[proofplan]
We construct the Deligne-Griffiths-Morgan-Sullivan zigzag through the subalgebra of $\partial$-closed forms. First we show that the inclusion of this subalgebra into the de Rham differential graded algebra is a quasi-isomorphism. Then we map the same subalgebra to $\partial$-cohomology and use the $\partial\bar{\partial}$ lemma, component by component in bidegree, to prove that this map is also a quasi-isomorphism. Finally, the cohomological form of the $\partial\bar{\partial}$ lemma identifies this terminal algebra with the usual complex de Rham cohomology algebra, giving formality.
[/proofplan]
[step:Define the intermediate differential graded algebra of $\partial$-closed forms]
Let $n:=\dim_{\mathbb C}X$. For integers $p,q$, let $A^{p,q}(X)$ denote the complex [vector space](/page/Vector%20Space) of smooth complex-valued differential forms of type $(p,q)$ on $X$, with $A^{p,q}(X)=0$ when either index is outside $\{0,\dots,n\}$. For each integer $k\geq 0$, let $A^k(X;\mathbb C)=\bigoplus_{p+q=k}A^{p,q}(X)$ denote the complex vector space of smooth complex-valued $k$-forms on $X$. By the [type decomposition of the exterior derivative](/theorems/7004), we have maps
\begin{align*}
\partial:A^{p,q}(X)\to A^{p+1,q}(X)
\end{align*}
and
\begin{align*}
\bar\partial:A^{p,q}(X)\to A^{p,q+1}(X)
\end{align*}
whose total-degree extensions satisfy $d=\partial+\bar\partial$ on $A^\bullet(X;\mathbb C)$. For each integer $k\geq 0$, define the complex vector subspace $K^k\subset A^k(X;\mathbb C)$ by
\begin{align*}
K^k:=\ker\left(\partial:A^k(X;\mathbb C)\to A^{k+1}(X;\mathbb C)\right).
\end{align*}
Set
\begin{align*}
K^\bullet:=\bigoplus_{k\geq 0}K^k.
\end{align*}
The restriction of $\bar\partial$ to $K^\bullet$ is well-defined. Indeed, if $\alpha\in K^k$, then $\partial\alpha=0$, and the Dolbeault relation
\begin{align*}
\partial\bar\partial+\bar\partial\partial=0
\end{align*}
gives
\begin{align*}
\partial(\bar\partial\alpha)=-\bar\partial(\partial\alpha)=0.
\end{align*}
Thus $\bar\partial\alpha\in K^{k+1}$.
The wedge product preserves $K^\bullet$. If $\alpha\in K^k$ and $\beta\in K^\ell$, then the graded derivation rule for $\partial$ gives
\begin{align*}
\partial(\alpha\wedge\beta)=(\partial\alpha)\wedge\beta+(-1)^k\alpha\wedge\partial\beta=0.
\end{align*}
Hence $\alpha\wedge\beta\in K^{k+\ell}$.
Therefore
\begin{align*}
(K^\bullet,\bar\partial,\wedge)
\end{align*}
is a differential graded algebra over $\mathbb C$. On $K^\bullet$, the [exterior derivative](/theorems/1525) satisfies
\begin{align*}
d\alpha=\partial\alpha+\bar\partial\alpha=\bar\partial\alpha
\end{align*}
for every $\alpha\in K^\bullet$. Hence the inclusion map
\begin{align*}
i:(K^\bullet,\bar\partial,\wedge)\to (A^\bullet(X;\mathbb C),d,\wedge)
\end{align*}
is a morphism of differential graded algebras.
[/step]
[step:Prove that the inclusion into the de Rham algebra is a quasi-isomorphism]
We use the cohomological consequences of the $\partial\bar{\partial}$ lemma from [citetheorem:8060] and [citetheorem:8061]. These results apply because $X$ is compact and satisfies the $\partial\bar{\partial}$ lemma by hypothesis.
First, every de Rham cohomology class has a representative lying in $K^\bullet$. Indeed, [citetheorem:8061] gives representatives of de Rham classes by forms whose pure-type components are both $\partial$-closed and $\bar\partial$-closed. Such a representative is in particular $\partial$-closed, hence lies in $K^\bullet$. Thus the induced map
\begin{align*}
H^\bullet(i):H^\bullet(K^\bullet,\bar\partial)\to H^\bullet(X;\mathbb C)
\end{align*}
is surjective.
Now suppose $\alpha\in K^k$ is $\bar\partial$-closed and that its image in de Rham cohomology is zero. Then $\partial\alpha=0$ by the definition of $K^k$, and $\bar\partial\alpha=0$ by hypothesis, so $d\alpha=0$. Since the de Rham class of $\alpha$ is zero, $\alpha$ is $d$-exact. By [citetheorem:8060], applied to the $d$-closed and $d$-exact form $\alpha$, there exists a form $\gamma\in A^{k-2}(X;\mathbb C)$ such that
\begin{align*}
\alpha=\partial\bar\partial\gamma.
\end{align*}
Define $\delta\in A^{k-1}(X;\mathbb C)$ by
\begin{align*}
\delta:=\partial\gamma.
\end{align*}
Then $\partial\delta=\partial^2\gamma=0$, so $\delta\in K^{k-1}$, and
\begin{align*}
\bar\partial\delta=\bar\partial\partial\gamma=-\partial\bar\partial\gamma=-\alpha.
\end{align*}
Thus $\alpha=\bar\partial(-\delta)$ is a $\bar\partial$-boundary in $K^\bullet$. This proves injectivity of $H^\bullet(i)$.
Therefore $i$ is a quasi-isomorphism of differential graded algebras.
[guided]
The inclusion map is not just an inclusion of graded algebras; it must preserve the differentials and induce an isomorphism on cohomology. The differential issue is handled by the defining condition $\partial\alpha=0$: for $\alpha\in K^\bullet$,
\begin{align*}
d\alpha=\partial\alpha+\bar\partial\alpha=\bar\partial\alpha.
\end{align*}
So the de Rham differential restricts to the $\bar\partial$ differential on $K^\bullet$.
For surjectivity on cohomology, take an arbitrary de Rham class in $H^k(X;\mathbb C)$. Since $X$ is compact and satisfies the $\partial\bar{\partial}$ lemma, [citetheorem:8061] supplies a representative whose components are $\partial$-closed and $\bar\partial$-closed. In particular, the total form is $\partial$-closed, so it belongs to $K^k$. Hence every de Rham cohomology class is hit by the inclusion.
For injectivity, let $\alpha\in K^k$ be a $\bar\partial$-closed element whose image in de Rham cohomology is zero. The condition $\alpha\in K^k$ means
\begin{align*}
\partial\alpha=0.
\end{align*}
The cocycle condition in $K^\bullet$ means
\begin{align*}
\bar\partial\alpha=0.
\end{align*}
Therefore
\begin{align*}
d\alpha=\partial\alpha+\bar\partial\alpha=0.
\end{align*}
The assumption that the image of $\alpha$ is zero in de Rham cohomology says that $\alpha$ is $d$-exact. By the cohomological form of the $\partial\bar{\partial}$ lemma, [citetheorem:8060], there exists $\gamma\in A^{k-2}(X;\mathbb C)$ such that
\begin{align*}
\alpha=\partial\bar\partial\gamma.
\end{align*}
Now define $\delta\in A^{k-1}(X;\mathbb C)$ by
\begin{align*}
\delta:=\partial\gamma.
\end{align*}
Because $\partial^2=0$, we have
\begin{align*}
\partial\delta=\partial^2\gamma=0,
\end{align*}
so $\delta\in K^{k-1}$. Using $\bar\partial\partial=-\partial\bar\partial$, we get
\begin{align*}
\bar\partial\delta=\bar\partial\partial\gamma=-\partial\bar\partial\gamma=-\alpha.
\end{align*}
Thus $\alpha=\bar\partial(-\delta)$ is already a boundary inside $K^\bullet$. This proves that the inclusion induces an injective map on cohomology, and hence the inclusion is a quasi-isomorphism.
[/guided]
[/step]
[step:Map the intermediate algebra to $\partial$-cohomology]
Define the total $\partial$-cohomology groups by
\begin{align*}
H^k_\partial(A^\bullet(X;\mathbb C)):=\frac{\ker(\partial:A^k(X;\mathbb C)\to A^{k+1}(X;\mathbb C))}{\operatorname{im}(\partial:A^{k-1}(X;\mathbb C)\to A^k(X;\mathbb C))}.
\end{align*}
Let
\begin{align*}
H^\bullet_\partial(A^\bullet(X;\mathbb C)):=\bigoplus_{k\geq 0}H^k_\partial(A^\bullet(X;\mathbb C)).
\end{align*}
Since $\partial$ is a graded derivation and $\partial^2=0$, the wedge product descends to a graded-commutative algebra structure on $H^\bullet_\partial(A^\bullet(X;\mathbb C))$.
Define a graded algebra map
\begin{align*}
\pi:K^\bullet\to H^\bullet_\partial(A^\bullet(X;\mathbb C))
\end{align*}
by
\begin{align*}
\pi(\alpha):=[\alpha]_\partial
\end{align*}
for each $\alpha\in K^\bullet$, where $[\alpha]_\partial$ denotes the $\partial$-cohomology class of the $\partial$-closed form $\alpha$. The map is multiplicative because the product on the target is induced by wedge product:
\begin{align*}
\pi(\alpha\wedge\beta)=[\alpha\wedge\beta]_\partial=[\alpha]_\partial\wedge[\beta]_\partial=\pi(\alpha)\wedge\pi(\beta).
\end{align*}
[/step]
[step:Show that the map to $\partial$-cohomology is a chain map]
The target differential on $H^\bullet_\partial(A^\bullet(X;\mathbb C))$ is zero. Therefore $\pi$ is a chain map precisely when
\begin{align*}
[\bar\partial\alpha]_\partial=0
\end{align*}
for every $\alpha\in K^\bullet$.
Let $\alpha\in K^k$. Decompose $\alpha$ into pure types:
\begin{align*}
\alpha=\sum_{p+q=k}\alpha^{p,q},
\end{align*}
where $\alpha^{p,q}\in A^{p,q}(X)$. Since $\partial\alpha=0$ and $\partial\alpha^{p,q}$ has type $(p+1,q)$, each component satisfies
\begin{align*}
\partial\alpha^{p,q}=0.
\end{align*}
For each pair $(p,q)$, the form $\bar\partial\alpha^{p,q}$ is $\bar\partial$-exact and is $\partial$-closed, because
\begin{align*}
\partial\bar\partial\alpha^{p,q}=-\bar\partial\partial\alpha^{p,q}=0.
\end{align*}
It is also $d$-closed, since its $\partial$ and $\bar\partial$ derivatives both vanish. By [citetheorem:8060], each $\bar\partial\alpha^{p,q}$ is $\partial$-exact. Hence their finite sum $\bar\partial\alpha$ is $\partial$-exact, so
\begin{align*}
[\bar\partial\alpha]_\partial=0.
\end{align*}
Thus $\pi$ is a morphism of differential graded algebras
\begin{align*}
\pi:(K^\bullet,\bar\partial,\wedge)\to (H^\bullet_\partial(A^\bullet(X;\mathbb C)),0,\wedge).
\end{align*}
[/step]
[step:Prove that the map to $\partial$-cohomology is a quasi-isomorphism]
We prove that $\pi$ induces an isomorphism on cohomology.
For surjectivity, let $[\eta]_\partial\in H^k_\partial(A^\bullet(X;\mathbb C))$, where $\eta\in A^k(X;\mathbb C)$ satisfies $\partial\eta=0$. The form $\bar\partial\eta$ is $\partial$-closed because
\begin{align*}
\partial\bar\partial\eta=-\bar\partial\partial\eta=0.
\end{align*}
It is $\bar\partial$-closed because $\bar\partial^2=0$, and it is $d$-exact because $d\eta=\partial\eta+\bar\partial\eta=\bar\partial\eta$. By [citetheorem:8060], applied componentwise to the pure-type components of $\bar\partial\eta$, there exists $\gamma\in A^{k-1}(X;\mathbb C)$ such that
\begin{align*}
\bar\partial\eta=\partial\bar\partial\gamma.
\end{align*}
Define $\eta_0\in A^k(X;\mathbb C)$ by
\begin{align*}
\eta_0:=\eta-\partial\gamma.
\end{align*}
Then $\partial\eta_0=\partial\eta-\partial^2\gamma=0$, so $\eta_0\in K^k$. Moreover,
\begin{align*}
\bar\partial\eta_0=\bar\partial\eta-\bar\partial\partial\gamma=\bar\partial\eta+\partial\bar\partial\gamma.
\end{align*}
Replacing $\gamma$ by $-\gamma$ in the preceding construction if necessary, we obtain $\bar\partial\eta= -\partial\bar\partial\gamma$, and hence $\bar\partial\eta_0=0$. Finally, $\eta_0$ represents the same $\partial$-cohomology class as $\eta$ because $\eta_0-\eta=-\partial\gamma$. Therefore
\begin{align*}
\pi(\eta_0)=[\eta_0]_\partial=[\eta]_\partial.
\end{align*}
Thus $H^\bullet(\pi)$ is surjective.
For injectivity, let $\alpha\in K^k$ be $\bar\partial$-closed and suppose that
\begin{align*}
\pi(\alpha)=[\alpha]_\partial=0.
\end{align*}
Then there exists $\theta\in A^{k-1}(X;\mathbb C)$ such that
\begin{align*}
\alpha=\partial\theta.
\end{align*}
Since $\alpha\in K^k$ and $\bar\partial\alpha=0$, the form $\alpha$ is both $\partial$-closed and $\bar\partial$-closed. It is also $\partial$-exact. Applying [citetheorem:8060] gives a form $\lambda\in A^{k-2}(X;\mathbb C)$ such that
\begin{align*}
\alpha=\partial\bar\partial\lambda.
\end{align*}
Set $\mu:=-\partial\lambda\in A^{k-1}(X;\mathbb C)$. Then
\begin{align*}
\partial\mu=-\partial^2\lambda=0,
\end{align*}
so $\mu\in K^{k-1}$, and
\begin{align*}
\bar\partial\mu=-\bar\partial\partial\lambda=\partial\bar\partial\lambda=\alpha.
\end{align*}
Thus $\alpha$ is a $\bar\partial$-boundary in $K^\bullet$. Hence $H^\bullet(\pi)$ is injective.
Therefore $\pi$ is a quasi-isomorphism.
[/step]
[step:Identify the terminal algebra with de Rham cohomology and conclude formality]
By [citetheorem:8061], the $\partial$-cohomology algebra $H^\bullet_\partial(A^\bullet(X;\mathbb C))$ is naturally isomorphic, as a graded $\mathbb C$-algebra, to the complex de Rham cohomology algebra $H^\bullet(X;\mathbb C)$. Under this identification, the differential on the terminal algebra is zero.
We have constructed a zigzag of differential graded algebra quasi-isomorphisms
\begin{align*}
(A^\bullet(X;\mathbb C),d,\wedge)\xleftarrow{\ i\ }(K^\bullet,\bar\partial,\wedge)\xrightarrow{\ \pi\ }(H^\bullet_\partial(A^\bullet(X;\mathbb C)),0,\wedge)\cong (H^\bullet(X;\mathbb C),0,\smile).
\end{align*}
This is precisely formality of $(A^\bullet(X;\mathbb C),d,\wedge)$ over $\mathbb C$. Therefore the complex de Rham differential graded algebra of $X$ is formal over $\mathbb C$.
[/step]
