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The First Reciprocity Law

May 7, 2014

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

A 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

From → Algebraic Geometry

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