I think this makes sense, particularly because these are meromorphic functions.
Iterating meromorphic functions tend to be a whole nother beast compared to iterating transcendental entire functions. If we want to go this route, we equally have an iteration for \(f^{\circ t}(z) = \lambda^t z\), this is holomorphic at \(\infty\) and at \(0\), where they are both fixed points. By which any conjugation using linear fractional transformation makes another solution. We can also probably map \(\infty \mapsto A\) and \(0 \mapsto B\), then we have a holomorphic iteration at both fixed points.
But! and this is the big but, this is an iteration of a meromorphic function, not an iteration of an entire function.
Things get a lot more complicated with meromorphic functions. I challenge you to find an entire function that can be iterated about two fixed points!
This reminds me of the problem, there are only two kinds of entire fractional iterations:
\[
\begin{align}
f^{\circ t}(z) = z_0 + \lambda^t(z-z_0)\\
g^{\circ t}(z) = z+tz_0\\
\end{align}
\]
But as soon as you relax entire into meromorphic, there are a bunch of linear fractional transformation solutions to a meromorphic fractional iteration.
So essentially, because you have added poles to your iteration, this greatly changes the problem. Milnor has nothing to say about this.
It's also important to note that this still isn't a counter example to the proposition. Because this iteration does not have the local expansion:
\[
\begin{align}
f^{\circ t}(z) &= \lambda^t z + O(z^2)\,\,\text{for}\,\,|z|<\delta\,\,\Re(t) > 0\\
f^{\circ t}(z) &= 1 + \mu^t (z-1) + O(z^2)\,\,\text{for}\,\,|z-1| < \delta\,\,\Re(t) > 0\\
\end{align}
\]
Because there are poles in \(f^{\circ t}\) at precisely:
\[
\begin{align}
(2^t-1)z &= -1\\
2^t - 1 &= -1/z\\
t &= \log(1-1/z)/\log(2) + k\pi i/\log(2)\,\,\text{for}\,\,k\in\mathbb{Z}\\
\end{align}
\]
So take \(\Im(z) > 0\) and \(|z| < \delta\), then there's a pole in t at about \(t \approx K \) for \(K\) arbitrarily large as \(z \to 0\) (choose k = -1). This means the above local expansion is invalid for \(z \approx 0\). I would call this a different kind of iteration than regular, simply because it doesn't have the taylor expansion at 0 that we typically see with regular iteration. But that's for you to decide.
To clarify, what I mean by no analytic iteration at both fixed points, I mean the above expansion is valid. And as I said before, this can't happen because the Schroder coordinate change will become discontinuous somewhere along a path to the other fixed point.
So this function is not holomorphic for \(\Re(t) > A\) for some \(A \in \mathbb{R}\) about \(z =0\) and about \(z=1\). I guess I should've clarified that's what I mean by no holomorphic iteration about two fixed points.
Also, I think this is a very important distinction we should be considering. Is it okay to let the regular iteration have poles? Is it still regular iteration if there is a pole. If the answer is yes, I suggest that we make a restriction to types of regular iteration, one without poles and one with poles.
For example, using your parabolic iteration, there are poles in \(t\) at precisely:
\[
t = -1/z\\
\]
Is that still allowed as a regular iteration? I've never seen the term regular iteration, which allows for poles, in either variable, but if so, then yes I'll withdraw the claim that holomorphic about two points, and change it to, a regular iteration in a half plane can't contain two fixed points. And additionally, recall that there are no poles in \(z\) or poles in \(t\). Noticeably, this function \(f^{\circ t}(z)\) is holomorphic on \(\mathbb{C}^2 /\{t = -1/z\}\) which is a very different type of domain than what you will see with entire functions. If we let \(z\) get arbitrarily large here, then the function is getting more and more volatile near \(t = 0\)--and similarly if you take very large iterates \(t >T\) then we shrink the domain of holomorphy about \(z=0\). A feature we'll never really see with entire functions.
To me, iteration about a fixed point, means that we have a fixed neighborhood \(N\) and a fixed domain \(D\) in \(\mathbb{C}\) such that this object is holomorphic; and D is closed under addition of elements; \(\{D,+\}\) is a semigroup. Notably, this parabolic iteration is \(\mathbb{R}\) analytic in \(t\), but isn't holomorphic for any strip \(|\Im(t)| < \delta\) and \(|z| < \delta\). You have to shrink \(\delta\) as \(t\) grows, and thereby you are making a more dynamic kind of iteration.
It doesn't have the form: \(f^{\circ t}(z) : D \times N \to N\); where we are mapping a semi group, but it does have the form \(f^{\circ t}(z) : \mathbb{R} \times N \to N\). This is then a real iteration in my book, and not a complex iteration.
Because yes, there are singularities with \(\sqrt{2}^z\)'s regular iteration about \(2\), but they are not poles. They are logarithmic singularities/ branching problems, and additionally locally about \(z \approx 2\) it's holomorphic in a half plane (which is a semi-group under \(+\)). Which is a standard feature of Schroder iteration. Poles require us to think of these things meromorphically, and that can change a lot of the rules of how we manipulate these things.
I think this is a very important problem that we'll have to clarify our language on. Neither of these iterations are holomorphic on a domain \(D \subset \mathbb{C}\) that is a semi-group under \(+\), while additionally \(\mathbb{R}^+\subset D\); while \(z \in N\), where \(N\) is a neighborhood about the fixed point \(p\); \(N = \{|z-p| < \delta\}\). That's the exception of these iterations, which breaks the law that you can't iterate about two fixed points; you aren't mapping a semi-group in the complex plane; you're mapping a semi group on \(\mathbb{R}\), or you are mapping a semigroup in \(\widehat{\mathbb{C}}\). Which is a very important distinction.
I think the really technical statement is: On a closed Riemann Surface (\widehat{\mathbb{C}}\)) we can iterate about multiple fixed points. On open Riemann surfaces (for example the open \(\mathbb{C}\))--we cannot iterate about two fixed points. And specifically in this case. This is how Milnor would handle this "counter-example".
Iterating meromorphic functions tend to be a whole nother beast compared to iterating transcendental entire functions. If we want to go this route, we equally have an iteration for \(f^{\circ t}(z) = \lambda^t z\), this is holomorphic at \(\infty\) and at \(0\), where they are both fixed points. By which any conjugation using linear fractional transformation makes another solution. We can also probably map \(\infty \mapsto A\) and \(0 \mapsto B\), then we have a holomorphic iteration at both fixed points.
But! and this is the big but, this is an iteration of a meromorphic function, not an iteration of an entire function.
Things get a lot more complicated with meromorphic functions. I challenge you to find an entire function that can be iterated about two fixed points!
This reminds me of the problem, there are only two kinds of entire fractional iterations:
\[
\begin{align}
f^{\circ t}(z) = z_0 + \lambda^t(z-z_0)\\
g^{\circ t}(z) = z+tz_0\\
\end{align}
\]
But as soon as you relax entire into meromorphic, there are a bunch of linear fractional transformation solutions to a meromorphic fractional iteration.
So essentially, because you have added poles to your iteration, this greatly changes the problem. Milnor has nothing to say about this.
It's also important to note that this still isn't a counter example to the proposition. Because this iteration does not have the local expansion:
\[
\begin{align}
f^{\circ t}(z) &= \lambda^t z + O(z^2)\,\,\text{for}\,\,|z|<\delta\,\,\Re(t) > 0\\
f^{\circ t}(z) &= 1 + \mu^t (z-1) + O(z^2)\,\,\text{for}\,\,|z-1| < \delta\,\,\Re(t) > 0\\
\end{align}
\]
Because there are poles in \(f^{\circ t}\) at precisely:
\[
\begin{align}
(2^t-1)z &= -1\\
2^t - 1 &= -1/z\\
t &= \log(1-1/z)/\log(2) + k\pi i/\log(2)\,\,\text{for}\,\,k\in\mathbb{Z}\\
\end{align}
\]
So take \(\Im(z) > 0\) and \(|z| < \delta\), then there's a pole in t at about \(t \approx K \) for \(K\) arbitrarily large as \(z \to 0\) (choose k = -1). This means the above local expansion is invalid for \(z \approx 0\). I would call this a different kind of iteration than regular, simply because it doesn't have the taylor expansion at 0 that we typically see with regular iteration. But that's for you to decide.
To clarify, what I mean by no analytic iteration at both fixed points, I mean the above expansion is valid. And as I said before, this can't happen because the Schroder coordinate change will become discontinuous somewhere along a path to the other fixed point.
So this function is not holomorphic for \(\Re(t) > A\) for some \(A \in \mathbb{R}\) about \(z =0\) and about \(z=1\). I guess I should've clarified that's what I mean by no holomorphic iteration about two fixed points.
Also, I think this is a very important distinction we should be considering. Is it okay to let the regular iteration have poles? Is it still regular iteration if there is a pole. If the answer is yes, I suggest that we make a restriction to types of regular iteration, one without poles and one with poles.
For example, using your parabolic iteration, there are poles in \(t\) at precisely:
\[
t = -1/z\\
\]
Is that still allowed as a regular iteration? I've never seen the term regular iteration, which allows for poles, in either variable, but if so, then yes I'll withdraw the claim that holomorphic about two points, and change it to, a regular iteration in a half plane can't contain two fixed points. And additionally, recall that there are no poles in \(z\) or poles in \(t\). Noticeably, this function \(f^{\circ t}(z)\) is holomorphic on \(\mathbb{C}^2 /\{t = -1/z\}\) which is a very different type of domain than what you will see with entire functions. If we let \(z\) get arbitrarily large here, then the function is getting more and more volatile near \(t = 0\)--and similarly if you take very large iterates \(t >T\) then we shrink the domain of holomorphy about \(z=0\). A feature we'll never really see with entire functions.
To me, iteration about a fixed point, means that we have a fixed neighborhood \(N\) and a fixed domain \(D\) in \(\mathbb{C}\) such that this object is holomorphic; and D is closed under addition of elements; \(\{D,+\}\) is a semigroup. Notably, this parabolic iteration is \(\mathbb{R}\) analytic in \(t\), but isn't holomorphic for any strip \(|\Im(t)| < \delta\) and \(|z| < \delta\). You have to shrink \(\delta\) as \(t\) grows, and thereby you are making a more dynamic kind of iteration.
It doesn't have the form: \(f^{\circ t}(z) : D \times N \to N\); where we are mapping a semi group, but it does have the form \(f^{\circ t}(z) : \mathbb{R} \times N \to N\). This is then a real iteration in my book, and not a complex iteration.
Because yes, there are singularities with \(\sqrt{2}^z\)'s regular iteration about \(2\), but they are not poles. They are logarithmic singularities/ branching problems, and additionally locally about \(z \approx 2\) it's holomorphic in a half plane (which is a semi-group under \(+\)). Which is a standard feature of Schroder iteration. Poles require us to think of these things meromorphically, and that can change a lot of the rules of how we manipulate these things.
I think this is a very important problem that we'll have to clarify our language on. Neither of these iterations are holomorphic on a domain \(D \subset \mathbb{C}\) that is a semi-group under \(+\), while additionally \(\mathbb{R}^+\subset D\); while \(z \in N\), where \(N\) is a neighborhood about the fixed point \(p\); \(N = \{|z-p| < \delta\}\). That's the exception of these iterations, which breaks the law that you can't iterate about two fixed points; you aren't mapping a semi-group in the complex plane; you're mapping a semi group on \(\mathbb{R}\), or you are mapping a semigroup in \(\widehat{\mathbb{C}}\). Which is a very important distinction.
I think the really technical statement is: On a closed Riemann Surface (\widehat{\mathbb{C}}\)) we can iterate about multiple fixed points. On open Riemann surfaces (for example the open \(\mathbb{C}\))--we cannot iterate about two fixed points. And specifically in this case. This is how Milnor would handle this "counter-example".
