1. Riemann Roch Theorem
Let X be a compact Riemann surface and Div(X) be its group of divisors. Two divisors D and D0 are called linearly equivalent, written D ∼ D0 if D − D0 is a principal divisor.
The group Div(X) divided by the subgroup of principal divisors is called the divisor class group of X, and is denoted by Cl(X).
A divisor D =P
P ∈XnPP is effective if nP ≥ 0 for all P. In this case, we write D ≥ 0.
It is obvious that the sum of two effective divisors is again effective. This notion induces a partial ordering on divisors: we say that D ≥ D0 if D − D0 is effective, i.e. D − D0≥ 0.
For a divisor D on X, we set
L(D) = {f ∈ K(X) : (f ) + D ≥ 0} ∪ {0}.
Here K(X) is the field of meromorphic functions of X. The dimension of L(D) is denoted by h0(D).
Lemma 1.1. Let D, D0 ∈ Div(X). Then D ≥ D0 if and only if L(D) ⊂ L(D0).
Proof. If f ∈ L(D0), then (f ) + D0 ≥ 0. Since D ≥ D0, D − D0 ≥ 0 and thus (the sum of two effective divisors is effective)
(f ) + D = (f ) + D0+ D − D0 ≥ 0.
Hence f ∈ L(D).
Lemma 1.2. We have L(0) = C and thus h0(0) = 1.
Proof. If f ∈ L(0) and f 6= 0, then ordPf ≥ 0 for all P ∈ X. This implies that f has no pole on X, and thus f is holomorphic on X. Therefore f is a constant function. This shows
that L(0) = C.
Proposition 1.1. If D is a divisor with deg D < 0, then L(D) = {0}.
Proof. Let f ∈ L(D). Assume that f is nonzero. By deg(f ) = 0 and (f ) + D ≥ 0, we have that
deg((f ) + D) = deg f + deg D = 0 + deg D = deg D ≥ 0
which leads to a contradiction to the assumption that deg D < 0. For every D ∈ Div(X), we set
ΩX(D) = {ω : ω is an abelian differential with (ω) ≥ D}.
We denote h1(D) = dim ΩX(D).
Proposition 1.2. For D ∈ Div(X), h0(D) and h1(D) depend only on the divisor classes of D. Furthermore, if ω is any nonzero abelian differential on X, then h1(D) = h0((ω) − D) for any given nonzero abelian differential ω.
Proof. Let us assume that D ∼ D0. Choose f ∈ K(X) so that D − D0 = (f ). Define a map T : L(D) → L(D0), h 7→ h · f.
Then T is an isomorphism of C-vector spaces whose inverse map L(D0) → L(D) is given by h0 7→ h0· (1/f ). Hence h0(D) = h0(D0).
Let us fix a nonzero abelian differential ω on X. Given any other abelian differential ω0, we can find a meromorphic function f such that ω0= f ω. We write f by ω0/ω. If ω0∈ Ω(D), (ω0) ≥ D. If f = ω0/ω, then
(f ) + (ω) − D = (f ω) − D = (ω0) − D ≥ 0.
1
2
This shows that ω0/ω ∈ L((ω) − D). This observation allows us to define a C-linear map S : Ω(D) → L((ω) − D), ω0 7→ ω0
ω.
Let us prove that S is an isomorphism. If ω0 ∈ ker S, then ω0/ω = 0. This implies that ω0 = 0. For any f ∈ L((ω) − D), we have (f ) + (ω) − D ≥ 0. Hence (f ω) ≥ D. This shows that S(f ω) = f for any f ∈ L((ω) − D), i.e. S is surjective. We prove that S is a linear isomorphism. Thus h1(D) = h0((ω) − D).
Finally, if D0P D, then
h1(D) = h0((ω) − D) = h0((ω) − D0) = h1(D0).
It follows from definition that Ω(0) coincides with the space of holomorphic differentials H0(X, ωX). We obtain the following result:
Proposition 1.3. We have Ω(0) = H0(X, ωX) and thus h1(0) = g.
We have already known that h0(0) = 1. Therefore h0(0) − h1(0) = 1 − g.
Theorem 1.1. (Riemann-Roch Theorem) Let X be a compact Riemann surface of genus g and D be a divisor on X. Then
h0(D) − h1(D) = deg D − g + 1.
