The class number 100 problem

Some time ago, Mark Watkins busted open the “class number n” problem for smallish n, finding all imaginary quadratic fields of class number at most 100 (the original paper is here) Although the paper describes the method in detail, it does not actually give the complete list of imaginary quadratic fields which occur (for fairly obvious reasons given the size of the list). I’ve occasionally wanted to consult the actual list, and most of the time I have just emailed Mark to find out the answer. But now it is available online! Here is the link. (Maybe someone could put this on the LMFDB?)

Consulting the table one immediately notices a number of beautiful facts, such as the fact that (Z/3Z)^3 does not occur as a class group. (Our knowledge of p-parts of class groups, following Gauss, Pierce, Helfgott, Venkatesh, and Ellenberg, is enough to show that (Z/2Z)^n and (Z/3Z)^n for varying n only occur finitely often [similarly these groups plus any fixed group A], but those results are not effective.) One also sees that D = – 5519 and D = -1842523 are the first and last IQF discriminants of class number 97. It’s the type of table that immediately bubbles up interesting questions which one can at least try to give heuristic guesstimates. For example, let mu(A) denote the number of imaginary quadratic fields with class group A. Can one give a plausible guess for the rough size of mu(A)? One roughly wants to combine the Cohen-Lenstra heuristics with the estimate h \sim \Delta^{1/2}. To do this, I guess one would roughly want to have an estimate for

\displaystyle{\sum_{x^{1/2 - \epsilon} < |A| \le x^{1/2 + \epsilon}} \frac{1}{|\mathrm{Aut}(A)|}}.

I wouldn’t be surprised if someone has already carried out this analysis (thought I don’t know any reference). As a specific example, what is the expected growth rate of mu(Z/qZ) over primes q? A related question: is there a finitely generated abelian group which provably does not occur as the first homology of a congruence arithmetic hyperbolic 3-manifold?

At any rate, this is a result that Gauss would have appreciated. Curiously enough, this paper was recently posted as an answer on to the (typically ridiculous as usual) MO question Which results from the last 30 years, in any area of mathematics, do you think are the most important ones? While I wouldn’t quite put it in that class, I do find it curious that it this answer (at the time of writing) has -4 votes on mathoverflow. Given the enormous crap that does receive positive votes, I suppose that such minus votes are not to be taken too seriously. I would, however, make the following claim: Watkins’ result is as least as interesting as any original number theory research that has appeared on MO (at least as far as anything I have seen).

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Central Extensions, Updated

I previously mentioned a problem concerning polynomials, whose motivation came from thinking about weight one forms and the inverse Galois problem for finite subgroups of \mathrm{GL}_2(\mathbf{C}). I still like the polynomial problem, but I realized that I was confused about the intended application. Namely, given a weight one form with projective image A_5, there is certainly a unique minimal lift up to twist, but the images of the twists also automatically have image given by a central extension A_5. So, just by twisting, one can generate all such groups as Galois groups by starting with a minimal lift. More prosaically, every central extension of A_5 by a cyclic group is either a quotient of A_5 \times \mathbf{Z} or of \widetilde{A}_5 \times \mathbf{Z} where \widetilde{A}_5 is the Darstellungsgruppe of A_5 (which is \mathrm{SL}_2(\mathbf{F}_5)). So, to solve the inverse Galois problem for central extensions of A_5, it suffices to solve it for \mathrm{SL}_2(\mathbf{F}_5). That is not entirely trivial, but it is true.

I still think it’s an interesting problem to determine which extensions of A_5 by cyclic groups occur as the Galois groups of minimally ramified up to twist extensions, but that is not the same as the inverse Galois problem.

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Correspondance Serre-Tate, Part I

Reading the correspondence between Serre and Tate has been as delightful as one could expect. What is very nice to see — although perhaps not so surprising — is the utter delight that both Serre and Tate find in discussing numerical examples. One of the beautiful aspects of number theory is that there is an abundance of examples, each of which exhibit both special cases of a vast general theory and yet each delighting with their own idiosyncracies: \mathbf{Q}(\sqrt{-23}), X_0(11), 691, 144169, etc. (It is precisely the absence of such examples, or at least any discussion of them, why geometric Langlands tends to leave me completely cold.) Take, for example, the following:

Letter from Tate to Serre, Dec 8, 1958:

Are you aware that the class number of the field of 97th  roots of 1 is divisible by 3457 and 118982593? And that 3457 = 36 * 96 + 1 and 118982593 = 1239402 * 96 + 1?

If reading that doesn’t give you just a little thrill, then you have no soul. Does it have any significance mathematically? The class number is large, of course, which relates to the fact (proved by Odlyzko) that there are only finitely many Galois CM fields with bounded class number. (The reason why one can access class numbers of CM fields F/F+ is that the unit group of F and F^+ are the same up to finite index, so the *ratio* of zeta values \zeta_{F}(1)/\zeta_{F^+}(1) is directly related to the minus part of the class group h^{-} uncoupled from any regulator term, so one can access this analytically.) Alternatively, one might be interested in the congruences of the primes q dividing the class number. In this case, we see a reflection of the conjectures of Cohen and Lenstra. Namely, we expect that there is a strong preference for the class group to be “more cyclic,” especially for larger primes. The class group also has an action of (\mathbf{Z}/97\mathbf{Z})^* which is cyclic of order 96. Since one expects the plus part h^{+} to be very small (and indeed in this case it is trivial), this means that complex conjugation should act non-trivially, which means that the group of order 96 should (at least) act through a quotient of order at least 32. So if the class group is actually cyclic, this forces the prime divisors q of h_F to be 1 mod 32, and even 1 mod 96 if the class group of F doesn’t secretly come from the degree 32 subfield of F (which it doesn’t). (Not entirely irrelevant is Rene Schoof’s nice paper on computing class groups of real cyclotomic fields.)

Both Serre and Tate are unfailingly polite to each other. As a running joke, the expression “talking through one’s hat” occurs frequently, as for example the letter of Nov 14, 1961, where the subtle issue of the failure of B \otimes_A C \rightarrow B \widehat{\otimes}_A C is discussed. (Another amusing snippet from that letter “Even G. himself makes mistakes when he thinks causally.”) The correspondence is also fascinating from the perspective of mathematical history — one sees the progress of many ideas as they are created, including the Honda-Tate theorem and the Tate conjecture over finite fields. The first time the latter appears (as a very special case) it actually turns out to be an argument of Mumford, who shows Tate an argument (using Deuring) why when two elliptic curves have the same zeta function they are isogenous. This elicits the following reaction from Tate:

Letter from Tate to Serre, May 9, 1962:

“Damn! The result is certainly new to me, and it frankly makes me mad that I never noticed it”

We have all been there, although, to be fair, most of us have the excuse of not being Tate!

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Today is Persiflage’s birthday, but let us not also forget to wish many happy returns to friends of the blog Ana and Vytas, who share the same birthday!

Here is how I plan to celebrate: I will spare you with the complete 12 hour 23 minute breakdown of my carefully curated iTunes play list, but suffice to say it is both suitably seasonal and indulgent. A few highlights: Spem in Alium, Lupu and Perahia playing Schubert’s Fantasia in F minor, Bach Cello Suites, A Musical Offering, and plenty of carols from King’s College Cambridge. There will be a trip to Intelligentsia (by Uber —  it is -17 C outside) and then a wander around the Chicago Art Institute. For lunch, smoked salmon and smoked eel (hat tip to Bao who pointed out the existence of a fine food store in duty free at Schiphol airport!) with Champagne, followed by left-over home made Coq au vin with orzo. For dinner, Foie Gras mi-cuit with a side of poached apples, together with Sauternes. And then maybe some vegetable and tofu stir fry to balance things out slightly, in order to justify finishing off the evening with some delicious moist fruit cake. The final indulgence: I plan to spend the afternoon at home drinking tea, reading the Serre-Tate correspondence, and looking out my window onto the lake (which right now has a beautiful cover of fog which my poor photography isn’t quite able to capture):

Fog on Lake

All of this, of course, with the best possible company imaginable for a quiet day of self-indulgence. Happy holidays!

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Artin No-Go Lemma

The problem of constructing Galois representations associated to Maass forms with eigenvalue 1/4 is, by now, a fairly notorious problem. The only known strategy, first explained by Carayol, is to first transfer the representation to a unitary group over an imaginary quadrtic field, where one can realize the corresponding transfer in the coherent cohomology of a related “Griffiths-Schmidt” variety X. Then one hopes to study the action of Hecke operators on this space and relate it to some (hopefully existing) rational structure on the cohomology. The wrinkle is that X is not algebraic but merely a complex manifold, so it’s not so easy to see how to impose any rational structure on the (higher) coherent cohomology. I have nothing intelligent to say about whether this approach will work. However, suppose one is as optimistic as possible, and thinks about what one might *hope* to be true — not only to construct Galois representations but also prove the converse (Artin). Then, following [CG], one might hope to find an *integral* structure on this cohomology (with interesting torsion) on which to study congruences and then glue together torsion classes using Taylor-Wiles to produce a patched complex of the right length. What is the invariant l_0 in this case? One might (generally) hope in this context to study conjugate self-dual representations

\rho: G_E \rightarrow \mathrm{GL}_3(A)

for an imaginary quadratic field E (in which p splits) for local Artinian rings (A,m) with A/m of characteristic p which are unramified at p. The difference in dimension between the ordinary local deformation ring and the unramified deformation ring appears to be 3, and thus we expect l_0 = 3. Correspondingly, we expect cohomology to occur in a range of cohomological degrees [q_0,q_0 + 3] for some q_0. Moreover, in the presence of cohomology in characteristic zero, we expect to see cohomologies in all such degrees. Yet this doesn’t happen for X; in fact, the cohomology only occurs (in characteristic zero) in degrees 1 and 2 (according to RLT). This suggests not only that we *won’t* be able to prove modularity using integral cohomology of X, but even that — in the most naive sense — we should not expect an integral structure at least with the usual properties. Namely, if we patch a complex of integral cohomology of length 1, then the corresponding patched modules in cohomology will be too big for any unramified deformation ring to act. So it appears that the best possible scenario is too good to be true.

On a different matter, there is another pressing issue I would like to bring to my readers. In the papers I have written with Matt and David (and some by myself), we have used the notation l_0 — which has its origins in the book of Borel and Wallach. There is, however, a competing notation in some of Akshay’s papers, namely \delta. One argument for the latter is that l_0 specifically comes from a particular calculation in (\mathfrak{g},K)-cohomology, and is not compatible with other situations in which one might want to consider the problem of cohomology in various degrees. (For example, for weight one modular forms, the Galois l_0 = 1 whereas GL(2)/Q has discrete series.) My argument is that there will never be any confusion when using l_0, and that it has the property of being unlikely to every conflict with any other notation. Moreover, the phenomenology in both coherent and Betti cohomology both depend on l_0 in exactly the same way. Dear reader: what is your opinion?

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Z_p-extensions of Number Fields, Part II

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

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

x^2 - D = 0.

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

x^2 - Dx = 1.

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

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

\prod (x - a_i) - 1 = 0

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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Z_p-extensions of Number Fields, Part I

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

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

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

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

Definition: The field E is rigid at p if

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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