A general form of a maximal totally isotropic subspace in the split octonion algebra

Question

$\DeclareMathOperator\Ker{Ker}$Let $\mathbb O'$ be the split octonion algebra over $\mathbb R$. For each nonzero divisor of zero $x\in \mathbb O'$ ($x \neq 0$, $N(x)=0$) the kernel of the left multiplication by $x$, $\Ker L_x=\{y\in \mathbb O': xy=0\}$ is a maximal totally isotropic subspace in $\mathbb O'$ and is equal to $\overline{x} \mathbb O'$ (since both $\Ker L_x$, $\overline{x} \mathbb O'$ are totally isotropic (hence with dimension $\leq 4$) and $\dim \Ker L_x+\operatorname{Im} L_x=8$). Similarly, $\Ker R_x=\{y\in \mathbb O': yx=0\}=\mathbb O' \overline{x}$ is maximal totally isotropic, where $R_x(y)=yx$ for $y\in \mathbb O'$.

My question: is it maybe true that for each maximally totally isotropic $V\subset \mathbb O'$ there is a nonzero zero's divisor $x$ in $\mathbb O'$ such that $V=x\mathbb O'$ or $V=\mathbb O' x$? Is $x$ unique up to nonzero real constant?

Answer 1

Yes to both questions!

Every maximal totally isotropic $V$ is of this form by Theorem 3 on page 164 of van der Blij, F.; Springer, T. A., Octaves and triality, Nieuw Arch. Wiskd., III. Ser. 8, 158-169 (1960). ZBL0127.11804.

The element $x$ is determined up to a non-zero constant by Theorem 4 on page 165.