Exercise 24 chapter 6

Vasco,

Here is my attempt. Confusing topic with all the orthogonal semicircles, reflections, translations, and rotations. I usually end up referring to [29], p. 163, or one of its equivalents. I'm casting about for rules of thumb.

nh.ch6.ex24.pdf (182.47 KB)

Edited following Vasco's comments.

Gary

Vasco,

Thanks for the corrections. Regarding your final question, I'm not sure of anything regarding that part. I'll study your answer.

Gary

Gary

In part (i) you have shown that the transformation is an opposite motion of the map, but the exercise requires proof that it is also an opposite motion of the PUHP.



Although your suggestion on page 2 of your document (about how part (ii) of the exercise might read), would work, I think it is unlikely, for the following reasons:

We can use any suitable expression for the Möbius transformation $M$ that we create, for example:
$M(z)=\frac{\alpha z+\beta}{\gamma z+\delta}$ where $\alpha,\beta,\gamma,\delta$ are real and $(\alpha\delta-\beta\gamma)>0$
This leads to the following expression for $\widetilde{\mathcal{M}}$
$\widetilde{\mathcal{M}}(z)=\frac{-\alpha\bar{z}-\beta}{\gamma\bar{z}+\delta}$.
We can then choose to define $a$ as $-\alpha$, $b$ as $-\beta$, $c$ as $\gamma$ and $d$ as $\delta$ and we can then write
$\widetilde{\mathcal{M}}(z)=\frac{a\bar{z}+b}{c\bar{z}+d}$ where $a,b,c,d$ are real and $(ad-bc)<0$.
This expression is then of the form we see in the exercise with just the $>$ replaced by $<$.

I also looked at how the fixed points of $M(z)$ and $\widetilde{\mathcal{M}}(z)$ and how they are related. I found that I agree with your conclusions that the fixed points are the same, its just that if the fixed points are complex and $\xi$ is a fixed point of one transformation, then $\bar{\xi}$ is the corresponding fixed point of the other. So we can say generally that if one transformation has fixed points $\xi_{\pm}$ then the other has corresponding fixed points $\bar{\xi}_{\pm}$. So if the fixed points are real then they are the same for both transformations, and if they are complex then corresponding fixed points are complex conjugates.

On page 3, near the bottom, you write "H-lines are invariants under a Möbius transformation". I think this is only true for elliptic Möbius transformations. Also, although your proof of the invariance of $L$ seems perfectly fine to me, it seems natural that having shown that $\widetilde{\mathcal{M}}(z)$ has fixed points on the horizon, we use this fact to show that an h-line through these points is invariant under $\widetilde{\mathcal{M}}(z)$.

In part (iv) I think we need to show from the results obtained in the previous parts that that $\widetilde{\mathcal{M}}(z)$ is always a glide reflection.

There are still one or two places where you have used $\mathcal{M}$ when I think it should be $M$.

Re-reading my solution I noticed a typo in the first line of part (ii) where I wrote "h-circle" instead of "h-line" - I have corrected and republished.

I've also made some changes to parts (i), (ii) and (iv) to try and make things clearer.

Vasco

PS In this post I have increased the font size by one notch - do you think it makes the maths easier to read?

Gary



No, it's because $\widetilde{\mathcal{M}}$ is a motion - see the last paragraph of page 35.

Vasco

Gary



I took your "interesting" comment as an expression of doubt and looked again at this and found that my conclusions were indeed incorrect. I have now done a complete re-think which may be of interest:

Fixed points of $M$ and $\widetilde{\mathcal{M}}$
Let's define the Möbius transformation $M(z)$ as

$M(z)=\dfrac{-az-b}{cz+d}$ where $a,b,c,d$ are real and $(ad-bc)<0$, and

define the opposite motion $\widetilde{\mathcal{M}}$ as

$\widetilde{\mathcal{M}}=\dfrac{a\bar{z}+b}{c\bar{z}+d}$ where $a,b,c,d$ are real and $(ad-bc)<0$.

If $\xi$ is a fixed point of $M$ then

$M(\xi)=\dfrac{-a\xi-b}{c\xi+d}=\xi$

We can also write

$\widetilde{\mathcal{M}}(\xi)=\dfrac{a\bar{\xi}+b}{c\bar{\xi}+d}$

We can write this last expression as

$\overline{\widetilde{\mathcal{M}}(\xi)}=\dfrac{a\xi+b}{c\xi+d}=-M(\xi)=-\xi$

We can write this as

$\widetilde{\mathcal{M}}(\xi)=-\bar{\xi}$

It follows that if $\xi$ is a fixed point of $M$ then it can only be a fixed point of $\widetilde{\mathcal{M}}$ if $\xi=-\bar{\xi}\Longrightarrow \xi+\bar{\xi}=0\Longrightarrow\text{Re}(\xi)=0\Longrightarrow \xi=i\alpha$ where $\alpha$ is real.
So the fixed points of $\widetilde{\mathcal{M}}$ can only be the same as the fixed points of $M$ if the fixed points of $M$ lie on the imaginary axis.
This makes sense because $f$ maps the imaginary axis to itself.

Vasco

Gary

Up early this morning Gary!
When you say the fixed points are equal you mean the fixed points for $M$ and $\widetilde{\mathcal{M}}$ are equal presumably, not that the two fixed points of $M$ are equal, a repeated root. Just checking!

Vasco

Vasco,

Yes, but I think a = d = 0 to resolve (27) with the quadratic of $\widetilde{\mathcal{M}}$ and with your finding that the two fixed points of each motion must lie on the imaginary axis if the fixed points of $\widetilde{\mathcal{M}}$ are to equal those of $\mathcal{M}$. If d $\ne$ 0, then $Re(\xi_\pm) \ne 0$ and $\xi_\pm \ne i\alpha$. Then where a = d = 0, $\widetilde{\mathcal{M}}$ reduces to $\frac{-b}{c\overline{z}}$ and $\xi_\pm  = \frac{\sqrt{-bc}}{2c} = \frac{i}{2}\sqrt{b/c}$, where b and c have equal signs. But this causes a problem. If the root is given the sign $\pm$, then there is a negative fixed point, which lies outside the UHP. So perhaps we can not actually have two equal fixed points between $\widetilde{\mathcal{M}}$ and $\mathcal{M}$ in the UHP. Perhaps we are allowed only one.

Gary