Proposition 1

Let $A(\Phi)$ be the subalgebra of $\mathscr{L}_A(E)$ generated by the direct similitudes of E.

a) The direct similitudes are the invertible elements of $A(\Phi)$. The algebra $A(\Phi)$ is a commutative algebra of degree 2 over A. When E contains no isotropic vector $\neq 0$, $A(\Phi)$ is a field, a quadratic extension of A; in the contrary case it is the direct sum of two fields isomorphic to A.

b) Let $(e_1, e_2)$ be an orthogonal basis of E; put $\alpha_i = \Phi(e_i, e_i)$ $(i = 1, 2)$ and $\delta = -\alpha_2/\alpha_1$. Then the matrices of the elements of $A(\Phi)$ with respect to this basis are the matrices of the form $\begin{pmatrix} a & \delta b \\ b & a \end{pmatrix}$ where $a \in A, b \in A$.

c) The space E is a monogeneous free $A(\Phi)$-module, generated by any non-isotropic vector.

Let us introduce, in fact, an auxiliary alternating bilinear form $B \neq 0$ on E; then B is non-degenerate. There exists therefore an endoNo. 1

morphism $\omega$ of E such that $\Phi(x, y) = B(\omega(x), y)$ for all $x, y$ in E. For every invertible endomorphism $u$ of E, one has

therefore, in order that $u$ be a direct similitude, it is necessary and sufficient that one have for all $x, y$ in E; since $\det u \neq 0$ and B is non-degenerate, this is equivalent to $u^{-1}\omega u = \omega$, or again to $u\omega = \omega u$. Let us take for B the alternating bilinear form whose matrix $S$ with respect to $(e_1, e_2)$ is $\begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix}$; denoting by $R$ and $W$ the matrices of $\Phi$ and of $\omega$ with respect to this basis, the relation $\Phi(x, y) = B(\omega(x), y)$ is written, by virtue of formula (47) of § 1, No. 10, $R = {}^tW.S$; writing this out shows that one has $W = \begin{pmatrix} 0 & \alpha_2 \\ -\alpha_1 & 0 \end{pmatrix}$. If $\begin{pmatrix} a & c \\ b & d \end{pmatrix}$ denotes the matrix of $u$ with respect to $(e_1, e_2)$, the relation $u\omega = \omega u$ is therefore equivalent to the relations $b\alpha_2 = -c\alpha_1, a\alpha_2 = d\alpha_2, a\alpha_1 = d\alpha_1$, that is to say to $a = d$ and $c = \delta b$; this proves that the matrices of the direct similitudes are the invertible matrices of the form $\begin{pmatrix} a & \delta b \\ b & a \end{pmatrix}$ ($a, b$ in A).

Now the endomorphisms of E whose matrices are of the form $\begin{pmatrix} a & \delta b \\ b & a \end{pmatrix}$ ($a, b$ in A) form a vector subspace of dimension 2 of $\mathfrak{L}_A(E)$, generated by 1 and by the endomorphism $\omega$; since $\omega^2$ is the homothety of ratio $-\alpha_1\alpha_2$, this subspace is the subalgebra $A(\Phi)$ of $\mathfrak{L}_A(E)$ generated by the direct similarities. The direct similarities are the invertible elements of $A(\Phi)$, that is to say those whose matrices satisfy $a^2 - \delta b^2 \neq 0$. The fact that the algebra $A(\Phi)$ is commutative is evident. Let us apply to it the results of Chap. II, § 7, No. 7: if $\delta$ is not a square in A, that is to say if no nonzero vector of E is isotropic, $A(\Phi)$ is a field; if, on the contrary, $\delta$ is a square in A, that is to say if E contains isotropic vectors $\neq 0$, $A(\Phi)$ is the direct sum of two fields isomorphic to A. This proves a) and b).

Finally any nonisotropic vector of E may be taken as first vector $e_1$ of an orthogonal basis $(e_1, e_2)$ (§ 6, No. 1); therefore its transforms $u(e_1)$ by the elements $u$ of $\Lambda(\Phi)$ are the vectors of the form $ae_1 + be_2$ ($a, b$ in $\Lambda$), that is to say all the vectors of $E$, since all the matrices $\begin{pmatrix} a & \delta b \\ b & a \end{pmatrix}$ are matrices of elements of $\Lambda(\Phi)$. In other words $E$ is a monogenic $\Lambda(\Phi)$-module, generated by any nonisotropic vector. One sees further that it is a free monogenic $\Lambda(\Phi)$-module, since $u(e_1) = ae_1 + be_2 = 0$ implies $a = b = 0$, hence $u = 0$. This proves c).