Chidambaram on genus two curves, I

Before we start, just to alert you to a minor blogpage design change: all the posts (including this one) which talk about my students work can be accessed in one place by clicking the “work of my students” tab just below the picture on the top of this page.
resume normal service.

Those who study elliptic curves certainly know that if you start with an elliptic curve \(E/\mathbf{Q}\), the \(p\)-torsion gives rise to a Galois representation:

\(\rho: G_{\mathbf{Q}} \rightarrow \mathrm{GL}_2(\mathbf{F}_p)\)

with cyclotomic determinant. Conversely, if \(p = 2,3,5\) then the converse is true, that is, any such Galois representation comes from an elliptic curve. Moreover, any such representation comes from an infinite number of curves which are parametrized by \(\mathbf{P}^1_{\mathbf{Q}}\). This is intimately related to the fact that the curves \(X(p)\) have genus zero for these \(p\).

What is also true is that, given any \(E\), one can write down explicit parametrizations of these families. This was done by Rubin and Silberberg for \(p=3,5\) around the time Fermat’s last theorem was proved. Indeed, the idea of passing between elliptic curves with the same mod-\(3\) Galois representation features prominently in Wiles’ argument.

One might ask what happens for higher genus. First of all, there is a geometric problem over the complex numbers: when is the moduli space \(\mathcal{A}_g(p)\) of PPAV of dimension \(g\) with full \(p\)-level structure a rational variety? It turns out the only possibilities when \(g > 1\) are \(p=2,3\) when \(g = 2\) and \(p=2\) when \(g = 3\). The case \((g,p) = (2,3)\) arose in my work with Boxer, Gee, and Pilloni (discussed here). In that paper, we proved a weaker version of the result above, namely the following:

Proposition: [BCGP] If \(\rho: G_{\mathbf{Q}} \rightarrow \mathrm{GSp}_4(\mathbf{F}_3)\) is any continuous representation with cyclotomic similitude character, then the corresponding twist \(\mathcal{A}_2(\rho)\) is unirational over \(\mathbf{Q}\) via a map of degree at most six. In particular, it has many rational points.

The degree six cover is not so mysterious. When \(g = 2\), PPAV are (more or less, the less being the Humbert surface) are Jacobians of genus two curves. So certainly birationally one can replace \(\mathcal{A}_2(\rho)\) by \(\mathcal{M}_2(\rho)\), the moduli of genus two curves whose Jacobian has the given \(3\)-torsion. The degree six cover is then the moduli of genus two curves with a fixed Weierstrass point, or, more prosaically, the genus two curves of the form:

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

whose Jacobian has the given \(3\)-torsion. (Any fixed Weierstrass point can be moved to \(\infty\), and then the \(x^4\) term can be surpressed by an obvious linear transformation.) This moduli space will be discussed in more detail in the next post. But for now, this leaves open the question of whether \(\mathcal{A}_2(\rho)\) itself is rational.

Over the complex nubmers, things are well understood. The space \(\mathcal{A}_2(3)\) has a number of compactifications, including the (singular) Satake compactification, and the various smooth toroidal compactifications. When \(g = 2\), things work out extra nicely: there is a somewhat canonical compactification \(\mathcal{A}^*_2(3)\) due to Igusa. It turns out that \(\mathcal{A}_2(\rho)\) is birational to a very nice \(3\)-fold known as the Burkhardt quartic. The Burkhardt quartic is given explicitly in \(\mathbf{P}^5\) by the equations:

\(\sigma_1 = x_0 + x_1 + x_2 + x_3 + x_4 + x_5 = 0\),
\(\sigma_4 = x_0 x_1 x_2 x_3 + \ldots + x_2 x_3 x_4 x_5 = 0.\)

Eliminating any variable using the first equation leads to a quartic in \(\mathbf{P}^4\), but this is the most symmetric presentation. This variety \(\mathcal{B}\) is singular and has \(45\)-nodes — a maximal number, as it turns out. Not surprisingly, it also has an action by automorphisms of the simple group \(G = \mathrm{PSp_4}(\mathbf{F}_3)\). Blowing up \(\mathcal{B}\) at these nodes gives the smooth variety \(\mathcal{A}^*_2(3)\).

Things are more subtle over \(\mathbf{Q}\). It turns out that for the trivial level \(3\) structure corresponding to the representation \((\mathbf{Z}/3 \mathbf{Z})^2 \oplus (\mu_3)^2\) with the obvious symplectic structure, the corresponding variety \(\mathcal{A}^*_2(3)\) is still rational (e.g. see here). But it is no longer so obvious whether the twists we are considering should be rational over \(\mathbf{Q}\) or not. (There are actually some twists of a different flavor which don’t have points, but all the ones we are considering do.) Note there is a big difference between what happens in higher dimensions and what happens in dimension one: In dimension one the only unirational smooth projective curve with a rational point is projective space itself, but this is completely false in higher dimensions (for example, take products of projective spaces).

We left the question of the rationality of \(\mathcal{A}_2(\rho)\) open in [BCGP]. But my student Shiva Chidambaram took up the question. The first question is how can you prove a smooth projective variety \(X\) is not rational over \(\mathbf{Q}\) assuming that it is rational over \(\mathbf{C}\) and has rational points. One obstruction was found by Manin. If \(X\) is projective space, then the geometric Picard group of \(X\) is \(\mathbf{Z}\). The Picard group does not always have to be \(\mathbf{Z}\) for a smooth rational variety, but Manin showed that, still assuming that \(X\) is smooth and projective, if it is birational to projective space then its (geometric) Picard group is similar to the trivial representation in a technical sense we now explain. Here we say that two \(\mathbf{Z}[G_{\mathbf{Q}}]\)-modules \(A\) and \(B\) (which are free finitely generated abelian groups) are similar if there are integral permutation representations \(P\) and \(Q\) of \(G_{\mathbf{Q}}\) such that

\(A \oplus P \simeq B \oplus Q.\)

(I think one should imagine a sequence of birational maps where one introduces (or removes) the class of some cycle and all of its conjugates.)

Now we can hope to apply this in practice if we can compute the \(G_{\mathbf{Q}}\) action on \(M = \mathrm{Pic}_{\overline{\mathbf{Q}}}(\mathcal{A}^*_2(\rho))\).

How might one go about computing \(M\)? First of all, consider the non-twisted space \(\mathcal{A}^*_2(\rho)\). Using the explicit geometry of this space, one can hope to go about computing the Neron-Severi group completely explicitly, together with the action of \(G = \mathrm{PSp}_4(\mathbf{F}_3)\). And this was indeed done by Hoffman and Weintraub (amongst other things) in this paper. In particular, they show that the cohomology of this variety is all torsion free, trivial in odd degrees, and satisfies

\(H^2(X,\mathbf{Z}) \simeq H^4(X,\mathbf{Z}) = \mathbf{Z}^{61}.\)

Moreover, the cohomology is entirely generated by cycles, and these cycles are all defined over \(E = \mathbf{Q}(\sqrt{-3})\) and can be written down explicitly, together with the corresponding intersection pairing, and the action of the group \(G = \mathrm{PSp}_4(\mathbf{F}_3)\) on these cycles is self-evident because of their geometric nature. Clearly the Neron-Severi group of any twist will also be \(\mathbf{Z}^{61}\), because the geometric object is the same — the only thing that will change is the Galois action. For this, it is more convenient to work over \(E = \mathbf{Q}(\sqrt{-3})\). In this case, the \(G_{E}\) action will be as follows: the projective image of \(\rho\) when restricted to \(G_E\) factors through \(\mathrm{PSp}_4(\mathbf{F}_3)\) given the assumption on the similitude character. Thus \(\rho\) gives a canonical map

\( G_E \rightarrow G,\)

and the action of \(G_E\) on \(M\) is simply the restriction of the action of \(G\). Manin’s obstruction says that for the variety to be rational over \(E\), the action of \(G_E\) has to factor through a representation similar to the trivial representation. But that depends only on the image \(H \subset G\). Thus the problem (at least in terms of when we can apply Manin’s criterion) is “reduced” to group theory.

Some more caveats: It turns out to be pretty hard to tell if a representation is similar to the trivial representation. There is one obstruction coming from cohomology: using Shapiro’s lemma, if \(H\) is acting on \(M\) by a permutation representation, then

\(H^1(P,M) = H^1(P,M^{\vee}) = 0, \quad \text{all} \ P \subset H.\)

But then it follows that the same is true of \(M\) is similar to a permutation representation. This gives a way to explicitly verify in some cases that \(M\) is not similar to a permutation representation by finding a subgroup \(P\) for which the group above is non-trivial. Moreover, computing cohomology is something that magma can do! So it remains to:

1. Explicitly translate the description of Hoffman-Weintraub into a presentation of \(\mathbf{Z}^{61}\) as a \(G = \mathrm{PSp}_4(\mathbf{F}_3)\)-representation.
2. Determine for what subgroups \(P\) of \(G\) one has \(H^1(P,M) = H^1(P,M^{\vee}) = 0\).
3. Deduce that \(\mathcal{A}^*_2(\rho)\) is not rational whenever the projective image contains such a \(P\) as above.

The conclusion:

Theorem (C-Shiva Chidambaram): For all but \(27\) of the \(116\) conjugacy classes of \(G\), the corresponding twist \(\mathcal{A}^*_2(\rho)\) is not rational over \(E = \mathbf{Q}(\sqrt{-3})\) and hence not rational over \(\mathbf{Q}\) either. In particular, if the projective image over \(E\) has order greater than \(20\), the twist is not rational.

You actually get something stronger from Manin’s criterion — if the variety becomes rational over some map of degree \(d\), then the cohomology of the modules must be annihilated by \(d\). From our computations we find, for example:

Theorem (C-Shiva Chidambaram): Suppose that \(\rho: G_{\mathbf{Q}} \rightarrow \mathrm{GSp}_4(\mathbf{F}_3)\) is surjective with cyclotomic similitude character. Then the minimal degree of any dominant rational map \(\mathbf{P}^3_{\mathbf{Q}} \rightarrow \mathcal{A}^*_2(\rho)\) is six.

Note that from the construction of [BCGP] we know that there is such a cover of degree six, so the six in this theorem is best possible! (It was good that the computation was consistent with the existence of this cover!)

It turns out that (in the surjective case) one can give a softer argument that only depends on the rational representation. The point is that there can still be an obstruction to a rational representation to being a difference of permutation representations. This is easy enough to compute using the character table; you take the group of all virtual representations over \(\mathbf{Z}\) and compute the subgroup of all induced representations. For \(G = \mathrm{PSp}_4(\mathbf{F}_3)\), this quotient, sometimes called the Burnside cokernel (at least this is what it is called in the magma documentation), turns out to be \(\mathbf{Z}/2\mathbf{Z}\) (magma computes it). It’s also not so hard to see that there exist subgroups \(G_{40}\) and \(G_{45}\) of the obvious index such that

\([H^2(X,\mathbf{Q})] = [G/G_{40}] + [G/G_{45}] – [\chi_{24}],\)

where \(\chi_{24}\) is the unique representation of \(G\) of dimension \(24\) which also happens to be defined over \(\mathbf{Q}\) and also generates the Burnside cokernel. On the other hand, this method this gives weaker results for subgroups of \(G = \mathrm{PSp}_4(\mathbf{F}_3)\) and even in the surjective case only shows the minimal cover has degree two rather than six.

A word on the actual computation: Shiva went off and did the task of converting the description in Hoffman–Weintraub into a form which could be used by magma. I also went off and tried to do this independently. We then both produced codes (mine much messier) which computed the cohomology of all the subgroups and arrived at completely different answers, which was a bit troubling. But then Shiva pointed out to me that magma automatically does something with matrices that converts right actions to left actions or something like that [could it really be that Magma treats matrices as acting from the right? that sounds crazy], and so his computation of \(H^1(P,M)\) was correct, but I was computing \(H^1(P,M^{\vee})\). But fortunately both are useful! (Of course, one could easily also extract that data from Shiva’s code which was much cleaner than mine.)