In this post, I will rely on this last post for results and notations, for example will also mean a compact Riemann surface. We will note the canonical bundle of , i.e. the bundle with sheaf of sections the holomorphic 1-forms on . We will also note the dimension of by which we really mean the j’th cohomology group of in the sheaf of holomorphic sections of . I’ll also write to mean where is the line bundle associated with the divisor , and sometimes the line bundle might get written accordingly. We will sometimes also suppress the and simply write for . Let us first state the Riemann-Roch theorem in the form that we aim to get to:

Theorem (Riemann-Roch):For any holomorphic line bundle , the following relation holds:.

We will freely use the fact that any line bundle on a compact Riemann surface actually comes from a divisor, which can be seen by Jacobi’s inversion theorem, and define the degree of simply as the degree of its associated divisor. To be more precise, we will do the proof for line bundles coming from generic divisors with the ‘s distinct. We will first prove that

which is relatively easy. The hard part will be

Theorem (Serre duality): There is a natural coupling which induces an isomorphism

In particular, this tells us that and Riemann-Roch follows.

**Proof of the first part:** We show the first part by induction on the degree of If , i.e. if is trivial, then by the maximum principle and . So

and we are done with the base case. Now, suppose that we can represent as for a divisor with the plus or minus sign in accordance with the sign of , and with all ‘s distinct.

Suppose by induction that the statement is true for . Then it suffices to show it is true for the line bundles associated to and for some divisor of degree not containing . In the case of positive degree, we use the short exact sequence

.

Note that this short exact sequence was also considered in my last post with trivial . Here we do the usual identification between sections with meromorphic functions such that and the first arrow corresponds to the inclusion of families of these spaces of meromorphic functions. We get a long exact sequence

hence adding alternating dimensions we find

.

Using our induction assumption, i.e. that , we deduce that

which is what we wanted.

For the second case, when , we use another but related short exact sequence:

.

A similar argument shows that

and we are done.

QED

Before attacking Serre duality, recall that for any line bundle there is an operator

where is the sheaf of smooth -forms with values in . It is the unique such operator satisfying Leibniz rule

and such that

is exact. It is called the **del-bar operator associated to **. Analogously to last post, we get a **Dolbeault isomorphism**

.

The identifications we discovered at the end of last post still hold in this context, with the necessary adjustments. Given two line bundles , this isomorphism permits us to define a **pairing**

by setting

.

In particular, with and , we get **Serre’s pairing**

where the last isomorphism is (which follows from the same argument as for ) composed with integration .

Theorem (Serre duality): This pairing is non-degenerate.

**Proof:** We will proceed by induction as in the proof of Riemann-Roch. We say that satisfies Serre duality if Serre’s pairing induces an isomorphism

and .

Note that satisfies Serre duality if and only if does.

The base case is the statement that the pairing given by is non-degenerate, which is just the isomorphism given by integration coupled with the fact that is injective.

Suppose now by induction that all line bundles with satisfy Serre’s duality. Like in Riemann-Roch, it suffices to show that and satisfy Serre duality (where with not containing ). We will now identify with meromorphic sections of with at worst a simple pole at . We thus get the short exact sequence

where the last arrow is (recall that the residue of a meromorphic function at is naturally an element of the holomorphic tangent space , as discussed in my last post and under our identifications, ). We obtain in this way our **first exact sequence**

.

Recall from last post that if for , then for an extension of in the neighborhood of .

Similarly, we see elements of as holomorphic sections of vanishing at , and we have the following short exact sequence of sheaves:

where now the last arrow is simply . This gives our **second exact sequence**

.

Dualizing the first exact sequence, we get the following diagram:

where the vertical arrows are given by Serre’s pairing. By the **Five Lemma **and by using our induction hypothesis, it suffices to show that this diagram is commutative. The only difficult parts are obviously the two squares in the middle.

Actually, to be exactly precise, we need to consider instead of just and similarly .

For **the third square**, consider (so is a holomorphic 1-form and a section of ). We need to see that for all , we have

.

But we have seen in last post that if near , then after identifying with , we have . Thus if , we find

which, since is supported near , by using Stokes theorem and Cauchy’s formula, is equal to

for a small circle around . But , so this shows the third square commutes.

**The fourth square** is similar. Take . Then unwinding the definition of , we see that

if in some coordinates around . Then under the Dolbeault isomorphism, this class corresponds to and it acts on by

where , for and . Like for the third square, this integral can be rewritten

for a small loop around and the residue of in the coordinate. But this is exactly what we needed since

.

In this post, will mean a compact Riemann surface (without boundary).

Let us start by considering a form of the **Mittag-Leffler problem**: Given a point , can we find a global meromorphic function such that and ? i.e. such that the only poles of is a simple pole at with residue in some coordinates. Since this would give a biholomorphic function , this is only possible if the genus of is zero, but let us see how this obstruction manifests itself.

We will consider a covering of where is a coordinate patch centered at with (holomorphic) coordinate , while is just . We can reformulate the question as follows: does there exist such that and ? NB: Here means the restriction .

Before going further, note that the **residue** of a meromorphic function is only invariantly defined as a **tangent vector**: if has a simple pole at , then we define, for ,

where is any extension of , and where

is a well-defined residue for any meromorphic 1-form defined around , where is a small loop around . Another way to see this is to take a holomorphic coordinate around and a meromorphic function with a simple pole at , say for holomorphic. Then if is another choice of coordinates, then so

for some holomorphic . In other words,

i.e.

is well-defined as an element of .

**First approach:** Our first approach to solving the Mittag-Leffler problem is with Cech cohomology: What we want is to find a global meromorphic function such that on we have . Thus we see , as a **Cech 1-cochain** in the sheaf of holomorphic functions and ask that on for , i.e. that defines a **co****boundary** in . Indeed, would then be holomorphic on with near .

Another way to realize this is via the following short exact sequence of sheaves:

where is the sheaf of meromorphic functions having at worst a simple pole at and where we consider as the associated skyscraper sheaf. From the induced long exact sequence we get

and in this formulation, finding a meromorphic function with at worst a simple pole at with residue is possible if and only if . But unwinding the definition of , we see that iff such that and in . Thus we can take and , and

.

We once again come to the conclusion that the problem is solvable exactly when is a coboundary.

**Second approach:** Consider a smooth cut-off function supported in and identically equal to 1 near . Then and the probem reduces to finding such that . Note that near so we can consider

as an element of , i.e. a -form on . Reformulating this, we can solve the problem exactly when .

To relate this to our first approach, we note the sheaf of -forms and use the short exact sequence of sheaves

which gives the exact sequence

.

In other words, this gives an isomorphism

Unwinding the definition of , we see that iff in , for .

This isomorphism is the simplest instance of the so-called **Dolbeault isomorphism** which is a similar isomorphism valid for arbitrary compact complex manifolds and complex vector bundle , where is the sheaf of holomorphic p-forms with values in .We can recap our discussion with the following:

Proposition:The obstruction to solving the Mittag-Leffler problem coming from our second approach, which is represented by the 1-cocycle , corresponds under Dolbeault’s isomorphism to , the obstruction from our first approach.

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 is surjective.

Theorem (Jacobi’s inversion):Let be a compact Riemann surface of genus and take any . Then for any , we can find points such that.

In other words, for a basis of , for all , there is and paths from to such that

for all .

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

.

**Lemma 1: **The set of effective divisors of degree on is a compact complex manifold.

**Proof of lemma 1:** Consider the action of the permutation group on

( times).

The quotient, denoted , inherits the quotient topology making a continuous map. Clearly, is in bijection with . Suppose for a moment that . We will write an element of as a sum to indicate the unimportance of the ordering (note that the ‘s are not necessarily distinct). We consider the map

that takes to the d-tuple consisting of the coefficents of the monic polynomial having as its roots, i.e

if

.

In other words, where is the ith elementary symmetric polynomial. By the fundamental theorem of algbera, is a bijection, and in fact a homeomorphism.

The difficulty is in seeing that is also continuous. To see this, take an open set and consider . In other words,

with .

We need to show that there is an open set . Write

which, when defined, is the number of zeros of the polynomial inside . We can choose an open set containing small enough so that is defined for all . Then since is continuous and takes only integer values, we have

for all .

But this means that for all , all the roots of are in so that as we wanted.

This gives the structure of a complex manifold of dimension , in fact biholomorphic to .

Now we put a similar complex manifold structure on for arbitrary . Consider and take holomorphic charts around on such that if and if . We obviously have an injective mapping

where , by doing as above;

.

The verifications that this gives a homeomorphism onto an open set of is just as in the earlier case. These maps thus provide with a holomorphic atlas.

QED

Fixing a base point we get a (holomorphic) injection

and thus holomorphic mappings

(mod )

by composing and . Jacobi’s inversion theorem says that is surjective.

**Lemma 2: ** Let be a holomorphic map between two compact connected complex manifolds of the same dimension. If is not everywhere singular, i.e. the Jacobian matrix is not identically zero, then is necessarily surjective.

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

Consider a volume form on . Since is not identically and since preserves the orientation (being holomorphic), we have

.

Since for any we have the volume form is exact in and

for some -form on . But then if , we have

,

which contradicts earlier considerations.

QED

To prove the theorem, we thus have to show that is not everywhere singular.

**Proof of the theorem:** At points such that the ‘s are distinct, the quotient map is locally a biholomorphism. So choosing disjoint charts in centered at , we get a chart

.

In such coordinates, for near , we have

mod .

So

.

But with near , this is

hence

The jacobian matrix of near is thus given by

.

It suffices to see that this matrix is of full rank for some choice of and basis . We simply do Gauss reduction: Choose such that . Then subtracting a multiple of to , we make it so that

.

The ‘s are still a basis of and we continue like this, choosing a such that etc… We eventually arrive at the form

which is of maximal rank since for all . This shows that 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 (i.e. with zero first Chern class) has the structure of a complex torus of dimension . Indeed, the group consisting of those (isomorphism classes of) holomorphic line bundles is isomorphic to , which by Abel’s and Jacobi’s theorem is isomorphic to .

Note that in fact the fibres of consist of projective spaces. Indeed, if , then by Abel’s theorem the fiber is the set of effective divisors linearly equivalent to , which corresponds to the projectivisation of . It can be shown that generically the fiber of is a point and that is a **birational map**.

In last post we proved Abel’s theorem and as a corollary we saw that any compact Riemann surface of genus (a **smooth** **elliptic curve**) is biholomorphic to a complex torus, the biholomorphism being given by the Abel-Jacobi mapping:

which is given by

(mod )

where some , is a holomorphic 1-form and is the **period lattice** for the two **period vectors** for giving a basis for .

In particular, every such Riemann surface has a **group structure **and we will see that this description of the group structure on an elliptic curve is the same as the usual one given for elliptic curves in given by a cubic polynomial. By the usual group structure I mean the following: recall that by Bézout’s theorem, if you intersect a line in with the zero locus of a cubic polynomial, you get (counting multiplicities) exactly 3 points. Thus to add 2 points , you consider the line between them and declare that if is the third point of intersection of this line and the elliptic curve.

After a preliminary discussion of projective embeddings of Riemann surfaces, we will see that in fact any genus 1 compact Riemann surface can be embedded in as the zero locus of some cubic polynomial. In particular when the initial surface is a complex torus , this embedding is given by the so-called **Weierstrass -functions.** We will then see that this embedding is actually an inverse to the Abel-Jacobi mapping.

First the remarks on projective embeddings of compact Riemann surfaces. Suppose is a holomorphic line bundle over a compact complex manifold and are a basis for the global holomorphic sections of . Suppose that there is no point on such that every sections in vanish simultaneously. Then for every , we have a non-zero vector

.

By choosing a trivialisation, we can consider this vector as sitting in . Of course this vector will depend on the choice of trivialisation of around but different choices of trivialisations will only change the vector by a complex multiple. We thus get a map

defined by .

**Kodaira’s embedding theorem** states that when is a **positive** line bundle (i.e. admits a connection of positive curvature), there exists a such that for , the map

is well-defined (the global sections of are never all vanishing) and is an embedding of .

We can in fact give a sharper version of Kodaira’s embedding theorem for compact Riemann surfaces (you can find a detailed discussion of this on pages 214, 215 of Griffiths-Harris):

Theorem:Let be a compact Riemann surface and a holomorphic line bundle. If , then is well-defined and an embedding.

In this proposition, the degree of a line bundle is just under the identification or equivalently, since if then the Poincaré dual , the degree is . Also, is the **canonical bundle of ** and .

The case that will interest us is when the genus of is . What we show is that in fact we can take . Indeed, since because is trivial ( is topologically a torus), for any the above theorem tells us that for the bundle , the map gives an embedding of into for If , then are 3 linearly independent global holomorphic sections so . On the other hand, a global section of corresponds to a meromorphic function on having poles only at and at worst of degree . Such a function is completely determined by the first four coefficients in its Laurent expension around :

(the difference of two such functions having all of these 4 numbers equal would be a global holomorphic function vanishing at hence everywhere). Moreover, two such functions having the same and must have the same because otherwise their difference would be a meromorphic function having only one pole at and of order 1, which is impossible if the genus is not zero. So we need at most 3 parameters to determine such a function, hence . This shows .

We can write this embedding explicitly as follows: by Kodaira’s vanishing theorem, we have so considering the long exact sequence associated to the short exact sequence

where is the skyscraper sheaf at , we find so there exists a meromorphic function on having a double pole at and no other poles. Since , there is also on a nonzero holomorphic 1-form . But as remarked earlier, since the canonical bundle is trivial, the divisor has zero Poincaré dual so is itself zero, i.e. is nowhere vanishing. We consider . It is a meromorphic form with only one pole at , of order 2, and since the sum of the residues of a meromorphic 1-form must be zero, .

So if is a local holomorphic coordinate near , we can write

near

after possibly multiplying by a constant and adding a constant. Considering also the meromorphic function , which is holomorphic except at where it has a triple pole, we get a global meromorphic function which is

near

for the right choice of ‘s. Since and are three linearly independant global sections of , they form a basis and thus the embedding is given by

.

We can use this to explicitly describe as the zero set of a polynomial in : around , we have

and

.

Thus the meromorphic function has at most a simple pole at and no other poles, so is constant. This tells us that is included in the zero locus of the polynomial which can be rewritten by making suitable linear changes in the coordinates by

for . Note that here is a set of affine coordinates in on the open set . Since there are both topological torus (the second by the degree-genus formula), they must be equal, i.e. the zero locus of this polynomial in is exactly .

We will show that in fact the above construction gives an inverse for the Abel-Jacobi map which realises such a non-singular planar curve as a complex torus.

Consider and as above, with their pole at . Take as a basis for and the coordinate on such that . In this context, the function is the so-called **Weierstrass -function**. Its derivative is denoted . If the Laurent expansions of around contained a term of odd degree, then killing the order 2 pole by we would obtain a noncontant holomorphic function on . So it has no term of odd degree and we can write

.

We find the relation

,

where and . The corresponding embedding is then

where ,

and is the zero locus of the polynomial

,

written in suitable affine coordinates. On the other hand, the Abel-Jacobi map from to is given by

(mod )

with let’s say . Indeed, so is a non-zero holomorphic 1-form on . Moreover, is actually inverse to the embedding because

.

We have thus found **an inverse for the Abel-Jacobi mapping of a smooth elliptic curve in **.

We are now in a position to discuss the **group structure** from different point of views. Any Riemann surface of genus inherits a group structure simply by letting

for

There is also a group structure induced by an embedding as above : if are colinear in . These are in fact the same. To see this, consider as embedded in , take 3 points and denote the corresponding points in the complex torus. Then by Abel’s theorem,

iff with .

But suppose for , i.e. with the line joining (in affine coordinates), we also have . Then

is a meromorphic function on the torus having divisor so the function has divisor . Thus if are collinear, we have and conversely, if there is a meromorphic function on such that , then is a meromorphic function with and this forces because otherwise we would have a meromorphic function having a simple pole at and no other poles.

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

This is a follow up to this post where I defined the Abel-Jacobi map of a compact Riemann surface of genus to its jacobian . My first goal will be to demonstrate Abel’s theorem, which goes like this:

Theorem (Abel):For and a basis for the space of holomorphic 1-forms, the divisor is principal, i.e. for some meromorphic function if and only if , i.e. iff.

As a corollary, we will have that is in fact a smooth analytic embedding of 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 from last post.

Proposition (First Reciprocity Law):Let be cycles inducing a canonical basis of and suppose are respectively differential of the first and third kind. Let be the poles of . Then the following relation holds:.

On the right hand side, the integral is taken on any path inside which is simply connected and is any base point.

**Proof:** The region is the standard polygonal representation of the Riemann surface , which is a -sided plane polygonal with sides labelled 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 -polygonal when ) but with more sides.

Anyways, since is simply connected, we can choose and define without ambiguity the function

for . Then is a holomorphic mapping on the closure of with . By construction, if and are two points of that are identified in on the cycle , then

and similarly for and that are identified in , we have

.

Using this, we find that

and similarly

.

Moreover, by the residue theorem, we have

.

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 . This will be a consequence of the

Corollary 1:For a non-zero holomorphic 1-form, we have.

**Proof:** Take . Then

for like above. So

.

So by integrating like above, we find

so letting gives the result since is positive.

QED

A **normalised**** basis** for with respect to the basis for will be a basis such that is if and if not. Corollary 1 permits us to always choose a normalised basis. Indeed, consider the linear mapping defined by

.

Then iff for all which by corollary 1 would mean .

Recall from last post the **period matrix** defined by where the columns are the vectors translated. By choosing a normalised basis, the period matrix becomes

for some matrix consisting of the **B-periods**.

Corollary 2 (Riemann’s bilinear relations):

1) First bilinear relation:If are two differentials of the first kind (i.e. holomorphic), then the reciprocity law tells us.

In particular, with a normalised basis,

,

i.e. is a symmetric matrix.

2) Second bilinear relation:Im, i.e. the matrix consisting of the imaginary parts of the coefficients of 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

.

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

.

QED

In the next couple of posts, I will write about Riemann surfaces and their Jacobian, following Griffiths and Harris. To each compact Riemann surface of genus , we will associate a -dimensional complex torus, which is a complex manifold of the form for a discrete full-rank integral lattice in .

Recall that by Dolbeault’s theorem, on a compact complex manifold there is an isomorphism

.

So on a Riemann surface of genus , taking -cohomology class we find that . Hence the dimension is

because

so since .

Let us first consider a genus compact Riemann surface, i.e. a torus. Then is 1-dimensional, let be a generator. If and are two points on the torus, of course the integral

is not well-defined as a complex number, it depends on the path chosen from to on which we integrate. Instead of fixing a path to make sense of this integral, we can change the space on which it lives to account for this indeterminacy: if are two paths from to , their difference will be a closed loop representing a cycle in . The idea is to view the integral as a point in the quotient .

More precisely, for a compact Riemann surface of genus , we consider cycles which give a basis for . We will suppose that is a **canonical** basis, i.e. that and for , where means the intersection product. This just means that intersects once positively and intersects no other cycles, think of the canonical cycles on the g-holed torus. The first half of the basis is the **-cycles** and the **-cycles** are the others, . Let’s also fix a basis for . Then the **periods** are the vectors

.

The **period matrix** is the matrix having the periods as columns:

.

Note that the periods are -linearly independent so they generate a full-dimensional **lattice**

.

To see this, recall that since , the forms generate . So if we had a relation with , this would give

and for all

so since the pairing is non-degenerate, this would give a relation , which is impossible since give a basis for the homology group.

The **Jacobian** of the Riemann surface is defined to be the **complex torus**

.

Intrinsically, is just the quotient where we identify with its image under the natural inclusion which takes a class of closed loops and maps it to the functional . This space is the right one to consider integrals on in the sense that the vector

is not well-defined as a point of but it is if we consider is as living on , i.e. **modulo the period lattice** .

After choosing a base point , we can define a mapping

by

(mod ).

In fact, given a **divisor** , we can extend this map by letting . In particular, for divisors of degree 0

,

the map is independent of the base point and

(mod ).

When considered on divisors of degree 0, this is the **Abel-Jacobi map**

.

The first goal will be to show **Abel’s theorem**, which says that the kernel of is exactly the divisors of meromorphic functions on . Recall the short exact sequence

where is the space of all meromorphic functions on . If we restrict this to and the corresponding flat line bundles , Abel’s theorem gives an injection

.

Our second goal will be to show **Jacobi’s inversion theorem** which essentially says that this injection is in fact an isomorphism. From this point of view, we see that the -dimensional torus naturally parametrizes flat line bundles on . In fact, **Torelli’s theorem** says that , considered as a **principally polarized abelian variety**, entirely determines the Riemann surface we started with. Jacobian varieties are thus key in the study of compact Riemann surfaces.

Let be a compact oriented Riemannian manifold of dimension with boundary . The aim of this note is to define the divergence and Laplacian operators on and to clarify the validy and meaning of various formulas such as integration by parts

or Green’s formula

which are well-known to hold for domains in .

The Laplace operator acting on functions defined over is usually defined by the simple formula

.

Obviously, this definition is no good to define anything on a manifold, so we formulate it in a more geometric way as

.

In a beginning calculus course, this expression is usually understood as some kind of matrix multiplication, and we formally see that it holds. But this equation tells us what the Laplacian really is. Indeed, the divergence of a vector field is a real-valued function that at a point measures the amount of infinitesimal dilation an infinitesimal object placed at would experience if it was to flow infinitesimally along the vector field. This is the geometric interpretation of the Laplacian. On a Riemannian manifold, we can define the gradient of a function by duality via its exterior derivative. Thus we only need a notion of divergence. From our informal discussion, it makes sense to define the divergence of a vector field on an oriented Riemannian manifold as the infinitesimal change of measurement of volumes when the volume form is flowing along , that is we want

where denotes the volume form associated to the metric and the one-parameter group of diffeomorphisms associated to . The right side of this equation is called the Lie derivative of along and is noted . Since the bundle of top forms of an oriented Riemannian manifold is trivialised by , i.e. every top form is of the form for a unique function , we actually get a well-defined function by the relation

,

given of course that there actually is a volume form, i.e. that our manifold is orientable. We thus define the Laplacian acting on smooth functions as where is the gradient of , caracterized by the relation

.

The only thing that remains undefined is the normal vector field . It is not hard to check that if all the transition functions of some atlas on have positive Jacobian determinants, then their restriction to the boundary also have positive Jacobian determinants, enough so that the boundary is also orientable. Supposing without loss of generality that all charts containing a point of the boundary have their image lying in only one half-space of , say , we can define a notion of “inward-pointing vector” tangent to the boundary. Indeed, at a point of the boundary, a tangent vector in will be called “inward-pointing” if its image in any chart is contained in the half-space . At each point of the boundary, we decompose orthogonally with the metric, and we define the normal unit inward-pointing vector field as the unit vector field lying in the second component of this decomposition that is everywhere inward-pointing. Note that this gives a canonical volume form on the boundary, given by where is the interior product.

Before we start to state and prove theorems, let’s recall Cartan’s formula which says that for every vector field and differential form ,

.

This identity tells us that because the volume form is of course closed. This will be useful when paired with Stoke’s theorem, which says that for all forms of degree on ,

where is the inclusion. We can now prove the generalization of Gauss’s divergence theorem:

Theorem 1:With the notation introduced above,for all vector fields .

**Proof: **By Cartan’s formula and Stoke’s theorem, we get

.

But if is an orthonormal basis of for , then we see that

is equal to

because and for all . Thus at all points of the boundary we have the equality

.

We can now conclude from the first equation in the proof.

QED

Recall that the integration by parts formula for functions follows from Leibniz rule and from the fundamental theorem of calculus:

.

To get the generalized version for oriented Riemannian manifolds, we need to establish the following Leibniz rule for the divergence operator:

Lemma:For all functions and vector fields ,.

**Proof: **This again follows from Cartan’ identity. Indeed, by the very definition of the interior product, we get that for any differential form so we get

.

But by the usual Leibniz rule for the exterior derivative, this is equal to

hence

so it suffices to see that . We proceed as in theorem 1. Let be an orthonormal basis of at some point . Then

which is equal to

.

This shows that

but is precisely .

QED

Coupled with the divergence theorem, this immediately yields the

Theorem:For all vector fields and smooth functions , there is an integration by parts formula.

Note that for a closed Riemannian manifold, with , this shows that

i.e. that minus the divergence operator is kind of a formal adjoint to the gradient operator. The so-called Green formulas are a simple application of integration by parts. Recall that the Laplacian of a smooth function is defined as and that is the inward-pointing vector field on the boundary. We will denote by .

Theorem: (Green formulas)For any two functions ,and hence

.

**Proof:** Integrating by parts, we get

hence the first formula. The second evidently follows from the first.

QED

As an immediate application, we can show the

Proposition:Consider the equation on a compact Riemannian manifold with boundary subject to either Dirichlet (i.e. ) or Neumann (i.e. ) boundary conditions. Then if a smooth solution exists, it is unique in the case of Dirichlet conditions and unique up to a constant in the case of Newmann conditions.

**Proof:** Consider two solutions and to the problem and consider their difference . Then since , it follows that . Integrating by parts, we thus find

so if either the Dirichlet or Neumann boundary conditions are satisfied, the integral on the right vanishes and we deduce that is a constant, i.e. that . In the case of Dirichlet boundary conditions, the function has to vanish on the boundary hence everywhere, so .

QED

In the same vein, if is a harmonic function on a closed Riemannian manifold (i.e. and ), then so the only harmonic functions on a closed Riemannian manifold are the constants.

From now on, let us denote by and by . Another immediate application of integration by parts yield results about the spectrum of :

Proposition:Let be a closed Riemannian manifold. Then (with our definition), the eigenvalues of the Laplacian are non-positive. Moreover, eigenfunctions corresponding to different eigenvalues are orthogonal for the inner product.

**Proof: **For the first assertion, suppose . Then integrating by parts gives

.

The second assertion follows from Green’s formula since if the eigenfunctions and correpond to eigenvalues and , then

.

QED

This last proposition is not surprising in view of the fact that integrating by parts actually shows that is formally self-adjoint with respect to the inner product on a closed manifold. Indeed, in this context Green’s formula reads The first assertion of this proposition is the reason why geometers often define the Laplacian as , in order to get a positive spectrum. Finally, from the first equation of the proof, we see that eigenvalues of (minus) the Laplacian satisfy

with is their associated eigenfunctions, which one could recognize as a renormalized Dirichlet energy functional. With this point of view, the variational min/max principle which says that the ‘st eigenvalue is given by the infimum of that functional over the functions orthogonal to the eigenfunctions associated to the first eigenvalues makes plenty of sense. This min/max principle in turn explains the link between the first eigenvalue of the Laplacian and the important Poincaré’s inequality, which says that for some constant , all smooth functions integrating to (i.e. orthogonal to the constants) satisfy

,

the optimal constant being exactly , attained by the first eigenfunctions of the Laplacian.

In this post I introduce the notion of a Fredholm operator between two Hilbert spaces, following part 1 of Booss & Bleecker’s book Topology and Analysis. These operators end up providing some very nice bridges between topological ideas and functional analytic ideas. For example the Atiyah-Jänich theorem says that the space of Fredholm operators on the complex Hilbert space represents the K functor in topological K-theory. It is also at the heart of index theory (see the Atiyah-Singer index theorem), where some deep analogies between ideas coming from the study of PDEs and ideas from geometry are established. In the hope of understanding this all one day, let’s start with those Fredholm operators.

We will place ourselves in the context of a complex Hilbert space having a countable orthonormal basis. Note that all such Hilbert spaces are isometrically isomorphic to . We will denote the Banach algebra of bounded linear operators on . A **Fredholm** operator is an operator with closed range and finite dimensional kernel and cokernel. We will denote the set of Fredholm operators by . The **index** of a Fredholm operator is defined as

.

For example, the situation for an operator between two finite dimensional spaces, the situation is quite simple: We have since and .

These operators originally appeared in the theory of integral equations. The condition of being Fredholm means that the equation has a finite dimensional space of solutions and that there are a finite number of linear relations we can impose on to make sure that is solvable (because has to be in the orthogonal complement of the cokernel).

We can also phrase the index of in terms of its adjoint. Recall that by Riesz’s representation theorem, for each we have the adjoint operator that satisfies, for all ,

such that is an isometry. It is an easy exercice to check that and thus . In fact, this holds for any bounded linear operator with closed range. We can then give an alternative definition: An operator is **Fredholm** if its image is closed and both and are finite dimensional. The **index** is then . Here is another example:

Proposition:The operator for a finite-rank operator (ie. an operator with finite dimensional image) is Fredholm with index 0.

**Proof:** We first show that is Fredholm. Let . We proceed in two steps: (1) show that and then (2) that . For (1), if , then for , we have

so . For (2), first note that is orthogonal to . To see this, take and . We get

as wanted.

Now define a linear map by . This is well defined since . Since , this map is injective and (2) follows.

Consider now the following diagram:

Clearly the rows are exact. To see that it is commutative and that the columns are well-defined, it is enough to show that . to see this, let and . Then

so . Now since is finite dimensional, we have and since , it’s index is also 0. Computing the alternating sum of the dimensions we get from the snake lemma, we finally conclude that so that .

QED

Fredholm operators are often encoutered in the study of PDE’s and more classically in the theory of integral equations. Since for , the equation is solvable for iff . In particular, if the index of is 0, i.e. , then we get the so-called **Fredholm alternative** : Either (1) the inhomogeneous equation has a unique solution for every or (2) the homogeneous equation has linearly independant solutions and there are such that if for then has a solution.

For example, if for some closed interval, then it can be shown that for ,

is a Fredholm operator of index 0. It follows that the equation

has a solution if and only if is -orthogonal to every solution of the homogeneous adjoint equation .

This principle can also be seen to be behind the fact that the laplace equation can be solved for on a compact manifold iff .

The fact that the integral equation above gives a Fredholm operator of index 0 is a consequence of the compactness of and of the

**Theorem **(Riesz, 1918): For any compact operator , the operator is Fredholm with index 0.

In the next post, I will prove this after introducing compact operators and then discuss a close relation between compact and Fredholm operators given by Atkinson’s theorem.

Hilbert’s theorem 90 is the 90’th theorem in Hilbert’s *Zahlbericht* (meaning number report according to google translate), which is a famous report on the state of algebraic number theory at the end of the nineteenth century. It is a basic theorem shedding some light on cyclic field extensions (a Galois extension with a cyclic Galois group) and it seems to lead to some really big ideas in algebraic number theory. As a corollary of this, we directly get a parametrization of the rational points on the circle, hence of integral pythagorean triples.

Given a finite extension , we can naturally consider as a -vector space. We can use this point of view to define a notion of norm in field extensions. For any , we define the **norm** of with respect to this extension as

where is multiplication by viewed as a -linear transformation on . It also makes sense to speak of the trace of an element with respect to a field extension and we actually get an analogous “additive” Hilbert 90 without further difficulty.

Take for example the complex numbers as an extension of the reals. If , then in the basis the application is represented by the matrix so we get . A similar argument shows that in a quadratic number field of the form for an integer which is not a square, we find .

**Lemma: **For a simple algebraic extension , we have where is the constant term of the minimal polynomial of in .

**Proof: **Let be the minimal polynomial of , say of degree and recall that can be seen as the field of -polynomials of degree less than where multiplication is done modulo . Then is identified with the class of . A nice basis for this space is . Here multiplication by acts on these vectors as (everything modulo ). But so multiplication by is represented by

.

This matrix clearly has determinant hence the assertion.

QED

Note that it also follows from the proof that . To get a better description of that norm, we need another lemma:

**Lemma:** For a finite extension and , let . Then .

**Proof:** If is a basis of as a vector space over and is the canonical basis of over , then for and is a basis for over . Ordering this basis by , the matrix of multiplication by is given by the block diagonal matrix

where is the matrix representing multiplication by in as in the first lemma.

QED

The analogous result for the trace is that . These two lemmas let us interpret the norm and trace in a clarifying way:

**Proposition:** For a Galois extension and , we have

and .

**Proof:** If are the roots of the minimal polynomial of whose constant term we denote by and is of degree over , then since , the preceding lemmas tell us that

.

Now, recall that acts transitively on the set of roots of since it is irreducible. Denoting by the stabilizer of a root for this action, we have and since there is only one orbit, all stabilizers are conjugate of each others so for all root . This shows that to each root of there corresponds at least elements of . Since , this correspondence is bijective and we have

as we wanted.

QED

We are now in a position to state and prove our main theorem, which characterizes unit norm elements in a cyclic extension:

**Theorem (Hilbert 90):** Let be a cyclic (Galois) extension of degree , take a generator for the Galois group and . Then if and only if for some .

One direction is immediate. Indeed, for any and , we have

(where the products are all taken over ). For the other direction, we need a lemma:

**Lemma:** If are pairwise distinct -automorphisms of , then they are linearly independant over .

**Proof of the lemma:** Let be a minimal linear relation among the automorphisms. Since they are pairwise distinct, there exists such that . Then for all , since , we have also that . We obtain in this way that for all ,

and

so subtracting give us which is a non-trivial relation (since we took so that ). But this relation is shorter than our initial relation, which we supposed to be minimal. This is a contradiction and the lemma is proven.

**Proof of the theorem:** Suppose and define as . Then it suffices to find a fixed point of . It is easy to see that for , we have

.

In particular, so if , we have for all . In particular, the order of is . Since moreover is -linear over , we get a representation by . From this point of view, it is easy to determine if has a fixed point. Indeed, for any group and any representation , recall that

is the projection of onto the trivial summand of the representation. So to show that a group action has a fixed point, it suffices to show that this map is not identically . But in our case we have

and since is a generator for the Galois group, the automorphisms are linearly independent by the lemma so is not identically zero because that would give us a non-trivial linear relation between them. This shows that for some and that .

QED

Note that the analogous result for the trace is that in a cyclic extension, is and only if . As a cool application of this theorem, we can give a parametrization of pythagorean triples. Indeed, consider the equation for not a square. Recall that the norm of is

(this is easy to see with the finer interpretation of the norm as a product over the Galois group). So rational solutions of our equation correspond exactly to the elements of unit norm in . By Hilbert’s theorem 90 and since conjugation generates the Galois group, these elements are those of the form

.

Since multiplying and by some rational number does not change the expression, we can take and as integers. Therefore, the rational solutions to the equation are

where . Letting , we find all the rational points on the unit circle, which give us the integer pythagorean triples by the correspondence .