2
$\begingroup$

I'm working on some presentation of the fundamental group of the complement of a complex lines arrangement in $\mathbb{C}^{2}.$ In particular, in Arvola's article "The fundamental group of the complement of an arrangement of complex hyperplanes" it is described an algorithm to compute this group via a certain diagram associated to the arrangement. This is the so called wiring diagram of the arrangement.

Moreover, using some related arguments, Cohen and Suciu in "The braid monodromy of plane algebraic curves and hyperplane arrangements" use wiring diagrams to compute the braid monodromy of a complex line arrangement in $\mathbb{C}^{2}.$ Particularly, the give an alternative presentation of the fundamental group of the complement which is (of course) isomorphic to the one provided by Arvola.

Similarly, in the recent arxiv preprint "On complex line arrangement and thei boundary manifolds," Florens, Guerville-Balle and Marco Buzurnaz give a third alternative presentation of the fundamental group of the complement which is based on some wiring diagrams algorithm, too.

I'm interested in the study of the groups built in these article which comes from an "abstract" wiring diagram, i.e., a graph with a sequence of crossings and braids. For more details, a full definition is given in page 299 of Cohen Suciu, "The braid monodromy of plane algebraic curves and hyperplane arrangements".

In order to study these groups, I have the following question. Maybe this is a well known result in hyperplane arrangements theory.

Given an "abstract" wiring diagrams $W,$ there always exists a complex lines arrangement $\mathcal{A}$ in $\mathbb{C}^{2}$ such that (up to a suitable choice of coordinates and projection) the wiring diagram associated to $\mathcal{A}$ is exactly $W$?

If my question is not completely clear, I'd like to discuss better about that.

$\endgroup$
1
$\begingroup$

Certainly not. Just take the wiring diagram of a "non-existant" arrangement (e.g., Desargues arrangement with one of the triple points perturbed into three double: this can be done in the realm of pseudo-holomorphic lines, so the diagram still makes sense).

$\endgroup$

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.