The de Rham complex of a formal category with divided powers

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

The cochains of a cosimplicial algebra form a differential graded algebra under the Alexander–Whitney product: a cochain of degree $p$ is pushed to the first $p+1$ vertices, one of degree $q$ to the last $q+1$, and the two are multiplied. For the nerve of a formal category with arrows $P$ and augmentation ideal $I$, the normalized cochains begin $A\to I\to I\otimes_AI$, and the tensor algebra of $I$ maps to them. Dividing $I$ by its second divided power gives a module of differentials $\Omega^1_R=I/I^{[2]}$, and the differential of the cochains descends to an antiderivation of $\Lambda\Omega^1_R$, the exterior derivative. This lesson proves these constructions, shows that the exterior derivative squares to zero in three cases (equal source and target maps on $A$, $\Omega^1_R$ generated by differentials of functions, $\Omega^1_R$ flat), and identifies the result for the Čech–Alexander complex of $A/K$ with the de Rham complex $\Omega^\bullet_{A/K}$, and for the part of the Čech–Alexander complex of a torsor under a group scheme invariant under the group with the de Rham complex of invariant differential forms.

Prerequisites. Formal categories with divided powers and the Čech–Alexander complex (its notation throughout: 1.4, Proposition 1.6 with (1.1)–(1.4), Lemmas 2.3, 2.4, Proposition 2.6, Definition 3.5, Example 3.6); De Rham complexes and Lie algebroids (Lemma 1.2, Propositions 2.2–2.4, Construction 6.1, Lemma 6.2, Proposition 6.3); Torsors, co-Lie complexes and the Atiyah class (Lemma 3.1, Theorem 2.5, Corollary 2.6). From the AI-integrated Stacks project: the module of differentials as the conormal module of the diagonal [Stacks, Tag 00RW] and divided powers [Tag 07GL]; flat base change and localization of divided power envelopes [Tag 07HD].

Conventions

The conventions are those of the lesson on formal categories with divided powers. For an affine $K$-category $C=(A,P,d^0,d^1,\varepsilon,\delta)$, the nerve $R=\operatorname{Ner}(C)$ has $R^0=A$ and $R^n=P\otimes_A\cdots\otimes_AP$. In $P\otimes_AP$ the first factor is an $A$-algebra through $d^0$ and the second through $d^1$; so $I$ is an $A$-bimodule, with the left action $a\cdot x=d^1(a)x$ and the right action $x\cdot a=x\,d^0(a)$, and $I\otimes_AI$ is formed with these actions. For a strict formal category everything below is applied levelwise.

1. The Alexander–Whitney algebra of a cosimplicial algebra

Let $R$ be a cosimplicial $K$-algebra, with cofaces $d^i$ and codegeneracies $s^i$. For an increasing map $\alpha\colon[p]\to[n]$ write $\alpha_*\colon R^p\to R^n$ for the induced ring map; $d^i=(\delta_i)_*$, where $\delta_i\colon[n]\to[n+1]$ omits $i$, and $s^i=(\sigma_i)_*$, where $\sigma_i\colon[n+1]\to[n]$ takes the value $i$ twice. The front and back maps are $\begin{aligned}\phi_{p,q}&\colon[p]\to[p+q],\ i\mapsto i,\\ \beta_{p,q}&\colon[q]\to[p+q],\ j\mapsto p+j.\end{aligned}$

Definition 1.1. The Alexander–Whitney algebra $\tilde R$ of $R$ is the cochain complex with $\tilde R^n=R^n$ and $d=\sum_{i=0}^{n+1}(-1)^id^i\colon R^n\to R^{n+1}$, with the product $x\cdot y=(\phi_{p,q})_*(x)\,(\beta_{p,q})_*(y)$ for $x\in R^p$ and $y\in R^q$, the product on the right being taken in $R^{p+q}$. The normalized complex $NR$ has $N^nR=\bigcap_{i=0}^{n-1}\ker(s^i\colon R^n\to R^{n-1})$.

Proposition 1.2. $\tilde R$ is a differential graded $K$-algebra: the product is associative and unital, with unit $1\in R^0$, and $\begin{aligned}d(x\cdot y)&=dx\cdot y\\&\quad+(-1)^px\cdot dy\end{aligned} \tag{1.1}$ for $x\in R^p$. The normalized complex $NR$ is a sub-differential graded algebra.

Proof. Associativity and unit. For $x\in R^p$, $y\in R^q$, $z\in R^r$, both $(x\cdot y)\cdot z$ and $x\cdot(y\cdot z)$ equal $(\phi_{p,q+r})_*(x)\,\gamma_*(y)\,(\beta_{p+q,r})_*(z),$ where $\gamma\colon[q]\to[p+q+r]$, $j\mapsto p+j$. This holds because the push-forwards are ring maps and $\phi_{p+q,r}\phi_{p,q}=\phi_{p,q+r}$, $\phi_{p+q,r}\beta_{p,q}=\gamma=\beta_{p,q+r}\phi_{q,r}$, $\beta_{p,q+r}\beta_{q,r}=\beta_{p+q,r}$. The unit: $(\phi_{0,q})_*(1)=1$ and $\beta_{0,q}=\mathrm{id}$.

Leibniz rule. Let $n=p+q$. For $0\leq i\leq p$, $\delta_i\phi_{p,q}=\phi_{p+1,q}\delta_i$ and $\delta_i\beta_{p,q}=\beta_{p+1,q}$, as $\delta_i$ shifts all values $\geq i$; so $d^i(x\cdot y)=(d^ix)\cdot y$. For $p+1\leq i\leq n+1$, $\delta_i\phi_{p,q}=\phi_{p,q+1}$ and $\delta_i\beta_{p,q}=\beta_{p,q+1}\delta_{i-p}$; so $d^i(x\cdot y)=x\cdot(d^{i-p}y)$. Moreover $(d^{p+1}x)\cdot y=x\cdot(d^0y)$, both being $(\phi_{p,q+1})_*(x)\,(\beta_{p+1,q})_*(y)$. Hence $\begin{aligned}d(x\cdot y)&=\sum_{i=0}^p(-1)^i(d^ix)\cdot y\\&\quad+\sum_{j=1}^{q+1}(-1)^{p+j}x\cdot(d^jy),\end{aligned}$ while $dx\cdot y+(-1)^px\cdot dy$ has the two further terms $(-1)^{p+1}(d^{p+1}x)\cdot y$ and $(-1)^px\cdot(d^0y)$, which cancel.

Normalized cochains. For $x\in N^nR$ and $j\leq n$, the cosimplicial identities $s^jd^i=d^is^{j-1}$ ($i<j$), $s^jd^j=s^jd^{j+1}=\mathrm{id}$ and $s^jd^i=d^{i-1}s^j$ ($i>j+1$) give $s^j(dx)=(-1)^jx+(-1)^{j+1}x=0$; so $NR$ is a subcomplex. For $x\in N^pR$, $y\in N^qR$ and $i<p+q$: if $i<p$, then $\sigma_i\phi_{p,q}=\phi_{p-1,q}\sigma_i$ and $\sigma_i\beta_{p,q}=\beta_{p-1,q}$, so $s^i(x\cdot y)=(s^ix)\cdot y=0$; if $i\geq p$, then $\sigma_i\phi_{p,q}=\phi_{p,q-1}$ and $\sigma_i\beta_{p,q}=\beta_{p,q-1}\sigma_{i-p}$, so $s^i(x\cdot y)=x\cdot(s^{i-p}y)=0$. $\square$

Example 1.3. For the nerve of an affine category, the front map of $x\in R^p$ appends $p$ copies of $1$ on the right and the back map of $y\in R^q$ prepends them on the left, by (1.2) of the lesson on formal categories. So the product is concatenation: $\begin{aligned}&(x_1\otimes\cdots\otimes x_p)\cdot(y_1\otimes\cdots\otimes y_q)\\&\quad=x_1\otimes\cdots\otimes x_p\otimes y_1\otimes\cdots\otimes y_q,\end{aligned}$ and for $a\in R^0=A$, $a\cdot x=d^1(a)x_1\otimes\cdots$ and $x\cdot a=x_1\otimes\cdots\otimes x_pd^0(a)$: the bimodule actions of the Conventions. In degrees $0$ and $1$, $\begin{aligned}da&=d^0(a)-d^1(a),\\ dx&=1\otimes x-\delta(x)\\&\quad+x\otimes1.\end{aligned} \tag{1.2}$

2. The normalized nerve and the tensor algebra

Proposition 2.1. Let $R$ be the nerve of an affine $K$-category. Then $N^0R=A$, $N^1R=I$, and $N^2R$ is the image of $I\otimes_AI$, which maps isomorphically onto it. More generally the image of $I^{\otimes n}=I\otimes_A\cdots\otimes_AI$ in $R^n$ lies in $N^nR$.

Proof. $N^1R=\ker(s^0)=\ker\varepsilon=I$. By (1.4) of the lesson on formal categories, $R^2=A\oplus I\oplus I\oplus(I\otimes_AI)$, the first factor decomposed as a right and the second as a left $A$-module. The codegeneracy $s^0$ deletes the first factor, $s^1$ the second, and $s^0$ is injective on $A\oplus I_{(2)}$ and zero on $I_{(1)}\oplus I\otimes_AI$, where $I_{(k)}$ is the copy in the $k$-th factor; similarly for $s^1$. So $\ker s^0\cap\ker s^1=I\otimes_AI$. In general, $s^i$ applies $\varepsilon$ to the factor $i+1$, which kills any tensor with that factor in $I$. $\square$

Construction 2.2. Let $T_A(I)=\bigoplus_{n\geq0}I^{\otimes n}$ be the tensor algebra of the $A$-bimodule $I$, a graded $K$-algebra, not commutative, with $I^{\otimes0}=A$. By Example 1.3, the restrictions $d_0\colon A\to I$ and $d_1\colon I\to I\otimes_AI$ of the differential (1.2) satisfy, by (1.1) and Proposition 2.1, $\begin{aligned}d_0(ab)&=d_0(a)\cdot b\\&\quad+a\cdot d_0(b),\\ d_1(a\cdot x)&=d_0(a)\otimes x\\&\quad+a\cdot d_1(x),\\ d_1(x\cdot a)&=d_1(x)\cdot a\\&\quad-x\otimes d_0(a).\end{aligned} \tag{2.1}$

Lemma 2.3. There is a unique $K$-antiderivation $d$ of degree $1$ of $T_A(I)$ restricting to $d_0$ and $d_1$. On $I^{\otimes n}$, $\begin{aligned}&d(x_1\otimes\cdots\otimes x_n)\\&=\sum_i(-1)^{i-1}x_1\otimes\cdots\\&\quad\otimes d_1x_i\otimes\cdots\otimes x_n.\end{aligned} \tag{2.2}$ Write $N'R$ for $T_A(I)$ with this antiderivation.

Proof. As in Proposition 2.2 of the lesson on de Rham complexes, the right side of (2.2) is additive in each variable, and it is balanced. Let $t=x_1\otimes\cdots\otimes x_i\otimes d_0(a)$ $\otimes x_{i+1}\otimes\cdots\otimes x_n$. Replacing $(x_i\cdot a,x_{i+1})$ by $(x_i,a\cdot x_{i+1})$ adds $(-1)^{i-1}t$ to the term $i$, by the third line of (2.1), and $(-1)^it$ to the term $i+1$, by the second; these cancel, and the other terms do not change. So (2.2) defines $d$ on $I^{\otimes n}$, first on the presheaf tensor products, then after sheafification. With $d=d_0$ on $A$, the rule $d(st)=ds\cdot t+(-1)^{m}s\cdot dt$ for $s$ of degree $m$ holds on products of pure tensors by (2.2), and for $s$ or $t$ in $A$ by the second and third lines of (2.1). Uniqueness: two such antiderivations agree on $A$ and $I$, which generate $T_A(I)$ (Lemma 1.2 (4) of the lesson on de Rham complexes, which does not use commutativity). $\square$

Proposition 2.4. The maps $I^{\otimes n}\to N^nR$ of Proposition 2.1 form a map $u\colon N'R\to NR$ of differential graded algebras, an isomorphism in degrees $\leq2$. If $d^0=d^1\colon A\to P$, then $u$ is an isomorphism.

Proof. By Example 1.3, $u$ is multiplicative. The maps $u\circ d$ and $d\circ u$ are $u$-antiderivations $T_A(I)\to NR$ of degree $1$ (by (1.1) for $d\circ u$), equal on $A$ and $I$ by construction, hence equal. Proposition 2.1 gives the degrees $\leq2$.

Let $d^0=d^1$. Then the left and right actions of $A$ on $P$ agree, and $P=d^0(A)\oplus I$ is a decomposition of bimodules. Decomposing every factor, $R^n=\bigoplus_{S\subseteq\{1,\ldots,n\}}I^{\otimes S},$ where $I^{\otimes S}$ has $I$ in the places of $S$ and $A$ elsewhere. The codegeneracy $s^i$ is zero on the summands with $i+1\in S$ and maps the others injectively, onto distinct summands of $R^{n-1}$. So $N^nR=\bigcap_i\ker s^i$ is the summand $I^{\otimes\{1,\ldots,n\}}$, which is the image of $I^{\otimes n}$, mapped injectively. $\square$

Without $d^0=d^1$, the bimodule $P$ need not split, and for $n\geq3$ the comparison of $I^{\otimes n}$ with $N^nR$ is not covered by these arguments.

3. The module of differentials and the exterior derivative

From here on $C$ is a PD formal category: $I$ has a divided power structure $\gamma$ for which $\delta$ is a PD morphism (Proposition 2.6 of the lesson on formal categories). It is affine or strict, or the PD completion of an affine or strict PD formal category $C_0$ (Proposition 2.7 there), which need not be strict. In the last case everything below is formed from $C_0$: the completion does not change the quotients $R/J^{[r]}$ or the maps between them induced by the structure maps, and the constructions below only involve these quotients for $r\leq3$. Example 3.6 applies the same description through the quotients to a $G$-invariant direct image.

Definition 3.1. The module of differentials of $R$ is $\Omega^1_R=I/I^{[2]}$. The left and right actions of $A$ on it agree: $d^0(a)-d^1(a)$ lies in $I$, and $I\cdot I\subseteq I^{[2]}$. So $\Omega^1_R$ is an $A$-module. Put $\Omega^\bullet_R=\Lambda_A\Omega^1_R$, and let $p\colon T_A(I)\to\Omega^\bullet_R$ be the map of graded $K$-algebras extending the projection $I\to\Omega^1_R$, $p(x_1\otimes\cdots\otimes x_n)=\bar x_1\wedge\cdots\wedge\bar x_n$; it is defined because the two actions of $A$ agree on $\Omega^1_R$.

Proposition 3.2. For the Čech–Alexander complex $R=\hat C_{A/K}$, $\Omega^1_R=\Omega^1_{A/K}$: the class of $d^0(a)-d^1(a)=1\otimes a-a\otimes1$ corresponds to $d_{A/K}a$.

Proof. Let $J\subseteq A\otimes_KA$ be the ideal of the diagonal and $D=D_{A\otimes A}(J)$, with $\bar J$. In the completion, $I/I^{[2]}$ is the pro-object of the $(\bar J/\bar J^{[r]})/(\bar J^{[2]}/\bar J^{[r]})$, constant equal to $\bar J/\bar J^{[2]}$ for $r\geq2$. The map $J/J^2\to\bar J/\bar J^{[2]}$ is an isomorphism. It is surjective: $\bar J$ is generated by the $\bar\gamma_n(x)$, $x\in J$ [Tag 07HA], and these lie in $\bar J^{[2]}$ for $n\geq2$. It has an inverse: the ideal $J/J^2$ of $A\otimes A/J^2$ has square zero, so $\gamma_n=0$ for $n\geq2$ is a divided power structure on it. The projection $A\otimes A\to A\otimes A/J^2$ therefore factors through a PD morphism $D\to A\otimes A/J^2$, which maps $\bar J^{[2]}$ to $(J/J^2)^{[2]}=0$ and induces $\bar J/\bar J^{[2]}\to J/J^2$, inverse to the first map on the generators. Finally $J/J^2\cong\Omega^1_{A/K}$ with $1\otimes a-a\otimes1\mapsto d_{A/K}a$ [Tag 00RW], on sections and after sheafification. $\square$

Lemma 3.3. $p(d(I^{[2]}))=0$, where $d\colon I\to I\otimes_AI$ is $d_1$.

Proof. Let $E=I\otimes_AI^{[2]}+I^{[2]}\otimes_AI$; then $p(E)=0$, as $p$ kills $I^{[2]}$. For a local section $x$ of $I$, $dx\in I\otimes_AI$ by Proposition 2.1, so by (1.2) $\delta x\equiv x\otimes1+1\otimes x$ modulo $I\otimes_AI$. For local sections $x,y$ of $I$, as $\delta$ is a ring map, $\begin{aligned}\delta(xy)&\equiv xy\otimes1+x\otimes y\\&\quad+y\otimes x+1\otimes xy\end{aligned}$ modulo $I\otimes I^2+I^2\otimes I\subseteq E$, and as $\delta$ is a PD morphism and $\gamma_n(x\otimes1+1\otimes x)=$ $\sum_{i+j=n}\gamma_i(x)\otimes\gamma_j(x)$, $\begin{aligned}\delta(\gamma_2x)&\equiv\gamma_2x\otimes1+x\otimes x\\&\quad+1\otimes\gamma_2x,\\ \delta(\gamma_nx)&\equiv\gamma_nx\otimes1+1\otimes\gamma_nx\\&\qquad(n\geq3)\end{aligned}$ modulo $E$; here $\gamma_n(\delta x)=\gamma_n(x\otimes1+1\otimes x+w)$ with $w\in I\otimes I$, and the terms involving $w$ lie in $E$. By (1.2), $\begin{aligned}d(xy)&\equiv-(x\otimes y+y\otimes x),\\ d(\gamma_2x)&\equiv-x\otimes x,\\ d(\gamma_nx)&\equiv0\quad(n\geq3)\end{aligned}$ modulo $E$, and $p$ kills the right sides: $\bar x\wedge\bar y+\bar y\wedge\bar x=0$ and $\bar x\wedge\bar x=0$. The ideal $I^{[2]}$ is generated as a sheaf of abelian groups by the products $a\,w$ with $a$ in $P$ and $w$ a product $xy$ or a $\gamma_n(x)$, $n\geq2$. By (0.1) of the lesson on formal categories, $a=d^1(\alpha)+x'$ with $\alpha\in A$ and $x'\in I$, and $x'w\in I^2$ is a sum of products $xy$. Finally $d(\alpha\cdot w)=d_0(\alpha)\otimes w+\alpha\cdot dw$ by (2.1), with $d_0(\alpha)\otimes w\in E$. $\square$

Proposition 3.4 (the exterior derivative). There is a unique $K$-antiderivation $d$ of degree $1$ of $\Omega^\bullet_R$ with $d\circ p=p\circ d$ on $N'R$. It satisfies $d^2=0$ on $A$, and $(\Omega^\bullet_R,d)$ is a de Rham complex over $A/K$ in each of the following cases:

$d^0=d^1\colon A\to P$;
$\Omega^1_R$ is generated as an $A$-module by the classes of the $d^0(a)-d^1(a)$;
$\Omega^1_R$ is a flat $A$-module.

Proof. By Lemma 3.3, $p\circ d_1\colon I\to\Omega^2_R$ factors through a $K$-linear map $\bar d_1\colon\Omega^1_R\to\Omega^2_R$; put $\bar d_0=p\circ d_0\colon A\to\Omega^1_R$. Applying $p$ to (2.1) gives $\bar d_0(ab)=a\,\bar d_0b+b\,\bar d_0a$ and $\bar d_1(a\bar x)=a\,\bar d_1\bar x+\bar d_0a\wedge\bar x$, and $\bar d_1\bar d_0=p\,d_1d_0=0$, because $N'R$ and $NR$ agree in degrees $\leq2$ (Proposition 2.4) and $NR$ is a complex. By Proposition 2.2 of the lesson on de Rham complexes, $(\bar d_0,\bar d_1)$ extends uniquely to a $K$-antiderivation $d$ of degree $1$ of $\Omega^\bullet_R$, and $d^2=0$ on $A$. The maps $d\circ p$ and $p\circ d$ are $p$-antiderivations of $T_A(I)$ that agree on $A$ and $I$, hence are equal; any $d$ with $dp=pd$ has these restrictions, which gives uniqueness. By the same proposition, $d^2=0$ once it vanishes on $\Omega^1_R$.

(1) By Proposition 2.4, $N'R\cong NR$ is a complex; so $d^2p=pd^2=0$, and $p$ is surjective.

(2) $d(a\,\bar d_0b)=\bar d_0a\wedge\bar d_0b$ and $d(\bar d_0a\wedge\bar d_0b)=0$, as $d^2=0$ on $A$; so $d^2$ vanishes on generators of $\Omega^1_R$.

(3) Let $P^1=P/I^{[2]}$, with the exact sequence $0\to\Omega^1_R\to P^1\xrightarrow{\ \varepsilon\ }A\to0, \tag{3.1}$ split as a right $A$-module sequence by $d^0$ and as a left one by $d^1$. Tensor products are over $A$. The inclusion $\Omega^1_R\otimes\Omega^1_R\to P^1\otimes\Omega^1_R$ is split, by the splitting of the first factor through $d^0$. Tensoring on the left with the flat module $\Omega^1_R$ keeps it injective, so $\Omega^{1\otimes3}_R\to\Omega^1_R\otimes P^1\otimes\Omega^1_R$ is injective. Further $\Omega^1_R\otimes P^1\otimes\Omega^1_R\to P^1\otimes P^1\otimes P^1$ is split injective, by the splittings of the first factor through $d^0$ and of the last through $d^1$. So $\Omega^1_R\otimes\Omega^1_R\otimes\Omega^1_R\longrightarrow P^1\otimes P^1\otimes P^1$ is injective. For a local section $x$ of $I$, the image of $d^2x\in I^{\otimes3}$ in $P^{\otimes3}=R^3$ is the image of $x$ under the square of the differential of $\tilde R$ (Proposition 2.4), which is zero. Its image in $P^1\otimes P^1\otimes P^1$ is therefore zero. This image is also that of the class of $d^2x$ in $\Omega^{1\otimes3}_R$, so that class is zero, and so is its image $p(d^2x)=d^2\bar x$ in $\Omega^3_R$. $\square$

The proposition does not say whether $d^2=0$ on $\Omega^\bullet_R$ for every PD formal category; the three cases cover the examples of this course.

Example 3.5. For the Čech–Alexander complex $R=\hat C_{A/K}$, Proposition 3.2 gives $\Omega^1_R=\Omega^1_{A/K}$ with $\bar d_0=d_{A/K}$, so case (2) applies. By Proposition 2.4 of the lesson on de Rham complexes, a de Rham complex on $\Omega^1_{A/K}$ extending $d_{A/K}$ is unique; so $(\Omega^\bullet_R,d)$ is the de Rham complex $\Omega^\bullet_{A/K}$. For $A=K[x]$ (Example 3.6 of the lesson on formal categories), $\Omega^1_R$ is free on the class $\bar\tau$ of $\tau=1\otimes x-x\otimes1$, and $\bar d_0(a)$ is the class of $a(x+\tau)-a(x)=\sum_{n\geq1}a^{(n)}(x)\tau^{[n]}$, which is $a'(x)\bar\tau$.

Example 3.6 (invariant differential forms). Let $G$ be a group scheme over a scheme $S$, flat and locally of finite presentation, $f\colon X\to Y$ a $G_Y$-torsor and $\varpi\colon Y\to S$ a morphism, with the nerve $a\colon N(G,X)\to Y$ and the de Rham complex $\omega^\bullet_{X/S}$ of invariant forms of Construction 6.1 and Proposition 6.3 of the lesson on de Rham complexes and Lie algebroids. For $n\geq0$ let $R_n$ be the Čech–Alexander complex of $\mathcal O_{N_n(G,X)}$ over the inverse image of $\mathcal O_{N_n(G)}$, on $N_n(G,X)_{\mathrm{Zar}}$, with arrows $P_n$ and augmentation ideal $I_n$.

The quotients $P_n/I_n^{[r]}$, and the corresponding quotients of the terms of the nerves, are quasi-coherent modules through each factor. For $\alpha\colon[m]\to[n]$, the square formed by $N(G,X)(\alpha)$ and $N(G)(\alpha)$ induces isomorphisms $N(G,X)(\alpha)^*(P_m/I_m^{[r]})\cong P_n/I_n^{[r]}$, compatible with all the structure maps and divided powers. So these quotients form cartesian quasi-coherent modules on $N(G,X)_{\mathrm{Zar}}$.
Their direct images under $a$ form the nerve of a PD-adic formal category $R_{X/Y}$ over $K=\varpi^{-1}\mathcal O_S$, with objects $a_*\mathcal O=\mathcal O_Y$: the $G$-invariant direct image of the Čech–Alexander complex of $X/S$.
$\Omega^1_{R_{X/Y}}=\omega^1_{X/S}$, and $(\Omega^\bullet_{R_{X/Y}},d)=(\omega^\bullet_{X/S},d)$ as de Rham complexes over $\mathcal O_Y/K$.
For $Y=S$ and $X=G$ acting on itself by left translations, $d^0=d^1$ on $\mathcal O_S$, case (1) of Proposition 3.4 applies, and $\Omega^\bullet_{R_{G/S}}$ is the complex $\omega^\bullet_{G/S}$ of left-invariant forms of $G$.

Proof. (1) Let $U=\operatorname{Spec}A$ be an affine open of $N_n(G,X)$ over an affine open $\operatorname{Spec}B$ of $N_n(G)$, $J$ the ideal of the diagonal of $A\otimes_BA$ and $D$ its divided power envelope. On $U$, $P_n/I_n^{[r]}$ is the sheaf associated with $D/\bar J^{[r]}$: for $g\in A$, the sections over $D(g)$ come from $A_g\otimes_BA_g$, whose envelope is the localization of $D$ at $g\otimes g$ [Tag 07HD], and modulo $\bar J^{[r]}$ inverting $g\otimes g$ is inverting $g$ through either factor, since $g\otimes1-1\otimes g\in\bar J$ and every $x\in\bar J$ satisfies $x^r=r!\,\gamma_r(x)\in\bar J^{[r]}$. So $P_n/I_n^{[r]}$ is quasi-coherent through each factor, and so are the quotients of the tensor products forming the nerve. The square of the statement is cartesian (Lemma 3.1 (1) of the lesson on torsors, as in the proof of Lemma 6.2 of the lesson on de Rham complexes and Lie algebroids), and $N(G)(\alpha)$ is flat, $G$ being flat over $S$. On affine pieces it is given by a flat map $B\to B'$ and $A'=A\otimes_BB'$; then $A'\otimes_{B'}A'=(A\otimes_BA)\otimes_BB'$ is flat over $A\otimes_BA$ and its diagonal ideal is generated by $J$, so its envelope is $D\otimes_BB'$, with the extended divided powers [Tag 07HD]. Hence $N(G,X)(\alpha)^*(P_m/I_m^{[r]})\cong P_n/I_n^{[r]}$, and the identifications are compatible with $d^0$, $d^1$, $\varepsilon$, $\delta$ and the divided powers, all being induced by the same maps of rings.

(2) By Theorem 2.5 and Corollary 2.6 (1) of the lesson on torsors, applied in degree $0$, $a^*$ is an equivalence from quasi-coherent $\mathcal O_Y$-modules onto cartesian quasi-coherent modules on $N(G,X)_{\mathrm{Zar}}$, with quasi-inverse $a_*$. It commutes with tensor products, so $a_*$ carries the tensor products forming the nerves of the $R_n$ to tensor products over $\mathcal O_Y$, and the module maps among them (products, cofaces, codegeneracies) to module maps. The divided powers $\gamma_m\colon I_n\to I_n$ are maps of sheaves of sets compatible with the transition maps, so $a_*$ carries them to maps $a_*I\to a_*I$ satisfying the same identities. Corollary 2.6 (1) there gives $a_*\mathcal O=\mathcal O_Y$. So the direct images form the nerve of a PD formal category, PD-adic because each $R_n$ is (Proposition 2.7 of the lesson on formal categories).

(3) By Proposition 3.2, $I_n/I_n^{[2]}=\Omega^1_{N_n(G,X)/N_n(G)}$, compatibly with the transition maps, so $\Omega^1_{R_{X/Y}}=a_*\Omega^1_{N(G,X)/N(G)}=\omega^1_{X/S}$, and exterior powers correspond (Proposition 6.3 (1) there). The exterior derivative of Proposition 3.4 is determined by the cochain differential of the nerve and the projection $p$, which $a^*$ carries to those of the $R_n$; by Example 3.5 these give the de Rham differentials of $N_n(G,X)/N_n(G)$, whose direct image is the differential of $\omega^\bullet_{X/S}$ (Proposition 6.3 (2) there).

(4) For $Y=S$, the map $a$ factors through $N(G)\to S$, so a section $s$ of $\mathcal O_S$ acts on $P_n$ through the inverse image of $\mathcal O_{N_n(G)}$, over which the tensor product is taken: $1\otimes s=s\otimes1$. $\square$

4. Exercises

Exercise 4.1. In Example 3.5 with $A=K[x]$, compute $d\tau^{[n]}\in N^2R$ for $n\geq1$, and check the three congruences in the proof of Lemma 3.3 for $x=\tau$.

Exercise 4.2. Let $A$ be a $K$-algebra and $P=A\langle t\rangle$, with $d^0=d^1$ the inclusion of $A$, $\varepsilon(t^{[n]})=0$ for $n\geq1$, and $\delta(t^{[n]})=\sum_{i+j=n}t^{[i]}\otimes t^{[j]}$. Show that this is a PD formal category, that $N^nR=I^{\otimes n}$, and that $\Omega^\bullet_R=\Lambda_A(A\bar t)$ with $d=0$.

Exercise 4.3. Let $R=C_{A/K}$ with $A=\mathbf Q[x]$, so that $R^n=\mathbf Q[x_0,\ldots,x_n]$, and let $\tau=x_1-x_0\in N^1R$. Show that $\tau\cdot\tau\neq0$ in $\tilde R^2$. Conclude that the Alexander–Whitney algebra is not graded commutative, in contrast with $\Lambda_A\Omega^1_R$, where odd elements square to zero.

5. Complete solutions

Solution 4.1. Here $R^1=A\langle\tau\rangle$ and $R^2=A\langle\tau_1,\tau_2\rangle$ with $\tau_1=\tau\otimes1$, $\tau_2=1\otimes\tau$. By (1.2) and Example 3.6 of the lesson on formal categories, $\begin{aligned}d\tau^{[n]}&=\tau_2^{[n]}-\textstyle\sum_{i+j=n}\tau_1^{[i]}\tau_2^{[j]}+\tau_1^{[n]}\\&=-\textstyle\sum_{i,j\geq1,\,i+j=n}\tau^{[i]}\otimes\tau^{[j]}.\end{aligned}$ So $d\tau=0$, $d\tau^{[2]}=-\tau\otimes\tau$ (exactly, not only modulo $E$), and for $n\geq3$ every term has $i\geq2$ or $j\geq2$, so lies in $E$. For $x=y=\tau$: $d(\tau^2)=d(2\tau^{[2]})=$ $-2\tau\otimes\tau=-(\tau\otimes\tau+\tau\otimes\tau)$.

Solution 4.2. The axioms (1.1) of the lesson on formal categories hold: $\delta$ is the PD morphism extending $t\mapsto t\otimes1+1\otimes t$, coassociative and counital (as in Exercise 4.1 there). By Proposition 2.6 there, the structure on $I=A\langle t\rangle_+$ makes $C$ a PD formal category, $\delta$ being a PD morphism. By Proposition 2.4, $N^nR=I^{\otimes n}$. Here $I^{[2]}$ is spanned by the $t^{[n]}$, $n\geq2$, so $\Omega^1_R=A\bar t$, free of rank $1$, and $\Omega^\bullet_R=A\oplus A\bar t$. Since $d^0=d^1$, $\bar d_0=0$; and $dt=1\otimes t-\delta t+t\otimes1=0$, so $\bar d_1\bar t=0$. Hence $d=0$.

Solution 4.3. By Example 1.7 of the lesson on formal categories, $R^n=A^{\otimes(n+1)}=\mathbf Q[x_0,\ldots,x_n]$, with $d^i$ inserting $1$, that is, renumbering the variables $x_j\mapsto x_{j+1}$ for $j\geq i$. The front map of $\tau\in R^1$ into $R^2$ is $d^2$, which keeps $x_1-x_0$, and the back map is $d^0$, which gives $x_2-x_1$. So $\tau\cdot\tau=(x_1-x_0)(x_2-x_1)\neq0$. In a graded commutative algebra, $\tau\cdot\tau=-\tau\cdot\tau$ for $\tau$ of degree $1$, so $2\tau\cdot\tau=0$, and over $\mathbf Q$ this would give $\tau\cdot\tau=0$.

Sources

The Alexander–Whitney algebra of a cosimplicial algebra, the comparison of the normalized nerve of a formal category with the tensor algebra of its augmentation ideal, the module of differentials $I/I^{[2]}$ and the exterior derivative, with the three cases of Proposition 3.4 and Examples 3.5 and 3.6, are due to Illusie, in his work on the cotangent complex.