Lemma 1

Let $f$ be an element of the dual $E^*$ of E. There exists one and only one linear mapping $i_f$ of T(E) into itself such that

for every $x \in E$ (where I denotes the identity mapping). The mapping $f \to i_f$ of $E^*$ into $\mathscr{L}(T(E))$ is linear. We have $i_f(T^n) \subset T^{n-1}$, $(i_f)^2 = 0$, and $i_f \circ i_g + i_g \circ i_f = 0$ for $f, g$ in $E^*$. The mapping $i_f$ is zero on the subalgebra of

T(E) generated by the kernel of f. The ideal I(Q) is stable under $i_f$; passing to the quotient, $i_f$ therefore defines a linear mapping (still denoted by $i_f$) of $C(Q)$ into itself.

Indeed formula (5) can be written

As (4) completely determines $i_f$ on $T^0$, and since (6) determines $i_f$ on $T^n$ if its values on $T^{n-1}$ are known, the uniqueness of $i_f$ is proved. On the other hand, for $x \in E$ and $u \in T^{n-1}$, the right-hand side of (6) is bilinear on $E \times T^{n-1}$; this proves the existence of $i_f$ by induction on $n$ (Chap. III, § 1, No. 2) and also proves, by induction on $n$, that $i_f(T^n) \subset T^{n-1}$. If $f = ag + bh$ (a, b in A, g, h in E*) it is clear that $ai_g + bi_h$ satisfies (4) and (5), hence is equal to $i_f$. We have

and, since $(i_f)^2(1) = 0$, it follows by induction on $n$ that $(i_f)^2$ is zero on $T^n$. Replacing $f$ by $f + g$, the relation $(i_f)^2 = 0$ gives $i_f \circ i_g + i_g \circ i_f = 0$. An analogous reasoning by induction to the preceding ones shows that $i_f$ is zero on the subalgebra generated by the kernel of $f$. Finally (6) entails that the set of elements $u$ of $I(Q)$ such that $i_f(u) \in I(Q)$ is a left ideal of $T(E)$; moreover, if $u = (x \otimes x - Q(x) \cdot 1) \otimes \nu$ ($x \in E,\ \nu \in T(E)$), we have

therefore $I(Q)$ is stable under $i_f$, and the last assertion follows. QED.

Let F be a bilinear form on E; in the remainder of this paragraph we shall denote by $i_x^F$ ($x \in E$) the mapping $i_f$ corresponding to the linear form $f : y \to F(x, y)$ on E.