1-5\sum_{\ell=0}^{m-1}e^{-2^{\ell}c_{1}\varepsilon^{2}N_{0}\hat{L}^{d}}-2e^{-2^{m}c_{1}\varepsilon^{2}N_{0}\hat{L}^{d}},$

as claimed. ∎

We now return to the main argument, i.e., the proof of Lemma 5.3. Note that
the assumptions on $\mathcal{G}_{2^{m-2}N_{0}}^{(1)}$ are verified in the base
case $m=3$ by Lemma 5.7. Hence, the inductive scheme applies and therefore
(5.558), (5.559) and (5.560) hold for $m=k$.

Step $k+1$. Recall that $n_{0}=N_{1}-2^{k}N_{0}$. Let

(5.575) $\displaystyle\tilde{g}_{n_{0},1}(\theta)=$
$\displaystyle\wedge^{d}M_{n_{0},E}(\theta)$ (5.576)
$\displaystyle\tilde{g}_{n_{0},4}(\theta)=$
$\displaystyle\wedge^{d}M_{n_{0},E}(\theta+(2N_{1}-n_{0})d\alpha+\kappa).$

By (5.558) with $m=k$, and recalling the definition of $g_{2^{k}N_{0},j}$ in
(5.545), we note that

(5.577)
$\displaystyle\log\|g_{2^{k}N_{0},3}(\theta)g_{2^{k}N_{0},2}(\theta)\|=$
$\displaystyle\log\|g_{2^{k-1}N_{0},4}(\theta)\cdots
g_{2^{k-1}N_{0},1}(\theta)\|$ (5.578) $\displaystyle\geq$ $\displaystyle
2^{k+1}N_{0}(1-3\varepsilon)\hat{L}^{d}-4\tilde{C}_{0}c_{0},$

where we used (5.551) in the last inequality. Recall also that

(5.579) $\displaystyle
M_{N_{1},E}(\theta+N_{1}d\alpha+\kappa)M_{N_{1},E}(\theta)=\tilde{g}_{n_{0},4}(\theta)g_{2^{k}N_{0},3}(\theta)g_{2^{k}N_{0},2}(\theta)\tilde{g}_{n_{0},1}(\theta).$

Let

(5.580) $\displaystyle\mathcal{G}_{2^{k}N_{0}}^{(1)}:=$
$\displaystyle\mathcal{G}_{2^{k-1}N_{0}}^{(1)}\cap\mathcal{G}_{n_{0}}\cap(\mathcal{G}_{n_{0}}-(2N_{1}-n_{0})d\alpha-\kappa)\cap$
(5.581)
$\displaystyle\qquad\cap\mathcal{G}_{n_{0}+2^{k}N_{0}}\cap(\mathcal{G}_{n_{0}+2^{k}N_{0}}-N_{1}d\alpha-\kappa)$

In this last step, we are going to apply Theorem 2.4 to the four matrices on
the right-hand side of (5.579). We now verify that the hypotheses are
satisfied for $\theta\in\mathcal{G}_{2^{k}N_{0}}^{(1)}$. By Lemma 2.2, one has
for $j=2,3$ that

(5.582) $\displaystyle
e^{2^{k}N_{0}(1-\varepsilon)\hat{L}^{d}}\leq\sigma_{1}(g_{2^{k}N_{0},j}(\theta))\leq
e^{2^{k}N_{0}(1+\varepsilon)\hat{L}^{d}},$

and for $j=1,4$ that

(5.583) $\displaystyle
e^{n_{0}(1-\varepsilon)\hat{L}^{d}}\leq\sigma_{1}(\tilde{g}_{n_{0},j}(\theta))\leq
e^{n_{0}(1+\varepsilon)\hat{L}^{d}}.$

We also have

(5.584)
$\displaystyle\|g_{2^{k}N_{0},2}(\theta)\tilde{g}_{n_{0},1}(\theta)\|\geq
e^{(2^{k}N_{0}+n_{0})(1-\varepsilon)\hat{L}^{d}},$ (5.585)
$\displaystyle\|\tilde{g}_{n_{0},4}(\theta)g_{2^{k}N_{0},3}(\theta)\|\geq
e^{(2^{k}N_{0}+n_{0})(1-\varepsilon)\hat{L}^{d}}.$

Similar to (5.518), for $j=2,3$,

(5.586) $\displaystyle\sigma_{2}(g_{2^{k}N_{0},j}(\theta))\leq
e^{2^{k}N_{0}(1-\varepsilon)(\hat{L}^{d-1}+\hat{L}_{d+1})}\cdot
e^{2^{k+2}N_{0}\varepsilon\hat{L}^{d-1}},$

and for $j=1,4$,

(5.587) $\displaystyle\sigma_{2}(\tilde{g}_{n_{0},j}(\theta))\leq
e^{n_{0}(1-\varepsilon)(\hat{L}^{d-1}+\hat{L}_{d+1})}\cdot
e^{2^{2}n_{0}\varepsilon\hat{L}^{d-1}}.$

Combining (5.577), (5.582), (5.583), (5.584), (5.586) and (5.587), we conclude
that for $j=2,3$

(5.588)
$\displaystyle\frac{\sigma_{1}(g_{2^{k}N_{0},j}(\theta))}{\sigma_{2}(g_{2^{k}N_{0},j}(\theta))}\geq
e^{2^{k+1}N_{0}(1-\varepsilon)\hat{L}_{d}}e^{-2^{k+2}N_{0}\varepsilon\hat{L}^{d-1}}=:(\tilde{\kappa}_{2^{k}N_{0}}^{(1)})^{-1},$

while for $j=1,4$, the following holds:

(5.589)
$\displaystyle\frac{\sigma_{1}(\tilde{g}_{n_{0},j}(\theta))}{\sigma_{2}(\tilde{g}_{n_{0},j}(\theta))}\geq
e^{2n_{0}(1-\varepsilon)\hat{L}_{d}}e^{-4n_{0}\varepsilon\hat{L}^{d-1}}\geq
e^{2^{k+1}N_{0}(1-\varepsilon)\hat{L}_{d}}e^{-2^{k+2}N_{0}\varepsilon\hat{L}^{d-1}}=(\tilde{\kappa}_{2^{k}N_{0}}^{(1)})^{-1},$

where we used $n_{0}\geq 2^{k}N_{0}$ as in (5.491). Moreover,

(5.590)
$\displaystyle\frac{\|g_{2^{k}N_{0},3}(\theta)g_{2^{k}N_{0},2}(\theta)\|}{\sigma_{1}(g_{2^{k}N_{0},3}(\theta))\cdot\sigma_{1}(g_{2^{k}N_{0},2}(\theta))}$
$\displaystyle\geq
e^{-2^{k+3}N_{0}\varepsilon\hat{L}^{d}}e^{-4\tilde{C}_{0}c_{0}}\geq
e^{-4N_{1}\varepsilon\hat{L}^{d}}e^{-4\tilde{C}_{0}c_{0}}=:\tilde{\varepsilon}_{2^{k}N_{0}}^{(1)}$
(5.591)
$\displaystyle\frac{\|g_{2^{k}N_{0},2}(\theta)\tilde{g}_{n_{0},1}(\theta)\|}{\sigma_{1}(g_{2^{k}N_{0},2}(\theta))\cdot\sigma_{1}(\tilde{g}_{n_{0},1}(\theta))}$
$\displaystyle\geq
e^{-2N_{1}\varepsilon\hat{L}^{d}}\geq\tilde{\varepsilon}_{2^{k}N_{0}}^{(1)}$
(5.592)
$\displaystyle\frac{\|\tilde{g}_{n_{0},4}(\theta)g_{2^{k}N_{0},3}(\theta)\|}{\sigma_{1}(\tilde{g}_{n_{0},4}(\theta))\cdot\sigma_{1}(g_{2^{k}N_{0},3}(\theta))}$
$\displaystyle\geq
e^{-2N_{1}\varepsilon\hat{L}^{d}}\geq\tilde{\varepsilon}_{2^{k}N_{0}}^{(1)},$

where we used $2^{k+1}N_{0}\leq N_{1}$ in the first inequality. Clearly,

(5.593)
$\displaystyle\tilde{\kappa}_{2^{k}N_{0}}^{(1)}\cdot(\tilde{\varepsilon}_{2^{k}N_{0}}^{(1)})^{-2}<c_{0}.$

Therefore, Theorem 2.4 implies that

(5.594)
$\displaystyle\log\|\tilde{g}_{n_{0},4}(\theta)g_{2^{k}N_{0},3}(\theta)g_{2^{k}N_{0},2}(\theta)\tilde{g}_{n_{0},1}(\theta)\|$
(5.595)
$\displaystyle\qquad\geq\log\|\tilde{g}_{n_{0},4}(\theta)g_{2^{k}N_{0},3}(\theta)\|+\log\|g_{2^{k}N_{0},3}(\theta)g_{2^{k}N_{0},2}(\theta)\|+\log\|g_{2^{k}N_{0},2}(\theta)\tilde{g}_{n_{0},1}(\theta)\|$
(5.596)
$\displaystyle\qquad-\log\|g_{2^{k}N_{0},3}(\theta)\|-\log\|g_{2^{k}N_{0},2}(\theta)\|-4\tilde{C}_{0}\tilde{\kappa}_{2^{k}N_{0}}^{(1)}\cdot(\tilde{\varepsilon}_{2^{k}N_{0}}^{(1)})^{-2}$
(5.597) $\displaystyle\qquad\geq
2N_{1}(1-\varepsilon)\hat{L}^{d}-2^{k+3}N_{0}\varepsilon\hat{L}^{d}-8\tilde{C}_{0}c_{0}$
(5.598) $\displaystyle\qquad\geq
2N_{1}(1-3\varepsilon)\hat{L}^{d}-8\tilde{C}_{0}c_{0}.$

Finally, to estimate the measure, recall $n_{0}\geq 2^{k}N_{0}$ and $N_{1}\geq
2^{k+1}N_{0}$ whence

(5.599) $\displaystyle\mathrm{mes}(\mathcal{G}_{2^{k}N_{0}}^{(1)})\geq$
$\displaystyle\mathrm{mes}(\mathcal{G}_{2^{k-1}N_{0}}^{(1)})-2e^{-c_{1}\varepsilon^{2}n_{0}}-2e^{-c_{1}\varepsilon^{2}N_{1}}$
(5.600) $\displaystyle\geq$ $\displaystyle
1-5\sum_{\ell=0}^{k-1}e^{-2^{\ell}c_{1}\varepsilon^{2}N_{0}\hat{L}^{d}}-2e^{-2^{k}c_{1}\varepsilon^{2}N_{0}\hat{L}^{d}}-2e^{-c_{1}\varepsilon^{2}n_{0}\hat{L}^{d}}-2e^{-c_{1}\varepsilon^{2}N_{1}\hat{L}^{d}}$
(5.601) $\displaystyle\geq$ $\displaystyle
1-5\sum_{\ell=0}^{\infty}e^{-2^{\ell}c_{1}\varepsilon^{2}N_{0}\hat{L}^{d}}>\frac{19}{20},$

where we used (5.560) with $m=k$ and (5.478). This proves the claimed result.
∎

As a corollary of Lemma 5.3, we now establish

######  Corollary 5.9.

Under the same conditions as in Lemma 5.3, for large enough
$\kappa_{0}$-admissible $n$, with $n\geq 8N_{0}$, and for any
$\theta\in(\tilde{\mathcal{B}}^{d}_{n,E,\varepsilon})^{c}$, we have that

(5.602) $\displaystyle|\langle v_{1,n}(\theta)\wedge\cdots\wedge
v_{d,n}(\theta),w_{1,n}(\theta)\wedge\cdots\wedge
w_{d,n}(\theta)\rangle|=|\det(\langle
v_{j,n}(\theta),w_{k,n}(\theta)\rangle)_{1\leq j,k\leq d}|\geq
e^{-6n\varepsilon\cdot\hat{L}^{d}}.$

###### Proof.

Towards a contradiction, suppose that for some
$\theta\in(\tilde{\mathcal{B}}^{d}_{n,E,\varepsilon})^{c}$ one has

(5.603) $\displaystyle|\langle v_{1,n}(\theta)\wedge\cdots\wedge
v_{d,n}(\theta),w_{1,n}(\theta)\wedge\cdots\wedge w_{d,n}(\theta)\rangle|$
(5.604) $\displaystyle\qquad\qquad=|\det(\langle
v_{j,n}(\theta),w_{k,n}(\theta)\rangle)_{1\leq j,k\leq d}|\leq
e^{-6n\varepsilon\cdot\hat{L}^{d}}.$

We expand

(5.605) $\displaystyle w_{1,n}(\theta)\wedge\cdots\wedge
w_{d,n}(\theta)=\sum_{1\leq j_{1}<...<j_{d}\leq
2d}a_{j_{1},...,j_{d}}(\theta)\cdot v_{j_{1},n}(\theta)\wedge\cdots\wedge
v_{j_{d},n}(\theta),$

in which

(5.606) $\displaystyle a_{j_{1},...,j_{d}}(\theta)=\langle
w_{1,n}(\theta)\wedge\cdots\wedge
w_{d,n}(\theta),v_{j_{1},n}(\theta)\wedge\cdots\wedge
v_{j_{d},n}(\theta)\rangle.$

For arbitrary $1\leq j_{1}<...<j_{d}\leq 2d$, there is the trivial bound

(5.607) $\displaystyle|a_{j_{1},...,j_{d}}(\theta)|\leq 1.$

By assumption (5.603), one has

(5.608) $\displaystyle|a_{1,...,d}(\theta)|\leq
e^{-6n\varepsilon\cdot\hat{L}^{d}}.$

We also require the following two uniform upper bounds.

######  Lemma 5.10.

For $n$ large enough, and for any $1\leq k\leq 2d-1$, $k\neq d$, one has that
uniformly in $\theta\in\mathbb{T}$,

(5.609) $\displaystyle\prod_{\ell=1}^{k}\sigma_{j_{\ell}}(M_{n,E}(\theta))\leq
e^{n\cdot(1+\varepsilon)\hat{L}^{d-1}}.$

Here $\varepsilon>0$ is arbitrary but fixed.

###### Proof.

Clearly, by Lemma 2.1,

(5.610)
$\displaystyle\prod_{\ell=1}^{k}\sigma_{j_{\ell}}(M_{n,E}(\theta))\leq\prod_{j=1}^{k}\sigma_{j}(M_{n,E}(\theta))\leq
e^{n(1+\varepsilon)\cdot\hat{L}^{k}}\leq
e^{n(1+\varepsilon)\cdot\hat{L}^{d-1}},$

where we used that for any $k\neq d$, $\hat{L}^{k}\leq\hat{L}^{d-1}$ due to
$\hat{L}_{d}>0$ and $\hat{L}_{d+1}<0$. ∎

######  Lemma 5.11.

For $(j_{1},...,j_{d})\neq(1,...,d)$, and for $n$ large enough,

(5.611) $\displaystyle\prod_{\ell=1}^{d}\sigma_{j_{\ell}}(M_{n,E}(\theta))\leq
e^{n(1+\varepsilon)\cdot\hat{L}^{d-1}}.$

uniformly in $\theta\in\mathbb{T}$.

###### Proof.

If $\sigma_{j_{d}}(M_{n,E}(\theta))<1$, then

(5.612)
$\displaystyle\prod_{\ell=1}^{d}\sigma_{j_{\ell}}(M_{n,E}(\theta))<\prod_{\ell=1}^{d-1}\sigma_{j_{\ell}}(M_{n,E}(\theta))\leq
e^{n(1+\varepsilon)\cdot\hat{L}^{d-1}},$

by Lemma 5.10. On the other hand, if $\sigma_{j_{d}}(M_{n,E}(\theta))\geq 1$,
then

(5.613)
$\displaystyle\prod_{\ell=1}^{d}\sigma_{j_{\ell}}(M_{n,E}(\theta))\leq\prod_{\ell=1}^{j_{d}}\sigma_{\ell}(M_{n,E}(\theta))\leq
e^{n(1+\varepsilon)\cdot\hat{L}^{d-1}},$

where we used $j_{d}>d$ and Lemma 5.10. ∎

By Lemmas 2.1, 2.2, 5.11 and estimates (5.607) and (5.603), one has for $n$
large enough,

(5.614)
$\displaystyle\|(\wedge^{d}M_{n,E}^{2}(\theta))(v_{1,n}(\theta)\wedge\cdots\wedge
v_{d,n}(\theta))\|$ (5.615) $\displaystyle=$
$\displaystyle\prod_{j=1}^{d}\sigma_{j}(M_{n,E}(\theta))\|(\wedge^{d}M_{n,E}(\theta)(w_{1,n}(\theta))\wedge\cdots\wedge
w_{d,n}(\theta))\|$ (5.616) $\displaystyle\leq$ $\displaystyle
e^{n(1+\varepsilon)\cdot\hat{L}^{d}}\sum_{1\leq j_{1}<...<j_{d}\leq
2d}|a_{j_{1},...,j_{d}}(\theta)|\prod_{\ell=1}^{d}\sigma_{j_{\ell}}(M_{n,E}(\theta))$
(5.617) $\displaystyle=$ $\displaystyle
e^{n(1+\varepsilon)\cdot\hat{L}^{d}}\cdot\left(|a_{1,...,d}(\theta)|\cdot\prod_{j=1}^{d}\sigma_{j}(M_{n,E}(\theta))+\sum_{(j_{1},...,j_{d})\neq(1,...,d)}|a_{j_{1},...,j_{d}}(\theta)|\cdot\prod_{\ell=1}^{d}\sigma_{j_{\ell}}(M_{n,E}(\theta))\right)$
(5.618) $\displaystyle\leq$ $\displaystyle
e^{n(1+\varepsilon)\cdot\hat{L}^{d}}\cdot\left(e^{n(1-5\varepsilon)\cdot\hat{L}^{d}}+C_{d}\cdot
e^{n(1+\varepsilon)\cdot\hat{L}^{d-1}}\right)$ (5.619) $\displaystyle\leq$
$\displaystyle e^{2n(1-2\varepsilon)\cdot\hat{L}^{d}},$

where we used (5.470). This leads to a contradiction with
$\theta\in(\tilde{\mathcal{B}}^{d}_{n,E,\varepsilon})^{c}$. ∎

Combining Corollary 5.9 with (5.475), we infer the following result.

######  Corollary 5.12.

Under the same conditions as in Lemma 5.3, for large enough
$\kappa_{0}$-admissible $n$, with $n\geq 8N_{0}$, and for any
$\theta\in(\tilde{\mathcal{B}}^{d}_{n,E,\varepsilon})^{c}\cap(\mathcal{B}_{n,E,\varepsilon}^{d})^{c}$,
one has

(5.620) $\displaystyle|\langle v_{d+1,n}(\theta)\wedge\cdots\wedge
v_{2d,n}(\theta),w_{d+1,n}(\theta)\wedge\cdots\wedge w_{2d,n}(\theta)\rangle|$
(5.621) $\displaystyle\qquad\qquad=|\det(\langle
v_{d+j,n}(\theta),w_{d+k,n}(\theta)\rangle)_{1\leq j,k\leq d}|\geq
e^{-6n\varepsilon\cdot\hat{L}^{d}}.$

The goal of the remainder of this section is to prove the following lemma.

######  Lemma 5.13.

Under the same conditions as in Lemma 5.3, for large enough
$\kappa_{0}$-admissible $n$, with $n\geq 8N_{0}$, for any
$\theta\in(\tilde{\mathcal{B}}^{d}_{n,E,\varepsilon})^{c}\cap(\mathcal{B}_{n,E,\varepsilon}^{d})^{c}$,
one has

(5.622) $\displaystyle|\det(M_{n,E}(\theta)-I_{2d})|\geq
e^{n(1-8\varepsilon)\cdot\hat{L}^{d}}.$

###### Proof.

Let $U_{v,n}(\theta)=(v_{1,n}(\theta),...,v_{2d,n}(\theta))$ and
$U_{w,n}(\theta)=(w_{1,n}(\theta),...,w_{2d,n}(\theta))$, and

$D_{n}(\theta)=\mathrm{diag}(\sigma_{1}(M_{n,E}(\theta)),...,\sigma_{2d}(M_{n,E}(\theta)))$

The singular value decomposition
$M_{n,E}(\theta)U_{v,n}(\theta)=U_{w,n}(\theta)D_{n}(\theta)$ yields

(5.623)
$\displaystyle\det(M_{n,E}(\theta)-I_{2d})=\det(D_{n}(\theta)-U_{w,n}^{*}(\theta)U_{v,n}(\theta)).$

Note that

(5.624) $\displaystyle
U_{w,n}^{*}(\theta)U_{v,n}(\theta)=\left(\begin{matrix}w_{1,n}^{*}(\theta)\\\
\vdots\\\
w_{2d,n}^{*}(\theta)\end{matrix}\right)\cdot(v_{1,n}(\theta),...,v_{2d,n}(\theta))$
$\displaystyle=\left(\langle
w_{j,n}(\theta),v_{k,n}(\theta)\rangle\right)_{1\leq j,k\leq
2d}=:Q_{n}(\theta).$

Hence,

(5.625)
$\displaystyle|\det(D_{n}(\theta)-U_{w,n}^{*}(\theta)U_{v,n}(\theta))|$
(5.626) $\displaystyle\leq$ $\displaystyle|\det(-Q_{n}(\theta))|$ (5.627)
$\displaystyle+\sum_{k=1}^{2d}\,\,\sum_{1\leq j_{1}<...<j_{k}\leq
2d}\,\,\prod_{\ell=1}^{k}\sigma_{j_{\ell}}(M_{n,E}(\theta))\cdot\left|\det(Q_{n}(\theta))_{\\{1,...,2d\\}\setminus\\{j_{1},...,j_{k}\\},\\{1,...,2d\\}\setminus\\{j_{1},...,j_{k}\\}}\right|,$

recall that for a matrix $M$, $M_{B_{1},B_{2}}$ refers to the submatrix with
row numbers in the set $B_{1}$ and column numbers taken from the set $B_{2}$.
By Hadamard’s inequality,

(5.628) $\displaystyle|\det
Q_{n}(\theta)|\leq\prod_{\ell=1}^{2d}\|(Q_{n}(\theta))_{\ell}\|\leq 1,$

and

(5.629)
$\displaystyle\left|\det(Q_{n}(\theta))_{\\{1,...,2d\\}\setminus\\{j_{1},...,j_{k}\\},\\{1,...,2d\\}\setminus\\{j_{1},...,j_{k}\\}}\right|\leq\prod_{\ell\notin\\{j_{1},..,j_{k}\\}}\|(Q_{n}(\theta))_{\ell}\|\leq
1$

where $(Q_{n}(\theta))_{\ell}$ refers to the $\ell$-th column of
$Q_{n}(\theta)$. Moreover, we used the following bound

(5.630) $\displaystyle\|(Q_{n}(\theta))_{\ell}\|\leq\|v_{\ell,n}(\theta)\|=1.$

Corollary 5.12 implies that

(5.631)
$\displaystyle\left|\det(Q_{n}(\theta))_{\\{d+1,...,2d\\},\\{d+1,...,2d\\}}\right|$
(5.632) $\displaystyle=$ $\displaystyle\left|\langle
w_{d+1,n}(\theta)\wedge\cdots\wedge
w_{2d,n}(\theta),v_{d+1,n}(\theta)\wedge\cdots\wedge
v_{2d,n}(\theta)\rangle\right|\geq e^{-6n\varepsilon\cdot\hat{L}^{d}}.$

Combining the estimates (5.628), (5.629), (5.631) with (5.625), and using
Lemmas 5.10, 5.11 and Corollary 5.12, we infer that for
$\theta\in(\tilde{\mathcal{B}}_{n,E,\varepsilon}^{d})^{c}\cap(\mathcal{B}^{d}_{n,E,\varepsilon})^{c}$

(5.633) $\displaystyle|\det(M_{n}(\theta)-I_{2d})|\geq$
$\displaystyle\prod_{j=1}^{d}\sigma_{j}(M_{n}(\theta))\cdot
e^{-6n\varepsilon\cdot\hat{L}^{d}}-C_{d}\,\sup_{k=0,\ldots,2d}\,\sup_{\begin{subarray}{c}1\leq
j_{1}<...<j_{k}\leq 2d\\\
(j_{1},...,j_{k})\neq(1,...,d)\end{subarray}}\,\prod_{\ell=1}^{k}\sigma_{j_{\ell}}(M_{n,E}(\theta))$
(5.634) $\displaystyle\geq$ $\displaystyle
e^{n(1-\varepsilon)\cdot\hat{L}^{d}}\cdot
e^{-6n\varepsilon\cdot\hat{L}^{d}}-C_{d}\,e^{n(1+\varepsilon)\cdot\hat{L}^{d-1}}$
(5.635) $\displaystyle\geq$ $\displaystyle
e^{n(1-8\varepsilon)\cdot(\hat{L}^{d}-\varepsilon)},$

where we used (5.470). This is the claimed result. ∎

Alternatively, Lemma 5.13 can also be proved via Schur complements. Lemma 3.2
follows immediately from combining Lemmas 5.1 and 5.13 with the measure
estimates of $\tilde{\mathcal{B}}^{d}_{n,E,\varepsilon}$ in Lemma 5.3 and
$\mathcal{B}^{d}_{n,E,\varepsilon}$ in Lemma 2.2. ∎

###  5.1. Proof of Lemma 3.5

Let $z=e^{2\pi i\theta}$ and $F_{E,n}(z)=f_{E,n}(\theta)$, and
$g_{E,n}(z)=z^{nd}F_{E,n}(z)$. It is easy to check that $|g_{E,n}(0)|=1$. Let
$\mathcal{C}$ be the unit circle. In the following, we use Herman’s
subharmonic argument:

(5.636)
$\displaystyle\frac{1}{nd}\int_{\mathbb{T}}\log|f_{E,n}(\theta)|\,\mathrm{d}\theta=$
$\displaystyle\frac{1}{nd}\int_{\mathcal{C}}\log|F_{E,n}(z)|\,\frac{\mathrm{d}z}{2\pi
iz}$ (5.637) $\displaystyle=$
$\displaystyle\frac{1}{nd}\int_{\mathcal{C}}\log|z^{nd}F_{E,n}(z)|\,\frac{\mathrm{d}z}{2\pi
iz}$ (5.638) $\displaystyle=$
$\displaystyle\frac{1}{nd}\int_{\mathcal{C}}\log|g_{E,n}(z)|\,\frac{\mathrm{d}z}{2\pi
iz}$ (5.639) $\displaystyle\geq$
$\displaystyle\frac{1}{nd}\log|g_{E,n}(0)|=0.$

It remains to recall that by Haro-Puig’s result, see Theorem 1.8, for
$E\in\sum_{\alpha,0}$ in the subcritical regime,

(5.640) $\displaystyle
0=L(\alpha,A_{E})=\hat{L}^{d}(\alpha,\hat{A}_{E})+\log|\hat{v}_{d}|.$

This proves Lemma 3.5. ∎

##  6\. Non-selfadjoint Haro-Puig formula

###  6.1. Proof of Corollary 1.12

We shall write $\hat{L}_{j}(\alpha,\hat{A}_{E}^{\varepsilon})$ as
$\hat{L}_{j}(\varepsilon)$ for simplicity. By Theorem 1.10 and (2.50) one has
that

(6.641) $\displaystyle L(\alpha,A_{E,\varepsilon})=$
$\displaystyle\sum_{j:\hat{L}_{j}(\varepsilon)\geq
0}\hat{L}_{j}(\varepsilon)+\log|\hat{v}_{-d}e^{2\pi d\varepsilon}|$ (6.642)
$\displaystyle=$ $\displaystyle\sum_{j:\hat{L}_{j}(0)\geq
2\pi\varepsilon}(\hat{L}_{j}(0)-2\pi\varepsilon)+\log|\hat{v}_{-d}|+2\pi
d\varepsilon$ (6.643) $\displaystyle=$ $\displaystyle
L(\alpha,A_{E,0})-\sum_{\begin{subarray}{c}1\leq j\leq d\\\
0\leq\hat{L}_{j}(0)<2\pi\varepsilon\end{subarray}}\hat{L}_{j}(0)+2\pi\varepsilon\cdot\\#\\{j:1\leq
j\leq d,\,0\leq\hat{L}_{j}<2\pi\varepsilon\\}$ (6.644) $\displaystyle=$
$\displaystyle L(\alpha,A_{E,0})+\sum_{\begin{subarray}{c}1\leq j\leq d\\\
0\leq\hat{L}_{j}(0)<2\pi\varepsilon\end{subarray}}(2\pi
d\varepsilon-\hat{L}_{j}(0))$

This can be further be simplified in the form:

(6.645) $\displaystyle L(\alpha,A_{E,\varepsilon})=$ $\displaystyle
L(\alpha,A_{E,0})+\int_{0}^{2\pi\varepsilon}\sum_{\begin{subarray}{c}1\leq
j\leq d\\\
0\leq\hat{L}_{j}(0)<2\pi\varepsilon\end{subarray}}\chi_{[\hat{L}_{j}(0),2\pi\varepsilon]}(y)\,\mathrm{d}y$
(6.646) $\displaystyle=$ $\displaystyle
L(\alpha,A_{E,0})+\int_{0}^{2\pi\varepsilon}\\#\\{j:1\leq j\leq
d,\,0\leq\hat{L}_{j}(0)<y\\}\,\mathrm{d}y.$

This implies

(6.647) $\displaystyle\kappa(\alpha,A_{E,\varepsilon})=\\#\\{j:1\leq j\leq
d,\,0\leq\hat{L}_{j}(0)\leq 2\pi d\varepsilon\\},$

whence, in particular,

(6.648) $\displaystyle\kappa(\alpha,A_{E,0})=\frac{1}{2}\cdot\\#\\{1\leq j\leq
2d\;:\;\hat{L}_{j}(0)=0\\}.$

This is the claimed result. ∎

The remainder of this section consists of a proof of Theorem 1.10.

###  6.2. Preparation

In the following, let $(\alpha,A)\in\mathbb{T}\times
C^{\omega}(\mathbb{T},\mathrm{Mat}(d,\mathbb{C}))$ be an analytic cocycle. The
following lemma and theorems from [AJS] play an important role in our proof.
All fractions $p/q$ are reduced (in lowest terms), and we may assume in
addition that $d|q$ (we will need this at some point below). Indeed, if
$|\frac{p_{n}}{q_{n}}-\alpha|\to 0$ as $n\to\infty$ for an irrational
$\alpha$, then also $|\frac{dp_{n}+1}{dq_{n}}-\alpha|\to 0$. No prime dividing
$d$ can divide $dp_{n}+1$, so the denominator remains divisible by $d$ after
removing common factors.

######  Definition 6.1.

[AJS] We say $(\alpha,A)$ is $k$-regular if $t\to L^{k}(\alpha,A(\cdot+it))$
is an affine function of $t$ in a neighborhood of $0$.

We will only use the definition of $1$-regular for the cocycle
$(\alpha,A_{E,\varepsilon})\in\mathbb{T}\times
C^{\omega}(\mathbb{T},\mathrm{Mat}(2,\mathbb{C}))$.

######  Lemma 6.1.

[AJS]*Lemma 5.1 If $L^{k}(\alpha,A)>-\infty$, then for arbitrary small
$\varepsilon_{0}>0$ and $p/q$ sufficiently close to $\alpha$, one has
uniformly for all $\theta$ that

(6.649) $\displaystyle L^{k}(p/q,A,\theta)\leq
L^{k}(\alpha,A)+\varepsilon_{0}.$

######  Theorem 6.2.

[AJS]*Theorem 5.2 Assume $L^{k}(\alpha,A)>-\infty$ and that $(\alpha,A)$ is
$k$-regular. Then for arbitrary small $\varepsilon_{0}>0$ and $p/q$
sufficiently close to $\alpha$, one has uniformly for all
$\theta\in\mathbb{T}$,

(6.650)
$\displaystyle|L^{k}(p/q,A,\theta)-L^{k}(\alpha,A)|\leq\varepsilon_{0}.$

######  Theorem 6.3.

[AJS]*Theorem 1.5 The functions
$(\mathbb{T},C^{\omega}(\mathbb{T},\mathrm{Mat}(d,\mathbb{C})))\ni(\alpha,A)\to
L_{k}(\alpha,A)\in[-\infty,\infty)$ are continuous at any
$(\alpha^{\prime},A^{\prime})$ with
$\alpha^{\prime}\in\mathbb{R}\setminus\mathbb{Q}$.

The case of $L_{k}=-\infty$ does not arise here. Since there are only finitely
many $\varepsilon$’s such that $(\alpha,A_{E,\varepsilon})$ is not
$1$-regular, it suffices to prove Theorem 1.10 for any $\varepsilon$ at which
$(\alpha,A_{E,\varepsilon})$ is $1$-regular. Also, due to (2.50), there are
only finitely many values of $\varepsilon$ for which some of the exponents
$\hat{L}_{j}(\alpha,\hat{A}_{E}^{\varepsilon})$ vanish. Hence it suffices to
assume that none of the
$\hat{L}_{E}^{\varepsilon}(\alpha,\hat{A}_{E}^{\varepsilon})$ are zero. The
identity (1.17) for arbitrary $\varepsilon$ will then follow from continuity
in $\varepsilon$ by Theorem 6.3. We will thus make the following assumption
below.

######  Assumption 6.4.

Let $\varepsilon$ be such that $(\alpha,A_{E,\varepsilon})$ is $1$-regular.

###  6.3. Proof of Theorem 1.10

The goal is to prove the following rational frequency $\alpha=p/q$ version of
Theorem 1.10. Let $k_{0}\in\\{1,...,2d\\}$ be such that

(6.651)
$\displaystyle\hat{L}_{1}(\alpha,\hat{A}_{E}^{\varepsilon})\geq\cdots\geq\hat{L}_{k_{0}}(\alpha,\hat{A}_{E}^{\varepsilon})>0>\hat{L}_{k_{0}+1}(\alpha,\hat{A}_{E}^{\varepsilon})\geq\cdots\geq\hat{L}_{2d}(\alpha,\hat{A}_{E}^{\varepsilon}),$

assuming that $\hat{L}_{1}(\alpha,\hat{A}_{E}^{\varepsilon})>0$. We allow for
all exponents to be positive here. We will deal with the remaining case
$\hat{L}_{1}(\alpha,\hat{A}_{E}^{\varepsilon})<0$ at the end of this section
(the vanishing case can be ignored, see above).

######  Lemma 6.5.

For any $\varepsilon$ such that $(\alpha,A_{E}^{\varepsilon})$ is $1$-regular
and for arbitrary small $\varepsilon_{0}$ such that

$0<\varepsilon_{0}<\frac{1}{10}\min(\hat{L}_{k_{0}}(\alpha,\hat{A}_{E}^{\varepsilon}),-\hat{L}_{k_{0}+1}(\alpha,\hat{A}_{E}^{\varepsilon})),$

the following holds for some constant $C>0$:

(6.652)
$\displaystyle\big{|}L^{1}(\alpha,A_{E,\varepsilon})-\hat{L}^{k_{0}}(\alpha,\hat{A}_{E}^{\varepsilon})-\log|\hat{v}_{d}e^{2\pi
d\varepsilon}|\big{|}\leq
C\varepsilon_{0}\left(L^{k_{0}}(\alpha,\hat{A}_{E}^{\varepsilon})+\big{|}\log|\hat{v}_{-d}e^{2\pi
d\varepsilon}|\big{|}\right).$

Clearly, Theorem 1.10 follows from taking $\varepsilon_{0}\to 0^{+}$ in Lemma
6.5. To prove Lemma 6.5, we consider the following three periodic operators
acting on $\ell^{2}(\mathbb{Z}_{q})$:

(6.653) $\displaystyle(H_{p/q,\theta,\varepsilon}u)_{[n]}$
$\displaystyle=u_{[n+1]}+u_{[n-1]}+v(\theta+i\varepsilon+np/q)u_{[n]},$
(6.654) $\displaystyle(\tilde{H}_{p/q,\theta,\varepsilon}u)_{[n]}$
$\displaystyle=e^{2\pi i\theta}u_{[n+1]}+e^{-2\pi
i\theta}u_{[n-1]}+v(np/q+i\varepsilon)u_{[n]},$

and

(6.655)
$\displaystyle(\hat{H}_{p/q,\theta,\varepsilon}\phi)_{[n]}=\sum_{|k|=1}^{d}\hat{v}_{k}e^{-2\pi
k\varepsilon}\phi_{[n-k]}+2\cos(2\pi(\theta+np/q))\phi_{[n]}.$

It is clear that we have

######  Fact 6.6.

$H_{p/q,0,\varepsilon}=\tilde{H}_{p/q,0,\varepsilon}$.

We will work with the Fourier transform on the cyclic group $\mathbb{Z}_{q}$.
We write the characters on this finite Abelian group in the form
$\chi_{m}(n)=e^{-2\pi imnp/q}$, Haar measure being counting measure. Note that
$np$ for $1\leq n\leq q$ is just a relabeling of the congruence classes modulo
$q$. The Fourier transform is thus defined in the form

(6.656) $\displaystyle(\mathcal{F}\phi)_{[n]}$
$\displaystyle=\frac{1}{\sqrt{q}}\sum_{m\in\mathbb{Z}_{q}}e^{2\pi
imnp/q}\phi_{[m]},$ (6.657) $\displaystyle(\mathcal{F}^{-1}u)_{[n]}$
$\displaystyle=\frac{1}{\sqrt{q}}\sum_{m\in\mathbb{Z}_{q}}e^{-2\pi
imnp/q}u_{[m]},$

and gives a unitary transformation on $\ell^{2}(\mathbb{Z}_{q})$. Lemma 6.7
establishes Aubry duality for the periodic operators by means of this Fourier
transform.

######  Lemma 6.7.

$\mathcal{F}\tilde{H}_{p/q,\theta,\varepsilon}\mathcal{F}^{-1}=\hat{H}_{p/q,\theta,\varepsilon}.$

###### Proof.

For simplicity, we write $\hat{H}_{p/q,\theta,\varepsilon}$ as $\hat{H}$. Then

(6.658) $\displaystyle(\mathcal{F}\hat{H}\mathcal{F}^{-1}u)_{[n]}$ (6.659)
$\displaystyle=$
$\displaystyle\frac{1}{\sqrt{q}}\sum_{m\in\mathbb{Z}_{q}}e^{2\pi
imnp/q}(\hat{H}\mathcal{F}^{-1}u)_{[m]}$ (6.660) $\displaystyle=$
$\displaystyle\frac{1}{\sqrt{q}}\sum_{m\in\mathbb{Z}_{q}}e^{2\pi
imnp/q}\left(\sum_{|k|=1}^{d}\hat{v}_{k}e^{-2\pi
k\varepsilon}(\mathcal{F}^{-1}u)_{[m-k]}+2\cos(2\pi(\theta+mp/q))(\mathcal{F}^{-1}u)_{[m]}\right)$
(6.661) $\displaystyle=$
$\displaystyle\frac{1}{q}\sum_{m,\ell\in\mathbb{Z}_{q}}\sum_{|k|=1}^{d}e^{2\pi
imnp/q}e^{-2\pi i\ell(m-k)p/q}\,\hat{v}_{k}\,e^{-2\pi k\varepsilon}u_{[\ell]}$
(6.662) $\displaystyle+\frac{1}{q}\sum_{m,\ell\in\mathbb{Z}^{q}}e^{2\pi
imnp/q}e^{-2\pi im\ell p/q}(e^{2\pi i(\theta+mp/q)}+e^{-2\pi
i(\theta+mp/q)})u_{[\ell]}$ (6.663) $\displaystyle=$
$\displaystyle\sum_{|k|=1}^{d}\hat{v}_{k}e^{-2\pi k\varepsilon}e^{2\pi
inkp/q}u_{[n]}+e^{2\pi i\theta}u_{[n+1]}+e^{-2\pi i\theta}u_{[n-1]}$ (6.664)
$\displaystyle=$ $\displaystyle v(np/q+i\varepsilon)u_{[n]}+e^{2\pi
i\theta}u_{[n+1]}+e^{-2\pi i\theta}u_{[n-1]}$ (6.665) $\displaystyle=$
$\displaystyle(\tilde{H}_{p/q,\theta,\varepsilon}u)_{[n]},$

as claimed. ∎

For $E\in\mathbb{C}$, let
$D_{p/q,E,\varepsilon}(\theta):=\det(H_{p/q,\theta,\varepsilon}-E)$ and define
$\tilde{D}$, $\hat{D}$ analogously for the operators $\tilde{H}$ and
$\hat{H}$, respectively. The Aubry duality Lemma 6.7 implies the following
invariance property.

######  Corollary 6.8.

For any $E\in\mathbb{C}$ and any $\theta\in\mathbb{T}$, we have

(6.666)
$\displaystyle\hat{D}_{p/q,E,\varepsilon}(\theta)=\tilde{D}_{p/q,E,\varepsilon}(\theta).$

By inspection we have

######  Fact 6.9.

For any $E\in\mathbb{C}$, $D_{p/q,E,\varepsilon}(\theta)$ is a
$(p/q)$-periodic (hence $(1/q)$-periodic) function in $\theta$. Analogous
results hold for $\tilde{D}$ and $\hat{D}$.

The following Chamber’s type formula is an easy consequence of Fact 6.9.

######  Lemma 6.10.

For any $E\in\mathbb{C}$, we have

(6.667)
$\displaystyle\hat{D}_{p/q,E,\varepsilon}(\theta)=\tilde{D}_{p/q,E,\varepsilon}(\theta)=a(p/q,E,\varepsilon)+2(-1)^{q+1}\cos(2\pi
q\theta),$

where $a(p/q,E,\varepsilon)$ is independent of $\theta$. This, combined with
Fact 6.6 implies that

(6.668)
$\displaystyle\hat{D}_{p/q,E,\varepsilon}(\theta)=\tilde{D}_{p/q,E,\varepsilon}(\theta)=D_{p/q,E,\varepsilon}(0)+(-1)^{q+1}(2\cos(2\pi
q\theta)-2).$

###### Proof.

The $\hat{D}=\tilde{D}$ part is due to Corollary 6.8. The term of highest
degree (in $e^{2\pi i\theta}$) in $\tilde{D}_{p/q,E,\varepsilon}(\theta)$
equals $(-1)^{q+1}e^{2\pi iq\theta}$, and that of lowest degree is
$(-1)^{q+1}e^{-2\pi iq\theta}$. The lemma now follows from Fact 6.9. ∎

Combining (6.668), Corollary 6.8 with Jensen’s formula implies the following.

######  Lemma 6.11.

Denote the two zeros of

$z^{2}+((-1)^{q+1}D_{p/q,E,\varepsilon}(0)-2)z+1=0$

by $z_{1},z_{2}$, respectively. Then for any $E\in\mathbb{C}$,

(6.669)
$\displaystyle\int_{\mathbb{T}}\log|\hat{D}_{p/q,E,\varepsilon}(\theta)|\,\mathrm{d}\theta=\log\max(|z_{1}|,|z_{2}|,1).$

Next, we compare the right-hand side of (6.669) to
$qL^{1}(\alpha,A_{E,\varepsilon})$ for all large $q$. Note that
$A_{E,\varepsilon}\in\mathrm{SL}(2,\mathbb{C})$ implies that
$L^{1}(\alpha,A_{E,\varepsilon})\geq 0$.

######  Lemma 6.12.

If $(\alpha,A_{E,\varepsilon})$ is $1$-regular, for some constant $C>0$, we
have

(6.670)
$\displaystyle\left|\frac{1}{q}\int_{\mathbb{T}}\log|\hat{D}_{p/q,E,\varepsilon}(\theta)|\,\mathrm{d}\theta-L^{1}(\alpha,A_{E,\varepsilon})\right|\leq
C\varepsilon_{0}$

provided $q$ is large.

###### Proof.

First, simple considerations yield

(6.671)
$\displaystyle\left|\max(|z_{1}|,|z_{2}|)-|D_{p/q,E,\varepsilon}(0)|\right|\leq
3.$

Hence by Lemma 6.11, we have

(6.672)
$\displaystyle\frac{1}{q}\log\left(\max(|D_{p/q,E,\varepsilon}(0)|-3,1\right)\leq\frac{1}{q}\int_{\mathbb{T}}\log|\hat{D}_{p/q,E,\varepsilon}(\theta)|\,\mathrm{d}\theta\leq\frac{1}{q}\log(|D_{p/q,E,\varepsilon}(0)|+3).$

Next, we relate $D_{p/q,E,\varepsilon}(0)$ to
$L^{1}(\alpha,A_{E,\varepsilon})$. Let

$|\lambda_{1,E,\varepsilon}(\theta)|\geq|\lambda_{1,E,\varepsilon}(\theta)|^{-1}=|\lambda_{2,E,\varepsilon}(\theta)|$

be the eigenvalues of $A_{q,E,\varepsilon}(p/q,\theta)$. Applying (2.26) to
$A=A_{E,\varepsilon}$, we have

(6.673) $\displaystyle
L^{1}(p/q,A_{E,\varepsilon},\theta)=\frac{1}{q}\log|\lambda_{1,E,\varepsilon}(\theta)|.$

The following lemma relates $D_{p/q,E,\varepsilon}(\theta)$ to
$A_{q,E,\varepsilon}(p/q,\theta)$.

######  Lemma 6.13.

We have for all $\theta\in\mathbb{T}$

(6.674)
$\displaystyle|D_{p/q,E,\varepsilon}(\theta)|=|\det(A_{q,E,\varepsilon}(p/q,\theta)-I_{2})|=|\lambda_{1,E,\varepsilon}(\theta)-1|\cdot|\lambda_{2,E,\varepsilon}(\theta)-1|.$

###### Proof.

The proof is very similar to that of Lemma 5.1. As in the previous sections,
we perform elementary row operations (which leave the determinant invariant),
on the $q\times q$ matrix

(6.675) $\displaystyle
H_{p/q,\theta,\varepsilon}-E=\left(\begin{matrix}v_{q-1}&1&&&&&1\\\
1&v_{q-2}&\ddots&&&&\\\ &\ddots&\ddots&&&&\\\ \\\ &&&&&\ddots&\\\
&&&&\ddots&\ddots&1\\\
1&&&&&1&v_{0}\end{matrix}\right)=:\left(\begin{matrix}\text{Row}_{1}\\\
\vdots\\\ \text{Row}_{q}\end{matrix}\right)$

where we write $v(\theta+i\varepsilon+jp/q)-E$ as $v_{j}$ for simplicity. We
also denote $A_{n,E,\varepsilon}(p/q,\theta+jp/q):=A_{n}(j)$, and

(6.676) $\displaystyle
A_{n}(j)=\left(\begin{matrix}A_{n}^{11}(j)\,&A_{n}^{12}(j)\\\
A_{n}^{21}(j)&A_{n}^{22}(j)\end{matrix}\right).$

Our calculations are based on the following recursive relations:

(6.677)
$\displaystyle\begin{cases}A_{k}^{11}(q-k)=-v_{q-k}A_{k-1}^{11}(q-k+1)+A_{k-1}^{12}(q-k+1)\\\
A_{k}^{12}(q-k)=-A_{k-1}^{11}(q-k+1)\\\ A_{1}^{11}(k)=-v_{k}\\\
A_{1}^{12}(k)=-1\end{cases}$

Replacing $\text{Row}_{1}$ with $\text{Row}_{1}-v_{q-1}\cdot\text{Row}_{2}$
yields

(6.678) $\displaystyle\text{Row}_{1}^{(1)}=$
$\displaystyle(\begin{matrix}0,&1-v_{q-1}v_{q-2},&-v_{q-1},&0,\cdots,&0,&1\end{matrix})$
(6.679) $\displaystyle=$
$\displaystyle(\begin{matrix}0,&-A_{2}^{11}(q-2),&-A_{2}^{12}(q-2),&0,\cdots,&0,&1\end{matrix})$

Replacing $\text{Row}_{1}^{(1)}$ with
$\text{Row}_{1}^{(1)}+A_{2}^{11}(q-2)\cdot\text{Row}_{3}$ yields

(6.680) $\displaystyle\text{Row}_{1}^{(2)}=$
$\displaystyle(\begin{matrix}0,&0,&-A_{2}^{12}(q-2)+v_{q-3}A_{2}^{11}(q-2),&A_{2}^{11}(q-2),&0,\dots,&0,&1\end{matrix})$
(6.681) $\displaystyle=$
$\displaystyle(\begin{matrix}0,&0,&-A_{3}^{11}(q-3),&-A_{3}^{12}(q-3),&0,\cdots,&0,&1\end{matrix})$

We repeat this process until

(6.682)
$\displaystyle\text{Row}_{1}^{(q-2)}=(\begin{matrix}0,\cdots&0,&-A_{q-1}^{11}(1),&1-A_{q-1}^{12}(1)\end{matrix}).$

Next, we carry out an analogous row reduction on the second row. Replacing
$\text{Row}_{2}$ with $\text{Row}_{2}-v_{q-2}\cdot\text{Row}_{3}$ yields

(6.683)
$\displaystyle\text{Row}_{2}^{(1)}=(\begin{matrix}1,&0,&-A_{2}^{11}(q-3),&-A_{2}^{12}(q-3)\,,&0,\cdots,&0\end{matrix}).$

We repeat this process until

(6.684)
$\displaystyle\text{Row}_{2}^{(q-3)}=(\begin{matrix}1,&0,\cdots,&0,&-A_{q-2}^{11}(1),&-A_{q-2}^{12}(1))\end{matrix}.$

Note that the first, second and last rows, respectively, of the matrix we
started with now read

(6.685) $\displaystyle\left(\begin{matrix}\text{Row}_{1}^{(q-2)}\\\
\text{Row}_{2}^{(q-3)}\\\
\text{Row}_{q}\end{matrix}\right)=\left(\begin{matrix}0&0&\cdots&0&-A^{11}_{q-1}(1)&1-A^{12}_{q-1}(1)\\\
1&0&\cdots&0&-A^{11}_{q-2}(1)&-A^{12}_{q-2}(1)\\\
1&0&\cdots&0&1&v_{0}\end{matrix}\right).$

We expand the determinant of the transformed matrix
$H_{p/q,\theta,\varepsilon}-E$ along the second column using the identities
$A_{q-2}^{11}(1)=A_{q-1}^{21}(1)$, $A_{q-2}^{12}(1)=A_{q-1}^{22}(1)$. This
yields

(6.686) $\displaystyle|D_{p/q,E,\varepsilon}(\theta)|=$
$\displaystyle\left|\det\left(\begin{matrix}0&-A_{q-1}^{11}(1)&1-A_{q-1}^{12}(1)\\\
1&-A_{q-2}^{11}(1)&-A_{q-2}^{12}(1)\\\ 1&1&v_{0}\end{matrix}\right)\right|$
(6.687) $\displaystyle=$
$\displaystyle\left|\det\left(\begin{matrix}0&-A_{q-1}^{11}(1)&1-A_{q-1}^{12}(1)\\\
1&-A_{q-1}^{21}(1)&-A_{q-1}^{22}(1)\\\ 1&1&v_{0}\end{matrix}\right)\right|$
(6.688) $\displaystyle=$
$\displaystyle\left|\det\left(\begin{matrix}0&0&-A_{q-1}^{11}(1)&1-A_{q-1}^{12}(1)\\\
0&1&-A_{q-1}^{21}(1)&-A_{q-1}^{22}(1)\\\ 1&0&0&-1\\\
0&1&1&v_{0}\end{matrix}\right)\right|$ (6.689) $\displaystyle=$
$\displaystyle\left|\det\left(\begin{matrix}\left(\begin{matrix}0&0\\\
0&1\end{matrix}\right)&-A_{q-1}(1)+\left(\begin{matrix}0&1\\\
0&0\end{matrix}\right)\\\ I_{2}&-(A_{1}(0))^{-1}\end{matrix}\right)\right|$
(6.690) $\displaystyle=$
$\displaystyle\left|\det\left(\left(\begin{matrix}0&0\\\
0&1\end{matrix}\right)-\left(A_{q-1}(1)-\left(\begin{matrix}0&1\\\
0&0\end{matrix}\right)\right)A_{1}(0)\right)\right|$ (6.691) $\displaystyle=$
$\displaystyle\left|\det(A_{q}(0)-I_{2})\right|,$

as claimed. ∎

By Assumption 6.4, $(\alpha,A_{E,\varepsilon})$ is 1-regular, hence by Theorem
6.2 and (6.673), for $p/q$ sufficiently close to $\alpha$,

(6.692)
$\displaystyle\left|\frac{1}{q}\log|\lambda_{1,E,\varepsilon}(0)|-L^{1}(\alpha,A_{E,\varepsilon})\right|\leq\varepsilon_{0}.$

This implies, by $\lambda_{1,E,\varepsilon}\lambda_{2,E,\varepsilon}=1$, that

(6.693)
$\displaystyle\left|\frac{1}{q}\log|\lambda_{2,E,\varepsilon}(0)|+L^{1}(\alpha,A_{E,\varepsilon})\right|\leq\varepsilon_{0}.$

Hence by Lemma 6.13, we have for all sufficiently large $q$

(6.694)
$\left|\frac{1}{q}\log|D_{p/q,E,\varepsilon}(0)|-L^{1}(\alpha,A_{E,\varepsilon})\right|\leq
C\varepsilon_{0}.$

Combining the above with (6.672) yields Lemma 6.12. ∎

Finally it remains to relate $\hat{D}_{p/q,E,\varepsilon}$ to the dual
Lyapunov exponents. As stated, the following argument requires $q$ to be
divisible by $d$ (see above why can assume this).

######  Lemma 6.14.

For $q=rd$ with $r\in\mathbb{N}$, we have

(6.695) $\displaystyle\log|\hat{D}_{p/q,E,\varepsilon}(\theta)|=$
$\displaystyle\log|\det(M_{r,E}^{\varepsilon}(p/r,\theta)-I_{2d})|+q\log|\hat{v}_{-d}e^{2\pi
d\varepsilon}|$ (6.696) $\displaystyle=$
$\displaystyle\log|\det(\hat{A}_{q,E}^{\varepsilon}(p/q,\theta)-I_{2d})|+q\log|\hat{v}_{-d}e^{2\pi
d\varepsilon}|.$

###### Proof.

(6.696) follows from (6.695), due to (2.48). The proof of (6.695) is similar
to those of Lemmas 6.13 and 5.1, and is based on elementary row operations.
Consider the operator

(6.697)
$\displaystyle\hat{H}_{p/q,\theta,\varepsilon}-E=\left(\begin{matrix}C_{r-1}&\tilde{B}&&&&&B\\\
B&C_{r-2}&\ddots&&&&\\\ &\ddots&\ddots&&&&\\\ \\\ &&&&\ddots&\ddots&\\\
&&&&\ddots&C_{1}&\tilde{B}\\\
\tilde{B}&&&&&B&C_{0}\end{matrix}\right)=:\left(\begin{matrix}\text{Row}_{1}\\\
\vdots\\\ \text{Row}_{r}\end{matrix}\right),$

where $C_{j}=C_{\varepsilon}(\theta+jp/r)-E$ and $B=B_{\varepsilon}$,
$\tilde{B}=\tilde{B}_{\varepsilon}$. Denote
$M_{r,E}^{\varepsilon}(p/r,\theta+jp/r)=:M_{r}(j)$, and

(6.698) $\displaystyle
M_{r}(j)=\left(\begin{matrix}M_{r}^{11}(j)&M_{r}^{12}(j)\\\
M_{r}^{21}(j)&M_{r}^{22}(j)\end{matrix}\right).$

We will use the following recursive relations repeatedly:

(6.699)
$\displaystyle\begin{cases}M_{r}^{11}(r-k)=-M_{r-1}^{11}(r-k+1)C_{r-k}B^{-1}+M_{r-1}^{12}(r-k+1)B^{-1}\\\
M_{r}^{12}(r-k)=-M_{r-1}^{11}(r-k+1)\tilde{B}\\\ M_{1}^{11}(k)=-C_{k}B^{-1}\\\
M_{1}^{12}(k)=-\tilde{B}\end{cases}$

Replacing $\text{Row}_{1}$ with
$\text{Row}_{1}-C_{r-1}B^{-1}\cdot\text{Row}_{2}$ yields

(6.700) $\displaystyle\text{Row}_{1}^{(1)}=$
$\displaystyle(\begin{matrix}0,&\tilde{B}-C_{r-1}B^{-1}C_{r-2},&-C_{r-1}B^{-1}\tilde{B},&0,\cdots,&0,&B\end{matrix})$
(6.701) $\displaystyle=$
$\displaystyle(\begin{matrix}0,&-M_{2}^{11}(r-2)B,&-M_{2}^{12}(r-2),&0,\cdots,&0,&B).\end{matrix}$

Replacing $\text{Row}_{1}^{(1)}$ with
$\text{Row}_{1}^{(1)}+M_{2}^{11}(r-2)\cdot\text{Row}_{3}$ yields

(6.702) $\displaystyle\text{Row}_{1}^{(2)}=$
$\displaystyle(\begin{matrix}0,&0,&-M_{2}^{12}(r-2)+M_{2}^{11}(r-2)C_{r-3},&M_{2}^{11}(r-2)\tilde{B},&0,\cdots,&0,&B\end{matrix})$
(6.703) $\displaystyle=$
$\displaystyle(\begin{matrix}0,&0,&-M_{3}^{11}(r-3)B,&-M_{3}^{12}(r-3),&0,\cdots,&0,&B\end{matrix}).$

Repeating this process we arrive at

(6.704)
$\displaystyle\text{Row}_{1}^{(r-2)}=(\begin{matrix}0,\cdots,&0,&-M_{r-1}^{11}(1)B,&B-M_{r-1}^{12}(1)\end{matrix}).$

Next we replace $\text{Row}_{2}$ with
$\text{Row}_{2}-C_{r-2}B^{-1}\cdot\text{Row}_{3}$, which yields

(6.705) $\displaystyle\text{Row}_{2}^{(1)}=$
$\displaystyle(\begin{matrix}B,&0,&\tilde{B}-C_{r-2}B^{-1}C_{r-3},&-C_{r-2}B^{-1}\tilde{B},&0,\cdots,&0\end{matrix})$
(6.706) $\displaystyle=$
$\displaystyle(\begin{matrix}B,&0,&-M_{2}^{11}(r-3)B,&-M_{2}^{12}(r-3),&0,\cdots,&0\end{matrix})$

Repeating this process we obtain

(6.707)
$\displaystyle\text{Row}_{2}^{(r-3)}=(\begin{matrix}B,&0,\cdots&0,&-M_{r-2}^{11}(1)B,&-M_{r-2}^{12}(1)\end{matrix})$

The first, second and last rows are

(6.708) $\displaystyle\left(\begin{matrix}\text{Row}_{1}^{(q-2)}\\\
\text{Row}_{2}^{(q-3)}\\\
\text{Row}_{r}\end{matrix}\right)=\left(\begin{matrix}0&0&\cdots&0&-M_{r-1}^{11}(1)B&B-M_{r-1}^{12}(1)\\\
B&0&\cdots&0&-M_{r-2}^{11}(1)B&-M_{r-2}^{12}(1)\\\
\tilde{B}&0&\cdots&0&B&C_{0}\end{matrix}\right)$

Expanding determinants and using that $M_{r-2}^{11}(1)=BM_{r-1}^{21}(1)$,
$M_{r-2}^{12}(1)=BM_{r-1}^{22}(1)$, we have

(6.709) $\displaystyle|\hat{D}_{p/q,E,\varepsilon}(\theta)|=$
$\displaystyle|\det
B|^{r-2}\cdot\left|\det\left(\begin{matrix}0&-M_{r-1}^{11}(1)B&B-M_{r-1}^{12}(1)\\\
B&-M_{r-2}^{11}(1)B&-M_{r-2}^{12}(1)\\\
\tilde{B}&B&C_{0}\end{matrix}\right)\right|$ (6.710) $\displaystyle=$
$\displaystyle|\det
B|^{r-2}\cdot\left|\det\left(\begin{matrix}0&-M_{r-1}^{11}(1)B&B-M_{r-1}^{12}(1)\\\
B&-BM_{r-1}^{21}(1)B&-BM_{r-1}^{22}(1)\\\
\tilde{B}&B&C_{0}\end{matrix}\right)\right|$ (6.711) $\displaystyle=$
$\displaystyle|\det
B|^{r-1}\cdot|\det\tilde{B}|\cdot\left|\det\left(\begin{matrix}0&-M_{r-1}^{11}(1)&B-M_{r-1}^{12}(1)\\\
B&-BM_{r-1}^{21}(1)&-BM_{r-1}^{22}(1)\\\
I_{2}&\tilde{B}^{-1}&\tilde{B}^{-1}C_{0}\end{matrix}\right)\right|$ (6.712)
$\displaystyle=$ $\displaystyle|\det
B|^{r-1}\cdot|\det\tilde{B}|\cdot\left|\det\left(\begin{matrix}0&0&-M_{r-1}^{11}(1)&B-M_{r-1}^{12}(1)\\\
0&I_{d}&-M_{r-1}^{21}(1)&-M_{r-1}^{22}(1)\\\ I_{d}&0&0&-B\\\
0&I_{d}&\tilde{B}^{-1}&\tilde{B}^{-1}C_{0}\end{matrix}\right)\right|$ (6.713)
$\displaystyle=$ $\displaystyle|\det
B|^{r-1}\cdot|\det\tilde{B}|\cdot\left|\det\left(\begin{matrix}\left(\begin{matrix}0&0\\\
0&I_{d}\end{matrix}\right)&-M_{r-1}(1)+\left(\begin{matrix}0&B\\\
0&0\end{matrix}\right)\\\ I_{2d}&-M_{1}(0)^{-1}\end{matrix}\right)\right|$
(6.714) $\displaystyle=$ $\displaystyle|\det
B|^{r-1}\cdot|\det\tilde{B}|\cdot|\det
M_{1}(0)^{-1}|\cdot\left|\det\left(\left(\begin{matrix}0&0\\\
0&I_{d}\end{matrix}\right)-M_{r}(0)+\left(\begin{matrix}0&B\\\
0&0\end{matrix}\right)M_{1}(0)\right)\right|$ (6.715) $\displaystyle=$
$\displaystyle|\det B|^{r}|\det(M_{r}(0)-I_{2d})|.$

This implies the claimed result by noting $|\det B|=|\hat{v}_{-d}e^{2\pi
d\varepsilon}|^{d}$. ∎

Taking Lemmas 6.12 and 6.14 into account, Lemma 6.5 will follow from the
following.

######  Lemma 6.15.

Under the assumptions of Lemma 6.5 one has

(6.716)
$\displaystyle\left|\frac{1}{q}\int_{\mathbb{T}}\log|\det(\hat{A}_{q,E}^{\varepsilon}(p/q,\theta)-I_{2d})|\,\mathrm{d}\theta-\hat{L}^{k_{0}}(\alpha,\hat{A}_{E}^{\varepsilon})\right|$
(6.717) $\displaystyle\qquad\leq
C\varepsilon_{0}\left(q^{-1}\varepsilon_{0}^{-\frac{1}{2}}+\hat{L}^{k_{0}}(\alpha,\hat{A}_{E}^{\varepsilon})+\left|\log|\hat{v}_{-d}e^{2\pi
d\varepsilon}|\right|\right)$

for all sufficiently large $q$.

###### Proof.

Let

$|\hat{\lambda}_{1,E,\varepsilon}(\theta)|\geq\cdots\geq|\hat{\lambda}_{2d,E,\varepsilon}(\theta)|$

be the eigenvalues of $\hat{A}_{q,E}^{\varepsilon}(p/q,\theta)$. We need the
following elementary estimate.

######  Lemma 6.16.

For any $1\leq m\leq 2d$, we have

(6.718)
$\displaystyle\hat{L}^{m}(p/q,\hat{A}_{E}^{\varepsilon},\theta)=\frac{1}{q}\sum_{j=1}^{m}\log|\hat{\lambda}_{j,E,\varepsilon}(\theta)|.$

###### Proof.

Note by (6.673) the claimed is true for $m=1$. Let

(6.719)
$\displaystyle\hat{A}_{q,E}^{\varepsilon}(p/q,\theta)=U_{q}^{-1}(\theta)B_{q}(\theta)U_{q}(\theta),$

be the Jordan form of $\hat{A}_{q,E}^{\varepsilon}(p/q,\theta)$, where we omit
the dependence of $U$ and $B$ on $E,\varepsilon$ for simplicity. Clearly
$U_{q},U_{q}^{-1},B_{q}$ are all continuous in $\theta$. Thus there exists
some constants $C_{m}>0$ such that

(6.720)
$\displaystyle\sup_{\theta\in\mathbb{T}}\max(\|\wedge^{m}U_{q}^{-1}(\theta)\|,\|\wedge^{m}U_{q}(\theta)\|)\leq
C_{m}.$

It is also easy to see that

(6.721)
$\displaystyle\|\wedge^{m}(B_{q}(\theta))^{k}\|=\prod_{j=1}^{m}|\hat{\lambda}_{j,E,\varepsilon}(\theta)|^{k}$

Combining (6.721) with (6.720), we have for any $k\geq 1$,

(6.722)
$\displaystyle\frac{1}{k}\log\|\wedge^{m}(\hat{A}_{q,E}^{\varepsilon}(p/q,\theta))^{k}\|$
(6.723) $\displaystyle\leq$
$\displaystyle\frac{1}{k}\left(\log\|\wedge^{m}U_{q}^{-1}(\theta)\|+\log\|\wedge^{m}(B_{q}(\theta))^{k}\|+\log\|(\wedge^{m}U_{q}(\theta))^{k}\|\right)$
(6.724) $\displaystyle\leq$
$\displaystyle\sum_{j=1}^{m}\log|\hat{\lambda}_{j,E,\varepsilon}(\theta)|+\frac{2C_{m}}{k},$

which implies

(6.725)
$\displaystyle\hat{L}^{m}(p/q,\hat{A}_{E}^{\varepsilon},\theta)\leq\frac{1}{q}\sum_{j=1}^{m}\log|\hat{\lambda}_{j,E,\varepsilon}(\theta)|.$

This other direction of the inequality can be proved similarly. ∎

######  Lemma 6.17.

If $p/q$ is sufficiently close to $\alpha$, then for any $1\leq
j_{1}<j_{2}<...<j_{m}\leq 2d$ such that $(j_{1},...,j_{m})\neq(1,...,k_{0})$,
one has

(6.726)
$\displaystyle\sup_{\theta\in\mathbb{T}}\frac{1}{q}\sum_{\ell=1}^{m}\log|\hat{\lambda}_{j_{\ell},E,\varepsilon}(\theta)|\leq\max(\hat{L}^{k_{0}-1}(\alpha,\hat{A}_{E}^{\varepsilon}),\hat{L}^{k_{0}+1}(\alpha,\hat{A}_{E}^{\varepsilon}))+\varepsilon_{0},$

and

(6.727)
$\displaystyle\sup_{\theta\in\mathbb{T}}\frac{1}{q}\sum_{j=1}^{k_{0}}\log|\hat{\lambda}_{j,E,\varepsilon}(\theta)|\leq\hat{L}^{k_{0}}(\alpha,\hat{A}_{E}^{\varepsilon})+\varepsilon_{0}.$

###### Proof.

Case 1. If $m\neq k_{0}$, we estimate by Lemmas 6.1 and 6.16 that uniformly in
$\theta$

(6.728)
$\displaystyle\frac{1}{q}\sum_{\ell=1}^{m}\log|\hat{\lambda}_{j_{\ell},E,\varepsilon}(\theta)|\leq\frac{1}{q}\sum_{j=1}^{m}\log|\hat{\lambda}_{j,E,\varepsilon}(\theta)|=\hat{L}^{m}(p/q,\hat{A}_{E}^{\varepsilon},\theta)\leq\hat{L}^{m}(\alpha,\hat{A}_{E}^{\varepsilon})+\varepsilon_{0}.$

Case 2.1. If $m=k_{0}$ and
$|\hat{\lambda}_{j_{k_{0}},E,\varepsilon}(\theta)|\leq 1$, we estimate

(6.729)
$\displaystyle\frac{1}{q}\sum_{\ell=1}^{k_{0}}\log|\hat{\lambda}_{j_{\ell},E,\varepsilon}(\theta)|\leq\frac{1}{q}\sum_{\ell=1}^{k_{0}-1}\log|\hat{\lambda}_{j_{\ell},E,\varepsilon}(\theta)|\leq\hat{L}^{k_{0}-1}(\alpha,\hat{A}_{E}^{\varepsilon})+\varepsilon_{0},$

where we used the bound in Case 1.

Case 2.2. If $m=k_{0}$ and
$|\hat{\lambda}_{j_{k_{0}},E,\varepsilon}(\theta)|>1$. We have
$|\hat{\lambda}_{j,E,\varepsilon}(\theta)|>1$ for any $1\leq j\leq k_{0}$.
Since $(j_{1},...,j_{k_{0}})\neq(1,...,k_{0})$, we have $j_{k_{0}}\geq
k_{0}+1$. We estimate, in this case,

(6.730)
$\displaystyle\frac{1}{q}\sum_{\ell=1}^{k_{0}}\log|\hat{\lambda}_{j_{\ell},E,\varepsilon}(\theta)|\leq\frac{1}{q}\sum_{j=1}^{j_{k_{0}}}\log|\hat{\lambda}_{j,E,\varepsilon}(\theta)|\leq\hat{L}^{j_{k_{0}}}(\alpha,\hat{A}_{E}^{\varepsilon})+\varepsilon_{0},$

where we used again the estimate in Case 1. Finally, note that due to the
definition of $k_{0}$, we have

(6.731) $\displaystyle\max_{\begin{subarray}{c}m=1,\ldots,2d\\\ m\neq
k_{0}\end{subarray}}\hat{L}^{m}(\alpha,\hat{A}_{E}^{\varepsilon})\leq\max(\hat{L}^{k_{0}-1}(\alpha,\hat{A}_{E}^{\varepsilon}),\hat{L}^{k_{0}+1}(\alpha,\hat{A}_{E}^{\varepsilon})),$

which proves (6.726). Finally, (6.727) follows directly from Lemmas 6.16 and
6.1. ∎

Let

(6.732) $\displaystyle\delta:=$
$\displaystyle\frac{1}{2}\left(\hat{L}^{k_{0}}(\alpha,\hat{A}_{E}^{\varepsilon})-\max(\hat{L}^{k_{0}-1}(\alpha,\hat{A}_{E}^{\varepsilon}),\hat{L}^{k_{0}+1}(\alpha,\hat{A}_{E}^{\varepsilon}))\right)$
(6.733) $\displaystyle=$
$\displaystyle\frac{1}{2}\min(\hat{L}_{k_{0}}(\alpha,\hat{A}_{E}^{\varepsilon}),-\hat{L}_{k_{0}+1}(\alpha,\hat{A}_{E}^{\varepsilon}))>5\varepsilon_{0},$

and introduce the good set for fixed $q$

(6.734)
$\displaystyle\mathcal{G}_{q}=\mathcal{G}:=\left\\{\theta\in\mathbb{T}:\frac{1}{q}\sum_{j=1}^{k_{0}}\log|\hat{\lambda}_{j,E,\varepsilon}(\theta)|>\hat{L}^{k_{0}}(\alpha,\hat{A}_{E}^{\varepsilon})-\delta\right\\}.$

######  Lemma 6.18.

The Lebesgue measure of $\mathcal{G}$ satisfies

(6.735) $\displaystyle|\mathcal{G}|\geq
1-\frac{2\varepsilon_{0}}{\delta+\varepsilon_{0}}.$

###### Proof.

By Lemma 6.17, we have

(6.736)
$\displaystyle\sup_{\theta}\frac{1}{q}\sum_{j=1}^{k_{0}}\log|\hat{\lambda}_{j,E,\varepsilon}(\theta)|\leq\hat{L}^{k_{0}}(\alpha,\hat{A}_{E}^{\varepsilon})+\varepsilon_{0}.$

Hence by the definition of $\mathcal{G}$ in (6.734) and (6.736), we have

(6.737)
$\displaystyle|\mathcal{G}|\cdot(\hat{L}^{k_{0}}(\alpha,\hat{A}_{E}^{\varepsilon})+\varepsilon_{0})+(1-|\mathcal{G}|)\cdot(\hat{L}^{k_{0}}(\alpha,\hat{A}_{E}^{\varepsilon})-\delta)\geq$
$\displaystyle\int_{\mathbb{T}}\hat{L}^{k_{0}}(p/q,\hat{A}_{E}^{\varepsilon},\theta)\,\mathrm{d}\theta$
(6.738) $\displaystyle=$
$\displaystyle\hat{L}^{k_{0}}(p/q,\hat{A}_{E}^{\varepsilon})>\hat{L}^{k_{0}}(\alpha,\hat{A}_{E}^{\varepsilon})-\varepsilon_{0}.$

This implies

(6.739) $\displaystyle|\mathcal{G}|\geq
1-\frac{2\varepsilon_{0}}{\delta+\varepsilon_{0}},$

as desired. ∎

On the good set we have the following approximation result.

######  Lemma 6.19.

If $p/q$ is sufficiently close to $\alpha$, then

(6.740)
$\displaystyle\frac{1}{q}\int_{\mathcal{G}}\sum_{j=1}^{2d}\log|\hat{\lambda}_{j,E,\varepsilon}(\theta)-1|\,\mathrm{d}\theta=\frac{1}{q}\int_{\mathcal{G}}\sum_{j=1}^{k_{0}}\log|\hat{\lambda}_{j,E,\varepsilon}(\theta)|\,\mathrm{d}\theta+O(1/q).$

###### Proof.

For each $\theta\in\mathcal{G}$, for $q$ large, we estimate by Lemma 6.17 that

(6.741)
$\displaystyle\prod_{j=1}^{2d}|\hat{\lambda}_{j,E,\varepsilon}(\theta)-1|\geq$
$\displaystyle\prod_{j=1}^{k_{0}}|\hat{\lambda}_{j,E,\varepsilon}(\theta)|-\sum_{\begin{subarray}{c}j_{1}<j_{2}<...<j_{m}\\\
(j_{1},...,j_{m})\neq(1,...,k_{0})\end{subarray}}\prod_{\ell=1}^{m}|\hat{\lambda}_{j_{\ell},E,\varepsilon}(\theta)|$
(6.742) $\displaystyle\geq$
$\displaystyle\frac{1}{2}\prod_{j=1}^{k_{0}}|\hat{\lambda}_{j,E,\varepsilon}(\theta)|.$

Analogously, for all $\theta\in\mathcal{G}$ we have the upper bound

(6.743)
$\displaystyle\prod_{j=1}^{2d}|\hat{\lambda}_{j,E,\varepsilon}(\theta)-1|\leq
2\prod_{j=1}^{k_{0}}|\hat{\lambda}_{j,E,\varepsilon}(\theta)|.$

The result follows by (6.741) and (6.743). ∎

Next, we estimate

######  Lemma 6.20.

Under the assumptions of Lemma 6.5,

(6.744)
$\displaystyle\left|\frac{1}{q}\int_{\mathcal{G}^{c}}\log|\det(\hat{A}_{q,E}^{\varepsilon}(p/q,\theta)-I_{2d})|\,\mathrm{d}\theta\right|\leq
C\varepsilon_{0}\left(q^{-1}\varepsilon_{0}^{-\frac{1}{2}}+\hat{L}^{k_{0}}(\alpha,\hat{A}_{E}^{\varepsilon})+\big{|}\log|\hat{v}_{d}e^{2\pi
d\varepsilon}|\big{|}\right).$

###### Proof.

By Lemma 6.17, we have

(6.745)
$\displaystyle|\det(\hat{A}_{q,E}^{\varepsilon}(p/q,\theta)-I_{2d})|\leq\sum_{1\leq
j_{1}<...<j_{m}\leq
2d}\,\,\prod_{\ell=1}^{m}|\hat{\lambda}_{j_{\ell},E,\varepsilon}(\theta)|\leq
C_{d}\exp(q\cdot(\hat{L}^{k_{0}}(\alpha,\hat{A}_{E}^{\varepsilon})+\varepsilon_{0})).$

This implies, after combining with (6.739),

(6.746)
$\displaystyle\frac{1}{q}\int_{\mathcal{G}^{c}}\log|\det(\hat{A}_{q,E}^{\varepsilon}(p/q,\theta)-I_{2d})|\,\mathrm{d}\theta\leq$
$\displaystyle|\mathcal{G}^{c}|\cdot(\hat{L}^{k_{0}}(\alpha,\hat{A}_{E}^{\varepsilon})+2\varepsilon_{0})$
(6.747) $\displaystyle\leq$ $\displaystyle
C\varepsilon_{0}(\hat{L}^{k_{0}}(\alpha,\hat{A}_{E}^{\varepsilon})+2\varepsilon_{0}).$

By Lemmas 6.14, 6.10, and (6.739), we have

(6.748)
$\displaystyle\frac{1}{q}\int_{\mathcal{G}^{c}}\log|\det(\hat{A}_{q,E}^{\varepsilon}(p/q,\theta)-I_{2d})|\,\mathrm{d}\theta$
(6.749) $\displaystyle=$
$\displaystyle\frac{1}{q}\int_{\mathcal{G}^{c}}\left(\log|\hat{D}_{p/q,E,\varepsilon}(\theta)|-q\log|\hat{v}_{-d}e^{2\pi
d\varepsilon}|\right)\,\mathrm{d}\theta$ (6.750) $\displaystyle=$
$\displaystyle\frac{1}{q}\int_{\mathcal{G}^{c}}\log|D_{p/q,E,\varepsilon}(0)+2(-1)^{q+1}\cos(2\pi
q\theta)|\,\mathrm{d}\theta-|\mathcal{G}^{c}|\cdot\log|\hat{v}_{-d}e^{2\pi
d\varepsilon}|$ (6.751) $\displaystyle\geq$
$\displaystyle-C\varepsilon_{0}\left|\log|\hat{v}_{-d}e^{2\pi
d\varepsilon}|\right|$

provided $|D_{p/q,E,\varepsilon}(0)|\geq 3$. On the other hand, if
$|D_{p/q,E,\varepsilon}(0)|\leq 3$, then

(6.752)
$\displaystyle\frac{1}{q}\int_{\mathcal{G}^{c}}\log|D_{p/q,E,\varepsilon}(0)+2(-1)^{q+1}\cos(2\pi
q\theta)|\,\mathrm{d}\theta\geq$ (6.753)
$\displaystyle-\frac{1}{q}|\mathcal{G}^{c}|^{\frac{1}{2}}\Big{(}\int_{\mathbb{T}}\big{(}\log|D_{p/q,E,\varepsilon}(0)+2(-1)^{q+1}\cos(2\pi\theta)|\big{)}^{2}\,\mathrm{d}\theta\Big{)}^{\frac{1}{2}}\geq-
Cq^{-1}\sqrt{\varepsilon_{0}},$

as desired. ∎

We also have

######  Lemma 6.21.

(6.754)
$\displaystyle\left|\frac{1}{q}\int_{\mathcal{G}^{c}}\sum_{j=1}^{k_{0}}\log|\hat{\lambda}_{j,E,\varepsilon}(\theta)|\,\mathrm{d}\theta-|\mathcal{G}^{c}|\cdot\hat{L}^{k_{0}}(\alpha,\hat{A}_{E}^{\varepsilon})\right|\leq
2\varepsilon_{0}.$

###### Proof.

For simplicity, we write
$g(\theta):=\frac{1}{q}\log\sum_{j=1}^{k_{0}}|\hat{\lambda}_{j,E,\varepsilon}(\theta)|$
and $\hat{L}^{k_{0}}:=\hat{L}^{k_{0}}(\alpha,\hat{A}_{E}^{\varepsilon})$. By
Lemmas 6.16 and 6.17, we have

(6.755)
$\displaystyle\hat{L}^{k_{0}}-\varepsilon_{0}\leq\int_{\mathbb{T}}g(\theta)\,\mathrm{d}\theta\leq\int_{\mathcal{G}^{c}}g(\theta)\,\mathrm{d}\theta+|\mathcal{G}|\cdot(\hat{L}^{k_{0}}+\varepsilon_{0}),$

which implies

(6.756)
$\displaystyle\int_{\mathcal{G}^{c}}(g(\theta)-\hat{L}^{k_{0}})\,\mathrm{d}\theta\geq-2\varepsilon_{0}.$

We also have by (6.739) and Lemma 6.17 that

(6.757)
$\displaystyle\int_{\mathcal{G}^{c}}(g(\theta)-\hat{L}^{k_{0}})\,\mathrm{d}\theta\leq\varepsilon_{0}\cdot|\mathcal{G}^{c}|\leq\frac{2\varepsilon_{0}^{2}}{\delta+\varepsilon_{0}}.$

This proves the claimed result. ∎

Finally, combining (6.740) with Lemmas 6.20 and 6.21, we have

(6.758)
$\displaystyle\left|\frac{1}{q}\int_{\mathbb{T}}\log|\det(\hat{A}_{q,E}^{\varepsilon}(p/q,\theta)-I_{2d})|\,\mathrm{d}\theta-\hat{L}^{k_{0}}(\alpha,\hat{A}_{E}^{\varepsilon})\right|$
(6.759) $\displaystyle\leq$
$\displaystyle\left|\frac{1}{q}\int_{\mathcal{G}}\sum_{j=1}^{2d}\log|\hat{\lambda}_{j,E,\varepsilon}(\theta)-1|\,\mathrm{d}\theta-\frac{1}{q}\int_{\mathcal{G}}\sum_{j=1}^{k_{0}}\log|\hat{\lambda}_{j,E,\varepsilon}(\theta)|\,\mathrm{d}\theta\right|$
(6.760)
$\displaystyle+\left|\frac{1}{q}\int_{\mathcal{G}^{c}}\log|\det(\hat{A}_{q,E}^{\varepsilon}(p/q,\theta)-I_{2d})|\,\mathrm{d}\theta\right|+\left|\frac{1}{q}\int_{\mathbb{T}}\sum_{j=1}^{k_{0}}\log|\hat{\lambda}_{j,E,\varepsilon}(\theta)|\,\mathrm{d}\theta-\hat{L}^{k_{0}}(\alpha,\hat{A}_{E,\varepsilon})\right|$
(6.761)
$\displaystyle+\left|\frac{1}{q}\int_{\mathcal{G}^{c}}\sum_{j=1}^{k_{0}}\log|\hat{\lambda}_{j,E,\varepsilon}(\theta)|\,\mathrm{d}\theta-|\mathcal{G}^{c}|\cdot\hat{L}^{k_{0}}(\alpha,\hat{A}_{E}^{\varepsilon})\right|+|\mathcal{G}^{c}|\cdot\hat{L}^{k_{0}}(\alpha,\hat{A}_{E}^{\varepsilon})$
(6.762) $\displaystyle\leq$ $\displaystyle
C\varepsilon_{0}(q^{-1}\varepsilon_{0}^{-\frac{1}{2}}+\hat{L}^{k_{0}}(\alpha,\hat{A}_{E}^{\varepsilon})+\big{|}\log|\hat{v}_{-d}e^{2\pi
d\varepsilon}|\big{|}).$

Combining Lemmas 6.12, 6.14, 6.15, we have proved Lemma 6.5. ∎

It remains to deal with the case
$\hat{L}^{1}(\alpha,\hat{A}_{E}^{\varepsilon})<0$. By Lemmas 6.1 and 6.15,
there exists $\delta>0$ so that uniformly in $\theta$, and provided $p/q$ is
close to $\alpha$, one has

$-\delta\geq\frac{1}{q}\sum_{j=1}^{m}\log|\hat{\lambda}_{j,E,\varepsilon}(\theta)|$

for all $1\leq m\leq 2d$. Hence, for all $1\leq j_{1}\leq\ldots\leq j_{m}\leq
2d$ we have

$\sup_{\theta}\prod_{\ell=1}^{m}|\hat{\lambda}_{j_{\ell},E,\varepsilon}(\theta)|\leq
e^{-\delta q}$

Therefore,

$\prod_{j=1}^{2d}|\hat{\lambda}_{j,E,\varepsilon}(\theta)-1|=1+O_{d}(e^{-\delta
q})$

and

$\frac{1}{q}\int_{\mathbb{T}}\log|\det(\hat{A}_{q,E}^{\varepsilon}(p/q,\theta)-I_{2d})|\,\mathrm{d}\theta=O_{d}(e^{-\delta
q})$

We now conclude Theorem 1.10 from Lemmas 6.12 and 6.14 as before.

##  Appendix A Proof of Theorem 1.2

The argument closely follows [AJ2], with a simplification from [Av2].
Throughout this section we write
$\hat{L}^{j}(\alpha,\hat{A}_{E}),\hat{L}_{j}(\alpha,\hat{A}_{E})$ as
$\hat{L}^{j}(\hat{A}_{E}),\hat{L}_{j}(\hat{A}_{E})$ for simplicity.

###  A.1. Preparation

######  Theorem A.1.

[AJ2]*Theorem 3.3 If $E\in\sigma(H_{\alpha,\theta})$, then there exist
$\theta\in\mathbb{R}$ and a bounded solution of
$\hat{H}_{\alpha,\theta}\hat{u}=E\hat{u}$ with $\hat{u}_{0}=1$ and
$|\hat{u}_{n}|\leq 1$.

Given Fourier coefficients $\hat{w}=(\hat{w}_{k})_{k\in\mathbb{Z}}$ and an
interval $I\subset\mathbb{Z}$, we let $w_{I}=\sum_{k\in I}\hat{w}_{k}e^{2\pi
ikx}$. The length of the interval $I=[a,b]$ is $|I|=b-a$. We will say that a
trigonometrical polynomial $p:\mathbb{T}\to\mathbb{C}$ has essential degree at
most $k$ if its Fourier coefficients outside an interval $I$ of length $k$
vanish. We let $\frac{p_{n}}{q_{n}}$ be the convergents of $\alpha$.

######  Theorem A.2.

[AJ2]*Theorem 6.1 Let $1\leq r\leq\lfloor q_{n+1}/q_{n}\rfloor$. If $p$ has
essential degree at most $k=rq_{n}-1$ and $x_{0}\in\mathbb{T}$, then for some
absolute constant $C$

(A.1) $\displaystyle\|p\|_{0}\leq Cq_{n+1}^{Cr}\sup_{0\leq j\leq
k}|p(x_{0}+j\alpha)|.$

###  A.2. Proof of Theorem 1.2

Fix $E\in\sigma(H_{\alpha,\theta})$, and let $\theta(E)$ be as in Theorem A.1.
Let $\varepsilon_{0}$ be such that
$0<\varepsilon_{0}<\frac{\hat{L}_{d}(\hat{A}_{E})}{30\hat{L}^{d}(\hat{A}_{E})+100C_{3}d}$
where $C_{3}$ is the absolute constant in Lemma 3.10. Finally, let
$\\{n_{j}\\}$ be the $\varepsilon_{0}$-resonances of $\theta(E)$ and $\hat{u}$
be the almost localized solution as in Theorem 3.8. Hence for
$3(|n_{j}|+1)<|n|<|n_{j+1}|/3$ and some $C_{2}=C_{2}(v,d)>0$, one has

(A.2) $\displaystyle|\hat{u}_{n}|\leq
C_{2}e^{-\frac{\hat{L}_{d}(\hat{A}_{E})}{10}|n|}.$

For simplicity we shall write $\theta(E)$ as $\theta$ for the remainder of
this appendix.

The cocycle $(\alpha,A_{E})$ being subcritical, which precisely means that
$L(\alpha,A_{E,\eta})=0$ for all $|\eta|\leq\hat{L}_{d}/(2\pi)$, implies via
Lemma 2.1 the following subexponential growth

(A.3) $\displaystyle\|A_{k,E}\|_{\frac{\hat{L}_{d}}{2\pi}}\leq
C_{\delta}e^{\delta|k|},$

for arbitrary $\delta>0$. Henceforth, we fix
$\delta=\min(\frac{\hat{L}_{d}(\hat{A}_{E})}{5000},\frac{\varepsilon_{0}}{25})$.
Theorem A.3 improves this subexponential bound to a polynomial one.

######  Theorem A.3.

For some absolute constants $C_{4}>0$, and
$C_{5}=C_{5}(\alpha,v,E,\varepsilon_{0})>0$,

(A.4) $\displaystyle\|A_{k,E}\|_{\frac{\delta}{2\pi}}\leq
C_{5}(1+|k|)^{C_{4}},$

for any $k\geq 1$.

###### Proof.

The proof proceeds by constructing an $\mathrm{SL}(2,\mathbb{C})$ almost
conjugacy of $A_{E}(x)$. Thus, let $n$ be such that

(A.5) $\displaystyle\begin{cases}9|n_{j}|<n<|n_{j+1}|/9,\text{ for some }j,\\\
n=rq_{m}-1<q_{m+1},\text{ for some }r,m\geq 1\end{cases}$

To see that such an $n$ exists, note that Lemma 3.7 implies that

(A.6) $\displaystyle\frac{|n_{j+1}|}{81|n_{j}|}>3,$

Moreover, for any $r,m\geq 1$, one has

(A.7) $\displaystyle\frac{(r+1)q_{m}-1}{rq_{m}-1}\leq 2.$

Let $I=[x_{1},x_{2}]:=[-\lfloor n/2\rfloor+1,\lfloor
n/2\rfloor]\cap\mathbb{Z}$, and define

(A.8) $\displaystyle U^{I}(x)=\left(\begin{matrix}e^{2\pi i\theta}u^{I}(x)\\\
u^{I}(x-\alpha)\end{matrix}\right).$

By direct computation one obtains

(A.9) $\displaystyle A_{E}(x)U^{I}(x)=e^{2\pi
i\theta}U^{I}(x+\alpha)+G^{I}(x),$

where

(A.10) $\displaystyle\|G^{I}(x)\|_{\frac{\hat{L}_{d}(\hat{A}_{E})}{40\pi}}\leq
C_{2}e^{-\frac{\hat{L}_{d}(\hat{A}_{E})}{40}n}.$

We select $U^{I}(x)$ as the first column of our putative conjugacy matrix. In
order to construct the second column, we first establish a lower bound on
$\|U^{I}(x)\|$.

######  Lemma A.4.

For $n>n_{0}(\alpha,v,E,\varepsilon_{0})$ satisfying (A.5), we have

(A.11)
$\displaystyle\inf_{|\mathrm{Im}(x)|\leq\frac{\hat{L}_{d}(\hat{A}_{E})}{40\pi}}\|U^{I}(x)\|\geq
e^{-2\delta n}.$

###### Proof.

If the lemma fails, then there exists $x_{0}$ with
$|\mathrm{Im}(x_{0})|\leq\frac{\hat{L}_{d}(\hat{A}_{E})}{40\pi}$ so that

(A.12) $\displaystyle\|U^{I}(x_{0})\|<e^{-2\delta n}.$

Iterating (A.9) yields

(A.13) $\displaystyle e^{2\pi i\ell\theta}U^{I}(x_{0}+\ell\alpha)=$
$\displaystyle A_{\ell,E}(x_{0})U_{I}(x_{0})$ (A.14)
$\displaystyle-\sum_{m=1}^{\ell}e^{2\pi
i(m-1)\theta}A_{\ell-m,E}(x_{0}+m\alpha)G^{I}(x_{0}+(m-1)\alpha).$

Combining the above with (A.3), (A.12) and (A.10), we infer that for
$0\leq\ell\leq n$,

(A.15)
$\displaystyle|u^{I}(x_{0}+\ell\alpha)|\leq\|U^{I}(x_{0}+\ell\alpha)\|\leq
C_{\delta}e^{\delta n}e^{-2\delta
n}+\sum_{m=1}^{\ell}C_{\delta}C_{2}e^{\delta(\ell-m)}e^{-\frac{\hat{L}_{d}(\hat{A}_{E})}{40}n}\leq
C_{\delta}C_{2}e^{-\delta n}.$

By Theorem A.2, we finally arrive at the uniform bound

(A.16) $\displaystyle\|u^{I}(x+i\mathrm{Im}(x_{0}))\|_{\mathbb{T}}\leq
Cq_{m+1}^{Cr}\sup_{0\leq\ell\leq n}|u^{I}(x_{0}+\ell\alpha)|\leq
C_{\delta}C_{2}e^{-\delta n/2},$

where we used that $q_{m+1}\leq C_{\delta}e^{\delta q_{m}/(2C)}$, see (3.77).
Clearly, (A.16) contradicts

$\int_{\mathbb{T}}u(x+i\mathrm{Im}(x_{0}))\,\mathrm{d}x=\hat{u}_{0}=1$

for $n$ large enough. ∎

The now invoke the following lemma to construct the second column of our
conjugacy matrix.

######  Lemma A.5.

[Av2]*Lemma 4.2 Let $V:\mathbb{T}\to\mathbb{C}^{2}$ be analytic in
$|\mathrm{Im}(x)|\leq\eta$. Assume $\delta_{1}<\|V(x)\|<\delta_{2}^{-1}$ holds
on $|\mathrm{Im}(x)|<\eta$. Then there exists
$M:\mathbb{T}\to\mathrm{SL}(2,\mathbb{C})$ analytic on $|\mathrm{Im}(x)|<\eta$
with the first column equals to $V$ and $\|M\|_{\eta}\leq
C\delta_{1}^{-2}\delta_{2}^{-1}(1-\log(\delta_{1}\delta_{2}))$.

Let $M^{I}(x)$ be the matrix obtained by means of Lemma A.5 whose first column
equals $U^{I}$. By Lemma A.4 and the fact that $|\hat{u}_{k}|\leq 1$, for any
$|\mathrm{Im}(x)|\leq\frac{\delta}{2\pi}$,

(A.17) $\displaystyle e^{-2\delta n}\leq\|U^{I}(x)\|\leq e^{\delta n}.$

By Lemma A.5,

(A.18)
$\displaystyle\|M^{I}\|_{\frac{\delta}{2\pi}}=\|(M^{I})^{-1}\|_{\frac{\delta}{2\pi}}\leq
e^{6\delta n}.$

and by (A.9),

(A.19) $\displaystyle
A_{E}(x)M^{I}(x)=M^{I}(x+\alpha)\left(\begin{matrix}e^{2\pi
i\theta}&a_{12}(x)\\\ 0&e^{-2\pi
i\theta}+a_{22}(x)\end{matrix}\right)+\left(\begin{matrix}G^{I}(x)&0\end{matrix}\right).$

Thus,

$\displaystyle M^{I}(x+\alpha)^{-1}A_{E}(x)M^{I}(x)=$
$\displaystyle\left(\begin{matrix}e^{2\pi i\theta}&a_{12}(x)\\\ 0&e^{-2\pi
i\theta}+a_{22}(x)\end{matrix}\right)+M^{I}(x+\alpha)^{-1}\left(\begin{matrix}G^{I}(x)&0\end{matrix}\right)$
(A.20) $\displaystyle=:$ $\displaystyle\left(\begin{matrix}e^{2\pi
i\theta}&a_{12}(x)\\\ 0&e^{-2\pi
i\theta}+a_{22}(x)\end{matrix}\right)+\left(\begin{matrix}a_{11}(x)&0\\\
a_{21}(x)&0\end{matrix}\right)$

In view of (A.10) and (A.18),

(A.21)
$\displaystyle\max(\|a_{11}\|_{\frac{\delta}{2\pi}},\|a_{21}\|_{\frac{\delta}{2\pi}})\leq\|M^{I}(x+\alpha)^{-1}\left(\begin{matrix}G^{I}(x)&0\end{matrix}\right)\|_{\frac{\delta}{2\pi}}\leq
e^{-\frac{\hat{L}_{d}(\hat{A}_{E})}{100}n},$

whence

(A.22)
$\displaystyle\|a_{12}\|_{\frac{\delta}{2\pi}}\leq\|(M^{I})^{-1}\|_{\frac{\delta}{2\pi}}\cdot\|A_{E}\|_{\frac{\delta}{2\pi}}\cdot\|M^{I}\|_{\frac{\delta}{2\pi}}+\max(\|a_{11}\|_{\frac{\delta}{2\pi}},\|a_{21}\|_{\frac{\delta}{2\pi}})\leq
C_{\delta}e^{12\delta n}.$

Passing to determinants in (A.2), and using (A.21), we conclude that

(A.23) $\displaystyle(e^{2\pi i\theta}+a_{11}(x))(e^{-2\pi
i\theta}+a_{22}(x))-a_{12}(x)a_{21}(x)=1,$

which implies

(A.24) $\displaystyle(e^{2\pi
i\theta}+a_{11}(x))a_{22}(x)=a_{12}(x)a_{21}(x)-e^{-2\pi i\theta}a_{11}(x).$

Combining the above equation with the bounds in (A.21) and (A.22) implies that

(A.25) $\displaystyle\|a_{22}\|_{\frac{\delta}{2\pi}}\leq
e^{-\frac{\hat{L}_{d}(\hat{A}_{E})}{200}n}.$

Let

(A.26)
$\displaystyle\Phi^{I}(x)=M^{I}(x)\left(\begin{matrix}e^{\frac{\hat{L}_{d}(\hat{A}_{E})}{400}n}&0\\\
0&e^{-\frac{\hat{L}_{d}(\hat{A}_{E})}{400}n}\end{matrix}\right).$

Then by (A.2), (A.21), (A.22) and (A.25),

(A.27)
$\displaystyle\Phi^{I}(x+\alpha)^{-1}A_{E}(x)\Phi^{I}(x)=\left(\begin{matrix}e^{2\pi
i\theta}&0\\\ 0&e^{-2\pi i\theta}\end{matrix}\right)+H(x),$

where

(A.28) $\displaystyle\|H\|_{\frac{\delta}{2\pi}}\leq$
$\displaystyle\max(\|a_{12}\|_{\frac{\delta}{2\pi}}e^{-\frac{\hat{L}_{d}(\hat{A}_{E})}{200}},\|a_{11}\|_{\frac{\delta}{2\pi}},\|a_{22}\|_{\frac{\delta}{2\pi}},\|a_{21}\|_{\frac{\delta}{2\pi}}e^{\frac{\hat{L}_{d}(\hat{A}_{E})}{200}n})$
(A.29) $\displaystyle\leq$ $\displaystyle C_{\delta}e^{12\delta
n}e^{-\frac{\hat{L}_{d}(\hat{A}_{E})}{200}n}\leq
e^{-\frac{\hat{L}_{d}(\hat{A}_{E})}{400}n}.$

Moreover, by (A.18) and (A.26),

(A.30)
$\displaystyle\|\Phi^{I}\|_{\frac{\delta}{2\pi}}=\|(\Phi^{I})^{-1}\|_{\frac{\delta}{2\pi}}\leq\|M^{I}\|_{\frac{\delta}{2\pi}}e^{\frac{\hat{L}_{d}(\hat{A}_{E})}{400}}\leq
e^{\frac{\hat{L}_{d}(\hat{A}_{E})}{200}}.$

In view of (A.27), this implies that for all $1\leq s\leq
e^{\frac{\hat{L}_{d}(\hat{A}_{E})}{800}n}$ one has

(A.31) $\displaystyle\|A_{s,E}(x)\|_{\frac{\delta}{2\pi}}\leq
2\|\Phi^{I}\|_{\frac{\delta}{2\pi}}\cdot\|(\Phi^{I})^{-1}\|_{\frac{\delta}{2\pi}}\leq
2e^{\frac{\hat{L}_{d}(\hat{A}_{E})}{100}n}.$

There exists an absolute constant $c_{4}>0$, such that for
$s>s(\alpha,v,E,\varepsilon_{0})$ large enough, there exists $n$, with
$9|n_{j}|<n<|n_{j+1}|/9$, and satisfying the requirement of Lemma A.4, such
that $c_{4}n<\frac{800}{\hat{L}_{d}(\hat{A}_{E})}\log s<n$. Thus by (A.31), we
finally obtain

(A.32) $\displaystyle\|A_{s,E}\|_{\frac{\delta}{2\pi}}\leq
C_{5}(1+|s|)^{8/c_{4}},$

as claimed. ∎

Next, let
$J=[-\lfloor|n_{j+1}|/9\rfloor,\lfloor|n_{j+1}|/9\rfloor]\cap\mathbb{Z}$, and
define

(A.33) $\displaystyle U^{J}(x)=\left(\begin{matrix}e^{2\pi i\theta}u^{J}(x)\\\
u^{J}(x-\alpha)\end{matrix}\right).$

By the almost localization Theorem 3.8, one has

(A.34) $\displaystyle A_{E}(x)U^{J}(x)=e^{2\pi
i\theta}U^{J}(x+\alpha)+G^{J}(x),\text{ with
}\|G^{J}\|_{\frac{\hat{L}_{d}}{32\pi}}\leq
e^{-\frac{\hat{L}_{d}(\hat{A}_{E})}{150}|n_{j+1}|},$

and

(A.35) $\displaystyle\|U^{J}\|_{\frac{\delta}{2\pi}}\leq e^{3\delta|n_{j}|}.$

Let $M_{J}(x)$ be the matrix with the first column equal to $U^{J}(x)$, and
the second column equal to $\overline{U^{J}(x)}$, respectively. By the
preceding,

(A.36) $\displaystyle
A_{E}(x)M_{J}(x)=M_{J}(x+\alpha)\left(\begin{matrix}e^{2\pi i\theta}&0\\\
0&e^{-2\pi i\theta}\end{matrix}\right)+H(x),$

where $H(x)=\left(\begin{matrix}G(x)&\overline{G(x)}\end{matrix}\right)$
satisfies $\|H\|_{\frac{\hat{L}_{d}}{32\pi}}\leq
e^{-\frac{\hat{L}_{d}(\hat{A}_{E})}{150}|n_{j+1}|}.$

We also define $\tilde{U}^{J}(x):=e^{\pi in_{j}x}U^{J}(x)$. When $n_{j}$ is
even, $\tilde{U}^{J}$ is well-defined on $\mathbb{T}$, and when $n_{j}$ is
odd, $\tilde{U}^{J}(x+1)=-\tilde{U}^{J}(x)$. By (A.34), with
$\tilde{\theta}:=\theta-n_{j}\alpha/2$,

(A.37) $\displaystyle A_{E}(x)\tilde{U}^{J}(x)=e^{2\pi
i\tilde{\theta}}\tilde{U}^{J}(x+\alpha)+\tilde{G}^{J}(x),\text{ where
}\|\tilde{G}^{J}\|_{\frac{\hat{L}_{d}(\hat{A}_{E})}{32\pi}}\leq
e^{-\frac{\hat{L}_{d}(\hat{A}_{E})}{200}|n_{j+1}|}.$

By Theorem 3.8, we also have

(A.38) $\displaystyle\|\tilde{U}^{J}\|_{\frac{\delta}{2\pi}}\leq
e^{4\delta|n_{j}|}.$

Let $\tilde{M}_{J}(x)$ be the matrix with columns $\tilde{U}^{J}(x)$ and
$\overline{\tilde{U}^{J}(x)}$. Clearly,

(A.39) $\displaystyle|\det\tilde{M}_{J}(x)|=e^{-2\pi n_{j}\mathrm{Im}(x)}|\det
M_{J}(x)|.$

######  Theorem A.6.

Let $L=\|2\theta-n_{j}\alpha\|_{\mathbb{T}}^{-1}$. Then for
$j>j_{0}(\alpha,v,E,\varepsilon_{0})$,

(A.40) $\displaystyle|\det M_{J}(x)|\geq L^{-5C_{4}},\text{ for
}|\mathrm{Im}(x)|\leq\frac{\delta}{2\pi},$

where $C_{4}$ is the constant in Theorem A.3.

###### Proof.

We first establish the following estimates on $L$.

######  Lemma A.7.

For $j>j_{0}(\alpha,\hat{L}_{d})$,

(A.41) $\displaystyle e^{\varepsilon_{0}|n_{j}|}\leq L\leq
e^{\frac{\hat{L}_{d}(\hat{A}_{E})}{4000C_{4}}|n_{j+1}|}.$

###### Proof.

For $\varepsilon_{1}=\frac{\hat{L}_{d}(\hat{A}_{E})}{4000C_{4}}$, we have by
(3.77) that for $n>N(\alpha,\hat{L}_{d}(\hat{A}_{E}))$,

(A.42) $\displaystyle C(\alpha,\hat{L}_{d})e^{-2\varepsilon_{1}|n_{j+1}|}$
$\displaystyle\leq
C(\alpha,\hat{L}_{d})e^{-\varepsilon_{1}|n_{j+1}-n_{j}|}\leq\|(n_{j+1}-n_{j})\alpha\|_{\mathbb{T}}$
(A.43) $\displaystyle\leq 2\|2\theta-n_{j}\alpha\|_{\mathbb{T}}=2L^{-1}\leq
2e^{-\varepsilon_{0}|n_{j}|}.$

This implies the claimed result. ∎

We now prove Theorem A.6 by contradiction, based on the next lemma.

######  Lemma A.8.

For any $\kappa\in\mathbb{C}$ such that $|\kappa|=1$,

(A.44)
$\displaystyle\inf_{x:|\mathrm{Im}(x)|\leq\frac{\delta}{2\pi}}\|\tilde{U}^{J}(x)-\kappa\cdot\overline{\tilde{U}^{J}(x)}\|\geq
L^{-4C_{4}}.$

###### Proof.

Suppose for some $x_{0}$, $|\mathrm{Im}(x_{0})|\leq\frac{\delta}{2\pi}$, and
some constant $|\kappa|=1$ we have

(A.45)
$\displaystyle\|\tilde{U}^{J}(x_{0})-\kappa\cdot\overline{\tilde{U}^{J}(x_{0})}\|<L^{-4C_{4}}.$

By (A.37), Theorem A.3 and Lemma A.7, for all $0\leq|\ell|\leq L^{2}$ one
concludes that

$\displaystyle\|\tilde{U}^{J}(x_{0}+\ell\alpha)e^{2\pi
i\ell\tilde{\theta}}-\kappa\cdot\overline{\tilde{U}^{J}(x_{0}+\ell\alpha)}e^{-2\pi
i\ell\tilde{\theta}}\|$ (A.46)
$\displaystyle\,\,\,\,\leq\left\|\sum_{m=0}^{\ell-1}A_{\ell-m,E}(x_{0}+m\alpha)\tilde{G}^{J}(x_{0}+m\alpha)-\kappa\sum_{m=0}^{\ell-1}A_{\ell-m,E}(x_{0}+m\alpha)\overline{\tilde{G}^{J}(x_{0}+m\alpha)}\right\|$
$\displaystyle\,\,\,\,+\|A_{\ell,E}(x_{0})\|L^{-4C_{4}}$ (A.47)
$\displaystyle\,\,\,\,\leq
C_{5}L^{2C_{4}+3}e^{-\frac{\hat{L}_{d}(\hat{A}_{E})}{200}|n_{j+1}|}+C_{5}L^{-2C_{4}}<L^{-C_{4}}.$

Setting $\ell=\ell_{0}=\lfloor L/2\rfloor$ in (A.2), we conclude that

(A.48) $\displaystyle\|\tilde{U}^{J}(x_{0}+\ell_{0}\alpha)e^{2\pi
i\ell_{0}\tilde{\theta}}-\kappa\cdot\overline{\tilde{U}^{J}(x_{0}+\ell_{0}\alpha)}e^{-2\pi
i\ell_{0}\tilde{\theta}}\|<L^{-C_{4}}.$

Writing $\ell_{0}=L/2+\delta$ with $|\delta|\leq 1/2$, and $2\theta-
n_{j}\alpha=m_{0}\pm\|2\theta-n_{j}\alpha\|_{\mathbb{T}}$, we infer that

(A.49) $\displaystyle 2\pi\ell_{0}\tilde{\theta}=\pi(L/2+\delta)(2\theta-
n_{j}\alpha)=\pi\ell_{0}m_{0}\pm\frac{\pi}{2}\pm\pi\delta
L^{-1}=\pi\ell_{0}m_{0}\pm\frac{\pi}{2}+O(L^{-1}).$

Plugging (A.49) into (A.48), and combining the resulting estimate with (A.38),
yields

$\displaystyle\|\tilde{U}^{J}(x_{0}+\ell_{0}\alpha)+\kappa\cdot\overline{\tilde{U}^{J}(x_{0}+\ell_{0}\alpha)}\|$
$\displaystyle\leq
L^{-C_{4}}+L^{-1}\|\tilde{U}^{J}\|_{\frac{\delta}{2\pi}}\leq
L^{-C_{4}}+L^{-1}e^{4\delta|n_{j}|}$ (A.50) $\displaystyle<L^{-1/2}.$

For $|\ell|\leq L^{1/2}$, we have in analogy to (A.49) that

(A.51) $\displaystyle 2\pi\ell\tilde{\theta}=\pi\ell m_{0}+O(L^{-1/2}).$

Plugging (A.51) into (A.37) yields

$\displaystyle\|\tilde{U}^{J}(x_{0}+\ell\alpha)-\kappa\cdot\overline{\tilde{U}^{J}(x_{0}+\ell\alpha)}\|$
$\displaystyle\leq
L^{-C_{4}}+L^{-1/2}\|\tilde{U}^{J}\|_{\frac{\delta}{2\pi}}\leq
L^{-C_{4}}+L^{-1/2}e^{4\delta|n_{j}|}$ (A.52) $\displaystyle<L^{-1/3}.$

Next, we let $\tilde{n}$ be such that

(A.53)
$\displaystyle\begin{cases}9|n_{j}|<\tilde{n}<\min(27|n_{j}|,|n_{j+1}|/9)\\\
\tilde{n}=rq_{m}-1<q_{m+1},\text{ for some }r,m\geq 1\end{cases}$

Let
$\tilde{I}=[-\lfloor\tilde{n}/2\rfloor+1,\lfloor\tilde{n}/2\rfloor]\cap\mathbb{Z}$
and define

(A.54) $\displaystyle U^{\tilde{I}}(x)=\left(\begin{matrix}e^{2\pi
i\theta}u^{\tilde{I}}(x)\\\ u^{\tilde{I}}(x-\alpha)\end{matrix}\right),$

as well as $\tilde{U}^{\tilde{I}}(x)=e^{\pi in_{j}x}U^{\tilde{I}}(x)$. Then by
Theorem 3.8, we have

(A.55)
$\displaystyle\|\tilde{U}^{J}-\tilde{U}^{\tilde{I}}\|_{\frac{\delta}{2\pi}}\leq
e^{-\frac{\hat{L}_{d}(\hat{A}_{E})}{2}|n_{j}|}$

Combining the above with (A.2) yields

(A.56)
$\displaystyle\|\tilde{U}^{\tilde{I}}(x_{0}+\ell\alpha)-\kappa\cdot\overline{\tilde{U}^{\tilde{I}}(x_{0}+\ell\alpha)}\|\leq
L^{-1/3}+2e^{-\frac{\hat{L}_{d}(\hat{A}_{E})}{2}|n_{j}|}.$

Denoting

(A.57) $\displaystyle g(x)=e^{-\pi n_{j}\mathrm{Im}(x_{0})}e^{2\pi
in_{j}x}\sum_{|k|\leq 9|n_{j}|}e^{-2\pi
k\mathrm{Im}(x_{0})}\left(\hat{u}_{k}e^{2\pi
ikx}-\kappa\,\overline{\hat{u}_{k}}e^{-2\pi ikx}\right),$

we obtain from (A.56) that

(A.58) $\displaystyle|g(\mathrm{Re}(x_{0})+\ell\alpha)|=$
$\displaystyle\left|e^{\pi
in_{j}(x_{0}+\ell\alpha)}u^{\tilde{I}}(x_{0}+\ell\alpha)-\kappa\;\overline{e^{\pi
in_{j}(x_{0}+\ell\alpha)}u^{\tilde{I}}(x_{0}+\ell\alpha)}\right|$ (A.59)
$\displaystyle\leq$
$\displaystyle\|\tilde{U}^{\tilde{I}}(x_{0}+\ell\alpha)-\kappa\cdot\overline{\tilde{U}^{\tilde{I}}(x_{0}+\ell\alpha)}\|\leq
L^{-1/3}+e^{-\frac{\hat{L}_{d}(\hat{A}_{E})}{2}|n_{j}|}$

for any $|\ell|\leq L^{1/2}$. Lemma A.7 implies that
$L^{1/2}>e^{\varepsilon_{0}|n_{j}|}>\tilde{n}$, whence by Theorem A.2,

(A.60) $\displaystyle\sup_{x\in\mathbb{T}}|g(x)|\leq
Cq_{m+1}^{Cr}(L^{-1/3}+e^{-\frac{\hat{L}_{d}}{2}|n_{j}|})<L^{-1/4}+e^{-\frac{\hat{L}_{d}(\hat{A}_{E})}{4}|n_{j}|}.$

Here we used that $q_{m+1}<C_{\varepsilon_{0}}e^{\varepsilon_{0}q_{m}/350C}$,
see (3.77), and $\tilde{n}=rq_{m}-1<27|n_{j}|$. This implies

(A.61)
$\displaystyle\sup_{x\in\mathbb{T}}\|\tilde{U}^{\tilde{I}}(x+\mathrm{Im}(x_{0}))-\kappa\cdot\overline{\tilde{U}^{\tilde{I}}}(x+\mathrm{Im}(x_{0}))\|\leq
2\sup_{x\in\mathbb{T}}|g(x)|\leq
2L^{-1/4}+2e^{-\frac{\hat{L}_{d}(\hat{A}_{E})}{4}|n_{j}|},$

and in particular

(A.62)
$\displaystyle\|\tilde{U}^{\tilde{I}}(x_{0}+\ell_{0}\alpha)-\kappa\cdot\overline{\tilde{U}^{\tilde{I}}(x_{0}+\ell_{0}\alpha)}\|\leq
2L^{-1/4}+2e^{-\frac{\hat{L}_{d}(\hat{A}_{E})}{4}|n_{j}|}.$

Combining (A.2) with (A.55), we have

(A.63)
$\displaystyle\|\tilde{U}^{\tilde{I}}(x_{0}+\ell_{0}\alpha)+\kappa\cdot\overline{\tilde{U}^{\tilde{I}}(x_{0}+\ell_{0}\alpha)}\|\leq
L^{-1/2}+e^{-\frac{\hat{L}_{d}(\hat{A}_{E})}{2}n}.$

Finally combining (A.63) with (A.62), we conclude that

(A.64) $\displaystyle\|\tilde{U}^{\tilde{I}}(x_{0}+\ell_{0}\alpha)\|\leq
2L^{-1/4}+2e^{-\frac{\hat{L}_{d}(\hat{A}_{E})}{4}|n_{j}|}.$

This leads to a contradiction with Lemma A.4 which implies

(A.65)
$\displaystyle\inf_{|\mathrm{Im}(x)|\leq\frac{\delta}{2\pi}}\|\tilde{U}^{\tilde{I}}(x)\|\geq
e^{-2\delta\tilde{n}}e^{-\pi|n_{j}\mathrm{Im}(x)|}\geq
e^{-55\delta|n_{j}|}>L^{-1/2}+e^{-\frac{\hat{L}_{d}(\hat{A}_{E})}{2}|n_{j}|},$

where we used Lemma A.7 and $\tilde{n}<27|n_{j}|$. ∎

Next, we return to the estimate of $\det M_{J}$. Combining (A.55) with (A.65),
we obtain

(A.66)
$\displaystyle\inf_{|\mathrm{Im}(x)|\leq\frac{\delta}{2\pi}}\left(\max(|e^{\pi
in_{j}x}u^{J}(x)|,|e^{\pi
in_{j}x}u^{J}(x-\alpha)|)\right)\geq\inf_{|\mathrm{Im}(x)|\leq\frac{\delta}{2\pi}}\|U^{J}(x)\|\geq$
$\displaystyle
e^{-55\delta|n_{j}|}-e^{-\frac{\hat{L}_{d}(\hat{A}_{E})}{2}|n_{j}|}$ (A.67)
$\displaystyle\geq$ $\displaystyle e^{-60\delta|n_{j}|}.$

For an arbitrary $x$ with $|\mathrm{Im}(x)|\leq\frac{\delta}{2\pi}$, suppose
we have (the other case is similar)

(A.68) $\displaystyle|e^{\pi in_{j}x}u^{J}(x)|\geq e^{-60\delta|n_{j}|}.$

Setting $\kappa:=\overline{e^{\pi in_{j}x}u^{J}(x)}/(e^{\pi
in_{j}x}u^{J}(x))$, which satisfies $|\kappa|=1$, one has

(A.69) $\displaystyle|\det\tilde{M}_{J}(x)|=$
$\displaystyle|\det(\begin{matrix}\tilde{U}^{J}(x),\,\,\,\overline{\tilde{U}^{J}(x)}-\kappa\cdot\tilde{U}^{J}(x)\end{matrix})|=|e^{\pi
in_{j}x}u^{J}(x)|\cdot\|\tilde{U}^{J}(x)-\overline{\kappa}\cdot\overline{\tilde{U}^{J}(x)}\|$
(A.70) $\displaystyle\geq$ $\displaystyle L^{-4C_{4}}e^{-60\delta|n_{j}|},$

where we used (A.68) and Lemma A.8 in the last line. The claimed result on
$|\det M_{J}(x)|$ follows by combining (A.39) with Lemma A.7. ∎

Next, we construct an $\mathrm{SL}(2,\mathbb{R})$ conjugacy. Let
$S(x)=\mathrm{Re}\,U^{J}(x)$ and $T(x)=\mathrm{Im}\,U^{J}(x)$, and define
$W_{1}(x)$ to be the matrix with columns $S(x)$ and $T(x)$. Thus

(A.71) $\displaystyle M_{J}(x)=W_{1}(x)\left(\begin{matrix}1\ &1\\\
i&-i\end{matrix}\right).$

Clearly (A.35) implies

(A.72) $\displaystyle\|W_{1}\|_{\frac{\delta}{2\pi}}\leq e^{3\delta|n_{j}|}.$

By Theorem A.6, $\det W_{1}(x)\neq 0$ for
$|\mathrm{Im}(x)|\leq\frac{\delta}{2\pi}$. Without loss of generality we
assume $\det W_{1}(x)>0$ on $\\{x:|\mathrm{Im}(x)|\leq\frac{\delta}{2\pi}\\}$
(otherwise we take $W_{1}(x)$ be the matrix with columns $S(x)$ and $-T(x)$).
By (A.36), we conclude that

(A.73)
$\displaystyle\|A_{E}(x)W_{1}(x)-W_{1}(x+\alpha)R_{-\theta}\|_{\frac{\hat{L}_{d}(\hat{A}_{E})}{32\pi}}\leq
e^{-\frac{\hat{L}_{d}(\hat{A}_{E})}{150}|n_{j+1}|}.$

Combining this estimate with (A.72), we infer that for
$|\mathrm{Im}(x)|\leq\frac{\delta}{2\pi}$,

(A.74) $\displaystyle\det W_{1}(x)=\det
W_{1}(x+\alpha)+O(e^{-\frac{\hat{L}_{d}(\hat{A}_{E})}{200}|n_{j+1}|}).$

By Theorem A.6, Lemma A.7 and (A.74), for all
$|\mathrm{Im}(x)|\leq\frac{\delta}{2\pi}$,

(A.75) $\displaystyle\det
W_{1}(x)=w_{0}+O(e^{-\frac{\hat{L}_{d}(\hat{A}_{E})}{300}|n_{j+1}|}),$

with $w_{0}\geq L^{-5C_{4}}$. Combining (A.73) with (3.7), (A.72), (A.75) and
Lemma A.7, we have

$\displaystyle\|W_{1}(x+\alpha)^{-1}A_{E}(x)W_{1}(x)-R_{-\theta}\|_{\frac{\delta}{2\pi}}\leq$
$\displaystyle\frac{1}{\inf_{|\mathrm{Im}(x)|\leq\frac{\delta}{2\pi}}\det
W_{1}(x)}\|W_{1}\|_{\frac{\delta}{2\pi}}e^{-\frac{\hat{L}_{d}(\hat{A}_{E})}{150}|n_{j+1}|}$
(A.76) $\displaystyle=$ $\displaystyle
O(L^{6C_{4}}e^{3\delta|n_{j}|}e^{-\frac{\hat{L}_{d}(\hat{A}_{E})}{150}|n_{j+1}|})=O(e^{-\frac{\hat{L}_{d}(\hat{A}_{E})}{200}|n_{j+1}|}).$

Define $W_{2}(x)=(\det W_{1}(x))^{-1/2}W_{1}(x)\in
C^{\omega}_{\frac{\delta}{2\pi}}(\mathbb{T},\mathrm{SL}(2,\mathbb{R}))$. By
(A.2) and (A.75), we have that for $|\mathrm{Im}(x)|\leq\frac{\delta}{2\pi}$,

(A.77) $\displaystyle W_{2}(x+\alpha)^{-1}A_{E}(x)W_{2}(x)=$
$\displaystyle\frac{(\det W_{1}(x+\alpha))^{1/2}}{(\det
W_{1}(x))^{1/2}}R_{-\theta}+O(e^{-\frac{\hat{L}_{d}(\hat{A}_{E})}{200}|n_{j+1}|})$
(A.78) $\displaystyle=$ $\displaystyle
R_{-\theta}+O(e^{-\frac{\hat{L}_{d}(\hat{A}_{E})}{200}|n_{j+1}|}).$

Taking (A.72), (A.75) and Lemma A.7 into account, one has

(A.79) $\displaystyle\|W_{2}\|_{\frac{\delta}{2\pi}}\leq
L^{6C_{4}}e^{3\delta|n_{j}|}\leq L^{7C_{4}}.$

The estimate of $\mathrm{deg}(W_{2})$, see Definition B.1, follows the same
Rouché argument as the proof of Theorem 3.5 in [AJ2]. In fact, in the
subcritical case, one should be able to obtain an improved bound
$|\mathrm{deg}(W_{2})|\lesssim\log|n_{j}|$. We leave those details to
interested readers.

##  Appendix B Proof of Theorem 1.6

###  B.1. Rotation number

######  Definition B.1.

Any $A:\mathbb{T}\to\mathrm{SL}(2,\mathbb{R})$ is homotopic to $x\to R_{nx}$
for some integer $n$, called the degree of $A$ and denoted
$\mathrm{deg}(A)=n$.

Assume that $A:\mathbb{T}\to\mathrm{SL}(2,\mathbb{R})$ is homotopic to the
identity. There exist $\psi:\mathbb{T}^{2}\to\mathbb{R}$ (lift of $A$) and
$v:\mathbb{T}^{2}\to\mathbb{R}^{+}$ such that

(B.1) $\displaystyle A(x)\left(\begin{matrix}\cos(2\pi y)\\\ \sin(2\pi
y)\end{matrix}\right)=v(x,y)\left(\begin{matrix}\cos(2\pi(y+\psi(x,y)))\\\
\sin(2\pi(y+\psi(x,y)))\end{matrix}\right).$

Let $\mu$ be any probability measure on $\mathbb{T}^{2}$ that is invariant
under the map

$T:(x,y)\to(x+\alpha,y+\psi(x,y)),$

projecting over the Lebesgue measure on the first coordinate. Then the number

(B.2)
$\displaystyle\rho(\alpha,A)=\int\psi\,\mathrm{d}\mu\qquad(\mathrm{mod}\mathbb{Z}),$

is independent of $\psi$ and $\mu$, and is called the fibered rotation number
of $(\alpha,A)$. It is clear from the definition that

(B.3) $\displaystyle|\rho(\alpha,A)-\theta|\leq C\|A-R_{\theta}\|_{0}.$

If the cocycles $(\alpha,A^{(1)})$ and $(\alpha,A^{(2)})$ are conjugate to
each, i.e., $B(x+\alpha)^{-1}A^{(1)}(x)B(x)=A^{(2)}(x)$ for some continuous
$B:\mathbb{T}\to\mathrm{PSL}(2,\mathbb{R})$ with $\mathrm{deg}(B)=k$, then

(B.4)
$\displaystyle\rho(\alpha,A^{(1)})=\rho(\alpha,A^{(2)})+\frac{k}{2}\alpha.$

The following relation between the rotation number on the one hand, and
integrated density of states (see (2.60)) on the other hand, is well-known:

(B.5) $\displaystyle\mathcal{N}(E)=1-2\rho(\alpha,A_{E}).$

###  B.2. Almost localization and almost reducibility

The following connection between localization and reduciblity was first
explored in [P], leading to the proof the Ten Martini problem for the almost
Mathieu operator for all Diophantine $\alpha$’s. The exact form cited below
appeared in [AJ2].

######  Theorem B.1.

[AJ2]*Theorem 2.5 For $\alpha\in\mathbb{R}\setminus\mathbb{Q}$, let
$H_{\alpha,\theta,v}$ be the Schrödinger operator with a real-valued analytic
potential $v$. Let $\theta,E\in\mathbb{R}$ be such that there exists a non-
trivial exponentially decaying solution of the dual operator
$\hat{H}\hat{u}=E\hat{u}$. Then

  * •

If $2\theta\notin\mathbb{Z}\alpha+\mathbb{Z}$, then there is a
$B:\mathbb{T}\to\mathrm{SL}(2,\mathbb{R})$ analytic such that
$B(x+\alpha)^{-1}A_{E}(x)B(x)=R_{\pm\theta}$.

  * •

If $2\theta-n\alpha\in\mathbb{Z}$ for some $n\in\mathbb{Z}$, then there exists
$B:\mathbb{T}\to\mathrm{PSL}(2,\mathbb{R})$ and $w:\mathbb{T}\to\mathbb{R}$
analytic such that

(B.6) $\displaystyle B(x+\alpha)^{-1}A_{E}(x)B(x)=\left(\begin{matrix}\pm
1\,&w(x)\\\ 0&\pm 1\end{matrix}\right).$

If moreover $\beta(\alpha)=0$, then $w$ can be chosen to be a constant.

In either case, $\rho(\alpha,A_{E})=\pm\theta+\frac{m}{2}\alpha$ for some
$m\in\mathbb{Z}$.

###  B.3. Proof of Theorem 1.6

By Theorem 1.2, we have the following spectral consequence, which is the
analogue of Theorem 4.1 in [AJ2].

######  Theorem B.2.

For $\beta(\alpha)=0$ and $E\in\Sigma_{\alpha,0}$, there exists a constant
$c>0$ (depending on $\hat{L}_{d}(\alpha,\hat{A}_{E})$) with the following
properties:

  1. (1)

If $\rho(\alpha,A_{E})$ is $c-$resonant, then there exists a sequence
$B^{(n)}:\mathbb{T}\to\mathrm{SL}(2,\mathbb{R})$ such that
$B^{(n)}(x+\alpha)^{-1}A_{E}(x)B^{(n)}(x)$ converges to a constant rotation
uniformly in $\\{x:|\mathrm{Im}(x)|<c\\}$.

  2. (2)

If $\rho(\alpha,A_{E})$ is not $c-$resonant and
$2\rho(\alpha,A_{E})\notin\mathbb{Z}\alpha+\mathbb{Z}$, then there exists
$B\in C^{\omega}_{c}(\mathbb{T},\mathrm{SL}(2,\mathbb{R}))$ such that
$B(x+\alpha)^{-1}A_{E}(x)B(x)$ is a constant rotation.

  3. (3)

If $2\rho(\alpha,A_{E})\in\mathbb{Z}\alpha+\mathbb{Z}$, then there exists
$B:\mathbb{T}\to\mathrm{PSL}(2,\mathbb{R})$ analytic such that
$B(x+\alpha)^{-1}A_{E}(x)B(x)$ is a constant.

###### Proof.

Let $\varepsilon_{0}:=\frac{\hat{L}_{d}(\alpha,\hat{A}_{E})}{50C_{3}}$ and
$\theta=\theta(E),\\{n_{j}\\},\delta,L_{j},W$ be defined as in Theorem 1.2.
First we show that if $\theta$ is resonant, then $\rho(\alpha,A_{E})$ must be
resonant. Thus suppose that $\theta$ is $\varepsilon_{0}$-resonant. By Theorem
1.2, for $j>j_{0}(\alpha,v,E)$ there exists $W_{j}\in
C^{\omega}_{\frac{\delta}{2\pi}}(\mathbb{T},\mathrm{SL}(2,\mathbb{R}))$ such
that

(B.7)
$\displaystyle\|W_{j}(x+\alpha)^{-1}A_{E}(x)W(x)-R_{-\theta}\|_{\frac{\delta}{2\pi}}\leq
e^{-\frac{\hat{L}_{d}(\alpha,\hat{A}_{E})}{300}|n_{j+1}|},$

and $|\mathrm{deg}(W_{j})|\leq C_{0}|n_{j}|$ with absolute constant $C_{0}>0$.
By (B.3), (B.4) and (B.7),

(B.8)
$\displaystyle|\rho(\alpha,A_{E})-\theta+\mathrm{deg}(W_{j})\,\alpha|\leq
Ce^{-\frac{\hat{L}_{d}(\alpha,\hat{A}_{E})}{300}|n_{j+1}|}.$

This implies, after combining the preceding with Lemma A.7, that

(B.9)
$\displaystyle\|2\rho(\alpha,A_{E})-(n_{j}-2\mathrm{deg}(W_{j}))\alpha\|_{\mathbb{T}}\geq$
$\displaystyle
L_{j}^{-1}-Ce^{-\frac{\hat{L}_{d}(\alpha,\hat{A}_{E})}{300}|n_{j+1}|}$ (B.10)
$\displaystyle\geq$ $\displaystyle
e^{-\frac{\hat{L}_{d}(\alpha,\hat{A}_{E})}{4000C_{4}}|n_{j+1}|}-Ce^{-\frac{\hat{L}_{d}(\alpha,\hat{A}_{E})}{300}|n_{j+1}|}>0,$

and

(B.11)
$\displaystyle\|2\rho(\alpha,A_{E})-(n_{j}-2\mathrm{deg}(W_{j}))\alpha\|_{\mathbb{T}}\leq$
$\displaystyle
L_{j}^{-1}+Ce^{-\frac{\hat{L}_{d}(\alpha,\hat{A}_{E})}{300}|n_{j+1}|}$ (B.12)
$\displaystyle\leq$ $\displaystyle
e^{-\varepsilon_{0}|n_{j}|}+Ce^{-\frac{\hat{L}_{d}(\alpha,\hat{A}_{E})}{300}|n_{j+1}|}$
(B.13) $\displaystyle\leq$ $\displaystyle
e^{-(2C_{0}+2)^{-1}\varepsilon_{0}|2\mathrm{deg}(W_{j})-n_{j}|}.$

Together with (3.78) this implies that $\rho(\alpha,A_{E})$ is
$(2C_{0}+2)^{-1}\varepsilon_{0}$-resonant at $n_{j}-2\mathrm{deg}(W_{j})$. Let
$c=(2C_{0}+2)^{-1}\varepsilon_{0},\frac{\delta}{2\pi}$. We divide the analysis
into three cases.

Case 1. If $\rho(\alpha,A_{E})$ is $c$-resonant and $\theta_{0}$ is
non-$\varepsilon_{0}$-resonant, then Theorem 1.7 implies that there is an
exponentially decaying eigenfunction. Theorem B.1 applies and yields the
claimed result.

Case 2. If $\rho(\alpha,A_{E})$ is $c$-resonant and $\theta_{0}$ is
$\varepsilon_{0}$-resonant, then (B.7) is the claimed result.

Case 3. If $\rho(\alpha,A_{E})$ is non-$c$-resonant, then by the arguments
above $\theta$ is non-$\varepsilon_{0}$ resonant. Theorem B.1 implies the
claimed result. ∎

Finally, Theorem 1.6 is the consequence of the following statement.

######  Theorem B.3.

For $\beta(\alpha)=0$, $|\lambda|<1$ and $|\nu|<\nu_{0}(\alpha,\lambda,w)$
sufficiently small. Suppose
$E\in\sigma(H^{\mathrm{PAMO}}_{\alpha,\theta,\nu,w})$ is such that
$\mathcal{N}(E)\in\mathbb{Z}\alpha+\mathbb{Z}$, then $E$ belongs to the
boundary of a component of
$\mathbb{R}\setminus\sigma(H^{\mathrm{PAMO}}_{\alpha,\theta,\nu,w})$.

###### Proof.

Due to the arguments in [P], it is enough to show that
$(\alpha,A^{\mathrm{PAMO}}_{E})$ is reducible. Suppose
$(\alpha,A^{\mathrm{PAMO}}_{E})$ is not reducible, then Theorem B.2 implies
that $\rho(\alpha,A_{E})$ is $c$-resonant for some $c>0$. But this contradicts
the property $2\rho(\alpha,A_{E})\in\mathbb{Z}\alpha+\mathbb{Z}$, which
follows from (B.5). ∎

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