In this post, I will rely on this last post for results and notations, for example $X$ will also mean a compact Riemann surface. We will note $K$ the canonical bundle of $X$, i.e. the bundle with sheaf of sections $\Omega^1$ the holomorphic 1-forms on $X$. We will also note $h^j(X,L)$ the dimension of $H^j(X,L)$ by which we really mean $H^j(X,\mathcal{O}(L))$ the j’th cohomology group of $X$ in the sheaf of holomorphic sections of $L$. I’ll also write $H^j(X,D)$ to mean $H^j(X,[D]$ where $[D]$ is the line bundle associated with the divisor $D$, and sometimes the line bundle $L_1 \otimes L_2$ might get written $L_1 + L_2$ accordingly. We will sometimes also suppress the $X$ and simply write $H^j(L)$ for $H^j(X,L)$. 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 $L \to X$, the following relation holds:

$h^0(X,L) - h^0(X,K-L) = \text{deg}(L) - g + 1$.

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 $L$ simply as the degree of its associated divisor. To be more precise, we will do the proof for line bundles coming from generic divisors $\sum p_i$ with the $p_i$‘s distinct. We will first prove that

$h^0(X,L) - h^1(X,L) = \text{deg}(L) - g + 1$

which is relatively easy. The hard part will be

Theorem (Serre duality): There is a natural coupling $H^0(K - L) \otimes H^1(L) \to \mathbb{C}$ which induces an isomorphism

$H^1(L) \cong H^0(K-L)^*.$

In particular, this tells us that $h^1(L) = h^0(K-L)$ and Riemann-Roch follows.

Proof of the first part: We show the first part by induction on the degree of $L = [D].$ If $\text{deg}(L) = 0$, i.e. if $L$ is trivial, then by the maximum principle $h^0(X,L) = h^0(X,\mathcal{O}) = 1$ and $h^1(X,\mathcal{O}) = h^{0,1}(X) = g$. So

$h^0(X,L) - h^1(X,L) = 1 -g = \text{deg}(L) - g + 1$

and we are done with the base case. Now, suppose that we can represent $L$ as $[D]$ for a divisor $D = \pm \sum_{i=1}^d p_i$ with the plus or minus sign in accordance with the sign of $d = \text{deg}(L)$, and with all $p_i$‘s distinct.

Suppose by induction that the statement is true for $|\text{deg}(L)| \leq d$. Then it suffices to show it is true for the line bundles associated to $D + p$ and $D - p$ for some divisor $D$ of degree $\pm d$ not containing $p$. In the case of positive degree, we use the short exact sequence

$0 \to \mathcal{O}(D) \to \mathcal{O}(D+p) \overset{\text{res}_p}\to T_pX \to 0$.

Note that this short exact sequence was also considered in my last post with trivial $D$. Here we do the usual identification between sections $H^0(X,\mathcal{O}([D]))$ with meromorphic functions $f$ such that $D + (f) \geq 0$ and the first arrow corresponds to the inclusion of families of these spaces of meromorphic functions. We get a long exact sequence

$0 \to H^0(D) \to H^0(D+p) \to T_pX \to H^1(D) \to H^1(D+p) \to 0$

hence adding alternating dimensions we find

$0 = h^0(D) - h^0(D+p) + 1 - h^1(D) + h^1(D+p)$.

Using our induction assumption, i.e. that $h^0(D) - h^1(D) = d-g+1$, we deduce that

$h^0(D+p) - h^1(D+p) = (d+1) - g + 1$

which is what we wanted.

For the second case, when $\text{deg}(L) = -k$, we use another but related short exact sequence:

$0 \to \mathcal{O}(D-p) \to \mathcal{O}(D) \overset{\text{ev}_p}\to \mathbb{C}_p \to 0$.

A similar argument shows that

$h^0(D-p) - h^1(D-p) = -(d+1) - g + 1$

and we are done.

QED

Before attacking Serre duality, recall that for any line bundle $L \to X$ there is an operator

$\overline{\partial}_L : A^{0,0}(L) \to A^{0,1}(L)$

where $A^{p,q}(L)$ is the sheaf of smooth $(p,q)$-forms with values in $L$. It is the unique such operator satisfying Leibniz rule

$\overline{\partial}_L(fs) = f\overline{\partial}_L(s) + (\overline{\partial}_Lf)s$

and such that

$0 \to \mathcal{O}(L) \to A^{0,0}(L) \overset{\overline{\partial}_L}\to A^{0,1}(L) \to 0$

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

$\Delta' : H^{0,1}_{\overline{\partial}_L}(X,L) = \text{coker}(\overline{\partial}_L : A^{0,0}(X,L) \to A^{0,1}(X,L)) \cong H^1(X,\mathcal{O}(L))$.

The identifications we discovered at the end of last post still hold in this context, with the necessary adjustments. Given two line bundles $L_1, L_2$, this isomorphism permits us to define a pairing

$H^0(L_1) \times H^1(L_2) \to H^1(L_1 \otimes L_2)$

by setting

$\alpha \otimes \beta \to \Delta'(\alpha (\Delta')^{-1}\beta)$.

In particular, with $L_1 = L^* \otimes K = K-L$ and $L_2 = L$, we get Serre’s pairing

$H^0(K-L) \otimes H^1(L) \to H^1(K) \cong \mathbb{C}$

where the last isomorphism is $H^1(X,K) \cong H^{1,1}_{\overline{\partial}}(X)$ (which follows from the same argument as for $H^{0,1}$) composed with integration $\int : H^{1,1}_{\overline{\partial}} \cong H^2(X,\mathbb{C}) \overset{\cong}\to \mathbb{C}$.

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

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

$H^0(K-L) \cong H^1(L)^*$     and     $H^0(L)^* \cong H^1(K-L)$.

Note that $L$ satisfies Serre duality if and only if $K-L$ does.

The base case is the statement that the pairing $\Omega^1(X) \otimes H^{0,1}(X) \to \mathbb{C}$ given by $\int \alpha \wedge \beta$ is non-degenerate, which is just the isomorphism $H^{1,1}(X) \cong \mathbb{C}$ given by integration coupled with the fact that $\Omega^1(X) \to H^{1,0}(X)$ is injective.

Suppose now by induction that all line bundles $L$ with $|\text{deg}(L)| \leq d$ satisfy Serre’s duality. Like in Riemann-Roch, it suffices to show that $L+p$ and $L-p$ satisfy Serre duality (where $L = [D]$ with $D$ not containing $p$). We will now identify $\mathcal{O}(L + p)$ with meromorphic sections of $L$ with at worst a simple pole at $p$. We thus get the short exact sequence

$0 \to \mathcal{O}(L) \to \mathcal{O}(L+p) \to L_p \otimes K_p^* \to 0$

where the last arrow is $\text{ev}_p \otimes \text{res}_p$ (recall that the residue of a meromorphic function at $p$ is naturally an element of the holomorphic tangent space $K^*_p$, as discussed in my last post and under our identifications, $\mathcal{O}(L+p) \subset \mathcal{O}(L) \otimes \mathcal{M}_X$). We obtain in this way our first exact sequence

$0 \to H^0(L) \to H^0(L +p) \to L_p \otimes K_p^* \overset{\Delta_1}\to H^1(L) \to H^1(L + p) \to 0$.

Recall from last post that if $l\otimes v \in L_p \otimes K_p^*$ for $v = \lambda\frac{\partial}{\partial z}$, then $\Delta_1(l\otimes v) = [\frac{\lambda}{z}\tilde{l}] \in H^1(L)$ for $\tilde{l}$ an extension of $l$ in the neighborhood $U_1$ of $p$.

Similarly, we see elements of $\mathcal{O}(K-L-p)$ as holomorphic sections of $K-L$ vanishing at $p$, and we have the following short exact sequence of sheaves:

$0 \to \mathcal{O}(K-L-p) \to \mathcal{O}(K-L) \to L_p^* \otimes K_p \to 0$

where now the last arrow is simply $\text{ev}_p \otimes \text{ev}_p$. This gives our second exact sequence

$0 \to H^0(K-L-p) \to H^0(K-L) \to L_p^* \otimes K_p \to \cdots$

$\cdots \to H^1(K-L-p) \to H^1(K-L) \to 0$.

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 $2\pi i {ev}_p$ instead of just $\text{ev}_p$ and similarly $2\pi i \text{res}_p^*$.

For the third square, consider $l^*\otimes \alpha \in H^0(K-L)$ (so $\alpha$ is a holomorphic 1-form and $l^*$ a section of $L^*$). We need to see that for all $f \otimes u \in L_p \otimes K_p^*$, we have

$2\pi i \langle \text{ev}_p(l^*\otimes \alpha), f \otimes u \rangle = \langle l^*, f \rangle \int_X \alpha \wedge \Delta_1(u)$.

But we have seen in last post that if $u = \lambda\frac{\partial}{\partial z}$ near $p$, then after identifying $H^1(X,\mathcal{O})$ with $H^{0,1}(X)$, we have $\Delta_1(u) = \overline{\partial}(\beta)\frac{\lambda}{z}$. Thus if $\alpha = g(z)dz$, we find

$\int_X \alpha \wedge \Delta_1(u) = \int_X \overline{\partial}(\beta)\frac{\lambda}{z}g(z)dz = \int_X d(\beta \frac{\lambda}{z}g(z)dz)$

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

$\int_{\gamma}\frac{\lambda}{z}g(z)dz = 2\pi i \lambda g(0)$

for $\gamma$ a small circle around $p$. But $\text{ev}_p(l^*\otimes \alpha)(f\otimes u) = \langle l^*, f \rangle g(0)\lambda$, so this shows the third square commutes.

The fourth square is similar. Take $l^* \otimes u^* \in L_p^* \otimes K_p$. Then unwinding the definition of $\Delta_2$, we see that

$\Delta_2(l^* \otimes u^*) = [l^* \otimes \lambda dz] \in H^1(K-L-p)$

if $u^* = \lambda dz$ in some coordinates around $p$. Then under the Dolbeault isomorphism, this class corresponds to $l^* \otimes \overline{\partial}(\beta)\lambda dz \in H^{0,1}(K-L-p)$ and it acts on $H^0(L+p)$ by

$\langle \Delta_2(l^* \otimes u^*), e \otimes f \rangle = \langle l^*, e \rangle \int_X\lambda f(z) \overline{\partial}(\beta)dz$

where $e \otimes f \in H^0(L+p)$, for $e \in \mathcal{O}(L)(X)$ and $f \in \mathcal{M}(X)$. Like for the third square, this integral can be rewritten

$\int_X \lambda f(z) \overline{\partial}(\beta)dz = \int_{\gamma}\lambda f(z)dz = \lambda 2\pi i a_{-1}$

for $\gamma$ a small loop around $p$ and $a_{-1}$ the residue of $f$ in the $z$ coordinate. But this is exactly what we needed since

$\langle l^* \otimes \text{res}_p^*(u^*) , e \otimes f \rangle = \langle l^*, e \rangle \langle u^*, \text{res}_p(f) \rangle = \langle l^*, e \rangle \lambda a_{-1}$.

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

Let us start by considering a form of the Mittag-Leffler problem: Given a point $p \in X$, can we find a global meromorphic function $f \in \mathcal{M}(X)$ such that $(f)_{\infty} = p$ and $\text{res}_p(f) = \lambda$? i.e. such that the only poles of $f$ is a simple pole at $p$ with residue $\lambda$ in some coordinates. Since this would give a biholomorphic function $f : X \to \mathbb{P}^1$, this is only possible if the genus of $X$ is zero, but let us see how this obstruction manifests itself.

We will consider a covering $\mathfrak{U} = U_1, U_2$ of $X$ where $U_1$ is a coordinate patch centered at $p$ with (holomorphic) coordinate $z$, while $U_2$ is just $X \setminus p$. We can reformulate the question as follows: does there exist $f \in \mathcal{M}(X)$ such that $f_1 \in \frac{\lambda}{z} + \mathcal{O}(U_1)$ and $f_2 \in \mathcal{O}(U_2)$? NB: Here $f_i$ means the restriction $f|_{U_i}$.

Before going further, note that the residue of a meromorphic function is only invariantly defined as a tangent vector: if $f \in \mathcal{M}(X)$ has a simple pole at $p$, then we define, for $\alpha_p \in T^*_pX$,

$\langle \tilde{\alpha}, \text{res}_p(f) \rangle = \text{res}_p(f\alpha)$

where $\tilde{\alpha}$ is any extension of $\alpha_p$, and where

$\text{res}_p(\alpha) = \frac{1}{2\pi i}\int_{\gamma}\alpha$

is a well-defined residue for any meromorphic 1-form $\alpha$ defined around $p$, where $\gamma$ is a small loop around $p$. Another way to see this is to take $z$ a holomorphic coordinate around $p$ and $f$ a meromorphic function with a simple pole at $p$, say $f(z) = \frac{\lambda}{z} + h(z)$ for $h$ holomorphic. Then if $\tilde{z} = c_1z + c_2z^2 + \cdots$ is another choice of coordinates, then $\frac{1}{z} = \frac{c_1}{\tilde{z}} + \frac{c_2z}{\tilde{z}} + \cdots$ so

$f(\tilde{z}) = \frac{c_1\lambda}{\tilde{z}} + g(\tilde{z})$

for some holomorphic $g$. In other words,

$(\text{res}_pf)_{\tilde{z}} = \left(\frac{d\tilde{z}}{dz}|_p\right)(\text{res}_pf)_{z}$

i.e.

$\text{res}_p(f) = \lambda\frac{\partial}{\partial z}$

is well-defined as an element of $T_pX$.

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 $f$ such that on $U_1 \cap U_2 = U_1 \setminus p$ we have $f = \frac{\lambda}{z}$. Thus we see $f \in C^1(\mathfrak{U},\mathcal{O})$, as a Cech 1-cochain in the sheaf of holomorphic functions and ask that $f = f_2 - f_1$ on $U_1 \setminus p$ for $f_i \in \mathcal{O}(U_i)$, i.e. that $f$ defines a coboundary in $H^1(X,\mathcal{O})$. Indeed, $f_2$ would then be holomorphic on $X \setminus p$ with $f_2 = \frac{\lambda}{z} + f_1$ near $p$.

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

$0 \to \mathcal{O} \to \mathcal{O}(p) \overset{\text{res}_p}\to T_pX \to 0$

where $\mathcal{O}(p)$ is the sheaf of meromorphic functions having at worst a simple pole at $p$ and where we consider $T_pX$ as the associated skyscraper sheaf. From the induced long exact sequence we get

$H^0(X,\mathcal{O}(p)) \to T_pX \overset{\Delta}\to H^1(X,\mathcal{O})$

and in this formulation, finding a meromorphic function with at worst a simple pole at $p$ with residue $v = \lambda \frac{\partial}{\partial z}$ is possible if and only if $\Delta(v) = 0 \in H^1(X,\mathcal{O})$. But unwinding the definition of $\Delta$, we see that $\Delta(v) = f$ iff $\exists f_i \in \mathcal{O}(p)(U_i)$ such that $\text{res}_p(f_1) = v$ and $f_1 - f_2 = f$ in $U_1 \setminus p$. Thus we can take $f_1 = \frac{\lambda}{z}$ and $f_2 = 0$, and

$\Delta(\lambda\frac{\partial}{\partial z}) = [\frac{\lambda}{z}] \in H^1(X,\mathcal{O})$.

We once again come to the conclusion that the problem is solvable exactly when $\frac{\lambda}{z}$ is a coboundary.

Second approach: Consider a smooth cut-off function $\beta$ supported in $U_1$ and identically equal to 1 near $p$. Then $\beta \frac{\lambda}{z} \in C^{\infty}(U_2)$ and the probem reduces to finding $g \in C^{\infty}(X)$ such that $\overline{\partial}(\beta \frac{\lambda}{z}) = \overline{\partial}g$. Note that $\overline{\partial}\beta = 0$ near $p$ so we can consider

$\Psi = \overline{\partial}(\beta)\frac{\lambda}{z}$

as an element of $A^{0,1}(X)$, i.e. a $(0,1)$-form on $X$. Reformulating this, we can solve the problem exactly when $[\Psi] = 0 \in H^{0,1}_{\overline{\partial}}(X)$.

To relate this to our first approach, we note $A^{p,q}$ the sheaf of $(p,q)$-forms and use the short exact sequence of sheaves

$0 \to \mathcal{O} \to A^{0,0} \overset{\overline{\partial}}\to A^{0,1} \to 0$

which gives the exact sequence

$H^0(X,A^{0,0}) \to H^0(X,A^{0,1}) \overset{\Delta'}\to H^1(X,\mathcal{O}) \to 0$.

In other words, this $\Delta'$ gives an isomorphism

$H^{0,1}_{\overline{\partial}}(X) = \text{coker} (\overline{\partial}:A^{0,0}(X) \to A^{0,1}(X)) \cong H^1(X,\mathcal{O}).$

Unwinding the definition of $\Delta'$, we see that $\Delta'(\overline{\partial}(f_i)) = f$ iff $f = f_2 - f_1$ in $U_1 \cap U_2$, for $f_i \in \mathcal{O}(U_i)$.

This isomorphism $H^{0,1}_{\overline{\partial}}(X) \cong H^1(X,\mathcal{O})$ is the simplest instance of the so-called Dolbeault isomorphism which is a similar isomorphism $H^{p,q}_{\overline{\partial}}(X,E) \cong H^q(X,\Omega^p(E))$ valid for arbitrary compact complex manifolds and complex vector bundle $E$, where $\Omega^p(E)$ is the sheaf of holomorphic p-forms with values in $E$.We can recap our discussion with the following:

Proposition: The obstruction to solving the Mittag-Leffler problem coming from our second approach, $[\Psi] \in H^{0,1}_{\overline{\partial}}(X)$ which is represented by the 1-cocycle $\overline{\partial}(\beta)\frac{\lambda}{z} \in A^{0,1}(X)$, corresponds under Dolbeault’s isomorphism to $[\frac{\lambda}{z}] \in H^1(X,\mathcal{O})$, 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 $\mu$ is surjective.

Theorem (Jacobi’s inversion): Let $S$ be a compact Riemann surface of genus $g$ and take any $p_0 \in S$. Then for any $\lambda \in J(S)$, we can find $g$ points $p_1, \dots, p_g \in S$ such that

$\mu\left(\displaystyle\sum_{i=1}^g(p_i-p_0)\right) = \lambda$.

In other words, for $\omega_1, \ldots, \omega_g$ a basis of $H^0(S,\Omega^1)$, for all $\lambda \in \mathbb{C}^g$, there is $p_1, \ldots, p_g \in S$ and paths $\alpha_i$ from $p_0$ to $p_i$ such that

$\displaystyle\sum_{i=1}^g \int_{\alpha_i}\omega_j = \lambda_j$     for all $1\leq j \leq g$.

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

$\mathcal{M}(S) \overset{()}\to \text{Div}_0(S) \overset{\mu}\to J(S) \to 0$.

Lemma 1: The set $S^{(d)}$ of effective divisors of degree $d$ on $S$ is a compact complex manifold.

Proof of lemma 1: Consider the action of the permutation group $\mathfrak{S}_d$ on

$S^d = S \times \cdots \times S$    ($d$ times).

The quotient, denoted $S^d/\mathfrak{S}_d = \text{Sym}^d(S)$, inherits the quotient topology making $\pi : S^d \to \text{Sym}^d(S)$ a continuous map. Clearly, $\text{Sym}^d(S)$ is in bijection with $S^{(d)}$. Suppose for a moment that $S = \mathbb{C}$. We will write an element of $\text{Sym}^d(\mathbb{C})$ as a sum $\lambda_1 + \cdots \lambda_d$ to indicate the unimportance of the ordering (note that the $\lambda_i$‘s are not necessarily distinct). We consider the map

$\varphi : \text{Sym}^d(S) \to \mathbb{C}^d$

that takes $\sum_{i=1}^d\lambda_i$ to the d-tuple $(a_1, \ldots, a_d) \in \mathbb{C}$ consisting of the coefficents of the monic polynomial having $\lambda_1, \ldots, \lambda_d$ as its roots, i.e

$\varphi(\lambda_1 + \cdots \lambda_d) = (a_1, \ldots, a_d)$

if

$z^d + a_1z^{d-1} + \cdots a_{d-1}z + a_d = (z-\lambda_1)\cdots (z-\lambda_d)$.

In other words, $\varphi(\{\lambda_i\}) = (\sigma_1(\{\lambda_i\}), \ldots, \sigma_d(\{\lambda_i\})$ where $\sigma_i$ is the ith elementary symmetric polynomial. By the fundamental theorem of algbera, $\varphi : \text{Sym}^d(\mathbb{C}) \to \mathbb{C}^d$ is a bijection, and in fact a homeomorphism.

The difficulty is in seeing that $\varphi^{-1}$ is also continuous. To see this, take an open set $\pi (U) \subset \text{Sym}^d(\mathbb{C})$ and consider $(a_1, \ldots, a_d) \in V := \varphi (\pi (U)) \subset \mathbb{C}^d$. In other words,

$p_0 := z^d + a_1z^{d-1} + \cdots + a_{d-1} + a_d = \displaystyle\prod_{i=1}^d(z-\lambda_i)$     with $\lambda_i \in U$.

We need to show that there is an open set $W \subset V$. Write

$N_U(p) = \dfrac{1}{2\pi\sqrt{-1}}\displaystyle\int_{\partial U} \frac{dp}{p}$

which, when defined, is the number of zeros of the polynomial $p$ inside $U$. We can choose an open set $W \subset \mathbb{C}^d$ containing $p_0$ small enough so that $N_U(p)$ is defined for all $p \in W$. Then since $p \mapsto N_U(p)$ is continuous and takes only integer values, we have

$N_U(p) = N_U(p_0) = d$    for all $p \in W$.

But this means that for all $p \in W$, all the roots of $p$ are in $U$ so that $p_0 \subset W \subset V$ as we wanted.

This gives $\text{Sym}^d(\mathbb{C})$ the structure of a complex manifold of dimension $d$, in fact biholomorphic to $\mathbb{C}^d$.

Now we put a similar complex manifold structure on $\text{Sym}^d(S)$ for arbitrary $S$. Consider $D = \sum_{i=1}^dp_i \in \text{Sym}^d(S)$ and take holomorphic charts $(U_i,z_i)$ around $p_i$ on $S$ such that $U_i \cap U_j = \emptyset$ if $p_i \neq p_j$ and $U_i = U_j, z_i = z_j$ if $p_i = p_j$. We obviously have an injective mapping

$\varphi_D : \pi(U_1 \times \cdots U_d) \to V_D \subset \mathbb{C}^d$

where $\pi : S^d \to \text{Sym}^d(S)$, by doing as above;

$\varphi_D(\sum_jz_j) = \left(\sigma_1(\{z_j\}), \ldots, \sigma_d(\{z_j\})\right)$.

The verifications that this gives a homeomorphism onto an open set of $\mathbb{C}^d$ is just as in the earlier case. These maps thus provide $\text{Sym}^d(S)$ with a holomorphic atlas.

QED

Fixing a base point $p_0 \in S$ we get a (holomorphic) injection

$\iota : S^{(d)} \hookrightarrow \text{Div}_0(S)$

$\iota : \displaystyle\sum_{i=1}^d p_i \mapsto \sum_{i=1}^d(p_i-p_0)$

and thus holomorphic mappings

$\mu^{(d)} : S^{(d)} \to J(S)$

$\mu^{(d)} : \displaystyle\sum_{i=1}^dp_i \mapsto \sum_{i=1}^d \left(\int_{p_0}^{p_i}\omega_1, \ldots, \int_{p_0}^{p_i}\omega_g\right)$    (mod $\Lambda$)

by composing $\mu$ and $\iota$. Jacobi’s inversion theorem says that $\mu^{(g)}$ is surjective.

Lemma 2:  Let $f : M \to N$ be a holomorphic map between two compact connected complex manifolds of the same dimension. If $f$ is not everywhere singular, i.e. the Jacobian matrix $J(f)$ is not identically zero, then $f$ is necessarily surjective.

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

Consider $\psi_N$ a volume form on $N$. Since $J(f)$ is not identically $0$ and since $f$ preserves the orientation (being holomorphic), we have

$\displaystyle\int_Mf^*\psi_N > 0$.

Since for any $q \in N$ we have $H^{2n}(N-\{q\},\mathbb{R}) = 0,$ the volume form is exact in $N - \{q\}$ and

$\psi_N = d\varphi$

for some $(2n-1)$-form $\varphi$ on $N - \{q\}$. But then if $q \notin f(M)$, we have

$\displaystyle\int_M f^*\psi_N = \int_{\partial M}df^*\varphi = 0$,

QED

To prove the theorem, we thus have to show that $\mu^{(g)}$ is not everywhere singular.

Proof of the theorem: At points $D = \sum_ip_i \in S^{(d)}$ such that the $p_i$‘s are distinct, the quotient map $\pi : S^d \to S^{(d)}$ is locally a biholomorphism. So choosing disjoint charts $(U_i,z_i)$ in $S$ centered at $p_i$, we get a chart

$\varphi : \pi(U_1 \times \cdots \times U_d) \to \mathbb{C}^d$

$\varphi(\sum_iq_i) = (z_1(q_1), \ldots, z_d(q_d))$.

In such coordinates, for $D' = \sum_iq_i$ near $D$, we have

$\mu^{(g)}(D') = \displaystyle\sum_{i=1}^g\left(\int_{p_0}^{z_i}\omega_1, \ldots, \int_{p_0}^{z_i}\omega_g \right)$     mod $\Lambda$.

So

$\dfrac{\partial}{\partial z^i}(\mu^{g}_j(D')) = \displaystyle\sum_{k=1}^g \dfrac{\partial}{\partial z^i}\left(\int_{p_0}^{z_k}\omega_j\right)$.

But with $\omega_j(z^i) = h_{ji}(z^i)dz^i$ near $p_i$, this is

$\dfrac{\partial}{\partial z^i} \displaystyle\int_{p_0}^{z_i}h_{ji}(z^i)dz^i$

hence

$\dfrac{\partial}{\partial z^i} (\mu^{(g)}_j(D')) = h_{ji}(z^i).$

The jacobian matrix of $\mu^{(d)}$ near $D$ is thus given by

$J(\mu^{(d)}) = \left(\begin{array}{ccc}h_{11} & \cdots & h_{1d}\\ \vdots && \vdots \\ h_{g1} & \cdots & h_{gd}\end{array}\right)_{g\times d}$.

It suffices to see that this matrix is of full rank for some choice of $D = \sum_{i=1}^d p_i$ and basis $\omega_1, \ldots, \omega_g$. We simply do Gauss reduction: Choose $p_1$ such that $\omega_1(p_1) \neq 0$. Then subtracting a multiple of $\omega_1(p_1)$ to $\omega_2, \ldots, \omega_g$, we make it so that

$\omega_2(p_1) = \cdots = \omega_g(p_1) = 0$.

The $\omega_i$‘s are still a basis of $\Omega^1(S)$ and we continue like this, choosing a $p_2$ such that $\omega_2(p_2) \neq 0$ etc… We eventually arrive at the form

$J(\mu^{(d)}) = \left(\begin{array}{cccccc}h_{11}&h_{12}&\cdots&h_{1g}&\cdots&h_{1d} \\ 0&h_{22}&\cdots&h_{2g}&\cdots&h_{2d} \\ \vdots&\vdots&\ddots&&& \\ 0&0&\cdots&h_{gg}&\cdots&h_{gd}\end{array}\right)$

which is of maximal rank since $h_ii(p_i) \neq 0$ for all $1 \leq i \leq g$. This shows that $\mu^{(g)} : S^{(g)} \to J(S)$ 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 $L \to S$ (i.e. with zero first Chern class) has the structure of a complex torus $\mathbb{C}^g/\Lambda$ of dimension $g$. Indeed, the group $\text{Pic}_0(S)$ consisting of those (isomorphism classes of) holomorphic line bundles is isomorphic to $\text{Div}_0/\mathcal{M}(S)$, which by Abel’s and Jacobi’s theorem is isomorphic to $J(S)$.

Note that in fact the fibres of $\mu^{(g)} : S^{(g)} \to J(S)$ consist of projective spaces. Indeed, if $\mu^{(g)}(D) = \lambda$, then by Abel’s theorem the fiber is $(\mu^{(g)})^{-1}(\mu(D)) = |D|,$ the set of effective divisors linearly equivalent to $D$, which corresponds to the projectivisation of $H^0(S,\mathcal{O}(D))$. It can be shown that generically the fiber of $\mu^{(g)}$ is a point and that $\mu^{(g)}$ is a birational map.

In last post we proved Abel’s theorem and as a corollary we saw that any compact Riemann surface $S$ of genus $g=1$ (a smooth elliptic curve) is biholomorphic to a complex torus, the biholomorphism being given by the Abel-Jacobi mapping:

$\mu|_S : S \overset{\cong}\longrightarrow \mathbb{C}/\Lambda$

which is given by

$\mu(p) = \displaystyle\int_{p_0}^p\omega$     (mod $\Lambda$)

where some $p_0 \in S$, $\omega \in \Omega^1(S)$ is a holomorphic 1-form and $\Lambda$ is the period lattice $\Lambda = \{\sum_{i=1}^2m_i\Pi_i \mid m_i \in \mathbb{Z}\}\subset \mathbb{C}$ for $\Pi_i$ the two period vectors $\Pi_i = \int_{\delta_i}\omega \in \mathbb{C}$ for $\delta_1, \delta_2$ giving a basis for $H_1(S,\mathbb{Z})$.

In particular, every such Riemann surface $S$ 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 $\mathbb{P}^2$ 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 $\mathbb{P}^2$ with the zero locus of a  cubic polynomial, you get (counting multiplicities) exactly 3 points. Thus to add 2 points $p_1, p_2$, you consider the line between them and declare that $p_1+p_2 + p_3 = 0$ if $p_3$ 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 $\mathbb{P}^2$ as the zero locus of some cubic polynomial. In particular when the initial surface is a complex torus $\mathbb{C}/\Lambda$, this embedding is given by the so-called Weierstrass $\wp$-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 $L \to M$ is a holomorphic line bundle over a compact complex manifold $M$ and $s_0, \ldots, s_N$ are a basis for $V = H^0(S,\mathcal{O}(L))$ the global holomorphic sections of $L$. Suppose that there is no point on $M$ such that every sections in $V$ vanish simultaneously. Then for every $p \in M$, we have a non-zero vector

$(s_0(p), \ldots, s_N(p)) \in (L_p)^{N+1}$.

By choosing a trivialisation, we can consider this vector as sitting in $\mathbb{C}^{N+1}$. Of course this vector will depend on the choice of trivialisation of $L$ around $p$ but different choices of trivialisations will only change the vector by a complex multiple. We thus get a map

$\iota_L : M \to \mathbb{P}^N$ defined by $\iota_L(p) = [s_0(p):\ldots:s_N(p)]$.

Kodaira’s embedding theorem states that when $L \to M$ is a positive line bundle (i.e. admits a connection of positive curvature), there exists a $k_0 \in \mathbb{N}$ such that for $k \geq k_0$, the map

$\iota_{L^k} : S \to \mathbb{P}^N$

is well-defined (the global sections of $\mathcal{O}(L)$ are never all vanishing) and is an embedding of $M$.

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 $S$ be a compact Riemann surface and $L \to S$ a holomorphic line bundle. If $\deg L > \deg K_S + 2$, then $\iota_L : S \to \mathbb{P}^N$ is well-defined and an embedding.

In this proposition, the degree of a line bundle is just $c_1(L)$ under the identification $H^2(S,\mathbb{Z}) \cong \mathbb{Z}$ or equivalently, since if $[\sum_ia_ip_i] = L$ then the Poincaré dual $\eta_D = c_1(L)$, the degree is $\sum_ia_i$. Also, $K_S \cong \bigwedge^{1,0}T^*S$ is the canonical bundle of $S$ and $N = \dim H^0(S,\mathcal{O}(L)) - 1$.

The case that will interest us is when the genus of $S$ is $g=1$. What we show is that in fact we can take $N=2$. Indeed, since $K_S = 0$ because $T^*S$ is trivial ($S$ is topologically a torus), for any $p \in S$ the above theorem tells us that for the bundle $L = [3p]$, the map $\iota_L$ gives an embedding of $S$ into $\mathbb{P}^N$ for $N = h^0(S, \mathcal{O}(L)) - 1.$ If $s \in H^0(S,\mathcal{O}([p]))$, then $s,s^2,s^3 \in H^0(S,\mathcal{O}(L))$ are 3 linearly independent global holomorphic sections so $N \geq 2$. On the other hand, a global section of $\mathcal{O}(L)$ corresponds to a meromorphic function on $S$ having poles only at $p$ and at worst of degree $3$. Such a function is completely determined by the first four coefficients in its Laurent expension around $p$:

$\dfrac{a_{-3}}{z^3} + \dfrac{a_{-2}}{z^2} + \dfrac{a_{-1}}{z} + a_0 + \cdots$

(the difference of two such functions having all of these 4 numbers equal would be a global holomorphic function vanishing at $p$ hence everywhere). Moreover, two such functions having the same $a_{-3}$ and $a_{-2}$ must have the same $a_{-1}$ because otherwise their difference would be a meromorphic function having only one pole at $p$ 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 $h^0(S,\mathcal{O}(L) \leq 3$. This shows $N=2$.

We can write this embedding explicitly as follows: by Kodaira’s vanishing theorem, we have $H^1(S,\mathcal{O}(p)) = 0$ so considering the long exact sequence associated to the short exact sequence

$0 \to \mathcal{O}(p) \to \mathcal{O}(2p) \overset{res}\to \mathbb{C}_p \to 0$

where $\mathbb{C}_p$ is the skyscraper sheaf at $p$, we find $H^0(S,\mathcal{O}(2p)) \to \mathbb{C} \to 0$ so there exists a meromorphic function $F$ on $S$ having a double pole at $p$ and no other poles. Since $h^0(S,\Omega^1) = g = 1$, there is also on $S$ a nonzero holomorphic 1-form $\omega$. But as remarked earlier, since the canonical bundle $K_S$ is trivial, the divisor $(\omega)$ has zero Poincaré dual so is itself zero, i.e. $\omega$ is nowhere vanishing. We consider $F\omega$. It is a meromorphic form with only one pole at $p$, of order 2, and since the sum of the residues of a meromorphic 1-form must be zero, $\text{res}_p(F\omega) = 0$.

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

$F(z) = \dfrac{1}{z^2} + a_1z + a_2z^2 + \cdots$    near $p$

after possibly multiplying by a constant and adding a constant. Considering also the meromorphic function $dF/\omega$, which is holomorphic except at $p$ where it has a triple pole, we get a global meromorphic function $F' = \lambda(dF/\omega) + \lambda'F + \lambda''$ which is

$F'(z) = \dfrac{1}{z^3} + a_1'z + a_2'z^2 + \cdots$   near $p$

for the right choice of $\lambda$‘s. Since $1, F$ and $F'$ are three linearly independant global sections of $\mathcal{O}([3p])$, they form a basis and thus the embedding $\iota_L : S \to \mathbb{P}^2$ is given by

$\iota_L(q) = [1:F(q):F'(q)]$.

We can use this to explicitly describe $S$ as the zero set of a polynomial in $\mathbb{P}^2$: around $p$, we have

$F'(z)^2 = \dfrac{1}{z^6} + \dfrac{c}{z^2} + \dfrac{a}{z} + \cdots$

and

$F(z)^3 = \dfrac{1}{z^6} + \dfrac{c'}{z^3} + \dfrac{c''}{z^2} + \dfrac{a'}{z} + \cdots$.

Thus the meromorphic function $F'(z)^2 + c'F'(z) - F(z)^3 + (c''-c)F(z)$ has at most a simple pole at $p$ and no other poles, so is constant. This tells us that $\iota_L(S)$ is included in the zero locus of the polynomial $y^2 + c'y = x^3 + ax + b$ which can be rewritten by making suitable linear changes in the coordinates by

$y^2 = x(x-1)(x-\lambda)$

for $\lambda \in \mathbb{C}$. Note that here $(x,y)$ is a set of affine coordinates in $\mathbb{P}^2$ on the open set $\{[z_0,z_1,z_2] \mid z_0 \neq 0\}$. 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 $\mathbb{P}^2$ is exactly $\iota_L(S)$.

We will show that in fact the above construction gives an inverse for the Abel-Jacobi map $\mu : \mathbb{P}^2 \supset S \overset{\cong}\to \mathbb{C}/\Lambda$ which realises such a  non-singular planar curve as a complex torus.

Consider $S = \mathbb{C}/\Lambda$ and $F,F'$ as above, with their pole at $p_0 \in S$. Take $\omega = dF/F'$ as a basis for $H^0(S,\Omega^1)$ and $z$ the coordinate on $S$ such that $\omega = dz$. In this context, the function $F$ is the so-called Weierstrass $\wp$-function. Its derivative $(\partial / \partial z)\wp = -2F'$ is denoted $\wp'$. If the Laurent expansions of $\wp$ around $p_0$ contained a term of odd degree, then killing the order 2 pole by $\wp(-z) - \wp(-z)$ we would obtain a noncontant holomorphic function on $S$. So it has no term of odd degree and we can write

$\wp(z) = \dfrac{1}{z^2} + az^2 + bz^4 + \cdots$

$\wp'(z) = -\dfrac{2}{z^3} + 2az + 4bz^3 + \cdots$

$\wp(z)^3 = \dfrac{1}{z^6} + \dfrac{3a}{z^2} + 3b + cz^2 + \cdots$

$\wp'(z)^2 = \dfrac{4}{z^6} - \dfrac{8a}{z^2} - 16b + c'z + \cdots$.

We find the relation

$\wp'^2 = 4\wp^3 - g_2\wp - g^3$,

where $g_2 = 20a$ and $g_3 = 28b$. The corresponding embedding is then

$\psi : \mathbb{C}/\Lambda \to \mathbb{P}^2$ where $\psi(z) = [1:\wp(z):\wp'(z)]$,

and $\psi(\mathbb{C}/\Lambda)$ is the zero locus of the polynomial

$y^2 = 4x^3 - g_2x - g_3$,

written in suitable affine coordinates. On the other hand, the Abel-Jacobi map from $\psi(S)$ to $\mathbb{C}/\Lambda$ is given by

$\mu(p) = \displaystyle\int_{p_0}^p\frac{dx}{y}$    (mod $\Lambda$)

with let’s say $p_0 = \psi(0)$. Indeed, $\psi^*(dx/y) = \wp'(z)dz/\wp'(z) = dz = \omega$ so $dx/y$ is a non-zero holomorphic 1-form on $S$. Moreover, $\mu|_S$ is actually inverse to the embedding $\psi$ because

$\mu(p) = \displaystyle\int_{\psi(0)}^p \frac{dx}{y} = \int_0^{\psi^{-1}(p)}dz = \psi^{-1}(p)$.

We have thus found an inverse for the Abel-Jacobi mapping of a smooth elliptic curve in $\mathbb{P}^2$.

We are now in a position to discuss the group structure from different point of views. Any Riemann surface $S$ of genus $g=1$ inherits a group structure simply by letting

$p_1+_{\mu}p_2 = \mu^{-1}(\mu(p_1) + \mu(p_2))$   for $p_1, p_2 \in S.$

There is also a group structure induced by an embedding $\psi : S \hookrightarrow \mathbb{P}^2$ as above : $p_1 +_{\psi} p_2 +_{\psi} p_3 = 0$ if $\psi(p_1),\psi(p_2),\psi(p_3)$ are colinear in $\mathbb{P}^2$. These are in fact the same. To see this, consider $S$ as embedded in $\mathbb{P}^2$, take 3 points $p_1,p_2,p_3 \in S$ and denote $z_i = \mu(p_i) \in \mathbb{C}/\Lambda$ the corresponding points in the complex torus. Then by Abel’s theorem,

$z_1 + z_2 + z_3 = 0$ iff $\exists f \in \mathcal{M}(S)$ with $(f) = p_1 + p_2 + p_3 - 3p_0 = 0$.

But suppose $p_1 +_{\psi} p_2 +_{\psi} p' = 0$ for $p' \in S$, i.e. with $A(x,y) = ax + by + c$ the line joining $p_1, p_2$ (in affine coordinates), we also have $A(p') = 0$. Then

$g(z) = A(\wp(z),\wp'(z)) = a\wp(z) + b\wp'(z) + c \in \mathcal{M}(\mathbb{C}/\Lambda)$

is a meromorphic function on the torus having divisor $(g) = z_1 + z_2 + \mu(z') - 3.0$ so the function $f = g\circ \mu \in \mathcal{M}(S)$ has divisor $(f) = p_1 + p_2 + p' - 3p_0$. Thus if $p_1, p_2, p_3$ are collinear, we have $p_1 +_{\mu} p_2 +_{\mu} p_3 = 0$ and conversely, if there is a meromorphic function $h$ on $S$ such that $(h) = p_1 + p_2 + p_3 - 3p_0$, then $h-g$ is a meromorphic function with $(h-g) = p_3 - p'$ and this forces $p_3 = p'$ because otherwise we would have a meromorphic function having a simple pole at $p'$ and no other poles.

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

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

In the next couple of posts, I will write about Riemann surfaces and their Jacobian, following Griffiths and Harris. To each compact Riemann surface $S$ of genus $g \geq 1$, we will associate a $g$-dimensional complex torus, which is a complex manifold of the form $\mathbb{C}^g/\Lambda$ for $\Lambda$ a discrete full-rank integral lattice in $\mathbb{C}^g$.

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

$H^q(M,\Omega^p) \cong H_{\bar{\partial}}^{p,q}(M)$.

So on a Riemann surface $S$ of genus $g \geq 1$, taking $\bar{\partial}$-cohomology class we find that $H^0(S,\Omega^1) \cong H^{1,0}_{\bar{\partial}}(S)$. Hence the dimension is

$\dim H^0(M,\Omega^1) = g$

because

$H^1(S,\mathbb{C}) \cong H^{1,0}(S) \oplus H^{0,1}(S)$

so $2h^{1,0}(S) = h^1(S) = 2 - \chi(S) = 2g$ since $\overline{H^{1,0}(S)} = H^{0,1}(S)$.

Let us first consider $C$ a genus $1$ compact Riemann surface, i.e. a torus. Then $H^0(S,\Omega^1)$ is 1-dimensional, let $\omega$ be a generator. If $p$ and $q$ are two points on the torus, of course the integral

$\displaystyle\int_q^p \omega$

is not well-defined as a complex number, it depends on the path chosen from $q$ to $p$ 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 $\gamma_1,\gamma_2$ are two paths from $q$ to $p$, their difference will be a closed loop representing a cycle in $H_1(C,\mathbb{Z})$. The idea is to view the integral as a point in the quotient $\mathbb{C}/H_1(C,\mathbb{Z})$.

More precisely, for a compact Riemann surface $S$ of genus $g$, we consider cycles $\delta_1, \ldots, \delta_{2g}$ which give a basis for $H_1(S,\mathbb{Z})$. We will suppose that $(\delta_i)$ is a canonical basis, i.e. that $[\delta_i] \cdot [\delta_{i+g}] = 1$ and $[\delta_i] \cdot [\delta_j] = 0$ for $j \neq i+g$, where $\cdot$ means the intersection product. This just means that $\delta_i$ intersects $\delta_{i+g}$ once positively and intersects no other cycles, think of the canonical cycles on the g-holed torus. The first half of the basis $\delta_1, \ldots, \delta_{g}$ is the $A$-cycles and the $B$-cycles are the others, $\delta_{g+1}, \ldots, \delta_{2g}$. Let’s also fix a basis $\omega_1, \ldots, \omega_g$ for $H^0(S,\Omega^1)$. Then the periods are the vectors

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

The period matrix is the $g \times 2g$ matrix $\Omega$ having the periods as columns:

$\Omega = (\Pi_1 \cdots \Pi_{2g})$.

Note that the periods are $\mathbb{R}$-linearly independent so they generate a full-dimensional lattice

$\Lambda = \{m_1\Pi_1 + \cdots + m_{2g}\Pi_{2g} \mid m_i \in \mathbb{Z} \}$.

To see this, recall that since $H^1(S,\mathbb{C}) = H^{1,0}(S) \oplus H^{0,1}(S)$, the forms $\{\omega_i, \overline{\omega}_i\}$ generate $H^1(S,\mathbb{C})$. So if we had a relation $\sum_ik_i\Pi_i = 0$ with $k_i \in \mathbb{R}$, this would give

$\displaystyle\sum_i k_i \displaystyle \int_{\delta_i} \omega_j = 0$ and $\displaystyle\sum_i k_i \displaystyle\int_{\delta_i} \overline{\omega}_i = 0$ for all $j$

so since the pairing $(\omega, \delta) \mapsto \int_{\delta} \omega$ is non-degenerate, this would give a relation $\sum_i k_i [\delta_i] = 0 \in H_1(S,\mathbb{R})$, which is impossible since $(\delta_i)$ give a basis for the homology group.

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

$J(S) = \mathbb{C}^g/\Lambda$.

Intrinsically, $J(S)$ is just the quotient $(H^0(S,\Omega^1)^*/H_1(S,\mathbb{Z})$ where we identify $H_1(S,\mathbb{Z})$ with its image under the natural inclusion which takes a class of closed loops $[\gamma]$ and maps it to the functional $\theta \mapsto \int_{\gamma} \theta$. This space is the right one to consider integrals on $S$ in the sense that the vector

$\left( \displaystyle\int_q^p \omega_1, \ldots, \displaystyle\int_q^p \omega_g \right)$

is not well-defined as a point of $\mathbb{C}^g$ but it is if we consider is as living on $J(S)$, i.e. modulo the period lattice $\Lambda$.

After choosing a base point $p_0 \in S$, we can define a mapping

$\mu : S \to J(S)$

by

$\mu(p) = \left( \displaystyle\int_{p_0}^p \omega_1, \ldots, \displaystyle\int_{p_0}^p \omega_g \right)$    (mod $\Lambda$).

In fact, given a divisor $D = \sum_i a_ip_i \in \text{Div}(S)$, we can extend this map by letting $\mu(D) = \sum_i a_i \mu(p_i)$. In particular, for divisors of degree 0

$D = \displaystyle\sum_i (p_i - q_i) \in \text{Div}_0(S)$,

the map $\mu$ is independent of the base point and

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

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

$\mu : \text{Div}_0(M) \to J(S)$.

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

$\mathcal{M}(S) \longrightarrow \text{Div}(S) \longrightarrow \text{Pic}(S) \to 0$

where $\mathcal{M}(S)$ is the space of all meromorphic functions on $S$. If we restrict this to $\text{Div}_0(S)$ and the corresponding flat line bundles $\text{Pic}_0(S)$, Abel’s theorem gives an injection

$\text{Pic}_0(S) \cong \text{Div}_0(S)/\mathcal{M}(S) \hookrightarrow J(S)$.

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 $g$-dimensional torus $J(S)$ naturally parametrizes flat line bundles on $S$. In fact, Torelli’s theorem says that $J(S)$, considered as a principally polarized abelian variety, entirely determines the Riemann surface $S$ we started with. Jacobian varieties are thus key in the study of compact Riemann surfaces.

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

$\int_M \langle \nabla f, X \rangle = - \int_M f \text{div}X + \int_{\partial M} f \langle X, \nu \rangle$

or Green’s formula

$\int_M (f \Delta g - g \Delta f) = - \int_{\partial M}\left(f\dfrac{\partial g}{\partial \nu} - g\dfrac{\partial f}{\partial \nu}\right)$

which are well-known to hold for domains in $\mathbb{R}^n$.

The Laplace operator acting on functions defined over $\mathbb{R}^n$ is usually defined by the simple formula

$\Delta f = \dfrac{\partial^2 f}{\partial (x^1)^2} + \cdots \dfrac{\partial^2 f}{\partial (x^n)^2}$.

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

$\Delta f = \text{div}(\text{grad}(f))$.

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 $p$ measures the amount of infinitesimal dilation an infinitesimal object placed at $p$ 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 $X$ on an oriented Riemannian manifold as the infinitesimal change of measurement of volumes when the volume form is flowing along $X$, that is we want

$(\text{div}X)d\mu = \left.\dfrac{d}{dt} \left(\Phi_t^{-1}\right)^*d\mu \right|_{t=0}$

where $d\mu$ denotes the volume form associated to the metric and $\Phi_t$ the one-parameter group of diffeomorphisms associated to $X$. The right side of this equation is called the Lie derivative of $d\mu$ along $X$ and is noted $\mathcal{L}_Xd\mu$. Since the bundle of top forms of an oriented Riemannian manifold is trivialised by $d\mu$, i.e. every top form is of the form $\alpha d\mu$ for a unique $C^{\infty}$ function $\alpha$, we actually get a well-defined function $(\text{div}X)$ by the relation

$(\text{div}X)d\mu = \mathcal{L}_Xd\mu$,

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 $\Delta = \text{div}(\text{grad}f)$ where $\text{grad}f$ is the gradient of $f$, caracterized by the relation

$g(\text{grad}f, X) = df(X)$.

The only thing that remains undefined is the normal vector field $\nu$. It is not hard to check that if all the transition functions of some atlas on $M$ 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 $\mathbb{R}^n$, say $\{e_1 \geq 0\}$, we can define a notion of “inward-pointing vector” tangent to the boundary. Indeed, at a point $p$ of the boundary, a tangent vector in $T_pM$ will be called “inward-pointing” if its image in any chart is contained in the half-space $\{e_1 > 0\}$. At each point of the boundary, we decompose $T_pM = T_p\partial M \oplus \mathbb{R}$ orthogonally with the metric, and we define the normal unit inward-pointing vector field $\nu$ 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 $d\tilde{\mu} = \iota_{\nu}d\mu$ where $\iota$ 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 $X$ and differential form $\omega$,

$\mathcal{L}_X\omega = d(\iota_X\omega) + \iota_Xd\omega$.

This identity tells us that $(\text{div}X)d\mu = \mathcal{L}_Xd\mu = d(\iota_Xd\mu)$ because the volume form $d\mu$ is of course closed. This will be useful when paired with Stoke’s theorem, which says that for all forms $\omega$ of degree $n-1$ on $M$,

$\int_M d\omega = \int_{\partial M} i^*\omega$

where $i : \partial M \to M$ is the inclusion. We can now prove the generalization of Gauss’s divergence theorem:

Theorem 1: With the notation introduced above,

$\int_M \text{div}Xd\mu = - \int_{\partial M} g(X,\nu)d\tilde{\mu}$

for all vector fields $X$.

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

$\int_M \text{div}Xd\mu = - \int_M d\left(\iota_X d\mu\right) = - \int_{\partial M} \iota_X d\mu$.

But if $(\nu, e_2, \ldots, e_n)$ is an orthonormal basis of $T_pM$ for $p \in \partial M$, then we see that

$(\iota_Xd\mu)(e_2, \ldots, e_n) = d\mu(X,e_2, \ldots, e_n)$

is equal to

$g(X,\nu)d\mu(\nu, e_2, \ldots, e_n) = g(X, \nu)d\tilde{\mu}(e_2, \ldots, e_n)$

because $X = g(X,\nu)\nu + \sum_{j=2}^ng(X,e_j)e_j$ and $d\mu(e_j, e_2, \ldots, e_n) = 0$ for all $2 \leq j \leq n$. Thus at all points of the boundary we have the equality

$\iota_Xd\mu = g(X,\nu)d\tilde{\mu}$.

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

QED

Recall that the integration by parts formula for functions $f : \mathbb{R} \to \mathbb{R}$ follows from Leibniz rule $(uv)' = u'v + v'u$ and from the fundamental theorem of calculus:

$u(1) - u(0) = \int_{[0,1]} (uv)' = \int_{[0,1]}u'v + v'u$.

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 $f \in C^{\infty}(M)$ and vector fields $X$,

$\text{div}(fX) = f(\text{div}X) + g(\text{grad}f,X)$.

Proof: This again follows from Cartan’ identity. Indeed, by the very definition of the interior product, we get that $\iota_{fX}\omega = f\iota_X\omega$ for any differential form $\omega$ so we get

$\mathcal{L}_{fX}\omega = d(\iota_{fX}\omega) + \iota_{fX}d\omega = d(f\iota_X\omega) + f\iota_Xd\omega$.

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

$df \wedge \iota_X \omega + fd (\iota_X \omega) + f\iota_X d\omega$

hence

$\mathcal{L}_{fX}\omega = f\mathcal{L}_X\omega + df\wedge \iota_X\omega$

so it suffices to see that $df \wedge \iota_Xd\mu = g(\text{grad}f,X)$. We proceed as in theorem 1. Let $e_1, \ldots, e_n$ be an orthonormal basis of $T_pM$ at some point $p \in M$. Then

$df \wedge \iota_Xd\mu (e_1, \ldots, e_n) = \sum_i (-1)^{i-1}df(e_i) \iota_Xd\mu(e_1, \ldots, \hat{e_i}, \ldots, e_n)$

which is equal to

$\sum_i (-1)^{i-1}df(e_i)d\mu(X,e_1,\ldots,\hat{e_i}, \ldots, e_n) = \sum_i df(e_i)X^id\mu (e_1, \ldots, e_n)$.

This shows that

$df \wedge \iota_X d\mu = \sum_i df(e_i)X^i d\mu$

but $\sum_i df(e_i)X^i$ is precisely $df(X) = g(\text{grad}f,X)$.

QED

Coupled with the divergence theorem, this immediately yields the

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

$\int_M g(\text{grad}f,X)d\mu = -\int_M f \text{div}X d\mu + \int_{\partial M} f \cdot g(X, \nu) d\tilde{\mu}$.

Note that for a closed Riemannian manifold, with $\langle X,Y \rangle = \int_M g(X,Y) d\mu$, this shows that

$\langle \text{grad} f, X \rangle = \langle f, -\text{div}X \rangle$

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 $f$ is defined as $\Delta f = \text{div}(\text{grad}f)$ and that $\nu$ is the inward-pointing vector field on the boundary. We will denote $g(\text{grad}f, \nu)$ by $\frac{\partial f}{\partial \nu}$.

Theorem: (Green formulas) For any two functions $u,v \in C^{\infty}(M)$,

$\int_M u\Delta v \; d\mu + \int_M g\left( \text{grad}u, \text{grad}v \right) d\mu = \int_{\partial M} u \dfrac{\partial v}{\partial \nu} \; d\tilde{\mu}$

and hence

$\int_M(u\Delta v - v\Delta u) \; d\mu = \int_\partial M \left(u\dfrac{\partial v}{\partial \nu} - v \dfrac{\partial u}{\partial \nu}\right)d\tilde{\mu}$.

Proof: Integrating by parts, we get

$\int_M g\left( \text{grad}u, \text{grad}v \right) d\mu = -\int_M u \text{ div}\left(\text{grad }v\right) d\mu + \int_{\partial M} u\dfrac{\partial v}{\partial \nu} \; d\tilde{\mu}$

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

QED

As an immediate application, we can show the

PropositionConsider the equation $\Delta f = 0$ on a compact Riemannian manifold with boundary $(M,g)$ subject to either Dirichlet (i.e. $f(x)|_{\partial M} = H(x)$) or Neumann (i.e. $\frac{\partial f(x)}{\partial \nu} = H(x)$) 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 $u$ and $v$ to the problem and consider their difference $w = u-v$. Then since $\Delta = 0$, it follows that $\int_M w\Delta w = 0$. Integrating by parts, we thus find

$\int_M g\left(\text{grad }w, \text{grad }w \right) d\mu = \int_{\partial M} w \dfrac{\partial w}{\partial \nu} \; d\tilde{\mu}$

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

QED

In the same vein, if $f$ is a harmonic function on a closed Riemannian manifold (i.e. $\Delta f = 0$ and $\partial M = \emptyset$), then $\int_M g(\text{grad }f, \text{grad }f)d\mu = 0$ so the only harmonic functions on a closed Riemannian manifold are the constants.

From now on, let us denote $\text{grad }f$ by $\nabla f$ and $\int_M g\left(\text{grad }f, \text{grad }f\right) d\mu$ by $\int_M |\nabla f|^2 d\mu$. Another immediate application of integration by parts yield results about the spectrum of $\Delta$:

Proposition: Let $(M,g)$ 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 $L^2$ inner product.

Proof: For the first assertion, suppose $\Delta u = \lambda u$. Then integrating by parts gives

$\lambda \int_M u^2 \; d\mu = \int_M u\Delta u \; d\mu = -\int_M |\nabla u|^2 \; d\mu \leq 0$.

The second assertion follows from Green’s formula since if the eigenfunctions $u$ and $v$ correpond to eigenvalues $\lambda$ and $\lambda'$, then

$\lambda \int_M uv \; d\mu = \int u\Delta v \; d\mu = \int_M v \Delta u \; d\mu = \lambda' \int_M uv \; d\mu$.

QED

This last proposition is not surprising in view of the fact that integrating by parts actually shows that $\Delta$ is formally self-adjoint with respect to the $L^2$ inner product on a closed manifold. Indeed, in this context Green’s formula reads $\langle \Delta u, v \rangle = \langle u, \Delta v \rangle.$ The first assertion of this proposition is the reason why geometers often define the Laplacian as $-\Delta$, in order to get a positive spectrum. Finally, from the first equation of the proof, we see that eigenvalues of (minus) the Laplacian satisfy

$\lambda = \dfrac{\int_M |\nabla u|^2 \; d\mu}{\int_M u^2 \; d\mu}$

with $u$ 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 $(k+1)$‘st eigenvalue is given by the infimum of that functional over the functions orthogonal to the eigenfunctions associated to the first $k$ 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 $C$, all smooth functions integrating to $0$ (i.e. orthogonal to the constants) satisfy

$||u||_{L^2} \leq C||\nabla u||_{L^2}$,

the optimal constant being exactly $\frac{1}{\sqrt{\lambda_1}}$, 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 $H$ a complex Hilbert space having a countable orthonormal basis. Note that all such Hilbert spaces are isometrically isomorphic to $l^2$. We will denote $B$ the Banach algebra of bounded linear operators on $H$. A Fredholm operator is an operator $T \in B$ with closed range and finite dimensional kernel and cokernel. We will denote the set of Fredholm operators by $F$. The index of a Fredholm operator $T \in F$ is defined as

$\text{ind } T := \dim \ker T - \dim \text{coker } T$.

For example, the situation for $T : V \to V'$ an operator between two finite dimensional spaces, the situation is quite simple: We have $\text{ind } T = \dim V - \dim V'$ since $\dim V = \dim \ker T + \dim \text{Im } T$ and $\dim V' = \dim \text{coker }T + \dim \text{Im }T$.

These operators originally appeared in the theory of integral equations. The condition of being Fredholm means that the equation $Tu = 0$ has a finite dimensional space of solutions and that there are a finite number of linear relations we can impose on $v$ to make sure that $Tu = v$ is solvable (because $v$ has to be in the orthogonal complement of the cokernel).

We can also phrase the index of $T$ in terms of its adjoint. Recall that by Riesz’s representation theorem, for each $T \in B$ we have the adjoint operator $T^* \in B$ that satisfies, for all $u,v \in H$,

$\langle u,T^*v \rangle = \langle Tu, v \rangle$

such that $T \mapsto T^*$ is an isometry. It is an easy exercice to check that $\text{im } T = (\ker T^*)^{\perp}$ and thus $\text{coker } T = \ker T^*$. In fact, this holds for any bounded linear operator with closed range. We can then give an alternative definition: An operator $T \in B$ is Fredholm if its image is closed and both $\ker T$ and $\ker T^*$ are finite dimensional. The index is then $\text{ind } T = \dim \ker T - \dim \ker T^*$. Here is another example:

Proposition: The operator $Id + P$ for $P : H \to H$ a finite-rank operator (ie. an operator with finite dimensional image) is Fredholm with index 0.

Proof: We first show that $Id + P$ is Fredholm. Let $h = \text{im }P$. We proceed in two steps: (1) show that $\ker (Id + P) \subset h$ and then (2) that $\dim \text{coker}(Id + P) \leq \dim h$. For (1), if $u \in \ker(Id + P)$, then for $w \in (\ker P^*)$, we have

$0 = \langle u + Pu, w \rangle = \langle u, w + P^*w \rangle = \langle u, w \rangle$

so $u \in (\ker P^*)^{\perp} = h$. For (2), first note that $\text{coker }(Id + P) = \ker (Id + P^*)$ is orthogonal to $\ker P$. To see this, take $u \in \ker(Id + P^*)$ and $w \in \ker P$. We get

$0 = \langle u + P^*u, w \rangle = \langle u, w + Pw \rangle = \langle u,w \rangle$

as wanted.

Now define a linear map $\varphi : \text{coker }(Id + P) \to \text{im }P$ by $\varphi(v + \text{im}(Id + P)) = Pv$. This is well defined since $P(\text{im}(Id + P)) \subset \text{im}(Id + P)$. Since $\text{coker}(Id + P) \perp \ker P$, 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 $(Id + P)(h) \subset h$. to see this, let $u \in h = (\ker P^*)^{\perp}$ and $w \in \ker P^*$. Then

$\langle u+Pu, w \rangle = \langle u,w \rangle + \langle u, P^*w \rangle = 0$

so $(Id + P)u \in (\ker P^*)^{\perp} = h$. Now since $h$ is finite dimensional, we have $\text{ind}(Id + P)|_h = \dim h - \dim h = 0$ and since $(Id+P)_{H/h} = Id_{H/h}$, it’s index is also 0. Computing the alternating sum of the dimensions we get from the snake lemma, we finally conclude that $\dim \ker(Id + P) = \dim\text{coker}(Id + P)$ so that $\text{ind }(Id + P) = 0$.

QED

Fredholm operators are often encoutered in the study of PDE’s and more classically in the theory of integral equations. Since $\text{im }T = (\ker T^*)^{\perp}$ for $T \in F$, the equation $Tu = v$ is solvable for $u$ iff $v \perp \ker T^*$. In particular, if the index of $T$ is 0, i.e. $\dim \ker T = \dim \ker T^*$, then we get the so-called Fredholm alternative : Either (1) the inhomogeneous equation $Tu = v$ has a unique solution for every $v \in H$ or (2) the homogeneous equation $Tu = 0$ has $k (> 0)$ linearly independant solutions and there are $w_1, \ldots, w_k$ such that if $\langle v, w_j \rangle = 0$ for $j = 1, \ldots, k$ then $Tu = v$ has a solution.

For example, if $K \in L^2(I \times I)$ for $I$ some closed interval, then it can be shown that for $\phi \in L^2(I)$,

$\phi \mapsto \phi + \int_I K(x,y)\phi(y)dy$

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

$\phi(x) + \int_I K(x,y) \phi(y) dy = h(x)$

has a solution if and only if $h(x)$ is $L^2$-orthogonal to every solution $\psi(x)$ of the homogeneous adjoint equation $u(x) + \int_I \overline{K(y,x)}u(y)dy = 0$.

This principle can also be seen to be behind the fact that the laplace equation $\Delta f = g$ can be solved for $f$ on a compact manifold iff $\int g = 0$.

The fact that the integral equation above gives a Fredholm operator of index 0 is a consequence of the compactness of $\phi \mapsto \int K(x,y) \phi(y)dy$ and of the

Theorem (Riesz, 1918): For any compact operator $Q \in K$, the operator $Id + Q$ 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 $k \hookrightarrow F$, we can naturally consider $F$ as a $k$-vector space. We can use this point of view to define a notion of norm in field extensions. For any $\alpha \in F$, we define the norm of $\alpha$ with respect to this extension as

$N_{F/k}(\alpha) = \det(m_{\alpha})$

where $m_{\alpha}$ is multiplication by $\alpha$ viewed as a $k$-linear transformation on $F$. 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 $z = a + bi \in \mathbf{C}$, then in the basis $(1, i)$ the application $m_z$ is represented by the matrix $\bigl( \begin{smallmatrix} a&-b \\ b & a \end{smallmatrix} \bigr)$ so we get $N_{\mathbf{C}/\mathbf{R}}(z) = a^2 + b^2$. A similar argument shows that in a quadratic number field of the form $\mathbf{Q}(\sqrt{-d})$ for $d$ an integer which is not a square, we find $N_{\mathbf{Q}(\sqrt{-d})/\mathbf{Q}}(a+b \sqrt{-d}) = a^2 + b^2d$.

Lemma: For a simple algebraic extension $k \hookrightarrow k(\alpha)$, we have $N_{k(\alpha)/k}(\alpha) = (-1)^da_0$ where $a_0$ is the constant term of the minimal polynomial of $\alpha$ in $k$.

Proof: Let $f_{\alpha}$ be the minimal polynomial of $\alpha$, say of degree $d$ and recall that $k(\alpha) \cong k[x]/(f_{\alpha})$ can be seen as the field of $k$-polynomials of degree less than $d$ where multiplication is done modulo $(f_{\alpha})$. Then $\alpha$ is identified with the class of $x$. A nice basis for this space is $(1, x, x^2, \ldots, x^{d-1})$. Here multiplication by $\alpha$ acts on these vectors as $1 \mapsto x, x \mapsto x^2, \ldots x^{d-1} \mapsto x^d$ (everything modulo $(f_{\alpha})$). But $x^d = -a_0 - a_1x - a_2x^2 - \cdots - a_{d-1}x^{d-1} \mod (f_{\alpha})$ so multiplication by $\alpha$ is represented by

$\left( \begin{smallmatrix} 0 & 0 & \cdots & 0 & -a_0 \\ 1 & 0 & \cdots & 0 & -a_1 \\ && \ddots && \\ 0 & 0 & \cdots & 1 & -a_{d-1} \end{smallmatrix} \right)$.

This matrix clearly has determinant $(-1)^da_0$ hence the assertion.

QED

Note that it also follows from the proof that $\text{tr}_{k(\alpha)/k}(\alpha) = -a_{d-1}$. To get a better description of that norm, we need another lemma:

Lemma: For $k \hookrightarrow F$ a finite extension and $\alpha \in F$, let $[F : k(\alpha)] = r$. Then $N_{F/k}(\alpha) = (N_{k(\alpha)/k} (\alpha))^r$.

Proof: If $(e_1, \ldots, e_r)$ is a basis of $F$ as a vector space over $k(\alpha)$ and $(1, x, x^2, \ldots, x^{d-1})$ is the canonical basis of $k(\alpha)$ over $k$, then $(e_ix^j)$ for $i=1, \ldots, r$ and $j=1, \ldots, d-1$ is a basis for $F$ over $k$. Ordering this basis by $(e_1x^1, e_1x^2, \ldots, e_1x^{d-1}, e_2x^1, \ldots, e_2x^{d-1}, \ldots, e_rx^1, \ldots, e_rx^{d-1})$, the matrix of multiplication by $\alpha$ is given by the block diagonal matrix

$\left(\begin{smallmatrix} A &&& \\ &A&& \\ & & \ddots & \\ &&&A \end{smallmatrix} \right)$

where $A$ is the matrix representing multiplication by $\alpha$ in $k(\alpha)$ as in the first lemma.

QED

The analogous result for the trace is that $\text{tr}_{F/k} (\alpha) = r(\text{tr}_{k(\alpha)/k}(\alpha))$. These two lemmas let us interpret the norm and trace in a clarifying way:

Proposition: For $k \hookrightarrow F$ a Galois extension and $\alpha \in F$, we have

$N_{F/k}(\alpha) = \displaystyle\prod_{\sigma \in Gal(F/k)} \sigma (\alpha)$     and      $\text{tr}_{F/k}(\alpha) = \displaystyle\sum_{\sigma \in Gal(F/k)} \sigma (\alpha)$.

Proof: If $\lambda_1, \ldots, \lambda_d$ are the roots of the minimal polynomial $f_{\alpha}(x)$  of $\alpha$ whose constant term we denote by $a_0$ and $F$ is of degree $r$ over $k(\alpha)$, then since $(-1)^da_0 = \prod_{i=1}^r \lambda_i$, the preceding lemmas tell us that

$N_{F/k}(\alpha) = \left(\displaystyle\prod_{i=1}^d \lambda_i \right) ^r$.

Now, recall that $Gal(F/k)$ acts transitively on the set of roots of $f_{\alpha}(x)$ since it is irreducible. Denoting by $\text{Stab}(\lambda)$ the stabilizer of a root $\lambda$ for this action, we have $\text{Stab}(\lambda) = Gal(F/k(\lambda))$ and since there is only one orbit, all stabilizers are conjugate of each others so $|Gal(F/k(\lambda))| = |Gal(F/k(\alpha))| = r$ for all root $\lambda$. This shows that to each root of $f_{\alpha}(x)$ there corresponds at least $r$ elements of $Gal(F/k)$. Since $|Gal(F/k)| = rd$, this correspondence is bijective and we have

$\displaystyle\prod_{\sigma \in Gal(F/k)} \sigma (\alpha) = \left( \displaystyle\prod_{i=1}^d \lambda_i \right)^r = N_{F/k}(\alpha)$

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 $k \hookrightarrow F$ be a cyclic (Galois) extension of degree $n$, take $\sigma$ a generator for the Galois group and $\alpha \in F$. Then $N_{F/k}(\alpha) = 1$ if and only if $\alpha = b/\sigma(b)$ for some $b \in F$.

One direction is immediate. Indeed, for any $\beta \in F$ and $\sigma \in Gal(F/k)$, we have

$N_{F/k}(\beta/\sigma(\beta)) = \prod (\tau(\beta)/\tau \circ \sigma (\beta)) = \prod \tau(\beta) / \prod \tau'(\beta) = 1$

(where the products are all taken over $Gal(F/k)$). For the other direction, we need a lemma:

Lemma: If $\varphi_1, \ldots, \varphi_m$ are pairwise distinct $k$-automorphisms of $F$, then they are linearly independant over $F$.

Proof of the lemma: Let $\varphi = \sum_{j=1}^rc_j \varphi_{i_j} = 0$ be a minimal linear relation among the automorphisms. Since they are pairwise distinct, there exists $\lambda \in F$ such that $\varphi_{i_1}(\alpha) \neq \varphi_{i_2}(\alpha)$. Then for all $x \in F$, since $\varphi(x) = 0$, we have also that $\varphi(\lambda x) = 0$. We obtain in this way that for all $x \in F$,

$0 = \varphi(\lambda x) = \displaystyle\sum_{j=1}^r c_j\varphi_{i_j}(\lambda) \varphi_{i_j}(x)$

and

$0 = \varphi_{i_1}(\lambda)\varphi(x) = \displaystyle\sum_{j=1}^rc_j \varphi_{i_1}(\lambda)\varphi_{i_j}(x)$

so subtracting give us $0 = \sum_{j=2}^rc_j(\varphi_{i_j}(\lambda) - \varphi_{i_1}(\lambda)) \varphi_{i_j}(x)$ which is a non-trivial relation (since we took $\lambda$ so that $\varphi_{i_2}(\lambda) \neq \varphi_{i_1}(\lambda)$). 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 $N_{F/k}(\alpha) = 1$ and define $\tau : F \to F$ as $\tau(b) = \alpha \sigma (b)$. Then it suffices to find a fixed point of $\tau$. It is easy to see that for $k \in \{1, \ldots, n\}$, we have

$\tau^k(b) = \alpha \sigma(\alpha)\sigma^2(\alpha) \cdots \sigma^{k-1}\sigma^k(b)$.

In particular, $\tau^n(b) = \prod_{i=1}^n\sigma^n(\alpha)\sigma^n(b) = N_{F/k}(\alpha)b$ so if $N_{F/k}(\alpha) = 1$, we have $\tau^n(b) = b$ for all $b \in F$. In particular, the order of $\tau$ is $n$. Since moreover $\tau$ is $k$-linear over $F$, we get a representation $\rho : \mathbf{Z}/n\mathbf{Z} \to GL(F)$ by $\rho(m) = \tau^m$. From this point of view, it is easy to determine if $\tau$ has a fixed point. Indeed, for any group $G$ and any representation $\phi : G \to GL(V)$, recall that

$p := \dfrac1{|G|} \displaystyle\sum_{g \in G} \phi(g)$

is the projection of $V$ 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 $0$. But in our case we have

\begin{aligned} p &= \frac1n(1_F + \tau + \tau^2 + \cdots + \tau^{n-1}) \\ &= \frac1n(1_F + \alpha \sigma+ \alpha \sigma(\alpha)\sigma^2 + \cdots + \alpha \sigma(\alpha)\cdots \sigma^{n-2}(\alpha)\sigma^{n-1}) \end{aligned}

and since $\sigma$ is a generator for the Galois group, the automorphisms $1_F, \sigma, \sigma^2, \ldots, \sigma^{n-1}$ are linearly independent by the lemma so $p$ is not identically zero because that would give us a non-trivial linear relation between them. This shows that $\tau(b) = b$ for some $b \in F$ and that $\alpha = b/\sigma(b)$.

QED

Note that the analogous result for the trace is that in a cyclic extension, $Tr(\alpha) = 0$ is and only if $\alpha = \beta - \sigma(\beta)$. As a cool application of this theorem, we can give a parametrization of pythagorean triples. Indeed, consider the equation $x^2 + dy^2 = 1$ for $d$ not a square. Recall that the norm of $\alpha = a + b\sqrt{-d} \in \mathbf{Q}(\sqrt{-d})$ is

$N_{\mathbf{Q}(\sqrt{-d})/\mathbf {Q}}(\alpha) = a^2 + db^2$

(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 $\mathbf{Q}(\sqrt{-d})$. By Hilbert’s theorem 90 and since conjugation generates the Galois group, these elements are those of the form

$\alpha = \dfrac{a+b\sqrt{-d}}{a-b\sqrt {-d}} = \dfrac{a^2 - db^2}{a^2 + b^2} + \dfrac{2ab}{a^2 + b^2}\sqrt{-d}$.

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

$(x,y) = \left(\dfrac{a^2 - db^2}{a^2 + b^2}, \dfrac{2ab}{a^2 + b^2}\right)$

where $(a,b) \in \mathbf{N}^2$. Letting $d=1$, we find all the rational points on the unit circle, which give us the integer pythagorean triples by the correspondence $(a/c, b/c) \mapsto (a,b,c)$.