Alexandra Florea on the average central value of hyperelliptic L-functions

Alexandra Florea, a student of Soundararajan, has a nice new paper up, which I heard about in a talk by Michael Rubinstein.  She computes the average of

$L(1/2, \chi_f)$

as f ranges over squarefree polynomials of large degree.  If this were the value at 1 instead of the value at 1/2, this would be asking for the average number of points on the Jacobian of a hyperelliptic curve, and I could at least have some idea of where to start (probably with this paper of Erman and Wood.)  And I guess you could probably get a good grasp on moments by imitating Granville-Soundararajan?

But I came here to talk about Florea’s result.  What’s cool about it is that it has the a main term that matches existing conjectures in the number field case, but there is a second main term, whose size is about the cube root of the main term, before you get to fluctuations!

The only similar case I know is Roberts’ conjecture, now a theorem of Bhargava-Shankar-Tsimerman and Thorne-Taniguchi, which finds a similar secondary main term in the asymptotic for counting cubic fields.  And when I say similar I really mean similar — e.g. in both cases the coefficient of the secondary term is some messy thing involving zeta functions evaluated at third-integers.

My student Yongqiang Zhao found a lovely geometric interpretation for the secondary term the Roberts conjecture.  Is there some way to see what Florea’s secondary term “means” geometrically?  Of course I’m stymied here by the fact that I don’t really know how to think about her counting problem geometrically in the first place.

Cold Topics Workshop

I was in Berkeley the other day, chatting with David Eisenbud about an upcoming Hot Topics workshop at MSRI, and it made me wonder:  why don’t we have Cold Topics workshops?  In the sense of “cold cases.”  There are problems that the community has kind of drifted away from, because we don’t really know how to do them, but which are as authentically interesting as they ever were.  Maybe it would be good to programatically focus our attention on those cold topics from time to time, just to see whether the passage of time has given us any new ideas, or cast these cold old problems in a new and useful light.

If this idea catches on, we could even consider having an NSF center devoted to these problems.  The Institute for Unpopular Mathematics!

What cold topics workshops would you propose to me, the founding director of the IUM?

Tagged ,

Idle question: are Kakeya sets winning?

Jayadev Athreya was here last week and reminded me about this notion of “winning sets,” which I learned about from Howie Masur — originally, one of the many contributions of Wolfgang Schmidt.

Here’s a paper by Curt McMullen introducing a somewhat stronger notion, “absolute winning.”

Anyway:  a winning set (or an absolute winning set) in R^n is “big” in some sense.  In particular, it has to have full Hausdorff dimension, but it doesn’t have to have positive measure.

Kakeya sets (subsets of R^n containing a unit line segment in every direction) can have measure zero, by the Besicovitch construction, and are conjectured (when n=2, known) to have Hausdorff dimension n.  So should we expect these sets to be winning?  Are Besicovitch sets winning?

I have no reason to need to know.  I just think these refined classifications of sets which are measure 0 yet still “large” are very interesting.  And for all I know, maybe there are sets where the easiest way to prove they have full Hausdorff dimension is to prove they’re winning!

Shin-Strenner: Pseudo-Anosov mapping classes not arising from Penner’s construction

Balazs Strenner, a Ph.D. student of Richard Kent graduating this year, gave a beautiful talk yesterday in our geometry/topology seminar about his recent paper with Hyunshik Shin.  (He’s at the Institute next year but if you’re looking for a postdoc after that…!)

A long time ago, Robert Penner showed how to produce a whole semigroup M in the mapping class group with the property that all but a specified finite list of elements of M were pseudo-Anosov.  So that’s a good cheap way to generate lots of certified pseudo-Anosovs in the mapping class group.  But of course one asks:  do you get all pA’s as part of some Penner semigroup?  This can’t quite be true, because it turns out that the Penner elements can’t permute singularities of the invariant folation, while arbitrary pA’s can.  But there are only finitely many singularities, so some power of a given pA clearly fixes the singularities.

So does every pA have a power that arises from Penner’s construction?  This is what’s known as Penner’s conjecture.  Or was, because Balazs and Hyunshik have shown that it is falsitty false false false.

When I heard the statement I assumed this was going to be some kind of nonconstructive counting argument — but no, they actually give a way of proving explicitly that a given pA is not in a Penner semigroup.  Here’s how.  Penner’s semigroup M is generated by Dehn twists Q_1, … Q_m, which all happen to preserve a common traintrack, so that there’s actually a representation

$\rho: M \rightarrow GL_n(\mathbf{R})$

such that the dilatation of g is the Perron-Frobenius eigenvalue $\lambda$ of $\rho(g)$.

Now here’s the key observation; there is a quadratic form F on R^n such that F(Q_i x) >= F(x) for all x, with equality only when x is a fixed point of Q_i.  In particular, this shows that if g is an element of M not of the form Q_i^a, and x is an arbitrary vector, then the sequence

$x, g x, g^2 x, \ldots$

can’t have a subsequence converging to x, since

$F(x), F(gx), F(g^2 x), \ldots$

is monotone increasing and thus can’t have a subsequence converging to F(x).

This implies in particular:

g cannot have any eigenvalues on the unit circle.

But now we win!  Because $\rho(g)$ is an integral matrix, so all the Galois conjugates of $\lambda$ must be among its eigenvalues.  In other words, $\lambda$ is an algebraic number none of whose Galois conjugates lie on the unit circle.  But there are lots of pseudo-Anosovs whose dilatations $\lambda$ do have Galois conjugates on the unit circle.  In fact, experiments by Dunfield and Tiozzo seem to show that in a random walk on the braid group, the vast majority of pAs have this property!  And these pAs, which Shin and Strenner call coronal, cannot appear in any Penner semigroup.

Cool!

Anyway, I found the underlying real linear algebra question very appealing.  Two idle questions:

• If M is a submonoid of GL_n(R) we may say a continuous real-valued function F on R^n is M-monotone if F(mx) >= F(x) for all m in M, x in R^n.  The existence of a monotone function for the Penner monoid is the key to Strenner and Shin’s theorem.  But I have little feeling for how it works in general.  Given a finite set of matrices, what are explicit conditions that guarantee the existence of an M-monotone function?  Nonexistence?  (I have a feeling it is roughly equivalent to M containing no element with an eigenvalue on the unit circle, but I’m not sure, and anyway, this is not a checkable condition on the generating matrices…)
• What can we say about the eigenvalues of matrices appearing in the Penner subgroup?  Balazs says he’ll show in a later paper that they can actually get arbitrarily close to the unit circle, which is actually not what I had expected.  He asks:  are those eigenvalues actually dense in the complex plane?

What I learned at the Joint Math Meetings

Another Joint Meetings in the books!  My first time in San Antonio, until last weekend the largest US city I’d never been to.  (Next up:  Jacksonville.)  A few highlights:

• Ngoc Tran, a postdoc at Austin, talked about zeroes of random tropical polynomials.  She’s proved that a random univariate tropical polynomial of degree n has about c log n roots; this is the tropical version of an old theorem of Kac, which says that a random real polynomial of degree n has about c log n real roots.  She raised interesting further questions, like:  what does the zero locus of a random tropical polynomial in more variables look like?  I wonder:  does it look anything like the zero set of a random band-limited function on the sphere, as discussed by Sarnak and Wigman?  If you take a random tropical polynomial in two variables, its zero set partitions the plane into polygons, which gives you a graph by adjacency:  what kind of random graph is this?
• Wait, I’ve got still one more thing about random graphs!  Russ Lyons gave a plenary about his work with Angel and Kechris about unique ergodicity of the action of the automorphism group of the random graph.  Wait, the random graph? I thought there were lots of random graphs!  Nope — when you try to define the Erdös-Renyi graph on countably many vertices, there’s a certain graph (called “the Rado graph”) to which your random graph is isomorphic with probability 1!  What’s more, this is true — and it’s the same graph — no matter what p is, as long as it’s not 0 or 1!  That’s very weird, but proving it’s true is actually pretty easy.  I leave it an exercise.
• Rick Kenyon gave a beautiful talk about his work with Aaron Abrams about “rectangulations” — decompositions of a rectangle into area-1 subrectangles.  Suppose you have a weighted directed graph, representing a circuit diagram, where the weights on the edges are the conductances of the corresponding wires.  It turns out that if you fix the energy along each edge (say, to 1) and an acyclic orientation of the edges, there’s a unique choice of edge conductances such that there exists a Dirichlet solution (i.e. an energy-minimizing assignment of a voltage to each node) with the given energies.  These are the fibers of a rational map defined over Q, so this actually gives you an object over a (totally real) algebraic number field for each acyclic orientaton.  As Rick pointed out, this smells a little bit like dessins d’enfants!  (Though I don’t see any direct relation.)  Back to rectangulations:  it turns out there’s a gadget called the “Smith Diagram” which takes a solution to the Dirichlet problem on the graph  and turns it into a rectangulation, where each edge corresponds to a rectangle, the area of the rectangle is the energy contributed by the current along that edge, the aspect ratio of the rectangle is the conductance, the bottom and top faces of the rectangle correspond to the source and target nodes, the height of a face is the voltage at that node, and etc.  Very cool!  Even cooler when you see the pictures.  For a 40×40 grid, it looks like this:

How many rational distances can there be between N points in the plane?

Terry has a nice post up bout the Erdös-Ulam problem, which was unfamiliar to me.  Here’s the problem:

Let S be a subset of R^2 such that the distance between any two points in S is a rational number.  Can we conclude that S is not topologically dense?

S doesn’t have to be finite; one could have S be the set of rational points on a line, for instance.  But this appears to be almost the only screwy case.  One can ask, more ambitiously:

Is it the case that there exists a curve X of degree <= 2 containing all but 4 points of S?

Terry explains in his post how to show something like this conditional on the Bombieri-Lang conjecture.  The idea:  lay down 4 points in general position.  Then the condition that the 5th point has rational distances from x1,x2,x3, and x4 means that point lifts to a rational point on a certain (Z/2Z)^4-cover Y of P^2 depending on x1,x2,x3,x4.  (It’s the one obtained by adjoining the 4 distances, each of which is a square root of a rational function.)

With some work you can show Y has general type, so under Lang its rational points are supported on a union of curves.  Then you use a result of Solymosi and de Zeeuw to show that each curve can only have finitely many points of S if it’s not a line or a circle.  (Same argument, except that instead of covers of P^2 you have covers of the curve, whose genus goes up and then you use Faltings.)

It already seems hard to turn this approach into a proof.  There are few algebraic surfaces for which we can prove Lang’s conjecture.  But why let that stop us from asking further questions?

Question:  Let S be a set of N points on R^2 such that no M are contained in any line or circle.  What is the maximal number of rational distance among the ~N^2 distances between points of S?

The Erdos-Ulam problem suggests the answer is smaller than N^2.  But surely it’s much smaller, right?  You can get at least NM rational distances just by having S be (N/M) lines, each with M rational points.  Can you do better?

How many points does a random curve over F_q have?

So asks a charming preprint by Achter, Erman, Kedlaya, Wood, and Zureick-Brown.  (2/5 Wisconsin, 1/5 ex-Wisconsin!)  The paper, I’m happy to say, is a result of discussions at an AIM workshop on arithmetic statistics I organized with Alina Bucur and Chantal David earlier this year.

Here’s how they think of this.  By a random curve we might mean a curve drawn uniformly from M_g(F_q).  Let X be the number of points on a random curve.  Then the average number of points on a random curve also has a geometric interpretation: it is

$|M_{g,1}(\mathbf{F}_q)|/|M_{g}(\mathbf{F}_q)|$

$|M_{g,2}(\mathbf{F}_q)|/|M_{g}(\mathbf{F}_q)|$?

That’s just the average number of ordered pairs of distinct points on a random curve; the expected value of X(X-1).

If we can compute all these expected values, we have all the moments of X, which should give us a good idea as to its distribution.  Now if life were as easy as possible, the moduli spaces of curves would have no cohomology past degree 0, and by Grothendieck-Lefschetz, the number of points on M_{g,n} would be q^{3g-3+n}.  In that case, we’d have that the expected value of X(X-1)…(X-n) was q^n.  Hey, I know what distribution that is!  It’s Poisson with mean q.

Now M_g does have cohomology past degree 0.  The good news is, thanks to the Madsen-Weiss theorem (née the Mumford conjecture) we know what that cohomology is, at least stably.  Yes, there are a lot of unstable classes, too, but the authors propose that heuristically these shouldn’t contribute anything.  (The point is that the contribution from the unstable range should look like traces of gigantic random unitary matrices, which, I learn from this paper, are bounded with probability 1 — I didn’t know this, actually!)  And you can even make this heuristic into a fact, if you want, by letting q grow pretty quickly relative to g.

So something quite nice happens:  if you apply Grothendieck-Lefschetz (actually, you’d better throw in Kai Behrend’s name, too, because M_g is a Deligne-Mumford stack, not an honest scheme) you find that the moments of X still agree with those of a Poisson distribution!  But the contribution of the tautological cohomology shifts the mean from q to q+1+1/(q-1).

This is cool in many directions!

• It satisfies one’s feeling that a “random set,” if it carries no extra structure, should have cardinality obeying a Poisson distribution — the “uniform distribution” on the groupoid of sets.  (Though actually that uniform distribution is Poisson(1); I wonder what tweak is necessary to be able to tune the mean?)
• I once blogged about an interesting result of Bucur and Kedlaya which showed that a random smooth complete intersection curve in P^3 of fixed degree had slightly fewer than q+1 points; in fact, about q+1 – 1/q + o(q^2).  Here the deviation is negative, rather than positive, as the new paper suggests is the case for general curves; what’s going on?
• I have blogged about the question of average number of points on a random curve before.  I’d be very interested to know whether the new heuristic agrees with the answer to the question proposed at the end of that post; if g is a large random matrix in GSp(Z_ell) with algebraic eigenvalues, and which multiplies the symplectic form by q, and you condition on Tr(g^k) > (-q^k-1) so that the “curve” has nonnegatively many points over each extension of F_q, does this give something like the distribution the five authors predict for Tr(g)?  (Note:  I don’t think this question is exactly well-formed as stated.)

Silas Johnson on weighted discriminants with mass formulas

My Ph.D. student Silas Johnson just posted his thesis paper to the arXiv, and it’s cool, and I’m going to blog about it!

How should you count number fields?  The most natural way is by discriminant; you count all degree-n number fields K with a given Galois group G in S_n and discriminant bounded in absolute value by B.  This gives you a value N_G(B) whose asymptotic behavior in B you might want to study.  The classical results and exciting new ones you’ve heard about — Davenport-Heilbron, Bhargava, and all that — generally concern counts of this kind.

But there are reasons to consider other kinds of counts.  I once had a bunch of undergrads investigate the behavior of N_3(X,Y), the number of cubic fields whose discriminant had squarefree part at most X and maximal square divisor at most Y.  This gives a more refined picture of the ramification behavior of the fields.  Asymptotics for this are still unknown!  (I would expect the main term to be on order $X Y^{1/2}$, but I don’t know what the secondary terms should be.)

One nice thing about the discriminant, though, is that it has a mass formula.  In brief:  a map f from Gal(Q_p) to S_n corresponds to a degree-n extension of Q_p, which has a discriminant (a power of p) which we call Disc(f).  Averaging Disc(f)^{-1} over all homomorphisms f gives you a polynomial in p^{-1}, which we call the local mass.  Now here’s the remarkable fact (shown by Bhargava, deriving from a formula of Serre) — there is a universal polynomial P(x) such that the local mass at p is equal to P(p^{-1}) for every P.  This is not hard to show for the tame primes p (you can see this point discussed in Silas’s paper or in the paper by Kedlaya I linked above) but that it holds for the wild primes is rather mysterious and strange.  And it certainly seems to ratify the idea that there’s something especially nice about the discriminant.  What’s more, this polynomial P is not just some random thing; it’s the product over p of P(p^{-1}) that gives the constant in Bhargava’s conjectural asymptotic for the number of number fields for degree n.

But here’s the thing.  If we replace G by a subgroup of S_n, there need not be a universal mass formula anymore.  Kedlaya (and Daniel Gulotta, in the appendix) show lots of examples.  The simplest example is the dihedral group of order 8.

All is not lost, though!  Wood showed in 2008 that you could fix this problem by replacing the discriminant of a D_4-extension with a different invariant.  Namely:  a D_4 quartic field M has a quadratic subextension L.  If you replace Disc(L/Q) with Disc(L/Q) times the norm to Q of Disc(L/M), you get a different invariant of M — an example of what Silas calls a “weighted discriminant” — and when you compute the local mass according to {\em this} invariant, you get a polynomial in p^{-1} again!

So maybe Wood’s modified discriminant, not the usual discriminant, is the “right” way to count dihedral quartics?  Does the product of her local masses give the right asymptotic for the number of D_4 extensions with Woodscriminant at most B?

It’s not at all clear to me how, if at all, you can cook up a modified discriminant for an arbitrary group G that has a universal mass formula.  What Silas shows is that having a mass formula is indeed special; when G is a p-group, there are only finitely many weighted discriminants that have one.  Silas thinks, and so do I, that this is actually true for every finite group G, and that some version of his approach will eventually prove this.  And he classifies all such weighted discriminants for groups of size up to 12; for any individual G, it’s a computation which can be made nicely algorithmic.  Very cool!

Squares and Motzkins

Greg Smith gave an awesome colloquium here last week about his paper with Blekherman and Velasco on sums of squares.

Here’s how it goes.  You can ask:  if a homogeneous degree-d polynomial in n variables over R takes only non-negative values, is it necessarily a sum of squares?  Hilbert showed in 1888 that the answer is yes only when d=2 (the case of quadratic forms), n=2 (the case of binary forms) or (n,d) = (3,4) (the case of ternary quartics.)  Beyond that, there are polynomials that take non-negative values but are not sums of squares, like the Motzkin polynomial

$X^4 Y^2 + X^2 Y^4 - 3X^2 Y^2 Z^2 + Z^6$.

So Greg points out that you can formulate this question for an arbitrary real projective variety X/R.  We say a global section f of O(2) on X is nonnegative if it takes nonnegative values on X(R); this is well-defined because 2 is even, so dilating a vector x leaves the sign of f(x) alone.

So we can ask:  is every nonnegative f a sum of squares of global sections of O(1)?  And Blekherman, Smith, and Velasco find there’s an unexpectedly clean criterion:  the answer is yes if and only if X is a variety of minimal degree, i.e. its degree is one more than its codimension.  So e.g. X could be P^n, which is the (n+1,2) case of Hilbert.  Or it could be a rational normal scroll, which is the (2,d) case.  But there’s one other nice case:  P^2 in its Veronese embedding in P^5, where it’s degree 4 and codimension 3.  The sections of O(2) are then just the plane quartics, and you get back Hilbert’s third case.  But now it doesn’t look like a weird outlier; it’s an inevitable consequence of a theorem both simpler and more general.  Not every day you get to out-Hilbert Hilbert.

Idle question follows:

One easy way to get nonnegative homogenous forms is by adding up squares, which all arise as pullback by polynomial maps of the ur-nonnegative form x^2.

But we know, by Hilbert, that this isn’t enough to capture all nonnegative forms; for instance, it misses the Motzkin polynomial.

So what if you throw that in?  That is, we say a Motzkin is a degree-6d form

expressible as

$P^4 Q^2 + P^2 Q^4 - 3P^2 Q^2 R^2 + R^6$

for degree-d forms P,Q,R.  A Motzkin is obviously nonnegative.

It is possible that every nonnegative form of degree 6d is a sum of squares and Motzkins?  What if instead of just Motzkins we allow ourselves every nonnegative sextic?  Or every nonnegative homogeneous degree-d form in n variables for n and d less than 1,000,000?  Is it possible that the condition of nonnegativity is in this respect “finitely generated?”

Breuillard’s ICM talk: uniform expansion, Lehmer’s conjecture, tauhat

Emmanuel Breuillard is in Korea talking at the ICM; here’s his paper, a very beautiful survey of uniformity results for growth in groups, by himself and others, and of the many open questions that remain.

He starts with the following lovely observation, which was apparently in a 2007 paper of his but which I was unaware of.  Suppose you make a maximalist conjecture about uniform growth of finitely generated linear groups.  That is, you postulate the existence of a constant c(d) such that, for any finite subset S of GL_d(C),  you have a lower bound for the growth rate

$\lim |S^n|^{1/n} > c(d)$.

It turns out this implies Lehmer’s conjecture!  Which in case you forgot what that is is a kind of “gap conjecture” for heights of algebraic numbers.  There are algebraic integers of height 0, which is to say that all their conjugates lie on the unit circle; those are the roots of unity.  Lehmer’s conjecture says that if x is an algebraic integer of degree n which is {\em not} a root of unity, it’s height is bounded below by some absolute constant (in fact, most people believe this constant to be about 1.176…, realized by Lehmer’s number.)

What does this question in algebraic number theory have to do with growth in groups?  Here’s the trick; let w be an algebraic integer and consider the subgroup G of the group of affine linear transformations of C (which embeds in GL_2(C)) generated by the two transformations

x -> wx

and

x -> x+1.

If the group G grows very quickly, then there are a lot of different values of g*1 for g in the word ball S^n.  But g*1 is going to be a complex number z expressible as a polynomial in w of bounded degree and bounded coefficients.  If w were actually a root of unity, you can see that this number is sitting in a ball of size growing linearly in n, so the number of possibilities for z grows polynomially in n.  Once w has some larger absolute values, though, the size of the ball containing all possible z grows exponentially with n, and Breuillard shows that the height of z is an upper bound for the number of different z in S^n * 1.  Thus a Lehmer-violating sequence of algebraic numbers gives a uniformity-violating sequence of finitely generated linear groups.

These groups are all solvable, even metabelian; and as Breuillard explains, this is actually the hardest case!  He and his collaborators can prove the uniform growth results for f.g. linear groups without a finite-index solvable subgroup.  Very cool!

One more note:  I am also of course pleased to see that Emmanuel found my slightly out-there speculations about “property tau hat” interesting enough to mention in his paper!  His formulation is more general and nicer than mine, though; I was only thinking about profinite groups, and Emmanuel is surely right to set it up as a question about topologically finitely generated compact groups in general.