The union thugs at the AMS

After twice failing to achieve a 2/3 majority on previous votes, the AMS busted up the filibuster and railroaded through their new fellows program. The initial implementation was based on having achieved one of several things, the easiest of which was to have given an invited AMS address. There was, however, another condition: you had to have been a full fee paying member of the AMS from 2010 onwards. I had given such an address, but I was not a member of the AMS – mostly out of indifference rather than anything else. The AMS had clearly accounted for this possibility; they were offering indulgences in exchange for cash: for a price, you could retroactively purchase AMS membership for 2010 and 2011 (this was in 2012) and all your sins would be forgiven. There was some ambiguity as to whether one’s invited address had to be before or after Jan 1 2012 (mine was after), and in the end, I did not pony up the cash, and I am not an AMS Fellow (nor, to take a selection of other members of my department, is Andrei Suslin or Kari Vilonen).

I don’t want to argue here about the merits of this particular program (see some opinions here), but I want instead to ask: should I become a member of the AMS? My department chair is (gently) applying some pressure for me to do so. The issues are mainly financial and moral. It is admissible for me to pay my dues with startup funds; this means that there is no immediate personal cost, although, as my startup funds are by no means infinite, this might eventually have a direct impact on my wallet. On the other hand, some have suggested that having such a fellowship will ultimately be better for one’s salary (this is hard to quantify, however). Note also that it is not so clear how easy it will be to become a member in the near future – there are presumably some really good mathematicians who have not given AMS invited addresses. Finally, there’s the moral argument: I feel a mild distaste concerning the notion that unless I pay up, the AMS will officially sanction my research output as not up to scratch. But maybe one could argue instead that the AMS does all sorts of wonderful things which mitigate such behavior. Please give me your thoughts!

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András Schiff Plays Bach

The first Bach CD I ever bought was András Schiff’s recording of the Well Tempered Clavier. My memory is a little hazy, but I think I paid almost $100 (AUS) for the 4-CD set (I ordered it through Allans Music on Collins Street: your typical 20th century classical music store which, in addition to selling CDs, charged exorbitant prices for sheet music which can now be legally downloaded in massive quantities here). I had a chance to see Schiff play live in 2000 (at the Barbican), but, after a tip off from the NY times, I found out he was also giving a Bach recital at Chicago Symphony Hall. His played the complete second book from the 48 (composed, incidentally, 20 years after the first). It was an intense performance that lasted almost three hours, and was one of the best concerts I’ve been to in a while (the audience, as might be expected from a solo Bach concert, was younger and better behaved than the usual crowd of wheezing octogenarians). Schiff eschews the use of the pedal in Bach completely, which perhaps is easier to do when you have such amazing legato.

It was a little tough finding the right video to post – most of the Well Tempered Clavier on Youtube by Schiff is from his 80’s recording, not his recent one. But then I found this snippet from a masterclass, and, although it’s not the WTC (it’s my favourite Partita), just from listening to the first five seconds of the Rondeau (about 30 seconds in) you get a sense of his magical touch. (Plus, you also get to see Fred Armisen in character as Harry Potter playing the piano!)

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The Bourgeois Pig is dead to me

While on the way to meet up with BF, I stopped by and ordered an 8 ounce latte at the Pig. The result: a foamy scalding hot disaster which brought great shame upon hipster cafes everywhere. It was so bad, I brought it back to the cafe without drinking it. Why would I ever go there again when Intelligentsia is only three minutes drive away? Pig, you are dead to me and I shall not return.

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How I deal with technical support

I usually manage to avoid having to make any of those annoying calls to banks, credit card companies, utility companies, etc. because DW does it all for me. However, there are times (for example, when things are in my name) in which I have to make the call. I have, however, a fairly successful strategy for making these calls run smoothly. Namely, I ask whether they think Sachin Tendulkar should retire or not. More than half the time, it leads to better service.

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Schoenberg

The best show on television…is surely classic arts showcase. It’s RAGE for classical music, by which I mean something like an eclectic classical music version of MTV (during the 80’s). Given the limited range of classical music that makes it onto the radio, it’s one of the few sources to discover “new” music. One clip that particularly electrified me was Christine Schäfer’s performance of Pierrot Lunaire: (apologies for the poor quality):

I thoroughly recommend the dvd which I own; it also contains a smashing performance of Schumann’s Dichterliebe. (Added personal bonus: the lyric Nach Bergamo, zur Heimat)

Sticking with the Schoenberg theme, here is Gould again playing the Gigue from Op.25. I think this counts as another successful use of video accompaniment; this time the clip is from the movie 32 short films about Glenn Gould (another personal favourite) (update: due to link rot, this is a different clip of the same recording).

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Jacobi by pure thought

JB asks whether there is a conceptual proof of Jacobi’s formula:

\(\Delta = q \prod_{n=1}^{\infty}(1 – q^n)^{24}\)

Here (to me) the best proof is one that requires the least calculation, not necessarily the “easiest.” Here is my attempt. We use the following property of \(\Delta\), which follows from its moduli theoretic definition: the only zero of \(\Delta\) is a simple zero at the cusp, moreover, the evaluation of \(\Delta\) on the Tate curve is normalized so that the leading coefficient is \(q\).

Let \(p\) be prime. I claim that

\(\Delta(\tau)^{p+1} \prod_{i=0}^{p-1} \zeta^i = \Delta(p \tau) \prod_{i=0}^{p-1} \Delta \left(\frac{\tau + i}{p}\right).\)

Observe that both these expressions are modular forms of level one and weight \((p+1)\) times the weight of \(\Delta\). One can prove this “by hand,” but also by noting that the RHS is equal to the norm of \(\Delta(p \tau)\) on \(X_0(p)\) down to \(X_0(1)\). On the other hand, the RHS also has a zero of order \(p+1\) at \(q = 0\), from which the result immediately follows, since the ratio will be holomorphic of weight zero. If one defines Hecke operators on \(q\)-expansions in the usual way, it also immediately follows that the logarithmic derivative \(q d/dq \log(\Delta)\) is (as a \(q\)-expansion) an eigenform for \(T_p\) of weight two with eigenvalue \(p+1\) for all primes \(p\). In fact, the same argument as above (with \(X_0(p)\) replaced by \(X_0(n)\)) implies that this derivative is also an eigenform for \(T_n\) with eigenvalue \(\sigma_1(n) = \displaystyle{\sum_{d|n} d}\). This is almost enough to determine the \(q\)-expansion uniquely: in particular, it implies that

\(q \cdot \frac{d}{dq} \log(\Delta) = 1 + m \cdot \sum_{n=1}^{\infty} \sigma_1(n) q^n\)

for some integer \(m\), from which it follows that

\(\Delta = q \prod_{n=1}^{\infty} (1 – q^n)^m.\)

To finish the argument, it suffices to check that \(m = 24\), or that \(\tau(2) =-24\). One way to do this is to note (by uniqueness) that \(\Delta\) is a Hecke eigenform, and then use the equation \(\tau(2) \tau(3) = \tau(6)\) which implies that \(m \in \{0,1,2,3,24\}\); the cases \(m = 1,2,3\) are then ruled out by the equations \(\tau(2) \tau(7) = \tau(14)\) and \(\tau(2) \tau(13) = \tau(26)\), and \(m = 0\) is ruled out by the fact that \(q\) is not a modular form. Curiously enough, this determines \(\Delta\) without ever using the fact that it has weight \(12\). Another (more traditional way) is to show that \(1728 \Delta = E^3_4 – E^2_6= q – 24 q^2 + \ldots\). Is there a way to do this final step by pure thought?

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En Passant

Several times in NYC, I’ve had the chance to visit Eataly, an italian food court/upscale delicatessen run by Mario Batali. You can either sit down at one of the various restaurants for an antipasto plate with a glass of wine, or grab a prosciutto fig sandwich (and a bottle of chinotto) to go. The good news is that Eataly is coming to Chicago.

There’s an updated version of BLGGT on the ArXiv. The numbering (at a quick glance) appears to be consistent with previous versions, although its now at least 20 pages longer. Perhaps (hint hint) one of my readers can update me with the contents?

The latest hot tip for a fields medal: Kai Wen Lan. Go Kai Wen! (In that first link, Kai Wen also exhibits the ability to morph into Shimura, a fields medal worthy feat itself.)

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Prokofiev

Glenn Gould is best known (deservedly) for his Bach, but there are other treasures in his discography, including Schoenberg and (perhaps surprisingly) Brahms. Here’s Gould playing some mean Prokofiev:

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Remarks on Buzzard-Taylor

Let \(\rho: G_{\mathbf{Q},S} \rightarrow \mathrm{GL}_2(\overline{\mathbf{Q}}_p)\) be continuous and unramified at \(p\). The Fontaine-Mazur conjecture predicts that \(\rho\) has finite image and is automorphic. Buzzard and Taylor proved this result under the assumption the natural assumption that \(\rho\) is odd, that \(\overline{\rho}\) is modular (now uneccessary), but also under the futher two assumptions:

  • \(\overline{\rho}\) is irreducible,
  • \(\overline{\rho}\) is \(p\)-distinguished.
  • (caveat: there are mild extra assumptions required when \(p = 2\).) The point of this post is to note that is seems possible to remove either of these conditions and to wonder whether both can be removed simultaneously.

    Suppose the second condition fails. One may enlarge the ordinary Hecke algebra \(\mathbf{T}\) to include the operator \(U_p\), call the resulting ring \(\widetilde{\mathbf{T}}\). After appropriate localizations, this is different from \(\mathbf{T}\) exactly when \(\overline{\rho}(\mathrm{Frob}_p)\) is a scalar. On the Galois side, let \(R_p\) denote the universal (framed) deformation ring. Then, for any lift of Frobenius, one can define the quadratic extension \(\widetilde{R}_p = R_p[\alpha]\) where \(\alpha\) is an eigenvalue of Frobenius. Fix a weight \(k \ge 2\). If \(R^{\mathrm{ord}}_p\) denotes the ordinary deformation ring, then there is a corresponding quotient \(\widetilde{R}^{\mathrm{ord}}_p\) of \(\widetilde{R}_p\) which records ordinary deformations together with the action of Frobenius on the unramified quotient. The \(\overline{\mathbf{Q}}_p\)-points of these local deformation rings are the same (since \(k \ge 2\)). The usual Taylor-Wiles-Kisin method produces an isomorphism of the type \(\widetilde{R}[1/p] = \widetilde{\mathbf{T}}[1/p]\), where \(\widetilde{R}\) is the global deformation ring which takes into account the extra data at \(p\). This isomorphism holds for all weights \(k \ge 2\), which is enough to get an isomorphism on the corresponding ordinary families.

    If \(\rho(\mathrm{Frob}_p)\) has distinct eigenvalues, one may now deduce Buzzard-Taylor (by the same argument as BT). If \(\rho(\mathrm{Frob}_p)\) is scalar, then one has to make a slight adjustment. To see what to do, note that if \(f(\tau)\) is the desired weight one form, then the old space \(f(\tau), f(p \tau)\) can no longer be diagonalized with respect to \(U_p\). Instead, it should give rise to a surjective map \(\psi: \widetilde{\mathbf{T}} \rightarrow \mathcal{O}[\epsilon]/\epsilon^2\) such that the image of \(\mathbf{T}\) is \(\mathcal{O}\). Conversely, if there is such a map \(\psi\), this produces the ordinary old forms necessary to recover \(f\) (both forms have the same Hecke eigenvalues away from \(p\), so one can determine the \(q\)-expansions). From the modularity result, it suffices to construct such a map on the Galois side. Yet this exists precisely because \(\rho\) comes with two distinct unramified quotients.

    DG and I used these these flavours of deformations rings for a somewhat different purpose (although we required more precise integral information concerning \(\widetilde{R}^{\mathrm{ord}}_p\) coming out of a very nice paper of Snowden). On the other hand, as far as the argument above goes, it was apparently known to RLT many years ago (as I learnt by chatting with TG whilst drinking an $8 can of Boddingtons in Toronto).

    Suppose one assumes instead that \(\overline{\rho}\) is reducible. Recall that one has maps \(R^{\mathrm{ps}} \rightarrow R\) and \(R^{\mathrm{ps}} \rightarrow \mathbf{T}\) for a suitable pseudo-deformation ring \(R^{\mathrm{ps}}\). In higher weights, Skinner-Wiles essentially prove that \(R[1/p]^{\mathrm{red}} = \mathbf{T}[1/p]\), which should be sufficient to construct the required overconvergent forms \(f_{\alpha}\) and \(f_{\beta}\) in weight one. While chatting with PA over espresso today, it also seems reasonable that (using appropriate framings, as above) one may generalize this to the case where \(\overline{\rho}\) is no longer \(p\)-distinguished, as long as the characteristic zero eigenvalues \(\alpha\) and \(\beta\) are distinct. The problem, however, with the \(\alpha = \beta\) case is that one needs to promote a non-reduced quotient \(R \rightarrow \mathcal{O}[\epsilon]/\epsilon^2\) to a map from \(\mathbf{T}\), and the methods of Skinner-Wiles in the reducible case only give information about the reduced quotients of \(R\). Is there any way around this?
    This seems (pretty close) to the only remaining obstruction for a complete solution to the weight one odd case of Fontaine-Mazur.

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    Winning the Chocolate

    Once a month, there is an (unrated) round robin blitz tournament at the Evanston chess club. The time controls are set on five minutes (a little slow for me, I usually play one minute games online), but it’s nice to play over the board for a change. The best players are rated around 2000, which means (for me at least) that every game is in play. There’s even a prize for the top two finishers (a chocolate bar), so the goal is to “win the chocolate.” So far, I’m 3 for 4 on the chocolate. Slightly amusingly, two of the players are dead ringers for Michael Schein and David Geraghty respectively (current record: 2:0 against MS and 1:1 against DG). As for my record against (actual) mathematicians, I’m 1:1 against Serre (I beat him playing the King’s gambit, then lost the second game playing the French), and I’ve won and lost many games of bughouse against Noam Elkies (to some extend this doesn’t count, since at bughouse it’s often more important how strong the weaker player is.)

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