Skip to content

Riemann-Roch and Serre duality on compact Riemann surfaces

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:

Serre_duality

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}.

Advertisement

The Mittag-Leffler problem and the Dolbeault isomorphism.

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.

Jacobi’s inversion theorem

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,

which contradicts earlier considerations.

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.

On the group structure of elliptic curves

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.

Abel’s Theorem

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

The First Reciprocity Law

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

The Abel-Jacobi map

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.

Integration by parts and Green’s formula on Riemannian manifolds

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.

First introduction to Fredholm operators.

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:

snake_fredholm

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.

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).