Startup culture, VC culture, and Mazurblogging

Those of us outside Silicon Valley tend to think of it as a single entity — but venture capitalists and developers are not the same people and don’t have the same goals.  I learned about this from David Carlton’s blog post.  Cathy O’Neil reposted it this morning.  It’s kind of cool that the three of us, who started grad school together and worked with Barry Mazur, are all actively blogging!  We just need to get Matt Emerton in on it and then we’ll have the complete set.  Maybe we could even launch a new blogging platform and call it mazr.  You want startup culture, I’ll give you startup culture!

“Deforming Galois Representations” is online, too

The hits just keep on coming, as Barry Mazur has now posted a scan of his paper, “Deforming Galois Representations,” from the long-unavailable Galois Groups over Q proceedings, on his homepage.  I didn’t link directly to the .pdf because there’s tons of other interesting stuff on Barry’s homepage to look at!

Diophantineness: Mazur-Rubin and Kollar

Last year I blogged about an argument of Bjorn Poonen, which shows that Hilbert’s tenth problem has a negative solution over the ring of integers O_K of a number field K whenever there exists an elliptic curve E/Q such that E(Q) and E(K) both have rank 1.  That is:  there’s no algorithm that tells you whether a given polynomial equation over O_K is solvable.  The idea is that under these circumstances one can construct a Diophantine model for Z inside O_K; one already knows (by Matijasevic, Robinson, etc.) that no algorithm can determine whether a polynomial equation over Z has a solution, and the same property is now inherited by the ring O_K.

The necessary fact about existence of low-rank elliptic curves over number fields (actually, not quite the fact Poonen asked for but something weaker that suffices) has now been proven, subject to a hypothesis on the finiteness of Sha, by Mazur and Rubin: see Theorems 1.11 and 1.12.  So, if you believe Sha to be finite, you believe that Hilbert’s tenth problem has a negative answer for the ring of integers of every number field.

The result of Mazur and Rubin is actually much more substantial than the corollary I mention here, giving for instance quite strong lower bounds on the number of twists of an elliptic curve E with specified 2-Selmer rank.  But I haven’t studied the argument sufficiently to say anything serious about what’s inside.

I recently returned from the “Spaces of curves and their interaction with Diophantine problems” conference at Columbia, where Janos Kollar discussed the question:  Which subsets S of C(t) are Diophantine?  That is, which have the property that they can be written as the set of s in C(t) such that $\exists x_1, x_2, ..., x_k: f(s,x_1, ... , x_k) = 0$ for some polynomial f in k+1 variables with coefficients in C(t).  Kollar explained how to prove that the polynomial ring C[t] is not Diophantine in C(t).  The idea is to show that any “sufficiently large” Diophantine subset S of C(t) contains functions whose denominators are essentially arbitrary; more precisely (but not completely precisely!) if X in Sym^d P^1 is the locus of degree-d denominators of elements of S, the Zariski closure of X needs to be — well, it doesn’t have to contain all degree-d polynomials, but it has to contain a set of the form $\{ FG^r\}$ as F,G range over polynomials of degrees s,t with s+rt = d.  In particular, it’s not possible for the denominator to be identically 1, as would be the case if S were C[t].  In fact, this argument shows that no finitely generated C-subalgebra of C(t) is Diophantine over C[t].

Open question:  is the localization of C(t) at t Diophantine over C(t)?

Update: When I first posted this I didn’t notice that Kollar’s result is already out, in the new journal Algebra and Number Theory, so you can go to the source for more details.  ANT, by the way, is a free electronically distributed journal with a terrific editorial board, and I highly recommend submitting there.

Non-simple abelian varieties in a family

Here’s a funny question. Let f in C[x] be a squarefree polynomial of degree at least 6. Let S be the set of complex numbers t such that the Jacobian of the hyperelliptic curve

$y^2 = f(x)(x-t)$

is not simple. Is S always finite? Even more, is there a bound on |S| which doesn’t depend on f, or depends only on the degree of f?

This question comes from the introduction to “Non-simple abelian varieties in a family: geometric and analytic approaches” , a new paper by me, Christian Elsholtz, Chris Hall, and Emmanuel Kowalski. In its original form this was a four-author, six-page paper — fortunately we’ve now added enough material to make the ratio a bit more respectable!

The paper isn’t about complex algebraic geometry at all — it explains how to get bounds on S when f has rational coefficients and t ranges over rational numbers, which is quite a different story. The point of the paper is partly to prove some theorems and partly to make a metamathematical point — that problems of this kind can be approached via either arithmetic geometry or analytic number theory, and that the two approaches have complementary strengths and weaknesses. Bounds from arithmetic geometry are stronger but less uniform; bounds from analytic number theory are weaker but have better uniformity.

Here’s my favorite example of this phenomenon. Let X be a smooth plane curve over Q of degree d at least 4. Then by Faltings’ Theorem we know that X has only finitely many rational points.

On the other hand, a beautiful theorem of Heath-Brown tells us that the number of rational points on X with coordinates of height at most B is at most C B^(2/d), for some constant C depending only on d. At first, this seems to give much less than Faltings. After all, as B gets larger and larger, the upper bound given by Heath-Brown gets arbitrarily large — whereas we know by Faltings that there are only finitely many points on the whole curve, no matter how large we allow the coordinates to be.

But note that the constant in Heath-Brown’s result doesn’t depend on the curve X. It is what’s called a uniform bound. Faltings’ theorem, by contrast, gives an upper bound on the number of points which depends very badly on the choice of X. Depending on what you’re trying to accomplish, you might be willing to sacrifice uniformity to get finiteness — or the reverse. But it’s best to have both options at hand.

Is it possible to have uniformity and finiteness simultaneously? Conjecturally, yes. Caporaso, Harris, and Mazur showed that, conditional on Lang’s conjecture, there is a constant B(g) such that every genus-g curve X/Q has at most B(g) rational points. The Caporaso-Harris-Mazur paper came out when I was in graduate school, and the idea of such a uniform bound was considered so wacky that CHM was thought of as evidence against Lang’s conjecture. Joe Harris used to wander around the department, buttonholing graduate students and encouraging us to cook up examples of genus-g curves with arbitrarily many points, thus disproving Lang. We all tried, and we all failed — as did many more experienced people. And nowadays, the idea that there might be a uniform bound for the number of rational points on a genus-g curve is considered fairly reputable, even among people who have their doubts about Lang’s conjecture. As far as I know, the world record for the number of rational points on a genus-2 curve is 588, due to Kulesz. Can you beat it?