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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?

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    $\begingroup$ @Wojowu There is actually a much easier proof that every genus 1 curve over a finite field has a point, and it generalizes to the statement that if $F$ Is a finite field and $A/F$ is a variety such that $A/\overline{F}$ is an abelian variety, then $A(F)\ne\emptyset$, so $A/F$ is itself already an abelian variety. (I have a recollection this may be due to Lang, but I could be wrong.) $\endgroup$
    – Joe Silverman
    Commented Aug 27, 2020 at 12:50
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    $\begingroup$ Have you considered the dimension 2 case? Clearly a necessary condition is that the Kodaira dimension be 0. There are just a few such classes of varieties, so you can check to see whether their discrete invariants suffice to pick out the abelian surfaces. $\endgroup$
    – Joe Silverman
    Commented Aug 27, 2020 at 12:52
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    $\begingroup$ I don't know about finite fields, but over $\mathbb C$ by arxiv.org/abs/math/9903184 and arxiv.org/abs/math/0011042, if $h^0(2K_X)=1$ and $h^0(\Omega ^1_X)=\dim X$, then $X$ is birational to an abelian variety. If $k$ is an algebraically closed field of char $p>0$, then related results are contained in arxiv.org/abs/1703.06631. $\endgroup$
    – Hacon
    Commented Aug 27, 2020 at 15:20
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    $\begingroup$ Sommese, Quaternion Manifolds (1975) link.springer.com/article/10.1007/BF01357140 constructed a complex 3-fold which is diffeomorphic to $(S^1)^6$ but not a complex torus. So there is grounds for caution. I found a readable description of the construction in the first pages of projecteuclid.org/euclid.kjm/1291041217 $\endgroup$
    – David E Speyer
    Commented Aug 27, 2020 at 18:16
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    $\begingroup$ Okay, I read further in the Catanese et al paper, and I retract my warning. According to Theorem 2.3, if $X$ is a compact complex manifold which (1) has the integer cohomology ring of a torus and (2) is bimeromorphic to a compact Kahler manifold, then $X$ is a complex torus. In particular, if (1) holds and $X$ is algebraic then Chow's lemma tells us it is birational to a projective variety, so this should show it is an abelian variety. That doesn't prove anything in char p, but it means that I no longer consider this a reason for caution. $\endgroup$
    – David E Speyer
    Commented Aug 27, 2020 at 19:41

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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.

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    $\begingroup$ the last sentence is incorrect. One can take $X$ to be a non-classical Enriques surface in characteristic $2$ arxiv.org/abs/1608.05198 $\endgroup$
    – Nguyen
    Commented Aug 28, 2020 at 7:01

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