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This question has been bogging me down lately. I'm not sure how to come up with an approach to tackle the proof exactly. I'm without a proof, butI think the result I'm searching for is true. Similarly, I do believe I'm not the only person to consider this question and I expect it is already answered somewhere in the vestibule of the internet.

The $n$'th super root is defined as the inverse to the function

$$^n x = F_n(x)=x^{x^{...n\,times...^x}}$$

so that if $\Psi_n(x)$ is the function in question then

$$^n \Psi_n(x) = F_n(\Psi_n(x)) = x$$

Now $F_n(x)$ depends on the branch of the logarithm chosen to define it, and therefore it is multivalued. We will stick to the principal branch of the logarithm as it simplifies the question (and I assume solving for other branches only requires a modification of the proof). It is obvious $F_n(x) : \mathbb{C}/ (0,-\infty) \to \mathbb{C}^{\times}$. It also follows that for $\Re(y) > 1$ there exists $x \in \mathbb{C}$ such that $y = F_n(x)$.

Now we can define an inverse through the analytic implicit function theorem granted that $\frac{d}{dx} F_n(x) \neq 0$ when $\Re(F_n(x)) > 1$--but showing this is rather daunting. It invovles showing when $\Re(F_n(x)) > 1$ we have

$$\frac{F_{n-1}(x)}{x} + \log(x)F'_{n-1}(x) \neq 0$$

which is rather daunting to say the least.

This brings me to my question which is more of a reference request than anything. Unless I'm completely missing something and the answer is right on my nose.

1.) Is $\Psi_n(x)$ holomorphic for $\Re(x) > 1$?

or

2.) What is the maximal domain $\Psi_n(x)$ is holomorphic for in $x$?

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    $\begingroup$ How about starting with $n=2$ ? $\endgroup$ Commented Aug 29, 2016 at 20:37
  • $\begingroup$ @Gerald Edgar Well with $n=2$ it is not very hard to construct an inverse in the right half plane, Namely $x^x = y$ then $\log(y) = x \log(x) = ue^u$ so then $u = W(\log(y))$ and $x = e^{W(\log(y))}$, where $W$ is Lamberts function. $\endgroup$
    – user78249
    Commented Aug 30, 2016 at 16:16

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This is not a complete answer, but will give you some pointers in any case.

In our paper Bifurcations in the space of exponential maps (Invent. math. (2009) 175, doi:10.1007/s00222-008-0147-5; see also arXiv:math/0311480) with Schleicher, we have to consider a similar question in order to control certain features of exponential parameter space. The key relevant result is Lemma 4.4.

For $\kappa,z\in\mathbb{C}$, define $E_{\kappa}(z):= e^z+\kappa$. Lemma 4.4 implies, in particular, that for all $m$ the function $$ \mathcal{E}_m\colon \kappa \mapsto E_{\kappa}^{\circ m}(\kappa)$$ has an inverse branch that is defined on a half-strip of the form $$ \{a+ib: a>R_m, |b|<\pi \}$$ and takes real numbers $>a$ to real numbers. Here the number $R_m$ increases rapidly with $m$.

(Remark. The result in Lemma 4.4 is more general, and applies around any "parameter ray" at an address $\underline{s}$; to obtain the special case above, use the sequence $\underline{s}=00000\dots$. Moreover, the number $n$ in Lemma 4.4 corresponds to $n=m+2$.)

To note the connection with your question, observe that $$ {\ }^n a = f_{\lambda}^{\circ(n+1)}(0),$$ where $$ f_{\lambda}(w) = e^{\lambda w},$$ and $e^\lambda=a$. Furthermore, if $e^{\kappa}=\lambda$, then $E_{\kappa}$ and $f_{\lambda}$ are conformally conjugate, with $z=\lambda w + \kappa$.

So the desired function is given in terms of $\mathcal{E}$ as the inverse of $$ a\mapsto \frac{\mathcal{E}_{n+1}(\kappa(a)) - \kappa(a)}{\lambda(a)},$$ with $\lambda(a)=\log(a)$ and $\kappa(a)=\log(\lambda)$.

While Lemma 4.4 in our paper does not directly imply that the inverse of this function also exists on a similar strip, the proof relies on the derivative of $\mathcal{E}_n$ being large, which will remain true under the above reparameterisation. So this is certainly true in your setting also.

I have not thought about this for a long time, but it seems reasonable that the proof might extend to show that the inverse exists on a right half-plane (we simply did not need this in our paper).

The proof will not give you an inverse on a right half-plane independently of $n$. For this - if true - you would likely need to use some information from the theory of parabolic explosion. (Actually, this is related to some work of mine with Benini, whose write-up is long overdue. Again, I am not quite sure whether it will give exactly what you want.)

Asking about the maximal domain on which the inverse exists seems like a hopeless question. First of all, it does not seem altogether likely that such a maximal domain would be unique.

EDIT. I just remembered that Baker and Rippon (Iteration of exponential functions, Ann. Acad. SCi. Fenn, 1984) already gave a version of the above argument in 1984. We needed more information, but that is irrelevant for your question. Moreover they use the parameterisation by $\lambda$, so their results are directly applicable in your setting. Having a quick look at their paper, the key point seems to be in Lemma 7.2. If I understand correctly, their result implies your map is defined on the intersection of a right half-plane with a sector. Again, by the nature of the argument, you will not get a fixed domain independent of $n$; for this you would need to pull back towards the parabolic point at $\kappa=-1$ resp. $\lambda=1/e$.

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  • $\begingroup$ Thanks a bunch! This helps. I'll look into parabolic explosion, maybe something will pop up related to my question. I was hoping the result wouldn't require too much deep mathematics, but I'm starting think I'm wrong by how many roadblocks I've hit. $\endgroup$
    – user78249
    Commented Aug 30, 2016 at 16:09
  • $\begingroup$ I've added a reference to Baker&Rippon in an edit; they actually work in the $\lambda$-plane, so their results are directly applicable. The arguments are a bit technical, but basically elementary. $\endgroup$ Commented Aug 30, 2016 at 16:24
  • $\begingroup$ However, in order to get all branches defined on a fixed domain, you will be getting closer and closer to the parabolic parameter at $\lambda=1/e$, and hence need some arguments involving Fatou coordinates. Getting a specific half-plane as you requested is likely to need, in addition, some specific estimates. But (if true) I see no reason this couldn't be done by using existing ideas. $\endgroup$ Commented Aug 30, 2016 at 16:26

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