_{s}\in\Lambda_{s}$ and $m_{s}\in\mathbb{Z}^{d_{s}^{\prime}}$.}$

One can handle the sum over $\lambda_{s}$ as in the previous proof using a
change of variables in $(v_{s,1},\ldots,v_{s,d_{s}^{\prime}})$ and the Siegel
formula for the affine lattices. When $d_{s}\geq 2$ and $d_{s}^{\prime}=0$,
the parameter $\lambda_{s}$ runs over $\Lambda_{s}$, and one can apply the
Siegel formula for the space of lattices. Ultimately, this gives the integral

$\int_{\mathbb{R}^{d_{1}}\times\cdots\times\mathbb{R}^{d_{k}}}\prod_{s=1}^{k}\max\big{(}1,\|x_{s}\|\big{)}^{-\ell}\,\omega\Big{(}\pi
L{\sum}_{j=1}^{k}\left<w_{j},x_{j}\right>\Big{)}\,d\rho_{1}(x_{1})\cdots
d\rho_{k}(x_{k}),$ (4.6)

where $\rho_{s}$ denotes the Lebesgue measure on $\mathbb{R}^{d_{s}}$ when
$d_{s}\geq 2$ and $\rho_{s}=\sum_{z\in\mathbb{Z}\backslash\\{0\\}}\delta_{z}$
when $d_{s}=1$. We write $x_{j}=(x_{j}^{(1)},\ldots,x_{j}^{(d_{j})})$.
Applying orthogonal changes of variables in (4.6), we may assume that

$\left<w_{j},x_{j}\right>=\|w_{j}\|\cdot|x_{j}^{(1)}|\quad\hbox{for
$j=1,\ldots,k$.}$

When $d_{s}=1$ for some $s$, we observe that for $\max_{j}|x_{j}|\geq 1$,

$\omega\Big{(}\pi
L{\sum}_{j=1}^{k}\|w_{j}\|\cdot|x_{j}|\Big{)}\ll(L\|w_{k}\|)^{-2}.$

This also implies the bound $O_{\ell}\big{(}(L\|w_{k}\|)^{-2}\big{)}$ for the
integral (4.6).

Now let us assume that $d_{s}\geq 2$ for all $s=1,\ldots k$. Let $C\geq 1$.
When we integrate over the domain defined by $|x_{j}^{(1)}|\leq C^{-1}$ for
all $j=1,\ldots,k$, we obtain the bound:

$\displaystyle\ll$ $\displaystyle\prod_{s=1}^{k}\int_{|x_{s}^{(1)}|\leq
C^{-1}}\max\big{(}1,\|x_{s}\|\big{)}^{-\ell}\,d\rho_{s}(x_{s})\ll_{\ell}C^{-k}$

when $\ell$ is sufficiently large. Also to estimate the integrals over the
domains definded by $|x_{j}^{(1)}|\geq C^{-1}$ for some $j$, we use the bound:

$\omega\Big{(}\pi
L{\sum}_{j=1}^{k}\|w_{j}\|\cdot|x_{j}|\Big{)}\leq\frac{C^{2}}{\pi^{2}L^{2}\|w_{j}\|^{2}},$

which leads to the bound

$\displaystyle\int_{\mathbb{R}^{d_{s}}}\max\big{(}1,\|x_{s}\|\big{)}^{-\ell}\omega\Big{(}\pi
L{\sum}_{j=1}^{k}\|w_{j}\|\cdot|x_{j}|\Big{)}\,d\rho_{s}(x_{s})\ll_{\ell}C^{2}(L\|w_{k}\|)^{-2}$

provided that $\ell$ is sufficiently large. Ultimately, we conclude that the
integral (4.6) is bounded up to a constant by

$C^{-k}+C^{2}(L\|w_{k}\|)^{-2}.$

with a parameter $C\geq 1$. Optimizing in $C$, we get the required bound. ∎

A simple modification of the previous argument also gives:

######  Corollary 4.5.

There exists $\delta>0$ such that for every $\xi\in\mathbb{R}$,
$\phi_{s},\psi_{s}\in C_{c}^{\infty}(Z_{s})$, $s=1,\ldots,k$, and $L>0$,

$\displaystyle\frac{1}{L^{2}}\int_{0}^{L}\int_{0}^{L}e^{i(u_{1}-u_{2})\xi}$
$\displaystyle\Big{(}\prod_{s=1}^{k}\mu_{s}\big{(}(\phi_{s}\circ\sigma_{s}(u_{1}))\cdot(\bar{\psi}_{s}\circ\sigma_{s}(u_{2}))\big{)}-\prod_{s=1}^{k}\mu_{s}(\phi_{s}^{\star}\cdot{\bar{\psi}_{s}^{\star}})\Big{)}\,du_{1}du_{2}$
$\displaystyle=$ $\displaystyle
O\Big{(}\max\big{(}1,L\|w_{k}\|\big{)}^{-\frac{2}{k(k+2)}}\,{\prod}_{s=1}^{k}\|\phi_{s}\|_{C^{\ell}}\|\psi_{s}\|_{C^{\ell}}\Big{)}$

uniformly on $\xi$, for sufficiently large $\ell$.

###### Proof.

It turns out that the previous proof requires only minor modifications. As
before, we obtain

$\displaystyle\frac{1}{L^{2}}\int_{0}^{L}\int_{0}^{L}e^{i(u_{1}-u_{2})\xi}\Big{(}\prod_{s=1}^{k}$
$\displaystyle\int_{(\mathbb{R}^{d_{s}}\times\mathbb{R}^{d_{s}^{\prime}})/\Lambda_{s,v_{s}}^{\perp}}\phi^{(\Lambda_{s},v_{s})}(x_{s}+u_{1}w_{s},y_{s})$
$\displaystyle\times{\bar{\psi}^{(\Lambda_{s},v_{s})}(x_{s}+u_{2}w_{s},y_{s})}\,dx_{s}dy_{s}-\prod_{s=1}^{k}\widehat{\phi}^{\Lambda_{s}}(0)\overline{\widehat{\psi}^{\Lambda_{s}}(0)}\Big{)}du_{1}du_{2}$
$\displaystyle=$
$\displaystyle{\sum_{\lambda_{1},\ldots,\lambda_{s}}}^{\\!\\!\\!\\!\\!\prime}\,\prod_{s=1}^{k}\widehat{\phi}^{\Lambda_{s}}(\lambda_{s})\overline{\widehat{\psi}^{\Lambda_{s}}(\lambda_{s})}\,\omega\Big{(}\pi
L{\sum}_{j=1}^{k}\left<(w_{j},0),\lambda_{j}\right>+\xi\Big{)},$

where the sum is taken over
$(\lambda_{1},\ldots,\lambda_{s})\neq(0,\ldots,0)$. Using the same argument as
above, we obtain the integrals

$\int_{\mathbb{R}^{d_{1}}\times\cdots\times\mathbb{R}^{d_{k_{1}}}}\prod_{s=1}^{k_{1}}\max\big{(}1,\|x_{s}\|\big{)}^{-\ell}\,\omega\Big{(}\pi
L{\sum}_{j=1}^{k_{1}}\left<w_{j},x_{j}\right>+\xi\Big{)}\,d\rho_{1}(x_{1})\cdots
d\rho_{k_{1}}(x_{k_{1}}),$

with $k_{1}\leq k$. Applying orthogonal changes of variables, we may assume
that

$\left<w_{j},x_{j}\right>=\hbox{sign}(\xi)\cdot\|w_{j}\|\cdot|x_{j}^{(1)}|\quad\hbox{for
$j=1,\ldots,k_{1}$,}$

so that we obtain the integral

$\int_{\mathbb{R}^{d_{1}}\times\cdots\times\mathbb{R}^{d_{k_{1}}}}\prod_{s=1}^{k_{1}}\max\big{(}1,\|x_{s}\|\big{)}^{-\ell}\,\omega\Big{(}\pi
L{\sum}_{j=1}^{k_{1}}\|w_{j}\|\cdot|x_{j}|+|\xi|\Big{)}\,d\rho_{1}(x_{1})\cdots
d\rho_{k_{1}}(x_{k_{1}}).$

Then the same argument works. ∎

######  Remark 4.6.

Let us consider a slight generalization of the above results. For $d\geq 2$
and $d_{0},d^{\prime}\geq 1$, we set

$H:=\left(\begin{tabular}[]{ccc}$I_{d_{0}}$&$\hbox{Mat}_{d_{0},d}(\mathbb{R})$&$\hbox{Mat}_{d_{0},d^{\prime}}(\mathbb{R})$\\\
0&$\hbox{SL}_{d}(\mathbb{R})$&$\hbox{Mat}_{d,d^{\prime}}(\mathbb{R})$\\\
0&0&$I_{d^{\prime}}$\end{tabular}\right)\quad\hbox{and}\quad\Delta:=H\cap\hbox{SL}_{d+d^{\prime}+d_{0}}(\mathbb{Z})$
(4.7)

and consider the homogeneous space

$Z:=H/\Delta.$ (4.8)

Let $Z^{\star}$ denote the homogeneous space obtained from $Z$ by removing the
first $d_{0}$ rows and the first $d_{0}$ columns. In fact, the space
$Z^{\star}$ is exactly as in Theorem 4.3. Namely, it can be parameterized by
tuples $(\Lambda,v)$ where $\Lambda$ is a unimodular lattice in
$\mathbb{R}^{d}$, and $v\in(\mathbb{R}^{d}/\Lambda)^{d^{\prime}}$. We have the
projection map

$\pi:Z\to Z^{\star}$

with the fibers

$\pi^{-1}(\Lambda,v)\simeq(\mathbb{R}^{d+d^{\prime}}/\Lambda_{v})^{d_{0}}.$
(4.9)

Let us take a non-zero element $w\in\hbox{Mat}_{d_{0},d}(\mathbb{R})\subset H$
such that all non-zero components of $w$ are contained in the $i_{0}$-th row.
We denote by $U_{w}$ the subgroup of $H$ whose Lie algebra is generated by
$E_{i_{0},j}$, $j=d_{0}+1,\ldots,d+d^{\prime}+d_{0}$. We note that $U_{w}$ is
normal in $H$. The homogeneous subspace $Z_{w}:=U_{w}\Delta$ of $Z$ is a
$(d+d^{\prime})$-dimensional torus and we denote by $\tau_{w}$ the probability
invariant measure on $Z_{w}$. For $\phi\in C_{c}(Z)$, we define
$\phi^{\star}\in C_{c}(Z)$ by

$\phi^{\star}(h\Delta):=\int_{Z_{w}}\phi(hz)\,d\tau_{w}(z),\quad h\in H.$

In this set-up, we can apply the above arguments, in particular, decompose the
restriction of the functions to the fibers (4.9) into Fourier series on
$(\mathbb{R}^{d+d^{\prime}}/\Lambda_{v})^{d_{0}}$. Because of our choice of
$w$, only the $i_{0}$-component is affected by the flow, so that the previous
computations require only minor modifications, and Theorems 4.3–4.4 and
Corollary 4.5 can be proved in this setting as above.

We also note that one can take a non-zero element
$w\in\hbox{Mat}_{d,d^{\prime}}(\mathbb{R})\subset H$ such that all non-zero
components of $w$ are contained in the $j_{0}$-th row. Clearly, our argument
applies in this case as well.

##  5\. A proposition about weights

Let $I\subset\\{1,\ldots,m+n\\}$ be an admissible subset, i.e. the sets

$I^{\prime}:=I\cap\\{1,\ldots,m\\}\quad\hbox{and}\quad
I^{\prime\prime}:=I\cap\\{m+1,\ldots,m+n\\}$

are both not-empty. Note that $I=I^{\prime}\sqcup I^{\prime\prime}$. We define
the subsemigroup of $A^{+}$ as

$A^{+}_{I}:=\\{t\in A^{+}:\,t_{i}=0\hbox{ for $i\notin I$}\\}.$

We consider the adjoint action of $A^{+}$ on $\hbox{Lie}(U)$. Then for $t\in
A^{+}$ and the elementary matrices $E_{ij}$ generating $\hbox{Lie}(U)$,

$[t,E_{ij}]=\alpha_{ij}(t)E_{ij},$

where the weights $\alpha_{ij}$ are given by $\alpha_{ij}(t)=t_{i}+t_{j}$.

We say that the weight $\alpha_{ij}$ (or the corresponding weight space
$\mathbb{R}E_{ij}$, or a non-zero vector there) is _mixing for $I$_ if $i\in
I^{\prime}$ and $j\in I^{\prime\prime}$. We note that this condition is
equivalent to $\mathbb{R}E_{ij}\subset\hbox{Lie}(S_{I})$, and thus Theorem 3.1
should explain the terminology.

######  Proposition 5.1.

Let $I_{1}\neq I_{2}$ be admissible subsets. Then there exists $\theta>0$ such
that for every $s\in A^{+}_{I_{1}}$ and $t\in A^{+}_{I_{2}}$, either there
exists a weight $\alpha$ mixing for $I_{1}$ such that

$\alpha(s-t)\geq\theta\,\min\big{(}\lfloor s\rfloor_{I_{1}},\lfloor
t\rfloor_{I_{2}}\big{)},$ (5.1)

or there exists a weight $\alpha$ mixing for $I_{2}$ such that

$\alpha(t-s)\geq\theta\,\min\big{(}\lfloor s\rfloor_{I_{1}},\lfloor
t\rfloor_{I_{2}}\big{)}.$ (5.2)

###### Proof.

For simplicity we write

$\Pi(s,t):=\min\big{(}\lfloor s\rfloor_{I_{1}},\lfloor
t\rfloor_{I_{2}}\big{)}$

and use the following notations:

$\displaystyle D_{1}\prec_{\varepsilon}D_{2}\;$
$\displaystyle\Longleftrightarrow\;D_{1}\leq D_{2}+\varepsilon\,\Pi(s,t),$
$\displaystyle D_{1}\succ_{\varepsilon}D_{2}\;$
$\displaystyle\Longleftrightarrow\;D_{1}\geq D_{2}-\varepsilon\,\Pi(s,t),$
$\displaystyle D_{1}\approx_{\varepsilon}D_{2}\;$
$\displaystyle\Longleftrightarrow\;|D_{1}-D_{2}|\leq\varepsilon\,\Pi(s,t).$

The proof is subdivided into several cases:

Case 1: $I_{1}^{\prime}\backslash I_{2}^{\prime}\neq\emptyset$ and
$I_{1}^{\prime\prime}\backslash I_{2}^{\prime\prime}\neq\emptyset$.

We take $i\in I_{1}^{\prime}\backslash I_{2}^{\prime}$ and $j\in
I_{1}^{\prime\prime}\backslash I_{2}^{\prime\prime}$. Then

$\alpha_{ij}(s-t)=s_{i}+s_{j},$

which immediately implies the proposition.

Case 2: $I_{2}^{\prime}\backslash I_{1}^{\prime}\neq\emptyset$ and
$I_{2}^{\prime\prime}\backslash I_{1}^{\prime\prime}\neq\emptyset$.

This can be handled as in Case 1.

The complement of Cases 1–2 can be subdivided into the following sub-cases:

  * •

Case 3: $I_{1}^{\prime}\backslash I_{2}^{\prime}=\emptyset$ and
$I_{2}^{\prime\prime}\backslash I_{1}^{\prime\prime}=\emptyset$
$\Longleftrightarrow$ $I_{1}^{\prime}\subseteq I_{2}^{\prime}$ and
$I_{2}^{\prime\prime}\subseteq I_{1}^{\prime\prime}$,

  * •

Case 4: $I_{1}^{\prime}\backslash I_{2}^{\prime}=\emptyset$ and
$I_{2}^{\prime}\backslash I_{1}^{\prime}=\emptyset$ $\Longleftrightarrow$
$I_{1}^{\prime}=I_{2}^{\prime}$,

  * •

Case 5: $I_{2}^{\prime}\backslash I_{1}^{\prime}=\emptyset$ and
$I_{1}^{\prime\prime}\backslash I_{2}^{\prime\prime}=\emptyset$
$\Longleftrightarrow$ $I_{2}^{\prime}\subseteq I_{1}^{\prime}$ and
$I_{1}^{\prime\prime}\subseteq I_{2}^{\prime\prime}$,

  * •

Case 6: $I_{1}^{\prime\prime}\backslash I_{2}^{\prime\prime}=\emptyset$ and
$I_{2}^{\prime\prime}\backslash I_{1}^{\prime\prime}=\emptyset$
$\Longleftrightarrow$ $I_{1}^{\prime\prime}=I_{2}^{\prime\prime}$.

We only discuss Cases 3–4 since the remaining cases can be treated similarly.

Let us fix $\varepsilon>0$, to be specified later, chosen independently of $s$
and $t$, but will depend on $m$ and $n$.

Case 3: $I_{1}^{\prime}\subseteq I_{2}^{\prime}$ and
$I_{2}^{\prime\prime}\subseteq I_{1}^{\prime\prime}$.

We may additionally assume that $I_{1}^{\prime}\subsetneq I_{2}^{\prime}$ and
$I_{2}^{\prime\prime}\subsetneq I_{1}^{\prime\prime}$ because otherwise we
land either in Case 4 or Case 6.

Take $i\in I_{1}^{\prime}$ and $k\in I_{1}^{\prime\prime}\backslash
I_{2}^{\prime\prime}$. If $(s_{i}+s_{k})-t_{i}\geq\varepsilon\,\Pi(s,t)$, we
can take the weight $\alpha_{ik}$ to obtain (5.1), so that we may assume that

$s_{i}+s_{k}\prec_{\varepsilon}t_{i}\quad\hbox{for all $i\in I_{1}^{\prime}$
and $k\in I_{1}^{\prime\prime}\backslash I_{2}^{\prime\prime}$}.$ (5.3)

Take $l\in I_{2}^{\prime}\backslash I_{1}^{\prime}$ and $j\in
I_{2}^{\prime\prime}$. If $(t_{l}+t_{j})-s_{j}\geq\varepsilon\,\Pi(s,t)$, we
can take the weight $\alpha_{lj}$ to obtain (5.2), so that we may assume that

$t_{l}+t_{j}\prec_{\varepsilon}s_{j}\quad\hbox{for all $l\in
I_{2}^{\prime}\backslash I_{1}^{\prime}$ and $j\in I_{2}^{\prime\prime}$}.$
(5.4)

When $t_{j}\succ_{\varepsilon}s_{j}$ for some $j\in I_{2}^{\prime\prime}$,
then it follows from (5.3) that for $i\in I_{1}^{\prime}$ and $k\in
I_{1}^{\prime\prime}\backslash I_{2}^{\prime\prime}$,

$(t_{i}+t_{j})-(s_{i}+s_{j})\succ_{\varepsilon}s_{k}+(t_{j}-s_{j})\succ_{\varepsilon}s_{k}\geq\Pi(s,t).$

Taking $\varepsilon<1/3$, we obtain (5.2) with the weight $\alpha_{ij}$. Now
we assume that

$t_{j}\leq s_{j}-\varepsilon\,\Pi(s,t)\quad\hbox{for all $j\in
I_{2}^{\prime\prime}$.}$ (5.5)

When $s_{i}\succ_{\varepsilon}t_{i}$ for some $i\in I_{1}^{\prime}$, then it
follows from (5.4) that for $j\in I_{2}^{\prime\prime}$ and $l\in
I_{2}^{\prime}\backslash I_{1}^{\prime}$,

$(s_{i}+s_{j})-(t_{i}+t_{j})\succ_{\varepsilon}t_{l}+(s_{i}-t_{i})\succ_{\varepsilon}t_{l}\geq\Pi(s,t).$

Taking $\varepsilon<1/3$, we obtain (5.1). Now we assume that

$s_{i}\leq t_{i}-\varepsilon\,\Pi(s,t)\quad\hbox{for all $i\in
I_{1}^{\prime}$.}$ (5.6)

From (5.5), we obtain

${\sum}_{j\in I_{2}^{\prime\prime}}t_{j}\leq{\sum}_{j\in
I_{2}^{\prime\prime}}s_{j}-|I_{1}^{\prime\prime}|\varepsilon\,\Pi(s,t)$ (5.7)

From (5.6), we obtain

${\sum}_{i\in I_{1}^{\prime}}s_{i}\leq{\sum}_{i\in
I_{1}^{\prime}}t_{i}-|I_{1}^{\prime}|\varepsilon\,\Pi(s,t).$ (5.8)

Finally, we use that

${\sum}_{i\in I_{1}^{\prime}}s_{i}={\sum}_{j\in
I_{1}^{\prime\prime}}s_{j}\quad\hbox{and}\quad{\sum}_{i\in
I_{2}^{\prime}}t_{i}={\sum}_{j\in I_{2}^{\prime\prime}}t_{j}.$

Then it follows from (5.7) and (5.8) that

${\sum}_{i\in I_{2}^{\prime}}t_{i}\leq{\sum}_{i\in
I_{1}^{\prime}}t_{i}-(|I_{1}^{\prime\prime}|+|I_{1}^{\prime}|)\varepsilon\,\Pi(s,t).$

Since $I_{1}^{\prime}\subsetneq I_{2}^{\prime}$, this is only possible when
$t_{i}=0$ for $i\in I_{2}^{\prime}\backslash I_{1}^{\prime}$ and $\Pi(s,t)=0$.
In this case, the proposition holds trivially.

Case 4: $I_{1}^{\prime}=I_{2}^{\prime}$.

Take $i\in I_{1}^{\prime}$ and $j\in I_{1}^{\prime\prime}\cap
I_{2}^{\prime\prime}$. If
$(s_{i}+s_{j})-(t_{i}+t_{j})\geq\varepsilon\,\Pi(s,t)$, we can take the weight
$\alpha_{ij}$ to obtain (5.1). Also if
$(t_{i}+t_{j})-(s_{i}+s_{j})\geq\varepsilon\,\Pi(s,t)$, we can take the weight
$\alpha_{ij}$ to obtain (5.2). Hence, it remains to consider the case when

$s_{i}+s_{j}\approx_{\varepsilon}t_{i}+t_{j}\quad\hbox{for all $i\in
I_{1}^{\prime}$ and $j\in I_{1}^{\prime\prime}\cap I_{2}^{\prime\prime}$}.$
(5.9)

Take $i\in I_{1}^{\prime}$ and $k\in I_{1}^{\prime\prime}\backslash
I_{2}^{\prime\prime}$. If $(s_{i}+s_{k})-t_{i}\geq\varepsilon\,\Pi(s,t)$, we
can take the weight $\alpha_{ik}$ to obtain (5.1), so that we may assume that

$s_{i}+s_{k}\prec_{\varepsilon}t_{i}\quad\hbox{for all $i\in I_{1}^{\prime}$
and $k\in I_{1}^{\prime\prime}\backslash I_{2}^{\prime\prime}$}.$ (5.10)

Take $i\in I_{1}^{\prime}$ and $l\in I_{2}^{\prime\prime}\backslash
I_{1}^{\prime\prime}$. If $(t_{i}+t_{l})-s_{i}\geq\varepsilon\,\Pi(s,t)$, we
can take the weight $\alpha_{il}$ to obtain (5.2), so that we may assume that

$t_{i}+t_{l}\prec_{\varepsilon}s_{i}\quad\hbox{for all $i\in I_{1}^{\prime}$
and $l\in I_{2}^{\prime\prime}\backslash I_{1}^{\prime\prime}$}.$ (5.11)

Let us suppose that $I_{1}^{\prime\prime}\backslash
I_{2}^{\prime\prime}\neq\emptyset$ and $I_{2}^{\prime\prime}\backslash
I_{1}^{\prime\prime}\neq\emptyset$. We write

$\Sigma_{1}:={\sum}_{i\in I_{1}^{\prime}}s_{i}={\sum}_{j\in
I_{1}^{\prime\prime}}s_{j}\quad\hbox{and}\quad\Sigma_{2}:={\sum}_{i\in
I_{2}^{\prime}}t_{i}={\sum}_{j\in I_{2}^{\prime\prime}}t_{j}.$

Recall that $I_{1}^{\prime}=I_{2}^{\prime}$. Let $k\in
I_{1}^{\prime\prime}\backslash I_{2}^{\prime\prime}$ and $l\in
I_{2}^{\prime\prime}\backslash I_{1}^{\prime\prime}$. Summing (5.10) and
(5.11), we obtain:

$\Sigma_{1}+|I_{1}^{\prime}|\,s_{k}\prec_{n\varepsilon}\Sigma_{2}\quad\hbox{and}\quad\Sigma_{2}+|I_{1}^{\prime}|\,t_{l}\prec_{n\varepsilon}\Sigma_{1}.$

Hence, it follows that

$|I_{1}^{\prime}|\,s_{k}+|I_{1}^{\prime}|\,t_{l}\prec_{2n\varepsilon}0.$

Taking $\varepsilon<1/(2n)$, we conclude that that $s_{k}=t_{l}=0$. Then
$\Pi(s,t)=0$ and the proposition follows trivially.

Now we can assume that either $I_{1}^{\prime\prime}\subset
I_{2}^{\prime\prime}$ and $I_{2}^{\prime\prime}\subset I_{1}^{\prime\prime}$.
We only discuss the case when $I_{1}^{\prime\prime}\subset
I_{2}^{\prime\prime}$ because the other case can be treated similarly. Since
$I_{1}\neq I_{2}$ and $I^{\prime}_{1}=I^{\prime}_{2}$, we also have
$I_{1}^{\prime\prime}\subsetneq I_{2}^{\prime\prime}$. It follows from (5.9)
that

$|I_{1}^{\prime\prime}|\sum_{i\in
I_{1}^{\prime}}s_{i}+|I_{1}^{\prime}|\sum_{j\in
I_{1}^{\prime\prime}}s_{j}\approx_{mn\varepsilon}|I_{1}^{\prime\prime}|\sum_{i\in
I_{1}^{\prime}}t_{i}+|I_{1}^{\prime}|\sum_{j\in I_{1}^{\prime\prime}}t_{j}.$

Therefore,

$\Sigma_{1}\approx_{mn\varepsilon}\frac{|I_{1}^{\prime\prime}|}{|I_{1}|}\sum_{i\in
I_{1}^{\prime}}t_{i}+\frac{|I_{1}^{\prime}|}{|I_{1}|}\sum_{j\in
I_{1}^{\prime\prime}}t_{j}=\Sigma_{2}-\frac{|I_{1}^{\prime}|}{|I_{1}|}\sum_{j\in
I_{2}^{\prime\prime}\backslash I_{1}^{\prime\prime}}t_{j}.$ (5.12)

Also from (5.11), we derive that

$\sum_{i\in
I_{1}^{\prime}}t_{i}+|I_{1}^{\prime}|t_{l}\prec_{n\varepsilon}\sum_{i\in
I_{1}^{\prime}}s_{i}=\Sigma_{1}$ (5.13)

for $l\in I_{2}^{\prime\prime}\backslash I_{1}^{\prime\prime}$. Combining
(5.12) and (5.13), we conclude that

$\Sigma_{2}+|I_{1}^{\prime}|t_{l}\prec_{(m+1)n\varepsilon}\Sigma_{2}-\frac{|I_{1}^{\prime}|}{|I_{1}|}\sum_{j\in
I_{2}^{\prime\prime}\backslash I_{1}^{\prime\prime}}t_{j},$

so that

$t_{\ell}\prec_{(m+1)n\varepsilon}0.$

When we take $\varepsilon<1/((m+1)n)$, this implies that $t_{\ell}=0$ for
$\ell\in I_{2}^{\prime\prime}\backslash I_{1}^{\prime\prime}$. Then
$\Pi(s,t)=0$ and the proposition follows trivially. This completes analysis of
Case 4. Since the other cases can be treated similarly, this concludes the
proof of the proposition. ∎

##  6\. Multiple equidistribution

In this section, we prove the following multiple equidistribution result,
which generalizes Theorem 1.3.

######  Theorem 6.1.

Let $I_{1},\ldots,I_{r}\subset\\{1,\ldots,m+n\\}$ be admissible subsets. There
exist $\ell\geq 1$ and $\delta_{r}>0$ such that for every $f\in
C_{c}^{\infty}(Y)$, $\varphi_{1},\ldots,\varphi_{r}\in C^{\infty}_{c}(X)$, and
$t_{1}\in A^{+}_{I_{1}}$, $\ldots$, $t_{r}\in A^{+}_{I_{r}}$,

$\displaystyle\int_{Y}f\left(\prod_{s=1}^{r}\varphi_{s}\circ
a(t_{s})\right)\,d\nu=$
$\displaystyle\left(\int_{Y}f\,d\nu\right)\prod_{s=1}^{r}\int_{X_{I_{s}}}\varphi_{s}\,d\mu_{I_{s}}$
$\displaystyle\quad+O\left(e^{-\delta_{r}\min\big{(}\lfloor
t_{1}\rfloor_{I_{1}},\ldots,\lfloor
t_{r}\rfloor_{I_{r}},\Delta(t_{1},\ldots,t_{r})\big{)}}\,\|f\|_{W}{\prod}_{s=1}^{r}\|\varphi_{s}\|_{C^{\ell}}\right).$

To simplify notation, we write

$D(t_{1},\ldots,t_{r}):=\min\big{(}\lfloor t_{1}\rfloor_{I_{1}},\ldots,\lfloor
t_{r}\rfloor_{I_{r}},\Delta(t_{1},\ldots,t_{r})\big{)}.$

When $I_{1}=\cdots=I_{r}$, Theorem 6.1 can be proved using the method of [5]
with minor modifications. Indeed, let us consider the span $W_{I}$ of mixing
directions $E_{ij}$, namely, with indices satisfying $i\in
I\cap\\{1,\ldots,m\\}$ and $j\in I\cap\\{m+1,\ldots,m+n\\}$. Since the adjoint
action of $A_{I}$ on $W_{I}$ is proper, we can run exactly the same argument
as in [5] by considering only one-parameter subgroups
$\\{\exp(uw)\\}_{u\in\mathbb{R}}$ with $w\in W_{I}$. Since those one-parameter
subgroups satisfy the required mixing estimate by Theorem 3.1, the original
argument is applicable. The base case with $r=1$, which is needed to start
this argument, was handled in Theorem 3.3.

Now we suppose that some of the sets $I_{s}$ are distinct. Then by Proposition
5.1, there exist a constant $\theta>0$, $s_{1}\neq s_{2}\in\\{1,\ldots,r\\}$,
and a direction $w=E_{i_{0}j_{0}}\in\hbox{Lie}(U)$ with the weight
$\alpha=\alpha_{i_{0}j_{0}}$ such that

$\alpha(t_{s_{2}}-t_{s_{1}})\geq\theta\,D(t_{1},\ldots.t_{r})$ (6.1)

and $w$ is mixing for $I_{s_{2}}$. We change the indexation so that

$\alpha(t_{1})\geq\alpha(t_{2})\geq\cdots\geq\alpha(t_{r}).$ (6.2)

Then $s_{1}>s_{2}$. It follows from (6.1) that there exists
$k=s_{2},\ldots,s_{1}-1$ such that

$\alpha(t_{k}-t_{k+1})\geq\theta\,D(t_{1},\ldots.t_{r})/r.$ (6.3)

From now on, we fix the direction $w=E_{i_{0}j_{0}}$ and the index $k$. It is
important for our argument later that for some $s=1,\ldots,k$, the direction
$w$ is mixing for $I_{s}$.

For smooth functions $f$ on $Y$, we have the uniformly convergent expansion as
in (2.4):

$f=\sum_{\xi\in\Xi_{Y}}\widehat{f}(\xi)\xi.$

Then

$\displaystyle\int_{Y}f\left(\prod_{s=1}^{r}\varphi_{s}\circ
a(t_{s})\right)\,d\nu=$
$\displaystyle\left(\int_{Y}f\,d\nu\right)\int_{Y}\left(\prod_{s=1}^{r}\varphi_{s}\circ
a(t_{s})\right)\,d\nu$ $\displaystyle\quad\quad\quad+\sum_{\xi\neq
1}\widehat{f}(\xi)\int_{Y}\xi\left(\prod_{s=1}^{r}\varphi_{s}\circ
a(t_{s})\right)\,d\nu.$ (6.4)

Now we proceed with estimating the integrals

$I(\xi):=\int_{Y}\xi\left(\prod_{s=1}^{r}\varphi_{s}\circ
a(t_{s})\right)\,d\nu\quad\quad\hbox{with $\xi\in\Xi_{Y}.$}$

We claim that for some $\ell\geq 1$ and $\delta_{r}>0$,

$\displaystyle I(1)$
$\displaystyle=\prod_{s=1}^{r}\int_{X_{I_{s}}}\varphi_{s}\,d\mu_{I_{s}}+O\left(e^{\delta_{r}D(t_{1},\ldots,t_{k})}{\prod}_{s=1}^{r}\|\varphi_{s}\|_{C^{\ell}}\right),$
$\displaystyle I(\xi)$
$\displaystyle=O\left(e^{\delta_{r}D(t_{1},\ldots,t_{k})}{\prod}_{s=1}^{r}\|\varphi_{s}\|_{C^{\ell}}\right),\quad\hbox{for
$\xi\neq 1$},$ (6.5)

uniformly on $\xi$. Once this claim is established, Theorem 6.1 follows
directly from (6.4). The proof proceeds by induction on $r$.

We use that the measure $\nu$ is invariant under the one-parameter subgroup
$\\{\exp(uw)\\}_{u\in\mathbb{R}}$, so that for any $u\in\mathbb{R}$,

$\displaystyle I(\xi)=$
$\displaystyle\int_{Y}\xi\circ\exp(uw)\left(\prod_{s=1}^{r}\varphi_{s}\circ
a(t_{s})\circ\exp(uw)\right)\,d\nu$ $\displaystyle=$
$\displaystyle\xi(\exp(uw))\int_{Y}\xi\left(\prod_{s=1}^{r}\varphi_{s}\circ\exp(uw^{(s)})\circ
a(t_{s})\right)\,d\nu,$

where

$w^{(s)}:=\operatorname{Ad}(a(t_{s}))w=e^{\alpha(t_{s})}w.$

In particular, it follows from (6.2) that

$\|w^{(1)}\|\geq\|w^{(2)}\|\geq\cdots\geq\|w^{(r)}\|.$

Integrating over $u\in[0,L]$, we deduce that for any $L>0$,

$\displaystyle I(\xi)$
$\displaystyle=\int_{Y}\xi\left(\frac{1}{L}\int_{0}^{L}\xi(\exp(uw))\prod_{s=1}^{r}\varphi_{s}\circ\exp(uw^{(s)})\circ
a(t_{s})\,du\right)\,d\nu.$ (6.6)

We choose the parameter $L$ so that $L\|w^{(s)}\|$ is ”large” for $s\leq k$
and ”small” for $s\geq k+1$. The exact value of $L$ will be given at the end
of the argument.

Step I: Controlling the components $\exp(uw^{(s)})$ with $s=k+1,\ldots,r$.

Let us consider a related integral

$\displaystyle I_{1}(\xi):=$
$\displaystyle\int_{Y}\xi\left(\frac{1}{L}\int_{0}^{L}\xi(\exp(uw))\prod_{s=1}^{k}\varphi_{s}\circ\exp(uw^{(s)})\circ
a(t_{s})\,du\right)\left(\prod_{s=k+1}^{r}\varphi_{s}\circ
a(t_{s})\right)\,d\nu.$ (6.7)

Since $|\xi|=1$, we obtain that

$\displaystyle|I(\xi)-I_{1}(\xi)|\leq$
$\displaystyle\prod_{s=1}^{k}\|\varphi_{s}\|_{C^{0}}$ (6.8)
$\displaystyle\quad\quad\times\sup_{u\in[0,L]}\Big{\|}\prod_{s=k+1}^{r}\varphi_{s}\circ\exp(uw^{(s)})\circ
a(t_{s})-\prod_{s=k+1}^{r}\varphi_{s}\circ a(t_{s})\Big{\|}_{C^{0}},$

and by the triangle inequality,

$\displaystyle\Big{\|}\prod_{s=k+1}^{r}\varphi_{s}\circ\exp(uw^{(s)})\circ
a(t_{s})-\prod_{s=k+1}^{r}\varphi_{s}\circ a(t_{s})\Big{\|}_{C^{0}}$ (6.9)
$\displaystyle=$
$\displaystyle\,\Big{\|}\prod_{s=k+1}^{r}\varphi_{s}\circ\exp(uw^{(s)})-\prod_{s=k+1}^{r}\varphi_{s}\Big{\|}_{C^{0}}$
$\displaystyle\leq\;$
$\displaystyle\sum_{l=k+1}^{r}\Big{\|}\Big{(}\prod_{s=k+1}^{l}\varphi_{s}\circ\exp(uw^{(s)})\Big{)}\prod_{s=l+1}^{r}\varphi_{s}-\Big{(}\prod_{s=k+1}^{l-1}\varphi_{s}\circ\exp(uw^{(s)})\Big{)}\prod_{s=l}^{r}\varphi_{s}\Big{\|}_{C^{0}}$
$\displaystyle\leq\;$
$\displaystyle\sum_{l=k+1}^{r}\big{\|}\varphi_{l}\circ\exp(uw^{(l)})-\varphi_{l}\big{\|}_{C^{0}}\prod_{k+1\leq
s\leq r,s\neq\ell}\|\varphi_{s}\|_{C^{0}}$ $\displaystyle\ll$ $\displaystyle
L\|w^{(k+1)}\|\,{\prod}_{s=k+1}^{r}\|\varphi_{s}\|_{C^{1}}.$

Therefore, we conclude that

$|I(\xi)-I_{1}(\xi)|\ll
L\|w^{(k+1)}\|\,{\prod}_{s=1}^{r}\|\varphi_{s}\|_{C^{1}}.$ (6.10)

Step II: Applying the inductive hypothesis.

We use notation: $I^{(1)}:=I\cup\\{m+1,\ldots,m+n\\}$ and
$I^{(2)}:=I\cup\\{1,\ldots,m\\}$

When $w=E_{i_{0}j_{0}}$ is not mixing with respect to a set $I$, it is
convenient to introduce a unipotent subgroup $U_{I,w}$, which is defined as
follows:

  * •

If $i_{0}\in I$ and $j_{0}\notin I$, the Lie algebra of $U_{I,w}$ is spanned
by $E_{ij_{0}}$ with $i\in I^{(2)}$.

  * •

If $i_{0}\notin I$ and $j_{0}\in I$, the Lie algebra of $U_{I,w}$ is spanned
by $E_{i_{0}j}$ with $j\in I^{(1)}$.

  * •

If $i_{0}\notin I$ and $j_{0}\notin I$, the Lie algebra of $U_{I,w}$ is
spanned by $E_{i_{0}j_{0}}$.

We note that the spaces $X_{I}$ are isomorphic to the spaces from Remark 4.6.
Indeed, in the first two cases, the action $\exp(\mathbb{R}w)$ on $X_{I}$
corresponds to one-parameter subgroups as in Remark 4.6. In the last case,
since $U_{I,w}$ lies in the center of $G_{I}$, the space $X_{I}$ is a bundle
with fibers $gU_{I,w}\Gamma$ and the action on the fibers is given by

$gz\mapsto g\exp(uw)z,\quad u\in\mathbb{R},\;z\in U_{I,w}\Gamma,$

so that we obtain the circle rotation which corresponds to the case $d=1$ in
Section 4. Hence, the results of Section 4 are applicable in our setting.

We denote by $\nu_{I,w}$ the invariant probability measure supported on
$U_{I,w}\Gamma$. When $w$ is not mixing for $I_{s}$, we set

$\varphi^{\star}_{s}(g\Gamma):=\int_{U_{I_{s},w}\Gamma}\varphi_{s}(gy)\,d\nu_{I_{s},w}(y).$

We note that $\varphi^{\star}_{s}$ is $\exp(\mathbb{R}w)$-invariant, and
$\int_{X_{I_{s}}}\varphi^{\star}_{s}\,d\mu_{I_{s}}=\int_{X_{I_{s}}}\varphi_{s}\,d\mu_{I_{s}}$.

When $w$ is mixing for $I_{s}$, we set

$\varphi^{\star}_{s}(g\Gamma):=\int_{X_{I_{s}}}\varphi_{s}\,d\mu_{I_{s}}.$

Let

$\displaystyle I_{2}(\xi):=$
$\displaystyle\int_{Y}\xi\left(\frac{1}{L}\int_{0}^{L}\xi(\exp(uw))\prod_{s=1}^{k}\varphi^{\star}_{s}\circ\exp(uw^{(s)})\circ
a(t_{s})\,du\right)\left(\prod_{s=k+1}^{r}\varphi_{s}\circ
a(t_{s})\right)\,d\nu$

and

$\displaystyle I_{3}(\xi):=$
$\displaystyle\int_{Y}\xi\left(\prod_{s=1}^{k}\varphi^{\star}_{s}\circ
a(t_{s})\right)\left(\prod_{s=k+1}^{r}\varphi_{s}\circ a(t_{s})\right)\,d\nu.$

Arguing as in Step I, we obtain that

$|I_{2}(\xi)-I_{3}(\xi)|\ll
L\|w^{(k+1)}\|\,{\prod}_{s=1}^{r}\|\varphi_{s}\|_{C^{1}}.$ (6.11)

Since the functions $\varphi_{s}^{\star}$ are $\exp(\mathbb{R}w)$-invariant,

$I_{2}(\xi)=\int_{Y}\xi\left(\frac{1}{L}\int_{0}^{L}\xi(\exp(uw))\prod_{s=1}^{k}\varphi^{\star}_{s}\circ
a(t_{s})\,du\right)\left(\prod_{s=k+1}^{r}\varphi_{s}\circ
a(t_{s})\right)\,d\nu.$

Let us introduce a function ${\Psi}_{\xi,L}:X^{k}\rightarrow\mathbb{R}$
defined by

$\displaystyle{\Psi}_{\xi,L}(\underline{x}):=$
$\displaystyle\frac{1}{L}\int_{0}^{L}\xi(\exp(uw))\Big{(}\prod_{s=1}^{k}\varphi_{s}(\exp(uw^{(s)})x_{s})-\prod_{s=1}^{k}\varphi^{\star}_{s}(\exp(uw^{(s)})x_{s})\Big{)}\,du$
(6.12) $\displaystyle=$
$\displaystyle\frac{1}{L}\int_{0}^{L}\xi(\exp(uw))\Big{(}\prod_{s=1}^{k}\varphi_{s}(\exp(uw^{(s)})x_{s})-\prod_{s=1}^{k}\varphi^{\star}_{s}(x_{s})\Big{)}\,du$

for $\underline{x}=(x_{1},\ldots,x_{k})\in X^{k}$. Setting

$\Phi:=\prod_{s=k+1}^{r}\varphi_{s}\circ a(t_{s})\quad\hbox{and}\quad
a(t):=(a(t_{1}),\ldots,a(t_{k})),$

we obtain that

$I_{1}(\xi)-I_{2}(\xi)=\nu\big{(}\xi\cdot(\Psi_{\xi,L}\circ
a(t))|_{diag(X^{k})}\cdot\Phi\big{)}.$

Hence, by Cauchy-Schwartz inequality,

$\displaystyle|I_{1}(\xi)-I_{2}(\xi)|$
$\displaystyle\leq\nu\big{(}(|\Psi_{\xi,L}|^{2}\circ
a(t))|_{diag(X^{k})}\big{)}^{1/2}\|\Phi\|_{C^{0}}$ (6.13)
$\displaystyle\leq\nu\big{(}(|\Psi_{\xi,L}|^{2}\circ
a(t))|_{diag(X^{k})}\big{)}^{1/2}{\prod}_{s=k+1}^{r}\|\varphi_{s}\|_{C^{0}}.$

Let $\mu^{(k)}:=\prod_{s=1}^{k}\mu_{I_{s}}$ be the measure on $X^{k}$. Using
that the inequality

$\alpha^{1/2}\leq|\alpha-\beta|^{1/2}+\beta^{1/2}\quad\hbox{ for all
$\alpha,\beta\geq 0$,}$

we see that

$\displaystyle\nu\big{(}(|\Psi_{\xi,L}|^{2}\circ
a(t))|_{diag(X^{k})}\big{)}^{1/2}\leq$
$\displaystyle\big{|}\nu\big{(}(|\Psi_{\xi,L}|^{2}\circ
a(t))|_{diag(X^{k})}\big{)}-\mu^{(k)}\big{(}|{\Psi}_{\xi,L}|^{2}\big{)}\big{|}^{1/2}$
(6.14)
$\displaystyle\quad\quad+\mu^{(k)}\big{(}|{\Psi}_{\xi,L}|^{2}\big{)}^{1/2}.$

We proceed to estimate the first term in (6.14). The function
$|\Psi_{\xi,L}|^{2}$ can be written as linear combination of functions of the
form

${\Omega}_{\xi,L}(\underline{x}):=\frac{1}{L^{2}}\int_{0}^{L}\int_{0}^{L}\xi((u_{1}-u_{2})w)\left(\prod_{s=1}^{k}\rho_{1,s}(\exp(u_{1}w^{(s)})x_{s})\rho_{2,s}(\exp(u_{2}w^{(s)})x_{s})\right)\,du_{1}du_{2},$

where $\rho_{i,s}$ is either $\varphi_{s}$, $\varphi^{\star}_{s}$,
$\bar{\varphi}_{s}$, $\bar{\varphi}^{\star}_{s}$. Let

$\omega^{u_{1},u_{2}}_{s}:=\big{(}\rho_{1,s}\circ\exp(u_{1}w^{(s)})\big{)}\cdot\big{(}\rho_{2,s}\circ\exp(u_{2}w^{(s)})\big{)}.$

Then

$\displaystyle\nu\big{(}(\Omega_{\xi,L}\circ
a(t))|_{diag(X^{k})}\big{)}=\frac{1}{L^{2}}\int_{0}^{L}\int_{0}^{L}\xi((u_{1}-u_{2})w)\int_{Y}\left(\prod_{s=1}^{k}\omega_{s}^{u_{1},u_{2}}\circ
a(t_{s})\right)\,d\nu\,du_{1}du_{2}.$

Since $k<r$, we can apply the inductive hypothesis to conclude that for some
$\ell\geq 1$ and $\delta_{k}>0$,

$\int_{Y}\left(\prod_{s=1}^{k}\omega_{s}^{u_{1},u_{2}}\circ
a(t_{s})\right)\,d\nu=\prod_{s=1}^{k}\int_{X_{I_{s}}}\omega_{s}^{u_{1},u_{2}}\,d\mu_{I_{s}}+O\left(e^{-\delta_{k}D(t_{1},\ldots,t_{k})}{\prod}_{s=1}^{k}\|\omega_{s}^{u_{1},u_{2}}\|_{C^{\ell}}\right).$

Since

$\displaystyle\mu^{(k)}\big{(}\Omega_{\xi,L}\big{)}=\frac{1}{L^{2}}\int_{0}^{L}\int_{0}^{L}\xi((u_{1}-u_{2})w)\left(\prod_{s=1}^{k}\int_{X_{I_{s}}}\omega_{s}^{u_{1},u_{2}}\,d\mu_{I_{s}}\right)\,du_{1}du_{2},$

it remains to estimate the error term. We have

$\displaystyle\|\omega_{s}^{u_{1},u_{2}}\|_{C^{\ell}}$
$\displaystyle\leq\big{\|}(\rho_{1,s}\circ\exp(u_{1}w^{(s)}))\cdot(\rho_{2,s}\circ\exp(u_{2}w^{(s)}))\big{\|}_{C^{\ell}}$
$\displaystyle\ll\big{\|}\rho_{1,s}\circ\exp(u_{1}w^{(s)})\big{\|}_{C^{\ell}}\cdot\big{\|}\rho_{2,s}\circ\exp(u_{2}w^{(s)})\big{\|}_{C^{\ell}}$
$\displaystyle\ll\max\big{(}1,L\|w^{(s)}\|\big{)}^{2\ell}\|\rho_{1,s}\|_{C^{\ell}}\|\rho_{2,s}\|_{C^{\ell}}$
$\displaystyle\leq\max\big{(}1,L\|w^{(s)}\|\big{)}^{2\ell}\|\varphi_{s}\|_{C^{\ell}}^{2}.$

Hence, we deduce that

$\displaystyle\Big{|}\nu\big{(}(\Omega_{\xi,L}\circ
a(t))|_{diag(X^{k})}\big{)}-$
$\displaystyle\mu^{(k)}\big{(}\Omega_{\xi,L}\big{)}\Big{|}$ $\displaystyle\ll$
$\displaystyle\max\big{(}1,L\|w^{(1)}\|\big{)}^{2k\ell}e^{-\delta_{k}D(t_{1},\ldots,t_{k})}{\prod}_{s=1}^{k}\|\varphi_{s}\|_{C^{\ell}}^{2}.$

Recalling that $|\Psi_{\xi,L}|^{2}$ is a linear combination of functions of
the form $\Omega_{\xi,L}$, we also conclude that

$\displaystyle\Big{|}\nu\big{(}(|\Psi_{\xi,L}|^{2}\circ
a(t))|_{diag(X^{k})}\big{)}-$
$\displaystyle\mu^{(k)}\big{(}|\Psi_{\xi,L}|^{2}\big{)}\Big{|}$ (6.15)
$\displaystyle\ll\max\big{(}1,L\|w^{(1)}\|\big{)}^{2k\ell}e^{-\delta_{k}D(t_{1},\ldots,t_{k})}{\prod}_{s=1}^{k}\|\varphi_{s}\|_{C^{\ell}}^{2}.$

Step III: Decorrelation estimates for the flows $\exp(uw^{(s)})$ with
$s=1,\ldots,k$.

Now we proceed with estimating the second term in (6.14). We write
${\Psi}_{\xi,L}$ as

${\Psi}_{\xi,L}(\underline{x})=\frac{1}{L}\int_{0}^{L}\xi(\exp(uw))\Theta_{u}(\underline{x})\,du$

with

$\Theta_{u}(\underline{x}):=\prod_{s=1}^{k}\varphi_{s}\big{(}\exp(uw^{(s)})x_{s}\big{)}-\prod_{s=1}^{k}\varphi^{\star}_{s}(x_{s}),$

so that

$\left|{\Psi}_{\xi,L}(\underline{x})\right|^{2}=\frac{1}{L^{2}}\int_{0}^{L}\int_{0}^{L}\xi(\exp((u_{1}-u_{2})w))\Theta_{u_{1}}(\underline{x})\overline{\Theta}_{u_{2}}(\underline{x})\,du_{1}du_{2}.$

We have

$\displaystyle\Theta_{u_{1}}(\underline{x})\overline{\Theta}_{u_{2}}(\underline{x})=$
$\displaystyle\prod_{s=1}^{k}\varphi_{s}\big{(}\exp(u_{1}w^{(s)})x_{s}\big{)}\bar{\varphi}_{s}\big{(}\exp(u_{2}w^{(s)})x_{s}\big{)}$
$\displaystyle\quad\quad-\prod_{s=1}^{k}\varphi_{s}\big{(}\exp(u_{1}w^{(s)})x_{s}\big{)}\bar{\varphi}^{\star}_{s}(x_{s})-\prod_{s=1}^{k}\varphi^{\star}_{s}(x_{s})\bar{\varphi}_{s}\big{(}\exp(u_{2}w^{(s)})x_{s}\big{)}$
$\displaystyle\quad\quad+\prod_{s=1}^{k}\varphi_{s}^{\star}(x_{s})\bar{\varphi}^{\star}_{s}(x_{s}).$

Since

$\int_{X_{I_{s}}}\big{(}\varphi_{s}\circ\exp(u_{1}w^{(s)})\big{)}\cdot\bar{\varphi}^{\star}_{s}\,d\mu_{I_{s}}=\int_{X_{I_{s}}}\varphi^{\star}_{s}\cdot\big{(}\bar{\varphi}_{s}\circ\exp(u_{2}w^{(s)})\big{)}\,d\mu_{I_{s}}=\int_{X_{I_{s}}}\varphi^{\star}_{s}\bar{\varphi}^{\star}_{s}\,d\mu_{I_{s}},$

we obtain that

$\displaystyle\mu^{(k)}\left(\Theta_{u_{1}}\cdot\overline{\Theta}_{u_{2}}\right)=\prod_{s=1}^{k}\mu_{I_{s}}\big{(}\varphi_{s}\circ\exp(u_{1}w^{(s)})\cdot{\bar{\varphi}_{s}\circ\exp(u_{2}w^{(s)})}\big{)}-\prod_{s=1}^{k}\mu_{I_{s}}\big{(}\varphi^{\star}_{s}\cdot\bar{\varphi}^{\star}_{s}\big{)}.$

We claim that

$\displaystyle\mu^{(k)}\big{(}|{\Psi}_{\xi,L}|^{2}\big{)}$
$\displaystyle=\frac{1}{L^{2}}\int_{0}^{L}\int_{0}^{L}\xi(\exp((u_{1}-u_{2})w))\mu^{(k)}\left(\Theta_{u_{1}}\cdot\overline{\Theta}_{u_{2}}\right)\,du_{1}du_{2}$
(6.16)
$\displaystyle\ll\max\big{(}1,L\|w^{(k)}\|\big{)}^{-c}\,{\prod}_{s=1}^{k}\|\varphi_{s}\|_{C^{\ell}}^{2}$

for some $c>0$. We write

$\\{1\ldots,k\\}=S_{mix}\sqcup S_{mix}^{c},$

where $S_{mix}$ consists of such indices $s$ that $w$ is a mixing direction
for $I_{s}$. By Corollary 3.2, for $s\in S_{mix}$,

$\displaystyle\mu_{I_{s}}\big{(}\varphi_{s}\circ\exp(u_{1}w^{(s)})\cdot{\bar{\varphi}_{s}\circ\exp(u_{2}w^{(s)})}\big{)}=$
$\displaystyle\mu_{I_{s}}\big{(}\varphi^{\star}_{s}\cdot\bar{\varphi}^{\star}_{s}\big{)}$
(6.17)
$\displaystyle\;+O\Big{(}\max\big{(}1,|u_{1}-u_{2}|\|w^{(s)}\|\big{)}^{-\tau}\,\|\varphi_{s}\|_{C^{\ell}}^{2}\Big{)}.$

Of course, in this case,
$\mu_{I_{s}}\big{(}\varphi^{\star}_{s}\cdot\bar{\varphi}^{\star}_{s}\big{)}=\mu_{I_{s}}(\varphi_{s})\mu_{I_{s}}(\bar{\varphi}_{s})$.

Let

$\Theta^{\star}_{u}(\underline{x}):=\prod_{s\in
S_{mix}^{c}}\varphi_{s}\big{(}\exp(uw^{(s)})x_{s}\big{)}\cdot\prod_{s\in
S_{mix}}\varphi^{\star}_{s}(x_{s})-\prod_{s=1}^{k}\varphi^{\star}_{s}(x_{s})$

and

${\Psi}^{\star}_{\xi,L}(\underline{x}):=\frac{1}{L}\int_{0}^{L}\xi(\exp(uw))\Theta^{\star}_{u}(\underline{x})\,du.$

As above, we obtain that

$\mu^{(k)}\big{(}|{\Psi}^{\star}_{\xi,L}|^{2}\big{)}=\frac{1}{L^{2}}\int_{0}^{L}\int_{0}^{L}\xi(\exp((u_{1}-u_{2})w))\mu^{(k)}\left(\Theta^{\star}_{u_{1}}\cdot\overline{\Theta}^{\star}_{u_{2}}\right)\,du_{1}du_{2},$

and

$\displaystyle\mu^{(k)}\left(\Theta^{\star}_{u_{1}}\cdot\overline{\Theta}^{\star}_{u_{2}}\right)=$
$\displaystyle\prod_{s\in
S_{mix}^{c}}\mu_{I_{s}}\big{(}\varphi_{s}\circ\exp(u_{1}w^{(s)})\cdot{\bar{\varphi}_{s}\circ\exp(u_{2}w^{(s)})}\big{)}\prod_{s\in
S_{mix}}\mu_{I_{s}}\big{(}\varphi^{\star}_{s}\cdot\bar{\varphi}^{\star}_{s}\big{)}$
$\displaystyle\quad\quad-\prod_{s=1}^{k}\mu_{I_{s}}\big{(}\varphi^{\star}_{s}\cdot\bar{\varphi}^{\star}_{s}\big{)}.$

It follows from (6.17) that

$\displaystyle\mu^{(k)}\left(\Theta_{u_{1}}\cdot\overline{\Theta}_{u_{2}}\right)=$
$\displaystyle\mu^{(k)}\left(\Theta^{\star}_{u_{1}}\cdot\overline{\Theta}^{\star}_{u_{2}}\right)$
$\displaystyle\quad+O\Big{(}\max\big{(}1,|u_{1}-u_{2}|\|w^{(k)}\|\big{)}^{-\tau}\,{\prod}_{s=1}^{k}\|\varphi_{s}\|_{C^{\ell}}^{2}\Big{)},$

so that using [5, Lemma 4.3], we conclude that

$\displaystyle\Big{|}\mu^{(k)}\big{(}|{\Psi}_{\xi,L}|^{2}\big{)}-\mu^{(k)}\big{(}|{\Psi}^{\star}_{\xi,L}|^{2}\big{)}\Big{|}$
$\displaystyle\leq\frac{1}{L^{2}}\int_{0}^{L}\int_{0}^{L}\big{|}\mu^{(k)}\left(\Theta_{u_{1}}\cdot\overline{\Theta}_{u_{2}}\right)-\mu^{(k)}\left(\Theta^{\star}_{u_{1}}\cdot\overline{\Theta}^{\star}_{u_{2}}\right)\big{|}\,du_{1}du_{2}$
$\displaystyle\ll\max\big{(}1,L\|w^{(k)}\|\big{)}^{-\tau}\,{\prod}_{s=1}^{k}\|\varphi_{s}\|_{C^{\ell}}^{2}.$

Finally, the estimate for
$\mu^{(k)}\big{(}|{\Psi}^{\star}_{\xi,L}|^{2}\big{)}$ reduces to analysis of
the integrals

$\displaystyle\frac{1}{L^{2}}\int_{0}^{L}\int_{0}^{L}\xi(\exp((u_{1}-u_{2})w))\Big{(}\prod_{s\in
S_{mix}^{c}}\mu_{I_{s}}\big{(}\varphi_{s}\circ\exp(u_{1}$ $\displaystyle
w^{(s)})\cdot{\bar{\varphi}_{s}\circ\exp(u_{2}w^{(s)})}\big{)}$
$\displaystyle-\prod_{s\in
S_{mix}^{c}}\mu_{I_{s}}\big{(}\varphi^{\star}_{s}\cdot\bar{\varphi}^{\star}_{s}\big{)}\Big{)}\,du_{1}du_{2}.$

In this case, according to Corollary 4.5 and Remark 4.6, we also have the
bound

$\ll\max\big{(}1,L\|w^{(k)}\|\big{)}^{-\tau^{\prime}}\,{\prod}_{s=1}^{k}\|\varphi_{s}\|_{C^{\ell}}^{2}$

for some $\tau^{\prime}>0$. This verifies (6.16).

Combining (6.13), (6.14), (6.15), and (6.16), we conclude that

$\displaystyle|I_{1}(\xi)-I_{2}(\xi)|\ll\Big{(}\max\big{(}1,L\|w^{(1)}\|\big{)}^{k\ell}$
$\displaystyle e^{-\delta_{k}D(t_{1},\ldots,t_{k})/2}$ (6.18)
$\displaystyle+\max\big{(}1,L\|w^{(k)}\|\big{)}^{-c/2}\Big{)}{\prod}_{s=1}^{r}\|\varphi_{s}\|_{C^{\ell}}.$

Step IV: Optimizing the parameters.

Combining the bound (6.18) with (6.10) and (6.11), we derive that

$\displaystyle\int_{Y}\xi\left(\prod_{s=1}^{r}\varphi_{s}\circ
a(t_{s})\right)\,d\nu=$
$\displaystyle\int_{Y}\xi\left(\prod_{s=1}^{k}\varphi^{\star}_{s}\circ
a(t_{s})\right)\left(\prod_{s=k+1}^{r}\varphi_{s}\circ a(t_{s})\right)\,d\nu$
$\displaystyle+O\Big{(}L\|w^{(k+1)}\|+\max\big{(}1,L\|w^{(1)}\|\big{)}^{k\ell}e^{-\delta_{k}D(t_{1},\ldots,t_{k})/2}$
$\displaystyle\quad\quad\quad\quad\quad+\max\big{(}1,L\|w^{(k)}\|\big{)}^{-c/2}\Big{)}{\prod}_{s=1}^{r}\|\varphi_{s}\|_{C^{\ell}}.$

Since $\|w^{(s)}\|=e^{\alpha(t_{s})}$, it remains to optimize

$L\,e^{\alpha(t_{k+1})}+\max\big{(}1,L\,e^{\alpha(t_{1})}\big{)}^{k\ell}\cdot
e^{-\delta_{k}D(t_{1},\ldots,t_{k})/2}+\max\big{(}1,L\,e^{\alpha(t_{k})}\big{)}^{-c/2}$

by choosing $L$ suitably. We recall that

$\alpha(t_{k}-t_{k+1})\geq\theta/r\,D(t_{1},\ldots.t_{r})$

We consider two cases:

Case A: $\alpha(t_{1}-t_{k})\leq\rho\,D(t_{1},\ldots,t_{r})$ with
$\rho:=\delta_{k}/(8k\ell)$.

In this case, we take

$L=e^{-\alpha(t_{k})}\cdot e^{\epsilon D(t_{1},\ldots,t_{r})}\quad\hbox{with
$\epsilon=\min(\theta/(2r),\rho)$.}$

Then

$\displaystyle L\,e^{\alpha(t_{k+1})}$
$\displaystyle=e^{\alpha(t_{k+1}-t_{k})}\cdot e^{\epsilon
D(t_{1},\ldots,t_{r})}\leq e^{-\epsilon\,D(t_{1},\ldots,t_{r})},$
$\displaystyle\max\big{(}1,L\,e^{\alpha(t_{1})}\big{)}^{k\ell}e^{-\delta_{k}D(t_{1},\ldots,t_{k})/2}$
$\displaystyle\leq e^{-(\delta_{k}/2-(\rho+\varepsilon)\cdot
k\ell)D(t_{1},\ldots,t_{k})}=e^{-\delta_{k}/4\,D(t_{1},\ldots,t_{r})},$
$\displaystyle\max\big{(}1,L\,e^{\alpha(t_{k})}\big{)}^{-c/2}$
$\displaystyle\leq e^{-\epsilon c/2\,D(t_{1},\ldots,t_{r})}.$

Therefore, we conclude that for some $\delta^{\prime}>0$,

$\displaystyle\int_{Y}\xi\left(\prod_{s=1}^{r}\varphi_{s}\circ
a(t_{s})\right)\,d\nu=$
$\displaystyle\int_{Y}\xi\left(\prod_{s=1}^{k}\varphi^{\star}_{s}\circ
a(t_{s})\prod_{s=k+1}^{r}\varphi_{s}\circ a(t_{s})\right)\,d\nu$
$\displaystyle\quad\quad\quad\quad\quad\quad+O\Big{(}e^{-\delta^{\prime}\,D(t_{1},\ldots,t_{r})}\Big{)}{\prod}_{s=1}^{r}\|\varphi_{s}\|_{C^{\ell}}.$

Furthermore, for some $s_{0}=1,\ldots k$, the direction $w$ is mixing for
$I_{s_{0}}$, so that

$\varphi^{\star}_{s_{0}}=\int_{X_{I_{s_{0}}}}\varphi_{s_{0}}\,d\mu_{I_{s_{0}}}.$

Then

$\displaystyle\int_{Y}\xi\left(\prod_{s=1}^{k}\varphi^{\star}_{s}\circ
a(t_{s})\right)\left(\prod_{s=k+1}^{r}\varphi_{s}\circ a(t_{s})\right)\,d\nu$
(6.19) $\displaystyle=$
$\displaystyle\int_{X_{I_{s_{0}}}}\varphi_{s_{0}}\,d\mu_{I_{s_{0}}}\cdot\int_{Y}\xi\left(\prod_{1\leq
s\leq k,s\neq s_{0}}\varphi^{\star}_{s}\circ
a(t_{s})\right)\left(\prod_{s=k+1}^{r}\varphi_{s}\circ a(t_{s})\right)\,d\nu,$

and the estimate follows by induction on the number of factors.

Case B: $\alpha(t_{1}-t_{k})>\rho\,D(t_{1},\ldots,t_{r})$ with
$\rho=\delta_{k}/(8k\ell)$.

In this case, there exists $k_{1}=1,\ldots k-1$ such that

$\alpha(t_{k_{1}}-t_{k_{1}+1})>\rho/(k-1)\cdot D(t_{1},\ldots,t_{r})$ (6.20)

Then we change our choice of $k$ to $k_{1}$ and apply the above argument with
$k_{1}$. Again we land either in Case A or Case B (with $k$ replaced by
$k_{1}$). In Case A, we get

$\displaystyle\int_{Y}\xi\left(\prod_{s=1}^{r}\varphi_{s}\circ
a(t_{s})\right)\,d\nu=$
$\displaystyle\int_{Y}\xi\left(\prod_{s=1}^{k_{1}}\varphi^{\star}_{s}\circ
a(t_{s})\right)\left(\prod_{s=k_{1}+1}^{r}\varphi_{s}\circ
a(t_{s})\right)\,d\nu$ (6.21)
$\displaystyle\quad+O\Big{(}e^{-\delta^{\prime}\,D(t_{1},\ldots,t_{r})}\Big{)}{\prod}_{s=1}^{r}\|\phi_{s}\|_{C^{\ell}}$

for some $\delta^{\prime}>0$. In Case B, we can replace $k_{1}$ by a smaller
index, so that a version of (6.20) still holds. Ultimately, for some choice of
$k_{1}\leq k$ we get the estimate (6.21), and it remains to analyze the latter
integral. If there exists $s\leq k_{1}$ such that $w$ is a mixing direction
for $I_{s}$, we can argue as in (6.19) to finish the proof. Otherwise, we may
assume that the functions $\varphi_{s}$, $s=1,\ldots,k_{1}$, are invariant
under the subgroup $U_{I_{s},w}$. Then (6.6) can be rewritten as

$\displaystyle I(\xi)$
$\displaystyle=\int_{Y}\xi\left(\frac{1}{L}\int_{0}^{L}\xi(\exp(uw))\prod_{s=1}^{k_{1}}\varphi_{s}\circ
a(t_{s})\prod_{s=k_{1}+1}^{r}\varphi_{s}\circ\exp(uw^{(s)})\circ
a(t_{s})\,du\right)\,d\nu.$

We note that the one-parameter subgroups $\exp(uw^{(s)})$ appear only in the
factors $s=k_{1}+1,\ldots,r$. In this case, our argument gives

$\displaystyle\int_{Y}\xi\left(\prod_{s=1}^{r}\varphi_{s}\circ
a(t_{s})\right)\,d\nu$ $\displaystyle=$
$\displaystyle\int_{Y}\xi\left(\prod_{s=1}^{k}\varphi^{\star}_{s}\circ
a(t_{s})\right)\left(\prod_{s=k+1}^{r}\varphi_{s}\circ a(t_{s})\right)\,d\nu$
$\displaystyle\quad\quad+O\Big{(}L\|w^{(k+1)}\|+\max\big{(}1,L\|w^{(k_{1}+1)}\|\big{)}^{(k-k_{1})\ell}e^{-\delta_{k-k_{1}}D(t_{k_{1}+1},\ldots,t_{k})/2}$
$\displaystyle\quad\quad\quad\quad\quad\quad\quad\quad+\max\big{(}1,L\|w^{(k)}\|\big{)}^{-c/2}\Big{)}{\prod}_{s=1}^{r}\|\varphi_{s}\|_{C^{\ell}}.$

As above we consider two cases:

  * •

Case A: $\alpha(t_{k_{1}+1}-t_{k})\leq\rho\,D(t_{1},\ldots,t_{r})$ with
$\rho:=\delta_{k-k_{1}}/(8(k-k_{1})\ell)$.

  * •

Case B: $\alpha(t_{k_{1}+1}-t_{k})>\rho\,D(t_{1},\ldots,t_{r})$.

In Case A, we complete the proof by induction on the number of factors using
that
$\varphi^{\star}_{s_{0}}=\int_{X_{I_{s_{0}}}}\varphi_{s_{0}}\,d\mu_{I_{s_{0}}}$
for some $s_{0}=k_{1}+1,\ldots k$.

In Case B, we argue as above to show that there exists
$k_{2}=k_{1}+1,\ldots,k$ such that

$\displaystyle\int_{Y}\xi\left(\prod_{s=1}^{r}\varphi_{s}\circ
a(t_{s})\right)\,d\nu=$
$\displaystyle\int_{Y}\xi\left(\prod_{s=1}^{k_{2}}\varphi^{\star}_{s}\circ
a(t_{s})\right)\left(\prod_{s=k_{2}+1}^{r}\varphi_{s}\circ
a(t_{s})\right)\,d\nu$
$\displaystyle\quad\quad\quad\quad+O\Big{(}e^{-\delta^{\prime}\,D(t_{1},\ldots,t_{r})}\Big{)}{\prod}_{s=1}^{r}\|\varphi_{s}\|_{C^{\ell}}$

for some $\delta^{\prime}>0$. Therefore, we can assume that all functions
$\varphi_{s}$ with $s=1,\ldots,k_{2}$ are are $U_{w,I_{s}}$-invariant.

Arguing this way, we eventually either land in Case A or reach the point when
$s_{0}\leq k_{2}$, so that we can complete the proof by induction on the
number of factors.

This completes the proof of Theorem 6.1.

##  7\. Quasi-independence for separated tuples

For $I\subset\\{1,\ldots,m+n\\}$ and $t\in A^{+}$, we define

$\lfloor t\rfloor_{I}:=\min_{i\in I}t_{i}\quad\textrm{and}\quad\lceil
t\rceil_{I}:=\max_{i\in I}t_{i}.$

When $I=\emptyset$, we set $\lfloor t\rfloor_{I}=\lceil t\rceil_{I}=0$. For
simplicity, we also write

$\lfloor t\rfloor:=\lfloor
t\rfloor_{\\{1,\ldots,m+n\\}}\quad\textrm{and}\quad\lceil t\rceil:=\lceil
t\rceil_{\\{1,\ldots,m+n\\}}.$

We fix $\ell\geq 1$ and $\delta>0$ such that Theorem 6.1 holds.

######  Lemma 7.1.

Let $\alpha,\beta\in(0,1)$ and let
$I_{1},\ldots,I_{r}\subseteq\\{1,\ldots,m+n\\}$ be admissible subsets. Then
for every $f\in C^{\infty}(Y)$, $\varphi_{1},\ldots,\varphi_{r}\in
C^{\infty}_{c}(X)$, and $t_{1},\ldots,t_{r}\in A^{+}$ such that

$\displaystyle\min\big{(}\lfloor t_{1}\rfloor_{I_{1}},\ldots,\lfloor
t_{r}\rfloor_{I_{r}}\big{)}$
$\displaystyle\geq\alpha\,\Delta(t_{1},\ldots,t_{r}),$ (7.1)
$\displaystyle\max\big{(}\lceil t_{1}\rceil_{I_{1}^{c}},\ldots,\lceil
t_{r}\rceil_{I_{r}^{c}}\big{)}$
$\displaystyle\leq\beta\,\Delta(t_{1},\ldots,t_{r}),$ (7.2)

we have

$\displaystyle\int_{Y}f\left(\prod_{s=1}^{r}\varphi_{s}\circ
a(t_{s})\right)\,d\nu=$
$\displaystyle\int_{Y}f\,d\nu\prod_{s=1}^{r}\int_{Y}\varphi_{s}\circ
a(t_{s})\,d\nu$
$\displaystyle\quad\quad\quad+O\left(e^{-(\delta\alpha-2r\ell\beta)\Delta(t_{1},\ldots,t_{r})}\,\|f\|_{W}{\prod}_{s=1}^{r}\|\varphi_{s}\|_{C^{\ell}}\right).$

###### Proof.

We write

$t_{s}=t^{\prime}_{s}+t^{\prime\prime}_{s}\quad\hbox{with $t^{\prime}_{s}\in
A^{+}_{I_{s}}$ and $t^{\prime\prime}_{s}\in A^{+}_{I_{s}^{c}}$.}$

Then

$\int_{Y}f\left(\prod_{s=1}^{r}\varphi_{s}\circ
a(t_{s})\right)\,d\nu=\int_{Y}f\left(\prod_{s=1}^{r}\psi_{s}\circ
a(t^{\prime}_{s})\right)\,d\nu,$

where

$\psi_{s}:=\varphi_{s}\circ a(t^{\prime\prime}_{s}).$

By Theorem 6.1,

$\displaystyle\int_{Y}f\left(\prod_{s=1}^{r}\psi_{s}\circ
a(t^{\prime}_{s})\right)\,d\nu=$
$\displaystyle\int_{Y}f\,d\nu\prod_{s=1}^{r}\int_{X_{I_{s}}}\psi_{s}\,d\mu_{I_{s}}$
(7.3) $\displaystyle\quad+O\left(e^{-\delta\min\big{(}\lfloor
t^{\prime}_{1}\rfloor_{I_{1}},\ldots,\lfloor
t^{\prime}_{r}\rfloor_{I_{r}},\Delta(t^{\prime}_{1},\ldots,t^{\prime}_{r})\big{)}}\,\|f\|_{W}{\prod}_{s=1}^{r}\|\psi_{s}\|_{C^{\ell}}\right).$

We claim that for any $1\leq s_{1},s_{2}\leq r$,

$\|t^{\prime}_{s_{1}}-t^{\prime}_{s_{2}}\|\geq\|t_{s_{1}}-t_{s_{2}}\|.$ (7.4)

Let us suppose that $\|t_{s_{1}}-t_{s_{2}}\|=|t_{s_{1},k}-t_{s_{2},k}|$ for
some $1\leq k\leq m+n$. Without loss of generally, we may assume that
$t_{s_{1},k}\geq t_{s_{2},k}$. When $k\in I_{s_{1}}\cap I_{s_{2}}$,

$|t_{s_{1},k}-t_{s_{2},k}|=|t^{\prime}_{s_{1},k}-t^{\prime}_{s_{2},k}|\leq\|t^{\prime}_{s_{1}}-t^{\prime}_{s_{2}}\|,$

so that (7.4) holds. When $k\in I_{s_{1}}\backslash I_{s_{2}}$, we also obtain
that

$|t_{s_{1},k}-t_{s_{2},k}|\leq
t_{s_{1},k}=|t^{\prime}_{s_{1},k}-t^{\prime}_{s_{2},k}|$

since $t^{\prime}_{s_{2},k}=0$. When $k\notin I_{s_{1}}$, it follows from
(7.2) that

$|t_{s_{1},k}-t_{s_{2},k}|\leq t_{s_{1},k}\leq\beta\,\|t_{s_{1}}-t_{s_{2}}\|.$

Since $\beta<1$, this case is not possible. This verifies (7.4).

In view of (7.4) and (7.1), the estimate (7.3) gives

$\displaystyle\int_{Y}f\left(\prod_{s=1}^{r}\psi_{s}\circ
a(t^{\prime}_{s})\right)\,d\nu$ $\displaystyle=$
$\displaystyle\int_{Y}f\,d\nu\prod_{s=1}^{r}\int_{X_{I_{s}}}\psi_{s}\,d\mu_{I_{s}}$
$\displaystyle\quad\quad+\left(e^{-\delta\min\big{(}\lfloor
t^{\prime}_{1}\rfloor_{I_{1}},\ldots,\lfloor
t^{\prime}_{r}\rfloor_{I_{r}},\Delta(t_{1},\ldots,t_{r})\big{)}}\,\|f\|_{W}{\prod}_{s=1}^{r}\|\psi_{s}\|_{C^{\ell}}\right)$
$\displaystyle=$
$\displaystyle\int_{Y}f\,d\nu\prod_{s=1}^{r}\int_{X_{I_{s}}}\psi_{s}\,d\mu_{I_{s}}+O\left(e^{-\delta\alpha\,\Delta(t_{1},\ldots,t_{r})}\,\|f\|_{W}{\prod}_{s=1}^{r}\|\psi_{s}\|_{C^{\ell}}\right).$

Since by Theorem 6.1 with $r=1$,

$\int_{Y}\psi_{s}\circ
a(t^{\prime}_{s})\,d\nu=\int_{X_{I_{s}}}\psi_{s}\,d\mu_{I_{s}}+O\left(e^{-\delta\lfloor
t^{\prime}_{s}\rfloor_{I_{s}}}\|\psi_{s}\|_{C^{\ell}}\right),$

we conclude that

$\prod_{s=1}^{r}\int_{X_{I_{s}}}\psi_{s}\,d\mu_{I_{s}}=\prod_{s=1}^{r}\int_{Y}\psi_{s}\circ
a(t^{\prime}_{s})\,d\nu+O\left(e^{-\delta\min\big{(}\lfloor
t^{\prime}_{1}\rfloor_{I_{1}},\ldots,\lfloor
t^{\prime}_{r}\rfloor_{I_{r}}\big{)}}\,{\prod}_{s=1}^{r}\|\psi_{s}\|_{C^{\ell}}\right),$

and

$\displaystyle\int_{Y}f\left(\prod_{s=1}^{r}\psi_{s}\circ
a(t^{\prime}_{s})\right)\,d\nu=$
$\displaystyle\int_{Y}f\,d\nu\prod_{s=1}^{r}\int_{Y}\psi_{s}\circ
a(t^{\prime}_{s})\,d\nu$ (7.5)
$\displaystyle\quad+O\left(e^{-\delta\alpha\,\Delta(t_{1},\ldots,t_{r})}\,\|f\|_{W}{\prod}_{s=1}^{r}\|\psi_{s}\|_{C^{\ell}}\right).$

where

$\prod_{s=1}^{r}\int_{Y}\psi_{s}\circ
a(t^{\prime}_{s})\,d\nu=\prod_{s=1}^{r}\int_{Y}\varphi_{s}\circ
a(t_{s})\,d\nu.$

Furthermore, it follows from (7.2) that

$\|\psi_{s}\|_{C^{\ell}}\ll e^{2\ell\lceil
t^{\prime\prime}_{s}\rceil_{I_{s}^{c}}}\|\varphi_{s}\|_{C^{\ell}}=e^{2\ell\lceil
t_{s}\rceil_{I_{s}^{c}}}\|\varphi_{s}\|_{C^{\ell}}\leq
e^{2\ell\beta\,\Delta(t_{1},\ldots,t_{r})}\|\varphi_{s}\|_{C^{\ell}}.$

This implies the result. ∎

##  8\. Proof of main theorem

We fix $\ell\geq 1$ and $\delta>0$ such that Theorem 6.1 holds and set

$\lambda:=\max\big{(}2,{4r\ell}/{\delta}\big{)}.$ (8.1)

Since the case $r=1$ is trivial, we may assume by induction that the theorem
holds for a smaller number of factors. Namely, for every $r_{1}\leq r-1$,
there exists $\eta(r_{1})>0$

$\displaystyle\int_{Y}\left(\prod_{s=1}^{r_{1}}\varphi_{s}\circ
a(t_{s})\right)\,d\nu=$
$\displaystyle\prod_{s=1}^{r_{1}}\int_{Y}\varphi_{s}\circ
a(t_{s})\,d\nu+O\left(e^{-\eta(r_{1})\,\Delta(t_{1},\ldots,t_{r})}{\prod}_{s=1}^{r_{1}}\|\varphi_{s}\|_{C^{\ell}}\right).$
(8.2)

Throughout the proof, we use constants $c_{1},\ldots,c_{r}$ such that

$c_{r}\geq c_{r-1}\geq\cdots\geq c_{1}>\max(m,n),$ (8.3)

which will be specified later.

We split the set of parameters $(t_{1},\ldots,t_{r})\in(A^{+})^{r}$ into
multiple domains defined in terms of the constants $c_{1},\ldots,c_{r}$. These
domains correspond to Cases $1^{\prime}_{k}$, $1^{\prime\prime}_{k}$, $2_{k}$
with $k=1,\ldots,r$, described below.

Case $1_{r}$: there exist admissible subsets
$I_{1},\ldots,I_{r}\subset\\{1,\ldots,m+n\\}$ such that

$\Delta(t_{1},\ldots,t_{r})\leq
c_{r}\min\big{(}\lfloor{t_{1}}\rfloor_{I_{1}},\ldots,\lfloor{t_{r}}\rfloor_{I_{r}}\big{)}.$

In this case, we distinguish two possibilities — Cases $1^{\prime}_{r}$ and
$1^{\prime\prime}_{r}$:

Case $1^{\prime}_{r}$: the lower bound

$\Delta(t_{1},\ldots,t_{r})\geq\lambda c_{r}\max\big{(}\lceil
t_{1}\rceil_{I_{1}^{c}},\ldots,\lceil t_{r}\rceil_{I_{r}^{c}}\big{)}$

holds, where $\lambda$ is defined in (8.1).

In this case, Lemma 7.1 is applicable with $\alpha=c_{r}^{-1}$ and
$\beta=(\lambda c_{r})^{-1}<c_{r}^{-1}$, and we conclude that

$\displaystyle\int_{Y}\left(\prod_{s=1}^{r}\varphi_{s}\circ
a(t_{s})\right)\,d\nu$ $\displaystyle=\prod_{s=1}^{r}\int_{Y}\varphi_{s}\circ
a(t_{s})\,d\nu+O\left(e^{-(\delta/c_{r}-2r\ell/\lambda
c_{r})\Delta(t_{1},\ldots,t_{r})}\,{\prod}_{s=1}^{r}\|\varphi_{s}\|_{C^{\ell}}\right)$
$\displaystyle=\prod_{s=1}^{r}\int_{Y}\varphi_{s}\circ
a(t_{s})\,d\nu+O\left(e^{-\delta/(2c_{r})\,\Delta(t_{1},\ldots,t_{r})}\,{\prod}_{s=1}^{r}\|\varphi_{s}\|_{C^{\ell}}\right).$

This gives the required estimate.

On the other hand, if the assumption in Case $1^{\prime}_{r}$ fails, we have:

Case $1^{\prime\prime}_{r}$: at least one of the sets

$J_{s}:=\big{\\{}j\in I_{s}^{c}\,:\,t_{s,j}>(\lambda
c_{r})^{-1}\Delta(t_{1},\ldots,t_{r})\big{\\}},\quad s=1,\ldots,r,$

is non-empty.

In this case, we introduce the new sets

$I^{\prime}_{s}:=I_{s}\sqcup J_{s},$

and note that

$\Delta(t_{1},\ldots,t_{r})\leq\lambda c_{r}\min\big{(}\lfloor
t_{1}\rfloor_{I_{1}^{\prime}},\ldots,\lfloor
t_{r}\rfloor_{I_{r}^{\prime}}\big{)}.$

This means that we are in case $1_{r}$ with $\lambda c_{r}$ instead of $c_{r}$
and $I_{s}^{\prime}$ instead of $I_{s}$. The key point in this argument is
that at least one of the sets $I^{\prime}_{s}$ is strictly larger than the
corresponding original sets $I_{s}$.

The new Case $1^{\prime}_{r}$ is that

$\lambda^{2}c_{r}\max\big{(}\lceil
t_{1}\rceil_{(I^{\prime}_{1})^{c}},\ldots\lceil
t_{r}\rceil_{(I^{\prime}_{r})^{c}}\big{)}\leq\Delta(t_{1},\ldots,t_{r}),$

which then implies as before that

$\displaystyle\int_{Y}\left(\prod_{s=1}^{r}\varphi_{s}\circ
a(t_{s})\right)\,d\nu=$ $\displaystyle\prod_{s=1}^{r}\int_{Y}\varphi_{s}\circ
a(t_{s})\,d\nu+O\left(e^{-\delta/(2\lambda
c_{r})\,\Delta(t_{1},\ldots,t_{r})}\,{\prod}_{s=1}^{r}\|\varphi_{s}\|_{C^{\ell}}\right).$

The new Case $1^{\prime\prime}_{r}$ is that at least one of the sets

$J^{\prime}_{s}:=\big{\\{}j\in(I^{\prime}_{s})^{c}\,:\,t_{s,j}>(1/\lambda^{2}c_{r})\Delta(t_{1},\ldots,t_{r})\big{\\}},\quad
s=1,\ldots,r,$

is non-empty. If this happens, we consider again the new sets

$I_{s}^{\prime\prime}:=I^{\prime}_{s}\sqcup J^{\prime}_{s}$

and note that

$\Delta(t_{1},\ldots,t_{r})\leq\lambda^{2}c_{r}\min\big{(}\lfloor
t_{1}\rfloor_{I_{1}^{\prime\prime}},\ldots,\lfloor
t_{r}\rfloor_{I_{r}^{\prime\prime}}\big{)}.$

We can now again restart Case $1_{r}$, but with $\lambda^{2}c_{r}$ instead of
$\lambda c_{r}$, and $I_{s}^{\prime\prime}$ instead of $I_{s}^{\prime}$. Again
at least one of the sets $I_{s}^{\prime\prime}$ has strictly increased. For
each such restart, we are either in the favourable Case $1_{r}^{\prime}$, when
we always get a suitable bound, or we are in the complementary Case
$1_{r}^{\prime\prime}$, which allows us to increase (at least one of) the
index sets, and give Case $1_{r}^{\prime}$ a new chance. Since there are at
most $r(m+n)$ indices in the union of the $r$ index sets, there can be at most
$r(m+n)$ restarts, after which we must reach Case $1_{r}^{\prime}$. Therefore,
the resulting bound after these iterations is:

$\displaystyle\int_{Y}\left(\prod_{s=1}^{r}\varphi_{s}\circ
a(t_{s})\right)\,d\nu=$ $\displaystyle\prod_{s=1}^{r}\int_{Y}\varphi_{s}\circ
a(t_{s})\,d\nu$
$\displaystyle\quad\quad+O\left(e^{-\delta/(2\lambda^{r(m+n)}c_{r})\,\Delta(t_{1},\ldots,t_{r})}\,{\prod}_{s=1}^{r}\|\varphi_{s}\|_{C^{\ell}}\right).$

Hence, in Case $1_{r}$ we always get a favourable bound.

The complement of Case $1_{r}$ is:

Case $2_{r}$: for all admissible
$I_{1},\ldots,I_{r}\subset\\{1,\ldots,m+n\\}$, we have

$\Delta(t_{1},\ldots,t_{r})>c_{r}\min\big{(}\lfloor
t_{1}\rfloor_{I_{1}},\ldots\lfloor t_{r}\rfloor_{I_{r}}\big{)}.$

For $t\in A^{+}$, we set

$\lceil t\rceil^{\prime}:=\min\big{(}\lceil t\rceil_{\\{1,\ldots,m\\}},\lceil
t\rceil_{\\{m+1,\ldots,m+n\\}}\big{)}.$

We recall that for $t\in A^{+}$, one has
$t_{1}+\cdots+t_{m}=t_{m+1}+\cdots+t_{m+n}$, so that

$\lceil t\rceil_{\\{1,\ldots,m\\}}\leq n\cdot\lceil
t\rceil_{\\{m+1,\ldots,m+n\\}}\quad\hbox{and}\quad m\cdot\lceil
t\rceil_{\\{1,\ldots,m\\}}\geq\lceil t\rceil_{\\{m+1,\ldots,m+n\\}}.$

Hence,

$\lceil t\rceil^{\prime}\geq\max(m,n)^{-1}\cdot\lceil t\rceil.$

By varying the sets $I_{s}$ over all possible two element admissible sets, we
see that Case $2_{r}$ can be equivalently formulated as:

$\Delta(t_{1},\ldots,t_{r})>c_{r}\min\big{(}\lceil
t_{1}\rceil^{\prime},\ldots\lceil t_{r}\rceil^{\prime}\big{)}.$

Then

$\Delta(t_{1},\ldots,t_{r})>c_{r}\max(m,n)^{-1}\min\big{(}\lceil
t_{1}\rceil,\ldots\lceil t_{r}\rceil\big{)}.$

This means that there exists an index $s$ such that

$\Delta(t_{1},\ldots,t_{r})>c_{r}\max(m,n)^{-1}\lceil t_{s}\rceil.$

Without loss of generality, we may assume that

$\Delta(t_{1},\ldots,t_{r})>c_{r}\max(m,n)^{-1}\lceil t_{r}\rceil.$ (8.4)

In this case, we introduce a function

$f_{r-1}:=\varphi_{r}\circ a(t_{r}).$

From (8.4),

$\big{\|}f_{r-1}|_{Y}\big{\|}_{W}\leq\|f_{r-1}\|_{C^{1}}\ll e^{2\lceil
t_{r}\rceil}\|\varphi_{r}\|_{C^{1}}\ll
e^{2\max(m,n)/c_{r}\,\Delta(t_{1},\ldots,t_{r})}\|\varphi_{r}\|_{C^{1}},$

and

$\int_{Y}\left(\prod_{s=1}^{r}\varphi_{s}\circ
a(t_{s})\right)\,d\nu=\int_{Y}f_{r-1}\left(\prod_{s=1}^{r-1}\varphi_{s}\circ
a(t_{s})\right)\,d\nu.$

Now we apply the above reasoning to the later integral and consider the
corresponding cases:

Case $1_{r-1}$: we have

$\Delta(t_{1},\ldots,t_{r})>c_{r}\max(m,n)^{-1}\lceil t_{r}\rceil,$

and there are admissible subsets
$I_{1},\ldots,I_{r-1}\subset\\{1,\ldots,m+n\\}$ such that

$\Delta(t_{1},\ldots,t_{r})\leq
c_{r-1}\min\big{(}\lfloor{t_{1}}\rfloor_{I_{1}},\ldots,\lfloor{t_{r-1}}\rfloor_{I_{r-1}}\big{)}.$

This case splits as above into two complementary subcases:

Case $1^{\prime}_{r-1}$: we have

$\Delta(t_{1},\ldots,t_{r})\geq\lambda c_{r-1}\max\big{(}\lceil
t_{1}\rceil_{I_{1}^{c}},\ldots,\lceil t_{r-1}\rceil_{I_{r-1}^{c}}\big{)}.$

As in Case $1^{\prime}_{r}$ above, we apply Lemma 7.1 with
$\alpha=c_{r-1}^{-1}$ and $\beta=(\lambda c_{r-1})^{-1}<c_{r-1}^{-1}$, and we
can conclude that

$\displaystyle\int_{Y}f_{r-1}\left(\prod_{s=1}^{r-1}\varphi_{s}\circ
a(t_{s})\right)\,d\nu$ $\displaystyle=$
$\displaystyle\left(\int_{Y}f_{r-1}\,d\nu\right)\prod_{s=1}^{r-1}\int_{Y}\varphi_{s}\circ
a(t_{s})\,d\nu$ $\displaystyle\quad+O\left(e^{-(\delta/c_{r-1}-2r\ell/\lambda
c_{r-1})\Delta(t_{1},\ldots,t_{r})}\,\big{\|}f_{r-1}|_{Y}\big{\|}_{W}{\prod}_{s=1}^{r-1}\|\varphi_{s}\|_{C^{\ell}}\right)$
$\displaystyle=$ $\displaystyle\prod_{s=1}^{r}\int_{Y}\varphi_{s}\circ
a(t_{s})\,d\nu$
$\displaystyle\quad+O\left(e^{-(\delta/(2c_{r-1})-2\max(m,n)/c_{r})\,\Delta(t_{1},\ldots,t_{r})}\,{\prod}_{s=1}^{r}\|\varphi_{s}\|_{C^{\ell}}\right).$

Case $1^{\prime\prime}_{r-1}$: at least one of the sets

$J_{s}:=\big{\\{}j\in I_{s}^{c}\,:\,t_{s,j}>(1/\lambda
c_{r-1})\Delta(t_{1},\ldots,t_{r})\big{\\}},\quad s=1\ldots,r-1,$

is non-empty.

As in Case $1^{\prime\prime}_{r}$, we define the new sets

$I^{\prime}_{s}:=I_{s}\sqcup J_{s},\quad s=1,\ldots,r-1,$

and note that

$\Delta(t_{1},\ldots,t_{r})\leq\lambda c_{r-1}\min(\lfloor
t_{1}\rfloor_{I_{1}^{\prime}},\ldots,\lfloor t_{r}\rfloor_{I_{r}^{\prime}}).$

Hence, we can now restart Case $1_{r-1}$ with $\lambda c_{r-1}$ instead of
$c_{r-1}$ and $I_{s}^{\prime}$ instead of $I_{s}$. We apply this argument
recursively. If eventually we arrive to Case $1^{\prime}_{r-1}$ and obtain the
bound:

$\displaystyle\int_{Y}\left(\prod_{s=1}^{r}\varphi_{s}\circ
a(t_{s})\right)\,d\nu=$ $\displaystyle\prod_{s=1}^{r}\int_{Y}\varphi_{s}\circ
a(t_{s})\,d\nu$
$\displaystyle+O\left(e^{-\big{(}\delta/(2\lambda^{r(m+n)}c_{r-1})-2\max(m,n)/c_{r}\big{)}\Delta(t_{1},\ldots,t_{r})}\,{\prod}_{s=1}^{r}\|\varphi_{s}\|_{C^{\ell}}\right).$

This provides a favourable estimate when

$\delta/(2\lambda^{r(m+n)}c_{r-1})-2\max(m,n)/c_{r}>0.$

The remaining case is:

Case $2_{r-1}$: we have

$\Delta(t_{1},\ldots,t_{r})>c_{r}\max(m,n)^{-1}\lceil t_{r}\rceil,$

and for all admissible $I_{1},\ldots,I_{r-1}\subset\\{1,\ldots,m+n\\}$,

$\Delta(t_{1},\ldots,t_{r})>c_{r-1}\min\big{(}\lfloor
t_{1}\rfloor_{I_{1}},\ldots\lfloor t_{r-1}\rfloor_{I_{r-1}}\big{)}.$

As above, the last inequality implies that

$\Delta(t_{1},\ldots,t_{r})>c_{r-1}\max(m,n)^{-1}\min\big{(}\lceil
t_{1}\rceil,\ldots\lceil t_{r-1}\rceil\big{)}.$

This means that there exists $1\leq s\leq r-1$ such that

$\Delta(t_{1},\ldots,t_{r})>c_{r-1}\max(m,n)^{-1}\lceil t_{s}\rceil.$

Without loss of generality, we may assume that

$\Delta(t_{1},\ldots,t_{r})>c_{r-1}\max(m,n)^{-1}\lceil t_{r-1}\rceil.$

Since $c_{r}\geq c_{r-1}$, we have

$\Delta(t_{1},\ldots,t_{r})>c_{r-1}\max(m,n)^{-1}\max\big{(}\lceil
t_{r-1}\rceil,\lceil t_{r}\rceil\big{)}.$

In this situation, we introduce the function

$f_{r-2}:=\big{(}\varphi_{r-1}\circ
a(t_{r-1})\big{)}\cdot\big{(}\varphi_{r}\circ a(t_{r})\big{)}$

and apply the argument as above to the product
$f_{r-2}\left(\prod_{s=1}^{r-2}\varphi_{s}\circ a(t_{s})\right)$. We treat
this product by considering Cases $1_{r-2}$ and $2_{r-2}$ defined as above.
Then we apply this reasoning again to deal with the remaining case and so on.

Generally, the outlined procedure will allow us to reduce the argument to the
situation when

$\Delta(t_{1},\ldots,t_{r})>c_{k+1}\max(m,n)^{-1}\max\big{(}\lceil
t_{k+1}\rceil,\ldots,\lceil t_{r}\rceil\big{)}.$ (8.5)

Then we introduce the function

$f_{k}:=\prod_{s=k+1}^{r}\varphi_{s}\circ a(t_{s}).$

It follows from (8.5) that

$\displaystyle\big{\|}f_{k}|_{Y}\big{\|}_{W}$
$\displaystyle\leq\prod_{s=k+1}^{r}\big{\|}\varphi_{s}\circ
a(t_{s})|_{Y}\big{\|}_{W}\leq\prod_{s=k+1}^{r}\|\varphi_{s}\circ
a(t_{s})\|_{C^{1}}\ll\prod_{s=k+1}^{r}e^{2\lceil
t_{s}\rceil}\|\varphi_{s}\|_{C^{1}}$ $\displaystyle\ll
e^{2(r-k)\max(m,n)/c_{k+1}\,\Delta(t_{1},\ldots,t_{k})}\prod_{s=k+1}^{r}\|\varphi_{s}\|_{C^{1}}.$

and

$\int_{Y}\left(\prod_{s=1}^{r}\varphi_{s}\circ
a(t_{s})\right)\,d\nu=\int_{Y}f_{k}\left(\prod_{s=1}^{k}\varphi_{s}\circ
a(t_{s})\right)\,d\nu.$

To analyze this integral, we consider the cases:

Case $1_{k}$: we have

$\Delta(t_{1},\ldots,t_{r})>c_{k+1}\max(m,n)^{-1}\,\max\big{(}\lceil
t_{k+1}\rceil,\ldots\lceil t_{r}\rceil\big{)},$

and there are admissible subsets $I_{1},\ldots,I_{k}\subset\\{1,\ldots,m+n\\}$
such that

$\Delta(t_{1},\ldots,t_{r})\leq
c_{k}\min\big{(}\lfloor{t_{1}}\rfloor_{I_{1}},\ldots,\lfloor{t_{k}}\rfloor_{I_{k}}\big{)}$

Case $1_{k}$ splits into two subcases:

Case $1^{\prime}_{k}$: we have

$\Delta(t_{1},\ldots,t_{r})\geq\lambda c_{k}\max\big{(}\lceil
t_{1}\rceil_{I_{1}^{c}},\ldots,\lceil t_{k}\rceil_{I_{k}^{c}}\big{)}.$

Applying Lemma 7.1 with $\alpha=c_{k}^{-1}$ and $\beta=(\lambda
c_{k})^{-1}<c_{k}^{-1}$, we conclude that

$\displaystyle\int_{Y}f_{k}\left(\prod_{s=1}^{k}\varphi_{s}\circ
a(t_{s})\right)\,d\nu$ $\displaystyle=$
$\displaystyle\left(\int_{Y}f_{k}\,d\nu\right)\prod_{s=1}^{k}\int_{Y}\varphi_{s}\circ
a(t_{s})\,d\nu$
$\displaystyle\quad\quad+O\left(e^{-(\delta/c_{k}-2r\ell/\lambda
c_{k})\Delta(t_{1},\ldots,t_{r})}\,\big{\|}f_{k}|_{Y}\big{\|}_{W}{\prod}_{s=1}^{k}\|\varphi_{s}\|_{C^{\ell}}\right)$
$\displaystyle=$
$\displaystyle\left(\int_{Y}f_{k}\,d\nu\right){\prod}_{s=1}^{k}\int_{Y}\varphi_{s}\circ
a(t_{s})\,d\nu$
$\displaystyle\quad\quad+O\left(e^{-\big{(}\delta/(2c_{k})-2(r-k)\max(m,n)/c_{k+1}\big{)}\,\Delta(t_{1},\ldots,t_{r})}\,{\prod}_{s=1}^{k}\|\varphi_{s}\|_{C^{\ell}}\right).$

Our inductive assumption (8.2), gives that

$\int_{Y}f_{k}\,d\nu=\prod_{s=k+1}^{r}\int_{Y}\varphi_{s}\circ
a(t_{s})\,d\nu+O\left(e^{-\eta(r-k)\,\Delta(t_{1},\ldots,t_{r})}\,{\prod}_{s=k+1}^{r}\|\varphi_{s}\|_{C^{\ell}}\right)$

for some $\eta(r-k)>0$. Therefore, the above estimate gives a favourable bound
provided that

$\delta/(2c_{k})-2(r-k)\max(m,n)/c_{k+1}>0.$

Case $1^{\prime\prime}_{k}$: at least one of the sets

$J_{s}:=\big{\\{}j\in I_{s}^{c}\,:\,t_{s,j}>(1/\lambda
c_{k})\Delta(t_{1},\ldots,t_{r})\big{\\}},\quad s=1\ldots,k$

is non-empty.

We introduce the new sets

$I^{\prime}_{s}:=I_{s}\sqcup J_{s},\quad s=1,\ldots,k,$

and note that

$\Delta(t_{1},\ldots,t_{r})\leq\lambda c_{k}\min\big{(}\lfloor
t_{1}\rfloor_{I_{1}^{\prime}},\ldots,\lfloor
t_{r}\rfloor_{I_{k}^{\prime}}\big{)}.$

We can restart Case $1_{k}$ with $\lambda c_{k}$ instead of $c_{k}$ and
$I_{s}^{\prime}$ instead of $I_{s}$. We apply this argument recursively. Then
eventually we arrive to Case $1^{\prime}_{k}$ at some point and we obtain the
bound:

$\displaystyle\int_{Y}f_{k}\left(\prod_{s=k+1}^{r}\varphi_{s}\circ
a(t_{s})\right)\,d\nu$ $\displaystyle=$
$\displaystyle\left(\int_{Y}f_{k}\,d\nu\right)\prod_{s=1}^{k}\int_{Y}\varphi_{s}\circ
a(t_{s})\,d\nu$
$\displaystyle\quad\quad+O\left(e^{-(\delta/(2\lambda^{r(m+n)}c_{k})-2(r-k)\max(m,n)/c_{k+1})\Delta(t_{1},\ldots,t_{r})}\,{\prod}_{s=1}^{r}\|\varphi_{s}\|_{C^{\ell}}\right),$

or otherwise we end up in Case $2_{k}$ below. We note that this bound is
favourable provided that

$\delta/(2\lambda^{r(m+n)}c_{k})-2(r-k)\max(m,n)/c_{k+1}>0.$ (8.6)

The remaining case is:

Case $2_{k}$: we have

$\Delta(t_{1},\ldots,t_{r})>c_{k+1}\max(m,n)^{-1}\max\big{(}\lceil
t_{k+1}\rceil,\ldots\lceil t_{r}\rceil\big{)},$

and for all admissible $I_{1},\ldots,I_{k}\subset\\{1,\ldots,m+n\\}$, we have

$\Delta(t_{1},\ldots,t_{r})>c_{k}\min\big{(}\lfloor
t_{1}\rfloor_{I_{1}},\ldots\lfloor t_{k}\rfloor_{I_{k}}\big{)}.$

As above, the last inequality implies that

$\Delta(t_{1},\ldots,t_{r})>c_{k}\max(m,n)^{-1}\min\big{(}\lceil
t_{1}\rceil,\ldots\lceil t_{k}\rceil\big{)}.$

This means that there exists $1\leq s\leq k$ such that

$\Delta(t_{1},\ldots,t_{r})>c_{k}\max(m,n)^{-1}\lceil t_{s}\rceil.$

Without loss of generality, we may assume that

$\Delta(t_{1},\ldots,t_{r})>c_{k}\max(m,n)^{-1}\lceil t_{k}\rceil.$

Since $c_{k+1}\geq c_{k}$, we have

$\Delta(t_{1},\ldots,t_{r})>c_{k}\max(m,n)^{-1}\max\big{(}\lceil
t_{k}\rceil,\ldots\lceil t_{r}\rceil\big{)}.$

We analyze this case by considering Cases $1^{\prime}_{k-1}$,
$1^{\prime\prime}_{k-1}$, $2_{k-1}$ and so on. Proceeding in this way, we end
up with the final case when

$\Delta(t_{1},\ldots,t_{r})>c_{1}\max(m,n)^{-1}\max\big{(}\lceil
t_{1}\rceil,\ldots\lceil t_{r}\rceil\big{)}.$ (8.7)

We observe that for some $1\leq s_{1},s_{2}\leq r$ and $1\leq k\leq m+n$,

$\Delta(t_{1},\ldots,t_{r})=\|t_{s_{1}}-t_{s_{2}}\|=|t_{s_{1},k}-t_{s_{2},k}|\leq\max\big{(}\lceil
t_{s_{1}}\rceil,\lceil t_{s_{2}}\rceil\big{)}.$

Since $c_{1}>\max(m,n)$, it follows that (8.7) is impossible. Hence, we have
covered all the cases.

It remains to justify the choice of parameters $c_{1},\ldots,c_{r}$. We recall
that they must satisfy (8.3) and (8.6). Therefore, we can choose them
recursively so that $c_{1}>\max(m,n)$ and

$c_{k+1}>\max\left(1,4(r-k)\max(m,n)\lambda^{r(m+n)}/\delta\right)c_{k}.$

This completes the proof of Theorem 1.4.

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