Classification of Principally polarized abelian surface.

Question

We know that a principally polarized abelian surface is either the Jacobian of a smooth curve of genus 2 or the canonically polarized product of two elliptic curves. Can anyone suggest me proof of that.

Answer 1

There is a beautiful theorem, called Matsusaka-Ran criterion, that can be used to give a direct proof of what you are saying. The Matsusaka-Ran criterion says the following

Let $X$ be a $g$-dimensional abelian variety with polarization $\mathcal{O}(D)$. Let $C=\sum_{i=1}^{l} r_iC_i$ be an effective algebraic 1-cycle such that $C$ generates $X$ and $(C\cdot D)=g$. Then $r_i=1$ and $C_i$ is smooth for each $i$ and there exists an isomorphism $\psi:J(C_1)\times...\times J(C_l) \simeq X$.

Now, let $X$ be an abelian surface with principal polarization $\mathcal{O}(D)$, so that in this case $D$ is an effective algebraic 1-cycle. Because of the fact that $\mathcal{O}(D)$ is ample, we can apply the Kodaira vanishing theorem which tells us that $0=H^i(X,\mathcal{O}(D)\otimes K_X)=H^i(X,\mathcal{O}(D))$ for $i>0$, where in the last passage we used the fact that the canonical bundle of abelian varieties is trivial. Thus $\chi(\mathcal{O(D)})=h^0(\mathcal{O}(D))=1$, where in the last passage we used the hypothesis that $\mathcal{O}(D)$ is a $\textit{principal}$ polarization.

The Riemann-Roch theorem for surfaces tells us that $2\chi(\mathcal{O}(D))=(D\cdot D)$: using what we have discovered above we can conclude that $(D\cdot D)=2$. Thus we are in position to apply the Matsusaka-Ran criterion.

If $D$ is irreducible, it tells us that $D$ is a smooth curve and $X=J(D)$. By the fact that dim$(J(D))=$ dim$(X)=2$ and that dim$(J(D))=g(D)$, we deduce that $D$ has genus two.

If $D$ is reducible, because of the dimension of $X$ the only possibility is that $X\simeq J(D_1)\times J(D_2)$, with $D=D_1+D_2$ and $J(D_i)$ of dimension one for $i=1,2$. This implies that the $D_i$ are both elliptic curves, so that $J(D_i)\simeq D_i$, thus $X\simeq D_1\times D_2$, and we are done.