Algebraic Geometry – Cohomology of Singular Projective Cubic Surface

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Let $X\subset \mathbb{P}_{\mathbb{C}}^3$ be a projective singular cubic surface with two singular points. Is the rationalcohomology of such objects known? As an example of the type of surfaces I'd be interested in one can take the hypersurface defined by the homogenous cubic $$x^2w+y^2w+z^2w+xyz-zw^2-w^3=0 .$$

In this case for example the singular points are given by $$([2: -\sqrt{3} :\sqrt{3} :1] , [2:\sqrt{3}:-\sqrt{3}:1] .$$

I think that the cohomology should be known for projective smooth cubic surfaces which should be realized as blowup of $\mathbb{P}^2_{\mathbb{C}}$ in six points but I'm not sure about singular ones.

Best Answer

By the classification theorem of cubic surfaces (p.6 in this paper), a cubic surface belongs to the following classes

  1. Has at worst ADE singularities.

  2. Has an elliptic singularity, i.e., the surface is cone over a smooth cubic curve.

  3. Non-normal or non-integral, and singular along a curve.

So if $X$ has two singularities, $X$ belongs to case 1 and all singularities are rational. We can compute the cohomology by the minimal resolution $\tilde{X}\to X$. The surface $\tilde{X}$ is called a weak del Pezzo surface, which is still blowup of 6 points on $\mathbb P^2$, but in less general positions, so $H^{2}(\tilde{X})=\mathbb Z^7$.

Now let's do some topology: Let $E$ be the exceptional divisor. Then the long exact sequence of the pair $(\tilde{X},E)$ reads $$H^1(E)\to H^2(X)\to H^2(\tilde{X})\xrightarrow{r} H^2(E),$$

where we used $H^*(\tilde{X},E)\cong H^*(X)$ because $\tilde{X}/E\cong X$ as CW complex.

$E$ is the disjoint union of two bunches of rational curves over the two singularities, so $H^2(E)$ has rank $\mu_1+\mu_2$, where $\mu_i$ is the Milnor number of the singularity. Also, $H^1(E)=0$ and $r$ is surjective, so

$$H^2(X)=\mathbb Z^{7-\mu_1-\mu_2}.$$

Cohomologies at other degrees are easy to compute.