Since my previous question

Hyperelliptic loci in Teichmueller spaces

resulted in two quick and helpful replies, let me ask another question in a similar vein:

A smooth compact complex curve is called trigonal, if it is a triple cover of the projective line. Let ${\mathcal X}_g$ be the trigonal locus in the moduli space of smooth genus $g$ curves (not sure what the standard notation is). What can be said about the topology of the preimage of ${\mathcal X}_g$ in the Teichmueller space? In particular, is it connected? If not, is there a description of its connected components in terms of mapping class groups similar to the one in the hyperelliptic case? Can one say anything reasonable about the fundamental groups of the connected components?


2 Answers 2


The argument I gave in that thread extends to a pretty wide class of surfaces with various symmetries. It's the same argument but there's some facts that are true for the hyperelliptic curves that are a little more fussy for curves with other symmetries. The line of reasoning I think is this:

Let $G$ be a finite subgroup of the diffeomorphism group of a surface $\Sigma_g$. Let $N(G)$ be the normalizer of $G$ in $Diff(\Sigma_g)$. So in most situations there's a short exact sequence $0 \to G \to \pi_0 N(G) \to \pi_0 Diff(\Sigma_g / G) \to 0$ (Birman-Hilden) where we're thinking of $\Sigma_g /G$ as an orbifold. Further there's a map $\pi_0 N(G) \to \pi_0 Diff(\Sigma_g)$. Provided the image of this map is the normalizer of $G$ in $\pi_0 Diff(\Sigma_g)$ (which according to Birman-Hilden is a pretty generic list of cases), then I think the argument goes through.

The Birman-Hilden reference. The case you're interested in is (?I think?) when $\Sigma_g / G$ is an orbifold whose underlying topological space is a sphere.

This is just off the top of my head and I'm likely glossing over some important details. I'll be happy to try and clarify.

So in general the lift of your moduli space is disconnected and its components are indexed by the cosets of the group $\pi_0 N(G)$ in the mapping class group. $\pi_0 N(G)$ is the hyperelliptic group in the case that $G = \mathbb Z_2$ is the hyperelliptic involution of your surface.

  • $\begingroup$ Ryan -- the coverings are not assumed to be regular i.e. there may be no $G$. (Which is why I don't see how to adapt the techniques mentioned in your and JSE's answers.) $\endgroup$
    – algori
    Feb 5, 2010 at 1:04
  • $\begingroup$ Ah, okay. For an irregular covering that'll take some more thought. $\endgroup$ Feb 5, 2010 at 1:27

The trigonal locus in the Teichmuller space - under mild hypotesis - is connected. The answer follows from one of the main results of


  • $\begingroup$ btw why did you wanted to know? I am curious $\endgroup$
    – IMeasy
    May 26, 2015 at 20:52
  • $\begingroup$ I think the result is essentially due to Hilden - look at the top of p. 8 of the paper, where the reference is given. $\endgroup$
    – Ian Agol
    May 26, 2015 at 21:09

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