## In Brief

The start of the academic year has a habit of bringing forth distractions, not least of all to someone as disorganized as me. So here are a few remarks in brief.

The class number of $\mathbf{Q}(\zeta_{151})^{+}$ is one.

John Miller, a student of Iwaniec at Rutgers, wrote the following nice paper, which improves upon a previous result of Schoof. One technique that is useful in computing the class numbers of fields with small discriminant is to make use of the Odlyzko bounds. Here’s a typical example. If $K =\mathbf{Q}(\zeta_{37})^{+}$, then the root discriminant of $K$ is $30$ or so. However, by consulting Odlzyko, one sees that any totally real field with this root discriminant has degree at most $40$. Hence the class number of $K$ is either one or two, and it is easy to rule out the second possibility by using genus theory. More generally, whenever one has an a priori bound on $h^{+}$, one can compute $h^{+}$ by relating $h^{+}$ to the index of the circular units (Schoof did this in a previous paper.) This trick only works if the root discriminant of the totally real field is at most $60$ (or so), which seems to prevent one from applying this to real cyclotomic fields for $p > 67$. (There’s always a bound on the class group by Minkowski, but that is a terrible bound.) The idea behind this paper is that Odlyzko’s bound can be improved if one in addition knows that certain primes of small norm are principal. And since one has explicit fields, it is possible to show that the relevant ideals are principal by exhibiting explicit elements with the appropriate norm. I can’t quite tell how lucky the author was to find such elements (he searches for cyclotomic elements expressible as a small number of roots of unity), but it works! Perhaps, a postiori, it is useful that these fields do actually turn out to have class number one.

Matisse cut-outs

The NYT reports on an exhibit of Matisse cut-outs at MoMA. I have a particular soft spot for these works. I was particularly struck by a cut-out I once saw at an exhibit at the Centre Pompidou, so much so that I painted a replica (almost full sized) on the wall of my rental apartment in Cambridge:

I’m not sure if this particular cut-out is at MoMA, though.

256

Thanks to DS, I was hooked on 2048 for far too long. I eventually got bored trying to get the 16384 tile, and moved on to the more compact 256 instead. The latter game is slightly less random in that only 2s are created on each turn. The highest possible score is 7172, which is obtained when one ends up with (in any configuration) the powers of $2$ from $2$ to $512$. Recently, I finally managed to complete the game:

Notice that the $512,256,128$ tiles are not along the same edge (I think it must be theoretically possible to finish in that way, but it would be harder). Unfortunately, having reached this point, it has not cured me of my addiction. Curse you, Savitt!

Stickelberger’s Theorem.

I proved Stickelberger’s theorem in class the other day — well, with one caveat. I proved that all the ideals $\mathfrak{q}$ of prime norm are annihilated by the Stickelberger ideal. This certainly implies the result, because the class group is generated by such ideals. This follows, for example, by the Cebotarev density theorem applied to the Hilbert class field (which was my argument in class). But then I worried that this was an anachronistic argument, and indeed Stickelberger’s theorem was a solidly 19th century result. So what did Stickelberger do?

Posted in Art, Mathematics, Waffle | | 2 Comments

## iTunes top Ten

tl;dr: lots of Bach, if you’re not into that sort of thing, at least check out Mel Brooks. And if you’re not into that *either*, well then I don’t know what’s wrong with you.

Following Jordan, here is a list of my top 10 iTunes tracks by play count. In reality, it’s more like the top 10 *albums*, because most of the works are spread out over multiple tracks.

Technically speaking, my most-played “song” on iTunes is “Ambient Waterfall Sounds for Ultimate Bedtime Relaxation, Deeply Lucid Dreams,” with 188869 plays. But that’s a little bit misleading. The track basically consists of white noise, but (slightly irritatingly to me) it varies slightly in tone and pitch over the 4 minutes 44 seconds of the track. So I rigged it to run on a 2-second loop, which i play overnight when trying to sleep at conferences.

#1. Musical Offering, 296 plays [the number of plays from each track is not constant, so I will just go by the highest number for each album]. My recording (by the Ensemble Sonnerie) is arranged for Oboe, Violin, Viola, Harpsichord, Flute, and Viol. There comes a certain point in the evening where the only possible music one wants to listen to the Musical Offering. And that time is 3AM. As you can see from the play count, I am up a lot at 3AM.

#2. The French Suites, Glenn Gould 215 plays. The French Suites are easy to play and even easier to listen to.

Then again, perhaps you would prefer the French suites Mel Brooks style:

#3. The Cello Suites, Yo Yo Ma (his second recording), 207 plays. OK, this is also something to listen to at 3AM.

#4. The Art of Fugue (Juilliard String Quartet) 200 plays. They built their own custom-made viola in order to avoid having to transpose the score up a fifth (which is what the Delme string quartet do in one of my other recordings of this piece). I highly recommend this recording from 1982. Unfortunately, I couldn’t find a clip on youtube.

#5. The Art of Fugue (Glenn Gould) 196 plays. My only regret is that it is not complete, as Gould only recorded the fugues I,II,IV,IX,XI,XIII and, of course, XIV. (There’s also a version by Gould on the organ on this recording which I don’t listen to). Instead of linking to this recording, let me link instead to a fascinating interview between Gould and Bruno Monsaingeon with Gould at the piano (I’ve linked to the the video at the end of the first (fairly ordinary and early) fugue:

#6. Inventions and Sinfonias (Andras Schiff) 151 plays. There’s a common theme to this list so far, and it is Bach. This CD holds a special place for me as it was my first piano Bach recording. Glenn Gould has an even more than usually idiosyncratic recording of these, so I more often turn to Schiff on this one.

#7. Ave Verum Corpus (Byrd, King’s Singers) 150 plays. We’ve finally broken out of Bach! The intimacy of this recording (for so few voices rather than a choir) is what appeals to me.

#8. Arias “Erbarme dich, mein Gott” and “Aus Liebe will mein Heiland sterben” from St Matthew Passion, 129 plays (Michael Chance/Ann Monoyios). I don’t usually have three hours to listen to the entire recording, but these Arias are certainly some of the highlights. Here’s Michael Chance in a different recording of the same aria:

#9. Gnossiennes #1-#3 Jean-Yves Thibaudet, 121 plays. Perhaps you would like to see (actors portraying) Wittgenstein, John Maynard Keynes, and Lydia Lopokova pretend to be the solar system? I actually saw this Derek Jarman film in the theatre with Patrick Emerton…

#10. An Die Musik (Schwarzkopf) 119 plays. Whenever I sing/play lieder, I always finish by singing this.
This isn’t my favourite recording, but it’s the oldest one I had. In fact, when it comes to Schwarzkopf singer lieder, my favourite performance is her singing of Litanei auf das Fest Allerseelen here:

I should note that these numbers are all coming from iTunes on my laptop. My listening habits are somewhat different on my iPod, where I’m much more likely to listen to (say) Beethoven or Mahler. As an indication of how long a time period these numbers represents, I gave the analogous numbers on Jordan’s Blog http://quomodocumque.wordpress.com/2010/07/01/real-life-rock-top-10/ in 2010. Interpolating between these two sources, it suggests I listen (at home) to either the Art of Fugue or the Musical Offering about once a week, which seems about right.

Posted in Music | | 3 Comments

## Applying for an NSF grant

It’s not easy to write a good grant proposal. But it can be even harder to write one for the first time, especially if you’re not quite sure who will be reading your proposal. So today, I want to say a little bit about how an NSF mathematics panel is run, and give you some idea of who your target audience should be.

Before I start, I want to include a pseudo-legal disclaimer. For fairly obvious reasons, you are not supposed to reveal that you served on any particular panel. But I am allowed to say that I have served on *some* panels, and there is enough uniformity in the process to make me confident that what I say should resemble your reality if you decide to apply. (Let me also mention that I had some help on this post from a friend [whom I shall refer to as the Hawk] who is much more of an NSF pro than I am. He made various corrections and suggestions on a first draft of this blog, and I even included a few of his remarks verbatim in the text.)

The NSF administers many different types of grants. I’m not just talking about graduate fellowships, postgraduate fellowships and research grants here. There are FRG grants, RTG grants, CAREER NSF grants, REUs, conference grants, and so on. However, for the purpose of this email, I want to concentrate on research grants.

The Mechanics: The panel is comprised of approximately 10 or so mathematicians, who consider approximately 40-50 or so proposals. About six weeks before the panel takes place, each panelist is given the list of proposals and asked to rank the proposals 1,2,3,C based on the following criteria:

1 = I feel comfortable reviewing this proposal

2 = I could review this proposal if necessary

3 = It would be very difficult for me to review this proposal

C = I have a conflict of interest with this proposal

The next step is that the panel meets at the NSF headquarters in Virginia, sometime between November and March. A typical panel may last 2.5 days. The panel is chaired by the relevant program officer and three or so other NSF employees (usually professional mathematicians who have taken a leave of absence for a two year position at the NSF), so there will typically be 14-15 people in a conference room, each with either their laptop or a supplied computer. The first 1.5 days of the panel consist of going through the files one by one. For each file, the three (or so) panelists who were assigned the proposal read out their review of the proposal. During this time, other panelists (especially those with some expertise) will also offer their opinions. During this period, anyone who is conflicted with the proposal has to leave the room. At the end of each discussion (which takes about 10 minutes), a yellow sticky sheet with the PI’s name has to be placed on a white board with three columns. The columns are officially designated as “strongly recommended for funding,” “recommended for funding if possible,” and “not recommended for funding.” The desired outcome is to have 10% of proposals in the first column, 30% in the second, and 60% in the third. Within the first two columns the names are ordered, although, during the process, certain proposals can float up or down as they are re-evaluated in light of other proposals. During each discussion, a panelist who was not assigned to read the proposal is assigned to be a scribe and record the highlights of each discussion. Each panelist is a scribe on 3-4 proposals.

The final step is for each panelist to write up a summary of the panel discussions for which they were a scribe, highlighting what the panel thought were the strengths and weaknesses of the proposal, indicating “which column” the panel placed the name, and reflecting the extent to which there was uniform agreement or not. Everyone then goes over these summaries to confirm that the summary does reflect the panel discussion. If you ever apply for a grant, you will be able to read this summary, together with the evaluation of the three members who read your proposal in depth. (The panelists assigned to your proposal have an opportunity to modify their evaluations during the meeting if they change their minds in light of the discussion.)

Then the panel ends; the panel has given the program officer a (roughly) ordered set of names, and it is up to the NSF to decide whom to fund. I’m not sure the extent to which the recommendations of the panel exactly mirror the actual results, although I suspect that it is quite close. I can imagine, however, that a programme officer feels that a certain proposal suffered because nobody on the particular panel was an expert in that area, and they may decide to send that proposal off for further review. The Hawk says: The actual results can deviate significantly from the advice of the panel.  I think it’s safe to say that the ‘highly recommended’ proposals always get funded. After that, there are various other objectives that the program officers are trying to achieve — gender diversity, racial diversity, support for young PIs, support for worthy PIs at undergraduate-only institutions. The panel list is typically the default in cases where none of those other objectives apply, though you can imagine reasons to deviate from it (e.g. you might not let the same person suffer the bad luck of being the first person after the cutoff two years running, you might support a proposal in a subdiscipline that has otherwise been shut out, etc). So in the ‘recommended’ zone there are certainly some inversions. It’s also not unheard of for a ‘not recommended’ proposal to end up being funded. One way this can happen is for the proposal to be looked at by a second (perhaps more appropriate) panel that likes the proposal much better. But also, the program officers can simply decide that the panel’s conclusions about a proposal were unjust for some reason, and raise the proposal up in the rankings.

How narrow is the focus of each panel? As I mentioned, there are approximately 40-50 proposals for each panel, of which maybe 15 are funded. So take the 80 or so people who are research active and applying for grants who are closest to you mathematically, and that gives you a rough idea. If you study Galois representations and modular forms, or Iwasawa theory, or the arithmetic of Shimura varieties, or arithmetic geometry of some kind, your proposal may well end up in the same panel as mine was (it can happen — as it did to me last year — that your proposal ends up being evaluated by *two* panels — this is possibly done in order to normalize the orderings in some way. Because I wasn’t there, I can’t quite tell what the difference was between the two panels). The Hawk says: This is the first time I’ve heard a suggestion that normalization is the reason that some proposals are looked at by two panels. I think this happens either because the program officers feel that the proposal straddles two panels to such an extent that they feel both opinions could be useful; or because the proposal has two very different parts that genuinely fit in separate panels; or because the assigned panel decided that there were parts of a proposal that they didn’t have the expertise to comment on, and so they suggest getting the input of another panel. On the other hand, I’m pretty sure that my proposal would not be on the same panel as someone like Ken Ono or Soundararajan. Could my proposal be on the same panel as Akshay’s? I’m not sure. I probably would have said no if my proposal didn’t end up on two panels last time. And Akshay is a collaborator of mine! So it’s pretty focused. On the other hand, there are certainly areas in each field which are smaller than others, and if you work in such a sub-field, then it’s more likely that the panelists will not be experts in your area.

Who serves on the panel? First, there are a few NSF rules which apply to panels. (update: this information was wrong — it turns out there are no formal NSF requirements for the constitution of any panel.) Beyond this formal requirement, who is a typical member of the panel? Well, of course, one goal of the program officer is to make the panel is not *too* uniform. But, for example, I would expect that there would always be at least one person on the committee who knows as much (say) about modular forms and Galois representations as I do. So if that is what you do, then you can be pretty sure that whomever that person is will be reading your file. But you can also be sure that someone who is *not* an expert will also be reading your file, perhaps someone in Iwasawa theory, say. And this already should give you a pretty good idea of your target audience. In other words, you have to do two things:

• You have to explain to Iwasawa theory person why the modularity theorems you are going to prove are interesting. When is math interesting? Well, there are plenty of ways it can be interesting. You may have an idea of how to apply previous machinery in a novel way. You may have an interesting application in mind. You may have a completely new approach to an old theorem. You may have a completely new idea on how to solve an open problem. This is what you want to get across when you are talking to Iwasawa theory person — to give a sense of why the general problem you are studying is interesting, and how you are going to make a contribution to that field.
• You have an easier job convincing me (or equivalent) why your modularity theorems are broadly interesting, but you still have to conveince me that you particular proposal is interesting. More importantly, you have to convince me that you can carry out your proposal successfully, or at least to the point of producing interesting mathematics. The Hawk says: I think it would be worth mentioning here the fine line between saying enough about how you intend to carry out your plans that the panel is convinced you can do it, and saying so much that they think you’ve done it already. I think new proposers often struggle on this point.

Of course, if you do something other than what I do, then replace “Iwasawa theory person” above by me or equivalent and “me” by someone with expertise in your field.

What should I take away from this? First up, I think that an NSF grant proposal is probably the most technical audience you will write for in a context that is not one of your research papers. So you don’t need (beyond a cursory mention) to say how modular forms played a role in the proof of Fermat’s Last Theorem which you might do (say) in a job application. Nor do you need to define the class group of a number field, or explain what a modular curve is. But, at the same time, and this is very important, it can still be incredibly useful to place your work in a broader context. For example, on my last NSF proposal, I started out by reminding the reader briefly how there are very general conjectures linking Galois representations coming from geometry to automorphic L-functions. I reminded the reader that special degenerate cases of this conjecture correspond to very classical objects like the Riemann zeta function. I then mention how the work of Wiles addresses the case when the representation comes from the cohomology of an elliptic curve over $\mathbf{Q}.$ Then I explain how all the generalizations of Wiles’ theorem share a common assumption, namely, that the Galois representations over $\mathbf{Q}$ that one can study by this method have the property that they are, up to a twist, self-dual. So already, in perhaps not much more than a half a page, I have given the context to explain how proving that a non-self-dual Galois representation is modular is “interesting.” Of course, then I have to go on an explain *how* I am going to say anything interesting about non-self-dual representations.

Do fat cats just get their grants without trying? Every proposal is evaluated on its merits, but of course “prior success” is taken into account when judging future chances of success, and so it should be. But if Peter Scholze (say, to take someone who is not in the US so I can use his name) sends in an application consisting solely of “I am working on several projects that I decline to disclose but that I expect to be of the same importance as my prior results,” he would not be funded. More realistically, I have heard that it has been the case that fields medalists have been turned down for grants, but because all grants that are turned down are never officially acknowledged, this is just hearsay. My feeling is that, on the whole, the panels do a pretty good job, and (apart from the occasional controversial case) there is more of a uniform agreement than you might guess. The Hawk brings up the key point that this opinion only concerns number theory panels. It may be the case (and I occasionally here rumours to this effect) that other areas are not run as well. I would also say that the fat cats (on the whole) seem to put as much effort into writing their NSF proposals as everyone else.

How can I compete with the fat cats given I’m only just starting out? This is taken into account. If you are at most 6 years from your PhD, your proposal is evaluated in that context; an effort is made to fund promising young people, and also people who have never received prior NSF support. That said, it’s not easy to get a grant the first time you apply coming straight out of a postdoctoral position.

What about broader impact? This is hard for younger people. But everyone on the panel realizes this and so the expectations are lower. You probably don’t have any grad students yet, so what can you say? Perhaps you have given expository talks at a workshop? Perhaps you have written up detailed notes on otherwise hard to access topics? Perhaps you have gone into the public schools in some hardscrabble inner surburban neighbourhood and and taught calculus? (Not the last one? Then don’t suggest that you might if there’s no reason to suspect that you have any previous inclination to do so.)

Don’t Imagine that you are going to be held account for what you say you are going to prove in future proposals. Future proposals will be evaluated on their own merits (as well as prior research), and nobody is going to know or remember what you said in your previous NSF grants. It’s expected that some of problems you are working on might not work out, and that you will have new ideas while working on the proposal.

Two further suggestions from the Hawk:

When will I hear back? Answer: who the hell knows. Usually within six months from the deadline, but not always, especially these days when the federal government is funded from continuing resolution to continuing resolution. If you hear in January, either you are Peter Scholze or it’s bad news. By May, no news is good news: you probably weren’t in the ‘not recommended’ pile, and they’re waiting to see how far they can stretch the money in the ‘recommended’ pile.

If I get the grant, how much money will I get? Answer: probably less than what you asked for in your budget, and if not, you probably didn’t budget enough. Less glib answer: the program officers do adjust the award sizes in order to hit their target funding rates. You shouldn’t fret that if you ask for too much and the person who’s next on the list asks for a lower number, that could hurt your chances. The natural followup: “If program officers have that kind of discretion, wouldn’t it be better if they gave smaller awards to more people?” You can certainly argue that in theory that might be better, but in practice the answer is emphatically no. DMS’s (DMS = division of mathematical sciences at the NSF) funding rate is already much higher than that of other divisions, as high as can politically be sustained within NSF. If the funding rate went up, DMS’s budget would be cut, and the rate would go back down again.

Do you have any other thoughts? The fact that approximately 30% percent of proposals get accepted is a fairly immutable law of nature. It is no doubt depressing to be continually rejected by the NSF, and good people simply stop applying, in some sense making it then harder for everyone else. If, for some reason, the number of applications suddenly doubled, it wouldn’t be the case that the success rate would halve, but more proposals would be awarded. So, there is a real sense in which the more people who apply the more grants are awarded.

Posted in Mathematics, Politics | Tagged , | 6 Comments

## The nearly ordinary deformation ring is (usually) torsion over weight space

Let $F/{\mathbf{Q}}$ be an arbitrary number field. Let $p$ be a prime which splits completely in $F$, and consider an absolutely irreducible representation:

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

which is unramified outside finitely many primes.

If one assumes that $\rho$ is geometric, then the Fontaine–Mazur conjecture predicts that $\rho$ should be motivic, and the Langlands reciprocity conjecture predicts that $\rho$ should be automorphic. This is probably difficult, so let’s make our lives easier by adding some hypotheses. For example, let us assume that:

• A: For all $v|p$, the representation $\rho|G_{v}$ is crystalline and nearly ordinary,
• B: The residual representation ${\overline{\rho}}$ has suitably big image (Taylor–Wiles type condition.)

Proving the modularity of $\rho$ under these hypotheses is still too ambitious — it still includes even icosahedral representations and Elliptic curves over arbitrary number fields. Natural further hypotheses to make include conditions on the Hodge–Tate weights and conditions on complex conjugation.

We prove the following:

Theorem I: Assume, in addition to conditions A and B+, that

• C: The Hodge-Tate weights $[a_v, b_v]$ at each $v|p$ are sufficiently generic,
• D: If $F$ is totally real, then there exists at least one infinite place such that $\rho$ is even.

Then $\rho$ does not exist.

The condition B+ (which will be defined during the proof) is more restrictive than the usual Taylor–Wiles condition — we shall see from the proof exactly what it entails. Condition C will also be explained — but let us note that, for any suitable method of counting, almost all choices of integers are generic, even after imposing some condition on the determinant (say $a_v + b_v$ is constant) to rule out stupidities.

One should think of this theorem as follows. If $F$ is totally real, then condition D should be sufficient to rule out the existence of any automorphic $\rho$ in regular weight, because (for motivic reasons) such representations should be totally odd. On the other hand, if $F$ is not totally real, then the weights of any motive (with coefficients) should satisfy a certain non-trivial symmetry property with respect to the action of complex conjugation. So, for example, if $F$ has signature $(1,2)$, then either condition C or D should be sufficient, but we will require both. In fact, even condition C is stronger than what should be necessary. In addition to assuming regularity at all primes, it amounts to (on the representation theoretic side) insisting that none of the $\mathrm{GL}_2(\mathbf{C})$ weights are fixed by any conjugate of complex conjugation, whereas a single such example should be enough for a contradiction.

Perhaps a useful way to think about Theorem I is to make the following comparison. Hida proves the following theorem:

Theorem [Hida]: The nearly ordinary Hida family for $\mathrm{SL}(2)/F$ is finite over weight space and has positive rank if and only if $F$ is totally real and the corresponding ${\overline{\rho}}$ is odd at all infinite places.

On the other hand, a consequence of Theorem I is:

Theorem II: The fixed determinant nearly ordinary deformation ring of a residual representation ${\overline{\rho}}$ satisfying condition B+ is finite over weight space and has positive rank if and only if $F$ is totally real and the corresponding ${\overline{\rho}}$ is odd at all infinite places.

In both cases, I am only considering the deformation rings up to twist — the deformation ring of the character is torsion over the corresponding weight space whenever $\mathcal{O}_F$ has infinitely many units. Also in both cases, it is of interest to determine the exact co-dimension of the ordinary family — this is a difficult problem, because strong enough results would allow you do deduce Leopoldt by considering induced representations.

OK, so what is the argument? If you have read some of my papers, you can probably guess.

Assume that $\rho$ exists. Let $U$ be the representation corresponding to $\rho$. Now replace $U$ by $V = {\mathrm{Sym}}^2(U)$. Now replace $V$ by the tensor induction:

$\displaystyle{ W = \bigotimes_{G_F/G_{{\mathbf{Q}}}} V}$

of dimension $3^{[F:{\mathbf{Q}}]}$. We now let C be the condition that $W$ has distinct Hodge–Tate weights. To see that this is generic, it really suffices to show that there is at least one choice of weights for which this is true. But one can let the weights of $U$ up to translation consist of the 2-uples $[0,1]$, $[0,3]$, $[0,9]$, etc. and then the weights of $W$ are, again up to translation, $[0,1,2,\ldots,3^{[F:{\mathbf{Q}}]} - 1].$ We now let B+ be the condition that the residual representation is absolutely irreducible, and that the prime $p > 2 \cdot 3^{[F:{\mathbf{Q}}]} + 1$. This is generically true, and amounts to saying that the conjugates of ${\overline{\rho}}$ under $G_{{\mathbf{Q}}}$ are sufficiently distinct. Since the dimension of $W$ is odd, and because it is essentially self-dual (exercise), orthogonal (obvious), nearly ordinary (by assumption), has distinct Hodge–Tate weights (by construction), satisfies the required sign condition (automatic in odd dimension), we deduce that it is potentially modular by [BLGGT]. In order to win, it suffices to show, by a theorem I made Richard prove, that the action of complex conjugation on $W$ has trace $\pm 1.$ However, this is equivalent to condition D (see below). QED.

One can relax condition B+ slightly by only inducing down to the largest totally real subfield of $F$. On the other hand, there are plenty of examples to which Theorem II applies. I think one can take any elliptic curve $E/F$ without CM and such that $j_E \in F$ does not lie in any subfield of $F$, and then take $p$ to be any sufficiently large ordinary prime which splits completely in $F$ (caveat emptor, I didn’t check this). Of course, the condition that $p$ splits isn’t really necessary either, I guess…

The second theorem follows along the exact same lines — the conditions are strong enough to ensure, using results of Thorne, that the nearly ordinary deformation ring of (the now residual) representation $W$ is finite over weight space, which translates back into finiteness of deformations of $U$ over weight space. The result is obvious if $F$ is totally real and ${\overline{\rho}}$ is odd. Otherwise, we choose a sufficiently generic point in weight space (in the sense of C), and then, by Theorem I, we see that the specialization of the nearly ordinary deformation ring at that point must be torsion.

It remains to compute the sign of $W$. This is an exercise in finite group theory, we only recall enough of the details for our purposes. Let $V$ be a representation of $H$ of dimension $d$. Consider the tensor induction:

$\displaystyle{\bigotimes_{G/H} \sigma V}.$

Let $T$ denote a set of representatives of right cosets of $H$ in $G$. Let $t g \in T$ denote the corresponding choice for the coset $Htg$. For $g \in G$, let $n(t)$ denote the size of the $\langle g \rangle$-orbit which contains $T$. If $g = c$ has order $2$, then either $n(t) = 1$ or $n(t) = 2$ Certainly

$t c^{n(t)} t^{-1} \in H, \quad t \in T.$

Let $T_0$ be a set of representatives for the $\langle g \rangle$ orbits on $T_0$. Then (proof omitted)

$\displaystyle{\phi^{\otimes G}(c) = \prod_{t \in T_0} \phi(t c^{n(t)} t^{-1})}.$

We observe that:

1. If $n(t) = 2$, then $\phi(t c^{n(t)} t^{-1}) = \phi(t t^{-1}) = \phi(1) = d$.
2. If $n(t) = 1$, then $tct^{-1} \in H$ and $\phi(t c t^{-1})$ is what it is. For example, it is $0,\pm 1$ if and only if $V$ is ${\mathrm{GL}}$-odd with respect to $tct^{-1}$.

Now suppose that $G = G_{{\mathbf{Q}}}$ and $H = G_{F}$. The elements $tct^{-1}$ are exactly the different complex conjugations of the representations of the conjugates of $H$. We deduce:

1. If $\dim(V)$ is even, then $W$ is ${\mathrm{GL}}$-odd if and only if there exists at least one real place of $F$ such that $V$ is ${\mathrm{GL}}$-odd.
2. If $\dim(V)$ is odd, then $W$ is ${\mathrm{GL}}$-odd if and only if $F$ is totally real and $V$ is ${\mathrm{GL}}$-odd at every real place.

Equivalently, a product of even integers can equal zero only if at least one of them is zero, and a product of odd integers can equal $\pm 1$ if and only if all of them are $\pm 1$.

## Tricky Fingers

How is one supposed to play this exactly?

One can neither can play a 14th in the right hand (my hands are not that big) nor play legato parallel 10ths in the left; hence some sort of arpeggiation is required. But I can’t quite seem to reproduce how Glenn Gould plays this measure. Then again, that phenomenon is not unique to this passage.

Posted in Music | | 3 Comments

## The Artin conjecture is rubbish

Let $\rho: G_{\mathbf{Q}} \rightarrow \mathrm{GL}_N(\mathbf{C})$ be a continuous irreducible representation. Artin conjectured that the L-function $L(\rho,s)$ is analytically continues to an entire function on $\mathbf{C}$ (except for the trivial representation where the is a simple pole at one) and satisfies a functional equation of a precise shape. Langlands later had the profound insight to link this conjecture to functoriality in the Langlands program, which would additionally imply that $\rho$ is automorphic which implies, inter alia, that $L(\rho,s) = L(\pi,s)$ for a cuspidal automorphic representation $\pi$ for $\mathrm{GL}(N)(\mathbf{Q})$.

This is a beautiful and fundamental conjecture. However, it does appear to be completely useless for any actual applications. The most natural application of Artin’s conjecture is to prove … the Cebotarev density theorem. This is why Cebotarev’s density theorem is so amazing! True, one can upgrade the error estimates if one knows Artin, but to do this one also has to know GRH. And if you know GRH, you are not too far away from Artin anyway, because then $L(\rho,s)$ at worst has poles on the critical strip, and so you can (essentially) get close to optimal bounds for Cebotarev anyway.

I thought a little bit about applications of Artin’s conjecture when I wrote a paper about it, but I came up empty. Then recently, I had occasion to look at my paper again, and found to my chagrin that when Springer made the final edit they lopped off a sentence in the statement of one of the main theorems. I guess that’s why the good people at Springer get paid the big bucks. (My best ever copy editing job, by the way, was for a paper in an AMS journal.) In a different direction, I guess it also reflects the deep study of this paper by people in the field that nobody has asked me about it. However, I did notice a statement in the paper that could be improved upon, which I will mention now.

To set the context, let $K^{\mathrm{gal}}/\mathbf{Q}$ be a Galois extension with Galois group $S_5$, and suppose that complex conjugation in this group is equal to $(12)(34)$. Now suppose that $\rho$ is a representation of $\mathrm{Gal}(K^{\mathrm{gal}}/\mathbf{Q})$. We already know that $L(s,\rho)$ is meromorphic, as proved by Brauer and Artin. One thing that can be proven is that, in the particular case above, $L(s,\rho)$ is holomorphic in a strip $\mathrm{Re}(s) > 1 - c$ for some constant $c > 0$ which I described as “ineffective.” But looking at it again, I realized that it is not ineffective at all, due to a result of Stark. What one actually shows is that if $L(s,\rho)$ has a pole in the strip $\mathrm{Re}(s) > 1 - c$, then there must also be another L-function for the same field which has a zero on the real line in this interval. Note that, again from by Artin, it is trivially the case that a pole of one L-function must come from the zero of another L-function, since the product of all such L-functions is the Dedekind zeta function. So the content here is that the offending pole has to be on the real line. One consequence is that, in any particular case, one can rigorously check that the L-function in question has no such zeros, and hence (combined with other results in this paper) that $\rho$ is automorphic. With help from Andrew Booker, I was able to compute one such example (Jo Dwyer has since gone on to compute a number of other examples.) On the other hand, back to the general case, we do have effective results for zeros on the real line! The result in the paper is stated in terms of the existence of a zero of $\zeta_H(s)$ for a certain subfield $H$ of $K^{\mathrm{gal}}$ of degree twelve. (The definition of $H$ was exactly what was swallowed up by Springer, so it’s not actually defined in the paper. To define it, note that $S_5$ has a faithful representation on six points. There is a degree six extension $E$ which is the fixed field of the stabilizer of a point; then $H$ is the compositum of $E$ and the quadratic extension inside $K^{\mathrm{gal}}.$) However, the actual argument produces a zero in an Artin L-factor of $\zeta_H(s)$ which is not divisible by the Dirichlet L-function for the quadratic character of $S_5.$ Stark shows (Some Effective Cases of the Brauer-Siegel Theorem) that such an L-function does not have Siegel zeros, and also gives an explicit estimate for the largest zero on the real line. In particular, for the $L(\rho,s)$ of interest, one deduces that they are analytic on the strip $\mathrm{Re}(s) > 1 - c$ where one can take

$\displaystyle{1 - c = 1 - \frac{1}{4 \log |\Delta_H|}}.$

The result of Stark, BTW, is why one could effectively solve the class number at most $X$ problem for totally complex CM fields which were not imaginary quadratic fields before Goldfeld–Gross–Zagier.

Posted in Mathematics | | 8 Comments

## K_2(O_F) for number fields F

Belabas and Gangl have a nice paper ( Generators and relations for $K_2({\mathcal{O}}_F)$, which can be found here) where they compute $K_2({\mathcal{O}}_E)$ for a large number of quadratic fields $E$. There main result is a method for proving upper bounds for $K_2({\mathcal{O}}_E)$ in a rigorous and computationally efficient way. Tate had previously computed these groups for small imaginary quadratic fields by hand;  — the problem is finding an efficient way to do this in general. (Brownkin and Gangl had previously found a non-rigorous way of computing these groups using $K_3({\mathcal{O}}_E)$ and regulator maps, but more on that later.) A good analogy to keep in mind is the problem of computing the class groups of imaginary quadratic fields. In the latter case, however, there are rigorous ways to determine whether an element in the class group is non-trivial, and this is missing from the computation of $K_2({\mathcal{O}}_E)$.  To produce lower bounds, [BG] use theorems of Tate and Keune to relate the $p$-primary part of $K_2({\mathcal{O}}_E)$ to class groups of $E(\zeta_p)$, which they can then compute in some cases. One nice example they give is

$K_2\left({\mathbf{Z}} \displaystyle{\left[\frac{1 + \sqrt{-491}}{2} \right]} \right) = {\mathbf{Z}}/13 {\mathbf{Z}}.$

Akshay and I used this as one of the examples in our paper; in our context, it implies that the order of the group

$H_1(\Gamma_0({\mathfrak{p}}),{\mathbf{Z}})$

is always divisible by $13$ where $\Gamma_0({\mathfrak{p}})$ is the congruence subgroup of $\mathrm{PGL}_2({\mathcal{O}}_E)$ for $E = {\mathbf{Q}}(\sqrt{-491})$ and ${\mathfrak{p}}$ is any prime — even though the group $H_1(\Gamma,{\mathbf{Z}}) = ({\mathbf{Z}}/2{\mathbf{Z}})^{26}$ is not so divisible. (Because we are talking about $\mathrm{PGL}$ rather than $\mathrm{PSL}$, the cusps are quotients of tori by involutions, so only contribute $2$-torsion to $H_1.$ This group is occasionally infinite; we use the convention that $\infty$ is divisible by $13.$) It’s always nice to see a theoretical argument come to life in an actual computation — fortunately, Aurel Page was kind enough to compute a presentation for $\Gamma$ in order for us to do this. Now that I think about it, this and many other interesting things didn’t make it into the submitted version of the paper; you’ll have to read the “directors cut” to learn about it.

Alexander Rahm pointed out to me that the computation of $K_2({\mathcal{O}}_E)$ we used was annotated with an asterisk in [BG], meaning that what was proved was only an upper bound. The issue is as follows. Let $p = 13$, and let $F = E(\zeta_p)$, let $G = {\mathrm{Gal}}(F/E) = ({\mathbf{Z}}/p {\mathbf{Z}})^{\times}$, and let ${\mathrm{Cl}}(F)$ denote the class group of $F$. What is required is to show, in light of Tate’s work on $K_2$, is that

$({\mathrm{Cl}}(F)[p])^{G = \chi^{-1}} \ne 0,$

where $\chi: G \rightarrow {\mathbf{F}}^{\times}_p$ is the cyclotomic character. The problem is that $F$ has degree $24$, and it is difficult to compute class groups explicitly in such cases. Let $H = {\mathrm{Gal}}(F/{\mathbf{Q}})$, so there is a canonical decomposition $H = G \times {\mathbf{Z}}/2{\mathbf{Z}}$. There are two extensions of $\chi$ to $H$, given (with some abuse of notation) by $\chi$ and $\chi \eta$, where $\eta$ is the non-trivial character of ${\mathrm{Gal}}(F/{\mathbf{Q}})$. The main conjecture of Iwasawa Theory (Mazur–Wiles) allows one to easily compute minus parts of class groups in terms of $L$-values without actually computing with explicit number fields. However, we should not expect this to help us here. Namely, it’s not hard to show that there is an isomorphism $({\mathrm{Cl}}({\mathbf{Q}}(\zeta_p))[p])^{G = \chi^{-1}} \simeq ({\mathrm{Cl}}(F)[p])^{H = \chi^{-1}}.$ However, the former is trivial by Herbrand’s theorem, because $B_2 = 1/6$ is not divisible by $13$. That leaves us with the problem of proving that $({\mathrm{Cl}}(F)[p])^{H = \chi^{-1} \eta} \ne 0,$ which is a statement about the class group of a totally real cyclotomic extension. Since $\chi \eta^{-1}$ is an even character, we get some savings by working in the totally real subfield $F^{+}$ of degree $12$. Now $\text{\texttt{pari}}$ happily tells me via $\text{\texttt{bnfinit}}$ and $\text{\texttt{bnfclgp}}$ that the class group of this field is ${\mathbf{Z}}/13 {\mathbf{Z}}$, so it looks like we are in good shape. However, $\text{\texttt{pari}}$ has the habit when computing class groups of assuming not only GRH but something stronger. What information does $\text{\texttt{bnfinit}}$ actually contain? It certainly gives, inter alia:

1. The Galois automorphisms of $F^{+}$, using $\text{\texttt{gal:=nfisisom(nf,nf)}}$.
2. A finite index subgroup $V$ of the unit group $U:={\mathcal{O}}^{\times}_{F^{+}}$, using $\text{\texttt{bnfinit[8][5]}}$.

Let me show how, just with this data, one can prove that the relevant part of the class group we are interested in is non-zero. BTW, if you tell $\text{\texttt{pari}}$: can you confirm this answer is really correct? (using $\text{\texttt{bnfcertify}}$) it complains, and says the following:

*** bnfcertify: Warning: large Minkowski bound: certification will be VERY long.
*** bnfcertify: not enough precomputed primes, need primelimit 59644617.

A rough guess (in part) as to what it might be doing: to compute all the invariants necessary for class field theory, one needs to know the full unit group. To do this, one can take the units $V$ found so far and saturate them in the entire unit group $U$. For each prime $q$, one can do this by taking representatives in $V/q V$ and determining whether or not they are perfect $q$th powers. By taking enough primes, on either rules out the existence of such an element, or finds a candidate $v \in V$ and then checks whether it is a $q$th power. On the other hand, from $V$, one can compute a pseudo-regulator $R_V$, which is related to the actual regulator $R_U$ by the unknown index. So to make this computation finite, it suffices to have some a priori bound on the regulator (to give an upper bound on the index), which will ultimately come down to some a priori bound on an $L$-value at one, which GRH probably tells you something useful about.

One can identify the automorphims of $F^{+}$ computed by $\text{\texttt{pari}}$ with the elements of the Galois group given by the corresponding quotient of $H = G \times {\mathbf{Z}}/2{\mathbf{Z}}$ by $(-1,1)$. This group is generated by the image of $\sigma = (2,1) = \mathrm{Frob}_2$, so it is enough to find the automorphism $\sigma$ such that $\sigma \theta - \theta^2 \equiv 0 \mod 2.$ View $\chi \eta$, a character of degree $12$, as being valued in ${\mathbf{F}}^{\times}_{13}$. Now choose a random unit, say $\text{\texttt{u:=bnf[8][5][6]}}$ (Warning! I have a feeling that $\text{\texttt{bnfinit}}$ does something different each time you run it, which means you might have to tweak the choice of index $6$ above if you are doing this at home. And by “you,” I really mean “me” in six months time. I guess I should also tell myself that the relevant $\text{\texttt{pari}}$ file is $\text{\texttt{\~{}fcale/Zagier/BG491}}$.) We may write down a second unit as follows:

$\displaystyle{{\epsilon} = \prod_{i=0}^{12} (\sigma^i(u))^{\chi \eta (\sigma^i)} \in ({\mathcal{O}}^{\times}_F/{\mathcal{O}}^{\times p}_F)^{H = \chi^{-1} \eta}}$

What we have done is apply the appropriate projector in the group ring ${\mathbf{F}}_{13}[H]$ to $u$. Naturally enough, we can lift ${\epsilon}$ to an actual unit in $F^{+}$.

Now choose an auxiliary prime $q$ which splits completely in $F$, say $q = 38299$. I chose this because it actually splits completely in ${\mathbf{Q}}(\zeta_{13 \cdot 491})$, which will make a computation below slightly easier. We reduce ${\epsilon}$ modulo a prime ${\mathfrak{q}}$ above $q$ in ${\mathcal{O}}_{F^+}$ and we find that

${\epsilon}^{(q-1)/13} \not \equiv 1 \mod {\mathfrak{q}}.$

What this last computation proves is that ${\epsilon}$ actually generates
$({\mathcal{O}}^{\times}_F/{\mathcal{O}}^{\times p}_F)^{H = \chi^{-1} \eta}$, which has dimension one by Dirichlet’s theorem. Note also that the inequality above does not depend on the choice of ${\mathfrak{q}}$ — any other choice is conjugate to ${\mathfrak{q}}$ which replaces ${\epsilon}$ by $\sigma {\epsilon}$ and the latter is a non-zero scalar multiple of the former modulo $13$th powers by construction.

On the other hand, let $\zeta$ be a primitive $13 \cdot 491$th root of unity. Then we may consider the projection of $1 - \zeta$ modulo $13$th powers to the $\chi^{-1} \eta$ eigenspace (the latter is naturally also a character on $F(\zeta_{491})$). Remember this eigenspace is generated by ${\epsilon}$. Take $q = 38299$ again, so $q - 1 = 13 \cdot 491 \cdot 6$. Then $2^{6}$ is a primitive $13 \cdot 491$th root of unity modulo $q$. On the other hand,

$\displaystyle{\left(\prod_{({\mathbf{Z}}/13 \cdot 491 {\mathbf{Z}})^{\times}} (1 -2^{6n})^{n \left(\frac{n}{491}\right)} \right)^{(q-1)/13} \equiv 1 \mod q}$

(The exponent of $(1 - 2^{6n})$ is just the value of $\chi \eta(n)$ — remember that the character gets inverted in the projection formula — and that $\eta^{-1} = \eta$.) This implies that the projection of $(1 - \zeta)$ to the $\chi^{-1} \eta$-eigenspace of units modulo $p = 13$ is trivial, because the image of ${\epsilon}^{(q-1)/13}$ computed above was not $1 \mod q$. The same is trivially true for the units in ${\mathbf{Q}}(\zeta_{491})$ and ${\mathbf{Q}}(\zeta_{13})$, because the projection of any unit in a subfield of $F$ can only be an eigenvalue for a character of the correponding quotient of the Galois group. In particular, if $C$ denotes the group of circular units, we have shown that the map

$(C/13 C)^{\chi^{-1} \eta} \rightarrow ({\mathcal{O}}^{\times}_F/{\mathcal{O}}^{\times 13}_F)^{\chi^{-1} \eta}$

is the zero map. This proves that the index of the circular units in the entire units is divisible by $13$. This is enough to prove that $13$ divides $h^{+}_F$, but even better, by the Gras conjecture (also proved by Mazur–Wiles, following Greenberg) it follows that the $\chi^{-1} \eta$-part of the class group is non-zero, and hence, given the previous upper bound, this gives a proof that

$K_2({\mathcal{O}}_E) = {\mathbf{Z}}/13 {\mathbf{Z}}.$

Further Examples: Let’s now look at other examples in [BG]. Consider the following example:

$\displaystyle{K_2\left( {\mathbf{Z}} \left[ \frac{1 + \sqrt{-755}}{2} \right] \right) =^{?} {\mathbf{Z}}/41 {\mathbf{Z}} \oplus {\mathbf{Z}}/2 {\mathbf{Z}}}.$

Let $F = E(\zeta_{41})$, and let $F^{+}$ be the totally real subfield of $F$ of degree $40$.  Well we certainly won’t be able to say so much about the class group of $F^{+}$. On the other hand, we can do the latter part of the computation, namely, testing that the $\chi^{-1} \eta$-eigenspace in the circular units looks like it has index divisible by $p$ in the entire units. For example,  if $q = 123821 = 1 + 41 \cdot 755 \cdot 4$, we can compute that

$\displaystyle{\left(\prod_{({\mathbf{Z}}/41 \cdot 755 {\mathbf{Z}})^{\times}} (1 -2^{n(q-1)/(41 \cdot 755)})^{n \left(\frac{n}{755}\right)} \right)^{(q-1)/13} \equiv 1 \mod q}$

For good measure, the same congruence holds for the next seven primes which split completely in $F(\zeta_{755})$. (One also has to check that the multiplicative order of $2$ for all these primes is co-prime to $41 \cdot 755$.) But, although this is compelling numerically, it doesn’t prove anything. If ${\epsilon}$ is a generator of $({\mathcal{O}}^{\times}_F/{\mathcal{O}}^{\times p}_F)^{\chi^{-1} \eta}$, it might be the case that ${\epsilon}^{(q-1)/p} \equiv 1 \mod {\mathfrak{q}}$ for ${\mathfrak{q}}$ above the first thousand primes of norm $q \equiv 1 \mod p$.  This would simply correspond to a certain ray class group being divisible by $p$. By Cebotarev, we know that we can find some prime $q$ for which this congruence does not hold, but explicit Cebotarev bounds tend to be rubbish in practice.

If we re-think our original computation, what we really want is a “generic” unit of $F^{+}$ in order to project. Since $F$ is abelian, we actually know how to compute a finite index subgroup of the unit group, namely, by projecting (via the norm map) the group of circular units from some cyclotomic overfield. Of course, this exactly won’t be good enough to find a candidate unit ${\epsilon}$. One approach is to take our lattice $V \subseteq U = {\mathcal{O}}^{\times}_{F^+}$ and saturate it. Now we only have to saturate it at $p = 41$. In fact, we only need to saturate the $\chi^{-1} \eta$ eigenspace, which is one dimensional. That is, it suffices to show that

$e_{\chi^{-1} \eta} N_{F(\zeta_{41})/F^{+}} (1 - \zeta)$

is a $p$th power in $F^{+}$. (Before taking the norm, the element is already in $F^{+}$ up to $p$th powers, and $[F(\zeta_{41}):F^{+}]$ has order prime to $41$.) But if I ask $\text{\texttt{pari}}$ to compute the following:

$\displaystyle{\left(\prod_{({\mathbf{Z}}/41 \cdot 755 {\mathbf{Z}})^{\times}} (1 -\zeta^n)^{n \left(\frac{n}{755}\right)} \right)^{(q-1)/13} \mod \Phi_{41 \cdot 755}(\zeta)}$

it complains and conks out. Well, probably René Schoof could do this computation, but let’s think about these things a little differently.

Higher Regulators: So far, we’ve been relying on the fact that the fields $E$ we are considering are abelian, in order to be able to explicitly write down some finite index subgroup of the full unit group using circular units. But what if we want to compute $K_2({\mathcal{O}}_E)$ for non-abelian fields $E$?  For this, I want to talk about an earlier paper of Gangl with Brownkin ( Tame and wild kernels of quadratic imaginary number fields…  oh bugger, this should also be cited as [BG].) Their approach is through the study of higher regulators. Borel constructs a higher regulator map for odd $K$-groups (the even ones are trivial after tensoring with ${\mathbf{Q}}$). For imaginary quadratic fields and $K_3$, this amounts to a map

$K_3(\mathcal{O}_E) \rightarrow (2 \pi i)^2 \cdot \mathbf{R},$

where the co-volume of the image is a rational multiple of $\zeta_E(2)$. The Quillen–Lichtenbaum conjecture predicts that the covolume differs exactly from $\zeta_E(2)$ by a factor coming from the torsion in $K_3({\mathcal{O}}_E)$, which has order dividing $24$, some slightly mysterious powers of $2$ which I will ignore, and — the most relevant term for us — the order of $K_2({\mathcal{O}}_E)$. Now the Quillen–Lichtenbaum conjecture is true. So how does this help to compute anything? Well, first one has to ask how to compute $K_3({\mathcal{O}}_E)$.  As an abelian group, it is easy to compute, but this is not enough to compute the regulator map. One could give explicit classes in $\pi_3(\mathrm{BGL}({\mathcal{O}}_E)^{+})$, of course, but that may not be the most practical approach. It turns out that the group $K_3$ is computable in a natural way because of its relation to the Bloch group $B(E)$, due to theorems of Bloch and Suslin. (That is, via the Hurewicz map we get classes in $H_3(\mathrm{GL}_N({\mathcal{O}}_E),\mathbf{Z})$ which turn out to be seen by $\mathrm{GL}_2.$) To recall, the Bloch group is defined as the quotient of the pre-Bloch group:

$\displaystyle{ \sum n_i [x_i], x_i \in E^{\times}, \ \text{such that} \ \sum n_i (x_i \wedge (1 - x_i)) = 0 \in \bigwedge^2 E^{\times}}$

by the $5$-term relation

$\displaystyle{[x] - [y] + \left[\frac{y}{x} \right] - \left[ \frac{1-y}{1-x} \right] + \left[ \frac{1 - y^{-1}}{1 - x^{-1}} \right] = 0, x,y \in E^{\times} \setminus 1}.$

Now the Bloch group admits a very natural regulator map

$B(E) \rightarrow {\mathbf{R}}^{r_2}$

(where $E$ has signature $(r_1,r_2)$) given by (under the various complex embeddings) the Bloch–Wigner dilogarithm

$D(z) = \mathrm{Im} (\mathrm{Li}_2(z)) + \arg(1-z) \log |z| \in \mathbf{R}.$

Now all of this is (almost) very computable. Namely, one can replace $E^{\times}$ by the $S$-units of ${\mathcal{O}}_E$ for some (as large as you can) set $S$, compute the pre-Bloch group, then do linear algebra to find the quotient. Since (roughly) $K_3({\mathcal{O}}_E) = {\mathbf{Z}}^{r_2} \oplus T$ for an easy to understand finite group $T$ which has order dividing $24$, as soon as one has a enough indepdenent elements in the Bloch group (which can be detected by computing $D(z)$) you can compute a group $B_S(E)$ which has finite index in $B(E)$. Moreover, the dilogarithm is also easy to compute numerically, and so one can compute a regulator $D_S(E)$ coming from the Bloch group. Now this regulator map is known to be rationally the same as Bloch’s regulator map (by Suslin and Bloch). Assuming this is also true integrally, we expect there to be a formula:

$\displaystyle{\frac{3 |d_E|^{3/2}}{\pi^2 D(E)} \cdot \zeta_E(2) =^{?} K_2({\mathcal{O}}_F)},$

at least for primes $p > 3$.  (The $3$ is coming from the torsion of $K_3$, and this formula is probably only true up to powers of $2$ — this formulation above comes from Brownkin–Gangl.) For $S$ big enough, $D_S(E)$ should stabilize to $D(E)$, which gives a method of computing the order of $K_2({\mathcal{O}}_E)$. This is what Brownkin and Gangl do. There are two issues which naturally one has to worry about. The first is that it’s not known that the regulator map coming from dilogarithms is the same on the nose as Bloch’s map. However, even granting this (and it should be true), this algorithm will not certifiably end, because one can never be sure that $D_S(E) = D(E)$. If you compare this to the computation that $\text{\texttt{pari}}$ is doing with the class group, the problem is that there is no a priori bounds on the size of the corresponding regulators. Well, I guess this algorithm can sometimes end, namely, when one can be sure if the indicated upper bound for $K_2({\mathcal{O}}_E)$ matches with a known lower bound. However, we are exactly in a situation in which we are trying to prove a lower bound. For example, when $E = {\mathbf{Q}}(\sqrt{-755})$, Brownkin and Gangl predict that $|K_2({\mathcal{O}}_E)| = 2 \cdot 41$ because, for a set of larger and larger primes $S$, the index formula above stabilizes. So, beyond the issue of relating two different higher regulator maps, we have the problem of determining whether a class in $K_3({\mathcal{O}}_E)$ is divisible by a prime $p$ or not. This seems harder than our previous problem of determining whether a unit was divisible or not! (To be fair, however, it seems impossible to find units in $E(\zeta_p)$ once $E$ is non-abelian and $p$ is in any sense large.)

Chern Class Maps: We want to understand whether a class in the Bloch group $B(E)$ or in $K_3({\mathcal{O}}_E)$ is divisible by $p$ or not. Instead of working over ${\mathbf{R}}$, another approach is to work modulo a prime $q$. (It may seem a little strange to work modulo $q$ to detect divisibility by $p$, but bear with me.) Soulé constructed certain Chern class maps, which include a map:

$c_2: K_3({\mathcal{O}}_E) = K_3(E) \rightarrow H^1(E,{\mathbf{Z}}_p(2)).$

These maps are the boundary map in the Atiyah–Hirzebruch spectral sequence for étale $K$-theory. Now compose this maps with the reduction modulo $p$ map. Then, after restricting to $F = E(\zeta_p)$, we may identify ${\mathbf{Z}}_p(2)/p$ with $\mu_p$, and so, by Kummer and Hilbert 90, we get a map:

$c_2: K_3(\mathcal{O}_E)/p \rightarrow H^1(F,\mathbf{Z}_p(2)/p) \simeq H^1(F,\mathbf{Z}_p(1)/p) = F^{\times}/F^{\times p}.$

Keeping track of the various identifications, the image lands in the $\chi^{-1}$ invariant subspace, where $\chi$ is the cyclotomic character of $G = {\mathrm{Gal}}(F/E)$.

Lemma Let $p > 3$ be a prime which is totally ramified in $E(\zeta_p)/E$ and suppose that $p$ does not divide the order of $K_2({\mathcal{O}}_E)$. Then the Chern class map induces an isomorphism

$({\mathbf{Z}}/p {\mathbf{Z}})^{r_2} = K_3({\mathcal{O}}_E)/p K_3({\mathcal{O}}_E) \rightarrow ({\mathcal{O}}^{\times}_F/{\mathcal{O}}^{\times p}_F)^{\chi^{-1}}.$

That is, the image of $c_2$ in $F^{\times}/F^{\times p}$ may be taken to land in the unit group, and the ranks of all the groups are the same and equal to $r_2$, the number of complex places of the field $E.$

This lemma follows from Quillen–Lichtenbaum, but it can also be proved directly from the surjectivity of the Chern class map as proved by Soulé, the known rank of $K_3 \otimes {\mathbf{Q}}$ by Borel, and some knowledge of the torsion of $K_3$ proved by Merkuriev and Suslin.  It turns out that the hypothesis on $K_2({\mathcal{O}}_E)$ is necessary not only for the proof but for the lemma to be true.

To detect whether a class in $K_3$ is divisible by $p$, it suffices to “compute” the Chern class map above and see whether it is zero. If one ever wants to compute anything, it makes sense to work with the Bloch group $B(E)$ instead. On the other hand, it seems hopeless to give a “concrete” map:

$B(E) \rightarrow F^{\times}/F^{\times p}.$

Even though one can write down elements in the first group somewhat explicitly, it’s hard to imagine a recipe that would produce explicit elements in $F^{\times}$ with the correct Galois action.

Instead, what we do is reduce modulo $q$ for some prime $q \equiv -1 \mod p$. That is, we pass from the Bloch group over $E$ (which will be generated by $S$ units for some $S$) to the Bloch group of the field ${\mathbf{F}}_q$. The construction over ${\mathbf{F}}_q$ is just the same. By a theorem Hutchinson, this group will have order $q+1$. The numerology here is intimately related to Quillen’s result that $K_3({\mathbf{F}}_q) = {\mathbf{Z}}/(q^2 - 1){\mathbf{Z}}$. Now there are some commutative diagrams one has to check commute here; I think the key point to keep in mind is that Quillen’s computation of $K_3({\mathbf{F}}_q)$ can already be realized in the cohomology group $H^3({\mathrm{SL}}_2({\mathbf{F}}_q),{\mathbf{Z}})$, and so the map of Bloch groups will be the same as the map on $K$-groups via comparison with the Hurewicz map.

Let’s choose a prime $q \equiv -1 \mod p$ which splits completely in $E(\zeta_p + \zeta^{-1}_p) \subset F = E(\zeta_p)$. So we have a map $B(E) \rightarrow B({\mathcal{O}}_E/{\mathfrak{q}}) \otimes {\mathbf{F}}_p = B({\mathbf{F}}_q) \otimes {\mathbf{F}}_p = {\mathbf{F}}_p.$ The Bloch group can be thought of in terms of (a quotient of a subgroup of) the free abelian group of elements of $\mathbf{P}^1(E)$, so there’s no issue about this reduction map. Moreover, given an element of the Bloch group, we can explicitly compute its image in the latter group. If this image is non-zero, that gives a certificate that the original element is not divisible by $p$. This will be enough to compute $K_2({\mathcal{O}}_E)$ as long as the Bloch regulator map agrees with the dilogarithm map.

This argument is still yoked to real regular maps. Let’s try to work entirely with $c_2$ and finite auxiliary primes $q \equiv - 1 \mod p$. Another manifestation of the map $B(E) \rightarrow B({\mathbf{F}}_q) \otimes {\mathbf{F}}_p$ is the map:

$c_2: K_3({\mathcal{O}}_E) \rightarrow {\mathcal{O}}^{\times}_F/{\mathcal{O}}^{\times p}_F \rightarrow ({\mathcal{O}}_F/{\mathfrak{Q}})^{\times} \otimes {\mathbf{F}}_p = {\mathbf{F}}_p,$

where ${\mathfrak{Q}}$ is a prime above ${\mathfrak{q}}$ in $F$.  Let’s go back to considering the case when $E$ is an imaginary quadratic field. The image of a generator of $K_3({\mathcal{O}}_E)$ will map exactly to a non-zero multiple of the non-trivial element unit ${\epsilon} \in F^{\times}/F^{\times p}$. If $K_2({\mathcal{O}}_E)$ is prime to $p$, it will even land in $({\mathcal{O}}^{\times}_F/{\mathcal{O}}^{\times p}_F)^{\chi^{-1}}$. The latter map is exactly computing (up to a non-zero scalar) ${\epsilon}^{(q-1)/p} \mod {\mathfrak{q}},$ and so, purely using the Bloch group, we can check whether this is trivial or not. In particular, given an element of the Bloch group $B(E)$ which (we think) is a generator, or at least not divisible by $p$, we can find a prime $q \equiv -1 \mod p$ such that the reduction to $B({\mathbf{F}}_q) \otimes {\mathbf{F}}_p$ is non-zero, which will imply that the image of $c_2$ is non-zero, which will imply that

${\epsilon}^{(q-1)/p} \not\equiv 1 \mod p.$

This gives an explicit value of $q$ for which this is true without ever having to compute ${\epsilon}$. For such a prime $q$, we can then check that the circular units project to the identity in this space, which will prove unconditionally that $K_2({\mathcal{O}}_E)$ is divisible by $p$. (Part of this computation assumed that $p$ did not divide $K_2({\mathcal{O}}_E)$, but that’s OK, because to prove that $p$ does divide this group we are allowed make that assumption anyway). Back to our example. We now want a prime $q \equiv -1 \mod 41$, which is also a square modulo $755$. We take $q = 163$. Now this is not the most attractive computation in the world, because the root of unity $\zeta$ of order $37 \cdot 755$ cuts out the extension ${\mathbf{F}}_{q^{300}}$, as we can see by computing the multiplicative order of $q = 163$ modulo $41 \cdot 5 \cdot 151$. Let’s do it in baby steps. By choosing a suitable prime ${\mathfrak{Q}}$ in $E(\zeta_{755})$, we can ensure that

$\zeta^{755} + \zeta^{-755} = \zeta_{41} + \zeta^{-1}_{41} \equiv 4 \mod {\mathfrak{Q}}.$

We write

$\zeta^{1510} - 4 \zeta^{755} + 1 = F(\zeta) G(\zeta) \mod 163,$

where $F(\zeta)$ is any of the four factors of degree $300$ (there are also two factors of degree $150$, and factors of degrees $2$, $4$, and $4$.) Now we want to compute, with $p = 41$, $q = 163$, and $r = 755$,

$\displaystyle{\eta:= \left(\prod_{({\mathbf{Z}}/p r {\mathbf{Z}})^{\times}} (1 -\zeta^n)^{n \left(\frac{n}{r}\right)} \right)^{(q^{300}-1)/p} \mod \mathfrak{Q} = (163,F(\zeta))}$

Of course, one should first reduce the exponents $\chi^{-1} \eta(n) = n (n/r)$ modulo $p = 41$ before taking the powers. (Actually, it’s probably kind of stupid to take a product over $\varphi(pr) = 24000$ different terms, and one can surely set this up much more effeciently, but whatever.) We find (drum roll) that:

$\eta \equiv 1 \mod 163.$

To finish, we have to take an element in the Bloch group $B({\mathcal{O}}_E)$ and show that it doesn’t vanish in $B({\mathcal{O}}_E/{\mathfrak{q}}) \otimes {\mathbf{F}}_p = B({\mathbf{F}}_{163}) \otimes {\mathbf{F}}_{41}$. At this point, I email Herbert (Gangl), and he sends me an email with the following beautiful element of $B(E)$, where $\alpha^2 = - 755$:

$\displaystyle{-8 \left[\frac{3 - \alpha}{10}\right] - 10 \left[\frac{7-\alpha}{10} \right] - 8 \left[\frac{3 - \alpha}{100} \right] + \ldots + 6 \left[\frac{7 \alpha + 221}{972} \right]}.$

(There are $114$ terms in all! This should be a generator of $B(E).$) Into my $\text{\texttt{magma}}$ programme it goes, which cheerily reports that the image of this element is non-zero in $B({\mathbf{F}}_{163}) \otimes \mathbf{F}_{41}$! So $K_2$ is really divisible by $41$. (You might question the veracity of my programme’s output, but more on that below.)

Stark and Beyond: Here are some more general remarks. Let’s still suppose that $E$ is imaginary quadratic. Take the image of a generator $[M]$ of $B(E)$, which is defined up to torsion and up to sign. The image of the Chern class map for some $p > 3$ and $p$ not dividing $K_2({\mathcal{O}}_E)$ gives a canonical unit in ${\mathcal{O}}^{\times}_F/{\mathcal{O}}^{\times p}_F$, where $F = E(\zeta_p)$. Let me a be a bit more careful here: by writing $F$ as $F = E(\zeta_p)$, we are choosing a root of unity (this unit depends on this choice). There’s also an automorphism of ${\mathrm{Gal}}(E/{\mathbf{Q}})$ which acts, but this changes the sign of $[M]$, so that is the same ambiguity we had before. What is this canonical unit? It is not just a circular unit, but a canonical one (modulo $p$th powers). What is it? More generally, when $r_2 = 1$, both $K_3$ and $(\mathcal{O}^{\times}_F/\mathcal{O}^{\times p}_F)^{\chi^{-1}}$ have rank $r_2 = 1$, so if $p$ is prime to $K_2(\mathcal{O}_E)$ we are generating canonical units. It’s tempting here to conjecture some relation to Stark units here, and in particular to the special value of $L(1,E,\chi^{-1})$, but let me say no more about this. When $r_2 > 1$, one is no longer in the Stark world, but there is still a canonical map from the Bloch group to the unit group (the group $\mathbf{Z}^{r_2}$ has no canonical generator when $r_2 > 1$ — but in the manifestation of this group as a Bloch group, one does have explicit elements.)

Actually, I haven’t even explained how to compute $c_2$. So far, I have only explained how to compute whether it is zero or not modulo $p$. To evaluate it exactly requires a further threading of the needle through the previous maps (on the Bloch group), and ultimately uses a test element coming from torsion in $B({\mathbf{Q}}(\zeta_p + \zeta^{-1}_p))$. Although this is somewhat delicate, and I have not yet proved all of the appropriate diagrams commute (blech), one can work with it in practice and it gives many consistency checks on all the computations. (So, for example, once one has the image of $c_2$, one can compute the reduction of the corresponding element in the Bloch group in $B({\mathbf{F}}_{q}) \otimes {\mathbf{F}}_p$ for one prime $q \equiv -1 \mod p$ knowing its image in the corresponding group for another such prime. Generating the same element of $\mathbf{F}_p$ for $p = 13$ and twenty different primes $q$ is pretty convincing.)

In fact, computing this map exactly is exactly the problem that I was thinking about in the first place. I did compute it explicitly for $K = {\mathbf{Q}}(\sqrt{-491})$ and $p = 13$ (and also $K = {\mathbf{Q}}(\sqrt{-571})$ for $p = 5$), and the image of a generator of the Bloch group is not a unit. Instead, it gives a generator of $\frak{a}^{13}$ for a non-trivial ideal in the class group ${\mathrm{Cl}}(F)$ of $F = E(\zeta_p)$, indeed, an element of ${\mathrm{Cl}}(F)[p]^{\chi^{-1} \eta}$. (In particular, it gives, having fixed a root of unity, a canonical element of this class group, which is also somewhat mysterious.) Let me also mention the Coates–Sinnott conjecture, higher Stickelberger elements, and work of Banaszak and Popescu which are closely related to the topics in this post (in particular, using Chern class maps to construct Euler systems generalizing the circular unit Euler system, although not so much question of identifying these elements in some explicit way — especially because much less is known about higher analogues of the Bloch group). But perhaps this is enough for now.