Post by Gary on Feb 13, 2017 13:10:20 GMT
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.
I don't agree that $a=0$ and $d=0$. Here is my analysis of the situation. It continues on from my previous post on this subject.
Because $M$ as defined above in my original post is not normalised, and (27) on page 152 only applies to a normalised $M$, let's derive the formula for the fixed points from scratch. We obtain the following formula
$\xi_{\pm}=\dfrac{-(a+d)\pm\sqrt{(a+d)^2-4bc}}{2c}$
So if the fixed point is of the form $i\alpha$, where $\alpha$ is real then $(a+d)$ must equal zero and so
$\xi_{\pm}=\pm\sqrt{-b/c}$
Since $(a+d)=0$, lets choose $a=1$ and $d=-1$. If we now choose $b=2$ and $c=1$ then we can see that $(-b/c)<0$ as required if we want fixed points of the form $i\alpha$.
Also we can see that the quantity $(ad-bc)=-1-2=-3<0$ as required for the matrix $M$.
We can therefore write our matrix as
$M(z)=\dfrac{-az-b}{cz+d}=\dfrac{-z-2}{z-1}$ and the fixed points of this should be
$\xi_{\pm}=\pm\sqrt{-b/c}=\pm i\sqrt{2}$.
Let's check this:
$M(i\sqrt{2})=\dfrac{-i\sqrt{2}-2}{i\sqrt{2}-1}=\dfrac{(-i\sqrt{2}-2)(i\sqrt{2}+1)}{(i\sqrt{2}-1)(i\sqrt{2}+1)}=i\sqrt{2}$
Similarly we find that $M(-i\sqrt{2})=\dfrac{i\sqrt{2}-2}{-i\sqrt{2}-1}=-i\sqrt{2}$.
So we have verified that the two fixed points of $M(z)=\dfrac{-z-2}{z-1}$ are $\pm i\sqrt{2}$.
Since $\widetilde{\mathcal{M}}(\xi)=-\bar{\xi}$ we can see that these same points are also fixed points of $\widetilde{\mathcal{M}}(z)$, as expected.
Note: If a quadratic with real coefficients has complex roots then they must be complex conjugates. So if one fixed point lies above the real axis, the other must lie below the real axis and vice versa.
Vasco
Another early morning here. You have saved me some work! Shortly after I wrote my last message, I realized that I was comparing apples and oranges (normalized $\mathcal{M}$ and unnormalized $\widetilde{\mathcal{M}}$. I sat down to the computer to redo it and found your message. I agree with your analysis with the one reservation that since the map is the UHP, there should be no fixed points with negative values on the imaginary axis, in which case only one fixed point of $\widetilde{\mathcal{M}}$ equals one fixed point of $\mathcal{M}$. Or do we permit the fixed points to lie below the horizon, rather like the traditional Pacific navigators who find their course on cloudy nights by reference to a fictional guide star?
Gary
