By Huang X., Yin W.
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Extra resources for A Bishop surface with a vanishing Bishop invariant
5, which we will prove in the next section, we also conclude that M is biholomorphically equivalent to Ms . 4 (f) is a simple consequence of the results in (a) and (e). 4. ,s−1 with infinitely many aks+ j = 0. Then for any N > s, M is not equivalent to the Bishop surface MN defined by ks+ j≤N w = H(N+1) (z, z) = zz + 2Re z + aks+ j z ks+ j . ,s−1 Here H(N+1) is the N th -truncation from the Taylor expansion of H at 0. In fact, M(N+1) is equivalent to M(N +1) with N > N if and only if aks+ j = 0 for any N < ks + j ≤ N .
15). Let z = z + f ∗ (z , w ) and w = w + g∗ (z , w ) be its inverse transformation. Notice that the coefficients of ( f ∗ , g∗ ) in its Taylor expansion up to degree, say m, are universal polynomial functions of the coefficients of ( f, g) up to degree m A Bishop surface with a vanishing Bishop invariant for any m. Hence we have the defining equation of M ∗ , the image of M, as follows: w + g∗ (z , w ) = H(z + f ∗ (z , w ), z + f ∗ (z , w )). Applying an implicit function theorem to solve for w and making use of the uniqueness of the graph function, we see that the coefficients in the Taylor expansion of H ∗ up to degree m must also be (possibly non-holomorphic) polynomial functions of the coefficients of H of degree not exceeding m in its Taylor expansion.
49) Then χ −1 (τ, τ0 (u)) = τ+τ0 (u) . 1+τ0 (u)τ Write √ z , u nor , τ0nor (u nor ) , σ Dnor(u nor ) (z) := χ (σnor)−1 √ u nor which is a conformal map from Dnor(u) to ∆, mapping Anor 0 (u nor ) to the origin. By the definition, −1 Pnor (t, u nor ) = σ Dnor(u nor ) tσ Dnor(u nor ) Anor 1 (u nor ) √ √ = u nor σnor (βnor (t, u nor ) , u nor ). 50) Similarly, we have ∗ Pnor (t, u ∗nor ) = ∗ ∗ u ∗nor σnor (βnor (t, u ∗nor ), u ∗nor ). 51) 1. −1 N F( N+1) ◦ Φ−1 . 51), we have F( N+1) ◦ Φ−1 1 (Pnor (t, u nor ), u nor ) ∗ ∗ ∗ N N−2 = Φ−1 2 (Pnor (t, u nor ), u nor ) + O |(Pnor (t, u nor ), u nor )| + u nor .