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''?



Identifying the canonical principal polarization of a Jacobian
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) $\ i^\*\circ p=-\mathrm{Id}$.


