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

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

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

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

    When we consider that $\displaystyle f$ 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 $\displaystyle \widetilde{f}: \overline{\Omega} \textrm{ to } \overline{\mathbb{D}}$, so the boundary of $\displaystyle \Omega$ maps to the boundary of $\displaystyle \mathbb{D}$. But, again, I don't see why the segment of $\displaystyle \Omega \cap \mathbb{R}$ should map into a line (or specifically into the real axis) in the disk so that we can conclude that the reflection $\displaystyle \rho_K(z)$ is in fact $\displaystyle \rho_\mathbb{R}(z)$.

    Thanks in advance for your consideration!
    Last edited by huram2215; May 8th 2010 at 04:30 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 $\displaystyle f$ is a Riemann map (conformal) from a simply connected domain $\displaystyle \Omega \varsubsetneqq \mathbb{C}$ to the open unit disk $\displaystyle \mathbb{D}$ with $\displaystyle z_0 \in \mathbb{R}, f(z_0) = 0, f'(z_0) > 0$ (i.e., real and positive). Also suppose that $\displaystyle \Omega$ is symmetric with respect to the real axis.

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

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

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

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