A Sylvester-Gallai configuration in the the complex projective plane is a finite number of $n\ge 2$ points in the complex projective plane such that there is no line through exactly two of them. Trivial examples are obtained by taking $n\ge 3$ points on the same line. There is also the classical Hessian configuration, obtained by taking the nine inflexion points of a smooth cubic.

Question. Is there any other example? Is there a way to classify these examples? I would be particularly interested in finding a non-collinear example with an even number of points. Does it exist?

Remark: In higher dimension, it is known that the configuration of points has to be coplanar. If the points have coordinates defined over $\mathbb{R}$, the Sylvester-Gallai theorem shows that any configuration as above is in fact collinear. Over finite fields one can of course find plenty of configurations by taking all points.

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    $\begingroup$ There might be useful information in Hirzebruch's "Algebraische Flachen und Geradenkonfigurationen". $\endgroup$ – inkspot Feb 20 '17 at 8:07

Yes, there are other Sylvester-Gallai configurations in $\mathbb{P}^2(\mathbb{C})$. Apart from the Hesse configuration (that contains $9$ points) the minimum number of points for a non-collinear configuration is $12$.

A configuration with $12$ points actually exists over any field $\mathbb{K}$ of characteristic different from $2$ and containing a square root of $-1$, as proven by Kelly and Nwankpa.

See the answers to the MathOverflow question The Sylvester-Gallai theorem over $p$-adic fields for references and more details.

  • $\begingroup$ Is this 12-point configuration just the dual of the Hesse configuration? $\endgroup$ – potentially dense Feb 20 '17 at 13:54
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    $\begingroup$ No. The dual of the Hesse configuration has $9$ lines, whereas the Sylvester-Gallai configuration of cardinality $12$ (which over $\mathbb{C}$ is the dual of the so-called Ceva configuration) has $16$ lines. See this paper by Dolgachev, Section 5: arxiv.org/abs/math/0304258 $\endgroup$ – Francesco Polizzi Feb 20 '17 at 14:57
  • $\begingroup$ OK, thanks for the extra infomation! $\endgroup$ – potentially dense Feb 20 '17 at 15:24
  • $\begingroup$ I'm confused about something. Are you sure that the Sylvester-Gallai configuration of cardinality 12 has 16 lines and not 19? Unless I'm misreading the Kelly-Nwankpa paper it seems that their two configurations of cardinality 12 both have 19 lines. $\endgroup$ – Timothy Chow Feb 22 '17 at 5:02
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    $\begingroup$ @Timoty Chow. You are right. We must also add to the $4^2=16$ points of the Ceva configuration the three coordinate vertices, obtaining a configuration with $19$ points and $12$ lines. The dual of this configuration is a Sylvester-Gallai configuration with $12$ points and $19$ lines. Thanks for the remark. $\endgroup$ – Francesco Polizzi Feb 22 '17 at 9:17

I asked around about this question a while ago and the best answer I got was from Konrad Swanepoel. There are the well-known "Fermat" examples

$$(x^n - y^n)(y^n - z^n)(z^n - x^n) = 0, \qquad n \ge 3.$$

The $3n$ lines here, together with the $n^2$ points of intersection and the 3 coordinate vertices, form a dual Sylvester–Gallai configuration. So these give rise to a Sylvester–Gallai configuration with $3n$ points and $n^2+3$ lines. If you want an even number of points, just take $n$ to be even.

There are only two other ("sporadic") examples known, one due to Klein and one due to Wiman, which are described for example in this paper. They have 21 and 45 points respectively so they don't give you what you want.

According to Swanepoel the complete classification is still an open problem, but he believes that there are at most finitely many more sporadic examples.

In 1973, Kelly and Nwankpa classified all Sylvester–Gallai designs (a more general concept than configurations) with at most 14 points. Swanepoel warned me that there are some errors in this paper, but if we take its results at face value then it shows that there are no other complex Sylvester–Gallai configurations this small. It should be possible to extend the computational search beyond 14 points but nobody seems have done so (or at least has not published the results). By considering complex reflection groups, a colleague of mine has performed an unpublished computation that, if correct, shows that there are no other examples with reflective symmetry in every line.

  • $\begingroup$ I don't fully understand the "Fermat" example: do you mean the line configuration given by this equation? $\endgroup$ – Victor Protsak Feb 21 '17 at 15:04
  • $\begingroup$ @VictorProtsak : I have edited my answer to clarify. $\endgroup$ – Timothy Chow Feb 22 '17 at 5:08

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