By William N Hess, Chris Davey

The forty ninth FG used to be despatched to Australia in early 1942 to assist stem the tide of eastern conquest in Java. Too overdue to avoid wasting the island, the crowd went into motion within the defence of Darwin, Australia, the place the Forty-NinersвЂ™ handful of P-40E Warhawks have been thrown into strive against along survivors from the defeated forces that had fled from the Philippines and Java. This booklet assesses the phenomenal functionality of the forty ninth FG, pitted opposed to improved eastern forces. via VJ-Day the gang had scored 668 aerial victories and gained 3 wonderful Unit Citations and ten crusade stars for its striking efforts.

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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 .