Derived Langlands

Although it has been in the air for some time, it seems as though ideas from derived algebraic geometry have begun to inform developments in the Langlands program. (A necessary qualifier: I am talking about reciprocity in the classical arithmetic Langlands program here.)

I want to describe a very simple instance of this which came up in Akshay’s MSRI talk which I linked to last time.

Start by fixing a global residual (GL-)odd Galois representation:

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

Let us suppose that \overline{\rho} is surjective. Associated to this representation is a fixed determinant unrestricted global deformation ring R, and a fixed determinant unrestricted local deformation ring which I will call S. (Apologies for the notation, but wordpress is not great with lots of subscripts.) I assume that the reader can make the appropriate adjustments to these definitions if the local representation is not irreducible by adding framings. One knows, at least if p > 2N+1, that the map

\mathrm{Spec}(R) \rightarrow \mathrm{Spec}(S)

is finite. Let us now suppose that \overline{\rho} restricted to G_{\mathbf{Q}_p} admits a crystalline lift of some regular weight; associated to this weight is a Kisin local deformation ring which I shall call T. If you like, you can even imagine that we are working in small weight so that T has nice properties; perhaps it is even smooth. Barry Mazur has made a conjecture for what the (relative) dimension of R should be over \mathbf{Z}_p. Namely, it should be given by the Euler characteristic of the adjoint representation, which is equal (see Def.2.1 and the subsequent comment) to

\dim B - \ell_0,

where B is a Borel of \mathrm{SL}_N(\mathbf{R}), and \ell_0 is the difference between the rank of \mathrm{SL}_N(\mathbf{R}) and \mathrm{SO}_N(\mathbf{R}). Of course, these quantities can easily be calculated explicity in this (or any) case; for \mathrm{SL}(N)/\mathbf{Q}, \ell_0 is the integer part of (N-1)/2. On the other hand, we may also compute the (relative) dimensions of the rings S and T, and we find that

\mathrm{dim}(S/\mathbf{Z}_p) = N^2 - 1, \quad \mathrm{dim}(T/\mathbf{Z}_p) = N(N-1)/2.

(The notation here means the relative dimension.) The Fontaine-Mazur sanity check is to see that, on the associated rigid analytic spaces, the assumption that R and T meet transversally inside S should imply that their intersection only has finitely many points. Indeed, we can compute that the expected dimension of the intersection is exactly:

N(N-1)/2 + \dim(B) - \ell_0 - (N^2 - 1) = - \ell_0.

When N = 2, we have \ell_0 = 0, and everything is as expected. However, as soon as N > 2, we have \ell_0 > 0, and so the expected dimension is negative. This says that regular algebraic automorphic forms for such N are much rarer beasts than their counterparts for N = 2, where modular forms are abundant. For example, it is not known if there exists a regular algebraic cusp form for \mathrm{GL}(N)/\mathbf{Q} giving rise to a \overline{\rho} as above for any N \ge 5. (Note that forms from smaller groups coming via functorial lifts will fail to give rise to representations with such large image.) Now all of this is a philosophy that has been known and exploited for some time. But suppose we actually try to interpret this heuristic a little more literally. For a start, we do expect that forms of characteristic zero do exist. This means that, in general, there are unlikely intersections of \mathrm{Spf}(R) and \mathrm{Spf}(T) inside \mathrm{Spf}(S). That is, R and T will not, in general, be transverse! This is exactly a context in where, to understand the intersection, it makes sense to introduce the derived world (see, for example, the introduction to DAG-V).

In the classical picture, to recover the usual minimal deformation ring, one considers the intersection X:= R \otimes_S T. However, science now tells us it is more natural to consider the derived tensor product

Y = R \otimes^{\mathbf{L}}_{S} T.

If \ell_0 = 0, then the cohomology of Y should exactly recover X in degree zero and be zero otherwise — that is, the classical context should be related to a completely transverse intersection, and we are still in the usual word of schemes (or even complete local Noetherian rings). However, this will (essentially) never happen when \ell_0 > 0. Classically, the ring X may be identified with the ring of endomorphisms generated by Hecke operators on a single extremal degree of cohomology. More generally, the cohomology of Y should be identified with the ring of Hecke operators acting on the cohomology now in degrees q_0, \ldots q_0 + \ell_0, where the notation is as in Borel-Wallach and is fixed for all time. In particular, the only context in which one should expect the intersection to be transverse (beyond \ell_0 = 0) is the case when \ell_0 = 1 and X is a finite ring (which can happen). Indeed, in such contexts, the cohomology over \mathbf{Z}_p also occurs exactly in one degree. It might be worth noting here that the ring S is not in general regular, and so Y, a priori, is not even bounded.

On the other hand, science also tells us that the complex Y has more information than its cohomology; and so one should really think of Y as the correct object. Unfortunately, I don’t have anything clever to say about derived arguments, but let me use the Fontaine–Mazur heuristic to extract some tiny amount of juice. Since I am not so DAGgy, I will only use algebra that goes back 30 years or more.

Instead of looking at the intersection of R and T inside the formal spectrum of S, let us look at their intersection over \mathbf{Z}_p. In this case, all the dimensions have been shifted by one, so, when \ell_0 = 0, their intersection should be infinite (that is, the length over \mathbf{Z}_p). This is obviously the case, because X (which by assumption exists) is flat over \mathbf{Z}_p and so automatically infinite. Well, obvious modulo Serre’s conjecture and Fontaine-Mazur for \mathrm{GL}(2)/\mathbf{Q}, at least. On the other hand, when \ell_0 > 1, then the dimensions don’t add up and the intersection multiplicity should be zero. Thus, assuming the dimensions of the rings are correct, we should have:

  1. If \ell_0 = 1, then the intersection multiplicity is finite and non-zero;
  2. If \ell_0 > 1, then the intersection multiplicity is zero.

In the latter case, we are invoking the conjecture of Serre (proved by Roberts and Gillet-Soulé) that the intersection multiplicity is zero when the dimensions are too small. (Heuristically, one can “move around” classes whose codimension is sufficiently large so they don’t intersect at all, although one cannot literally do this in a local ring!) Actually, even this is a lie, because S is not necessarily regular, as mentioned above, but pretend for this paragraph that it is. Serre’s fomula tells us that the intersection multiplicity is given by the Euler characteristic of the complex. Let me now suppose that Y has no characteristic zero cohomology. For example, we could be working in a weight corresponding to a (strongly) acyclic local system (as long as \ell_0 \ne 0). What we want to compute is the alternating product of the cohomology groups, which should be equal to the alternating product of the integral cohomology groups (everything localized at the appropriate maximal ideal) of the corresponding congruence subgroup of \mathrm{GL}_N(\mathbf{Z}). Yet this product (without localizing at a maximal ideal) is basically equal to the Reidemeister torsion, which is equal to the analytic torsion, which always vanishes when \ell_0 > 1! Under our finiteness conditions, when \ell_0 = 1, all the cohomology occurs in just a single degree, and so the multiplicity is just the length of X. But when \ell_0 > 1, this gives (the expected) refinement of the vanishing of analytic torsion after localizing at a non-Eisenstein maximal ideal, which was one of the questions implicitly raised in the last blog post. To be more precise about our assumptions and conclusions, we have:

Proposition: Let \mathfrak{m} be a non-Eisenstein maximal ideal of the cohomology of a congruence subgroup of \mathrm{SL}_N(\mathcal{O}_F) for a CM number field F, and let \overline{\rho} be the corresponding Galois representation. Assume that:

  1. The required assumptions in C-G hold: (vanishing outside degrees q_0, \ldots, q_0 + \ell_0 after localization at \mathfrak{m}, the representation \overline{\rho} has big image, local-global compatibility, etc.);
  2. The cohomology localized at \mathfrak{m} vanishes in characteristic zero.

Then the alternating product of the orders of the cohomology groups localized at \mathfrak{m} is non-zero if and only if \ell_0 = 1.

One issue with proving this directly using the above argument is that we don’t actually know the dimension of R in general. So, instead of working with Y, we work instead with the output of the Taylor-Wiles method as in C-G, namely:

\displaystyle{Y_{\infty} = R_{\infty} \otimes^{\mathbf{L}}_{S_{\infty}} S_{\infty}/\mathfrak{a}}

Here S_{\infty} is an Iwasawa algebra of diamond operators of dimension q + \ell_0, the ideal \mathfrak{a} is the augmentation ideal with S_{\infty}/\mathfrak{a} = \mathbf{Z}_p, and R_{\infty} is a patched minimal deformation ring of dimension q. These are relative dimensions, so the transverse case over \mathbf{Z}_p corresponds exactly to the case when \ell_0 = 1. We note here that the ring S_{\infty} is regular, so we are in the appropriate context of Serre’s multiplicity formula, and the result follows. (Exercise: we are only using a very special case of the vanishing claim for intersection multiplicities when one component is \mathbf{Z}_p inside S_{\infty}; the vanishing should be easy to prove directly in this case.) This is good, because it gives a purely Galois theoretic proof (well, really only a heuristic because of all the conjectures one needs to assume) of a result (vanishing of analytic torsion) which is not at all obvious. (Well, not quite; the result is localized at a maximal ideal — which one can’t do analytically — but it only applies to non-Eisenstein maximal ideals.) At any rate, thinking through this example after Akshay’s talk has convinced me that this derived perspective is a very good one.

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