Integral Hasse Principle and Strong Approximation for Markoff Surfaces

Abstract

We study the failure of the integral Hasse principle and strong approximation for Markoff surfaces, as studied by Ghosh and Sarnak, using the Brauer-Manin obstruction. Contents1. Introduction 1 2. Geometry of affine cubic surfaces 4 3. Geometry of projective Markoff surfaces 8 4. Brauer group of affine Markoff surfaces 9 5. The Brauer-Manin obstruction 13 References 27 Theorem 1.3. Assume that m ∈ Z is such that U m has a Brauer-Manin obstruction to the integral Hasse principle. ThenThis qualitative statement shows that there is a Brauer-Manin obstruction to the integral Hasse principle for at most O(B 1/2 ) of the surfaces U m for m ∈ Z with |m| ≤ B. A more in-depth analysis of the Brauer-Manin obstruction allows us to prove that not only is m − 4 a product of small primes times a square, but all the prime divisors of m − 4 must satisfy very strong congruence conditions. This allows us to show the improved upper bound O(B 1/2 /(log B) 1/2 ), which is sharp by the following theorem.Theorem 1.4. We have Geometry of affine cubic surfacesBy an affine cubic surface, we mean an affine surface of the formwhere f is a polynomial of degree of 3. The closure of U in P 3 is a cubic surface S. The complement H = S \ U is a hyperplane section on S. Much of the geometry of U can be understood in terms of the geometry of S and H. We begin with some basic remarks.2.1. Basic geometry.Lemma 2.1. Let S be a smooth cubic surface over a field k, let H be a hyperplane section and set U = S \ H. Then O(U)

Brauer-Manin obstruction for Markoff surfaces

“…We then show that the counterexamples to the integral Hasse principle for U m in [10] may all be explained by a combination of integral Brauer-Manin obstruction and reduction theory. We increase the stock of such counterexamples, thus leading to an improvement on a counting result in [15]. In §6, we prove that strong approximation never holds for Markoff type surfaces.…”

“…• Assume p = 5 and ord 5 (d) = 1. One can use the lifting of smooth points of U m (Z/5) as in [15,Proposition 5.7] to show that B can take all possible values over U m (Z 5 ) except (0, 0, 0). We prove (0, 0, 0) ∈ B(U m (Z 5 )).…”

“…This work was started in Beijing in November 2017 and posted on arXiv in August 2018. In a preprint posted on arXiv in July 2018, D. Loughran and V. Mitankin [15] have made an independent study. With the restrictions m, d, md not squares, they independently solve problem (B).…”

“…Their paper also solves Problem (A), produces some more types of counterexamples, and gives an asymptotic lower bound for the number of integers m giving rise to such counterexamples. Our stock of counterexamples enables us to produce a slightly better asymptotic lower bound than [15,Theorem 1.5].…”

“…This finiteness result, which is in the spirit of the finiteness of exceptional spinor classes in the study of the representation of an integer by a ternary quadratic form (see [6,Remark 7.11]), explains why the examples in [10] based on the quadratic reciprocity law were of a rather special type. It is used in [15] to show that there are indeed far less values of m with Brauer-Manin counterexamples than the number of values of m predicted by [10] for counterexamples to the integral Hasse principle.…”