By Abrashkin V.A.

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**Example text**

For every holomorphic one form ω on H(t) , ω Proof: 2 ≥ | log t| 2π ω A The projection (z1 , z2 ) ∈ H(t) −→ z1 ∈ z1 t ≤ |z1 | ≤ 1 from H(t) to the annulus t ≤ |z1 | ≤ 1 is a global coordinate. 4 Denote by RX the period matrix of the marked Riemann surface (X; A1 , B1 , · · ·) on which there is an exhaustion function with finite charge. Let tj ∈ (0, 1) , j ≥ 1 . Suppose that for all j ≥ 1 there exists an injective , holomorphic map φj : H(tj ) −→ X such that tj e−iθ tj eiθ , φj 0 ≤ θ ≤ 2π is homologous to Aj and φk H(tk ) ∩ φ H(t ) = ∅ for all k = .

In ZZ∞ with only a finite number of nonzero components. 9. Consequently, 0 = X ωk ∧ ω = ∞ ωk i=1 Ai Bi ω − ωk Bi ω Ai = Rk, − R ,k since ωk ∧ ω = 0 . 9. 3 Fix 0 < t < 1 . Let A = √ t eiθ , √ −iθ te 0 ≤ θ ≤ 2π be the oriented waist on the model handle H(t) . For every holomorphic one form ω on H(t) , ω Proof: 2 ≥ | log t| 2π ω A The projection (z1 , z2 ) ∈ H(t) −→ z1 ∈ z1 t ≤ |z1 | ≤ 1 from H(t) to the annulus t ≤ |z1 | ≤ 1 is a global coordinate. 4 Denote by RX the period matrix of the marked Riemann surface (X; A1 , B1 , · · ·) on which there is an exhaustion function with finite charge.

21 A Riemann surface with finite ideal boundary has a finite number of ideal boundary points. The “infinitely branching” surface below has no ends and hence no ideal boundary points, but still does not have finite ideal boundary. 22 If X is a hyperelliptic surface and τ is the hyperelliptic projection from X to IP1 (C) \ I , then I is in one to one correspondence with the ideal boundary points of X . Let E be an end of the Riemann surface X . A function f on E has a limit w at the ideal boundary point represented by E if for every ε > 0 there is an end E ⊂ E such that sup x∈E f (x) − w ≤ ε If the intersection form on H1 (E, ZZ) vanishes, the end E is called planar.