Using "conformal" to mean a holomorphic bijection, the Riemann Mapping theorem guarantees the existence of a conformal map from the upper half-plane $\mathbb{H}=\{z=x+iy\in\mathbb{C}:y>0\}$ to the interior of any polygon $P$, and I know that such maps are given by a basic transformation of a Schwarz-Christoffel integral $$ S(z)=\int_0^z\frac{d\zeta}{(\zeta-A_1)^{\beta_1}\dots(\zeta-A_n)^{\beta_n}}$$ where the $A_k$'s are distinct and increasing, $\beta_k<1$ for each $k$, and $1<\sum_{k=1}^n\beta_k\leq 2$. More precisely, if $F:\mathbb{H}\to\Omega$ is conformal where $\Omega$ is the interior of a polygon, then $F(z)=c_1S(z)+c_2$ for some Schwarz-Christoffel integral $S(z)$ and $c_1,c_2\in\mathbb{C}$.

I'm interested in the reverse of this idea that conformal mappings onto polygons are essentially given by Schwarz-Christoffel integrals. That is, are Schwarz-Christoffel integrals conformal in general? From Stein and Shakarchi's Complex Analysis, Proposition 4.1 on p. 236, I know that a Schwarz-Christoffel integral maps to a polygon with vertices $S(A_k)$ (and possibly $S(\infty)$, if $\sum_{k=1}^n\beta_k<2$). As the text states, this polygon does not have to be a simple polygon, and so it's easy to see that $S(z)$ is not necessarily conformal for the case when it maps the real line to a self-intersecting curve.

However, the book goes further to state (without proof) that even when the mapped polygon is simple, $S$ still might not be conformal. Are there any illustrative examples of Schwarz-Christoffel integrals that map the real line to simple polygons but are not conformal?


The answer depends on the missing lower bound on $\beta_k$. The Schwarz-Chrisoffel formula requires $-1\le\beta_k<1$.

The integrand in the formula has some constant argument along the boundary segment $[A_{k-1},A_k]$. Along the next segment $[A_k,A_{k+1}]$ it has a new direction. The change in the argument is $\beta_k\pi$. This will be the external angle of the resulting polygon. So we need $-1<\beta_k<1$ (or $-1\le\beta_k<1$ if we allow the polygon to turn back by $180^\circ$).

If $\beta_k<-1$ then the map will not be conformal around $A_k$. Take a small a small semicircle around $A_k$; it will be mapped to the a disk slice with central angle $(1-\beta_k)\pi>2\pi$, so the function cannot be injective anymore.

Now let us see what happens if we have $-1<\beta_k<1$ for $k=1,\ldots,n$.

To obtain a bounded polygon, we need $1<\sum\beta_k$. Then the integral in the formula has the magnitude order $|\zeta|^{-\sum\beta_k}$, so the function provided has some finite limit $P_0$ at $\infty$. The images of the boundary segments $[A_k,A_{k+1}]$ are always line segments, because the argument of the integrand changes only at the points $A_k$. Therefore, the image of the boundary line is a closed polygon $P_0P_1\ldots P_n$ where $P_0$ is the image of $\infty$.

The angle change (oriented external angle) at vertex $P_k$ of the polygon is $\beta_k\pi$ for $k=1,\ldots,n$. The angle change at $P_0$ is in $(-\pi,+\pi)$; if the polygon is simple then the sum of all changes must be $\pm2\pi$. Due to $\sum\beta_k>1$, the sum cannot be negative, so the sum if the angle changes is $2\pi$ and therefore the polygon has positive orientation. (This also shows that positively oriented simple polygons are possible only for $1<\sum\beta_k<3$.)

If $-1<\beta_k<1$, $1<\sum\beta_k$ and the polygon is simple then it follows from the argument principle that the map is conformal: every interior point of the polygon is the image of a single point in the upper half-plane and vice versa.

  • $\begingroup$ Is there really a simple way to apply the argument principle? Stein and Shakarchi goes to some length to prove the elliptic integral $\int_0^z\frac{d\zeta}{(1-\zeta^2)^{\frac{1}{2}}(1-k^2\zeta^2)^{\frac{1}{2}}}$ is conformal onto a rectangle even though it satisfies the conditions you state. $\endgroup$ – Michael M May 16 '15 at 1:28
  • 1
    $\begingroup$ Since the function has a finite limit at infinity, we can consider the real line the closed boundary of the upper half plane; then the argument principle works. (If you find this approach is too wild, take a composition with the map $i\frac{1-z}{1+z}$ that maps the unit disk to the upper hall plane. The composition maps the unit circle to the rectangle; then the argument principle shows that is is conformal.) $\endgroup$ – Géza Kós May 17 '15 at 11:49

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.