TITLE: Identifying the canonical principal polarization of a Jacobian
QUESTION [5 upvotes]: 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''?

REPLY [9 votes]: 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) $\ i^*\circ p=-\mathrm{Id}$.