# Abel’s Theorem

In the first post of this series, I defined the **Abel-Jacobi map** from a compact Riemann surface of genus to its jacobian . Recall that the jacobian of is defined to be the complex torus where is the **period lattice**

for the **period vectors**

associated to some basis of the space of holomorphic 1-forms, and where are cycles giving a canonical basis of . For a divisor of degree 0, the Abel-Jacobi map is then

(mod ).

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

Theorem (Abel):Let and suppose is a basis of . Denote by the meromorphic functions on . Then for some if and only if .

**Proof:** The only if part is quite easy. Suppose is the divisor associated to a meromorphic function , i.e. for respectively the zeros and poles of . The idea is to view as a holomorphic function

.

Then has a well-defined **degree** which is the number of points (counted with multiplicity) in a fiber . Let’s note the divisor which is the fiber of at . Then the degree of is zero for all , and

(mod )

for some chain in with . Since the set of points in vary analytically with ( is a locally with ), we obtain a holomorphic function

by defining

.

But since is simply connected, it lifts to a holomorphic function which must be constant by the maximum principle. Hence itself is constant, which means in particular that , i.e. that

.

Since is additive and , this tells us that when , which is what we wanted to show.

The converse is harder. Given a divisor of degree , where and the are all distinct, such that , we search for a meromorphic function such that . We first reduce the problem to the existence of a certain differential of the third kind:

**Lemma 1:** There is an with if and only if there is a meromorphic 1-form such that

**1.** ,

i.e. has a simple pole exactly at the zeros and poles of ;

**2.** and ,

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

**3.** for any loop in .

The correspondence being given by

and

for any .

**Proof of ****lemma 1:**** **Let be such a meromorphic function and define as above. Then since is locally given by for some , we have locally, with . So has no zero and a simple pole at every zero and pole of , with residue at a point equal to the order of at that point. Conditions 1. and 2. are thus satisfied for this . To see that the periods of are integers, we write

for a loop in where are small loops around and . Then

.

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

for some not in . Note that condition 3. insures that this function is well defined. Near one of ‘s simple pole , we can write

for some never-vanishing holomorphic function around . Then for sufficiently near , we have in these coordinates

which gives

for never vanishing holomorphic functions around . Similarly around a point we find and we conclude that

,

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 on and complex numbers such that , there exists a meromorphic 1-form with only simple poles having its poles exactly at the points with residue at .

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

where is the sheaf of meromorphic 1-forms having only simple poles exactly at the points , i.e. if , then , and where is the skyscraper sheaf around . By **Kodaira-Serre duality** (see for example p. 153 in Griffiths & Harris, replacing with the trivial line bundle),

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

i.e.

so the residue map has 1-dimensional cokernel, i.e. the image of is of codimension at most 1. But if , then by the residue theorem hence the image of by the residue map, which is a linear subspace of codimension at most 1, is contained in the hyperplane which is of codimension 1. Hence these two sets are equal, i.e.

which is exactly what the lemma says.

QED

**Back to the proof of the theorem:** Given our divisor of degree 0, this last lemma tells us we can find a meromorphic 1-form satisfying conditions 1. and 2. of the first lemma. All that remains to be done is to show that we can perturb this such that its periods are integers, i.e. such that it satisfies condition 3. of the lemma. Since the biholomorphism class of is independant of the choice of basis for and for , we may take a canonical basis and normalised with respect to this basis, i.e. such that

for .

Recall from last post that such a choise of basis for is possible as a consequence of the reciprocity law. Let satisfy conditions 1. and 2. and denote its periods by

().

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

.

Then still satisfies conditions 1. and 2. because the ‘s are holomorphic, but clearly for , we have

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

.

So since for and the basis is normalised, by condition 2. we find

for all

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

by hypothesis, there exists a cycle

with

such that

for all

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

for all .

The periods of are thus

and

for . We can now correct for it to satisfy conditions 1. 2. and having integral B-periods by taking

.

Indeed, for , the periods of are

and

.

But by Riemann’s first bilinear relation, the expression in parentheses above vanishes for all , so . We have thus found a meromorphic 1-form 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 , the Abel-Jacobi map gives a holomorphic embedding of into (i.e. it is injective and has maximal rank 1 everywhere)

**Proof:** To see that is injective, suppose that for two distinct points , i.e. that . Then by Abel’s theorem, there is a meromorphic function having divisor . We see this meromorphic function a holomorphic function

.

Recall that any holomorphic function between two compact Riemann surfaces is in fact a branched cover. In particular, has a **degree** (the degree is the number of points in the fibers of , which, counting with multiplicity, does not depend on the point in the image). Since , we have . But this would mean that for every . So would be a biholomorphism, which constradicts the fact that the genus is not .

To see that has maximal rank everywhere, we just compute its differential. Let be a holomorphic coordinate centered at and write . Writing the basis of holomorphic one-forms in this chart as , we have

so

.

Since is a basis for , they never all vanish simultaneously, so has maximal rank 1 everywhere.

QED

**Corollary:** Every smooth Riemann surface of genus one is biholomorphic to a complex torus .

**Proof:** This is immediate from last corollary: if , the Abel-Jacobi map gives a biholomorphism

.

QED

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