This is a follow up to this post where I defined the Abel-Jacobi map $\mu$ of a compact Riemann surface $S$ of genus $g \geq 1$ to its jacobian $J(S)$. My first goal will be to demonstrate Abel’s theorem, which goes like this:

Theorem (Abel): For $D = \sum_i (p_i - q_i) \in \text{Div}_0(S)$ and $\omega_1, \ldots, \omega_g$ a basis for $\Omega^1(S)$ the space of holomorphic 1-forms, the divisor $D$ is principal, i.e. $D = (f)$ for some meromorphic function $f \in \mathcal{M}(S)$ if and only if $\mu(D) = 0 \in J(S)$, i.e. iff

$\mu(D) = \displaystyle\sum_i \left( \displaystyle\int_{q_i}^{p_i} \omega_1, \ldots, \displaystyle\int_{q_i}^{p_i} \omega_g \right) \in \Lambda$.

As a corollary, we will have that $\mu : S \to J(S)$ is in fact a smooth analytic embedding of $S$ into its Jacobian. Before proving the theorem, we will establish in this post a reciprocity law relating the periods of a holomorphic 1-form and a meromorphic 1-form having only simple poles. These two types of 1-forms are classically called differentials of the first and third kind, respectively. Recall the notion of canonical basis for $H_1(S,\mathbb{Z})$ from last post.

Proposition (First Reciprocity Law):Let $\delta_1, \ldots, \delta_{2g}$ be cycles inducing a canonical basis of $H_1(S,\mathbb{Z})$ and suppose $\omega, \eta$ are respectively differential of the first and third kind. Let $\{s_{\alpha}\}$ be the poles of $\eta$. Then the following relation holds:

$\displaystyle\sum_{i=1}^g\left(\displaystyle\int_{\delta_i}\omega\displaystyle\int_{\delta_{g+i}}\eta - \displaystyle\int_{\delta_i}\eta\displaystyle\int_{\delta_{g+i}}\omega\right) = 2\pi\sqrt{-1}\displaystyle\sum_{\alpha} \text{res}_{s_{\alpha}}(\eta)\displaystyle\int_{s_0}^{s_{\alpha}}\omega$.

On the right hand side, the integral is taken on any path inside $S_0 := S \backslash (\delta_1 \cup \ldots \cup \delta_{2g})$ which is simply connected and $s_0 \in S_0$ is any base point.

Proof: The region $S_0$ is the standard polygonal representation of the Riemann surface $S$, which is a $4g$-sided plane polygonal with sides labelled $\delta_1, \delta_{g+1}, \delta_1^{-1}, \delta_{g+1}^{-1}, \delta_2,$ etc…  with the corresponding orientation. This is hard to convey without drawing pictures so you should refer to some book if you’ve never seen this, maybe a book which treats the classification of closed surfaces like Munkres. Think of the standard way to represent a torus as a quotient of the square (which is a $4g$-polygonal when $g=1$) but with more sides.

Anyways, since $S_0$ is simply connected, we can choose $s_0 \in S_0$ and define without ambiguity the function

$\pi(s) = \displaystyle\int_{s_0}^s \omega$

for $s \in \overline{S_0}$. Then $\pi$ is a holomorphic mapping on the closure of $S_0$ with $d\pi = \omega$. By construction, if $p \in \delta_i$ and $p' \in \delta_i^{-1}$ are two points of $\overline{S_0}$ that are identified in $S$ on the cycle $\delta_i$, then

$\pi(p') - \pi(p) = \displaystyle\int_p^{p'} \omega = \int_p^{\delta_i(1)} \omega + \int_{\delta_{g+i}} \omega + \int_{\delta_i(1)}^{p'}\omega = \int_{\delta_{g+i}} \omega$

and similarly for $p \in \delta_{g+i}$ and $p' \in \delta_{g+i}^{-1}$ that are identified in $S$, we have

$\pi(p') - \pi(p) = -\displaystyle\int_{\delta_i} \omega$.

Using this, we find that

$\displaystyle\int_{\delta_i + \delta_i^{-1}} \pi\eta = \int_{\delta_i}\pi\eta + \int_{\delta_i^{-1}}\left(\pi(p) + \int_{\delta_{g+i}}\omega\right)\eta = -\int_{\delta_{g+i}}\omega\int_{\delta_i}\eta$

and similarly

$\displaystyle\int_{\delta_{g+i} + \delta_{g+i}^{-1}}\pi\eta = \int_{\delta_i}\omega\int_{\delta_{g+i}}\eta$.

Moreover, by the residue theorem, we have

$\displaystyle\int_{\partial S_0} \pi\eta = 2\pi\sqrt{-1}\displaystyle\sum_{\alpha}\text{res}_{s_{\alpha}}(\pi\eta) = 2\pi\sqrt{-1}\sum_{\alpha}\text{res}_{s_{\alpha}}(\eta)\int_{s_0}^{s_{\alpha}}\omega$.

Equating these two last equations, we obtain the first reciprocity law.

QED

A first consequence of this result is that it permits us to choose a very nice basis for the holomorphic 1-forms $\Omega^1(S)$. This will be a consequence of the

Corollary 1: For $\omega \in \Omega^1(S)$ a non-zero holomorphic 1-form, we have

$\sqrt{-1}\displaystyle\sum_{i=1}^g\left(\int_{\delta_i}\omega\overline{\int_{\delta_{g+i}}\omega} - \int_{\delta_{g+i}}\omega\overline{\int_{\delta_i}\omega}\right) > 0$.

Proof: Take $\omega, \omega' \in \Omega^1(S)$. Then

$d(\pi\omega') = d\pi \wedge \omega = \omega \wedge \omega'$

for $\pi = \int_{s_0}^s\omega$ like above. So

$\displaystyle\int_{\partial S_0} \pi\omega' = \int_S \omega \wedge \omega'$.

So by integrating like above, we find

$\displaystyle\int_S\omega \wedge \omega' = \sum_{i=1}^g\left(\int_{\delta_i}\omega\overline{\int_{\delta_{g+i}}\omega} - \int_{\delta_{g+i}}\omega\overline{\int_{\delta_i}\omega}\right)$

so letting $\omega = \omega'$ gives the result since $\omega \wedge \overline{\omega}$ is positive.

QED

normalised basis $\omega_1, \ldots, \omega_g$ for $\Omega^1(S)$ with respect to the basis $\delta_1, \ldots, \delta_{2g}$ for $H_1(S,\mathbb{Z})$ will be a basis such that $\int_{\delta_i}\omega_j$ is $1$ if $i=j$ and $0$ if not. Corollary 1 permits us to always choose a normalised basis. Indeed, consider the linear mapping $\psi : \Omega^1(S) \to (\mathbb{C} \{\delta_1, \ldots, \delta_g\} )^*$ defined by

$\psi(\theta) = \left(\sigma = \displaystyle\sum_{i=1}^ga_i\delta_i \mapsto \int_{\sigma}\theta\right)$.

Then $\theta \in \ker\psi$ iff $\int_{\delta_i}\theta = 0$ for all $i = 1, \ldots, g$ which by corollary 1 would mean $\theta = 0$.

Recall from last post the $g \times 2g$ period matrix defined by $\Omega = (\Pi_1 \cdots \Pi_{2g})$ where the columns $\Pi_i$ are the vectors $(\int_{\delta_i}\omega_1, \ldots, \int_{\delta_i}\omega_g)$ translated. By choosing a normalised basis, the period matrix becomes

$\Omega = \left( I_{g \times g} \: Z \right)$

for some $g \times g$ matrix $Z$ consisting of the B-periods.

Corollary 2 (Riemann’s bilinear relations):

1) First bilinear relation: If $\omega, \eta \in \Omega^1(S)$ are two differentials of the first kind (i.e. holomorphic), then the reciprocity law tells us

$\displaystyle\sum_{i=1}^g\left(\int_{\delta_i}\omega\int_{\delta_{g+i}}\eta - \int_{\delta_{g+i}}\omega\int_{\delta_i}\eta\right) = 0$.

In particular, with $\omega_1, \ldots, \omega_g$ a normalised basis,

$\displaystyle\int_{\delta_{g+i}}\omega_j = \int_{\delta_{g+j}}\omega_i$,

i.e. $Z$ is a symmetric matrix.

2) Second bilinear relation: Im$(Z) > 0$, i.e. the matrix consisting of the imaginary parts of the coefficients of $Z$ is positive definite.

Proof: The first bilinear relation is immediate from the first reciprocity law. For the second, we proceed as in the proof of the first corollary and write

$0 < \sqrt{-1}\displaystyle\int_S\omega_i \wedge \overline{\omega_j} = \sum_{k=1}^g\sqrt{-1}\left(\int_{\delta_k}\omega_i\overline{\int_{\delta_{g+k}}\omega_j} - \int_{\delta_{g+k}}\omega_i\overline{\int_{\delta_k}\omega_j}\right)$.

But by the first bilinear relation this last expression is equal to

$\sqrt{-1}\left(\overline{\int_{\delta_{g+i}}\omega_j} - \int_{\delta_{g+i}}\omega_j \right) = 2\text{Im}\left(\int_{\delta_{g+i}}\omega_j\right)$.

QED