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Let $X$ be a curve over an algebraically closed field $k$ (even over $k = \mathbb{C}$ if you want), let $J = Pic^0_{X/k}$ be its Jacobian, let $P \in X(k)$ be a point, and let $i \colon X \hookrightarrow J$ be the closed immersion that sends $Q$ to the divisor class of $[Q] - [P]$. How does the map $$Pic^0(i): Pic^0_{J/k} \rightarrow J$$ compare with the canonical principal polarization $$p \colon J \rightarrow Pic^0_{J/k}$$ induced by the $\Theta$-divisor? Are $Pic^0(i)$ and $p$ inverse to each other?

I came across Lemma 6.9 in the following notes http://www.jmilne.org/math/xnotes/JVs.pdf by James Milne, and, if I understand correctly, this lemma claims that $Pic^0(i) = -p^{-1}$. I am very confused about the minus sign though: if $X = J$ is an elliptic curve and $P = 0$ is the origin, aren't both $Pic^0(i)$ and $p$ ``the identity''?

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    $\begingroup$ An elliptic curve has a canonical autoduality (if we maintain a clear distinction between an elliptic curve and its dual, as is always good to do; e.g., it removes any statement about a map being "the identity"), but apart from the appeal of the language of divisors is there a reason to prefer that one over its negative? The approach in Mumford's book, which uses line bundles rather than divisors, identifies a good property: the pullback along $(1,p)$ of the Poincare bundle should be ample! For that the autoduality in basic texts on elliptic curves is not the good one (it gives anti-ample). $\endgroup$
    – nfdc23
    Mar 14 '16 at 5:15
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The answer depends very much on how you define the homomorphism $p$ associated to the polarization. My personal choice is $p(a)=\mathcal{O}(\Theta _a-\Theta )$, where $\Theta $ is a theta divisor and $\Theta _a:=\Theta +a$. Realizing $\Theta $ as $i(\mathrm{Sym}^{g-1}C)$, you find that $i^*\Theta _a$ is the divisor $Q_1+\ldots +Q_g$, where the $Q_i$ are the solutions of $h^0(Q_i+(g-2)P-a)\geq 1$, or equivalently $h^0(K-(g-2)P+a-Q_i)\geq 1$; this implies
$i^*\Theta _a=K-(g-2)P+a\ $ for general $a$, hence for all $a$, and therefore $i^*p(a)=i^*\Theta _a-i^*\Theta=a$, so that $i^*\circ p$ is the identity.

Now Mumford in Abelian varieties, and others following him, take the opposite sign for $p\,$, which leads to the formula (unfortunate in my view) $\ i^*\circ p=-\mathrm{Id}$.

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    $\begingroup$ The motivation for Mumford's approach is a property that doesn't refer to divisors (and so better-suited to the relative situation): inspired by the dictionary between quadratic forms and symmetric bilinear forms, Mumford's notion of polarization $f:A \rightarrow A^{\vee}$ is a map symmetric with respect to double duality such that the pullback of the Poincare bundle $P_A$ along $(1,f):A \rightarrow A \times A^{\vee}$ is ample (akin to positive-definte quadratic form). For a divisor $D$, $t_a^{\ast}(O(D))=O(t_{-a}(D))$. So the sign is caused by the language of divisors. $\endgroup$
    – nfdc23
    Mar 14 '16 at 5:22

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