(z)$ given by . $\Delta(z)$ is real on $\partial{{\mathbb{D}}}$ so $${\label}{3.10a} 
\Delta (z) = {\overline}{\Delta(1/\bar z)}$$

$\Delta(z)\in (-2,2)$ only if $z=e^{i\theta}$ and there are $p$ roots, each of ${\text{\rm{Tr}}}(T_p(z)) \mp 2z^{p/2}=0$, that is, $p$ solutions of $\Delta(z) =\pm 2$. These alternate on the circle at points $+2, -2, -2, +2, +2, -2, -2, \dots$, so we pick $${\label}{3.11} 
0\leq x_1 < y_1 \leq x_2 < y_2 \leq \cdots < y_p \leq 2\pi$$ where $e^{ix_j}, e^{iy_j}$ are solutions of $\Delta(z) =\pm 2$.

The bands $${\label}{3.12} 
B_j =\{e^{i\theta}\mid x_j \leq \theta \leq y_j\}$$ are precisely the points where $\Delta(z)\in [-2,2]$. In between are the gaps $${\label}{3.13} 
G_j \{e^{i\theta}\mid y_j < \theta < x_{j+1}\}$$ where $x_{p+1} = x_1 + 2\pi$. Some gaps can be closed, that is, $G_j$ is empty (i.e., $y_j =x_{j+1}$).

We also see that on ${{\mathbb{C}}}\backslash\text{bands}$, ${\lvert\Gamma_+\rvert} > {\lvert\Gamma_-\rvert}$, so the Lyapunov exponent is given by $${\label}{3.14} 
\lim_{n\to\infty}\, \tfrac{1}{n}\, \log \|T_n(z)\| = \tfrac12 \, \log {\lvertz\rvert} + 
\tfrac{1}{p}\, \log {\lvert\Gamma_+(z)\rvert} \equiv \gamma(z)$$

If we remove the bands from ${{\mathbb{C}}}$, holds. Moreover, $\Gamma_+(z)$ has an analytic continuation to the Riemann surface, ${{\mathcal S}}$, of $[\prod_{\text{open gaps}}
(z-e^{iy_{j+1}})(z-e^{ix_j})]^{1/2}$. The genus of ${{\mathcal S}}$, $\ell\leq p-1$, where $\ell+ 1$ is the number of 