## Jacquet-Langlands and an old R=T conjecture

This is part 2 of a series of posts on R=T conjectures for inner forms of GL(2). (See here for part 1).

I feel that I should preface this post with the following psychological remark. Occasionally you have the germ of an idea at the back of you mind that you sense is in conflict with your world view. Perhaps you try subconsciously to banish it from your mind, or perhaps you are drawn towards it. But inevitably, the idea breaks through your consciousness and demands to be addressed. The game is now winner-takes-all — either you can defeat the challenge to your world view, or you will be swallowed up by this new idea an emerge a new person. This is how I came face to face with the non-trivial multiplicities in cohomology for non-split forms of GL(2) over an imaginary quadratic field. Part of me somehow, unconsciously, worried about the conflict between extra multiplicities on the one hand and, on the other hand, the “numerical” equality between the space of “newforms” on the split side with the corresponding space on the non-split side (this equality is not known for each maximal ideal of the Hecke algebra, but rather the “averaged” version over all maximal ideals is the topic of my paper with Akshay). Then, earlier this week, I turned my face directly towards the problem and admitted its existence, which lead to the previous post. But now… there may be a way to defeat the beast after all!

Here is the issue. I talked last time about two types of local framed Steinberg deformation rings at l=1 mod p. The first was defined by imposing conditions on characteristic polynomials, but the second was a more restrictive quotient which demanded the existence of an eigenvalue which was genuinely equal to 1. This modification seemed to pass some consistency checks, and more importantly resolved the compatibility issue between having both the equality |M| = |M’| but also having M be cyclic whilst M’ was not. Then I went away for a few days and was distracted by other math, until I flew back to Chicago this evening. While on the plane, I tried to flesh out the argument a little more by writing down more carefully what these two deformation rings R (and its smaller quotient R’) were like. And here’s the problem. It started to seem as though this quotient R’ didn’t really exist — after all, demanding the existence of an eigenvector without pinning it down in the residual representation is a dangerous business, and runs into exactly the same issues one sees when trying to give an integral definition of the ordinary deformation ring for l=p. Then I thought a little more about the ring R, and it turns out that, for all the natural integral framed deformation rings one writes down, the ring R is a Cohen-Macaulay normal integral domain! In particular, since R’ has to be of the same dimension of R, this pretty much forces R to equal R’. So it seems that my last post is completely bogus.

So what then is going on? When you have eliminated the impossible, whatever remains, however improbable, must be the truth. It is impossible that R does not equal $\mathbf{T},$ so I can only conclude the improbable — that even when the representation rhobar is unramified at l and the image of Frobenius at l under rhobar is scalar, the multiplicity on the quaternionic side ramified at l will still have multiplicity one. In other words, the local multiplicity behavior will be sensitive to the archimedean places. This is not what I would (or did) guess, but I cannot see another way around it. So, at the very least, we should investigate this assumption more closely.

Let’s talk about two situations where multiplicity two occurs. The first is in the Jacobian J_1(Np) for mod-p representations which are ramified at p. In this case, the source of multiplicities is coming from the fact that the local deformation ring R is Cohen-Macaulay but not Gorenstein. On the other hand, the stucture of the Tate module is well understood to be of the form $\mathbf{T} \oplus \mathrm{Hom}(\mathbf{T},\mathbf{Z}_p),$ and so the multiplicity can (ultimately) be read off from the dualizing module of R. This is what happens in my paper with David Geraghty. The second, which is something I should have paid more attention to last time, is in the work of Jeff Manning (I can’t find a working link to either the paper or to Jeff!). The setting of Manning’s work is precisely as above: one has l=1 mod p and one is looking at the cohomology of an inner form of GL(2)/F. The only difference is that F is totally real and the geometric object is a Shimura curve. The corresponding local deformation ring R — which is basically the corresponding ring R above — is Cohen-Macaulay but not Gorenstein. On the other hand, one doesn’t now know what the structure of the Jacobian is as a module over the Hecke ring. Manning’s idea is to exploit the fact that, in his setting, the module M is reflexive (and generically of rank one), and then by studying the class group of R, pin down M exactly. But here is the thing. The reflexivity of M is coming, ultimately, from the fact that the cohomology group H^1 for Shimura curves is self-dual. And this is fundamentally not the case for these inner forms for GL(2) over an imaginary quadratic field, where the cohomology is spread between H^1 and H^2. So this is where the archimedean information can change the structure. At this point, I am pretty much obligated to make the following conjecture.

Conjecture: For inner forms of GL(2) over an imaginary quadratic field, and for a minimal rhobar which is irreducible and finite flat at primes dividing p > 2, the multiplicity of rhobar in cohomology is one. Moreover, the correpsonding module M’ of this cohomology group localized at this maximal ideal is isomorphic (as R-modules and so as Z_p-modules) to the space of newforms on the split side, as defined in the last post.

To put it another way, in Example 2 of the previous post, I am now forced to say that $M' = \mathbf{Z}/p^2\mathbf{Z}$ rather than $(\mathbf{Z}/p\mathbf{Z})^2.$

To reiterate from last time — perhaps this conjecture is worth a computation!

I guess we shall have to wait a few days to see whether there will be a part 3!

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### 8 Responses to Jacquet-Langlands and an old R=T conjecture

1. AP says:

Very interesting conjectures! I am trying to run some computational experiments. I downloaded some elliptic curves on imaginary number fields from the LMFDB and looked for some that satisfy your conditions. Unfortunately, the conditions p|N(Q)-1 and p^2||aQ-N(Q)-1 and the representation being irreducible are not very often satisfied together for small primes Q. I have started some computations for p=5 but they will take some time to complete (I had to go to arithmetic 3-orbifolds of volume in the ~4000). Would you be afraid of data for p=3, given that it could be an orbifold prime and I am computing group homology? (there would be smaller examples, and maybe more importantly, there would be more of them) An alternative: if one uses number fields of higher degree, there could be more examples of small volume, but one would not be able to compute on the split side. Is it still possible to check the conjecture in this case?

2. Compute first, ask questions later!

Here is the example I considered checking, but it may be better to reverse engineer it a little.

Let E be the elliptic curve

$E: y^2 - \theta x y + y = x^3 + (\theta - 1) x^2,$

where $\theta^2 = - 2.$ Let $\pi$ be the prime $1 + \theta.$
Then

$H_1(\Gamma_0(\pi \cdot \frak{p}),\mathbf{Z}_3) = \mathbf{Z}_3$

where $\mathfrak{p} = (2 \theta + 3).$ (This is our example 7.3.1 on page 176 of Short Cave). So now we want to find primes $\mathfrak{q}$ of norm l = 1 mod 3 such that:

1. $l \equiv 1 \mod 3,$
2. $a_{\frak{q}} = 1 + l \mod 9,$
3. $|E[3](F_{\frak{q}})| = 9,$

(One doesn’t have to do any of this computation in the presence of characteristic zero forms, that just helps explicitly to compute the mod-3 representation. As you say, it’s a little annoying to find primes that satisfy the last condition, because finding primes which split completely is a little rare in such a large field by Cebotarev, not to mention that splitting completely has a tendency to lose that Cebotarev race even with the appropriate handicap). In this case, one wants the polynomial

(-6*y^7 – 68/3*y^6 – 51*y^5 + 306*y^4 + 1107*y^3 – 162*y^2 – 3402*y – 3159)*t + (y^8 + 4*y^7 + y^6 – 39*y^5 – 117*y^4 + 1035*y^3 + 3564*y^2 + 2673*y – 2673)

to split completely. Then one wants to compute at the quaternion algebra split at $\pi$ and at $\mathfrak{q}$ and look at level $\Gamma_0(\mathfrak{p}).$ The problem, of course, is that there could be other random rubbish which turns up. (Unless — has someone finally written a program that computes Hecke operators??) For example, the prime $\frak{p}$ has norm 17, and so should contribute at least a copy of Z/(18Z) to the non-split side which is Eisenstein. But if it happens to be the case that there are no Eisenstein deformations and no other mod-3 representations and no further deformations of rhobar and the cohomology on the non-split side looks like (Z/9Z) + (Z/9Z) at the prime 3… then this new (or old) guess is supported.

Of course, even computing at this level requires finding the presentation for the non-split guy, which maybe Aurel Page could do, but you want to be sure it works before computing a new one every time. So it might be better to choose a non-split form of GL(2)/F and then reverse engineer an appropriate rhobar.

As to your explicit question — I think one really only has to compute on the non-split side — namely, write down a quaternion algebra ramified at P and Q (with N(Q)=1 mod 3, say) and then write down a mod-3 representation which is Steinberg at P and scalar at Q, and then hope that (after accounting for the Eisenstein part which one can predict in advance) the cohomology is cyclic as a Z_p-module.

3. AP says:

Oh, I see, I was using more strict condition on the example so I got larger ones.

I found a prime of norm 283 that satisfies your requirements. I computed the H_1 of the suitable unit group of the quaternion algebra, with coefficients in R=Z/3^20 (to emulate Z_3).
The H_1 is isomorphic to R/3 + R/3 + R/(3^3) + R + R. Modulo 3, the eigenvalue system corresponding to E has multiplicity 2 (the corresponding module is free of rank 2 under the Hecke algebra). We have H_1(Gamma)/m^2 isomorphic to R/3 + R/3, where m is the maximal ideal corresponding to that eigenvalue system.

I will write code to setup a more systematic computation and get more examples, but I might not have time to do that for a week.

To answer your question about code computing Hecke operators: I have written such code several years ago, but it is ugly and difficult to use and buggy in some places, so I did not release it. Tidying it up would be useful but would take time. Yes, that’s lame, sorry!

• OK, first of all I have to admit that when I read your first comment I somehow read the name as “AV” rather than “AP”, thus making my references to computations that I only know that Aurel Page can do slightly silly.

Second, and more importantly, I think your computations precisely validate my conjecture (guess), assuming I am reading what you say correctly. We are looking at the module

M = H_1(Gamma,Z_3)

localized at the maximal ideal m. (I’m not sure if Gamma is ramified at the primes above 17 and 283 with level 3 structure, or is ramified at the primes above 3 and 283 with level 17 structure, although it shouldn’t make any difference). I want to conjecture that M is a cyclic T module, where T is the Hecke algebra and T/m = F_3. By Nakayama, M will be cyclic exactly when

dim_(F_3)(M/m) = 1.

What I read from your computation is that M/m^2 has length 2. As long as T =/= F_3 (equivalently, as long as m =/= m^2), this implies that M/m has length one.

The other computation you do is to say that M/3 *has* multiplicity 2. But this is not a problem! This will just reflect that the image of T acting on M/3 is not Gorenstein, and so is not isomorphic to its dual. The correct “multiplicity” computation for mod-p representations in this sense should be on H^1(Gamma,F_3), and this is isomorphic to

(M/3)^vee = Hom(M/3,F_3).

But now

(M/3)^vee[m] = Hom(M/mM,F_3) = 1-dimensional.

• AP says:

Ah sorry, what I wrote was not very clear. First, Gamma is ramified at the primes above 3 and 283 with level 17 structure, although as you say it should not make a difference. The sentence starting with “modulo 3” was supposed to be interpreted as saying that M/3=M/m is free of rank 2 under T/3. This is already enough to conclude that M is not cyclic as a T-module. Then, to give more precise information, I computed an approximation to M_m by computing successive quotients M/m^k, but actually this stabilises already at M/m^2 which is isomorphic to R/3+R/3.
If I understand things correctly, this goes in favor of the conjecture of the first post, not that of the second one, although maybe we should compute a few more examples.

• AP says:

What confuses me is that v_3(aQ-Norm(Q)-1)=3. Don’t you predict that |M_m|=3^3 in this case?

4. OK, a few possible things, some of which may be related:

In the original example, the prime above 17 should be (2theta – 3) not (2theta + 3), there is a typo in the paper (note to self before proofs arrive). So one should be looking at Q ramified at (1 + theta) and (11theta + 9) and level Gamma_0(2theta – 3) just to be sure.

I’m still confused what you wrote, but you are saying M/m = M/m^2 = (R/p)^2?

If v_3(aQ-Norm(Q)-1)=3, then |M_m| should have order *at least 3^3* — but it could be higher if there is more relevant cohomology on the split side at level Gamma_0((1 + theta)*(2theta – 3)*(11theta + 9)) (what happens there?

Maybe you can list the first few Hecke eigenvalues to check it agrees with the relevant form? (but I thought you must have done this to distinguish m from the other maximal ideals…