TITLE: If it quacks like an abelian variety over a finite field
QUESTION [14 upvotes]: Consider smooth projective varieties over a finite field. If a curve "looks like" an elliptic curve (i.e. has genus $1$) then it can be made into an elliptic curve.
Is there something similar in higher dimensions, i.e. if we fix the values of some invariants can we guarantee that the variety admits an algebraic group structure? What is sufficient for threefolds?

REPLY [8 votes]: One possible answer to this could be Lang's theorem: it says that if $G/\mathbb{F}_q$ is a smooth connected algebraic group, then $H^1(\mathbb{F}_q,G)$ is trivial, or otherwise put every $G$-torsor has an $\mathbb{F}_q$-rational point.
This generalizes your example: if $X/\mathbb{F}_q$ is a smooth projective variety such that $X_{\bar{\mathbb{F}}_q}$ is isomorphic to an abelian variety, then $X$ is a torsor under its Albanese variety $A = Alb(X)$.
If $X$ has dimension $1$ then requiring the genus of $X$ to be $1$ is enough.
If $X$ has dimension $2$ then by the classification of surfaces it is enough to assume that e.g. the canonical bundle of $X$ is trivial and $X$ is not simply connected.
Edit: as pointed out below, the previous sentence is correct only if the characteristic of $\mathbb{F}_q$ is $\neq 2,3$. If we additionally assume that the second $l$-adic Betti number equals $6$, then we get a criterion in all characteristics.