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!
Maybe folks should start holding “research” office hours?
Daniel Litt, for his office hours, has the advantage that, because the questions grad students have are generally easier – they often want to know the basics of a field – he can give a reasonable answer in many more fields of arithmetic geometry than he does research in and could be called an expert on.
For instance, when I stopped by recently, someone asked about the eigencurve, and he and I were both able to give helpful information even though we are not any kinds of experts and someone like our illustrious host would be much more qualified.
Questions from already established researches are likely to be substantially more specialized. So you would need more experts there to have a lively session. Also, everyone is hopefully an expert in something, so it might make sense to have a roundtable format where everyone asks and answers than an office hours.
I think such a thing would be good, but it might be hard to organize outside the discussion section of a conference.
\(g = 2\) is special, for \(g > 2\) the Picard group of \(\mathcal{M}_g\) is indeed \(\mathbf{Z}\) (at least over \(\mathbb{C}\)) (see, e.g., Albarello–Cornalba “Picard groups of the moduli spaces of curves“. The paper of Arsie–Vistoli “Stacks of cyclic covers of projective spaces” contains a computation of the Picard group of the moduli stack of hyperelliptic curves from which it follows that \(\mathrm{Pic}(\mathcal{M}_2) = \mathbf{Z}/10 \mathbf{Z}\).
Excellent! So that surely means that \(\omega\) is the generator \(10\) and that \(\Delta\) as a Siegel modular form has weight \(10\). So now I guess I am implicitly suggesting that \(\omega\) admits a square root over \(\mathcal{M}^{w}_2\). Naïvely I would have even guess that there is an \(\mathcal{O}(1)\) on \(\mathbf{P}(4,6,8,10) \setminus (\Delta = 0)\) which had order \(40\) and may even generate \(\mathrm{Pic}(\mathcal{M}^{w}_2)\) but stacky weighted projective spaces confuse me.
I don’t see any conceptual way of showing that a square root of \(\omega\) (which is indeed a generator of \(\mathrm{Pic}(\mathcal{M}_2)\)) exists on \(\mathcal{M}_2^w\), but it should be easy to compute the Picard group of \(\mathbf{P}(4,6,8,10) \backslash (\Delta = 0)\). In general, if \(X\) is say a smooth variety and \(G\) is an algebraic group acting on \(X\), then the Picard group of the quotient stack \([X/G]\) can be computed as follows:
Let \(V\) be any linear representation of \(G\) on which the action is sufficiently free–to compute the Picard group it suffices to assume that the codimension of the non-free locus is at least two–and let \(U\) be the locus on which the action is free. Then \(\mathrm{Pic}([X/G]) = \mathrm{Pic}((X \times U)/G)\) where the action is the diagonal action. (Of course, the point here is that \((X \times U)/G\) is a smooth variety.)
If \(G = \mathbf{G}_m\), then one can just take \(V = \mathbf{A}^2\) with the action \(\lambda\cdot(x,y) = (\lambda x, \lambda y)\). In the case of weighted projective space one may even replace \(X = \mathbf{A}^n – 0\) by \(\mathbf{A}^n\) when computing \(\mathrm{Pic}(X \times U/\mathbf{G}_m)\) so the quotient is an affine bundle over \(\mathbf{P}^1\). In your example, I think for any irreducible \(\Delta\) (with the given homogeneity properties) one can then see that the Picard group is indeed \(\mathbf{Z}/(40)\).
Sorry again. I did a search and read that $latex $ was the right thing to use to make latex work on wordpress…
Actually with the relevant plugin I am using instead of “$latex” you start math formulas with \ followed by (, and then end them with \ followed by ).
Ooh, which plugin is this?
It’s MathJax-LaTeX. Unfortunately, this is not available on wordpress blogs hosted on wordpress.org (which is free) which means that I one has to pay $100 a year (or so) to use it. Indeed, this is precisely why I moved this blog, so I could use this instead of the rather crappy LaTeX version on the freely hosted site.
ps. I guess it is incumbent on me to remark that you are not the David Roberts who is the coauthor on the paper discussed above. That’s clear to you and to anyone who clicks on your username, but it did confuse me temporarily.