445 research outputs found

    Local rigidity for hyperbolic groups with Sierpi\'nski carpet boundaries

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    Let GG and G~\tilde G be Kleinian groups whose limit sets SS and S~\tilde S, respectively, are homeomorphic to the standard Sierpi\'nski carpet, and such that every complementary component of each of SS and S~\tilde S is a round disc. We assume that the groups GG and G~\tilde G act cocompactly on triples on their respective limit sets. The main theorem of the paper states that any quasiregular map (in a suitably defined sense) from an open connected subset of SS to S~\tilde S is the restriction of a M\"obius transformation that takes SS onto S~\tilde S, in particular it has no branching. This theorem applies to the fundamental groups of compact hyperbolic 3-manifolds with non-empty totally geodesic boundaries. One consequence of the main theorem is the following result. Assume that GG is a torsion-free hyperbolic group whose boundary at infinity \dee_\infty G is a Sierpi\'nski carpet that embeds quasisymmetrically into the standard 2-sphere. Then there exists a group HH that contains GG as a finite index subgroup and such that any quasisymmetric map ff between open connected subsets of \dee_\infty G is the restriction of the induced boundary map of an element hHh\in H.Comment: 14 page

    Maps conjugating holomorphic maps in C^n

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    If f is a bijection from C^n onto a complex manifold M, which conjugates every holomorphic map in C^n to an endomorphism in M, then we prove that f is necessarily biholomorphic or antibiholomorphic. This extends a result of A. Hinkkanen to higher dimensions. As a corollary, we prove that if there is an epimorphism from the semigroup of all holomorphic endomorphisms of C^n to the semigroup of holomorphic endomorphisms in M, or an epimorphism in the opposite direction for a doubly-transitive M, then it is given by conjugation by some biholomorphic or antibiholomorphic map. We show also that there are two unbounded domains in C^n with isomorphic endomorphism semigroups but which are neither biholomorphically nor antibiholomorphically equivalent.Comment: 10 page
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