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Let $X\subseteq\mathbb{P}^n(\mathbf{C})$ be a complete intersection (smooth if you want).

Q: Is there a good reference which gives (and proves in enough details) an explicit description of the graded ring $H^*(X,\mathbb{Z})$ (or $H^*(X,\mathbb{Q})$)?

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  • $\begingroup$ Have a look at Dimca's book "Singularities and topology of hypersurfaces", Chapter 5. $\endgroup$ Commented Nov 1, 2011 at 18:42
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    $\begingroup$ Another reference is SGA VII, Exp. XI. $\endgroup$
    – user19475
    Commented Feb 7, 2012 at 17:50
  • $\begingroup$ (for étale cohomology) $\endgroup$
    – user19475
    Commented Feb 7, 2012 at 17:52

2 Answers 2

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One can compute the Betti numbers of a smooth complete intersection $X$ of multidegree $d=(d_1,\dotsc,d_r)$ by induction on $r$. The case $r=0$ corresponds to $\mathbb{P}^n(\mathbb{C})$.

A smooth complete intersection of $X$ multidegree $(d_1,\dotsc, d_r)$ is a section of the positive line bundle $\mathcal{O}(d_1)$ over a smooth complete intersection of $Y$ multidegree $(d_2,\dotsc, d_r)$. By the Lefschetz hyperplane theorem we have $H^i(X,\mathbb{Q})\cong H^i(Y,\mathbb{Q})$ for $i\leq \dim Y-2=\dim X -1$. Using the induction hypothesis and the Poincaré duality we can compute all $H^i(X,\mathbb{Q})$ apart from $H^{\dim X}(X,\mathbb{Q})$.

But we can also compute the Euler characteristic of $X$ as follows. The normal bundle of $X$ is $\mathcal{O}(d_1)\oplus\dotsb\oplus\mathcal{O}(d_r)$, which extends to $\mathbb{P}^n$. The Chern class of $T\mathbb{P}^n$ is $(1+H)^{n+1}$ where $H$ is the class of the hyperplane section. So the Chern class of $TX$ is $$\frac{(1+H)^{n+1}}{\prod_{i=1}^{r}(1+d_i H)}$$ restricted to $X$. So the Euler characteristic of $X$ is $d_1\dotsm d_r$ times the coefficient of $H^{\dim X}$ in the above expression. Now we can compute the rank of $H^{\dim X}(X,\mathbb{Q})$.

Moreover, from the above and the universal coefficients formula it follows that $H^*(X,\mathbb{Z})$ is torsion free. Indeed, we may assume that $H_{<\dim X}(X,\mathbb{Z})$ and $H^{<\dim X}(X,\mathbb{Z})$ are torsion free by induction hypothesis and hence so are $H_{>\dim X}(X,\mathbb{Z})$ and $H_{>\dim X}(X,\mathbb{Z})$ by the Poincaré duality; moreover $H^{\dim X}(X,\mathbb{Z})=\operatorname{Hom} (H^{\dim X}(X,\mathbb{Z}),\mathbb{Z})$ is torsion free and so is $H_{\dim X}(X,\mathbb{Z})$ by the Poincaré duality again.

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  • $\begingroup$ I kind of new (vaguely) how to compute the Betti numbers, but I would like also to know the ring structure. $\endgroup$ Commented Nov 1, 2011 at 23:38
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    $\begingroup$ Hugo -- so would I. For hypersurfaces this should be feasible though, at least with real coefficients, since only the middle degree cohomology is different from the cohomology of the projective space, and we can compute the signature since we know the Chern classes. $\endgroup$
    – algori
    Commented Nov 2, 2011 at 0:06
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If $X$ is smooth then Lefschetz' hyperplane theorem and Poincaré duality yield that $H^i(X,\mathbb{Z})=H^i(\mathbb{P}^n,\mathbb{Z})$ for $i=0,\dots, 2\dim X$, $i\neq \dim X$. (This is proven in Dimca's book on topology and geometry of singular hypersurfaces. In this book there is also a proof for the fact that if $X$ is singular then $H^i(X,\mathbb{Z})=H^i(\mathbb{P}^n,\mathbb{Z})$ for $i=0,\dots, 2\dim X$, $i\neq \dim X,\dim X+1,\dots ,\dim X+\dim X_\text{sing}+1$.)

In the smooth case the primitive cohomology group $H^n(X,\mathbb{C})_\text{prim}$ can be studied by Cayley's trick: Let $c=n-\dim X$. Then Cayley's trick produces a hypersurface $Y$ in a $\mathbb{P}^{c-1}$-bundle over $\mathbb{P}^n$ such that you can represent classes in $H^n(X,\mathbb{C})$ by residues of differential forms on $\mathbb{P}^{c-1}\setminus Y$. (If $c=1$ then $X=Y$.) This is worked out in detail by several people in the 90s (one of them is Dimca, but there are also papers on this issue by other authors).

The singular case is less well understood and still a subject of present day research. Dimca's book is a nice introduction to this subject.

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