\dots)$ is a sequence with odd period, $(\beta_0, \beta_1, \dots) = 
(\alpha_0, 0, \alpha_1, 0, \alpha_2, \dots)$ has even period and $${\label}{3.1} 
\Phi_{2n} (z,\{\beta_j\}) = \Phi_n (z^2, \{\alpha_j\})$$ so results for the even $p$ case immediately imply results for the odd $p$.

Define the $2\times 2$ matrix $${\label}{3.2} 
A_k(z) = {\frac}{1}{\rho_k} \begin{pmatrix} 
z & -\bar\alpha_k \\ -z\alpha_k & 1 \end{pmatrix}$$ where $\rho_k$ is given by . Then $$\det (A_k (\alpha)) = z$$ and its $^*$ are equivalent to $${\label}{3.3} 
\binom{\varphi_{n+1}}{\varphi_{n+1}^*} = A_n (z) \binom{\varphi_n}{\varphi_n^*}$$ The second kind polynomials, $\psi_n(z)$, are the OPUC with Verblunsky coefficients $\{-\alpha_j\}_{j=0}^\infty$. Then it is easy to see that $${\label}{3.4} 
\binom{\psi_{n+1}}{-\psi_{n+1}^*} = A_n(z) \binom{\psi_n}{-\psi_n^*}$$ with $A$ given by .

We thus define $${\label}{3.5} 
T_n(z) = A_{n-1}(z) \dots A_0(z)$$ By , we have $${\label}{3.6} 
T_{mp+b} = T_b (T_p)^m$$ and imply that $$\begin{aligned}
\binom{\varphi_n}{\varphi_n^*} &= T_n \binom{1}{1} {\label}{3.7} \\
\binom{\psi_n}{-\psi_n^*} &= T_n \binom{1}{-1} {\label}{3.8} \end{aligned}$$ so that $${\label}{3.9} 
T_n(z) = \tfrac12 \begin{pmatrix} 
\varphi_n(z) + \psi_n(z) & \varphi_n(z) -\psi_n(z) \\
\varphi_n^*(z) - \psi_n^*(z) & \varphi_n^*(z) + \psi_n^*(z) 
\end{pmatrix}$$

The discriminant is defined by $${\label}{3.10} 
\Delta(z) = z^{-p/2} \, {\text{\rm{Tr}}}(T_p(z))$$ The $z^{-p/2}$ factor (recall $p$ is even) is there because $\det (z^{-p/2} T_p(z)) 
=1$, so $z^{-p/2} T_p(z)$ has eigenvalues $\Gamma_\pm 