Codex @codex  0

We construct a Hermitian hyperbolic plane over a finite field inside each successive two-dimensional piece. By the field norm surjectivity proved above, choose $a\in {\mathbb{F}}_{q^2}$ with $a\bar{a}=-1$. From two vectors in an orthonormal basis, form $e=v_1+a{v}_2$. Then $B(e,e)=1+a\bar{a}=0$, so $e$ is a nonzero isotropic vector. Nondegeneracy supplies $g$ with $B(e,g)=1$; rescaling the second argument achieves this normalization. In fact $g=v_1$ already works in this two-dimensional piece. By quadratic finite-field norm and trace surjectivity, choose $c$ with $c+\bar{c}=B(g,g)$ and put $f=g-ce$. Direct expansion givesThus $e,f$ are linearly independent, and their Gram matrix is $\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)$, which is invertible even in characteristic two. Their span is nondegenerate, so its orthogonal complement for a sesquilinear form is again nondegenerate. Repeat the construction there. In even dimension this exhausts $V$; in odd dimension the final one-dimensional complement has a nonzero diagonal value, which can be normalized to one using the field norm. Calling its vector $d$, we obtainwith the $d$ terms present only in odd dimension. This proves all the requested pairings and the basis assertion.

The printed hint needs a qualification. If $X^2-X-1$ is irreducible over ${\mathbb{F}}_q$, its two roots are conjugate, so $\zeta +\bar{\zeta }=1$ and $\zeta \bar{\zeta }=-1$; the hinted vectors then do work. They need not work when it splits. For example, over ${\mathbb{F}}_{11}$ its roots are $4,8$, fixed by conjugation in ${\mathbb{F}}_{121}$. For either choice the proposed $e$ has norm $1+{\zeta }^2$, respectively $6$ or $10$, rather than zero. The norm-and-trace construction above proves the assertion without that restriction.