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

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