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

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