Division by two in the jacobian Let $X$ be an hyperelliptic complex curve of genus $3$, $J$ its jacobian,
and $P,Q$ two Weierstrass points on $X$. Can one write explicitly
a class $x$ in $J$ such that $2x = P-Q$?
 A: Expanding on my comment above, here is what the algorithm in my paper gives for a genus 3 Jacobian. We write the curve as
$$ y^2 = f(x) = x(x-a_1^2) \cdots (x - a_6^2) $$
and take $P = (0,0)$ and $Q = \infty$, the point at infinity.
We cannot directly apply Cor. 5.4 (it requires the point to
have non-zero $y$-coordinate), but we can still run the
algorithm. The result is that the point on the Jacobian
with Mumford representation
$$ (x^3 - \sigma_2 x^2 + \sigma_4 x - \sigma_6,
    (\sigma_1 \sigma_2 - \sigma_3) x^2
    - (\sigma_1 \sigma_4 - \sigma_5) x + \sigma_1 \sigma_6) $$
is one of the ``halves'' of $P - Q$. Here $\sigma_j$ denotes
the $j$th elementary symmetric polynomial in the $a_i$.
The other solutions are obtained by changing the signs
of some of the $a_i$.
(Let $(a(x), b(x))$ be the Mumford representation above.
The the point is represented by the divisor
$D - (\deg a) \cdot\infty$, where $D$ is the sum of points
$(\xi, b(\xi))$, where $\xi$ runs through the roots of $a$,
counted with multiplicities.)
