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If $X$ is a compact K\"ahler manifold, then the $\partial\overline{\partial}$-lemma states that the inclusion of the subcomplex $(\ker\partial, d)$ into the complex $(A^{*,*}(X),d)$ of smooth complex differential forms on $X$ is a quasi-isomorphism.

I'm interested in an algebraic analogue of this (if it exists). The setting I have in mind is a smooth projective manifold $X$ over a characteristic zero algebraically closed field $\mathbb{K}$. Then a natural substitute for $(A^{*,*}(X),d)$ would be $\mathbf{R}\Gamma\Omega^*_X$, the derived global sections of the algebraic de Rham complex of $X$. The de Rham differential $d_{dR}: \Omega^*_X \to \Omega^{*+1}_X$ should induce a de Rham differntial $d_{dR}:\mathbf{R}\Gamma\Omega^*_X\to \mathbf{R}\Gamma\Omega^*_X$ which is "half" of the differential of $\mathbf{R}\Gamma\Omega^*_X$ (the other half coming from the Cech differential in the relaization of $\mathbf{R}\Gamma\Omega^*_X$ as the total complex of the Cech-de Rham bicomplex). So my candidate for replacing $(\ker\partial, d)$ is $(\ker d_{dR},d)$.

question: is the inclusion $(\ker d_{dR},d)\subseteq (\mathbf{R}\Gamma\Omega^*_X,d)$ a quasi-isomorphism? I know that the Hodge-to-de Rham spectral sequence abutting to the cohomology of $\mathbf{R}\Gamma\Omega^*_X$ degenerates at $E_1$, which is I'd like to think as an algebraic analogue of the $\partial\overline{\partial}$-lemma, but when I try to fix the details of this I am not able to see things as clearly as I'd like to, so I suspect there's something I'm missing.

Any suggestion, hint, proof, counterexample, or reference is welcome.

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I believe the "horizontal" differentials of the $E_1$ term of the Hodge-to-de Rham spectral sequence are exactly what you call de Rham differentials since the de Rham complex is quasi-isomorphic to the structure sheaf and $E_1$ is induced by the filtration bête, so in $E_1$ one direction is the de Rham differential and the other is the Čech. Then the fact that the spectral sequence degenerates at $E_1$ seems to imply exactly what you want.

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I think the answer to the question is yes. More precisely, let $X$ be a smooth compact complex algebraic variety and suppose $(F^{\ast,\ast},d',d'')$ is a bounded below double complex of $\Gamma$-acyclic sheaves (for the complex or Zariski topology) such that there is a map $\Omega_X^\ast\to F^{\ast,0}$ which is a resolution of $\Omega_X^\ast$. Let $R\Gamma\Omega_X^\ast$ be the complex of global sections of $s[F^{\ast,\ast}]$, the total complex of $F^{\ast,\ast}$. Then $(\ker d',d'')\subset (R\Gamma\Omega^\ast_X,d)$ is a quasi-isomorphism. Here is how I would try to prove this.

There are two left exact functors from the category $Kom^+(Sh(X),\mathbf{C}-Vect)$ of bounded below complexes of sheaves of $\mathbf{C}$-vector spaces on a topological space $X$ to the category $Kom^+(\mathbf{C}-Vect)$ of bounded below complexes of $\mathbf{C}$-vector spaces: one is the global sections functor $A:F^\ast\mapsto \Gamma(F^\ast)$ and the other, $B$, is the composition of $A$ and the functor that takes a complex $K^\ast$ of vector spaces to $\bigoplus_i \ker (K^i\to K^{i+1})$ with zero differential. Both the source and the target of $A$ and $B$ are abelian categories and there is a natural transformation from $B\to A$.

Now take $X$ to be a smooth complex analytic manifold and take $F^\ast$ to be the de Rham complex. The Dolbeault resolution of $F^\ast$ (considered as an object in $$Kom^+(Kom^+(Sh(X),\mathbf{C}-Vect))$$ with differential $\bar\partial$) is acyclic for $A$. This follows from the fact that the sheaves of smooth differential forms are soft. So $R^\ast A$ is represented in $K^+(Kom^+(\mathbf{C}-Vect))$, the homotopy category of $Kom^+(\mathbf{C}-Vect)$, by the complex of the global sections of the Dolbeault resolution with differential $\bar \partial$. Since $B$ is the composition of $A$ and some functor from vector spaces to vector spaces, one can compute $R^\ast B$ by taking the Dolbeault resolution and then taking the kernel of $\partial$.

The derived category of $Kom^+(\mathbf{C}-Vect)$ is equivalent to the $D^+(\mathbf{C}-Vect)$; on the level of complexes the equivalence takes a double complex $\in Kom^+(Kom^+(\mathbf{C}-Vect))$ to its total complex. Taking the compositions of $R^\ast A$ and $R^\ast B$ with this equivalence we get functors $$D^+(Sh(X),\mathbf{C}-Vect)\to D^+(\mathbf{C}-Vect)$$ and a natural transformation between then (it is induced by the transformation $B\to A$). Applied to $\Omega^\ast_X$ the value of one of the functirs is $R\Gamma\Omega_X^\ast$ and the value of the other is what is denoted $(\ker d_{dR}, d)$ in the above posting. When $X$ is algebraic and compact and the functors are applied to $\Omega^\ast_X$, this transformation is an isomorphism due to the classical $\partial\bar\partial$-lemma.

This proves the above claim for the complex topology on $X$. I think by GAGA the same should be true for the Zariski topology as well, but one should check the details.

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