Proposition 4

a) When $\omega^2 = \delta$ is not a square in $A$ (that is to say, when $E$ does not contain any isotropic right), the mapping $t$ of $S^+/H$ into $\tilde{A}$ is a bijection.

b) When $\delta$ is the square of an element $\gamma$ of $A$, the mapping $t$ is a bijection of $S^+/H$ onto $\tilde{A}$ with the elements $1/\gamma$ and $-1/\gamma$ removed.

c) Denoting $S^+/H$ additively, one has, for $\varphi, \varphi'$ in $S^+/H$

(2)

when $t(\varphi)$ and $t(\varphi')$ are finite and $1 + t(\varphi)t(\varphi')$ is $\neq 0$.

In fact, since $S^+/H$ is a set of lines (with 0 omitted) of $A(\Phi)$ considered as a vector plane over $A$, $t$ is injective. On the other hand, for an element $a + b\omega$ ($a \in A, b \in A$) of $A(\Phi)$ to be a direct similitude, it is necessary and sufficient that it be invertible, that is to say, that one have $N(a + b\omega) = a^2 - \delta b^2 \neq 0$, or again $(b/a)^2 \neq 1/\delta$; this proves the assertions of surjectivity in a) and b). Finally, the product of the similitudes $1 + t(\varphi)\omega$ and $1 + t(\varphi')\omega$ is the similitude $1 + \delta t(\varphi)t(\varphi') + (t(\varphi) + t(\varphi'))\omega$, which proves c).