$t$ and $\bar t$ are $\mathbb{Q}$-linearly independent

Let $K \subseteq \mathbb{C}$ be a subfield and $t \in \mathbb{C}$ transcendental over $K$ with $t\bar t \in K$. If $a, b \in \mathbb{Q}$ satisfy

then $a = 0$ and $b = 0$.

Why. Transcendence gives $t \neq 0$, and $\bar t = (t\bar t)/t$. Substituting and clearing denominators turns the relation into $a\,t^{2} + b\,(t\bar t) = 0$. If $a \neq 0$ this expresses $t^{2}$ as an element of $K$ times $t \bar t$, hence puts $t^{2}$ — and with it $\bar t^{2} = (t\bar t)^{2}/t^{2}$, hence the cross term $t^{2}+\bar t^{2}$ — inside $K$, which is impossible for transcendental $t$. So $a = 0$, and then $b\,\bar t = 0$ with $\bar t \neq 0$ gives $b = 0$.

Role. This is what makes the coordinate map $(a,b) \mapsto a t + b\bar t$ injective, so that "a function of the coordinates" is a well-defined function on the rational plane. Every statement of the model phrased in coordinates — the norm form, the coordinate action of the involution, the formal exponential $\mathrm{Exp}_0$ — depends on it. It is the model's analogue of the axis lemma.