## 24.31 Resolutions of differential graded algebras

This section is the analogue of Differential Graded Algebra, Section 22.38.

Let $(\mathcal{C}, \mathcal{O})$ be a ringed site. As in Remark 24.23.5 consider a sheaf of graded sets $\mathcal{S}$ on $\mathcal{C}$. Let us think of the $r$-fold self product $\mathcal{S} \times \ldots \times \mathcal{S}$ as a sheaf of graded sets with the rule $\deg (s_1 \cdot \ldots \cdot s_ r) = \sum \deg (s_ i)$. Here given local sections $s_ i \in \mathcal{S}(U)$, $i = 1, \ldots , r$ we use $s_1 \cdot \ldots \cdot s_ r$ to denote the corresponding section of $\mathcal{S} \times \ldots \times \mathcal{S}$ over $U$. Let us denote $\mathcal{O}\langle \mathcal{S} \rangle $ the free graded $\mathcal{O}$-algebra on $\mathcal{S}$. More precisely, we set

\[ \mathcal{O}\langle \mathcal{S} \rangle = \mathcal{O} \oplus \bigoplus \nolimits _{r \geq 1} \mathcal{O}[\mathcal{S} \times \ldots \times \mathcal{S}] \]

with notation as in Remark 24.23.5. This becomes a sheaf of graded $\mathcal{O}$-algebras by concatenation

\[ (s_1 \cdot \ldots \cdot s_ r) (s'_1 \cdot \ldots \cdot s'_{r'}) = s_1 \cdot \ldots s_ r \cdot s'_1 \cdot \ldots \cdot s'_{r'} \]

We may endow $\mathcal{O}\langle \mathcal{S} \rangle $ with a differential by setting $\text{d}(s) = 0$ for all local sections $s$ of $\mathcal{S}$ and extending uniquely using the Leibniz rule although it is important to also consider other differentials.

Indeed, suppose that we are given a system of the following kind

for $i = 0, 1, 2, \ldots $ sheaves of graded sets $\mathcal{S}_ i$,

for $i = 0, 1, 2, \ldots $ maps

\[ \delta _{i + 1} : \mathcal{S}_{i + 1} \longrightarrow \mathcal{A}_ i = \mathcal{O}\langle \mathcal{S}_0 \amalg \ldots \amalg \mathcal{S}_ i\rangle \]

of sheaves of graded sets of degree $1$ whose image is contained in the kernel of the inductively defined differential on the target.

More precisely, we first set $\mathcal{A}_0 = \mathcal{O}\langle \mathcal{S}_0 \rangle $ and we endow it with the unique differential satisfying the Leibniz rule where $\text{d}(s) = 0$ for any local section $s$ of $\mathcal{S}$. By induction, assume given a differential $\text{d}$ on $\mathcal{A}_ i$. Then we extend it to the unique differential on $\mathcal{A}_{i + 1}$ satisfying the Leibniz rule and with

\[ \text{d}(s) = \delta (s) \]

where $\delta (s) = \delta _ j(s)$ if $s$ is in the summand $\mathcal{S}_ j$ of $\mathcal{S}_0 \amalg \ldots \amalg \mathcal{S}_{i + 1}$. This makes sense exactly because $\delta (s)$ is in the kernel of the inductively defined differential.

Lemma 24.31.1. In the situation above the differential graded $\mathcal{O}$-algebra

\[ \mathcal{A} = \mathop{\mathrm{colim}}\nolimits \mathcal{A}_ i \]

has the following property: for any morphism $(f, f^\sharp ) : (\mathop{\mathit{Sh}}\nolimits (\mathcal{C}'), \mathcal{O}') \to (\mathop{\mathit{Sh}}\nolimits (\mathcal{C}), \mathcal{O})$ of ringed topoi, the pullback $f^*\mathcal{A}$ is flat as a graded $\mathcal{O}'$-module and is K-flat as a complex of $\mathcal{O}'$-modules.

**Proof.**
Observe that $f^*\mathcal{A} = \mathop{\mathrm{colim}}\nolimits f^*\mathcal{A}_ i$ and that

\[ f^*\mathcal{A}_ i = \mathcal{O}'\langle f^{-1}\mathcal{S}_0 \amalg \ldots \amalg f^{-1}\mathcal{S}_ i\rangle \]

with differential given by the inductive procedure above using $f^{-1}\delta _{i + 1}$. Thus it suffices to prove that $\mathcal{A}$ is flat as a graded $\mathcal{O}$-module and is K-flat as a complex of $\mathcal{O}$-modules. For this it suffices to prove that each $\mathcal{A}_ i$ is flat as a graded $\mathcal{O}$-module and is K-flat as a complex of $\mathcal{O}$-modules, compare with Lemma 24.23.3.

For $i \geq 1$ write $\mathcal{S} = \mathcal{S}_0 \amalg \ldots \amalg \mathcal{S}_ i$ so that we have $\mathcal{A}_ i = \mathcal{O}\langle \mathcal{S} \rangle $ as a graded $\mathcal{O}$-algebra. We are going to construct a filtration of this algebra by differential graded $\mathcal{O}$-submodules.

Set $W = \mathbf{Z}_{\geq 0}^{i + 1}$ considered with lexicographical ordering. Namely, given $w = (w_0, \ldots w_ i)$ and $w' = (w'_0, \ldots , w'_ i)$ in $W$ we say

\[ w > w' \Leftrightarrow \exists j,\ 0 \leq j \leq i : w_ i = w'_ i,\ w_{i - 1} = w'_{i - 1},\ \ldots , \ w_{j + 1} = w'_{j + 1},\ w_ j > w'_ j \]

and so on. Suppose given a section $s = s_1 \cdot \ldots \cdot s_ r$ of $\mathcal{S} \times \ldots \times \mathcal{S}$ over $U$. We say that the *weight of $s$ is defined* if we have $s_ a \in \mathcal{S}_{j_ a}(U)$ for a unique $0 \leq j_ a \leq i$. In this case we define the weight

\[ w(s) = (w_0(s), \ldots , w_ i(s)) \in W,\quad w_ j(s) = |\{ a \mid j_ a = j\} | \]

The weight of any section of $\mathcal{S} \times \ldots \times \mathcal{S}$ is defined locally. The reader checks easily that we obtain a disjoint union decompostion

\[ \mathcal{S} \times \ldots \times \mathcal{S} = \coprod \nolimits _{w \in W} \left( \mathcal{S} \times \ldots \times \mathcal{S}\right)_ w \]

into the subsheaves of sections of a given weight. Of course only $w \in W$ with $\sum _{0 \leq j \leq i} w_ j = r$ show up for a given $r$. We correspondingly obtain a decomposition

\[ \mathcal{A}_ i = \mathcal{O} \oplus \bigoplus \nolimits _{r \geq 1} \bigoplus \nolimits _{w \in W} \mathcal{O}[\left(\mathcal{S} \times \ldots \times \mathcal{S}\right)_ w] \]

The rest of the proof relies on the following trivial observation: given $r$, $w$ and local section $s = s_1 \cdot \ldots \cdot s_ r$ of $\left(\mathcal{S} \times \ldots \times \mathcal{S}\right)_ w$ we have

\[ \text{d}(s) \text{ is a local section of } \mathcal{O} \oplus \bigoplus \nolimits _{r' \geq 1} \bigoplus \nolimits _{w' \in W,\ w' < w} \mathcal{O}[\left(\mathcal{S} \times \ldots \times \mathcal{S}\right)_{w'}] \]

The reason is that in each of the expressions

\[ (-1)^{\deg (s_1) + \ldots + \deg (s_{a - 1})} s_1 \cdot \ldots s_{a - 1} \cdot \delta (s_ a) \cdot s_{a + 1} \cdot \ldots \cdot s_ r \]

whose sum give the element $\text{d}(s)$ the element $\delta (s_ a)$ is locally a $\mathcal{O}$-linear combination of elements $s'_1 \cdot \ldots \cdot s'_{r'}$ with $s'_{a'}$ in $\mathcal{S}_{j'_ a}$ for some $0 \leq j'_{a'} < j_ a$ where $j_ a$ is such that $s_ a$ is section of $\mathcal{S}_{j_ a}$.

What this means is the following. Suppose for $w \in W$ we set

\[ F_ w \mathcal{A}_ i = \mathcal{O} \oplus \bigoplus \nolimits _{r \geq 1} \bigoplus \nolimits _{w' \in W,\ w' \leq w} \mathcal{O}[\left(\mathcal{S} \times \ldots \times \mathcal{S}\right)_{w'}] \]

By the observation above this is a differential graded $\mathcal{O}$-submodule. We get admissible short exact sequences

\[ 0 \to \mathop{\mathrm{colim}}\nolimits _{w' < w} F_{w'}\mathcal{A}_ i \to F_ w\mathcal{A}_ i \to \bigoplus \nolimits _{r \geq 1} \mathcal{O}[\left(\mathcal{S} \times \ldots \times \mathcal{S}\right)_ w] \to 0 \]

of differential graded $\mathcal{A}$-modules where the differential on the right hand side is zero.

Now we finish the proof by transfinite induction over the ordered set $W$. The differential graded complex $F_0\mathcal{A}_0$ is the summand $\mathcal{O}$ and this is K-flat and graded flat. For $w \in W$ if the result is true for $F_{w'}\mathcal{A}_ i$ for $w' < w$, then by Lemmas 24.23.3, 24.23.2, and 24.23.6 we obtain the result for $w$. Finally, we have $\mathcal{A}_ i = \mathop{\mathrm{colim}}\nolimits _{w \in W} F_ w\mathcal{A}_ i$ and we conclude.
$\square$

Lemma 24.31.2. Let $(\mathcal{C}, \mathcal{O})$ be a ringed site. Let $(\mathcal{B}, \text{d})$ be a differential graded $\mathcal{O}$-algebra. There exists a quasi-isomorphism of differential graded $\mathcal{O}$-algebras $(\mathcal{A}, \text{d}) \to (\mathcal{B}, \text{d})$ such that $\mathcal{A}$ is graded flat and K-flat as a complex of $\mathcal{O}$-modules and such that the same is true after pullback by any morphism of ringed topoi.

**Proof.**
The proof is exactly the same as the first proof of Lemma 24.23.7 but now working with free graded algebras instead of free graded modules.

We will construct $\mathcal{A} = \mathop{\mathrm{colim}}\nolimits \mathcal{A}_ i$ as in Lemma 24.31.1 by constructing

\[ \mathcal{A}_0 \to \mathcal{A}_1 \to \mathcal{A}_2 \to \ldots \to \mathcal{B} \]

Let $\mathcal{S}_0$ be the sheaf of graded sets (Remark 24.23.5) whose degree $n$ part is $\mathop{\mathrm{Ker}}(\text{d}_\mathcal {B}^ n)$. Consider the homomorphism of differential graded modules

\[ \mathcal{A}_0 = \mathcal{O}\langle \mathcal{S}_0 \rangle \longrightarrow \mathcal{B} \]

where map sends a local section $s$ of $\mathcal{S}_0$ to the corresponding local section of $\mathcal{A}^{\deg (s)}$ (which is in the kernel of the differential, so our map is a map of differential graded algebras indeed). By construction the induced maps on cohomology sheaves $H^ n(\mathcal{A}_0) \to H^ n(\mathcal{B})$ are surjective and hence the same will remain true for all $i$.

Induction step of the construction. Given $\mathcal{A}_ i \to \mathcal{B}$ denote $\mathcal{S}_{i + 1}$ the sheaf of graded sets whose degree $n$ part is

\[ \mathop{\mathrm{Ker}}(\text{d}_{\mathcal{A}_ i}^{n + 1}) \times _{\mathcal{B}^{n + 1}, \text{d}} \mathcal{B}^ n \]

This comes equipped with a canonical map

\[ \delta _{i + 1} : \mathcal{S}_{i + 1} \longrightarrow \mathcal{A}_ i \]

whose image is contained in the kernel of $\text{d}_{\mathcal{A}_ i}$ by construction. Hence $\mathcal{A}_{i + 1} = \mathcal{O}\langle \mathcal{S}_0 \amalg \ldots \mathcal{S}_{i + 1}\rangle $ has a differential exteding the differential on $\mathcal{A}_ i$, see discussion at the start of this section. The map from $\mathcal{A}_{i + 1}$ to $\mathcal{B}$ is the unique map of graded algebras which restricts to the given map on $\mathcal{A}_ i$ and sends a local section $s = (a, b)$ of $\mathcal{S}_{i + 1}$ to $b$ in $\mathcal{B}$. This is compatible with differentials exactly because $\text{d}(b)$ is the image of $a$ in $\mathcal{B}$.

The map $\mathcal{A} \to \mathcal{B}$ is a quasi-isomorphism: we have $H^ n(\mathcal{A}) = \mathop{\mathrm{colim}}\nolimits H^ n(\mathcal{A}_ i)$ and for each $i$ the map $H^ n(\mathcal{A}_ i) \to H^ n(\mathcal{B})$ is surjective with kernel annihilated by the map $H^ n(\mathcal{A}_ i) \to H^ n(\mathcal{A}_{i + 1})$ by construction. Finally, the flatness condition for $\mathcal{A}$ where shown in Lemma 24.31.1.
$\square$

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