## Lieblich’s counter counterexample example

Max Lieblich gave a great talk at WAGS yesterday about something that looks like a counterexample to the Hasse principle, but secretly isn’t!  All mistakes in this summary are my own.  For a more authoritative take on the material below, see Max’s recent arXiv preprint.

The counterexample Max countered is the central simple algebra A over the field Q(t) obtained as the tensor product of the two generalized quaternion algebras $(17,t)$ and $(13,6(t-1)(t-11))$.  This algebra has index 4, which is to say (if I understand correctly) that its Brauer class isn’t the cup product of two elements in Q(t)^*.  On the other hand, it turns out that $A \otimes \mathbf{Q}_v$ has index 1 or 2 — that is, it’s either trivial or a cup product — for all places v of Q.

This seems like an example of a situation where there’s no Hasse principle; A fails to be a cup product despite the fact that A is a cup product over every completion of Q.  But the truth, as Max explained, is more complicated.

We know (and if we don’t, we read A Course in Arithmetic to remember) that the quaternion algebra (a,b) is trivial (i.e., has index 1) over a field k precisely when the conic $x^2 - ay^2 - bz^2$ has a k-rational point.  You might ask whether there’s a similar criterion for the tensor product of two quaternion algebras to have index 2.  It turns out that (at least over Q(t)) there is indeed a variety that does this trick — it’s a coarse moduli space M parametrizing certain twisted vector bundles on — well, not quite P^1/Q, but a certain orbifold version of P^1 with a bunch of stacky points with inertia Z/2Z.  And the assertion that A has index 4 is equivalent to the assertion that M(Q) is empty.

To really talk about what M is would take us too far afield; I just want to record Max’s observation that the definition of M is truly global, in the sense that the scheme $M_{\mathbf{Q}_v}$ is not determined by $A \otimes \mathbf{Q}_v$.  In particular, the fact that $A \otimes \mathbf{Q}_v$ has index 2 doesn’t imply that M has a $\mathbf{Q}_v$-point.  And indeed, in the case at hand, M has no points over $\mathbf{Q}_{17}$.  So there is, after all, a local obstruction to A having index 2; but it’s a local obstruction which, it seems, can’t be seen except in this rather intricate geometric way.

It makes you wonder what should actually be meant by “Hasse principle.”  Suppose, for instance, you had some class C of varieties X/Q, and suppose you had some construction which attached to each X in C a variety Y/Q such that X(Q) is empty if and only if Y(Q) is empty.   Now one way to prove X(Q) empty would be to prove that Y(Q) was empty, which you could in turn prove if you knew that Y(Q_v) was empty for some place v.  Do you consider this a local obstruction to rational points on X?

## 4 thoughts on “Lieblich’s counter counterexample example”

1. The descent formulation of the Brauer-Manin obstruction for curves works like this: (see Stoll’s paper for details). Given a curve X/Q you can look at Y/Q which is a disjoint union of twists of an etale cover of X (so Y is reducible). Often, when X(Q) is empty, (perhaps always?) you can find Y such that for each component Y_i of Y there is v_i with Y_i(Q_{v_i}) empty. So is not quite what you are asking but is close. I consider this a local obstruction.

2. JSE says:

Ah, that’s a good point, and an extra wrinkle — since Y itself will of course have points over every Q_v.

3. I would also say it’s a local obstruction, provided the construction X —> Y is suitably natural (so you’re not allowed to peek at X(Q) to construct Y…). It’s not clear how easy this would be to formulate precisely…

4. […] thought it was awesome that Max Lieblich gave a talk called “A counter-counterexample example,” but perhaps it’s even better that he’s giving two talks at Stanford this week, the […]