Well the periodic points can actually be seen in my graph!
If you see the sinusoidal curve, you're basically just finding the points \(\text{tet}_{\eta_-}(z_0)\) such that \(\text{tet}_{\eta_-}(z_0 + 2)\) are the same value. And again, since this is very sinusoidal, you're bound to get many of these points of arbitrary period \(n\). And additionally, since the sinusoid is small and restrictive about \(1/e\), we can expect the periodic points to sort of "coallesce" about the fixed point (\(1/e\)).
Iterating about periodic points is pretty standard for the repelling case, (a bit more complicated with neutral). The thing is, that the local iterations can never contain the periodic points.
If you define an Abel function \(\alpha\), then \(\alpha(f^{\circ 2}(z_0)) = \alpha(z_0) = \alpha(z_0) + 2\) so that we must have \(z_0\) is a pole. You can see this in my graph too, where the function is not injective, and therefore no inverse function exists on the real line and it blows up precisely at these points. If you take the Schroder route, than we get the same problem as with root two about the attracting or repelling fixed point. You can do an iteration about \(z_0\), but then it disagrees with an iteration about any other periodic/repelling point.
So for example, if I take:
\[
f^{\circ 2}(z) = z_0 + \lambda^2(z-z_0) + O(z-z_0)^2\\
\]
For \(f(z) = \eta_-^z\), then we get a schroder function (we know that \(|\lambda| \neq 0,1\)),
\[
\Psi(f^{\circ 2}(z)) = \lambda^2 \Psi(z)\\
\]
Then there exists exactly TWO squareroot functions:
\[
g(z) = \Psi^{-1}(\pm\lambda\Psi(z))\\
\]
in a neighborhood of \(z_0\). NEITHER OF THESE SQUARE ROOT FUNCTIONS ARE \(\eta_-^z\). That's simply because \(\eta_-^z\) doesn't have a fixed point at \(z_0\).
This goes for any local iteration about periodic points, you cannot, for lack of a better word, have both \(z_0\) and \(f(z_0)\) be in one iteration.
To over explain, if \(f^{\circ s}(z) : \mathcal{A} \to \mathcal{A}\) for some domain \(\mathcal{A}\) such that \(z_0,z_1 \in \mathcal{A}\), then:
\[
f^{\circ 2k}(z_0) = z_0\,\,\,\text{and}\,\,f^{\circ 2k+1}(z_0) = f(z_0)\\
\]
This implies two damning things. Which is pretty much exactly what happens with the square root two case. We have an iteration about \(2\), and we have an iteration about \(4\), we can't have an iteration about both at the same time. If we did, we'd be crossing from an attracting fixed point's immediate basin, into the Julia set... And no such iteration can exist. In this case either \(\log_{\eta_-}\) is attracting at both \(z_0,z_1\) or \(f\) is, and therefore there's a Julia set between them.
So in this case, it would imply that the julia set \(\mathcal{J}\) of \(f\) intersects with \(\mathcal{A}\). Now we may get lucky and \(\mathcal{A}\) is all Julia set, but that still poses a very big problem--the Julia set of \(\log_{\eta_-}\) separates the basin about both, and will therefore make a discontinuity as we move \(z\) from \(z_0 \to z_1\). \(f^{\circ s}\) can only be holomorphic on \(\mathcal{A}\) excluding the periodic points. This is again, because the Abel function is undefined here in \(z\). We'd end up with \(f^{\circ s + \alpha(z)}(1)\) which is a singularity at these periodic points.
Essentially, LOCAL iteration is precisely that, local to a fixed point/ periodic point. And cannot include other fixed points/periodic points, without producing a discontinuity. Exactly like with \(\sqrt{2}\) about \(2\) and \(4\).
AFAIK There is no way to avoid this though, even through Kneser. Kneser (for example base \(e\)), does not produce an iteration about ANY PERIODIC OR FIXED POINT. It is singular at every one. (I mean this as \(\exp^{\circ s}(z) = \text{tet}_K(s+\text{slog}_K(z))\) isn't holomorphic at any fixed/periodic point of \(\exp\)) Though it does have a regular looking structure at \(L^{\pm}\), but on different domains depending on \(s\), so the point still stands. There can be 1 fixed point, 0 fixed points, in an iteration, never more.
So, In doing a Kneser iteration as you're proposing, we'd be iterating \(f^{\circ 2}(z)\) about the conjugate fixed points. This could absolutely produce an iteration \(F(s)\) which would be holomorphic in the upper/lower half plane. The trouble...? \(F(1/2 + F^{-1}(z)) \neq f(z)\). You'd be iterating \(f^{\circ 2}\) and you'd essentially lose all information about \(f\).
Additionally, it wouldn't make sense because \(F\) would send to the periodic points in the upper/lower half planes, this is not a fixed point of \(\log_{\eta_-}(z)\), so asymptotically it's a bit meaningless.
You would have to do a very very weird crescent iteration, that I can't even imagine, it would be very non-standard. And would ultimately not be subject to the \(lim_{\Im(z) \to \pm \infty}\) condition we're used to seeing, as this would mean we're iterating \(f^{\circ 2}\).
If you see the sinusoidal curve, you're basically just finding the points \(\text{tet}_{\eta_-}(z_0)\) such that \(\text{tet}_{\eta_-}(z_0 + 2)\) are the same value. And again, since this is very sinusoidal, you're bound to get many of these points of arbitrary period \(n\). And additionally, since the sinusoid is small and restrictive about \(1/e\), we can expect the periodic points to sort of "coallesce" about the fixed point (\(1/e\)).
Iterating about periodic points is pretty standard for the repelling case, (a bit more complicated with neutral). The thing is, that the local iterations can never contain the periodic points.
If you define an Abel function \(\alpha\), then \(\alpha(f^{\circ 2}(z_0)) = \alpha(z_0) = \alpha(z_0) + 2\) so that we must have \(z_0\) is a pole. You can see this in my graph too, where the function is not injective, and therefore no inverse function exists on the real line and it blows up precisely at these points. If you take the Schroder route, than we get the same problem as with root two about the attracting or repelling fixed point. You can do an iteration about \(z_0\), but then it disagrees with an iteration about any other periodic/repelling point.
So for example, if I take:
\[
f^{\circ 2}(z) = z_0 + \lambda^2(z-z_0) + O(z-z_0)^2\\
\]
For \(f(z) = \eta_-^z\), then we get a schroder function (we know that \(|\lambda| \neq 0,1\)),
\[
\Psi(f^{\circ 2}(z)) = \lambda^2 \Psi(z)\\
\]
Then there exists exactly TWO squareroot functions:
\[
g(z) = \Psi^{-1}(\pm\lambda\Psi(z))\\
\]
in a neighborhood of \(z_0\). NEITHER OF THESE SQUARE ROOT FUNCTIONS ARE \(\eta_-^z\). That's simply because \(\eta_-^z\) doesn't have a fixed point at \(z_0\).
This goes for any local iteration about periodic points, you cannot, for lack of a better word, have both \(z_0\) and \(f(z_0)\) be in one iteration.
To over explain, if \(f^{\circ s}(z) : \mathcal{A} \to \mathcal{A}\) for some domain \(\mathcal{A}\) such that \(z_0,z_1 \in \mathcal{A}\), then:
\[
f^{\circ 2k}(z_0) = z_0\,\,\,\text{and}\,\,f^{\circ 2k+1}(z_0) = f(z_0)\\
\]
This implies two damning things. Which is pretty much exactly what happens with the square root two case. We have an iteration about \(2\), and we have an iteration about \(4\), we can't have an iteration about both at the same time. If we did, we'd be crossing from an attracting fixed point's immediate basin, into the Julia set... And no such iteration can exist. In this case either \(\log_{\eta_-}\) is attracting at both \(z_0,z_1\) or \(f\) is, and therefore there's a Julia set between them.
So in this case, it would imply that the julia set \(\mathcal{J}\) of \(f\) intersects with \(\mathcal{A}\). Now we may get lucky and \(\mathcal{A}\) is all Julia set, but that still poses a very big problem--the Julia set of \(\log_{\eta_-}\) separates the basin about both, and will therefore make a discontinuity as we move \(z\) from \(z_0 \to z_1\). \(f^{\circ s}\) can only be holomorphic on \(\mathcal{A}\) excluding the periodic points. This is again, because the Abel function is undefined here in \(z\). We'd end up with \(f^{\circ s + \alpha(z)}(1)\) which is a singularity at these periodic points.
Essentially, LOCAL iteration is precisely that, local to a fixed point/ periodic point. And cannot include other fixed points/periodic points, without producing a discontinuity. Exactly like with \(\sqrt{2}\) about \(2\) and \(4\).
AFAIK There is no way to avoid this though, even through Kneser. Kneser (for example base \(e\)), does not produce an iteration about ANY PERIODIC OR FIXED POINT. It is singular at every one. (I mean this as \(\exp^{\circ s}(z) = \text{tet}_K(s+\text{slog}_K(z))\) isn't holomorphic at any fixed/periodic point of \(\exp\)) Though it does have a regular looking structure at \(L^{\pm}\), but on different domains depending on \(s\), so the point still stands. There can be 1 fixed point, 0 fixed points, in an iteration, never more.
So, In doing a Kneser iteration as you're proposing, we'd be iterating \(f^{\circ 2}(z)\) about the conjugate fixed points. This could absolutely produce an iteration \(F(s)\) which would be holomorphic in the upper/lower half plane. The trouble...? \(F(1/2 + F^{-1}(z)) \neq f(z)\). You'd be iterating \(f^{\circ 2}\) and you'd essentially lose all information about \(f\).
Additionally, it wouldn't make sense because \(F\) would send to the periodic points in the upper/lower half planes, this is not a fixed point of \(\log_{\eta_-}(z)\), so asymptotically it's a bit meaningless.
You would have to do a very very weird crescent iteration, that I can't even imagine, it would be very non-standard. And would ultimately not be subject to the \(lim_{\Im(z) \to \pm \infty}\) condition we're used to seeing, as this would mean we're iterating \(f^{\circ 2}\).
