(Joint work with Tobias Kaiser; this is a repost)

We are interested in holomorphic continuations of one-variable functions definable in $\Ranexp$. Since $\exp$ and $\log$ are two crucial functions definable in $\Ranexp$, the natural domain on which to consider holomorphic continuations of all definable functions is the Riemann surface of the logarithm $$\LL:= (0,\infty) \times \RR$$ with its usual covering map $$\pi:\LL \into \CC \setminus {0}$$ defined by $\pi(r,\theta) = re^{i\theta}$. We let $$\left|(r,\theta)\right|:= r$$ be the **modulus** and $$\arg(r,\theta):= \theta$$ be the **argument** of $(r,\theta) \in \LL$. We also denote by $\Log:\LL \into \CC$ the biholomorphic map $$\Log x:= \log|x| + i \arg x,$$ and by $\Exp:\CC \into \LL$ its inverse.

Finally, we set $$d(x,y):= \max{||x|-|y||, |\arg x -\arg y|}$$ and $$B(x,s):= \set{y \in \LL:\ d(x,y) \lt s}.$$

**Remarks**

- One drawback of $\LL$ is that addition does not extend globally on $\LL$. However, the map $\Log$ gives $\LL$ the structure of a holomorphic manifold, thereby giving meaning to the term “holomorphic function” on $\LL$.
- Since $\pi = \exp \circ \Log$, one might simply work in the “logarithmic chart” on $\CC$ instead of in $\LL$. However, describing the continuation properties of functions in the logarithmic chart really means describing the continuation properties of their compositions with $\log$, which is not what we want in the end. On the other hand, we do need to work in the logarithmic chart of $\CC$ whenever we have to compute with derivatives of functions on $\LL$.
- While $\LL$ and $\Log$ are clearly definable, the projection $\pi$ is not; indeed, any restriction of $\pi$ to a domain in $\LL$ on which $\arg x$ is unbounded is not definable.

**Example:** Holomorphic continuation on $\LL$ of power functions.

For real $r\gt 0$, we define the **power function** $\p_r:\LL \into \LL$ by $$\p_r(s,\theta):= (s^r,r\theta)$$ and the **scalar multiplication** $\m_r:\LL \into \LL$ by $$\m_r(s,\theta):= (rs,\theta).$$ Note that these continuations are definable and injective, hence biholomorphic.

**Definition**

For $a\gt 0$ we define the **half-plane** $$H_\LL(a):= \set{x \in \LL:\ |x| \gt a}$$ and the **sector** $$S_\LL(a):= \set{x \in \LL:\ |\arg x| \lt a}.$$ Note that the restriction $$\pi_0:= \pi\rest{S_\LL(\pi)}:S_\LL(\pi) \into \CC \setminus (-\infty,0]$$ is biholomorphic and definable.

**Example:** Holomorphic continuations on $\LL$ of $\log$ and $\exp$.

If $\log$ also denotes the standard branch of the logarithm on $\CC \setminus (-\infty,0]$, then we have $$\Log(x) = \log(\pi(x)), \quad\text{for } x \in S_\LL(\pi).$$ So we define $\llog:\LL \setminus (0,1) \into S_\LL(\pi)$ by $$\llog(x) := (\pi_0)^{-1}(\Log(x)),$$ and we define $\eexp: S_\LL(\pi) \into \LL \setminus (0,1)$ by $\eexp := \llog^{-1}$. Note in particular that $$\llog\left(H_\LL(1)\right) = S_\LL(\pi/2).$$ Note again that these continuations are biholomorphic and definable.

A general observation about holomorphic continuations on $\LL$ arises out of the existence of branches of $\log$ (see e.g. Theorem 13.11 in Rudin’s book):

**Lemma***Let $U \subseteq \CC \setminus {0}$ be a domain and $f:U \into \CC \setminus{0}$ be holomorphic, and assume that $\pi^{-1}(U)$ is simply connected. Then $f$ lifts to a unique holomorphic $\f:\pi^{-1}(U) \into \LL$ such that $f \circ \pi = \pi \circ \f$.*

**Examples** arising from the lemma:

- $f \in \CC[X,1/X]$ and $U= D(a)$, where $a\gt 0$ is such that $Z(f):= \set{z \in \CC:\ f(z) = 0} \subseteq B_a(0)$ and $$D(a):= \set{z \in \CC:\ |z|\gt a}.$$
- For $a \in \RR$, the injective map $t_a: D(|a|) \into \CC \setminus \{0\}$ defined by $t_a(z):= z+a$. Note that, in this case, the corresponding lifting even extends to a biholomorphic map $$\t_a:\LL \setminus \pi^{-1}(-a) \into \LL \setminus \pi^{-1}(a)$$ with compositional inverse $\t_{-a}$.
- $f \in \Pc{R}{X}$ is nonzero, and $a>0$ is sufficiently small such that $F(z) \ne 0$ for $z \in U:= B_a(0) \setminus \{0\}$. In this situation, if $f(0) = 0$, then $$\left|\f(x)\right| \to 0 \text{ as } |x| \to 0;$$ while if $f(0) = a \ne 0$, then $$d\left(\f(x),a\right) \to 0 \text{ as } |x| \to 0.$$

**Final remark**

The holomorphic continuations discussed here are not always definable. For instance, the translation $\t_a$ in Example 2 above is not definable because, for instance, it maps the line ${\arg x = 2a}$ to an infinitely oscillating curve.