De Rham complexes and Lie algebroids

Written by Claude Opus 5.5 (Anthropic) on 22 October 2026. Self-checked by the writing AI. Public domain (CC0).

Let $K\to A$ be a map of sheaves of commutative rings on a topos. A de Rham complex over $A/K$ is an $A$-module $E$ together with a $K$-linear antiderivation $d$ of degree $1$ of the exterior algebra $\Lambda_AE$ with $d^2=0$; the de Rham complex $\Omega^\bullet_{A/K}$ is the initial one. The dual module $L=\mathcal Hom_A(E,A)$ then carries a bracket and a map $\theta$ to the derivations of $A$, defined by Cartan's formulas $i_xd+di_x=\theta_x$ and $\theta_xi_y-i_y\theta_x=i_{[x,y]}$: a Lie algebroid over $A/K$. This lesson constructs both structures and proves their properties. A pair of first-order operators extends uniquely to an antiderivation; a quotient of $\Omega^1_{A/K}$ carries a de Rham complex exactly when an explicit condition holds, and some quotients carry none. The differential is given by the Chevalley–Eilenberg formula. When $E$ is locally free of finite type, the bracket satisfies the Jacobi identity exactly when $d^2=0$, and de Rham complexes and Lie algebroids on locally free modules of finite type are anti-equivalent. For a torsor under a flat group scheme, the invariant differential forms form a de Rham complex; its Lie algebroid fits into the Atiyah sequence, whose kernel is the Lie algebra of the group twisted by the torsor.

Prerequisites. Torsors, co-Lie complexes and the Atiyah class (Conventions, Definition 1.1, Lemma 1.2, Proposition 1.3, Corollary 2.6, Lemma 3.1, Theorem 3.3, Proposition 4.1); The adjoint co-Lie complex of a group scheme (the maps $c(\rho)$ and formula (2.1) before Lemma 2.3, and Proposition 2.4); The cotangent complex (Theorem 6.1: $H^0(L_{B/A})=\Omega_{B/A}$). From the AI-integrated Stacks project: exterior algebras of sheaves of modules [Stacks, Tag 01CF, Tag 01CI], differentials of sheaves of rings [Tag 04BL, Tag 08TP], the de Rham complex of a ring map [Tag 0FKF] and of a ringed space [Tag 0FKM], and differentials of morphisms of schemes [Tag 00RS, Tag 01UT, Tag 01UX, Tag 01V0, Tag 02G1, Tag 02K4].

Conventions

$T$ is a topos, $K$ a sheaf of commutative rings on $T$, and $A$ a commutative $K$-algebra, that is, a sheaf of commutative rings with a ring map $K\to A$. A local section of a sheaf is a section over an object $U$ of $T$; equalities of maps of sheaves are checked on local sections. Modules are sheaves of modules. Tensor products, exterior powers and the module of differentials $\Omega^1_{A/K}$ [Tag 08TP] are the sheaves associated with the corresponding presheaf constructions; a map out of one of them is defined by maps out of the presheaf construction that are compatible with restrictions, followed by sheafification.

The exterior algebra $\Lambda_AE=\bigoplus_{n\geq0}\Lambda^n_AE$ of an $A$-module $E$ is the quotient of the tensor algebra of $E$ over $A$ by the two-sided ideal generated by the local sections $x\otimes x$ of $E\otimes_AE$, as for ringed spaces [Tag 01CF]. Its degree $n$ part is the quotient of $E^{\otimes n}$ by the submodule generated by the tensors $x_1\otimes\cdots\otimes x_n$ with $x_i=x_{i+1}$ for some $i$. It is strictly graded commutative: $\omega\eta=(-1)^{pq}\eta\omega$ for $\omega$ of degree $p$ and $\eta$ of degree $q$, and $\omega^2=0$ for $\omega$ of odd degree. Moreover $\Lambda^0_AE=A$, $\Lambda^1_AE=E$, and every local section of $\Lambda^n_AE$ is locally a finite sum of products $ax_1\wedge\cdots\wedge x_n$. An $A$-linear map $u\colon E\to F$ induces the map $\Lambda u\colon\Lambda_AE\to\Lambda_AF$ of graded $A$-algebras. For an $A$-module $E$, $E^\vee=\mathcal Hom_A(E,A)$, and $x(e)$ is the value of $x\in E^\vee$ at $e\in E$.

1. Graded antiderivations

Definition 1.1. A graded $K$-algebra is a sheaf $B=\bigoplus_{n\in\mathbf Z}B^n$ of graded $K$-modules with a $K$-bilinear multiplication $B^m\times B^n\to B^{m+n}$, not necessarily associative or commutative. For an integer $k$, a $K$-antiderivation of degree $k$ of $B$ is a $K$-linear map $f\colon B\to B$ with $f(B^n)\subseteq B^{n+k}$ such that $f(xy)=f(x)y+(-1)^{kn}xf(y)$ for all $n$, local sections $x$ of $B^n$ and $y$ of $B$. For $k$ even, $f$ is also called a derivation. More generally, if $\varphi\colon B\to B'$ is a map of graded $K$-algebras, a $\varphi$-antiderivation of degree $k$ is a $K$-linear map $\delta\colon B\to B'$ of degree $k$ with $\delta(xy)=\delta(x)\varphi(y)+(-1)^{kn}\varphi(x)\delta(y)$. For maps $f$, $g$ of degrees $p$, $q$ put $[f,g]=fg-(-1)^{pq}gf. \tag{1.1}$

Lemma 1.2. Let $f$, $g$, $h$ be $K$-antiderivations of $B$ of degrees $p$, $q$, $r$.

$[f,g]$ is a $K$-antiderivation of degree $p+q$, and $[g,f]=-(-1)^{pq}[f,g]$.
(Jacobi identity) $[f,[g,h]]=$ $[[f,g],h]+(-1)^{pq}[g,[f,h]]$.
If $p$ is odd, then $f^2$ is a derivation of degree $2p$, $[f,f]=2f^2$, and $[f,[f,g]]=[f^2,g]$.
Let $\varphi\colon B\to B'$ be a map of graded $K$-algebras and $C\subseteq B$ a graded subsheaf such that the only graded subsheaf of $B$ that contains $C$ and is closed under products is $B$. Two $\varphi$-antiderivations of the same degree that agree on $C$ are equal. This applies to $B=\Lambda_AE$ and $C=A\oplus E$.

Proof. (1) For $x$ of degree $n$, $\begin{aligned}fg(xy)&=fg(x)y+(-1)^{p(q+n)}g(x)f(y)\\&\quad+(-1)^{qn}f(x)g(y)\\&\quad+(-1)^{(p+q)n}x\,fg(y),\end{aligned}$ and $gf(xy)$ is given by the same formula with $(f,p)$ and $(g,q)$ exchanged. In $fg(xy)-(-1)^{pq}gf(xy)$ the terms in $f(x)g(y)$ have the coefficients $(-1)^{qn}$ and $(-1)^{pq+q(p+n)}=(-1)^{qn}$, and those in $g(x)f(y)$ the coefficients $(-1)^{pq+pn}$ and $(-1)^{pq+pn}$, so they cancel, leaving $[f,g](x)y+(-1)^{(p+q)n}x[f,g](y)$. The second assertion is (1.1).

(2) Expanding, $\begin{aligned}[f,[g,h]]&=fgh-(-1)^{qr}fhg\\&\quad-(-1)^{pq+pr}ghf\\&\quad+(-1)^{pq+pr+qr}hgf,\end{aligned}$ while $[[f,g],h]+(-1)^{pq}[g,[f,h]]$ is the sum of $\begin{aligned}&fgh-(-1)^{pq}gfh\\&\quad-(-1)^{pr+qr}hfg\\&\quad+(-1)^{pq+pr+qr}hgf\end{aligned}$ and $\begin{aligned}&(-1)^{pq}gfh-(-1)^{pq+pr}ghf\\&\quad-(-1)^{qr}fhg+(-1)^{pr+qr}hfg,\end{aligned}$ which is the same.

(3) For $x$ of degree $n$, $\begin{aligned}f^2(xy)&=f^2(x)y+xf^2(y)\\&\quad+\bigl((-1)^{p(p+n)}+(-1)^{pn}\bigr)\\&\qquad\cdot f(x)f(y),\end{aligned}$ and $(-1)^{p(p+n)}=-(-1)^{pn}$ since $p$ is odd. Next $[f,f]=f^2-(-1)^{p^2}f^2=2f^2$. Finally $(-1)^{p(p+q)}=-(-1)^{pq}$, so $\begin{aligned}[f,[f,g]]&=f(fg-(-1)^{pq}gf)\\&\quad+(-1)^{pq}(fg-(-1)^{pq}gf)f\\&=f^2g-gf^2=[f^2,g],\end{aligned}$ $f^2$ having even degree.

(4) The difference $\delta$ of the two maps is a $\varphi$-antiderivation vanishing on $C$. Its kernel is a graded subsheaf, $\delta$ being homogeneous, and it is closed under products, since $\delta(xy)=\delta(x)\varphi(y)\pm\varphi(x)\delta(y)$. So it is $B$. In $\Lambda_AE$, a graded subsheaf closed under products and containing $A$ and $E$ contains all products $ax_1\wedge\cdots\wedge x_n$, hence all local sections. $\square$

2. De Rham complexes

Definition 2.1. A de Rham complex over $A/K$ is an $A$-module $E$ together with a $K$-antiderivation $d$ of degree $1$ of $\Lambda_AE$ such that $d^2=0$; it is written $(\Lambda E,d)$. A morphism $(\Lambda E,d)\to(\Lambda F,d)$ is an $A$-linear map $u\colon E\to F$ with $\Lambda u\circ d=d\circ\Lambda u$. When $A=K$ one speaks of a de Rham complex over $A$.

For every $K$-antiderivation $d$ of degree $1$ of $\Lambda_AE$, $d(1)=d(1\cdot1)=2d(1)$, so $d(1)=0$, and $d$ vanishes on the image of $K$. Its restriction $d|_A\colon A\to E$ is a $K$-derivation: $d(ab)=(da)b+a\,db=a\,db+b\,da$. If $A=K$, then $d|_A=0$.

Proposition 2.2 (extending first-order operators). Let $E$ be an $A$-module and $d_0\colon A\to E$, $d_1\colon E\to\Lambda^2_AE$ $K$-linear maps such that $\begin{aligned}d_0(ab)&=a\,d_0b+b\,d_0a,\\ d_1(ax)&=a\,d_1x+d_0a\wedge x\end{aligned} \tag{2.1}$ for local sections $a,b$ of $A$ and $x$ of $E$.

There is a unique $K$-antiderivation $d$ of degree $1$ of $\Lambda_AE$ with $d|_A=d_0$ and $d|_E=d_1$. For $n\geq1$ it is given by (2.2) below.
$d^2=0$ if and only if $d_1\circ d_0=0$ and $d\circ d_1=0$, where $d(x\wedge y)=d_1x\wedge y-x\wedge d_1y$ on $\Lambda^2_AE$.

\begin{aligned}&d(ax_1\wedge\cdots\wedge x_n)\\&=d_0a\wedge x_1\wedge\cdots\wedge x_n\\&\quad+a\sum_i(-1)^{i-1}x_1\wedge\cdots\\&\qquad\wedge d_1x_i\wedge\cdots\wedge x_n.\end{aligned} \tag{2.2}

Proof. (1) Uniqueness is Lemma 1.2 (4). For existence, note that $d_0(1)=0$ by (2.1), so $d_0$ vanishes on the image of $K$. For $n\geq1$ and local sections $x_1,\ldots,x_n$ of $E$ over $U$ put $\begin{aligned}&\delta(x_1,\ldots,x_n)\\&\quad=\sum_{i=1}^n(-1)^{i-1}x_1\wedge\cdots\wedge x_{i-1}\\&\qquad\wedge d_1x_i\wedge x_{i+1}\wedge\cdots\wedge x_n,\end{aligned}$ a section of $\Lambda^{n+1}_AE$ over $U$. It is additive in each variable, and (2.1) gives, for a local section $a$ of $A$, $\begin{aligned}&\delta(x_1,\ldots,ax_i,\ldots,x_n)\\&\quad=a\,\delta(x_1,\ldots,x_n)\\&\qquad+d_0a\wedge x_1\wedge\cdots\wedge x_n,\end{aligned} \tag{2.3}$ because the additional term $(-1)^{i-1}x_1\wedge\cdots\wedge x_{i-1}\wedge$ $d_0a\wedge x_i\wedge\cdots\wedge x_n$ equals $d_0a\wedge x_1\wedge\cdots\wedge x_n$, $d_0a$ passing $i-1$ elements of degree $1$. The right side of (2.3) does not depend on $i$, so $\delta$ is balanced over $A$ and defines an additive map $\delta'$ on the presheaf tensor power $E^{\otimes n}$, with $\delta'(at)=a\,\delta'(t)+d_0a\wedge\bar t$, where $\bar t$ is the image of $t$ in $\Lambda^n_AE$. If $x_i=x_{i+1}=x$, the terms $i$ and $i+1$ of $\delta$ cancel, because $x\wedge d_1x=d_1x\wedge x$, $d_1x$ having degree $2$; the other terms contain $x\wedge x=0$. So $\delta'$ vanishes on the tensors with two equal neighbouring factors, and on their multiples $at$, since $\bar t=0$ for them. These generate the kernel of $E^{\otimes n}\to\Lambda^n_AE$, so $\delta'$ induces, after sheafification, a $K$-linear map $D_n\colon\Lambda^n_AE\to\Lambda^{n+1}_AE$ with $\begin{aligned}D_n(a\omega)&=a\,D_n\omega\\&\quad+d_0a\wedge\omega.\end{aligned} \tag{2.4}$ $K$-linearity holds because $d_0$ vanishes on $K$. Let $d=d_0$ on $\Lambda^0=A$ and $d=D_n$ on $\Lambda^n$, $n\geq1$. Then (2.2) holds and $d|_E=d_1$. To see that $d$ is an antiderivation it suffices, by biadditivity, to take $\omega=ax$ and $\eta=by$ with $x=x_1\wedge\cdots\wedge x_p$, $y=y_1\wedge\cdots\wedge y_q$ (read as $1$ for $p=0$ or $q=0$). Write $Dx=\delta(x_1,\ldots,x_p)$, so that $D(x\wedge y)=Dx\wedge y+(-1)^px\wedge Dy$, the summands at the positions of $y$ carrying the extra sign $(-1)^p$. Then $\begin{aligned}d(\omega\eta)&=d_0(ab)\wedge x\wedge y\\&\quad+ab\,Dx\wedge y\\&\quad+(-1)^pab\,x\wedge Dy,\\ d\omega\wedge\eta&=b\,d_0a\wedge x\wedge y\\&\quad+ab\,Dx\wedge y,\\ (-1)^p\omega\wedge d\eta&=a\,d_0b\wedge x\wedge y\\&\quad+(-1)^pab\,x\wedge Dy,\end{aligned}$ and $d_0(ab)=a\,d_0b+b\,d_0a$.

(2) By Lemma 1.2 (3), $d^2$ is a derivation of degree $2$; by Lemma 1.2 (4) it vanishes if and only if it vanishes on $A$ and on $E$. $\square$

Proposition 2.3 (the de Rham complex of $A/K$). There is a unique $K$-linear map $d_1\colon\Omega^1_{A/K}\to\Omega^2_{A/K}=\Lambda^2_A\Omega^1_{A/K}$ with $d_1(a\,db)=da\wedge db$, where $d=d_{A/K}$. The antiderivation that extends $(d_{A/K},d_1)$ by Proposition 2.2 is a de Rham complex $\Omega^\bullet_{A/K}$ over $A/K$, and $\begin{aligned}&d(a_0\,da_1\wedge\cdots\wedge da_n)\\&\quad=da_0\wedge da_1\wedge\cdots\wedge da_n.\end{aligned} \tag{2.5}$ For every de Rham complex $(\Lambda E,d)$ over $A/K$ there is a unique morphism $\Omega^\bullet_{A/K}\to(\Lambda E,d)$. Its component of degree $1$ is the $A$-linear map $u\colon\Omega^1_{A/K}\to E$ with $u\circ d_{A/K}=d|_A$. For a morphism of schemes $f\colon X\to S$, $\Omega^\bullet_{X/S}=\Omega^\bullet_{\mathcal O_X/f^{-1}\mathcal O_S}$ on $X_{\mathrm{Zar}}$ is the de Rham complex of $X$ over $S$ of [Tag 0FKM].

Proof. $\Omega^1_{A/K}$ is generated as a sheaf of abelian groups by the local sections $a\,db$, which gives uniqueness. For a ring map $R\to P$, the de Rham complex of $P$ over $R$ [Tag 0FKF] has a map $\Omega^1_{P/R}\to\Omega^2_{P/R}$, $a\,db\mapsto da\wedge db$, compatible with ring maps. Applied to $K(U)\to A(U)$ and composed with the maps to $\Lambda^2_A\Omega^1_{A/K}$, these maps form a map from the presheaf $U\mapsto\Omega^1_{A(U)/K(U)}$ to the sheaf $\Omega^2_{A/K}$, hence $d_1$ on its sheafification $\Omega^1_{A/K}$ [Tag 08TP]. The rule (2.1) holds on the generators, $d_1(ab\,dc)=$ $(a\,db+b\,da)\wedge dc=$ $a\,d_1(b\,dc)+da\wedge b\,dc$, hence everywhere by additivity. Since $d_1(da)=d_1(1\cdot da)=d(1)\wedge da=0$, formula (2.2) gives (2.5) by induction on $n$. Then $d_1\circ d=0$, and $d(d_1(b\,dc))=$ $d(db\wedge dc)=$ $d_1(db)\wedge dc-db\wedge d_1(dc)=0$, so $d^2=0$ by Proposition 2.2 (2).

Let $(\Lambda E,d)$ be a de Rham complex. $d|_A$ is a $K$-derivation, so $u$ exists and is unique [Tag 04BL]. The maps $d\circ\Lambda u$ and $\Lambda u\circ d$ from $\Omega^\bullet_{A/K}$ to $\Lambda E$ are $\Lambda u$-antiderivations of degree $1$. They agree on $A$, where both are $d|_A$. On the generators $b\,d_{A/K}c$ of $\Omega^1_{A/K}$, with $u(b\,d_{A/K}c)=b\,dc$, the first gives $d(b\,dc)=db\wedge dc+b\,d^2c=db\wedge dc$, and the second gives $\Lambda u(d_{A/K}b\wedge d_{A/K}c)=db\wedge dc$. By Lemma 1.2 (4) they are equal, so $u$ is a morphism. A morphism $v$ satisfies $v(d_{A/K}a)=d(a)$, so $v=u$. For schemes, (2.5) is the rule that defines the differentials of [Tag 0FKM]. $\square$

Proposition 2.4 (quotients of $\Omega^1_{A/K}$). Let $\pi\colon\Omega^1_{A/K}\to E$ be a surjective map of $A$-modules with kernel $N$, and $d_0=\pi\circ d_{A/K}$.

There is at most one de Rham complex $(\Lambda E,d)$ over $A/K$ with $d|_A=d_0$.
It exists if and only if $\Lambda^2\pi(d\omega)=0$ for every local section $\omega$ of $N$, where $d$ is the differential of $\Omega^\bullet_{A/K}$. If $N$ is generated as an $A$-module by local sections $\omega_j$, it suffices that $\Lambda^2\pi(d\omega_j)=0$ for all $j$.
If it exists, $\Lambda\pi\colon\Omega^\bullet_{A/K}\to(\Lambda E,d)$ is the morphism of Proposition 2.3. It is surjective, and its kernel is the ideal generated by $N$.

A $K$-derivation $d_0\colon A\to E$ arises in this way exactly when $E$ is generated as an $A$-module by the local sections $d_0a$: the map $\pi$ with $\pi\circ d_{A/K}=d_0$ is then surjective.

Proof. (1) Two such complexes $d$, $d'$ agree on $A$. For local sections $a,b$ of $A$, $d(a\,d_0b)=d_0a\wedge d_0b+a\,d^2b=$ $d_0a\wedge d_0b$, and the same holds for $d'$. The elements $a\,d_0b=\pi(a\,db)$ generate $E$ as a sheaf of abelian groups, so $d=d'$ on $E$, and $d=d'$ by Lemma 1.2 (4).

(2), (3) For every surjective map of $A$-modules $M\to M'$ with kernel $N'$, the map $\Lambda_AM\to\Lambda_AM'$ is surjective, and its kernel is the ideal $J'$ generated by $N'$. Indeed it kills $J'$, and the induced map $\Lambda_AM/J'\to\Lambda_AM'$ has an inverse: the map of $A$-algebras $\Lambda_AM'\to\Lambda_AM/J'$ that extends $M'\cong M/N'$, which exists because $\Lambda_AM/J'$ is strictly graded commutative. Let $J\subseteq\Omega^\bullet_{A/K}$ be the ideal generated by $N$; so $\Lambda\pi$ is surjective with kernel $J$. If $(\Lambda E,d)$ exists, $\Lambda\pi$ is the morphism of Proposition 2.3, its degree $1$ component $\pi$ satisfying $\pi\circ d_{A/K}=d_0$; so $\Lambda^2\pi(d\omega)=d(\pi\omega)=0$ for $\omega$ in $N$. Conversely, assume the condition. $J$ is generated as a sheaf of abelian groups by the products $\omega\wedge\eta$ with $\omega$ in $N$, and $d(\omega\wedge\eta)=d\omega\wedge\eta-\omega\wedge d\eta$, where $d\omega$ lies in $\ker\Lambda^2\pi=J\cap\Omega^2_{A/K}$. So $d(J)\subseteq J$, and $d$ induces a $K$-antiderivation of degree $1$ with square zero on $\Omega^\bullet_{A/K}/J\cong\Lambda_AE$, extending $\pi\circ d_{A/K}=d_0$. If the $\omega_j$ generate $N$, a local section of $N$ is locally a sum of terms $a\omega_j$, and $d(a\omega_j)=da\wedge\omega_j+a\,d\omega_j$ with $\Lambda^2\pi(da\wedge\omega_j)=\pi(da)\wedge\pi(\omega_j)=0$. The last assertion is the universal property of $\Omega^1_{A/K}$. $\square$

Example 2.5. The condition in Proposition 2.4 (2) can fail. Let $T$ be the punctual topos, $K$ a nonzero ring, $A=K[x,y]$, so that $\Omega^1_{A/K}$ is free on $dx$ and $dy$, and let $E=\Omega^1_{A/K}/A\,y\,dx$, with the projection $\pi$ and $d_0=\pi\circ d_{A/K}$. Write $\bar e_x=d_0x$, $\bar e_y=d_0y$; they generate $E$, and $y\bar e_x=0$. By the description of exterior algebras of quotients in the proof above, $\Lambda^2_AE$ is $A\,dx\wedge dy$ modulo $A\,y\,dx\wedge dy$, that is, $A/yA$ with the basis $\bar e_x\wedge\bar e_y$. Suppose that some $K$-antiderivation $d$ of degree $1$ of $\Lambda_AE$ restricts to $d_0$ on $A$. Then $\begin{aligned}0&=d(y\bar e_x)=d_0y\wedge\bar e_x+y\,d\bar e_x\\&=-\bar e_x\wedge\bar e_y+y\,d\bar e_x.\end{aligned}$ As $y$ annihilates $\Lambda^2_AE\cong A/yA$, this says $\bar e_x\wedge\bar e_y=0$, which is false. So no antiderivation of degree $1$ extends $d_0$, let alone a de Rham complex, although $E$ is generated by $d_0A$; in the terms of Proposition 2.4, $\Lambda^2\pi(d(y\,dx))=-\bar e_x\wedge\bar e_y\neq0$. Exercise 7.1 treats a quotient where the condition holds.

3. Lie algebroids

The tangent sheaf is $T_{A/K}=\mathcal Hom_A(\Omega^1_{A/K},A)$, identified with the sheaf of $K$-derivations of $A$ by $x\mapsto x\circ d_{A/K}$ [Tag 04BL]. The commutator $[x,y]=xy-yx$ of two $K$-derivations is a $K$-derivation; it makes $T_{A/K}$ a sheaf of $K$-Lie algebras, and $[x,ay]=x(a)y+a[x,y]$.

Definition 3.1. A Lie algebroid over $A/K$ (also Lie algebra over $A/K$, or Lie–Rinehart algebra) is an $A$-module $L$ with a $K$-bilinear bracket $L\times L\to L$ that makes the underlying sheaf of $K$-modules a sheaf of $K$-Lie algebras, that is, $[x,x]=0$ and $[x,[y,z]]=[[x,y],z]+[y,[x,z]],$ together with an $A$-linear map $\theta\colon L\to T_{A/K}$, $x\mapsto\theta_x$, the anchor, that is a homomorphism of $K$-Lie algebras, such that $[x,ay]=\theta_x(a)y+a[x,y] \tag{3.1}$ for local sections $x,y$ of $L$ and $a$ of $A$. A morphism $(L,\theta)\to(L',\theta')$ is an $A$-linear map $v\colon L\to L'$ compatible with the brackets such that $\theta'\circ v=\theta$.

$T_{A/K}$ with the commutator and $\theta=\mathrm{id}$ is a Lie algebroid. If $A=K$, then $T_{A/K}=0$, and a Lie algebroid over $A/K$ is a sheaf of $K$-Lie algebras.

Lemma 3.2. Let $L$ be an $A$-module with a $K$-bilinear alternating bracket and an $A$-linear map $\theta\colon L\to T_{A/K}$ satisfying (3.1). Put $J(x,y,z)=$ $[x,[y,z]]-[[x,y],z]-[y,[x,z]]$. Then $\begin{aligned}&J(x,y,az)=aJ(x,y,z)\\&\quad+\bigl([\theta_x,\theta_y]-\theta_{[x,y]}\bigr)(a)\,z.\end{aligned}$ If the Jacobi identity holds and $L$ is locally free of finite type, then $\theta$ is a homomorphism of Lie algebras.

Proof. By (3.1), $\begin{aligned}[x,[y,az]]&=\theta_x\theta_y(a)z+\theta_y(a)[x,z]\\&\quad+\theta_x(a)[y,z]+a[x,[y,z]],\\ [[x,y],az]&=\theta_{[x,y]}(a)z+a[[x,y],z],\\ [y,[x,az]]&=\theta_y\theta_x(a)z+\theta_x(a)[y,z]\\&\quad+\theta_y(a)[x,z]+a[y,[x,z]],\end{aligned}$ and the alternating sum is the formula. If $J=0$, then $([\theta_x,\theta_y]-\theta_{[x,y]})(a)z=0$ for all local sections $z$. Locally $L$ is free of some rank $r$; for $r=0$ there is nothing to prove, and for $r\geq1$ take $z$ a member of a basis. $\square$

4. The Lie algebroid of a de Rham complex

In this section $E$ is an $A$-module, $L=E^\vee$, and $d$ is a $K$-antiderivation of degree $1$ of $\Lambda_AE$; the condition $d^2=0$ is assumed only where stated.

Construction 4.1 (interior products). For a local section $x$ of $L$ over $U$, let $i_x$ be the unique $K$-antiderivation of degree $-1$ of $\Lambda_AE|_U$ with $i_x(e)=x(e)$ for local sections $e$ of $E$. It is given by $\begin{aligned}&i_x(e_1\wedge\cdots\wedge e_n)\\&=\sum_j(-1)^{j-1}x(e_j)\\&\quad\cdot e_1\wedge\cdots\wedge\widehat{e_j}\\&\qquad\wedge\cdots\wedge e_n.\end{aligned} \tag{4.1}$ Indeed, the right side is $A$-multilinear and alternating in $(e_1,\ldots,e_n)$: if $e_k=e_{k+1}$, the terms $j=k$ and $j=k+1$ cancel and the others contain $e_k\wedge e_{k+1}=0$. So it defines an $A$-linear map $\Lambda^n_AE\to\Lambda^{n-1}_AE$, and the antiderivation rule on products of decomposable elements follows from (4.1), the terms coming from the second factor carrying the sign $(-1)^p$ when the first has degree $p$. Uniqueness is Lemma 1.2 (4); $i_x$ vanishes on $A$ for degree reasons. For local sections $\omega$ of $\Lambda^n_AE$ and $x_1,\ldots,x_n$ of $L$ put $\omega(x_1,\ldots,x_n)=i_{x_n}\cdots i_{x_1}\omega,$ a local section of $A$.

Lemma 4.2. Let $x$, $y$ be local sections of $L$ and $a,b$ of $A$.

$i_x$ is $A$-linear, $i_{ax+by}=a\,i_x+b\,i_y$, $i_x^2=0$ and $i_xi_y+i_yi_x=0$.
Every $K$-antiderivation of degree $-1$ of $\Lambda_AE|_U$ is $i_x$ for a unique local section $x$ of $L$ over $U$.
$(e_1\wedge\cdots\wedge e_n)(x_1,\ldots,x_n)=$ $\det\bigl(x_j(e_i)\bigr)_{i,j}$. In particular $\omega(x_1,\ldots,x_n)$ is $A$-multilinear and alternating in the $x_j$.

Proof. (1) $i_x(a\omega)=i_x(a)\omega+a\,i_x\omega=a\,i_x\omega$. The maps $i_{ax+by}$ and $a\,i_x+b\,i_y$ are antiderivations of degree $-1$ that agree on $E$, hence equal. By Lemma 1.2 (3), $i_x^2$ is a derivation of degree $-2$, and by Lemma 1.2 (1), $[i_x,i_y]=i_xi_y+i_yi_x$ is an antiderivation of degree $-2$; both vanish on $A$ and $E$ for degree reasons, so both are zero by Lemma 1.2 (4).

(2) Such a map $\delta$ vanishes on $A$, so $\delta(ae)=a\,\delta(e)$ on $E$, and $x=\delta|_E$ is a local section of $L$. Then $\delta=i_x$ by uniqueness, and $x$ is determined by $i_x|_E=x$.

(3) By (4.1) for $x=x_1$ and induction on $n$, the left side is the expansion of the determinant along its first column. $\square$

Construction 4.3 (anchor and bracket). For a local section $x$ of $L$ put $\theta_x=[d,i_x]=d\,i_x+i_x\,d, \tag{4.2}$ a $K$-derivation of degree $0$ of $\Lambda_AE|_U$ (Lemma 1.2 (1)). On $A$, $\theta_x(a)=i_x(da)=x(da)$; this defines $\theta\colon L\to T_{A/K}$. If $u\colon\Omega^1_{A/K}\to E$ is the $A$-linear map with $u\circ d_{A/K}=d|_A$, then $\theta_x=x\circ u$ in $T_{A/K}=\mathcal Hom_A(\Omega^1_{A/K},A)$: the anchor is the transpose $u^\vee$. For local sections $x,y$ of $L$, $[\theta_x,i_y]=\theta_xi_y-i_y\theta_x$ is a $K$-antiderivation of degree $-1$, so by Lemma 4.2 (2) $[\theta_x,i_y]=i_{[x,y]} \tag{4.3}$ for a unique local section $[x,y]$ of $L$. Evaluating (4.3) on a local section $e$ of $E$ and using (4.2) gives the Maurer–Cartan formula $\begin{aligned}[x,y](e)&=\theta_x\bigl(y(e)\bigr)-\theta_y\bigl(x(e)\bigr)\\&\quad-(de)(x,y).\end{aligned} \tag{4.4}$ Indeed $\theta_x(i_ye)=\theta_x(y(e))$, and $i_y\theta_xe=$ $i_yi_x\,de+i_y\,d(x(e))=$ $(de)(x,y)+\theta_y(x(e))$.

Proposition 4.4.

$\theta$ is $A$-linear; the bracket is $K$-bilinear and alternating and satisfies (3.1). For local sections $\omega$ of $\Lambda^n_AE$ and $x_0,\ldots,x_n$ of $L$, the Chevalley–Eilenberg formula (4.5) below holds.
If $d^2=0$, then $[d,\theta_x]=0$ and $\theta_{[x,y]}=[\theta_x,\theta_y]$ as derivations of $\Lambda_AE$, and $(L,[\ ,\ ],\theta)$ is a Lie algebroid over $A/K$.

Here, writing $x_{\hat\imath}$ for the sequence $x_0,\ldots,x_n$ with $x_i$ omitted and $x_{\hat\imath\hat\jmath}$ for the sequence with $x_i$ and $x_j$ omitted, $\begin{aligned}&(d\omega)(x_0,\ldots,x_n)\\&=\sum_i(-1)^i\theta_{x_i}\bigl(\omega(x_{\hat\imath})\bigr)\\&\quad+\sum_{i<j}(-1)^{i+j}\\&\qquad\cdot\omega\bigl([x_i,x_j],x_{\hat\imath\hat\jmath}\bigr).\end{aligned} \tag{4.5}$

Proof. (1) As operators on $\Lambda_AE$, $\theta_{ax}=d(a\,i_x)+a\,i_xd=$ $a\,\theta_x+da\wedge i_x(\,\cdot\,)$; since $i_x$ vanishes on $A$, $\theta_{ax}(b)=a\,\theta_x(b)$, so $\theta$ is $A$-linear. The maps $x\mapsto i_x$ and $x\mapsto\theta_x$ are $K$-linear, so the bracket is $K$-bilinear. It is alternating by (4.4), since $(de)(x,x)=0$ (Lemma 4.2 (3)). As $\theta_x$ is a derivation of degree $0$, $\begin{aligned}i_{[x,ay]}&=[\theta_x,a\,i_y]=\theta_x(a)\,i_y+a[\theta_x,i_y]\\&=i_{\theta_x(a)y+a[x,y]},\end{aligned}$ and Lemma 4.2 (2) gives (3.1).

For (4.5) we use induction on $n$. For $n=0$, $(da)(x_0)=i_{x_0}da=\theta_{x_0}(a)$. Let $n\geq1$. By (4.2), $i_{x_0}d\omega=\theta_{x_0}\omega-d\,i_{x_0}\omega$, so $\begin{aligned}&(d\omega)(x_0,\ldots,x_n)\\&=i_{x_n}\cdots i_{x_1}\theta_{x_0}\omega\\&\quad-(d\,i_{x_0}\omega)(x_1,\ldots,x_n).\end{aligned}$ By (4.3), $i_{x_k}\theta_{x_0}=\theta_{x_0}i_{x_k}-i_{[x_0,x_k]}$; moving $\theta_{x_0}$ to the left past $i_{x_1},\ldots,i_{x_n}$ gives $\begin{aligned}&i_{x_n}\cdots i_{x_1}\theta_{x_0}\omega=\theta_{x_0}\bigl(\omega(x_1,\ldots,x_n)\bigr)\\&\quad-\sum_{k=1}^n\omega(x_1,\ldots,[x_0,x_k],\ldots,x_n),\end{aligned}$ with $[x_0,x_k]$ in the place of $x_k$. Moving it to the front costs $(-1)^{k-1}$, which produces the terms $i=0$ of the second sum of (4.5). The form $\eta=i_{x_0}\omega$ of degree $n-1$ has $\eta(y_1,\ldots,y_{n-1})=\omega(x_0,y_1,\ldots,y_{n-1})$. By induction, $\begin{aligned}&(d\eta)(x_1,\ldots,x_n)\\&=\sum_{i\geq1}(-1)^{i-1}\theta_{x_i}\bigl(\omega(x_{\hat\imath})\bigr)\\&\quad+\sum_{1\leq i<j}(-1)^{i+j}\omega\bigl(x_0,[x_i,x_j],\ldots\bigr),\end{aligned}$ and exchanging $x_0$ with $[x_i,x_j]$ changes the sign. Subtracting gives the remaining terms of (4.5).

(2) By Lemma 1.2 (3) for the odd map $d$, $[d,\theta_x]=[d,[d,i_x]]=[d^2,i_x]=0$. Lemma 1.2 (2) for $d$, $\theta_x$, $i_y$, of degrees $1$, $0$, $-1$, gives $\begin{aligned}\theta_{[x,y]}&=[d,i_{[x,y]}]=[d,[\theta_x,i_y]]\\&=[[d,\theta_x],i_y]+[\theta_x,[d,i_y]]\\&=[\theta_x,\theta_y].\end{aligned}$ On $A$ this says that $\theta$ is a homomorphism of Lie algebras. For the Jacobi identity, Lemma 1.2 (2) for $\theta_x$, $\theta_y$, $i_z$ gives $\begin{aligned}i_{[x,[y,z]]}&=[\theta_x,[\theta_y,i_z]]\\&=[[\theta_x,\theta_y],i_z]+[\theta_y,[\theta_x,i_z]]\\&=[\theta_{[x,y]},i_z]+[\theta_y,i_{[x,z]}]\\&=i_{[[x,y],z]}+i_{[y,[x,z]]},\end{aligned}$ which is the Jacobi identity, by Lemma 4.2 (2). $\square$

Proposition 4.5 (functoriality). Let $u\colon E\to F$ be an $A$-linear map and $d$, $d$ antiderivations of degree $1$ of $\Lambda_AE$, $\Lambda_AF$ with $\Lambda u\circ d=d\circ\Lambda u$. Then $u^\vee\colon F^\vee\to E^\vee$, $x\mapsto x\circ u$, is compatible with the brackets and anchors of Construction 4.3. In particular, a morphism of de Rham complexes induces a morphism of Lie algebroids, and this defines a functor from the opposite of the category $DR(A/K)$ of de Rham complexes over $A/K$ to the category $LA(A/K)$ of Lie algebroids over $A/K$.

Proof. For a local section $x$ of $F^\vee$, $\Lambda u\circ i_{u^\vee x}$ and $i_x\circ\Lambda u$ are $\Lambda u$-antiderivations of degree $-1$ that agree on $A$ and on $E$, so they are equal (Lemma 1.2 (4)). With (4.2) this gives $\Lambda u\circ\theta_{u^\vee x}=\theta_x\circ\Lambda u$, and in degree $0$, where $\Lambda u$ is the identity of $A$, $\theta_{u^\vee x}=\theta_x$ on $A$. With (4.3), $\begin{aligned}\Lambda u\circ i_{[u^\vee x,u^\vee y]}&=\Lambda u\circ[\theta_{u^\vee x},i_{u^\vee y}]\\&=[\theta_x,i_y]\circ\Lambda u\\&=i_{[x,y]}\circ\Lambda u,\end{aligned}$ and on a local section $e$ of $E$ this reads $[u^\vee x,u^\vee y](e)=[x,y](u(e))$, that is, $[u^\vee x,u^\vee y]=u^\vee[x,y]$. $\square$

Example 4.6. For $\Omega^\bullet_{A/K}$, $L=T_{A/K}$ and $u=\mathrm{id}$, so $\theta=\mathrm{id}$. By (4.4) with $e=db$, $[x,y](db)=x(y(b))-y(x(b))$, since $d(db)=0$; the elements $db$ generate $\Omega^1_{A/K}$, so the bracket is the commutator of derivations. Hence the anchor of any de Rham complex $(\Lambda E,d)$ is the morphism of Lie algebroids $E^\vee\to T_{A/K}$ induced, through Proposition 4.5, by the morphism $\Omega^\bullet_{A/K}\to(\Lambda E,d)$ of Proposition 2.3.

5. The locally free case

Lemma 5.1. Let $E$ be locally free of finite type and $L=E^\vee$.

$L$ is locally free of finite type, and $E\to L^\vee$, $e\mapsto(x\mapsto x(e))$, is an isomorphism.
The map sending $\omega$ to the function $(x_1,\ldots,x_n)\mapsto\omega(x_1,\ldots,x_n)$ is an isomorphism from $\Lambda^n_AE$ onto the sheaf of alternating $A$-multilinear maps $L^n\to A$.
The pairing $\langle\omega,x_1\wedge\cdots\wedge x_n\rangle=\omega(x_1,\ldots,x_n)$ identifies $\Lambda^n_AL$ with $(\Lambda^n_AE)^\vee$.

Proof. All three are local, so let $E$ be free with basis $e_1,\ldots,e_r$ and $x_1,\ldots,x_r$ the dual basis of $L$. (1) is clear. For increasing multi-indices $I$, $J$ of length $n$, Lemma 4.2 (3) gives $e_I(x_J)=1$ if $I=J$ and $0$ otherwise, where $e_I=e_{i_1}\wedge\cdots\wedge e_{i_n}$. An alternating multilinear map is determined by its values on the $x_J$, and the $e_I$ form a basis of $\Lambda^n_AE$, which gives (2); (3) follows in the same way, the $x_J$ forming a basis of $\Lambda^n_AL$. $\square$

Theorem 5.2 (Jacobi identity and $d^2=0$). Let $E$ be locally free of finite type and $d$ a $K$-antiderivation of degree $1$ of $\Lambda_AE$. The bracket of Construction 4.3 on $L=E^\vee$ satisfies the Jacobi identity if and only if $d^2=0$. In that case $(L,[\ ,\ ],\theta)$ is a Lie algebroid over $A/K$.

Proof. If $d^2=0$, Proposition 4.4 (2) applies. Conversely, put $D=d^2$, a derivation of degree $2$ (Lemma 1.2 (3)), and assume the Jacobi identity. Two identities hold without any assumption on $d$. First, $[d,\theta_x]=[d,[d,i_x]]=[D,i_x]$ by Lemma 1.2 (3), so the computation in the proof of Proposition 4.4 (2) gives $\begin{aligned}&\theta_{[x,y]}-[\theta_x,\theta_y]\\&\quad=[[D,i_x],i_y].\end{aligned} \tag{5.1}$ Second, Lemma 1.2 (2) gives $i_{[x,[y,z]]}-i_{[y,[x,z]]}=[[\theta_x,\theta_y],i_z]$, while $i_{[[x,y],z]}=[\theta_{[x,y]},i_z]$, so with $J$ as in Lemma 3.2, $\begin{aligned}&i_{J(x,y,z)}\\&\quad=-\bigl[[[D,i_x],i_y],i_z\bigr].\end{aligned} \tag{5.2}$

$D$ vanishes on $A$. By Proposition 4.4 (1), Lemma 3.2 applies, and $L$ is locally free of finite type, so $\theta_{[x,y]}=[\theta_x,\theta_y]$ on $A$. For a local section $a$ of $A$, $[D,i_x](a)=-i_x(Da)$, and $[[D,i_x],i_y](a)=$ $i_y\bigl([D,i_x](a)\bigr)=$ $-(Da)(x,y)$, $[D,i_x]$ having odd degree. By (5.1), $(Da)(x,y)=0$ for all $x,y$, so $Da=0$ by Lemma 5.1 (2).

$D$ vanishes on $E$. As $D$ vanishes on $A$, $[D,i_x]$ vanishes on $A$ and sends $e$ to $-i_xDe$; then $[[D,i_x],i_y]$ vanishes on $A$ and sends $e$ to $-i_yi_xDe$; and $[[[D,i_x],i_y],i_z](e)=$ $i_zi_yi_xDe=(De)(x,y,z)$. By (5.2) and the Jacobi identity, $(De)(x,y,z)=0$ for all $x,y,z$, so $De=0$ by Lemma 5.1 (2). Hence $D=0$ by Lemma 1.2 (4). $\square$

Local freeness cannot be omitted: Exercise 7.2 gives $E$ with $E^\vee=0$, so that the Jacobi identity holds trivially, and an antiderivation with $d^2\neq0$.

Theorem 5.3 (de Rham complexes and Lie algebroids). Let $DR_{\mathrm{lf}}(A/K)$ and $LA_{\mathrm{lf}}(A/K)$ be the full subcategories of de Rham complexes $(\Lambda E,d)$, resp. Lie algebroids $L$, with $E$, resp. $L$, locally free of finite type. The functor of Proposition 4.5 restricts to an equivalence $DR_{\mathrm{lf}}(A/K)^{\mathrm{op}}\longrightarrow LA_{\mathrm{lf}}(A/K).$ A quasi-inverse sends $L$ to $\Lambda_A(L^\vee)$ with the differential given by (4.5); in degrees $0$ and $1$, $\begin{aligned}(da)(x)&=\theta_x(a),\\ (d\alpha)(x,y)&=\theta_x\bigl(\alpha(y)\bigr)\\&\quad-\theta_y\bigl(\alpha(x)\bigr)\\&\quad-\alpha([x,y]).\end{aligned} \tag{5.3}$

Proof. By Lemma 5.1 (1) the functor $\Phi$ of Proposition 4.5 sends $DR_{\mathrm{lf}}$ to $LA_{\mathrm{lf}}$.

The quasi-inverse $\Psi$. Let $L$ be in $LA_{\mathrm{lf}}(A/K)$ and $E=L^\vee$, locally free of finite type. Identify $L$ with $E^\vee$ by Lemma 5.1 (1). Define $d_0a$ by $(d_0a)(x)=\theta_x(a)$, a local section of $E$ because $\theta$ is $A$-linear; $d_0$ is a $K$-derivation, each $\theta_x$ being one. Define $d_1\alpha$ by the second line of (5.3). It is alternating, $\alpha([x,x])=0$, and $A$-bilinear: by (3.1), $[bx,y]=-[y,bx]=-\theta_y(b)x+b[x,y]$, so $\begin{aligned}&(d_1\alpha)(bx,y)-b\,(d_1\alpha)(x,y)\\&\quad=-\theta_y(b)\alpha(x)+\theta_y(b)\alpha(x)=0.\end{aligned}$ By Lemma 5.1 (2), $d_1\alpha$ is a local section of $\Lambda^2_AE$. Rule (2.1) holds: $d_1(a\alpha)(x,y)-a\,(d_1\alpha)(x,y)=$ $\theta_x(a)\alpha(y)-\theta_y(a)\alpha(x)=$ $(d_0a\wedge\alpha)(x,y)$ by Lemma 4.2 (3). Proposition 2.2 gives an antiderivation $d^L$ of degree $1$ of $\Lambda_AE$. By Construction 4.3 its anchor is $x\mapsto(a\mapsto(d_0a)(x))=\theta$, and by (4.4) its bracket is $[x,y]'(\alpha)=$ $\theta_x(\alpha(y))-\theta_y(\alpha(x))-(d_1\alpha)(x,y)=$ $\alpha([x,y])$: the given ones. The bracket of $L$ satisfies the Jacobi identity, so $(d^L)^2=0$ by Theorem 5.2. Put $\Psi(L)=(\Lambda_AL^\vee,d^L)$. For a morphism $v\colon L\to L'$, the maps $\Lambda(v^\vee)\circ d^{L'}$ and $d^L\circ\Lambda(v^\vee)$ are $\Lambda(v^\vee)$-antiderivations, and their values on $A$ and on $L'^\vee$, evaluated on local sections of $L$, are $\theta'_{vx}(a)=\theta_x(a)$ and $\begin{aligned}&\theta'_{vx}(\alpha(vy))-\theta'_{vy}(\alpha(vx))-\alpha([vx,vy])\\&\quad=\theta_x(\alpha(vy))-\theta_y(\alpha(vx))-\alpha(v[x,y]).\end{aligned}$ So they are equal by Lemmas 5.1 (2) and 1.2 (4), and $v^\vee$ is a morphism $\Psi(L')\to\Psi(L)$.

$\Phi\Psi\cong\mathrm{id}$. We have just seen that the bracket and anchor of $\Phi\Psi(L)$ on $L^{\vee\vee}$ are those of $L$ under the biduality isomorphism, which is natural in $L$.

$\Psi\Phi\cong\mathrm{id}$. Let $(\Lambda E,d)$ be in $DR_{\mathrm{lf}}$, $L=E^\vee$, and $\beta\colon E\to L^\vee$ the biduality isomorphism. The maps $\Lambda\beta\circ d$ and $d^L\circ\Lambda\beta$ are $\Lambda\beta$-antiderivations. On $A$, $(d^La)(x)=\theta_x(a)=x(da)=\beta(da)(x)$. On $E$, by (4.4), $\begin{aligned}(d^L\beta e)(x,y)&=\theta_x(y(e))-\theta_y(x(e))\\&\quad-[x,y](e)=(de)(x,y),\end{aligned}$ and $(\Lambda^2\beta\,de)(x,y)=(de)(x,y)$ by Lemma 4.2 (3). So the two maps agree by Lemmas 5.1 (2) and 1.2 (4), and $\beta$ is an isomorphism $(\Lambda E,d)\to\Psi\Phi(\Lambda E,d)$, natural in $(\Lambda E,d)$. $\square$

Remark 5.4 (Lie algebras). Let $A=K$, so that $T_{A/K}=0$. For a sheaf of $K$-Lie algebras $L$ that is locally free of finite type, the de Rham complex $\Psi(L)$ has the differential $\begin{aligned}&(d\omega)(x_0,\ldots,x_n)\\&\quad=\sum_{i<j}(-1)^{i+j}\omega\bigl([x_i,x_j],x_{\hat\imath\hat\jmath}\bigr),\end{aligned}$ by (4.5): it is the Chevalley–Eilenberg cochain complex of $L$ with coefficients in $K$. By Lemma 5.1 (3) its transpose is a complex on $\Lambda^\bullet_KL$, and by (4.5) its differential $\partial\colon\Lambda^{n+1}_KL\to\Lambda^n_KL$ is $\begin{aligned}&\partial(x_0\wedge\cdots\wedge x_n)\\&=\sum_{i<j}(-1)^{i+j}\\&\quad\cdot[x_i,x_j]\wedge x_{\hat\imath\hat\jmath},\end{aligned} \tag{5.4}$ where $x_{\hat\imath\hat\jmath}$ is the product $x_0\wedge\cdots\wedge x_n$ with $x_i$ and $x_j$ omitted, since both sides pair in the same way with every $\omega$: this is the Chevalley–Eilenberg chain complex, or Koszul complex, of $L$.

6. Invariant differential forms on torsors

In this section $G$ is a group scheme over a scheme $S$, flat and locally of finite presentation, $f\colon X\to Y$ is a $G_Y$-torsor, and $\varpi\colon Y\to S$ is a morphism. The notation is that of Torsors, co-Lie complexes and the Atiyah class: $N(G,X)$ is the nerve of the action, with the levels $N_n(G,X)=G^n\times_SX$ and the faces $\mathrm{act},\mathrm{pr}\colon G\times_SX\to X$ in degree $1$; $N(G)=N(G,S)$, $N(G,Y)=N(G)\times_SY$, and $a\colon N(G,X)\to Y$. A sheaf $F$ on $N(G,X)_{\mathrm{Zar}}$ is a family of sheaves $F_n$ on the $N_n(G,X)_{\mathrm{Zar}}$ with transition maps $N(G,X)(\alpha)^{-1}F_m\to F_n$ for $\alpha\colon[m]\to[n]$. A section of $a_*F$ over an open $V\subseteq Y$ is a compatible family of sections $s_n$ of $F_n$ over $a_n^{-1}(V)$. Such a family is determined by $s_0$, which has the same images under the two faces $N_1(G,X)\rightrightarrows N_0(G,X)$. Conversely every such $s_0$ extends: any two vertex maps $[0]\to[n]$ factor through one map $[1]\to[n]$, so the pullbacks of $s_0$ along all vertex maps agree.

Construction 6.1. For $n\geq0$ let $\Omega^\bullet_n$ be the de Rham complex of $N_n(G,X)$ over $N_n(G)$, and $\Omega'^\bullet_n$ that over $N_n(G,Y)$ (Proposition 2.3). For $\alpha\colon[m]\to[n]$, the commutative squares formed by $N(G,X)(\alpha)$ and $N(G)(\alpha)$, resp. $N(G,Y)(\alpha)$, with the projections induce maps of sheaves of differential graded algebras $N(G,X)(\alpha)^{-1}\Omega^\bullet_m\to\Omega^\bullet_n$, resp. on the $\Omega'^\bullet$: in each degree the exterior power of the map of differentials, compatible with $d$ by (2.5). This gives sheaves of differential graded algebras $\Omega^\bullet_{N(G,X)/N(G)}$ and $\Omega^\bullet_{N(G,X)/N(G,Y)}$ on $N(G,X)_{\mathrm{Zar}}$, and the quotient maps of the triangles $N_n(G,X)\to N_n(G,Y)\to N_n(G)$ give a map from the first to the second. Put $\begin{aligned}\omega^\bullet_{X/S}&=a_*\Omega^\bullet_{N(G,X)/N(G)},\\ \omega^\bullet_{X/Y}&=a_*\Omega^\bullet_{N(G,X)/N(G,Y)}.\end{aligned}$ Explicitly, a section of $\omega^p_{X/S}$ over $V$ is a $p$-form $\sigma$ of $X$ over $S$ on $f^{-1}(V)$ whose images under the maps $\mathrm{act}^\natural,\mathrm{pr}^\natural$ to the $p$-forms of $G\times_SX$ over $G$ are equal, $\mathrm{act}$ and $\mathrm{pr}$ lying over the structure map $G\to S$. These are the left-invariant $p$-forms of $X$ over $S$. For $Y=S$ and $X=G$, acting on itself by left translations, $f$ is the structure map, which has the unit section, and $(g,x)\mapsto(gx,x)$ is an automorphism of $G\times_SG$: $G$ is a $G_S$-torsor, and $\omega^\bullet_{G/S}$ is the complex of left-invariant differential forms of $G$.

Lemma 6.2. Every $\Omega^p_{N(G,X)/N(G)}$ and $\Omega^p_{N(G,X)/N(G,Y)}$ is a cartesian quasi-coherent module on $N(G,X)_{\mathrm{Zar}}$, and it is the $p$-th exterior power of the corresponding module of degree $1$.

Proof. The levels are quasi-coherent [Tag 01UT], and exterior powers of the modules of differentials by definition. For $\alpha\colon[m]\to[n]$, the squares formed by $N(G,X)(\alpha)$, $N(G,Y)(\alpha)$ and by $N(G,Y)(\alpha)$, $N(G)(\alpha)$ are cartesian (Lemma 3.1 (1) of the lesson on torsors), hence so is their composite. Differentials commute with base change [Tag 01V0], so the transition maps of the modules of degree $1$ are isomorphisms, and exterior powers commute with pullback [Tag 01CI]. $\square$

Proposition 6.3.

The $\omega^p_{X/S}$ and $\omega^p_{X/Y}$ are quasi-coherent $\mathcal O_Y$-modules, the counit maps $a^*\omega^p_{X/S}\to\Omega^p_{N(G,X)/N(G)}$ and $a^*\omega^p_{X/Y}\to\Omega^p_{N(G,X)/N(G,Y)}$ are isomorphisms, and so are the product maps $\Lambda^p_{\mathcal O_Y}\omega^1\to\omega^p$. In particular $f^*\omega^p_{X/S}\cong\Omega^p_{X/S}$ and $f^*\omega^p_{X/Y}\cong\Omega^p_{X/Y}$. Moreover $\omega^1_{X/S}\cong H^0(\ell_n(X/S))$ for every $n\leq0$ and $\omega^1_{X/Y}\cong H^0(\ell_{X/Y})$, through the isomorphisms of Theorem 3.3 (2), (1) of the lesson on torsors.
$(\omega^\bullet_{X/S},d)$ is a de Rham complex over $\mathcal O_Y/\varpi^{-1}\mathcal O_S$, $(\omega^\bullet_{X/Y},d)$ is a de Rham complex over $\mathcal O_Y/\mathcal O_Y$, and the map $\omega^\bullet_{X/S}\to\omega^\bullet_{X/Y}$ induced by Construction 6.1 is a surjective morphism of de Rham complexes over $\mathcal O_Y/\varpi^{-1}\mathcal O_S$. The degree $1$ component of the morphism $\Omega^\bullet_{Y/S}\to\omega^\bullet_{X/S}$ of Proposition 2.3 is the pullback of forms, $db\mapsto d(f^\sharp b)$.
An isomorphism $\varphi\colon X'\to X$ of $G_Y$-torsors induces, by pullback of forms, isomorphisms of de Rham complexes $\omega^\bullet_{X/S}\cong\omega^\bullet_{X'/S}$ and $\omega^\bullet_{X/Y}\cong\omega^\bullet_{X'/Y}$. For an open $V\subseteq Y$, the restrictions to $V$ are the complexes of the $G_V$-torsor $X_V\to V$.

Proof. (1) Let $\mathcal E=H^0(\ell_n(X/S))$, quasi-coherent by Theorem 3.3 (2) of the lesson on torsors. The functor $a^*$ is exact, the $a_n$ being flat, so $a^*\mathcal E\cong H^0(t_{[n}L_{N(G,X)/N(G)})$, which is $\Omega^1_{N(G,X)/N(G)}$ levelwise (Theorem 6.1 of The cotangent complex; $n\leq0$). Hence $a^*\Lambda^p\mathcal E\cong\Lambda^pa^*\mathcal E\cong\Omega^p_{N(G,X)/N(G)}$ by Lemma 6.2. By Corollary 2.6 (1) of the lesson on torsors, applied to $\Lambda^p\mathcal E$ in degree $0$ and followed by $H^0$, the unit $\Lambda^p\mathcal E\to a_*a^*\Lambda^p\mathcal E\cong\omega^p_{X/S}$ is an isomorphism. It sends $e_1\wedge\cdots\wedge e_p$ to the product of the images of the $e_i$; for $p=1$ it is an isomorphism $\mathcal E\cong\omega^1_{X/S}$, and through it the product maps are isomorphisms. By the triangle identity, the counit composed with $a^*$ of the unit is the identity of $a^*\Lambda^p\mathcal E$, so the counit is an isomorphism. At the level $0$, $a^*$ is $f^*$. The same argument with $\ell_{X/Y}$ and Theorem 3.3 (1) of the lesson on torsors applies to $\omega^\bullet_{X/Y}$.

(2) The differentials of the $\Omega^\bullet_n$ commute with the transition maps, so $a_*$ gives a differential $d$ on $\omega^\bullet_{X/S}$ with $d^2=0$, an antiderivation for the products, which are computed levelwise. It is linear over $\varpi^{-1}\mathcal O_S$, since the differential of $\Omega^\bullet_n$ is linear over the inverse image of $\mathcal O_{N_n(G)}$. By Corollary 2.6 (1) there, $\omega^0_{X/S}=a_*\mathcal O=\mathcal O_Y$, and by (1) $\omega^\bullet_{X/S}=\Lambda_{\mathcal O_Y}\omega^1_{X/S}$. The differentials of the $\Omega'^\bullet_n$ are linear over the inverse image of $\mathcal O_{N_n(G,Y)}$, which contains that of $\mathcal O_Y$; so $d$ on $\omega^\bullet_{X/Y}$ is $\mathcal O_Y$-linear and vanishes on $\mathcal O_Y$. The levelwise maps of Construction 6.1 are maps of differential graded algebras, the identity in degree $0$, so their direct image is a morphism of de Rham complexes over $\mathcal O_Y/\varpi^{-1}\mathcal O_S$. By (1), $a^*$ of it is in degree $1$ the levelwise surjection of differentials; $f$ is flat and surjective (Proposition 1.3 (1) there), and the cokernel is quasi-coherent with $f^*$ of it zero, so it vanishes. Exterior powers of a surjection are surjective. Finally, for a local section $b$ of $\mathcal O_Y$, $d(b)$ is the invariant form $d(f^\sharp b)$, and $u\circ d_{Y/S}=d|_{\mathcal O_Y}$ characterizes $u$.

(3) The isomorphisms $N(G,\varphi)\colon N(G,X')\to N(G,X)$, levelwise $\mathrm{id}\times\varphi$, lie over $N(G,Y)$ and $N(G)$, so they induce isomorphisms of the relative de Rham complexes, compatible with the transition maps, hence of their direct images. For $V\subseteq Y$, $a^{-1}(V)=N(G,X_V)$. $\square$

Proposition 6.4 (the Lie algebra of $G$). Let $t_G=\mathcal Hom_{\mathcal O_S}(\omega^1_{G/S},\mathcal O_S)$ with the bracket of Proposition 4.4 for the de Rham complex $\omega^\bullet_{G/S}$ over $\mathcal O_S$: a sheaf of $\mathcal O_S$-Lie algebras, with $\theta=0$. Let $f\colon G\to S$ be the structure map. For a local section $x$ of $t_G$ over $V\subseteq S$, let $D_x$ be the derivation of $\mathcal O_{G_V}$ over $f^{-1}\mathcal O_V$ whose $\mathcal O_G$-linear map $\Omega^1_{G/S}\to\mathcal O_G$ is $f^*x$, through $f^*\omega^1_{G/S}\cong\Omega^1_{G/S}$. Write $\langle D,\sigma\rangle$ for the value of a derivation on a $1$-form.

$\langle D_x,\sigma\rangle=f^\sharp(x(\sigma))$ for invariant $\sigma$, and $x\mapsto D_x$ is injective and $\mathcal O_S$-linear. Its image consists of the left-invariant derivations, those $D$ with $\langle D,\sigma\rangle$ in $f^\sharp\mathcal O_S$ for all local invariant $1$-forms $\sigma$.
$D_{[x,y]}=D_x\circ D_y-D_y\circ D_x$.
Restriction to the unit section $e$ is an isomorphism $\omega^1_{G/S}\cong e^*\Omega^1_{G/S}=\omega_G$. If $\omega_G$ is finite locally free, for instance if $G$ is smooth over $S$ or $S$ is the spectrum of a field, then $\omega^\bullet_{G/S}$ is the Chevalley–Eilenberg cochain complex of $t_G$, and its transpose is the Chevalley–Eilenberg chain complex (5.4) of $t_G$.

Proof. (1) The isomorphism $f^*\omega^1_{G/S}\cong\Omega^1_{G/S}$ of Proposition 6.3 (1) is the counit, which sends $f^*\sigma$ to $\sigma$; this gives the formula. Since $fe=\mathrm{id}_S$, $f^\sharp$ is injective, so $D_x=0$ forces $x=0$. If $D$ is left-invariant, $x(\sigma)=(f^\sharp)^{-1}\langle D,\sigma\rangle$ defines an $\mathcal O_S$-linear $x$, and $D$, $D_x$ agree on the invariant forms, which generate $\Omega^1_{G/S}$ as an $\mathcal O_G$-module; so $D=D_x$.

(2) The derivations $D_x$, $D_y$ are local sections of the Lie algebroid $T_{\mathcal O_G/f^{-1}\mathcal O_S}$ of $\Omega^\bullet_{G/S}$, whose bracket is the commutator (Example 4.6). For an invariant $\sigma$, (4.4) in $\Omega^\bullet_{G/S}$ gives $\begin{aligned}\langle[D_x,D_y],\sigma\rangle&=D_x\langle D_y,\sigma\rangle-D_y\langle D_x,\sigma\rangle\\&\quad-(d\sigma)(D_x,D_y)\\&=-(d\sigma)(D_x,D_y),\end{aligned}$ since derivations over $f^{-1}\mathcal O_S$ kill $f^\sharp\mathcal O_S$. By Proposition 6.3 (1), $d\sigma$ is locally $\sum_kc_k\,\alpha_k\wedge\beta_k$ with $c_k$ in $\mathcal O_S$ and $\alpha_k$, $\beta_k$ invariant, so by Lemma 4.2 (3) and (1), $(d\sigma)(D_x,D_y)=f^\sharp\bigl((d\sigma)(x,y)\bigr)$, the right side computed in $\omega^\bullet_{G/S}$. There (4.4) with $\theta=0$ reads $[x,y](\sigma)=-(d\sigma)(x,y)$. So $\langle[D_x,D_y],\sigma\rangle=\langle D_{[x,y]},\sigma\rangle$ for all invariant $\sigma$, and the two derivations are equal.

(3) The restriction is $e^*$ of the isomorphism $f^*\omega^1_{G/S}\cong\Omega^1_{G/S}$, as $e^*f^*=\mathrm{id}$. If $G$ is smooth, $\Omega^1_{G/S}$ is finite locally free [Tag 02G1], hence so is $\omega_G$. Over a field, $\Omega^1_{G/S}$ is of finite type, $G$ being locally of finite presentation, so $\omega_G$ is a finite-dimensional vector space. When $\omega_G$ is finite locally free, Lemma 5.1, (4.5) with $\theta=0$ and Remark 5.4 apply. $\square$

Theorem 6.5 (the Atiyah sequence). Let $t_{X/S}=\mathcal Hom(\omega^1_{X/S},\mathcal O_Y)$, $t_{X/Y}=\mathcal Hom(\omega^1_{X/Y},\mathcal O_Y)$ and $T_{Y/S}=\mathcal Hom(\Omega^1_{Y/S},\mathcal O_Y)$.

The sequence of the maps of Proposition 6.3 (2), $\begin{aligned}&\Omega^1_{Y/S}\to\omega^1_{X/S}\\&\quad\to\omega^1_{X/Y}\to0,\end{aligned} \tag{6.1}$ is exact. Its pullback by $f$ is the exact sequence $f^*\Omega^1_{Y/S}\to\Omega^1_{X/S}\to\Omega^1_{X/Y}\to0$ [Tag 01UX].
The transposed sequence $\begin{aligned}&0\to t_{X/Y}\to t_{X/S}\\&\quad\xrightarrow{\theta}T_{Y/S}\end{aligned} \tag{6.2}$ is exact, and its second map is the anchor of the Lie algebroid $t_{X/S}$ of $\omega^\bullet_{X/S}$ over $\mathcal O_Y/\varpi^{-1}\mathcal O_S$. Its first map is a morphism of Lie algebroids from the Lie algebroid of $\omega^\bullet_{X/Y}$, whose anchor is zero and whose bracket is $\mathcal O_Y$-bilinear; it identifies $t_{X/Y}$ with the kernel of $\theta$.
If $G$ is smooth over $S$, then (6.1) is also exact on the left, $\omega^1_{X/Y}$ is finite locally free, and (6.2) is also exact on the right.
Let $V\subseteq Y$ be open, $s$ a section of $f$ over $V$ and $\psi_s\colon G_V\to X_V$, $h\mapsto hs$. Pullback along $\psi_s$ is an isomorphism of de Rham complexes $\omega^\bullet_{X/Y}|_V\cong\omega^\bullet_{G_V/V}$ over $\mathcal O_V$, where $G_V$ is viewed as a group scheme over $V$; its transpose is an isomorphism of $\mathcal O_V$-Lie algebras $t_{G_V}\cong t_{X/Y}|_V$. For $h\in G(V)$, the isomorphisms for $hs$ and $s$ differ by the automorphism $\rho_h^*$ of $\omega^\bullet_{G_V/V}$, $\rho_h$ being the right translation by $h$. Through restriction to the unit section, $\omega^1_{G_V/V}\cong\omega_G\otimes\mathcal O_V$, the map $\rho_h^*$ is $c(\operatorname{int}(h^{-1}))$.

So $t_{X/Y}$ is the Lie algebra of $G_Y$ twisted by $X$ through the transposes of the $c(\operatorname{int}(h^{-1}))$, that is, through the adjoint action (Proposition 2.4 of The adjoint co-Lie complex of a group scheme).

Proof. (1) By Proposition 6.3 (1), (2), $f^*$ of (6.1) is the displayed sequence of differentials: $f^*$ of the first map is the pullback of forms, and $f^*$ of the second is the quotient map. It is exact [Tag 01UX]. Since $f$ is flat and surjective, $\mathcal O_{Y,f(x)}\to\mathcal O_{X,x}$ is faithfully flat for every point $x$ of $X$, and exactness of a sequence of $\mathcal O_Y$-modules can be checked after $f^*$.

(2) $\mathcal Hom(-,\mathcal O_Y)$ is left exact. By Construction 4.3 the anchor of $t_{X/S}$ is the transpose of the degree $1$ component of $\Omega^\bullet_{Y/S}\to\omega^\bullet_{X/S}$, which is the first map of (6.1) by Proposition 6.3 (2). The second map of (6.1) is the degree $1$ component of a morphism of de Rham complexes over $\mathcal O_Y/\varpi^{-1}\mathcal O_S$, so its transpose is a morphism of Lie algebroids (Proposition 4.5). On $\omega^\bullet_{X/Y}$, $d$ vanishes on $\mathcal O_Y$, so its anchor $x\mapsto x(d\,\cdot\,)$ is zero, and its bracket is $\mathcal O_Y$-bilinear by (3.1). Exactness identifies $t_{X/Y}$ with $\ker\theta$.

(3) By Proposition 4.1 (1) of the lesson on torsors, $f$ is smooth, so $0\to f^*\Omega^1_{Y/S}\to\Omega^1_{X/S}\to\Omega^1_{X/Y}\to0$ is exact [Tag 02K4] and $\Omega^1_{X/Y}$ is finite locally free [Tag 02G1]. The argument of (1) gives exactness on the left. The module $\omega^1_{X/Y}$ is quasi-coherent with $f^*\omega^1_{X/Y}\cong\Omega^1_{X/Y}$, and finite locally free modules descend along the fppf covering $\{X\to Y\}$ [Tag 05B2]. A short exact sequence with finite locally free quotient is locally split, so its transpose is exact.

(4) By Lemma 1.2 (2) of the lesson on torsors, $\psi_s$ is an isomorphism of $G_V$-torsors from the trivial torsor. By Proposition 6.3 (3), pullback along $\psi_s$ is an isomorphism of de Rham complexes $\omega^\bullet_{X_V/V}\cong\omega^\bullet_{G_V/V}$. Here $N(G,G_V)$ and $N(G,V)$ are the nerves of $G_V$ acting on itself and of $G_V$, formed over $V$, so $\omega^\bullet_{G_V/V}$ is the complex of left-invariant forms of the group scheme $G_V$ over $V$. Its transpose is an isomorphism of Lie algebras by Proposition 4.5. Next, $\psi_{hs}(k)=khs=\psi_s(\rho_h(k))$, and $\rho_h$ is an automorphism of the trivial torsor, as it commutes with left translations; so $\psi_{hs}^*=\rho_h^*\circ\psi_s^*$, with $\rho_h^*$ an automorphism of de Rham complexes by Proposition 6.3 (3).

Let $\sigma$ be a left-invariant $1$-form of $G_V$ over $V$. Pull back the equality $\mathrm{act}^\natural\sigma=\mathrm{pr}^\natural\sigma$ along $j\colon G_V\to G\times_SG_V$, $k\mapsto(h^{-1},k)$, which lies over the point $h^{-1}\colon V\to G$; since $\mathrm{act}\circ j$ is the left translation $L_{h^{-1}}$ and $\mathrm{pr}\circ j=\mathrm{id}$, this gives $L_{h^{-1}}^*\sigma=\sigma$. Hence $\rho_h^*\sigma=\rho_h^*L_{h^{-1}}^*\sigma=\operatorname{int}(h^{-1})^*\sigma,$ as $L_{h^{-1}}\circ\rho_h=\operatorname{int}(h^{-1})$, and since $\operatorname{int}(h^{-1})$ fixes the unit section, $e_V^*(\rho_h^*\sigma)=c(\operatorname{int}(h^{-1}))(e_V^*\sigma)$ with $c(\rho)=e_V^*(d\rho)$. $\square$

7. Exercises

Exercise 7.1. Let $K$ be a ring, $A=K[x,y]$ and $E=\Omega^1_{A/K}/A\,dx$. Show that the derivation $d_0=\pi\circ d_{A/K}$ extends to a unique de Rham complex over $A/K$, and that this complex is $\Omega^\bullet_{A/K[x]}$. Describe it explicitly.

Exercise 7.2. Let $K=\mathbf Z$, $A=\mathbf Z[t]$ and $E=(A/2A)\bar e_1\oplus(A/2A)\bar e_2$. Show that $E^\vee=0$, that $\Lambda^2_AE$ is free over $A/2A$ on $\bar e_1\wedge\bar e_2$, and that there is a $K$-antiderivation $d$ of degree $1$ of $\Lambda_AE$ with $d(t)=\bar e_1$ and $d^2(t)=\bar e_1\wedge\bar e_2$. Conclude that Theorem 5.2 fails without the local freeness of $E$.

Exercise 7.3. Let $G=GL_{n,S}$, with the matrix of coordinates $g=(g_{ij})$, and let $\omega=g^{-1}dg$ be the matrix of $1$-forms $\omega_{ij}=\sum_k(g^{-1})_{ik}\,dg_{kj}$. Show that the $\omega_{ij}$ are left-invariant and form a basis of $\omega^1_{G/S}$, that $d\omega=-\omega\wedge\omega$, and that $x\mapsto(x(\omega_{ij}))_{i,j}$ is an isomorphism of Lie algebras from $t_G$ onto the $n\times n$ matrices over $\mathcal O_S$ with the commutator $xy-yx$. Compute $D_x(g)$.

8. Complete solutions

Solution 7.1. Here $N=A\,dx$ is generated by $dx$, and $d(dx)=0$ in $\Omega^2_{A/K}$. By Proposition 2.4 there is a unique de Rham complex $(\Lambda E,d)$ with $d|_A=d_0$. The module $\Omega^1_{A/K[x]}$ is the quotient of $\Omega^1_{A/K}$ by the image of $\Omega^1_{K[x]/K}$, that is, by $A\,dx$ [Tag 00RS], so $E=\Omega^1_{A/K[x]}$ and $d_0=d_{A/K[x]}$. The complex $\Omega^\bullet_{A/K[x]}$ is a de Rham complex over $A/K[x]$, hence over $A/K$, with differential $d_0$ on $A$; by uniqueness it is $(\Lambda E,d)$. Explicitly $E$ is free on $\bar e_y=d_0y$, $\Lambda^2_AE=0$, and the complex is $A\to A\bar e_y$, $p\mapsto(\partial p/\partial y)\,\bar e_y$.

Solution 7.2. $E$ is killed by $2$ and $A$ is torsion free, so every $A$-linear map $E\to A$ is zero. By the description of exterior algebras of quotients in the proof of Proposition 2.4, applied to $Ae_1\oplus Ae_2\to E$, $\Lambda^2_AE$ is $A\,e_1\wedge e_2$ modulo $2A\,e_1\wedge e_2$ (the images of $2e_1\wedge e_2$ and $e_1\wedge2e_2$), that is, $(A/2A)\,\bar e_1\wedge\bar e_2$. Let $d_0\colon A\to E$, $d_0p=(p'\bmod2)\,\bar e_1$, with $p'$ the derivative; it is a $\mathbf Z$-derivation. On the free module $Ae_1\oplus Ae_2$ put $\begin{aligned}d_1(p_1e_1+p_2e_2)&=d_0p_1\wedge\bar e_1+d_0p_2\wedge\bar e_2\\&\quad+\bar p_1\,\bar e_1\wedge\bar e_2.\end{aligned}$ It is additive, and it vanishes on $2pe_1$ and $2pe_2$, every term being a multiple of $2$ in $\Lambda^2_AE$, which is killed by $2$. So it defines $d_1\colon E\to\Lambda^2_AE$, and $d_1(ax)=a\,d_1x+d_0a\wedge x$ holds by construction. Proposition 2.2 gives $d$, with $d(t)=\bar e_1$ and $d^2(t)=d_1(\bar e_1)=\bar e_1\wedge\bar e_2\neq0$. The bracket on $E^\vee=0$ satisfies the Jacobi identity, but $d^2\neq0$.

Solution 7.3. $\Omega^1_{G/S}$ is free on the $dg_{ij}$, and $\omega=g^{-1}dg$ is obtained from $dg$ by the invertible matrix $g^{-1}$, so the $\omega_{ij}$ form a basis of $\Omega^1_{G/S}$. In $\Omega^1_{G\times_SG/G}$, where the first factor $h$ is constant, $\mathrm{act}^\natural\omega=(hg)^{-1}d(hg)=$ $g^{-1}h^{-1}h\,dg=\mathrm{pr}^\natural\omega$, so the $\omega_{ij}$ are invariant. The map $\mathcal O_S^{n^2}\to\omega^1_{G/S}$ that they define becomes the isomorphism onto $\Omega^1_{G/S}$ after $f^*$ (Proposition 6.3 (1)); $f$ being flat and surjective, it is an isomorphism, as in the proof of Theorem 6.5 (1). From $g^{-1}g=1$, $d(g^{-1})=-g^{-1}\,dg\,g^{-1}$, and by (2.5) $\begin{aligned}d\omega&=d(g^{-1})\wedge dg\\&=-g^{-1}dg\,g^{-1}\wedge dg=-\omega\wedge\omega,\end{aligned}$ that is, $d\omega_{ij}=-\sum_k\omega_{ik}\wedge\omega_{kj}$. For $x$, $y$ in $t_G$ write $x_{ij}=x(\omega_{ij})$. By (4.4) with $\theta=0$ and Lemma 4.2 (3), $\begin{aligned}[x,y](\omega_{ij})&=-(d\omega_{ij})(x,y)\\&=\sum_k(x_{ik}y_{kj}-y_{ik}x_{kj})\\&=(xy-yx)_{ij}.\end{aligned}$ The $\omega_{ij}$ form a basis, so $x\mapsto(x_{ij})$ is an isomorphism of Lie algebras onto the matrices with the commutator. Finally $\langle D_x,\omega\rangle=x$ by Proposition 6.4 (1), and $dg=g\,\omega$, so $D_x(g)=g\,x$: $D_x$ is the left-invariant vector field with value $x$ at the unit.

Sources

De Rham complexes over $A/K$, the Lie algebroid of a de Rham complex, the equivalence of Theorem 5.3 and the de Rham complexes of invariant forms on torsors, with their Atiyah sequence, are due to Illusie, in his work on the cotangent complex. Proposition 2.4 is the criterion of Stacks, Tag 07HY for rings, extended to topoi and with its converse. Lie algebroids over $A/K$ are also called Lie–Rinehart algebras; formula (4.5) is the Chevalley–Eilenberg formula, and (6.1) is the algebraic form of the sequence introduced by Atiyah for principal bundles.