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Let $X$ be a Riemann Surface with genus $g$, $S^g(X)$ be the symmetric power of $X$ (which is naturally identified with the set of effective divisors of degree $g$). Let $A$ be the Abel-Jacobi map from $S^g(X)$ to $Jac(X)$. We know that for any $D \in S^g(X)$, $A^{-1}A(D)$ is given by the linear system $|D|$.

By computing the rank of $A$, we know that $A$ has constant rank along $A^{-1}A(D)$, but does this (maybe also using the Rank Theorem for holomorphic maps) imply $A^{-1}A(D)$ is a submanifold of $S^g(X)$?

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You will find a positive answer (and a precise description of the structure of the Abel-Jacobi map) in the reference given by Vivek Shende in his answer to the following related MO post : mathoverflow.net/questions/68687/… –  Olivier Benoist Oct 15 '13 at 11:41
    
It seems that there is no discussion about whether the fiber is a submanifold of $S^g(X)$(i.e., a subvariety without singularities). –  qcji Oct 15 '13 at 14:03
    
In the last statement (Theorem 4) of this paper, it is shown that the symmetric power of $X$ may be identified with the projectivization of a coherent sheaf on the jacobian of $X$. As a consequence, the scheme-theoretical fibers of the Abel-Jacobi map are projective spaces, hence smooth varieties. –  Olivier Benoist Oct 15 '13 at 16:42

1 Answer 1

Every fiber of $A$ is a projective space. If $V = H^0(X, \mathcal{O}(D))$, then $A^{-1}(A(D))$ is naturally identified with $\mathbb{P}(V)$. A divisor $E$ in $A^{-1}(A(D))$ is the zero locus of a nonzero function $f$ in $V$, and $f$ is uniquely determined up to scaling. So this gives a bijection between the points of $\mathbb{P}(V)$ and $A^{-1}(A(D))$.

To prove they are isomorphic as schemes, you should check that, for any $\mathbb{C}$-algebra $R$, the $R$-valued points of $\mathbb{P}(V)$ and $A^{-1}(A(D))$ are equal. There are probably references for this, but I don't know them

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