# Fundamental group of $M_g^\circ$

Let's work over the complex numbers $$\mathbb{C}$$. Let $$g\geq3$$ be an integer. Let $$\mathcal{M}_g$$ be moduli stack of smooth genus $$g$$ curves. Let $$M_g$$ be the corresponding coarse moduli scheme. They share an open subscheme $$M_g^\circ$$ parametrizing automorphism-free smooth genus $$g$$ curves.

What is the fundamental group $$\pi_1(M_g^\circ)$$? How does it compare to the following notions:

(1) The orbifold mapping class group $$\pi_1(\mathcal{M}_g)$$,

(2) The topological fundamental group $$\pi_1(M_g)$$,

(3) The mapping class group of a genus $$g$$ surface, $$\mathrm{Mod}_g$$?

Except in a few trivial cases, the locus of curves which have an extra automorphism will have codimension greater than one in $$\mathcal M_g$$. When that happens, the fundamental group of $$\mathcal M_g$$ must equal the fundamental group of $$\mathcal M_g^{\circ}$$. (To see this, one can pass to the $$3$$-torsion cover of $$\mathcal M_g$$, which is a smooth scheme, and use the fact that removing a codimension two subscheme of a smooth scheme does not affect $$\pi_1$$.) So it equals cases 1 and 3.

• For completeness: the locus of curves with automorphisms is of codimension $g-2$. – Dan Petersen Sep 10 '19 at 20:26
• What happens when $g=3$? – Kevin Casto Sep 10 '19 at 20:42
• @KevinCasto I don't know except it seems that, because the hyperelliptic locus has nontrivial class in $H^2$, the abelianization of $\pi_1$ does not increase except possibly by some torsion when it is removed. – Will Sawin Sep 10 '19 at 21:11
• @WillSawin Thanks! Do we know if (1) coincides with (2)? (Since $M_g$ is normal, finite etale covers are determined by integral closure in the function field extension?) – Qixiao Sep 10 '19 at 22:45
• @Qixiao As Paul Johnson has already explained, $\pi_1(M_g)$ is trivial while $\pi_1(\mathcal M_g)$ is not. The normalcy is completely correct, but the issue is simply that some finite extensions of the function field are ramified as covers of $M_g$ but unramified as covers of $\mathcal M_g$. – Will Sawin Sep 10 '19 at 22:55

Since $$\mathcal{M}_g=[\mathcal{T}_g/\text{Mod}_g]$$, and $$\mathcal{T}_g$$ Teichmueller space is contractible, (1) and (3) are going to be isomorphic.

This was mentioned previously on MathOverflow here, or can be found for instance on page 361 of Farb and Margalit "A primer on Mapping Class Groups".

Meanwhile, (2), the fundamental group of the coarse moduli space $$\pi_1(M_g)$$, is trivial, as described by Andy Putnam's answer to a question here

So, we really just need to compare $$\pi_1(M^\circ_g)$$ to $$\pi_1(\mathcal{M}_g)$$. The inclusion map will give us a homomorphism from $$\pi_1(M^\circ_g)\to \pi_1(\mathcal{M}_g)$$, and presumably we can say something nice about this but I don't know anything off the top of my head and have run out of time to research it...

• I dream that someday mathematicians will learn how to spell my name... – Andy Putman Sep 10 '19 at 21:13
• @AndyPutman Thanks for the explanation on triviality of $\pi_1(M_g)$ in that answer! – Qixiao Sep 11 '19 at 1:01
• @AndyPutman Argh how embarrassing. So sorry. Question was interesting to me because it was on the edge of what I knew, but had already spent too long tracking down answers I did find to pay attention to basic things like formatting or getting people's names right. – Paul Johnson Sep 11 '19 at 14:47
• @PaulJohnson: Don't worry, you're not the first to make that mistake... – Andy Putman Sep 11 '19 at 18:25