By Chen G.
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Additional resources for A Characterization of Alternating Groups by the Set of Orders of Maximal Abelian Subgroups
9 Let X be a noncompact, parabolic Riemann surface. Let f be a nonconstant meromorphic function on X with 1 ≤ m < ∞ (counted with multiplicity) poles that is bounded on the complement of each neighborhood of its poles. Then, nf (a) ≤ m for all a ∈ C , and the set a ∈ C nf (a) < m has Lebesque measure zero. 43 Proof: Let h be a harmonic exhaustion function on X . It can be assumed, by scaling, that the restriction of h to h−1 [1, ∞) is harmonic. For all r > 0 and a ∈ IP1 (C) , (counting with multiplicity) set x ∈ X f (x) = a and h(x) ≤ r n(a, r) = v(r) = f∗ Φ 1 π x∈X | h(x) ≤ r where, Φ is the volume form of the Fubini-Study metric on IP1 (C) .
It is possible to choose a nonzero η ∈ Ω(X) with η(x1 ) = 0 , since dimC Ω(X) ≥ 2 . Let m ≥ 0 be the order to which η vanishes at x2 . Observe that, by the first part of the proposition, m is even when x2 is a ramification point of τX . Set ω(x) = τ (x)−τ (x1 ) m τ (x)−τ (x2 ) m τ (x)−τ (x1 ) 2 τ (x)−τ (x2 ) η(x) , x2 is not a ramification point η(x) , x2 is a ramification point The proof that ϕX is immersive goes in the same way. 19 (i) Unravelling the definitions, the injectivity of ϕX is equivalent to the statement that for all x1 , x2 ∈ X and coordinates z1 , z2 in neighbourhoods of x1 , x2 , there exists a constant ω ω λ such that dz1 (x1 ) = λ dz2 (x2 ) for all ω ∈ Ω(X) if and only if τ (x1 ) = τ (x2 ) .
Suppose Xt ⊂ X , t ≥ 0 is a family of compact submanifolds with piecewise smooth boundary such that Xs ⊂ Xt for all t > s > 0 . If there exists a linear density g with L(g:Xs ,Xt )2 t→∞ A(g;Xt \Xs ) lim = ∞ for some fixed s > 0 , then X is parabolic. 15B] . 6 There is an exhaustion function with finite charge on the Riemann surface X without boundary if and only if X is parabolic. Proof: Suppose, h is an exhaustion function with finite charge on X . As in Section 2, Xt = h−1 ([0, t]) , for all t ≥ 0 , and the linear density gh on X associated to h is gh (x) = dh ⊗ dh + ∗dh ⊗ ∗dh 41 Observe that for any path γ joining Γs to Γt , (dh)2 + (∗dh)2 ≥ = length of γ with respect to gh γ γ |dh| ≥ t − s and consequently, L(gh ; Xs , Xt ) ≥ t − s Fix s > 0 .