Bourbaki

Proposition 8

Tag 0DOI, permanent. Cite it as [Bourbaki, Tag 0DOI].

Algebra, chapter IX, § 10. Angles

Suppose that A is a maximal ordered field, and that the form Φ\Phi is positive. Let D1,D2,D1′,D2′D_1, D_2, D'_1, D'_2 be four lines (resp. half-lines) with origin 0 of E. A necessary and sufficient condition that the angles (D1,D2^)(\widehat{D_1, D_2}) and (D1′,D2′^)(\widehat{D'_1, D'_2}) be equal is that the angles (D1,D1′^)(\widehat{D_1, D'_1}) and (D2,D2′^)(\widehat{D_2, D'_2}) be equal.

Let us prove the necessity of the stated condition. Let u be a direct similitude (resp. a rotation) such that u(D1)=D1′u(D_1) = D'_1 and u(D2)=D2′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(D1)=D2v(D_1) = D_2. Since the group S+S^+ (resp. O+O^+) is commutative, we have D2′=u(v(D1))=v(u(D1))=v(D1′)D'_2 = u(v(D_1)) = v(u(D_1)) = v(D'_1), and this shows that (D1,D1′^)=(D2,D2′^)(\widehat{D_1, D'_1}) = (\widehat{D_2, D'_2}). The sufficiency is deduced from the necessity by interchanging D2D_2 and D1′D'_1.

It follows from prop. 8 that to every angle (D1,D2^)(\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+/HS^+/H (resp. O+O^+), namely the class mod. H of the direct similitudes φ\varphi (resp. the rotation φ\varphi) such that u(D1)=D2u(D_1) = D_2 for any representative (D1,D2)(D_1, D_2) of the angle (D1,D2^)(\widehat{D_1, D_2}). Thus there has been defined a canonical bijection hh (resp. h′h') of the set A0\mathfrak{A}_0 of angles of lines (resp. A\mathfrak{A} of angles of half-lines) onto S+/HS^+/H (resp. O+O^+); in particular, for every φ∈A\varphi \in \mathfrak{A}, one says that h(φ)h(\varphi) is the rotation of angle φ\varphi. We shall transport to A0\mathfrak{A}_0 and A\mathfrak{A}, by means of h−1h^{-1} and of h′−1{h'}^{-1}, the structures of commutative groups of S+/HS^+/H and of O+O^+, and we shall denote additively the groups A0\mathfrak{A}_0 and A\mathfrak{A} thus obtained. If D,D′,D′′D, D', D'' denote lines (resp. half-lines) with origin 0 of E, we have by definition

(9)(D,D′′^)=(D,D′^)+(D′,D′′^);(relation de Chasles) (9) \quad (\widehat{D, D''}) = (\widehat{D, D'}) + (\widehat{D', D''}) \tag{relation de Chasles};

whence one deduces

(10)(D,D^)=0,(D,D′^)=−(D′,D^). (10) \quad (\widehat{D, D}) = 0, \qquad (\widehat{D, D'}) = - (\widehat{D', D}).

Bourbaki puts the proof directly after the statement, so what is above is the statement and its proof together.

Cited by

Nothing.

Cites

Nothing.

Translated with machine help, not yet reviewed by a person.