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.
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4$\begingroup$ What exactly do you mean by ordinary cohomology? $\endgroup$– MohanCommented Apr 26, 2018 at 22:24
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4$\begingroup$ Perhaps ordinary does not add anything to the statement. I just mean why "bother" with the hypercohomology. What problem would one encounter by defining the de Rham cohomology of $X/k$ by the cohomology of the complex as usual. $\endgroup$– Alekos RobotisCommented Apr 26, 2018 at 22:26
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10$\begingroup$ First of all, as pointed out in other comments it doesn't make sense to consider 'usual cohomology' because the result won't be a vector space. You are probably asking why we define de Rham cohomology as hypercohomology of the de Rham complex rather than homology of the complex of its global sections. The main reason is that we want this invariant to be useful. For example, we want the algebraic de Rham cohomology of a smooth finite type $\mathbf C$ scheme to coincide with ($\mathcal C^{\infty}$ or holomorphic) de Rham cohomology of the associated analytic space. $\endgroup$– gdbCommented Apr 27, 2018 at 1:45
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12$\begingroup$ There is no chance to have a result of this sort for the naive definition of the algebraic de Rham cohomology. For instance, by the very definition $\mathrm H^i_{dR-naive}(X)=0$ for $i>\operatorname{dim} X$ but we know (by the comparison of smooth de Rham cohomology with topological cohomology) that $\mathrm H^{2d}_{dR-\mathcal C^{\infty}}(X^{an})=\mathbf C$ for a smooth proper $\mathbf C$-scheme of dimension $d$. Also, you can see by hands that the naive de Rham complex doesn't give you the correct answer for $\mathbf P^1_{\mathbf C}$ (it never gives you cor. ans. for proper sm. schemes) $\endgroup$– gdbCommented Apr 27, 2018 at 1:51
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11$\begingroup$ However, it is worth mentioning that the naive definition of de Rham cohomology does give you the right answer in the affine case because each term $\Omega^i_{X/k}$ is coherent. Thus the algebraic de Rham complex is a complex of $\Gamma$-acyclic sheaves. In this case you can use spectral sequence of a filtered complex to argue that $\mathrm H^i_{dR}(X)=\mathrm H^i_{dR-naive}(X)$ provided that $X$ is an affine scheme. $\endgroup$– gdbCommented Apr 27, 2018 at 1:56
1 Answer
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 $$\mathbb{H}^i(X,\Omega_X^\bullet)\cong H^i(X^{an},\mathbb{C})$$
