For u ∉ Q̄ and real algebraic s with s² = uū, the cyclic space with double poles at ±s carries no 2×3 configuration whose ratio is a Möbius function of u

Let $u \in \mathbb{C} \setminus \overline{\mathbb{Q}}$ and let $s \in \overline{\mathbb{Q}}$ be real with $s^2 = u\bar u$ (so $\bar u = s^2/u$). With $v = u/s$, put $G_1 = 1/(v - 1) + 2/(v + 1)$ and $G_2 = 1/(v - 1)^2 + 1/(v - 1) - 2/(v + 1)^2 + 2/(v + 1)$, and let $W$ be the $\overline{\mathbb{Q}}$-span of $1, u, \bar u, G_1, G_2$ (poles at $u = \pm s$, each a double pole). Then there are no free $x_1, x_2$ and free $y_1, y_2, y_3$ with all six products in $W$ and $x_2/x_1 = (\alpha u + \beta)/(\gamma u + \delta)$, $\alpha, \beta, \gamma, \delta \in \overline{\mathbb{Q}}$, $\alpha\delta - \beta\gamma \neq 0$.

So, like R14's $W_1$, it is none of R11's real pencil spaces.

Proof. First, if $t$ and $ut$ both lie in $W$, then $t \in \overline{\mathbb{Q}} + \overline{\mathbb{Q}}\bar u$. Write $t = c_1 + c_2u + c_3\bar u + c_4G_1 + c_5G_2$, and $ut$ likewise, and multiply each by $u(u^2 - s^2)^2$, which turns them into polynomials of degree $6$ in $u$ (as in DiazModulus.cyclic_pair_space_excludes_squares_and_reciprocals). Then $u \cdot u(u^2 - s^2)^2t = u(u^2 - s^2)^2 \cdot ut$ is an identity of degree $7$ over $\overline{\mathbb{Q}}$ at the transcendental $u$. Its $u^7$ coefficient gives $c_2 = 0$, and two combinations of the eight coefficient equations give $16s^6c_4 = 0$ and $16s^6c_5 = 0$. (In terms of poles: in $W$ the leading polar coefficients at $s$ and $-s$ are in the ratio $1 : -2$ for double poles and $1 : 2$ for simple ones, and multiplication by $u$ changes the sign of that ratio.) Now let free $x$ and $y$ have $x_iy_j \in W$ and $x_2(\gamma u + \delta) = x_1(\alpha u + \beta)$ with $\alpha\delta - \beta\gamma \neq 0$. Then $c = \alpha x_1 - \gamma x_2 \neq 0$ by the freeness of $x$, the three numbers $t_j = cy_j$ are free and lie in $W$, and $ut_j = \delta x_2y_j - \beta x_1y_j \in W$. So three free numbers lie in the two-dimensional $\overline{\mathbb{Q}} + \overline{\mathbb{Q}}\bar u$, which is impossible.

Novelty. Not found in the sources read; routine.