[Math] Why Use Hypercohomology When Defining the de Rham Cohomology of a Smooth Scheme over $k$

Question

Hopefully this question is of an appropriate level for this site: I'm reading some notes by Claire Voisin titled Géométrie Algébrique et Géométrie Complexe. Let $X$ be a smooth $k-$scheme. In these notes, one constructs the (algebraic) de Rham complex
$$ 0\xrightarrow{}\mathscr{O}_X\xrightarrow{d}\Omega_{X/k}\xrightarrow{d}\Omega_{X/k}^2\xrightarrow{d}\cdots \xrightarrow{d}\Omega_{X/k}^n\xrightarrow{}0$$
where $n=\dim(X)$. Now, one constructs the algebraic de Rham cohomology by way of using the hypercohomology of the complex, $H^l_{dR}(X/k):=\mathbb{H}^l(\Omega_{X/k}^{\cdot})$. My question is not regarding the construction of the hypercohomology. I am just wondering why one should think to not use the ordinary cohomology of the complex in this case.

Answer 1

This is pretty much explained in the comments, but let me put it into an answer. One wants algebraic de Rham cohomology to be isomorphic to the usual de Rham cohomology (using $C^\infty$ forms) when $k=\mathbb{C}$, and have similar properties when $k$ is an arbitrary field of characteristic zero. From this point of view, as Grothendieck observed in the 1960's, the definition using hypercohomology is the correct one to use. To understand why, one should observe that a quasi-isomorphism of (bounded below) complexes induces an isomorphism of hypercohomologies. The holomorphic Poincar\'e lemma, says that on the complex manifold $X^{an}$, $\mathbb{C}_{X^{an}}$ and $\Omega_{X^{an}}^\bullet$ are quasi-isomorphic and so have the same hypercohomologies. We are halfway there. Now a (not completely straightforward) application of GAGA shows that $$\mathbb{H}^i(X,\Omega_X^\bullet)\cong H^i(X^{an},\mathbb{C})$$