Abel map on the symmetric product: why it is locally boundedness and holomorphicitiy I'm studying Jacobian Varieties from Griffith's Introduction to Algebraic Curves and I have a problem with this: let $u$ be the Abel map
\begin{align*}
u\colon  C^{(d)} && \longrightarrow && J(C)\\
\sum{n_i p_i} && \longmapsto && \sum n_i \left( \int_{q}^{p_i}\omega_1,...,\int_{q}^{p_i}\omega_g \right)
\end{align*}
with $C^{(d)}$ the $d$-symmetric product and $J(C)$ the Jacobian of the curve $C$. Why it is holomorphic and locally bounded near a generic divisor $D=p_1 + ... + p_d\in C^{(d)}$ (generic means that $p_i$ are distinct).
More in general, I need to understand why the Abel map is holomorphic on the $d$-symmetric product of $C$.
Thanks for your help!
 A: Question:  "I need to understand why the Abel map is holomorphic on the d-symmetric product of C."
Special case: If $C:=E$ is an elliptic curve, it is known that $Sym^d(E) \cong \mathbb{P}(F^*)$ is the projective bundle of a locally trivial $\mathcal{O}_{J(E)}$-module
$F$ on $J(E)\cong E$ and the projective bundle $\mathbb{P}(F^*)$ is a complex manifold with the canonical map $\pi: \mathbb{P}(F^*) \rightarrow J(E)$ a holomorphic map. The projection map $\pi$ "commutes" with your map $u$ under this isomorphism. In fact it holds more generally by  "Geometry of Algebraic Curves" by Arbarello-Cornalba-Grifiths-Harris. You find some references at the MO site:
https://mathoverflow.net/questions/68687/symmetric-powers-of-a-curve-projective-bundle-over-jacobian-and-the-relative?rq=1
Speculation: In the book you are reading they are constructing a structure as complex manifold on $Sym^d(C)$ using the elementary symmetric functions. If $t,x_1,..,x_n$  are independent variables $f(t)$ the polynomial
$$f(t):=(t-x_1)\cdots (t-x_n):=t^n-s_1(x_i)t^{n-1}+s_2(x_i)t^{n-2}+\cdots +(-1)^ns_n(x_i)$$
has the elementary symmetric functions $s_1,..,s_n$ as coefficients. There is an action of the symmetric group $S_n$ on the polynomial ring $A:=\mathbb{Z}[x_1,..,x_n]$ and the invariant ring $A^{S_n}\cong \mathbb{Z}[s_1,..,s_n]$ is the ring generated by these functions. These functions are algebraically independent over $\mathbb{Z}$. Use these functions to construct a structure as complex manifold on $Sym^n(C)$. You must construct an atlas $U_i$ on $Sym^n(C)$ and local coordinates on $U_i$ and prove that the transition maps on $U_i\cap U_j$ are holomorphic. You prove existence of a map
$$u: Sym^n(C) \rightarrow J(C)$$
which is "generically holomorphic" and then the RE theorem proves the result.
An alternative construction using symmetric products: If $C \subseteq \mathbb{P}^n_k$ is a smooth quasi projective curve over the complex numbers $k$, you may cover the product $C^{\times n}$ by open affine $S_n$-invariant subschemes $U_i:=Spec(A_i)$ where $A_i$ is a one dimensional regular ring over $k$. You must prove the invariant ring
$$B_i:=(A_i^{\otimes_k^n})^{S_n} \subseteq A_i^{\otimes_k^n}$$
is a regular ring and that $V_i:=Sym^n(U_i):=Spec(B_i)$ is an open cover of $Sym^n(C)$. Since $V_i$ is a regular $n$-dimensional algebraic variety over $k$ it follows $V_i$ has the structure of a complex manifold and since $\cup V_i = Sym^n(C)$ is an open cover it follows $Sym^n(C)$ is a complex manifold.
You must give a construction of the map
$$u: Sym^n(C) \rightarrow J(C)$$
and if $u$ is algebraic it follows $u$ is holomorphic. In the case when $Sym^n(C)$ and $J(C)$ are projective it follows $u$ is algebraic iff $u$ is holomorphic.
The definition in the book uses complex analysis and the map $u$ is holomorphic iff for any open set $V \subseteq J(C)$ and any $s\in \mathcal{O}_{J(C)}(V)$ it follows
$$H1.\text{  }s \circ u_V \in \mathcal{O}_{Sym^n(C)}(u^{-1}(V)),$$
hence you must give an explicit construction of atlases for $Sym^n(C)$ and $J(C)$ and verify condition $H1$. With the construction above you get an atlas $V_i$ of $Sym^n(C)$.
Example: If $A:=\mathbb{C}[x]$ and $C:=Spec(A)$ it follows $Sym^n(C)\cong \mathbb{A}^n_{\mathbb{C}}$ and there is a canonical projection map
$$\pi: C^{\times n} \rightarrow Sym^n(C)$$
defined by
$$\pi(x):=(s_1(x),..,s_n(x)).$$
The "discriminant locus" of the map $\pi$ is a hypersurface $D_n \subseteq Sym^n(C)$ and the complement $U:=Sym^n(C)-D_n$ has the property that the induced map
$$\pi_U: \pi^{-1}(U) \rightarrow U$$
has $\#\pi^{-1}(x):=n!$ - the cardinality of the fiber is constant. The map $\pi_U$ is a "principal $S_n$-bundle", hence there is a local trivialization $W_i$ of $\pi_U$ with
$$\pi_U^{-1}(W_i) \cong W_i \times S_n$$
and where the induced map is the projection map. But $W_i \times S_n \cong W_i \times \cdots \times W_i$ is a product and the projection map is a holomorphic map. Hence it seems $\pi_U$ is holomorphic. A similar result holds for any affine smooth curve $C$. Generically the map
$$\pi: C^{\times n} \rightarrow Sym^n(C)$$
is a principal $S_n$-bundle. There is a discriminant hypersurface $D_n \subseteq Sym^n(C)$ whose complement $U_n:=Sym^n(C)-D_n$ has the property that
$$\pi_{U_n}:\pi^{-1}(U_n) \rightarrow U_n$$
is a principal $S_n$-bundle. The locus $U_n$ are divisors $D:=p_1+\cdots +p_n$ on $C$ with $p_i$ distinct. The map $\pi_{U_n}$ is locally a projection map.
Note: The Jacobian variety $J(C)$ is related to the Picard group $Pic(C)$ of line bundles on $C$: Over a number field $K$ you may define the Jacobian variety as $Pic^0(C)$ - the abelian variety parametrizing degree $0$ line bundles on $C$. Hence over other fields you need line bundles in order to define $Jac(C)$.
