Chidambaram on genus two curves, II

We now continue a series of posts on the work of my student Shiva Chidambaram. (Click here for part I.) Today I would like to discuss another project with Shiva that was also joint with David Roberts (no, not David Roberts).

We saw last time that the moduli spaces \(\mathcal{A}_2(\rho)\) and \(\mathcal{M}_2(\rho)\) are not in general rational over \(\mathbf{Q}\). On the other hand, the degree six cover \(\mathcal{M}^w_2(\rho)\) is always rational. So the next question is: what is an explicit parametrization? Slightly differently, start with a genus two curve with a Weierstrass point

\(y^2 = x^5 + a x^3 + b x^2 + c x + d\)

Problem: Parametrize all other genus two curves with a Weierstrass point which have the same \(3\)-torsion representation.

It might be worth briefly revisiting the argument from [BCGP] that such a parameterization exists. The key point is that there is an birational map

\(\mathcal{M}^{w}_2(3) \rightarrow \mathbf{P}^3\)

which is \(\mathrm{PSp}_4(\mathbf{F}_3)\)-equivariant. This allows one to show that the corresponding twists are Brauer-Severi varieties, and then deduce they are rational by the same group theoretic trick which appears in this paper of Shepherd-Barron and Richard Taylor. More explicitly, there are maps

\( H^1(\mathbf{Q},\mathrm{GL}_4(\overline{\mathbf{Q}}))
\rightarrow H^1(\mathbf{Q},\mathrm{PGL}_4(\overline{\mathbf{Q}})) \rightarrow H^2(\mathbf{Q},\overline{\mathbf{Q}}^{\times})\)

Here the LHS is trivial by Hilbert 90. One shows, using the fact that the Darstellungsgruppe of \(\mathrm{PSp}_4(\mathbf{F}_3)\) is \(\mathrm{Sp}_4(\mathbf{F}_3)\), that the cocycle corresponding to any Galois representation \(\rho: G_{\mathbf{Q}} \rightarrow \mathrm{GSp}_4(\mathbf{F}_3)\) with cyclotomic determinant lifts in an explicit way to a cocycle in the LHS and hence is trivial. The problem is to bridge the gap between a theoretical argument that a cocycle is trivial and a way to produce an equation of the corresponding twist. That amounts to the problem of taking a cocycle

\(Z^1(\mathbf{Q},\mathrm{GL}_4(\overline{\mathbf{Q}}))\)

and writing it as a coboundary. Before going further, it’s worth pointing out that the case of \((g,p) = (2,3)\) is very similar (but more complicated) than the case of \((g,p) = (1,3)\). In the latter case, one has that

\(\mathrm{dim} H^0(X(3),\omega) = 2,\)

and this simple equality leads to an idenfication of \(X(3)\) with \(\mathrm{Proj} H^0(X(3),\omega)\). So, let’s talk about the problem of parametrizing elliptic curves as a warm-up case. If you start with a curve

\(E: y^2 = x^3 + a x + b\)

for which (for convenience of exposition) you assume \(\rho_{E,3}\) surjective, then the splitting field \(L/\mathbf{Q}\) is a \(\mathrm{GL}_2(\mathbf{F}_3)\) extension \(L\) which contains \(F = \mathbf{Q}(\sqrt{-3})\). There is an isomorphism of \(H = \mathrm{SL}_2(\mathbf{F}_3)\)-modules

\(L = F[H]\).

The group \(H\) admits a certain specific \(2\)-dimensional representation \(V\), and the representation \(\rho\) can be interpreted as giving an explicit map

\(V \rightarrow L.\)

OTOH, the identification above means there is an inclusion \(V^2 \rightarrow L\) because \(\mathrm{dim}(V) = 2.\) The problem of solving Hilbert 90 (ignoring a certain descent from \(F\) to \(\mathbf{Q}\)) then becomes the question of finding the “other” extension. Now if you can write \(L\) down you can do this by linear algebra. But even in any specific example, \(L\) has degree 48, and computations with fields of that size can be pretty formidable and are at the limit of what one can do with explicit number fields in (say) pari/gp or magma.

Of course, one wants to do this over the field \(\mathbf{Q}(a,b)\) with general parameters in order to have the general formula. The extension \(L\), for example, is the Galois closure of the equation

\(27 y^8 + 216 b y^6 – 18 \Delta y^4 – \Delta^2 = 0,\)

but you probably don’t want to even write down a polynomial of degree 48 in these general variables which gives \(L\), let alone try to compute the Galois action. We did succeed in solving this problem by a certain amount of trickery — working in special cases and making the right ansatz for the general case. There were many intermediate formulas which involved polynomials with (say) 100 terms, but the final answer turns out to be perhaps surprisingly simple, namely, the general equation is given by

\(y^2 = x^3 + A(a,b,s,t) x + B(a,b,s,t),\)

where

\(
\begin{eqnarray*}
3A(a,b,s,t) & = & 3 a s^4 +18 b s^3 t -6 a^2 s^2 t^2 -6 a b s t^3 -(a^3+9
b^2) t^4, \\
9B(a,b,s,t) & = & 9 b s^6-12 a^2 s^5 t-45 a b s^4 t^2-90 b^2 s^3 t^3 + 15 a^2 b s^2 t^4 \\
&& \qquad -2 a
(2 a^3+9 b^2 ) s t^5 -3 b (a^3+6 b^2 ) t^6.
\end{eqnarray*}
\)

Here \([s,t]\) is the \(\mathbf{P}^1\) parameter. Curiously enough this exact formula was also found before in the literature. That reflects something a little surprising about this equation. The moduli space we are looking for is a \(\mathbf{P}^1\), and this has many automorphisms. On the other hand, we are starting with an \(E\) so we have a fixed point normalized here to be \([1,0]\). But the projective line with one fixed point still has many automorphisms! However, it turns out that there is some extra hidden structure which gives rise to a second canonical point normalized here as \([0,1]\), which is why different people would possibly end up with the same equation independently. (\(\mathbf{P}^1\) with two fixed points still has a \(\mathbb{G}_m\)’s worth of automorphisms, but an informal consideration of the integral structure can be used to pin this down further.) The map which takes one \(E\) and spits out the other point therefore ends up giving a canonical rational map on \(\mathbf{P}^1_j\) which has the property that it preserves the (projective) \(3\)-torsion representation. Explicitly it is given by:

\(\displaystyle{j \mapsto \frac{(6912 – j)^3}{27 j^2}}\)

I wonder if this has interesting dynamical properties?

The computation above was not so easy, even though the answer turned out to be simple enough. But for \((g,p) = (2,3)\) things are looking pretty bad. First of all, the extension \(L\) now has degree \(103680\), which one is not going to write down explicitly. Even the analogue of the degree \(8\)-polynomial above is a degree \(40\) polynomial in \(x^6\) with \(1673\) terms.

Despite that, we found the answer:

Theorem: [C, Shiva Chidambaram, David Roberts] There exist (and we compute) explicit polynomials \(A,B,C,D\) in \(\mathbf{Q}[a,b,c,d,s,t,u,v]\) which specialize to \(a,b,c,d\) at \([s,t,u,v]=[1,0,0,0]\) such that

\(y^2 = x^5 + A x^3 + B x^2 + C x + D\)

is the general genus two curve with a rational Weierstrass point and fixed \(3\)-torsion representation.

Even though we find the simplest form of these polynomials, they turn out to be quite big. As in, the number of monomial terms they contain are \(14671\), \(112933\), \(515454\), and \(1727921\) respectively. The text files were so big that I ran into space problems on my university account! (OK, so it’s only 200MB or so, but that’s a big text file!)

The reason such a computation is ultimately possible relates to an accidental fact that is common between the two cases, namely, that the groups \(\mathrm{SL}_2(\mathbf{F}_3)\) and \(\mathrm{Sp}_4(\mathbf{F}_3) \times \mathbf{Z}/3\mathbf{Z}\) are two of the 37 exceptional complex reflection groups as determined by Shephard and Todd. The story is explained in our paper so I won’t discuss it here, but it might be worth mentioning two further facts:

The first is that these methods can also deal (in principle) with an analogue of this problem for \(g = 3\) and \(p = 2\). Just as with \(g = 2\), the moduli space which admits an equivariant birational map to \(\mathbf{P}^6\) is not \(\mathcal{M}_3(2)\) but once more a finite cover, and this cover does not correspond to any level structure but rather some cover coming genuinely from the mapping class group. This picture relates to the isomorphism \(\mathbf{Z}/2 \mathbf{Z} \times \mathrm{Sp}_6(\mathbf{F}_2) \simeq W(E_7)\), another exceptional complex reflection group. There is even a less analogous version for \(g = 4\) and \(p = 2\) related to the fact that the largest complex reflection group \(W(E_8)\) admits a description \(W(E_8) \simeq 2.\mathrm{O}^{+}_8(\mathbf{F}_2):2\), and the projective version of this group \(\mathrm{O}^{+}_8(\mathbf{F}_2):2\) is a subgroup of \(\mathrm{Sp}_8(\mathbf{F}_2)\), although of genuine index (\(136\)) rather than as an isomorphism, which is the main reason why this is a little different to the other cases. We estimated that an explicit version of the last moduli problem would involve polynomials with approximately 100 trillion terms, so needless to say we did not try to compute it.

Second, there is an interesting story concerning the auxiliary copy of \(\mathbf{Z}/3 \mathbf{Z}\) that turns up in the \(g = 2\) setting. The formulas that we write down actually correspond not only to projective spaces \(\mathbf{P}^1\) and \(\mathbf{P}^3\) but actually to affine spaces \(\mathbf{A}^2\) and \(\mathbf{A}^4\) which represent moduli problems related to the complex reflection groups. In these affine families, not only is the representation corresponding to \(\mathrm{ker}(\rho)\) fixed, but the splitting field of \(X^3 – \Delta\) also remains unchanged. When \(g = 1\), this is not surprising, because the \(S_3\) extension comes from the map \(\mathrm{GL}_2(\mathbf{F}_3) = \widetilde{S_4} \rightarrow S_4 \rightarrow S_3\). On the other hand, that’s obviously not happening in the genus two case where the group is almost simple. This is a little peculiar! However, it related to the fact that the splitting field of \(X^3 – \Delta\) for genus two curves depends on the Weierstrass equation. If you scale the Weierstrass equation by (\(x,y) \mapsto (t^2 x,t^5 y)\), this sends \(\Delta \rightarrow t^{40} \Delta\). So the affine equation represents a moduli space for some larger group which disappears when considering the equation projectively, and you can always normalize your Weierstrass equation so that \(\Delta\) is a perfect cube.