The iterational paradise of fractional linear functions

[bo198214]

Taking conjugations of \(\lambda^t z\) by linear fractional transformations, will probably already supply that.

This really raises a point that always puzzled me: In the calculations of the linear fractional iterations there naturally always occur the powers of *two* Eigenvalues \(\lambda_1^t\) and \(\lambda_2^t\). I was wondering how they relate to the *one* \(\lambda\) you are talking about and how they relate to the two fixed points and their derivations. So I write it down to make it clear for me:

So we have a linear fractional function \(\phi\) and its matrix representation \(M\). \(M\) can be diagonalized in two ways with \(\lambda_1,\lambda_2\) being the Eigenvalues of the characteristic polynomial:
\begin{align}
M &= A_1 D_1 A_1^{-1}, \quad\quad D_1=\left(\begin{matrix}\lambda_1 &0\\0 &\lambda_2\end{matrix}\right)\\
M &= A_2 D_2 A_2^{-1}, \quad\quad D_2=\left(\begin{matrix}\lambda_2 &0\\0 &\lambda_1\end{matrix}\right)
\end{align}

And because these are only permutations in columns and rows we know that both iterations are the same
\[ M^t = A_1 D_1^t A_1^{-1} = A_2 D_2^t A_2^{-1} \]

\(A_i\) and \(D_i\) correspond to the linear fractional functions \(\mu_i\) and \(\ell_i\)
\[ \phi = \mu_i \circ \ell_i \circ \mu_i^{-1}, \quad\quad\ell_i(z) = c_i z \]
\(\mu_i\) are (obviously?) the mappings which map \(\{0,\infty\}\) to the fixed points \(\{z_1,z_2\}\). Say \(\mu_1(0)=z_1\), \(\mu_2(0)=z_2\).
From this viewpoint we also can derive equality: knowing that there can only be one Schröder iteration at a fixed point and the iterations are analytic at both fixed points and that's why the iterations are equal.

And it turns now out how the fixed point derivations are connected with the Eigenvalues: 
\(c_1 = \frac{\lambda_1}{\lambda_2}\) and \(c_2 = \frac{\lambda_2}{\lambda_1}\) EDIT: and it makes totally sense that it is a ratio, because the matrix \(M\) is only determined up to a constant a. If \(M\) corresponds to \(\phi\) then also \(aM\) corresponds to \(\phi\).

Though we still have to show that \(c_i\) are really the fixed point derivations of \(\phi\):

\begin{align}
\phi' &= \mu'\circ \ell\circ\mu^{-1} \cdot \ell'\circ \mu^{-1} \cdot \frac{1}{\mu'\circ \mu^{-1}}\\
\phi'(z_i) &= \mu_i'(\ell_i(0)) \cdot \ell_i'(0) \cdot \frac{1}{\mu_i'(0)} = \ell_i'(0) = c_i
\end{align}

So the \(c_1=\phi'(z_1)\) and \(c_2=\phi'(z_2)\) are indeed the derivatives at both fixed points and they are reciprocal.

So in my constructed parabolic plosion one could express \(q_t(s)\) also in terms of one fixed point derivative \(d=\frac{1-s}{1+s}\):
\[ q_t(s) = \frac{(1+s)^t - (1-s)^t}{s((1+s)^t+(1-s)^t} = \frac{1-d^t}{s\left(1+d^t\right)} \]

And in this form it is quite natural to show periodicity: Because \(d^t\) has obviously period \(2\pi i/\log(d)\) and we showed already that \(\log(d)\) is purely imaginary for s imaginary. Even that is natural to show with this simpler form of the derivative d (which I didn't see in my previous post).