Edit: In my original framing of this question it was not so clear what I was looking for, so this is basically a re-write.

I feel that I should already know the answer to this, but it never sits quite right in my head.

When dealing with Gromov-Witten theory, one is obviously interested in part in the moduli stack of stable curves, $\overline{M}_{g,n}$ (actually, we probably should really care about the stack of pre-stable curves, but that's not going to be hugely relevant for my case).

When dealing with orbifold Gromov-Witten theory, we instead need to consider the moduli stack of twisted stable curves, which are those curves that may have isotropy at the marked points and nodes. For the problems that I am interested in, it is usually the case that every marked point has the same isotropy---in particular, this is $\mathbb{Z}/2$. Let us denote then the stack of twisted stable curves whose marked points all have $\mathbb{Z}/r$ isotropy as $\overline{M}_{g,n}^{\mathbb{Z}/r}$.

There is an obvious map $\overline{M}_{g,n}^{\mathbb{Z}/r} \to \overline{M}_{g,n}$ given by taking the coarse moduli space.

My question is: What is the relationship between these two stacks?

Edit further: Some further questions that I feel that are relevant.

The moduli spaces $\overline{M}_{g,n}$ have a lot of structure maps between them. For example, they have forgetful maps $\overline{M}_{g,n+1} \to \overline{M}_{g,n}$, gluing maps, etc. They also have a bunch of canonically defined bundles over them, such as the cotangent line bundles, the Hodge bundles, etc. In particular, the divisor classes of the cotangent line bundles can be computed recursively given a few base cases by pulling them back via the forgetful maps.

How does these work for the moduli spaces $\overline{M}_{g,n}^{\mathbb{Z}/r}$? From what I understand, the $\psi$-classes come from the coarse curve, and so all of the expected relations should still hold. In particular, for $g = 0$ we can describe the $\psi$-classes purely combinatorially in terms of partitioning the marked points among different components of our curve. Is this picture correct?

Perhaps a better question: Is there a good reference which covers this?


2 Answers 2


I'm a bit confused as to what you're after, here.

You're first question seems to be "Which is the correct moduli space to look at?" -- and from my perspective you seem to answer this yourself in your edit. The correct moduli space is $\overline{\mathcal{M}}_{g,n}(\mathcal{B}\mathbb{Z}_2)$.

To build on this, by this "$\overline{\mathcal{M}}^{\mathbb{Z}/2}_{g,n}$" you seem to be just adding orbifold marked points, is kind of a silly thing to consider. It should really look like exactly like the usual moduli space of curves, except the universal family over it will have some orbifold structure. But this orbifold structure on the universal family is all very boring -- you've just added some orbifold structure along smooth, disjoint divisors, and so nothing much has happened at all. (The technical way stacky people describe adding orbifolds on divisors is as a root stack). But really, nearly everything about this $\overline{\mathcal{M}}^{\mathbb{Z}/2}_{g,n}$ will be just like $\overline{\mathcal{M}}_{g,n}$. The $\psi$ classes will look just like the usual $\psi$ classes except for a factor of 1/2, the hodge bundle $\mathbb{E}$ will be exactly the same...

Now, $\overline{\mathcal{M}}_{g,n}(\mathcal{B}\mathbb{Z}_2)$ and relatives are subtly different. In general, these will be finite covers of $\overline{\mathcal{M}}_{g,n}$ -- so the deformation theory will basically be the same. The interesting bits happen now when you get nodal curves, as you sometimes have to put orbifold structure at the nodes, and it is all a little subtle. Your universal family now is a lot richer -- you now have a universal non-orbifold curve of some genus $g^\prime$ found by Riemann Hurwitz, with an action of $G$, and the quotient of that is the genus $g$ orbifold curve. The $\psi$ classes of either universal curve is essentially the same (except maybe a factor of $1/r$, where $r$ is the size of the orbifold structure). But the Hodge bundle $\mathbb{E}$ is much richer now as we can take differential forms on the higher genus curve, and the group $G$ will act on it, and we can decompose it into isotypical components and get many bundles...

But as you say, I feel like this is stuff you "know". If there's a really specific thing you want to know, let's try to get on that, but otherwise, to help you with the unease I'd just reiterate that you never really want to think about this "$\overline{\mathcal{M}}^{\mathbb{Z}/2}_{g,n}$" at all, but want to use $\overline{\mathcal{M}}_{g,n}(\mathcal{B}\mathbb{Z}_2)$.

  • $\begingroup$ I think that I framed this question poorly. I'm not so explicitly interested in the moduli space of hyperelliptic curves thought of as twisted stable maps into $B\mathbb{Z}/2$, but really the difference between the moduli space of stable curves and the moduli space of stable twisted curves. I will edit the question. $\endgroup$
    – Simon Rose
    Oct 24, 2014 at 8:10

If you want to express the stack of hyperelliptic curves in terms of a genus zero moduli problem, you have a few options:

  1. You may view a hyperelliptic curve equipped with its hyperelliptic involution as a $\mathbb{Z}/2\mathbb{Z}$-torsor over a twisted genus zero curve with étale $\mathbb{Z}/2\mathbb{Z}$-gerbes attached to the $2g+2$ disjoint sections. Then you can write this space as a component of the Hom stack $\underline{\operatorname{Hom}}(\mathcal{M}_{0,2g+2}^{tw,2},B(\mathbb{Z}/2\mathbb{Z}))$ corresponding to representable objects. When $g>1$, this is the Abramovich-Vistoli twisted stable maps stack $\mathcal{K}_{0,2g+2}(B(\mathbb{Z}/2\mathbb{Z}))$, also written as $\mathcal{B}^{bal}_{0,2g+2}(\mathbb{Z}/2\mathbb{Z})$.

  2. You may view a hyperelliptic curve as a ramified double cover of a marked genus zero curve. Then you can use the Abramovich-Corti-Vistoli stack $\mathcal{A}dm_{0,2g+2}(\mathbb{Z}/2\mathbb{Z})$ of admissible $\mathbb{Z}/2\mathbb{Z}$-covers.

These stacks are equivalent (ACV Theorem 4.3.2), but you have to forget the markings (or the distinguished choice of hyperelliptic involution) to get an honest stack of hyperelliptic curves.

The coarse moduli map induces a morphism between stack of twisted curves and the stack of marked curves. The main difference between them is the deformation theory at the boundary. If you have a versal deformation space $\operatorname{Spec} R$ for nodal curves, with smooth divisors $D_i$ defined by elements $t_i \in R$ parametrizing deformations where a nodal point $q_i$ remains a node, then the versal deformation space for twisted nodal curves has the form $$\operatorname{Spec} R[z_1,\ldots,z_n]/(t^{r_1}-z_1,\ldots,t^{r_n}-z_n)$$ where $r_i$ is the inertia at $q_i$. This is described (essentially verbatim) in the first section of Olsson's "On log twisted curves".

  • $\begingroup$ As I commented on the other response (although this is a good answer!), I think I framed the question poorly. I'm more interested in the difference between the moduli space of stable curves and the moduli space of stable twisted curves, which was (it seems) not clear from the question. $\endgroup$
    – Simon Rose
    Oct 24, 2014 at 8:11

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