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 [108] 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.