naive de Rham cohomology fails for singular varieties

Question

Let $X$ be a variety over a field $k$ of characteristic zero. If $X$ is smooth, algebraic de Rham cohomology defined as $$ H^n_{dR}(X / k)=\mathbb{H}^n(X, \Omega^\bullet_{X/k})\qquad (\star) $$ is a good cohomological theory, for instance, is isomorphic tensor $\mathbb{C}$ to the singular cohomology of $X(\mathbb{C})$ once one chooses and embedding $k \hookrightarrow \mathbb{C}$.

It is "well known" that this fails for singular $X$ unless one considers more sophisticated definitions of de Rham cohomology, but I had difficulties finding a counterexample of, say, a singular variety $X$ such that the dimension of $(\star)$ is different from the dimension of singular cohomology. After a while, I found a paper by Arapura-Kang where the example of the plane curve $$ x^5+y^5+x^2y^2=0 $$ is presented. The proof is not so easy. Does anybody know a simpler example?

Answer 1

The nonexactness of the naïve analytic de Rham complex for complex analytic spaces is discussed in the following papers.

H.J. Reiffen, Das Lemma von Poincaré fȕr holomorphs Differential-formen auf komplexen Räumen, Math Z. 101 (1967).

M. Sebastiani, Preuve d'une Conjecture de Brieskorn, Manuscripta math. 2 (1970).

K. Saito, Quasihomogene isolierte Singularitäten von Hyperflächen, Invent. math. 14 (1971).

There are many isolated hypersurface singularities which are not quasihomogeneous (in the sense that there are no local analytic coordinates in which the hypersurface is locally defined by a quasihomogeneous polynomial) and for any such one can globalize it to a projective complex hypersurface with one singular point, and by the above in this case the Euler characteristic of the singular cohomology differs from the Euler characteristic of the naïve de Rham cohomology.