(Joint work with Tobias Kaiser)

Recall from this post that not all germs in $\H$ have a holomorphic extension that maps definable real domains to definable real domains. In fact, the extension $\t_a$ of the translation $t_a$, for $a\gt 0$, does not even map real domains to real domains.

So, in order to describe the extension properties of general $f \in \H$, we need to relax the notion of (definable) real domain, while keeping in mind the scale provided by the angular level.

For $\,k \in \NN \cup \{-1\}$, we set $$\H_k := \set{h \in \H_{\gt 0}:\ \alevel(h) = k}.$$

**Definition**

Let $\lambda \in \ZZ$, $U,V \subseteq \LL$ be domains and $\phi:U \into V$.

- $U$ is a
**$k$-domain**if there exist $\,f,g \in \H_k$ such that $U_f \subseteq U \subseteq U_g$. - $\phi$ has
**complex level $\lambda$**if, for $k \ge \lambda -1$ and every $k$-domain $U’ \subseteq U$, the image $\phi(U)$ is a $(k-\lambda)$-domain.

**Example**

Let $w \in \E^\circ$ of level $l$, and set $\eta:= \max\{l,0\}$. Combining the proof of the corollary in this post with the corollary of this post, it follows that there exists an $(\eta-1)$-domain $U \subseteq \LL$ and an injective holomorphic extension $\w:U \into \LL$ such that $\w$ has complex level $l$.

It is this way of describing the extension properties of words in $\E^\circ$ that works for general definable $f$:

**Extension Theorem***Let $\,f \in \I$ and set $\eta:=\max\{\eh(f),0\}$ and $\lambda:= \level(f) \le \eta$. Then there exist an $(\eta-1)$-domain $U$, and $(\eta-\lambda-1)$-domain $V$ and a biholomorphic extension $\f: U \into V$ of $\,f$ of complex level $\lambda$.*

How optimal are these holomorphic extensions? Using Ilyashenko’s Phragmén-Lindelöf principle (see Section 3.1C in this book) and the Extensions Theorem, we obtain a converse of the latter:

**Proposition***Let $\,f \in I$ and $\,\eta \in \NN$, and assume that $\,f$ has a holomorphic extension $\,\f:U \into H_{\LL}(a)$, where $U \subseteq \LL$ is an $(\eta-1)$-domain and $a > 0$. Then $\eh(f) \le \eta$.*

This brings us right back to the Hardy field $\H$ of $\Ranexp$: it easily follows from the definition of $\eh$ that, for $f \in \I$, we have $$\tag{1} \eh(f) \ge \eh(\lm(f)) = \level(f).$$ The above proposition provides an *upper bound* on $\eh(f)$.

To see how one can use this upper bound, let $\,f \in \I$, set $\eta:= \max\{\eh(f),0\}$ and $\lambda:= \level(f)$, and let $\,\f:U \into \LL$ be an injective, holomorphic extension of $\,f$ obtained from the Extension Theorem. Then $V:= \f(U)$ is an $(\eta-\lambda-1)$-domain, and clearly $\,\f^{-1}:V \into U$ is a biholomorphic extension of $\,f^{-1}$. Thanks to the above proposition, we can say more:

**Corollary 1***$\eh\left(f^{-1}\right) \le \eta-\lambda$. In particular, after replacing $U$ by a smaller $(\eta-1)$-domain if necessary, $\,\f^{-1}$ has complex level $-\lambda$.*

This corollary is of particular interest for the following germs:

**Definition**

A germ $\,f \in \I$ is **simple** if $\level(f) = \eh(f)$.

Clearly, all pure germs are simple.

**Corollary 2***Let $\,f,g \in \I$ be simple. Then $$\tag{2} \eh\left(f \circ g^{-1}\right) \le \max\{\level(f)-\level(g),0\}.$$*

This last corollary implies that $\,f \circ g^{-1}$ not only has an “obvious” holomorphic extension obtained by composing an extension of $\,f$ with an extension of $\,g^{-1}$ (which would be defined on a $(\level(g)-1)$-domain), but actually has one defined on a possibly larger domain. This observation provides the basis for a construction of quasianalytic classes based on monomials in $\H$, a topic to be discussed in a future set of posts.

Let me finish with one last corollary and an open question: combining (1) and (2), we obtain

**Corollary 3***The set of all simple germs of level 0 is a subgroup of $\I$ under composition.*

In fact, the same argument shows that if $\,f,g \in \I$ are simple and $\level(f) \ge \level(g)$, then $\,f \circ g^{-1}$ is simple. But what happens if $\level(f) \lt \level(g)$?

**Open question**

Is the set of all simple germs a subgroup of $\I$ under composition?

Maybe this question can be answered in the affirmative by iterating the passage from $\CC$ to $\LL$, thereby defining $k$-domains for all $k \in \ZZ$, and by correspondingly restating the Extension Theorem…

Actually, I take the last remark back—because either the iteration suggested in the remark does not lead to a better extension theorem, or the set of all simple germs is not a group under composition (or both)! If $f$ is a simple germ of level $gt 0$, and if $f^{-1}$ had a biholomorphic extension of complex level $-level(f)$ on some $(-level(f)-1)$-domain (however these are defined), then $f$ would have an extension to a $(-1)$-domain. But then $level(f) le 0$ by the above, a contradiction.

So maybe it’s time to look for a counterexample showing that the set of all simple germs is not a group under composition…?