4
$\begingroup$

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 ?

$\endgroup$
  • 1
    $\begingroup$ So are you assuming $r = N - 2$ since you say you get a surface? $\endgroup$ – Dori Bejleri Dec 15 '13 at 0:23
8
$\begingroup$

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

$$ \chi^m(V_n^{d_1,\ldots, d_r}) = (-1)^mh^{m,m}(V_n^{d_1,\ldots, d_r}) \enspace \enspace \text{for} \enspace \enspace 2m = n $$

$\endgroup$
  • $\begingroup$ So, this computation based on Lefschetz hyperplane theorem? $\endgroup$ – Li Yutong Dec 15 '13 at 2:30
  • $\begingroup$ 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. $\endgroup$ – Dori Bejleri Dec 15 '13 at 2:55
1
$\begingroup$

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 $P^3$.

$\endgroup$
  • 1
    $\begingroup$ Can you give an explicit example of this phenomena? $\endgroup$ – 54321user Aug 12 '17 at 23:04
  • $\begingroup$ 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. $\endgroup$ – Maxim Leyenson Jan 16 at 3:40
0
$\begingroup$

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:

Link, at the GitLab

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.