Möbius transformation with $B(-i) = i,B(i) = 2i$ and maps real line to real line

Question

Given $B$ Möbius transformation such that $B(-i) = i, B(i) = 2i$ and $B(\{z\in \mathbb{C} : Im z = 0 \} \cup \{ \infty\}) = \{z\in \mathbb{C} : Im z = 0 \} \cup \{ \infty\}$

Does such transformation exist and if it does map $\{z\in \mathbb{C} : Re z = 0 \}$ with it.

As I understand if cline $k_1$ maps to cline $k_2$ under a Möbius transformation then points symmetric under inversion on $k_1$ map to points inverted onto $k_2$.

$i$ is symmetric to $-i$ on line $Im z =0$ since on lines inversion is reflection?

But since $B(i)$ and $B(-i)$ aren't symmetric to line $Im z = 0$ such transformation doesn't exist.

I'm not sure if I'm correct, the inversion symmetry is confusing. Does this mean assuming $B(-i) = i$ we'd need $B(i) = -i$ ? Would it be possible to find the transformation in this case?

Answer 1

I belive the answer to this is that no such map exists. In the following denote for any subset $\Omega \subseteq \mathbb C$, $\Omega^{\ast}$ as the complex conjugation of this set.

Supose now that such a map exists. Without loss of generality we can ignore the value taken at infinty, since either it is mapped to the real line and to itself and in both cases this does not really matter for the following argument.

Main Idea: Analytic Contiuation are unique

Now consider $B \colon \mathbb C \to \mathbb C$ and denote by $B\vert_{\mathbb H} \colon \mathbb H \to \mathbb C$ the restriction of $B$ to the upper half plane. Then since this restriction maps a $I \subseteq \mathbb R$ to the real line (note that here one of real values could be mapped to infinity but this does not matter since you dont need to map the full real axis to the whole real axis), applying the Schwarz reflection principle yields that there exists an unique analytic extension $\tilde{B} \colon \mathbb H \cup \mathbb H^{\ast} \cup I \to \mathbb C$ such that one has $$ \tilde{B}(z) = \overline{B(\overline{z})} \quad z \in \mathbb H^{\ast}. $$ Hence from our assumptions $\tilde{B}(-i) = \overline{B(i)} = \overline{2i} = -2i \neq i=B(i)$, but since analytic continuation is unique, it must be that $\tilde{B} = B$, hence we arrive at a contradiction.

Answer 2

The answer is no, there are two ways to prove it.

Geometric argument: Moebious transformations maps circles in circles, which are eventually straight lines in the model $\mathbb{P}^1(\mathbb{C})=\mathbb{C}\cup\{\infty\}$.

If $B(\infty)=\infty$, the straight line through $-i$, $i$ and $\infty$ is preserved, hence $0$ is sent in a point $ix$, $x\in(1,2)$, that is $\mathbb{R}$ is not preserved.

If $B(\infty)=0$, the straight line $i\mathbb{R}$ is preserved, hence $B(0)=\infty$. It follows that $B(z)=\pm1/z$, hence $B(i)\ne2i$.

Finally, if $B(\infty)\in\mathbb{R}^*$, the straight line through $-i$, $i$ and $\infty$ is sent to a circle in $\mathbb{C}$, which crosses $i\mathbb{R}$ in $B(-i)=i$ and $B(i)=2i$, and $\mathbb{R}$ in $B(\infty)$ and $B(0)$. Moreover, since Moebius transformations are conformal maps (namely they preserve angles), such a circle has to intersect orthogonally $\mathbb{R}$. It is an easy exercise of Euclidean geometry to show that such a circle does not exists.

Algebraic argument: $B$ can be written as $$B(z)=\frac{az+b}{cz+d}\,,\qquad\begin{pmatrix} a & b\\ c & d\end{pmatrix}\in\mathbb{P}\mathrm{SL}_2(\mathbb{C}).$$

It is a classical result, but it is not difficult to check, that $B$ preserves $\mathbb{R}\cup\{\infty\}$ if and only if $B\in\mathbb{P}\mathrm{SL}_2(\mathbb{R})$. In particular, we can assume $\det B=1$.

By hypothesis \begin{align} &\frac{ai+b}{ci+d}=B(i)=2i=2B(-i)=2\frac{-ai+b}{-ci+d}\\ \iff& (ai+b)(-ci+d)=2(-ai+b)(ci+d)\\ \iff&(ac+bd+i(\underbrace{ad-bd}_{=1})=2(ac+bd-i(\underbrace{ad-bd}_{=1}))\\ \iff& 3i=ac+bd\in\mathbb{R}, \end{align} which is absurd.

EDIT: another direct argument is to use invariancy of cross ratio by Moebious transformation.

Answer 3

Your map $B$ cannot exist for basic topological reasons. When you look at $\mathbb C_\infty = \mathbb C\cup\big\{\infty\big\}$, $B$ gives a homeomorphism from $\mathbb C_\infty$ to itself and the map restricts to a homeomorphism from $\mathbb C-\mathbb R$ to itself. This has 2 components, where $i$ is in one of them and $-i$ is in the other.

But under the image of $B$ these two points are in the same component which is impossible-- e.g. $g:\mathbb C-\mathbb R\longrightarrow \mathbb Z$, where $g(z)=-1$ if $\Im(z)\lt 0$ and $g(z)=1$ if $\Im(z)\gt 0$ then $g\circ B^{-1}$ would be a continuous integer valued function that is not constant on the upper half plane (check at $i$ and $2i$).