Let $\kappa$ be the distribution of $(m_j)_{j \ge 0}$. Let $X = \Gamma^{2\mathbb{N}}$ and $Y = \Gamma^{2\mathbb{N}+1} \times [0, 1]^\mathbb{N}$. Note that $\Omega$ naturally identifies with $\bar{X} \times Y$. Let $\mu_Y = \nu_s^{\otimes \mathbb{N}} \otimes \mathcal{U}^{\otimes \mathbb{N}}$. For all $x \in X$, let $\mathbb{P}_x = \delta_x \otimes \mu_Y$. The fact that $(\gamma_{2k})_{k \in \mathbb{N}}$ is independent of the joint data of $(\gamma_{2k+1})_{k \in \mathbb{N}}$ and $(\tau_k)_{k \in \mathbb{N}}$ means that there is a probability distribution $\mu_X$ on $X$ such that:

$$ \mathbb{P} = \mu_X \otimes \mu_Y = \int_{x \in X} \mathbb{P}_x d\mu_X(x). $$

Then by Lemma 4.12, applied to $\mathbb{P}_x$ for all $x$, we have:

$$ ((\gamma_{2k})_{k \ge 0}, (m_j)_{j \ge 0}) \sim \int_{x \in X} (\delta_x \otimes \kappa) d\mu_X(x) = \mu_X \otimes \kappa, $$

Which means that the sequences $(\gamma_{2k})_{k \ge 0}$ and $(m_j)_{j \ge 0}$ are independent.

Let us now prove that each $p_k$ has a finite exponential moment. Let $x$ be a random variable that takes values in $\{1\} \cup \mathbb{Z}_-$ such that:

$$ \mathbb{P}(x=1) = 1-2\rho \quad \text{and} \quad \forall k < 0, \mathbb{P}(x=k) = 2\rho \frac{1-3\rho}{\rho} \left( \frac{\rho}{1-2\rho} \right)^k. $$

Then $\mathbb{E}(x) = \frac{(1-2\rho)(1-5\rho)}{1-3\rho} > 0$ and $x$ has finite exponential moment. Hence, by Lemma 5.2, we have $m_j \to +\infty$ almost surely. Moreover there exist constants $C, \beta > 0$ such that:

$$ \forall j \ge 0, \mathbb{P}(m_j \le 1) \le C e^{-\beta t}. $$

Hence by taking the sum, we get:

$$ \forall t \ge 0, \mathbb{P}(\exists j \ge t, m_j \le 1) \le \frac{C}{1-e^{-\beta}} e^{-\beta t}. $$

Then by (64), we have:

$$ \forall t \ge 0, \mathbb{P}(\bar{p}_3 \ge 2j+1) \le \frac{C}{1-e^{-\beta}} e^{-\beta t}. $$

This yields (78).

<figure><img src="image_2.png" /></figure>

Now let us prove the Markov property, which implies points (8), (6) and (11).

**Lemma 4.14 (Markov Property).** Let $\nu_u$ be any probability distribution on $\Gamma$ and let $\mathbb{P} := (\nu_u \otimes \nu_s)^{\otimes \mathbb{N}} \otimes \mathcal{U}^{\otimes \mathbb{N}}$. Then, for all $k \ge 0$, the data of $(\tilde{\gamma}_{k'})_{k' \le 2k}$ is independent of the data of $(\tilde{\gamma}_{k'})_{k' \ge 2k+2}$ and we have:

$$ (79) \quad \forall A \in \mathcal{A}_\Gamma, \mathbb{P}(\gamma_{2k+1}^p \in A | (\tilde{\gamma}_{k'}^p)_{k' \ne 2k+1}) = \frac{\nu_s\{s \in A | \gamma_{2k}^p A s A \gamma_{\bar{p}_{2k+2}}\}}{\nu_s\{s \in \Gamma | \gamma_{2k}^p A s A \gamma_{\bar{p}_{2k+2}}\}}. $$
Therefore $(\tilde{\gamma}_{2k+2}^p)_{k \ge 0}$ is i.i.d. and independent of $\tilde{\gamma}_0^p$.

*Proof.* Let $S$ be as in Lemma 4.11. For all $k \in \mathbb{N}$, we write $\mathcal{P}_k'$ for the $\sigma$-algebra generated by the joint data of $\overline{p}_{2k+1}$ and $(\tilde{\gamma}_{k'}^p)_{k' \le 2k}$. We remind that for all $j$, we write $\mathcal{P}_j$ for the $\sigma$-algebra generated by $(\gamma_n)_{n \le 2j}$ and $(\tau_k)_{k < j}$.

First we show that for all $k \ge 0$, the conditional distribution of $T_{\Omega}^{\frac{\overline{p}_{2k+2}}{2}}$ with respect to $\mathcal{P}_k'$ is almost surely equal to $\frac{1(S)}{\mathbb{P}(S)}\mathbb{P}$. Let $l, k \ge 0$. We have:

$$ (\overline{p}_{2k+1} = 2l + 1) = (m_l = k) \cap A_l \cap T_{\Omega}^{-l-1}(S) $$

by (66) in Lemma 4.11. The event $(m_l = k)$ only depends on $\mathcal{P}_l$, which is independent of $(T_{\Omega}^{l+1})^*\mathcal{A}_{\Omega}$. Moreover, we have $\mathbb{P}(A_l | \mathcal{P}_l \lor (T_{\Omega}^{-l-1})^*\mathcal{A}_{\Omega}) = 1 - 2\rho$ so $A_l$ is independent of the joint data of $\mathcal{P}_l$ and $(T_{\Omega}^{-l-1})^*\mathcal{A}_{\Omega}$. Reciprocally $(T_{\Omega}^{l+1})^*\mathcal{A}_{\Omega}$ is independent of the joint data of $\mathcal{P}_l$ and $A_l$. Therefore the conditional distribution of $T_{\Omega}^{l+1}$ with respect to the $(\overline{p}_{2k+1} = 2l + 1)$ and $\mathcal{P}_{\frac{\overline{p}_{2k+1}-1}{2}}$ is almost surely equal to the conditional distribution of $T_{\Omega}^{l+1}$ with respect to $T_{\Omega}^{-l-1}(S)$, which is equal to $\frac{1(S)}{T_{\Omega}^{l+1} \mathbb{P}(S)} T_{\Omega}^{l+1} \mathbb{P}$. Moreover, we have $\frac{1(S)}{T_{\Omega}^{l+1} \mathbb{P}(S)} T_{\Omega}^{l+1} \mathbb{P} = \frac{1(S)}{\mathbb{P}(S)}\mathbb{P}$ because $\mathbb{P}$ is $T$-invariant. We conclude by integrating over all possible values for $l \ge 0$.

By (66) and because $\gamma_{2l+1}$ is independent of $\mathcal{P}_l \lor (T_{\Omega}^{-l-1})^*\mathcal{A}_{\Omega}$ by construction of $\mathbb{P}$, we know that the conditional distribution of $\gamma_{2l+1}$ with respect to $(\overline{p}_{2k+1} = 2l + 1)$ and $\mathcal{P}_l \lor (T_{\Omega}^{-l-1})^*\mathcal{A}_{\Omega}$ is almost surely equal to its conditional distribution with respect to $A_l$, which is characterized by (79). We conclude by integrating over all possible values of $l$ to prove (79).

By Lemma 4.11, we have $p_{2k+2+i} = p_i \circ T_{\Omega}^{\frac{\overline{p}_{2k+2}}{2}}$ for all $k, i \ge 0$. From that we deduce that $\tilde{\gamma}_{2k+2+i}^p = \tilde{\gamma}_{2k+2+i}^p \circ T_{\Omega}^{\frac{\overline{p}_{2k+2}}{2}}$. Therefore, the data of $(\tilde{\gamma}_{2k+2+i}^p)_{i \ge 0}$ is independent of $\mathcal{P}_k'$ and its distribution is $(\otimes_{i \ge 0} \tilde{\gamma}_i^p)_* (\frac{1(S)}{\mathbb{P}(S)}\mathbb{P})$. This distribution odes not depend on $k$ so the sequence $(\tilde{\gamma}_{2k+2}^p)_{k \ge 0}$ is identically distributed. Moreover, for all $k \ge 0$, the data of $(\tilde{\gamma}_{2k}^p)_{k' \le k}$ is $\mathcal{P}_k'$-measurable so it is independent of $\tilde{\gamma}_{2k+2}^p$. Therefore, the random variables $(\tilde{\gamma}_{2k}^p)_{k \ge 0}$ are globally independent. $\square$

Finally, we prove the recursive alignment property, which implies points (9) and (10).

**Lemma 4.15.** With the Notations of Definitions 4.9 and 4.10. For all $k \ge 0$, we have $\gamma_{2k+1}^p \mathbb{A} \gamma_{\overline{p}_{2k+2}}$ and there is a measurable function defined on $\Omega'$ and taking values in the space of finite families of odd integers $1 = c_1^k < c_2^k \dots < c_{jk}^k = p_{2k+2}$
such that for all $1 \le i < j_k$, we have:

$$ (80) \qquad \gamma \bar{p}_{2k+2} \cdots \gamma \bar{p}_{2k+2} + c_i^k - 1 \mathbb{A} \gamma \bar{p}_{2k+2} + c_i^k \mathbb{A} \gamma \bar{p}_{2k+2} + c_i^k + 1 \cdots \gamma \bar{p}_{2k+2} + c_{i+1}^k - 1. $$

*Proof.* Let $l_k := \frac{\bar{p}_{2k+1}-1}{2} = \max\{j | m_j \le k\}$ and $j_k := \#\{j > l_k | m_j = k+1\}$. Both these integers are well defined and measurable on $\Omega' = (m_j \to +\infty)$. Moreover $j_k > 0$ because $m_{l_k+1} = k+1$. We define by induction $b_1^k := l_k+1$ and:

$$ b_{i+1}^k := \min\{j > b_i^k | m_j = k+1\}. $$

Let $c_i^k = 2b_i^k + 1 - \bar{p}_{2k+2}$ for all $1 \le i \le j_k$. For $i=0$, we have $A_{l_k}$ by (66) in Lemma 4.11 so $\gamma_{2k+1}^p \mathbb{A} \gamma_{\bar{p}_{2k+2}}$.

Let $1 \le i < j_k$. By maximality of $l_k$, we have $m_{b_i^{k+1}} \ge k+1 = m_{b_i^k}$.

Therefore, we have $A_{b_i^k}$ so $\gamma_{2k+2}^{p b_i^k} \mathbb{A} \gamma_{2b_i^{k+1}}$. Then by (65), and because $b_i^k \ge l_k$, we have $\bar{p}_{2k+2}^{b_i^k} = \bar{p}_{2k+2}$ so $\gamma_{2k+2}^{p b_i^k} = \gamma_{\bar{p}_{2k+2}} \cdots \gamma_{2b_i^k} = \gamma_{\bar{p}_{2k+2}} \cdots \gamma_{\bar{p}_{2k+2}+c_i^k-1}$. This proves the first alignment condition in (80).

Since $m_{b_{i+1}^k} = k+1$ and $m_{b_{i+1}^k-1} > k$, we can not have $m_{b_{i+1}^k} > m_{b_{i+1}^k-1}$.

Therefore $m_{b_{i+1}^k} < m_{b_{i+1}^k-1}$. Hence, we have $\gamma_{2k+3}^{p b_{i+1}^k-1} \mathbb{A} \gamma_{\bar{p}_{2k+4}}^{b_{i+1}^k-1} \cdots \gamma_{2b_i^k}$. Moreover, we have $m_j > k+1$ for all $b_i^k < j < b_{i+1}^k$ so $\bar{p}_{2k+3}^{b_{i+1}^k-1} = 2b_i^k + 1$ by (65). Therefore $\gamma_{\bar{p}_{2k+4}}^{b_{i+1}^k-1} \cdots \gamma_{2b_i^k} = \gamma_{\bar{p}_{2k+2}+c_i^k+1} \cdots \gamma_{\bar{p}_{2k+2}+c_{i+1}^k-1}$. This proves the second alignment condition in (80). $\square$

Now to prove the main theorem, we consider a random sequence $(p_k)_{k \ge 0}$ given by applying the pivot algorithm to the random sequence $(\gamma_n)_{n \ge 0} = (\gamma_n^{\hat{v}})_{n \ge 0}$. The above results are all we need to prove points (5) to (11). Let us give more details.

*Proof of Theorem 4.7.* Let $(\hat{v}_k)_{k \ge 0}$ be as in Lemma 4.8 and let $\hat{v} = \hat{v}^{(3,1)^{\otimes N}}$. Let $(\tau_k)_{k \ge 0} \sim \mathcal{U}^{\otimes N}$ be independent of the joint data of $(s_k)_k$, $(u_k)_k$, $(w_k)_k$ and $(v_k)_k$ and therefore independent of $(\hat{\gamma}_k^{\hat{v}})_{k \ge 0}$. Let $\nu_u$ be the distribution law of $\gamma_0^{\hat{v}}$. Then by Lemma 4.8, we have $(\gamma_k^{\hat{v}})_{k \ge 0} \sim (\nu_u \otimes \nu_s)^{\otimes N}$. Let $(\tau_k)_{k \ge 0} \in [0, 1]^N$ be uniformly distributed, globally independent and independent of Let $(p_k)_{k \ge 0}$ be the length of the pivotal blocks associated to the sequence $(\gamma_k^{\hat{v}})_{k \ge 0}$ with weights $(\tau_k)_{k \ge 0}$. We have already proven points (1) to (4) in Lemme 4.8. Let us now prove points (5) to (11).

We know by Corollary 4.13 that the data of $(p_k)_{k \ge 0}$ is independent of $(\gamma_{2k}^{\hat{v}})_{k \ge 0}$. Moreover, by Lemma 4.8, the joint data of $(\hat{\gamma}_{2k}^{\hat{v}})_{k \ge 0}$, $(v_k)_{k \ge 0}$, $(w_k)_{k \ge 0}$
and $(u_k)_{k \ge 0}$ determines $(\gamma_{2k}^{\hat{v}})_{k \ge 0}$ and is independent of the joint data of $(\gamma_{2k+1}^{\hat{v}})_{k \ge 0}$ and $(\tau_k)_{k \ge 0}$ so its conditional distribution with respect to the joint data of $(\gamma_k^{\hat{v}})_{k \ge 0}$ and $(\tau_k)_{k \ge 0}$ only depends on $(\gamma_{2k}^{\hat{v}})_{k \ge 0}$. Note also that the sequence $(p_k)_k$ is given by a measurable function of $(\gamma_k^{\hat{v}})_{k \ge 0}$ and $(\tau_k)_{k \ge 0}$. Therefore the joint data of $(\tilde{\gamma}_{2k}^{\hat{v}})_{k \ge 0}$, $(v_k)_{k \ge 0}$ and $(w_k)_{k \ge 0}$ is independent of $(p_k)_{k \ge 0}$. This proves (5).

Let us now prove points (6) and (8). For all $k$, we have $p_{2k+1} = 1$ by construction and $\bar{p}_{2k+1}$ is odd by (64) in Lemma 4.11. For $k = 0$, we have $p_0 = \bar{p}_1$ which is odd. For $k \ge 1$, we have $p_{2k} = \bar{p}_{2k+1} - \bar{p}_{2k-1} - 1$, which is odd. So all the $p_k$'s are odd and therefore positive. Then by Lemma 4.14, we have (8), which states that $(\tilde{\gamma}_{2k+2}^p)_k$ is i.i.d. and independent of $\tilde{\gamma}_0^p$ then taking the push-forward by $L^{\otimes N}$, we get that $(p_{2k+2})_k$ is i.i.d. and independent of $p_0$.

Point (7), follows directly from (78) in Corollary 4.13 because $\bar{p}_3 = p_0 + p_2 + 1$.

Points (9) and (10) follow directly from Lemma 4.15.

Point (11) follows from (79) in Lemma 4.14 and from the fact that the conditional distribution of $\tilde{\gamma}_{2k+1}^p$ with respect to $(\gamma_k^{\hat{v}})_{k \ge 0}$ and $(\tau_k)_{k \ge 0}$ only depends on $\gamma_{2k+1}^p$. $\square$

## 5. PROOF OF THE RESULTS

In this section, we use Theorem 1.6 to prove the other results of this paper. In this discussion, $(E, \|\cdot\|)$ is a Euclidean, Hermitian or ultra-metric vector space over a local field $(\mathbb{K}, |\cdot|)$. Unless specified otherwise, $\nu$ is a strongly irreducible an proximal probability distribution over $\text{End}(E)$. The constants $0 < \varepsilon \le 1/2$ and $m$ are as in Theorem 1.6. On a probability space $(\Omega, \mathcal{P})$, we define two random sequences $(\gamma_n)_{n \ge 0} \sim \nu^{\otimes N}$ and $(p_n)_{n \ge 0}$ that satisfy the conclusion of Theorem 1.6.

Point (3) tells us that $p_{2k+1} = m$ almost surely and for all $k \ge 0$. So one should keep in mind that the data of the sequence $(p_n)_{n \ge 0}$ is determined by the data of $(\tilde{\gamma}_{2k}^p)_{k \ge 0}$.

### 5.1. Preliminaries about one-sided large deviations.
In this Paragraph we give statements and proofs of folklore results about sequences that satisfy large deviations inequalities.

**Definition 5.1 (One-sided large deviations).** Let $(\bar{x}_n) \in \mathbb{R}^N$ be a random sequence and $\lambda \in \mathbb{R} \cup \{+\infty\}$. We say that $(\bar{x}_n)$ satisfies large deviations inequalities below the speed $\lambda$ if:

$$ (81) \qquad \forall \alpha < \lambda, \exists C, \beta > 0, \forall n, \mathbb{P}(\bar{x}_n \le \alpha n) \le Ce^{-\beta n}. $$
Note that (81) admits the following equivalent reformulations:

$$
\begin{align*}
& \forall \alpha < \lambda, \exists C, \beta > 0, \forall n_0, \mathbb{P}(\exists n \ge n_0, \bar{x}_n \le \alpha n) \le Ce^{-\beta n}, \\
& \forall \alpha < \lambda, \exists n_0, \beta > 0, \forall n \ge n_0, \mathbb{P}(\bar{x}_n \le \alpha n) \le e^{-\beta n}, \\
& \forall \alpha < \lambda, \limsup_{n_0 \to +\infty} \log \mathbb{P}(\exists n \ge n_0, \bar{x}_n \le \alpha n) < 0, \\
& \forall \alpha < \lambda, \exists \beta > 0, \mathbb{E}(\exp(\beta \max\{n | \bar{x}_n \le \alpha n\})) < +\infty.
\end{align*}
$$

**Lemma 5.2 (Classical large deviations inequalities).** Let $(x_n) \in \mathbb{R}^N$ be a random i.i.d. sequence and let $\beta_0 > 0$ be a constant. Assume that $\mathbb{E}(e^{-\beta_0 x_0}) < +\infty$. Then the random sequence $(\bar{x}_n)_n$ satisfies large deviations inequalities below the speed $\mathbb{E}(x_0) \in \mathbb{R} \cup \{+\infty\}$.

*Proof.* The first step is to prove the following:

$$
(82) \qquad \forall \alpha < \mathbb{E}(x_0), \exists \beta > 0, \mathbb{E}(e^{-\beta x_0}) < e^{-\beta \alpha}.
$$

We prove it using the monotonous convergence theorem. For all $t \in \mathbb{R}$, the map $\beta \mapsto \frac{1-e^{-\beta t}}{\beta}$ is non-increasing on $(0, +\infty)$ and $\lim_{\beta \to 0} \frac{1-e^{-\beta t}}{\beta} = t$. Moreover $\frac{1-e^{-\beta x_0}}{\beta}$ is in $L^1$, therefore $\mathbb{E}(\frac{1-e^{-\beta x_0}}{\beta})$ is non-increasing and admits $\mathbb{E}(x_0)$ as its limit. Hence, for all $\alpha < \mathbb{E}(x_0)$ there exists $\beta > 0$ such that $\mathbb{E}(\frac{1-e^{-\beta x_0}}{\beta}) > \alpha$, and by linearity $\mathbb{E}(e^{-\beta x_0}) < 1 - \beta\alpha \le e^{-\beta\alpha}$. This proves (82).

Let $\alpha < \mathbb{E}(x_0)$ and let $\beta > 0$ be as in (82). By independence, we have
$\mathbb{E}(e^{-\beta \bar{x}_n}) = \mathbb{E}(e^{-\beta x_0})^n$ for all $n$. Then by Markov inequality, we have:

$$
\mathbb{P}(\bar{x}_n \le \alpha n) \le \frac{\mathbb{E}(e^{-\beta \bar{x}_n})}{e^{\beta \alpha n}} = \left( \frac{\mathbb{E}(e^{-\beta x_0})}{e^{\beta \alpha}} \right)^n . \quad \square
$$

**Lemma 5.3.** Let $(\bar{x}_n)_{n \ge 0} \in \mathbb{R}^N$ and $(l_n)_{n \ge 0} \in \mathbb{N}^N_{\ge 0}$ be random sequences and let $\lambda \le +\infty$ and $C_0, \beta_0 > 0$ be constants. Assume that $(\bar{x}_n)_n$ satisfies large deviations inequalities below the speed $\lambda$ and that $\mathbb{E}(e^{\beta_0 l_n}) \le C_0$ for all $n$. Then $(\bar{x}_n - l_n)_n$ satisfies large deviations inequalities below the speed $\lambda$.

*Proof.* Let $\alpha < \alpha' < \lambda$. By assumption, there exist constants $C', \beta' > 0$ such that $\mathbb{P}(\bar{x}_n \le \alpha' n) \le C' e^{-\beta' n}$ for all $n$. Let $C', \beta'$ be such constants. Moreover, by Markov's inequality, we have $\mathbb{P}(l_n \ge (\alpha' - \alpha)n) \le C_0 e^{-\beta_0(\alpha' - \alpha n)}$ for all $n$. Therefore, we have:

$$
\forall n, \mathbb{P}(\bar{x}_n - l_n \le \alpha n) \le C_0 e^{-\beta_0(\alpha' - \alpha n)} + C' e^{-\beta' n}. \quad \square
$$
**Lemma 5.4.** Let $(\bar{x}_n)_{n \ge 0} \in \mathbb{R}^N$ and $(\bar{w}_n)_{n \ge 0} \in \mathbb{N}^N$ be random sequences and let $0 \le \delta, \lambda \le +\infty$ be constants. Assume that $(\bar{x}_n)_n$ satisfies large deviations inequalities below the speed $\lambda$ and that $(\bar{w}_n)_n$ satisfies large deviations inequalities below the speed $\delta$. Then $(\bar{x}_{\bar{w}_n})_n$ satisfies large deviations inequalities below the speed $\lambda\delta$.

*Proof.* Let $\alpha < \lambda\delta$ and let $0 < \alpha_0 < \lambda$ and $0 < \alpha_1 < \delta$ be such that $\alpha \le \alpha_0\alpha_1$. Let $C_0, \beta_0, C_1, \beta_1 > 0$ be such that $\mathbb{P}(\bar{x}_n \le \alpha_0 n) \le C_0 e^{-\beta_0 n}$ and $\mathbb{P}(\bar{w}_n \le \alpha_1 n) \le C_1 e^{-\beta_1 n}$ for all $n \ge 0$. Then for all $n \ge 0$, we have:

$$ \begin{aligned} \mathbb{P}(\bar{x}_{\bar{w}_n} \le \alpha n) &\le \mathbb{P}(\bar{w}_n \le \alpha_1 n \cup \exists k \ge \alpha_1 n, \bar{x}_k \le \alpha_0 k) \\ &\le C_1 e^{-\beta_1 n} + \sum_{k \ge \alpha_1 n} C_0 e^{-\beta_0 k} \le C_1 e^{-\beta_1 n} + \frac{C_0}{1 - e^{-\beta_0}} e^{-\beta_0 \alpha_1 n}. \quad \square \end{aligned} $$

**Lemma 5.5.** Let $(\bar{x}_n)_{n \ge 0}$ be a random sequence of non-negative numbers that satisfies large deviations inequalities above a speed $\lambda$. Let $w$ be a random non-negative integer that has a finite exponential moment. Then without any independence assumption, the random variable $\bar{x}_w$ has a finite exponential moment.

*Proof.* Let $t \ge 0$, we have $(\bar{x}_w \ge t) \subset (\bar{x}_{\lfloor t/(2\lambda) \rfloor} \ge t) \cup (w > \lfloor t/(2\lambda) \rfloor)$. The probability of the first term decreases exponentially fast in $t$ by the large deviations inequalities and the probability decreases exponentially fast in $t$ because $w$ has a finite exponential moment. $\square$

## 5.2. Law of large numbers and large deviations inequalities for the pivot extraction.

In this paragraph we prove the following Lemma. This is the only point where we use Kingman's ergodic Theorem. However the use is non-obvious since $\sigma$ is not sub multiplicative in general.

In this paragraph, we use points (2) and (4) from Theorem 1.6 in a non-obvious way. One should remember that these points are a reformulation of the fact that for all $k \ge 0$, the random sequence $(\tilde{\gamma}_{j+2k+2}^p)_{j \ge 0}$ is independent of $(\tilde{\gamma}_j^p)_{j \le 2k}$ and its distribution does not depend on $k$.

**Lemma 5.6 (Escape speed and large deviations inequalities for the aligned extraction).** Let $(\gamma_n)_{n \ge 0}$ be a random sequence in End($E$). Let $(p_k)_{k \ge 0}$ be a random sequence of integers. Let $0 < \varepsilon \le 1/2$. Assume that points (2), (3) and (4) from Theorem 1.6 hold. Then there exists a constant $\lambda' \in (0, +\infty]$ such that for all $i_0 \ge 1$, we have almost surely:

$$ (83) \qquad \lim_{j \to +\infty} \frac{-\log \sigma (\gamma_{2i_0}^p \cdots \gamma_{2j}^p)}{j - i_0} = \lambda'. $$
Moreover, for all $\alpha < \lambda'$, there exist constant $C, \beta > 0$ such that for all $i \le j$, we have:

$$ (84) \qquad \mathbb{P}(-\log \sigma(\gamma_{2i}^p \cdots \gamma_{2j}^p) \le \alpha(j-i)) \le Ce^{-\beta(j-i)}. $$

*Proof.* First we claim that for all $i \le j < k$, we have the following sub-additivity result:

$$ (85) \qquad \log \sigma(\gamma_{2i}^p \cdots \gamma_{2k}^p) \le \log \sigma(\gamma_{2i}^p \cdots \gamma_{2j}^p) + \log \sigma(\gamma_{2j+2}^p \cdots \gamma_{2k}^p) - \log(3). $$

This is a consequence of Lemma 2.11 with $f = \gamma_{2i}^p \cdots \gamma_{2j}^p$, $g = \gamma_{2j+1}^p$ and $h = \gamma_{2j+2}^p \cdots \gamma_{2k}^p$. By (3), we have $\sigma(g) \le \varepsilon^6/48$ and by (8), we have $fA^{\frac{\varepsilon^6}{2}}gA^{\frac{\varepsilon^6}{2}}h$ so $\sigma(fgh) \le \sigma(f)\frac{\varepsilon^6}{48}\sigma(h)\frac{64}{\varepsilon^4}$ and $\varepsilon \le \frac{1}{2}$ so $4\varepsilon^2/3 \le 1/3$, which proves (85).

By (2) and (4), the map $T: \Gamma^N \to \Gamma^N; (g_k)_{k \ge 0} \mapsto (g_{k+2})_{k \ge 0}$ preserves the measure $\mu$ defined as $(\gamma_{k+2}^p)_{k \ge 0} \sim \mu$. We claim that $\mu$ is in fact ergodic for $T$. Simply note that for all $i > 1$, the data of $(\gamma_k^p)_{2 \le k \le 2i}$ is independent of $(\gamma_k^p)_{k \ge 2i+2} = T^i((\gamma_k^p)_{k \ge 2})$. Let $A' \subset \Gamma^N$ be $T$-invariant and let $A = ((\gamma_k^p)_{k \ge 2} \in A')$, that way $\nu(A') = \mathbb{P}(A)$. We claim that $\mu(A') \in \{0, 1\}$, which is equivalent to saying that $A$ is independent of itself. Then for all $i > 0$, we have $A = ((\gamma_k^p)_{k \ge 2} \in T^{-i}(A')) = ((\gamma_k^p)_{k \ge 2i+2} \in A')$ Therefore for all $i > 0$, $A$ is independent of the data of $(\gamma_k^p)_{2 \le k \le 2i}$ is independent of $A$. Hence $A$ is independent of the joint data of $(\gamma_k^p)_{k \ge 2}$ and therefore of itself, which proves that $\mu$ is ergodic for $T$.

For all $n \ge 0$, let $\phi_n := -\log \sigma(\gamma_2^p \cdots \gamma_{2n+2}^p)$. Then for all $m, n \ge 0$ by (85) with $i = 1, j = m+1$ and $k = m+n+1$, we have $\phi_{m+n} \ge \phi_m + \phi_n \circ T^m$. Note also that $\phi_n \ge 0$ for all $n$. Therefore, by Kingman's sub-additive ergodic Theorem[Kin68], there exists a constant $\lambda' = \sup \frac{\mathbb{E}(\phi_n)}{n} = \lim \frac{\mathbb{E}(\phi_n)}{n} \in [0, +\infty]$ such that $\frac{\phi_n}{n} \to \lambda'$ almost surely. Not also that $\lambda' > 0$ because $\phi_1 \ge \log(3)$ by (85).

Let us now prove (84). Let $\alpha < \lambda'$ and let $n \in \mathbb{N}$ be such that $\alpha < \lambda_n = \frac{\mathbb{E}(-\log \sigma(\gamma_2^p \cdots \gamma_{2n}^p))}{n}$. For all $k, r \ge 0$, and by (85), we have:

$$ (86) \qquad -\log \sigma(\gamma_2^p \cdots \gamma_{2kn+r}^p) \le \sum_{j=0}^{k-1} -\log \sigma(\gamma_{2jn+2}^p \cdots \gamma_{2(j+1)n}^p). $$

Moreover, for all $i$, the data of $(\gamma_k)_{k \le 2i}$ is independent of the data of $(\gamma_k)_{k \ge 2i+2}$ so the sequence $(-\log \sigma(\gamma_{2jn+2}^p \cdots \gamma_{2(j+1)n}^p))_{j \ge 0}$ is i.i.d. hence, by Lemma 5.2, for all $\alpha' < \mathbb{E}(-\log \sigma(\gamma_2^p \cdots \gamma_{2n}^p))$ there exist constants $C, \beta > 0$ such that:

$$ \forall k \in \mathbb{N}, \mathbb{P}\left(\sum_{j=0}^{k-1} -\log \sigma(\gamma_{2jn+2}^p \cdots \gamma_{2(j+1)n}^p) \le \alpha'k\right) \le Ce^{-\beta k}. $$
Then by (86), we have:

$$ \forall m, \mathbb{P}(-\log \sigma(\gamma_2^p \cdots \gamma_{2m}^p) \le \alpha'[m/n]) \le Ce^{-\beta[m/n]} \le Ce^{\beta}e^{-\frac{\beta}{n}m}. $$

Moreover, we may assume that $n\alpha < \alpha'$ hence, there exists $m_0$ such that $\alpha m \le \alpha'[m/n]$ for all $m \ge m_0$, which concludes the proof. □

**5.3. The bi-lateral pivoting technique.** In this Paragraph, we use Point (5) from Theorem 1.6. We remind that (5) says that for all $k \ge 0$, the conditional distribution of $\gamma_{2k+1}^p$ with respect to $(\tilde{\gamma}_{k'}^p)_{k' \ne 2k+1}$ is almost surely $\frac{1}{4}$-Schottky for the binary relation $\mathbb{A}^\epsilon$. It is equivalent to saying that for all random $g$, which is given by a measurable function of $(\tilde{\gamma}_{k'}^p)_{k' \ne 2k+1}$, we have $\mathbb{P}(\gamma_{2k+1}^p \mathbb{A}^\epsilon g | (\tilde{\gamma}_{k'}^p)_{k' \ne 2k+1}) \ge 3/4$ and $\mathbb{P}(g \mathbb{A}^\epsilon \gamma_{2k+1}^p | (\tilde{\gamma}_{k'}^p)_{k' \ne 2k+1}) \ge 3/4$ almost surely.

Let us now describe the left, the right and the cyclical pivoting techniques, which respectively allow us to control the behaviour of the lines, of the columns and of the top eigenspace of $\bar{\gamma}_n$.

**Lemma 5.7 (Pivoting technique).** Let $(\gamma_n)_{n \ge 0}$ be a random sequence in $\text{End}(E)$. Let $(p_k)_{k \ge 0}$ be a random sequence of integers such that $p_{2k+1}$ is equal to a non-random positive constant for all $k$. Let $(\tau_k)_{k \ge 0}$ be independent an uniformly distributed and independent of the joint data of $(\tilde{\gamma}_k^p)_{k \ge 0}$. Let $0 < \epsilon \le 1/2$. Assume that points (2), (5) and (6) from Theorem 1.6 hold. For all $f \in E^* \cup \text{End}(E)$, there exists a random integer $l^f \sim \mathcal{G}_{1/4}$, that is independent of $(\tilde{\gamma}_{2k}^p)_{k \ge 0}$ and such that:

$$ (87) \qquad f \gamma_0^p \cdots \gamma_{2lf}^p \mathbb{A}^\epsilon \gamma_{2lf+1}^p $$

Let $h \in E \cup \text{End}(E)$. For all $n \in \mathbb{N}$, let $q_n = \max\{k \ge 0 | \bar{p}_{2k} \le n\}$. Then for all integer $n$, there exists two random integer $r_n^h \sim \mathcal{G}_{1/4}$ and $c_n \sim \mathcal{G}_{1/2}$, that are independent of $(\tilde{\gamma}_{2k}^p)_{k \ge 0}$ and such that:

$$ (88) \qquad \gamma_{2q_n-2r_n^h-1}^p \mathbb{A}^\epsilon \gamma_{\bar{p}_{2q_n-2r_n^h}} \cdots \gamma_{n-1}^p h \quad \text{or} \quad r_n^h \ge q_n, $$

$$ (89) \qquad \left\{ \begin{array}{l} \gamma_{2q_n-2c_n-1}^p \mathbb{A}^\epsilon \gamma_{\bar{p}_{2q_n-2c_n}} \cdots \gamma_{n-1}^p \gamma_0^p \cdots \gamma_{2c_n}^p \\ \text{and } \gamma_{\bar{p}_{2q_n-2c_n-1}}^p \cdots \gamma_{n-1}^p \gamma_0^p \cdots \gamma_{2c_n}^p \mathbb{A}^\epsilon \gamma_{2c_n+1}^p \end{array} \right. \quad \text{or} \quad 2c_n+1 \ge q_n, $$

Moreover, there exist constants $C, \beta > 0$ such that $\mathbb{E}(e^{\beta \bar{p}_{2lf}}) \le C$ for all $f$ and $\mathbb{E}(e^{\beta(n-\bar{p}_{2q_n-2r_n^h})}) \le C$, for all $n$ and all $h$ and $\mathbb{E}(e^{\beta(n-\bar{p}_{2q_n-2c_n-2}+\bar{p}_{2c_n})}) \le C$.
*Proof.* Let $f \in E^* \cup \text{End}(E)$. We construct $l^f$ as:

$$l^f := \min \left\{ k \in \mathbb{N} \mid \tau_k < \frac{3}{4} \frac{\mathbb{1}\left(f\gamma_0^p \cdots \gamma_{2lf}^p A^\epsilon \gamma_{2lf+1}^p\right)}{\mathbb{P}\left(f\gamma_0^p \cdots \gamma_{2lf}^p A^\epsilon \gamma_{2lf+1}^p \middle| (\tilde{\gamma}_{2k}^p)_{k \ge 0}\right)} \right\}.$$

Then we have (87). By (5), we have $\mathbb{P}(f\gamma_0^p \cdots \gamma_{2k}^p A^\epsilon \gamma_{2k+1}^p \mid (\tilde{\gamma}_{k'}^p)_{k' \ne 2k+1}) \ge 3/4$ for all $k$, so $\mathbb{P}(l^f = k \mid (\tilde{\gamma}_{k'}^p)_{k' \ne 2k+1}, l^f \ge k) = \frac{3}{4}$. Therefore, the conditional distribution of $l^f$ with respect to $(\tilde{\gamma}_{2k}^p)_{k \ge 0}$ is $\mathcal{G}_{1/4}$. Moreover the $p_k$'s have a bounded exponential moment and are independent so $\bar{p}_{2lf}$ has a finite exponential moment by Lemmas 4.2 4.4.

Let $h \in E \cup \text{End}(E)$ and let $n \in \mathbb{N}$. Let us construct $r_n^h$. Since $(p_{2k+1})_{k \ge 0}$ is non-random, the data of $q_n$ is measurable for the $\sigma$-algebra generated by $(\tilde{\gamma}_{2k}^p)_{k \ge 0}$. For all $k \ge 0$, we write:

$$A_{n,k}^h = (k \ge q_n) \cap \left( \tau_k < \frac{3}{4} \right) \\ \cup (k < q_n) \cap \left( \tau_k < \frac{3}{4} \frac{\mathbb{1}\left(\gamma_{2q_n-2r_n^h-1}^p A^\epsilon \gamma_{\bar{p}_{2q_n-2r_n^h}} \cdots \gamma_{n-1} h\right)}{\mathbb{P}\left(\gamma_{2q_n-2k-1}^p A^\epsilon \gamma_{\bar{p}_{2q_n-2k}} \cdots \gamma_{n-1} h \mid (\tilde{\gamma}_{2j}^p)_{j \ge 0}\right)} \right)$$

Then by (5), we have $\mathbb{P}(A_{n,k}^h \mid q_n, (\tilde{\gamma}_j^p)_{j \ne 2q_n-2k-1}, (\tau_j)_{j \ne k}) = \frac{3}{4}$. So the $A_{n,k}^h$ are globally independent and independent of $(\tilde{\gamma}_{2j}^p)_{j \ge 0}$. Now let $r_n^h = \min\{k \ge 0 \mid A_{n,k}^h\}$, then the conditional distribution of $r_n^h$ with respect to $(\tilde{\gamma}_{2k}^p)_{k \ge 0}$ is equal to $\mathcal{G}_{1/4}$.

First, we claim that we have $\mathbb{P}(n - \bar{p}_{2q_n-2k} \ge t) \le t\mathbb{P}(\bar{p}_{2k+2} \ge t)$ for all $k, n, t \ge 0$. Let $k, n, t \ge 0$. We have:

$$\begin{aligned} \mathbb{P}(n - \bar{p}_{2q_n-2k} = t) &= \sum_{i \ge 0} \mathbb{P}(\bar{p}_{2i} = n - j \cap \bar{p}_{2i+k} \le n \cap \bar{p}_{2i+2k+2} > n) \\ &\le \sum_{i \ge 0} \mathbb{P}(\bar{p}_{2i} = n - j \cap \bar{p}_{2i+2k+2} > n) \end{aligned}$$

Hence, by independence of the sequence $(p_i)_{i \ge 0}$, the events $(\bar{p}_{2i} = n - t)$ and $(p_{2i} + \cdots + p_{2i+2k+1} > j)$ are independent for all $i, t, n$. Note also that the sequence $\bar{p}_{2i}$ is almost surely increasing, therefore $\sum_{i \ge 0} \mathbb{P}(\bar{p}_{2i} = n - t) \le 1$ for all
$n, t$. Therefore, we have:

$$
\begin{align*}
\mathbb{P} (n - \bar{p}_{2q_n-2k} = t) &\le \sum_{i \ge 0} \mathbb{P} (\bar{p}_{2i} = n-j) \mathbb{P} (p_{2i} + \dots + p_{2i+2k+1} > j) \\
&\le \max_i \mathbb{P} (p_{2i} + \dots + p_{2i+2k+1} > t)
\end{align*}
$$

Let $C, \beta > 0$ be such that $\mathbb{E}(e^{\beta p_{2k}}) \le C$ for all $k \ge 0$. We remind that $p_{2k+1} = m$
almost surely and for all $k$ by (3). Then by independence, for all $i, k \ge 0$, we
have $\mathbb{E}(e^{\beta(p_{2i}+\cdots+p_{2i+2k+1})}) \le (e^{\beta m}C)^k$ and by Markov's inequality, we have:

$$
\forall i, j, k \ge 0, \mathbb{P} \left( p_{2i} + \dots + p_{2i+2k+1} \ge j + k \frac{\beta m + \log(2C)}{\beta} \right) \le 2^{-k} e^{-\beta j}.
$$

As a consequence, for $C' = m + \frac{\log(2C)}{\beta}$, we have:

$$
\mathbb{P} (n - \bar{p}_{2q_n - 2k} \geq kC') \leq \frac{2^{-k}}{1 - e^{-\beta}}
$$

Moreover, by independence of $r_n^h$ with $(p_i)_{i \ge 0}$, we have:

$$
\begin{align*}
(n - \bar{p}_{2q_n - 2r_n^h} \ge t) &= \sum_{k \ge 0} \mathbb{P}(n - \bar{p}_{2q_n - 2k} \ge t) \mathbb{P}(r_n^h = k) \\
&\le \mathbb{P}(r_n^h \ge \lfloor t/C' \rfloor) + \mathbb{P}(n - \bar{p}_{2q_n - 2\lfloor t/C' \rfloor} \ge t) \\
&\le 4^{-\lfloor t/C' \rfloor} + 2^{-\lfloor t/C' \rfloor}.
\end{align*}
$$

Let us now construct $c_n$. We use the same technique as for the construction
of $l^f$ and $r_n^h$. For all $n, k \ge 0$, we write:

$$
S'_{n,k} = (2k+1 \le q_n) \cup (2k+1 < q_n) \cap \left( \gamma_{2q_n-2k-1}^p A^\epsilon \gamma_{\bar{p}_{2q_n-2k}} \cdots \gamma_{n-1} \gamma_0^p \cdots \gamma_{2k}^p \right) \\
\cap \left( \gamma_{\bar{p}_{2q_n-2k-1}} \cdots \gamma_{n-1} \gamma_0^p \cdots \gamma_{2k}^p A^\epsilon \gamma_{2k+1}^p \right).
$$

Then for all $n, k$, by (5), we have the rough estimate:

$$
\mathbb{P} \left( S'_{n,k} \mid (\tilde{\gamma}_{k'}^p)_{k' \notin \{2k+1, 2q_n-2k-1\}}, 2k+1 < q_n \right) \ge \frac{1}{2}
$$

Now for all $n, k$ we write:

$$
S_{n,k} = \left( \tau_k < \frac{1(S'_{n,k})}{2\mathbb{P}\left( S'_{n,k} \mid q_n, (\tilde{\gamma}_{k'}^p)_{k' \notin \{2k+1, 2q_n-2k-1\}} \right)} \right)
$$
and define Let $c_n := \min\{k \ge 0 | S_{n,k}\}$ for all $n$. That way, we have $\mathbb{P}(S_{n,k})^{1/2}$ for all $n, k$. Moreover, the $(S_{n,k})_{k \ge 0}$ are independent and their joint data is independent of $(\tilde{\gamma}_{2j}^p)_{j \ge 0}$. Therefore $c_n \sim \mathcal{G}_{1/2}$ and $c_n$ is independent of $(\tilde{\gamma}_{2j}^p)_{j \ge 0}$. With the same proof as for $r_n^h$, we have:

$$ \forall t \ge 0, \quad (n - \bar{p}_{2q_n-2c_n-2} \ge t) \le 2^{1-\lfloor t/C' \rfloor} + 2^{-\lfloor t/C' \rfloor}. $$

so $n - \bar{p}_{2q_n-2c_n-2}$ has a bounded exponential moment that does not depend on $n$ and by Lemmas 4.2 and 4.4, $\bar{p}_{2c_n}$ also has a bounded exponential moment that does not depend on $n$. Therefore, their sum also has, which concludes the proof. $\square$

Note that in Lemma 5.7, we treat $\mathbb{A}^\epsilon$ as an abstract binary relation, so it could be reformulated using the abstract notations of Theorem 4.7. In fact all of the results of this article are consequences of analogous results in term of abstract alignment. These results, though interesting by themselves, are needlessly technical to formulate for the purpose of the present article.

Given $i$ and $j$ two random integers and $(\gamma_n)$ a random sequence, we use the convention $\gamma_i \cdots \gamma_j = \text{Id}$ on the set $(j < i)$.

**Lemma 5.8.** Let $(\gamma_n)_{n \ge 0} \in \text{End}(E)^\mathbb{N}$ and $(p_n)_{n \ge 0} \in \mathbb{N}^\mathbb{N}$ be random sequences, let $m \ge 1$ and let $0 < \epsilon \le \frac{1}{2}$. Assume that points (2) to (8) in Theorem 1.6 hold. Let $\lambda'$ be as in Lemma 5.6. Let $f \in E^* \cup \text{End}(E)$ and let $h \in E \cup \text{End}(E)$. Let $l^f$, $(q_n)_{n \ge 0}$ and $(r_n^h)_{n \ge 0}$ be as in Lemma 5.7. Let $\lambda := \frac{\lambda'}{m+\mathbb{E}(p_2)}$. Then the random sequences $(-\log \sigma(\gamma_{2lf+2}^p \cdots \gamma_{2q_n-2r_n^h-2}^p))_{n \ge 0}$ and $(-\log \sigma(\gamma_{2c_n+2}^p \cdots \gamma_{2q_n-2c_n-2}^p))_{n \ge 0}$ both satisfy large deviations inequalities below the speed $\lambda$.

*Proof.* First let us show that $(q_n)_{n \ge 0}$ satisfies large deviations inequalities below the speed $\frac{1}{m+\mathbb{E}(p_2)}$. Let $\alpha > m + \mathbb{E}(p_2)$. The sequence $(\bar{p}_{2k})_{k \ge 0}$ is non-decreasing so for all $n$, we have:

$$ \mathbb{P}(q_n \le n/\alpha) = \mathbb{P}\left(\bar{p}_{2\lfloor n/\alpha \rfloor} > n\right) \le \mathbb{P}\left(\bar{p}_{2\lfloor n/\alpha \rfloor} > \alpha\lfloor n/\alpha \rfloor\right) $$

Moreover, the sequence $(p_{2k+2} + m)_{k \ge 0}$ is i.i.d and $p_2$ has a finite exponential moment by Lemma 5.2, the sequence $(\bar{p}_{2k+2} - p_0 - m)$ satisfies large deviations inequalities below the speed $-m - \mathbb{E}(p_2)$ and by Lemma 5.3, the sequence $(\bar{p}_{2k+2})$ also does. Therefore, there exists constants $C, \beta > 0$ such that $\mathbb{P}(\bar{p}_{2k+2} \ge \alpha k) \le Ce^{-\beta k}$ for all $k$. Let $C, \beta$ be such constants and $n \in \mathbb{N}$, for $k = \lfloor n/\alpha \rfloor$, we have:

$$ \mathbb{P}(q_n \le n/\alpha) \le Ce^{-\beta\lfloor n/\alpha \rfloor}. $$
Therefore $(q_n)_{n \ge 0}$ satisfies large deviations inequalities below the speed $\lambda'' := \frac{1}{m+E(p_2)}$. By Lemma 5.3, the sequences $(q_n - r_n^h - 1)_{n \ge 0}$ and $(q_n - c_n - 1)_{n \ge 0}$ also do.

Now let $\alpha < \lambda = \lambda'\lambda''$. Let $0 < \delta < \lambda''/2$ and let $0 < \alpha_1 \le \lambda'$ be such that $\alpha < (\lambda'' - 2\delta)\alpha_1$. By the large deviations inequalities, there exist constants $C_0, \beta_0$ such that the following assertions hold:

$$ (90) \qquad \forall n, \mathbb{P} (q_n - r_n^h - 1 \le (\lambda'' - \delta)n) \le C_0 e^{-\beta_0 n}, $$

$$ (91) \qquad \forall n, \mathbb{P} (q_n - c_n - 1 \le (\lambda'' - \delta)n) \le C_0 e^{-\beta_0 n}. $$

Moreover, since $c_n$ and $l^f$ both follow a geometric law, we have constants $C_0, \beta_0 > 0$ such that the following assertions hold:

$$ (92) \qquad \forall n, \mathbb{P} (l^f + 1 \ge \delta n) \le C_0 e^{-\beta_0 n}, $$

$$ (93) \qquad \forall n, \mathbb{P} (c_n + 1 \ge \delta n) \le C_0 e^{-\beta_0 n}. $$

By Lemma 5.6, there exists constants $C_1, \beta_1 > 0$ such that:

$$ (94) \qquad \forall 1 \le i \le j, \mathbb{P} \left( -\log \sigma \left( \gamma_{2i}^p \cdots \gamma_{2j}^p \right) \le \alpha_1 (i-j) \right) \le C_1 e^{-\beta_1 (j-i)}. $$

Let $C_0, \beta_0 > 0$ and $C_1, \beta_1$ be such that (90), (91), (92), (93) and (94) hold. We have:

$$ (95) \qquad \left( -\log \sigma(\gamma_{2lf+2}^p \cdots \gamma_{2q_n-2r_n^h-2}^p) \le \alpha n \right) \subset \left( q_n - r_n^h - 1 \le (\lambda'' - \delta)n \right) \\ \cup \left( l^f + 1 \ge \delta n \right) \cup \left( \exists i \le \delta n, \exists j \ge \lambda'' - \delta n, -\log \sigma(\gamma_{2i}^p \cdots \gamma_{2j}^p) \le \alpha_1(i-j) \right) $$

Then by (95), we have:

$$ \begin{align*} \mathbb{P} \left( -\log \sigma(\gamma_{2lf+2}^p \cdots \gamma_{2q_n-2r_n^h-2}^p) \le \alpha n \right) &\le 2C_0 e^{-\beta_0} + \sum_{i \le \delta n, j \ge (\lambda'' - \delta)n} C_1 e^{-\beta_1(j-i)} \\ &\le 2C_0 e^{-\beta_0} + \sum_{i \le \delta n} \frac{C_1}{1 - e^{-\beta_1}} e^{-\beta_1((\lambda'' - \delta)n - i)} \\ &\le 2C_0 e^{-\beta_0} + \frac{C_1}{(1 - e^{-\beta_1})^2} e^{-\beta_1(\lambda'' - 2\delta)n}. \end{align*} $$

Therefore, the random sequence $\left(-\log \sigma(\gamma_{2lf+2}^p \cdots \gamma_{2q_n-2r_n^h-2}^p)\right)_n$ satisfies large deviations inequalities below the speed $\lambda$. The exact same reasoning works for $\left(-\log \sigma(\gamma_{2c_n+2}^p \cdots \gamma_{2q_n-2c_n-2}^p)\right)_n$. $\square$
5.4. **Quantitative estimates for the coefficients and spectral radius.** In this section, we use Lemma 5.7 and point (7) in Theorem 1.6 to prove Theorem 1.13. More precisely, we will prove the following results. First let us remind that given $g \in \text{GL}(E)$, we write $N(g) := \log \|g\|\|g^{-1}\|$.

**Theorem 5.9.** Let $\nu$ be a strongly irreducible and proximal probability distribution on $\text{GL}(E)$. Let $(\gamma_n)_{n \ge 0} \sim \nu^{\otimes N}$. Let $f \in E^* \setminus \{0\}$ and let $v \in E \setminus \{0\}$. There exist universal constants $C, D, \beta > 0$ that do not depend on $f$ or $v$ and such that:

$$ (96) \quad \mathbb{P} \left( -\log \left( \frac{|f \bar{\gamma}_n v|}{\|f\| \|\bar{\gamma}_n\| \|v\|} \right) \ge t \right) \le C e^{-\beta n} + \sum_{k \ge 1} C e^{-\beta k} k N_* \nu(t/k - D, +\infty). $$

*Proof.* Let $(p_n)_{n \ge 0}$ and $K \subset \text{GL}(E)$ be as in Theorem 1.6. Let $l_f$ be as in Lemma 5.7 and for all $n$, let $q_n$ and $r_n^v$ be as in Lemma 5.7. Then by Lemma 5.7, there exist constants $C, \beta > 0$ such that for all $k$ and all $n$, we have:

$$ (97) \quad \mathbb{P} \left( n - \bar{p}_{2q_n-2r_n^v} + \bar{p}_{2lf+1} = k \right) \le C e^{-\beta k}. $$

The first issue is that the joint data of $r_n^v$ and $l_f$ is not independent of $(\tilde{\gamma}_{2k}^p)_{k \ge 0}$ in general. However, there exist constants $C, \beta > 0$ such that for all $n$, we have:

$$ (98) \quad \mathbb{P} \left( l^f \ge \lfloor q_n/2 \rfloor \text{ or } r_n^v - 1 \ge \lfloor q_n/2 \rfloor \right) \le C e^{-\beta n} $$

Moreover, on the intersection of the level sets of $(\tilde{\gamma}_{2k}^p)_{k \ge 0}$ with the event $S_n^{f,v} := (l^f < \lfloor q_n/2 \rfloor) \cap (r_n^v + 1 < \lfloor q_n/2 \rfloor)$, the random integer $l^f$ satisfies a condition that only depends on the data of $(\gamma_{2k+1}^p)_{k \le \lfloor q_n/2 \rfloor}$ and $r_n^v$ satisfies a condition that only depends on the data of $(\gamma_{2k+1}^p)_{k > \lfloor q_n/2 \rfloor}$. Moreover, the data of $(\gamma_{2k+1}^p)_{k \le \lfloor q_n/2 \rfloor}$ and $(\gamma_{2k+1}^p)_{k > \lfloor q_n/2 \rfloor}$ are almost surely independent with respect to the $\sigma$-algebra generated by $(\tilde{\gamma}_{2k}^p)_{k \ge 0}$, therefore, we may assume that $l^f$ and $r_n^v$ are almost surely independent with respect to the same $\sigma$-algebra. Moreover, they are both independent of said $\sigma$-algebra by lemma 5.7 so their joint data is independent of the $\sigma$-algebra generated by $(\tilde{\gamma}_{2k}^p)_{k \ge 0}$.

Let us now give an upper bound on $-\log \left( \frac{|f \bar{\gamma}_n v|}{\|f\| \|\bar{\gamma}_n\| \|v\|} \right)$. By Definition of $l^f$ and $r_n^v$, we have $f \gamma_0^p \cdots \gamma_{2lf}^p A^\epsilon \gamma_{2lf+1}^p$ and $\gamma_{2q_n-2r_n-1}^p A^\epsilon \gamma_{\bar{p}_{2q_n-2r_n}}^p \cdots \gamma_{n-1}^p v$. Therefore, on the set $S_n^{f,v}$, by point (8) in Theorem 1.6, we have:

$$ f \gamma_0 \cdots \gamma_{\bar{p}_{2lf+1}-1} A^{\frac{\epsilon}{2}} \gamma_{\bar{p}_{2lf}} \cdots \gamma_{n-1} v \quad \text{and} \quad \gamma_{\bar{p}_{2lf}} \cdots \gamma_{\bar{p}_{2q_n-2r_n-1}} A^{\frac{\epsilon}{2}} \gamma_{\bar{p}_{2q_n-2r_n}} \cdots \gamma_{n-1} v $$
Therefore, we have:

$$|f \overline{\gamma}_n v| \ge \left\| f \gamma_0 \cdots \gamma \overline{\rho}_{2lf+1} - 1 \right\| \frac{\varepsilon}{2} \left\| \gamma \overline{\rho}_{2lf} \cdots \gamma \overline{\rho}_{2q_n-2r_n} - 1 \right\| \frac{\varepsilon}{2} \left\| \gamma \overline{\rho}_{2q_n-2r_n} \cdots \gamma_{n-1} v \right\|$$

Therefore, we have:

$$\frac{|f \overline{\gamma}_n v|}{\|f\| \|\overline{\gamma}_n\| \|v\|} \ge \frac{\varepsilon^2}{4} \frac{\left\| f \gamma_0 \cdots \gamma \overline{\rho}_{2lf+1} - 1 \right\|}{\|f\| \left\| \gamma_0 \cdots \gamma \overline{\rho}_{2lf+1} - 1 \right\|} \frac{\left\| \gamma \overline{\rho}_{2q_n-2r_n} \cdots \gamma_{n-1} v \right\|}{\left\| \gamma \overline{\rho}_{2q_n-2r_n} \cdots \gamma_{n-1} \right\| \|v\|}$$

Moreover, we have $\|fg\| \|g^{-1}\| \ge \|f\|$ for all $f, g$. Therefore:

$$(99) \quad -\log \left( \frac{|f \overline{\gamma}_n v|}{\|f\| \|\overline{\gamma}_n\| \|v\|} \right) \le |\log(\varepsilon^2/4)| + \sum_{k=0}^{\overline{p}_{2lf+1}-1} N(\gamma_k) + \sum_{k=\overline{p}_{2q_n-2r_n}^{v}}^{n-1} N(\gamma_k).$$

Let us write $I_n^{f,v}$ for the set $\{0, \dots, \overline{p}_{2lf+1}-1\} \cup \{\overline{p}_{2q_n-2r_n^v}, n-1\}$. Note that $\#I_n^{f,v} = n - \overline{p}_{2q_n-2r_n^v} + \overline{p}_{2lf+1} \ge 1$.

Let us now give a probabilistic estimate for $\sum_{k \in I_n^{f,v}} N(\gamma_k) + |\log(\varepsilon^2/4)|$. Let $B = \max_{s \in K} N(s)$. Then, for all $t \ge B$, we have $N^{-1}(t, +\infty) \cap K = \emptyset$. Hence, by point (7) in Theorem 1.6, we have:

$$\forall t \ge B, \forall k, \mathbb{P} \left( N(\gamma_k) > t \mid (p_j)_{j \ge 0} \right) \le 2N_*\nu(t, +\infty).$$

Note that for all $k$, on the measurable set $\text{odd}_p(k) = (\exists j, \overline{p}_{2j+1} \le k < \overline{p}_{2j+2})$, we have $\gamma_k \in K$ by point (3) in Theorem 1.6 so $\mathbb{P}(N(\gamma_k) > B \mid (p_j)_{j \ge 0}) = 0$ and on the set $\text{even}_p(k) = (\exists j, \overline{p}_{2j} \le k < \overline{p}_{2j+1})$, the data of $l^f$ and $r_n^v$ is independent of $\gamma_k$ so we have:

$$(100) \quad \forall t \ge B, \forall k \le n, \mathbb{P} \left( N(\gamma_k) > t \mid (p_j)_{j \ge 0}, l^f, r_n^v \right) \le 2N_*\nu(t, +\infty).$$

Note moreover that for all $t \ge B$, we have $N_*\nu(t, +\infty) \le N_*\nu(t - B, +\infty)$ and for all $t < B$, we have $N_*\nu(t - B, +\infty) = 1$. Therefore, for all $t \ge 0$, by (99) we have:

$$\begin{align*}
\mathbb{P} \left( -\log \left( \frac{|f \overline{\gamma}_n v|}{\|f\| \|\overline{\gamma}_n\| \|v\|} \right) > t \cap S_n^{f,v} \right) &\le \mathbb{P} \left( \sum_{k \in I_n^{f,v}} N(\gamma_k) > t - |\log(\varepsilon^2/4)| \right) \\
&\le \mathbb{P} \left( \exists k \in I_n^{f,v}, N(\gamma_k) > \frac{t - |\log(\varepsilon^2/4)|}{\#I_n^{f,v}} \right) \\
&\le \sum_{k=1}^{\infty} \mathbb{P} \left( \#I_n^{f,v} = k \right) 2kN_*\nu \left( \frac{t - |\log(\varepsilon^2/4)|}{k} - B, +\infty \right).
\end{align*}$$
By (97), there exist constants $C, \beta$ such that $2\mathbb{P}(\#I_n^{f,v} = k) \le Ce^{-\beta k}$. Without loss of generality, we may also assume that $\mathbb{P}(S_n^{f,v}) \ge 1 - Ce^{-\beta n}$. To conclude, let $D = B + |\log(\varepsilon^2/4)|$ and we have (96). $\square$

One may note that the law of $(p_k)$ only depends on the constants $\alpha$ and $m$ defined in Corollary 3.17. Therefore, the constants $C$ and $\beta$ defined in Theorem 5.9 depend on $\nu$ only trough $\alpha$ and $m$ and the constant $D$ only depends on $\nu$ through $\varepsilon$ and $K$.

We get a stronger result for the spectral radius because it is pinched by the singular values i.e., $s_d \le \rho_1 \le s_1$. This allows us to have a uniform control over the spectral radius at all times, even when $2c_n + 1 \ge q_n$.

**Theorem 5.10.** Let $\nu$ be a strongly irreducible and proximal probability distribution on GL($E$). Let $(\gamma_n)_{n \ge 0} \sim \nu^{\otimes N}$. There exist constants $C, D, \beta > 0$ that depend only on $\alpha$ and $m$ and such that:

$$ (101) \qquad \mathbb{P}\left(-\log\left(\frac{\rho_1(\gamma_n)}{\|\bar{\gamma}_n\|}\right) > t\right) \le \sum_{k=1}^{\infty} Ce^{-\beta k}kN_*\nu(t/k - D, +\infty). $$

*Proof.* Let $(p_n)_{n \ge 0}$ and $K \subset \text{GL}(E)$ be as in Theorem 1.6. For all $n$, let $q_n$ and $c_n$ be as in Lemma 5.7. Then by Lemma 5.7, there exist constants $C, \beta > 0$ such that for all $k$ and all $n$, we have:

$$ (102) \qquad \mathbb{P} \left( n - \bar{p}_{2c_n-2c_n} + \bar{p}_{2c_n} = k \right) \le Ce^{-\beta k}. $$

Let us now give an upper bound on $-\log(\rho_1(\gamma_n)/\|\bar{\gamma}_n\|)$. Let $S_n = (2c_n + 1 < q_n)$ By definition of $c_n$, and by point (8) in Theorem 1.6, on $S_n$ for:

$$ g_n := \gamma_{\bar{p}_{2c_n+1}} \cdots \gamma_{n-1} \gamma_0 \cdots \gamma_{\bar{p}_{2c_n+1}-1}, $$

we have $g_n A^{\frac{\varepsilon}{4}} g_n$. Let us give more details. Write for all $n$:

$$ f_n := \gamma_{\bar{p}_{2c_n-2c_n-1}} \cdots \gamma_{n-1} \gamma_0^p \cdots \gamma_{2c_n}^p $$

$$ h_n := \gamma_{\bar{p}_{2c_n-2c_n}} \cdots \gamma_{n-1} \gamma_0^p \cdots \gamma_{2c_n}^p $$

$$ z_n := \gamma_{2c_n+1}^p \cdots \gamma_{2q_n-2c_n-2}^p $$

We have $\gamma_{2q_n-2c_n-1}^p A^{\varepsilon} h_n$ and $\sigma(\gamma_{2q_n-2c_n-1}^p) \le \varepsilon^6/48$ by point (3) in Theorem 1.6. Hence $\sigma(f_n) < \varepsilon^4/48$ by Lemma 2.8. Moreover, we have $f_n A^{\varepsilon} \gamma_{2c_n+1}^p$ We apply point (8) to both aforementioned conditions, and we get $f_n A^{\frac{\varepsilon}{2}} z_n A^{\frac{\varepsilon}{2}} f_n$. Moreover $\gamma_{2c_n+1}^p A^{\frac{\varepsilon}{2}} \gamma_{2c_n+2}^p \cdots \gamma_{2q_n-2c_n-2}^p$ so $\sigma(z_n) \le \varepsilon^4/12$ by Lemma 2.8. Then by Lemma 2.9 applied to $z_n$ and $f_n$, we have $z_n f_n A^{\frac{\varepsilon}{4}} z_n f_n$.
Then by (32) in Lemma 2.17, we have:

$$ \rho_1(g_n) \geq \frac{\varepsilon}{4} \|g_n\| \geq \frac{\varepsilon^2 \|z_n\|}{8 \|f_n\|} $$

Moreover, $g_n$ is conjugated to $\bar{\gamma}_n$ so $\rho_1(g_n) = \rho_1(\bar{\gamma}_n)$. Therefore, we have on $S_n$:

$$ \begin{aligned} -\log \left( \frac{\rho_1(\bar{\gamma}_n)}{\|\bar{\gamma}_n\|} \right) &\leq |\log(\varepsilon^2/8)| + N(f_n) \\ &\leq |\log(\varepsilon^2/8)| + \sum_{k=0}^{\bar{p}_{2c_n+1}-1} N(\gamma_k) + \sum_{k=\bar{p}_{2q_n-2c_n-1}}^{n-1} N(\gamma_k). \end{aligned} $$

Outside of $S_n$, we have $-\log(\frac{\rho_1(\bar{\gamma}_n)}{\|\bar{\gamma}_n\|}) \leq N(\bar{\gamma}_n) \leq \sum_{k=0}^{n-1} N(\gamma_k)$. Outside of $S_n$, let $J_n := \{0, \dots, n-1\}$ and on $S_n$, let $J_n := \{0, \dots, \bar{p}_{2c_n+1}-1\} \cup \{\bar{p}_{2q_n-2c_n-1}, \dots, n-1\}$. Then, for all $t \geq 0$, we have:

$$ \begin{aligned} \mathbb{P} \left( -\log \left( \frac{\rho_1(\gamma_n)}{\|\bar{\gamma}_n\|} \right) > t \right) &\leq \mathbb{P} \left( \sum_{j \in J_n} N(\gamma_j) > t - |\log(\varepsilon^2/8)| \right) \\ &\leq \sum_{k=0}^{n} \mathbb{P} \left( \max_{j \in J_n} N(\gamma_j) > \frac{t - |\log(\varepsilon^2/8)|}{k} \cap \# J_n = k \right) \end{aligned} $$

By Lemma 5.7, there exist constants $C, \beta > 0$ such that $2\mathbb{P}(\# J_n = k) \leq C e^{-\beta k}$ for all $k \leq n$. Moreover, with the same reasoning as in the proof of (100) in Theorem 5.9, there exists a constant $B$ such that:

$$ \forall k \leq n, \mathbb{P}(N(\gamma_k) > t | J_n) \leq 2N_*\nu(t - B, +\infty). $$

From the above, we deduce (101) for $D = B + |\log(\varepsilon^2/8)|$, with the same reasoning as in the proof of Theorem 5.9. $\square$

*Proof of Theorem 1.13.* To derive (1.13) from Theorems 5.9 and 5.10, we simply need to prove that for all distribution $\eta \neq \delta_0$ on $\mathbb{R}_{\geq 0}$ and for all $C, D, \beta > 0$, there exist constants $C', \beta' > 0$ such that:

$$ \forall t \geq 0, \sum_{k \geq 1} C e^{-\beta k} k \eta(t/k - D, +\infty) \leq \sum_{k \geq 0} C' e^{-\beta' k} \eta(t/k, +\infty). $$

Indeed, if $N_*\nu = \delta_0$, then $\nu$ is supported on $\mathbb{K}^\times \mathcal{O}(E)$ and therefore $\rho_1 = \rho_2 = \dots = \rho_d$ on $\Gamma_\nu$, which contradicts the proximality.
Let $\eta = N_*\nu$ and let $C, \beta > 0$. First let $C_1, \beta_1 > 0$ be such that $Ce^{-\beta k}k \le C_1e^{-\beta_1 k}$ for all $k$. Moreover, for all $t \ge 2Dk$, we have $t/k - D \ge t/(2k)$. Hence, for all $t > 0$:

$$ \sum_{k \ge 0} Ce^{-\beta k}k\eta(t/k - D, +\infty) \le \sum_{1 \le k \le t/(2D)} C_1e^{-\beta_1 k}\eta(t/(2k), +\infty) + \sum_{k > t/(2D)} C_1e^{-\beta_1 k} \\ \le \sum_{k=1}^{+\infty} 2C_1e^{-\beta_1 k}\eta(t/k, +\infty) + \frac{C_1}{1 - e^{-\beta_1}}e^{-\beta_1[t/(2D)]} $$

Let $\delta > 0$ be such that $\eta(\delta, +\infty) > 0$ and let $\alpha = \eta(\delta, +\infty)$. Then, for all $C_2, \beta_2 > 0$, we have:

$$ \sum_{k=1}^{+\infty} C_2e^{-\beta_2 k}\eta(t/k, +\infty) \ge C_2e^{-\beta_2[t/\delta]}\alpha $$

Let $\beta_2 = \delta\beta_1/(2D)$ and $C_2 = \frac{C_1e^{\beta_1+\beta_2}}{\alpha(1-e^{-\beta_1})}$. Then for all $t \ge 0$, we have $C_2e^{-\beta_2[t/\delta]}\alpha \ge \frac{C_1}{1-e^{-\beta_1}}e^{-\beta_1[t/(2D)]}$. Moreover, we may assume that $2D \ge 1$ without loss of generality so $\beta_2 \le \beta_1$ and $C_2 \ge C_1$. Therefore, for all $t \ge 0$, we have:

$$ \sum_{k=1}^{+\infty} Ce^{-\beta k}k\eta(t/k - D, +\infty) \le \sum_{k=1}^{+\infty} 3C_2e^{-\beta_2 k}\eta(t/k, +\infty). \quad \square $$

Note that (96) and (101) do not give a satisfying result in case $N_*\nu$ has a finite exponential moment. More precisely, there is no constant $\alpha > 0$ such that $\sum_k e^{-k}e^{-t/k} \le Ce^{-\alpha t}$. However, we know by the work of Guivarc'h and Lepage that when $N_*\nu$ has a finite exponential moment, the invariant measure in Hölder regular, which implies that the logarithm of the coefficients and of the spectral radius both satisfy large deviations inequalities above and below the top Lyapunov exponent. This is because when we use point (7) in Theorem 1.6, we forget about the fact that the random variables $N(\gamma_k)$ are independent. In fact they are not conditionally independent with respect to the data of $(p_k)_{k \ge 0}$.

To get a sharper estimate using the pivoting technique, we simply ignore point (7) and use Lemmas 5.2 and 5.5 instead.

**5.5. Law of large numbers for the singular gap.** This paragraph is dedicated to the proof of Theorem 1.7. Let $\nu$ be strongly irreducible and proximal on End$(E)$. Let $(\gamma_n)_{n \ge 0} \sim \nu^{\otimes \mathbb{N}}$. We remind that Theorem 1.7 tells us that $\frac{-\log \sigma(\bar{\gamma}_n)}{n}$ converges almost surely to a non-random positive limit that we denote by $\lambda_{1-2}(\nu)$ and (6) tells us that the random sequence $(-\log \sigma(\bar{\gamma}_n))_n$ satisfies large deviations
inequalities below the speed $\lambda_{1-2}(\nu)$. The proof of (6) is pretty straightforward from Lemma 5.8 and it implies that $\liminf_{\frac{-\log \sigma(\bar{\gamma}_n)}{n} \ge \lambda_{1-2}(\nu)}$ almost surely. The hard part is to prove that $\limsup_{\frac{-\log \sigma(\bar{\gamma}_n)}{n} \le \lambda_{1-2}(\nu)}$ from the fact that $\lambda_{1-2}(\nu) < +\infty$ only. For the proof of the almost sure convergence, we use the following Lemma.

**Lemma 5.11.** Let $\nu$ be a strongly irreducible and proximal probability distribution on End($E$). Let $(\gamma_n) \sim \nu^{\otimes\mathbb{N}}$. Let $0 < \varepsilon \le 1/2$ be as in Theorem 1.6. There exist constants $l_0 \in \mathbb{N}$, and $0 < \beta < 1$ such that:

$$ (103) \qquad \forall f \in E \cup \text{End}(E), \mathbb{P}\left(\forall n \ge l_0, f\mathbb{A}^{\frac{\varepsilon}{2}\bar{\gamma}_n} > \beta\right). $$

*Proof.* Let $m, \alpha \in (0, 1)$ and $\nu_s$ be as in Corollary 3.17 for $\delta(\varepsilon) = \frac{\varepsilon^4}{12}$ and $\rho = 1/4$. Let $\mu$ be as in Theorem 1.6 and let $(\gamma_{-m}, \dots, \gamma_{-1}, (\gamma_n)_{n \ge 0}, (p_n)_{n \ge 0}, (\tau_k)_{k \ge 0}) \sim \nu^{\otimes m} \otimes \mu \otimes \mathcal{U}^{\otimes \mathbb{N}}$ and let $p_0 = m$. For all $n \ge 0$, we write $q_n := \max\{k \ge 0 | \bar{p}_{2k+1} \le n+m\}$. Then, by Lemma 5.7, for all $n$, there exists a random integer $r_n = r_n^{\text{id}} \sim \mathcal{G}_{1/4}$ such that $\gamma_{2q_n-2r_n-1}^p \gamma_{2q_n-2r_n}^{\bar{\gamma}_{2q_n-2r_n}} \cdots \gamma_{n-1}^p$ or $r_n \ge q_n$. Then by (8), for all $n$ such that $r_n < q_n$, we have $\gamma_0^p \gamma_1^p \mathbb{A}^{\frac{\varepsilon}{2}} \gamma_{\bar{p}_2} \cdots \gamma_{n-1}$. Moreover, by Lemma 5.8, the probability of having $q_n \le r_n$ decreases exponentially fast so there exists an integer $l_0$ such that $\mathbb{P}(\exists n \ge l_0 - m, q_n \le r_n) \le \frac{1}{2}$. Let $l_0$ be such an integer. Let:

$$ A_1 = \left( \forall n \ge l_0, \gamma_0^p \gamma_1^p \mathbb{A}^{\frac{\varepsilon}{2}} \gamma_{\bar{p}_2} \cdots \gamma_{n-m-1} \right). $$

The event $A_1$ is independent of $(\gamma_{-m} \cdots \gamma_{-1})$ and by the above argument, it has probability at least $1/2$.

Let $f \in E \cup \text{End}(E)$. We have $\nu^{*m} \ge \alpha\nu_s$ by Corollary 3.17 and $\nu_s$ is $\frac{1}{4}$-Schottky so:

$$ \mathbb{P} \left( f \mathbb{A}^{\varepsilon} \gamma_{-m} \cdots \gamma_{-1} \mathbb{A}^{\varepsilon} \gamma_0^p \gamma_1^p \mid (\tilde{\gamma}_k^p)_{k \ge 0} \right) = \nu^{*m} \left\{ \begin{array}{l} s \mid f \mathbb{A}^{\varepsilon} s \mathbb{A}^{\varepsilon} \gamma_0^p \gamma_1^p \\ \ge \alpha \nu_s \left\{ s \mid f \mathbb{A}^{\varepsilon} s \mathbb{A}^{\varepsilon} \gamma_0^p \gamma_1^p \right\} \ge \alpha/2. \end{array} \right. $$

Moreover $A_1$ only depends on $(\tilde{\gamma}_k^p)_{k \ge 0}$. Therefore, we have:

$$ \mathbb{P} \left( \forall n \ge l_0, f \mathbb{A}^{\varepsilon} \gamma_{-m} \cdots \gamma_{-1} \mathbb{A}^{\varepsilon} \gamma_0^p \gamma_1^p \mathbb{A}^{\frac{\varepsilon}{2}} \gamma_{\bar{p}_2} \cdots \gamma_{n-m-1} \right) \ge \alpha/4. $$

Moreover, we have $\gamma_0^p \mathbb{A}^{\frac{\varepsilon}{2}} \gamma_1^p$ so $\sigma(\gamma_0^p \gamma_1^p) \le \varepsilon^4/12$ and by construction, we have $\sigma(\gamma_{-m} \cdots \gamma_{-1}) \le \varepsilon^2/12$. Then by Lemma 2.9 applied twice, we have the alignment
$f\mathbb{A}^{\frac{\epsilon}{2}}\gamma_{-m}\cdots\gamma_{n-m-1}$ as soon as $f\mathbb{A}^{\epsilon}\gamma_{-m}\cdots\gamma_{-1}\mathbb{A}^{\epsilon}\gamma_{0}^{p}\gamma_{1}^{p}\mathbb{A}^{\frac{\epsilon}{2}}\gamma_{\bar{p}_{2}}\cdots\gamma_{n-m-1}$. Hence, we have:

$$ \mathbb{P}\left(\forall n \ge l_0, f\mathbb{A}^{\frac{\epsilon}{2}}\gamma_{-m} \cdots \gamma_{n-m-1}\right) \ge \alpha/4 $$

and $(\gamma_{n-m})_{n \ge 0} \sim \nu^{\otimes \mathbb{N}}$, which proves (103) for $\beta = \frac{\alpha}{4}$. $\square$

Note that we can not derive Lemma 5.11 from Lemma 5.7 alone because, we do not prove anywhere that $f\gamma_{0}^{p}\cdots\gamma_{2lf}^{p} \ne 0$ with positive probability. A reformulation of Lemma 5.11 is that the distribution of $\bar{\gamma}_n$ is $(1-\beta)$-Schottky for all $n \ge l_0$ (for that we need to apply Lemma 5.11 to both $\nu$ and $^t\nu$ but we get the same $\beta$ and $l_0$, because Corollary 3.17 is symmetric). Let us now give a full proof of Theorem 1.7. We remind that it states that there exists a constant $0 < \lambda_{1-2}(\nu) \le +\infty$ such that $\frac{-\log \sigma(\bar{\gamma}_n)}{n} \to \lambda_{1-2}(\nu)$ almost surely and (6) states that $(-\log \sigma(\bar{\gamma}_n))_{n \ge 0}$ satisfies large deviations inequalities below the speed $\lambda_{1-2}(\nu)$.

*Proof of Theorem 1.7.* Let $0 < \varepsilon \le \frac{1}{2}$, let $m \in \mathbb{N}_{\ge 1}$ and let $\mu$ be as in Theorem 1.6. Let $(\gamma_n)_{n \ge 0}$, $(p_n)_{n \ge 0}$, $(\tau_n)_{n \ge 0} \sim \mu \otimes \mathcal{U}^{\otimes \mathbb{N}}$. Let $\lambda'$ be as in Lemma 5.6 and let $\lambda_{1-2}(\nu) := \frac{\lambda'}{m+\mathbb{E}(p_2)}$. We start with the proof of (6). For all $n \ge 0$, let $q_n$, and $r_n = r_n^{\text{Id}}$ be as in Lemma 5.7. We claim that:

$$ (104) \qquad \sigma(\bar{\gamma}_n) \le \sigma(\gamma_0^p \cdots \gamma_{2q_n-2r_n-2}^p). $$

Indeed, we have $\gamma_{2q_n-2r_n-2}^p \mathbb{A}^{\epsilon} \gamma_{\bar{p}_{2q_n-2r_n}} \cdot \gamma_{n-1}$ by (8) in Theorem 1.6, we have $\gamma_0^p \cdots \gamma_{2q_n-2r_n-2}^p \mathbb{A}^{\frac{\epsilon}{2}} \gamma_{\bar{p}_{2q_n-2r_n-1}} \cdot \gamma_{n-1}$. By Lemma 2.8, we have $\sigma(\gamma_{\bar{p}_{2q_n-2r_n-1}} \cdots \gamma_{n-1}) \le \varepsilon^4/48$ and by Lemma 2.8 again, we have $\sigma(\bar{\gamma}_n) \le \sigma(\gamma_0^p \cdots \gamma_{2q_n-2r_n-2}^p) \varepsilon^2/12$.

Then by Lemma 5.8 ($-\log \sigma(\bar{\gamma}_n)$) satisfies large deviations inequalities below the speed $\lambda_{1-2}(\nu)$. Then, by Borel Cantelli's Lemma, we have almost surely $\liminf_n \frac{1}{n} \log \sigma(\bar{\gamma}_n) \ge \lambda_{1-2}(\nu)$. This concludes the proof when $\lambda_{1-2}(\nu) = +\infty$. Otherwise, we have to prove that $\limsup_n \frac{-\log \sigma(\bar{\gamma}_n)}{n} \le \lambda_{1-2}(\nu)$. We do it by contraposition. By Lemma 5.6, we have $\frac{-\log \sigma(\gamma_2^p \cdots \gamma_{2k}^p)}{k} \to \lambda'$. almost surely. For all $n \ge 0$, we define $k_n := \min\{k \ge 0 \mid \bar{p}_{2k} > n\} = q_n + 1$. We have $k_n \to +\infty$ almost surely and $\bar{p}_{2k_n-2} \le n \le \bar{p}_{2k_n}$. Hence, by the law of large numbers, we have $\frac{k_n}{n} \to \frac{1}{\mathbb{E}(p_2)+m}$ almost surely. Then by Lemma 5.6, we have $\frac{-\log \sigma(\gamma_2^p \cdots \gamma_{2k_n}^p)}{n} \to \lambda_{1-2}(\nu)$. In other words, for all $\lambda'' > \lambda_{1-2}(\nu)$ we have:

$$ (105) \qquad \lim_{n_0 \to +\infty} \mathbb{P} \left( \exists n \ge n_0, -\log \sigma(\gamma_2^p \cdots \gamma_{2k_n}^p) > \lambda'' n \right) = 0. $$
Let $\lambda'' > \lambda_{1-2}(\nu)$ and assume by contraposition that for all $n_0$, we have:

$$ \mathbb{P}(\exists n \ge n_0, -\log \sigma(\bar{\gamma}_n) > \lambda''n) \ge \delta. $$

Let $l_0, \beta > 0$ be as in Lemma 5.11. Assume that (103) holds for both $\nu$ and for the push-forward ${}^t\nu$ of $\nu$ by the transposition map. Let $l_1$ be such that $\mathbb{P}(p_0 + p_1 \ge l_1) \le \beta^2\delta/3$ and let $n_0$ be such that:

$$ \mathbb{P}(\exists n \ge n_0, -\log \sigma(\gamma_2^p \cdots \gamma_{2k_n}^p) > \lambda''(n - 2l_0 - l_1) - \log(32\epsilon^{-4})) \le \beta^2\delta/3. $$

Such a $n_0$ exists because $\lim_{n} \frac{-\log \sigma(\gamma_2^p \cdots \gamma_{2k_n}^p) + C}{n-C} = \lambda_{1-2}(\nu)$ almost surely and for all constant $C$. Let:

$$ n_1 := \min \{n \ge n_0, -\log \sigma(\gamma_{l_0+l_1} \cdots \gamma_{l_0+l_1+n-1}) > \lambda''n\}. $$

That way, we have $\mathbb{P}(n_1 \le +\infty) \ge \delta$. Let $g_1 := \gamma_{l_0+l_1} \cdots \gamma_{l_0+l_1+n-1}$. Let us show that:

(106) $$ \mathbb{P} \left( \forall i < l_1, \forall j \ge n_1 + l_1 + 2l_0, \gamma_i \cdots \gamma_{l_1+l_0-1} \mathbb{A}^{\frac{\epsilon}{2}} g_1 \mathbb{A}^{\frac{\epsilon}{2}} \gamma_{l_0+l_1+n} \cdots \gamma_{j-1} \right) \ge \delta\beta^2. $$

First note that, the data of $(\gamma_k)_{k<l_0+l_1}$ is independent of $g_1$ and $(^t\gamma_k)_{k<l_0+l_1} \sim {}^t\nu^{\otimes l_0+l_1}$. Then, by (103) applied to ${}^t\nu$, we have:

(107) $$ \mathbb{P} \left( \forall i' \ge l_0, {}^t g_1 \mathbb{A}^{\frac{\epsilon}{2}t} \gamma_{l_0+l_1-1} \cdots {}^t \gamma_{l_0+l_1-i'} \middle| n_1, g_1 \right) \ge \beta. $$

Moreover $l_0 + l_1 + n_1$ is a stopping time in the sense that the event $(l_0 + l_1 + n_1 = n)$ only depends on $(\gamma_k)_{k<n}$. Therefore, the conditional distribution of $(\gamma_{k+n_1+l_0+l_1})_{k \ge 0}$ with respect to $(\gamma_k)_{k<l_0+l_1+n_1}$ is $\nu^{\otimes \mathbb{N}}$ and therefore, we have:

(108) $$ \mathbb{P} \left( \forall j' \ge l_0, g_1 \mathbb{A}^{\frac{\epsilon}{2}} \gamma_{l_0+l_1+n_1} \cdots \gamma_{l_0+l_1+n_1+j'-1} \middle| (\gamma_k)_{k<l_0+l_1+n_1} \right) \ge \beta. $$

Then we compose (108) with (107) and with the fact that $n_1$ exists with probability at least $\delta$ and we get (106). Now note that (106) combined with the fact that $\bar{p}_2 < l_1$ outside of an event of probability $1 - \beta^2\delta/3$ implies that:

$$ \mathbb{P} \left( \gamma_{\bar{p}_2} \cdots \gamma_{l_1+l_0-1} \mathbb{A}^{\frac{\epsilon}{2}} g_1 \mathbb{A}^{\frac{\epsilon}{2}} \gamma_{l_0+l_1+n_1} \cdots \gamma_{\bar{p}_{2k_2l_0+l_1+n_1}-1} \right) \ge 2\delta\beta^2/3. $$

Then by Lemma 2.11, we have:

$$ \mathbb{P} \left( \sigma(\gamma_{\bar{p}_2} \cdots \gamma_{\bar{p}_{2k_2l_0+l_1+n_1}-1}) \le 32\epsilon^{-4} \sigma(g_1) \right) \ge 2\delta\beta^2/3. $$

Moreover, we have $-\log \sigma(g_1) > \lambda''n_1$ and $n_1 \ge n_0$. Therefore:

$$ \mathbb{P}(\exists n \ge n_0, -\log \sigma(\gamma_2^p \cdots \gamma_{2k_n}^p) > \lambda''(n - 2l_0 - l_1) - \log(32\epsilon^{-4})) \ge 2\beta^2\delta/3, $$

which contradicts the definition of $n_0$. $\square$

<figure><img src="image_3.png" /></figure>
**5.6. Contraction property.** In this section, we prove and give more details on Theorems 1.8 and 1.10.

**Definition 5.12** (The space of contracting sequences). Let $E$ be a Euclidean, Hermitian or ultra-metric vector space. We define the set of contracting sequences:

$$ \Omega'(E) := \left\{ \gamma \in \mathrm{End}(E)^{\mathbb{N}} \mid \begin{array}{l} \forall n \in \mathbb{N}, \bar{\gamma}_n \neq \{0\} \quad \text{and} \quad \exists l^{\infty} \in \mathrm{P}(E), \\ \forall \varepsilon \in (0, 1), \lim_{n \to \infty} \max_{u \in U^{\varepsilon}(\bar{\gamma}_n)} d(l^{\infty}, [u]) = 0 \end{array} \right\}. $$

We define $T := \mathrm{End}(E)^{\mathbb{N}} \to \mathrm{End}(E)^{\mathbb{N}}$ to be the Bernoulli shift and we define $l^{\infty} : \Omega'(E) \to \mathrm{P}(E)$ to be the only map such that:

$$ \forall \gamma \in \Omega'(E), \lim_{n \to +\infty} \max_{u \in U^{\varepsilon}(\bar{\gamma}_n) \setminus \{0\}} d(l^{\infty}(\gamma), [u]) = 0. $$

We define the shift-invariant space of contracting sequences as:

$$ \Omega(E) := \left\{ \gamma \in \bigcap_{k=0}^{+\infty} T^{-k}\Omega'(E) \mid \forall k \in \mathbb{N}, \gamma_k l^{\infty}(T^{k+1}\gamma) = l^{\infty}(T^k\gamma) \right\}. $$

Note that the space $\Omega(E)$ defined in Definition 5.12 is measurable and $T$-invariant. Note that $l^{\infty}$ is measurable because given $(e_1, \dots, e_d)$ a basis of $E$, for $\varepsilon$ small enough and for all $g \in \mathrm{End}(E)$ there exists at least one index $1 \le i \le d$ such that $e_i \in V^{\varepsilon}(g)$. So we can define $l^{\infty}(\gamma) = \lim_n[\bar{\gamma}e_{i_n}]$ with $i_n = \min(\arg\max_j \|\bar{\gamma}_n e_j\|)$. Moreover $l^{\infty}$ is $T$-equivariant on $\Omega(E)$ in the sense that $l^{\infty}(\gamma) = \gamma_0 l^{\infty}(T\gamma)$.

Note also that $\Omega'(E) \neq \Omega(E)$. For example let $E = \mathbb{R}^2$, and let $\pi_1$ and $\pi_2$ be the orthogonal projections onto the first and second coordinates. If $\gamma_0 = \pi_1$ and $\gamma_k = \pi_1 + 2\pi_2$ for all $k \ge 1$, then $\bar{\gamma}_n = \pi_1$ for all $n \ge 1$ and $[\gamma_k \cdots \gamma_n] \xrightarrow{m \to \infty} [\pi_2]$ for all $k \ge 1$. So $l^{\infty}(\gamma)$ is the first coordinate axis and $l^{\infty}(T\gamma)$ is the second coordinate axis. Hence $\gamma_0 l^{\infty}(T\gamma) = [0] \neq l^{\infty}(\gamma)$ so $\gamma \in \Omega'(E) \setminus \Omega(E)$.

Let us prove the following theorem, which trivially imply Theorem 1.8 but also gives more precision. The fact that the limit is given by a measurable map that does not depend on $\nu$ is non trivial. Indeed, there is no vector $v \in E$ such that $v \notin \ker(\nu)$ for all strongly irreducible and proximal $\nu$.

**Theorem 5.13.** Let $\nu$ be a strongly irreducible and proximal probability distribution on $\mathrm{End}(E)$. Let $\gamma = (\gamma_n)_{n \in \mathbb{N}} \sim \nu^{\otimes \mathbb{N}}$. Then $\gamma \in \Omega(E)$ almost surely. Let $\alpha < \lambda_{1-2}(\nu)$. There exist constants $C, \beta > 0$ such that:

$$ (109) \quad \mathbb{P} \left( \exists u \in U^1(\bar{\gamma}_n) \setminus \{0\}, \text{d}([u], l^{\infty}(\gamma)) \ge \exp(-\alpha n) \right) \le C \exp(-\beta n). $$
Moreover, for all $v \in E$, we have:

$$ (110) \quad \mathbb{P}(d([\bar{\gamma}_n v], l^\infty(\gamma)) \ge \exp(-\alpha n) \mid \bar{\gamma}_n v \ne 0) \le C \exp(-\beta n). $$

*Proof.* Let $0 < \varepsilon \le 1$ and $\mu$ be as in Theorem 1.6. Let $((\gamma_n)_{n \ge 0}, (p_n)_{n \ge 0}) \sim \mu$. Let $v \in E \setminus \{0\}$. For all $n \in \mathbb{N}$, let $q_n$ and $r_n^v$ and $r_n^{\text{Id}}$ be as in Lemma 5.7. Let $\alpha < \sigma(\nu)$ and let $C, \beta > 0$ be as in Lemma 5.8, i.e., such that:

$$ (111) \quad \forall n, \forall h, \mathbb{P}\left(-\log \sigma\left(\gamma_0^p \cdots \gamma_{2q_n-2r_n^h-2}^p\right) \le \alpha n\right) \le Ce^{-\beta n}. $$

For all $n$, let $j'_n = \min\{2q_n - 2r_n^v, 2q_n - 2r_n^{\text{Id}}\}$. Then we have:

$$ (112) \quad \forall n, \mathbb{P}\left(\exists k \ge n, -\log \sigma\left(\bar{\gamma}_{j'_n}^p\right) \le \alpha n\right) \le \frac{2C}{1-e^{-\beta}} e^{-\beta n}. $$

Let $j_n = \min_{k \ge n} j'_k$ for all $n$. By (112), the random integer $n_\alpha = \max\{n \mid \sigma(\bar{\gamma}_{j_n}^p) \ge e^{-\alpha n}\}$ has a finite exponential moment. By point (8) in Theorem 1.6, we have the alignment $\bar{\gamma}_{j_n}^p A_\frac{\varepsilon}{2} \gamma_{\bar{p}_{j_n}} \cdots \gamma_{k-1} h$ for all $k \ge n$ and $h \in \{v, \text{Id}\}$. Then by Lemma 2.8 and (22), for all $u \in U^1(\bar{\gamma}_{j_n}^p) \setminus \{0\}$ and for all $u' \in [\bar{\gamma}_k v] \cup U^\varepsilon(\bar{\gamma}_k) \setminus \{0\}$ for $k \ge n$, we have:

$$ (113) \quad \mathrm{d}([u], [u']) \le \frac{2}{\varepsilon}\sigma(\bar{\gamma}_{j_n}^p) \le \frac{4e^{-\alpha n}}{\varepsilon} + 1_{n<n_\alpha}. $$

Let $u_n^{\varepsilon'} \in U^{\varepsilon'}(\bar{\gamma}_n) \setminus \{0\}$ for all $n \in \mathbb{N}$ and all $\varepsilon > 0$. Note that (113) holds for all $0 < \varepsilon' < \varepsilon$ Therefore, for all $n_0 \le n \le k$, we have $\mathrm{d}(u_n, u_k) \le 4e^{-\alpha n}/\varepsilon'$, by (113). Therefore $(u_n^{\varepsilon'})_{n \ge 0}$ a Cauchy sequence on the full measure set $(n_\alpha < +\infty)$ so $(\gamma_n)_{n \ge 0} \in \omega'(E)$ on $(n_\alpha < +\infty)$ and therefore $\nu^{\otimes \mathbb{N}}(\Omega'(E)) = 1$. Since (113) holds for all $\alpha < \lambda_{1-2}$, the random sequence $(-\log \mathrm{d}([u_n^1], l^\infty(\Omega)))_{n \ge 0}$ satisfies large deviations inequalities below the speed $\lambda_{1-2}$, which implies (109). When $v \notin \ker(\nu)$, the random sequence $(-\log \mathrm{d}([\bar{\gamma}_n v], l^\infty))_{n \ge 0}$ satisfies large deviations inequalities below the speed $\lambda_{1-2}$, which implies (110). Then, we have $[\bar{\gamma}_n v] \to l^\infty(\gamma)$ and by the same reasoning applied to $(\bar{\gamma}_n)_{n \ge 1}$, we have $[\gamma_1 \cdots \gamma_n v] \to l^\infty(T\gamma)$. Hence, we have $\gamma_0 l^\infty(T\gamma) = l^\infty$ almost surely so $\nu^{\otimes \mathbb{N}}(\Omega) = 1$ By the same argument, for all $\alpha < \lambda_{1-2}$, there exist constants $C', \beta' > 0$ such that:

$$ \mathbb{P}(d([\bar{\gamma}_n v], l^\infty(\gamma)) \ge \exp(-\alpha n) \cap \bar{\gamma}_n v \ne 0) \le C' \exp(-\beta' n). $$

By Lemma 3.4, there exists $\delta > 0$ such that $\mathbb{P}(\bar{\gamma}_n v \ne 0) \ge \delta$ for all $n$ and all $v$, which implies (110). $\square$
*Proof of Theorem 1.10.* We want to use Lemma 2.17. For that, let $0 < \varepsilon$ and let $(\gamma_n), (p_n) \sim \mu$ be as in Theorem 1.6 and for all $n$, let $q_n$ and $c_n$ be as in Lemma 5.7. When $2c_n + 1 < q_n$, let:

$$ 
\begin{align*}
f_n &:= \gamma \bar{p}_{2q_n-2c_n-1} \cdots \gamma_{n-1} \gamma_0^p \cdots \gamma_{2c_n}^p \\
h_n &:= \gamma \bar{p}_{2q_n-2c_n} \cdots \gamma_{n-1} \gamma_0^p \cdots \gamma_{2c_n}^p \\
z_n &:= \gamma_{2c_n+1}^p \cdots \gamma_{2q_n-2c_n-2}^p.
\end{align*}
$$

By definition of $c_n$, we have $\gamma_{2q_n-2c_n-1}^p A^\varepsilon h_n$ so by point (3) in Theorem 1.6 and Lemma 2.8, we have $\sigma(f_n) \le \varepsilon^4/48$ and by point (8) in Theorem 1.6, we have $z_n A^{\frac{\varepsilon}{2}} f_n$. By definition of $c_n$ again, we have $f_n A^\varepsilon \gamma_{2c_n+1}^p$ so by point (8) in Theorem 1.6, we have $f_n A^{\frac{\varepsilon}{2}} z_n$. Moreover, we have $\gamma_{2c_n+1}^p A^{\frac{\varepsilon}{2}} \cdots \gamma_{2q_n-2c_n-2}^p$ by Lemma 2.9 so $\sigma(z_n) \le \varepsilon^4/12$ by Lemma 2.8. Then by Lemma 2.9, we have $f_n z_n A^{\frac{\varepsilon}{4}} f_n z_n$ and $\sigma(f_n z_n) \le 4\varepsilon^{-2}\sigma(f_n)\sigma(z_n)$. Hence, by (33) in Lemma 2.17, we have:

$$ \frac{\rho_2}{\rho_1}(\bar{\gamma}_n) = \frac{\rho_2}{\rho_1}(f_n z_n) \le \frac{64\sigma(f_n z_n)}{\varepsilon^2} \le \sigma(\gamma_{2c_n+2}^p \cdots \gamma_{2q_n-2c_n-2}^p) $$

And when $2c_n + 1 \ge q_n$, we still have $\frac{\rho_2}{\rho_1}(\bar{\gamma}_n) \le 1$. Then by Lemma 5.8, for $\lambda_{1-2} = \log(\rho_1/\rho_2)$, the random sequence $(\lambda_{1-2}(\bar{\gamma}_n))_{n \ge 0}$ satisfies large deviations inequalities below the speed $\lambda_{1-2}(\nu)$.

Now we prove the second part of (10), namely that the random sequence $(-\log d(E^+(\bar{\gamma}_n), l^\infty(\gamma))_{n \ge 0}$ satisfies large deviations inequalities below the speed $\lambda_{1-2}(\nu)$. Let $n \ge 0$ and assume that $2c_n + 1 < q_n$. By point 8, we have $\bar{\gamma}_{2q_n-2c_n-1}^p = \gamma_0^p \cdots \gamma_{2q_n-2c_n-2}^p A^{\frac{\varepsilon}{2}} f_n$. Then by Lemma 2.15, we have $\bar{\gamma}_{2q_n-2c_n-1}^p A^{\frac{\varepsilon}{4}} (f_n z_n)^k$ for all $k \ge 0$ moreover, for all choice of $e_n \in E^+(f_n z_n) \setminus \{0\}$ and $e'_n \in E^+(z_n^t f_n) \setminus \{0\}$, we have $[(f_n z_n)^m]_{k \to \infty} \to [e_n e'_n]$. Moreover $A^\varepsilon$ is bi-homogeneous and closed so $\bar{\gamma}_{2q_n-2c_n-1}^p A^{\frac{\varepsilon}{4}} e_n e'_n$ and therefore $\bar{\gamma}_{2q_n-2c_n-1}^p A^{\frac{\varepsilon}{4}} e_n$.

Moreover $\bar{\gamma}_n$ and $f_n z_n$ are semi-conjugated by $\bar{\gamma}_{2q_n-2c_n-1}^p$ so $E^+(\bar{\gamma}_n) = \bar{\gamma}_{2q_n-2c_n-1}^p E^+(f_n z_n) = K \bar{\gamma}_{2q_n-2c_n-1}^p e_n$. Therefore, $E^+(\bar{\gamma}_n) \subset U^{\frac{\varepsilon}{4}}(\bar{\gamma}_{2q_n-2c_n-1}^p)$. Moreover $\bar{\gamma}_{2q_n-2c_n-1}^p A^{\frac{\varepsilon}{2}} \gamma_{2c_n+1}^p \cdots \gamma_k^p$ for all $k$ and $A^{\frac{\varepsilon}{2}}$ is closed, so going to the limit we have $l^\infty((\gamma_{n+\bar{p}_{2q_n-2c_n-1}})_{n \ge 0}) \subset U^{\frac{\varepsilon}{4}}(\bar{\gamma}_{2q_n-2c_n-1}^p)$. Then by Lemma 2.5 and by triangular inequality, we have:

$$ d(E^+(\bar{\gamma}_n), l^\infty) \le \frac{6\sigma(\bar{\gamma}_{2q_n-2c_n-1}^p)}{\varepsilon} \le \sigma(\gamma_{2c_n+2}^p \cdots \gamma_{2q_n-2c_n-2}^p) $$

We conclude using Lemma 5.8 again. $\square$

<figure><img src="image_4.png" /></figure>
A direct consequence of Theorem 1.10 is that $\liminf_n \frac{1}{n} \log \frac{\rho_1}{\rho_2}(\bar{\gamma}_n) \ge \lambda_{1-2}(\nu)$ almost surely. However, for $d \ge 3$ and without moment condition, we do not know anything about the upper limit. The issue is that the pivoting technique does not tell us much about $\rho_2(\bar{\gamma}_n)$ due to the fact that $\Lambda_*^2 \nu$ may not be strongly irreducible. Using a reduction method as we did in the proof of Corollary 1.15 and the concentration results of (12) in Theorem 1.13, one may conjecture that we have a convergence in law for the distribution of $\frac{1}{n} \log \frac{\rho_1}{\rho_2}(\bar{\gamma}_n)$ to the Dirac measure at $\lambda_{1-2}(\nu)$ in the invertible case. This weak convergence result may or may not be optimal.

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INSTITUT DENIS POISSON, UNIVERSITÉ DE TOURS, FACULTÉ DES SCIENCES ET TECHNIQUES BÂTIMENT
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