Proof. Note that it suffices to prove this for \(\mathcal{C}=\mathcal{O}_{\infty}^n\), since \[\Vert \mathbf{v}\Vert_{\mathcal{C}}=\inf\left\{r\geq 0:\mathbf{v}\in x^{\log_qr}h\mathcal{O}_{\infty}^n\right\}=\inf\left\{r\geq 0:h^{-1}\mathbf{v}\in x^{\log_qr}\mathcal{O}_{\infty}^n\right\}=\Vert h^{-1}\mathbf{v}\Vert.\] By [23], there exists an \(\mathcal{R}\)-basis \(\{\mathbf{v}_1, \ldots, \mathbf{v}_n\}\) for \(\Lambda\) such that \[\Vert \mathbf{v}_i\Vert=\lambda_i(\Lambda),\quad i=1,\dots,n,\] where \(\lambda_i(\Lambda)\) is the \(i\)-th successive minima. Moreover, by combining Theorem 9 and [23], we obtain that \(\{\mathbf{v}_1,\dots,\mathbf{v}_n\}\) is an orthogonal \(\mathcal{R}\)-basis of \(\Lambda\). Denote by \(\mathbf{v}=\sum_{i=1}^n c_i\mathbf{v}_i\) (\(c_i\in \mathcal{K}_\infty\)). It follows that \[\begin{align} \mathbf{u}=\sum_{i=1}^n a_i\mathbf{v}_i\in \Lambda\cap B(\mathbf{v}, q^t) \quad &\Leftrightarrow\quad \left\|\sum_{i=1}^n (a_i-c_i)\mathbf{v}_i\right\|\le q^t\\ &\Leftrightarrow\quad \max_{i=1,\ldots, n} |a_i-c_i|\cdot \lambda_i(\Lambda) \le q^t. \end{align}\] As a consequence, \(\vert a_i-c_i\vert\leq \frac{q^t}{\lambda_i(\Lambda)}\) for every \(i\). If \(t\) is chosen so that \(q^t>\vert c_i\vert\lambda_i(\Lambda)\) for every \(i\), we have \(\mathbf{u}\in B(\mathbf{v},q^t)\) if and only if \(\vert a_i\vert\leq \frac{q^t}{\lambda_i(\Lambda)}\) for every \(i\). In this case, we have \[\#\Lambda\cap B(\mathbf{v},q^t)=\#\left\{(a_1,\dots,a_n)\in \mathcal{R}^n:\vert a_i\vert\leq \frac{q^t}{\lambda_i(\Lambda)}\right\}=\prod_{i=1}^n\left\lceil\frac{q^{t+1}}{\lambda_i(\Lambda)}\right\rceil.\] ◻

Proof. If \(\lambda_n(\Lambda)\leq q^{t+1}\), then, \(\left\lceil\frac{q^{t+1}}{\lambda_i(\Lambda)}\right\rceil=\frac{q^{t+1}}{\lambda_i(\Lambda)}\) for every \(i=1,\dots,n\). Hence, by Lemma 1, \[\label{eqn:LambdacapB40044q94t41SmallLambda95n} \#\Lambda\cap B(0,q^t)=\prod_{i=1}^n\frac{q^{t+1}}{\lambda_i(\Lambda)}=\frac{q^{(t+1)n}}{\det(\Lambda)}=q^{(t+1)n}\leq q^{t(n+1)}\alpha(\Lambda).\tag{8}\] On the other hand, if \(\lambda_n(\Lambda)>q^{t+1}\), then there exists some \(j=1,\dots,n-1\), such that \(\lambda_j(\Lambda)\leq q^{t+1}<\lambda_{j+1}(\Lambda)\). Let \(\mathbf{v}_1,\dots,\mathbf{v}_n\in \Lambda\) be a linearly independent set of vectors satisfying \(\Vert \mathbf{v}_i\Vert=\lambda_i(\Lambda)\). Then, by Theorem 9 and Lemma 1, \[\label{eqn:LambdacapB40044q94t41} \#\Lambda\cap B(0,q^t)=\prod_{i=1}^j\frac{q^{t+1}}{\lambda_i(\Lambda)}=\frac{q^{(t+1)j}}{\Vert \mathbf{v}_1\wedge \dots \wedge \mathbf{v}_j\Vert}\leq q^{n(t+1)}\alpha(\Lambda).\tag{9}\] For [lem:SchmidtLem], it suffices to prove that \(\widetilde{f}(\Lambda)\leq C\alpha(\Lambda)\) for \(f=\mathbf{1}_{B(\mathbf{u},q^t)}\) for \(t\in\mathbb{Z}\), since any \(f\in C_c^{\infty}(\mathcal{K}_{\infty}^n)\) can be approximated by sums of finitely many indicator functions. Note that in this case, we have \[\label{eqn:tilde40f41Count} \widetilde{f}(\Lambda)=\#\Lambda\cap B(\mathbf{u},q^t)=\#\{\mathbf{v}\in \Lambda:\Vert \mathbf{v}-\mathbf{u}\Vert\leq q^{t}\}.\tag{10}\] Note that if \(\Vert \mathbf{u}\Vert\leq q^t\), then \(\Vert \mathbf{v}-\mathbf{u}\Vert\leq q^t\) if and only if \(\Vert \mathbf{v}\Vert\leq q^t\). Therefore, in this case, by Lemma 1, the right hand side of 10 is equal to \[\label{eqn:BallCntAlphaBND} \#\Lambda\cap B(0,q^t)=\prod_{i=1}^n\left\lceil\frac{q^{t+1}}{\lambda_i(\Lambda)}\right\rceil\leq q^{nt}\alpha(\Lambda).\tag{11}\] If \(\Vert \mathbf{u}\Vert>q^t\), then \(\Vert \mathbf{v}-\mathbf{u}\Vert\leq q^t\) implies that \(\Vert \mathbf{v}\Vert=\Vert \mathbf{u}\Vert\). As a consequence, 8 , 9 , and 10 imply that \[\widetilde{f}(\Lambda)\leq \#\Lambda\cap \left(B(0,\Vert \mathbf{u}\Vert)\setminus B(0,q^{-1}\Vert \mathbf{u}\Vert)\right)=\prod_{i=1}^n\left\lceil \frac{q\Vert \mathbf{u}\Vert}{\lambda_i(\Lambda)}\right\rceil-\prod_{i=1}^n\left\lceil\frac{\Vert \mathbf{u}\Vert}{\lambda_i(\Lambda)}\right\rceil\leq q^n\Vert \mathbf{u}\Vert^n\alpha(\Lambda).\] ◻

Proof of Theorem 7. Note that \(\alpha\) is \(\operatorname{SL}_n(\mathcal{O}_{\infty})\)-invariant. Thus, by 7 and Theorem 10, for any \(\Lambda\in \mathcal{L}_n\), we have \[\label{eqn:alphaCompDiag} \alpha(\Lambda)=\alpha\left(\operatorname{diag}\{x^{\log_q\lambda_1(\Lambda)},\dots,x^{\log_q\lambda_n(\Lambda)}\}\mathcal{R}^n\right)=\max_{j=0,\dots,n}\prod_{1\leq i\leq j}\lambda_i(\Lambda)^{-1}.\tag{12}\] Let \[A_0=\left\{a=\operatorname{diag}\{x^{k_1},\dots,x^{k_n}\}: k_i\leq k_{i+1},k_i\in \mathbb{Z},\,\,\,\forall i=1,\dots,n-1\right\}\subseteq A.\] By Theorem 10, the set \(\operatorname{SL}_n(\mathcal{O}_{\infty})A_0\) contains a fundamental domain for the right action of \(\operatorname{SL}_n(\mathcal{R})\) on \(\operatorname{SL}_n(\mathcal{K}_{\infty})\). Thus, it suffices to compute \[\label{eqn:IntDecAlpha} \begin{align} \int_{\mathcal{L}_n}\alpha_j^m\mathrm{d}m_{\mathcal{L}_n}&=\int_{\operatorname{SL}_n(\mathcal{O})A_0}\alpha_j^m(g\mathcal{R}^n)dm_{\mathrm{G}}(g)\\ &=\int_{\operatorname{SL}_n(\mathcal{O}_{\infty})}\int_{A_0}\alpha_j^m(oa\mathcal{R}^n)\mathrm{d}m_{\operatorname{SL}_n(\mathcal{O})}(o)\mathrm{d}m_A(a)\\ &\leq C\int_{A_{0}}\alpha_j^m(a\mathcal{R}^n)\mathrm{d}m_A(a), \end{align}\tag{13}\] where the last inequality stems from compactness of \(\operatorname{SL}_n(\mathcal{O}_{\infty})\). Hence, by 6 , it suffices to compute \[\label{eqn:alphaAInteg} \int_{-\infty}^{\infty}\cdots\int_{-\infty}^{\infty}\frac{dk_1\dots d_{k_{n-1}}}{q^{\sum_{i=1}^jmk_i+2\sum_{i=1}^{n-1}(n-i)k_i}}.\tag{14}\] For \(a=\operatorname{diag}\{x^{k_1},\dots,x^{k_n}\}\in A_0\), denote \(\beta_i=k_{i+1}-k_i\geq 0\), so that \(k_{i+1}=k_1+\sum_{j=1}^i\beta_j\). Then, \(k_1=-\sum_{i=1}^{n-1}k_{i+1}=-(n-1)k_1-\sum_{i=1}^{n-1}\sum_{j=1}^i\beta_j\), so that \(k_1=-\frac{1}{n}\sum_{j=1}^{n-1}(n-j)\beta_j\), and for every \(i=1,\dots,n-1\), we have \[\begin{align} \label{eqn:k95i431} k_{i+1}=-\frac{1}{n}\sum_{j=1}^{n-1}(n-j)\beta_j+\sum_{j=1}^i\beta_j=\frac{1}{n}\sum_{j=1}^ij\beta_j-\frac{1}{n}\sum_{j=i+1}^{n-1}(n-j)\beta_j. \end{align}\tag{15}\] Hence by 15 , the exponent of the denominator in 14 is equal to \[\begin{align} \label{eqn:denomAlphaIntExp} &\frac{1}{n}\left[\sum_{i=1}^jm\left(\sum_{\ell=1}^i\ell\beta_{\ell}-\sum_{\ell=i+1}^{n-1}(n-\ell)\beta_{\ell}\right)+\frac{2}{n}\sum_{i=1}^{n-1}(n-i)\left(\sum_{\ell=1}^i\ell\beta_{\ell}-\sum_{\ell=i+1}^{n-1}(n-\ell)\beta_{\ell}\right)\right]\\ &=\frac{1}{n}m\left[\sum_{\ell=1}^j\beta_{\ell}\left(n+\ell(j+2-n)\right)-\sum_{\ell=j+1}^{n-1}(n-\ell)\beta_{\ell}\right]\\ &+\frac{2}{n}\sum_{\ell=1}^{n-1}\beta_{\ell}\left[\ell\sum_{i=\ell}^{n-1}(n-i)-(n-\ell)\sum_{i=1}^{\ell-1}(n-i)\right]\\ &=\frac{1}{n}m\left[\sum_{\ell=1}^j\beta_{\ell}\left(n+\ell(j+2-n)\right)-\sum_{\ell=j+1}^{n-1}(n-\ell)\beta_{\ell}\right]+\frac{1}{n}\sum_{\ell=1}^{n-1}\beta_{\ell}\left(3\ell^2+2(n-1)\ell+n^2\right)\\ &=\frac{1}{n}\sum_{\ell=1}^j\beta_{\ell}\left(mn+m\ell(j+2-n)+3\ell^2+2(n-1)\ell+n^2\right)\\ &+\frac{1}{n}\sum_{\ell=j+1}^{n-1}\beta_{\ell}\left(3\ell^2+(2n-2+m)\ell+n(n-m)\right). \end{align}\tag{16}\] Note that since \(m\leq n-1\), then \(3\ell^2+(2n-2+m)\ell+n(n-m)\geq 3\ell^2+n\geq 1\) for every \(\ell=j+1,\dots,n-1\). Moreover, for \(\ell=1,\dots,j\), the coefficient of \(\beta_{\ell}\) in the second to last line of 16 is equal to \[\begin{align}\label{eqn:ellSmallExp} &mn(1-\ell)+m\ell(j+2)+2(n-1)\ell+n^2+3\ell^2\\ &=m\ell(j-n)+mn+2m\ell+2(n-1)\ell+n^2+3\ell^2\\ &\geq (n-1)(2\ell-m)+n^2+3\ell^2+2m\ell+mn\\ &\geq (n-1)(2-m)+n^2+3\ell^2+2m\ell+mn\\ &\geq n\left(n-m+2\right)+3\ell^2+2m\ell+mn>0. \end{align}\tag{17}\] As a consequence of 16 and 17 , there exist \(c_1,\dots,c_{n-1}\geq 1\), such that 14 is equal to \[\int_0^{\infty}\cdots \int_0^{\infty}\frac{d\beta_1\dots d\beta_{n-1}}{q^{\sum_{i=1}^n c_i\beta_i}}<\infty.\] Thus, for every \(j=1,\dots,n-1\), and for every \(m=1,\dots, n-1\), the integral \(\int_{\mathcal{L}_n}\alpha_j^mdm_{\mathcal{L}_n}\) converges. Since \(\alpha^m(\Lambda)=\max_{j=1,\dots,n-1}\alpha_j^m(\Lambda)\), the integral \(\int_{\mathcal{L}_n}\alpha^mdm_{\mathcal{L}_n}\) converges for every \(m=1,\dots,n-1\). ◻

Proof. Assume that \(H\) acts on a locally compact homogeneous space \(X\) with finitely many orbits, \(X_1,\dots, X_k\), and \(\nu\) is an \(H\)-invariant measure on \(X\). Then, \(\nu\) can be decomposed into a sum \(\nu=\sum_{i=1}^m\nu_i\), where each \(\nu_i\) is of the form \(\nu_i(E):=\nu(E\cap X_i)\) and is an \(H\)-invariant measure supported on \(X_i\). In this case, the \(\operatorname{SL}_n(\mathcal{K}_{\infty})\)-action on \(\mathcal{K}_{\infty}^n\) has two orbits, which are \(\mathcal{K}_{\infty}^n\setminus \{0\}\) and \(\{0\}\) (see [20] for more details). An invariant measure supported on \(\mathcal{K}_{\infty}^n\setminus\{0\}\) is measure theoretically isomorphic to a scalar multiple of the Lebesgue measure, and an invariant measure supported on \(\{0\}\) is a scalar multiple of the Dirac-delta mass at \(0\). Hence, there exist \(c_0,c_1\geq 0\), such that \(\nu=c_0\delta_0+c_1\operatorname{Vol}\). ◻

Proof of Theorem 4. By Lemma [cor:LattPointCnt][lem:SchmidtLem] and Theorem 7, the function \(\widetilde{f}:\mathcal{L}_n\rightarrow \mathbb{C}\) is integrable. It follows from the Riesz-Markov-Kakutani representation theorem that the linear functional \(f\mapsto \int_{\mathcal{L}_n}\widetilde{f}dm_{\mathcal{L}_n}\) corresponds to some \(\mathrm{G}\)-invariant Radon measure \(\nu\), i.e., \[\int_{\mathcal{L}_n}\widetilde{f}dm_{\mathcal{L}_n}=\int_{\mathcal{K}_{\infty}^n}fd\nu =c_0f(0)+c_1\int_{\mathcal{K}_\infty^n} f d\mathrm{Vol}\] for some \(c_0,\;c_1\in [0,\infty)\), where the last equality follows from Lemma 4.

To compute \(c_1\), consider the sequence of test functions \(f_m=\frac{1}{q^{mn}}\mathbf{1}_{x^m\mathcal{O}_{\infty}^n}\), where \(m\in \mathbb{N}\). By Lemma 1, we have \[\begin{align} \label{eqn:STLargeBalls} \int_{\mathcal{L}_n}\widetilde{f}_m dm_{\mathcal{L}_n}=\frac{1}{q^{mn}}\int_{\mathcal{L}_n}\left(\#\left(\Lambda\cap x^m\mathcal{O}_{\infty}^n\right)-1\right)\: dm_{\mathcal{L}_n}(\Lambda). \end{align}\tag{18}\] Note that if \(\Lambda\in \mathcal{L}_n^{\le {q^{m+1}}}\), then, \(\lambda_n(\Lambda)\leq q^{m+1}\), so that \[\widetilde{f_m}(\Lambda)=\frac{1}{q^{mn}}\prod_{i=1}^n\left\lceil\frac{q^{m+1}}{\lambda_i(\Lambda)}\right\rceil-\frac{1}{q^{mn}}=\frac{q^{(m+1)n}}{q^{mn}\det(\Lambda)}-\frac{1}{q^{mn}}=q^{n}-\frac{1}{q^{mn}}.\] Observe that for any \(\Lambda\in \mathcal{L}_n^{> q^{m+1}}\), where \(m\in \mathbb{N}\), \[\lambda_1(\Lambda) < \frac{1}{q^{(m+1)/(n-1)}} \quad \text{and hence} \quad \alpha(\Lambda) > q^{(m+1)/(n-1)}.\]

Moreover, in the spirit of the proof of Lemma 3, one can take a uniform constant \(C>0\) for which \(\widetilde{f_m}\le C\alpha\) for all \(m\in \mathbb{N}\). It follows that \[\begin{align} \int_{\mathcal{L}_n} \widetilde{f_m} dm_{\mathcal{L}_n} &= \int_{\mathcal{L}_n^{\le q^{m+1}}} \widetilde{f_m} dm_{\mathcal{L}_n}+\int_{\mathcal{L}_n^{> q^{m+1}}} \widetilde{f_m} dm_{\mathcal{L}_n}\\ &= \left(q^n-\frac{1}{q^{mn}}\right) m_{\mathcal{L}_n}\left(\mathcal{L}_n^{\le q^{m+1}}\right)+\int_{\mathcal{L}_n^{> q^{m+1}}} \widetilde{f_m} dm_{\mathcal{L}_n}. \end{align}\] Thus by Lemma 7, \[\begin{align} &\left|\int_{\mathcal{L}_n} \widetilde{f_m} dm_{\mathcal{L}_n} - \left(q^n-\frac{1}{q^{mn}}\right) m_{\mathcal{L}_n}\left(\mathcal{L}_n^{\le q^{m+1}}\right)\right| \le C\int_{\mathcal{L}_n^{> q^{m+1}}} \alpha\: dm_{\mathcal{L}_n}\\ &= C\int_{\mathcal{L}_n^{> q^{m+1}}} \alpha^{1.5}\cdot \alpha^{-0.5} dm_{\mathcal{L}_n} \le \frac{C}{q^{(m+1)/2(n-1)}} \int_{\mathcal{L}_n^{> q^{m+1}}} \alpha^{1.5} dm_{\mathcal{L}_n} \rightarrow 0 \end{align}\] as \(m\rightarrow \infty\). By Mahler’s compactness criterion [22] (see also [18]), \(\lim_{m\rightarrow \infty}m_{\mathcal{L}_n}\left(\mathcal{L}_n^{\le q^{m+1}}\right)=m_{\mathcal{L}_n}(\mathcal{L}_n)=1\). On the other hand, \[\int_{\mathcal{L}_n} \widetilde{f_m} dm_{\mathcal{L}_n} = c_1\int_{\mathcal{K}_{\infty}^n}f_md\operatorname{Vol}+c_0f_m(0)=\frac{c_1q^{mn}+c_0}{q^{mn}}\rightarrow c_1\] as \(m\rightarrow \infty\). Thus, \(c_1=q^n\).

Now, consider the sequence of test functions \(h_m=\mathbf{1}_{x^{-m}\mathcal{O}_{\infty}^n}\), \(m\in \mathbb{N}\), to compute \(c_0\). By Lemma 1, \[\begin{align} \label{eqn:STSmallBalls} \int_{\mathcal{L}_n}\widetilde{h}_mdm_{\mathcal{L}_n}=\int_{\mathcal{L}_n}\left(\#\left(\Lambda\cap x^{-m}\mathcal{O}_{\infty}^n\right)-1\right)dm_{\mathcal{L}_n}(\Lambda). \end{align}\tag{19}\] Note that \(\lambda_1(\Lambda)\geq q^{-(m-1)}\) if and only if \[\#\left(\Lambda\cap x^{-m}\mathcal{O}_{\infty}^n\right)=1.\] Hence, \[\begin{align} 0\le \int_{\mathcal{L}_n} \widetilde{h_m} dm_{\mathcal{L}_m} &=\int_{\{\Lambda\in \mathcal{L}_n: \alpha_1(\Lambda)\ge q^{(m-1)}\}} \widetilde{h_m} dm_{\mathcal{L}_m}\\ &\le C \int_{\{\Lambda\in \mathcal{L}_n: \alpha_1(\Lambda)\ge q^{(m-1)}\}} \alpha\:dm_{\mathcal{L}_m}\\ &\le \frac{C}{q^{(m-1)/2}} \int_{\{\Lambda\in \mathcal{L}_n: \alpha_1(\Lambda)\ge q^{(m-1)}\}} \alpha^{1.5} dm_{\mathcal{L}_n}, \end{align}\] where \(C>0\) is the same constant for functions \(f_m\) as before. By Lemma 7 again, the above quantity converges to zero as \(m\) goes to \(\infty\).

On the other hand, \[\int_{\mathcal{L}_n} \widetilde{h_m} \:dm_{\mathcal{L}_m}= q^n\int_{\mathcal{K}_{\infty}^n}h_md\operatorname{Vol}+c_0h_m(0)=q^{-(m-1)n}+c_0\rightarrow c_0\] as \(m\rightarrow \infty\). Thus \(c_0=0\) and one obtains the integral formula 3 . ◻

Proof. Firstly, it is easy to verify that \(\Psi(D,Q)\) is an additive group, satisfying \[Q(\mathcal{R}^n)^r\subseteq \Psi(D,Q)\subseteq (\mathcal{R}^n)^r.\] Hence, \(\Psi(D,Q)\) is a lattice in \((\mathcal{R}^n)^r\). By taking the projection \(\pi_Q: (\mathcal{R}^n)^r\rightarrow(\mathcal{R}^n/Q\mathcal{R}^n)^r\), one obtains that \[\operatorname{covol}(\Psi(D,Q))=\frac{\#(\mathcal{R}^n/Q\mathcal{R}^n)^r}{\#\left\{(\mathbf{v}_1,\dots,\mathbf{v}_r)\in (\mathcal{R}^n/Q\mathcal{R}^n)^r:\frac{1}{Q}(\mathbf{v}_1,\dots,\mathbf{v}_r)D\in (\mathcal{R}^n)^k\right\}}=\frac{\vert Q\vert^{nr}}{N(D,Q)^n}.\] ◻

Proof of Theorem 12. For \(1\le r \le k\) and \(Q\in \mathcal{R}_{\text{monic}}\), fix \(D\in \mathfrak D^k_{r,Q}\), and let \(\Phi(D,Q)\) and \(\Psi(D,Q)\) be as in 22 and 25 respectively.

For a compactly supported, continuous, and bounded function \(H:\left(\mathcal{K}_{\infty}^n\right)^r\rightarrow \mathbb{R}_{\geq 0}\), define \[\begin{align} \phi_H(g\Gamma)&=\sum_{(\mathbf{w_1},\dots,\mathbf{w}_r)\in \Phi(D,Q)}H\left(g\mathbf{w}_1,\dots,g\mathbf{w}_r\right),\quad \forall g\Gamma\in \mathcal{L}_n.\\ \end{align}\] Then equation 24 follows from the following formula \[\label{eq:32H-integral} \int_{\mathcal{L}_n} \phi_H(g\Gamma) dm_{\mathcal{L}_n}(g\Gamma) =\frac{N(D,Q)^r}{|Q|^{nr}} \int_{(\mathcal{K}^n)^r} H(\mathbf{w}_1, \ldots, \mathbf{w}_r) d\mathbf{w}_1 \cdots d\mathbf{w}_r,\tag{26}\] when taking \(H=H_F\) defined by \[H_F(\mathbf{w}_1, \ldots, \mathbf{w}_r) :=F\left(\frac{1}{Q} (\mathbf{w}_1, \ldots, \mathbf{w}_r)D\right).\]

Let \(E=\{(\mathbf{w}_1, \ldots, \mathbf{w}_r)\in \mathcal{K}_\infty^r: \mathrm{rk}(\mathbf{w}_1, \ldots, \mathbf{w}_r)\le r-1\}\). Note that \((\mathcal{K}^n_\infty)^r\setminus E\) is locally compact Hausdorff space and consists of a single orbit of the \(\mathrm{G}\)-action defined by \(g.(\mathbf{w}_1, \ldots, \mathbf{w}_r)=(g\mathbf{w}_1, \ldots, g\mathbf{w}_r)\) for all \(g\in \mathrm{G}\) and \((\mathbf{w}_1, \ldots, \mathbf{w}_r)\in (\mathcal{K}^n_\infty)^r\). Moreover, \(E\) is locally diffeomorphic to \(K^{(r-1)(n+1)}_\infty\). Since \(r\le n-1\), we have \((r-1)(n+1)<nr\), and the unique \(\mathrm{G}\)-invariant measure \(m_{(\mathcal{K}_\infty^n)^r\setminus E}:=\left.m_{(\mathcal{K}_\infty^n)^r}\right\vert_{(\mathcal{K}_\infty^n)^r\setminus E}\) on \((\mathcal{K}_\infty^n)^r\setminus E\) is measure-theoretically equal to the Haar measure \(m_{(\mathcal{K}_\infty^n)^r}\).

First, we claim that for \(H\in C_c\big((\mathcal{K}_\infty^n)^r\setminus E\big)\), the integral \(\int_{\mathcal{L}_n} \phi_H(g\Gamma) dm_{\mathcal{L}_n}\) is finite: one can take a compactly supported, bounded, continuous function \(h:\mathcal{K}_{\infty}^n\rightarrow \mathbb{R}_{\geq 0}\), satisfying \(H(\mathbf{v}_1,\dots,\mathbf{v}_r)\leq h(\mathbf{v}_1)\cdots h(\mathbf{v}_r)=:h^{\otimes r}(\mathbf{v}_1,\dots,\mathbf{v}_r)\). As a consequence, \(\phi_H(\Lambda)\leq \big(\widetilde{h}(\Lambda)\big)^r\) for every \(\Lambda\in \mathcal{L}_n\). Hence, by Theorem 7 and Lemma [cor:LattPointCnt] [lem:SchmidtLem], there exists \(C_h\in \mathbb{R}_{>0}\), such that \[\int_{\mathcal{L}_n} \phi_H \: dm_{\mathcal{L}_n}\leq \int_{\mathcal{L}_n}\widetilde{h}^r\:dm_{\mathcal{L}_n}\leq C_h\int_{\mathcal{L}_n}\alpha^rdm_{\mathcal{L}_n}<\infty.\] Thus by Riesz–Kakutani–Markov theorem, together with the fact that the positive linear functional \[H\in C_c\big((\mathcal{K}_\infty^n)^r\setminus E\big)\mapsto \int_{\mathcal{L}_n} \phi_H \:dm_{\mathcal{L}_n}\] is \(\mathrm{G}\)-invariant, there exists \(c>0\) such that for any \(H\in C_c\big((\mathcal{K}_\infty^n)^r\setminus E\big)\), \[\label{eqn:inthat40F4161cint40F41} \int_{\mathcal{L}_n}\phi_H\: dm_{\mathcal{L}_n}=c\int_{\left(\mathcal{K}_{\infty}^n\right)^r} H\: dm_{\left(\mathcal{K}_{\infty}^n\right)^r}.\tag{27}\]

Second, let us compute the constant \(c>0\) for the functions in \(C_c\left((\mathcal{K}^n_\infty)^r\setminus E\right)\). For each \(t\in \mathbb{N}\), consider \(B_t=B_{(\mathcal{K}_{\infty}^n)^r}(0,q^t)\setminus N_{1}(E)\subseteq (\mathcal{K}_{\infty}^n)^r\), where \(N_{1}(E)\) is the \(1\)-neighborhood of \(E\). Define a function \(H_t\) on \((\mathcal{K}^n_\infty)^r\setminus E\) as \(H_t=\frac{1}{q^{trn}} \mathbf{1}_{B_t}.\) Observe that \[\begin{gather} m_{(\mathcal{K}_\infty^n)^r} \left(B_{(\mathcal{K}_{\infty}^n)^r}(0,q^t)\cap N_{1}(E)\right)=O\left(q^{t(r-1)(n+1)}\right),\\ \#\left(\Lambda\cap B_{(\mathcal{K}_{\infty}^n)^r}(0,q^t)\cap N_{1}(E)\right)=O_\Lambda\left(q^{t(r-1)(n+1)}\right) \end{gather}\] for any lattice \(\Lambda\in \mathcal{L}_n\). More precisely, the implied constant in the second Big-O depends on the successive minima \(\lambda_1(\Lambda),\cdots ,\lambda_n(\Lambda)\) of the lattice \(\Lambda\). It follows that \[\label{eqn:H95tInt} \lim_{t\rightarrow \infty}\int_{(\mathcal{K}_{\infty}^n)^r} H_t\:dm_{(\mathcal{K}_{\infty}^n)^r} =\lim_{t\rightarrow \infty}\int_{(\mathcal{K}_{\infty}^n)^r}\frac{1}{q^{trn}}\mathbf{1}_{B_{(\mathcal{K}_{\infty}^n)^r}(0,q^t)}\:dm_{(\mathcal{K}_{\infty}^n)^r}=1.\tag{28}\] Moreover, for every \(\Lambda \in \mathcal{L}_n\), \[\label{eqn:phi95H95t}\begin{align} \phi_{H_t}(\Lambda) &=\frac{1}{q^{nrt}} \#\left(g\Phi(D,Q)\cap B_t\right) =\frac{1}{q^{nrt}} \#\left(g\Psi(D,Q)\cap B_t\right)\\ &=\frac{1}{q^{nrt}} \#\left(g\Psi(D,Q) \cap (B_{(\mathcal{K}_\infty^n)^r}(0, q^t)\setminus N_{1}(E))\right)\\ &\underset{t\rightarrow \infty}{\longrightarrow} \frac{1}{q^{nrt}} \#(g\Psi(D,Q)\cap B_{(\mathcal{K}_\infty^n)^r}(0, q^t)), \end{align}\tag{29}\] where the limit is uniform on any compact subset of \(\mathcal{L}_n\). By Lemma 1 and Lemma 5, the right hand side of 29 converges to \(N(D,Q)^n/|Q|^{nr}\) as \(t\rightarrow \infty\), uniformly on any compact subset of \(\mathcal{L}_n\). Thus, by 27 and 28 , we have \[c=\lim_{t\rightarrow \infty} c\int_{\left(\mathcal{K}_{\infty}^n\right)^r} H_t\: dm_{\left(\mathcal{K}_{\infty}^n\right)^r}=\lim_{t\rightarrow \infty} \int_{\mathcal{L}_n} \phi_{H_t}\: dm_{\mathcal{L}_n}=\frac{N(D,Q)^n}{|Q|^{nr}}.\]

Finally, we show that the identity 27 extends to functions in \(C_c\left((\mathcal{K}^n_\infty)^r\right)\). For any given \(F\in C_c\left((\mathcal{K}^n_\infty)^{r}\right)\), consider the sequence of functions \(F_t \in C_c\left((\mathcal{K}^n_\infty)^{r}\setminus E\right)\) defined by \[F_t=F\mid^{}_{(\mathcal{K}_\infty^n)^r\setminus N_{\frac{1}{q^{-t}}}(E)},\] where \(N_{\frac{1}{q^{-t}}}(E)\) is the \(\frac{1}{q^{-t}}\)-neighborhood of \(E\). Note that \((F_t)\nearrow F\), and thus there exists a constant \(C=C_F>0\) such that \[\label{eqn:phi60alpha94r} \phi_{F_t}\le \phi_F\le C\alpha^r, \quad\forall t\in \mathbb{N}.\tag{30}\] Since \(m_{(\mathcal{K}_\infty^n)^r} (\mathop{\mathrm{supp}}F \cap N_{\frac{1}{q^{-t}}}(E))=O_F(q^{-t(r-1)(n+1)})\), it follows that \[\lim_{t\rightarrow \infty} \int_{(\mathcal{K}_\infty^n)^r} F_t \: dm_{(\mathcal{K}_\infty^n)^r}=\int_{(\mathcal{K}_\infty^n)^r} F\: dm_{(\mathcal{K}_\infty^n)^r}.\] Hence, it suffices to show that \[\label{eqn:IntConverge}\lim_{t\rightarrow \infty}\int_{\mathcal{L}_n} \phi_{F_t} dm_{\mathcal{L}_n}=\int_{\mathcal{L}_n} \phi_{F} dm_{\mathcal{L}_n}.\tag{31}\] Since \(F\) is compactly supported, for any \(\Lambda=g\mathcal{R}^n\in \mathcal{L}_n\), there is \(t_0>1\) depending on \(\|g\|,\;\|g^{-1}\|\) (thus uniform on any compact set on \(\mathcal{L}_n\)) such that if \(t\ge t_0\), we have \(\phi_{F_t}=\phi_F\), that is, \(\mathop{\mathrm{supp}}F \cap N_{\frac{1}{q^{-t}}}(E) \cap g\Phi(D,Q)=\emptyset\).

Therefore, for any \(\varepsilon>0\), there exists \(\ell=\ell_{\varepsilon}\in \mathbb{N}\) for which \(\int_{\mathcal{L}_n^{> q^{\ell}}} \alpha^{r+\frac{1}{2}} dm_{\mathcal{L}_n}<\frac{\varepsilon}{C}\). Hence, by 30 and [lem:UM], for all sufficiently large \(t\), we have \[\left|\int_{\mathcal{L}_n} \phi_F \: dm_{\mathcal{L}_n} -\int_{\mathcal{L}_n} \phi_{F_t} \: dm_{\mathcal{L}_n}\right|\leq \lim_{t\rightarrow \infty} \int_{\mathcal{L}_n^{> q^{\ell}}} \left|\phi_F- \phi_{F_t}\right| dm_{\mathcal{L}_n} <\frac{\varepsilon}{C}\cdot C=\varepsilon.\] Thus, 31 holds, so that all functions \(F\in C_c((\mathcal{K}_{\infty}^n)^r)\) satisfy 27 . ◻

Proof. For each \(Q\in \mathcal{R}_{\text{monic}}\), the matrix \(D\in \mathfrak D^2_{1,Q}\) is of the form \(\begin{pmatrix} Q & a\end{pmatrix}\), where \(a\in \mathcal{R}\setminus \{0\}\) is coprime with \(Q\). In view of Theorem 11, it suffices to compute \(N(D,Q)\). Note that a vector \(\mathbf{v}\in \mathcal{K}_{\infty}\) satisfies \(\left(\mathbf{v},\frac{a}{Q}\mathbf{v}\right)=\frac{1}{Q}\mathbf{v}\begin{pmatrix} Q&a \end{pmatrix}\in \mathcal{R}^2\) if \(\mathbf{v}\in \mathcal{R}\) and \(a\mathbf{v}\in Q\mathcal{R}\). Since \(\mathrm{gcd}(a,Q)=1\), then, \(\mathbf{v}\in Q\mathcal{R}\). Hence, \[\begin{align} N(D,Q)=\#Q\mathcal{R} \cap \mathcal{R}_{\leq \mathrm{deg}(Q)}=\#(Q\mathbb{F}_q)=q. \end{align}\] Thus, Equation 32 follows directly from Theorem 11. ◻

Proof of Theorem 5. By Lemma 6, it suffices to prove that for a measurable set \(B\subseteq \mathcal{K}_\infty^n\), \[\label{eqn:var}\begin{align} \operatorname{Var}(X_{B}) =\sum_{Q\in \mathcal{R}_{\text{monic}}} \sum_{\scriptsize \begin{array}{c} a\in \mathcal{R}\setminus \{0\}\\ \mathrm{gcd}(a,Q)=1 \end{array}} \frac{q^n}{|Q|^n} \int_{\mathcal{K}_\infty^n} X_B\left(\mathbf{v}\right)X_B\left(\frac{a}{Q}\mathbf{v}\right) d\mathrm{Vol}(\mathbf{v}) \end{align}\tag{33}\] is bounded by \(C_n\mathrm{Vol}(B)\), where \(C_n\) is the constant given as in 4 . Note that \[\label{eqn:VolBND}\begin{align} \int_{\mathcal{K}_{\infty}}X_B(\mathbf{v})X_B\left(\frac{a}{Q}\mathbf{v}\right)d\operatorname{Vol}(\mathbf{v}) &=\operatorname{Vol}\left(B\cap \frac{Q}{a}B\right)\\ &\leq \begin{cases} \operatorname{Vol}(B)& \text{if }\mathrm{deg}(Q)\ge \mathrm{deg}(a);\\[0.05in] \left|\frac{Q}{a}\right|^n\operatorname{Vol}(B)& \text{if }\mathrm{deg}(Q)<\mathrm{deg}(a). \end{cases}\end{align}\tag{34}\] Thus, one can consider three cases which are \(\mathrm{deg}(Q)\ge \mathrm{deg}(a)\), \(\mathrm{deg}(Q)=\mathrm{deg}(a)\), and \(\mathrm{deg}(Q)<\mathrm{deg}(a)\). For the last case, one can switch the roles of \(Q\) and \(a\) by the following equality that holds for any integrable function \(f:\mathcal{K}_\infty^n \rightarrow \mathbb{C}\): \[\label{eqn:symmetry}\begin{align} \frac{1}{|Q|^n} \int_{\mathcal{K}^n_\infty} f\left(\mathbf{v}\right)f\left( \frac{a}{Q} \mathbf{v}\right) d\mathrm{Vol}(\mathbf{v}) &=\int_{\mathcal{K}^n_\infty} f\left(Q\mathbf{v}\right)f\left( a\mathbf{v} \right) d\mathrm{Vol}(\mathbf{v})\\ &=\frac{1}{|a|^n} \int_{\mathcal{K}^n_\infty} f\left(\frac{Q}{a} \mathbf{v}\right)f\left(\mathbf{v}\right)d\operatorname{Vol}(\mathbf{v}). \end{align}\tag{35}\] Moreover, since \(\{a\in \mathcal{R}: \mathrm{deg}(a)=d\}=\mathbb{F}_q^{\times} \times \{a\in \mathcal{R}_{\text{monic}}: \mathrm{deg}(a)=d\}\) (\(d\neq 0\)), by changing the order of summation of \(Q\) and \(a\) in 33 , it follows that 33 is bounded above by \[\label{eqn:sumVOlBND} q^n\sum_{d=0}^{\infty}\frac{1}{q^{nd}} \sum_{\scriptsize \begin{array}{c} Q\in \mathcal{R}_{\text{monic}}\\ \mathrm{deg}(Q)=d\end{array}}\left(q\times \sum_{\scriptsize \begin{array}{c} a\in \mathcal{R}_{< d}\\ \mathrm{gcd}(a,Q)=1\end{array}}\operatorname{Vol}(B)+\sum_{\scriptsize\begin{matrix} a\in \mathcal{R}_{=d}\\ \mathrm{gcd}(a,Q)=1 \end{matrix}} \operatorname{Vol}(B)\right).\tag{36}\] We now bound the first internal summand in 36 . Recall the Euler \(\varphi\)-function for a function field defined as \[\varphi(Q)=\#\{a\in \mathcal{R}_{<\mathrm{deg}(Q)}: \mathrm{gcd}(a,Q)=1\},\] and that by [28], we have \[\sum_{Q\in \mathcal{R}_{\text{monic}}\cap \mathcal{R}_{=d}}\varphi(Q)=q^{2d}(1-q^{-1}).\] To count the number of \(a\in \mathcal{R}\) with \(\vert a\vert=q^m\), where \(m\ge d\) (we only need the case when \(m=d\)) and \(a\) is coprime with \(Q\), we use the inclusion-exclusion principle. Let \(P_1, \cdots, P_{\ell}\) be irreducible factors of \(Q\). It holds that \[\begin{align} \label{eqn:CoprimeGenDeg} &\#\{\vert a\vert=q^m:\mathrm{gcd}(a,Q)=1\} =\#\mathcal{R}_{=m}\setminus \bigcup_{i=1}^{\ell}P_i\mathcal{R}\\ &=q^m(q-1)+\sum_{k=1}^{\ell}(-1)^k\sum_{1\leq i_1<\dots<i_k\leq \ell}\#\mathcal{R}_{=m}\cap P_{i_1}\cdots P_{i_k}\mathcal{R}\\ &=q^m(q-1)\left[1+\sum_{k=1}^{\ell}(-1)^k\sum_{1\leq i_1<\cdots <i_k\leq \ell}\frac{1}{\vert P_{i_1}\cdots P_{i_k}\vert}\right]\\ &=q^m(q-1)\prod_{ \scriptsize \begin{array}{c} P|Q\\ P\text{ irred.}\end{array}}\left(1-\frac{1}{\vert P\vert}\right) =q^m(q-1)\frac{\varphi(Q)}{\vert Q\vert}, \end{align}\tag{37}\] where the last line is obtained through [28]. Thus, the fact that \(n\geq 3\) implies that \[\begin{align} \label{eqn:smallA} \eqref{eqn:sumVOlBND} =q^n(2q-1)(1-q^{-1})\sum_{d=0}^{\infty}\frac{\operatorname{Vol}(B)}{q^{nd}}q^{2d}=(2q-1)(1-q^{-1})\frac{q^{2n-2}}{q^{n-2}-1}\operatorname{Vol}(B). \end{align}\tag{38}\]  ◻

Proof of Proposition 6. When \(\|\alpha\|\le r\), by Lemma 7, we have \(B(\alpha, r)=B(0, r)\). Thus we may assume that \(\alpha\) is the origin. In this case, every intersection \(B(0,r)\cap B\left(0, \left|\frac{Q}{a}\right| r\right)\) is non-empty. Moreover, the inequality in 34 is an equality for every pair of coprime polynomials \((a,Q)\). Thus, by following the proof of Theorem 5, we have \[\operatorname{Var}(X_{B(\alpha, r)}) =\operatorname{Var}(X_{B(0, r)}) =C_n.\] Now, assume that \(\|\alpha\|>r\). By Lemma 7, \(B(\alpha,r)\cap \frac{Q}{a}B(\alpha,r)=B(\alpha,r)\cap B\left(\frac{Q}{a}\alpha,\left|\frac{Q}{a}\right|r\right)\neq \emptyset\) if and only if \(\left|1-\frac{Q}{\alpha}\right|\cdot \Vert \alpha\Vert\leq \max\left\{1,\left|\frac{Q}{a}\right|\right\}r\), i.e., \[\label{eqn:admissRange} B(\alpha,r)\cap B\left(\frac{Q}{a}\alpha,\left|\frac{Q}{a}\right|r\right)\neq \emptyset \quad\Leftrightarrow\quad r<\Vert \alpha\Vert\leq \frac{\max\left\{|a|,\left|{Q}\right|\right\}}{\left|a-{Q}\right|}r.\tag{39}\] Observe that if \(|a|\neq \left|{Q}\right|\), then [lem:UM61] implies that the right hand side of 39 is equal to \(r\), so that \(B(\alpha,r)\cap B\left(\frac{Q}{a}\alpha,\left|\frac{Q}{a}\right|r\right)=\emptyset\). Therefore, the only case in which \(B\left(\alpha,r\right)\cap B\left(\frac{Q}{a}\alpha,\left|\frac{Q}{a}\right|r\right)\neq \emptyset\) is when \(\left|a-Q\right|\le\frac{|Q|}{\|\alpha\|/r}<\vert Q\vert\).

Let \(t\ge 1\) be a positive integer such that \(\frac{\|\alpha\|}{r} = q^t>1\). By 36 , for \(Q\in \mathcal{R}_{\text{monic}}\) with \(\mathrm{deg}(Q)=d\), we would like to count \[\label{eqn:124a-Q12460q94d-tCoPrime} \#\left\{a\in \mathcal{R}_{=d}: |a-Q|\le q^{d-t},\;\mathrm{gcd}(a,Q)=1 \right\}.\tag{40}\] First, note that 40 is zero whenever \(\mathrm{deg}(Q)<t\), since \(|a-Q|\ge 1\) provided that \(a\neq Q\). For \(Q\in \mathcal{R}_{\text{monic}}\) with \(d=\mathrm{deg}(Q)\ge t\), denote the distinct irreducible factors of \(Q\) by \(P_1,\ldots, P_{\ell}\), where \(\ell\ge 1\). It follows that for \(1\le i_1 < \ldots < i_k\le \ell\), \[\begin{align} \#\left\{a\in \mathcal{R}: |a-Q|\le q^{d-t} \right\}\cap P_{i_1}\cdots P_{i_k}\mathcal{R} &=\#\left\{a'\in \mathcal{R}: \left|a'-\frac{Q}{P_{i_1}\cdots P_{i_k}}\right|\le \frac{q^{d-t}}{P_{i_1}\cdots P_{i_k}}\right\}\\ &=\frac{q^{d-t+1}}{|P_{i_1}\cdots P_{i_k}|}, \end{align}\] provided that we put \(a=P_{i_1}\cdots P_{i_k}a'\in P_{i_1}\cdots P_{i_k}\mathcal{R}\) and noting that \(Q/(P_{i_1}\cdots P_{i_k})\) is again a monic polynomial. Using the inclusion-exclusion principle in a similar manner as in 37 , we obtain that \[\begin{align} &\#\left\{a\in \mathcal{R}: |a-Q|\le q^{d-t}\;\text{and}\; \mathrm{gcd}(a,Q)=1 \right\}\\ &=\#\left\{a\in \mathcal{R}: |a-Q|\le q^{d-t}\right\} +\sum_{k=1}^\ell (-1)^k \sum_{1\le i_1 < \cdots < i_k \le \ell} \#\left\{a\in P_{i_1}\cdots P_{i_k} \mathcal{R}: |a-Q|\le q^{d-t}\right\}\\ &=q^{d-t+1}\left(1+\sum_{k=1}^\ell (-1)^k \sum_{1\le i_1< \cdots < i_k\le \ell} \frac{1}{|P_{i_1} \cdots P_{i_k}|}\right) =q^{d-t+1} \frac{\varphi(Q)}{|Q|}\\ &=\frac{\varphi(Q)}{q^{t-1}}. \end{align}\]

Consequently, \[\begin{align} \operatorname{Var}(X_{B}) &=\sum_{d=t}^\infty \frac{q^n}{q^{dn}} r^n \sum_{\scriptsize \begin{array}{c} Q\in \mathcal{R}_{\text{monic}}\\ \mathrm{deg}(Q)=d\end{array}} q^{-t+1}\varphi(Q)\\ &=q^{n-t+1}r^n(1-q^{-1})\sum_{d=t}^\infty \frac{1}{q^{(n-2)d}} =\frac{q^{2n-1}}{q^{n-2}-1}(1-q^{-1})\left(\frac{r}{\Vert \alpha\Vert}\right)^{n-1}\mathrm{Vol}(B). \end{align}\] ◻

Proof. Equation 41 follows directly from Lemma 8 (see the proof of [8] for details). ◻

Proof of Theorem 15. Since smooth functions can be approximated with sums of finitely many characteristic functions, we may assume that \(h\) is the characteristic function of a Borel set \(A\subseteq \mathcal{O}^n\setminus \mathfrak{m}^n\) satisfying \[A+x^{-K}\mathcal{O}^n=A.\] By defining \(x^{-t}\mathbf{v}=\mathbf{w}\), we have \[\label{eqn321:32volume32estimate1} \int_{\mathcal{K}_\infty^n} h(x^{-t}\mathbf{v}) \chi^{}_I(Q_0(\mathbf{v})) d\mathbf{v} =q^{nt}\int_{\mathcal{K}_\infty^n} h(\mathbf{w}) \chi^{}_I(x^{2t}Q_0(\mathbf{w})) d\mathbf{w}.\tag{46}\] Set \[C_t(A,I)=\left\{\mathbf{w}\in \mathcal{K}_\infty^n: \mathbf{w} \in A\;\text{and}\; x^{2t}Q_0(\mathbf{w})\in I\right\},\] so that \[\eqref{eqn321:32volume32estimate1}= q^{nt} \mathrm{Vol}(C_t(A,I)).\]

The asymptotic formula 44 follows if we define \[J_{h,Q_0}=\lim_{\ell\rightarrow \infty} \frac{\#[A\cap \mathrm{Cone}(Q_0)]_\ell}{q^{(n-1)\ell}},\] where \(J_{h,Q_0}\) is well-defined for general \(h\in C_c(\mathcal{O}^n_\infty\setminus x^{-1}\mathcal{O}^n)\). ◻

Proof of Theorem 13. We may assume that \(h\) is the characteristic function of a Borel set \(A\subseteq \mathcal{K}_\infty^n\) satisfying that \[A+ x^{-K}\mathcal{O}^n=A\] and that \(\mathop{\mathrm{supp}}h\subseteq x^R\mathcal{O}^n\setminus x^{r-1}\mathcal{O}^n\). For each \(r\le z \le R\), let \[h_z(\mathbf{v})=h(\mathbf{v})\cdot \chi^{}_{x^z\mathcal{O}^n\setminus x^{z-1}\mathcal{O}^n}(\mathbf{v}).\] Define \(h'_z(\mathbf{v}):=h_z(x^{z}\mathbf{v})\) for all \(\mathbf{v}\in \mathcal{K}_\infty^n\) so that \(\mathop{\mathrm{supp}}h'_z\subseteq \mathcal{O}^n\setminus \mathfrak{m}^n\) for all \(r\le z\le R\). It follows that \[\begin{align} \int_{\mathcal{K}_\infty^n} h(x^{-t}\mathbf{v}) \chi_I(Q_0(\mathbf{v}))d\mathbf{v} &=\int_{\mathcal{K}_\infty^n} \sum_{z=r}^R h_z(x^{-t}\mathbf{v}) \chi_I (Q_0(\mathbf{v})) d\mathbf{v}\\ &=\sum_{z=r}^R q^{zn}\int_{\mathcal{K}_\infty^n} h'_z(x^{-t}\mathbf{v})\chi^{}_{I_z}(Q_0(\mathbf{v})) d\mathbf{v},\\ \end{align}\] where \(I_z=x^{-2z}a+x^{b-2z}\mathcal{O}\subset \mathcal{K}_\infty\). Moreover, for each \(z\), \[\begin{gather} h'_z(\mathbf{v}+x^{-K-z} \mathcal{O}^n) =h_z(x^{z}\mathbf{v} + x^{-K}\mathcal{O}^n) =h_z(x^{z}\mathbf{v}) =h'_z(\mathbf{v}),\quad\forall \mathbf{v}\in \mathcal{K}^n_\infty.\\ \end{gather}\] Applying 44 to each \(h'_z\), if \(2t>\max\{-2z+B, N+B-b, -z+B+K\}\), we have \[\int_{\mathcal{K}_\infty^n} h'_z(x^{-t}\mathbf{v}) \chi^{}_{I_z}(Q_0(\mathbf{v})) d\mathbf{v} =J_{h'_z,Q_0}q^{b-2z}q^{(n-2)t}.\] Thus, if \(2t>\max\{-2r+B, N+B-b,-3r+B+K\}\), one has \[\int_{\mathcal{K}_\infty^n} h(x^{-t}\mathbf{v}) \chi_I(Q_0(\mathbf{v}))d\mathbf{v} =\sum_{z=r}^R q^{zn} J_{h'_z,Q_0}q^{b-2z}q^{(n-2)t}.\] In conclusion, if \(2t>\max\{-2r+B, N+B-b,-3r+B+K\}\), the asymptotic formula 44 holds if we take \[\label{eqn:J95Q} J_{h,Q_0}=\sum_{z=r}^R q^{z(n-2)}J_{h'_z,Q_0}.\tag{50}\] Let \(Q=Q_0\circ g\) for some \(g\in \mathrm{SL}_n(\mathcal{K}_\infty)\). Let \(h=\chi^{}_{g(\mathcal{O}^n\setminus x^{-1}\mathcal{O}^n)}\), so that for \[t_Q=\frac{1}{2}\max\{-2r+B,N+B-b,-3r+B+K\}\in \mathbb{N}\] and \(J_{Q}>0\) defined as in 50 , for every \(t\ge t_Q\), we have \[\mathrm{Vol}\left\{\mathbf{v}\in \mathcal{K}_\infty^n: \|\mathbf{v}\|=q^t,\; Q(\mathbf{v}) \in I \right\} =J_{Q}\mathrm{Vol}(I) q^{(n-2)t}.\] Let \[\mathrm{V}_{t_Q}:= \mathrm{Vol}\left\{\mathbf{v}\in \mathcal{K}_\infty^n: \|\mathbf{v}\|\le q^{t_Q},\; Q(\mathbf{v})\in I\right\}.\] For \(T> t_Q\), it follows that \[\begin{align} &\mathrm{Vol}\left\{\mathbf{v}\in \mathcal{K}_\infty^n: \|\mathbf{v}\|\le q^{t} ,\; Q(\mathbf{v})\in I\right\}\\ &=\mathrm{Vol}\left\{\mathbf{v}\in \mathcal{K}_\infty^n: q^{t_Q}<\|\mathbf{v}\|\le q^{t} ,\; Q(\mathbf{v})\in I\right\}+\mathrm{V}_{t_Q}\\ &=\sum_{t=t_Q+1}^T J_Q\mathrm{Vol}(I)q^{(n-2)t}+\mathrm{V}_{t_Q}\\ &=\frac{q^{n-2}J_Q}{q^{n-2}-1}\cdot \mathrm{Vol}(I)q^{(n-2)T}+\left(\mathrm{V}_{t_Q}-\frac{q^{n-2}J_Q}{q^{n-2}-1}\cdot \mathrm{Vol}(I)q^{(n-2)t_Q}\right) \end{align}\] Thus, we obtain the volume formula by taking \(c_Q=\frac{q^{n-2}J_Q}{q^{n-2}-1}\). ◻

Proof. For a fixed compact set \(\mathcal{C}\subseteq \mathrm{G}\) and \(t\in \mathbb{N}\), define \(\mathcal{B}_t\subseteq \mathcal{C}\) by \[\mathcal{B}_t=\left\{g\in \mathcal{C}:D\left(g\mathcal{R}^n,A_{g,I,t}\right)\geq \operatorname{Vol}(A_{g,I,t})^{\delta}\right\}.\] We prove that \(m_{\mathrm{G}}\left(\limsup_{t\rightarrow \infty}\mathcal{B}_t\right)=0\), so that for \(m_\mathrm{G}\)-almost every \(g\in \mathcal{C}\), we have \(D\left(g\mathcal{R}^n,A_{g,I_t,t}\right)<\operatorname{Vol}(A_{g,I_t,t})^{\delta}\) for every sufficiently large \(t\). Since this holds for any compact set \(\mathcal{C}\), this concludes the proof. Note that \[m_{\mathrm{G}}\left(\limsup_{t\rightarrow \infty}\mathcal{B}_t\right)=\lim_{m\rightarrow \infty}m_{\mathrm{G}}\left(\bigcup_{t\geq m}\mathcal{B}_t\right)\leq \lim_{m\rightarrow\infty}\sum_{t=m}^{\infty}m_{\mathrm{G}}\left(\mathcal{B}_t\right).\] We now prove that \(\sum_{t>0}m_{\mathrm{G}}(\mathcal{B}_t)<\infty\), so that the Borel Cantelli lemma implies that \(m_{\mathrm{G}}(\limsup \mathcal{B}_t)=0\). By Remark 18, there exist \(h_1,\dots,h_{\eta}\in \mathcal{C}\) such that \(\mathcal{C}\subseteq \bigcup_{i=1}^{\eta}h_i\operatorname{SL}_n(\mathcal{O})\). Thus, for every \(g\in \mathcal{C}\), there exists \(i\in\left\{1,\dots, \eta\right\}\), such that \(g\in h_i\operatorname{SL}_n(\mathcal{O})\). For \(i=1,\dots, \eta\), define \[\mathcal{B}_{t,i}:=\left\{g\in h_i\operatorname{SL}_n(\mathcal{O}):D\left(g\mathcal{R}^n,A_{h_i,I,t}\right)\geq \operatorname{Vol}(A_{h_i,I,t})^{\delta}\right\}.\] By Lemma 8 and Theorem 13, there exists \(t_{h_i,I}\), such that for every \(t>t_{h_i,I}\), we have \[\begin{align} \label{eqn:m40B95t44i41Bnd} m_{\mathrm{G}}(\mathcal{B}_{t,i})\ll_{h_i}\operatorname{Vol}(A_{h_i,I_t,t})^{1-2\delta} \ll_{h_i,I} q^{t(n-2)(1-2\delta)}\\ \end{align}\tag{51}\] for every large enough \(t\). Since \(\frac{1}{2}<\delta<1\), we have \(\sum_{t=0}^{\infty}m_{\mathrm{G}}(\mathcal{B}_{t,i})<\infty\). Since \(\mathcal{C}\subseteq \bigcup_{i=1}^{\eta}h_i\operatorname{SL}_n(\mathcal{O})\), it follows that \(\sum_{t=0}^{\infty}m_{\mathrm{G}}(\mathcal{B}_t)<\infty\), so that \(m_{\mathrm{G}}(\limsup_{t\rightarrow \infty}\mathcal{B}_t)=0\). ◻

Proof of Theorem 1. Let \(\frac{1}{2}<\delta<1\). By Theorem 13, Remark 18, and Theorem 19, for \(m_\mathrm{G}\) almost every \(g\), for \(Q=Q_0\circ g\), and for every sufficiently large \(t\), we have \[\begin{align} \label{eqn:N95Q40I95t44t41Cnt} &\left|N_Q(I,t)-c_Qq^{(n-2)t+n}m_{\mathcal{K}_{\infty}}(I)\right|\\ &\leq D(g\mathcal{R}^n,A_{g,I,t})+q^n\left|\operatorname{Vol}(A_{g,I,t})-c_Qq^{t(n-2)}m_{\mathcal{K}_{\infty}}(I)\right|\\ &<\operatorname{Vol}(A_{g,I,t})^{\delta}+O_{Q,I}(1)\\ &\ll \left(c_Qm_{\mathcal{K}_{\infty}}(I)q^{t(n-2)}+O_{Q,I}(1)\right)^{\delta}+O_{Q,I}(1)\\ &=O_{Q,I}\left(q^{\delta t(n-2)}m_{\mathcal{K}_{\infty}}(I)^{\delta}\right). \end{align}\tag{52}\]  ◻

Proof of Theorem 16. We need to prove that for any \(\mathbf{v}_1,\mathbf{v}_2\in \mathcal{K}_{\infty}^n\) with \(\Vert \mathbf{v}_i\Vert=q^z\) and \(Q(\mathbf{v}_i)=\alpha\) for \(i=1,2\), there exists \(k\in \mathrm{K}\) such that \(k\mathbf{v}_1=\mathbf{v}_2\). It suffices to assume that \(z=0\) and \(\mathbf{v}_1=\mathbf{e}_1\) and \(\mathbf{v}_2=\mathbf{v}\in \mathcal{O}_{\infty}^n\setminus \mathfrak{m}^n\), since all other cases follow directly from this case. We may assume that \(Q(y_1,\dots,y_n)\) has one of the following forms: \[\begin{align} u_1y_1^2+\dots+u_iy_i^2+x^{-1}(u_{i+1}y_{i+1}^2+\dots+u_ny_n^2)\text{ or }\\ y_1y_2+u_3y_3^2+\dots+u_iy_i^2+x^{-1}(u_{i+1}y_{i+1}^2+\dots+u_ny_n^2), \end{align}\] where \(u_i\in \mathcal{O}_{\infty}\setminus \mathfrak{m}\). Then for any \(\mathbf{w}\in \mathcal{K}_{\infty}^n\), we have \(Q(\mathbf{w})=\mathbf{w}^tY\mathbf{w}\), where \(Y=\begin{pmatrix} Y'&\\ &x^{-1}Y'' \end{pmatrix}\), where \(Y'\in \operatorname{GL}_i(\mathcal{K}_{\infty}),Y''\in \operatorname{GL}_{n-i}(\mathcal{K}_{\infty})\) are nondegenerate modulo \(x^{-1}\). The goal is to construct a matrix \(k\in \mathrm{K}\) satisfying

1. \(k\mathbf{e}_1=\mathbf{v}\) and

2. \(k^tYk=Y\),

where the second condition ensures that \(k\in \mathrm{K}\). Since \(\mathcal{O}_{\infty}\) is the inverse limit of \(\mathcal{R}_x/x^{-j}\mathcal{R}_x\), where \(\mathcal{R}_x:=\mathbb{F}_q[x^{-1}]\), we construct a sequence \(k^{(j)}\in \operatorname{GL}_n(\mathcal{R}_x/x^{-(j+1)}\mathcal{R}_x)\) satisfying

1. \(k^{(j)}\mathbf{e}_{1}=\mathbf{v}\mod x^{-(j+1)}\),

2. \({k^{(j)}}^t Y k^{(j)}=B\mod x^{-(j+1)}\), and

3. \(k^{(\ell)}=k^{(j+1)}\mod x^{-(j+1)}\) for every \(\ell \leq j\).

Then \(\lim_{\leftarrow} k_{\ell}\) will be an element satisfying [cond:Transitive] and [cond:StabQ]. In this case, we write \(k^{(j)}=\sum_{i=0}^jk_ix^{-i}\), where \(k_i\in \mathbb{F}_q\), and \(\mathbf{v}=\sum_{j=0}^{\infty}x^{-j}\mathbf{v}_j=(v_1,\dots,v_n)\), where \(\mathbf{v}_j\) is the projection of \(\mathbf{v}\) on \(\mathcal{R}_x/x^{-j}\mathcal{R}_x\).

5.0.0.1 Step 1: \(j=0\).

Let \(k^{(0)}=k_0=\begin{pmatrix} A_0&B_0\\ C_0&D_0 \end{pmatrix}\). Then, \(x^{-1}Y''\equiv 0\mod x^{-1}\), so that the matrix \(k_0\) should satisfy the following \[\begin{align} \label{eqn:WittStep0} \begin{pmatrix} Y'&\\ &0 \end{pmatrix}\equiv \begin{pmatrix} A_0^t&C_0^t\\ B_0^t&D_0^t \end{pmatrix}\begin{pmatrix} Y'&\\ &0 \end{pmatrix}\begin{pmatrix} A_0&B_0\\ C_0&D_0 \end{pmatrix}\mod x^{-1}\\ =\begin{pmatrix} A_0^tY'A_0&A_0^tY'B_0\\ B_0^tY'A_0&B_0^tY'B_0 \end{pmatrix}\mod x^{-1}. \end{align}\tag{53}\] Let \(\operatorname{pr}_i:(\mathcal{R}_x/x^{-1}\mathcal{R}_x)^n\rightarrow (\mathcal{R}_x/x^{-1}\mathcal{R}_x)^i\) be the projection onto the first \(i\) coordinates, that is \(\operatorname{pr}_i(\alpha_1,\dots,\alpha_n)=(\alpha_1,\dots,\alpha_i)\). By assumption, \(Q(\operatorname{pr}_i(\mathbf{e}_1))=Q(\operatorname{pr}_i(\mathbf{v}_0))\mod x^{-1}\). Note that \(\mathcal{R}_x/x^{-1}\mathcal{R}_x\cong \mathbb{F}_q\). Therefore, by applying Witt’s theorem for finite fields [29] to \(\left((\mathcal{R}_x/x^{-1}\mathcal{R}_x)^i,Y'\right)\), we obtain an isometry \(A_0\), whose first column is \(\operatorname{pr}_i(\mathbf{v}_0)\), and satisfies \(A_0^tY'A_0=Y'\mod x^{-1}\). Since \(A_0^tY'\) is invertible, by 53 , \(B_0=0\). Similarly to [11], at this step, we cannot determine \(C_0\) and \(D_0\), as they do not appear in the right hand side of 53 .

5.0.0.2 Step 2: \(j=1\).

Let \(k^{(1)}=\begin{pmatrix} A_0+xA_1&B_0+xB_1\\ C_0+xC_1&D_0+xD_1 \end{pmatrix} \in \mathcal{R}_x/x^{-2}\mathcal{R}_x\) be a matrix whose first column is \(\mathbf{v}_0+x^{-1}\mathbf{v}_1\) and satisfies \((k^{(1)})^tYk^{(1)}=Y\). Hence, \[\begin{align} A_0^tY'A_0+x(A_1^tY'A_0+A_0^tY'A_1+C_0^tY''C_0)\equiv Y'\mod x^{-2},\\ A_0^tY'B_0^t+x\left[A_1^tY'B_0+A_0^tY'B_1+C_0^tY''D_0\right]\equiv 0\mod x^{-2},\\ B_0^tY'B_0+x\left[B_1^tY'B_0+B_0^tY'B_1+D_0^tY''D_0\right]\equiv xY''\mod x^{-2}. \end{align}\] Since \(B_0=0\) and \(A_0^tY'A_0=Y'+xZ_1^{11}\mod x^2\) for some symmetric matrix \(Z_1^{11}\), then, we have \[\begin{align} \label{eqn:x94123-1125CONDS} Z_1^{11}+A_1^tY'A_0+A_0^tY'A_1+C_0^tY''C_0\equiv 0 \mod x^{-1},\\ A_0^tY'B_1+C_0^tY''D_0\equiv 0 \mod x^{-1},\\ D_0^tY''D_0\equiv Y''\mod x^{-1}. \end{align}\tag{54}\] Let \(C_0\in M_{n-i\times i}(\mathcal{R}_x/x^{-1}\mathcal{R}_x)\) be some matrix whose first column is \((v_{i+1},\dots,v_n)\). By Witt’s theorem for finite fields [29], there exists a matrix \(D_0\in M_{n-i\times n-i}(\mathcal{R}_x/x^{-1}\mathcal{R}_x)\) satisfying \(D_0^tY''D_0=Y''\mod x^{-1}\). Hence, it suffices to prove that there exists \(A_1\in M_{i\times i}(\mathcal{R}_x/x^{-1}\mathcal{R}_x)\) such that \(Z_1^{11}+A_1^tY'A_0+A_0^tY'A_1+C_0^tY''C_0\equiv 0\mod x^{-1}\).

5.0.0.3 Step 3

We now prove the following claim to show that \[Z_1^{11}+A_1^tY'A_0+A_0^tY'A_1+C_0^tY''C_0\equiv 0 \mod x^{-1}\] has a solution \(A_1\).

Hence, by applying Claim 20, with \(C=-Z_1^{11}-C_0^tY''C_0\), the equation \(Z_1^{11}+A_1^tY'A_0+A_0^tY'A_1+C_0^tY''C_0\equiv 0\mod x^{-1}\) has a solution \(A_1\), where \(Z_{1}^{11},Y',Y'',A_0,C_0\) and the first column of \(A_1\) are all known.

5.0.0.4 Step 4: General Case

Assume that there exists some solution \(k=\sum_{j=0}^{\infty}k_jx^{-j}\) which satisfies [cond:Transitive] and [cond:StabQ]. Since \(k^tYk=Y\), \[\begin{align} \begin{pmatrix} Y'&\\ &x^{-1}Y'' \end{pmatrix}=\sum_{j=0}^{\infty}x^{-j}\left(\sum_{i=0}^j\begin{pmatrix} A_i^t&C_i^t\\ B_i^t&D_i^t \end{pmatrix}\begin{pmatrix} Y'&\\ &x^{-1}Y'' \end{pmatrix}\begin{pmatrix} A_i&B_i\\ C_i&D_i \end{pmatrix}\right)\\ =\sum_{j=0}^{\infty}x^{-j}\left(\sum_{i=0}^j\begin{pmatrix} A_i^tY'A_{j-i}&A_i^tY'B_{j-i}\\ B_i^tY'A_{j-i}&B_i^tY'B_{j-i} \end{pmatrix}\right)\\ +\sum_{j=0}^{\infty}x^{-(j+1)}\left(\sum_{i=0}^j\begin{pmatrix} C_i^tY''C_{j-i}&C_i^tY''D_{j-i}\\ D_i^tY''C_{j-i}&D_i^tY''D_{j-i} \end{pmatrix}\right). \end{align}\] We now find \(A_j,B_j,C_j,D_j\) inductively using step 3. Let \(C_{j-1}\) be a matrix whose first column equal to \((v_{j-1}^{(i)},\dots,v_{j-1}^{(n)})\), where \(\mathbf{v}=\sum_{\ell=0}^{\infty}x^{-\ell}\mathbf{v}_\ell\) and \(\mathbf{v}_\ell=(v_\ell^{(1)},\dots,v_{\ell}^{(n)})^t\in \mathbb{F}_q^n\). Then by step 3 and the fact that \(\mathbf{v}^tY\mathbf{v}=\sum_{k=0}^jx^{-k}\left(\sum_{i=0}^k\mathbf{v}_i^tY\mathbf{v}_{k-i}\right)\mod x^{-(j+1)}\), there exist \(A_j,D_{j-1},B_j\) which satisfy the following equations: \[\begin{align} A_0^tY'A_j+A_j^tY'A_0\equiv -\sum_{i=1}^{j-1}A_i^tY'A_{j-i}-\sum_{i=0}^{j-1}C_i^tY''C_{j-i-1}+Z_j^{11}\mod x^{-1}\\ D_{j-1}^tY''D_0+D_0^tY''D_{j-1}=-\sum_{i=1}^{j-1}\left(D_i^tY''W_{j-i-1}+B_i^tY'B_{j-i}\right)+Z_n^{22}\mod x^{-1}\\ A_0^tYB_j=-\sum_{i=0}^{j-1}\left(A_i^tY'B_{j-i}+C_i^tY''D_{j-i-1}\right)+Z_j^{12}\mod x^{-1}, \end{align}\] where \(Z_j^{11},Z_j^{22},\) and \(Z_j^{12}\) are obtained from step 2 and step \(j-1\). Hence, there exists \(k\in \mathrm{K}\) such that \(k^tYk=Y\) and \(k\mathbf{e}_1=\mathbf{v}\). ◻