Proof. By definition, $a=(a_{ij})_{1\leqslant i,j\leqslant n}\in \operatorname{GL}(n,{\mathcal{O}}_{F})$ implies that $a_{ij}\in {\mathcal{O}}_{F}$ and $|\text{det}(a)|=1$ . Now take any $x\in \operatorname{Mat}(n,\unicode[STIX]{x1D71B}{\mathcal{O}}_{F})$ . First, we have $a+x\in \operatorname{Mat}(n,{\mathcal{O}}_{F})$ . Second, write $x=\unicode[STIX]{x1D71B}y$ with $y\in \operatorname{Mat}(n,{\mathcal{O}}_{F})$ . By definition, there exists $z\in {\mathcal{O}}_{F}$ , such that $\det (a+x)=\det (a)+\unicode[STIX]{x1D71B}z$ . Since $|\unicode[STIX]{x1D71B}z|\leqslant q^{-1}$ , by ultrametricity, we obtain $|\text{det}(a+x)|=|\text{det}(a)+\unicode[STIX]{x1D71B}z|=1$ , whence $a+x\in \operatorname{GL}(n,{\mathcal{O}}_{F})$ and the set on the right-hand side of (16) is contained in $\operatorname{GL}(n,{\mathcal{O}}_{F})$ . Conversely, since ${\mathcal{C}}_{q}$ is a complete set of representatives of the cosets of $\unicode[STIX]{x1D71B}{\mathcal{O}}_{F}$ in ${\mathcal{O}}_{F}$ , for any $A\in \operatorname{GL}(n,{\mathcal{O}}_{F})$ , there exists a unique $t\in \operatorname{GL}(n,{\mathcal{C}}_{q})$ , such that $A\equiv t(\operatorname{mod}\unicode[STIX]{x1D71B}{\mathcal{O}}_{F})$ . This completes the proof of (16).

Recalling that

$$\begin{eqnarray}\displaystyle \#\text{GL}(n,{\mathcal{C}}_{q})=\#\text{GL}(n,\mathbf{F}_{q})=\mathop{\prod }_{j=0}^{n-1}(q^{n}-q^{j}), & & \displaystyle \nonumber\end{eqnarray}$$

we arrive at (17). ◻

Proof. The proof follows immediately from Proposition 2.2 and the formula (18) for the normalized Haar measure on $\operatorname{GL}(n,{\mathcal{O}}_{F})$ .◻

Proof. Since $F$ is non-dyadic, we have $|2|=1$ . For any $x=\unicode[STIX]{x1D6FC}^{2}\in ({\mathcal{O}}_{F}^{\times })^{2}$ , since ${\mathcal{C}}_{q}$ is a complete set of representatives for ${\mathcal{O}}_{F}/\unicode[STIX]{x1D71B}{\mathcal{O}}_{F}$ , there exists $a\in {\mathcal{C}}_{q}^{\times }$ , such that $x\equiv a(\operatorname{mod}\unicode[STIX]{x1D71B}{\mathcal{O}}_{F})$ , that is,

(22) $$\begin{eqnarray}\displaystyle |x-a|<1. & & \displaystyle\end{eqnarray}$$

Take any $b\in {\mathcal{O}}_{F}^{\times }$ such that $|b-x|<1=|2\unicode[STIX]{x1D6FC}|^{2}$ . Then the polynomial $P_{b}(X)=X^{2}-b\in {\mathcal{O}}_{F}[X]$ satisfies

(23) $$\begin{eqnarray}\displaystyle |P_{b}(\unicode[STIX]{x1D6FC})|<|P_{b}^{\prime }(\unicode[STIX]{x1D6FC})|^{2}. & & \displaystyle\end{eqnarray}$$

By Hensel’s lemma (see Cassels [Reference CasselsCas86, pp. 49–51]), the inequality (23) implies that there exists $\unicode[STIX]{x1D6FD}\in F$ such that

(24) $$\begin{eqnarray}\displaystyle P_{b}(\unicode[STIX]{x1D6FD})=\unicode[STIX]{x1D6FD}^{2}-b=0\quad \text{and}\quad |\unicode[STIX]{x1D6FD}-\unicode[STIX]{x1D6FC}|\leqslant \frac{|P_{b}(\unicode[STIX]{x1D6FC})|}{|P_{b}^{\prime }(\unicode[STIX]{x1D6FC})|}<|P_{b}^{\prime }(\unicode[STIX]{x1D6FC})|=1. & & \displaystyle\end{eqnarray}$$

In particular, we have

(25) $$\begin{eqnarray}\displaystyle \{b\in {\mathcal{O}}_{F}^{\times }:|b-x|<1\}=x+\unicode[STIX]{x1D71B}{\mathcal{O}}_{F}\subset ({\mathcal{O}}_{F}^{\times })^{2}. & & \displaystyle\end{eqnarray}$$

Combining (22) and (25), we get $a\in ({\mathcal{O}}_{F}^{\times })^{2}$ . Hence

$$\begin{eqnarray}\displaystyle ({\mathcal{O}}_{F}^{\times })^{2}\subset \bigsqcup _{a\in ({\mathcal{C}}_{q}^{\times })^{2}}(a+\unicode[STIX]{x1D71B}{\mathcal{O}}_{F}). & & \displaystyle \nonumber\end{eqnarray}$$

Conversely, since $({\mathcal{C}}_{q}^{\times })^{2}\subset ({\mathcal{O}}_{F}^{\times })^{2}$ , for any $a\in ({\mathcal{C}}_{q}^{\times })^{2}$ , replacing $x$ by $a$ in the above argument, the inclusion (25) implies $a+\unicode[STIX]{x1D71B}{\mathcal{O}}_{F}\subset ({\mathcal{O}}_{F}^{\times })^{2}$ . Hence

$$\begin{eqnarray}\displaystyle \text{}\bigsqcup _{a\in ({\mathcal{C}}_{q}^{\times })^{2}}(a+\unicode[STIX]{x1D71B}{\mathcal{O}}_{F})\subset ({\mathcal{O}}_{F}^{\times })^{2}.\Box & & \displaystyle \nonumber\end{eqnarray}$$

Proof. Since ${\mathcal{T}}$ is a complete set of representatives for the quotient $F/({\mathcal{O}}_{F}^{\times })^{2}$ , it suffices to show that any symmetric matrix $A\in \operatorname{Sym}(n,F)$ is diagonalizable. Assume that $A$ is not a zero matrix. We claim that, up to passing $A$ to $gAg^{t}$ for some $g\in \operatorname{GL}(n,{\mathcal{O}}_{F})$ , we may assume that the $(1,1)$ -coefficient of $A$ has maximal absolute value.

Case 1: Assume first that there exists $1\leqslant i_{0}\leqslant n$ , such that $|A_{i_{0}i_{0}}|=\max _{1\leqslant i,j\leqslant n}|A_{ij}|$ . If $i_{0}=1$ , then there is nothing to prove. Otherwise, take $g=M_{(1i_{0})}$ , where $M_{(1i_{0})}$ is the permutation matrix associated with the transposition $(1i_{0})$ . Then the matrix $gAg^{t}$ has $A_{i_{0}i_{0}}$ as its $(1,1)$ -coefficient.

Case 2: Now assume that there exists $i_{0}<j_{0}$ such that

(27) $$\begin{eqnarray}\displaystyle |A_{i_{0}j_{0}}|=\max _{1\leqslant i,j\leqslant n}|A_{ij}|>\max _{1\leqslant i\leqslant n}|A_{ii}|. & & \displaystyle\end{eqnarray}$$

Let us do operations on the submatrix indexed by $\{i_{0},j_{0}\}\times \{i_{0},j_{0}\}$ as follows:

$$\begin{eqnarray}\displaystyle \left[\begin{array}{@{}cc@{}}2A_{i_{0}j_{0}}+A_{i_{0}i_{0}}+A_{j_{0}j_{0}} & A_{i_{0}j_{0}}+A_{j_{0}j_{0}}\\ A_{i_{0}j_{0}}+A_{j_{0}j_{0}} & A_{j_{0}j_{0}}\end{array}\right]=\left[\begin{array}{@{}cc@{}}1 & 1\\ 0 & 1\end{array}\right]\left[\begin{array}{@{}cc@{}}A_{i_{0}i_{0}} & A_{i_{0}j_{0}}\\ A_{i_{0}j_{0}} & A_{j_{0}j_{0}}\end{array}\right]\left[\begin{array}{@{}cc@{}}1 & 0\\ 1 & 1\end{array}\right]. & & \displaystyle \nonumber\end{eqnarray}$$

Since $F$ is non-dyadic and non-Archimedean, (27) implies

$$\begin{eqnarray}\displaystyle |2A_{i_{0}j_{0}}+A_{i_{0}i_{0}}+A_{j_{0}j_{0}}|=|A_{i_{0}j_{0}}+A_{j_{0}j_{0}}|=|A_{i_{0}j_{0}}|. & & \displaystyle \nonumber\end{eqnarray}$$

This shows that we can reduce the second case to the first case where a diagonal coefficient has maximal absolute value.

Now assume that

$$\begin{eqnarray}\displaystyle A=\left[\begin{array}{@{}cc@{}}x & c^{t}\\ c & A_{1}\end{array}\right],\quad c\in F^{n-1}, & & \displaystyle \nonumber\end{eqnarray}$$

such that $x$ attains the maximal absolute value of all coefficients of $A$ . Then $x^{-1}c\in {\mathcal{O}}_{F}^{n-1}$ and we have

$$\begin{eqnarray}\displaystyle \left[\begin{array}{@{}cc@{}}1 & 0\\ -x^{-1}c & 1\end{array}\right]\left[\begin{array}{@{}cc@{}}x & c^{t}\\ c & A_{1}\end{array}\right]\left[\begin{array}{@{}cc@{}}1 & -x^{-1}c^{t}\\ 0 & 1\end{array}\right]=\left[\begin{array}{@{}cc@{}}x & 0\\ 0 & A_{1}-x^{-1}cc^{t}\end{array}\right]. & & \displaystyle \nonumber\end{eqnarray}$$

By continuing the above procedure on the submatrix $A_{1}-x^{-1}cc^{t}$ , we prove finally that $A$ is diagonalizable.◻

Proof. First assume that $y\in \unicode[STIX]{x1D71B}^{-l}{\mathcal{O}}_{F}$ . Then for any $x\in \unicode[STIX]{x1D71B}^{l}{\mathcal{O}}_{F}$ , we have $xy\in {\mathcal{O}}_{F}$ . Consequently, by (28), we have $\unicode[STIX]{x2A0D}_{\unicode[STIX]{x1D71B}^{l}{\mathcal{O}}_{F}}\unicode[STIX]{x1D712}(xy)\,dx=1.$ Now assume that $y\notin \unicode[STIX]{x1D71B}^{-l}{\mathcal{O}}_{F}$ . Since the Haar measure $d\text{vol}$ on $F$ is invariant under the multiplication action by any element $u\in {\mathcal{O}}_{F}^{\times }$ , without loss of generality, we may assume that $y=\unicode[STIX]{x1D71B}^{k}$ with $k\leqslant -l-1$ . By (33), we have

$$\begin{eqnarray}\displaystyle \unicode[STIX]{x2A0D}_{\unicode[STIX]{x1D71B}^{l}{\mathcal{O}}_{F}}\unicode[STIX]{x1D712}(xy)\,dx=\unicode[STIX]{x2A0D}_{\unicode[STIX]{x1D71B}^{l}{\mathcal{O}}_{F}}\unicode[STIX]{x1D712}(\unicode[STIX]{x1D71B}^{k}x)\,dx=\unicode[STIX]{x2A0D}_{{\mathcal{O}}_{F}}\unicode[STIX]{x1D712}(\unicode[STIX]{x1D71B}^{k+l}z)\,dz=0. & & \displaystyle \nonumber\end{eqnarray}$$

This completes the proof of (34). ◻

Proof. Write $X=(X_{i}^{(1)})_{i=1}^{m}$ . It suffices to prove that for any $A\in \operatorname{GL}(m,{\mathcal{O}}_{F})$ and any $y=(y_{1},\ldots ,y_{m})\in F^{m}$ , we have

(35) $$\begin{eqnarray}\displaystyle \mathbb{E}[\unicode[STIX]{x1D712}(X\cdot y)]=\mathbb{E}[\unicode[STIX]{x1D712}((AX)\cdot y)]. & & \displaystyle\end{eqnarray}$$

By the independence between $X_{i}^{(1)}:i=1,\ldots ,m$ and Lemma 3.3, we have

$$\begin{eqnarray}\displaystyle \mathbb{E}[\unicode[STIX]{x1D712}(X\cdot y)]=\mathop{\prod }_{j=1}^{n}\mathbb{E}[\unicode[STIX]{x1D712}(X_{j}y_{j})]=\mathop{\prod }_{j=1}^{n}\unicode[STIX]{x1D7D9}_{{\mathcal{O}}_{F}}(y_{j})=\unicode[STIX]{x1D7D9}_{{\mathcal{O}}_{F}^{m}}(y). & & \displaystyle \nonumber\end{eqnarray}$$

Similarly,

$$\begin{eqnarray}\displaystyle \mathbb{E}[\unicode[STIX]{x1D712}((AX)\cdot y)]=\mathbb{E}[\unicode[STIX]{x1D712}(X\cdot (A^{t}y))]=\unicode[STIX]{x1D7D9}_{{\mathcal{O}}_{F}^{m}}(A^{t}y)=\unicode[STIX]{x1D7D9}_{{\mathcal{O}}_{F}^{m}}(y). & & \displaystyle \nonumber\end{eqnarray}$$

Hence we get (35). The proof of Lemma 3.4 is completed. ◻

Proof of Proposition 3.1.

It suffices to prove that the following probability measures

$$\begin{eqnarray}\displaystyle {\mathcal{L}}[X_{i}^{(1)}Y_{j}^{(1)}]_{i,j\in \mathbb{N}}\quad \text{and}\quad {\mathcal{L}}([Z_{ij}]_{i,j\in \mathbb{N}}) & & \displaystyle \nonumber\end{eqnarray}$$

are $\operatorname{GL}(\infty ,{\mathcal{O}}_{F})\times \operatorname{GL}(\infty ,{\mathcal{O}}_{F})$ -invariant. The invariance of both measures follows immediately from Lemma 3.4.◻

Proof. Let $\unicode[STIX]{x1D707}\in {\mathcal{P}}_{\text{inv}}(\operatorname{Mat}(\mathbb{N},F))$ . For any $\unicode[STIX]{x1D70E}\in S(\infty )$ , the associated permutation matrix $M_{\unicode[STIX]{x1D70E}}$ , defined by

$$\begin{eqnarray}\displaystyle M_{\unicode[STIX]{x1D70E}}(i,j):=1_{\unicode[STIX]{x1D70E}(i)=j}, & & \displaystyle \nonumber\end{eqnarray}$$

is an element in $\operatorname{GL}(\infty ,{\mathcal{O}}_{F})$ . By the invariance of $\unicode[STIX]{x1D707}$ under the multiplication by all permutation matrices $M_{\unicode[STIX]{x1D70E}},\unicode[STIX]{x1D70E}\in S(\infty )$ on left and on right, it is easy to see that $\unicode[STIX]{x1D6F9}(\unicode[STIX]{x1D707})\in {\mathcal{P}}_{\text{inv}}^{S(\infty )}(F^{\mathbb{N}})$ .

Now we show that the map (36) is injective. By the definition of pushforward map, the Fourier transform of $\unicode[STIX]{x1D6F9}(\unicode[STIX]{x1D707})$ is given as follows: for any $x_{1},\ldots ,x_{r}\in F$ ,

(37) $$\begin{eqnarray}\displaystyle \widehat{\unicode[STIX]{x1D6F9}(\unicode[STIX]{x1D707})}((x_{1},\ldots ,x_{r},0,0,\ldots ))=\widehat{\unicode[STIX]{x1D707}}(\operatorname{diag}(x_{1},\ldots ,x_{r},0,0,\ldots )). & & \displaystyle\end{eqnarray}$$

By Remark 2.9, $\widehat{\unicode[STIX]{x1D707}}$ is determined by

$$\begin{eqnarray}\displaystyle \widehat{\unicode[STIX]{x1D707}}(\operatorname{diag}(x_{1},\ldots ,x_{r},0,0,\ldots )),\quad x_{1},\ldots ,x_{r}\in F. & & \displaystyle \nonumber\end{eqnarray}$$

Now, the equality (37) implies that $\widehat{\unicode[STIX]{x1D707}}$ and hence $\unicode[STIX]{x1D707}$ itself is determined uniquely by $\unicode[STIX]{x1D6F9}(\unicode[STIX]{x1D707})$ . The injectivity of the map (36) is proved.◻

Proof of Theorem 3.5.

By construction, for any $\Bbbk \in \unicode[STIX]{x1D6E5}$ , the measure $\unicode[STIX]{x1D6F9}(\unicode[STIX]{x1D707}_{\Bbbk })$ is Bernoulli. Indeed, the definition

$$\begin{eqnarray}\unicode[STIX]{x1D707}_{\Bbbk }={\mathcal{L}}\biggl(\biggl[\mathop{\sum }_{n:k_{n}>k}\unicode[STIX]{x1D71B}^{-k_{n}}X_{i}^{(n)}Y_{j}^{(n)}+\unicode[STIX]{x1D71B}^{-k}Z_{ij}\biggr]_{i,j\in \mathbb{N}}\biggr)\end{eqnarray}$$

implies that the diagonal marginal measure $\unicode[STIX]{x1D6F9}(\unicode[STIX]{x1D707}_{\Bbbk })$ is

$$\begin{eqnarray}\unicode[STIX]{x1D6F9}(\unicode[STIX]{x1D707}_{\Bbbk })={\mathcal{L}}\biggl(\biggl(\mathop{\sum }_{n:k_{n}>k}\unicode[STIX]{x1D71B}^{-k_{n}}X_{i}^{(n)}Y_{i}^{(n)}+\unicode[STIX]{x1D71B}^{-k}Z_{ii}\biggr)_{i\in \mathbb{N}}\biggr),\end{eqnarray}$$

and our original assumption, that all $X_{i}^{(n)},Y_{i}^{(n)},Z_{ii}$ are independent sampled with respect to the normalized Haar measure on the compact additive group ${\mathcal{O}}_{F}$ , implies that the measure $\unicode[STIX]{x1D6F9}(\unicode[STIX]{x1D707}_{\Bbbk })$ is Bernoulli. Consequently, by the De Finetti Theorem, $\unicode[STIX]{x1D6F9}(\unicode[STIX]{x1D707}_{\Bbbk })$ is $S(\infty )$ -ergodic. Therefore, by Remark 3.7,

$$\begin{eqnarray}\displaystyle \unicode[STIX]{x1D6F9}(\unicode[STIX]{x1D707}_{\Bbbk })\in {\mathcal{P}}_{\text{erg}}^{S(\infty )}(F^{\mathbb{N}})=\text{Ext}({\mathcal{P}}_{\text{inv}}^{S(\infty )}(F^{\mathbb{N}})). & & \displaystyle \nonumber\end{eqnarray}$$

It is clear that for any convex subset $C$ of $\text{Ext}({\mathcal{P}}_{\text{inv}}^{S(\infty )}(F^{\mathbb{N}}))$ , we have

$$\begin{eqnarray}C\cap \text{Ext}({\mathcal{P}}_{\text{inv}}^{S(\infty )}(F^{\mathbb{N}}))\subset \text{Ext}(C).\end{eqnarray}$$

Taking $C=\unicode[STIX]{x1D6F9}({\mathcal{P}}_{\text{inv}}(\operatorname{Mat}(\mathbb{N},F)))$ , we see that

(38) $$\begin{eqnarray}\displaystyle \unicode[STIX]{x1D6F9}(\unicode[STIX]{x1D707}_{\Bbbk })\in \text{Ext}(\unicode[STIX]{x1D6F9}({\mathcal{P}}_{\text{inv}}(\operatorname{Mat}(\mathbb{N},F)))). & & \displaystyle\end{eqnarray}$$

By Lemma 3.6, $\unicode[STIX]{x1D6F9}$ is an affine embedding. Hence

$$\begin{eqnarray}\displaystyle \text{Ext}(\unicode[STIX]{x1D6F9}({\mathcal{P}}_{\text{inv}}(\operatorname{Mat}(\mathbb{N},F))))=\unicode[STIX]{x1D6F9}(\text{Ext}({\mathcal{P}}_{\text{inv}}(\operatorname{Mat}(\mathbb{N},F)))). & & \displaystyle \nonumber\end{eqnarray}$$

The relation (38) implies $\unicode[STIX]{x1D707}_{\Bbbk }\in \text{Ext}({\mathcal{P}}_{\text{inv}}(\operatorname{Mat}(\mathbb{N},F)))$ . Since indecomposability implies ergodicity, we get the desired relation $\unicode[STIX]{x1D707}_{\Bbbk }\in {\mathcal{P}}_{\text{erg}}(\operatorname{Mat}(\mathbb{N},F))$ . ◻

Proof. For any $g\in \operatorname{GL}(n,{\mathcal{O}}_{F})$ , the linear map

(39) $$\begin{eqnarray}\displaystyle \left.\begin{array}{@{}rcl@{}}\operatorname{Sym}(n,{\mathcal{O}}_{F})\ & \longrightarrow \ & \operatorname{Sym}(n,{\mathcal{O}}_{F})\\ S\ & \mapsto \ & gSg^{t}\end{array}\right. & & \displaystyle\end{eqnarray}$$

is invertible. Clearly, if we use the group identification

$$\begin{eqnarray}\operatorname{Sym}(n,{\mathcal{O}}_{F})\simeq {\mathcal{O}}_{F}^{(n^{2}+n)/2},\end{eqnarray}$$

then the linear map (39) is represented by an invertible matrix from $\operatorname{GL}((n^{2}+n)/2,{\mathcal{O}}_{F})$ . Hence by Remark 2.1, it preserves the normalized Haar measure.◻

Proof of Proposition 3.8.

It suffices to prove the $\operatorname{GL}(\infty ,{\mathcal{O}}_{F})$ -invariance of the following probability measures on $\operatorname{Sym}(\mathbb{N},F)$ :

$$\begin{eqnarray}\displaystyle {\mathcal{L}}([X_{i}^{(1)}X_{j}^{(1)}]_{i,j\in \mathbb{N}})\quad \text{and}\quad {\mathcal{L}}([H_{ij}]_{i,j\in \mathbb{N}}). & & \displaystyle \nonumber\end{eqnarray}$$

The $\operatorname{GL}(\infty ,{\mathcal{O}}_{F})$ -invariance of ${\mathcal{L}}([X_{i}^{(1)}X_{j}^{(1)}]_{i,j\in \mathbb{N}})$ follows immediately from Lemma 3.4, while the $\operatorname{GL}(\infty ,{\mathcal{O}}_{F})$ -invariance of ${\mathcal{L}}([H_{ij}]_{i,j\in \mathbb{N}})$ follows from Lemma 3.9.◻

Proof. The proof is similar to that of Lemma 3.6. ◻

Proof of Theorem 3.10.

The proof is similar to that of Theorem 3.5 by using Lemma 3.11 instead of Lemma 3.6. ◻

Proof. Let $|x|=q^{\ell }$ . Then there exists $u\in {\mathcal{O}}_{F}^{\times }$ , such that $x=\unicode[STIX]{x1D71B}^{-\ell }u$ . By rotation invariance, $\unicode[STIX]{x1D6E9}(x)=\unicode[STIX]{x1D6E9}(\unicode[STIX]{x1D71B}^{-\ell })$ . Now by Lemma 3.3,

$$\begin{eqnarray}\displaystyle \unicode[STIX]{x1D6E9}(\unicode[STIX]{x1D71B}^{-\ell }) & = & \displaystyle \mathbb{E}[\unicode[STIX]{x1D712}(\unicode[STIX]{x1D71B}^{-\ell }X_{1}^{(1)}Y_{1}^{(1)})]=\mathbb{E}(\mathbb{E}[\unicode[STIX]{x1D712}(\unicode[STIX]{x1D71B}^{-\ell }X_{1}^{(1)}Y_{1}^{(1)})\mid Y_{1}^{(1)}])\nonumber\\ \displaystyle & = & \displaystyle \mathbb{E}(\unicode[STIX]{x1D7D9}_{{\mathcal{O}}_{F}}(\unicode[STIX]{x1D71B}^{-\ell }Y_{1}^{(1)}))=\mathbb{P}(|Y_{1}^{(1)}|\leqslant q^{-\ell })=q^{-\ell \unicode[STIX]{x1D7D9}_{\{\ell \geqslant 1\}}}.\Box \nonumber\end{eqnarray}$$

Proof of Proposition 4.1.

The identity (41) follows from the independence between all diagonal coefficient of $M_{\Bbbk }$ . So we only need to prove the identity (40).

First assume that $\Bbbk =(k_{n})_{n\in \mathbb{N}}\in \unicode[STIX]{x1D6E5}$ is such that $\lim _{n\rightarrow \infty }k_{n}=-\infty$ . By the independence between all $X_{1}^{(n)}$ and $Y_{1}^{(n)},n\in \mathbb{N}$ , we have

$$\begin{eqnarray}\displaystyle \widehat{\unicode[STIX]{x1D707}_{\Bbbk }}(\unicode[STIX]{x1D71B}^{-\ell }e_{11}) & = & \displaystyle \mathbb{E}\biggl[\unicode[STIX]{x1D712}\biggl(\operatorname{tr}\hspace{-0.8pt}\biggl(\mathop{\biggl[\mathop{\sum }_{n=1}^{\infty }\unicode[STIX]{x1D71B}^{-k_{n}}X_{i}^{(n)}Y_{j}^{(n)}\biggr]}\nolimits_{i,j\in \mathbb{N}}\unicode[STIX]{x1D71B}^{-\ell }e_{11}\biggr)\biggr)\biggr]\nonumber\\ \displaystyle & = & \displaystyle \mathbb{E}\biggl[\unicode[STIX]{x1D712}\biggl(\mathop{\sum }_{n=1}^{\infty }\unicode[STIX]{x1D71B}^{-k_{n}-\ell }X_{1}^{(n)}Y_{1}^{(n)}\biggr)\biggr]=\mathop{\prod }_{n=1}^{\infty }\mathbb{E}[\unicode[STIX]{x1D712}(\unicode[STIX]{x1D71B}^{-k_{n}-\ell }X_{1}^{(n)}Y_{1}^{(n)})].\nonumber\end{eqnarray}$$

By Lemma 4.2, we get

$$\begin{eqnarray}\displaystyle \widehat{\unicode[STIX]{x1D707}_{\Bbbk }}(\unicode[STIX]{x1D71B}^{-\ell }e_{11}) & = & \displaystyle \mathop{\prod }_{n=1}^{\infty }\exp (-\text{log}\,q\cdot (k_{n}+\ell )\unicode[STIX]{x1D7D9}_{\{k_{n}+\ell \geqslant 1\}})\nonumber\\ \displaystyle & = & \displaystyle \exp \biggl(-\text{log}\,q\cdot \mathop{\sum }_{n=1}^{\infty }(k_{n}+\ell )\unicode[STIX]{x1D7D9}_{\{k_{n}+\ell \geqslant 1\}}\biggr).\nonumber\end{eqnarray}$$

Now assume that there exists $m\in \mathbb{N}\cup \{0\}$ and $k\in \mathbb{Z}$ , such that

$$\begin{eqnarray}\displaystyle k_{1}\geqslant \cdots \geqslant k_{m}>k\quad \text{and}\quad k_{n}=k\text{ for any }n\geqslant m+1. & & \displaystyle \nonumber\end{eqnarray}$$

By previous computation and the formula (32) for the characteristic functions of convolutions of probability measures, we only need to consider the case when $\Bbbk =(k_{n})_{n\in \mathbb{N}}$ is such that

$$\begin{eqnarray}\displaystyle k_{n}=k\in \mathbb{Z}\quad \text{for any }n\in \mathbb{N}. & & \displaystyle \nonumber\end{eqnarray}$$

In this case, $\unicode[STIX]{x1D707}_{\Bbbk }={\mathcal{L}}(\unicode[STIX]{x1D71B}^{-k}Z)$ with $Z$ an infinite random matrix sampled uniformly from $\operatorname{Mat}(\mathbb{N},{\mathcal{O}}_{F})$ . Hence by Lemma 3.3, we obtain

$$\begin{eqnarray}\displaystyle \widehat{\unicode[STIX]{x1D707}_{\Bbbk }}(\unicode[STIX]{x1D71B}^{-\ell }e_{11})=\mathbb{E}[\unicode[STIX]{x1D712}(\unicode[STIX]{x1D71B}^{-\ell -k}Z_{11})]=\unicode[STIX]{x1D7D9}_{{\mathcal{O}}_{F}}(\unicode[STIX]{x1D71B}^{-\ell -k})=\unicode[STIX]{x1D7D9}_{\{k+\ell \leqslant 0\}}. & & \displaystyle \nonumber\end{eqnarray}$$

But if $k_{n}=k$ for any $n\in \mathbb{N}$ , we have

(43) $$\begin{eqnarray}\displaystyle \unicode[STIX]{x1D7D9}_{\{k+\ell \leqslant 0\}}=\exp \biggl(-\text{log}\,q\cdot \mathop{\sum }_{n=1}^{\infty }(k_{n}+\ell )\unicode[STIX]{x1D7D9}_{\{k_{n}+\ell \geqslant 1\}}\biggr). & & \displaystyle\end{eqnarray}$$

This proves the identity (40) in the second case and we complete the proof of Proposition 4.1. ◻

Proof. Let $\Bbbk =(k_{j})_{j\in \mathbb{N}}\in \unicode[STIX]{x1D6E5}$ . Since $\Bbbk$ is a non-increasing sequence, we have

$$\begin{eqnarray}\displaystyle |a\cdot \unicode[STIX]{x1D71B}^{-k_{1}}|\geqslant |a\cdot \unicode[STIX]{x1D71B}^{-k_{2}}|\geqslant \cdots \geqslant |a\cdot \unicode[STIX]{x1D71B}^{-k_{j}}|\cdots \,. & & \displaystyle \nonumber\end{eqnarray}$$

Then either there exists $j_{0}\in \mathbb{N}$ , such that $|a\cdot \unicode[STIX]{x1D71B}^{-k_{j}}|\leqslant 1$ for all $j\geqslant j_{0}$ or $|a\cdot \unicode[STIX]{x1D71B}^{-k_{j}}|>1$ for all $j\in \mathbb{N}$ . Consequently, the infinite product (50) either is a finite product or equals $0$ . The identity (51) follows immediately form (49).◻

Proof of Proposition 4.4.

The identity (45) follows from the independence between all diagonal coefficient of $S_{h}$ . So we only need to prove the identity (44).

Case 1: $h=(-\infty ;\Bbbk ,\Bbbk ^{\prime }),\Bbbk \in \unicode[STIX]{x1D6E5}[-\infty ],\Bbbk ^{\prime }\in \unicode[STIX]{x1D6E5}^{\sharp }[-\infty ]$ .

In this case, we have

$$\begin{eqnarray}\displaystyle \unicode[STIX]{x1D708}_{h}={\mathcal{L}}(W_{\Bbbk }+\unicode[STIX]{x1D700}W_{\Bbbk ^{\prime }})={\mathcal{L}}(W_{\Bbbk })\ast {\mathcal{L}}(\unicode[STIX]{x1D700}W_{\Bbbk ^{\prime }}). & & \displaystyle \nonumber\end{eqnarray}$$

Thus for proving (45), it suffices to prove it for the probability measures ${\mathcal{L}}(W_{\Bbbk })$ and ${\mathcal{L}}(\unicode[STIX]{x1D700}W_{\Bbbk ^{\prime }})$ . For instance, we have

$$\begin{eqnarray}\displaystyle \widehat{{\mathcal{L}}(W_{\Bbbk })}(xe_{11}) & = & \displaystyle \mathbb{E}\biggl[\unicode[STIX]{x1D712}\biggl(\operatorname{tr}\hspace{-0.8pt}\biggl(\mathop{\biggl[\mathop{\sum }_{n=1}^{\infty }\unicode[STIX]{x1D71B}^{-k_{n}}X_{i}^{(n)}X_{j}^{(n)}\biggr]}\nolimits_{i,j\in \mathbb{N}}xe_{11}\biggr)\biggr)\biggr]\nonumber\\ \displaystyle & = & \displaystyle \mathbb{E}\biggl[\unicode[STIX]{x1D712}\biggl(\mathop{\sum }_{n=1}^{\infty }\unicode[STIX]{x1D71B}^{-k_{n}}(X_{1}^{(n)})^{2}x\biggr)\biggr]=\mathbb{E}\biggl[\mathop{\prod }_{n=1}^{\infty }\unicode[STIX]{x1D712}(\unicode[STIX]{x1D71B}^{-k_{n}}(X_{1}^{(n)})^{2}x)\biggr].\nonumber\end{eqnarray}$$

Then by dominated convergence theorem and the independence between all $X_{1}^{(n)},n\in \mathbb{N}$ , we get

$$\begin{eqnarray}\displaystyle \widehat{{\mathcal{L}}(W_{\Bbbk })}(xe_{11})=\mathop{\prod }_{n=1}^{\infty }\mathbb{E}[\unicode[STIX]{x1D712}(\unicode[STIX]{x1D71B}^{-k_{n}}(X_{1}^{(n)})^{2}x)]=\mathop{\prod }_{n=1}^{\infty }\unicode[STIX]{x1D703}(\unicode[STIX]{x1D71B}^{-k_{n}}x). & & \displaystyle \nonumber\end{eqnarray}$$

Similar computation works for ${\mathcal{L}}(\unicode[STIX]{x1D700}W_{\Bbbk ^{\prime }})$ .

Case 2: There exists $k\in \mathbb{Z}$ and $h=(k;\Bbbk ,\Bbbk ^{\prime }),\Bbbk \in \unicode[STIX]{x1D6E5}[k],\Bbbk ^{\prime }\in \unicode[STIX]{x1D6E5}^{\sharp }[k]$ .

In this case, we have

$$\begin{eqnarray}\displaystyle \unicode[STIX]{x1D708}_{h} & ={\mathcal{L}}(W_{\Bbbk }+\unicode[STIX]{x1D700}W_{\Bbbk ^{\prime }}+\unicode[STIX]{x1D71B}^{-k}H)={\mathcal{L}}(W_{\Bbbk })\ast {\mathcal{L}}(\unicode[STIX]{x1D700}W_{\Bbbk ^{\prime }})\ast {\mathcal{L}}(\unicode[STIX]{x1D71B}^{-k}H). & \displaystyle \nonumber\end{eqnarray}$$

By (32), for proving (45), it suffices to prove it for the probability measures

$$\begin{eqnarray}\displaystyle {\mathcal{L}}(W_{\Bbbk }),\quad {\mathcal{L}}(\unicode[STIX]{x1D700}W_{\Bbbk ^{\prime }})\quad \text{and}\quad {\mathcal{L}}(\unicode[STIX]{x1D71B}^{-k}H). & & \displaystyle \nonumber\end{eqnarray}$$

By the computation in Case 1, we only need to verify (45) for ${\mathcal{L}}(\unicode[STIX]{x1D71B}^{-k}H)$ . A simple computation yields the desired identity

$$\begin{eqnarray}\displaystyle \widehat{{\mathcal{L}}(\unicode[STIX]{x1D71B}^{-k}H)}(xe_{11})=\mathbb{E}[\unicode[STIX]{x1D712}(\operatorname{tr}(\unicode[STIX]{x1D71B}^{-k}Hxe_{11}))]=\mathbb{E}[\unicode[STIX]{x1D712}(\unicode[STIX]{x1D71B}^{-k}H_{11}x)]=\unicode[STIX]{x1D7D9}_{{\mathcal{O}}_{F}}(\unicode[STIX]{x1D71B}^{-k}x).\Box & & \displaystyle \nonumber\end{eqnarray}$$

Proof. We need to show that if $\Bbbk =(k_{j})_{j\in \mathbb{N}}$ and $\widetilde{\Bbbk }=(\widetilde{k}_{j})_{j\in \mathbb{N}}$ are two distinct elements in $\unicode[STIX]{x1D6E5}$ , then there exists $\ell \in \mathbb{Z}$ , such that

(52) $$\begin{eqnarray}\displaystyle \mathop{\sum }_{j=1}^{\infty }(k_{j}+\ell )\unicode[STIX]{x1D7D9}_{\{k_{j}+\ell \geqslant 1\}}\neq \mathop{\sum }_{j=1}^{\infty }(\widetilde{k}_{j}+\ell )\unicode[STIX]{x1D7D9}_{\{\widetilde{k}_{j}+\ell \geqslant 1\}}. & & \displaystyle\end{eqnarray}$$

By assumption, there exists $j_{0}\in \mathbb{N}$ , such that

(53) $$\begin{eqnarray}\displaystyle k_{j}=\widetilde{k}_{j}\quad \text{for any }1\leqslant j<j_{0}\quad \text{and}\quad k_{j_{0}}\neq \widetilde{k}_{j_{0}}. & & \displaystyle\end{eqnarray}$$

By symmetry, let us assume that $k_{j_{0}}>\widetilde{k}_{j_{0}}$ . Under this assumption (whether $\widetilde{k}_{j_{0}}$ equals to $-\infty$ or not), we will have $k_{j_{0}}\in \mathbb{Z}$ . Now by taking $\ell =1-k_{j_{0}}\in \mathbb{Z}$ , we have

$$\begin{eqnarray}\displaystyle \left\{\begin{array}{@{}l@{}}k_{j}+\ell =\widetilde{k}_{j}+\ell \geqslant 1\quad \text{for any }1\leqslant j<j_{0},\quad \\ k_{j_{0}}+\ell =1\quad \text{and}\quad \widetilde{k}_{j_{0}}+\ell \leqslant 0.\quad \end{array}\right. & & \displaystyle \nonumber\end{eqnarray}$$

Consequently,

$$\begin{eqnarray}\mathop{\sum }_{j=1}^{\infty }(k_{j}+\ell )\unicode[STIX]{x1D7D9}_{\{k_{j}+\ell \geqslant 1\}}\geqslant \mathop{\sum }_{j=1}^{j_{0}}(k_{j}+\ell )=1+\mathop{\sum }_{j=1}^{j_{0}-1}(k_{j}+\ell ),\end{eqnarray}$$

while

$$\begin{eqnarray}\mathop{\sum }_{j=1}^{\infty }(\widetilde{k}_{j}+\ell )\unicode[STIX]{x1D7D9}_{\{\widetilde{k}_{j}+\ell \geqslant 1\}}=\mathop{\sum }_{j=1}^{j_{0}-1}(\widetilde{k}_{j}+\ell )=\mathop{\sum }_{j=1}^{j_{0}-1}(k_{j}+\ell ).\end{eqnarray}$$

Thus we prove that the inequality (52) holds for $\ell =1-k_{j_{0}}$ .◻

Proof of Proposition 5.1.

Proposition 5.1 follows from Proposition 4.1, Lemma 5.2 and the fact any probability measure on $\operatorname{Mat}(\mathbb{N},F)$ is uniquely determined by its characteristic function.◻

Proof of Proposition 5.3.

Let $h=(k;\Bbbk ,\Bbbk ^{\prime })$ and $\widetilde{h}=(\widetilde{k};\widetilde{\Bbbk },\widetilde{\Bbbk }^{\prime })$ be two elements in $\unicode[STIX]{x1D6FA}$ such that $\unicode[STIX]{x1D708}_{h}=\unicode[STIX]{x1D708}_{\widetilde{h}}$ . By Proposition 4.4, this is equivalent to the following identity: for any $x\in F$ ,

(54) $$\begin{eqnarray}\displaystyle \unicode[STIX]{x1D7D9}_{{\mathcal{O}}_{F}}(\unicode[STIX]{x1D71B}^{-k}x)\cdot \mathop{\prod }_{n=1}^{\infty }\unicode[STIX]{x1D703}(\unicode[STIX]{x1D71B}^{-k_{n}}x)\mathop{\prod }_{n=1}^{\infty }\unicode[STIX]{x1D703}(\unicode[STIX]{x1D700}\unicode[STIX]{x1D71B}^{-k_{n}^{\prime }}x)=\unicode[STIX]{x1D7D9}_{{\mathcal{O}}_{F}}(\unicode[STIX]{x1D71B}^{-\widetilde{k}}x)\cdot \mathop{\prod }_{n=1}^{\infty }\unicode[STIX]{x1D703}(\unicode[STIX]{x1D71B}^{-\widetilde{k}_{n}}x)\mathop{\prod }_{n=1}^{\infty }\unicode[STIX]{x1D703}(\unicode[STIX]{x1D700}\unicode[STIX]{x1D71B}^{-\widetilde{k}_{n}^{\prime }}x).\quad & & \displaystyle\end{eqnarray}$$

Recall the identity (49). By taking the modulus and square of both sides of (54) and substituting $x=\unicode[STIX]{x1D71B}^{-\ell }u$ with $\ell \in \mathbb{Z}$ and $u\in {\mathcal{O}}_{F}^{\times }$ , we obtain

(55) $$\begin{eqnarray}\displaystyle & & \displaystyle \unicode[STIX]{x1D7D9}_{\{k+\ell \leqslant 0\}}\cdot q^{-\hspace{-0.9pt}\mathop{\sum }_{n=1}^{\infty }(k_{n}+\ell )\unicode[STIX]{x1D7D9}_{\{k_{n}+\ell \geqslant 1\}}-\mathop{\sum }_{n=1}^{\infty }(k_{n}^{\prime }+\ell )\unicode[STIX]{x1D7D9}_{\{k_{n}^{\prime }+\ell \geqslant 1\}}}\nonumber\\ \displaystyle & & \displaystyle \quad =\unicode[STIX]{x1D7D9}_{\{\widetilde{k}+\ell \leqslant 0\}}\cdot q^{-\hspace{-0.9pt}\mathop{\sum }_{n=1}^{\infty }(\widetilde{k}_{n}+\ell )\unicode[STIX]{x1D7D9}_{\{\widetilde{k}_{n}+\ell \geqslant 1\}}-\mathop{\sum }_{n=1}^{\infty }(\widetilde{k}_{n}^{\prime }+\ell )\unicode[STIX]{x1D7D9}_{\{\widetilde{k}_{n}^{\prime }+\ell \geqslant 1\}}}.\end{eqnarray}$$

Claim 1: $k=\widetilde{k}$ .

Indeed, if $k=-\infty$ , then the left-hand side of the identity (55) never vanishes. Consequently, so does the right-hand side. It follows that $\widetilde{k}=-\infty$ . If $k\in \mathbb{Z}$ . Then the left-hand side of the identity (55) vanishes at $\ell =1-k$ . Consequently, so does the right-hand side of (55) vanishes both at $\ell =1-k$ . It follows that $\widetilde{k}+1-k>0$ or equivalently $\widetilde{k}\geqslant k$ . By symmetry, we have $k=\widetilde{k}$ .

Claim 2: $(\Bbbk ,\Bbbk ^{\prime })=(\widetilde{\Bbbk },\widetilde{\Bbbk }^{\prime })$ .

For simplifying notation, let us define $\Bbbk ^{\ast },\widetilde{\Bbbk }^{\ast }\in \unicode[STIX]{x1D6E5}$ as follows: if $k=\widetilde{k}=-\infty$ , then set $\Bbbk ^{\ast }:=\Bbbk ,\widetilde{\Bbbk }^{\ast }:=\widetilde{\Bbbk }$ ; if $k=\widetilde{k}\in \mathbb{Z}$ , then both $\Bbbk$ and $\widetilde{\Bbbk }$ are finite sequences in $\mathbb{Z}_{{>}k}$ , set $\Bbbk ^{\ast }$ and $\widetilde{\Bbbk }^{\ast }$ by adding infinitely many $k$ . Clearly, for proving $(\Bbbk ,\Bbbk ^{\prime })=(\widetilde{\Bbbk },\widetilde{\Bbbk }^{\prime })$ , it suffices to prove that $(\Bbbk ^{\ast },\Bbbk ^{\prime })=(\widetilde{\Bbbk }^{\ast },\widetilde{\Bbbk }^{\prime })$ . By Remark 5.4, it suffices to prove that for any $l\in \mathbb{Z}$ ,

(56) $$\begin{eqnarray}\displaystyle \left.\begin{array}{@{}l@{}}\#\{n\in \mathbb{N}\mid k_{n}^{\ast }=l\}=\#\{n\in \mathbb{N}\mid \widetilde{k}_{n}^{\ast }=l\},\\ \#\{n\in \mathbb{N}\mid k_{n}^{\prime }=l\}=\#\{n\in \mathbb{N}\mid \widetilde{k}_{n}^{\prime }=l\}.\end{array}\right. & & \displaystyle\end{eqnarray}$$

Applying (43) to $\unicode[STIX]{x1D7D9}_{\{k+\ell \leqslant 0\}}$ and $\unicode[STIX]{x1D7D9}_{\{\widetilde{k}+\ell \leqslant 0\}}$ , we may write (54) as

(57) $$\begin{eqnarray}\displaystyle q^{-\hspace{-0.9pt}\mathop{\sum }_{n=1}^{\infty }(k_{n}^{\ast }+\ell )\unicode[STIX]{x1D7D9}_{\{k_{n}^{\ast }+\ell \geqslant 1\}}-\mathop{\sum }_{n=1}^{\infty }(k_{n}^{\prime }+\ell )\unicode[STIX]{x1D7D9}_{\{k_{n}^{\prime }+\ell \geqslant 1\}}}=q^{-\hspace{-0.9pt}\mathop{\sum }_{n=1}^{\infty }(\widetilde{k}_{n}^{\ast }+\ell )\unicode[STIX]{x1D7D9}_{\{\widetilde{k}_{n}^{\ast }+\ell \geqslant 1\}}-\mathop{\sum }_{n=1}^{\infty }(\widetilde{k}_{n}^{\prime }+\ell )\unicode[STIX]{x1D7D9}_{\{\widetilde{k}_{n}^{\prime }+\ell \geqslant 1\}}}.\qquad & & \displaystyle\end{eqnarray}$$

By Lemma 5.2 and Remark 5.4, the equality (57) implies that for any $l\in \mathbb{Z}$ , we have

(58) $$\begin{eqnarray}\displaystyle \#\{n\in \mathbb{N}\mid k_{n}^{\ast }=l\}+\#\{n\in \mathbb{N}\mid k_{n}^{\prime }=l\}=\#\{n\in \mathbb{N}\mid \widetilde{k}_{n}^{\ast }=l\}+\#\{n\in \mathbb{N}\mid \widetilde{k}_{n}^{\prime }=l\}. & & \displaystyle\end{eqnarray}$$

The identity (58) implies in particular that the two identities in (56) hold or are violated simultaneously. Now assume by contradiction that there exists $l_{0}\in \mathbb{Z}$ , such that the identities in (56) are violated. Obviously, such $l_{0}$ verifies

$$\begin{eqnarray}\displaystyle k<l_{0}\leqslant \max \{k_{n},\widetilde{k}_{n},k_{n}^{\prime },\widetilde{k}_{n}^{\prime }\}<+\infty . & & \displaystyle \nonumber\end{eqnarray}$$

Now let $l_{\max }\in \mathbb{Z}$ be the largest $l_{0}$ such that the identities in (56) are violated. Substituting $x=\unicode[STIX]{x1D71B}^{l_{\max }-1}u$ with $u\in {\mathcal{O}}_{F}^{\times }$ into the identity (54), we obtain

(59) $$\begin{eqnarray}\displaystyle & & \displaystyle \mathop{\prod }_{n_{1}=1}^{\infty }\unicode[STIX]{x1D703}(\unicode[STIX]{x1D71B}^{-k_{n_{1}}}\unicode[STIX]{x1D71B}^{l_{\max }-1}u)\mathop{\prod }_{n_{2}=1}^{\infty }\unicode[STIX]{x1D703}(\unicode[STIX]{x1D700}\unicode[STIX]{x1D71B}^{-k_{n_{2}}^{\prime }}\unicode[STIX]{x1D71B}^{l_{\max }-1}u)\nonumber\\ \displaystyle & & \displaystyle \quad =\mathop{\prod }_{m_{1}=1}^{\infty }\unicode[STIX]{x1D703}(\unicode[STIX]{x1D71B}^{-\widetilde{k}_{m_{1}}}\unicode[STIX]{x1D71B}^{l_{\max }-1}u)\mathop{\prod }_{m_{2}=1}^{\infty }\unicode[STIX]{x1D703}(\unicode[STIX]{x1D700}\unicode[STIX]{x1D71B}^{-\widetilde{k}_{m_{2}}^{\prime }}\unicode[STIX]{x1D71B}^{l_{\max }-1}u).\end{eqnarray}$$

By the assumption of $l_{\max }$ , we know that for any $l>l_{\max }$ ,

(60) $$\begin{eqnarray}\displaystyle \left.\begin{array}{@{}l@{}}\#\{n\in \mathbb{N}\mid k_{n}^{\ast }=l\}=\#\{n\in \mathbb{N}\mid \widetilde{k}_{n}^{\ast }=l\},\\ \#\{n\in \mathbb{N}\mid k_{n}^{\prime }=l\}=\#\{n\in \mathbb{N}\mid \widetilde{k}_{n}^{\prime }=l\}.\end{array}\right. & & \displaystyle\end{eqnarray}$$

Since $l_{\max }>k$ , by definition of $\Bbbk ^{\ast }$ and $\widetilde{\Bbbk }^{\ast }$ we also have for any $l>l_{\max }$ ,

(61) $$\begin{eqnarray}\displaystyle \left.\begin{array}{@{}l@{}}\#\{n\in \mathbb{N}\mid k_{n}^{\ast }=l\}=\#\{n\in \mathbb{N}\mid k_{n}=l\},\\ \#\{n\in \mathbb{N}\mid \widetilde{k}_{n}^{\ast }=l\}=\#\{n\in \mathbb{N}\mid \widetilde{k}_{n}=l\}.\end{array}\right. & & \displaystyle\end{eqnarray}$$

By Proposition 4.5, the function $\unicode[STIX]{x1D703}$ never vanishes. Hence by the identities (60) and (61), we can remove simultaneously all those terms concerning $k_{n_{1}}>l_{\max },k_{n_{2}}^{\prime }>l_{\max }$ and $\widetilde{k}_{m_{1}}>l_{\max },\widetilde{k}_{m_{2}}^{\prime }>l_{\max }$ from both sides of identity (59). Again by Proposition 4.5, for any $k_{n_{1}}<l_{\max },k_{n_{2}}^{\prime }<l_{\max }$ and $\widetilde{k}_{m_{1}}<l_{\max },\widetilde{k}_{m_{2}}^{\prime }<l_{\max }$ , we have

$$\begin{eqnarray}\displaystyle \unicode[STIX]{x1D703}(\unicode[STIX]{x1D71B}^{-k_{n_{1}}}\unicode[STIX]{x1D71B}^{l_{\max }-1}u)=\unicode[STIX]{x1D703}(\unicode[STIX]{x1D700}\unicode[STIX]{x1D71B}^{-k_{n_{2}}^{\prime }}\unicode[STIX]{x1D71B}^{l_{\max }-1}u)=\unicode[STIX]{x1D703}(\unicode[STIX]{x1D71B}^{-\widetilde{k}_{m_{1}}}\unicode[STIX]{x1D71B}^{l_{\max }-1}u)=\unicode[STIX]{x1D703}(\unicode[STIX]{x1D700}\unicode[STIX]{x1D71B}^{-\widetilde{k}_{m_{2}}^{\prime }}\unicode[STIX]{x1D71B}^{l_{\max }-1}u)=1. & & \displaystyle \nonumber\end{eqnarray}$$

Consequently, we may remove simultaneously all those terms concerning $k_{n_{1}}<l_{\max },k_{n_{2}}^{\prime }<l_{\max }$ and $\widetilde{k}_{m_{1}}<l_{\max },\widetilde{k}_{m_{2}}^{\prime }<l_{\max }$ from both sides of identity (59) as well. Then we arrive at the identity

(62) $$\begin{eqnarray}\displaystyle \unicode[STIX]{x1D703}(\unicode[STIX]{x1D71B}^{-1}u)^{D}\cdot \unicode[STIX]{x1D703}(\unicode[STIX]{x1D700}\unicode[STIX]{x1D71B}^{-1}u)^{D^{\prime }}=\unicode[STIX]{x1D703}(\unicode[STIX]{x1D71B}^{-1}u)^{\widetilde{D}}\cdot \unicode[STIX]{x1D703}(\unicode[STIX]{x1D700}\unicode[STIX]{x1D71B}^{-1}u)^{\widetilde{D^{\prime }}}, & & \displaystyle\end{eqnarray}$$

where

$$\begin{eqnarray}\displaystyle D & := & \displaystyle \#\{n\in \mathbb{N}\mid k_{n}=l_{\max }\}\quad \text{and}\quad D^{\prime }:=\#\{n\in \mathbb{N}\mid k_{n}^{\prime }=l_{\max }\},\nonumber\\ \displaystyle \widetilde{D} & := & \displaystyle \#\{n\in \mathbb{N}\mid \widetilde{k}_{n}=l_{\max }\}\quad \text{and}\quad \widetilde{D}^{\prime }:=\#\{n\in \mathbb{N}\mid \widetilde{k}_{n}^{\prime }=l_{\max }\}.\nonumber\end{eqnarray}$$

By definitions for $\unicode[STIX]{x1D6E5}[k]$ and $\unicode[STIX]{x1D6E5}^{\sharp }[k]$ , we must have

(63) $$\begin{eqnarray}\displaystyle D,\quad \widetilde{D}\in \mathbb{N}\cup \{0\}\quad \text{and}\quad D^{\prime },\widetilde{D^{\prime }}\in \{0,1\}. & & \displaystyle\end{eqnarray}$$

The identity (58) now implies that

(64) $$\begin{eqnarray}\displaystyle D+D^{\prime }=\widetilde{D}+\widetilde{D^{\prime }}. & & \displaystyle\end{eqnarray}$$

By definition of $l_{\max }$ , we have $D\neq D^{\prime }$ . Without loss of generality, we may assume that $D^{\prime }=0$ and $\widetilde{D^{\prime }}=1$ . Then the identity (62) becomes

(65) $$\begin{eqnarray}\displaystyle \unicode[STIX]{x1D703}(\unicode[STIX]{x1D71B}^{-1}u)^{D}=\unicode[STIX]{x1D703}(\unicode[STIX]{x1D71B}^{-1}u)^{\widetilde{D}}\cdot \unicode[STIX]{x1D703}(\unicode[STIX]{x1D700}\unicode[STIX]{x1D71B}^{-1}u). & & \displaystyle\end{eqnarray}$$

But now $D=\widetilde{D}+1$ and since $\unicode[STIX]{x1D703}$ never vanishes, the identity (65) is equivalent to

$$\begin{eqnarray}\displaystyle \unicode[STIX]{x1D703}(\unicode[STIX]{x1D71B}^{-1}u)=\unicode[STIX]{x1D703}(\unicode[STIX]{x1D700}\unicode[STIX]{x1D71B}^{-1}u). & & \displaystyle \nonumber\end{eqnarray}$$

Since $|\unicode[STIX]{x1D71B}^{-1}u|>1,\text{ord}_{F}(\unicode[STIX]{x1D71B}^{-1}u)=-1\equiv 1(\operatorname{mod}2)$ and $L_{2}(\unicode[STIX]{x1D700})=-1$ , by Proposition 4.5, we have $\unicode[STIX]{x1D703}(\unicode[STIX]{x1D700}\unicode[STIX]{x1D71B}^{-1}u)=-\unicode[STIX]{x1D703}(\unicode[STIX]{x1D71B}^{-1}u)$ . Consequently, we would have $\unicode[STIX]{x1D703}(\unicode[STIX]{x1D71B}^{-1}u)=\unicode[STIX]{x1D703}(\unicode[STIX]{x1D700}\unicode[STIX]{x1D71B}^{-1}u)=0$ . This contradicts to the non-vanishing property of $\unicode[STIX]{x1D703}$ . Hence we complete the proof of Claim 2.

Combining Claims 1 and 2, we complete the proof of Proposition 5.3. ◻

Proof of Theorem 7.1.

Fix $n,r\in \mathbb{N}$ and fix the two diagonal matrices $D$ and $A$ . Let $T=T(n),T^{\prime }=T^{\prime }(n)$ be two independent copies of random matrices sampled uniformly from the finite set $\operatorname{GL}(n,{\mathcal{C}}_{q})$ , and let $V=V(n),V^{\prime }=V^{\prime }(n)$ be two independent random matrices sampled uniformly from $\operatorname{Mat}(n,\unicode[STIX]{x1D71B}{\mathcal{O}}_{F})$ and independent of $T,T^{\prime }$ . By Proposition 2.3, we have

$$\begin{eqnarray}\displaystyle \int _{{\mathcal{K}}(n)}\unicode[STIX]{x1D712}(\operatorname{tr}(g_{1}Dg_{2}A))\,dg_{1}\,dg_{2}=\mathbb{E}[\unicode[STIX]{x1D712}(\operatorname{tr}((T+V)D(T^{\prime }+V^{\prime })A))]. & & \displaystyle \nonumber\end{eqnarray}$$

Since the transposed random matrix $(T^{\prime }+V^{\prime })^{t}$ and the original random matrix $T^{\prime }+V^{\prime }$ have the same distribution, we have

$$\begin{eqnarray}\displaystyle & & \displaystyle \mathbb{E}[\unicode[STIX]{x1D712}(\operatorname{tr}((T+V)D(T^{\prime }+V^{\prime })A))]\nonumber\\ \displaystyle & & \displaystyle \quad =\mathbb{E}[\unicode[STIX]{x1D712}(\operatorname{tr}((T+V)D(T^{\prime }+V^{\prime })^{t}A))]\nonumber\\ \displaystyle & & \displaystyle \quad =\mathbb{E}\biggl[\unicode[STIX]{x1D712}\biggl(\mathop{\sum }_{i=1}^{r}\mathop{\sum }_{j=1}^{n}(T_{ij}+V_{ij})(T_{ij}^{\prime }+V_{ij}^{\prime })a_{i}x_{j}\biggr)\biggr]\nonumber\\ \displaystyle & & \displaystyle \quad =\frac{1}{[\#\text{GL}(n,{\mathcal{C}}_{q})]^{2}}\mathop{\sum }_{t,t^{\prime }\in \operatorname{GL}(n,{\mathcal{C}}_{q})}\mathbb{E}\biggl[\unicode[STIX]{x1D712}\biggl(\mathop{\sum }_{i=1}^{r}\mathop{\sum }_{j=1}^{n}(t_{ij}+V_{ij})(t_{ij}^{\prime }+V_{ij}^{\prime })a_{i}x_{j}\biggr)\biggr].\nonumber\end{eqnarray}$$

By Lemma 7.3, we obtain

$$\begin{eqnarray}\displaystyle & & \displaystyle \mathbb{E}[\unicode[STIX]{x1D712}(\operatorname{tr}((T+V)D(T^{\prime }+V^{\prime })A))]\nonumber\\ \displaystyle & & \displaystyle \quad =\frac{1}{[\#S(r\times n)]^{2}}\mathop{\sum }_{M,M^{\prime }\in S(r\times n)}\mathbb{E}\biggl[\unicode[STIX]{x1D712}\biggl(\mathop{\sum }_{i=1}^{r}\mathop{\sum }_{j=1}^{n}(M_{ij}+V_{ij})(M_{ij}^{\prime }+V_{ji}^{\prime })a_{i}x_{j}\biggr)\biggr].\nonumber\end{eqnarray}$$

For simplifying notation, for any pair of rectangular matrices $M,M^{\prime }\in \operatorname{Mat}(r\times n,{\mathcal{C}}_{q})$ , we write

$$\begin{eqnarray}\displaystyle F(M,M^{\prime }):=\mathbb{E}\biggl[\unicode[STIX]{x1D712}\biggl(\mathop{\sum }_{i=1}^{r}\mathop{\sum }_{j=1}^{n}(M_{ij}+V_{ij})(M_{ij}^{\prime }+V_{ji}^{\prime })a_{i}x_{j}\biggr)\biggr]. & & \displaystyle \nonumber\end{eqnarray}$$

Then we may write

$$\begin{eqnarray}\displaystyle \mathbb{E}[\unicode[STIX]{x1D712}(\operatorname{tr}((T+V)D(T^{\prime }+V^{\prime })A))]=\frac{1}{[\#S(r\times n)]^{2}}\underbrace{\biggl[\mathop{\sum }_{M,M^{\prime }\in \operatorname{Mat}(r\times n,{\mathcal{C}}_{q})}F(M,M^{\prime })+{\mathcal{E}}_{1}\biggr]}_{\text{denoted}~\text{by}~I}, & & \displaystyle \nonumber\end{eqnarray}$$

where

$$\begin{eqnarray}\displaystyle {\mathcal{E}}_{1}:=-\mathop{\sum }_{M,M^{\prime }\in \operatorname{Mat}(r\times n,{\mathcal{C}}_{q})\,\setminus \,S(r\times n)}F(M,M^{\prime }). & & \displaystyle \nonumber\end{eqnarray}$$

Note that if $M_{ij}$ and $V_{ij}$ are sampled independently and uniformly from the finite set ${\mathcal{C}}_{q}$ and from the compact additive group $\unicode[STIX]{x1D71B}{\mathcal{O}}_{F}$ respectively, then $M_{ij}+V_{ij}$ is uniformly distributed on ${\mathcal{O}}_{F}$ . It follows that

$$\begin{eqnarray}\displaystyle \mathop{\prod }_{i=1}^{r}\mathop{\prod }_{j=1}^{n}\unicode[STIX]{x1D6E9}(a_{i}x_{j})=\frac{1}{[\#\text{Mat}(r\times n,{\mathcal{C}}_{q})]^{2}}\mathop{\sum }_{M,M^{\prime }\in \operatorname{Mat}(r\times n,{\mathcal{C}}_{q})}F(M,M^{\prime }). & & \displaystyle \nonumber\end{eqnarray}$$

Consequently,

$$\begin{eqnarray}\displaystyle \mathbb{E}[\unicode[STIX]{x1D712}(\operatorname{tr}((T+V)D(T^{\prime }+V^{\prime })A))] & = & \displaystyle \frac{I}{[\#\text{Mat}(r\times n,{\mathcal{C}}_{q})]^{2}}+{\mathcal{E}}_{2}\nonumber\\ \displaystyle & = & \displaystyle \mathop{\prod }_{i=1}^{r}\mathop{\prod }_{j=1}^{n}\unicode[STIX]{x1D6E9}(a_{i}x_{j})+\frac{{\mathcal{E}}_{1}}{[\#\text{Mat}(r\times n,{\mathcal{C}}_{q})]^{2}}+{\mathcal{E}}_{2},\nonumber\end{eqnarray}$$

where

$$\begin{eqnarray}\displaystyle {\mathcal{E}}_{2}:=\frac{I}{[\#S(r\times n)]^{2}}-\frac{I}{[\#\text{Mat}(r\times n,{\mathcal{C}}_{q})]^{2}}. & & \displaystyle \nonumber\end{eqnarray}$$

Now let us estimate these error terms. By the obvious estimate $|F(M,M^{\prime })|\leqslant 1$ , we have

$$\begin{eqnarray}\displaystyle |{\mathcal{E}}_{1}|\leqslant [\#(\operatorname{Mat}(r\times n,{\mathcal{C}}_{q})\,\setminus \,S(r\times n))]^{2}. & & \displaystyle \nonumber\end{eqnarray}$$

Note that $|I|\leqslant [\#S(r\times n)]^{2}$ . Hence

$$\begin{eqnarray}\displaystyle |{\mathcal{E}}_{2}| & = & \displaystyle |I|\biggl(\frac{1}{[\#S(r\times n)]^{2}}-\frac{1}{[\#\text{Mat}(r\times n,{\mathcal{C}}_{q})]^{2}}\biggr)\nonumber\\ \displaystyle & {\leqslant} & \displaystyle \biggl(\frac{\#\text{Mat}(r\times n,{\mathcal{C}}_{q})-\#S(r\times n)}{\#\text{Mat}(r\times n,{\mathcal{C}}_{q})}\biggr)^{2}.\nonumber\end{eqnarray}$$

Taking (75) into account, we get

$$\begin{eqnarray}\displaystyle & & \displaystyle \biggl|\mathbb{E}[\unicode[STIX]{x1D712}(\operatorname{tr}((T+V)D(T^{\prime }+V^{\prime })A))]-\mathop{\prod }_{i=1}^{r}\mathop{\prod }_{j=1}^{n}\unicode[STIX]{x1D6E9}(a_{i}x_{j})\biggr|\nonumber\\ \displaystyle & & \displaystyle \quad \leqslant \frac{|{\mathcal{E}}_{1}|}{[\#\text{Mat}(r\times n,{\mathcal{C}}_{q})]^{2}}+|{\mathcal{E}}_{2}|\nonumber\\ \displaystyle & & \displaystyle \quad \leqslant 2\biggl(\frac{\#\text{Mat}(r\times n,{\mathcal{C}}_{q})-\#S(r\times n)}{\#\text{Mat}(r\times n,{\mathcal{C}}_{q})}\biggr)^{2}\nonumber\\ \displaystyle & & \displaystyle \quad \leqslant 2\biggl(1-\mathop{\prod }_{w=0}^{r-1}(1-q^{w-n})\biggr)^{2}.\Box \nonumber\end{eqnarray}$$

Proof. By inequalities (73) and (74), we have

$$\begin{eqnarray}\displaystyle & & \displaystyle \int _{{\mathcal{K}}(n)}\unicode[STIX]{x1D712}(\operatorname{tr}(g_{1}Dg_{2}A))\,dg_{1}\,dg_{2}=\mathop{\prod }_{i=1}^{r}\mathop{\prod }_{j=1}^{n}\unicode[STIX]{x1D6E9}(a_{i}x_{j})+\unicode[STIX]{x1D700}_{0}\nonumber\\ \displaystyle & & \displaystyle \quad =\mathop{\prod }_{i=1}^{r}\biggl(\underbrace{\int _{{\mathcal{K}}(n)}\unicode[STIX]{x1D712}(a_{i}\cdot \operatorname{tr}(g_{1}Dg_{2}e_{11}))\,dg_{1}\,dg_{2}+\unicode[STIX]{x1D700}_{i}}_{=\mathop{\prod }_{j=1}^{n}\unicode[STIX]{x1D6E9}(a_{i}x_{j})}\biggr)+\unicode[STIX]{x1D700}_{0},\nonumber\end{eqnarray}$$

with the errors $\unicode[STIX]{x1D700}_{0},\unicode[STIX]{x1D700}_{1},\ldots ,\unicode[STIX]{x1D700}_{r}$ controlled by

$$\begin{eqnarray}\displaystyle |\unicode[STIX]{x1D700}_{0}|\leqslant 2\biggl(1-\mathop{\prod }_{w=0}^{r-1}(1-q^{w-n})\biggr)^{2}\quad \text{and}\quad |\unicode[STIX]{x1D700}_{i}|\leqslant 2q^{-2n},\quad i=1,\ldots ,r. & & \displaystyle \nonumber\end{eqnarray}$$

Using the elementary inequalities $|\!\prod _{j=1}^{n}\unicode[STIX]{x1D6E9}(a_{i}x_{j})|\leqslant 1$ and by a simple computation, we get

$$\begin{eqnarray}\displaystyle & & \displaystyle \biggl|\int _{{\mathcal{K}}(n)}\unicode[STIX]{x1D712}(\operatorname{tr}(g_{1}Dg_{2}A))\,dg_{1}\,dg_{2}-\mathop{\prod }_{i=1}^{r}\int _{{\mathcal{K}}(n)}\unicode[STIX]{x1D712}(a_{i}\cdot \operatorname{tr}(g_{1}Dg_{2}e_{11}))\,dg_{1}\,dg_{2}\biggr|\nonumber\\ \displaystyle & & \displaystyle \quad \leqslant \unicode[STIX]{x1D700}_{0}+\unicode[STIX]{x1D700}_{1}+\cdots +\unicode[STIX]{x1D700}_{r}\nonumber\\ \displaystyle & & \displaystyle \quad \leqslant 2\biggl(1-\mathop{\prod }_{w=0}^{r-1}(1-q^{w-n})\biggr)^{2}+2rq^{-2n}.\Box \nonumber\end{eqnarray}$$

Proof. Fix $n,r\in \mathbb{N}$ and fix the two diagonal matrices $D$ and $A$ . Let $T=T(n)$ be a random matrix uniformly distributed on the finite set $\operatorname{GL}(n,\mathbf{F}_{q})$ , and let $V=V(n)$ be a random matrix uniformly distributed on $\operatorname{Mat}(n,\unicode[STIX]{x1D71B}{\mathcal{O}}_{F})$ and independent of $T$ . By Proposition 2.3, we have

$$\begin{eqnarray}\displaystyle \int _{\operatorname{GL}(n,{\mathcal{O}}_{F})}\unicode[STIX]{x1D712}(\operatorname{tr}(gDg^{t}A))\,dg=\mathbb{E}\biggl[\unicode[STIX]{x1D712}\biggl(\mathop{\sum }_{i=1}^{r}\mathop{\sum }_{j=1}^{n}a_{i}x_{j}(T_{ij}+V_{ij})^{2}\biggr)\biggr]. & & \displaystyle \nonumber\end{eqnarray}$$

By similar arguments as in the proof of Theorem 7.1, we may get the desired inequality (77). The second inequality (78) follows immediately by taking $e=1$ and $a_{1}=a$ .◻

Proof. The proof is similar to that of Theorem 7.4. ◻

Proof. Let $\unicode[STIX]{x1D707}\in \mathscr{ORB}_{\infty }(\operatorname{Mat}(\mathbb{N},F))$ . Then by definition, there exists an increasing sequence $(n_{k})_{k\in \mathbb{N}}$ in $\mathbb{N}$ and a sequence $(\unicode[STIX]{x1D707}_{n_{k}})_{k\in \mathbb{N}}$ of orbital measures with $\unicode[STIX]{x1D707}_{n_{k}}\in \mathscr{ORB}_{n_{k}}(\operatorname{Mat}(n_{k},F))$ , such that

(80) $$\begin{eqnarray}\displaystyle \unicode[STIX]{x1D707}_{n_{k}}\Longrightarrow \unicode[STIX]{x1D707}\quad \text{as }k\rightarrow \infty . & & \displaystyle\end{eqnarray}$$

Take any $x_{1},\ldots ,x_{r}\in F$ . By the inequality (76), we have

(81) $$\begin{eqnarray}\displaystyle \lim _{k\rightarrow \infty }\biggl|\widehat{\unicode[STIX]{x1D707}_{n_{k}}}(\operatorname{diag}(x_{1},\ldots ,x_{r},0,\ldots ))-\mathop{\prod }_{j=1}^{r}\widehat{\unicode[STIX]{x1D707}_{n_{k}}}(x_{j}e_{11})\biggr|=0. & & \displaystyle\end{eqnarray}$$

Combining (80) and (81), we get the desired identity (79).

By Vershik’s Theorem 6.3 and Proposition 6.8, to obtain (81), we only need to prove the inclusion