For Elliptic curves over a finite field, there is a very useful characterization of ordinary elliptic as those with commutative, quadratic endomorphism rings and of supersingular curves as those with Endomorphism ring a non commutative division algebra of rank $4$.

**Question:** Is there any such characterization in the case of (simple) Abelian surfaces over a finite field? How about in higher dimensions?

There are three possibilities for supersingularity in terms of the p-adic Tate module - it can be $0,1$ or $2$ dimensional (over $\mathbb Z_p)$.

Similarly, there are three possibilities for the endomorphism ring- it can be a commutative (order in a) number ring of dimension $4$ or a non commutative division algebra of order $8$ or $16$ (dimensions over $\mathbb Z$). In the $8$ dimensional case, the center is a quadratic number ring and in the $16$ dimensional case, the center is $\mathbb Z$.

**What I know:** This mathoverflow question says that ordinary abelian surfaces over finite fields are always commutative. Why is this true and is this also true for higher dimensional abelian varieties?

In the case that the p-adic Tate module is one dimensional, we know that the endomorphism ring has to act faithfully on it and hence it has to be commutative.

In the case of a $16$ dimensional algebra, since the Frobenius is in the center and the center is $\mathbb Z$, the frobenius and it's dual are both just multiplication by some power of $p$ and so the surface is supersingular.

Conversely, I can show that in the supersingular case, the algebra is definitely not commutative.