Category Archives: math

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


What about


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.)


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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!





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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?”






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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


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.







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August linkdump

  • The company that makes OldReader, the RSS reader I fled to after the sad demise of Google Reader, is from Madison!  OK, Middleton.  Still part of Silicon Isthmus.
  • I never new that Mark Alan Stamaty, one of my favorite cartoonists, did the cover of the first They Might Be Giants album.
  • Hey I keep saying this and now Allison Schrager has written an article about it for Bloomberg.  Tenure is a form of compensation.  If you think tenure is a bad way to pay teachers, and that compensation is best in the form of dollars, that’s fine; but if California pretends that the elimination of tenure isn’t a massive pay cut for teachers, they’re making a basic economic mistake.
  • New “hot hand” paper by Brett Green and Jeffrey Zweibel, about the hot hand for batters in baseball.  They say it’s there!  And they echo a point I make in the book (which I learned from Bob Wardrop) — some of the “no such thing as the hot hand” studies are way too low-power to detect a hot hand of any realistic size.
  • Matt Baker goes outside the circle of number theory and blogs about real numbers, axioms, and games.  Daring!  Matt also has a very cool new paper with Yao Wang about spanning trees as torsors for the sandpile group; but I want that to have its own blog entry once I’ve actually read it!
  • Lyndon Hardy wrote a fantasy series I adored as a kid, Master of the Five Magics.  I didn’t know that, as an undergrad, he was the mastermind of the Great Caltech Rose Bowl Hoax.  Now that is a life well spent.
  • Do you know how many players with at least 20 hits in a season have had more than half their hits be home runs?  Just two:  Mark McGwire in 2001 and Frank Thomas in 2005.
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Grothendieck’s parents

From “Who is Alexander Grothendieck?  Anarchy, Mathematics, Spirituality, Solitude,” by Winfried Scharlau (trans. Melissa Schneps)

If one is to believe the account given in Eine Frau, Sascha saw Hanka’s photograph by chance, probably one of the photographs that still exist today, and immediately informed the dismayed husband: “I will take your wife away!”  A few days later Hanka appeared, still rather weak from her abortion — and it was love at first sight.”

Bounded rank was probable in 1950

Somehow I wrote that last post about bounded ranks without knowing about this paper by Mark Watkins and many other authors, which studies in great detail the variation in ranks in quadratic twists of the congruent number curve.  I’ll no doubt have more to say about this later, but I just wanted to remark on a footnote; they say they learned from Fernando Rodriguez-Villegas that Neron wrote in 1950:

On ignore s’il existe pour toutes les cubiques rationnelles, appartenant a un corps donné une borne absolute du rang. L’existence de cette borne est cependant considérée comme probable.

So when I said the conventional wisdom is shifting from “unbounded rank” towards “bounded rank,” I didn’t tell the whole story — maybe the conventional wisdom started at “bounded rank” and is now shifting back!

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Show your work

Here’s another comment on that New York Times piece:

“mystery number game …. ‘I’m thinking of a mystery number, and when I multiply it by 2 and add 7, I get 29; what’s the mystery number?’ “

See, that’s what I mean, the ubiquitous Common Core approach to math teaching these days wouldn’t allow for either “games” or “mystery”: they would insist that your son provide a descriptive narrative of his thought process that explains how he got his answer, they would insist on him drawing some matrix or diagram to show who that process is represented pictorially.

And your son would be graded on his ability to provide this narrative and draw this diagram of his thought process, not on his ability to get the right answer (which in child prodigies and genius, by definition, is out of the ordinary, probably indescribable).

Actually, I do often ask CJ to talk out his process after we do a mystery number.  I share with the commenter the worry of slipping into a classroom regime where students are graded on their ability to recite the “correct” process.  But in general, I think asking about process is great.  For one thing, I learn a lot about how arithmetic facility develops in the mind.  I asked CJ the other night how many candies he could buy if each one cost 7 cents and he had a dollar.  He got the right answer, 14, not instantly but after a little thought.  I asked him how he got 14 and he said, “Three 7s is 21, and five 21s is a dollar and five cents, so 15 candies is a little too much, so it must be 14.”

How would you have done it?

Rank 2 versus rank 3

One interesting feature of the heuristics of Garton, Park, Poonen, Wood, Voight, discussed here previously: they predict there are fewer elliptic curves of rank 3 than there are of rank 2.  Is this what we believe?  On one hand, you might believe that having three independent points should be “harder” than having only two.  But there’s the parity issue.  All right-thinking people believe that there are equally many rank 0 and rank 1 elliptic curves, because 100% of curves with even parity have rank 0, and 100% of curves with odd parity have rank 1.  If a curve has even parity, all that has to happen to force it to have rank 2 is to have a non-torsion point.  And if a curve has odd parity, all that has to happen to force it to have rank 3 is to have one more non-torsion point you don’t know about it.  So in that sense, it seems “equally hard” to have rank 2 or rank 3, given that parity should be even half the time and odd half the time.

So my intuition about this question is very weak.  What’s yours?  Should rank 3 be less common than rank 2?  The same?  More common?

Are ranks bounded?

Important update, 23 Jul:  I missed one very important thing about Bjorn’s talk:  it was about joint work with a bunch of other people, including one of my own former Ph.D. students, whom I left out of the original post!  Serious apologies.  I have modified the post to include everyone and make it clear that Bjorn was talking about a multiperson project.  There are also some inaccuracies in my second-hand description of the mathematics, which I will probably deal with by writing a new post later rather than fixing this one.

I was only able to get to two days of the arithmetic statistics workshop in Montreal, but it was really enjoyable!  And a pleasure to see that so many strong students are interested in working on this family of problems.

I arrived to late to hear Bjorn Poonen’s talk, where he made kind of a splash talking about joint work by Derek Garton, Jennifer Park, John Voight, Melanie Matchett Wood, and himself, offering some heuristic evidence that the Mordell-Weil ranks of elliptic curves over Q are bounded above.  I remember Andrew Granville suggesting eight or nine years ago that this might be the case.  At the time, it was an idea so far from conventional wisdom that it came across as a bit cheeky!  (Or maybe that’s just because Andrew often comes across as a bit cheeky…)

Why did we think there were elliptic curves of arbitrarily large rank over Q?  I suppose because we knew of no reason there shouldn’t be.  Is that a good reason?  It might be instructive to compare with the question of bounds for rational points on genus 2 curves.  We know by Faltings that |X(Q)| is finite for any genus 2 curve X, just as we know by Mordell-Weil that the rank of E(Q) is finite for any elliptic curve E.  But is there some absolute upper bound for |X(Q)|?  When I was in grad school, Lucia Caporaso, Joe Harris, and Barry Mazur proved a remarkable theorem:  that if Lang’s conjecture were true, there was some constant B such that |X(Q)| was at most B for every genus 2 curve X.  (And the same for any value of 2…)

Did this make people feel like |X(Q)| was uniformly bounded?  No!  That was considered ridiculous!  The Caporaso-Harris-Mazur theorem was thought of as evidence against Lang’s conjecture.  The three authors went around Harvard telling all the grad students about the theorem, saying — you guys are smart, go construct sequences of genus 2 curves with growing numbers of points, and boom, you’ve disproved Lang’s conjecture!

But none of us could.

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