Bourbaki

Proposition 1

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

Algebra, chapter IX, § 9. Algèbres de Clifford

*Let f be a linear mapping of E into an algebra D over A such that f(x)2=Q(x).1f(x)^2 = Q(x).1 for all x∈Ex \in E. There exists one and only one homomorphism fˉ\bar{f} of C(Q)C(Q) into D such that f=fˉ∘ρQf = \bar{f} \circ \rho_Q.

The uniqueness of fˉ\bar{f} follows from the fact that C(Q)C(Q) is generated by ρQ(E)\rho_Q(E). Let hh be the unique homomorphism of T(E)T(E) into D which extends ff (hh is defined by h(x1⊗⋯⊗xn)=f(x1)…f(xn)h(x_1 \otimes \cdots \otimes x_n) = f(x_1) \ldots f(x_n)). We have

h(x⊗x−Q(x).1)=(f(x)2−Q(x)).1=0, h(x \otimes x - Q(x).1) = (f(x)^2 - Q(x)).1 = 0,

and consequently hh vanishes on I(Q)I(Q) and defines by passing to the quotient the homomorphism fˉ\bar{f} in question.

Prop. 1 expresses that C(Q)C(Q) is the solution of a universal mapping problem (Sets, Chap. IV, § 3, No. 1).

Take in particular for D the opposite algebra of C(Q)C(Q) and for ff the mapping ρ\rho; prop. 1 entails that there exists one and only one anti-automorphism β\beta of C(Q)C(Q) whose restriction to ρ(E)\rho(E) is the identity; it is called the principal antiautomorphism of C(Q)C(Q). It is clear that β2=1\beta^2 = 1.

D’autre part let Q′Q' be a quadratic form on an A-module E′E' and ff a linear mapping of EE into E′E' such that Q′∘f=QQ' \circ f = Q. We have ρQ(f(x))2=Q′(f(x)).1=Q(x).1\rho_Q(f(x))^2 = Q'(f(x)).1 = Q(x).1, and consequently there exists one and only one homomorphism C(f)C(f) of C(Q)C(Q) into C(Q′)C(Q') such that C(f)∘ρQ=ρQ′∘fC(f) \circ \rho_Q = \rho_{Q'} \circ f. If ff is the identity, C(f)C(f) is the identity; if Q′′Q'' is a quadratic form on an A-module E′′E'' and gg a linear mapping of E′E' into E′′E'' such that Q′′∘g=Q′Q'' \circ g = Q', we have C(g∘f)=C(g)∘C(f)C(g \circ f) = C(g) \circ C(f). When E′E' is a submodule of EE and ff the canonical injection of E′E' into EE (so that Q′Q' is the restriction of QQ to E′E'), we say that C(f)C(f) is the canonical homomorphism of C(Q′)C(Q') into C(Q)C(Q).

Let us take in particular Q′=QQ' = Q and for ff the mapping x→−xx \to -x; we see that there exists one and only one automorphism α\alpha of C(Q)C(Q) such that α∘ρ=−ρ\alpha \circ \rho = -\rho; it is called the principal automorphism of C(Q)C(Q). It is clear that α2=1\alpha^2 = 1, and that the restriction of α\alpha to C+C^+ (resp. C−C^-) is the identity (resp. the mapping u→−uu \to -u).

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

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