1
$\begingroup$

I have a very specific question. How does one check the following map ${\mathbb C}^n-\bigcup_{i\neq j} \{z \mid z_i= \pm z_j\}\to {({\mathbb C}^*)}^{n-1}-\bigcup_{i\neq j}\{w \mid w_i= w_j\}$ defined by $(z_1,z_2,\dotsc , z_n)\mapsto (z_n^2-z_1^2,\dotsc , z_n^2-z_{n-1}^2)$ is a locally trivial fibration? This is stated in the paper of E. Brieskorn ‘Sur les groupes de tresses’.

Actually, in the paper there are few other above type maps are defined and mentioned to be locally trivial fibration. Probably, there is some standard technique to check. Any hint, how to prove it?

Improve this question
$\endgroup$
5
  • $\begingroup$ I would try doing it in two steps, first $\mathbb{C}^n \setminus \{ z_i = \pm z_j \} \to \mathbb{C}^n \setminus \{ z_i = z_j \}$, $(z_1,\dots) \mapsto (z_1^2, \dots)$ looks like a covering space of degree $2$. Then you have the usual Fadell–Neuwirth fibration which forgets a point in a configuration space (see their paper for the proof that it's a fibration). Of course writing this I now realize that the first map has a problem at $z_i = 0$ so maybe it's more complicated. $\endgroup$
    – Najib Idrissi
    Commented Jul 4, 2018 at 7:44
  • $\begingroup$ Thanks for your response! Fadell-Neuwirth considers projection maps. Do you mean the same for the second map. In the above case it is different. $\endgroup$
    – RKS
    Commented Jul 4, 2018 at 9:09
  • $\begingroup$ I know it's different, what I'm seeing is that I expect the same techniques to work. Your map is like translation + symmetry followed by projection. $\endgroup$
    – Najib Idrissi
    Commented Jul 4, 2018 at 9:51
  • $\begingroup$ It has non-compact fiber, how it is proper? $\endgroup$
    – RKS
    Commented Jul 5, 2018 at 4:12
  • $\begingroup$ Sorry I must have misread your map $\endgroup$
    – Phil Tosteson
    Commented Jul 5, 2018 at 14:40

0

Reset to default

You must log in to answer this question.

Browse other questions tagged .