Real Multivalued/Parametrized Iteration Groups

[tommy1729]

This was the original comments by myself and MphLee when Leo first came on the scene. This is all Riemann surface territory. Riemann surface stuff gets wild when you talk about iterations. These are well known constructions; but it classifies itself under "dynamics on a Riemann surface". It behaves very similar to the normal dynamics we have; which is why I keep mentioning Milnor. We either have \(\mathbb{C}\) (Euclidean) or \(\mathbb{D}\) (A simply connected domain) or \(\widehat{\mathbb{C}}\) (The Riemann Surface). These are just three Riemann surfaces that we use as our base, especially with most of the work done here.

Choosing an arbitrary Riemann surface \(S\)--which can be as wild as possible, and taking iterates of \(f^{\circ n} : S \to S\)--we get everything Leo is talking about. The trouble is, talking about it as multivalued functions won't give you all the juice that Riemann surfaces will give you. It's like drinking orange juice compared to eating oranges. Whether we process our oranges first (Project the Riemann surface into the space of multivalued functions), or we just eat oranges (Prove everything with Riemann surfaces, that Milnor and most of complex dynamics set up).

If you want to go down the Riemann surface route though. We should refer to this more broadly. Where the preimage of the \(\beta\) tetration of \(\eta^-\) is Riemann surface, and constructing an action from this riemann surface to \(\mathbb{C}\) creates a multivalued semi group at \(\eta^-\).

This becomes:

\[
f^{\circ t}(z) = F\left(t+\mu(\mathcal{F}(z))\right)\\
\]

Where \(F\) is the beta iteration with period \(2 \pi i/\lambda\), and \(\mu : S \to \mathbb{C}\) and \(\mathcal{F}(z) = \{y \in \mathbb{C}\,|\, F(z) = y\}\). Then we just note that \(\{\forall z \mathcal{F}(z)\} = S\); and we are just choosing a projection schema (a multivalued function).

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Let's go down the rabbit hole, bo! I loved Leo's original thesis. And I loved the idea. The trouble I had, was that it was the image, and not the Riemann surface preimage.

let me get this straith

\(\mathcal{F}\) send you to the right complex value and mu sends you to the right branch ?

regards

tommy1729