# Analytic continuation of Schwarz-Christoffel mappings?

I have not studied carefully the topic of analytic continuation. Here is a question that popped up in my mind.

Let $$f: \mathbb{D} \to \Omega$$, where $$\mathbb{D}$$ is the open unit disk, and $$\Omega$$ is the interior of a polygon be a Schwarz-Christoffel mapping. Then $$f$$ extends continuously from $$\overline{\mathbb{D}}$$ into $$\overline{\Omega}$$, and maps the unit circle continuously onto the polygon $$\partial \Omega$$.

My question is this. Can one analytically continue $$f$$? If so, what would be its maximal domain of definition? What type of singularities would lie at the preimages of the polygon under $$f$$?

• Thank you. I feel one can do better with Riemann surfaces to deal with branch points, right? – Malkoun Dec 27 '18 at 12:30
• @reuns, that sounds interesting. So there could be several analytic continuations? What are you using, which result please? Is there like a one-to-one correspondence between the analytic continuations of $f^{-1}$ and some subgroups of the monodromy group perhaps (similar to covering spaces)? Also, why do you write $f^{-1}$ and not $f$, is there a specific reason please? – Malkoun Dec 27 '18 at 12:46
• See how it works for the upper half plane to a triangle. – reuns Dec 27 '18 at 12:54

My other answer is a little sloppy. I will probably delete it once thisone is completed.

I'll look at a biholomophic map from the upper half-plane to a polygon.

• Let $$a_1,\ldots,a_n \in (0,1)^n, \sum_m a_m < n-1$$ and $$b_1<\ldots < b_n\in \mathbb{R}$$

$$f'(z) =\prod_{m=1}^n (z-b_m)^{a_m-1}, \qquad f(z) = \int_0^z f'(s)ds$$

In $$\int_0^z f'(s)ds$$ we need to choose a curve $$0 \to z$$. When integrating $$f'$$ we continue it analytically along the curve.

All the analytic continuations of $$f'$$ and $$f$$ are locally analytic away from $$b_1,\ldots,b_n$$.

• As $$|z| \to \infty$$, $$f'(z) = O(z^{-1-\varepsilon})$$ thus $$\lim_{R \to \infty} \int_0^t f'(R e^{it}) d(R e^{it}) = 0$$ and $$f$$ is continuous at $$\infty$$.

Look at the upper semi-disks $$D_R = \{ z, Im(z) > 0, |z|< R\}$$ and the branch of $$f$$ analytic on $$Im(z) >0$$ and continuous on $$Im(z) \ge 0$$. Then $$f(\partial D_R)$$ is a closed curve and $$\lim_{R \to \infty} f(\partial D_R) = f(\mathbb{R})$$ is the boundary of the polygon $$P$$ with vertices $$f(b_0),f(b_1),\ldots,f(b_m)$$ where $$f(b_0)= f(+\infty) = f(i\infty) = f(-\infty)$$. Also the $$\pi a_m$$ are the angles at those vertices.

• Since $$f'$$ doesn't vanish on $$Im(z) > 0$$ then $$f$$ maps biholomorphically $$Im(z) > 0$$ to the interior of $$P$$.

• Rotation around a branch point : given a branch of $$f$$ analytic around $$b_m+z$$, continue analytically $$f(b_m+ze^{2i\pi t})$$ from $$t=0$$ to $$t=1$$ and set $$f(b_m+ze^{2i\pi}) =\lim_{t \to 1} f(b_m+ze^{2i\pi t})$$.

• For $$|z| < \min_{l \ne m} |b_l-b_m|$$ then $$f'(b_m+z e^{2i \pi}) = f'(b_m+z ) e^{2i \pi a_m}$$ thus $$f(b_m+ze^{2i\pi}) = f(b_m+z)+\lim_{\epsilon \to 0} \int_z^{\epsilon z} f'(b_m+s)ds\qquad\qquad\qquad\qquad \\ \qquad\qquad\qquad\qquad + \int_0^{2\pi} f'(b_m+\epsilon z e^{it}) d(\epsilon e^{it})+\int_{\epsilon z}^{z} f'(b_m+se^{2\pi})ds$$ $$= f(b_m+z)+ \int_z^0 f'(b_m+s)ds+\int_0^{z} f'(b_m+se^{2\pi})ds$$ $$= f(b_m)+e^{2i \pi a_m} (f(b_m+z)-f(b_m))$$

• Fix some branch $$\tilde{f}$$ of $$f$$ and let $$G$$ be the group generated by the affine transformations $$g_m(f(z))= e^{2i \pi a_m} f(z) + (1-e^{2i \pi a_m})\tilde{f}(b_m)$$.

Then the monodromy $$M$$ group of $$f$$ is the group generated by the affine transformations $$\sigma_{m,s}(f(z)) = e^{2i \pi a_m} f(z) + (1-e^{2i \pi a_m})s(\tilde{f}(b_m))$$ for each $$m$$ and each $$s \in G$$.

• How do we get that for some choices of $$a_m,b_m$$ then $$f^{-1}$$ is periodic, or doubly-periodic, or is some sort of tilling of the plane ?

For example with $$n=2,a_1 =a_2=1/2, b_1=1,b_2 = -1$$ then $$f(z) = \text{arcosh}(z)$$ and $$f^{-1}(z) = \cosh(z)$$ is $$2i\pi$$ periodic because rotating around $$z=1$$ then rotating in the other direction around $$z=-1$$ yields $$f(z) \mapsto -f(z)+2 f(1) \mapsto -(-f(z)+2 f(1))-2 (-f(-1)+2 f(-1))= f(z)+C$$ where $$C=-2 f(1)-2f(-1)=2i\pi$$

so $$f^{-1}(f(z)) = f^{-1}(f(z)+C)$$ and $$f^{-1}$$ is $$C= 2i\pi$$ periodic.

• If $f_1$ and $f_2$ are two branches of $f$, then isn't $f_1(b_m) = f_2(b_m)$? Does one really need to fix a branch in your second-to-last paragraph? – Malkoun Dec 30 '18 at 4:16
• Finally, why are there two parameters $m$ and $s$ for the generators of the monodromy group? Isn't just $m$ enough? – Malkoun Dec 30 '18 at 4:17
• Thank you for your nice answer though. – Malkoun Dec 30 '18 at 4:19