algebras to quadratic forms.

Recall that for any quadratic form $\varphi$ of dimension $8$ with $\operatorname{disc}\varphi=1$ the even Clifford algebra decomposes into a direct product of central simple $F$-algebras of degree $8$, $$C_0(\varphi)\simeq C_+(\varphi)\times C_-(\varphi).$$ The canonical involution $\tau_0$ on $C_0(\varphi)$ restricts to orthogonal involutions $\tau_+$, $\tau_-$ on $C_+(\varphi)$ and $C_-(\varphi)$, and we have $$(C_+(\varphi),\tau_+)\simeq(C_-(\varphi),\tau_-).$$ It is easily checked that $(C_+(\varphi),\tau_+)$ is totally decomposable.

\[totdec.prop\] For every central simple algebra of degree $8$ with totally decomposable orthogonal involution $(A,\sigma)$, there is a quadratic form $\varphi$ with $\dim\varphi=8$ and $\operatorname{disc}\sigma=1$ such that $$(A,\sigma)\simeq(C_+(\varphi),\tau_+).$$ The form $\varphi$ is uniquely determined by $(A,\sigma)$ up to similarity. Moreover,

-   the algebra $A$ is split if and only if $\varphi$ is a multiple of a $3$-fold Pfister form; in that case $(A,\sigma)\simeq\operatorname{Ad}_\varphi$;

-   the algebra $(A,\sigma)$ contains ${(Q,{\overline{\rule{2.5mm}{0mm}\rule{0mm}{4pt}}})}$ if and only if $\varphi$ contains a subform similar to ${{\left\langle{1,-a,-b}\right\rangle}}$.

Furthermore, the following conditions are equivalent:

1.  $(A,\sigma)$ is isotropic;

2.  $(A,\sigma)$ is hyperbolic;

3.  $\varphi$ is isotropic.

Since $(A,\sigma)$ is totally decomposable, it follows from [@KMRT (42.11)] that one of the components $C_+(A,\sigma)$ of the Clifford algebra is split. The canonical involution $\sigma_+$ on $C_+(A,\sigma)$ is orthogonal, hence there is a quadratic form $\varphi$ of 