The Stacks project

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

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

  2. 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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