Which hypersurfaces in $\mathbb{P}^n$ are abelian varieties? 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.
 A: 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:


*

*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.

*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 $\binom{2d+1}{d}$.
