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Theorem.
Let $f$ be a function that is biholomorphic on both of the initial regions $D_1$ and $D_2$ that share as boundaries the same conjugate pair of fixed points of f. Also let f have no other fixed points. Let there be two Abel functions of f, $A_1$ and $A_2$, that are biholomorphic on these initial regions and satisfy A(d) = c.
For each f, there exists exactly one biholomorphism $A_{cont}(z)$ on a single simply connected open set $C \supseteq D_1,\,D_2$ such that $\forall z \in D_i\, A_{cont}(z) = A_i (z).$ (i is an index that can be 1 or 2) i.e. there is an analytic continuation, and it's unique.

Proof.
1.
Let $D_1, D_2$ be disjoint, simply connected domains that have as boundaries:
(1) $\partial_1 D,\, \partial_2 D \not \subset D$, disjoint curves which are homeomorphic to (0,1);
(2) $L$ and $\bar{L}$, which are boundaries of, but not contained in, $\partial _1 D_1, \, \partial_2 D_1,\, \partial_1 D_2, \,\partial_2 D_2$.

2.
Let $f$ be a biholomorphism on $S := \lbrace z:|\Im(z)| < \Im(L) + \epsilon i \rbrace$, where $\epsilon>0$, (tried to make the domain of biholomorphism into an open set) that:
(1) bijects $\partial_1 D$ to $\partial_2 D$;
(2) has a conjugate pair of fixed points $L$ and $\bar{L}$;
(3) has no other fixed points in the domain of biholomorphy.

3.
(1) Let $A_1$, a biholomorphism on $D_1$, and $A_2$, a biholomorphism on $D_2$, both satisfy $A(f(z))=A(z)+1$ for all applicable z. (for all z such that A(f(z)) is defined)
(2) Let $A(d)=c$ for some $d \in S$.

to be continued
(06/20/2009, 07:27 PM)Tetratophile Wrote: [ -> ]1. .. 2. ... 3. ... 4.

I dont see where it shows that $A_1(z)=A_2(z)$ for $z\in D_1\cap D_2$.
btw, why does kouznetsov have IMAGINARY infinities $i\infty$ and $-i\infty$ as the values of the super logarithm at the fixed points? The choice seems arbitrary. I think it should be a kind of negative infinity because of how you get to the fixed points. exp^(some complex number) (1) = i or some other nonreal complex number, take the logarithm of that number an infinite number of times and you get the fixed point. for some complex c, corresponds to a complex iteration plus a negative infinity itterations of exp: $\exp^{-\infty+c}(1).$ We lose injectivity, but it's a fair price.

(06/21/2009, 08:19 AM)bo198214 Wrote: [ -> ]Is the proof already finished?
I dont see where it shows that $A_1(z)=A_2(z)$ for $z\in D_1\cap D_2$.

no, it isn't finished.... It just lists all i know about the situation. I don't know how to proceed. I just have a hunch that the neighborhood of the fixed points might be important (the only thing the domains D1 and D2 share as part of their boundaries is the fixed point pair, and both D1 and D2 include a subset of the neighborhood); if I can show that the Abel functions are analytic and take the same values in all directions in the deleted neighborhood of the fixed points, I will have established the existence of a unique analytic continuation of both A1 and A2.

I have attached a drawing to summarize the problem and the proposed proof technique.

[attachment=496]
(06/21/2009, 01:57 PM)Tetratophile Wrote: [ -> ]btw, why does kouznetsov have IMAGINARY infinities $i\infty$ and $-i\infty$ as the values of the super logarithm at the fixed points? The choice seems arbitrary.
Infinity can be imagined as a point on the extended complex plane, which is a sphere.
You can regard infinity as any other point on the complex plane.
The tetrational is not holomorphic (not even continuous) at infinity.
However one can approach infinity from different angles/sectors where the limit may be a single value.
E.g. if you approach infinity along the real axis $z \to \infty$ the limit is infinity.
If you however approach infinity along the imaginary axis $z\to i \infty$ then the limit is $L$. This is writtten exactly as:
$\lim_{x\to\infty} f(x+iy) = \infty$ for each real $y$.
$\lim_{y\to\infty} f(x+iy) = L$ for each real $x$.
$\lim_{y\to -\infty} f(x+iy) = L^\ast$ for each real $x$.

Quote:no, it isn't finished.... It just lists all i know about the situation. I don't know how to proceed. I just have a hunch that the neighborhood of the fixed points might be important (the only thing the domains D1 and D2 share as part of their boundaries is the fixed point pair, and both D1 and D2 include a subset of the neighborhood);

I guess its about deforming one initial region into the other initial region, and continuing the one Abel function to the other region. Then we have reduced the problem to both Abel function having the same domain, to which we can apply the theorem.

For different fixed point pairs this would not be possible.

Quote: if I can show that the Abel functions are analytic and take the same values in all directions in the deleted neighborhood of the fixed points,
No, they are not analytic in the punctured neighborhood. Or in other words the fixed points are not isolated singularities. They are branching points, we need cut-line ending in the fixed point, to have it holomorphic.

Your drawing however is I also see the situation (except holomorphy at the fixed points).
(06/21/2009, 05:18 PM)bo198214 Wrote: [ -> ]Infinity can be imagined as a point on the extended complex plane, which is a sphere.
You can regard infinity as any other point on the complex plane.
The tetrational is not holomorphic (not even continuous) at infinity.
However one can approach infinity from different angles/sectors where the limit may be a single value.

I said that the superlogarithm should be negative infinity at the fixed points, instead of infinity*i or = -infinity*i, because the fixed point is approached via a complex iteration, and then infinite negative iteration of exp, ie iteration of log. (fixed points are repelling in graph of exp) Or leave it an essential singularity if there is no way to complex-iterate exp to get an imaginary number.

'bo198214 Wrote:I guess its about deforming one initial region into the other initial region
How do we go about doing that? It is not given whether the Abel functions are holomorphic outside of their domains.
(06/21/2009, 05:28 PM)Tetratophile Wrote: [ -> ]I said that the superlogarithm should be negative infinity at the fixed points,

No, its not. The super-exponential approaches L for $z=x+iy$, $y\to+\infty$. Thatswhy the super-logarithm has positively unbounded imaginary part, if we approach the upper fixed point in the initial region, which one could interpret as $+i\infty$. And it has negatively unbounded imaginary when we approach $L^\ast$ in the initial region which we could interpret as $-i\infty$. However strictly considered there is no $\pm i\infty$ as value of a limit.

Quote: instead of infinity*i or = -infinity*i, because the fixed point is approached via a complex iteration, and then infinite negative iteration of exp, ie iteration of log.
If you approach the fixed point via repeated logarithm, then you dont stay inside the initial region.

Quote:
'bo198214 Wrote:I guess its about deforming one initial region into the other initial region
How do we go about doing that? It is not given whether the Abel functions are holomorphic outside of their domains.
If you go slightly to the right of $\partial_1 D_1$ having a curve $\gamma$ inside $D_1$ then you can continue $A_1$ to the region between $\partial_2 D_1$ and $F(\gamma)$, i.e. $A_1(F(z)):=A_1(z)+1$ where we use the already established values of $A_1$ on the region between $\partial_1 D_1$ and $\gamma$.

This works if $F$ is bijective in the considered area. For example $F=\exp$ is bijective in the strip along the real axis $\{z: -\pi < \Im(z) < \pi \}$, which includes the fixed points $L$ and $L^\ast$.
(06/21/2009, 06:04 PM)bo198214 Wrote: [ -> ]
(06/21/2009, 05:28 PM)Tetratophile Wrote: [ -> ]
'bo198214 Wrote:I guess its about deforming one initial region into the other initial region
How do we go about doing that? It is not given whether the Abel functions are holomorphic outside of their domains.
If you go slightly to the right of $\partial_1 D_1$ having a curve $\gamma$ inside $D_1$ then you can continue $A_1$ to the region between $\partial_2 D_1$ and $F(\gamma)$, i.e. $A_1(F(z)):=A_1(z)+1$ where we use the already established values of $A_1$ on the region between $\partial_1 D_1$ and $\gamma$.

This works if $F$ is bijective in the considered area. For example $F=\exp$ is bijective in the strip along the real axis $\{z: -\pi < \Im(z) < \pi \}$, which includes the fixed points $L$ and $L^\ast$.

Of course, this procedure might not help if the derivatives at the fixed points are positive reals (or are 0). In that case, near the fixed points, the expanded $A_1$ won't be any closer to $A_2$ than the original one was.
Please go back up to the beginning of the proof;
I made the conditions a lot stronger: changed the holomorphy condition for $f$ to BIholomorphy on a strip with infinite range of real parts), and added "f has no other fixed points", because I thought other fixed points would mess things up.
Proof continued...
4.
Consider a simple curve $\gamma \subset D_1$ that also has $L$ and $\bar{L}$ as non-inclusive boundaries.
(1) Since $A_1$ is biholomorphic on $\gamma$, $A_1$ will still be biholomorphic on the curve $f(\gamma)$, because if $A_1(z)$ is biholomorphic, so is $A_1(z)+1=A_1(f(z)) \forall z \in D_1$.
(2) We can do this for every curve in $D_1$ that has $L$ and $\bar{L}$ as boundaries, to extend the domain of $A_1$
(3) We can repeat (4.2) as many times as needed to get to $D_2$. (we can do the above the other way around, using $f^{-1}$ to get from $D_2$ to $D_1$, because f is BIholomorphic) Then $A_1$ is biholomorphic where $A_2$ was defined to be biholomorphic.* Then we know, by the theorem that was proven on the other day, that they are the same biholomorphism. So we know that there exists a single open set C that includes $D_1 \cup D_2$ where not only the biholomorphism $A_{cont}$that maps d to c exists, but also $A_1 = A_2 = A_{cont} \forall z \in C.$

$\mathcal{Q. E. D.}$

Corollary.
There exists a unique superlogarithm for $b>e^{1/e}$ that uniquely bijects holomorphically each simple initial region of arbitrary "width" in an open set C that: (1) contains in its boundary fixed points of exp_b; (2) does not include branch cuts of the superlogarithm; to its resp. vertically infinite strip of arbitrary width in $A( C )$.

*Now I don't know if $A_1 = A_2$ in $D_2$. It must have something to do with the condition $A(d) = c$. For your theorem to apply, I need to prove that both A1 and A2 are equal in some neighborhood of d.
(06/21/2009, 11:22 PM)BenStandeven Wrote: [ -> ]Of course, this procedure might not help if the derivatives at the fixed points are positive reals (or are 0). In that case, near the fixed points, the expanded $A_1$ won't be any closer to $A_2$ than the original one was.

Right, this would make up an extra case; also it needs to be investigated whether this case allows an initial region at all. However for $\exp_b$, $b>e^{1/e}$ this is already satisfied.
(06/22/2009, 12:37 AM)Tetratophile Wrote: [ -> ]and added "f has no other fixed points", because I thought other fixed points would mess things up.
Well, but $\exp$ *has* more fixed points. In every strip $2\pi i k < \Im(z) < 2\pi i (k+1)$ there is a fixed point of $\exp$.

Quote:(3) We can repeat (4.2) as many times as needed to get to $D_2$.

No, we can not.
1. The curve $f(\gamma)$ may get out of the strip of bijectivity of $f$ what indeed happens for $f=\exp$. (plot the curves!)
2. $f(\gamma)$ may hit $D_1$, i.e. the extension of $A_1$ overlaps with the original domain. If $\gamma$ meets $L$ at an angle $\alpha$ then $f(\gamma)$ meets $L$ at an angle $\alpha+\arg(f'(L))$. So the image curves rotate around L. This includes also the comment of Ben, that if $\arg(f'(L))=0$, i.e. if $f'(L))$ is real, then then the images do not rotate, and hence one would never reach $D_2$ if it hits $L$ in a different angle.
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