Report From Berkeley

My recent trip to Berkeley did not result in a chance to test whether the Cheeseboard pizza maintained its ranking, but did give me the opportunity to attend the latest Bay Area Number Theory and Algebraic Geometry day, on a (somewhat disappointingly) rainy Saturday in Evans Hall. The weather was somewhat better on Sunday, however, allowing myself to make the trip to Mint Plaza for the following cup, which should bear some resemblance to the banner picture on this site. (Unfortunately, they were no longer serving their mini-Brioche buns.)

Blue Bottle Coffee

But now on to the good stuff, a report on some of the talks:

Jaclyn Lang gave a talk on her work concerning the image of big Galois representions. The setup is roughly as follows. Let

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

be an absolutely irreducible odd Galois representation over a finite field (hence modular). Suppose this Galois representation was the residual representation associated to an ordinary modular form that lived inside a Hida family that was smooth over weight space. Then one might expect that the corresponding representation

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

to have image which was as big as possible, namely, containing \mathrm{SL}_2(\mathbf{Z}_p[[T]]). This can’t always be the case, however; for example, the residual representation (or the entire family) could be dihedral. However, if the residual representation contains \mathrm{SL}_2(\mathbf{F}_p), and one additionally assumes that the image of inertia at p is sufficiently large, then this is indeed the case (probably this assumes that the residue characteristic is at least 5). There have been a number of generalizations of this result due to Hida and others which improves the result by weaken the various hypotheses; for example, allowing coefficients, allowing the residual representation to be dihedral, and weakening the ramification hypothesis at p. For these results, one can’t expect that the image is full, but rather that the image contains an appropriate congruence subgroup of \mathrm{SL}_2. I like to think of this as follows: at classical specializations, one knows that the image (if it is not CM or weight one) will have open image; the results of this talk and of previous work show that index can be controlled in families. Actually, this is not quite true, because another obstruction to having open image even classically is the existence of inner twists. The main result of the talk was to deal with this issue of inner twists, and hence also allow for a generalization of the results not only to smooth Hida families but to any irreducible component of any Hida family. (More details to be found here.)

A natural question: one output of Lang’s result is to give an ideal \mathfrak{b} of the Hida family for which the image of these Galois representations contains the \mathfrak{b}-congruence subgroup (after accounting for inner twists). In characteristic zero, my impression from the talk was that one can identify the support of this ideal as coming from CM points and classical weight one modular forms. On the other hand, apparently there is also a version of this result in the reducible case (due to Hida and with extra hypotheses); in that case the zeros should correspond to the reducible locus, or equivalently, the zeroes of the p-adic zeta function. However, a stronger result is true, namely, that \mathfrak{b} can essentially be identified with this p-adic zeta function. So, returning back to the residually irreducible case, the natural question is: can the support of \mathfrak{b} contain the prime p?

Kestutis Cesnavicius gave a talk on the Manin-Stevens and Manin constants for elliptic curves, with emphasis on the prime p=2. He raised the following question: Suppose that N is odd. Is there a surjection from the space of weight 2 classical modular cusp forms of level \Gamma_0(N) with coefficients in \mathbf{Z}_2 to the space of weight 2 Katz cusp forms of the same level over \mathbf{F}_2? The issue here is that the latter space is really the cohomology of the associated stack, not the course moduli space. Unfortunately, this question distracted me a little as I tried to find a counter-example (I failed). A result of Serre and Carayol basically implies that the result can only fail after localizing at a non-Eisenstein maximal ideal \mathfrak{m} of the Hecke algebra \mathbf{T} if the corresponding representation \overline{\rho}_{\mathfrak{m}} is induced from \mathbf{Q}(\sqrt{-1}). (Analogously, for p =3, when the representation is induced from \mathbf{Q}(\sqrt{-3}).) This is related to the classic failure of the first version of Serre’s conjecture for p =3 at level \Gamma_1(13). However, as Serre quickly realized, this failure ultimately comes from a failure to lift mod-p forms as above, except in this case from the intermediate curve X_H(13), not from X_0(13). I ultimately convinced myself that lifting was always possible unless \mathfrak{m} was not only Eisenstein but also the ideal containing T_{p} for all odd p not dividing N. I think this must be related to Ken’s result on component groups of Neron models, and how the non-Eisenstein parts arise for X_H(N) but not for X_0(N) or X_1(N). (More details here.)

The final speaker of the day was Daquin Wan. The key question that arose in his talk was the following. Suppose that D(k,T) is the characteristic power series of the U operator on the space of overconvergent p-adic modular forms in integral weight k. Can one show that

L(k,T) = \displaystyle{\frac{D(k+2,T)}{D(k,pT)}}

has infinitely many zeros and infinitely many poles? One actually has to assume that k \ne 0 here, since otherwise the result is false, as this will be a polynomial of dimension the space of weight two forms. One feels that p-adic Langlands should be able to say enough about slopes in these weights to obtain a contradiction, but I don’t unfortunately see how to do it. The main point of the talk was two-fold. There is an argument of Coleman that shows that D(k,T) is not itself a polynomial. This argument can be generalized to prove that L(k,T) is not a rational function. Second, the product L(k,T) L(-k,p^k T) is actually a rational function because of the properties of the theta operator. So one deduces that at least one of these functions had infinitely many poles and the other had infinitely many zeroes. This also relies on a previous result of Wan that these functions are meromorphic. (Oh, I should mention that this was joint work with … and here I didn’t take notes for a talk two weeks ago … Liang Xiao? Please correct me if I’m wrong)

(Romyar Sharifi also talked, but since I am actively trying to understand something about that talk on a more technical level, so I will have to return to a discussion of it later.)

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Tensor Products

Let W be an irreducible representation of a finite group G. Say that W is tensor indecomposable if any isomorphism W = U \otimes V implies that either U or V is a character.
In conversations with Matt and Toby (which permeate the rest of this post as well), the following problem came up:

Question: Let G be a finite group. Let V be an irreducible representation of G. Is there a unique decomposition

V = V_1 \otimes V_2 \ldots \otimes V_k

of V as a tensor product of tensor indecomposable representations (up to re-ordering and twist)? I don’t think this can be too hard, but I confess I don’t see how to do it. (Since we didn’t really need this, we didn’t think about it too hard.)
(edit: when I say I don’t think this can be too hard, I don’t mean to imply that I think it is true; just that I think either a proof or counterexample should not be too hard to find — hopefully not both.)

One can ask an analogous problem for Lie groups. Actually, the problem for Lie algebras is actually quite simple (and the answer is positive). It is related to the following:

Let V and W be irreducible non-trivial representations of a simple Lie group \mathfrak{g}. Then V \otimes W is reducible.

Proof: Assume that V \otimes W is irreducible. In particular, it is determined by its highest weight. Let the highest weights of V and W be \lambda and \mu respectively. Then the highest weight of V \otimes W must be \lambda + \mu. But now, by the Weyl character formula, we deduce that

\displaystyle{1 = \frac{\dim(V) \dim(W)}{\dim(V \otimes W)} = \prod_{\Phi^{+}} \frac{ \langle \rho,\alpha \rangle \langle \rho + \lambda + \mu, \alpha \rangle} { \langle \rho + \lambda,\alpha \rangle \langle \rho + \mu, \alpha \rangle}}.

The product term can also be written as:

\displaystyle{1 + \frac{ \langle \lambda,\alpha \rangle \langle \mu, \alpha \rangle} { \langle \rho + \lambda,\alpha \rangle \langle \rho + \mu, \alpha \rangle}}.

In particular, since the pairing is non-negative between positive roots and highest weights, we deduce a contradiction unless

\langle \lambda,\alpha \rangle \langle \mu, \alpha \rangle = 0

for all \alpha \in \Phi^{+}. The assumption that \mathfrak{g} is irreducible, however, is equivalent to saying that \Phi^{+} has a maximal root \beta, and for such a maximal root, we have

\langle \lambda, \beta \rangle = 0 \Rightarrow \langle \lambda, \alpha \rangle = 0, \ \forall \alpha \in \Phi^{+} \Leftrightarrow \lambda = 0.

Note that this lemma is actually a special case of a theorem of Rajan, who proved that, for simple \mathfrak{g}, the factors of a (not necessarily irreducible) tensor product are determined by the representation. In particular, the tensor product of two non-trivial irreducible representations cannot be irreducible.

The problem with the initial question is that it’s hard to construct tensor products of irreducible representations which are irreducible. Or rather, it is easy, simply by taking U \otimes V where U is an irreducible representation of H and V is an irreducible representation of G and the tensor is irreducible for H \times G. Yet the interesting case is something closer to assuming that U and V are faithful. Actually, this motivates the following question:

Question: Do there exist irreducible non-trivial representations U and V of a finite simple non-abelian group G such that U \otimes V is irreducible?

The argument above for Lie groups suggests that this may not happen for Chevelley groups (although it certainly doesn’t prove this). It also suggests (relating the representation theory of A_n and S_n to \mathrm{GL}_n) that it doesn’t happen for the alternating groups either. It almost surely doesn’t happen for the sporadic groups either. So my guess that the answer to the problem above is no, and that this is probably known, and probably requires classification. (Please comment if you know the answer.) Actually, this also reminds me of a similar problem which I think is open.

Question: Fix N. Does there exist a non-trivial representation V of a finite group G of dimension N such that \mathrm{Hom}^0(V,V) (of dimension N^2 - 1) is irreducible?

This question came up in my paper with Barry, where I was surprised to find very few examples. I seem to remember that the Mathieu group M_{12} has an 11-dimensional representation whose corresponding 120 dimensional adjoint is irreducible. Can one classify all such examples coming from simple groups?

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Due Process

When I was a senior (’93), my high school began a program of inviting “distinguished” visitors to give talks to a selected number of students, several of whom were also invited to lunch with the speaker. I was fortunate that the program was run by Lawrence (Larry) Doolan, head of maths, which meant that I got more than my fair share of free lunches. I’m not sure how many talks there were, but I particularly remember talks by John Halfpenny (who gave, perhaps unexpectantly, a thoughful and nuanced take on labour unions), Meg Lees (ugh, no wonder the Democrats no longer exist as a party), Michael Dukakis, and also someone who I believe was a judge on the International Court of Justice. I remember at the time (over lunch) questioning the latter gentleman what the ICJ (and UN more generally) was planning on doing about the ongoing crisis in Yugoslavia. He responded (somewhat peevishly in my memory) that this was an internal matter which was not within the juristriction of any UN courts. To my young mind, he certainly seemed to display a perverse attitude which elevated process far above justice. It was therefore satisfying today to see that the International Criminal Court (which did not exist until 2002) sentence Radovan Karadzic to 40 years in prison. I mean, it’s cetainly at least 20 years too late, but better than nothing. (Slobodan Milošević, on the other hand, got off lightly.)

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Ventotene, Part II

I promised to return to a more mathematical summary of the conference in Ventotene, and indeed I shall do so in the next two posts.

One of the themes of the conference was bounding the order of the torsion subgroup in arithmetic lattices. Tsachik Gelander gave a number of talks (in part) on the seven author paper. One nice result was a uniform bound of the shape

\log |H^*(\Gamma,\mathbf{Z})^{\mathrm{tors}}| < C \cdot \mathrm{Vol}(\Gamma),

where \Gamma ranges (say) over all lattices in \mathrm{SL}_n(\mathbf{R}) for a fixed n \ge 3. (The key result here is the uniformity — this result is much easier for covers of a fixed manifold.)

Two natural questions that came up (in conversation at least) during the conference are as follows:

  1. Can one do better in low degree?
  2. What is the true expectation for the size of this group for (say) congruence subgroups of \mathrm{SL}_n(\mathbf{Z})?

Let’s consider the first question. For (congruence) subgroups of \mathrm{SL}_n(\mathbf{Z}), one can certainly say quite a bit more. For example, H^1 is essentially trivial, by the congruence subgroup property. However, in the stable range of cohomology (in particular, when the completed cohomology groups become stable), the groups H^* are finite over \mathbf{Z}_p, and so contribute very little. One does, at least, have the following soft arguments for general groups.

Proposition: Let G be a semi-simple group over \mathbf{Q} with \mathbf{Q}-rank r = r_{\mathbf{Q}}. Then \widetilde{H}_i is a torsion \Lambda = \mathbf{Z}_p[[G(\mathbf{Z_p})]]-module for i <  r_{\mathbf{Q}}.

The proof is as follows: the boundary terms are also torsion, so it suffices to show that all the \widetilde{H}^{BM}_i in the appropriate range are also torsion, where we consider Borel-Moore homology. Assume otherwise. Let \dim G(\mathbf{R})/K(\mathbf{R}) = d. From the spectral sequence \mathrm{Ext}^i(\widetilde{H}^{BM}_j,\Lambda) \Rightarrow \widetilde{H}_{d-i-j}, we deduce that there is at least one i < r_{\mathbf{Q}} such that \widetilde{H}_{d-i} \ne 0. Yet the homological dimension of \Gamma \backslash G(\mathbf{R})/K(\mathbf{R}) is, by Borel-Serre, equal to d - r_{\mathbf{Q}}, and so all the homology in these degrees (and hence certainly the completed homology) vanishes.

One can do better in certain algebraic cases, where one can deduce vanishing of the completed cohomology in certain degrees by perfectoid technology (as in Corollary 4.2.3 of Scholze’s paper).

The answer to the second question, even conjecturally, is more mysterious. There are some speculations related to this question in Bergeron-Venkatesh. But it seems a little tricky to formulate a precise guess (for a good upper bound).

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Best seat in the house

I recently attended an András Schiff recital at Chicago Symphony Center (not the first such Schiff performance I have been to). The conceit of this concert was the “last sonatas,” a performance of the last sonatas of Beethoven, Haydn, Mozart, and Schubert respectively. (This was actually the third in a series of concerts, the previous two consisting of the antepenultimate and penultimate sonatas respectively of the same four composers.) I was a little bit disorganized about buying tickets — I only heard about the concert on the radio the week before — but this had a positive effect: the tickets were almost sold out, so the CSO introduced extra “stage seating,” which is exactly what it sounds like. I literally had the best seat in the house, roughly located in the middle of where the first violins would be. I swear that Andras Schiff looked directly into my eyes for one of his bows. (To be fair, hew bowed quite a few times — there were about 6 curtain calls in between two encores: a middle movement of the penultimate Schubert sonata and the aria to the Goldberg variations.) If you ever get a chance to purchase stage tickets at the CSO, I strongly recommend it!

Yes, this is really my seat

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Hilbert Modular Forms of Partial Weight One, Part III

My student Richard Moy is graduating!

Richard’s work has already appeared on this blog before, where we discussed his joint work with Joel Specter showing that there existed non-CM Hilbert modular forms of partial weight one. Today I want to discuss a sequel of sorts to that paper, which also forms part of Richard’s thesis (I should note that he already has five publications and will have 7 or 8 papers by the time he graduates.) The starting observation is as follows. Fix a real quadratic field F. From the perspective of Galois representations, the Hilbert modular forms of partial weight one fall under the case \ell_0 = 1 in the notation of my paper with David Geraghty (this is in the context of coherent cohomology). To orient the reader, let us discuss three classes of such forms:

  1. Hilbert modular forms of weight [2k+1,1] for a real quadratic field F.
  2. Regular algebraic cuspidal automorphic forms for \mathrm{GL}(3)/\mathbf{Q}.
  3. Regular algebraic cuspidal automorphic forms for \mathrm{GL}(2)/F for an imaginary quadratic field F.

Suppose one fixes a tame level N and then looks at the space of such forms as the weights vary. In both of the latter cases, the problem has been raised (or even conjectured, for N = 1 and \mathrm{GL}(3) by Ash and Pollack here), of whether all but finitely many such forms arise via functoriality from a smaller group. More explicitly, one can ask whether:

  1. If G = \mathrm{GL}(2)/F, then all but finitely many cuspidal regular algebraic forms of conductor N either arise (up to twist) via base change from \mathrm{GL}(2)/\mathbf{Q}, or are induced from a quadratic CM extension E/F.
  2. If G = \mathrm{GL}(3)/\mathbf{Q}, then all but finitely many cuspidal regular algebraic forms of conductor N arise up to twist as the symmetric square of a form from \mathrm{GL}(2)/\mathbf{Q}.

Naturally enough, one can make the same conjecture whenever \ell_0 > 0, appropriately formulated. There does not seem to be any case of this conjecture which is known, although there are analogous results (where one fixes the weight and varies the level) in both weight one (where it is almost trivial) and for imaginary quadratic fields (in the work of Calegari-Dunfield and Boston-Ellenberg). Still, the conjectures in varying weight seem pretty hard even for N = 1. In that context, Richard proves the following nice complementary pair of theorems below. Let F = \mathbf{Q}(\sqrt{7}). The field F has narrow class number 2 and there is a unique odd everywhere unramified quadratic character \chi of G_F with fixed field E = F(\sqrt{-1}).

Theorem I (Moy) Let F and \chi be as above. Every Hilbert modular form over F of weight [2k+1,1] and level N = 1 is CM, and in particular is induced from E.

Theorem II (Moy) Let F and \chi be as above. Let M be a strongly compatible family of two dimensional Galois representations of F with determinant \chi, level N = 1, and Hodge–Tate weights [0,0] and [k,-k]. Then M is induced from E.

Theorem I is almost an immediate consequence of Theorem II, with the caveat that one doesn’t quite have complete local-global compatibility for partial weight one modular forms (though results and methods of Luu, Jorza, and Newton get close). Theorem II on the other hand is a consequence of the following:

Theorem III (Moy) Let F and \chi be as above. Let

\rho: G_F \rightarrow \mathrm{GL}_2(\overline{\mathbf{Q}}_3)

be a continuous irreducible representation with determinant \chi that is unramified at all finite places except for one prime v|3. Then \rho is induced from a character of G_E.

The argument in this case is (roughly) the following. Using a Tate-style argument (with discriminant bounds), one proves that the residual representation \overline{\rho} must have semi-simplification \chi \oplus 1. The restriction of \rho to G_E then has the property that its image is pro-3 and unramified outside the fixed prime v|3. Yet one shows by a class field theory computation that the largest abelian 3-extension unramified outside v|3 is cyclic, which (by consideration of the Frattini quotient) immediately implies that the image of \rho restricted to G_E factors through a cyclic quotient as well, and one is done.

Note that to deduce Theorem I, one first has to prove (using a congruence argument) that at the other prime w|3, either:

  1. The representation \rho is unramified at w,
  2. The representation \rho restricted to D_w has unramified semi-simplification. In particular, the generalized eigenvalues of \mathrm{Frob}_w for \overline{\rho} are both the same.

To finish, one rules out the second possibility by computing all the modular residual representations explicitly by doing computations in low weight (this can ultimately be reduced to a computation on the definite quaternion side, although Richard had to write his own programs to do this since the current magma implementation required trivial character for non-parallel weight.)

It is true that these arguments will not suffice for the more general conjecture, but then, I haven’t seen a viable strategy to prove those conjectures either!

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Ventotene, part I

I recently returned from Ventotene, an Island some 40 miles west of Naples, where I attended a very pleasant conference on Manifolds and Groups. I have several mathematical thoughts on the conference, but for today I will content myself to a brief description of some of the extra-curricular activities related to spending a week on an Italian Island. The participants of the conference were scattered over 5 of the Island’s numerous hotels. My particular hotel, the Hotel Borgo Cacciatori, had the feature that it was roughly a mile walk from either downtown or the conference center (the Island itself is not two miles in length). This was good for two reasons. First, it provided the more sedentary amongst us to get some exercise, secondly, being near the highest point of the island, it afforded a nice view of the ocean:


Finally, on the two evenings with no cloud cover, the walk back to the hotel provided a spectacular view of the Milky Way.

The food options included all manner of interesting melanzane based dishes, some good antipasti:

and, of course, some very tasty pork with crackling:


The end of the week saw the start of the feast of Santa Candida, the Island’s patron saint, which included the ceremonial release of a balloon (of which I have a wonderful video which can not be embedded in this post). Some number-theoretic Proustian character was also notable…

Kowalki's Blog

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