For u ∉ Q̄, real algebraic s with s² = uū, |λ| = 1 and real f, the span of 1, u, ū, f, f/(s + λu) is stable under conjugation and carries a 2×3 configuration

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 $s = \pm|u|$). Let $\lambda \in \overline{\mathbb{Q}}$ with $|\lambda| = 1$, let $f \in \mathbb{C}$ be real, put $g = f/(s + \lambda u)$, and let $W$ be the $\overline{\mathbb{Q}}$-span of $1, u, \bar u, f, g$. Suppose these five numbers are free over $\overline{\mathbb{Q}}$. Then $W$ is stable under complex conjugation, and the products of $(1, u)$ and $(u^{-1}, 1, g)$, two free families, all lie in $W$.

This is the family $W(\lambda, f)$ of the research notes: the conjugation-stable five-dimensional spaces containing $H_0 = \overline{\mathbb{Q}} + \overline{\mathbb{Q}}u + \overline{\mathbb{Q}}\bar u$ that carry a $2 \times 3$ configuration $(1, u) \otimes (y_1, y_2, y_3)$ (after rescaling). It is not a geometric progression in general, so it is a separate case from the involution spaces of R9.

Proof. $u \neq 0$, $s \neq 0$ and $s + \lambda u \neq 0$ since $u$ is transcendental; $\bar u = s^2/u$. Using $\lambda\bar\lambda = 1$, $\bar g = f/s - g$ and $ug = (f - sg)/\lambda$, so conjugation maps each generator into $W$, and $x = (1, u)$, $y = (u^{-1}, 1, g)$ has its six products $u^{-1} = \bar u/s^2, 1, g, u, 1, ug$ in $W$. Both families are free: $u \notin \overline{\mathbb{Q}}$, and $(u^{-1}, 1, g)$ is a rescaled subfamily of the five free generators.

Novelty. Not found in the sources read; it is short.