By Richard Henry Stoddard
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19) is equivalent to any hypersurface of the form x0 = P(z, z) + ReQ(z), where Q is a holomorphic polynomial, and therefore we can assume that P does not contain harmonic terms. Spherical hypersurfaces of this kind are described in the following proposition. 1. 19), where P does not contain harmonic terms. Assume further that there exists an open subset of M which is spherical. Then P(z, z) = ± |R(z)|2 , where R(z) is a holomorphic polynomial. Proof. Let W be a spherical subset of M and U its projection to the z-coordinate.
Zn ) in Cn+1 such that V = Im z j = 0, j = 0, . . , n , we identify V with Rn+1 by means of the coordinates x j := Re z j , j = 0, . . , n. Thus, we always regard MR as a hypersurface in Rn+1 and represent the tube hypersurface M as M = MR + iRn+1 . Note that for Z := (z0 , z) any transformation of Cn+1 of the form Z → Z + ib, b ∈ Rn+1 is a CR-automorphism of M. We let x := (x1 , . . , xn ), X := (x0 , x) and consider the map Π : Cn+1 → Rn+1 , Z → X. Clearly, if M is a tube hypersurface, we have Π −1 (MR ) = M.
Fn ) has a zero lying in this subspace. Thus, we can assume Eγ = 0 for γ = k + 1, . , n. 8). 5). 1 but also for a majority of results in the rest of the book. 1 The above proof is due to M. A. Mishchenko. The original proof given in  for the strongly pseudoconvex case seems to be incomplete. 5) with respect to xν and the (α , ν )th equation with respect to xβ . Clearly, we have ∂ Fαβ ∂ Fαν ≡ . 10) involve some first- and second-order partial derivatives of the function F. 10) the second-order partial derivatives of F by the corresponding expressions in terms of its first-order partial derivatives.