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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.

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    $\begingroup$ At least in characteristics zero, Lefshetz hyperplanes theorem, together with $h^{1,0}\neq 0$ for abelian varieties shows that $n=2$ is the only case. $\endgroup$
    – Lev Borisov
    Feb 27, 2016 at 2:25
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    $\begingroup$ At least over $\mathbb{C}$, there are many obstructions, even just topologically - for example all of the Chern classes also need to vanish, and a straightforward computation with the Euler sequence and the Lefschetz hyperplane theorem shows that this only happens when $n = 2, d = 3$. This computation also shows that smooth hypersurfaces almost always have nonzero Euler characteristic (for example the degree $4$ hypersurface in $\mathbb{P}^3$ is a K3 surface and has Euler characteristic $24$) so maybe we can use etale cohomology or something in other characteristics. $\endgroup$
    – Qiaochu Yuan
    Feb 27, 2016 at 2:44
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    $\begingroup$ Indeed I thought it was known that a $d$-dimensional abelian variety doesn't even embed smoothly into ${\bf CP}^{\,2d}$ except for $d=1$ (cubic curves) and $d=2$ (Horrocks-Mumford surfaces). $\endgroup$
    – Noam D. Elkies
    Feb 27, 2016 at 2:47
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    $\begingroup$ @QiaochuYuan the Lefschetz hyperplane theorem and facts about Chern classes hold in etale cohomology so this all works great in characteristic $p$. $\endgroup$
    – Will Sawin
    Feb 27, 2016 at 5:14
  • $\begingroup$ One can calculate the étale cohomology of a smooth hypersurface using purity/Gysin sequence and compare them with the cohomology groups of projective space (see e.g. Milne's lecture notes, chapter 16). The étale cohomology ring of an abelian variety is an exterior algebra over $H^1$. $\endgroup$
    – user19475
    Feb 1, 2018 at 13:53

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

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