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COMPLEX GEOMETRY Course notes

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1<br />

b g<br />

a g<br />

a 1<br />

Since S is oriented we have that there exists an intersection pairing in H 1 (any two loops, or elements in H 1 ,<br />

are homotopic to loops which are transversal). Note that a and b are the same element in H 1 if and only<br />

if a + (−b) = a − b = ∂U, for some open set U ⊆ S, i.e., a is homologous to b. Here −b de<strong>notes</strong> reverse<br />

orientation. So if a is homotopic to b then they are homologous.<br />

Consider the following picture:<br />

chart<br />

−→ b<br />

y<br />

C<br />

p<br />

−→ a<br />

x<br />

We have −→ a ∧ p<br />

−→ b = kx ∧p y, and denote<br />

and<br />

(a, b) p = sign(k)<br />

(a, b) = ∑ p∈a∩b (a, b) p<br />

Since the previous sum does not depend on the choice of the prepresentative, we have a well defined map<br />

( , ) : H 1 × H 1 −→ Z<br />

If S is a Riemann surface of genus g, we have in terms of a basis that<br />

(a i , b j ) = (b i , b j ) = 0, for every i, j,<br />

(a i , b j ) = −(b j , a i ) = δ ij .<br />

28

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