指标定理（五） (9-2)

ⱼ₌₁

₂ₖ xⱼ

∏ ────────) [X]

ⱼ₌₁ (eˣʲ/² – eˉˣʲ/²)²

₂ₖ xⱼ(eˣʲ/²＋eˉˣʲ/²)

＝(∏ ────────) [X]

ⱼ₌₁ (eˣʲ/² – eˉˣʲ/²)

₂ₖ xⱼ

＝(∏ ───────) [X]

ⱼ₌₁ tanh(xⱼ/2)

1 ₂ₖ 2xⱼ

＝─ (∏ ────) [X]

2²ᵏ ⱼ₌₁ tanh(xⱼ)

₂ₖ xⱼ

＝(∏ ──── ) [X].

ⱼ₌₁ tanh(xⱼ)

Consequently，since p₁(lⱼ)＝ –c₂ (lⱼ ⨂ ℂ)＝ –c₂(lⱼ ⨁ ˉlⱼ)＝ –c₁(lⱼ)c₁(ˉlⱼ)＝c₁(lⱼ)²，

22

L(TX)[X]＝∏ L(1＋p₁(lⱼ))[X]

₂ₖ √p₁(lⱼ)

＝∏ ───── [X]

ⱼ₌₁ tanh(√p₁(lⱼ))

₂ₖ xⱼ

＝(∏ ──── ) [X]＝indₜ(d＋d*).

ⱼ₌₁ tanh(xⱼ)

5 Riemann-Roch Theorem

5.1 Divisors on Riemann surfaces

Recall that a Riemann surface M is a one-dimensional complex manifold，and a diυisor is a mapping D：X → ℤ， such that in ∀compact K ⊂ X，there are only finitely many points where D takes nonzero values. Divisors form an abelian group Diυ(X). One can define a divisor (f) for a meromorphic function f (or a meromorphic one-form) in the obvious way：(f)(P)＝k＞0 if f has a zero of order k at P，and (f)(P)＝k＜0 if f has a pole of order –k at P. We require that f is never identically zero in any open set.

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