Remark 60.6.8. Let $A \to B$ be a ring map. Let $\Omega _{B/A} \to \Omega $ be a quotient satisfying the assumptions of Algebra, Lemma 10.132.1. Let $M$ be a $B$-module. A connection is an additive map
\[ \nabla : M \longrightarrow M \otimes _ B \Omega \]
such that $\nabla (bm) = b \nabla (m) + m \otimes \text{d}b$ for $b \in B$ and $m \in M$. In this situation we can define maps
\[ \nabla : M \otimes _ B \Omega ^ i \longrightarrow M \otimes _ B \Omega ^{i + 1} \]
by the rule $\nabla (m \otimes \omega ) = \nabla (m) \wedge \omega + m \otimes \text{d}\omega $. This works because if $b \in B$, then
\begin{align*} \nabla (bm \otimes \omega ) - \nabla (m \otimes b\omega ) & = \nabla (bm) \wedge \omega + bm \otimes \text{d}\omega - \nabla (m) \wedge b\omega - m \otimes \text{d}(b\omega ) \\ & = b\nabla (m) \wedge \omega + m \otimes \text{d}b \wedge \omega + bm \otimes \text{d}\omega \\ & \ \ \ \ \ \ - b\nabla (m) \wedge \omega - bm \otimes \text{d}(\omega ) - m \otimes \text{d}b \wedge \omega = 0 \end{align*}
As is customary we say the connection is integrable if and only if the composition
\[ M \xrightarrow {\nabla } M \otimes _ B \Omega ^1 \xrightarrow {\nabla } M \otimes _ B \Omega ^2 \]
is zero. In this case we obtain a complex
\[ M \xrightarrow {\nabla } M \otimes _ B \Omega ^1 \xrightarrow {\nabla } M \otimes _ B \Omega ^2 \xrightarrow {\nabla } M \otimes _ B \Omega ^3 \xrightarrow {\nabla } M \otimes _ B \Omega ^4 \to \ldots \]
which is called the de Rham complex of the connection.
Comments (2)
Comment #4223 by Dario Weißmann on
Comment #4401 by Johan on
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