We classify moduli spaces of arrangements of 10 lines with quadruple points. We show that moduli spaces of arrangements of 10 lines with quadruple points may consist of more than 2 disconnected components, namely 3 or 4 distinct points. We also present defining equations to those arrangements whose moduli spaces are still reducible after taking quotients of complex conjugations.2 sifying moduli spaces of arrangements of 10 lines. Section 3 shows that moduli spaces of arrangements with multiple points of high multiplicity are most likely irreducible. Section 4 and Section 5 deal with arrangements of 10 lines with a quadruple point. All possible arrangements of 10 lines with quadruple point whose moduli spaces are reducible, for instance, consisting of 2 points, 3 points, 4 points, or 2 one-dimensional components, can be found there.Acknowledgements: This work was partially supported by the Oswald Veblen Fund and by the Minerva Foundation of Germany. The authors thank M. Falk and the referee for their comments and suggestions that helped improve the clarity of the paper.
PreliminariesLet A = {L 1 , L 2 , · · · , L n } be a line arrangement in CP 2 . We say a singularity of L 1 ∪ L 2 ∪ · · · ∪ L n is a multiple point of A, if it has multiplicity at least 3.Definition 2.1. A line arrangement A is said to be C ≤3 if all multiple points of A are on at most three lines, say L 1 , L 2 and L 3 . A line arrangement is called simple C ≤3 if it is C ≤3 and one of the following conditions holds:Theorem 3.1. Let A be an arrangement of 10 lines. If there is a multiple point of multiplicity ≥ 6, then the moduli space M A is irreducible.Proof. Assume that L 1 ∩L 2 ∩· · ·∩L 6 = ∅. It is easy to check that at least one of the six lines contains at most two multiple points. By Lemma 2.3 and classification of arrangements of 9 lines (see [Ye11] Proposition 3.3), we see that M A is irreducible.When the highest multiplicity is 5, we notice that there is a case that the moduli space is reducible. However, a close look (see Remark 3.3) shows that the fundamental groups are still isomorphic.Theorem 3.2. Let A be a non-reductive arrangement of 10 lines with a quintuple point and no multiple points of higher multiplicities. Then A contains a Falk-Sturmfels arrangement as a sub-arrangement.Proof. By Lemma 2.4 and Theorem 2.5, we have the following inequality n 3 +n 4 ≤ 140−44n 5 9 .Since n 5 ≥ 1, thus n 3 + n 4 ≤ 10. On the other hand, there must be at least 11 − n 5 multiple points so that each line will pass through at least 3 multiple points. Therefore, 11 − n 5 ≤ n 3 + n 4 . The two inequalities together tell us that n 5 = 1 and n 3 + n 4 = 10. Apply Lemma 2.4 again, and we see that n 4 ≤ 1.
In the study of line arrangements, searching for minimal examples of line arrangements whose fundamental groups are not combinatorially invariant is a very interesting and hard problem. It was known that such a minimal arrangement must have at least 9 lines. In this paper, we extend the number to 10 by a new method. We classify arrangements of 9 projective lines according to the irreducibility of their moduli spaces and show that fundamental groups of complements of arrangements of 9 projective lines are combinatorially invariant. The idea and results have been used to classify arrangements of 10 projective lines.2000 Mathematics Subject Classification. 14N20, 32S22, 52C35.
Abstract. We introduce an algorithm that exploits a combinatorial symmetry of an arrangement in order to produce a geometric reflection between two disconnected components of its moduli space. We apply this method to disqualify three real examples found in previous work by the authors from being Zariski pairs. Robustness is shown by its application to complex cases, as well.
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