2
$\begingroup$

Consider the moduli space $$ \overline{M}_{0,4}(B\mathbb{Z}/2) $$ This has virtual (and real) dimension one. In a certain sense this moduli space paramaterizes "genus 1 hyperelliptic curves"; that is, given a family of objects in this moduli space, by pulling back the universal family $pt \to B\mathbb{Z}/2$ we obtain a genus one curve with marked 2-torsion points. In particular, there is a morphism $$ \overline{M}_{0,4}(B\mathbb{Z}/2) \to \overline{M}_{1,4} $$ whose image is those genus 1 curves whose four marked points are exactly the 2-torsion points.

Because of this map, we can pull back the Hodge bundle to $\overline{M}_{0,4}(B\mathbb{Z}/2)$, and so we can consider the integral $$ \int_{\overline{M}_{0,4}(B\mathbb{Z}/2)} \lambda_1 $$

Question What is the value of this integral?

Some thoughts: We know that $\int_{\overline{M}_{1,1}} \lambda_1 = \frac{1}{24}$. Since the Hodge bundle pulls back via the forgetful maps (forgetting marked points), it seems that we should end up with either $3!\frac{1}{24} = \frac{1}{4}$ (choosing our favourite marked point, mapping to $\overline{M}_{1,1}$) or $4!\frac{1}{24} = 1$ (forgetting all the marked points).

I suppose another way of phrasing this: What is the degree of the map $\overline{M}_{0,4}(B\mathbb{Z}/2) \to \overline{M}_{1,1}$?

$\endgroup$
4
$\begingroup$

You are right that it's equivalent to compute the degree of the forgetful map. The degree is $3!$, since it's the quotient by $S_3$ permuting the last three markings. If you divide by $S_4$ you get the moduli stack of genus one curves with a distinguished degree $2$ map to $\mathbf P^1$, which is not the same as $\overline M_{1,1}$.

In fact there is a canonical isomorphism of stacks $\overline M_{0,4}(B\mathbf Z/2) \cong X(2)$, where $X(2)$ is the modular curve for the full level 2 congruence subgroup, parametrizing elliptic curves with a basis of their 2-torsion. This is easy to show on the interior. To get the result also on the boundary, use the modular interpretation of the cusps of $X(2)$ described in Deligne--Rapoport. For any $T \to X(2)$, you get a family $E \to T$ of generalized elliptic curves, and inversion in the group structure defines an admissible double cover $E \to E/\langle \pm 1\rangle$ branched over four points where $E/\langle \pm 1\rangle$ is a rational curve, hence a $T$-point of $\overline M_{0,4}(B\mathbf Z/2)$.

$\endgroup$
  • $\begingroup$ Hmmm. Now I'm wondering if maybe I phrased the question wrong. Technically, the integral that I'm interested in isn't over $\overline{M}_{0,4}(B\mathbb{Z}/2)$, but in a certain sense over $\mathfrak{M}_{0,4}$. $\endgroup$ – Simon Rose Jan 15 '15 at 8:43
  • $\begingroup$ That is, over the stack of pre-stable stacky curves (whose marked points all have $\mathbb{Z}/2$ stabilizers). $\endgroup$ – Simon Rose Jan 15 '15 at 8:44

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.