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Given a (complex) abelian variety $A$ of a fixed dimension $g$, let $d(A)$ be the dimension of the smallest complex projective space it embeds into.

Is $d(A)$ uniform over all abelian varieties of a fixed $g$? Or are there special ones that embed into even smaller projective spaces?

Can $d(A)$ be computed explicitly? I am particularly interested in the case $g = 2$.

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Recall that any smooth projective variety of dimension $g$ embeds into $\mathbf{P}^{2g+1}$. Consider now an abelian variety $A$ of dimension $g$ which embeds into $\mathbf{P}^{2g}$. Van de Ven proves (essentially by applying the self-intersection formula to the normal bundle of $A$ in $\mathbf{P}^{2g}$) that the degree of $A$ in $\mathbf{P}^{2g}$ is given by ${2g+1\choose g}$, and notes that the Riemann-Roch theorem implies that the degree has to be divisible by $g!$. This is possible only if $g=1$ or $g=2$. Of course elliptic curves embed into $\mathbf{P}^2$. One cannot embed abelian surfaces into $\mathbf{P}^3$, and if an abelian surface embeds into $\mathbf{P}^4$, then its degree is $10$. Abelian surfaces with this property exist (Mumford-Horrocks surfaces). This is all folklore. Maybe less known is the fact that Comessatti proved in 1919 (see this paper of Lange for a modern account) that for some curves of genus $2$ the Jacobian embeds into $\mathbf{P}^4$. More precisely: if $C$ is a curve of genus $2$ and $J(C)$ contains a curve $D$ with self-intersection $2$ and $C\cdot D=3$, then $J(C)$ embeds into $\mathbf{P}^4$ (with embedding given by $|C+D|$). But these should also be of the Mumford-Horrocks type (Theorem 5.2 in the paper of Mumford and Horrocks says that any abelian surface in $\mathbf{P}^4$ is projectively equivalent to one of theirs).

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  • $\begingroup$ Is it known what is the smallest degree that the image of $\mathbf{A}$ in $\mathbf{P}^{2g+1}$ can be? $\endgroup$ – Kim Jan 21 at 23:32
  • $\begingroup$ @Kim Possibly, but not to me. $\endgroup$ – Einfacher Schreiberling Jan 22 at 10:29

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