In the first post of this series, I defined the Abel-Jacobi map $\mu : \text{Div}_0(S) \to J(S)$ from a compact Riemann surface $S$ of genus $g \geq 1$ to its jacobian $J(S)$. Recall that the jacobian of $S$ is defined to be the complex torus $\mathbb{C}^g/\Lambda$ where $\Lambda$ is the period lattice

$\Lambda = \left\{\displaystyle\sum_{i=1}^{2g}\alpha_i\Pi_i \mid \alpha_i \in \mathbb{Z}\right\}$

for $\Pi_i$ the period vectors

$\Pi_i = \left( \displaystyle\int_{\delta_i}\omega_1, \ldots, \int_{\delta_i}\omega_g\right) \in \mathbb{C}^g$

associated to some basis $\omega_1, \ldots, \omega_g$ of the space $\Omega^1(S)$ of holomorphic 1-forms, and where $\delta_1, \ldots, \delta_{2g}$ are cycles giving a canonical basis of $H_1(S,\mathbb{Z})$. For $D = \sum_{i=1}^d(p_i-q_i)$ a divisor of degree 0, the Abel-Jacobi map is then

$\mu(D) = \displaystyle\sum_{i=1}^d\left(\int_{q_i}^{p_i}\omega_1, \ldots, \int_{q_i}^{p_i}\omega_g\right)$    (mod $\Lambda$).

The goal of this post of to prove the following result:

Theorem (Abel): Let $D \in \text{Div}_0(S)$ and suppose $\omega_1, \dots, \omega_g$ is a basis of $\Omega^1(S)$. Denote by $\mathcal{M}(S)$ the meromorphic functions on $S$. Then $D = (f)$ for some $f \in \mathcal{M}(S)$ if and only if $\mu(D) = 0 \in J(S)$.

Proof: The only if part is quite easy. Suppose $D$ is the divisor associated to a meromorphic function $f \in \mathcal{M}(S)$, i.e. $D = (f) = (f)_0 - (f)_{\infty}$ for $(f)_0, (f)_{\infty}$ respectively the zeros and poles of $f$. The idea is to view $f$ as a holomorphic function

$f : S \to \mathbb{CP}^1$.

Then $f$ has a well-defined degree $d$ which is the number of points (counted with multiplicity) in a fiber $f^{-1}(p)$. Let’s note $D(t) = f^{-1}(t)$ the divisor which is the fiber of $f$ at $t \in \mathbb{CP}^1$. Then the degree of $(D(t) - dp_0)$ is zero for all $p_0 \in S$, and

$\mu(D(t) - dp_0) = \left(\displaystyle\int_{\sigma}\omega_1, \ldots, \int_{\sigma}\omega_g\right)$    (mod $\Lambda$)

for $\sigma$ some chain in $S$ with $\partial \sigma = D(t) - dp_0$. Since the set of points in $f^{-1}(t)$ vary analytically with $t$ ($f$ is a locally $z^k$ with $1 \leq k \leq d$), we obtain a holomorphic function

$\varphi : \mathbb{CP}^1 \to J(S)$

by defining

$\delta(t) = \mu(D(t) - dp_0)$.

But since $\mathbb{CP}^1$ is simply connected, it lifts to a holomorphic function $\tilde{\varphi} : \mathbb{CP}^1 \to \mathbb{C}^g,$ which must be constant by the maximum principle. Hence $\varphi$ itself is constant, which means in particular that $\varphi(0) = \varphi(\infty)$, i.e. that

$\mu(D(0)) - \mu(D(\infty)) = 0$.

Since $\mu$ is additive and $D(0) - D(\infty) = (f)_0 - (f)_{\infty} = (f)$, this tells us that $\mu(D) = 0$ when $D = (f)$, which is what we wanted to show.

The converse is harder. Given a divisor $D = \sum_i(a_ip_i-b_iq_i)$ of degree $0$, where $a_i, b_i \in \mathbb{Z}_{>0}$ and the $p_i,q_i$ are all distinct, such that $\mu(D) = 0$, we search for a meromorphic function $f \in \mathcal{M}(S)$ such that $(f) = D$. We first reduce the problem to the existence of a certain differential of the third kind:

Lemma 1: There is an $f \in \mathcal{M}(S)$ with $(f) = D$ if and only if there is a meromorphic 1-form $\eta \in (\Omega^1 \otimes \mathcal{M})(S)$ such that

1. $(\eta)_{\infty} = -\left(\displaystyle\sum_i(p_i+q_i)\right)$,

i.e. $\eta$ has a simple pole exactly at the zeros and poles of $f$;

2. $\text{res}_{p_i}(\eta) = \dfrac{a_i}{2\pi\sqrt{-1}}$      and      $\text{res}_{q_i}(\eta) = \dfrac{-b_i}{2\pi\sqrt{-1}}$,

i.e. the residues at those poles are given by the order of $f$;

3. $\int_{\gamma}\eta \in \mathbb{Z}$ for any loop $\gamma$ in $S \backslash \{p_i, q_i\}$.

The correspondence being given by

$f \mapsto \eta = \dfrac{1}{2\pi\sqrt{-1}}\dfrac{df}{f}$

and

$\eta \mapsto f(p) = \exp\left(2\pi\sqrt{-1}\displaystyle\int_{p_0}^p\eta\right)$

for any $p_0 \in S$.

Proof of lemma 1: Let $f$ be such a meromorphic function and define $\eta$ as above. Then since $f$ is locally given by $f(z) = z^k$ for some $k \in \mathbb{Z}$, we have $df/f = k/z$ locally, with $k = \text{ord}(f)$. So $df/f$ has no zero and a simple pole at every zero and pole of $f$, with residue at a point equal to the order of $f$ at that point. Conditions 1. and 2. are thus satisfied for this $\eta$. To see that the periods of $\eta$ are integers, we write

$[\gamma] = \displaystyle\sum_iw_i[C_{p_i}] + w_i'[C_{q_i}]$

for a loop $\gamma$ in $S \backslash \{p_i,q_i\}$ where $C_{p_i},C_{q_i}$ are small loops around $p_i$ and $q_i$. Then

$\displaystyle\int_{\gamma}\eta = \dfrac{1}{2\pi\sqrt{-1}}\displaystyle\sum_i\left(w_i\int_{C_{p_i}}d\log f + w_i' \int_{C_{q_i}}d\log f\right) = \sum_i w_i + w_i' \in \mathbb{Z}$.

Conversely, given a meromorphic 1-form $\eta$ satisfying conditions 1., 2. and 3., we let $f$ be the function

$f(p) = \exp\left(2\pi\sqrt{-1}\displaystyle\int_{p_0}^p\eta\right)$

for some $p_0 \in S$ not in $\{p_i,q_i\}$. Note that condition 3. insures that this function is well defined. Near one of $\eta$‘s simple pole $p_i$, we can write

$\eta = \dfrac{a_i}{2\pi\sqrt{-1}}\dfrac{dz}{z} + h(z)dz$

for $h(z)$ some never-vanishing holomorphic function around $p_i$. Then for $p_0^i$ sufficiently near $p_i$, we have in these coordinates

$f(z) = \exp\left(2\pi\sqrt{-1}\left[\displaystyle\int_{p_0}^{p_0^i}\eta + \int_{p_0^i}^z\frac{a_i}{2\pi\sqrt{-1}}\frac{dz}{z} + \int_{p_0}^zh(z)dz\right]\right)$

which gives

$f(z) = H_1(z)\exp\left(a_i\displaystyle\int_{p_0}^zd\log (z)\right) = H_2(z)\exp\left(a_i\log (z)\right) = z^{a_i}H(z)$

for $H_1, H_2, H$ never vanishing holomorphic functions around $p_i$. Similarly around a point $q_i$ we find $f(z) = z^{-b_i}H(z)$ and we conclude that

$(f) = \displaystyle\sum_i(a_ip_i - b_iq_i)$,

concluding the proof of the lemma

QED

To construct such a meromorphic 1-form, we will use another lemma:

Lemma 2: Given a finite number of points $\{p_i\}$ on $S$ and complex numbers $a_i \in \mathbb{C}$ such that $\sum_ia_i = 0$, there exists a meromorphic 1-form $\eta$ with only simple poles having its poles exactly at the points $p_i$ with residue $a_i$ at $p_i$.

Proof of lemma 2: We consider the exact sequence of sheaves

$0 \longrightarrow \Omega^1 \longrightarrow \Omega^1(\sum_ip_i) \overset{\text{res}}\longrightarrow \bigoplus_i\mathbb{C}_{p_i} \longrightarrow 0$

where $\Omega^1(\sum_ip_i)$ is the sheaf of meromorphic 1-forms having only simple poles exactly at the points $p_i$, i.e. if $D = \sum_ip_i$, then $\Omega^1(\sum_ip_i) = \Omega^1 \otimes_{\mathcal{O}_S} \mathcal{O}([D])$, and where $\mathbb{C}_{p_i}$ is the skyscraper sheaf around $p_i$. By Kodaira-Serre duality (see for example p. 153 in Griffiths & Harris, replacing $E$ with the trivial line bundle),

$H^1(S,\Omega^1) \cong H^0(S,\mathcal{O}) \cong \mathbb{C}$

where the last isomorphism comes from the maximum principle. Then the long exact sequence gives the exact sequence

$H^0(S,\Omega^1(\sum_ip_i)) \overset{\text{res}}\longrightarrow H^0(S,\bigoplus_i\mathbb{C}_{p_i}) \longrightarrow H^1(S,\Omega^1)$

i.e.

$H^0(S,\Omega^1(\sum_ip_i)) \overset{\text{res}}\longrightarrow \bigoplus_i \mathbb{C} \to \mathbb{C}$

so the residue map has 1-dimensional cokernel, i.e. the image of $H^0(S,\Omega^1(\sum_ip_i))$ is of codimension at most 1. But if $\eta \in H^0(S,\Omega^1(\sum_ip_i))$, then by the residue theorem $\sum_i \text{res}_{p_i}(\eta) = 0$ hence the image of $H^0(S,\Omega^1(\sum_ip_i))$ by the residue map, which is a linear subspace of codimension at most 1, is contained in the hyperplane $\{\sum_ia_i = 0\} \subset \bigoplus_i\mathbb{C}$ which is of codimension 1. Hence these two sets are equal, i.e.

$\text{res}\left(H^0(S,\Omega^1(\sum_ip_i))\right) = \{(a_i)_i \in \bigoplus_i\mathbb{C} \mid \sum_ia_i = 0\}$

which is exactly what the lemma says.

QED

Back to the proof of the theorem: Given our divisor $D = \sum_i(a_ip_i - b_iq_i)$ of degree 0, this last lemma tells us we can find a meromorphic 1-form $\eta$ satisfying conditions 1. and 2. of the first lemma. All that remains to be done is to show that we can perturb this $\eta$ such that its periods are integers, i.e. such that it satisfies condition 3. of the lemma. Since the biholomorphism class of $J(S)$ is independant of the choice of basis $\delta_1, \ldots, \delta_{2g}$ for $H_1(S,\mathbb{Z})$ and $\omega_1, \ldots, \omega_g$ for $\Omega^1(S)$, we may take $(\delta_i)$ a canonical basis and $(\omega_i)$ normalised with respect to this basis, i.e. such that

$\displaystyle\int_{\delta_i}\omega_i = \delta_{ij}$      for $1 \leq i,j \leq g$.

Recall from last post that such a choise of basis for $\Omega^1(S)$ is possible as a consequence of the reciprocity law. Let $\eta$ satisfy conditions 1. and 2. and denote its periods by

$N^i = \displaystyle\int_{\delta_i}\eta$      ($i=1, \ldots, 2g$).

By correcting $\eta$ with a linear combinations of the $\omega_i$‘s, we can suppose its A-periods vanish. Indeed, take

$\eta' = \eta - \displaystyle\sum_{i=1}^g N^i \omega_i$.

Then $\eta'$ still satisfies conditions 1. and 2. because the $\omega_i$‘s are holomorphic, but clearly for $i = 1, \ldots, g$, we have $\int_{\delta_i}\eta' = 0.$

The game is now to add an integral linear combination of the $\omega_i$‘s to $\eta$ to make its B-periods integers. By the reciprocity law,

$\displaystyle\sum_{i=1}^g \left(N^{i+g}\int_{\delta_i}\omega_j - N^i\int_{\delta_{i+g}}\right) = 2\pi\sqrt{-1}\sum_{p \text{ pole of } \eta}\text{res}_{p}(\eta) \int_{p_0}^p\omega_j$.

So since $N^i = 0$ for $i \geq g$ and the basis $(\omega_i)$ is normalised, by condition 2. we find

$N^{j+g} = \displaystyle\sum_ia_i\int_{p_0}^{p_i}\omega_j - b_i\int_{p_0}^{q_i}\omega_j = \sum_i\int_{q_i}^{p_i}\omega_j$    for all $j = 1, \ldots, g$

for a proper choice of path in this last integral (take  path from $q_i$ to $p_i$ circling the right amount of times along the $\delta_j$‘s to incorporate the $a_i,b_i$‘s in the integral). Let us denote by $\gamma_i$ those paths on which we integrate in this last expression. Since

$\mu(D) = \displaystyle\sum_i\left(\int_{\gamma_i}\omega_1, \ldots, \int_{\gamma_i}\omega_g\right) \in \Lambda \subset \mathbb{C}^g$

by hypothesis, there exists a cycle

$\sigma \sim \displaystyle\sum_{k=1}^{2g}m_k\delta_k$        with $m_k \in \mathbb{Z}$

such that

$\displaystyle\sum_i\int_{\gamma_i}\omega_j = \int_{\sigma}\omega_j$      for all $j=1, \ldots, g$

(this is the definition of being 0 in the jacobian). Then we have

$N^{g+j} = \displaystyle\sum_i\int_{\alpha_i}\omega_j = \int_\sigma\omega_j$     for all $j = 1, \ldots, g$.

The periods of $\eta$ are thus

$N^i = 0$

and

$N^{g+i} = \displaystyle\sum_{k=1}^{2g}m_k\int_{\delta_k}\omega_i = m_i + \sum_{k=1}^gm_{g+k}\int_{\delta_{g+k}}\omega_i$

for $i=1, \ldots, g$. We can now correct $\eta$ for it to satisfy conditions 1. 2. and having integral B-periods by taking

$\eta' = \eta - \displaystyle\sum_{k=1}^gm_{g+k}\omega_k$.

Indeed, for $i=1, \ldots, g$, the periods of $\eta'$ are

$N'^i = -m_{g+1}$

and

$N'^{g+i} = N^{g+i} - \displaystyle\sum_{k=1}^gm_{g+k}\int_{\delta_{g+i}}\omega_k = m_i + \sum_{k=1}^gm_{g+k}\left(\int_{\delta_{g+k}}\omega_i - \int_{\delta_{g+i}}\omega_k\right)$.

But by Riemann’s first bilinear relation, the expression in parentheses above vanishes for all $1 \leq i,k \leq g$, so $N'^{g+i} = m_i$. We have thus found a meromorphic 1-form $\eta$ satisfying conditions 1, 2 and 3 of lemma 1, concluding the proof of Abel’s theorem

QED

Corollary: For a compact Riemann surface of genus $g \geq 1$, the Abel-Jacobi map gives a holomorphic embedding of $S$ into $J(S)$ (i.e. it is injective and has maximal rank 1 everywhere)

Proof: To see that $\mu|_S$ is injective, suppose that $\mu(p) = \mu(q)$ for two distinct points $p,q \in S$, i.e. that $\mu(p-q) = 0 \in J(S)$. Then by Abel’s theorem, there is a meromorphic function $f \in \mathcal{S}$ having divisor $(f) = p-q$. We see this meromorphic function a holomorphic function

$f : S \to \mathbb{CP}^1$.

Recall that any holomorphic function between two compact Riemann surfaces is in fact a branched cover. In particular, $f$ has a degree $d$ (the degree is the number of points in the fibers of $f$, which, counting with multiplicity, does not depend on the point in the image). Since $f^{-1}(\infty) = \{q\}$, we have $\text{deg}(f) = 1$. But this would mean that $\#f^{-1}(p) = 1$ for every $p \in \text{Im}(f) = \mathbb{CP}^1$. So $f$ would be a biholomorphism, which constradicts the fact that the genus is not $0$.

To see that $\mu$ has maximal rank everywhere, we just compute its differential. Let $z$ be a holomorphic coordinate centered at $p \in S$ and write $\mu = (\mu_1, \ldots, \mu_g)$. Writing the basis of holomorphic one-forms in this chart as $\omega_j = \eta_jdz$, we have

$\mu_j(z) = \displaystyle\int_{p_0}^p\omega_j + \int_0^z\eta_j(z)dz$

so

$\dfrac{\partial \mu_j}{\partial z} = \eta_j(z)$.

Since $\omega_1, \ldots, \omega_g$ is a basis for $\Omega^1(S)$, they never all vanish simultaneously, so $\mu|_S$ has maximal rank 1 everywhere.

QED

Corollary: Every smooth Riemann surface $T$ of genus one is biholomorphic to a complex torus $\mathbb{C}/\Lambda$.

Proof: This is immediate from last corollary: if $g=1$, the Abel-Jacobi map gives a biholomorphism

$\mu : T \to J(T) \simeq \mathbb{C}/\Lambda$.

QED