TITLE: Can we count isogeny classes of abelian varieties?
QUESTION [15 upvotes]: Let's fix a finite field F and consider abelian varieties of dimension g over F.  Can we say how many isogeny classes there are?  Is it even clear that there's more than one isogeny class?  For g=1, and some fixed F with characteristic not 2 or 3, we could probably just write down all the Weierstrass equations and count isogeny classes by brute force, but is there a cleaner way to do it?

REPLY [5 votes]: Let Fq be a finite field.  Two elliptic curves E1 and E2 defined over Fq are Fq-rationally isogenous iff # E1(Fq) = # E2(Fq).
(==>) Let \varphi be an isogeny, take l a prime which is prime to q*(degree of \varphi).  Then \varphi induces a Frobenius-equivariant isomorphism on the l-adic Tate modules, so the characteristic polynomials of Frobenius agree.  Since
E(Fq) = q+1 - trace(Frob),
this implies that the two elliptic curves have the same number of points.
(<==) Similarly, if the elliptic curves have the same number of rational points, their Frobenius traces aq are the same, so their characteristic polynomials of Frobenius are both T2 - aq T + q2.  It follows from Honda-Tate theory that they have the same number of points.
Since it is known exactly what the possibilities for #E(Fq) are in terms of q -- this the Hasse-Deuring-Waterhouse theorem -- it seems that it is known exactly how many Fq-rational isogeny classes there are.  For instance, when q = p is prime, the theorem asserts that any integer N satisfying the Weil bounds
p+1 - 2\sqrt{p} < N < p+1 + 2\sqrt{p}
occurs as #E(Fp) for some E/Fp, so it seems that there are exactly
1 + 2*floor(2\sqrt{p})
Fp-rational isogeny classes.
For abelian varieties of dimension g, to determine the Frobenius polynomial and hence the Fq-rational isogeny class, you need to know not just #A(Fq) but #A(Fqi) for 1 <= i <= g.
The answer for isogeny over the algebraic closure is different -- let me know if you're interested in that.