## Poonen-Rains and lines on a quadric surface

One feature of the Poonen-Rains heuristics that might seem strange at first is that the dimension of the Selmer group isn’t 0 almost all  the time.  This is by contrast with the Cohen-Lenstra heuristic, where the p-torsion in the class group is indeed trivial about 1-1/p of the time.  Instead, the Poonen-Rains heuristics predict that the p-Selmer rank is 0 about half the time and 1 about half the time, with about 1/p’s worth of measure devoted to ranks 2 or higher.  Of course, given that we expect a random elliptic curve to have Mordell-Weil rank 1 half the time, it would be bad news for their heuristic if it predicted a lower frequency of positive Selmer rank!

But why is the intersection of two maximal isotropic subspaces 1 half the time and 0 half the time?  You can get a nice picture of what’s going on by thinking about the case of  a quadratic form Q in 4 variables.  The vanishing of the quadratic form cuts out a quadric surface in P^3.  A maximal isotropic subspace is a 2-dimensional space on which Q vanishes — in other words, a line on the quadric.  The intersection of two maximal isotropics is o-dimensional if the corresponding lines are disjoint, 1-dimensional if the lines intersect at a point, and 2-dimensional when the lines coincide.  So what’s the probability that two random lines on the quadric intersect?  The key point is that there are two families of lines.  If L1 and L2 come from different families, they intersect; if they come from the same family, they’re disjoint (except in the unlikely event they coincide.)  So there you go — the intersection of the maximal isotropics is split 50-50 between 0-dimensional and 1-dimensional.  More generally, the variety of maximal isotropic subspaces in an even-dimensional orthogonal space has two components, and this explains the leading term of Poonen-Rains.

It would be interesting to understand how to describe the “two types of maximal isotropics” in the infinite-dimensional F_p-vector space considered by Poonen-Rains, and to understand why the two maximal isotropics supplied by a given elliptic curve lie in the same family if and only if the L-function of E has even functional equation, which should lead one to expect that Sel_p(E) has even rank  (or even, thanks to recent progress on the parity conjecture by Nekovar, Kim, los Dokchitsers, etc., implies that Sel_p(E) has even rank, subject to finiteness of Sha.)

## Can the trace hear the shape of its field?

The most natural arithmetic invariant of a number field K is its discriminant D_K, an integer congruent to either 0 or 1 mod 4 whose prime factors are precisely the primes where K/Q is ramified.  Oftentimes D_K is a squarefree, in which case it’s just the product of the ramified primes; even if not, the multiplicity of a prime factor of D_K can be described quite cleanly in terms of the p-inertia subgroup of Gal(K/Q).

The situation is especially handsome for quadratic field.  The discriminants of quadratic fields are just those integers congruent to 0 or 1 mod 4 which have no square factor larger than 4.  Better still, the discriminant specifies the field uniquely!  So in order to describe a quadratic field it suffices to write down a single integer.

Life gets worse in higher degree.  There can be lots of number fields with the same discriminant D.  For example:  if D is squarefree and K is a cubic field with discriminant D, then the Galois closure L of K is an unramified (Z/3Z)-extension of the quadratic field M with discriminant D.  So if the ideal class group of M has a lot of (Z/3Z) in it, there are going to be a lot of cubic fields with discriminant D!

Just how bad is this multiplicity?  It’s widely believed that, for every n, there are at most D^eps number fields of discriminant D.  But I think nobody has a good idea about how to prove this, even for n=3.

So it’s naturally interesting to ask whether there are other invariants which might uniquely specify a number field.  Here’s one natural candidate.  The ring of integers O_K is a free rank n Z-module, endowed with a natural quadratic form q(x,y) = Tr_{K/Q}(xy), called the trace form.  The discriminant of this trace form is, up to known factors, the discriminant of K.  So you can think of the isomorphism class of the trace form as a refinement of the discriminant.  The question is:  is it such a good refinement that it actually specifies the field?

My former student Guillermo Mantilla-Soler, now at UBC,  just posted a preprint offering the first real insight into this question, which he colorfully phrases “Can the trace hear the shape of its field?”  He shows that the answer to the original question is no:  for instance, he displays two non-isomorphic quintic fields of discriminant 34129 which have isomorphic trace form.  More generally, he gives a necessary condition which I would expect is satisfied by examples in every degree (though it might be hard to prove this.)

But in some sense this is a local issue; the fields in these examples are not totally real, so the trace forms aren’t definite, and so, as Guillermo observes, it suffices to show the forms lie in the same spinor genus.  In the definite case, it’s “harder” for quadratic forms to be isomorphic.  Are there two non-isomorphic totally real number fields with isomorphic trace forms?  Guillermo includes the results of a fairly large computer search which finds no examples in degree < 10 and discriminant < 10^9.

## The different does not have a canonical square root

Just wanted to draw attention to this very nice exchange on Math Overflow.   Matt Emerton remarks that the different of a number field is always a square in the ideal class group, and asks:  is there a canonical square root of the ideal class of the different?

What grabs me about this question is that the word “canonical” is a very hard one to define precisely.   Joe Harris used to give a lecture called, “The only canonical divisor is the canonical divisor.”  The difficulty around the word “canonical” is what gives the title its piquancy.

Usually we tell students that something is “canonical” if it is “defined without making any arbitrary choices.”  But this seems to place a lot of weight on the non-mathematical word “arbitrary.”

Here’s one way to go:  you can say a construction is canonical if it is invariant under automorphisms.  For instance, the abelianization of a group is a canonical construction; if f: G_1 -> G_2 is an isomorphism, then f induces an isomorphism between the abelianizations.

It is in this sense that MathOverflow user “Frictionless Jellyfish” gives a nice proof that there is no canonical square root of the different; the slick cnidarian exhibits a Galois extension K/Q, with Galois group G = Z/4Z, such that the ideal class of the different of K has a square root (as it must) but none of its square roots are fixed by the action of G (as they would have to be, in order to be called “canonical.”)  The different itself is canonical and as such is fixed by G.

But this doesn’t seem to me to capture the whole sense of the word.  After all, in many contexts there are no automorphisms!  (E.G. in the Joe Harris lecture, “canonical” means something a bit different.)

Here’s a sample question that bothers me.  Ever since Gauss we’ve known that there’s a bijection between the set of proper isomorphism classes of primitive positive definite binary quadratic forms of discriminant d and the ideal class group of a quadratic imaginary field.

Do you think this bijection is “canonical” or not?  Why?