Here are some algebraic geometry musings related to the last post, most of which is hopefully correct. Everything below is secretly over \(\mathbf{Z}[1/6]\) but I think one may as well think about what is happening over \(\mathbf{C}\). Warning: I don’t know any algebraic geometry, please correct me if you see any nonsense.
As mentioned in the last post, if you fix a \(3\)-torsion representation with cyclotomic determinant and look at the corresponding moduli space of elliptic curves with this \(3\)-torsion, you get a \(\mathbf{P}^1\) (at least accounting for cusps). A natural followup question is: what geometric object do you get over the stack \(\mathcal{A}_{1} = \mathcal{M}_{1,1}\)?
Thinking about stacks in the most naive way, we just consider
\(y^2 = x^3 + a x + b\)
for \((a,b)\) in \(\mathbf{P}(4,6)\) minus \(\Delta = 0\) in the stacky sense. But just thinking about this as an elliptic curve over \(\mathbf{Q}(a,b)\), you can write down:
\(y^2 = x^3 + A x + B\)
where
\(\begin{eqnarray*}
3A(a,b,s,t) & = & 3 a s^4 +18 b s^3 t -6 a^2 s^2 t^2 -6 a b s t^3 -(a^3+9
b^2) t^4, \\
9B(a,b,s,t) & = & 9 b s^6-12 a^2 s^5 t-45 a b s^4 t^2-90 b^2 s^3 t^3 + 15 a^2 b s^2 t^4 \\
&& \qquad -2 a
(2 a^3+9 b^2 ) s t^5 -3 b (a^3+6 b^2 ) t^6,
\end{eqnarray*}
\)
Now one thing you notice straight away about these equations is that they change when one replaces \(a,b\) by \(a \lambda^4, b \lambda^6\), namely:
\(A(\lambda^4 a, \lambda^6 b,s,t) = A(a,b, \lambda s, \lambda^3 t)\)
and the same equation holds for \(B\). That is, the parametrization of \(\mathbf{P}^1\) changes, and so the family is not literally projective space over this stack. Of course, if
\(\Delta(a,b) = 16(-4 a^3 – 27 b^2),\)
then
\(\Delta(a \lambda^4,b \lambda^6) = \lambda^{12} \Delta(a,b),\)
where \(\Delta\) trivializes \(\omega^{12}\). In order to remove the ambiguity, one can then define
\(\displaystyle{A^*(a,b,s,t) = A \left(a,b, \frac{s}{\Delta^{1/12}},\frac{t}{\Delta^{3/12}}\right)}\)
and similarly with \(B^*\), then the equation is well defined, at least after addressing the issue of taking 12th roots correctly. This suggests that after pulling back to the space where you adjoin \(\Delta^{1/12}\) you get projective space, but that the original space is not projective space at all but maybe something like the projective bundle
\(\mathrm{Proj}(\mathcal{O}_X \oplus \omega^2) = \mathrm{Proj}(\omega \oplus \omega^3)\)
where \(\omega\) is the usual line bundle which has order \(12\) in the Picard group of \(\mathcal{A}_{1}\).
Something very similar happens for the equations for families of fixed three torsion over \(\mathcal{M}^{w}_2\), the moduli stack of genus two curves with a fixed Weierstrass point. In this case, the base looks like
\(y^2 = x^5 + a x^3 + b x^2 + c x + d\)
or \(\mathbf{P}(4,6,8,10)\) minus \(\Delta = 0\). (You need to be a little bit more careful at the prime \(5\).) Here the corresponding identity for \(A,B,C,D\) is
\(A(\lambda^4 a,\lambda^6 b,\lambda^8 c,\lambda^{10} d,s,t,u,v)
= A(a,b,c,d,\lambda s, \lambda^7 t,\lambda^{13} u,\lambda^{19} v)\)
and
\(\Delta(\lambda^4 a,\lambda^6 b,\lambda^8 c,\lambda^{10} d) = \lambda^{40} \Delta(a,b,c,d)\).
So now one wants to trivialize the family by taking the cover with various roots of \(\Delta\), including \(\Delta^{1/20}\). Except now I don’t really know what the Picard group of \(\mathcal{M}^{w}_2\) is. Somehow I first assumed that the Picard group would be the same as that of the corresponding moduli space of abelian surfaces \(\mathcal{A}^{w}_2\), and since \(\Delta\) seems to give a trivialization of some power of the determinant bundle it should be related to torsion in \(H_1(\Gamma,\mathbf{Z})\) for the corresponding congruence subgroup \(\Gamma\) of \(\mathrm{Sp}_4(\mathbf{Z})\). But because of the congruence subgroup property, presumably \(H_1(\mathrm{Sp}_4(\mathbf{Z}),\mathbf{Z})\) is equal to \(\mathbf{Z}/2 \mathbf{Z}\), and that’s not going to change by taking the map to \(S_6 = \mathrm{PSp_4(\mathbf{F}_2)}\) and taking the pre-image of \(S_5\). But it is pure folly to imagine the Picard group of \(\mathcal{M}^{w}_2\) and \(\mathcal{A}^{w}_2\) coincide. The latter contains an extra divisor, the Humbert divisor, consisting of direct sums of elliptic curves. Moreover, (I guess) the Siegel modular form corresponding to \(\Delta\) is probably very close to the Igusa form, which vanishes not only at the cusp but also along the Humbert divisor. So the line bundle \(\omega\) on \(\mathcal{A}_2\) has infinite order even though its pullback to \(\mathcal{M}_2\) does not because \(\Delta\) itself is giving a trivialization of some power of \(\omega\). So it is indeed plausible that abelianization of the corresponding (index five subgroup of) the \(g = 2\) Torelli group has \(20\)-torsion. One way to try to compute this is to explicitly compute the abelianization of the corresponding cover of the mapping class group (I guess there are explicit presentations?). So the first question is can someone confirm that \(\mathrm{Pic}(\mathcal{M}^{w}_2)\) does indeed have \(20\)-torsion? If only there was someone in my department who could prime me on the properties of mapping class groups… Actually, Andrew Putman is probably the obvious person to ask. The second problem is confirm that the family explicitly computed in the last post does indeed coincide with \(\mathrm{Proj}(\mathcal{O}_X \oplus \omega^6 \oplus \omega^{12} \oplus \omega^{18})\).
I confess my efforts to do a literature search in this case have not been very thorough. In my mind I somehow thought that the Picard group of the stack \(\mathcal{M}_g\) (for \(g \ge 2\)) was \(\mathbf{Z}\), but that is transparently false, at least for \(g = 2\). I got as far as doing a google search for Picard groups of moduli stacks and found a few pages of notes written by Daniel Litt. So I naturally zoomed in to Daniel Litt’s office hours once after he advertised them on twitter… but I soon realized that it would take too long to explain and he had better things to do like explaining modular forms to his students… so here it is now in blog form!