We study the limiting distribution of dense orbits of a lattice subgroup $\Gamma\leq\SL(m+1,\mathbb{R})$ acting on $H\backslash\SL(m+1,\mathbb{R})$, with respect to a filtration of growing norm balls. The novelty of our work is that the groups $H$ we consider have infinitely many non-trivial connected components. For a specific such $H$, the homogeneous space $H\backslash G$ identifies with $X_{m,m+1}$, a moduli space of rank $m$-discrete subgroups in $\mathbb{R}^{m+1}$. This study is motivated by the work of Shapira-Sargent who studied random walks on $X_{2,3}$.

Consider the moduli space $X_{m,m+1}$ of homothety classes of rank-two discrete subgroups with orientation in $\R^{3}$. The group $\SL(m+1,\R)$ acts transitively on $X_{m,m+1}$, so that $X_{m,m+1}\cong H\backslash \SL(m+1,\R)$, where $H\leq\SL(m+1,\R)$ is a closed subgroup. It turns out that for each $x_{0}\in X_{m,m+1}$ and each lattice subgroup $\Gamma \leq \SL(m+1,\R)$, the orbit $ x_{0}\cdot \Gamma $ is dense in $X_{m,m+1}$. We compute the limiting distribution of those orbits with respect to norm balls, where the norm is given by $\|g\|^2=\sum g_{i,j}^2$, $g\in\SL(m+1,\R)$, the sum of squares of the entries of $g$. Namely, we consider $\Gamma_{T}=\{\gamma\in\Gamma : \|\gamma\| \leq T \}$, and we show that for all $\varphi\in C_c(X_{m,m+1})$

\to\infty} \frac{1}{\#\Gamma_{T}}\sum_{\gamma\in \Ga_T}\varphi(x_{0}\cdot \gamma)=c_{\Ga}\int_{X_{m,m+1}}\varphi(x)d\nu_{x_{0}}(x)$$

for some explicit measure $\nu_{x_0}$ on $X_{m,m+1}$ which depends on $x_{0}$ and constant $c_{\Ga}$ which depends on $\Gamma$. To prove our result, we use the general recipe for applying the duality principle developed by Gorodnik and Weiss which reduces the above problem to computation of volumes of growing skewed balls in H and proving ergodic theorems of the left H-action on $\SL(m+1,\R)/\Gamma$. The main novelty of our work is that we apply the duality principle in the case that H has infinitely many connected components.

§ INTRODUCTION

A well studied topic in homogeneous dynamics concerns the limiting distributions of dense orbits of a lattice $\Ga\leq G$ acting on a homogeneous space $H\backslash G$.
An notable basic result in this line of research is the work of Ledrappier <cit.> which gives the limiting distribution of the $\Ga\leq\SL(2,\R)$-dense orbits in $ \begin{bmatrix}
\end{bmatrix}\backslash \SL(2,\R)\cong\R^2\smallsetminus \textbf{0}$. An important aspect of <cit.> is the method of reducing the question on the limiting distribution of $\Ga$ orbits in $H\backslash G$ to a question on $H$ orbits in $G/\Ga$. This kind of idea, which we refer to as the duality principle, has appeared extensively in the literature, and we refer the reader to <cit.> for a detailed exposition. In this paper, we continue the above line of research, and the main novelty of our work is that the subgroup $H\leq \SL(m+1,\R)$ we consider has infinitely many non-trivial connected components. In this regard, we note that general results for a connected $H$ were obtained in <cit.>, and results for $H$ being a lattice were given in <cit.>.  Our method will also exploit the duality principle through a general theorem of <cit.>.

We now proceed to describe our results, and then we will explain our motivation to consider this particular setting. In what follows, $G:=\SL(m+1,\R)$, $\Ga\leq G$ is a lattice, and
\begin{equation}\label{eq: def of general H}
H:=\left \{\begin{bmatrix}
t^{-\frac{1}{m}}q & 0 \\
v &  t
\end{bmatrix}:t> 0, q \in \De, v \in \R^m \right\},
\end{equation}
where $\De\leq\SL(m,\R)$ is a lattice. In this setting, we have the following observation:

For each $x_0\in H\backslash G$, the orbit $x_0\Ga$  is dense.

This statement follows from our Theorem <ref>, but we briefly sketch a simple proof  –  A $\Ga$-orbit passing through $Hg$ is dense in $H\backslash G$ if and only if the “dual" $H$-orbit passing through $g\Ga$ is dense in $G/\Ga$. It's well known that each orbit of the connected component of $H$ is dense, see e.g. Proposition 1.5 of <cit.>.

We will be considering the filtration of the dense orbits given by the Hilbert-Schmidt norm:
$$\|g\|=\sqrt{\text{Trace}(^tgg)}=\sqrt{\sum_{ij}g_{ij}^2},~g\in G.$$
Namely, we denote
\begin{equation}\label{eq:def of Ga T}
\Ga_T:=\{\gamma \in \Ga:\|\gamma\|\leq T\},
\end{equation}

and we consider the following probability measures on $H\backslash G$:

$$\mu_{T,x_0}:=\frac{1}{\#\Ga_T}\sum_{\gamma\in\Ga_T}\de_{x_0.\gamma},~T>0,x_0\in H\backslash G.$$

We will prove that the above measures converge as $T\to\infty$ to the explicit measure   $\tilde{\nu}_{x_0}$ which we describe now.
Consider the  group $P$ containing $H$, which is given by
\begin{equation*}
P:=\left \{\begin{bmatrix}
t^{-\frac{1}{m}}\eta & 0 \\
v &  t
\end{bmatrix}:t> 0, \eta \in \SL(m,\R), v \in \R^m \right\}.
\end{equation*}
The space $H\backslash G$ is naturally a fiber-bundle over $P\backslash G$ with respect to the natural map sending

and the fibers are isomorphic to $\Delta\backslash\SL(m,\R)$. We note that the base space $P\backslash G$ is identified with $\Sph^{m}$ the unit sphere in $\R^{m+1}$ via the orbit map of the right action $w.g:=\frac{w (^tg^{-1})}{\|w (^tg^{-1})\|},~g\in G,~w\in\Sph^{m}$. Thus we may view $H\backslash G$ as a fiber bundle over the sphere. Notice that $\SO(m+1,\R)$ acts transitively on $P\backslash G$ (with respect to the natural action), and we denote by $\mu_{P.\SO(m+1,\R)}$ the right $\SO(m+1,\R)$-invariant probability on $P\backslash G=P.\SO(m+1,\R)$. Now we define measures on each fiber
Using the Iwasawa decomposition, we consider
$$x_0=:H \begin{bmatrix} \text{g}_0  & 0 \\ \textbf{v}_0 & \nicefrac{1}{\det(\text{g}_0)} \end{bmatrix}\rho_0,$$
where $|\det(\text{g}_0)|=1$ and $\rho_0\in\SO(m+1,\R)$, and we define the function
\begin{equation}\label{ef:def of Phi_x_0}
\Phi_{x_0}\left(H\begin{bmatrix} \eta  & 0 \\ v & 1 \end{bmatrix}\rho\right)=\sum_{q\in \Delta}\frac{1}{\|\text{g}_0^{-1}q \eta\|^{m^2}}.
\end{equation}

Here $\rho\in\SO(m+1,\R)$ and $\eta\in \SL(m,\R)$. The expression on the right is well-defined as the Hilbert-Schmidt norm is bi-$\SO(m+1,\R)$ invariant. The infinite sum is convergent due to Lemma <ref> below. We naturally identify the “standard" fiber $\pi^{-1}(P)=H\backslash P$ with $\Delta\backslash \SL(m,\R)$ by mapping $\De\eta\mapsto H\begin{bmatrix} \eta  & 0 \\ 0 & 1 \end{bmatrix}$, which allows to define the $\SL(m,\R)$-invariant measure $\mu_{H.\SL(m,\R)}$  on $H\backslash P$ with the choice of normalization such that the measure
\begin{equation}\label{eq:def of measures on fibers}
\nu_{x_0,\pi^{-1}(P)}(f):=\int_{\pi^{-1}(P)}f(y)\Phi_{x_0}(y)d\mu_{H.\SL(m,\R)}(y),~f\in C_c(H\backslash G)
\end{equation}
is a probability measure. On any other fiber $\pi^{-1}(P\rho),~\rho\in\SO(m+1,R)$, we define the pushed measure
where the pushforward is via the right translation by $\rho.$ Having defined the above measures on the base space and the fibers, we can now define our probability measure on $H\backslash G$ to be
\begin{equation}\label{eq:def of nu tilde x0}
\tilde\nu_{x_0}(f)=\int_{P.\SO(m+1,\R)}\nu_{x_0,\pi^{-1}(b)}(f)d\mu_{P.\SO(m+1,\R)}(b),~f\in C_c(H\backslash G).
\end{equation}

Let $\Ga\leq G$ be a lattice and fix ${x_0}\in H\backslash G$. Then, $\mu_{T,{x_0}}$ converges in the weak-* topology to $\tilde\nu_{x_0}$ as $T\to\infty$.

§.§ Main motivating example - unimodular $m$-lattices in $(m+1)$-space

The main motivation behind this project was to study the limiting distribution of orbits of a lattice $\Ga\leq G=\SL(m+1,\R)$ acting on a moduli space of discrete groups. Such a research direction was originally suggested by U. Shapira as natural continuation of the work <cit.>  which considers random walks on the space of rank 2 discrete subgroups in $\R^3$ (see also the more recent work in this direction <cit.> which generalizes <cit.>). We now describe our result in this setting, which is a particular case of Theorem <ref> with $H$ as in (<ref>) such that $\De=\SL(m,\Z)$.

In what follows, $m$ is a natural number strictly larger than 1.

We say that $\La\subset\R^{m+1}$ is an $m$-lattice if $\La$ is the $\Z$-Span of a tuple of linearly independent vectors  ${v_1,v_2,...,v_m}\in\R^{m+1}$, that is,
We define the covolume function:
$$\cov(\Lambda):=\sqrt{\det(\langle v_i,v_j\rangle)},$$

which is the area of a fundamental parallelogram of $\Lambda$.
We call an $m$-lattice $\La$  unimodular if $\cov(\La)=1$, and we consider the moduli space of unimodular $m$-lattices in $(m+1)$-space with a marked orthogonal vector:

Compared to the space of (non-marked) unimodular $m$-lattices
the space $X_{m,m+1}$ adds only slightly more information – for each $\La$ there are only two choices of an orthogonal vector $w\in\Sph^m$. All the results below imply directly the analogous results of $L_{m,m+1}$.

We define a right $\SL(m+1,\R)$ action on $X_{m,m+1}$ by
\begin{equation} \label{action of SL on X_m,3}
(\La,w). g:=\left(\frac{\La g}{\sqrt{\cov(\La g)}}, \frac{w (^tg^{-1})}{\|w (^tg^{-1})\|} \right),~g\in \SL(m+1,\R),~w\in \mathbb{S}^m,
\end{equation}
where $\La g$ and $w (^tg^{-1})$ are defined using the usual matrix multiplication, and  where $\|\cdot\|$ is the usual Euclidean norm. This action is transitive, and the stabilizer subgroup of the base point $\left(\text{Span}_{\Z}\{e_1,\cdots,e_m\},e_{m+1}\right)$ is $H$ as in (<ref>) with $\De=\SL(m,
\Z)$. This gives a natural identification $X_{m,m+1}\cong H\backslash G$ and the corresponding action of $\Ga$ on $ H\backslash G$ is the natural one.
$$x_0:=(\La_0,w_0)\in X_{m,m+1},$$
consider the probability measures
$$\mu_{T,x_0}:=\frac{1}{\#\Ga_T}\sum_{\gamma\in \Ga_T}\delta_{x_0.\gamma},~T>0,$$
where $\Ga_T$ is as in (<ref>).

When $m=1$ the space $X_{m,m+1}$ is  identified with $\Sph^1$ the unit circle in $\R^2$. The limiting distribution of $\mu_{T,x_0}$ in the case of $m=1$ was obtained in <cit.>,

Having the identification $X_{m,m+1}\cong H\backslash G$, we obtain via Theorem <ref> that the limiting measure of the measures $\mu_{T,x_0}$ as $T\to\infty$ is $\tilde{\nu}_{x_0}$ defined in (<ref>). In the following we interpret this limiting measure in terms of the $m$-lattices. The projection to the sphere $\mathbb{S}^m$ defined by
endows $X_{m,m+1}$ with a fiber-bundle structure over $\Sph^{m}$, where the fibers are isomorphic to

The fiber $\pi_{\perp}^{-1}(w)$ may be thought of as the collection of unimodular $m$-lattices orthogonal to $w\in \Sph^{m}$.
We now give an interpretation for the measures on $\nu_{x_0,w}$ on those fibers. We first consider the function $\Phi_{x_0}$ which was defined in (<ref>). For an operator T from a hyperplane $U\subset \R^{m+1}$ to another hyperplane $V\subset \R^{m+1}$, we define

\begin{equation*}
\|\text{T}\|^2_{\text{HS}}:=\sum_{i=1}^m\|\text{T}u_i\|^2,
\end{equation*}

where $\{u_1,u_2,...,u_m\}$ is an orthonormal basis of $U$, and where the norm on the right hand side is the usual Euclidean norm on $\R^{m+1}$. We note that this norm is independent of the choice of an orthonormal basis $\{u_1,u_2,...,u_m\}$.
For an ordered tuple  $B=(u_1,u_2,...,u_m)$ of linearly independent vectors in $\R^{m+1}$ we define the linear map $\text{T}_B:\text{Span}_{\R}  \{e_1,e_2,...,e_m\} \to\text{Span}_{\R}\{u_1,u_2,...,u_m\}$, by sending $e_1\mapsto u_1,...,e_m\mapsto u_m$. Now fix unimodular $m$-lattice $\Lambda_0\subset\R^{m+1}$, and let $\mathscr{B}_0$ be an ordered tuple of linearly independent vectors forming a $\Z$-basis for $\La_0 $. Then for $x_0=(\La_0,w_0)$ the function $\Phi_{x_0}$ in (<ref>)  becomes
\begin{equation*}
\Phi_{x_0} (\La,w):=\sum_{\Span_\Z{\mathscr{B}}=\La}\frac{1}{\|\text{T}_{\mathscr{B}}\circ \text{T}_{\mathscr{B}_0}^{-1}\|_{\text{HS}}^{m^2}}.
\end{equation*}

By identifying $\pi_\perp^{-1}(e_{m+1})$ with $\SL(m,\Z)\backslash \SL(m,\R)$, we obtain the $\SL(m,\Z)\backslash \SL(m,\R)$-invariant measure $\mu_{e_{m+1}}$ supported on $\pi_\perp^{-1}(e_{m+1})$ scaled such that the measure $\nu_{x_0,e_{m+1}}$ defined by
$$\nu_{x_0,e_{m+1}}(f):=\int_{\pi^{-1}_\perp(e_{m+1})}f(\La,e_{m+1})\Phi_{x_0}(\La,e_{m+1})d\mu_{e_{m+1}}(\La),~f\in C_c(X_{m,m+1}),$$
is a probability measure. Next, for any other $w\in \Sph^{m}$, we choose $\rho_w\in \SO(m+1,\R)$ such that $w=e_{m+1} \rho_w$, and we define,
which is the push-forward of the right translation by $\rho_w$ via the right action of $\SO(m+1,\R)$ on $X_{m,m+1}$ defined in (<ref>). Note that $\nu_{x_0,w}$ is independent of the choice of $\rho_w$. Finally, we define $\tilde\nu_{x_0}$ by

Let $\Ga\leq\SL(m+1,\R)$ be a lattice and fix ${x_0}\in X_{m,m+1}$. Then, $\mu_{T,{x_0}}$ converges in the weak-* topology to $\tilde\nu_{x_0}$ as $T\to\infty$. In other words,
for all $f\in C_c(X_{m,m+1})$, we have

\begin{equation*}
\lim_{T
\to\infty} \frac{1}{\#\Gamma_{T}}\sum_{\gamma\in \Ga_T}f(x_{0}. \gamma)=\int_{X_{m,m+1}}f(x)d \tilde\nu_{x_{0}}(x).
\end{equation*}

An example for $\Ga=\SL(m+1,
\Z)$. We would like to point out a classical result in number theory that  is related to the orbit $x_0.\Ga$ where $\Ga=\SL(m+1,\Z)$ and $x_0=(\text{Span}_\Z\{e_1,...,e_m\},e_{m+1}) \in X_{m,m+1}$.
The collection of $m$-lattices
consists of all primitive (maximal) rank $m$-discrete subgroups of $\Z^{m+1}$. Notice that here the $m$-lattices $\text{Span}_\Z\{e_1,...,e_m\}\gamma$ are without the covolume normalization, and their covolume can be arbitrarily large. It's natural to consider the equivalence relation saying that $m$-lattices are equivalent if they differ by a rotation or by a dilation.  If $\La\leq \R^{m+1}$ is an $m$-lattice, then it's equivelence class $[\La]$ can be naturally represented by a point in $ \text{PGL}(m,\Z)\backslash \text{PGL}(m,\R)/\text{PO}(m,\R)$, where $\text{PGL}$ denotes the projective linear group and where $\text{PO}$ denotes the projective orthogonal group. A particular consequence of a classical result of Schmidt <cit.>, who generalized the works of Roelcke <cit.> and Maass <cit.>, states that the uniform probability measures on
$$\left\{[\La]:\La\in \Z_{m,m+1}\text{ and Cov}(\La)\leq T\right\}$$
converge to the natural measure on $\text{PGL}(m,\Z)\backslash \text{PGL}(m,\R)/\text{PO}(m,\R)$ as $T\to\infty$

Using our Theorem <ref>, we can deduce the limiting measure of the following different sequence of probability measures on $\Z_{m,m+1}$ given by $\nu_T:=\frac{1}{\#\Ga_T}\sum_{\gamma\in\Ga_T}\delta_{[\La_0\gamma]},$ where $\La_0=\text{Span}_\Z\{e_1,...,e_m\}.$ To obtain the limiting measure, we note that the equivalence relation modulo rotations and dilations defines a continuous map $[\cdot]:X_{m,m+1}\to \text{PGL}(m,\Z)\backslash \text{PGL}(m,\R)/\text{PO}(m,\R)$. Pushing forward the  measure $\tilde{\nu}_{x_0}$ from Theorem  <ref> via this map gives the limiting measure of $\nu_T$, as $T\to\infty.$

Next, we recall the definition of the shape of lattices, a notion that was extensively studied in e.g <cit.>, which refined the classical work of Schmidt <cit.>.
We view $\R^{m+1}$ as row vectors, and for a unimodular $m$-lattice $\La\subset\R^{m+1}$, let $w\in\mathbb{S}^{m}$ be such that $w\perp \La$. We choose a $\rho \in \SO(m+1,\R)$ such that $w \rho=e_{m+1}:=(0,\dots,0,1)$, and we define the shape of the pair $(\La,w)$ as

\begin{equation}\label{definition of shape}
\s(\La,w):=\La \rho\begin{bmatrix}
\SO(m,\R) & 0 \\
0 &  1
\end{bmatrix},
\end{equation}

which is independent of the choice of $\rho$. By identifying the unimodular $m$-lattices $\La\subset e_{m+1}^\perp$ with $X_m:=\SL(m,\Z)\backslash \SL(m,\R)$, we view $\s(\Lambda,w)$ as a point in $X_m\slash \SO(m,\R)$. Note that in general

A more intrinsic definition of a shape of a discrete subgroup, see e.g. <cit.>, is defined by the equivalence class under the equivalence relation of scaling and rotations. When defining shape in this way, one gets a point in $X_m/\text{O}(m,\R)$, and it captures slightly less information. Our definition (<ref>) is mainly motivated by the definition in <cit.>, which produces a point in the more familiar space $X_m/\SO(m,\R)$.

We consider
and note that $\s$ defined in (<ref>) yields a map
$$\s:X_{m,m+1}\to X_m/\SO(m,\R).$$

TODO: write how it can be proven and give reference to Dani

We observe that that the push-forward of $\tilde{\nu}_{x_0}$ by $\s$ is given by

Let $\Ga\leq\SL(m+1,\R)$ be a lattice and fix ${x_0}\in X_{m,m+1}$. Then, the probability measures on $X_m\slash \SO(m,\R)$ given by

converge in the weak-* topology to the probability measure $\s_*\nu_{x_0,e_{m+1}},$ as $T\to\infty$.

§.§ Paper organization

As was mentioned above, to prove our main result we verify the conditions of a theorem due to Gorodnik and Weiss, see Section 2.5 in <cit.>. Those conditions are certain volume estimate for the left Haar measure of "skew"-norm balls in $H$, and ergodic theorems of those $H$-balls acting on $G/\Ga$. The organization of the paper is as follows: Section <ref> states and proves the aforementioned volume estimates, and Section <ref> states and proves the aforementioned ergodic theorems. Finally, Section <ref> applies the theorem of Gorodnik-Weiss.

§ NOTATIONS AND CONVENTIONS

Throughout this paper, for functions $f$ and $g$ we say that as $x\to \al$ where $\al\in \R\cup\{\infty\}$ that $f(x)=O(g(x))$ or $f(x)\ll g(x)$ if there is some $C>0$ such that $|f(x)|\le C |g(x)|$ for all $x$ sufficiently close to $\al$; by $f(x)\asymp g(x)$ we mean $f(x)\ll g(x)$ and $g(x) \ll f(x)$; by $f(x)\sim g(x)$ we mean $\lim_{x\to \al}\left|\frac{f(x)}{g(x)} \right|=L$ for $L\neq0$; by $f(x)=o(g(x))$ we mean $\lim_{x\to\al}f(x)/g(x)=0.$

§ VOLUME ESTIMATES OF EXPANDING SKEW BALLS IN $H$

We begin by describing the left invariant measure on $H$. We consider

\begin{equation}\label{eq:def of the subgroups UADelta}
I_m & 0 \\
\mathbb{\R}^m & 1
\end{bmatrix}, A=\left \{\begin{bmatrix}
t^{-\frac{1}{m}}I_m & 0 \\
0 & t
\end{bmatrix}:t>0 \right\},\tilde{\De}:=\begin{bmatrix}
\De & 0 \\
0 &  1
\end{bmatrix},
\end{equation}

where $\De\leq\SL(m,\R)$ is a lattice, and we denote

\begin{equation}\label{eq:definition of u_v a_t and qtilde}
I_m & 0 \\
v & 1
\end{bmatrix},
t^{-\frac{1}{m}}I_m & 0 \\
0 & t
\end{bmatrix},
\tilde{q}:=\begin{bmatrix}
q & 0 \\
0 &  1
\end{bmatrix},
\end{equation}

for $q\in \Delta,t>0, v \in \R^m$. Any element in $x\in H$ can be represented as $x=u_v a_t\tilde{q} $  uniquely in $v\in\R^{m},t>0,q\in \De$, and we have

Namely, $H$ is the semi-direct product

$$H\cong \R^{m}\rtimes(\R_{>0} \times \Delta),$$

and it follows that a left Haar measure on $H$ is given by

\begin{align}\label{eq:Haar measure on $H$}
\int_H f(x)d\mu(x)
=&\sum_{q\in \Delta}\int_0^{\infty}\int_{\R^m} f\left(u_v a_t\tilde{q}\right) dv \frac{1}{t^{m+2}} dt,
\end{align}

where the measures $dv$  and $dt$ denote the Lebesgue measures on $\R^m$ and $\R$ correspondingly.

To check the left invariance, fix $q_0 \in \SL_{m,1}(\Z)$, $a_{t_0}:=\begin{bmatrix}
t_0^{-\frac{1}{2}}I_2 & 0 \\
0 & t_0
\end{bmatrix}, t_0\ne 0$ and $u_{v_0}:=\begin{bmatrix}
I_2 & 0 \\
v_0 & 1
\end{bmatrix}$, $v_0:=(x_0,y_0)\in \R^m$, then

\begin{align*}
u_{v_0} a_{t_0} q_0 \cdot u_v a_t q
=& u_{v_0} [a_{t_0} q_0 u_v (a_{t_0} q_0)^{-1}] a_{t_0} q_0 a_t q \\
=& u_{v_0} u_{t_0^{\frac{3}{2}} vq_0^{-1}}a_{t_0} a_t q_0 q \\
=& u_{v_0+t_0^{\frac{3}{2}} vq_0^{-1}}a_{t_0 t} q_0 q
\end{align*}

It follows that

\begin{align*}
& \int_{\SL_{m,1}(\Z)}\int_0^{\infty}\int_{\R^m} L_{u_{v_0} a_{t_0} q_0}[f(uaq) ] dx dy \frac{1}{t^{m+2}}dt dq \\
=&  \int_{\SL_{m,1}(\Z)}\int_0^{\infty}\int_{\R^m} f(u_{v_0} a_{t_0} q_0uaq)  dx dy \frac{1}{t^{m+2}}dt dq \\
=&  \int_{\SL_{m,1}(\Z)}\int_0^{\infty}\int_{\R^m} f(u_{v_0+t_0^{\frac{3}{2}} vq_0^{-1}}a_{t_0 t} q_0 q) dx dy \frac{1}{t^{m+2}}dt  dq
\end{align*}

By the change of variable

\begin{equation*}
v:=v\mapsto v':=(x',y')=v_0+t_0^{\frac{3}{2}}vq_0^{-1}, t\mapsto t':=t_0t, q \mapsto q':=q_0 q,
\end{equation*}

and noticing that $q_0$ (with determinant $1$) does not contribute to the Jacobian, the last line of the equation above becomes

\begin{equation*}
\int_{\SL_{m,1}(\Z)}\int_0^{\infty}\int_{\R^m} f(u_{v'}a_{t'}q') (t_0^{-\frac{3}{2}})^2 dx' dy' \frac{t_0^4}{t'^4}d(\frac{t'}{t_0})  dq'=    \int_{\SL_{m,1}(\Z)}\int_0^{\infty}\int_{\R^m} f(u_{v'}a_{t'}q') dx' dy' t'^4 dt'  dq'
\end{equation*}

This verifies the invariance of measure. Finally, by the uniqueness of Haar measure, the measure $dq$ on $\SL_{m,1}(\Z)$ is the counting measure and therefore the out-most integral reduces to the sum.

Throughout this paper, $\|M\|$ for a matrix $M$ denotes it's Hilbert-Schmidt norm.
Following <cit.>, for a subgroup $L\leq G$,  $g_1,g_2 \in G$ and $T>0$, we define the so-called “skewed balls" as follows (defined slightly differently from the definition in <cit.>):

\begin{equation*}
L_T[g_1,g_2]:=\{h\in L: \|g_1^{-1}hg_2\|<T \},
\end{equation*}

and we denote

\begin{equation}\label{meaning of the subscript T}
\end{equation}

For $g_1, g_2 \in G$, by using Iwasawa decomposition, we denote $k_1,k_2 \in \SO(m+1,\R)$, $\text{g}_1,\text{g}_2\in \GL(m,\R)$, and $\text{v}_1, \text{v}_2\in \R^m$ by

\begin{equation}\label{matrix representation of g1 and g2}
g_1^{-1}=:k_1\begin{bmatrix} \text{g}_1 & 0 \\\text{v}_1 & \det(\text{g}_1)^{-1} \end{bmatrix}, ~g_2=:\begin{bmatrix} \text{g}_2 & 0 \\\text{v}_2 & \det(\text{g}_2)^{-1} \end{bmatrix} k_2.
\end{equation}

We note that the above $k_i,\text{g}_j$, are not unique, but due to $\SO(m+1,\R)$-invariance, all  their appearances below are well defined regardless of their choices.

The key volume estimates we prove in this section are given in the following proposition.
Let $\Gamma(z)$ be the classical Gamma function, and let
$$C(m,g_1,g_2):=\frac{m\pi^\frac{m}{2}\Ga(\frac{m^2}{2})}{2\Ga(\frac{m^2}{2}+\frac{m}{2}+1)}\left|\frac{\det(\emph{\text{g}}_1)^m}{ \det(\emph{\text{g}}_2)}\right|.$$
There exists $\kappa>0$ such that for a fixed bounded subset $B\subseteq \SL(m+1,\R)$ it holds for all $g_1,g_2\in B$ that
\begin{equation}\label{eq:the estimate for H skew balls}
\mu(H_T[g_1,g_2])=\left(C(m,g_1,g_2)\sum_{q\in \Delta}\frac{1}{\|\emph{\text{g}}_1q\emph{\text{g}}_2\|^{m^2}}\right)T^{m(m+1)}+O(T^{m(m+1)-\kappa}),
\end{equation}

as $T\to\infty$. The implied constant depends on $B$ only.

We note that  the sum $\sum_{q\in \Delta}\frac{1}{\|\text{g}_1q\text{g}_2\|^{m^2}}$ is convergent due to Lemma <ref> below.

Statements of this form are known in great generality for a connected $H$, see e.g. Theorems 7.17–7.18 in <cit.>. Here, since our $H$ has infinitely many connected components, the volume of the skew-ball is a sum over the volumes of skew-balls of the connected component, see (<ref>) below.  Our main challenge here is to quantify the dependence of the error term of the main term appearing of the volume of each skew-ball of the connected component, see Lemma <ref> below. Proving such a result for more general norms is the main obstacle for  generalizing our results for more general norms.

As an immediate corollary, we  obtain the following statements which are key requirements for the method of <cit.>.

(Uniform volume growth for skewed balls in $H$, property D1 in <cit.>)
For any bounded subset $B\subset G$ and any $\e>0$, there are $T_0$ and $\de>0$ such that for all $T>T_0$ and all $g_1,g_2 \in B$ we have:

\begin{equation*}
\mu \left(H_{(1+\de)T}[g_1,g_2] \right)\le (1+\e)\mu \left(H_T[g_1,g_2] \right).
\end{equation*}

Corollary <ref> follows since Proposition <ref> states that
$$\mu\left(H_T[g_1,g_2]\right)\sim (\text{constant depending on $g_1,g_2$})\times T^{m(m+1)},$$
and importantly, this estimate is uniform when $g_1,g_2$ are restricted to a compact set.
Next we have:

For any $g_1,g_2 \in G$. the limit

\begin{equation}\label{eq:alpha}
\omega(g_1,g_2):=\lim_{T\to \infty}\frac{\mu \left(H_T[g_1,g_2] \right)}{\mu \left(H_T \right)}=\left|\frac{\det(\emph{\text{g}}_1)^m}{ \det(\emph{\text{g}}_2)}\right|\frac{\sum_{q\in \Delta}\frac{1}{\|\emph{\text{g}}_1q\emph{\text{g}}_2\|^{m^2}}}{\sum_{q\in \Delta}\frac{1}{\|q\|^{m^2}}},
\end{equation}

exists and is positive and finite.

The importance of the explicit expression (<ref>) is that it allows  to deduce, via Thoerem 2.3 of <cit.>, the explicit limiting distribution of Theorem <ref>.

We now detail the structure of the proof of Proposition <ref>. Consider for $q\in\De$

\begin{equation}\label{definition of V q T}
V_{q,T}[g_1,g_2]:=\{h\in U A \tilde{q}: \|g_1^{-1}hg_2\|<T \}.
\end{equation}

Since $H=\bigsqcup_{q \in \Delta}UA\tilde{q}$, we have

\begin{equation}\label{decomposition of H_T into V_T}
H_T[g_1,g_2]=\bigsqcup_{q \in \Delta}V_{q,T}[g_1,g_2],
\end{equation}

and in particular

\begin{align}\label{eq:haar measure of skewed H ball as a sum of skewed balls on connected component}
\mu \left(H_T[g_1,g_2] \right)
=& \sum_{\substack{q\in \Delta}} \mu \left(V_{q,T}[g_1,g_2]\right).
\end{align}

Now denote $V:=UA$,  (the connected component of $H$) and observe that $V_{q,T}[g_1,g_2]$ is the following translated skewed ball of the group $V$:

More explicitly, note that

$$V_T[g_1,\tilde{q}g_2]=\{u_va_t:t>0,v\in\R^m\text{ and }\|(g_1^{-1})a_tu_v(\tilde{q}g_2)\|\leq T\},$$

and observe that by  (<ref>),
$$\mu\left(V_{q,T}[g_1,g_2]\right)=\iint\limits_{V_T[g_1,\tilde{q}g_2]}  dv \frac{1}{t^{m+2}} dt.$$
When the norm of $q$ is large we have that  $\mu \left(V_{q,T}[g_1,g_2]\right)$, is small (see Lemma <ref>), and the summation over such terms in (<ref>) doesn't contribute to the main term of (<ref>). In fact, we partition the sum into two parts

$$ \left(\sum_{\substack{q\in \Delta,\\ \|\text{g}_1 q \text{g}_2\|< \sqrt{T}}}+\sum_{\substack{q\in \Delta,\\ \sqrt{T}\leq\|\text{g}_1 q \text{g}_2\|}} \right)\mu\left(V_{q,T}[g_1,g_2]\right),$$

and we prove the following lemma.

If $g_1,g_2$ vary in a bounded set $B$, then

$$\sum_{\substack{q\in \Delta,\\ \sqrt{T}\leq\|\text{g}_1 q \text{g}_2\|}} \mu\left(V_{q,T}[g_1,g_2]\right)=O(T^{m(m+1)-m/2}),$$
where the implied constant depends on $B$ only.

For the rest of the terms, we have the following key estimate.

There exist $\ka_1,\ka_2>0$ which can be small as we wish and explicitly determined, such that whenever $g_1,g_2$ vary in a bounded set $B$, and $q\in\De$ such that $\|\text{g}_1q\text{g}_2\|\leq \sqrt{T}$, it holds that

\begin{equation*}
\mu\left(V_{q,T}[g_1,g_2]\right)=\left(C(m,g_1,g_2)\frac{1}{\|\emph{\text{g}}_1q\emph{\text{g}}_2\|^{m^2}}\right)T^{m(m+1)}+O\left(\frac{T^{m(m+1)-\ka_1}}{\|q\|^{m^2}}+\frac{T^{m+\ka_2\frac{m^2}{m+1}}}{\|q\|^{\frac{m^2}{m+1}}}\right),
\end{equation*}

where the implied constant depends on $B$ only.

We prove Proposition <ref> in Section <ref>.

For each point $Hg_0\in H\backslash G$, there exists a measure $\nu_{Hg_0}$ (not $G$-invariant) on $H\backslash G$ such that for any compactly supported function $\varphi \in C_c(H\backslash G)$, we have

\begin{equation}
\lim_{T\to \infty}\frac{1}{\mu(H_T)}\int_{G_T}\varphi(Hg_0g)dg=\int_{H\backslash G} \varphi d\nu_{Hg_0}.
\end{equation}

§.§ Structure of the paper and further work

* The outline of our proof of Theorem <ref> is as follows. We will first give an estimate for the volume of $H_T[g_1,g_2]$ (Section 2). This estimate will help us to show the unipotent invariance of limiting measure (Section 3), which in turn opens the gate to the application of Ratner theory. In particular, a dichotomy theorem (Theorem <ref>) due to Shah plays the central role here. To exclude the first outcome of the dichotomy theorem, we use a lemma (Lemma <ref>) on the expansion of the norms of vectors under the action of unipotent subgroups. This allows us to prove the non-escape of mass on the space of one-point compactification of $G/\Ga$ and the $G$-invariance will follow from Ratner's measure classification theorem (<cit.>, <cit.>).

* In Section 5 we prove Theorem <ref> as a consequence of the duality principle stated in Theorem 2.2 to Corollary 2.4 in <cit.>.

TODO:write here that our method of proof gives a more general result... sl Z can be replaced with other lattices.We remark that despite that the theorem is stated from the case of rank-$m$ lattices in $\R^{m+1}$. Our work in progress is to generalize the results to more general classes of moduli spaces of discrete subgroup of the Euclidean space. The main difficulty lies in the estimtate of the volumes of balls as well as the expansion of norms under unipotent actions.
TODO:add small section with notations for the asymptotic symbols etc
The structure of the rest of the section is as follows. First, in Section  <ref> we prove the main Proposition <ref> by assuming Lemmata <ref> - <ref>. Finally Section <ref>  is dedicaded for the proof of Lemmata <ref> - <ref>.

§.§ Proof of Proposition <ref> based on Lemmata <ref> - <ref>

The following lemma is essentialy all we need to get Proposition <ref> from Lemmata <ref> - <ref>.

Let $\De\leq \SL(m,\R)$ by a lattice, and fix an integer $m\geq2$. Then, for $\sigma>-m(m-1)$ it holds that

\begin{equation*}
\sum_{q\in \De_T}\|q\|^\sigma\ll T^{m(m-1)+\sigma},~\text{as }T\to\infty,
\end{equation*}

and if $\sigma<-m(m-1)$,
the sum above converges with the following tail estimate:
\begin{equation*}
\sum_{q\in \De_T}\|q\|^\sigma=I_\sigma+ O(T^{m(m-1)+\sigma}),~\text{as }T\to\infty.
\end{equation*}

Our method of proof doesn't give explicitly the constants $I_{\sigma}$.

By <cit.>, there is $C>0$ (an explicit constant), such that

\begin{equation}\label{eq:DRS asymp}
\#\De_s=Cs^{m(m-1)}+o(s^{m(m-1)}),~\text{as }s\to\infty.
\end{equation}

This proves the statement in case that $\sigma=0$. Now, for $\sigma\in\R$

\begin{align*}
\sum_{q\in \De_T}\|q\|^\sigma
&= \sum_{q\in \De_T}\int_0^{\|q\|^{\sigma}}1dt\\
&= \sum_{q\in \De_T}\int_0^{\infty}\mathbf{1}_{[t<\|q\|^{\sigma}]}dt\\
&=\int_0^{\infty}\sum_{q\in \De_T}\mathbf{1}_{[t<\|q\|^{\sigma}]}dt.
\end{align*}

Consider the case that $\sigma>0$. We have

\begin{align*}
\int_0^{\infty}\sum_{q\in \De_T}\mathbf{1}_{[t<\|q\|^{\sigma}]}dt&=\int_0^{T^{\sigma}}\#\{q\in\De:t<\|q\|^\sigma<T^\sigma\}dt \\
&\underbrace{\ll}_{\eqref{eq:DRS asymp}}T^{m(m-1)+\sigma}.
\end{align*}

Next, consider $-m(m-1)\ne \sigma<0$. We have

\begin{align*}
\int_0^{\infty}\sum_{q\in \De_T}\mathbf{1}_{[t<\|q\|^{\sigma}]}dt&=\int_{0}^{\infty}\#\{q\in\De:\|q\|<\min(t^{\frac{1}{\sigma}},T)\} dt\\
&=\int_{T^{\sigma}}^{\infty}\#\{q\in\De:\|q\|<t^{\frac{1}{\sigma}}\}dt+ \int_0^{T^{\sigma}}\#\{q\in\De:\|q\|<T\}dt\\
&\underbrace{=}_{\eqref{eq:DRS asymp}}\int_{T^{\sigma}}^{\infty}\#\{q\in\De:\|q\|<t^{\frac{1}{\sigma}}\}dt+ CT^{m(m-1)+\sigma}+o(T^{m(m-1)+\sigma}),~\text{as }T\to\infty.
\end{align*}

To prove the claim, it remains to treat the integral $\int_{T^{\sigma}}^{\infty}\#\{q\in\De:\|q\|<t^{\frac{1}{\sigma}}\}dt$. Since $\|g\|\geq1$ for all $g\in\SL(m,\R)$, we have
By (<ref>), there is $C'>0$ such that
$$\#\{q\in\De:\|q\|<t^{\frac{1}{\sigma}}\}\leq C't^{\frac{m(m-1)}{\sigma}},~\text{for all }t\in(0,1).$$

Suppose that $-m(m-1)<\sigma<0$. Then

\begin{align*}
\int_{T^{\sigma}}^{1}\#\{q\in\De:\|q\|<t^{\frac{1}{\sigma}}\}dt\leq& C'\int_{T^\sigma}^1 t^{\frac{m(m-1)}{\sigma}}dt\\
&\ll T^{m(m-1)+\sigma}.
\end{align*}

It's left to treat the case that $\sigma<-m(m-1)$. Since  $\int_0^1t^\frac{m(m-1)}{\sigma}dt$ converges, we have by dominated convergence that
for some  constant $=I_{\sigma}$. Finally,

\begin{align*}
\int_{T^{\sigma}}^{1}\#\{q\in\De:\|q\|<t^{\frac{1}{\sigma}}\}dt=&I_{\sigma}-\int^{T^{\sigma}}_{0}\#\{q\in\De:\|q\|<t^{\frac{1}{\sigma}}\}dt\\
\leq & I_{\sigma}+C'\left(\int^{T^{\sigma}}_{0}t^\frac{m(m-1)}{\sigma}dt\right)\\
\end{align*}

For the following, we note the basic fact – when $g_1,g_2$ vary in a bounded set $B$, there exists constants $C^+,C^->0$ such that

\begin{equation}\label{eq:lower bound on A_m}
C^-\|q\|\leq \|\text{g}_1 q \text{g}_2\|\leq C^+\|q\|,
\end{equation}

where $\text{g}_1,\text{g}_2$ are given by (<ref>).

We assume that $g_1,g_2$ vary in a fixed bounded set $B$. We have:

\begin{align}
\mu \left(H_T[g_1,g_2] \right)
=& \sum_{q\in \Delta} \mu \left(V_{q,T}[g_1,g_2]\right)\nonumber\\
\underbrace{=}_{\text{Lemma \ref{lem:bound on skewed V vol} }}&~\sum_{\substack{q\in \Delta,\nonumber\\ \|\text{g}_1q\text{g}_2\|< \sqrt{T}}}\mu \left(V_{q,T}[g_1,g_2]\right)+O(T^{m(m+1)-m/2})\nonumber\\
\underbrace{=}_{\text{Lemma \ref{lem:main estimate for large V skewed balls } }}&\left(C(m,g_1,g_2)\sum_{\substack{q\in \Delta,\\ \|\text{g}_1 q \text{g}_2\|<\sqrt{T}}}\frac{1}{\|\text{g}_1q\text{g}_2\|^{m^2}}\right)T^{m(m+1)}\nonumber\\
&+O\left(\sum_{\substack{q\in \Delta,\nonumber\\ \| q\|<\sqrt{T}/C^-}}\frac{T^{m(m+1)-\ka_1}}{\|q\|^{m^2}}+\frac{T^{m+\ka_2\frac{m^2}{m+1}}}{\|q\|^{\frac{m^2}{m+1}}}\right).
\end{align}

We now replace the summation $\sum_{\substack{q\in \Delta,\\ \|\text{g}_1 q \text{g}_2\|<\sqrt{T}}}$ in the last expression by $\sum_{q\in \Delta}$:

\begin{align}
\left(C(m,g_1,g_2)\sum_{\substack{q\in \Delta,\\ \|\text{g}_1 q \text{g}_2\|<\sqrt{T}}}\frac{1}{\|\text{g}_1q\text{g}_2\|^{m^2}}\right)T^{m(m+1)}&\nonumber\\
=\left(C(m,g_1,g_2)\sum_{q\in \Delta}\frac{1}{\|\text{g}_1q\text{g}_2\|^{m^2}}\right)T^{m(m+1)}+O&\left(\sum_{\substack{q\in \Delta,\\ \| q \|\geq \sqrt{T}/C^+}}\frac{1}{\|q\|^{m^2}}\right)T^{m(m+1)}\nonumber
\end{align}

\begin{align}
\underbrace{=}_{\text{Lemma } \ref{lemma on the integral and summation in a ball}}\left(C(m,g_1,g_2)\sum_{q\in \Delta}\frac{1}{\|\text{g}_1q\text{g}_2\|^{m^2}}\right)T^{m(m+1)}&+O\left(T^{m(m+1)-m/2}\right).\nonumber
\end{align}

Finally, we estimate the terms in the big-O notation above by again applying Lemma <ref>:

\begin{align}
\sum_{\substack{q\in \Delta,\nonumber\\ \| q\|<\sqrt{T}/C^-}}\frac{T^{m(m+1)-\ka_1}}{\|q\|^{m^2}}=O(T^{m(m+1)-\ka_1}),
\end{align}

\begin{align}
\sum_{\substack{q\in \Delta,\nonumber\\ \| q\|<\sqrt{T}/C^-}} \frac{T^{m+\ka_2\frac{m^2}{m+1}}}{\|q\|^{\frac{m^2}{m+1}}}=O(T^{0.5m(m+1)-\frac{(0.5-\ka_2)m^2}{m+1}}).
\end{align}

The proof is complete by combining all the above estimates.

§.§ Proving Lemmata <ref> - <ref>

We recall (<ref>):

\begin{equation*}
g_1^{-1}=:k_1\begin{bmatrix} \text{g}_1 & 0 \\\text{v}_1 & \det(\text{g}_1)^{-1} \end{bmatrix}, ~g_2=:\begin{bmatrix} \text{g}_2 & 0 \\\text{v}_2 & \det(\text{g}_2)^{-1} \end{bmatrix} k_2.
\end{equation*}

By bi-$\SO(m+1,\R)$ invariance, we have for $q\in\De$

\begin{align*}
V_T[g_1,\tilde{q}g_2]&=\left\{u_va_t:v\in\R^m,t>0,\left\|g_1^{-1}u_va_t(\tilde{q}g_2) \right\|\leq T\right\}\\ &=\left\{u_va_t:v\in\R^m,t>0,\left\|\begin{bmatrix} \text{g}_1 & 0 \\\text{v}_1 & \det(\text{g}_1)^{-1} \end{bmatrix}u_va_t\begin{bmatrix}
q & 0 \\
0 &  1
\end{bmatrix} \begin{bmatrix} \text{g}_2 & 0 \\\text{v}_2 & \det(\text{g}_2)^{-1} \end{bmatrix}\right\|\leq T\right\},
\end{align*}

where $u_v\text{ and }a_t$ are given by (<ref>). We will now describe the above skew-ball as a family of ellipsoids.
A computation shows

\begin{align*}
\begin{bmatrix} \text{g}_1 & 0 \\\text{v}_1 & \det(\text{g}_1)^{-1} \end{bmatrix}u_va_t\begin{bmatrix}
q & 0 \\
0 &  1
\end{bmatrix}& \begin{bmatrix} \text{g}_2 & 0 \\\text{v}_2 & \det(\text{g}_2)^{-1} \end{bmatrix}\\
t^{-\frac{1}{m}}\text{g}_1q \text{g}_2 & 0 \\
t^{-\frac{1}{m}}\text{v}_1q \text{g}_2+\det(\text{g}_1)^{-1}(t^{-\frac{1}{m}}vq \text{g}_2 +  t\text{v}_2) &  \det(\text{g}_1\text{g}_2)^{-1}t
\end{bmatrix}.
\end{align*}

Upon taking the sum of squares, we conclude that $\|g_1^{-1}u_va_t(\tilde{q}g_2)\| \le T$ is equivalent to

\begin{equation}\label{equation defining the ball of integration}
\|\text{v}_1q \text{g}_2+\det(\text{g}_1)^{-1} vq \text{g}_2+\det(\text{g}_1)^{-1} t^{\frac{1}{m}+1}\text{v}_2\|^2
\le -\det(\text{g}_1\text{g}_2)^{-2} t^{\frac{2}{m}+2}+t^{\frac{2}{m}}T^2-\|\text{g}_1 q \text{g}_2\|^2.
\end{equation}

\begin{equation}\label{eq:def of R_q,T}
R_{q,T}(t):=-\det(\text{g}_1\text{g}_2)^{-2} t^{\frac{2}{m}+2}+t^{\frac{2}{m}}T^2-\|\text{g}_1 q \text{g}_2\|^2,
\end{equation}

and note that the solutions in $v\in\R^m$ for  (<ref>)  define an interior of an ellipsoid

\begin{equation}\label{eq:def of D}
D_{q,T,t}:=\{v\in\R^m:\|\text{v}_1q \text{g}_2+\det(\text{g}_1)^{-1} vq \text{g}_2+\det(\text{g}_1)^{-1} \text{v}_2t^{\frac{1}{m}+1}\|^2 \leq R_{q,T}(t)\}.
\end{equation}
We have

\begin{equation}\label{eq:volume of D_gamma,t}
\vol(D_{q,T,t})= \begin{cases}
v_{m}\left|\frac{\det(\text{g}_1)^m}{\det(\text{g}_2)}\right|\left(R_{q,T}(t) \right)^{\frac{m}{2}}& \text{if } R_{q,T}(t)>0, \\
0 & \text{otherwise.}
\end{cases}
\end{equation}

where $\vol(D_{q,T,t})$ is the Lebesgue measure of the ellipse $D_{q,T,t}$. Here $v_m=\frac{\pi^{\frac{m}{2}}}{\Ga(\frac{m}{2}+1)}$ denotes the volume of Euclidean ball of radius one in $\R^m$.
We conclude that

\begin{equation}\label{eq:the measure of V skew ball}
\mu \left(V_{q,T}[g_1,g_2]\right):=v_{m}\left|\frac{\det(\text{g}_1)^m}{\det(\text{g}_2)}\right|\int_{\{t>0:R_{q,T}(t)>0\}} \frac{\left(R_{q,T}(t) \right)^{\frac{m}{2}}}{t^{m+2}}dt.
\end{equation}

As we will show now, the set $\{t>0:R_{q,T}(t)>0\}$ is either an interval $(\al_{q,T},\be_{q,T}),$ or  empty when the norm of $q$ is large. In particular, we will give estimates of the roots $\al_{q,T},\be_{q,T}$ which are essential in our proof.

Consider for $T>0$,
\begin{equation}\label{eq:maximal value}
M_{T}:=\frac{m}{m+1}\left(\frac{1}{m+1}\right)^{\frac{1}{m}} \frac{T^{2+\frac{2}{m}}}{|\det(\text{g}_1\text{g}_2)^{-1}|^{\frac{2}{m}}},
\end{equation}

\begin{equation}\label{eq:critical value where M_T is attained}
\theta_T:=\frac{1}{\sqrt{1+m}} \frac{T}{|\det(\text{g}_1\text{g}_2)^{-1}|}.
\end{equation}

* $\{t>0:R_{q,T}(t)>0\}\neq\emptyset$ if and only if $\|\text{g}_1q\text{g}_2\|^2< M_T.$

* Suppose that $\{t>0:R_{q,T}(t)>0\}\neq\emptyset$. Then $\{t>0:R_{q,T}(t)>0\}=(\al_{q,T},\be_{q,T})$, and the following estimates hold:

\begin{equation}\label{bounds on roots}
0<\al_{q,T}<\theta_T<\be_{q,T}<\sqrt{m+1} \theta_T,
\end{equation}

\begin{equation}\label{range for a}
\al_{q,T}^{\frac{2}{m}}\in
\left(\frac{\|\text{g}_1 q \text{g}_2\|^2}{T^2}, \frac{(m+1)\|\text{g}_1 q \text{g}_2\|^2}{mT^2}\right),
\end{equation}

and moreover, if $g_1,g_2$ vary in a bounded set $B$, and $q\in\De$ such that $\|\text{g}_1q\text{g}_2\|\leq \sqrt{T}$

\begin{equation}\label{eq:asymp for a when A small then sqrt T}
\al_{q,T}^{\frac{2}{m}}=\frac{\|\text{g}_1 q \text{g}_2\|^2}{T^2}+O\left(\frac{1}{T^{m+3}}\right).
\end{equation}

First, observe that the maximal value of $-\det(\text{g}_1\text{g}_2)^{-2} t^{\frac{2}{m}+2}+t^{\frac{2}{m}}T^2$ for $t>0$ is (<ref>). Since $R_{q,T}(t)=-\det(\text{g}_1\text{g}_2)^{-2} t^{\frac{2}{m}+2}+t^{\frac{2}{m}}T^2-\|\text{g}_1 q \text{g}_2\|^2$, we conclude that $\{t>0:R_{q,T}(t)>0\}\neq\emptyset$ if only if $\|\text{g}_1q\text{g}_2\|^2< M_T.$

Next, observe that (<ref>) is the  unique positive critical point of $R_{q,T}(\cdot)$. More precisely, $R_{q,T}(\cdot)$  is monotonic increasing in $(0,\theta_T)$ and monotonic decreasing in $(\theta_T,\infty)$. Thus, $R_{q,T}(\cdot)$ has two positive roots $\al_{q,T},\be_{q,T}$,and the bounds (<ref>) are concluded  by noting that $R_{q,T}(0)<0$ and that $R_{q,T}(\sqrt{m+1}\theta_T)<0$.

It remains to prove the more precise bounds for the root $\al_{q,T}$. By rearranging terms in
$$-\det(\text{g}_1\text{g}_2)^{-2} \al_{q,T}^{\frac{2}{m}+2}+\al_{q,T}^{\frac{2}{m}}T^2-\|\text{g}_1 q \text{g}_2\|^2=0,$$
we get that

\begin{equation}\label{eq:recursive form of a}
\al_{q,T}^{\frac{2}{m}}=\frac{\|\text{g}_1 q \text{g}_2\|^2}{T^2-\det(\text{g}_1\text{g}_2)^{-2} \al_{q,T}^2},
\end{equation}

and by using that $\al_{q,T}<\theta_T$, see (<ref>), we conclude that

\begin{equation*}
\al_{q,T}^{\frac{2}{m}}=\frac{\|\text{g}_1 q \text{g}_2\|^2}{T^2-\det(\text{g}_1\text{g}_2)^{-2} \al_{q,T}^2}\in
\left(\frac{\|\text{g}_1 q \text{g}_2\|^2}{T^2}, \frac{(m+1)\|\text{g}_1 q \text{g}_2\|^2}{mT^2}\right).
\end{equation*}

In particular, it follows that

\begin{equation}
\al_{q,T}\asymp \frac{\|\text{g}_1 q \text{g}_2\|^m}{T^m}. \label{estimate for a in all range}
\end{equation}

Finally, assume that $\|\text{g}_1 q \text{g}_2\|\leq\sqrt{T}$ and that $g_1,g_2$ vary in a bounded set. Then

\begin{align*}
\al_{q,T}^{\frac{2}{m}}=&\frac{\|\text{g}_1q\text{g}_2\|^2}{T^2}\frac{1}{1-\frac{\det(\text{g}_1\text{g}_2)^{-2}\al_{q,T}^2}{T^2}}\\
=&\frac{\|\text{g}_1 q \text{g}_2\|^2}{T^2}\left(1+\frac{\det(\text{g}_1\text{g}_2)^{-2}\al_{q,T}^2}{T^2}+\cdots \right)\\
=&\frac{\|\text{g}_1 q \text{g}_2\|^2}{T^2}+O\left(\frac{1}{T^{m+3}}\right).\tag{using \eqref{range for a} with $\|\text{g}_1 q \text{g}_2\|\leq\sqrt{T}$}
\end{align*}
where in the last line we also used the fact that for $g_1,g_2$ that vary in a bounded set it holds that $\det(\text{g}_1\text{g}_2)^{-2}$ is bounded above (here $\text{g}_i$ are given by (<ref>)).

We now conclude that whenever $\|\text{g}_1q\text{g}_2\|^2\leq M_T,$ we have  by (<ref>) that

\begin{equation}\label{eq:measure of V skewed balls as integral}
\mu \left(V_{q,T}[g_1,g_2] \right)
= \frac{\pi^{\frac{m}{2}}}{\Ga(\frac{m}{2}+1)}\left|\frac{\det(\text{g}_1)^m}{\det(\text{g}_2)}\right|\int_{\al_{q,T}}^{\be_{q,T}} \frac{(R_{q,T}(t))^{\frac{m}{2}}}{t^{m+2}}dt.
\end{equation}

§.§.§ Proving Lemma <ref>

The proof of Lemma <ref> consists of the following bound and on the summation Lemma <ref>.

It holds that

\begin{align}
\mu(V_{q,T}[g_1,g_2])
\le & v_m\left|\frac{\det(\text{g}_1)^m}{\det(\text{g}_2)}\right|\frac{T^{m(m+1)}}{\|\text{g}_1 q \text{g}_2\|^{m^2}}\label{eq:bound on measure of Vga T} \\
\leq&v_m\left|\frac{\det(\text{g}_1)^m}{\det(\text{g}_2)}\right| \|\text{g}_1^{-1}\|^{m^2}\|\text{g}_2^{-1}\|^{m^2}\frac{T^{m(m+1)}}{\|q\|^{m^2}}.\label{eq:bound on volume of connected component}
\end{align}

In particular, when $g_1,g_2$ vary in a compact subset of $\SL(m+1,\R)$, it holds that

$$\mu(V_{q,T}[g_1,g_2])\leq T^{m(m+1)}O\left(\frac{1}{\|q\|^{m^2}}\right).$$

We have

\begin{align*}
&\int_{\al_{q,T}}^{\be_{q,T}} \frac{(-\det(\text{g}_1\text{g}_2)^{-2}t^{\frac{2}{m}+2}+t^{\frac{2}{m}}T^2-\|\text{g}_1 q \text{g}_2\|^2)^{\frac{m}{2}}}{t^{m+2}}dt\\
\le &\int_{\al_{q,T}}^{\be_{q,T}} \frac{(t^{\frac{2}{m}}T^2)^{\frac{m}{2}}}{t^{m+2}}dt\\
=& T^m\int_{\al_{q,T}}^{\be_{q,T}}\frac{1}{t^{m+1}}\\
\leq& T^m\frac{1}{m\al_{q,T}^m}.
\end{align*}

Using (<ref>),

$$\frac{1}{\al_{q,T}^m}\leq\frac{T^{m^2}}{\|\text{g}_1 q \text{g}_2\|^{m^2}}\leq T^{m^2}\frac{\|\text{g}_1^{-1}\|^{m^2}\|\text{g}_2^{-1}\|^{m^2}}{\|q\|^{m^2}},$$

which proves the claim upon recalling (<ref>).

In the following  $g_1,g_2$ vary in a bounded set. By Lemma <ref>,

\begin{align*}
\sum_{\substack{q\in \Delta,\\ \sqrt{T}<\|\text{g}_1 q \text{g}_2\|}}\mu(V_{q,T}[g_1,g_2])\le   T^{m(m+1)}O\left(\sum_{\substack{q\in \Delta,\\ \sqrt{T}<\|\text{g}_1 q \text{g}_2\|}} \frac{1}{\|q\|^{m^2}}\right).
\end{align*}

The proof is complete by recalling (<ref>) and by applying Lemma <ref>.

We now return to $\mu \left(H_T[g_1,g_2] \right)$. We conclude by (<ref>) that

\begin{equation}\label{eq:measure of H skewed balls as sums of integrals}
\mu \left(H_T[g_1,g_2] \right)
=\sum_{\substack{q\in \Delta,\\ \|\text{g}_1 q \text{g}_2\|< \sqrt{M_T}}} \frac{v_m}{|\det(\text{g}_1)^{-1}|^m |\det(\text{g}_2)|}\int_{\al_{q,T}}^{\be_{q,T}} \frac{(-\det(\text{g}_1\text{g}_2)^{-2} t^{\frac{2}{m}+2}+t^{\frac{2}{m}}T^2-\|\text{g}_1 q \text{g}_2\|^2)^{\frac{m}{2}}}{t^{m+2}}dt,
\end{equation}

where $M_T$ is given by (<ref>). We split the sum into two parts as

\begin{equation}\label{eq:truncation of A_m<sqrt T}
\left(\sum_{\substack{q\in \Delta,\\ \|\text{g}_1 q \text{g}_2\|< \sqrt{T}}}+\sum_{\substack{q\in \Delta,\\ \sqrt{T}<\|\text{g}_1 q \text{g}_2\|< \sqrt{M_T}}} \right) \frac{v_m}{|\det(\text{g}_1)^{-1}|^m |\det(\text{g}_2)|}\int_{\al_{q,T}}^{\be_{q,T}} \frac{(-\det(\text{g}_1\text{g}_2)^{-2} t^{\frac{2}{m}+2}+t^{\frac{2}{m}}T^2-\|\text{g}_1 q \text{g}_2\|^2)^{\frac{m}{2}}}{t^{m+2}}dt.
\end{equation}

We will now estimate the second sum. The second sum is relatively easier to estimate, and it will eventually follow that it is of lower order in $T$ compared to the first sum.

If $g_1,g_2$ vary in a bounded set, then

\begin{equation}
\sum_{\substack{q\in \Delta,\\ \sqrt{T}<\|\text{g}_1 q \text{g}_2\|< \sqrt{M_T}}}  \frac{v_m}{|\det(\text{g}_1)^{-1}|^m |\det(\text{g}_2)|}\int_{\al_{q,T}}^{\be_{q,T}} \frac{(-\det(\text{g}_1\text{g}_2)^{-2} t^{\frac{2}{m}+2}+t^{\frac{2}{m}}T^2-\|\text{g}_1 q \text{g}_2\|^2)^{\frac{m}{2}}}{t^{m+2}}dt=O(T^{m(m+1)-m/2})
\end{equation}

In the following  $g_1,g_2$ vary in a bounded set. By Lemma <ref>,

\begin{align}
&\sum_{\substack{q\in \Delta,\\ \sqrt{T}<\|\text{g}_1 q \text{g}_2\|< \sqrt{M_T}}}\int_{\al_{q,T}}^{\be_{q,T}} \frac{(-\det(\text{g}_1\text{g}_2)^{-2}t^{\frac{2}{m}+2}+t^{\frac{2}{m}}T^2-\|\text{g}_1 q \text{g}_2\|^2)^{\frac{m}{2}}}{t^{m+2}}dt\\
\le &  T^{m(m+1)}O\left(\sum_{\substack{q\in \Delta,\\ \sqrt{T}<\|\text{g}_1 q \text{g}_2\|}} \frac{1}{\|q\|^{m^2}}\right).
\end{align}

We have
$$\|\text{g}_1 q \text{g}_2\|=\|\text{g}_1q\text{g}_2\|\leq\|\text{g}_1\|\|q\|\|\text{g}_2\|,$$
where $\|\text{g}_1\|,\|\text{g}_2\|$ are bounded above since they depend continuously on $g_1,g_2$. Then, for some $C>0$, we conclude that
$$O\left(\sum_{\substack{q\in \Delta,\\ \sqrt{T}<\|\text{g}_1 q \text{g}_2\|}} \frac{1}{\|q\|^{m^2}}\right)=O\left(\sum_{\substack{q\in \Delta,\\ C\sqrt{T}<\|q\|}} \frac{1}{\|q\|^{m^2}}\right).$$

By Lemma <ref> with $n=m$ and $\sigma=-m^2$, it follows that
$\sum_{\substack{q\in \Delta}} \frac{1}{\|q\|^{m^2}}$ converges
with the tail estimate

$$\sum_{\substack{q\in \Delta,\\ \sqrt{T}<\|q\|}} \frac{1}{\|q\|^{m^2}}=O(\sqrt{T}^{-m}),$$

which proves our claim.

§.§.§ Proving Lemma <ref>

Meanwhile by Taylor expansion,

\begin{align}
a=&\frac{\|\text{g}_1 q \text{g}_2\|^m}{T^m}\left(\frac{1}{1-\frac{\det(\text{g}_1)^{-1}^2\det(\text{g}_2)^{-1}^2 a^2}{T^2}}\right)^{\frac{m}{2}} \nonumber\\
=&\frac{\|\text{g}_1 q \text{g}_2\|^m}{T^m}\left(1+\frac{\det(\text{g}_1\text{g}_2)^{-2} a^2}{T^2}+\cdots \right)^{\frac{m}{2}} \label{taylor expansion for a}
\end{align}

when $\|\text{g}_1 q \text{g}_2\|\leq T$ we have that it follows from (<ref>) that

\begin{equation}\label{estimate for a in smaller range taylor expansion}
a=\frac{\|\text{g}_1 q \text{g}_2\|^m}{T^m}\left(1+\det(\text{g}_1\text{g}_2)^{-2} O_{g_1,g_2}\left(\frac{1}{T^2} \right)\right)^{\frac{m}{2}}, \text{ whenever }\|\text{g}_1 q \text{g}_2\|\le T.
\end{equation}

In this section we assume throughout that $\|\text{g}_1q\text{g}_2\|\leq\sqrt{T}$. For $\e_1\in(0,1)$ we define $\de=\de(T,q,\e_1)$ by

\begin{equation}\label{eq:definition of delta}
\de:=\frac{\|\text{g}_1 q \text{g}_2\|}{T^{1+\e_1}},\end{equation}
we define $\al_\de=\al_\de(q,T,\e_1)$ by
\begin{equation}\label{eq:defintion of a_delta}
\al_\de^{\frac{1}{m}}:= \al_{q,T}^{\frac{1}{m}}+\de,
\end{equation}
and for $\e_2\in(0,1)$ we let $\la=\la(q,T,\e_2)$

\begin{equation}\label{eq:definition of c}
\la:=\left(\frac{\|\text{g}_1 q \text{g}_2\|}{T^{\e_2}}\right)^{\frac{m}{m+1}}.
\end{equation}
By the bounds (<ref>) and (<ref>), it follows that when  $\|\text{g}_1 q\text{g}_2\|\leq\sqrt{T}$ and $g_1,g_2$ vary in a compact set, there exists a uniform $T_0$ such that for all $T\geq T_0,$
$$\al<\al_{\de}< \lambda<\be.$$
In what follows the reader should keep in mind that $T\geq T_0$. We consider the following partition of the integral appearing in  (<ref>),

\begin{align}
&\int_{\al_{q,T}}^{\be_{q,T}} \frac{(R_{q,T}(t))^{\frac{m}{2}}}{t^{m+2}}dt
=\left(\int_{\al_{q,T}}^{\al_{\de}(q,T,\e_1)}+\int_{\al_{\de}(q,T,\e_1)}^{\la(q,T,\e_2)}+\int_{\la(q,T,\e_2)}^{\be_{q,T}}\right) \frac{(R_{q,T}(t))^{\frac{m}{2}}}{t^{m+2}}dt\label{eq:partition of integral into four part}
\end{align}

In the following Lemmata we will estimate each of those integrals, and Lemma <ref> will follow by combining those estimates. The main term will come from  the integral in the range $\int_{\al_{\de}(q,T,\e_1)}^{\la(q,T,\e_2)}$.

Suppose that $\|\text{g}_1 q \text{g}_2\|\leq\sqrt{T}$ and that $g_1,g_2$ vary in a bounded set of $\SL(m+1,\R)$. Fix $\e_1\in(0,1)$. Then

\begin{align}\label{eq:first integral}
\int_{\al_{q,T}}^{\al_{\de}(q,T,\e_1)} \frac{(R_{q,T}(t))^{\frac{m}{2}}}{t^{m+2}}dt =O\left(\frac{T^{m(m+1)-\e_1\frac{m+2}{2}}}{\|q\|^{m^2}}\right)
\end{align}

Note since $\al_{q,T}$ is a root, we have that $R_{q,T}(t)$ is small in the range of the integral (<ref>). In fact, we will show that it's small enough compared with $t^{-m+2}$ which is large in that range. In the following, to ease the reading, we will omit from the notations the dependencies on $q$, $T$ and $\e_1$. We substitute the variable $s=t^{\frac{1}{m}}$ in (<ref>):

\begin{align*}
f(s):=R_{q,T}(s)=-\det(\text{g}_1\text{g}_2)^{-2} s^{2m+2}+T^2s^2-\|\text{g}_1 q \text{g}_2\|^2.
\end{align*}

We apply Taylor expansion to $f(s)$ in the range $s\in[\al^{\frac{1}{m}},\al_\de^{\frac{1}{m}}]$
at $s=\al^{\frac{1}{m}}$ which gives

\begin{align}
f(\al_\de^\frac{1}{m})&=f(\al^{\frac{1}{m}}+\de)=\underbrace{f(\al^{\frac{1}{m}})}_{=0}+f'(\al^{\frac{1}{m}})\de+\frac{f''(\xi)}{2}\de^2\nonumber\\&= [-(2m+2)\det(\text{g}_1\text{g}_2)^{-2} \al^\frac{2m+1}{m}+2T^2\al^\frac{1}{m}]\de+\frac{f''(\xi)}{2}\de^2\nonumber\\
&\ll T^2\al^\frac{1}{m}\de+f''(\xi)\de^2\label{eq:bound by taylor}
\end{align}

where $\xi\in [\al^\frac{1}{m},\al_\de^{\frac{1}{m}}]$. In the last line we used that for bounded $g_1,g_2$, $|\det(\text{g}_1\text{g}_2)^{-1}|$ is bounded above and away from zero. We not bound the second derivative $f''(\xi)$ over $[\al^\frac{1}{m},\al_\de^{\frac{1}{m}}]$:

\begin{align}
=&|-(2m+2)(2m+1)\det(\text{g}_1\text{g}_2)^{-2} \xi^{2m}+2T^2|\nonumber\\
\le & (2m+2)(2m+1)\det(\text{g}_1\text{g}_2)^{-2} \al_\de^\frac{2m}{m}+2T^2\nonumber \\
\ll & T^2.
\label{eq:bound on sectond der}\end{align}
Thus by (<ref>) and (<ref>),

\begin{equation}
f(\al_\de^{\frac{1}{m}})\ll T^2\al_\de^\frac{1}{m}\de.\label{eq:bound for f(al_de 1 ove m)}
\end{equation}
We note that for all large $T$ (independently of $q$ and $\e_1$) $f$ is monotonically increasing on the interval $(\al^\frac{1}{m},\al_\de^{\frac{1}{m}})$. To see this, recall that $\theta_T^\frac{1}{m}\sim T^\frac{1}{m}$ is a critical point for $f(s)$,  $f(s)$ is monotonically increasing in $(0,\theta_T^\frac{1}{m})$ (see Lemma <ref> and it's proof), and $\al_\de=o(1)$. Then,

\begin{align*}
\int_{\al}^{\al_{\de}} \frac{(R_{q,T}(t))^{\frac{m}{2}}}{t^{m+2}}dt
\le&\frac{1}{\al^{m+2}}\int_{\al}^{\al_{\de}} (R_{q,T}(t))^{\frac{m}{2}}dt\\
=&\frac{1}{\al^{m+2}}\int_{\al^\frac{1}{m}}^{\al_{\de}^\frac{1}{m}} f(s)^{\frac{m}{2}}ms^{m-1}ds
\end{align*}

\begin{align*}
\ll&\frac{1}{\al^{m+2}}(f(\al_\de^{\frac{1}{m}})^\frac{m}{2}\al_\de^{\frac{m-1}{m}})\de\tag{monotonicity of $f(s)$}\\
\ll&\frac{1}{\al^{m+1+\frac{1}{m}}}f(\al_\de^{\frac{1}{m}})^\frac{m}{2}\de\tag{$\al_\de\ll\al$}
\\
\ll&\frac{\left\{T^2\al_\de^{\frac{1}{m}}\right\}^{\frac{m}{2}}}{\al^{m+1+\frac{1}{m}}}\cdot\delta^{\frac{m+2}{2}}\tag{by \eqref{eq:bound for f(al_de 1 ove m)}}\\
\ll& \frac{\left\{T^2\al^{\frac{1}{m}}\right\}^{\frac{m}{2}}}{\al^{m+1+\frac{1}{m}}}\cdot\delta^{\frac{m+2}{2}} \tag{$\al_\de\ll\al$}
\end{align*}

Using the bounds of $\al$ in (<ref>) and by applying the definition of $\delta$ in (<ref>), we get
\begin{equation*}
\frac{\left\{T^2\al^{\frac{1}{m}}\right\}^{\frac{m}{2}}}{\al^{m+1+\frac{1}{m}}}\cdot\delta^{\frac{m+2}{2}}\ll\frac{T^{m(m+1)-\e_1\frac{m+2}{2}}}{\|\text{g}_1 q \text{g}_2\|^{m^2}},
\end{equation*}

which concludes our proof.

We proceed to treat the middle integral, which will be the main term in (<ref>).

Let $\|\text{g}_1 q \text{g}_2\|\leq\sqrt{T}$, and assume that $g_1,g_2$ vary in a bounded set of $\SL(m+1,\R)$. Suppose that $0<\e_1<2\e_2<1$, and let
\begin{equation*}
\e:=\min\{\e_1,2\e_2-\e_1,1+\frac{1-2\e_2}{m+1}\}.
\end{equation*}
\begin{align}
\cdot \frac{T^{m(m+1)}}{\|\text{g}_1 q \text{g}_2\|^{m^2}}+O\left(\frac{T^{m(m+1)-\e}}{\|q\|^{m^2}}\right). \label{eq:main term estimate for integral in a delta to c}
\end{align}

To ease the reading, we will omit from some notations the dependencies on $q$, $T$ and $\e_1,\e_2$.
We rewrite our integral (<ref>) as

\begin{align*}
\int_{\al_{\de}}^{\la}\frac{(-\det(\text{g}_1\text{g}_2)^{-2} t^{\frac{2}{m}+2}+t^{\frac{2}{m}}T^2-\|\text{g}_1 q \text{g}_2\|^2)^{\frac{m}{2}}}{t^{m+2}}dt&\\
=\int_{\al_\de}^\la\frac{(t^{\frac{2}{m}}T^2-\|\text{g}_1 q \text{g}_2\|^2)^{\frac{m}{2}}}{t^{m+2}}
& \left(1-\frac{\det(\text{g}_1\text{g}_2)^{-2}t^{\frac{2}{m}+2}}{t^{\frac{2}{m}}T^2-\|\text{g}_1 q \text{g}_2\|^2} \right)^{\frac{m}{2}}dt.
\end{align*}

We now show that in the range $t\in(\al_\de,\la)$ it holds that

\begin{equation}\label{much less than 1}
\frac{\det(\text{g}_1\text{g}_2)^{-2}t^{\frac{2}{m}+2}}{t^{\frac{2}{m}}T^2-\|\text{g}_1 q \text{g}_2\|^2}= O\left(\frac{1}{T^{2\e_2-\e_1}}\right).
\end{equation}

Recall that we assume here $\|\text{g}_1 q \text{g}_2\|\leq\sqrt{T}$, and as $g_1,g_2$ vary in a bounded set, $|\det(\text{g}_1\text{g}_2)^{-1}|$ will be bounded above. We have for $t\in(\al_\de,\la)$

\begin{align*}
\frac{\det(\text{g}_1\text{g}_2)^{-2}t^{\frac{2}{m}+2}}{t^{\frac{2}{m}}T^2-\|\text{g}_1 q \text{g}_2\|^2}
\ll &  \frac{\la^{2\frac{m+1}{m}}}{(\al^\frac{1}{m}+\de)^2T^2-\|\text{g}_1 g \text{g}_2\|^2}  \\
\ll & \frac{\la^{2\frac{m+1}{m}}}{O(T^{-m-3})+2\al^\frac{1}{m}\de T^2+\de^2 T^2} \tag{by \eqref{eq:asymp for a when A small then sqrt T}}\\
\ll &\frac{\la^{2\frac{m+1}{m}}}{\al^\frac{1}{m}\de T^2} \\
\leq & \frac{(\nicefrac{\|\text{g}_1 q \text{g}_2\|}{T^{\e_2}})^2}{ \frac{\|\text{g}_1 q \text{g}_2\|}{T}\frac{\|\text{g}_1 q \text{g}_2\|}{T^{1+\e_1}}T^2}=O\left(\frac{1}{T^{2\e_2-\e_1}}\right)\tag{using \eqref{range for a} for bounding $\al$}.
\end{align*}

Now in view of the estimate $(1+x)^{\alpha}\sim 1+\alpha x$ as $x\to0$, we conclude that

\begin{align}
\int_{\al_\de}^\la \frac{(t^{\frac{2}{m}}T^2-\|\text{g}_1 q \text{g}_2\|^2)^{\frac{m}{2}}}{t^{m+2}}
\left(1-\frac{\det(\text{g}_1\text{g}_2)^{-2}t^{\frac{2}{m}+2}}{t^{\frac{2}{m}}T^2-\|\text{g}_1 q \text{g}_2\|^2} \right)^{\frac{m}{2}}dt\nonumber\\=\left(\int_{\al_\de}^\la \frac{(t^{\frac{2}{m}}T^2-\|\text{g}_1 q \text{g}_2\|^2)^{\frac{m}{2}}}{t^{m+2}}
dt\right)& \left(1+O\left(\frac{1}{T^{2\e_2-\e_1}}\right)\right). \label{eq:estimate for the fraction in the integral a_de to c}
\end{align}

We shall now estimate $\int_{\al_\de}^\la \frac{(t^{\frac{2}{m}}T^2-\|\text{g}_1 q \text{g}_2\|^2)^{\frac{m}{2}}}{t^{m+2}}

\begin{align}
&\int_{\al_\de}^\la \frac{(t^{\frac{2}{m}}T^2-\|\text{g}_1 q \text{g}_2\|^2)^{\frac{m}{2}}}{t^{m+2}}dt\nonumber\\
=&\int_{\al_\de}^\la\frac{T^m}{t^{m+1}}\left(1-\frac{\|\text{g}_1 q \text{g}_2\|^2}{T^2 t^{\frac{2}{m}}}\right)^{\frac{m}{2}}dt \nonumber\\
=&\frac{m}{2} \frac{T^{m(m+1)}}{\|\text{g}_1 q \text{g}_2\|^{m^2}}\int^{\nicefrac{\|\text{g}_1 q \text{g}_2\|^2}{T^2 \al_{\de}^{\frac{2}{m}}}}_{\nicefrac{\|\text{g}_1 q \text{g}_2\|^2}{T^2 \la^{\frac{2}{m}}}}
u^{\frac{m^2}{2}-1}(1-u)^{\frac{m}{2}}du. \label{upper and lower limits of the integral}
\end{align}
It remains to prove the estimate

\begin{align}
\int^{\nicefrac{\|\text{g}_1 q \text{g}_2\|^2}{T^2 \al_{\de}^{\frac{2}{m}}}}_{\nicefrac{\|\text{g}_1 q \text{g}_2\|^2}{T^2 \la^{\frac{2}{m}}}}
.\label{eq:estimate of the diffence for the beta integral}
\end{align}
where in the last equality we used the classical formula for the beta function. Notice that by  (<ref>), by (<ref>) and by (<ref>) the proof is complete.

To prove (<ref>) it suffices to estimate the differences between the endpoints of the corresponding integral, namely it suffices to estimate $\left|{1}-{\frac{\|\text{g}_1 q \text{g}_2\|^2}{T^2 \al_{\de}^{\frac{2}{m}}}}\right|$ and $\left|\frac{\|\text{g}_1 q \text{g}_2\|^2}{T^2\la^{\frac{2}{m}}}-0\right|=\frac{\|\text{g}_1 q \text{g}_2\|^2}{T^2\la^{\frac{2}{m}}}.$
We have

\begin{align*}
\left|{1}-{\frac{\|\text{g}_1 q \text{g}_2\|^2}{T^2 \al_{\de}^{\frac{2}{m}}}}\right|=&\left|\frac{T^2\al_\de^\frac{2}{m}-\|\text{g}_1 q \text{g}_2\|^2}{T^2\al_\de^\frac{2}{m}}\right|\\
= &\left|\frac{(T^2\alpha^\frac{2}{m}-\|\text{g}_1 q \text{g}_2\|^2)+(2\al^\frac{1}{m}\de+\de^2)T^2}{T^2\al_\de^\frac{2}{m}}\right|\tag{using definition of $\al_\de$, see \eqref{eq:defintion of a_delta}}\\
= &\frac{\left|O(\nicefrac{1}{T^{m+1}})+(2\al^\frac{1}{m}\de+\de^2)T^2\right|}{T^2\al_\de^\frac{2}{m}}\tag{see  \eqref{eq:asymp for a when A small then sqrt T}}\\
\leq & \frac{\left|O(\nicefrac{1}{T^{m+1}})+(2\al^\frac{1}{m}\de+\de^2)T^2\right|}{\left|T^2\al^\frac{2}{m}\right|}\\
= & O\left(\frac{1}{T^{\e_1}}\right)\tag{since $T^2\al^\frac{2}{m}\asymp \|\text{g}_1q\text{g}_2\|^2$ and $(2\al^\frac{1}{m}\de+\de^2)T^2=O(\frac{\|\text{g}_1q\text{g}_2\|^2}{T^{\e_1}})$},
\end{align*}

\begin{align}
\left|{1}-{\frac{\|\text{g}_1 q \text{g}_2\|^2}{T^2 \al_{\de}^{\frac{2}{m}}}}\right|=&\left|{1}-\frac{\|\text{g}_1 q \text{g}_2\|^2}{T^2a^{\frac{2}{m}}(1+a^{-\frac{1}{m}}\de)^2}\right|\\
= &\left|{1}-\frac{\|\text{g}_1 q \text{g}_2\|^2}{T^2\left(\frac{\|\text{g}_1 q \text{g}_2\|^2}{T^2}+O(\frac{1}{T^{m+1.5}})\right)(1+a^{-\frac{1}{m}}\de)^2}\right|\\
= & \left|{1}-\frac{\|\text{g}_1 q \text{g}_2\|^2}{T^2\left(\frac{\|\text{g}_1 q \text{g}_2\|^2}{T^2}+O(\frac{1}{T^{m+1.5}})\right)}(1+O(a^{-\frac{1}{m}}\de))\right|\\
= & \left|{1}-\left(1+O(\frac{1}{\|\text{g}_1 q \text{g}_2\|^2T^{m-0.5}})\right)\left(1+O(\frac{T}{\|\text{g}_1 q \text{g}_2\|}\frac{\|\text{g}_1 q \text{g}_2\|}{T^{1+\e_1}})\right)\right|\\
= & O\left(\frac{1}{T^{\e_1}}\right).
\end{align}

and finally

\begin{equation*}
\frac{\|\text{g}_1 q \text{g}_2\|^2}{T^2\la^{\frac{2}{m}}}=\frac{\|\text{g}_1 q \text{g}_2\|^2}{T^2\left(\nicefrac{\|\text{g}_1 q \text{g}_2\|}{T^{\e_2}}\right)^{\frac {2}{m+1}}}=\frac{\|\text{g}_1 q \text{g}_2\|^{2-\frac{2}{m+1}}}{T^{2-\frac{2\e_2}{m+1}}}\underbrace{=}_{\|\text{g}_1 q \text{g}_2\|\leq\sqrt{T}}O\left(\frac{1}{T^{1+\frac{1-2\e_2}{m+1}}}\right).
\end{equation*}

We now turn to our estimate for the last integral.

Let $\|\text{g}_1 q \text{g}_2\|\leq\sqrt{T}$, and assume that $g_1,g_2$ vary in a bounded set of $\SL(m+1,\R)$. Fix $\e_2\in(0,1)$. Then
\begin{equation*}
\int_{\la(q,T,\e_2)}^{\be_{q,T}} \frac{(R_{q,T}(t))^{\frac{m}{2}}}{t^{m+2}}dt=T^{m+\e_2\frac{ m^2}{m+1}}O\left(\frac{1}{\|q\|^{\frac{m^2}{m+1}}}\right).
\end{equation*}

Again, to ease the reading, we will omit from the notations the dependencies on $q$,$T$ and $\e_2$.
We have

\begin{align*}
&\int_\la^\be \frac{(-\det(\text{g}_1\text{g}_2)^{-2} t^{\frac{2}{m}+2}+t^{\frac{2}{m}}T^2-\|\text{g}_1 q \text{g}_2\|^2)^{\frac{m}{2}}}{t^{m+2}}dt\\
\le &\int_\la^\be \frac{(t^{\frac{2}{m}}T^2)^{\frac{m}{2}}}{t^{m+2}} dt =\int_\la^\be \frac{T^m}{t^{m+1}}dt \\
\le & \frac{T^m}{m\la^m}=\frac{T^m}{m(\|\text{g}_1 q \text{g}_2\|/T^{\e_2})^{\frac{m^2}{m+1}}}=\frac{T^{m+\frac{\e_2 m^2}{m+1}}}{m\|\text{g}_1 q \text{g}_2\|^{\frac{m^2}{m+1}}}
\end{align*}

§ EQUIDISTRIBUTION ALONG SKEWED $H$-BALLS

Recall $G=\SL(m+1,\R)$, and $\Ga\leq G$ is a lattice. The second important ingredient required for the method of <cit.> are the following ergodic theorems.  For  $x\in X,g_1, g_2\in G$  and $T>0$, we consider the measure defined by the integral

\begin{equation}\label{definition of measure funtional}
\mu_{T,g_1,g_2}(F):=\frac{1}{\mu \left(H_T[g_1,g_2] \right)}\int_{H_T[g_1,g_2]}F(h^{-1}x)d\mu(h),~F\in C_c(G/\Ga).
\end{equation}

Let $\mu_X$ be the  $G$-invariant probability measure on $X=G/\Ga$. For all $x\in X,g_1, g_2\in G$  and all $F\in C_c(X)$

\begin{equation}
\lim_{T\to \infty} \mu_{T,g_1,g_2}(F)=\mu_X(F).
\end{equation}

TODO: explain more in detail hwo the ergodic theorem is proved, that it generalizes to other norms, and that it's not possible to use Margulis's banana trickThe above ergodic theorem is proven in Section <ref>. In an overview, we will establish unipotent invariance of the limiting measure, which opens the way to apply the celebrated results of Ratner (see e.g. <cit.>), combined with the results of Shah (see e.g. <cit.>) on the behaviour of polynomial orbits. The main technical result we prove is a certain divergence of polynomial maps in representation space, where the domain of the maps "shrinks" (see Lemma <ref>).
We start by reducing Theorem <ref> to the  equidistribution statement along skewed balls of the connected component of $H$. Similarly to the above, consider for $x\in X,g_1,g_2\in G$ and $T>0$ the measures

\begin{equation}
\label{eq:measures along the connected component} \mu^\circ_{T,g_1,g_2}(F):=\frac{1}{\mu(V_T[g_1,g_2])}\int_{V_T[g_1,g_2]}F(a_t^{-1}u_{v}^{-1}x)dv\frac{1}{t^{m+2}}dt,
\end{equation}

where $V_T[g_1,g_2]$ is the skew-ball of the identity component of $H$.

For all $x\in X, g_1,g_2\in G$ and all $F\in C_c(X)$ it holds that

In view of (<ref>), we can decompose $\mu_{T,g_1,g_2}(F)$ as the following convex linear combination:
\begin{align*}
\mu_{T,g_1,g_2}(F)
=\sum_{\substack{q\in \Delta}}\frac{\mu \left(V_{q,T}[g_1,g_2] \right)}{\mu \left(H_T[g_1,g_2] \right)}\frac{1}{\mu \left(V_{q,T}[g_1,g_2] \right)}\int_{V_{q,T}[g_1,g_2]} F(\tilde{q}^{-1}a_t^{-1}u_v^{-1}x) \frac{1}{t^{m+2}}dv dt
\end{align*}

Recall that
\begin{equation*}
\mu(V_{q,T}[g_1,g_2])=\mu(V_T[g_1,\tilde{q} g_2]),
\end{equation*}
and observe that
\begin{equation}
\frac{1}{\mu \left(V_{q,T}[g_1,g_2] \right)}\int_{V_{q,T}[g_1,g_2]} F(\tilde{q}^{-1}a_t^{-1}u_v^{-1}x) \frac{1}{t^{m+2}}dv dt=\mu_{T,g_1,\tilde{q} g_2}^\circ(L_{\tilde{q}^{-1}}(F)).
\end{equation}
By assuming Theorem <ref>, we have for all $q\in\Delta$,
$$\lim_{T\to\infty}\mu_{T,g_1,\tilde{q} g_2}^\circ(L_{\tilde{q}^{-1}}(F))=\mu_X(L_{\tilde{q}^{-1}}(F))\underbrace{=}_{\text{invariance}}\mu_X(F).$$
We denote
\begin{equation*}
c_{q,T}:=\frac{\mu \left(V_{q,T}[g_1,g_2] \right)}{\mu \left(H_T[g_1,g_2] \right)}.
\end{equation*}

Then clearly,
$$\sum_{q\in\Delta}c_{q,T}=1,~\forall T,\text{ and }c_{q,T}\leq 1,~\forall q,T.$$

Importantly, by Lemma <ref> and by (<ref>) there is $C>0$ such that for all $T>0$

By Lemma <ref>  the sum $\sum_{q\in\Delta}\frac{1}{\|q\|^{m^2}}$ converges. Then for arbitrary small $\e>0$, we may find $N_\e$ such that
$$\sum_{\substack{q\in\Delta,\\\|q\|>N_\e}}c_{q,T}\leq \e,~ \forall T>0.$$

\begin{align*}
|\mu_{T,g_1,g_2}(F)-\mu_X(F)|=&\left|\sum_{q\in\Delta}c_{q,T}(\mu_{T,g_1,\tilde{q} g_2}^\circ(F)-\mu_X(F))\right|\\
\leq &\sum_{q\in\Delta}c_{q,T}\left|\mu_{T,g_1,\tilde{q} g_2}^\circ(F)-\mu_X(F)\right|\\
\leq&\sum_{\substack{q\in\Delta,\\\|q\|\leq N_\e}}\left|\mu_{T,g_1,\tilde{q} g_2}^\circ(F)-\mu_X(F)\right|+2\e\|F\|_\infty.
\end{align*}

Then, by assuming Theorem <ref>, we get that
which concludes our proof.

In the rest of the section we will be proving Theorem <ref>, and we now explain the main ideas. In an overview, we prove unipotent invariance of the limiting measure and apply the linearisation technique, which is by now a standard technique in homogeneous dynamics. More specifically, let $\overline{X} = \overline{G/\Ga}$ denote the one-point compactification of $X=G/\Ga$. Then, by the Banach-Alaoglu theorem, there is a weak-* limit $\eta$ of the measures $\mu^\circ_{T,g_1,g_2},~ T>0$, and $\eta$ is a probability measure on $\overline{X}$. In Section <ref>, we prove that $\eta$ is invariant under $U=\begin{bmatrix}
\end{bmatrix}$. The celebrated  Ratner theorems on measure rigidity <cit.> state that any $U$-invariant and ergodic measure on $G/\Ga$ is supported on a closed orbit having finite measure. In particular, this allows to describe the ergodic decomposition of the $U$-invariant measure $\eta$, and reduces the problem to proving that there is no escape of mass, namely $\eta(\infty)=0$, and to showing that the $\eta$-measure of proper closed orbits is zero. To prove the later fact, notice that the measures $\mu^\circ_{T,g_1,g_2}$ involve integration along families of polynomial trajectories. This allows to utilize deep results due to Shah on the behavior of polynomial trajectories in $G/\Ga$, see <cit.> and <cit.> (generalizing  the celebrated results of Dani-Margulis <cit.>) building on the linearisation technique. The essential challenge in applying the theorems of Shah is proving a certain expansion property of the acting group in representation space. We note that our apporach is similar to the one taken by Gorodnik in  <cit.>. The essential difference between our approach compared to <cit.> is in our treatment of the expansion in the representation space.

Notice that our measures are explicitly given by
$$\mu^\circ_{T,g_1,g_2}(F)= \frac{1}{\mu \left(V_{T}[g_1,g_2] \right)} \int_{\al_T}^{\be_T} \int_{D_{T,t}} F \left(a_t^{-1} u_v^{-1} g_1\Gamma \right) dv \frac{1}{t^{m+2}}dt,~F\in C_c(G/\Ga).$$
Now $U$ is the horosphere for $a_t^{-1}$ which expands as $t\to 0.$ Then one might hope to apply the classical Margulis thickening trick (see Margulis's thesis
<cit.>) in order to prove the equidistribution of $\int_{D_{T,t}} F \left(a_t^{-1} u_v^{-1} g_1\Gamma \right) dv$ as $t\to0$. This approach fails since the ellipsoids $D_{T,t}\subseteq\R^{m+1}$ shrink as $t\to0$. Moreover, the integration above is essentially supported on the range where $t$ is small, see (<ref>).

§.§ The $U$-invariance of the measure along skewed balls of $V$

The main goal here is to prove that any weak-* limit of $\mu_{T,g_1,g_2}^\circ$  is $U:=\begin{bmatrix}
I_m & 0 \\
\mathbb{\R}^m & 1
\end{bmatrix}$-invariant. For any $u_0\in U$ and $F\in C_c(X)$ consider $L_{u_0}(F)(x):=F(u_0^{-1}x)$.

For all $g_1,g_2\in G$ and all $F\in C_c(X)$ it holds that

\begin{equation*}
\lim_{T\to \infty}\left(\mu^\circ_{T,g_1,g_2}(F)-\mu^\circ_{T,g_1,g_2}(L_{u_0} (F)\right)=0.
\end{equation*}

The following lemma will be needed:

Let $f:\R^{d}\to \R$ be a bounded continuous function. Let $E\subset \R^{d}$ be an ellipsoid with surface area $S$, then for $y\in \R^{d}$ we have the estimate:

\begin{equation*}
\left|\int_{E}[f(v)-f(y+v)]dv\right| \ll \|f\|_\infty \|y\|S,
\end{equation*}

where the implied constant depends on $d$ only.

An elementary geometric argument (thickening the boundary) implies that
$$\vol(E\triangle (E-y))\ll \|y\|S,$$
where $S$ is the surface area of the ellipsoid, and Vol stands for the Lebesgue volume. Alternatively, see <cit.> for a much more general theorem implying this fact. Now

\begin{align*}
\left|\int_{E}[f(v)-f(y+v)]dv\right|
= &\left|\int_{E}f(v)dv - \int_{E}f(v)dv\right|\\
\le & \int_{E\triangle
\ll & \|f\|_\infty\|y\|S, \tag{$f$ is bounded}
\end{align*}

Recall that
$$V_T[g_1,g_2]=\{u_va_t:t>0,~v\in D_{T,t}\},$$

\begin{equation}\label{eq:ellipsoid for connected component}
D_{T,t}:=\{v\in\R^m:\|\text{v}_1 \text{g}_2+\det(\text{g}_1)^{-1} v \text{g}_2+\det(\text{g}_1)^{-1} \text{v}_2t^{\frac{1}{m}+1}\|^2 \leq R_{T}(t)\},
\end{equation}
and where
\begin{equation}\label{eq:radius for connected component}
R_T(t):=-\det(\text{g}_1\text{g}_2)^{-2} t^{\frac{2}{m}+2}+t^{\frac{2}{m}}T^2-\|\text{g}_1 \text{g}_2\|^2,
\end{equation}
see (<ref>), (<ref>) (with $q=I_m$). Here $\text{g}_i,\text{v}_j$ are defined by (<ref>).  We denote $\al_T:=\al_{I_m,T}$ and $\be_T:=\be_{I_m,T}$ the roots $R_T(t)$, see Lemma <ref> (again, with $q=I_m$).
Let $F\in C_c(X)$ and fix $u_{v_0}\in U$. We have

\begin{align*}
&\left| \int_{V_{T}[g_1,g_2]} F(h^{-1}x) d\mu(h)-\int_{V_{T}[g_1,g_2]} F(u_{v_0}^{-1}h^{-1}x) d\mu(h) \right |\\
=&\left|\int_{\al_{T}}^{\be_{T}}\int_{D_{T,t}} \left(F( a_t^{-1} u_v^{-1} x) - F(u_{v_0}^{-1} a_t^{-1} u_v^{-1}x) \right) dv \frac{1}{t^{m+2}}dt\right |
\end{align*}

\begin{align*}=&\left|\int_{\al_{T}}^{\be_{T}} \int_{D_{T,t}} \left(F( a_t^{-1} u_v^{-1} x) - F(a_t^{-1} [a_t u_{v_0}^{-1}a_t^{-1}]  u_v^{-1}x) \right) dv \frac{1}{t^{m+2}}dt\right |\\
=&\left|\int_{\al_{T}}^{\be_{T}} \int_{D_{T,t}} \left(F( a_t^{-1} u_{-v} x) - F(a_t^{-1} u_{-t^{\frac{1}{m}+1}v_0-v}  x) \right) dv \frac{1}{t^{m+2}}dt\right |.
\end{align*}

For the use of Lemma <ref>, notice that the surface area of the ellipsoids $D_{T,t}$ is $\asymp (R_T(t))^{\frac{m-1}{2}}$. We have

\begin{align*}
&\left| \int_{\al_T}^{\be_T}  \int_{D_{T,t}} \left(F( a_t^{-1} u_{-v} g_1\Gamma) - F(a_t^{-1} u_{-t^{\frac{1}{m}+1}v_0-v}  g_1\Gamma) \right) dv \frac{1}{t^{m+2}}dt \right|\\
\ll &  \int_{\al_T}^{\be_T}   \|-t^{\frac{1}{m}+1}v_0\|(-\det(\text{g}_1\text{g}_2)^{-2} t^{\frac{2}{m}+2}+t^{\frac{2}{m}}T^2-\|\text{g}_1  \text{g}_2\|^2)^\frac{m-1}{2} \frac{1}{t^{m+2}}dt  \tag{by Lemma \ref{lemma on symmetric difference}}\\
\ll &  \int_{\al_T}^{\be_T}   \frac{(t^{\frac{2}{m}}T^2)^\frac{m-1}{2}}{t^{m-\frac{1}{m}+1}}dt \\
\ll& T^{m-1}\frac{1}{\al_T^{m-1}}\\
\ll& T^{m-1}\frac{1}{1/T^{m(m-1)}}=T^{m^2-1}. \tag{by \eqref{range for a}}
\end{align*}

After normalization by $V_{T}[g_1,g_2]\asymp T^{m(m+1)}$, this term goes to zero as $T\to \infty$.

Recall that
$$V_T[g_1,g_2]=\{u_va_t:t>0,~v\in D_{T,t}\},$$

\begin{equation}
D_{T,t}:=\{v\in\R^m:\|\text{v}_1 \text{g}_2+\det(\text{g}_1)^{-1} v \text{g}_2+\det(\text{g}_1)^{-1} \text{v}_2t^{\frac{1}{m}+1}\|^2 \leq R_{T}(t)\},
\end{equation}
and where
\begin{equation}
R_T(t):-\det(\text{g}_1\text{g}_2)^{-2} t^{\frac{2}{m}+2}+t^{\frac{2}{m}}T^2-\|\text{g}_1 \text{g}_2\|^2,
\end{equation}
see (<ref>), (<ref>) (with $q=I_m$).  We denote $\al_T:=\al_{I_m,T}$ and $\be_T:=\be_{I_m,T}$ the roots $R_T(t)$, see Lemma <ref> (again, with $q=I_m$). Let $F\in C_c(X)$ and fix $u_{v_0}=\begin{bmatrix}
I_m & 0 \\
v_0 & 1
\end{bmatrix}\in U$. We have

\begin{align*}
&\left| \int_{V_{T}[g_1,g_2]} F(h^{-1}x) d\mu(h)-\int_{V_{T}[g_1,g_2]} F(u_{v_0}^{-1}h^{-1}x) d\mu(h) \right |\\
=&\left|\int_{\al_{T}}^{\be_{T}}\int_{D_{T,t}} \left(F( a_t^{-1} u_v^{-1} x) - F(u_{v_0}^{-1} a_t^{-1} u_v^{-1}x) \right) dv \frac{1}{t^{m+2}}dt\right |\\
=&\left|\int_{\al_{T}}^{\be_{T}} \int_{D_{T,t}} \left(F( a_t^{-1} u_v^{-1} x) - F(a_t^{-1} [a_t u_{v_0}^{-1}a_t^{-1}]  u_v^{-1}x) \right) dv \frac{1}{t^{m+2}}dt\right |\\
=&\left|\int_{\al_{T}}^{\be_{T}} \int_{D_{T,t}} \left(F( a_t^{-1} u_{-v} x) - F(a_t^{-1} u_{-t^{\frac{1}{m}+1}v_0-v}  x) \right) dv \frac{1}{t^{m+2}}dt\right |\\
=&\left| \left( \int_{\al_T}^{c_T} + \int_{c_T}^{\be_T}  \right) \int_{D_{T,t}} \left(F( a_t^{-1} u_v^{-1} x) - F(a_t^{-1} u_{-t^{\frac{1}{m}+1}v_0-v}  x) \right) dv \frac{1}{t^{m+2}}dt\right |,
\end{align*}

where the auxiliary parameter $c_T$ is chosen by
\begin{equation}\label{the truncation c}
\end{equation}

To estimate the integral $\int_{c_T}^{\be_T}$ we use the trivial bound

\begin{align*}
&\left|  \int_{D_{T,t}} \left(F( a_t^{-1} u_{-v} g_1\Gamma) - F(a_t^{-1} u_{-t^{\frac{1}{m}+1}v_0q^{-1}-v}  g_1\Gamma) \right) dv  \right|\\
\le &   \int_{D_{T,t}} 2\|F\|_\infty dv
\ll  (-\det(\text{g}_1\text{g}_2)^{-2} t^{\frac{2}{m}+2}+t^{\frac{2}{m}}T^2-\|\text{g}_1  \text{g}_2\|^2)^{\frac{m}{2}} \tag{by \eqref{eq:volume of D_gamma,t}},
\end{align*}

so that the integral $\int_{c_T}^{\be_T}$ will be bounded by

\begin{align*}
&\int_{c_T}^{\be_T} \frac{(-\det(\text{g}_1\text{g}_2)^{-2} t^{\frac{2}{m}+2}+t^{\frac{2}{m}}T^2-\|\text{g}_1  \text{g}_2\|^2)^{\frac{m}{2}}}{t^{m+2}}dt\\
\leq & \int_{c_T}^{\be_T}\frac{(t^\frac{2}{m}T^{2})^\frac{m}{2}}{t^{m+2}}dt\\
\ll & {T^{\frac{m^2}{2}+m}}\tag{using \eqref{the truncation c}}.
\end{align*}

After normalization by $V_{T}[g_1,g_2]\asymp T^{m(m+1)}$, this integral vanishes as $T\to \infty$.

Now it remains to estimate the integral $\int_{\al_T}^{c_T}$. For that we use Lemma <ref>. Notice that the surface area of the ellipsoids $D_{T,t}$ is $\asymp (R_T(t))^{\frac{m-1}{2}}$. We have

\begin{align*}
&\left| \int_{\al_T}^{c_T}  \int_{D_{T,t}} \left(F( a_t^{-1} u_{-v} g_1\Gamma) - F(a_t^{-1} u_{-t^{\frac{1}{m}+1}v_0-v}  g_1\Gamma) \right) dv \frac{1}{t^{m+2}}dt \right|\\
\ll &  \int_{\al_T}^{c_T}   \|-t^{\frac{1}{m}+1}v_0\|(-\det(\text{g}_1\text{g}_2)^{-2} t^{\frac{2}{m}+2}+t^{\frac{2}{m}}T^2-\|\text{g}_1  \text{g}_2\|^2)^\frac{m-1}{2} \frac{1}{t^{m+2}}dt  \tag{by Lemma \ref{lemma on symmetric difference}}\\
\ll &  \int_{\al_T}^{c_T}   t^{\frac{1}{m}-m-1} (t^{\frac{2}{m}}T^2)^\frac{m-1}{2} dt \\
\ll& T^{m-1}\frac{1}{\al_T^{m-1}}\\
\ll& T^{m-1}\frac{1}{1/T^{m(m-1)}}=T^{m^2-1}. \tag{by \eqref{range for a}}\\
\end{align*}

Again, this term goes to zero as $T\to \infty$, after normalization by $V_{T}[g_1,g_2]\asymp T^{m(m+1)}$.

§.§ Expanding property of $UA$ in representations of $\SL(m+1,\R$)

In order to apply the results of Shah, we need to establish that vectors expand by the action of $V=UA$ in a representation of $G=\SL(m+1,\R)$. We denote by $\V_G$ a representation of $G$ and we let
$$\V_G=\V_0\oplus \V_1,$$
where $\V_0$ is the space of $\SL(m+1,\R)$-fixed vectors and $\V_1$ is it's invariant complement. Suppose that $\V_1$ is not trivial, and let $\Pi$ be the projection of $\V_G$ onto $\V_1$ with kernel $\V_0$. Our main result in this section is the following.
Fix $g_0\in G$, let $K\subseteq \R^m$ be a bounded set, and let $\chi\in(0,1)$. Then for any $\nu>0$, there exist $t_0>0$ such that for any $0<t<t_0$, any $\xi\in K$ and any $\textbf{x}\in \V_G$ such that $\|\Pi(\textbf{x})\|\geq r$, it holds

\begin{equation}\label{expansion inequality for applying shah dichotomy}
\sup_{v\in \text{B}\left(t^{\frac{1-\chi}{m}}\right)+\xi} \|a_t^{-1}u_v^{-1}g_0.\textbf{x}\| >\nu,
\end{equation}

where $\text{B}(\beta)$ is the ball of radius $\beta$ in $\R^m$ centered at the origin.

The idea behind the proof of the lemma to decompose the elements  $a_t^{-1}u_v^{-1}$ into  $m$ elements  each belonging to a subgroup isomorphic to $\SL(2,\R)$. That's useful since we can use the explicit description of irreducible representations  of $\SL(2,\R)$ to prove an   expansion of the norm for each term in the decomposition. We note that a similar approach was used in <cit.>.

We now give the decomposition. Let  $E_{i,j}$ be the $(m+1)\times(m+1)$ matrix where the $(i,j)$-th entry is $1$ while all other entries are zero. Consider the following copy of $\SL(2,\R)$ in $\SL(m+1,\R)$

\begin{equation}\label{eq:j-th copy of SL2 in SLm+1}
\SL^{(j)}(2,\R):=\{I_{m+1}+(a-1)E_{jj}+bE_{m+1,j}+cE_{j,m+1}+(d-1)E_{m+1,m+1}:ad-bc=1\},
\end{equation}

for $j=1,2,...,m$. We will denote
\begin{equation*}
\pi^{(j)}\begin{bmatrix}
\end{bmatrix}:=I_{m+1}+(a-1)E_{jj}+bE_{m+1,j}+cE_{j,m+1}+(d-1)E_{m+1,m+1}.
\end{equation*}

We have the following observation which we leave the reader to verify.

Let $\sigma:\{1,2,\dots m\}\to \{1,2,\dots,m\}$ be a permutation. Then it holds that
\begin{equation}\label{eq:decompostion of AU to SL2 mats}
\begin{bmatrix}
\end{bmatrix}
\begin{bmatrix}
\end{bmatrix}\right),
\end{equation}

for $a_t,u_v$ as in (<ref>).

The decomposition (<ref>) reduces the proof of  Lemma <ref> to the study of the expansion of elements of the form
t^{\frac{1}{m}} & 0 \\
0 & t^{-\frac{1}{m}}
\end{bmatrix}\begin{bmatrix}
1 & 0 \\
y & 1
\end{bmatrix}$$
in representations of $\SL(2,\R)$.

§.§.§ Expansion in $\SL(2,\R)$-irreducible representations

We now briefly recall some basic facts on $\SL(2,\R)$-irreducible representations.

Recall that if $\pi$ is an $(n+1)$-dimensional irreducible representation of $\SL(2,\R)$ and $\pi'$ is the induced Lie algebra representation, then there exists a basis $v_0,v_1,...,v_n$ such that

\begin{align*}
&\pi'(H)(v_i)=(n-2i)v_i, i=0,1,...,n;\\
&\pi'(X)(v_i)=i(n-i+1)v_{i-1}, i=0,1,...,n;\\
&\pi'(Y)(v_i)=v_{i+1}, i=0,1,...,n~(v_{n+1}=0).
\end{align*}

$H = \begin{bmatrix}
1 & ~~0\\
0 & -1
\end{bmatrix},
X = \begin{bmatrix}
0 & 1\\
0 & 0
\end{bmatrix}$, and
$Y = \begin{bmatrix}
0 & 0\\
1 & 0
\end{bmatrix}$ form a generating set of $\frak{sl}(2,\R)$.

Under the matrix Lie group-Lie algebra correspondence,

\[\pi \left(\begin{bmatrix}
t^{\frac{1}{m}} & 0\\
0 & t^{-\frac{1}{m}}
\end{bmatrix} \right)=\exp(\pi'(\log(t^{\frac{1}{m}})H))\]
has the matrix representation

\begin{equation}\label{matrix rep of a_t under the basis}
\begin{bmatrix}
t^{\frac{n}{m}} & & &\\
& t^{\frac{n-2}{m}} & & \\
& & \ddots &\\
& & & {t}^{-\frac{n}{m}}
\end{bmatrix}
\end{equation}

\[\pi \left(\begin{bmatrix}
1 & 0\\
-y & 1
\end{bmatrix} \right)=\exp(\pi'(-yY))\]

has the matrix representation

\begin{equation}\label{matrix rep of u_y under the basis}
\begin{bmatrix}
1 & & &\\
p_{1}(y) & 1 & & \\
\vdots   & \ddots & \ddots &\\
p_{n}(y) & \cdots &  p_{1}(y)  &  1
\end{bmatrix}
\end{equation}

where $p_{l}(y)=\frac{(-y)^l}{l!},~l=1,...,n$.
Observe that the line
$$W:=\R v_n$$
is the subspace of fixed vectors of $\pi \left(\begin{bmatrix}
1 & 0\\
-y & 1
\end{bmatrix}\right),y\in\R.$ This space is important for us since it's the  eigenspace of the matrices $\pi \left(\begin{bmatrix}
t^{\frac{1}{m}} & 0\\
0 & t^{-\frac{1}{m}}
\end{bmatrix}\right)$ which corresponds to the largest eigenvalue as $t\to 0$.
We denote by $\pr_W$ the orthogonal projection on $W$ with respect to some scalar product on $\V_n$.

The following lemma is a very special case of <cit.> (which is essentially <cit.>). In our setting the proof simplifies considerably and we prove it for completeness.

Let $\pi:\SL(2,\R)\to \V_n$ be the $(n+1)$-dimensional irreducible representation of $\SL(2,\R)$.   Fix a bounded interval I. Then there exists a constant $C>0$ such that for any $\beta \in (0,1)$, $\tau\in \emph{\text{I}}$ and $\textbf{x}\in \V_n$,

\begin{equation*}
\sup_{y\in([0,\beta]+\tau)}\left\|\pr_W\left(\pi \left(\begin{bmatrix}
1 & 0\\
-y & 1
\end{bmatrix}\right)\textbf{x}\right)\right\| \ge C \beta^n\|\textbf{x}\|.
\end{equation*}

First, note that it's enough to prove that there exists $C_0>0$ such that:

\begin{equation}\label{eq:expansion of the norm removing tau}
\sup_{y\in[0,\beta]}\left\|\pr_W\left(\pi \left(\begin{bmatrix}
1 & 0\\
-y & 1
\end{bmatrix}\right)\textbf{x}\right)\right\|\geq C_0\|\textbf{x}\|.
\end{equation}

In fact, by assuming (<ref>), we get

\begin{align*}
\sup_{y\in([0,\beta]+\tau)}\left\|\pr_W\left(\pi \left(\begin{bmatrix}
1 & 0\\
-y & 1
\end{bmatrix}\right)\textbf{x}\right)\right\| =&\sup_{y\in[0,\beta]}\left\|\pr_W\left(\pi \left(\begin{bmatrix}
1 & 0\\
-y & 1
\end{bmatrix}\right)\pi \left(\begin{bmatrix}
1 & 0\\
-\tau & 1
\end{bmatrix}\right)\textbf{x}\right)\right\|\\
&\geq C_0\left\|\pi \left(\begin{bmatrix}
1 & 0\\
-\tau & 1
\end{bmatrix}\right)\textbf{x}\right\|\\
&\geq C_0 C_1 \|\textbf{x}\|,
\end{align*}

where $C_1$ depends on I.
Now, in the coordinates of the basis $\{v_0,...,v_n\}$, upon recalling (<ref>) we see

\begin{align*}
\pr_W\left(\pi \left(\begin{bmatrix}
1 & 0\\
-y & 1
\end{bmatrix}\right)\textbf{x}\right)&=\begin{bmatrix}
1 & & &\\
p_{1}(y) & 1 & & \\
\vdots   & \ddots & \ddots &\\
p_{n}(y) & \cdots &  p_{1}(y)  &  1
\end{bmatrix}\begin{bmatrix}
\vdots\\
\end{bmatrix}\\
\vdots
\\
\end{bmatrix},
\end{align*}

where $p_{\textbf{x}}(y)=c\sum_{k=0}^n\frac{(-1)^k}{k!}x_{n-k}y^k$ (here $c$ is some constant depending on chosen scalar product for the orthogonal projection $\text{pr}_W.$)  Note the general fact that there exists $C_2>0$ such that for any polynomial $p(x)=a_nx^n+...+a_0$ of degree at most $n$
$$\sup_{s\in[0,1]}|p(s)|\geq C_2\|\textbf{a}\|,$$

where $\textbf{a}$ is the vector whose entries are the coefficients of $p(x)$. Thus we may conclude:
$$\sup_{y\in[0,\beta]}|p_{\textbf{x}}(y)|=\sup_{s\in[0,1]}\left|\sum_{k=0}^n\frac{(-1)^k}{k!}\beta^k x_{n-k}s^k\right|\gg \beta^n\|\textbf{x}\|,$$

which proves the lemma.

The above lemma implies the following statement, which is an analogue of Lemma <ref> for $\SL(2,\R)$.

Suppose that $\pi:\SL(2,\R)\to\V$ is a representation, fix an interval I and let $\chi\in (0,1)$. Then, for all $t\in(0,1)$ the following holds.

*  If $\pi$ is irreducible, then for all $\tau\in\emph{\text{I}}$ it holds that
\begin{equation}\label{eq:expansion of sl(2,r) reps of y in t interval}
\sup_{y\in[0,t^{\frac{1-\chi}{m}}]+\tau}\left \|\pi\left(\begin{bmatrix}
t^{\frac{1}{m}} & 0 \\
0 & t^{-\frac{1}{m}}
\end{bmatrix}
\begin{bmatrix}
1 & 0 \\
-y & 1
\end{bmatrix}\right)\textbf{x} \right \|\gg t^{-\frac{\chi}{m}}\|\textbf{x}\|,
\end{equation}
for all $\emph{\textbf{x}}\in\V$, where the implied constant depends on $\pi$ only.

*  If $\pi$ is any representation, then for all $\tau\in \emph{\text{I}}$
\begin{equation}\label{eq:expansion of sl(2,r) in not necessarily irreducible}
\sup_{y\in[0,t^{\frac{1-\chi}{m}}]+\tau}\left \|\pi\left(\begin{bmatrix}
t^{\frac{1}{m}} & 0 \\
0 & t^{-\frac{1}{m}}
\end{bmatrix}
\begin{bmatrix}
1 & 0 \\
-y & 1
\end{bmatrix}\right)\textbf{x} \right \|\gg \|\textbf{x}\|,
\end{equation}

for all $\emph{\textbf{x}}\in\V$, where the implied constant depends on $\pi$ only.

In the following, for convenience, we will omit the representation symbol $\pi$. Suppose that $\V$ is the $n+1$'th irreducible representation. Then for all $y\in \R$ and all $\textbf{x}\in \V$

\begin{align}
& \left \|\begin{bmatrix}
t^{\frac{1}{m}} & 0 \\
0 & t^{-\frac{1}{m}}
\end{bmatrix}
\begin{bmatrix}
1 & 0 \\
-y & 1
\end{bmatrix}.\textbf{x} \right \| \nonumber \\
\ge & C \left \| \pr_W\left(\begin{bmatrix}
t^{\frac{1}{m}} & 0 \\
0 & t^{-\frac{1}{m}}
\end{bmatrix}
\begin{bmatrix}
1 & 0 \\
-y & 1
\end{bmatrix}.\textbf{x} \right) \right \| \tag{by the boundedness of the linear operator $\pr_W$}\\
= & C \left \| \begin{bmatrix}
t^{\frac{1}{m}} & 0 \\
0 & t^{-\frac{1}{m}}
\end{bmatrix}.
\pr_W\left(
\begin{bmatrix}
1 & 0 \\
-y & 1
\end{bmatrix}.\textbf{x} \right) \right \| \tag{since the matrix \eqref{matrix rep of a_t under the basis} commutes with $\pr_W=\pr_{\R e_n}$}\\
=& C t^{-\frac{n}{m}} \left \|
\pr_W\left(\begin{bmatrix}
\end{bmatrix}\textbf{x} \right) \right \| \tag{see the last component of \eqref{matrix rep of a_t under the basis}}
\end{align}

Then, part 1 of our corollary follows by Lemma <ref>. To conclude part 2, note that we may always decompose $\V$ as $\V_0\oplus\V_1$ where $\V_0$ is the space of fixed vectors, and $\V_1$ is an invariant complement. By further decomposing $\V_1$ into irreducible representations, and applying part 1 of the corollary in each component, we get the result.

§.§.§ Proving Lemma <ref>

Recall $\V_G=\V_0\oplus \V_1$ where $\V_0$ is the space of $\SL(m+1,\R)$-fixed vectors, and $\V_1$ is it's invariant complement. Note that in order to prove Lemma <ref>, it's enough to verify that for a fixed $r>0$, a bounded set $K\subset \R^m$ and $\chi\in(0,1)$ it holds that
\begin{equation}\label{eq:what we need to show for lem 3.6}
\min_{\textbf{x}\in \V_1,~\|\textbf{x}\|\geq r}\sup_{v\in \text{B}\left(t^{\frac{1-\chi}{m}}\right)+\xi} \|a_t^{-1}u_v^{-1}\textbf{x}\|\gg t^{-\chi/m},~\forall\xi\in K
\end{equation}

where $t\in(0,1)$. As $\V_1$ is invariant by the induced action of $\SL^{(j)}(2,\R)$ (see (<ref>)) for $j=1,2,...,m$, we get the decompositions

\begin{align}
\V_1 &= \V_{0}^{(j)} \oplus \V_{1}^{(j)},j=1,2,...,m,
\label{decompositive of vs w.r.t. SL2}
\end{align}
where $\V_0^{(j)}$ is the subspace of fixed vectors of the action of $\SL^{(j)}(2,\R)$ and $\V_1^{(j)}$ is its invariant complement.
For any $\textbf{x}\in \V_1$ we write

\begin{equation}    \textbf{x}=\textbf{x}_{0}^{(j)}+\textbf{x}_{1}^{(j)}, j=1,2,...,m.
\end{equation}

where $\textbf{x}_{i}^{(j)} \in \V_{i}^{(j)}, i=1,2,...,m$ from above.

It holds that
$\inf_{\textbf{x}\in \V_1, \|\textbf{x}_1\|\ge r} \left( \|\textbf{x}_{1}^{(1)}\|+\cdots+\|\textbf{x}_{1}^{(m)}\| \right) = Lr$ for some $L>0$

We recall that $UA$ (see (<ref>)) is an  epimorphic subgroup, cf. <cit.>. This simply means that if $\textbf{x}\in \V_1$ is fixed by $UA$, then it's fixed by all of $\SL(m+1,\R)$. But in such case, $\textbf{x}=0$, since $\V_1$ is an invariant complement of the fixed vectors. In particular, this implies that $\bigcap_{i=1}^m\V_{0}^{(i)}=\{0\}$ using the decomposition in Lemma <ref>. Since the sphere of radius $1$ is compact and as $\bigcap_{i=1}^m\V_{0}^{(i)}=\{0\}$, we have
$$\inf_{\textbf{x}_1\in \V_1, \|\textbf{x}_1\|= 1} \left( \|\textbf{x}_{1}^{(1)}\|+\cdots+\|\textbf{x}_{1}^{(m)}\| \right) = L > 0.$$

By scaling, it is easy to see for $r \ge 1$,

\begin{align*}
\inf_{\textbf{x}_1\in \V_1, \|\textbf{x}_1\|= r} \left( \|\textbf{x}_{1}^{(1)}\|+\cdots+\|\textbf{x}_{1}^{(m)}\| \right)
%    =& \frac{r}{1}\inf_{w_1\in V_1, \|w_1\|= r} \left( \|w_{1}^{(1)}\|+\|w_{1}^{(2)}\| \right) \\
= & r \inf_{\textbf{x}_1\in \V_1, \|\textbf{x}_1\|= 1} \left( \|\textbf{x}_{1}^{(1)}\|+\cdots+\|\textbf{x}_{1}^{(m)}\| \right)  \\
=& Lr>0.
\end{align*}

Now take $\textbf{x}\in \V_1$ and fix $j$ such that $\|\textbf{x}^{(j)}_1\|\geq Lr$. We pick a permutation $\sigma:\{1,...,m\}\to\{1,...,m\}$ such that $\sigma(m)=j$. Using Lemma <ref> we write
\begin{align}\label{eq:chose permutation for j with large component}
\begin{bmatrix}
\end{bmatrix}
\begin{bmatrix}
\end{bmatrix}\right)
\pi^{(j)}
\left(
\begin{bmatrix}
\end{bmatrix}
\begin{bmatrix}
\end{bmatrix}\right)\textbf{x}.
\end{align}

For a bounded set $K\subseteq \R^m$ there exist intervals $\text{I}_1,...,\text{I}_m$ such that
$$\left(\left[0,\frac{1}{\sqrt{m}}t^{\frac{1-\chi}{m}}\right]+\text{I}_1 \right)\times...\times \left(\left[0,\frac{1}{\sqrt{m}}t^{\frac{1-\chi}{m}}\right]+\text{I}_m \right)\subseteq \text{B}\left(t^{\frac{1-\chi}{m}}\right)+K.$$
Then (<ref>) is obtained by applying Lemma <ref> as follows: We decompose $\V_1^{(j)}$ into $\SL^{(j)}(2,\R)$-irreducible representations, and we assume (without loss of generality, as all norms are equivalent) that our norm $\|\cdot\|$ on $\V_1$ is obtained by taking the sup norm with respect to a basis of $\V_1$ composed out of a basis for $\V_0^{(j)}$ and bases for each of the $\SL^{(j)}(2,\R)$-irreducible spaces.  Then, using Lemma <ref>,(<ref>), we have for all $\tau_j\in \text{I}_j$  that

\begin{equation*}
\sup_{v_j\in[0,t^{\frac{1-\chi}{m}}]+\tau_j}\left \|\pi^{(j)}\left(\begin{bmatrix}
t^{\frac{1}{m}} & 0 \\
0 & t^{-\frac{1}{m}}
\end{bmatrix}
\begin{bmatrix}
1 & 0 \\
-y & 1
\end{bmatrix}\right)\textbf{x} \right \|\gg t^{-\frac{\chi}{m}}r,
\end{equation*}

and by further applying Lemma <ref>,(<ref>) when taking the supremum of (<ref>) over the parameters $\left(\left[0,\frac{1}{\sqrt{m}}t^{\frac{1-\chi}{m}}\right]+\text{I}_1 \right)\times...\times \left(\left[0,\frac{1}{\sqrt{m}}t^{\frac{1-\chi}{m}}\right]+\text{I}_m \right)$, we obtain (<ref>).

As before we make the substitution $s=t^{\frac{1}{m}}$. So the equation above becomes

\begin{equation}
\sqrt{-\det(\text{g}_1\text{g}_2)^{-2}s^{2+2m}+s^2T^2-\|\text{g}_1 q \text{g}_2\|^2} \gtrsim
\end{equation}

Put $s_a=a^{\frac{1}{m}}$ and $s=s_a+z$ with $z\ge \de:=\frac{\|\text{g}_1 q \text{g}_2\|}{T^{1+\e}}$. $z$ is treated as a variable. When $t\in (a_{\de}, c)$, we have $z\in (\de,c^{\frac{1}{m}}-s_a)$. Recall from (<ref>) we used the truncation

\begin{equation}
c=\left(\frac{\|\text{g}_1 q \text{g}_2\|}{T^{3\e}}\right)^{\frac{m}{m+1}}.
\end{equation}

Since $s_a\asymp \frac{\|\text{g}_1 q \text{g}_2\|}{T}$ is a root of $-\det(\text{g}_1\text{g}_2)^{-2}s^{2+2m}+s^2T^2-\|\text{g}_1 q \text{g}_2\|^2$, we have

\begin{align}
&\left(t^{-\frac{1}{m}}\sqrt{-\det(\text{g}_1\text{g}_2)^{-2}t^{\frac{2}{m}+2}+t^{\frac{2}{m}}T^2-\|\text{g}_1 q \text{g}_2\|^2}\right)^2\\
=&\left(s^{-1}\sqrt{-\det(\text{g}_1\text{g}_2)^{-2}s^{2+2m}+s^2T^2-\|\text{g}_1 q \text{g}_2\|^2}\right)^2\\
\end{align}

Note on the right hand side of the last line, the dominating term is

which is a monotonically increasing function on $(\de, c^{\frac{1}{m}}-s_a)$ and the minimal value is when $z$ approaches $\de$. But

\begin{equation}
\lim_{T\to \infty}f_T(\de)=\infty.
\end{equation}

This justifies that (<ref>) is indeed expanding on $(a_{\de},c)$ as $T\to \infty$.

§.§ The non-escape of mass

Let $\frak g$ be the Lie algebra of $G=\SL(m+1,\R)$. For positive integers $d$ and $n$, denote by
$\mathcal {P}_{d,n}(G)$ the set of functions $\varphi: \R^n \to G$ such that for any $\textbf{a}, \textbf{b} \in \R^n$, the map

\begin{equation*}
\R\ni \tau \mapsto \Ad(\varphi(\tau \textbf{a} +\textbf{b}))\in \GL(\frak{g})
\end{equation*}

is a polynomial of degree at most $d$ when the above linear transformation is represented by some basis of $\frak{g}$.

Let $\V_G = \sum_{i=1}^{\dim \frak g} \bigwedge^i \frak{g}$. We consider the natural action of $G$ on $\V_G$ induced from the Ad-representation.  Fix a norm $\|\cdot\|$ on $\V_G$. For a Lie subgroup H of $G$ with Lie algebra $\frak{h}$, take a unit vector $p_\text{H} \in \bigwedge^{\dim \frak h} \frak{h}$.

There exist closed subgroups $U_i (i=1,2,...,l)$ such that each $U_i$ is the unipotent radical of a parabolic subgroup, $U_i\Ga$ is compact in $X=G/\Ga$ and for any $d,n\in \N$, $\e,\de>0$, there exists a compact set $C \subset G/\Ga$ such that for any $\varphi \in \mathcal{P}_{d,n}(G)$ and a bounded open convex set $D\subset \R^n$, one of the following holds:

* there exist $\gamma\in \Ga$ and $i=1,...,l$ such that $\sup_{v\in D}\|\varphi(v)\gamma.p_{U_i}\| \le \de$;

* $\vol(v\in D:\varphi(v)\Ga \notin C)< \e \vol(D)$, where $\vol$ is the Lebesgue measure on $\R^n$.

Fix  $g_1,g_2\in\SL(m+1,\R)$. We shall consider the family of polynomial maps  appearing in the integration along the measures $\mu^\circ_{T,g_1,g_2}$ which are given by
\begin{equation}\label{eq:definition of q t v}
\varphi_t(v):= a_t^{-1} u_v^{-1} g_0,
\end{equation}

where $g_0\in \SL(m+1,\R)$,  $t>0$ and $v\in\R^m$.
For each fixed $t$, the map $\varphi_t(\cdot)$ is in $\mathcal{P}_{2,m}(G)$.
Using Lemma <ref>, we will show that $\varphi_t(v)$ fails the condition 1 of the Theorem <ref> in the range appearing in the integration along $\mu^\circ_{T,g_1,g_2}$. This will be the key fact that allows to prove the non-escape of mass and also the equidistribution, see Section <ref>.

We have the following statement which will enable to verify the requirement of Lemma <ref> for a lower bound of the norm of the projection to the space of non-fixed vectors of the vectors $\gamma.p_{U_i},\gamma\in\Ga,i=1,...,l.$

Denote by $\mathcal{H}_{\Ga}$ the family of all proper closed connected subgroups H of $G$ such that $\Ga \cap \text{H}$ is a lattice in H, and $\Ad(\text{H} \cap \Ga)$ is Zariski-dense in $\Ad(\text{H})$. We have the following important theorem:

The set $\mathcal{H}_{\Ga}$ is countable. For any $\emph{\text{H}}\in \mathcal{H}_{\Ga}, \Ga.p_\emph{\text{H}}$ is discrete.

Consider the finite set of unipotent radicals of parabolic subgroups, denoted $U_1,...,U_l$ appearing in Theorem <ref>.  By Theorem <ref>, we have that  $\Ga.p_{U_i}$ is discrete in $\V_G$ for all $i$.

Write $\V_G=\V_0 \bigoplus \V_1$, where $\V_0$ is the space of vectors fixed by $G$ and $\V_1$ is its $G$-invariant complement. Recall that $\Pi$ denotes the projection of $\V_G$ onto $\V_1$ with kernel $\V_0$. By Lemma 17 in <cit.> it holds that $\Pi(\Ga.p_{U_i})$ is discrete for all $i$. Since $p_{U_i}$ is not fixed by $G$ (this is because the action is through conjugation and $U_i$'s are not normal subgroups in the simple group $G$), $\Pi(\Ga.p_{U_i})$ does not contain $0$. So it follows that

\begin{equation}\label{inf of the projection is at least r}
\inf_{\textbf{x}\in \cup_{i=1}^l \Ga.p_{U_i}} \|\Pi(\textbf{x})\|:=r>0.
\end{equation}

§.§.§ Proof of $\eta{(\infty)}=0$

Therefore, for any $x\in \cup_{i=1}^r \Ga.p_{U_i}$,

\begin{equation}
\lim_{t\to 0} \sup_{v\in D(\frac{1}{2}\st^{\frac{1-\chi}{m}})} \|a_t^{-1}u_v^{-1}g_1.x\| =\infty.
\end{equation}

Recall that $\eta$ is a weak-* limit of the measures

\begin{align*}
\mu^\circ_{T,g_1,g_2}(F):=& \frac{1}{\mu \left(V_{T}[g_1,g_2] \right)} \int_{\al_T}^{\be_T} \int_{D_{T,t}} F \left(a_t^{-1} u_v^{-1} g_0\Gamma \right) dv \frac{1}{t^{m+2}}dt,~F\in C_c(G/\Ga),
\end{align*}

where $0<\al_T<\be_T$ are the two positive roots of (<ref>) (see Lemma <ref> with $q=I_m$), and $D_{T,t}$ is the ellipsoid (<ref>). We note that the ellipsoid $D_{T,t}$ is centered at

\begin{equation}\label{eq:def of translation}
\xi_T(t):=\det(\text{g}_1)\text{v}_1 -\text{v}_2\text{g}_2^{-1}t^{\frac{1}{m}+1} \in \R^m.
\end{equation}
Then  $D_{T,t}$ contains the ball $\text{B}(\rho_T(t))+\xi_T(t)$ with the same center whose radius $\rho_T(t)$ is equal to the shortest radius of the ellipse, and $\rho_T$ has the estimate:

\begin{equation}\label{eq:radius of the contained ball}
\rho_T(t):=C_{g_1,g_2}\sqrt{R_T(t)}=C_{g_1,g_2}\sqrt{ -\det(\text{g}_1\text{g}_2)^{-2} t^{\frac{2}{m}+2}+t^{\frac{2}{m}}T^2-\|\text{g}_1 \text{g}_2\|^2},
\end{equation}

for some constant $C_{g_1,g_2}>0$. The  displacement  $\xi_T(t)$ has two parts: one part ($-\det(\text{g}_1)\text{v}_1$) is constant and the other part ($-\text{v}_2\text{g}_2^{-1}t^{\frac{1}{m}+1}$) involves $t$ but is  bounded when $t\in(0,1)$. To use Lemma <ref> in our proof for $\eta(\infty)=0$, we require that $\rho_{T}(t)>t^{\frac{1-\chi}{m}}$. But, when $t$ approaches to either $\al_T$ or to $\be_T$, $\rho_{T}(t)$ becomes too small. Thus we use the following truncation – for a fixed  $F\in C_c(G/\Ga)$, $\e_1,\e_2\in(0,1)$ there is $\ka=\ka(\e_1,\e_2)>0$ such that
\begin{equation}\label{eq:truncated integral}
\mu^\circ_{T,g_1,g_2}(F)= \frac{1}{\mu \left(V_{T}[g_1,g_2] \right)} \int_{\al_{\de}(T,\e_1)}^{\la(T,\e_2)} \int_{D_{T,t}} F \left(a_t^{-1} u_v^{-1} g_0\Gamma \right) dv \frac{1}{t^{m+2}}dt+O(T^{-\ka}).
\end{equation}

The above estimate follows by Lemmata <ref> and <ref> together with the estimate of $\mu(V_T[g_1,g_2])\asymp T^{m(m+1)}$.
Here $\al_{\de}\text{ and }\la$ are given by (<ref>) and (<ref>) with $q=I_m.$

Fix $\chi\in(0,1)$. Then, there exist $0<\e_1<\e_2<1$ such that for all $T$ large enough and all $t\in(\al_{\de}(T,\e_1),\la(T,\e_2))$, it holds that

\begin{equation*}
\rho_{T}(t)>t^{\frac{1-\chi}{m}}.
\end{equation*}

For all $T$ large enough its holds that for all $t\in(\al_{\de,T},\la_T)$

\begin{equation}
\sqrt{ -\det(\text{g}_1\text{g}_2)^{-2} t^{\frac{2}{m}+2}+t^{\frac{2}{m}}T^2-\|\text{g}_1 \text{g}_2\|^2}\geq t^{\frac{1-\chi}{m}}
\end{equation}

To prove the inequality, we bound from below the left hand side, and bound from above the right hand side. For the right hand side, by recalling  (<ref>) defining  $\la$, we get for $t\in(\al_{\de}(T,\e_1),\la(T,\e_2))$ that

\begin{equation}\label{eq:asymp for righthand side}
\frac{1}{T^\frac{\e_2(1-\chi)}{m+1}}\asymp \la(T,\e_2)^{\frac{1-\chi}{m}}\geq t^\frac{1-\chi}{m}.
\end{equation}
We now estimate $\rho_{T}(t)=C_{g_1,g_2}\sqrt{R_T(t)}$ from below. Recall that $R_T(t)$ is monotonically increasing in $(0,\theta_T)$  and  $\theta_T\asymp T$ (see the proof of Lemma <ref>). Since $\al_{\de}(T,\e_1)\text{ and }\la(T,\e_2)$ converge to zero as $T\to\infty$, have that $(\al_{\de}(T,\e_1),\la(T,\e_2))\subseteq(0,\theta_T)$. Thus $\forall t\in(\al_{\de}(T,\e_1),\la(T,\e_2))$ it holds
\begin{align}
\sqrt{R_T(t)}\geq\sqrt{R_T(\al_\de(T,\e_1))}=&\sqrt{-\det(\text{g}_1\text{g}_2)^{-2} \al_{\de}(T,\e_1)^{\frac{2}{m}+2}+\al_{\de}(T,\e_1)^{\frac{2}{m}}T^2-\|\text{g}_1 \text{g}_2\|^2}\nonumber
\end{align}

We now determine the asymptotics of the terms below the above square-root. Upon recalling the definition of $\al_{\de}$ in (<ref>) and using the estimate (<ref>), we see that

$$\al_{\de}(T,\e_1)^{\frac{2}{m}+2}\asymp \al_{T}^{\frac{2}{m}+2}\asymp \frac{1}{T^{2+2m}}$$

For the second term,

$$\al_{\de}(T,\e_1)^\frac{2}{m}=\left(\al_T^{\frac{1}{m}}+\frac{\|\text{g}_1 \text{g}_2\|}{T^{1+\e_1}}\right)^2\underbrace{=}_{\eqref{eq:asymp for a when A small then sqrt T}}\|\text{g}_1 \text{g}_2\|^2\left(\frac{1}{T^2}+\frac{2}{T^{2+\e_1}}\right)+O\left(\frac{1}{T^{2+2\e_1}}\right),$$
which implies that

$$\al_{\de}(T,\e_1)^{\frac{2}{m}}T^2-\|\text{g}_1 \text{g}_2\|^2\asymp \frac{1}{T^{\e_1}}$$

\begin{equation}\label{eq:asymp for lefthand side}
\sqrt{R_T(\al_\de(T,\e_1))}\asymp  \frac{1}{T^{\e_1/2}}.
\end{equation}

We are free to choose $\e_1$ and $\e_2$ in the interval $(0,1)$ as we wish. In particular, we as well may assume that
With the latter choice,  we obtain our claim by the estimates (<ref>) and (<ref>).

We are now ready to prove $\eta(\infty)=0.$

By the above Lemma <ref> and Lemma <ref>, we conclude that the first outcome in Theorem <ref> for the translated ellipsoids $D=D_{T,t}$ fails in the range $t\in(\al_\de(T,\e_1),\la(T,\e_2))$ for all $T$ large enough. Here $\e_1,\e_2$ are fixed such that Lemma <ref> holds.
Thus, by the second outcome of Theorem <ref>, for an  $\e>0$  there is a compact $C\subset G/\Ga$  such that

\begin{equation}\label{eq:small proporotion of visits outside a compact C}
\vol(v\in D_{T,t}:\varphi_t(v)\Ga \notin C)< \e \vol(D_{T,t}),~\forall t\in(\al_\de(T,\e_1),\la(T,\e_2)).
\end{equation}

Now fix $x=g_0\Ga$, and let $F\in C_c(G/\Ga)$ with $\mathbf{1}_C \le F \le 1$. As $T\to \infty$,

\begin{align*}
\mu_{T,g_1,g_2}^\circ(F)=&\frac{1}{\mu \left(V_{T}[g_1,g_2] \right)}\int_{V_{T}[g_1,g_2]} F( h^{-1}g_0\Gamma) d\mu(h)\\
=& \frac{1}{\mu \left(V_{T}[g_1,g_2] \right)} \int_{\al_{\de}(T,\e_1)}^{\la(T,\e_2)} \int_{D_{T,t}} F \left(\varphi_t(v)\Ga \right) dv \frac{1}{t^{m+2}}dt+O(T^{-\ka})\tag{by \eqref{eq:truncated integral}}\\
\ge& \frac{1}{\mu \left(V_{T}[g_1,g_2] \right)} \int_{a_{\de}(T,\e_1)}^{\la(T,\e_2)} \int_{D_{T,t}} \mathbf{1}_C \left(\varphi_t(v)\Ga \right) dv \frac{1}{t^{m+2}}dt +O(T^{-\ka})\\
\ge& \frac{1}{\mu \left(V_{T}[g_1,g_2] \right)} \int_{a_{\de}(T,\e_1)}^{\la(T,\e_2)}  (1-\e)\vol(D_{T,t}) \frac{1}{t^{m+2}}dt+O(T^{-\ka}) \tag{by \eqref{eq:small proporotion of visits outside a compact C}}\\
=&(1-\e)\frac{1}{\mu \left(V_{T}[g_1,g_2] \right)} \int_{\al_T}^{\be_T} \vol(D_{T,t}) \frac{1}{t^{m+2}}dt +O(T^{-\ka})\tag {by Lemmata \ref{lemma for integral from a to a delta} and \ref{lem:integral from c to b}}\\
\end{align*}

\begin{equation*}
\liminf_{T\to \infty}\mu_{T,g_1,g_2}^\circ(F)\ge 1-\e
\end{equation*}

so that

\begin{equation*}
\eta(\infty)\le\limsup_{T\to \infty}\mu_{T,g_1,g_2}^\circ(\text{support}(F)^c) \le \e.
\end{equation*}

Since $\e$ is arbitrary $\eta(\infty)=0$.

§.§ Zero measure for the singular set – concluding equidistribution

Recall $U=\begin{bmatrix}
I_m & 0\\
\R^m & 1
\end{bmatrix}$. For a closed subgroup
H of $G$, denote

\begin{align*}
&N(\text{H},U):=\{g\in G:Ug\subset g\text{H}\},\\
&S(\text{H},U):=\cup_{\text{H}'\subsetneq \text{H}, \text{H}'\in \mathcal{H}_{\Ga}}N(\text{H}',U).
\end{align*}

\begin{equation}
Y:=\bigcup_{\text{H}\in \mathcal{H}_{\Ga}}N(\text{H},U)\Ga=\bigcup_{\text{H}\in \mathcal{H}_{\Ga}}[N(\text{H},U)-S(\text{H},U)]\Ga \subset G/\Ga,
\end{equation}

where $\mathcal{H}_
{\Ga}$ was defined above Theorem <ref>.
The equality holds since for any $g\in N(\text{H},U)$, if $g$ is also in $S(\text{H},U)$, then $g$ must belong to $N(\text{H}',U)$ for some $\text{H}'\subsetneq H$ (note for Lie subgroups, this condition means $\dim \text{H}' < \dim \text{H}$) and $\text{H}'\in \mathcal{H}_{\Ga}$. Since $\text{H}'$ has strictly lower dimension, by repeating this argument we see eventually, $g$ will fall into some $N(\tilde{\text{H}}, U)-S(\tilde{\text{H}}, U)$ (with $S(\tilde{\text{H}}, U)$ possibly empty when $\tilde{\text{H}}$ has minimal dimension).

By Theorem 2.2 of <cit.>, in order to show that $\eta$ is the $G$-invariant probability measure $\mu_X$, it is sufficient to prove the following lemma.

By the discreteness of $\mathcal{H}_{\Ga}$ (Theorem <ref>), it suffices to show that for each fixed $\text{H}\in\mathcal{H}_{\Ga}$ it holds

\begin{equation*}
\eta([N(\text{H},U)-S(\text{H},U)]\Ga)=0.
\end{equation*}
Since $[N(\text{H},U)-S(\text{H},U)]\Ga$ is a countable union of compact subsets in $G/\Ga$ (See <cit.> Proposition 3.1), it suffices to show $\eta(C)=0$ for any compact subset $C$ of $[N(\text{H},U)-S(\text{H},U)]\Ga$.

The main tool in the proof of the latter statement is the following consequence of Proposition 5.4 in <cit.>. We use in the following the same notations as in Section <ref>.

Let $d,n \in \N, \e>0, \emph{\text{H}}\in \mathcal{H}_{\Ga}$. For any compact set $C \subset [N(\emph{\text{H}}, U)-S(\emph{\text{H}},U)]\Ga$, there exists a compact set $F \subset \V_G$ such that for any neighborhood $\Phi$ of $F$ in $\V_G$, there exists a
neighborhood $\Psi$ of $C$ in $G/\Ga$ such that for any $\varphi \in \mathcal{P}_{d,n}(G)$ and a bounded open convex set
$D \subset \R^n$, one of the following holds:

* There exist $\gamma\in \Ga$ such that $\varphi(D)\gamma.p_\emph{\text{H}} \subset \Phi$.

* $\vol(t\in D: \varphi(t)\Ga \in \Psi)<\e \vol(D)$, where $\vol$ is the Lebesgue measure on $\R^n$.

Fix $\e>0$ and $\text{H}\in\mathcal{H}_\Ga$. Recall that $\varphi_t(\cdot)$ which was defined in (<ref>) is in $\mathcal{P}_{2,m}$. Let $C \subset [N({\text{H}}, U)-S({\text{H}},U)]\Ga$ be a compact set and  take a compact set $F\subset \V_G$ satisfying the outcome of Theorem <ref>. By the same argument as in the <ref> with Lemma <ref> applied, for $t\in (\al_{\de}(T,\e_1), \la(T,\e_2))$ for all $T$ large enough, the first outcome of Theorem <ref> fails for $\varphi_t(D_{t,T})$. Then, for $t\in (\al_{\de}(T,\e_1), \la(T,\e_2))$ for all $T$ large,

\begin{equation}\label{eq:visits to Psi is negligble}
\vol(v\in D_{T,t}:\varphi_t(v)\Ga \in \Psi)<\e \vol(D_{T,t}),
\end{equation}

where $C\subseteq\Psi$ is the compact neighborhood from Theorem <ref>.
Now let $f\in C_c(G/\Ga)$ be such that $\mathbf{1}_C \le f \le 1$ and $\text{support}(f)\subset \Psi$, it follows that

\begin{align*}
\eta(C)\leq& \limsup_{T\to \infty}\mu_{T,g_1,g_2}^\circ(f)\\
=& \limsup_{T\to \infty} \frac{1}{\mu \left(V_{T}[g_1,g_2] \right)} \int_{\al_T}^{\be_T} \int_{D_{T,t}} f \left(\varphi_t(v)\Ga \right) dv\frac{1}{t^{m+2}}dt \\
=& \limsup_{T\to \infty} \frac{1}{\mu \left(V_{T}[g_1,g_2] \right)} \int_{\al_{\de}(T,\e_1)}^{\la(T,\e_2)} \int_{D_{T,t}} f \left(\varphi_t(v)\Ga \right) dv \frac{1}{t^{m+2}}dt \tag{by Lemmata \ref{lemma for integral from a to a delta} and \ref{lem:integral from c to b}}\\
\le& \limsup_{T\to \infty} \frac{1}{\mu \left(V_{T}[g_1,g_2] \right)} \int_{\al_{\de}(T,\e_1)}^{\la(T,\e_2)} \int_{D_{T,t}} \mathbf{1}_{\Psi} \left(\varphi_t(v)\Ga \right) dv \frac{1}{t^{m+2}}dt \\
\le& \limsup_{T\to \infty} \frac{1}{\mu \left(V_{T}[g_1,g_2] \right)} \int_{\al_{\de}(T,\e_1)}^{\la(T,\e_2)} \e \vol(D_{T,t}) \frac{1}{t^{m+2}}dt \\
\le & \e.
\end{align*}

Since $\e>0$ is arbitrary, $\eta(C)=0$ and thus $\eta(Y)=0$.

§ PROOF OF THEOREM <REF>: THE LIMITING MEASURE ON $H\BACKSLASH G$.

We need to prove for all $\varphi\in C_c(X_{m,m+1})$, we have

\begin{equation}
\lim_{T
\to\infty} \frac{1}{\#\Gamma_{T}}\sum_{\gamma\in \Ga_T}\varphi(x_{0}\cdot \gamma)=\int_{X_{m,m+1}}\varphi(x)d\tilde\nu_{x_{0}}(x)
\end{equation}

for some  measure $\tilde\nu_{x_0}$ on $X_{m,m+1}$ which depends on $x_{0}$.

The first thing we need to do is to construct a measure $\nu_{x_0}$ on $H\backslash G$. Since $G$ is unimodular while $H$ is not (the measure $\mu$ we defined is not right invariant), there is no $G$-invariant Haar measure on $H\backslash G$.

Nevertheless, it is still possible to find a lift $Y\subset G$ of $H\backslash G$ such that $H \times Y \to G$ is a Borel isomorphism and decompose the Haar measure on $G$ to obtain a measure on $Y$. Here

We follow Section 2.5 of <cit.>.
First, we provide an explicit measurable section
$$\sigma:H\backslash G \to Y\subset G,$$
and then we provide a measure $\nu_Y$ on $Y$ such that $dg=d\mu d\nu_Y$, where $d\mu$ is the left Haar measure on $H$ given by (<ref>) and $dg$ is the Haar measure on $G$ normalized such that $\vol(G/\Ga)=1$

To define the section,  let $\mathcal{F}_m \subset \begin{bmatrix}
\SL(m,\R) & 0\\
0 & 1
\end{bmatrix}$ denote a measurable fundamental domain of

\begin{equation*}
\begin{bmatrix}
\Delta & 0\\
0 & 1
\end{bmatrix} \bigg\backslash
\begin{bmatrix}
\SL(m,\R) & 0\\
0 & 1
\end{bmatrix} \bigg\slash  \begin{bmatrix}
\SO(m,\R) & 0\\
0 & 1
\end{bmatrix}
\cong
\Delta\backslash \SL(m,\R) /\SO(m,\R),
\end{equation*}

and consider the product
$$Y:=\mathcal{F}_m\cdot \SO(m+1,\R).$$

The product map $H \times Y \to G$ is a Borel isomorphism. In particular $\sigma:H\backslash G\to Y$ defined by
$$\sigma(Hg)=y,\text{ where $g=hy,\text{ for }h\in H,y\in Y$}$$
is a well defined section identifying measurably $H\backslash G$ with $Y$.

The surjectivity is clear from the block-wise Iwasawa decomposition. We only verify the injectivity here:

Suppose $h_1g_1s_1=h_2g_2s_2$ where $h_i\in H, g_i\in \mathcal{F}_m$ and $s_i\in \SO(m+1,\R)$. Then

* & 0 \\
* & *
\end{bmatrix}\ni g_2^{-1} h_2^{-1}h_1 g_1=s_2 s_1^{-1} \in \SO(m+1,\R),$$

and it follows from the definition of $\SO(m+1,\R)$ that both sides of the above equality must lie in $\begin{bmatrix}\SO(m,\R) & 0\\0 & 1 \end{bmatrix}$. So $h_2^{-1}h_1=g_2s_2s_1^{-1}g_1^{-1}\in \begin{bmatrix}\SL(m,\R) & 0\\0 & 1 \end{bmatrix}$. But $\begin{bmatrix}\SL(m,\R) & 0\\0 & 1 \end{bmatrix} \cap H= \begin{bmatrix}\Delta & 0\\0 & 1 \end{bmatrix}$. Hence $q g_1s_1=g_2s_2$ for some $q \in \begin{bmatrix}
\Delta & 0 \\0 & 1
\end{bmatrix}$. It follows that $g_2^{-1}q g_1=s_2 s_1^{-1}\in \begin{bmatrix}\SL(m,\R) & 0 \\0 & 1 \end{bmatrix} \cap \SO(m+1,\R)=\begin{bmatrix}\SO(m,\R) & 0 \\0 & 1 \end{bmatrix}$. Hence $g_1=g_2, s_1=s_2$ by the definition of fundamental domain, and $h_1=h_2$.

Next, let $P:=
\left\{ u_va_t\tilde{\text{g}}:\text{g}\in\SL(m,\R),~v\in\R^m,~t>0 \right\},$ where $u_v,a_t$ are given by (<ref>), and $\tilde{\text{g}}:=\begin{bmatrix}
\text{g}&0\\0&1
\end{bmatrix}$. On $P$ we define the Haar measure  $dp=dv \frac{dt}{t^{m+2}} d\text{g}$, where $d\text{g}$ is the Haar measure on $\SL(m,\R)$ defined through standard Iwasawa decomposition, under which $\SO(m,\R)$ has volume $1$. We note the formula
\begin{equation}
\int_{\SL(m,\R)}f(\text{g})d\text{g}=\int_{\mathcal{F}_m}\sum_{q\in\Delta}\int_{\SO(m,\R)}f(q\text{g}\rho')d\rho' d\text{g}.\label{eq:unfolding formula on the double coset space}
\end{equation}

Let $d\rho$ the  Haar measure probability measure on $\SO(m+1,\R)$, and consider the measure  $\nu_Y$ on $Y$ defined by  $d\nu_Y:=d\rho d\text{g}$. Then
\begin{equation} \label{unfolding haar measure on G using section}
dg=d\mu d\nu_Y,
\end{equation}
where $\mu$ is (<ref>), and $dg$ is the Haar measure on $G=\SL(m+1,\R)$ such that $\vol(G/\Ga)=1$.

by using Theorem 8.32 of <cit.>, we unfold a Haar measure on $G$  as follows

\begin{align*}
&\int_G f(g)dg\\
=&\int_{\SO(m+1,\R)} \int_P f(p\rho) dp d\rho \\
=&\int_{\SO(m+1,\R)}\int_{\SL(m,\R)} \int_0^{\infty}\int_{\R^m} f(u_v a_t \tilde{\text{g}} \rho)    dv \frac{dt}{t^{m+2}}d\text{g} d\rho \tag{by definition of $dp$}\\
=&\int_{\SO(m+1,\R)}\sum_{q \in \Delta}\int_{\mathcal{F}_m}\int_{\SO(m,\R)}  \int_0^{\infty}\int_{\R^m} f(u_v a_t \tilde{q}\tilde{\text{g}}\tilde{\rho'} \rho)    dv \frac{dt}{t^{m+2}}d\text{g}  d\rho'd\rho\tag{by formula \eqref{eq:unfolding formula on the double coset space}}\\
=& \int_{\SO(m+1,\R)}\sum_{q \in \Delta}\int_{\mathcal{F}_m} \int_0^{\infty}\int_{\R^m}  f(u_v a_t \tilde{q} \tilde{\text{g}}\rho)    dv \frac{dt}{t^{m+2}}d\text{g}d\rho\tag{invariance of $d\rho$}\\
=& \int_{\SO(m+1,\R)}\sum_{q \in \Delta} \int_0^{\infty}\int_{\R^m} \int_{\mathcal{F}_m}  f(u_v a_t \tilde{q} \tilde{\text{g}}\rho)  d\text{g}  dv \frac{dt}{t^{m+2}}d\rho\\
=&\int_H \int_{\SO(m+1,\R)}\int_{\mathcal{F}_m}  f(h \tilde{\text{g}}\rho)  d\text{g} d\rho  d\mu(h)\\
%   =&\text{Vol}(\SO(m,\R))^2 \int_H \int_{\frac{\mathcal{F}_m \times S}{K}}  f(h gK) d(gK)  dh
%   =&\text{Vol}(\SO(m,\R))^2\int_H \int_{\mathcal{F}_m S}  f(h g) dg  dh\\
=& \int_H\left( \int_{\mathcal{F}_m}\int_{\SO(m+1,\R)}  f(h \tilde{\text{g}}\rho)  d\rho d\text{g}\right)  d\mu(h),
\end{align*}

which proves the statement.

Fix $Hg_0\in H\backslash G$. By identifying $H \backslash G$ with $Y$, we define a measure on $H \backslash G$ via

\begin{equation}\label{relation between measures on H G and Y}
\end{equation}

where $\omega(\cdot,\cdot)$ is given in (<ref>).

In view of Theorem 2.2 and Corollary 2.4 in <cit.>, an immediate consequence of Theorem <ref> is the following

Fix $g_0\in G$. For any compactly supported $\varphi \in C_c(H\backslash G)$,

\begin{equation*}
\lim_{T\to \infty}\frac{1}{\mu(H_T)}\sum_{\gamma \in \Ga_T}\varphi(Hg_0.\gamma)= \int_{H\backslash G}\varphi(Hg)d\nu_{Hg_0}(Hg).
\end{equation*}

Now we would like to replace the normalization factor $\mu(H_T)$ by $\# \Ga_T$.
By <cit.>,

\begin{equation*}
\lim_{T\to \infty}\frac{\vol(G_T)}{\# \Ga_T}=1.
\end{equation*}

By formula A.1.15 in <cit.> and by Proposition <ref>, the limit
$$L:=\lim_{T\to \infty}\frac{\mu(H_{T})}{\vol(G_T)}$$
exists. Thus we conclude

\begin{equation*}
\lim_{T\to \infty}\frac{\mu(H_{T})}{\#\Ga_T}=\lim_{T\to \infty}\frac{\mu(H_{T})}{\vol(G_T)}\frac{\vol(G_T)}{\#\Ga_T}=\lim_{T\to \infty}\frac{\mu(H_{T})}{\vol(G_T)}=L,
\end{equation*}

which shows that
$$\lim_{T\to \infty}\frac{1}{\#\Ga_T}\sum_{\gamma \in \Ga_T}\varphi(Hg_0.\gamma)= L\int_{H\backslash G}\varphi(Hg)d\nu_{Hg_0}(Hg).$$

Our goal now will be to show that $L\nu_{Hg_0}(H\backslash G)=1$. We first have the following lemma.

It holds that $\nu_{H}(H\backslash G)<\infty$ and   $\nu_{Hg_0}(H\backslash G)=\nu_{H}(H\backslash G)$ for all $Hg_0\in H \backslash G.$

For $g_0,y \in Y$, we consider the decompositions

\begin{equation} \label{decompositions of g_0 and y in Y}
g_0:=\begin{bmatrix} \text{g}_0 & 0 \\0 & 1 \end{bmatrix}\rho_0,~ y:=\begin{bmatrix} \text{g} & 0 \\0 & 1 \end{bmatrix} \rho,
\end{equation}

where $\text{g}_0,\text{g}\in\mathcal{F}_m$ and $\rho_0,\rho\in\SO(m+1,\R).$ Hence $\alpha$ in (<ref>) takes the form: $ \omega(g_0,y)
= \frac{\sum_{q\in \Delta}\frac{1}{\|\text{g}_0^{-1}q \text{g}\|^{m^2}}}{\sum_{q\in \Delta}\frac{1}{\|q \|^{m^2}}}.$
Using the $\SO(m+1,\R)$-invariance of the Hilbert-Schmidt norm, we compute

\begin{align*}\label{folding trick}
\nu_{Hg_0}(H\backslash G)
%   =& \vol(\SO(m+1,\R)) \int_{\mathcal{F}_m}2\pi dg\\
=&\frac{1}{\sum_{q\in \Delta}\frac{1}{\|q \|^{m^2}}} \int_{\SO(m+1,\R)}\int_{\mathcal{F}_m}\sum_{q\in \Delta}\frac{1}{\|\text{g}_0^{-1}q \text{g}\|^{m^2}} d\text{g} d\rho \tag{definition of $d\nu_Y$}\\
=&\frac{1}{\sum_{q\in \Delta}\frac{1}{\|q \|^{m^2}}} \int_{\SL(m,\R)}\frac{1}{\|\text{g}_0^{-1}\text{g}\|^{m^2}}  d\text{g}\tag{by formula \eqref{eq:unfolding formula on the double coset space}}\\
=&\frac{1}{\sum_{q\in \Delta}\frac{1}{\|q \|^{m^2}}} \int_{\SL(m,\R)}\frac{1}{\|\text{g}\|^{m^2}}  d\text{g}\tag{invariance of $d\text{g}$}\\
=&\nu_{H}(H\backslash G).
\end{align*}

It remains to verify that $\int_{\SL(m,\R)}\frac{1}{\|\text{g}\|^{m^2}}  d\text{g}<\infty.$ By monotone convergence, it's enough to prove that $\int_{\SL(m,\R)_T}\frac{1}{\|\text{g}\|^{m^2}}  d\text{g}$ converges as $T\to\infty$, where $\SL(m,\R)_T=\{\text{g}\in\SL(m,\R):\|\text{g}\|\leq T\}.$  For that we use a Fubini trick, similarly to the proof of Lemma <ref>:

\begin{align*}
\int_{\SL(m,\R)_T}\frac{1}{\|\text{g}\|^{m^2}}  d\text{g}&=\int_{\SL(m,\R)_T}\int_0^\infty1_{\{t\leq\|\text{g}\|^{-m^2}\}} dt d\text{g}\\
&=\int_0^\infty \int_{\SL(m,\R)_T}1_{\{t\leq\|\text{g}\|^{-m^2}\}}d\text{g} dt\\
&=\int_0^1 \vol\{\text{g}:\|\text{g}\|\leq \text{min}(T,t^{-\frac{1}{m^2}})\} dt,\tag{using that $\|\text{g}\|\geq 1,\forall\text{g}$}
\end{align*}
where  $\vol$ stands for the $\SL(m,\R)$-Haar volume. The convergence is established by using formula A.1.15 in <cit.> which gives $\vol\{\SL(m,\R)_s\}\sim s^{m(m-1)}.$
Finally, we prove

$\nu_{Hg_0}(H\backslash G)=\nu_{H}(H\backslash G)=\lim_{T\to \infty}\frac{\vol(G_T)}{\mu(H_T)}=\frac{1}{L}$.

Since $\nu_H$ is a Radon measure, for any $\e>0$, we can choose $f_{\e}\in C_c(H\backslash G)$ with support $B_{\e}$  such that

\begin{equation*}
\int_{Y}|f_{\e}(Hy)-1|\omega(e,y)d\nu_Y(y)=\int_{H\backslash G}|f_{\e}(Hg)-1|d\nu_H(Hg)\le \e.
\end{equation*}

As in <cit.>, we observe that
\begin{align*}
\frac{1}{\mu(H_T)}\int_{G_T}|f_{\e}(Hg)-1|dg
=&\frac{1}{\mu(H_T)}\int_Y |f_{\e}(Hy)-1|\mu(\{h:\|hy\|<T\})d\nu_{Y}(y)\\
=&\int_Y |f_{\e}(Hy)-1| \frac{\mu(H_T[e,y])}{\mu(H_T)}d\nu_{Y}(y)
\end{align*}

Recall that $\lim_{T\to\infty}\frac{\mu(H_T[e,y])}{\mu(H_T)}=\omega(e,y).$
We will use below the dominated convergence theorem, and for that we will now show that the integrand $\frac{\mu(H_T[e,y])}{\mu(H_T)}$ is bounded by a function in $L^1(Y)$ for large $T$.
By (<ref>) and by (<ref>), we see

\begin{equation*}
\frac{\mu(H_T[e,y])}{\mu(H_T)} \ll \sum_{q\in \Delta}\frac{1}{\|q \text{g}\|^{m^2}},\text{ where $y=\begin{bmatrix} \text{g} & 0 \\0 & 1 \end{bmatrix} \rho,~\text{g}\in \SL(m,\R),\rho\in\SO(m+1,\R)$}.
\end{equation*}

Since $\int_{\SL(m,\R)}\frac{1}{\|\text{g}\|^{m^2}}d\text{g}$ converges, it follows that $\Psi(y):=\sum_{q\in \Delta}\frac{1}{\|q \text{g}\|^{m^2}}\in L^{1}(Y)$ (more precisely, to see that this is a $L^1$ function, use (<ref>)).

By the dominant convergence theorem, the second term satisfies

\begin{equation*}
\lim_{T\to \infty}\int_{Y} |f_{\e}(Hy)-1| \frac{\mu(H_T[e,y])}{\mu(H_T)}d\nu_{Y}(y)= \int_{Y} |f_{\e}(Hy)-1| d\nu_{H}(y)\le \e.
\end{equation*}

Therefore, by triangular inequality

\begin{align*}
&\limsup_{T\to \infty}\left|\frac{\vol(G_T)}{\mu(H_T)}-\nu_H(H\backslash G)\right|\\
\le & \limsup_{T\to \infty}\left|\frac{\vol(G_T)}{\mu(H_T)}-\frac{1}{\mu(H_T)}\int_{G_T}f_{\e}(Hg)dg\right|\\
&+\limsup_{T\to \infty}\left|\frac{1}{\mu(H_T)}\int_{G_T}f_{\e}(Hg)dg-\int_{H\backslash G}f_{\e}(Hg)d\nu_H(g) \right|\\
&+\limsup_{T\to \infty}\left|\int_{H\backslash G}f_{\e}(Hg)dg-\nu_H(H\backslash G) \right|\\
\le& \e+0+\e \tag{the middle term vanishes because of Theorem 2.3, \cite{Gorodnik2004DistributionOL}}
\end{align*}

Now let $\e\to 0$, and this finishes the proof.

Therefore, $\tilde\nu_{Hg_0}:=L\nu_{Hg_0}$ is a probability measure, and we conclude

\begin{equation*}
\lim_{T\to \infty}\frac{1}{\# \Ga_T}\sum_{\gamma \in \Ga_T}\varphi(Hg_0.\gamma)= \int_{H\backslash G}\varphi(Hg)d\tilde \nu_{Hg_0}(g).
\end{equation*}

Concluding the proof of Theorem <ref>.
We express $\tilde\nu_{Hg_0}$ as
$$d\tilde\nu_{Hg_0}=\Phi_{Hg_0}(y)dg d\rho,$$
$$\Phi_{Hg_0}(y):=\sum_{q\in \GL(m,\Z)}\frac{1}{\|\text{g}_1^{-1}q \text{g}_2\|^{m^2}},$$
$\Phi_{Hg_0}(y)dg$ is a probability measure on $\mathcal{F}_m$, and $d\rho$ is a probability measure on $\SO(m+1,\R)$. Now we interpret the coefficient $\Phi_{Hg_0}(y)$
in terms of rank-two discrete subgroups in $\R^{m}$. Let $(\La_0,w_0)$ and $(\La,w)$ denote the oriented rank-two discrete subgroups of $\R^{m}$ corresponding to $g_0$ and $y$, respectively. Let $e_1,e_2,...,e_m$ be the canonical basis of $\R^m$, written in terms of row vectors:

Then by (<ref>) it follows that
$$\mathscr{B}_0:=\{(e_1\text{g}_1,0)\rho_0,...,(e_m\text{g}_1,0)\rho_0 \},$$
$$\mathscr{B}:=\{(e_1q \text{g}_2,0)\rho,...,(e_mq \text{g}_2,0)\rho\}$$
form  $\Z$-bases of $\La_0$ and $\La$, respectively.

Consider the following linear maps between two-dimensional subspaces of $\R^{m}$:

\begin{equation}
T_{\mathscr{B}_0}: \Span_{\R}\{(e_1,0),...,(e_m,0)\} \to \Span_{\R}\{\mathscr{B}_0\}, (e_i,0)\mapsto (e_i\text{g}_1,0)\rho_0, i=1,2,...,m
\end{equation}

\begin{equation}
T_{\mathscr{B}}: \Span_{\R}\{(e_1,0),...,(e_m,0)\} \to \Span_{\R}\{\mathscr{B}\}, (e_i,0)\mapsto (e_iq \text{g}_2,0)\rho, i=1,2,...,m.
\end{equation}

Recall from the introduction (see (<ref>)) the Hilbert-Schmidt norm   $\|T\|_{\text{HS}}$ of an operator $T$ from the two-dimensional subspace $U\subset \R^{m}$ to the two-dimensional subspace $V\subset \R^{m}$, which is computed by choosing an orthonormal basis $\{u_1,u_2,...,u_m\}$ of $U$ and evaluating

\begin{equation}
\|T\|^2_{\text{HS}}:=\sum_{i=1}^m\|Tu_i\|^2,
\end{equation}

where the norm on the right hand side is the usual Euclidean norm on $\R^{m}$. Using the orthonormal basis $\{(e_1,0)\rho_0 , (e_2,0)\rho_0,...,(e_m,0)\rho_0\}$ of $\Span_\R\{\mathcal{B}_0\}$, we get
$$\|T_{\mathscr{B}}\circ T_{\mathscr{B}_0}^{-1}\|_{\text{HS}}^2=\|\text{g}_1^{-1}q \text{g}_2\|^2,$$

which shows

\begin{equation}\label{intrinsic interpretation of the sum}
\Phi_{g_0}(y)=\sum_{q\in \GL(m,\Z)}\frac{1}{\|\text{g}_1^{-1}q \text{g}_2\|^{m^2}}=\sum_{\Span_{\Z}\mathscr{B}=\La}\frac{1}{\|T_{\mathscr{B}}\circ T_{\mathscr{B}_0}^{-1}\|_{\text{HS}}^{m^2}}=\Psi_{\La_0}(\La),
\end{equation}

where $\Psi_{\La_0}(\La)$ defined in (<ref>).
It's now straight forward to verify that the measure $\tilde\nu_{x_0}$ for $x_0=(\Span_\Z\{e_1,e_2,...,e_m\},e_{m+1})\cdot g_0$ defined before Theorem <ref> is identified with $\tilde\nu_{H_{g_0}}$, which we leave to the reader to verify.

ν̃_x_0=ν̃_Hg_0$ under the identification of $X_m,m+1$ with $H\G$ and $x_0$ with $Hg_0$.

Let $X_m=Δ\(m,)$. For any $f∈C_c(X_m,m+1)$, we have
\begin{align}
\tilde\nu_{x_0}(f)
:=&\int_{\mathbb S^2}\int_{\pi_{\perp}^{-1}(w)}f((\La,w),\rho_w) \Phi_{\La_0}(\La)d((\rho_w)_*\mu_{e_{m+1}})(\La)d\mu_{\mathbb S^2}(w)\\
=&\int_{\mathbb S^2}\int_{\pi_{\perp}^{-1}(e_{m+1})}f((\La,e_{m+1}),\rho_w) \Psi_{\La_0}(\La)d\mu_{e_{m+1}}(\La)d\mu_{\mathbb S^2}(w)\\
=&\int_{\mathbb S^2}\int_{X_m}f(\Z^2\times 0,\eta\rho_w) \Psi_{\La_0}(\Z^2\times 0)d\eta d\mu_{\mathbb S^2}(w)\\
=&\frac{1}{\vol(\SO(m,\R))}\int_{\mathbb S^2}\int_{\SO(m,\R)}\int_{X_m}f(\Z^2\times 0,\eta k \rho_w) \Psi_{\La_0}(\Z^2\times 0)d\eta dk d\mu_{\mathbb S^2}(w)\\
=&\frac{1}{\vol(\SO(m,\R))}\int_{ \SO(m+1,\R)}\int_{X_m}f(\Z^2\times 0,\eta\rho) \Psi_{\La_0}(\Z^2\times 0)d\eta d\rho\\
=&\frac{1}{\vol(\SO(m,\R))}\int_{ \SO(m+1,\R)}\int_{\mathcal F_2}\int_{\SO(m,\R)}f(\Z^2\times 0,\eta\rho k) \Psi_{\La_0}(\Z^2\times 0)d\eta d\rho dk\\
=&\int_{ \SO(m+1,\R)}\int_{\mathcal F_2}f(\Z^2\times 0,\eta\rho) \Psi_{\La_0}(\Z^2\times 0)d\eta d\rho
\end{align}

The last line is the same as $∫_H\Gφ(Hg)dν̃_Hg_0(g)=∫_H\Gφ(Hg)dc̃_ν_Hg_0(g)$ (note that both $ν̃_x_0$ and $ν̃_Hg_0$ are probability measures).
This proves Theorem \ref{equidistribution result on  G mod H}.
\fi
%Since we are integrating with respect to the variable $y$. It is an interesting question how $\omega(g_1,y)$ depends $g_1$.

\section*{Acknowledgements}
The authors would like to thank Nimish Shah for helpful discussions and generous sharing of his ideas leading to the proof of Lemma \ref{our lemma expansion inequality for applying shah dichotomy}. We thank Uri Shapira for many helpful discussions and encouragement. We thank Alex Gorodnik and Barak Weiss for their comments on a previous version of our manuscript. We would also like to thank Osama Khalil for discussions related to this work. Finally, we would like to thank an anonymous referee for their careful reading and valuable feedback which helped us improve many places in this paper. This work has received funding from the European Research Council (ERC) under the European Union’s Horizon 2020 Research and Innovation Program, Grant agreement No. 754475.

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