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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.

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  • $\begingroup$ The cohomology is easy to compute, but it will depend of the type of the singular points. $\endgroup$
    – abx
    Commented Sep 9, 2021 at 15:01
  • $\begingroup$ If we look at the one I wrpte what should come out? $\endgroup$ Commented Sep 9, 2021 at 16:09
  • $\begingroup$ Can you say which points are singular? I could probably compute this but I figure you must already know... $\endgroup$
    – Will Sawin
    Commented Sep 9, 2021 at 17:31
  • $\begingroup$ Sorry I'm editing! This was actually a key point! $\endgroup$ Commented Sep 9, 2021 at 17:42
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    $\begingroup$ The surface is smooth by my computation. $\endgroup$
    – AG learner
    Commented Sep 9, 2021 at 20:35

1 Answer 1

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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.

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    $\begingroup$ The singularities could also have type $\mathrm{D}$, so the components of the divisor are not necessarily chains of curves. $\endgroup$
    – Sasha
    Commented Sep 10, 2021 at 7:04
  • $\begingroup$ @Sasha Thanks! I have edited accordingly. $\endgroup$
    – AG learner
    Commented Sep 10, 2021 at 22:14
  • $\begingroup$ I'm sorry but what is the Milnor number of a singularity? $\endgroup$ Commented Sep 12, 2021 at 14:22
  • $\begingroup$ @TommasoScognamiglio It's the dimension of the Jacobi ring $\mathbb C[x,y,z]/(\partial f/\partial x,\partial f/\partial y,\partial f/\partial z)$, where $f$ has an isolated singularity at 0. For ADE singularities, it's the number of the irreducible components on exceptional divisor of the minimal resolution. E.g., $A_1$ is 1, $D_4$ is 4, $E_6$ is 6. (So it is also equal to the substript index). $\endgroup$
    – AG learner
    Commented Sep 12, 2021 at 16:46

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