Chidambaram on Galois representations (not) associated to abelian varieties over Q

Posted on October 13, 2020 by Persiflage

Today’s post is about a new paper by my student Shiva. Suppose that \(A/\mathbf{Q}\) is a principally polarized abelian variety of dimension \(g\) and \(p\) is a prime. The Galois representation on the \(p\)-torsion points \(A[p]\) gives rise to a Galois representation:

\(\rho: G_{\mathbf{Q}} \rightarrow \mathrm{GSp}_{2g}(\mathbf{F}_p)\)

with the property that the similitude character coincides with the mod-\(p\) cyclotomic character. A natural question to ask is whether the converse holds. Namely, given such a representation as above with the constraint on the similtude character, does it necessarily come from an abelian variety (principally polarized or not)?

When \(g=1\), the answer is that all such representations come from elliptic curves when \(p \le 5\), but that for \(p \ge 7\) there exist representations for any \(p\) which do not. For \(p \le 5\), more is true: the twisted modular curves \(X(\rho)\) all are isomorphic to \(\mathbf{P}^1\). When \(p \ge 7\), the curves \(X(\rho)\) are of general type, so one might expect a “random” such example to have no rational points. Dieulefait was the first person to find explicit representations (for any such \(p\)) which do not come from elliptic curves (and there is a similar result in my paper here). Both of these arguments exploit the Hasse bound. Namely, if

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

is unramified at \(l \ne p \ge 5\) and \(\rho\) comes from \(E/\mathbf{Q}\), then \(E\) must have either good or multiplicative reduction at \(l\). But this puts a constraint on the possible trace of Frobenius at the prime \(l\). For \(l = 2\), for example, this leads to explicit examples of non-elliptic mod-\(p\) representations for \(p \ge 11\). The case \(p = 7\), however, requires a different argument. More generally, while the Hasse argument does generalize to larger \(g\), it only works when \(p\) is large compared to \(g\). On the other hand, the Siegel modular varieties \(\mathcal{A}_g(p)\) of principal level \(p\) are rational over \(\mathbf{C}\) for only very few values of \(g\) and \(p\). Indeed, they are rational only for

\((g,p) = (1,2), (1,3), (1,5), (2,2), (2,3), (3,2)\)

whereas \(\mathcal{A}_g(p)\) turns out to be of general type for all other such pairs. When \((g,p)\) is on this list, then, as discussed in these posts, the twists \(\mathcal{A}_g(\rho)\) can all be shown to be unirational over \(\mathbf{Q}\) and so any such representation \(\rho\) does indeed come from infinitely many (principally polarized) abelian varieties.

Thus one is left to consider all the remaining pairs. This is exactly the question resolved by Shiva:

Theorem [Chidambaram]: Suppose that \((g,p)\) is not one of the six pairs above such that \(\mathcal{A}_g(p)/\mathbf{C}\) is rational. Then there exists a representation:

\(\rho: G_{\mathbf{Q}} \rightarrow \mathrm{GSp}_{2g}(\mathbf{F}_p)\)

with cyclotomic similitude character which does not come from an abelian variety over \(\mathbf{Q}\).

Shiva’s argument does not use the Weil bound. Instead, the starting point for his argument is based on the following idea. Start by assuming that \(\rho\) comes from an abelian variety \(A\). Suppose also that \(\rho\) is ramified at \(v \ne p\) and the image of the inertia group at \(v\) contains an element of order \(n\) for some \((n,p) = 1\). Using this, one deduces (using independence of \(p\) arguments) that

\(|\mathrm{Sp}_{2g}(\mathbf{F}_l)| = l^{g^2} \prod_{m=1}^{g} l^{2m} – 1\)

is divisible by \(n\) for all large enough primes \(l\), and hence divides the greatest common divisor \(K_g\) of all these orders. This is actually a very restrictive condition on \(n\). For example, using Dirichlet’s theorem, the number \(K_g\) is only divisible by primes at most \(2g+1\). But now if the order of the group \(\mathrm{Sp}_{2g}(\mathbf{F}_p)\) for any particular \(p\) is divisible by a prime power \(n\) with \(n\) not dividing \(K_g\), then one can hope to construct a mod-\(p\) Galois representation whose inertial image at some prime \(v\) has order divisible by this \(n\), and this representation cannot come from an abelian variety over \(\mathbf{Q}\).

The good news is that one can show that (most) symplectic groups have orders divisible by large primes using Zsigmondy’s theorem. Combined with a few extra tricks and calculations for some boundary cases, the groups \(\mathrm{Sp}_{2g}(\mathbf{F}_p)\) contain elements of “forbidden” orders exactly when one is not in the case of the six exceptional pairs \((g,p)\). Note that Zsigmondy’s theorem already arises in the literature in this context in order to understand prime factors of the (corresponding) simple groups.

So now one would be “done” if one could (for example) solve the inverse Galois problem for \(\mathrm{GSp}_{2g}(\mathbf{F}_p)\) with local conditions. The inverse Galois problem is solved for these groups, but only because there is an obvious source of such representations coming from abelian varieties. Of course, these are precisely the representations Shiva wants to avoid.

Instead Shiva looks for solvable groups inside \(\mathrm{GSp}_{2g}(\mathbf{F}_p)\) containing elements of order \(n\) for suitable large prime powers \(n\). Note that the obvious thing would simply be to take the cyclic group generated by the element of the corresponding order. The problem is that there is no way to turn the corresponding representation into a Galois representation whose similitude character is cyclotomic. The groups Shiva actually uses are constructed as follows. Start by finding prime powers \(n | p^{m} + 1\) for some \(m \le g\), then embed the non-split Cartan subgroup of \(\mathrm{SL}_2(\mathbf{F}_{p^m})\) into \(\mathrm{GSp}_{2g}(\mathbf{F}_p)\), and then consider the normalizer of this image. One finds a particularly nice metabelian subgroup whose similitude character surjects onto \(\mathbf{F}^{\times}_p\). Shiva then has to prove the existence of a number field whose Galois group is this metabelian extension with the desired ramification properties at some auxiliary prime \(v\) but also crucially satisfying the cylotomic similitude character condition. This translates into a (typically) non-split embedding problem — such problems can be quite subtle! Shiva solves it by a nice trick where he relates the obstruction to a similar one which can be shown to vanish using methods related to the proof of the Grunwald-Wang Theorem. Very nice! In retrospect, the case of \(g = 1\) and \(p = 7\) in my original paper is a special example of Shiva’s argument, except it falls into one of the “easy” cases where the relevant metabelian extension actually is a split extension over the cyclotomic field. In general, this only happens when the the maximum power of \(2\) dividing \(g\) is strictly smaller than the maximum power of \(2\) dividing \(p-1\) which is automatic when \(g\) is odd. (The case when \(p = 2\) is easier because the cyclotomic similitude character condition disappears!)

Posted in Mathematics, Students, Work of my students | Tagged , , , , , , , , | 5 Comments

Tips on becoming a computational number theorist

Posted on June 16, 2020 by Persiflage

How would you advise a student who has talents in both the computational and theoretical aspects of algebraic number theory? There is no hard border between computational and theoretical algebraic number theory, but there is a definite computational number theory community with its own norms and expectations. While I interact with this world, I am not a part of it, and so I don’t necessarily have the best practical advice to offer such a student. So instead of just guessing, I emailed a few people (including Drew Sutherland and John Voight) who graciously sent me a number of suggestions. Below is a lightly edited version of Drew’s email (incorporating some of John’s responses), posted with permission: I think there is some great advice here for both students and advisors! One comment I would like to amplify is the idea of spending time engaging with experts in the field. This is a useful suggestion for anyone in mathematics. It can certainly be terrifying for a fresh graduate student to put themselves out there and talk to the famous experts, but that of course is why it is important for senior people to make the effort to interact with junior people. (No doubt some areas of mathematics have more approachable famous people than others; computational number theory seems to be a pretty friendly place.) Finally, it is incumbent upon me as an advisor to note that my student Shiva Chidambaram (expected graduation 2021) will indeed be giving a talk at ANTS this year. If you are (virtually) attending ANTS this year, please consider both going to this talk and also saying hello to Shiva!

  1. Learn how to use Magma, Pari/GP, and Sage well. Each of these tools has particular strengths and there will almost certainly come a time when you want to be able to take direct advantage of them. In addition to your own research, if you are marketing yourself as a computational person you need to be prepared for random computational requests that may fall well outside your mathematical expertise; if you know the right button to push you can still come out looking like a genius.


  2. Should I be using a low level programming language like C/C++ and learning how to exploit libraries like GMP, Pari, FLINT, or NTL?
    This might eventually become important, but unless you have a project that would benefit from some serious computational horsepower, I wouldn’t burn a lot of time writing (or learning to write) low level code right now. But if you do have such a project (or have an advisor who can suggest one), then by all means, swing for the fences! Nothing better demonstrates your computational chops than computing something no one else has been able to compute. But I should note that this doesn’t necessarily require writing any low level code, a better algorithm that cleverly leverages functionality built in to Magma, Pari/GP, or Sage may achieve the same result.


  3. Make it your goal to deliver a talk at a major workshop or conference focused on computational number theory or explicit methods, such as the Algorithmic Number Theory Symposium (ANTS). Your future employer probably won’t be represented there, but you will get a chance to join a network of people who can really help you, both with your immediate job search and in the longer term. ANTS VIII was my first mathematics conference, and looking at the participant list twelve years later I can count at least twenty names of people who have played a direct role in my career as co-authors, co-editors, co-organizers, or mentors in some form. Ideally your talk will also lead to or be based on a publication, but making connections is really the crucial thing when you are getting started, I can trace eight of the papers I wrote in the years immediately following ANTS VIII directly to conversations or people I met there.

    It’s worth noting that as a young computational number theorist you are an ideal collaborator; you have the ability to make the mathematical ideas of others explicit in a way they may not be able to, and you likely have more time to devote to new projects than more senior collaborators (they will have other projects in progress, students to supervise, and myriad administrative obligations). But to make this work in your favor you have to network, and you need to be able to take a project and run with it.

  4. In order to achieve (3) you need an interesting research result to present. A good way to get started is to attend a project-based summer school or research workshop (not a conference), ideally one that has a mix of graduate students, post-docs, and more senior people. There are usually a few of these each year, often at places like MSRI or ICERM; one to keep an eye out for is the IAS/PCMI summer program on Number Theory Informed by Computation, which was meant to take place this summer but is now likely to happen in the summer of 2021 or 2022.


  5. Create a professional webpage that showcases your work and research interests, especially things you have computed. Don’t rely on the profile page you might have on your institution’s web site; you will soon be leaving and you need to establish your own brand. Every mathematician looking to attract invitations to give talks and eventually a job offer should do this, but if you are going to sell yourself as someone with computational expertise it is especially important that you have a credible web presence.


  6. Deposit your code and results on GitHub. Here I am simply going to repeat the excellent advice of my colleague John Voight: “this signals a level of seriousness, transparency, and a desire to share that is important in a collaborative working environment”.


  7. Prove theorems! If you are going to call yourself a mathematician this should go without saying, but it is worth emphasizing. As John Voight has also noted, proving theorems about your algorithms not only establishes results that may be of independent mathematical interest, it makes your algorithms better. For me, writing a paper is always (at least) a four step process: (a) come up with a new idea for computing something new, or computing something old in a new way; (b) implement the algorithm to see if it actually works and return to step (a) if it does not; (c) state and prove precise theorems showing that the algorithm does what it is supposed to do, and that it does so efficiently; (d) re-implement the algorithm to reflect all the ideas necessary to complete step (c), which will often make the code both clearer and faster.
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Families of Hilbert Modular Forms of Partial Weight One.

Posted on June 3, 2020 by Persiflage

Today I would like to talk about a beautiful new theorem of my student Eric Stubley (see also here). The first version of Eric’s result assumed (unknown) cases of the general Ramanujan conjecture for Hilbert modular forms, and relied on a beautiful idea due to Hida. The final argument, however, is unconditional, and goes beyond Hida’s ideas in a way (I hope) that he would be delighted to see.

Suppose that \(F\) is a real quadratic field in which \(p = vw\) splits. If \(f\) is a Hilbert modular form of (paritious) weight \((1,2k+1)\) and level prime to \(p\), then the corresponding Galois representation (really only defined up to twist):

\(\rho_f: G_F \rightarrow \mathrm{GL}_2(\overline{\mathbf{Q}}_p)\)

has the property that, for exactly one of the places \(v|p\), the restriction \(\rho_f |_{G_v}\) is unramified. Forms of partial weight one are slippery objects — one can construct such forms which are CM, but the existence of any such form which is not CM was open until an example was found by my students Richard Moy and Joel Specter (see here, here, and here). They behave in many ways like tempered cohomological automorphic forms for groups without discrete series, more specifically Bianchi modular forms or cohomological forms for \(\mathrm{GL}(3)/\mathbf{Q}\). In each of these cases, the invariant \(l_0\) as considered in Calegari-Geraghty (see for example section 2.8 of this paper) is equal to \(1\). Following work of Ash-Stevens and Calegari-Mazur, one might consider whether or not \(f\) deforms into a family of classical forms. For example, the form \(f\) will be ordinary at \(v\), and so it lives in a Hida family \(\mathcal{H}\) over \(\Lambda = \mathbf{Z}_p[[\mathcal{O}^{\times}_v(p)]] \simeq \mathbf{Z}_p[[T]]\) where we keep the weight and level at \(w\) fixed and consider (nearly) ordinary forms at \(v\). The specialization of this family to regular paritious weights will give a space of classical Hilbert modular forms. What can one say about the other specializations in partial weight one?

Theorem [Stubley]: only finitely many partial weight one specializations of the one variable \(v\)-adic Hida family \(\mathcal{H}\) associated to \(f\) are both classical and not CM.

This gives a completely general rigidity result for all partial weight one Hilbert modular forms in the split case. Over the past decade or so, the prevailing philosophy is that the only algebraic automorphic forms which are not exceedingly rare are either those coming from automorphic forms which are discrete series at infinity, or come from such forms on lower rank groups by functoriality. In this setting, this predicts that non-CM forms of partial weight one should be rare. It might even be plausible to conjecture that, up to twisting, there are only finitely many such forms of fixed tame level. However, such conjectures are completely open, and Stubley’s result is one of the first general theorems which points in that direction. (Stronger results for very specific \(F\) and \(p\) and tame level were obtained by Richard Moy and are discussed in some of the links above.)

One way to think about this theorem is in terms of the Galois representation associated to \(\mathcal{H}\). Assume for convenience of exposition that the family is free of rank one over \(\Lambda\). The Galois representation \(\rho_f\) extends to a family:

\(\rho: G_F \rightarrow \mathrm{GL}_2(\mathbf{Z}_p[[T]])\)

where \(\Lambda = \mathbf{Z}_p[[T]]\) represents weight space, so \(T = 0\) corresponds to the original specialization, and \(T = \zeta – 1\) for a \(p\)-power root of unity \(\zeta\) corresponds to a specialization to partial weight one with non-trivial level structure at \(v\). These representations are all nearly ordinary at \(v\). Is it possible that they could be split locally at \(v\) for infinitely many specializations to partial weight one? Since a non-zero Iwasawa function has only finitely many zeros, this would actually force the local representation to split for all \(T.\) Moreover, it should imply (and does in many cases) that the specializations \(T = \zeta – 1\) are all classical by modularity lifting theorems. Thus, by Stubley’s theorem, this can only happen when the family \(\rho\) is CM. In particular, Stubley’s result implies a theorem (assuming some Taylor-Wiles hypothesis) that a family of Galois representations which is (say) nearly ordinary at \(w\) of fixed weight and level and nearly ordinary at \(v\) is locally split at \(v\) if and only if it is CM.

Experts should recognize the similarity between the Galois theoretic version of Stubley’s theorem and the work of Ghate-Vatal, who prove that an ordinary family over \(\mathbf{Q}\) cannot be locally split unless it is CM. The main ingredient in their proof is the fact that there are only finitely many weight one forms of fixed tame level (up to twist) which are not CM, since these correspond either to \(A_4, S_4, A_5\) extensions of \(\mathbf{Q}\) unramified outside a fixed set of primes, which are clearly finite, or real multiplication forms, whose finiteness comes down to the finiteness of the ray class group of conductor \(N \mathfrak{p}^{\infty}\) for a split prime \(\mathfrak{p}\) in a real quadratic field. However, the analogous statement for partial weight one forms is completely open as mentioned above, so Stubley’s theorem requires a quite different argument.

Before discussing the proof, we first need to discuss a result of Hida (see this paper) about fields of definition of ordinary forms in families. Consider an ordinary family over \(\mathbf{Q}\), and consider specializations in some fixed weight, amounting (with some normalization) to specializing \(T\) to \(\zeta – 1\) for a \(p\)th power root of unity. The coefficient field will automatically contain \(\mathbf{Q}(\zeta)\). Suppose that for any prime \(q\), the degrees \([\mathbf{Q}(a_q,\zeta):\mathbf{Q}(\zeta)]\) are bounded for infinitely many specializations. Then Hida proves the family has to be a CM family. Let \(\alpha_q\) be one of the corresponding Frobenius eigenvalues. Hida’s key insight is to note that \(\alpha_q\) is a Weil number, and that Weil numbers over extensions of \(\mathbf{Q}(\zeta)\) of uniformly bounded degree are extremely restricted, and in particular given an infinite collection of such numbers then infinitely many of them have to be of the form \(\alpha \zeta\) for a fixed \(\alpha\). Using a rigidity lemma fashioned for this very purpose, he then deduces that \(\alpha_q\) in the Iwasawa algebra more or less has to equal \(\alpha (1+T)^s\) for some \(s \in \mathbf{Z}_p\), and this puts enough restrictions on \(a_q\) for him to be able to deduce the family is CM.

Stubley’s first idea is to use Hida’s result in the context of partial weight one forms. The key fact that is different in partial weight one is that when \(a_{v} \ne 0\), the form \(f\) is automatically ordinary at \(v\), and hence the \(\mathrm{Gal}(\overline{\mathbf{Q}}/\mathbf{Q}(\zeta))\) conjugates of \(f\) will still be ordinary at \(v\)! This is completely false in regular weights. However, in partial weight one, the only possible (finite) slope of any form at a split prime is \(0\). As a consequence, the boundedness assumption of Hida’s theorem is always going to be satisfied, because all of the conjugates have to lie on one of the finitely many Hida families which all have bounded rank over \(\Lambda\).

There is, however, a problem. Hida’s argument crucially uses the fact that \(\alpha_q\) is a Weil number, which uses the Ramanujan conjecture for forms of regular weight. The Ramanujan conjecture is completely open for partial weight one forms, since we have no idea how to prove they occur motivically (nor prove modularity of their symmetric powers). This is where Stubley’s second idea comes in. Instead of the Ramanujan conjecture, one does have standard bounds on the coefficients \(a_q\). This is not enough to deduce that \(\alpha_q\) has the form \(\alpha \zeta\) for some fixed \(\alpha\). Instead, Stubley shows that it does allow one to show that the trace of \(a_q\) (together with the trace of any if its powers) to \(\mathbf{Q}(\zeta)\) (which has uniformly bounded degree) can be written as a finite sum of roots of unity where the number of terms does not depend on \(\zeta\). Again for convenience of exposition and to avoid circumlocutions with traces, let us suppose that the rank of the Hida algebra is one and so \(\mathbf{Q}(\zeta,f) = \mathbf{Q}(\zeta)\). Then Eric shows that infinitely many of the \(a_q\) satisfy:

\(a_q = \alpha_1 \zeta_1 + \alpha_2 \zeta_2 + \ldots + \alpha_N \zeta_N\)

for varying \(p\)-power roots of unity \(\zeta_i\), but where \(\alpha_i\) and \(N\) are fixed. Then Stubley proves a new rigidity theorem in this context (not unrelated to results of Serban) showing that one must have an equality

\(a_q = \alpha_1 (1+T)^{s_1} + \alpha_2 (1+T)^{s_2} + \ldots + \alpha_N (1 + T)^{s_n}\)

over the Iwasawa algebra. This is probably enough to show the family has to be CM using ideas similar to Hida, but even that is not necessary — by using this formula for specializations in regular weight one deduces that the \(\alpha_i\) are in \(\overline{\mathbf{Q}}\), and then applying Hida’s theorem in this fixed regular weight one deduces that the family is CM.

Stubley’s theorem is the first result that gives general theoretical evidence towards the conjecture (if one is so bold to make such a conjecture) that there are only finitely many non-CM partial weight one forms of fixed tame level up to twist. It also shows that certain \(v\)-ordinary deformations of a non-CM partial weight one form \(f\) will not be classical. But there is also a second possible way to deform \(f\), namely, to vary the weight at \(w|p\) instead (or as well). For example, if the form \(f\) was also ordinary at \(w|p\), one could look at the ordinary at \(w\) Hida family. One might also conjecture that this family only contains finitely many non-CM points, but this is still open. (Boxer has raised this question.) I think this is an interesting but very hard question!

Posted in Mathematics, Students, Work of my students | Tagged , , , , , , , , , , , | 2 Comments

Picard Groups of Moduli Stacks update

Posted on May 27, 2020 by Persiflage

A tiny update on this post. I was chatting with Benson and realized that I may as well ask him directly for a presentation of the mapping class group of a genus two surface. Perhaps unsurprisingly, it can be found in his book with Dan Margalit (see page 122 of their book which might be downloadable from a Russian website) and is given as follows:

\(G \simeq \langle a_1,a_2,a_3,a_4,a_5 | \ [a_i,a_j] \ \text{for $|i-j|>1$}, a_i a_{i+1} a_i = a_{i+1} a_i a_{i+1},\)
\( (a_1 a_2 a_3)^4 = a^2_5, [(a_5 a_4 a_3 a_2 a_1 a_1 a_2 a_3 a_4 a_5),a_1], (a_5 a_4 a_3 a_2 a_1 a_1 a_2 a_3 a_4 a_5)^2, \rangle.\)

The next task is to find the representation

\(G \rightarrow \mathrm{Sp}_4(\mathbf{Z}) \rightarrow \mathrm{Sp}_4(\mathbf{F}_2) \simeq S_6\)

and then take the index \(6\) preimage \(\Gamma \subset G\) of the \(S_5 \subset S_6\) corresponding to fixing a Weierstrass point. Note there are two conjugacy classes of \(S_5\), the correct one is the one whose restriction to \(A_5\) still acts absolutely irreducibly on \((\mathbf{F}_2)^4\). Then one can use Reidemeister-Schreier to compute a presentation of \(\Gamma\) and then compute \(H_1(\Gamma,\mathbf{Z})\). This is all good in theory, and Farb-Margalit does have a chapter on the symplectic representation, but actually having to read the book in detail to extract the precise symplectic representation sounded like too much work, especially since all of this is ultimately just for a two sentence comment in a paper that might be removed for space reasons anyway. So instead I just fired up magma with the representation \(G\) and asked it to find *all* index six subgroups. It turns out that there are only two of them (up to conjugation), which must come exactly from the two subgroups of \(S_5 \subset S_6\). The abelianization of one is \(\mathbf{Z}/10 \mathbf{Z} \simeq G^{\mathrm{ab}}\), but the other group is

\(\Gamma = \langle a_1,a_2,a_3,a_4 \rangle\), and one finds that \(H_1(\Gamma,\mathbf{Z}) \simeq \mathbf{Z}/40 \mathbf{Z}.\)

Hence this is (in light of the previous dicussion) the correct subgroup, and this (unsurprisingly although not entirely independently) confirms the analysis of naf in the comments. Now I suspect that if you think a little harder than I am prepared to do (or if you just know a little bit more than me), you might be able to see directly from the definition of the \(a_i\) that \(a_1,a_2,a_3,a_4\) fix a Weierstrass point; if you are such a person please make a comment!

Posted in Mathematics | Tagged , , , , | 2 Comments

Panel Discussion: Mathematics Research Online

Posted on May 13, 2020 by Persiflage

Odds are high that virtual conferences in mathematics are here to stay. It seems crucial, therefore, to think long and hard about ways to make them work for all participants. We need to have conversations as a community to better understand how to make this happen. Andrew Sutherland and Bianca Viray are organizing a panel (virtual of course!) on this very topic one week from today (May 20th), with a number of panelists who have already had experience running virtual workshops. I encourage you to join in and learn from there experience, but also to add your own voice to the conversation.

For the link, click here!

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Picard Groups of Moduli Stacks

Posted on April 30, 2020 by Persiflage

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!

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Chidambaram on genus two curves, II

Posted on April 24, 2020 by Persiflage

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.

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En Passant VIII

Posted on April 23, 2020 by Persiflage

I just found out that Lucien Szpiro recently passed away. I met him only once in late 2018 when I gave the joint NY number theory at CUNY. When I arrived at my hotel (the Gregory) around 10pm, I was somewhat shocked to find that nobody had made me a reservation, and the hotel was completely booked out. I was tempted to try the Langham on the opposite side of the street, but thought that might probably bankrupt the CUNY seminar budget for the next few years, so instead I started wandering the streets of NYC trying to find a more modest hotel with a vacancy (without any success — I’m not sure if there was something going on or whether it’s always hard to find a hotel room in NYC in mid-October). With a little help from JC back in Chicago with a reliable internet connection and access to hotels.com, I eventually found the one remaining room in a Best Western a few streets away for about $450 a night. When Lucien found out about what happened the next day, he vigorously tried to persuade me to accept a personal check from him to cover my costs before the reimbursement process could be sorted out. Though I insisted on declining his generous offer, a few days later I received a personal check from him in the mail, and this time I cashed it. And it was just as well, because CUNY did deny my reimbursement request for going over the daily limit for hotels, before finally [with a little more help from locals] “generously” agreeing to a “one time exception” to cover their own incompetence. I was then finally able to pay Szpiro back via Venmo (at least through the intermediary of Alex Gamburd, friend of the blog). Although this was our only ever interaction, it certainly left a very positive impression on me.

Tim Brooke-Taylor also recently died. If you grew up in Australia in the 1980s then there’s a good chance you have a deep place in your heart for the Goodies. (Further evidence: see this week’s Australian letter in the New York Times.) In addition to the absurdist humor, there were also plenty of 70s British cultural and political references which were somewhat lost on a 10 year old kid in 80s Melbourne; I wonder what a 10 year old kid in 2020s Chicago would think?

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En Passant VII

Posted on April 19, 2020 by Persiflage

Idle question which has surely been asked (and answered!). If \(X^{+}_{\mathrm{nsp}}(p)\) is the modular curve corresponding to the normalizer of the non-split Cartan, then one reason it is hard to find all rational points is that the all factors of the Jacobian have positive rank (probably contingent on BSD). Is the same true for \(X^{+}_{\mathrm{nsp}}(pq)\)?

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Chidambaram on genus two curves, I

Posted on April 15, 2020 by Persiflage

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.)

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