## 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 cappuccino, Coffee, Food | | 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)).

## Prime divisors of polynomials

A heuristic model from the last post suggests that the “expected” order of the Galois group associated to a weight one modular form of projective type $A_5$ is infinite. And when one tries to solve the inverse Galois problem for central extensions of this group, one is lead to problems concerning the prime divisors of polynomials and their properties modulo 2. But I don’t know how to answer this type of problems! Here is an analogous question that seems a little tricky to me:

Problem: Show that there are infinitely many integers $n$ such that all the odd prime divisors of $n^2 + 1$ are of the form 5 modulo 8.

To make the problem slightly easier, one can replace (5 modulo 8) by not (1 modulo 2^m) for any fixed m.

Is this an open problem?

Posted in Mathematics | | 4 Comments

## Central Extensions and Weight One Forms

As mentioned in the comments to the last post, Kevin Buzzard and Alan Lauder have made an extensive computation of weight one modular forms in characteristic zero (see also here). Thinking about what that data might contain, I wondered about the following question: what are the images of the Galois representations associated to the weight one forms of type $A_5$?

Let us take a step back. Consider a projective representation

$\psi: G_{{\mathbf{Q}}} \rightarrow {\mathrm{PGL}}_2({\mathbf{C}})$

with image $A_5$, and assume that it is odd. (That is, complex conjugation has order $2.$) According to Tate, there exists a lift

$\rho: G_{{\mathbf{Q}}} \rightarrow {\mathrm{GL}}_2({\mathbf{C}}).$

This lift is unique up to twisting. Since the Schur multiplier $H_2(A_5,{\mathbf{Z}})$ of $A_5$ is ${\mathbf{Z}}/2{\mathbf{Z}}$, there is a unique minimal lift up to twist whose image is a central extension ${\widetilde{A}_5}$ by a cyclic group $\Delta$ of $2$-power order. Note that $\Delta$ is not trivial, since $A_5$ does not have any two-dimensional representations. If $|\Delta| = 2$, then the determinant of the corresponding 2-dimensional representation of ${\widetilde{A}_5}$ is trivial, which contradicts the assumption that $\psi$ is odd. (Equivalently, there is an obstruction at $\infty$ to lifting to the central extension by ${\mathbf{Z}}/2{\mathbf{Z}}.$) Hence $4$ divides $|\Delta|.$ What is the expected distribution of $\Delta$ as one runs over all odd $A_5$-extensions?

My first guess (without any prior thought or computation) was that this might obey some form of Cohen–Lenstra heuristic, suitably interpreted.

Note that the image of the determinant has order $|\Delta|/2.$ The corresponding determinant representation is a character of ${\mathbf{Q}}$ of $2$-power order. Since ${\mathbf{Q}}$ has trivial class number, the order $|\Delta|/2$ is equal to the maximal ramification degree $e_p$ of this representation over all primes $p.$

Over ${\mathbf{Q}}$, Tate’s lifting theorem has the following stronger form: one may choose a lift $\rho_p$ of $\psi_p := \psi|_{D_p}$ and insist that $\rho|_{I_p} = \rho_p | I_p$; that is, they agree on inertia. This is essentially a consequence of the fact that ${\mathbf{Q}}$ has trivial class group. For convenience, suppose that $\psi$ is unramified at $2$ and $3.$ Suppose that $\psi$ is ramified at $p.$ There are three possibilities:

1. The image of $\psi_p$ at a ramified prime $p$ is cyclic of order 2, 3, or 5.
2. The image of $\psi_p$ at a ramified prime $p$ is $S_3.$
3. The image of $\psi_p$ at a ramified prime $p$ is ${\mathbf{Z}}/2{\mathbf{Z}} \oplus {\mathbf{Z}}/2{\mathbf{Z}}.$

For a fixed $p$, let $\epsilon$ denote the Teichmuller lift of the mod-$p$ cyclotomic character. (Fix an isomorphism of $\mathbf{C}$ with $\overline{\mathbf{Q}}_p$ for all $p.$)

Let us consider the three cases in turn.

In the first case, the image factors through a cyclic quotient. One may thus take $\rho_p$ to be a direct sum which, on inertia, has the shape $\chi \oplus 1$ up to twist. By comparing this to the projective representation, we see that $\chi$ has order 2, 3, or 5, and so, after finding the twist such that the determinant has $2$-power order, we see that $e_p = 1$ or $e_p = 2.$

In the second case, the lift on inertia is (up to twist) of the form:

$\omega^{r}_2 \oplus \omega^{pr}_2$

for some $r.$ Since the order of $\omega_2$ is $p^2 - 1$, the order of the ratio is

$\displaystyle{\frac{p+1}{\gcd(r,p+1)}}.$

which must be equal to 3. It follows that $r$ is even. Yet the determinant is equal to

$\displaystyle{ \omega^{(p+1)r}_2 = \epsilon^{r}},$

Since $r$ is even, we see that, after twisting, we may take $e_p = 1.$

Finally, in the third case, the lift is of the form:

$\omega^{r}_2 \oplus \omega^{pr}_2$

for some $r.$ We now find that the order of the ratio of these characters is

$\displaystyle{\frac{p+1}{\gcd(r,p+1)}}.$

which must be equal to 2, and the determinant is $\epsilon^r.$ If $r$ is even, then, as above, we may twist so that $e_p = 1.$ Hence, the only way that the image after minimal twist does not have $|\Delta| = 4$ is if we are in this third situation with $p \equiv 1 \mod 4$, with $r$ odd, and then (after twisting) we find that $e_p$ is the largest power of $2$ dividing $p - 1.$

(I confess that I originally forgot the fact that the third possibility could occur, and was only after noticing that this seemed to imply the inverse Galois problem was false thought a little bit more about the possibilities.)

To summarize:

Lemma Assume that $\psi$ is unramified at $2$ and $3$ and has projective image $A_5$, and a lift with image $\widetilde{A}_5$ with minimal kernel. Then order of $\Delta$ is $4$ unless there exists a prime $p \equiv 1 \mod 4$ such that the image of the decomposition group at $p$ under $\psi$ is ${\mathbf{Z}}/2{\mathbf{Z}} \oplus {\mathbf{Z}}/2 {\mathbf{Z}}.$ In this case, we have $\Delta$ to be twice the largest power of $2$ dividing $p -1$ for all such primes $p.$

Let $\Delta(\psi)$ denote the corresponding power of $2.$

We see that $\Delta(\psi)$ is determined by purely local phenomena. This still doesn’t quite answer what the distribution of the extension $\widetilde{A}_5$ will be. However, I imagine that Bhargava style heuristics should certainly be able to predict the ratio of $A_5$ with $\Delta(\psi) = 2^n.$ Does anyone have a sense of how easy this might be to prove? Or, much more modestly, how easy it would be to compute from these quite precise heuristics the exact predicted distribution of the central extensions of $A_5$ coming from weight one modular forms?

(I confess, it is not even obvious to me from this construction how to prove that all central extensions $\widetilde{A}_5$ occur as Galois groups — but I presume this is known, and hopefully one of my readers can provide a reference.)

(According to KB, BTW, all the $A_5$ representations with $N \le 1500$ have $\Delta = 4.$)

## LMFDB!?

The LMFDB has gone live!

I previously expressed on this blog a somewhat muted opinion about certain aspects of the website’s functionality, and it seems that my complaints have mainly been addressed in the latest version. On the other hand, this live version has been released with a certain amount of fanfare, not to mention press releases. (Is AIM involved? yes it is.) First of all, there’s something about press releases in science (or mathematics) which I find deeply troubling. What is the point of such a press release? To drum up future funding? To generate mainstream press articles and thus communicate a sense of wonder and amazement to the public, who rarely get to glimpse the excitement of modern mathematics in action? Hopefully it’s the later which is true. But if so, does it really require that we stretch the truth about what we are doing in order to generate such excitement? (To see that the answer to the latter question is no, one need only look back on Scientific American articles on mathematics and physics from the ‘60s and ‘70s.) To be fair, I should also link to a more modest description of the project here. But then again, I dare you to click on the following website.

But back to the topic at hand. I have asked the opinion of at least three mathematicians about either the LMFDB or on computational aspects of the Langlands Programme more generally. For reasons of anonymity, I will not mention their names here. The first comment addresses a widely circulated quote from John Voight in the press release: “Our project is akin to the first periodic table of elements.” One source offered the following take on this (literally copied from my email and modified only by adjusting the spelling of the Langlands Programme to the preferred Canadian spelling):

The periodic table was a fantastic synthesis of decades, maybe even centuries, of empirical investigation, that led to profound new theoretical insights into chemistry and the physical world. The LMFDB is a rag-tag assortment of empirical facts in the Langlands Programme which lag far behind the theoretical advances of the past decade.

A second source, speaking more broadly on the question of computations in the Langlands programme, wondered if there had been any fundamental discoveries made via numerical computations since the BSD conjecture. While these are certainly pretty strong statements, I feel comfortable agreeing at least that, in our field, the theory is way in advance of the computations. We can prove potential modularity theorems for self-dual representations of any dimension, and yet it’s barely possible to compute even weight zero forms for U(3) (David Loeffler did some computations once). In part, I think this actually provides justification for effort into understanding how to actually compute these objects. After all, a really nice aspect of number theory is having a collection of beautiful concrete examples, from X_0(11) to Q(sqrt(-23)) (take that, Geometric Langlands!). Yet these statements also provide an alternate framework with which to view the LMFDB, which is less glorious than the press releases suggest. To continue with the periodic table analogy, the LMDFB is less a construction of a periodic table, but more a collection of sample elements from the periodic table neatly contained in small glass vials. Let me make the following point clear: some of the samples took a great deal of effort and ingenuity to extract. But it’s not entirely obvious to me how easy it will be to actually use the data to do fundamental new mathematics. This was the main point of my third commentator: the problem is that, if one actually wants to undertake a fairly serious computation, the complete set of data one has to compute will either not be available on the LMFDB (however big the LMDFB grows), or not available in a format which is at all practical to extract for actually doing computations. So, in the end, if you need some serious data, you are probably going to have to compute it again yourself. In some cases you may be able to do this, and in other cases not.

This leads to probably the most frustrating thing about the website from my perspective. I thought quite a bit about what the most useful format for some of the data might be (for me). Here is a typical thing that I might want to do: find a Hecke eigenform of some particular weight and level, and then compute information about the mod-p representations for various primes p (as well as congruences between forms). This is a little tricky to do at the moment, in part because the data for the coefficients is given in terms of a primitive element in the Hecke field (often the eigenvalue of T_2), and then the order generated by this element has (almost always) huge index (divisible by many small primes) inside the ring of integers, which makes computing the reductions slightly painful. There are certainly ways to address this specific problem and incorporate such functionality into the website. But it ultimately would be silly to customize the LMFDB for my particular needs. Instead, in the end, what I think would actually be most useful would be if the webpages on modular forms included enough pari/gp/sage/magma code to allow me to go away and compute the q-expansion myself. This is why I think that, even within computational number theory, the impact of programs like pari/sage/mamga will be far greater than the LMFDB.

If I think of the three most serious computations I have been involved with recently, they include partial weight one Hilbert modular forms, non-liftable classes of low weight Siegel modular forms, and Artin L-functions of S_5 extensions. The first required customized programs in magma (some written in part by John Voight), and the tables of higher weight HMFs in the LMFDB would not have been of any use. The computations of low weight Siegel modular forms in finite characteristic, which are absolutely terrifying, require completely custom computations (which, by the way, are completely beyond my capability of doing and involve getting Cris Poor and David Yuen to do them). Finally, the computation which would be the closest to an off the shelf computation was proving that the Artin L function $L(\rho_5,s)$ of some $S_5$ extension provably had no zero in $[0,1].$ However, the LMFDB tables only go up to degree four representations. Fortunately I knew that Andy Booker was an expert in this sort of thing and he did it for me. (Then again, even if the data in this case was included, it’s totally unclear to me the extent to which the data in the tables has been “proved.”) And my point here is not to complain about the lack of certain computations being in the LMFDB, just to caution against the idea that its existence will be particularly useful for future computations.

The bigger point, however, is surely this: Is hype in science or mathematics a necessary evil to generate public enthusiasm, or is it an ultimately corrosive influence?

Posted in Mathematics, Rant | | 31 Comments