Riemann-Roch定理 (6-5)

For example, letS be the Riemann sphere. The meromorphic function are just the rational functions (any meromorphic function on the complex plane that isn't rational has an essential singularity at ∞ ). The functions having no poles except at ∞ are the polynomials, by Liouville's Theorem. The space L(m∞) (here, we are taking only one point, namely, p₁＝∞ ) is the set of functions that are holomorphic except at ∞ , where they're alowed to have a zero of order at most m . That's just the set of polynomials of degree ≤ m . So ℓ(m∞)＝m＋1 , for the functions 1，z，z²，· · ·，zᵐ form a basis. Here is a second example. The space L(1∞ – 1(1＋i)) ( so p₁＝∞，m₁＝1， and p₂＝1＋i，m₂＝–1 ) consist of those meromorphic funcitions with a zero at 1＋i and, at worst, a pole of order 1 at ∞ . This is a subspace of L(1∞) , and so it consist of just those functions of the form c(z – 1 – i) ; hence, ℓ(1∞ – 1(1＋i))＝1 . Similarly, L(2∞ – 1(1＋i)) is the space of all those functions of the form (z – 1 – i)(αz＋b) , so the dimension ℓ(2∞ – 1(1＋i))＝2 . Generally, L(m₁p₁＋· · ·＋mₙpₙ) is the space of meromorphic functions of the form (z – p₁)⁻ᵐ¹ (z – p₂)⁻ᵐ² · · · (z – pₙ)⁻ᵐⁿ (c₀zᵈ＋· · ·＋cd)；

where d＝m₁＋· · ·＋mₙ if d ≥ 0 . (If d＜0， the only function satisfying the conditions is the function identically 0 .) So, on the Riemann sphere we always get ℓ(m₁p₁＋· · ·＋mₙpₙ)＝m₁＋· · ·＋mₙ＋1＝d＋1

unless that's negative, in which case we get ℓ(m₁p₁＋· · ·＋mₙpₙ)＝0.

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