## Bush, the Messiah (and Emma Kirkby)

The title of this post is, in part, a public service message to use the Oxford comma. However, there is a thread (in my mind) linking the three titular subjects. The connection between George Bush and the Messiah is not an obvious one, I admit, but hear me out. When I think of Bush, I think of the phrase “either you’re with us, or your against us.” I then always associate this phrase with “If God be for us, who can be against us?.” Was Bush consciously echoing the King James Bible? To me, of course, the latter phrase does not recall the Bible but rather Handel’s Messiah. All of which is a roundabout way of saying that this is another music post, with (who else) but Emma Kirkby performing with Christopher Hogwood and the Academy of Ancient Music (I have a recording on CD by the same ensemble which sounds to have been made contemporaneously with the video). The vintage of the haircuts is more H.W. than G., however.

## Finiteness of the global deformation ring over local deformation rings

(This post is the result of a conversation I had with Matt). Suppose that

$\overline{\rho}: G_{F} \rightarrow \mathrm{GL}_n(\mathbf{F})$

is a continuous mod-$p$ absolutely irreducible Galois representation. For now, let’s assume that $F/F^{+}$ is a CM field, and $\overline{\rho}$ is essentially self-dual and odd. Associated to this representation is a global deformation ring $R$ (of essentially self-dual representations) consisting of representations with no local restriction at primes dividing $p$ and the condition of being unramified at primes away from $p$. One also has a (collection of) local (unrestricted) deformation rings for the set of primes $v|p$, combining to give a ring $R^{\mathrm{loc}}$. Let us also assume that $\overline{\rho}$ has suitably big image (for example, its restriction to $F(\zeta_p)$ is adequate). Then we have:

Proposition: The map $R^{\mathrm{loc}} \rightarrow R$ is finite.

(Matt and Vytas prove this in the modular (odd) case when $n = 2$ and $F = \mathbf{Q}$, although I’m not sure whether the paper exists yet [actually, I'm pretty sure it doesn't]. Possibly if I was listening closer to Matt’s talk at Fields I might have remembered the argument, since I vaguely think it came up there, although possibly only briefly.)

Here one has to be a little careful defining deformation rings in the local case, of course (for those worried by such issues, simply choose suitable framings). To prove this, it suffices to prove the result after base change, so we may assume that $\overline{\rho}$ is unramified at all primes, and completely trivial at all primes dividing $p$. By Nakayama’s lemma, the problem above reduces to the following:

Proposition: Let $F^{\mathrm{ur}}$ be the maximal extension of $F$ unramified everywhere. Let $\Gamma$ be the Galois group of $F^{\mathrm{ur}}$ over $F$. Then $\Gamma$ does not admit a continuous essentially self-dual representation:

$\Gamma \rightarrow \mathrm{GL}_n(A)$

such that $A$ is a complete local Notherian $\mathbf{F}$-algebra of positive dimension.

This is a special case of the generalization of the unramified Fontaine-Mazur conjecture due to Boston. Recall that the group $\Gamma$ may be infinite (Golod-Shafarevich), but that Fontaine-Mazur predicts that the image of any such representation into any characteristic zero $p$-adic analytic group has finite image. Boston conjectured that the same finiteness would hold for homomorphisms of $\Gamma$ into $\mathrm{GL}_n(A)$ for rings like $A = \mathbf{F}[[T]]$. It turns out that even though the Fontaine-Mazur conjecture is hard, when $A$ has characteristic $p$ the conjecture is amenable to modularity lifting theorems by comparison to a new deformation ring in regular weight.

The proof is as follows:

Step 1: Using lifting theorems (Theorem 4.3.1 from BLGGT), we may assume, after a finite base change, that $\overline{\rho}$ is potentially ordinarily modular of level one for some regular weight $w$.

Step 2: Using minimal modularity theorems in the ordinary case (Section 10 from Thorne’s Jussieu paper, or Theorem 2.2.2 of BLGGT, both using work of Geraghty), deduce that the minimal weight $w$ ordinary deformation ring $S$ is finite over $W(\mathbf{F})$, and hence that $S/p$ is finite over $\mathbf{F}$. Strictly speaking, theorems of this kind are required to prove the previous result.

Step 3: Note that the minimal everywhere unramified deformations of $\overline{\rho}$ (i.e., the deformations coming from $\Gamma$) of characteristic $p$ are all ordinary of weight $w$, because everything unramified is ordinary, and in characteristic $p$ any two weights are the same. Hence $R/p$ is a quotient of $S/p$, from which it follows from the finiteness of $S$ that $R$ is also finite.

While I am using the latest modularity lifting theorems here, weaker versions for $n=2$ with some local assumptions on $\overline{\rho}$ follow from 90′s era technology (say Taylor’s Remarks on a conjecture of Fontaine and Mazur paper from 2000, or even earlier if one assumes residual modularity).

Via the usual argument, this result also applies to even Galois representations $\overline{\rho}: G_{\mathbf{Q}} \rightarrow \mathrm{GL}_2(\mathbf{F})$ with large image. In particular, the unramified deformation rings in these cases will be finite over $W(\mathbf{F})$, and there will be at most finitely many counter examples to the unramified Fontaine-Mazur conjecture in characteristic zero for a fixed residual representation. One can also apply it to many classes of higher dimensional non-self dual representations by taking irreducible summands of $\rho \otimes \rho^{\vee}$. For example, one can take any representation of $\mathbf{Q}$ whose image contains $\mathrm{SL}_n(\mathbf{F}_p)$ if $n$ is even, since then the associated $(n^2 - 1)$-dimensional representation $\mathrm{Ad}^0(\overline{\rho})$ restricted to an auxiliary CM field is irreducible, odd, self-dual, and adequate for large enough $p$. Similar remarks apply to representations over an arbitrary field $F$ with generic enough image by taking the tensor induction down to $\mathbf{Q}$.

If one starts allowing ramification at auxiliary primes, things become a little harder. One fix is to build the auxiliary primes into the local deformation ring $R^{\mathrm{loc}}$, although this might be considered cheating. The problem is that one cannot deduce (in general) that more general ordinary deformation rings $S$ are finite in the non-minimal situation. Although perhaps one can get by with the Taylor trick in some contexts. One should be OK with $\mathrm{GL}_2$ by Ihara’s Lemma.

Posted in Mathematics | | 3 Comments

## Michael Pollan is not a scientist

Michael Pollan is popular because he is an engaging speaker who spins a narrative about food that dovetails with the political inclinations of his audience. He has a degree in English, and, as far as I know, no scientific training whatsoever, but yet, he commands an enourmous amount of space in the New York Times and other liberal media to pontificate about nutritional science. Why does anyone take him seriously?

I don’t see any reason why I should care what Pollan thinks I should be eating. Science reporting should consist of a reporter explaining the consensus opinion (or otherwise) of scientists, not a dilettante pedalling an Alice Waters based cult dressed up as homespun wisdom. Let me be clear that I am not claiming anything he says in particular is wrong, I’m just feel that most of his conclusions are not arrived at in any scientific way, and the reason he has such a following is that his voice resonates with the intuition of self-indulgent (relatively) highly paid and well educated liberal elites (a class which I include myself). I avoid processed food, I seek out organic produce [for certain foods when it makes an appreciable difference in taste] (well, to be honest, it’s not usually me who does the food shopping because when I’m in charge I usually forget half the ingredients), and I almost always eat home-cooked meals with relatively little meat and plenty of fresh vegetables; and I do this for reasons of culture, taste, socioeconomic status, and because my wife wants me to be healthy. I pretty much agree with a lot of Pollan says (in the brief interviews I’ve seen him give), but what’s to stop him deciding (if he hasn’t already) that genetically modified foods are rubbish based on his own oversimplified philosophy rather than what science has to say? Or that vaccines are dangerous because his grandmother didn’t get them?

## Equidistribution of Heegner Points

I saw a nice talk by Matt Young recently (joint work with Sheng-Chi Liu and Riad Masri) on the following problem.

For a fundamental discriminant $|D|$ of an imaginary quadratic field $F$, one has $h_D$ points in $X_0(1)(\mathbf{C})$ with complex multiplication by the ring of integers of $F$. Choose a prime $q$ which splits in $F = \mathbf{Q}(\sqrt{-|D|})$. One obtains a set of $2 h_D$ points in $X_0(q)(\mathbf{C})$, given explicitly as follows:

$\mathbf{C}/\mathfrak{a} \mapsto \mathbf{C}/\mathfrak{a} \mathfrak{q}^{-1}$

for $\mathfrak{a}$ in the class group and $\mathfrak{q}$ one of the two primes above $q$ in $F$. The complex points $X_0(q)(\mathbf{C})$ can be thought of as being tiled by $q+1$ copies of the fundamental domain $\Omega$ in the upper half plane.

Problem: How large does $D$ have to be to guarantee that every one of the $q+1$ copies of $\Omega$ contains one of the $2 h_K$ CM points by $\mathcal{O}_F$?

This is the question that Young and his collaborators answer. Namely, one gets an upper bound of the shape $|D| < O(q^{m + \epsilon})$ (with some explicit $m$, possibly 20), the point being that this is a polynomial bound. Note that this proof is not effective, since it trivially gives a lower bound on the order of the class group which is a power bound in the discriminant, and no such effective bounds are known.

I idly wondered during the talk about the following "mod-$p$" version of this problem. To be concrete, suppose that $p = 2$ (the general case will be similar). We now suppose that $D$ is chosen so that $2$ is inert in $F$. Then all the $h_K$ points in $X_0(1)(\overline{\mathbf{F}}_2)$ are supersingular, which means that they all reduce to the same curve $E_0$ with $j$-invariant $1728$. Now, as above, choose a prime $q$ which splits in $F$. The pre-image of $j=1728$ in $X_0(q)(\overline{\mathbf{F}}_2)$ consists of exactly $q+1$ points.

Problem: How large does $|D|$ have to be to ensure that these points all come from the reduction of one of the $2 h_K$ CM points by $\mathcal{O}_F$ as above?

Since $E_0$ is supersingular, we know that $\mathrm{Hom}(E_0,E_0)$ is an order in the quaternion algebra ramified at $2$ and $\infty$. In fact, it is equal to the integral Hamilton quaternions $\mathbf{H}$. If $E$ and $E'$ are lifts of $E_0$, then there is naturally a degree preserving injection:

$\mathrm{Hom}(E,E') \rightarrow \mathrm{Hom}(E_0,E_0) = \mathbf{H}.$

The degree on the LHS is the degree of an isogeny, and it is the canonical norm on the RHS.
In particular, if $E = \mathbf{C}/\mathfrak{a}$ and $E' = \mathbf{C}/\mathfrak{a} \mathfrak{q}^{-1}$, then one obtains a natural map:

$\psi_{\mathfrak{a}}: \mathfrak{q}^{-1} \simeq \mathrm{Hom}(E,E') \rightarrow \mathbf{H}$

preserving norms. The norm map on $\mathfrak{q}^{-1}$ is $N(x)/N(\mathfrak{q}^{-1})$. The image of the natural $q$ isogeny is simply $\psi_{\mathfrak{a}}(1)$, whose image has norm $q$. Hence the problem becomes:

Problem: If one considers all the $2 h_K$-maps:

$\psi_{\mathfrak{a}}: \mathfrak{q}^{-1} \rightarrow \mathbf{H}, \qquad \psi_{\mathfrak{a}}: \overline{\mathfrak{q}}^{-1} \rightarrow \mathbf{H},$

do the images of $1$ cover the $q+1$ elements of $\mathbf{H}$ of norm $q$?

Given a field $F$ in which $2$ is inert, it wasn’t obvious how to explicitly write down the maps $\psi_{\mathfrak{a}}$, but this problem does start to look similar in flavour to the original one. Moreover, to make things even more similar, in the original formulation over $\mathbf{R}$ one can replace modular curves by definite quaternion algebras ramified at (say) $2$ and $q$, and then the Archimidean problem now also becomes a question of a class group surjecting onto a finite set of supersingular points. In fact, this Archimedean analogue may well be *equivalent* to the $\mod 2$ version I just described! Young told me that his collaborators had mentioned working with various quotients coming from quaternion algebras as considered by Gross, which I took to mean the finite quotients coming from definite quaternion algebras as above. Hence, with any luck, they will provide an answer this problem.

## Exercise concerning quaternion algebras

Here’s a fun problem that came up in a talk by Jacob Tsimerman on Monday concerning some joint work with Andrew Snowden:

Problem: Let $D/\mathbf{Q}(t)$ be a quaternion algebra such that the specialization $D_t$ splits for almost all $t$. Then show that $D$ itself is split.

As a comparison, if you replace $\mathbf{Q}$ by $\overline{\mathbf{Q}}$, then although the condition that $D_t$ splits becomes empty, the conclusion is still true, by Tsen’s theorem.

This definitely *feels* like the type of question which should have a slick solution; can you find one?

Posted in Mathematics | | 2 Comments

## Catalan’s Constant and periods

There is a 60th birthday conference in honour of Frits Beukers in Utrech in July; I’m hoping to swing by there on the way to Oberwolfach. Thinking about matters Beukers made me reconsider an question that I’ve had for while.

There is a fairly well known explanation of why $\zeta(3)$ should be irrational (and linearly independent of $\pi^2$) in terms of Motives. There is also a fairly good proof that $\zeta(3) \ne 0$ in terms of the non-vanishinjg of Borel’s regulator map on $K_5(\mathbf{Z})$. (I guess there are also more elementary proofs of this fact.) A problem I would love to solve, however, is to show that, for all primes $p$, the Kubota-Leopoldt $p$-adic zeta function $\zeta_p(3)$ is non-zero. Indeed, this is equivalent to the injectivity of Soule’s regulator map

$K_5(\mathbf{Z}) \otimes \mathbf{Z}_p \rightarrow K_5(\mathbf{Z}_p).$

(Both these groups have rank one, and the cokernel is (at least for $p > 5$) equal to $\mathbf{Z}_p/\zeta_p(3) \mathbf{Z}_p$ by the main conjecture of Iwasawa theory.) It is somewhat of a scandal that we can’t prove that $\zeta_p(3)$ is zero or not; it rather makes a mockery out of the idea that the “main conjecture” allows us to “compute” eigenspaces of class groups, since one can’t even determine if there exists an unramified non-split extension

$0 \rightarrow \mathbf{Q}_p(3) \rightarrow V \rightarrow \mathbf{Q}_p \rightarrow 0$

or not. Well, this post is about something related to this but a little different. Namely, it is about the vaguely formed following question:

What is the relationship between a real period and its $p$-adic analogue?

Since one number is (presumably) in $\mathbf{R} \setminus \mathbf{Q}$ and the other in $\mathbf{Q}_p \setminus \mathbf{Q}$, it’s not entirely clear what is meant by this. So let me give an example of what I would like to understand. One could probably do this example with $\zeta(3)$, but I would prefer to consider the “simpler” example of Catalan’s constant. Here

$G = \displaystyle{\frac{1}{1} - \frac{1}{3^2} + \frac{1}{5^2} - \frac{1}{7^2} \ldots } = L(\chi_4,2) \in \mathbf{R},$

is the real Catalan’s constant, and

$G_2 = L_2(\chi_4,2) \in \mathbf{Q}_2$

is the $2$-adic analogue. (There actual definition of the Kubota-Leopoldt zeta function involves an unnatural twist so that one could conceivably say that $L_2(\chi_4,2) = 0$ and that the non-zero number is $\zeta_2(2)$, but this is morally wrong, as the examples below will hopefully demonstrate. Morally, of course, they both relate to the motive $\mathbf{Q}(2)(\chi_4)$.)

So what do I mean is the “relation” between $G$ and $G_2$. Let me give two relations. The first is as follows. Consider the recurrence relation (think Apéry/Beukers):

$n^2 u_n = (4 - 32 (n-1)^2) u_{n-1} - 256 (n-2)^2 u_{n-2}.$

It has two linearly independent solutions with $a_1 = 1$ and $a_2 = -3$, and $b_1 = -2$ and $b_2 = 14$. One fact concerning these solutions is that $b_n \in \mathbf{Z}$, and $a_n \cdot \mathrm{gcd}(1,2,3,\ldots,n)^2 \in \mathbf{Z}.$ Moreover one has that:

$\displaystyle{ \lim_{n \rightarrow \infty} \frac{a_n}{b_n}} = G_2 \in \mathbf{Q}_2.$

The convergence is very fast, indeed fast enough to show that $G_2 \notin \mathbf{Q}$. What about convergence in $\mathbf{R}$, does it converge to the real Catalan constant? Well, a numerical test is not very promising; for example, when $n = 40000$ one gets $0.625269 \ldots$, which isn’t anything like $G = 0.915966 \ldots$; for contrast, for this value of $n$ one has $a_n/b_n - G_2 = O(2^{319965})$, which is pretty small. There are, however, two linearly independent solutions over $\mathbf{R}$ given analytically by

$\displaystyle{\frac{(-16)^n}{n^{3/2}} \left( 1 + \frac{5}{256} \frac{1}{n^2} - \frac{903}{262144} \frac{1}{n^4} + \frac{136565}{67108864} \frac{1}{n^6} - \frac{665221271}{274877906944} \frac{1}{n^8} + \ldots \right)},$

\begin{aligned} \frac{(-16)^n \cdot \log n}{n^{3/2}} \left( 1 + \frac{5}{256} \frac{1}{n^2} - \frac{32261}{7864320} \frac{1}{n^4} + \frac{136565}{67108864} \frac{1}{n^6} - \frac{665221271}{274877906944} \frac{1}{n^8} + \ldots \right)\\ +\frac{(-16)^n}{n^{3/2}} \left( -\frac{1}{768} \frac{1}{n^2} + \frac{32261}{7864320} \frac{1}{n^4} - \frac{30056525}{8455716864} \frac{1}{n^6} + \frac{1778169492137}{346346162749440} \frac{1}{n^8} + \ldots \right) \end{aligned},

from which one can see that $a_n/b_n$ must converge very slowly, and indeed, one has (caveat: I have some idea on how to prove this but I’m not sure if it works or not):

$\displaystyle{\frac{a_n}{b_n} = G - \frac{1}{(0.2580122754655 \ldots) \cdot \log n + 0.7059470639 \ldots}}$

So one has a naturally occurring sequence which converges to $G$ in $\mathbf{R}$ and $G_2$ in $\mathbf{Q}_2$. So that is some sort of “relationship” alluded to in the original question. Here’s another connection. Wadim Zudilin pointed out to me the following equality of Ramanujan:

$\displaystyle{G = \frac{1}{2} \sum_{k=0}^{\infty} \frac{4^k}{(2k + 1)^2 \displaystyle{\binom{2k}{k}}}} \in \mathbf{R}.$

This sum also converges $2$-adically. So, one can naturally ask whether

$\displaystyle{G_2 =^{?} \frac{1}{2} \sum_{k=0}^{\infty} \frac{4^k}{(2k + 1)^2 \displaystyle{\binom{2k}{k}}}} \in \mathbf{Q}_2.$

(It seems to be so to very high precision.) These are not random sums at all. Indeed, they are equal to

$\displaystyle{ \frac{1}{2} \cdot F \left( \begin{array}{c} 1,1,1/2 \\ 3/2,3/2 \end{array} ; z \right)}$

at $z = 1$. Presumably, both of these connections between $G$ and $G_2$ must be the same, and must be related to the Picard-Fuchs equation/Gauss-Manin connection for $X_0(4)$. This reminds me of another result of Beukers in which one compares values of hypergeometric functions related to Gauss-Manin connections and elliptic curves, and finds that they converge in $\mathbf{R}$ and $\mathbf{Q}_p$ for various $p$ to algebraic (although sometimes different!) values. Of course, things are a little different here, since the values are (presumably) both transcendental. Yet it would be nice to understand this better, and see to what extent there is a geometric interpretation of (say) the non-vanishing of $L_p(\chi,2)$ for some odd quadratic character $\chi$. Of course, one always has to be careful not to accidentally prove Leopoldt’s conjecture in these circumstances.