I am trying to improve my moderate knowledge of moduli spaces/stacks by examining the moduli stack of stable double covers of $\mathbb{P}^1$ with one marked point. My idea is to ignore the stack structure at first, construct a moduli space and take the stack structure finally into account by displaying the stack as an algebraic groupoid $[M/G]$, with the action of $G$ on $M$ counting in automorphisms of the domain if necessary. A naive first approach is to describe the branching locus together with the marked point via $\mathbb{P}^1 \times \mathbb{P}^1 \times \mathbb{P}^1$ , letting the first and second component describe the branch points and the third component describe the marked point.

So far I ignore the fact that the marked point is only on one sheet of the domain, is it legit?

Now when it comes to taking care of automorphisms I struggle. The generic case (three distinct points) should only possess a trivial automorphism group, but degenerate cases (e.g. the marked point coinciding with one of the branch points) hold automorphisms (e.g. the automorphism swapping the sheets of the domain).

Are there further automorphisms to be considered and how do I arrive at a global presentation of the form $[M/G]$ (is it even possible)? Last but not least I am not sure if I have to adjust the unnatural labels of the branch points by a $\mathbb{Z}_2$ action arriving at something like $\mathbb{P}^1 \times \mathbb{P}^1 \times \mathbb{P}^1/\mathbb{Z}_2$.

I am thankful for any advice.


1 Answer 1


It's very important to define things carefully the problem of interest before constructing the moduli space / stack. In the most general setting you want to carefully define the moduli functor, but you should always at least have a precise idea of what the field-valued points are.

For instance, you don't specify how many branch points / what genus you want the covers to have. I guess you want two branch points, and genus zero.

You also have to define the stability condition for a cover. The most reasonable choice is probably Kontsevich's definition of stable maps, where double cover is defined to be any stable map of degree 2, but you could also take a double cover to literally be a two-to-one map with no constant components. This is no problem in the genus zero case because there are no stable covers with constant component.

Your construction is completely valid for the coarse moduli space, although of course you are right that you have to quotient by $\mathbb Z/2$ switching the two marked points. However, taking that quotient in stacks gives the wrong answer, because it implies you have extra automorphism when the two points coincide. Assuming you interpret this as a (stable) nodal curve, that extra automorphism is meaningless. Instead it's better to use the isomorphism $(\mathbb P^1 \times \mathbb P^1 ) / (\mathbb Z/2) = \mathbb P^2$, which you can see explicitly by viewing a degree $2$ branch divisor as a line inside the space of sections of $\mathcal O(2)$.

The moduli stack of double covers of $\mathbb P^1$ is not quite $\mathbb P^2$, because we have to account for the hyerelliptic involution. In fact, it is the quotient of $\mathbb A^3 - \{0\}$ by $\mathbb G_m$, where $\mathbb G_m$ acts via scalar multiplication by the square. To see this, represent a double cover as the relative Spec of $\mathcal O_{\mathbb P^1} + \mathcal O_{\mathbb P^1}(-1)$, with multiplication given by $(a_1,b_1)\times(a_2,b_2) \to (a_1a_2 + f b_1b_2, a_1b_2 + a_2b_1)$ for $f$ a nonzero section of $\mathcal O_{\mathbb P^1}(2)$. This defines a double cover, every double cover can be defined this way, and equivalences between the double covers defined by $f_1$ and $f_2$ are parameterized by $\lambda \in \mathbb G_m$ with $f_2=\lambda^2 f_1$. Locally, this is equal to the quotient of $\mathbb P^2$ by $ \pm 1 \subset \mathbb G_m$, but not globally (as I earlier claimed).

To construct the moduli stack in this setting, you have to take the universal family of double covers over this stack. In other words, you take the universal family of double covers over $\mathbb A^3 - \{0\}$, constructed by exactly this relative Spec, and quotient by this $\mathbb G_m$ action. This works because in general, the moduli stack of varieties with one marked point is the universal family over the moduli stack of varieties. For curves, this is true with Deligne-Mumford or Kontsevich stability conditions even if the marked point touches a singularity, although this case requires blowing up the universal family of curves over the moduli space of curves with marked points to produce the universal family of curves with marked points.

With greater numbers of branch points, this won't quite work, as when three or more branch points coincide the double cover is no longer stable, and you will need to do some blow-up, or other tricks, to deal with it.

  • $\begingroup$ Thanks a lot for your detailed answer. In deed i am interested in genus zero covers/ 2 branch points ( it is an equivalence due to Riemann-Hurwitz, isn't it?). Furthermore I had Kontsevich stability in mind. Am I getting it right that by exchanging $\mathbb{P}^1 \times \mathbb{P}^1$ with $\mathbb{P}^2$ I avoid having to deal with the two marked branch points globally on the whole stack? Could you explain in detail why the action of $\mathbb{Z}/2$ is trivial on $\mathbb{P}^2$? Last but not least, do I have to pay special attention to the case when all three points coincide? $\endgroup$
    – gmp
    Mar 3, 2018 at 20:19
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    $\begingroup$ @gmp Yes it's equivalent. Yes, $\mathbb P^2$ is the moduli space of degree $2$ divisors on $\mathbb P^1$, so doing it this way you don't have to deal with two marked branch points (which is not the right choice anyways). I was wrong about the $\mathbb Z/2$ quotient, it's actually a gerbe, I will edit to explain. The way the stability conditions are set up, you don't have to worry too much about that - though if another marked point was there it would require some attention. When the two branch points collide, that forms a node. When the node would be marked, you blow it up. $\endgroup$
    – Will Sawin
    Mar 3, 2018 at 20:55
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    $\begingroup$ @gmp But there is still exactly one stable curve, with no extra automorphisms or trickery, that corresponds to that blown-up curve. One has to check something more carefully at the level of functors and stacks but this has all been worked out. $\endgroup$
    – Will Sawin
    Mar 3, 2018 at 20:56
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    $\begingroup$ @gmp The double cover necessarily is a finite flat map to $\mathbb P^1$, hence the relative spectrum of a vector bundle, which must be rank two. This vector bundle admits an automorphism, the hypereliptic/Galois involution, with eigenvalues $1$ and $-1$. The eigenvalue $1$ part must be equal to the base ring, and has degree $0$. The eigenvalue $-1$ part must have degree $-1$ to make the arithmetic genus match up. I have written don the most general multiplication law for such a sum. I don't know a more concrete presentation than the relative spectrum here. $\endgroup$
    – Will Sawin
    Mar 8, 2018 at 7:07
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    $\begingroup$ @gmp I guess an alternate description is that it is the locus in $\mathbb A^3 - 0 $ times the space of pairs of a point $x \in \mathbb P^1$ and a section of $\mathcal O(1)$ at that point (i.e. literally a line bundle over $\mathbb P^1) where the sections satisfies $s^2=f(x)$ with $f$ the section of $\mathbb A^3-0$. $\endgroup$
    – Will Sawin
    Mar 8, 2018 at 7:10

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