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Let ${\cal M}_g$ be the moduli space of smooth complex genus $g$ curves, let ${\cal H}_g\subset {\cal M}_g$ be the hyperelliptic locus and set ${{\cal H}}'_g$ to be the preimage of ${\cal H}_g$ in the Teichmueller space.

While working on a problem I arrive at two results that can't be reconciled unless ${\cal H}'_3$ is disconnected.

While it seems a bit strange to me that ${\cal H}'_3$ should be disconnnected, I don't see why it should't be. So I'd like to ask whether this is known or known to be false.

[sorry, had to cut this into small paragraphs, otherwise the tex part wouldn't show properly.]

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  • $\begingroup$ I'd appreciate it if you could define the hyperelliptic locus, if the definition is not too long. Thank you! $\endgroup$
    – Ilya Grigoriev
    Commented Feb 4, 2010 at 2:07
  • $\begingroup$ $\mathcal H_g \subset \mathcal M_g$ presumably are the surfaces that admit an involution with $2g+2$ fixed points. $\endgroup$
    – Ryan Budney
    Commented Feb 4, 2010 at 2:22

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That $\mathcal H'_g$ can't be connected for $g \geq 3$, isn't this just the "lift to Teichmuller space" of the result that the mapping class group isn't the hyperelliptic group?

In particular, the path components of $\mathcal H'_g$ are indexed by the cosets of the hyperelliptic group in the mapping class group. No?

edit: Anweshi, I'm not sure how you're thinking about Teichmuller space but the answer to your question can be seen in many ways, you don't have to use the language of orbifolds, it's just a convienient container. In my mind I suppose I think of a path in Teichmuller space as a motion of the surface -- make this concrete using Fenchel-Nielsen coordinates, for example. So if you have a path that connects one point to another there is an associated diffeomorphism of the surfaces that stretches/twists the metric appropriately and matches up the markings of the surfaces. So if you go between two points in your $\mathcal H_g'$ covering the same point in $\mathcal H_g$ the relating diffeomorphism is in the hyperelliptic group (since the hyperelliptic group is a subgroup of the mapping class group). This is how you `see' the cosets of the hyperelliptic group in the mapping class group as indexing $\pi_0 \mathcal H'_g$.

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  • $\begingroup$ What makes this a silly reason? $\endgroup$
    – JSE
    Commented Feb 4, 2010 at 2:22
  • $\begingroup$ Er, sorry that's just how I talk. What I mean by that is this is a literal translation of the fact that the hyperelliptic group isn't the whole mapping class group, lifted up to Teichmuller space. I'll edit my comment to make it seem less annoying. $\endgroup$
    – Ryan Budney
    Commented Feb 4, 2010 at 2:29
  • $\begingroup$ I'll answer on Anweshi's behalf: thanks, Ryan, that was illuminating. $\endgroup$
    – algori
    Commented Feb 4, 2010 at 14:29
  • $\begingroup$ Oh my, I think I'm confusing your user name with Anweshi's. Sorry! $\endgroup$
    – Ryan Budney
    Commented Feb 4, 2010 at 18:03
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    $\begingroup$ Anweshi also has some interest in Teichmuller theory. So thanks are due from me too. :) $\endgroup$
    – Anweshi
    Commented Feb 6, 2010 at 23:32
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Teichmuller space is the universal cover of $M_g$. Thus, if $X$ is a locus in $M_g$, the preimage of $X$ in Teichmuller space is connected if and only if the induced map

$$\pi_1(X) \to \pi_1(M_g)$$

is surjective. In your case, you are asking: is the hyperelliptic mapping class group in genus $3$ the whole of the genus $3$ mapping class group $\Gamma_3$? No, it isn't: the hyperelliptic mapping class group is the centralizer of an involution in $\Gamma_3$.

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    $\begingroup$ This is morally correct, but Teichmuller space is not quite the universal cover of M_g because the action of the mapping class group on Teichmuller space is not free. It is thus a universal cover only in the sense of orbifolds, but everything still goes through like you'd hope. Of course, I'm sure JSE knows this, but this kind of thing can be confusing to a beginner. $\endgroup$
    – Andy Putman
    Commented Feb 4, 2010 at 2:29
  • $\begingroup$ Thanks, JSE. What I don't quite get is the following: the Teichmueller space is not exactly the universal cover of the moduli space. So why does the "$\pi_1$'s map surjectively" criterion still work? $\endgroup$
    – algori
    Commented Feb 4, 2010 at 2:37
  • $\begingroup$ Sorry! I indeed had the orbifold pi_1 in mind for both M_g and H_g. So the Teichmuller space IS the universal cover of the moduli space; not the coarse moduli space but the actual moduli space, which is an orbifold. $\endgroup$
    – JSE
    Commented Feb 4, 2010 at 2:41
  • $\begingroup$ The category of orbifolds is set up so that you can perform this sort of thing. Without mentioning orbifolds explicitly, you can do the following. Let M_g(L) be the moduli space of curves with level L structures, ie T_g/Mod_g(L) where Mod_g(L) is the subgroup of the mapping class group Mod_g that acts trivially on H_1(S_g;Z/L). For L>=3, the action of Mod_g(L) on T_g is free, so T_g really is the universal cover of M_g(L). Let X(L) be the pullback of the hyperelliptic locus to M_g(L). If X(L) is disconnected, then we are done. Otherwise, we can apply Jordan's argument to X(L). $\endgroup$
    – Andy Putman
    Commented Feb 4, 2010 at 2:44
  • $\begingroup$ Dear JSE -- many thanks for the answer, but I've decided to throw in a couple of dollars. Hope you'll forgive me. $\endgroup$
    – algori
    Commented Jul 26, 2011 at 12:15
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There's a slight issue I believe with the other answers. If we consider moduli space as an orbifold (of complex dimension $3g-3$), and the hyperelliptic locus an immersed suborbifold (of complex dimension $2g-1$ or so), then we may (essentially) identify the hyperelliptic locus with the orbifold of $2g+2$ points on $S^2$, obtained by quotienting each Riemann surface by the hyperelliptic involution. However, how does one know that this space doesn't "cross" itself? Imagine by analogy an immersed geodesic curve on a hyperbolic surface, such that each complementary component is a disk: the preimage in the universal cover is connected, when taken as a union of geodesics, even though each geodesic lift is embedded.

This sort of crossing does not occur for the hyperelliptic locus. If two branches of the hyperelliptic universal cover in Teichmuller space were to intersect, then there would be a single Riemann surface fixed by two distinct hyperelliptic involutions. But a hyperelliptic involution fixes precisely the $2g+2$ Weierstrauss points of the surface, and is therefore uniquely determined, a contradiction. So in fact the hyperelliptic locus is "embedded", in the sense that each lift corresponds to a fixed set of a hyperelliptic involution, and distinct hyperelliptic involutions give distinct components in Teichmuller space.

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