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Jan 22, 2021 at 10:37 comment added Igor Sikora Thank you very much!
Jan 21, 2021 at 14:33 comment added Sam Hopkins @IgorSikora: That's correct. The paper Adrien linked to is very nice and explains everything in the case you care about (Type D).
Jan 21, 2021 at 9:31 comment added Adrien I did :) arxiv.org/abs/math/0210438 this is definitely not the only paper explaining this, but it does focus on the type D case.
Jan 21, 2021 at 8:19 comment added Igor Sikora @Adrien You probably missed adding a link to the paper :)
Jan 21, 2021 at 8:18 comment added Igor Sikora @SamHopkins I think so. If A type is "classical" braid group, then the complement of associated arrangement is a configuration space of points in $\mathbb{C}$. So $\pi_1$ of this space is a pure braid group. Is that correct?
Jan 20, 2021 at 19:57 comment added Adrien @IgorSikora The introduction of this paper sketch the answer to your question and provides references. This is also explained in Brieskorn's paper, in case you read french.
Jan 20, 2021 at 15:18 comment added Sam Hopkins Here is definition of the associated braid group for arbitrary type Coxeter groups: en.wikipedia.org/wiki/Artin%E2%80%93Tits_group. The pure braid group is by definition the kernel of the canonical homomorphism from the braid group to its Coxeter group. In order to understand why this pure braid group is the fundamental group of the associated Coxeter hyperplane arrangement, it helps if you understand the Type A case first. Do you understand the Type A case?
Jan 20, 2021 at 15:10 comment added Igor Sikora @Adrien - the very basics, i.e. the definition of pure braid group of type D (or any other type) and why this is a fundamental group of this arrangement.
Jan 20, 2021 at 15:07 comment added Adrien Right, type D sorry, this is still handled in Brieskorn paper. $\pi$ is then called the pure braid group of type D. This is the kernel of the natural map from the full braid group of type D to the corresponding Coxeter group, where the former has a presentation obtained from the standard Coxeter presentation of the Coxeter group by removing the torsion relations. There is quite a lot of literature on those, I'm not sure what exactly you'd like to know.
Jan 20, 2021 at 15:02 comment added Andreas Blass I also think it's type D. Type B would also include the coordinate hyperplanes, i.e., the polynomial would have each $x_i$ as a factor.
Jan 20, 2021 at 12:55 comment added Igor Sikora @Adrien And what is $\pi$ in this situation? And isn't it of type D?
Jan 20, 2021 at 12:34 comment added Adrien This is the arrangement underlying the braid/Coxeter group of type B. That this is a $K(\pi,1)$ is due to Brieskorn in that case maths.ed.ac.uk/~v1ranick/papers/brieskorn8.pdf.
Jan 20, 2021 at 12:17 history asked Igor Sikora CC BY-SA 4.0