In this post, I will discuss Jacobi’s inversion theorem. It is a follow up in a series of posts about Riemann surfaces, the first of which being this one. This theorem tells us in what sense the Abel-Jacobi map $\mu$ is surjective.

Theorem (Jacobi’s inversion): Let $S$ be a compact Riemann surface of genus $g$ and take any $p_0 \in S$. Then for any $\lambda \in J(S)$, we can find $g$ points $p_1, \dots, p_g \in S$ such that

$\mu\left(\displaystyle\sum_{i=1}^g(p_i-p_0)\right) = \lambda$.

In other words, for $\omega_1, \ldots, \omega_g$ a basis of $H^0(S,\Omega^1)$, for all $\lambda \in \mathbb{C}^g$, there is $p_1, \ldots, p_g \in S$ and paths $\alpha_i$ from $p_0$ to $p_i$ such that

$\displaystyle\sum_{i=1}^g \int_{\alpha_i}\omega_j = \lambda_j$     for all $1\leq j \leq g$.

With this theorem coupled with Abel’s theorem, we will have completely proved the exactness of

$\mathcal{M}(S) \overset{()}\to \text{Div}_0(S) \overset{\mu}\to J(S) \to 0$.

Lemma 1: The set $S^{(d)}$ of effective divisors of degree $d$ on $S$ is a compact complex manifold.

Proof of lemma 1: Consider the action of the permutation group $\mathfrak{S}_d$ on

$S^d = S \times \cdots \times S$    ($d$ times).

The quotient, denoted $S^d/\mathfrak{S}_d = \text{Sym}^d(S)$, inherits the quotient topology making $\pi : S^d \to \text{Sym}^d(S)$ a continuous map. Clearly, $\text{Sym}^d(S)$ is in bijection with $S^{(d)}$. Suppose for a moment that $S = \mathbb{C}$. We will write an element of $\text{Sym}^d(\mathbb{C})$ as a sum $\lambda_1 + \cdots \lambda_d$ to indicate the unimportance of the ordering (note that the $\lambda_i$‘s are not necessarily distinct). We consider the map

$\varphi : \text{Sym}^d(S) \to \mathbb{C}^d$

that takes $\sum_{i=1}^d\lambda_i$ to the d-tuple $(a_1, \ldots, a_d) \in \mathbb{C}$ consisting of the coefficents of the monic polynomial having $\lambda_1, \ldots, \lambda_d$ as its roots, i.e

$\varphi(\lambda_1 + \cdots \lambda_d) = (a_1, \ldots, a_d)$

if

$z^d + a_1z^{d-1} + \cdots a_{d-1}z + a_d = (z-\lambda_1)\cdots (z-\lambda_d)$.

In other words, $\varphi(\{\lambda_i\}) = (\sigma_1(\{\lambda_i\}), \ldots, \sigma_d(\{\lambda_i\})$ where $\sigma_i$ is the ith elementary symmetric polynomial. By the fundamental theorem of algbera, $\varphi : \text{Sym}^d(\mathbb{C}) \to \mathbb{C}^d$ is a bijection, and in fact a homeomorphism.

The difficulty is in seeing that $\varphi^{-1}$ is also continuous. To see this, take an open set $\pi (U) \subset \text{Sym}^d(\mathbb{C})$ and consider $(a_1, \ldots, a_d) \in V := \varphi (\pi (U)) \subset \mathbb{C}^d$. In other words,

$p_0 := z^d + a_1z^{d-1} + \cdots + a_{d-1} + a_d = \displaystyle\prod_{i=1}^d(z-\lambda_i)$     with $\lambda_i \in U$.

We need to show that there is an open set $W \subset V$. Write

$N_U(p) = \dfrac{1}{2\pi\sqrt{-1}}\displaystyle\int_{\partial U} \frac{dp}{p}$

which, when defined, is the number of zeros of the polynomial $p$ inside $U$. We can choose an open set $W \subset \mathbb{C}^d$ containing $p_0$ small enough so that $N_U(p)$ is defined for all $p \in W$. Then since $p \mapsto N_U(p)$ is continuous and takes only integer values, we have

$N_U(p) = N_U(p_0) = d$    for all $p \in W$.

But this means that for all $p \in W$, all the roots of $p$ are in $U$ so that $p_0 \subset W \subset V$ as we wanted.

This gives $\text{Sym}^d(\mathbb{C})$ the structure of a complex manifold of dimension $d$, in fact biholomorphic to $\mathbb{C}^d$.

Now we put a similar complex manifold structure on $\text{Sym}^d(S)$ for arbitrary $S$. Consider $D = \sum_{i=1}^dp_i \in \text{Sym}^d(S)$ and take holomorphic charts $(U_i,z_i)$ around $p_i$ on $S$ such that $U_i \cap U_j = \emptyset$ if $p_i \neq p_j$ and $U_i = U_j, z_i = z_j$ if $p_i = p_j$. We obviously have an injective mapping

$\varphi_D : \pi(U_1 \times \cdots U_d) \to V_D \subset \mathbb{C}^d$

where $\pi : S^d \to \text{Sym}^d(S)$, by doing as above;

$\varphi_D(\sum_jz_j) = \left(\sigma_1(\{z_j\}), \ldots, \sigma_d(\{z_j\})\right)$.

The verifications that this gives a homeomorphism onto an open set of $\mathbb{C}^d$ is just as in the earlier case. These maps thus provide $\text{Sym}^d(S)$ with a holomorphic atlas.

QED

Fixing a base point $p_0 \in S$ we get a (holomorphic) injection

$\iota : S^{(d)} \hookrightarrow \text{Div}_0(S)$

$\iota : \displaystyle\sum_{i=1}^d p_i \mapsto \sum_{i=1}^d(p_i-p_0)$

and thus holomorphic mappings

$\mu^{(d)} : S^{(d)} \to J(S)$

$\mu^{(d)} : \displaystyle\sum_{i=1}^dp_i \mapsto \sum_{i=1}^d \left(\int_{p_0}^{p_i}\omega_1, \ldots, \int_{p_0}^{p_i}\omega_g\right)$    (mod $\Lambda$)

by composing $\mu$ and $\iota$. Jacobi’s inversion theorem says that $\mu^{(g)}$ is surjective.

Lemma 2:  Let $f : M \to N$ be a holomorphic map between two compact connected complex manifolds of the same dimension. If $f$ is not everywhere singular, i.e. the Jacobian matrix $J(f)$ is not identically zero, then $f$ is necessarily surjective.

Proof of lemma 2: This is immediate from the proper mapping theorem which says that if $V \subset M$ is an analytic subvariety, then $f(V)$ is an analytic subvariety of $N$. In this case, $f(M)$ would be a compact subvariety containing an open set which would mean $f(N) = M$. Griffiths-Harris presents a more elementary proof which does not use the rather deep proper mapping theorem:

Consider $\psi_N$ a volume form on $N$. Since $J(f)$ is not identically $0$ and since $f$ preserves the orientation (being holomorphic), we have

$\displaystyle\int_Mf^*\psi_N > 0$.

Since for any $q \in N$ we have $H^{2n}(N-\{q\},\mathbb{R}) = 0,$ the volume form is exact in $N - \{q\}$ and

$\psi_N = d\varphi$

for some $(2n-1)$-form $\varphi$ on $N - \{q\}$. But then if $q \notin f(M)$, we have

$\displaystyle\int_M f^*\psi_N = \int_{\partial M}df^*\varphi = 0$,

QED

To prove the theorem, we thus have to show that $\mu^{(g)}$ is not everywhere singular.

Proof of the theorem: At points $D = \sum_ip_i \in S^{(d)}$ such that the $p_i$‘s are distinct, the quotient map $\pi : S^d \to S^{(d)}$ is locally a biholomorphism. So choosing disjoint charts $(U_i,z_i)$ in $S$ centered at $p_i$, we get a chart

$\varphi : \pi(U_1 \times \cdots \times U_d) \to \mathbb{C}^d$

$\varphi(\sum_iq_i) = (z_1(q_1), \ldots, z_d(q_d))$.

In such coordinates, for $D' = \sum_iq_i$ near $D$, we have

$\mu^{(g)}(D') = \displaystyle\sum_{i=1}^g\left(\int_{p_0}^{z_i}\omega_1, \ldots, \int_{p_0}^{z_i}\omega_g \right)$     mod $\Lambda$.

So

$\dfrac{\partial}{\partial z^i}(\mu^{g}_j(D')) = \displaystyle\sum_{k=1}^g \dfrac{\partial}{\partial z^i}\left(\int_{p_0}^{z_k}\omega_j\right)$.

But with $\omega_j(z^i) = h_{ji}(z^i)dz^i$ near $p_i$, this is

$\dfrac{\partial}{\partial z^i} \displaystyle\int_{p_0}^{z_i}h_{ji}(z^i)dz^i$

hence

$\dfrac{\partial}{\partial z^i} (\mu^{(g)}_j(D')) = h_{ji}(z^i).$

The jacobian matrix of $\mu^{(d)}$ near $D$ is thus given by

$J(\mu^{(d)}) = \left(\begin{array}{ccc}h_{11} & \cdots & h_{1d}\\ \vdots && \vdots \\ h_{g1} & \cdots & h_{gd}\end{array}\right)_{g\times d}$.

It suffices to see that this matrix is of full rank for some choice of $D = \sum_{i=1}^d p_i$ and basis $\omega_1, \ldots, \omega_g$. We simply do Gauss reduction: Choose $p_1$ such that $\omega_1(p_1) \neq 0$. Then subtracting a multiple of $\omega_1(p_1)$ to $\omega_2, \ldots, \omega_g$, we make it so that

$\omega_2(p_1) = \cdots = \omega_g(p_1) = 0$.

The $\omega_i$‘s are still a basis of $\Omega^1(S)$ and we continue like this, choosing a $p_2$ such that $\omega_2(p_2) \neq 0$ etc… We eventually arrive at the form

$J(\mu^{(d)}) = \left(\begin{array}{cccccc}h_{11}&h_{12}&\cdots&h_{1g}&\cdots&h_{1d} \\ 0&h_{22}&\cdots&h_{2g}&\cdots&h_{2d} \\ \vdots&\vdots&\ddots&&& \\ 0&0&\cdots&h_{gg}&\cdots&h_{gd}\end{array}\right)$

which is of maximal rank since $h_ii(p_i) \neq 0$ for all $1 \leq i \leq g$. This shows that $\mu^{(g)} : S^{(g)} \to J(S)$ is not everywhere singular, so it is surjective by lemma 2.

QED

Putting everything together, we showed that the set of (isomorphism classes) of topologically trivial holomorphic line bundles $L \to S$ (i.e. with zero first Chern class) has the structure of a complex torus $\mathbb{C}^g/\Lambda$ of dimension $g$. Indeed, the group $\text{Pic}_0(S)$ consisting of those (isomorphism classes of) holomorphic line bundles is isomorphic to $\text{Div}_0/\mathcal{M}(S)$, which by Abel’s and Jacobi’s theorem is isomorphic to $J(S)$.

Note that in fact the fibres of $\mu^{(g)} : S^{(g)} \to J(S)$ consist of projective spaces. Indeed, if $\mu^{(g)}(D) = \lambda$, then by Abel’s theorem the fiber is $(\mu^{(g)})^{-1}(\mu(D)) = |D|,$ the set of effective divisors linearly equivalent to $D$, which corresponds to the projectivisation of $H^0(S,\mathcal{O}(D))$. It can be shown that generically the fiber of $\mu^{(g)}$ is a point and that $\mu^{(g)}$ is a birational map.