# 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?

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 $$(*)$$.

• many thanks for the explanation
– john
Jun 21 '19 at 9:23
• @Sasha can you provide some reference for this implication "closed embedding => blow up"?. I can't figure out why this map is actually a blow up and not some other map satisfying given properties (e.g. weighted blow up)?. Sep 11 '20 at 22:06

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.

• many thanks for the explanation and reference.
– john
Jun 21 '19 at 9:22