Joint work with A. Skorobogatov.

Let be an indefinite division quaternion algebra over (indefinite means that ). Let be a maximal order in . Let be the discriminant of (the product of primes that do not split ), and suppose . Let be the union of the upper and lower half planes and consider the Shimura curve

Then is a compact Riemann surface with a canonical model over .

Let be a prime, . Let , be the inverse images in of the usual subgroups of . Let , . Then is a Galois covering, with Galois group , which is unramified if such that , .

- Jordan and Livne gave a complete description of points of over
local fields.
- Jordan gave some results on points of over imaginary quadratic
fields. For example, if ,
, then
does not satisfy the Hasse Principle.
- Skorobogatov and Siksek showed that if does not have a -rational divisor class of degree 1 and is finite, then the failure of the Hasse Principle for is accounted for by the Manin obstruction.

Consider now the covering of curves over , as defined above. We want to find an imaginary quadratic field and and such that for any twist of by a character , but . Since, as in Skorobogatov's talk, , this shows that the Hasse principle for fails and provides a cohomological obstruction that explains this failure.

Suppose first that for a prime dividing . Then has bad reduction at . Furthermore, it is known (Jordan-Livne-Varshavsky) that if is inert. So we suppose that all primes dividing are inert in .

Now consider what happens at . We have the usual model of with two components , intersecting at supersingular points defined over . We assume again that is inert. It can be shown that Supersingular Points. This then shows that if is inert, for .

Finally, we consider places which do not divide . If is ramified at such a place , . So we need only consider characters unramified outside . If we suppose further

- that is prime to for any
- is prime to

Now, and have good reduction outside . To deal with places , we count points to show that the curve has points over (which is or according as is split or inert in ), then lift them using Hensels Lemma.

The point counts make use of the following trace formulas:
then

where is Frobenius and . We get , and if , such that

Putting these together we get the following proposition:

We can now find counterexamples to the
Hasse principle as follows:

- Choose , with and .
- Choose such that is coprime to .
- Let be a prime such that . Call good if there exists such that condition above is satisfied, bad otherwise. (The set of bad primes is finite.) Now find a good such that () is satisfied.
- Now choose a field imaginary quadratic such that are inert in , bad primes are inert, primes from Step are split or ramified, and is prime to . Ono can prove that if there is one satisfying these conditions there are infinitely many.

Example: , , .

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