By James R. Munkres

A readable creation to the topic of calculus on arbitrary surfaces or manifolds. obtainable to readers with wisdom of easy calculus and linear algebra. Sections contain sequence of difficulties to augment concepts.

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Then, by a suitable β choice of τ = φ(0) and Dt φ(0), one can achieve that E (m+1) (0) < 0 and E (j ) (0) = 0 for 1 ≤ j ≤ m. Proof Set N := L − 1, M := L − (α + β + 1) = N − (α + β), hence L − 1 = α + β + M. By Leibniz’s formula, DtN {[Zˆ w · Zˆ w ]φ} = N−β N α=0 β=0 N! N −β−α ˆ β Zw ) · (Dtα Zˆ w )Dt φ. (N − β − α)! 1 The Strategy of the Proof 39 we can use Leibniz’s formula to compute E (L) (t) from E (L) (t) = 2 Re S1 wDtN {[Zˆ w (t) · Zˆ w (t)]φ(t)} dw. We choose L := m + 1; then L ≥ 5 as we have assumed m ≥ 4.

Now, if we know that the image under Xˆ of a small neighbourhood U of 0 is an analytic regular (embedded) surface S, then w = 0 is a false branch point and it is not hard to see that w = 0 is also analytically false. To this end, let Y : U → S be a C 2,α smooth regular conformal parametrization of f . Then ϕ := Y −1 ◦ X is conformal and we may presume holomorphic. Since Xˆ has a branch point of order n, ϕ locally has the form ϕ(z) = an+1 wn+1 + · · · , where an+1 = 0. Therefore, there is a holomorphic function ψ defined on a neighbourhood V ⊂ U of 0 such that ϕ = ψ n+1 and we may assume ψ : V → ψ(V ) is biholomorphic.

By Leibniz’s formula, DtN {[Zˆ w · Zˆ w ]φ} = N−β N α=0 β=0 N! N −β−α ˆ β Zw ) · (Dtα Zˆ w )Dt φ. (N − β − α)! 1 The Strategy of the Proof 39 we can use Leibniz’s formula to compute E (L) (t) from E (L) (t) = 2 Re S1 wDtN {[Zˆ w (t) · Zˆ w (t)]φ(t)} dw. We choose L := m + 1; then L ≥ 5 as we have assumed m ≥ 4. 1) where the terms J1 , J2 , J3 are defined as follows: Set ˆ T α,β := w(Dtα Z(0)) w Dt φ(0). 2) Then, J1 := 4 Re S1 ˆ w τ ) dw ˆ [DtL−1 Z(0)] w · (w X + 4 · (L − 1) Re L−3 +4 M> 12 (L−1) S1 ˆ [DtL−2 Z(0)] w f dw (L − 1)!