Over an algebraically closed field $k$, which smooth hypersurfaces $X \subset \mathbb{P}^n$ are abelian varieties? 


If $n=2$, then the smooth hypersurfaces of degree 3 (i.e. elliptic curves) are abelian varieties. Furthermore, the degree $d$ of the hypersurface $X$ must be $n+1$ (so that the canonical bundle is trivial), but I doubt that this condition is sufficient.



Which hypersurfaces in $\mathbb{P}^n$ are abelian varieties?
Really this is mostly just consolidating what has been (implicitly) said in the comments and cleaning it up a bit (e.g. using the Chow ring instead of singular cohomology), but might as well make it an answer...


Let $A$ be an abelian variety of dimension $d$. We prove that $A$ cannot embed in $\mathbb{P}^{2d-1}$ and can only embed in $\mathbb{P}^{2d}$ if $d=1$ or $2$. The algebraic version of Whitney embedding shows that $A$ can be embedded into $\mathbb{P}^{2d+1}.$ I believe the second half of that was first proved in *On the embedding of abelian varieties in projective spaces*, Van de Ven. The only facts we use about abelian varieties are the following:


1. The tangent bundle of an abelian variety is trivial. This is intuitively because the tangent spaces at different points on $A$ can be identified by a translation.
2. Any divisor $D$ on an abelian variety has $D^n$ divisible by $n!$. This is easy to see in characteristic 0 and follows in general from e.g. Hirzebruch-Riemann-Roch.


Now take an embedding $f$ of $A$ into $\mathbb{P}^n$ and let $h$ be the pullback of the class of a hyperplane in Chow. We have the exact sequence $0\rightarrow T\_A\rightarrow f^\*T\_{\mathbb{P}^n}\rightarrow N\_{A/\mathbb{P}^n}\rightarrow 0.$ Taking Chern classes, we see that $c(N\_{A/\mathbb{P}^n})=(1+h)^{n+1},$ and in particular, $c\_d(N\_{A/\mathbb{P}^n})=\binom{n+1}{d}h^d\neq 0.$ This immediately shows that $N\_{A/\mathbb{P}^n}$ has rank at least $d$, and hence that $n$ must be at least $2d$.


Assume $n=2d$. Then $c\_d(N\_{A/\mathbb{P}^n})$ is the self intersection number of $A$ (times the class of a point). If $h^d$ is $k$ times the class of a point, then the self intersection number of $A$ is $\binom{n+1}{d}k.$ But then $A$ also has degree $k$, and so the self intersection number of $A$ is also $k^2.$ This shows $k=\binom{n+1}{d}$, and so by fact 2, we see that $\binom{2d+1}{d}$ must be divisible by $d!.$ This can explicitly be checked to be false for low $d$ not equal to $1,2$, and for $d$ at least $8$, $d!$ is larger than $\binom{2d+1}{d}$.


