Everett W. Howe, Enric Nart, and Christophe Ritzenthaler: Jacobians in isogeny classes of abelian surfaces over finite fields, Ann. Inst. Fourier (Grenoble) 59 (2009) 239–289, MR 2010b:11064.

(A preprint and an official version are available online.)

We give a complete answer to the question of which polynomials occur as the characteristic polynomials of Frobenius for genus-2 curves over finite fields. Here is our main theorem. (Note that we assume we are given the Weil polynomial of an isogeny class of abelian surfaces, so to apply the theorem you have to know how to use the Honda-Tate theorem to determine what those polynomials are. We explain how to do this in an appendix to the paper.)

Theorem. Let f = x4 + ax³ + bx² + aqx + be the Weil polynomial of an isogeny class A of abelian surfaces over Fq, where q is a power of a prime p.

• Suppose that A contains a product of elliptic curves, so that f can be written as a product f = (sx + q) (tx + q) where the two factors are the Weil polynomials of isogeny classes of elliptic curves over Fq and where we may assume that |s| ≥ |t|. Then A does not contain a Jacobian if and only if the conditions in one of the rows of Table 1 are met.
• Suppose that A is simple. Then A does not contain a Jacobian if and only if the conditions in one of the rows of Table 2 are met.

Table 1. Conditions that ensure that the split isogeny class with Weil polynomial (sx + q)(tx + q) does not contain a Jacobian. Here we assume that |s| ≥|t|.
p-rank of A Condition on p and q Conditions on s and t
|st| = 1
2 s = t and − 4q∈{−3,−4,−7}
2 q = 2 |s| = |t| = 1 and st
1 q square = 4q and st squarefree
1 p > 3
1 p = 3 and q nonsquare = t² = 3q
0 p = 3 and q square st is not divisible by 3√q
0 p = 2 is not divisible by 2q
0 q = 2 or q = 3 s = t
0 q = 4 or q = 9 = = 4q

Table 2. Conditions that ensure that the simple isogeny class with Weil polynomial x4 + ax³ + bx² + aqx + does not contain a Jacobian.
p-rank of A Condition on p and q Conditions on a and b
b = q and b < 0 and all prime divisors of b are 1 mod 3
2 a = 0 and b = 1 − 2q
2 p > 2 a = 0 and b = 2 − 2q
2 p ≡ 11 mod 12 and q square a = 0 and b = −q
0 p = 3 and q square a = 0 and b = −q
0 p = 2 and q nonsquare a = 0 and b = −q
0 q = 2 or q = 3 a = 0 and b = −2q