It is known (Scwhartz-Pick) that holomorphic functions are distance non-increasing in the hyperbolic metric.Does it imply that they preserve convexity in the hyperbolic metric? How about the converse?
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No. Take a convex set and use an exponential function $e^{2\pi i z/c}$, where $c$ is the difference between two points in the set. The image of the set is no longer simply connected, and thus cannot be convex in a Riemannian manifold with no closed geodesics like the hyperbolic disc (which I assume is what you are talking about here). You may need to scale the exponential function down by a constant to fit in the disc, this is obviously immaterial |
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@Will gave an excellent answer, but also, a mapping preserving convexity must map (hyperbolic) straight lines to (hyperbolic) straight lines (since the images of both the half-spaces need to be convex), and the only analytic functions which do that are Mobius transformations, so the answer is NO almost always. |
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