# Appendix: loose ends

In this appendix to the Fields 2022 module, I briefly discuss a couple of loose ends left in this post.

First, let $f = (f_0, \dots, f_k) \in \U^{k+1}$ be such that $f_0 > \cdots > f_k$. We saw that if $f_0 \succ \cdots \succ f_k$, then our construction yields a qaa field $(K_f, e^x \circ \langle f \rangle, T_f)$. Is the condition $f_0 \succ \cdots \succ f_k$ really necessary?

### Exercise 1

There exist $l \le k$ and $g_0 \succ \cdots \succ g_l$ in $\langle f \rangle$ such that, with $g = (g_0, \dots, g_l)$, we have $\langle g \rangle = \langle f \rangle$.

In particular, the sets $e^x \circ \langle f \rangle$ and $e^x \circ \langle g \rangle$ of convergent transmonomials are the same. Moreover, our construction is independent of the particular $g$ obtained in the previous exercise:

### Proposition

Let $g$ and $h$ be two tuples obtained from Exercise 1. Then $(\K_g,e^x \circ \langle g \rangle, T_g)$ and $(\K_h, e^x \circ \langle h \rangle, T_h)$ are the same.

In view of this proposition, our construction assigns a unique qaa field $(K_f, e^x \circ \langle f \rangle, T_f)$ to every tuple $f$ in $\U$, and the direct limit of these qaa fields is again $(\K, \la, T)$.

Second, how do we get the Admissibility Theorem from the Continuation Theorem?

The Continuation Theorem as stated here is not actually the full statement needed in order to be able to prove it by induction on terms. What we do get from our statement is the following corollary concerning complex valued holomorphic continuations (where $H(a) = \{z \in \CC:\ \re(z) > a\}$):

### Corollary 1

Let $f \in \I$ be such that $\eh(f) \le -1$. Then there exist $a \ge 0$ and a holomoprhic continuation $\f:H(a) \into \CC$ that maps every standard power domain into every standard power domain; in particular, $e^{-f}$ is bounded on every standard power domain. $\qed$

We also obtained the following from the Uniqueness Principle:

### Corollary 2

Let $g > f$ be simple germs in $\I$. Then $\eh(f \circ g^{-1}) \le 0$. \qed

So to prove the Admissibility Theorem, we need to work extra in the case $\eh(f \circ g^{-1}) = 0$.

### Example

Corollary 1 fails for $f = x^2$: if $U \subseteq \CC$ is a standard power domain, then $U^2 \nsubseteq H(a)$ for every $a \in \RR$; i.e., $e^{-x^2}$ is unbounded on $U$. Indeed, Corollary 1 holds for $f = x^r$ if and only if $r \le 1$.

The full statement of the Continuation Theorem gives the following strengthening of Corollary 1:

### Complex Continuation Theorem (CCT)

Let $f \in \I$ be such that $\eh(f) \le 0$.

1. There exist $a \ge 0$ and a holomorphic continuation $\f:H(a) \into \CC$ such that $\frac1\f$ is bounded.
2. $|\f(z)| \to \infty$ as $|z| \to \infty$ in $H(a)$.
3. Assume in addition that $f \prec x^2$. Then $\f(H(a)) \subseteq \CC \setminus (-\infty,0]$, the map $\f$ is bijolomorphic onto $\f(H(a))$, and for $z \in H(a)$ we have $$\sign(\arg\f(z)) = \sign(\arg z) = \sign(\im z) = \sign(\im\f(z)).$$

In the Admissibility Theorem, we are dealing with germs of the form $f \circ g^{-1}$ in place of the $f$ in the CCT, where $g > f$. So we need to obtain the desired holomorphic continuation properties for $f$ as in the CCT satisfying $f < x$. To do so, we distinguish three cases:

• Case 1: there exists $\epsilon \in (0,1)$ such that $f \prec x^\epsilon$.
• Case 2: $x^\epsilon \prec f \prec x$ for every $\epsilon \in (0,1)$.
• Case 3: $f \asymp x$.

To illustrate, let’s consider Case 1. For $\alpha \in (0,\pi/2)$ set $S(\alpha):= \{z \in \CC:\ |\arg z| < \alpha\}.$

### Lemma

Assume that $f \prec x^\epsilon$ for some $\epsilon \in (0,1)$. Then there exists $a \ge 0$ such that $\f(H(a)) \subseteq S(\epsilon \frac\pi2)$.

Proof. Set $h:= \frac{x^\epsilon}f$. Then $h \in \I$, and since $f < x$, we also have $h \prec x^2$. Since $\eh(f) \le 0$ and $\eh(x^\epsilon) \le 0$, we also have $\eh(h) \le 0$. So by the CCT, there exist $a>0$ and a biholomorphic continuation $\h:H(a) \into \h(H(a))$ such that, for $z \in H(a)$, we have $\sign(\arg\h(z)) = \sign(\arg z)$.

After increasing $a$ if necessary, we have $\h = \frac{z^\epsilon}\f$, where $z^\epsilon$ denotes the principle branch of this power function. So if $\arg(z)>0$, we get from the above that $$0 < \arg \frac{z^\epsilon}{\f(z)} = \epsilon\arg(z) – \arg(\f(z)).$$ The case $\arg(z) < 0$ is handled similarly, which proves the lemma. $\qed$

The other two cases are a bit more elaborate, but they are still handled using similar reductions to statement 3 of the CCT.

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