Let $X$ be a hyperelliptic curve and let $i:X\to X$ denote the hyperelliptic involution. Once we fix a point $x_0\in X$ we get the AbelJacobi map $AJ:X\to J$ where $J$ denotes the Jacobian variety. Now the Jacobian is also equipped with an involution, namely $x\mapsto x^{1}$. Is it possible to choose the base point $x_0$ in such a way that the restriction of the involution on the Jacobian is the involution on $X$.
2 Answers
Yes, pick $x_0$ to be a Weierstrass point, i.e. a fixed point of the hyperelliptic involution.
Let $\sigma$ denote the hyperelliptic involution on $X$. Under the AbelJacobi map we have $x \mapsto [xx_0]$ and $\sigma(x) \mapsto [\sigma(x)  x_0]$. Now $[x + \sigma(x)  2x_0]$ is the divisor of a function, since it is the pullback under the hyperelliptic map of a degree zero divisor on $\mathbf P^1$. Hence $AJ(x)$ and $AJ(\sigma(x))$ are inverses.
Another way to think of this is the following (at least over $\mathbb{C}$).
Consider the diagram $$ \begin{array}{ccccc} X & \xrightarrow{AJ \times AJ \circ \sigma} & J \times J & & \\\\ \downarrow & & \downarrow & & \\\\ \mathbb{P}^1 & \to & Sym^2J & \xrightarrow{+} & J \end{array}. $$ Note that the composition along the bottom row must be constant, as there are no nontrivial maps from $\mathbb{P}^1$ to any Abelian variety.
What this tells you is that $f(x) = AJ(x) + AJ\big(\sigma(x)\big)$ is constant. So if you translate it (i.e. choose a different basepoint): $$ \tilde{AJ}(x) = AJ(x)  \frac{f(x)}{2} $$ then you find that the corresponding $\tilde{f}(x) = 0$ for all $x$. That is, the two involutions agree with each other as desired.

$\begingroup$ There seems to be a small argument missing here, since not every translation on $\mathrm{Pic}^0$ comes from a different choice of basepoint. $\endgroup$ Commented May 13, 2011 at 6:55

$\begingroup$ If $x_0$ is chosen to be a Weierstrass point then the image of $x_0$ (and hence all of $\mathbb{P}^1$) in $J$ is 0, so there is no need to translate. $\endgroup$ Commented May 13, 2011 at 14:40