4
$\begingroup$

Let $\mathcal{M}_{0,n}$ denote the moduli stack of $\mathbb{P}^1$'s equipped with $n$ distinct sections. My understanding is that $\mathcal{M}_{0,3}$ is a point, so $\mathcal{M}_{0,4}$ is a scheme, isomorphic to $\mathbb{P}^1 - \{0,1,\infty\}$. This has fundamental group $F_2$ (the free group of rank 2), and so $\mathcal{M}_{0,5}$ is a fibration over $\mathbb{P}^1 - \{0,1,\infty\}$ with fibers isomorphic to 4-times punctured $\mathbb{P}^1$'s.

My question is - is there a good description of the fundamental groups of $\mathcal{M}_{0,n}$? (say, over $\overline{\mathbb{Q}}$). Are they ever trivial?

$\endgroup$
7
  • $\begingroup$ I think they are n-stranded braid groups. $\endgroup$ Commented Sep 25, 2016 at 23:01
  • 2
    $\begingroup$ The ordered versions are the pure braid groups. $\endgroup$
    – HJRW
    Commented Sep 25, 2016 at 23:12
  • 1
    $\begingroup$ One description is given by the Birman exact sequence, which says the kernel of the forget-a-strand map is free. $\endgroup$
    – HJRW
    Commented Sep 25, 2016 at 23:14
  • 2
    $\begingroup$ Do you actually mean the profinite completion of $F_2$? $\endgroup$
    – HeinrichD
    Commented Sep 26, 2016 at 0:31
  • 1
    $\begingroup$ The fundamental group of the moduli space $M_{g,n}$ is the mapping class group of a genus $g$ surface with $n$ punctures. So you are asking about the mapping class group of an $n$ times punctured sphere. It is related both to the usual pure braid group and to the spherical braid group. $\endgroup$ Commented Sep 26, 2016 at 16:33

1 Answer 1

7
$\begingroup$

Over an algebraically closed field of characteristic zero, a choice of complex analytification gives you an equivalence between finite étale covers of the moduli scheme $\mathcal{M}_{0,n}$ and finite covers of the configuration space of $n$ ordered points on the 2-sphere.

We can compute the fundamental group of the latter space by fixing the $n$th point at infinity, and considering configurations of $n-1$ points on the plane, up to rotations, translations, and dilations. If we ignore the rotation ambiguity, the configuration space has fundamental group given by the pure braid group on $n-1$ strands, which has a description in terms of an iterated semi-direct product of free groups.

Adding back in the rotation ambiguity amounts to quotienting by the global "total rotations", which form a central copy of $\mathbb{Z}$. The moduli scheme therefore has fundamental group isomorphic to the profinite completion of $P_{n-1}/\mathbb{Z}$.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .