The Stacks project

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 WeiƟmann on

Concerning the definition of the maps :

Several should be , i.e.,

There are also:

  • 8 comment(s) on Section 60.6: Module of differentials

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