It is said in many papers that the hyperelliptic locus $\mathcal{H}_g\subseteq \mathcal{M}_g$ is a $K(\pi,1)$. (in the sense of orbifolds). This is justified by saying that it can be constructed as an orbifold quotient of a contractible analytic subspace of the Teichmuller space by the hyperelliptic mapping class group. Apparently, this is "well-known from Teichmuller theory", but I can't find a precise reference (I am more of an algebraic geometer). Are you aware of a place where this is written up?
3
-
3$\begingroup$ Let $X = (\mathbb{P}^1\setminus \{0,1,\infty\})^{d}$, where $d = \dim \mathcal{H}_g$. Let $U\subset X$ be the complement of the diagonals $\Delta_{ij}$. Then there is a finite etale morphism $U\to \mathcal{H}_g$ (in the category of algebraic stacks). Thus, don't we now reduce to showing that $U$ is a $K(\pi,1)$? $\endgroup$– Ariyan JavanpeykarCommented Mar 21, 2018 at 21:22
-
1$\begingroup$ Yes that is one way, but is it easy to see that $U$ is a $K(\pi,1)$? Most importantly, I was hoping for a proof that would simultaneously show that the orbifold fundamental group is the hyperelliptic mapping class group. $\endgroup$– F. GermanoCommented Mar 21, 2018 at 21:37
-
4$\begingroup$ The space $U$ in Ariyan's comment is just $M_{0,2g+2}$. So it is a $K(\pi,1)$ because its universal cover is a Teichmuller space. One can also use this description to see that the fundamental group is the hyperelliptic mapping class group, given that the latter is a Z/2 central extension of the mapping class group of a sphere with 2g+2 unordered punctures. $\endgroup$– Dan PetersenCommented Mar 22, 2018 at 0:05
Add a comment
|