Naive question on the Jacobian of a curve Let $X$ be a smooth, projective curve of genus $g \ge 2$. We know that the Jacobian $J(X)$ of the curve is a principally polarized abelian variety. The principal polarization is induced by the intersection form (or cup-product) on $H^1(X,\mathbb{Z})$. My question is: Is this the only principal polarization on $J(X)$? In other words, is there another unimodular, alternating, non-degenerate bilinear form on $H^1(X,\mathbb{Z})$ different from the cup-product (this will induce a different principal polarization on $J(X)$)?
I would think this is very basic, but I am not able find a good literature for this question. Any hint/reference will be most welcome.
 A: It is possible for the Jacobian's  of non-isomorphic curves to be isomorphic as abelian varieties, but obviously, not as principally polarized abelian varieties.  This paper https://arxiv.org/pdf/math/0304471.pdf by Howe gives examples.  Also from the same paper - "it has been known since the late 1800s that distinct curves
can have isomorphic unpolarized Jacobians. I"
A: If the Neron-Severi group of an abelian variety is $\mathbb{Z}$, then a principal polarisation, if it exists, is unique. This is the generic case, as well as for generic jacobians.
To find counter-examples, one can look for abelian varieties with an automorphism which acts non-trivially on the Neron-Severi group. The example suggested by abx has this form.
A: The answer is no. Your question is equivalent to the question that is there only one curve on an abelian variety? I mean that let $C$ be a smooth, genus 2 curve on an abelian surface $A$. Then $J_C\simeq A$. Let $P_A$ be the set of isomorphism classes of smooth genus 2 curves on $A$. Then your question is that $|P_A|=1$? The answer is no. Moreover, it is unbounded. Hayashida and Nishi(1965,1968) raised this question and gave some partial results, i.e., formulae for $|P_A|$. Recently, E. Kani(2014,2016) gave some answer for this question. Check their related papers.
