An arithmetic Zariski pair of line arrangements with non-isomorphic fundamental group

Abstract

Abstract. In a previous work, the third named author found a combinatorics of line arrangements whose realizations live in the cyclotomic group of the fifth roots of unity and such that their non-complex-conjugate embedding are not topologically equivalent in the sense that they are not embedded in the same way in the complex projective plane. That work does not imply that the complements of the arrangements are not homeomorphic. In this work we prove that the fundamental groups of the complements are not isomorphic. It provides the first example of a pair of Galois-conjugate plane curves such that the fundamental groups of their complements are not isomorphic (despite the fact that they have isomorphic profinite completions).

Configurations of points and topology of real line arrangements

“…They can be realized with equations which are Galois-conjugated in Q(ζ 5 ), where ζ 5 is a primitive fifth-root of unit. Moreover, their fundamental groups are non-isomorphic as is later proved by Artal, Cogolludo, Marco and the first author [ABCAGBMB17]. Once again, these arrangements are not rational.…”

“…L 8 , L 13 }, {L 5 , L 10 , L 11 }, {L 5 , L 10 , L 13 }, {L 6 , L 10 , L 11 }, {L 7 , L 8 , L 13 }.4.1.2. Realization over a number field.A Zariski pair is said to be arithmetic if the equations of the arrangements are Galois-conjugated in a number field (as the example in[GB16,ABCAGBMB17]). Such pairs are particular since they cannot be distinguished by algebraic arguments.…”

On the non-connectivity of moduli spaces of arrangements: the splitting-polygon structure

“…This means that, if it exists an isomorphism between π 1 (CP 2 \ ML i,j k ) and π 1 (CP 2 \ ML i ,j k ), for i, j, i , j ∈ {1, 2}, then it induces the identity on the Abelianization of these groups. By applying the Alexander Invariant test of level 2 (described in [4,5]), we obtain the following theorem. We don't give here more details about the proof, since the strategy is the exactly the same as in [4,5,13], and since the author used the same program to perform the computation.…”

“…By applying the Alexander Invariant test of level 2 (described in [4,5]), we obtain the following theorem. We don't give here more details about the proof, since the strategy is the exactly the same as in [4,5,13], and since the author used the same program to perform the computation. We refer to these articles for more details (the construction of the Alexander Invariant test is done in [4] while a Sagemath code is given in [5]).…”

“…As mentioned in the abstract, the questions related to the combinatorial nature of some properties of a hyperplane arrangement are numerous in the literature. If some of them have been solved by the affirmative, as for the number of chambers of a real arrangement [27], the cohomology ring of the complement [19], the rank of the lower central series quotients of the fundamental group of its complement [10] or the deletion and addition-deletion theorems of free arrangements [2,1]; some others obtained a negative answer, as for the embedded topology of a complex arrangements or the fundamental group of its complements, see [21,12,5,13], (also negative for the smaller class of real complexified arrangements [3,14]), the torsion of the lower central series quotients [8] or the existence of unexpected curves [14]. Naturally, the number of problems which are still open (or conjectural) is larger; like the famous Terao's conjecture [24,20], the combinatorial nature of the characteristic varieties [15] or of the homology of the Milnor fiber [11,Problem 4.5], to name some but a few.…”