How does one check that the following space is aspherical? $X_n=\{(x_1,x_2,\ldots , x_n)\in {(\mathbb C^*)}^n\ |\ x_i\neq x_j\ and\ x_ix_j\neq 1\ for\ i\neq j\}$.

One way I can think of is to give a map to $Y_{n-1}=\{(y_1,y_2,\ldots , y_{n-1})\in {(\mathbb C^*)}^{n-1}\ |\ y_i\neq y_j\ for\ i\neq j\}$ which is a locally trivial fibration. After replacing $\mathbb C$ by $\mathbb R$ and for $n=2$, I drew a picture and it seems a suitable map should be $(x_1,x_2,\ldots , x_n)\mapsto (\frac{x_i-x_n}{2},\ldots , \frac{x_{n-1}-x_n}{2})$.

But then how does one check that this map is a locally trivial fibration? $Y_n$ is aspherical by taking projection and applying long exact homotopy sequence and induction. Using a result of Fadell-Neuwirth the projections in the case of $Y_n$ are locally trivial fibrations.

  • The space in question is an orbit configuration space $F_{C_2}(\mathbb{C}^*,n)$, where the cyclic group $C_2$ acts by $x\mapsto x^{-1}$. Xicotencatl showed that orbit configuration spaces $F_G(M,n)$ have Fadell-Neuwirth fibrations, but only in the case when $G$ acts freely on $M$. This makes me think you should try to prove non-asphericity by looking for a sphere in $X_n$ enclosing a singular point, such as $(1,1)$. – Mark Grant Jul 13 at 10:17
  • Thanks a lot for the reference of Xicotencatl. I believe that there is some kind of method to check whether such an algebraically defined map (when fiber is non-compact) is a locally trivial fibration? But no luck so far! – Roushan Jul 16 at 6:53

Your Answer

 
discard

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Browse other questions tagged or ask your own question.