F eb 1 99 6 EXAMPLES OF DOMAINS WITH NON - COMPACT AUTOMORPHISM GROUPS

We give an example of a bounded, pseudoconvex, circular domain in Cn for any n ≥ 3 with smooth real-analytic boundary and non-compact automorphism group, which is not biholomorphically equivalent to any Reinhardt domain. We also give an analogous example in C, where the domain is bounded, non-pseudoconvex, is not equivalent to any Reinhardt domain, and the boundary is smooth real-analytic at all points except one. Let D be a bounded or, more generally, a hyperbolic domain in C. Denote by Aut(D) the group of biholomorphic self-mappings of D. The group Aut(D), with the topology given by uniform convergence on compact subsets of D, is in fact a Lie group [Kob]. A domain D is called Reinhardt if the standard action of the n-dimensional torus T on C, zj 7→ ejzj , φj ∈ R, j = 1, . . . , n, leaves D invariant. For certain classes of domains with non-compact automorphism groups, Reinhardt domains serve as standard models up to biholomorphic equivalence (see e.g. [R], [W], [BP], [GK1], [Kod]). It is an intriguing question whether any domain in C with non-compact automorphism group and satisfying some natural geometric conditions is biholomorphically equivalent to a Reinhardt domain. The history of the study of domains with non-compact automorphism groups shows that there were expectations that the answer to this question would be positive (see [Kra]). In this note we give examples that show that the answer is in fact negative. While the domain that we shall consider in Theorem 1 below has already been noted in the literature [BP], it has never been proved that this domain is not biholomorphically equivalent to a Reinhardt domain. Note that this domain is circular, i.e. it is invariant under the special rotations zj 7→ ezj , φ ∈ R, j = 1, . . . , n. Our first result is the following Theorem 1. There exists a bounded, pseudoconvex, circular domain Ω ⊂ C with smooth real-analytic boundary and non-compact automorphism group, which is not biholomorphically equivalent to any Reinhardt domain. Proof. Consider the domain Ω = {|z1| + |z2| + |z3| + (z2z3 + z3z2) < 1}. Mathematics Subject Classification: 32A07, 32H05, 32M05