# Smooth curves of odd genus

Let $$C$$ be a smooth curve of genus $$g$$ and $$J_C$$ its intermediate Jacobian. Recall that $$J_C$$ is a ppav of dimension $$g$$.

Fixing a point $$p\in C$$, one can define the Abel-Jacobi map $$a\colon C\longrightarrow J_C$$ in the usual way and we know that the class of the image $$a(C)$$ is equal to $$\theta^{g-1}/(g-1)!$$. (Here and later, the image is always taken with the reduced scheme structure.) Analogously, one can define a map $$a_{g-1}$$ from the symmetric product $$C^{(g-1)}$$ to $$J_C$$ and in this case the image $$a_{g-1}(C^{(g-1)})$$ has the same class of the $$\theta$$ (both follows from the Poincare' Formula).

If now the genus $$g=2d+1$$ is odd, one can define a choice-point-free analog of the map $$a_{2d}$$ by considering the difference map $$b_d\colon C^{(d)}\times C^{(d)}\longrightarrow J_C$$ defined by sending $$((x_1,\cdots,x_d),(y_1,\cdots,y_d))$$ to the equivalence class of the divisor $$(x_1+\cdots+x_d)-(y_1+\cdots+y_d)$$.

My main question is: Is the image $$b_d(C^{(d)}\times C^{(d)})$$ in the same class of $$\theta$$?

I do not see why it should not, but on the other hand I find it tricky to prove that it is. Looking at the proof of Poicare' Formula for example, one of the main ingredients is the remark (which follows from Abel's Theorem) that $$a_{2d}$$ is $$1-1$$ onto its image: the rest of the proof is a local computation that seems to carry on without problems. So I was led to think that $$[b_d(C^{(d)}\times C^{(d)})]=k\theta$$, where $$k$$ is the degree of $$b_d$$. Is it true? And if it is, what is the degree of $$b_d$$?

I did not find any reference to this difference map anywhere, is there any?

P.S.: in particular, my curves are hyper-elliptic, so I am happy to restrict to this case if it helps; but by duality with the sum map, I do not expect such a dichotomy... In the hyper-elliptic case, using the existence of the $$g^1_2$$, the sum map $$a_{2d}$$ is as well choice-point-free, but I am mainly interested in the difference map.