## Me versus Magnus

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.

## Pseudo-representations and the Eisenstein Ideal

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.

## Erdős Number 3!

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.

Posted in Waffle | | 6 Comments

## Film Criticism

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:

## A non-liftable weight one form modulo p^2

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

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

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.

## Report from MSRI and Berkeley

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

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

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:

Aeropress at Blue Bottle

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.

Posted in Uncategorized | | 12 Comments