COMPLEX GEOMETRY Course notes

Then we have that each (a i , b j ) is a skew-linear pairing which is unimodular (det = 1).
a i
b i
a i 0 1
b i −1 0
det
1
Hence H 1 (S; Z) = Hom(H 1 (S; Z), Z) =: H 1 (S; Z).
Let ω ∈ M ′ (S) be a closed meromorphic differential. Recall that ω is exact if and only if ω = df for some
meromorphic function f ∈ M(S). Can we choose a function f ∈ M(S) such that f = ∫ z
ω? As an exercise,
p
think if it is possible that Res p (ω) = 0 for every p ∈ S. Consider the period homomorphism
Π ω : H 1 (S; Z) −→ C
∫
[γ] ↦→ ω, γ is a loop.
γ
This map is well defined. For if γ = γ ′ in H 1 then γ − γ ′ = ∂u, and by the Stokes Theorem we have
∫
γ ω = ∫ γ ′ ω. Also, Π ω is a C-linear map.
By the de Rham Theorem, we have an isomorphism HdR 1 (Z) ∼ = H 1 (S; Z) ⊗ C if Z is compact.
isomorphism is given by
[ω] ∈ HdR(Z) 1 ↦→ Π ω
Such an
Corollary 2.10.1. dim C HdR 1 (S) = 2g.
Proof: We have
2 − 2g = χ(S)
= ∑ (−1) i h i dR(S), where h i dR (S) = dim(H1 dR (S)),
= h 0 dR(S) − h 1 dR(S) + h 2 dR(S)
= 1 − dim C HdR(S) 1 + 1.
Theorem 2.10.1. The de Rham ismomorphism carries wedge product of forms defined by
isomorphically to the (cup) product on H 1 (S).
HdR 1 × HdR 1 −→ C
∫
([ω 1 ], [ω 2 ]) ↦→ ω 1 ∧ ω 2
S
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