# Is the moduli space of graphs simply connected?

The moduli space of graphs $MG_n$ is the quotient of Culler-Vogtmann's outer space $X_n$ by the action of $\mathrm{Out}(F_n)$. It can be thought of as the space of metric graphs homotopy equivalent to a wedge of $n$ circles with all vertices of valence $\geq 3$. A metric graph can be thought of as simply a graph with an assignment of positive edge lengths to all edges, and when an edge length goes to $0$, the edge itself also contracts, yielding a new graph type in $MG_n$. One also often assumes that the sum of edge lengths is $1$, since this condition yields a homotopy equivalent space.

One of the useful properties of $MG_n$ is that it is a rational $K(\pi,1)$ for $\mathrm{Out}(F_n)$: $H_*(MG_n;\mathbb Q)\cong H_*(\mathrm{Out}(F_n);\mathbb{Q})$. If $MG_n$ were an actual $K(\pi,1)$, the fundamental group would be $\mathrm{Out}(F_n)$ and that would be that, but it seems in practice that $MG_n$ is fairly highly connected, and in particular simply connected. $MG_2$ is certainly contractible. I'm not sure about $MG_3$; I would guess it's contractible. I suspect that $MG_4\simeq S^4$.

My question is whether $MG_n$ is indeed simply connected. I am unaware of any particular consequences this would have, but it also seems we ought to know such a basic property about such a fundamental space.

Yes, it is. If $G$ is a discrete group acting on a simply-connected simplicial complex $X$, then a theorem of M. A. Armstrong says that there is a short exact sequence

$$1 \longrightarrow H \longrightarrow G \longrightarrow \pi_1(X/G) \longrightarrow 1,$$

where $H$ is the subgroup of $G$ generated by elements that fix a point in $X$. The original reference is

M. A. Armstrong, On the fundamental group of an orbit space, Proc. Cambridge Philos. Soc. 61 (1965), 639–646.

A. Putman, Obtaining presentations from group actions without making choices, Algebr. Geom. Topol. 11 (2011), 1737-1766.

which proves a more specific result and contains a streamlined version of Armstrong's proof in Section 3.1 (which is really just a meditation on the usual facts about covering spaces, which the above restricts to if the action is free and thus $H = 1$).

In any case, if we apply this with $G = \text{Out}(F_n)$ and $X = X_n$ (Outer Space), then we see that $H = G$ since $G$ is generated by torsion elements, which fix points of Outer space. The quotient is thus simply-connected, as desired.

A similar argument shows that the (ordinary, non-orbifold) fundamental group of the moduli space of curves is trivial. This was originally proved (by this argument) in

Maclachlan, Colin, Modulus space is simply-connected, Proc. Amer. Math. Soc. 29 1971 85–86.

• Thanks Andy. That is a very nice short exact sequence. I wasn't expecting anything quite so elegant. – Jim Conant Nov 17 '15 at 23:23
• @JimConant: Glad to help. It's one of my favorite lesser-known results in elementary algebraic topology! – Andy Putman Nov 17 '15 at 23:32
• @JimConant: By the way, you might be interested in the following paper of mine, which uses the above exact sequence to construct generators for $\text{IA}_n$: M. Day, A. Putman, The complex of partial bases for $F_n$ and finite generation of the Torelli subgroup of $\text{Aut}(Fn)$, Geom. Dedicata 164 (2013), 139-153. – Andy Putman Nov 17 '15 at 23:40
• It is not quite obvious that the mapping class group is generated by elements which fix something - I think this is the nontrivial part of Maclachlan's result. – Igor Rivin Nov 18 '15 at 0:01
• @IgorRivin: Right, one has to prove that the mapping class group is generated by torsion elements (which Maclachlan does basically by observing that you can write the Lickorish generators as products of conjugates of some specific torsion elements that had previously been produced by Birman). – Andy Putman Nov 18 '15 at 0:16