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Question: Let $D_2=\bar D(2,1)$ and $D_{-2}=\bar D(-2,1)$ be the closed disks of radius $1$ centered at $z=2$ and $z=-2$ in the complex plane, respectively. Set $X= \mathbb C-\{D_2 \cup D_{-2} \}$, and suppose $$f:X \rightarrow X,$$ is analytic, 1-1, and satisfies $f(X)=X.$ Show that $f$ is a Möbius transformation.

Attempt(s):

(1) I was thinking this could be done by contradiction. Here is a vague sketch of what I had in mind. Assume it's not a Möbius transformation. Viewing $f$ as a map on the Riemann sphere/extended complex plane minus these two disks, either $\infty\rightarrow \infty$ or some point on the boundary of one of the two disks is mapped to $\infty$. (If both "large complex numbers" and points near the boundary headed to $\infty$ we would not have injectivity. If neither, then we wouldn't have surjectivity.)

If all points on the boundary of the disks do not go to $\infty$, then we can extend $f$ into the disks... (I get lost here) then somehow need to show that this maps $D_{-2}$ to $D_2$ and $D_2$ into $D_{-2}$, or fixes them both.

What I would like to end up with is some automorphism of the Riemann sphere which must be a Möbius transformation.

(2) Use symmetry and the Schwarz reflection principle somehow, but I don't know that the real line is mapped into real line.

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  • $\begingroup$ By "bilinear transformation", do you mean Möbius transformation [aka fractional linear transformation]? $\endgroup$ – Daniel Fischer Jun 27 '15 at 21:12
  • $\begingroup$ yes, i guess that's more standard term. i've changed it, $\endgroup$ – Ashley Jun 27 '15 at 21:18
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    $\begingroup$ Okay. Point 1, $\infty$ is an isolated singularity of $f$. What type can it be? [The answer will come out to be "a simple pole".] Point 2, $X\cup \{\infty\}$ is conformally equivalent to an annulus. Point 3, every automorphism of an annulus is a Möbius transformation. $\endgroup$ – Daniel Fischer Jun 27 '15 at 21:33
  • $\begingroup$ Actually, it's probably a tiny bit easier to start with point 2. $\endgroup$ – Daniel Fischer Jun 27 '15 at 21:41
  • $\begingroup$ yeah point 2 is nice...didn't see that at all. okay, I guess it's a bit of work to show that all automorphisms of annulus are mobius, but maybe thats "well-known"? $\endgroup$ – Ashley Jun 27 '15 at 21:52
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The key is that the set $Y = X\cup \{\infty\}$ is conformally equivalent to an annulus.

Let's find a Möbius transformation that maps $Y$ to an annulus with centre $0$. First, it is trivial to find a Möbius transformation mapping one of the two disks to the closed unit disk $\overline{\mathbb{D}}$, let's take the translation $z\mapsto z+2$ mapping $D_{-2}$ to $\overline{\mathbb{D}}$. Then $D_2$ is mapped to a closed disk in the exterior of the unit disk, and $Y$ to the remainder of the exterior. Composing with inversion in the unit circle then produces a Möbius transformation mapping $Y$ conformally to $\mathbb{D}\setminus \overline{D}$ for some disk $D$. Since the intersections of the circle $\lvert z-2\rvert = 1$ with the real axis are mapped to $\frac{1}{3}$ and $\frac{1}{5}$ by the Möbius transformation $T_1 \colon z \mapsto \frac{1}{z+2}$, and Möbius transformations preserve angles, the centre of $D$ must lie on the real axis, and thus it follows that $D = \bigl\{ z : \bigl\lvert z-\frac{4}{15}\bigr\rvert = \frac{1}{15}\bigr\}$. Next we need an automorphism of the unit disk mapping $D$ to a disk with centre $0$. Knowing the form of automorphisms of the unit disk, we look for an $r \in \bigl(\frac{1}{5},\frac{1}{3}\bigr)$ such that $T_2 \colon z \mapsto \frac{z-r}{1-rz}$ satisfies $T_2\bigl(\frac{1}{5}\bigr) = - T_2\bigl(\frac{1}{3}\bigr)$. Solving the arising quadratic equation yields $r = 2 - \sqrt{3}$. Calculating the composition, we find that

$$T_3 \colon z \mapsto (2-\sqrt{3}) \frac{\sqrt{3}-z}{\sqrt{3}+z}$$

is a Möbius transformation mapping $Y$ to the annulus

$$A_1 = \{z : (2-\sqrt{3})^2 < \lvert z\rvert < 1\}.$$

Doing a little further normalisation,

$$T \colon \frac{z-\sqrt{3}}{z+\sqrt{3}}$$

maps $Y$ to the annulus

$$A = \{ z : 2-\sqrt{3} < \lvert z\rvert < 2+\sqrt{3}\},$$

and $T(\infty) = 1$, so $X$ is mapped biholomorphically to $A\setminus \{1\}$.

Then $g = T\circ f \circ T^{-1}$ is a biholomorphic map of $A\setminus \{1\}$ with itself. Since $A$ is bounded, $1$ is a removable singularity of $g$, and after removing it, we have $g(1) \in \overline{A}$. Since $g$ is not constant, the open mapping theorem tells us that $g(1) \notin \partial A$, and also that $g(1) \neq 1$ is impossible, since if we had $1\neq g(1) = g(z_0)$ for $z_0\in A\setminus \{1\}$, $g$ could not be injective. Thus $g$ is an automorphism of the annulus $A$ with fixed point $A$. Since all automorphisms of $A$ are of the form $z \mapsto e^{i\varphi} z$ or $z\mapsto e^{i\varphi}/z$ for some $\varphi\in \mathbb{R}$, the only automorphisms of $A$ with fixed point $1$ are the identity and the inversion $z \mapsto \frac{1}{z}$. That means that $f$ is either the identity or the negation $z \mapsto -z$, in particular, it is a Möbius transformation.

It remains to see that the automorphisms of annuli with centre $0$ are all of the form $z \mapsto a\cdot z$ or $z \mapsto a/z$ for some $a\in \mathbb{C}\setminus \{0\}$. We show a little more, that every biholomorphic map between annuli with centre $0$ is of that form. From that, it follows that two annuli $A_1 = \{ z : \rho_1 < \lvert z\rvert < \rho_2\}$ and $A_2 = \{ z : r_1 < \lvert z\rvert < r_2\}$ are biholomorphically equivalent if and only if $\frac{\rho_2}{\rho_1} = \frac{r_2}{r_1}$. We assume that both radii of an annulus are strictly positive and finite. The extension to annuli where the inner radius is $0$ or the outer radius is $+\infty$ but not both is straightforward (it is in fact simpler). The case $\mathbb{C}\setminus \{0\}$ is special, it is of course not biholomorphically equivalent to any annulus with at least one strictly positive and finite radius.

So suppose $h_0\colon A_1 \to A_2$ is biholomorphic. Let $C = \{ z : \lvert z\rvert = \sqrt{\rho_1 \rho_2}\}$ be the circle of symmetry of $A_1$. Since $A_1\setminus C$ is a disjoint union of two annuli, $A_2 \setminus h_0(C)$ is - topologically - also a disjoint union of two annuli, so $\gamma\colon t\mapsto h_0(\sqrt{\rho_1\rho_2}e^{it}),\, t\in [0,2\pi]$ is a Jordan curve with $n(\gamma, 0) = \pm 1$. If $n(\gamma,0) = -1$, compose $h_0$ with $z \mapsto \frac{\rho_1\rho_2}{z}$ and call the resulting map $h$, otherwise let $h = h_0$. Then $h$ maps the annulus $\{ z : \rho_1 < \lvert z\rvert < \sqrt{\rho_1\rho_2}\}$ to the part of $A_2$ between the circle $\{ z : \lvert z\rvert = r_1\}$ and $h(C)$.

If $(z_n)$ is a sequence in $A_1$ with $\lvert z_n\rvert \to \rho_1$, then $\bigl(h(z_n)\bigr)$ has a convergent subsequence by the Bolzano-Weierstraß theorem. We can assume that the full sequence $\bigl(h(z_n)\bigr)$ converges, say to $w$. Now it is impossible that $w \in A_2$, for then we'd have $w = h(z_\ast)$ for some $z_\ast \in A_1$, and we could find an open neighbourhood $U$ of $z_\ast$ and an $n_0 \in \mathbb{N}$ such that $z_n \notin U$ for $n \geqslant n_0$. But by the open mapping theorem $V = h(U)$ is then an open neighbourhood of $w$, and we have $h(z_n) \in V$ for all large enough $n$ since $h(z_n) \to w$. But that contradicts the injectivity of $h$. So we must have $w \in \partial A_2$. Since for all large enough $n$ we have $\lvert z_n\rvert < \sqrt{\rho_1\rho_2}$, it follows that $h(z_n)$ is in the interior of $h(C)$ for all large enough $n$, and thus $\lvert w\rvert = r_1$. Essentially the same argument shows that $\lvert h(z_n)\rvert \to r_2$ whenever $\lvert z_n\rvert \to \rho_2$.

Now we have shown that the premises of the (general) Schwarz reflection principle are fulfilled, so by reflecting in the inner boundary circles, we can extend $h$ to a biholomorphic mapping

$$h_1 \colon \biggl\{ z : \frac{\rho_1^2}{\rho_2} < \lvert z\rvert < \rho_2\biggr\} \to \biggl\{ z : \frac{r_1^2}{r_2} < \lvert z\rvert < r_2\biggr\}.$$

Continuing the extension to larger annuli by reflection in the inner boundary circle, we obtain a biholomorphic map

$$h_\omega \colon \{ z : 0 < \lvert z\rvert < \rho_2\} \to \{ z : 0 < \lvert z\rvert < r_2\},$$

and we know all such maps are of the form $z \mapsto az$ for an $a\in \mathbb{C}$ with $\lvert a\rvert = \frac{r_2}{\rho_2}$. Since $h_\omega$ maps the circle with radius $\rho_1$ to the circle with radius $r_1$, it follows that $\frac{r_2}{\rho_2} = \lvert a\rvert = \frac{r_1}{\rho_1}$, which shows that the condition $\frac{r_2}{r_1} = \frac{\rho_2}{\rho_1}$ is necessary for the existence of a biholomorphic map between $A_1$ and $A_2$. The sufficiency is clear, if the ratios of the radii are equal, $z \mapsto \frac{r_2}{\rho_2}z$ is a biholomorphic map between the two annuli.

Since we might at the beginning have composed our original $h_0$ with an inversion, it follows that $h_0$ either had the form $z \mapsto az$ or $z \mapsto \frac{a\rho_1\rho_2}{z}$ for some $a$ with $\lvert a\rvert = \frac{r_2}{\rho_2}$.

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