On Superspecial abelian surfaces over finite fields III

Xue, Jiangwei; Yu, Chia-Fu; Zheng, Yewei

doi:10.1007/s40993-021-00303-8

Cited by 3 publications

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References 27 publications

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“…The calculation of class numbers of certain complicated orders arising in Section 5 is postponed to Section 6. The handful of cases where the assumption fails will be treated in an upcoming paper [28], where the ramification requires much greater care.…”

Section: Explicit Formulas Formentioning

confidence: 99%

On superspecial abelian surfaces over finite fields II

Xue¹,

Yang²,

Yu³

2020

J. Math. Soc. Japan

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Extending the results of [24, Asian J. Math.], in [26, Doc. Math. 21, 2016] we calculated explicitly the number of isomorphism classes of superspecial abelian surfaces over an arbitrary finite field of odd degree over the prime field Fp. A key step was to reduce the calculation to the prime field case, and we calculated the number of isomorphism classes in each isogeny class through a concrete lattice description. In the present paper we treat the even degree case by a different method. We first translate the problem by Galois cohomology into a seemingly unrelated problem of computing conjugacy classes of elements of finite order in arithmetic subgroups, which is of independent interest. We then explain how to calculate the number of these classes for the arithmetic subgroups concerned, and complete the computation in the case of rank two. This complements our earlier results and completes the explicit calculation of superspecial abelian surfaces over finite fields.Date: October 4, 2018. 2010 Mathematics Subject Classification. 11R52, 11G10. 1 This is well known when the ground field k is algebraically closed, and is proved in [29] for arbitrary k 1 Lemma 6.1. Suppose that S × ⊆ N ( R), the normalizer of R in B × . Then the suborder R J := x Rx −1 ∩ B of S J is independent of the choice of x ∈ B × for J, and.Proof. Suppose that J = x ′ S for x ′ ∈ B × as well. Then there exists u ∈ S × such that x ′ = xu. Since S × ⊆ N ( R), we havewhich proves the independence of R J of the choice of x. If I is a locally principal right R-ideal such that IS = J, then R J = O l (I), the associated left order of I.Conjugating by x ∈ B × on the right hand side of (6.3), we obtainThe assumption S × ⊆ N ( R) also implies that R × S × , and hence R × J S × J and R × J S × J . The left action of S × J on the quotient group S × J / R × J factors through S × J /R × J ⊆ S × J / R × J , and its orbits are the right cosets of S × JRemark 6.2. The condition S × ⊆ N ( R) implies that R × S × . However, the converse does not hold in general. It is enough to provide a counterexample locally Corollary 6.3. Keep the notation and assumption of Lemma 6.1. If the natural homomorphism S ×Proof. It is enough to show that π is injective. The surjectivity of S × J → S × J / R × J implies that the monomorphism S × J /R × J ֒→ S × J / R × J is an isomorphism, and hence |π −1 ([J])| = [ S × J / R × J : S × J /R × J ] = 1.

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Section: Explicit Formulas Formentioning

confidence: 99%