## A strange continuity

Returning to matters OPAQUE, here is the following problem which may well now be approachable by known methods.

Let me phrase the conjecture in the case when the prime p = 2 and the level N = 1.

As we know from Buzzard-Kilford, in every classical weight $\kappa$ “close enough” to the boundary of weight space, the slopes of the space of overconvergent forms are given by the arithmetic progression $nt$ where t depends only on the 2-adic valuation of $\kappa(5) - 1.$ Now, for each of these overconvergent forms, one obtains a Galois representation

$\rho_{n}: G_{\mathbf{Q}} \rightarrow \mathrm{GL}_2(\overline{\mathbf{Q}}_2)$

for every positive integer n. This gives a map from the integers $\mathbf{N}$ considered as a discrete set to $\mathrm{Spf}(R)$ for a deformation ring R (there is only one residual representation in this setting. Yes, it is residually reducible, but ignore this for the moment).

Problem: Show that this map from $\mathbf{N}$ extends to a continuous map from $\mathbf{Z}_2.$

I’ve never done any computations in these weights, but my spidey senses says it should be true. Naturally, one should also try to work out the most precise statement where N and p are now arbitrary.

I don’t have any sense about is whether, for a fixed weight $\kappa,$ there is actually a representation

$\rho: G_{\mathbf{Q}} \rightarrow \mathrm{GL}_2(\mathcal{O} [[T]])$

(for some $\mathcal{O}$ containing enough roots of unity) whose specialization to $T = n$ for a non-negative integer n is $\rho_n,$ or whether the continuity is not so strong. That might be interesting to check.

More natural questions:

1. Once one has the correct formulation in fixed weight $\kappa,$ explain what happens over the entire boundary, and at the halo.

2. I’m pretty sure that $\rho_0$ will just be the Eisenstein series, or more accurately the Galois representation $1 \oplus \chi,$ where $\chi$ is determined from $\kappa$ in some easy way involving normalizations which I don’t want to get wrong. But what is $\rho_{-1}?$ I’m not sure if it is interesting or not. But is there any way of parametrizing this family of Galois representations so that the potentially crystalline points transparently correspond to non-negative integers?

All of this is just to say that, even for N = 1 and p = 2, there’s a lot we don’t know about the eigencurve over the boundary of weight space.

Posted in Uncategorized | | 2 Comments

## Hausdorff Trimester: May 4-August 21, 2020

This post is to let everyone know that there will be a trimester at the Hausdorff institute in 2020 organized by Ana Caraiani, Laurent Fargues, Peter Scholze, and myself on “The Arithmetic of the Langlands Program“. The dates for both the summer school and the conferences have now been set: mark your calendars!

## Irregular Lifts, Part II

This is the global counterpart to the last post. I was going to write this post in a more general setting, but the annoyances of general reductive groups got the better of me.

Suppose we fix the following:

1. A number field F and a prime p > 2.
2. A conjugacy class of involutions (possibly trivial) c_v of $\mathfrak{gl}_n$ for all real places of F.

Then does every representation

$\overline{\rho}: G_F \rightarrow \mathrm{GL}_n(\overline{\mathbf{F}}_p),$

with complex conjugation acting on the adjoint by c_v for each real place of F

1. Have a de Rham lift?
2. Have a de Rham lift of non-regular weight?

I have basically come to the conclusion that the answer to this question is, almost always, no. Namely, the only time the answer is yes is when F is totally real and all the complex conjugations are totally odd. (With one caveat that comes later.)

Most of the theoretical evidence — slim that it is — is in favour of this minimalist conjecture. Namely:

• When n = 2 and F = Q and c is non-scalar, there is a global obstruction to lifting to a weight one modular form, since the image of such forms is a finite subgroup of GL_2(C), and this can already be precluded from making the image of $\overline{\rho}$ contain a large Borel subgroup.
• When n = 2 and p is totally split in F, there are also even local obstructions to lifting to non-regular weight. (There may be local obstructions in other cases as well, although I’m not sure.)
• When F has a real place, the usual conjectures imply that, when c_v is not the “odd” involution, there are obstructions to lifting to regular weights. In the extreme case when c_v is trivial, all lifts should be of trivial weight, and then one can prevent this happening by local (or conjecturally global) reasons similar to those mentioned above.

One can also extend this conjecture to other settings, where one still might conjecture the answer is always no unless one is in a context (regular weight) where $l_0 = 0.$

One caveat is that the case of GL(1) doesn’t quite work out. In this case, oddness is automatic and regularity is automatic, but even when F is not totally real there still exist lifts. I think this is too degenerate to really be so persuasive.

The first real case of this conjecture is when F is an imaginary CM field, and then the claim is that there should be representations

$\overline{\rho}: G_F \rightarrow \mathrm{GL}_2(\overline{\mathbf{F}}_p),$

with no de Rham lifts. To be honest, I don’t have anything intelligent to say about how to prove this, I merely wanted to put on the record that I think I used to believe that such lifts might always exist, and now I’m willing to go on the record and conjecture that they don’t always exist. And, as I tell my group theory class, half the battle to answering a question in mathematics is determining what you think the right answer should be!

## Irregular Lifts, Part I

This post motivated in part by the recent preprint of Fakhruddin, Khare, and Patrikis, and also by Matt’s number theory seminar at Chicago this week. (If you are interested in knowing what the calendar is for the Chicago number theory seminar this quarter, then that makes two of us. Actually, if you are giving a number theory seminar at Chicago this quarter, please leave a comment on this post with the day you are visiting, because several readers of this blog would be interested in finding out who is coming and what they are talking about.)

Let

$\overline{\rho}: G_{\mathbf{Q}_p} \rightarrow \mathrm{GL}_n(\overline{\mathbf{F}}_p)$

be a continuous representation. We now know, by the work of Emerton-Gee, that this representation admits a lift to characteristic zero representation of regular weight which is de Rham (and is even potentially diagonalizable).

On the other hand, can it be the case that there do not exist any de Rham lifts in non-regular weight?

In the most extreme case, where we demand that all the Hodge–Tate weights are zero, then there are obstructions to lifting. In this case, the image of inertia on any lift must have finite image, but the image of inertia of $\overline{\rho}$ may already be sufficiently large to preclude this possibility. (This was exploited in the proof of Theorem 5.1 here.) So this answers the case when n = 2.

But what happens (for example) for n > 2 and HT weights = [0,…,0,1]? Or even n = 2 and replacing $\mathbf{Q}_p$ by a finite extension K? The first remark is that even when the residual image lands inside the Borel, there will certainly be obstructions to finding lifts inside the Borel, which means that inductive arguments will not be sufficient. On the other hand, this definitely smells like a tractable problem.

I offer an Aperol Spritz to an answer to this problem — let me do so even in the constrained version in weight [0,0,1] and $K = \mathbf{Q}_p.$

## More or less OPAQUE

I recently talked with Lynnelle Ye (a soon to be graduating student of Mark Kisin) for a few hours about her thesis and related mathematics. In her thesis, she generalizes (in part) the work Liu-Wan-Xiao on the boundary (halo) of the eigencurve to unitary groups. One of her main results gives a precise asymptotic growth rate of the Newton Polygon of U_p as one moves towards the boundary.

Turning this around, this leads to estimates for the function $N_{\lambda}(X)$ which counts the number of eigenvalues $\lambda$ of U_p (with multiplicity) of valuation at most $X.$

I have always had a soft spot for counting slopes, although I haven’t really done anything in this business for many years. It is already interesting to estimate this growth function for classical overconvergent modular forms in the centre of weight space. Precise estimates were first obtained by Wan in his work on the Gouvea-Mazur conjectures.

Suppose we fix a tame level \Gamma, and let X = X(\Gamma) denote the relevant modular curve. Then it turns out that, conjecturally at least, that:

$\displaystyle{N_{\lambda}(X) \sim^{?} \frac{\mathrm{Vol}(X_0(p))}{4 \pi} X.}$

But this is precisely the growth estimate in Weyl’s law for the Laplacian on X_0(p)! This suggests an analogy between the spectrum of the compact operator U_p in the p-adic case and the spectrum of the Laplacian operator in the complex case which was first suggested to me by Don Blasius and which I always hoped but never quite managed to extract anything from (see section 5 of these notes, which also contain more precise details about Wan’s results and related results towards the conjecture above, as well as many further speculations on Overconvergent P-Adic Quantum Unique Ergodicity, if you were wondering about the title).

What growth rate should one expect for the Unitary group U(n)? Lynnelle exploits the fact (as do Liu-Wan-Xiao) that one can work on a compact form of the group which is zero dimensional. However, the eigenvariety is (or should be) essentially the same as the eigenvariety for other forms of the group. Following the analogy above, we can consider the growth rate of Weyl’s law for U(n-1,1), which, since the Shimura variety for U(n-1,1) has complex dimension n-1, grows like $X^{n-1}.$ However, the exponent in Lynnelle’s work turns out to be

$X^{n(n-1)/2}.$

If I understood correctly, this one can even predict (if not prove) by simply counting the dimension of certain classical spaces of regular algebraic automorphic forms as one ranges over local systems of appropriate weights (proving it requires more work, of course). However, this seems to spoil the very precise (up to the level of constants) analogy for the complex dimension n = 1 case above. Is there something one can do to massage these results so they look more similar or was the n = 1 case simply misleading?

Posted in Mathematics | | 3 Comments

## Mazur’s Program B on Abelian Surfaces

In the book “More mathematical people,” there is an interview with Robin Wilson with the following quote:

At the meal I found myself sitting next to Alistair Cooke who was very charming, and absolutely fascinating to listen to. The very next Sunday when I was back in England I turned on his “Letter from America” on the radio — he started off by saying, “I went to a very boring dinner at the White House. There was no one interesting to talk to.” That amused me a lot.

So let me start off by saying that even though this post is about one or two things I learnt at Oberwolfach, it is deliberately not about anything I learnt in the talks, lest my choosing some talks over others leading to false inferences on what I thought interesting. For example, the title of this post alludes to David Zureick-Brown’s talk, which I will not mention again.

Let g be a non-negative integer and p a prime. Suppose one starts with a representation

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

with (say) cyclotomic similitude character. To avoid later circumlocutions, let me (most of the time) assume it is absolutely irreducible. One can ask whether this representation arises infinitely often from the p-torsion on an abelian variety — perhaps additionally assuming that it does arise from at least one such variety, or perhaps not.

This problem is very well-studied in the case g = 1, where we know that the answer is positive exactly for the primes p = 2, 3, and 5, where the corresponding moduli space X(p) has genus zero, and the associated twists X(rho) are Brauer-Severi varieties that also turn out to be rational over Q under the given hypothesis on the similitude character. When p > 5, the curves X(p) have genus at least 3, and so their twists always have at most finitely many points over any field, by Faltings. So there is certainly a satisfactory answer in this case. (Of course, there are many more subtle versions of this question — for example, replacing “infinitely” by “at least twice” — and those variations are open in general.)

If we move to genus g = 2, then the case of p = 2 is also straightforward — the 2-torsion of the Jacobian of y^2 = f(x) for a degree 6 polynomial with Galois group G just comes from the isomorphism $G \hookrightarrow S_6 \simeq \mathrm{GSp}_4(\mathbf{F}_2).$ (One needs to be a little bit careful here because the outer automorphism of S_6 means there are two non-conjugate such maps and one has to choose the right one.) Given that one can write down families of sextics where the etale Q-algebra Q[x]/f(x) is constant, it’s easy to see that the answer is positive in this case without any restrictions. For example, given an $S_6$ extension, there’s a six dimensional family of polynomials one can write down whose splitting field generically gives this extension, and so after accounting for the action of $\mathrm{PGL}_2$ on the roots, this still gives a three-dimensional rational family of genus two curves whose two torsion comes from this extension.

In my paper with Boxer, Gee, and Pilloni (coming soon!), will also give a similarly conclusive answer for p = 3, although there are some unexpected surprises, as well as some complementary results recently proved by my student Shiva Chidambaram. But more on this in a post coming up soon!

When p > 3, then the corresponding 3-folds obtained by taking full level p-structure of the corresponding Siegel 3-fold $\mathcal{A}_2$ are of general type. (Note that it is essentially known when $\mathcal{A}_g(n)$ is either geometrically rational of general type, see for example Theorem II.2.1 and the surrounding comments in this paper.) Of course, unlike the case of curves, varieties of dimension greater than one of general type can have many rational points. For example, it’s obvious that there are many abelian surfaces over Q whose 5-torsion has the form $(\mathbf{Z}/5 \mathbf{Z} \oplus \mu_5)^2,$ because one can take A to be E + E where E is an elliptic curve whose 5-torsion has the form $(\mathbf{Z}/5 \mathbf{Z} \oplus \mu_5),$ and there are infinitely many such E because the classical modular curve of full level 5 is rational over Q. To put it a different way, the 3-fold A_2(5) corresponding to abelian surfaces with fixed 5-torsion will contain a number of rational Shimura subvarieties coming both from Hilbert modular surfaces and from modular curves, even though it itself is of general type. This can happen even if the mod-p representation rho is irreducible. For example, given an elliptic curve over a quadratic field K/Q, there will once more be a rational curve of elliptic curves with the same mod-5 representation, and so the restriction of scalars will give a rational curve on some twist A_2(rho) of A_2(5). On the other hand, one might at least start off by making the following naive minimal guess.

Question Suppose that $\rho$ is surjective for g = 2 and p >= 5. Then are there only finitely many points on $\mathcal{A}_2(\rho)?$

An even more extreme version of this question would be to ask if there is at most one such point. This seems a little unlikely even by comparison with the case of g=1. I learnt the following nice example talking to John Cremona during the hike through the Black Forest: for g=1 and p=7 and varying E, the twist X(E[7]) has genus 3 (it is a twist of the Klein quartic). This twist is still geometrically a plane quartic. By considering the tangent to the point of X(E[7]) corresponding to E, the line has two further intersections with the curve, and one obtains two further points over X(E[7]) which now (in general) lie over a quadratic extension. But one can parametrize the E for which these points are actually *rational* and this turns out to be the rational cover of the $j$-line corresponding to asking that the invariant c_4 is a square. So there are infinitely many elliptic curves A (even with A[7] surjective) for which there exist at least a pair of non-isogenous elliptic curves A,B with A[7]=B[7] as symplectic Galois representations. So a better question is the following:

Question Can one find examples of non-isogenous abelian surfaces A and B with A[5]=B[5] and such that the corresponding representation has a surjective Galois representation?

This is the type of question where it is useful to have Andrew Sutherland nearby with a laptop. Within an hour or two, he sent me the following examples:

\displaystyle{\begin{aligned} C_1: y^2 = & \ -120x^6-264x^5+186x^4+276x^3-201x^2+24x \\ C_2: y^2 = & \ 16x^5 - 33x^4 + 60x^3 - 42x^2 + 36x - 9 \end{aligned}}

both of conductor $2^{10} \cdot 3^7$ with surjective and isomorphic mod-5 Galois representations which are not isogenous. Nice!

Naturally, the question turned to the existence of a pair with A[7] = B[7]. That proved a tougher challenge, but not an insurmountable one, and here is such a pair (again found by Andrew the same day):

\displaystyle{\begin{aligned} C_1: y^2 + (x^3 + x) y = & \ -x^6+2x^4+2x^3+16x^2+4x+16 \\ C_2: y^2 + (x^2) y = & \ 14x^5-44x^4+46x^3-23x^2+12x-3 \end{aligned}}

this time of conductor $2^7 \cdot 3^2 \cdot 7^4.$

Any guesses as to whether there are any such pairs for $p = 11?$ I’m not sure I have any idea.

Other news from Oberwolfach:

I do appreciate being invited to the Oberwolfach conference on computational number theory — it pushes me outside my usual range of interests. It’s also the conference I have attended most often, now 8 times since 2003, although even that is far fewer than some of the regular participants. The conference is also chance to see a bunch of people I pretty much never get to see anywhere else. Even better, they are all nice enough to still invite me after this post. On the other hand, every time I give a talk I think that this is the time that I finally have something interesting to say to this audience, and it never quite seems to work out that way. I was certainly convinced that this was going to be the year, but then during my talk I managed to catch three people asleep in the front row. To be fair, it was the third last talk of the conference. On the other hand, Mike Bennett talked directly after me and completely failed to rise to my level of soporificness, despite his best efforts and his own predictions he would do otherwise.

There was a lunar eclipse on the final night of our stay. Most of us took to the roof to observe it, but the tall mountains of the Schwarzwald obscured our view until the final moment. Mike Bennett took the following photo, which he describes as the “best of a bad bunch”

For comparison, here is my best photo of the same scene:

Finally, if you are travelling to Oberwolfach during the summer, you mind will naturally turn to the question of whether you can enjoy the warm evenings by sipping on an Aperol spritz or two. Certainly they have lots of sparkling water, and so if Champagne (or equivalent) is available, you will easily be able to set yourself up simply by bringing along a bottle of Aperol. You might then consider trying to determine in advance whether they actually have any sparkling wine. Let me inform you, gentle reader, that the answer to that question is a definite yes, although you may have to look at the alcohol supply near the lecture theatre rather than near the dining room:

Alas, my own intelligence was not up to snuff, as my informant on Facebook who was visiting the previous week was unable to locate this bottle.

There can be only one way to end this post (mostly only relevant to those brought up in Australia in the early 80s or perhaps in England a decade or so earlier:

Now for a walk in the Black Forest:

(10:40, 11:00, 18:52, 22:06 for relevant times if the timestamp link doesn’t work for you…)

## Bristol 2005

(This is really just a supplement to this post.)

The AIM model of conferences encourages real time collaboration, which is unusual as far as mathematical conferences go. But the ne plus ultra of such a conference (among those I have attended) was not at AIM at all, but rather organized by the University Bristol (Although to be fair, I believe it was organized specifically by Brian Conrey). The mission was to take a group of mathematicians and have them work on a very specific problem (which we were not told about in advance). The result: we failed to solve it (c’est la vie). On the other hand, I met a bunch of interesting mathematicians for the first time. My records are spotty, but I did manage to dig out some (poorly executed) photographic evidence from the time, which I present to you below.

The conference was actually located in Clifton rather than Bristol. It didn’t look much like its namesake Clifton Hill (in Melbourne) to me.

Emmanuel Kowalski and Mark Watkins heading towards the Clifton Suspension Bridge. (Check out the snazzy red suitcase!) The conference centre was located in an old manor (Burwalls house, now apparently sold by the University of Bristol to a developer) which is visible in the photo as the orange brick building to the left.

I can’t quite tell if the red suitacase has now transformed into a red backpack or if this is a different day and my fellow blogger has a predilection for vermillion satchels.

Akshay Venkatesh (I’m not going to comment on the hair colour.)

Soundararajan (see comment above)

Elon Lindenstrauss and Erez Lapid

Ben Green and Jon Keating

Brian Conrey and David Farmer.

This collection of photos is definitely incomplete: attending but missing from the photos includes William Stein (who I’m pretty sure was there) and Andy Booker and Sally Koutsoliotas (who were both definitely there) [also Mike Rubinstein]. I think there were a few more local Bristol people as well.

Other things I learnt at this conference: the naive Ramanujan conjecture is false for GSp(4), pork pies are pretty much best avoided, and Collins’s 628.