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Math Help - Conformal Reflection across Real Axis

  1. #1
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    Conformal Reflection across Real Axis

    I have an exercise and an approach to it's solution with a gap in it. The exercise comes from Ahlfors Complex Analysis text in Ch. 6 on Conformal mappings.

    Suppose that f is a Riemann map (conformal) from a simply connected domain  \Omega \varsubsetneqq \mathbb{C} to the open unit disk \mathbb{D} with z_0 \in \mathbb{R}, f(z_0) = 0, f'(z_0) > 0 (i.e., real and positive). Also suppose that  \Omega is symmetric with respect to the real axis.

    Claim:  f(\overline{z}) = \overline{f(z)}.

    Scratch Work: For the moment assume f is a mobius map and recall that mobius tranforms map Riemann circles to Riemann circles (including "lines"). Define \rho_\mathbb{R}(z) be a reflection across the real axis and \rho_K(z) be the reflection about the circle f(\mathbb{R}). Form the function g(z) = f^{-1} \circ \rho_K \circ f \circ \rho_\mathbb{R}(z) which maps from \Omega to  \Omega. Since z_0 \in \Omega \cap \mathbb{R} and \Omega is open, we can get (a,b) \subseteq \Omega \cap \mathbb{R}. If w \in (a,b) then g(w) = w because \rho_\mathbb{R}(w) = w and f(w) \in f(\mathbb{R})\Rightarrow \rho_K(f(w)) = f(w). So g(w) = f^{-1} \circ f(w) = w. Therefore, we have that g is analytic (since conformal) and equals the identity map on  \Omega. Then, analytic extension gives us that g(z) is the identity map, i.e. g(z) = z. So f \circ \rho_\mathbb{R}(z) = \rho_K \circ f(z) \Rightarrow f(\overline{z}) = \rho_K(f(z)).

    So there is the first gap; I don't see how to find that  f(\overline{z}) = \overline{f(z)} as claimed. If it is true, it must be that K = \mathbb{R} so that f( (a,b) ) \subseteq \mathbb{R}, but I'm not seeing it.

    When we consider that f is a general conformal map, there is a similar gap. In this case it is not necessarily true that the real line maps to a Riemann "circle". We have the Carathedory-Osgood extension that gives a homeomorphic extension \widetilde{f}: \overline{\Omega} \textrm{ to } \overline{\mathbb{D}}, so the boundary of \Omega maps to the boundary of \mathbb{D}. But, again, I don't see why the segment of \Omega \cap \mathbb{R} should map into a line (or specifically into the real axis) in the disk so that we can conclude that the reflection \rho_K(z) is in fact \rho_\mathbb{R}(z).

    Thanks in advance for your consideration!
    Last edited by huram2215; May 8th 2010 at 05:30 AM. Reason: clarify notation
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  2. #2
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    Quote Originally Posted by huram2215 View Post
    I have an exercise and an approach to it's solution with a gap in it. The exercise comes from Ahlfors Complex Analysis text in Ch. 6 on Conformal mappings.

    Suppose that f is a Riemann map (conformal) from a simply connected domain  \Omega \varsubsetneqq \mathbb{C} to the open unit disk \mathbb{D} with z_0 \in \mathbb{R}, f(z_0) = 0, f'(z_0) > 0 (i.e., real and positive). Also suppose that  \Omega is symmetric with respect to the real axis.

    Claim:  f(\overline{z}) = \overline{f(z)}.

    Scratch Work: For the moment assume f is a mobius map and recall that mobius tranforms map Riemann circles to Riemann circles (including "lines"). Define \rho_\mathbb{R}(z) be a reflection across the real axis and \rho_K(z) be the reflection about the circle f(\mathbb{R}). Form the function g(z) = f^{-1} \circ \rho_K \circ f \circ \rho_\mathbb{R}(z) which maps from \Omega to  \Omega. Since z_0 \in \Omega \cap \mathbb{R} and \Omega is open, we can get (a,b) \subseteq \Omega \cap \mathbb{R}. If w \in (a,b) then g(w) = w because \rho_\mathbb{R}(w) = w and f(w) \in f(\mathbb{R})\Rightarrow \rho_K(f(w)) = f(w). So g(w) = f^{-1} \circ f(w) = w. Therefore, we have that g is analytic (since conformal) and equals the identity map on  \Omega. Then, analytic extension gives us that g(z) is the identity map, i.e. g(z) = z. So f \circ \rho_\mathbb{R}(z) = \rho_K \circ f(z) \Rightarrow f(\overline{z}) = \rho_K(f(z)).

    So there is the first gap; I don't see how to find that  f(\overline{z}) = \overline{f(z)} as claimed. If it is true, it must be that K = \mathbb{R} so that f( (a,b) ) \subseteq \mathbb{R}, but I'm not seeing it.

    When we consider that f is a general conformal map, there is a similar gap. In this case it is not necessarily true that the real line maps to a Riemann "circle". We have the Carathedory-Osgood extension that gives a homeomorphic extension \widetilde{f}: \overline{\Omega} \textrm{ to } \overline{\mathbb{D}}, so the boundary of \Omega maps to the boundary of \mathbb{D}. But, again, I don't see why the segment of \Omega \cap \mathbb{R} should map into a line (or specifically into the real axis) in the disk so that we can conclude that the reflection \rho_K(z) is in fact \rho_\mathbb{R}(z).
    In Ahlfors's book this is an exercise on the uniqueness statement in the Riemann mapping theorem. Given the map f:\Omega\to\mathbb{D}, define g(z) = \overline{f(\overline{z})}. Check that g:\Omega\to\mathbb{D} is analytic and bijective, with g(z_0) = 0 and g'(z_0)>0, and deduce from the uniqueness that g = f. (It then follows that f must be real-valued on the real axis.)
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