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.