TITLE: A map from a symmetric product of a curve to its Jacobian
QUESTION [5 upvotes]: Let $C$ be a smooth projective curve over an algebraically closed field $k$, of genus $g$.
It is well known that, after fixing a point $p_0$, the map $C^{(n)}\to J$ sending $\{a_1,\dots,a_n\}$ to $[a_1+\dots+a_n-np_0]$, from the n-th symmetric product of the curve to its Jacobian, is an algebraic projective bundle (for $n>2g-2$).
Consider the map $C^{(n)}\times C^{(n)}\to J$, sending $(\{a_1,\dots,a_n\},\{b_1,\dots,b_n\})$ to $[a_1+\dots+a_n-b_1-\dots-b_n]$.
Is this map also a fiber bundle?

REPLY [5 votes]: Let $J_n = Pic(C)_n$ --- the moduli space of line bundles of degree $n$ on $C$. Then there is a map
$$
C^{(n)} \to J^n,\qquad \{a_1,\dots,a_n\} \mapsto O(a_1+\dots+a_n).
$$
This is a slightly more canonical version of the map you considered, in particular it is a projective bundle for $n > 2g - 2$.
The map you are interested in can be written as the composition
$$
C^{(n)} \times C^{(n)} \to J_n \times J_n \cong J_n \times J_{-n} \to J_0.
$$
Here the first map is the product of to projective bundles (its fiber is a product of two projective spaces), the second is an isomorphism (given by dualization of a line bundle in the second factor), and the third is a (trivial) abelian fibration.
So, altogether, the fibers of your map are fiber products of two projective bundles over an abelian variety.