Theta-divisor and special divisors I'm studying the Jacobi inversion problem. Could you help me with the next question?
Let us consider an algebraic curve $C$ and it's Jacobian $J(C)$. For a initial point $P_0$ we can construct the Abel map $\mathcal{A}\colon C \mapsto J(C)$.
We denote by $K$ the vector of Riemann constants of $\mathcal{A}$. These constants depend on the choice of initial point.   
One can define the Theta-divisor $\Theta$ in $J(C)$ as the set of zeros of $\theta(z-K)$. Here $\theta(z)$ is theta-function of $C$. 
Is it true that the image $\mathcal{A}(D)$ of any special (not general) divisor $D$ belongs to $\Theta$?
Thanks.
 A: There is a relationship between $\Theta$-divisors and global sections of line bundles on $Pic_{g-1}(C)$. But the relationship is not obvious and may not be what you wanted. 
Let $C$ be a Riemann Surface, let $Pic_{g-1}(C)$ be the Picard variety of line bundles of degree $g-1$ on $C$. Let $E$ be a divisor of degree $r+g-1$. Then $O_{X}(D)\in Pic_{g-1}(X)$ can be viewed as $O_{X}(E-P_1\cdots -P_r)$. Therefore by fixing $E$, we can define a map 
$$
\phi: X^{r}\rightarrow Pic_{g-1}(X)\cong Jac(X)
$$
And the pull-back of $O_{J}(-\Theta)$ via $\phi$ is precisely the line bundle whose fibre defines the so-called "Faltings volume" of $\det (O_{X}(D))$. A rather dense proof can be found in Lang's book, Chapter V. 
A: The answer is yes.
It is known that $\Theta$ can be parametrized by effective divisors of degree $g-1$ via $\Theta = \{A(D') + K \mid D' \in C^{[g-1]} \}$. If $deg(D) \leq g-1$ then being special means up to equivalence $D$ is an effective divisor. Then $A(D) = A(D')$ where $D' \in C^{[g-1]} $ is defined by adding to $D$ an appropirate number of the reference point $P_0$. If $\deg(D) > g$, then being special means we have at least $\deg(D) - (g-1)$ degrees of freedom to vary the points defining $D$. We then take advantage of these degrees of freedom to move at least $\deg(D) - (g-1)$ points defining $D$ to $P_0$, and again obtain $A(D) = A(D')$ for some $D' \in C^{[g-1]}$.
