Abel-Jacobi map on symmetric product of genus 4 curve Suppose $C$ is a genus $4$ smooth projective curve over complex numbers. 
The Abel-Jacobi map from $Sym^4(C)$ to $Jac(C)$ is birational.
Is this a blow-up along a surface or a curve. Can one determine this surface or curve?
 A: In any genus, it is true that the Abel-Jacobi map $\operatorname{Sym}^g(C)\rightarrow \operatorname{Jac}^g(C)$ is the blow up of the integral subvariety $W^1_g$ of $\operatorname{Jac}^g(C)$ parametrizing line bundles $L$ with $h^0(L)\geq 2$. This is a result of Muñoz-Porras, On the structure of the birational Abel morphism, 
Math. Ann. 281 (1988), no. 1, 1-6.
A: The locus in $Jac(C) = Pic^4(C)$ over which the map is not an isomorphism is the locus of degree 4 line bundles with $h^0 \ge 2$. By Serre duality this is the same as the locus of degree 2 line bundles with $h^0 \ge 2$, i.e., the image of $Sym^2(C)$ under the map 
$$
Sym^2(C) \to Pic^4(C),
\qquad 
(P,Q) \mapsto K_C - P - Q.\tag{$*$}
$$
If $C$ is not hyperelliptic, this map is a closed embedding and thus 
$$
Sym^4(C) \cong Bl_{Sym^2(C)} Jac(C).
$$
If $C$ is hyperelliptic, $Sym^2(C)$ contains a smooth rational curve (the hyperelliptic linear system) that is contracted by the map $(*)$, and the fiber of the Abel-Jacobi map over the corresponding point of $Jac(C)$ is $\mathbb{P}^2$. Most probably, the Abel-Jacobi map is still the blowup of the singular surface, the image of $Sym^2(C)$ under $(*)$.
