Section 10.132 (0FKF): The de Rham complex

10.132 The de Rham complex

Let $A \to B$ be a ring map. Denote $\text{d} : B \to \Omega _{B/A}$ the module of differentials with its universal $A$ -derivation constructed in Section 10.131. Let $\Omega _{B/A}^ i = \wedge ^ i_ B(\Omega _{B/A})$ for $i \geq 0$ be the $i$ th exterior power as in Section 10.13. The de Rham complex of $B$ over $A$ is the complex

$\Omega _{B/A}^0 \to \Omega _{B/A}^1 \to \Omega _{B/A}^2 \to \ldots$

with $A$ -linear differentials constructed and described below.

The map $\text{d} : \Omega ^0_{B/A} \to \Omega ^1_{B/A}$ is the universal derivation $\text{d} : B \to \Omega _{B/A}$ . Observe that this is indeed $A$ -linear.

For $p \geq 1$ we claim there is a unique $A$ -linear map $\text{d} : \Omega _{B/A}^ p \to \Omega _{B/A}^{p + 1}$ such that

10.132.0.1

$\begin{equation} \label{algebra-equation-rule} \text{d}\left(b_0\text{d}b_1 \wedge \ldots \wedge \text{d}b_ p\right) = \text{d}b_0 \wedge \text{d}b_1 \wedge \ldots \wedge \text{d}b_ p \end{equation}$

Recall that $\Omega _{B/A}$ is generated as a $B$ -module by the elements $\text{d}b$ . Thus $\Omega ^ p_{B/A}$ is generated as an $A$ -module by the elements $b_0\text{d}b_1 \wedge \ldots \wedge \text{d}b_ p$ and it follows that the map $\text{d} : \Omega ^ p_{B/A} \to \Omega ^{p + 1}_{B/A}$ if it exists is unique.

Construction of $\text{d} : \Omega _{B/A}^1 \to \Omega _{B/A}^2$ . By Definition 10.131.2 the elements $\text{d}b$ freely generate $\Omega _{B/A}$ as a $B$ -module subject to the relations $\text{d}a = 0$ for $a \in A$ and $\text{d}(b' + b'') = \text{d}b' + \text{d}b''$ and $\text{d}(b'b'') = b'\text{d}b'' + b''\text{d}b'$ for $b', b'' \in B$ . Hence to show that the rule

$\sum b'_ i \text{d}b_ i \longmapsto \sum \text{d}b'_ i \wedge \text{d}b_ i$

is well defined we have to show that the elements

$b\text{d}a, \quad \text{and}\quad b\text{d}(b' + b'') - b\text{d}b' - b\text{d}b'' \quad \text{and}\quad b\text{d}(b'b'') - bb'\text{d}b'' - bb''\text{d}b'$

for $a \in A$ and $b, b', b'' \in B$ are mapped to zero. This is clear by direct computation using the Leibniz rule for $\text{d}$ .

Observe that the composition $\Omega ^0_{B/A} \to \Omega ^1_{B/A} \to \Omega ^2_{B/A}$ is zero as $\text{d}(\text{d}(b)) = \text{d}(1 \text{d}b) = \text{d}(1) \wedge \text{d}(b) = 0 \wedge \text{d}b = 0$ . Here $\text{d}(1) = 0$ as $1 \in B$ is in the image of $A \to B$ . We will use this below.

Construction of $\text{d} : \Omega _{B/A}^ p \to \Omega _{B/A}^{p + 1}$ for $p \geq 2$ . We will show the $A$ -linear map

$\gamma : \Omega ^1_{B/A} \otimes _ A \ldots \otimes _ A \Omega ^1_{B/A} \longrightarrow \Omega _{B/A}^{p + 1}$

defined by the formula

$\omega _1 \otimes \ldots \otimes \omega _ p \longmapsto \sum (-1)^{i + 1} \omega _1 \wedge \ldots \wedge \text{d}(\omega _ i) \wedge \ldots \wedge \omega _ p$

factors over the natural surjection $\Omega ^1_{B/A} \otimes _ A \ldots \otimes _ A \Omega ^1_{B/A} \to \Omega ^ p_{B/A}$ to give the desired map $\text{d} : \Omega ^ p_{B/A} \to \Omega ^{p + 1}_{B/A}$ . According to Lemma 10.13.4 the kernel of $\Omega ^1_{B/A} \otimes _ A \ldots \otimes _ A \Omega ^1_{B/A} \to \Omega ^ p_{B/A}$ is generated as an $A$ -module by the elements $\omega _1 \otimes \ldots \otimes \omega _ p$ with $\omega _ i = \omega _ j$ for some $i \not= j$ and $\omega _1 \otimes \ldots \otimes f\omega _ i \otimes \ldots \otimes \omega _ p - \omega _1 \otimes \ldots \otimes f\omega _ j \otimes \ldots \otimes \omega _ p$ for some $f \in B$ . A direct computation shows the first type of element is mapped to $0$ by $\gamma$ , in other words, $\gamma$ is alternating. To finish we have to show that

$\gamma ( \omega _1 \otimes \ldots \otimes f\omega _ i \otimes \ldots \otimes \omega _ p) = \gamma ( \omega _1 \otimes \ldots \otimes f\omega _ j \otimes \ldots \otimes \omega _ p)$

for $f \in B$ . By $A$ -linearity and the alternating property, it is enough to show this for $p = 2$ , $i = 1$ , $j = 2$ , $\omega _1 = b \text{d}b'$ and $\omega _2 = c \text{d} c'$ for $b, b', c, c' \in B$ . Thus we need to show that

$\begin{align*} & \text{d}(fb) \wedge \text{d}b' \wedge c \text{d}c' - fb \text{d}b' \wedge \text{d}c \wedge \text{d}c' \\ & = \text{d}b \wedge \text{d}b' \wedge fc\text{d}c' - b \text{d}b' \wedge \text{d}(fc) \wedge \text{d}c' \end{align*}$

in other words that

$(c \text{d}(fb) + fb \text{d}c - fc \text{d}b - b \text{d}(fc)) \wedge \text{d}b' \wedge \text{d}c' = 0.$

This follows from the Leibniz rule. Observe that the value of $\gamma$ on the element $b_0\text{d}b_1 \otimes \text{d}b_2 \otimes \ldots \otimes \text{d}b_ p$ is $\text{d}b_0 \wedge \text{d}b_1 \wedge \ldots \wedge \text{d}b_ p$ and hence (10.132.0.1) will be satisfied for the map $\text{d} : \Omega ^ p_{B/A} \to \Omega ^{p + 1}_{B/A}$ so obtained.

Finally, since $\Omega ^ p_{B/A}$ is additively generated by the elements $b_0\text{d}b_1 \wedge \ldots \wedge \text{d}b_ p$ and since $\text{d}(b_0\text{d}b_1 \wedge \ldots \wedge \text{d}b_ p) = \text{d}b_0 \wedge \ldots \wedge \text{d}b_ p$ we see in exactly the same manner that the composition $\Omega ^ p_{B/A} \to \Omega ^{p + 1}_{B/A} \to \Omega ^{p + 2}_{B/A}$ is zero for $p \geq 1$ . Thus the de Rham complex is indeed a complex.

Given just a ring $R$ we set $\Omega _ R = \Omega _{R/\mathbf{Z}}$ . This is sometimes called the absolute module of differentials of $R$ ; this makes sense: if $\Omega _ R$ is the module of differentials where we only assume the Leibniz rule and not the vanishing of $\text{d}1$ , then the Leibniz rule gives $\text{d}1 = \text{d}(1 \cdot 1) = 1 \text{d}1 + 1 \text{d}1 = 2 \text{d}1$ and hence $\text{d}1 = 0$ in $\Omega _ R$ . In this case the absolute de Rham complex of $R$ is the corresponding complex

$\Omega _ R^0 \to \Omega _ R^1 \to \Omega _ R^2 \to \ldots$

where we set $\Omega ^ i_ R = \Omega ^ i_{R/\mathbf{Z}}$ and so on.

Suppose we have a commutative diagram of rings

$\xymatrix{ B \ar[r] & B' \\ A \ar[r] \ar[u] & A' \ar[u] }$

There is a natural map of de Rham complexes

$\Omega ^\bullet _{B/A} \longrightarrow \Omega ^\bullet _{B'/A'}$

Namely, in degree $0$ this is the map $B \to B'$ , in degree $1$ this is the map $\Omega _{B/A} \to \Omega _{B'/A'}$ constructed in Section 10.131, and for $p \geq 2$ it is the induced map $\Omega ^ p_{B/A} = \wedge ^ p_ B(\Omega _{B/A}) \to \wedge ^ p_{B'}(\Omega _{B'/A'}) = \Omega ^ p_{B'/A'}$ . The compatibility with differentials follows from the characterization of the differentials by the formula (10.132.0.1).

Lemma 10.132.1. Let $A \to B$ be a ring map. Let $\pi : \Omega _{B/A} \to \Omega$ be a surjective $B$ -module map. Denote $\text{d} : B \to \Omega$ the composition of $\pi$ with the universal derivation $\text{d}_{B/A} : B \to \Omega _{B/A}$ . Set $\Omega ^ i = \wedge _ B^ i(\Omega )$ . Assume that the kernel of $\pi$ is generated, as a $B$ -module, by elements $\omega \in \Omega _{B/A}$ such that $\text{d}_{B/A}(\omega ) \in \Omega _{B/A}^2$ maps to zero in $\Omega ^2$ . Then there is a de Rham complex

$\Omega ^0 \to \Omega ^1 \to \Omega ^2 \to \ldots$

whose differential is defined by the rule

$\text{d} : \Omega ^ p \to \Omega ^{p + 1},\quad \text{d}\left(f_0\text{d}f_1 \wedge \ldots \wedge \text{d}f_ p\right) = \text{d}f_0 \wedge \text{d}f_1 \wedge \ldots \wedge \text{d}f_ p$

Proof. We will show that there exists a commutative diagram

$\xymatrix{ \Omega _{B/A}^0 \ar[d] \ar[r]_{\text{d}_{B/A}} & \Omega _{B/A}^1 \ar[d]_\pi \ar[r]_{\text{d}_{B/A}} & \Omega _{B/A}^2 \ar[d]_{\wedge ^2\pi } \ar[r]_{\text{d}_{B/A}} & \ldots \\ \Omega ^0 \ar[r]^{\text{d}} & \Omega ^1 \ar[r]^{\text{d}} & \Omega ^2 \ar[r]^{\text{d}} & \ldots }$

the description of the map $\text{d}$ will follow from the construction of the differentials $\text{d}_{B/A} : \Omega ^ p_{B/A} \to \Omega ^{p + 1}_{B/A}$ of the de Rham complex of $B$ over $A$ given above. Since the left most vertical arrow is an isomorphism we have the first square. Because $\pi$ is surjective, to get the second square it suffices to show that $\text{d}_{B/A}$ maps the kernel of $\pi$ into the kernel of $\wedge ^2\pi$ . We are given that any element of the kernel of $\pi$ is of the form $\sum b_ i\omega _ i$ with $\pi (\omega _ i) = 0$ and $\wedge ^2\pi (\text{d}_{B/A}(\omega _ i)) = 0$ . By the Leibniz rule for $\text{d}_{B/A}$ we have $\text{d}_{B/A}(\sum b_ i\omega _ i) = \sum b_ i \text{d}_{B/A}(\omega _ i) + \sum \text{d}_{B/A}(b_ i) \wedge \omega _ i$ . Hence this maps to zero under $\wedge ^2\pi$ .

For $i > 1$ we note that $\wedge ^ i \pi$ is surjective with kernel the image of $\mathop{\mathrm{Ker}}(\pi ) \wedge \Omega ^{i - 1}_{B/A} \to \Omega _{B/A}^ i$ . For $\omega _1 \in \mathop{\mathrm{Ker}}(\pi )$ and $\omega _2 \in \Omega ^{i - 1}_{B/A}$ we have

$\text{d}_{B/A}(\omega _1 \wedge \omega _2) = \text{d}_{B/A}(\omega _1) \wedge \omega _2 - \omega _1 \wedge \text{d}_{B/A}(\omega _2)$

which is in the kernel of $\wedge ^{i + 1}\pi$ by what we just proved above. Hence we get the $(i + 1)$ st square in the diagram above. This concludes the proof. $\square$

Comments (7)

Comment #4933 by Théo de Oliveira Santos on February 19, 2020 at 02:23

Very minor typo: "The map $\mathrm{d}\colon\Omega^0_{B/A}\rightarrow\Omega^1_{B/A}$ is the univeral derivation $\mathrm{d}\colon B\rightarrow\Omega_{B/A}$ ".

Comment #5200 by Johan on June 10, 2020 at 17:03

Thanks and fixed here.

Comment #7458 by ? on June 27, 2022 at 13:48

?The most difficult part is omitted?

Comment #7459 by Johan on June 28, 2022 at 13:43

The omitted part is something about exterior powers and has nothing to do with differentials. It just says that given a ring $R$ , an $R$ -module $M$ , and an integer $p \geq 1$ the map $M \times \ldots \times M \to \wedge^p_R(M)$ , $(m_1, \ldots, m_p) \mapsto m_1 \wedge \ldots \wedge m_p$ is the universal map which is $R$ -linear in each entry and alternating. This belongs in Section 10.13.

Comment #7610 by Stacks Project on August 14, 2022 at 08:48

Tried to improve this a little bit. See here.

Comment #8617 by Brian Nugent on August 16, 2023 at 16:47

Minor typo, in the paragraph below equation 10.132.0.1, I think you mean "generated by elements of the form" and not "generated by the element"

Comment #9421 by Stacks project on June 05, 2024 at 16:59

Fixed here.

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The Stacks project

10.132 The de Rham complex

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