# Square root of biholomorphic mappings between multiply connected domains

I've been looking in the literature for a reference on the following, but without success :

Let $\Omega_1$ be a bounded finitely connected domain in the complex plane. Suppose that the boundary of $\Omega_1$ consists of pairwise disjoint piecewise analytic Jordan curves.

A repeated application of the Riemann mapping theorem gives a biholomorphic mapping $F: \Omega_1 \rightarrow \Omega_2$, where $\Omega_2$ is a bounded finitely connected domain whose boundary consists of pairwise disjoint analytic Jordan curves. See for example Ahlfors, 3rd edition, p.252.

It is well known that the Riemann map of a Jordan domain extends to a homeomorphism on the closure. Furthermore, it also extends analytically across any analytic arc on the boundary (Ahlfors p.235).

Since $F$ is a composition of Riemann maps, it extends to a continuous function in $\overline{\Omega_1}$, the closure of $\Omega_1$, and also analytically across the boundary $\partial \Omega_1$, except maybe at the "corners". My question is the following :

Is it true that $F'$ has an analytic square root in $\Omega_1$, i.e. $F'=G^2$ where $G$ is analytic in $\Omega_1$ and continuous in $\overline{\Omega_1}$, except maybe at the "corners" of $\partial \Omega_1$?

Any reference on this and biholomorphic mappings between multiply connected domains is welcome.

Thank you, Malik

• You have a reference for $F$ extending analytically across the boundary except possibly at corners? Meanwhile, what happens when $\Omega_1$ is a concentric slit region ( Ahlfohrs pages 247-249) which entails no loss of generality? – Will Jagy Apr 10 '12 at 20:29
• @Will Jagy : I've seen a proof of this somewhere, but I forgot where and can't find the reference... I only need a weaker form of that result though, so I'll edit the question accordingly. As for concentric slit regions, if I understand correctly what is meant by that, $\Omega_1$ is not of this form. I suppose that the boundary of $\Omega_1$ consists of a finite number of pairwise disjoint curves, and that each of these curves is simple, closed and piecewise analytic. – Malik Younsi Apr 11 '12 at 13:54

Proposition. If $\Omega\subseteq \mathbb C$ is a domain and $f:\Omega\to\mathbb C$ an injective holomorphic map, then $\sqrt{f'}$ has a holomorphic branch in $\Omega$.
Proof. Since $f'$ does not vanish in $\Omega$, the function $\sqrt{f'}$ admits analytic continuation along any path. It remains to show that for every closed smooth curve $\gamma$ in $\Omega$ the change of $\frac{1}{2\pi}\arg f'$ along $\gamma$ (henceforth denoted $\operatorname{ind}_\gamma f'$) is an even integer. By cutting $\gamma$ in pieces and smoothing them up, we reduce the problem to simple closed smooth curve $\gamma$.
The tangent vector $\gamma'$ rotates once as it travels around $\gamma$: that is, $$\operatorname{ind}_\gamma \gamma'\in\{-1,1\}\tag1$$ Indeed, (1) is true for a circle and invariant under diffeotopy. Applying (1) to $f\circ \gamma$ (which is also a simple closed curve), and using the relation $\arg (f\circ\gamma)'=(\arg f')\circ \gamma+\arg \gamma'$, we obtain $$\operatorname{ind}_\gamma f' + \operatorname{ind}_\gamma\gamma'\in\{-1,1\}\tag2$$ From (1) and (2), $$\operatorname{ind}_\gamma f' \in\{-2,0,2\}\tag3$$ which completes the proof. $\Box$
The continuity of $\sqrt{f'}$ up to the boundary (except at corners) is a local issue, which is resolved by reflecting $f$ across the relevant boundary arc.