Jacquet-Langlands and a new R=T conjecture

It is somewhat mysterious how one should formulate the Jacquet-Langlands correspondence integrally, particularly in the presence of torsion classes. Even the classical case has many subtleties including for example some results in this paper of Ribet.

In the case of imaginary quadratic fields, Akshay and I observed a number of new pathologies that don’t occur in the classical case. One of the confusing aspects was how to define a “space of newforms” which might match (in some vague sense) the cohomology of some inner form. I want to discuss here a new conjecture which is very speculative and for which I have absolutely no computational evidence. It started off as a troubling example in my mind where things seemed to go wrong in the setting of my work with Akshay, and this is the result of me trying to put down those concerns in written form. My guiding principle is that R=T in every situation, so if this doesn’t seem to work, you have to find the right definition of R (or T).

Let F be a fixed imaginary quadratic field, say of class number 1, and let P and Q be primes (of residue characteristic different from p). Suppose that

\displaystyle{H_1(\Gamma_0(P),\mathbf{Z}_p)_{\mathfrak{m}} = \mathbf{Z}_p},

where localization is done with respect to a non-Eisenstein maximal ideal of the Hecke algebra (assume all Hecke algebras are anemic for now). It can (and does) totally happen that one might have

\displaystyle{H_1(\Gamma_0(PQ),\mathbf{Z}_p)_{\mathfrak{m}} = \mathbf{Z}^2_p},

That is, at level PQ there are two old forms but nothing new either in characteristic zero or at the torsion level. In this setting, there are apparently no “newforms” of level PQ, and so one might predict that, on the quaternionic side ramified at PQ, there is no cohomology at this maximal ideal. This is certainly true in characteristic zero by classical Jacquet-Langlands. But it is false integrally! In particular, suppose that the corresponding residual representation

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

has the property that the image of Frobenius at Q has eigenvalues with ratio N(Q). Then one indeed expects a contribution on the non-split side. Akshay and I managed to find an interpretation of this result by giving a “better” definition of the space of newforms as the cokernel of a transfer map:

\Phi^{\vee}: \displaystyle{H_1(\Gamma_0(P),\mathbf{Z}_p)^2_{\mathfrak{m}} \rightarrow H_1(\Gamma_0(PQ),\mathbf{Z}_p)_{\mathfrak{m}}},

and this can have interesting torsion even in the context above. In fact, by a version of Ihara’s Lemma, one can (and we did) compute that the order of the cokernel in this case will be exactly the order of

\mathbf{Z}_p/(a^2_Q - 1 - N(Q)) \mathbf{Z}_p,

and (again in this precise setting) Akshay and I predicted that this should have the same order as the corresponding localization at the same maximal ideal on the non-split side. (In the Eisenstein case, this is not true, and one sees contributions from various K_2 groups). We even prove a few theorems which prove results of this form taking a product over all maximal ideals of the Hecke algebra.

But even in this example, something a little strange can happen. In particular, I want to argue in this post that there are two natural definitions of the appropriate global deformation ring, and in order to have a consistent theory, one should consider both of them. To remind ourselves, we now have two modules, one, defined in terms of the cokernel above, call it M, and then the cohomology localized at the appropriate maximal ideal on the non-split side, which we call M’.

What should we predict about M? The first prediction is that the image of the Hecke algebra should be precisely the universal deformation ring R_Q which records deformations that are Steinberg at Q (and what they should be at the other places). But what does Steinberg at Q even mean for torsion representations? There are basically two types of guesses for the corresponding local deformation ring, and correspondingly two guesses for the associated global deformation ring.

  1. A deformation ring defined in terms of characteristic polynomials. In particular, the maximal quotient of R_Q which corresponds to classes unramified at Q is the unramified deformation ring where the characteristic polynomial of Frob_Q is (X-1)(X-N(Q)).
  2. A more restrictive ring in which (on this same unramified quotient) the image of Frob_Q must actually fix a line.

These certainly will have the same points in characteristic zero, but they need not a priori coincide integrally. And this will save us below.

Returning to the corresponding global deformation rings (which should be framed, but now ignore the framing), call the corresponding rings R_Q and R’_Q. There is a surjection from R_Q to R’_Q.

Now we make the following conjecture on the smell of an oily rag:

Conjecture: The Hecke action on M has image R_Q while the Hecke action on M’ has image R’_Q.

I base this conjecture entirely on the following thought experiment.

Let’s suppose for convenience that N(Q) is not -1 mod p. This implies that a_Q is congruent to precisely one of 1+N(Q) or its negative — assume the former. Then the “space of newforms” M as we define it (under all the hypotheses above) will be actually be isomorphic to

\mathbf{Z}_p/(a^2_Q - 1 - N(Q)) \mathbf{Z}_p =:\mathbf{Z}_p/\eta \mathbf{Z}_p,

because one of the factors will be a direct summand. (The case when N(Q) = -1 mod p is no problem but one has to break things up more using the Hecke operator at U_Q which I am ignoring.) So the claim in this case is that R_Q is isomorphic to this ring. What about R’_Q? Let us consider two possibilities.

(added: Note that if N(Q) =/= 1 mod p then R_Q=R’_Q, so we are assuming that N(Q)=1 mod p in the examples below.)

Example 1: Suppose that a_Q – 1 – N(Q) is exactly divisible by p^2, and that

\rho(\mathrm{Frob}_Q) = \left(\begin{matrix} 1 & 1 \\ 0 & 1 \end{matrix}\right) \mod p.

In this case, the non-split property implies that the corresponding matrix modulo p^2 will always have 1 as an eigenvalue, so the prediction is that R_Q = R’_Q.

Example 2: Suppose that a_Q – 1 – N(Q) is exactly divisible by p^2, and that

\rho(\mathrm{Frob}_Q) = \left(\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}\right) \mod p.

In this case, the split condition and the assumption that a_Q – 1 – N(Q) is exactly divisible by p^2 force the lift to be of the form

\rho(Frob_Q) = \left(\begin{matrix} 1 + a p & p b \\ 0 & 1 + c p \end{matrix}\right) \mod p^2.

where a and c are non-zero. In particular, 1 will never be an eigenvalue. So in this case, one predicts that R_Q = Z/p^2Z but R’_Q = Z/pZ.

So how do we see this in terms of R=T and Jacquet-Langlands and our Conjecture above? First of all, my paper with Akshay suggests indeed that |M’|=|M|= p^2, and certainly M’ should be an R_Q-module. But now the following should happen:

  1. In Example 1, we should have multiplicity one, and so M’ should be free of rank 1 over R_Q = R’_Q.
  2. In Example 2, we should have multiplicity two, following Ribet (Helm, Cheng, Manning…), since multiplicities should be determined by local conditions, and in particular multiplicities should arise exactly when primes which ramify in the quaternion algebra are split and such that the image of the corresponding Frobenius is scalar. Hence M’ should be free of rank 2 over R’_Q in this case.

In particular, the Hecke action on M’ should factor through R’_Q in both cases, and R_Q does not act faithfully. Perhaps this conjecture is worth a computation!

Update: Read Part 2 of this series.

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4 Responses to Jacquet-Langlands and a new R=T conjecture

  1. JT says:

    I am a bit confused by your examples. Shouldn’t Frob_Q have eigenvalues 1, N(Q)?

    • I guess I didn’t say this, but the only interesting case is when N(Q) = 1 mod p, since if the eigenvalues are distinct mod p then one should immediately have R_Q = R’_Q, so one is certainly assuming this in the examples. (I might just edit this in now to make it more explicit.)

  2. Pingback: Jacquet-Langlands and an old R=T conjecture | Persiflage

  3. html-pro says:

    I think you might want to terminate the boldface at the end of the Update.

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