Elementary Class Groups Updated

Posted on June 10, 2017 by Persiflage

In a previous post, I gave a short argument showing that, for odd primes p and N such that \(N \equiv -1 \mod p,\) the p-class group of \(\mathbf{Q}(N^{1/p})\) is non-trivial. This post is just to remark that the same argument works under weaker hypotheses, namely:

Proposition: Assume that N is p-power free and contains a prime factor of the form \(q \equiv -1 \mod p,\) and that p is at least 5. Then the p-class group of \(K = \mathbf{Q}(N^{1/p})\) is non-trivial.

The proof is pretty much the same. If N has a prime factor of the form \(1 \mod p,\) then the genus field is non-trivial. Hence we may assume there are no such primes, from which it follows that \(H^1_S(\mathbf{F}_p)\) has dimension one and \(H^2_S(\mathbf{F}_p)\) is trivial, where S denotes the set of primes dividing Np. The prime q gives rise to a non-trivial class \(b \in H^1_S(\mathbf{F}_p(-1))\) which is totally split at p (this requires that p be at least 5), and the field K itself gives rise to a class \(a \in H^1_S(\mathbf{F}_p(1)).\) But now the vanishing of H^2 implies that \(a \cup b = 0\) and hence there exists a representation of G_S of the form:

\(\rho: G_S \rightarrow \left( \begin{matrix} 1 & a & c \\ 0 & \chi^{-1} & b \\0 & 0 & 1 \end{matrix} \right),\)

where \(\chi\) is the mod-p cyclotomic character. The class c gives the requisite extension (after possibly adjusting by a class in the one-dimensional space \(H^1_S(\mathbf{F}_p)).\) The main point is that the image of inertia at primes away from p is tame and so cyclic, but any unipotent element of \(\mathrm{GL}_3(\mathbf{F}_p)\) has order p if p is at least three. This ensures c is unramified over K away from the primes above p. On the other hand, the class \(b\) is totally split at p. This implies that the class c is locally a homomorphism of the Galois group of \(\mathbf{Q}_p,\) and so after modification by a multiple of the cyclotomic class in \(H^1_S(\mathbf{F}_p)\) may also be assumed to be unmarried at p. The fact that \(b \ne 0\) ensures that \(c \ne 0,\) and moreover the fact that p is at least 5 implies that the kernel of c is distinct from that of a, completing the proof. (This result was conjectured in the paper Class numbers of pure quintic fields by Hirotomo Kobayashi, which proves the claim for \(p = 5.)\)

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Thoughts on Paris

Posted on May 25, 2017 by Persiflage

Nothing is more pretentious or annoying than when an American offers, uninvited, their opinions of Paris. Here, then, are some of mine.

  • Starting the day with a two hour lecture on elliptic integals:OUI: Who does not get a slight frisson upon seeing the identity$$ \displaystyle{\frac{j}{16} = (\sqrt{k})^4 + \left(\frac{1}{\sqrt{k}}\right)^4
    + \left(\frac{1 – \sqrt{k}}{1 + \sqrt{k}}\right)^4 + \left(\frac{1 + \sqrt{k}}{1 – \sqrt{k}}\right)^4
    + \left(\frac{1 – \sqrt{-k}}{1 + \sqrt{-k}}\right)^4 + \left(\frac{1 + \sqrt{-k}}{1 – \sqrt{-k}}\right)^4 + 42}.$$

    where j is the modular invariant and k is the usual parameter of elliptic integrals, given in terms of theta functions as \(\theta^2_2/\theta^2_3\) where \(\theta_2 = \sum q^{(n+1/2)^2}\) and \(\theta_3 = \sum q^{n^2}.\)

  • Starting the morning with a croissant:NON: There are decent enough croissants available, but in the general spectrum of correctly proportioning one’s caloric intake, there are better choices.
  • Starting the morning with a Kouign Amann:OUI…ET NON: Yes, I did wake up at 6:45 so I could bike to Blé Sucré and have a Kouign Amann before they were sold out. It was indeed good. But it still didn’t live up to the buttery sugary indulgences I had in Brittany. Calling on Jacques Tillouine to organize another conference in Roscoff!
  • Using Vélibs (the Paris bikeshare program):OUI: Travelling by bike, especially from my location at Paris 7, was extremely convenient, not to mention very pleasant in the clear 70 degree days with a light breeze that were pretty much a constant throughout my stay. The bike paths were excellent, and rarely required having to get too close to cars. But even on-the-road traffic (for example, cycling around the place de la Bastille) was less stressful than it can sometimes be in Chicago or London. The Velib stations themselves were not perfect: there were a number of times the internet connection was down, or the machine inexplicably returned to the initial screen or gave some other error (the “you already have a bike rented” being the most disturbing one), or the closest stations were either all full or empty depending on whether you were trying to return or rent a bike, but this type of thing seems to happen for many such programs. Extra points for the baskets on the front of the bikes which were extremely useful. Also points for being so much cheaper than Divvy: I had about three weeks of use for 24E, wheras in Chicago the cheapest option would have been to get a $100 yearly membership.
  • Going anywhere by car:NON: Traffic was terrible. Fortunately, I mostly avoided having to be in a car. We did go by bus to the Paris Mosque. We ended up being stuck in one stretch of road for about 10 minutes, at a point where the alternative would have been a very pleasant (and less than 10 minute) walk through the jardin de plantes.
  • The Gardens at Giverny: OUI: I had to choose a day exursion for my young charges, and I was very happy with this choice. Admittedly, a Parisian local described my choice as “American,” so make of that what you will.
  • Lunch with Clozel: OUI: I didn’t have much time for socializing on this trip, but I did get to have a very pleasant lunch with Laurent. If you leave this off your itenerary, you haven’t seen Paris!
  • Orange SIM cards: NON: My phone would randomly claim that I had used up all my data, and I would hae to turn it off and start it again before it would work. It was truly the worst SIM card I have ever used in Europe. I strongly recommend using anyone but Orange.
  • Third Wave Coffee: OUI…ET NON: It is well known that the French have mastered all aspects of cafe culture except making drinkable coffee. But I was very interested to see how much of the third wave had infiltrated into Paris. Here’s a breakdown of the third wave places I visited in order of preference: Telescope, Boot (Right Bank and Left Bank — the Right Bank store is much smaller and has wifi, the Left Bank is bigger and does not), Coffee Cuillier, Fragments, Strada (two locations), Le Peleton Cafe, Ten Belles, and Passager, although the gap between almost all of these was close to non-existent and I would revisit any of them if I was in the neighbourhood. (I had a very pleasant stay at Passager working on my laptop outside. I stayed there for so long I very nearly forgot to pay for my coffee when I left.) Given the weather and general ambience, the general experience of biking to these cafes and then sitting down for a flat white (or equivalent) or espresso was overall very pleasant. On the other hand, I would rate the coffee at these places as generally fine but not great. Many of these places seem to be run (or staffed) by Australians, which is no surprise. (As mentioned previously, Australians have also done wonders for coffee in London.)
  • Background music in cafes: NON: There seems to be some sort of cultural time warp, with Paris 7 students consisting of skateboarding dudes smoking and wearing ’80s fashion. The music in the cafes is similarly pretty bad. Of course, YMMV.
  • Restaurants:My restaurant list is somewhat longer than my cafe list, and I have a detailed set of notes, but I would say the best overall meal was at La Bourse et La Vie. For those on a budget looking for a cheap place to have a light lunch, I strongly recommend Canard & Champagne. Other notable courses: a rendition of vitello tonnato at Paul Bert, a light egg tapas dish whose name I don’t remember at Sourire tapas françaises, a fluffy squid dish which tasted like liquid quiche at Semilla, seared Foie Gras at Domaine De Lintellac, and a few more.
  • The weather in May:OUI: It poured the first day or so, and threatened in the forecast to rain quite frequently. But future forecasts faded, and for almost the entire three weeks, it was pretty close to a blissful 70 degrees, clear, with a slight breeze. Perfect!
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Who proved it first?

Posted on April 25, 2017 by Persiflage

During Joel Specter’s thesis defense, he started out by remarking that the \(q\)-expansion:

\(\displaystyle{f = q \prod_{n=1}^{\infty} (1 – q^n)(1 – q^{23 n}) = \sum a_n q^n}\)

is a weight one modular forms of level \(\Gamma_1(23),\) and moreover, for \(p\) prime, \(a_p\) is equal to the number of roots of

\(x^3 – x + 1\)

modulo \(p\) minus one. He attributed this result to Hecke. But is it really due to Hecke, or is this more classical? Let’s consider the following claims:

  1. The form \(f\) is a modular form of the given weight and level.
  2. If \(p\) is not a square modulo 23, then \(a_p = 0\).
  3. If \(p\) is a square modulo 23, and \(x^3 – x + 1\) has three roots modulo \(p,\) then \(a_p= 2.\)
  4. If \(p\) is a square modulo 23, and \(x^3 – x + 1\) is irreducible modulo \(p,\) then \(a_p = -1.\)

At when point in history could these results be proved?

Let’s first start with Euler, who proved that

\(\displaystyle{ \prod_{n=1}^{\infty} (1 – q^n) = \sum_{-\infty}^{\infty} q^{(3n^2+n)/2} (-1)^{n}} \)
Using this, one immediately sees that

\(\displaystyle{f = \sum \sum q^{\frac{1}{24} \left((6n+1)^2 + 23 (6m+1)^2 \right)} (-1)^{n+m}}\)

This exhibits \(f\) as a sum of theta series. With a little care, one can moreover show that

\(\displaystyle{2f = \sum \sum q^{x^2 + x y + 6 y^2} – \sum \sum q^{2 x^2 + x y + 3 y^2}}.\)

This is not entirely tautological, but nothing that Gauss couldn’t prove using facts about the class group of binary quadratic forms of discriminant \(-23.\) The fact that \(f\) is a modular form of the appropriate weight and level surely follows from known results about Dedekind’s \(\eta\) function, which covers (1). From the description in terms of theta functions, the claim (2) is also transparent. So what remains? Using elementary number theory, we are reduced to showing that a prime \(p\) with \((p/23) = +1\) is principal in the ring of integers of \(\mathbf{Q}(\sqrt{-23})\) if and only if \(p\) splits completely in the Galois closure \(H\) of \(x^3 – x + 1.\)

Suppose that \(K = \mathbf{Q}(\sqrt{-23}) \subset H.\) What is clear enough is that primes \(p\) with \((p/23) = + 1\) split in \(K,\) and those which split principally can be represented by the form \(x^2 + xy + 6y^2\) in essentially a unique way up to the obvious automorphisms. Moreover, the class group of \(\mathrm{SL}_2(\mathbf{Z})\) equivalent forms has order \(3,\) and the other \(\mathrm{GL}_2(\mathbf{Z})\) equivalence class is given by \(2x^2 + xy + 3y^2.\) In particular, the primes which split non-principally in \(K\) are represented by the binary quadratic form \(2 x^2 + xy + 3y^2\) essentially uniquely. From Minkowski’s bound, one can see that \(H\) has trivial class group. In particular, if \(x^3 – x + 1\) has three roots modulo \(p,\) then the norm of the corresponding ideal to \(K\) is also principal and has norm \(p = x^2 + xy + 6y^2.\) This is enough to prove (3).

So the only fact which would not obviously be easy to prove in the 19th century is (4), namely, that if \(p = x^2 + xy + 6y^2,\) then \(p\) splits completely in \(H\). The most general statement along these lines was proved by Furtwängler (a student of Hilbert) in 1911 — note that this is a different (and easier?) statement than the triviality of the transfer map, which was not proved until 1930 (also by Furtwängler), after other foundational results in class field theory had been dispensed with by Tagaki (another student of Hilbert!). Yet we are not dealing with a general field, but the much more specific case of an imaginary quadratic field, which had been previously studied by Kronecker and Weber in connection with the Jugendtraum. I don’t know how much Kronecker could actually prove about (for example) the splitting of primes in the extension of an imaginary quadratic field given by the singular value \(j(\tau).\) Some of my readers surely have a better understanding of history than I do. Does this result follow from theorems known before 1911? Who proved it first?

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Return to Northwestern

Posted on April 21, 2017 by Persiflage

Today I returned to Northwestern, taking part in both the communal faculty lunch and the post prandial espresso. Jared Wunsch worked his magic on the Silvia (Rancilio) to pour one of the best espressos I have had in a very long time (Metropolis beans). It has inspired me to put aside my recent flirtations with pourovers to return to the pure espresso form. The question is: what are the reliable cafes in Chicago that don’t overextract their espresso in order to serve milky drinks?

Oh, and other news: Joel Specter succesfully had his thesis defense. Congratulations Joel! You have now gained couch privileges.

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Me versus Magnus

Posted on April 2, 2017 by Persiflage

I’ve recently been distracting myself with the new Magnus Carlsen “app,” a free chess app for the iPhone whose distinguishing feature is that it tries to play “like” Magnus did at various ages. It does seem like a very tricky problem to make a computer “play like a human,” and this app does not entirely succeed in this endeavor. I’ve spent most of my time playing Magnus “aged 10.” My record against Magnus aged 10 is decidedly in the negative (although I definitely have a few wins), in part because I play fairly quickly, and in part because Magnus aged 10 is pretty tricky. One can get a good sense of exactly how tricky from the following videos:

(Does this mean my Magnus Carlsen number is two? Hmmm, probably not.) At this age, Magnus does not resign. He is also is not very inclined to accept draw offers, as the following screenshot indicates:

I haven’t yet attempted to play Magnus aged 11, but apparently Vlad won against Magnus 11 on either the very first or second try, which suggests that he may have been going easy on me when we played blitz at Northwestern.

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Pseudo-representations and the Eisenstein Ideal

Posted on March 29, 2017 by Persiflage

Preston Wake is in town, and on Tuesday he gave a talk on his recent joint work with Carl Wang Erickson. Many years ago, Matt and I studied Mazur’s Eisenstein Ideal paper from the perspective of Galois deformation rings. Using some subterfuge (involving a choice of auxiliary ramification line at the prime \(N\) following an idea of Mark Dickinson), we proved an \(R = \mathbf{T}\) theorem. One satisfactory aspect (to us, at least) of our paper was that we were able to reconstruct from a purely Galois theoretic perspective some of the thorny geometric issues in Barry’s paper, particularly at the prime 2. Another problem of Barry’s that we studied was the question of determining for which N and p the cuspidal Hecke algebra was smooth (equivalently, whether the cuspidal Hecke algebra completed at a maximal Eisenstein ideal \(\mathfrak{m}\) of residue characteristic p was equal to Z_p). Our theorem showed this was equivalent to the existence of certain Galois deformations to GL_2(F[e]/e^3). Although we were able to give a precise account of what happens for p = 2, for larger p we could only prove the following:

Theorem Let p > 3 be prime, and let N = 1 mod p be prime. If the rank of the cuspidal Hecke algebra of level \(\Gamma_0(N)\) localized at the Eisenstein prime is greater than one, then

\(K = \mathbf{Q}(N^{1/p})\)

has non-cyclic p-class group.

Note that there is always trivial p-torsion class in the class group of K coming from the degree p extension inside the Nth roots of unity. In our paper, we speculated that this was actually an equivalence. To quote the relevant passage:

We expect (based on the numerical evidence) that the condition that the class group of K has p-rank [at least] two is equivalent to the existence of an appropriate group scheme, and thus to [the rank being greater than one].

Not a conjecture, fortunately, as it turns out to be false, already for p = 7 and N = 337. Oops! In fact, this had already been observed by Emmanuel Lecouturier here. Wake and Wang Erickson, however, give a complete characterization of when the rank is greater than one, namely

Theorem [Wake, Wang Erickson] Let \(a \in H^1(\mathbf{Z}[1/Np],\mathbf{F}_p(1))\) be the Kummer class corresponding to N. Let \(b \in H^1(\mathbf{Z}[1/Np],\mathbf{F}_p(-1))\) be the (unique up to scalar) non-trivial class which is unramified at p. Then the rank of the Hecke algebra is greater than one if and only if the cup product \(a \cup b\) vanishes.

They prove many other results in their paper as well. The main theoretical improvement of their method over the old paper was to work with pseudo-representations rather than representations. On the one hand, this requires some more technical machinery, in particular to properly define exactly what it means for a pseudo-representation to be finite flat. On the other hand, it avoids certain tricks that Matt and I had to make to account properly for the ramification at N as well as to make the deformation problem representable. Our methods would never work as soon as N is (edit: not) prime, whereas this is not true for the new results of W-WE. In particular, there is real hope that there method can be applied to much more general N.

Let me also note that Merel in the ’90s found a completely different geometric characterization of when the cuspidal Hecke algebra had rank bigger than one; explicitly, for p > 3 and N = 1 mod p, it is bigger than one when the slightly terrifying expression:

\(\displaystyle{\prod_{i=1}^{(N-1)/2} i^i}\)

is a pth power modulo N. So now there are a circle of theorems relating three things: vanishing of cup products, ranks of Eisenstein Hecke algebras, and Merel’s invariant above. It turns out that one can directly relate Merel’s invariant to the cup product using Stickelberger’s Theorem. On the other hand, Wake and Wang Erickson also have a nice interpretation of the expression above as it relates to Mazur-Tate derivatives (possibly this observation is due to Akshay), and they also prove some nice results in this direction. And I haven’t even mentioned their other results relating to higher ranks and higher Massey products, and many other things. Lecouturier’s paper is also a good read, and considers the problem from another perspective.

In Preston’s talk, he sketched the relatively easy implication that the vanishing of the cup product \(a \cup b\) above implies that the class group of Q(N^{1/p}) has non-cyclic p-part. The main point is that the vanishing of cup products is exactly what is required for a certain extension problem, and in particular the existence of a Galois representation of the form:

\(\left( \begin{matrix} 1 & a & c \\ 0 & \chi^{-1} & b \\0 & 0 & 1 \end{matrix} \right),\)

where \(\chi\) is the mod-p cyclotomic character. The class c gives the requisite extension (after some adjustment). Curiously enough, both the classes a and b exist for primes N = -1 mod p. On the other hand, the corresponding H^2 group vanishes in this case, and so the pairing is always zero. Hence one deduces the following very curious corollary:

Theorem: Let p > 3, and let N = – 1 mod p be prime. Then the class number of \(\mathbf{Q}(N^{1/p})\) is divisible by p.

Question: Is there a direct proof of this theorem? In particular, is there an easy way to contruct the relevant unramified extension of degree p for all such primes N? I offer a beer to the first satisfactory answer.

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Erdős Number 3!

Posted on March 26, 2017 by Persiflage

My chances at this point of writing a paper with Erdős are probably very small. My chances of writing a paper with one of Erdős’ collaborators is also quite small. I had assumed that I had not even met — let alone talked math — with anyone with Erdős number one, but consulting with this link shows this is definitely false; I had a number of delightful chats with Andrew Granville at Oberwolfach in 2015. Still, given my interests, I imagine I am destined to always have an Erdős Number at least three. And now I have achieved that lower bound, having just written a paper with Don Zagier (and Stavros Garoufalidis).

I first met Zagier in 1993, during my last year of high school. He was the Mahler lecturer, a position which carries the responsibility of giving as many lectures as one can on mathematics (and number theory in particular) all around Australia. My brother encouraged me to play truant from school and sneak into a colloquium talk by Zagier, who talked (with characteristic enthusiasm) about Ramanujan’s Delta function and the Birch–Swinnerton-Dyer conjecture. My brother also introduced me (this very same day) to Matthew Emerton, who talked to me about math for three hours; in particular he talked about elliptic curves and Mazur’s theorem on the possible torsion subgroups over \(\mathbf{Q}.\) So it was a very auspicious day for me indeed! At the time, I was enthralled by Edwards’ book on the Riemann Zeta function and was expecting to become an analytic number theorist. But on that day, I completely abandoned those plans and decided to do algebraic number theory instead.

Zagier gave another lecture the next day (which I also skipped school to see). This time it was on volumes of hyperbolic manifolds and their relationship with the dilogarithm and the Bloch group. It is remarkably pleasing then to now — almost 24 years later — write a paper with Don and Stavros which is related to the theme of both those talks, namely the Bloch group and modularity.

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Film Criticism

Posted on March 22, 2017 by Persiflage

You know that feeling you get when you want to understand the precise conjectural relationship between the cohomology of arithmetic varieties and Galois representations?

I finally get it. Du musst Caligari werden! Oh, and if you think I’m crazy, it’s not me; it’s you.

Edit: Found a frame from the English version:

English Version

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A non-liftable weight one form modulo p^2

Posted on March 10, 2017 by Persiflage

I once idly asked RLT (around 2004ish) whether one could use Buzzard-Taylor arguments to prove that any representation:

\(\rho: \mathrm{Gal}(\overline{\mathbf{Q}}/\mathbf{Q}) \rightarrow \mathrm{GL}_2(\mathbf{Z}/p^2 \mathbf{Z})\)

which was unramified at p and residually irreducible (and modular) was itself modular (in the Katz sense). Galois representations of this flavour are obviously something I’ve thought about (and worked on with David G) quite a lot since then. But I have never actually seen any examples of mod p^2 forms which didn’t lift to characteristic zero. I asked George Schaeffer about it once, but his computations were only set up to detect the primes for which non-liftable forms existed rather than to compute the precise structure of the torsion in H^1(X,omega). But just today I stumbled across an example in relation to a pairing I learned about from Akshay (which I will tell you all about some other time).

The particular form (or rather pair, since it comes with a twist by the nebentypus character) occurs at level \(\Gamma_0(103) \cap \Gamma_1(3),\) and is defined over the ring \(\mathbf{Z}/11^2 \mathbf{Z}.\) It doesn’t lift to a weight one form mod 11^3. The nebentypus character is the only one it could be at this level and weight: the odd quadratic character of conductor 3. When I looked again at Schaeffer’s thesis, he does indeed single out this particular level as a context where computations suggested their might exist a mod p^2 form. (Literally, he says that a computation “seems to imply the existence” of such a form.) I guess this remark was not in any previous versions of the document I had, so I hadn’t seen it. Here are the first few terms of the q-expansion(s):

\(g = q + 16 q^2 + 20 q^3 + 15 q^4 + 58 q^5 + 78 q^6 + 22 q^7 + \ldots + 91 q^{11} + \ldots + 104 q^{103} + \ldots \)

\(f = q + 105 q^2 + 115 q^3 + 15 q^4 + 63 q^5 + 96 q^6 + 22 q^7 + \ldots \)

Some remarks. Note that the coefficients of g and f satisfy \(a(g,n) = \chi(n) a(f,n)\) for all \((n,3) = 1\) and where \(\chi\) is the quadratic character of conductor 3 (the nebentypus character). On the other hand, at the prime 3, we have

$latex \rho_f = \left( \begin{matrix} \chi \psi^{-1} & 0 \\ 0 & \psi \end{matrix} \right), \qquad
\rho_g = \chi \otimes \rho_f = \left( \begin{matrix} \chi \psi & 0 \\ 0 & \psi^{-1} \end{matrix} \right),$

and so the eigenvalue of U_3 is the image of Frobenius at 3 under \(\psi\) or \(\psi^{-1},\) and hence satisfies the equality

\(a(g,3)a(f,3) = \psi(\mathrm{Frob}_3) \psi^{-1}(\mathrm{Frob}_3) = 20 \cdot 115 = 1 \mod 121.\)

I was temporarily confused about the fact that \(a_q = 1 + q\) for the Steinberg prime \(q = 103\) rather than \(a_q = \pm 1,\) and thought for a while I had made an error or mathematics was wrong. But then I realized this was weight one not weight two, and so one should have instead that \((a_q)^2 = q^{-1}\) (note that \(\chi(103) = 1.\)) And it just so happens that the equation

\((1+q)^2 = q^{-1}\)

in a weird coincidence has a solution very close to 103 (this is a solution mod 11^3, in fact). It’s easy enough to see that the image of rho and its twist contains \(\mathrm{SL}_2(\mathbf{Z}/11^2 \mathbf{Z})\) with index two, and so has degree 3513840. (At this level, the only real alternative is that the form is Eisenstein, which it isn’t.) The root discriminant is not particularly small, it is

\(103^{1 – 1/11^2} \cdot 3^{1/2} = 171.6970 \ldots\)

Finally, the Frobenius eigenvalues at the prime p = 11 are distinct, which is easy enough to see because otherwise the coefficient of q^11 would have to be twice the squareroot of chi(11) = -1, which isn’t even a square mod 11.

Perhaps there’s not too much more to say about this particular example, but I was happy to come across it, nonetheless. Well, perhaps I should also say that I computed this example in SAGE, as I slowly wean myself off magma dependency.

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Report from MSRI and Berkeley

Posted on March 7, 2017 by Persiflage

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:

Babette

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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