I was reading, some answers about fractional linear transformations and find this old question that was never answer and I think is a nice question. How do you prove it?

We define a Möbius transformation through: $$z\rightarrow \frac{az+b}{cz+d}, ad-bc\neq0, a,b,c,d\in \mathbb{C}$$ and extend to the Riemann sphere as follows: if $c=0$, $T(\infty)=\infty$, and if $c\neq0$, $T(\infty)=a/c$ and $T(-d/c)=\infty$. Show that the Möbius transformations are continuous in that area with the chordal metric. We define the chordal metric thus: $$d_{c}(z_{1},z_{2})= \begin{cases} \frac{2|z_{1}-z_{2}|}{\sqrt{1+|z_{1}|^2}\sqrt{1+|z_{2}|^2}}& \text{if $z_{1},\,z_{2}\neq \infty$}\\ \frac{2}{\sqrt{1+|z_{1}|^2}} & \text{if $z_{2}= \infty$.} \end{cases}$$


Under the stereographic projection, we can identify points on the riemman sphere $\hat{\mathbb{C}}=\mathbb{C}\cup\{\infty\}$ with points on the unit sphere, $S^2$. If two points $x,y\in\hat{\mathbb{C}}$ are mapped to $p,q\in S^2$ respectivley, the chordal metric $d(x,y)=|p-q|$ with the usual euclidean metric on $\mathbb{R}^3$; for a proof you can look at these great notes (which also explain and prove the formulas for the stereographic projection, which I'll use implicitly later on).

Denote the stereographic projection by $\pi:\hat{\mathbb{C}}\to S^2$. As the distance is taken with respect to points on $S^2$, the statement is equivalent to saying that, for a given mobius transformation $M:\hat{\mathbb{C}}\to\hat{\mathbb{C}}$, the function $\pi\circ M\circ\pi^{-1}$ is continuous with respect to the euclidean distance. We can write every mobius transformation as a composition of linear maps $z\to az$ (Denoted $H_a$), an inversion $z\to 1/z$ (denoted $I$) and a translation $z\to z+b$ (Denoted $M_b$). So, if we know that $\pi\circ H_a\circ \pi^{-1}$,$\pi\circ M_b\circ \pi^{-1}$, and $\pi\circ I\circ \pi^{-1}$ are continuous with respect to the euclidean distance- we may finish, cause we have:

$$\pi\circ (M\circ N)\circ \pi^{-1}=\pi\circ M\circ \pi^{-1}\circ\pi\circ N\circ \pi^{-1}$$

And the composition of continuous maps is continuous. So we only need to check any of these cases. Right now I'm short in time so I'll give only the example of $I$, I hope to expand tomorrow on how to do the other cases, but they can be done similarly. Maybe, also, I'll have a way with less calculations - I'm sure there's a much more elegant to do it.

So, fot the $I$ case- write a point on $S^2$ as $(t,u,v)$. Then $\pi^{-1}(t,u,v)=\frac{t+iu}{1-v}$. (Ignore now from the point $(0,0,1)$ - the north pole - which is mapped to $\infty$). Now, $I(\pi^{-1}(t,u,v))=\frac{1-v}{t^2+u^2}(t-iu)$. Now applying $\pi$ gives the point $(2\cdot\frac{(1-v)t}{t^2+u^2+(1-v)^2},-2\cdot\frac{(1-v)u}{t^2+u^2+(1-v)^2},\frac{(1-v)^2 t^2+(1-v)^2 u^2-(t^2+u^2)^2}{(1-v)^2 t^2+(1-v)^2 u^2+(t^2+u^2)^2})=(t, - u,-v) $. Tracking where the point $\infty$ goes, we see that $(0,0,1)$ goes to $(0,0,-1)$. This, therefore, is a continuous mapping (it is continuous in any of its coordinates- the case where we approach to the north pole also works). Geometrically, this is a rotation around the x-axis, so this is another way to see that it is continuous.

EDIT: I've been trying to think about ways to complete the other cases without messy calculations (which can be done). I came up with a nice interpretation only for special kind of the linear maps- you can decompose these further, to multiplication by an element of the form $z\to e^{i\theta}z$ for real $\theta$, and $z\to az$ for $a>0$. If you think about it, the first kind of translations correspond to rotations around the z-axis: if you rotate the plane, the projection will just rotate.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.