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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2 Answers
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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
answered Jun 11, 2012 at 16:31
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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.
answered Jun 11, 2012 at 17:10
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$\begingroup$ If the function is bijective, one does not even need analyticity, according to the reference: Demirel, Oğuzhan; Soytürk Seyrantepe, Emine A characterization of Möbius transformations by use of hyperbolic regular polygons. J. Math. Anal. Appl. 374 (2011), no. 2, 566–572 They prove that a continuous bijection of the unit disk is Moebius if and only if it preserves hyperbolic regular polygons. $\endgroup$ Commented Jun 13, 2012 at 17:35