Proposition 14

Tag 0DT1, permanent. Cite it as [Bourbaki, Tag 0DT1].

Algebra, chapter X, § 2. Complexes de A-modules

(i) There exists a unique mapping $k$ -linéaire $\nabla$ from the right $\Omega_{A/k}$ -module $M \otimes_A \Omega_{A/k}$ into itself, graded of degree 1, which extends $\nabla^0$ in degree 0 and satisfies the identity :

(14)

\nabla(x \omega) = (\nabla x) \omega + (-1)^p x(d\omega) \quad \text{for} \quad x \in M \otimes_A \Omega^p_{A/k},\ \omega \in \Omega_{A/k}.

(ii) The composite mapping $\nabla \circ \nabla$ is $\Omega_{A/k}$ -linéaire ; in particular the mapping $R = \nabla^1 \circ \nabla^0$ from $M$ into $M \otimes_A \Omega^2_{A/k}$ is $A$ -linéaire, and we have

\nabla \circ \nabla(m \otimes \omega) = R(m).\omega \quad \text{for} \quad m \in M,\ \omega \in \Omega_{A/k}.

The homomorphism $R$ is sometimes called curvature homomorphism of the connection $\nabla^0$ ; if it is zero, the pair $(M \otimes_A \Omega_{A/k}, \nabla)$ is a complex, also called de Rham complex of $(M, \nabla^0)$ over $k$ .

Let us prove (i). The uniqueness of $\nabla$ is obvious. Let us define a $k$ -homomorphisme $\overline{\nabla}$ from $M \otimes_k \Omega_{A/k}$ into $M \otimes_A \Omega_{A/k}$ by

\overline{\nabla}(m \otimes_k \omega) = (\nabla^0 m) \omega + m \otimes d\omega \quad \text{for} \quad m \in M,\ \omega \in \Omega_{A/k}.

It follows from (13) that $\overline{\nabla}(am \otimes \omega) = \overline{\nabla}(m \otimes a \omega)$ , so that one obtains by passing to the quotient a $k$ -homomorphisme $\nabla$ from $M \otimes_A \Omega_{A/k}$ into itself, graded of degree 1, extending $\nabla^0$ in degree 0. Let us verify (14): we have for $m \in M$ , $\alpha \in \Omega^p_{A/k}$ , $\omega \in \Omega_{A/k}$ :

\begin{align*} \nabla((m \otimes \alpha).\omega) &= \nabla(m \otimes (\alpha \wedge \omega)) = \nabla^0(m).(\alpha \wedge \omega) + m \otimes d(\alpha \wedge \omega) \\ &= \nabla^0(m) \alpha.\omega + (m \otimes d\alpha) \omega + (-1)^p (m \otimes \alpha) d\omega \\ &= (\nabla(m \otimes \alpha)) \omega + (-1)^p (m \otimes \alpha) d\omega \end{align*}

which proves (14) for

x = m \otimes \alpha

; the general case follows by linearity.

Let us prove (ii). Let $x \in M \otimes_A \Omega_{A/k}^p, \omega \in \Omega_{A/k}$ ; by repeated application of (14), we obtain :

\nabla \circ \nabla(x \omega) = \nabla(\nabla(x) \omega) + (-1)^p \nabla(x d \omega) = (\nabla \circ \nabla(x)) \omega + (-1)^{p+1} \nabla(x)(d \omega) + (-1)^p \nabla(x)(d \omega) = (\nabla \circ \nabla(x)) \omega,

which proves the first assertion of (ii); the others follow immediately.

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

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