Lemma 2

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

Algebra, chapter IX, § 9. Algèbres de Clifford

There exists one and only one linear mapping $\lambda_F$ of $T(E)$ into itself such that

\begin{align*} \lambda_F(1) &= 1 \\ \lambda_F \circ e_x &= (e_x + i_x^F) \circ \lambda_F \qquad (x \in E). \end{align*}

For any $f \in E^*$ , we have

\lambda_F \circ i_f = i_f \circ \lambda_F.

Indeed, formula (8) is equivalent to

\lambda_F(x \otimes u) = x \otimes \lambda_F(u) + i_x^F(\lambda_F(u)) \quad (x \in E, u \in T(E)).

Comme (7) completely determines $\lambda_F$ on $T^0$ , and since (10) determines $\lambda_F$ on $T^n$ if its values on $T^{n-1}$ are known, the uniqueness of $\lambda_F$ is demonstrated. On the other hand, for $x \in E$ and $u \in T^{n-1}$ , the right-hand side of (10) is bilinear on $E \times T^{n-1}$ ; this demonstrates the existence of $\lambda_F$ by induction on $n$ . It remains to prove (9), which we shall do by induction; both members of (9) are zero on $T^0$ ; suppose (9) to be true on $\sum_{h=0}^{n-1} T^h$ ; then, for $x \in E$ and $u \in T^{n-1}$ , we have:

\begin{align*} (\lambda_F \circ i_f)(x \otimes u) &= (-\lambda_F \circ e_x \circ i_f + f(x)\lambda_F)(u) \\ &= -(e_x + i_x^F) \circ \lambda_F \circ i_f(u) + f(x)\lambda_F(u) \\ &= -(e_x + i_x^F) \circ i_f \circ \lambda_F(u) + f(x)\lambda_F(u) \\ &= (i_f \circ e_x \circ \lambda_F - f(x)\lambda_F + i_f \circ i_x^F \circ \lambda_F + f(x)\lambda_F)(u) \\ &= (i_f \circ (e_x + i_x^F) \circ \lambda_F)(u) = (i_f \circ \lambda_F)(x \otimes u), \end{align*}

whence our last assertion.

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

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