Timeline for Hyperelliptic locus is a $K(\pi,1)$

5 events

when toggle format	what		by	license	comment
Mar 22, 2018 at 0:05	comment	added	Dan Petersen		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.
Mar 21, 2018 at 21:37	comment	added	F. Germano		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.
Mar 21, 2018 at 21:22	comment	added	Ariyan Javanpeykar		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 $K(\pi,1)$?
Mar 21, 2018 at 20:59	review	First posts
Mar 21, 2018 at 21:44
Mar 21, 2018 at 20:56	history	asked	F. Germano	CC BY-SA 3.0