03/16/2023, 03:56 AM
This reduces into a problem with riemann mappings.
We can write tommy's equation as:
\[
f(g(f^{-1}(z)) = g(z)\\
\]
These equations are only solvable if \(g : D \to D\). Then we need that \(f : D \to D\). The only domain \(g = \exp\) takes to itself is \(\mathbb{C}\). Trying to create conjugations on the entire complex plane is a null effort. There exists no conjugations in this case. The cases where you can create conjugations on all of \(\mathbb{C}\), always require a smaller domain \(D\) that satisfies this.
If \(D\) were simply connected and biholomorphic to the unit disk; then we can write the solutions for:
\[
f(g(f^{-1}(z))) = h(z)\\
\]
Very simply for all \(f,g,h\). But as \(\exp(z)\) only fixes the complex plane, this isn't possible; unless you introduce a lot of branching talk.
So, Daniel is right in saying that \(z \mapsto z\) is the trivial solution, but it's also the only solution on \(\mathbb{C}\). To get a different solution, we can take something like Kneser's:
\[
f(z) = \text{sexp}_K(\text{slog}_K(z) + 1/2)\\
\]
But this is not holomorphic on \(\mathbb{C}\). And we get a chicken and the egg situation... Do we solve tetration, or do we solve this conjugation first?
I'm not sure how helpful it is to look at this first; rather than looking at iterates which appear more natural; but I could be wrong...
Regards, James
We can write tommy's equation as:
\[
f(g(f^{-1}(z)) = g(z)\\
\]
These equations are only solvable if \(g : D \to D\). Then we need that \(f : D \to D\). The only domain \(g = \exp\) takes to itself is \(\mathbb{C}\). Trying to create conjugations on the entire complex plane is a null effort. There exists no conjugations in this case. The cases where you can create conjugations on all of \(\mathbb{C}\), always require a smaller domain \(D\) that satisfies this.
If \(D\) were simply connected and biholomorphic to the unit disk; then we can write the solutions for:
\[
f(g(f^{-1}(z))) = h(z)\\
\]
Very simply for all \(f,g,h\). But as \(\exp(z)\) only fixes the complex plane, this isn't possible; unless you introduce a lot of branching talk.
So, Daniel is right in saying that \(z \mapsto z\) is the trivial solution, but it's also the only solution on \(\mathbb{C}\). To get a different solution, we can take something like Kneser's:
\[
f(z) = \text{sexp}_K(\text{slog}_K(z) + 1/2)\\
\]
But this is not holomorphic on \(\mathbb{C}\). And we get a chicken and the egg situation... Do we solve tetration, or do we solve this conjugation first?
I'm not sure how helpful it is to look at this first; rather than looking at iterates which appear more natural; but I could be wrong...
Regards, James