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Why is there only one canonical principal polarization per Jacobian? I don't yet see why it is true, but I have seen "the canonical polarization" stated many times. This is perhaps a naive question, but I am nevertheless confused.

Let us say we have two different curves $C$, $C'$ (over a perfect field $k$), such that $A:= Jac(C) \simeq Jac(C')$ as unpolarized abelian varieties.

The canonical principal polarization (thought of as a symplectic form on $Jac(C)$) from $C$ associated to $Jac(C)$ comes from the intersection form $Q: H_1(C; \mathbb{Z}) \otimes H_1(C; \mathbb{Z}) \to\mathbb{Z}$. This is because $H^2(Jac(C); \mathbb{Z}) \simeq Hom(\bigwedge H_1(C; \mathbb{Z}); \mathbb{Z})$.

Let us say that $C$ and $C'$ are not isomorphic, but they have isomorphic unpolarized Jacobians $A:= Jac(C) \simeq Jac(C')$ (such curves exist, e.g. this paper). Then, $C'$ must produce a different canonical principal polarization of $A$ than $C$, otherwise $C$ and $C'$ would be isomorphic by Torelli's theorem.

I have heard that the canonical principal polarization may be equivalently defined by the polarization associated to the Poincare correspondence on $Jac(C) \times Jac(C) \simeq Jac(C') \times Jac(C')$, which indicates there should only be one canonical principal polarization on $Jac(C) \simeq Jac(C')$. I do not understand (and cannot locate written) why the associated polarization of this correspondence is principal.

Here are my questions:

Let $A$ be an abelian variety in the Jacobian locus.

  1. Is there only one canonical principal polarization of $A$?

  2. If so, why is there only one canonical principal polarization of $A$, especially if $A$ is the Jacobian of multiple non-isomorphic curves?

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  • $\begingroup$ As soon as one chooses a rational point on $C$, one obtains a principal polarization on $Jac(C)$ in a manner that requires no additional choices. Is it possible that your curves $C$ already have a natural choice of basepoint (eg, they are elliptic curves)? $\endgroup$
    – KReiser
    Jul 1, 2018 at 23:42
  • $\begingroup$ Does the principal polarization depend on the choice of rational point? No, these are genus 2 curves. $\endgroup$
    – Catherine Ray
    Jul 2, 2018 at 4:06
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    $\begingroup$ The polarization doesn't depend on the choice of a rational point. Milne's notes, section 6 (see jmilne.org/math/xnotes/JVs.pdf) provide an explanation. It seems that Howe's construction picks a particular principal polarization - not necessarily the canonical one, and there's no reason for there to even be a map of polarized abelian varieties between the two of them. Perhaps this potential explanation for q2 is just (incorrect) speculation, though. Howe's original 1996 paper may prove instructive: sciencedirect.com/science/article/pii/S0022314X96900268 $\endgroup$
    – KReiser
    Jul 2, 2018 at 8:06
  • $\begingroup$ @KReiser: Thank you, I looked over this orginal paper, but I am still flummoxed regarding the difference between the principal polarization $a_C$ of a $A \simeq Jac(C)$ coming from a curve $C$, and the canonical principal polarization $a$ of $A$. I understand how $a_C$ is constructed via the intersection form of $C$, and by this paper of Howe, it need not coincide with $a$. So how is $a$ constructed? $\endgroup$
    – Catherine Ray
    Jul 4, 2018 at 20:35

1 Answer 1

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There is only a canonical principal polarization on $A$ if a curve is specified. More precisely, to get a canonical principal polarization, it is not sufficient to know that $A$ is in the essential image of the functor $\text{Jac}: \text{RiemSurf} \to \text{AbVar}_{/\mathbb{C}}$. We may only specify a canonical principal polarization with respect to $C$ if we know that $A = \text{Jac}(C)$.

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