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.


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"

  • $\begingroup$ @agniesky: Are you saying that the Jacobian of a fixed curve can have more than one principal polarization? That is my question. $\endgroup$ – Jana Aug 6 '18 at 19:37
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    $\begingroup$ Yes, that is exactly what the answer says. To take another example, it is well-known that the Jacobian $J$ of the Klein quartic is isomorphic to $E^3$, where $E$ is the elliptic curve with complex multiplication by $i$; thus $J$ admits another principal polarization, which is reducible. $\endgroup$ – abx Aug 6 '18 at 19:50
  • $\begingroup$ @abx: Is there any condition under which an abelian variety has an unique principal polarization? $\endgroup$ – Jana Aug 6 '18 at 20:40
  • $\begingroup$ @Jana In general it is difficult to compute the number of ppal polarizations on an ab. variety. Generically, if it has one, it is the only one. But also, for any $N$ there exists $N$ genus 2 curves non isomorphic with all jacobians isomorphic. $\endgroup$ – Xarles Aug 6 '18 at 22:20

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