Suppose we have a smooth complete intersection of hypersurfaces with degrees $d_1,...,d_r$ in some $\mathbb{P}^N$. This should be a surface and in certain situations a surface of general type. What can one say about the Hodge diamond? Or what is its Grothendieck group ?
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1So are you assuming $r = N - 2$ since you say you get a surface? – Dori Bejleri Dec 15 '13 at 00:23
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This question is addressed in Appendix I of Topological Methods in Algebraic Geometry. Let $V_n^{d_1, \ldots, d_r}$ denote the complete intersection of $r$ generic hypersurfaces of degrees $d_1, \ldots d_r$ in $\mathbb{P}^{n + r}$. Let
$$ \chi_y(V_n^{d_1,\ldots, d_r}) = \sum_{p,q \geq 0} (-1)^qh^{p,q}(V_n^{d_1,\ldots, d_r})y^p = \sum_{p \geq 0} \chi^p(V_n^{d_1,\ldots, d_r})y^p $$
where $y$ is an indeterminate and $h^{p,q}$ are the Hodge numbers and
$$ \chi^p(V_n^{d_1,\ldots, d_r}) = \sum_{q \geq 0} (-1)^qh^{p,q}(V_n^{d_1,\ldots, d_r}). $$
Then Theorem 22.1.1 of the above reference says that
$$ \sum_{n \geq 0} \chi_y(V_n^{d_1,\ldots, d_r}) z^{n+r} = \frac{1}{(1-z)(1 + zy)}\prod_{i=1}^r\frac{(1 + zy)^{d_i}-(1-z)^{d_i}}{(1+zy)^{d_i}+y(1-z)^{d_i}}. $$
This let's you compute the numbers $\chi^p(V_n^{d_1,\ldots, d_r})$ which aren't exactly the Hodge numbers. However, the next Theorem makes it possible to find the actual Hodge numbers from this data. Thereom 22.1.2 in the same section says that
$$ h^{p,q}(V_n^{d_1,\ldots, d_r}) = \delta_{p,q} \enspace \enspace \text{for} \enspace \enspace p + q \neq n, $$
$$ \chi^p(V_n^{d_1,\ldots, d_r}) = (-1)^{n-p}h^{p,n-p}(V_n^{d_1,\ldots, d_r}) + (-1)^p \enspace \enspace \text{for} \enspace \enspace 2p \neq n $$
and
$$ \chi^m(V_n^{d_1,\ldots, d_r}) = (-1)^mh^{m,m}(V_n^{d_1,\ldots, d_r}) \enspace \enspace \text{for} \enspace \enspace 2m = n $$
Dori Bejleri
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Yes. In particular, the computations of theorem 22.1.2 use Lefschetz hyperplane theorem. The argument for packaging them together into the generating function of theorem 22.1.1 uses Grothendieck-Riemann-Roch to relate that generating function to the Chern classes. – Dori Bejleri Dec 15 '13 at 02:55
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The Grothendieck group/ring tensor $Q$ is the same as the Severi-Chow group/ring tensor $Q$, which varies a lot when you vary your hypersurfaces. This happens already in the case of algebraic surfaces in $P^3$.
Maxim Leyenson
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Well, take a look of surfaces of degree 4 in P^3. Configurations of curves on them can be vastly different (for example, some of them contain lines, (which are rigid -- can not be moved), while most surfaces of degree 4 have no lines on them. Thus their Picard (= Neron-Severi, in this case) lattices will be diffenet, and thus their K-groups are different. – Maxim Leyenson Jan 16 '19 at 03:40
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I have written a small (Python) program which computes Hodge numbers of hypersurfaces;
It is very easy to modify to work for complete intersections, too.
It contains references (to Hirzebruch and Deligne) in the comments section, and also a couple of examples in dimensions two and three:
Maxim Leyenson
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