Conformal Reflection across Real Axis

• May 7th 2010, 02:38 PM
huram2215
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 $\rho_\mathbb{R}(z)$.

• May 8th 2010, 09:30 AM
Opalg
Quote:

Originally Posted by huram2215
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.)