Report from MSRI and Berkeley

Having attended last Friday’s academic sponsors’ day at MSRI, I can provide a little more context concerning the issues expressed last time.

But first, it’s time for the current edition of NAME AND SHAME:

There are currently 105 universities which contribute a yearly amount of money to MSRI. Although the amount of money is not insubstantial (about $5000 per institution), it sends a clear message to those outside mathematics (= potential donors) that we as a community value the job that MSRI does. (As a certain dean at Harvard once explained to me, nobody wants to make large donations to an institution unless they feel part of a larger movement, which requires a high rate of participation amongst everyone else.) Since the main focus of MSRI is generally as a research institution, you would certainly expect that (say) that all the top 25 ranked graduate programs in the US would be sponsors of MSRI. But this is not true! How could this be, you ask? I imagine the reason is that this funding is at the discretion of the chair, and you (as a department member) might not even think to ask if your institution is an academic sponsor of MSRI. So who are the miscreants who have skirted their obligations? The guilty party: Brown! Why is Brown not an institutional member of MSRI? Is it because of their precipitous drop in the USA rankings as an undergraduate programme over the last few decades? Is it because — let’s face it — things haven’t been the same since the halcyon days of the (admittedly rather sexy) undergraduate class of 2001? Is it because their endowment is so low that their check to MSRI bounced? Let your disappointment be known by emailing the chair, Jeffrey Brock. (Oh, and NYU is not a member also.)

OK, now back to the meeting. I learned a few things. First (and it’s a little hard to determine exactly what this means given the fungibility of money), the summer graduate schools are exactly paid for by contributions from academic sponsors. In this light, the issue of fairness is even more acute than I previously realized. Second, there is already a committee in place which is very much aware of the issues relating to mixed levels of backgrounds and is trying to find ways to address it. They seem to have a few good ideas (in particular, making clear before hand what the expectations will be, in order for universities to self-select appropriate students as well as give students with weaker backgrounds information on what they should learn about before hand), but I agree there’s no simple fix.

Naturally, of course, you also want the update on what’s going in the Berkeley culinary scene, or at least what can be ascertained from by a casual and infrequent visitor.

Andronico’s (also known as Astrinomico’s) has been replaced by Safeway. The decline in quality is immediately apparent — although to be honest, this judgement is mostly based from the brief look I had at the current Champagne selections.

Cafe Rouge has closed! I think Dipankar introduced me to this store. I had a great cassoulet there once.

Babette still does a very good coffee and pastry, although there was a charm to the previous outdoor space which seems to have been a little lost in the move. The new Blue Bottle Cafe is very shiny and has the very good manners to open early (e.g. early enough to get two cortados before going up the hill to MSRI).
I also made it (as is my habit) to visit Blue Bottle on Mint Plaza on my way to the airport for breakfast for some aeropress coffee as well as another cordato:


Cortado at Babette

Blue Bottle Aeropress

Aeropress at Blue Bottle

Blue Bottle Cortado

Cortado at Blue Bottle (these are both from the Mint Plaza store in SF)


Besides hitting the cheeseboard (more of a miss this time, too much potato) and Babette for lunch, I only really had one free evening, but I did manage to also visit Cesar’s, Gregoire’s, and Chez Panisse for a progressive dinner of fino Sherry, lamb pasta (delicious), and a pear galette with Madeira. Arthur Ogus also made a cameo appearance on his bike picking up takeout at Gregoire’s.

Pear Galette

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Complain about MSRI day

I will be heading off later this week to the Academic Sponsors’ Day at MSRI, going as Shmuel’s proxy for uchicago. I’m not sure to what extent (if any) there is for me to make policy suggestions, but any comments you would like me to pass on to MSRI management would be appreciated. Post them below!

There is one program at MSRI that (from my mostly second hand observations) seems as though it could do with some improvement. MSRI regularly holds a number of summer schools. For example, this summer, Kevin Buzzard is giving an introduction to the theory of automorphic forms. I would have loved to been able to go to such talks when I started out as an graduate student, and I really wanted to send some of my current students to this. The problem? Firstly, there are a severely limited number of places. I’m not entirely sure I understand this, but I can imagine a few reasons. However, MSRI (apparently) goes to extreme lengths to be as “equitable” as possible in admitting people from as many different sponsor schools as possible. The result? I am told (internally) that the University of Chicago will be at most able to send two or three students in total to the seven or so different programs available. The result is that the only people uchicago sends to these programs are basically advanced students who are about to graduate and aren’t in any sense the target audience. On the other hand, these programs also tend to admit students from schools with much weaker backgrounds who aren’t even comfortable with basic concepts from algebraic number theory. This seems to be a very stupid way to choose participants for any program, even if it is purely in the name of fairness. What does a Harvard number theory student have to do to be able to attend an introductory course on automorphic forms — prove the Sato-Tate conjecture?

Apparently at least one lecturer plans to give their 20 courses in the order 1,11,2,12,3,13,etc in order to try to please at least half the people half the time.

One wag suggested that schools from weaker programs should consider it in their best interest to only admit students from places like Harvard, since then at least the programs would be training their future professors ([Caveat: I heard this second hand, so it may well have been in jest]). Presumably the reason behind MSRI’s policy is that the spoils of MSRI programs should go equally to (students of) universities that fund MSRI (sssuming they contribute a similar amount.) However, there seems to me to be a very natural alternative solution to this problem, namely, to continue having “advanced” summer courses but also introduce some deliberately introductory courses tailored to people from schools with less background. Students from schools with more advanced programs could be barred from the lower level introductory programs (since they could be more easily reproduced locally) and then, when it comes to the more advanced topics, there wouldn’t be the restriction to admit at most one student from each school. In particular, the selection criteria should concentrate (in part) on who stands to get the most out of each program.

Please add any further suggestions or complaints below

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Virtual coherent cohomology

I gave a talk yesterday where I attempted to draw parallels between the cohomology of (arithmetic) 3-manifolds and weight one modular forms. It was natural then to think about whether there was an analogue of the virtual Betti number conjecture. Recall the following:

Theorem [Agol] Let M be a compact hyperbolic 3-manifold. Then \mathrm{dim} H^1(N,\mathbf{Q}) is unbounded as N ranges over all finite covers N \rightarrow M.

(There’s an analogous version for finite volume hyperbolic manifolds with cusps.) What is the corresponding conjecture in coherent cohomology? Here is a first attempt at such a question.

Question: Let X be a proper smooth curve of genus g \ge 2 defined over \mathbf{Q}. Let \mathscr{L} denote a line bundle such that \mathscr{L}^{\otimes 2} = \Omega^1_X. As one ranges over all (finite etale) covers \pi: Y \rightarrow X, are the groups

H^0(Y,\pi^* \mathscr{L})

of unbounded dimension?

One might ask the weaker question as to whether there is a cover where this space has dimension at least one (and in fact this is the first question which occurred to me). However, there are some parity issues. Namely, Mumford showed the dimension of H^0(X, \mathscr{L}) is locally constant in (X,\mathscr{L}), and this dimension is odd for precisely 2^{g-1}(2^g + 1) choices of \mathscr{L} (there are 2^{2g} such choices and the choices are a torsor for 2-torsion in the corresponding Jacobian). But I think this means that one can always make \pi^* \mathscr{L} effective for some degree 2 cover, and thus produce at least one dimensions worth of sections. For example, when g = 1, then \Omega^1_X = \mathcal{O}_X, and \mathscr{L} = \mathcal{O}_X has global sections whereas the other square-roots correspond literally to 2-torsion points. But those sections become trivial after making the appropriate 2-isogeny.

Another subtlety about this question which is worth mentioning is that I think the result will have to be false over the complex numbers, hence the deliberate assumption that X was defined over \mathbf{Q}, or at least over a number field. Specifically, I think it should be a consequence of Brill-Noether theory that the set of X in \mathscr{M}_g such that

\mathrm{dim} H^0(Y, \pi^* \mathscr{L}) > 1

for any choice of \mathscr{L} and any cover \pi: Y \rightarrow X of degree bounded by a fixed constant D will be a finite union of proper varieties of positive dimension. And now the usual argument shows that, as D increases, any countable union of varieties cannot exhaust \mathscr{M}_g. But it can, of course, exhuast all the rational points, and even all the algebraic points.

There’s not really much evidence in favor of this question, beyond the following three very minor remarks.

  1. The only slightly non-trivial case one can say anything about is when X is a Shimura curve over \mathbf{Q}, and then the answer is positive because there exist lots of weight one forms (which one can massage to have the right local structures after passing to a finite cover).
  2. The analogy between H^0(X,\mathscr{L}) and H^1(M,\mathbf{Q}) is fairly compelling in the arithmetic case, so why not?
  3. There doesn’t seem to be any a priori reason why the virtual Betti number conjeture itself was true, and it is certainly false in for related classes of groups (groups with the same number of generators and relations, word hyperbolic groups), so, by some meta-mathematical jiu-jitsu, one can view the lack of a good heuristic in the hyperbolic case as excusing any real heuristic in the coherent case.
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The class number 100 problem

Some time ago, Mark Watkins busted open the “class number n” problem for smallish n, finding all imaginary quadratic fields of class number at most 100 (the original paper is here) Although the paper describes the method in detail, it does not actually give the complete list of imaginary quadratic fields which occur (for fairly obvious reasons given the size of the list). I’ve occasionally wanted to consult the actual list, and most of the time I have just emailed Mark to find out the answer. But now it is available online! Here is the link. (Maybe someone could put this on the LMFDB?)

Consulting the table one immediately notices a number of beautiful facts, such as the fact that (Z/3Z)^3 does not occur as a class group. (Our knowledge of p-parts of class groups, following Gauss, Pierce, Helfgott, Venkatesh, and Ellenberg, is enough to show that (Z/2Z)^n and (Z/3Z)^n for varying n only occur finitely often [similarly these groups plus any fixed group A], but those results are not effective.) One also sees that D = – 5519 and D = -1842523 are the first and last IQF discriminants of class number 97. It’s the type of table that immediately bubbles up interesting questions which one can at least try to give heuristic guesstimates. For example, let mu(A) denote the number of imaginary quadratic fields with class group A. Can one give a plausible guess for the rough size of mu(A)? One roughly wants to combine the Cohen-Lenstra heuristics with the estimate h \sim \Delta^{1/2}. To do this, I guess one would roughly want to have an estimate for

\displaystyle{\sum_{x^{1/2 - \epsilon} < |A| \le x^{1/2 + \epsilon}} \frac{1}{|\mathrm{Aut}(A)|}}.

I wouldn’t be surprised if someone has already carried out this analysis (thought I don’t know any reference). As a specific example, what is the expected growth rate of mu(Z/qZ) over primes q? A related question: is there a finitely generated abelian group which provably does not occur as the first homology of a congruence arithmetic hyperbolic 3-manifold?

At any rate, this is a result that Gauss would have appreciated. Curiously enough, this paper was recently posted as an answer on to the (typically ridiculous as usual) MO question Which results from the last 30 years, in any area of mathematics, do you think are the most important ones? While I wouldn’t quite put it in that class, I do find it curious that it this answer (at the time of writing) has -4 votes on mathoverflow. Given the enormous crap that does receive positive votes, I suppose that such minus votes are not to be taken too seriously. I would, however, make the following claim: Watkins’ result is as least as interesting as any original number theory research that has appeared on MO (at least as far as anything I have seen).

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Central Extensions, Updated

I previously mentioned a problem concerning polynomials, whose motivation came from thinking about weight one forms and the inverse Galois problem for finite subgroups of \mathrm{GL}_2(\mathbf{C}). I still like the polynomial problem, but I realized that I was confused about the intended application. Namely, given a weight one form with projective image A_5, there is certainly a unique minimal lift up to twist, but the images of the twists also automatically have image given by a central extension A_5. So, just by twisting, one can generate all such groups as Galois groups by starting with a minimal lift. More prosaically, every central extension of A_5 by a cyclic group is either a quotient of A_5 \times \mathbf{Z} or of \widetilde{A}_5 \times \mathbf{Z} where \widetilde{A}_5 is the Darstellungsgruppe of A_5 (which is \mathrm{SL}_2(\mathbf{F}_5)). So, to solve the inverse Galois problem for central extensions of A_5, it suffices to solve it for \mathrm{SL}_2(\mathbf{F}_5). That is not entirely trivial, but it is true.

I still think it’s an interesting problem to determine which extensions of A_5 by cyclic groups occur as the Galois groups of minimally ramified up to twist extensions, but that is not the same as the inverse Galois problem.

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Correspondance Serre-Tate, Part I

Reading the correspondence between Serre and Tate has been as delightful as one could expect. What is very nice to see — although perhaps not so surprising — is the utter delight that both Serre and Tate find in discussing numerical examples. One of the beautiful aspects of number theory is that there is an abundance of examples, each of which exhibit both special cases of a vast general theory and yet each delighting with their own idiosyncracies: \mathbf{Q}(\sqrt{-23}), X_0(11), 691, 144169, etc. (It is precisely the absence of such examples, or at least any discussion of them, why geometric Langlands tends to leave me completely cold.) Take, for example, the following:

Letter from Tate to Serre, Dec 8, 1958:

Are you aware that the class number of the field of 97th  roots of 1 is divisible by 3457 and 118982593? And that 3457 = 36 * 96 + 1 and 118982593 = 1239402 * 96 + 1?

If reading that doesn’t give you just a little thrill, then you have no soul. Does it have any significance mathematically? The class number is large, of course, which relates to the fact (proved by Odlyzko) that there are only finitely many Galois CM fields with bounded class number. (The reason why one can access class numbers of CM fields F/F+ is that the unit group of F and F^+ are the same up to finite index, so the *ratio* of zeta values \zeta_{F}(1)/\zeta_{F^+}(1) is directly related to the minus part of the class group h^{-} uncoupled from any regulator term, so one can access this analytically.) Alternatively, one might be interested in the congruences of the primes q dividing the class number. In this case, we see a reflection of the conjectures of Cohen and Lenstra. Namely, we expect that there is a strong preference for the class group to be “more cyclic,” especially for larger primes. The class group also has an action of (\mathbf{Z}/97\mathbf{Z})^* which is cyclic of order 96. Since one expects the plus part h^{+} to be very small (and indeed in this case it is trivial), this means that complex conjugation should act non-trivially, which means that the group of order 96 should (at least) act through a quotient of order at least 32. So if the class group is actually cyclic, this forces the prime divisors q of h_F to be 1 mod 32, and even 1 mod 96 if the class group of F doesn’t secretly come from the degree 32 subfield of F (which it doesn’t). (Not entirely irrelevant is Rene Schoof’s nice paper on computing class groups of real cyclotomic fields.)

Both Serre and Tate are unfailingly polite to each other. As a running joke, the expression “talking through one’s hat” occurs frequently, as for example the letter of Nov 14, 1961, where the subtle issue of the failure of B \otimes_A C \rightarrow B \widehat{\otimes}_A C is discussed. (Another amusing snippet from that letter “Even G. himself makes mistakes when he thinks causally.”) The correspondence is also fascinating from the perspective of mathematical history — one sees the progress of many ideas as they are created, including the Honda-Tate theorem and the Tate conjecture over finite fields. The first time the latter appears (as a very special case) it actually turns out to be an argument of Mumford, who shows Tate an argument (using Deuring) why when two elliptic curves have the same zeta function they are isogenous. This elicits the following reaction from Tate:

Letter from Tate to Serre, May 9, 1962:

“Damn! The result is certainly new to me, and it frankly makes me mad that I never noticed it”

We have all been there, although, to be fair, most of us have the excuse of not being Tate!

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Today is Persiflage’s birthday, but let us not also forget to wish many happy returns to friends of the blog Ana and Vytas, who share the same birthday!

Here is how I plan to celebrate: I will spare you with the complete 12 hour 23 minute breakdown of my carefully curated iTunes play list, but suffice to say it is both suitably seasonal and indulgent. A few highlights: Spem in Alium, Lupu and Perahia playing Schubert’s Fantasia in F minor, Bach Cello Suites, A Musical Offering, and plenty of carols from King’s College Cambridge. There will be a trip to Intelligentsia (by Uber —  it is -17 C outside) and then a wander around the Chicago Art Institute. For lunch, smoked salmon and smoked eel (hat tip to Bao who pointed out the existence of a fine food store in duty free at Schiphol airport!) with Champagne, followed by left-over home made Coq au vin with orzo. For dinner, Foie Gras mi-cuit with a side of poached apples, together with Sauternes. And then maybe some vegetable and tofu stir fry to balance things out slightly, in order to justify finishing off the evening with some delicious moist fruit cake. The final indulgence: I plan to spend the afternoon at home drinking tea, reading the Serre-Tate correspondence, and looking out my window onto the lake (which right now has a beautiful cover of fog which my poor photography isn’t quite able to capture):

Fog on Lake

All of this, of course, with the best possible company imaginable for a quiet day of self-indulgence. Happy holidays!

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