# Limit theorems for a strongly irreducible product of independent random matrices under optimal moment assumptions

Axel Péneau

► To cite this version:

Axel Péneau. Limit theorems for a strongly irreducible product of independent random matrices under optimal moment assumptions. 2024. hal-04438391v3

HAL Id: hal-04438391

https://hal.science/hal-04438391v3

Preprint submitted on 24 Feb 2025

HAL is a multi-disciplinary open access
archive for the deposit and dissemination of sci-
entific research documents, whether they are pub-
lished or not. The documents may come from
teaching and research institutions in France or
abroad, or from public or private research centers.

L'archive ouverte pluridisciplinaire HAL, est
destinée au dépôt et à la diffusion de documents
scientifiques de niveau recherche, publiés ou non,
émanant des établissements d'enseignement et de
recherche français ou étrangers, des laboratoires
publics ou privés.

<figure><img src="image_1.png" /></figure>

Distributed under a Creative Commons Attribution 4.0 International License
# LIMIT THEOREMS FOR A STRONGLY IRREDUCIBLE PRODUCT OF INDEPENDENT RANDOM MATRICES UNDER OPTIMAL MOMENT ASSUMPTIONS

AXEL PÉNEAU

**ABSTRACT.** Let $\nu$ be a probability distribution over the semi-group of square matrices of size $d \ge 2$. We assume that $\nu$ is proximal, strongly irreducible and that $\nu^{*n}\{0\} = 0$ for all integers $n \in \mathbb{N}$. We consider the sequence $\overline{\gamma}_n := \gamma_0 \cdots \gamma_{n-1}$ for $(\gamma_k)_{k \in \mathbb{N}}$ independent of distribution law $\nu$. Let $s_1 \ge s_2 \ge \dots s_d$ be the singular values given by the Cartan projection. We show that $(\log \frac{s_1}{s_2}(\overline{\gamma}_n))_{n \in \mathbb{N}}$, escapes to infinity linearly and satisfies exponential large deviations inequalities below its escape rate. This is an extension the simplicity of the top Lyapunov exponent result, proven by Guivarc'h and Raugi under a first moment assumption. We also show that the image of a generic line by $\overline{\gamma}_n$ as well as its eigenspace of maximal eigenvalue both converge to the same random line $l^\infty$ at an exponential speed.

If we moreover assume that $\nu$ is supported on the group of invertible matrices and that the push-forward distribution $N_*\nu$ is $L^p$ for $N: g \mapsto \log \|g\| \|g^{-1}\|$ and for some $p > 0$, then we show that $-\log d(l^\infty, H)$ is uniformly $L^p$ for all proper subspace $H \subset \mathbb{R}^d$. For $p=1$, we moreover show that the rescaled logarithm of each coefficient of $\overline{\gamma}_n$ almost surely converges to the top Lyapunov exponent. This is an extension of results by Benoist and Quint which were themselves improvements of historic results by Guivarc'h, Lepage and Raugi, first proven under exponential moment assumptions.

To prove these results, we do not rely on the existence of the stationary measure nor on the existence of the Lyapunov exponent. Instead we describe an effective way to group the i.i.d. factors into i.i.d. random words that are aligned in the Cartan decomposition. We moreover have an explicit control over the moments.

## 1. INTRODUCTION

**Notable results.** Let $d \ge 2$. To a random independent and identically distributed sequence $(\gamma_n)_{n \ge 0} \in \text{GL}(\mathbb{R}^d)$, we associate the random walk $(\overline{\gamma}_n)_{n \ge 0}$ defined as $\overline{\gamma}_n := \gamma_0 \cdots \gamma_{n-1}$ for all $n \ge 0$. The random matrix $\overline{\gamma}_n$ is then a product

<footnote>Date: February 24, 2025.</footnote>

<footnote n="2020">2020 Mathematics Subject Classification. Primary 60B20, 60F10, 37H15, 60B15, 60F15, 60F25.</footnote>
of $n$ independent random $d \times d$ matrices. We are interested in quantitative probabilistic estimates for $\bar{\gamma}_n$ that hold for all time $n$ as well as qualitative asymptotic results that hold almost surely for the sequence $(\bar{\gamma}_n)_{n \ge 0}$. We write $^t e_i \bar{\gamma}_n e_j$ for the entry of $\bar{\gamma}_n$ at the coordinates $(i, j)$. We write $\rho_1(\bar{\gamma}_n) \ge \rho_2(\bar{\gamma}_n) \ge \dots \ge \rho_d(\bar{\gamma}_n) \ge 0$ for the spectral values of $\bar{\gamma}_n$ i.e., the moduli of the eigenvalues. We say that a matrix $h$ is proximal when $\rho_1(h) > \rho_2(h)$, which implies that the eigenspace of $h$ associated to its top eigenvalue is a line and that $\rho_1(h)^{-n}h^n$ converges to a rank one projection or equivalently that the projective class $[h^n]$ converges to the projective class of a rank one projection.

First let us state two notable results of the present article to highlight the strength and versatility of the pivoting technique. A notable qualitative result is the following theorem.

**Theorem A (Law of large numbers for the coefficients).** Let $(\gamma_n)_{n \ge 0}$ be a sequence of independent and identically distributed invertible square matrices of size $d \ge 2$. Assume the following:

* $\mathbb{E}(\log^+ \|\gamma_0\|) < +\infty$ and $\mathbb{E}(\log^+ \|\gamma_0^{-1}\|) < +\infty$.

* For all $N \ge 1$ and for all $\{0\} \subsetneq V_1, \dots, V_N \subsetneq \mathbb{R}^d$, we have $\mathbb{P}(\gamma_0(\bigcup V_i) \subset \bigcup V_i) < 1$.

* There exists $n \ge 1$ such that $\mathbb{P}(\rho_1(\bar{\gamma}_n) > \rho_2(\bar{\gamma}_n)) > 0$.

Then there exists a non-random constant $\lambda_1$ such that for all $1 \le i, j \le d$, we have almost surely:

$$ \lim_{n \to +\infty} \frac{\log |^t e_i \bar{\gamma}_n e_j|}{n} = \lim_{n \to +\infty} \frac{\log \rho_1(\bar{\gamma}_n)}{n} = \lambda_1 $$

Theorem A is notable for answering the long-standing conjecture of convergence of the coefficients under $L^1$ moment assumption. From the work of Furstenberg and Kesten [FK60], it was already known with the first moment assumption only that $\frac{\log \|\bar{\gamma}_n\|}{n} \to \lambda_1$. We also know that $\lambda_1 = \lim_{n \to +\infty} \frac{\mathbb{E}(\log \|\bar{\gamma}_n\|)}{n}$. This is simply a particular case of Kingman's sub-additive ergodic Theorem [Kin68]. Weaker versions of Theorem A were stated by Guivarc'h and Lepage [Gui80] [LP82] with an exponential moment condition i.e., assuming that $\mathbb{E}(\|\gamma_0\|^\beta) < +\infty$ and $\mathbb{E}(\|\gamma_0^{-1}\|^\beta) < +\infty$ for some $\beta > 0$ and by Xiao, Gramma and Liu [XGL21] using the work of Benoist and Quint [BQ16c] with an $L^2$ moment condition i.e., assuming that $\mathbb{E}(\log^2 \|\gamma_0\|) < +\infty$ and $\mathbb{E}(\log^2 \|\gamma_0^{-1}\|) < +\infty$.

Let us now state a notable quantitative result of the present article. Given $x \in \mathbb{R}^d$, we write $[x]$ for the projective class of $x$ the space of projective classes, that we denote by $\mathbb{P}(\mathbb{R}^d)$, is a compact manifold so all distances are equivalent, let us pick $d$ to be such a distance. Given $g \in \text{GL}(\mathbb{R}^d)$, we write $s_1(g) \ge s_2(g) \ge$
$\cdots \ge s_d(g)$ for the singular values of $g$. It is known from Oseledets multiplicative ergodic Theorem that given $(\gamma_n)_{n \ge 0}$ i.i.d. that satisfy the $L^1$ moment assumption, there exists a Lyapunov spectrum $\lambda_1 \ge \lambda_2 \ge \cdots \ge \lambda_d$ such that $\frac{s_k(\bar{\gamma}_n)}{n} \to \lambda_k$ almost surely and for all $k \in \{1, \dots, d\}$. This implies that $\frac{\log \frac{s_1}{s_2}(\bar{\gamma}_n)}{n} \to \lambda_1 - \lambda_2$. In [GR89], Guivarc'h and Raugi show that with the algebraic assumptions of Theorem A, we moreover have $\lambda_1 > \lambda_2$.

Once we get rid of the moment assumption, the Lyapunov spectrum does not make sense. However, with the same algebraic assumptions as in [GR89], we have the following result.

**Theorem B (Exponential contraction without moment condition).** Let $(\gamma_n)_{n \ge 0}$ be a sequence of independent and identically distributed invertible square matrices of size $d \ge 2$. Assume the following:

* For all $N \ge 1$ and for all $\{0\} \subset V_1, \dots, V_N \subset K^d$, we have $\mathbb{P}(\gamma_0(\bigcup V_i) \subset \bigcup V_i) < 1$.

* There exists $n \ge 1$ such that $\mathbb{P}(\rho_1(\bar{\gamma}_n) > \rho_2(\bar{\gamma}_n)) > 0$.

Then there exists a constant $0 < \lambda_{1-2} \le +\infty$ such that $\frac{1}{n} \log \frac{s_1}{s_2}(\bar{\gamma}_n) \to \lambda_{1-2}$ almost surely. Moreover, for all $\alpha < \lambda_{1-2}$, there exist constants $C, \beta > 0$ such that $\mathbb{P}(\log \frac{s_1}{s_2}(\bar{\gamma}_n) \le \alpha n) \le Ce^{-\beta n}$ for all $n$ and for all $x, y \in \mathbb{R}^d \setminus \{0\}$ and for all $0 \le n \le m$, we have:

$$ (C) \qquad \mathbb{P}(d([\bar{\gamma}_n x], [\bar{\gamma}_m y]) \ge e^{-\alpha n}) \le Ce^{-\beta n}. $$

Let us now introduce some definitions and give more complete and precise statements in a more general setting. Namely we replace $\mathbb{R}$ with any local field and do not assume that the matrices are invertible in Theorem B.

In Paragraph 1.1 we remind some standard notations for products of random matrices and extend the notions of strong irreducibility and proximality to the case of non-invertible matrices. In Paragraph 1.2, we state the main theorem of this paper, which is the main tool in the proof of all the other results. In Paragraph 1.3, we define the limit line and state the exponential contraction results, from which Theorem B follows. In Paragraph 1.4, we explain how these contraction results imply contraction on the space of flags and formulate a conjecture on the Poisson boundary. In Paragraph 1.5, we state the quantitative regularity result from which Theorem A follows and we reformulate these results as regularity results for the invariant measure, in the fashion of the regularity results of Benoist and Quint [BQ16b].
1.1. **Preliminary definitions.** Consider $\mathbb{K}$ a local field i.e., a field endowed with an absolute value $|\cdot| = \mathbb{K} \to \mathbb{R}_{\ge 0}$ that makes it locally compact. Let $d \ge 2$ be fixed and let $E = \mathbb{K}^d$, when $\mathbb{K}$ is Archimedean i.e., $\mathbb{K} = \mathbb{R}$ or $\mathbb{K} = \mathbb{C}$, we endow $E$ with the Euclidean (real) or Hermitian (complex) norm $\|\cdot\| : (x_1, \dots, x_d) \mapsto \sqrt{|x_1|^2 + \dots + |x_d|^2}$, otherwise, we endow $E$ with the ultra-metric norm $\|\cdot\| : (x_1, \dots, x_d) \mapsto \max\{|x_1|, \dots, |x_d|\}$.

For the rest of this paper, we fix a local field $(\mathbb{K}, |\cdot|)$ and a Euclidean, Hermitian or ultra-metric space $(E, \|\cdot\|)$ of dimension $d \ge 2$.

Given $1 \le k \le d$, we write $\wedge^k E$ for the $k$-th exterior product of $E$, which is simply the dual space of the space of $k$-linear alternate forms. For all $j, k \ge 1$ such that $j + k \le d$, we have a bi-linear map $\wedge : \wedge^j E \times \wedge^k E \to \wedge^{j+k} E$. There is also a canonical norm $\|\cdot\|$ on $\wedge^k E$ for all $k$, which is characterized by the fact that for all $j, k$, for all $x \in \wedge^j E$ and for all $y \in \wedge^k E$, we have $\|x \wedge y\| \le \|x\|\|y\|$ and this inequality is sharp in the sense that there exists $y' \in \wedge^k E$ such that $x \wedge y = x \wedge y'$ and such that $\|x \wedge y'\| = \|x\|\|y'\|$.

To be more explicit, the norm on $\wedge^k E$ is the Euclidean (resp. Hermitian, resp. ultra-metric) norm associated to the basis $(e_{i_1} \wedge \dots \wedge e_{i_k})_{1 \le i_1 < \dots < i_k \le d}$, where $(e_i)_{i \le i \le d}$ is the canonical basis of $E$.

We write $\text{End}(E)$ for the semi-group of linear endomorphisms of $E$, which is naturally identified with the set of square matrices of size $d$ with coefficients in $\mathbb{K}$. We write $\text{GL}(E)$ for the group of linear automorphisms of $E$.

Given $g \in \text{End}(E)$, we write $s_1(g) \ge s_2(g) \ge \dots \ge s_d(g) \ge 0$ for the singular values of $g$ counted with multiplicity i.e., the absolute values of the coefficients of the diagonal matrix in the polar decomposition<fnref n="1" />.

The first singular value $s_1(g)$ is simply the operator norm of $g$, that we also denote by $\|g\| = \max_{x \in E \setminus \{0\}} \|gx\|/\|x\|$ and we have the following characterization for the higher order singular values:

$$ \forall g \in \text{End}(E), \forall 2 \le k \le d, s_k(g) = \frac{\| \wedge^k g \|}{\| \wedge^{k-1} g \|}, $$

where $\wedge^k g \in \text{End}(\wedge^k E)$ is the linear map characterized by $\wedge^k g : x_1 \wedge \dots \wedge x_k \mapsto gx_1 \wedge \dots \wedge gx_k$.

Given $g \in \text{End}(E)$, we write $\rho_1(g) \ge \rho_2(g) \ge \dots \ge \rho_d(g)$ for the spectral values of $g$ i.e., the moduli of its eigenvalues counted with multiplicity, or dynamically speaking $\rho_k(g) = \lim_n (s_k(g^n))^{1/n}$. We say that $g$ is proximal if $\rho_1(g) > \rho_2(g)$.

<footnote n="1">¹We remind that the Cartan, or polar, decomposition Theorem tells us that for all matrix $g \in \text{End}(\mathbb{K}^d)$, there exist three matrices $k, a, k'$, with $k, k'$ being isometries and $a$ a diagonal matrix such that $g = kak'$. Moreover, the multi-set of absolute values of the coefficients of $a$ does not depend on the choice of $k$ and $k'$.</footnote>
Given a probability measure $\nu$ on $\text{End}(E)$, we write $(\gamma_n)_{n \ge 0} \sim \nu^{\otimes \mathbb{N}}$ to introduce a sequence of random variables, that are defined on the same probability space $(\Omega, \mathbb{P})$, that are independent and that all have distribution law $\nu$. For the rest of the discussion, we will assume that we are given a probability measure $\nu$ on $\text{End}(E)$ and a random sequence $(\gamma_n)_{n \ge 0} \sim \nu^{\otimes \mathbb{N}}$. We will moreover assume that $\nu$ is strongly irreducible and proximal in the following sense.

**Definition 1.1 (Proximality).** Let $\nu$ be a probability distribution on $\text{End}(E)$. We say that $\nu$ is proximal if both of the following conditions are satisfied:

$$ (1) \qquad \forall n \in \mathbb{N}, \nu^{*n}\{0\} = 0. $$

$$ (2) \qquad \exists n \in \mathbb{N}, \nu^{*n}\{\gamma \in \text{End}(E) \mid \rho_1(\gamma) > \rho_2(\gamma)\} > 0, $$

The condition (1) is trivially satisfied when $\nu$ is supported on the set $\text{GL}(E)$ of invertible matrices. We show in Lemma 3.2 that we have a $0-1$ law for $\nu^{*n}\{0\}$ in the sense that we either have $\nu^{*n}\{0\} = 0$ for all $n$ or there exists $C, \beta > 0$ such that $\nu^{*n}\{0\} \ge 1 - Ce^{-\beta n}$ for all $n \ge 0$. So when the second condition for proximality is not satisfied, the sequence $(\bar{\gamma}_n)_{n \ge 0}$ is almost surely stationary to $0$ for $(\gamma_n) \sim \nu^{\otimes \mathbb{N}}$ and the random first time $n_0$ such that $\bar{\gamma}_{n_0} = 0$ has a finite exponential moment.

**Definition 1.2 (Strong irreducibility).** Let $\nu$ be a probability distribution on $\text{End}(E)$. We say that $\nu$ is irreducible if:

$$ \forall f \in E^* \setminus \{0\}, \forall v \in E \setminus \{0\}, \exists n \ge 0, \nu^{*n}\{\gamma \in \text{End}(E) \mid f\gamma v \ne 0\} > 0. $$

We say that $\nu$ is strongly irreducible if:

$$ \forall N \ge 1, \forall (f_1, \dots, f_N) \in (E^* \setminus \{0\})^N, \forall v \in E \setminus \{0\}, \exists n \ge 0, $$

$$ (3) \qquad \nu^{*n} \left\{ \gamma \in \text{End}(E) \left| \prod_{i=1}^{N} f_i \gamma v \neq 0 \right. \right\} > 0, $$

$$ \text{and} \quad \forall N \ge 1, \forall f \in E^* \setminus \{0\}, \forall (v_1, \dots, v_N) \in (E \setminus \{0\})^N, \exists n \ge 0, $$

$$ (4) \qquad \nu^{*n} \left\{ \gamma \in \text{End}(E) \left| \prod_{j=1}^{N} f_j \gamma v_j \neq 0 \right. \right\} > 0. $$

Note that the above conditions are symmetric in the sense that saying that a measure $\nu$ is strongly irreducible and/or proximal is equivalent to saying that the transpose $^t\nu \in \text{End}(E^*)$ is.

Guivarc'h and Lepage have shown that, under the assumption that $\nu$ gives measure 1 to the set of invertible matrices, the conditions (3) and (4) are equivalent. Moreover, under the proximality assumption, the strong irreducibility is
equivalent to the fact that there is no non-trivial union of linearly independent subspaces of $E$ that is invariant under the action of $\Gamma_\nu$.

Let $\Gamma_\nu$ be the smallest closed semi-group such that $\nu(\Gamma_\nu) = 1$. Note that conditions (2), (3) and (4) are actually conditions on $\Gamma_\nu$ rather than on $\nu$ itself. Indeed, saying that an open set has positive $\nu^{*n}$-measure for some $n$ is equivalent to saying that said open set intersects $\Gamma_\nu$.

Note that without the assumption that $\nu(\mathrm{GL}(E)) = 1$, conditions (3) and (4) are not equivalent. For example: let $\nu$ be the convolution of the normalized Lebesgue measure on the unit sphere in $E = \mathbb{R}^d$ with the uniform measure on a basis $\{w_1, \dots, w_d\}$ of $E^*$; then $\Gamma_\nu = \{uw_i | u \in E, \|u\| \le 1, 1 \le i \le d\} \cup \{\mathrm{Id}\}$ does not preserve any finite union of proper subspaces because $\Gamma_\nu \mathbb{K}v = E$ for all $v \ne 0$ and (3) is satisfied for the same reason, yet $\nu$ is not irreducible in the sense of Definition 1.2 because (4) is not satisfied for $2 \le N \le d$, for $f = w_1$ and for $(v_1, \dots, v_d) \in E^d$ being the dual basis of $(w_1, \dots, w_d)$ (i.e., $w_i v_j = \mathbb{1}_{i=j}$ for all $i, j$).

Note that condition (1) is not an algebraic condition. Indeed, if we consider $\nu$ to be a Gaussian distribution over $\mathrm{End}(E)$, then we have $\Gamma_\nu = \mathrm{End}(E)$ and $\nu(\mathrm{GL}(E)) = 1$ so $\nu$ satisfies conditions (2), (1), (3) and (4), hence it is strongly irreducible and proximal. On the other hand, if $\nu$ is the barycentre of a Gaussian distribution with the Dirac measure at 0, then we still have $\Gamma_\nu = \mathrm{End}(E)$, so conditions (2), (3) and (4) are satisfied, but $\nu^{*n}\{0\} = 1 - 2^{-n}$ for all $n$ so (1) is not satisfied.

## 1.2. Alignment and pivotal extraction.
First let us introduce a useful notation for alignment of matrices.

**Definition 1.3 (Coarse alignment of matrices).** Let $g \in E^* \cup \mathrm{End}(E)$ and let $h \in E \cup \mathrm{End}(E)$. Let $0 < \varepsilon \le 1$, we say that $g$ is aligned with $h$ and write $g\mathbb{A}^\varepsilon h$ if we have:

$$ (5) \qquad \|gh\| \ge \varepsilon\|g\|\|h\|. $$

**Definition 1.4 (Contraction coefficient).** Let $g \in \mathrm{End}(E)\setminus\{0\}$. We write $\sigma(g) = \frac{s_2(g)}{s_1(g)} = \frac{\|g\|}{\|g\|^2}$. We endow $\mathbb{N}$ with the addition map to make it a semi-group. That way, given $(w_k)_{k \ge 0} \in \mathbb{N}^\mathbb{N}$, we have $\overline{w}_k = \sum_{0 \le j < k} w_j$ for all $k \ge 0$.

**Definition 1.5 (Power notation for extractions).** Let $\Gamma$ be a semi-group. Let $(\gamma_n)_{n \ge 0} \in \Gamma^\mathbb{N}$ and let $(w_k)_{k \ge 0} \in \mathbb{N}^\mathbb{N}$ For all $k \ge 0$, we write:

$$ \tilde{\gamma}_k^w := (\gamma_{\overline{w}_k}, \gamma_{\overline{w}_{k+1}}, \dots, \gamma_{\overline{w}_{k+1-1}}) $$

$$ \gamma_k^w := \gamma_{\overline{w}_k} \cdots \gamma_{\overline{w}_{k+1-1}}. $$
Note that given a sequence $(\gamma_n)_{n \ge 0}$ is a semi-group and a sequence of positive integers $(w_k)_{k \ge 0}$, the data of the sequence of words $(\tilde{\gamma}_k^w)_{k \ge 0}$ is equivalent to the joint data of $((\gamma_n)_{n \ge 0}, (w_k)_{k \ge 0})$. The main result of the present paper, from which we deduce the other results, is the following.

**Theorem 1.6.** Let $\mathbb{K}$ be a local field and let $d \ge 2$. Let $\nu$ be a strongly irreducible and proximal probability distribution over $\Gamma = \text{Mat}_{d \times d}(\mathbb{K}) = \text{End}(\mathbb{K}^d)$ or $\Gamma = \text{GL}(\mathbb{K}^d)$. There exist constants $0 < \varepsilon \le 1/2$ and $m \in \mathbb{N}$, a compact $K \subset \Gamma$ and a probability distribution $\mu$ on $\Gamma^\mathbb{N} \times \mathbb{N} \setminus \{0\}^\mathbb{N}$ such that given $((\gamma_n)_{n \ge 0}, (p_n)_{n \ge 0}) \sim \mu$, the following assertions hold.

(1) We have $(\gamma_n)_{n \ge 0} \sim \nu^{\otimes \mathbb{N}}$.

(2) The sequence $(\tilde{\gamma}_{2k}^p)_{k \ge 1}$ is i.i.d. and independent of $\tilde{\gamma}_0^p$.

(3) For all $k \ge 0$, we have $\tilde{\gamma}_{2k+1}^p \in K^m$ and $\sigma(\gamma_{2k+1}^p) \le \frac{\varepsilon^6}{48}$ almost surely.

(4) For all $k \ge 0$, the conditional distribution of $\tilde{\gamma}_{2k+1}^p$ with respect to the data of $(\tilde{\gamma}_{k'}^p)_{k' \ne 2k+1}$ is given by a function of $(\tilde{\gamma}_{2k}^p, \tilde{\gamma}_{2k+2}^p)$ that does not depend on $k$.

(5) Almost surely, and for all $k \ge 0$, we have $\mathbb{P}(g\mathbb{A}^\varepsilon \gamma_{2k+1}^p \mid (\tilde{\gamma}_{k'}^p)_{k' \ne 2k+1}) \ge 3/4$ for all $g$ and $\mathbb{P}(\gamma_{2k+1}^p \mathbb{A}^\varepsilon h \mid (\tilde{\gamma}_{k'}^p)_{k' \ne 2k+1}) \ge 3/4$ for all $h$.

(6) There exists constants $C, \beta > 0$ such that $\mathbb{E}(e^{\beta p_n}) \le C$ for all $n \ge 0$.

(7) For all $A \subset \Gamma \setminus K$, we have $\mathbb{P}(\gamma_n \in A \mid (p_k)_{k \in \mathbb{N}}) \le 2\nu(A)$.

(8) For all $i \le j \le k$ such that $i < k$, for all $f \in E^* \cup \text{End}(E)$ such that $g\mathbb{A}^\varepsilon \gamma_i^p$ and for all $h \in E \cup \text{End}(E)$ such that $\gamma_{k-1}^p \mathbb{A}^\varepsilon h$, we have $g\gamma_i^p \cdots \gamma_{j-1}^p \mathbb{A}^{\frac{\varepsilon}{2}} \gamma_j^p \cdots \gamma_k^p h$.

Note that (3) implies that $p_{2k+1} = m$ almost surely and for all $k$ and (2) implies that $(p_{2k})_{k \ge 1}$ is i.i.d..

**1.3. Contraction results without moment assumptions.** In this section $E$ is a Euclidean, Hermitian or ultra-metric vector space of dimension $d \ge 2$. The measure $\nu$ is assumed to be strongly irreducible and proximal but we may have $\nu(\text{GL}(E)) < 1$.

**Theorem 1.7 (Contraction).** Let $\nu$ be a proximal and strongly irreducible probability distribution on $\text{End}(E)$. Let $(\gamma_n) \sim \nu^{\otimes \mathbb{N}}$. Write $\bar{\gamma}_n := \gamma_0 \cdots \gamma_{n-1}$ for all $n$. Then there exists a positive constant $\lambda_{1-2}(\nu) \in (0, +\infty]$ such that almost surely $\frac{-\log \sigma(\bar{\gamma}_n)}{n} \to \lambda_{1-2}(\nu)$. Moreover, we have the following large deviations inequalities:

$$ (6) \quad \forall \alpha < \lambda_{1-2}(\nu), \exists C, \beta > 0, \forall n \in \mathbb{N}, \mathbb{P}(-\log \sigma(\bar{\gamma}_n) \le \alpha n) \le Ce^{-\beta n}. $$
The quantitative estimates in (6) are new even in the setting of [GR86]. In fact they are key to our approach and easier to prove that the qualitative convergence result. We deduce the qualitative convergence from the strong quantitative estimates.

We denote by $P(E)$ the projective space associated to $E$ i.e., the set of vector lines in $E$. Write $[\cdot] : E \setminus \{0\} \to P(E)$ for the projection map. We endow $P(E)$ with the metric:

$$ (7) \qquad \mathrm{d} : ([x], [y]) \mapsto \frac{\|x \wedge y\|}{\|x\|\|y\|} = \min_{c \in \mathbb{K}} \frac{\|x - cy\|}{\|x\|}. $$

**Theorem 1.8 (Quantitative convergence of the image).** Let $\nu$ be a strongly irreducible and proximal probability distribution on End($E$). Let $(\gamma_n) \sim \nu^{\otimes N}$. There exists a random line $l^\infty \in P(E)$ such that for all $\alpha < \lambda_{1-2}(\nu)$, there exist constants $C, \beta > 0$, such that:

$$ (8) \quad \forall v \in E \setminus \{0\}, \forall n \in \mathbb{N}, \mathbb{P}(d([\bar{\gamma}_n v], l^\infty) \ge \exp(-\alpha n) | \bar{\gamma}_n v \ne 0) \le C e^{-\beta n}, $$

Let us now prove how Theorem 1.8 implies Theorem B.

*Proof of Theorem B.* Let $\nu$ be a strongly irreducible and proximal probability distribution on GL($E$). Let $l^\infty$ be as in Theorem 1.8. Let $x, y \in E \setminus \{0\}$ and let $0 \le n \le m$. Let $0 < \alpha < \lambda_{1-2}(\nu)$ and let $C, \beta > 0$ be as in Theorem 1.8. Let $C' := \max\{2, 2C\}$ and let $\beta' := \min\{\alpha, \beta\}$. By sub-additivity of $\mathbb{P}$, we have:

$$ \mathbb{P}(d([\bar{\gamma}_n x], l^\infty) \ge e^{-\alpha n} \cup d([\bar{\gamma}_m y], l^\infty) \ge e^{-\alpha m}) \le C e^{-\beta n} + C e^{-\beta m}. $$

Moreover, we have $e^{-\alpha n} + e^{-\alpha m} \le C'e^{-\beta'n}$ and $C e^{-\beta n} + C e^{-\beta m} \le C'e^{-\beta'n}$. Then by triangular inequality, we have:

$$ \mathbb{P} \left( d([\bar{\gamma}_n x], [\bar{\gamma}_m y]) \ge C' e^{-\beta'n} \right) \le C' e^{-\beta'n}. \quad \square $$

We assumed with the proximality condition that $\mathbb{P}(\bar{\gamma}_n = 0) = 0$ for all $n$. However, given a non-random vector $v$, we may have $\mathbb{P}(\bar{\gamma}_n v = 0) > 0$ when $\bar{\gamma}_n$ is not invertible. We show in Lemma 3.4 that the set $\ker(\nu)$ of vectors $v$ such that $\sup_n \mathbb{P}(\bar{\gamma}_n v = 0) > 0$ is a countable union of proper subspaces of $E$ and we show that that $\mathbb{P}(\bar{\gamma}_n v = 0)$ is bounded above by a constant that does not depend on $n$ or $v$ but only on $\nu$. Therefore (8) is not trivial for any $v$ because $\mathbb{P}(\bar{\gamma}_n v \ne 0)$ is bounded below by a positive constant. Note that (8) is non-trivial as long as $\mathbb{P}(\bar{\gamma}_n = 0) < 1$ for all $n$ and in this case it is an interesting open question to ask whether it remains true or not with the same algebraic assumptions on $\Gamma_\nu$.
When $v$ is such that $\sup_n \mathbb{P}(\bar{\gamma}_n v = 0) = 0$ (which is true for $v$ in a dense and uncountable subset of $P(E)$ by Lemma 3.4) (8) implies that $[\bar{\gamma}_n v] \to l^\infty$ almost surely.

We define $\xi_\nu^\infty$ to be the distribution of $l^\infty$. Then $\xi_\nu^\infty$ is the only $\nu$-stationary measure on $P(E)$ in the sense that $\nu * \xi_\nu^\infty = \xi_\nu^\infty$. A direct consequence of Theorem 1.8 is the following result, which we can see as an exponential mixing result for the Wasserstein metric on the projective space.

**Corollary 1.9 (Exponential mixing).** Let $\nu$ be a strongly irreducible and proximal probability distribution on End($E$). There exists a unique $\nu$-stationary probability distribution $\xi_\nu^\infty$ on $P(E)$. Moreover, there exist constants $C, \beta$ such that for all probability distribution $\xi$ on $P(E) \setminus \ker(\nu)$ and for all Lipschitz function $f: P(E) \to \mathbb{R}$ with Lipschitz constant $\lambda(f)$, we have:

$$ (9) \quad \forall n \in \mathbb{N}, \quad \left| \int_{P(E)} f d\xi_\nu^\infty - \int_{P(E)} f d\nu^{*n} * \xi \right| \le \lambda(f) C \exp(-\beta n). $$

*Proof.* Let $0 < \alpha < \lambda_{1-2}(\nu)$ and let $C, \beta > 0$ be constants that satisfy (8) in Theorem 1.8. Let $[x], (\gamma_n)_n \sim \xi \otimes \nu^{\otimes N}$. Since $\mathbb{P}(x \in \ker(\nu)) = 0$, we have $\bar{\gamma}_n x \ne 0$ almost surely and for all $n$. Let $l^\infty$ be as in Theorem 1.8. Then $\mathbb{E}(f(l^\infty)) = \int f d\xi_\nu^\infty$ for all $f$. Moreover, for all $n$, we have $\mathbb{P}(d([\bar{\gamma}_n x], l^\infty) \le e^{-\alpha n}) \le Ce^{-\beta n}$. Moreover the distance map takes values in $[0, 1]$ so $\mathbb{E}(d([\bar{\gamma}_n x], l^\infty)) \le Ce^{-\beta n} + e^{-\alpha n} \le C'e^{-\beta'n}$ for $C' = C + 1$ and $\beta' = \min\{\beta, \alpha\}$. Then, by convexity of the absolute value, we have $|\mathbb{E}(f[\bar{\gamma}_n x]) - \mathbb{E}(f(l^\infty))| \le \mathbb{E}(|f[\bar{\gamma}_n x] - f(l^\infty)|) \le \mathbb{E}(\lambda(f)d([\bar{\gamma}_n x], l^\infty)) \le \lambda(f)C'e^{-\beta'n}$. $\square$

Note that saying that $\xi$ is supported on $P(E) \setminus \ker(\nu)$ is not very restrictive because any measure that gives measure 0 to all hyperplanes would satisfy that condition. However, $\xi_\nu^\infty$ itself may give positive measure to some hyperplanes. For example if $\nu$ is the barycentre of the Haar measure on the group of isometries and a Dirac mass $\delta_\pi$ at a projection endomorphism $\pi$, then $\xi_\nu^\infty$ is the average of the isometry-invariant measure and of the Dirac mass on the image of $\pi$. In particular $\xi_\nu^\infty$ gives positive measure to any hyperplane that contains the image of $\pi$.

Let $g$ be a square matrix such that $\rho_1(g) > \rho_2(g)$. Then the top eigenvalue of $g$ is simple and in $\mathbb{K}$. We write $E^+(g)$ for the associated eigenspace, which is a $\mathbb{K}$-line.

**Theorem 1.10 (Contraction properties for the dominant eigenspace).** Let $E$ be a Euclidean space and let $\nu$ be a proximal and strongly irreducible probability
distribution on End($E$). Let $(\gamma_n) \sim \nu^{\otimes \mathbb{N}}$. Then for all $\alpha < \lambda_{1-2}(\nu)$, there exist constants $C, \beta > 0$ such that for all $n > 0$, we have:

$$ (10) \qquad \mathbb{P} \left( \rho_2(\bar{\gamma}_n) \ge e^{-\alpha n} \rho_1(\bar{\gamma}_n) \cup d(E^+(\bar{\gamma}_n), l^\infty) \ge e^{-\alpha n} \right) \le C e^{-\beta n}. $$

Theorem 1.10 implies that we have $E^+(\bar{\gamma}_n) \to l^\infty$ almost surely and that $\liminf_n \log(\rho_1(\bar{\gamma}_n)/\rho_2(\bar{\gamma}_n)) \ge \lambda_{1-2}(\nu)$ almost surely. The question whether this $\liminf$ is actually a limit or not remains open when $\nu$ is not in $L^1$ (See corollary 1.15 for the $L^1$ case). In dimension 2, we have $\log \frac{\rho_1}{\rho_2} \le \log \frac{s_1}{s_2}$ so we know that $\frac{1}{n} \log \frac{\rho_1}{\rho_2}(\bar{\gamma}_n) \to \lambda_{1-2}(\nu)$ almost surely by Theorem 1.7.

Note that if $\nu$ is supported on $\text{GL}(E)$, then $\ker(\nu) = \{0\}$. The existence and uniqueness of the stationary measure are well known in this case. This was in fact the first step towards the formalization of boundary theory by Furstenberg [Fur73]. Even in this case, with the pivoting technique, we get regularity results for the stationary measure which are better than the ones obtained using ergodic theory.

## 1.4. Convergence in the space of flags and Poisson boundary.
In [GR86], Guivarc'h and Raugi derive contraction results on the flag space for totally strongly irreducible measures (i.e., the push-forward $\bigwedge_*^k \nu$ is strongly irreducible for all $k < d$). In this paragraph, we note point out that the same strategy works without moment conditions.

We write $\text{Gr}_k(E)$ for the set of subspaces of $E$. For all $k$, we write $\text{Gr}_k(E)$ for the set of subspaces of $E$ of dimension $k$. Given $\Theta \subset \{1, \dots, d-1\}$, we write $\text{Fl}_\Theta(E)$ for the set of partial flags of type $\Theta$, which is the space of totally ordered$^2$ collections of subspaces of $E$ whose set of dimensions (i.e., its image by $\dim : \text{Gr}(E) \to \{1, \dots, d\}$) is $\Theta$. Given $F$ a partial flag of type $\Theta$ and $k \in \Theta$, we write $F_k \in \text{Gr}_k(E)$ for the element of $F$ that has dimension $k$. For all $g \in \text{End}(E)$, we write $F^{\text{eigen}}(g)$ for the collection for $\lambda \ge 0$ of the sums of the eigenspaces of $g$ associated to an eigenvalue of modulus at least $\lambda$. Write $\Theta(g)$ for the type of $F^{\text{eigen}}(g)$, note that $\Theta(g) = \{k | \rho_k > \rho_{k+1}\}$.

The following straightforward corollary of Theorems 1.8 and 1.10 can be interpreted as an extension of Oseledets multiplicative ergodic theorem in the i.i.d. case without the $L^1$ moment condition.

**Corollary 1.11** (Contraction property in the flag space.). Let $\nu$ be a probability measure on End($E$). Let $\Theta(\nu)$ be the set of indices $k < d$ such that $\bigwedge_*^k \nu$ is proximal. Assume that $\bigwedge_*^k \nu$ is strongly irreducible for all $k \in \Theta(\nu)$. Let

<footnote>$^2$For the inclusion order relation.</footnote>
$(\gamma_n)_{n \ge 0} \sim \nu^{\otimes \mathbb{N}}$. Then there exists a measurable map $F^{\infty}$ that is defined $\nu^{\otimes \mathbb{N}}$ almost everywhere and such that for all flag $F \in \text{Fl}_{\Theta(\nu)}(E)$, we have $\bar{\gamma}_n F \to F^{\infty}(\gamma)$ almost surely. Moreover, we have $F^{\text{eigen}}(\bar{\gamma}_n) \to F^{\infty}(\gamma)$ almost surely. Moreover, for all $i \le i < j \le d$, we have $\lim_n \frac{1}{n} \log \frac{s_i}{s_j}(\bar{\gamma}_n) = \lambda_{i-j}(\nu)$ almost surely, with $\lambda_{i-j}(\nu) := \sum_{k \in \Theta(\nu), i \le k < j} \lambda_{1-2}(\wedge_*^k \nu)$.

**Proof.** We apply Theorems 1.8 and 1.10 to the random sequence $(\wedge_*^k \bar{\gamma}_n)_{n \ge 0}$ for all $k \in \Theta(\nu)$. We get that $E^+(\wedge_*^k \bar{\gamma}_n)$ converges almost surely to a random projective class in $\text{P}(\wedge_*^k E)$, let us call $l^+(\wedge_*^k \gamma)$ said projective class. Now we need to show that $l^+(\wedge_*^k \gamma)$ is pure in the sense that it is of the form $\wedge_*^k V$ for $V \in \text{Gr}_k(E)$. We know that $E^+(\wedge_*^k g) = \wedge_*^k F_k^{\text{eigen}}(g)$ for all $g \in \text{End}(E)$ such that $\rho_k(g) > \rho_{k+1}(g)$. Moreover, the set of pure projective classes is closed so $l^+(\wedge_*^k \gamma) = \lim_n E^+(\wedge_*^k \bar{\gamma}_n)$ is a pure projective class. Write $l^+(\wedge_*^k \gamma) = \wedge_*^k F_k^{\infty}(\gamma)$. Note that, the inclusion relation is closed in $\text{Gr}(E)$ so the family $(F_k^{\infty}(\gamma))_{k \in \Theta(\nu)}$, which is a limit of partial flags of type $\Theta(\nu)$ is nested and therefore a partial flag. $\square$

The above corollary implies that there exists a unique $\nu$ invariant measure on $\text{Fl}_{\Theta(\nu)}(E)$. In the proximal case (i.e., $1 \in \Theta(\nu)$), the invariant measure on $\text{P}(E)$ is the push-forward of the invariant measure on $\text{Fl}_{\Theta(\nu)}(E)$ by the map $F \mapsto F_1$. We may wonder whether the space $\text{Fl}_{\Theta(\nu)}(E)$ endowed with said invariant measure is itself the push-forward of the unique invariant measure on a bigger $\Gamma_\nu$-space (i.e., a compact space endowed with a continuous action of $\Gamma_\nu$). In other words, we wonder if $\text{Fl}_{\Theta(\nu)}(E)$ is isomorphic to the Poisson boundary of the Markov chain $(\bar{\gamma}_n)_{n \ge 0}$. This is not true in general for arithmetic reasons<fnref n="3" /> but we do not have any counterexamples in the case where $\Gamma_\nu$ is discrete. Using the pivoting technique, one should be able to prove the following result.

**Conjecture 1.12.** Let $\nu$ be a probability measure on $\text{End}(E)$. Assume that $\wedge_*^k \nu$ is strongly irreducible for all $k \in \Theta(\nu)$. Assume also that $\Gamma_\nu$ is discrete and that $\nu$ has finite entropy i.e., $\sum_{g \in \text{End}(E)} \nu\{g\} \log \nu\{g\} > -\infty$. Then the Poisson boundary of $\nu$ is isomorphic to $\text{Fl}_{\Theta(\nu)}(E)$ endowed with the probability distribution $F_*^\infty \nu^{\otimes \mathbb{N}}$.

Kaimanovich's entropy criterion tells us that this is equivalent to saying that the entropy of the random walk $(\bar{\gamma}_n)_{n \ge 0}$ conditioned to the value of the limit flag $F^{\infty}(\gamma)$ is zero. A similar result is shown in [CFFT22] for random walks on hyperbolic groups.

<footnote n="3">³If $\nu$ is a measure that generates $\text{GL}(\mathbb{Q}^d)$ for example, then there exists a unique invariant measure on the product $\prod_p \text{Fl}(\mathbb{Q}_p^d) \times \text{Fl}(\mathbb{R}^d)$.</footnote>
1.5. **Regularity results with optimal moment assumptions.** In This Paragraph, $E$ is a Euclidean, Hermitian or ultra-metric vector space of dimension $d \ge 2$ and $\nu$ is a probability distribution over $GL(E)$. For all $g \in GL(E)$, we write $N(g) = \log \|g\|\|g^{-1}\|$. We assume that $\nu$ is strongly irreducible and proximal in the sense of Definitions 1.1 and 1.2. Since $\nu$ is supported on $GL(E)$, these notions are equivalent to the usual notions of proximality and strong irreducibility introduced by Guivarc'h and Lepage.

Under these purely algebraic assumption, we give concentration results for the distribution of the coefficients and of the spectral radius that depend on the tail of $N_*\nu$. From these results, and from Theorem 1.8, we deduce regularity results for the invariant measure.

**Theorem 1.13** (Probabilistic estimates for the coefficients and the spectral radius). Let $\nu$ be strongly irreducible and proximal over $GL(E)$. There exist constants $C, \beta > 0$ such that for all $f \in E^* \setminus \{0\}$, all $v \in E \setminus \{0\}$, for all $n \in \mathbb{N}$, for $\bar{\gamma}_n \sim \nu^{*n}$ and for all $t \in \mathbb{R}_{\ge 0}$, we have:

$$ (11) \quad \mathbb{P} \left( \log \frac{\|f\| \|\bar{\gamma}_n\| \|v\|}{|f\bar{\gamma}_n v|} > t \right) \le Ce^{-\beta n} + \sum_{k=1}^{+\infty} Ce^{-\beta k} N_* \nu (t/k, +\infty). $$

Moreover:

$$ (12) \quad \forall t \ge 0, \quad \mathbb{P} \left( \log \frac{\|\bar{\gamma}_n\|}{\rho_1(\bar{\gamma}_n)} > t \right) \le \sum_{k=1}^{+\infty} Ce^{-\beta k} N_* \nu (t/k, +\infty). $$

The term $Ce^{-\beta n}$ in (11) naturally appears in the proof. This term is annoying because it prevents us from giving a bound on the moments of $-\log \frac{|f\bar{\gamma}_n v|}{\|f\|\|\bar{\gamma}_n\|\|v\|}$. However on can not get rid of it, to give a clear example, let $\nu$ be a lazy measure that gives measure 1/2 to the identity matrix, then given $fv = 0$, we have $\mathbb{P}(f\bar{\gamma}_n v = 0) \ge 1 - 2^{-n}$ for all $n$.

Note also that (11) implies that for all non-random sequence $\alpha_n \to 0$ and for all $1 \le i, j \le \dim(E)$, the sequence $\left( \frac{\| (\bar{\gamma}_n)_{i,j} \|^\alpha_n}{\|\bar{\gamma}_n\|^\alpha_n} \right)_n$ converges weakly in distribution to the Dirac measure at 1, without any moment assumption. The same holds for $\rho_1$ by (12). This observation implies that any weak convergence result that holds for $\log \|\bar{\gamma}_n\|$ immediately holds for $\log |(\bar{\gamma}_n)_{i,j}|$ and $\log \rho_1(\bar{\gamma}_n)$.

Given $C, \beta > 0$ and $\nu$ a probability measure on $GL(E)$, we denote by $\zeta_\nu^{C,\beta}$ the probability distribution on $\mathbb{R}_{\ge 0}$ characterized by the fact that for all $t \ge 0$, we have:

$$ (13) \quad \zeta_\nu^{C,\beta}(t, +\infty) = \min \left\{ 1, \sum_{k=1}^{+\infty} C \exp(-\beta k) N_* \nu (t/k, +\infty) \right\}. $$
Note that $\sum_{k=1}^{+\infty} C \exp(-\beta k) < +\infty$ so $\sum_{k=1}^{+\infty} C \exp(-\beta k) N_* \nu (t/k, +\infty)$ goes to 0 as $t \to +\infty$ by the monotonous convergence Theorem. Moreover, since everything is non-negative, we have:

$$ (14) \qquad \begin{aligned} \mathbb{E}(\zeta_{\nu}^{C,\beta}) &= \int_{0}^{\infty} \zeta_{\nu}^{C,\beta}(t, +\infty)dt \\ &\le \int_{0}^{\infty} \sum_{k=0}^{+\infty} Ce^{-\beta k} N_{*}\nu(t/k, +\infty)dt \\ &\le \sum_{k=0}^{+\infty} Ce^{-\beta k} \int_{0}^{\infty} kN_{*}\nu(t, +\infty)dt \end{aligned} \\ \mathbb{E}(\zeta_{\nu}^{C,\beta}) \le \frac{C}{e^{\beta} - 2 + e^{-\beta}} \mathbb{E}(N_{*}\nu). $$

Let us now derive Theorem A from 1.13.

*Proof of Theorem A.* By (14), we have $\mathbb{E}(\zeta_{\nu}^{C,\beta}) < +\infty$. Then by Theorem 1.13, for all $\varepsilon > 0$, we have:

$$ \sum_{n \in \mathbb{N}} \mathbb{P} \left( \log \frac{\|f\| \|\bar{\gamma}_n\| \|v\|}{|f \bar{\gamma}_n v|} \ge n\varepsilon \right) \le \sum_{n \in \mathbb{N}} C e^{-\beta n} + \sum_{n \in \mathbb{N}} \zeta_{\nu}^{C,\beta}(n\varepsilon, +\infty) \\ \le \frac{C}{1 - e^{-\beta}} + \varepsilon^{-1} \mathbb{E}(\zeta_{\nu}^{C,\beta}) < +\infty. $$

Then by Borel-Cantelli's Lemma, we have $n^{-1} \log \frac{\|f\| \|\bar{\gamma}_n\| \|v\|}{|f \bar{\gamma}_n v|} \to 0$ almost surely. Moreover, by Furstenberg and Kesten [FK60, Theorem 1], we have $n^{-1} \log \|\bar{\gamma}_n\| \to \lambda_1(\nu)$. The same proof works for $\rho_1$ because the term $C e^{-\beta n}$ is positive. $\square$

The following Corollary is about the regularity of the stationary measure. The formulation (16) is analogous to the regularity result for the stationary measure on hyperbolic groups [BQ16c, Proposition 5.1]. This is also an improvement of the regularity results for the invariant measure on the projective space stated in [BQ16b, Proposition 4.5].

Given $V \in \text{Gr}(E)$, and $l \in \text{P}(E)$, we write $\text{d}(\text{P}(V), l) := \min_{v \in V \setminus \{0\}} \text{d}([v], l)$ and given $0 < r \le 1$, we define $\mathcal{N}_r(V) := \{l \in \text{P}(E) \mid \text{d}(\text{P}(V), l) < r\}$. Given $p > 0$, we say that a random variable $x \in \mathbb{R}_{\ge 0}$ has a finite $\text{L}^p$ moment if $\mathbb{E}(x^p) < +\infty$ and we say that $x$ has a weak $\text{L}^p$ moment if $\sup_{t \ge 0} t^p \mathbb{P}(x > t) < +\infty$. By integration by parts, we have $\mathbb{E}(x^p) = \int_0^\infty pt^{p-1} \mathbb{P}(x > t)dt$ for all $p > 0$. So with the same computations as for (14), for all $p > 0$, the (weak) $\text{L}^p$ moment of $\zeta_{\nu}^{C,\beta}$ is bounded above by the (weak) $\text{L}^p$ moment of $N_*\nu$ times a constant that depends on $C, \beta$ and $p$.
**Corollary 1.14** (Regularity of the measure). Let $E$ be a Euclidean vector space. Let $\nu$ be a strongly irreducible and proximal probability measure on $GL(E)$. Let $C, \beta$ be as in Theorem 1.13 and let $\xi_{\nu}^{\infty}$ be the only $\nu$-stationary measure on $P(E)$. Then we have:

$$ (15) \quad \forall V \in \text{Gr}(E) \setminus \{E\}, \forall 0 < r \le 1, \xi_{\nu}^{\infty}(\mathcal{N}_r(V)) \le \zeta_{\nu}^{C,\beta}(|\log(r)|, +\infty). $$

Let $p > 0$. If we assume that $N_*\nu$ has a finite $L^p$ moment, then there exists a constant $C'$ such that:

$$ (16) \quad \forall V \in \text{Gr}(E) \setminus \{E\}, \quad \int_{l \in P(E)} |\log d(P(V), l)|^p d\xi_{\nu}^{\infty}(l) \le C'. $$

If we assume that $N_*\nu$ has a weak $L^p$ moment, then there exists a constant $C'$ such that:

$$ (17) \quad \forall V \in \text{Gr}(E) \setminus \{E\}, \forall 0 < r < 1, \xi_{\nu}^{\infty}(\mathcal{N}_r(V)) \le C' |\log(r)|^{-p}. $$

*Proof.* First note that (16) and (17) follow directly from (15). Let us prove (15). Let $f \in E^* \setminus \{0\}$ be such that $fV = \{0\}$ and let $u \in E \setminus \{0\}$ be constant. Let $\gamma = (\gamma_n)_{n \ge 0} \sim \nu^{\otimes \mathbb{N}}$. We claim that:

$$ (18) \quad \xi_{\nu}^{\infty}(\mathcal{N}_r(V)) \le \limsup_{n \to +\infty} \mathbb{P} (|f \bar{\gamma}_n u| < r \|f\| \|\bar{\gamma}_n\| \|u\|). $$

Let $l^{\infty}$ be as in Theorem 1.8. We have $l^{\infty} \sim \xi_{\nu}^{\infty}$ so $\xi_{\nu}^{\infty}(\mathcal{N}_r(V)) = \mathbb{P}(l^{\infty} \in \mathcal{N}_r(V))$. Moreover, by Theorem 1.8, we have $[\bar{\gamma}_n u] \to l^{\infty}$ almost surely and $\mathcal{N}_r(V)$ is open, so assuming that $l^{\infty} \in \mathcal{N}_r(V)$ implies that there exists $n_0$ such that $[\bar{\gamma}_n u] \in \mathcal{N}_r(V)$ for all $n \ge n_0$. Therefore, we have $\xi_{\nu}^{\infty}(\mathcal{N}_r(V)) \le \lim_n \mathbb{P}([\bar{\gamma}_n u] \in \mathcal{N}_r(V))$.

Moreover, we have $\mathcal{N}_r(V) \subset \{[v] \mid |fV| < r \|f\| \|v\|\}$ so $([\bar{\gamma}_n u] \in \mathcal{N}_r(V)) \subset (|f\bar{\gamma}_n u| < r \|f\| \|\bar{\gamma}_n u\|)$. Moreover $\|\bar{\gamma}_n u\| \le \|\bar{\gamma}_n\| \|u\|$ so we have (18). Then by Theorem 1.13, we have $\limsup_n \mathbb{P} (|f\bar{\gamma}_n u| < r \|f\| \|\bar{\gamma}_n\| \|u\|) \le \zeta_{\nu}^{C,\beta}(|\log(r)|, +\infty)$. $\square$

Note also that by reducing to the strongly irreducible and proximal case with the same reasoning as in [GR86], we can get rid of the algebraic assumptions for the law of large numbers on the spectral radius. Let us write $\lambda = \log \rho$.

**Corollary 1.15** (Law of large numbers for the spectral radius without algebraic assumptions). Let $\nu$ be any measure on $GL(E)$. Let $(\gamma_n)_{n \ge 0} \sim \nu^{\otimes \mathbb{N}}$. Assume that $\mathbb{E}(\log^+ \|\gamma_0\|) < +\infty$ and that $\mathbb{E}(\log^+ \|\gamma_0^{-1}\|) < +\infty$. Then for all $1 \le k \le d$, we have almost surely:

$$ (19) \quad \lim_{n \to +\infty} \frac{\lambda_k(\bar{\gamma}_n)}{n} = \lambda_k(\nu). $$
*Proof.* We remind that by definition $\lambda_k(\nu) = \lim_{n \to +\infty} \frac{\log s_k(\bar{\tau}_n)}{n}$. Let $\Gamma_\nu$ be the closed semi-group generated by $\nu$.

Assume that $\Gamma_\nu$ is irreducible and that $\lambda_1(\nu) > \lambda_2(\nu)$. Then by [GR85], $\Gamma_\nu$ is actually strongly irreducible and proximal so $\frac{\lambda_1(\bar{\tau}_n)}{n} \to \lambda_1(\nu)$ almost surely by Theorem A.

Now let us reduce the problem to the above case. First note that for all $g$, by sub-multiplicativity of the norm, we have $\rho_1(g) \le s_1(g) = \|g\|$ the same holds for $g^{-1}$ so $s_d(g) \le \rho_d(g)$ and therefore $s_d(g) \le \rho_k(g) \le s_1(g)$ for all $1 \le k \le d$. Hence, if we assume that $\lambda_1(\nu) = \lambda_d(\nu)$, then (19) holds. Note also that (19) is simply the law of large numbers when $d=1$.

Assume that $\lambda_1(\nu) > \lambda_2(\nu)$ and that $\Gamma_\nu$ is not irreducible. Then $\Gamma_\nu$ is conjugated to a semi-group of block-wise upper triangular matrices whose diagonal blocks are all irreducible. Moreover, we know that the Lyapunov spectrum of $\nu$ is the union of the Lyapunov spectrums of the diagonal blocks and the spectrum of each element $g \in \Gamma_\nu$ is the union of the spectrums of the diagonal blocks of $g$. Then for all $g \in \Gamma_\nu$, we write $\pi(g)$ for the restriction of $g$ to the diagonal block on which $\lambda_1(\nu)$ appears. Then $\pi_*\nu$ is strongly irreducible by assumption and $\lambda_1(\pi_*\nu) = \lambda_1(\nu)$ by assumption, moreover $\lambda_2(\pi_*\nu) = \lambda_j(\nu)$ for some $j \ge 2$ so $\lambda_2(\pi_*\nu) < \lambda_1(\pi_*\nu)$. Therefore, we have $\frac{\lambda_1\pi(\bar{\tau}_n)}{n} \to \lambda_1(\nu)$ almost surely. Moreover, we have $\rho_1\pi(g) \le \rho_1(g)$ for all $g \in \Gamma_\nu$ so we have $\frac{\lambda_1(\bar{\tau}_n)}{n} \to \lambda_1(\nu)$ almost surely.

Now assume that $\lambda_1(\nu) = \lambda_2(\nu) > \lambda_d(\nu)$. Let $j$ be the smallest integers such that $\lambda_j(\nu) > \lambda_{j+1}(\nu)$. Then $\lambda_1(\Lambda^j \nu) = j\lambda_j(\nu) > \lambda_2(\Lambda^j \nu) = (j-1)\lambda_j(\nu) + \lambda_{j+1}(\nu)$. Therefore, by the above argument, we have $\frac{\lambda_1(\Lambda^j \bar{\tau}_n)}{n} \to j\lambda_j(\nu)$ almost surely. Therefore, we have $\frac{\lambda_1(\bar{\tau}_n) + \dots + \lambda_j(\bar{\tau}_n)}{n} \to j\lambda_1(\nu)$ almost surely. Moreover, we always have $\limsup_n \frac{\lambda_k(\bar{\tau}_n)}{n} \le \lambda_1(\nu)$ for all $k$. Therefore, we have $\frac{\log \rho_1(\bar{\tau}_n)}{n} \to \lambda_1(\nu)$ almost surely.

In conclusion, without any algebraic assumptions on $\nu$, we have $\frac{\log \rho_1(\bar{\tau}_n)}{n} \to \lambda_1(\nu)$ almost surely. Then we apply the formula $\rho_k(g) = \rho_1(\Lambda^k g)/\rho_1(\Lambda^{k-1} g)$ to prove (19) for all $k \ge 2$. $\square$

**1.6. Background.** The study or products of random matrices bloomed with the eponym article [FK60] where Furstenberg and Kesten construct an escape speed for the logarithm of the norm using the sub-additivity. This proof was generalized by Kingman's sub-additive ergodic Theorem [Kin68]. This article followed the works of Bellman [Bel54] who showed the almost sure convergence of the rescaled logarithms of coefficients as well as a central limit theorem for one specific example. In [FK60] Furstenberg and Kesten show that we have a law of large numbers for the norm under a strong L$^1$ moment condition for $\log \| \cdot \|$.
For matrices that have positive entries and under an $L^\infty$ moment condition, they show that moreover, we have a law of large numbers for the coefficients (entries) and under an additional $L^{2+\delta}$ moment assumption, they show that we have a central limit Theorem. These works on matrices inspired the theory of measurable boundary theory for random walks on groups [Fur73]. In [BL85], Bougerol and Lacroix give an overview of the field of study with applications to quantum physics.

In [GR86], Guivarc'h and Raugi show a qualitative version of Theorem 1.7: in the case when $\nu$ is proximal and strongly irreducible, the two top Lyapunov exponents are distinct. In [GR89] the same authors show that we have almost sure convergence of the limit flag for totally strongly irreducible distributions. In [GM89] Goldsheid and Margulis show that the distribution $\nu$ is proximal and totally strongly irreducible when the support of $\nu$ generates a Zariski-dense subgroup of $SL(E)$.

In [BQ16a] Yves Benoist and Jean-François Quint give an extensive state of the art overview of the field of study with an emphasis on the algebraic properties of semi-groups. Later, in [XGL21] Xiao, Grama and Liu use [BQ16b] to show that coefficients satisfy a law of large numbers under some technical $L^2$ moment assumption. We can also mention [GQX20] and [XGL22] that give other probabilistic estimates for the distribution of the coefficients. The strong law of large numbers and central limit-theorem for the spectral radius were proven by Aoun and Sert in [Aou20] and in [AS21] under an $L^2$ moment assumption.

The importance of alignment of matrices was first noted in [AMS95] along with the importance of Schottky sets. Those notions were then used by Aoun in [Aou11] where he uses it to show that independent draws of an irreducible random walk that has finite exponential moment generate a free group outside of an exponentially rare event (note that the pivoting technique allows us to drop the finite exponential moment assumption). In [CDJ16] and [CDM17], Cuny, Dedecker, Jan and Merlevède give KMT estimates for the behaviour of $(\log \|\bar{\gamma}_n\|)_{n \in \mathbb{N}}$ under $L^p$ moment assumptions for $p > 2$.

The main difference between these previous works and this paper is that the measure $\nu$ has to be supported on the General Linear group $GL(E)$ for the above methods to work. Indeed, they rely on the existence of the stationary measure $\xi_\nu^\infty$ on $P(E)$, which is a consequence of the fact that $GL(E)$ acts continuously on $P(E)$, which is compact. Some work has been done to study non-invertible matrices in the specific case of matrices that have real positive coefficients. In [FK60], Furstenberg and Kesten show limit laws for the coefficients under an $L^\infty$ moment assumption, in [Muk87] and [KS84] Mukherjea, Kesten and Spitzer show some limit theorems for matrices with non-negative entries that are later improved by
Hennion in [Hen97] and more recently improved by Cuny, Dedecker and Mer-
levède in [CDM23].

In [LP82], Le Page shows the exponential mixing property by exhibiting a spectral gap for the action of $\nu$ on the projective space under some moments assumptions on $\nu$. The large deviations inequalities were already known for the norm in the specific case of distributions having finite exponential moment by the works of Sert [Ser18].

1.7. **About the pivoting technique.** In the second part of this article we mainly use the tools introduced in [Gou22], some of them having been introduced or used in former works like [BMSS20] where Adrien Boulanger, Pierre Mathieu, Cagri Sert and Alessandro Sisto state large deviations inequalities from below for random walks in discrete hyperbolic groups or [MS20] where Mathieu and Sisto show some bi-lateral large deviations inequalities in the context of distributions that have a finite exponential moment. In [Gou22] Sébastien Gouëzel uses the pivoting technique in the setting of hyperbolic groups to get large deviations estimates below the escape speed and to show the continuity of the escape speed. For us, the most interesting part of Gouëzel's work is the "toy model" described in section 2. In [Cho24a] [Cho24b] [Cho24c] Inhyeok Choi applies the pivoting technique to show results that are analogous to the ones of Gouëzel for the mapping class group of an hyperbolic surface as well as other groups that satisfy some contraction property. In [CFFT22], Chawla, Forghani, Frisch and Tiozzo use another view of the pivoting technique and the results of [Gou22] to show that the Poisson boundary of random walk with finite entropy on a group that has an acylindrical action on an hyperbolic space is in fact the Gromov Boundary of said space. I believe that similar method can be used to describe the Poisson boundary of a totally strongly irreducible random walk that has finite entropy, in the sense of Conjecture 1.12.

1.8. **Structure of this paper.** Theorem 1.6 is the main result of this paper, form which we deduce Theorems 1.7, 1.8, 1.10 and 1.13. In Section 2 of this article, we state some local-to-global properties for alignment of matrices. In Section 3, we state some preliminary results about random products of non-invertible matrices.

In Section 4, we state Theorem 4.7, which is an abstract version of the construction of the pivoting extraction and prove Theorem 1.6 as a corollary of that statement We prove Theorem 4.7 using an improved version of the pivoting technique described in [Gou22, section 2]. In the present article, we use Theorem 4.7 to study a strongly irreducible and proximal product of random matrices. However, this result can be applied to wide range of random walks on groups.
Namely Gromov hyperbolic groups as in [Gou22], but also mapping class groups, relatively hyperbolic groups, group of outer morphisms of a free group or cat(0) groups as in [Cho24a] [Cho24b] [Cho24c].

In Section 5, we use Theorem 1.6 as a black-box. We prove technical Lemmas from which Theorems 1.7 1.8, 1.7 and 1.13 follow. The proof of Theorem 1.7 is the most technical and relies on Kingman's sub-additive ergodic Theorem so we prove Theorem 1.13 first as its proof relies on simple probabilistic arguments.

## 2. LOCAL-TO-GLOBAL PROPERTIES FOR THE ALIGNMENT OF MATRICES

In this section, we describe the geometry of the monoid $\Gamma := \text{End}(E)$ for $E$ a Euclidean space. We can think of $K = \mathbb{R}$ but all the proofs work the same when $K = \mathbb{C}$ or when $K$ is a ultra-metric local field.

Given $E$ a $K$-vector space, we will identify $E$ with $\text{Hom}(K, E)$. Note that up to choosing a canonical basis for all Euclidean spaces, linear maps between Euclidean spaces can be seen as matrices. Moreover, vectors and linear form can also be seen as matrices.

We want to translate ideas of hyperbolic geometry into the language of products of endomorphisms. The idea is to exhibit a local-to global property in the same fashion as [Can84, Theorem 4]. That way we can adapt the arguments of [Gou22] to the setting of products of random matrices.

This strategy is similar to the one used by Yves Benoist in [Ben97] to study asymptotic properties of linear groups.

**2.1. Alignment and contraction coefficient.** We remind that we defined the notation $A^\epsilon$ in Definition 1.3, given $g$ and $h$ two matrices such that the product $gh$ is well defined (i.e., $g$ has as many columns as $h$ has rows) and $0 < \epsilon < 1$, we write $gA^\epsilon h$ as a short notation for $\|gh\| \ge \epsilon\|g\|\|h\|$. Note that $g$ may have a single line and $h$ may have a single column, in which case $gh$ is a $1 \times 1$ square matrix i.e., a scalar. We remind the reader that given $x, y$ two vectors, we have the characterization $\|x \wedge y\| = \min_{a \in K} \|x - ya\|\|y\|$. Therefore, given $h \in \text{Hom}(E, F)$, we have $\|h \wedge h\| = \max_{x, y \in E \setminus \{0\}} \min_{a \in K} \frac{\|h(x - ya)\|\|h(y)\|}{\|x\|\|y\|}$.

**Definition 2.1 (Singular gap).** Let $E, F$ be Euclidean vector spaces and $h \in \text{Hom}(E, F) \setminus \{0\}$. We define the first singular gap of $h$ as:

$$ \sigma(h) := \frac{\|h \wedge h\|}{\|h\|^2}. $$

Given a matrix $h$ of size $d \times d$, the quantity $\sigma(h)$ is in the interval $[0, 1]$ and it is equal to the ratio $\frac{s_2(h)}{s_1(h)}$ where $s_1(h) \ge s_2(h) \ge \dots \ge s_d(h) \ge 0$ are
the singular values of $h$ i.e., the coefficients of the diagonal matrix in the Cartan decomposition of $h$.

**Definition 2.2** (Distance between projective classes). Let $E$ be a Euclidean, Hermitian or ultra-metric space. We denote by $P(E)$ the projective space of $E$ i.e., the set of lines in $E$, endowed with the following distance map:

$$ (20) \qquad \forall x, y \in E \setminus \{0\}, \quad d([x], [y]) := \frac{\|x \wedge y\|}{\|x\|\|y\|} = \min_{a \in \mathbb{K}} \frac{\|x - ya\|}{\|x\|}. $$

**Lemma 2.3** (Lipschitz property for the norm cocycle). Let $E$ and $F$ be two Euclidean, Hermitian or ultra-metric spaces and let $f \in \text{End}(E) \setminus \{0\}$. Let $x, y \in E \setminus \{0\}$, we have:

$$ (21) \qquad \left| \frac{\|fx\|}{\|f\|\|x\|} - \frac{\|fy\|}{\|f\|\|y\|} \right| \le d([x], [y]). $$

*Proof.* Let $f \in \text{Hom}(E, F)$ and let $x, y \in E \setminus \{0\}$. We show that $\|fx\| \le \|fy\| + \|f\|\|x\|d([x], [y])$, which implies (21) since $x$ and $y$ play symmetric roles. Let $c \in \mathbb{K}$ be such that $\|x - yc\| = \min_{a \in \mathbb{K}} \|x - ya\|$. Then by Definition 2.2, we have $d([x], [y]) = \|x - yc\|/\|x\|$. Moreover $\|yc\| \le \|x\|$ by property of the orthogonal projection in the Euclidean and Hermitian cases and by ultra-metric inequality otherwise. By triangular inequality and by homogeneity of the norm, we have:

$$ \|fx\| \le \|fyc\| + \|f(x - yc)\| \le \|fy\| + \|f\|\|x - yc\|. \quad \square $$

We remind that given $g$ and $h$ two matrices such that the product $gh$ is well defined and given $0 < \varepsilon \le 1$, we write $gA^{\varepsilon}h$ when $\|gh\| \ge \varepsilon\|g\|\|h\|$. We also remind that given $h \in \text{Hom}(E, F)$, we write $^t h \in \text{Hom}(F^*, E^*)$ for the map $f \mapsto fh$, we call it the transpose of $h$.

**Definition 2.4.** Let $E, F$ be Euclidean spaces and $h \in \text{Hom}(E, F) \setminus \{0\}$. Let $0 < \varepsilon \le 1$. We define $V^{\varepsilon}(h) := \{x \in E; \|hx\| \ge \varepsilon\|h\|\|x\|\}$ and $U^{\varepsilon}(h) := h(V^{\varepsilon}(h))$ and $W^{\varepsilon}(h) := U^{\varepsilon}(^t h)$.

Note that for all $h$ and all $0 \le \varepsilon \le 1$ the sets $V^{\varepsilon}(h)$, $U^{\varepsilon}(h)$ and $W^{\varepsilon}(h)$, are homogeneous in the sense that they are invariant under scalar multiplication.

Note also that for $h$ an endomorphism of rank one, and for all $0 < \varepsilon \le 1$, the cone $U^{\varepsilon}(h)$ is the image of $h$ so it has diameter 0 in the projective space.

The idea to have in mind is that given $h$ a matrix that has a large singular gap $U^{\varepsilon}(h)$ will have a small diameter in the following sense.
**Lemma 2.5.** Let $E, F$ be Euclidean spaces, let $h \in \operatorname{Hom}(E, F) \setminus \{0\}$ and let $0 < \varepsilon \le 1$. Let $u \in U^1(h) \setminus \{0\}$ and let $u' \in U^\varepsilon(h) \setminus \{0\}$. Then we have:

$$ (22) \qquad d([u], [u']) \le \frac{\sigma(h)}{\varepsilon}. $$

*Proof.* Let $v \in V^1(h)$ and let $v' \in V^\varepsilon(h)$ be such that $u = hv$ and $u' = hv'$. Then we have $u \wedge u' = \wedge^2 h(v \wedge v')$ so:

$$ \|u \wedge u'\| \le \|\wedge^2 h\| \|v \wedge v'\|. $$

Now saying that $v \in V^1(h)$ and $v' \in V^\varepsilon(h)$ means that $\|u\| = \|h\|\|v\|$ and $\|u'\| \ge \varepsilon\|h\|\|v'\|$. Hence:

$$ \|u\|\|u'\| \ge \varepsilon\|h\|^2\|v\|\|v'\|. $$

Then by taking the quotient of the above inequalities, we have:

$$ \frac{\|u \wedge u'\|}{\|u\|\|u'\|} \le \frac{\|h \wedge h\|}{\varepsilon\|h\|^2} \frac{\|v \wedge v'\|}{\|v\|\|v'\|} \le \frac{\|h \wedge h\|}{\varepsilon\|h\|^2}. $$

By definition, the term on the left is $d([u], [u'])$ and the term on the right is $\frac{\sigma(h)}{\varepsilon}$. $\square$

Lemma 2.5 tells us that the projective image of $U^\varepsilon(h)$ has diameter at most $\varepsilon$ as long as $\sigma(h) \le \varepsilon^2$. With the toy model analogy, the condition $\sigma(h) \le \varepsilon^2$ will play the role of the condition for word to be non-trivial. We will extensively use the following simple remarks.

**Lemma 2.6.** Let $g$ and $h$ be non-zero matrices such that the product $gh$ is well defined and let $0 < \varepsilon \le 1$. We have $gA^\varepsilon h$ if and only if $^t hA^{\varepsilon t} g$. Moreover $\sigma(^t h) = \sigma(h)$.

*Proof.* This is a consequence of three well known facts. One is that we have $\|h\| = \|^t h\|$ for all homomorphism $h$. One way of seeing that is to notice that the operator norm admits the following (obviously symmetric) characterization:

$$ (23) \qquad \forall E, \forall F, \forall h \in \operatorname{Hom}(E, F), \|h\| = \max_{\substack{f \in F^* \\ v \in E \setminus \{0\}}} \frac{|fhv|}{\|f\|\|v\|}. $$

The second fact is that $^t(gh) = {}^t h {}^t g$. It implies that for all non trivial $g, h$, we have $\frac{\|gh\|}{\|g\|\|h\|} = \frac{\|^t h {}^t g\|}{\|^t h\|^t \|g\|}$. The third fact is that $^t h \wedge {}^t h = {}^t (h \wedge h)$. It implies that $\sigma(^t h) = \sigma(h)$. $\square$
**Lemma 2.7.** Let $g$ and $h$ be non-zero matrices such that the product $gh$ is well defined and let $0 < \varepsilon \le 1$. If there exist $u \in U^1(h) \setminus \{0\}$ and $w \in W^1(g) \setminus \{0\}$ such that $\frac{|wu|}{||u||||w||} \ge \varepsilon$, then $g\mathbb{A}^\varepsilon h$. If $g\mathbb{A}^\varepsilon h$, then there exist $u \in U^\varepsilon(h) \setminus \{0\}$ and $w \in W^\varepsilon(g) \setminus \{0\}$ such that $w\mathbb{A}^\varepsilon u$ and $g\mathbb{A}^\varepsilon u$ and $w\mathbb{A}^\varepsilon h$.

*Proof.* Let $u \in U^1(h) \setminus \{0\}$ and $w \in W^1(g) \setminus \{0\}$. Assume that $\frac{|wu|}{||u||||w||} \ge \varepsilon$. Let $f \in V^1(tg)$ and let $v \in V^1(h)$ be such that $w = fg$ and $u = hv$. We have $\frac{|fghv|}{||fg||||hv||} \ge \varepsilon$, therefore $\frac{|fghv|}{||f||||g||||h||||v||} \ge \varepsilon$, so $\frac{||gh||}{||g||||h||} \ge \varepsilon$, which means that $g\mathbb{A}^\varepsilon h$. This proves the first implication of Lemma 2.7.

Now assume that $g\mathbb{A}^\varepsilon h$. Let $f \in E^* \setminus \{0\}$ and let $v \in E \setminus \{0\}$ be such that $|fghv| = \|f\|\|gh\|\|v\|$. Then $\frac{|fghv|}{||f||||g||||h||||v||} = \frac{||gh||}{||g||||h||} \ge \varepsilon$. Let $u := hv$ and let $w := fg$. Then we have:

$$ \varepsilon \le \frac{|wu|}{||w||||u||} \cdot \frac{||fg||||hv||}{||f||||g||||h||||v||} = \frac{|fgu|}{||f||||g||||u||} \cdot \frac{||hv||}{||h||||v||} = \frac{|whv|}{||w||||h||||v||} \cdot \frac{||fg||}{||f||||g||} $$

All the factors are in $[0, 1]$ so $\frac{|wu|}{||w||||u||} \ge \varepsilon$ hence $w\mathbb{A}^\varepsilon u$, and $\frac{||qu||}{||g||||u||} \ge \frac{|fgu|}{||f||||g||||u||} \ge \varepsilon$ hence $g\mathbb{A}^\varepsilon u$, and $\frac{||wh||}{||w||||h||} \ge \frac{|whv|}{||w||||h||||v||} \ge \varepsilon$ hence $w\mathbb{A}^\varepsilon h$. Moreover $\frac{||hv||}{||h||||v||} \ge \varepsilon$, hence $v \in V^\varepsilon(h)$ so $u \in U^\varepsilon(h)$, and $\frac{|fg|}{||f||||g||} \ge \varepsilon$ so $w \in W^\varepsilon(g)$. This proves the second implication of Lemma 2.7. $\square$

**Lemma 2.8.** Let $g$ and $h$ be non-zero matrices such that the product $gh$ is well defined and let $0 < \varepsilon \le 1$. Assume that $g\mathbb{A}^\varepsilon h$. Then one has:

$$ (24) \qquad \sigma(gh) \le \varepsilon^{-2}\sigma(g)\sigma(h). $$

Moreover, for every non-zero vectors $u \in U^1(g) \setminus \{0\}$, and $u' \in U^1(gh) \setminus \{0\}$, we have:

$$ (25) \qquad d([u], [u']) \le \frac{\sigma(g)}{\varepsilon}. $$

*Proof.* Note that the norm of the $\wedge$ product is sub-multiplicative because it is a norm so:

$$ (26) \qquad \|gh \wedge gh\| \le \|g \wedge g\| \|h \wedge h\|. $$

So if we do $(5)^2/\log(26)$ we find (24).

Now to prove (25), we only need to show that $U^1(gh) \subset U^\varepsilon(g)$ and use (22) from Lemma 2.5. Indeed, consider $v \in V^1(gh)$, then one has $\|ghv\| \ge \varepsilon\|g\|\|h\|\|v\| \ge \varepsilon\|g\|\|hv\|$ which means that $hv \in V^\varepsilon(g)$, therefore $ghv \in U^\varepsilon(g)$. Therefore, we have $U^1(gh) \subset U^\varepsilon(g)$. Moreover, by (22) in Lemma 2.5, we have $d([u], [u']) \le \frac{\sigma(g)}{\varepsilon}$ for all $u \in U^1(g)$ and all $u' \in U^\varepsilon(g)$, which proves (25). $\square$
2.2. **Local-to-global properties for the alignment.** In this paragraph, we prove local to global properties for the alignment relation under a contraction condition. Note that given a family $g_0, \dots, g_n$ of matrices of rank 1 such that $g_1 \cdots g_n \neq 0$, for all $k$ and for all $0 < \varepsilon \le 1$, the following local and global alignment relations $g_0 \cdots g_{k-1}A^\varepsilon g_k \cdots g_n$ and $g_{k-1}A^\varepsilon g_k$ and $g_0 \cdots g_{k-1}A^\varepsilon g_k$ and $g_{k-1}A^\varepsilon g_k \cdots g_n$ are equivalent. Under a condition on $\sigma$ the local and global conditions are almost equivalent in the following sense.

**Lemma 2.9 (Transmission of the alignment).** Let $f, g, h$ be non-zero matrices such that the product $fgh$ is well defined and let $0 < \varepsilon \le 1$. Assume that $\sigma(g) \le \varepsilon^2/4$ and that $fA^\varepsilon gA^{\frac{\varepsilon}{2}}h$. Then $fA^{\frac{\varepsilon}{2}}gh$.

*Proof.* Let $u \in U^1(gh)$. Then we have $\|ghu\| = \|gh\|\|u\| \ge \frac{\varepsilon}{2}\|g\|\|h\|\|u\|$ so $\|ghu\| \ge \frac{\varepsilon}{2}\|g\|\|hu\|$ and therefore $u \in U^{\frac{\varepsilon}{2}}(g)$. Let $v \in V^1(fg) \setminus \{0\}$. Then we have $\|fgv\| \ge \|f\|\|g\|\|v\|\varepsilon$ so $gv \in U^\varepsilon(g) \setminus \{0\}$. Let $u' \in U^1(g) \setminus \{0\}$. By Lemma 2.5, we have $d([gv], [u']) \le \frac{\varepsilon}{4}$. and by Lemma 2.8, we have $d([u], [u']) \le \frac{\varepsilon}{4}$. Hence, by triangular inequality, we have $d([u], [gv]) \le \frac{\varepsilon}{2}$. Then by Lemma 2.3, we have $\frac{\|fu\|}{\|f\|\|u\|} \ge \frac{\|fgv\|}{\|f\|\|gv\|} - \frac{\varepsilon}{2}$. Moreover $\frac{\|fu\|}{\|f\|\|u\|} \ge \frac{\varepsilon}{2}$ and $u \in U^1(gh)$ hence $fA^{\frac{\varepsilon}{2}}gh$. $\square$

**Remark 2.10 (The ultra-metric case is easier).** Let $\mathbb{K}$ be a ultra-metric locally compact field and let $f, g, h$ be matrices with entries in $\mathbb{K}$ such that the product $f, g, h$ is non trivial. Let $0 < \varepsilon \le 1$ be so that $\sigma(g) < \varepsilon^2$ and $gA^\varepsilon h$. Then $fA^\varepsilon gh$ if and only if $fA^\varepsilon g$. This simple remark allows us to get rid of the multiplicative constants in the results of this section for the ultra-metric case.

**Lemma 2.11.** Let $f, g$ and $h$ be non-zero matrices such that the product $fgh$ is well defined and let $0 < \varepsilon \le 1$. Assume that $fA^\varepsilon gA^\varepsilon h$ and that $\sigma(g) \le \varepsilon^4/4$. Then $\sigma(fgh) \le 4\varepsilon^{-4}\sigma(f)\sigma(g)\sigma(h) \ge 4\varepsilon^{-4}\sigma(g)$.

*Proof.* By (24) in Lemma 2.8, we have $\sigma(gh) \le \varepsilon^{-2}\sigma(g)\sigma(h)$. Moreover, by Lemma 2.9, we have $fA^{\frac{\varepsilon}{2}}gh$ so $\sigma(fgh) \le 4\varepsilon^{-2}\sigma(f)\sigma(gh) \le 4\varepsilon^{-4}\sigma(f)\sigma(g)\sigma(h)$. We conclude using the fact that $\sigma$ takes values in $[0, 1]$. $\square$

**Lemma 2.12 (Contraction property for aligned chains).** Let $E$ be a Euclidean vector space, let $0 < \varepsilon \le 1$ let $n \ge 1$. Let $g_0, \dots, g_n$ be non-zero matrices such that the product $g_0 \cdots g_n$ is well defined. Assume that for all $k \in \{0, \dots, n-1\}$,
we have $\sigma(g_k) \ge \varepsilon^2/8$ and $g_k A^\varepsilon g_{k+1}$. Then we have $g_0 A^{\frac{\varepsilon}{2}} g_1 \cdots g_n$ and:

$$ (27) \qquad \|g_0 \cdots g_n\| \ge \left(\frac{\varepsilon}{2}\right)^n \prod_{j=0}^{n} \|g_j\| \quad \text{and} $$

$$ (28) \qquad \sigma(g_0 \cdots g_n) \le 2^{2n} \varepsilon^{-2n} \prod_{j=0}^{n} \sigma(g_j). $$

**Proof.** We prove by induction on $1 \le k \le n$ that $g_{k-1}A^{\frac{\varepsilon}{2}}g_k \cdots g_n$. For $k = n$, this is a direct consequence of the fact that $g_{n-1}A^\varepsilon g_n$. Let $1 \le k < n$ and assume that $g_k A^{\frac{\varepsilon}{2}} g_{k+1} \cdots g_n$. Then by Lemma 2.9 applied to $f = g_{k-1}$, $g := g_k$ and $h := g_{k+1} \cdots g_n$, we have $g_{k-1}A^{\frac{\varepsilon}{2}}g_k \cdots g_n$.

Therefore, for all $0 \le k < n$, we have $\|g_k \cdots g_n\| \ge \frac{\varepsilon}{2}\|g_k\|\|g_{k+1} \cdots g_n\|$. Then by induction on $k$, we have $\|g_k \cdots g_n\| \ge (\varepsilon/2)^{n-k} \|g_k\|\|g_{k+1} \cdots g_n\|$, for all $k$. Taking $k = 0$, we have (27).

Now by (24), we have $\sigma(g_k \cdots g_n) \le \frac{4}{\varepsilon^2}\sigma(g_k)\sigma(g_{k+1} \cdots g_n)$ for all $0 \le k < n$. Then by induction, we have $\sigma(g_k \cdots g_n) \le (4/\varepsilon^2)^{n-k}\sigma(g_k)\cdots\sigma(g_n)$ for all $0 \le k < n$. Therefore we have (28). $\square$

**Corollary 2.13 (Limit line).** Let $0 < \varepsilon \le 1$, let $E$ be a Euclidean space and let $(g_n)_{n \ge 0}$ be a sequence in End($E$). Assume that for all $n \ge 0$, one has $g_n A^\varepsilon g_{n+1}$ and $\sigma(g_{n+1}) \le \varepsilon^2/8$. Then there exists $l^\infty \in P(E)$ such that:

$$ (29) \qquad \forall n, \forall u_n \in U^1(g_0 \cdots g_{n-1}) \setminus \{0\}, \quad d([u_n], l^\infty) \le \frac{2}{\varepsilon} \sigma(g_0 \cdots g_{n-1}). $$

**Proof.** For all $k > 0$, let $u_k \in U^1(g_0 \cdots g_{k-1}) \setminus \{0\}$. Let $0 < k < m$. By Lemma 2.15 applied to $(g_0, \dots, g_{m-1})$, we have $(g_0 \cdots g_{k-1})A^{\frac{\varepsilon}{2}}(g_k \cdots g_{m-1})$. Hence, by Lemma 2.8, we have:

$$ (30) \qquad d([u_k], [u_m]) \le \frac{2\sigma(g_0 \cdots g_{k-1})}{\varepsilon}. $$

And by (28) in Lemma 2.12, we have $\sigma(g_0 \cdots g_{k-1}) \le 2^{-k+1}$. Hence, the sequence $([u_k])_{k \ge 1}$ is a Cauchy sequence in $P(E)$, therefore it has a limit. Let $l_\infty = \lim_{m}[u_m]$. Then by taking $m \to +\infty$ in (30), we have:

$$ d([u_k], l^\infty) \le \frac{2\sigma(g_0 \cdots g_{k-1})}{\varepsilon}. $$

To conclude, note that for all $n$, if we only change the value of $u_n$ in the sequence $(u_k)_{k \ge 0}$ then the limit $l_\infty$ does not change and (2.2) still holds for $k = n$ as long as $u_n \in U^1(g_0 \cdots g_{n-1}) \setminus \{0\}$. $\square$
Let us now state more technical results on local to global properties for the alignment.

**Lemma 2.14 (Rigidity of the alignment).** Let $f, g_1, g_2, h$ be non-zero matrices such that the product $fg_1g_2h$ is well defined and let $0 < \varepsilon \le 1$. Assume that $\sigma(g_i) \le \varepsilon^2/12$ for $i = 1, 2$ and that $f\mathbb{A}^{\frac{\varepsilon}{2}}g_1\mathbb{A}^{\varepsilon}g_2\mathbb{A}^{\frac{\varepsilon}{2}}h$. Then $fg_1\mathbb{A}^{\frac{\varepsilon}{2}}g_2h$.

*Proof.* The proof is similar to the proof of 2.9. Let $w \in W^1(fg_1)$ and $u \in U^1(g_2h)$. Then $w \in W^{\frac{\varepsilon}{2}}(fg_1)$ and $u \in U^{\frac{\varepsilon}{2}}(g_2)$. Let $w' \in W^{\varepsilon}(g_1)$ and $u' \in U^{\varepsilon}(g_2)$ be such that $w'\mathbb{A}^{\varepsilon}u'$. Such $u', w'$ exist by Lemma 2.7. Moreover, we have $d([w], [w']) \le \frac{3\sigma(g_1)}{\varepsilon} \le 3\varepsilon/12 = \varepsilon/4$ by Lemma 2.5 applied to $^t g_1$ and by triangular inequality. With the same argument we have $d([u], [u']) \le \varepsilon/4$. Then by the transpose of Lemma 2.3, we have $w\mathbb{A}^{\varepsilon-\frac{\varepsilon}{4}}u'$ and Lemma 2.3 again, we have $w\mathbb{A}^{\frac{\varepsilon}{2}}u$. Hence, by Lemma 2.7, we have $fg_1\mathbb{A}^{\frac{\varepsilon}{2}}g_2h$. $\square$

**Lemma 2.15 (local-to-global property for the alignment).** Let $g_0, \dots, g_n$ be non-zero matrices such that the product $g_0 \cdots g_n$ is well defined. Let $0 < \varepsilon \le 1$. Assume that for every $k \in \{1, \dots, n-1\}$ we have $\sigma(g_i) \le \varepsilon^2/12$. Assume also that $g_0\mathbb{A}^{\varepsilon}g_1\mathbb{A}^{\varepsilon}\cdots\mathbb{A}^{\varepsilon}g_n$ i.e., for all $k \in \{0, \dots, n-1\}$, we have $g_k\mathbb{A}^{\varepsilon}g_{k+1}$. Then for all $k \in \{1, \dots, n\}$, we have $g_0 \cdots g_{k-1}\mathbb{A}^{\frac{\varepsilon}{2}}g_k \cdots g_n$.

*Proof.* The case $k = 1$ is a reformulation of Lemma 2.12 and the case $k = n$ is the transpose of the case $k = 1$.

Let $2 \le k \le n - 1$. First by Lemma 2.9 applied to $(g_k, \dots, g_n)$, we have $g_k\mathbb{A}^{\frac{\varepsilon}{2}}g_{k+1} \cdots g_n$. Again by Lemma 2.9 applied to $(^t g_{k-1}, \dots, ^t g_0)$, we have $g_0 \cdots g_{k-2}\mathbb{A}^{\frac{\varepsilon}{2}}g_{k-1}$. Moreover, we have $g_{k-1}\mathbb{A}^{\varepsilon}g_k$ an $\sigma(g_{k-1})$, $\sigma(g_k) \le \varepsilon^2/12$. Therefore, by Lemma 2.14, we have $g_0 \cdots g_{k-1}\mathbb{A}^{\frac{\varepsilon}{2}}g_k \cdots g_n$. $\square$

Lemmas 2.15 and corollary 2.13 give a good intuition on how local alignment properties of a sequence of matrices are linked to contraction properties for the action on the projective space.

However we will have to deal with a more intricate notion of half local half global alignment. Let us give a more complete version of Lemma 2.15, which is exactly what we need for the proof of (8) in Theorem 1.6.

**Lemma 2.16.** Let $0 < \varepsilon \le 1$ and let $n \ge 0$. Let $g_{-1}, g_0, g_1, \dots, g_{2n}$ be non-zero matrices and assume that the product $g_{-1} \cdots g_{2n}$ is well defined. Assume that $g_{-1}\mathbb{A}^{\varepsilon}g_0$, that $\sigma(g_0) \le \varepsilon^2/12$, that for all $1 \le i \le n$, we have $\sigma(g_{2i-1}) \le \varepsilon^6/48$ and that for all $0 \le i < n$, we have:

$$ (31) \qquad g_0 \cdots g_{2i} \mathbb{A}^{\varepsilon} g_{2i+1} \mathbb{A}^{\varepsilon} g_{2i+2}. $$

Then $g_{-1}\mathbb{A}^{\frac{\varepsilon}{2}}(g_0 \cdots g_{2n})$ and $\sigma(g_0 \cdots g_{2n}) \le \sigma(g_0)$.
Let $f, h$ be non-zero matrices such that the products $fg_{-1}$ and $g_n h$ are well defined. Assume moreover that $\sigma(g_{-1}) \le \varepsilon^6/48$, that $f A^{\frac{\varepsilon}{2}} g_{-1}$ and that $g_0 \cdots g_n A^{\frac{\varepsilon}{2}} h$. Then we have $fg_{-1} A^{\frac{\varepsilon}{2}} g_0 \cdots g_{2n} h$.

*Proof.* Let $0 \le i < n$. By Lemma 2.11 and by (31), we have:

$$ \sigma(g_0 \cdots g_{2i+2}) \le 4\varepsilon^{-4}\sigma(g_0 \cdots g_{2i})\sigma(g_{2i+1}) \le \sigma(g_0 \cdots g_{2i})\varepsilon^2/12. $$

By induction on $i$, we have $\sigma(g_0 \cdots g_{2i}) \le \sigma(g_0)(\varepsilon^2/12)^i \le (\varepsilon^2/12)^{i+1}$. Moreover, by Lemma 2.9, and by (31) we have $g_0 \cdots g_{2i} A^{\frac{\varepsilon}{2}} g_{2i+1} g_{2i+2}$. Moreover, by (25) in Lemma 2.8, for all $u_i \in U^+(g_0 \cdots g_{2i}) \setminus \{0\}$ and all $u_{i+1} \in U^+(g_0 \cdots g_{2i+2}) \setminus \{0\}$, we have:

$$ d([u_i], [u_{i+1}]) \le \frac{2\sigma(g_0 \cdots g_{2i})}{\varepsilon} \le \frac{\varepsilon}{6}(\varepsilon^2/12)^i. $$

Then by triangular inequality, for all $u_0 \in U^1(g_0)$, we have:

$$ d([u_0], [u_i]) \le \frac{\varepsilon}{4} \sum_{k=0}^{i-1} (\varepsilon^2/12)^k \le \frac{\varepsilon/6}{1 - \varepsilon^2/12} \le \frac{2\varepsilon}{11}. $$

Now let $u \in U^\varepsilon(u_0)$ be such that $g_{-1} A^\varepsilon u$, as in Lemma 2.7 applied to $g_{-1} A^\varepsilon g_0$. Then by Lemma 2.5, we have $d([u], [u_0]) \le \varepsilon/8$ and by triangular inequality, we have $d([u], [u_n]) \le \varepsilon/12 + 2\varepsilon/11 \le 3\varepsilon/11 \le \varepsilon/2$. Hence, by Lemma 2.3, we have $g_{-1} A^{\frac{\varepsilon}{2}} u_n$ with $u_n \in U^+(g_0 \cdots g_{2n}) \setminus \{0\}$ and therefore $g_{-1} A^{\frac{\varepsilon}{2}} (g_0 \cdots g_{2n})$.

Now assume that $\sigma(g_{-1}) \le \varepsilon^6/48$ and let $f, h$ be such that $f A^{\frac{\varepsilon}{2}} g_{-1}$ and $g_0 \cdots g_n A^{\frac{\varepsilon}{2}} h$. When $n=0$, we have $fg_{-1} A^{\frac{\varepsilon}{2}} g_0 \cdots g_{2n} h$ by Lemma 2.14. Assume now that $n \ge 1$. Then, we have $\sigma(g_0 \cdots g_n) \le (\varepsilon^2/12)^{n+1} \le \varepsilon^4/144$. Let $u' \in U^1(g_0 \cdots g_n h)$. By Lemma 2.8, we have:

$$ d([u_n], [u']) \le \frac{2\sigma(g_0 \cdots g_n)}{\varepsilon} \le \frac{\varepsilon^3}{72}. $$

Therefore $d(u_n, u) \le \frac{\varepsilon^3}{72} + \frac{\varepsilon}{12} + \frac{2\varepsilon}{11}$ by triangular inequality. Now let $w \in W^\varepsilon(g_{-1})$ be such that $w A^\varepsilon u$, let $w' \in W^1(fg_{-1})$ and let $w'' \in W^1(g_{-1})$. By Lemmas 2.5, we have $d([w], [w'']) \le \frac{\varepsilon^5}{48}$, by Lemma 2.8, we have $d([w'], [w'']) \le \frac{2\varepsilon^5}{48}$ and by triangular inequality, we have $d([w], [w']) \le \frac{3\varepsilon^5}{48}$. Then by Lemma 2.3, we have $\frac{|w'u|}{\|w'\|\|u\|} \ge \varepsilon - \frac{\varepsilon^5}{16}$ and by 2.3 again, we have $\frac{|w'u'|}{\|w'\|\|u'\|} \ge \varepsilon - \frac{\varepsilon^5}{16} - \frac{\varepsilon^3}{72} - \frac{\varepsilon}{12} - \frac{2\varepsilon}{11} \ge \varepsilon/2$. $\square$

2.3. **Link between singular values and eigenvalues.** In this short section, we prove the following lemma.
**Lemma 2.17.** Let E be a Euclidean space and let $0 < \varepsilon \le 1$. Let g be an endomorphism such that $\sigma(g) \le \varepsilon^2/8$ and $gA^\varepsilon g$. Then g is proximal and we have the following:

$$ (32) \qquad \rho_1(g) \ge \frac{\varepsilon}{2} \|g\| $$

$$ (33) \qquad \frac{\rho_2(g)}{\rho_1(g)} \le \frac{4\sigma(g)}{\varepsilon^2} $$

$$ (34) \qquad \forall u \in U^1(g), d([u], E^+(g)) \le \frac{2\sigma(g)}{\varepsilon}. $$

*Proof.* Consider $(g_k)_{k \ge 0}$ to be the sequence of copies of g. First, by Lemma 2.12, we have $\sigma(g^n) \le \left(\frac{4\sigma(g)}{\varepsilon^2}\right)^n$ and $\|g^n\| \ge \left(\frac{\varepsilon}{2}\|g\|\right)^n$. Therefore, we have $\inf_n \|g^n\|^{1/n} \ge \frac{\varepsilon}{2}\|g\|$. Moreover, by the spectral theorem, this inferior limit is in fact an honest limit and it is equal to $\rho_1(g)$, which proves (32).

To get (34), we apply Corollary 2.13. We get a line $l^\infty$ such that for any sequence $u_n \in U^1(g^n) \setminus \{0\}$, we have $[u_n] \to l^\infty$. Let $l^\infty$ be such a line and $(u_n)_{n \ge 0}$ be such a sequence. Moreover, for all $u \in U^1(g)$, we have $d([u], l^\infty) \le \frac{2\sigma(g)}{\varepsilon}$. Now we only need to show that $l^\infty = E^+(g)$. Let $e \in E^+(g)$. Then we have $\|ge\| = \rho_1(g)\|e\|$ so by (32), $e \in V^{\frac{\varepsilon}{2}}(g)$. Moreover, e is an eigenvector so $e \in Kge$ and as a consequence $e \in U^{\frac{\varepsilon}{2}}(g)$. This reasoning holds for all powers of g so we have $e \in U^{\frac{\varepsilon}{2}}(g^n)$ for all $n$. Moreover, $\sigma(g^n) \to 0$ by (33) so by Lemma 2.5, the projective diameter of $U^{\frac{\varepsilon}{2}}(g^n)$ goes to zero so $[u_n] \to [e]$ and therefore $l^\infty = [e]$. $\square$

### 3. RANDOM PRODUCTS AND EXTRACTIONS

This section is dedicated to the proof of basic results for products of random matrices. First we describe the essential kernel of a random walk on the semi-group of non-invertible matrices. Then we construct the Schottky measure for a strongly irreducible and proximal random walk.

For the remaining of the discussion, E denotes a Euclidean, Hermitian of ultra-metric vector space, $\nu$ a probability measure on End(E) and $(\gamma_n)_{n \ge 0} \sim \nu^{\otimes N}$ a random sequence. In Paragraph 3.1, we describe the probabilistic behaviour of the kernel $\ker(\bar{\gamma}_n)$ in the case $\nu(\mathrm{GL}(E)) < 1$. In the second paragraph, we exhibit a mixing property in the strongly irreducible and proximal case. This mixing property is linked with the notion of Schottky measures and is what allows us to prove Theorem 1.6.
We use the following notations. Given $\Gamma$ a metric semi-group with a unit $\mathbf{1}_{\Gamma}$, we write $\tilde{\Gamma} := \bigsqcup_{n \ge 0} \Gamma^n$ for the set of words on the alphabet $\Gamma$ and write $\odot$ for the concatenation product. We write $()$ for the empty word.

As a disjoint union of finite products of metric space, $\tilde{\Gamma}$ is endowed with a natural metrizable topology<fnref n="4" /> and a Borel $\sigma$-algebra. We also endow $\tilde{\Gamma}$ with two continuous semi-group morphisms:

$$ L: \begin{array}{ccc} (\tilde{\Gamma}, \odot) & \to & (\mathbb{N}, +) \\ (\gamma_0, \dots, \gamma_{n-1}) & \mapsto & n \end{array} $$

$$ \Pi: \begin{array}{ccc} (\tilde{\Gamma}, \odot) & \to & (\Gamma, \cdot) \\ (\gamma_0, \dots, \gamma_{n-1}) & \mapsto & \gamma_0 \cdots \gamma_{n-1}. \end{array} $$

Given $(\tilde{\gamma}_i)_{i \ge 0} \in \tilde{\Gamma}^N$ a sequence which is not stationary to the empty word, we write $\odot_{i=0}^{+\infty} \tilde{\gamma}_i \in \Gamma^N$ for the infinite concatenation, in other words, for all $k \ge 0$ and for $0 \le n < \sum_{i=0}^k L(\tilde{\gamma}_i)$, the entry of the sequence $\odot_{i=0}^{+\infty} \tilde{\gamma}_i$ of index $n$ is the $(1+n)$-th letter of the word $\tilde{\gamma}_0 \odot \cdots \odot \tilde{\gamma}_k$.

Given $(w_k)_{k \ge 0} \in \mathbb{N}^N$, we write $\overline{w}_k := w_0 + \cdots + w_{k-1}$ for all $k \in \mathbb{N}$ and given $(\gamma_n)_{n \ge 0} \in \Gamma^N$, we write $\gamma_k^w := \gamma_{\overline{w}_k} \cdots \gamma_{\overline{w}_{k+1-1}}$ and $\tilde{\gamma}_k^w := (\gamma_{\overline{w}_k}, \dots, \gamma_{\overline{w}_{k+1-1}})$ for all $k \in \mathbb{N}$. That way, for all sequence $(w_k)_{k \ge 0} \in \mathbb{N}^N N$ that is not stationary to 0 and for all sequence $(\gamma_n)_{n \ge 0} \in \Gamma^N$, we have $\odot_{k=0}^{+\infty} \tilde{\gamma}_k^w = (\gamma_n)_{n \ge 0}$.

Given $\tilde{\kappa}_1$ and $\tilde{\kappa}_2$ two probability distributions on $\tilde{\Gamma}$, we write $\tilde{\kappa}_1 \odot \tilde{\kappa}_2$ for the convolution product i.e., the distribution of $\tilde{\gamma}_1 \odot \tilde{\gamma}_2$ when $(\tilde{\gamma}_1, \tilde{\gamma}_2) \sim \tilde{\kappa}_1 \otimes \tilde{\kappa}_2$. Given $(\tilde{\kappa}_i)_{i \ge 0}$ a family of probability distributions not stationary to $\delta_0$, we write $\odot_{i=0}^{+\infty} \tilde{\kappa}_i$ for the infinite convolution i.e., the distribution of $\odot_{i=0}^{+\infty} \tilde{\gamma}_i$, when $(\tilde{\gamma}_n) \sim \otimes \tilde{\kappa}_i$, which is a measure on $\Gamma^N$. If we have $\tilde{\kappa}_i = \tilde{\kappa}$ for all $i$ and for a given measure $\kappa \ne \delta_0$, then we write $\tilde{\kappa}^{\odot N}$ as short for $\odot_{i=0}^{+\infty} \tilde{\kappa}_i$.

**3.1. Rank and essential kernel of a probability distribution.** In this paragraph, we describe the probabilistic behaviour of the kernel of a product of i.i.d. random matrices. This allows us to study products of random matrices that are not almost surely invertible and not asymptotically almost surely zero. Given $h$ a linear map, we denote by rk($h$) the rank of $h$ i.e., the dimension of the image of $h$.

<footnote n="4">⁴Several metric classes give the same topology on $\tilde{\Gamma}$, given a sequence $(\gamma_n)_{n \ge 0}$, the sequence $((\gamma_0, \dots, \gamma_n))_{n \in \mathbb{N}}$ may or may not be a Cauchy sequence depending on the choice of metric. This choice does not matter because the sequence is divergent anyway. For the sake of intuition, it is better to have a complete locally compact metric. Given $k \le n$ and given two words $\tilde{\gamma} := (\gamma_0, \dots, \gamma_{k-1})$ and $\tilde{\gamma}' := (\gamma'_0, \dots, \gamma'_{n-1})$, we set $d_{\tilde{\Gamma}}(\tilde{\gamma}, \tilde{\gamma}') = \sum_{i=0}^{k-1} d_{\Gamma}(\gamma_i, \gamma'_i) + \sum_{i=k}^{n-1} (1 + d_{\Gamma}(\mathbf{1}_{\Gamma}, \gamma'_i))$, that way closed bounded sets are compact.</footnote>
We say that a probability measure $\nu$ is supported on a set $S$ if $\nu(S) = 1$, we write $\text{supp}(\nu)$ for the minimal closed set of full $\nu$-measure. That way, saying that a measure $\nu$ may be supported on $\text{GL}(E)$ does not mean that its support in $\text{End}(E)$ is included in $\text{GL}(E)$.

In this paragraph $E$ will be a finite dimensional vector space of dimension $d \ge 2$ over an unspecified local field but the results hold for any measurable field.

**Definition 3.1** (Rank of a distribution). Let $\nu$ be a probability measure on $\text{End}(E)$. We define the eventual rank of $\nu$ as the largest integer $\lfloor k \rfloor(\nu)$ such that:

$$ (35) \qquad \forall n \ge 0, \nu^{*n}\{\gamma \in \text{End}(E) \mid \text{rk}(\gamma) < \lfloor k \rfloor(\nu)\} = 0. $$

**Lemma 3.2** (Eventual rank of a distribution). Let $\Gamma := \text{End}(E)$. Let $\nu$ be a probability measure on $\Gamma$ such that $\lfloor k \rfloor(\nu) < d$. There exists a probability measure $\tilde{\kappa}$ on $\tilde{\Gamma}$ such that $\tilde{\kappa}^{\otimes N} = \nu^{\otimes N}$ and $\Pi_*\tilde{\kappa}$ is supported on the set of endomorphisms of rank equal to $\lfloor k \rfloor(\nu)$. Moreover $L_*\tilde{\kappa}$ has finite exponential moment.

*Proof.* Given a non-random sequence $\gamma = (\gamma_n)_{n \ge 0} \in \Gamma^N$, the sequence $(\text{rk}(\overline{\gamma}_n))_n$ is a non-increasing sequence of non-negative integers so it is stationary. Write $r_\gamma$ for the limit of $(\text{rk}(\overline{\gamma}_n))_{n \in \mathbb{N}}$. Then for all sequence $\gamma \in \Gamma^N$, there is an integer $n' \ge 1$ such that $\text{rk}(\overline{\gamma}_n) = r_\gamma$ for all $n \ge n'$. Write $n_\gamma$ for the minimal such $n'$. Note that $\gamma \mapsto r_\gamma$ and $\gamma \mapsto n_\gamma$ are measurable maps.

Now let $(\gamma_n)_{n \in \mathbb{N}} \sim \nu^{\otimes N}$ be a random sequence. We define $\tilde{\kappa}$ to be the distribution of $(\gamma_0, \dots, \gamma_{n_\gamma-1})$. Note that for all $k$, the event $(n_\gamma = k)$ is measurable for the $\sigma$-algebra generated by $(\gamma_0, \dots, \gamma_{k-1})$, which is independent of $(\gamma_{n+k})_{n \ge 0}$. Therefore, the conditional distribution of $(\gamma_{n+n_\gamma})_{n \in \mathbb{N}}$ with respect to $(\gamma_0, \dots, \gamma_{n_\gamma-1})$ is $\nu^{\otimes N}$. Hence, we have $\tilde{\kappa} \odot \nu^{\otimes N} = \nu^{\otimes N}$, moreover $\tilde{\kappa} \ne \delta_0$ (because $\text{rk}(\Pi()) = \text{dim}(E)$) so $\tilde{\kappa}^{\otimes N} = \nu^{\otimes N}$.

Moreover $\Pi_*\tilde{\kappa}$ is the distribution of $\overline{\gamma}_{n_\gamma}$ which has rank $r_\gamma$ and $L_*\tilde{\kappa}$ is the distribution of $n_\gamma$. Therefore, we only need to show that $r_\gamma$ is almost surely constant and that $n_\gamma$ has finite exponential moment.

Let $r_0$ be the essential lower bound of $r_\gamma$ i.e., the largest integer such that $\mathbb{P}(r_\gamma \ge r_0) = 1$. Let $n_0 \in \mathbb{N}$ be such that $\mathbb{P}(\text{rk}(\overline{\gamma}_{n_0}) = r_0) > 0$ and write $\alpha := \mathbb{P}(\text{rk}(\overline{\gamma}_{n_0}) = r_0)$. We claim that such an integer $n_0$ exists. Indeed, by minimality, we have $\mathbb{P}(r_\gamma > r_0) < 1$ so $\mathbb{P}(r_\gamma = r_0) > 0$, which means that $\mathbb{P}(\text{rk}(\overline{\gamma}_{n_\gamma}) = r_0) > 0$. Let $n_0$ to be such that $\mathbb{P}(n_\gamma \le n_0 \cap r_\gamma = r_0) > 0$. Such an $n_0$ exists, otherwise $n_0$ would be almost surely infinite, which is absurd.

Now since the sequence $(\gamma_n)$ is i.i.d, we have $\mathbb{P}(\text{rk}(\gamma_{k'n_0} \cdots \gamma_{(k'+1)n_0-1}) = r_0) > 0$ for all $k \in \mathbb{N}$. Moreover these events are independents so for all $k \in \mathbb{N}$, we have:

$$ \mathbb{P}(\forall k' < k, \text{rk}(\gamma_{k'n_0} \cdots \gamma_{(k'+1)n_0-1}) > r_0) = (1 - \alpha)^k. $$
Now note that the rank of a product is bounded above by the rank on each of its factor so:

$$ 
\begin{aligned}
\forall k \in \mathbb{N}, \mathbb{P}(\mathrm{rk}(\bar{\gamma}_{kn_0}) > r_0) &\le (1-\alpha)^k \\
\forall n \in \mathbb{N}, \mathbb{P}(\mathrm{rk}(\bar{\gamma}_{\lfloor \frac{n}{n_0} \rfloor n_0}) > r_0) &\le (1-\alpha)^{\lfloor \frac{n}{n_0} \rfloor}
\end{aligned}
$$

Now note that for all $n \in \mathbb{N}$, we have $\lfloor \frac{n}{n_0} \rfloor \ge \frac{n}{n_0} - 1$ and $\lfloor \frac{n}{n_0} \rfloor n_0 \le n$ so:

$$ \forall n \in \mathbb{N}, \mathbb{P}(\mathrm{rk}(\bar{\gamma}_n) > r_0) \le (1-\alpha)^{\frac{n}{n_0}-1}. $$

Let $C = \frac{1}{1-\alpha}$ and $\beta = \frac{-\log(1-\alpha)}{n_0} > 0$. Then for all $n$, we have $\mathbb{P}(\mathrm{rk}(\bar{\gamma}_n) > r_0) \le C \exp(-\beta n)$. Note that for all $n$, we have $\mathbb{P}(\mathrm{rk}(\bar{\gamma}_n) > r_0) \ge \mathbb{P}(r_\gamma > r_0)$ so $\mathbb{P}(r_\gamma > r_0) = 0$, which means that $r_0 = r_\gamma$. Hence $\mathbb{P}(\mathrm{rk}(\bar{\gamma}_n) > r_0) = \mathbb{P}(n < n_\gamma)$ for all $n \in \mathbb{N}$, so $\mathbb{P}(n_\gamma > n) \le C \exp(-\beta n)$, which means that $n_\gamma$ has finite exponential moment. $\square$

**Definition 3.3** (Essential kernel). Let $\nu$ be a probability distribution on $\mathrm{End}(E)$. We define the essential kernel of $\nu$ as:

$$ (36) \quad \underline{ker}(\nu) := \{v \in E \mid \exists n \ge 0, \nu^{*n}\{h \in \mathrm{End}(E) \mid hv = 0\} > 0\}. $$

**Lemma 3.4.** Let $E$ be a Euclidean space of dimension $d \ge 2$ and let $\nu$ be a probability distribution on $\mathrm{End}(E)$. There exists a probability distribution $\kappa$ on $\mathrm{End}(E)$ which is supported on the set of rank $\underline{rk}(\nu)$ endomorphisms and such that:

$$ (37) \quad \underline{ker}(\nu) = \underline{ker}(\kappa) = \{v \in E \mid \kappa\{h \in \mathrm{End}(E) \mid hv = 0\} > 0\} $$

$$ (38) \quad \forall v \in E, \lim_{n \to +\infty} \nu^{*n}\{h \mid hv = 0\} = \sup_{n \ge 0} \nu^{*n}\{h \mid hv = 0\} = \kappa\{h \mid hv = 0\}. $$

Moreover, there exists a constant $\alpha < 1$ such that:

$$ (39) \quad \forall v \in E, \sup_{n \in \mathbb{N}} \nu^{*n}\{h \in \mathrm{End}(E) \mid hv = 0\} \in [0, \alpha] \cup \{1\}. $$

Moreover, the set:

$$ (40) \quad \left\{ v \in E \left| \sup_{n \in \mathbb{N}} \nu^{*n}\{h \in \mathrm{End}(E) \mid hv = 0\} = 1 \right. \right\} $$

is a subspace of $E$ which is $\nu$-almost surely invariant.
Proof. Let $(\gamma_n)_{n \ge 0} \sim \nu^{\otimes \mathbb{N}}$. We define the random integer:

$$n_0 := \min \{n \in \mathbb{N} \mid \mathrm{rk}(\gamma_{n-1} \cdots \gamma_0) = \underline{\mathrm{rk}}(\nu)\}.$$

Let $g := \gamma_{n_0-1} \cdots \gamma_0$ and let $\kappa$ be the distribution of $g$. Then by Lemma 3.2 applied to the transpose of $\nu$, the random integer $n_0$ has finite exponential moment. Let $v \in E$, and let $n \in \mathbb{N}$. We claim that:

$$(41) \qquad \nu^{*n}\{h \in \mathrm{End}(E) \mid hv = 0\} = \mathbb{P}(\gamma_{n-1} \cdots \gamma_0 v = 0) \le \mathbb{P}(gv = 0).$$

When $n < n_0$, we factorize $g = \gamma_{n_0-1} \cdots \gamma_n \gamma_{n-1} \cdots \gamma_0$ so $\mathrm{ker}(\gamma_{n-1} \cdots \gamma_0) \subset \mathrm{ker}(g)$, therefore the event $\gamma_{n-1} \cdots \gamma_0 v = 0$ implies that $gv = 0$. When $n \ge n_0$, by factorization, we have $\mathrm{ker}(\gamma_{n-1} \cdots \gamma_0) \supset \mathrm{ker}(g)$ and by definition of $n_0$, we have $\mathrm{rk}(\gamma_{n-1} \cdots \gamma_0) = \mathrm{rk}(g)$ therefore $\mathrm{ker}(\gamma_{n-1} \cdots \gamma_0) = \mathrm{ker}(g)$ so $\gamma_{n-1} \cdots \gamma_0 v = 0$ if and only if $gv = 0$. In both cases $(\gamma_{n-1} \cdots \gamma_0 = 0) \subset (gv = 0)$, which proves (41). A direct consequence of (41) is that:

$$(42) \qquad \underline{\mathrm{ker}}(\nu) \subset \{v \in E \mid \kappa\{h \in \mathrm{End}(E) \mid hv = 0\} > 0\}.$$

Let $v$ be such that $\mathbb{P}(gv = 0) > 0$. By the above argument, we have:

$$\forall n, \mathbb{P}(\gamma_{n-1} \cdots \gamma_0 v = 0) \ge \mathbb{P}(gv = 0) - \mathbb{P}(n_0 > n).$$

Moreover $\mathbb{P}(n_0 > n) \to 0$, so we have (38) by double inclusion. Therefore, there exists an integer $n \in \mathbb{N}$ such that $\mathbb{P}(\gamma_{n-1} \cdots \gamma_0 v = 0) > 0$. This proves (37) by double inclusion.

Let us prove (39). Let $V = \{v \in E \mid \mathbb{P}(gv = 0) = 1\}$. First note that $V$ is a subspace of $E$ and that $gV = \{0\}$ with probability 1. Let

$$(43) \qquad \alpha := \sup_{n \in \mathbb{N}, v \in E \setminus V} \nu^{*n}\{h \in \mathrm{End}(E) \mid hv = 0\}.$$

Assume by contradiction that $\alpha = 1$. Let $(v_n)_{n \ge 0}$ be a non-random sequence in $E \setminus V$ such that $\mathbb{P}(gv_n = 0) \ge 1 - 2^{-n}$ for all $n$. Then we have $\sum_{n=0}^{+\infty} \mathbb{P}(gv_n \ne 0) < +\infty$. Therefore, by Borel-Cantelli's Lemma, the set $\{n \in \mathbb{N} \mid gv_n \ne 0\}$ is almost surely finite. Let $m_0 := \max\{n \in \mathbb{N} \mid gv_n \ne 0\}$ and let $V' := \bigcap_{m \to +\infty} \langle (v_n)_{n \ge m} \rangle$. Then since $E$ is finite dimensional, the sequence of subspaces $\langle (v_n)_{n \ge m} \rangle_{m \ge 0}$ is stationary i.e., there is an integer $m \in \mathbb{N}$ such that $v_n \in V'$ for all $n \ge m$. Then on the set $(m_0 < +\infty)$, we have $g(V') \subset \langle (gv_n)_{n \ge m_0} \rangle = \{0\}$ so $V' \subset V$, which is absurd.
Let us prove that the set described in (40) is $\nu$-almost-surely invariant. Assume by contradiction that $\mathbb{P}(\gamma_0(V) = V) \neq 1$. Let $v \in V$ be such that $\mathbb{P}(\gamma_0 v \notin V) > 0$. Then for all $n \in \mathbb{N}$, we have:

$$ \begin{aligned} \mathbb{P}(\gamma_n \cdots \gamma_0 v \neq 0) &\geq \mathbb{P}(\gamma_0 v \notin V) \mathbb{P}(\gamma_n \cdots \gamma_1 \gamma_0 v \neq 0 | \gamma_0 v \notin V) \\ &\geq \mathbb{P}(\gamma_0 v \notin V) (1 - \alpha) > 0, \end{aligned} $$

which is absurd because $\mathbb{P}(\gamma_n \cdots \gamma_0 v \neq 0) \to 0$.

□

**Proposition 3.5.** Let $\nu$ be a probability distribution on $\text{End}(E)$. The set $\underline{\text{ker}}(\nu)$ is a countable union of subspaces of $E$ that each have dimension at most $\text{dim}(E) - \text{rk}(\nu)$.

*Proof.* Let $d' := \text{dim}(E) - \text{rk}(\nu)$. For all $k \in \{0, \dots, \text{dim}(E)\}$, we denote by $\text{Gr}_k(E)$ the set of subspaces of $E$ of dimension $k$. Let $\kappa$ be as in Lemma 3.4 First we show that $\underline{\text{ker}}(\nu)$ is included in a countable union of subspaces of dimension exactly $d'$. Given $\alpha > 0$, we define:

$$ K_{\alpha} := \{x \in E \mid \kappa\{h \in \text{End}(E) \mid hx = 0\} \geq \alpha\} $$

Note that we have:

$$ \underline{\text{ker}}(\nu) = \bigcup_{m \in \mathbb{N}} K_{2^{-m}}, $$

so we only need to show that $K_{2^{-m}}$ is included in a countable union of subspaces for all $m \geq 0$.

Let $m \in \mathbb{N}$, we claim that $K_{2^{-m}}$ is included in a union of at most $\binom{d'2^m}{d'}$ subspaces of $E$ of dimension $d'$. Let $g \sim \kappa$, write $\alpha := 2^{-m}$ and assume that $K_{\alpha} \neq \{0\}$.

Let $N$ be an integer and let $(x_1, \dots, x_N) \in K_{\alpha}$. Assume that for all $1 \leq i_1 < \dots < i_k \leq N$ with $k \leq d' + 1$, the space $\langle x_{i_1}, \dots, x_{i_k} \rangle$ has dimension exactly $k$. In this case, we say that the family $(x_i)_{1 \leq i \leq N}$ is in general position up to $d'$. We claim that in this case:

$$ (44) \qquad N \leq \frac{d'}{\alpha}. $$

To all index $i \in \{1, \dots, N\}$, we associate a random integer variable $a_i := 1_{gx_i=0} \in \{0, 1\}$ i.e., such that $a_i = 1$ when $g(x_i) = 0$ and $a_i = 0$ otherwise. Note that $\text{ker}(g)$ has dimension at most $d'$ almost surely. As a consequence, for all $\leq i_1 < \dots < i_{d'+1} \leq N$, we have $\text{dim}\langle x_{i_j} \rangle_{1 \leq j \leq d'+1} > \text{dim}(\text{ker}(g))$ almost surely. Hence $\mathbb{P}(\langle x_{i_j} \rangle \subset \text{ker}(g)) = 0$ so $gx_{i_j} \neq 0$ for at least one index $j$. This means
that, with probability 1, the random set of indices $\{1 \le i \le N \mid gx_i = 0\}$ does not admit any subset of size $d' + 1$ so it has cardinal at most $d'$. In other words, $\sum_{i=1}^N a_i \le d'$ almost surely. Now, note that by definition of $K_\alpha$, we have $E(a_i) = \mathbb{P}(gx_i = 0) \ge \alpha$ for all $i \in \{1, \dots, N\}$. Hence $N\alpha \le \sum_{i=1}^N E(a_i) \le d'$, which proves (44).

Now we want to construct an integer $N \le d'/\alpha$ and a family $(x_1, \dots, x_N) \in K_\alpha$ that is in general position up to $d'$, and such that:

$$ (45) \qquad K_\alpha \subset \bigcup_{1 \le i_1 < \dots < i_{d'} \le N} \langle x_{i_j} \rangle_{1 \le j \le d'}. $$

We do it by induction. Since we assumed that $K_\alpha \ne \{0\}$, there exists a non-zero vector $x_1 \in K_\alpha$. Let $j \ge 1$ and assume that we have constructed a sequence $(x_1, \dots, x_j) \in K_\alpha^j$ that is in general position up to $d'$. If we have:

$$ K_\alpha \subset \bigcup_{1 \le i_1 < \dots < i_{d'} \le j} \langle x_{i_j} \rangle_{1 \le j \le d'}. $$

then we write $N := j$ and the algorithm ends as (45) is satisfied. Otherwise, we take:

$$ x_{j+1} \in K_\alpha \setminus \left( \bigcup_{1 \le i_1 < \dots < i_{d'} \le j} \langle x_{i_1}, \dots, x_{i_{d'}} \rangle \right). $$

Then we have constructed a family $(x_1, \dots, x_{j+1}) \in K_\alpha$ that is in general position up to $d'$ and we can proceed to the next step of the induction. This process reaches the stopping condition after at most $\lfloor \frac{d'}{\alpha} \rfloor$ steps by (44).

Now for all $m$ we choose a family $(V_1^m, \dots, V_{\binom{d'2m}{d'}}^m) \in \text{Gr}_{d'}(E)^{\binom{d'2m}{d'}}$ such that $K_{2-m} \subset \bigcup_{j=1}^{\binom{d'2m}{d'}} V_j^{m,n}$ and we have:

$$ \underline{\text{ker}}(\nu) \subset \bigcup_{\substack{m \in \mathbb{N}, j \in \mathbb{N}, \\ 1 \le j \le \binom{d'2m}{d'}}} V_j^m. $$

This proves that $\underline{\text{ker}}(\nu)$ is included in a countable union of subspaces of dimension exactly $d'$.

Now we will show that $\underline{\text{ker}}(\nu)$ is in fact equal to a countable union of subspaces. Let $g \sim \kappa$ and let $K := \underline{\text{ker}}(\nu)$. Let $(V_k)_{k \in \mathbb{N}} \in \text{Gr}_{d'}(E)^\mathbb{N}$ be such
that $K \subset \bigcup V_k$. We will construct a family $(V_{k_0, \dots, k_j})_{\substack{0 \le j \le d', \\ (k_0, \dots, k_j) \in \mathbb{N}^{j+1}}}$ such that $(V_{k_0})_{k_0 \in \mathbb{N}} = (V_k)_{k \in \mathbb{N}}$, and such that for all $j \in \{1, \dots, d'\}$, we have:

$$ (46) \qquad K \subset \bigcup_{(k_0, \dots, k_j) \in \mathbb{N}^{j+1}} V_{k_0, \dots, k_j}, $$

and for all multi-index $(k_0, \dots, k_j) \in \mathbb{N}^{j+1}$, we have $V_{k_0, \dots, k_j} \subset V_{k_0, \dots, k_{j-1}}$, with equality if and only if $V_{k_0, \dots, k_{j-1}} \subset K$. Then we have:

$$ (47) \qquad K = \bigcup_{(k_0, \dots, k_{d'}) \in \mathbb{N}^{d'+1}} V_{k_0, \dots, k_{d'+1}}. $$

Indeed, for all $(k_0, \dots, k_{d'}) \in \mathbb{N}^{d+1}$, we either have $V_{k_0, \dots, k_{d'}} \subset K$ or $V_{k_0} \supsetneq V_{k_0, k_1} \supsetneq \dots \supsetneq V_{k_0, \dots, k_{d'}}$. In the second case, we have $\dim(V_{k_0, \dots, k_{d'}}) \le \dim(V_{k_0}) - d'$ so $V_{k_1, \dots, k_{d'}} = \{0\}$, which is a contradiction because $0 \in K$ by definition.

We do it by induction. Let $0 \le c \le d'$. Assume that we have constructed a family $(V_{k_0, \dots, k_j})_{\substack{0 \le j \le c, \\ (k_0, \dots, k_j) \in \mathbb{N}^{j+1}}}$, such that for all $j \in \{0, \dots, c\}$, we have:

$$ (48) \qquad K \subset \bigcup_{(k_0, \dots, k_j) \in \mathbb{N}^j} V_{k_1, \dots, k_j}, $$

and such that for all $j \in \{1, c\}$ and all $(k_0, \dots, k_j) \in \mathbb{N}^{j+1}$, we have $V_{k_0, \dots, k_j} \subset V_{k_0, \dots, k_{j-1}}$, with equality if and only if $V_{k_0, \dots, k_{j-1}} \subset K$.

Let $(k_0, \dots, k_c) \in \mathbb{N}^{c+1}$ be a multi-index such that $V_{k_1, \dots, k_c} \not\subset K$. Then we have almost surely $g(V_{k_0, \dots, k_c}) \ne \{0\}$ so the restriction of $h$ to $V_{k_0, \dots, k_c}$ has rank at least 1 almost surely. By the previous argument, the set:

$$ K \cap V_{k_0, \dots, k_c} = \{x \in V_{k_0, \dots, k_c} \mid \mathbb{P}(hx = 0) > 0\} $$

is included in the union of a countable family of subspaces of $V_{k_0, \dots, k_c}$ that have dimension $\dim(V_{k_0, \dots, k_c}) - 1$. For all multi-index $(k_0, \dots, k_c) \in \mathbb{N}^{c+1}$ such that $V_{k_1, \dots, k_c} \not\subset K$, we define $(V_{k_0, \dots, k_{c+1}})_{k_{c+1} \in \mathbb{N}}$ to be such a family. For every other multi-index $(k_0, \dots, k_c) \in \mathbb{N}^{c+1}$ and for all $k_{c+1} \in \mathbb{N}$, we define $V_{k_0, \dots, k_{c+1}} := V_{k_0, \dots, k_c}$. $\square$

**Remark 3.6.** Let $(\gamma_n)_{n \ge 0}$ be a random i.i.d. sequence in End($E$). Then for every $x \in E$, the sequence $\mathbb{P}(\bar{\gamma}_n x = 0)$ is non-decreasing and its limit is positive if and only if $x \in \ker(\nu)$.

*Proof.* For all $n$, we have $\mathbb{P}(\bar{\gamma}_n x = 0) = \mathbb{P}(\gamma_{n-1} \cdots \gamma_0 x = 0)$, which is trivially non-decreasing. Then the second statement is a reformulation of the definition of $\ker(\nu)$. $\square$
3.2. **Rank one boundary of a semi-group.** Given a subset A of a topological space X, we denote by $cl_X(A)$ the closure of A in X. Note that saying that an endomorphism has rank one is equivalent to saying that it is the product of a non-trivial vector on the left by a non-trivial linear form on the right.

Given a probability measure $\nu$ on a topological space $X$, we denote by $\text{supp}_X(\nu)$ or simply $\text{supp}(\nu)$ the smallest closed subset of $X$ on which $\nu$ is supported. Then $\text{supp}(\nu)$ is characterized by the fact that it is closed and for all open $U \subset X$, we have $\nu(U) > 0$ if and only if $U \cap \text{supp}(\nu) \neq \emptyset$.

We remind that given $E$ a vector space and $u \in E \setminus \{0\}$, we denote by $[u]$ the projective class of $u$ and we denote by $\text{P}(E)$ the projective space of $E$.

In this paragraph, $E$ is a vector space of dimension $d \ge 2$ over a local field $(\mathbb{K}, |\cdot|)$). Then $E$ is naturally endowed with a metric topology that does not depend on the choice of the norm and that makes the projective space $\text{P}(E)$ compact for the quotient topology. The results of this section are well known in the invertible case and rely on this compactness.

**Definition 3.7 (Rank one boundary).** Let $\Gamma < \text{End}(E)$ be a sub-semi-group. Let $\bar{\Gamma} := \text{cl}_{\text{End}(E)}(\mathbb{K}\Gamma)$. We denote by $\partial\Gamma$ the rank-one boundary of $\Gamma$, defined as:

$$ (49) \qquad \partial\Gamma := \{[\gamma] \mid \gamma \in \bar{\Gamma}, \text{rk}(\gamma) = 1\} $$

We define the left and right boundaries of $\Gamma$ as:

$$ \partial_u \Gamma := \{[hv] \mid [h] \in \partial \Gamma, v \in E \setminus \ker(h)\} \subset \text{P}(E) $$

$$ \partial_w \Gamma := \{[fh] \mid [h] \in \partial \Gamma, f \in E^* \setminus \ker(h^*)\} \subset \text{P}(E^*) $$

**Definition 3.8 (Range and boundary of a distribution).** Let $E$ be a Euclidean space and let $\nu$ be a probability measure on $\text{End}(E)$. We denote by $\Gamma_\nu$ the range of $\nu$ defined as the smallest closed sub-semi-group of $\text{End}(E)$ that has measure 1 for $\nu$. We define $\partial\nu := \partial\Gamma_\nu$ and $\partial_u\nu := \partial_u\Gamma_\nu$ and $\partial_w\nu := \partial_w\Gamma_\nu$.

Let us now show that the strong irreducibility in the sense of Definition 1.2 implies strong irreducibility in the usual sense. However, these notions are not equivalent, as mentioned in paragraph 1.1.

**Lemma 3.9 (Invariant subspaces and irreducibility).** Let $E$ be a Euclidean space. Let $\nu$ be a probability measure on $\text{End}(E)$. Let $S := \text{supp}(\nu)$. Let $V \subset E$ be a proper non-trivial subspace. If $SV \subset V$ then $\nu$ is not irreducible.

Let $N \ge 1$ and let $V_1, \dots, V_n \subset E$ be proper non-trivial subspaces. If $\bigcup_{i=1}^N S \cdot V_i \subset \bigcup_{i=1}^N V_i$ then $\nu$ is not strongly irreducible in the sense of Definition 1.2.
Proof. Let $\Gamma = \bigcup_{n \in \mathbb{N}} S^n = \bigcup_{n \in \mathbb{N}} \Pi(S^n)$ be the semi-group generated by $S$. Note that for all $n \in \mathbb{N}$, one has $\nu^{*n}(\Gamma) = 1$. Let $V \subset E$ be a proper non-trivial subspace such that $SV \subset V$. The fact that $V$ is a proper subspace implies that there exists a linear form $f \in E^* \setminus \{0\}$ such that $V \subset \ker(f)$. The fact that $V$ is not trivial implies that there exists a vector $v \in V \setminus \{0\}$. Let $f$ and $v$ be as above. We have $\Gamma \cdot v \subset V$, hence $f\gamma v = 0$ for all $\gamma \in \Gamma$ so $\nu$ is not irreducible.

Let $N \ge 1$ and let $V_1, \dots, V_n \subset E$ be a family of proper non-trivial subspaces such that $\bigcup_{i=1}^N S \cdot V_i \subset \bigcup_{i=1}^N V_i$. Let $f_1, \dots, f_n \in E^* \setminus \{0\}$ be such that $V_i \subset \ker(f_i)$ for all $i \in \{1, \dots, N\}$. Let $v \in V_1 \setminus \{0\}$. Then one has $\Gamma \cdot v \subset \bigcup V_i$, hence $\prod_{i=1}^N f_i \gamma v = 0$ for all $\gamma \in \Gamma$ so $\nu$ is not strongly irreducible. $\square$

We call irreducible semi-group a semi-group $\Gamma \subset \text{End}(E)$ that has no non-trivial invariant subspace. Note that given $\Gamma$ a semi-group, $\bar{\Gamma}$ is also a semi-group

**Lemma 3.10.** Let $E$ be a Euclidean space and let $\Gamma < \text{End}(E)$ be an irreducible semi-group. Then we have a factorization:

$$ (50) \qquad \partial\Gamma = \{[uw] \mid [u] \in \partial_u\Gamma, [w] \in \partial_w\Gamma\}. $$

Proof. Note that the space $\text{P}(\text{End}(E))$ is metrizable so the closure is characterized by sequences. Let $\pi$ be a rank one endomorphism. Saying that $[\pi] \in \partial\Gamma$ is equivalent to saying that there is a sequence $(\gamma_n) \in (\Gamma \setminus \{0\})^\mathbb{N}$ such that $[\gamma_n] \to [\pi]$. Note also that the product map is continuous so $\bar{\Gamma}$ is a semi-group. Therefore, for all $[\pi_1], [\pi_2] \in \partial\Gamma$, and for all $g \in \Gamma$ such that $\pi_1 g \pi_2 \ne 0$, we have $[\pi_1 g \pi_2] \in \partial\Gamma$.

Let $v_1 \in \partial_u\Gamma$ and let $f_2 \in \partial_v\Gamma$. Let $v_2 \in E$ and $f_1 \in E^*$ be such that $[v_1 f_1] \in \partial\Gamma$ and $[v_2 f_2] \in \partial\Gamma$. By definition of the irreducibility, there is an element $g \in \Gamma$ such that $f_1 g v_2 \ne 0$. Let $g$ be such an element. Then we have $v_1 f_1 g v_2 f_2 \ne 0$ hence $[v_1 f_1 g v_2 f_2] \in \partial\Gamma$. Moreover $[v_1 f_1 g v_2 f_2] = [v_1 f_2]$ because $f_1 g v_2$ is a scalar, therefore $[v_1 f_2] \in \partial\Gamma$. $\square$

**Lemma 3.11** (Characterisation of proximality). Let $\nu$ be a probability measure on $\text{End}(E)$. Let $(\gamma_n) \sim \nu^{\otimes N}$. Assume that $\nu$ is irreducible and that $\text{rk}(\nu) \ge 1$. Then the following assertions are equivalent:

(1) There exists a constant $b > 0$ such that $\sigma(\bar{\gamma}_n) \ge b$ almost surely and for all $n \in \mathbb{N}$.

(2) $\partial\nu = \emptyset$.

(3) We have $\rho_1(\bar{\gamma}_n) = \rho_2(\bar{\gamma}_n)$ almost surely and for all $n \in \mathbb{N}$.

Proof. Let $S = \text{supp}(\nu)$ as a warm-up, we prove that (1) is equivalent to (2). It is a direct consequence of the two following facts:

* The map $\sigma$ continuous on $\text{End}(E) \setminus \{0\}$ and it only depends on the projective class.
• a linear map $h \neq 0$ has rank one if and only if $\sigma(h) = 0$.

By continuity of $\sigma$ (1) is equivalent to saying that there exists $b$ such that $\sigma(S^{\cdot n} \setminus \{0\}) \subset [b, 1]$ for all $n \in \mathbb{N}$, by homogeneity, this is equivalent to saying that $\sigma(\mathbb{K}S^{\cdot n} \setminus \{0\}) \subset [b, 1]$ for all $n \in \mathbb{N}$, which in turn is equivalent to saying that $\sigma(\bigcup_n \mathbb{K}S^{\cdot n} \setminus \{0\}) \subset [b, 1]$, which again by continuity is equivalent to saying that $\sigma(\bar{\Gamma}_\nu \setminus \{0\}) \subset [b, 1]$. In conclusion (1) is equivalent to saying that $0 \notin \sigma(\bar{\Gamma}_\nu \setminus \{0\})$ which means that there are no rank one matrices in $\bar{\Gamma}_\nu$, which is equivalent to (2).

Let us now prove that (1) is equivalent to (3) by double contraposition. The proof relies on the fact that for all $1 \le j \le d$, the map $\rho_i : \text{End}(E) \to \mathbb{R}_{\ge 0}$ is continuous and homogeneous. Hence (3) is equivalent to saying that $\rho_1(g) = \rho_2(g)$ for all $g \in \bar{\Gamma}_\nu$. Assume that there exist $g \in \bar{\Gamma}_\nu$ such that $\rho_1(g) > \rho_2(g)$, which contradicts (3). Then we have $\sigma(g^k) \to 0$, which contradicts (1)

Reciprocally, let $[\pi] \in \partial\nu$. By irreducibility, there exists $g \in \bar{\Gamma}_\nu$ such that $g(\text{im}(\pi)) \not\subset \ker(\pi)$. Since $g\pi$ has rank one, we have $\rho_2(g\pi) = 0$ and $\rho_1(g\pi) = \frac{\|g\pi g\pi\|}{\|g\pi\|} > 0$. Therefore, we have $\rho_1(g\pi) > \rho_2(g\pi)$ and $g\pi \in \bar{\Gamma}_\nu$, which contradicts (3). $\square$

**3.3. Construction of the Schottky measure.** In this paragraph $E$ is a vector space of dimension $d \ge 2$ over a local field $(\mathbb{K}, |\cdot|)$). We endow $E$ with a norm $\|\cdot\|$ that is Euclidean when $\mathbb{K} = \mathbb{R}$, Hermitian when $\mathbb{K} = \mathbb{C}$ and ultra-metric otherwise. We endow $\text{End}(E)$ with the operator norm. We recall the notation introduced in Definition 1.3. Given $0 < \varepsilon \le 1$ and $g, h$ two matrices, we write $g\mathbb{A}^\varepsilon h$ when $\|gh\| \ge \varepsilon\|g\|\|h\|$.

Remind that a measurable binary relation over a measurable space $\Gamma$ is a subset $\mathbb{A} \subset \Gamma \times \Gamma$ that is measurable for the product $\sigma$-algebra. Given $g, h \in \Gamma$, we write $g\mathbb{A}h$ to say that $(g, h) \in \mathbb{A}$. Given $S, T \subset \Gamma$, we write $SAT$ to say that $S \times T \subset \mathbb{A}$. Given $n \in \mathbb{N}$ and $(g_0, \dots, g_n) \in \Gamma^{n+1}$, we write $g_0\mathbb{A}\dots\mathbb{A}g_n$ to say that $g_i\mathbb{A}g_{i+1}$ for all $i \in \{0, \dots, n-1\}$.

**Definition 3.12.** Let $\Gamma$ be a measurable space, let $\mathbb{A}$ be a measurable binary relation on $\Gamma$ and let $0 \le \rho < 1$. Let $\nu_s$ be a measure on $\Gamma$. We say that $\nu_s$ is $\rho$-Schottky for $\mathbb{A}$ if:

$$ \forall h \in \Gamma, \nu_s\{\gamma \in \Gamma | \neg h\mathbb{A}\gamma\} \le \rho \quad \text{and} \quad \nu_s\{\gamma \in \Gamma | \neg \gamma\mathbb{A}h\} \le \rho. $$

**Remark 3.13.** Given $\Gamma$ be a measurable space, given $\mathbb{A} \subset \mathbb{A}'$ two measurable binary relations on $\Gamma$, given $0 \le \rho \le \rho' < 1$ and given $\nu$ be a measure on $\Gamma$, if $\nu_s$ is $\rho$-Schottky for $\mathbb{A}$, then it is also $\rho'$-Schottky for $\mathbb{A}'$.
To construct a Schottky measure, we will approximate a uniform distribution over a set of rank 1 matrices that are in general position i.e., the sum of k distinct matrices in said set has rank exactly k for all $k \le d$.

**Lemma 3.14.** Let $\Gamma < \text{End}(E)$ be a strongly irreducible and proximal semi-group and let $0 < \rho \le 1$. There exist an integer $N \in \mathbb{N}$, a constant $\varepsilon > 0$ and a family $([\pi_1], \dots, [\pi_N]) \in \partial\Gamma^N$ such that:

$$ \forall h \in \text{End}(E) \setminus \{0\}, \ \#\{k \mid \pi_k A^\varepsilon h\} \ge (1-\rho)N \text{ and } \#\{k \mid h A^\varepsilon \pi_k\} \ge (1-\rho)N. $$

*Proof.* Let $d := \dim(E)$. Let $m \ge 0$. Assume that we have constructed a family $([u_1], \dots, [u_m]) \in \partial_u \Gamma^m$ that is in general position in the sense that for all $k \le d$, and for all $1 \le i_1 < \dots < i_k \le m$, we have $\dim(\langle u_{i_1}, \dots, u_{i_k} \rangle) = k$. Let:

$$ u_{m+1} \in \{u \in E \mid [u] \in \partial_u \Gamma\} \setminus \bigcup_{\substack{k \le d-1, \\ 1 \le i_1 < \dots < i_k \le m}} \langle u_{i_j} \rangle_{1 \le j \le k}. $$

Such a $u_{m+1}$ exists because $\{u \in E \mid [u] \in \partial_u \Gamma\} \cup \{0\}$ is $\Gamma$-invariant by definition and it is not $\{0\}$ by Lemma 3.11. Hence $\{u \in E \mid [u] \in \partial_u \Gamma\}$ cannot be included in $\bigcup_{i_1 < \dots < i_k} \langle u_{i_j} \rangle_{1 \le j \le k}$, which is a finite union of hyperplanes. Then $[u_{m+1}] \in \partial \Gamma$ and we can easily check that $([u_1], \dots, [u_{m+1}])$ is in general position.

let $N := \lfloor \frac{d-1}{\rho} \rfloor$. Let $([w_1], \dots, [w_N]) \in \partial_u \Gamma^N$ be in general position and let $([w_1], \dots, [w_N]) \in \partial_w \Gamma^N$ be in general position. We can construct the family $(u_i)_{1 \le i \le N}$ by induction using the above construction. To construct $(w_i)_{1 \le i \le N}$, note that the transpose semi-group $^t\Gamma := \{^t\gamma \mid \gamma \in \Gamma\} \subset \text{End}(E^*)$ is also strongly irreducible and proximal and $\partial_w \Gamma = \partial_u {}^t\Gamma$.

For all $k \in \{1, \dots, N\}$, we define $\pi_k := u_k w_k$. Then $([\pi_1], \dots, [\pi_N]) \in \partial \Gamma^N$ by Lemma 3.10. With this formulation, one can check that $gA^\varepsilon \pi_k$ if and only if $gA^\varepsilon u_k$ and $\pi_k A^\varepsilon h$ if and only if $w_k A^\varepsilon h$ (simply note that $\|\pi_k\| := \|u_k\|\|w_k\|$).

Let $h \in \text{End}(E) \setminus \{0\}$. Let $I^h := \{i \mid u_i \in \ker(h)\}$. Since the family $([u_1], \dots, [u_N])$ is in general position and $\langle u_i \rangle_{i \in I} \subset \ker(h)$, we have $\#I^h \le d-1$. Let $J^h := \{j \mid w_j \in \ker(h^t)\}$. By the same argument, we have $\#J^h \le d-1$. For all $h \in \text{End}(E) \setminus \{0\}$, let:

$$ \psi(h) := \max_{\substack{I, J \subset \{1, \dots, N\}, \\ \#I \le d-1, \#J \le d-1}} \min \left\{ \min_{i \notin I} \frac{\|hu_i\|}{\|h\|\|u_i\|}, \min_{j \notin J} \frac{\|w_jh\|}{\|w_j\|\|h\|} \right\}. $$

By the previous argument, one has $\psi(h) > 0$ for all $h$. Moreover the maps $h \mapsto \frac{\|hv\|}{\|h\|\|v\|}$ and $\frac{\|fh\|}{\|f\|\|h\|}$ are continuous for all $v \in E \setminus \{0\}$ and all $f \in E^* \setminus \{0\}$. Hence $\psi$ is continuous. Moreover, $\psi$ is invariant by scalar multiplication so there
is a continuous map $\phi : \mathrm{P}(\mathrm{End}(E)) \to (0,1]$ such that $\phi([h]) = \psi(h)$ for all $h \in \mathrm{End}(E) \setminus \{0\}$. The projective space $\mathrm{P}(\mathrm{End}(E))$ is compact so $\phi(\mathrm{P}(\mathrm{End}(E)))$ is compact. Let $\varepsilon$ be its lower bound.

Now let $h \in \mathrm{End}(E) \setminus \{0\}$. The set of indices $\{k | \pi_k A^\varepsilon h\}$ is $\{k | w_k A^\varepsilon h\}$, which has cardinality at least $N - d + 1 \ge (1 - \rho)N$. Indeed, since $\psi(h) \ge \varepsilon$, a sufficient condition to have $w_j A^\varepsilon h$ is for $j$ not to be included in a set $J$ that realises the maximum in the definition of $\psi(h)$. By the same argument $\#\{k | h A^\varepsilon \pi_k\} \ge N - d + 1$. $\square$

**Corollary 3.15.** Let $\nu$ be a strongly irreducible and proximal probability measure on $\mathrm{End}(E)$, let $0 < \rho \le 1$ and let $\delta : (0,1) \to (0,1)$. There exist an integer $N$, two constants $\alpha', \varepsilon \in (0,1)$, a family $(n_k)_{1 \le k \le N} \in \mathbb{N}^N$ and a family $(S'_k)_{1 \le k \le N}$ of measurable subsets of $\Gamma$ such that $\nu^{*n_k}(S'_k) \ge \alpha'$ for all $1 \le k \le N$ and such that:

$$ (51) \quad \forall h \in \mathrm{End}(E), \#\{k | S'_k A^{2\varepsilon} h\} \ge (1-\rho)N \quad \text{and} \quad \#\{k | h A^{2\varepsilon} S'_k\} \ge (1-\rho)N, $$

$$ (52) \quad \forall 1 \le j \le N, \#\{k | S'_k A^\varepsilon S'_j\} \ge (1-\rho)N \quad \text{and} \quad \#\{k | S'_j A^\varepsilon S'_k\} \ge (1-\rho)N, $$

$$ (53) \quad \forall 1 \le k \le N, \forall s' \in S'_k, \sigma(s') \le \delta(\varepsilon). $$

*Proof.* Let $N \in \mathbb{N}$, let $\varepsilon > 0$ and let $([\pi_1], \dots, [\pi_N]) \in \partial\nu^N$ be as in Lemma 3.14. To all $k \in \{1, \dots, N\}$, we associate the open set:

$$ S'_k := \left\{ s \in \mathrm{End}(E) \setminus \{0\} \left| \min_{\substack{c \in \mathbb{K}, \\ \|c\pi_k\| \ge \|s\|}} \frac{\|s - c\pi_k\|}{\|s\|} \le \varepsilon, \sigma(s) \le \delta(\varepsilon) \right. \right\}. $$

Now let $k \in \{1, \dots, N\}$ and $h \in \mathrm{End}(E)$. Assume that $h A^{3\varepsilon} \pi_k$. We claim that for all $s' \in S'_k$, we have $h A^{2\varepsilon} s'$. Let $s' \in S'_k$ and let $c \in \mathbb{K}$ be such that $\|s' - c\pi_k\| < \varepsilon\|s'\|$ and $\|c\pi_k\| \ge \|s\|$. The left multiplication by $h$ is $\|h\|$-Lipschitz on $\mathrm{End}(E)$ so $\|hs' - ch\pi_k\| \le \varepsilon\|h\|\|s'\|$. We assumed $h A^{3\varepsilon} \pi_k$, which means that $\|h\pi_k\| \ge 3\varepsilon\|h\|\|c\pi_k\|$. Hence by triangular inequality, we have $\|hs'\| \ge 3\varepsilon\|h\|\|c\pi_k\| - \|h\|\|s'\| \ge 2\varepsilon\|h\|\|s'\|$. Moreover, for all $h$, there exist at least $N(1-\rho)$ indices $k$ such that $h A^{3\varepsilon} \pi_k$ and therefore such that $h A^{2\varepsilon} S_k$. The same argument works for the left alignment, which proves (51).

Now we show (52). Let $j, k \in \{1, \dots, N\}$ be such that $\pi_j A^{3\varepsilon} \pi_k$ and let $s'_j \in S'_j$ and $s'_k \in S'_k$. Then by the above argument, applied to $h = s'_j$, we have $s'_j A^{2\varepsilon} \pi_k$ and by the same argument applied to the transpose, we have $s'_j A^\varepsilon s'_k$. Moreover, for all $j$, we have $N(1-\rho)$ indices $k$ such that $\pi_j A^{3\varepsilon} \pi_k$ and vice versa by Lemma 3.14, which proves (52).

Let us now show the existence of a family $(n_k) \in \mathbb{N}^N$ such that $\nu^{*n_k}(S'_k) > 0$ for all $k$, to conclude, we simply need to take $\alpha = \min_k \nu^{*n_k}(S'_k)$. Let $k \in$
$\{1, \dots, N\}$. The interior of $S'_k$ contain $\pi_k$. It means that the interior of $S'_k$ intersects $\bar{\Gamma}_\nu = \text{cl}(\mathbb{K}\Gamma_\nu)$ and $S'_k$ is homogeneous so it intersects $\Gamma_\nu$. Hence, there exists an integer $n_k$ such that the interior of $S'_k$ intersects $\text{supp}(\nu^{*n_k})$. Therefore, $\nu^{n_k}(S'_k) > 0$ by characterization of the support. $\square$

**Lemma 3.16.** Let $E$ be a Euclidean space, let $\nu$ be a strongly irreducible and proximal probability measure on $\text{End}(E)$, let $0 < \rho < 1$ and let $f = (0, 1) \to (0, 1)$. There exist an integer $N$, two constants $\alpha, \varepsilon \in (0, 1)$, an integer $m$ and a family $(S_k)_{1 \le k \le N}$ of measurable subsets of $\text{End}(E)$ such that $\nu^{*m}(S_k) \ge \alpha$ for all $1 \le k \le N$ and such that:

$$ (54) \quad \forall h \in \text{End}(E), \#\{k \mid S_k A^\varepsilon h\} \ge (1-\rho)N \quad \text{and} \quad \#\{k \mid h A^\varepsilon S_k\} \ge (1-\rho)N, $$

$$ (55) \quad \forall 1 \le k \le N, \forall s \in S_k, \sigma(s) \le \delta(\varepsilon). $$

*Proof.* Without loss of generality, we may assume that $\rho < \frac{1}{3}$ and that $\delta(\varepsilon) \le \varepsilon^2/12$. Let $N \in \mathbb{N}$, let $\alpha', \varepsilon \in (0, 1)$, let $(n_k)$ and let $(S'_k)$ be as in Corollary 3.15. To all index $k \in \{1, \dots, N\}$, we associate two indices $i_k, j_k \in \{1, \dots, N\}$ such that $S'_k A^\varepsilon S'_i$ and $S'_i A^\varepsilon S'_j$ and $S'_j A^\varepsilon S'_k$. By (52) in Corollary 3.15, such indices $i_k, j_k$ exist because:

$$ \#\{(i, j) \mid S'_i A^\varepsilon S'_k A^\varepsilon S'_i A^\varepsilon S'_j A^\varepsilon S'_k A^\varepsilon S'_j\} \ge (1-2\rho)N(1-3\rho)N > 0. $$

Hence the set of all possible values for $i_k, j_k$ is non-empty. Let $m'$ be the smallest common multiple of $\{n_k + n_{i_k} \mid 1 \le k \le N\}$, let $m''$ be the smallest common multiple of $\{n_k + n_{j_k} \mid 1 \le k \le N\}$ and let $m = m' + m''$. Let $k \in \{1, \dots, N\}$. We define $p_k := \frac{m'}{n_k+n_{i_k}}$ and $q_k := \frac{m''}{n_k+n_{j_k}}$ and:

$$ S_k := (S'_k \cdot S'_{i_k})^{p_k} \cdot (S'_{j_k} \cdot S'_k)^{q_k} = \Pi \left( (S'_k \times S'_{j_k})^{p_k} \times (S'_{j_k} \times S'_k)^{q_k} \right). $$

Then we have $\nu^{*m}(S_k) \ge (\nu^{n_k}(S'_k)\nu^{n_{i_k}}(S'_{i_k}))^{p_k} (\nu^{n_{i_k}}(S'_{i_k})\nu^{n_{j_k}}(S'_k))^{q_k} \ge \alpha'^{2p_k+2q_k} \ge \alpha'^{2m} =: \alpha$.

Now let $h \in \text{End}(E) \setminus \{0\}$ and let $k \in \{1, \dots, N\}$. Assume that $h A^{2\varepsilon} S'_k$, then by Lemma 2.15 and because we assumed that $\delta(\varepsilon) \le \varepsilon^2/12$, we have $h A^\varepsilon S_k$. If we instead assume that $S'_k A^{2\varepsilon} h$, then by the same argument, we have $S_k A^\varepsilon h$. Hence, we can derive (54) in Lemma 3.16 from (51) in Corollary 3.15.

Moreover, by Lemma 2.8 and because we assumed that $\delta(\varepsilon) \le \varepsilon^2/8$, for all $k$ and for all $s \in S_k$, we have $\sigma(s) \le \delta(\varepsilon)^{2p_k+2q_k} 2^{4p_k+4q_k-4}\varepsilon^{-4p_k-4q_k+2} \le \delta(\varepsilon)$, which proves (55). $\square$

Now let us prove the main result of this section. In fact, for the rest of the discussion, we will not directly use the formal definitions of strong irreducibility
and proximality. Instead, we simply think of a strongly irreducible and proximal measure as a measure that satisfies the conclusions of Corollary 3.17 for $\rho = \frac{1}{6}$ and $\delta : \varepsilon \mapsto \varepsilon^6/48$.

**Corollary 3.17.** Let $E$ be a Euclidean space, let $\Gamma \in \{\text{End}(E), \text{GL}(E)\}$ and let $\nu$ be a strongly irreducible and proximal probability measure on $\Gamma$, let $0 < \rho < 1$ and let $\delta : (0, 1) \to (0, 1)$. There exists an integer $m$, two constants $\alpha, \varepsilon \in (0, 1)$ and a probability measure $\tilde{\nu}_s$ on $\Gamma^m$ such that:

(1) The measure $\Pi_*\tilde{\nu}_s$ is $\rho$-Schottky for $\mathbb{A}^\varepsilon$.

(2) The measure $\tilde{\nu}_s$ is absolutely continuous with respect to $\nu^{\otimes m}$ in the sense that $\alpha\tilde{\nu}_s \le \nu^{\otimes m}$.

(3) We have $\sigma_*\Pi_*\tilde{\nu}_s[0, \delta(\varepsilon)] = 1$ i.e., for all $(s_1, \dots, s_m)$ in the support of $\tilde{\nu}_s$, we have $\sigma(s_1 \cdots s_m) \ge \delta(\varepsilon)$.

(4) The support of $\tilde{\nu}_s$ is compact in $\Gamma^m$.

*Proof.* Let $m, N \in \mathbb{N}$, $\alpha, \varepsilon \in (0, 1)$ and $(S_k)_{1 \le k \le N}$ be as in Lemma 3.16. Define $f : \Gamma^m \to \mathbb{R}_{\ge 0}$ as:

$$f := \sum_{k=1}^{N} \frac{\mathbb{1}_{S_k} \circ \Pi}{\nu^{*m}(S_k)}.$$

Then $f \le \frac{N}{\alpha}$ because we assumed that $\nu^{*m}(S_k) \ge \alpha$ for all $k$. Moreover, we can check that $\int_{\Gamma^m} f d\nu^{\otimes m} = N$ and for all $k \in \{1, \dots, N\}$ we have $\int_{\Pi^{-1}(S_k)} f d\nu^{\otimes m} \ge 1$. Let:

$$\tilde{\nu}_s := \frac{f \nu^{\otimes m}}{\int_{\Gamma^m} f d\nu^{\otimes m}} \quad \text{and} \quad \nu_s := \Pi_* \tilde{\nu}_s.$$

Then $\frac{N}{\alpha}\nu_s \le \nu^{*m}$ by definition. Moreover, for all $I \subset \{1, \dots, N\}$, we have $\nu_s(\bigcup_{i \in I} S_i) \ge \frac{\#I}{N}$. Hence $\nu_s$ is $\rho$-Schottky by (54). Moreover $\nu_s$ is supported on $\bigcup S_k$ and $\sigma(S_k) \subset [0, \delta(\varepsilon)]$ for all $k$ and as a consequence $\sigma_*\nu_s[0, \delta(\varepsilon)] = 1$.

Now it remains to prove that we can moreover assume that $\tilde{\nu}_s$ has compact support. Let $\beta \in (0, 1)$. Since $\Gamma$ is $\sigma$-compact, so $\Gamma^m$ also is, hence there exists a compact $K \subset \Gamma^m$ such that $\tilde{\nu}_s(K) > 1 - \beta$. Let $\tilde{\nu}_s^K := \frac{1_K \tilde{\nu}_s}{\tilde{\nu}_s(K)}$. Let $\nu_s^K := \Pi_* \tilde{\nu}_s^K$. Then $\nu_s^K(\bigcup_{i \in I} S_i) \ge \frac{1}{N} - \beta$ for all $I \subset \{1, \dots, N\}$. Hence $\nu_s^K$ is $(\rho + \beta)$-Schottky. Moreover $\alpha(1 - \beta)\tilde{\nu}_s \le \nu^{\otimes m}$. This is true for all $0 < \beta < \frac{1}{3} - \rho$ so $\rho + \beta$ can take any value in $(0, 1)$ and we always have $\alpha(1 - \beta) > 0$. $\square$
4. CONSTRUCTION OF A GLOBALLY ALIGNED EXTRACTION

In this section, we use Corollary 3.17 to study a product of i.i.d. random
matrices whose distribution that we call $\nu$ is strongly irreducible and proximal.
Let $\alpha, m, \rho, \varepsilon$ and $\nu_s$ be as in Corollary 3.17 for $\delta : \varepsilon \mapsto \varepsilon^6/32$ and let $\mathbb{A} = \mathbb{A}^\varepsilon$.

The idea starts with the following observation. Since we have a measure $\tilde{\nu}_s \le \alpha^{-1}\nu^{\otimes m}$, then $\nu^{\otimes m}$ is a convex combination of $\tilde{\nu}_s$ and of an other measure, that we call $\tilde{\kappa}$ and know nothing about. This means that we can construct a random sequence $(\gamma_n) \sim \nu^{\otimes \mathbb{N}}$ by alternating randomly between independent $\tilde{\nu}_s$ distributed blocks of length $m$ and unknown blocks of length $m$. From such a decomposition, we construct another decomposition that alternates non-randomly between random blocks of random length and random $\tilde{\nu}_s$ distributed blocks. These blocks are independent. Using the properties of $\nu_s := \Pi_*\tilde{\nu}_s$, we also make sure that each block have the inner structure of being a concatenation of three blocs $\tilde{u}_0 \odot \tilde{s}_1 \odot \tilde{u}_2$ with $u_0 \mathbb{A} s_1 \mathbb{A} u_2$ (removing the $\tilde{\cdot}$ means taking the product) and $s_1$ is in the support of $\nu_s$, which means that $\sigma(s_1)$ is small enough to apply Lemma 2.11.

Starting from this sequence alternating between words whose product are
close to rank one matrices and $\tilde{\nu}_s$ distributed words, we use an inductive construc-
tion to create a sequence of blocks whose products form an aligned sequence. The
issue is that with this construction the lengths of the blocks are not stopping times
and the blocks are not independent.

A crucial technical point is to make sure that each step of the construction
is independent of the contents of each $\tilde{\kappa}$ distributed block. Allowing us to have
control over the tail of the distribution of each block and therefore over the
moments.

In this section, we will use the following notations for usual probability
distributions. Given a parameter $\alpha \in (0, 1)$, we write $\mathcal{B}_\alpha := \alpha\delta_1 + (1-\alpha)\delta_0$ for
the Bernoulli law of parameter $\alpha$ and $\mathcal{G}_\alpha := \sum_{k \ge 0} \alpha^k (1-\alpha)\delta_k$. We write $\mathcal{U}$ for the
uniform probability distribution on $[0, 1]$ i.e., the Lebesgue measure. Given an
integer $m$, we will write $m \times_*$ and $m+_*$ for the push forward operators associated
to the multiplication by $m$ and the addition of $m$ respectively.

**Definition 4.1 (Power convolution).** Let $\Gamma$ be a metric semi-group. We write $*$ for the convolution product of probability distributions on $\Gamma$. Let $\nu$ be a probability distribution on $\Gamma$. Let $\eta$ be a probability distribution on $\mathbb{N}$. We define the probability distributions $\nu^{*\eta}$ on $\Gamma$ and $\nu^{\otimes\eta}$ on $\bar{\Gamma}$ by:

$$
(56) \qquad \nu^{*\eta} = \sum_{k \in \mathbb{N}} \eta\{k\} \nu^{*k} \quad \text{and} \quad \nu^{\otimes\eta} = \sum_{k \in \mathbb{N}} \eta\{k\} \nu^{\otimes k}.
$$
In other words, given an i.i.d. sequence of integers, matrices or words: $(\gamma_n)_{n \in \mathbb{N}} \sim \nu^{\otimes \mathbb{N}}$ and a random sequence of integers $(w_k)_{k \ge 0} \sim \eta^{\otimes \mathbb{N}}$ that are independent, and using the notation of Definition 1.5, we have $(\tilde{\gamma}_k^w)_{k \ge 0} \sim (\nu^{\otimes \eta})^{\otimes \mathbb{N}}$ and $(\gamma_k^w)_{k \ge 0} \sim (\nu^{*\eta})^{\otimes \mathbb{N}}$. Let us give a few basic results to understand the notation of power convolution.

**Lemma 4.2.** Let $\nu$ be a probability distribution on $(\mathbb{R}_{\ge 0}, +)$ and $\eta$ be a probability distribution on $(\mathbb{N}, +)$. Let $(x_n)_{n \ge 0}$ be a random sequence and let $w$ be a random non-negative integer. Assume that $\mathbb{P}(w \ge k) \le \eta(\mathbb{N}_{\ge k})$ for all $k \ge 0$ and assume that $\mathbb{P}(x_n > t | x_0, \dots, x_{n-1}, w) \le \nu(t, +\infty)$ almost surely, for all $t \ge 0$ and all $n$. Then we have $\mathbb{P}(\bar{x}_w > t) \le \nu^{*\eta}(t, +\infty)$ for all $t \ge 0$.

*Proof.* The proof is straightforward using a monotonous coupling. Let $(u_k)_{k \ge -1} \sim \mathcal{U}^{\otimes \mathbb{N}}$. Let $p = \min\{k \in \mathbb{N}_{\ge 0} | \eta(\mathbb{N}_{\ge k}) \le u_{-1}\}$. For all $n \ge 0$, write $y_n = \min\{t \ge 0 | \nu(t, +\infty) \le u_k\}$. That way $\bar{y}_p \sim \nu^{*\eta}$. Let $w' = \min\{k \in \mathbb{N}_{\ge 0} | \mathbb{P}(w \ge k) \le u_{-1}\}$. Then $w'$ has the same distribution law as $w$ and $w' \le p$ for all values of $(u_k)_{k \ge -1} \in (0, 1)^{\mathbb{N}}$. We define a sequence $(x'_n)_{n \ge 0}$ that has the same law as $(x_n)_{n \ge 0}$ by induction on $n$ with the following formula:

$$ \forall n \in \mathbb{N}, x'_n = \min \{t \ge 0 \mid \mathbb{P}(x_n > t \mid \forall k < n, x_k = x'_k \cap w = w') \le u_k\}. $$

that way, we have $\mathbb{P}(x'_n > t | x'_0, \dots, x'_{n-1}, w') = \mathbb{P}(x_n > t | x_0, \dots, x_{n-1}, w)$ almost surely<fnref n="5" /> and for all $t \ge 0$ so the joint distribution of $(w', (x'_n)_{n \ge 0})$ is equal to the joint distribution of $(w, (x_n)_{n \ge 0})$. Moreover, we have $x'_n \le y_n$ for all $n$ so $\bar{x}'_n \le \bar{y}_n$ for all $n \ge 0$. Moreover the sequence $(\bar{x}'_n)$ and $(\bar{y}_n)$ are both non-decreasing so $\bar{x}'_{w'} \le \bar{y}_p$. $\square$

The following Lemmas illustrate how well the notion of finite exponential moment behaves under power convolution.

**Lemma 4.3.** Let $0 < \alpha, \beta < 1$, we have $(1 +_* \mathcal{G}_{1-\alpha})^{*(1+_* \mathcal{G}_{1-\beta})} = (1 +_* \mathcal{G}_{1-\alpha\beta})$.

*Proof.* Let $(a_n)_{n \ge 1}$, $(b_n)_{n \ge 1} \sim \mathcal{B}_{\alpha}^{\otimes \mathbb{N}} \otimes \mathcal{B}_{\beta}^{\otimes \mathbb{N}}$. Then $(a_n b_n) \sim \mathcal{B}_{\alpha\beta}^{\otimes \mathbb{N}}$ For all $n \ge 1$, let $\bar{x}_n$ be the $n$-the element of $\{n | a_n = 1\}$. Then $(x_n)_{n \ge 0} \sim (1 +_* \mathcal{G}_{1-\alpha})^{\otimes \mathbb{N}}$. Moreover, by independence, $(b_{\bar{x}_n}) \sim \mathcal{B}_{\beta}^{\otimes \mathbb{N}}$ and $(b_{\bar{x}_n})_{n \ge 1}$ is independent of $(x_n)_{n \ge 0}$. Let $y \ge 1$ be the smallest integer such that $b_{\bar{x}_y} = 1$, then $y \sim (1 +_* \mathcal{G}_{1-\beta})$ and $y$ is independent of $(x_n)$ so $\bar{x}_y \sim (1 +_* \mathcal{G}_{1-\alpha})^{*(1+_* \mathcal{G}_{1-\beta})}$ and $\bar{x}_y$ is also the first index $n \ge 1$ such that $a_n b_n = 1$ so $\bar{x}_y \sim (1 +_* \mathcal{G}_{1-\alpha\beta})$. $\square$

<footnote n="5">⁵For all $t \ge 0$, one can choose a map $x'_0, \dots, x'_{n-1}, w' \mapsto \mathbb{P}(x_n > t | \forall k < n, x_k = x'_k \cap w = w') \le \nu(t, +\infty)$ that satisfies the integral formula for conditional probabilities because $\mathbb{R}_{\ge 0}^n \times \mathbb{N}$ is second countable. However this map is not unique. Yet, two realization of this map match on a set that has measure 1 for the distribution of $(x_0, \dots, x_{n-1}, w)$ and by induction on $n$, this set has measure 1 for the distribution of $(x'_0, \dots, x'_{n-1}, w')$ so the equivalence class of $x'_n$ as a random variable is well defined and $(x'_0, \dots, x'_n, w')$ has the same distribution law as $(x_0, \dots, x_n, w)$, which concludes the induction.</footnote>
**Lemma 4.4.** Let $\nu$ be a probability distribution on $\mathbb{R}$ and $\eta$ be a probability distribution on $\mathbb{N}$. Assume that $\nu$ and $\eta$ both have a finite positive exponential moment, then $\nu^{*\eta}$ also has. Let $w \sim \eta$, $x \sim \nu$ and $y \sim \nu^{*\eta}$. Given $\alpha, \beta > 1$ such that $\mathbb{E}(\beta^x) \le \alpha$, we have $\mathbb{E}(\beta^y) \le \mathbb{E}(\alpha^w)$. Hence, there exists $\beta > 1$ such that $\mathbb{E}(\beta^y) < +\infty$.

*Proof.* We first show the second result. Let $(w, (x_n)_{n\ge 0}) \sim \eta \otimes \nu^{\otimes \mathbb{N}}$ and let $\alpha, \beta > 1$. Assume that $\mathbb{E}(\beta^{x_n}) \le \alpha$ for all $n$. Then by independence, we have $\mathbb{E}(\beta^{\bar{x}_w}) = \sum_k \mathbb{P}(w=k)\mathbb{E}(\beta^{\bar{x}_k}) \le \sum_k \mathbb{P}(w=k)\alpha^k = \mathbb{E}(\alpha^w)$.

To conclude, note that the fact that $\eta$ has a finite exponential moment means that there exists $\alpha > 1$ such that $\mathbb{E}(\alpha^w) < +\infty$. The fact that $\nu$ has a finite exponential moment means that there exists $\beta_0 > 1$ such that $\mathbb{E}(\beta_0^{x_0}) < +\infty$. By Jensen's inequality and by concavity of the power function, we have $\mathbb{E}(\beta^{x_0}) \le \mathbb{E}(\beta_0^{x_0})^{\frac{\log(\beta)}{\log(\beta_0)}}$ for all $1 < \beta < \beta_0$ so there exists $\beta > 1$ such that $\mathbb{E}(\beta^{x_0}) \le \alpha$. $\square$

**4.1. Statement of the result and motivation.** In this paragraph, we give more detail on the probabilistic idea behind the construction. The following lemma is simple and not really new but central in the pivoting technique, we state it using the power notation for extractions in order to get familiar with this notation.

**Lemma 4.5.** Let $\Gamma$ be a metric space. Let $\kappa, \eta$ be probability distributions on $\Gamma$ and let $\alpha \in (0, 1)$. Let $\nu := \alpha\eta + (1-\alpha)\kappa$. Let $(u_n)_{n \sim \kappa} \otimes \mathbb{N}$ and let $(s_n)_{n \sim \eta} \otimes \mathbb{N}$ be independent of the data of $(u_n)_n$. Let $(w_{2n})_{n \in \mathbb{N}} \sim \mathcal{G}_{\alpha}^{\otimes \mathbb{N}}$ be independent of the joint data of $(u_n)_n$ and $(s_n)_n$ and let $w_{2k+1} = 1$ for all $k \in \mathbb{N}$. For all $k \in \mathbb{N}$, we define:

$$ \tilde{g}_{2k} := \tilde{u}_{2k}^w = (u_{\overline{w}_{2k}}, \dots, u_{\overline{w}_{2k+1}-1}) = (u_{\overline{w}_{2k}}, \dots, u_{\overline{w}_{2k}+w_{2k}-1}) \\ \tilde{g}_{2k+1} := \tilde{s}_{2k+1}^w = (s_{\overline{w}_{2k+1}}, \dots, s_{\overline{w}_{2k+2}-1}) = (s_{\overline{w}_{2k}}). $$

Let $(\gamma_n)_n = \bigoplus_{k=0}^{+\infty} \tilde{g}_k$ i.e., $\tilde{\gamma}_k^w = \tilde{g}_k$ for all $k \in \mathbb{N}$. Then $\gamma_n \sim \nu^{\otimes \mathbb{N}}$.

*Proof.* Let $(u_n)_n$, $(s_n)_n$ and $(w_k)_k$ be as in Lemma 4.5. To all integer $n$, we associate a random number $i_n$, we write $i_n = 1$ when there is an integer $k$ such that $\overline{w}_{2k} = n$ and 0 otherwise. We first show that $(i_n)_n \sim \mathcal{B}_{\alpha}^{\otimes \mathbb{N}}$. For that, we first observe that by definition of $\mathcal{G}_{\alpha}$, we have:

$$ \forall t, k \in \mathbb{N}, \mathbb{P}(w_{2k} = t | w_{2k} \ge t) = \alpha. $$

Moreover, the $(w_{2k})_k$ are independent so:

(57) \qquad $\forall t, k \in \mathbb{N}, \mathbb{P}(w_{2k} = t | w_{2k} \ge t, (w_{2k'})_{k'<k}) = \alpha.$
Now for all $n \in \mathbb{N}$, we write $q_n := \max\{k \in \mathbb{N} | \overline{w}_{2k} < n\}$. Then for all $k \in \mathbb{N}$, we have $q_n = k$ if and only if $\overline{w}_{2k} < n$ and $\overline{w}_{2k+2} \ge n$. Note that by construction, $\overline{w}_{2k+2} = \overline{w}_{2k} + w_{2k} + 1$ so we can decompose the event ($q_n = k$) into the intersection of $\overline{w}_{2k} \le n$ (which is in the $\sigma$-algebra generated by $(w_{2k'})_{k'<k}$) and $w_{2k} \ge n - \overline{w}_{2k} - 1$. Moreover, we have $i_n = 1$ if and only if $\overline{w}_{2q_n+2} = n$, which is equivalent to say that $w_{2q_n} = n - \overline{w}_{2q_n} - 1$. Note also that in (57) one can replace the constant parameter $t$ by a random parameter $t_n^k := n - \overline{w}_{2k} - 1$ which can be expressed in terms of $(w_{2k'})_{k'<k}$ and then use (57) on the (non-negative) level sets of $t_n^k$. Hence we have:

$$ \forall k, n \in \mathbb{N}, \mathbb{P}(w_{2k} = n - \overline{w}_{2k} - 1 | w_{2k} \ge n - \overline{w}_{2k} - 1 \ge 0, (w_{2k'})_{k'<k}) = \alpha, $$

where $(w_{2k} \ge n - \overline{w}_{2k} - 1 \ge 0)$ is equivalent to ($q_n = k$) and is the intersection of $(t_n^k \ge 0)$, which is in the $\sigma$-algebra generated by $(w_{2k'})_{k'<k}$ with the event $(w_{2k} \ge t_n^k)$ and under that condition, the event $(w_{2k} = n - \overline{w}_{2k} - 1)$ is equivalent to ($i_n = 1$). The above is true for all $k$ in the support of $q_n$, so we rephrase it as:

$$ \forall n \in \mathbb{N}, \mathbb{P}(i_n = 1 | (w_{2k'})_{k'<q_n}) = \alpha. $$

Moreover, the data of $(i_n)_{n'<n}$ is determined by (and actually equivalent to) the joint data of $q_n$ and $(w_{2k'})_{k'<q_n}$. In conclusion, $\mathbb{P}(i_n = 1 | i_0, \dots, i_{n-1}) = \alpha$ for all $n \in \mathbb{N}$, which characterizes the fact that $(i_n)_{n \sim \mathcal{B}^{\otimes \mathbb{N}}}$.

Note moreover that the data of $(i_n)_n$ is determined by the data of $(w_k)_k$ so it is independent of the joint data of $(u_n)_n$ and $(s_n)_n$. Since all three sequences are i.i.d. and by global independence, the sequence $(u_n, s_n, i_n)_{n \in \mathbb{N}}$ is also i.i.d. Moreover we can also check that for all $n \in \mathbb{N}$, we have $\gamma_n = s_n$ when $i_n = 1$ and $\gamma_n = u_n$ when $i_n = 0$, meaning that $\gamma_n$ is given the image of $(u_n, s_n, i_n)$ by a measurable function so $(\gamma_n)$ is i.i.d. and we only need to check that $\gamma_0 \sim \nu$. Let $A \subset \Gamma$ be measurable. Since $(i_0 = 1)$ and $(i_1 = 0)$ are disjoint events, we have:

$$ \begin{aligned} \mathbb{P}(\gamma_0 \in A) &= \mathbb{P}(\gamma_0 \in A \cap i_0 = 1) + \mathbb{P}(\gamma_0 \in A \cap i_0 = 0) \\ &= \mathbb{P}(s_0 \in A \cap i_0 = 1) + \mathbb{P}(u_0 \in A \cap i_0 = 0) \end{aligned} $$

So by independence:

$$ \begin{aligned} \mathbb{P}(\gamma_0 \in A) &= \mathbb{P}(s_0 \in A)\mathbb{P}(i_0 = 1) + \mathbb{P}(u_0 \in A)\mathbb{P}(i_0 = 0) \\ &= \alpha\eta(A) + (1-\alpha)\kappa(A) = \nu(A). \end{aligned} \quad \square $$

**Corollary 4.6.** Let $E$ be a Euclidean vector space and let $\nu$ be a probability distribution on $\text{End}(E)$. Let $\alpha \in (0, 1)$ and let $m \in \mathbb{N}$. Let $\tilde{v}_s$ be a probability distribution on $\text{End}(E)^m$ such that $\alpha \tilde{v}_s \le \nu^{\otimes m}$ and let $\tilde{\kappa} = \frac{1}{1-\alpha}(\nu^{\otimes m} - \alpha \tilde{v}_s)$. Let
$(u_k)_k \sim \tilde{\kappa}^{\otimes \mathbb{N}}$, let $(s_k)_k \sim \tilde{\nu}_s^{\otimes \mathbb{N}}$ and let $(w_{2k})_{k \in \mathbb{N}} \sim \mathcal{G}_{\alpha}^{\otimes \mathbb{N}}$ be globally independent random sequences defined on the same probability space. Let $w_{2k+1} = m$ for all $k \in \mathbb{N}$ and let $(\tilde{g}_n)$ be the random sequence defined as $\tilde{g}_{2k} = \tilde{u}_{2k}^w$ and $\tilde{g}_{2k+1} = \tilde{s}_{2k+1}^w$ for all $k \in \mathbb{N}$. Let $(\gamma_n)_{n \in \mathbb{N}} \sim \bigodot_{k=0}^{+\infty} \tilde{g}_k$. Then $\tilde{g}_k = \gamma_k^{mw}$ for all $k \in \mathbb{N}$ and $(\gamma_n)_n \sim \nu^{\otimes \mathbb{N}}$.

**Proof.** We apply Lemma 4.5 to $\nu^{\otimes m} = \alpha \tilde{\nu}_s + (1-\alpha) \tilde{\kappa}$. Note that $(\tilde{u}_k^m)_k \sim \tilde{\kappa}^{\otimes \mathbb{N}}$ and $(\tilde{s}_k^m)_k \sim \tilde{\nu}_s^{\otimes \mathbb{N}}$. Let $\tilde{g}_{2k} = \tilde{u}_{2k}^m$ and $\tilde{g}_{2k+1} = \tilde{s}_{2k+1}^m$ for all $k \in \mathbb{N}$. Then $\tilde{g}_k = \bigodot_k \tilde{g}_k$ for all $k$ and $\bigodot_{k=0}^{+\infty} \tilde{g}_k = (\tilde{\gamma}_k^m)_k$ because each letter of each $\tilde{g}_k$ is a word of length $m$. So Lemma 4.5 tells us that $(\tilde{\gamma}_k^m)_k \sim (\nu^{\otimes m})^{\otimes \mathbb{N}}$ and $(\tilde{\gamma}^m)_k^w = \tilde{g}_k$ for all $k \in \mathbb{N}$ and as a consequence $(\gamma_n)_n \sim \nu^{\otimes \mathbb{N}}$ and $\tilde{g}_k = \gamma_k^{mw}$ for all $k$. $\square$

We call the $\tilde{\kappa}^{\otimes \mathbb{N}}$-distributed sequence $u$ for "unknown" because we do not assume anything about $\tilde{\kappa}$ and we call the $\tilde{\nu}_s^{\otimes \mathbb{N}}$-distributed sequence $s$ for Schottky because in practice we need the product $\nu_s = \Pi_* \tilde{\nu}_s$ to be $\rho$-Schottky for a binary relation $\mathcal{A}$ and for a constant $\rho < 1/5$.

With the notation of Definition 4.1, the words $\tilde{g}_{2k}$'s defined in Lemma 4.5 have distribution law $\tilde{\kappa}^{\otimes \mathcal{G}_{\alpha}}$. Let us now give a fully detailed statement of the main result of this section. We give an abstract statement for topological semi-groups for further applications as mentioned in the introduction but in this article, we only use Theorem 4.7 with $\Gamma = \text{End}(E)$.

**Theorem 4.7 (Pivot extraction).** Let $\Gamma$ be a metric semi-group endowed with a measurable binary relation $\mathcal{A}$. Let $\alpha \in (0,1)$, let $\rho \in (0,1/5)$ and let $m \ge 1$. Let $\tilde{\nu}_s$ be a probability distribution on $\Gamma^m$ such that $\alpha \tilde{\nu}_s \le \nu^{\otimes m}$ and let $\tilde{\kappa} := \frac{1}{1-\alpha}(\nu^{\otimes m} - \alpha \tilde{\nu}_s)$. Let $\nu_s = \Pi_* \tilde{\nu}_s$ and assume that $\nu_s$ is $\rho$-Schottky for $\mathcal{A}$. Then there exist random sequences $(u_k)_{k \in \mathbb{N}}$, $(s_k)_{k \in \mathbb{N}}$, $(w_{2k})_{k \in \mathbb{N}} \sim \tilde{\kappa}^{\otimes \mathbb{N}} \otimes \tilde{\nu}_s^{\otimes \mathbb{N}} \otimes \mathcal{G}_{\alpha}^{\otimes \mathbb{N}}$ and two random sequences $(v_k)_{k \in \mathbb{N}}$ and $(p_k)_{k \in \mathbb{N}}$ all defined on the same probability space such that if we write $\tilde{w} := mw$ and $(\gamma_n)_{n \in \mathbb{N}} := \bigodot_{k=0}^{\infty} \tilde{u}_{2k}^{\tilde{w}} \tilde{s}_{2k+1}^{\tilde{w}}$ and $\tilde{v} := \tilde{w}^v$, $\hat{v}_{2k} := \tilde{v}_{4k} + \tilde{v}_{4k+1} + \tilde{v}_{4k+2}$ and $\hat{v}_{2k+1} = \tilde{v}_{4k+3}$ for all $k$ (or in compact notations $\hat{v} = \tilde{v}^{(3,1)^{\otimes \mathbb{N}}}$) and $\tilde{p} := \hat{v}^p$, then the following assertions hold:

(1) The data of $(v_k)_k$ is independent of the joint data of $(\tilde{u}_k)_k$ and $(w_k)_k$.

(2) For all $k \in \mathbb{N}$, we have $v_{4k+1} = v_{4k+2} = v_{4k+3} = 1$ and $(\frac{v_{4k}-1}{2})_{k \in \mathbb{N}} \sim \mathcal{G}_{2\rho}^{\otimes \mathbb{N}}$.

(3) For all $k \in \mathbb{N}$, we have $\gamma_{4k}^{\tilde{v}} \mathcal{A} \gamma_{4k+1}^{\tilde{v}} \mathcal{A} \gamma_{4k+2}^{\tilde{v}}$.

(4) We have $(\tilde{\gamma}_{4k+3}^{\tilde{v}})_{k \in \mathbb{N}} \sim \tilde{\nu}_s^{\otimes \mathbb{N}}$ and the sequence $((\tilde{\gamma}_{4k}^{\tilde{v}}, \tilde{\gamma}_{4k+1}^{\tilde{v}}, \tilde{\gamma}_{4k+2}^{\tilde{v}}))_{k \in \mathbb{N}}$ is i.i.d. and independent of $(\tilde{\gamma}_{4k+3}^{\tilde{v}})_{k \in \mathbb{N}}$.

(5) The data of $(p_k)_k$ is independent of the joint data of $(\tilde{s}_k)_k$ and $(w_k)_k$ and $(v_k)_k$ and $(\tilde{\gamma}_k^{\tilde{v}})_{k \in \{0,1,2\} + 4\mathbb{N}}$.

(6) Each $p_k$ is a positive odd integer, $p_{2k+1} = 1$ for all $k$ and $(p_{2k+2})_k$ is i.i.d. and independent of $p_0$.
(7) The distribution laws of $p_0$ and $p_2$ only depend on $\rho$ and have a finite exponential moment.

(8) The sequence $(\tilde{\gamma}_{2k+2}^{\tilde{p}})_{k \ge 0}$ is i.i.d. and independent of $\tilde{\gamma}_0^{\tilde{p}}$.

(9) For all $k \in \mathbb{N}$, we have $\gamma_{2k}^{\tilde{p}} A \gamma_{2k+1}^{\tilde{p}}$ almost surely.

(10) Almost surely and for all $k \in \mathbb{N}$, we have $\gamma_{2k+1}^{\tilde{p}} A \gamma_{p_{2k+2}}^{\tilde{p}}$ and there is a family of odd integers $1 = c_1^k < c_2^k < \dots < c_{j_k}^k = p_{2k+2}$ such that for all $1 \le i < j_k$, we have:

$$ (58) \qquad \gamma_{\tilde{p}_{2k+2}}^{\tilde{v}} \cdots \gamma_{\tilde{p}_{2k+2}+c_i^k-1}^{\tilde{v}} A \gamma_{\tilde{p}_{2k+2}+c_i^k}^{\tilde{v}} \cdots \gamma_{\tilde{p}_{2k+2}+c_{i+1}^k-1}^{\tilde{v}} $$

(11) For all $k \in \mathbb{N}$, the conditional distribution of $\tilde{\gamma}_{2k+1}^{\tilde{p}}$ with respect to the joint data of $(\tilde{\gamma}_{k'}^{\tilde{p}})_{k' \ne 2k+1}$ and $(\tilde{u}_k)_k$ and $(w_k)_k$ and $(v_k)_k$ is the normalized restriction of $\tilde{v}_s$ to the measurable set:

$$ (59) \qquad C_k := \Pi^{-1} \left\{ \gamma \in \Gamma \mid \gamma_{2k}^{\tilde{p}} A \gamma_{2k+1}^{\tilde{p}} A \gamma_{\tilde{p}_{2k+2}}^{\tilde{v}} \right\}. $$

Let us quickly break down what Theorem 4.7 says. Note that writing $(\gamma_n)_{n \in \mathbb{N}} := \bigodot_{k=0}^{\infty} \tilde{u}_{2k}^w \tilde{s}_{2k+1}^w$ is simply a compact notation to say that $(\gamma_n)_{n \in \mathbb{N}}$ is so that $\tilde{\gamma}_{2k}^{mw} = \tilde{u}_{2k}^w$ and $\tilde{\gamma}_{2k+1}^{mw} = \tilde{s}_{2k+1}^w$ for all $k \in \mathbb{N}$. The notation $\tilde{v} := \tilde{w}^v$ simply means that $\tilde{v}_k = m(w_{\tilde{v}_k} + \dots + w_{\tilde{v}_{k+1}-1})$.

Let us point out a few facts that follow directly from Theorem 4.7. For all $k \in \mathbb{N}$, we have $\tilde{\gamma}_{2k}^{\tilde{v}} = \tilde{\gamma}_{4k}^{\tilde{v}} \odot \tilde{\gamma}_{4k+1}^{\tilde{v}} \odot \tilde{\gamma}_{4k+2}^{\tilde{v}}$ and $\tilde{\gamma}_{2k+1}^{\tilde{v}} = \tilde{\gamma}_{4k+3}^{\tilde{v}}$, by definition of $\tilde{v}$ and by the first part of (2). So (4) implies that the sequence $(\tilde{\gamma}_{2k}^{\tilde{v}})_k$ is i.i.d. and independent of $(\tilde{\gamma}_{2k+1}^{\tilde{v}})_{k \in \mathbb{N}} \sim \tilde{v}_s^{\otimes \mathbb{N}}$.

It is also important to note that the set $C_k$ defined in (11) has $\tilde{v}_s$ measure at least $1 - 2\rho$ by the Schottky property of $v_s$. So the product of the normalised restriction of $\tilde{v}_s$ to $C_k$ is $\frac{\rho}{1-2\rho}$-Schottky.

On point that is really important to note is that the joint data of the random sequences $(p_k)_k$, $(v_k)_k$ and $(w_k)_k$ is independent of the random sequence $(\tilde{u}_k)_k$. If we moreover assume that $\tilde{v}_s$ has compact support, this fact allows us to bound the tail of the conditional distribution of $\gamma_n$ for a given $n \ge 0$ with respect to the joint data of $(p_k)_k$, $(v_k)_k$ and $(w_k)_k$ by the tail distribution of the marginals of $\tilde{\kappa}$. For example, if we assume that $\mathbb{E}(N(\gamma_0))$ is finite for a given continuous and sub-additive map $N$ and write $B := \max\{N(s_k) \mid 0 \le k < m, (s_0, \dots, s_{m-1}) \in \text{supp}(\tilde{v}_s)\}$, then we have a uniform bound over $\mathbb{E}(N(\gamma_n) \mid (p_k)_k, (v_k)_k, (w_k)_k) \le C$ with $C = \frac{1}{1-\alpha}\mathbb{E}(N(\gamma_0)) + B$, which in turn gives a uniform bound over $\mathbb{E}(N(\gamma_k^{\tilde{p}})) \le C\mathbb{E}(\tilde{p}_k)$ for all $k \in \mathbb{N}$. This argument is the main ingredient in the proof of Theorem 1.13.

Last but not least, Theorem 4.7 is the key in the proof of Theorem 1.6. Let us prove Theorem 1.6 using Theorem 4.7. More precisely, we will show that the
random sequence $(\check{p}_k)$ from Theorem 4.7 satisfies the conclusion of Theorem 1.6 when $\tilde{\nu}_s$ is as in Corollary 3.17.

*Proof of Theorem 1.6.* Let $E$ be a Euclidean, Hermitian or ultra-metric over a local field $\mathbb{K}$. Let $\Gamma = \text{End}(E)$ or $\Gamma = \text{GL}(E)$ and let $\nu$ be a probability measure on $\Gamma$. Let $(\gamma_n) \sim \nu^{\otimes\mathbb{N}}$. Let $\delta : \varepsilon \mapsto \varepsilon^6/48$. Let $m$, let $0 < \alpha, \varepsilon \le 1/2$, let $m \ge 1$ and let $\tilde{\nu}_s$ be as in Corollary 3.17.

Let $\mathbb{A} = \mathbb{A}^{\varepsilon}$, let $\rho = 1/6$ and let $\check{p}$ be as in Theorem 4.7. Let $\mu$ be the joint distribution of $((\gamma_n)_{n\ge 0}, (\check{p}_k)_{k\ge 0})$.

Point (1) in Theorem 1.6 states that $(\gamma_n) \sim \nu^{\otimes\mathbb{N}}$, this is a consequence of Lemma 4.5.

Point (2) states that the sequence $(\gamma_{2k}^{\check{p}})_{k\ge 1}$ is i.i.d. and independent of $\gamma_0^{\check{p}}$. This is exactly what point (8) says in Theorem 4.7.

Let $K \subset \Gamma$ be the smallest compact set such that $\nu_s(K^m) = 1$. Point (3), which states that $\tilde{\gamma}_{2k+1}^{\check{p}} \in K^m$ and $\sigma(\gamma_{2k+1}^{\check{p}}) \le \varepsilon^6/48$ almost surely and for all $k \ge 0$ is then a consequence of point (11) in theorem 4.7, which implies that the distribution of $\tilde{\gamma}_{2k+1}^{\check{p}}$ is absolutely continuous with respect to $\tilde{\nu}_s$ for all $k \ge 0$.

Point (4) states that the conditional distribution of $\tilde{\gamma}_{2k+1}^{\check{p}}$ with respect to $(\tilde{\gamma}_{k'}^{\check{p}})_{k' \ne k+1}$ only depends on $\tilde{\gamma}_{2k}^{\check{p}}$ and $\tilde{\gamma}_{2k+2}^{\check{p}}$. It is again a consequence of (11) in Theorem 4.7. Indeed, even though $\gamma_{\check{p}_{2k+2}}^{\hat{v}}$ does not explicitly depend of $\tilde{\gamma}_{2k+2}^{\check{p}}$, we derive from points (4), (5) and (6) in Theorem 4.7 that the data of $(p_k)_{k\ge 0}$ is independent of the data of $(\tilde{\gamma}_k^{\hat{v}})_{k\ge 0}$ and that $(p_{2k})_{k\ge 2}$ is i.i.d. and independent of $p_0$. From that we deduce that the sequence $(\gamma_{\check{p}_{2k+2}}^{\hat{v}}, \tilde{\gamma}_{2k+2}^{\check{p}})_{k\ge 0}$ is i.i.d. and independent of $\tilde{\gamma}_0^{\check{p}}$. Hence, by (11), applied to all $k' \ne k$ and by reciprocity of the relative independence relation, the conditional distribution of $\gamma_{\check{p}_{2k+2}}^{\hat{v}}$ with respect to $(\tilde{\gamma}_{k'}^{\check{p}})_{k' \ne 2k+1}$ only depends on $\tilde{\gamma}_{2k+2}^{\check{p}}$. Then by (11) applied to $k$, the conditional distribution of $\gamma_{2k+1}^{\check{p}}$ with respect to $(\tilde{\gamma}_{k'}^{\check{p}})_{k' \ne 2k+1}$ only depends on $\tilde{\gamma}_{2k}^{\check{p}}$ and $\tilde{\gamma}_{2k+2}^{\check{p}}$.

Point (5) states that the conditional distribution of $\gamma_{2k+1}^{\check{p}}$ with respect to $(\tilde{\gamma}_{k'}^{\check{p}})_{k' \ne 2k+1}$ is almost surely $\frac{1}{4}$-Schottky for $\mathbb{A}$. This comes from (11) again. By the Schottky property, we have $\tilde{\nu}_s(C_k) \ge 4/6$. Therefore, the product of the normalized restriction of $\tilde{\nu}_s$ to $C_k$, is bounded above by $6\nu_s/4$, which is $\frac{1}{4}$-Schottky.

Point (6) in Theorem 1.6 states that each $\hat{p}_k$ has a finite exponential moment. Points (6) and (7) in Theorem 4.7 imply that the $p_k$'s have a bounded exponential moment. Point (2) implies that the $v_k$'s have a bounded exponential moment and the $w_k$'s have a bounded exponential moment by definition. By (1), $(v_k)_k$ is independent of $(w_k)_k$ so the $\hat{v}_k$'s have a bounded exponential moment
by Lemma 4.4. By (5) $(p_k)_k$ is independent of $(\hat{v}_k)_k$ so the $\check{p}_k$'s have a bounded exponential moment by Lemma 4.4 again.

Point (7) states that for all $A \subset \Gamma \setminus K$, we have for all $n$ and almost surely $\mathbb{P}(\gamma_n \in A | (p_k)_{k \in \mathbb{N}}) \le 2\nu(A)$. First note that data of $(w_k)_k$ determines whether $\gamma_n = u_k$ or $\gamma_n = s_k$ for all $n$. Hence, for all $n$, we have almost surely $\mathbb{P}(\gamma_n \in A | (w_k)_{k \in \mathbb{N}}) = \mathbb{1}(s_n = \gamma_n)\mathbb{P}(s_n \in A) \le \nu(A)/(1-\alpha) \le 2\nu(A)$. On the set $(\gamma_n = u_k)$, we have $\mathbb{P}(\gamma_n \in A | (w_k)_{k \in \mathbb{N}}) = 0$ so $\mathbb{P}(\gamma_n \in A | (w_k)_{k \in \mathbb{N}}, (p_k)_{k \in \mathbb{N}}) = 0$. On the set $(\gamma_n = s_k)$, we have $\mathbb{P}(\gamma_n \in A | (w_k)_{k \in \mathbb{N}}, (p_k)_{k \in \mathbb{N}}) = \mathbb{P}(s_n \in A | (w_k)_{k \in \mathbb{N}}, (p_k)_{k \in \mathbb{N}}) = \mathbb{P}(s_n \in A)$ by the independence assumption (5).

Let us now prove point (8) using Lemmas (2.11) and 2.16. These points are purely geometric properties on the support of $\nu$. The proof relies on points (9) and (10) in Theorem 4.7, which are also almost surely properties (though not properties of the support because we do not assume $A$ to be closed in general). Point (8) states that for all $i < j \le k$, for all $f\mathbb{A}^\varepsilon\gamma_i^{\check{p}}$ and for all $\gamma_k^{\check{p}}\mathbb{A}^\varepsilon h$, we have $f\gamma_i^{\check{p}} \cdot \gamma_{j-1}^{\check{p}}\mathbb{A}^{\frac{\varepsilon}{2}}\gamma_j^{\check{p}} \cdot \gamma_k^{\check{p}}h$. We will prove it by induction but first let us show how Lemma 2.16 applies.

By (3) and by Lemma 2.11, we have $\sigma(\gamma_{2k}^{\hat{v}}) \ge 4\varepsilon^{-4}\sigma(\gamma_{4k+1}^{\check{v}}) = \varepsilon^2/12$ for all $k \ge 0$. Let $k \ge 0$. By point (10) in Theorem 4.7, the sequence $(g_{-1}, \dots, g_{2i_j})$ defined as $g_{-1} = \gamma_{2k+1}^{\check{p}}$, $g_0 = \gamma_{\frac{\hat{v}}{\mathbb{P}_{2k+2}}, 2k+2}^{\check{p}}$, and for all $1 \le i \le j_k$, $g_{2i+1} = \gamma_{\frac{\hat{v}}{\mathbb{P}_{2k+2}+c_i}, 2k+2}^{\check{p}}$ and $g_{2i+2} = \gamma_{\frac{\hat{v}}{\mathbb{P}_{2k+2}+c_i+1}, 2k+2}^{\check{p}} \cdots \gamma_{\frac{\hat{v}}{\mathbb{P}_{2k+2}+c_{i+1}-1}, 2k+2}^{\check{p}}$ satisfies the hypotheses of Lemma 2.16. The alignment condition comes from (58) and the condition on $\sigma(g_{2i+1})$ comes from the fact that $c_i$ is odd so $\gamma_{\frac{\hat{v}}{\mathbb{P}_{2k+2}+c_i}, 2k+2}^{\check{p}}$ is almost surely in the support of $\nu_s$, on which $\sigma \le \varepsilon^6/48$. Therefore, we have $\gamma_{2k+1}^{\check{p}}\mathbb{A}^{\frac{\varepsilon}{2}}\gamma_{2k+2}^{\check{p}}$ and $\sigma(\gamma_{2k+2}^{\check{p}}) \le \varepsilon^2/8$ almost surely and for all $k$. Moreover, for all $g\mathbb{A}^{\frac{\varepsilon}{2}}\gamma_{2k+1}^{\check{p}}$ and $\gamma_{2k+2}^{\check{p}}\mathbb{A}^{\frac{\varepsilon}{2}}h$, we have $\gamma_{2k+1}^{\check{p}}\mathbb{A}^{\frac{\varepsilon}{2}}\gamma_{2k+2}^{\check{p}}h$.

Let $k \ge 1$ and let $\gamma_k^{\check{p}}\mathbb{A}^{\frac{\varepsilon}{2}}h$. We show that then $\gamma_{k-1}^{\check{p}}\mathbb{A}^{\frac{\varepsilon}{2}}\gamma_k^{\check{p}}h$. When $k$ is odd by point (9) in Theorem 4.7, we have $\gamma_{k-1}^{\check{p}}\mathbb{A}^\varepsilon\gamma_k^{\check{p}}$ and $\sigma(\gamma_k^{\check{p}}) \le \varepsilon^6/48 \le \varepsilon^2/8$. By Lemma 2.9, we have $\gamma_{k-1}^{\check{p}}\mathbb{A}^{\frac{\varepsilon}{2}}\gamma_k^{\check{p}}h$. When $k$ is even, by Lemma 2.16, we have $\gamma_{k-1}^{\check{p}}\mathbb{A}^{\frac{\varepsilon}{2}}\gamma_k^{\check{p}}h$. By induction, we have $\gamma_j^{\check{p}}\mathbb{A}^{\frac{\varepsilon}{2}}\gamma_{j+1}^{\check{p}}\gamma_k^{\check{p}}h$ for all $j \le k$.

Let $i \ge 0$ and let $f\mathbb{A}^{\frac{\varepsilon}{2}}\gamma_i^{\check{p}}$. We show that then $f\gamma_{i+1}^{\check{p}}\mathbb{A}^{\frac{\varepsilon}{2}}\gamma_{i+1}^{\check{p}}$ When $i$ is odd, by Lemma 2.16, we have $f\gamma_i^{\check{p}}\mathbb{A}^{\frac{\varepsilon}{2}}\gamma_{i+1}^{\check{p}}$. When $i$ is even, by Lemma 2.9, we have $f\gamma_i^{\check{p}}\mathbb{A}^{\frac{\varepsilon}{2}}\gamma_{i+1}^{\check{p}}$. By induction, we have $f\gamma_i^{\check{p}} \cdots \gamma_{j-1}^{\check{p}}\mathbb{A}^{\frac{\varepsilon}{2}}\gamma_j^{\check{p}}$ for all $j \ge i$.

Then for all $i < j < k$, we have $f\gamma_i^{\check{p}} \cdots \gamma_{j-2}^{\check{p}}\mathbb{A}^{\frac{\varepsilon}{2}}\gamma_{j-1}^{\check{p}}$ and $\gamma_j^{\check{p}}\mathbb{A}^{\frac{\varepsilon}{2}}\gamma_{j+1}^{\check{p}} \cdots \gamma_{k-1}^{\check{p}}h$. Then, by Lemma 2.14 when $j$ is even and Lemma 2.16 when $j$ is odd, we have $f\gamma_i^{\check{p}} \cdot \gamma_{j-1}^{\check{p}}\mathbb{A}^{\frac{\varepsilon}{2}}\gamma_j^{\check{p}} \cdot \gamma_k^{\check{p}}h$.

In the case $i = j$, we have $f\mathbb{A}^\varepsilon\gamma_j^{\check{p}}\mathbb{A}^{\frac{\varepsilon}{2}}\gamma_{j+1}^{\check{p}} \cdots \gamma_{k-1}^{\check{p}}h$. So by Lemma 2.9, we have $f\mathbb{A}^\varepsilon\gamma_j^{\check{p}} \cdots \gamma_{k-1}^{\check{p}}h$. The same argument works when $j = k$, which concludes the proof of point (8). $\square$
Now we need to prove Theorem 4.7 for the above to make sense.

4.2. **First step of the proof of Theorem 4.7.** In this paragraph, we will use the notations of Theorem 4.7. The first step of the proof is the construction of the sequence $(v_k)_{k \in \mathbb{N}}$ that satisfies points (1) to (4).

The naive idea is to construct it using a stopping time. We know that the $(\gamma_{2k+1}^{mw})$ are i.i.d. $\rho$-Schottky and their joint data is independent of $(\gamma_{2k}^{mw})_{k \in \mathbb{N}}$ so for all $n \in \mathbb{N}$, we have:

$$ (60) \qquad \mathbb{P} \left( \gamma_0^{mw} \cdots \gamma_{2n}^{mw} \mathbb{A} \gamma_{2n+1}^{mw} \mathbb{A} \gamma_{2n+2}^{mw} \mid \gamma_0^{mw}, \ldots, \gamma_{2n}^{mw} \right) \geq 1 - 2\rho. $$

Let $n_0$ to be the smallest integer such that $\gamma_0^{mw} \cdots \gamma_{2n_0}^{mw} \mathbb{A} \gamma_{2n_0+1}^{mw} \mathbb{A} \gamma_{2n_0+2}^{mw}$. We could define $v_0 := 2n_0 + 1$, that way $\gamma_0^{\tilde{v}} = \gamma_0^{mw} \cdots \gamma_{2n_0}^{mw}$ and $\gamma_1^{\tilde{v}} = \gamma_{2n_0+1}^{mw}$ and $\gamma_2^{\tilde{v}} = \gamma_{2n_0+1}^{mw}$ so we have (3). Then $v_0$ would have a finite exponential moment by (60) and by a stopping time argument, the data of $(\gamma_{2k+v_0+2}^{mw})_{k \in \mathbb{N}}$ would be independent of $(\tilde{\gamma}_0^{mw}, \ldots, \tilde{\gamma}_{v_0+1}^{mw})$ so $\tilde{\gamma}_3^{\tilde{v}} = \tilde{\gamma}_{2k+v_0+2}^{mw}$ has law $\tilde{\nu}_s$, and is independent of the data of $(\tilde{\gamma}_0^{\tilde{v}}, \tilde{\gamma}_1^{\tilde{v}}, \tilde{\gamma}_2^{\tilde{v}})$ and the sequence $(\gamma_{2k\tilde{\nu}_4}^{mw})_{k \in \mathbb{N}}$ is independent of $(\tilde{\gamma}_0^{\tilde{v}}, \ldots, \tilde{\gamma}_3^{\tilde{v}})$ and has the same distribution as $(\gamma_{2k}^{mw})_{k \in \mathbb{N}}$ so we may the reiterate the process.

The advantage of this naive approach is that the sequence $(v_k)_k$ can be computed explicitly from the data of $(\gamma_{2k}^{mw})_{k \in \mathbb{N}}$. The issue is that the data of $(v_k)_{k \in \mathbb{N}}$ is not independent of the data of $(\tilde{\gamma}_{2k}^{mw})_{k \in \mathbb{N}}$ so we would not have the same control over the tail of the distribution of $\gamma_0^{\tilde{v}}$.

Luckily, this correlation between $(v_k)_{k \in \mathbb{N}}$ and $(\tilde{\gamma}_{2k}^{mw})_{k \in \mathbb{N}}$ is easy to get rid of. A quick observation is that if we had:

$$ (61) \qquad \mathbb{P} \left( \gamma_0^{mw} \cdots \gamma_{2n}^{mw} \mathbb{A} \gamma_{2n+1}^{mw} \mathbb{A} \gamma_{2n+2}^{mw} \mid (\gamma_k^{mw})_{k \neq 2n+1} \right) = 1 - 2\rho, $$

almost surely and for all $n$, then the conditional distribution of $n_0$ with respect to the data of $(\tilde{\gamma}_{2k}^{mw})_{k \in \mathbb{N}}$ would be almost surely equal to $\mathcal{G}_{2\rho}$, giving us the independence of $v_0$ with the data of $(\tilde{\gamma}_{2k}^{mw})_{k \in \mathbb{N}}$. However (61) has no reason to be satisfied as the Schottky property only gives us an inequality. Let $A_n := \gamma_0^{mw} \cdots \gamma_{2n}^{mw} \mathbb{A} \gamma_{2n+1}^{mw} \mathbb{A} \gamma_{2n+2}^{mw}$. A sure-fire way to make (60) into an equality is to artificially remove mass from $A_n$. Consider a family $(\tau_n)_{n \in \mathbb{N}} \in [0, 1]^\mathbb{N}$ uniformly distributed an independent of the joint data of all the formerly defined random variables and then replace the event $A_n$ with:

$$ (62) \qquad A'_n := \left( \tau_n < \frac{(1-2\rho)1_{A_n}}{\mathbb{P}(A_n \mid (\gamma_k^{mw})_{k \neq 2n+1})} \right). $$

We can then check that $A'_n \subset A_n$ and $\mathbb{P}(A_n \mid (\gamma_k^{mw})_{k \neq 2n+1}) = 1 - 2\rho$ almost surely and for all $n \in \mathbb{N}$. In some sense, asking moreover that $\tau_n < \frac{(1-2\rho)}{\mathbb{P}(A_n \mid (\gamma_k^{mw})_{k \neq 2n+1})}$
when $A_n$ holds is like inflicting a penalty to $A_n$ to lower its probability of success and make the $(A'_j)_j$ i.i.d. and independent of the data of $(\tilde{\gamma}_{2k}^{mw})_k$.

The notion of independence is central in the following discussion. We remind that we say that the joint data of a first family of random variables is independent of the joint data of a second family of random variables if for all measurable event A defined in term of the first family of random variables and B defined in term of the second family, we have $P(A \cap B) = P(A)P(B)$. This is a symmetric notion which is equivalent to saying that the conditional distribution of the first family of random variables with respect to the $\sigma$-algebra generated by the second one is almost surely constant. Let us now give a proof of the first step of 4.7.

**Lemma 4.8.** Let $\Gamma$ be a metric semi-group endowed with a measurable binary relation $\Lambda$. Let $\alpha \in (0,1)$, let $\rho \in (0,1/5)$ and let $m \in \mathbb{N}$. Let $\tilde{\nu}_s$ be a probability distribution on $\Gamma^m$ such that $\alpha\tilde{\nu}_s \le \nu^{\otimes m}$ and let $\tilde{\kappa} = \frac{1}{1-\alpha}(\nu^{\otimes m} - \alpha\tilde{\nu}_s)$. Let $\nu_s = \Pi_*\tilde{\nu}_s$ and assume that $\nu_s$ is $\rho$-Schottky for $\Lambda$. Then there exist random sequences $(u_k)_k, (s_k)_k, (w_{2k})_k \sim \tilde{\kappa}^{\otimes \mathbb{N}} \otimes \tilde{\nu}_s^{\otimes \mathbb{N}} \otimes \mathcal{G}_{\alpha}^{\otimes \mathbb{N}}$ and two random sequences $(v_k)_k$ and $(p_k)_k$ such that if we write $(\gamma_n)_{n \in \mathbb{N}} := \bigodot_{k=0}^{\infty} \tilde{u}_{2k}^{w} \tilde{s}_{2k+1}^{w}$ and $\tilde{w} := mw$ and $\tilde{v} := \tilde{w}^v$, then the following assertions hold:

(1) The data of $(v_k)_k$ is independent of the joint data of $(\tilde{u}_k)_k$ and $(w_k)_k$.

(2) For all $k \in \mathbb{N}$, we have $v_{4k+1} = v_{4k+2} = v_{4k+3} = 1$ and $(\frac{v_{4k}-1}{2})_{k \in \mathbb{N}} \sim \mathcal{G}_{2\rho}^{\otimes \mathbb{N}}$.

(3) For all $k \in \mathbb{N}$, we have $\gamma_{4k}^{\tilde{v}} A \gamma_{4k+1}^{\tilde{v}} A \gamma_{4k+2}^{\tilde{v}}$.

(4) We have $(\tilde{\gamma}_{4k+3})_{k \in \mathbb{N}} \sim \tilde{\nu}_s^{\otimes \mathbb{N}}$ and the sequence $((\tilde{\gamma}_{4k}^{\tilde{v}}, \tilde{\gamma}_{4k+1}^{\tilde{v}}, \tilde{\gamma}_{4k+2}^{\tilde{v}}))_{k \in \mathbb{N}}$ is i.i.d. and independent of $(\tilde{\gamma}_{4k+3})_{k \in \mathbb{N}}$.

**Proof.** Let $\Omega = \Gamma^{\mathbb{N}} \times \Gamma^{\mathbb{N}} \times \mathbb{N}^{\mathbb{N}} \times [0,1]^{\mathbb{N}}$. Let $\mu := \tilde{\kappa}^{\otimes \mathbb{N}} \otimes \tilde{\nu}_s^{\otimes \mathbb{N}} \otimes (\mathcal{G}_{\alpha} \otimes \delta_1)^{\otimes \mathbb{N}} \otimes \mathcal{U}^{\otimes \mathbb{N}}$. To all $j \in \mathbb{N}$, we associate the $\sigma$-algebra $\mathcal{C}_j$, which is the pull-back of the Borel $\sigma$-algebra by the map:

$$ (u_n)_n, (s_n)_n, (w_k)_k, (\tau_k)_k \mapsto (u_n)_{n<m\bar{w}_{2j}}, (s_n)_{n<m\bar{w}_{2j}}, (w_k)_{k<2j}, (\tau_k)_{k<j} $$

Let $T: \Omega \to \Omega$ be the map:

$$ (u_n)_n, (s_n)_n, (w_k)_k, (\tau_k)_k \mapsto (u_{n+m\bar{w}_2})_n, (s_{n+m\bar{w}_2})_n, (w_{k+2})_k, (\tau_{k+1})_k. $$

First we check that $T$ is measure preserving and that for all $j \in \mathbb{N}$, the push-forward of the Borel $\sigma$-algebra by $T^j$ is independent of $\mathcal{C}_j$. Let:

$$ Q: (u_n)_n, (s_n)_n, (w_k)_k, (\tau_k)_k \mapsto ((\tilde{u}_{2k}^{mw} \odot \tilde{u}_{2k+1}^{mw}, \tilde{s}_{2k}^{mw} \odot \tilde{s}_{2k+1}^{mw}, w_{2k}, w_{2k+1}, \tau_k))_{k \in \mathbb{N}}. $$
Note that $Q$ is one to one, that $T$ is conjugated to a simple shift by $Q$, and $(C_j)_j$ is the pull-back of the cylinder filtration by $Q$. Moreover

$$Q_*\mu = \left( \sum_{k \in \mathbb{N}} \frac{\alpha^k}{1-\alpha} \tilde{\kappa}^{\odot k+1} \otimes \tilde{\nu}_s^{\odot k+1} \otimes \delta_k \otimes \delta_1 \otimes \mathcal{U} \right)^{\otimes \mathbb{N}}$$

So $Q_*\mu$ is shift-invariant. Let $j \in \mathbb{N}$. Note that:

$$T^j : (u_n)_n, (s_n)_n, (w_k)_k, (\tau_k)_k \mapsto (u_{n+m\bar{w}_{2j}})_n, (s_{n+m\bar{w}_{2j}})_n, (w_{k+2j})_k, (\tau_{k+j})_k.$$

Let $(u_n)_n, (s_n)_n, (w_k)_k, (\tau_k)_k \sim \mu$. We need to show that the joint conditional distribution with respect to $C_j$ of $(u_{n+m\bar{w}_{2j}})_n, (s_{n+m\bar{w}_{2j}})_n, (w_{k+2j})_k, (\tau_{k+j})_k$ is almost surely equal to $\mu$.

It is a well known fact that since $(\tau_k)_k$ is i.i.d. the conditional distribution with respect to $(\tau_k)_{k<j}$ of $(\tau_{k+j})_k$ is almost surely equal to $\mathcal{U}^{\otimes \mathbb{N}}$. Since $\tilde{\kappa}$ and $\tilde{\nu}_s$ are supported on $\Gamma^m$, the joint conditional distribution of $(u_{n+mi})_n, (s_{n+mi})_n$ with respect to the joint data of $(u_n)_{n \le mi}, (s_n)_{n \le m_i}$ is almost surely equal to $\tilde{\kappa}^{\odot \mathbb{N}} \otimes \tilde{\nu}_s^{\odot \mathbb{N}}$ for all $i \in \mathbb{N}$. As a consequence, the joint conditional distribution with respect to the joint data of $(w_k)_k$ and $(u_n)_{n<m\bar{w}_{2j}}, (s_n)_{n<m\bar{w}_{2j}}$ of the random sequences $(u_{n+m\bar{w}_{2j}})_n, (s_{n+m\bar{w}_{2j}})_n$ is almost surely equal to $\tilde{\kappa}^{\odot \mathbb{N}} \otimes \tilde{\nu}_s^{\odot \mathbb{N}}$. Moreover the conditional distribution of $(w_{k+2j})_k$ with respect to $(w_k)_{k<2j}$ is almost surely equal to $(\mathcal{G}_\alpha \otimes \delta_1)^{\odot \mathbb{N}}$ by the same argument.

By independence we can add that the conditional distribution of $(w_{k+2j})_k$ with respect to the joint data of $(u_n)_{n<m\bar{w}_{2j}}, (s_n)_{n<m\bar{w}_{2j}}$ and $(w_k)_{k<2j}$ is almost surely equal to $(\mathcal{G}_\alpha \otimes \delta_1)^{\odot \mathbb{N}}$. Hence, the joint conditional distribution of $(u_{n+m\bar{w}_2})_n, (s_{n+m\bar{w}_2})_n$ and $(w_{k+2j})_k$ with respect to the joint data of $(u_n)_{n<m\bar{w}_{2j}}, (s_n)_{n<m\bar{w}_{2j}}$ and $(w_k)_{k<2j}$ is almost-surely equal to $\tilde{\kappa}^{\odot \mathbb{N}} \otimes \tilde{\nu}_s^{\odot \mathbb{N}} \otimes (\mathcal{G}_\alpha \otimes \delta_1)^{\odot \mathbb{N}}$. To conclude note that the data of $(u_{n+m\bar{w}_2})_n, (s_{n+m\bar{w}_2})_n, (w_{k+2j})_k), (u_n)_{n<m\bar{w}_{2j}}, (s_n)_{n<m\bar{w}_{2j}}, (w_k)_{k<2j}, (\tau_k)_{k<j}$ and $(\tau_{k+j})_k$ are globally independent so the joint data of $(u_{n+m\bar{w}_{2j}})_n, (s_{n+m\bar{w}_{2j}})_n, (w_{k+2j})_k$ and $(\tau_{k+j})_k$ is independent of the joint data of $(u_n)_{n<m\bar{w}_{2j}}, (s_n)_{n<m\bar{w}_{2j}}, (w_k)_{k<2j}$ and $(\tau_k)_{k<j}$ and its distribution is the product $\tilde{\kappa}^{\odot \mathbb{N}} \otimes \tilde{\nu}_s^{\odot \mathbb{N}} \otimes (\mathcal{G}_\alpha \otimes \delta_1)^{\odot \mathbb{N}} \otimes \mathcal{U}^{\otimes \mathbb{N}} = \mu$. This proves the claim.

Let $\omega := (u_n)_n, (s_n)_n, (w_k)_k, (\tau_k)_k \sim \mu$ and let $(\gamma_n)_{n \in \mathbb{N}} := \bigoplus_{k=0}^{\infty} \tilde{u}_{2k}^w \tilde{s}_{2k+1}^w$. We construct the sequence $(v_k)_k$ by induction. For all $j \in \mathbb{N}$, we write:

$$A_j := (\gamma_0^{\tilde{w}} \cdots \gamma_{2j}^{\tilde{w}} \mathcal{A} \gamma_{2j+1}^{\tilde{w}} \mathcal{A} \gamma_{2j+2}^{\tilde{w}}) = (u_0^{\tilde{w}} s_1^{\tilde{w}} \cdots u_{2j}^{\tilde{w}} \mathcal{A} s_{2j+1}^{\tilde{w}} \mathcal{A} u_{2j+2}^{\tilde{w}})$$

$$A'_j := \left( \tau_j < \frac{(1-2\rho)1_{A_j}}{\mathbb{P}(A_j | (u_n)_n, (\tilde{s}_k^w)_{k \neq 2j+1}, (w_k)_k, (\tau_k)_{k \neq 2j+1})} \right)$$
Then we define $j_0 := \min\{j \in \mathbb{N} | A'_j\}$, which is a measurable function $j_0 : \Omega \to \mathbb{N}$ and for all $k \in \mathbb{N}$, we write $j_k := j_0 \circ T^{\bar{v}_{4k}/2}$ and $v_{4k} = 2j_k + 1$ and $v_{4k+1} = v_{4k+2} = v_{4k+3} = 1$. We can easily check that $A'_j \subset A_j$ and by definition of $j_0$, $A_{j_0}$ holds almost surely for all $j \in \mathbb{N}$. Hence $T^{-\bar{v}_{4k}/2}(A_{j_k})$ holds almost surely and for all $k$ and by definition:

$$ 
\begin{align*}
T^{-\bar{v}_{4k}/2}(A_{j_k}) &= (\gamma_{\bar{v}_{4k}}^{\bar{w}} \cdots \gamma_{\bar{v}_{4k+2j_k}}^{\bar{w}} A \gamma_{\bar{v}_{4k+2j_k+1}}^{\bar{w}} A \gamma_{\bar{v}_{4k+2j_k+2}}^{\bar{w}}) \\
&= (\gamma_{\bar{v}_{4k}}^{\bar{w}} \cdots \gamma_{\bar{v}_{4k}+v_{4k}-1}^{\bar{w}} A \gamma_{\bar{v}_{4k+1}}^{\bar{w}} A \gamma_{\bar{v}_{4k+2}}^{\bar{w}}) \\
&= (\gamma_{4k}^{\bar{v}} A \gamma_{4k+1}^{\bar{v}} A \gamma_{4k+2}^{\bar{v}})
\end{align*}
$$

so (3) holds.

Let us now prove (1) (2) and (4). For this proof, we introduce useful notations. For all $j \in \mathbb{N}$, we write $\mathcal{F}_j$ for the $\sigma$-algebra generated by the joint data of $(u_n)_n$, $(\tilde{s}_k^w)_{k \neq 2j+1}$, $(w_k)_k$, $(\tau_k)_{k \neq j}$, we write $\mathcal{C}'_j = \mathcal{C}_{j+2} \cap \mathcal{F}_j$ and we define

$$ P_j := \mathbb{P}(A_j | (u_n)_n, (\tilde{s}_k^w)_{k \neq 2j+1}, (w_k)_k, (\tau_k)_{k \neq 2j+1}) = \mathbb{P}(A_j | \mathcal{F}_j) $$

Note that $A_j \in \mathcal{C}_{j+2}$, so we have almost surely $P_j = \mathbb{P}(A_j | \mathcal{C}'_j)$, which is a $\mathcal{C}_{j+2}$-measurable function, and $\tau_j$ is $\mathcal{C}_{j+2}$-measurable therefore $A'_j \in \mathcal{C}_{j+2}$. From that we deduce that the event $(j_0 = j)$ is $\mathcal{C}_{j+2}$-measurable for all $j \in \mathbb{N}$. In other words $j_0$ is a stopping time for the filtration $(\mathcal{C}_{j+2})_{j \in \mathbb{N}}$ so $T^{j_0+2}(\omega)$ is independent of $\mathcal{C}_{j_0+2}$ and the data of $(\tilde{\gamma}_k^{\tilde{v}})_{k < 4}$ is $\mathcal{C}_{j_0+2}$-measurable. Therefore, the sequence $((\tilde{\gamma}_{4k}^{\tilde{v}}, \dots, \tilde{\gamma}_{4k+3}^{\tilde{v}}))_{k \in \mathbb{N}}$ is i.i.d.. Moreover, for all $j \in \mathbb{N}$, the events $(A'_{j'})_{j' \le j}$ and the data of $(\tilde{\gamma}_0^{\tilde{w}}, \dots, \tilde{\gamma}_{2j+3}^{\tilde{w}})$ are measurable for $\mathcal{F}_{j+1}$, which is independent of $\tilde{s}_{2j+3}^w$. Therefore, $\tilde{\gamma}_3^{\tilde{v}} = \tilde{s}_{2j_0+3}^w$ is independent of the joint data of $j_0$ and $(\tilde{\gamma}_0^{\tilde{v}}, \tilde{\gamma}_1^{\tilde{v}}, \tilde{\gamma}_2^{\tilde{v}})$. This concludes the proof of (4).

We have $A'_j \subset A_j$ because if $\tau_j < \frac{(1-2\rho)1_{A_j}}{P_j}$, then $1_{A_j} > 0$. Let us check that for all $j \in \mathbb{N}$, we have $\mathbb{P}(A'_j | \mathcal{C}'_j) = 1 - 2\rho$. Let $j \in \mathbb{N}$ be fixed. First note that $\tilde{s}_{2j+1}^w$ is independent of $\mathcal{F}_j$ and has distribution law $\nu_s$ which is $\rho$-Schottky for $A$ so $P_j \ge 1 - 2\rho$ almost surely. Moreover $\tau_j$ is uniformly distributed and independent of $\mathcal{F}_j$ so:

$$ \mathbb{P} \left( \tau_j < \frac{(1 - 2\rho)}{P_j} \mid \mathcal{F}_j \right) = \frac{(1 - 2\rho)}{P_j}. $$

Note also that $A'_j = A_j \cap (\tau_j < \frac{(1-2\rho)}{P_j})$ and $A_j$ is independent of the joint data of $\tau_j$ and $\mathcal{F}_j$ so:

$$ \mathbb{P}(A'_j | \mathcal{F}_j) = \mathbb{P}(A'_j | C'_j) \mathbb{P} \left( \tau_j < \frac{(1 - 2\rho)}{P_j} \mid \mathcal{F}_j \right) = P_j \frac{(1 - 2\rho)}{P_j} = 1 - 2\rho. $$
Note moreover that for all $j' < j$, the event $A_j'$ is $\mathcal{F}_j$-measurable. Therefore, $\mathbb{P}(A_j | (A_j)_{j'<j}) = (1-2\rho)$ for all $j \in \mathbb{N}$ so $j_0 \sim \mathcal{G}_{2\rho}$, so $\frac{v_0-1}{2} \sim \mathcal{G}_{2\rho}$. Moreover $(v_{4k})_{k \in \mathbb{N}}$ is i.i.d. so $(\frac{v_{4k}-1}{2})_{k \in \mathbb{N}} \sim \mathcal{G}_{2\rho}^{\otimes \mathbb{N}}$, which proves (2). Moreover, the fact that $A_j'$ is independent of $\mathcal{F}_j$ for all $j \in \mathbb{N}$ implies that the joint data of $(A_j')_{j \in \mathbb{N}}$ is independent of $\mathcal{F} := \bigcap_j \mathcal{F}_j$. Therefore $j_0$ is independent of the joint data of $(u_n)_n$ and $(w_n)_n$. Moreover $T^{\tilde{v}_{4k/2}}$ is measure-preserving and the data of $T^{\overline{v}_{4k/2}}(\omega)$ is independent of $C_{\overline{v}_{4k/2}}$. So for all $k \in \mathbb{N}$, $j_k$ is independent of the joint data of $(u_{n+\tilde{v}_{4k/2}})_n$ and $(w_{n+\overline{v}_{4k/2}})_n$ and the joint data of $j_i$ with $(u_{n+\tilde{v}_{4k/2}})_n$ and $(w_{n+\overline{v}_{4k/2}})_n$ is independent of $C_{\overline{v}_{4k/2}}$. As a consequence, $j_k$ is independent of the joint data of $(u_{n+\tilde{v}_{4k/2}})_n$ and $(w_{n+\overline{v}_{4k/2}})_n$ and $C_{\overline{v}_{4k/2}}$ and therefore independent of the joint data of $(j_{k'})_{k'<k}$ and $\mathcal{F}$. Then by induction, the joint data of $(j_k)_k$ is independent of $\mathcal{F}$, which proves (1) and concludes the proof of Lemma 4.8. $\square$

This first part looks a bit technical but we simply used the theory of stopping times for i.i.d. sequences, which is a classical tool in probability theory. See it as a warm-up for the next part of the proof of Theorem 4.7 (namely the construction of $(p_k)_k$), which is the more interesting part.

**4.3. Pivot algorithm.** In this paragraph, we describe an algorithm to construct the sequence $(p_k)$ of Theorem 4.7 from the random sequence $(\gamma_k^{\tilde{v}})_{k \in \mathbb{N}}$ with $v$ as constructed in Lemma 4.8 and from a random sequence of penalties $(\tau_k)_k \sim \mathcal{U}^{\otimes \mathbb{N}}$, which is independent of all previously defined random variables. We start by defining the algorithm, then introduce useful notations, then illustrate how the algorithm works on a toy model example, then prove a non-probabilistic renewal Lemma from which we deduce results on the probabilistic behaviour of the pivot algorithm. In the end, we give a full proof of Theorem 4.7.

**Definition 4.9 (Pivoting technique with penalty).** Let $\Gamma$ be a metric semi-group$^6$ endowed with a measurable binary relation $\triangle$. Let $0 < \rho < 1/5$ and let $\nu_s$ be a probability distribution on $\Gamma$ that is $\rho$-Schottky for $\mathbb{A}$. Let $(\gamma_n) \in \Gamma^{\mathbb{N}}$ and $(\tau_n) \in [0, 1]^\mathbb{N}$. To $(\gamma_n)$ and $(\tau_n)$ we associate a family $(p_k^j)_{j,k \in \mathbb{N}}$ that we define by induction on $j$ by the following algorithm. For $j = 0$, we set $p_k^0 = 1$ for all $k \in \mathbb{N}$. Let $j \ge 0$, we assume that we have constructed a sequence of odd integers $(p_k^j)_{k \ge 0}$ such that that $p_{2k+1}^j = 1$ for all $k$, such that there is an integer $m_j$ such that $\overline{p}_{2m_j+1}^j = 2j+1$ and such that $p_{2k}^j = 1$ for all $k > m_j$.

<footnote>$^6$We call metric semi-group a metric space endowed with a continuous and associative product map.</footnote>
• If $\gamma_{2m_j}^{p_j} \mathbb{A} \gamma_{2j+1}^{p_j} \mathbb{A} \gamma_{2j+2}^{p_j}$ and:

$$ \tau_j < \frac{1 - 2\rho}{\nu_s \left\{ s \in \Gamma \mid \gamma_{2m_j}^{p_j} \mathbb{A} s \mathbb{A} \gamma_{2m_j+2}^{p_j} \right\}}, $$

we set $m_{j+1} = m_j + 1$ and $p_k^{j+1} = p_k^j$ for all $k \in \mathbb{N}$.

• Otherwise, and if there exists $0 \le k < m_j$ such that:

$$ \tau_{\frac{\bar{p}_{2k+1}-1}{2}} < \frac{(1 - 3\rho)\mathbb{1}\left(\gamma_{2k+1}^{p_j} \mathbb{A} \gamma_{\bar{p}_{2k+2}}^{p_j} \cdots \gamma_{2j+2}\right)}{\nu_s \left\{ s \in \Gamma \mid \gamma_{2k}^{p_j} \mathbb{A} s \mathbb{A} \gamma_{\bar{p}_{2k+2}}^{p_j} \cap s \mathbb{A} \gamma_{\bar{p}_{2k+2}}^{p_j} \cdots \gamma_{2j+2} \right\}}, $$

we define $m_{j+1}$ to be the largest such $k$ and we set $p_k^{j+1} = p_k^j$ for all $k < 2m_{j+1}$, we set $p_{2m_{j+1}}^{j+1} = 2j + 3 - \bar{p}_{2m_{j+1}}^j$ and we set $p_k^j = 1$ for all $k > 2m_{j+1}$.

• Otherwise, we set $p_0^{j+1} = 2j + 3$ and $p_k^{j+1} = 1$ for all $k > 0$.

We fix a metric semi-group $\Gamma$, endow it with its Borel $\sigma$-algebra $\mathcal{A}_\Gamma$ and fix a measurable binary relation $\mathbb{A}$. We fix a constant $0 < \rho < 1/5$ and a probability distribution $\nu_s$ on $\Gamma$ that is $\rho$-Schottky for $\mathbb{A}$. Let $\Omega := \Gamma^\mathbb{N} \times [0, 1]^\mathbb{N}$ and let $\mathcal{A}_\Omega$ be the infinite product of the Borel $\sigma$-algebras on $\Gamma$ and $[0, 1]$. For all $n \in \mathbb{N}$, we write $\gamma_n : \Omega \to \Gamma$ for the map $((\gamma'_k)_k, (\tau'_k)_k) \mapsto \gamma'_n$ and $\tau_n : \Omega \to \Gamma$ for the map $((\gamma'_k)_k, (\tau'_k)_k) \mapsto \tau'_n$. Then for all $n \in \mathbb{N}$, the maps $\gamma_n$ and $\tau_n$ are continuous and therefore measurable. Note that the algorithm describes in Definition 4.9 associates a family $(p_k^j)_{j,k \ge 0}$ to all pair of sequences $((\gamma'_k)_k, (\tau'_k)_k) \in \Omega$ so it describes a function $(p_k^j)_{j,k \ge 0} : \Omega \to \mathbb{N}^{\mathbb{N} \times \mathbb{N}}$.

**Definition 4.10.** Let $(p_k^j)_{j,k \ge 0} : \Omega \to (\mathbb{N}^\mathbb{N})^\mathbb{N}$ be as in Definition 4.9. Let $\Omega' \subset \Omega$ be the set of entries $(\gamma_n)_n, (\tau_n)_n$ such that the sequence $(p_k^j)_{j \ge 0}$ is stationary for all $k \ge 0$. For all $k \ge 0$, we define $p_k : \Omega' \to \mathbb{N}$ by $p_k = \lim_j p_k^j$. For all $k \ge 0$, we say that $\bar{p}_{2k+1}$ is the $k$-th pivotal time of the sequence $(\gamma_k)_{k \ge 0}$ with weights $(\tau_k)_{k \ge 0}$.

For convenience, we use the following notations for this section. For all $j \ge 0$, let:

$$ A_j := \left( \tau_j < \frac{(1 - 2\rho)\mathbb{1}\left(\gamma_{2m_j}^{p_j} \mathbb{A} \gamma_{2m_j+1}^{p_j} \mathbb{A} \gamma_{2m_j+2}^{p_j}\right)}{\nu_s \left\{ s \in \Gamma \mid \gamma_{2m_j}^{p_j} \mathbb{A} s \mathbb{A} \gamma_{2m_j+2}^{p_j} \right\}} \right). $$
Note that for all $j$, we have $A_j = (m_{j+1} = m_j + 1)$. We say that the Pivot algorithm goes forward at step $j$ when $A_j$ is satisfied, otherwise we say that it backtracks. For all $j, k \ge 0$, let:

$$B_{j,k} := (k < m_j) \cap \left( \tau_{\frac{\overline{p}_{2k+1}-1}{2}} < \frac{(1-3\rho)\mathbb{1}\left(\gamma_{2k+1}^{p_j} \mathbb{A} \gamma_{\overline{p}_{2k+2}}^{p_j} \cdots \gamma_{2j+2}\right)}{\nu_s \left\{ s \mid \gamma_{2k}^{p_j} \mathbb{A} s \mathbb{A} \gamma_{\overline{p}_{2k+2}}^{p_j} \cap s \mathbb{A} \gamma_{\overline{p}_{2k+2}}^{p_j} \cdots \gamma_{2j+2} \right\}} \right).$$

Note that for all $j, k$, the event $B_{j,k}$ implies that $m_{j+1} \ge k$ and in fact we have:

$$(63) \quad \forall j \ge 0, \forall \omega \in \Omega \setminus A_j, m_{j+1}(\omega) = \max\{k \ge 0 \mid \omega \in B_{j,k} \text{ or } k = 0\}.$$

Let us illustrate what the pivot algorithm does on the first 4 steps of a toy model example. Let $\Gamma = F_6 = \langle a, b, c \rangle$, let $\ell: \Gamma \to \mathbb{N}$ be the word length associated to the set of generators $S = \{a, b, c, a^{-1}, b^{-1}, c^{-1}\}$ let $\mathbb{A} = \{(f, g) \mid \ell(fg) = \ell(f) + \ell(g)\}$ and let $\nu_s$ be the uniform distribution on $S$ and let $\rho = 1/6$. One can easily check that $\nu_s$ is $\frac{1}{6}$-Schottky for $\mathbb{A}$. Let us start with:

$$(\gamma_0 | [\gamma_1, \tau_0], \dots, \gamma_8, \dots) = (a^9 | [a, \tau_0], aba, [c, \tau_1], b^{-5}, [b^{-1}, \tau_2], b^{269}, [c^{-1}, \tau_3], a, \dots)$$

We mark with brackets the candidate pivotal times, which are simply the oddly indexed times and to each of them, we associate a penalty $\tau_k$. We denote by a vertical bar the position $m_j$. We start with $m_0 = 0$. At time 1, we check whether $\gamma_0 \mathbb{A} \gamma_1 \mathbb{A} \gamma_2$, this is true because the word $a^9 aba$ is reduced. Moreover, we need to check whether $\tau_0 < 4/5$ because $\nu_s\{s \mid \gamma_0 \mathbb{A} s \mathbb{A} \gamma_2\} = \nu_s(S \setminus \{a^{-1}\}) = 5/6$. Assume that $\tau_0 < 4/5$. Then at time $j = 1$, we have $m_1 = 1$ and:

$$\begin{aligned} & \left( \gamma_0^{p_j}, [\gamma_1^{p_j}, \tau_{(\overline{p}_3^0-1)/2}], \dots, \gamma_{2m_j}^{p_j} \mid [\gamma_{2j+1}, \tau_j], \gamma_{2j+2}, \dots \right) \\ & = (ac^9, [b, \tau_0], aba \mid [c, \tau_1], b^{-5}, \dots) \end{aligned}$$

Now we check whether $\gamma_{2m_1}^{p_1} = \gamma_2 \mathbb{A} \gamma_3 \mathbb{A} \gamma_4$, this is true because $abacb^{-5}$ is reduced. Moreover, there is no condition to check on $\tau_1$ because $\nu_s\{s \mid \gamma_2 \mathbb{A} s \mathbb{A} \gamma_4\} = \nu_s(S \setminus \{a^{-1}, b\}) = 4/6$. Then, at time $j = 2$, we have $m_2 = 2$ and:

$$\begin{aligned} & \left( \gamma_0^{p_j}, \dots, \gamma_{2m_j}^{p_j} \mid [\gamma_{2j+1}, \tau_j], \gamma_{2j+2}, \dots \right) \\ & = (ac^9, [b, \tau_0], aba, [c, \tau_1], b^{-5} \mid [b^{-1}, \tau_2], b^{269}, \dots). \end{aligned}$$

Here, we do not have $b^{-1} \mathbb{A} b^{269}$ so we have to backtrack. First, we check whether $c \mathbb{A} b^{263}$, this is true, moreover, there is no condition to check on $\tau_1$ because $\nu_s\{s \mid \gamma_2 \mathbb{A} s \mathbb{A} \gamma_4 \cap s \mathbb{A} \gamma_4 \gamma_5 \gamma_6\} = \nu_s(S \setminus \{a^{-1}, b, b^{-1}\}) = 3/6$. We would have had
to check conditions on $\tau$ in two cases: either the alignment condition with the product $\gamma_4\gamma_5\gamma_6$ is redundant with one of the former alignment conditions (in this case, the first letter would have to be a or $b^{-1}$) or when the product $\gamma_4\gamma_5\gamma_6$ is trivial. Then at time $j = 3$, we have backtracked to $m_3 = 1$ again and we have:

$$ (\gamma_0^{p_j}, \dots, \gamma_{2m_j}^{p_j} \mid [\gamma_{2j+1}, \tau_j], \gamma_{2j+2}, \dots) = (ac^9, [b, \tau_0], abacb^{263} \mid [c^{-1}, \tau_3], a, \dots). $$

We check whether $abacb^{263}A c^{-1}A a$, this is true. Moreover $b \neq a$ so there is no condition to check on $\tau_3$. At time $j = 4$, we have $m_4 = 2$ and:

$$ (\gamma_0^{p_j}, \dots, \gamma_{2m_j}^{p_j} \mid \dots) = (ac^9, [b, \tau_0], abacb^{263}, [c^{-1}, \tau_3], a \mid \dots). $$

Note that this situation is undistinguishable of the one we had at time $j = 2$.

The reason why we need to check conditions on $\tau_k$ is to make sure that the Markov chain $(m_j)_{j \ge 0}$ is independent of the joint data of $(\gamma_{2k})_{k \ge 0}$ when $(\gamma_{2k+1}) \sim \nu_s^{\otimes N}$. In fact the law of $(m_j)_{j \ge 0}$ only depends on $\rho$ and satisfies a renewal condition in the sense that for all $n$, the renewal event $(\forall k > n, m_k > m_n)$ is independent the joint data of $(\gamma_k)_{k \le 2j}$ and $(m_j)_{j \le n}$ i.e., the past. Let us now remind some standard notations and give more details on what this means exactly.

We use the following notations. Given $k \ge 0$, given $F(X_1, \dots, X_k)$ a formula and given $f_1, \dots, f_k$ a family of functions on $\Omega$, we write $(F(f_1, \dots, f_k))$ for the set $\{\omega \in \Omega \mid F(f_1(\omega), \dots, f_k(\omega))\}$. Given $A \subset \Omega$, we say that the formula $F(f_1, \dots, f_k)$ is satisfied on $A$ when $A \subset (F(f_1, \dots, f_k))$. Given $f: \Omega \to \mathbb{N}$ and $T: \Omega \to \Omega$, we write $T^f: \Omega \to \Omega$; $\omega \mapsto T^f(\omega)$. Note that if $T$ and $f$ are measurable then $T^f$ is.

For now $\Omega$ is nothing more that a metric space endowed with its Borel $\sigma$-algebra, in particular there is no notion of almost-sure properties on $\Omega$. Let:

$$ T_{\Omega}: \Omega \to \Omega; (\gamma_n)_{n \ge 0}, (\tau_n)_{n \ge 0} \mapsto (\gamma_{n+2})_{n \ge 0}, (\tau_{n+1})_{n \ge 0}. $$

That way, for all $k \ge 0$, we have $\tau_k = \tau_0 \circ T_{\Omega}^k$ and $\gamma_{2k} = \gamma_0 \circ T_{\Omega}^k$ and $\gamma_{2k+1} = \gamma_1 \circ T_{\Omega}^k$.

The following Lemma is a non-probabilistic toolbox whose role is to break down in details what the pivot algorithm does and introduce useful notations for the rest of the discussion.

**Lemma 4.11.** The functions $(p_k^j)_{j,k \ge 0} : \Omega \to (\mathbb{N}^N)^N$ defined in Definition 4.9 is measurable. Moreover, the set $\Omega'$ and the function $(p_k)_{j \ge 0} : \Omega' \to \mathbb{N}^N$ are measurable. Let $(m_j)_{j \ge 0} : \Omega \to \mathbb{N}^N$ be as in Definition 4.9. Then $\Omega' = (\lim_j m_j = +\infty)$ and on $\Omega'$, we have:

$$ (64) \qquad \forall k \ge 0, \quad \frac{\overline{p}_{2k+1} - 1}{2} = \max\{j \in \mathbb{N} \mid m_j \le k\}. $$
More precisely, we show that:

$$ (65) \quad \forall j, k \ge 0, \quad \frac{\bar{p}_{2k+1}^j - 1}{2} = \max\{j' \le j \mid m_{j'} \le k\} + (k - m_j)^+. $$

Then $\Omega'$ is $T_{\Omega}^{\frac{p_0+1}{2}}$ invariant. Moreover, for all $k \ge 0$, we have $(T_{\Omega}^{\frac{p_0+1}{2}})^k = T_{\Omega}^{\frac{\bar{p}_{2k}}{2}}$ and $p_{2k+2i} = p_{2i} \circ T_{\Omega}^{\frac{\bar{p}_{2k}}{2}}$ for all $i, k \ge 0$. Moreover, there is a measurable subset $S \subset \Omega$ such that on $\Omega$ we have:

$$ (66) \quad \forall j, k \ge 0, \quad (\bar{p}_{2k+1} = 2j+1) = (m_j = k) \cap A_j \cap T_{\Omega}^{-j-1}(S). $$

More precisely, we show that there exists a non-increasing sequence of measurable sets $(S^{j'})_{j' \ge 0}$ such that:

$$ (67) \quad \forall j, j', k, \quad (\bar{p}_{2k+1}^{j'+j} = 2j+1) = (m_j = k) \cap A_j \cap T_{\Omega}^{-j-1}(S^{j'}). $$

*Proof.* Let $(p_k^j)_{j,k \in \mathbb{N}} : \Omega \to (\mathbb{N}^\mathbb{N})^\mathbb{N}$ be as in Definition 4.9. We write:

$$ S := \bigcap_{j=0}^{+\infty} \left( A_j \cup \bigcup_{k=0}^{+\infty} B_{j,k} \right). $$

First we claim that for all $j, k \in \mathbb{N}$, we have:

$$ (68) \quad (m_{j+1} = k) = A_j \cap (m_j = k-1) \sqcup B_{j,k} \setminus A_j \setminus \bigcup_{k'>k} B_{j,k'} \sqcup (k=0) \setminus A_j \setminus \bigcup_{k'} B_{j,k'}. $$

This is simply a reformulation of (63).

Now we prove (65). First note that we have $m_{j+1} \le m_j + 1$ for all $j \ge 0$ because $m_{j+1} = m_j + 1$ on $A_j$ and $m_{j+1} \le m_j$ outside of $A_j$. We show by induction that for all $j \ge 0$, we have:

$$ (69) \quad \bar{p}_{2m_j+1}^j = 2j+1 \quad \text{and} \quad \forall k > 2m_j, \quad p_k^j = 1. $$

This is true for $j = 0$ because $\bar{p}_1^0 = p_0^0 = 1$. Assume that $\bar{p}_{2m_j+1}^j = 2j+1$ for an integer $j \ge 0$. Then on $A_j$, we have $\bar{p}_{2m_{j+1}+1}^{j+1} = \bar{p}_{2m_j+3}^j = \bar{p}_{2m_j+3}^j = \bar{p}_{2m_j+1}^j + p_{2m_j+1}^j + p_{2m_j+2}^j = 2j+3$. Outside of $A_j$, we explicitly set $p_k^{j+1} = 1$ for all $k > 2m_{j+1}$. Moreover, $\bar{p}_{2m_{j+1}+1}^{j+1} = \bar{p}_{2m_j+1}^{j+1} + p_{2m_j+1}^{j+1} = 2j+3$. This conclude the proof of (69). Let us quickly show that for all $j \ge 0$, we have:

$$ (70) \quad \forall k < 2m_{j+1}, \quad p_k^{j+1} = p_k^j. $$
This is true for all $k$ on $A_j$. Outside of $A_j$, we explicitly set $p_k^{j+1} = p_k^j$ for all $k < m_{j+1}$. An immediate consequence of (69) and (70) is that $p_{2k+1}^j = 1$ for all $j$ and all $k$.

Let $j \ge k \ge 0$. On the set $(k \le m_j)$, let $l_k^j := \max\{j' \le j \mid m_{j'} \le k\}$. Then we have $m_{l_k^j} = k$ so we have $\overline{p}_{2k+1}^{l_k^j} = 2l_k^j + 1$ by (70), when $l_k^j = j$, we have (65), otherwise we have $m_{j'+1} > k$ for all $j' \in \{l_k^j, \dots, j-1\}$ so by (70) we have $p_k^j = p_{k'}^{l_k^j}$ for all $k' \le 2k$. Therefore $\overline{p}_{2k+1}^j = 2l_k^j + 1$, which proves (65) on the set $(k \le m_j)$. On $(k \ge m_j)$, we have $\overline{p}_{2k+1}^j = 2j+1 + p_{2m_{j+1}} + \dots + p_{2k} = 2j+1+2(k-m_j)$, which concludes the proof of (65).

Note also that (70) and (69) imply that for all $k$, the sequence $(\overline{p}_k^j)_{j \ge 0}$ is non-decreasing so it has a limit in $\mathbb{N} \cup \{+\infty\}$. Let $\overline{p}_k$ be the limit of $\overline{p}_k^j$ for all $k$. Then, by taking the limit $j \to +\infty$ in (65), we get (64). Moreover (65) and the fact that for all $j$, we have $p_{2k+1}^j = 1$ for all $k$ and $p_{2k}^j = 1$ for all $k > m_j$ tells us that the data of $(p_k^j)_{k \ge 0}$ is determined by the data of $(m_{j'})_{j' \le j}$.

We show by induction that the data of $(m_{j'})_{j' \le j}$ is $\mathcal{P}_j$-measurable. Let $j \ge 0$ and assume that the data of $(m_{j'})_{j' \le j}$ is $\mathcal{P}_j$-measurable. Then the data of $(p_k^j)_{k \ge 0}$ is also $\mathcal{P}_j$-measurable by (65). Moreover the data of $(\gamma_0, \dots, \gamma_{2j})$ is $\mathcal{P}_j$ measurable by definition so $\gamma_{2m_j}^{p_j}$ is $\mathcal{P}_j$-measurable. Moreover $\gamma_{2m_{j+1}}^{p_j} = \gamma_{2j+1}$ and $\gamma_{2m_{j+2}}^{p_j} = \gamma_{2j+2}$ by (70) therefore $A_j$ is $\mathcal{P}_{j+1}$-measurable. Then for all $k \ge 0$, the set $(k < m_j)$ is $\mathcal{P}_j$-measurable by assumption and on this set, we have $\overline{p}_{2k+1} - 12 < j$ by (65) so $C_{j,k}$ is $\mathcal{P}_{j+1}$-measurable. Therefore, by (68), the data of $m_{j+1}$ is $\mathcal{P}_{j+1}$-measurable, which concludes the induction argument. Therefore the functions $(p_j^k)_{j,k \ge 0}$, $(p_k)_{k \ge 0}$ and $(m_j)_{j \ge 0}$ are measurable.

Now let us prove (66), which states that for all $j \ge 0$, the set $(\overline{p}_{2k+1} = 2j+1)$ is equal to $(m_j = k) \cap T_{\Omega}^{-j-1}(S) \cap A_j$. By (64), we know that $(p_0 = 2j+1)$ is equal to $(m_j = 0) \cap (\forall j' > j, m_{j'} \ge m_j)$. Moreover, the set $(\forall j' > j, m_{j'} > m_j)$ is the intersection of $A_j$ with the set:

$$S_j := \bigcap_{j'=j+1}^{\infty} (m_{j'+1} \ge m_{j+1}) = \bigcap_{j'=j+1}^{+\infty} \left( A_{j'} \cup \bigcup_{k=1}^{+\infty} B_{j',k} \cap (k \ge m_{j+1}) \right).$$

We claim that $S_j \cap A_j = T_{\Omega}^{-j-1}(S) \cap A_j$. First note that (69) implies that $\gamma_{2m_j+2+k}^{p_j} = \gamma_{2j+2+k}$ for all $k \ge 0$. Moreover, $A_j$ implies that $\gamma_{2m_{j+1}+k}^{p_{j+1}} = \gamma_{2j+2+k}$ for all $k \ge 0$. For all $j' \ge 0$, let $S_{j'} := \bigcap_{j''=0}^{j'} (A_{j''} \cup \bigcup_{k=0}^{+\infty} B_{j'',k})$ and for all $j' > j$, let $S_j' := \bigcap_{j''=j+1}^{j'} (A_{j'} \cup \bigcup_{k=1}^{+\infty} B_{j'',k} \cap (k \ge m_{j+1}))$. Note that for all
$j \le j'$, we have $S_j^{j'} \subset S_j^{j'+1}$. We claim that for all $j' > j$, we have $S_j^{j'} \cap A_j = T_{\Omega}^{-j-1}(S^{j'-j-1}) \cap A_j$ and on $S_j^{j'} \cap A_j$, we have:

$$ (71) \quad m_{j'+j+1} = m_{j'} \circ T_{\Omega}^{j+1} + m_{j+1}, \quad \text{and} \quad \forall k \ge 0, \quad p_{2k+2m_{j+1}}^{j'} = p_{2k}^{j'-j-1} \circ T_{\Omega}^{j+1}. $$

For $j' = j + 1$, we have $S^0 = A_0$ and $S_j^{j+1} = A_{j+1}$. Moreover, on $A_j$, we have $(\gamma_{2m_{j+1}}^p, \gamma_{2m_{j+1}+1}^p, \gamma_{2m_{j+1}+2}^p) = (\gamma_{2j+2}, \gamma_{2j+2}, \gamma_{2j+4}) = (\gamma_0^{p_0}, \gamma_1^{p_1}, \gamma_2^{p_2}) \circ T_{\Omega}^{j+1}$, which proves the claim. Now let $j' > j$ and assume that the claim holds for $j'$. We want to show that it also holds for $j' + 1$. First, the fact that for all $k \ge 0$, we have $p_{2k+2m_{j+1}}^{j'} = p_{2k}^{j'-j-1} \circ T_{\Omega}^{j+1}$ and that $\bar{p}_{2m_{j+1}}^{j'} = 2j+2$ implies that $\gamma_{2k+2m_{j+1}}^{p'j'} = \gamma_{2k}^{p'j'-j-1} \circ T_{\Omega}^{j+1}$ for all $j', k$. Therefore, for all $k \ge 0$, on the set $(k > m_j) \cap A_j$, we have $B_{j'+1,k} = T_{\Omega}^{-j-1}(B_{j'-j,k-m_j})$ and $A_{j'+1} \cap A_j = T_{\Omega}^{-j-1}(A_{j'-j}) \cap A_j$. Therefore, we have $S_j^{j'+1} \cap A_j = T_{\Omega}^{-j-1}(S^{j'-j}) \cap A_j$ and on $S_j^{j'+1} \cap A_j$, we have $m_{j'+1} = m_{j'-j} \circ T_{\Omega}^{j+1} + m_{j+1}$. This concludes the induction argument which proves (67).

Taking the limit $j' \to +\infty$, we have $S_j \cap A_j = T_{\Omega}^{-j-1}(S) \cap A_j$, which concludes the proof of (66). From (71), we deduce that:

$$ (72) \qquad p_{2k+2i}^{j'} = p_{2i}^{j' - \frac{\bar{p}_{2k}^j}{2}} \circ T_{\Omega}^{\frac{\bar{p}_{2k}^j}{2}} $$

on the set $(m_j \ge k)$ for all $i, j, k \ge 0$. Indeed, (72) is trivial for $k = 0$ and on $(m_{j'} \ge k)$, let $j < j'$ be maximal and such that $m_{j+1} = k$, then $\frac{\bar{p}_{2k}}{2} = j + 1$ by (65). By taking the sum of (72) on $i$, we deduce that:

$$ (73) \qquad \bar{p}_{2k+2i}^{j'} = \bar{p}_{2i}^{j' - \frac{\bar{p}_{2k}^j}{2}} \circ T_{\Omega}^{\frac{\bar{p}_{2k}^j}{2}} + \bar{p}_{2k}^{j'} $$

on the set $(m_j \ge k)$ for all $i, j, k \ge 0$. Therefore, on $\Omega' = (\forall k, \bar{p}_k < +\infty)$, for all $i, k \ge 0$ and for all $j' \ge \frac{\bar{p}_{2k}}{2}$ we have $\bar{p}_{2i}^{j'} \circ T_{\Omega}^{\frac{\bar{p}_{2k}}{2}} \le \bar{p}_{2k+2i} < +\infty$. Hence $T_{\Omega}^{\frac{\bar{p}_{2k}}{2}}(\Omega') \subset \Omega'$. $\square$

Now we can properly study the probabilistic behaviour of the pivoting algorithm. For all $k \in \mathbb{N}$, we write $\mathcal{P}_j$ for the $\sigma$-algebra on $\Omega$ generated by $(\gamma_k)_{k \le 2j}$ and $(\tau_k)_{k < j}$. Note that $m_j$ is $\mathcal{P}_j$-measurable for all $j$. First let us describe the distribution law of $(m_j)_{j \ge 0}$.

**Lemma 4.12.** Let $\mathbb{P}$ be a probability distribution on $\Omega$ such that $(\gamma_{2k})_{k \ge 0}$ is independent of the joint data of $(\gamma_{2k+1})_{k \ge 0}$ and $(\tau_k)_{k \ge 0}$ and such that $(\gamma_{2k+1})_{k \ge 0}$,
$(\tau_k)_{k \ge 0} \sim \nu_s^{\otimes \mathbb{N}} \otimes \mathcal{U}^{\otimes \mathbb{N}}$. Then for all $j, k \in \mathbb{N}$, we have:

$$ (74) \qquad \mathbb{P}(m_{j+1} = m_j + 1 | m_0, \dots, m_j) = (1 - 2\rho) $$

$$ (75) \qquad \mathbb{P}(m_{j+1} < m_j - k | m_0, \dots, m_j) = 2\rho \left( \frac{\rho}{1-2\rho} \right)^k \mathbb{1}(k < m_j). $$

*Proof.* Let $(A_j)_{j \ge 0}$ and $(B_{j,k})_{j,k \ge 0}$ be as defined above below Definition 4.10. First note that by construction of $\mathbb{P}$, the conditional distribution of $\gamma_{2j+1}$, $\tau_j$ with respect to $\mathcal{P}_j \lor (T_{\Omega}^{j+1})^* \mathcal{A}_{\Omega}$ is constant equal to $\nu_s \otimes \mathcal{U}$. Moreover the random variable $\gamma_{2m_j}^{p_j}$ is $\mathcal{P}_j$-measurable and $\gamma_{2m_j+1}^{p_j}$ is equal to $\gamma_{2j+1}$. Therefore, we have:

$$ \mathbb{P} \left( \gamma_{2m_j}^{p_j} \mathbb{A} \gamma_{2m_j+1}^{p_j} \mathbb{A} \gamma_{2m_j+2}^{p_j} \mid \mathcal{P}_j, (T_{\Omega}^{j+1})^* \mathcal{A}_{\Omega} \right) = \nu_s \left\{ s \in \Gamma \mid \gamma_{2m_j}^{p_j} \mathbb{A} s \mathbb{A} \gamma_{2m_j+2}^{p_j} \right\}. $$

Moreover, $\tau_j$ is independent of $\mathcal{P}_j \lor T_{\Omega}^{j+1*} \mathcal{A}_{\Omega}$ by construction. Therefore, for all $t_j : \Omega \to [0, 1]$ that is $\mathcal{P}_j \lor T_{\Omega}^{j+1*} \mathcal{A}_{\Omega}$-measurable, we have:

$$ \mathbb{P} \left( \tau_j < t_j \mid \mathcal{P}_j, T_{\Omega}^{j+1*} \mathcal{A}_{\Omega}, \gamma_{2j+1} \right) = t_j. $$

Let $t_j := \frac{1-2\rho}{\nu_s \{s \mid \gamma_{2m_j}^{p_j} \mathbb{A} s \mathbb{A} \gamma_{2m_j+2}^{p_j}\}}$. The Schottky property for $\nu_s$ guarantees us that $t_j \le 1$. Moreover $\gamma_{2m_j}^{p_j}$ is $\mathcal{P}_j$-measurable and $\gamma_{2m_j+2}^{p_j} = \gamma_{2j+2}$ is $T_{\Omega}^{j+1*} \mathcal{A}_{\Omega}$-measurable, $t_j$ is a function of those two so it is $\mathcal{P}_j \lor T_{\Omega}^{j+1*} \mathcal{A}_{\Omega}$-measurable. we get that:

$$ (76) \qquad \mathbb{P} \left( A_j \mid \mathcal{P}_j, T_{\Omega}^{j+1*} \mathcal{A}_{\Omega} \right) = 1 - 2\rho. $$

Note that it is the Schottky property for $\nu_s$ that guarantees us that $t_j \le 1$. This proves (74).

Now we prove (75). Let $j \ge 0$. For all $k \ge 0$ let $l_k^j := \max\{j' \le j \mid m_{j'} \le k\}$. Note that $l_k$ is measurable for the algebra generated by $(m_0, \dots, m_j)$ and when $k \le m_j$, we have $l_k^j = \frac{\bar{p}_{2k+1}^j - 1}{2}$ by (65). Then by (67), we know that for all fixed $l$, the event $l_k^j = l$ factorizes as

$$ (l_k^j = l) = (m_l = k) \cap A_l \cap T_{\Omega}^{-l-1}(S^{j-l}) $$

So the distribution of $(\gamma_{2l+1}, \tau_l)$ with respect to the event $(l_k^j = l)$ is almost surely equal to its distribution with respect to $A_l$. Let:

$$ C_{l,k}^j = \left( \tau_l < \frac{(1-3\rho)\mathbb{1}(\gamma_{2l+1}\mathbb{A}\gamma_{2l+2}\cdots\gamma_{2j+2}) \mathbb{1}(\gamma_{2k}^{p_k}\mathbb{A}\gamma_{2l+1}\mathbb{A}\gamma_{2l+2})}{\nu_s\{s \mid \mathbb{A}\gamma_{2l+2}\cdots\gamma_{2j+2} \cap \gamma_{2k}^{p_k}\mathbb{A} s \mathbb{A}\gamma_{2l+2}\}} \right) $$
We claim that $C_{l,k}^j \cap (l_k^j = l) \subset A_l$. It is obvious for the indicator function part because $\gamma_{2k}^{p_j} = \gamma_{2m_l}^{p_l}$ when $(l_k^j = l)$ by (70) and (69) and $\gamma_{2l+1}, \gamma_{2l+2} = \gamma_{2m_l+1}, \gamma_{2m_l+2}^{p_l}$ so we only need to prove that:

$$ \nu_s\{s \mid s \mathbb{A}\gamma_{2l+2} \cdots \gamma_{2j+2} \cap \gamma_{2k}^{p_j} \mathbb{A} s \mathbb{A}\gamma_{2l+2}\} \ge \frac{1-3\rho}{1-2\rho} \nu_s\{s \mid \gamma_{2k}^{p_j} \mathbb{A} s \mathbb{A}\gamma_{2l+2}\} $$

Let $A = \{s \mid \gamma_{2m_j}^{p_j} \mathbb{A} s \mathbb{A} \gamma_{2m_j+2}^{p_j}\}$ and $B = \{s \mid s \mathbb{A} \gamma_{2l+2} \cdots \gamma_{2j+2}\}$. Then $\nu_s(A) \ge 1-2\rho$ and $\nu_s(B) \ge 1-\rho$ by the Schottky property so $\nu_s(A \cap B) \ge \nu_s(A) - \rho = \nu_s(A) \left(1 - \frac{\rho}{\nu_s(A)}\right) \ge \nu_s(A) \frac{1-3\rho}{1-2\rho}$. This proves that $C_{l,k}^j \cap (l_k = l) \subset A_l$. Moreover $\mathbb{P}(C_{l,k} \mid \mathcal{P}_l \lor (T_\Omega^{l+1})^* \mathcal{A}_\Omega) = 1-3\rho$ because $\gamma_{2l+1}$ is $\nu_s$ distributed. Therefore, we have $\mathbb{P}(C_{l,k}^j \mid \mathcal{P}_l, (T_\Omega^{l+1})^* \mathcal{A}_\Omega, A_l) = \frac{1-3\rho}{1-2\rho}$. Moreover, the data of $m_0, \dots, m_l$ is $\mathcal{P}_l$-measurable and $m_{i+l+1} = m_i \circ T_\Omega^{l+1} + m_l + 1$ on $A_l$ and for all $i \ge 0$ by (71). Note also that, on the level sets of $(m_0, \dots, m_j)$ and for all $k < m_j$, we have:

$$ (77) \qquad (m_{j+1} < m_j - k) = \Omega \setminus A_j \setminus \bigcup_{k'=m_j-k}^{m_j-1} C_{l_{k'},k'}^j. $$

Moreover the $(l_{k'})_{0 \le k' < m_j}$ are distinct by definition and on each level set of $(m_0, \dots, m_j)$, the event $C_{l_{k'},k'}^j$ is independent of $(\mathcal{P}_{l_k} \lor (T_\Omega^{l_{k'+1}})^* \mathcal{A}_\Omega)$ has probability $\frac{1-3\rho}{1-2\rho}$. Therefore, on each level set of $(m_0, \dots, m_j)$ and for all $k \le m_j$, the event $\bigcap_{k'=m_j-k}^{m_j-1} (\Omega \setminus C_{l_{k'},k'}^j)$ has conditional probability $(\frac{\rho}{1-2\rho})^k$ and is independent of $A_j$, from that we deduce (75). To conclude, we trivially have $\mathbb{P}(m_j < m_j - k) = 0$ for all $k \ge m_j$. $\square$

Now we use Lemmas 4.11 and 4.12 to prove points (5), (6), (7), (8), (9), (10) and (11) in Theorem 4.7. First we prove the finite exponential moment result, which implies points (5) and (7).

**Corollary 4.13.** Let $\mathbb{P}$ be a probability distribution on $\Omega$ such that $(\gamma_{2k})_{k \in \mathbb{N}}$ is independent of the joint data of $(\gamma_{2k+1})_{k \in \mathbb{N}}$ and $(\tau_k)_{k \in \mathbb{N}}$ and $((\gamma_{2k+1})_{k \in \mathbb{N}}, (\tau_k)_{k \in \mathbb{N}}) \sim \nu_s^{\otimes \mathbb{N}} \otimes \mathcal{U}^{\otimes \mathbb{N}}$. Then the data of $(m_j)_{j \ge 0}$ is independent of the data of $(\gamma_{2k})_{k \ge 0}$. Moreover $m_j \to +\infty$ almost surely and there exist constants $C, \beta > 0$ such that:

$$ (78) \qquad \forall k, t \in \mathbb{N}, \mathbb{P}(\bar{p}_3 \ge t) \le C e^{-\beta t}. $$

*Proof.* First we prove the independence result. Note that (74) and (75) together with the fact that $m_0 = 0$ determine the distribution law of the sequence $(m_j)_{j \ge 0}$.