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
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 , .
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
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:
Putting these together we get the following proposition:
We can now find counterexamples to the Hasse principle as follows:
Example: , , .
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