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).

The first assertion follows from the fact that the algebra $A(\Phi)$ is commutative. Since $E$ is a monogenic free $A(\Phi)$-module generated by $x$ (resp. $y$), there exists one and only one element $u$ (resp. $u'$) of $A(\Phi)$ such that $y = u(x)$ (resp. $x = u'(y)$); hence $x = u(u'(x))$, and $uu'$ is the identity; this shows that $u$ is invertible, and is therefore a direct similitude.

The first assertion follows from Corollary 1. The latter also shows that there exists one and only one direct similarity $u$ such that $y = u(x)$; since $\Phi(u(x), u(x)) = \Phi(x, x)$, the multiplier of $u$ is equal to 1, and $u$ is therefore a rotation.

This follows from Corollary 1 and from the fact that $H$ leaves every line of $E$ globally invariant.

In fact 1 and $-1$ are the only elements of $H \cap O^+$.

In fact the relation $u/\overline{u} = 1$ is equivalent to $u = \overline{u}$, that is to say to $u \in A^* = H$; hence $H$ is the kernel of $u \to u/\overline{u}$. Since $N(u/\overline{u}) = 1$, $u/\overline{u}$ is a rotation (Remark 1). It remains to prove that every element $\varphi$ of $O^+$ is of the form $u/\overline{u}$ ($u \in S^+$). If $1 + \varphi$ is invertible, one may take $u = 1 + \varphi$, since the relation $N(\varphi) = \varphi \overline{\varphi} = 1$ implies $1 + \varphi = \varphi(1 + \overline{\varphi})$. Otherwise, one has $N(1 + \varphi) = (1 + \varphi)(1 + \overline{\varphi}) = 0$, that is to say, on putting $\varphi = a + b\omega$ ($a \in A,\ b \in A,\ \omega^2 = \delta \in A$), $2(1 + a) = 0$, whence $a = -1$; now the relations $a = -1$ and $N(\varphi) = a^2 - \delta b^2 = 1$ imply $b = 0$, whence $\varphi = -1$; since $\overline{\omega} = -\omega$, it is enough, in this case, to take $u = \omega$.

When $A(\Phi)$ is a field, Proposition 3 is a particular case of Hilbert's theorem (Chapter V, § 11, No. 5, Theorem 3).

In fact, if $\overline{\varphi}$ is a rotation, we have $\overline{\varphi} = \varphi^{-1}$, whence $d(i(\varphi)) = \varphi/\overline{\varphi} = \varphi^2$. On the other hand, if $\varphi \in S^+/H$, $\varphi$ is the class mod. $H$ of a direct similarity $u$, and $d(\varphi) = u/\overline{u} = u^2 / N(u)$ is congruent to $u^2$ mod. $H$; whence the second formula.

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).

(3)

(4) (5)

Relation (3) expresses in fact that $N(c(\theta) + s(\theta)\omega) = 1$. For (4) and (5) it is enough to calculate, in $A(\Phi)$, the product of the rotations $c(\theta) + s(\theta)\omega$ and $c(\theta') + s(\theta')\omega$.

(6)

In fact $\varphi$ is the class mod. $H$ of the similitude $1 + t\omega$, and the rotation $d(\varphi)$ is therefore $(1 + t\omega)^2/N(1 + t\omega) = (1 + \delta t^2 + 2t\omega)/(1 - \delta t^2)$ (Prop. 3), which proves (6).

(7)

For every element $\varphi$ of $S^+/H$ such that $t(\varphi)$ is finite and $1 + \delta t(\varphi)^2 \neq 0$, one has (8)

In fact this follows immediately from Prop. 6 and from the corollary to Prop. 3.

We have already seen that the homomorphisms in question are injective, and it is enough to prove that the first is surjective. Let $(e_1, e_2)$ be an orthonormal basis of E, and let $\omega$ be the direct similarity such that $\omega(e_1) = e_2$ (cor. 1 of prop. 1, No. 1); one then has $\omega^2 = -1$ (prop. 1, b)). Given an arbitrary direct similarity u = a + bw (a \in A,\ b \in A),\ \text{one has}\ N(u) = a^2 + b^2 > 0,\ \text{and there exists one and only one rotation contained in the same half-line of } A(\Phi) \text{ as } u, \text{ namely } (a^2 + b^2)^{1/2}\ u.\ \text{QED.}

The hypotheses imply in fact that E contains no isotropic lines. Our assertion then follows from cor. 1 of prop. 1 (No. 1).

Henceforth we shall suppose that A is a maximal ordered field, and that the form $\Phi$ is positive. In the set of pairs $(D_1, D_2)$ of lines (resp. half-lines with origin 0) of E, the relation “there exists a direct similitude (resp. a rotation) u such that $u(D_1) = D'_1$ and $u(D_2) = D'_2$” is an equivalence relation between the pairs $(D_1, D_2)$ and $(D'_1, D'_2)$. The equivalence class of the pair $(D_1, D_2)$ is called, by definition, the angle of the lines (resp. half-lines) $D_1, D_2$ (taken in this order); it is denoted by $(\widehat{D_1, D_2})$.

Let us prove the necessity of the stated condition. Let u be a direct similitude (resp. a rotation) such that $u(D_1) = D'_1$ and $u(D_2) = D'_2$. There exists, by cor. 3 of prop. 1, a direct similitude (resp. by the cor. of prop. 7, a rotation) v such that $v(D_1) = D_2$. Since the group $S^+$ (resp. $O^+$) is commutative, we have $D'_2 = u(v(D_1)) = v(u(D_1)) = v(D'_1)$, and this shows that $(\widehat{D_1, D'_1}) = (\widehat{D_2, D'_2})$. The sufficiency is deduced from the necessity by interchanging $D_2$ and $D'_1$.

It follows from prop. 8 that to every angle $(\widehat{D_1, D_2})$ of lines (resp. half-lines) with origin 0 of E, there is canonically associated a well determined element of $S^+/H$ (resp. $O^+$), namely the class mod. H of the direct similitudes $\varphi$ (resp. the rotation $\varphi$) such that $u(D_1) = D_2$ for any representative $(D_1, D_2)$ of the angle $(\widehat{D_1, D_2})$. Thus there has been defined a canonical bijection $h$ (resp. $h'$) of the set $\mathfrak{A}_0$ of angles of lines (resp. $\mathfrak{A}$ of angles of half-lines) onto $S^+/H$ (resp. $O^+$); in particular, for every $\varphi \in \mathfrak{A}$, one says that $h(\varphi)$ is the rotation of angle $\varphi$. We shall transport to $\mathfrak{A}_0$ and $\mathfrak{A}$, by means of $h^{-1}$ and of ${h'}^{-1}$, the structures of commutative groups of $S^+/H$ and of $O^+$, and we shall denote additively the groups $\mathfrak{A}_0$ and $\mathfrak{A}$ thus obtained. If $D, D', D''$ denote lines (resp. half-lines) with origin 0 of E, we have by definition

whence one deduces

Since $\mathfrak{A}_0, S^+/H, \mathfrak{A}$ and $O^+$ are isomorphic (Prop. 3), it is enough to give the proof for $O^+$, that is to say, to show that there are exactly $n$ rotations $\varphi$ such that $\varphi^n = 1$. Now, since $\mathbf{A}$ is a maximal ordered field and $\mathbf{A}(\Phi)$ is an extension field of degree 2 of $\mathbf{A}$ (Prop. 1 a)), $\mathbf{A}(\Phi)$ is an algebraically closed field (Chapter VI, § 2, No. 6, Theorem 3). Hence the $n$-th roots of unity in $\mathbf{A}(\Phi)$ form a cyclic group of order $n$ (Chapter V, § 11, No. 1, Theorem 1). Since one has $N(u) = u \overline{u} \geqslant 0$ for every $u \in \mathbf{A}(\Phi)$, the relation $u^n = 1$ implies that one has $N(u) = 1$, hence that $u$ is a rotation (No. 1, Remark 1). This proves our assertion.

Finally, we shall suppose that the plane $E$ is oriented.

The case where $u$ is a homothety is trivial. In the contrary case one has $x \wedge u(x) \neq 0$ for all $x \neq 0$; let $x, y$ be two vectors of $E$ ($x \neq 0, y \neq 0$); there exists $\varphi \in S^+$ such that $y = \varphi(x)$, whence $y \wedge u(y) = \varphi(x) \wedge u \varphi(x) = \varphi(x) \wedge \varphi(u(x)) = (\det \varphi)(x \wedge u(x))$; taking an orthonormal basis of $E$, one sees that $\det \varphi$ is positive (Proposition 1 b)); whence our assertion.

This being so, among the two generators $\omega$ of $\mathbf{A}(\Phi)$ such that $\omega^2 = -1$, there exists one and only one such that the bivector $x \wedge \omega(x)$ is positive for all $x \in E$. It is this generator that we shall choose in order to define the functions $c_w, s_w$ and $t_w$ (No. 2). Let $h$ and $h'$ be the canonical bijections defined above from the group $\mathfrak{A}_0$ of angles of lines onto $S^+/H$ and from the group $\mathfrak{A}$ of angles of half-lines onto $O^+$. The composite mappings $t_w \circ h$ of $\mathfrak{A}_0$ into the projective field $\tilde{\mathfrak{A}}$, $c_w \circ h'$ and $s_w \circ h'$ of $\mathfrak{A}$ into the field $A$ are denoted respectively by $\operatorname{tg}$, $\cos$ and $\sin$, and are called the tangent, cosine and sine functions. The mapping $\varphi \to 1/\operatorname{tg} \varphi$ of $\mathfrak{A}_0$ into $\tilde{\mathfrak{A}}$ is denoted by $\operatorname{cotg}$ and is called the cotangent function. One says that the sine, tangent and cotangent functions are the trigonometric functions. The composite mappings $\operatorname{tg} \circ p$ and $\operatorname{cotg} \circ p$, where $p$ denotes the canonical homomorphism of $\mathfrak{A}$ onto $\mathfrak{A}_0$ (Remark 2 above), are again denoted by $\operatorname{tg}$ and $\operatorname{cotg}$ by abuse of language.

Formulas (2), (8), (3), (4), (5) and (7) of No. 2 give, since one has here $\delta = -1$

for $\varphi, \varphi'$ in $\mathfrak{A}_0$;

for $\theta, \theta'$ in $\mathfrak{A}$. On the other hand one has, by definition or as an easy consequence of the preceding formulas:

for $\theta \in \mathfrak{A}$.

Given two non-zero vectors $x, y$ of $E$, the angle of these two vectors (taken in this order) is defined to be, and is denoted by $\widehat{(x, y)}$, the angle of the half-lines to which they belong. For every vector $x$ of $E$ the length of $x$ is defined to be, and is denoted by $|x|$, the element $\Phi(x, x)^{1/2}$ of $A$.

In fact, since the vectors $x' = x/|x|$ and $y' = y/|y|$ both have length 1, there exists one and only one rotation $\varphi$ such that $\varphi(x') = y'$ (No. 1, Cor. 2 of Prop. 1). If one writes $\varphi = a + b \omega$ ($a, b$ in $\mathbf{A}$), then by definition one has $a = \cos (\widehat{x, y})$ and $b = \sin (\widehat{x, y})$. The relation $y' = \varphi(x') = ax' + b \omega(x')$ gives $\Phi(x', y') = a \Phi(x', x') = a$ since $x'$ and $\omega(x')$ are orthogonal (Remark 2 of No. 1); this proves (19). On the other hand this relation also gives $x' \wedge y' = bx' \wedge \omega(x') = b.e$ by the definition of the extension $\Phi_{(2)}$ of $\Phi$ to $\bigwedge^2 \mathbf{E}$ ($§ 1,$ No. 9, formula (37)) and the choice of $\omega$; this proves (20).