endomorphisms of the Jacobian of a curve Let $C$ be a smooth, projective curve of genus >1 over the complex numbers and let $J(C)$ be its Jacobian. The Torelli theorem relates the automorphisms of $C$ to the automorphisms of $J(C)$. Precisely, $Aut(J(C))$ is isomorphic to $Aut(C)$ if $C$ is hyperelliptic and to $Aut(C) \times \mathbb{Z}/2$ otherwise. 
Is there something similar for the endomorphisms of $J(C)$. My guess is that one has to relate them to correspondences on the curve, but I have been unable to find a reference.  
 A: You can relate $\mathrm{End}(JC)$ to correspondences on $C$, but this is essentially tautological : one defines the group $\mathrm{Corr}(C)$ as being $\mathrm{Pic}(C\times C)/(pr_1^*\mathrm{Pic(C)}+pr_2^*\mathrm{Pic}(C))$, then there is a natural isomorphism $\mathrm{Corr}(C)\rightarrow \mathrm{End}(JC)\ $ $-$ see for instance Birkenhake-Lange, Complex Abelian Varieties, Theorem 11.5.1. Unfortunately, as I said, this is essentially tautological; I don't think there is a result similar to the Torelli theorem, in particular taking into account the polarization.
A: For a principally polarized abelian variety $A$, there is an embedding $\text{NS}(A)\otimes\mathbb{Q}\hookrightarrow\text{End}(A)\otimes\mathbb{Q}$ and a Rosatti involution on $\text{End}(A)\otimes\mathbb{Q}$ so that the image of $\text{NS}(A)\otimes\mathbb{Q}$ is the subset fixed by Rosatti. (For what it's worth, this gives $\text{NS}(A)\otimes\mathbb{Q}$ the structure of a Jordan algebra.) If we take $A=\text{Jac}(C)$, this "reduces" the problem to computing the Neron-Severi group of $\text{Jac}(C)$, and one might then use the birational map $C^g\to\text{Jac}(C)$. But in the end, this simply reinforces the answer given by abx, namely that the structure of $\text{End}(\text{Jac}(C))$ depends on the higher-dimensional geometry of products of $C$.
