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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.

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.

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 cohomology. I am just wondering why one should think to not use the ordinary cohomology of the complex in this case.

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.

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 cohomology. I am just wondering why one should think to not use the ordinary cohomology of the complex in this case.

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 cohomology. I am just wondering why one should think to not use the ordinary cohomology of the complex in this case.