5
$\begingroup$

I am aware of the following examples of normal surfaces in $\mathbb P^3$ that are projectively isomorphic to their dual varieties:

  1. the smooth quadric;

  2. Kummer surfaces;

  3. The surface with the equation $x_0^3=x_1x_2x_3$ (in homogeneous coordinates).

What else is known? The base field is algebraically closed of characteristic zero.

Thank you in advance, Serge

Improve this question
$\endgroup$

1 Answer 1

Reset to default
7
$\begingroup$

In his paper [Some invariants for conics and their applications, Publ. RIMS (Kyoto Univ.) 19 (1983), 1139-1151] Naruki gives an example of a self-dual quartic surface in $\mathbb{P}^3$ with three singular points of type $A_3$ and seven points of type $A_1$ (i.e., ordinary double points).

The paper can be dowloaded here.

Improve this answer
$\endgroup$
3
  • 2
    $\begingroup$ As $3\cdot 3 + 7 = 16$ one could imagine this surface is a degeneration of Kummer surfaces. Is it really the case? $\endgroup$
    – Sasha
    Nov 16, 2013 at 7:36
  • 1
    $\begingroup$ Nice observation. Naruki's quartic is obtained by taking the double cover of $\mathbb{P}^2$ branched over a configuration of three irreducible conics in special position. On the other hand, Kummer's branch locus is given by six distinct lines, all tangent to the same conic. It does not seem to me that this can degenerate to Naruki's branch, but perhaps there is some degeneration between the two surfaces not coming from the double cover construction. Maybe, in order to have more insight, one should compute some topological invariant of Naruki's quartic (for instance, the fundamental group). $\endgroup$
    – Francesco Polizzi
    Nov 16, 2013 at 7:50
  • $\begingroup$ Well, regarding my precedent comment, since all the singularities of Naruki's quartic $S$ are Rational Double Points, by the simultaneous resolution of RDP it follows that the desingularization of $S$ is diffeomorphic to a smooth quartic, i.e. a $K3$ surface. So I think that $S$ itself is simply connected. $\endgroup$
    – Francesco Polizzi
    Nov 16, 2013 at 9:32

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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