Midlife crisis

Plein Air is certainly the best cafe in Hyde Park. (Arguably Build Coffee is fine as well, but they are only open about 5 hours a week.) But it is also true to say that Plein Air is (at best) pretty inconsistent when it comes to espresso drinks; some baristas are definitely better than others, but frequently the result is honestly pretty disappointing. As a daily ritual, I really would hope for a lot more. So what better time to (finally) get serious about making coffee myself.

Thus the latest addition to my office: a Silvia Pro X, a Baratza Sette 270 grinder, shims added, and some other accoutrements, perhaps most gratuitously a Luna Acaia scale. A few days in and I’m already very pleased; not only is the taste better than anything else one can get in Hyde Park, but even the ritual of waiting for the machine to warm up is not at all an unpleasant experience. Also, as far as midlife crises go, neither very expensive nor excessively time consuming!

I’ve always been a little resistant to going down this path — my experiences with Jared Wunsch’s Silvia at Northwestern suggested a certain learning curve was required, and I would not call myself mechanically adept. But a few things have changed. First of all, modern espresso had moved a bit more towards “science” than “art” over the past few years: a meticuluous approach via weighting, ratios, tamping, the science of extraction, etc. have made consistency and reproducibility much more possible. Add to that a souped up version of the Silvia with more controls to make things easier, and an infinite number of online resources (James Hoffmann youtube videos) available online.

My intent it to experiment with milk drinks at some point, but for now I’m quite happy just concentrating on espressos. Arriving soon: extra cups for visitors!

Here is the setup. Those with a keen eye will note the novel application of potential modularity…

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The horizontal Breuil-Mezard conjecture

Postdoc hiring season will be upon us soon! I have two excellent graduate students who will be applying for academic jobs soon, Chengyang Bao and Andreea Iorga. I have mentioned Chengyang’s first project before here and an introduction to the results in Andreea’s thesis is here. Today I wanted to talk about Chengyang’s thesis.

Fix a local mod-p representation, say

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

given on inertia by \(\omega_2 \oplus \omega^p_2\). Associated to this residual representation is a Kisin deformation ring \(R\) corresponding to fixed determinant crystalline lifts of weights \([0,k-1]\), for some fixed positive integer \(k \equiv 2 \bmod (p-1)\). The special fibres \(R/p\) of these rings have dimension one, and so, if one denotes their maximal ideal by \(\mathfrak{m})\), then the Hilbert series

\(H_k(x) = \sum \dim(\mathfrak{m}^k/\mathfrak{m}^{k+1}) X^k\)

has the form

\(H_k(x) = \displaystyle{\frac{P_k(x)}{1 – x}}\)

where \(P_k(x)\) is a polynomial. The Hilbert-Samuel multiplicities of these rings are given by the Breuil-Mezard conjecture (also proved by Kisin). These numbers are explicitly given by the values \(P_k(1)\).

It seems quite surprising that understanding the seemingly simple number \(P_k(1)\) is so intimately linked to the proof of the Fontaine-Mazur conjecture. At the same time, we know very little about these rings \(R\) (or their special fibres) when \(k\) is large.

In the example above, the unrestricted (fixed determinant) local deformation ring \(R^{\mathrm{loc}}\) is is formally smooth of dimension three over \(\mathbf{Z}_p\). Although the rings \(R/p\) only have dimension one, one expects that for larger and larger \(k\) they start to “fill out” the unrestricted deformation ring. It is natural to wonder: how fast does this happen?

More explicitly, the Hilbert-Samuel series of the special fibre of the unrestricted deformation ring with fixed determinant is

\(\displaystyle{\frac{1}{(1-x)^3}} = 1 + 3 x + 6 x^2 + 10 x^3 + \ldots \)

So one can ask: what weight does one have to go to to see all three dimensions of the tangent space? How far does one have to go to see all of \(R^{\mathrm{loc}}/(p,\mathfrak{m}^n)\)?

This was the thesis problem of Chengyang Bao, which grew out of (in part) questions arising during her work here. In this particular case, it seems that one has to go to weight \(k = p^2 + 1\) to see the entire tangent space. Actually, an even more basic question (which also came up here), is whether there exists surjective map

\(R_{k+p-1}/p \rightarrow R_k/p,\)

this seems very tricky and is still open (but Chengyang’s work stongly suggests that it is true).

One of the difficulties in this project is that close to nothing was known about the rings \(R\) for \(k\) anything larger than \(k = 2p\) or so (although there has certainly been quite a bit of work understanding the link between \(a_p\) and the residual representation, including work of Buzzard-Gee, Sandra Rozensztajn, and many others using p-adic Langlands for larger \(k\)).

Chengyang’s approach was, perhaps surprisingly, to use global methods. The basic summary of the Taylor-Wiles method as formulated by Kisin is that via patching one finds that a patched Hecke ring may be identified with a power series ring over \(R\). By reverse engineering this, if one finds a residual representation with sufficiently nice global properties, one can use explicit Taylor-Wiles primes to get arbitarily close approximations to the Kisin deformation ring \(R\). One of the tricks here is to be able to do this in a way where one can work efficiently after fixing a residual representation and then increasing the weight.

By doing these computations, Chengyang generated lots of explicit data about these rings \(R\) from which one can start making conjectures. I said before how the Breuil-Mezard conjecture amounts to predicting the value of \(P_k(x)\) at \(x = 1\). Chengyang has, at least in this particular case, been able to formulate an exact conjectural answer for the *entire* polynomial \(P_k(x)\). As a consequence, one can read off from this the answer of how large a weight one has to go to see all the directions in \(R^{\mathrm{loc}}/\mathfrak{m^n}\). I mentioned before that for \(n=2\) the answer is \(p^2+1\). and my guess was that the answer in general might be of order \(p^n\). But somehow the conjectural answer (up to constants which depend on \(p\)) turns out to be of order \(O(n^2)\), which is honestly completely different from anything that I would have guessed. I think of this conjecture as a new “horizonal Breuil-Mezard conjecture.” But really, it’s only half a conjecture; the hope is that one can understand and interpret Chengyang’s conjecture on the \(\mathrm{GL}_2(\mathbf{Q}_p)\)-side, and working this out is an exciting problem.

At the same time, there are lots of other things one can start to guess from looking at these explicit rings. Chengyang has a precise conjecture which says when the rings \(R\) in this setting are complete intersections or Gorenstein, and it also seems that they are always Cohen-Macaulay.

Even though we “know” \(p\)-adic Langlands for \(\mathrm{GL}_2(\mathbf{Q}_p)\) much better than in any other situation, there seems to be a real opportunity here to tease out many more precise and explicit conjectures from Chengyang’s work, and really to discover new phenomena which have hitherto never been noticed because computations of these rings has been so limited. (Another basic question: how many components does the generic fibre of \(R\) have in terms of \(k\)?).

I recommend that anyone interested in more details read Chenyang’s research statement!

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Magma Instability

I had occasion to return to some magma scripts I wrote in 2012. I the script used a number of pre-computed auxiliary files with computations, and was a little complicated, but didn’t use anything particularly complicated. So I was really surprised to run them in 2023 and find that they no longer worked. That is, they compiled, but the results they gave were different (and also incompatible with the truth). It was quite confusing to understand what has gone wrong, but eventually I traced it to the following. Early on in the file one has (having defined \(t\) as a variable using code that’s easy to write but which is somehow causing issues with wordpress):

F := NumberField(t^2 – 5);
ZF:=Integers(F);

So far, so good. But later on, the script called upon elements of \(F\) of the form \(\texttt{ZF![a,b]}\). But it turns out that if \(\texttt{x:=ZF[0,1]}\) that

\(x^2 + x = 1\)

in 2012, but

\(x^2 – x = 1\)

in 2023; that is, \(x\) was replaced by \(-x\), which is not an automorphism. I have no idea how or why that changed, but it certainly broke everything and took several days to fix. Annoying!

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Clozel 70, Part II

Many years ago, Khare asked me (as I think he asked many others at the time) whether I believed their existed an irreducible motive \(M\) over \(\mathbf{Z}\) (so good reduction everywhere) with Hodge-Tate weights \([0,1,2,\ldots,n-1]\) for any \(n > 1\). (Here the Motive is allowed to have coefficients.) When \(n=2\), the answer is no. Assuming all conjectures, such an \(M\) must be modular associated to a cusp form of weight \(2\) and level one, but no such cuspform exists. But the answer is also no unconditionally (for any notion of motive), and this fact is intertwined with the (inductive) proof by Khare and Wintenberger of Serre’s Conjecture. The hope might be that if no such motive existed for all \(n\), it could serve as the inductive basis for a more general form of Serre’s Conjecture.

My response at the time was that I guessed that no such motive existed for any \(n\). I generally feel that my intuition is quite good in these matters, so it was surprising to learn some time later a convincing meta-argument that such motives should really exist. This idea, which I can’t now remember whether I learned from Chenevier or Clozel, is related to trying to construct such forms which are in addition self-dual and so come from a classical group. In favourable situations, there exists a compact inner form on this group, so that computing these forms “reduces” to computing on certain finite sets. One such finite set turns out to be the set of positive definite lattices of discriminant one and dimension \(n\). As is well-known, they only occur in dimensions a multiple of 8. For \(n=8\) there is just \(E_8\), and for \(n=16\) there are two, and for \(n=24\) there turn out to be exactly \(24\), as classified by Niemeier, and which include the famous Leech lattice whose automorphism group is a central extension of the first sporadic group discovered Conway. Easier to compute is the weighted sum of such lattices by automorphisms; for \(n=24\), for example, this weighted sum is

\( \displaystyle{\frac{1027637932586061520960267}{129477933340026851560636148613120000000}}\)

which is very small, and of course is related to the fact that these lattices are quite symmetric. For \(n = 32\), however, the weighed sum is bigger than \(10^7\), and so there are lots of lattices. You might then think that the existence of these lattices (even just \(E_8\) when \(n=8\)) implies the existence of automorphic forms which then should give rise to the desired automorphic forms on \(\mathrm{GL}(n)\). But there are issues. One concerns the technical issue of trasfering forms between groups which is of course a subtle problem. But there is another. A form which is cuspidal on some group need no longer be cuspidal after transferring to \(\mathrm{GL}(n)\). So to see which forms are cuspidal you really need to do a computation. But these objects are of large complexity — already computing Hecke actions on supersingular points for \(X_0(11)\) is a non-trivial exercise; here the objects involve lattices of enormously high dimension. Chenevier and his co-authors, (including Lannes, Renard, and Taïbi) have done a remarkable job understanding what is going on here. The most basic example of the type of theorem they prove is as follows. When \(n = 16\) so there are two lattices; one can try to compute the action of a Hecke operator \(T_p\), and it turns out (see for example Theorem A here) that the answer involves Ramanujan’s function \(\tau(p)\). But this also tells you that the transfer to \(\mathrm{GL}_{16}\) will have some explicit isobaric decomposition corresponding to twists of the modular form \(\Delta\), and in particular the associated \(\pi\) will clearly not be cuspidal.

At the same time, there are some automorphic arguments which show that cusp forms of level one (and cohomologically trivial weight) cannot exist. Here the idea goes back to the (automorphic) proof of lower bounds for discriminants of number fields by Stark and Odlzyko. The idea in that case to use the explicit formula for \(\zeta_{K}(s)\) to construct an expression (and in particular the normalized version \(\Lambda_K(s)\) which satisfies the functional equation and involves \(N^s\) where \(N\) is the level which is directly related to the discriminant of \(K\)) and ultimately arrive at some expression which is provably non-negative unless the root discriminant of \(K\) is larger than some explicit constant (minus some explicit \(o(1)\) depending on the degree). Mestré generalized this argument to automorphic forms corresponding to other Motives, in particular proving that, assuming conjectures of Langlands type, that there did notexist any abelian varieties over \(\mathbf{Z}\) of dimension at least one (which was proved unconditionally by Fontaine) but also that the conductor of such an abelian variety had to be at least \(10^g\). This was then later generalized by Fermigier (a student of Mestré) and then by Stephen Miller (Rutgers!) to prove that there are no automorphic forms \(\pi\) for \(\mathrm{GL}_n/\mathbf{Q}\) of level one which are cohomological for the trivial representation when \(n < 27\). These are exactly the forms associated (conjecturally) to the motives of weight \([0,1,\ldots,n-1]\). Returning to the conference at Orsay: Chenevier gave a talk on understanding automorphic representations \(\pi\) of level one and low motivic weight, and once again raised the automorphic version of Khare’s question. Now I have known about this question for a long time, but somehow being reminded of a problem can sometimes be the spark to help one think about the question again.

Correcting what was a past failing of my own intuition, I was very happy that George Boxer, Toby Gee, and I were able to come up with a very simple argument to answer both questions; there does exist a compatible family of crystalline Galois representations with Hodge-Tate weights \([0,1,2,\ldots,n-1]\) for some \(n\); for example one can take \(n=105\). Moreover, this compatible system is even automorphic and associated to a cuspidal
\(\pi\) of level one and cohomological weight zero for \(\mathrm{GL}_n\). (With work, it is even “motivic” in the sense that the compatible system can be found inside an explicit algebraic variety over \(\mathbf{Q}\), so it is in particular also pure.) Now while the argument is very simple, it must also be said that is uses some extremely hard theorems; for a start, it uses both the full modularity lifting results of BLGGT (Barnet-Lamb, Gee, Geraghty, Taylor), following Clozel-Harris-Taylor and many others, *and* it uses the even more recent full symmetric power functoriality result for classical modular forms by Newton and Thorne. (Since the paper is only nine pages and the proof only half of that, I won’t explain it here.)

One would still like to prove, of course, that there are a huge number of self-dual forms for all sufficiently large \(n\). And one can naturally ask what is the smallest such \(n\), which we now know satisfies \(27 \le n \le 105\). The expectation is certainly that \(n\) is probably close to around \(32\). It would be nice to know!

Of course, there is an endless list of other tricky problems one can pose of this form. For example, does there exist a regular motive (with coefficients) over \(\mathbf{Z}\) with Hodge-Tate weights \([0,1,\ldots,n-1]\) for some \(n\) which is *not* essentially self-dual?

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Clozel 70, Part I

I recently returned home from a trip to Paris for Clozel’s 70th birthday conference. Naturally I stayed in an airbnb downtown, and the RER B gods smiled on me with a hassle free commute for the entire week. Tekés was an interesting find, a fun (and surprisingly cheap) Israeli vegetarian restaurant right near where I was staying. But surely the food highlight of the week was the lunch spreads during the conference at Orsay — certainly the best conference food I’ve ever had! Great vegetarian food with amazing eggplant dishes, feta, figs, all the good stuff. (Rumor was it was chosen by Valentin Hernandez and paid for by Vincent Pilloni; I’m not sure if that’s true but a great job all round.)

There were quite a number of interesting talks, although as mentioned before I don’t like singling out because that sometimes seems like an implicit criticism of the other talks. But a few thoughts spurred by some of the talks (which you can more or less guess if you wanted to), some of which were already raised by others in conversation during the conference:

(Global) modules for Galois deformation rings. As a result of Taylor-Wiles patching, one usually constructs a CM-module \(M_{\infty}\) defined over some local deformation ring \(R\). Often quite a bit of mileage can be gained by exploiting the ring theoretic structure of \(R\) to conclude something about the module \(M_{\infty}\) and vice versa. Perhaps the ur-version of this argument is Diamond’s argument showing (in some circumstances) that the formal smoothness of \(R\) implies that \(M_{\infty}\) is free. A more recent example is in the work of Jeff Manning where he exploits the geometry of some particular \(R\) (by relating to a more geometric situation where one can perhaps understand the Picard group) to restrict the possible \(M_{\infty}\) to a very small number of possibilities from which one can then get some mileage. But one question raised is the extent to which this one can always do this. As one considers more and more complicated \(R\), is there some constraint on possible \(R\) which means that there are only going to ever exist a small number of faithful maximal rank one CM-modules \(M\), or are there going to be situations where \(R\) is very complicated and one can’t hope classify all such \(M\), but only (for some mysterious situation) a very small number of them turn up in global situations. Note that globally there is often a few possibilities of the type of cohomology one patches, and even for \(\mathrm{GL}(2)\) you can be in situations where you can force \(M\) to be free or self-dual by working in coherent cohomology or etale cohomology respectively and these modules are not always the same.

The work of Arthur: (Some) experts are at the point where they no longer expect Arthur to publish proofs of results he has claimed, leaving a huge gap in the literature. The summary seems to be that many very smart people are putting lots of effort into filling in some of these details, and that this seems to require new arguments. For example, it seems to be the case that one of Arthur’s proposed inductive arguments will not (at least naively) work. The mathematical community should be immensely grateful to people working on this!

Shimura Varieties: I once joked that today’s generation is more likely to learn about Galois representations associated to elliptic curves and modular curves before learning any class field theory. Well that generation has passed! We may be approaching a moment where people learn about Shimura varieties without ever thinking about modular curves, let alone getting close and personal with \(X_0(11)\). (Note: I don’t think that RLT knows the genus of \(X_0(11)\) and that never stopped him, of course.) Some people can look at the abstract definition of a Shimura variety and then start proving things; I am certainly not one of those people. Fortunately there are still many interesting open questions even about classical modular forms.

A result of Garland (at least) two talks reminded me of a vanishing result of Garland. Suppose that \(\Gamma\) is (for example) a lattice in \(\mathrm{SL}_n(\mathbf{Q}_p)\). Then the cohomology (with coefficients in \(\mathbf{Q}\)) vanish in positive degrees below \((n-1)\). But I think that much more should be true, namely, that the cohomology should all have a “trivial” Hecke action in the expected ways, i.e. the completed cohomology groups should all be finite in this range, as they are for \(\mathrm{SL}_n(\mathbf{Z})\) (more or less, let’s not be precise about ranges and what eactly is known). It feels like conjectures of this sort are not completely out of reach. Is this too optimistic? This is already an interesting problem in the case of \(H^2(\Gamma,\mathbf{F}_p)\).

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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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