## Z_p-extensions of Number Fields, Part II

This is continuation of the last post. We claimed there that we were going to deform a totally real number field of degree n into a field with signature (r,s) with r+2s = n, and pass information about Leopoldt’s conjecture from one field to the other.

How does one “deform” a number field? One natural way is to think of a finite etale map of varieties X->Y defined over Q, and then consider the fibres. More prosaically, write down some family of polynomials and then vary the coefficients. Most of the time, the unit group doesn’t behave so well in such families. For example, consider the equation:

$x^2 - D = 0.$

If one varies D, even with some local control at primes dividing infinity (that is, keeping D positive), then it is not at all clear how the fundamental unit varies. In fact, one knows that the height of the fundamental unit is very sensitive to the size of the class number, which changes somewhat irregularly with D. On the other hand, consider the equation:

$x^2 - Dx = 1.$

Here one is in much better shape: as D varies, the element x will always be a unit, and moreover always generates a finite index subgroup of the full unit group. How might one use this for arguments concerning Leopoldt’s conjecture? The idea is to consider (as above) a family of number fields in which some finite index subgroup of the full unit group is clearly visible, and is deforming “continuously” in terms of the parameters. Then, by Krasner’s Lemma, we see that Leopoldt’s conjecture for one number field (and a fixed prime p) will imply the same for all sufficiently close number fields. To start, however, one needs to have such nice families.

Ankely-Brauer-Chowla Fields: One nice family of number fields that deforms nicely is the class of so-called Ankeny-Brauer-Chowla fields (from their 1956 paper):

$\prod (x - a_i) - 1 = 0$

It is manifestly clear that in the field Q(x), the elements x – a_i are all units, and that (generically) there is only one multiplicative relation, namely that the product over all such units is trivial. In this way, we get a family of number fields (with generic Galois group) S_n and with a family of units generating a free abelian group of rank n-1. With a little tweak, we can also ensure that the prime p splits completely. Concretely, consider the equations

$\prod (X - a_i) - \prod (X - b_i) = 1,$

where X is a formal variable. The corresponding variety Y is connected of dimension n, and the projection to A^n given by mapping to {b_i} is a finite map, and so, generically, the values of b_i on Y are all distinct. In particular, for sufficiently large primes p, Y has points over F_p where all the b_i are distinct modulo p. Fix such a point {a_i,b_i} over F_p. Lift the a_i in F_p to arbitrary integers in Z. Then, by Hensel’s lemma, there exist p-adic integers v_i congruent to b_i mod p such that

$\prod (x - a_i) - 1 = \prod (x - v_i),$

and so p splits completely in our field as long as the a_i satisfy some suitable non-empty congruences mod p.

Deforming the signature: Suppose we assume that, for a fixed choice A = (a_1,..,a_n), the corresponding field F satisfies Leopoldt’s conjecture. Then we see that, in a sufficiently small neighbourhood of A, we obtain many other fields which are totally real with Galois group S_n that also satisfy Leopoldt’s conjecture. On the other hand, our goal is to study fields of signature (r,s) with r+2s=n. So we want to deform our fields to have non-trivial signature. I learnt this trick by reading a paper of Bilu: we deform the fields in a slightly different way, by making the replacement

$(x - a_i)(x - a_j) \Rightarrow (x - a_i)(x - a_j) + u,$

where u has very small p-adic valuation, and yet is a large positive integer. The corresponding field no longer has n obvious units (whose product is one), but now only n-1 obvious units (whose product is one), where one of the units is now the quadratic polynomial above. On the other hand, one can also see that the signature of the number field is now (n-2,1). So we still have a nice finite index subgroup of the unit group. Moreover, p-adically, if our original units are written as {u_i}, then we get (p-adically) something very close (by Krasner), except now u_i and u_j have been replaced by u_i + u_j. By combining other pairs of units in the same way, we can reduce the signature to (r,s) with any r+2s = n and still have a nice p-adically continuous finite index family of global units.

Proposion Suppose that Leopoldt’s conjecture holds for the original field K at p. Then, by deforming suitably chosen pairs of roots, we obtain a (infinitely many) fields L with Galois group S_n and signature (r,s) with r+2s=n such that, for a choice of r+s-1 primes above p in L, the p-adic regulator of the units at those r+s-1 primes is non-zero.

As a consequence, for that choice of r+s-1 primes, the corresponding maximal Z_p-extension has rank zero. This proves that (L,p) is rigid for this choice of S. However, since S_n is n-transitive, the same result applies for any such choice of r+s-1 primes. It’s an elementary lemma to see that this also implies the result for sets S which are either larger or smaller than r+s-1.

The argument is exactly as you expect: Given the original field K, the assumption of Leopoldt’s conjecture for K implies that at least one of the corresponding (r+s-1) x (r+s-1) minors must be non-zero. We then deform the field globally so that the corresponding units in L of signature (r,s) are related to this minor, which (by Krasner) will still be non-zero. QED

Questions: The starting point of this construction was the assumption that K satisfied Leopoldt’s conjecture. Can one prove this directly? That is, can one find a choice of a_i such that the field

$\prod (x - a_i) - 1 = 0.$

satisfies Leopoldt at p? This seems quite plausible, after all, we have seen above that there are n nice units of finite index in the unit group whose regulator varies p-adically. So, it suffice to show that the regulator is not zero in the entire family. This certainly seems like an easier problem, because it’s easier to prove a function is non-zero rather than the special value of a function (for example, by looking at the derivative). Still, I confess that I don’t know how to prove this.

## Z_p-extensions of Number Fields, Part I

In the next few posts, I want to discuss a problem that came up when I wrote a paper with Barry Mazur. We had a few observations and remarks that we discussed as part of a possible sequel but which we never wrote up(*); mostly because we never could quite prove what we wanted to prove. But some of those remarks might be worth sharing.

The basic problem is as follows. Let E/Q be a number field of signature (r,s). Let p be a prime that splits completely in E (this is not strictly necessary, but it makes things cleaner). Let S be a set of primes above p. If S includes all the primes above p, then the Leopoldt Conjecture for E and p is the statement that

$r_S:=\mathrm{dim}_{\mathbf{Q}_p} \mathrm{Gal}(E^S/E)^{\mathrm{ab}} \otimes \mathbf{Q} = 1+s.$

The question is then to predict what happens when S is a strict subset of the primes above p. This leads to the following minimalist definition:

Definition: The field E is rigid at p if

$r_S:=\mathrm{dim}_{\mathbf{Q}_p} \mathrm{Gal}(E^S/E)^{\mathrm{ab}} \otimes \mathbf{Q} = \begin{cases} \#S - (r+s-1), & \#S \ge r + s - 1, \\ 0, & \text{otherwise}. \end{cases}$

Note that, for any field E, the right hand side is always a lower bound. So rigid pairs (E,p) are those which have no “unexpected” Z_p-extensions. If E is totally real, the Leopoldt Conjecture at p is equivalent to E being rigid. However, one does not predict that all fields E are rigid. The following is elementary:

Proposition If E is a totally imaginary CM field, then complex conjugation acts naturally on the set S. There are inequalities $r_S \ge [E^+:\mathbf{Q}] + 1$ if S consists of all primes above p, and

$r_S \ge \frac{1}{2} \# (S \cap c S)$

otherwise. If Leopoldt’s conjecture holds, then these inequalities are equalities.

It follows that if E is a CM field of degree at least 4, then E is not rigid for any prime p, because when S consists of two primes conjugate to each other under complex conjugation, then

$r_S \ge 1 > 2 - (r+s-1) = 2 - s.$

The “extra” extensions are coming from algebraic Hecke characters. Our expectation is that this is the only reason for a pair (E,p) to be rigid. For example:

Conjecture: Suppose that E does not contain a totally imaginary CM extension F of degree at least 4. Then (E,p) is rigid for any prime p that splits completely in E.

(When I say conjecture here, I really mean a guess; it could be false for trivial reasons.) Naturally these conjectures are hard to prove, since they imply Leopoldt’s Conjecture. Even if one assumes Leopoldt’s Conjecture, this conjecture still seems tricky. It makes sense, however, to see what can be proven under further “genericity” hypotheses on the image of the global units inside the local units. To this end, let me recall the Strong Leopoldt Conjecture which Barry and I formulated our original paper. Let F/Q be the splitting field of E/Q. Let G be the Galois group of F/Q. There is a G-equivariant map

$\mathcal{O}^{\times}_F \otimes \mathbf{Q}_p \rightarrow \prod_{v|p} \mathcal{O}^{\times}_{F,v} \otimes \mathbf{Q}_p.$

The right hand side is isomorphic as a G-module to $\mathbf{Q}_p[G]$. However, more is true; for a fixed prime v|p, there is an isomophism

$\mathbf{Q}_p[G] = \mathbf{Q}[G] \otimes \mathbf{Q}_p$

which is well-defined up to a scalar in $\mathbf{Q}_p$ coming from a choice of p-adic logarithm for the given place at p. It makes sense to talk about a rational subspace V of the right hand side, namely, a space of the form $V = V_{\mathbf{Q}} \otimes \mathbf{Q}_p$ for some $V_{\mathbf{Q}} \subset \mathbf{Q}[G].$ The strong Leopoldt conjecture asserts that the intersection of the global units wich such a rational subspace is as small as it can possibly be subject to the constraints of the G-action on both V and the units, together with Leopoldt’s conjecture that the map from the units tensor $\mathbf{Q}_p$ is injective. Let H = \Gal(F/E). By inflation-restriction, there is an isomorphism

$H^1_S(E,\mathbf{Q}_p) = H^1_T(F,\mathbf{Q}_p)^{H},$

where the subscript denotes classes “unramified outside S,” and where T denotes the set of primes in F above S. By class field theory, this may be identified with the H-invariants of the cokernel of the map

$\mathcal{O}^{\times}_F \otimes \mathbf{Q}_p \rightarrow \prod_{T} \mathcal{O}^{\times}_{F,v} \otimes \mathbf{Q}_p.$

The cokernel is larger than expected if and only if the kernel is bigger than expected. In particular, $r_S = \mathrm{dim} H^1_S(E,\mathbf{Q}_p)$ is bigger than expected only if

$\left( \mathcal{O}^{\times}_F \otimes \mathbf{Q}_p \bigcap \prod_{\neg T} \mathcal{O}^{\times}_{F,v} \otimes \mathbf{Q}_p \right)^{H}$

is bigger than expected. Note that the product over any subset T of primes in the right hand side is a rational subspace. Certainly the Strong Leopoldt Conjecture determines the dimension of the intersection $U \cap V$ of the unit group with a rational subspace. What is slightly less clear is that the intersection $(U \cap V)^{H}$ for any subgroup $H$ is also determined by the strong Leopoldt Conjecture, but this is true (and we prove it). As a consequence, one has:

Lemma: Assuming the Strong Leopoldt Conjecture, the dimension $r_S$ depends only on G, H, and S.

This “reduces” the computation of r_S to an intersection problem in a certain Grassmannian. But this is a computation we were never really able to do!

This is the problem: One knows very well the structure of the unit group of F as a G-module. So to compute the relevant intersections, one only has to compute the intersection with a “generic” rational subspace. Paradoxically, it seems very difficult in general to give explicit examples of rational subspaces which are generic enough to obtain the correct minimal value. So while, for formal reasons, almost any rational subspace will do, none of the nice subspaces which allow us to compute the intersection tend to be good enough.

Instead, to compute these intersections, we somewhat perversely look at actual number fields and their unit groups. This seems like a bad idea, since even verifying Leopoldt for a particular K and p is not so easy to do. So instead, we start with a totally real number field K of a certain form. Then, under the assumption of Leopolodt’s conjecture we can (non-constructively) find subspaces of rational subspaces V which provably minimize various intersections $\mathrm{dim}(W \cap V)$ for various unit-like submodules W. We then deform the field K to other fields L of different signature, and use this construction (as well as the Strong Leopoldt Conjecture) to make deductions about L. In the next post, we explain how this led Barry and me to a proof of the following:

Theorem: Let E/Q be a degree n field with whose Galois closure F has Galois group G = S_n. Assume the Strong Leopoldt Conjecture. Then (E,p) is rigid for any prime p which splits completely in p.

I will explain the details next time. But to unwind the serpentine argument slightly, we do not prove the result by finding rational subspaces in $\mathbf{Q}_p[G]$ whose intersection with the units of F has the a dimension which we can compute to be the expected value, but only rational subspaces whose dimension we can compute contingent on Leopoldt’s conjecture for some auxiliary totally real number field. In other words, we would like to compute the generic dimension of some intersection inside some G-Grassmannian, a problem which has nothing to do with number theory, and we compute it using Leopoldt’s conjecture. More next time!

(* never wrote up = actually written up in a pdf file on my computer somewhere)

Posted in Mathematics | | 1 Comment

## Coffee Passport

Having completed my visits to 22 Chicago coffee shops on the Chicago indie coffee passport tour, here are my notes together with a (rough) ranking in four categories from best to worst, taking into account a number of factors, including location.

In the rotation:

• Cafe Integral:
Artisan/scientific approach to coffee that interweaves the hipster with the pretentious. excellent 6 ounce cappuccino. A keeper.
• Heritage Outpost:
As good if not better than Heritage General. I have returned several times, although not since I moved downtown.
• Heritage General:
All the hipster accoutrements, including an in-the-store bike store. Croissant was absolutely rubbish, coffee was excellent. Will (and have) returned.
• Metropolis:
Excellent as expected; consistently good, have returned several times.

Good, but not in convenient (for me) location:

• Buzz Killer Espresso:
Nice atmosphere (modern decor (black, straight lines), pretentious vibe). Though it was my third drink of the day (on my Wicker Park Tour), it served a more than decent cortado.
• Groundswell Coffee:
I really liked this place; good coffee and decent food as well. Not really anywhere near where I go nowadays, but worth visiting.
• Caffe Streets:
The first stop on my day in Wicker Park.
Nice latte. The internet somewhat bizarrely allowed LT6 websites to load, however, so I couldn’t stay so long. OTOH, the barista had a top tip for lunch (Antique Taco) which made up for that.

Might be worth a second chance:

• Cup & Spoon:
An acceptable latte; would give a second chance if it was in a more convenient location.
• Bru Chicago:
Had a 90’s brewed awaking vibe: decent espresso, but would probably to go Buzz Killer across the street. Next to an great second hand book store where I picked up a copy of B.F.Skinner’s autobiography. It was only later that I realized it was part 2 of 3, and now I’m left searching for the other two volumes in the same edition.
• Ellipsis:
Opposite my favorite Thai restaurant in Chicago (Thai Spice.) I ordered an almond croissant (I feel that I should make at least one genuine purchase) which was absolutely terrible. Coffee was quite decent, however.
• Big Shoulders Cafe:
This cafe falls into this category only because it’s “in the middle of nowhere.” I had a cortado, which showed promise.
• Brew Brew :
Opposite Kosciusko park (but probably not the same Kosciusko that Melbournians would think of.) Coffee decent, not ideal texture. Had a donut. Music was turned up and sucked (even headphones blaring Schumann were not sufficient).
• Everybody’s Coffee:
My timing was poor, they were closing early for some reason, so I only got a rushed coffee to go; I’ve heard decent things about this place, but I probably won’t be returning.
• Sol Cafe:
I’ve heard about this place for a while; it was less than 5 minutes drive from my old house in Evanston. Yet I never managed to make it there until my visit with coffee passport. I had a latte; the server told the barista that I was a “passport” customer, which may (or may not) have led to a coffee with one fewer shot; the result was a pleasant enough mild coffee, but nothing amazing. They had some interesting looking donuts. (Update, I have since tried buttermilk donuts; they are an abomination.)
• HERO: Dearborn:
Curiously better than the other HERO store below. 8 ounce latte. Mildy offensive popular music.

Don’t Bother:

• Atomix Cafe:
Friendly staff; happy to substitute a 12 ounce latte for an 8 ounce latte. So it was a little sad that, after the first sip, I had to toss it into the trash. The main culprit, as usual, was over-frothed milk.
• HERO (Roscoe):
Double Espresso to go, discretely emptied on the sidewalk after the first sip.
• Wired:
Like a 00’s London cafe, and not in a good way. (In other words, before the Australians came and upped the game.) Latte fail.
• Nitecap Coffee Bar:
When the servings offered on indie passport are the “special” flavoured lattes, you know you are in trouble. After some negotiation, they agreed to make a plain latte. It went down with some difficulty (I was in polite company and there was no convenient cuspidor); overly frothed milk.
• Spoken Cafe:
This is what passes for coffee in the suburbs, I guess. Not much memory of this place besides a terrible croissant and the note “better than starbucks.” Damned with faint praise.
• Awake Cafe:
I would have preferred to be asleep. No substitutions, forced to get
a 12 ounce gingerbread latte. Went straight into the trash after the first sip. (wisely got it to go).
• Bourgeious Pig:
Is there any more to add than this? Undrinkable.
Posted in Coffee | | 1 Comment

## Serre 1: Calegari 0

I just spent a week or so trying to determine whether Serre’s conjecture about the congruence subgroup property was false for a very specific class of S-arithmetic groups. The punch line, perhaps not surprisingly, was that I had made an error. I should note that I was pretty skeptical during the entire endeavour, so the final resolution was not a surpise, but there were still a few interesting twists along the way. (Thanks to Matt for some informative chats along the way.)

Let’s start by recalling Ribet’s proof of (what is one of many statements known as) Ihara’s lemma. Let $\Gamma$ be a congruence subgroup of $\mathrm{SL}_2(\mathbf{Z})$ of level prime to q. There is a congruence subgroup $\Gamma_0(q)$ defined in the usual way, where $q|c.$ However, there is also a second copy $\Gamma^0(q)$ of this group inside $\Gamma$ with $q|b.$ (Well, there are $q+1$ copies of this group, but let’s just consider these two for the moment.) The two groups are conjugate inside $\mathrm{GL}_2(\mathbf{Q}),$ but not inside $\Gamma$. An argument of Serre now shows that the amalgam of $\Gamma$ with itself along these groups (identified by conjugation by $[q,0;0,1])$ is the congruence subgroup $\Gamma[1/q]$ of $\mathrm{SL}_2(\mathbf{Z}[1/q]).$ That is, the congruence subgroup where the local conditions away from q are the same as $\Gamma.$ The Lyndon long exact sequence associated to an amalgam of groups shows that there is an exact sequence:

$H_1(\Gamma_0(q),A) \rightarrow H_1(\Gamma,A)^2 \rightarrow H_1(\Gamma[1/q],A) \rightarrow 0,$

for any trivial coefficient system $A.$ Now the group $\Gamma[1/q]$ satisfies the congruence subgroup property, so the group on the right is easily seem to be finite and Eisenstein. By duality, there is also a map

$H_1(\Gamma,A)^2 \rightarrow H_1(\Gamma_0(q),A),$

and the composition of this map with the projection above is a matrix with determinant $T^2_q - (1+q)^2.$ A bookkeeping argument now gives Ribet’s famous level raising theorem (taking coefficients $A = \mathbf{F}_p.$)

Fred Diamond and Richard Taylor generalized this theorem by replacing the modular curve with both definite and indefinite quaternion algebras. The actual theorem itself at this point is probably quite easily to prove by the K-W method, but that’s not relevant here. Instead, let’s think a little about the proof. The more difficult and interesting case is when $\Gamma$ comes from the norm one units in an indefinite quaternion algebra, which we consider from now on (the case of Shimura curves over $\mathbf{Q}.)$ Morally, the proof should be exactly the same. The only wrikle is that the corresponding group $\Gamma[1/q]$ is notoriously not known to satisfy the congruence subgroup property, although Serre conjectures that it does. Diamond and Taylor instead argued in the following way. (Let us specialize to the case of trivial weight, which is the only relevant case here.) Suppose that p is a prime greater than two and different from q. Then instead of working with Betti cohomology, one can instead, via a comparison theorem, use de Rham cohomology. The Hodge filtration consists of two pieces, one of which is $H^0(X,\Omega^1),$ and the other is $H^1(X,\mathcal{O}_X).$ They then investigate the kernel of the map:

$H^1(X,\Omega^1)^2 \rightarrow H^1(X_0(q),\Omega^1)$

where everything is now over $\mathbf{F}_p.$ Here the two maps are the two pullbacks under the two projections $X_0(q) \rightarrow X.$ They now show that element in the kernel gives rise to a differential $\omega$ which vanishes at all the supersingular points or does not vanish at all. The first is impossible by a degree argument when $p > 3,$ and the second is always impossible. They conclude that, returning to etale cohomology, any kernel of the map

$H^1(X,\mathbf{F}_p)^2 \rightarrow H^1(X_0(q),\mathbf{F}_p)$

must lie entirely in one filtered piece, from which they deduce it must be Eisenstein. But let’s look at this argument a little more closely. Even in Ribet’s case, the conclusion is really much stronger than level raising for non-Eisenstein primes; there is a very precise description of the kernel (or cokernel in homology) in terms of the homology of $\Gamma$ coming from congruence quotients, which one can compute quite explicitly. So Ribet’s theorem also gives level raising for Eisenstein representations in some contexts. In particular, for a suitable choice of congruence subgroup (with $p > 2)$ one can make the group $H_1(\Gamma[1/q],\mathbf{F}_p)$ vanish identically. Let’s now return to the argument of Diamond and Taylor when $p = 3.$ All the comparison theorems are still valid, so the only issue is that the map

$H^1(X,\Omega^1)^2 \rightarrow H^1(X_0(q),\Omega^1)$

does have a kernel, namely, if one takes the “Hasse Invariant” $A$ which vanishes to degree one at all supersingular points, then the two pullbacks of $A$ to $X_0(q)$ coincide up to a scalar, and so the kernel is at least one dimensional. In fact, the argument of Diamond-Taylor shows that the kernel is at most one dimensional. But what does this mean in the proof of Ihara’s Lemma? It means that, assuming $X$ has good reduction at the prime $p = 3,$ the level raising map always has a kernel, and thus $H_1(\Gamma[1/q],\mathbf{F}_3)$ is always non-trivial.

This now seems suspicious: all we need to do is find a quaternion algebra which doesn’t have any congruence homology of degree $3.$ If the quaternion algebra $D/\mathbf{Q}$ is ramified at a prime $r,$ then the congruence homology coming from this prime (for $p \ne 2 r)$ is a subgroup of the norm one elements of $\mathbf{F}^{\times}_{r^2},$ which has order $r + 1.$ So it makes sense to take a quaternion algebra ramified at $7 \cdot 13,$ since these are the two smallest primes different from 3 which are congruent to 1.

Because this seemed to contradict Serre’s conjecture, I decided for fun to explicitly compute a presentation for the amalgam $\Gamma[1/2]$ to help work out what was going on. To first start, one needs a presentation for $\Gamma.$ John Voight (friend of the blog) has written a very nice magma package to do exactly this. (More precisely, it’s trivial to write down a presentation — $\Gamma$ is torsion free, and hence a surface group $\pi_1(\Sigma_g)$ for a genus $g$ one can compute via other means to be $g = 7;$ the point is that one also wants an explicit representation as well as an explicit identification with the norm one units of the correponding quaternion algebra.)

I then took an embarassingly long time to compute the subgroup $\Gamma_0(2).$ The main issue was finding a suitable element in $D$ to play the role of $\eta = [2,0;0,1]$ in $M_2(\mathbf{Q}).$ There certainly exists such a unit in $D \otimes \mathbf{Q}_2,$ so in real life one just has to find an actual norm 2 unit which is sufficiently close 2-adically to this. However, I am absolute rubbish at mathematica and so repeatedly made the following error: when you define suitable quaternions $i, j, k$ in $D \otimes_{\mathbf{Q}} E$ for some quadratic splitting field $E/\mathbf{Q},$ and then compute with the matrix $a + b i + c j + d k,$ mathematica helpfully interprets “$a$” here as $[a,a;a,a]$ rather than a multiple of the identity, a programming decision which makes a lot of sense, said no one ever. I did this more times than I care to admit. Then, using John’s program, one can find the subgroup $\Gamma_0(2),$ and then write down a presentation for the amalgam by conjugating this subgroup by $\eta$ and identifying the correpsonding elements via a solution to the word problem as words in the original generators, and then substitute the names for these generators for the second copy of $\Gamma.$ The result is a group with $14 + 14$ generators and $2 + 38$ relations (corresponding to the 2 surface relations and the fact that $\Gamma_0(2)$ has $3(14 - 2) + 2 = 38$ generators.) Finally, one takes this group, plugs it into magma, and finds:

$\mathtt{AbelianQuotientInvariants(G);}$
$\mathtt{> [168]}$

There are known congruence factors coming from $7+1$ and $13 + 1,$ but here one sees that the factor of three survives!

And then, shortly after this point, I realized that $\mathrm{SL}_2(\mathbf{F}_3)$ has a quotient of order $3,$ because it is $A_4.$ So that degree three quotient is congruence after all… Oops! Still, it’s nice to see that mathematics is consistent.

However, at this point one might just ask why can’t one replace the quaternion algebra $D/\mathbf{Q}$ by (say) a real quadratic field in which $3$ is unramified and inert. Serre got away with it above because $\mathrm{SL}_2(\mathbf{F}_3)$ is solvable, but $\mathrm{SL}_2(\mathbf{F}_9)$ has the good manners not to have any such quotients. So why can’t one now run the same argument as above and disprove Serre’s conjecture? That’s a good question, and the entire argument works, up to the issue of defining the Hasse invariant. Quaternion algebras over fields other than $\mathbf{Q}$ are a bit of a disaster, because they don’t have nice moduli theoretic descriptions. That doesn’t mean they don’t have Hasse invariants, however. But now what happens, which at this point in the game I suspected but was confirmed and explained to be by George Boxer (Keerthi also suggested a computation which would lead to the same conclusion): the Hasse invariant is no longer a section of $\Omega^1 = \omega^{\otimes 2} = \omega^{p-1},$ but rather a section of $\omega^{p^2 - 1},$ and this has too large a degree to contribute to the cohomology of $\Omega^1.$ Since $2^2 - 1 > 2,$ it still has too large a degree when $p = 2,$ which is good, because otherwise working at this prime could have given rise to a counter-example to Serre’s conjecture because $\mathrm{SL}_2(\mathbf{F}_4) = A_5$ is perfect. (One would have to be slightly more careful with p=2 about comparison theorems, but at least one is dealing with curves.) So the conclusion is that Serre’s conjecture still stands, but only because various Hasse invariants in low weight are exactly accounted for by the solvability of $\mathrm{SL}_2(\mathbf{F})$ when $|\mathbf{F}| = 2,3.$

(Also, completely randomly and apropos of nothing, this link is now the top hit on the web to the search “Fred Diamond’s Beard.”)

## Champagne Marxists

One wonders whether it is only the US in which highly educated white liberal college students from the upper reaches of the socio-economic strata support the extremely regressive Bernie Sanders-style policy of spending a trillion dollars on free education for all^* (* = all people like themselves), but I was pleased to note during my recent visit to King’s College London that even more extreme positions exist. Here is the following sign located on what appeared to be a graduate student working room in the school of social science & public policy at King’s College London. (I promise that, given the other political statements posted, that this is meant in all seriousness.)

Posted in Politics | | 7 Comments

The Tip: I received a hot tip to visit Sawada Coffee, proclaimed (by a facebook friend of my source) as “the best coffee in Chicago.” Here is a review.

First impressions: Somewhat of a skater-hipster vibe. It turned out that Sawada shares its space with Green Street Meats:

The barista didn’t know the difference between a wet and a dry cappuccino, which was not reassuring. Nor was the fact that the smallest sized serving was 8oz rather than 5 or 6 (or 5.5). Still, I ordered away, and was slightly reassured when a different barista actually made my drink, and produced the following very acceptable looking cup:

So at least the milk foaming and pouring skills seemed up to par. (The claim to fame of Sawada seems to be the green tea latte, but a cafe should stand or fall on its espresso and cappuccino.)

The taste: Hmmm. Terrible espresso shot. Probably the most accurate description was that it tasted like the coffee one gets in Germany. This is not a point in its favor.

General Ambience: Terrible music, blaring far too loudly over the bass-heavy speakers. (The sound was something like Santana meets rockabilly, but even that description makes it sound better than it was.) Even the St Matthew Passion (full volume with earbuds) wasn’t enough to mask the sound.

Verdict: Erbarm dich, mein Gott!

Posted in Food, Coffee | | 3 Comments

## fivethirtyeight.com doesn’t understand mathematics

I happened to find myself browsing fivethirtyeight.com a few weeks ago, where I came across a column known as the “riddler.” The particular problem of the week (see here) was to answer the following:

Problem: It’s Friday and that means it’s party time! A group of N people are in attendance at your shindig, some of whom are friends with each other. (Let’s assume friendship is symmetric — if person A is friends with person B, then B is friends with A.) Suppose that everyone has at least one friend at the party, and that a person is “proud” if her number of friends is strictly larger than the average number of friends that her own friends have. (A competitive lot, your guests.)

Importantly, more than one person can be proud. How large can the share of proud people at the party be?

If you take a completed graph and omit one edge, then $N - 2$ people know everyone and have $N$ friends, and the remaining two people have $N - 1$ friends. In this case, there are $N - 2$ proud people. On the other hand, at least one person has a smaller or equal number of friends than everybody else, and so they can’t have more friends than any of their other friends let alone their average number. So at least one person is not proud. Hence the real content of the question is to determine whether there can be $N - 1$ proud people. The answer to this question (which is no) is harder than coming up with the example but neither terribly difficult nor particularly interesting. The absolute shocker, however, is that the riddler’s “solution” (scroll to the bottom of the page) to the puzzle is merely to exhibit the example above with $N - 2$ proud people. There’s not any hint that an argument is required to show that $N - 1$ is not possible. I nearly choked on my cappuccino when reading this. (You could try to argue that the formulation of the problem allows for some wiggle room: one is asked to find the highest proportion of people with the indicated property, and so imaging that N is not fixed, one might claim it merely suffices to show that the limit is 1 as N goes to infinity. But I don’t buy this.) Click and Clack would never have made this mistake.

Maybe the author of the riddler was aware of this issue or maybe they weren’t. But the whole point about online media is that it doesn’t require dumbing down the message to reach the right audience. I may well agree that in this (or other) particular cases, the technical details of a correct solution may be a little annoying, but in that case it is OK as long as:

1. One provides a link with the full argument, and crucially:
2. One makes it very clear in the main text that there is something left to do to fully answer the question.

Not mentioning that there is any issue at all represents a failure in what I hoped the website fivethirtyeight.com was supposed to represent. Instead of hiring technical people who can write and training them in journalism, they appear to have simply hired journalists with a fairly mixed level of technical expertise. That’s a missed opportunity.

(I did look up the person responsible for the riddler website, and they appear not to have any scientific training. Rather, they were trained as an economist. At my institution, no less (hmmm)).