Kouign-Amann, Chicago Tasting

A free morning on the north side this week meant a chance for a bike ride and a new cafe; nothing out of the ordinary. But this time I prepared an itinerary to hit up some of the most highly rated Kouign-Amann in Chicago. First stop, the Good Ambler, then on to Aya Pastry, and then to the Publican bakery; one order of Kouign-Amann at each stop! Then onto Metric coffee for a (good) cortado and a Kouign-Amann taste off:

Video Introduction

First of all; these were all good pastries! But none were in the neighbourhood of transcendent. Here’s a little more detail.

Good Ambler: Very crunchy and caramelized on the outside. I noted the inside was “bready”. I guess this is supposed to contrast not only with a cake but also with “buttery”. This was perhaps a complaint one could have made about each of them. I imagine that the relative ratio of both butter and sugar used in these versions is less than in the original. I also wonder about the relative quality of the butter. What’s the best butter one can get in Chicago?

Publican: From my notes: “If I had to, I could definitely eat all of this, but I won’t”. I think that same comment could honestly have been applied to all of them. Now that doesn’t sound very enthusiastic, but for context the amount of food I typically consume before lunch is half a piece of toast (and this was probably before 10:00AM). From the picture below, the interior was certainly in the ballpark of a croissant.

Aya: This was much denser than the other two, although again the closest point of comparison for the inside was a still sweet croissant, and while not a million miles away from the reality is not what I am going for. Both the Aya and the Publican had a deliberate layer of caramelization on the bottom, but in both cases this was very thin and didn’t make that much difference to the overall tast.

To get more of a hint of what I am looking for in my dreams, I think this video hopefully conjures up some idea (19:03 in to the video). It almost makes me want to try (again) to make it myself!

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Haebler and Gilberto

Two obituaries in the NYT within one week for musicians in my music collection.

I can’t quite say that Mozart is my composer of choice, especially when it comes to the piano, although I could listen to Mitsuko Uchida play all day. But every now and again, Mozart knocks one out of the park. The G minor piano quartet, for example. While that quartet has a great arrangement for two pianos, here’s a Mozart fugue actually originally written for two pianos; here is my recording, performed by Ingrid Haebler and Ludwig Hoffmann

I once had some Brazilian students surprised during office hours that I was listening to Getz/Gilberto (I was surprised by their surprise in turn). But at any rate this is what I now assume all Dutch TV is like (featuring Astrud Gilberto):

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Google and Franck

I have a google play (which plays streaming music) and it’s really terrible for classical music. If you choose virtually any piece of classical music and then flip forward three songs you invariably end up with Claire de Lune or the Moonlight sonata. For science, I just tested this again right now, and here are the unfiltered results:

  1. Initial Selection: Art of Fugue
  2. Next Piece: Toccata and Fugue in D minor (ha ha; semantically similar I guess)
  3. Next Piece: First movement, Moonlight sonata
  1. Initial Selection: Brahms, Op 119 no.1
  2. The initial selection continued with Op 119 no.2 on the same track; I don’t think that counts
  3. Prelude and Fugue in B flat minor #22 from Well-tempered Clavier Book 2. Pretty good! Mind you, when I asked google what it was playing to confirm, it said “Country road, by John Denver.” In fact, this appears to happen when I ask about any piece of music at all.
  4. Moonlight sonata (but third movement!)
  5. What sounds like cheap Debussy but google play still claims is John Denver.
  6. Another Brahms intermezzo! (Op 118)
  7. Flight of the Valkyries (blech)
  1. Initial Selection: Schubert Op 929, second movement.
  2. The Swan (arranged for violin and Piano)
  3. Schubert Op 929, second movement. Hmmm.
  4. The Nutcracker suite (perhaps complete? from the beginning recorded live.)
  5. Air on a G-string.

I say: hypothesis confirmed.

Some other strange results: I once asked for the Franck violin sonata and somehow got the piano accompaniment without the violin part. Actually that may have been the most interesting selection it ever played.

Speaking of Franck, I only recently learnt that he composed his violin sonata when he was 64. It’s not so usual for great composers to write great music late in their lives (Art of Fugue, unfinished when Bach was 65), but a little unusual, I think, for one’s most famous piece of music (perhaps by some distance) to be composed relatively late in life. I hope I prove my most famous theorem when I’m 64!

My go to recording is by Perlman and Ashkenazy; here they are (very young; from 1968, apparently) in the recording studio (apparently I am unable to embed this video, but I promise this is a real and interesting link!)

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Quadratic Reciprocity

I accidentally proved quadratic reciprocity in class today, or at least three quarters of a proof. Can you finish it off? Here’s the proof: start with a real quadratic field K, and the sequence

1 \rightarrow \mathcal{O}^{\times}_K \rightarrow K^{\times} \rightarrow K^{\times}/\mathcal{O}^{\times} \rightarrow 1

then take cohomology. If P_{K} is the group of principal ideals of K, then from Hilbert Theorem 90 you deduce that

P^{G}_K/P_{\mathbf{Q}} \simeq H^1(G,\mathcal{O}^{\times}_K).

If I_K is the group of all ideals of K, the left hand side is a subgroup of I^G_K/P_{\mathbf{Q}} which is a product of groups of order two for each ramified prime. Now if K = \mathbf{Q}(\sqrt{pq}) with p \equiv q \equiv 3 \bmod 4, there is no unit of norm -1 because (-1/p) = (-1/q) = -1, so you deduce that the cohomology of the unit group has order 4 and so from order considerations that the prime ideals of norm p and q are principal. Write \mathfrak{p} = (\alpha) and \mathfrak{q} = (\beta). One has x and y with

x^2 – p q y^2 = N(\alpha)

Here N(\alpha) = \pm p. But the right hand side must be a square modulo q, and so N(\alpha) = p if (p/q)=+1 and -p if (p/q)=-1. Equivalently, N(\alpha) = (p/q)p, and similarly N(\beta) = (q/p)q. But \mathfrak{p} \mathfrak{q} = (\sqrt{pq}), so \alpha \beta = \varepsilon \sqrt{pq} for some unit \varepsilon, and since all units in K have norm one, it follows that

pq(p/q)(q/p) = N(\alpha \beta) = N(\varepsilon) N(\sqrt{p q}) = – p q,

which is quadratic reciprocity! There is a similar argument for p \equiv 1 \bmod 4 and q \equiv -1 \bmod 4 (though now using facts about (2/p) rather than (-1/p)). However, it is not so clear if one can prove the case p \equiv q \equiv 1 \bmod 4 using this argument. Is there one? Of course the challenge here is to “only” use Hilbert 90 and unique factorization of ideals, but never to make any arguments about how primes are splitting.

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En Passant VIII

Coffee in Hyde Park: I had the misfortune of being stuck in down town Hyde Park needing a coffee without access to a car. I started at Sip and Savour where I ordered a cortado. They did not know what a cortado was. I paused, considered the situation, and slowly walked out. Attempt number two was at a cafe which was nominally a Peets but was actually a “Capital One” cafe. They did have a fine looking Marzocco machine which was a good sign, but they also had a very bored looking barista and my attempt to order a cortado lead to strike two: they also had no idea what I was taking about. I then peeked into Philz Coffee which I know is not really an espresso place; indeed they did not seem to have an espresso machine at all. I considered just getting a regular coffee, but I didn’t like the vibe and ended up at “Cafe 53.” Here I somehow ended up with two small (to go) espresso cups, one consisting of a very watered down and undrinkable espesso, and the other filled with frothed milk. I politely put them on the table, worked on my laptop for 30 minutes, then discretely through them in the trash. Thank god for Plein Air near campus!

The quest for a good Kouign-Amann: In 2009 I went to a conference in Roscoff organized by Tilouine and Hida (mentioned previously here). I have a number of fond memories from that conference: The mathematical discussions; the tides; the speaker who wore loose hanging pants and then kept bending over very low to the ground to write on the extremities of the fairly low white boards to reveal a generous plumber’s crack to those of us fortunate to be sitting in the front row; but also the food. The culinary highlight for me (gizzards aside) was surely the Kouign-Amann, a super indulgent buttery cake which is a speciality of Brittany and of which I got to try several versions (all excellent). Little did I know how hard it would be to reproduce the same dish anywhere else, where very few people seemed to have heard of it. The less said about my attempts to make it myself the better. Now enter Dominique Ansel (spit). This third-rate fraud is probably best known for inventing the cronut, sold at his trendy eponymous bakery in NYC which took a certain section of the US culinary world by storm around 2013. But for me he will be eternally known as the chef who introduce the “Kouign-Amann” to the US. I use quotes here deliberately, since Ansel’s version is something resembling a slightly caramelized croissant and nothing than anything I had had in Brittany. I mean the guy has probably not been west of Rouen let alone actually gone to Brittany. It’s as if a Mexican chef in Baja boiled a tortilla with a hole cut out of it and called it a bagel. Of course, bakeries the entire world over started selling Kouign-Amanns based not on the original but on copies of Ansel’s version, which was a sad simulacrum of the original. I have now been on an almost 15-year quest around the world to find a cake sold by that name which in some way was as amazing as the first ones I tried but to no avail. From East coast to West coast, from Blé Sacré in Paris to, most recently, an hour waits in line at Lune croissanterie on Collins street. Mostly the results have been bad, but even when they have been quite decent they have not been Kouign-Amanns. The quest continues…

Chess: For those watching Ding-Nepo, what a match it has been so far! One thing I have appreciated about watching live on youtube is that the commentators (including Caruana and Giri) analyze the positions in real time instead of using stockfish. That helps you distinguish from an advantage which is obvious to a super GM from an advantage which is obvious only to a super computer! (By the time it is obvious to me the players have already resigned). I’m looking forward to tomorrow.

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Deciphering Quanta

Sometimes it is claimed that Quanta articles are so watered down of mathematical content that they become meaningless. That presents a challenge: do I understand the quanta article on my own work?

Here goes:

New Proof Distinguishes Mysterious and Powerful ‘Modular Forms’

I can confirm that I did not see this article in any form before it appeared. Overall I would say that it is faithful to the facts and I can interpret what everything means. I’m not quite sure why there is an Alex Kontorovich explainer about the Langlands Program in there but why not?

I did, however, have to stop and wonder when I saw the following picture:

This appeared with the caption *Congruence modular forms (left) have additional structure that noncongruence modular forms (right) lack.* OK, here is the challenge: can I work out what this picture actually represents?

In both pictures, there is clearly some color gradient between yellow and blue, and this has discontinuities along some regions we call L and R for left and right. These are also clearly pictures inside \mathbf{H}^2 in the Poincaré disc model.

Here L looks at least superficially like a fundamental domain for some Fuchsian group. The rays of L going towards the point i on the boundary do look like geodesics in the hyperbolic metric (which are circular arcs meeting the boundary at right angles). There is a corresponding invariance for the figure by the parabolic element (with some cusp width) through i. Consider a conformal map from the upper half plane to this model which sends \infty to i:

Using the standard conformal map with the upper half plane:

\phi: \mathbf{H} \rightarrow D(0,1), \quad z \mapsto \displaystyle{\frac{1 + i z}{1 + z}}

From the picture, we see the images of the geodesics from the infinite cusp to n \alpha for n odd. All parabolic elements are conjugate, but if we are to guess what \Gamma is by looking at the picture then working out \alpha for this specific model is important. I tried to eyeball it for a while before printing it out and trying to compute the angle using continued fractions, which wasn’t so accurate but gave \tan \theta \sim 1/(1 + 1/3) with \theta in the mid 30s. Then using some angle tools in keynote it came out to somewhere between 36 and 37 degrees, closer to the latter. So maybe it was exactly a 10th root of unity, which would make \alpha live in a degree 4 field, or (much better!) maybe \alpha = 1/2 in which case the angle is 36.8698 \ldots, and then L (or rather \phi(L) is invariant under exactly

A = \left( \begin{matrix} 1 & 1 \\ 0 & 1 \end{matrix} \right).

Maybe I should have guessed this in retrospect! Let’s look at the geodesic from i \infty to 1/2, which in the picture goes from i to 4/5 – 3i/5. There appears to be a point with 4-fold symmetry. That suggests invariance under some order four element corresponding to an elliptic point. But this is a little worrying, since \mathrm{PSL}_2(\mathbf{Z}) does not have any elements of order 4! Now there are some ways around this. For example, this could be the level set of a function which satisfies an extra invariance property under the normalizer of some congruence subgroup, and so does not itself come from an element of \mathrm{SL}_2(\mathbf{Z}) but \mathrm{GL}_2(\mathbf{Q})^{+}. For example, the Fricke involution on X_0(N) is \tau \mapsto -1/N \tau which corresponds to. a matrix in \mathrm{GL}_2({\mathbf{Q}}) with determinant N, although that has order 2. Any element of order 4 has to have eigenvalues of the form \alpha and \alpha i, with \alpha + i \alpha \in {\mathbf{Z}}, and thus eigenvalues 1+i and 1-i up to rational multiples. That means the determinant should be 2 times a square. The unique element such element up to conjugacy is

S = \left( \begin{matrix} 1 & 1 \\ 1 & -1 \end{matrix} \right)

which has order 4 and fixes i. If we want to fix a point on the geodesic 1/2 + i t, we can just make a hyperbolic scaling and then conjugate to get the matrix

S_t = \left( \displaystyle{ \begin{matrix} 1 – 1/2t & t + 1/4t \\ -1/t & 1 + 1/2t \end{matrix} } \right).

Here I started to run into a problem because it’s hard to approximate t to any precision from the picture, although t \sim 0.1 numerically. There are a few natural integral matrices one can write down but it wasn’t clear what was going on.

Next I turned to the picture around the cusp -i. This looks kind of weird because the biggest curves decidedly do not look like geodesics. But if we ignore this, we might decide that -i is another cusp. It’s also disturbing that the fundamental domain appears to contain enough of the boundary to suggest that \Gamma is a thin group, but let’s ignore this as well. So what is the cusp width (with out normalizations) at -i? Here numerically I simply get nonsense.because if there really were geodesics around i translated by a fixed parabolic element and the first one was of the size indicated in the picture then there would be many fewer visible ones. As a typical example, one should expect a picture like this:

which doesn’t look anything like the original picture.

So I’m ready to give up now. What about the picture on the right? Well here I don’t even know how to begin. There is not any obvious evidence of parabolic elements, which are not only present in congruence subgroups of \mathrm{SL}_2(\mathbf{Z}) but of any non-congruence subgroup as well. Perhaps there is a cusp at i with some large cusp width. There also seem to be singularities inside the disc. But that just suggests this might be a level set of a meromorphic form. But why choose a meromorphic form rather than one that is holomorphic away from the cusps? I wonder if that gives hints about the first picture; perhaps the transition from yellow to blue occurs when \mathrm{Im}(f) goes from positive to negative, and where it might be 0 along some geodesic arc (For example: the Fourier coefficients are rational so the form is real along the purely imaginary axis) but the form can also have imaginary part 0 along some non-geodesic regions and that is what one is seeing around -i. That might allow one to reinterpret the graphs as something related to \mathrm{Im}(f). This suggests that the point on the left that appears to have degree four instead is related to a “local” symmetry coming from a vanishing point of the derivative of the modular function, but I’m just guessing now.

I should probably spend no more time on this, so instead I open up speculations to the comments!

Full Disclosure The picture comes with an attribution to David Lowry-Duda but I did not try to follow that lead.

Update: Perhaps it was not as obvious as intended, but the request for speculation was intended to be “how do you interpret this from the picture alone” not “send me a link” which is something which I deliberately avoided trying to do. (I will no doubt look at the way they are generated at some point in more detail.) A brief look at the LMFDB link suggests that my second explanation is pretty close, although the graph on the left is of the argument of a modular function (or rather the modular form \Delta) rather than the imaginary part. But the guess as to the explanation for the “order four” symmetry is correct, it is explained by the vanishing of \Delta'(\tau) or equivalently the vanishing of E_2(\tau) at a point on the geodesic 1/2 + i t. The picture on the right is still a mystery, however, and original speculations are still welcome.

Update II: So here is an new example. Like the graphs above, it plots the argument of a two modular forms on (different) finite index subgroups of \mathbf{SL}_2(\mathbf{Z}). I have normalized both pictures so that the cusp width at the cusp i is the same in both cases (under the normalizations above invariant under \tau \rightarrow \tau + 3; making the cusp width too small makes the behavior at tau = i \infty dominate the picture as the covolume goes up. There are dome differences with this example however:

  1. I have used (holomorphic) modular functions rather than modular forms.
  2. For these particular choices of functions, they are non-zero everywhere.
  3. I made the picture myself so it is not as pretty as the ones above.
  4. The functions I chose have the property that they are real precisely on geodesics, and thus the singularities here do form a tesselation.


Now one of the forms is defined on a congruence subgroup and has a Fourier series in \mathbf{Z}[[q]], the other has a Fourier series in \mathbf{Q}[[q]] but not in \mathbf{Q} \otimes \mathbf{Z}[[q]] and is only defined on a non-congruence subgroup. But which is which? The pictures here are clearly much more similar than the pair of pictures above!

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Boxes for Boxer update

As noted in this post, exactly 42 reprints of [BCGP] were recovered in January of 2022 from boxes left out in the snow outside Eckhart Hall addressed to George Boxer. As mentioned there, the packaging (5 boxes of 8 plus a further smaller box with 2) suggested that there was a “missing box” with 8 more copies of the paper, and that one should be on the lookout at Hyde Park used book stores for the remaining copies. These reprints are after all roughly of a similar scarcity to the extant Gutenberg Bibles. Well there has been an update! My student Abhijit noted that 7 copies had been left out in the hall as an offering for graduate students (and that he had secured a copy). I initially assumed that several of my copies had escaped my hands but they all seem to be accounted for (there were 39 in my office, at least one at home, and at least one I gave to TG). So the missing box did exist after all! I still haven’t determined what happened to the box. There could have been a reprint thief who, out of guilt or madness (depending on how much of it they tried to read), was driven them to give them to the world as an attempt of repentance. Then again, the box may have ended up in someone’s hands much more unwittingly and they had only just discovered the contents. Or finally it could have been a janitor who dumped them out from a dusty closet where they had inadvertently been placed. The plot, as they say, thickens! Anonymous tips from uchicago graduate students welcome in the comments.

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What the slopes are

Let f be a classical modular eigenform of weight k, for example, f = \Delta. The Ramanujan conjecture states that the Hecke eigenvalues a_p satisfy the bound

|a_p| \le 2 p^{(k-1)/2}.

A slightly fancier but cleaner way of saying this is as follows. Associated to f of weight k, level N prime to p and finite order Nebentypus character \chi is a polynomial

X^2 – a_p X + p^{k-1} \chi(p).

(For \Delta one has \chi(p) = 1 for all p, but in general it can be some other root of unity.) This is the characteristic polynomial of Frobenius on the \ell-adic representation associated to f for \ell \ne p, or the characteristic polynomial of crystalline Frobenius for \ell = p. The Ramanujan conjecture is now that the Frobenius eigenvalues \alpha_p and \beta_p satisfy

|\alpha_p| = |\beta_p| = p^{(k-1)/2}.

This conjecture was famously proved by Deligne. Having determined the complex valuation of these eigenvalues, one might ask about their \ell-adic valuations as well. If \ell \ne p, then since the polynomial above has constant term prime to \ell, then \alpha_p and \beta_p are \ell-adic units. So the remaining case is p = \ell.

When p = \ell, the p-adic valuation of the two roots \alpha_p and \beta_p certainly satisfy v(\alpha_p), v(\beta_p) \ge 0 and v_p(\alpha) + v_p(\beta) = k-1. Given f, we call these valuations the *slopes* of f. The first observation is that the slopes depend on more than just the weight k. For example, suppose that f is the weight 2 eigenform associated to a (modular) elliptic curve E with good reduction at p, then either E has ordinary reduction, in which case the slopes are 0 and 1, or E has supersingular reduction, and the slopes are both 1/2.

Instead of fixing f and varying p, one can fix both p and the tame level N and ask how the slopes vary as the weight changes. For example if N=1 and p=2, then the first interesting case is when k=12 and f = \Delta. In this case \tau(2) = 24, and the two slopes are 3 and 11-3 = 8.

I’ve already tried to suggest by analogy to the Ramanujan conjecture why determining the p-adic valuations (the slopes) might naturally be an interesting question. But let me mention two other natural reasons. The first is that, from p-adic Hodge Theory, the crystalline eigenvalues \alpha_p and \beta_p determine the restriction of the p-adic Galois representation associated to f to the local Galois group at p. The valuation of these slopes while containing less information than the eigenvalues themselves still tell you a lot about the p-adic representation, although determining exactly what is still a question of active and open interest. Secondly, as Coleman observed (following Hida in the case when one of the slopes is 0), one can deform the Galois representations associated to these finite slope forms into continuous p-adic families of Galois representations associated to cuspidal eigenforms, which leads to the story of the eigencurve of Coleman-Mazur and beyond.

Gouvêa was one of the first people to undertake a numerical study of the roots.
In this paper where the slopes are, Gouvêa observed a number of interesting behavior of the slopes which were all somewhat mysterious. First, the slopes were almost all integers. This is not surprising when a_p(f) \in \mathbf{Z}, but in general a_p(f) will be a random algebraic integer. The slopes also seemed to be distributed in a number of surprisings way. For example, although the slopes in weight k associated to f add up to k-1, they tended to be concentrated in the intervals [0,(k-1)/(p+1)] and [p(k-1)/(p+1),k-1].

Kevin Buzzard went one step further and looked more closely at the slopes when N=1 and p=2. Ostensibly, according to Kevin, this was to test William Stein’s latest magma code for bugs! Kevin found that the slopes in this case satisfied a much more regular pattern. As mentioned above, when k=12, there is a single eigenform whose smallest slope is 3. When k = 12 + 64, there are six eigenforms whose smallest slopes are 3, 7, 13, 15, 17, 25, and when k = 12 + 2^{10}, there are 86 forms whose slopes are

3, 7, 13, 15, 17, 25, 29, 31, 33, 37, 47, 49, 51, 57 \ldots

where all of these sequences are given explicitly by the 2-adic valuation of 2 ((3n)!/n!)^2. Note that the Gouvêa-Mazur conjecture says that these sequences should have some initial segment in common, but in fact the Gouvêa-Mazur conjecture only implies that the slopes in weights 16 + 2^6 and 16 + 2^{10} should be the same up to slope 6, and in practice they agree ridiculously further than this. This was all very mysterious. Kevin found a general algorithm which conjecturally computed by an inductive procedure all the slopes in all weights for a fixed tame level N. (In what I always regarded as a missed opportunity, he did not call the paper “What the slopes are”).

In fact, Kevin’s conjectural answer required an assumption on N and p which for p>2 was equivalent to asking that the local residual representations associated to all low-weight forms are locally reducible. For N=1, the first case for which this does not happen is p = 59, and this story is related to our counterexample to the original form of the Gouvêa-Mazur conjecture.

Kevin Buzzard was a speaker at the Arizona Winter School in 2001, and for his project he outlined a special case of his conjecture corresponding to overconvergent modular forms of weight k = 0. Of course there are no classical modular cuspidal forms of weight k = 0, but by Coleman’s theory there is a direct link between the questions about classical forms in all weights and overconvergent forms of finite slope in all weights. Kevin’s problem in its original formulation is described here.

I was a graduate student at the time, and although I wasn’t actually assigned to Kevin’s group, I did see his problem and had an idea which more or less amounted to the idea of using a slightly different explicit basis for this space than Kevin had considered and for which the U_2 operator has a much nicer explicit form. In fact this basis was related to computations I had done in the summer of 93-94 and shortly afterwards during my first year at Melbourne uni, using what Matthew Emerton and I had come to definitely know as the “f” function

f = q \prod_{n=1} (1 + q^{n})^{24}.

(When It came to writing my paper with Kevin, I still felt strongly enough to insist we call it by this letter.)

Kevin and I put quite a bit of effort into proving his conjecture for more general k, still with N=1 and p=2, but only succeeded in a few cases, including k = -12, but also k = -84 which was somewhat randomly chosen as a case where our approach failed but only by a little bit which could be overcome. There are a few other cases where X_0(p) has genus zero and one can do something similar, but otherwise very little progress was made on these conjectures in this situation.

There are a plethora of related conjectures which also came out (at least indirectly) from similar calculations. For example, Kevin’s algorithm certainly always produces integers, so, in light of the p-adic theory, there is the natural question of whether a crystalline representation with Hodge-Tate weights [0,k-1] for k even whose residual representation is reducible always has finite slope, Then there are questions of the exact relationship between the slope and the Galois representation, and so on.

Concerning the exact slopes themselves, perhaps the biggest advance over time were refinements of Kevin’s conjecture. The Ghost Conjecture, formulated by Bergdall and Pollack, is some “master conjecture” of a combinatorial nature generalizing a number of previous conjectures concerning the slopes of classical overconvergent modular forms over the centre of weight space. (Although, perhaps amusingly, it’s not clear that these conjectures do actually imply Buzzard’s original conjecture.)

Now cut to the present day. In a recent preprint, Ruochuan Liu, Nha Xuan Truong, Liang Xiao, Bin Zhao have now proved all of these conjectures, at least up to some genericity hypotheses (excluding the case N=1 and p=2!). The authors certainly employ technologies that didn’t exist 20 years ago (p-adic Langlands, for example), but that was not the only obstruction to previous progress: the paper contains a number of very original and clever ideas. Very amazingly and satisfyingly, it resolves a large number of the open problems discussed above, including Gouvêa’s conjectures about the distribution of slopes, the integrality properties of slopes for locally reducible representations, and even a version of the Gouvêa-Mazur conjecture. It is also satisfying that the arguments use p-adic local Langlands, given that some of the initial computations of slopes served at least in part as inspiration for some aspects of this program. I myself have not really worked on this circle of problems for almost 20 years but I am still very happy to see these questions answered!

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Check the arXiV regularly!

In a previous post, I discussed a new result of Smith which addressed the following question: given a measure \mu on \mathbf{R} supported on some finite union of intervals \Sigma, under what conditions do there exist polynomials of arbitrarily large degree whose roots all lie in \Sigma whose distribution (in the limit) converge to \mu?

A natural generalization is to replace \Sigma by a subset of \mathbf{C} subject to certain natural constrains, including that \mu is invariant under complex conjugation. I decided that this had a chance of being a good thesis problem and scheduled a meeting with one of my graduate students to discuss it. Our meeting was scheduled at 11AM. Then, around 9:30AM, I read my daily arXiv summary email and noticed the preprint (https://arxiv.org/abs/2302.02872) by Orloski and Sardari solving this exact problem! There are a number of other very natural questions along these lines of course, so this was certainly excellent timing. When I chatted with Naser over email about this, he mentioned he had become interested in this problem (in part) by reading my blog post!

There is of course a general danger of giving my students problems related to my blog posts, and indeed I have refrained from posting a number of times on possible thesis problems, but in this case everything turned out quite well.

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Report from Australia, Part I, Coffee

My travel often involves making some effort to find good local coffee. From Palo Alto to Portland, a little effort finds quality cafes with reliable espresso drinks. How does the rest of the world then stack up with Australia, the acknowledged home of coffee?

My first stop was Sydney, where my airbnb was conveniently located a stone’s throw from Skittle Lane Coffee. Many other cafes were on my list (Cabrito Coffee Traders, St Dreux Espresso Bar, and so on). What consistently stood out was not necessarily how far above in quality the coffee was from elsewhere in the world, but the sheer consistency of a cup at almost any completely random cafe in Sydney, and there are a *lot* of cafes. I mean, walking down a city street and finding three different cafes in a row was a common sight.

After a while, it began to dawn on me that I am not a coffee snob; it’s just that the rest of the world sucks when it comes to consistent espresso drinks. There are some places which can make a decent cup on occasion, but when the baristas are under pressure (especially with long lines of customers) their technique falters and they screw up the microfoam. New York City is the absolute worst in this regard — full of hipster baristas with beards to match making subpar coffee when they are rushed. So as my trip continued, I could relax and order coffee almost anywhere (country cafes, even Sydney airport). It helped that almost every single cafe had a La Marzocco coffee machine, a serious 20K piece of equipment.

All this is not to say that some coffee places were better than others. My favourite was probably Poolhouse coffee, (whose product has replaced the old coffee picture on my webpage) but this was just one great option among many.

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