If P(u)/Q(u), of degree 2, maps g₀, g₁ into span(1, u, u⁻¹, g₀, g₁) of dimension 5, that span is complete, two-orbit or cyclic

Let $u \in \mathbb{C}$ be transcendental, that is, $u \notin \overline{\mathbb{Q}}$ (the algebraic numbers); spans and linear independence are over $\overline{\mathbb{Q}}$. Let $P, Q \in \overline{\mathbb{Q}}[X]$ be coprime with $\max(\deg P, \deg Q) = 2$, and let $g_0, g_1 \in \mathbb{C}$ be such that $S = \mathrm{span}(1, u, u^{-1}, g_0, g_1)$ has dimension $5$ and $(P(u)/Q(u))\,g_j \in S$ for $j = 0, 1$. Write $P_c = P(u) - cQ(u)$. Then one of the following holds:

• (scalar) for some $c \in \overline{\mathbb{Q}}$, $Q(u)R(u)/(uP_c) \in S$ for every $R \in \overline{\mathbb{Q}}[X]$ of degree at most $2$;
• (two-orbit) for some $c_1 \neq c_2$ in $\overline{\mathbb{Q}}$ and $R_1, R_2 \in \overline{\mathbb{Q}}[X]$ of degree at most $2$, $S = \mathrm{span}(1, u, u^{-1}, Q(u)R_1(u)/(uP_{c_1}), Q(u)R_2(u)/(uP_{c_2}))$;
• (cyclic) for some $c \in \overline{\mathbb{Q}}$ and $R_0, R_1 \in \overline{\mathbb{Q}}[X]$ of degree at most $2$, $S = \mathrm{span}(1, u, u^{-1}, Q(u)R_0(u)/(uP_c), Q(u)(Q(u)R_0(u) + P_cR_1(u))/(uP_c^2))$.

Proof. Coprimality and degree give $Q \neq 0$ and $P - cQ \neq 0$, so $Q(u) \neq 0$ and $P_c \neq 0$. Let $k = P(u)/Q(u)$. DiazModulus.quadratic_complement_multiplication_matrix gives $T$ over $\overline{\mathbb{Q}}$ and $R_0, R_1$ of degree at most $2$ with $kg_j = R_j(u)/u + \sum_iT_{ij}g_i$. With $G(x) = x_0g_0 + x_1g_1$ and $H(x) = x_0R_0 + x_1R_1$ this reads $kG(x) = H(x)(u)/u + G(Tx)$, so $Tx = cx + y$ gives $G(x) = Q(u)(H(x)(u) + uG(y))/(uP_c)$. Apply DiazModulus.qbar_matrix_fin_two_eigen_or_jordan to $T$. Eigenvectors $v, w$ for $c_1 \neq c_2$: $G(v)$, $G(w)$ have the two-orbit form with $R_1 = H(v)$, $R_2 = H(w)$, and as $v, w$ span $\overline{\mathbb{Q}}^2$, $\mathrm{span}(1, u, u^{-1}, G(v), G(w)) = S$. A Jordan pair: $G(v)$ as above and $G(w) = Q(u)(Q(u)H(v)(u) + P_cH(w)(u))/(uP_c^2)$, the cyclic form, with the same span. Scalar $T = cI$: multiplication by $k - c = P_c/Q(u)$ sends $g_0, g_1, Q(u)/u$ to $R_0(u)/u, R_1(u)/u, (P - cQ)(u)/u$, which are independent because $g_0, g_1, Q(u)/u$ are (the five generators of $S$ are), so they span the three-dimensional $H_0$. Every $R(u)/u$ is then $(k - c)(\alpha_0g_0 + \alpha_1g_1 + \alpha_2Q(u)/u)$, and $Q(u)R(u)/(uP_c) = \alpha_0g_0 + \alpha_1g_1 + \alpha_2Q(u)/u \in S$.

Novelty. Not found in the sources read; routine.